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Full text of "Physics - Principles with Applications"

PHYSICS 

PRINCIPLES with APPLICATIONS 

SIXTH EDITION 



DOUGLAS C. GIANCOLT 




Upper Saddle River, New Jersey 07458 



I.ilnaiy of ( (infjrcss t alaloginn-in-Fuhlieation Data 

Gianeoli. Douglas C. 
Physics ; principles with applications / Douglas C. Gianeoli.-- 6lh cd. 

p. em. 
Includes index. 

IS LIN Q-13-0352S7-S (vol.2 pbk. : alk. paper) — ISBN 0- 1 3-06(Xi20-u (full buok vol I & 2. raw bound. : alk. paper) - 
ISBN 0-13-035256- X (Vol I reprint, pbk.. ; alk. paper) — [SEN 0-13-18*61-2 (Nasla edition : alt. paper) — 
ISBN [1-13-191 1 S3- X (International edition : alk. paper) 

1. Physics, I, Title. 



QC23.G;W 2<XLS 
530— dc22 



umomx, 



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Cover Photo: K2, the world's second highest summit, H61 1 m (see pp. 10-11) as seen from Concordia. It is said to he the most 
difficult of the world's highest peaks (over 8000 m— see Table 1 -6 and Example 1 -3), and was first climbed by 
Lino Lacedelli and Achille Compagnoni in 1954 (see page 438). (Art Wolfe/Getty Images. Inc.) 




©2005, 1998, 1<M5. 199 1, 1985, ivsoby Douglas CGianeoli 
Published by Pearson Education. Inc. 
Pe arson Prentice Hall 
Pearson Education, Inc. 
Upper Saddle River. NJ 07458 

All rights reserved. No part of this hook may he reproduced, in any form 
or by any means, without permission in writing from the publisher, 

Pearson Prentice Hall' is a trademark of Pearson Education, Inc. 

Photo credits appear on page A-57, which constitutes 

a continuation of the copyright page. 

Printed in the United States of America 
10 9 8 7 6 5 



isbn n-i^-nbatED-rj 



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TM 



MasteringPhysics Constants 

Note: The value of a constant given in the text of an item takes precedence over the values listed below. 
Physical Constants 



Fundamental Constants 



Quantity 



Symbol 



A |) proving life Value 



1 until E ]lo<4 Viilu l ,t 



Spaed of light in vacuum 
<Ji;i\iuni(>llnl tiiil^UuiL 
Avogiidro's number 
Gas toiislcnil 

Boltzniann s constant 
Charge kin electron 

SteLnn-Rollzmarm constani 
Permittivity of free space 
Pei mealiility of free spare 
Planck's ronslatit 
Electron rest mass 

Proton rest mass 

Neutron rest mass 

Atomic mass unit (1 it J 



(■ 


3.00 lO*n V \ 


o 


6.6? x l^'N-mVkg 2 


Na 


6,02 x lfr^niol -1 


It 


8.314 J/moI'K = L.99cal/mol'K 




= 0.0821 Latm/mol-K 


I: 


1.38 x 10~ 23 J/K 


(- 


1.60 x itf 19 c 


i- 


5,67 x l<T s W/nr-K 4 


e - (l/^ fl ) 


S.85 x irr ]2 c 2 /N'm 2 


m 


4tt x lO- v Tm/A 


h 


6.63 X KT^J-s 


m,. 


9.11 x l(r ?! kg = 0. 0005+9 u 




= 0511 MeV/t 2 


»ip 


1 .6726 X 10 _27 kg - 1 .00725 u 




= 93S.3 MeV/t 2 


m n 


1,6749 x lQ~ i7 kg = 1.008665 u 



= 939.6 McV/f 2 
1.6605 x 10~ 27 kg = 931.5 MeV/c 2 



2.99792458 X 10 s m/s 
6.6742(10) x 10-"NmVkg 2 
6.022 1 Jl 5(10) X 10 23 mot -1 
8.314472(15) J /mol-K 

1 3S06505( 24 ) X 1 0~ 23 J/K 



-H . 



1.60217653(14) x 10 
5.670400(40) x lO _B W/nr-K 4 
S.S54IS7S17... X 10" 12 C 2 /N'm 2 

1.2566370614... x Kr s T-in/A 
6.6260693(11) x 10 _34 J-s 
9,1093826(16) x I0 - " 11 kg 

= 54857990945(24) X lO^u 
1,67262171(29) X 10~ 27 kg 

= 1 . 007276466 8S( 13 )u 
1.67492725(29) x l0 _3T Iqj 

= 1.00866491560(55) u 
1 .66053886(28) x 10 -37 kii 

= 931 49+043(80) Me V/c 2 



' CO DATA ( ]2/<J3). Pater J. Moist and Barry N.Taylor National Institute of Standards, and Technology. Numbers ut parent lieses indicate one-slandard- 
deviafioii experimental uncertainties in final digit?. Values without parentheses are exact [i.e.. defined quantities). 



Other Useful Data 



The Greek Alphabet 



Joule equivalent (1 cal) 


4.1S6J 


Alpha 


Absolute zero (OK) 


-273,15°C 


Beia 


Acceleration due to gravity 




Gamma 


ttl Earth's surface (uvg.) 


9.S0m/s 2 (- g) 


Delta 


Speed of sound in air (20'C) 


3+3 m/s 


Epsilnn 


Density of ;ur (dry) 


129kg/m 3 


Zeln 


Earth: Mass 


5.98 x lfjr*kg 


Liu 


Radius (mean) 


6.38 X IfFkm 


Tlietii 


Mooil: Mass 


7.35 x 10" ky 


lota 


Radius (mean) 


1.74 x itfkm 


Kappa 


Sun: Mass 


1.99 X lO^kg 


Lambda 


Radius (mean) 


6.96 x 10? km 


Mu 


Earth-Sun distance (mean} 


149,6 x If/Vkm 






Earth -Moon distance (mean) 


384 X 10 3 km 





A 


■:< 


B 


P 


r 


V 


A 


(i 


F 


,■■ 


Z 


- 


H 


■■ 


u 


f 


I 


L 


K 


KT 


A 


A 


M 


.'■■ 



Nu 


\ 


j- 


Xi 


= 


. 


O micron 


O 





[ s i 


i: 


~ 


Rlw 


p 


•■' 


Sigma 


V 


1! 


Tau 


I 


~ 


( "pstlon 


Y 


r 


Phi 


+ 


4>, <p 


Chi 


X 


X 


Psi 


'■]' 


'■■■ 


Ome^a 


n 


<■■ 



Values of Some Numbers 



77 = 3.1415927 
e = 2.7182818 



V2 = 14142136 
V'I= 1.7320508 



in 2 =0,6931472 
In 10 = 2.3025 S51 



log [of = 0.+342945 
1 rad = 57.2957795" 



TM 



MasteringPhysics Tools 

Periodic Table 



Main groups 

1A 
] 
1 
1 H 



Main groups 



3 

2 Li 

6*11 
II 

3 Na 

22,{J$S>4 



r 



2A 

2 

4 
Be 

ftOI2IS 

12 r 

Mg 



3B 

3 



41J 
■J 



Metals Metalloids 



transition metals 



No n metals 



5B 

.5 



6B 
6 



7B 
7 



i 

8 



8B 
9 



10 



IB 

I [ 



19 
K 

JWjffJR? 

37 
Rb 

fti.4678 


20 
Ca 

warn 
38 
Sr 

47.62 


21 

Sc 

44.9.559 

39 
Y 


22 
Ti 

47.SA7 


23 
V 

50.9415 


24 
Cr 

54. 9961 


25 

Mil 

54.9340 


26 

Fe 

55-845 


27 

Co 

S&9332 


28 

Ni 

5M4J4 


29 
Cu 

63.546 


40 
Zr 

* 1.224 


41 

Nb 

92.9*64 


42 
Mo 

9S.*1 


43 
Tc 


44 
Ru 

101.07 


45 
Rb 

102.90*5 


46 
Pd 

106.42 


47 
Ag 

107.S642 


55 
Cs 

131WBJ 


56 

Ba 

137.337 


71 

Lu 

174.947 

iol 

Lr 

i»n i] 


72 
Hf 

I7S.49 


73 

Ta 

lSfli.9479- 


74 
W 

1S3.S4 


75 

Re 

140.20? 


76 
Os 

l».W 


77 
lr 

192,217 


78 
Pt 

I95.07S 


79 

An 


87 
Fr 

122102] 


88 
Ka 

I22WIJ] 


104 

Rf 

1361.111 


105 
Db 

126111! 


106 

[266.12) 


!07 
Bh 

1 264.121 


I0S 

Hs 

1 269.131 


109 

Ml 
[2MLI4 1 


110 
[2TJ.IS| 


111 

Kg 

[272, IS] 



2B 
12 
30 
Zn 

*5,3? 
48 

Cd 

1 12.411 

80 
Hg 

200.59 
I 12 



3A 
13 
5 
B 

10.9 ! I 



13 
A) 

24.9&IS 
31 

Ga 

49 
In 

114.81* 

81 

TI 

204.3833 

113 



4A 
14 
6 
C 

12.0107 
14 

Si 

24jMJ5 



5 A 

15 

7 

N 

I4jOMT 

15 
P 

30,5738 



6A 

lft 
8 

o 

16 

s 

32.0SS 



32 


33 


34 


Ge 


AS 


Sc 


7J.A4 


74.9216 


2*-% 

52 


50 


51 


Sn 


Sb 


Te 


i 4.710 


131.760 


127.60 


82 


83 


84 


Pb 


Bi 


Po 


'207 2 


wwhh 


|2C(S.<JS| 



114 M5 116 



7A 
17 

9 
F 

IK.W&4 

17 
CI 

11 4,** 
35 
Br 

79.9*1 

53 
I 

1 2&.e*Ji 
85 
At 



8A 
IS 

2 
He 

4.OM60 

10 

Ne 

20.179? 
18 

Ar 

39.948 

36 
Kr 

43.80 

54 
Xe 

131.293 

86 
Rn 

[I23.02J 



(277 1 [2841 |:*9|i [2Rlt| 1 292! 





57 58 


59 
Pr 


60 


61 


62 


63 


64 


65 


(56 


67 


68 


69 


70 


Lanthanitle series 


La Ce 


Nd 


Pm 


Sm 


hu 


Gd 


Tb 


D.V 


Hd 


Er 


Tm 


Yb 




I34.9QS5 140.116 


I40.W77 


144.24 


EI45I 


1 J0.3S 


151.964 


1*7.25 


158.9233 


162.30 


I64.93U3 


167.259. 


164.9342 


173.04 




S9 90 


91 


92 


93 


94 


95 


96 


97 


98 


99 


100 


101 


!()2 


Actinidc series 


Ac Th 


Pa 


U 


Np 


Pu 


Am 


Cm 


Bk 


cr 


Es 


Em 


Md 


No 




IZ2TJ03.I [232.0381 


231 .M» 


23S.02S* 


|237.«| 


1244.06! 


[2*J,«] 


[247,97| 


[24J.QJ] 


[25!, m 


1252 m\ 


(257, I0t 


(258,10] 


[!?■>. Hi 



Contents 




Kinematics in Two 
Dimensions: Vectors 



45 



Applications List xii 

Preface xiv 

Available Supplements and Media xxii 

Notes to Students (and Instructors) 

on the Format xxiv 

Color Use: Vectors, Fields, and Svmbols xxv 

^ Introduction, Measurement, 

J_ Estimating 1 

1-1 The Nature of Science 1 

1-2 Physics and its Relation to Other Fields 3 

1-3 Models, Theories, and Laws 4 
1-4 Measurement and Uncertainty; 

Significant Figures 5 

1-5 Units, Standards, and the ST System 8 

1-6 Converting Units 10 

1-7 Orderof Magnitude: Rapid Estimating 12 

*l-8 Dimensions and Dimensional Analysis 14 

SUMMARY 15 QUESTIONS 16 
PROBLEMS 16 GENERAL PROBLEMS 17 

- Describing Motion: Kinematics 

2 in One Dimension 19 

2-1 Reference Frames and Displacement 20 

2-2 Average Velocity 21 

2-3 Instantaneous Velocity 23 

2-4 Acceleration 23 

2-5 Mot i on at C on s tan t A ccc 1 e ra t ion 26 

2-6 Solving Problems 28 

2-7 Falling Objects 31 

*2-8 Graphical Analysis of Linear Motion 36 

SUMMARY 38 QUESTIONS 38 

PROBLEMS 39 GENERAL PROBLEMS 42 



3-1 
3-2 
3-3 

3-4 
3-5 
3-6 

*3-7 
*3-8 



5- 

*5- 

5- 



Vectors and Scalars 45 

Addition of Vectors — Graphical Methods 46 
Subtraction of Veclors, and 

Multiplication of a Vector by a Scalar 48 

Adding Vectors by Components 49 

Projectile Motion 54 
Solving Problems Involving Projectile 

Motion 56 

Projectile Motion Is Parabolic 62 

Relative Velocity 62 

SUMMARY 64 QUESTIONS 65 
PROBLEMS 65 GENERAL PROBLEMS 69 



Dynamics: Newton's Laws 
oh Motion 



72 



4-1 Force 72 

4-2 Newton's First Law of Motion 73 

4-3 Mass 75 

4-4 Newton's Second Law of Motion 75 

4-5 Newton's Third Law of Motion 77 

4-6 Weight — the Force of Gravity; and the 

Normal Force 80 

4-7 Solving Problems with Newton's Laws; 

Free-Body Diagrams 84 

4-8 Proble m s I n vol vi ng Frict i on , I ncl in es 90 

4-9 Problem Solving — A General Approach 96 

SUMMARY 96 QUESTIONS 97 
PROBLEMS 98 GENERAL PROBLEMS 103 



_ Circular Motion; 
3 Gravitation 



106 



5-1 Kinematics of Uniform Circular Motion 106 

5-2 Dynamics of Uniform Circulai Motion 10y 
5-3 Highway Curves, Banked 

and Un banked 112 

*5-4 Nonuniform Circular Motion 115 

*5-5 Centrifugation 1 16 

5-6 Newton's Law of Universal Gravitation 1 17 

5-7 Gravity Near the Earth's Surface; 

Geophysical Applications 121 

8 Satellites and "Weightlessness" 122 

9 Kepler's Laws and Newton's Synthesis 125 

10 Types of Forces in Nature 128 

SUMMARY 128 QUESTIONS 129 

PROBLEMS 130 GENERAL PROBLEMS 133 



Work and Energy 



136 



6-1 Work Dune by a Constant Force 
*6-2 Work Done by a Varying Force 
6-3 Kinetic Energy, and the Work-Energy 

Principle 
6-4 Potential Energy 

6-5 Conservative and Nonconservative 
Forces 

6-6 Mechanical Energy and Its 

Conservation 
6-7 Problem Solving Using Conservation 

of Mechanical Energy 
6-8 Other Forms of Energy; 

Energy Transformations and the 

Law of Conservation of Energy 

6-9 Energy Conservation with Dissipative 
Forces: Solving Problems 

6-10 Power 

SUMMARY 160 QUESTIONS 160 
PROBLEMS 162 GENERAL PROBLEMS 



Linear Momentum 



7-1 
7-2 

7-3 
7-4 

7-5 
7-6 

*7-7 

7-8 

-7-9 

: -7-]i) 



8 



137 

141 

141 
144 

148 

149 

150 



155 

156 
158 



165 



167 



Momentum and Its Relation to Force 

Conservation of MomenLum 

Collisions and Impulse 

Conservation of Energy and Momentum 

in Collisions 

Elastic Collisions in One Dimension 

Inelastic Collisions 

Collisions in Two or Three Dimensions 

Center of Mass (CM) 

CM for the Human Body 

Center of Mass and Translational 

Motion 

SUMMARY 187 QUESTIONS 187 
PROBLEMS 188 GENERAL PROBLEMS 



168 

170 
173 

175 
176 
178 
179 
182 
184 

185 



192 



Rotational Motion 



194 



8-1 Angular Quantities 195 

8-2 Constant Angular Acceleration 201 

8-3 Rolling Motion (Without Slipping) 202 

8-4 Torque 203 
8-5 Rotational Dynamics; Torque and 

Rotational Inertia 206 

8-6 Solving Problems in Rotational 

Dynamics 208 

8-7 Rotational Kinetic Energy 210 

8-8 Angular Momentum and Its 

Conservation 213 

s 8-9 Vector Nature of Angulai Quantities 215 
SUMMARY 217 QUESTIONS 217 

PROBLEMS 219 GENERAL PROBLEMS 223 




9-1 

9-2 
*9-3 

9-4 
*9-5 
*9-6 
*9-7 



10 



Static Equilibrium; 
Elasticity and Fracture 



226 



The Conditions for Equilibrium 227 

Solving Statics Problems 229 

Applications to Muscles and Joints 234 

Stability and Balance 236 

Elasticity; Stress and Strain 237 

Fracture 241 
Spanning a Space: 

A rch es a n d D o m es 243 

SUMMARY 246 QUESTIONS 246 
PROBLEMS 247 GENERAL PROBLEMS 252 



Fluids 



255 



10-1 
10-2 
10-3 

1 0-4 

10-5 
10-6 

10-7 
10-8 

10-9 
10-10 



* 10-11 
*10-12 

*10-13 
*10-14 



Phases of Matter 255 

Density and Specific Gravity 256 

Pressure in Fluids 257 

Atmospheric Pressure and 

Gauge Pressure 259 

Pascal's Principle 260 

Measurement of Pressure; Gauges 

and the Barometer 260 

Buoyancy and Archimedes 1 Principle 263 

Fluids in Motion; Flow Rate and 

the Equation of Continuity 268 

Bernoulli's Equation 270 
Applications of Bernoulli's Principle: 
from Torricelli to Airplanes, Baseballs, 

andTIA 272 

Viscosity 274 
Flow in Tubes: Poise ui lie's Equation, 

Blood Flow 275 

Surface Tension and Capillarity 276 

Pumps, and the Heart 278 

SUMMARY 279 QUESTIONS 280 
PROBLEMS 28! GENERAL PROBLEMS 284 



vi CONTENTS 



11 



Vibrations and Waves 



286 



Ll-l 
11-2 
11-3 
11-4 
11-5 
11-6 
11-7 
11-8 

Ll-9 
* 1 1 - 1 

11-11 
11-12 
11-13 

•11-14 

♦11 

*Ll 



■ 15 

■ in 



Simple Harmonic Motion 287 

Energy in the Simple Harmonic Oscillator 289 

The Period and Sinusoidal Nature of SHM 292 

The Simple Pendulum 2% 

Damped Harmonic Motion 298 

Forced Vibrations; Resonance 299 

Wave Motion 300 
Types of Waves: Transverse and 

Longitudinal 303 

Energy Transported by Waves 305 

Intensity Related to Amplitude 

and Frequency 306 

Reflection and Transmission of Waves 307 

Interference; Principle of Superposition 308 

Standing Waves; Resonance 310 

Refraction 312 

Diffraction 313 

Mathematical Representation of 

a Traveling Wave 314 

SUMMARY 315 OUESTIONS 316 
PROBLEMS 317 GENERAL PROBLEMS 320 



^ ^ Temperature and 
13 Kinetic Theory 



352 



12 



Sound 



322 



12-1 

12-2 

+ 12-3 

12-4 

*12-5 

12-6 

12-7 

•12-8 

+ 12-9 



Characteristics of Sound 322 

Intensity of Sound: Decibels 325 

The Ear and Its Response; Loudness 328 

Sources of Sound: Vibrating Strings and 

Air Columns 329 

Quality of Sound, and Noise; 

Superposition 334 

In terference of Sound Waves; Beats 335 

Doppler Effect 338 

Shock Waves and the Sonic Boom 342 
Applications: Sonar, Ultrasound, and 

Medical Imaging 343 

SUMMARY 345 OUESTFONS 346 
PROBLEMS 347 OENERAL PROBLEMS 349 




13-1 

13-2 

+ 13-3 

13-4 
*13-5 
13-6 
13-7 
13-8 
13-9 

13-10 

*13-11 
*13-12 
*13-13 
+ 13-14 



14 



Atomic Theory of Matter 

Temperature and Thermometers 

Thermal Equilibrium and the Zeroth 

Law of Th e r m ody n a m i cs 

Thermal Expansion 

Thermal Stresses 

The Gas Laws and Absolute Temperature 

The Ideal Gas Law 

Problem Solving with the Ideal Gas Law 

Ideal Gas Law in Terms of Molecules: 

Avogadro's Number 

Kinetic Theory and the Molecular 

Interpretation of Temperature 

Distribution of Molecular Speeds 

Real Gases and Changes of Phase 

Vapor Pressure and Humidity 

Diffusion 

SUMMARY 378 OUESTIONS 379 

PROBLEMS 380 OLNLRAL PROBLLMS 



352 

354 

357 
357 
361 
361 
3h3 
364 

366 

367 
371 
371 
373 
376 



382 



Heat 



384 



14-1 Heat as Energy Transfer 385 

14-2 Internal Energy 386 

14-3 Specific HeaL 387 

14-4 Calorimetry — Solving Problems 388 

14-5 Latent Heat 391 

14-6 Heat Transfer: Conduction 395 

14-7 Heat Transfer: Convection 397 

14-8 Heat Transfer: Radiation 399 

SUMMARY 403 OUESTIONS 403 
PROBLEMS 404 OENERAL PROBLEMS 406 

15 The Laws of Thermodynamics 408 

15-1 The First Law of Thermodynamics 409 

15-2 Thermodynamic Processes and the First Law 410 

*15-3 Human Metabolism and the First Law 414 
15-4 Second Law of Thermodynamics — 

Introduction 415 

15-5 Heat Engines 416 

15-6 Refrigerators, Air Conditioners, and Heat 

Pumps 42 1 

15-7 Entropy and the 

Second Law of Thermodynamics 424 

15-8 Order to Disorder 426 

15-9 Unavailability of Energy; Heat Death 426 

*15-10 Evolution and Growth; "Time's Arrow" 427 

+ 15-1 1 Statistical Interpretation of Entropy and 

the Second Law 428 

+ 15-12 Thermal Pollution and Global Warming 430 
SUMMARY 432 OUESTIONS 433 
PROBLEMS 433 GENERAL PROBLEMS 436 



CONTENTS vii 




18 



Electric Currents 



493 



16 



Electric Charge and 
Electric Field 



439 



16-1 

16-2 
16-3 
16-4 
16-5 
16-6 

16-7 

16-8 

16-9 

*16-10 

*16-11 

*16-12 



17 



Static Electricity; Electric Charge and 
Its Conservation 
Electric Charge in the Atom 
Insulators and Conductors 
Induced Charge; the Electroscope 
Coulomb's Law 

Solving Problems Involving Coulomb's 
Law and Vectors 

The Electric Field 

Field Lines 

Electric Fields and Conductors 

Gauss s Law 

Electric Forces in Molecular Biology: 
DNA Structure and Replication 

Photocopy Machines and Computer 
Printers Use Electrostatics 

SUMMARY 463 QUESTIONS 464 
PROBLEMS 465 GENERAL PROBLEMS 468 



460 



462 



Electric Potential 



470 



17-1 Electric Potential Energy and Potential 

Difference 470 
17-2 Relation between Electric Potential and 

Electric Field 474 

17-3 Equipotential Lines 474 

17-4 The Electron Volt, a Unit of Energy 476 

17-5 Electric Potential Due to Point Charges 476 

*17-6 Potential Due to Electric Dipole; 

Dipole Moment 479 

17-7 Capacitance 480 

17-S Dielectrics 482 

17-9 S t o r age of E 1 ect rie En e rgy 484 

*17-10 Cathode Ray Tube: TV and Computer 

Monitors, Oscilloscope 485 

* 17-1 1 The Electrocardiogram (ECG or EKG) 487 

SUMMARY 488 QUESTIONS 488 
PROBLEMS 489 GENERAL PROBLEMS 491 



18-1 


The Electric Battery 


494 


18-2 


Electric Current 


496 


18-3 


Ohm's Law: Resistance and Resistors 


498 


18-4 


Resistivity 


5IK) 


18-5 


Electric Power 


502 


18-6 


Power in Household Circuits 


505 


18-7 


Alternating Current 


51 K! 


♦18-8 


Microscopic View of Electric Current 


509 


*18-9 


S u pe rcon duct i vi t y 


510 


*18-10 


Electrical Conduction in the Human 






Nervous System 


510 




SUMMARY 514 QUESTIONS 514 






PROBLEMS 515 GENERAL PROBLEMS 


518 


19 


DC Circuits 


520 



440 


19-1 


441 


19-2 


441 


19-3 


442 


*19-4 


444 






19-5 


447 




450 


19-6 


454 




45b 


19-7 


457 


*19-8 



EMF and Terminal Voltage 520 

Resistors in Series and in Parallel 522 

Kirchhoff s Rules 528 
EMFs in Series and in Parallel; 

Charging a Battery 532 
Circuits Containing Capacitors in Series 

and in Parallel 533 

RC Circuits — Resistor and Capacitor in 
Series 535 

Electric Hazards 538 

Ammeters and Voltmeters 541 

SUMMARY 545 QUESTIONS 545 
PROBLEMS 547 GENERAL PROBLEMS 551 



20 



Magnetism 



554 



20-1 Magnets and Magnetic Fields 554 

20-2 Electric Currents Produce Magnetic 

Fields 557 

20-3 Force on an Electric Current in „ 

a Magnetic Field; Definition of B 558 

20-4 Force on Electric Charge Moving in 

a M agn e ti c Fi e Id 560 

20-5 Magnetic Field Due to a Long 

Straight Wire 563 

20-6 Force between Two Parallel Wires 565 

20-7 Solenoids and Electromagnets 567 

*2Q-S Ampere's Law 568 

*20-9 Torque on a Current Loop; 

Magnetic Moment 570 

* 20- 1 A ppl i ca ti on s : G alvan o m c tc is. M o to is. 

Loudspeakers 571 

*20-ll Mass Spectrometer 572 

20-12 Ferromagnelism: Domains and 

Hysteresis 573 

SUMMARY 575 QUESTIONS 576 
PROBLEMS 577 GENERAL PROBLEMS 581 



viii CONTENTS 



,_ ^ Electromagnetic Induction 
Ll and Faraday's Law 



584 



21-1 
21-2 
21-3 
21-4 

21-5 
*21-6 

21-7 
21-8 

*21-9 

•21-10 

*21-11 

+21-12 

*21-13 

*21-14 



22 



Induced EMF 584 

Faraday's Law of Induction; Lenz's Law 586 
EMF Induced in a Moving Conductor 590 
Changing Magnetic Flux Produces an 
Electric Field 
Electric Generators 

Back EMF and Counter Torque; 
Eddy Currents 

Transformers and Transmission of Power 595 

Applications of Induction: Sound Systems, 

Computer Memory, Seismograph, GFCI 

Inductance 

Energy Stored in a Magnetic Field 

LR Circuit 

AC Circuits and Reactance 

LRC Scries AC Circuit 

Resonance in AC Circuits 

SUMMARY 608 QUESTIONS 609 
PROBLEMS 610 GENERAL PROBLEMS 613 



591 

592 
593 



598 
600 

602 
602 
603 
606 
60S 



Electromagnetic Waves 



615 



22- 

22- 



22-4 

5 
6 



*22- 
*22- 



*22-7 



23 



Changing Electric Fields Produce 

Magnetic Fields; Maxwell's Equations 616 

Pro duct i on of E le ct romagn e tic Wa ves 617 

Light as an Electromagnetic Wave and 

the Electromagnetic Spectrum 619 

Measuring the Speed of Light 622 

Energy in EM Waves 623 
Momentum Transfer and Radiation 

Pressure 625 

Radio and Television; Wireless 
Communication 626 

QUESTIONS 629 

GENERAL PROBLEMS 631 



SUMMARY 629 
PROBLEMS 629 



Light: Geometric Optics 



632 



23-1 The Ray Model of Light 632 
23-2 Reflection; Image Formation by a 

Plane Mirror 633 
23-3 Formation of Images by Spherical 

Mirrors fi35 

23-4 Index of Refraction 642 

23-5 Refraction: Sncll's Law M2 

23-fi Total Internal Reflection; Fiber Optics 645 

23-7 Thin Lenses; Ray Tracing 647 

23-8 The Thin Lens Equation; Magnification 650 

*23-9 Combinations of Lenses 654 

*23- 10 The Lensmaker's Equation 656 

SUMMARY 656 QUESTIONS 657 
PROBLEMS 658 GENERAL PROBLEMS 662 



-'■'•|\ins.ni of pari [u) of Hp. , 
I Hewed Kt ihi nsMf (win! »iih «h(J 
1v object WfetBWh ;il ihe eye is much ldrp:A 
insular mf nffinlion or nmeniltiiiK potter. W. ■ 
'.if the an£t« Subtended by an phjevl when usirvg rhi 
U5in£ I he unaided eye. wilh ihi; ybjecl. 41C [he 
i .N' ■ 2^ cm fni a nnmial ijyitj: 



iU ii i^il ww 
_.il* -I- .■■ Ihr li|i;i. 

ilt.-l |i UKdi ITil 1 
Vinctl a£ I he rnln' 
\i I i i. - . . 

IS' a! lliu cvf 



115-11 

I ■ *A ■!■.. !■'. I 

mull nfl :iniL 
\ye ••!- .11 i ihe 



(Mil 



24 



where f) and r arc shown in Fig. 25-liS. We: esin •u.ji\n 
k-nglh bf ncilirifc thai «" = h/N (Fig, 25-lfth) and r 
where i"i is Ehc height uf the uhjecl and we jissume I ho 
fl" equal iheir Nihes artd laflflaiibs. If Iht tye i* rtHasttJ 
iniiipt; friM be At infinily and the ohjtct will be prvri' 
Fir- 25- 17. Then &«/ and fl' - A//. Tlius 

Um *-NL m E. r c 

# A/A' / L N 

* sec Chat the shurteT the focal lenclh of Ihj 
: mjipxiificaticm of i\ given It 
nit vtHir^yt; ncs it F(kli<.».-s 



The Wave Nature of Light 664 



24-1 

*24-2 

24-3 

24-4 
24-5 
24-6 

*24-7 
24-8 

*24-9 
24-10 

*24-ll 

+24-12 



25 



Waves Versus Particles; 

Huygens* Principle and Diffraction 665 
Huygens T Principle and the Law 

of Refraction 666 
Interference — Young's Double-Slit 

Experiment 668 

The Visible Spectrum and Dispersion 671 

Diffraction by a Single Slit or Disk 673 

Diffraction Grating 676 

The Spectrometer and Spectroscopy 678 

Interference by Thin Films 679 

Michelson Interferometer 684 

Polarization 684 

Liquid Crystal Displays (LCD) 688 
Scattering of Light by the Atmosphere 690 

SUMMARY 690 OUESIIONS 691 

PR URL EMS 692 GENERAL PROBLEMS 694 



Optical Instruments 



696 



25-1 Cameras, Film and Digital 697 

25-2 The Human Eye; Corrective Lenses 70 1 

25-3 Magnifying Glass 704 

25-4 Telescopes 706 

+25-5 Compound Microscope 708 

+25-6 Aberrations of Lenses and Mirrors 710 
25-7 Limits of Resolution; Circular 

Apertures 711 
25-8 Resolution of Telescopes and 

Microscopes; the 1 Limit 714 
25-9 Resolution of the Human Eye and 

U sef ul M agn i f ica tio n 715 

+25-10 Specialty Microscopes and Contrast 716 

+25-11 X-Rays and X-Ray Diffraction 717 
*25-l2 X-Ray Imaging and Computed 

Tom ogra ph y ( CT S e an ) 718 

SUMMARY 721 OUESTIONS 722 
PROBLEMS 722 GENERAL PROBLEMS 725 



CON II NFS ix 



- g. The Special Theory of 
Zo Relativity 



726 



_ o Quantum Mechanics of 
Zo Atoms 



786 



26-1 Galilean-Newtonian Relativity 
26-2 Postulates of the Special Theory 

of Relativity 
26-3 Simultaneity 

26-4 Time Dilation and the Twin Paradox 
26-5 Length Contraction 
*26-6 Four-Dimensional Space-Time 
26-7 Relativistic Momentum and Mass 
26-8 The Ultimate Speed 
26-9 E = m c 2 ; Mass and Energy 
26-10 Relativistic Addition of Velocities 
26-1 1 The Impact of Special Relativity 

SUMMARY 749 QUESTIONS 750 
PROBLEMS 751 GENERAL PROBLEMS 



'hi 



727 


28-1 




28-2 


730 




731 


28-3 


734 


*28-4 


740 
742 


28-5 


742 


28-6 


743 
744 
748 
748 


28-7 

28-.S 

*28-9 

*28-10 




*28-11 




*28-l2 



_ _ Early Quantum Theory and 

Z 7 Models of the Atom 754 

27-1 Discovery and Properties of the Electron 754 

27-2 Planck's Quantum Hypothesis; 
Blackbody Radiation 

27-3 Photon Theory of Light and the 

Photoelectric Effect 
27-4 Ene rgy, Mass, and Momen turn of a 

Photon 
+27-5 Compton Effect 
27-6 Photon Interactions; Pair Production 
27-7 Wave-Particle Duality; the Principle of 

Complementarity 
27-8 Wave Nature of Matter 
*27-9 Electron Microscopes 
27-10 Early Models of the Atom 
27-11 Atomic Spectra: Key to the Structure 

of the Atom 
27-12 The Bohr Model 
27-13 de Broglie's Hypothesis Applied to Atoms 780 

SUMMARY 781 QUESTIONS 782 
PROBLEMS 782 GENERAL PROBLEMS 784 




Quantum Mechanics — A New Theory 787 
The Wave Function and Its Interpretation; 

the Double-Slit Experiment 787 

The Hciscnbcrg Uncertainty Principle 789 
Philosophic Implications; Probability 

versus Determinism 792 

Quantum-Mechanical View of Atoms 794 
Quantum Mechanics of the Hvdrugen 

Atom; Quantum Numbers 794 

Complex Atoms; the Exclusion Principle 797 

The Periodic Table of Elements 798 

X-Ray Spectra and Atomic Number 800 

Fluorescence and Phosphorescence 802 

Lasers 803 

Holography 806 

SUMMARY 807 QUESTIONS 808 
PROBLEMS 809 GENERAL PROBLEMS 810 



29 



Molecules and Solids 



812 



756 


*29-l 
*?9-2 


Bonding in Molecules 
Potential-Energy Diagrams for Molecul 


812 

les 815 


758 


*29-3 


Weak (van der Waals) Bonds 


817 




*29-4 


Molecular Spectra 


821 


762 


*29-5 


Bonding in Solids 


824 


763 


*29-6 


Band Theory of Solids 


825 


764 


*29-7 


Semiconductors and Doping 


827 




*29-S 


Semiconductor Diodes 


828 


765 


*?9-9 


Transistors and Integrated Circuits 


830 


766 




SUMMARY 831 QUESTIONS 832 




768 




PROBLEMS 832 GENERAL PROBLEMS 


833 


769 










30 


Nuclear Physics and 




771 

'771 


Radioactivity 


835 



30-1 Structure and Properties of the Nucleus 835 

30-2 Binding Energy and Nuclear Forces 838 

30-3 Radioactivity 841 

30-4 Alpha Decay 842 

30-5 Beta Decay 845 

30-6 Gamma Decay 847 
30-7 Conservation of Nucleon Number and 

Other Conservation Laws 848 

30-8 Half-life and Rate of Decay 848 
30-9 Calculations Involving Decay Rates 

and Half-Life 850 

30-10 Decay Series 852 

30-11 Radioactive Dating 853 

*30-12 Stability and Tunneling 855 

30-13 Defection of Radiation 856 

SUMMARY 858 QUESTIONS 859 

PROBLEMS 860 GENERAL PROBLEMS 861 



x CONTENTS 



— ^ Nuclear Energy; Effects and 
JL Uses of Radiation 863 



31- 

31- 
31- 
31- 



31-5 
+31-6 
*31-7 

*31-8 
31-9 



1 Nuclear Reactions and the 
Transmutation of Elements 863 

2 Nuclear Fission; Nuclear Reactors 866 

3 Nuclear Fusi on 87 1 

4 Passage of Radiation Through Matter; 
Radiation Damage 876 

5 Measurement of Radiation — Dosimetry 877 
Radiation Therapy 880 

Tracers and Imaging in Research and 

Medicine SSO 

Emission Tomography 881 

Nuclear Magnetic Resonance (NMR) 

and Magnetic Resonance Imaging (MRI) 882 

SUMMARY 885 QUESTIONS 885 
PROBLEMS 886 GENERAL PROBLEMS 887 




32 



Elementary Particles 



889 



32-1 High-Energy Particles and Accelerators 890 
32-2 Beginnings of Elementary Particle 

Physics — Particle Exchange 895 

32-3 Particles and Antiparticles 898 

32-4 Particle Interactions and 

Conservation Laws 898 

32-5 Neutrinos— Recent Results 900 

32-6 Particle Classification 90 1 

32-7 Particle Stability and Resonances 902 
32-8 Strange Particles? Charm? 

Maybe a New Model Is Needed! 903 

32-9 Quarks 904 
32-10 The ''Standard Model"; Quantum 

Chromo dynamics (QCD) and the 

Electroweak Theory 906 

32-11 Grand Unified Theories 908 

32-12 Strings and Supersymmetry 910 

SUMMARY 910 QUESTIONS 91 I 
PROBLEMS 911 GENERAL PROBLEMS 913 













33 



Astrophysics 
and Cosmology 



914 



33-1 Stars and Galaxies 

33-2 Stellar Evolution: The Birth and Death 

of Stars 
33-3 Distance Measurements 
33-4 General Relativity: Gravity and the 

Curvature of Space 
33-5 The Expanding Universe: 

Redshilt and Huhhle's Law 

33-6 The Big Bang and the Cosmic 
Microwave Background 

33-7 The Standard Cosmological Model: 
the Early History of the Universe 

33-8 Dark Matter and Dark Energy 

33-9 Large-Scale Structure of the Universe 

33-10 Finally... 

SUMMARY 943 OUESTIONS 944 
PROBLEMS 944 GENERAL PROBLEMS 945 

Appendices 



915 

918 
924 

926 

930 

933 

936 
939 
942 
942 



A 

A-i 

A-2 

A-3 
A-4 
A-5 
A-6 
A-7 
A-8 

B 

C 

n 

E 



Mathematical Review A-i 

Relationships, Proportionality, and Equations A-I 

Exponents A -2 

Powers of 10. or Exponential Notation A-3 

Algebra A-3 

The Binomial Expansion A-fi 

Plane Geometry A-7 

Trigonometric Functions and Identities A-8 

Logarithms A- 10 

Selected Isotopes A- 12 

Rotating Frames of Reference; 

Inrrtial Forces; Corioi is Effect A-16 

Molar Specific Heats for Gases, and 

the eolt partition of energy a-20 



Galilean and Lorentz 
Transformations 



Answers to Odd -Numbered Problems 

Index 

Photo Credits 



A-23 

A-27 
A-42 

A-57 



CON II N"S xi 



Applications to Biology and Medicine 



Chapter 1 




Estimating number of heartbeats 


in a lifetime 


13 


Chapter 4 




How we walk 


-N 


Chapter 5 




Centrifuge 


116,201 


Chapter 7 




Don't break a leg 


174 


Body parts center of mass 


184 


Chapter S 




Biceps torque 


205,221 


Chapter 9 




Teeth straightening 


2.27 


Forces in muscles and joints 


234 


Muscle insertion and lever arm 


234 


Spine, back pain 


235 


Body balance 


236 


Chapter III 




Body suspension in water 


255 


Bbod circulation 


269 


Blood loss to brain — T1A 


273 


Blood flow and heart disease 


275 


Insect on water surface 


276 


Heart as a pump 


278 


Blood pressure 


278 


Chapter 11 




Spider web 


293 


Echolocation in whales, bats 


304 


Chapter 12 




Wide range of human hearing 


325. 329 


Human ear and its sensitivity 


328 



341 
344 

360 
367 

374, 395 
378 

386 
399 
400 

402 

414 



Doppler blood-flow meter 

and other medical uses 
Ultrasound medical imaging 
Chapter 13 
Life under ice 
Molecules in one breath 
Evaporation cools 
Diffusion in living organisms 
Chapter 14 
Working off Calories 
Convection by blood 
Humans' radiative heal loss 
Medical thermography 
Chapter 15 

Energ}' in the human body 
Biological evolution and development 427 
Chapter 16 

Cells: electric forces plus kinetic theory 460 
DNA structure and replication 460 

Chapter 17 

Dipoles in molecular biology 
Capacitor burn or shock 
Heart defibrillator 
Electrocardiogram (ECG) 
Chapter IS 
Electrical conduction in the 

human nervous system 

Chapter 19 

Heart pacemaker 

Electric shock, grounding, and safety 539 

Chapter 21 

EM blood flow measurement 590 

Ground fault circuit interrupters 599 



4 Ml 

485 
485 

487 



510 



.\-8 



Pacemaker 


600 


Chapter 22 

Optical tweezers 


fOn 


Chapter 23 

Medical endoscopes (fiber optics) 646 


Chapter 2? 

Human eye 


701 


Corrective lenses 


702 


Contact lenses 


703 


Seeing under water 


704 


Light microscopes 

Resolution of human eye- 


708 
713 


X-ray diffraction in biology 


718 


X-ray images 


718 


CAT scans 


719 


Chapter 27 

Photosynthesis 


763 


Measuring tone density 


764 


Electron microscope, STM, AFM 


768.769 


Chapter 28 




Medical uses of lasers 


80." 


Chapter 29 
Activation energy. ATP 


817 


Weak bonds in cells 


818 


Protein synthesis 


820 


Chapter 31 

Biological radiation damage 
Radiation dosimetry 


876 
877 


Radiation therapy 

Tracers in medicine and biology 


880 
SK0 


Medical imaging PET and SPET 


881,882 


NMR imaging (MRI) 


883 



Applications to Other Fields and Everyday Life 



Chapter 1 




Chapter 6 




The 80QfJ-m peaks 


10 


Car Stopping distance a v 2 


144 


Estimating volume of lake 


12 


Roller coaster 


SI, 157 


Estimating height by 




Pole vault 


152 


triangulation 


13 


Dart gun 


1 53 


Chapter 2 




Car power 


159 


Airport runway design 


27 


I .ever 


If- 2 


Car safety — air bags 


29 


Chapter 7 




Braking distances 


30 


Tennis serve 


169,173 


Rapid transit 


42 


Gun recoil 


172 






Rockets 


172.186 


Chapter 3 
Kicked football 


5S.61 


High jump 


185 


Ballsporis 66,67,70,71 


Chapter 8 

Hard drive and bit speed 


200 


Chapter 4 




Rotating skater, diver 


214 


Rocket acceleration 


78 


Neutron star collapse 


215 


What force accelerates car? 


79 


Chapter 9 

Lever 




ElevatoT and counterweight 


SS 


229 


Mechanical advantage of pulley 


89 


Cantilever 


231 


Rock elimbing 


102.105 


Reinforced & pres tressed 




Chapter 5 




concrete 


242 


Skidding on a curve 


113 


Tragic collapse 


242 


Antilock bTakes 


113 


Arches and domes 


243 


Banked curves 


114 


Chapter 10 




Geophysics applications 


122 


Car brakes, hydraulic lift 


260 


Artificial Earth satellites 


122 


Hydrometer 


266 


Geosynchronous satellites 


123 


Airplane wings, lift 


272 


Weightlessness 


124 


Sailing against the wind 


273 



A baseball curve 


273 


Surface tension, capillarity 


277 


Soaps and detergents 


277 


Pumps 


278 


Chapter 11 




Pendulum clock 


297 


Shock absorbers, building dampers 


298 


Resonant bridge collapse 


299 


Earthquakes ' 3U4.305.306.3I3 


Chapter 12 




Lightning, distance to strike 


323 


Autofocusing camera 


324 


Musical instruments, stringed and wind 329 


Wind noise 


334 


Tuning with beats 


337 


Doppler effect, weather forecasting 


341 


Redshift in cosmology 


342 


Sonic boom 


342 


Sonar 


343 


Chapter 13 




Expansion joints 


354 


Opening a tight lid 


359 


Gas lank overflow 


359 


Highway buckling 


361 


Mass (and weight) of air in a room 


365 


Pressure in a hot tire 


366 


Chemical reactions, temperature- 




dependence 


371 



xii APPLICATIONS 



Superfluidity 




373 


Humidity, weather 


375,3 


Thermostat 




379 


Chapter 14 






Heat loss through windows 




396 


Thermal windows 




397 


R-v allies of thermal insulation 




397 


How clothes insulate 


397.399 


Convective house heating 




398 


Convection on a steep hike 




398 


Radiation from the Sun 


4i) 1 


,402 


Astronomy — size of a Star 




4P J 


Chapter 15 






Steam engine 




1 1 (- 


Internal combustion engine 




417 


Refrigerator 




121 


Air conditioner 




477 


Heat pump 




423 


SEER rating 




423 


Thermal pollution, global warm 


i'v: 


430 


Energy resources 




430 


Chapter 16 






Electrical shielding, safety 




457 


Photocopy machines 




462 


Laser printers and inkjet printers 


463 


Chapter 17 






Capacitors in camera flashes. 






backups, surge protectors, 






memory, keyboards 480, 481 


,482,4 


Super high capacitance 




4SJ 


CRT: TV and computer monitors 


486 


Oscilloscope 




list- 


Photocell 




_p:- 


Chapter 18 






Loudspeaker wires 




SOI 


Resistance thermometer 




502 


Heating element, lightbulb filament 


503 


Why bulbs burn out when fh"st 






turned on 




503 


A lightning bolt 




504 


Household circuits 




505 


Fuses and circuit breakers 




505 


Shorts & safety 




506 


Extension cords 




506 


Hair dryer 




508 


Superconductors 




510 


Chapter 19 






Car battery charging 




532 


Jump starling a car 




532 


Blinking flashes windshield wipers 


537 


Electric hazards 




538 



Ground wires and plugs 540 

Leakage currents 541 

Downed power lines 541 
Digital &. analog meters 541 . 544 

Meter connection, corrections 543-544 

Condenser microphone 546 

Chapter 20 

Compass use. magnetic 

declination 556 

Aurora borealis 563 

Electromagnets and solenoids 567 

Solenoid switching 567 

Magnetic circuit breakers 567 
Motors 571 , 572 

Loudspeaker 572 

M ass spec t rome te r 572 

Electromagnetic pumping 576 

Relay 577 

Chapter 21 

Induction stove 588 

Generators^ car alternators 592 

Motor start-up current 593 

Motor over! oad 5 94 

Eddy current damping 594 

Airport metal detector 595 

Radio transformers 5 L >6 

Electric power transmission 597 

Magnetic microphone 598 

Read/write on tape and disks 598 

Digital coding 598 

Credi t card swipe 599 

Seismograph 599 
Ground fault circuit interrupters 599 

Capacitors as filters 605 

Electric resonance 608 

Chapter 22 

AM and FM transmission 627 

Tuning a station 627 

Antennas 628 
Cell phones, remote control, cable 

and satellite TV 628 

Chapter 23 

How tall a mirror do you need 635 

Where you can see yourself in a 

concave mirror 639 
Curved mirror uses 635, 640, 641 

Optical illusions 643 

Apparent water depth 644 
Fiber optics in telecommunications 646 
Where you can see a lens image 649 



Chapter 24 




Highway mirages 


667 


Rainbows and diamonds 


672 


Spectroscopic analysis 


679 


Soap bubbles and oil films 


679 


I. ens coatings 


682 


Polaroids 


685 


Seeing into the river 


687 


Liquid crystal displays (LCDs] 


688 


Why the sky is blue, sunsets 




are red, clouds are white 


690 


Chapter 25 




Digital cameras, CCD, artifacts 


697 


Camera adjustments 


698 


Telescopes 


706 


Microscopes 


70S 


Hubble space telescope 


713 


Telescope resolution 


714 


Special microscopes 


716 


Uses of X-ray diffraction 


?:« 


Chapter 26 




Global positioning system (GPS) 


739 


Chapter 27 




Photocells, photodiodes 


762 


Chapter 28 




Fluorescence analysis 


802 


Fluorescent lighlbulbs 


803 


Laser uses 


805 


CD, DVD, barcodes 


806 


Holography 


M:>n 


(hauler 29 




Semiconductor diodes, transistors 828. 830 


ReetifieT circuits 


829 


LED displays: photodiodes 


830 


Integrated circuits 


83: 


Chapter 30 




Smoke detectors 


844 


Carbon- 14 dating 


853 


Archeological and geological dating 


854 


Oldest Earth rocks and earliest life 


855 


Chapter 31 




Nuclear power plants 


869 


Manhattan Project 


87 : 


Radon gas pollution 


878 


Chapter 33 




Star evolution 


933 


Supernovae 


935 


Star distances 


936 


Black holes 


94 : 


Evolution of universe 


948 



Problem Solving Boxes 



Chapter 2 

Problem Solving 28 

Chapter 3 

Problem Solving: Adding Vectors 53 

Problem Solving: Projectile Motion 56 

Chapter 4 

Problem Solving: Newton's Laws: 

Free -Body Diagrams 85 

Problem Solving: In General 96 

Chapter 5 

Problem Solving: Uniform Circular 
Motion 112 

Chapter fi 

Pro blem S ol vi ng: Work 1 39 



Problem Solving: Conservation of 
Energy 157 

Chapter 7 

Problem Solving: Momentum 

Conservation and Collisions 181 

Chapter 8 

Problem Solving: Rotational Motion 209 

Chapter 9 

Problem Solving: Statics 230 

Chapter 14 

Pro b lem Sol vi n g: Ca lori m etry 394 

Chapter 15 

Problem Solving: Thermodynamics 432 



Chapter 16 

Problem Solving: Electrostatics: 

Electric Forces and Electric Fields 454 

Chapter 19 

Problem Solving: Kirchhoff's Rules 530 

Chapter 20 

Problem Solving: Magnetic Kields 562 

Chapter 21 

Problem Solving: I .enz's Law 588 

Chapter 23 

Problem Solving: Spherical Mirrors 64 1 

Prob lem Sol vi n g: Thin Lense s 651 

Chapter 24 

Problem Solving: Interference 683 

APPLICATIONS xiii 



Preface 

See the World through Eyes that Know Physics 

This book is written for students, Tt has been written to give students a thorough 
understanding of the basic concepts of physics in all its aspects, from mechanics to 
modern physics. It aims to explain physics in a readable and interesting manner that 
is accessible and cleaT, and to teach students by anticipating their needs and 
difficulties without oversimplifying. A second objective is to show students how- 
useful physics is in their own lives and future professions by means of interesting 
applications. In addition, much effort has gone into techniques and approaches 
foT solving problems. 

This textbook is especially suited for students taking a one year introductory 
course in physics that uses algebra and trigonometry but not calculus. Many of 
these students are majoring in biology or (pre)medicine. and others may be in 
architecture, technology, or the earth or environmental sciences. Many applications 
to these fields are intended to answer that common student query: "Why must T 
study physics?" The answer is that physics is fundamental to a full understanding 
of these fields, and here they can see how. Physics is all about us in the everyday 
world. Tt is the goal of this book to help students "see the world through eyes that 
know physics," 
n e w ► Some of the new features in this sixth edition include: ( I ) in-text Exercises for 

students to check their understanding; (2) new Approach paragraphs for worked- 
out Examples; (3) new Examples that stcp-by-step follow each Problem Solving 
Box; (4) new physics such as a rigorously updated Chapter 33 on cosmology and 
astrophysics to reflect the latest results in today's "Cosmological Revolution"; and 
(5) new applications such as detailed physics-based descriptions of liquid crystal screens 
(LCD), digital cameras (with CCD), and expanded coverage of electrical safety 
and devices. These and other new aspects are highlighted below. 

Phvsics and How to Understand It 



I have avoided the common, dry, dogmatic approach of treating topics formally 
and abstractly first, and only later relating the material to the students' own 
experience. My approach is to recognize that physics is a description of reality and 
thus to start each topic with concrete observations and experiences that students 
can directly relate to. Then we move on to the generalizations and more formal 
treatment of the topic, Not only does this make the material more interesting and 
easier to understand, but it is closer to the way physics is actually practiced. 

A major effort has been made to not throw too much at students reading the 
first few chapters. The basics have to be learned first; many aspects can come later, 
when the students are more prepared. If we don't overwhelm students with too 
much detail, especially at the start, maybe they can find physics interesting, fun, and 
helpful — and those who were afraid may lose their fear. 

The great laws of physics are emphasized by giving them a tan-coloTed screen 
and a marginal note in capital letters enclosed in a rectangle. All important 
equations are given a number to distinguish them from less useful ones, To help 
make cleaT which equations are general and which are not, the limitations of 
important equations are given in brackets next to the equation, such as 



x, 



+ vjj t + j at:. [constant acceleration] 



Malkeinalics can be an obstacle to student understanding. I have aimed at 
including all steps in a derivation. Important mathematical tools, such as addition 



XIV 



of vectors and trigonometry, are incorporated in the text where first needed, so 
Ihey come with a context rather than in a scary introductory Ch a pter, Appendices 
contain a Teview of algebra and geometry (plus a few advanced topics; rotating 
reference frames, inertiai forces, Coriolis effect; heat capacities of gases and 
equipartilion of energy; Lorentz transformations). sSysteme International (SI) units 
are used throughout, Other metric and British units aTe defined for informational 
purposes 

Chapter I is not a throwaway, H is fundamental to physics to realize that every 
measurement has an uncertainty, and how significant figures are used to reflect 
that. Converting units and being able to make rapid estimates are also basic. The 
cultural aspects at the start of Chapter 1 broaden a person's understanding of the 
world hut do not have to be covered in class. 

The many applications sometimes serve only as examples of physical principles. 
Others are treated in depth. They have been carefully chosen and integrated into 
the text So as not to interfere with the development of the physics, but rather to 
illuminate it. To make it easy to spot the applications, a Physics Applied marginal 
note is placed in the margin. 

Color is used pcdagogically to bring out the physics, Different types of vectors 
are given different colors (see the chart on page xxv).This book has been printed 
in 5 colors (5 passes through the presses) to provide better variety and definition 
for illustrating vectors and other concepts such as fields and rays. The photographs 
opening each Chapter, some of which have vectors superimposed on them, have 
been chosen so that the accompanying caption can be a sort of summary of the 
Chapter. 

Some of the new aspects of physics and pedagogy in this sixth edition are: 

Cosmologicui Revolution: The latest results in cosmology and astrophysics are 
presented with the generous help of top experts in the field. We give readers 
the latest results and interpretations from the present, ongoing "Golden Age 
of Cosmology." 

Greater clarity: No topic, no paragraph in this book was overlooked in the 
search to improve the clarity of the presentation. Many changes and clarifi- 
cations have been made, hoth small and not so small. One goal has been to 
eliminate phrases and sentences that may slow down the principle argument: 
keep to the essentials at first, give the elaborations later. 

Vector notation, arrows: The symbols for vector quantities in the text and 4 new 
Figures now have a tiny arrow over them, so they are similar to what a 
professor writes by hand in lecture. The letters are still the traditional boldface: 
thus v for velocity, P for force. 

Exercises within the text, for students to check their understanding. Answers 4 new 
are given at the end of the Chapter, 

Siep-hy Step Examples, after a Problem Solving Rox, as discussed on 4 new 
page xvii. 

Conceptual Examples are not a new feature, but there are some new ones. 

Examples modified: more math steps are spelled out, and many new Examples 
added: see page xvii. 

Page layout: Complete Derivations. Even more than in the previous edition, 
serious attention has been paid to how each page is formatted. Great effort has 
been made to keep important derivations and arguments on facing pages. 
Students then don't have to turn back and forth. Throughout the book readers 
see before them, on two facing pages, an important slice of physics. 

Subheads: Many of the Sections within a Chapter arc now divided into 4NEW 
subsections, thus breaking up the topics into more manageable "bites." They 
allow "pauses" for the students to rest or catch their breath. 



PREFACE xv 



Marginal notes; Caution. Margin notes, in blue, point out main topics acting 
as a sort of outline and as an aid to find topics in review. They also point out 
NEW*- applications and problem -solving hints. A new type, labeled CAUTION, points 

out possible misunderstandings discussed in the adjacent text, 

Deletions. To keep the book from being too long, and also to reduce the burden 
on students in more advanced topics, many topics have been shortened or 
streamlined, and a few dropped. 



New Physics Topics and Major Revisions 

Here is a list of major changes or additions, but there arc many others: 

Symmetry used more, including for solving Problems 
N £ w ► Dimensional analysis, optional (Ch. I ) 

More graphs in kinematics (Ch, 2) 

Engine efficiency (Chs. 6, 15) 

Work -energy principle, and conservation of energy: new subsection (Ch, 6); 
carried through in thermodynamics (Ch. 15) and electricity (Ch. 17) 
n e w *■ FoTce on tennis ball by racket (Ch. 7) 

N E w ► Airplane wings, curve balls, sailboats, and other applications of Bernoulli's 

principle: improved and clarified with new material (Ch. 10) 

Distinguish wave interference in space and in time (beats) (Ch. 1 1 ) 

Dopplcr shift for light (Ch. 12 now, as well as Ch, 33) 
n e w ► Giant star radius (Ch. 14) 

First law of thermodynamics rewritten and extended, connected better to 
work-energy principle and energy conservation (Ch. 15) 

Energy resources shortened (Ch. 15) 
N E w ► SEER rating (Ch. 15) 

N E w .► Separation of charge in nonconductors (Ch, 16) 

n ew ► Gausss law, optional (Ch. 16) 

new*- Photocopiers and computer printers (Ch. 16) 

Electric force and field directions emphasized more (Chs, 16, 17) 

Electric potential related better to work, more detail (Ch. 17) 
n e w * Dielectric effect on capacitor with and without connection to voltage plus 

other details (Ch. 17) 
N E w *■ Parallel-plate capacitor derivation, optional (Ch. 1 7) 

n e w * Electric hazards, grounding, safety, current interrupters: expanded with much 

new material (Chs, 17, 18, 19 especially, 20,21) 
N E w ► Electric current, misconceptions discussed in Chapter 18 

Superconductivity updated (Ch. 18) 

Terminal voltage and emf reorganized, with more detail (Ch. 19) 

Magnetic materials shortened (Ch.20) 
n e w ► Right-hand rules summarized in a Table (Ch. 20) 

Faraday's and Lenz's laws expanded (Ch.2l) 

AC circuits shortened (Ch. 21 ), displacement current downplayed (Ch. 22) 
N E w *■ Radiation pressure and momentum of EM waves (Ch, 22) 

new*- Where to see yourself in a mirror; where you can actually see a lens image (Ch. 23) 

new* Liquid crystal displays (LCD) (Ch, 24) 

N E w *■ Physics behind digital cameras and CCD (Ch. 25) 

NEW*- Seeing under water (Ch. 25) 

Relativistic mass redone (Ch.26) 
new* Revolutionary results in cosmology: flatness and age of universe, WMAR 

SDSS, dark matter, and dark energy (Ch. 33) 
new*- Specific heats of gases, equipartition of energy (Appendix) 



xvi PRMAU 



Problem Solving, with New and Improved Approaches 

Being able to solve problems is a valuable technique in general. Solving problems 
is also an effective; way to understand the physics more deeply. Here are some of 
the ways this book uses to help students become effective problem solvers. 

Problem Solving Boxes f about 20 of them, are found throughout the book 
(there is a list on p. xiii.). Each one outlines a step-by-step approach to solving 
problems in general, or specifically for the material being covered. The 
best students may find these "boxes" unnecessary (they can skip them), but 
many students may find it helpful to be reminded of the general approach and 
of steps they can take to get started. The general Problem Solving Box in 
Section 4-9 is placed there, after students have had some experience wrestling 
with problems, so they may be motivated to read it with close attention. Section 
4-9 can be covered earlieT if desired. Problem Solving Boxes htc not intended 
to be a prescription, but ratheT a guide. Hence they sometimes follow the 
Examples to serve as a summary for future use. 

Problem Solving Sections (such as Sections 2-6, 3-6, 4-7, 6-7, 8-6, and 13-8) 
arc intended to provide extra drill in areas where solving problems is especially 
important, 

Eva mphs: Wo i ked-out Exam pies, each with a title for easy reference, fall into 
four categories; 

( 1 ) The majority are regul ar worked-out Examples that serve as "practice 
problems," New ones have been added, a few old ones have been dropped, 
and many have been reworked lo provide greater clarity, more math steps, 
more of "why we do it this way," and with the new Approach paragraph 
more discussion of the reasoning and approach. The aim is to "think aloud" 
with the Students, leading them to develop insight The level of the Worked- 
out Examples for most topics increases gradually, with the more compli- 
cated ones being on a par with the most difficult Problems at the end of 
each Chapter. Many Examples provide relevant applications to various 
fields and to everyday life. 

(2) Step-hy-step Examples: After many of the Problem Solving Boxes, the 4 new 
next Example is done Step-by-step following the Steps of the. preceding Box, 

just to show students how the Box can be used. Such solutions are long and 
can be redundant, so only one of each type is done in this manner. 

(3) Estimating Examples, roughly 10% of the total, arc intended to 
develop the skills for making ordcr-of-magnitudc estimates, even when the 
data are scarce, and even when you might never have guessed that any 
result was possible at all. See, for example, Section I -7, Examples I -6 to I -9. 

(4) Conceptual Examples? Each is a brief Socratic question intended to 
stimulate student response before reading the Response given. 

APPROACH paragraph: WoTked-out numerical Examples now all have a 4 new 

short introductory paragraph before the Solution, outlining an approach and 

the steps we can take to solve the given problem. 

NOTE: Many Examples now have a brief "note" after the Solution, some- <new 

times remarking on the Solution itself, sometimes mentioning an application, 

sometimes giving an alternate approach to solving the problem, These new 

Note paragraphs let the student know the Solution is finished, and now we 

mention a related issue (s). 

Additional Examples: Some physics subjects require many different worked- 4 new 

out Examples to clarify the issues. Bui so many Examples in a row can be 

overwhelming to some students. In those places, a subhead 'Additional 

Exaniple(s)" is meant to suggest to students that they could skip these in a 

first reading. When students include them during a second reading of the 

Chapter, they can give power to solve a greater range of Problems, 

Exercises within the text, after an Example or a derivation, which give students 4 NEW 

a chance to see if they have understood enough to answer a simple question 

or do a simple calculation. Answers are given at the bottom of the last page 

of each Chapter. PREFACE xvii 



Problems at the end of each Chapter have been increased in quality and quan- 
lily. Some old ones have been replaced or rewritten to make them clearer, 
and/or have had their numerical values changed. Each Chapter contains a 
large group of Problems arranged by Section and graded according to 
(approximate) difficulty: level T Problems are simple, designed to give students 
confidence; level TI are "normal" Problems, providing more of a challenge and 
often the combination of two different concepts; level Til are the most complex 
and are intended as "extra credit" Problems that will challenge even superior 
students. The arrangement by Section number is to help the instructors choose 
which material they want to emphasize, and means that those Problems depend on 
material up to and including that Section: earlier material may also he relied upon. 
General Problems are unranked and grouped together at the end of each 
Chapter, accounting for perhaps 30% of all Problems, These are not neces- 
sarily more difficult, but they may be more likely to call on material from 
earlier Chapters. They are useful for instructors who want to give students a 
few Problems without the clue as to what Section must be referred to or how 
hard they are. 

OuestioaSf also at the cud of each Chapter, arc conceptual. They help students 
to use and apply the principles and concepts, and thus deepen their under- 
standing (or let them know they need to study more). 

Assigning Problems 

T suggest that instructors assign a significant numbei of the level T and level II Problems, 
as well as a small number of General Problems, and reserve level III Problems only as 
"extra credit" to stimulate the best students. Although most level T problems may seem 
easy, they help to build self-confidence — an important part of learning, especially in 
physics. Answers to odd-numbered Problems are given in the back of the book. 

Organization 

The general outline of this new edition retains a traditional order of topics: 
mechanics (Chapters 1 to 9); fluids, vibrations, waves, and sound (Chapters 10 to 12); 
kinetic theory and thermodynamics (Chapters 13 to 1 5); electricity and magnetism 
(Chapters 16 to 22); light (Chapters 23 to 25); and modern physics (Chapters 26 to 
33), Nearly all topics customarily taught in introductory physics courses are 
included here. 

The tradition of beginning with mechanics is sensible because it was devel- 
oped first, historically, and because so much else in physics depends on it. Within 
mechanics, there are various ways to order topics, and this book allows for considerable 
flexibility. I prefer to cover statics after dynamics, partly because many students 
have trouble with the concept of force without motion. Furthermore, statics is a 
special case of dynamics — we study statics so that we can prevent structures from 
becoming dynamic (falling down). Nonetheless, statics (Chapter 9) could be 
covered earlier after a brief introduction to vectors, Another option is light, which 
T have placed after electricity and magnetism and EM waves. But light could be 
treated immediately after waves (Chapter 1 1), Special relativity (Chapter 26) could 
be treated along with mechanics, if desired — say, after Chapter 7. 

Not every Chapter need be given equal weight, Whereas Chapter 4 or Chap- 
ter 2 [ might require I j to 2 weeks of coverage. Chapter 1 2 or 22 may need only 
j week or less, Because Chapter 1 1 covers standing waves, Chapter 12 could be 
left to the students to read on their own if little class time is available. 

The book contains more material than can be covered in most one-year courses. 
Yet there is great flexibility in choice of topics, Sections marked with a star (*) 
are considered optional. They contain slightly more advanced physics material 
(perhaps material not usually covered in typical courses) and/or interesting 
applications.They contain no material needed in later Chapters, except perhaps in 
later optional Sections. Not all unstarred Sections must be covered; there remains 
considerable flexibility in the choice of material. For a brief course, all optional 
material could be dropped, as well as major parts of Chapters 10, 12, 19,22, 28,29, 
32, and 33, and perhaps selected parts of Chapters 7, 8, 9, 15,21,24,25, and 31, 
xviii PREFACE Topics not covered in class can be a resource to students for later study. 



New Applications 



Relevant applications of physics to biology and medicine, as well as to architecture, 
otheT fields, and everyday life, have always been a strong feature of this book, and 
continue to be. Applications are interesting in themselves, plus they answer the 
students' question, "Why must T study physics'?" New applications have been added. 
Here are a few of the new ones (see list after Table of Contents, pages xii 
and xiii). 

Digital cameras, charge coupled devices (CCD) (Ch. 25) 

Liquid Crystal Displays (LCD) (Ch. 24) 

Electric safety, hazards, and various types of current interrupters and circuit 

breakers (Chs. 17,18,19,20,21) 
Photocopy machines (Ch. 16) 
Inkjet and Laser printers (Ch. 16) 
World's tallest peaks (unit conversion, Ch. I ) 
Airport metal detectors (Ch. 21) 
Capacitor uses (Ch. 1 7) 
Underwater vision (Ch. 25) 
SEER rating (Ch. 15) 
Curve ball (Ch. 10) 
Jump starting a car (Ch. 19) 

RC circuits in pacemakers, turn signals, wipers (Ch. 19) 
Digital voltmeters (Ch. 19) 



< A LL 
ARE 
NEW 



Thanks 

Over 50 physics professors provided input and direct feedback on every aspect 
of the text: organization, content, figures, and suggestions for new Examples and 
Problems. The reviewers for this sixth edition are listed below. I owe each of them 
a debt of gratitude: 



Zaven Allounian (McGill University) 

David Amadio (Cypress Falls Senior High School) 

Andrew Bacher (Indiana University) 

Rama Ban si I (Boston University) 

Mitchell C. Be gel man (University of Colorado) 

Cornelius Bennhold (George Washington University) 

Mike Burger (Indiana University) 

George W. Brandenburg (Harvard University) 

Robert Coakley (University of Southern Maine) 

Renee D, Diehl (Penn State University) 

Kalbryn Dimiduk (University of New Mexico) 

Leroy W. Dubcck (Temple University) 

Andrew Duffy (Boston University) 

John J. Dykla (Loyola University Chicago) 

John Essick (Reed College) 

David Faust (Ml. Hood Community College) 

Gerald Feldman {George Washington University) 

Frank A. Ferrone (Drexel University) 

Alex Filippenko (University of California, Berkeley) 

Richard Firestone (Lawrence Berkeley Lab) 

Theodore Gotis (Oakton Community College) 

J Frik HendTickson (University of Wisconsin, Fau Claire) 

Laurent Hodges (Iowa State University) 

Brian Houser (Eastern Washington University) 

Brad Johnson (Western Washington University) 

Randall S. Jones (Loyola College of Maryland) 

Joseph A, Keane (St, Thomas Aquinas College) 

Arthur Kusowsky (Rutgers University) 

Amitabh Lath (Rutgers University) 



Paul L. Lee (California State University, Northridge) 

Jerome R. Long (Virginia Tech) 

Mark Lucas (Ohio University) 

Dan Maclsaac (Northern Arizona University) 

William W MeNairy (Duke University) 

Laszlo Mihaly (SUNY Stony Brook) ' 

Peter J. M oh r (N I ST) 

Lisa K, Morris (Washington State University) 

Paul Morris (Abilene Christian University) 

Hon-Kic Ng (Florida State University) 

Mark Oreglia (University of Chicago) 

Lyman Page (Princeton University) 

Bruce Partridge (Haverford College) 

R, Daryl Pedigo (University of Washington) 

Robert Pelcovits (Brown University) 

Alan Pepper (Campbell School, Adelaide, Australia) 

Kevin T. Pitts (University of Illinois) 

Steven Pollock (University of Colorado, BouldeT) 

W. Steve (Juon (Ventura College) 

Michele Rallis (Ohio State University) 

James J. Rhyne (University of Missouri, Columbia) 

Paul L. Richards (University of California, Berkeley) 

Dennis Rioux (University of Wisconsin, Oshkosh) 

Robert Ross (University of Detroit. Mercy) 

Roy S. Rubins (University of Texas. Arlington) 

Wolfgang Rueckner (Harvard University Fxtension) 

Randall .1. Scalise (Southern Methodist University) 

Arthur G. Schmidt (North western University) 

Cindy Schwarz (Vassar College) 



PREFACE xix 



Bartlett M. Sheinberg (Houston Community College) 

J. L. Shinpaugh (East Carolina University) 

Ross L. Spencer (Brigham Young University) 

Mark Sprague (East Carolina University) 

Michael G, Strauss (University at Oklahoma) 

Chun Fu Su (Mississippi State University) 

Ronald G.Taback (Youngstown State University) 

Leo H.Takahashi (Pennsylvania State University, Beaver) 



Ed Try on (Hunter College, NYC) 

Raymond C Turner (Clemson University J 

Robert C. Webb (Texas A&M University) 

Arthur Wiggins (Oakland Community College) 

Stanley Wojcicki (Stanford University) 

Edward 1. Wright (University of California, Los Angeles) 

Andrzej Zieminski (Indiana University) 



I am grateful also to those other physicist reviewers of earlier editions: 



David B. Aaron (South Dakota State University) 

Narahari Aehar (Memphis State University) 

William T. Achor (Western Maryland College) 

Arthur All (College of Great Falls) 

John Anderson (University of Pittsburgh) 

Sutahash Antani (Edgewood College) 

Atam P. Ai'ya (West Virginia University) 

Sirus Aiyainejad (Eastern Illinois University) 

Charles R, Bacon ( Herri s Slate University) 

Arthur Ballato (Brookhaven National Laboratory) 

David E Baiuion (Chemeketa Community College) 

Gene Barnes (California State University. Sacramento) 

Isaac Bass 

Jacob Becher (Old Dominion University) 

Paul A. Bender (Washington State University) 

Michael S. Berger (Indiana University) 

Donald E. Bowen (Stephen F. Austin University) 

Joseph Boyle (Miami-Dade Community College) 

Peter Brancazio (Brooklyn College. CUNY) 

Michael ¥.. Browne (University of Idaho) 

Michael Broyles (Collin County Community College) 

Anthony Buffa (California Polytechnic State University) 

David Bush ne 1 1 (Northern Illinois University) 

Neal M, Cason (University of Notre Dame) 

H, R, Chandrasekhar (University of Missouri) 

Ram D. Chaudhari (SUNY, Oswego) 

K. Kelvin Cheng (Texas Tech University) 

Lowell O. Christensen (American River College) 

Mark W. Piano ClaTk (Doane College) 

Irvine G. Clator (UNC, Wilmington) 

Albert C. Claus (I .oyola University of Chicago) 

Scott Cohen (Portland State University) 

Lawrence Coleman (University of California. Davis) 

Lattie Collins (East Tennessee State University) 

Sally Daniels (Oakland University) 

Jack E. Denson (Mississippi State University) 

Waren Deshotels (Marquette University) 

Eric Dieti (California State University, Chico) 

Frank Drake (University of California. Santa Cruz) 

Paul Draper (University of Texas. Arlington) 

Miles X Dresser (Washington State University) 

Ryan Droste (The College of Charleston) 

F. Eugene Dunnam (University of Florida) 

Len Feuerhelm (Oklahoma Christian University) 

Donald Foster (Wichita Stale University) 

Gregory E. Francis (Montana State University) 

Philip Gash (California State University, Chico) 

J. David Gavenda (University of Texas. Austin) 

Simon George (California State University, Long Beach) 

James Gerhart (University of Washington) 

Bernard Gerslman (Florida International University) 

Charles Glashausser (Rutgers University) 

Grant W. Hart (Brigham Young University) 

Hershel J. Hatisman (Ohio State University) 

Melissa Hill (Marquette University) 

Mark Hillery (Hunter College) 

Hans Hoehheimer (Colorado State University) 

Joseph M. Hoffman (Frostburg State University) 



Peter H off man-Pi nther (University of Houston. Downtown) 

Alex Holloway (University of Nebraska. Omaha) 

Fred W, Inman (Mankato State University) 

M. Azad Nan (SUNY. Potsdam) 

James P Jacobs (University of Montana) 

Larry D, Johnson (Northeast Louisiana University) 

Gordon Jones (Mississippi State University) 

Rex Joyner (Indiana Institute of Technology) 

Sina David Kaviani (El Camino College) 

Kirby W Kemper (Florida State University) 

Sanford Kern (Colorado State University) 

James E. Kettler (Ohio University. Eastern Campus) 

James R. Kirk (Edinboro University of Pennsylvania) 

Alok Kuman (SUNY, Oswego) 

Sung Kyu Kim (Macalester College) 

Amer Lahamer (Berea College) 

Clement Y. Lam (North Harris College) 

David Lamp (Texas Tech University) 

Peler Landry (McGill University) 

Michael I.ieber (University of Arkansas) 

Bryan H. Long (Columbia State College) 

Michael C. Lo Presto (Henry Ford Community College) 

James Madsen (University of Wisconsin, River Falls) 

Ponn Mahes (Winthrop University) 

Robert H- March (University of Wisconsin, Madison) 

David Markowit? (University of Connecticut) 

Daniel J. McLaughlin (University of Hartford) 

E. R. Menzel (Texas Tech University) 

Robert Messina 

David Mills (College of the Redwoods) 

George K. Miner (University of Dayton) 

Victor Montemeyer (Middle Tennessee State University) 

Marina Morrow (Lansing Community College) 

Ed Nelson (University of Iowa) 

Dennis Nemeschansky (USC) 

Gregor Novak (Indiana University/Purdue University) 

Roy I. Peterson (University of Colorado, Boulder) 

Frederick M. Phelps (Central Michigan University) 

Brian L. Pickering (Laney College) 

T. A. K, Pillai (University of Wisconsin. La Crosse) 

John Polo (Edinboro University of Pennsylvania) 

Michael Ram (University of Buffalo) 

John Reading (Texas A&M University) 

David Reid (Eastern Michigan University) 

Charies Richardson (University of Arkansas) 

William Riley (Ohio State University) 

Larry Rowan (University of North Carolina) 

D. Lee Rutledge (Oklahoma State University) 

Hajime Sakai (University of Massachusetts. Amherst) 

Thomas Sayetta (East Carolina University) 

Neil Schiller (Ocean County College) 

Ann Schmiedekamp (Pennsylvania State University, Ogont?) 

Juergen Schroeer (Illinois State University) 

Mark Semon (Bates College) 

James R Sheerin (Eastern Michigan University) 

Eric Sheldon (University of Massachusetts, Lowell) 

K.Y. Shen (California State University, Long Beach) 

Marc Sher (College of William and Mary! 



xx PREFACE 



Joseph Shinar (Iowa State University) 

Thomas W. Si] is (Wilbur Wright College) 

Anthony A. Siluidi (Kent State University) 

Michael A. Simon (Housatonic Community College) 

Upindranalh Singh (Hmbry-Riddle) 

Michael 1. Sobel (Brooklyn College) 

Donald Sparks (I.oS Angeles Pierce College) 

Thor F, Sinimherg (New Mexico State University) 

James F, Sullivan (University of Cincinnati) 

Kenneth Swinney (Bevill Stale Community College) 

Harold F.Taylor (Stockton State University) 

John R.Teggins (Auburn University at Montgomery) 

Colin Terry (Ventura College) 

Michael Thoennessen (Michigan State University) 

Kwok YeungTsang (Georgia Institute of Technology) 



Jagdish K.Tuli (Rrookhaven National Laboratory) 

Paul Urone (CSU", Sacramento) 

Linn D.Van Woerkom (Ohio State University) 

S, L, Varghese (University of South Alabama) 

fear] WalkeT (Cleveland Slate University) 

Robert A. Walking (University of Southern Maine) 

Jai-Ching Wang (Alabama A&M University) 

Thomas A.Weber (Iowa State University) 

John C, Wells (Tennessee Technological) 

Garelh Williams (San Jose State University) 

Wendall S- Williams (Case Western Reserve University) 

Jerry Wilson (Metropolitan State College at Denver) 

Lowell Wood (University of Houston) 

David Wright (Tidewater Community College) 

Peter Zimmerman (Louisiana State University) 



I owe special thanks to Profs, Bob Davis and J.Erik Hendrickson for much valuable 
input, and especially tot working out all the Problems and producing the Solutions 
Manual with solutions to all Problems and Questions, as well as for providing the 
answers to odd-numbered Problems at the end of this book. Thanks as well to the 
team [hey managed (Profs. David Curotl, Bryan Long, and Richard Louie) who 
also worked out all the Problems and Questions, each checking the others. 

T am grateful to Profs, RobeTt Coakley, Lisa Morris, KalhTyn Dimiduk, Robert 
Pelcovits, Raymond Turner, Cornelius Bennhold, GeTald Feldman, Alan Pepper, 
Michael Strauss, and Zaven Allounian, who inspired many of the Examples, 
Questions, Problems, and significant clarifications. 

Chapter 33 on Cosmology and Astrophysics absorbed more time by far than 
any other Chapter because of the very recent, and ongoing, "revolutionary" results 
that [ wanted to present. I was fortunate to receive generous input from some of 
the top experts in the field, to whom T owe a debt of gratitude: Paul Richards and Alex 
Filippenko (U.C. Berkeley). Lyman Page (Princeton and WMAP), Edward Wright 
(UC.L.A. and WMAP), Mitchell Begelman (U. Colorado), Bruce Partridge 
(Haverford College), Arthur Kosowsky (Rutgers), and Michael Strauss (Princeton 
and SDSS). 

I especially wish to thank Profs. Howard Shugart, Chris McKee, and many 
others at the University of California, Berkeley, Physics Department for helpful 
discussions, and for hospitality. Th an ks also to Prof. Tito Arccchi and others at the 
Istituto Nazionale di Ottica, Florence, Italy, 

Finally, I am most grateful to the many people at Prentice Hall with whom I 
worked on this project, especially to the highly professional and wonderfully ded- 
icated Paul Corey, Karen Karliti, and Susan Fisher. The final responsibility for all 
errors lies with mc. I welcome comments, corrections, and suggestions 1 as soon as 
possible to benefit students for the next reprint, 

D.C.G. 



' Please send t<i: 

email: pliysies.servicet'Jpvreiiliall.coJn 
or by postal service: Physics Udirur 

Prentiec Hall Inc. 

One Lake Street 

Unf-a Saddle River, NJ 07458 



PRFfAC! xxi 



Available Supplements and Media 

MasteringPhysics™ (www.masteringphysks.cum) 

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Sliicienl Pockt-I Companion (II- 13-1 1.15249-7) 

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Mathematics fur College Physics (0-13- 141427-5) 

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XXII 



Instructor's Solutions Manual 

(Volume l:u-I.Vu3S237-3, Volume Il:tt-I3-I41545-X> 

by Bob Davis (Taylor University) and J. Erik Hendrickson (University of Wisconsin- 

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XXIII 



NOTES TO STUDENTS (AND INSTRUCTORS) ON THE FORMAT 

1. Sections marked with a star (*) arc considered optional. They can be omitted without 
interrupting the main flow of topics. No later material depends on them except possibly 
later starred Sections. They may be fun to read. 

2. The customary conventions are used: symbols for quantities (such as m for mass) are 
italicized, whereas units (such as m for meter) are not italicized. Symbols for vectors 
arc shown in boldface with a small arrow above: F. 

3. Few equations are valid in all situations. Where practical, the limitations of important 
equations aTe slated in square brackets next to the equation. The equations that 
represent the great laws of physics are displayed with a tan background, as are a few 
other indispensable equations. 

4. The number of significant figures (Section I -4) should not be assumed to be greater or less 
than given: if a numbeT is slated as (say) 6, with its units.it is mean t to be 6 and not 6.0 ot 6.00. 

5. At the end of each Chapter is a set of Questions that students should attempt to answer 
(to themselves at least). These arc followed by Problems which are ranked as level I, 
II, or III, according to estimated difficulty, with level I Problems being the easiest. Level 
II are normal Problems, and level III are for "extra credit." These ranked Problems are 
arranged by Section, but Problems for a given Section may depend on earlieT material 
as well. There follows a group of General Problems, which are not arranged by Section 
nor ranked as to difficulty. Questions and Problems that relate to optional Sections are 
starred (*), Answers to odd-numbered Problems arc given at the end of the book. 

6. Being able to solve problems is a crucial part of learning physics, and provides a 
powerful means for understanding the concepts and principles. This book contains 
many aids to problem solving: (a) worked-out Examples and their solutions in the text 
(set off with a vertical blue line in the margin) which should be studied as an integral 
part of the text; (b) some of the worked-out Examples arc Estimation Examples, which 
show how rough or approximate results can be obtained even if the given data are 
sparse (see Section 1 -7); (c) special "Problem Solving Boxes"' placed throughout the 
text to suggest a step- by-step approach to problem solving for a particular topic — but 
don't get the idea that every topic has its own "techniques," because the basics remain 
the same; some of these "Boxes" arc followed by an Example that is solved by explic- 
itly following the suggested Steps; (d) Special prohlem-Solving Sections; (e) "Problem 
Solving" marginal notes (see point 9 below) which refer to hints for solving problems 
within the text; (f) Exercises within the text that you should work out immediately, 
and then check your response against the answer given at the bottom of the last page 
of that Chapter; (g) the Problems themselves at the end of each Chapter (point 5 above), 

7. Conceptual Examples are conceptual rather than numerical. Each poses a question ot 
two, which hopefully starts you to think and come up with a response. Give yourself a 
little time to come up with your own response before reading the Response given. 

8. "Additional Examples" subheadings contain Examples that you could skip on a first 
reading, in case you arc feeling overwhelmed. But a day or two later, when you read 
the Chapter a second time, try to work through these Examples too because they can 
give you more poweT in doing a wide range of Problems. 

9. Margin notes: brief notes in the margin of almost every page are printed in blue and 
arc of five types: (a) ordinary notes (the majority) that serve as a sort of outline of the 
text and can help you later locate important concepts and equations: (b) notes that 
refer to the great laws and principles of physics, and these are in capital letters and in 
a box foT emphasis; (c) notes that refer to a problem -solving hint or technique treated 
in the text, and these say "Problem Solving"; (d) notes that refer to an application 
of physics in the text or an Example, and these say "Physics Applied"; (e) "Caution" 
notes that point out a possible misconception spelled out in the adjacent text. 

10, This book is printed in full color — but not simply to make it more attractive. The coIot 
is used above all in the Figures, to give them greater clarity foT our analysis. The Table 
on the next page is a summary of which colors are used for the different kinds of 
vectors, for field lines, and for other symbols and objects. These colors arc used 
consistently throughout the book. 

11, Math review, plus some additional topics, are found in Appendices. Useful data, con- 
version factors, and math formulas are found inside the front and back covers, 



XXIV 



usn or color 



Vcitirrs 



A general vector 

resultant vector (sum) is slightly thicker 

components oi' any vector arc dashed 
Displacement f I>, F> 
Velocity (v) 
Acceleration (it) 
V,:io: (I' I 

Force on second or 

third object in same iigurc ^— ^~ 

MmrrenlniTi Lpnrwrv) 
Angular momentum j L) 
Angular velocity iai) 
Tortiuc (T) 
Electric lield i El 
Magnetic field (B) 



Electricity and magnetism 



Electric circuit symbols 



Electric field lines 

Fqui potential linen 

Magnetic field lines • ■— 

Electric charge (+) * '"' 

Flee trie charge (- J - (VI 




Optics 



Other 



Ljght rays 




Object 


i 


Real image 
(dashed) 


* 

■ 
■ 
■ 


Virtual image 


§ 



(dashed rind paler) '. 



Energy level 
(atom, etc .J 
Measurement lines 

Palh [if a moving 
object 

Direction of motion 
or current 



■I.O111-H 



XXV 




This- computer-enhanced photo of Fiarth was taken from 

about 36,000 kin away. North and South America are 

clearly visible below the clouds. From this 

distance the "sky" is black'. (Why we see blue 

sky from Farth is discussed in Chapter 24.) 

We start this Chapter by learning some 

basics about science and its theories, 

and about measurement and units. 

We also learn how to make quick 

estimates. 



CHAPTER 



1 



Introduction, 
Measurement Estimating 



Physics is the most basic of the sciences. It deals with the behavior and 
structure of matter. The field of physics is usually divided into classical 
physics which includes motion, fluids, heat, sound, light, electricity, and 
magnetism; and modem physics which includes the topics of relativity, atomic 
structure, condensed matter, nuclear physics, elementary particles, and cosmology 
and astrophysics. We will coveT all these topics in this book, beginning with 
motion (or mechanics, as it is often called). But before we begin on the physics 
itself, we take a brief look at how this overall activity called "science," including 
physics, is actually practiced. 

The Nature of Science 

The principal aim of all sciences, including physics, is generally considered lo be 
the search for order in our observations of the world around ns. Many people 
think that science is a mechanical process of collecting facts and devising theo- 
ries. But it is not so simple. Science is a creative activity that in many respects 
resembles other creative activities of the human mind. 



FIGURE 1-1 Aristotle is the 
central figure (dressed in blue) at 
the top of the stairs (the figure next 
ti.i him is Plato) in this famous 
Renaissance portrayal of The 
School of Athens, painted by 
Raphael around 15 10- Also in this 
painting, considered one of the 
great masterpieces in art, are Fuelid 
(drawing a circle at the lower right), 
Ptolemy (extreme right with globe), 
Pythagoras. Socrates, and Diogenes. 




Observation ami experiment 



Motion is as natural as ri'st 



Theories 



Testing a theory 



Theory acceptance 



One important aspect of science is observation of events, which includes the 
design and carrying out of experiments, But observation requires imagination, 
for scientists can never include everything in a description of what they observe. 
Hence, scientists must make judgments about what is relevant in their observa- 
tions and experiments. Consider, for example, how two great minds, Aristotle 
(384-322 B.C.; Fig. 1-1) and Galileo (1564-1642; Fig, 2-17), interpreted motion 
along a horizontal surface. Aristotle noted that objects given an initial push 
along the ground (or on a tabletop) always slow down and stop. Consequently, 
Aristotle argued that the natural state of an object is at rest. Galileo, in his reexam- 
ination of horizontal motion in the early 1600s, imagined that if friction could be 
eliminated, an object given an initial push along a horizontal surface would 
continue to move indefinitely without stopping, He concluded that for an object to 
be in motion was just as natural as for it to be at rest. Ry inventing a new approach, 
Galileo founded our modern view of motion (Chapters 2, 3, and 4), Galileo made 
this intellectual leap conceptually, without actually eliminating friction. 

Observation, with careful experimentation and measurement, is one side of 
the scientific process. The other side is the invention or creation of theories to 
explain and OTder the observations. Theories are never derived directly from 
observations, Observations may help inspire a theory, and theories are accepted 
or rejected based on observation and experiment. 

Theories are inspirations that come from the minds of human beings. For 
example, the idea that matter is made up of atoms (the atomic theory) was not 
arrived at by direct observation of atoms — we can't see atoms directly. Rather, 
the idea sprang from creative minds. The theory of relativity, the electromag- 
netic theory of light, and Newton's law of universal gravitation were likewise 
the result of human imagination. 

The great theories of science may he compared, as creative achievements, 
with great works of art or literature, But how does science differ from these other 
creative activities? One important difference is that science requires testing of its 
ideas or theories to sec if their predictions are borne out by experiment, Btit 
theories are not "proved" by testing, First of all, no measuring instrument is 
perfect, so exact confirmation cannot be possible. Furthermore, it is not possible 
to test a theory for every possible set of circumstances, Hence a theory can 
never be absolutely "proved,"' Indeed, the history of science tells us that long- 
held theories arc sometimes replaced by new ones. 

A new theory is accepted by scientists in some cases because its predictions 
are quantitatively in better agreement with experiment than those of the older 



2 CHAPTER 1 Introduction, Measurement, Estimating 




FIGURE 1 -2 (a) Ptolemy's geocentric view of the universe. Note at the center the four elements of the 
ancients: Earth, water, air (clouds around the Earth), and fire; then the circles, with symbols, for the Moon, 
Mercury, Venus, Sun, Mars, Jupiter, Saturn, the fixed stars, and the signs of the zodiac (b) An early repre- 
sentation of Copemicus's heliocentric view of the universe with the Sun at the center, (See Chapter 5.) 



theory. But in many cases, a new theory is accepted only if it explains a greater 
Hinge of phenomena than docs the older one. Copemicus's Sun-centered theory 
of the universe (Fig. 1— 2b), for example, was originally no more accurate than 
Ptolemy's Earth-centered theory (Fig. I -2a) for predicting the motion of heav- 
enly bodies (Sun, Moon, planets). But Copemicus's theory had consequences 
that Ptolemy's did not, such as predicting the moon like phases of Venus. A 
simpler and richer theory, one which unifies and explains a greater variety of 
phenomena, is more useful and beautiful to a scientist. And this aspect, as well 
as quantitative agreement, plays a major role in the acceptance of a theory. 

An important aspect of any theory is how well it can quantitatively predict 
phenomena, and from this point of view a new theory may often seem to be only 
a minor advance over the old one. For example, Einstein's theory of relativity 
gives predictions that diffeT very little from the older theories of Galileo and 
Newton in nearly all everyday situations, fts predictions are better mainly in the 
extreme case of very high speeds close to the speed of light. But quantitative 
prediction is not the only important outcome of a theory, Out view of the world 
is affected as well. As a result of Einstein's theory of relativity, for example, our 
concepts of space and lime have been completely altered, and we have come to 
see mass and energy as a single entity (via the famous equation F. = mc 2 ). 

Physics and its Relation to Other Fields 

For a long time science was more or less a united whole known as natural 
philosophy. Not until a century or two ago did the distinctions between physics 
and chemistry and even the life sciences become prominent. Indeed, the sharp 
distinction we now sec between the arts and the sciences is itself but a few 
centuries old, It is no wonder then that the development of physics has both 
influenced and been influenced by other fields. For example, the notebooks 
(Pig. 1-3) of Leonardo da Vinci, the great Renaissance artist, researcher, and 
engineer, contain the first references to the forces acting within a structure, a 
subject we consider as physics today; but then, as now, it has great relevance to 
architecture and building. 



FIGURE 1-3 Studies on the forces 

in structures by l.eonardu da Vinci 
(1452-1519). 



■■> e a- 

h ' X 


"1 





> a ■ n 

£7 «^ 




-a, 
u 



SECTION 1-2 Physics and its Relation to Other Fields 3 




FIGURE 1-4 fa) '["his Roman aqueduct was built 2000 years ago and still stands, (b) Collapse of the 
Hartford Civic Center in 1 978, just two years after it was built. 



Physics applies la many fields 



Models 



Early work in electricity that led to the discovery of the electric battery and 
electric current was done by an eighteenth -century physiologist, Luigi Galvani 
< 1737— 1 798). He noticed the twitching of frogs' legs in response to an electric 
spark and later that the muscles twitched when in contact with two dissimilar 
metals (Chapter IS). At first this phenomenon was known as "animal elec- 
tricity," but it shortly became clear that electric current itself could exist in the 
absence of an animal. 

Physics is used in many fields, A zoologist, for example, may find physics 
useful in understanding how prairie dogs and other animals can live underground 
without suffocating. A physical therapist will do a more effective job if aware of 
the principles of center of gravity and the action of forces within the human body, 
A knowledge of the operating principles of optical and electronic equipment is 
helpful in a variety of fields. Life scientists and architects alike will be interested 
in the nature of heat loss and gain in human beings and the resulting comfort or 
discomfort. Architects themselves may not have to calculate, for example, the 
dimensions of the pipes in a heating system or the forces involved in a given 
structure to determine if it will remain standing (Fig. 1-4). But architects must 
know the principles behind these analyses in order to make realistic designs and 
to communicate effectively with engineering consultants and other specialists, 
From the aesthetic or psychological point of view, too, architects must be aware 
of the forces involved in a structure — for instability, even if only illusory, can be 
discomforting to those who must live or work in the structure. 

The list of ways in which physics relates to other fields is extensive, Tn the 
Chapters that follow we will discuss many such applications as we carry out out 
principal aim of explaining basic physics- 

Models, Theories, and Laws 

When scientists are trying to understand a particular set of phenomena, they 
often make use of a model. A model, in the scientific sense, is n kind of analogy 
or mental image of the phenomena in terms of something else we are already 
familiar with. One example is the wave model of light. We cannot see waves of 
light as we can water waves, But it is valuable to think of light as if it were made 
up of waves, because experiments indicate that light behaves in many respects 
as water waves do. 



4 CHAPTER 1 Introduction, Measurement, Estimating 



The purpose of a model is to give us an approximate mental ot visual 
picture — something to hold onto — when we cannot see or understand what 
actually is happening. Models often give us a deepeT understanding; the analogy 
to a known system (for instance, water waves in the above example) can suggest 
new experiments to perform and can provide ideas about what oilier related 
phenomena might occur. 

You may wonder what the difference is between a theory and a model. 
Usually, a model is relatively simple and provides a structural similarity to the 
phenomena being studied. A theory is broader, more detailed, and can give 
quantitatively testable predictions, often with great precision. It is important, 
however, not to confuse a model or a theory with the real system or the 
phenomena themselves. 

Scientists give the title law to certain concise but general statements about 
how nature behaves (that energy is conserved, for example). ,Sometimes the 
statement takes the form of a relationship or equation between quantities (such 
as Newton's second law. F - ma). 

To be called a law, a statement must be found experimentally valid over a 
wide range of observed phenomena, For less general statements, the term 
principle is often used (such as Archimedes' principle). 

Scientific laws are different from political laws in that the latter are 
prescriptive: they tell us how we ought to behave. Scientific laws arc descriptive: 
they do not say how nature should behave, but rather arc meant to describe 
how nature does behave. As with theories, laws cannot be tested in the infinite 
variety of cases possible. So wc cannot be sure that any law is absolutely true. 
We use the term "law" when its validity has been tested over a wide range of 
cases, and when any limitations and the range of validity are clearly understood. 

Scientists normally do their work as if the accepted laws and theories were 
true. But they are obliged to keep an open mind in case new information should 
alter the validity of any given law or theory. 



Theories (vs. models) 



Linrs 



;n ui 



principles 




Measurement and Uncertainty; 
Significant Figures 



Tn the quest to understand the world around us, scientists try to work out rela- 
tionships among physical quantities that can be measured. 

Uncertainty 

Accurate, precise measurements are an important part of physics. But no 
measurement is absolutely precise. There is an uncertainty associated with every 
measurement. Among the most important sources of uncertainty, other than 
blunders, are the limited accuracy of every measuring instrument and the 
inability to read an instrument beyond some fraction of the smallest division 
shown. For example, if you were to use a centimeter ruler to measure the width 
of a board (Fig. 1-5), the result could be claimed to be preeise to about 0.1 cm 
(I mm), the smallest division on the ruler, although half of this value might be a 
valid claim as well. The reason for this is that it is difficult for the observer to 
estimate between the smallest divisions. Furthermore, the ruler itself may not 
have been manufactured to an accuracy any better than this,' 1 

'There is a technical difference hetvvcen "precision" and "accuracy." Precision in a strict sense refers 
to the repeatability of [he measurement using a given instrument. For example, if yoj measure the 
width of a board many limes, gelling results like 8.81 cm, 8.85 cm. 8,78cm, 8_82cm (estimating 
between the 0.1-em marks as best as possible each time J, you could say rhe measurements give a 
prexinon a bit better than (J.I em. Accuracy refers to how close a measurement is to the true value. 
For example, if the ruler shown in Fig. I 5 was manufactured with a 2'A. error, the accuracy of its 
measurement of the board's width (about K.8 cm) would be about 2'X of ti.8em, or about ± 0.2 cm. 
Estimated uncertainty is meant to take both accuracy and precision into account. 



Every measurement has an uncertainty 

FIGURE 1-5 Measuring the width 
of a board with a centimeter ruler. 
Accuracy is about ± I mm. 




SECTION 1-4 Measurement and Uncertainty; Significant Figures 5 



Stating the uncertainly 



Assumed uncertainty 



When giving the Tesult of a measurement, it is important to state the esti- 
mated uncertainty in the measurement. For example, the width of a board might 
be written as 8.8 ± 0.1 cm. The ± 0.1 cm ("plus or minus 0.1 cm") represents the 
estimated uncertainty in the measurement, so thai the actual width most likely 
lies between 8.7 and 8.9 cm. The percent uncertainty is simply the ratio of the 
uncertainty to the measured value, multiplied by 100- For example, if the measure- 
ment is 8.8 and the uncertainty about 0.1 cm, the percent uncertainty is 



S* l(10 « 



I 1 : 



where = means "is roughly equal to." 

Often the uncertainty in a measured value is not specified explicitly. Tn such 
cases, the uncertainty is generally assumed to be one or a few units in the last 
digit specified. For example, if a length is given as 8.3 em, the uncertainty is 
assumed to be about 0. 1 em or 0.2 cm, It is important in this case that you do not 
write 8.80 cm. for this implies an uncertainty on the order of 0.01 cm; it assumes 
that the length is probably between 8.79 cm and 8.81 em, when actually you 
believe it is between 8.7 and 8.9 cm. 



Which digits are significant? 



*» PROBLEM SOLVING 

Number of significant figures in final 

result should hi' same as least 

significant input value 



6 CHAPTER 1 



CONCEPTUAL EXAMPLE 1-1 | Is the diamond yours? A friend asks to 
borrow your precious diamond for a day to show her family. You are a bit 
worried, so you carefully have your diamond weighed on a scale which reads 
8. 17 grams. The scale's accuracy is claimed to be ± 0,05 gram. The next day you 
weigh the returned diamond again, getting 8.09 grams, Ts this your diamond? 

RESPONSE The scale readings are measurements and do not necessarily give 
the "true" value of the mass. Each measurement could have been high or low 
by up to O.OSgram or so, The actual mass of your diamond lies most likely 
between 8. 12 grams and 8.22 grams. The actual mass of the returned diamond 
is most likely between 8.04 grams and 8. 14 grams, These two ranges overlap, so 
there is not a strong reason to doubt that the returned diamond is yours, at 
least based on (he scale readings. 



Significant Figures 

Tie number of reliably known digits in a number is called the number of 
significant figures- Thus there are four significant figures in the number 23.21 cm 
and two in the number 0,062cm (the zeros in the latteT are merely place holders 
that show where the decimal point goes). The number of significant figures may 
not always be clear. Take, for example, the number 80, Are there one ot two 
significant figures? If we say it is about 80 km between two cities, there is only- 
one significant figure (the 8) since the zero is merely a place holder, Tf it is 
exactly 80 km within an accuracy of 1 or 2 km, then the 80 has two significant 
figures." If it is precisely 80 km, to within ± 0.1 km, then we write 80.0 km. 

When making measurements, or when doing calculations, you should avoid 
the temptation to keep more digits in the final answer than is justified. For 
example, to calculate the area of a rectangle 11.3 cm by 6.8 cm, the result of 
multiplication would be 76.84 cm 2 . But this answer is clearly not accurate to 
0.01 enr, since (using the outer limits of the assumed uncertainty for each 
measurement) the result could be between 1 1 .2 cm X 6.7 cm = 75.04 cm 2 and 
11.4 cm X 6.9 cm = 78.66 cm 2 . At best, we can quote the answer as 77 cm 2 , 
which implies an uncertainty of about I or 2 cm 2 . The other two digits (in the 
number 76.84 enr) must be dropped since they arc not significant, As a rough 
general rule (i.e., in the absence of a detailed consideration of uncertainties), we 
can say that the final result of a multiplication or division should have only as 
many digits us the number with the least number of significant figures used in the 
calculation, In our example, 6,8cm has the least number of significant figures, 
namely two. Thus the result 76,84 cm 2 needs to be rounded off to 77 cm 2 . 

f ff the SO has two significant figures, some people prefer to write it HO., with a decimal point. This is 
not usually done, so the number of significant figures in fit) can he ambiguous unless something is 
said about it suth as "about" (meaning &i ± 10), or "very nearly" or "pretistly" (meaning 80 + 1). 



I EXERCISE A The area of a rectangle 4.5 cm by 3-25 cm is correctly given by 
| («) 14.625 cittj (b) 14.63 enr; (c) 14.6 cm 3 ; (d) 15 cm'. 

When adding or subtracting numbers, the final result is no more accurate 
than the least accurate number used. For example, the result of subtracting 0.57 
from 3.6 is 3.0 (and not 3.03). 

Keep in mind when you use a calculator that all the digits it produces may 
not be significant. When you divide 2,0 by 3.0, the proper answer is 0.67, and not 
some such thing as 0.666666666. Digits should not be quoted in a result, unless 
they are truly significant figures. However, to obtain the most accurate result, 
you should normally keep one or more extra significant figures throughout a 
calculation, and round off only in the final result, (With a calculator, you can 
keep all its digits in intermediate results,) Note also that calculators sometimes 
give too few significant figures. For example, when you multiply 2.5 X 3.2, a 
calculator may give the answer as simply S. But the answer is good to two 
significant figures, so the proper answer is 8.0. See Fig. 1-6. 

| EXERCISE B Do 0.00324 and 0,00056 have the same number of significant figures? 

Be careful not to confuse significant figures with the number of decimal places. 

EXERCISE C For each of the following numbers, state the number of significant 
figures and the- number of decimal places: («) 1 .23: (b) 0.12;!; (<:) 0.0123. 



CONCEPTUAL EXAMPLE 1-2 | Significant figures. Using a protractor 
(Fig, I -7), you measure an angle to be 30", (a) How many significant figures 
should you quote in this measurement? (b) Use a calculator to find the cosine 
of the angle you measured. 

RESPONSE (a) Tf you look at a protractor, you will see that the precision with 
which you can measure an angle is about one degree (certainly not 0.I 3 ). So you 
can quote two significant figures, namely, 30 u (not 30.0"). (b) Tf you enteT cos 30° in 
your calculator, you will get a number like 0.866025403, However, the angle you 
entered is known only to two significant figures, so its cosine is coTTeetly given 
by 0.87; i.e., you must round your answer to two significant figures. 
NOTE We discuss trigonometric functions like cosine in Chapter 3. 



Scientific Notation 

We commonly write numbers in "powers of ten,' or "scientific" notation — for 
instance 36,900 as 3.69 X 10*, or 0.0021 as 2.1 X 10 \ One advantage of scien- 
tific notation (discussed in Appendix A) is that it allows the number of significant 
figures to be clearly expressed. For example, it is not clear whether 36,900 has 
three, four, or five significant figures. With powers of ten notation the ambiguity 
can be avoided: if the number is known to an accuracy of three significant 
figures, we write 3.69 x 10*, but if it is known to four, we write 3. 690 x 10*. 

* Percent Error 

The significant figures rule is only approximate, and in some cases may under- 
estimate the precision of the answer, Suppose for example we divide 97 by 92: 
97 

Both 97 and 92 have two significant figures, so the rule says to give the answer as 1.1. 
Yet the numbers 97 and 92 both imply an uncertainty of ± I if no other uncer- 
tainty is stated. Now 92 ± I and 97 ± 1 both imply an accuracy of about \% 
(1/92 = 0.01 = 1%). But the final result to two significant figures is 1.1, with an 
implied uncertainty of ±0.1, which is an uncertainty of 0,1/1.1 ~ 0.1 ~ 10%. 
In this case it is better to give the answer as 1.05 (which is three significant 
figures), Why'.' Because 1.05 implies an uncertainty of ±0.01 which is 
0.01/1.05 =0.01 ~ |%, just like the uncertainty in the original numbers 92 
and 97. 

SUGGESTION: Use the significant figures rule, but consider the % uncertainty 
too, and add an extra digit if it gives a more realistic estimate of uncertainty. 



,05 * II. 



CAUTION 



Calculators err with significant figures 

*+ PROBLEM SOLVING 
Report only the proper number of 
significant figures in the final result 
Keep extra digits during 
the calculation. 




si ^j taW 

noiiEica 

DQQQI3 
DBQB1 

DDian 




■ p ld'=i 

B El SI Et E3 

^□[3 2) O CD 

DDDBD 

□ b a a : ! 
Baa 3 3 ' 

B B ■ a 2 

(a) (b) 

FIGURE 1-6 'Ihcsc two calculators 
show the wrong number of significant 
figures. In (a), 20 was divided by 3-0- 
The correct final result would be 0.67. 
In (h), 2.5 was multiplied by 3.2. The 
correct result is 8.0. 

FIG U RE 1 -7 Ex ample 1 -2. A 
protractor used to measure an angle. 

m 




SECTION 1-4 Measurement and Uncertainty; Significant Figures 7 



Standard of length {meter) 



FIGURE 1-8 Some lengths: 

(a) viruses; (about lfT 7 m long) 
attacking a cell: (b) Ml- Everest's 
height is on the order of 10* m 

(885U m. to lit precise). 




Units, Standards, and the SI System 

The measurement of any quantity is made relative to a particular standard or 
unit, and this unit must be specified along with the numerical value of the quan- 
tity. FOr example, we can measure length in units such as inches, feet, or miles, or 
in the metric system in centimeters, meters, or kilometers. To specify that the 
length of a particular object is 18.6 is meaningless. The unit must be given; for 
clearly, 18.6 meters is very different from 18.6 inches or 18.6 millimeters. 

For any unit we use, such as the meter for distance or the second for time, 
we need to define a slarjdarrJ which defines exactly how long one meter or one 
second is, It is important that standards be chosen that arc readily reproducible 
so that anyone needing to make a very accurate measurement can refer to the 
standard in the laboratory. 

Length 

The first truly international standard was the meter (abbreviated m) established 
as the standard of length by the French Academy of Sciences in the 1 790s. The 
standard meter was originally chosen to be one ten-millionth of the distance from 
the Earth's equator to either pole," and a platinum rod to represent this length 
was made. (One meter is, very roughly, the distance from the tip of your nose to 
the tip of your finger, with arm and hand stretched out to the side.) In 1889, the 
meter was defined more precisely as the distance between two finely engraved 
marks on a particular bar of platinum-iridium alloy. In I960, to provide greater 
precision and reproducibility, the meter was redefined as 1,650,763.73 wave- 
lengths of a particular orange light emitted by the gas krypton-86. In 1983 the 
meter was again redefined, this time in terms of the speed of light (whose best 
measured value in terms of the older definition of the meter was 299,792,458 m/s, 
with an uncertainty of I m/s). The new definition reads: "The meter is the length 
of path traveled by light in vacuum during a time interval of 1/299,792,458 of a 
second." % 

British units of length (inch, foot, mile) are now defined in terms of the 
meter. The inch (in,) is defined as precisely 2,54 centimeters (cm; 1 cm - 0,01 m). 
Other conversion factors are given in the Table on the inside of the front cover 
of this book. Table 1-1 presents some typical lengths, from very small to very 
large, rounded off to the nearest power of ten. See also Fig. I -8. [Note that the 
abbreviation for inches (in.) is the only one with a period, to distinguish it from 
the word "in".! 



TABLE 1-1 Some Typical L 


engths or 


Distances 


[order of magnitude) 


Length (or Distance) 




Meters (approximate) 


Neutron or proton (radius) 






10 ^ 


m 


Atom 






ln -iu 


m 


Virus [see Fig.. 1-Ra] 






lO" 7 


in 


Sheet of paper ( thickness) 






1(T 4 


m 


Finger width 






irr 2 


m 


Football field length 






10* 


m 


Height of Ml Everest [see Fig. 1- 


-Sb| 




10 4 


in 


Earth diameter 






10 7 


m 


Earth to Sun 






10 11 


m 


Earth to nearest star 






10" 


in 


Earth to nearest galaxy 






10" 


m 


Earth to farthest galaxy visible 






10 26 


in 



'Modern measurements of the Earth's circumference reveal thai [he intended length is off by about 
one-fiftieth of IX Not bud! 

? 'l"he new definition of the meter has the effect of giving the speed of light the exact value of 
2W.792.45 8 m/s. 



8 CHAPTER 1 Introduction, Measurement, Estimating 



TABLE 1-2 Some Typical Tim 


e Intervals 




Time Interval 




Seconds {approximate) 


Lifetime of very unstable subatomic 


particle 


irr z 


l S 


Lifetime of radioactive elements 




Ltr z 


L StC.10 2S 5 


Lifetime of muofl 




iir h 


s 


Time between human heartbeats 




I0 (h 


s(=ls) 


One day 




to 5 


s 


One year 




3 x It) 7 


s 


Human life span 




2 X ll/ h 


s 


Length of recorded history 




10" 


s 


Humans on Earth 




I0 14 


s 


Life on Earth 




It) 17 


s 


Age of Universe 




I0 ls 


s 



TABLE 1-3 Some 


Masses 




Objeel 


Kih 


jgrams (approximate) 


Electron 






10" 3 " kg 


Proton, neutron 






10~ 27 kg 


DNA molecule 






KT 17 kg 


Bacterium 






10-" kg 


Mosquito 






i<r 5 k g 


Plum 






10 ' kg 


Human 






10 2 kg 


Ship 






10 s kg 


Earth 




6 x 10 24 kg 


Sun 




2 x 


10» kg 


Galaxy 






10 41 kg 



Time 

The standard unit of time is the second (s). For many years, the second was 
defined as I /86,400 of a mean solar day. The standard second is now defined 
more precisely in terms of the frequency of radiation emitted by cesium atoms 
when they pass between two particular states. [Specifically, one second is 
defined as the time required for 9,192,631,770 periods of this radiation. | There are, 
by definition, 60 s in one minute (min) and 60 minutes in one hour (h). Table 1 -2 
presents a range of measured time intervals, rounded off to the nearest power of ten. 

Mass 

The standard unit of mass is the kilogram (kg), The standard mass is a particular 
platinum-indium cylinder, kept at the International Bureau of Weights and 
Measures near Paris, France, whose mass is defined as exactly 1 kg. A range of 
masses is presented in Table 1-3. [For practical purposes, 1 kg weighs about 
2.2 pounds on Earth.] 

When dealing with atoms and molecules, we usually use the unified atomic 
mass unit (u). In terms of the kilogram, 

lu = 1.6605 X l(T 2T kg. 

The definitions of other standard units for other quantities will be given as 
we encounter them in later Chapters. 

Unit Prefixes 

In the metric system, the larger and smaller units are defined in multiples of 10 
from the standard unit, and this makes calculation particularly easy. Thus 
1 kilometer (km) is 1000 m, 1 centimeter is Tro m > I millimeter (mm) is ui^ni or 
37} cm, and so on. The prefixes "centi," "kilo," and others are listed in Table 1-4 
and can be applied not only to units of length, but to units of volume, mass, or 
any other metric unit. For example, a centiliter (cL) is 755 liter (L), and a kilo- 
gram (kg) is 1 000 grams (g). 

Systems of Units 

When dealing with the laws and equations of physics it is very important to use a 
consistent set of units. Several systems of units have been in use over the years. 
Today the most important is the Systeme International (French for International 
System), which is abbreviated SI. In ST units, the standard of length is the meter, 
the standard for time is the second, and the standard for mass is the kilogram. 
This system used to be called the MKS (meter-kilogram-second) system. 

A second metric system is the cgs system, in which the centimeter, gram, 
and second are the standard units of length, mass, and time, as abbreviated in 
the title. The British engineering system takes as its standards the foot for 
length, the pound for force, and the second for time, 

SI CIIJN 1-5 Units, Standards, and :he SI System 



TABLE 1-4 
Metric (SI) Prefi 


xes 




Prefix 


Abbrevi 


Btfon 


Value 


yotta 




Y 




lfJ 24 


zetta 




Z 




10 21 


exa 




F 




10 1S 


peta 




P 




10 15 


Cera 




"1 




10 12 


giga 
mega 




G 




10' 


M 




10* 


kilo 




k 




id 3 


hecto 




h 




10 2 


dc-ka 




da 




HI 1 


deci 




d 




10" 1 


centi 




c 




HI" 2 


milli 




m 




10" 3 


micro' 




^ 




10 -5 


nana 




n 




10-* 


pi to 




P 




10" 12 


fern to 




f 




10 -15 


alto 




a 




10" ls 


zepto 




Z 




10" 21 


yocto 




y 




10" 24 


' ti is the CjcC'Ck 1 


ettei ' 


'ran." 




^* PRO 


B t 1 


■ M 


5 O 


t V 1 N G 



Always use a consistent set of units 



SI units 



TABLE 1-5 
and Units 


SI Base Quantities 


Quantity 


Unit 


L'nit 

A 1 1 HV- 

viatiun 


Length 


meter 


m 


Time 


second 


s 


Vl.iss 


kilogram 


kg 


Flcctric 
current 


ampere 


A 


Temperature 


fcelvin 


K 


Amount 
of substance mole 


mol 


Luminous 
in tensity 


candela 


cd 



SI units are the princip;il ones used today in scientific wot k. We will there- 
fore use SI units almost exclusively in this book, although we will give the cgs 
and British units for various quantities when introduced. 

Base vs. Derived Quantities 

Physical quantities can be divided into two categories: base quantities and 
derived quantities. The corresponding units for these quantities are called base 
units and derived units. A base quantity must be defined in terms of a standard. 
Scientists, in the interest of simplicity, want the smallest number of base quanti- 
ties possible consistent with a full description of the physical world. This number 
turns out to be seven, and those used in the SI arc given in Table I -5. All other 
quantities can be defined in ternvs of these seven base quantities, 7 and hence arc 
referred to as derived quantities, An example of a derived quantity is speed, 
which is defined as distance divided by the time it takes to travel that distance. A 
Tabic inside the front cover lists many derived quantities and their units in terms 
of base units. To define any quantity, whether base or derived, we can specify a 
rule or procedure, and this is called an operational definition. 



Converting Units 



Any quantity we measure, such as a length, a speed, or an electric current, 
consists of a number and a unit. Often we are given a quantity in one set of 
units, but we want it expressed in another set of units. For example, suppose we 
measure that a table is 21.5 inches wide, and we want to express this in centi- 
meters, Wc must use a conversion factor, which in this case is 



FIGURE 1-9 the world's second 
highest peak, K2. whose summit is 
considered the most difficult of the 
8,000-ers, K2 is seen here from the 
north (China). Our cover shows 
K2 from the south (Pakistan). 
Example 1-3. 




(%) PH 



YSICS APPLIED 



The world 's tallest peaks 



1 in. = 2.54 cm 
or, written another way, 

1 = 2.54 cm/in. 

Since multiplying by one does not change anything, the width of our table. 

in cm, is 

21.5 inches - (21.5%*.} X (2,54-^j - 54.6 cm. 

Note how the units (inches in this case) cancelled out, A Table containing 
many unit conversions is found inside the front cover of this book. Let's take 
some Examples- 



EXAMPLE 1-3 



The 8000-m peaks. The fourteen tallest peaks in the world 
(Fig. 1-9 and Table 1-6) are referred to as "eight-thousanders," meaning their 
summits are over 8000 m above sea level. What is the elevation, in feet, of an 
elevation of 8000 m? 

APPROACH We need simply to convert meters to feet, and we can start with 
the conversion factor I in. = 2.54 cm, which is exact. That is, I in. = 2.5400 cm 
to any number of significant figures. 

SOLUTION One foot is 12 in,, so we can write 

1ft = (12^)|2,54^M = 30.48 em = 03048 m. 
The units cancel (colored slashes), and our result is exact. We can rewrite this 



''ll'ic only exceptions arc for angle (radians — see Chapter 8) and solid angle (sreradian). No general 
agreement lias been reached as to whether these are hase or derived quantities. 



10 CHAPTER 1 Introduction, Measurement, Estimating 



equation to find the numbeT of feet in I meter: 
I ft 



1 m 



3.28084 ft. 



0.3048 
We multiply this equation by 8,000.0 (to have five significant figures); 

8,000.0 m = (8,000.0 m.)( 3.28084 — J = 26,247 ft. 

An elevation of 8000 m is 26.247 ft above sea level. 
NOTE We could have done the conversion all in one line: 



8000 m = (8000 ~m.) 



I in^ \ { I ft \ 



100 JHT ' 

IX j\ 2,54 jsnr/ 1.12 -rn^j 



26,247 ft. 



The key is to multiply conversion factors, each equal to one (- 1.0000), and to 
make sure the units cancel. 



EXERCISE D There are only 14 eight-thousand-meter peaks in the world (see 
Example 1-3) and their names and elevations are given in Table 1-6. They are all in 
the Himalaya mountain range in India, Pakistan. Tibet, and China. Determine the 
elevation of the world's three highest peaks in feet. 



EXAMPLE 1-4 



Area of a semiconductor chip. A silicon chip has an area 
of 1. 25 square inches. Express this in square centimeters. 

APPROACH We use the same conversion factor, 1 in. = 2.54 cm, but this 
time we have to use it twice. 

SOLUTION Because 1 in. = 2.54 cm, then I in. 2 = (2.54 cm ) : = 6.45 cm 2 . So 

1.25 in. 2 - (1.25 in. 2 ) (z54 ^V - (l.25W)(fi.4S^) - S.06cm 2 . 



EXAMPLE 1-5 



Speeds. Where the posted speed limit is 55 miles per hour 
(mi/h or mph), what is this speed (a) in meters per second (m/s) and (b) in 
kilometers per hour (km/h)? 

APPROACH Wc again use the conversion factor I in, =2.54 cm, and we 
recall that there arc 5280 ft in a mile and 12 inches in a foot; also, one hour 
contains (60min/h) X (60s/min) - 3600 s/h. 

SOLUTION (a) We can write 1 mile as 



1 mi = (5280 if) 12 



"m. 



2.54 



jaBtf\ 



\ 1 00 .CRT j 



1609 m. 



if A "hwl . IOOjbrt. 
Note that each conversion factor is equal to one. We also know that I hour 
contains 3600 s, so 

/'__"mL\/. „ m \( 1 Jr 



55 



= 55 



1609 



m \f ljf \ _ 
TBi 1 1 3600 s / " 



h V Jf 7 \ TBi ,/ V 3600 s , 

where we rounded off to two significant figures. 
(b) Now we use I mi = 1609 m = 1.609 km; then 

km \ .... km 



25 



ni 



55 ™ = ss^b) ( 1. fio, J™ ) = sg i™ 
h V h A -vol.} h 



NOTE These unit conversions are very handy. You can always look them up in 
the Table inside the front cover. 



EXERCISE E Would a driver traveling at 15 m/s in a 35 mi/h zone be exceeding the 
Speed limit? 

When changing units, you can avoid making an error in the use of conver- 
sion factors by checking that units cancel out properly. For example, in our 
conversion of 1 mi to 1609 m in Example I -5(a), if we had incorrectly used the 
factor (nf^p 1 ) instead of ( m' L) "' m )i- the meter units would not have cancelled out; we 
would not have ended up with meters. 



TABLE 1-6 The 


8000-m Peaks 


Peak 


Height (in) 


Mt. Everest 


8850 


K2 


8611 


Kangehenjunga 


8586 


Lhotse 


8516 


Makalu 


8462 


Cho Oyu 


8201 


Dhaulagiri 


8167 


Manaslu 


8156 


Nan git Partial 


8125 


Aiinapurna 


8091 


Gasherbrum I 


8068 


Broad Peak 


8047 


Gasherbrum II 


8035 


Shisha Pangma 


8013 



t onversion factors - '. 



— ■ PROBLEM SOLVING 

Unit conversion is wrong ,■■' 
units do not cant ei 



SECTION 1-6 Converting Units 11 



PROBLEM SOLVING 

How to make a rough estimate 



PHY5IC5 APPLIED 

Estimating the volume 

{or mass) of a lake; see also 

Fig. I -10 



Order of Magnitude: Rapid Estimating 

We are sometimes interested only in an approximate value for a quantity. This 
might be because an accurate calculation would take more time than it is worth or 
would require additional data that are not available. Tn other cases, we may want to 
make a rough estimate in order to check an accurate calculation made on a calcu- 
lator, to make sure that no blunders were made when the numbers were entered. 

A rough estimate is made by rounding off all numbers to one significant 
figure and its power of 10, and after the calculation is made, again only one signif- 
icant figure is kept. Such an estimate is called an order-of-magnitude estimate and 
can be accurate within a factor of 10, and often better, In fact, the phrase "order 
of magnitude" is sometimes used to refer simply to the power of 10. 

To give you some idea of how useful and powerful rough estimates can be, 
let us do a few "worked- out Examples." 



EXAMPLE 1-6 l^kmiJ 



Volume of a lake. Estimate how much 
watei there is in a particular lake, Fig. 1- 10a, which is roughly circular, about 
1 km across, and you guess it has an average depth of about 10 ra. 

APPROACH No lake is a perfect circle, nor can lakes be expected to have a 
perfectly flat bottom. We are only estimating here. To estimate the volume, we 
can use a simple model of the lake as a cylinder: wc multiply the average 
depth of the lake times its roughly circular surface area, as if the lake were a 
cylinder (Fig. 1 - 10b). 

SOLUTION The volume V of a cylinder is the product of its height h times 
the area of its base: V - hirr 2 , where r is the radius of the circular base/ The 
radius r is jkm = 500 m, so the volume is approximately 

V = finr 2 ffi(10m) X (3) X (5 X 10 2 m) 2 « 8 X 10" m J « lfj 7 m\ 

where it was rounded off to 3. So the volume is on the order of l(fm 3 , ten 
million cubic meters. Because of all the estimates that went into this calcula- 
tion, the order-of-magnitude estimate (10' m 1 ) is probably better to quote than 
the 8 X I0 6 m 1 figure, 

'Formulas like this for volume, area, etc., are found inside the back cover of this book. 





FIGURE 1-10 Example 1-6. (a) How much 
water is in this lake? (Photo is of one of the Rac 
Lakes in the Sierra Nevada of California.) 
(I>) Model of the lake as a cylinder. [We could go 
one step further and estimate thy mass or weight 
of this lake. We will see later that water has a 
density of 1000 kg/m 3 . so this lake has a mass of 
about (l0 3 kg/m 3 )(lQ 7 m 3 ) * It) 10 kg. which is 
about 10 billion kg or 10 million metric tons. 
(A metric ton is 1000 kg, about 2200 lbs, slightly 
larger than a British ton, 2000 lbs.)] 



12 CHAPTER 1 Introduction, Measurement, Estimating 



NOTE To express out result in U.S. gallons, we see in ihe Table on the inside 
front cover thai 1 liter = 10 'm' & j gallon. Hence, Ihe lake contains 
(l0 7 m 3 )(l gallon/4 X 10" 3 m 3 ) « 2 X 10' ) gallons of water. 



EXAMPLE 1-7 MailflEH^ 



Thickness of a page. Estimate the thick- 
ness of a page of this book. 

APPROACH At first you might think that a special measuring device, a 
micrometer (Fig. 1-1 1), is needed to measure the thickness of one page since 
an ordinary ruler clearly won't do. But we can use a trick or, to put it in 
physics terms, make use of a symmetry: we can make the reasonable assump- 
tion that all the pages of this book are equal in thickness. 
SOLUTION We can use a ruler to measure hundreds of pages at once, ft you 
measure the thickness of the first 500 pages of this book (page 1 to page 500), 
you might get something like 1.5 cm. Note that 500 pages counted front and 
back is 250 separate pieces of paper. So one page must have a thickness of about 

1.5 cm , „ , „ , 

— *s 6 x 10 - 1 cm - 6 x 10 3 mm, 

250 pages 

or less than a tenth of a millimeter (0. 1 mm). 



■* PROBLEM SOLVING 

Use symmetry when possible 



EXAMPLE 1-8 




ESTIMATE - ! Total number of heartbeats. Estimate the 

total number of beats a typical human heart makes in a lifetime, 

FIGURE 1-11 Example 1-7. A 

APPRO AGH A typical resting heart rate is 70 beats/min. But during exercise micrometer, which is used for 

it can be a lot higher. A reasonable average might be 80 beats/min, measuring small thicknesses. 

SOLUTION If an average person lives 70 years ~ 2 X iCs (sec Tabic 1-2), 



Si I 



beats 



als\/ 1 min\. 



x 10 s s) * 3 X 10* 



or 3 trillion, 



Now let's take a simple Example of how a diagram can be useful for 
making an estimate, ft cannot be emphasized enough how important it is to 
draw a diagram when trying to solve a physics problem. 



EXAMPLE 1-9 



ESTIMATE 



1 



J Height by triangulation. Estimate the height 
12, by "triangulation," with the help of a bus-stop 



of the building shown in Fig. 
pole and a friend. 

APPROACH By standing your friend next to the pole, you estimate the height 
of the pole to be 3 m. You next step away from the pole until the top of the 
pole is in line with the top of the building, Fig. I -12a. You are 5 ft 6 in. tall, so 
youT eyes are about 1 ,5 m above the ground, Your friend is talleT, and when 
she stretches out her amis, one hand touches you, and the other touches the 
pole, so you estimate that distance as 2 m (Fig. I -12a). You then pace off the 
distance from the pole to the base of the building with big, I -m-long, steps, and 
you get a total of 16 steps or 16 m. 

SOLUTION Now you draw, to scale, the diagram shown in Fig. I -12b using 
these measurements, You can measure, right on the diagram, the last side of 
the triangle to be about x - 13 m. Alternatively, you can use similar triangles 
to obtain the height x: 



FIGURE 1-12 Fxamplc 

Diagrams are really usefu 



UO 



1-9. 



1.5 m 



L 



.1 in 



1 -12 nil- 



1.5 m 
2m 



18 m 



St) 



Finally you add in your eye height of 

result: the building is about 15 m tall. 



I3lm. 
1 ,5 m above the ground to gel your final 




18m 



SECTION 1-7 Order of Magnitude: Rapid Estimating 13 



Another technique for estimating, this one made famous by Enrico Fermi 
to his physics students, is to estimate the number of piano tuners in a city, say, 
Chicago oi" San Francisco. To gel a rough order-of-magnitude estimate of the 
number of piano tuners today in San Francisco, a city of about 700,000 inhabi- 
pkOblim solving lams, we can proceed by estimating the numbeT of functioning pianos, how 
Estimating hot:- tmin\ piano often each piano is tuned, and how many pianos each tuner can tune- To esti- 
titners there are in a city rnate the number of pianos in San Francisco, we note that ceriainly not 
everyone lias a piano. A guess of 1 family in 3 having a piano wouid correspond 
to I piano per 12 persons, assuming an average family of 4 persons. As an ordeT 
of magnitude, let's say 1 piano per 10 people. This is certainly more reasonable 
than I per 100 people, or 1 per every person, so let's proceed with the estimate 
that 1 person in If) has a piano, or about 70,000 pianos in San Francisco. Now a 
piano tuner needs an hour or two to tune a piano. So let's estimate that a tuner 
can tune 4 or 5 pianos a day. A piano ought to be tuned every 6 months or a 
year — let's say once each year. A piano tuner tuning 4 pianos a day, 5 days a 
week. 50 weeks a year can tune about 1000 pianos a year. So San Francisco, with 
its (very ) roughly 70,000 pianos, needs about 70 piano tuners. This is, of course, 
only a rough estimate. 1 ft tells us that there must be many more than 10 piano 
tuners, and surely not as many as 1000. If you were estimating the number of ear 
mechanics, on the other hand, your estimate would be rather different! 

Dimensions and Dimensional Analysis' 

When we speak of the dimensions of a quantity, we are referring to the type of 
units or base quantities that make it up. The dimensions of area, for example, 
arc always length squared, abbreviated [L 2 ], using square brackets; the units 
can be square meters, square feet, cm 2 , and so on, Velocity, on the other hand, 
can be measured in units of km/h, m/s, or mi/h, but the dimensions arc always 
a length [L] divided by a time [T]; that is, [L/T]. 

The formula for a quantity may be different in different cases, but the 
dimensions remain the same. For example, the area of a triangle of base b and 
height k is A - \bh, whereas (he area of a circle of radius r is A — trr 2 . The 
formulas are different in the two cases, but the dimensions of area in both cases 
are the same: \L 2 ]. 

When we specify the dimensions of a quantity, we usually do so in terms of 
base quantities, not derived quantities. For example, force, which we will see 
later has the same units as mass \M] times acceleration [L/T 2 ], has dimensions 
of \ML/T 2 \. 

Dimensions can be used as a help in working out relationships, and such a 
Dimensional anaiysh procedure is referred to as dimensional analysis.* One useful technique is the 
use of dimensions to check if a relationship is incorrect- A simple rule applies 
here: we add ot subtract quantities only if they have the same dimensions (we 
don't add centimeters and hours). This implies that the quantities on each side 
of an equals sign must have the same dimensions. (Tn numerical calculations, the 
units must also be the same on both sides of an equation.) 

Fot example, suppose you derived the equation v - v,-, + jat 2 , where v is 
the speed of an object after a time (, v tl is the object's initial speed, and the 
object undergoes an acceleration a. Let's do a dimensional check to see if this 
equation is correct; note that numerical factors, like the j here, do not affect 

'A check of the San Francisco Yellow Pages (done after this calculation) reveals about SO listings. 
Haeli tit these listings may employ more than one tuner, but on the other hand, each may also do 
repairs as well as tuning, fn any case, our estimate is reasonable. 

*Some Sections of this book, such as this one, may be considered optivnal at the discretion of the 
instructor. Sec the Preface for more details. 

s 'l~he techniques described in the nest few paragraphs may seem more meaningful after you have 
studied a few Chapters of this boot. Reading this Section now will give you an overview of the 
subject, and you can then return to it later as needed. 

14 CHAPTER 1 Introduction, Measurement, Estimating 



dimensional checks. We write a dimensional equation as follows, remembering 
thai the dimensions of speed are [I-/T] and (as we shall see in Chapter 1) the 
dimensions of acceleration are [/./T 2 ]; 



% + w, 



+ [/.]. 



The dimensions are incorrect: on the right side, we have the sum of quantities 
whose dimensions are not the same. Thus we conclude that an error was made 
in the derivation of the original equation. 

If such a dimensional check does come out correct, it does not prove that 
the equation is correct. For example, a dimensionless numerical factor (such as i 
ot 2-n) could be wrong. Thus a dimensional check can only tell you when a rela- 
tionship is wrong, ft can't tell you if it is completely light, 

Dimensional analysis can also be used as a quick check on an equation you 
are not sure about. For example, suppose that you can't rememher whether the 
equation for the period T (the time to make one back-and-forth swing) of a 
simple pendulum of length / is T = 7-n\/7jg or T = 2irV g/i , where g is the 
acceleration due to gravity and. like all accelerations, has dimensions [L/T : ]. 
(Do nut worry ahout these formulas — the correct one will be derived in 
Chapter 1 1: what we are concerned about here is a person's forgetting whether it 
contains t/g or g/i.) A dimensional check shows that the former (l/g) is correct: 



\T] 






V[t 2 ] - \n 



whereas the latter (#//) is not: 



\T\ 



+ /M3 



r 



\ in \[T 2 } 



i 
in' 



Note that the constant 2tt has no dimensions and so can't be checked using 
dimensions. 



| Summary 



[I Tie Summary thai appears at the end of each Chapter in [his 
book gives a brief overview of the main ideas of the Chapter. 
The Summary cannot serve to give an understanding of the 
material, which can be accomplished only by a detailed 
reading of the Chapter.] 

Physics, like- other sciences, is a creative endeavor. It is 
not simply a collection of facts. Important theories are created 
with the idea of explaining observations. To be accepted, theo- 
ries are "tested" by comparing their predictions with the 
results of actual experiments. Note that, in general, a theory 
cannot be "proved" in an absolute sense. 

Scientists often devise models of physical phenomena. A 
model is a kind of picture or analogy that helps to describe 
the phenomena in terms of something we already know, A 
theory, often developed from a model, is usually deeper and 
more complex than a simple model. 

A scientific law is a concise statement, often expressed in 
the form of an equation, which quantitatively describes a wide 
range of phenomena, 

Measurements play a crucial role in physics, but can 
never be perfectly precise. It is important to specify the 



uncertainty of a measurement either by staling it directly 
using the ± notation, and/or by keeping only the correct 
number of significant figures. 

Physical quantities are always specified relative to a 
particular standard or unit, and Che unit used should always 
be stated. The commonly accepted set of units today is 
the Sy slime International (SI), in which the standard units 
of length, mass, and time are the meter, kilogram, and 
second, 

When converting units, check all conversion factors for 
correct cancellation of units. 

Making rough, order-of-magnitude estimates is a very 
useful technique in science as well as in everyday life. 

["The dimensions of a quantity refer to the combination 
of hase quantities that comprise it. Velocity, for example, has 
dimensions of [length/time] or [I./T\. Working with only the 
dimensions of the various quantities in a given relationship 
(this technique is called dimensional analysis) make it 
possible to check a relationship for correct form.| 



Summary 15 



I Questions 



1. What are Che merits and drawbacks of using a person's 
foot as a standard? Consider both («) a particular 
person's foot, and (h) any person's foot, Keep in mind 
that it is advantagous that fundamental standards he 
accessible (easy to compare to), invariable (do not 
change), indestructible, and reproducible. 

2. When traveling a highway in the mountains, you may see 
elevation signs that read "91 4 m (3000 ft)." Critics of the 
metric system claim that such numbers show the metric 
system is more complicated. How would you alter such signs 
to be more consistent with a switch to the metric system? 

3. Why is it incorrect to think that the more digits you 
represent in your answer, the more accurate it is? 

4. What is wrong with this road sign: 

Memphis 7 mi (11.263 km)? 

5. For an answer to be complete, the units need to be speci- 
fied. Why? 



6. Discuss how the notion of symmetry could be used to 
estimate the number of marbles in a 1-liter jar. 

7. You measure the radius of a wheel to he 4.16 cm. If you 
multiply by 2 to get the diameter, should you write the 
result as 8 cm ot as 8,32 cm? Justify your answer. 

8. Express the sine of 30,0" with the correct number of 
significant figures. 

9. A recipe for a souffle specifies that the measured ingredi- 
ents must be exact, or the souffle will not rise. The recipe 
calls for 6 large eggs. Ilie size of "large" eggs can vary by 
10%, according to the USDA specifications. What does 
this tell you about how exactly you need to measure the 
other ingredients? 

10. List assumptions useful to estimate the number of car 
mechanics in (Yr) San brancisco. (b) your hometown, and 
tli en make the estimates. 



| Problems 



[The Problems at the end of each Chapter are ranked I, II, or III 
according to estimated difficulty, with (I) Problems being easiest. 
Level (III) Problems are meant mainly as a challenge for the best 
students for "extra credit." The Problems are arranged hy Sections, 
meaning that the reader should have read up to and including 
that Section, but not only that Section — Problems often depend 
on earlier material. Fach Chapter also has a group of General 
Problems that are not arranged by Section and not ranked, | 

1-4 Measurement, Uncertainty, Significant Figures 
(Note- In Problems, assume a number like 6.4 is accurate 

to ±0.]; and 950 is =10 unless 950 is said to be "precisely 1 ' or 
"very nearly" 950. in which case assume 950 ± 1.) 

1. (I)"lne age of the universe is thought to be about 14 
billion years. Assuming two significant figures, write this 
in powers of ten in {#) years, {b) seconds 

2. (I) How many significant figures do each of the following 
numbers have- (a) 214. (b) 81.60, (c) 7.03, (rf) 0,03. 
(e) 0,0086. (/) 3236. and (#) 8700? 

3. (I) Write the following numbers in powers of ten nota- 
tion: (a) 1.156. (b) 21.8, (c) 0.0068, (<i) 27.635, (e) 0-219, 
and (/') 444. 

4. (I) Write out the following numbers in full with the 
correct number of zeros: (a) 8.69 x 10 4 , (b) 9.1 X 1()\ 
(c) 8.8 x lfT 1 , (d) 4.76 x 10 2 , and fe) 3.62 x 10" 5 . 

5. (II) What, approximately, is the percent uncertainty for 
the measurement given as 1 .57 m~? 

6. (II) What is the percent uncertainty in the measurement 
3.76 ± 0.25 in? 

7. (II) Time intervals measured with a stopwatch typically have 
an uncertainty of about 0.2s, due to human reaction time at 
the start and slop moments. What is the percent uncertainly 
of a handtimed measurement of (a) 5 s, (fe) 50s. (t) 5 min? 

8. (II) Add (9.2 X 10-* s) + (8.3 X lQ 4 s) + (O.00S X 10 6 s). 

9. (II) Multiply 2.079 X 10 2 m by 0.082 X 10 ', taking into 
account significant figures. 

1(1. (Ill) What is the area, and its approximate uncertainty, of 
a circle of radius 3.8 x 10* cm? 



11. (Ill) What, roughly, is the percent uncertainty in the 
volume of a spherical beach ball whose radius is r = 
2.86 ± 0.09 m? 

1-5 and 1-6 Units, Standards, SI, Converting Units 

12. (I) Write the following as full (decimal) numbers with 
standard units: (a) 2866mm, (b) S5,uV, (c) 760 mg, 
(rf) 60.0 ps, (e ) 22.5 fm. (/) 2.50 gigavolts. 

13. (I) Express the following using the prefixes of Table 1-4: 
(,v) 1 x 10" volts, (b) 2 x 10"* meters, (c) 6 x K) 3 days, 
(rf) 18 X I0 2 bucks, and (e) 8 X 10"* pieces. 

14. (I) Determine your own height in meters, and your mass 
in kg. 

15. (I) The Sun, on average. Is 93 million miles from Earth. 
How many meters is this? Fxpress (a) using powers of 
ten. and (6) using a metric prefix, 

16. (II) What is the conversion factoT between (a) ft 2 and yd 2 , 
{!>) in 2 and ft 2 ? 

17. (II) An airplane travels at 950 km/h. How long does it 
take to travel ] .00 km? 

18. (II) A typical atom has a diameter of about 
1,0 X 10" l[l m, (a) What is this in inches? (ft) Approx- 
imately how many atoms are there along a 1.0-cni line? 

19. (II) Fxpress the following sum with the correct number of 
significant figures: 1.80m + 142.5cm + 5.34 X 10 Vm. 

211. (II) Determine the conversion factor between (a) km/h 
and mi/h, (b) m/s and ft/ a. and (c) km/h and m/s. 

21. (II) How much longer (percentage) is a one-mile race 
than a 1500-m race ("the metric mile")? 

22. (II) A light-year is the distance light travels in one yeaT 
(at speed =2,998 X ]Q s m/s). (n) How many meters are 
there in 1,00 light-year? (A) An astronomical unit (AC) is 
the average distance from the Sun to Farth, 
1,50 X 10 s km. How many AU are theTe in 1.00 light- 
year? (c) What is the speed of light in AU/h? 

23. (Ill) The diameter of the Moon is 3480 km. («) What is 
the surface area of the Moon? (h) How many limes largeT 
is the surface area of the Earth? 



16 CHAPTER 1 Introduction, Measurement, Estimating 



1-7 Order of Magnitude Estimating 

(Note: Remember that for rough estimates, only round numbers 
are needed both as input to calculations and as final results.) 

24. (I) Estimate the order of magnitude (power of ten) of: 
(a) 2800. (b) 86,30 x IU 2 . (c) 0,0076, and (rf) 15.0 x 10 s . 

25. (II) Estimate how many books tan be shelved in a college 
library with 3500 square meters of floor space. Assume 8 
shelves high, having books on both sides, with corridors 1.5 m 
wide. Assume books are about the size of this one, on average. 

'fi. (IT) Estimate how many hours it would take a runner to 
run (at lOkm/h) across the United States from New York 
to California. 

27. (11) Rstimatc how long it would take one person to mow 
a football field using an ordinary home lawn mower 
(Fig. 1 -13), Assume the mower moves with a 1 kin/h 
*.peed, and has a 0.5 m width. 




28. (II) Estimate the number of liters of water a human 
drinks in a lifetime. 

29. (II) Make a rough estimate of the volume of your body 

(in cm 1 ). 

30* (II) Make a rough estimate, for a typical suburban house, 
of the % of its outside wall aTea that consists of window 
area, 

31. (Ill) The rubber worn from tires mostly enters the atmos- 
phere as particulate pollution. Estimate how much rubber 
(in kg) is put into the air in the United States every year. 
To get started, a good estimate for a tire tread's depth is 
1 cm when new. and the density of rubber is about 
1200kg/nr\ 

1-8 Dimensions 

* 32. (II) The speed, ?;. of an object is given by the equation 

v = /If' - Bt. where t refers to lime. What are the 

dimensions of A and B'l 

* 33. (II) Three students derive the following equations in which 

x refers to distance traveled. « the speed, a [he acceleration 
(m/s 2 ). and t the time, and the subscript („) means a quan- 
tity at time / = 0: (a) v = IK 2 -I- 2a/. (h) v = %f + lar 1 , 
and (c) A" = Hjf + 2ar. Which of these could possibly be 
correct according to a dimensional check? 



FIGURE 1-13 Problem 27 



| General Problems 



34. Global positioning satellites (GPS) can he used to deter- 
mine positions with great accuracy. The system works by 
determining the distance between the observer and each 
of several satellites orbiting Earth. If one of the satellites 
is at a distance of 20,0(10 km from you, what percent accu- 
racy in the distance is required if we desire a 2-meter 
uncertainty'? How many significant figures do we need to 
have in the distance'? 

35. Computer chips (Fig, 1-14) are etched on circular silicon 
wafers of thickness 0-60 mm that are sliced from a solid 

cylindrical silicon crystal of 
length 30 cm, If each wafer can 
hold IUU chips, what is the 
maximum number of chips 
that can be produced from one 
entire cylinder'? 

FIGURE 1-14 Problem 35. 
The wafer held by the hand 
(above) is shown below, 
enlarged and illuminated by 
colored light. Visible btc rows 
of integrated circuits (chips). 

36. {a) How many seconds are there in 1 .(JO year? (b) How 
many nanoseconds are there in 1,00 yeaT? (c) How many 

years are there in 1 ,00 second? 




37. A typical adult human lung contains about 300 million 
tiny cavities called alveoli. Estimate the average diameter 
of a single alveolus. 

,ltf. One hectare is defined as I0 4 m : . One acre is 4 x lO^ft 2 . 
How many acres are in one hectare? 

39. Use Table 1-3 to estimate the total number of protons or 
neutrons in (a) a bacterium, (b) a DMA molecule, (c) the 
human body, (d) our Galaxy. 

4il Rstimatc the number of gallons of gasoline consumed by the 
total of all automobile drivers in the United States, per yeaT. 

41. Estimate the number of gumballs in the machine of 
Fig 1-15, 



FIGURE 1-15 

Problem 41, Estimate the 
number of gumballs in 
the machine. 




General Problems 17 



42. An average family of four uses roughly 1200 liters (about 
300 gallons) of water peT day. (One liter = 1 000 cm 3 .) 
How much depth would a lake lose per year if it 
uniformly covered an area of 50 square kilometers and 
supplied a local town with a population of 40,000 people? 
Consider only population uses, and neglect evaporation 
and so on, 

43. How big is a ton? That is. what is the volume of some- 
thing thai weighs a Ion? To be specific, estimate the diam- 
eter of a 1-ton rock, but first make a wild guess: will it he 
1 ft across, 3 ft, or the si7.e of a car? [Hint Rock has mass 
per volume about 3 times that of water, which is 1 kg per 
liter (10 3 cm 3 ) or 62 lb per cubic font 1 

44. A heavy rainstorm dumps 1.0 cm of rain on a city 5 km 
wide and 8 km long in a 2-h period. How many metric 
tons (1 metric Ion = 10" l kg) of water fell on (he city? 
[1 cm 3 of water has a mass of 1 gram = 10 _3 kg, ] How 
many gallons of water was this? 

45. Hold a pencil in front of your eye at a position where 
its blunt end just blocks out the Moon (Fig. 1-16). 
Make appropriate measurements lo estimate the diameter 
of the Moon, given that the Earth-Moon distance is 
3.8 x Ul s km. 




FIGURE 1-16 Problem 45. 

How big is the Moon? 



46. Estimate how many days it would take to walk around 
the world, assuming 10 h walking per day at 4 km/h, 

47. Noah's ark was ordered to be 300 cubits long, 50 cubits 
wide, and 30 cubits high. "Hie cubit was a unit of measure 
equal to the length of a human forearm, elbow to the tip 
of the longest finger. Express the dimensions of Noah's 
ark in meters, and estimate its volume (m '). 

4Jf. One liter (1000 em 3 ) of oil is spilled onto a smooth lake. If 
the oil spreads out uniformly until it makes an oil slick 
just one molecule thick, with adjacent molecules just 
touching, estimate the diameter of the oil slick. Assume 
the oil molecules have a diameter of 2 x 10 '"m. 



49. Jc-an camps beside a wide river and wonders how wide it 
is, She spots a large rock on the bank directly across from 
her. She then walks upstream until she judges that the 
angle between her and the rock, which she can still see 
clearly, is now at an angle of 30" downstream (Fig. 1-17), 
Jean measures her stride to be about one yard long. The 
distance back to her camp is 120 strides. About how faT 
across, both in yards and in meters, is the river? 



v 

\ 
\ 
\ 

&. N 

"l ?.i "i Strides* 



FIGURE 1-17 
Prublcm 49. 



50. A watch manufacturer claims that its watches gain or 
lose no more than 8 seconds in a year. How accurate is 
this watch, expressed as a percentage? 

51. The diameter of the Moon is 3480 km. What is the volume 
of the Moon? How many Moons would be needed to 
create a volume equal to that of Earth' 

52. An angstTom (symbol A) is a unit of length, defined as 
10 _lo m, which is on the order of the diameter of an atom, 
[a) How many nanometers are in 1,0 angstrom? (/>) How 
many femtometers or fermis (the common unit of length 
in nuclear physics) are in 1.0 angstrom? (c) How many 
angstroms are in 1.0 meter? (d) How many angstroms are 
in 1,0 light-year (see Problem 22)? 

53. Determine the percent uncertainty in Q. and in sin tf, when 
{a) B = 15.0 ,J ± 0-5", (h) ft = 75.0' 5 ± 0.5", 

54. If you began walking along one of Earth's lines of 
longitude and walked until you had changed latitude by 
1 minute of arc (theTe are 60 minutes per degree), how faT 
would J on have walked (in miles)? This distance is called 
a "nautical mile." 



Answers to Exercises 

A: (d). 

B: No: 3, 2. 

C: All three have three significant figures, although the 

number of decimal places is (a) 2. (i) 3, (c) 4, 



D: Ml. Everest. 29,035 ft; K2. 28.251 ft: Kangchenjunga. 

28,169 ft. 
E: No: 15m/s = 34mi/h. 



18 CHAPTER 1 Introduction, Measurement, Estimating 




A high-speed car has released a parachute to reduce its 
speed quickly. The directions of the car's velocity arid 
acceleration are shown by the green (?) and gold (a) 
arrows. Motion is described using the concepts of velocity 
and acceleration. We see here 
that the acceleration a can 
sometimes be in the opposite 
direction from the velocity v. 
We will also examine in detail 
motion with constant acceler- 
ation, including the vertical 
motion of objects failing 
under gravity, 



CHAPTER 



2 



Describing Motion: 
Kinematics in One Dimension 



T~~ he motion of objects — baseballs, automobiles, joggers, and even the Sun 
and Moon — is an obvious part of everyday life. "Ft was not until the 
sixteenth and seventeenth centuries that our modern understanding of 
motion was established. Many individuals contributed to this understanding, 
particularly Galileo Galilei (1564-1642) and Isaac Newton (1642-1727). 

The study of the motion of objects, and the related concepts of force and 
energy, form the field called mechanics. Mechanics is customarily divided into 
two parts: kinematics, which is the description of how objects move, and 
dynamics, which deals with force and why objects move as they do. This Chapter 
and the next deal with kinematics. 

For now we only discuss objects that move without rotating (Fig. 2- la). 
Such motion is called translational motion. In this Chapter we will be 
concerned with describing an object that moves along a straight-line path, 
which is one-dimensional translational motion. Tn Chapter 3 we will describe 
translational motion in two (or three) dimensions along paths that are not 
straight. (We discuss rotation, as in Fig, 2- lb, in Chapter 8.) 

We will often use the concept, or model, of an idealized particle which is 
considered to be a mathematical point and to have no spatial extent (no si^e). 
A particle can undergo only translational motion. The particle model is useful 
in many real situations where we are interested only in translational motion 
and the object's size is not so significant. For example, we might consider a 
billiard ball, ur even a spacecraft traveling toward the Moon, as a particle for 
many purposes. 




4 



§ m 



(j) 



(b) 



FIGURE 2-1 The pinecone in (a) 
undergoes pure translation as it 

falls, whereas in (b) it is rotating as 
well as translating. 



19 



All measurements are 
made relative to 

a frame of reference 



Reference Frames and Displacement 

Any measurement of position, distance, or speed must be made with respect to 
a reference frame, or frame of reference. For example, while you aTe on a tTain 
traveling at 80km/h, suppose a person walks past you toward the front of the 
train at a speed of. say, 5 km/h (Fig. 2-2). This 5 km/h is the person's speed with 
respect to the train as frame of reference. With respect to the ground, that 
person is moving at a speed of 80 km/h + 5 km/h = 85 km/h. It is always 
important to specify the frame of reference when stating a speed. In everyday 
life, we usually mean "with respect to the Earth" without even thinking about it. 
but the reference frame must be specified whenever there might be confusion. 



FIGURE 2-2 A person walks toward 
the from of a train at 5 km/h. The Irain 
is moving 80 kni/h with respect to the 

ground SO the- walking person's Speed, 

relative lo the ground, is 85 km/h. 




■ y 



FIGURE 2-3 Standard SCI of xy 
coordinate axes. 



Displacement 



<j> C A U T I Q IM 



The displacement may not equal 
the total distance traveled 



When Specifying the motion of an object, it is important to specify not Only 
the speed but also the direction of motion. Often we can specify a direction by 
using north, east, south, and west, and by "up'' and "down," In physics, we often 
draw a set of coordinate axes, as shown in Fig. 2-3. to represent a frame of 
reference. Wc can always place the origin 0. and the directions of the Jt and v 
axes, as we like for convenience. The x and v axes are always perpendicular to 
each other. Objects positioned to the right of the origin of coordinates (0) on 
the x axis have an _t coordinate which we usually choose to be positive; objects 
to the left of then have a negative x coordinate, The position along the v axis 
is usually considered positive when above U, and negative when below 0, 
although the reverse convention can be used if convenient. Any point on the 
plane can be specified by giving its x and y coordinates. In three dimensions, a z 
axis perpendicular to the x and y axes is added. 

For one-dimensional motion, we often choose the x axis as the line along 
which the motion takes place. Then the position of an object at any moment is 
given by its x coordinate. If the motion is vertical, as for a dropped object, wc 
usually use the y axis. 

We need to make a distinction between the distance an object has traveled 
and its displacement, which is defined as the change in position of the object. 
Tit at is, displacement is how far the object is from its starting point. To see the 
distinction between total distance and displacement, imagine a person walking 
70 m lo the east and then turning around and walking back (west) a distance of 
30 m (see Fig. 2-4). The total distance traveled is 100 m. but the displacement is 
only 40 m since the person is now only 40 m from the starting point. 



FIGURE 2-4 A person walks 70 m east, 
then 30 m west. The total distance traveled 
is 100 m (path is shown dashed in black); 
bm the displacement, shown as 
a blue arrow, is 40 in to the east. 



West 



70 m 



40 m 30 in 
Displacement 



tiast 



Displacement is a quantity that has both magnitude and direction. Such 
quantities are called vectors, and are represented by arrows in diagrams. For 
example, in Fig. 2-4, the blue arrow represents the displacement whose magni- 
tude is 40 m and whose direction is to the right (east). 



20 CHAPTER 2 Describing Motion: Kinematics in One Dimension 



We will deal with vectors more fully in Chapter 3. For now, we deal 
only with motion in one dimension, along ;i line- Tn this case, vectors 
which point in one direction will have a positive sign, whereas vectors that 
point in the opposite direction will have a negative sign, along with their 
magnitude. 

Consider the motion of an object over a particular lime interval, Suppose that 
at some initial time, call it f, , the object is on the x axis at the position X\ in the 
coordinate system shown in Fig. 2-5. At some later time, t 2 , suppose the object 
has moved to position a 3 . The displacement of out object is x 2 - X\ , and is repre- 
sented hy the arrow pointing to the right in Fig. 2-5, It is convenient to write 

Ax = x 1 - x, , 






10 20 30 40 
Distance Cm) 



FIGURE 2-5 The arrow represents 
the displacement x 2 — X\ . Distances 
are in meters. 



where the symbol A (Greek letter delta) means "change in." Then Ax means 
"the change in x," or "change in position, 11 which is the displacement. Note that 
the "change in" any quantity means the final value of that quantity, minus the 
initial value. 

Suppose a | = 10.0 m and x 2 = 30.0 m. Then 



i means final value 
minus initial value 



Ax — x 7 



jti = 30.0 m - 10.0 tn = 20.0 m, 



so the displacement is 20.0 m in the positive direction, as in Fig. 2-5. 

Now consider an object moving to the left as shown in Fig. 2-6. Here the 
object, say, a person, starts at x v = 30,0 m and walks to the left to the point 
jf 2 = 1 0.0 m. In this case 



Ax 



10.0 m - 30.0 m 



-20.0 m, 



and the blue arrow representing the vector displacement points to the left- The 
displacement is 20.0 m in the negative direction. This example illustrates that for 
one-dimensional motion along the x axis, a vector pointing to the right has a 
positive sign, whereas a vectOT pointing to the left has a negative sign. 



FIGURE 2-6 For the displacement 
Ajc = x 2 ~ xi = 1 0.0 m - 30.0 in. 
the displacement vector points to 
the left. 



M 



"fr" I T fr 



10 20 30 40 
Distance (m) 



Average Velocity 



Consider a Tacing sprinter, a galloping horse, a speeding Ferrari, or a rocket shol 
off into space. The most obvious aspect of IheiT motion is how fast they are 
moving, which brings us to the idea of speed and velocity, 

The term 'speed" refers to how far an object travels in a given time interval, 
regardless of direction. If a car travels 240 kilometers (km) in 3 hours (h), we 
say its average speed was 80 km/h. In general, the average speed of an object is 
defined as the lolal distance traveled along its path divided by the time it takes to 
travel this distance: 



average speed = 



distance traveled 
time elapsed 



(2—1) Average speed 



The terms "velocity" and "speed" are often used interchangeably in ordi- 
nary language, Rut in physics we make a distinction between the two. Speed is 
simply a positive number, with unils. Velocity, on the other hand, is used to 
signify both the magnitude (numerical value) of how fast an object is moving 
and also the direction in which it is moving. (Velocity is therefore a vector.) 
There is a second difference between speed and velocity: namely, the average 
velocity is defined in terms of displacement, rather than total distance traveled: 

displacement final position - initial position 

average velocity = = 

time elapsed time elapsed 



Velocity 



Average xxlocity 



SECTION 2-2 Average Velocity 21 



^ CAUTION 



Average speed is not necessarily 

equal to the magnitude of the 

average velocity 



Average velocity 



» PROBLEM SOLVING 

+ or — sign can signify the direction 
for linear motion 



Average speed and average velocity have the same magnitude when the 
motion is all in one direction, In other cases, Ihey may differ; recall the walk we 
described earlier, in Fig. 2-4, where a peison walked 70 m east and then 30 m west. 
The total distance traveled was 70 m + 30 m = 100 m, but the displacement was 
40 m. Suppose this walk took 70s to complete. Then the average speed was: 

distance 100 m . , 

~ ', T - ~^7, — - I -4111/5. 

time elapsed 70 s 

The magnitude of the average velocity, on the other hand, was: 

displacement 40 m 



70s 



0.57 m/s. 



time elapsed 

This difference hetween the speed and the magnitude of the velocity can occur 
when we calculate average values. 

To discuss one-dimensional motion of an ohject in general, suppose that at 
some moment in time, call it f, , the object is on the x axis at position x, in a 
coordinate system, and at some later time, / 2 , suppose it is at position x 2 . The 
elapsed time is f 2 - /, ; during this time interval the displacement of our ohject 
is Ax = x 2 — Xj . Then the average velocity, defined as the displacement divided 
by the elapsed time, can be written 

x 2 - Xy Ax ,„ -. 

v = — = — ■ 2-2 

where v stands for velocity and the baT (~) over the v is a standard symbol 
meaning "average."' 

The elapsed lime, or time interval, t 2 — r, , is the time that lias passed 
during our chosen period of observation. 

For the usual case of the +x axis to the right, note that if x? is less than x { , 
the object is moving to the left, and then Ax — jt 2 — jt| is less than zero. The 
sign of the displacement, and thus of the average velocity, indicates the direc- 
tion: the average velocity is positive for an object moving to the right along the 
+x axis and negative when the object moves to the left. The direction of the 
average velocity is always the same as the direction of the displacement. 



H h 



Finish Start 
U,i (x,l 



t [x 



10 20 30 40 50 60 
Distance (m) 



FIGURE 2-7 Example 2-1. 
A person runs from x\ = 50.0 m 

to .V2 = 30.5 m. The displacement 
is -19.5 m. 



EXAMPLE 2-1 



Runner's average velocity. The position of a runner as a 
function of time is plotted as moving along the x axis of a coordinate system. 
During a 3.00-s time interval, the runner's position changes from x t - 50.0m 
to x z - 30.5 m, as shown in Fig 2-7, What was the runner's average velocity? 

APPROACH We want to find the average velocity, which is the displacement 
divided by the elapsed time. 

SOLUTION The displacement is Ax = x 2 - x { = 30.5 m - 50.0 m = -19.5m. 
The elapsed time, or time interval, is At = 300 s, The average velocity is 



v 



A', 

a7 



- 1 9.5 m 

3.00 s 



= -6.50 m/s. 



The displacement and average velocity are negative, which tells us that the 
runner is moving to the left along the x axis, as indicated by the arrow in Fig. 2-7. 
Thus we can say that the runner's average velocity is 6.50 m/s to the left. 



EXAMPLE 2-2 



Distance a cyclist travels. How far can a cyclist travel in 
2.5 h along a straight road if her average velocity is 1 8 km/b? 

APPROACH We are given the average velocity and the time interval (= 2,5 h), 
We want to find the distance traveled, so we solve Eq. 2-2 for Ax. 
SOLUTION We rewrite Eq. 2-2 as Ax - T>Ac, and find 
Ax = vAt= (lSkm/h)(2.5h) = 45 km. 



22 CHAPTER 2 Describing Motion: Kinematics in One Dimension 



Instantaneous Velocity 



If you drive a car 150 km along a straight road in one direction for 2.0 h, the magni- 
tude of your average velocity is 75 km/h. It is unlikely, though, that you were 
moving at precisely 75 km/h at every instant. To deal with this situation we need 
the eoneept of instantaneous velocity, which is the velocity at any instant of time. 
(Its magnitude is the number, with units, indicated hy a speedometer; Fig. 2-8.) 
More precisely, the instantaneous velocity at any moment is defined as the average 
velocity daring an infmitesimally short time internal. That is, starting with Eq. 2-2, 

_ Ar 

M 

we define instantaneous velocity as the average velocity as we let A/ become 
extremely small, approaching zero. We can write the definition of instantaneous 
velocity, v, for one-dimensional motion as 



v - km -t— 



FIG U RE 2 -8 Car speedom e I er 
showing mi/h in white, and km/h in 
orange. 



(2-3) Instantaneous velocity 



The notation lim^.,,, means the ratio A.v/Af is to be evaluated in the limit of Af 
approaching zero. 

For instantaneous velocity we use the symbol v, whereas for average 
velocity we use v, with a baT. In the rest of this book, when we use the term 
"velocity, 1 it will refer to instantaneous velocity. When we want to speak of the 
average velocity, we will make this clear hy incl tiding the word "average." 

Note that the instantaneous speed always equals the magnitude of the 
instantaneous velocity. Why? Because the distance and the magnitude of 
the displacement beeome the same when they become infinitesimally small. 

If an object moves at a uniform (that is, constant) velocity during a partic- 
ular time interval, then its instantaneous velocity at any instant is the same as its 
average velocity (see Fig. 2 -9a), But in many situations this is not the case. For 
example, a car may start from rest, speed up to 50 km/h, remain at that velocity 
for a time, then slow down to 20 km/h in a traffic jam, and finally stop at its 
destination after traveling a total of 15 km in 30min,This trip is plotted on the 
graph of Fig. 2 -9b. Also shown on the graph is the average velocity (dashed 
line), which is li = Ajr/Af = 15 km/0.50 h = 30 km/h. 



FIGU RE 2-9 Velocity of a car as a 
function of time: (a) at constant 
velocity; (b) with varying velocity. 



£60 

^40 



Q 20 



0.1 0.2 0.3 0.4 0.5 

{'■>) Time(b) 



2 fid 



HA 

£ oT — i h 



Average velocity 



0.1 0.2 0.3 0.4 0.5 

(h) Timeth) 



Acceleration 



An object whose velocity is changing is said to be accelerating. For instance, a 
car whose velocity increases in magnitude from zero to 80 km/h is accelerating. 
Acceleration specifies how rapidly the velocity of an object is changing. 

Average acceleration is defined as the change in velocity divided by the 
time taken to make this change; 



average acceleration 



change of velocity 
time elapsed 



In symbols, the average acceleration, a, during a time interval A/ = t 2 - t t over 
which the velocity changes by An = v 2 — V] , is defined as 



a 



u-u 



Av 
At 



(2-4) A ret a»e ttcivivt Mum 



Acceleration is also a vector., but for one-dimensional motion, we need only use 
a plus or minus sign to indicate direction relative to a chosen coordinate system. 



SECTION 2-4 Acceleration 23 



I ns tt a aaneous acceieratioi i 



At it's/means v = U 



The instantaneous acceleration, a, can he defined in analogy to instanta- 
neous velocity, for any specific inslanl: 



Km 
ii-to At 



(2-5) 



Here iw is the very small change in velocity during the very short time interval Af . 



EXAMPLE 2-3 



Average acceleration. A car accelerates along a straight 
road from rest to 75 km/h in 5.0 s, Fig. 2-10. What is the magnitude of its 
average acceleration? 

APPROACH Average acceleration is the change in velocity divided by elapsed 
lime, 50 s, The caT starts from rest, so t> L = 0, The final velocity is v 2 = 75 km/h, 
SOLUTION From Eq, 2-4, the average acceleration is 



wi 75km/h - Okm/h 



= 15 



km/h 



t 2 - l, 5.0 s s 

This is read as "fifteen kilometers per hour per second" and means that, on 
average, the velocity changed by 1 5 km/h during each second. That is, assuming 
the acceleration was constant, during the first second the car's velocity increased 
from zero to 15 km/h. During the next second its velocity increased by another 
15 km/h, reaching a velocity of 30 km/h at / - 2.0 s, and so on. See Fig. 2-10, 
NOTE Our result contains two different time units: hours and seconds, We 
usually prefer to use only seconds. To do so we can change km/h to m/s (see 
Section 1-6, and Example 1-5): 

75km/h = (75^)( l0, 



Then 



Jec I \ 1 It™, 

21 m/s - 0.0 m/s 



A 3600 s^ 



21 m/s. 



5.0 s 



= 4.2 



m/s 



= 4.2 ■ 



We almost always write the units for acceleration as m/s 2 (meters per second 
squared), as we just did, instead of m/s/s. This is possible because: 

m/s 



m 

8' s 



m 



S S'S s* 

According to the calculation in Example 2-3, the velocity changed on the average 
by 4.2 m/s during each second, for a total change of 21 m/s over the 5.0s, 



FIGURE 2-10 Fxample 2-3. The 
car is shown at the start with 
i>i = at i\ = 0. I "he tar is shown 
three more times, at f = 1,0 s, 
t = 2.0 s, and at the end of our 
lime interval, r; = 5.0 s. We 
assume the acceleration is constant 
and equals 15 km/h/s. The green 
arrows represent the velocity 
vectors: the length of each arrow 
represents (he magnitude of the 
velocity at that moment. The accel- 
eration vector is the orange arrow. 
Distances are not to scale. 



I, =0 
i?l = 



Acceleration 

,, km/h 

a - 15 -j— 



al f- 1.0 s 

v - 15 km/h 



at r = 2.0 s 

= 3D km/h 



at i = r ; = 5,0 $ 
r- r 2 = 7s fcm/h 



-<P 



24 CHAPTER 2 Describing Motion: Kinematics in One Dimension 



Note thill acceleration tells us how quickly the velocity changes, whereas velocity ^ > CAUTION 

tells lis how quickly the position changes. Distinguish velocity from acceleration 



CONCEPTUAL EXAMPLE 2-4 | Velocity and acceleration. (a) If the veloc- 



ity of an object is zero, does it mean that the acceleration is zero? (b) If the 
acceleration is zero, does it mean that the velocity is zero? Think of some examples, 

RESPONSE A zero velocity does not necessarily mean that the accelera- 
tion is zero, nor does a zero acceleration mean that the velocity is zero. 
(a) For example, when you put your foot on the gas pedal of your car 
which is at rest, the velocity starts from zero but the acceleration is nol 
zero since the velocity of the car changes. (How else could your car starl 
forward if its velocity weren't changing — that is, accelerating?) {b) As you 
cruise along a straight highway at a constant velocity of lOOkm/h, your 
acceleration is zero: a = 0, i> ^ 0. 



4>_ 



CAUTION 



// v or a is zero, is the other zero too? 



EXERCISE A A car is advertised to go from zero to 60mi/h in 60s What does this say 
about the car; (a) it is fast (high speed): or (A) it accelerates well? 



EXAMPLE 2-5 



Car slowing down. An automobile is moving to the right 
along a straight highway, which we choose to be the positive* axis (Fig. 2-11). 
Then the driver puts on the brakes. If the initial velocity (when the driver hits 

the brakes) is v t - 15.0 m/s, and it takes 5.0 s to slow down to v 2 - 5.0 m/s, 
what was the cars average acceleration? 

APPROACH We are given the initial and final velocities and the elapsed time, 
so we can calculate a using Eq, 2-4, 

SOLUTION We use Eq. 2-4 and call the initial time /, - 0; then f 2 - 5,0 s. 
(Note that our choice of fj - doesn't affect, the calculation of a because 
only A/ = l 2 - f| appears in Eq. 2-4,) Then 

5.0 m/s - 1 5.0 m/s 



5.0 s 



= - 2.0 m/s . 



The negative sign appears because the final velocity is less than the initial 
velocity. In this case the direction of the acceleration is to the left (in the 
negative x direction) — even though the velocity is always pointing to the right. 
We say that the acceleration is 2.0 m/s 2 to the left, and it is shown in Fig. 2-1 1 

as an orange arrow. 



Acceleration 

ill f, = 

D] - 15.0 m/s fl = -2.0mA 2 



at t 2 = 5.0 5 

th = 5.1) in/* 



FIGURE 2-11 Fxample2-5, 
showing the position of the car at 
times; t t and t 2 , as well as the car's 
velocity represented by the green 
arrows. The acceleration vector 
(orange) points to the left as the 
ear slows down while moving to 
the right. 



Deceleration 

When an object is slowing down, we sometimes say it is decelerating. Rul he <T > CAUTION 

careful: deceleration does not mean that the acceleration is necessarily negative. 

For an object moving to the right along the positive x axis and slowing down 

(as in Fig. 2-1 1), the acceleration is negative. But the same car moving to the 

left (decreasing x"), and slowing down, has positive acceleration that points to 

the light, as shown in Fig. 2- 1 2. We have a deceleration whenever the 

magnitude of the velocity is decreasing, and then the velocity and acceleration 

point in opposite directions. 



Deceleration means the magnitude of 
the velocity is decreasing; it does not 
necessarily mean a is negative 



i = -5.0 m/s Vt = -15.0 m/s FIGURE 2-12 The car of Example 2-5, now moving to the left and decelerating. 



-0- 



The acceleration is 

v 2 ~ U[ -5.0 m/s - (-15.0 m/s) -5,0m/s + 15.0 m/s 
a = — — — = — = — = +2.0 m/s. 



A.' 



5.0 s 



5.0 s 



EXERCISE B A car moves along the x axis. What is the sign of the car's acceleration if 
it is moving in the positive x direction with {a) increasing speed or {b) decreasing 
speed? What is the sign of the acceleration if the car moves in the negative direction 
with (c) increasing speed or (d) decreasing speed? 



SECTION 2-4 Acceleration 25 



Motion at Constant Acceleration 

Many practical situations occur in which the acceleration is constant or nearly 
Let a = constant constant. We now examine this situation when the magnitude of the accelera- 
tion is constant and the motion is in a straight line. Tn this case, the 
instantaneous and average accelerations are equal. 

We now use our definitions of velocity and acceleration to derive a set of 
extremely useful equations that relate jt, v, a, and / when a is constant, allowing 
us to determine any one of these variables if we know the others. 

To simplify our notation, let us take the initial time in any discussion to be 

zero, and we call it f () : r L = t, } - 0. (This is effectively starting a stopwatch at /,,.) 

xinit = 0) = ,Y|| We can then let r 2 = ( be the elapsed time. The initial position (jcJ and the 

v (;u i - 0; - in initial velocity (v L ) of an object will now be represented by x u and t^, since they 

i ri;«)vii iinw represent x and v at t = 0. At time t the position and velocity will be called x 

and a (rather than x 2 and v 2 ). The average velocity during the time interval 

t - f (l will be (Eq. 2-2) 



i, : , 



in 



v related to a and t 
(a = constant, t = elapsed time) 



^ CAUTION 



A verage velocity, but only if 
a = constant 



x related to a ami t 
(a = constant) 



V =- 

I ~ 'a t 

since we chose t„ = 0. The acceleration, assumed constant in time, is (Eq. 2-4) 

V - Vq 
a = 

A common problem is to determine the velocity of an object. afteT any 
elapsed time (, when we are given the object's constant, acceleration, We can 
solve such problems by solving for v in the last equation to obtain: 

v - v„ + at. [constant acceleration! (2-6) 

Eor example, it may be known that the acceleration of a particular motor- 
cycle is 4.0m/s ; , and we wish to determine how fast it will be going after an 
elapsed time t - 6.0 s when it starts from rest (» = at r = 0). At / = 6.0 s, 
the velocity will be v = at = (4,0m/s 2 )(6,0s} = 24m/s, 

Next, let us sec how to calculate the position of an object after a time t 
when it is undergoing constant acceleration, The definition of average velocity 
(Eq. 2-2) is I> = (jt - Jf )/f, which we can rewrite as 

x = x + vt, (2-7) 

Because the velocity increases at a uniform rate, the average velocity,!;, will be 
midway between the initial and final velocities: 



i> — — 



[constant acceleration! (2— S) 

(Careful: Eq. 2-8 is not necessarily valid if the acceleration is not constant.) We 
combine the last two Equations with Eq. 2-6 and find 

„ + v 

X = Xf, + Vt = X(, + 

= x + 



Vq + i*, + at \ 

2 r 



or 



x + v^t + {at 2 . 



[constant acceleration] (2-9) 

Equations 2-6, 2-8, and 2-9 are three of the four most useful equations for 
motion at constant acceleration. We now derive the fourth equation, which is 
useful in situations where the time ( is not known. We begin with Eq. 2-7 and 
substitute in Eq. 2-8: 



x, I rl 



v -^y 



26 CHAPTER 2 Describing Motion: Kinematics in One Dimension 



Next we solve Eq. 2-6 for t, obtaining 
v - Vg 



i - 



a 



and substituting this into the previous equation we have 



'.., 



v + Vr, 






'Ti 



2fl 



Wc solve this for v 2 and obtain 

v 1 = vl + 2a(x - jf„) 



[constant acceleration] (2-10) 

which is the useful equation we sought. 

We now have four equations relating position, velocity, acceleration, and 
time, when the acceleration a is constant. We collect these kinematic equations 
here in one place for future reference (the tan background screen emphasizes 
theii" usefulness): 



\a — constant| (2-lla) 
[a = constant] (2-llb) 
[a = constant] (2-llt) 



V 


- 


Vt> 


+ 


at 




X 


= 


*0 


+ 


v v t + 


w 


V 2 


= 


«a 


+ 


2a(x - 


- Jfll) 


V 


- 


V 


1 
2 


% 





\a - constant I (2-lld) 



These useful equations are not valid unless a is a constant. In many cases we can set 
x it - 0, and this simplifies the above equations a bit. Note that x represents posi- 
tion, not distance, that x — x n is the displacement, and that t is the elapsed time. 



EXAMPLE 2-6 



Runway design. You are designing an airport foT small 
planes. One kind of airplane that might use this airfield must reach a speed 
before takeoff of at least 27.8 m/s (100 km/h), and can accelerate at 2,00 m/s 2 . 
(a) Tf the runway is 150 m long, can this airplane reach the required speed for 
take off? (h) If not, what minimum length must the runway have? 

APPROACH The plane's acceleration is given as constant (a - 2,00 m/s 2 ), so 
we can use the kinematic equations for constant acceleration. In (a), we are 
given that the plane can travel a distance of 150 m.The plane starts from Test, 
so V(, — and we take x u - 0. We want to find its velocity, to determine if it 
will be at least 27.8 m/s. We want to find v when we are given: 



KllilV. I! 



Wauled 



v, - (1 
«0 = 
-t = 150 m 

a = 2.U0 m/s : 



SOLUTION (a) Of the above four equations, Eq. 2-1 le will give us o when we 
know v lt . a,x, and x n : 

•ii 2 = vl + 2e(x - jf ) 

= + 2(2.0m/s 2 )(l50m) = 6QOm 2 /s 2 

v = \/600 m 2 /s 2 = 24.5 m/s, 

This runway length is not sufficient. 

(b) Now we want to find the minimum length of runway, x - x„, given 

v - 27.8 m/s and a - 2.00 m/s 2 . So we again use Eq. 2—1 Ic, but rewritten as 

(27.8 m/s) 2 - 



(x - x ) = 



ir - »s 



2a 2(2.0 m/s 2 ) 

A 200- m runway is more appropriate for this plane. 



193 m. 



■ ■ related to aandx 
(a = constant) 



Kinei untie eq t lotions 
for constant acceleratit >n 
(we'll use them a lot) 



®_ 



PHYSICS APPLIED 



Airport design 



■» PROBLEM SOLVING 

Equations 2 i I are rath! only when 
sht acceleration is constant, which we 
assume in this Example 



SECTION 2-5 Motion at Constant Acceleration 27 



Solving Problems 



Before doing more worked-out Examples, let us look at how to approach prohlem 
solving. First, it is important to note that physics is not a collection of equations to 
be memorized. (In fact, rather than memorizing the very useful Eqs. 2- IK it is 
better to understand how to derive them from the definitions of velocity and accel- 
eration as we did above.) Simply searching for an equation that might work can 
lead you to a wrong result and will surely not help you understand physics. A better 
approach is to use the following (rough) procedure, which wc put in a special "Box." 
(Other such Problem Solving Boxes, as an aid, will be found throughout the book.) 



PROBLEM SOLVING 



1. Read and reread the whole problem carefully 
before trying to strive it. 

2. Decide what object (or objects) you are going to 
study, and for what time interval. You can often 
choose the initial time to be t — 0. 

3. Draw a diagram or picture of the situation, with 
coordinate axes wherever applicable. [You can 
place the origin of coordinates and the axes wher- 
ever you like to make your calculations easier. You 
also choose which direction is positive and which is 
negative, Usually wc choose the x axis to the right 
as positive.] 

4. Write down what quantities are "known" or 
"given," and then what you want to know. Consider 
quantities both at the beginning and at the end of 
the chosen time interval, You may need to "trans- 
late" stated language into physical tenus, such as 
"Starts from rest' 1 means v u = 0- 

5. Think about which principles of physics apply in 
this problem. Use common sense and your own 
experiences. Then plan an approach. 

6. Consider which equations (and/or definitions) relate 
the quantities involved. Before using them, be sure 
their range of validity includes your problem (for 
example, Eqs. 2- 1 1 arc valid only when the 
acceleration is constant). If you find an applicable 



equation that involves only known quantities and 
one desired unknown, solve the equation alge- 
braically for the unknown. In many instances 
several sequential calculations, or a combination of 
equations, may be needed. It is often preferable to 
solve algebraically for the desired unknown before 
putting in numerical values. 

Carry out the calculation if it is a numerical 
problem. Keep one or two extra digits during the 
calculations, but round off the final answer(s) to the 
correct number of significant figures (Section 1 -4). 
Think carefully about the result you obtain: Is it 
reasonable? Does it make sense according to your 
own intuition and experience? A good check is to 
do a rough estimate using only powers of ten, as 
discussed in Section 1-7. Often it is preferable to 
do a rough estimate at the start of a numerical 
problem because it can help you focus your atten- 
tion on finding a path toward a solution. 
A very important aspect of doing problems is keeping 
track of units, An equals sign implies the units on 
each side must be the same, just as the numbers must. 
If the units do not balance, a mistake has no doubt 
been made, This can serve as a check on your solution 
(but it only tells you if you're wrong, not if you're 
right). And: always use a consistent set of units. 



PROBLEM SOLVING 
"Starting from rest" means 
» = at f = [i.e., i;,j =0] 



FIGURE 2-13 Is am pie 2-7. 



a = 2.00 m/i 1 



»0=D 



a = 2.00 m/s? 



r- 

x - 

.10,0 m 



EXAMPLE 2-7 



Acceleration of a car. How long does it take a car to cross 
a 30.0-m-wide intersection after the light turns green, if the car accelerates 
from rest at a constant 2.00 m/s 2 ? 

APPROACH We follow the Problem Solving Box, step by step, 
SOLUTION 

1. Reread the problem. Be sure you understand what it asks for (here, a time 
period). 

2. The object under study is the car. We need to choose the time interval 
during which we look at the car's motion: we choose ( - 0, the initial time, 
to be the moment the car starts to accelerate from Test (t\ } - 0); the time t is 
the instant the car has traveled the full 30.0-m width of the intersection, 

3. Draw a diagram: the situation is shown in Fig. 2-13, where the car is shown 
moving along the positive x axis. We choose x n = at the front bumper of 
the car before it starts to move. 



28 CHAPTER 2 Describing Motion: Kinematics in One Dimension 



4, The "knoHns" and the "wanted" are shown in the Table in the margin, and 
we choose x = 0- Note that "starting from rest" means v = at i — 0; 
that is, Vo = 0- 

5, The physics: the motion takes place at constant acceleration, so we can use 
the kinematic equations, Eqs. 2—11. 

6, Equations: we want to find the time, given the distance and acceleration; 
Eq. 2-1 lb is perfect since the only unknown quantity is t. Setting Vq = 
arid J*,, = in Eq. 2-1 lb (jt = x n + v n I + { at 1 ), wc can solve for /: 

,2 



A - -ill 

Zx 
a 



** = — , 



so 



r = 



7. The cakulaliim: 



2x 



t = \! — 



2(30,0 m) 

2.00 m/s 2 



5.48 s, 



This is our answer. Note that the units come out Correctly. 

8, We can check the reasonableness of the answer by calculating the final velocity 
1.1 = at = (2.00 m/s 2 ) (5.48 s) = 10.96 m/s, and then finding x = x n + at - 
+ | (10.% m/s + 0)(5.48s) = 30.0 m, which is our given distance. 

9. We checked the units, and they came out perfectly (seconds). 

NOTE In steps 6 and 7, when we took the square root, we should have written 
t— + y/2x/a = ± 5.48s. Mathematically there are two solutions. But the 
second solution, t = -5.48 s, is a time before our chosen time interval and 
makes no sense physically. We say it is "unphysical" and ignore it. 



We explicitly followed the Steps of (he Problem Solving Box in Example 2-7. 
In upcoming Examples, we will use our usual 'approach" and "solution" to avoid 
being wordy. 



EXAMPLE 2-8 



ESTIMATE | Air bags, Suppose you want to design an air- 
bag system that can protect the driver in a head-on collision at a speed of 
lOOkm/h (60mph). Estimate how fast the air bag must inflate (Fig. 2-14) to 
effectively protect the driver. How docs the use of a scat belt help the driver? 

APPROACH We assume the acceleration is roughly constant, so we can use 
Eqs. 2-11. Both Eqs, 2-1 la and 2-1 lb contain r, our desired unknown. They 
both contain a, so we must first find a, which we can do using Eq. 2-1 1c if we 
know the distance x over which the car crumples, A rough estimate might be 
about 1 meter. We choose the time interval to start at the instant of impact with 
the car moving at v a = KJ0km/h, and to end when the car comes to Test 
(v = 0) after traveling I m. 

SOLUTION We convert the given initial speed to SI units: lOQkm/h = 
100 x I0 1 m/3600s = 28 m/s. We then find the acceleration from Eq. 2-1 Ic: 



<■:, 



(28 m/s) 3 



= -39€ m/s 2 . 



2x 2.0 m 

This enormous acceleration takes place in a time given by (Eq. 2-1 la): 

v ~ q, 0-28 m/s 

t = = r = 0.07 s. 

a -390 m/s" 

To be effective, the air bag would need to inflate faster than this. 

What docs the air bag do? It spreads the force over a large area of the 
chest (to avoid puncture of the chest by the steering wheel). Ibe seat belt 
keeps the person in a stable position against the expanding air bag. 



Known 



Wanted 



-T() " 


- 


.V 


= 30.0 m 


;',■ 


- 2.00 m/s 2 


!'■, . 


- 



I 



•* PROBLEM SPUING 
Check your answer 



JPHYSICS APPLIED 
Car mfety — air bags 




FIGURE 2-14 An ah hag 
deploying on impact. Example 2-8. 



SECTION 2-6 Solving Problems 29 



FIGURE 2-15 Rxample 2-9: stop- 
ping distance for a braking car. 




Travel during 
reaction time 



v = cor s tan l 
S = 0.50 s 
a = 



1 4 m/s 



Travel durintt 
braking 



v decreases from 14 m/s to zero 
a = - 6.0 m/s- 



PHYSICS APPLIED 
Braking distances 



Pari I: Reaction time 



Known 



Wanted 



r = 0.50 s 

v (> = 14 m/s 
I? = 14 m/s 
w = 

x<, = 



Prj/v 2.- Braking 



Known 



Wanted 



,v ( , = 7-0 m 
Vci =14 m/s 

t> = 

(T = -6,0 m/s 2 



FIGURE 2 


-16 Example 2-9. 


Graph of 1 


I VS- t- 


14- 
P- 




i?v 


3T0 

s 8- 
i 6- 

4- 




i \\ 

| /=0.5s^S. 

i ^v. 
1 ^s. 


i.) 




15 1.0 1.5 2.0 2.5 
t(&) 



EXAMPLE 2-9 IMMOHIi 



Iraki ng distances. Estimate the minimum 
stopping distance for a car, which is important for traffic safety and traffic 
design. The problem is best dealt with in two parts, two separate time intervals, 
(I) The first time interval begins when the driver decides to hit the brakes, and 
ends when the foot touches the brake pedal. This is the "reaction time" during 
which the speed is constant, so a - 0. (2) The second time interval is the 
actual braking period when the vehicle slows down (a # 0) and comes to a 
stop. The stopping distance depends on the reaction time of the driver, the 
initial speed of the ear (the final speed is zero), and the acceleration of the car, 
For a dry road and good tires, good brakes can decelerate a car at a rate of 
about 5 m/s 2 to 8 m/s J . Calculate the total stopping distance for an initial 
velocity of 5Gkm/h (14 m/s ~ 31 mi/h) and assume the acceleration of the 
car is -6,0 m/s 2 (the minus sign appears because the velocity is taken to be in 
the positive * direction and its magnitude is decreasing), Reaction time for 
normal drivers varies from perhaps 0,3 s to about 1,0s; take it to be 0,50s 

APPROACH During the "reaction time," part (1), the car moves at constant 
speed of 14 m/s, so a - 0. Once the brakes are applied, part (2), the acceler- 
ation is a - —6.0 m/s 2 and is constant over this time interval. For both parts 
a is constant, so we can use Eqs. 2-11. 

SOLUTION Part (1). Wc take x yi = for the first part of the problem, in 
which the car travels at a constant speed of 14 m/s during the time interval when 
the driver is reacting (0,50 s). See Fig, 2-15 and the Table in the margin. To find x, 
the position of the car at f = 0.50 s (when the brakes arc applied), we cannot 
use Eq. 2- 1 1 c because x is multiplied by a, which is zero. But Eq, 2- 1 1 b works; 

x = vat + = (I4m/s)(0.50s) = 7.0 m. 
Thus the car travels 7,0 m during the driver's reaction time, until the moment 
the brakes are applied. We will use this result as input to part (2). 

Part (2). Now we consider the second time interval, during which the 
brakes are applied and the car is brought to rest. We have an initial position 
jf = 7,0 m (result of part ( I)), and other variables are shown in the Table in 
the margin. Equation 2-1 la doesn't contain x\ Eq. 2-1 lb contains x but also 
the unknown t. Equation 2-1 Ic, v 2 — vl = 2a(x - jc^), is what wc want; after 
setting x = 7.0 m, wc solve for jj, the final position of the car (when it stops): 



x - x ( , + 



v 1 - *£ 



= 7.0 m + 



?.tt 
0- (14 m/s) 2 



2( -6.0 m/s 2 ) 
= 7.0 m + 16 m = 23 m. 



7.0 m 



-l%m 2 .,V 
-12 m/s 3 



The car traveled 7,0 m while the driver was reacting and another 16m during the 
braking period before coming to a stop. The total distance traveled was then 23 m. 
Figure 2-16 shows a graph of v vs. t: v is constant from t — to / - 0.50s 
and decreases linearly, to zero, after t - 0.50 s. 

NOTE From the equation above for x, wc sec that the stopping distance after you 
hit the brakes (= x - x it ) increases with the square of the initial speed, not just linearly 
with speed. If you arc traveling twice as fast, it takes four times the distance to stop, 



30 CHAPTER 2 Describing Motion: Kinematics in One Dimension 



The analysis of motion we have been discussing in this Chapter is basically 
algebraic. It is sometimes helpful to use a graphical interpretation as well; see 
the optional Section 2-8. 



Falling Objects 



One of the most common examples of uniformly accelerated motion is that of an 
object allowed to fall freely near the Earth "s surface. That a falling object is 
accelerating may not be obvious at first And beware of thinking, as was widely 
believed until the time of Galileo (Fig, 2-17), that heavier objects fall faster than 
lighter objects and that the speed of fall is proportional to how heavy the object is. 

Galileo's analysis of falling objects made use of his new and creative tech- 
nique of imagining what would happen in idealized (simplified) cases. For free 
fall, he postulated that all objects would fall with the rame constant acceleration 
in the absence of air or other resistance. He showed that this postulate predicts 
that for an object falling from rest, the distance traveled will be proportional to 
the square of the time (Fig. 2-18); that is, d x t 2 . We can see this from 
Eq. 2-1 lb, but Galileo was the first to derive this mathematical relation. 
[Among Galileo's great contributions to science was to establish such mathe- 
matical relations, and to insist on specific experimental consequences that could 
be quantitatively checked, such as d tx i 2 .] 

Ti support his claim that falling ohjects increase in speed as they fall, Galileo 
made use of a clever argument: a heavy stone dropped from a height of 2 m will 
drive a stake into the ground much further than will the same stone dropped from 
a height of only 0.2 m. Clearly, the stone must be moving faster in the former case. 

As we saw, Galileo also claimed that all objects, light or heavy, fall with the 
same acceleration, at least in the absence of air, If you hold a piece of paper 
horizontally in one hand and a heavier object — say, a baseball — in the other, 
and release them at the same time as in Fig, 2- 19a, the heavier object will reach 
the ground first, But if you repeat the experiment, this time crumpling the paper 
into a small wad (see Fig. 2-1%), you will find that the two objects reach the 
floor at nearly the same time. 

Galileo was sure that air acts as a resistance to very light objects that, have a 
large surface area. But in many circumstances this air resistance is negligible. In a 
chamber from which the air has been removed, even light objects like a feather or 
a horizontally held piece of paper will fall with the same acceleration as any other 
object (see Fig. 2-20). Such a demonstration in vacuum was not possible in 
Galileo's time, which makes Galileo's achievement all the greater. Galileo is often 
called the "father of modern science," not only for the content of his science 
(astronomical discoveries, inertia, free fall), but also for his style or approach to 
science (idealization and simplification, mathemati'/ation of theory, theories that 
have testable consequences, experiments to test theoretical predictions). 




n/ 



FIGURE 2-19 (a) A ball 
and a light piece of paper are 
d rupped at the same time, 
(b) Repeated, with the paper 
wadded up. 




FIGURE 2-20 A ruck 
and a feather are dropped 

simultaneously (a) in air, 
(b) in a vacuum. 




FIGURE 2-17 Galileo Galilei 

(1564-1642). 

<j > CAUTION 

The speed of a falling object is NOT 
proportional t» its mass or weight 



FIGURE 2-1S Multiftash 

photograph of a falling apple, at 
equal lime intervals, line apple falls 
farther during each successive 
interval, which means it is 
accelerating, 




< 



. 



f 



Air-filled tube 

(a) 



Evacuated tube 
(b) 



SECTION 2-7 Falling Objects 31 



f. 'ml ilea's hypothesis: free fa!! is tit 
constant necetei filial! •:■ 



Acceleration due to gravity 



^ PROBLEM SOLVING 
Yi.isi choose v it.' be positii e 
either tip or down 

"Drop" means vq = 

FIGURE 2 21 Rx ample 2- 1 0. 
(a) An object dropped from a lower 

fails with progressively greater 
speed and covers greater distance 
with each successive second. (See 
also Fig. 2-18.) (b) Graph of y vs. t. 

Acceleration 

due to 
gravity 

(After 1 .00 s) 

v 2 = ly.ft m 

(After 2.00 s) 



V3 = 44. 1 m 
'(After 3.00 s) 




Galileo's specific contribution to our understanding of the motion of falling 
objects can be summarized as follows; 

at a given location on the Earth and in the absence of air resistance, all 
objects fall with the same constant acceleration. 

We call this acceleration the acceleration due lo gravity tin the Earth, arid we 
give it the symbol jf. Its magnitude is approximately 

g - 9.80 m/s 2 . |at surface of Earth] 

In British units g is about 32 ft/s 2 . Actually, g varies slightly according to latitude 
and elevation, but these variations are so small that we will ignore them for most 
purposes. The effects of air resistance are often small, and we will neglect them I'ot 
the most part. However, air resistance will be noticeable even on a reasonably 
heavy object if the velocity becomes large.* Acceleration due to gravity is a 
Vector, as is any acceleration, and its direction is toward the center of the Earth. 

When dealing with freely falling objects we can make use of Eqs. 2-11, 
where for a we use the value of g given above. Also, since the motion is vertical 
we will substitute v in place of x, and y<> in place of x (l . We take y n = unless 
otherwise Specified. If is arbitrary whether 'we choose y to be positive in the 
upward direction or in the downward direction; but we must be consistent about 
if throughout a problem s solution. 

M 31fA fl I a I ^a ['■ Falling from a tower. Suppose that a ball is dropped 
(v a = 0) from a tower 70,0 rn high, flow far will the ball have fallen after a 
time fj = 1.00 s, t 2 = 2.00 s, and r 3 = 3.00 s? 

APPROACH Let us take y as positive downward. We neglect any air resistance. 
Thus the acceleration is a = g = +9,80 m/s 2 , which is positive because we 
have chosen downward as positive. We set i;„ = and y, - 0. We want to find 
the position y of the ball after three different time intervals. Equation 2- 1 lb, 
with x replaced by y, relates the given quantities (t, a, and v„) to the unknown v. 
SOLUTION We set t - t, - 1.00s in Eq. 2-1 lb: 

y\ = v a t, + jar? 

= + {at] = |(9,80 m/V)( 1,00 s) 2 = 4,90 m. 
The ball has fallen a distance of 4,90m dining the time interval / = to 
/, = 1. 00s. Similarly, after 2.00s (= i 2 ), the ball's position is 

y 2 = jat] = ^(9.80m/s 2 )(2.00s) 2 = 19.6 m. 
Finally, after 3.00s (= f_,), the ball's position is (see Fig. 2-21) 

y, ={at\ = +(9.80 m/s 2 ) (3.00 s) 2 = 44.1 m. 
NOTE Whenever we say "dropped," we mean v a - 0. 



EXAMPLE 2-11 



Thrown down from a tower. Suppose the hall in 
Example 2-10 is thrown downward with an initial velocity of 3,00 m/s, instead 
of being dropped, (a) What then would be its position after 1 .00 s and 2.00 s? 
(b) What would its speed be after 1, 00s and 2,00s? Compare with the speeds 
of a dropped ball. 

APPROACH We can approach this in the same way as in Example 2-10, 
Again we use Eq. 2-1 lb, but now <; n is not zero, it is v Q = 3,00 m/s. 
SOLUTION (a) At t = 1.00s, the position of the ball as given by Eq. 2-1 lb is 

y = v a ( + Ut 2 = (3.00 m/s )( 1. 00s) + £(9.80 m/s 2 )(l.Q0 s) 2 = 7.90 m, 



At i = 2.00 s, (time interval t = to t = 2.00 s), the position is 

y = i*,r + W = (3.00m/s)(2.00s) + £(9.80m/s 2 )(2.00 s) 2 = 25.6 m. 
As expected, the ball falls farther each second than if it were dropped with v n - 0, 



'The speed of an object falling in air (or other fluid) does no[ increase indefinitely. If [he object falls 
far enough, it will reach a maximum velocity called the terminal velocity due to air resistance. 



32 CHAPTER 2 Describing Motion: Kinematics in One Dimension 



(b) The velocity is obtained from Eq. 2- 1 1 a: 
v - v a + at 
= 3.00 m/s + (9.80 m/s 2 )( 1.00 s) = 12.8 m/s [at t, = 1.00 s] 
= 3.00 m/s + (9.80 m/s 2 )(2.00 s) = 22.6 m/s. [at i 2 = 2.00 s] 

In Example 2-10, when the ball was dropped {v tt - 0), the first term (t\ h ) in 

these equations was zero, so 
v = + ai 
= (9.80 m/V)( 1 .00 s) = 9.80 m/s [at r, = 1 .00 s] 

= (9.80 m/s 2 )(2.00 s) = 19.6 m/s. [at t 2 = 2,00 s] 

NOTE For both Examples 2-10 and 2-1 1, the speed increases linearly in time by 
9.80 m/s during each second. But the speed of the downwardly thrown ball at any 
moment is always 3.00 m/s (its initial speed) higher than that of a dropped ball. 



EXAMPLE 2-12 



Ball thrown upward, I. A person throws a ball upward 
into the air with an initial velocity ol 15.0 m/s. Calculate (a) how high it goes, 
and (b) how long the ball is in the air before it eomes back to his hand. 

APPROACH We are not concerned here with the throwing action, but only with 
the motion of the ball after it leaves the thrower's hand (Fig. 2-22) and until it 
comes back to his hand again. Let us choose y to be positive in the upward direc- 
tion and negative in the downward direction. (This is a different convention from 
that used in Examples 2-10 and 2-11, and so illustrates our options.) The accel- 
eration due to gravity will have a negative sign, a = —g - -9. SO m/s 2 . As the 
ball rises, its speed decreases until it reaches the highest point (B in Fig. 2-22), 
where its speed is zero for an instant; then it descends, with increasing speed, 
SOLUTION (a) We consider the time interval from when the ball leaves the 
thrower's hand until the ball reaches the highest point. To determine the 
maximum height, wc calculate the position of the ball when its velocity equals 
zero (v — at the highest point). At / = (point A in Fig. 2-22) we have 
y a - 0, v n = 15.0 m/s, and a = -9.30 m/s 2 , At time / (maximum height), 
v = 0, a - -9.S0m/s 2 , and wc wish to find y. We use Eq. 2-1 1 e, replacing x 
with y: v 1 = v\ + lay. We solve this equation for y: 

- ( 15.0 m/s) 2 



7 



11.5 m. 



l>i 2[ -9.80 m/s : ) 

The ball reaches a height of 1 1.5 m above the hand. 

(b) Now we need to choose a different time interval to calculate how long the 
ball is in the air before it returns to his hand. We could do this calculation in 
two parts by first determining the time required for the ball to reach its highest 
point, and then determining the time it takes to fall back down. However, it is 
simpler to consider the time interval for the entire motion from A to B to C 
(Fig. 2-22) in one step and use Eq. 2- 1 1 b- We can do this because > (or x) 
represents position ot displacement, and not the total distance traveled. Thus, at 
both points A and C, y = 0, We use Eq. 2-1 lb with a = -9,80 m/s 2 and find 

y = v u t + jar 

= (15.0 m/s )t + j(-9.80m/s 2 > 2 . 
This equation is readily factored (we factor out one t): 

(1 5.0 m/s - 4.90 m/s 2 i)t = 0. 
There are two solutions: 

15.0 m/s 



ii 



and 



3.06 s. 



4.90 m/s" 

The first solution (r. — 0} corresponds to the initial point (A) in Fig. 2-22, when the 
ball was first thrown from y - 0. The second solution, t — 3.06 s, corresponds 
to point C, when the ball has returned to y - 0, Thus the ball is in the air 3.06s. 




FIG U RE 2 -22 A n o hject t h rown 
into the air leaves the thrower's hand 
at A, reaches its maximum height at 
B, and returns to the original position 

at C. Examples 2-12, 2-13, 2-14. 
and 2-15. 



SECTION 2-7 Falling Objects 33 



<j > C A U T I M 



Quadratic equations have two 

solutions. Sometimes only one 

■■ am tponds to reality, 

sometimes both 



<| ) caution 



( Velocity and acceleration are not 
always in the same direction; the 

acceleration (of gravity) always 

points Jo ivn 

(2) a =£ even at the highest point 

of a trajectory 




FIG U RE 2-22 ( Repeated for 
Hxamples 2- 1 3. 2- 14. and 2- 15.) 



Note the symmetry: the speed at any 

height is the ■iunte tt'ilnl gone.; u;> as 

when coming down 

(but the direction is opposite) 



We did not consider Ihe throwing action in this Example. Why? Because during 
the throw, the thrower's hand is touching the ball and accelerating the ball at a 
rate unknown to us — the acceleration is not g. We consider only the time when 
the ball is in the air and the acceleration is equal to g. 

Every quadratic equation (where the variable is squared) mathematically 
produces two solutions- In physics, sometimes only one solution corresponds to 
the real situation, as in Example 2-7, in which case we ignore the "unphysical" 
solution. But in Example 2-12, both solutions to our equation in I 1 aTe 
physically meaningful: r = and t = 3.06s. 



EXAMPLE 2-13 1 Two possible misconceptions. Give 



examples to show the error in these two common misconceptions: (I) that accel- 
eration and velocity are always in the same direction, and (2) that an object 
thrown upward has zero acceleration at the highest point (B in Fig. 2-22). 

RESPONSE Both are wrong. (I) Velocity and acceleration are not necessarily in 
the same direction. When the ball in Example 2- 12 is moving upward, its velocity 
is positive (upward), whereas the acceleration is negative (downward), (2) At the 
highest point (B in Fig. 2-22), the ball lias zero velocity for an instant. Ts the 
acceleration also zero at this point? No, The velocity near the top of the arc 
points upward, then becomes zero (for zero time) at the highest point, and then 
points downward. Gravity does not stop acting, so a = —g = -9.8(1 m/s 2 even 
there. Thinking that a = at point B would lead to the conclusion that upon 
reaching point B. the ball would stay there: if the acceleration (= rate of change of 
velocity) were zero, the velocity would stay zero at the highest point, and the ball 
would stay up there without falling. Tn sum, the acceleration of gravity always 
points down toward the Earth, even when the object is moving up. 



EXAMPLE 2-14 



Ball thrown upward, II. Let us consider again the ball 
thrown upward of Example 2-12, and make nxvre calculations. Calculate (a) how 
much time it takes for the ball to reach the maximum height (point B in Fig. 2-22), 

and (b) the velocity of the ball when it returns to the thrower's hand (point C). 

APPROACH Again we assume the acceleration is constant, so Eqs. 2-1 1 are valid. 
We have the height of 1 1.5 m from Example 2-12. Again we take y as positive upward 
SOLUTION (a) Wc consider the time interval between the throw (t - 0, 
i\) = 15.0 m/s) and the top of the path (y = + 1 1.5 m), v - 0, and wc want 
to find t. The acceleration is constant at a - -g = -9.80 m/s 2 . Both 
Eqs. 2-1 la and 2-1 lb contain the time t with other quantities known. Let us 
use Eq, 2-1 la with a — —9.80 m/s 2 , v - 15.0 m/s, and v — 0: 

v = % + at; 



setting -tt 



and solving for r gives 
Vq 1 5.0 m/s 






-9.80 m/s 






1 .53 s. 



This is just half the time it takes the ball to go up and fall back to its original 
position [3.06 s, calculated in part (b) of Example 2- 12]. Thus it takes the same 
time to reach the maximum height as to fall back to the starting point. 
(b) Now we consider the time interval from the throw (r = 0, v = 1 5,0 m/s) 
until the ball's return to the hand, which occurs at / = 3.06s (as calculated 
in Example 2-12), and we want to find v when t - 3.06s: 

v - ii,, + at - 15.0 m/s - (9.80 m/s 2 ) (3.06 s) - -15.0 m/s. 

NOTE The ball has the same magnitude of velocity when it returns to the 
starting point as it did initially, but in the opposite direction (this is the 
meaning of the negative sign). Thus, as we gathered from paTt (a), the motion 
is symmetrical about the maximum height. 



34 CHAPTER 2 Describing Motion: Kinematics in One Dimension 



EXERCISE C Two balls arc thrown from a cliff. Oric is thrown directly Up, the Other 
directly down. Both balls have the same initial speed, and both hit the ground below 
the cliff. Which ball hits the ground at the greater speed: (a) the ball thrown upward. 
(b) the ball thrown downward, or (t) both the same? IgnoTe air resistance. [Hint: See 
the result of Example 2-14, part (fc).| 

The acceleration of objects such as rockets and fast airplanes is often 
given as a multiple of g = 9.80 m/s 2 . For example, a plane pulling out of a 
dive and undergoing 3.00 gs would have an acceleration of (3.00}(9,80m/s ; ) - 
29.4 m/s 2 . 

| EXERCISE D If a car is said to accelerate at 0.50 g, what is its acceleration in tn/s : ? 

Additional Example — Using the Quadratic Formula 



EXAMPLE 2-15 



Ball thrown upward. III. For the ball in Example 2-14, 
calculate at what time / the ball passes a point 8.00 m above the person's hand. 

APPROACH We choose the time interval from the throw (f = 0, v (i = 15.0 m/s) 
until the time t (to be determined) when the ball is at position v = 8.00m, 
using Eq. 2-1 lb. 

SOLUTION We want /, given y = 8.00 m, y = 0, »„= 15.0 m/s, and 
a = -9.80 m/s 2 . We use Eq. 2-1 lb: 

v = yo + <v + ^tit 2 
8,00m - + (15.0 m/s) ( + |( -9.80 m/s 2 )/ 2 . 

To solve any quadratic equation of the form at 1 + bt + c = 0, where a, 6, 
and c arc constants (a is not acceleration here), we use the quadratic formula 
(sec Appendix A -4): 



t = 



b ± \/b 2 - 4<Tc 
la 



We rewrite our v equation just above in standard form, at 2 + bt + c — 0: 

(4.90 m/s 2 ) t 2 - (15-0 m/s) t + (8.00m) = 0. 

So the coefficient a is 4.90 m/s 2 , b is - 15.0 m/s, and c is 8,00 m, Putting these 
into the quadratic formula, we obtain 



15.0 m/s ± V(l5.0m/s) 2 - 4(4,90 m /V)(8,00 m) 
2(4,90 m/s 2 ) 

which gives us t - 0.69 s and i = 2.37s. Are both solutions valid? Yes, 
because the ball passes y = 8.00m when it goes up {( = 0.69s) and again 
when it comes down (r = 2.37 s). 

For some people, graphs can be a help in understanding. Figure 2-23 shows 
graphs of y vs. / and v vs. / for the ball thrown upward in Fig. 2-22, incorporating 
the results of Examples 2-12, 2-14, and 2-15. We shall discuss some useful 
properties of graphs in the next Section. 

We will use the word "vertical" a lot in this book, What docs it mean? (Try 
to respond before reading on.) Vertical is defined as the line along which an 
object falls. Or, if you put a small sphere on the end of a string and let it hang, 
the string represents a vertical line (sometimes called a plumb line). 

| EXERCISE E What does horizontal mean? 



Acceleration expressed in g's 



*» PROBLEM SOLVING 
Using the quadratic formula 

\1 
10 

s 

I 6 









. 


.. 




3 m 


\> 










.' 


.53 s\ 








p- 1 = 

f o.6y s 


1=1 

5.37 s 

















































.5 2 2.5 3 

Ms) 



v5 



1 



9,5 
(a) 
20 

l> 

[0 

1 5 
5 

-5 

-10 

-15 

-20 

0.5 I 1.5 2 2.5 3 3.5 

(b) f(s) 

FIGURE 2-23 Graphs of (a) y vs. r. 

(b) v vs. / for a ball thrown upward, 
Examples 2-12, 2-14, and 2-15. 



















































1 - 


= 1..' 


3 s 





























































SECTION 2-7 Falling Objects 35 



>(] - 






lll- 




v 






J 1 1 m 


9 30- 


/ ! A ' = 




4 20- 

■z 

| lO- 


/ l 1.0 s 




il 


£- 1 1 1 


1 1- 



1.0 



2.0 3,0 4,0 
Time, t (s) 



5 



FIGURE 2-24 

vs. time for an 
uniform veloci 



Graph of position 
object moving at a 
ly of 1 1 m/s. 



Velocity - slopeofx vs. I ?rap/i 



.S/(j,'jf of a t urve 



Graphical Analysis of Linear Motion 

Figure 2-9 showed the graph of the velocity of a car versus time for two cases 
of linear motion: (a) constant velocity, arid (b) a particular case in which the 
magnitude of the velocity varied. Tt is also useful to graph, or '"plot," the posi- 
tion x (or y) as a function of time, as we did in Fig. 2-23a. The time t is 
considered the independent variable and is measured along the horizontal axis. 
The position, x, the dependent variable, is measured along the vertical axis, 

Let us make a graph of x vs. (, and make the choice that at t = 0, the 
position is jfii = 0. First we consider a car moving at a constant velocity of 
40 km/h, which is equivalent to 1 1 m/s. Equation 2-1 lb tells us x = vt, and we 
sec that x increases by 1 1 m every second. Thus, the position increases linearly in 
time, so the graph of x vs. f is a straight line, as shown in Fig. 2-24, Each point 
on this straight line tells us the car's position at a particular time. For example, 
at r - 3.0 s, the position is 33 m, and at ( = 4.0 s, x = 44 m, as indicated by 
the dashed lines. The small (shaded) triangle on the graph indicates the slope of 
the straight line, which is defined as the change in the dependent variable (Ax) 
divided by the corresponding change in the independent variable (At): 

i ir 

slope ——■ 

At 

We see, using the definition of average velocity (Eq. 2-2), that the dope of the x 
vs. t graph is equal to the velocity. And, as can be seen from the small triangle on 
the graph, A.v/Ar = ( 1 1 m)/(1,Qs) = 1 1 m/s, which is the given velocity. 

The slope of the x vs. : graph is everywhere the same if the velocity is 
constant, as in Fig, 2-24. But if the velocity changes, as in Fig, 2-25a, the slope of 
the x vs. t graph also varies. Consider, for example, a car that (1) accelerates 
uniformly from Test to 15 m/s in 15s, afteT which (2) it remains at a constant 
velocity of 15 m/s for the next 5.0s; (3) dining the following 5.0s, the car slows 
down uniformly to 5,0 m/s, and then (4) remains at this constant velocity. This 
velocity as a function of time is shown in the graph of Fig. 2-25a. To construct 
the x vs. f gTaph, we can use Eq. 2-1 lb {x — x a + Vnt + i<tt 2 } with constant 
acceleration for the interval f = to t = 15 s and for i = 20s to i - 25 s; 
for the constant velocity period I = 15 s to t = 20 s, and after t = 25 s, we 
set a - 0. The result is the .v vs. t graph of Fig. 2-25b. 

From the origin to point A, the x vs. / graph (Fig. 2-25bJ is not a straight line, 
hut is curved. The slnpe of a curve at any point is defined as the slope of the 
tangent to the curve at that point. (The tangent is a straight line drawn so it 
touches the curve only at that one point, but does not pass across or through the 
curve.) For example, the tangent to the x vs. t curve at the time ( = 10.0s is 
drawn on the graph of Fig. 2-25b. A triangle is drawn with At chosen to be 4.0 s; 

'Some Sections of [his book, such as this one. may be considered optional ill Ihe discretion of [he 

instructor. See the Preface for more details, 



FIGURE 2-25 (a) Velocity vs. time and (h) displacement vs. time for an object with variable velocity. (See text.) 

300 r- 



■>" 


15.0 






■= 




*> 


10,0 


£■ 




■J 




-; 


5.0 



- 




A 




15 




(2) 




— 


A\) 






\ (4) 




i i 


1 




C D 

1 1 1 



10 10.0 15,0 20.0 25,0 
(a) Time, i (s) 



=-0.0 



s<;\ 



iiu 





X i Ax = 


= 45m 




^fTangeut 




^%t 
1 1 


\jA*=40m 
-4..:i', 
ill 1 


1 


10.0 


15,0 200 
Time, t (s) 


25.0 



30,0 



36 CHAPTER 2 Describing Motion: Kinematics in One Dimension 



Ax can be measured off the graph for this chosen At and is found to be 40m. 
Thus, the slope of the curve al t = 10,0 s, which equals the instantaneous 
velocity at that instant, is v = Ax/ At = 40m/4.0s = 10 m/s. 

In the region between A and R (Fig, 2-25b) the x vs. t graph is a straight line 
because the slope (equal to the velocity) is constant. The slope can be measured 
using the triangle shown for the time interval between t = 17s and t = 20 s, 
where the increase in x is 45 m: Ax! At = 45 m/XOs = 15 m/s. 

The slope of an .v vs. / graph at any point is Ax /At and thus equals the 
velocity of the object being described at that moment. Similarly, the slope at any 
point of a ■» vs. / graph is Av/At and so (by Eq. 2-4) equals the acceleration at 
that moment. 

Suppose we were given the x vs. / graph of Fig. 2-25b. We could measure 
the slopes at a number of points and plot these slopes as a function of time. 
Since the slope equals the velocity, we could thus reconstruct the v vs. ( graph! 
In other words, given the graph of x vs. (, we can determine the velocity as a 
function of time using graphical methods, instead of using equations. This tech- 
nique is particularly useful when the acceleration is not constant, for then 
Eqs, 2-1 1 cannot be used, 

It, instead, wc arc given the v vs. t graph, as in Fig. 2-25a, wc can determine the 
position, a-, as a function of time using a graphical procedure, which wc illustrate by 
applying it to the v vs. t graph of Fig. 2-25a. Wc divide the total time interval into 
subintcrvals, as shown in Fig. 2 -26a, where only six are shown (by dashed vertical 
lines). In each interval, a horizontal dashed line is drawn to indicate the average 
velocity during that time interval. For example, in the first interval, the velocity 
increases at a constant rate from zero to 5.0 m/s, so v - 2.5 m/s; and in the 
fourth interval the velocity is a constant 15 m/s, so v - 15 m/s (no horizontal 
dashed line is shown in Fig. 2-26a since it coincides with the curve itself). The 
displacement (change in position) during any subinterval is Ax - vAt. Thus 
the displacement during each subinterval equals the product of v and Al, which is 
just the area of the rectangle (height X base - I> X if), shown shaded in rose, for 
that interval. The total displacement after 25 s, say, will be the sum of the areas of 
the first five rectangles. 

If the velocity varies a great deal, it may be difficult to estimate v from the 
graph. To reduce this difficulty, we can choose to divide the time interval into 
many more — but narrower — subintervals of time, making each A; smaller as 
shown in Fig. 2-26b, More intervals give a better approximation, Ideally, we 
could let A( approach zero; this leads to the techniques of integral calculus, 
which we don't discuss here. The result, in any case, is that the total displacement 
between any two times is equal to the area under the vvs.t graph between these 
two times , 



EXAMPLE 2-16 



Displacement using v vs, t graph. A space probe accele- 
rates uniformly from 50 m/s at t - to 150 m/s at t — 10s. How far did it 
move between ( - 2.0 s and ( - 6,0 s? 

APPROACH A graph of it vs. t can be drawn as shown in Fig. 2-27. We need to 
calculate the area of the shaded region, which is a trapezoid, The area will be 
the average of the heights (in units of velocity) times the width (which is 4.0 s). 
SOLUTION The acceleration is a = ( 150 m/s - 50m/s)/10s = 10 m/s 2 . Using 
Eq. 2-1 1 a, or Fig. 2-27, at f = 2.0 s, v = 70 m/s; and at f = 6.0 s, v = 110 m/s. 
Thus the area, (v X At), which equals Ax, is 

'70 m/s + 110 m/s V 
— — 1(4.0 s) = 360 m. 



Ax 



NOTE For this case of constant acceleration, we could use Eqs. 2- 1 1 and we 
would get the same result, 




10 15 20 25 30 
Time, ( (S) 

(b) 

FIGURE 2-26 Determining the 

displacement from the graph of 
v vs. t is done by calculating areas. 



Displacement = area 

under v w tgrttph 



FIGURE 2-27 Example 2-16. The 

shaded area represents the displace- 
ment during the time interval 

t = 2.0 s to I = 6-0 s. 



E 




Tn cases where the acceleration is not constant, the area can be obtained by- 
counting squares on graph paper, 



♦SECTION 2-8 Graphical Analysis of Linear Motion 37 



I Summary 



[The Summary that appears at the end of each Chapter in this; 
book gives a brief overview of the main ideas of the Chapter, 
The Summary ainnoi serve to give an understanding of the 
material, which can be accomplished only by a detailed 
reading of the Chapter,] 

Kinematics deals with the description of how objects 
move. The description of the motion of any object must 
always be given relative to some particular reference frame. 

The displacement of an object is the change in position of 
(he object, 

Average speed is the distance traveled divided by the 
elapsed time ot lime interval, if. the lime period over which 
we choose to make our observations. An object's average 
velocity over a particular time interval \t is its displacement 
Av during that lime interval, divided by if: 



where Av is the change of velocity during the time interval Af. 
Instantaneous acceleration is the average acceleration taken 
over an infinitesimally short time interval. 

If an object has position x (1 and velocity ty, at time i = 
and moves in a straight line with constant acceleration, the 
velocity v and position x at a later time f are related to the 
acceleration a. Ihe initial position ,t ( |. and Ihe initial velocity 
% by Eqs. 2-11: 



v = v,) + at, 



ir = ■)>?. 



V(, + M* ~ X<] 



x = jt ( j +■ i\ t t + \at 2 , 
v + % 



(2-U) 



v - 



Ax 

if" 



12-2) 



The instantaneous velocity, whose magnitude is the same 
as the instantaneous speed, is defined as the average velocity 
taken over an infinitesimally short lime interval. 

Acceleration is the change of velocity per unit time, An 
object's average acceleration over a time interval A/ is 



Objects that move vertically near the surface of the 
Earth, either falling or having been projected vertically up ot 
down, move with the constant downward acceleration due to 
gravity, whose magnitude is g - 9.80 m/s : if air resistance 
can be ignored. We can apply Eqs. 2-11 for constant accelera- 
tion to objects that move up or down freely near the Farth's 
m.i riacc 

[*The slope of a curve at any point on a graph is the 
slope of the tangent to the curve at that point. If the graph is 
x vs. f , the slope is Ax/ At and equals the velocity at that point, 
The area under a v vs. I graph equals the displacement 
between any two chosen limes,] 



| Questions 



7. 

8. 
id. 



12. 



Does a car speedometer measure speed, velocity, or both? 
Can an object have a varying speed if its velocity is 
constant? If yes, give examples. 

When an object moves with constant velocity, does its 
average velocity during any time interval differ from its 
instantaneous velocity at any instant? 
In drag racing, is it possible for the car with the greatest 
speed crossing the finish line to lose the race? Explain. 
If one object has a greater speed than a second object, 
does the first necessarily have a greater acceleration? 
Explain, using examples. 

Compare the acceleration of a motorcycle lhal accelerates 
from 80 km/h to 90 km/h with the acceleration of a bicycle 
dial accelerates from rest to 10 km/h in die same time. 
Can an object have a northward velocity and a southward 
acceleration? Explain, 

Can the velocity of an object he negative when its 
acceleration is positive? What about vice versa? 
Give an example where both the velocity and accelera- 
tion are negative, 

Two cars emerge side by side from a tunnel. Car A is 
traveling with a speed of 60 km/h and has an accelera- 
tion of 40 km/h/min, CaT B has a speed of 40 km/h and 
has an acceleration of 60 km/h/min. Which car is 
passing the otheT as they come out of the tunnel? 
Explain your reasoning. 

Can an object be increasing in speed as its acceleration 
decreases? If so, give an example, If not, explain, 
A baseball player hits a foul ball straight up into the air. It 
leaves the bal with a speed of 120 km/h, In the absence of 
air resistance, how fast will the ball be traveling when the 
catcher catches it? 



13. As a freely falling object speeds up. what is happening to 
its acceleration due to gravity — does i( increase, decrease, 
or stay the same? 

14. How would you estimate the maximum height you could 
throw a ball vertically upward' How would you estimate 
the maximum speed you could give it? 

15. You travel from point A to point B in a car moving at a 
constant speed of 70 km/h. Then you travel the same 
dis lance from point B to another poinl C. moving al a 
constant speed of 90 km/h, Is your average speed for the 
entire trip from A to C 80 km/h? Explain why or why not. 

16. In a lecture demonstration, a 3.0-m-long vertical string 
with ten bolts lied to it al equal intervals is dropped from 
the ceiling of the lecture hall, The string falls on a tin 
plate, and the class hears the clink of each bolt as it hits 
the plate, The sounds will not occur al equal lime 
intervals. Why? Will the time between clinks increase or 
decrease near the end of the fall? How could the bolls be 
tied so that the clinks occur at equal intervals? 

17. Which one of these motions is not at constant acceleration: 
a rock falling from a cliff, an elevator moving from the 
second floor to the fifth floor making stops along the way. 
a dish resting on a table? 

18. An object that is thrown vertically upward will return to its 
original position with the same speed as it had initially if air 
resistance is negligible, tf air resistance is appreciable, will this 
result be altered, and if so. how? [Hint: The acceleration due to 
wr resistance is always in a direction opposite to the motion.] 

19. Can an object have zero velocity and nonzero accelera- 
tion at the same time? Give examples. 

20. Can an object have zero acceleration and nonzero 
velocity al (he same lime? Give examples. 



38 CHAPTER 2 Describing Motion: Kinematics in One Dimension 



'21. Describe in words the motion plotted in Fig. 2-28 in * 22, Describe in words the motion of the object graphed in 
terms of w, fl, etc. [Hint: First try to duplicate the motion Fig, 2-29, 

plotted by walking or moving your hand | 



20 






































\ 


i 


"■ 10 




















V 
























II 























in 



20 30 

MS) 



40 



50 




10 20 30 40 50 60 70 80 90 100 1 10 120 
FIGURE2-29 Question22. Prob] ems 49 and 54. 



FIGURE 2-28 Question 2] , Problems 50, 51 , and 55 



| Problems 



[The- Problems at the end of each Chapter are ranked I, II, or 
Til according to estimated difficulty, with (I) Problems being 
easiest. Level III are meant as challenges for the best 
students. The Problems art arranged by Section, meaning that 
the reader should have read up to and including that Section, 
but not only that Section — Problems often depend on earlier 
material, Finally, there is a set of un ranked "General Prob- 
lems" not arranged by Section number. | 

2-1 to 2-3 Speed and Velocity 

1. (I) What must be your car's average speed in order to 
travel 235 km in 325 h? 

2. (I) A bird can fly 25 km/h. How long does it take to fly 
15 km? 

i. (I) ff you are driving 1 1(1 km/h along a straight road and 
you look to the side for 2.0 s, how far do you travel during 
this inattentive period? 

4. (I) Convert 35 mi/h to (a) km/h. (h) m/s, and (c) ft/s. 

5. (I) A rolling ball moves from x l = 3.4cm to .v; = -4.2 cm 
during the time from 1] = 3.0 s to t 2 = 6.1 s. What is its 
average velocity? 

6. (II) A particle at f| = -2.0s is at .\\ = 3.4cm and at 
t 2 = 4,5 s is at x 2 = 8,5 cm, What is its average velocity? 
Can you calculate its average speed from these data'? 

7. (II) You are driving home from school steadily at 95 km/h 
for 130 km. It then begins to rain and yon slow to 65 km/h. 
You arrive home after driving 3 hours and 20 minutes. 
(a) How far is your hometown from school"? (b) What was 
your average speed? 

8. (II) According to a rule -of -thumb, eveTy five seconds 
between a lightning flash and the following thunder gives 
the distance to the flash in miles. Assuming that the flash 
of light arrives in essentially no time at all. estimate the 
Speed of sound in m/s from this rule 

9. (II) A person jogs eight complete laps around a quarter- 
mile track in a total lime of 12.5 min, Calculate (a) the 
average speed and (b) the average velocity, in m/s. 

UK (II) A horse canters away from its trainer in a straight 
line, moving llfim away in 14.0s. It then turns abruptly 
and gallops halfway back in 4,8 s, Calculate (a) its average 
speed and (b) its average velocity for the entire trip, using 
"away from the iTainer" as the positive direction, 



II 



(II) Two locomotives approach each other on parallel 
tracks. Bach has a speed of 95 km/h with respect lo the 
ground. If they are initially 8.5 km apart, how long will it 
be before they reach each other? (See Fig. 2-30). 



8.5 km 



95 km/h 



B = 
95 km/h 




FIGURE 2-30 Problem II. 



12. (II) A car traveling 88 km/h is 1 10m behind a truck trav- 
eling 75 km/h. How long will it take the car to reach 
the truck? 

13. (II) An airplane travels 3100 km at a speed of 790 km/h. 
and then encounters a tailwind that boosts its speed to 
W0 km/h for the next 2800 km. What was the total 
time for the trip? What was the average speed of the 
plane for this trip? \Hinl: Think carefully before using 
Fq. 2-lld| 

14. (II) Calculate the average speed and average velocity of a 
complete round-trip in which the outgoing 250 km is 
covered at 95 km/h. followed by a 1.0-hour lunch break. 
and the return 250 km is covered at 55 km/h. 

is. (Ill) A bowling ball traveling with constant speed hits the 
pins at the end of a bowling lane 16,5m long. The bowler 
hears the sound of the hall hitting the pins 2.50 s after the 
ball is released from his hands. What is the speed of 
the ball? The speed of sound is 340 m/s, 

2-4 Acceleration 

lf>- (I) A sports car accelerates from rest to 95 km/h in 6.2s. 

What is its average acceleration in m/s 2 ? 

17. (I) A sprinter accelerates from rest to 10,0 m/s in 1.35 s. 
What is her acceleration (a) in m/s'. and (b) in km/h 2 ? 



Problems 39 



IS. (II) At highway speeds, a particular automobile is capable 
of an acceleration of about 1.6 m/s". At this rate, how long 
does it take to accelerate from SO km/It to 110 km/h? 

19, (II) A sports car moving at constant speed [ravels 110 m 
in 5.0 s. If it then brakes and comes to a stop in 4,0 s, what 
is its acceleration in m/s 21 '? Express the answer in terms of 
"gV where 1.00 £ = 9,80 m/s 2 . 

211. (Ill) The position of a racing car, which starts from rest at 
( = and moves in a straight line, is given as a function 
of time in the following Table. Estimate («) its velocity 
and (b) ils acceleration as a function of time. Display each 
in a Table and on a graph. 

l(s) 0.25 0.50 0.75 1.00 1.50 2.00 2.50 

x(m) 0.11 0.46 1.06 1.94 4.62 K.55 13.79 

77s) Jm 3iso 4joo 4^ sm sTso 6m 

.¥(m) 20.36 28.31 37.65 48.37 60.30 73.2ft 87.1ft 

2-5 and 2-6 Motion at Constant Acceleration 

21. (I) A car accelerates from 13 m/s to 25 m/s in 6.0s. What 
was ils acceleration? How far did it travel in this lime? 
Assume constant acceleration, 

22. (I) A car slows down from 23 m/s to rest in a distance of 
85 m. What was its acceleration, assumed constant? 

23. (I) A light plane must reach a speed of 33 m/s for takeoff. 
How long a runway is needed if the (constant) accelera- 
tion is 3.0 m/s 2 ? 

24. (II) A world-class sprinter can buTst out of the blocks to 
essentially top speed (of about 1 1.5 m/s) in the first 15.0 in 
of the race. What is the average acceleration of this 
sprinter, and how long does it take her to reach that speed? 

25. (II) A caT slows down uniformly from a speed of 21.0 m/s 
to rest in 6.00 s, How far did it travel in that time'? 

26. (II) In coming to a stop, a car leaves skid marks 92 m long 
on the highway, Assuming a deceleration of 7.00 m/s 2 . 
estimate the speed of the car just before braking. 

27. (II) A car traveling 85 km/h strikes a tree. The front end 
of the car compresses and the driver comes to rest after 
traveling 0.80 m. What was the average acceleration of the 
driver during the collision? Express the answer in terms 
of "gX" where 1 .00 g = 9.80 m/s 2 . 

28. (II) Determine the stopping distances for a car with an initial 
speed of 95 km/h and human reaction time of 1.0 s for an 
acceleration (a) a = -4-0 m/s 2 ; (b)a - -8,0 m/s 2 . 

29. (Ill) Show that the equation for the stopping distance 
of a car is d$ = v^t^ — hj t /(2a). where t\j is the initial 
speed of the car. f R is the driver's reaction time, and a is 
the constant acceleration (and is negative). 

341. (Ill) A ear is behind a truck going 25 m/s on the highway, 
The car's driver looks for an opportunity to pass, guessing 
that his car can accelerate at 1.0 m/s 2 . He gauges that he 
has to cover the 20-m length of the truck, plus 10 m clear 
room at the rear of the truck and 10 m more at the front of 
it. In the oncoming lane, he sees a car approaching, prob- 
ably also traveling at 25 m/s. He estimates that the car is 
about 400 m away. Should he attempt the pass? Give details 

31. (Ill) A runner hopes to complete the 10,000-m run in less 
than 30.0min. After exactly 27.0 min, there are Still 1100 m 
to go. The runner must then accelerate at 0,20 m/s~ for 
how nianv seconds in order to achieve the desired time? 



32, (in) A person driving her car at 45 km/h approaches an 

intersection just as the traffic light turns yellow. She 
knows that the yellow light lasts only 2.0 s before turning 
red, and she is 2.8 m away from the near side of the inter- 
section (Fig. 2-31 ). Should she try to stop, or should she 
speed up to cross the intersection before the light turns 
red? The intersection is 15 m wide. Her car's maximum 
deceleration is -5.8 m/s\ whereas it can accelerate Irom 
45 km/h to 65 km/h in 6,0 s, IgnoTe the length of heT caT 
and her reaction time. 



DDDDDD 
DDDDDD 
□ □ □ 







FIGURE 2-31 Problem 32. 

2-7 Falling Objects [neglect air resistance] 

33. (I) A stone is dropped from the top of a eliff. II hits the 
ground below after 3.25 s, How high is the cliff? 

34. (I) If a car rolls gently (i^ = 0) off a vertical cliff, how 
long does it take it to reach 85 km/h? 

35. (I) Estimate (a) how long it took King Kong to fall 
straight down from the top of the Empire State Building 
(380 m high), and (b) his velocity just before "landing"? 

36. (II) A baseball is hit nearly straight up into the air with a 
speed of 22 m/s. {a) How high does il go? (b) How long is 
it in the a It? 

37. (II) A ballplayer catches a ball 3.0 s after throwing it 
vertically upward. With what speed did he throw it. and 
what height did il Teach? 

38. (II) An object starts from rest and falls undeT the influ- 
ence of gravity. Draw graphs of (a) its speed and (6) the 
distance it has fallen, as a function of time from ( = to 
: - 5.00 s. Ignore air resistance. 

39. (II) A helicopter is ascending vertically with a speed of 
5.20 m/s. At a height of 125 m above the Earth, a package 
is dropped irom a window, How much lime does il take 
for the package to reach the ground';' [Him: The 
package's initial speed equals the helicopter's. | 

411. (II) For an object falling freely from rest, show that the 
dislanee traveled during each successive second increases 
in the ratio of successive odd integers (1, 3. 5, etc.). This 
was first shown by Galileo. See Kigs, 2-18 and 2-21, 

41. (II) If air resistance is neglected, show (algebraically) that 
a ball thrown vertically upward with a speed v,, will have 
the same speed, 1^. when it comes back down to the 
starting point. 

42. (II) A stone is thrown vertically upward with a speed of 
18.0 m/s, (a) How fast is it moving when it reaches a 
height of ll-0m? {b) How long is required to reach this 
height? (c) Why are there two answers to (h)'' 

43. (III) Estimate the time between each photoflash of the 
apple in Fig, 2-18 (or number of pholnf lashes per second). 
Assume the apple is about 10 em in diameter. [Hint: Use 
two apple positions, but not the unclear ones at the top.] 



40 CHAPTER 2 Describing Motion: Kinematics in One Dimension 



44. (Ill) A falling stone takes 0.28 s to travel past a window 

2.2 m tall (Fig. 2-32), From what height above the top of 
the window did the stone fall? 



2.2 m 



To travel 
this 
> distance 

took 
0.28 6 



FIGURE 2-32 
Problem 44. 



4?. 



(Ill) A rock is dropped from a sea cliff, and the sound of 

it striking the ocean is heard 3.2s later. If the speed 

of sound is 340 m/s, how high is the cliff'; 1 
46. (Ill) Suppose you adjust your garden hose nozzle for a hard 

stream of water. You point the nozzle vertically upward at 
a height of 1, 5m above the 
ground (Fig. 2-33). When you 
quickly move the nozzle away 
from the vertical, you hear the 
water striking the ground nest 
to you for another 2.0 s. What 
is the water speed as it leaves 
the nozzle? 
1.5 m 



FIGURE 2-33 

Problem 46. 



41, (III) A stone is thrown vertically upward with a speed of 
12.0 m/s from the edge of a cliff 70.0 m high (Fig. 2-34). 

r{a] How much later does it 
reach the bottom of the cliff? 
(h) What is its speed just 
before hitting? (c) What total 
distance did it travel? 




I 









I 



t 



FIGURE 2-34 
Problem 47. 



48. (Ill) A baseball is seen to pass upward by a window 28 m 
above the street with a vertical speed of 13 m/s. If the ball 
was thrown from the street, {a) what was its initial speed, 
(b) what altitude does it reach, (c) when was it thrown. 
and (d) when does it reach the street again? 

2-8 Graphical Analysis 

: 49. (I) Figure 2-29 shows the velocity of a train as a function 
of time, (a) At what time was its velocity greatest? 
(b) During what periods, if any, was the velocity constant? 
(t) During what periods if any. was the acceleration 
constant? (d) When was the magnitude of the accelera- 
tion greatest? 

50. (II) Tli e position of a rabbit along a straight tunnel as a 
function of time is plotted in Fig. 2-28 What is its instan- 
taneous velocity (n) at r = 10.0 s and (6) at I - 30.0s? 
What is its average velocity (c) between t = and 
( = 5,0 s, (rf) between ; = 25. (J s and t = 30,0 s, and 
(e) between f = 40.0s and I = 50.0 s? 

1 51. (II) In Fig. 2-2S, (a) during what time periods, if any, is 
the velocity constant? (b) At what time is the velocity 
greatest? (c) At what time, if any. is the velocity zero? 
((0 Does the object move in one direction or in both 
directions during the time shown? 

■ 52. (II) A certain type of automobile can accelerate approxi- 
mately as shown in the velocity -time graph of Fig. 2-35. 
(The shoTt Hal spots in the curve represent shifting of the 
gears.) (a) Estimate the average acceleration of the car in 

second gear and in fourth gear, (b) Estimate how far the 
caT traveled while in fourth gear. 




10 20 30 



as) 



FIGURE 2-35 Problems 52 and 53. The velocity of an auto- 
mobile as a function of time, starting from a dead stop. The 
jumps in the curve represent gear shifts. 



* Si. (II) Estimate the average acceleration of the car in the 

previous Problem (Fig. 2-35) when it is in (a) first. 
(b) third, and (c) fifth gear, (rf) What is its average accel- 
eration through the first four gears? 

* 54. (II) In Fig. 2-29, estimate the distance the object traveled 

during (n) the first minute, and (b) the second minute, 

* 55. (II) Construct the v vs. t graph for the object whose 

displacement as a function of time is given by Fig, 2-28. 



P rob I ems 41 



56. (II) Figure 2-36 is a position versus time graph for the 
motion of an object along the _r axis. Consider the time 
interval from A to B. (a) Is the object moving in the posi- 
tive or negative direction? (b) Is the object speeding up 
or slowing down? (c) Is the acceleration of the object 
positive or negative? Now consider the time interval from 
D to E, {(f) Is the object moving in the positive or nega- 
tive direction? (e) Is the object speeding up or slowing 
down? if) Is the acceleration of the object positive or 
negative? (g) Finally, answer these same three questions 
for the time interval from C to D. 



50 



20 



B 15 



■ . _ 














\ 












\ 








E 
\ 






\ 






I 






\ 




s 


/ 






c 




\ 

D 





r is j 



FIGURE 2-36 
Problem 56 



| General Problems 



57. A person jumps from a fourth -story window 150 m above a 
firefighter's safety net. The survivor stretches the net I. Urn 
before coming to rest. Fig. 2-37. (a) What was the average 
deceleration experienced by the survivor when she was 

slowed to rest by the net? 
(hi) What would you do 
to make it "safer" (that 
is, to generate a smaller 
deceleration): would you 
stiffen or loosen the net? 
Explain. 




FIGURE 2-37 

LO m Problem 57. 
i 



58. The acceleration due to gravity on the Moon is about one- 
sixth what it is on Earth- I tan object is thrown vertically 
upward on the Moon, how many times higher will it go 
than it would on Earth, assuming the same initial velocity? 

59. A person who is properly constrained by an over-the- 
shoulder seat belt has a good chance of surviving a car 
collision if the deceleration does not exceed about 30 "gY" 
(1.0 g = 9.8 m/s 2 ). Assuming uniform deceleration of 
this value, calculate the distance over which the front end 
of the car must be designed to collapse if a crash brings 
the car to rest from lOOkm/h. 

611. Agent Bond is standing on a bridge. 12 m above the road 
below, and his pursuers are getting too close for comfort. He 
spots a flatbed track approaching at 2? m/s. which he 
measures by knowing that the telephone poles the truck is 
passing, are 25 m apart in this country. The bed of the [Tuek 
is 1.5 m above the road, and Bund quickly calculates how 
many poles away the truck should be when he jumps down 
from the bridge onto the truck to make his getaway, How 
many poles is it? 



61. Suppose a car manufacOjrer tested its cars for front-end 
collisions by hauling them up on a crane and dropping 
them from a certain height, (ti) Show that the speed just 
before a car hits the ground, after falling from rest a 
vertical distance H. is given by \'2gH .What height corre- 
sponds to a collision at (b) 60 km/h? (c) 100km/h7 

62. Every year the Earth travels about 10 L) km as it orbits the 
Sun. What is Earth's average speed in km/h? 

63. A 95-m-long train begins uniform acceleration from rest. 

The front of the train has a speed of 25 m/s when it 
passes a railway worker who is standing 180 m from 
wheTe the front of the train started. What will be the 
speed of the last ear as it passes the worker? (See 
Fig, 2-38.) 



95 m 



r = 25 m/s 



r 




FIGURE 2-38 Problem 63. 

64. A person jumps off a diving board 4.0 m above the 
water's surface into a deep pool, The person's downward 
motion stops 2.0 m below the surface of the water. Esti- 
mate the average deceleration of the person while under 
the water. 

65. In the design of a rapid transit system, it is necessary to 
balance the average speed of a train against the distance 

between stops. The more stops there are, the slower the 
train's average speed. To get an idea of this problem, 
calculate the time it takes a train to make a 9.0-km trip in 
two situations: (a) the stations at which the trains must 
stop are 1.8km apart (a total of 6 stations, including those 
at the ends): and (b) the stations are 30 km apart (4 stations 
total). Assume that at each station the train accelerates at a 
rate of 1 ,1 m/s 2 until it reaches 90 km/h. then stays at this 
speed until its brakes are applied for arrival at the next 
station, at which time it decelerates at -2.0 m/s 2 . Assume 
it stops at each intermediate station for 20s. 



42 CHAPTER 2 Describing Motion: Kinematics in One Dimension 



6f>. Pelicans tuck their wings and free fall straight down when 
diving for fish. Suppose a pelican starts its dive from a 
height of 16,0 m and cannot change its path ones 
committed. If it takes a fish 0.20s to perform evasive 
action, at what minimum height must it spot the pelican 
to escape? Assume the fish is at the surface of the water. 

67, In putting, the force with which a golfer strikes a ball is 
planned so that the ball will stop within some small 
distance of the cup, say, 1.0m long or short, in case the 
putt is missed, Accomplishing this from an uphill lie (that 
is. putting downhill, see Fig. 2-39) is more difficult than 
from a downhill lie, To see why. assume that on a partic- 
ular green the ball decelerates constantly at 2.0 m/s 2 
going downhill, and constantly at 3.0 m/s 2 going uphill. 
Suppose we have an uphill lie 7.0 m from the cup, Calcu- 
late the allowable range of initial velocities we may 
impart to the ball so that it stops in the range l.Om short 
to 1.0 m long of the cup, Do the same for a downhill lie 
7.1.) m from the cup. What in your results suggests that the 
downhill putt is more difficult.' 




FIGURE 2-39 Problem 67. Golf on Wednesday morning, 

nH. A fugitive tries to hop on a freight train traveling at a 
constant speed of 6.0 m/s, Just as an empty box car passes 
him, the fugitive starts from rest and accelerates at 
a = 4.0 m/s- to his maximum speed of 8-0 m/s. (ti) How 
long does it take him to catch up to the empty box car? 

(b) What is the distance traveled to reach the box car? 

69. A stone is dropped from the roof of a high building. A 
second stone is dropped i .50 s later. How far apart are the 
stones when the second one has Teached a speed of 
12.0 m/s? 

70 1 . A race car driver must average 200.0 km/h over the 
course of a time trial lasting ten laps. If the first nine laps 
were done at 198.0 km/h, what average speed must be 
maintained for the last lap? 

71. A bicyclist in the Tour de Fiance crests a mountain pass 

as he moves at 18 km/h. At the bottom. 4.0 km farther, his 
speed is 75 km/h. What was his average acceleration 
(in m/s-) while riding down the mountain? 

72, Two children are playing on two trampolines. The first 
child can bounce up one-and-a-half times higher than the 

second child. The initial speed up of the second child is 
5.0 m/s, {a} Find the maximum height the second child 
reaches, (b) What is the initial speed of the first child? 

(c) How long was the first child in the air? 



73, An automobile traveling % km/h overtakes a 1. 10-km-long 
train traveling in the same direction on a track parallel to 
the road. If the train's speed is 75 km/h. how long does it 
take the car to pass it, and how far will the car have traveled 
in this time? See Fig, 2-40, What are the results if the car 
and train are traveling in opposite directions? 



1. 10 km 



v = 75 km/h 

»* • * , « 

— _ i i - 95 km/h 

FIGURE 2-40 Problem 73. 

74, A baseball pitcher throws a baseball with a speed of 
44 m/s, In throwing the baseball, the pitcher accelerates 

the ball through a displacement of about 3.5 m. from 
behind the body to the point where it is released 
(Fig. 2-41). Estimate the average acceleration of the ball 
during the throwing motion. 



3.5 m 




FIGURE 2-41 
Problem 74, 



75, A rocket rises vertically, from rest, with an acceleration of 
3,2 m/s - until it runs out of fuel at an altitude of 1200 m. 
After this point, its acceleration is that of gravity, down- 
ward, (a) What is the velocity of the rocket when it runs 
out of fuel? (b) How long does it take to reach this point? 

(c) What maximum altitude does the rocket reach? 

(d) How much time (total) does it take to reach maximum 
altitude? (c) With what velocity does the rocket strike the 
EaTth? (/) How long (total) is it in the air? 

76. Consider the street pattern shown in Fig. 2-42. Each 
intersection has a traffic signal, and the speed limit is 
50 km/h. Suppose you are driving from the west at the 
Speed limit. When you are 10 m from the first intersection, 
all the lights turn green. 'Hie lights are- green for 13 s each. 
(a) Calculate the time needed to reach the third stoplight. 
Can you make it through all three lights without stop- 
ping? (A) Another car was stopped at the first light when 
all the lights turned green. It can accelerate at the rate of 
2,0 m/s 2 to the speed limit, Can the second car make it 
through all three lights without stopping? 



West i 

L _ C -I 



■ W ' ' f 



Et?E 



Yi.ur 
car 



10 ra 



Speed limit 
SO kmrti 



!5 m 15 m 



70. 



I last 

n . t 



15m 



FIGURE 2-42 Problem 76. 

General Problems A3 



77. A police car at rest, passed by a speeder traveling at a 
constant 12Qkm/b, takes off in hot pursuit, The police 
officer catches up to the speeder in 750 m. maintaining a 
constant acceleration. (17) Qualitatively plot the- position 
vs. time graph for both cars from the police ear's start to 
the catch-up point. Calculate (b) how long it took the 
police officer to overtake the speeder, (c) ihe required 
police car acceleration, and (d) the speed of the police car 
at the overtaking point. 

7K. A stone is dropped from the roof of a building: 2.00 s after 
that, a second stone is thrown straight down with an initial 
speed of 25.0 m/s, and the two stones land at the same 
time, (n) How long did it lake the first stone to reach the 
ground? (6) How high is the building? (c) What are the 
speeds of the two stones just before they hit the ground? 

79. Two stones are thrown vertically up at the same time. The 
first stone is thrown with an initial velocity of 11.0 m/s 
from a 12th-floor balcony of a building and hits the 
ground after 4.5 s. With what initial velocity should the 
second stone be thrown from a 4 Ih -floor balcony so that it 
hits the ground at the same time as the first stone? Make 
simple assumptions, like equal-height floors. 

tftl. If there were no air resistance, how long would it take a 
free-falling parachutist to fall from a plane at 3200 m to 
an altitude of 350 m, where she will pull her ripcord? 
What would her speed be at 350m? (In reality, the air 
resistance will restrict her speed to perhaps 150km/h.) 

81. A fast-food restaurant uses a conveyor belt to send the 
burgers through a grilling machine. If the grilling machine 
is 1 ,1 m long and the burgers require 2.5 min to cook, how 
fast must the conveyor belt travel? If the hurgers are 
spaced 15 cm apart, what is the rate of burger production 
(in burgers/miii)? 



82. Bill can throw a ball vertically at a speed 1.5 times faster 
than Joe can. How many times higher will Bill's ball go 
than Joe's? 

83. You stand at the top of a cliff while your friend stands on 
the ground below you. You drop a ball from rest and see 
that it takes 1.2s for the ball to hit the ground below, 
Your friend then picks up the ball and throws it up to 
you, such that it just comes to rest in your hand. What is 
the speed with which your friend threw the ball? 

84. Two students are asked to find the height of a particular 
building using a barometer. Instead of using the barom- 
eter as an altitude-measuring device, they take it to the 
roof of the building and drop it off. timing its fall. One 
student reports a fall time of 2,0 s. and the olheT. 2.3 5. 
How much difference does the 0.3 s make for the esti- 
mates of the building's height? 

■ 85. Figure 2-43 shows the position vs. time graph for two 
bicycles. A and B. (a) Is there any instant al which the 
two bicycles have the same velocity? (b) Which bicycle 
has the larger acceleration? (c) Al which instanl(s) are 
the bicycles passing each other? Which bicycle is passing 
the other? (rf) Which bicycle has the highest instanta- 
neous velocity? (e) Which bicycle has the higher average 
velocity? 




FIGURE 2-43 Problem 85. 



Answers to Exercises 

A: (/>). 

Bi (a) +:(b) -:(c) -:(d) +. 



D: 4.9 m/s 2 . 

E; That plane on which a smooth ball will not roll: or 
perpendicular to vertical, 



44 CHAPTER 2 Describing Motion: Kinematics in One Dimension 




This multiflash photograph of a ping pong ball shows examples 
of motion in two dimensions The arcs of the ping pong ball 
are parabolas that represent "projectile motion." Galileo 

analyzed projectile motion 
into its horizontal and 
vertical components; the gold 
arrow represents the down- 
ward acceleration of gravity, 
g. We will discuss how to 
manipulate vectors and how 
to add them. Besides analyzing 
projectile motion, we will also 
see how to work with relative 
velocity. 



CHAPTER 



3 



Kinematics in Two Dimensions; 

Vectors 



In Chapter 2 wc dealt with motion along a straight line. We now consider the 
description of the motion of objects that move in paths in two (or three) 
dimensions. In particular, wc discuss an important type of motion known as 
projectile motion: objects projected outward near the surface of the Earth, such 
as struck baseballs and golf balls, kicked footballs, and other projectiles. Before 
beginning our discussion of motion in two dimensions, we first need to present a 
new tool — vectors — and how to add them. 

Vectors and Scalars 

We mentioned in Chapter 2 that the term velocity refers not only to how fast 
something is moving but also to its direction. A quantity such as velocity, which 
has direction as well as magnitude, is a vector quantity. Other quantities that are 
also vectors are displacement, force, and momentum. However, many quantities 
have no direction associated with them, such as mass, time, and temperature. 
They are specified completely by a number and units. Such quantities are called 
scalar quantities. 



45 




FIGURE 3-1 Car traveling on a 
Toad. [Tie green arrows represent 
the velocity vector at each position. 



FIGURE 3-2 Combining vectors 
in one dimension. 



Resultant = 1 4 km (east) 



I I I ■ I |i| — I — x (km) 
8 km 6 km East 



Resultant = 1 km least) 
J_ 6 km 







I I n I — I — I — I — I — x (km) 
8 km East 

(hi 



Drawing a diagram of a particular physical situation is always helpful in 
physics, and this is especially true when dealing with vectors. On a diagram, each 
vector is represented by an arrow. The arrow is always drawn so that it points in 
the direction of the vector quantity it represents. The length of the arrow is 
drawn proportional to the magnitude of the vector quantity. For example, in 
Fig, 3-1, green arrows have been drawn representing the velocity of a car at 
various places as it rounds a curve. The magnitude of the velocity at each point 
can be read off Fig. 3-1 by measuring the length of the corresponding arrow 
and using the scale shown (I cm = 90km/h). 

When we write the symbol for a vector, we will always use boldface type, with 
a tiny arrow over the symbol. Thus for velocity we write v. If we are concerned 
only with the magnitude of the vector, we will write simply v, in italics, as we do 
for other symbols. 

Addition of Vectors— Graphical Methods 

Because vectors are quantities that have direction as well as magnitude, they 
must be added in a special way. In this Chapter, we will deal mainly with 
displacement vectors, for which we now use the symbol D, and velocity vectors, v. 
But the results will apply for other vectors we encounter later. 

Wc use simple arithmetic for adding scalars. Simple arithmetic can also be 
used for adding vectors if they arc in the same direction. For example, if a 
person walks 8 km cast one day, and 6 km cast the next day, the person will 
be 8 km + 6 km = 14 km cast of the point of origin, Wc say that the net or 
resultant displacement is 14 km to the cast (Fig. 3-2a). If, on the other hand, the 
person walks 8 km east on the first day, and 6 km west (in the reverse direction) 
on the second day, then the person will end up 2 km from the origin (Fig, 3-2b), 
so the resultant displacement is 2 km to the east. In this case, the resultant 
displacement is obtained by subtraction: 8 km — 6 km - 2 km. 

But simple arithmetic cannot be used if the two vectors are not along the same 
line. For example, suppose a person walks 10,0 km east and then walks 5,0 km 
north. These displacements can be represented on a graph in which the positive 
y axis points north and the positive x axis points east, Fig. 3-3. On this graph, we 
draw an arrow, labeled C t , to represent the displacement vector of the 10,0-km 
displacement to the east. Then we draw a second arrow, D 2 , to represent the 
5.0-km displacement to the north. Both vectors are drawn to scale, as in Fig. 3-3. 

After taking this walk, the person is now 1 0.0 km east and 5.0 km north of 
the point of origin. The resultant displacement is represented by the arrow 
labeled R in Fig. 3-3. Using a ruler and a protractor, you can measure on this 
diagram that [he person is I 1.2 km from the origin at an angle 6 = 11" north of 
east. In other words, the resultant displacement vector has a magnitude of 
1 1.2km and makes an angle B = 27 u with the positive x axis. The magnitude 
(length) of D| t can also be obtained using the theorem of Pythagoras in this 
case, since D\ , D t , and D K fomi a right triangle with D R as the hypotenuse. Thus 



I\ = \/D] + D 2 . = \/( 10.0 km) 2 + (5.0 km) 3 = \/ 125 km 2 = 1 1.2 km. 

You can use the Pythagorean theorem, of course, only when the vectors are 
perpendicular to each other. 



FIGURE 3-3 A person walks 10,0 km east and then 

5-0 km north. These two displacements are represented by 
the vectoTS D| and D; , which are shown as arrows, Tlie 
resultant displacement vector, D R , which is the vector sum 
of fi| and f>2 - is also shown. Measurement on the graph 
with ruler and protractor shows that D R has a magnitude of 
1 1 .2 km and points at an angle (i = 27" north of cast. 



y (km) 
North 



14- 
2 



West 



V? 







f- x (km) 
East 



South 



46 CHAPTER 3 Kinematics in Two Dimensions; Vectors 



The resultant displacement vector, D K , is Ihe sum of the vectors B l and D 2 
That is, 

6 K -fi, + 6 1 . 

This is a vector equation. An important featuie of adding two vectors that are 
not along the same line is that the magnitude of the resultant vector is not equal 
to the sum of the magnitudes of the two separate vectors, but is smaller than 
their sum: 



D R < A + P 2 , 



[vectors not along the same line| 



Vector equation 



In our example (Fig. 3-3), D K - 1 1.2 km, whereas D l + Eh equals IS km. Note 
also that we cannot set D K equal to 1 1.2 km, because we have a vector equation and 
1 1,2 km is only a part of the resultant vector, its magnitude. We could write some- 
thing like this, though: 5 I4 - 0, + D, - (11.2 km, 27" N of E). 

EXERCISE A Under what conditions can the magnitude of ihe resultant vector above 
be D R = D, + ft? 

Figure 3-3 illustrates the general rules for graphically adding two vectors 
together, no matter what angles they make, to get their sum. The rules arc as 
follows: 



1. On a diagram, draw one of the vectors — call it D t — to scale. 

2. Next draw the second vectoT, t) 2 , to scale, placing its tail at the tip of the 
first vector find being sure its direction is correct. 

3. The arrow drawn from the tail of the first vector to the tip of the second 
vector represents the sum, or resultant, of the two vectors, 

The length of the resultant vector represents its magnitude. Note that vectors 
can be translated parallel to themselves (maintaining the same length aod 
angle) to accomplish these manipulations. The length of the resultant can be 
measured with a ruler and compared to the scale. Angles can be measured with 
a protractor, This method is known as the tail-to-tip method or adding vectors. 
It is not important in which order the vectors arc added, For example, a 
displacement of 5.0 km north, to which is added a displacement of 10.0 km cast, 
yields a resultant of 1 1.2 km and angle = 27' (sec Fig. 3-4), the same as 
when they were added in reverse order (Fig, 3-3). That is, 



Tail-to-tip ttnehod 

of 

adding vectors 



V, + V, = V, + V, 



WlMI 



The tail-to- tip method of adding vectors can be extended to three ot more 
vectors. The resultant is drawn from the tail of the first vector to the tip of the 
last one added. An example is shtiwn in Fig. 3-5; the three vectors could repre- 
sent displacements (northeast, south, west) or perhaps three forces. Check for 
yourself that you get the same resultant no matter in which order you add the 
three vectors. 




x (km) 
Gasi 



South 



FIGURE 3-4 [f the vectors are 
added in reverse order, the resultant 
is the same. (Compare to Fig. 3-3.) 



FIGURE 3-5 The resultant of thTee vectors: V R = \\ + \ 2 + \\ . 

V 



A 



+ v, 




SECTION 3-2 Addition of Vectors— Graphical Methods 47 



Parallelogram iitclluttl of 
adding vectors 



A second way to add two vectors is the parallelogram method. Ft is fully 
equivalent to the lail-ki-lip method, In this method, the two vectors are drawn 
starting from a common origin, and a parallelogram is constructed using these 

two vectors as adjacent sides as shown in Fig. 3 -6b. The resultant is the diagonal 
drawn from the common origin. In Fig. 3-6a, the tail-to-tip method is shown, 
and it is clear that both methods yield the same result, 



V, 



FIGURE 3-6 Vector addition by 
two different methods, (a) and (b) 
Pan (c) is. incorrect. 



^ CAUTION 



Be surf to use the correct dutgortai 
on parallelogram to get the resultant 




y (a) Tail-to-tip 



(b) Parallelogram 



(c) Wrong 



It is a common error to draw the sum vector as the diagonal running 
between the tips of the two vectors, as in Fig. 3-6c. This is incorrect: it does not 
represent the sum of the two vectors. (In fact, it represents their difference, 
V 2 — V, , as we will see in the next Section.) 



CONCEPTUAL EXAMPLE 3-1 I Range of vector lengths Suppose two 



vectors each have length 3.0 units. What is the range of possible lengths for the 
vector representing the sum of the two? 

RESPONSE The sum can take on any value from fi.O (= 3.0 + 3.0) where the 
vectors point in the same direction, to (=3.0 — 3.0) when the vectors are 
antiparallcl. 



// 



FIGURE 3-7 The negative of a 
vector is a vector having the same 
length but opposite direction. 




EXERCISE B If the two vectors of Conceptual Example 3-1 are perpendicular to each 
other, what is the resultant vector length? 

Subtraction of Vectors, and 

Multiplication of a Vector by a Scalar 

Given a vector V, we define the negative of this vectoT ( - V) to be a vectoT with 
the same magnitude as V but opposite in direction, Fig. 3-7. Note, however, that 
no vector is ever negative in the sense of its magnitude: the magnitude of every 
vector is positive. Rather, a minus sign tells us about its direction. 

We can now define the subtraction of one vector from another: the differ- 
ence between two vectors V 2 - V, is defined as 

v 2 - v, = % + (-v,). 

That is, the difference between two vectors is equal to the sum of the first plus 
the negative of the seeond. Thus our rules for addition of vectors can be applied 
as shown in Fig. 3-8 using the tail-to-tip method. 



FIG U RE 3 -8 Sub tract i ng two 

vectors: V, - V, . 



/- -^ -V 



-V, 



-V, 



-^sy* 



48 CHAPTER 3 Kinematics in Two Dimensions; Vectors 



A vector V can be multiplied by ;i scalaT c. We define their product so that 
cV has the same direction as V and has magnitude cV. Thai is, multiplication of 
a vector by a positive scalaT c changes the magnitude of the vector by a factor c 
but doesn't alter the direction, Tf c is a negative scalar, the magnitude of the 
product cV is still cV (without the minus sign), but the direction is precisely 
opposite to that of V. See Fig, 3-9 , 




-2.0 V 



FIGURE 3-9 Multiplying a vector V by a scalar c 
gives a vector whose magnitude is c times greater 
and in the same direction as V (or opposite direction 
if c is negative). 



Adding Vectors by Components 

Adding vectors graphically using a mler and protractor is often not sufficiently 
accurate and is not useful for vectors in three dimensions. We discuss now a 
more powerful and precise method for adding vectors. Rut do not forget 
graphical methods — they aTe always useful for visualizing, for checking your 
math, and thus for getting the correct Tesult. 

Consider first a vector V that lies in a particular plane. It can be expressed 
as the sum of two other vectors, called the components of the original vector. 
The components are usually chosen to he along two perpendicular direc- 
tions. The process of finding the components is known as resolving the vector 
into its components. An example is shown in Fig. 3-10; the vector V could be 
a displacement vector that points at an angle & = 30" north of east, where 
we have chosen the positive x axis to be to the east and the positive y axis 
north. This vector V is resolved into its .v and y components by drawing 
dashed lines out from the tip (A) of the vector (lines AH and AC) making 
them perpendicular to the x and y axes, Then the lines OB and DC repre- 
sent the x and y components of V, respectively, as shown in Fig, 3- 10b. 
These vector components arc written V s and V y . We generally show vector 
components as arrows, like vectors, but dashed, The scalar components, V x 
and V y , ate numbers, with units, that are given a positive or negative sign 
depending on whether they point along the positive or negative x or y axis. 



As can be seen in Fig, 3- 
addine vectors. 



10, V, + Vj, = V by the parallelogram method of 



Resolving a vector into components 



North 




North 



U) 




East 



FIGURE 3-10 Resolving a vectoT V into its 
components along an arbitrarily chosen set of 
x and y axes. The components, once found, 
themselves represent the vector. That is, the 
components contain as much information as 
the vector itself, 



Space is made up of three dimensions, and sometimes it is necessary to 
resolve a vector into components along three mutually perpendicular directions. 
In rectangular coordinates the components are \ x , V t ,, and V-. Resolution of a 
vector in three dimensions is merely an extension of the above technique. We 
will mainly be concerned with situations in which the vectors arc in a plane and 
two components are all that are necessary. 

To add vectors using the method of components, we need to use the 
trigonometric functions sine, cosine, and tangent, which we now review. 



SECTION 3^ Adding Vectors by Components 49 



FIGURE 3-11 Starting with an angle # as 
in (a), we can construct right triangles of 
different sizes, (h) and (c), but the ratio of 
the lengths of the sides does not depend on 

the size of the triangle. 




j£ 



(a) 



(hi 




(c) 



Given any angle 0, as in Fig, 3- 1 la, a right triangle tan be constructed by 
drawing a line perpendicular to either of its sides, as in Fig. 3-1 lb. The longest 
side of a right triangle, opposite the right angle, is called the hypotenuse, which 
we label h. The side opposite the angle B is labeled o, and the side adjacent is 
labeled a. We let h, o, and a represent the lengths of these sides, respectively. 
We now define the three trigonometric functions, sine, cosine, and tangent 
(abbreviated sin, cos, tan), in terms of the right triangle, as follows: 



Trigonometric 

functions 

defined 




WW 





V 


COS 


= 


v* 
1 


tan = 




V* = 


v\ 


i r- 
y 



FIGURE 3-12 Finding the 

components of a vector using 
trigonometric functions. 



I omponents 



sin# = 



cos 8 = 



tanft 



side opposite o 

hypotenuse h 

side adjacent a 

hypotenuse h 

side opposite o 

side adjacent a 



(3-1) 



Tf we make the triangle bigger, but keep the same angles, then the ratio of the 
length of one side to the other, or of one side to the hypotenuse, remains 
the same. That is, in Fig. 3-1 1c we have: afh =a'/h'\ o/h =o'/k'; and 
o/a - o' fa'. Thus the values of sine, cosine, and tangent do not depend on 
how big the triangle is, They depend only on the size of the angle, The values of 
sine, cosine, and tangent for different angles can be found using a scientific 
calculator, or from the Table in Appendix A, 
A useful trigonometric identity is 

sin 1 8 + cos 2 6 = 1 (3-2) 

which follows from the Pythagorean theorem (o 2 + a 2 - h 1 in Fig. 3-1 l).That is: 

' + a 1 ft 1 



sin 2 8 + cos 2 8 = — + — = 

h 



77 



lr 






(See also Appendix A for other details on trigonometric functions and identities.) 
The use of trigonometric functions for finding the components of a vector is 
illustrated in Fig, 3-12, where a vector and its two components are thought of as 
making up a right triangle. We then see that the sine, cosine, and tangent are as 
given in the Figure, Tf we multiply the definition of stn# - V y /V by Von both 
sides, we get 



V„ = V sin 8. 



•it a Similarly, from the definition of cos 6, wc obtain 

vector V x = V cos 8, 



(3-3a) 



(3-31,1 



Note that 8 is chosen (by convention) to be the angle that the vector makes 
with the positive jt axis. 

Using Eqs. 3-3, we can calculate V x and V y for any vector, such as that illus- 
trated in Fig, 3-10 or Fig. 3-12. Suppose V represents a displacement of 500 m 



50 CHAPTER 3 Kinematics in Two Dimensions; Vectors 




l$.= Vsm0 = 



250 m 



V T =Vcos0=433m 

I' = VV? + V'J = 500 in 



FIGURE 3-13 (a) Vector V repre- 
sents a displacement of 500 m at a 30 " 
angle north of east. (I>) Tlie compo- 
nents of V are V x and V v , whose 
magnitudes are given on the right, 



in a direction 30 u north of east, as shown in Fig. 3- 1 3. Then V — 500 m. From 
acalculatoi orTtbles, sin30 u = 0.500 and cos30 u = 0.866. Then 



Fcosfl = (500 m) (0.866) = 433 m. (east), 
Ksintf = (500 m) (0.500) = 250m (north). 
There arc two ways to specify a vector in a given coordinate system: 



V, 



1, We can give its components, V x and V y . 

2, We can give its magnitude V and the angle f> it makes with the positive x axis. 

We can shift from one description to the other using Eqs. 3-3, and, for the 
reverse, by using the theorem of Pythagoras + and the definition of tangent: 

V = y/vl + V y 



tan e = 



11 



(3-ini 
(3-4b) 



as can be seen in Fig. 3-12, 

We can now discuss how to add vectors using components. The first step is 
to resolve each vector into its components. Next we can see, using Fig. 3-14, 
that the addition of any two vectors V, and V 2 to give a resultant, 
V = V, + V 2 , implies that 

V - V, + v, 

* lx 2x (3-5) 

V y = V ly + V 2y . 

That is, the sum of the x components equals the x component of the resultant, 
and similarly for y. That this is valid can be verified by a careful examination of 
Fig, 3-14. But note that we add all the x components together to get the x 
component of the resultant; and we add all the y components together to get the 
v component of the resultant. We do not add x components to y components. 

If the magnitude and direction of the resultant vector are desired, they can 
be obtained using Eqs. 3-4. 



: In three dimensions, Iht theorem of Pythagoras becomes V = \/V\ — V% + V\ 
component along the third, or z, axis. 



where V- is the 



Two ways 
to specify 

a vector 



( tlllipnlU HIS 

related to 
magnitude and 
direction 



Adding vectors 

analytically 

■'■, components) 




FIGURE 3-14 lhe components of V = V, + Vj 

are V x = V ix + V 2x and V, = V iy + V ly . 



SECTION 3-4 Adding Vectors by Components 51 



Ciitiur ti( rf.i.r'i tan simplify 
effort needed 



The components of a given vector will be different for different choices of 
coordinate axes, The choice of coordinate axes is always arbitrary. You can often 
reduce the work involved in adding vectors by a good choice of axes — foT 
example, by choosing one of the axes to be in the same direction as one of the 
vectors. Then that vector will have only one nonzero component. 




FIGURE 3-15 Example 3-2. 

(a) The two displacement vectors, 
D| and Dj , (b) D 2 is resolved into 
its components, (c) D| and D. are 
added graphically to obtain 
the resultant D. The component 
method of adding the vectors is 
explained in (he Example. 



EXAMPLE 3-2 



Mail carrier's displacement. A rural mail carrier leaves 
the post office and drives 22.0 km in a northerly direction. She then drives in a 
direction 60.tr south of cast for 47.0 km (Fig. 3- 1 5a). What is her displacement 
from the post office? 

APPROACH We Tesolve each vector into its x and y components, We add the 
x components together, and then the v components together, giving us the x 
and y components of the resultant, We choose the positive x axis to be east 
and the positive y axis to be north, since those are the compass directions used 
on most maps, 

SOLUTION Resolve each displacement vector into its components, as shown 
in Fig, 3- 1 5b. Since D, has magnitude 22.0 km and points north, it has only a y 
component: 

D lx = 0, D u . = 22.0 km, 

fi 2 has both x and y components; 

D, x = + (47.0 km) (cos 60") = +(47.0km)(0.500) = +23.5 km 



D 



(47.0 km) (sin 60") = -(47.0km)(0.866) = -40.7 km. 



Notice that D 2} is negative because this vector component points along the 
negative v axis. The resultant vector, D, has components: 

A - Aj + Ai - km + 23.5 km - +23.5 km 
D } = D ly + D lv = 22.0 km + (-40.7 km) = -18.7 km. 
This specifies the resultant vectoT completely: 
D z = 23.5 km, D y = -18.7 km. 

We can also specify the resultant vector by giving its magnitude and angle 
using Eqs. 3-4; 



D = \/D\ + D 2 = V(23.5 km) 2 + (-18.7 km) 2 = 30.0 km 



tantf = 



El 
A 



-18.7 km 

23.5 km 



= -0.796, 



A calculator with an imv tan, an arc tan, or a tan" 1 key gives 6 = 
tan _1 ( -0.796} = -38.5°. The negative sign means ft = 38.5° below the r axis, 
Fig. 3- 15c. So, the resultant displacement is 30.0 km directed at 38.5' in a 
southeasterly direction. 

NOTE Always be attentive about the quadrant in which the resultant vector 
lies. An electronic calculator does not fully give this information, but a good 
diagram does. 



PROBLEM SOLVING 
Identify she correct quadrant by 
drawing a careful diagram 



The signs of trigonometric functions depend on which "quadrant" the angle 
falls in: for example, the tangent is positive in the first and third quadrants (from 
0" to 90°, and I80 : ' to 270'), but negative in the second and fourth quadrants; see 
Appendix A-7. The best way to keep track of angles, and to check any vector 
result, is always to draw a vector diagram, A vector diagram gives you something 
tangible to look at when analyzing a problem, and provides a check on the results, 

The following Problem Solving Box should not he considered a prescrip- 
tion. Rather it is a summary of things to do to get you thinking and involved in 
the problem at hand. 



52 CHAPTER 3 Kinematics in Two Dimensions; Vectors 



PROBLEM SOLVING 



Adding Vectors 



Here is a brief summary of how to add two or more 
vectors using components: 

1. Draw a diagram, adding the vectors graphically hy 
cither the parallelogram or tail-to-tip method. 

2. Chouse x and y axes. Choose them in a way, if possible, 
that will make your work easier. (For example, choose 
one axis along the direction of one of the vectors so 
that vector will have only one component.) 

3. Resolve each vector into its x and y components, 
showing each component along its appropriate 
(x or y) axis as a (dashed) arrow. 

4. Calculate each component (when not given) using 
sines and cosines, ff tf, is the angle that vector 
V) makes with the positive a- axis, then: 

V lx = Vlcosfl, , F lv = F,sin0|. 



Pay careful attention to signs: any component that 
points along the negative x or v axis gets a — sign. 
Add the x components together to get the x 
component of the resultant. Ditto for v: 

Kt = Vix + Vix + an y others 
V y = V u + V 2y + any others. 
This is the answer: the components of the resultant 
vector. Check signs to sec if they fit the quadrant 
shown in your diagram (point I above). 
If you want to know the magnitude and direction of 
the resultant vector, use Eqs. 3-4: 

V 

V my/vl + Vl, tan 0-7T- 

*x 

The vector diagram you already drew helps to 
obtain the correct position (quadrant) of the angle ft. 



EXAMPLE 3-3 



Three short trips. An airplane trip involves three legs, 
with two stopovers, as shown in Fig. 3- 16a. The first leg is due east for 620 km; 
the second leg is southeast (45° } for 440 km; and the third leg is at 53° south of 
west, for 550 km, as shown. What is the plane's total displacement? 

APPROACH We follow the steps in the above Problem Solving Box. 
SOLUTION 

1. Draw a diagram such as Fig. 3-l6a, where D t , D 2 , and D\, represent the 
three legs of the trip, and D K is the plane's total displacement. 

2. Choose axes: Axes are also shown in Fig, 3- 1 6a. 

3. Resolve components: It is imperative to draw a good figure. The components 
are drawn in Fig. 3- 16b- Instead of drawing all the vectors starting from a 
common origin, as we did in Fig. 3- 15b, here we draw them "tail-to-tip" 
style, which is just as valid and may make it easier to see. 

4. Calculate the components: 

D,:Du = +D|CosG" = D, = 620 km 

D ly = + D,sinO° = km 

B 2 :D 2 . t - +D,cos45 u - +(440 km) (0.707) - +311 km 

Djy = -D,sin45" = -(440 km) (0.707) = -311 km 

I),: f\ x = -D,cos53 u = -(550 km) (0.602) = -331 km 

O,, = -OjsinSS 11 = -(550 km) (0.799) = -439 km. 

We have given a minus sign to each component that in Fig, 3-l6b points in 
the —,v or — y direction, The components are shown in the Table in the margin. 

5. Add the components; We add the x components together, and we add the y 
components together to obtain the x and y components of the resultant: 

Pt = ®ix + P.< + Aix = 620 km + 311 km - 331 km = 600 km 
D y = D ly + D 2v + D iy = km - 31 1 km - 439 km = -750 km. 

The x and y components are 600 km and —750 km, and point respectively to 
the east and south. This is one way to give the answer. 

6. Magnitude and direction: We can also give the answer as 

D K = \/Di + D\ = \/(n00)' + (-750) 2 km 
A -750 km 



tan 6 = -r- 



-1.25, 



sofl 



960 km 

-51°. 




FIGURE 3-16 



Vector 


Cumpuiii' 


llt> 




x (km) 


V (kill) 


l> 


62U 





IJ 


311 


-311 


L> 


-331 


-439 


■>R 


60U 


-750 



600 km 

Thus, the total displacement has magnitude 960 km and points 51° below 
the x axis (south of east), as was shown in our original sketch, Fig. 3- 1 6a. 

SECTION 3-4 Adding Vectors by Components 53 




Projectile Motion 



FIGURE 3-17 This strobe photo- 
graph of a hall making a series of 
bounces shows the characteristic 
"parabolic" path of projectile motion. 



Horizonial and 

vertical motion 

anal) zed separately 



Tn Chapter 2, we studied the motion of objects in one dimension in terms of 
displacement, velocity, and acceleration, including purely vertical motion of 
falling bodies undergoing acceleration due to gravity. Now we examine the more 
general motion of objects moving through the air in two dimensions near the 
Earth's surface, such as a golf ball, a thrown or hatted baseball, kicked footballs, 
and speeding bullets. These are all examples of projectile motion (see Pig. 3-17), 
which wc can describe as taking place in two dimensions. Although air resistance 
is often important, in many cases its effect can be ignored, and we will ignore it 
in the following analysis. Wc will not be concerned now with the process by 
which the object is thrown or projected. Wc consider only its motion after it has 
been projected, and before it lands or is caught — that is, wc analyze our 
projected object only when it is moving freely through the air under the action 
of gravity alone. Then the acceleration of the object is that due to gravity, which 
acts downward with magnitude # — 9.80 m/s\ and we assume it is constant/ 

Galileo was the first to describe projectile motion accurately. He showed 
that it could be understood by analyzing the horizontal and vertical components 
of the motion separately. For convenience, we assume that the motion begins at 
time t - at the origin of an xy coordinate system (so x$ = y (} = 0). 



FIGURE 3-1 3 Projectile motion of a small ball 
projected horizontally. The dashed Mack line 

represents the path of the object. The velocity vector v 
at each point is in the direction of motion and thus 
is tangent to the path. The velocity vectors are green 
arrows, and velocity components are dashed. 
(A vertically falling object starting at the same point 
is shown at the left for comparison: v y is the same for 
the falling object and the projectile.) 



Pntjjeetile 
motion 



? is tangent to the path 



Vertical motion 
(tjy = constant = —g) 




Lei us look at a (tiny) ball rolling off the end of a horizontal table with an 
initial velocity in the horizontal (x) direction, v xf) . See Fig, 3- IS, where an object 
falling vertically is also shown for comparison. The velocity vector v at each 
instant points in the direction of the ball's motion at that instant and is always 
tangent to the path. Following Galileo's ideas, we treat the horizontal and vertical 
components of the velocity, t> x and v v , separately, and we can apply the kinematic 
equations (Eqs, 2- 1 la through 2-1 1 c) to the x and v components of the motion. 

First we examine the vertical ( y) component of the motion. At the instant 
the ball leaves the table's top (t = 0), it has only an x component of velocity. 
Once the ball leaves the table (at i - 0), it experiences a vertically downward 
acceleration #, the acceleration due to gravity. Thus t\. is initially zero (t), H) = 0) 
but increases continually in the downward direction (until the ball hits the 
ground). Let us take y to be positive upward. Then a.. = — g, and from 



Eq, 2-1 la we can write t\. = 
ment is given by y = — f gr 1 . 



gt since wc set VyQ = 0. The vertical displacc- 



' 'II 115 restricts us to objects whose disrance Traveled and maximum height above the liarth sec small 
compared to the F.arth's radius ((i4Q0kni), 



54 CHAPTER 3 Kinematics in Two Dimensions; Vectors 




FIGURE 3-19 Multiple-exposure photograph 

showing positions of two balls at equal time Intervals. 
One ball was dropped from rest at the same time the 
other was projected horizontally outward. The vertical 
position of each ball is seen to be the same. 



In the horizontal direction, on the other hand, there is no acceleration (we 
are ignoring air resistance). So the horizontal component of velocity, v x , 
remains constant, equal to its initial value, » j0 , and thus has the same magni- 
tude at each point on the path. The horizontal displacement is then given by 
x - !'jM t. The two vector components, % and \ y , can be added vectorially at any 
instant to obtain the velocity v at that time (that is, for each point on the path), 
ass shown in Fig. 3-18, 

One result of this analysis, which Galileo himself predicted, is that an object 
projected horizontally will reach (lie ground in the same time as an object 
dropped vertically. This is because the vertical motions are the same in both 
cases, as shown in Fig. 3- IS, Figure 3-19 is a multiple-exposure photograph of 
an experiment that confirms this 

I EXERCISE C Two balls having different speeds roll off the edge of a horizontal table 
| at the same time, Which hits the flooT sooner the faster ball or the slower one' 

If an object is projected at an upward angle, as in Fig. 3-20, the analysis is 
similar, except that now there is an initial vertical component of velocity, v yU . 
Because of the downward acceleration of gravity, v y gradually decreases with 
time until the object reaches the highest point on its path, at which point 
v y - 0. Subsequently the object moves downward (Fig. 3-20) and i\. increases 
in the downward direction, as shown (that is, becoming more negative), As 
before, v x remains constant. 



Horizontal motion 
n, = 0, t), = constant) 



Object projected upward 




V 


= Oat this point 











\ \v 



a = g 




FIG U RE 3 - 20 Pa th o f a p roj c c ti 1 a 
fired with initial velocity * () at 
angle D to the horizontal, Path is 
shown in black, the velocity vectors 
are green arrows, and velocity 
components are dashed, 



SECTION 3-5 Projectile Motion 55 



PROBLEM SOLVING 
Choice of time interval 



Solving Problems Involving Projectile Motion 

We now work through several Examples of projectile motion quantitatively. We 
use the kinematic equations (2- 1 la through 2-1 Ic) separately for the vertical and 
horizontal components of the motion. These equations are shown separately for 
the x and y components of the motion in Table 3-1, for the general case of two- 
dimensional motion at constant acceleration. Note that x and y are the respective 
displacements, that v x and i\. arc the components of the velocity, and that 
a x and a K . arc the components of the acceleration, each of which is constant. 
The subscript means "at / = 0." 



TABLE 3-1 General Kinematic Equations for Constant Acceleration 
in Two Dimensions 


x component (horizontal) 




'■' component (vertical) 


V x = VxD + il x ! 


(Eq. 2-lla) 


v y = v w +- a y t 


X = -t<j + Vxat + jfljtf 2 


(Eq.2-llb) 


y = ft + Vyat + ka f t 2 


»* - «jifl + 2 <J*(- V ~ *o) 


(Eq.2-llc) 


Vy = ">o + 2a y(y ~ yu) 



We cati simplify these equations for the case of projectile motion because we 

can set a x = 0. See Table 3-2, which assumes y is positive upward, so a y = —g 
= —9,80 m/s ! . Note that if 9 is chosen relative to the + x axis, as in Fig. 3-20, then 

''in — VoCosB, and v^ = %ma6, 

Tn doing Problems involving projectile motion, we must consider a time interval foT 
which out chosen object is in the air, influenced only by gravity. We do not consider 
the throwing (or projecting) process, nor the time alter the object lands or is caught, 
because then otheT influences act on the object, and we can no longer set a = g. 

TABLE 3-2 Kinematic Equations for Projectile Motion 
{y positive upward; a, = 9, a f = -g = -S.gOm/s 1 ) 



Horizontal Motion 

(flj = 0, t! 5 = constant} 



Vertical Motion* 

(o v = — g = constant ) 



v x — V ,ii 
,v = Xo + Vat 



(Eq. 2-lla) 


Vy = Bjfl - gt 




(Eq.2-llb) 


y = ye + Vyat - 


Igt 2 


(Eq.2-llc) 


4 = «*» - 2 s(y 


-*) 



' II v is taken positive downward, the minus (-] signs in front of£ become + signs. 



PROBLEM SOLVING 



Projectile Motion 



Our approach to solving problems in Section 2-6 also 
applies here, Solving problems involving projectile 
motion can require creativity, and cannot be done just 
by following some rules. Certainly you must avoid just 
plugging numbers into equations that seem to "work." 

1. As always, read carefully; choose the object (or 
objects) you arc going to analyze. 

2. Draw a careful diagram showing what is happening 
to the object. 

3. Choose an origin and an xy coordinate system, 

4- Decide on the time interval, which for projectile 
motion can only include motion under the effect of 
gravity alone, not throwing or landing. The time 
interval must be the same for thex andy analyses. The 
x and y motions are connected by the common time. 



5. Examine the horizontal (x) and vertical (y) motions 
separately. If you are given the initial velocity, you 
may want to resolve it into its % and y components. 

6. List the known and unknown quantities, choosing 
a x — and a y - —g or + g, where- g - 9.80 m/s\ 
and using the + or — sign, depending on whether 
you choose y positive down or up. Remember that 
v s never changes throughout the trajectory, and 
that Vy - at the highest point, of any trajectory 
that returns downward. The velocity just before 
landing is generally not zero. 

7. Think for a minute before jumping into the equations. 
A little planning goes a long way. Apply the relevant 
equations (Table 3-2), combining equations if 
necessary. Y<m may need to eomhine components of 
a vector to get magnitude and direction (Eqs. 3-4). 



56 CHAPTER 3 Kinematics in Two Dimensions; Vectors 



EXAMPLE 3^ 



Driving off a cliff. A movie stunt driver on a motorcycle 
speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave 
the cliff top to land on level ground below, 90.0 m from the base of the cliff 
where the cameras arc? Ignore air resistance. 

APPROACH We explicitly follow the steps of the Problem Solving Box, 

SOLUTION 

1. and 2. Read, choose the object, and draw a diagram. Our object is the motor- 
cycle and driver, taken as a single unit- The diagram is shown in Fig, 3-21. 

3. Choose a coordinate system. We choose the v direction to be positive 
upward, with the top of the cliff as y tl = 0. The ,t direction is horizontal 
with x l} - at the point where the motorcycle leaves the cliff. 

4. Choose a time interval. We choose our time interval to begin (i - 0) just as 
the motorcycle leaves the cliff top at position x u - 0, v M - 0; our time 
interval ends just before the motorcycle hits the ground below, 

5. Examine x and y motions. In the horizontal (jt) direction, the acceleration 
a x - 0, so the velocity is constant The value of x when the motorcycle 
reaches the ground is x - +90,0 m. In the vertical direction, the accelera- 
tion is the acceleration due to gravity, a y - — g — -9.80 m/s 2 . The value 
of v when the motorcycle reaches the ground is y - —50,0 m. The initial 
velocity is horizontal and is our unknown, !!„,; the initial vertical velocity is 
zero, Vyt) = 0. 

6. List knowns and unknowns. See the Table in the margin. Note that in addition 
to not knowing the initial horizontal velocity v xU (which stays constant until 
landing), we also do not know the time / when the motorcycle reaches the 
ground. 

7. Apply relevant equations. The motorcycle maintains constant v 3 as long as 
it is in the air. The time it stays in the air is determined by the y motion- 
when it hits the ground. So we first find the time using the y motion, and 
then use this time value in the x equations. To find out how long it takes the 
motorcycle to reach the ground below, we use Eq. 2-Mb (Table 3-2) for 
the vertical (>■) direction with y„ - and i>,„ - 0: 





y = 


>}) + Vyt 


i + 


iv= 




= 


0+0 


+ 


k-gy 


(II 


V 


-!*». 






We solve 


for 


and set 


v - 


- -50.0 m 



I 



2(- 50.0 m) 



\' ~g \f -9.80 m/s 2 



3.19 s. 



To calculate the initial velocity, v xlj , we again use Eq. 2-1 lb, but this time 
for the horizontal (i) direction, with a x - and jt u - 0: 

* = -*!) + V m t + ka x i 2 
= + v M t + 



or 



Then 



x 

t 



90.0 m 
3.19 s 



2S.2 m/s. 



which is about 100 km/h (roughly 60 mi/h). 

NOTE In the time interval of the projectile motion, the only acceleration is g 
in the negative y direction. The acceleration in the x direction is zero. 



f 


+ X 

~. a = B 

s 

s 
\ 

\ 
\ 

v = -50.0ni\ 
"-4 


«■ r-i. 
(50.0 m 


r— 




mo m - ■ — h 



FIGURE 3-21 Example 3-4 



Known 






Unknown 


.T„ = JUj = 






v m 


x = 90.0 m 






t 


v = -50.0 m 








a x - 








«> = -g = " 


9.80 


m/s 2 




,:, , - 









SECTION 3-6 Solving Problems Involving Projectile Motion 57 



FIGURE 3-22 Example 3-5. 



y v = Oat I his point 




a = g 



:*-jc 



^PHYSICS APPLIED 
Sports 



EXAMPLE 3-5 



A kicked football. A football is kicked at an angle 
8„ = 37.0" wilh a velocity of 20.0 m/s, as shown in Fig. 3-22. Calculate (a) the 
maximum heigh i, (b) the lime of travel before Ihe football hits the ground, 
(c) how far away it hits the ground, (d) the velocity vector at the maximum 
height, and (e) the acceleration vector at maximum height. Assume the ball 
leaves the foot at ground level, and ignore air resistance and rotation of the ball. 

APPROACH This may seem difficult at first because there aTe so many ques- 
tions. But we can deal with them one at a lime. We take the y direction as posi- 
tive upward, and treat the x and v motions separately. The total time in the air is 
again determined by the y motion. The x motion occurs at constant velocity. The 
v component of velocity varies, being positive (upward) initially, decreasing to 
zero at the highest point, and then becoming negative as the football falls, 
SOLUTION Wc resolve the initial velocity into its components (Fig, 3-22): 

V M = Vo COS 37,0° = (20.0 m/s) (0.799) = lfi.Om/s 

Vya = Vo sin 37.0° = (20.0 m/s) (0.602) = 12.0 m/s. 

(a) We consider a time interval that begins just after the football loses contact 
with the foot until it reaches its maximum height. During this time interval, the 
acceleration is g downward. At the maximum height, the velocity is horizontal 
(Fig. 3-22), so v y = 0; and this occurs at a time given by v v = t\ H) - gt with 
v, = (see Eq. 2-1 la in Table 3-2). Thus 



«,<> (I2.0m/s) 



£ (9.80 m/s 2 ) 
From Eq. 2- 1 lb, with y\, - 0, wc have 



1 .22s. 



V = VyQt 



W 



= (l2.0m/s)(U2s) - 1(9.80 m/s 2 )( 1.22s) 2 = 735 m. 
Alternatively, wc could have used Eq. 2-1 1c, solved for y, and found 



y 



2g 



(12.0 m/s) 2 - (0m/s) 2 
2(9.80 m/s 2 ) 



- 7.35 m. 



The maximum height is 7.35 m. 

(b) To find the time it takes for the ball to return to the ground, we consider a 
different time interval, starting at. the moment the ball leaves the foot 
(f - 0, v M - 0) and ending just before the ball touches the ground (v - 
again). We can use Eq. 2-1 lb with y„ — and also set y - (ground level): 

v = y« + v f ~ iS* z 

= + (12.0 m/s)/ - |(9.S0m/s 2 )i 2 . 
This equation can be easily factored; 

[Jt(9.80m/s 2 )(- ]2,0m/s]jr - 0. 

There are two solutions, ( = (which corresponds to the initial point, y M ), and 

2(1 2,0 m/s) 

f = ~r -4- = 2-45 s, 

(9.80 m/s 2 ) 

which is the total travel time of the football. 



58 CHAPTER 3 Kinematics in Two Dimensions; Vectors 



NOTE The lime t = 2.45s for the whole trip is double the time to reach 
the highest point, calculated in (a). That is, the lime lo go up equals the lime to 
come hack down lo the same level, but only in the absence of aiT resistance, 
(c) The total distance traveled in the .v direction is found by applying 
Eq, 2-1 lb with x a - 0, a x - G, v M - 16.0 m/s: 

* = *W = (I6.0m/s)(2.45s) = 39.2 m. 

(ci) At the highest point, there is no vertical component to the velocity. There 

is only the horizontal component (which remains constant throughout the 

flight), so v = v M = b () cos 31.(1" - 16.0 m/s. 

(e) The acceleration vector is the same at the highest point as it is throughout 

the flight, which is 9.80 m/s 2 downward. 

NOTE We treated the football as if it were a particle, ignoring its rotation. We 

also ignored air resistance, which is considerable tin a rotating football, so our 

results are not verv accurate. 



EXERCISE D Two balls are thrown in the air at different angles, but each reaches the 

same height. Which ball remains in the air longer: the one thrown at Che steeper angle 
or the one thrown at a shallower angle? 



CONCEPTUAL EXAMPLE 3-6 | Where does the apple land? A child sits 



upright in a wagon which is moving to the right at constant speed as shown in 
Fig. 3-23, The child extends her hand and throws an apple straight upward 
(from her own point of view, Fig. 3-23a), while the wagon continues to travel 
forward at constant speed. If air resistance is neglected, will the apple land 
(a) behind the wagon, (b) in the wagon, or (c) in front of the wagon? 

RESPONSE The child throws the apple straight up from her own reference 
frame with initial velocity V ri) (Fig, 3 -23a). Bui when viewed by someone on 
the ground, the apple also has an initial horizontal component of velocity 
equal to the speed of the wagon, V, . Thus, to a person on the ground, the 
apple will follow the path of a projectile as shown in Fig. 3-23b. The apple 
experiences no horizontal acceleration, so v«, will stay constant and equal to 
the speed of the wagon. As the apple follows its arc, the wagon will be directly 
under the apple at all times because they have the same horizontal velocity. 
When the apple conies down, it will drop right into the outstretched hand of 
the child. The answer is (b). 



Time up = time down 



i 




(a) Wagon reference frame 



4 









*,o 




(b) Ground reference frame 
FIGURE 3-23 Example 3-(i. 



CONCEPTUAL EXAMPLE 3-7 | The wrong strategy. A boy on a small 



hill aims his water-balloon slingshot horizontally, straight at a second boy 
hanging from a tree branch a distance d away. Fig. 3-24. At the instant the 
water balloon is released, the second boy lets go and falls from the tree, hoping 
to avoid being hit. Show that he made the wrong move, (He hadn't studied 
physics yet.) f gnore air resistance. 

RESPONSE Both ihe water balloon and the boy in ihe tTee start falling at the 
same instant, and in a lime t they each fall the same vertical distance v = \gt 2 , 
much like Fig. 3-19. In the time it takes the water balloon to travel the hori- 
zontal distance d, Ihe balloon will have the same y position as Ihe falling boy- 
Splat. If the boy had stayed in the tree, he would have avoided the humiliation. 




y = 



FIGURE 3-24 Example 3-7. 



SECTION 3-6 Solving Problems Involving Projectile Motion 58 



EXERCISE E A package is dropped from a plane flying at constant velocity parallel to 
the ground, If air resistance is ignored, the package will (a) fall behind the plane, 
(ft) remain directly below the plane until hitting the ground, (l) move ahead of the 
plane, or (d) it depends on the speed of the plane. 



Horizontal range of'a projectile 



v = again here 
(where ,r=#) 




60° 



(h) 

FIGURE 3-25 Example 3-8. 

(a) ITie range R of a projectile; 

(b) there are generally two angles O 
that will give the same range. Can 
you show that if one angle is 6 n \ . 
the other is U2 = 90" - B m ? 



Level range formula 
[y ifinaf) = ya] 



EXAMPLE 3-8 



Level horizontal range, (a) Derive a formula for the hori- 
zontal range R of a projectile in terms of its initial velocity <>,, and angle M .The 
horizontal range is defined as the horizontal distance the projectile travels 
before returning to its original height (which is typically the ground); that is, 
y (final) = y„. See Fig. 3- 25a. (b) Suppose one of Napoleon's cannons bad a 
muzzle velocity, o„, of 60,0 m/s. At what angle should it have been aimed 
(ignore air resistance) to strike a target 320m away? 

APPROACH The situation is the same as in Example 3-5, except we are not 
now given numbers in (a). We will algebraically manipulate equations to 
obtain our result. 

SOLUTION (a) We set *„ = and y (h = at i = 0. After the projectile 
travels a horizontal distance R, it returns to the same level, y - 0, the final 
point. We choose our time interval to start (t - 0) just after the projectile is 
fired and to end when it returns to the same vertical height. To find a general 
expression for R, we set both y — and y„ — in Eq, 2-1 lb for the vertical 
motion, and obtain 



so 



V - }\l + Vyut + lilyt 2 



- + iw - kgt 1 . 



We solve for t, which gives two solutions; t — and ( — Iv^Jg. The first 
solution corresponds to the initial instant of projection and the second is the 
time when the projectile returns to y - 0. Then the range, ft, will be equal to x 
at the moment t has this value, which we put into Eq, 2-1 lb for the horizontal 
motion {x - v xi) t, with jc u - 0).Thus we have: 



Vrat 



2"f ya \ 2v xB Vyo 2vl sin n cos ft,, 



I J 



1 'I 



= v.; M cos 6 a and v >v = U,, sin & u . This is the result 



where we have written u 

we sought. It can be rewritten, using the trigonometric identity 2 sin cos 6 = 

sin 2ft (Appendix A or inside the rear cover): 



R = 



vf, sin 2fl„ 



[y = V.)] 



We see that the maximum range, for a given initial velocity v u , is obtained 
when sin 2d takes on its maximum value of 1.0, which occurs for 28 a - 90°; so 

9 — 45" for maximum range, and R mM = iftfg. 

[When air resistance is important, the range is less for a given <;„, and the 
maximum range is obtained at an angle smaller than 45".] 
NOTE The maximum range increases by the square of v„, so doubling the 
muzzle velocity of a cannon increases its maximum range by a factor of 4. 
{b) We put R - 320 m into the equation we just derived, and (assuming, 
unrealistically, no au resistance) we solve it to find 

Rg (320m)(9.80m/s 2 ) 



sin 20, 



<v 



(60.0 m/s) 2 



0.871. 



We want to solve for an angle 8 lt that is between 0" and 90°, which means 2fl n 
in this equation can be as large as 180". Thus, 20 o = 60.6° is a solution, but 



60 CHAPTER 3 Kinematics in Two Dimensions; Vectors 



2* = 180" - 60.6° = 119.4° is also a solution (see Appendix A-7). In 
general we will have two solutions (see Fig, 3-25b), which in the present case 
are given by 

8„ - 30.3° or 59.7°, 

Either angle gives the same range. Only when sin 20 (l = 1 (so f) a = 45") is 
there a single solution (that is, both solutions arc the same). 



Additional Example: Slightly more Complicated, hut Fun 



EXAMPLE 3-9 



A punt. Suppose the football in Example 3-5 was a punt 
and left the punters foot at a height of 1 .00 m above the ground. How far did 



0, V'n 



il. 



the football travel before hitting the ground? Set jv n 

APPROACH The x and y motions are again treated separately. But we 
cannot use the range formula from Example 3-8 because it is valid only it 
y (final) - y,, which is not the case here, Now we have y« — 0, and the foot- 
ball hits the ground where y = — 1 .00 m (see Fig, 3-26). We choose our time 
interval to start when the ball leaves his foot {( = 0, _y () = 0, jc,, = 0) and end 
just before the ball hits the ground (y = - 1 .00 m). We can get x from 
Eq, 2-1 lb, x = v x „t, since we know [hat v M) = 16,0 m/s from Example 3-5. 
Rut first we must find t, the time at which the ball hits the ground, which we 
obtain from they motion, 

SOLUTION With y = -1.00 m and » y) = 12.0 m/s (see Example 3-5), we 

use the equation 



is**, 



v v„ - l\,,l 

and obtain 

-1.00m =0 + (12.0m/s)t - (4.90 m/s 2 > 2 . 

Wc rearrange this equation into standard form so we can use the quadratic 
formula (Appendix A -4; also Example 2-15): 

(4.90 m/s 2 V 1 - (12,0 m/s}* - (1.00m) = 0. 

Using the quadratic formula gives 

1 2.0 m/s ± V'( 12.0 m/s) 2 - 4(4,90 m,V)( - 1 .00 m ) 



i 



2.53 s or 



2(4.90 m/s 2 ) 
-0.081 s. 



The second solution would correspond to a time prior to the kick. so it doesn't 
apply. With ( - 2.53 s for the time at which the ball touches the ground, the 
horizontal distance the ball traveled is (using v x0 - 16.0 m/s from Example 3-5): 



tW 



(I6.0m/s)(2.53s) = 40.5 m. 



Our assumption in Example 3-5 that the ball leaves the foot at ground level 
results in an underestimate of about 1.3 m in the distance traveled. 



\fi 


j PHYS 


i C 5 


A P 


PLIED 




Sports 










PROBLEM 


SOLVING 



Do not use any formula unless you 
iii-i sure its range oj validity fits the 
problem. The range formula does not 
apply here because y *■ y\\ 




. y = 







Cuo Lin J 



FIGURE 3-26 Example 3-9: the 
football leaves the punters foot at 
y = 0. and reaches the ground 
where y = —1.00 m. 



SECTION 3-6 Solving Problems Involving Projectile Motion 61 




Projectile Motion Is Parabolic 



We now show that the path followed by any projectile is a paranoia, if we ignore 
air resistance and assume that g is constant. To show this, we need to find v as a 
function of x by eliminating / between the two equations for horizontal and 
vertical motion (Eq. 2-1 lb), and we set x = y () = 0: 



x = v M t 

y = V' 



$8? 



From the first equation, we have t - x/v m , and we substitute this into the 
second one to obtain 



hi 



X 



2i4> 



Tf we write v a , = i?ocos0 n and v yi , = I'nsinfy,, we can also write 



y = (tan 8,)jc - 



A . 



\2vIcq& 2 8 0/ 

Tn either case, we see that y as a function of .v has the form 

y = Ax - #jt, 

where 4 and B aTe constants for any specific projectile motion, This is the well- 
known equation for a parabola. See Figs. 3-17 and 3-27. 

The idea that projectile motion is parabolic was, in Galileo's day, at the 
forefront of physics research. Today we discuss it in Chapter 3 of introductory 

physics! 



Relative Velocity 



FIG U RE 3 -27 Rs amp Its of proj ee- 
tile motion — sparks (small hoi 
glowing pieces of metal), water, and 
fireworks, All exhibit the parabolic 
path characteristic of projectile 
motion, although the effects of air 
resistance can be seen to alter the 
path of some trajectories. 



PROBLEM SOLVING 

Subscripts for adding velocities: 

first subscript for the object; 

second subscript for the reference 

frame 



We now consider how observations made in different reference frames are 
related to each other. For example, consider two trains approaching one 
another, each with a constant speed of 80 km/h with respect to the Earth. 
Observers on the Earth beside the tracks will measure 80 km/h for the speed of 
each train. Observers on either of the trains (a different reference frame) will 
measure a speed of 160 km/h for the other train approaching them. 

Similarly, when one car traveling 90 km/h passes a second ear traveling in 
the same direction at 75 km/h, the first car has a speed relative to the second 
car of 90 km/h - 75 km/h = 15 km/h. 

When the velocities are along the same line, simple addition or subtrac- 
tion is sufficient to obtain the relative velocity. But if they arc not along the 
same line, we must use vector addition. We emphasize, as mentioned in 
Section 2-1, that when specifying a velocity, it is important to specify what the 
reference frame is. 

When determining relative velocity, it. is easy to make a mistake by adding 
or subtracting the wrong velocities. It is important, therefore, to draw a 
diagram and use a careful labeling process. Each velocity is labeled by two 
subscripts: the first refers to the object, the second to the reference frame in 
which if has this velocity. For example, suppose a boat is to cross a river to the 
opposite side, as shown in Fig, 3-28. We let v BW be the velocity of the Boat 
with respect to the Water. (This is also what tire boat's velocity would be rela- 
tive to the shore if the wateT were still.) Similarly, v bi is the velocity of the 
Boat with respect to the Shore, and v ws is the velocity of the WaleT with 
respect to the Shore (this is the river current). Note that v RW is what the boat's 
motor produces (against the water), whereas v, ls is equal to v BW plus the effect 
of the current, v w $. Therefore, the velocity of the boat relative to the shore is 



62 CHAPTER 3 Kinematics in Two Dimensions; Vectors 



(see vector diagram, Fig. 3-28) 



v KS 



*ws ■ 



(3-6) Follow the subscripts 



By writing the subscripts using this convention, we sec that the inner 
subscripts (the two W's) on the right-hand side of Eq. 3-6 are the same, 
whereas the outer subscripts on the right of Eq. 3-6 (the B and the S) are the 
same as the two subscripts for the sum vector on the left, v^ s , By following this 
convention (first subscript for the object, second for the reference frame), one 
can write down the correct equation relating velocities in different reference 
frames." Equation 3-6 is valid in general and can be extended to three or more 
velocities. For example, if a fisherman on the boat walks with a velocity v,.-^ rela- 
tive to the boat, his velocity relative to the shore is v hS - V KB + V BW + v ws . 
The equations involving relative velocity will be correct when adjacent inner 
subscripts are identical and when the outermost ones correspond exactly to the 
two on the velocity on the left of the equation. But this works only with plus 
signs (on the right), not minus signs. 

Tt is often useful to remember that for any two objects or reference frames, 
A and B, the velocity of A relative to B has the same magnitude, but opposite 
direction, as the velocity of B relative to A; 

%a = -v A „. (3-7) 

For example, if a train is traveling lOOkm/h relative to the Earth in a certain 
direction, objects on the Earth (such as trees) appear to an observer on the 
train to be traveling lODkm/h in the opposite direction. 



CONCEPTUAL EXAMPLE 3-1 I Crossing a river. A man in a smal 



motor boat is trying to cross a river that flows due west with a strong current. 
The man starts on the south bank and is trying to reach the north bank 
directly north from his starting point. Should he (a) head due north, (b) head 
due west, (c) head in a northwesterly direction, (d) head in a northeasterly 
direction? 

RESPONSE Tf the man heads straight across the riveT, the current will drag 
the boat downstream (westward). To overcome the liver's westward current, 
the boat must acquire an eastward component of velocity as well as a north- 
ward component. Thus the boat must (rf) head in a northeasterly direction (see 
Fig. 3-28). The actual angle depends on the strength of the current and how 
fast the boat moves relative to the water. Tf the current is weak and the motor 
is strong, then the boat can head almost, but not quite, due north. 




FIGURE 3-28 To move directly 
across the river, the boat must head 
upstream at an angle 6, Velocity 
vectors are shown as green arrows: 

V[js = velocity of Boat with respect 

to the Shore, 
vuw = velocity of Boat with Tespect 

to the Water, 
?ws = velocity of Water with 

respect to Che Shore (river 

current). 



EXAMPLE 3-11 



Heading upstream. A boat's speed in still water is 
Tf the boat is to travel directly across a river whose current 
1 .20 m/s, at what upstream angle must the boat head? (See 



c uw - I.S5m/s. 
has speed ih# s = 
Fig. 3-29.) 

APPROACH We reason as in Example 3-I0, and use subscripts as in Eq. 3-6. 

Figure 3-29 has been drawn with v, ts , the velocity of the Boat relative to the 

Shore, pointing directly across the river since this is how the boat is supposed 

to move. (Note that v BS = v HVV + v ws .) To accomplish this, the boat needs to 

head upstream to offset the current pulling it downstream, 

SOLUTION Vector v BW points upstream at an angle B as shown. From the 

diagram, 

i>ws _ 1.20 m/s 
'%w 1 .85 ni /s 

Thus - 40,4°, so the boat must head upstream at a 40,4° angle. 



FIGURE 3-29 Example 3-1 1 . 



sin0 = 



= 0.6486. 



'We Ihus would know by inspection thai (for trample) the equation Vjjvu = %j +- ? W j ' s wrong; 
the inner subscripts are not the same, and the outer ones on the right are nor the same as the 
subscripts on the left. 




'SECTION 3-8 Relative Velocity 63 




FIGURE 3-30 Example 3- 1 2. A 

hoat heading directly across a river 
whose current moves at 1.20 m/s. 



across the river. The same boat (ub W = 
across the river whose current is still 1.20 m/s, 



EXAMPLE 3-12 



Heading 

1 .85 m/s) now heads directly 
(a) What is the velocity (magnitude and direction) of the boat relative to the 
shore? (b) It the river is 1 1 m wide, how long will it take to cross and how far 
downstream will the boat be then? 

APPROACH The boat now heads directly across the river and is pulled down- 
stream by the current, as shown in Fig. 3-30. The boat's velocity with respect to 
the shore, v BS , is the sum of its velocity with respect to the water, v BVl , plus the 
velocity of the water with Tespecl to the shoTe, v ws : 

*us = v BW + v ws , 

just as before, 

SOLUTION (a) Since V| lw is perpendicular Lo v w $, we can get Vr$ using the 
theorem of Pythagoras: 

';■■., 



V»bw + »ws - V'( 1-&5 m/s) 2 + (1,20 m/s) 2 - 2.21 m/s. 

We can obtain the angle (note how 8 is defined in the diagram) from: 

tantf = *W*W = (1.20 m/s)/(l.SS m/s) = 0.6486. 

A calculator with an inv tan, an arc tan, or a tan -1 key gives 6 = 
Lan" 1 (0.6486) = 33.0". Note thai this angle is not equal lo Ihe angle calculated 
in Example 3-11. 

(b) The travel time for the boal is determined by Ihe time it takes lo cross Ihe 
river. Given Ihe rivers width D = 110 m, we can use the velocity component 
in the direction of O, yi| tw = D/f, Solving for f, we get t = 110 m/1.85 m/s = 
60 s. The boat will have been carried downstream, in this time, a distance 

d = t% s ( = (1.20 m/s) (60s) = 72 m. 

NOTE There is no acceleration in this Example, so the motion involves only 
constant velocities (of the boat ot of the river). 



| Summary 



A quantity such as velocity, that has both a magnitude and a 
direction, is called a vector. A quantity such as mass, that has 
only a magnitude, is called a scalar. 

Addition of vectors can be done graphically by placing 
the tail of each successive arrow at the tip of the previous one, 
The sum, or resultant vector, is the arrow drawn from the tail 
of the first vector lo the tip of the last vector. Two vectors can 
also he added using the parallelogram method 

Vectors can be added more accurately by adding their 
components along chosen axes with the aid of trigonometric 
functions. A vector of magnitude V making an angle 6 with 
the x axis has components 



V x =Vcas0. V„=Vsmtt. 



(3-3) 



Given the components, we can find a vector's magnitude and 
direction from 



v = vV| + v|, 



tan ft 



V, 



(3-4) 



Projectile motion is the motion of an object in an arc 

near the Earth's surface under the effect of gravity alone. It 
can be analyzed as two separate motions if air resistance can 
be ignored, 'Ifie horizontal component of motion is at 
constant velocity, whereas the vertical component is at 
constant acceleration, g, just as for a body falling vertically 
under the action of gravity, 

[*The velocity of an object relative to one frame of refer- 
ence can be found by vector addition if its velocity relative to 
a second frame of reference, and the relative velocity of the 
two reference frames, arc known. | 



64 CHAPTER 3 Kinematics in Two Dimensions; Vectors 



I Questions 



1. One car travels due east at 40 km/h. and a second ear 
travels mirth at 40 km/h Arc their velocities equal? 
Explain. 

2. Can you give several examples of an object's motion in 
which a great distance is traveled but the displacement is 
7cro? 

3. Can the displacement vector for a particle moving in two 
dimensions ever be longer than the length of path trav- 
eled by the particle over the same time interval? Can it 
ever he less? Discuss. 

4. During baseball practice, a hatter hits a very high fly hall 
and then runs in a straight line and catches it Which had 
the greater displacement, the hatter or the ball? 

5. If V = Vj + V 2 , is V necessarily greater than V, and/or 
Kj? Discuss. 

6. Iwo vectors have length V { - 3.5 km and V? = 4.0 km. 
What are the maximum and minimum magnitudes of their 
vector sum? 

7. Can two vectors of unequal magnitude add up to give the 
zero vector? Can three unequal vectors? UndeT what 
conditions? 

8» Can the magnitude of a vector ever (a) be equal to one of 
its components, or (ft) be less than one of its components? 

9. Can a particle with constant speed be accelerating? What 
if it has constant velocity? 

10. A child wishes to determine the speed a slingshot imparts 
to a rock. How can this be done using only a meter stick, 
a rock, and the slingshot? 

11. It was reported in World War I thai a pilot flying at an 
altitude of 2 km caught in his hare hands a bullet fired at 
the plane! Using the fact that a bullet slows down consid- 
erably due to air resistance, explain how this incident 
occurred. 

12. At some amusement parks, to get on a moving "car" the 
riders first hop onto a moving walkway and then onto the 
cars themselves, Why is this done? 



13. If you are riding on a train that speeds past another train 
moving in the same direction on an adjacent track, it 
appears that the other train is moving backward, Why? 

14. Tf you Stand motionless under an umbrella in a rainstorm 
wheTe the drops fall vertically, you remain relatively dry. 
However, if you start running, the rain hegins to hit your 
legs even if they remain under the umbrella. Why? 

15. A person sitting in an enclosed train car, moving at 
constant velocity, throws a ball straight up into the air in 
her reference frame, (a) Where does the ball land? What 

is your answer if the car (b) accelerates, (c) decelerates, 
(ti) rounds a curve, (tf) moves with constant velocity but is 
open to the air? 

16. Two roweTS. who can row at the same speed in still water, 
set off across a river at the same time, One heads straight 
across and is pulled downstream somewhat by the 
current, The other one heads upstream at an angle so as 
to arrive at a point opposite the starling point. Which 
rower reaches the opposite side first? 

17. How do you think a baseball player -judges'' the flight of 
a fly ball? Which equation in this Chapter becomes part 
of the player's intuition? 

18. In archery, should the arrow be aimed directly at the 
target? How should your angle of aim depend on the 
distance to the target? 

19. A projectile is launched at an angle of 30° to the hori- 
zontal with a speed ol 30m/s, How does the- horizontal 
component of its velocity 1 .0 s after launch compare with 
its horizontal component of velocity 2.0 s after launch? 

20. Two cannonballs, A and B, are fired from the ground with 
identical initial speeds, but with A larger than H-& . 
(a) Which cannonball reaches a higher elevation? (ft) Which 
stays longer in the air? (c) Which travels farther? 



| Problems 



3-2 to 3-4 Vector Addition 

1. (I) A car is driven 215 km west and then $5 km south west- 
What is the displacement of the car from the point of 
origin (magnitude and direction)? Draw a diagram. 

1. (I) A delivery truck travels 18 blocks north. 10 blocks 
east, and 16 blocks south. What Is its final displacement 
from ihe origin? Assume ihe blocks are equal length. 

3. (I) Show that the vector labeled "incorrect" in Fig. 3-6c is 
actually the difference of the two vectors. Is it Yj - Vj , 
OT Vj - V 2 ? 

4. (I) If V x = 6.80 units and V y 7.40 units, determine 

the magnitude and direction of V, 

5. (II) Graphically determine the resultant of the following 
three vector displacements: (1) Mm, 25" north of east: 
(2) 48 m. 33° east of north: and (3) 22 m. 56° west of south. 



6, (II) The components of a vector V can be written 
(F, , V\.. V^). What are the components and length of a vector 
which is the sum of the two vectors. Vj and V 2l whose 
components are (8-0. -3.7.0.0) and (3.9, -8.1.-4-4)? 

7, (II) V is a vector 14.3 units in magnitude and points al an 
angle of 34.8" above the negative x axis, (j?) Sketch this 
vector, (ft) Find V x and V y . (c) Use V x and V y to obtain 
(again) the magnitude and direction of V. \.\'oie; Pan (c) 
is a good way to check if you've resolved your vector 
correctly.] 

K, (II) Vector Vj is 6 6 units long and points along the nega- 
tive x axis, Vector V; is 8.5 units long and points at +45" 
to the positive x axis, (a) What are the x and y compo- 
nents of each vector? (ft) Determine the sum Yj + \\ 
(magnitude and angle), 



Problems 65 



(II) An airplane is traveling 735 km/h in a direction 41.5' : 
wesl of north (Fig. 3-31 ). (a) Find the components of the 
velocity vector in the northerly and westerly directions, 
(ft) How far north and how far west has the plane trav- 
eled after 3.00 h? 



N 



(735 km/h) 



W 




FIGURE 3- 
Prohlem 9. 



.11 



UK (II) Three vectors are shown in Fig. 3-32. Their 
magnitudes are given in arbitrary units. Determine the 
sum of the three vectors. Give the resultant in terms of 
(a) components, (ft) magnitude and angle with the .t axis. 




FIGURE 3-32 Problems 1 0. 11. 12. 13, and 14. 
Vector magnitudes are given in arbitrary units. 

11. (II) Determine the vector A — C. given the Vectors A and 
C in Fig. 3-32. 

12. (II) (a) Given the vectors A and B shown in Fig. 3-32. 
determine B - A. (b) Determine A - I) without using 
yorjT answer in (a). Then compare your results and see if 
they are opposite. 

U. (II) For the vectors given in Fig. 3-32, determine 
(a) A - B + C. (ft) A + B - C. and (c) C - A - B. 

M, (II) For the vectors shown in Fig, 3-32. determine 
(a) B - 2A, (ft) 2A - 3B + 2C. 

IS. (II) The summit of a mountain, 2450 m above base camp, 
is measured on a map to be 4580 m horizontally from the 
camp in a direction 32.4" west of north. What are the 
components of the displacement vector from camp to 
summit? What is its magnitude"? Choose the ,v axis east, y 
axis north, and z axis up, 



Hi. (II) You are given a vector in the xy plane that has a 

magnitude of 70,0 units and a _y component of -55.0 units. 
What are the two possibilities for its .v component? 

3-5 and 3-6 Projectile Motion (neglect air resistance) 

17. (I) A tiger leaps horizontally from a 6.5-m-high rock with 
a speed of 3.5 m/s, How far from the base of the rock will 
she land? 

18. (I) A diveT running 1.8 m/s dives out horizontally from 
the edge of a vertical cliff and 3.0 s later reaches the water 
below, How high was the dill, and how 1st from its base 
did the diver hit the water? 

{'). (II) A fire hose held near the ground shoots water at a 
speed of 6.8 m/s. At what angle(s) should the nozzle point 
in order that the water land 2.0 m away (Fig. 3-33)? Why 
are there two different angles? Sketch the two trajectories. 



A 






:.() m 



20. 



FIGURE 3-33 Problem 19. 

(II) Romeo is chucking pebbles gently up to Juliet's 
window, and he wants the pebbles to hit the window with 
only a horizontal component of velocity. He is standing at 
the edge of a rose garden 4.5 m below her window and 
5.0 m from the base of the wall (Fig, 3-34), How fast are 
the pebbles going when they hit her window? 



I r in / 






5.0 m 



FIGURE 3-34 

Problem 20- 



21. (II) A ball is thrown horizontally from the roof of a 
building 45-0 m tall and lands 24.0 m from the base. What 
was the ball's initial speed? 

22. (II) A football is kicked at ground level with a speed of 
1S.0 m/s at an angle of 35.0" to the horizontal. How much 
later does it hit the ground? 

23. (II) A ball thrown horizontally at 22,2 m/s from the roof 
of a building lands 36.0 m from the base of the building. 
How tall is the building? 



66 CHAPTER 3 Kinematics in Two Dimensions; Vectors 



24. (II) An athlete executing a long jump leaves the ground 
at a 28.0° angle and travels 7.&0m. («) What was the 
takeoff speed? (h) If this speed were increased by just 
5-0%, how much longer would the jump be? 

35. (M) Determine how much farther a person can jump on 
the Moon as compared to the Earth if the takeoff speed 
and angle are the same. '1 rie acceleration due to gravity 
on the Moon is one-sixth what it is on Earth. 

26. (II) A hunter aims directly at a target (on the same level) 
75.0 m away (a) If the bullet leaves the gun at a speed of 
180 m/s. by how much will it miss Ihe target? (b) At what 
angle should the gun l>e aimed so as to hit the target? 

27. (II) The pilot of an airplane traveling ISOkm/h wants to 
drop supplies lo flood victims isolated on a patch of land 
160 m below. The supplies should be dropped how many 

seconds before the plane is directly overhead? 

38. (II) Show lhal the speed with which a projectile leaves 
the ground is equal to its speed just before it strikes the 
ground at the end of its journey, assuming the firing level 
equals Che landing level. 

29. (II) Suppose the kick in Example 3-5 is attempted 3G.0m 
from the goalposts, whose crossbar is 3 .00m above the 
ground. If the football is directed correctly between the 
goalposts, will it pass over the bar and be a field goal? 
Show why or why not. 

3tt. (II) A projectile is fired with an initial speed of 65.2 m/s 
at an angle of 34.5 cr above the horizontal on a long flat 
firing range. Determine (a) the maximum height reached 
by the projectile, (b) the total time in the air. (<::) the total 
horizontal distance covered (that is, the range), and 
(4) the velocity of the projectile 1 ,50 s after tiring. 

31. (II) A projectile is shot from the edge of a cliff 125 m 
above ground level with an initial speed of 65.0 m/s at an 
angle of 37.0" with the horizontal, as shown in Fig. 3-35. 
(a) Determine the lime taken by the projectile to hit point 
P at ground level, (h) Determine the range X of the projec- 
tile as measured from the base of the clitT. At the instant 
just before the projectile hits point P. find (c) the horizontal 
and the vertical components of its velocity, (d) the magni- 
tude of the velocity, and (c) the angle made by the velocity 
vector with the horizontal. (J) Find the maximum height 
above the cliff top reached by the projectile. 



.12. (II) A shotputter throws the shot with an initial Speed of 
15.5 m/s at a 34.0 :! angle to the horizontal. Calculate the 
horizontal distance traveled by the shot if it leaves the 
athlete's hand at a height of 2.20 m above the ground, 

33. (II) At what projection angle will the range of a projectile 

equal its maximum height? 

34. (Ill) Revisit Conceptual Example 3-7. and assume that the 
boy with the slingshot is below the boy in the tree 
(Fig. 3-36), and so aims upward, directly at the boy in the 
tree, Show that again the boy in the tree makes the wrong 
move by letting go at the moment the water balloon is shot. 




FIGURE 3-36 Problem 34. 

35. (Ill) A rescue plane wants to drop supplies to isolated 
mountain climbers on a rocky ridge 235 m below. If the 
plane is traveling horizontally with a speed of 250 km/h 
(69.4 m/s), (a) how far in advance of the recipients (hori- 
zontal distance) must the goods be dropped (Fig. 3-37a)? 
(b) Suppose, instead, that the plane releases the supplies a 
horizontal distance of 425 m in advance of the mountain 
climbers- What vertical velocity (up or down) should the 
supplies be given so that they arrive precisely at the 
climbers' position (Fig. 3-37b)? (<:) With what Speed do 
the supplies land in the latter case? 




H g = 65.0 m/s 




h- (25 in 



r^*' ■ ■ *^ 



235 m 



(a) 



"Dropped" 



■f 




' "- -» Thrown upward? 



(b) 



«35 m Thrown dowtiward? 
1 — _425m— 



S 

\ 

si 



FIGURE 3-35 Problem 31. 



FIGURE 3-37 Problem 35. 



Problems 67 



: - 3-8 Relative Velocity 

* M. (I) A person going for a morning jug on the deck of a 

cruise ship is running toward the bow (front) of the ship 
at 2.2 tn/s while the ship is moving ahead at 7.5 m/s. What 
is the velocity of the jogger relative to the water? Later, 
the jogger is moving toward the stern (rear) of the ship, 
What is the jogger's velocity relative to the water now? 

* 37. (II) Huck Finn walks at a speed of 0.60 m/s across his raft 

(that in, he walks perpendicular to the raft's motion rela- 
tive to the shore). The raft is traveling down the Missis- 
sippi River at a speed of 1 .70 m/s relative to the river 
hank (Fig. 3-38), What is Huck's velocity (speed and 
direction) relative to the river bank? 




: 4X 



44. 



(II) Determine the speed of the boat with respect to the 
shore in Example 3-11. 

(II) A passenger on a boat moving at 1.50 tn/s on a still 
lake walks up a flight of stairs at a speed of 0.50 m/s 
(Fig. 3-39), The stairs are angled at 45 :s pointing in the 
direction of motion as shown. What is the velocity of the 
passenger relative to the water? 




FIGURE 3-3B Problem 37. 



FIGURE 3-39 Problem 44. 

* 45. (II) A motorboat whose speed in still water is 2.60 m/s 
must aim upstream at an angle of 2.8.5" (with respect to a 
line perpendicular to the shore) in ordeT to travel directly 
across the stream, (a) What is the speed of the current? 
(b) What is the resultant speed of the boat with respect to 
the shore? (See Fig, 3-28.) 

*4fi. (II) A boat, whose speed in still water is 1.70 m/s, must 
cross a 260-m-wide river and arrive at a point 110m 

upstream from where it starts (Fig. 3-40). To do so, the 
pilot must head the boat at a 45" upstream angle, What is 

the speed of the river's current? 



48 38. (II) You are driving south on a highway at 25 m/s 
(approximately 55mi/h) in a snowstorm. When you last 
stopped, you noticed that the snow was coming down 
vertically, but it is passing the windows of the moving car 
at an angle of 30" to the horizontal. Estimate the speed of 
the snowflakes relative to the car and relative to the 
ground. 

* 39. (II) A boat can travel 2.30 m/s in still water, («) If the 

boat points its prow directly across a stream whose 
current is 1.20 m/s, what is the velocity (magnitude and 
direction) of the boat relative to the shoTe? (b) What will 
be the position of Che boat, relative to its point of origin, 
after 3.00 s 1 (See Fig. 3-30.) 

*40. (II] Two planes approach each other head-on, Each has a 
speed of 785 km/h, and they spot each other when they 
are initially 1 1 .0 km apart, How much time do the pilots 
have to take evasive action'? 

* 41. (II) An airplane is heading due south at a speed of 

600 km/h, If a wind begins blowing from the southwest at 
a speed of 100 km/h (average), calculate: (a) the velocity 
(magnitude and direction) of the plane relative to the 
ground, and (b) how far from its intended position will it 
be after lOmin if the pilot takes no corrective action. 
\Hirir. First draw a diagram, | 

*42. (II) In what direction should the pilot aim the plane in 
Problem 41 so that it will flv due south? 




FIGURE 3-40 Problem 46. 

* 47. (II) A swimmer is capable of swimming 0.45 m/s in still 

water, (a) II she aims her body directly across a 75-m- 
wide river whose current is 0,40 m/s, how far downstream 
(from a point opposite her starting point) will she land? 
(ri) How long will it lake heT to reach the other side? 

* 48. (II) (a) At what upstream angle must the swimmer in 

Problem 47 aim. if she is to arrive at a point directly 
across the stream? (b) How long would it take her? 



CHAPTER 3 Kinematics in Two Dimensions; Vectors 



'■ 4<J, (III) An airplane whose air speed is 620 km/h is supposed 
to fly in a straight path 35.0" north of east But a steady 
95 km/h wind is blowing from [he north. In what direc- 
tion should the plane head? 

: SO. (Ill) An unmarked police car, traveling a constant 95 km/h, 
is passed hy a speeder traveling 145 km/h. Precisely 1 ,00 s 
alter the speeder passes, the policeman steps on the accel- 
erator, If the police car's acceleration is 2,00 m/s 2 , how 
much timt elapses after the police car is passed until 
it overtakes the speeder (assumed moving at constant 
speed)'? 

SI. (Ill) Assume in Problem 50 that the speeder's speed is 
not known. If the police car accelerates uniformly as 

given above, and overtakes the speeder after 7. DO s, what 
was the speeder's speed? 



; 52. (NT) Two cars approach a street corner at righ 
each other (Fig. 3-41). Car 1 travels at a speed 
Earth t; Jt = 35 km/h. and car 2 at w 3H = 
What is the Tela live velocity of car 1 as seen 
What is the velocitv of car 2 relative to car 1? 



t angles to 

relative to 

= 55 km/h. 

by car 11 



G 



>) 



FIGURE 3-41 

Problem 52. 




| General Problems 



Si, William Tell must split the apple atop his son's head from 
a distance of 27m. When William aims directly at the 
apple, the arrow is horizontal. At what angle must he aim 
it to hit the apple if the arrow travels at a speed of 35 m/s? 

54. A plumber steps out of his truck, walks 50 m east and 
25 m south, and then takes an elevator 10 m down into the 
subbasement of a building where a bad leak is occurring. 
What is the displacement of the plumber relative to his 
truck? Give your answer in components, and also give the 
magnitude and angles with the x axis in the vertical and 
horizontal planes. Assume x is east, v is north, and z is up. 

55. On mountainous downhill roads, escape routes are some- 
times placed to the side of the road for trucks whose 
brakes might fail. Assuming a constant upward slope of 
32", calculate the horizontal and vertical components 
of the acceleration of a truck that slowed from 120 km/h 
to rest in 6.0 s, See Fig. 3-42. 




>fl. What is the y component of a vector (in the .vv plane) 
whose magnitude is 88.5 and whose x component is 75.4? 
What is the direction of this vector (angle it makes with 
the x axis)? 

57. Raindrops make an angle H with the vertical when viewed 
through a moving train window (Fig, 3-43). If the speed 
of the train is i> r . what is the speed of the raindrops in the 
reference frame of the Barlh in which they are assumed 
to fall vertical] v? 



5K. 



5 U . 




TIGURE 3-42 Problem 55. 



FIGURE 3-43 Problem 57. 

A light plane is headed due south with a speed of 
155 km/h relative to still air. After 1, 00 hour, the pilot 
notices that they have covered only 125 km and their 
direction is not south but southeast (45.0'' ). What is the 
wind velocity? 

A car moving at 95 km/h passes a 1 ,00-km-long train 
traveling in the same direction on a track that is parallel 
to the road. If the speed of the train is 75 km/h. how long 
does it take the car to pass the train, and how far will the 
car have traveled in this time? What are the results if the 
caT and tTain are instead traveling in opposite directions? 



General Problems 69 



(in. An Olympic lung jumper is capable of jumping 8-0 m_ 
Assuming his horizontal speed is 9.1 m/s as he leaves the 
ground, how long Is he in the air and how high does he 
go? Assume that he lands standing upright — thai is. the 
same way he left the ground. 

61. Apollo astronauts took a "nine iron" to the Moon and hit 
a golf hall ahout ISO m! Assuming that the swing, launch 
angle, and so on. were the same as on Earth where the 
same astronaut could hit it only 35 m, estimate the accel- 
eration due to gravity on the surface of the Moon. 
(Neglect aiT resistance in both cases, but on the Moon 
there is none!) 

62. When Babe Ruth hit a homer over the 7.5-m-high right- 
field fence 95 m from home plate, roughly what was the 
minimum speed of the ball when it left the hat? Assume 
the ball was hit 1.0m above the ground and its path 
initially made a 38" angle with the ground, 

63. 'lire cliff divers of Acapulco push off horizontally from 

rock platforms about 35 m above the water, but they must 
clear rocky outcrops at water level that extend out into the 
water 5,0 m from the base of the cliff directly under their 
launch point. See Fig. 3-44. What minimum pushoff speed 
is necessary to clear the rocks? How long are they in 
the air? 




65. Spymasler Paul, flying a constant 2l5km/h horizontally 
in a low-flying helicopter, wants to drop secret documents 
into his contact's open car which is traveling 155 km/ h on 
a level highway 78,0m below, At what angle (to the hori- 
zontal) should the car be in his sights when the packet is 
released (Fig, 3-46)? 



— *- 2l5km/h 




66. 



I55km/h 



TIGURE 3-46 Problem 65. 



The speed of a boat in still water is v. The boat is to make 

a round trip in a river whose current travels at speed u. 
Derive a formula for the time needed to make a round trip 
of total distance D if the boat makes the round trip by 
moving (a) upstream and back downstream, (b) directly 
across the river and back. We must assume u < t>\ whv? 



FIGURE 3-44 Problem 63. 



A. 




195 m 



64. At serve, a tennis player aims to hit the ball horizontally. 
What minimum speed is required Fot the ball to clear the 
0.90-m-high net about 15.0 m from the server if the ball is 
"launched" from a height of 2.50 m? Where will the ball 
land if it just clears the net (and will it be "good" in the 
sense that it lands within 7.0 m of the net)? How long will 
it be in the air? See Fis«. 3-45, 



FIGURE 3-47 Problem 67. 

67. A projectile is launched from ground level to the top of a 
cliff which is 195 m away and 155 m high (see Fig, 3-47), 
If the projectile lands on top of the cliff 7.6 s after it is 
fired, find the initial velocity of the projectile (magnitude 
and direction), Neglect air resistance, 



FIGURE 3-45 Problem 64. 



? S() Ml 

a 



15.0 m 



7.(1 rn 



70 CHAPTER 3 Kinematics in Two Dimensions; Vectors 



6H. fa) A skier is accelerating down a 30-0" hill at 1-80 m/s 2 
(Fig. 3-48). What is the vertical component of her accel- 
eration? (A) How long will it take her to reach the bottom 
of the hill, assuming she starts from rest and accelerates 
uniformly, if the elevation change is 335 m? 




FIGURE 3-48 Problem 68. 

69. A basketball leaves a player's hands at a height of 2.10m 
above the floor, The basket is 2.60 m above the floor, The 
player likes to shoot the ball at a 38.0" angle, [f the shot is 
made from a horizontal distance of 1 1 -00 m and must be 
accurate to ± 0.22 m (horizontally), what is the range of 
initial speeds allowed to make the basket? 

7tt, A high diver leaves the end of a 5.0-m-high diving board 
and strikes the wateT 1 .3s later, 3.0 m beyond the end of the 
board. Considering the diver as a particle, determine 
(a) her initial velocity, \„. (b) the maximum height reached, 
and (c) the velocity Vj with which she enters the water. 

71. A stunt driver wants to make his ear jump over eight ears 
parked side by side below a horizontal ramp (Fig. 3-49). 
(d) With what minimum speed must he drive off the hori- 
zontal ramp? "Hie vertical height of the ramp is 1.5 m 
above the cars, and the horizontal distance he must clear 
is 20m. (6) ff the ramp is now tilted upward, so that 
"takeoff angle" is 10° above the horizontal, what is the 
new minimum speed? 



J* 

T 



20 m 



1.5 m 



'Must clear 
ihi*, point! 



72. A batter hits a fly ball which leaves the bat O.sXlm above 
the ground at an angle of 61 <: with an initial speed of 
28 m/s heading toward centerfield. Ignore air resistance. 
(a) How faT from home plate would the ball land if not 
caught? (b) The ball is caught by the centerfielder who, 
starting at a distance of 105 m from home plate, runs 
straight toward home plate al a constant speed and makes 
the catch at ground level. Find his speed. 

73, Al t = a batter hits a baseball with an initial speed of 
32 m/s at a 55" angle to the horizontal. An outfielder is 
85 m from the batter at i = 0, and, as seen from home 
plate, the line of sight to the outfielder makes a horizontal 
angle of 22° with the plane in which the ball moves (see 
Fig. 3-50). What speed and direction must the fielder take 
in order to catch the ball at the same height from which it 
was struck? Give angle with respect to the outfielder's 
line of sight to home plate. 



Fielder runs 

to here fruni here 




FIGURE 3-50 Problem 73. 

74. A ball is shot from the top of a building with an initial 
velocity of 18 m/s at an angle (t = 42" above the hori- 
zontal, (a) What are the .v and y components of the initial 
velocity? (h) Tf a nearby building is the same height and 
55 m away, how far below the top of the building will the 
ball strike the nearby building? 

75. You buy a plastic dart gun, and being a clever physics 
student you decide to do a quick calculation to find its 
maximum horizontal range. You shoot the gun straight Up, 
and it lakes 4,0 s for Ihe dart to land back at the barrel. 
What is the maximum horizontal range of your gun? 



FIGURE 3-49 Problem 71 



Answers to Exercises 

A: When the two vectors D\ and Di point in the same 

direction. 
B: 3V2 = 4,24. 
C: They hit at the same time. 



D: Both balls reach the same height; therefore they are in 

the air for the same length of time. 
E: (b\ 



General Problems 71 



This airplane is Ulking off- Tl is 

accelerating, increasing in speed 

rapidly, To do so, a force must be 

exerted on h according lo Newton's 

second law, SF = ma. What exerts this 

farce? The two jet engines of this plane exert a 

strong force on the gases they push out toward the 

rear of the plane (labeled P c ,v)- According to Newton's third 

law, these ejected gases exert an equal and opposite farce on 

the airplane in the forward direction, It is these "reaction" 

farces exerted on the plane by the gases, labeled Fpq , 

that accelerate the plant forward- 




CHAPTER 




U0z?*Z£ 




:j* s? 



Dynamics: Newton's Laws of Motion 



We have discussed how motion is described in terms of velocity and 
acceleration. Now we deal with the question of why objects move as 
they do: What makes an object at rest begin to move? What causes 
an object to accelerate or decelerate? What is involved when an object moves in 
a circle? We can answer in each case that a force is required. In this Chapter, we 
will investigate the connection between force and motion, which is the subject 
called dynamics. 

We begin with intuitive ideas of what a force is. and then discuss Newton's 
three laws of motion. We next look at several types of force, including friction 
and the force of gravity. We then apply Newton's laws to real problems. 

Force 

Intuitively, we experience force as any kind of a push or a pull on an object. 

When you push a stalled car or a grocery cart (Fig. 4-1). you arc exerting a 
FIGURE 4-1 A force exerted on a force on it. When a motor lifts an elevator, or a hammer hits a nail, or the wind 
grocery carl — in this case exerted blows the leaves of a tree, a foTce is being exerted. We say that an object falls 
by a child. because of ibt force of gravity. 




72 




FIGURE 4-2 A spring scale used 
to measure a force. 






Tf an object is at rest, to start it moving requires force — that is, a force is 
needed to accelerate an object from zero velocity to a nonzero velocity. For an 
object already moving, if you want to change its velocity — either in direction or 
in magnitude — again a force is required. In other words, to accelerate an object, 
a foTce is required. 

One way to measure the magnitude (or strength) of a force is to use a 
spring scale (Fig. 4-2). Normally, such a spring scale is used to find the weight of Measuring force 
an object; by weight we mean the force of gravity acting on the object 
(Section 4-6). The spring scale, once calibrated, can be used to measure other 
kinds of forces as well, such as the pulling force shown in Fig, 4-2. 

A force exerted in different directions has a different effect. Clearly, force has 
direction as well as magnitude, and is indeed a vector that follows the rules of 
vector addition discussed in Chapter 3, We can represent any force on a diagram by 
an arrow, just as we did with velocity. The direction of the arrow is the direction of 
the push or pull, and its length is drawn proportional to the magnitude of the force. 



Newton's First Law of Motion 



What is the relationship between force and motion? Aristotle (384-322 B.C.) Aristotle 
believed that a force was required to keep an object moving along a horizontal 
plane. To Aristotle, the natural state of an object was at rest, and a force was 
believed necessary to keep an object in motion. Furthermore, Aristotle argued, 
the greater the force on the object, the greater its speed. 

Some 2000 years later, Galileo disagreed: ITc maintained that it is just as Galileo 
natural for an object to be in motion with a constant velocity as it is for it. to 
be at test. 

To understand Galileo's idea, consider the following observations involving 
motion along a horizontal plane. To push an object with a rough surface along a 
tabletop at constant speed requires a certain amount of force. To push an 
equally heavy object with a very smooth surface across the table at the same 
speed will require less force. If a layer of oil or other lubricant is placed 
between the surface of the object and the table, then almost no force is required 
to move the object. Notice that in each successive step, less force is required. As 
the next step, we imagine that the object does not rub against the table at all — 
or there is a perfect lubricant between the object and the table — and theorize 
that once started, the object would move across the table at constant speed with 
no force applied. A steel ball bearing rolling on a hard horizontal surface 
approaches this situation. So does a puck on an air table, in which a thin layer of 
air reduces friction almost to zero, 

It was Galileo's genius to imagine such an idealized worid — in this case, one 
where there is no friction — and to see that it could lead to a more accurate and 
rieher understanding of the real world. This idealization led him to his remark- 
able conclusion that if no force is applied to a moving object, it will continue to 
move with constant speed in a straight line. An object slows down only if a force 
is exerted on it. Galileo thus interpreted friction as a force akin to ordinary i-i-u tion ,c a force 
pushes and pulls. 



SECTION 4-2 Newton's First Law of Motion 73 



To push an object across a table aL constant speed requires a force from 
your hand that can balance out the force of friction (Fig, 4-3)- When ihe object 
moves at constant speed, your pushing I'oTce is equal in magnitude to the fric- 
tion force, but these two forces are in opposite directions, so the net force on the 
object (the vector sum of the two forces) is zero. This is consistent with Galileo's 
viewpoint, for the object moves with constant speed when no net force is 
exerted on it. 



FIGURE 4-3 F represents the force applied 
by the person and F|> represents (he forte of 
friction. 




NEWTON'S FIRST LAW 
OF MOTION 



Inertia 



FIGURE 4-4 

Isaac Newton (1642-1727). 




•'.' tertia ! referen ce frames 



Upon this foundation laid by Galileo, Isaac Newton (Fig. 4-4) built his 
great theory of motion, Newton's analysis of motion is summarized in his 
famous "three laws of motion/' In his great work, the Principia (published in 
1 687), Newton readily acknowledged his debt to Galileo. In fact, Newton's first 
la« of motion is dose to Galileo's conclusions, It states that 

Every object continues in its state of rest, or of uniform velocity in a straight 
line, as long as no net force acts on il. 

The tendency of an object to maintain its state of rest or of uniform motion in a 
straight line is called inertia. As a result. Newton's first law is often called the 
law of inertia. 



CONCEPTUAL EXAMPLE 4-1 Newton's first law, A school bus comes 



to a sudden stop, and all of the backpacks on the floor start to slide forward, 
What force causes them to do that? 

RESPONSE It isn't "force" that does it. The backpacks continue their state of 
motion, maintaining their velocity (friction may slow them down), as the 
velocity of the bus decreases. 



Inertial Reference Frames 

Newton's first law docs not hold in every reference frame, For example, if your 
reference frame is fixed in an accelerating car, an object such as a cup resting on 
the dashboard may begin to move toward you (it stayed at rest as long as the car's 
velocity remained constant). The cup accelerated toward you, but neither you nor 
anything else exerted a force on it in that direction. Similarly, in the reference 
frame of the bus in Example 4-1, there was no force pushing the backpacks 
forward. In accelerating reference frames. Newton's first law does not hold. Refer- 
ence frames in which Newton's first law does hold are called inertial reference 
frames (the law of inertia is valid in them). For most purposes, we can usually 
assume that reference frames fixed on the Earth are inertial frames, (This is not 
precisely true, due to the Earth's rotation, but usually if is close enough.) Any 
reference frame that moves with constant velocity (say, a car or an airplane) relative 
to an inertial frame is also an inertial reference frame. Reference frames where the 
law of inertia does not hold, such as the accelerating reference frames discussed 
above, are called nonincrtbil reference frames. How can we be sure a reference 
frame is inertial or not? By checking to see if Newton's first law holds. Thus 
Newton's first law serves as the definition of inertial reference frames. 



74 CHAPTER 4 Dynamics: Newton's Laws of Motion 



Mass 



Newton's second law, which we come to in the next Section, makes use of the 
concept of mass. Newton used the term mass as a synonym for quantity of matter. 
This intuitive notion of the mass of an object is not very precise because the 
concept "quantity of matter" is not very well defined. More precisely we can say 
that mass is a measure of the inertia of an object. The more mass an ohject has, 
the greater the force needed to give it a particular acceleration. It is harder to 
start it moving from rest, or to stop it when it is moving, or to change its velocity 
sideways out of a straight-line path, A truck has much more inertia than a baseball 
moving at the same speed, and it requires a much greater force to change the truck's 
velocity at the same rate as the ball's. The truck therefore has much more mass. 

To quantify the concept of mass, we must define a standard. In SI units, the 
unit of mass is the kilogram (kg) as we discussed in Chapter I, Section 1-5. 

The terms mass and weight arc often confused with one another, but it is 
important to distinguish between them. Mass is a property of an object itself (a 
measure of an object's inertia, or its "quantity of matter"). Weight, on the other 
hand, is a force, the pull of gravity acting on an object. To see the difference, suppose 
we take an object to the Moon, The object will weigh only about one-sixth as much 
as it did on Earth, since the force of gravity is weaker. But its mass will be the same. 
It will have the same amount of matter as on Earth, and will have just as much 
inertia — for in the absence of friction, it will be just as hard to start it moving on the 
Moon as on Earth, or to stop it once it is moving. (More on weight in Section 4-6.) 



Mass as inertia 



CAUTION 



Distinguish mass from weight 



Newton's Second Law of Motion 



Newton's first law states that if no net force is acting on an object at rest, the 
object remains at rest; or if the object is moving, it continues moving with 
constant speed in a straight line. But what happens if a net force is exerted on an 
object? Newton perceived that the object's velocity will change (Fig. 4-5). A net 
force exerted on an object may make its velocity increase. Or, if the net force is 
in a direction opposite to the motion, the force will reduce the objects velocity. 
If the net force acts sideways on a moving object, the direction of the object's 
velocity changes (and the magnitude may as well). Since a change in velocity is 
an acceleration (Section 2-4), we can say that a net force causes acceleration. 

What precisely is the relationship between acceleration and force? Everyday 
experience can suggest an answer. Consider the foTce required to push a cart when 
friction is small enough to ignore. (If there is friction, consider the net force, which 
is the force you exert minus the force of friction.) Now if you push with a gentle 
but constant force for a certain period of lime, you will make the cart accelerate 
from rest up to some speed, say 3 km/h. Tf you push with twice the force, the cart 
will reach 3 km/h in half the time. The acceleration will be twice as great. If you 
triple the I'oTce, the acceleration is tripled, and so on. Thus, the acceleration of an 
object is directly proportional' to the net applied force. But the acceleration 
depends on the mass of the object as well. Tf you push an empty grocery cart with 
the same force as you push one that is filled with groceries, you will find that the 
full cart accelerates more slowly. The greater the mass, the less the acceleration for 
the same net force. The mathematical relation, as Newton aTgued, is that the accel- 
eration of an ohjeet is inversely proportional to its mass. These relationships are 
found to hold in general and can be summarized as follows: 

The acceleration of an object is directly proportional to the net force acting 
on it. and is inversely proportional to its mass. The direction of the acceler- 
ation is in the direction of the net force acting on the object. 

This is Newton's second law of motion. 




FIGURE 4-5 The bobsled 

accelerates because the team exerts 
a force. 



NEWTON'S SECOND LAW 
Of- MOTION 



'A review of proportionality is given in Appendix A. at tlie hack of this book. 



SECTION 4-4 Newton's Second Law of Motion 75 



Net forte 



N E W 1 ON 'S S E( O N P LAW 
OF MOTION 



Force defined 



Unit of force: 
the newton 



*» PROBLEM SOLVING 


Use a ctjtisiuetii \c-7 oj mats 


TABLE 4-1 

Units for Mass and Force 


System 


Mass Force 


SI 


kilogram newton (N) 

(kg) (=kg-m/s 2 ) 


cgs 


gram (g) dyne 

(= g ■ cm/s 2 ) 


British 


slug pound (lb) 


Conversion factors: [dyne = lf)" v N; 



Newton's second law can be written as an equation; 
m 



a 



where a stands for acceleration, m for the mass, and 2F* for the net force on the 
object. The symbol 2 (Greek "sigma") stands for "sum of; P stands for force, 
so XF" means the vector sum of all forces acting on the object, which we define 
as the net forte. 

We rearrange this equation to obtain the familiar statement of Newton's 
second law; 



2F = ma. 



(4-1) 



Newton's second law relates the description of motion to the cause of motion, 
foTce. ft is one of the most fundamental relationships in physics, From Newton's 
second law we can make a more precise definition of force as an action capable 
of accelerating an object. 

Every force F is a vector, with magnitude and direction. Equation 4-1 is a 
vector equation valid in any inertial reference frame. It can be written in 
component form in rectangular coordinates as 



2F, 



■I 7 . 



XF, 



Tf the motion is all along a line (one-dimensional), we can leave out the 
subscripts and simply write XF = ma. 

In SI units, with the mass in kilograms, the unit of force is called the newton (N). 
One newton, then, is the force required to impart an acceleration of 1 m/s 2 to a 
mass of 1 kg. Thus I N = I kg- m/s 2 . 

In cgs units, the unit of mass is the gram (g) as mentioned earlier.' The unit 
of force is the dyne, which is defined as the net force needed to impart an accel- 
eration of 1 cm/s ; to a mass of 1 g. Thus 1 dyne = 1 g-cm/s 2 . ft is easy to show 
that I dyne = l<T ? N. 

In the British system, the unit of force is the pound (abbreviated lb), where 
1 lb = 4.44822 N ~ 4.45 N, The unit of mass is the dug, which is defined as 
that mass which will undergo an acceleration of I ft/s 2 when a force of 1 lb is 
applied to it. Thus 1 lb = 1 slug ft/s 2 . Table 4-1 summarizes the units in the 
different systems. 

It is very important that only one set of units be used in a given calculation 
or problem, with the SI being preferred. If the force is given in, say, newtons, 
and the mass in grams, then before attempting to solve for the acceleration in ST 
units, we must change the mass to kilograms. For example, if the force is given 
as 2.0 N along the x axis and the mass is 500 g, we change the latter to 0.50 kg, 
and the acceleration will then automatically come out in m/s 2 when Newton's 
second law is used (we set I N - I kg- m/s 2 ): 



<tx = 



y.F x 2.0 N 2.0 kg -m/s 2 



a i 



0.50 kg 0.50 kg 



= 4.0 m/V, 



EXAMPLE 4-2 ■Jail^WJ 



Force to accelerate a fast car. Estimate 
the net force needed to accelerate (a) a 1000-kg car at jg; (b) a 200-g apple at 
the same rate. 

APPROACH We can use Newton's second law to find the net force needed (ot 
each object, because we are given the mass and the acceleration. This is an 
estimate (the { is not said to be precise) so we round off to one significant 
figure, 

: Bc cartful not to confuse g for gram with g for the acceleration due to gravity. The latter is always 
italicized (or boldface when a vector). 



76 CHAPTER 4 Dynamics: Newton's Laws of Motion 



SOLUTION (a) The car's acceleration is a = jg = 4(9.8 m/s 2 ) w 5m/sl w e 
use Newton's second law to get the net force needed to achieve this acceleration: 

XF = ma ~ (1000 kg)(5 m/s 2 ) = 5000 N. 

(If you aTe used to British units, to get an idea of what a 5000-N force is, you 
can divide by 4,45 N/lb find get a force of about 10001b) 
(b) Fot the apple, m = 200 g = 0.200 kg, so 

XF = ma ~ (0,200 kg)(5 m/s 2 ) = I N, 



EXAMPLE 4-3 



Force to slop a car. What average net force is required to 
bring a 1500-kg car to rest from a speed of 100 ktn/h within a distance of 55 m? 

APPROACH We can use Newton's second law, SF = ma, to determine the 
force if we know the mass and acceleration of the car, We are given the mass, 
but we will have to calculate the acceleration a. We assume the acceleration is 
constant, so we can use the kinematic equations, Fqs. 2-1 1, to calculate it. 



Bo=lOOkm/li o=0 



i r 

'. (1 *=55m 

SOLUTION We assume the motion is along the +x axis (Fig. 4-6). We are 
given the initial velocity t> - lOOkm/h -28 m/s (Section 1-6), the final 
velocity v — 0, and the distance traveled x - x u — 55 m, From Eq. 2-1 1 c, 
we have 

v 1 = v\ + 2a{x - x n ), 

so 

v'-vl 0- (28 m/s} 2 

= -7.1 m/s . 



FIGURE 4-6 Example 4-3. 

i" (rli) 



2(x - _i- ) 2(55 m) 

The net force requited is then 

EF = ma = (1500 fcg)(-7. 1 m/s 2 ) = -1.1 X 10 4 N. 

The force must be exerted in the direction opposite to the initial velocity, 
which is what the negative sign means. 

NOTE When we assume the acceleration is constant, even though it may not 
be precisely true, we are determining an "average" acceleration and we obtain 
an "average"' net force (or vice versa). 



Newton's second law, like the first law, is valid only in inertial reference 
frames (Section 4-2). Tn the noninertial reference frame of an accelerating car, 
for example, a cup an the dashboard starts sliding — it accelerates — even though 
the net force on it is zero; thus EF = ma doesn't work in such an accelerating 
reference frame. 

Newton's Third Law of Motion 

Newton's second law of motion describes quantitatively how forces affect A force is exerted on an object and is 

motion. But where, we may ask, do forces come from? Observations suggest exerted by another object 

that a force applied to any object is always applied by another object. A horse 

pulls a wagon, a person pushes a grocery cart, a hammer pushes on a nail, a 

magnet attracts a paper clip, In each of these examples, a force is exerted on one 

object, and that force is exerted by another object. For example, the force 

exerted on the nail is exerted by the hammer. 

SECTION 4-5 Newton's Third Law of Motion 77 



N E WI () ,V 'S THIRD LA W 
OF MOTION 



But NewLon realized that things are not so one-sided. True, the hammer 
exerts a force on the nail (Fig, 4-7). Bui the nail evidently exerts a force back 
on Ihe hammer as well, for the hammer's speed is rapidly reduced to zero upon 
contact Only a strong force could cause such a rapid deceleration of the 
hammer. Thus, said Newton, the two objects must be treated on an equal basis. 
The hammer exerts a force on the nail, and the nail exeTts a force back on the 
hammer. This is the essence of Newton 1 s third law of motion: 

Whenever one object exerts a force on a second object, the second exerts 

mi equal force in the opposite direction on the first. 



This law is sometimes paraphrased as "to every action there is an equal and 
<i> caution opposite reaction." This is perfectly valid. But to avoid confusion, it is very 

Action and reaction forces act important to remember that the "action" force and the "reaction" force are 
on different objects acting on different objects. 




FIGURE 4-7 A hammer striking a 
nail. The hammer exerts a force on 
the nail and the nail exerts a force 
back on the hammer. The latter 
force decelerates the hammer and 
brings it to rest. 

FIG U RE 4 -9 A n example of 
Newton's third law: when an ice 
skater pushes against the wall, the 
wall pushes back and this force 
causes her to accelerate awav. 




Rocket acceteratit >n 



Force exerted 

■ ■I hnild 

by desk 




Batce BiBCte 
on di'sk by h;ind 



FIGURE 4-8 II your hand 
pushes against the edge of a desk 
(the force vector is shown in red), 
the desk pushes back against vout 
hand (this force vector is shown 
in a different color, violet, 
to remind us that this force 
acts on a different object) 



78 CHAPTER 4 Dynamics: IM 



As evidence for the validity of Newton's third law, look at your hand when 
you push against the edge of a desk. Fig. 4-8, Your hand's shape is distorted, 
clear evidence that a force is being exerted on it. You can see the edge of the 
desk pressing into your hand. You can even feel the desk exerting a force on 
your hand; it hurts! The harder you push against the desk, the harder the desk 
pushes hack on your hand. (You only feel forces exerted on you; when you exert 
a force on another object, what you feel is that object pushing back on you.) 

As another demonstration of Newton's third law, consider the ice skater 
in Fig. 4-9, There is very little friction between her skates and the ice, so 
she will move freely if a force is exerted on her. She pushes against the 
wall: and then she starts moving backward. The force she exerts on the wall 
cannot make her start moving, for that force acts on the wall. Something 
had to exert a force on her to start her moving, and that force could only have 
been exerted by the wall. The force with which the wall pushes on her is, by 
Newton's third law, equal and opposite to the force she exerts on the wall. 

When a person throws a package out of a boat (initially at rest), the boat 
starts moving in the opposite direction. The person exerts a force on the 
package. The package exerts an equal and opposite force back on the person, 
and this force propels the person (and the boat) backward slightly. 

Rocket propulsion also is explained using Newton's third law (Fig, 4-10). A 
common misconception is that rockets accelerate because the gases rushing out 
the back of the engine push against the ground ot the atmosphere. Not true. 
What happens, instead, is that a rocket exerts a strong force on the gases, 
expelling them; and the gases exert an equal and opposite force on the rocket. U 
is Ihis latter force that propels Ihe rocket forward — the force exeUed on ihe 
rocket by Ihe gases. Thus, a space vehicle is maneuvered in empty space by 
firing ils rockets in the direction opposite to that in which it needs to accelerate, 
When the rocket pushes on the gases in one direction, the gases push back on 
the rocket in the opposite direction, 

ewton's Laws of Motion 




FIGURE 4-10 Another example of Newton's third law: 
the launch of a rocket, The rocket engine pushes the gases 
downward, and Che gases exert an equal and opposite force 
upward on the rocket, accelerating it upward- (A rocket 
does no! accelerate as a result of its propelling gases 
pushing against the ground.) 



Consider how we walk. A person begins walking by pushing with the foot 
backward against the ground. The ground then exerts an equal and opposite 
force forward on the person (Fig. 4-11), and it is this force, on the person, that 
moves the person forward. (If you doubt this, try walking normally where there 
is no friction, such as on very smooth slippery ice.) Tn a similar way, a bird flies 
forward by exerting a backward force on the air, but it is the air pushing 
forward on the bird's wings that propels the bird forward. 



CONCEPTUAL EXAMPLE 4-4 What exerts the force on a car? What 



We tend to associate forces with active objects such as humans, animals, 
engines, or a moving object like a hammer, Tt is often difficult to see how an 
inanimate object at rest, such as a wall or a desk, or the wall of an ice rink 
(Fig. 4-9), can exert a force. The explanation is that every material, no matter 
how hard, is elastic (springy), at least to some degree. A stretched rubber band 
can exert a force on a wad of paper and accelerate it to fly across the room. 
Other materials may not stretch as readily as rubber, but they do stretch or 
compress when a force is applied to them. And just as a stretched rubber band 
exerts a force, so docs a stretched (or compressed) wall, desk, or car fender, 

From the examples discussed above, we can sec how important it is to 
remember on what object a given force is exerted and by what object that force 
is exerted, A force influences the motion of an object only when it is applied on 
that object. A force exerted by an object docs not influence that same object; it 
only influences the other object on which it is exerted. Thus, to avoid confusion, 
the two prepositions on and by must always be used — and used with care. 

One way to keep clear which force acts on which object is to use double 
subscripts. For example, the force exerted on the Person by the Ground as the 
person walks in Fig. 4- 1 1 can be labeled F^ . And the force exerted on the 
ground by the person is P GJ >. By Newton's third law 



r CP 



-Fro 



How itt'iM icalk 

FIG U RE 4- 1 1 We can walk 

forward because, when one foot 
pushes backward against the 
ground, the ground pushes forward 
on that toot. I ne two forces shown 

act on different objects. 



makes a cat go forward? 

RESPONSE A common answer is that the engine makes the car move 
forward. But it is not so simple. The engine makes the wheels go around. But if 
the tires are on slick ice or deep mud, they just spin. Friction is needed. On 
solid ground, the tires push backward against the ground because of friction. 
By Newton's third law, the ground pushes on the tires in the opposite direc- 
tion, accelerating the car forward. 



Horizontal 

force exerted 
on the ground 
by person's 

fool ' 



I'd* 




Horizontal 
force exerted 
on the 

person's fool 
b}' Che ground 



Inanimate objects can exert a force 
(due in elasticity) 



*+ PROBLEM SOLVING 
For each force, be dear on which object 

,-.' i ,-. n \ : n id bj 1 1 'hit h ob i< % t it i s i a ei a id 

2F = ma applies only to forces acting 
on an object. 



(4-2) 

F CiP and Fj, Ci have the same magnitude (Newton's third law), and the minus sign 
reminds us that these two forces arc in opposite directions. 

Note carefully that the two forces shown in Fig. 4-11 act on different 
objects — hence we used slightly different colors for the vector arrows repre- 
senting these forces. These two forces would never appear together in a sum 
of forces in Newton's second law, £p - ma. Why not? Because they act on 
different objects: a is the acceleration of one particular object, and SP must 
include only the forces on that one object. 



NEWTON'S THIRD LAW 
OF MOTION 



SECTION 4-5 79 



FIGURE 4-12 Example 4-5. 

showing only horizontal forces. 
Seventy-year-old Michelangelo has 
selected a fine black of marhle for 
his next sculpture. Shown here is his 
assistant pulling it on a sled away 
from the quarry. Forces on the assis- 
tant are shown as red (magenta) 
arrows. Forces on the sled are purple 
arrows. Forces acting on the ground 
are orange arrows. Act ion -re action 
forces that are equal and opposite 
are labeled by the same subscripts 
hut reversed (such as Tq a and Fag) 
and aTe of different colors because 
they act on different objects. 



PROBLEM SOLVING 
A study of Newton's second and 
third laws 



Force on 
assistant 
exerted 



by sled 



#> 




Force on 

assistant 
exerted 

by ground 

FIGURE 4-13 Rxample 4-5. The 
horizontal forces on the assistant. 



Force on sled 
exerted by 
assistant 



Force on 
assistant 

exerted 
hv sled 




Friction Force on 

farce on ground 

sled exerted exerted 

by ground by sled 



Force on 

ground 

exerted 

by assistant 



Force on 

assistant 
exerted 
by ground 



] Third law clarification. Michelangelo's 



assistant has been assigned the task of moving a block of marble using a sled 
(Fig. 4- 1 2). He says to his boss, "When T exerL a forward force on the sled, the 
sled exerts an equal and opposite force backward. So how can I eveT start it 
moving? No matteT how hard I pull, the backward reaction force always 
equals my forward force, so the net force must be zero. I'll never be able to 
move this load." Ts this a case of a little knowledge being dangerous? Explain. 

RESPONSE Yes, Although it is true that the action arid reaction forces are equal 
in magnitude, the assistant has forgotten that they are exerted on different 
objects. The forward ("action") force is exerted by the assistant on the sled 
(Fig. 4-12), whereas the backward "reaction'" force is exerted by the sled on the 
assistant. To determine if the assistant moves or not, we must consider only 
the forces on the assistant and then apply 2F = ma, where XF is the net force 
on the assistant, a. is the acceleration of the assistant, and in is the assistant's mass 
There are two forces on the assistant that affect his forward motion; they are 
shown as bright red (magenta) arrows in Figs. 4-12 and 4-13: they are (1) the 
horizontal force P AG exerted on the assistant by the ground (the harder he pushes 
backward against the ground, the harder the ground pushes forward on him — 
Newton's third law), and (2) the force tf AS exerted on the assistant by the sled, 
pulling backward on him; see Fig, 4-13, If he pushes hard enough on the ground, 
the force tin him exerted by the ground, F ACi , will be larger than the sled pulling 
back, F AS , and the assistant accelerates forward (Newton's second law). The sled, 
tin the other hand, accelerates forward when the force on // exerted by the assis- 
tant is greater than the frictional force exerted backward on it by the ground (that 
is, when F SA has greater magnitude than F SCi in Fig. 4-12). 



Using double subscripts to clarify Newton's third law can become cumber- 
some, and we won't usually use them in this way. Nevertheless, if there is any 
confusion in your mind about a given force, go ahead and use them to identify 
on what object and by what object the force is exerted. We will usually use a 
single subscript referring to what exerts the force on the object being discussed, 

EXERCISE A \ massive truck collides head-on with a small sports car. (a) Which 
vehicle experiences the greater force of impact? (ft) Which experiences the greater 
acceleration? (t) Which of Newton's laws is useful to obtain the correct answer? 



4-6 



Weight — the Force of Gravity; 
and the Normal Force 



As we saw in Chapter 2, Galileo claimed that all objects dropped near the surface 
of the Earth will fall with the same acceleration, g, if air resistance can be neglected. 
The force that causes this acceleration is called the force of gravity or gravitational 
force. What exerts the gravitational force on an object? It is the Earth, as we will 



SO CHAPTER 4 Dynamics: Newton's Laws of Motion 



discuss in Chapter 5, and the force acts vertically 1 downward, toward the center of 
the Earth. Let us apply Newton's second law to an object of mass m falling due to 
gravity; for the acceleration, a, we use the downward acceleration due to gravity, g. 
Thus, the gravitational force on an object, F& , can be written as 

Pc = '"I- (4-3) Weight - gravitational foac 

The direction of this force is down toward the center of the Earth. The magni- 
tude of the force of gravity on an object is commonly called the object's weight. 

Tn SI units, # = 9.80 m/s 2 = 9.80 N/kg,' so the weight of a 1 .00-kg mass on 
Earth is 1 .00 kg X 9.80 m/s 2 = 9.80 N. We will mainly he concerned with the 
weight of objects on Earth, but we note that on the Moon, on other planets, or 
in space, the weight of a given mass will be different than it is on Earth. For 
example, on the Moon the acceleration due to gravity is about one-sixth what it 
is on Earth, and a 1.0-kg mass weighs only 1.7 N, Although wc will not use 
British units, we note that for practical purposes on the Earth, a mass of I kg <|> CAJTION 
weighs about 2.2 lb. (On the Moon, I kg weighs only about 0.4 lb.) Mass vs. weighs 

The force of gravity acts on an object when it is falling. When an object is at 
rest on the Earth, the gravitational force on it does not disappear, as we know if 
wc weigh it on a spring scale, The same force, given by Eq. 4-3, continues to act. 
Why, then, doesn't the object move? From Newton's second law, the net force 
on an object that remains at rest is zero, There must be another force on the 
object to balance the gravitational force. For an object resting on a table, the 
table exerts this upward force; see Fig. 4- 14a. The table is compressed slightly 
beneath the object, and due to its elasticity, it pushes up on the object as shown. 
The force exerted by the table is often called a contact force, since it occurs Contact force 
when two objects are in contact. (The force of your hand pushing on a cart is 
also a contact force.) When a contact force acts perpendicular to the common 
surface of contact, it is referred to as the normal force ("normal" means perpen- Normal force 
dicular); hence it is labeled P N in Fig. 4- 1 4a. 

FIGURE 4-1* (is) The net force on an 
object at rest is zero according to Newton's 
second law. Therefore the downward force of 
gravity (Frj) on an object must be balanced 
by an upward force (the normal force Fis) 
exerted by the table in this case, (b) F^ is the 
foTce exerted on the table by the statue and 
is the_reactk)Ti force to F K per Newton's third 
law, (P'k is shown in a different color to 
remind us it acts on a different object,) The 
reaction to F fi is not shown. 

The two forces shown in Fig. 4- 14a are both acting on [he statue, which 

remains at rest, so the vector sum of these two forces must be zero (Newton's ^t > CAUTION 

second law). Hence F G and F N must be of equal magnitude and in opposite Weight and normal force are not 

directions. But they are not the equal and opposite forces spoken of in Newton's action-reaction pairs 

third law. The action and reaction forces of Newton's third law act on different 

objects, whereas the two forces shown in Fig. 4- 14a act on the tame object. For 

each of the forces shown in Fig. 4- 14a, we can ask, ''What is the reaction force?" 

The upward force, F N , on the statue is exerted by the table. The reaction to this 

force is a force exerted by the statue downward on the table. Tt is shown in 

Fig. 4- 14b, where it is labeled F^-, This force, Fi,-, exerted on the table by the 

statue, is the reaction force to F N in accord with Newton's third law. What about 

the other force on the statue, the force of gravity F Ci exerted by the Earth? Can 

you guess what the reaction is to this force? Wc will sec in Chapter 5 that the 

reaction force is also a gravitational force, exerted on the Earth by the statue. 

' The concept of ""vertical" is lied to gravity. The best definition of vertical is [hat it is [he direction in 
which objects fall, A surface that is "" horizontal." on the other hand, is a surface cm which a round 
object won't start rolling: gravity has no effect- Horizontal is perpendicular "J vcrcical- 
1 Since I N = I kg- m/s 2 (Section 4-4), I m/s 2 = i N/kg. 

SECTION 4-6 Weight— the Force of Gravity; and the Normal Force 81 





'■'-. 





^- 
















^m 








f 


M 








1 


i) 


»'g 







FIGURE 4-15 Example 4-6. 
(a) A lU-kg gift box is at rest on 
a table, (b) A person pushes down 
on the hox with a force of 4(1.0 N. 
(e) A person pulls upward on the 
box with a force of 40,0 N. The forces 
are all assumed to act along a line: 
they are shown slightly displaced in 
order to be distinguishable. Only 
forces acting on the box are shown. 

^ CAUTION 

The normal force if not 
necessarily equal to the weight 



EXAMPLE 4-6 



Weight, normal force, and a box. A friend has given you 
a special gift, a box of mass 10.0 kg with a mystery surprise inside. The hux is 
resting on the smooth (frictionlcss) horizontal surface of a table (Fig. 4-15a), 
(a) Determine the weight of the box and the normal force exerted on it by the 
table, (b) Now your friend pushes down on the box with a force of 40.0 N, as in 
Fig. 4- 15b. Again determine the normal force exerted on the box by the table, 
(c) If your friend pulls upward on the box with a force of 40.0 N (Fig. 4- 15c), 
what now is the normal force exerted on the box hy the tabic ? 

APPROACH The box is at rest on the table, so the net force on the box in 
each case is zcto (Newton's second law). The weight of the box equals mg in 
all three cases. 

SOLUTION (a) The weight of the box is mg = (10.0 kg}(9.80m/s 2 ) = °S.O N, 
and this force acts downward. The only other force on the box is the normal 
force exerted upward on it by the lable, as shown in Fig. 4- 15a. We chose the 
upward direction as the positive v direction; then the net force 2F,, on the box 
is iiF,, = F N — mg. The box is at rest, so the net force on it must be zero 
(Newton's seeond law, X/%. = ma y , and a y = 0).Thus 

F x - mg = 0, 

and we have in this case 



2F y 



F N = mg. 

The normal force on the box, exerted by the table, is 98,0 N upward, and has 
magnitude equal to the box's weight. 

(b) Your friend is pushing down on the box with a force of 40.0 N. So instead 
of only two forces acting on the box, now there are three forces acting on the 
box, as shown in Fig, 4- 15b. The weight of the box is still mg - 98.0 N. The 
net force is ~2F y - F N - mg - 40.0 N, and is equal to zero because the box 
remains at rest. Thus, since a — 0, Newton's second law gives 

SF, = F s - mg -40.0N = 0. 

We solve this equation for the normal force; 

F N = mg + 40.0 N = 98.0 N + 40.0 N = 138.0 N, 

which is greater than in (a). The tahle pushes hack with more force when a person 
pushes down on the box. The normal force is not always equal to the weight! 

(c) The box's weight is still 98.0 N and acts downward. The force exerted by 
your friend and the normal force both act upward (positive direction), as 
shown in Fig. 4- 1 5c. The box doesn't move since your friend's upward force is 
less than the weight. The net force, again set to zero in Newton's second law 
because a - 0, is 



in 



SF> - F y - mg + 40.0N - 0, 
40.0 N - 9-8.0 N- 



40.0 N -58.0N. 



F N - mg 

The table does not push against the full weight of the box because of the 
upward pull exerted by your friend. 

NOTE The weight of the box (= mg) does not change as a Tesult of vout 
friend's push or pull. Only the normal force is affected. 



CAUTION 



The normal force. V\ . is not 

lift fi ,■,(":',■''. ri'rtirat 



Recall that the normal force is elastic in origin (the table in Fig. 4- 1 5 sags 
slightly under the weight of the box). The normal force in Example 4-6 is 
vertical, perpendicular to the horizontal table. The normal force is not always 
vertical, however, When you push against a vertical wall, for example, the 
normal force with which the wall pushes back on you is horizontal. For an 
object on a plane inclined at an angle to the horizontal, such as a skier or cai on 
a hill, the normal force acts perpendicular to the plane and so is not vertical. 



82 CHAPTER 4 Dynamics: Newton's Laws of Motion 



EXAMPLE 4-7 



Accelerating the box. What happens when a person pulls 
upward an the box in Example 4-6 (c) with a force equal to, or greater than, 
the box's weight, say F v - 100.0 N rather than the 40,0 N shown in Fig, 4- 15c? 

APPROACH We can start just as in Example 4-6, but be ready for a surprise. 
SOLUTION The nel force on the box is 

SFj. = F N - mg + F r 

= F N - 98.0 N + 100.0 N, 

and if we set this equal to zero (thinking the acceleration might be zero), we 
would get F H = -2.0 N, This is nonsense, since the negative sign implies F N 
points downward, and the table surely cannot putt down on the box (unless 
there's glue on the table). The least F s can be is zero, which it will be in this 
case. What really happens here is that the box accelerates upward because the 
net force is not zero. The net force (setting the normal force F N - 0) is 

XF,. = F P - mg = 100.0N - 98.0 N 
= ION 

upward. See Fig. 4-16. We apply Newton's second law and see that the box 
moves upward with an acceleration 



XF 



m 



2.0 N 
10.0 kg 
0.20 m/s z . 



F P (l(X).(IN) 




(98.0 N) 



FIGURE 4-16 Example 4-7. The 
box accelerates upward because 
Fp > mg. 



Additional Example 



EXAMPLE 4-8 



Apparent weight loss. A 65-kg woman descends in an 
elevator thai briefly accelerates at 0.20g downward when leaving a floor. 
She stands on a scale that Teads in kg. (a) During this acceleration, what is her 
weight and what does the scale read? (b) What does the scale read when the 
elevator descends at a constant speed of 2.0 m/s? 

APPROACH Figure 4- 1 7 shows all the forces thai act on the woman (and 
only those that act on her). The direction of the acceleration is downward, 
which we take as positive. 

SOLUTION (d) From Newton's second law, 

2F = ma 
mg -F N = m(0.20g). 

We solve for F x : 

F N = mg - 0.20ms = O.SOmj?, 

and it acts upward. The normal force f N is the force the scale exerts on the 
person, and is equal and opposite to the force she exerts on the scale: 
F' s - O.SOwg downward. Her weight (force of gravity on her) is still 
mg - (65 kg}(9,8 m/s 2 ) - 640 N, But the scale, needing to exert a force of 
only OMOmg, will give a reading of 0.80m = 52 kg. 

(ft) Now there is no acceleration, h = 0, so by Newton's second law, 
mg - F N = and F N = mg. The scale reads her lure mass of 65 kg. 

NOTE The scale in (a) may give a reading of 52 kg (as an "apparent 
mass"), but her mass doesn't change as a result of the acceleration: it stays 

at 65 kg. 



FIGURE 4-17 




SECTION 4-6 Weight— the Force of Gravity; and the Normal Force 83 




FIG U RE 4 - 1 8 [ a) ' I'wo forces. F A 

and Fu , exerted by workers A and 
B, act on a crate (b) The sum, or 
resultant, of F A and F|j is Fr. 



FIGURE 4-19 Example 4-9: Two 
force vectors act on a boat, 




la) 




PROBLEM SOLVING 

Free-body diagram 



Identifying every force 




Solving Problems with Newton's Laws: 
Free-Body Diagrams 

Newton's second law tells us thai the acceleration of an object is proportional 
to the net force acting on the object- The net force, us mentioned earlier, is the 
vector sum of all forces acting on the object. Indeed, extensive experiments 
have shown that forces do add together as vectors precisely according to the 
Tules we developed in Chapter 3. Fot example, in Fig. 4-18, two forces of equal 
magnitude (I00N each) are shown acting on an object at right angles to each 
other. Intuitively, we can see that the ohject will start moving at a 45° angle and 
thus the net force acts at a 45" angle. This is just what the rules of vector addi- 
tion give. F rom the theorem of P ythagoras, the magnitude of the resultant force 
is F K = V(100N) 2 + (ICON J 2 = 141 N. 



EXAMPLE 4-9 



Adding force vectors. Calculate the sum of the two 
forces exerted on the boat by workers A and B in Fig. 4- 19a. 

APPROACH We add force vectors like any other vectors as described in 
Chapter 3. The first step is to choose an xy coordinate system, as in Fig. 4- 19a, 
and then resolve vectors into their components. 

SOLUTION The two force vectors are shown resolved into components in 
Fig. 4-1%, We add the forces using the method of components. The compo- 
nents of F A are 

Fax ~ f A cas45.G u - (40,0N)(0.707) - 28.3 N, 
F A) . = ^sin45,0 u = (40,0 N) (0.707) = 28.3 N. 

The components of P Li are 

^kr = + F U cos 37.0° = + (30.0 N) (0,799) = +24.0 N, 



'":■ 



■ F„ sin 37.0° = - (30.0 N) (0.602} = - 18.1 N. 



Fuy is negative because it points along the negative v axis. The components of 
the resultant force arc (see Fig. 4- 19c) 

Fiu = F Ax + F Kx = 2S.3 N + 24.0 N = 52.3 N, 

FRy ~ F Ay + Fay = 28.3 N - 18.1 N - I0.2N. 

To find the magnitude of the resultant force, we use the Pythagorean theorem: 



Fr = V'FL + F\ y = V(S2.3) J + (10.2) 2 N = 53.3 N, 

The only remaining question is the angle f) that the net force F K makes with 
the x axis. We use: 



tan(? 



F Vl 



10.2 N 

52.3 N 



- 0.195, 



and tan '(0.195) = 11.0". The net force on the boat has magnitude 53.3 N 
and acts at an 1 1.0 (> angle to the x axis. 



When solving problems involving Newton's laws and force, it is very impor- 
tant to draw a diagram showing all the forces acting an each object involved. 
Such a diagram is called a free-body diagram, or force diagram: choose one 
ohject, and draw an arrow to represent each force acting an it. Include every 
force acting on that object. Do not show forces that the chosen object exerts on 
other objects. To help you identify each and every force that is exerted on your 
chosen object, ask yourself what other objects could exert a force on it. If your 
problem involves more than one object a separate free-body diagram is needed 
for each object. 



84 CHAPTER 4 Dynamics: Newton's Laws of Motion 



MmUm 



Mot km 



Motion 



(a) f Po 



J 



(b) 



L 




FIGURE 4-20 Example 4-10. 
Which is the Correct free-body 
diagram for a hockey puck sliding 

HLTuss frictionless ice? 



(<0 ' F rr 



CONCEPTUAL EXAMPLE 4-10 | The hockey puck. A hockey puck is 
sliding at constant velocity across a flat horizontal ice surface that is assumed 
to be frictionless. Which of the sketches in Fig, 4-20 is the correct free-body 
diagram for this puck? What would your answer be if the puck slowed down? 

RESPONSE Did you choose (a)? Tf so, can you answer the question; what 
exerts the horizontal force labeled Fqn the puck? Tf you say that if is the force 
needed to maintain the motion, ask yourself: what exerts this force? 
Remember that another object must exert any force — and there simply isn't 
any possibility here. Therefore, (a) is wrong. Besides, the force F in Fig. 4-20a 
would give rise to an acceleration by Newton's second law, It is (If) that is 
correct, as long as there is no friction. No net force acts on the puck, and the 
puck slides at constant velocity across the ice. 

In the real world, where even smooth lee exerts at least a tiny friction 
force, then (c) is the correct answer. The tiny friction force is in the direction 
opposite to the motion, and the puck's velocity decreases, even if very slowly. 



Here now is a brief summary of how to approach solving problems 
involving Newton's laws. 



PROBLEM SOLVING 



Newton's Laws; Free-Body Diagrams 



1. Draw a sketch of the situation, 

2. Consider only one object (at a time), and draw a 
free-body diagram for that object, showing all the 
forces acting on that object. Include any unknown 
forces that you have to solve for. Do not show any 
forces that the chosen object exerts on other objects. 
Draw the arrow for each force vector reasonably 
accurately for direction and magnitude. Label each 
force, including forces you must solve for, as to its 
source (gravity, person, friction, and so on). 

Tf several objects are involved, draw a free- 
body diagram foT each object separately, showing 
all the forces acting on thai object (and only forces 
acting on that object). For each (and every) foTce, 
you must be dear about: on what object that force 



acts, and by what object that force is exerted. Only 
forces acting on a given object can be included in 
2F" - ma for that object. 

Newton's second law involves vectors, and it is 
usually important to resolve vectors into compo- 
nents. Choose .v and y axes in a way that simplifies 
the calculation. For example, it often saves work if 
you choose one coordinate axis to be in the direc- 
tion of the acceleration. 

For each object, apply Newtim's second law to the x 
and v components separately. That is, the x compo- 
nent of the net force on that object is related to 
the x component of that object's acceleration: 
EF-t = ma x , and similarly for the y direction. 
Solve the equation or equations for the unknown(s), 



Force arrow placement 
ok diagrams 



This Problem Solving Box should not be considered a prescription. Rather it is 
a summary of things to do that will start you thinking and getting involved in 
the problem at hand. 

When we arc concerned only about translational motion, all the forces on a 
given object can be drawn as acting at the center of the object, thus treating the 
object as a point particle. 1 lowevcr, for prohlems involving rotation or statics, the 
place where each force acts is also important, as we shall see in Chapters S and 9. 

In the Examples that follow, we assume that all surfaces are very smooth so 
that friction can be ignored. (Friction, and Examples using it, are discussed in 
Section 4-8.) 

SECTION 4-7 Solving Problems with Newton's Laws: Free-Body Diagrams 85 




(a) 



i ^ 



froii 



7 



(b) 



wj 



(el 



FIGURE 4-21 (a) Pulling the bust. 
Example 4- 1 1 : (b) is the tree-body 
diagram for the hox, and (c) is the 
tree -body diagram considering all 
the forces to act at a point (transla- 
tional motion only, which is what we 
have here). 



PROBLEM SOLVING 

C ofids can pull has can't push: 
tension exists throughout a cord 



EXAMPLE 4-11 



Pulling the mystery box. Suppose a friend asks to 
examine the 10,0-kg box you were given (Example 4-6, Fig. 4-15), hoping to 
guess what is inside; and you respond, "Sure, pull the box over to you." She 
then pulls the box by the attached cord, as shown in Fig. 4-21 a, along the 
smooth surface of the table, Trie magnitude of the force exerted by the person 
is F P = 40.0 N, and it is exerted at a 30.0' angle as shown, Calculate (a) the 
acceleration of the box, and (b) the magnitude of the upward force F s exerted 
by the table on the box. Assume that friction can be neglected. 

APPROACH We follow the Problem Solving Box on the previous page. 
SOLUTION 

1. Draw a sketch: The situation is shown in Fig, 4-21 a; it shows the box and 
the force applied by the person, F v , 

2. Free-body diagram: Figure 4-2 lb shows the free-body diagram of the box, 
To draw it correctly, we show at! the forces acting on the box and only the 
forces acting on the box, They arc: the force of gravity mg; the normal force 
exerted by the table F x ; and the force exerted by the person F P . We are 
interested only in translational motion, so wc can show the three forces 
acting at a point. Fig, 4-2 1 c. 

3. Choose axes and resolve vectors: We expect the motion to be horizontal, so 
we choose the x axis horizontal and the y axis vertical, The pull of 400 N 
has components 

F P , = (40.0 N)(cos 30,0°) = (40.0 N) (0.866) = 34.6 N, 

F Ps , - (40.0N}(sin30.0 u ) - (40.0 N) (0.500) - 20.0 N. 

In the horizontal (x) direction, T^ N and ing have zero components. Thus the 
horizontal component of the net force is fp f , 

4. (a) Apply Newton's second law to determine the x component of the 
acceleration: 

Fpx ~ »Wx ■ 

5. (a) Solve: 

F Pr (34.6 N) 



a r = 



= 3.46 m/s 2 



(10.0 kg) 

The acceleration of the box is 3.46 m/s 2 to the right. 
(b) Next we want to find F N . 

4. (b) Apply Newton's second law to the vertical (v) direction, with upward as 
positive: 

2Fj. - ma v 
F N - mg + F,, } = ma y . 

5. (b) Solve: We have mg - (10.0 kg)(9.8Q m/s 2 ) - 98.0 N and, from point 3 
above, F Vy - 20.0 N. Furthermore, since F Vy < mg, the box does not move 
vertically, so a y - 0. Thus 

F^ - 98.0 N + 20.0 N = 0, 
so 

F^ = 78.0 N. 
NOTE F N is less than mg: the table does not push against the full weight of the 
box because part of the pull exerted by the person is in the upward direction, 



Tension in a Flexible Cord 

When a flexihle cord pulls on an object, the cord is said to he under tension, and 
the foree it exerts on the object is the tension F T . If the cord has negligible mass. 
the force exerted at une end is transmitted undiminished to each adjacent piece of 
cord along the entire length to the other end, Why? Because SF = ma = for 
the cord if the cord's mass m is zero (or negligible) no matter what a is. 1 lence the 
forces pulling on the cord at its two ends must add up to zero (F T and — F T ). Note 
that flexible cords and strings can only pull. They can't push because they bend. 



86 CHAPTER 4 Dynamics: Newton's Laws of Motion 



FIGURE 4-22 Example 4-12. (a) Two boxes, A and B. 
are connected hy a cord. A person pulls horizontally on 
box A with forte F P = 40.0 N. (b) Free-body diagram 
for box A, (c) Free-body diagram for box B. 



u 




BoxB 



Our next Example involves two boxes connected by a cord, We can refer to 
this group of objects as a system. A system is any group of one or more objects 
we choose to consider and study. 



EXAMPLE 4-12 



Two boxes connected by a cord. Two boxes, A and B, 
are connected hy a lightweight cord and are resting on a smooth (friction less) 
table. The boxes have masses of 1 2.0 kg and 10.0 kg. A horizontal force F P of 
40.0 N is applied to the 10.0-kg box, as shown in Fig. 4— 22a. Find («) the accel- 
eration of each box, and (b) the tension in the cord connecting the boxes. 

APPROACH We streamline our approach by not listing each step. We have two 
hoxes so we need to draw a free-body diagram for each box. To draw them 
correctly, we must consider the forces on each box by itself, so that Newton's 
second law can be applied to each. The person exerts a force F r on box A. 
Box A exerts a force F T on the connecting cord, and the cord exerts an opposite 
hut equal magnitude force F r back on box A (Newton's third law). These two 
horizontal forces on box A are shown in Fig, 4-22b, along with the force of 
gravity /n A g downward and the normal force F AN exerted upward by the table. 
The cord is light, so we neglect its mass. The tension at each end of the cord is 
thus the same. ] Icncc the cord exerts a force F T on the second box. Figure 4-22c 
shows the forces on box B, which are P, , m u g, and the normal force F UN . 
There will be only horizontal motion. We take the positive .v axis to the right. 
SOLUTION (a) We apply 2F, = ma x to box A: 

Sfj. - F r — F T - m A a A , |box A| 

For box B, the only horizontal force is F , , so 

EF A = F T = m B a B . [boxB] 

The boxes are connected, and if fhe cord remains taut and doesn't stretch, then 
the two boxes will have the same acceleration a, Thus a A = fl u = a- We are 
given m A - 10.0 kg and w B = 12.0 kg. We can add the two equations above 
to eliminate an unknown [F T ) and obtain 



Box A 



(m A + m u )o - Ff, - F, + F v — F r 



or 



u 



Fp 



m A + m B 
This is what we sought. 



40.0 N 
22.0 kg 



1 .82 m/s 1 . 



Alternate Solution We would have obtained the same result had we 
considered a single system, of mass m A + m lt ,, acted on hy a net horizontal 
force equal to F f . (The tension forces F, would then be considered internal to 
the system as a whole, and summed together would make zero contribution to 
the net force on the whole system.) 
(b) From the equation above for box B (F T - wj u a u ), the tension in the cord is 

Fj = m K n = (12.0 kg)( 1.82 m/s 2 ) = 21.8 N. 

Tnus, F r is less than />(= 40.0 N), as we expect, since F T acts to accelerate only /n u . 

NOTE ft might be templing to say that the force the person exerts, F P , acts not 
only on box A but also on box B. It doesn't. F v acts only on box A. Tt affects box B 
via the tension in the cord, F T . which acts on hox B and accelerates it. 




(b) 



KN 



Y in 



m Bfi 
(C) 



*+■ PROBLEM SOLVING 
An alternate analysis 



<k 



CAUTION 



For any object, use only 
the forces on that object in 
calculating T.F = ma 



SECTION 4-7 Solving Problems with Newton's Laws: Free-Body Diagrams 87 



PHYSICS APPLIED 
Elevator (asArwood's machine) 




'".£ . CeHfflerweighi 

I mc =lOO0ke 



. U - 
11 f 















'"fR 



w c g 



fb) 



(c) 



FIGURE 4-23 Example 4-11 
(a) At wood's machine in the form 
of an elevator-counterweight 
system, (b) and (c) Free-body 
diagrams for the two objects. 



*+ PROBLEM SOLVING 
Check your result 6j seeing if it 

U-tJik.\ in siliiitliuns tl'itcic ih< uastr-ir 
is easily guessed 



Additional Examples 

Here are some more worked-out Examples to give you practice in solving a 
wide range of Problems. 



EXAMPLE 4-13 



Elevator and counterweight (Atwood's machine!, 
A system of two objects suspended over a pulley by a flexible cable, as shown 
in Fig, 4- 23a, is sometimes referred to as an Atwood's machine. Consider the 
real-life application of an elevator (m E ) and its counterweight (m c ). To mini- 
mize the work done by the motor to raise and lower the elevator safely, w L 
and m c are similar in mass. We leave the motor out of the system for this 
calculation, and assume that the cables mass is negligible and that the mass of 
the pulley, as well as any friction, is small and ignorable. These assumptions 
ensure that the tension F T in the cable has the same magnitude on both sides 
of the pulley. Let the mass of the counterweight be m € - 1000 kg. Assume 
the mass of the empty elevatoT is 850 kg, and its mass when carrying four 
passengers is m E = 1150 kg, For the laller case (wi E = 1150 kg), calculate 

(a) the acceleration of the elevator and (b) the tension in the cable. 

APPROACH Again we have two objects, and we will need to apply Newton's 
second law to each of them separately. Faeh mass has two forces acting on it: 
gravity downward and the cable tension pulling upward, F, . Figures 4-23b 
and e show the free-body diagrams for the elevator (m F J and for the counter- 
weight (wi(J. The elevator, being the heavier, will accelerate downward, 
whereas the counterweight will accelerate upward. The magnitudes of their 
accelerations will be equal (we assume the cable doesn't stretch). For the 
counterweight, m c g = (1000 kg} (9,80 m/s : ) = 9800 N, so F r must be greater 
than 9800 N (in order that m c will accelerate upward), For the elevator, 
m E # = (ll50kg)(9.3Gm/s 2 ) = 1 1,300 N, which must have greater magnitude 
than F T so that m^ accelerates downward. Thus our calculation must give F r 
between 9800 N and 1 1,300 N, 

SOLUTION (a) To find F r as well as the acceleration a, we apply Newton's 
second law, SF = ma, to each object. We take upward as the positive v 
direction for both objects. With this choice of axes, a c = a because m c accel- 
erates upward, and « E = -a because m E accelerates downward. Thus 

F, - m^.g = «1f.(?k = — my a 
P\ ~ m c8 — mc a c ~ +tn c a. 
We can subtract the first equation from the second to get 

Wa ~ f»c)g ~ ('"I- + "'c)ii 

where a is now the only unknown. We solve this foT a: 

m v - m,- 1150 kg - 1000 kg 

a = — g = x = 0,070 a = 0.68 m/s 2 , 

m E + m € . 1150 kg + 1000 kg * * ' 

The elevator (m^) accelerates downward (and the counterweight m c upward) 
at a - 0.070^ -0.68 m/s 2 . 

(b) The tension in the cable F T can be obtained from eitheT of the two 
SF = ma equations, setting a = 0.070# = 0.68 m/V: 

F y = m E g - m E a = m E (g - a) 

= 1150 kg (9.80 m/s 2 - 0.68 m/s 2 ) = 10,500 N, 
F v = m c g + m c a = m c (g + a) 

= 1000 kg (9,80 m/s : + 0,68 m/s : ) = 10,500 N, 

which are consistent. As predicted, our result lies between 9800 N and 1 1 ,300 N. 
NOTE We can check our equation for the acceleration a in this Example by 
noting that if the masses were equal (m E = m c ), then our equation above for 
a would give a - 0, as we should expect. Also, if one of the masses is zero 
(say, m c = 0), then the other mass (m L ^ 0) would be predicted by our 
equation to accelerate at a — g, again as expected. 



S3 CHAPTER 4 Dynamics: Newton's Laws of Motion 



CONCEPTUAL EXAMPLE 4-14 | The advantage of a pulley. A mover is 



trying tu lift a piano (slowly) up to a second-story apartment (Fig, 4-24), He is 
using a rope looped over two pulleys as shown. What force must he exert on 
the rope to slowly lift the piano's 2000-N weight? 

RESPONSE The magnitude of the tension force F, within the rope is the 
same at any point along the rope if we assume we can ignore its mass. First 
notice the forces acting on the lower pulley at the piano. The weight of the 
piano pulls down on the pulley via a short cable. The tension in the rope, 
looped through this pulley, pulls up twice, once on each side of the pulley. Let 
us apply Newton's second law to the pulley-piano combination (of mass m): 

2F r - trig — ma. 

To move the piano with constant speed (set a - in this equation) thus 
requires a tension in the rope, and hence a pull on the rope, of F r = rng/2. 
The mover can exert a force equal to half the piano's weight, We say the pulley 
has given a mechanical advantage of 2, since without the pulley the mover 
would have to exert twice the force. 




FIGURE 4-24 Example 4-14, 



EXAMPLE 4-15 



Getting the car out of the mud. finding her car stuck 
in the mud, a bright graduate of a good physics course tics a strong rope to the 
back bumper of the car, and the other end to a boulder, as shown in 
Fig. 4-25a. She pushes at the midpoint of the rope with her maximum effort, 
which she estimates to be a force /> '■=> 300 N. The car just begins to budge 
with the rope at an angle (see the Figure), which she estimates to be 5°. With 
what force is the rope pulling on the car? Neglect the mass of the rope, 

APPROACH First, note that the tension in a rope is always along the rope. 
Any component perpendicular to the rope would cause the rope to bend or 
buckle (as it does here where F v acts) — in other words, a rope can support a 
tension force only along its length. Let F liH and F CK be the forces on the boulder 
and on the car, exerted via the tension in the rope, as shown in Fig. 4-25a. Let us 
choose to look at the forces on the tiny section of Tope where she pushes, The free- 
body diagram is shown in Fig. 4-25b, which shows F f as well as the tensions in the t 



rope (note that we have used Newton's third law: F, m = 



Fur — 



■Fcr) 



CM)- 

At the moment the car budges, the acceleration is still essentially zero, so a = 0. 
SOLUTION For the x component of 2? - ma - on that small section of 
rope (Fig, 4-25b), we have 

2,F t - F Kli cosf) — F KC cos& - 0. 

Hence F Rli = F K £ , and these forces represent the magnitude of the tension in the 
rope, call it F y i then we can write F r - F Kli - F RC . In the v direction, the forces 
acting are F?, and the components of F Kli and F KC that point in the negative y 
direction (each equal to F , sin 8). So for the y component of SP - ma, we have 



**, = 



i> - 2F, sin 9 = 0. 
We solve this for F r , and insert ft - 5" and F v 
F., 300 N 



300 N, which were given: 



F, 



1700N, 



2 sin 2 sin 5° 

When out physics graduate exerted a force of 300 N on the rope, the force 
produced on the car was 1700 N, She was able to magnify her effort almost six 
times using this technique! 

NOTE Notice the symmetry of the problem, which ensures that F RB = F RC . 
NOTE Compare Figs. 4-25a and b. Notice that we cannot write down 
Newton's second law using Fig. 4 -25 a because the force vectors are not acting 
on the same object, Tt is only by choosing a tiny section of rope as our object, 
and using Newton's third law (in this case, the boulder and the car pulling back 
on the rope with forces F R]i and F R( ), that all forces apply to the same object. 



How to gel out 
of the mud 



FIGURE 4-25 Example 4-15. 
(a) Getting a car out of the mud, 
showing the forces on the bouldeT, 

on the car, and exerted hy the 
person, (b) The free -body diagram: 
fortes on a small segment of rope. 




*» PROBLEM SOLVING 
Use any symmetry present to simplify 

a problem 



SECTION 4-7 Solving Problems with Newton's Laws: Free-Body Diagrams 89 




Problems Involving Friction, Inclines 



FIGURE 4-26 An object moving 
to the riglit on a table or floor. "The 
two surfaces in contact are rough, at 
least on a microscopic scale, 



K&tetit friction 



FIGURE 4-27 When an object 

is pulled by an applied force (F A ) 
along a surface, the force of friction 
Fly opposes the motion. The magni- 
tude- of Ff r is proportional to the 
magnitude of the normal foTce (Fv). 











! ^ 








M 






i% 


w 




J- 




H 




J\ 














»'g 







Friction 

Until row we have ignored friction, buL it must be taken into account in most 
practical situations, Friction exists between two solid surfaces because even the 
smoothest looking surface is quite rough on a microscopic scale, Fig. 4-26. When 
we try to slide an object across another surface, these microscopic bumps impede 
the motion. Exactly what is happening at the microscopic level is not yet fully 
understood. It is thought that the atoms on a bump of one surface may come so 
close to the atoms of the other surface that attractive electric forces between the 
atoms can i; bond" as a tiny weld between the two surfaces, Sliding an object 
across a surface is often jerky, perhaps due to the making and breaking of these 
bonds. Even when a round object rolls across a surface., there is still some fric- 
tion, called rolling friction, although it is generally much less than when an object 
slides across a surface. We focus now on sliding friction, which is usually called 
kiiieiii' friction {kinetic is from the Greek for "moving"). 

When an object slides along a tough surface, the force of kinetic friction acts 
opposite to the direction of the object's velocity. The magnitude of the force of 
kinetic friction depends on the nature of the two sliding surfaces. For given 
surfaces, experiment shows that the friction force is approximately proportional 
to the normal force between the two surfaces, which is the force that either 
object exerts on the other, perpendicular to their common surface of contact (see 
Fig. 4-27), The force of friction between hard surfaces in many cases depends 
very little on the total surface area of contact: that is, the friction force on this 
book is roughly the same whether it is being slid on its wide face or on its spine, 
assuming the surfaces have the same smoothness. We consider a simple model of 
friction in which we make this assumption that the friction force is independent 
of area. Then we write the proportionality between the friction force F r , and the 
normal force /\ as an equation by insetting a constant of proportionality, ju, k : 

This relation is not a fundamental law; it is an experimental relation between 
the magnitude of the friction force F tr , which acts parallel to the two surfaces, 
and the magnitude of the normal force F N , which acts perpendicular to the 
surfaces. It is not a vector equation since the two forces have directions perpen- 
dicular to one another. The term ,u. k is called the coefficient of kinetic friction, and 
its value depends on the nature of the two surfaces, Measured values for a 
variety of surfaces are given in Table 4-2. These arc only approximate, however, 
since il depends on whether the surfaces arc wet or dry, on how much they have 
been sanded or rubbed, if any burrs remain, and other such factors. But ^ k is 
roughly independent of the sliding speed, as well as the area in contact. 



90 CHAPTER 4 



TABLE 4-2 Coefficients of Friction 1 


Surfaces 


Coefficient uf 
Static l'rictiun./.i, 


Coefficient «r 
Kinetic friction. ;i i 


Wood on wood 


0.4 


0.2 


Ice on ice 


0.1 


003 


Metal on metal (lubricated) 


0.15 


0.07 


Steel on steel (untubrieated) 


0.7 


0,6 


Rubber on dry concrete 


1.0 


0.8 


Rubber on wet concrete 


0.7 


0.5 


Rubber on other solid surfaces 


1-4 


1 


Teflon* on teflon in air 


0.04 


0.04 


Teflon on steel in air 


0.04 


0.O4 


Lubricated ball bearings 


<0.01 


<0.0l 


Synovial joints (in human limbs) 


0.01 


0.01 


' Values arc approximate and intended only 


at a guide 





What we have been discussing up lo now is kinetic friction, when one object 
slides over ;motheT. There is also static friction, which refers to a force parallel 
to the two surfaces thai can arise even when they are not sliding. Suppose an 
object such as a desk is resting on a horizontal floor. Tf no horizontal force is 
exerted on the desk, there also is no friction force. But now suppose you try to 
push the desk, and it doesn't move. You are exerting a horizontal force, but the 
desk isn't moving, so there must be another force on the desk keeping it from 
moving (the net force is zero on an object that doesn't move). This is the force 
of static friction exened by the floor on the desk. If you push with a greater 
force without moving the desk, the force of static friction also has increased. Tf 
you push hard enough, the desk will eventually start to move, and kinetic fric- 
tion takes over. At this point, you have exceeded the maximum force of static 
friction, which is given by F (r (max) = /i s F N , where fa is the coefficient of static 
friction (Table 4-2). Since the force of static friction can vary from zero to this 
maximum value, we write 

F\ r. s ^ s ^ n ■ 
You may have noticed that it is often easier to keep a heavy object sliding 
than it is to start it sliding in the first place. This is consistent with ju^ generally 
being greater than jt k (see Table 4-2). 



EXAMPLE 4-16 



Friction: static and kinetic. Our 10.0-kg mystery box 
rests on a horizontal floor. The coefficient of static friction is ji^ = 0.40 and 
the coefficient of kinetic friction is ii k = 0,30, Determine the force of friction, 
F fr , acting on the box if a horizontal external applied force F A is exerted on it 
of magnitude: (a) 0, (h) 1 N, (c) 20 N, (d) 38 N, and (e) 40 N. 

APPROACH We don't know, right off, if we are dealing with static friction or 
kinetic friction, nor if the box remains at rest or accelerates. We need to draw 
a free-body diagram, and then determine in each case whether or not the box 
will move, by using Newton's second law. The forces on the box are gravity >ng, 
the normal force exerted by the floor P N , the horizontal applied force F A , and 
the friction force F B ., as shown in Fig, 4-27- 

SOLUTION The free-body diagram of the box is shown in Fig, 4-27. In the 
vertical direction there is no motion, so Newton's second law in the vertical 
direction gives SF, = ma y — 0, which tells us F N - mg - 0. Hence the 
normal force is 

F N - mg - (I0.0kg)(9.8m/s 2 ) - 9SN. 

(a) Since no external force F A is applied in this first case, the box doesn't 
move, and F ir - 0. 

(b) The force of static friction will oppose any applied force up lo a maximum of 

j^F N = (0.40)(98N) = 39 N. 
When the applied force is F A - ION, the box will not move. Since 
2f. - F A - Fir ~ 0, then F r , - ION. 

(c) An applied force of 20 N is also not sufficient to move the box. Thus 
F fr - 20 N to balance the applied force, 

(d) The applied force of 38 N is still not quite large enough to move the box; 
so the friction force has now increased to 38 N to keep the box at rest. 

(<?) A force of 40 N will start the box moving since it exceeds the maximum 
force of static friction, ii s F N = (0.40)(9SN) = 39 N. Instead of static friction, 
we now have kinetic friction, and its magnitude is 

F (r = ^Fn = (0.30)(98N) = 29 N 

now a net (horizontal) force on 



There is 

F = 40 N 



a 



the box 

29 N = UN, so the box will accelerate at a rate 
^F UN 
in 10 kg 



of magnitude 



1 . 1 m/s 2 



Static friction 




40 50 60 

Applied force, F A \ 



■■< f \ 



no 
"motion" 



-h 



- sliding - 



as long as the applied force is 40 N. Figure 4-28 shows a graph that summa 

rizes this Example, 

SECTION 4-8 



FIG U RE 4-28 E xam pie 4- 1 6 . 

Magnitude of the force of friction as 
a function of the external force 
applied to an object initially at rest. 
As the applied force is increased in 
magnitude, Ihe force of static fric- 
tion increases linearly to just match 
it. until the applied force equals 
/i s F K . If the applied force increases 
further, the object will begin lo 
move, and the friction force drops 
to a roughly constant value charac- 
teristic of kinetic friction. 



Friction, Inclines 91 




FIGURE 4-29 Example 4-17. 



FIGURE 4-30 Example 4-1 8. 



FIGURE 4-31 Example 4-19, 




CONCEPTUAL EXAMPLE 4-17 | A box against a wall. You can hold a 



box against a rough wall (Fig. 4-29) and prevent it from slipping down by 
pressing hard horizontally, How docs the application of a horizontal force 

keep an ohject from moving vertically? 

RESPONSE This won't wotIc well if the wall is slippery. You need friction. 
Even then, if you don't press hard enough, the box will slip. The horizontal 
force you apply produces a normal force on the box exerted by the wall. The 
force of gravity mg, acting downward on the box, can now be balanced by an 
upward friction force whose magnitude is proportional to the normal force. 
The harder you push, the greater F N is and the greater F h can be. If you don't 
press hard enough, then mg > ^ F N and the box begins to slide down. 



Additional Examples 

Here are some more worked-out Examples that can help you for solving Problems 



CONCEPTUAL EXAMPLE 4-18 | To push or to pull a sled? Your little 



sister wants a ride on her sled. If you are on flat ground, will you exert less 
force if you push her or pull her? See Figs. 4-30a and b. Assume the same 
angle B in each case. 

RESPONSE Let us draw fTee-body diagrams for the sled-sisteT combination, 
as shown in Figs. 4-30c and d. They show, for the two cases, the forces exerted 
by you, F (an unknown), by the snow, F v and F fl , and gravity mg. (a) If you 
push her, and > 0, there is a vertically downward component to your force. 
Hence the normal force upward exerted by the ground (Fig. 4-30c) will be 
larger than mg (where m is the mass of sister plus sled), (ft) If you pull her. yoUT 
force has a vertically upward component, so the normal force F N will be less 
than mg, Fig. 4-30d. Because the friction force is proportional to the normal 
force, F[j will be less if you pull her. So you exert less force if you pull her. 






EXAMPLE 4-19 



Pulling against friction, A 10,0-kg box is pulled along a 
horizontal surface by a force F P of 40.0 N applied at a 30XT angle. This is like 
Example 4- 1 1 except now there is friction, and we assume a coefficient of 
kinetic friction of 0.30. Calculate the acceleration. 

APPROACH The free-body diagram is like that in Fig. 4-21, but with one 
more force, that of friction; see Fig. 4-3 1 . 

SOLUTION The calculation for the vertical (y) direction is just the same 
as in Example 4-11, where we saw that F Pv = 20.0 N, F Vx = 34.6 N, and the 
normal force is F s = 78,0 N. Now we apply Newton's second law for the 
horizontal (v) direction (positive to the right), and include the friction force: 

The friction force is kinetic as long as F : , = ^F M is less than F rx (= 34.6 N}, 

which it is: 

F fr = MkF N = (0.30)(78.0N) = 23.4 N. 



92 CHAPTER 4 Dynamics: Newton's Laws of Motion 



Hence Ihe box doe* accelerate; 

F Vx - F lr 34,6 N 



23.4 N 



flr = 



I.I m/s 2 



n 10.0 kg 

In the absence of friction, as we saw in Example 4-11, the acceleration would 
he much greater than this. 

NOTE Our final answer has only two significant figures because our least 
significant input value (fi k = 0.30) has two. 



| EXERCISE B If jttj, F N were greater than fj, v , what would you conclude? 



EXAMPLE 4-20 



Two boxes and a pulley. In Fig. 4 -32a, two boxes are 
connected by a coTd running over a pulley, The coefficient of kinetic friction 
between box A and the table is 0.20. We ignore the mass of the cord and pulley and 
any friction in the pulley, which means we can assume that a force applied to one 
end of the cord will have the same magnitude at the other end. We wish to find the 
acceleration, a, of the system, which will have the same magnitude for both boxes 
assuming the cord doesn't stretch. As box R moves down, box A moves to the right. 

APPROACH We need a free -body diagram for each box. Figs. 4 -32b and c, so 
we can apply Newton's second law to each. The forces on box A aTe the pulling 
force of the cord F T , gravity m A g, the normal force exerted by the table F s , 
and a friction force exerted by the table F h \ the forces on box B are gravity 
wug, and the cord pulling up, F T . 

SOLUTION Box A does not move vertically, so Newton's second law tells us 
the normal force just balances the weight, 

F N = m A g = (5.0kg)(9.8m/s 2 ) = 49 N. 

In the horizontal direction, there are two forces on box A (Fig, 4-32b); Fj, the 
tension in the coTd (whose value we don't know), and the force of friction 

F k -^F N - (0.20}(49N)-9.8N. 

The horizontal acceleration is what wc wish to find; we use Newton's second 
law in the x direction, 2F A _< - nt^a^, which becomes (taking the positive 
direction to the right and setting a Aj . = a): 

EF A < = F, - Ff, = m A a. [box A] 

Next consider box B. The force of gravity m^g = (2.0 kg)(9.8m/s 2 ) = 19.6N 
pulls downward; and the cord pulls upward with a force F T . So we can write 
Newton's second law for box B (taking the downward direction as positive): 

2F Ut , - m a g — F, _ >n u a. |box B| 

[Notice that if a =t 0, then F T is not equal to tn^g.] 

We have two unknowns, a and F, , and wc also have two equations, We 
solve the box A equation for F r : 

F f = Ff r + >n A ti, 
and substitute this into the box R equation: 

m B g — F [r — m A a - m^a. 
Now wc solve for a and put in numerical values: 



a = 



mug -Fir I9.6N-9.BN 



= 1.4 m/s 2 , 



m A + m B 5.0 kg + 2.0 kg 

which is the acceleration of box A to the right, and of box B down. 
If we wish, we can calculate F, using the first equation: 

F f = F h + m h a = 9.8 N + (5.Qkg}(l.4m/s 2 ) = 17 N. 

NOTE Box B is not in free fall, ft does not fall at a = g because an addi- 
tional force, F T , is acting upward on it. 



FIG U RE 4-32 E xam pie 4 - 2U . 

5-0 kg 



(a) 

tv 



2.0 kg 



■ 



«a8 



(O 



<j> CAUTION 



"'b£ 



Tension m a cord supporting a jailing 
object may not equal object's weight 



SECTION 4-8 Problems Involving Friction, Inclines 93 



<c> 




FIGURE 4-33 Forces on an object 
sliding down an incline 



F<j-Bll 



\ 



PROBLEM SOLVING 

Goodchoice of coordinate system 
simplifies the calculation 



Inclines 

Now we consider what happens when an object slides down an incline, such 
as a hill or ramp. Such problems are interesting because gravity is the accel- 
erating force, yet the acceleration is not vertical. Solving problems is usually 
easier if we choose the xy coordinate system so the x axis points along the 
incline and the y axis is perpendicular to the incline, as shown in Fig. 4-33. 
Note also that the normal force is not vertical, but is perpendicular to the 
sloping surface of the plane in Fig. 4-33. 



EXERCISE C Is the gravitational force always perpendicular to an inclined plane? Is it 
always vertical? 



EXERCISE Is the nonnal force always perpendicular to an inclined plane? Is it 
always vertical? 



PHYSICS APPLIED 



S« ' 



EXAMPLE 4-21 



The skier. The skier in Fig, 4-34 has just begun 
descending the 30" slope. Assuming the coefficient of kinetic friction is 0.10, 
calculate (a) her acceleration and (6) the speed she will teach after 4.0 s. 

APPROACH We choose the x axis along the slope, positive pointing down- 
slope in the direction of the skier's motion. The y axis is perpendicular to 
the surface as shown. The forces acting on the skier are gravity, F = mg, which 
points vertically downward (not perpendicular to the slope), and the two 
forees exerted on her skis by the snow — the normal force perpendicular to the 
snowy slope (not vertical), and the friction force parallel to the surface. These 
three forces are shown acting at one point in Fig. 4-34h, for convenience, and 
is our free-body diagram for the skier. 






(a) 



FIGURE 4-34 Example 4-21 . A skier descending a slope; F^ = wig is the force 

of gravity (weight) on the ikier. 



Id 



94 CHAPTER 4 Dynamics: Newton's Laws of Motion 



SOLUTION We have to resolve only one vector inlo components, the weight P G , 
and its components are shown as dashed lines in Fig. 4-34c, To be general, we 
use t) rather than 30 J for now. We use the definitions of sine ("side opposite") 
and cosine ("side adjacent") to obtain the components; 

Fcx = m g sin e - 
F G} . - -mg cos Q. 

where F Cy is in the negative v direction. 

(a) To calculate the skier's acceleration down the hill, rt r , we apply Newton's 

second law to the x directum: 

2F t - ma x 

mg sin 8 — ju, k F]m - ma x 

where the two forces are the x component of the gravity force ( + x direction) 
and the friction force (~x direction). We want to find the value of u x , but we 
don't yet know F N in the last equation. Let's see if we can get /\ from the 
v component of Newton's second law: 



1,F V - ma v 

F N - mg cos 9 - ma K . 



I) 



where we set a y - because there is no motion in the v directum (perpen- 
dicular to the slope). Thus we can solve for F N : 

F N - mg cos 

and we can substitute this into our equation above for ma x : 

tng sin B — ^(nig cos 6) - ma x . 

There is an m in each term which can be canceled out. Thus (setting — 30^ 
and ju k - 0.10): 

a x = g sin 30° - tit g cos 30" 

= 0.50g - (O.IO)(0.866)g = 0.41g. 

The skier's acceleration is 0.41 times the acceleration of gravity, which in 

numbers is a - (0.4l)(9.8m/s 2 ) - 4.0 m/s 2 . It is interesting that the mass 

canceled out here, and so we have the useful conclusion that the acceleration 

doesn't depend on the mass. That such a cancellation sometimes occurs, and *» problem solving 

thus may give a useful conclusion as well as saving calculation, is a big It i: 

advantage of working with the algebraic equations and putting in the only ctt rite end 

numbers only at the end. 

(b) The speed after 4.0s is found, since the acceleration is constant, by using 

Eq.2-lla; 

is = v,V, + at 

= + (4.0m/s 2 )(4.0s} = I6m/s, 

where we assumed a start from rest. 



Tn problems involving a slope or 'inclined plane," it is common to make an 
error in the direction of the normal force or in the direction of gravity. The 
normal force is not vertical in Example 4-21. It is perpendicular to the slope <ir 
plane. And gravity is not perpendicular to the slope or plane — gravity acts verti- 
cally downward toward the center of the Earth. 



<S>_ 



CAUTION 



Diret tions of gravity and the normal 

forca 



SECTION 4-8 Problems Involving Friction, Inclines 95 



Problem Solving— A General Approach 

A basic part of a physics course is solving problems effectively. The approach 
discussed here, though emphasizing Newton's laws, can be applied generally for 
other topics discussed throughout this boot. 



PROBLEM SOLVING 



In General 



L Read and reread written problems carefully. A 
common error is to skip a word or two when 
reading, which can completely change the meaning 
of a problem. 

2. Draw an accurate picture or diagram of the situa- 
tion, (This is probably the most overlooked, yet 
most crucial, part of solving a problem. ) Use arrows 
to represent vectors such as velocity or force, and 
label the vectors with appropriate symbols. When 
dealing with forces and applying Newton's laws, 
make sure to include all forces on a given object, 
including unknown ones, and make clear what 
forces act on what object (otherwise you may make 
an error in determining the net force on a particular 
object), A separate free-body diagram needs to be 
drawn for each object involved, and it must show 
all the forces acting on a given object (and only on 
that object). Do not show forces that act on other 
objects, 

3. Choose a convenient xy coordinate system (one 
that makes vout calculations easier, such as one 
axis in the direction of the acceleration). Vectors 
are to be resolved into components along the coor- 
dinate axes, When using Newton's second law, apply 
SF = ma separately to x and y components, 
remembering that x direction forces are related to 
a x , and similarly for y. ff more than one object is 
involved, you can choose different (convenient) 
coordinate systems for each. 

4. List the knowns and the unknowns (what you are 
trying to determine), and decide what you need 
in order to find the unknowns, For problems in 
the present Chapter, we use Newton's laws. 
More generally, it may help to see if one or more 
relationships (or equations) relate the unknowns 
ta the knowns. But be sure each relationship is 
applicable in the given case. Tt is very important 



to know the limitations of each formula or 
relationship — when it is valid and when not. In 
this book, the more general equations have been 
given numbers, but even these can have a limited 
range of validity (often stated in brackets to the 
right of the equation). 

5. Try to solve the problem approximately, to sec if it 
is doable (to check if enough information has been 
given) and reasonable, Use your intuition, and 
make rough calculations — sec "Order of Magnitude 
Estimating" in Section 1-7. A rough calculation, or 
a reasonable guess about what the range of final 
answers might be, is very useful, And a rough calcu- 
lation can be checked against the final answer to 
catch errors in calculation, such as in a decimal 
point or the powers of 10. 

fi. Solve the. prohlem, which may include algebraic 
manipulation of equations and/or numerical calcu- 
lations. Recall the mathematical rule that you need 
as many independent equations as you have 
unknowns; if you have three unknowns, for 
example, then you need three independent equa- 
tions. It is usually best to work out the algebra 
symbolically before putting in the numbers. Why? 
Because (a) you can then solve a whole class of 
similar problems with different numerical values; 
(b) you can check your result for cases already 
understood (say, 6 - D or 90") ; (c) there may be 
cancellations or other simplifications; (d) there is 
usually less chance for numerical error; and (e) you 
may gain better insight into the problem. 

7. Be sure to keep track of units, for they can serve as 
a check (they must balance on both sides of any 
equation). 

8. Again consider if your answer is reasonable. The use 
of dimensional analysis, described in Section 1-8, 
can also serve as a check for many problems. 



| Summary 



Newton's three laws of motion art the basic classical laws 
describing motion, 

Newton's first law [the law of inertia) states that if the 
net force on an object is zero, an object originally at rest 
remains at rest, and an object in motion remains in motion in 
a straight line with constant velocity. 



Newknt's second law states that the acceleration of an 
object is directly proportional lo the net force acting on it. 
and inversely proportional to its mass! 

£F = ma. (4-1) 

Newton's second law is one of the most important and funda- 
mental laws in classical physics. 



96 CHAPTER 4 Dynamics: Newton's Laws of Motion 



Newton's third law states that whenever one object 
exerts a force on a second object, the second object always 
exerts a force on the first object which is equal in magnitude 
but opposite in direction: 

Fab = -Sua <4-2> 

where ¥$ A is the force on object B exerted by object A, 

The tendency of an object to resist a change in its motion 
is called inertia. Mass is a measure of the inertia of an object. 
Weight refers to the gravitational force on an object, and 
is equal to the product of the object's mass m and the acceler- 
ation of gravity g: 

F c , = fflg. (4-3) 

Force, which is a vector, can be considered as a push or 
pull: or. from Newton's second law, force can he defined as an 



action capable of giving rise to acceleration. The net forte on 
an object is the vector sum of all forces acting on it. 

When two objects slide over one another, the force of 
friction that each object exerts on the olheT can be written 
approximately as 5j- r = ju.^ F K , where F^ is the normal force 
(the force each object exerts on the other perpendicular to 
their contact surfaces), and ^ is the coefficient of kinetic 
friction, [f the objects are at rest relative to each other, then 
Ff r is just laTge enough to hold them at rest and satisfies the 
inequality F( f < fi s F N , where /i s is the coefficient of static 
friction. 

For solving problems involving the forces on one or moTe 
objects, it is essential to draw a free-body diagram for each 
object, showing all the forces acting on only that object. 
Newton's second law can be applied to the vector compo- 
nents for each object. 



[ Questions 



Why does a child in a wagon seem to fall backward when 
you give the wagon a sharp pull forward? 

A box rests on the (friction less) bed of a truck. The truck- 
driver starts the truck and accelerates forward. The box 
immediately starts to slide toward the rear of the truck- 
bed. Discuss the motion of the box, in terms of Newton's 
laws, as seen (a) by Mary standing on the ground beside 
the truck, and (b) bv Chris who is riding on the truck 
(Fig. 4-35). 




FIGURE 4-35 Quest ion 2. 

3. If the acceleration of an object is zero, are no forces 
acting on it? Explain, 

4. Only one force acts on an object. Can the object have 
zero acceleration? Can it have zero velocity'? Explain, 

5. When a golf ball is dropped to the pavement.it bounces 
back up. (a) Is a force needed to make it bounce back up? 
(/>) If so, what exerts the force? 

6. If you walk along a log floating on a lake, why does the log 
move in the opposite direction? 

7. Why might your foot hurl if you kick a heavy desk ot a 
wall? 

8. When you are running and want to stop quickly, you must 
decelerate quickly, (a) What is the origin of the force that 
causes you to Stop? (b) Estimate (using your own experi- 
ence) the maximum Tate of deceleration of a person 
running at top speed to come to rest. 



A stone hangs by a fine thread from the ceiling, and a 
section of the same thread dangles from the bottom of 
the stone (Fig. 4-36), If a person gives a sharp pull on the 
dangling thread, where is the thread likely to break: 
below the stone or above it? What if the person gives a 
slow and steady pull? Explain your answers. 



i 



FIGURE 4-36 Question <J. 



LI) 



II 



The force of gravity on a 2-kg rock is twice as great as 
that on a I -kg rock. Why then doesn't the heavier rock 
fall fasteT? 

Would a spring scale carried to the Moon give accurate 
results if the scale had been calibrated (a) in pounds, or 
(A) in kilograms? 

12. You pull a hox with a constant force across a frictionless 
table using an attached rope held horizontally. If you now 
pull the rope with the same force at an angle to the hori- 
zontal (with the box remaining flat on the table), does the 
acceleration of the box (a) remain the same, (b) increase, 
or (c) decrease? Explain. 

13, When an object falls freely under the influence of gravity 
there is a net force nig exerted on it by the Earth. Yet by 
Newton's third law the object exerts an equal and oppo- 
site force on the Earth Why doesn't the Earth move? 
Compare the effort (or force) needed to lift a lt)-kg object 
when you are on the Moon with the force needed to lift it 
on Earth. Compare the force needed to throw a 2-kg object 
horizontally with a given speed on the Moon and on Earth. 



14 



Questions 97 



IS, According to Newton's third law, each team in a tug of 
war (Fig. 4-37) pulls with equal foTce on the other team, 
What, then, determines which team will win? 




FIGURE 4-37 Question 15. A tug of war. Describe 
the forces on each of the teams, and on the rope. 

16. A person exerts an upward force of 40 N to hold a hag of 
groceries. Describe the "reaction" force (Newton's third 
law) by staling {a) its magnitude, (b) its direction, (c) on 
what object it is exerted, and (d) by what object it is exerted. 



17. When you Stand Still on the ground, how large a force 
does the ground exert on you? Why doesn't this force 

make you rise up into the air? 

18. Whiplash sometimes results from an automobile accident 
when the victim's car is struck violently from the rear. 
Explain why the head of the victim seems to be thrown 
backward in this situation. Is it really? 

19. A heavy crate rests on the bed of a flatbed truck. When 
the truck accelerates, the crate remains wheTe it is on the 
truck, so it, too. accelerates. What force causes the crate to 
accelerate? 

20. A block is given a push so that it slides up a ramp. After 
the block reaches its highest point., it slides back down 
but the magnitude of its acceleration is less on the 
descent than on the ascent. Why? 

21. What would your bathroom scale read if you weighed 
yourself on an inclined plane? Assume the mechanism 
functions properly, even at an angle. 



Problems 



4-4 to 4-6 Newton's Laws, Gravitational Force, 
Mormat Fores 

1. (I) What force is needed to accelerate a child on a sled 

(total mass = 60.0 kg) at 1 .25 m/s 2 ? 

2. (I) A net force of 265 N accelerates a bike and rider at 
2.30 m/s 2 . What is the mass of the bike and rider 
together? 

3. (I) How much tension must a rope withstand if it is used 
to accelerate a 960- kg car horizontally along a frict ion less 
surface at 1 .20 m/s 2 ? 

4. (I) What is the weight of a 76-kg astronaut (a) on Earth. 
(b) on the Moon (g = 1 .7 m/s 2 ), (c) on Mars {g = 3-7 m/s 2 ), 
(d) in outer space traveling with constant velocity? 

5. (II) A 20.0-kg box rests on a table, {a) What is the weight 
of the box and the normal foTce acting on it? (b) A 
10,0-kg box is placed on top of the 20.0-kg box, as shown 
in Fig. 4-38. Determine the normal force that the tabic 
exerts on the 20.0-kg box and the normal force that the 
20.0-kg box exerts on the 10.0-kg box. 




FIGURE 4 -3 a Problem! 

6. (II) What average force is required to stop an 1100-kgcar 
in 8.0 s if the car is traveling at 95 km/h? 



7. (II) What average force is needed to accelerate a 
7-00-gram pellet from rest to 125 m/s ovct a distance of 
0,80(1 m along the barrel of a rifle? 

8, (II) A fisherman yanks a fish vertically out of the waleT 
with an acceleration of 2,5 m/s" using very light fishing 
line that has a breaking strength of 22 N. The fisherman 
unfortunately loses the fish as the line snaps, What can 
you say alxint the mass of the fish? 

■X (tl) A 0.140-kg baseball traveling 35.0 m/s strikes the 
catcher's mitt, which, in bringing the ball to rest, recoils 
backward 11.0 em, What was the average force applied by 
the ball on the glove? 
111. (II) How much tension must a rope withstand if it is used 
to accelerate a 1200-kg ear vertically upward al 0.80 m/s 2 ? 

11. (II) A particular race car can cover a quarter-mile track 
(402 m) in 6.40 s starting from a standstill. Assuming the 
acceleration is constant, how many "gV does the driveT 
experience? If the combined mass of the driver and race 
car is 485 kg, what horizontal force must the road exert on 
the tires? 

12. til) A 12.0-kg bucket is lowered vertically by a rope in 
which there is 163 N of tension at a given instant, What is 
the acceleration of the bucket? Is it up or down? 

13. (II) An elevator (mass 4850kg) is to be designed so that 
the maxim um acceleration is 0,0680^. What are the 
maximum and minimum forces the motor should exert on 
the supporting cable? 

14. (II) A 75-kg petty thief wants to escape from a third-story 
jail window, Unfortunately, a makeshift rope made of 
sheets tied together can support a mass of only 58 kg. 
How might the thief use this "rope" to escape? Give a 
quantitative answer. 

15. (II) A person stands on a hathroom scale in a motionless 
elevator. When the elevator begins to move, the scale 
briefly reads only 0.75 of the person's regular weight, 
Calculate the acceleration of the elevator, and find the 
direction of acceleration. 



98 CHAPTER 4 Dynamics: Newton's Laws of Motion 



If,. 



17. 



(II) The cable supporting a 2125-kg elevator has a 
maximum strength of 21 ,750 N. What maximum upward 
acceleration can it give the elevator without breaking? 
(II) (a) What is the acceleration of two falling sky divers 
(mass 132 kg including parachute) when the upward 
force of air resistance is equal to one-fourth of their 
weight? (h) After popping open the parachute, the divers 
descend leisurely to the ground at constant speed. What 
now is the force of air resistance on the sky divers and 
their parachute? See Fig. 4-39. 




FIGURE 4-39 Problem 17. 

IH. (Ill) A person jumps from the roof of a house 3-9-m high. 
When he strikes the ground below, he bends his knees so 
that his torso decelerates over an approximate distance of 
0.70m. If the mass of his torso (excluding legs) is 42kg. 
find (a) his velocity just before his feet strike the ground. 
and (b) the average force exerted on his torso by his legs 
during deceleration. 

4-7 Newton's Laws and Vectors 

19, (T) A box weighing 77.0 N rests on a table. A rope tied to 
the bos runs vertically upward 
over a pulley and a weight is 
hung from the other end 
(Fig. 4-40). Determine the 
force that the table exerts on 
the box if the weight hanging 
on the other side of the pullev 
weighs (a) 30.0 N, (t>) 60.0 N. 
and (c) 90.0 N. 




FIGURE 4-40 
Problem 19. 



2(1. (1) Draw the free-body diagram for a basketball player 
(a) just before leaving the 
ground on a jump, and 
(6) while in the air. See 
Fig. 4-41 . 



FIGURE 4-41 

Problem 20. 




21. (I) Sketch the free-body diagram of a baseball (a) at the 
moment it is hit by the bat, and again (b) after it has left 
the bat and is flying toward the outfield. 

22. (I) A 650-N foTce acts in a northwesterly direction, A 
second 650-N force must l>e exerted in what direction so 
that the resultant of the two fortes points westward? 
Illustrate your answer with a vector diagram. 

23. (11) Arlene is to walk across a "high wire 1 ' strung horizontally 
between two buildings 1 0.0 in apart. The sag in the rope when 
she is at the midpoint is 10.0 : . as shown in Fig. 4-42. If her 
mass is 50,0 kg. what is the tension in the rope at this point? 




FIGURE 4-42 Problem 23. 

24. (II) The two forces Fj and F 2 shown in Fig. 4 -43 a and b 
(looking down) act on a 27.0-kg object on a fnclionless 
tabletop. If Fj = 10.2 N and F 2 = 16.0 N, find the net 
force on the object and its acceleration for (a) and (b). 



Vil 




(a) 
FIGURE 4-43 Problem 24. 



(b) 



25. 



(II) One 3.2-kg paint bucket is hanging by a massless eord 
from another 3.2-kg paint 
bucket, also hanging by a mass- 
less coTd, as shown in Fig. 4-44, 
(tf) If the buckets are at rest. 
what is the tension in each cord? 
(b) If the two buckets are pulled 
upward with an acceleration of 



1,60 m/r 
calculate 
cord- 



by the upper cord, 
the tension in each 




FIGURE 4-44 
Problem 25. 



Problems 99 



26. (IT) A person pushes a l4.Q-kg lawn mower at constant 
speed with a force of F = 88,0 N directed along the handle, 
which is at an angle of 45.0 C to the horizontal (Fig. 4—45). 
(a) Dmw the tree-bod}' diagram showing all fortes acting 
on the mower. Calculate (6) the horizontal friction 
force on the mower, then (c) the normal force everted 
vertically upward on the mower by the ground, (d) What 
force must the person exert on the lawn mower to accel- 
erate it from rest to 1 ,5 m/s in 2.5 seconds, assuming the 
same friction force? 



2M. 




FIGURE 4-45 Problem 20. 

27. (II) Two snowcats tow a housing unit to a new location at 
McMurdo Base. Antarctica, as shown in Fig. 4-46. The sum 
of the forces F A and F B exerted on the unit by the hori- 
zontal cables is parallel to the line L. and F A = 4500 N, 
Determine F u and the magnitude of F A + F u . 




Top view 



FIGURE 4-46 
Problem 27. 



2& 



(II) A train locomotive is pulling two cars of the same 
mass behind it. Fig, 4-47. Determine the ratio of the 
tension in the coupling between the locomotive and the 
first car (F||), to that between the first car and the second 
car {Fj 2 ), for any nonzero acceleration of the train. 

Cur 2 p_ Car l 



M\. 



31 



(II) A window washer pulls herself Upward using the bucket- 
pulley apparatus shown in Fig. 4-48. {a) How hard must she 
pull downward to raise herself slowly at constant speed? 
(b) II she increases this force by 15%, what will her accelera- 
tion be? The mass of the person plus the bucket is 65 kg. 




FIGURE 4-48 
Problem 29. 

(II) At the instant a race began, a 65-kg sprinter exerted a 
force of 720 N on the starting block at a 22° angle with 
respect to the ground, (a) What was the horizontal acceler- 
ation of the sprinter? (b) It' the force was exerted lor 0,32 s, 
with what speed did the sprinter leave the starting block' 
(II) Figure 4-49 shows a block (massm A ) on a smooth 
horizontal surface, connected by a thin cord that passes 
over a pulley to a second block (mj). which hangs verti- 
cally, (a) Draw a free -body diagram for each block, 
showing the force of gravity on each, the force (tension) 
exerted by the cord, and any normal force, (b) Apply 
Newton's second law to find formulas for the acceleration 
ol the system and foT the tension in the cord- Ignore fric- 
tion and the masses of the pulley and cord. 






m n 



FIGURE 4-49 
Problem 31. Mass m A 

rests on a smooth 
horizontal surface, 
m K hangs vertically. 



(II) A pair ol fuzzy dice is hanging by a siring from j'out 
rearview mirror. While you are accelerating from a stop- 
light to 28 m/s in 6.0 s, what angle H does the string make 
with the vertical? See Fig. 4-50, 



f"IGURE4-47 I'roblciTi 2«. 




FIGURE 4-50 
Problem 32. 



100 CHAPTER 4 Dynamics: Newton's Laws of Motion 



33. (Ill) Three blocks on a friction less horizontal surface are 
in contact with each other, as shown in Fig. 4-51. A force F 
is applied to Mock A (mass tn A ). (a) Draw a free-body 
diagram tor each block, Determine (b) the acceleration of 
the system (in terms of m A , m u , and m f ), (c) the net force 
on each block, and (rf) the force of contact that each block 
exerts on its neighbor, (c) If m A — m^ = ffif = 12.0 kg and 
F = Qfi.ON, give numerical answers to (b), (c), and (d). 
Do your answers make sense intuitively? 




FIGURE 4-51 Problem 33. 



34. ( 



III) The two masses shown in Fig, 4-52 are each initially 
1.80m above the ground, and the massless frictioniess 
pulley is 4.8 m above the ground. What maximum height 
docs the lighter object reach after the system is released? 
[Hint: First determine the acceleration of the lighter mass 
and then its velocity at the moment the heavier one hits 
the ground, This is its "launch" speed. Assume it doesn't 
hit the pulley,] 







W 










4> 


ii 


1 : 

1.80 m 


..: 


kj 


> 


... 

u 


1 

kg 




1 




FIGURE 


4 


_l 


)2 




Pruble 


1134 



35. (Ill) Suppose two boxes on a frictioniess table are 
connected by a heavy cord of mass 1,0 kg, Calculate the 
acceleration of each Ixix and the tension at each end of 

the cord, using the free-body diagrams shown in Fig. 4-53. 
Assume F P = 40,0 N. and ignore sagging of the cord. 
Compare your results to Example 4-12 and Fig. 4-22. 



4-8 Newton's Laws with Friction; Inclines 

36. (I) If the coefficient of kinetic friction between a 35-kg 
crate and the floor is 0,30. what horizontal force is 
required to move the crate at a steady speed across the 
floor? What horizontal force is required if ^t is zero? 

37. (I) A force of 48,0 N is required to start a 5.0-kg box 
moving across a horizontal concrete floor, (a) What is the 
coefficient of static friction between the box and the floor? 
(b) If the 48.0-N force continues, the box accelerates at 
0.70 m/s 2 . What is the coefficient of kinetic friction? 

38. (I) Suppose that you are standing on a train accelerating 
at 0.20g. What minimum coefficient of static friction must 
exist between your feet and the floor if you are not lo 
slide? 

3'). (I) What is the maximum acceleration a car can undergo 
if the coefficient of sialic friction between the tires and 
the ground is 0.80? 

411. (II) The coefficient of static friction between hard rubber 
and normal street pavement is about 0,8. On how steep a 
hill (maximum angle) can you leave a car parked? 

41. (II) A 15.0-kg box is released on a 32' ; incline and accel- 
erates down the incline at 0,30 m/s 2 . Find the friction 
force impeding its motion. What is the coefficient of 
kinetic friction? 

42. (II) A car can decelerate at — 4.80 m/s 2 without skidding 
when coming to rest on a level road. What would its 
deceleration be if (he road were inclined at 13" uphill? 
Assume the same static friction coefficient. 

43. (II) (a) A box sits at rest on a rough 30" inclined plane. 
Draw the free-body diagram, showing all the forces acting 
on the box. (b) How would the diagram change if the box 
weTe sliding down the plane? (c) How would it change if 
the box were sliding up the plane after an initial shove? 

44. (II) Drag-race tires in contact with an asphalt surface 
have a very high coefficient of static friction, Assuming a 
constant acceleration and no slipping of tires, estimate the 
coefficient of static friction needed for a drag racer lo 
cover 1.0 km in 12 s, starting from rest. 

45. (II) The coefficient of kinetic friction for a 22-kg bobsled 
on a track is 0.10. What force is required lo push it down 
a fi.fT incline and achieve a speed of 60 km/h at the end 
of 75 m? 

4<>. (II) Fot the system of Fig. 4-32 (Example 4-20) how 
large a mass would box A have to have to prevent any 
motion from occurring? Assume jtt s = 0.30. 

47. (II) A box is given a push so that it slides across the floor. 
How far will it go. given that the coefficient of kinetic 
friction is U.2U and (he push imparts an initial speed of 
4,(1 m/s? 



i~r 



12.0 kg 



c> 



BT 



'TB 



Cord 



1 .0 kg 



<b) 



t v 




■m 



'"A " 

1 0.0 kg 



■v 



.jt 



(c) 



FIGURE 4-53 Problem 35. Free-body diagrams for two boxes on a table connected by a heavy 
cord, and being pulled to the right as in Fig. 4-22a, Vertical forces, F N and F^ . are not shown. 



Problems 101 



4K. (II) Two crates, of mass 75 kg and HO kg, art in contact 
and at rest on a horizontal surface (Fig, 4-54). A 620- N 
force is exerted on the 75-kg crate. If the coefficient of 
kinetic friction is 0.15. calculate (if) the acceleration ol the 
system, and (b) the force that each crate exerts on the 
other, (c .-) Repeat with the crates reversed 




FIGURE 4-54 Problem 4S. 

49. (II) A flatbed truck is carrying a heavy crate. The coeffi- 
cient of static friction between the crate and the bed of 
the truck is 0.75. What is the maximum rate at which the 
driver can decelerate and still avoid having the crate slide 
against the cab of the truck? 

5(1. (II) On an icy day. you Worry about parking your car in 
your driveway, which has an incline of 12". Your neighbor's 
driveway has an incline of 9.0", and the driveway across 
the street is at 6.0 <: . "The coefficient of static friction 
l>elween tire rubber and ice is 0.15. Which driveway(s) 
will be safe to park in? 

51. (II) A child slides down a slide with a 28" incline, and at 
the bottom her speed is precisely hall what it would have 
l>een if the slide had been friction I ess. Calculate the coef- 
ficient of kinetic friction between the slide and the child, 

52. (II) The carton shown in Fig. 4-55 lies on a plane tilted at 
an angle B - 22,0" to the horizontal, with ^ - 0,12, 
(a) Determine the acceleration of the carton as it slides 
down the plane, (b) If the carton starts from rest 9.30 m 
up the plane from its base, what will be the carton's speed 
when it reaches the bottom of the Incline? 




53. 



FIGURE 4-55 Carton on inclined plane 
Problems 52 and 53. 

(II) A carton is given an initial speed of 3,0 m/s up the 
22,0" plane shown in Fig, 4-55, ia) How far up the plane 
will it go? (b) How much time elapses before it returns to 
its starting point? Ignore friction 



54. (II) A roller coaster reaches the top of the steepest hill 
with a speed of 6.0 km/h. It then descends the hill, which is 
at an average angle of 45 :I and is 45.0m long. Estimate Its 
speed when it Teaches the bottom, Assume fxv - 0. 18. 

55. (II) An 18.0-kg box is released on a 37.0" incline and 
accelerates down the incline at 0.270 m/s 2 . Find the fric- 
tion force impeding Its motion. How laTge is the coeffi- 
cient of kinetic friction? 

56. (II) A small box is held in place against a rough wall by 
someone pushing on it with a force directed upward at 
28" above the horizontal, "lfie coefficients of static and 
kinetic friction between the box and wall are 0.40 and 
030, respectively. The box slides down unless the applied 
force has magnitude 13 N, What is the mass of the box? 

57. (II) Piles of snow on slippery roofs can become 
dangerous projectiles as they melt. Consider a chunk of 
snow at the ridge of a roof with a pitch of 30' ! . {a) What is 
the minimum value of the coefficient of static friction that 
will keep the snow from sliding down? (b) As the snow 
begins to melt, the coefficient of static friction decreases 
and the snow eventually slips. Assuming that the distance 
from the chunk to the edge of the roof is 5.0 m and the 
coefficient of kinetic friction is 0.20. calculate the speed of 
the snow chunk when it slides off the roof, (c) If the edge 
of the roof is 10,0 m above ground, what is the speed of 
the snow when it hits the ground? 

5N. (Ill) {a) Show that the minimum stopping distance for an 
automobile traveling at speed v is equal to v 2 f2/isg, 
wheTe ^ is the coefficient of static friction between the 
tires and the road, and g is the acceleration of gravity. 
(b) What is this distance for a 1200- kg car traveling 
95 km/h if Ms = 0-75? 

59. (III) A coffee cup on the dashboard of a car slides 
forward on the dash when the driver decelerates from 
45 km/h to rest in 3.5 s or less, but not if he decelerates in 
a longer time, What is the coefficient of static friction 
between the cup and the dash? 

60. (Ill) A small block of mass hi is given an initial speed v M 
up a ramp inclined at angle # to the horizontal. It travels a 
distance d up the ramp and comes to rest. Determine a 
formula for the coefficient of kinetic friction between 
block and ramp. 

61. (Ill) The 75-kg climber in Fig, 4-56 is supported in the 
"chimney" by the friction forces exerted on his shoes and 
back. The static coeffi- 
cients of friction between 
his shoes and the wall, and 
between his back and the 
wall, are 0.80 and 0.60, 
respectively, What is the 
minimum normal force he 
must exert? Assume the 
walls are vertical and that 
friction forces are both at 
a maximum. Ignore his 
gTip on the rope. 



FIGURE 4-56 

Problem 61. 




102 CHAPTER 4 Dynamics: Newton's Laws of Motion 



62. (HI) Roxes arc moved on a conveyor belt from where they 
are filled to the packing station 11.0m away. The belt is 
initially stationary and must finish with zero speed. The most 
Tapid transit is accomplished if the belt accelerates tor half 
the distance, then decelerates for the final half of the trip. If 
the coefficient of static friction between a box and the belt is 
0.60, what is the minimum transit time for each box? 

fix (111) A block (masswi) lying on a frictionlcss inclined 
plane is connected to a mass rri; by a massless cord passing 
over a pulley, as shown in Fig. 4-57. (<j) Determine a 
formula for the acceleration of the system of the two blocks 
in teims of ni[ , mj, ft and g, (b) What conditions apply to 
masses itij and m 2 for the acceleration to be in one direc- 
tion (say. mi down the plane), or in the opposite direction? 




64. (Ill) (<i) Suppose the coefficient of kinetic friction between 
m i and the plane in Fig. 4-57 is ^ = 0.15, and that 
m, = m 2 = 2.7 kg, As m 2 moves down, determine the 
magnitude of the acceleration of mi and m>, given = 25 <: - 
(b) What smallest value of fif, will keep this system from 
accelerating? 

65, (III) A bicyclist of mass 65 kg (including the bicycle) can 
coast down a 6.0 :: ' hill at a steady speed of CO km/h because 
of air resistance, How much force must l>e applied to climb 
the hill at the same speed and same air resistance? 



FIGURE 4-57 
Problems 63 
and 64, 



| General Problems 



6& According to a simplified model of a mammalian heart, at 
each pulse approximately 20 g of blood is accelerated from 
0.25 m/s to 0.35 m/s during a period of 0.10 s, What is the 
magnitude of the force exerted by the heart muscle? 

67. A person has a reasonable chance of surviving an auto- 
mobile crash if the deceleration is no more than 30 "g's." 
Calculate the force on a 70-kg person undergoing this 
acceleration. What distance is traveled if the person is 
brought to rest at this rate from 100 km/h'? 

68. (a) If the horizontal acceleration produced by an earth- 
quake is a, and if an object is going to "hold its place" on 
the ground, show that the coefficient of static friction with 
the ground must be at least m = afg. (h) The famous 
Loma Prieta earthquake thai slopped the 1 989 World Series 
produced ground accelerations of up to 4.0 m/s - in the San 
Francisco Bay Area. Would a chair have started to slide on 
a linoleum floor with coefficient of static friction U-25? 

69. An 1 150-kg car pulls a 450-kg trailer. 'Hie car exerts a hori- 
zontal force of 3.8 X 10'' N against the ground in order to 
accelerate. What force does the car exert on the trailer? 
Assume an effective friction coefficient of 0,15 for the trailer. 

7tt. Police investigators, examining the scene of an accident 
involving two cars measure 72-m-long skid marks of one 
of the caTS, which nearly came to a stop before colliding. 
The coefficient of kinetic friction between rubber and the 
pavement is about 0.80. Fstimate the initial speed of that 
car assuming a level road. 

71. A car starts Tolling down a l-in-4 hill (1-tn^l means that 
for each 4 m traveled along the road, the elevation change 
is 1 m). How fast is it going when it reaches the bottom 
after traveling 55 m? (a) [gnaTe friction, (b) Assume an 
effective coefficient of friction equal to U.10. 



72. A 2.0-kg purse is dropped from the top of the 1. caning 
Tower of Pisa and falls 55m before reaching the 
ground with a speed of 29 m/s, What was the average 
force of air resistance? 

73. A cyclist is coasting at a steady speed of 12 m/s but enters 
a muddy stretch where the effective coefficient of friction 
is 0,(50, Will the cyclist emerge from the muddy stretch 
without having to pedal If the mud lasts for 11 m? If so, 
what will be the speed upon emerging? 

74. A city planner is working on the redesign of a hilly 
portion of a city. An important consideration is how steep 
the roads can be so that even low-powered cars can get 
up the hills without slowing down. A particular small car, 
with a mass of 1 100 kg. can accelerate on a level Toad 
from rest to 21 m/s (75 km/h) in 14.0 s. Using these data, 
calculate the maximum steepness of a hill. 

75. Francesca, who likes physics experiments, dangles her 
watch from a thin piece of String while the jetliner she is in 
takes off from JFK Airport (Fig. 4-58). She notices that 
the string makes an angle of 25° with respect to the 
vertical as the aircraft accelerates for 

takeoff, which takes ahout 18 s, Esti- 
mate the takeoff speed of the 
aircraft. 



FIGURE 4 53 

Problem 75. 




General Problems 103 



76. 



77. 



n. 



A 28-0-kg block is connected to an empty ] .35-kg bucket 
by a cord running over a frictionless pulley (Fig. 4-59), 
The coefficient of static friction between the table and the 
block is 0.450 and the coefficient of kinetic friction 
between the table and the block is 0.320. Sand is gradu- 
ally added to the bucket until the system just begins to 
move, (a) Calculate the mass of sand added to the bucket, 
(ft) Calculate the acceleration of the system. 



28.0 kg 



SO. 




78. 



FIGURE 4-59 Problem 76. 

In the design of a supermarket, there are to be several 
ramps connecting different parts of the store. Customers 
will have to push grocery carts up the ramps and it is 
obviously desirable that this not be too difficult/Hie engi- 
neer has done a survey and found that almost no one 
complains if the force directed up the ramp is no more than 
20 N. Ignoring friction, at what maximum angle 6 should the 
ramps be built, assuming a full 30-kg grocery cart? 

((?) What minimum force F is needed to lift the piano 
(mass M) using the pulley apparatus shown in Fig, 4-60? 
(r>) Determine the tension in each section of rope: Fn , 
Ff2 * ^T3 ■ and r^ , 




FIGURE 4-60 
Problem 78, 



A jet aircraft is accelerating at 3-5 m/s 2 at an angle of 45' : 
above the horizontal. What is the total force that the 
cockpit seat exerts on the 75 -kg pilot? 



In the design process for a child-restTaint chair, an engi- 
neer considers the following set of conditions: A 12-kg 
child is riding in the chair, which is securely fastened to 
the seat of an automobile (Fig, 4-61 J. Assume the auto- 
mobile is involved in a head-on collision with anotheT 
vehicle. The initial speed i\ t of the car is 45 km/h, and this 
speed is reduced to zero during the collision time of 
0.20 s. Assume a constant car deceleration during the 
collision and estimate the net horizontal force F lhal the 
straps of the restraint chair must exert on the child in 
order to keep her fixed to the chair. Treat the child as a 
particle and state any additional assumptions made 
during your analysis. 




FIGURE 4-61 Problem 80. 

A 7650-kg helicopter accelerates upward at 0.80 m/s~ 
while I if dug a 1250-kg frame at a construction site. Fig. 4-62. 
(a) What is the lift force exerted by the air on the heli- 
copter rotors? (b) What is the tension in the cable (ignore 
its mass) that connects the frame to the helicopter? 
(c) What force does the cable exert on the helicopter? 



X. 



Ft 




FIGURE 4-62 Problem SI. 

52. A super high-speed 12-car Italian train has a mass of 
660 metric tons (660,000 kg), It can exert a maximum 
force of 400 kN horizontally against the tracks, whereas 
at maximum velocity (300 km/h). it cxctis a foTce of 
about 150 kN. Calculate (a) its maximum acceleration, 
and (b) estimate the force of air resistance at top 
speed. 

53. A 65-kg ice skater coasts with no effort for 75 m until she 
slops. If the coefficient of kinetic friction between her 
skates and the ice is ^i. k = 0.10, how fast was she moving 
at the start of her coast? 



1(M CHAPTER 4 Dynamics: Newton's Laws of Motion 



N4. Two ruck climbers. Rill and Karen, vise safety ropes of 
similar length, Karen's rope is more elastic, called a 
dynamic rope by climbers. Bill has a sunk rope, not 
recomm ended foT safely purposes in pro climbing, Karen 
falls freely about 2,0 m and then the rope stops her over a 
distance of I .Cm (Fig. 4-63). (a) Fsti mate, assuming that 
the force is constant, how large a force she will feel from, 
the rope. (Express the result in multiples of her weight.) 
(b) In a similar fall. Bill's rope stretches by 30cm only. 
How many times his weight will the rope pull on him? 
Which climber is more likelv to be hurt"? 




-' V* " *■■ 



FIGURE 4-63 
Problem 84 



Ss. A fisherman in a boat is using a "10-lh test'' fishing line. 
This means that the line can exert a force of 45 N without 
breaking (1 lb = 4.45 N). (a) How heavy a fish cart the 
fisherman land if he pulls the fish up vertically at constant 
speed? (b) If he accelerates the fish upward at 2.0m/s~, 
what maximum weight fish can he land? (c) Is it possible 
to land a 15-lb trout on 10-lb lest line? Why or why not? 



86, An elevator in a tall building is allowed to reach a 

maximum speed of 3.5 m/s going down, What must the 
tension be in the cable to stop this elevator over a distance 
of 2.6m if the elevator has a mass of 13UU kg including 
occupants'' 

87, Two boxes, m\ = 1.0 kg with a coefficient of kinetic fric- 
tion of 0.10, and m 2 = 2,0 kg with a coefficient of 0.20, 
are placed on a plane inclined at 9 = 30", (a) What 
acceleration does each ho* experience? (/>) If a taut string 
is connected to the boxes (Fig. 4-64), with m-> initially 
farther down the slope, what is the acceleration of each 
box? (c) If the initial configuration is reversed with nt\ 
starting lower with a taut string, what is the acceleration 
of each box? 







FIGURE 4-64 Problem 87. 

88. A 75-0-kg person stands on a scale in an elevator. What 
does the scale read (in N and in kg) when the elevator is 
(w) at rest, (ft) ascending at a constant speed of 3.0 m/s, 
(c) falling at 3-0m/s, ((0 accelerating upward at 3.0 m/s 2 , 
(e ) accelerating downward at 3.0 m/sr? 

89. Three mountain climbers who are roped together aTe 

ascending an icefield inclined at 21.0° to the horizontal. 
The last climber slips, pulling the second climber off his 
feet. The first climber is able to hold them both, ff each 
climber has a mass of 75 kg, calculate the tension in each 
of the two sections of rope between the llmee climbers. 
Ignore friction between the ice and the fallen climbers. 



Answers to Exercises 

A.: (a) "The same; (ft) the sports cart (c) third law for part (a), 

second law for part (b). 
B: The force applied by the person is insufficient to keep the 

box moving. 



C: No: yes. 
11; Ycs:no. 



General Problems 105 



The astronauts in the upper left of 
this photo are working on the space 
shuttle- As they orbit the Earth— at 
a rather high speed — they experi- 
ence apparent weightlessness. The 
Moon, in the background, also is 
orbiting the Earth at high speed. 
Both the Moon and the space 
shuttle move in nearly circular 
orbits, and each undergoes a 
centripetal acceleration. What keeps 
the Moon and the space shuttle (and 
its astronauts) from moving off in a 
straight line away from Earth? It is 
the force of gravity. Newton's law of 
universal gravitation states that all 
objects attract all other objects with 
a force proportional to then - masses 
and inversely proportional to the 
square of the distance between 
them, 



CHAPTER 



5 







Circular Motion; Gravitation 



FIGURE 5-1 A small object 
moving in a circle, showing how the 
velocity changes. At each point, the 
instantaneous velocity is in a direc- 
tion tangent to the circular path. 




An object moves in a straight line if the net force on it acts in the direction 
of motion, or the net force is zero. If the net force acts at an angle to the 
direction of motion at any moment, then the object moves in a curved 
path. An example of the latter is projectile motion, which we discussed in Chapter 3. 
Another important case is that of an object moving in a circle, such as a ball at 
the end of a string revolving around one's head, or the nearly circular motion of 
the Moon about the Earth. 

In this Chapter, we study the circular motion of objects, and how Newton's 
laws of motion apply. We also discuss how Newton conceived of another great 
law by applying the concepts of circular motion to the motion of the Moon and 
the planets. This is the law of universal gravitation, which was the capstone of 
Newton's analysis of the physical world. 

Kinematics of Uniform Circular Motion 

An object that moves in a circle at eons.tant speed v is said to experience 
uniform circular motion. The magnitude of the velocity remains constant in this 
ease, but the direction of the velocity continuously changes as the ohject moves 
around the circle (Fig. 5-1). Because acceleration is defined as the rate of 



106 



change of velocity, a change in direction of velocity constitutes an acceleration, 
just as a change in magnitude of velocity does, Thus, an object revolving in a 
circle is continuously accelerating, even when the speed remains constant 
(if] = Vj = v). We now investigate this acceleration quantitatively, 
Acceleration is defined as 

v, - V] A* 

9 = — = . 

Af 



Af 

where Av is the change in velocity during the short time interval Af. We will 
eventually consider the situation in which Af approaches zero and thus obtain 
the instantaneous acceleration, But for purposes of making a clear drawing. 
Fig. 5-2, we consider a nonzero time interval. During the time interval Af, the 
particle in Fig. 5-2a moves from point A to point B, covering a distance A/ 
along the arc which subtends an angle AW. The change in the velocity vector is 
v 2 — V, = Av, and is shown in Fig. 5-2b, 

If we let Af be very small (approaching zero), then A/ and AS are also 
very small, and v, will be almost parallel to v,; Av will be essentially perpen- 
dicular to them (Fig. 5-2c). Thus Av points toward the center of the circle. 
Since 5, hy definition, is in the same direction as Av, it too must point toward 
the center of the circle. Therefore, this acceleration is called centripetal 
acceleration ('"center-pointing" acceleration) or radial acceleration (since it is 
directed along the radius, toward the center of the circle), and we denote it 
by a H . 

We next determine the magnitude of the centripetal (radial) acceleration, a K . 
Because CA in Fig 5-2a is perpendicular to v, , and CB is perpendicular to v 2 , it 
follows that the angle AS, defined as the angle between CA and CB, is also the 
angle between v t and v 2 . Hence the vectors f,,v,, and Av in Fig, 5-2b fornv a 
triangle that is geometrically similar' to triangle CAB in Fig, 5-2a. If we take AS 
to be very small (letting Af be very small) and setting v = v t - v 2 because the 
magnitude of the velocity is assumed not to change, we can write 

Av Al 
v r 

This is an exact equality when Af approaches zero, for then the arc length Al 
equals the cord length AB, We want to find the instantaneous acceleration, so 
we let At approach zero, write the above expression as an equality, and then 
solve for Av: 

Av - - Al. 
r 

To get the centripetal acceleration, o K ,we divide Av by Af: 

_ &v = v A[ 
" K Af r Af' 

But Al/At is just the linear speed, v, of the object, so 



FIGURE 5-2 Determining the 
change in velocity. Av, for a particle 

moving in a circle. ITie length Al is the 
distance alona the arc, from A to G. 





<it 



(5-1) Centripetal (radial) acceleration 



Equation 5-1 is valid even when v is not constant. 

To summarize, an object mooing in a circle of radius r at constant speed v 
has an acceleration whose direction is toward the center of the circle and whose 
magnitude is a R = v 2 /r. It is not surprising that this acceleration depends on if 
and r. The greater the speed v, the faster the velocity changes direction; and the 
larger the radius, the less rapidly the velocity changes direction. 



<fc. 



CAUTION 



In uniform circular motion, the fpeed is 
constant, !>ut the lucfitrttiiuii is not zero 



' Appendix A contains a review of geometry. 



SECTION 5-1 Kinematics of Uniform Circular Motion 107 



^ CAUTION 



The direction of motion (V) and 

the mi deration f aj art not in 

thi same direction; instead, a -L V 



The acceleration vector points toward the center of the circle. Rut the 
velocity vectOT always points in the direction of motion, which is tangential to 
the circle. Thus the velocity and acceleration vectors are perpendicular to each 
other at every point in the path for uniform circular motion (Fig- 5-3), This is 
another example that illustrates the erroT in thinking that acceleration and 
velocity are always in the same direction. For an object falling vertically, a and v 
are indeed parallel. But in circular motion, a and v are perpendicular, not 
parallel (nor were they parallel in projectile motion. Section 3-5). 



FIGURE 5-3 For uniform 
circular motion, a is always 
perpendicular lo v. 



Period and frequency 




Circular motion is often described in terms of the frequency /, the number 
of revolutions per second. The period T of an object revolving in a circle is the 
time required for one complete revolution. Period and frequency are related by 



'■f 



(5-2) 



Foi" example, if an object revolves at a frequency of 3 rev/s, then each revolution 
takes js. For an object revolving in a circle (of circumference 2t7t) at constant 
speed v, we can write 

27rr 
v - -p. 

since in one revolution the object travels one circumference. 



EXAMPLE 5-1 



Acceleration of a revolving ball. A 150-g ball at the end 
of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in 
Fig. 5- 1 or 5-3, The ball makes 2.00 revolutions in a second. What is its centripetal 
acceleration? 

APPROACH The centripetal acceleration is a R = v 2 fr, We are given r, and 
we can find the speed of the ball, v, from the given radius and frequency. 
SOLUTION If the ball makes two complete revolutions per second, then the 
ball travels in a complete circle in a time interval equal to 0-500 s„ which is its 
period 7*. The distance traveled in this time is the circumference of the circle, 
27rr, where r is the radius of the circle. Therefore, the ball has speed 

2nr 2(3.14)(0.600m) 
v = 



T (0.500 s) 

The centripetal acceleration is 

_ v 2 _ (7.54 m/s) 2 

" R " r (0.600 m) 



= 7.54 m/s. 



94,7 m/s 2 



I EXERCISE A If the string is doubled in length to 1 ,20 m but all else stays the same, by 
what factor will the centripetal acceleration change? 

' Differences in [he final digit ean depend on whether you keep all digits in your calculator for i- 
(which gives a R = y4,7m/s ? ).or if you use o = 7.54 m/s in which case you get a R = 94.Bm/s 2 . Both 
results are valid since our assumed accuracy is about ± 0.1 m/s (see Section 1-4). 



108 CHAPTERS Circular Motion; Gravitation 



EXAMPLE 5-2 



Moon's centripetal acceleration. The Moon's nearly 
circular orbit ahuut the Earth has a radius of about 384,000 km and a period T 
of 27.3 days. Determine the acceleration of the Moon toward the Earth. 

APPROACH Again we need to find the velocity v in order to find a R , We will 
need to convert to SI units to get » in m/s, 

SOLUTION In one orbit around the Earth, the Moon travels a distance 2nr, 
where r — 3,&4 x 10 s m is the radius of its circular path. The time required 
for one complete orbit is the Moon's period of 27,3 d. The speed of the Moon 
in its orbit about the Earth is v - ItrrjT. The period T in seconds is 
T - (27,3 d) (24.0 h/d)(3600s/h) - 2.36 x 10" s. Therefore, 



a K - — - 



r 
r 



(2irr} 2 47r 2 r 4ir J (3.84 X I0*m) 
T 2 r T 2 (2.36 X ltf's) : 

= 0.00272 m/s 2 = 2.72 x KT'm/s 2 , 



We can write this acceleration in terms of # = 9.80 m/s 2 (the acceleration of 
gravity at the Earth's surface) as 



a = 2.72 X I0" 3 m/s : 



9.80 m/s' 



= 2.78 X UT 4 g. 



NOTE The centripetal acceleration of the Moon, a = 2.78 X 10 4 #. is not <j > CAUTION 



the acceleration of gravity for objects at the Moon's surface due to the 
Moon's gravity. Rather, it is the acceleration due to the Earth's gravity for 
any object (such as the Moon) that is 384.000 km from the Earth. Notice how 
small this acceleration is compared to the acceleration of objects near the 
Earth's surface, 



Distinguish Mot 
gravity on objects m itssurface, 
from Earth 's gravity acting 
on Moon (litis Example) 



Dynamics of Uniform Circular Motion 

According to Newton's second law (SF = ma), an object that is accelerating 
must have a net force acting on it. An object moving in a circle, such as a ball 
on the end of a string, must therefore have a force applied to it to keep it 
moving in that circle, That is, a net force is necessary to give it centripetal 
acceleration. The magnitude of the required force can be calculated using 
Newton's second law for the radial component, £F K - rna R , where a R is the 
centripetal acceleration, a K - v 2 /r, and YF tl is the total (or net) force in the 
radial direction: 



2F K = ma R = m 



[circular motion] (5-3) 



For uniform circular motion (v = constant), the acceleration is a R , which is 
directed toward the center of the circle at any moment Thus the net force too 
must be directed toward the center of the circle (Fig. 5-4). A net force is neces- 
sary because otherwise, if no net force were exerted on the object, it would not 
move in a circle but in a straight line, as Newton's first law tells us. The direction 
of the net force is continually changing so that it is always directed toward the 
center of the circle. This force is sometimes called a centripetal ('"pointing 
toward the center") force. But be aware that "centripetal force" does not indi- 
cate some new kind uf force. The term merely describes the direction of the net 
force needed to provide a circular path: the net force is directed toward the 
circle's center. The force must he applied by other objects. For example, to swing 
a ball in a circle on the end of a string, you pull on the string and the string 
exerts the force on the hall. (Try it.) 



Force is needed to provide 
centripetal acceleration 




FIGURE 5-4 A force is required 
to keep an object moving in a circlc- 
If the speed is constant, the force is 
directed toward the circle's center. 



<j> CAUTION 

Centripetal force is not a new 

kind of force 

: !■ ■■<■>■■. force must be exerted 

by an object) 



SECTION 5-2 Dynamics of Uniform Circular Motion 109 



<t> C A U T I O 



There -s no >■;■<;■ t?xlr;j:i%n; •(>>'■<■' 



S Force tin bill] N 




FIG U RE 5 -S Swi n gi ng a ball on 
the end of a string, 



FIGURE 5-6 If centrifugal force 
existed, the revolving ball would fly 
outward as in (a) when released, In 

fact.it flies off tangential]} 1 as in (b). 
For example, in (c) sparks fly in 
straight lines tangentially from the 
edge of a rotating grinding wheel. 



/ \ 

f \ 

( DOESN'T \ 

/ HAPPEN 



U] 





» 



There is a common misconception that an object moving in a circle has an 
outward force acting on if, a so-called centrifugal ("center- fleeing") force. This 
is incorrect: there is no outward force on the revolving object. Consider, for example, 
a person swinging a ball on the end of a string around her head (Fig, 5-5). If you 
have ever done this yourself, you know that you feel a force pulling outward on 
your hand- The misconception arises when this pull is interpreted as an 
outward "centrifugal" force pulling on the ball that is transmitted along the 
string to your hand. This is not what is happening at all. To keep the ball 
moving in a circle, you pull inwardly on the string, and the string exerts this 
force on the ball The ball exerts an equal and opposite force on the string 
(Newton's third law), and {his is the outward force your hand feels (see Fig. 5-5), 

The force on the ball is the one exerted inwardly on it by you, via the 
string. To see even more convincing evidence that a "centrifugal force" does 
not act on the ball, consider what happens when you let go of the string. If a 
centrifugal force were acting, the ball would fly outward, as shown in 
Fig. 5-6a. But it doesn't; the ball flies off tangentially (Fig. 5-6b), in the direc- 
tion of the velocity it had at the moment it was released, because the inward 
force no longer acts. Try it and sec ! 



EXAMPLE 5-3 B^. T^J 



Force on revolving ball [horizontal). 

Estimate the force a person must exert on a string attached to a 0. 150-kg ball 
to make the ball revolve in a horizontal circle of radius 0.600m. The ball 
makes 2,00 revolutions per second (T - 0.500 s), as in Example 5-1. 

APPROACH First we need to draw the free-body diagram for the ball. The 
forces acting on the ball are the force of gravity, mj* downward, and the 
tension force P, that the string exerts toward the hand at the center (which 
occurs because the person exerts that same force on the string). The free -body 
diagram for the ball is as shown in Fig. 5-7. The ball's weight complicates 
matters and makes it impossible to revolve a ball with the cord perfectly hori- 
zontal. We assume the weight is small, and put <$> ~ in Fig. 5-7, Thus P T will act 
nearly horizontally and, in any case, provides the force necessary to give the 
ball its centripetal acceleration. 

SOLUTION We apply Newton's second law to the radial direction, which we 
assume is horizontal: 

(SF) R - ma K , 

where a R = v 2 /r and v = 2-irrjT = 2jt(0.600 m)/(0.500 s) = 7.54 m/s. Thus 

~ 2 (7,54 m/s) 2 



'': 



W 



(0.150 kg) 



(0.600 m) 



« I4N. 



NOTE We keep only two significant figures in the answer because mg = 
(0.150 kg)(9.80 m/s 2 ) = 1.5N, being about ^ of our result, is small but not so 
small as to justify stating a more precise answer since we ignored the effect of mg. 

NOTE To include the effect of wig, resolve F T in Fig. 5-7 into components, 
and set the horizontal component of F, equal to mv 2 /r and its vertical 
component equal to mg. 




( c ) FIGURE 5-7 Example 5-3. 

110 CHAPTER 5 Circular Motion; Gravitation 



EXAMPLE 5-4 



Revolving ball (vertical circle!. A 0. 1 5 0-kg ballon the end 
of a 1.10-m-long curd (negligible mass) is swung in a vertical circle, 
(a) Determine the minimum speed the ball must have at the top of its arc so that 
the ball continues moving in a circle, (b) Calculate the tension in the cord at the 
bottom of the arc, assuming the ball is moving at twice the speed of part (a). 

APPROACH The ball moves in a vertical circle and is not undergoing uniform 
circular motion. The radius is assumed constant, but the speed v changes 
because of gravity. Nonetheless, Eq. 5-1 is valid at each point along the circle, 
and we use it at points 1 and 2. The free-body diagram is shown in Fig. 5-8 for 
both positions I and 2, 

SOLUTION (a) At the top (point I), two forces act on the ball: wig, the force 
of gravity, and F TI , the tension force the cord exerts at point I. Both act 
downward, and their vector sum acts to give the ball its centripetal accelera- 
tion a R . We apply Newton's second law, for the vertical direction, choosing 
downward as positive since the acceleration is downward (toward the center): 



(2F) R 



on,. 



F, t + trig - m 



[at top] 



From this equation we can see that the tension force F ri at point I will get larger 
if v, (hall's speed at top of circle) is made larger, as expected. But we are asked 
for the minimum speed to keep the ball moving in a circle. The eord will remain 
taut as long as there is tension in it. But if the tension disappears (because >\ is 
too small) the cord can go limp, and the ball will fall out of its circular path. Thus, 
the minimum speed will occur if F, , = 0, for which we have 



mg -■ 
We solve for i\ : 



[minimum speed at top] 



v, = Vgr = \/(9.80m/s-)(U0m) = 3.28 m/s. 

This is the minimum speed at the top of the circle if the ball is to continue 
moving in a circular path. 

(b) When the ball is at the bottom of the circle (point 2 in Fig. 5-8), the cord 
exerts its tension force F, 2 upward, whereas the force of gravity, mg, still acts 
downward. So we apply Newton's second law, this time choosing upward as 
positive since the acceleration is upward (toward the center): 

(2F) R = ma R 

■■ 

F t2 ~ mg = m — [at bottom] 

The speed v 2 is given as twice that in (a), namely 6.56 m/s. We solve for F ri : 

F\i = m— + >ng 

(6.56 m/s) 2 



- (0.150 kg) 



(1.10m) 



+ (0.150 kg)(9.80 m/s : ) - 7.34 N. 



EXERCISE B In a tumble dryer, the speed of the drum should be just large enough so 
that the clothes are carried nearly to the top of the drum and then fall away, rather 
than being pressed against the drum for the whole revolution Determine whether this 
speed will be different for heavier wet clothes than foT lighter dry clothes. 

EXERCISE C A rider on a Ferris wheel moves in a vertical chcle of radius r at constant 
speed v (Fig. 5-9). Is the normal force that the seat exerts on the rider at the top of 
the wheel (a) less than, (h) more than, or (c) the same as, the force the seat exerts at the 
bottom of the wheel? 



"fh, 



t Tl 



t 

VS.* 



"■'K 

FIGURE 5-8 Example 5-4. Free- 
body diagrams for positions l and 2. 

< ord tension ami gravity together 
provide centripetal acceleration 



Gravity provides 
centripetal acceleration 



String tension and gravity 
acting in opposite directions 
provide centripetal acceleration 

FIGURE 5-9 Exercise C. 




SECTION 5-2 Dynamics of Uniform Circular Motion 111 




FIGURE 5-10 Example 5-5. 



CONCEPTUAL EXAMPLE 5-5 | Tetherball. The game of tetherball is 



played with a ball tied to a pole with a string. After the hall is struck, it 
revolves around the pole as shown in Pig. 5-10. In what direction is the accel- 
eration of the ball, and what force causes the acceleration? 

RESPONSE If the ball revolves in a horizontal plane as shown, then the 
acceleration points horizontally toward the center of the ball's circular path 
(not toward the lop of the pole), The force responsible for the acceleration 
may not be obvious at first, since there seems to be no force pointing directly 
horizontally- But it is the net force (the sum of wg and F T here) Lhat must point 
in the direction of the acceleration. The vertical component of the string 
tension, F Ty , balances the ball's weight, wig. The horizontal component of the 
string tension, F Tj ., is the force that produces the centripetal acceleration 
toward the center. 



PROBLEM SOLVING 



Uniform Circular Motion 



1, Draw a free -body diagram, showing all the forces 
acting on each object under consideration. Be sure 
you can identify the source of each force (tension 
in a cord, Earth's gravity, friction, normal force, and 
so on). Don't put in something that doesn't belong 
(like a centrifugal force). 

2. Determine which of the forces, or which of their 
components, act to provide the centripetal accelera- 
tion — that is, all the forces or components that act 



radially, toward or away from the center of the 
circular path. The sum of these forces (or compo- 
nents) provides the centripetal acceleration. 

Choose a convenient coordinate system, preferably 
with one axis along the acceleration direction. 
Apply Newton's second law to the Tadial component: 



(EF) R = ma ]t = m ■ 



[radial direction] 



PHYSICS APPLIED 

Driving around a curve 



Highway Curves, Banked and Unbanked 

An example of circular dynamics tieeurs when an automobile rounds a curve, say 
to the left. In such a situation, you may feel that you are thrust outward toward the 
right side door. But there is no mysterious centrifugal force pulling on you. 
What is happening is that you tend to move in a straight line, whereas the car 
has begun to follow a curved path. To make you go in the curved path, the seat 
(friction) or the door of the car (direct contact) exerts a force on you 
(Fig. 5-1 1). The car also must have a force exerted on it toward the center of 
the curve if it is to move in that curve. On a flat road, this force is supplied by 
friction between the tires and the pavement. 




FIGURE 5-11 Hie road exerts an 

inward force (friction against the 
tires) on a car to make it move in a 
circle. The car exerts an inward 
force on the passenger. 



Force on car 
(sum of friction forces 
acting on each tire) 




Tendency lor 
passenger to 
go straight 



passenger 



112 CHAPTER 5 Circular Motion; Gravitation 



If the wheels and tires of the car are rolling normally without slipping or 
sliding, the bottom of the tire is at rest against the Toad at each instant; so the 
friction force the Toad exeTts on the tires is static friction. But if the static friction 
force is not great enough, as under icy condition s, sufficient friction force cannot 
be applied and the car will skid out of a circular path into a more nearly straight 
path. See Fig. 5-12. Once a car skids or slides, the friction force becomes kinetic 
friction, which is less than static friction. 



EXAMPLE 5-6 



Skidding on a curve. A 1000 -kg car rounds a curve on a 
flat road of radius 50 m at a speed of 50 km/h ( 14 m/s). Will the car follow the 
curve, or will it skid? Assume; (a) the pavement is dry and the coefficient of 
static friction is /i, = 0.60; (6) the pavement is icy and ^i s = 0.25. 

APPROACH The forces on the car are gravity mg downward, the normal 
force F N exerted upward by the toad, and a horizontal friction force due to the 
road. They are shown in Fig. 5-13, which is the free-body diagram for the car. 
The car will follow the curve if the maximum static friction force is greater 
than the mass times the centripetal acceleration. 

SOLUTION In the vertical direction there is no acceleration. Newton's second 
law tells us that the normal force F s on the car is equal to the weight mg since 
the road is flat: 

f N - mg - ( 1000 kg}(9.8 m/s 2 ) - 9800 N. 

In the horizontal direction the only force is friction, and we must compare it to 
the force needed to produce the centripetal acceleration to see if it is suffi- 
cient. The net horizontal force required to keep the car moving in a circle 
around the curve is 



(2F) R = ma R = my=(\( 



(I4m/s) 2 

k S> tw / =3W0 N. 
(50 m) 



Now we compute the maximum total static friction force (the sum of the fric- 
tion forces acting on each of the four tires) to see if it can be large enough to 
provide a safe centripetal acceleration. For (a), jj, s = 0.60, and the maximum 
friction force attainable (recall from Section 4-8 that F (t =£ fa F N J is 

(*"ftW - P*F N - (0.60)(9800N) - 5900 N. 

Since a force of only 3900 N is needed, and that is, in fact, how much will be 
exerted by the road as a static friction force, the ear can follow the curve. But 
in (b) the maximum static friction force possible is 

(ffrUx = M^n = (0.25)(9S00N) = 2500 N. 

The car will skid because the ground cannot exert sufficient force (3900 N is 
needed) to keep it moving in a curve of radius 50 m at a speed of 50 km/h. 



The possibility of skidding is worse if the wheels lock (stop rotating) when the 
brakes are applied too hard. When the tires are rolling, static friction exists. 
But if the wheels lock (stop rotating), the tires slide and the friction force, 
which is now kinetic friction, is less. More importantly, the direction of the 
friction force changes suddenly if the wheels lock. Static friction can point 
perpendicular to the velocity, as in Fig. 5- 1 3b, but if the car slides, kinetic 
friction points opposite to the velocity. The force no longer points toward the 
center of the circle, and the car cannot continue in a curved path (see 
Fig. 5-12). Even worse, if the road is wet or icy, locking of the wheels occurs 
with less force on the brake pedal since there is less road friction to keep the 
wheels turning rather than sliding. Antilock brakes (ABS) are designed to 
limit brake pressure just before the point where sliding would occur, by means 
of delicate sensors and a fast computer. 




FIGURE 5-12 Race tar heading 
into a curve, From the tire marks we 

see that most cars experienced a 
sufficient friction foTce to give them 
the needed centripetal acceleration 
for rounding the aiTve safely. But. 
we also see tire tracks of cars on 
which there was not sufficient 
force — and which followed more 
nearly straight-line paths. 



FIGURE 5-13 Example 5-6. Forces 

on a ear rounding a curve on a flat 
road, (a) Front view, (h) top view. 



4 



(a) 



Pti = '«S 




(h) 



C&L 



PHYSICS APPLIED 



Antilock brakes 



SECTION 5-3 Highway Curves, Banked and Unbanked 113 



@ PHYSICS APPLIED 

Banked curves 




FIGURE 5-14 Normal force on a 
car rounding a banked curve. 
resolved into its horizontal and 
vertical components. The centripetal 
acceleration is horizontal (not 
parallel to the sloping road). The 
friction force on the tires, not 
si: own, could point up or down 
along the slope, depending on the 
ear's speed, The friction force will 
he zero for one particular speed, 



> CAUTION 

F N is not alway i equal to mg 



Horizontal component of normal force 

urn .'i ) provide cenlripetid two ieviiiiu!} 

{friction i.t desired to be zero — 
otherwise it too would consribute) 



Blinking angle (friction not needed) 



The banking of curves can reduce Lhe chance of skidding. The normal force 
exerted by a banked road, acting perpendicular to the mad, will have a compo- 
nent toward the center of the circle (Fig. 5-14), thus reducing the reliance on 
friction. For a given banking angle B, there will be one speed for which no fric- 
tion at all is required. This will be the case when the horizontal component of the 
normal force toward the center of the curve, F s sin B (see Fig, 5- 1 4), is just equal 
to the force required to give a vehicle its centripetal acceleration — that is, when 



F v sin0 = m — 
r 



[no friction required] 



The banking angle of a road, ft, is chosen so that this condition holds for a 
particular speed, called the "design speed." 



EXAMPLE 5-7 



Banking angle, (a) For a car traveling with speed v 
around a curve of radius r t determine a formula for the angle at which a road 
should be banked so that no friction is required, (b) What is this angle for an 
expressway off-ramp curve of Tadius 50 m at a design speed of 50km/h? 

APPROACH Even though the road is banked, the car is still moving along a 
horizontal circle, so the centripetal acceleration needs to be horizontal. We 
choose our x and v axes as horizontal and vertical so that a R , which is hori- 
zontal, is along the x axis. The forces on the caT are the Earth's gravity mg 
downward, and the normal force F N exerted by the road perpendicular to its 
surface, See Fig. 5— 14, where the components of F N are also shown. We don't 
need to consider the friction of the road because we are designing a road to be 
banked so as to eliminate dependence on friction, 
SOLUTION (a) For the horizontal direction, SF K = ma K gives 



F N sin B = 



mvr 
r 



Since there is no vertical motion, the y component of the acceleration is zero. 
y gives us 

Fy, cos 8 - mg 



so XF,. = h 



f). 



Thus, 



l\ 



mg 
costf 



[Note in this case that F s s: nig since costf =s ].] 

We substitute this relation for F N into the equation for the horizontal motion. 



Fvsinfl = m — - 



and obtain 



or 



mg 

cos B 



sin B 



mg tan ft — m — * 



so 



tantf 



rg 



This is the formula for the banking angle 0: no friction needed at speed v. 
(b) For r = 50m and v = 50km/h (or I4m/s), 
'14m/s) 2 



tan e = 



50m)(9.Sm/s 2 ) 



= 0.40, 



so B = 22 <J . 



Ilfl CHAPTER 5 Circular Motion; Gravitation 



EXERCISE D Tu negotiate an unbanked curve at a faster speed, a driver puts a couple 
of sand hags in his van aiming to increase the force of friction between the tires and the 
road. Will the sand bags help? 

EXERCISE E Can a heavy truck and a small ear travel safely at the same speed around 
an icy, banked-curve road? 

*( ] Nonuniform Circular Motion 

Circular motion at constant speed occurs when the net force on an object is 
exerted toward the center of the circle. Tf the net force is not directed toward 
the center but is at an angle, as shown in Fig. 5-l5a, the force has two compo- 
nents. The component directed toward the center of the circle, F K , gives rise to 
the centripetal acceleration, a R , and keeps the object moving in a circle. The 
component tangent to the circle, F UfI , acts to increase (or decrease) the speed, 
and thus gives rise to a component of the acceleration tangent to the circle, 
fli an . When the speed of the object is changing, a tangential component of 
force is acting. 



\ ' I 




/ [/__+Pr \ / Z_J 9r \ FIGURE 5-15 The speed of an object moving 

I ' 1 in a circle changes if the force on it has a tangen- 



^Lan 



1 tial component, F Un Part (a) shows the force F 

, \ / audits vector components: paTt (b) shows the 

\ / \ / acceleration vector and its vector components. 

\ / \ / 

(a) (b) 

When you first start revolving a ball on the end of a string around your 
head, you must give it tangential acceleration. You do this by pulling on the 
string with youT hand displaced from the center of the circle. In athletics, a 
hammer thrower accelerates the hammer tangenlially in a similar way so that it 
reaches a high speed before Telease. 

The tangential component of the acceleration, a lari , is equal to the rate of 
change of the magnitude of the objects velocity; 

A v.) 

a7" 

The radial (centripetal) acceleration arises from the change in direction of the 
velocity and, as we have seen (Eq. 5-1), is given by 

vr 

The tangential acceleration always points in a direction tangent to the circle, and 
is in the direction of motion (parallel to v, which is always tangent to the circle) 
if the speed is increasing, as shown in Fig. 5- 1 5b. If the speed is decreasing, a Ia]1 
points antiparallcl to v. In cither case, a lan and a K arc always perpendicular to 
each other; and their directions change continually as the object moves along its 
circular path. The total vector acceleration si is the sum of these two: 

a — a lull + a R- 

Since a K and a la „ are always perpendicular to each other, the magnitude of a at 
any moment is 



♦SECTION 5-4 Nonuniform Circular Motion 115 



EXAMPLE 5-8 



Two components of acceleration. A Taee car starts from 
rest in the pit area and accelerates at a uniform rate to a speed of 35m/s in 
1 1 s, moving on a circular track of radius 500 m. Assuming constant tangential 
acceleration, find (a) the tangential acceleration, and (b) the radial accelera- 
tion, at the instant when the speed is v = 15 m/s. 

APPROACH The tangential acceleration relates to the change in speed of the 
car, and can be calculated as a tm = AvfAt. The centripetal acceleration 

relates to the change in the direction of the velocity vector and is calculated 
using a R = v z /r. 

SOLUTION (a) During the ll-s time interval, we assume the tangential 
acceleration a un is constant, Tts magnitude is 

Ac (35 m/s - Om/s) 
A7 = _ 



tflsin — 



II s 



= 3.2 m/s 2 . 



(b) When v - 15 m/s, the centripetal acceleration is 



a K = 



V 2 (15 m/s) 2 
(500 m) 



= 0.45 m/s 2 . 



EXERCISE F When the speed of the race car in Example 5-8 is 30 m/s. how are 
(a) a u „ and (/>) a K changed? 

These concepts can be used for an object moving along any curved path, 
such as that shown in Fig. 5-16, We can treat any portion of the curve as an arc 
of a circle with a radius of curvature r. The velocity at any point is always 
tangent to the path. The acceleration can be written, in general, as a vector sum 
of two components: the tangential component a lm - Av/At, and the radial 
(centripetal) component n K = v 2 /r l 



FIGURE 5- IB Object following a curved path (solid line). 
At point P the path has a radius of curvature r. The object 
has velocity v, tangential acceleration a^,, (the object is 
increasing in speed), and radial (centripetal) acceleration a K 
(magnitude Mr = V*/r) which points toward the center of 
curvature C. 




PHYSICS APPLIED 

Centrifuge 



Centrifugal on 



A useful device that nicely illustrates circular motion is the centrifuge, ot the 
very high speed ultracentrifuge. These devices are used to sediment materials 
quickly or to separate materials. Test tubes are held in the centrifuge rotor, 
which is accelerated to very high rotational speeds: see Fig. 5-17, where one test 
tube is shown in two positions as the rotor turns. The small green dot represents 
a small particle, perhaps a macro molecule, in a fluid-filled test tube. When the 
tube is at position A and the rotor is turning, the particle has a tendency to 
move in a straight line in the direction of the dashed arrow. But the fluid, 
resisting the motion of the particles, exerts a centripetal force that keeps the 
particles moving nearly in a circle. Usually, the resistance of the fluid (a liquid, a 
gas, or a gel, depending on the application) does not quite equal rm> 2 /r, and the 
particles eventually reach the bottom of the tube. The purpose of a centrifuge is 
to provide an "effective gravity" much larger than normal gravity because of the 
high rotational speeds, thus causing more rapid sedimentation. 



116 CHAPTER 5 Circular Motion; Gravitation 



EXAMPLE 5-9 



Ultracentrifuge. The rotor of an ultracentrifuge rotates at 
50,000 rpm (revolutions per minute). The top of a 4.00-em-long test tube 
(Fig, 5-17) is 6.00ctn from the rotation axis and is perpendicular to it. 
The bottom of the tube is 1 0.00 cm from the axis of rotation, Calculate the 
centripetal acceleration, in "gV at the top and the bottom of the tube, 

APPROACH We can calculate the centripetal acceleration from a n = v 2 /r. 

We divide by g = 9.80 rn/s : to find a n in g\ 

SOLUTION At the top of the tube, a particle revolves in a circle of circumfer- 
ence 2irr, which is a distance 

2tti- = (2tt) (0.0600 m) = 0.377 m per revolution. 

It makes 5,00 x 10 4 such revolutions each minute, or, dividing by 60s/min, 
833 rev/s.The time to make one revolution, the period T. is 



T = 



I 



= 1.20 X 10 - ' s/rev. 



(833 rev/s) 
The speed of the particle is then 

2vr ( 0.377 m /rev 



V = 



r V 1.20 x 10 Vrev 
The centripetal acceleration is 

(3.14 X 10 z m/s) 2 



= 3.14 X 10 2 m/s. 



"< 



0.0600 m 



- 1.64 x IO ft m/s 2 , 



which, dividing by g = 9.80 m/s 2 , is 1.67 x \Q? g'&. 

At the bottom of the tube (V = 0.1000 m), the speed is 



V = 



Then 



«■< 



2vr (277)(0.l000m) 
7" " 1.20 X 10 'Vrcv 

v 2 (523.6 m/s) : 



= 523.6 m/s. 



(0.1000 m) 



-2.74 x lO'm/s 2 
= 2.80 X 10 5 S \ 



oi 280,000 g\ 



Force exerted 
by liquid 




FIGURE 5-17 Two positions of a 
rotating test tube in a centrifuge 
(top view). A I A. the green dot 
represents a macromolecule or 
Other particle being sedimented- It 
would lend lo follow the dashed 
line, heading toward the bottom of 
(he lube, bill the fluid resists this 
motion by exerting a force on the 
particle as shown at point B. 



Newton's Law of Universal Gravitation 

Besides developing the three laws of motion, Sir Isaac Newton also examined 
the motion of the planets and the Moon. Tn particular, he wondered about the 
nature of the force that must act to keep the Moon in its nearly circular orbit 
around the Earth. 

Newton was also thinking about the problem of gravity. Since falling objects 
accelerate, Newton had concluded that they must have a force exerted on them, 
a force we call the force of gravity. Whenever an object has a force exerted on 
it, that force is exerted by some other object. But what exerts the force of 
gravity? Every object on the surface of the Earth feels the force of gravity, and 
no matter where the object is, the Iotcc is directed toward the center of the 
Earth (Fig. 5- IS). Newton concluded thai it must be Ihe Earth itself that exeTts 
the gravitational force on objects at its surface. 

According to legend, Newton noticed an apple drop from a tree. He is said 
to have been struck with a sudden inspiration; If gravity acts at the tops of trees, 
and even at the tops of mountains, Ihen perhaps it acts all the way to the Moon! 



Fl G U R E S- 1 8 Any whe re o n 
fciarlh. whether in Alaska. Peru, or 
Australia, the force of gravity acts 

downward toward the Farth's center. 






y% 



i 



SECTION 5-6 Newton's Law of Universal Gravitation 117 



Tke Moon's 

acceleration 

toward Earth 



Willi this idea that it is the Earth's gTavily that holds Ihe Moor in iLs orbit, 
Newton developed his great theory of gravitation, But there was controversy at 
the time. Many thinkers had trouble accepting the idea of a force "acting at a 
distance" Typical forces act through contact — youT hand pushes a carl and pulls 
a wagon, a bat hits a ball, and so on. Bui gravity acts without contact, said 
Newton: the Earth exeTts a force on a falling apple and on the Moon, even 
though there is no contact, and the two objects may even be very far apart. 

Newton set about determining the magnitude of the gravitational force that 
the EaTth exerts on the Moon as compared to the gravitational force on objects 
at the Earth's surface. The centripetal acceleration of the Moon, as we calcu- 
lated in Example 5-2, is a R = 0.00272 m/s 2 . In terms of the acceleration of 
gravity at the Earth's surface, # = 9.80 m/s 2 , 

0.00272 m/s 2 



Ik " 



9.8 m/s 2 



I 
3600 



That is. the acceleration of the Moon toward the Earth is about ^ as great as 
the acceleration of objects at the Earth's surface. The Moon is 384,000 km from 
the Earth, which is about 60 times the Earth's radius of 6380 km. That is, the 
Moon is 60 times farther from the Earth's center than are objects at the Earth's 
surface. But 60 x 60 = 60 2 = 3600. Again that number 3600. Newton 
concluded that the gravitational force exerted by the Earth on any object 
decreases with the square of its distance r from the Earth's center: 



force of gravity <x 



I 



The Moon is 60 Earth radii away, so it feels a gravitational force only — : = j^j 
times as strong as an equal mass would at the Earth's surface. 

Newton realized that the force of gravity on an object depends not only tin 
distance but also on the ohject's mass, Tn fact, it is directly proportional to its 
mass, as wc have seen. According to Newton's third law, when the Earth exerts 
its gravitational force on any object, such as the Moon, that object exerts an 
equal and opposite force on the Earth (Fig. 5-19). Because of this symmetry, 
Newton reasoned, the magnitude of the force of gravity must be proportional to 
both the masses. Thus 



F x 



m F m 



OI>j 



where m E is the mass of the Earth, m iWj the mass of the other object, and r the 
distance from the Earth's center to the center of the other object. 



FIGURE 5-19 The gravitational force one 
object exerts on a second object is directed 
toward the first object, and (by Newton's third 
law) is equal and opposite to the force exerted 
by the second object on the first. 



Earth 



Moon Q) 

/ Gravitational 
r force exerted uit 
Moon by Earth 




Gravitational force 
exerted on Earth 
hy the Mooti 



Newton went a step further in his analysis of gravity, In his examination of 
the orbits of the planets, he concluded that the force required to hold the planets 
in their orbits around the Sun seems to diminish as the inverse square of their 
distance from the Sun. This led him to believe that it is also the gravitational 
force that acts between the Sun and each of the planets to keep them in theiT 
orbits. And if gravity acts between these objects, why not between all objects? 



118 CHAPTER 5 Circular Motion; Gravitation 



Thus he proposed his law of universal gravitation, which we can slate as follows: 

Every particle in the universe attracts every other particle with a force that 
is proportional to the product of their masses and inversely proportional to 
the square of the distance between them. This force acts along the line 
joining the two particles. 

The magnitude of the gravitational force can he written as 



G 



m t wi 3 



(5-4) 



where m, and m 2 are the masses of the two particles, r is the distance between 
them, and G is a universal constant which must be measured experimentally 
and has the same numerical value for all objects. 

The value of G must be very small, since we are not aware of any force of 
attraction between ordinary-sized objects, such as between two baseballs, The 
force between two ordinary objects was first measured by Henry Cavendish in 
179S, over 100 years after Newton published his law. To detect and measure the 
incredibly small force between ordinary objects, he used an apparatus like that 
shown in Fig. 5-20, Cavendish confirmed Newton's hypothesis that two objects 
attract one another, and that Eq. 5-4 accurately describes this force, In addition, 
because Cavendish could measure F, m t , m 2 , and r accurately, he was able to 
determine the value of the constant G as well. The accepted value today is 

G = 6.67 x l(T ll N-m7kg 2 . 

[Strictly speaking, Eq. 5-4 gives the magnitude of the gravitational knee 
that one particle exerts on a second particle that is a distance r away. For an 
extended object (that is, not a point), we must consider how to measure the 
distance r. This is often best done using integral calculus, which Newton himself 
invented. Newton showed that for two uniform spheres, Eq. 5-4 gives the 
correct force where r is the distance between their centers. When extended 
objects are small compared to the distance between them (as for the Earth-Sun 
system), little inaccuracy results from considering them as point particles.] 



EXAMPLE 5-10 ■tf.-UhmU 



Can you attract another person gravita- 
tionally? A 50-kg person and a 75-kg person arc sitting on a bench, Estimate 
the magnitude of the gravitational force each exerts on the other, 

APPROACH This is an estimate: we let the distance between the people be 



:m, and round off G to 1 ~ L " N-m : /kg 2 . 
SOLUTION We use Eq. 5-4: 



I G 



. ra, m 2 (I0" ia N-m 2 /kg 2 )(50 kg) (75 kg) 



~ ia~*N, 



(0.5 m) 2 
which is unnoticenbly small unless very delicate instruments are used. 



NEWTON'S 

LAW 

OF 

UNIVERSAL 

GRAVITATION 




Light 
source 
narrow beam) 



FIGURE 5-20 Schematic diagram of Cavendish's apparatus. 
Two spheres are attached to a lightweight horizontal rod. 
which is suspended at its center hv a thin fiber. When a third 
sphere labeled A is brought close to one of (he suspended 
spheres, the gravitational force causes the latter to move, and 
this twists the fiber slightly. The tiny movement is magnified by 
the use of a narrow light beam directed at a mirror mounted 
on the fiber, The beam reflects onto a scale. Previous determi- 
nation of how large a force will twist the fiber a given amount 
then allows one to determine the magnitude of the gravita- 
tional force between two objects. 



SECTION 5-6 Newton's Law of Universal Gravitation 119 



^y 



Motion 




FIGURE 5-21 Rxample5-ll. 



EXAMPLE 5-11 



Spacecraft at 2r E . What is the force of gravity acting on 
a 2000-kg spacecraft when it orbits two Earth radii from the Earth's center 
(that is, a distance r E = 6380 km above the Earth's surface. Fig. 5-21)? The 
mass of the Earth is M E = 5.98 X I0 34 kg. 

APPROACH We could plug alt the numbers into Eq. 5-4, but there is a 
simpler approach. The spacecraft is twice as far from the Earth's center as 
when it is at the surface of the Earth. Therefore, since the force of gravity 
decreases a* the square of the distance (and p = 5), the force of gravity on the 
satellite will be only one -fourth its weight at the Earth's surface. 

SOLUTION At the surface of the Earth, F c - mg. At a distance from the 
Earth's center of 2r ]i , F G is 5 as great: 



Fc ~ \mg - { (2000 kg}(9.80 m/s 2 ) 
= 4900 N. 



FIGURE 5-22 Example 5-12. 
Orientation of Sun (S). Earth (E), 
and Moon (M ) at right angles to 
each other (not to scale). 



MiMin 



hJ 



ML 



Earth 



O 



Sun 



EXAMPLE 5-12 



Force on the Moon. Find the net force on the Moon 

(m M = 7,35 x IO 22 kg) due to the gravitational attraction of both the Earth 
(m E = 5.98 X I0 24 kg) and the Sun (m s = I.99 X 10 ;5 "kg), assuming they are 
at right angles to each other as in Fig, 5-22. 

APPROACH The forces on our object, the Moon, are the gravitational force 
exerted on the Moon by the Earth F MH and that exerted by the Sun F MS , as 
shown in the free-body diagram of Fig, 5-22. We use the law of universal 
gravitation to find the magnitude of each force, and then add the two forces 
as vectors. 

SOLUTION The Earth is 3,84 x l(f km - 3.84 x I0 s m from the Moon, so 
f ME (the gravitational force on the Moon due to the Earth) is 



f'vii-- — 



(6.67 x IO""N-mVkg 2 )(7.35 x I0 22 kg)(5,98 x I0"kg) 



(3.84 X I0 s m) : 



= 1. 99 X 10 2L1 N, 



The Sun is 1 .50 X I if km from the Earth and the Moon, so F M5 (the gravita- 
tional force on the Moon due to the Sun) is 



(6.67 X IO-"N-m 2 /kg 2 )(7.35 X I0 22 kg)(t.99 X l0 M kg) 



'MS 



(1.50 x lfj"m) 2 



4.34 x l0 2l 'N. 



The two forces act at Tight angles in the case we are considering (Fig. 5-22), 
so we can apply the Pythagorean theorem to find the magnitude of the 
total force: 



F = V'0.99 X I0 20 N) 2 + (4.34 X !0 2t, N) 2 = 4.77 x I0 2IJ N. 
The force acts at an angle (Fig. 5-22) given by 6 = tan"' (1.99/4,34) = 24,6° 



CAUTION 



Distinguish between 

Newton '$ second law and 

the fywofami <t rsa I gra vitetwn 



Don't confuse the law of universal gravitation with Newton's second law of 
motion, SF = >na. Tie former describes a particular force, gravity, and how its 
strength varies with the distance and masses involved- Newton's second law, on 
the other hand, relates the net foTce on an object (i.e., the vector sum of all the 
different forces acting on the object, whatever their sources) to the mass and 
acceleration of that object. 



120 CHAPTER 5 Circular Motion; Gravitation 




Gravity Near the Earth's Surface; 
Geophysical Applications 

When Eq. 5-4 is applied lo Ihe gravitational force between the Earth and an 
object at its surface, m becomes the mass of the Earth m^,rrt 2 becomes the 
mass of the object m, and r becomes the distance of the object from the Earths 
centeT, which is the Tadius of the EaTlh r v . This force of gravity due to the 
Earth is the weight of the object, which we have been writing as mg. Thus, 



mg 



a- 



We can solve this for g, the acceleration of gravity at the Earth's surface: 



(5 -S ) gin terms of G 



Thus, the acceleration of gravity at the surface of the Earth,?,', is determined by 
m K and r E . (Don't confuse G with g: they arc very different quantities, but are 
related by Eq, 5-5.) 

Until G was measured, the mass of the Earth was not known. But once G 
was measured, Eq. 5-5 could be used to calculate the Earth's mass, and 
Cavendish was the first to do so. Since g = 9.80 m/s 2 and the radius of the 
Earth is r L = 6.3S x 10" m, then, from Eq. 5-5, we obtain 

(9.80 m/s 2 )( 6.38 X 10" m^ 



d 



m E = 



gn 



6.67 x IO- ll N-m 2 /kg 2 



= 5-98 x 1 2A kg 



for the mass of the Earth, 

Equation 5-5 can be applied to other planets, where g, m, and r would refer 

to that planet, 



EXAMPLE 5-13 



ESTIMATE - ] Gravity on Everest. Estimate the effective 
value of g on the top of M t. Everest, 8850 m (29,035 ft) above sea level. That is, 
what is the acceleration due to gravity of objeets allowed to fall freely at this 
altitude? 

APPROACH The force of gravity (and the acceleration due to gravity g) 
depends on the distance from the center of the Earth, so there will be an effec- 
tive value g' on top of Mt. Everest which will be smaller than g at sea level. We 
assume the Earth is a uniform sphere (a reasonable "estimate"). 
SOLUTION We use Eq. 5-5, with r K replaced by r = 6380 km + 8.9 km = 
6389 km = 6.389 x I0 6 m: 

,Ti K (6.67 x 10 " N>m7kg 2 )(5.98 x 1 i4 kg) 

g = r ' — = ; — — j = 9.77 m/s-, 

r (6.389 X 10* m) 2 

which is a reduction of about 3 parts in a thousand (0.3%). 

NOTE This is an estimate because, among other things, we ignored the mass 
accumulated under the mountaintop. 



Note that Eq. 5-5 does not give precise values for g at different locations 
because the Earth is not a perfect sphere. The Earth not only has mountains and 
valleys, and bulges at the equator, but also its mass is not distributed precisely 
uniformly (see Table 5-1). The Earth's rotation also affects the value of g. 
However, for most practical purposes, when an object is near the Earth's surface, 
we will simply use g = 9.80 m/s 3 and write the weight of an object as mg- 

'That the distance is measured from the Earth's center docs not imply that the force of gravity 
somehow emanates from that one point. Rather, all parts of the Earth attract gravitatioiially, but the 
net effect is a force acting toward the F.arth's center. 



<$ > C A U T I O M 
Disii^gaisk G fronts 



Mass of the Earih 



TABLE 5-1 

Acceleration Due to Gravity 

at Various Locations on Earth 


Location 


Elevation 
(m> 


S 
(m/s ) 


New York 







9.803 


San Francisco 







9,800 


Denver 


1650 




9.796 


Pikes Peak 


4300 




9.789 


Sydney, 
Australia 







9.79S 


Equator 


o 




9,780 


North Pole 
(calculated) 


D 




9.832 



SECTION 5-7 Gravity Near the Earth's Surface; Geophysical Applications 121 



PHYSICS APPLIED 



Geology- 



-mineral arid oil 
exploration 



The value of g can vary locally on the Earth's surface because of the presence 
of irregularities and rocks of different densities- Such variations in g, known as 
"gravity anomalies," are very small — on the OTder of I part per I0 n or 1 7 in the 
value of g. Rut they can be measured by "gravimeters" which delect variations in g 
to 1 part in I CI 1 *. Geophysicisls use such measurements as paTt of their investiga- 
tions into the structure of the Earth's crust, and in mineral and oil exploration. 
Mineral deposits, for example, often have a greater density than does surrounding 
material- Because of the greater mass in a given volume, g can have a slightly 
greater value on top of such a deposit than at its flanks. "Salt domes,'' under which 
petroleum is often found, have a lower than average density; searches for a slight 
reduction in the value of g in certain locales have led to the discovery of oil. 



PHYSICS APPLIED 

Artificial Earth satellites 



Satellites and "Weightlessness' 



Satellite Motion 

Artificial satellites circling the Earth are now commonplace (Fig, 5-23). A satel- 
lite is put into orbit by accelerating it to a sufficiently high tangential speed with 
the use of rockets, as shown in Fig. 5-24. If the speed is too high, the spacecraft 
will not be confined by the Earth's gravity and will escape, never to return. If 
the speed is too low, it will return to Earth. Satellites arc usually put into circular 
(or nearly circular) orbits, because such orbits require the least takeoff speed. 




27,000 krn/h 
circular 



30,000 kin/h 

L'lliplilJIll 



. 40.000 krn/h 
,~***i-^ I I escape 




FIGURE 5-23 A satellite circling the F.arth. 



FIGURE 5-24 Artificial satellites launched at 
different speeds, 



It is sometimes asked: "What keeps a satellite up?" The answer is: its high 
speed. If a satellite stopped moving, it would fall directly to Earth. But at the 
very high speed a satellite has, it would quickly fly out into space (Fig. 5-25) if 
it weren't for the gravitational force of the Earth pulling it into orbit, In fact, a 
satellite is falling (accelerating toward Earth), but its high tangential speed 
keeps it from hitting Earth. 

Without 
gravity 



WithN 



FIGURE 5-25 A moving satellite "falls" out of a 

straight-line path toward the HaTth. 




122 CHAPTER 5 Circular Motion; Gravitation 



PHYSICS APPLIED 



EXAMPLE 5-14 



Geosynchronous satellite. A geosynchronous satellite 
is one that stays above the same point on the Earth, which is possible only if it 
is above a point on the equator. Such satellites are used for TV and radio 
transmission, for weather forecasting, and as communication relays. Determine 

(a) the height above the Earth's surface such a satellite must orbit, and 

(b) such a satellite's speed, (c) Compare to the speed of a satellite orbiting 
200 km above Earth's surface, 

APPROACH To remain above the same point on Earth as the Earth rotates, 
the satellite must have a period of one day, We can apply Newton's second law, 
F = ma, where a = v 2 /r if we assume the orbit is circular, 

SOLUTION (a) The only force on the satellite is the force of universal gravita- 
tion. So Eq. 5-4 gives us the force F, which we insert into Newton's second law: 

F - ma 

G 5 — - w Sal — [satellite equation | 

r '' 

This equation has two unknowns,;' and v. But the satellite revolves around the 
Earth with the same period that the Earth rotates on its axis, namely once in 
24 ho uts. Thus the speed of the satellite must be 

1-srr 

v = -r—- 
T 

where T - I day - (24h)(3600 s/h} - 86,400s. We substitute this into the 
"satellite equation" above and obtain (after canceling m Sal on both sides) 

m E _ (2nrf 

After cancelling an r, we can solve for r~: 

Gm B T 2 (6.67 x 10 " N-mVkg 2 ){5.9S x 10 M kg)(86,400s) 2 



4tt 2 4tt 2 

= 7.54 X 1 » 22 m \ 

Taking the cube root, we get r = 4.23 X 10 7 m, or 42,300 km from the Earth's 
center. We suhtract the Earth's radius of 6380 km to find that a geosynchronous 
satellite must orhit about 36,000 km (about 6 r e ) above the Earth's surface. 
(b) We solve for v in the satellite equation given in part (a): 



[Gn^, 1(6.67 X 10- JI N'm : /kg : )(5.98 X I0 24 kg) 

if - -J \ = — : 3070 m/s. 

V r V (4.23 x 10 ? m) 

We get the same result if we use v = 2-n-r/T. 

(t) The equation in part (bj for v shows v en \f~\jr. So for r = r K + h = 

6380 km + 200 km = 6580 km, we get 

[7 /(42.300 km) 

^^ r (3070W^^ = ™ m /s. 

NOTE The center of a satellite orbit is always at the center of the Earth; so it is 
not possible to have a satellite orbiting above a fixed point on the Earth at any 
latitude other than 0". 



EXERCISE G Two satellites orbit the Earth in circular orbits of the same radius. One 
satellite is twice as massive as the other. Which of the following statements is true 
about the speeds of these satellites'? (a) The heavier satellite moves twice as fast as the 
lighter one, (b) The two satellites have the same speed, (c) "I he lighter satellite moves 
twice as fast as the heavier one. (d) The heavier satellite moves four times as fast as the 
lighter one. 



SECTION 5-8 Satellites and "Weightlessness" 123 



FIGURE 5-26 (a) An object in an 
elevator a I rest exerts a forte on a 
spring scale equal to its weight. 
(It) In an elevator accelerating 
upward at \g. the object's apparent 
weight is 1 i times larger than its 
true weight, (c) In a freely falling 
elevator, the object experiences 
"weightlessness": the scale reads zera 




(b} 1 f a-ifK (up) (c) f a = j?(ik>wn> 



"Weightlessness" in a 

fatting elevator 



FIGURE 5-27 Experiencing 
weightlessness on F.arth. 




Weightlessness 

People and other objects in a satellite circling the Earth arc said to experience 
apparent weightlessness. Let us first look at a simpler case, that of a falling 
elevator, fn Fig. 5-26a, an elevator is at rest with a bag hanging from a spring 
scale. The scale reading indicates the downward force exerted on it by the bag, 
Tbis force, exerted on the scale, is equal and opposite to the force exerted by 
the scale upward on the bag, and we call its magnitude w. Two forces act on the 
bag: the downward gravitational force and the upward force exerted by tbe scale 
(Newton's third law) equal to w. Because the bag is not accelerating, when we 
apply 2F - ma to the bag in Fig. 5-26a we obtain 

ib - mg — 0, 

where mg is the weight of the bag. Thus, w = mg, and since the scale indicates 
the force w exerted on it by the bag, it registers a force equal to the weight of 
the bag, as we expect. 

ff, now, the elevator has an acceleration, a, then applying XF = ma to the hag, 
we have 

w - mg - ma. 

Solving for w, wc have 

IP = mg + ma, [a is + upward] 

We have chosen the positive direction up. Thus, if the acceleration a is up, a is 
positive; and the scale, which measures w, will read more than mg. We call w the 
apparent weight of the bag, which in this case would be greater than its actual 
weight (mg). Ff the elevator accelerates downward, a will be negative and u\ the 
apparent weight, will be less than mg. The direction of the velocity v doesn't 
matter. Only the direction of the acceleration a. influences the scale reading. 
Suppose, for example, the elevator's acceleration is jg upward; then we find 

w - mg + m{\g) - \mg. 

That is, the scale reads I { times the actual weight of the bag (Fig. 5-26b). The 
apparent weight of the bag is 1 j times its real weight. The same is true of the 
person: her apparent weight (equal to the normal force exerted on her by the 
elevator floor) is I \ times her real weight. We can say that she is experiencing 
1 j#'s, just as astronauts experience so many #'s at a rocket's launch. 

If, instead, the elevator's acceleration is a - -jg (downward), then 
ic = mg — jmg - j mg, That is, the scale reads half the actual weight. If the 
elevator is in free fall (for example, if the cables break), then a = -g and 
w - mg — mg - 0. The scale reads zero. See Fig. S-26c, The bag appears 
weightless. If the person in the elevator accelerating at -g let go of a pencil, say, 
it would not fall to the floor. True, the pencil would be falling with acceleration g. 
But so would the floor of the elevator and the person. The pencil would hover 
Tight in front of the person. This phenomenon is called apparent weightlessness 
because in the reference frame of the person, objects don't fall or seem to have 
weight — yet gravity does not disappear. It is still acting on the object, whose 



12fl CHAPTER 5 Circular Motion; Gravitation 



weight is still mg. The objects seem weightless only because the elevator is in 
free tall, and there is no contact force to make us feel the weight. 

The "weightlessness" experienced by people in a satellite orbit close to the 
Earth is the same apparent weightlessness experienced in a freely falling 
elevator. Ft may seem strange, at first, to think of a satellite as freely falling. But 
a satellite is indeed falling toward the Earth, as was shown in Fig. 5-25, The 
force of gravity causes it to "fall" out of its natural straight line path. The accel- 
eration of the satellite must be the acceleration due to gravity at that point, 
since the only force acting on it is gravity. Thus, although the force of gravity 
acts on objects within the satellite, the objects experience an apparent weight- 
lessness because they, and the satellite, are all accelerating as in free fall. 

Figure 5-27 shows some examples of "free fall; 1 or apparent weightlessness, 
experienced by people on Earth for hrief moments. 

A different situation occurs when a spacecraft is out in space far from the 
Earth, the Moon, and other attracting objects. The force of gravity due to the 
Earth and other heavenly bodies will then be quite small because of the distances 
involved, and people in such a spacecraft will experience real weightlessness. 

Kepler's Laws and Newton's Synthesis 

More than a half century before Newton proposed his three laws of motion and 
his law of universal gravitation, the German astronomer Johannes Kepler 
(1571-1630) had worked out a detailed description of the motion of the planets 
about the Sun: three empirical findings that we now refer to as Kepler's laws of 
planetary motion. They are summarized as follows, with additional explanation 
in Figs. 5-28 and 5-29. 

Kepler's first law: The path of each planet about the Sun is an ellipse with 
the Sun at one focus (Fig. 5-28). 

Kepler's second law: Each planet moves so that an imaginary line drawn 
from the Sun to the planet sweeps out equal areas in equal periods of time 
(Fig. 5-29). 

Kepler's third law. The ratio of the squares of the periods 7" of any two planets 
revolving about the Sun is equal to the ratio of the cubes of their mean 
distances a from the Sun: (7"|/T 2 ) 2 = (*iA;) 3 . [Actually, s is the semimajor 
axis, defined as half the long (major) axis of the orbit, as shown in Fig. 5-28. 
We can also call it the mean distance of the planet from the Sun,| Present-day 
data are given in Table 5-2: see the last column, 

Kepler arrived at his laws through careful analysis of experimental data. 
Fifty years later, Newton was able to show that Kepler's laws could he derived 
mathematically from the law of universal gravitation and the laws of motion. 
Newton also showed that for any reasonable form for the gravitational farce 
law, only one that depends on the inverse square of the distance is fully consis- 
tent with Kepler's laws. He thus used Kepler's laws as evidence in favor of his 
law of universal gravitation, Eq. 5-4. 



TABLE 5-2 


Planetary Data Applied 


to Kepler's Third Law 




Planet 


Mean Distance 

from Sun, s 

(111 6 km) 


Period, T 
(Earth years) 


i-VT 2 
(lO^kmVy 2 ) 


Mercury 


57.9 


0.241 


3.34 


Venus 


108.2 


0.615 


3.35 


Earth 


149.6 


1.0 


3.35 


Mars 


227.9 


1.88 


3.35 


Jupiter 


778.3 


11.86 


3.35 


Saturn 


1427 


29.5 


3,34 


Uranus 


2870 


84.0 


3.35 


Neptune 


4497 


165 


3.34 


Pluto 


5900 


24S 


3.34 



"Weightlessness" in u satellite 



Planet 




FIG U RE 5 -28 ( a ) Kepler 's first 
krai. An ellipse is a closed curve 
such [hat the sum of the distances 
from any point P on the curve to 
two fixed points (called the foci. F| 
and F 2 ) remains constant. That is, 
the sum of the distances, 
H[ P +■ ¥2 P. is the same for all 
points on the curve, A circle is a 
special ease of an ellipse in which 
the two foci coincide, at the center 
of the circle. 

FIGURE 5-29 Kepters second taw. 
The two shaded regions have equal 
areas. 'Ifie planet moves from point 1 
to point 2 in the same time as it takes 
to move from point 3 to point 4. 
Planets move fastest in that part of 
their orbit where they are closest to 
the Sun. Exaggerated scale. 




^SECTION 5-9 125 



Derivation of We will derive KepleT's third law for the special case of a circular orbit. 

Kepters third kiw (Most planetary orbits are close to a circle.) First, we write Newton's second law 
of motion, J,F - ma. For F we use the gravitational force (Eq. 5-4) between 
the Sun and a planet of mass m, , and for a the centripetal acceleration, (i ! /r,We 
assume the mass of the Sun, M s , is much greater than the mass of its planets. 
Then 

Y.F = ma 



G — -., — = m 
n 



vi 



Here r , is the distance of one planet from the Sun, and i\ is its average speed in 
orbit; M s is the mass of the Sun, since it is the gravitational attraction of the Sun 
that keeps each planet in its orbit. The period 7", of the planet is the time 
required for one complete OTbit, a distance equal to the circumference of 
its orbit, 2irr, . Thus 

27TJ"| 



Kepler's third itnr 



<I>- 



CAUTION 



Compare orbits of objects 
onlv around the same tenter 



We substitute this formula torn into the equation above; 

A-n 3 r. 



i ■ 



a— ^ = 

r\ 

We rearrange this to get 
T\ _ 4tt : 

We derived this for planet 1 (say, Mars). The same derivation would apply for a 

second planet (say, Saturn) orbiting the Sun, 



(S-6a) 



>: 



4ir 2 
GM< 



where 7, and t 2 arc the period and orbit radius, respectively, for the second 
planet. Since the right sides of the two previous equations arc equal, we have 
T\fr\ = Tjfrl or, rearranging, 



%y 



I 



(5-fih) 



which is Kepler's third law. 

The derivations of Eqs. 5 -6a and 5-6b (Kepler's third law) compared 
two planets revolving around the Sun; but they are general enough to be 
applied to other systems. For example, we could apply Eq. 5-6a to our Moon 
revolving around Earth (then M$ would be M v , the mass of the Earth). Or we 
eould apply Eq. 5-6b to compare two moons revolving around Jupiter. But 
Kepler's third law applies only to objects orhiting the same attracting center. 
Do not use Eq. 5-6b to compare, say, the Moon's orbit around the Earth to 
the orbit of Mars around the Sun hecause they depend on different 
attracting centers. 

In the following Examples, we assume the orbits are circles, although it is 
not quite true in general. 



EXAMPLE 5-15 



Where is Mars? Mars' period (its "year") was noted 
by Kepler to be about 687 days (Earth days), which is (687 d/365 d) - 1.88 yr. 
Determine the distance of Mars from the Sun using the Earth as a reference. 

APPROACH We know the periods of Earth and Mars, and the distance from 
the Sun to Earth. We can use Kepler's third law to obtain the distance 
from the Sun to Mars. 



126 CHAPTERS Circular Motion; Gravitation 



SOLUTION The period of the Earth is 7^ = 1 yr, and the distance of Earth 
from the Sun is r ts = 1,50 x 10" m. From Kepler's third law (Eq, 5-6b); 



r MS 

>ks 



1 .88 yr 
I yr 



1.52. 



So Mars is 1,52 times the Earth's distance from the Sun, or 2,28 x 10" m. 



EXAMPLE 5-16 



The Sun's mass determined. Determine the mass of 
the Sun given the Earth's distance from the Sun as r ES = 1.5 X IO Ll m. 

APPROACH Equation 5-6a relates the mass of the Sun M s to the period and 
distance of any planet. We use the Earth. 

SOLUTION The Earth's period is 7 K = I yr = (36S|d)(24 h/d)(3600s/h) = 
3.16 X 10 7 s. We solve Eq. 5-6a for M s ; 



■V/, 



4jt 2 (I.5 x 10" m) 3 



A—l 3 

4n n-s = 

G7| (6.67 x 10 _11 N-m 2 /kg 2 )(3.1rj x I0 7 sf 



2.0 x 1 ;! " kg, 



Accurate measurements on the orbits of the planets indicated that they did 
not precisely follow Kepler's laws. For example, slight deviations from perfectly 
elliptical orbits were observed. Newton was aware that this was to be expected 
because any planet would be attracted gravitationally not only by the Sun but 
also (to a much lesser extent) by the other planets. Such deviations, or 
perturbations, in the orbit of Saturn were a hint that helped Newton formulate 
the law of universal gravitation, that all objects attract gravitationally. Observa- 
tion of other perturbations later led to the discovery of Neptune and Pluto. 
Deviations in the orbit of Uranus, for example, could nut all he accounted for 
by perturbations due to the other known planets. Careful calculation in the 
nineteenth century indicated that these deviations could be accounted for if 
another planet existed farther out in the solar system. The position of this planet 
was predicted from the deviations in the orbit of Uranus, and telescopes focused 
on that region of the sky quickly found it; the new planet was called Neptune. 
Similar but much smaller perturbations of Neptune's orbit led to the discovery 
of Pluto in 1930. 

Starting in the mid-1990s, planets revolving about distant stars (Pig, 5-30) 
were inferred from the regular "wobble'" of each star due to the gravitational 
attraction of the revolving planet (s). 

The development by Newton of the law of universal gravitation and the 
three laws of motion was a major intellectual achievement: with these laws, he 
was able to describe the motion of objects on Earth and in the heavens. The 
motions of heavenly bodies and objects on Earth were seen to follow the same 
laws. For this reason, and also because Newton integrated the results of earlier 
scientists into his system, we sometimes speak of Newton's synthesis. 



K&L 



PHYSICS APPtlED 



Determining the 
Sun's mass 



Perturbations and 

discovery of planets 



Planets around 
other stars 

Newton s 
synthesis 



(a) 



San 




47 
(b) Ursae 

Majoris ' 






Jupiter 



M\ 



Planet 

© 

3M } 



FIG U RE 5-30 Our solar system (a) 

is. compared to recently discovered 
planets orbiting (V) the star 47 Ursae 
Majoris and (c) the star Upsilon 
Andromcdae with at least three 
planets, Afj is the mass of Jupiter, 
(Sizes not to scale) 



■;,) 



Upsilon 
Aiidroinedae 




0JM, im } 



4M, 



♦SECTION 5-9 Kepler's Laws and Newton's Synthesis 127 



Causality The laws formulated by Newton are referred to as causal laws. By causality 

we mean ihe idea that one occurrence can cause another. When a rock strikes a 
window, we infer lhal the Tock causes the window to break. This idea of "cause 
and effect" relates to Newton's laws: Ihe acceleration of an object was seen to 
be caused by Ihe net force acting on it. 



5-10 



Types of Forces in Nature 



Eiectroweak and GUT 



Everyday forces 
are gravity and 
electromagnetic 



We have already discussed that Newton's law of universal gravitation, Eq. 5-4, 
describes how a particular type of force — gravity — depends on the masses of the 
objects involved and the distance between them, Newton's second law, 2P - ma, 
on the other hand, tells how an object will accelerate due to any type of force, 
But what are the types of forces that occur in nature besides gravity? 

Tn the twentieth century, physicists came to recognize four fundamental 
forces in nature: ( 1 ) the gravitational force; (2) the electromagnetic force (we 
shall see later that electric and magnetic forces are intimately related); (3) the 
strong nuclear foTce; and (4) the weak nuclear force. In this Chapter, we 
discussed the gravitational force in detail, The nature of the electromagnetic 
force will be discussed in Chapters 16 to 22. The strong and weak nuclear forces, 
which are discussed in Chapters 30 to 32, operate at the level of the atomic 
nucleus; although they manifest themselves in such phenomena as radioactivity 
and nuclear energy, they are much less obvious in our daily lives. 

Physicists have been working on theories that would unify these four 
forces — that is. to consider some or all of these forces as different manifesta- 
tions of the same basic force. So far, the electromagnetic and weak nuclear 
forces have been theoretically united to form eiectroweak theory, in which the 
electromagnetic and weak forces are seen as two different manifestations of a 
single eiectroweak force. Attempts to further unify the forces, such as in /•rand 
unified theories (GUT), arc hot research topics today. 

But where do everyday forces fit into this scheme:' Ordinary forces, other 
than gravity, such as pushes, pulls, and other contact forces like the normal force 
and friction, are today considered to be due to the electromagnetic force acting 
at the atomic level, For example, the force your fingers exert on a pencil is the 
result of electrical repulsion between the outer electrons of the atoms of your 
finger and those of the pencil. 



| Summary 



An object moving in a circle of radius r with constant speed V 
is said to be in uniform circular motion. It has a centripetal 
acceleration u R that is directed radially toward the center of 
[he circle (also called radial acceleration), and has magnitude 



r 



(5-1) 



The direction of the velocity vector and that of the accelera- 
tion a R are continually changing in direction, but are perpen- 
dicular to each other at each moment. 

A force is needed to keep a particle revolving in a circle, 
and the direction of this force is toward the center of the 
circle. This force may be due to gravity, to tension in a cord, to 
a component of the normal force, to another type of force, or 
to a combination of forces. 

|*When the speed of circular motion is not constant, the 
acceleration has two components, tangential as well as 
centripetal] 

Newton's law of universal gravitation states that even 1 
particle in the universe attracts every other particle with a force 



proportional to the product of their masses and inversely 
proportional to the square of the distance he t ween them: 



, mi "?; 



(S 4) 



The direction of this force is along the line joining the two 

particles, It is this gravitational force that keeps the Moon 
revolving around the barlh. and the planets revolving around 
the Sun. 

Satellites revolving around the Farth are acted on by 
gravity, but "stay up" because of their high tangential speed. 

[*Newton's three laws of motion, plus his law of universal 
gravitation, constituted a wide-ranging theory of the universe. 
With them, motion of objects on Earth and in the heavens 
could be accurately described. And they provided a theoret- 
ical base for Kepler's laws of planetary motion. | 

The four fundamental forces in nature are (1) the gravi- 
tational force, (2) electromagnetic force, (3) Strong nuclear 
force, and (4) weak nuclear force, The first two fundamental 
forces are responsible for nearly all "everyday'" forces. 



128 CHAPTERS Circular Motion; Gravitation 



I Questions 



I. Sometimes people say that water is removed from clothes 
in a Spin dryer by centrifugal force throwing the water 
outward, What is wrong with this statement? 

1. Win the acceleration of a car be the same when the car 
travels around a sharp curve at a constant fit) km/h as 
when it travels around a gentle curve at Ihe same speed? 
Explain. 

3*. Suppose a car moves at constant speed along a hilly road. 
Where does the car exert the greatest and least forces on 
the road; (a) at the lop of a hill. (6) at a dip between two 
hills, (c) on a level stretch near the bottom of a hill? 

4. Describe all the forces acting on a child riding a horse on 
a merry-go-round. Which of these forces provides the 
centripetal acceleration of the child? 

5. A bucket of water can be whirled in a vertical circle 
without the water spilling out, even at the top of the circle 

when the bucket is upside down. Explain. 

6. How many "accelerators" do you have in your car? There 
are at least three controls in the car which can be used to 

cause the car to accelerate. What are they? What acceler- 
ations do they produce"? 

7. A child on a sled comes flying over the crest of a small 
hill, as shown in Fig, 5-31, His sled does not leave the 
ground (he does not achieve "air"'), but he feels the 
normal force between his chest and the sled decrease as 
he goes over the hill, Explain this decrease using 
Newton's second law. 



UK 



II. 




,1 



FIGURE 5-31 Ouestion7. 

Why do bicycle riders lean inward when rounding a curve 
at high speed? 

Why do airplanes bank when they turn? How would you 
compute the banking angle given its speed and radius 
of the turn? 

A girl is whirling a ball on a siring around her head in a 
horizontal plane. She wants to let go at precisely the right 
time so that the ball will hit a target on the other side of 
the yard, When should she lei go of the siring? 

Does an apple exert a gravitational force on the Earth? If 
so, how large a force? Consider an apple [a) attached to a 
tree, and (b) falling. 



12. If the Earth's mass were double what it is, 
would the Moon's orbit be different? 



in what ways 



13. Which pulls harder gravitationally, the Earth on the Moon, 
or the Vloon on the Earth? Which accelerates more? 

14. The Sun's gravitational pull on the Earth is much larger 
than Ihe Moon's, Yet the Moon's is mainly responsible for 
the tides. Explain, \ffint: Consider the difference in gravi- 
tational pull from one side of the Earth to the other. | 

15. Will an object weigh more at the equator or at the poles? 
What two effects aTe at work? Do they oppose each other? 

16. The gravitational force on the Moon due to the Earth is 
only about half the force on the Moon due to the Sun. 
Why isn't the Moon pulled away from the Earth? 

17. Is Ihe centripetal acceleration of Mars in its orbit around 
the Sun larger or smaller than the centripetal acceleration 
of the Earth? 

IS. Would it require less speed to launch a satellite 
(a) toward the east or (6) toward the west? Consider the 
Earth's rotation direction. 

19. When will your apparent weight be the greatest, as 
measured by a scale in a moving elevator: when the 
elevator (a) accelerates downward, (b) accelerates upward, 
(t) is in free fall, (d) moves upward at constant speed? In 
which case would your weight be the least? When would it 
be the same as when you are on the ground? 

2U, What keeps a satellite up in its orbit around the Earth? 

21. Astronauts who spend long periods in outer space could 
be adversely affected by weightlessness. One way to simu- 
late gravity is to shape the spaceship like a cylindrical 
shell that rotates, with the astronauts walking on the 
inside surface (Fig. 5-32). Explain how this simulates 
gravity. Consider (it) how objects fall, (h) the force we feel 
on our feet, and (c) any other aspects of gravity you can 
think of. 




FIGURE 5-32 Question 21 and Problem 45 

Questions 129 



22. Explain how a runner experiences "free fall" or "apparent 
weightlessness" between steps. 
* 23. The Earth moves faster in its orbit around the Sun in 
January than in July. Is. the Earth closer to the Sun in 
January or in July? Fxplain. [Note: This is not much of a 
factor in producing the seasons — the main factor is the 
tilt of the Earth's axis relative to the plane of its orbit- J 



a: 24, The mass of Pluto was not known until it was discovered 
to have a moon, Explain how this discovery enahled an 
estimate of Pluto's mass. 



| Problems 



5-1 to 5-3 Uniform Circular Motion; Highway Curves 

1. (I) A child sitting 1.10 m from the center of a merry-go- 
round moves with a speed of 1.25 m/s. Calculate (a) the 
centripetal acceleration of the child, and (b) the net hori- 
zontal force exerted on the child (mass = 250 kg). 

2. (I) A jet plane traveling 1890 km/h (525 m/s) pulls out of 
a dive by moving in an arc of radius 6,00 km. What is the 
plane's acceleration in g's? 

3. (I) Calculate the centripetal acceleration of the Earth in 
its orbit around the Sun, and the net force exerted on the 
Earth. What exerts this force on the Earth? Assume that 
the Earth's orbit is a circle of radius 1 .50 x 1U IL m. [Hint: 
see the Tables inside the front cover of this book.] 

4. (T) A horizontal force of 21UN is exerted on a 2.0-kg 
discus as it rotates uniformly in a horizontal circle (at 
arm's length) of radius 0.90m. Calculate the speed of the 
discus. 

5. (II) Suppose the space shuttle is in orbit 400 km from the 
Earth's surface, and circles the Earth about once every 

90 minutes. Find the centripetal acceleration of the space 
shuttle in its OTbit. Express your answer in terms of g, the 
gravitational acceleration at the Earth's surface. 

6. (II) What is the magnitude of the acceleration of a speck 
of clay on the edge of a potter's wheel turning at 45rpm 
(revolutions per minute) if the wheel's diameter is 32 cm? 

7. (II) A ball on the end of a string is revolved at a uniform 
rate in a vertical circle of radius 72.0cm, as shown in 
Fig. 5-33. If its speed is 4.00 m/s and its mass is (J. 300 kg, 
calculate the tension in the string when the ball is (a) at 
the top of its path, and (b) at the bottom of its path. 




in 



K. (II) A O-45-kg ball, attached to the end of a horizontal 
cord, is rotated in a circle of radius 1,3 m on a (riclionless 
horizontal surface. If the cord will break when the tension 
in it exceeds 75 N. what is the maximum speed the ball 
can have? 

y. (II) What is the maximum speed with which a IU5U-kg car 
can round a turn of radius 77 m on a flat road if the coef- 
ficient of static friction between tires and road Is 0.80? Is 
this result independent of the mass of the car? 

(II) How large must the coefficient of static friction be 
between the tires and the road if a car is to round a level 
curve of radius 85 m at a speed of 95 km/h? 

11. (II) A device for training astronauts and jet fighter pilots 
is designed to rotate a trainee in a horizontal circle of 
radius 12.0 m. If the force felt by the trainee on heT back 
is 7.S5 times her own weight, how fast is she rotating? 
Express your answer in both m/s and rev/s, 

12. (II) A coin is placed 11.0 cm from the axis of a rotating 
turntable of variable speed, When the speed of the 
turntable is slowly increased, the coin remains fixed on 
the turntable until a rate of 36rpm is reached and the 
coin slides off. What is the coefficient of static friction 
between the coin and the turntable? 

13. (II) At what minimum speed must a roller coaster be 
traveling when upside down at the top of a circle 
(Fig. 5-34) so that the passengers will not fall out? 
Assume a radius of curvature of 7.4 m, 




14. 



FIGURE 5-33 Problem 7. 



FIGURE 5-34 Problem 13. 

(II) A sports car of mass 950 kg (including the driver) 
crasses the rounded top of a hill (radius = 95 m) at 22 m/s. 
Determine (a) the normal force exerted by the road on 
the car, (A) the normal force exerted by the car on the 
72- kg driver, and (r) the car speed at which the normal 
force on the driver equals zero, 



130 CHAPTER 5 Circular Motion; Gravitation 



16 



(II) How many revolutions- per minute would a 
15-m-diameter Ferris wheel need to make for the passen- 
gers to feel "weightless" at the topmost point? 
(II) A bucket of mass 2.00 kg is whirled in a vertical circle 
of radius l -10m At the lowest point of its motion the 
tension in the rope supporting the bucket is 25,0 N. (a) Find 
the speed of the bucket, (h) How fast must the bucket 
move at the top of the circle so that the rope does not go 
slack? 

17. (II) How fast (in rpm) must a centrifuge rotate- if a 
particle 9.00 cm from the axis of rotation is to experience 
an acceleration of I 15.000 gY? 

18. (II) In a "Rotor-ride" at a carnival, people are rotated in a 
cylindrical ly walled "'room.'" (See Fig. 5-35.) The room 
Tadius is 4.(5 m, and the rotation frequency is 0.50 revolu- 
tions per second when the floor drops out. What is the 
minimum coefficient of static friction so that the people 
will not slip down? People on this ride say they were 
"pressed against the wall." Is there really an outward 
force pressing them against the wall'? If so. what is its 
source? If not, what is the proper description of their situ- 
ation (besides "scary")? \Hint. First draw the free-body 
diagram for a person, | 




FIGURE 5-35 Problem 18, 

19. (II) A flat puck (mass M) is rotated in a circle on a fric- 
tionless air-hockey tahletop, and is held in this ortrit by a 
light cord connected to a dangling block (mass m) 
through a central hole as shown in Fig. 5-36, Show that 
the speed of the puck is given by 



j mgR 
V M 




, M - -if 



"*) 



I 



TF 



I 




FIGURE 5-36 Problem 19. 



2tt. (II) Redo Example 5-3. precisely this time, by not 
ignoring the weight of the ball which revolves on a string 
0-600 m long. In particular, find the magnitude of Ft. and 
the angle it makes with the horizontal. {Hint; Set the hori- 
7ontal component of F T equal to ma R : also, since there is 
no vertical motion, what can you say about the vertical 
component of Fj ?| 

21. (Ill) If a curve with a radius of 88 m Is perfectly banked 
for a car traveling 75 km/h, what must be the coefficient 
of static friction for a car not to skid when traveling at 
95 km/h? 



22. (Ill) A i200-kg car rounds a curve of radius 67 m banked 
at an angle of 12°. If the car is traveling at 95 km/h, will a 
friction force be required? If so, how much and in what 
direction? 

2X (III) Two blocks, of masses mi and m 3 , are connected to 
each other and to a central post by cords as shown in 
Fig. 5-37. They rotate about the post at a frequency / 
(revolutions per second) on a frictionlcss horizontal surface 
at distances r { and r 2 from the post. Derive an algebraic 
expression for the tension in each segment of the cord. 



/ /■ 

I f 
t \ 

\ X 

\ 

N 




FIGURE 5-37 Problem 23. 

24. (Ill) A pilot performs an evasive maneuver by diving verti- 
cally al 3 10 m/s, If he can withstand an acceleration of 
9,0 g's without blacking out, at what altitude must he hegin 
to pull out of the dive to avoid crashing into the sea? 

' 5-4 Nonuniform Circular Motion 

* 25. (I) Determine the tangential and centripetal components 

of the net force exerted on the car (by the ground) in 
Example 5-8 when its speed is 15 m/s. The car's mass is 
1100 kg. 

* 26. (II) A car at the Indianapolis 500 accelerates uniformly 

from the pit area, going from rest to 320 km/h in a semi- 
circular arc with a radius of 220 m. Determine the tangen- 
tial and radial acceleration of the car when it is halfway 
through the turn, assuming constant tangential accelera- 
tion. If the curve were flat, what would the coefficient of 
static friction have to be between the tires and the road to 
provide this acceleration with no slipping or skidding? 

*■ 27. (Ill) A particle revolves in a horizontal circle of radius 
2,90m, At a particular instant, its acceleration is 1,05 m/s 2 . 
in a direction that makes an angle of 32.0" to its direction 
of motion. Determine its speed {a) at this moment, and 
(b) 2.00 s later, assuming constant tangential acceleration. 

5-6 and 5-7 Law of Universal Gravitation 

28. (I) Calculate the force of Earth's gravity on a spacecraft 
12,800 km (2 Earth radii) above the Earth's surface if its 
mass is 1350 kg. 

29. (I) At the surface of a certain planet, the gravitational accel- 
eration g has a magnitude of 12.0 m/s 2 . A 21.0-kg brass ball 
is transported to this planet. What is (a) die mass of the 
brass ball on the Earth and on the planet, and (h) the 
weight of the brass ball on the Earth and on the planet? 

311, (II) Calculate the acceleration due to gravity on the 
Moon The Moon's radius is 1,74 X 10 fl m and its mass is 
7,35 x 10 22 kg. 



Problems 131 



II. 

.12. 
33. 

34. 

Mi. 

If,. 
37. 

IS. 



39. 



4(1. 



(II) A hypothetical planet has a radius 1.5 times that of 

Earth, hut has the same mass. What is the acceleration 

due to gravity near its surface? 

(II) A hypothetical planet has a mass 1,66 times thai of 

Earth, but the same radius. What is g near its surface? 

(I I J Two objects attract each other gravitation a My with a 

force of 2.5 X 10 ll 'N when they are 0.25 m apart. Their 

total mass is 4.0 kg. Find their individual masses 

(II) Calculate the effective value of g, the acceleration of 

gravity, at (a) 3200m, and (b) 3200 km, above the Earth's 

surface-. 

(II) What is the distance from the Earth's center to a point 

outside the Earth where the gravitational acceleration due 

to the Earth is -fa of its value at the Earth's surface? 

(II) A certain neutron star has five times the mass of our 
Sun packed into a sphere about 10 km in radius, Estimate 
the surface gravity on this monster. 

(II) A typical white-dwarf star, which once was an 
average star like our Sun but is now in the last stage of its 
evolution, is the size of our Moon but has the mass of our 
Sun. What is the surface gravity on this star? 

(II) You are explaining why astronauts feel weightless 
while orbiting in the space shuttle. Your friends respond 
that they thought gravity was jusl a lot weakeT up there. 
Convince them and yourself that it isn't so by calculating 
the acceleration of gravity 250 km above the Earth's 
surface in terms of g. 

(II) Four 9.5-kg spheres are located at the corners of a 
square of side 0,60 m. Calculate the magnitude and direc- 
tion of the total gravitational force exerted on one sphere 
by the other three. 

(II) Every few hundred years most of the planets line up 
on the same side of the Sun. Calculate the total force on 
the Earth due to Venus. Jupiter, and Saturn, assuming all 
four planets are in a line (Fig. 5-3fi)_ The masses are 
M v = 0.8I5M K . Wj = 3I8M H . M s = 95.1 M K , and their 
mean distances from the Sun are 108, 150, 778. and 
1430 million km, respectively. What fraction of the Sun's 
force on the Earth is this? 






# 






San 



FIGURE 5-33 Problem 40. (Not to scale) 

41. (II) Given that the acceleration of gravity at the surface 
of Mars is 0.38 of what it is on Earth, and that Mars' 
radius is 3400 km. determine the mass of Mare, 

42. (Ill) Determine the mass of the Sun using the known 
value for the period of the Earth and its distance from the 
Sun. [Note: Compare your answer to that obtained using 
Kepler's laws. Example 5-16.] 

5-8 Satellites; Weightlessness 

43. (I) Calculate the speed of a satellite moving in a stable 

circular orbit about the Earth at a height of 3600 km. 

44. (I) The space shuttle releases a satellite into a circular 
orbit 650 km above the Earth. How fast must the shuttle 
be moving (relative to Earth) when the release occurs? 



45. (II) At what rate must a cylindrical spaceship rotate if 
occupants are to experience simulated gravity of 0.60 g? 
Assume the spaceship's diameter is 32 m, and give your 
answer as the lime needed for one revolution. (See Ques- 
tion 21, Fig 5-32) 

46. (II) Determine the time it takes for a satellite to orbit the 
Earth in a circular "near-Earth" orbit. A "near-Earth" 
orbit is one at a height above the surface of the Earth 
which is very small compared to the radius of the Earth, 
Does your result depend on the mass of the satellite? 

47. (II) At what horizontal velocity would a satellite have to 
be launched from the top of Ml. Everest to be placed in a 
circular orbit around the Earth? 

48. (II) During an Apollo lunar landing mission, the 
command module continued to orbit the Moon at an alti- 
tude of about 1 00 km, How long did i( take to go around 
the Moon once? 

49. (II) The rings of Saturn are composed of chunks of ice 
that orbit the planet- The inner radius of the rings is 
73.000 km, while the outer radius is 170,000km. Find the 
period of an orbiting chunk of ice at the inner radius and 
the period ol a chunk at the outer radius, Compare youT 
numbers with Saturn's mean rotation period of 10 hours 
and 39 minutes. Tie mass of Saturn is 5.7 X 10 2(> kg, 

511. (II) A Ferris wheel 24.0 in in diameter rotates once every 
15.5s (see Fig. 5-9). What is the ratio of a person's 
apparent weight to her real weight (n) at the top, and 
(b) at the bottom? 

51. (II) What is the apparent weight of a 75-kg astronaut 
4200km from the center of the Earth's Moon in a 
space vehicle (a) moving at constant velocity, and 
(b) accelerating toward the Moon at 2.9 m/s 2 ? State the 
"direction" in each case. 

52, (II) Suppose that a binary-star system consists of two 
stars of equal mass. They are observed to be separated by 
360 mi I lion km and take 5.7 Earth years to orbit about a 
point midway between them. What is the mass of each? 

53, (II) What will a spring scale read for the weight ol a 55-kg 
woman in an elevatoT that moves (a) upward with constant 
speed of 6.0 m/s, (b) downward with constant speed of 
6,0 m/'s, (e) upward with acceleration of 0.33 g, 
(rf) downward with acceleration 0.33 g, and (*?) in free fall? 

54. (II) A 17.0-kg monkey hangs from a cord suspended from 
the ceiling of an elevator. The cord can withstand a 
tension of 220 N and breaks as the elevator accelerates, 
What was the elevator's minimum acceleration (magni- 
tude and direction)? 

5R (III) (a) Show that if a satellite orbits very near the surface 
of a planet with period T, the density (mass/volume) of the 
planet isp = m/V = "iir/GT 2 . (/>) Estimate the density of 
the Earth, given thai a satellite near the surface orbits with 
a period of about R5 min. 

■ 5-9 Kepler's Laws 

* 5(>. (I) Use Kepler's laws and the period of the Moon (27.4 d) 

to determine the period of an artificial satellite orbiting 

very near the Earth's surface, 

* 57. (I) The asteroid learns, though only a few hundred meters 

across, orbits the Sun like the planets. Its period is 410 d, 
What is its mean distance from the Sun? 



132 CHAPTER 5 Circular Motion; Gravitation 



* sK. (I) Neptune is an average distance of 4.5 x 10 km from the 
Sun. Estimate the length of the Neptunian year given that 
the Earth is 1 .50 x 10* km from the Sun on the average. 

*S9. (II) Halley's comet orbits the Sun roughly once every 
76 years. It comes very close to the surface of the Sun 
on its closest approach (Fig. 5-39). Estimate the greatest 
distance of the comet from the Sun. Is it still "in" the 
Solar System? What planet's orbit is nearest when it is out 
there? [Hiitt:Th& mean distance s in Kepler's third law 
is half the sum of the nearest and farthest distance from 
the Sun.l 



Halley's comet \ 




Sun - > 




___ r^ 


FIGURE 5-39 


___ _^ -— "™ 


Problem 59. 



*6<r. (II) Our Sun rotates about the center of the Galaxy 
[M == 4 x 10 41 kg) at a distance of about 3 x 10 4 light- 
years (1 ly = 3 X 10 s m/s X 3.16 X 10 7 s/y X 1 y). What 
is the period of our oTbital motion about the center of 
the Galaxy? 

(II) Table 5-3 gives the mass, period, and mean distance 
for the four largest moons of Jupiter (those discovered by 
Galileo in 1609). (a) Determine the mass of Jupiter using 
the data for Io. (h) Determine the mass of Jupiter 
using data for each of the other three moons. Art the 
results consistent? 



M. 



TABLE 5- 


-3 Principal M 


oons of Jupiter 


Mi.l.i 


Mass 
(kg) 


Mean distance 
Period from Jupiter 
(Earth days) (km) 


Io 


8.9 X IO 22 


1.77 422 X IO 3 


Europa 


4.9 X IO 22 


3.55 671 X 10-' 


Ganvmede 


15 X IO 22 


7.16 1070 x 10 1 


Callisto 


11 X IO 22 


16.7 1883 X 10 :! 



* 62. (II) Determine the mass of the Farth from the known 

period and distance of the Moon. 

* 63. (II) Determine the mean distance from Jupiter for each of 

Jupiter's moons, using Kepler's third law. Use the distance 
of Io and the periods given in Table 5-3. Compare to the 
values in the Table. 

* 64. (II) The asteroid belt between Mars and Jupiter consists 

of many fragments (which some space scientists think 
came from a planet that once orbited the Sun but was 
destroyed), (a) If the center of mass of the asteroid belt 
(where the planet would have been) is about three limes 
farther from the Sun than the Earth is. how long would it 
have taken this hypothetical plane! to orbit the Sun? 
(b) Can we use these data to deduce the mass of this 
planet? 

* 65. (IU) A science -fiction tale describes an artificial "planet" 

in the form of a band completely encircling a sun 
(Fig. 5-40). The inhabitants live on the inside surface 
(where it is always noon). Imagine thai this sun is exactly 
like our own, that the distance to the band is the same as 
the Earth -Sun distance (to make the climate temperate), 
and that the ring rotates quickly enough to produce an 
apparent gravity of j» as on Earth. What will be the period 
of revolution, this planet's year, in Earth days? 




FIGURE 5-40 
Problem 65. 



| General Problems 



64 Tar? an plans to cross a gorge by swinging in an arc 
from a hanging vine 
(Fig. 5-41). If his 
arms are capable of 
exerting a force of 
1400 N on the vine, 
what is the maximum 
speed he can tolerate 
at the lowest point of 
his swing? His mass 
is 80 kg, and the vine 
is 5.5 m long, 



FIGURE 5-41 

Problem 66. 




67. How far above the Earth's surface will the acceleration of 

gravity be half what it is on the surface? 

68. On an ice rink, two skaters of equal mass grab hands and 
spin in a mutual circle once every 2.5 s. If we assume their 
arms are each 0.80 m long and their individual masses are 

60-0 kg, how hard are they pulling on one another? 

69. Because the Earth rotates once per day. the apparent 
acceleration of gravity at the equator is slightly less than 
it would be if the Earth didn't rotate, Estimate the magni- 
tude of this effect. What fraction of # is this? 

70. At what distance from the Earth will a spacecraft trav- 
eling directly from the Earth to the Moon experience 
zero net force because the Farth and Moon pull with 
equal and opposite forces? 

71. You know your mass is 65 kg, but when you stand on a 

bathroom scale in an. elevator, it says your mass is 82 kg. 
What is the acceleration of the elevator, and in which 
direction? 



General Problems 133 



12. A projected space station consists of a circular tube that 
will rotate ahout its center (like a tubular hicycle lire) 
(Fig, 5-42). The circle formed by the tube has a diameter 
of about 11 km, What must be the rotation speed (revolu- 
tions per day) if an effect equal to gravity al the surface 
of the Earth (l.Og) is to be felt? 




FIGURE 5-42 

Problem 72. 



73. A jet pilot takes his aircraft in a vertical loop (Fig 5-43). 
(a) If (he jet is moving at a speed of 1300 km/h al the 
lowest point of the loop, determine the minimum radius 
of the circle so that the centripetal acceleration al the 
lowest point does not exceed 6.0j»'s. (ft) Calculate the 78-kg 
pilot's effective weight (the force with which the seat 
pushes up on him) at the bottom of the circle, and (c) al 
the top of the circle (assume the same speed). 




FIGURE 5-43 

Problem 73, 



74. Derive a formula for the mass of a planet in terms of its 
radius r, the acceleration due to gravity at its surface gp, 
and the gravitational constant G. 

75. A plumb bob (a mass m hanging on a string) is deflected 
from the vertical by an angle H due to a massive mountain 
nearby (Fig. 5-44), (a) Find an approximate formula for If 
in terms of the mass of the mountain, m M , the distance to 
its center. f\j, and the radius and mass of the Earth. 
(A) Make a rough estimate of the mass of Mt. Everest. 
assuming it has the shape of a cone 4000 m high and base 
of diameter 4000m, Assume its mass per unit volume is 
3000 kg per m 3 . (c) Estimate the angle B of the plumb boh 
if it is 5 km from the center of Vlt. Everest. 




76, A curve of radius 67 m is banked for a design speed of 
95 km/h, If the coefficient of static friction is 0,30 (wet 
pavement), at what range of speeds can a car safely 
handle (he curve'? 

77. How long would a day be if the Earth were rotating so fast 
that objects at the equator were apparently weightless 1 ? 

78, Two equal-mass stars maintain a constant distance apart 
of 8.0 X IO l!J m and rotate about a point midway between 
th^in at a rate of one revolution ever}' I2.6yr. (a) Why 
don't the two stars crash into one anotheT due to the 
gravitational force between them? (b) What must be the 
mass of each star? 

79. A train traveling at a constant speed rounds a curve of 
radius 235m. A lamp suspended from the ceiling swings 
out lo an angle of 17-5" throughout the curve. What is the 
speed of the train? 

NIL Jupiter is about 320 times as massive as the Rarth. Thus, it 
has been claimed thai a person would be crushed by the 
force of gravity on a planet the size of Jupiter since 
people can't survive more than a few g*s. Calculate the 
number of g's a person would experience at the equator 
of such a planet- Use the following data for Jupiter: 
mass = 1.9 X 10 27 kg. equatorial radius = 7.1 X 10 4 km. 
rotation period = 9 hr 55 min. Take the centripetal accel- 
eration into account, 

HI. Astronomers using the Hubhle Space Telescope deduced 
the presence of an extremely massive core in the distant 
galaxy M87, so dense that it could be a black hole (from 
which no light escapes). They did this by measuring the 
speed of gas clouds orbiting the core to be 780 km/'s at a 
distance of 60 light-years (5,7 x 10 l7 m) from the core. 
Deduce the mass of the cotc. and compare it to the mass 
of our Sun. 

82. A car maintains a constant speed v as it traverses the hill 
and valley shown in Fig. 5-45. Both the hill and valley 
have a radius of curvature R. (a) How do the normal 
forces acting on the car at A, B, and C compare 1 ? (Which 
is largest ? Smallest?) Explain, (ft) Where would the driver 
feel heaviest? Lightest? Explain, (c) How fast can the caT 
go without losing contact with the road at A? 




****» 



FIGURE 5-44 Problem 75. 



FIGURE 5-45 Problem 82. 

83, The Navstar Global Positioning System (GPS) utilizes 
a group of 24 satellites orbiting the Earth. Using "trian- 
gulation" and signals transmitted by these satellites, the 
position of a receiver on the Earth can be determined 
to within an accuracy of a few centimeters. The satellite 
orbits are distributed evenly around the Earlh. with 
four satellites in each of six orbits, allowing continuous 
navigational "fixes." The satellites orbit at an altitude 
of approximately 11, 000 nautical miles [1 nautical mile = 
1.852km = 6076 ft], (a) Determine the speed of each 
satellite, (h) Determine (he period of each satellite. 



13fl CHAPTER 5 Circular Motion; Gravitation 



84. The Near Earth Asteroid Rendezvous (NEAR), after 
(raveling 2.1 billion km. is meant to orbit the asteroid 
Eros at a height of ahout 15 km. Eros is roughly 
40 km X 6 km X 6 km. Assume Eros has a density 
(mass/volume) of about 2.3 x lO-'kg/url (a) What will 
be the period of \'EAR as it orbits Eros? (ft) If Eros 
weTe a sphere with the same mass and density, what 
would its radius be? (l) What would g be at the surface 
of this spherical Pros? 

85. You are an astronaut in the space shuttle pursuing a satel- 
lite in need of repair. You are in a circular orbit of the 
same radius as the satellite (400 km above the Earth), but 
25 km behind it. (a) How long will it take to overtake the 
satellite if you reduce your orbital radius by t.Okm? 
(i>) By how much must you reduce your orbital radius to 
catch up in 7-0 hours'? 

* 86. The comet Hale-Bopp has a period of 3000 years. 

(a) What is its mean distance from the Sun? (/>) At its 
closest approach, the come! is about I A.U, from the Sun 
(I A.U. = distance from Earth to the Sun). What is the 
farthest distance? (c) What is the ratio of the speed at the 
closes! point to the speed al the farthest point? \Hiul: Use 
Kepler's second law and estimate areas by a triangle (as 
in Kg, 5-29, but smaller distance travelled: see also Hint 
for Problem 5°.| 

87. Estimate what the value of G would need to be if you 
could actually "feel" yourself gravitalionally attracted to 
someone near you. Make reasonable assumptions, like 
F=1N. 

* 88. The Sun rotates around the center of the Milky Way Galaxy 

(Fig, 5-46) at a distance of about 30,000 light-years from the 
center (l ly = 9.5 X lG L> m). If it takes ahout 200 million 
years to make one rotation, estimate the mass of our 
Galaxy, Assume that the mass distribution of our Galaxy is 
concentrated mostly in a central uniform sphere. If all the 
stars had about the mass of our Sun (2 X lO^'kg), how 
many stars would there be in our Galaxy? 



89, Four l o-kg masses are located at the comers of a square 
0,50 m on each side. Find the magnitude and direction of 
the gravitational force on a fifth 1.0-kg mass placed at the 
midpoint of the bottom side of the square. 

911. A satellite of mass 5500 kg orbits the Earth (mass = 
6.0 x ]0 24 kg) and has a period of 6200s. Find (a) the 
magnitude of the Earth's gravitational force on the satel- 
lite. (6) the altitude of the satellite. 

91. What is the acceleration experienced by the tip of the 
1 5-cm-kmg sweep second hand on your wrist watch? 

92. While fishing, you get bored and star! to swing a sinker 
weight around in a circle below you on a 0.25-m piece of 
fishing line. The weight makes a complete circle every 
0,50 s. What is the angle that the fishing line makes with 
the vertical? [Hint See Fig 5-10-| 

93. A circular curve of radius R in a new highway is designed 
so that a car traveling at speed no can negotiate the turn 
safely on glare ice (zero friction). If a car travels loo 
slowly, then it will slip toward the center of the circle. If it 
travels too fast, then it will slip away from the center of 
the circle. If the coefficient of static friction increases, a 
car can stay on the road while traveling at any speed 
within a Tange from v w \„ to t; 1TlaY . Derive formulas for 
b iiiLii a "d "mas as functions of p^ Vq, and R. 

94. Am trait's high speed train, the Acela. utilizes tilt of the 
caTS when negotiating curves. 'I rie angle of lilt is adjusted 
so that the main force exerted on the passengers, to 
provide the centripetal acceleration, is the normal force. 
The passengers experience less friction force against the 
seat, thus feeling more comfortable. Consider an Acela 
train that rounds a curve with a radius of 620 m at a speed 
of 160km/h (approximately lt")()mi/h). (/<) Calculate the 
friction force needed on a train passenger of mass 75 kg if 
the track is not banked and the train does not tilt. 
(ft) Calculate the friction force on the passenger if the 
train lilts lo its maximum lilt of 8.0" toward Ihe center of 
the curve. 



Sun 

\ 



30,000 ly- 



FIGURE 5-46 Problem 88. F.dge-on view of our Galaxy. 



Answers to Exercises 



A: A factor of two (doubles). 

B: Speed is independent of the mass of the clothes. 

O. (a). 

U: No 



E; Yes. 

F: (ft) No change: (ft) four times larger. 

G: (b). 



General Problems 135 



This baseball pitcher is about 10 accelerate the 
baseball lo a high velocity by exerting a force on it. He 
will be doing work on the ball as he exerts Ihe force over a 
displacement of perhaps several meters, from behind his. 
head until he releases the ball with arm outstretched in 
front of him, The total work done on the ball will be equal to 
the kinetic energy (\mv z ) acquired by the ball, a result 
known as the work-energy principle. 




Displacement 




Work and Energy 



Until now we have been studying the translations! motion of an object in 
terms of Newton's three laws of motion. In this analysis, force has 
played a central role as the quantity determining the motion, In this 
Chapter and the next, we discuss an alternative analysis of the translational 
motion of objects in terms of the quantities energy and momentum. The signifi- 
cance of energy and momentum is that they are conserved. That is, in quite 
general circumstances they remain constant, That conserved quantities exist 
gives us not only a deeper insight into the nature of the world, but also gives us 
another way to approach solving practical problems. 

The conservation laws of energy and momentum are especially valuable in 
dealing with systems of many objects, in which a detailed consideration of the 
forces involved would be difficult or impossible. These laws are applicable to a 
wide range of phenomena, including the atomic and subatomic worlds, where 
Newton's laws do not apply. 



136 



This Chapter is devoted to the very important concepts of work and energy. 
These two quantities are scalars and so have no direction associated with them, 
which often makes them easier to work with than vector quantities, 



Work Done by a Constant Force 

The word work has a variety of meanings in everyday language. Rut in physics, 
wtiTk is given a very specific meaning to describe what is accomplished when 
a force acts on an object, and the object moves through a distance. Specifically, 
the work done on an object by a constant force (constant in both magnitude 
and direction) is defined to be the product of the magnitude of the displacement 
times the component of the force parallel to the displacement. Tn equation form, 
we can write 

W - F { \d, 

where Fj| is the component of the constant force F parallel to the displacement d. 
We can also write 



W - Fd cos 6, 



«5-l) 



where F is the magnitude of the constant force, d is the magnitude of the 
displacement of the ohjeet, and & is the angle between the directions of the 
force and the displacement (Fig. 6-1). The costf factor appears in Eq. 6-1 
because F cos 8 (= F\\) is the component of F that is parallel to d. Work is a 
scalar quantity — it has only magnitude, which can be positive or negative. 

Let us first consider the ease in which the motion and the force arc in the 
same direction, so 9 = and cos# = 1; in this case, W - Fd. For example, if 
you push a loaded grocery cart a distance of 50 m by exerting a horizontal force 
of 30 N on the cart, you do 30 N X 50 m = 1 500 N ■ m of work on the cart. 

As this example shows, in SI units work is measured in newt on- mete rs (N ■ m). 
A special name is given to this unit, the joule (J): 1J = INm. In the 
cgs system, the unit of work is called the erg and is defined as 
1 erg - 1 dyne- cm. Tn British units, work is measured in foot-pounds, Tt is easy 
to show that I J - I0 7 erg - 0.7376 ft* lb. 



Work 



Work defined 
(for constant force) 



Units far work: the italic 



FIGURE 6-1 A person pulling a crate along the floor. The work done by the force F is W 
where d is the displacement. 



Fdcosti. 




SECTION 6-1 Work Done by a Constant Force 137 



FIGURE 6-2 The person does no 
work on the bag of groceries since 
F P is perpendicular to the displace- 
ment d. 




<S>- 



CAUTION 



Forte without work 



<^ 



CAUTION 



State that work is done 
on why. on object 



A force can be exerted on an object and yet do no work, For example, if 
you hold a heavy bag of groceries in your hands at rest, you do no work on it, 
You do exert a force on the bag, but the displacement of the bag is zero, so the 
work done by you on the bag is W = 0, You need both a force and a displace- 
ment to do work, You also do no work on the bag of groceries if you carry it as 
you walk horizontally across the floor at constant velocity, as shown in 
Fig. 6-2. No horizontal force is required to move the bag at a constant velocity, 
The person shown in Fig, 6-2 docs exert an upward force F,, on the bag equal 
to its weight. But this upward force is perpendicular to the horizontal 
displacement of the bag and thus has nothing to do with that motion. Hence, 
the upward foTce is doing no work. This conclusion comes from our definition of 
work, Eq. 6-1: W — 0, because — 90" and cos 90" - 0. Thus, when a 
particular force is perpendicular to the displacement, no work is done by that 
force. (When you start or stop walking, there is a horizontal acceleration and 
you do briefly exert a horizontal force, and thus do work on the bag.) 

When we deal with work, as with force, it is necessary to specify whether you 
are talking about work done by a specific object or done on a specific object. It is 
also important to specify whether the work done is due to one particular force 
(and which one), or the total (net) work done by the net force on the object. 



EXAMPLE 6-1 



Work done on a crate, A person pulls a 50-kg CTate 40m 



along a horizontal floor by a constant force F v = I00N, which acts at a 
37 !J angle as shown in Fig, 6-3- The Hoot is rough and exerts a friction force 
Ff,. = SON. Determine (a) the work done by each force acting on the crate, 
and (b) the net work done on the crate. 

APPROACH We choose our coordinate system so that x can be the vector that 
represents the 40-m displacement (that is, along the ,t axis). Four forces act on the 
CTate, as shown in Fig. 6-3: the force exerted by the person P P ; the friction force P fr 
due to the floor; the crate's weight mj>; and the normal force ? s - exerted upward 
by the floor. The net foTce on the crate is the vector sum of these four forces. 
SOLUTION (a) The work done by the gravitational and normal knees is zero, 
since they are perpendicular to the displacement x (6 = 90 J in Eq. 6- 1): 

W G = nigx eos 90" = 

W N = F N x cos 90° = 0. 
The work done by f f . is 

W r = Fy,x con 8 = (100N)(40m)cos37 D = 3200 J. 
The work done by the friction force is 

W tt = F r ,xeosl80" = (50 N)(40m)(- I } = -2000 J. 

The angle between the displacement x and the force F\ r is 180° because they 
point in opposite directions. Since the force of friction is opposing the motion 

(and cos 180" - —1), the work done by friction on the crate is negative. 



138 CHAPTER 6 Work and Energy 




x (40 m) 



— 



FIGURE 6-3 Example 6-1. A 
50-kg crate is polled along a floor. 



(b) The net work can be calculated in two equivalent ways: 

(1) The net work done on an object is the algebraic sum of the work done by 

each force, since work is a scalar: 



1 * i 



w c + w s + w v + w b 



+ + 3200 J 
1 200 J. 



2000 J 



(2) The net work can also be calculated by first determining the net force 
on the object and then taking its component along the displacement: 
{F n ci)i = />cos# - F( r . Then the net work is 

W^l = (F as t)xX = {F T co6$ - F k )x 

= (100Ncos37° - 50N)(40m} 
U'i'LT. 

In the vertical ( y) direction, there is no displacement and no work done. 



IV net ss the work done by all the 

forces acting on the object 



Tn Example 6-1 we saw that friction did negative work. Tti general, the 
work done by a force is negative whenever the force (or the component of the 
force, f||) acts in the direction opposite to the direction of motion. Also, we can 
see that when the work done by a force on an object is negative, that force is 
trying to slow the object down (and would slow it down if that were the only 
force acting). When the woik is positive, the force involved is trying to speed up 
the object, 

EXERCISE A A box is dragged across a flt»T by a force F P which makes an angle # 
with the horizontal as in Fig. 6-1 or 6-3. If the magnitude of F,, is held constant but 
the angle 6 is increased, the work done by F P (a) remains the same; (b) increases: 
(c) decreases: (d) first increases, then decreases. 



<| ) CAUTION 
Negative work 



PROBLEM SOLVING 



Work 



1, Draw a ftcc-body diagram showing all the forces 
acting on the object you choose to study. 

2, Choose an xy coordinate system, If the object is in 
motion, it may be convenient to choose one of the 
coordinate directions as the direction of one of the 
forces, ot as the direction of motion. [Thus, for an 
object on an incline, you might choose one coordi- 
nate axis to be parallel to the incline.] 

3, Apply Newton's laws to determine any unknown 
forces, 



Find the work done by a specific force on the 
object by using TV = Fa" COS (9 for a constant force. 
Note that the work done is negative when a force 
tends to oppose the displacement. 
To find the net work done on the object, either 
(a) find the work done by each force and add the 
results algebraically; or (b) find the net force on the 
object, Foet, and then use it to find the net work 
done, which for constant net force is: 



TV 

* r rii 



F nBt d cos 0, 



SECTION 6-1 Work Done by a Constant Force 139 




w 




II V 




(c) 
FIGURE 6-4 Example 6-2. 



> PROBLEM SOLVING 

H ark done by gravity depends on 

the height of the hilt and 

not on (he tingle of incline 



EXAMPLE 6-2 



Work on a backpack, (a) Determine the work a hiker 
must do an a 15,0-kg backpack to carry it up a hill of height h = 10.0 m, as 
shown in Fig. 6-4a. Determine also (b) the work done by gravity on the back- 
pack, and (c) the net work done on the backpack. For simplicity, assume the 
motion is smooth and at constant velocity (i.e., acceleration is negligible). 

APPROACH We explicitly follow the Problem Solving Box step by step, 
SOLUTION 

1. Draw a free-body diagram. The forces on the backpack are shown in 
Fig. f>-4b: the force of gravity, mg, acting downward; and F H , the force the 
hiker must exert upward to support the backpack, Since we assume there is 
negligible acceleration, horizontal forces on the backpack are negligible. 

2. Choose a coordinate system. We are interested in the vertical motion of the 
backpack, so we choose the v coordinate as positive vertically upward, 

3. Apply Newton's laws, Newton's second law applied in the vertical direction 
to the backpack gives 

£F V = ma y 



Hence, 



F n - mg - 0. 



F li =mg = (15.0kg)(9.80m/s 2 ) = 147 N. 



4. Find the work done by a specific force, (a) To calculate the work done by 
the hiker on the backpack, we write Cq. 6- 1 as 

W H = Fn(rfcosW), 

and wc note from Fig. 6-4a that dcosB = h. So the work done by the 
hiker is 

W H = Fn(d cos D) = F H h = mgh 

- (l47N)(lQ.0m) - 1470 J. 

Note that the work done depends only on the change in elevation and not 
on the angle of the hill, (i. The hiker would do the same work to lift the pack 
vertically the same height h. 
(6) The work done by gravity on the backpack is (from Eq. 6-1 and Fig. 6-4c) 

W c = F Q dcm(WQ° - 9). 

Since cos(l80 a — B) — -cos 8, we have 

W G = F G d{-cos8) = mg{-dcw0) 
- -mgh 
= -(15.0 kg)(9.S0m/s 2 )( 10.0m} = -1470 J. 

NOTE The work done by gravity (which is negative here) doesn't depend on the 
angle of the incline, only on the vertical height k of the hill. This is because 
gravity acts vertically, so only the vertical component of displacement contributes 
to work done. 

5. Find the net work done, (a) The net work done on the backpack is 
Wnei = 0, since the net force on the backpack is zero (it is assumed not to 
accelerate significantly). We can also determine the net work done by 
adding the work done by each force: 



»',. 



W f . + 1V H = -1470 J + 1470 J = 0. 



NOTE Even though the net wink done by all the fences on the backpack is 
zero, the hiker does do work on the backpack equal to 1 4701 



140 CHAPTER 6 Work and Energy 



CONCEPTUAL EXAMPLE 6-3 I Does the Earth do work on the Moon? 



Moon 



The Moon revolves around the Earth in a nearly circular orbit, kept there hy 
the gravitational force exerted by the Earth. Does gravity do (a) positive 
work, (b) negative work, or (c) no work on the Moon? 

RESPONSE The gravitational force exerted by the Earth on the Moon 
(Fig. 6-5) acts toward the Earth and provides its centripetal acceleration, inward 
along the radius of the Moon's orbit. The Moon's displacement at any moment 
is tangent to the circle, in the direction of its velocity, perpendicular to the radius 
and perpendicular to the force of gravity. Hence the angle 6 between the force 
f c and the instantaneous displacement of the Moon is 90", and the work done 
by the Earth's gravity on the Moon as it orbits is therefore zero (cos90 u = 0). 
This is why the Moon, as well as artificial satellites, can stay in orbit without 
expenditure of fuel; no net work needs to be done against the force of gravity. 



Work Done by a Varying Force 



If the force acting on an object is constant, the work done by that force can be 
calculated using Eq, 6-1. Rut in many cases, the force varies in magnitude or 
direction during a process. For example, as a rocket moves away from Earth, 
work is done to overcome the force of gTavity, which varies as the inverse 
square of the distance from the Earth's center. Other examples are the force 
exerted by a spring, which increases with the amount of stretch, or the work 
done by a varying force in pulling a box or cart up an uneven hill 

The work done by a varying force can be determined graphically. The 
procedure is like that for determining displacement when the velocity is known 
as a function of time (Section 2-8). To determine the work done by a variable 
force, we plot F\ {= F cos 8, the component of F parallel to the direction of 
million at any point) as a function of distance d, as in Fig. 6-6a. We divide the 
distance into small segments Ad. For each segment, we indicate the average of 
F\\ by a horizontal dashed line. Then the work done for each segment is 
A IV = F\\ Ad, which is the area of a rectangle Ad wide and F\ high. The total 
work done to move the object a total distance d - d B - d A is the sum of the 
areas of the rectangles (five in the case shown in Fig. 6-6a). Usually, the average 
value of fj| for each segment must be estimated, and a reasonable approxima- 
tion of the work done can then be made. If we subdivide the distance into many 
more segments, Ad can be made smaller and our estimate of the work done 
would be more accurate. In the limit as Ad approaches zero, the total area of 
the many narrow rectangles approaches the area under the curve. Fig. 6-6b. 
That is, the work done by a variable force in moving an object between tivo 
points is equal to the area under the F\ vs. d curve between those two points. 

Kinetic Energy, and the Work-Energy Principle 

Energy is one of the most important concepts in science. Yet we cannot give a 
simple general definition of energy in only a few words. Nonetheless, each specific 
type of energy can be defined fairly simply. In this Chapter, we define transla- 
tional kinetic energy and some types of potential energy. In later Chapters, we will 
examine other types of energy, such as that related to heat (Chapters 14 and 15). 
The crucial aspect of all the types of energy is that the sum of all types, the total 
energy, is the same after any process as it was before: that is, the quantity "energy" 
is a conserved quantity. 

For the purposes of this Chapter, we can define energy in the traditional 
way as "the ability to do work." This simple definition is not very precise, nor is 
it really valid for all types of energy. 1 It is valid, however, for mechanical energy 
which wc discuss in this Chapter, and it serves to underscore the fundamental 

'Energy associated with heal is often no[ available to do work, as wc will discuss in Chapter 15. 



/'" T* N 




Earth 



FIGURE 6-5 Example 6-3. 



FIGURE 6-6 The work done by a 
force Fcan be calculated by taking: 
(a) the sum of the areas of the 
rectangles; (b) the area under the 
curve of fj vs. d. 




T 100 -- 



Distance, d 
(b) 



SECTION 6-3 Kinetic Energy, and the Work-Energy Principle 141 



FIGURE 6-7 A constant net force 
F nc[ accelerates a bus from speed vy 
to speed v 2 over a displacement rf.The 

net work done is W nrt = F nL . t d. 




Kinetic energy defined 



WORK-ENERG Y PRINCIPLE 



WO RK-ENERG Y PRINCIPE E 



CAUTION 



Work energy valid oniy for net work 



connection between work and energy. We now define and discuss one of the 
basic types of energy, kinetic energy, 

A moving object can do work on another ohject it strikes. A flying cannon- 
ball does work on a brick wall it knocks down; a moving hammer does work on 
a nail it drives into wood. Tn either ease, a moving ohject exerts a force on a 
second object which undergoes a displacement. An object in motion has the 
ability to do work and thus can he said to have energy. The energy of motion is 
called kinetic energy, from the Greek word kinetikos, meaning "motion." 

To obtain a quantitative definition for kinetic energy, let us consider a rigid 
object of mass m that is moving in a straight line with an initial speed i\ . To 
accelerate it uniformly to a speed v 2 , a constant net force F r(:l is exerted on it 
parallel to its motion over a displacement d. Fig. 6-7. Then the net work done on 
the object is 1V ]1!;1 = F aLl d. We apply Newton's second law, F ni:l = ma, and use 
Eq. 2-1 Ic, which wc now write as v\ — v] + lad, with i\ as the initial speed 
and v 2 the final speed, We solve for a in Eq. 2-1 1c. 

2 2 

vi - v\ 
a = ^d— 
then substitute this into F ncl = ma, and determine the work done: 



W = 



or 



F nct d = mad = 


HI 1 


, 2d 


I 1 1 ! 

j mv 2 — j w'i- 







d = m 



v- r. 



W nA = Imei - \tm& (6-2) 

We define the quantity j mv 2 to be the translational kinetic energy (KE) of the 
object: 

ke = | wit? 2 . (6— 3> 

(We call this "translationaF kinetic energy to distinguish it from rotational 
kinetic energy, which we will discuss in Chapter 8.) Equation 6-2, derived 
here for one-dimensional motion with a constant force, is valid in general for 
translational motion of an ohject in three dimensions and even if the force varies, 
Wc can rewrite Eq. 6-2 as: 

IV 

L> i 

or 



= KF, 



K I , 



W 

rT Til 



Akc. 



(6-4) 



Equation 6-4 (or Eq. 6-2) is an important result known as the work-energy 
principle. It can be stated in words: 

The net work done on An object is equal lo the change in the object's 
kinetic energy. 

Notice that we made use of Newton's second law, F I1C1 - ma, where F^ is the 
net force — the sum of all forces acting on the object. Thus, the work-energy 
principle is valid only if Wis the net work done on the object — that is, the work 
done by all forces acting on the object. 

The work-energy principle is a very useful reformulation of Newton's laws, 
It tells us that if (positive) nel work TV is done on an object, the object's kinetic 
energy increases by an amount W. The principle also holds true for the TeveTse 
situation: if the net work W done on an object is negative, the object's kinetic 



142 CHAPTER 6 Work and Energy 



energy decreases by an amount IV. That is, a net force exerted on an object 
opposite to the object's direction of motion decreases its speed and its kinetic 
energy. An example is a moving hammer (Fig. 6-8) striking a nail. The net force 
on the hammer (-F in Fig, 6-8, where F is assumed constant for simplicity) acts 
toward the left, whereas the displacement d of the hammer is toward the right. 
So the net work done on the hummer, W h = {F)(d)(co$ 180°} = — Fd, is 
negative and the hammer's kinetic eneTgy decreases (usually to zero). 

Figure 6-8 also illustrates how energy can be considered the ability to do 
work. The hammer, as it slows down, does positive work on the nail: if the nail 
exerts a force -F on the hammer to slow it down, the hammer exerts a farce 
+ F tin the nail (Newton's third law) through the distance d. Hence the work 
done on the nail by the hammer is W„ - (+F)(+d) = Fd and is positive. We 
also see that W„ = Fd = -W h : the wort done on the nail W„ equals the 
negative of the work done on the hammer. That is, the decrease in kinetic 
energy of the hammer is equal to the work the hammer can do on another 
object — which is consistent with energy being the ability to do work. 

Whereas the translational kinetic energy (= jtms 1 ) is directly proportional 
to the mass of the object, it is proportional to the square of the speed. Thus, if 
the mass is doubled, the kinetic energy is doubled. But if the speed is doubled, 
the object has four times as much kinetic energy and is therefore capable of 
doing four times as much work. 

Let us summarize the relationship between work and kinetic energy 
(Eq. 6-4): if the net work W done on an object is positive, then the object's 
kinetic energy increases. If the net work W done on an object is negative, its 
kinetic energy decreases. If the net work done on the object is zero, its kinetic 
energy remains constant (which also means its speed is constant). 

Because of the direct connection between work and kinetic energy 
(Eq. 6-4), eneTgy is measured in the same units as work: joules in ST units, ergs 
in the cgs, and foot-pounds in the British system. Like work, kinetic energy is a 
scalar quantity. The kinetic energy of a group of objects is the sum of 
the kinetic energies of the individual objects. 



f* 



'i 



, , A 

-F ' f/H 

{on hammer) fori nail) 






FIGURE 6-8 A moving hammer 
strikes a nail and comes to rest. The 
hammer exerts a force F on the 
nail: the nail exerts a force —F on 
the hammer (Newton's third law), 
'the work done on the nail by the 
hammer is positive (W n = Fd > 0). 
The work done on (he hammer by 
the nail is negative (W h = -Fd). 



//TV,,.., > 0. KE increases 
J/Wnet < 0, K£ decreases 



Energy units: 
the joule 



EXAMPLE 6-4 



KE and work done on a baseball. A 145-g baseball is 
thrown so that it acquires a speed of 25m/s. (a) What is its kinetic energy? 
(b) What was the net work done on the ball to make it reach this speed, if it 
started from rest? 

APPROACH Wc use the definition of kinetic energy, Eq. 6-3, and then the 

work -energy principle, Eq. 6-4, 

SOLUTION (a) The kinetic energy of the ball after the throw is 

ten = \mv 2 = |(0.I45 kg)(25m/s) z = 45.1. 

(b) Since the initial kinetic energy was zero, the net work done is just equal to 
the final kinetic energy, 45 J. 



EXAMPLE 6-5 



Work on a car, to increase its ke. I low much net work is 
required to accelerate a 1000-kg car from 20m/s to 30m/s (Fig, 6-9)? 

APPROACH To simplify a complex situation, let us treat the car as a particle 
or simple rigid object. We can then use the work-energy principle. 
SOLUTION The net work needed is equal to the increase in kinetic energy: 

W „ _ „ - 1 ~..,2 _ i 



Kl., 



Kl i 



7 

imvi 



■ m: 



- HlOOOkg)(30m/s} 2 - i(1000kg)(20 m/s) 2 - 2.5 X Iff J. 

NOTE You might be tempted to work this Example by finding the force and 
using Eq. 6-1. That won't work, however, because we don't know how far or 
for how long the car was accelerated, fn fact, a large knee could be acting for 
a small distance, or a small force could be acting over a long distance; both 
could give the same net work. 




Example 6-5. 

r-r - ?0 ni/s 



-€^-l~{> 



SECTION 6-3 Kinetic Energy, and the Work-Energy Principle 143 



(a, 



FIGURE 6-10 Example 6-6. (h) 



i'i = fit) kmrti 



U, = t) 



d (d = 20 in! 



I, = l20km/h 



_£ "& 



Bj = 

*y^"#~i 



d W = ?) 



PHYSICS APPLIED 

( li'.i\ stopping distance oc 
initial speed squared 



CONCEPTUAL EXAMPLE 6-6 | Work to stop a car. A car traveling 



fid Icm/h can brake to a stop within a distance d of 20 m (Fig. 6- Ifla). if the car 
is going twice as fast, I20km/h, what is its stopping distance (Fig. 6-10b)? 
Assume the maximum braking force is approximately independent of speed. 

RESPONSE Since the stopping force F is approximately constant, the work 
needed to stop the car, Fd, is proportional to the distance traveled. We apply 
the work-energy principle, noting that F and d are in opposite directions and 
that the final speed of the car is zero: 



VF nel = Fd cos 180" = -Fd. 



Then 



■Fd = Ake = 2 mv \ ~ l mv 

- - [mv\. 



■ mi , 
t_™2 



Thus, since the force and mass arc constant, we sec that the stopping distance, </, 
increases with the square of the speed: 

d ex v 2 . 

If the cars initial speed is doubled, the stopping distance is (2) 2 = 4 times as 
great, or 80 m. 



I EXERCISE B Can kinetic energy ever be negative? 



Potential enei \ ; 



Gravitational Pi 



Potential Energy 



We have just discussed how an object is said to have energy by virtue of its 
motion, which we call kinetic energy. But it is also possible to have potential 
energy, which is the energy associated with forces that depend on the position 
or configuration of an object (or objects) relative to the surroundings. Various 
types of potential energy (pe) can be defined, and each type is associated with a 
particular force, 

The spring of a wind-up toy is an example of an object with potential 
energy. The spring acquired its potential energy because work was done on it by 
the person winding the toy. As the spring unwinds, it exerts a force and does 
work to make the toy move. 

Perhaps the most common example of potential energy is gravitational 
potential energy. A heavy brick held high in the air has potential energy 
because of its position relative to the Earth. The raised brick has the ability to 
do work, for if it is released, it will fall to the ground due to the gravitational 
force, and can do work on, say, a stake, driving it into the ground. Let us seek 
the form for the gravitational potential energy of an object near the surface of 
the Earth, For an object of mass m to be lifted vertically, an upward force at 
least equal to its weight, mg, must be exerted on it, say by a person's hand. 



144 CHAPTER 6 Work and Energy 



To lift it without acceleration a vertical displacement of height h, from 
position Vi to v'2 in Fig, 6-11 (upward direction chosen positive), a person 
must do wotIc equal to the product of the needed external force, F CK , = mg 
upward, and the vertical displacement /i.That is. 



W«t = 



F cxt d cos U = mgh 



<6-Sa) 
v ( to v, , and docs work on 



Gravity is also acting on the object as it moves from 
it equal to 

W G = F G dcase = mgh COS ISO , 

where B = 180° because F c and d point in opposite directions. So 

W G - -mgh 

= -™g{yi - vi). 



<6-5b> 



If we now allow the object to start from rest and fall freely under the action of 
gravity, it acquires a velocity given by v 2 = 2gh (Eq. 2-1 Ic) aflei" falling a 
height h- It then has kinetic energy jmv 2 = jm(2gft) = mgfh and if il strikes a 
stake it can do work on the stake equal to mgh (woTk-energy principle). Thus, to 
raise an object of mass m to a height h requires an amount of work equal to mgh 
(Eq. 6-5a). And once at height h, the object has the ability to do an amount of 
work equal to mgh. 

We therefore define the gravitational potential energy of an object, due to 
Earth's gravity, as the product of the object's weight mg and its height y above 
some reference level (such as the ground): 



IT 



•.■''''■ 



mgy. 



The higher an object is above the ground, the moTe gravitational potential 
energy it has, We combine Eq. 6-5a with Eq. 6-6: 

W,-*i - mg{yj - Vj) 

W^xi — PE 2 - PE i = Ape. (6-7a) 

That is, the work done by an external force to move the object of mass m from 
point I to point 2 (without acceleration) is equal to the change in potential 
energy between positions I and 2. 

Alternatively, we can write the change in potential energy, Apf., in terms of 
the work done by gravity itself: starting from Eq. 6-5b, we obtain 

W Cr = -mjfCvj - y,) 

W = -(pe, - PC,) = 



Ape. 



(6-7b) 



That is, the work done by gravity as the object of mass m moves from point 1 to 
point 2 is equal to the negative of the difference in potential energy between 
positions I and 2. 

Potential energy belongs to a system, and not to a single object alone. 
Potential energy is associated with a force, and a force on one object is always 
exerted by some other object. Thus potential energy is a property of the system 
as a whole. For an object raised to a height v above the Earth's surface, the 
change in gravitational potential energy is mgy. The system here is the object 
plus the Earth, and properties of both arc involved: object (m) and Earth (g). 

Gravitational potential energy depends on the vertical height of the object 
above some reference level (Eq. 6-6). In some situations, you may wonder from 
what point to measure the height y. The gravitational potential energy of a 
book held high above a table, for example, depends on whether we measure y 
from the top of the table, from the floor, or from some other reference point. 



d i^CKl 


m 


JL 


" f c- 



(exerted 

by hand) 



yi — 

FIGURE 6- 11 A person exerts an 
upward force F L . xl = mg to lift a 
brick from y\ to yi- 



(6-6) Gravitatiotwi pe 



<k 



CAUTION 



Potential energy belongs to n system, 
not to a single object 



SECTION 6-4 Potential Energy 145 



X> CAUTION 
Change in PEis what is 
physically meaningful 



Grav. pi depends on vertical height 



What is physically important in any situation is the change in potential energy, 
Apr, because that is what is related to the work done, Eqs. 6-7; ;md it is Apr 
that can he measured. We can thus choose to measure y from any reference 
point that is convenient, but we must choose the reference point at the start 
and be consistent throughout a calculation. The change in potential energy 
between any two points does not depend on this choice, 

An important result we discussed earlier (see Example 6-2 and Fig. 6-4) 
concerns the gravity force, which does work only in the vertical direction; the 
work done by gravity depends only on the vertical height h, and not on the path 
taken, whether it be purely vertical motion or, say, motion along an incline. 
Thus, from Eqs, 6-7 we see that changes in gravitational potential energy 
depend only tin the change in vertical height and not on the path taken. 




FIGURE 6-12 Example 6-7. 



PE tiffined in general 



EXAMPLE 6-7 



Potential energy changes for a roller coaster. A 
1000-kg roller-coaster caT moves from point I, Fig, 6-12. to point 2 and then to 
point 3, (a) What is the gravitational potential energy at 2 and 3 relative to 
point I? That is, take y = at point 1. (fr) What is the change in potential 
energy when the ear goes from point 2 to point 3? (c) Repeat parts (a) and (A), 
but take the reference point (y = 0) to be at point 3. 

APPROACH We are interested in the potential energy of the car- Earth 
system, Wc take upward as the positive y direction, and use the definition of 
gravitational potential energy to calculate PE. 

SOLUTION (a) We measure heights from point 1, which means initially that 
the gravitational potential energy is zero. At point 2, where y 2 - 10 m, 

pe ? - mgyi - (10QQkg}(9.8m/s 2 )(l0m) - 9.8 x 1 4 J. 

At point 3, y\ - - 1 5 m, since point 3 is below point 1 . Therefore, 

pe, - mgy 3 - (100Qkg)(9.8m/s 2 )(-l5m) - -1.5 x 10 s J. 

(b) fn going from point 2 to point 3, the potential energy change (pEniiai ~~ pt iniiiai) ' s 

pe., - pe 2 - (-1.5 x 10 s J) - (9.8 x I0 4 J) 
= -2.5 X I0 ? J. 

The gravitational potential energy decreases by 2.5 X 10 s J. 

(c) In this instance, y, = + 15 m at point I, so the potential energy initially (at 
point 1 ) is 

pe, = ( 1000 kg)(9.8m/s 2 )( 15 m) = 1.5 X 10 5 J. 

At point 2, yj - 25 m, so the potential energy is 

pej - 2.5 x I0 5 J. 

At point 3, Vj - 0, so the potential energy is zero. The change in potential 
energy going from point 2 to point 3 is 



pe, - pe, - - 2.5 x 1 s J 
which is the same as in part (&). 



-2.5 x 10 s J, 



There are other kinds of potential energy besides gravitational. Each form 
of potential energy is associated with a particular force, and can be defined 
analogously to gravitational potential energy. In general, the change in potential 
energy associated with a particular force is equal to the negative of the work 
done by that force if the object is moved from one point to a second point (as in 
Eq. 6-7b for gravity). Alternatively, because of Newton's third law, we can 
define the change in potential energy ax the Work required of an external force to 
move the object 'without acceleration between the two points, as in Eq. 6-7a. 



146 CHAPTERS Work and Energy 



^wwww 



(a) 




* (b) 




[ (c) 



FIGURE 6-13 (a) A spring can store 
energy (elastic PE) when compressed as in (b) 
and can do work when released (c). 



We now consider another type of potential energy, thai associated with 
elastic materials. This includes a great variety of practical applications. Consider 
the simple coil spring shown in Fig. 6- 1 3. The spring has potential energy when 
compressed (or stretched), for when it is released, it can da work on a ball as 
shown. Tti hold a spring either stretched or compressed an amount x from its 
natural (unstretched) length requires the hand to exert a force on the spring, />, 
that is directly proportional to v. That is, 

Fp = kx, 

where A: is a constant, called the spring stiffness constant, and is a measure of the 
stiffness of the particular spring. The stretched or compressed spring exerts a 
force F s in the opposite direction on the hand, as shown in Fig. 6- 14: 

F s = -kx. (6-8) 

This force is sometimes called a "restoring force' 1 because the spring exerts its 
force in the direction opposite the displacement (hence the minus sign), acting 
to return it to its natural length. Equation 6-8 is known as the spring equation 
and also as H««ke's law, and is accurate for springs as long as x is not too great. 
To calculate the potential energy of a stretched spring, let us calculate the 
work required to stretch it (Fig. 6— 14b). We might expect to use Eq. 6-1 for 
the work done tin it, W = Fx, where x is the amount it is stretched from its 
natural length. But this would be incorrect since the force F P (- kx) is not 
constant but varies over this distance, becoming greater the more the spring is 
stretched, as shown graphically in Fig. 6-15, So let us use the average force, F. 
Since Fp varies linearly — from zero at the unstretched position to kx when 
stretched to x — the average force is F = ([0 + kx] - \kx, where x here is 
the final amount stretched (shown as x { in Fig. 6- IS for clarity). The work 
done is then 

W = Fx = {{kx){x) = \kx z . 

Hence the elastic potential energy is proportional to the square of the amount 
stretched: 



;v of eisak <tpnn$ 



clastic te = \kx l . 



If a spring is compressed a distance x from its natural length, the average force is 
again F - {kx, and again the potential energy is given by Eq. 6-9. Thus x 
can be either the amount compressed or amount stretched from the spring's 
natural length. ' Note that for a spring, wc choose the reference point for zero PE 
at the spring's natural position. 

{ Wc ran also obtain F.q. d-l using Section f>-2_ The work done, and hence Apr, equals- the area 
under the F vs. x graph of Fig. 6 15. This area is a triangle (colored in Fig. 6 15 ) of altitude kx and 
base x. and hence of area (for a Iriangle) equal to j(t.v)(.v) = j&v 2 . 




FIGURE 6-14 (a) Spring in natural 
(unstretched) position, (b) Spring is 
stretched by a person exerting a 
force Fp to the right (positive direc- 
tion), The spring pulls back with a 
force F s . where F s = ~kx- 
(c) Person compresses the spring 
[x < 0) by exerting a force F P to 
the left; the spring pushes back with 
a foTce F$ = — kx, wheTe Fg > 
because X < 0. 



(6-9) Elastic PE 




FIGURE 6-15 As a spring is stretched (ot compressed). 

the force needed increases linearly as x increases: graph 
of F = kx vs. .v from x = to x — xr. 



SECTION 6-4 Potential Energy 147 



TABLE 6-1 Conservative 
and Nonconservative Forces 


Cimservative 
Forces 


Nonconservative 

Forces 


Gravitational 


Friction 


Elastic 


Air resistance 


Electric 


Tension in cord 


MotoT or rocket 
propulsion 


Push or pull by 
a person 



In each of the above examples of potential energy — from a brick held at a 
height y, to a stretched or compressed spring — an object has the capacity ot 
potential to do work even though it is not yet actually doing it. These examples 
show that energy can be stored, for later use, in the form of potential energy 
(Fig. 6-13, for instance, for a spring). 

Note that there is a single universal formula foT the Uanslalional kinetic 
energy of an object, jflitr, but there is no single formula for potential energy. Instead, 
the mathematical form of the potential eneTgy depends on the force involved. 

Conservative and Nonconservative Forces 

The work done against gravity in moving an object from one point to another 
does not depend on the path taken. For example, it takes the same work 
(= m#y) to lift an object of mass m vertically a certain height as to carry it up 
an incline of the same vertical height, as in Fig. 6-4 (see Example 6-2). Forces 
such as gravity, for which the work done does not depend on the path taken but 
only on the initial and final positions, are called conservative fortes. The elastic 
force of a spring (or other clastic material) in which F = -kx, is also a conser- 
vative force, An object that starts at a given point and returns to that same point 
under the action of a conservative force has no net work done on it because the 
potential energy is the same at the start and the finish of such a round trip, 

Friction, on the other hand, is a nonconservative forte since the work it docs 
depends on the path, For example, when a crate is moved across a floor from one 
point to another, the work done depends on whether the path taken is straight, or is 
curved or zigzag. As shown in Fig. 6-16, if a crate is pushed from point I to point 2 
along the longer semicircular path rather than along the straight path, more work 
is done against friction. That is because the distance is greater and, unlike the 
gravitational force, the friction force is always directed opposite to the direction of 
motion. (The cosf? term in Eq. 6- 1 is always cos 180° - —I at all points on the 
path for the friction force.) Thus the work done by friction in Fig. 6-16 does not 
depend only on points 1 and 2, Other forces that are nonconservative include the 
force exerted by a person and tension in a rope (see Table 6- 1 ). 



FIGURE 6-16 A crate is pushed across 

the floor from position 1 to position 2 via 
two paths, one straight and one curved. 

The friction force is always in the direc- 
tion exactly opposed to the direction of 
motion. Hence, for a constant magnitude 
friction force, lVj r = -F lr rf, so if d is 
greater (as for the curved path), then W is 
greater- Hie work done does not depend 
only on points 1 and 2. 




pi: a defined only for a conservative force 
There is no pi: for friction 



Because potential energy is energy associated with the position or configu- 
ration of objects, potential energy can only make sense if it can be stated 
uniquely for a given point. This cannot be done with nonconservative forces 
since the work done depends on the path taken (as in Fig. 6-16). Hence, 
potential energy can be defined only for a conservative force. Thus, although 
potential energy is always associated with a force, not all forces have a potential 
energy. For example, there is no potential energy for friction. 

EXERCISE C An object acted on by a constant force F moves from point 1 to point 2 
and back again. The work done by the force F in this round trip is 60 J. Can you deter- 
mine from this information if f-'is a conservative or nonconservative force? 

We can now extend the work-energy principle (discussed in Section 6-3) to 
include potential energy. Suppose several forces act on an object which can 
undergo ttanslational motion. And suppose only some of these forces are 



148 CHAPTER 6 Work and Energy 



conservative. We write the total (net) woTk W ni . t as a sum of the work done by 
conservative foTces, W c , and the work dure by non conservative forces, VK NC ; 

1*^ = W c + W NC . 

Then, from the work-energy principle. Eq. 6-4, we have 

W asl = Akf. 

W c + W sc = Akf. 

where Ake - ke 2 - ke l . Then 

W NC = Akf. - W c . 

Work done by a conservative force can be written in terms of potential energy, 
as we saw in Eq. 6-7b for gravitational potential energy; 



W, 



Ape. 



We combine these last two equations: 
W NC - Ake + Apk. 



(6-10) 



Thus, the work W yi done by the rtonconsermtive forces acting on an object is 
equal to the total change in kinetic and potential energies. 

It must be emphasized that alt the forces acting on an object must be 
included in Eq, 6-10, cither in the potential energy term on the right (if it is a 
conservative force), or in the work term on the left (but not in both!). 



WORK-ENERG Y PRINCIPLE 
(general form) 



Mechanical Energy and Its Conservation 

If only conservative forces arc acting in a system, wc arrive at a particularly 
simple and beautiful relation involving energy. 

When no non conservative forces are present, then W m - = in Eq. 6-10, 
the general form of the work-energy principle, Then we have 



Ake + Ape = 



or 



[con 
for 



se native 
forces only 



(kE 2 - Kb,) + (Ft, - PE|) - 0, 



conservative 
forces only 



(6-lla) 



(6-llb) 



We now define a quantity E, called the total mechanical energy of our system, 

as the sum of the kinetic and potential energies at any moment: 

F = KE + PR. 
Now we can rewrite Eq. 6- 1 1 b as 

f conservative 

KE; + PE, = KE, + PE, ,. , 

[ forces only 
or 



(6- 12a) 



Ej = Ei = constant. 



conservative 
forces only 



(6- 12b) 



Equations 6-12 express a useful and profound principle regarding the total 
mechanical energy of a system — namely, that it is a conserved quantity. The total 
mechanical energy E remains constant as long as no nonconservative forces act: 
(ke + pe) at some initial time 1 is equal to the (ke + pe) at any later time 2. 

To say it another way, consider Eq. 6-lla which tells us Ape = -Ake; that 
is, if the kinetic energy ke of a system increases, then the potential energy pe 
must decrease by an equivalent amount to compensate. Thus, the total, ke + pe, 
remains constant: 

If only conservative forces are acting, the total mechanical energy of a 
system neither increases nor decreases in any process. It stays constant — it 
is conserved. 

This is the principle of conservation of mechanical energy for conservative forces. 



Total mechanical energy defined 



CONSERVATION OF 

M E CHA NICA L EN ER G Y 



CONSERVATION OF 
MECHANICAL ENERGY 



SECTION 6-6 Mechanical Energy and Its Conservation 149 




V, = h 



all ke 



v,=0 



FIGURE 6-17 As it falls, the rock's 
potential energy changes to kinetic 

energy. 



Vtnisa'<\ni<ni t'f tiitdtitincu! ciwi^y 
when only gravity acts 



FIGURE 6-18 F.ncrgy buckets (for 
Hxample 6-8). Kinetic energy is red 
and potential energy is blue. The 
total (ke + pv.) is [he siinic for the 
three points shown. "The speed at 
y = t), just before the rock hits 
the ground, is 

\/2(9.8 m/s 2 )(3.() m) = 7.7 m/s. 



W = 



3.0 m 



ll Jj_ 



6,3 m/s 



W = 1.0 m 



7.7 m/s c 



H 




Iii the next Section we shall see the great usefulness of the conservation of 
mechanical energy principle in a variety of situations, and how il is often easier 
to use than the kinematic equations or Newton's laws. After that we will discuss 
how other forms of energy can be included in the general conservation of 
energy law that includes energy associated with n on conservative forces. 

Problem Solving Using 
Conservation of Mechanical Energy 

A simple example of the conservation of mechanical energy (neglecting air resis- 
tance) is a rock allowed to fall under gravity from a height ft above the ground, as 
shown in Fig. 6-17, Tf the rock starts from rest, all of the initial energy is potential 
energy. As the rock falls, the potential energy decreases (because y decreases), but 
the rock's kinetic energy increases to compensate, so that the sum of the two remains 
constant. At any point along the path, the total mechanical energy is given by 

E - Kt + pt - \mv l + mgy 

where y is the rock's height above the ground at a given instant and v is its 
speed at that point. If we let the subscript 1 represent the rock at one point 
along its path (for example, the initial point), and the subscript 2 represent it at 
some other point, then we can write 

total mechanical energy at point I - total mechanical energy at point 2 

or (sec also Eq. 6- 12a) 



■ m 



v\ + mg\\ = j tnvl + mg v? . 



[grav. ee only] (6-13) 

Just before the rock hits the ground, where we chose _y — 0, all of the initial 
potential energy will have been transformed into kinetic energy. 



EXAMPLE 6-8 



Falling rock. If the original height of the rock in Fig. 6-17 



is v, = h = 3.0 m, calculate the rocks speed when it has fallen to 1.0 m above 

the ground. 

APPROACH One approach is to use the kinematic equations of Chapter 2. 
Let us instead apply the principle of conservation of mechanical energy, 
Eq. 6-13, assuming that only gravity acts on the Tock. We choose the ground as 
our reference level ( y = 0), 

SOLUTION At the moment of release (point 1) the rock's position is 
y, - 3.0 m and it is at rest: v t - 0. We want to find v 2 when the rock is at 
position y? - 1.0 m. Equation 6-13 gives 

jmuf + mgyi = jmv\ + mgy 2 . 

The »i's cancel out; setting i\ = and solving for vj we find 

v$ = 2g(y, - y 2 ) 



and 



= 2(9.8 m/s 2 )[ (3.0 m) - (1.0 m)] = 39.2 m 2 /s 2 , 



■»■> = V 3 9,2 m/s = 6.3 m/s. 
The rock's speed 1.0 m above the ground is 6.3 m/s downward. 
NOTE The velocity at point 2 is independent of the rock's mass. 



EXERCISE D Solve IsMmple 6-8 by using the work-energy principle applied to the rock, 
without the concept of potential energy. Show all equations you use. starting with Eq. 6-4. 

A simple way to visualize energy conservation is with an "energy bucket" as 
shown in Fig. 6-18. At each point in the fall of the rock, for example, the 
amount of kinetic energy and potential energy arc shown as if they were two 
differently colored materials in the bucket. The total amount of material in the 
bucket (- total mechanical energy) remains constant. 



150 CHAPTER 6 Work and Energy 



Equation 6-13 can be applied to any object moving without friction under 
the action of gravity. For example, Fig, 6-19 shows a roller-coaster car starting 
from rest at the top of a hill, and coasting without friction to the bottom and up 
the hill on the other side/ Initially, the car has only potential energy. As it coasts 
down the hill, il loses potential energy and gains in kinetic energy, hut the sum of 
the two remains constant, At the bottom of the hill it has its maximum kinetic 
energy; as it climbs up the other side, the kinetic energy changes back to potential 
energy. When the car comes to rest again, all of its energy will be potential energy- 
Given that the potential energy is proportional to the vertical height, energy 
conservation tells us that (in the ahsence of friction) the car comes to rest at a 
height equal to its original height. If the two hills are the same height, the car will 
just barely reach the top of the second hill when it stops. If the second hill is lower 
than the first, not all of the car's kinetic energy will be transformed to potential 
energy and the car can continue over the top and down the other side. If instead 
the second hill is higher, the car will only reach a height on it equal to its original 
height on the first hill. This is true (in the absence of friction) no matter how steep 
the hill is, since potential energy depends only on the vertical height (Eq. 6-6), 




FIGURE 6-19 A roller-coaster car 

moving without friction illustrates 
the conservation of mechanical 

energy. 



Grav. pf, depends on vertical height, 

not path length (F.fj. f>-6) 



EXAMPLE 6-9 



Roller coaster speed using energy conservation. 
Assuming the height of the hill in Fig, 6-19 is 40 m, and the roller-coaster car 
starts from rest at the top, calculate (a) the speed of the roller-coaster ear at 
the bottom of the hill, and (b) at what height it will have half this speed. Take 
y = at the bottom of the hill. 

APPROACH We choose point 1 to be where the car starts from rest (u, - 0) at 
the top of the hill (y, - 40 m). Point 2 is the bottom of the hill, which we choose 
as our reference level, so y, - 0. We use conservation of mechanical energy. 

SOLUTION (a) We use Eq. 6- 1 .3 with v, - and y, - 0. Then 
jjra>i + mgy L = hm>\ + mgy? 
m gyi = 3»iw|. 
The w's cancel out and, setting y, = 40 m, we find 



v 2 = \/2^ = V '2(9.8 m/s 2 )( 40 m) = 28 m/s. 

(b) We again use conservation of energy, 

\mv\ + mgy\ = \mv\ + mgy 2 , 

but now (ij = 14 m/s (half of 2-8 m/s) and y? is unknown. We cancel the m\ 
set V[ = 0, and solve for y^: 



"■': 



- - 1 = 30 m. 
2g 



That is, the car has a speed of 14 m/s when it is 30 vertical meters above the 
lowest point, both when descending the left-hand hill and when ascending the 
right-hand hill. 

NOTE The mathematics of this Example is almost the same as that in 
Example 6-8. But there is an important difference between them. Example 6-8 
could have been solved using force, acceleration, and the kinematic equations 
(Eqs. 2-1 1). But here, where the motion is not vertical, that approach would have 
been too complicated, whereas energy conservation readily gives us the answer. 



'The forces on the car arc gravity, the norma] force exerted hy the track, and friction (here, assumed 
zero). The normal force acts perpendicular to the track, and so is always perpendicular to the 
motion and does no work, lluis W^,- = (J in Eq. 6-10 (so mechanical energy is conserved) and we 
can use Lu,, 6-1.3 with the potential energy being only gravitational potential energy. We will see 
how to deal with friction, for which W sc 4- U, in Section fi-9. 



SECTION 6-7 Problem Solving Using Conservation of Mechanical Energy 151 




CONCEPTUAL EXAMPLE 6-10 1 Speeds on two water slides. Two 



FIGURE 6-20 



water slides at a pool are shaped differently, but have the same length arid 
start at the same height h (Fig. 6-20). Two riders, Paul and Kathleen, start 
from rest at the same time on different slides- (a) Which rider, Paul or Kath- 
leen, is traveling faster at the bottom? (b) Which rider makes it to the bottom 
first? Ignore friction, 

RESPONSE (a) Each rider's initial potential energy mgh gets transformed to 
kinetic energy, so the speed v at the bottom is obtained from \nw 2 = mgh. 
The mass cancels and so the speed will be the same, regardless of the mass of 
the rider. Since they descend the same vertical height, they will finish with the 
same speed. 

(b) Note that Kathleen is consistently at a lower elevation than Paul at any 
instant, until the end. This means she has converted her potential energy to kinetic 
energy earlier. Consequently, she is traveling faster than Paul for the whole trip, 
except toward the end where Paul finally gets up to the same speed. Since she 
was going faster for the whole trip, and the distance is the same, Kathleen gets 
to the bottom first. 



EXERCISE E Two balls are released from the- same height above the floor. Ball A tails 
freely through the air, whereas ball B slides On a curved frkiionless track to the Hour. 
How do the speeds of the balls compare when they reach the floor? 



«* PROBtEM SOLVING 
Whether m use energy, or Neirttm \- laws? 



Ig J F H V S I C S APPLIED 

Sports 

FIGURE 6-21 Transformation of 
energy during a pole vault. 




You may wonder sometimes whether to approach a problem using work 
and energy, or instead to use Newton's laws. As a rough guideline, if the foree(s) 
involved arc constant, cither approach may succeed. If the forces arc not 
constant, and/or the path is not simple, energy may be the surest approach. 

There are many interesting examples of the conservation of energy in 
sports, such as the pole vault illustrated in Fig. 6-21. We often have to make 
approximations, but the sequence of events in broad outline for the pole vault is as 
follows. The initial kinetic energy of the running athlete is transformed into elastic 
potential energy of the bending pole and, as the athlete leaves the ground, 
into gravitational potential energy. When the vaulter reaches the top and the 
pole has straightened out again, the energy has all been transformed into 
gravitational potential energy (if we ignore the vaulter's low horizontal speed 
over the bar). The pole docs not supply any energy, but it acts as a device to 
store energy and thus aid in the transformation of kinetic energy into gravita- 
tional potential energy, which is the net result- The energy required to pass 
over the bar depends on how high the center of mass (cm) of the vaulter 
must be raised. By bending their bodies, pole vaultcrs keep their cm so low 
that it can actually pass slightly beneath the bar (Fig, 6-22), thus enabling 
them to cross over a higher bar than would otherwise be possible. (Center of 
mass is covered in Chapter 7.) 



FIGURE 6-22 By heading their bodies, pole vauhers. can keep 
their center of mass so low that it may even pass below the bar. 
By changing their kinetic energy (of running) into gravitational 
potential energy (= tngy) in this way. vaulters can cross over a 
higher bar than if the change in potential energy were accom- 
plished without carefully bending the body. 




152 CHAPTERS Work and Energy 



As another example of the conservation of mechanical energy, let us consider 
an object of mass in connected to a horizontal spring whose own mass can be 
neglected and whose spring stiffness constant is fc.The mass m has speed v at any 
moment. The potential energy of the system (object plus spring) is given by 
Fq, 6-9, pf = jkx 1 , where x is the displacement of the spring from its unstretched 
length, If neither friction nor any otheT foTce is acting, conservation of mechanical 
energy tells us that 



,mv\ 



+ jkx] — jtnvj + jkxj, 



[elastic PE only| (6-14) 



where the subscripts 1 and 2 refer to the velocity and displacement at two 

different moments. 



( 'onservation of mechanical energy 
when PI is elastic 



EXAMPLE 6-11 



Toy dart gun. A dart of mass 0.100 kg is pressed against 
the spring of a toy dart gun as shown in Fig. 6-23a. The spring (with spring 
stiffness constant k - 250 N/m) is compressed 6.0 cm and released. If the dart 
detaches from the spring when the spring reaches its natural length (jr - 0), 
what speed does the dart acquire? 

APPROACH The dart is initially at rest (point 1), so ke, = 0. We ignore friction 
and use conservation of mechanical energy; the only potential energy is elastic. 

SOLUTION We use Cq. 6-14 with point I being at the maximum compres- 
sion of the spring, so v, - (dart not yet released) and x t - —0.060 m. 
Point 2 we choose to be the instant the dart flies off the end of the spring 
(Fig, 6-23b), so x z - and we want to find v 2 . Thus Eq, 6-14 can be written 



Then 



+ \kx 2 . = kmv\ + 0. 



kin 



(250 N/m) (-0.060 m) 2 



(0.1 00 kg) 



9.0 m z /s 2 



so lh 



Vti 



3.0 m/s. 



NOTE In the horizontal direction, the only force on the dart (neglecting 
friction) was the force exerted by the spring. Vertically, gravity was counterbal- 
anced by the normal force exerted on the dart by the gun barrel. After it 
leaves the barrel, the dart will follow a projectile's path under gravity. 




u) 



FIGURE 6-23 Example 6-11. (a) A dart is 
pushed against a spring, compressing it 6.0 cm. 
The daTt is (hen released, and in (b) il leaves 

the spring at velocity -i>, . 



- 



■ 



' I 



(fa) 



( < 



i- 6.0 cm - 



i£&ki4k£tffc 



- 



f-i 



SECTION 6-7 Problem Solving Using Conservation of Mechanical Energy 153 



C ttiiscrvaikm of energy: 
gravity and elastic pn 



Additional Example 

The next Example shows how to solve a problem involving two types of 
potential energy. 



EXAMPLE 6-12 



Two kinds of PE. A ball of mass m = 2.60 kg, starting 
from rest, falls a vertical distance h = 55.0 em before striking a vertical coiled 
spring, which it compresses an amount Y — 15,0cm (Fig. 6-24). Determine 
the spring stiffness constant of the spring. Assume the spring has negligible 
mass, and ignore air resistance. Measure all distances from the point where the 
ball first touches the uncompressed spring (y - at this point). 

APPROACH The forces acting on the ball are the gravitational pull of the 
Earth and the elastic force exerted by the spring, Roth forces are conservative, 
so we can use conservation of mechanical energy, including both types of 
potential energy. We must be careful, however: gravity acts throughout the fall 
(Fig. 6-24), whereas the elastic force does not act until the ball touches the 
spring (Fig, 6-24b). We choose y positive upward, and v = at the end of 
the spring in its natural (uncompressed) state. 

SOLUTION We divide this solution into two parts, (An alternate solution follows.) 
Part 1: Let us first consider the energy changes as the ball falls from a height 
y'i - k - 0,55 m, Fig. 6-24a, to y, - 0, just as it touches the spring. Fig. 6-24b, 
Our system is the ball acted on by gravity plus the spring, which up to this point 
doesn't do anything. Thus 

jmv] + mgV] = \m\i\ + wgy 3 

+ mgh - \mv\ + 0. 



Wc solve for t> 2 = \/2gh = \/2(9.80m/s r ](0.550m) = 3,2S3m/s « 3.28 m/s. 
This is the speed of the ball just as it touches the top of the spring, Fig, 6-24b, 
Part 2: Let's sec what happens as the ball compresses the spring. Figs, 6-24b to c, 
Now there arc two conservative forces on the ball — gravity and the spring force. 
So our conservation of energy equation becomes 

£(ball touches spring) = £(spring compresses) 

\mv\ + >ngy 2 + jky\ - {mv\ + ingyj + {kyj- 

We take point 2 to be the instant when the ball just touches the spring, so y z - 
and U; - 3.283 m/s (keeping an extra digit for now). We take point 3 to be 
when the ball comes to rest (for an instant) and the spring is fully compressed, 
so v 3 - and y, - —Y - -0.150 m (given). Substituting into the above 
energy equation, we get 

\mv\ + + - - mgY + ^kY 2 . 

We know m, v 2 , and Y, so we can solve for !c. 

jib* 
(2.60 kg) 



k = 



+ m g Y\ = ^[f$ + 2gY\ 



(0.150 m) ? 
which is the result we sought. 



[(3,283 m/s) 2 + 2(9,80 m/s 2 )(0. 150 m}] = l590N/m, 



FIGURE 6-24 f -*;iniplc 6 12. 



y = Q- 



illi 



T" 

A. 



; - V | - It 



y=j, = Q 



v = v, = -y 



(a) I 



(b)' 



U) 



151 CHAPTER 6 Work and Energy 



Alternate Solution Instead of dividing the solution into two parts, we can do 
it all at once. After all, we get to choose what two points are used on the left 
and right of the energy equation. Let us write the energy equation for points I 
and 3 (Fig, 6-24). Point 1 is the initial point just before the ball starts to fall 
(Fig, 6-24a), so if] = 0, Vj = h = 0.550 m; and point 3 is when the spring is 
fully compressed (Fig, 6— 24c), so V\ = 0, Vi = — Y — —0,150 m, The forces on 
the ball in this process are gravity and (at least pan of the time) the spring. So 
conservation of energy tells us 

\rnvl + mgvi + i^(0) 2 - 2 mv l + m SYi + ?kyl 
+ mgh +0 = - mgY + {kY J 

where we have set y = for the spring at point I because it is not acting and 
is not compressed or stretched at point I . We solve for fc 



*» PROBLEM SOLVING 
Alternate Solution 



k = 



2mg(h + Y} 2(2.60 kg}(9.80m/s 2 ) (0.550 m + 0.150m} 



Y 2 (0.150 m) 2 

just as in our first method of solution. 



= l590N/m 




Other Forms of Energy; Energy Transformations 
and the Law of Conservation of Energy 

Besides the kinetic energy and potential energy of ordinary objects, other forms 
of energy can be defined as well. These include electric energy, nuclear energy, 
thermal energy, and the chemical energy stored in food and fuels. With the advent 
of the atomic theory, these other forms of eneTgy have come to be considered as 
kinetic or potential energy at the atomic or molecular level, For example, 
according to the atomic theory, thermal energy is the kinetic eneTgy of rapidly 
moving molecules — when an object is heated, the molecules that make up the 
object move faster. On the other hand, the energy stored in food and fuel such as 
gasoline is potential energy stored by virtue of the relative positions of the atoms 
within a molecule due to electric forces between the atoms (referred to as chem- 
ical bonds), For the energy in chemical bonds to be used to do work, it must be 
released, usually through chemical reactions. This is analogous to a compressed 
spring which, when released, can do work. Electric, magnetic, and nuclear energies 
also can be considered examples of kinetic and potential (or stored) energies. We 
will deal with these other forms of energy in detail in later Chapters. 

Energy can be transformed from one form to another, and we have already 
encountered several examples of this. A rock held high in the air has potential 
energy; as it falls, it loses potential energy, since its height above the ground 
decreases. At the same time, it gains in kinetic energy, since its velocity is 
increasing. Potential energy is being transformed into kinetic energy. 

Often the transformation of energy involves a transfer of energy from one 
object to another. The potential energy stored in the spring of Fig. 6- 13b is 
transformed into the kinetic energy of the ball, Fig, 6- 13c, Water at the top of a 
dam has potential energy, which is transformed into kinetic energy as the water 
falls. At the base of the dam, the kinetic energy of the water can be transferred 
to turbine blades and further transformed into electric energy, as we shall sec in 
a later Chapter. The potential energy stored in a bent bow can be transformed 
into kinetic energy of the arrow (Pig. 6-25). 

Tn each of these examples, the transfer of energy is accompanied by the 
performance of work. The spring of Fig, 6-13 does work on the ball. Water does 
work on turbine blades. A bow does work on an arrow. This observation gives us 
a further insight into the relation between work and energy: work is done when 
energy is transferred from one object to another.'' A person throwing a ball or 
pushing a grocery cart provides another example. The work done is a manifesta- 
tion of energy being transferred from the person (ultimately derived from the 
chemical energy of food) to the ball or cart, 

'If the objects are at different temperatures, heat can flow between tlicm instead, i.>r in addition. See 
Chapters 14 and IS 




FIGURE B-25 Potential energy of 
a bent bow about to be transformed 
into kinetic energy of an arrow. 



Work is done when energy is 
transferred from one object to another 



SECTION 6-8 155 



LAW OF 

CONSERVATION 

OF ENERGY 



One of the great Tesults of physics is that whenever energy is transferred ot 
transformed, il is found that no energy is gained ot losl in the process. 

This is the law of conservation of energy, one of the most important princi- 
ples in physics; it can be stated as; 

The total energy is neither increased nor decreased in any process. Energy 
can be transformed from one form to another, and transferred from one 
object to another, but the total amount remains constant. 

We have already discussed the conservation of energy for mechanical systems 
involving conservative forces, and we saw how it could be derived from 
Newton's laws and thus is equivalent to them. But in its full generality, the 
validity of the law of conservation of energy, encompassing all forms of energy- 
including those associated with no n conservative forces like friction, rests on 
experimental observation. Even though Newton's laws are found to fail in the 
submicrtiscopie world of the atom, the law of conservation of energy has been 
found to hold in every experimental situation so far tested. 



Dissipative forces 



Conservation of energy 
with gravity mid friction 




Energy Conservation with 
Dissipative Forces: Solving Problems 

In our applications of energy conservation in Section 6-7, we neglected friction, 
a nonconservative force, But in many situations it cannot be ignored. In a real 
situation, the roller-coaster car in Fig. 6-19, lor example, will not in fact reach 
the same height on the second hill as it had on the first hill because of friction, 
In this, and in other natural processes, the mechanical energy (sum of the 
kinetic and potential energies) docs not remain constant but decreases. Because 
frictional forces reduce the mechanical energy (but not the total energy), they 
are called dissipative forces. Historically, the presence of dissipative forces 
hindered the formulation of a comprehensive conservation of energy law until 
well into the nineteenth century. It was only then that heat, which is always 
produced when there is friction (try rubbing your hands together), was inter- 
preted in terms of energy. Quantitative studies by nineteenth-century scientists 
(discussed in Chapters 14 and IS) demonstrated that if heat is considered as a 
transfer of energy (thermal energy), then the total energy is conserved in any 
process, For example, if the roller-coaster car in Fig, 6-19 is subject to frictional 
forces, then the initial total energy of the car will be equal to the kinetic plus 
potential energy of the car at any subsequent point along its path plus the 
amount of thermal energy produced in the process, The thermal energy 
produced by a constant friction force F fr is equal to the work done by friction. 
We now apply the general form of the work-energy principle, Eq, 6- 1 0: 



n\ 



Ake + Ape, 



We can write W NC - —F if d, where d is the distance over which the friction 
force acts. (P and d are in opposite directions, hence the minus sign.) Thus, 



with Kt 



01 



\mv and Pb - mgy, we have 

-F tr d - jmvl - jmv\ + mgy 2 - mg}\ 



\mv\ + mgy] = jtnvj + >ng\'2 + Ffrd, 



gravity and 
friction acting 



(G-15) 



156 CHAPTER 6 



where d is the distance along the path traveled by the object in going from 
point I to point 2, Equation 6-15 can be seen to be Eq, 6-13 modified to 
include friction. It can be interpreted in a simple way: the initial mechanical 
energy of the car (point 1) equals the (reduced) final mechanical energy of the 
car plus the energy transformed by friction into thermal energy. 

When other forms of energy are involved, such as chemical or electrical 
energy, the total amount of energy is always found to be conserved. Hence the 
law of conservation of energy is believed to be universally valid. 



Work-Energy versus Energy Conservation 

The work-energy principle ;ind ihe law of conservation of energy are basically 
equivalent. The difference between Ihem is in how you use them, and in particular 
on your choice of the system under study. If you choose as your system one or 
more objects on which external forces do work, then you must use the work- 
energy principle: the work done by the external forces on your system equals the 
total change in energy of your chosen system. 

On the other hand, if you choose a system on which no external forces do 
work, then you can apply conservation of energy to that system. 

Consider foT example, a spring connected to a block on a frictionless table 
(Fig. 6-26). If you choose the block as your system, then the work done on the 
block by the spring equals the change in kinetic energy of the block: the work- 
energy principle. (Energy conservation does not apply to this system — the 
block's energy changes.) If instead you choose the block plus the spring as your 
system, no external forces do work (since the spring is part of the chosen 
system), To this system you can apply conservation of energy: if you compress 
the spring and then release it, the spring still exerts a force on the block, but the 
subsequent motion can be discussed in terms of kinetic energy (jtnv 2 ) plus 
potential energy (j&A- 2 ), whose total remains constant. 

Conservation of energy applies to any system on which no work is done by 
external forces. 





-jft 


WVS 


d m 







FIGURE 6-26 A spring connected 
to a block on a frictionless table, If 
you choose your system lo be (he 
block plus spring, then 

F = jwt> 2 + \kx 2 
is conserved. 



PROBLEM SOLVING 



Conservation of Energy 



1. Draw a picture of the physical situation. 

2. Determine ihe system for which energy will be 
conserved: the object or objects and the forces acting. 

3. Ask yourself what quantity you are looking for, and 
decide what are the initial (point I) and final 
(point 2) positions. 

4. If the object under investigation changes its height 
during the problem, then chouse a reference frame 
with a convenient y = level for gravitational 
potential energy: the lowest point in the problem is 
often a good choice, 

If springs arc involved, choose the unstretched 
spring position to be x (or y) = 0. 



5, Apply conservation of energy. If no friction or 
other oonconservative forces act, then conservation 
of mechanical energy holds; 

KG, + PE, = KEj + PE 2 . 

If friction or other nonconservativc forces arc present, 
then an additional term (W NC ) will be needed: 

VY NC = Akr + Ape. 

To be sure which sign to give W Nr , you can use 
your intuition: is the total mechanical energy 
increased or decreased in the process? 

6. Use the equation(s) you develop to solve for the 
unknown quantity. 



EXAMPLE 6-13 



Friction on the roller coaster. The roller-coaster car in 
Example 6-9 reaches a vertical height of only 25 m on the second hill before 
coming to a momentary stop (Fig. 6-27), It traveled a total distance of 400 m. Esti- 
mate the average friction force (assume constant) on the car, whose mass is 1000 kg. 

APPROACH We explicitly follow the Problem Solving Box step by step. 
SOLUTION I. Draw a Picture. See Fig. 6-27. 

2. The system. The system is the roller-coaster car (and the Earth since it exerts the 
gravitational force). The forces acting on the car arc gravity and friction. (The 
normal force also acts on the car, but does no work, so it does not affect the energy.) 

3. Choose initial and final positions. We take point I to be the instant when 
the car started coasting (at the top of the first hill), and point 2 to be the 
instant it stopped 25 m up the second hill. 

4. Choose a reference frame. We choose the lowest point in the motion to be 
y = for the gravitational potential energy. 

5. Apply conservation of energy. There is friction acting on the caT, so we use 
conservation of energy in the form of Eq. 6-15, with vy = 0, V| = 40 m, 
v 2 = 0'. v i = 25m, and d = 400m. Thus 

0+ (1000kg)(<5.8m/s : )(40m) = + ( 1000kg)(9.8m/s 2 )(25m} + F, r (400m). 

6. Solve. We can solve this equation foT F r , : F ir - 370 N. 



FIG U RE 6-27 Ex am p le 6- 1 3. 

Because of friction, a roller coaster 
car does not reach the original 
height on the second hill. 




v - 



SECTION 6-9 157 



Problem solving is not ;i process that can be done by following a set of 
Tules. The Problem Solving Box on page 157 is thus nol a prescription, but is a 
summary of steps to help you gel started in solving problems involving energy. 



Power defined 



Average power 



Power units: the watt 



t'ht> lit/rit'ptucer 



<£■ 



CAUTION 



OisiragwisSi between 
power and energy 



1 Power 



Power is defined as the rate at which work is done. Average power equals the 
work dtme divided by the time to do it. Power can also be defined as the rate at 
which energy is transformed. Thus 



P — average power = 



work energy transformed 



time 



time 



(6-16) 



The poweT of a horse refers to how much work it can do per unit time. The 
power rating of an engine refers to how much chemical ot electrical energy can 
be transformed into mechanical energy per unit time, Tn ST units, power is 
measured in joules per second, and this unit is given a special name, the watt 
(W); I W = 1 J/s. We are most familiaT with the watt foT electrical devices: the 
Tate at which an electric lightbulb or heater changes electric energy into light ot 
thermal energy; but the watt is used for other types of energy transformations 
as well. In the British system, the unit of power is the foot-pound per second 
(ft-lb/s). For practical purposes, a larger unit is often used, the horsepower 
One horsepower' (hp) is defined as 550 ft ■ Ib/s, which equals 74fi W. 

To see the distinction between energy and power, consider the following 
example. A person is limited in the work he or she can do, not only by the total 
energy required, but also by how fast this energy is transformed: that is, by power. 
For example, a person may he ahle to walk a long distance or climb many flights 
of stairs before having to stop because so much energy has been expended. On 
the other hand, a person who runs very quickly upstairs may fall exhausted after 
only a flight or two, He or she is limited in this case by power, the rate at which 
his or her body can transform chemical energy into mechanical energy. 




FIGURE 6-Z8 Example 6-14, 



EXAMPLE 6-14 



Stair-climbing power. A 6(J-kg jogger runs up a long 
flight of stairs in 4.0s (Fig. 6-28). The vertical height of the stairs is 4.5m. 
(a) Estimate the jogger's power output in watts and horsepower, (b) How 
much energy did this require? 

APPROACH The work done by the jogger is against gravity, and equals 
IV = mgy. To get her power output we divide W by the time it took, 

SOLUTION (a) The average power output was 



~P=™ 
t 



mgy (60kg)(9.8m/s : )(4.5m) 
( 4.0 s 



660 W. 



Since there are 746 W in I hp, the joggeT is doing work at a Tate of just under I hp 
A human cannot do work at this rate for very long. 

(h) The energy required is E = Pi (Eq. 6- 16). Since P = 660 W = 660 J/s, 
then E = (660 J/s) (4.0 s) = 2600 J. This result equals W = nigy. 

NOTE The person had to transform more energy than this 2600 J. The total 
energy transformed by a person or an engine always includes some thermal 
energy (recall how hot you get running up stairs). 



f Tlic unit was chosen by James Watt {1736-1819), who needed a way to specify tlie power of his 
newly developed steam engines. He found hy experiment that a good horse can work all day at an 
average rate of about 360 fl-lb/s. So as not to be accused of exaggeration in Ihc sale of his steam 
engines, he multiplied Ibis by 1 j when he defined the hp. 



158 CHAPTER 6 Work and Energy 



Automobile engines do work to overcome the force of friction (including 
air resistance), to climb hills, and to accelerate, A car is limited by the rate at 
which it can do work, which is why automobile engines are rated in horsepower. 
A car needs power most when climbing hills and when accelerating, In the next 
Example, we will calculate how much power is needed in these situations for a 
car of reasonable size, Even when a car travels on a level road at constant 
speed, it needs some power just to do work to overcome the retarding forces of 
internal friction and air resistance, These forces depend on the conditions and 
speed of the caT T but aTe typically in the range 400 N to 1000 N. 

It is often convenient to write power in terms of the net force F applied to 
an ohjeet and its speed v. This is readily done since P = W/t and W = Fd, 
where d is the distance traveled. Then 



1 ! 



Fv, 



(6-17) 



where v - djt is the average speed of the object. 



EXAMPLE 6-15 



Power needs of a car. Calculate the power required of a 
1400-kg car under the following circumstances: (a) the car climbs a 10" hill (a 
fairly steep hill) at a steady 80 km/h; arid (b) (he ear accelerates along a level 
road from 90 to 1 10 km/h in 6.0 s to pass another car. Assume the retarding 
force on the car is F R = 700 N throughout. Sec Fig, 6-29. 

APPROACH First we must be careful not to confuse F„ , which is due to air 
resistance and friction that retards the motion, with the force F needed to 
accelerate the car, which is the friction al force exerted hy the road on the 
tires — the reaction to the motor-driven tires pushing against the road. We 
must determine the latter force F before calculating the power. 
SOLUTION (a) To move at a steady speed up the hill, the ear must, by 
Newton's seeond law, exert a force F equal to the sum of the retarding force, 
TOON, and the component of gravity parallel to the hill, wig sin I0' : . Thus 

F = 700 N + mg sin 10" 

= 700 N + (1400kg)(9.80m/s 2 )(0.174) = 3100 N. 

Since T) - SO km/h = 22m/s and is parallel to F, then (Eq. 6-17) the power is 

p = Fv = (3100 N)(22 m/s) = 6,80 X 10 4 W = 91 hp. 

(b) The car accelerates from 25.0 m/s to 30.6 m/s (90 to 1 10 km/h). Thus the ear 
must exert a force that overcomes the 700-N retarding force plus that required to 
give it the acceleration 

(30.6 m/s -ZS.0m/s) 



6.0 s 



= 0,93 m/s : . 



We apply Newton's second law with x being the direction of motion: 

ms x = y.F x = F — F R . 

Then the force required, F, is 

F = ma x + F K 

= (1400 kg)(0.93 m/s 2 ) + 700N 

= 1300 N + 700 N = 2000 N. 

Since P = Fv, the required power increases with speed and the motor must 
be able to provide a maximum power output of 

P - (2000 N)(30.6 m/s) - 6.12 X I0 4 W - 82 hp. 

NOTE Even taking into account the fact that only 60 to 80% of the engine's 
power output reaches the wheels, it is clear from these calculations that an 
engine of 100 to 150 hp is quite adequate from a practical point of view, 




FIGURE 6-29 Example 6- 1 5a, 
Calculation of power needed for a 
car to climb a hill. 



SECTION 6-10 Power 158 



Efficiency 



We mentioned in Example 6-15 that only pari of Ihe energy output of a caT 
engine reaches the wheels- Not only is some energy wasted in getting from 
the engine to the wheels, in the engine itself much of the input energy (from the 
gasoline) does not do useful work. An important characteristic of all engines is 
their overall efficiency e, defined as the ratio of the useful power output of the 
engine, P™ 



< )ut , to the power input, P ln : 
P:„ ' 



e = 



The efficiency is always less than 1 .0 because no engine can create energy, and 
no engine can even transform energy from one form to another without some 
energy going to friction, thermal energy, and other nonuseful forms of energy. 
For example, an automobile engine converts chemical energy released in the 
burning of gasoline into mechanical energy that moves the pistons and eventu- 
ally the wheels. But nearly K5% of the input energy is "wasted" as thermal 
energy that goes into the cooling system or out the exhaust pipe, plus friction in 
the moving parts. Thus car engines are roughly only about 15% efficient. We will 
discuss efficiency in detail in Chapter 15. 



[ Summary 



Wurk is done on an ohject by a force when the object moves 
through a distance d. If the direction of a constant force f- 
makes an angle ft with the direction of motion, the work done 
by this force is 



W = Fd cm 9. 



(6-1) 



Energy can be defined as the ability to do work. In SI 

units, work and energy are measured in joules (1 J = 1 N*m). 
Kinetic energy (KE) is energy of motion. An object of 
mass m and speed if has translational kinetic energy 

kf = jmv 1 . H>-3) 

Potential energy (pe) is energy associated with forces 
that depend on the position or configuration of objects, Grav- 
itational potential energy is 

P% av = mgy, (*-*) 

where v is the height of the object of mass m above an arbi- 
trary reference point. Elastic potential energy is given by 

elastic PC = ikx 1 ifi-9) 

for a stretched ot compressed spring, where .v is the displace- 
ment from the un stretched position and k is the spring stiff- 
ness constant, Other potential energies include chemical, 



electrical, and nuclear energy. The change in potential energy 
when an object changes position is equal to the external work 
needed to take the object from one position to the other. 

The work-energy principle states that the net work done 
on an object (by the net force) equals the change in kinetic 
energy of that object: 



WV-t = iKE = \mv$ 



(6-2, 6-4> 



The law of conservation of energy states that eneTgy can 

be transformed from one type to another, but the total energy 
remains constant. It is valid even when friction is present. 
since the heat generated can he considered a form of energy 
transfer, When only conservative, forces act. the total mechan- 
ical energy is conserved: 

KT-. + i-r. = con stan t- 

When non conservative forces such as friction act, then 



W NC = &ke + Ape. 



(6-IO> 



where W^c is the work done by nonconservative forces, 

Power is defined as the rate at which work is done, or the 
rate at which energy is transformed. The SI unit of power is 
the watt flW = I J/s). 



| Questions 



In what ways is the word "work" as used in even-day 

language the same as that defined in physics? In what 

ways is it different'? Give examples of both, 

Can a centripetal force ever do work on an object? 

Explain. 

Can the normal force on an object ever do work? Explain. 

A woman swimming upstream is not moving with respect 
to the short. Is she doing any work? If she stops swim- 
ming and merely floats, is work done on her? 



5, Is the work done hy kinetic friction forces always nega- 
tive? [Hint: Consider what happens to the dishes when 
you pull a tablecloth out from under them,| 

6, Why is it tiring to push hard against a solid wall even 
though you arc doing no work? 

7, You have two springs that are identical except that 
spring ] is stiffer than spring 2 (ky > fcj). On which spring 
is more work done (a) if they are stretched using the 
same force. (6) if they are stretched the same distance? 



160 CHAPTER 6 Work and Energy 



A hand exerts a constant horizontal force on a block that 
is free to slide on a friction [ess surface (Fig. 6-30). The 
hlocfc Marts from rest at point A. and by the time it has 
traveled a distance d to point B it is traveling with speed i>r . 
When the block has traveled another distance d to point C, 
will its speed be greater than, less than, or equal to 2*%? 
Explain your reasoning. 






Itt, 



[i 



FIGURE 6-30 

Question 8. 



By approximately how much dt>es your gravitational poten- 
tial energy change when you jump as high as you can"? 

In Fig. 6-31. water balloons 
are tossed from the roof of 
a building, all with the same 
speed but with different 
launch angles Which one has 
the highest speed on impact? 
Ignore air resistance. 




11. 



FIGURE 6-31 

Question 10, 



A pendulum is launched from a point that is a height h 
above its lowest point in two different ways (Fig. 6-32). 
During both launches, the pendulum is given an initial 
speed of 3.0 m/s. On the first launch, the initial velocity of 
the pendulum is directed upward along the trajectory, and 
on the second launch it is directed downward along the 
trajectory. Which launch will cause it to swing the largest 
angle from the equilibrium position? Explain. 




First launch) 



(Second launch) 



FIGURE 6-32 Question II. 



12, A coil spring of mass m rests upright on a table. If you 
compress the spring by pressing down with your hand and 

then release it, can the spring leave the table? Explain, 
using the law of conservation of energy. 



IX A bowling ball is hung from the ceiling by a Steel wiTe 
(Fig. 6-33). The instructor pulls the ball back and stands 
against the wall with the ball against his nose, To avoid 
injury the instructor is supposed to release Ihe ball 
without pushing it. Why? 




FIGURE 6-33 
Question 13. 

14. What happens to the gravitational potential energy when 
wateT at the top of a waterfall falls to the pool below? 

15. Describe the energy transformations when a child hops 
around on a pogo stick- 

16. Describe the energy transformations that lake place when 
a skier starts skiing down a hill, but after a time is brought 
to rest by Striking a snowdrift. 

17. A child on a sled (total mass m) starts from Test at the top 
of a hill of height h and slides down. Does the velocity at 
the bottom depend on the angle of the hill if [a) it is icy and 
there is no friction, and (ft) there is friction (deep snow)? 

18. Seasoned hikers prefer to step over a fallen log in their 
path rather than stepping on top and jumping down on 
the other side. Fxplain. 

19. Two identical arrows, one with twice the speed of the other, 
are fired into a bale of hay. Assuming the hay exerts a 
constant frictional force on the arrows, the faster arrow will 
penetrate how much farther than the slower arrow? Fxplain. 

2(1, Analyze the motion of a simple swinging pendulum in 
terms of energy, (a) ignoring friction, and (b) taking fric- 
tion into account. Explain why a grandfather clock has to 
be wound up, 

21. When a "superbair is dropped, can it rebound to a height 
greater than its original height? Explain. 

22. Suppose you lift a suitcase from the floor to a table, The 
work you do on the suitcase depends on which of the 
following: (a) whether you lift it straight up or along a 
more complicated path, (ft) the time it takes, (c) the 
height of the table, and (d) the weight of the suitcase? 

23. Repeat Question 22 for the power needed rather than the 
work. 

24. Why is it easier to climb a mountain via a zigzag trail than 
to climb straight up? 

25. Recall from Chapter 4. Example 
4-14, that you can use a pulley and 
ropes to decrease the force needed 
to raise a heavy load (see Fig. 6-34). 
But for every meleT the load is 
raised, how much rope must be 
pulled up? Account for this, using 
energy concepts, 

FIGURE 6-34 

Question 25. 




Questions 161 



I Problems 



6-1 Work, Constant Force 

1. (I) How much work is dune by the gravitational force 
when a 265-kg pile driver falls 2,80 m? 

2. (I) A 65.0-kg firefighter climbs a flight of stairs 20,0 m 
high. How much work is required? 

3. (I) A 1300-N crale rests on (he floor. How much work is 
required to move it at constant s|ieed (a) 4.0 ra along the 
floor against a friction force of 230 N. and (ft) 4.0 m vertically? 

J. (I) How much work did the movers do (horizontally) 
pushing a 160-kg crate 10.3 m across a rough floor 
without acceleration, if the effective coefficient of friction 
was 0,50? 

5. (II) A box of mass 5.0kg is accelerated from rest across a 
floor at a rate of 2.0 m/'s 2 for 7.0 s. Find the net work done 
on the box. 

6. (II) Eight books, each 4.3 cm thick with mass 1.7 kg, lie 
flat on a table. How much work is required to stack them 
one on lop of another? 

7. (II) A lever such as that shown in Fig, 6-35 can be used 
to lift objects we might not otherwise l>e able to lift. .Show 
that the ratio of output force. Fq , to input foTce. f-) . is 
related to the lengths /j and Iq from the pivot point by 
Fq/F] = Ij/Iq (ignoring friction and the mass of the 
lever), given that the work output equals work input. 




)C 




tf" T 



(b) 



FIGURE 6-35 
Problem 7. 

A simple lever, 



S. (II) A 330-kg piano slides 3.6 m down a 28' ; incline and is 
kepi from accelerating b>' a man who is pushing back on 
it parallel to the incline (Fig. 6-36). The effective coeffi- 
cient of kinetic friction is 0.40. Calculate: (n) the foTce 
exerted by the man, (ft) the work done by 
the man on the piano, (c) the work 
done by the friction force, 
(d) the work done by the 
force of gravity, and 
(c) the- net work done on 
the piano. 



FIGURE 6-36 

Problem 8, 




y. (II) (17) Find the force required to give a helicopter of mass M 
an acceleration of 0.10 g upward, (ft) Find the work done by 
this force as the helicopter moves a distance h upward. 
10, (II) What is the minimum work needed to push a 950- kg 
car 810 m up along a 9.0" incline? [a) Ignore friction, 
(ft) Assume the effective coefficient of friction retarding 
the car is 0,25, 

* 6-2 Work, Varying Force 

* LI. (II) In Fig. 6-fja, assume the distance axis is linear and 

that d.\ = 10-0 m and d fi = 35.0 m. Estimate the work 
done by force Fin moving a 2,80-kg object from d A to rf B . 

* 12. (II) The force on an object, acting along the x axis, varies 

as shown in Fig. 6-37. n>etermi«e the work done by this 
force to move the object (a) from x = 0.0 to x = 10.0 m. 
and (ft) from jc = 0.0 to x = 15.0m. 



400 
300 + 

- 100 
k 

-I TO 

-200 




I I I I I I I I I I I I I I I -t (m) 



K 




FIGURE 6-37 
Problem 12, 



* 13. (II) A spring has k = 88N/m. Use a graph to determine 

the work needed to stretch it from x — 3,8 cm to 
x = 5.8 cm, where x is the displacement from its 
unstretched length. 

* L4. (II) The net force exerted on a particle acts in the +.v 

direction, Its magnitude increases linearly from zero at 
X = 0. to 24.UN at x = 3.0 m. H remains constant at 
24.0 N from x = 3,0 m to x = 8,0 m. and then decreases 
linearly to zero at .v = 13.0 m. Determine the work done 
to move the particle from x = to ,v = 13.0 m graphi- 
cally by determining the area under the f 7 , vs. x graph. 

6-3 Kinetic Energy; Work-Energy Principle 

15. (I) At room temperature, an oxygen molecule, with 
mass of 5,31 X 10 kg. typically has a kk of about 
6.21 X 10~ :l J. How fast is the molecule moving? 

16. (I) (a) If the KE of an arrow is doubled, by what factor has 
its speed increased? (ft) If its speed is doubled, by what 
factor does its kc increase? 

17. (I) How much work is required to slop an electron 
(m = 9.11 x 10 ^ kg) which is moving with a speed of 
1.90 x 10*m/s? 

LN. (I) How much work must be done to stop a 1 250-kg car 
traveling at 105 km/h? 

1^. (II) An SS-g arrow is fired from a bow whose siring exerts 
an average force of HON on the arrow over a distance of 
78 cm. What is the speed of the arrow as it leaves the bow? 

2(1. (II) A baseball (m = I40g) traveling 32m/s moves a 
fielder's glove backward 25 cm when the ball is caught. What 
was the average force exerted by the ball on the glove? 

21. (II) If the speed of a car is increased by 50%. by what factor 
will its minimum braking distance be increased, assuming 
all else is the same? Ignore the driver's reaction time. 



162 CHAPTER 6 Work and Energy 



22. (II) At an accident scent on a level road, investigators 
measure a car's skid mark to he 88 m long. The accident 
occurred on a rainy day. and the coefficient of kinetic fric- 
tion was. estimated to be 0,42- Use these data to determine 
the speed of the car when the driveT slammed on (and 
locked) the hrakes. (Why does the car's mass not matter?) 

23. (II) A softball having a mass of 0-25kg is pitched at 
95 km/h, By the time it reaches the plate, it may have 
slowed by 107s. Neglecting gravity, estimate the average 
force of air resistance during a pitch, if the distance 
between the plate and the pitcher is about 15 m. 

24. (II) How high will a 1,85-kg rock go if thrown straight up 
by someone who does 800 J of work on it? Neglect air 
resistance. 

25. (Ill) A 285-kg load is lilted 22.0m vertically with an 
acceleration a = 0.160g by a single cable. Determine 
(n) the tension in the cable, (b) the net work done on the 
load, (t) the work done by the cable on the load, (ti) the 
work done by gravity on the load, and (<?) the final speed 
of the load assuming it started from rest, 

6-4 and 6-5 Potential Energy 

26. (I) A spring has a spring stiffness constant, k. of 440 N/m. 

How much must this spring be stretched to store 25 J of 
potential energy? 

27. (I) A 7.0-kg monkey swings from one branch to another 
1.2 m higher. What is the change in potential energy? 

2&. (I) By how much does the gravitational potential energy 
of a 64-kg pole vauller change if his center of mass rises 
alxjut 4.0 m during the jump? 

29. (11) A 1200-kg car rolling on a horizontal surface has 
speed v = 65 km/h when it strikes a horizontal coiled 
spring and is brought to rest in a distance of 2.2 m. What 
is the spring stiffness constant of the spring? 

30. (II) A l.&O-m tall person lifts a 2.10-kg book from the 
ground so it is 2.20 in above the ground. What is the poten- 
tial energy of the book relative to (a) the ground, and 
(£>) the top of the person's head? (c) How is the work done 
by the person related to the answers in parts (a) and (/>)? 

31. (II) A 55-kg hiker starts at an elevation of 1600 m and 
climbs to the top of a 3300-m peak, (a) What is the hiker's 
change in potential energy? (6) What is the minimum 
work required of the hiker? (l) Can the actual work done 
be more than this? Explain why. 

32. (II) A spring with k = 53 N/m hangs vertically nest to a 
mler, Hie end of the spring is next to the 15-em mark on 
the ruler. If a 2,5-kg mass is now attached to the end of 
the Spring, where will the end of the Spring line Up with 
the ruler marks? 

6-6 and 6-7 Conservation of Mechanical Energy 
Si. (I) .lane, looking for Tarzan, is running at top speed 
(5.3 m/s) and grabs a vine hanging vertically from a (all 
tree in the jungle. How high can she swing upward'; 1 Does 
the length of the vine affect your answer? 

34. (I) A novice skier, starting from rest, slides down a fric- 
tionless 35.0" incline whose vertical height is 185 m. How 
fast is she going when she reaches the bottom? 

35. (I) A sled is initially given a shove up a frictionless 28,0''' 
incline. It reaches a maximum vertical height 1.35 m 
higher than where it started. What was its initial speed? 



36. (II) In the high jump, Fran's kinetic energy is transformed 
into gravitational potential energy without the aid of a 
pole. With what minimum speed must Fran leave the 
ground in order to lift her center of mass 2.10 m and cross 
the bar with a speed of 0.70 m/s? 

37. (II) A 65-kg trampoline artist jumps vertically upward 
from the top of a platform 
with a speed of 5 m/s. 

(a) How fast is he going as 
he lands on the trampoline, 
3.0 m below (Fig. 6-38)? 

(b) If the trampoline behaves 
like a Spring with spring stiff- 
ness constant 6.2 X \Q A N/m, 
how far does he depress it? 

FIGURE 6-38 
Problem 37, 

38. (II) A projectile is fired at an upward angle of 45,0 t: from 
the top of a 265- m cliff with a speed of 1S5 m/s. What will 
be its speed when it strikes the ground below? (Use 
conservation of energy.) 

39. (II) A vertical spring (ignore its mass), whose spring stiff- 
ness constant is 950 N/m, is attached to a table and is 
compressed down 0.150 m. {a) What upward speed can it 
give to a 0.30-kg ball when released? (h) How high above 
its original position (spring compressed) will the ball fly? 

4U. (II) A block of mass m slides without friction along the 
looped track shown in Fig. 6-39. If tht block is to remain 
on the track, even at the top of the circle (whose radius 
is r), from what minimum height h must it be released? 





FIG URE 6-39 Problems 40 and 75. 

41. (II) A block of mass m is attached to the end of a spring 
(spring stiffness constant k). Fig, 6-40, The block is given 
an initial displacement .v (hl after which it oscillates back 
and forth. Write a formula for the total mechanical 
energy (ignore friction and the mass of the spring) in 
terms of .v (( , position .(, and speed v. 



VHQ 




/ 


/ 


GQQQSQQQQOfi 






JS" 




< I 


y^ 







FIGURE 6-40 Problems 41, 55, and 56. 

42. (II) A 62-kg bungee jumper jumps from a bridge. She is tied 
to a bungee cord whose unstretched length is 12 m, and falls 
a total of 31 m. (a) Calculate the spring stiffness constant k 
of the bungee cord, assuming Hooke's law applies. 

(b) Calculate the maximum acceleration she experiences. 



Problems 163 




43. (II) The roller-coaster car shown in Fig. 6-41 is dragged 

up to point 1 where it is released from rest. Assuming no 
friction, calculate the speed at points 2, 3. and 4, 



FIGURE 6-41 
Problems 43 

and 53. 



44. (II) A 0.40- kg ball is thrown with a speed of I2m/s at an 
angle of 33'. (a) What is its speed at its highest point, and 

(b) how high does it go? (Use conservation of energy, and 
ignoTe ah~ resistance.) 

45. (Ill) An engineer is designing a Spring to be placed at the 
bottom of an elevator shaft, If the elevator cable should 
break when the elevator is at a height It above the top of 
the spring, calculate the value that the spring stiffness 
constant k should have so that passengers undergo an 
acceleration of no more than 5.0 # when brought to rest. 
Let A/ be the total mass of the elevator and passengers. 

46. (Ill) A cyclist intends to cycle up a 7.8 U hill whose vertical 
height is 150 m. Assuming the mass of bicycle plus cyclist 
is 75 kg, {a) calculate how much work must be done 
against gravity, (b) If each complete revolution of the 
pedals moves the hike 5.1 m along its path, calculate the 
average force that must be exerted on the pedals tangent 
to their circular path. Neglect work done by friction and 
other losses. The pedals turn In a circle of diameter 36 cm, 

fi-8 and 6-9 Law of Conservation of Energy 

47. (I)'lwo railroad ears, each of mass 7650 kg and traveling 
95 km/b in opposite directions collide head-on and come to 
rest. How much thermal energy is produced in this collision? 

48. (II) A 21,7-kg child descends a slide 3.5 m high and reaches 
the hottom with a speed of 2.2 m/s. How much thermal 
energy due to friction was generated in this process? 

4D. (II) A ski starts from rest and slides down a 22" incline 
75 m long, (a) If the coefficient of friction is 0,090, what is 
the ski's speed at the base of the incline? (b) If the snow 
is level at the foot of the incline and has the same coeffi- 
cient of friction, how faT will the ski travel along the 
level'? Use energy methods. 

5(1. (II) A 145-g baseball is dropped from a tree 13.0 m above 
the ground, [a) With what speed would it hit the ground if 
air resistance could be ignored? (b) If it actually hits the 
ground with a speed of 8.00 m/s. what is the average force 
of air resistance exerted on it? 

51. (I I J You drop a ball from a height of 2.0 m, and it bounces 
back to a height of 1.5 m. (a) What fraction of its initial 
energy is lost during the bounce? (b) What is the ball's 
speed just as it leaves the ground after the bounce? 

(c) Where did the energy go7 

52. (II) A 1 1 0-kg crate, starting from rest, is pulled across a floor 
with a constant horizontal force of 350 N. For the first 15 m 
the flooT is frictionless, and for the next 15 m the coefficient 
of friction is 0.30. What is the final speed of the crate? 

53. (II) Suppose the Toller coasteT in Kig, 6-41 passes point 1 
with a speed of 1.70 m/s. If the average force of friction is 
equal to one-fifth of its weight, with what speed will it 
reach point 2? The distance traveled is 45.0 m, 



54, (II) A skier traveling 12.0 m/s reaches the loot ol a steady 
upward lfi.0 13 incline and glides 12.2 m up along this slope 
before coming to rest, What was the average coefficient ol 
friction? 

55. (Ill) A 0.620-kg wood block is firmly attached to a very- 
light horizontal spring (t = l80N/m) as shown in 
Fig. 6-40. It is noted that the block-spring system, when 
compressed 5.0 cm and released, stretches out 2,3 cm 
beyond the equilibrium position before stopping and 
turning back. What is the coefficient of kinetic friction 
between the block and the table? 

56. (Ill) A 280-g wood block is firmly attached to a very light 
horizontal spring. Fig. 6-40. The block can slide along a 
table where the coefficient of friction is 0,30, A lorce ol 
22 N compresses the spring 18 cm. If the spring is released 
from this position, how far beyond its equilibrium posi- 
tion will it stretch at its first maximum extension? 

57, (HI) Rarly test flights for the space shuttle used a "glider" 
(mass of 980 kg including pilot) that was launched horizon- 
tally at 500 km/h from a height of 3500 m.The glider even- 
tually landed at a speed of 200 km/h, (<r) What would its 
landing speed have been in the absence of aiT resistance? 
lb) What was the average force of air resistance exerted on 
it if it came in at a constant glide of 10° to the Earth? 

6-10 Power 
58. 



59. 

66. 
61. 

62. 



63. 



64. 
65. 



66. 



67, 



(I) How long will it take a 1750-W motor to lift a 3l>kg 

piano to a sixth-story window 16.0 m above? 

(I) If a car generates 18 hp when traveling at a steady 

88 km/h, what must be the average force exerted on the 

car due to friction and air resistance? 

(I) A 1400-kg sports car accelerates from rest to 95 km/h in 

7.4 s. What is the average power delivered by the engine? 

(I) (a) Show that one British horsepower (550ft-lb/s) is 
equal to 746 W, (b) What is the horsepower rating of a 
75 -W light bulb? 

(II) Electric energy units are often expressed in the form 
of "kilowatt-hours," (a) Show that one kilowatt-houT 
(kWh) is equal to 3.6 x io* J. {b) If a typical family of 
four uses electric energy at an average rate of 520 W, how 
many kWh would their electric bill be for one mouth, and 
(t) how many joules would this be? (d) At a cost of $0.12 
per kWh, what would their monthly bill be in dollars? 
Does the monthly bill depend on the rate at which they 
use the electric energy? 

(II) A driver notices that her 1150-kg car slows down 
from &5 km/h to 65 km/h in about 6.0 s on the level when 
it Is in neutral. Approximately what power (watts and hp) 
is needed to keep Ihe car traveling at a constant 75 km/h? 
(II) How much work can a 3 0-hp motor do in 1 .0 h? 
(II) A shot-putter accelerates a 7.3-kg shot from rest to 
14 m/s. If this motion takes 1 .5 s. what average power was 
developed? 

(II) A pump is to lift 18.0 kg of water per minute through 
a height of 3.60 m. What output rating (watts) should the 
pump motoT have'? 

(II) During a workout, the football players at State U. ran 
up Ihe stadium stairs in 66 s. The slarrs are 140 m long and 
inclined at an angle of 32°, If a typical player has a mass 
of 95 kg, estimate the average power output on the way 
up, Ignore friction and air resistance. 



161 CHAPTER 6 Work and Energy 



fiH. (II) How fast must a cyclist climb a 6-0" hill to maintain a 
power output of 0,25 lip? Neglect work done hy friction, 
and assume the mass of cyclist plus bicycle is 68kg. 

69. (II) A 12(X)-kg car has a maximum power output of 
120hp. How steep a hill can it climb at a constant speed 
of 75 km/h if the frictional forces add up to 650 N7 



71 L (II) What minimum horsepower must a motor have to be 
able to drag a 310-kg box along a level floor at a speed of 
1 ,20 m/'s if the coefficient of friction is 0.45? 

71. (Ill) A hicyclist coasts down a 7.0° hill at a steady speed 
of 5.0 m/s. Assuming a total mass of 75 kg (bicycle plus 
rider), what must be the cyclist's power output to climb 
the same hill at the same speed' 1 



| General Problems 



72. Designers of today's cars have built "5mi/h (8 km/h) 
bumpers" thai are designed to compress and rebound elasli- 
cally without any physical damage at speeds below 8 km/h. 
If the material of the bumpers permanently deforms after a 
compression of 1 .5 cm. but remains like an elastic spring up 
to that point, what must the effective spring stiffness 
constant of the bumper be. assuming the ear has a mass of 
1300 kg and is tested by ramming into a solid wall? 

7,*. In a certain library the first shelf is 10.0cm off the ground, 
and the remaining four shelves are each spaced 30-0 cm 
above the previous one. If the average hook has a mass of 
1 .5 kg with a height of 21 cm. and an average shelf holds 25 
books, how mueh work is required to fill all the shelves. 
assuming the books are all laying flat on the floor to start? 
A film of Jesse Owens's famous long jump (Fig. 6-42) in 
(he 1936 Olympics shows thai his center of mass Tose 
1.1 m from launch point to the top of the arc, What 
minimum speed did he need at launch if he was traveling 
at 6.5 m/s at the top of the arc? 



74, 








FIGURE 6-42 
Problem 74. 

75. The block of mass m sliding without friction along the 
looped track shown in Fig. 6-39 is to remain on the track 
at all times, even at the veTy top of the loop of radius r. 
(a) In terms of the given quantities, determine the 
minimum release height h (as in Problem 40). Next, if the 
actual Te lease height is 2A, calculate (6) the normal force 
exerted by the track at the bottom of the loop, (c) the 
normal foTce exerted by the tTack at the top of the loop, 
and (d) the normal force exerted hy the track after the 
block exits the loop onto [he fla! section. 

76. An airplane pilot fell 370 m after jumping from an aircraft 
without his parachute opening. He landed in a snowbank, 
creating a craleT 1 . 1 m deep, but survived with only minor 
injuries. Assuming the pilot's mass was 78 kg and his 
terminal velocity was 35 m/s, estimate (a) the work done 
by the snow in bringing him to Test: (b) the average force 
exerted on him by the snow to stop him; and (c) the work 
done on him by air resistance as he fell. 



1- 


—/.- 


* 


T 




i, 


\ 








V 

\ 




h 


\ 








\ 








\ 










^ 


Peg J 






^ 






^ 


-**T 








~ 



77. A ball is attached to a horizontal cord of length L whose 
other end is fixed (Fig. 6-43). («) If the ball is released, what 
will be its speed at the lowest point of its path? {b) A peg is 

located a distance h directly 
below the point of attach- 
ment of the cord. If 
h = 0.80/.. what will be 
the speed of the ball when it 
reaches the top of its 
circular path about the peg? 
1 

FIGURE 6-43 
Problem 77. 

78. A 65 -kg hiker climbs to the top of a 3700 -m- high moun- 
tain. The climb is made in 5.0 h starting at an elevation of 
2300m. Calculate {a) the work done by the hiker against 
gravity, (if) the average power output in watts and in 
horsepower and (t) assuming the body is 15% efficient, 
what rate of energy input was required. 

79. An elevator cable breaks when a 920-kg elevator is 28 tn 
above a huge spring (k = 2.2 x 10 ? N/m) at the bottom 
of the shaft. Calculate {a) the work done by gravity on the 
elevator before it hits the spring, (b) the speed of 
the elevatoT just before striking the spring, and (c) the 
amount the spring compresses (note that work is dune by 
both the spring and gravity in this pari). 

811. Squaw Valley ski area in California claims that its lifts can 
move 47.000 people per hour, If the average lift carries 
people about 200 m (vertically) higher, estimate the 
power needed. 

SI. Water flows (■» ~ 0) over a dam at the rate of 650kg/s 
and falls vertically 81 m before striking the tiubine blades. 
Calculate {a) the speed of the water just before striking the 
turbine blades (neglect air resistance), and (b) the rate at 
which mechanical energy is transferred to the turbine 
blades, assuming 58% efficiency, 

82. Show that on a roller coaster with a circular vertical loop 
(Fig. 6-44), the difference in your apparent weight at the 
top of the circular loop and the bottom of the circular loop 
is 6g's — that is, six times your weight. Ignore friction. Show 
also that as long as your speed is above the minimum 
needed, this answer doesn't depend on the size 
of the loop or how fast you go 
through it. 



FIGURE 6-44 
Problem 82. 




General Problems 165 



K3. (a) If the human body could convert a candy bar directly 
into work, how high could an 82- kg man climb a ladder if 
he were fueled by one bar (= 1100 fcJ)? (h) If the man 
then jumped off the ladder, what will be his speed when 
he reaches the bottom? 

84. A projectile is fired at an upward angle of 4S,U t: from the 
top of a 165-m cliff with a speed of 1 75 m/s. What will be 
its speed when it strikes the ground below? (Use conser- 
vation of energy and neglect air resistance.) 

US. If you stand on a bathroom scale, the spring inside the 
scale compresses 0.60 mm. and it tells you your weight is 
710 N. Now if you jump on the scale from a height of 
l.O m, what does the scale read at its peak? 

&6. A 65-kg student runs at 5.0 m/s, grabs a rope, and swings 
out over a lake (Fig. 6-45). He releases the rope when his 
velocity is zero, (nj What is the angle ft when he releases 
the rope'? (b) What is the tension in the rope just before he 
releases it? (c) What is the maximum tension in the rope? 




FIGURE 6-45 

Problem 86. 



87. In the rope climb, a 72-kg athlete climbs a vertical 
distance of 5-0 m in 9.0 s. What minimum power output 
was used to accomplish this feat? 

98. Some electric-power companies use water to store energy. 
Water is pumped by reversible turbine pumps from a low 
to a high reservoir. To store the energy produced in 
1.0 hour by a 120-MW (120 x lU ft W) electric-power 
plant, how many cubic meters of water will have to be 
pumped from the lower to the upper reservoir? Assume 
the upper reservoir is 520 m above the lower and we can 
neglect the small change in depths within each. Water has 
a mass of 1 000 kg for cvctv 1 .(.) m\ 

89. A spring with spring stiffness constant k is cut in half, 
What is the spring stiffness constant for each of the two 
resulting springs? 



911. A 6-0-kg block is pushed 8. Urn up a rough 37" inclined 
plane by a horizontal foTce of 75 N, If the initial speed 
of the block is 2.2 m/s up the plane and a constant 
kinetic friction force of 25 N opposes the motion. 
calculate (a) the initial kinetic energy of the block; 
(fr) the work done by the 75-N force; (t) the work done 
by the friction foTce; (rf) the work done by gravity; 
(f) the work done by the normal force; (f) the final 
kinetic energy of the block. 

91. If a 1500-kg car can accelerate from 35 km/h to 55 km/h 
in 3.2 s, how long will it take to accelerate from 55 km/h 
to 75 km/h? Assume the poweT stays the same, and 
neglect frictional losses. 

92. In a common test for cardiac function (the "stress test"), 
the patient walks on an inclined treadmill (Fig. 6-46). 
Estimate the power required from a 75 -kg patient when 
the treadmill is sloping at an angle of 15 <: and the velocity 
is 3.3 km/h. (How does this power compare to the power 
rating of a lightbulb?) 




FIGURE 6-46 Problem 92. 

93, (<j) If a volcano spews a 50U-kg rock vertically upward a 
distance of 500 m, what was its velocity when it left the 
volcano? (b) If the volcano spews the equivalent of 1000 
rocks of this size every minute, what is its power output? 

94. Water falls onto a water wheel from a height of 2.0 m at a 
rate of 95 kg/s. (a) If this water wheel is set up to provide 
electricity output, what is its maximum power output? 
(b) What is the speed of the water as it hits the wheel? 



Answers to Exercises 

A: (c). 

B: No. because the speed v would be the square root of a 
negative number, which is not real. 

O It is nonconservative. because for a conservative force 

W = in a round trip. 



D: W'jk-i = Ake , where W ne t = mg{y\ - »_) and 

Am - -.mtr, 
E: Equal speeds. 



mv\ = hrmA- Then v\ - 2g(\\ - .vj). 



166 CHAPTER 6 Work and Energy 



Conservation of linear momentum is another of the great conservation laws of physics Colli- 
sions, as between billiard or pool balls, illustrate this vector law very nicely: the total vector 
momentum before the collision equals the total vector momentum just after the colli- 
sion. In this photo, the moving cue ball strikes the 11 ball at rest. Roth balls move 
after the collision, at angles, but the sum of their vector momenta equals the 
initial momentum of the incoming cue ball. 

We will consider both elastic collisions (where kinetic energy is also 

conserved) and inelastic collisions. We also examine the 

'H|*i (before) concept of center of mass, and how it can make the study of 

complex motion more readily analyzed and understood, 





CHAPTER 



Linear Momentum 



The law of conservation of energy, which we discussed in the previous 
Chapter, is one of several great conservation laws in physics. Among the 
other quantities found to he conserved arc linear momentum, angular 
momentum, and electric charge. Wc will eventually discuss all of these because 
the conservation laws arc among the most important ideas in science. In this 
Chapter, we discuss linear momentum, and its conservation. The law of conser- 
vation of momentum is essentially a reworking of Newton's laws that gives us 
tremendous physical insight and problem-solving power. 

We make use of the laws of conservation of linear momentum and of 
energy to analyze collisions. Indeed, the law of conservation of momentum is 
particularly useful when dealing with a system of two or more objects that 
interact with each other, such as in collisions. 

Our focus up to now has been mainly on the motion of a single object, often 
thought of as a "panicle" in the sense that we have ignored any rotation or 
internal motion. In this Chapter we will deal with systems of two or more 
objects, and toward the end of the Chapter, the concept of center of mass. 



167 



Lmeur ttuimeftittm tlt'fiitctl 



Units of momentum 



NEWTON'S SECOND LAW 



Momentum and Its Relation to Force 

The linear momentum (or '■momentum" for short) of an object is defined as the 
product of its mass and its velocity. Momentum (plural is momenta) is repre- 
sented by the symbol p. If we let m represent the mass of an object and v repre- 
sent its velocity, then its momentum p is defined as 

p = mv. (7-1) 

Velocity is a vector, so momentum too is a vector. The direction of the 
momentum is the direction of the velocity, and the magnitude of the momentum 
is p - mv. Because velocity depends on the reference frame, so does momentum; 
thus the reference frame must be specified. The unit of momentum is that of 
mass X velocity, which in ST units is kg- m/s. There is no special name for this unit. 
Everyday usage of the term momentum is in accord with the definition above, 
According to Eq. 7- 1 , a fast-moving car has more momentum than a slow-moving 
car of the same mass; a heavy track has more momentum than a small car moving 
with the same speed. The more momentum an object has, the harder it is to stop 
it, and the greater effect it will have if it is brought to rest by striking another 
object. A football player is more likely to be stunned if tackled by a heavy oppo- 
nent running at top speed than by a lighter or slower-moving faekleT. A heavy, 
fast-moving truck can do more damage than a slow-moving motorcycle. 

EXERCISE A Can a small sports ear ever have the same momentum as a large sport- 
utility vehicle with three times the sports car's mass? Explain. 

A force is required to change the momentum of an object, whether it is to 
increase the momentum, to decrease it, or to change its direction. Newton orig- 
inally stated his second law in terms of momentum (although he called the 
product mv the ''quantity of motion"). Newton's statement of the second law of 
motion, translated into modern language, is as follows: 

The rate of change of momentum of an object is equal to the net force 
applied to it. 

We can write this as an equation. 



NEWTON'S SECOND LAW 



Sf- 



Ap 

At' 



(7-2) 



C A U T I O M 



The change in the momentum vet tor 
is in the direction of the net force 



Vewion's second law 
for constant mass 



where 2F is the net force applied to the object (the vector sum of all forces 
acting on it) and Ap is the resulting momentum change that occurs during the 

time interval 1 At. 

We can readily derive the familiar form of the second law, XI* 1 = ma, from 
Eq. 7-2 for the case of constant mass. Tf v, is the initial velocity of an object and 
v, is its velocity after a time interval At has elapsed, then 

Ap _ mv ; - mv, _ m(v 2 - v,) 
At ' At At 

Av 

Af ' 



XF 



m 



By definition, a = AvfAt, so 

y.f = ma. [constant mass] 

Newton's statement, Eq. 7-2, is moTe general than the more familiar version 
because it includes the situation in which the mass may change. A change in 
mass occurs in certain circumstances, such as for rockets which lose mass as they 
burn fuel, and also in the theory of relativity (Chapter 26). 

'Normally we think of Af as being a small time interval, If it is not small, then liq. 7-2 is valid if 2.K 
is constant during that time interval, or if SF is the average net force during that time interval. 



168 CHAPTER 7 Linear Momentum 



EXAMPLE 7-1 



ESTIMATE | Force of a tennis serve. For a top player, a 
tennis ball may leave the racket on the serve with a speed of 55 m/s (about 
I20 mi/h), Fig. 7- 1. If the ball has a mass of 0.060 kg and is in contact with the 
racket for about 4 ms (4 X 10 _3 s), estimate the average force on the ball. 
Would this force be large enough to lift a 60-kg person? 

APPROACH The tennis ball is hit when its initial velocity is very nearly zero at 
the top of the throw, so wc take i>, = 0. We use Newton's second law, Cq. 7-2, 
to calculate the force, ignoring all other forces such as gravity in comparison to 
that exerted by the tennis racket. 
SOLUTION The force exerted on the ball by the racket is 

Ap mih - rnv\ 
" = ~r~ = — "~ : 

where Uj = 55 m/s, v l = 0, and At = 0,004 s. Thus 
Ap (0.060 kg)(55 m/s) - 
F ~ If- 



0.004 s 



~ ma n. 



This is a large force, larger than the weight of a 60-kg person, which would 

require a force mg = (60 kg)(9.8 m/s 2 ) » 600 N to lift. 

NOTE The force of gravity acting on the tennis ball is mg - (0.060 kg)(9,8 m/s 3 ) 

- 0.5° N, which justifies our ignoring it compared to the enormous force the 

racket exerts, 

NOTE High-speed photography and radar can give us an estimate of the 

contact time and the velocity of the ball leaving the racket. But a direct 

measurement of the force is not practical. Our calculation shows a handy 

technique for determining an unknown force in the Teal world. 



EXAMPLE 7-2 



Washing a car: momentum change and force. Water 
leaves a hose at a rate of 1 ,5 kg/s with a speed of 20 m/s and is aimed at the 
side of a car, which slops it, Fig. 7-2. (Thai is, we ignore any splashing back.) 
What is Ihe force exerted by the water on the car? 

APPROACH The water leaving Ihe hose has mass and velocity, so il has a 
momentum p,,^. When the water hits the car, the water loses this 
momentum (pfj ril | = 0). We use Newton's second law in the momentum form, 
Fq. 7-2, to find the force that the car exerts on the water to stop it. By 
Newton's third law, the force exerted by the water on the car is equal and 
opposite. We have a continuing process: 1 .5 kg of water leaves the hose in each 
1.0-s time interval. So let us choose At = 1.0 s, and m = 1.5 kg in Fq. 7-2. 
SOLUTION We take the x direction positive to the right. In each 1.0-s time 
interval, water with a momentum of p x = mv x = ( 1.5 kg)(20m/s) = 30 kg- m/s 
is brought to rest when it hils Ihe car. The magnitude of the force (assumed 
constant) that ihe car must exert to change Ihe momentum of the walei by this 
amount is 

c &P Pfiim] - Pinilkil - 30 kg ■ m/s 

F - 47-- 



M 



1.0 s 



-3DN. 



The minus sign indicates that the force on the water is opposite to the water's 
original velocity. The car exerts a force of 30 N to the left to stop the water, so 
by Newton's third law. the water exerts a force of 30 N to the right on the car. 
NOTE Keep track of signs, although common sense helps too. The water is moving 
lo Ihe right, so common sense Lells us the force on Ihe car must be to the righl. 

EXERCISE B If the water splashes hack from the car in Example 7-2, would the force 
on Ihe car be larger or smaller? 




FIGURE 7-1 Example 7-1. 



Measuring force 



FIGURE 7-2 Example 7-2. 




SECTION 7-1 Momentum and Its Relation to Force 168 



=jr~\ '"A v A '"H VBAfc 



=® 






m A v A/"N^ ^=*k" I U v 13 



! ® 



A u , a 



FIGURE 7-3 Momentum is 
conserved in a collision of two balls, 
labelled A and B- 



t VSERVATiON Oh' MOMENTUM 
(for nco object n 'Hiding) 



Momentum i onseroation related 

■'■■■ \c>i-l<m\l/i<r\ 



FIGURE 7-4 Forces on the balls 
during the collision of Fig. 7-3. 




Conservation of Momentum 



The concept of momentum is particularly important because, under certain 
circumstances, momentum is a conserved quantity. Consider, for example, the 
head-on collision of two billiard balls, as shown in Fig. 7-3, We assume the net 
external force on this system of two balls is zero — that is, the only significant forces 
during the collision are the forces that each ball exerts on the other. Although the 
momentum of each of the two balls changes as a result of the collision, the sum of 
their momenta is found to be the sanvc before as after the collision. If m A v A is the 
momentum of ball A and m B v u the momentum of ball B, both measured just 
before the collision, then the total momentum of the two balls before the collision is 
the vector sum m A v A + rti u v u . Immediately after the collision, the balls each 
have a different velocity and momentum, which we designate by a "prime" on the 
velocity: m A v' A and m a v' a . The total momentum after the collision is the vector 
sum m A v' A + m^ v' a . No matter what the velocities and masses arc, experiments 
show that the total momentum before the collision is the same as afterward, 
whether the collision is head-on or not, as long as no net external force acts: 



momentum before - momentum after 
wi A v A + m n v B = tn A v\ + m n v^. 



(7-3) 



That is, the total vector momentum of the system of two colliding balls is 
conserved: it stays constant. 

Although the law of conservation of momentum was discovered experimen- 
tally, it is closely connected to Newton's laws of motion and they can be shown 
to be equivalent, We will do a derivation for the head-on collision illustrated in 
Fig. 7-3. We assume the force F that one ball exerts on the other during 
the collision is constant oveT the brief time interval of the collision Af. 
We use Newton's second law as expressed in Eq. 7-2, and rewrite it by multi- 
plying both sides by Af; 



Ap = FA/. 



(7 4) 



We apply this to ball B alone, noting that the force P UA on ball B exerted by ball 
A during the collision is to the right ( + x direction — see Fig, 7-4): 

Apu = F BA Af 
m B f' B - m H v B = F BA if. 

By Newton's third law; the force P AU on ball A due to ball B is P A u - -P^ A and 
acts to the left. Then applying Newton's second law in the same way to ball A yields 

Ap A -P,Mjif 
or 

m A v' A - m A v A - F AU dr 

= -PliA^. 

We combine these two Ap equations (theiT right sides differ only by a minus sign): 
fn A y' A - ffl A v A = -(i« B vjj - w B v,i) 



or 



»*A*A + «*fl*B = "! A v' A + IHhVJj 



which is Eq. 7-3, the conservation of momentum, 

The above derivation can be extended to include any numbeT of interacting 
objects. To show this, we let p in Eq, 7-2 represent the total momentum of a 
system — that is, the vector sum of the momenta of all objects in the system. (Fot 
our two-object system above, p = wj A v A + w b *b) If the net force 2F on the 
system is zero [as it was above for our two-object system, F + (-F) = 0,] then 



170 CHAPTER 7 Linear Momentum 



from Eq, 7-2, Ap = F At = 0, so the total momentum doesn't change. Tims 
the general statement of the law of conservation of momentum is 

The total mo men turn of an isolated system of ohjecls remains constant. 

By a system, we dimply mean a set of objects that we choose, and which may 
interact with each other. An isolated system is one in which the only (significant) 
forces are those between the objects in the system, The sum of all these 
"internal" forces within the system will be zero because of Newton's third law. If 
there arc external forces — by which we mean forces exerted by objects outside the 
system — and they don't add up to zero (vcctorially), then the total momentum of 
the system won't be conserved. However, if the system can be redefined so as to 
include the other objects exerting these forces, then the conservation of 
momentum principle can apply. For example, if we take as our system a rock 
falling under gravity; the momentum of this system (the rock) is not conserved: an 
external force, the force of gravity exerted by the Earth, is acting on it and changes 
its momentum. However, if we include the Earth in the system, the total 
momentum of rock plus Earth is conserved. (This means that the Earth comes up 
to meet the rock. But the Earth's mass is so great, its upward velocity is very tiny.) 



EXAMPLE 7-3 



Railroad cars collide: momentum conserved. A L0,0f)0-kg 
railroad ear, A, traveling at a speed of 24,0 m/s strikes an identical car, B, at 
rest. Tf the ears lock together as a result of the collision, what is their common 
speed just afterward? See Fig. 7-5. 

APPROACH We choose our system to be the two railroad cars, We consider a very 
brief time interval, from just before the collision until just afteT, so that external 
forces such as friction can be ignored, Then we apply conservation of momentum, 
SOLUTION The initial total momentum is 

initial = '«A' f A + ^B^B = '«A»A 

because car B is at rest initially (v tt - 0), The direction is to the right in the + x 
direction. After the collision, the two cars become attached, so they will have 
the same speed, call it v' . Then the total momentum after the collision is 

/Tin a I = ("?A + "!«)»'. 

We have assumed there are no external forces, so momentum is conserved: 

Piniliji] — PVnvii 

m A v A = (w A + m K )■!■'. 
Solving for v', we obtain 

tn A { 10,000 kg 



*A 



1(24.0 m/s) = 12.0 m/s, 



m A + m B ' \ 10,000 kg + 10,000 kg, 

to the right, Their mutual speed after collision is half the initial speed of car A. 

NOTE We kept symbols until the very end, so we have an equation we can use 
in other (related) situations. 



LAW OF CONSERVATION 
OF MOMENTUM 



Systems 
Isolated system 



A 



T^ 



n A = 24.0 m/s 



'□□□□ 
J B 



BE" 





(at rest) 



T= 1=1 1=1 T=l 1=T =1 =1 =1 =1 l=r 

(a) Before collision 



— i=i — i=i— 



FIGURE 7-5 Fxamplc 7-3. 



J b d b d 



J 



J b a b d 



(b) Alter collision 



r 



SECTION 7-2 Conservation of Momentum 171 



EXERCISE C In 



w 



m A = nj n , so in the- last equation. 
What result do you get if (a) m B = 3tn A , 



PHYSICS APPLIED 

Racket propulsion 



< |> CAUTION 

A rocket pushes on the gases 

released h} the ftwt, not on the 
Earth or other objects 



F.xamplc 7-3, 

i/^ m A + m a) = !■ Hence 
(6) m B is much larger than m A (m K » m A }. and (t) m H <$< tn A ? 

As long as no external forces act on our chosen system, conservation of 
momentum is valid. In Ihe real world, external forces do act: friction on billiard 
balls, gravity acting on a baseball, and so on. So it may seem that conservation 
of momentum cannot be applied. Ot can it? In a collision, the force each object 
exerts on the other acts only oveT a very brief time interval, and is very strong, 
When a racket hits a tennis ball (or a bat hits a baseball), both before and after 
the "collision" the hall moves as a projectile under the actum of gravity and air 
resistance. During the brief time of the collision, however, when the racket hits 
the ball, external forces (gravity, air resistance) are insignificant compared to 
the collision forces that the racket and ball exert on each other. So if we 
measure the momenta just before and just after the collision, we can apply 
momentum conservation with high accuracy. 

The law of conservation of momentum is particularly useful when we are 
dealing with fairly simple systems such as colliding objects and certain types of 
"explosions"'. For example, rocket propulsion, which we saw in Chapter 4 can be 
understood on the basis of action and reaction, can also be explained on the 
basis of the conservation of momentum. We can consider the rocket and fuel as 
an isolated system if it is far out in space (no external forces). In the reference 
frame of the rocket, the total momentum of rocket plus fuel is zero. When the 
fuel burns, the total momentum remains unchanged: the backward momentum 
of the expelled gases is just balanced by the forward momentum gained by the 
rocket itself (see Fig, 7-6). Thus, a rocket can accelerate in empty space, There 
is no need foT the expelled gases to push against the Earth or the air (as is 
sometimes erroneously thought). Similar examples of (nearly) isolated systems 
where momentum is conserved are the recoil of a gun when a bullet is fired, and 
the movemen t of a rowboat just after a package is thrown from it. 



FIGURE 7-6 (a) A rocket, containing fuel, at rest in some 
reference frame, (b) In the same reference frame, the rocket 
fifes and gases are expelled at high speed out the rear. The 
total vector momentum, p slls -I- pr^-tti , remains zero. 



*< 



p = 



EXAMPLE 7-4 



FIGURE 7-7 bxample 7-4. 



(a) Bulbre shooiing (* test) 



ft 



Pii 



(b) After shooting 



P^as "roctitl 

Rrfle recoil. Calculate the recoil velocity of a S.O-kg rifle 
that shoots a 0.020-kg bullet at a speed of 620 m/s, Fig, 7-7. 

APPROACH Our system is the rifle and the bullet, both at rest initially, just 
before the trigger is pulled. The trigger is pulled, an explosion occurs, and we 
look at the rifle and bullet just as the bullet leaves the barrel. The bullet moves 
to the right ( + x), and the gun recoils to the left. During the very short time 
interval of the explosion, we can assume the external farces are small 
compared to the forces exerted by the exploding gunpowder. Thus we can 
apply conservation of momentum, at least approximately. 
SOLUTION Let subscript B represent the bullet and R the rifle; the final velocities 
are indicated by primes. Then momentum conservation in the x direction gives 

momentum before = momentum after 

»"b% + mR"% = "Vwii + tn R v' R 
+ - m B i;jj + m R v' K 

so 

m B i>B (0.020 kg}(620 m/s) 



"H 



-2.5 m/s. 



m K (5.0 kg) 

Since the rifle has a much larger mass, its (recoil) velocity is much less than 
that of the bullet, The minus sign indicates that the velocity (and momentum) 
of the rifle is in the negative x direction, opposite to that of the bullet. 



172 CHAPTER 7 Linear Momentum 



CONCEPTUAL EXAMPLE 7-5 | Falling on ot off a sled, (a) An empty 



sled is sliding on friction I ess ice when Susan drops vertically from a tree above 
onto the sled. When she lands, does the sled speed up, slow down, or keep the 
same speed? (£>) Later. Susan falls sideways off the sled. When she drops off, 

does the sled speed up, slow down, or keep the same speed? 

RESPONSE (a) Because Susan falls vertically onto the sled, she has no initial 
horizontal momentum. Thus the total horizontal momentum afterward equals the 
momentum of the sled initially. Since the mass of the system (sled + person) has 
increased, the speed must decrease. 

(b) At the instant Susan falls off, she is moving with the same horizontal speed 
as she was while on the sled. At the moment she leaves the sled, she has the 
same momentum she had an instant before. Because momentum is conserved, 
the sled keeps the same speed. 



Collisions and Impulse 



Collisions are a common occurrence in everyday life: a tennis racket or a baseball 
bat striking a ball, billiard balls colliding, a hammer hitting a nail. When a collision 
occurs, the interaction between the objects involved is usually far stronger than 
any interaction between our system of objects and their environment. We can 
then ignore the effects of any other forces during the brief time interval of 
the collision, 

During a collision of two ordinary objects, both objects are deformed, often 
considerably, because of the large forces involved (Fig. 7-8). When the collision 
occurs, the force usually jumps from zero at the moment of contact to a very 
large force within a very short time, and then rapidly returns to zero again, A 
graph of the magnitude of the force that one objeet exerts on the other during a 
collision, as a function of time, is something like the red curve in Fig. 7-9. The 
time interval At is usually very distinct and very small. 

From Newton's second law, Eq. 7-2, the net force on one object is equal to 
the rate of change of its momentum: 



F- 



Ap 

if' 



(We have written F instead of ~£F for the net force, which we assume is entirely 
due to the brief but large average force that acts during the collision.) This 
equation applies to each of the two objects in a collision. We multiply both sides 
of this equation by the time interval if, and obtain 



F if - ip. 



(7-5) 



The quantity on the left, the product of the force F times the time if over which 
the force acts, is called the impulse: 

Impulse = F if. 

We sec that the total change in momentum is equal to the impulse. The concept 
of impulse is useful mainly when dealing with forces that act during a short tinve 
interval, as when a bat hits a baseball. The force is generally not constant, and 
often its variation in time is like that graphed in Figs, 7-9 and 7-10. Wc can often 
approximate such a varying force as an average force F acting during a time 
interval if, as indicated by the dashed line in Fig. 7-10. F is chosen so that the 
area shown shaded in Fig. 7-10 (equal to F X A/) is equal to the area under the 
actual curve of F vs. f, Fig, 7-9 (which represents the actual impulse), 

EXERCISE D Suppose Fig. 7-9 illustrates the force on a golf ball vs. the time when the 
ball hits a wall. How would the shape of this curve change if a softer rubber ball with 
the same mass and speed hit the same wall? 




FIGURE 7-8 Tennis racket striking 
a ball. Both the hall and the racket 
strings are deformed due to the large 
force each exerts on the other. 



FIGURE 7-9 Force as a function 
of time during a typical collision. 




Time, l 



FIGURE 7-10 ITi e average foTce 
F acting over an interval of time At 
gives the same impulse (F At) as 
the actual force. 




SECTION 7-3 Collisions and Impulse 173 



( fy PHYSICS APPLIED 

How not to break a leg 




v = 



v = 0- 

F1GURE 7-1 1 F.xamplc 7-6. Time 
interval Af during which [he 
impulse acts. 



PROBLEM SOLVING 

i ret body diagrams art 
always useful f 



EXAMPLE 7-6 



Bend your knees when landing, (a) Calculate the 
impulse experienced when a 70-kg person lands on firm ground after jumping 
from a height of 3.0 m. {b) Estimate the average force exerted on the person's 
feet by the ground if the landing is stiff-legged, and again (c) with bent legs. 
With stiff legs, assume the body moves 1.0 cm during impact, and when the 
legs arc bent, about 50 cm. 

APPROACH We consider the short time interval that starts just before the 
person hits the ground and ends when he is brought to rest. During this time 
interval, the ground exerts a force on him and gives him an impulse which 
equals his change in momentum (Eq. 7-5). For part (a) we know his final speed 
(zero, when he comes to Test), but we need to calculate his "initial' 1 speed just 
before impact with the ground. The latter is found using kinematics and his drop 
from a height of 3-0 m. Then Eq. 7-5 gives us FAf, fn parts (b) and (c) we 
calculate how long, Af, it Lakes him to slow down as he hits the ground, using 
kinematics, and then obtain F because we know F\{. 

SOLUTION (a) First we need to determine the velocity of the person just before 
striking the ground, which we do by considering the earlier time period between 
the initial jump from a height of 3.0 m until just before he touches the ground, 
The person falls under gravity, so we can use the kinematic Eq. 2- lie, 



v 1 = vl + 2a(y 



veil with a 
2s(y„ - y) 



-g and i\i = 0, so 



or 



v - \/2,?U - y) - \/2(9.8m/s 2 )(3.0m) - 7.7 m/s. 

This v = 7.7 m/s is his speed just before hiLting the ground, and so it is the 
initial speed for the short time interval of the impact with the ground, Af, 
Now we can determine the impulse by examining this brief time interval as the 
person hits the ground and is brought to rest (Fig. 7-1 1), We don't know Fand 
thus can't calculate the impulse FAf directly; but we can use Eq. 7-5: the 
impulse equals Lhe change in momentum of the object 

FAf - Ap - m An 

= (70 kg)(0 - 7.7 m/s) = -540 N -s. 

The negative sign tells us that the force is opposed to the original (downward) 
momentum; that is, the force acts upward. 

(b) In coming to rest, the person decelerates from 7.7 m/s to zero in a distance 
cl = 1.0 cm = 1.0 x I0~ 3 m. If we assume the upward force exerted on him by 
the ground is constant, then the average speed during this brief period is 

(7.7 m/s + 0m/s} 



= 



= 3.9 m/s. 



Thus the collision with the ground lasts for a time interval (recall the definition 

of Speed, I> = d/Al): 



d 

Af = - 

v 



(1.0 x 10 _2 m) 



2.6 X |(J- s. 



2.6 x 10 *s, 



(3.9 m/s) 

Since the magnitude of the impulse is F Af = 540 N-s, and Af 
the average net force F on the person has magnitude 

_ 540 Ns 

F ~ TZ T7TT- ~ 2- 1 x & 5 N. 

2.6 X 10" 3 s 

We are almost there. F equals the vector sum of the average force upward on 
the legs exerted by the ground, F iTj , which we take as positive, plus the down- 
ward force of gravity, -mg (see Fig. 7-12): 

F = F grd - mg. 

Since mg = (70 kg) (9 .8 m/s : ) = 690 N, then 



'Vu 



F + mg = (2.1 x 10 s N) + (0.690 x 10' N) ~ 2.1 x 10 s N. 



17fl CHAPTER 7 Linear Momentum 



(c) This is jusl like pari (b), except d = 0.50 m, so 



and 



At = 


d _ 

V 


0.50 m 

3.9 m/s 


= 0.13 s 


F = 


540 N ■ s , 
— — — = 4.2 X 10* N. 



0.13 s 

The upward force exerted on the person's feel, by ihe ground is, as in part (£>); 

Fpd - F+ mg - (4.2 X 10 3 N) + (0.69 X 10 1 N) - 4.9 X 10' N. 

Clearly, the force on the feet and legs is much less now with the knees bent, find 
the impulse occurs over a longer time interval. In fact, the ultimate strength of 
the leg bone (sec Chapter 9, Table 9-2) is not great enough to support the 
force calculated in part (b), so the leg would likely break in such a stiff 
landing, whereas it probably wouldn't in part (c) with bent legs. 




EXERCISE E In pari (b) of Example 7-6. we calculated the force exerted by the 
ground on the person during the collision, F gT(J . Was F gr j much greater than the 
"external" force of gravity on the person? By what factor? 

Conservation of Energy and Momentum in 
Collisions 

During most collisions, we usually don't know how the collision force varies 
over time, and so analysis using Newton's second law becomes difficult <ir 
impossible. But by making use of the conservation laws for momentum and 
energy, we can still determine a lot about the motion after a collision, given the 
motion before the collision, We saw in Section 7-2 that in the collision of two 
objects such as billiard balls, the total momentum is conserved. If the two 
objects arc very hard and no heat or other form of energy is produced in the 
collision, then kinetic energy is conserved as well. By this we mean that the sum 
of the kinetic energies of the two objects is the same after the collision as 
before. For the brief moment during which the two objects are in contact, some 
(or all) of the energy is stored momentarily in the form of elastic potential 
energy. But if we compare the total kinetic energy just before the collision with 
the total kinetic energy just after the collision, they are found to be the same. 
Such a collision, in which the total kinetic energy is conserved, is called an 
elastic collision. If we use the subscripts A and B to represent the two objects, 
we can write the equation for conservation of total kinetic energy as 

total kk before - total ke after 



\m x v\ + |m B v| 



jm A i/l + {m B v$. [elastic collision] (7-6) 

Here, primed quantities (') mean after the collision and unprimed mean before 
the collision, just as in Eq, 7-3 for conservation of momentum. 

At the atomic level the collisions of atoms and molecules are often elastic. 
But in the "macroscopic"' world of ordinary objects, an elastic collision is an 
ideal that is never quite reached, since at least a little thermal energy (and 
perhaps sound and other forms of energy) is always produced during a collision. 
The collision of two hard elastic balls, such as billiard balls, however, is very 
close to being perfectly elastic, and we often treat it as such. 

We do need to remember that even when the kinetic energy is not 
conserved, the total energy is always conserved. 

Collisions in which kinetic energy is not conserved are said to be inelastic 
collisions. The kinetic energy that is lost is changed info other forms of energy, 
often thermal energy, so that the total energy (as always) is conserved, Tn this case, 

ke a + KE K = ke' a + ke' b + thermal and other forms of energy. 

See Fig. 7-13, and the details in its caption. 




FIGURE 7-12 Example 7-6. 
When the person lands on the 
ground, the average net force 
during impact is F = F^j — mg, 
where F gr d is the force the ground 
exerts upward on the person. 



FIGURE 7-13 Two equal-mass 
objects (a) approach each other 
with equal speeds, (b) collide, and 
then (c) bounce off with equal 
speeds in the opposite directions if 
the collision is elastic, or (d) bounce 
hack much less or not at all if the 
collision is inelastic. 



'I!. 



(a) Approach 

40 

(hi Collision 

(C> if l-I Jslie 



v' R 



(d) I inelastic 



SECTION 7-4 Conservation of Energy and Momentum in Collisions 175 



*h 



■:;,! 



(b) 

FIG U RE 7- 1 4 Two *m a 1 1 obj ec t s 
of masses m A and m^ , (a) before 
Ihe collision and (b) after the 
Collision. 



Relative speeds (one dimension only) 



Elastic Collisions in One Dimension 

We now apply the conservation laws fur momentum and kinetic energy to an 
elastic collision between two small objects that collide head-on, so all the 
motion is along a line. Let us assume that the two objects are moving with 
velocities v; A and v^ along the x axis before the collision. Fig. 7- 14a. After the 
collision, their velocities are v' A and o B , Fig. 7- 14b. For any v > 0, the object is 
moving to the right (increasing a), whereas for v < 0, the object is moving to 
the left (toward decreasing values of x). 

From conservation of momentum, we have 

m A t> A + m u % = m A v' A + m B i^. 

Because the collision is assumed to be elastic, kinetic energy is also conserved: 

?m A v\ + jm b vl - jm A v£ + $m B v%- 

We have two equations, so we can solve for two unknowns. If we know the masses 
and velocities before the collision, then we can solve these two equations for the 
velocities after the collision, v\ and v' n . We derive a helpful Tesull by rewriting 
the momentum equation as 

m A (v A - v' A ) = m ]t [v'„ - t) R ), (i) 

and we rewrite the kinetic energy equation as 

m A (v A - v'l) = m B (v£ - v|). 
Noting that algebraically (a - b){a + h) - a 1 - b 1 , we write this last equation as 

m A {v A - v\)(v A + v' A ) - tn B {v' B - v B ){v' u + v b }. <ii> 

We divide Eq. (ii) by Eq. (i), and (assuming v A t- v' A and v B ^ v'^) obtain 

V A + U A = V' B + V B . 

We can rewrite this equation as 

D A - v B = 'Jjj - «a 

or 

v A — Vr = ~U\\ ~ v'm)- [head-on elastic collision] (7-7j 

This is an interesting result: it tells us that for any elastic head-on collision, the 
relative speed of the two objects after the collision has the same magnitude (but 
opposite direction) as before the collision, no matter what the masses are. 

Equation 7-7 was derived from conservation of kinetic energy for elastic 
collisions, and can be used in place of it. Because the v's are not squared in 
Eq. 7-7, it is simpler to use in calculations than the conservation of kinetic 
energy equation (Eq, 7-6) directly. 



EXAMPLE 7-7 



Pool or billiards. Billiard hall A of mass m moving with 
speed v collides head-on with ball B of equal mass at rest (v a = 0). What are 
the speeds of the two balls after the collision, assuming it is elastic? 

APPROACH There are two unknowns, v' A and v' a , so wc need two indepen- 
dent equations. We focus on the tinvc interval from just before the collision 
until just after, No net external force acts on our system of two balls {mg and 
the normal force cancel), so momentum is conserved. Conservation of kinetic 
energy applies as well because the collision is clastic, 

SOLUTION Given v A = v and w u = 0, and tn A = /n^ = m, then conserva- 
tion of momentum gives 

tno = mv' A + mi) B 

or, since the w's cancel out, 

V = V' A + l/ K . 

We have two unknowns (v' A and v B ) and need a second equation, which could 



176 CHAPTER 7 Linear Momentum 



be the conservation of kinetic energy or the simpleT Eq. 7-7 we derived from iL: 

«a - v a - v'u - v' A , or v - v' a - v' A 

since » A - c and ufo - 0. We subtract v - itjj - u A from our momentum 
equation (v - v' A + v' B ) and obtain 

Hence i,' A = 0. We can now solve for the other unknown (i;^) since » = v' K - v' A : 

v' B = v + b a = y + = 'it. 
To summarize, before the collision we have 

i> B - 



v'b - v. 



and after the collision 

That is, ball A is brought to rest by the collision, whereas ball B acquires the 
original velocity of ball A. See Fig. 7-15. 

NOTE Our result is often observed by billiard and pool players, and is valid 
only if the two balls have equal masses (and no spin is given to the halls). 



EXAMPLE 7-8 



A nuclear collision. A proton (p) of mass 1. 01 u (unified 
atomic mass units) traveling with a speed of 3,60 x 10 J m/s lias an elastic 
head-on collision with a helium (He) nucleus {tn He = 4.00 u) initially at Test. 
What are the velocities of the proton and helium nucleus after the collision? 
(As mentioned in Chapter 1, I u = 1,66 x 10 "kg, but we won't need l^' s 
fact.) Assume the collision takes place in nearly empty space. 

APPROACH Like Example 7-7, this is an elastic head-on collision, but now 
the masses of our two-particle system are not equal. The only external force is 
Earth's gravity, but it is insignificant compared to the strong force during the 
collision. So again we use the conservation laws of momentum and of kinetic 
energy, and apply them to our system of two particles, 

SOLUTION Let the proton (p) be paTticle A and the helium nucleus (He) be 
particle R. We have v% = v^ = and v A = v p = 3-60 x 1 0* m/s. We want to 
find the velocities v' p and v^ after the collision. From conservation of momentum, 

mpUp + - m p v p + m, „(>;,,, 

Because the collision is elastic, the kinetic energy of our system of two parti- 
cles is conserved and we can use Eq. 7-7, which becomes 

Vp ~ 
Thus 



i .■ 



p ' 



"p = '-'He ~ Op * 

and substituting this into our momentum equation displayed above, we get 

rap tip = TfifV'n,. - m p v p + m^v^. 
Solving for yj k , , we obtain 

2m p Up 2(1.01 u)(3,60 x 10* m/s) 



'iW = 



m p + »i Hc 



5.01 u 



= 1,45 x 10 J m/s. 



The other unknown is v' p , which wc can now obtain from 

*> P = i'hc - » P = (1-45 x 10* m/s) - (3.60 X It) 4 m/s) = -2.15 X I0 4 m/s. 

The minus sign for w p tells us that the proton reverses direction upon collision, 
and wc sec that its speed is less than its initial speed (sec Fig. 7-16), 

NOTE This result makes sense: the lighter proton would be expected to "bounce 
back" from the moTe massive helium nucleus, but not with its full original velocity 
as from a rigid wall (which corresponds to extremely large, ot infinite, mass). 



>>^ 



FIGURE 7-15 In this multiflash 
photo of a head-on collision 
between two balls of equal mass, the 
white cue ball is accelerated from 
Test by the cue stick and then 
strikes the red ball, initially at rest- 
The while ball slops in its tracks, 
and the (equal mass) red ball moves 
off with the same speed as the white 
hall had before the collision. See 
Example 7-7. 



Ik 
(a) 



(b) 



■ it. 



FIGURE 7-16 Example 7-8: 
(a) before collision, (h) after 
collision. 



SECTION 7-5 Elastic Collisions in One Dimension 177 



I Inelastic Collisions 



Completely inelastic collision 



Collisions in which kinetic energy is not conserved are called inelastic collisions. 
Some of the initial kinetic energy is transformed into other types of energy, such 
as thermal or potential energy, so the total kinetic energy after the collision is 
less than the total kinetic energy hefore the collision. The inverse can also 
happen when potential energy (such as chemical or nuclear) is released, in 
which case the total kinetic energy after the interaction can be greater than the 
initial kinetic energy. Explosions arc examples of this type. 

Typical macroscopic collisions arc inelastic, at least to some extent, and often 
to a large extent. It two objects stick together as a result of a collision, the colli- 
sion is said to be completely inelastic. Two colliding balls of putty that stick 
together or two railroad cars that couple together when they collide arc examples 
of completely inelastic collisions, The kinetic energy in some cases is all trans- 
formed to other forms of energy in an inelastic collision, but in other cases only 
part of it is. Tn Example 7-3, for instance, we saw that when a traveling railroad 
cat collided with a stationary one, the coupled cars traveled off with some kinetic 
energy. In a completely inelastic collision, the maximum amount of kinetic energy 
is transformed to other forms consistent with conservation of momentum. Even 
though kinetic energy is not conserved in inelastic collisions, the total energy is 
always conserved, and the total vector momentum is also conserved. 



EXAMPLE 7-9 



Railroad cars again. For the completely inelastic collision 
of two railroad cars that we considered in Example 7-3, calculate how much 
of the initial kinetic energy is transformed to thermal or other forms of energy, 

APPROACH The railroad ears stick together after the collision, so this is a 
completely inelastic collision. By subtracting the total kinetic energy after the 
collision from the total initial kinetic energy, we can find how much energy is 
transformed to other types of energy. 

SOLUTION Before the collision, only car A is moving, so the total initial 
kinetic energy is 

2m A i>i - | ( 10,000 kg}(24.0 m/s) 2 - 2.88 x 10" J. 

After the collision, both cars are moving with a speed of 12,0 m/s, by conser- 
vation of momentum (Example 7-3), So the total kinetic energy afterward is 

k (20,000 kg) (12.0 m/s) 2 = 1,44 X 10 ft J. 
Hence the energy transformed to other forms is 

(2.88 X 10° J) - (1.44 X 10* J) - 1.44 X 10* J, 
which is just half the original kinetic energy. 



Ballistic pendulum 



EXAMPLE 7-10 



Ballistic pendulum. The ballistic pendulum is a device 
used to measure the speed of a projectile, such as a bullet. The projectile, of 
mass m, is fired into a large block (of wood or other material) of mass M, 
which is suspended like a pendulum. (Usually, M is somewhat greater than m.) 
As a result of the collision, the pendulum and projectile together swing up to a 
maximum height A, Fig. 7- 17. Determine the relationship between the initial 
horizontal speed of the projectile, v, and the maximum height ft. 

APPROACH We can analyze the process by dividing it into two parts or two 
time intervals: (1) the time interval from just before to just after the collision 
itself, and (2) the subsequent time interval in which the pendulum moves from 
the vertical hanging position to the maximum height h- 



178 CHAPTER 7 Linear Momentum 



In part (I), Fig. 7- 1 7a, we assume the collision time is very short, so the projec- 
tile comes to rest in the block before the block has moved significantly from its 
position directly below its support. Thus there is effectively no net external force, 
and we can apply conservation of momentum to this completely inelastic collision. 
In part (2), Fig. 7- 1 7b, the pendulum begins to move, subject to a net external 
force (gravity, tending to pull it back to the vertical position); so for part (2), 
we cannot use conservation of momentum. But we can use conservation of 
mechanical energy because gravity is a conservative force (Chapter 6). The 
kinetic energy immediately after the collision is changed entirely to gravita- 
tional potential energy when the pendulum reaches its maximum height, h. 
SOLUTION In part (I) momentum is conserved: 



total p before = total p after 
mv = (m + M)v', 



(i> 



where v' is the speed of the block and embedded projectile just after the 
collision, before they have moved significantly. 

In part (2), mechanical energy is conserved. We choose y = when the 
pendulum hangs vertically, and (hen v = h when the pendulum-projectile 
system reaches its maximum height. Thus we write 

(kf. + pr) just after collision = (kf. + pr) at pendulum's maximum height 



or 

\{m + M)v' 2 + - + (m + M)gh, 

We solve for v': 

v' =s/lgh. 
Inserting this result for v' into Eq, (i) above, and solving for v, gives 



<"> 



m + M 

v - v 



m + M 



!)! 



Vigh, 



which is our final result. 

NOTE The separation of the process into two parts was crucial. Such an analysis 
is a powerful problem -solving tool. But how do you decide how to make such a 
division? Think about the conservation laws. They are your tools. Start a 
problem by asking yourself whether the conservation laws apply in the given 
situation. Here, we determined that momentum is conserved only during the 
brief collision, which we called part (1). But in part (1), because the collision is 
inelastic, the conservation of mechanical energy is not valid. Then in pari (2), 
conservation of mechanical energy is valid, but trot conservation of momentum. 
Note, however, that if (here had been significant motion of the pendulum 
during the deceleration of the projectile in the block, then there would have 
been an external force (gravity) during the collision, so conservation of 
momentum would not have been valid in part ( I ). 



Collisions in Two or Three Dimensions 

Conservation of momentum and energy can also be applied to collisions in two 
ot three dimensions, where the vector nature of momentum is especially impor- 
tant. One common type of non-head-on collision is that in which a moving 
object (called the "projectile") strikes a second ohjeet initially at rest (the 
"target"). This is the common situation in games such as billiards and pool, and 
for experiments in atomic and nuclear physics (the projectiles, from radioactive 
decay or a high-energy accelerator, strike a stationary target nucleus; Fig. 7-18). 



m 



M 



(a) 



r»~bi 



M + m 

FIGURE 7-17 Ballistic pendulum. 
Example 7-10, 



*■» PROBLEM SOLVING 

I ; \f the conservation laws to 
anal) ze a problem 



FIG URE 7-18 A recen t color- 
enhanced version of a cloud-chamber 
photograph made in the early days 
( 1920s) of nuclear physics. Green 
lines are paths of helium nuclei (He) 
coming from the left. One He, high- 
lighted in yellow, strikes a proton of 
the hydrogen gas in the chamber, and 
bod: scatter at an angle; the scattered 
proton's path is shown in red. 




*SECTION 7-7 Collisions in Two or Three Dimensions 179 



FIGURE 7-19 Object A. the projectile, collides 
with object B, the target. After the collision, they 
move off with m omenta p A and p' K at angles 
fl A and &' B . 



'AV^, 



/ Y*. 



_c 



p. 



I> \ 



^r 



LeHUt-rrni 



Figure 7-19 shows the incoming projectile, m A , heading along the x axis 
toward the target object, m B , which is initially at rest, If these arc billiard balls, 
m A strikes nijj and they go off at the angles &' A and flfj, respectively, which arc 
measured relative to m A 's initial direction (the x axis). 1 

Let us apply the law of conservation of momentum to a collision like that of 
Fig. 7-19. We choose the xy plane to be the plane in which the initial and final 
momenta lie, Momentum is a vector, and because the total momentum is 
conserved, its components in the x and y directions also arc conserved, The 
x component of momentum conservation gives 

Pax + P$x = P',\x + Pax 

or, with p Bx - m H v nx = 0, 

m A » A = m A u A cos9 A + m e .v' B ca$,8' B , (7-8aJ 

where the primes ('} refer to quantities after the collision. Because there is no 
motion in the y direction initially, the y component of the total momentum is 
zero before the collision. The y component equation of momentum conserva- 
tion is then 



or 



p. conserved 



P*. j + PBy ~ P\y + Phy 

= m A v' A sin B' A + m n v' n Sm ^'n ■ 



<7-8b) 



(A> 




Kl 



FIGURE 7-20 Example 7- 1 1 



EXAMPLE 7-11 



Billiard ball collision in 2-D. Billiard ball A moving with 
speed D A = 3.0 m/s in the +x direction (Fig. 7-20) strikes fin equal -mass ball 
B initially al rest, The two balls are observed to move off at 45 u to the x axis, 
ball A above the x axis and ball B below. That is, 6 A = 45° and 6' B = -45" in 
Fig. 7-20, What are the speeds of the two balls after the collision? 

APPROACH There is no net external force on our system of two balls. 
assuming the table is level (the normal force balances gravity). Thus 
momentum conservation applies, and we apply it to both the x and y compo- 
nents using the jrv coordinate system shown in Fig. 7-20. We get two equa- 
tions, and we have two unknowns. v' A and v' B . From symmetry we might guess 
that the two balls have the same speed. But let us not assume that now. Even 
though we aren't told whether the collision is clastic or inelastic, wc can still 
use conservation of momentum. 

SOLUTION We apply conservation of momentum, Eqs, 7-8a and h, and we 
solve for « A and v'q ■ We are given m A = m B ( = m), so 

(for x) mv A = mu A cos(45 u ) + mv' a cos(-45 u ) 
and 

(for y) - mv' A sin (45° ) + mv' u sin ( -45° ), 

The m's cancel out in both equations (the masses are equal). 

'Hit: objects may begin to deflect even Ixifore they touch if electric, magnetic, or nuclear forces act 
between tliem. You might think, for example, of two magnets oriented so that tliey repel each other: 
when one moves toward the other, the second moves away before the first one touches it. 



180 CHAPTER 7 Linear Momentum 



The second equation yields [recall that sin(-fl) - — sinS|: 

sin 45 ' J 



sin(45 :i ) 



-sin 45° 



So they do have equal speeds as we guessed at first. The x component equation 
gives [recall that cos(— 0) = COS &]'. 



s.o 



"a = W A cos(45 Q ) + »f,cos(45°) = 2i^cos(45 u ), 
v A 3.0 m/s 



«a = % = 



2 ras (45°) 2(0-707} 



= 2.1 m/s. 



NOTE When we have two independent equations, we can solve for, at most, 
two unknowns. 



EXERCISE F Make a calculation to see if kinetic energy was conserved in the collision 

of Example 7-11. 

Tf we know that a collision is elastic, we can also apply conservation of 
kinetic energy and obtain a third equation in addition to Eqs. 7-8a and b: 

KE A + KE B = KR A + KR{j 

or, for the collision shown in Fig. 7-20, 

|m A ^ = 2 m A v A + 2 m n v 'i' [elastic collision] (7-fte) 

If the collision is clastic, we have three independent equations and can solve for 
three unknowns. If we arc given m A , m B , v A (and v a , if it is not zero), we 
cannot, for example, predict the final variables, ti^ulj,^, and ft|j, because 
there are four of them. I lowever, if wc measure one of these variables, say 0' A , 
then the other three variables (v' A , v' a , and B' B ) are uniquely determined, and we 
can determine them using Eqs. 7-8a, b, c, 

A note of caution; Eq. 7-7 does not apply for two-dimensional collisions. Tt 
works only when a collision occurs along a line. 



KE conserved 



<P- 



CAUTION 



Equation 7-7 applies only in !-D 



PROBLEM SOLVING 



Momentum Conservation and Collisions 



1. Choose your system, If the situation is complex, 
think about how you might break it up into separate 
parts when one or more conservation laws apply. 

2. Consider whether a significant net external Force 
acts on your chosen system; if it docs, be sure the 
time interval At is so short that the effect on 
momentum is ncglible. That is, the forces that act 
between the interacting objects must be the signifi- 
cant ones if momentum conservation is to be used. 
[Note: Tf this is valid for a portion of the problem, 
you can use momentum conservation only for that 
portion. | 

3. Draw a diagram of the initial situation, just before 
the interaction (collision, explosion) takes place, 
and represent the momentum of each object with 
an arrow and a label. Do the same for the final situ- 
ation, just after the interaction. 

4. Choose a coordinate system and "+" and "-" 
directions. (For a head-on collision, you will need 
only an jc axis.) It is often convenient to choose 



the +jf axis in the direction of one object's initial 
velocity. 

5, Apply the momentum conservation equation(s): 

total initial momentum - total final momentum. 

You have one equation for each component 
(x, y, z): only one equation for a head-on collision. 
[Don't forget that it is the lotai momentum of the 
system that is conserved, not the momenta of indi- 
vidual objects.] 

6. If the collision is elastic, you can also write down a 
conservation of kinetic energy equation: 

total initial ke = total final ke, 

[Alternately, you could use Eq. 7-7: u A - Vq = 
Ax ~ "a . if the collision is one dimensional (head-on).] 

7, Solve for the unknown(s). 

8. Cheek your work, check the units, and ask yourself 
whether the results are reasonable. 



^SECTION 7-7 Collisions in Two or Three Dimensions 181 



1 Center of Mass (CM) 



Momentum is a powerful concept not only for analyzing collisions hut also for 
analyzing the transnational motion of real extended objects, Until now, whenever 
we have dealt with the motion of an extended object (that is, an object that has 
size), we have assumed that it could be approximated as a point particle or that it 
undergoes only translational motion. Real extended objects, however, can undergo 
rotational and other types of motion as well. For example, the diver in Fig. 7-2 1 a 
undergoes only translational motion (all parts of the object follow the same path), 
whereas the diver in Fig. 7-2 lb undergoes both translational and rotational 
motion, We will refer to motion that is not pure translation as genera! motion. 



FIGURE 7-21 The motion of the diver is 
pure translation in (a), but is translation plus 
rotation in (b). The black dot represents the 
diver's cm at each moment. 






(a) 



I I'M 



Observations indicate that even if an object rotates, ot several parts of a system 

of objects move relative to one another, there is one point that moves in [he same 

path that a particle would move if subjected to the same net force. This point is 

Cento- of mass called the tenter nf mass (abbreviated cm). The general motion of an extended 

General motion object (or system of objects) can be considered as the sum of the translational 

motion of the cm, plus rotational, vibrational, or other types of motion about the ow. 

As an example, consider the motion of the center of mass of the diver in 
Fig. 7-2 1; the ™ follows a parabolic path even when the diver rotates, as shown in 
Fig. 7-2 lb. This is the same parabolic path that a projected particle follows when 
acted on only by the force of gravity (that is, projectile motion). Other points in the 
rotating diver's body, such as her feet or head, follow more complicated paths. 

Figure 7-22 shows a wrench acted on by zero net force, translating and 
rotating along a horizontal surface, Note that its cm, marked by a red cross, 
moves in a straight line, as shown by the dashed white line. 



FIGURE 7-22 Translation plus 

rotation: a wrench moving over a 
horizontal surface. The cm, 
marked with a red cross, moves 
in a straight line. 



As*. 



■*">~(F « 



'7TT\ 



182 CHAPTER 7 



We will show in Section 7-10 that the important properties of the CM follow 
from Newton's laws if the cm is defined in the following way. We can consider any 
extended object as being made up of many tiny particles. Rut first we consider a 
system made up of only two particles (or small objects), of masses m A and h? b , 
We choose a coordinate system so that both particles lie on the x axis at positions 
_x A and ,v u , Fig. 7-23. The center of mass of this system is defined to be at the 

Linear Momentum 



position .v CM , given by 



XrM 



'"a + »h 



M 



x coordinate of cm (2 panicles) 



where M = m A + m s is the total mass of the system. The center of mass lies 
on the line joining wi A and m a . If the two masses are equal {m A = w B = wi), 
then x CM is midway between them, since in this case 

_ m(x A + x h ) _ Ua + jt b ) 
Xcu 2m ~ 2 

If one mass is greater than the other, say, m A > «i B , then the cm is closer to the 
larger mass. If all the mass is concentrated at ,r B , say, so m A - 0, then 
x cu = (0- t a + WuJTuJ/tO + wj b ) = * B , as we would expect. 

If there are more than two particles along a line, there will be additional 
terms: 



> 








— *A-+ 


X H -l 


m A 

*cm - 




■■" M 



FIGURE 7-23 The cenleT of mass 

of a two-particle system lies on Che 
line joining the two masses. Here 
m A > m-2 , so the CM is closer to 
m A than to m u , 



*CH 



A/ 



m A + m a + m c + ■■■ 
where M is the total mass of all the particles. 

■ =HfJ fi I ii * *flM CM of three guys on a raft. Three people of roughly 
equal masses m on a lightweight (air- filled) banana boat sit along the .v axis at 
positions x A - LOm.Jtij - 5.0 m, and x c - 6,0 m, measured from the left-hand 
end as shown in Fig. 7-24. Find the position of the cm. Ignore mass of boat. 

APPROACH We are given the mass and location of the three people, so we 
use three terms in Eq. 7-9a. We approximate each person as a point particle. 
Equivalent])', the location of each person is the position of that person's own cm. 
SOLUTION We use Eq. 7-9a with three terms: 

mx A + mx u + mxc w(jta + x a + x c) 



(7-9a) a t oordinate of cm (many particles) 



1.0 m 5.0 m 6.0 m x 

FIGURE 7-24 Example 7-12. 



m + m + m 3m 

(1.0 m + 5.0 m + 6.0 m) 12.0 m 



4,0 m, 



3 3 

The cm is 4.0 m from the left-hand end of the boat, 

NOTE The coordinates of the cm depend on the reference frame or coordinate 
system chosen. But the physical location of the cm is independent of that choice. 



EXERCISE G Calculate the CM of the three people in Example 7-12 taking the origin 
at the driver (x (: = 0) on the right Is the physical location of the cm the same? 

Tf the particles are spread out in two or three dimensionSi then we must specify 
not only the x coordinate of the cm (jr (M ), but also the v and z coordinates, which will 
be given hy formulas like Eq. 7-9a. For example, the y coordinate of the cm will be: 



A-M 



in-., ■ n}„ ! ■■• 



™a,Va + W'B v» + 
M 



A concept similar to center of mass is center of gravity (CG).The CG of an object 
is that point at which the force of gravity can be considered to act, The force of gravity 
actually acts on all the different parts or particles of an object, but for purposes of 
determining the translational motion of an object as a whole, we can assume that 
the entire weight of the object (which is the sum of the weights of all its parts) acts 
at the CG, There is a conceptual difference between the center of gravity and the 
center of mass, but for nearly all practical purposes, they are at the same point* 

It is often easier to determine the cm or cc of an extended object experimen- 
tally rather than analytically. If an object is suspended from any point.il will swing 
(Fig. 7-25) unless it is placed so its cc> lies on a vertical line directly below the 
point from which it is suspended. If the object is two dimensional, or has a plane of 
symmetry, it need only be hung from two different pivot points and the respective 
vertical (plumb) lines drawn. Then the center of gravity will be at the intersection 

■'I'here would be a difference between the cm find cc only in the unusual ease of an object so large 
that the acceleration due to gravity, g, was different at different parts of the object, 

SECTION 



(7-9b) ', ; twttiiititi >..'! m 
Center of gravity 



FIG UR E 7 -25 The force of gravi ty, 
considered to act at the cg, causes this 
object to rotate about the pivot point: 
if the cg were on a vertical line 
directly below the pivot, 
the object would remain at rest. 



Pivot point 




7-8 Center of Mass (CM) 183 




of the two lines, as in Fig. 7-26. If the object, doesn't have a plane of symmetry, 
the CO with respect to the third dimension is found by suspending the object 
from at least three points whose plumb lines do not lie in the same plane. Fot 
symmetrically shaped objects such as uniform cylinders (wheels), spheres, and 
rectangular solids, the cm is located at the geometric center of the object. 

Fot some objects, the cm may actually lie outside the object. The cm of a 
donut, for example, lies at the center of the hole. 



FIGURE 7-26 Finding the cg. 



CM for the Human Body 



Tf we have a group of extended objects, each of whose cm is known, we can find 
the cm of the group using Eqs. 7-9a and b. As an example, we consider the human 
body, Tabic 7-1 indicates the cm and hinge points (joints) for the different 
components of a "representative" person. Of course, there arc wide variations 
among people, so these data represent only a very rough average. The numbers 
represent a percentage of the total height, which is regarded as 100 units; similarly, 
the total mass is 100 units, For example, if a person is 1.70 m tall, his or her 
shoulder joint would be ( 1 .70 m }(8 1 .2/ 1 00) = 1.38m above the floor. 



TABLE 7-1 


Center of Mass of Parts of Tvpica 
(full height and mass = 100 units) 


1 Human Body 






Distance Above Floor 
of Hinge Points { c k ) 


Nin^t' Points 1 •) 
(Joints) 




Center of Mass 
{% Height Above 


Floor) 


Percent 
Mass 


91. 2 




Base of skull 






Head 


93.5 


6.9 


81. 2 




Shoulder joint 




:) 


Trunk and neck 


71.1 


46.1 


i»lhr'™r ftl 1 L_ 


s?\ 


Upper arms 


717 


6.6 










t W 


Lower arms 


55.3 


4.2 


52.1 




Hip joint 


7$ 


Hands 


43.1 


1.7 


1 : 


: 


Upper legs (thighs) 


42.5 


21.5 


28.5 




Knee joint 










I .ower legs 


18.2 


9.6 


4.0 




Ankle joint 


A 




Feet 

Body cm 


1.8 

= 58.0 


3.4 
10U.0 



FIGURE 7-27 Example 7-13: finding 

the cm of a leg in two different positions 
using percentages from Table 7-1 . 
(® represents die calculated cm). 




EXAMPLE 7-13 



A leg's cm. Determine the position of the cm of a whole 
leg (a) when stretched out, and (6) when bent at 90"'. See Fig. 7-27. Assume 
the person is I 70 m tall. 

APPROACH Our system consists of three objects: upper leg, lower leg, and 
foot. The location of the CM of each object, as well as the mass of each, is given 
in Table 7-1, where they are expressed in percentage units. To express the 
results in meters, these percentage values need to be multiplied by (1.70 m/100). 
When the leg is stretched out, the problem is one dimensional and we can 
solve for the x coordinate of the cm. When the leg is bent, the problem is two 
dimensional and we need to find both the x and y coordinates. 
SOLUTION (a) We determine the distances from the hip joint using Table 7-1 
and obtain the numbers (%) shown in Fig. 7-27a. Using Cq. 7 -9a, we obtain 

(2I.5}(9.6) + (9.6)(33.9) + (3.4}(50.3) 

x vm = TT^ 7T, T~, = 20.4 units. 

21.5 +9.6 + 3.4 

Thus, the center of mass of the leg and foot is 20.4 units from the hip joint, or 
52.1 - 20.4 - 31.7 units from the base of the foot. Since the person is 1.70 m 
tall, this is (1.70m) (3 1.7/ 1 00) -0.54 m above the bottom of the foot. 
(b) We use an xy coordinate system, as shown in Fig. 7-27b. First, we calculate 
how far to the right of the hip joint the CM lies, accounting for all three parts: 

(2I.5}(9.6) + (9.6)(23.6) + (3.4)(23.6) 



■f-r\a 



21.5 + 9.6 + 3.4 



- 1 4.9 units. 



184 CHAPTER 7 Linear Momentum 



For our 1.70-m-talI person, this is (1.70 m){ 14.9/100) = 0.25 m from the hip 
joint. Next, we calculate the distance, v CM , of the cm above the floor: 

(3.4)(1.8) + (9.6)(18,2) + (2L5)(28.5) 



VVv. = 



21.5 + 9.6 + 3.4 



= 23.0 units. 



or (1.70 m)(23. 0/100) = 0.39 m. Thus, the cm is located 39 cm above the floor 

and 25 cm to the right of the hip joint. 

NOTE The cm actually lies outside the body in (h). 




Knowing the CM of the body when it is in various positions is of great use in figure 7-28 A hieh iuniner's cm 

studying body mechanics. One simple example from athletics is shown in may actually pass beneath the bar. 
Fig. 7-28. H high jumpers can gel into the position shown, their cm can pass 

below the bar which their bodies go over, meaning that for a particular takeoff (a ) physics applied 
speed, they can clear a higher bar. This is indeed what they try to do, flitf* ia-nm:." 



7-10 



Center of Mass and Translational Motion 



As mentioned in Section 7-8, a major reason for the importance of the concept 
of center of mass is that the motion of the CM for a system of particles (or an 
extended object) is directly related to the net force acting on the system as a 
whole. We now show this, taking the simple ease of one-dimensional motion 
(x direction) and only three particles, but the extension to more objects and to 
three dimensions follows the same lines. 

Suppose the three particles lie on the .v axis and have masses wj a , wi B , m c , 
and positions Jt A , j: b . j: r . From Eq. 7-9a for the center of mass, we can write 



Af*ai = '"a-* a + m B x B + mexe, 



(i) 



where M = m A + m B + m c is the total mass of the system. If these particles 
arc in motion (say. along the x axis with velocities e a , i> h , and v c , respectively), 
then in a short time interval if they each will have traveled a distance 

Aj a = a a - jr A = v A Ar 

Ax B = x'u — X a = I'u Af 

Ax c = x'c- - Xc — w c Af, 

where x' A , x B , and x' c represent their new positions after time interval Af , The 
position of the new cm is given by 

Mx'ai = ™a 1 a + m B x' B + "if if ■ {") 

If we subtract from this equation (ii) the previous cm equation (i), we get 

MAj cm = wi A Ajt A + m E \x B + m c Ax c . 
During time interval At, the center of mass will have moved a distance 

A-V(M = -t" (:M _ Jr (:Sll = I'cm Af , 

where u rM is the velocity of the center of mass. We now substitute the relations 
for all the Ajt's into the equation just before the last oner 

Mv rK Af = m A v A A' + "in"(-'n ^ f + mcVc Af. 

We cancel At and get 

Mv m = m A v A + OT D t ! n + rrtci'c- (7-10) 

Since m A v A + m B i' B + m c i>c is the sum of the momenta of the particles of 
the system, it represents the total momentum of the system. Thus we see from 
Eq. 7-U) that the total (linear) momentum of a system of particles is equal to the 
product of the total muss M and the velocity of the center of mass of the system. 
Or, the linear momentum of an extended object is the product of the objects mass 
and the 'velocity of its cm. 



Total momentum, and 
velocity of cm 



•SECTION 7-10 Center of Mass and Franslational Motion 185 



Newton's second taw 

for a system of panicles 
oi-an ■. xtended object 



If forces are acting on the particles, then the panicles may be accelerating, 
Tn a shoTt lime interval Af, each particle's velocity will change by an amount 

& v a = °a Af, A !> b = a B At, Av c - a v At, 

Tf we now use the same reasoning as we did to derive Eq. 7-10, we obtain 



Ma (: , 



m A a A + '''u^U + Wlcftc- 



According to Newton's second law, m h a A = F A , m B a B = F B , and m c a c = t- 
where F A , F„ , and F c aTe the net forces on the three particles, respectively. Thus 
we get for the system as a whole Ma^ = F A + F K + F c , or 

Wflc« " f B L. t - (7-11) 

That is, /fe .sum of till the forces acting on the system is equal to the tola! muss of 
the system times the acceleration of its center of mass. This is Newtwn's second 
law for a system of particles, and it also applies to an extended object (which 
can be thought of as a collection of particles). Thus we conclude that the center 
of mass of a system of particles for of an extended object) iL'ith total mass M 
moves tike a single particle of mass M acted on by the same net external force. 
That is, the system moves as if all its mass were concentrated at the center of 
mass and all the external forces acted at that point. We can thus treat the trans- 
lational motion of any object or system of objects as the motion of a particle 
(see Figs. 7-21 and 7-22), This result simplifies our analysis of the motion of 
complex systems and extended objects. Although the motion of various parts of 
the system may be complicated, wc may often be satisfied with knowing the 
motion of the center of mass. This result also allows us to solve certain types of 
problems very easily, as illustrated by the following Example. 



CONCEPTUAL EXAMPLE 7-141 A Iwo-stage rocket. A rocket is shot 
into the air as shown in Fig. 7-29. At the moment the rocket reaches its 
highest point, a horizontal distance d from its starting point, a prearranged 
explosion separates it into two parts of equal mass. Part I is stopped in midair 
by the explosion, and it falls vertically to Earth. Where does part II land? 
Assume g = constant. 

RESPONSE After the rocket is fired, the path of the CM of the system 
continues to follow the parabolic trajectory of a projectile acted on by only a 
constant gravitational force. The CM will thus land at a point 2d from the 
starting point. Since the masses of f and II are equal, the cm must be midway 
between them at any time. Therefore, part II lands a distance 3d from the 
Starting point, 

NOTE If part T had been given a kick up or down, instead of merely falling, 
the solution would have been more complicated. 



EXERCISE H \ woman stands up in a rowboat and walks from one end of the boat to 
the other, How does the boat move, as seen from the shoTe? 



FIGURE 7-29 bxample 7- 14, 



■ 


„- 1 11 -. 

1 ■*** 

1 

1 

a.\ a 

1 

_ 1 

t 


"""*"• --Art 

\ 

\ 
\ 


■^ 




\ 



186 CHAPTER 7 Linear Momentum 



I Summary 



The momentum, p, of an object is defined as the product of its 

mass times its velocity, 

p = tttf. [1-1) 

In terms of momentum. Newton's second law can be 
written as 



SF = — -■ 
if 



(7-2) 



That is, the rate of change of momentum equals the net 
applied force, 

The law of conservation of momejiium states that the total 
momentum of an isolated system of objects remains constant. 
An isolated system is one on which the net external force is zero. 

The law of conservation of momentum is very useful in 
dealing with collisions. In a collision, two (or more) objects 
interact with each other over a very short time interval, and the 
forces between them during this time interval are very large. 

The impulse of a force on an object is defined as Fit. 
where F is the average force acting during the (usually short) 
lime interval if. The impulse is equal to the change in 
momentum of the object: 

Impulse = F if = ip, (7-5) 

Total momentum is conserved in any collision as long as 
any net external force is zero or negligible. If m A v A and m B v B 



are the momenta of two objects before the collision and m A v' A 

and i?i B TJ) are their momenta after, then momentum conserva- 
tion tell us thai 

"»AVi. + m B % = m A v' A + m u v' s (7-3) 

for this two-object system, 

Total energy is also conserved, but this may not be helpful 
in problem solving unless the only type of energy transformation 
involves kinetic energy. In that case kinetic energy is conserved 
and the collision is called an elastic collision, and we can write 

|m A HA + jm B Ds = ^m^v'i + lin n v'i. (7-6) 

If kinetic energy is not conserved, the collision is called 
inelastic. A completely inelastic collision is one In which the 
colliding objects Stick together after the collision. 

The center of mass (CM) of an extended object (or group 
of objects) is that point at which the net force can be consid- 
ered to act. for purposes of determining the translations] 
motion of the object as a whole. "The jr component of the CM 
for objects with mass m A ,m B is given by 

Ifl/yX/y ■+ H1,|.T B + ■■■ 

X(<a = 7 7" (7-9a) 

m A + m n H 

[*The complete motion of an object can be described as 
the iTanslalional motion of its center of mass plus rotation (or 
other internal motion) about its center of mass,] 



| Questions 



1. We claim that momentum is conserved, yet most moving 
objects eventually slow down and stop. Explain. 

2. When a person jumps from a tree to the ground, what 
happens to the momentum of the person upon striking 
the ground? 

3. When you release an in Hated but untied balloon, why 
does it fly across the room'? 

4 It is said that in ancient times a rich man with a bag of 
gold coins froze to death while stranded on a frozen lake. 
Because the ice was frictionless, he could not push himself 
to shore. What could he have done to save himself had he 
not been so miserly? 

5. How can a rocket change direction when it is far out in 
space and is essentially in a vacuum',' 

6. According to Eq. 7-5, the longer the impact time of an 
impulse, the smaller the force can be foT the same 
momentum change, and hence the smaller the deforma- 
tion of the object on which the force acts. On this basis, 
explain the value of air bags, which are intended to inflate 
during an automobile collision and reduce the possibility 
of fracture or death, 

7. Cars used to be built as rigid as possible to withstand 
collisions. Today, though, cars are designed to have 
"crumple zones" that collapse upon impact. What is the 
advantage of this new design? 

& Why can a batter hit a pitched baseball further than a ball 

tossed in the air by the batter? 
9. Is it possible for an object to receive a larger impulse 

from a small force than from a large force? Explain. 
Itt. A light object and a heavy object have the same kinetic 

energy. Which has the greater momentum? Explain. 
1 1, Describe a collision in which all kinetic eneTgy is lost- 



12. At a hydroelectric power plant, water is directed at high 
speed against turbine blades on an axle that turns an elec- 
tric generator. For maximum power generation, should 
the turbine blades be designed so that the water is 
brought to a dead slop, or so thai the water rebounds? 

13. A squash ball hits a wall at a 45" angle as 
shown in Fig. 7-30. What is the direction 
(a) of (he change in momentum of the 
ball, (t>) of the force on the wall? 



FIGURE 7-30 
Question f 3. 

14. A Superball is dropped from a height h onto a hard steel 
plate (fixed to the Earth), from which it Tebounds at very 
nearly its original s|ieed. (a) Is the momentum of the ball 
conserved during any part of this process? (b) If we 
consider the ball and Earth as our system, during what parts 
of the process is momentum conserved? (c) Answer part (h) 
for a piece of putty that falls and sticks to the steel plate. 

15. Why do you tend to lean backward when carrying a heavy 
load in your arms? 

16. Why is the cm of a 1 -m length of pipe at its mid-point, 
wheTeas this is not true for your arm or leg? 

:E 17. Show on a diagram how your cm shifts when you change 

from a lying position to a sitting position. 
:e 18. If only an external force can change the momentum of 

the center of mass of an object, how can the internal force 

of an engine accelerate a car? 
* 19. A rocket following a parabolic path through the air 

suddenly explodes into many pieces. What can you say 

about the motion of this system of pieces? 



Questions 187 



I Problems 



7-1 and 7-2 Momentum and Its Conservation 

1. (I) What is the magnitude of the momentum of a 28-g 
sparrow flying with a speed of 8-4 m/s? 

2. (I) A constant friction force of 25 N acts on a 65-kg skier 
for 20 s, What is the skieCs change in velocity? 

3. (II) A 0.145-kg baseball pitched at 39.0 m/s is hit on a 
horizontal line drive straight back toward the pitcher at 
52,0 m/s. If the contact time between bat and ball is 
3-00 x i0~ 3 s, calculate the average force between the 
ball and bat during contact. 

4. (II) A child in a boa! throws a 6,40-kg package out hori- 
zontally with a speed of 1 0.0 m/s. Fig. 7-31. Calculate the 
velocity of the boat immediately after, assuming it was 
initially at Test. The mass of the child is 26.0 kg. and thai 
of the boat is 45.0 kg. Ignore water resistance. 




FIGURE 7-31 Problem 4. 

5. (II) Calculate the force exerted on a rocket, given that the 
propelling gases are expelled at a Tate of 1500 kg/s with, a 
speed of 4.0 x in 4 m/s (at the moment of takeoff), 

ft. (II) A 95-kg halfback moving at 4.1 m/s on an apparent 
breakaway for a touchdown is tackled from behind. When 
he was tackled by an 85-kg cornerback running at 5.5 m/s 
in the same direction, what was their mutual speed imme- 
diately after the tackle? 

7. (II) A 12.600-kg railroad car travels alone on a level fric- 
tionless track with a constant speed of 1 8,0 m/s. A 
53504; g load, initially at rest, is dropped onto the car. 
What will be the car's new speed? 

(II) A 93O0-kg boxcar traveling at 15.0 m/s strikes a second 
boxcar at rest- The two stick together and move off with a 
speed of 6,0 m/s, What is the mass of the second car? 

'). (II) During a Chicago storm, winds can whip horizontally 
at speeds of 100 km/h. If the air strikes a person at the 
rate of 40 kg/s per square meter and is brought to rest, 
estimate the force of the wind on a person. Assume 
the person is 1 .50 m high and 0.50 m wide. Compare to the 
typical maximum force of friction (^ = 1.0) between the 
person and the ground, if the person has a mass of 70 kg. 

10. (II) A 38O0-kg open railroad car coasts along with a 
constant speed of 8.60 m/s on a level track. Snow begins 
to fall vertically and fills the car at a rate of 3.50 kg/min, 
Ignoring friction with the tracks, what is the speed of the 
car after 9O.0 min"? 



8 



IL. (II) An atomic nucleus initially moving at 420 m. s emits 
an alpha panicle in the direction of its velocity, and the 
remaining nucleus slows to 350 m/s. If the alpha particle 
has a mass of 4.0 u and the original nucleus has a mass of 
222 u, what speed does the alpha particle have when it is 
emitted? 

12. (II) A 23-g bullet traveling 230 m/s penetrates a 2.0-kg 
block of wood and emerges cleanly at 170 m/s, If the 
block is stationary on a frictlonless surface when hit, how 
fast does it move after the bullet emerges? 

13. (Ill) A 975-kg two-stage rocket is traveling at a speed of 
5.80 x 10 s m/s with respect to Earth when a pre-designed 
explosion separates the rocket into two sections of equal 
mass that then move at a speed of 2.20 x 10 3 m/s relative 
to each other along the original line of motion, {a) What 
aTC the speed and direction of each section (relative to 
Earth) after the explosion',' (h) How much energy was 
supplied by the explosion? [Hint. What is the change In 
ke as a result of the explosion?] 

L4. (Ill) A rocket of total mass 3180kg is traveling in outer 
space with a velocity of 115 m/s, To alter its course by 
35-0", its rockets can be fired briefly in a direction 
perpendicular to its original motion. If the rocket gases 
aTe expelled at a speed of 1750 m/s, how much mass must 
be expelled? 

7-3 Collisions and Impulse 

15. (II) A golf ball of mass 0.045kg is hit off the tee at a 
speed of 45 m/s. The golf club was in contact with the ball 
for 3.5 x 10~ 3 s. Find (a) the impulse imparted to the golf 
ball, and {b) the average force exerted on the ball by the 
golf club. 

16. (II) A 124^g hammer strikes a nail at a velocity of 8,5 m/s 
and comes to rest in a time interval of 8,0 ms, (a) What is 
the impulse given to the nail'? (b) What is the average 
force acting on the nail? 

17. (II) A tennis ball of mass m = 0.060 kg and 
speed v = 25 m/s strikes a wall at a 45 u 
angle and rebounds with the same 
speed at 45° (Fig. 7-32). What is 
the impulse (magnitude and 'V^ 



'A 



direction) given to the ball? 



45" 



45 



FIGURE 7-32 
Problem 17. 



18. (II) You are the design engineer in charge of the crashworthi- 
ness of new automobile models. Cars are tested by smashing 
them into fixed, massive barriers at 50 km/h (30mph), 
A new model of mass 1500 kg takes 0.15 s from the time 
of impact until it is brought to rest- (a) Calculate the 
average force exerted on the car by the barrier, 
(h) Calculate the average deceleration of the car. 



188 CHAPTER 7 Linear Momentum 



19, (I I) A 95-kg fullback is running a! 4.0 m/s. to (he east arid 
is stopped in 0,75 s hy a head-on tackle hy a tackier 
Tunning due west. Calculate (a) the original momentum 
of the fullback, (b) the impulse exerted on the fullback, 
(c) the impulse exerted on the tackier, and (rf) the 
average forte exerted on Ihe tackier. 

2<k (II) Suppose the force acting on a tennis hah (mass 
0.060 kg) points in the +x direction and is given by the 
graph of Fig. 7-33 as a function of lime. Use graphical 
methods to estimate {a) the total impulse given the 
ball, and (b) the velocity of the ball after being struck, 
assuming the ball is being served so it is nearly at 
rest initiallv. 



300 














































-. 200 

z 














































100 














































"t 
























1 








0. 


:l 
(s 


> 






0. 


02 



FIGURE 7-33 

Problem 20. 

21. (Ill) From what maximum height can a 75-kg person 
jump without breaking the lower leg bone of either leg? 
Ignore air resistance and assume the cm of the person 
moves a distance of 0.60 m from the standing to the 
seated position (that is. in breaking the fall). Assume the 
breaking strength (force per unit area) of bone is 
170 X lQ°N/m , and its smallest cross-sectional area is 
2.5 X 10 _4 nr, [Hint. Do not try this experimentally.] 

7-4 and 7-5 Elastic Collisions 

22. (II) A ball of mass 0.440 kg moving east (+jc direction) 
with a speed of 3,30 m/s collides head -on with a 
0,220-kg ball at rest. If the collision is perfectly elastic, 
what will be the speed and direction of each ball after 
the collision'? 

23. (II) A 0-450-kg ice puck, moving east with a speed of 
3.00 m/s, has a head-on collision with a 0,900-kg puck 
initially at rest. Assuming a perfectly elastic collision, 
what will be the speed and direction of each object after 
the collision? 

24. (II) Two billiard balls of equal mass undergo a perfectly 
elastic head-on collision, If one ball's initial speed was 
2.00m/s, and the other's was 3.00 m/s in the opposite 
direction, what will be their speeds after the collision? 

25. (II) A 0.060-kg tennis ball, moving with a speed of 
2.50 m/s. collides head-on with a 0.090-kg ball initially 
moving away from it at a speed of 1.15 m/s. Assuming a 
perfectly elastic collision, what are the speed and direc- 
tion of each ball after the collision? 

26. (II) A softball of mass 0.220 kg that is moving with a 
speed of 8,5 m/s collides head-on and elastically with 
another ball initially at rest. Afterward the incoming soft- 
ball bounces backward with a speed of 3.7 m/s. Calculate 
(a) the velocity of the target ball after the collision, and 
(6) the mass of the target ball. 



27. (II) Two bumper cars in an amusement park ride collide 
elastically as one approaches the other directly from the 
rear (Fig. 7-34), Car A has a mass of 450 kg and car G 
550kg. owing to differences in passenger mass. If car A 
approaches at 4.50 m/s and car B is moving at 3.70 m/s, 
calculate (a) their velocities after the collision, and (6) the 
change in momentum of eaeh- 



'"a- 

450 kg 



550 kg 



Ui 



(bj 



v A = '-'11 = 

4.50 m/s 3.70 m/s 



'', 



FIGURE 7-34 

Problem 27; 
(a) before colli- 
sion, (b) after 
collision. 



28 



(II) A 0.280-kg croquet ball makes an elastic head-on colli- 
sion with a second ball initially at rest. The second ball moves 
off with half the original speed of the first ball, (a) What is the 
mass of the second ball? (b) What fraction of the original 
kinetic energy ( Ake/ke) gets transferred to the second ball? 
2°. (Ill) In a physics lab, a cube slides down a frictionless incline 
as shown in Fig. 7-35, and elastically strikes another cube at 
the bottom that is only one-half its mass. If the incline is 30 cm 
high and the table is 90 cm off the floor, where does each 
cube land? [Hint: Roth leave the incline moving horiTontally.j 




90 cm 



FIGURE 7-35 
Problem 29. 



JO. (Ill) Take the general case of an object of mass i» A and 
velocity ua elastically striking a stationary (z>n = 0) object 
of mass mjj head-on. (a) Show that the final velocities v\ 

and v' n arc given by 

(m A - m B \ 



\ jti a +- m B 
f 2ffl A \ 
\m A +- m B / 



(b) What happens in the extreme case when m A is much 
smaller than m B ? Cite a common example of this, 
(e) What happens in the extreme case when m A is much 
larger than tn B ? Cite a common example of this, (rf) What 
happens in Ihe case when m,\ - "?n ? Cite a common example. 

7-6 Inelastic Collisions 

31. (I) In a ballistic pendulum experiment, projectile 1 results 

in a maximum height h of the pendulum equal to 2.6 cm. 
A second projectile causes the the pendulum to swing 
twice as high, hi =5.2 em. The second projectile was 
how many times faster than the first? 



Problems 189 



35 



32. (II) A 28-g rifle bullet traveling 230 m/s buries itself in a 
3.6 -kg pendulum hanging on a 2.8-m-long string, which 
makes [he pendulum swing upward in an are. Determine 
the vertical and horizontal components of the pendulum's 
displacement. 

35. (II) (a) Derive a formula for the fraction of kinetic 
energy lost, AKE/Kr., for the ballistic pendulum collision 
of Example 7-10. (b) Evaluate for hi = 14.0 g and 
M = 380". 

34. (II) An internal explosion breaks an object, initially at 
rest, into two pieces, one of which has 1 .5 times the mass 
of the other. If 7500 J were released in the explosion, how 
much kinetic energy did each piece acquire? 
(II) A 920-kg sports car collides into the rear end of a 
2300-kg SUV stopped at a red light. The bumpers lock, 
the brakes are locked, and the two cars skid forward 2,8 m 
before stopping. The police officer, knowing that the coef- 
ficient of kinetic friction between tires and road is 0-80, 
calculates the speed of the sports car at impact, What was 
that speed? 

(If) A ball is dropped from a height of 1.50 m and 
rebounds to a height of 1.20 m. Approximately how many 
rebounds will the ball make before losing 90'??. of its 
energy? 

37. (It) A measure of inelasticity in a head-on collision of two 
objects is the coefficient of restitution, c, defined as 

e = 

i>u ~ w A 

where v"a - v(j is the relative velocity of the two objects 
after the collision and vn — v& is them relative velocity 
before it. (a) Show that e = I for a perfectly elastic colli- 
sion, and t = for a completely inelastic collision, (b) A 
simple method for measuring the coefficient of restitution 
for an object colliding with a very hard surface like steel 
is to drop the object onto a heavy steel plate, as shown in 
Fig. 7-36. Determine a formula for e in terms of the orig- 
inal height h and the maximum height h' reached after 
one collision, 



3d 



o 



o 



u 



FIGURE 7-36 
of restitution. 



Problem 37. Measurement of the coefficient 



38. (II) A wooden block is cut into two pieces, one with three 
times the mass of the other. A depression is made in both 
faces of the cut. so that a firecracker can be placed in it 
with the block reassembled. The reassembled block is set 
on a rough-surfaced table, and the fuse is lit. When the 
firecracker explodes, the two blocks separate and slide 
apart. What is the ratio of distances each block travels? 



,iy. (Ill) A 15-0-kg object moving in the +x direction at 
5.5 m/s collides head-on with a 10.0-kg object moving in 
the —.v direction at 4,0 m/s. Find the final velocity of each 
mass if: (a) the objects stick together: (6) the collision is 
elastic: (c) the 1 5,0- kg object Is at rest after the collision; 
(rf) the 10.0-kg object is at rest after the collision; (e) the 
15.0-kg object has a velocity of 4.0 m/s in the —x direc- 
tion after the collision. Are the results in (c), (d), and (e) 
"reasonable"? Explain. 

* 7-7 Collisions in Two Dimensions 

* 41). (II) A radioactive nucleus at rest decays into a second 

nucleus, an electron, and a neutrino. The electron and 

neutrino are emitted at right angles and have momenta of 
9.30 X 10" 23 kg 'm/s and 5.40 X I0 _M kg-m/s. respec- 
tively. What are the magnitude and direction of the 
momentum of the second (recoiling) nucleus? 
"41. (II) An eagle (m A = 4.3 kg) moving with speed 
U A = 7.8 m/s is on a collision course with a second eagle 
(m R = 5.6 kg) moving at Vg = 10.2 m/s in a direction 
perpendicular to the first, After they collide, they hold 
onto one another. In what direction, and with what speed, 
are they moving after the collision? 

* 42. (II) Billiard ball A of mass m A = 0.400 kg moving with 

speed v A = 1,80 m/s strikes ball B. initially at rest, of 
mass ffin = 0.500 kg. As a result of the collision, ball A is 
deflected off at an angle of 30.0° with a speed 
v' A = 1.10 m/s. (a) Taking the x axis to be the original 
direction of motion of ball A. write down the equations 
expressing the conservation of momentum for the compo- 
nents in the .v and y directions separately, (b) Solve these 
equations for the speed wjj and angle Sg of ball B. Do 
not assume the collision is elastic. 
"43. (Ill) After a completely inelastic collision between two 
objects of equal mass, each having initial speed v, the two 
move off together with speed v/3. What was the angle 
between their initial directions? 

* 44. (Ill) Two billiard balls of equal mass move at right angles 

and meet at the origin of an xy coordinate system, Ball A 
is moving upward along the y axis at 2.0 m/s. and ball B is 
moving to the right along the x axis with speed 3.7 m/s 
After the collision, assumed elastic, ball R is moving along 
the positive y axis (Fig, 7-37), What is the final direction 
of ball A and what are their two speeds? 



= 3.7 m/s 



■+x 



® 



v A = 2.0 m/s 



FIGURE 7-37 

Problem 44. (Ball A after 
the collision is not shown.) 



: 4S. 



(Ill) A neon atom (m = 20.0 u) makes a perfectly elastic 
collision with another atom at rest, After the impact, the 
neon atom travels away at a 55.6° angle from its original 
direction and the unknown atom travels away at a -500" 
angle. What is the mass (in u) of the unknown atom? 
[Hint. You can use the law of sines.| 



190 CHAPTER 7 Linear Momentum 



7-8 Center of Mass 

46. (I) Find the center of mas* of the three-mass system shown 
in Kg. 7-38, Specify relative to the left-hand 1.00-kg mass. 



49 



511 



51. 



1. 00 kg 



0.50 IT1 



1.50 kg 1. 1 kg 



0.25 in 



FIGURE 7-38 
Problem 46. 



47, (I) The distance between a carbon atom (m c = I2u) and 
an oxygen atom (mo = 16 u) in the CO molecule is 



1.13 X 



in. How far from the carbon atom is the 



center of mass of the molecule'? 
48. (I) The cm of an empty 1050-kg ear is 2.50 m behind the 
front of the car. How far from the front of the car will the 
cm be when two people sit in the front seat 2.S0m from 
the front of [he ear. and IhTee people sit in [he back seat 
3.90 m from the front? Assume that each person lias a 
mass of 70,0 kg, 

(II) A square uniform raft, 18 m by 1 8 m, of mass 6800 kg, 
is used as a ferryboat, ff thTee cars, each of mass 1200 kg, 
occupy its NE, SE, and SW corners, determine the cm of 
the loaded ferryboat. 

(II) Three cubes, of sides / M , 24,. and 3I (I . are placed nest to 
one another (in contact) with their centers along a straight 
line and the I = 24, cube in the center (Fig. 7-39). What is 
the position, along this line, of the cm of this system? 
Assume the cubes are made of the same uniform material. 



FIGURE 7-39 


-- 


— - — 


Problem 50. 


- 2h - 



3/n 



(II) A (lightweight) pallet has a load of identical cases of 

tomato paste (see Fig. 7-40), each of which is a cube of 
length /. Find the center of gravity in the horizontal plane, 
so that the crane operator can pick up the load without 

lipping it. 



FIGURE 7-40 Problem 51 



FIGURE 7-41 Problem 52 



* 7-9 cm for the Human Body 

* 53. (I) Assume that your proportions are the same as those in 

Table 7-1, and calculate the mass of one of your legs. 

* 54. (I) Determine the CM of an outstretched arm using 

Table 7-1. 

* 55. (II) Use Fable 7-1 to calculate the position of the cm of 

an arm bent at a right angle. Assume that the person is 
155 em tall. 

'* 5f>. (II) When a high jumper is in a position such that his 
arms and legs are hanging vertically, and his trunk and 

head are horizontal, calculate how far below the torso's 
median line the CM will be. Will this cm be outside the 

body? Use Table 7-1. 

* 7-10 cm and Translation a I Motion 

* 57. (II) The masses of the Earth and Moon are 5.98 X 10 24 kg 

and 7,35 x 10 n kg, respectively, and their centers are 
separated by 3.84 x 10 s m. (a) Where is the cm of this 
system located' (fc) What can you say about the motion 
of the Earth -Moon system about the Sun, and of the 
tjarch and Moon separately about the Sun? 

* 58. (II) A 55-kg woman and an 80-kg man stand 10.0 m apart 

on frictionless ice, (a) How faT from the woman is their 
CM? (/>) If each holds one end of a rope, and the man 
pulls on the rope so that he moves 2.5m, how- far from 
the woman will he be now? (c) How far will the man have 
moved when he collides with the woman? 

* 59. (II) A mallet consists of a uniform cylindrical head of 

mass 2.00 kg and a diameter 0.0800 m mounted on a 
uniform cylindrical handle of mass 0-500 kg and length 
0,240 m, as shown in Fig, 7-42. If this mallet is tossed, 
spinning, into the air, how far above the bottom of 
the handle is the point that will follow a parabolic 
trajectory? 





52. (Ill) A uniform circular plate of radius 2R has a circular 
hole of radius R cut out of it. The center C of the smaller 
circle is a distance 0.80/? from the center C of the larger 
circle. Fig. 7-41 .What is the position of the center of mass 
of the plate? \llml: Try subtraction.) 



24.0 cm 



8.00 cm 
FIGURE 7-42 Problem 59. 

fill. (II) (vt) Suppose that in Example 7-14 (Fig. 7-29), 
m„ = 3/ni. Where then would w TT land? (b) What if 
m\ — 3m u 7 

61. (Ill) A helium balloon and its gondola, of mass M, are in 
the air and stationary with respect to the ground. A 
passenger, of mass m. then climhs out and slides down a 
rope with speed w, measured with respect to the balloon. 
With what s|wed and direction (relative to Earth) does 
the balloon then move? What happens if the passenger 
stops? 



Problems 191 



I General Problems 



62. A 0.145-kg baseball pitched horizontally at 35.0 m/s 
strikes a bat and is popped straight up to a height of 
55,6 m. Tf the contact time is 1,4ms, calculate the average 
force on the ball during the contact. 

63. A rocket of mass m traveling with speed %\, along the 
x axis suddenly shoots out fuel, equal to one-third of its 
mass, parallel to the y axis (perpendicular to the rocket 
as seen from the ground) with speed 2t%. Give the 
components of the final velocity of the rock el. 

: 64. A novice pool player is faced with the comer pocket shot 
shown in Fig. 7-43, Relative dimensions are also shown. 
Should the player be worried about this being a "scratch 
shot." in which the cue ball will also fall into a pocket? 
Give details. 



FT 
■,'3.0 
- 



-4.0- 



O Cue ball 



FIGURE 7-43 
Problem 64. 



65. A 140-kg astronaut (including space suit) acquires a 
speed of 2.50 m/s by pushing off with his legs from an 
1800-kg space capsule, (a) What is the change in speed 
of the space capsule? (b) If the push lasts 0.40 s, what is 
the average force exerted on the astronaut by the space 
capsule? As the reference frame, use the position of the 
space capsule before the push. 

66. Two astronauts, one of mass 60 kg and the other 80kg. are 
initially at rest in outer space. They then push each other 
apart, How far apart are they when the lighter astronaut 
has moved 12 m? 

67. A ball of mass m makes a head-on elastic collision with a 
second hall (at rest) and rehounds in the opposite direc- 
tion with a speed equal to one-fourth its original speed. 
What is the mass of the second ball? 

68. You have been lured as an expert witness in a court case 
involving an automobile accident. The accident involved 
car A of mass 1900 kg which crashed into stationary car B 
of mass 1100 kg, The driver of car A applied his brakes 
15 m before he crashed into car E. After the collision. 
car A slid 18 m while car B slid 30 m. The coefficient of 
kinetic friction between the locked wheels and the road 
was measured to be 0.60. Show that the driver of car A 
was exceeding the 55-mph (90 km/h) speed limit before 
applying the brakes. 

69. A golf ball rolls off the top of a flight of concrete steps of 
total vertical height 4.00 m. The ball hits four times on the 
way down, each time striking the horizontal pan of a 
different step 1.0m lower. If all collisions are perfectly 
elastic, what is the bounce height on the fourth bounce 
when the ball reaches the bottom of the stairs? 



7(1, A bullet is fired vertically into a 140-kg block of wood at 

rest directly above it. If the bullet has a mass of 2.9.0 g and 
a speed of 510 m/s, how high will the block rise after the 
bullet becomes embedded in it? 

71. A 25-g bullet strikes and becomes embedded in a 1.35-kg 
block of wood placed on a horizontal surface just in front 
of the gun. If the coefficient of kinetic friction between 
the block and the surface is 0.25, and the impact drives 
the block a distance of 9,5 m before it comes to rest, what 
was the muzzle speed of the bullet? 

72. Two people, one of mass 75 kg and the other of mass 
60 kg. sit in a rowboat of mass SO kg. With the boat 
initially at rest, the two people, who have been silting at 
opposite ends of the hoat 3.2 m apart from each other, 
now exchange seats. How far and in what direction will 
the boat move? 

73. A meteor whose mass was about 1.0 x 10 s kg struck the 
Earth (m t = 6,()x I0 24 kg) with a speed of about 15 km /s 
and came to rest in the Earth, (a) What was the Earth's 
recoil speed? (6) What fraction of the meteor's kinetic 
energy was transformed to kinetic energy of the Earth? 
(e) By how much did the Earth's kinetic energy change as 
a result of this collision? 

74. An object at rest is suddenly broken apart into two frag- 
ments by an explosion. One fragment acquires twice the 
kinetic energy of the other. What is the ratio of their masses? 

75. The force on a bullet is given by the formula 
F = 580 - (1.8 x 10 s )f over the time interval I = to 
t = 3.0 x 10 _i s. In this formula. ( is in seconds and F is 
in newtons. (a) Plot a graph of F vs. r for t = to 
; = 3.0 ms. (b) Estimate, using graphical methods, the 
impulse given the bullet, (c) If the bullet achieves a speed 
of 220 m/s as a result of this impulse, given to it in the 
barrel of a gun, what must its mass be? 

76. Two balls, of masses m A = 40 g and m a = 60 g, are 
suspended as shown in Fig. 7-44. The lighter ball is pulled 
away to a 60' : angle with the vertical and released, 
{a) What is the velocity of the lighter ball before impact? 
(h) What is the velocity of each ball after the elastic colli- 
sion? (c) What will be the maximum height of each ball 
after the elastic collision? 




] A " T U 



FIGURE 7-44 

Problem 76- 



77. An atomic nucleus at rest decays radioactivcly into an 
alpha particle and a smaller nucleus. What will be the 
speed of this recoiling nucleus if the speed of the alpha 
particle is 3.8 X 10 m/s? Assume the recoiling nucleus 
has a mass 57 times greateT than that of the alpha particle, 



192 CHAPTER 7 Linear Momentum 



u 



^T\30° 



Skeet ^- 

} D = 200 m/8 

Pellet 



I— —it H 



FIGURE 7-45 Problem 78. 



7H. A 0.25- kg skcct (clay target) is fired at an angle of 30" to 
the horizon with a speed of 25 m/s (Eg. 7 — 45). When it 
reaches the maximum height, it is hit from below by a 15-g 
pellet traveling vertically upward at a speed of 200 m/s. 
The pellet is embedded in the skeet. (a) How much higher 
did the skeet go up? (b) How much extra distance, A,v, 
does the skeet travel because of the collision? 

79, A Hock of mass m = 2,20 kg slides down a 30.0" incline 
which is 3 60 m high At the bottom, it strikes a block of mass 
M = 7.00 kg which is at rest on a hori7jontal surface. 
Fig, 7-46. (Assume a smooth transition at the bottom of the 
incline.) If the collision is elastic, and friction can be ignored, 
determine (a) the speeds of the two blocks after the collision, 
and (b) how far back up the incline the smaller mass will go, 



SI- The gravitational slingshot effect. Figure 7-47 shows the 
planet Saturn moving in the negative x direction at its 
orbital speed (with respect to the Sun) of 9.6 km/s. The 
mass of Saturn is 5.69 x 10^ kg. A spacecraft with mass 
825 kg approaches Saturn. When far from Satum.it moves in 
the +x direction at 10.4 km/s. The gravitational attraction 
of Saturn (a conservative force) acting on the spacecraft 
causes it to swing around the planet (orbit shown as dashed 
line) and head off in the opposite direction. F.stimate the 
final speed of the spacecraft after it is far enough away to 
he considered free of Saturn's gravitational pull. 



<■ = WA kmh 



3.6 m 



30"/ 



\ »lL 



FIGURE 7-46 Problems 79 and 80. 

S«. In Problem 79 (Fig. 7-46), what is the Upper limit on mass 
m if it is to rebound from M h slide up the incline, stop, 
slide down the incline, and collide with M again? 



''Wn = -' M >' !m ' ! i 



FIGURE 7-47 Problem 81. 




Answers to Exercises 

A: Yes. if the sports car's speed is three times greater. 

B: Larger. 

C: (a) 6.0 m/s: (ft) almost zero; (c) almost 24.0 m/s. 

D: "1'he curve would be wider and less high, 



E: Yes. by 300 times. 

F: Yes. ke was conserved. 

G: .v^y = -2.0 m; yes. 

H: l"he boat moves in the opposite direction. 



General Problems 193 



You too can experience rapid 
rotation — if your stomach can 
take the high angular velocity 
and centripetal acceleration 
of some of the faster amuse- 
ment park rides. If not, try 
the slower merry-go-round or 
Ferris wheel. Rotating carnival 
rides have rotational ki: as 
well as angular momentum. 



CHAPTER 



8 




m&sii 



Rotational Motion 



Until now, wc have been concerned mainly with translations! motion. We 
discussed the kinematics and dynamics of translational motion (the 
role of force), and the energy and momentum associated with it. In this 
Chapter we will deal with rotational motion. Wc will discuss the kinematics of 
rotational motion and then its dynamics (involving torque), as well as rotational 
kinetic energy and angular momentum (the rotational analog of linear 
momentum). We will find many analogies with translational motion, which will 
make our study easier. Our understanding of the world around us will be 
increased significantly — from rotating bicycle wheels and compact disks to 
amusement park tides, a spinning skater, the rotating Earth, and a centrifuge— 
and there may be a few surprises. 

We will consider mainly the rotation of rigid objects A rigid object is an 
object with a definite shape that doesn't change, so that the particles composing 
it slay in fixed positions relative to one another. Any Teal object is capable of 
vibrating or deforming when a force is exerted on it. But. these effects are often 
very small, so the concept of an ideal rigid object is very useful as a good 
approximation. 



194 



Angular Quantities 



We saw in Chapter 7 (Section 7-8) that the motion of a rigid object can he 
analyzed as the translations! motion of the object's center of mass, plus rota- 
tional motion about its center of mass. We have already discussed translational 
motion in detail, so now we focus on purely rotational motion. By purely rota- 
tional motion, we mean that all points in the object move in circles, such as the 
point P in the rotating wheel of Fig. 8-1, and that the centers of these circles all 
lie on a line called the axis of rotation. In Fig. 8-1 the axis of rotation is perpen- 
dicular to the page and passes through point O. 

Every point in an object rotating about a fixed axis moves in a circle 
(shown dashed in Fig, 8- 1 for point P) whose center is on the axis and whose 
radius is /-, the distance of that point from the axis of rotation. A straight line 
drawn from the axis to any point sweeps out the same angle in the same time. 

To indicate the angular position of a rotating object, or how far it has 
rotated, we specify the angle 8 of some particular line in the object (ted in Fig. 8- 1 ) 
with respect to a reference line, such as the x axis in Fig. 8-1. A point in the object, 
such as P in Fig. 8-1, moves through an angle 8 when it travels the distance / 
measured along the circumference of its circular path. Angles are commonly 
measured in degrees, but the mathematics of circular motion is much simpler if 
we use the radian for angular measure. One radian (abbreviated tad) is defined 
as the angle subtended by an arc whose length is equal to the radius. For 
example, in Fig. 8- lb, point P is a distance r from the axis of rotation, and it has 
moved a distance / along the arc of a circle. The arc length / is said to "subtend" 
the angle 0. T< I - r, then 8 is exactly equal to 1 rad. Tn radians, any angle is 
given by 

I 
& = -- (8-la) 



*\ — J( 




FIGURE 8-1 Looking at a wheel 

(hat is rotating counterclockwise 
about an axis through the wheel's 
center at O (axis perpendicular to the 
page). Each point, such as point P. 
moves in a circular path; I is the 
distance P travels as the wheel 
rotates through the angle 0, 



h in radians 



where r is the radius of the circle, and / is the are length subtended by the angle 
specified in radians. If / = r, then ft = I rad. 

The radian is dimension less since it is the ratio of two lengths. Nonetheless 
when giving an angle in radians, we always mention rad to remind us it is not 
degrees. It is often useful to rewrite Eq. 8- 1 a in terms of arc length /: 

/ = rB. (8-lb) 

Radians can be related to degrees in the following way. In a complete circle 
there are 36CF, which must correspond to an arc length equal to the circumfer- 
ence of the circle, I — litr. Thus 8 — l/r - lirr/r - 2 57 rad in a complete 
circle, so 

360° = 2w rad. 

One radian is therefore 360 < 72tt « 360 s /6-28 ** 57,3", An object that makes 
one complete revolution (rev) has rotated through 360", or 2-n radians' 

1 rev = 360° = 2tt rad, 



/ rad: arc length = radius 



Conversion, degrees to rad 
I rad a 57.3° 



EXAMPLE 8-1 



Bike wheel. A bike wheel rotates 4,50 revolutions. How 
many radians has it rotated? 

APPROACH All we need is a straightforward conversion of units using 

1 revolution = 360" = 2tt rad = 6,28 rad- 

SOLUTION 

4.50 revolutions = (4.50 rev )( 2?r^M = 9.00<rrad = 28.3 rad. 



SECTION 8-1 Angular Quantities 195 




Chord 



(a) 



J- Art: Icntnh 



[|v. 



FIGURE 8-2 (a) Example 8-2. 
(b) For small angles, arc length and 
(he choKl length (straight line) »Te 
neatly equal. 



Angular 

displacement {rati) 

FIGURE 8-3 A wheel rotates 
from (a) initial position ff t to 
(b) final position (f z .Thc angular 
displacement is AS = &i ~~ "l - 




Angular 

i riihis; 



EXAMPLE 8-2 



Birds of prey — in radians. A particular bird's eye can 
just distinguish objects that subtend an angle no smaller than about 
3 X 10~ 4 rad. (a) How many degrees is this? (b) How small an object can the 
bird just distinguish when flying at a height of 100 m (Fig. 8-2a)7 

APPROACH For (a) we use the relation 360 :| = 2tt rad. For (b) we use 
Eq. 8- lb, / = r&, to find the are length, 
SOLUTION (a) We convert 3 X 10 4 rad to degiees: 



(3 x IfT^rad 



f \2lTTSid/ 



0.0I7 :J . 



(b) We use Eq. S— lb, I = rB. For small angles, the arc length / and the chord 
length are approximately 1 the same (Fig. S-2b). Since r = 100 m and 
6 = 3 x 10" d rad. we find" 



/ - (100m)(3 x 10 *rad) - 3 x 10 



in 



3 cm. 



A bird can distinguish a small mouse (about 3 cm long) from a height of 
1 00 m. That is good eyesight. 

NOTE Had the angle been given in degrees, we would first have had to 
convert it to radians to make this calculation. Equation 8-1 is valid only if the 
angle is specified in radians, Degrees (or revolutions) won't work. 



To describe rotational motion, we make use of angular quantities, such as 
angular velocity and angular acceleration. These are defined in analogy to the 
corresponding quantities in linear motion, and are chosen to describe the 
to la ting object as a whole, so they are the same for each point in the rotating object 
Each point in a rotating object may also have translational velocity and acceleration, 
but they have different values for different points in the object. 

When an object, such as the bicycle wheel in Fig. 8-3, rotates from some 
initial position, specified by B t , to some final position, ft 2 , its angular 
displacement is 

Ad = 6 2 - B[- 

The angular velocity (denoted by tu, the Greek lowercase letter omega) is 
defined in analogy with linear (translational) velocity that was discussed in 
Chapter 2. Instead of linear displacement, we use the angular displacement. 
Thus the average angular velocity is defined as 

_ A0 

to = — * (8-2a) 

M 

where A0 is the angle through which the object has rotated in the time interval Af. 
We define the instantaneous angular velocity as the very small angle A#, 
through which the object turns in the very short time interval Af: 

tr> = lim (»-2b) 

Angular velocity is generally specified in radians per second (rad/s), Note that 
all points in a rigid object rotate with the same angular velocity, since every posi- 
tion in the object moves through the same angle in the same time interval, 

An object such as the wheel in Fig. 8-3 can rotate about a fixed axis either 
clockwise or counterclockwise. The direction can be specified with a + or - sign, 
just as we did in Chapter 2 for linear motion along the + x or — x axis. The usual 
convention is to choose the angular displacement AB and angular velocity to as 
positive when the wheel rotates counterclockwise. If the rotation is clockwise, 
then 9 would decrease, so Afl and w would be negative. 4 

' Lven for an angle as large as 15". the error in making this estimate is only I ST.. bur for larger angles 

the error increases rapidly, 

*Tlic vector nature of angular velocity and other angular quantities is discussed in Section S-f (optional). 



196 CHAPTER 8 Rotational Motion 



Angular acceleration (denoted by uc, the Greek lowercase letter alpha), in 
analogy to linear acceleration, is defined as the change in angular velocity 
divided by the time required to make this change. The average angular 
acceleration is defined as 



_ « 2 

IT 



At 



Aw 

aT 



<8-3a) 



where Uj is the angular velocity initially, and a> 2 is the angular velocity after a 
time interval At. Instantaneous angular acceleration is defined in the usual way 

as the limit of this ratio as Ar approaches zero: 



Aw 
a — hm -7— ■ 
ir->o At 



(8-3b) 



Since tit is the same for all points of a rotating object, Eq. 8-3 tells us that a also 
will be the same for all points, Thus, <w and a are properties of the rotating 
object as a whole. With w measured in radians per second and / in seconds, a 
will be expressed as radians per second squared (rad/s 2 ). 

Each point or particle of a rotating rigid object has, at any moment, a 
linear velocity v and a linear acceleration a. We can relate the linear quantities 
at each point, » and a, to the angular quantities of the rotating object, 
a and a. Consider a point P located a distance / from the axis of rota- 
tion, as in Fig. 8-4. If the object rotates with angular velocity <v, any point 
will have a linear velocity whose direction is tangent to its circular path. The 
magnitude of that point's linear velocity is v - A!/ At. From Eq. 8- lb, a 
change in rotation angle AS (in radians) is related to the linear distance traveled by 
At — r A8, I lencc 

A/ AG 

v - — — r—— 
At At 



or 



ra. 



(8-4) 



Thus, although w is the same for every point in the rotating object at any 
instant, the linear velocity v is greater for points farther from the axis (Fig. 8-5). 
Note that Eq. 8-4 is valid both instantaneously and on the average. 



Angular 
acceleration 



Linear and angular 
velocity related 




FIGURE 8-4 A point P on a rotating wheel 
has a linear velocity v at a«v moment. 




FIGURE 8-5 A wheel rotating uniformly counterclockwise. 
Two points on the wheel, at distances r A and r fi from the 
center, have the same angular velocity id because they travel 
through the same angle in the same time interval. But the 
two points have different linear velocities because they travel 
different distances in the same time interval. Since r u > r A , 
then Wp > t) A [v = ran). 



SECTION 8-1 Angular Quantities 197 



CONCEPTUAL EXAMPLE 8-3 J Is the lion fasteT than the horse? On a 



rotating carousel or merry-go-round, one child sits on a horse near the outer 
edge and another child sits on a lion halfway out from the center, (a) Which 
child has the greater linear velocity? (b) Which child has the greater angular 
velocity? 

RESPONSE (a) The linear velocity is the distance traveled divided by the 
time interval In one rotation the child on the outer edge travels a kingeT 
distance than the child near the center, but the time interval is the same Iot 
both. Thus the child at the outer edge, on the horse, has the greater linear 
velocily. 

(£>) The angular velocity is the angle of rotation divided by the time interval, In 
one rotation both children rotate through the same angle (360 y = 2v radians). 
The two children have the same angular velocity. 



Tangential acceleration 



If the angular velocity of a rotating object changes, the object as a whole— 
and each point in it — has an angular acceleration. Each point also has a linear 
acceleration whose direction is tangent to that point's circular path. We use 
Eq. 8-4 (v - no) to show that the angular acceleration a is related to the 
tangential linear acceleration ff lai] of a point in the rotating object by 



"lan — 



Aw 
to 



Aii> 
~At 



or 



(8-5) 



Tn this equation, r is the radius of the circle in which the particle is moving, and 
the subscript "tan" in a lmi stands for "tangential," 

The total linear acceleration of a point is the vector sum of two 
components: 



= a r . 



+ a'R, 



Cewipetal 
etixrleration 



where the radial* component, a R , is the radial or "centripetal" acceleration and 
its direction is toward the center of the point's circular path; see Fig. 8-6. We 
saw in Chapter 5 (Eq. 5-1) that a R - v 2 /r, and we can rewrite this in terms 
of tit using Eq. 8-4: 



'■'i: 



(fw) : 



(8-61 



Thus the centripetal acceleration is greater the farther you arc from the axis of 
rotation: the children farthest out on a carousel feel the greatest acceleration, 

FIGURE 8-6 On a rotating wheel Equations 8-4, 8-5, and 8-6 relate the angular quantities describing the rota- 

whose angular speed is increasing, a tion of an object to the linear quantities for each point of the object. Tabic 8- 1 

point P has both tangential and radial summarizes these relationships. 

(centripetal) components of linear 

acceleration. (See also Chapter 5.) 




TABLE 8- 


-1 


Linear and Rotational Quantities 




Lintiir 




Type 


Rotational 


Relation 


X 




displacement 


(t 


,v = r» 


V 




velocity 


to 


v = rca 


«Lan 




acceleration 


rt 


fftan = ™ 



"■Radial" means along the radius. — Tlia: is, towaid or away from ilic center or avis. 



198 CHAPTER 8 Rotational Motion 



EXAMPLE 8-4 



Angular and linear velocities and accelerations. A 
carousel is initially at rest. At / = it is given a constant angular acceleration 
a - Q.060 rad/s 2 , which increases its angular velocity tor 8.0 s. At t - 8,0 s, 
determine tire following quantities: (a) the angular velocity of the carousel; 
(b) the linear velocity of a child (Fig. S-7a) located 2.5 tn from the center, 
point P in Fig. 8-7b; (c) the tangential (linear) acceleration of that child; (d) the 
centripetal acceleration of the child; and (e) the total linear acceleration of 
the child. 

APPROACH The angular acceleration m is constant, so we can use Eq. 8-3a 

to solve for a) after a time t = 8,0 s. With this tit and the given a, we 
determine the other quantities using the relations we just developed, 
Eqs. 8-4, 8-5, and 8-6. 

SOLUTION {a) Equation 8-3a tells us 



At 
We are given At = 8.0 s, a = 0.060 rad/s : , and u, = 0. Solving for ia 2 , we get 

<1>2 — <D\ + H if 

- + (0.060 rad/s 2 )(8.0s) - 0.48 rad/s. 

During the 8.0-s interval, the carousel has accelerated from <y L = (rest) to 

£i> 2 = 0.48 rad/s. 

(b) The linear velocity of the child with r = 2.5 m at time / = 8.0 $ is found 
using Eq. 8-4: 

v - no - (2.5 m) (0.48 rad/s) - 1.2 m/s. 

Note that the "rad" has been dropped here because it is dimensionless (and 
only a reminder) — it is a ratio of two distances, Eq. 8- lb. 

(c) The child's tangential acceleration is given by Eq. 8-5: 

fltan = ra = (2.5 m)(0.060 rad/s 2 ) = 0.15 m/s : , 

and it is the same throughout the 8.0-s acceleration interval. 

(d) The child's centripetal acceleration at t = 8.0 s is given by Eq, 8-6: 



a„ = 



v 1 (1.2 m/s) 2 



(2.5 m) 



= 0.58 m/s'. 



(e) The two components of linear acceleration calculated in parts (c) and (d) 
are perpendicular to each other. Thus the total linear acceleration at / = 8.0 s 
has magnitude 



a ~ V<4n + ffii 



= v'(0-"5m/^) 2 + (0.58 m/s 2 ) 2 = 0.60 m/s 
Its direction (Fig. 8-7b) is 

'0.1 5 m/s 2 



B = t 



an-(^) 



tan 



0.58 m/s ; 



0.25 rad, 



so B & 1 5" 



NOTE The linear acceleration is mostly centripetal, keeping the child moving 
in a circle with the carousel. The tangential component that speeds up the 
motion is smaller. 





FIGURE 8-7 Example 8-4. The 

total acceleration vector 

3 = "inn + Sr- ac ' = S-0s. 



SECTION 8-1 Angular Quantities 199 



Frequency 



Period 



PHYSICS APPLIED 

Hard drive 
and hit speed 



We can relate the angular velocity m to the frequency of rotation, /. The 
frequency is the number of complete revolutions (rev) per second, as we saw 
in Chapter 5. One revolution (of a wheel, say) corresponds to an angle of 
2- radians, and thus 1 rev/s = 2jj- rad/s, Hence, in general, the frequency / 
is related to the angular velocity w by 



/ = 



■■- 



oi 



= 2irf. 



(8-7) 



The unit for frequency, revolutions per second (rev/s), is given the special name 
the hertz (Hz). That is 

1 Hz = I rev/s, 

Note that "revolution" is not really a unit, so we can also write 1 Hz = 1 s _l . 

The time required for one complete revolution is called the period T, and it 
is related to the frequency by 

1 



T = 



i 



(8-S> 



If a particle rotates at a frequency of three revolutions per second, then the 
period of each revolution is }s, 

EXERCISE A In Example 8-4, we found that the carousel, after 8.0 s, rotates at an 
angular velocity w = 0.48 rad/s. and continues to do so after : = 8,0 s because the 
acceleration ceased. What are the frequency and period of the carousel? 



EXAMPLE 8-5 



Hard drive. The platter of the hard drive of a computer 
rotates at 7200 rpm (revolutions per minute = rev/min). (a) What is the 
angular velocity of the platter? (b) If the reading head of the drive is located 
3.00 cm from the rotation axis, what is the linear speed of the point on the 
platter just below it? (c) If a single bit requires 0.50 /j,m of length along the 
direction of motion, how many bits per second can the writing head write 
when it is 3.00 cm from the axis? 

APPROACH We use the given frequency / to find the angular velocity w of 
the platter and then the linear speed of a point on the platter (v - fw). The 
bit rate is found by dividing the linear speed by the length of one bit 

(■» = distance/time), 

SOLUTION (a) First we find the frequency in rev/s, given / - 7200 rev/min: 



/ = 



(7200 rev/min) 



120 rev/s = 120 I Iz. 



(60s/min) 
Then the angular velocity is 

<i> - 2irf - 754 rad/s. 

(b) The linear speed of a point 3.00 cm out from the axis is given by Eq. 8-4: 

v - r«i - (3,00 x 10 2 m)(754 rad/s) - 22.6 m/s. 

(c) Each bit requires 0.50 X 10 h m, so at a speed of 22.6 m/s, the number of 
bits passing the head per second is 



22.6 m/s 



0.50 x 10 "m/bit 
or 45 megabits/s (Mbps). 



45 x 10'' bits per second. 



200 CHAPTER 8 Rotational Motion 



Constant Angular Acceleration 



In Chapter 2, we derived the useful kinematic equations (Eqs. 2-11) that relate 
acceleration, velocity, distance, and time for the special case of uniform linear 
acceleration. Those equations were derived from the definitions of linear 
velocity and acceleration, assuming constant acceleration. The definitions of 
angular velocity and angular acceleration are the same as those for their linear 
counterparts, except that 9 has replaced the linear displacement x, a has 
replaced i>, and a has replaced a. Therefore, the angular equations for constant 
angular aeteleralion will he analogous to Eqs. 2- 1 1 with x replaced by tf, v by w, 
and a by a, and they can be derived in exactly the same way. We summarize 
them here, opposite their linear equivalents (we've chosen x B = 0, and 6 U = 
at the initial time ( = 0): 



Angular 



J. in Lit r 



it> = <if t ) + tit 

9 = toe t + jtj( ! 

(o 1 = (of) + 2a9 

_ <1) + (tfi;; 



v = Va + at 
x = v a s + \at l 
xr = Vq + 2ax 

_ V + V u 



V 



[constant fjf, ff] (8-9j») Kinematic equations 

[constant of, a] («-9b) /,„■ constant 

[constant a, a] (8-9c) angular acceleration 

[constant a, a] (j{-9d) (.v (J = 0. 6 U = 0) 



Note that to tl represents the angular velocity at t - 0, whereas 8 and w represent 
the angular position and velocity, respectively, at time t. Since the angular accel- 
eration h constant, tit = a. 



EXAMPLE 8-6 



Centrifuge acceleration. A centrifuge rotor is accelerated 
from rest to 20,000 rpm in 30s. (a) What is its average angular acceleration? 
(fe) Through how many revolutions has the centrifuge rotor turned during its 
acceleration period, assuming constant angular acceleration? 

APPROACH To determine a - Ato/At, we need the initial and final angular 
velocities, For (b), we use Eqs. S-9 (recall that one revolution corresponds to 
9 - 2ir rad). 
SOLUTION (a) The initial angular velocity is <a = 0. The final angular velocity is 

(20.000 rev/min) 
(o = lirf = (27rrad/rev) ^——. — , . , = 2100rad/s. 

(60 s/ mm) 

Then, since a = Ato/At and Ar = 30s, we have 

<o — (o (l 21 00 rad/s - . , 

a = = = 70 rad/s'. 

Af 30 s 

That is, every second the rotor's angular velocity increases by 70 rad/s, or by 
(70/2tt) = II revolutions per second. 

(b) To find (i we could use either Eq, g-9b or 8-9c, or both to check our 
answer. The former gives 

ft = + f(7Qrad/s z )(30s) Z = 3.15 X 10 4 rad, 

where we have kept an extra digit because this is an intermediate result. To 
find the total number of revolutions, we divide by 2tt rad/rev and obtain 

3.15 x 10 4 rad en lft3 

— = 5.0 X 1 0' rev. 

2jt rad/ rev 

NOTE Let us calculate 9 using Eq. 8-9c: 

(2100 rad/s) 1 - n 



/.' 



of — w\ 



2a 2(70 rad/s 2 ) 

which cheeks our answer perfectly. 



3.15 x I0 4 rad 



SECTION 8-2 Constant Angular Acceleration 201 




FIGURE 8-8 (a) A wheel Rilling 
to the right. Its center C moves with 

velocity v. Point P is at rest at this 
instant, (b) Ine same wheel as seen 
from a reference frame in which the 
axle of the wheel C is at rest — that 
is, we are moving to the right with 
velocity v relative to the ground. 
Point P. which was at rest in (a), here 
in (b) is moving to the left with 
velocity -v as shown. (See also 
Section 3-8 on relative velocity.) 



Rolling Motion {Without Slipping) 

The. rolling motion of a ball or wheel is familiar in everyday life: a ball rolling 
across the floor, or the wheels and tires of a car or bicycle rolling along the pave- 
ment. Rolling without slipping is readily analyzed and depends on static friction 
between the rolling object and the ground. The friction is static because the rolling 
object's point of contact with the ground is at rest at each moment. 

Rolling without slipping involves both rotation and translation. There is 
then a simple relation between the linear speed v of the axle and the angular 
velocity to of the rotating wheel or sphere: namely, v = rcj (where r is the 
radius) as wc now show, Figure 8-8a shows a wheel rolling to the right without 
slipping. At the moment shown, point P on the wheel is in contact with the 
ground and is momentarily at test. The velocity of the axle at the wheel's center C 
is v, In Fig, S-8b we have put ourselves in the reference frame of the wheel — 
that is, we arc moving to the right with velocity v relative to the ground, In this 
reference frame the axle C is at rest, whereas the ground and point ? are 
moving to the left with velocity — v as shown. Here we are seeing pure rotation. 
So we can use Eq. 8-4 to obtain v - rw, where r is the radius of the wheel. 
This is the same v as in Fig. 8-8a, so we see that the linear speed v of the axle 
relative to (he ground is related to the angular velocity to by 



no. 



[rolling without slipping] 



This relationship is valid only if there is no slipping. 



EXAMPLE 8-7 



Bicycle. A bicycle slows down uniformly from t\, = 8.40 m/s 
to rest over a distance of 115 m, Fig. H-9. Each wheel and tire has an overall 
diameter of 68.0em. Determine (a) the angular velocity of the wheels at the 
initial instant (t = 0); (6) the total number of revolutions each wheel rotates 
before coming to rest; (c) the angular acceleration of the wheel; and (J) the 
time it took to come to a stop, 

APPROACH We assume the bicycle wheels roll without slipping and the tire is 
in firm contact with the ground- The speed of the bike v and the angulaT 
velocity of the wheels <i> are related by v = rm. The bike slows down 
uniformly, so the angular acceleration is constant and we can use Eqs. 8-9. 

SOLUTION (a) The initial angular velocity of the wheel, whose radius is 34,0cm, is 

% 8.40 m/s 
r 



(Bfl = 



0.340 m 



= 24.7 rad/s. 



(b) Tn coming to a stop, the hike passes over 115 m of ground- Tie circumference 
of the wheel is 2ttt, so each revolution of the wheel corresponds to a distance 
traveled of 2irr = (2ir) (0.340m). Thus the number of revolutions the wheel 
makes in coming to a stop is 

1 15 m 115 m 



2jrr (2 jt) (0,340 m) 



53.8 rev. 



FIGURE 8-9 FxamplcS-7. 




Bike as seen from the around at t = 



202 CHAPTER 8 Rotational Motion 



(c) The angular acceleration of the wheel can be obtained from Eq. 8-9c, for 
which we set w = and w y = 24,7 rad/s. Because each revolution corre- 
sponds to 2t7 radians of angle, then = 2?r rad/rev X 53. 8 rev (=338 rad} and 



<y, 



- (247 rad/s) 2 



= -0.902 rad/s ! . 



2B 2(2tt rad/rcv)(53.S rev) 

(rf) Equation 8 -9 a or b allows us to solve for the time. The first is easier: 



t = 



- 24.7 rad/s „„ , 

7T = 27.4 s, 

-0.902 rad/s 2 



NOTE When the bike lire completes one revolution, the bike advances 
linearly a distance equal to the outer circumference (2ur) of the tire, as long 
as there is no slipping ot sliding. 



Torque 



We have so far discussed rotational kinematics — the description of rotational 
motion in terms of angle, angular velocity, and angular acceleration. Now we 
discuss the dynamics, or causes, of rotational motion. Just as we found analogies 
between linear and rotational motion for the description of motion, so rota- 
tional equivalents for dynamics exist as well. 

To make an object start rotating about an axis clearly requires a force. But 
the direction of this force, and where it is applied, are also important. Take, for 
example, an ordinary situation such as the overhead view of the door in 
Fig. 8-I0. If you apply a force F A to the door as shown, you will find that the 
greater the magnitude. F A , the more quickly the door opens. But now if you 
apply the same magnitude force at a point closer to the hinge — say, F B in 
Fig. 8-10 — the door will not open so quickly. The effect of the force is less: 
where the force acts, as well as its magnitude and direction, affects how quickly 
the door opens. Indeed, if only this one force acts, the angular acceleration of 
the door is proportional not only to the magnitude of the force, but is also 
directly proportional to the perpendicular distance from the axis of rotation to 
the line along which the force acts. This distance is called the lever arm, or 
moment arm, of the force, and is labeled r A and r u for the two forces in Fig. 8- 1 0. 
Thus, if r A in Fig, 8- 1 is three times larger than r u , then the angular acceleration 
of the door will be three times as great, assuming that the magnitudes of the 
forces are the same. To say it another way, if r A — 3r a , then F B must be three 
times as large as F A to give the same angular acceleration. (Figure 8- 1 1 shows 
two examples of tools whose long lever arms are very effective,) 



— »hH 



T 



S? 



M fj 



FIGURE 8-10 Applying the same 
forte with different lever arms, r& 
and r u , If r A = 3r u , then to create 
the same effect (angular accelera- 
tion), F% needs to be three times F^ , 
or F A = 5 F H . 

Lever arm 





FIGURE 3-11 (a) A plumber can exert greater 
torque using a wrench with a long lever arm. 
(b) A tire iron too can have a long lever arm. 



SECTION 8-4 Torque 203 



Tort/in' defined 



in) 




>&< 



Of 



\ = ^r_ 






(b) 

FIGURE 8-12 (a) Furees acting ai 
different angles at the doorknob, 
fb) The lever arm is defined as the 
perpendicular distance from the axis 
of rotation (the hinge) to the line of 
action of the force, 



FIGURE 8-13 

Torque = r ± F = rh' L . 



Axis of 

rotation 



Point of 
■y. application 

A^s.^ uf force 




(a) 



-^ 






■ii 



-i 



ih] 



Magnitude of a torque 



The angular acceleration, then, is proportional to the product of the force 
times the lever arm, This product is called the moment of the force about the 
axis, or, more commonly, it is called the torque, and is represented by t (Greek 
lowercase letter Urn). Thus, the angular acceleration u of an object is directly 
proportional to the net applied torque t: 

a cc t, 

and wc sec that it is torque that gives rise to angular acceleration. This is the 
rotational analog of Newton's second law for linear motion, a cc F. 

We defined the lever arm as the perpendicular distance from the axis of 
rotation to the line of action of the force — that is, the distance which is perpen- 
dicular both to the axis of rotation and to an imaginary line drawn along the 
direction of the force. We do this to take into account the effect of forces acting 
at an angle. It is clear that a force applied at an angle, such as F c in Fig. 8—12, 
will be less effective than the same magnitude force applied perpendicular to 
the door, such as P A (Fig. 8- 12a). And if you push on the end of the door so that 
the force is directed at the hinge (the axis of rotation), as indicated by f D , the 
door will not rotate at all. 

The lever arm iot a force such as f c is found by drawing a line along the 
direction of f c (this is the "line of action" of P c ). Then we draw another line, 
perpendicular to this line of action, that goes to the axis of rotation and is 
perpendicular also to it. The length of this second line is the lever aim iot F c and is 
labeled r c in Fig. 8- 12b- The lever arm is perpendicular both to the line of 
action of the force and, at its other end, perpendicular to the rotation axis. 

The magnitude of the torque associated with F c is then r c F c . Tiis short leveT 
ami j- c and the corresponding smaller torque associated with F c is consistent 
with the observation that F c is less effective in accelerating the door than is F A . 
When the lever arm is defined in this way, experiment shows that the relation 
a ex t is valid in general, Notice in Fig. 8-12 that the line of action of the force 
Fr> passes through the hinge, and hence its lever arm is zero. Consequently, zero 
torque is associated with F n and it gives rise to no angular acceleration, in 
accord with everyday experience. 

In general, then, we can write the magnitude of the torque about a given axis as 

t = r ± F, (8- lua) 

where r ± is the lever arm, and the perpendicular symbol (_L) reminds us that wc 
must use the distance from the axis of rotation that is perpendicular to the line 
of action of the force (Fig. S— 13a). 

An equivalent way of determining the torque associated with a force is to 
Tesolve the force into components parallel and perpendicular to the line that 
connects the axis to the point of application of the force, as shown in Fig, 8- 13b. The 
component Fj| exerts no torque since it is directed at the rotation axis (its 
moment arm is zero). Hence the torque will be equal to F ± times the distance r 
from the axis to the point of application of the force: 

t = rF - (8- 10b) 

That this gives the same result as Eq. 8- 10a can be seen from the relations 
F = F sin 6 and r ± = r sin B- [Note that B is the angle between the directions 
of F and r (radial line from the axis to the point where F acts)]. So 

t = rF sin S (8-10e) 

in either case. We can use any of Cqs. 8- 10 to calculate the torque, whichever is 
easiest. 

Since torque is a distance times a force, it is measured in units of m ■ N in SI 
units, 1 cm dyne in the cgs system, and ft lb in the English system. 

'Note that the units for torque arc the same as those for energy. We write the unit for torque here 
as m-N {in St J to distinguish it from energy (N-m) because the two quantities are very different. 
An obvious difference is that energy is a scalar, whereas torque has a direction and is a vector. ITie 
special name joule (I J= IN m)is used only for energy (and for work), never for torque. 



204 CHAPTER 8 Rotational Motion 



EXAMPLE 3-8 



Biceps torque. The hieeps muscle exerts a vertical force (a) 
on the lower arm, bent as shown in Figs. 8- 14a and b. For each case, calculate 
the torque about the axis of rotation through the elbow joint, assuming the 
muscle is attached 5.0cm from the elbow as shown, 

APPROACH The force is given, and the lever arm in fa) is given. Tn (b) we 
have to take into account the angle to get the lever arm. 
SOLUTION (a) F - 700 N and r_ - 0.050 m, so 

t = r_F = (0.050 m)(700N} = 35m -N. 

(b) Beeause the arm is at an angle below the horizontal, the lever arm is shorter 
(Fig, 8-14c) than in part (a): r, = (0.050 m) (sin 60"),, where 6 = 60" is the 
angle between F and r. F is still 700 N, so 

t - (0.050 m) (0.866) (700 N) = 30m-N, 

The arm can exert less torque at this angle than when it is at 90". Weight machines 
at gyms are often designed to take this variation with angle into account. 
NOTE In (b), we could instead have used t - rF ± . As shown in Fig. S-14d, 
F L - F sin 60°. Then t - rF L - rFsmO - (0.050 m }( 700 N)( 0.866) gives the 
same result. 



EXERCISE B Two forces (F n = 20 N and F A - 30 N) are applied to a meter stick which 
can rotate about its left end. Fig. 8-15. Force Fjj is applied perpendicularly at the 
midpoint- Which force exerts the greater torque? 



■ 



■ 



Axis 



I' M 



p^< 



FIGURE 8-15 Exercise B. 



When more than one torque acts on an object, the angular acceleration a 
is found to be proportional to the net torque, If all the torques acting on an 
object tend to rotate it in the same direction about a fixed axis of rotation, the 
net torque is the sum of the torques But if, say, one torque acts to rotate an 
object in one direction, and a second torque acts to rotate the object in the 
opposite direction (as in Fig. 8-16), the net torque is the difference of the two 
torques. We normally assign a positive sign to torques that act to rotate the 
object counterclockwise, and a negative sign to torques that act to rotate 
the object clockwise. 



EXAMPLE 8-9 



Torque on a compound wheel. Two thin disk-shaped 
wheels, of radii r A = 30 cm and r u = 50 cm, are attached to each other on an 
axle that passes through the center of each, as shown in Fig. 8-16. Calculate 
the net torque on this compound wheel due to the two forces shown, each of 
magnitude 50 N. 

APPROACH The force F A acts to rotate the system counterclockwise, whereas 
F u acts to rotate it clockwise. So the two forces act in opposition to each other. 
We must choose one direction of rotation to be positive — say, counterclock- 
wise, Then F A exerts a positive torque, t a = r A F A , since the lever arm is t A . 
F u , on the other hand, produces a negative (clockwise) torque and does not 
act perpendicular to r u , so we must use its perpendicular component to 
calculate the torque it produces: t u — —r u F tiL - —r u F a sin 0, where — 60°. 
(Note that 9 must be the angle between ? u and a radial line from the axis.) 
SOLUTION The net torque is 

t = r A F A - r n F K sin 60 ,J 

= (0.30m)(50N) - (0.50 m)(50N)(0,866) = -6.7 m-N. 

This net torque acts to accelerate the rotation of the wheel in the clockwise direction. 

NOTE The two forces have the same magnitude, yet they produce a net 
torque because their lever arms are different, 



Axis at 
elbow 

(b) 




A\K 




(c) Fi 




(J.) « X v 


A&is J- 






Axis 


t 
/Pi 


FIGURE 8- 


14 


l : x 


ample S-S. 





FIGURE 8-16 Example 8-9. The 

torque due to F A tends to acceleralu 
the wheel counterclockwise, 
whereas the torque due to l"n tends 

to accelerate the wheel clockwise, 




SECTION 8-4 Torque 205 




* Forces that Act to Tilt the Axis 

We are considering only rotation about a fixed axis, and so we consider only 
forces that act in a plane perpendicular to the axis of rotation. If there is a force 
(or component of a force) acting parallel to the axis of rotation, it will tend to 
tilt the axis of rotation — the component Fy in Fig, 8-17 is an example- Since we 
are assuming the axis remains fixed in direction, either there can be no such 
forces or else the axis must he mounted in bearings or hinges that hold the axis 
fixed. Thus, only a force, or component of a force (F, in Fig. 8-17), in a plane 
perpendicular to the axis will give rise to rotation about the axis, and it is only 
these that we consider. 



FIGURE 8-17 Only the compo- 
nent of F that acts in the plane 
perpendicular to the rotation axis. F, . 
acts to turn the wheel about the axis, 
The component parallel to the 
axis. F , would tend to move the axis 
itself, which we assume is held fixed. 



FIGURE 8-1 

in a circle of 
point- 



8 A mass m rotating 
radius r ahout a fixed 



-J 

j: 




Rotational Dynamics; Torque and 
Rotational Inertia 



We have discussed that the angular acceleration ut of a rotating object is propor- 
tional to the net torque t applied to it: 

where wc write St to remind us 1 that it is the net torque (sum of all torques 
acting on the object) that is proportional to a. This corresponds to Newton's 
second law for tra relational motion, uocSf, but here torque has taken the 
place of force, and, correspondingly, the angular acceleration a takes the place 
of the linear acceleration a. In the linear case, the acceleration is not only 
proportional to the net force, but it is also inversely proportional to the inertia 
of the object, which we call its mass, m. Thus we could write a — SF/m. But 
what plays the role of mass for the rotational case? That is what we now set out 
to determine. At the same time, we will see that the relation « <x St follows 
directly from Newton's second law, SF - ma. 

We first consider a very simple case: a particle of mass m rotating in a circle 
of Tadius r at the end of a string or rod whose mass we can ignore compared to m 
(Fig. 8-18), and we assume a single force Facts on m as shown. The torque that 
gives rise to the angular acceleration is t - rF. If we use Newton's second law 
foT linear quantities, SF = ma, and Eq, 8-5 relating the angular acceleration 
to the tangential lineaT acceleration, <j lan = ra, then we have 

F - ma 
- mra. 

When we multiply both sides of this equation by r, we find that the torque 
t = rF is given by 

t = mra. [single particle] (8-11) 

Here at last we have a direct relation between the angular acceleration and the 
applied torque t The quantity irir 2 represents the rotational inertia of the 
particle and is called its moment of inertia. 

Now let us consider a rotating rigid object, such as a wheel rotating about an 
axis through its center, which could be an axle. We can think of the wheel as 
consisting of many particles located at various distances from the avis of rotation. 
We can apply Eq. 8-1 1 to each particle of the object, and then sum over all the 
particles. The sum of the various torques is just the total torque, St, so we obtain: 

Sr - (Sw 2 )a (8-12) 

where we factored out tx because it is the same for all the particles of the object. 
The sum Sm/- 2 represents the sum of the masses of each particle in the object 
multiplied by the square of the distance of that particle from the axis of 

: Recall from Chapter 4 that 2 (Greek letter sigma) means "sum of." 



206 CHAPTER 8 Rotational Motion 



rotation. If we assign each particle a number (1,2,3 ), then Swr 2 = 

ni\r\ + m 2 rj + m^rj, + ■-. This quantity is called the moment of inertia (or 

(8-13) Mouii'ill of inertia 



rotational inertia) /of the object: 

/ - Smr 2 — m x r] + mjr\ + *■• 
Combining Eqs. 8-12 and 8-13, we can write 



Ex = Ic 



(8-14) 



This is the rotational equivalent of Newton's second law, Tt is valid for the rota- 
tion of a rigid object about a fixed axis, 1 

We see that the moment of inertia, /, which is a measure of the rotational 
inertia of an object, plays the same role for rotational motion that mass does for 
translational motion. As can be seen from Eq. 8-13, the rotational inertia of an 
object depends not only on its mass, but also on how that mass is distributed 
with respect to the axis. For example, a large-diameter cylinder will have greater 
rotational inertia than one of equal mass but smaller diameter (and therefore 
greater length). Fig, 8-19. The former will be harder to start rotating, and 
harder to stop. When the mass is concentrated farther from the axis of rotation, 
the rotational inertia is greater. For rotational motion, the mass of an object 
cannot be considered as concentrated at its center of mass, 



NEWTON'S SECOND LAW 
FOR ROTATION 






CAUTION 



Mass can aut be considered 
concentrated m ( w for rotational 

morion 





FIGURE 8-19 A large-diameter cylinder 
has greater rotational inertia than one of 
smaller diameter hut equal mass 



EXAMPLE 8-10 



Two weights on a bar: different axis, different /. Two 
small "weights," of mass 5,0 kg and 7.0 kg, are mounted 4.0 m apart on a light 
rod (whose mass can be ignored), as shown in Fig. 8-20. Calculate the moment 
of inertia of the system (a) when rotated about an axis halfway between the 
weights, Fig. 8-20a. and (b) when rotated about an axis 0.50 m to the left of the 
5.0-kg mass (Fig. 8-20b). 

APPROACH In each case, the moment of inertia of the system is found by 
summing over the two parts using Eq. 8- 1 3. 

SOLUTION (a) Roth weights are the same distance, 2.0 m, from the axis of 
rotation, Thus 

/ = 2mr = (5,0kg)(2.0m) 2 + (7.0kg)(2.0m) : 
= 20kgm 3 + 2Skgm 3 = 48kgm ! . 

(b) The 5,0-kg mass is now 0,50 m from the axis, and the 7,0-kg mass is 4.50 m 
from the axis. Then 

/ - Imr 2 - (5.0 kg) (0.50 m) 2 + (7.0kg)(4.5m) 2 
1.3kg-nr + 142 kg -nr - 143 kg -nr. 

NOTE This Example illustrates two important points. First, the moment of 
inertia of a given system is different for different axes of rotation. Second, we 
sec in part (6) that mass close to the axis of rotation contributes little to the total 
moment of inertia; here, the 5.04cg object contributed less than 1% to the total. 



'liquation 8-14 is also valid when the object is translating with acceleration, as long as / and a are 
tabulated about the cenrer of mass of the object, and the rotation axis through the Cm doesn't 
change direct km. 



FIGURE 8-20 Example 8-10: 
calculating the moment of inertia. 



-4,0 ui- 
I 



5.0 k- 



0,50 m 



I 5.0 kg 

I 

V,:r; 



I 
I 

Axis 
00 



-4,0 m- 



(b) 



CAUTION 



7,0 kg 



7.0 kg 



I depends on axis of rotation 
and on distribution of mass 



SECTION 8-5 Rotational Dynamics; Torque and Rotational Inertia 207 



FIGURE 8-21 Moment!; of inertia 






Location 






Moment of 


for various objects of uniform 




Object 


of axis 




inertia 


composition, 








Axis 






(a) 


Thin lump. 

ruililK A' 


Through 

center 


<sL 


-", 


SfflP 
















(b) 


Thin hoop, 
radius R 
width W 


Through 

central 
diameter 


* 


A\ ■ 

■ 

Axis 


i.Wfl- i ^.v/w- 




(c) 


Solid cylinder, 

r\ulillS K 


Through 

center 


a:: 




>Jf> 



(rl) Hollow cylinder, 

inner radius fl[ 
outer radius R 2 



<e> 



Uniform sphere, 
radius R 



Through 

center R. 



Through 
center 




;AW? ' *J) 



iMR 2 



(f) 1 .on |> uniform rod. Through 

lenerh J. center 



Uis 



* L 



,MI? 



(g) Lang uniform rod. Through 

length L end 

(h) Rettangukr Through 

thin plate. center 
length L, width W 




\ml* 



iM(/,2i W-) 



For most ordinary objects, the mass is distributed continuously, and the 
calculation of the moment of inertia, 2mr : , can be difficult. Expressions can, 
however, be worked out (using calculus) for the moments of inertia of regularly 
shaped objects in terms of the dimensions of the objects. Figure S— 21 gives 
these expressions for a number of solids rotated about the axes specified. The 
only one for which the result is obvious is that for the thin hoop or ring rotated 
about an axis passing through its center perpendicular to the plane of the hoop 
(Fig. 8-2 la). For this object, all the mass is concentrated at the same distance 
from the axis, fi.Thus 2mr 2 - (Sm)S 2 - A/ft ! , where M is the total mass of 
the hoop. 

When calculation is difficult, /can be determined experimentally by measuring 
the angular acceleration a about a fixed axis due to a known net torque, St, and 
applying Newton's second law, / - 2r/of, Eq, 8-14. 

Solving Problems in Rotational Dynamics 

When working with torque and angular acceleration (Eq. 8-14), it is important 
to use a consistent set of units, which in ST is: a in rad/s 2 ; t in m - N; and the 
moment of inertia, /, in kg • nr. 



208 CHAPTER 8 Rotational Motion 



PROBLEM SOLVING 



Rotational Motion 



1. As always, draw a clear and complete diagram. 

2. Choose the object or objects that will be the system 
to be studied, 

3. Draw a free-body diagram for the object under 
consideration (or for each object, if moTe than 
one), showing only (and fill) the forces acting on 
that object and exactly where they act, so you can 
determine the torque due to each. Gravity acts at 
the CG of the object (Section 7-8). 

4. Identify the axis of rotation and determine the 
torques about it. Choose positive and negative 



directions of rotation (counterclockwise and clock- 
wise), and assign the correct sign to each torque, 

5. Apply Newton's second law for rotation, St = la. 
If the moment of inertia is not given, and it is not 
the unknown sought, you need to determine it first. 
Use consistent units, which in SI are: a in rad/s J ; 
t in m-N; and I in kg* m 2 . 

6. Also apply Newton's second law for translation, 
XF = mS, and other laws or principles as needed. 

7. Solve the resulting equation(s) for the unknown(s). 

8. Do a rough estimate to determine if your answer is 
reasonable. 



EXAMPLE 8-11 



A heavy pulley. A 15. ON force (represented by F r ) is 
applied to a cord wrapped around a pulley of mass M - 4.00kg and radius 
R - 33.0 cm, Fig. 8-22. The pulley accelerates uniformly from rest to an angular 
speed of 30.0 rad/s in 3,00 s. If there is a fractional torque r fr - 1,10 m. -N at the 
axle, determine the moment of inertia of the pulley. The pulley rotates about its center. 
APPROACH We follow the steps of the Problem Solving Box explicitly. 
SOLUTION 
T. Draw a diagram. The pulley and the attached cord are shown in Fig. 8-22. 

2. Choose the system: the pulley. 

3. Draw a free-body diagram. The force that the cord exerts on the pulley is 
shown as F, in Fig. 8-22. The friction force is also shown, but wc arc given 
only its torque. Two other forces could be included in the diagram: the force 
of gravity mg down and whatever force keeps the axle in place. They do not 
contribute to the torque (their lever arms are zero) and so are not shown. 

4. Determine the torques. The cord exerts a force r? r that acts at the edge of 
the pulley, so its lever arm is R. The torque exerted by the cord equals RF r 
and is counterclockwise, which we choose to be positive. The frictional 
torque is given as T rc =1.10 m-N; it opposes the motion and is negative. 

5. Apply Newton's second law fur rotation. The net torque is 

St - RF T - r : , 

= (0.330 m)( 15.0 N) - 1.10 m-N = 3.85 m-N. 

The angular acceleration a is found from the given data that it takes 3.0s to 
accelerate the pulley from rest to w = 30.0 rad/s: 

Aw 30.0 rad/s - 
a = — = —± = 10.0 rad/s 2 . 

Af 3.00 s 

We can now solve for [ in Newton's second law (see step 7). 

6. Other calculations: None needed. 

7. Solve for unknowns. We solve for / in Newton's second law for rotation 
and insert our values for St and a; 

_ Xt 3.85 m N 

/ — 

Of 



la, 



0.385 kg -m 2 



10.0 rad/s 2 

8, Do a rough estimate. We can do a rough estimate of the moment of inertia 
by assuming the pulley is a uniform cylinder and using Fig. 8-2 1c: 

I = ^MR 2 = |(4.00kg)(0.330m) ! = 0.218 kg-ra ! . 

This is the same order of magnitude as our result, but numerically somewhat 
less. This makes sense, though, because a pulley is not usually a uniform 
cylinder but instead has more of its mass concentrated toward the outside 
edge. Such a pulley would be expected to have a greater moment of inertia 
than a solid cylinder of equal mass; a thin hoop, Fig, S-2la, ought to 
have a greater /than our pulley, and indeed it does: / - MR 2 - 0.436 kg -m 2 . 




FIGURE 8-22 Example 8-11. 



Usefulness and power 
of rough estimates 



SECTION 8-6 209 




FIG URE 8-23 Example 8-12. 

(a) Pulley and tailing bucket ol 
mass m. (b) Free -body diagram for 
(he bucket, 



Additional Example — a bit more challenging 



EXAMPLE 8-12 



Pulley and bucket. Consider again the pulley in 
Fig. 8-22 and Example 8-11. But this time, instead of a constant 15.0- N force 
being exerted on the cord, we now have a bucket of weight w; = 15.0 N (mass 
m = w/g = 1 .53 kg) hanging from the cord. See Fig. 8-23a. We assume the 
cord has negligible mass and does not stretch or slip on the pulley. Calculate the 
angular acceleration a of the pulley and the linear acceleration a of the bucket, 
APPROACH This situation looks a lot like Example S- 1 1, Fig. 8-22. But there 
is a big difference: the tension in the cord is now an unknown, and it is no 
longer equal to the weight of the bucket if the bucket accelerates. Our system 
has two parts: the bucket, which can undergo translational motion (Fig. S-23b 
is its free-body diagram); and the pulley. The pulley docs not translate, but it 
can rotate. We apply the rotational version of Newton's second law to the 
pulley, St — la, and the linear version to the bucket, 2F - ma. 
SOLUTION Let F^ be the tension in the cord. Then a force F T acts at the edge of the 
pulley, and we apply Newton's second law; Eq. 8-14, for the rotation of the pulley: 

/(i = St = RF r - Tf r , [pulley] 

Next we look at the (linear) motion of the bucket of mass m- Figure 8-23b, the 
free- body diagram for the bucket, shows that two forces act on the bucket: 
the force of gravity mg acts downward, and the tension of the cord F T pulls 
upward. Applying Newton's second law, SF = ma, for the bucket, we have 
(taking downward as positive): 

mg - F r — ma. | bucket] 

Note that, the tension F t , which is the force exerted on the edge of the pulley, 
is not equal to the weight of the bucket (- mg - 15.0 N). There must be a net 
force on the bucket if it is accelerating, so F l: < mg. We can also see this from 
the last equation above, F T - mg - ma. 

To obtain «, we note that the tangential acceleration of a point on the 
edge of the pulley is the same as the acceleration of the bucket if the cord 
doesn't stretch or slip. Hence we can use Eq. 8-5, a lm = a= Ru. Substituting 
F T = mg - ma = mg - mfttx into the first equation above (Newton's second 
law for rotation of the pulley), we obtain 

In = St = RF T - Tf r = R(mg — mRtu} - Tf, = mgR - mR 2 a - Tr r , 
a appears in the second term on the right, so we bring that term to the left side 
and solve for a: 

mgR - t & 

a — z-' 

/ + mR'- 

The numerator (mgR - T fr ) is the net torque, and the denominator (/ + mR 2 } 

is the total rotational inertia of the system. Then, since / = 0.385 kg-m 2 , 

m = 1.53 kg, and T fr = l.lOnvN (from Example 8-1 1), 

(15.0N)(0.330m) - 1.10m -N 

a = — ; : — rz — = 6-98 rad/V, 

0.385 kg ■ m 2 + ( I .53 kg)(0.330 m) 2 

The angular acceleration is somewhat less in this case than the 10,0 rad/s 2 of 
Example 8-11. Why? Because F r (= mg — ma) is less than the 15.D-N weight 
of the bucket, mg. The linear acceleration of the bucket is 
a = Ra = (0.330 m)(6.9S rad/s 2 ) = 2.30 m/s 2 . 
NOTE The tension in the cord F v is less than mg because the bucket accelerates. 



Rotational Kinetic Energy 



210 CHAPTER 8 



The quantity j«n>* is the kinetic energy of an object undergoing translational 
motion. An object rotating about an axis is said to have rotational kinetic 
energy. By analogy with translational kinetic energy, we would expect this to be 
given by the expression j/w 2 , where fis the moment of inertia of the object and 
w is its angular velocity. We can indeed show that this is true. 
Rotational Motion 



Consider any rigid rotating object as made up of many tiny particles, each 
of mass m, f f we lei r represent the distance of any one particle from the axis of 
rotation, then its linear velocity is v - ru. The total kinetic energy of the 
whole object will be the sum of the kinetic energies of all its particles: 

ke = 1(\mv 2 } - S(fwirV) 

We have factored out the ^ and the to 1 since they are the same for every particle 
of a rigid object. Since S«ir ! - /, the moment of inertia, we see that the 
kinetic energy of a rigid rotating object is, as expected, 

rotational ke = jlar. (R-1S) 

The units are joules, as with all other forms of energy. 

An object that rotates while its center of mass (cm) undergoes translational 
motion will have both translational and rotational kinetic energies. Equation 8-15 
gives the rotational kinetic energy if the rotation axis is fixed. If the object is 
moving (such as a wheel rolling down a hill), this equation is still valid as long as 
the rotation axis is fixed in direction. Then the total kinetic energy is 

ke = \Mv} K + j/ rM <«r, (8-16) 

where i> CH is the linear velocity of the center of mass, / CM is the moment of 
inertia about an axis through the center of mass, to is the angular velocity about 
this axis, and M is the total mass of the object. 



EXAMPLE 8-13 



Sphere rolling down an incline. What will be the speed 
of a solid sphere of mass M and radius R when it reaches the bottom of an 
incline if it starts from rest at a vertical height Wand rolls without slipping? 
See Fig. 8-24. (Assume plenty of static friction, which does no work, so no 
slipping takes place.) Compare your result to that for an object sliding down a 
friction less incline, 

APPROACH Wc use the law of conservation of energy with gravitational poten- 
tial energy, now including rotational kinetic energy as well as translational ke. 
SOLUTION The total energy at any point a vertical distance y above the base 
of the incline is 

'tMv 2 + i/ cm m : + Mgy, 

where v is the speed of the center of mass, and Mgy is the gravitational PE. 
Applying conservation of energy, we equate the total energy at the top 
( y = H, v = 0, <i> = 0) to the total energy at the bottom (y = 0}: 

+ + MgH = ^Mo 2 + j/ CM u- + 0. 

The moment of inertia of a solid sphere about an axis through its center of 
mass is / CH - ^MR 2 , Fig. 8-21e. Since the sphere rolls without slipping, we 
have w - v/R (recall Fig. 8-8), Hence 



MgH = \Mv* + j{s mr7 ) 
Canceling the M's and R's, we obtain 

+ \y = g H 

or 



R 



v = V^gH. 

We can compare this result, for the speed of a rolling sphere to that for 
an object sliding down a plane without rotating and without friction, 
{tnv 2 - mg H (se e our energy equation above, removing the rotational term). 
Then v — ^JlgH , which is greater than our result. An object sliding without 
friction or rotation transforms its initial potential energy entirely into transla- 
tional KE (none into rotational KE),so the speed of its center of mass is greater. 
NOTE Our result for the rolling sphere shows (perhaps surprisingly) that v is 
independent of both the mass M and the radius R of the sphere. 



Rotational KE 



Tom! KE (translation + rotation) 




FIGURE S-24 A sphere rolling 
down a hill has both translational 
and rotational kinetic energy. 
Example 8-13, 

*» PROBLEM SOLVING 
Rotational energy adds 

to other forms of energy 
to get the total energy 
which is conserved 



SECTION 8-7 Rotational Kinetic Energy 211 




Mi .i;i 

f~\ Era pi y c;m 

Solid cylinder (D-ecll) 
Sphere (marble) 

Box (sliding) 



FIGURE 8-25 Example 8-14. 



FIGURE 8-26 A sphere rolling to 
the right on a plane surface. The 
point in contact with the ground at 
any moment, point P. is momen- 
tarily at rest. Point A to the left of P 
is moving nearly vertically upward 
at the instant shown, and point B to 
the right is moving nearly vertically 
downward. An instant laleT, point B 
will touch the plane and be at rest 
momentarily. Thus no work is done 
by the force of static friction, 

Sphere, niiling to the righl — *■ 



P B 
1 -*-- t 



<fc- 



CAUTION 



Rolling objects go slower 
than sliding objects because of 
rotational KE, not because of friction 



CONCEPTUAL EXAMPLE 8-14 I Who's fastest? Several objects roll 



without slipping down an incline of vertical height //, all starting from rest at 
the same moment. The objects arc a thin hoop (or a plain wedding band), a 
spherical marble, a solid cylinder (a D-cell battery), and an empty soup can. In 
addition, a greased box slides down without friction. In what order do they 
reach the bottom of the incline? 

RESPONSE The sliding box wins because the potential energy loss (MgfF) is 
transformed completely into translation a I ke for the box, whereas for rolling 
objects the initial I'K is shared between translational and rotational kinetic 
energies, and so their linear speed is less, For each of the rolling objects we can 
state that the loss in potential energy equals the increase in kinetic energy: 



MgH 



L 2 Mv 2 + {l^o/ 



For all our rolling objects, the moment of inertia / (M is a numerical factor 
times the mass M and the radius R 1 (Fig. 8-2 1 ). The mass M is in each term, so 
the translational speed o doesn't depend on M\ nor does it depend on the 
radius R since tr> = v/R, so R 2 cancels out for all the rolling objects, just as in 
Example 8- 13. Thus the speed v at the bottom depends only on that numerical 
factor in / t:M which expresses how the mass is distributed, The hoop, with all its 
mass concentrated at radius R (/, - M = MR 7 }, has the largest moment of inertia; 
hence it will have the lowest speed and will arrive at the bottom behind the 
D-cell (/ (M = jAf/? 1 ), which in turn will be behind the marble (J, M = I MR 2 ). 
The empty can, which is mainly a hoop plus a small disk, has most of its mass 
concentrated at R: so it will be a bit faster than the pure hoop but slower than 
the D-cell. Sec Fig. 8-25. 

NOTE As in Example 8-13, the speed at the bottom does not depend on the 
object's mass M or radius fi, but. only on its shape (and the height of the hill FJ). 



If iheTe had been little or no static friction between the rolling objects and the plane 
in these Examples, the round objects would have slid rather than rolled, or a 
combination of both. Static friction must be present to make a round object roll, We 
did not need to take friction into account in the energy equation because it is static 
friction and does no work — the point of contact of the sphere at each instant does 
not slide, but moves perpendicular to the plane (first down and then up as shown 
in Fig, 8-26) as the sphere rolls- Thus, no work is done by the static friction 
force because the force and the motion (displacement) are perpendicular. The 
reason the rolling objects in Examples 8-13 and 8-14 move down the slope 
more slowly than if they were sliding is not because friction is doing work. 
Rather, it is because some of the gravitional pe is converted to rotational kh, 
leaving less for the translational ke. 

Work Done by Torque 

The work done on an object rotating about a fixed axis, such as the pulleys in 
Figs. 8-22 and 8-23, can be written using angular quantities. As shown in 
Fig. 8-27, a force F exerting a torque t - rF on a wheel docs work 
W = F&l in rotating the wheel a small distance A/ at the point of application of F, 



FIGURE 8-27 Torque 7 - rF does 
work when rotating a wheel equal to 
W = FU = FrAfl = tM. 




212 CHAPTER 8 Rotational Motion 



The wheel has rotated through a small angle AS 

W = FAl = FrAB. 
Since 7 = rF, then 

W = tM 



Al/r (Eq. S-l).IIcncc 



Angular momentum 



(8-17) Work done by torque 

is the work done by the torque t when rotating the wheel through an angle M. 
Finally, power P is the rate work is done: P = W/Ai = tAB/Ai = ™. 

Angular Momentum and Its Conservation 

Throughout this Chapter we have seen that if we use the appropriate angular 
variables, the kinematic and dynamic equations for rotational motion are analo- 
gous to those for ordinary linear motion. We saw in the previous Section, for 
example, that rotational kinetic energy can be written as 5/w 2 , which is analo- 
gous to the translational kinetic energy, \mv 2 . In like manner, the linear 
momentum, p - mv, has a rotational analog. It is called angular momentum. L. 
For an object Totaling about a fixed axis, it is defined as 

L = /u, <N 18) 

where / is the moment of inertia and <w is the angular velocity about the axis of 
rotation. The SI units for L are kg-m 2 /s, which has no special name, 

We saw in Chapter 7 (Section 7-1 ) that Newton's second law can be written 
not only as S/ 7 = ma but also more generally in terms of momentum (Eq. 7-2), 
"LF = Ap/Al. In a similar way, the rotational equivalent of Newton's second 
law, which we saw in Eq. 8-14 can be written as St = la, can also be written 
in terms of angular momentum; 
AL 
Af ' 

where St is the net torque acting to rotate the object, and Af. is the change in 
angular momentum in a time interval At, Equation 8-14, St = /<*, is a special 
case of Eq. 8-19 when the moment of inertia is constant. This can be seen as 
follows. If an object has angular velocity w„ at time t = 0, and angular velocity 
(!) after a time interval Ar. then its angular acceleration (Eq. 8-3) is 

Ail) 11} — M|| 

u — —— - — 

if A! 

Then from Eq. 8-19, we have 

AL _ Ioj 

A( ~ 



St = 



(8-19) 



NEWTON'S SECOND LAW 
I OR ROTATION 



St = 



/(!>() 



I {to - d} ) Ail) 



= I 



At 



= In 



At At 

which is Eq. 8-14. 

Angular momentum is an important concept in physics because, under 
certain conditions, it is a conserved quantity. We can sec from Eq. 8-19 that if 
the net torque St on an object is zero, then AL/Al equals zero. That is, L does 
not change. This is the law of conservation of angular momentum for a rotating 
object: 

The total angular momentum of a rotating object remains constant if the 
nel torque acting on it is zero. 

The law of conservation of angular momentum is tine of the great conservation 
laws of physics, along with energy and linear momentum. 

When there is zero net torque acting on an object, and the object is rotating 
about a fixed axis or about an axis through its center of mass whose direction 

doesn't ehange, we can write 

/« = /,]» = constant, 

/(, and Mi, are the moment of inertia and angular velocity, respectively, about that 
axis at some initial time (/ - 0), and I and <u are their values at some other time. 



CONSERVATION OF 
ANGULAR MOMENTUM 



SECTION 8-8 Angular Momentum and Its Conservation 213 



1 large, 
ft} small 



/ sma.IL, 
tt> large 




(a) 



(b) 



FIGURE 8-28 A skater doing a 
spin on ice, illustrating conservation 
of angular momentum. In (a), / is. 
large and a> is small; in (b). / is 
smaller so w is larger. 

(^ ) PHYSICS APPLIED 

Spiiis in figure skating and diving 



FIGURE 8-29 A diver rotates 
faster when arms and legs are tucked 
in than when they are outstretched. 

Angular momentum is conserved. 

I 




FIGURE 8-30 Example 8-15. 




TTie parts of the object may alter their positions relative to one another so i 
changes, Bui then w changes as well to ensure that the product /'•> remains constant 
Many interesting phenomena can be understood on the basis of conserva- 
tion of angular momentum. Consider a skater doing a spin on the lips of heT 
skates, Fig. 8-25. She rotates at a relatively low speed when her arms aTe 
outstretched; when she brings her aims in close to her body, she suddenly spins 
much faster. From the definition of moment of inertia, / = Xjw 2 , it is cleaT 
that when she pulls her aims in closer to the axis of rotation, r is reduced for the 
arms, so her moment of inertia is reduced. Since the angular momentum Im 
remains constant (we ignore the small torque due to friction), if / decreases, 
then the angular velocity u must increase. Tf the skater reduces her moment of 
inertia by a factor of 2, she will then rotate with twice the angular velocity. 

EXERCISE C When a spinning figure skater pulls in her arms, her moment of inertia 
decreases; to conserve angular momentum, her angular velocity increases. Does heT 
rotational kinetic energy also increase? If so, where does the energy come from? 

A similar example is the diver shown in Fig. 8-29, The push as she leaves 
the board gives her an initial angular momentum about her center of mass. 
When she curls herself into the tuck position, she rotates quickly one or more 
times. She then stretches out again, increasing her moment of inertia which 
reduces the angular velocity to a small value, and then she enters the water. The 
change in moment of inertia from the straight position to the tuck position can 
be a factor of as much as 3j. 

Note that for angular momentum to be conserved, the net torque must be 
zero, but the net force docs not necessarily have to be zero, The net force on the 
diver in Fig. 8-29, for example, is not zero (gravity is acting), but the net torque 
on her is zero because the force of gravity acts at her center of mass. 



EXAMPLE 8-15 



Object rotating on a string of changing length. A 
small mass m attached to the end of a string revolves in a circle on a friction- 
less tabletop. The other end of the string passes through a hole in the table 
(Fig. 8-30). Initially, the mass revolves with a speed v, = 2.4 m/s in a circle of 
radius r, = 0.80 m. The string is then pulled slowly through the hole so that 
the radius is reduced to r 2 = 0.48 m. What is the speed, v 2 , of the mass now ? 

APPROACH There is no net torque on the mass m because the force exerted 
by the string to keep it moving in a circle is exerted toward the axis; hence the 
lever arm is zero. We can thus apply conservation of angular momentum. 
SOLUTION Conservation of angular momentum gives 

Our small mass is essentially a particle whose moment of inertia about the 
hole is / = mr 1 (Section S-5, Eq. 8- 1 1 ), so we have 



mrfwj = mrjoij, 



or 



Then, since v 



r-,tiii 



r«j, we can write 

■ 
1 
r- 



f 2 <U] 






(2.4 m/s) 



0.80 m 

0.48 m 



4.0 m/s. 



The speed increases as the radius decreases. 



EXERCISE The speed of mass m in Example 8- 
incieased. Where did the energy come from? 



15 increased, so its kinetic energy 



Zlfl CHAPTER 8 Rotational Motion 



EXAMPLE 8-16 



ESTIMATE | Star collapse. Astronomers often detect 
stars that are rotating extremely rapidly, known as neutron stars. These stars 
arc believed to have formed from the inner core of a larger star that collapsed, 
due to its own gravitation, to a star of very small radius and very high density. 
Before collapse, suppose the core of such a star is the size of our Stm 
(R ~ 7 X 10 5 km) with mass 2.0 times as great as the Sun, and is rotating at a 
speed of I.O revolution every 10 days. If it were to undergo gravitational 
collapse to a neutron star of radius 10 km, what would its rotation speed be? 
Assume the star is a uniform sphere at all times. 

APPROACH The star is isolated (no external forces), so we can use conserva- 
tion of angular momentum for this process. 

SOLUTION From conservation of angular momentum, 

where the subscripts 1 and 2 refer to initial (normal star) and final (neutron 
staT), respectively. Then, assuming no mass is lost in the process. 



<u 2 



hi 



\M 2 R\ 



R[ 
Rl 



The frequency / - <a>/2tt, so 



(1*2 ft | 



2ir Ri 



7 x 10*km V 
10 km 



1.0 rev 



10d(24h/d)(3600s/h) 



6 X lO^rev/s. 



) PHYSICS APPLIED 

XcttUtiti star 



Vector Nature of Angular Quantities 

Up to now we have considered only the magnitudes of angular quantities such 
as w, a, and L. But they have a vector aspect too, and now we consider the 
directions, In fact, we have to define the directions for rotational quantities, and 
we take first the angular velocity, &». 

Consider the rotating wheel shown in Fig. 8-3 1 a. The linear velocities of 
different particles of the wheel point in all different directions. The only unique 
direction in space associated with the rotation is along the axis of rotation, 
perpendicular to the actual motion. We therefore choose the axis of rotation to 
be the direction of the angular velocity vector, w. Actually, there is still an 
ambiguity since <o could point in cither direction along the axis of rotation (up 
or down in Fig, S-31a).Thc convention we use, called the right-hand rule, is the 
following: When the fingers of the right hand are curled around the rotation 
axis and point in the direction of the rotation, then the thumb points in the 
direction of to. This is shown in Fig, S-31b. Note that tit points in the direction 
a right-handed screw would move when turned in the direction of rotation. 
Thus, if the rotation of the wheel in Fig. 8-3 lb is counterclockwise, the direc- 
tion of w is upward. If the wheel rotates clockwise, then w points in the opposite 
direction, downward. Note that no part of the rotating object moves in the 
direction of w. 

Tf the axis of rotation is fixed, then w can change only in magnitude, Thus 
a - Aw/ \i must also point along the axis of rotation. If the rotation is coun- 
terclockwise as in Fig, 8-3 la, and if the magnitude 10 is increasing, then a points 
upward; but if w is decreasing (the wheel is slowing down), a points downward. 
If the rotation is clockwise, « will point downward if w is increasing, and point 
upward if w is decreasing. 



FIGURE 3-31 (a) 
(b) Right -hand rule 
the direction of tii. 



Rotating wheel. 

for obtaining 




(a) 



(b) 



High 
hand 
rule 



"SECTION 8-9 Vector Nature of Angular Quantities 215 



Angular momentum, like linear momentum, is a vector quantity. For a 
symmetrical object rotating about a symmetry axis (such as a wheel, cylinder, 
hoop, or sphere), we can write the vector angular momentum as 

L = lih, (8-2*) 

The angular velocity vector w (and therefore also L) points along the axis of 
rotation in the direction given by the right-hand rule (Fig, 8-3 lb). 



FIGURE 8-32 (a) A person standing on a circular platform, 
initially at rest, begins walking along the edge at speed I'.The 
platform, assumed lo be mounted on friction-free bearings, 
begins rotating in the opposite direction, so that the total 
angular momentum remains zero, as shown in (b). 



I m 




CM 




, J 



(a) 



"person 



'platform 



(b) 



FIGURE 8-33 Example 8-17. 




The vector nature of angular momentum can be used to explain a number 
of interesting (and sometimes surprising) phenomena. For example, consider a 
person standing at rest on a circular platform capable of rotating without fric- 
tion about an axis through its center (that is, a simplified merry-go-round), [f 
the person now starts to walk along the edge of the platform, Fig. 8-32a, the 
platform starts rotating in the opposite direction. Why? One way to look at it is 
that the person's foot exerts a force on the platform. Another way to look at 
it (and this is the most useful analysis here) is that this is an example of the 
conservation of angular momentum. If the person starts walking counter- 
clockwise, the person's angular momentum will point upward along the axis 
of rotation (remember how wc defined the direction of w using the right- 
hand rule). The magnitude of the person's angular momentum will be 
L - fas - (mr 2 )(vfr), where v is the person's speed (relative to Earth, not to 
the platform), / is his distance from the rotation axis, m is his mass, and mr 2 is 
his moment of inertia if we consider him a particle (mass concentrated at one 
point). The platform rotates in the opposite direction, so its angular momentum 
points downward. If the initial total angular momentum of the system (person 
and platform) was zero (person and platform at rest), it will remain zero after 
the person starts walking. That is, the upward angular momentum of the person 
just balances the oppositely directed downward angular momentum of the 
platform (Fig. 8-32b),so the total vector angular momentum remains zero. Even 
though the person exerts a force (and torque) on the platform, the platform 
exerts an equal and opposite torque on the person. So the net torque on the 
system of person plus platform is zero (ignoring friction), and the total angulaT 
momentum remains constant. 



v 



CONCEPTUAL EXAMPLE 8-17 | Spinning bicycle wheel. Your physics 



teacher is holding a spinning bicycle wheel while he stands on a stationary 
friction I ess turntable (Fig 8-33). What will happen if the teacher suddenly 
flips the bicycle wheel over so that it is spinning in the opposite direction? 

RESPONSE We consider the system of turntable, teacher, and bicycle wheel. 
The total angular momentum initially is L vertically upward. That is also what 
the system's angular momentum must be afterward, since L is conserved 
when there is no net torque. Thus, if the wheel's angular momentum after 
being flipped over is — L downward, then the angular momentum of teacher 
plus turntable will have to be +2L upward. Wc can safely predict that the 
teacher will begin spinning around in the same direction the wheel was 
spinning originally. 



216 CHAPTER 8 Rotational Motion 



I Summary 



When a rigid object rotates about a fixed axis, each point of 
the object moves in a circular path. Lines drawn perpendicu- 
larly from the rotation axis to various points in [he object all 
sweep out the same angle B in any given time interval. 

Angles are conveniently measured in radians, where one 
radian is the angle subtended by an arc whose length is equal 
to the radius, or 

2 it rad = 360" 
I rad = 573". 

Angular velocity, im, is defined as the rate of change of 
angular position: 



It 



(8-2) 



All parts of a rigid object rotating about a fixed axis have the 
same angular velocity at any instant, 

Angular acceleration, a. is defined as the rate of change 
of angular velocity: 






(K-3> 



'I ne linear velocity v and acceleration a of a point fixed at 
a distance r from the axis of rotation are related to to and a hy 

v = r«, (8-4) 

"tan = ra - (8-») 

a R = arr, (8-6) 

where a tan and a R are the tangential and radial (centripetal) 
components of the linear acceleration, respectively. 
The frequency / is related to &s by 

i.> = 2-n-f. f8-7) 

and to the period T by 

7" = 1// (8-8) 

The equations describing uniformly accelerated rotational 
motion {a = constant) have the same form as for uniformly 
accelerated linear motion: 



OJrj + at, 

fljg + 2t.a. 



B = a\-,t + kat; 

_ m + mo (8-9) 



The dynamics of rotation is analogous to the dynamics of 
linear motion. Force is replaced by torque t, which is defined as 
the product of force times lever arm (perpendicular distance 
from the line of action of the force to the axis of rotation): 



Mass is replaced by moment of inertia /. which depends 
not only on the mass of the object, but also on how the mass 
is distributed about the axis of rotation. Linear acceleration is 
replaced by angular acceleration. The rotational equivalent of 

Newton's second law is then 



= la. 



(8-14) 



The rotational kinetic energy of an object rotating about 
a fixed axis with angular velocity at is 



i i 7 
kl = - tu> . 



(8-15) 



For an object both translating and rotating, the total 
kinetic energy is the sum of the translations! kinetic energy of 
the object's center of mass plus the rotational kinetic energy 
of the object about its center of mass: 



KE = |M1)q, + 3^-ytU 1 



[8- 1 A) 



as long as the rotation axis is fixed in direction, 

The angular momentum L of an object about a fixed 
rotation axis is given by 

L = fa>. (8-18) 

Newton's second law, in terms of angular momentum, is 



~ T= aT 



(8-19) 



If the net torque on the object is zero. AL/Ar = 0. so 
L = constant. This is the law of conservation of angular 
momentum for a rotating object, 

The following Table summarizes angular (or rotational) 

quantities, comparing them to their translation a I analogs, 



Translation 


Rotation 


Connection 


'. 


9 


x = rB 


r 


at 


D = no 


a 


ft 


a — fa 


in 


/ 


I = Sffl/ 2 


F 


T 


r = i F sin 


KE = kmv 1 


1 I 2 




p = inv 


L = 1m 




W = Fd 


W = T» 




£F = ma 


•£.r = la 




— AP 


AL 





T = rF sin B - r L F = rF L . 



<8-io> 



IF = 



M 



2t = 



a; 



| Questions 



A bicycle odometer (which measures distance traveled) is 
attached near the wheel hub and is designed ioT 27-inch 
wheels. What happens if you use it on a bicycle with 
24-inch wheels'? 

Suppose a disk rotates at constant angular velocity. Hoes a 
point on the rim have radial and/ or tangential accelera- 
tion 1 ? If the disk's angular velocity increases uniformly. 



does the point have radial and/or tangential acceleration"? 
For which cases would the magnitude of either component 
of linear acceleration change'? 

3, Could a nonrigid body be described by a single value of 
the angular velocity w? Explain. 

4. Can a small force ever exert a greater torque than a larger 
force? Explain. 



Questions 217 



If a force P acts on an object such that its lever arm is 
zero, does it have any effect on the object's motion? Explain, 
Why is it more difficult to do a sit-up with your hands 
behind your head than when your arms ate stretched out 
in front of you'.' A diagram may help you to answer this, 
A 21 -speed bicycle has seven sprockets at the rear wheel 
and three at the pedal cranks. In which gear is it harder to 
pedal, a small Tear sprocket ot a large Tear sprocket? 
Why? In which gear is it harder to pedal, a small front 
sprocket or a large front sprocket? Why? 
Mammals that depend on being able to run (as: have 
slender lower legs with flesh and muscle concentrated 
high, close to the body (Fig. 8-34). On the basis of rota- 
tional dynamics, explain why this distribution of mass is 
advantageous. 




FIGURE 8-34 Question 8. A gazelle. 



FIGURE 8-35 Question 9. 




9. Why do tightrope walkers (Fig. 8-35) cany a long, narrow beam? 

10. If the net force on a system is zero, is the net torque also 
zero? If the net torque on a system is zero, is the net 
force zero? 

11. Two inclines have the same height but make different 
angles with the horizontal. The same steel ball is rolled 
down each incline, Qn which incline will the speed of the 
ball at the bottom be greater? Explain. 

12. Two solid spheres simultaneously start rolling (from rest) 
down an incline. One sphere has twice the radius and 
twice the mass of the other. Which reaches the bottom of 
the incline first? Which has the greater speed there? 
Which has the greater total kinetic energy at the bottom? 



IX 

M. 

15. 

n., 

17. 

1ft 

19. 

*20. 
*21. 



*22. 
*23. 



A sphere and a cylinder have the same radius and the same 
mass. They start from rest at the top of an incline. Which 
reaches the bottom first? Which has the greater speed at 
<be bottom? Which has the greater lolal kinetic energy at 
the bottom? Which has the greater rotational KC? 

We claim that momentum and angular momentum are 

conserved. Yet most moving or rotating objects eventually 

slow down and stop. Explain. 

If there were a great migration of people toward the Earth's 

equator, how would this affect thy length of the day? 

Can the diver of Kig, 8-29 do a somersault without having 

any initial rotation when she leaves the board? 

The moment of inertia of a rotating solid disk about an 
axis through its center of mass is \ MR 2 (Fig. 8-21 c). 
Suppose instead that the axis of rotation passes through a 
point on the edge of the disk. Will the moment of inertia 
be the same, larger, or smaller? 

Suppose you are sitting on a rotating stool holding a 2-kg 
mass in each outstretched hand. If you suddenly drop the 
masses, will your angular velocity increase, decrease, ot 
stay the same? Explain. 

Two spheres look identical and have the same mass. 
However, one is hollow and the other is solid. Describe an 
experiment to determine which is which. 
In what direction is the Earth's angular velocity vector as it 
rotates daily about its axis? 

The angular velocity of a wheel rotating on a horizontal 
axle points west. In what direction is the linear velocity of 
a point on the top of the wheel? If the angular accelera- 
tion points east, describe the tangential linear acceleration 
of this point at the top of the wheel. Is the angular speed 
increasing or decreasing? 

Suppose you are standing on the edge of a large freely rotating 
turntable. What happens if you walk toward the center? 
A shortstop may leap into the air to catch a ball and throw 
it quickly. As he throws the ball, the upper part of his body 
rotates. If you look quickly you will notice that his hips and 
legs rotate in the opposite direction (Fig. 8-36). Fxplain. 




*24 



FIGURE 8-36 

Question 23. A 
shortstop in the air. 
throwing the ball. 

On the basis of the law of conservation of angulaT 
momentum, discuss why a helicopter must have more than 
one rotor (or propeller). Discuss one or more ways the 
second propeller can operate to keep the helicopter stable, 



218 CHAPTER 8 Rotational Motion 



I Problems 



3-1 Angular Quantities 

1. (I) Express the following angles in radians: [a) 30°, 

(6) 57", (c) 90". (d) 360", and (e) 420< : . Give as numerical 
values and as fractions of it. 

2. (I) Eclipses happen on Earth because of an amazing coin- 
cidence. Calculate, using the information inside the Front 
Cover, the angular diameters (in radians) of the Sun and 
the Moon, as seen on Earth 

X (I) A laser beam is directed at the Moon, 380,000 km from 
Earth, "The beam diverges at an angle # (Fig. 8-37) of 
1.4 x 10 _;i rad. What diameter spot will it make on the 
Moon? 




FIGURE 8-37 Problem 3. 

4. (I) The blades in a blender rotate at a rale of 6500 rpm- 
When the motor is turned off during operation, the blades 
slow to rest in 3.0 s. What is the angular acceleration as 
the blades slow down? 

5. (II) A child rolls a ball on a level floor 3.5 m to another 
child. If the ball makes 150 revolutions, what is its diameter? 

6. (II) A bicycle with tires 68cm in diameter travels 8-0 km. 
How many revolutions do the wheels make? 

7. (II) (a) A grinding wheel 0.35m in diameter rotates at 
5500 rpm. Calculate its angular velocity in rad/s. (b) What 
are the linear speed and acceleration of a point on the 
edge of the grinding wheel? 

8. (II) A rotating merry-go-round makes one complete 
revolution in 4.0 s (Fig- 8-38). (a) What is the linear speed 
of a child seated 1.2 m from the center? (h) What is her 
acceleration (give components)? 




FIGURE 8-38 Problem 8, 



'). (II) Calculate the angular velocity of the Earth {a} in its 

orbit around the Sun, and (b) about its axis. 
III, (II) What is the linear speed of a point {a) on the equator, 
(b) on the Arctic Circle (latitude 6fi.5 c N), and (c) at a 
latitude of 45,0' J N. due to the Earth's rotation? 

11. (II) How fast (in Tpm) must a centrifuge rotate if a 
particle 7.0 cm from the axis of rotation is to experience 
an acceleration of lOQ.OOOg's? 

12. (II) A 70-cm -diameter wheel accelerates uniformly about 
its center from 130 rpm to 280 rpm in 4.0 S- Determine 
(a) its angular acceleration, and (b) the radial and tangen- 
tial components of the linear acceleration of a point on the 
edge of the wheel 2,0 s afleT it has started accelerating, 

13. (II) A turntable of radius R] is turned by a circular rubber 
roller of radius R 2 in contact with it at their outer edges. 
What is the ratio of their angular velocities, wj/wj? 

14. (Ill) In traveling to the Moon, astronauts aboard the 
Apollo spacecraft put themselves into a slow rotation to 
distribute the Sun's energy evenly. At the start of their 
trip, they accelerated from no rotation to 10 revolution 
every minute during a 12-min time interval. ITie space- 
craft can be thought of as a cylinder with a diameter of 
8.5 m. Determine (a) the angular acceleration, and (b) the 
radial and tangential components of the linear accelera- 
tion of a point on the skin of the ship 5. (J min after it 
started this acceleration. 

8-2 and 8-3 Constant Angular Acceleration, Rolling 

t5. (I) A centrifuge accelerates uniformly from rest to 
15,000 rpm in 220 s. Through how many revolutions did it 
turn in this time? 

Ifi. (I) An automobile engine slows down from 4500 rpm to 
1200rpm in 2,5s, Calculate (a) its angular acceleration, 
assumed constant, and (6) the total number of revolutions 
the engine makes in this time. 

17, (I) Pilots can be tested for the stresses of flying high- 
speed jets in a whirling "human centrifuge," which takes 
1.0 min to turn through 20 complete revolutions before 
reaching its final speed, (a) What was its angular acceler- 
ation (assumed constant), and (b) what was its final 
angular speed in Tpm? 

18, (II) A wheel 33 cm in diameter accelerates uniformly from 
240 rpm to 360 rpm in 6-5 S- How far will a point on the 
edge of the wheel have traveled in this time? 

V). (II) A cooling fan is turned off when it is running at 
850rev/min, It turns 1500 revolutions before it comes to a 
stop, (a) What was the fan's angular acceleration, assumed 
constant? (A) How long did it take the fan to come to a 
complete stop? 

20. (II) A small rubber wheel is used to drive a large pottery 
wheel, and they btc mounted so that their circular edges 
touch, The small wheel has a radius of 2,0cm and acceler- 
ates at the rate of 7.2 rad/s J , and it is in contact with the 
pottery wheel (radius 25.0 cm) without slipping. Calculate 

(a) the angular acceleration of the pottery wheel, and 

(b) the time it takes the pottery wheel to reach its 
required speed of 65 rpm. 



Problems 219 



21. (IT) The tires of a car mate 65 revolutions as the car 
reduces its speed uniformly from 95km/h to 45km./h, 
The tires have a diameter of 0.80 m. (a) What was the 
angular acceleration of the tires? (b) If the car continues 
to decelerate at this rate, how much more time is required 
for it to stop? 

3-4 Torque 

22. (I) A 55-kg person riding a bike puts all her weight on 
each pedal when climbing a hill. The pedals rotate in a 
circle of radius 17 cm. its') What is the maximum torque 
she exerts? (h) How could she exert more torque'.' 

23. (I) A person exerts a force of 55 N on the end of a door 
74 cm wide. What is the magnitude of the torque if the 
force is exerted (a) perpendicular to the door, and (As) at a 
45 ,: angle to the face of the door? 

24. (II) Calculate the net torque about the axle of the wheel 
shown in Fig. 8-39. Assume that a friction torque of 
0,40 m -N opposes the motion. 



j 2KN 



I8N 




FIGURE 8-39 

Problem 24. 



2S. (II) Two blocks, each of mass m. are attached to the ends 
of a massless rod which pivots as shown in Fig. 8-40. 
Initially the rod is held in the horizontal position 
and then released. Calculate the magnitude and direction 
of the net torque on this system. 




FIGURE 8-40 Problem 25. 



26, (II) The bolts on the cylinder head of an engine Tequire 
tightening to a torque of 88 m - N. If a wrench is 28 cm 
long, what force perpendicular to the wrench must the 
mechanic exert at its end'' If the six-sided holt head is 
15 mm in diameter, estimate the force applied near each 
of the six points by a socket wrench (Fig. 8-41). 



I5mmj 




28 em 



r * I., l! 

FIGURE 8-41 Problem 26. 




8-5 and 8-6 Rotational Dynamics 

27, (I) Determine the moment of inertia of a lOS-kg sphere 
of radius 0.648 m when the axis of rotation is through its 
center. 

28. (I) Calculate the moment of inertia of a bicycle wheel 
66.7 cm in diameter. The rim and tire have a combined 
mass of l .25 kg. The mass of the hub can be ignored 
(why?). 

2% (II) A small 650-gram ball on the end of a thin, light rod 
is rotated in a horizontal circle of radius 1.2 m. Calculate 
(a) the moment of inertia of the ball about the center of 
the circle, and (b) the torque needed to keep the ball 
rotating at constant angular velocity if air resistance 
exerts a force of 0020 N on the ball. Ignore the rod's 
moment of inertia and air resistance. 

3(1. (II) A potter is shaping a bowl on a potter's wheel 
rotating at constant angular speed (Fig, 8-42). Ine fric- 
tion force between her hands and the clay is 1 ,5 N total. 
(«) How large is her torque on the wheel, if the diameter 
of the bowl is 12 cm? (b) How long would it take for the 
potter's wheel to stop if the only torque acting on it is due 
to the potter's hand? The initial angular velocity of the 
wheel is 1 ,6 rev/s, and the moment of inertia of the wheel 
and the bowl is 0.11 kg-m 2 . 




FIGURE 8-42 Problem 30. 

(II) Calculate the moment of inertia of the array of 
point objects shown in Kg. 8-43 about (a) the vertical 
axis, and (b) the horizontal axis. Assume m = 1 ,8 kg, 
M = 3.1 kg, and the objects are wired together by very 
light, rigid pieces of wire. The airay is rectangular and is 
split through the middle by the horizontal axis, (c) About 
which axis would it be harder to accelerate this array? 



' 



I*- 
m 



■0.50 m 



I y 



1.50 in 



► 



0.50 m 



1 



m 



_<jAxis_ 



w 



M 



FIGURE 8-43 Problem 31. 



220 CHAPTER 8 Rotational Motion 



32. (N) An oxygen molecule consists of two oxygen atoms 
whose total mass is 5,3 X 10 _Ml kg and whose moment of 
inertia about an axis perpendicular to the line joining the 
two atoms, midway between them, is 1.9 x 10~ 4f, kgm 2 . 
From these data, estimate the effective distance between 
the atoms. 

33. (II) To get a flat, uniform cylindrical satellite spinning at 
the correct Tate, engineer's fire four tangential rockets as 
shown in Fig. 8-44. If the satellite has a mass of 3600 kg 
and a radius of 4.0 m, what is the required steady force of 
each rocket if the satellite is to reach 32 rpm in 5,0 min? 




FIGURE 8-44 

Problem 33. 



34. (II) A grinding wheel is a uniform cylinder with a radius of 
8.50 cm and a mass of 0-580 kg Calculate (a) its moment of 
inertia about its center, and (b) the applied torque needed 
to accelerate it from rest to 1500 rpm in 5 .00s if it is known 
to slow down from 151)0 rpm to rest in 55.0 s. 

35. (11) A softball player swings a bat, accelerating it from rest 



to 3,0 rev/s 



a time of 0,20 s. Approximate 



the hat as a 2.2-kg uniform rod of length 0.95 m, and 
compute the torque the player applies to one end of it. 
3d. (II) A leenageT pushes langenlially on a small hand- 
dTiven merry-go-round and is able to accelerate it from 
Test to a frequency of 15 rpm in 1 0-0 S- Assume the merry- 
go-round is a uniform disk of radius 2.5 m and has a mass 
of 760 kg, and two children (each with a mass of 25 kg) sit 
opposite each other on the edge. Calculate the torque 
Tequired to produce the acceleration, neglecting frictional 
torque. What force is required at the edge? 

37. (11) A centrifuge rotor rotating at 10,300 rpm Is shut off 
and is eventually bro Light uniformly to rest by a frictional 
torque of 1 .20 m j N. If the mass of the rotor is 4.80 kg and 
it can be approximated as a solid cylinder of radius 
0.0710m, through how many revolutions will the rotor 
turn before coming to rest, and how long will it take? 

38. (II) The forearm in Fig, 8-45 accelerates a 3.6-kg ball at 
7.1.) m/s 2 by means of the triceps muscle, as shown. Calcu- 
late (a) the torque needed, and (6) the force that must be 
exerted by the triceps muscle. Ignore the mass of the arm. 



31 cm 




■■'/ // ' 

\\is of rotation 

2.? ^ in" - - . Cat elbow ) 



Triceps 
muscle 



39. (II) Assume that a l-00-kg ball is thrown solely by the 
action of the forearm, which rotates about the elbow joint 
under the action of the triceps muscle. Fig. 8-45. The ball 
is accelerated uniformly from Test to lO.Om/s in 0,350s. at 
which point it is released. Calculate (a) the angular acceler- 
ation of the arm, and (b) the force required of the triceps 
musele, Assume thai the forearm has a mass of 3.70 kg and 
rotates like a uniform rod ahout an axis at its end. 

4(J. (II) A helicopter rotor blade can be considered a long thin 
rod. as shown in Fig. 8-46. (a) If each of the three rotor heli- 
copter blades is 3.75 m long and has a mass of I Ml kg. calcu- 
late the moment of inertia of the three rotor blades about 
the axis of rotation, (b) How much torque must the motor 
apply to bring the blades up to a speed of 5.0 rev/s in 8.0s? 



41 




m = 1 60 kg 



FIGURE 8-46 

Problem 40 




FIGURE 8-45 
Problems 38 and 39. 



(Ill) An Atwofid's machine consists of two masses, Mi| and 
Hi 2 , which are connected by a massless inelastic cord that 
passes over a pulley. Fig. 8-47. If the pulley has radius R 
and moment of inertia ! about 
its axle, determine the accelera- 
tion of the masses ffi[ and m 2 . 
and compare to the situation in 
which the moment of inertia of 
the pulley is ignored. | Hint: The 
tensions Fj] and Ff2 are not 
equal. We discussed this situa- 
tion in Example 4-13. assuming 
/ = foT the pulley. | 

FIGURE 8-47 
Problems 41 and 49. 

A [wood's machine. 

42. (Ill) A hammer thrower accelerates the hammer 
(mass = 7.30 kg) from rest within four full turns (revolu- 
tions) and releases it at a speed of 28,0 m/s. Assuming a 
uniform rate of increase in angular velocity and a hori- 
zontal circular path of radius 1 .20 m. calculate (a) the 
angular acceleration, (b) the (linear) tangential accelera- 
tion, (c) the centripetal acceleration just before Telease. 
(d) the net force being exerted on the hammer by the 
athlete just before release, and (e) the angle of this force 
with respect to the radius of the circular motion. 

8-7 Rotational Kinetic Energy 

43. (I) A centrifuge rotor has a moment of inertia of 
3,75 X 10 : kg- m\ How much energy Is required to bring 
it from rest to 8250 rpm? 

44. (II) An automobile engine develops a torque of 280 m • N at 
3800 rpm. What is the power in watts and in horsepower? 

45. (II) A bowling ball of mass 7.3 kg and radius 9.0 cm rolls 
without slipping down a lane at 1.3 m/s. Calculate its total 
kinetic energy. 



Problems 221 



46. (II) Estimate the kinetic energy of the Farth with respect 
to the Sun as the sum of two terms, (a) that due to its daily 
rotation about its axis, and (A) that due to its yearly revolu- 
tion about the Sun. [Assume the Earth is a uniform sphere 
with mass = 6,0 X 10 24 kg and radius = 6.4 X 10* m, 
and is 1 .5 x 10 s km from the Sun.| 

47. (II) A merry-go-round has a mass of 1 640 kg and a radius 
of 7.50 m, How much net work is required to accelerate it 
from rest to a rotation rate of 100 revolution per 8 00 s? 
Assume it is a solid cylinder. 

48. (II) A sphere of radius 20.0 cm and mass 1 ,80 kg starts 
from rest and rolls without slipping down a 300" incline 
that is 10.0 m long, {a) Calculate its translational and rota- 
tional speeds when it reaches the bottom, (b) What is the 
ratio of translational to rotational ke at the bottom? 
Avoid putting in numbers until the end so you can 
answer: (c) do your answers in («) and (b) depend on the 
radius of the sphere or its mass? 

49. (Ill) Two masses. m\ = 18,0 kg and in; = 26.5 kg, are 
connected by a rope that hangs over a pulley (as in 
Fig. 8-47), The pulley is a uniform cylinder of radius 
0.260m and mass 7.50kg. Initially, hi, is on the ground 
and mj rests 300m above the ground If the system is 
now released, use conservation of energy to determine 
the speed of m 2 just before it strikes the ground. Assume 
the pulley is frietionless. 

50. (Ill) A 2.30-m-long pole is balanced vertically on its lip, It 
starts to fall and its lower end does not slip. What will be 
the speed of the upper end of the pole just before it hits 
the ground? [Hinl: Use conservation of energy.] 

8-8 Angular Momentum 

51. (I) What is the angular momentum of a 0,21 0-kg ball 
rotating on the end of a thin string in a circle of radius 
1. 10m at an angular speed of 10.4rad/s? 

52. (I) (a) What is the angular momentum of a 2.8-kg 
uniform cylindrical grinding wheel of radius IS cm when 
rotating at 1500 rpm? (b) How much torque is required to 
stop it in 6,0 s? 

53. (II) A person stands, hands at his side, on a platform that 
is rotating at a rate of 1 ,30 rev/s. If he raises his arms to a 
horizontal position, Fig, 8-48, the speed of rotation 
decreases to 0.80 rev/s. (a) Why? (b) By what factor has 
his moment of inertia changed? 




FIGURE 8-48 
Problem 53. 



54. (II) A diver (such as the one shown in Fig. 8-29) can 
reduce her moment of inertia by a factor of about 3.5 
when changing from the straight position to the tuck posi- 
tion. If she makes 2.0 rotations in ]_5s when in the tuck- 
position, what is her angular speed (rev/s) when in the 
straight position? 



55. (II) A figure skater can increase her spin rotation rate 
from an initial rate of 1,0 rev every 2,0 s to a final rate of 
3.0 rev/s. If her initial moment of Inertia was 4.6 kg- in 2 , 
what is her final moment of inertia? How does she physi- 
cally accomplish this change? 

56. (II) A potter's wheel is rotating around a vertical axis 
through its center at a frequency of 1-5 rev/s. The wheel 
can be considered a uniform disk of mass 5.0 kg and 
diameter 0.40 m. The potter then throws a 3. 1 -kg chunk of 
clay, approximately shaped as a flat disk of radius 8-0 cm, 
onto the center of the rotating wheel, What is the 
frequency of the wheel after the clay sticks to it? 

57. (II) (a) What is the angular momentum of a figure skater 
spinning at 3,5 rev/s with arms in close to her body. 
assuming her to be a uniform cylinder with a height of 
1.5 m, a radius of 1 5 cm. and a mass of 55 kg? (b) How 
much torque is required to slow her to a stop in 5.0 s, 
assuming she does not move her arms? 

58. (II) Determine the angular momentum of the Earth 
(a) about its rotation axis (assume the Earth is a uniform 
Sphere), and (/>) in its orbit around the Sun (treat the 
Earth as a particle orbiting the Sun). The Farth has 
mass = 6.0 x lO^kg and radius = 6.4 x 10 fl m. and is 
1.5 X 10? km from the Sun. 

59. (II) A nonrotating cylindrical disk of moment of inertia / 
is dropped onto an identical disk rotating at angular 
speed to, Assuming no external torques, what is the final 
common angular speed of the two disks? 

nil. (II) A uniform disk turns at 2.4 rev/s around a frietionless 
spindle, A nonrotating rod, of the same mass as the disk 
and length equal to the disk's diameter, Is dropped onto 
the freely spinning disk. Fig. 8-49. They then both turn 
around the spindle with their centers superposed, What is 
the angular frequency in rev/s of the combination? 




FIGURE 8-49 

Problem 60. 



61 



(II) A person of mass 75 kg stands at the center of a 
rotating merry-go-round platform of radius 3.0m and 
moment of inertia 920 kg- m 2 .The platform rotates without 
friction with angular velocity 2.0rad/S- The person walks 
radially to the edge of the platform, (a) Calculate the 
angular velocity when the person reaches the edge. 
(b) Calculate the rotational kinetic energy of the system of 
platform plus person before and afteT the person's walk. 
62. (II) A 4.2-m-diameter merry-go-round is rotating freely 
with an angular velocity of 0.80 rad/s. Its total moment of 
inertia is I76Ukg'm 2 . Four people standing on the 
ground, each of mass 65 kg, suddenly step onto the edge 
of the merry-go -round. What is the angular velocity of the 
merry-go-round now? What if the people were on it 
initially and then jumped off in a radial direction (relative 
to the merry-go-round)? 



222 CHAPTER 8 Rotational Motion 



fix (II) Suppose our Sun eventually collapses into a white 
dwarf, losing about half its mass in the process, and 
winding up with a radius 1.0% of its existing radius. 
Assuming the lost mass carries away no angular 
momentum, what would the Sun's new rotation rate be? 
(Take thy Sun's current period to be about 30 days.) What 
would be- its final, kc in terms of its initial KH of today? 

64. (Ill) Hurricanes can involve winds in excess of 120km/h 
at the outer edge. Make a crude estimate of (a) the 
energy, and (b) the angular momentum., of such a hurri- 
cane, approximating it as a rigidly rotating uniform 
cylinder of air (density Ukg/m 1 ) of radius 100 km and 
height 4.0 km, 

65. (Ill) An asteroid of mass 1.0 x I0 3 kg. traveling at a 
speed of 30 km/s relative to the Earth, hits the Earth at 
the equator tangentially, and in the direction of Farth's 
rotation, Use angular momentum to estimate the 
percent change in the angular speed of the Earth as a result 
of the collision. 



1 8-9 Angular Quantities as Veetors 

B 66. (II) a person stands on a platform, initially at rest, that 
tan rotate freely without friction. 'ITie moment of inertia 
of the person plus the platform is / P . The person holds a 
spinning bicycle wheel with its axis horizontal, ITie wheel has 
moment of inertia / w and angular velocity (u^ . What will 
be the angular velocity &> P of the platform if the person 
moves the axis of the wheel so that it points (a) vertically 
upward, (b) at a diT angle to the vertical, (c) vertically 
downward? (d) What will <<> P be if the person reaches up 
and stops the wheel in part (a)1 

* 67. (Ill) Suppose a 55-kg person stands at the edge of a 6,5-m 
diameter merry-go-round tumtahle that is mounted on 
frictbnless bearings and has a moment of inertia of 
1700 kg-m 2 , The turntable is at Test initially, but when the 
person begins cunning at a speed of 3.8 m/s (with respect 
to the turntable) around its edge, the turntable begins to 
rotate in the opposite direction. Calculate the angular 
velocity of the turntable. 



| General Problems 



6N. 



A large spool of rope rolls on the ground with the end of 
the rope lying on the top edge of the spool. A person 
grabs the end of the rope and walks a distance L, holding 
onto it. big, 8-50, 'ITie spool rolls behind the person 
without slipping. What length of rope unwinds from the 
spool? How far does the spool's center of mass move? 





FIGURE 8-50 
Problem 68. 



ti/i. The Moon orbits the Harth such that the same side always 
faces the Earth. Determine the ratio of the Moon's spin 
angular momentum (about its own axis) to its orbital 
angular momentum. (In the latter case, treat the Moon as 
a particle orbiting the Earth.) 

7(1. A cyclist accelerates from rest at a rate of l 00 m/s*. How 
fast will a point on the rim of the tire (diameter = 6S cm) 
at the top be moving after 3.0 s? [Hint: At any moment, 
the lowest point on the tire is in contact with the ground 
and is at rest — see Fig, 8-51.] 




LOO m/s 3 



^Thi* puini nit tire 
at rest momfiirearily 



71. A 1 .4-kg grindstone in the shape of a uniform cylinder of 
radius 0.2D m acquires a rotational rate of 1800 rev/s from 
rest over a 6.0-s interval at constant angular acceleration. 
Calculate the torque delivered by the motor. 

72. (a) A yo-yo is made of two solid cylindrical disks, each of 
mass 0.050 kg and diameter 0.075 m. joined by a (concen- 
tric) thin solid cylindrical hub of mass 0.0050 kg and 
diameter 001 0m. Use conservation of energy to calculate 
the linear speed of the yo-yo when it reaches the end of 
its I.O-m-long string, if it is released from rest. (6) What 
fraction of its kinetic energy is rotational? 

73. (a) For a bicycle, how is the angular speed of the rear wheel 
(« n ) related to that of the pedals and front sprocket (up). 
Fig. 8-52? That is. derive a formula for io^/u>y , Let JVp and 
jV r be the number of teeth on the front and rear sprockets, 
respectively. The teeth are spaced equally on all sprockets 
so that the chain meshes properly, (h) Evaluate the ratio 
Mtt/eoy when the front and rear sprockets have 52 and 13 
teeth, respectively, and (e) when they have 42 and 28 teeth. 




FIGURE 8-51 Problem 70. 



FIGURE 8-52 Problem 73. 



General Problems 223 



74. Suppose a Star the size of our Sun, but with mass ISO limes 
as gTeat. were rotating at a speed of '1.0 revolution every 
12 days. If it were to undergo gravitational collapse to a 
neutron star of radius 1 1 km. losing three-quarters of its 
mass in the process, what would its rotation speed he? 
Assume that the star is a uniform sphere at all times, and 
that the lost mass carries off no angulaT momentum. 

75. One possibility for a low-pollution automobile is for it to 
use energy stored in a heavy rotating flywheel. Suppose 
such a ear has a total mass of 1 400 kg. uses a uniform 
cylindrical flywheel of diameter 1.50 m and mass 240 kg. 
and should be able to travel 350 km without needing 
a flywheel "spinup." (a) Make reasonable assumptions 
(average fractional retarding force = 450 N, twenty accel- 
eration periods from rest to 95 km/h, equal uphill and 
downhill, and that energy can be put back into the 
flywheel as the car goes downhill), and show that the 
total energy needed to be stored in the flywheel is about 
1.7 x 10 s J. (h) What is the angular velocity of the 
flywheel when it has a full "energy charge"? (c) About 
how long would it take a 150-hp motor to give the 
flywheel a full energy charge before a trip? 

76. Figure 8-53 illustrates an H 2 molecule The O-H 
bond length is 0.96 nni and the H-O-H bonds make an 
angle of 104°, Calculate the moment of inertia for the H 2 
molecule about an axis passing through the center of the 
oxygen atom (a) |>erpendicular to the plane of the 
molecule, and (b) in the plane of the molecule, bisecting 
the H-O-H bonds. 




FIGURE 3-53 
Problem 76, 



77. A hollow cylinder (hoop) is rolling on a horizontal surface 
at speed v = 3,3 m/s when it reaches a 15° incline, 
(a) How far up the incline will it go? (b) How long will it 
be on the incline before it arrives back at the bottom? 

7it. A uniform rod of mass M and length L can pivot freely 
(i.e., we ignore friction) about a hinge attached to a wall. 
as in Fig. 8-54. The rod is held horizontally and then 
released. At the moment of release, determine (a) the 
angular acceleration of the rod, and (b) the linear acceler- 
ation of the tip of the rod. Assume that Che force of 
gravity acts at the center of mass of the rod, as shown. 
|f/wi:"SeeFig.8-2lg.| 




7SL A wheel of mass M has radius R- U is standing vertically 
on the floor, and we want to exert a horizontal force Fat 
its axle so that it will climb a step against which it rests 
(Fig- 8-55), The slep has height h. where h < R. What 
minimum force F is needed? 




80. 



FIGURE 3 -55 Problem 79. 



A bicyclist traveling with speed v = 4.2 m/s on a flat 
road is making a turn with a radius r = 6.4 m. The forces 
acting on the cyclist and cycle are the normal force (F v ) 
and friction foTce (F[ r ) exerted by the road on the tires, 
and mg, the total weight of the cyclist and cycle (see 
Fig, 8-56). (s) Explain carefully why the angle % the 
bicycle makes with the vertical (Fig. 8-56) must be given 
by tan B = F( T /F K if the cyclist is to maintain balance. 
(b) Calculate ff for the values given, (c) If the coefficient 
of static friction between tires and road is /i s = 0.70, 
what is the minimum turning radius? 




FIGURE 8-54 

Problem 78. 



FIGURE 8-56 Problem 80. 

SI. Suppose David puts a 0.50-kg rock into a sling of length 
1,5 m and begins whirling the rock in a nearly horizontal 
circle above his head, accelerating it from rest to a rate of 
I20rpm after 5,0 s, What is the torque required to achieve 
this feat, and where does the torque come from? 

82. Model a figure skater's body as a solid cylinder and her 
anus as thin rods, making reasonable estimates for the 
dimensions. Then calculate the ratio of the angular speeds 
for a spinning skater with outstretched arms, and with 
arms held tightly against her body. 



224 CHAPTER 8 Rotational Motion 



XA. You arc designing a dutch assembly which consists 
of two cylindrical plates, of mass M,\ = CO kg and 
M B = 9.0 kg. with equal radii R = 0.60 m. They are 
initially separated (Fig. 8-57). Plate Ma is accelerated from 
Test to an angular velocity o> { = 7,2 rad/s in time 
if = 2.0s. Calculate {a) the angular momentum of M A . 
and (b) the torque required to have accelerated Af A from 
rest to (i>j . (l-) Plate M B , initially at rest but free to rotate 
without friction, is allowed to fall vertically (or pushed by a 
spring), so it is in firm contact with plate M A (their contact 
surfaces are high-friction). Before contact, M A was rotating 
at constant »] , After contact, at what constant angular 
velocity co 2 do the two plates rotate? 




FIGURE 8-57 
Problem 83. 



84. 



85. 



A marble of mass m and radius r rolls along the looped 
rough track of Fig. 8-58. What is the minimum value of 
the vertical height h that the marble must drop if it is to 
reach the highest point of the loop without leaving the 
track 1 ? Assume r <£. R, and ignore frictional losses, 

Repeat Problem 84. but do not assume t « R. 




FIG URE 8-58 Problems 84 and 85. 

8fi. The tires of a car make 85 revolutions as the car reduces 
its speed uniformly from 9IJ.0km/h to 60.0 km/h. The 
tires have a diameter of 0.90m, (a) What was the angular 

acceleration of each tire? (b) If the car continues to 
decelerate at this rate, how much more time is required 
for it to stop? 



Answers to Exercises 



A: / = 0.076 Hz: T = 13 s. 

B: F A . 

D Yes: she does wotk to pull in her arms. 



D: Work was done in pulling the string and decreasing the 
circle's radius. 



General Problems 225 



Our whole built environment, 
from modem bridges to 
skyscrapers, has required 

architects and engineers to 
determine the forces and 
stresses within these struc- 
tures. The object is to keep 
these structures static — that 
is. not in motion, especially 
not falling down. 

The study of statics applies 
equally well to the human 
body, including balance, the 
forces in muscles, joints, and 
bones, and ultimately the 
possibility of fracture. 



CHAPTER 



9 





^ 


m 




. 








|V»I-W 


( ..t 1 i.iviwv- % -^ 


H^^^pG^^P^H 


. ,, >'■■■' - - 


fm^- — 




.^i^^^^^M 


ixi 








gpfY^ !*g 




7 ■ i *j=~ ■ ■■ •:•■ 






■. » - ki j' . p •"-'W* " *^*1 . >-» ^ V^: 




• 











Static Equilibrium; 
Elasticity and Fracture 



This Chapter deals with 
forces within objects ai rest 



In this Chapter, we will study a special ease in mechanies — when the net farce 
and the net torque on an abject, or system of objects, are both zero. In this 
case both the linear acceleration and the angular acceleration of the ohject 
or system are zero. The object is either at rest, or its center of mass is moving at 
constant velocity. We will he concerned mainly with the first situation, in which 
the object or objects are all at rest. Now, you may think that the study of objects 
at rest is not very interesting since the objects will have neither velocity nor 
acceleration, and the net force and the net torque will be zero, But this docs not 
imply that no forces at all act on the objects, In fact it is virtually impossible to 
find an object on which no forces act. Just how and where these forces act can 
be very important, both for buildings and other structures, and in the human body. 

Sometimes, as we shall sec in this Chapter, the forces may be so great that 
the object is seriously deformed, or it may even fracture (break) — and avoiding 
such problems gives this field of statics even greater importance. 

Statics is concerned with the calculation of the forces acting on and within 
structures that are in equilibrium. Determination of these forces, which occupies 
us in the first part of this Chapter, then allows a determination of whether the 
structures can sustain the forces without significant deformation or fracture, 
subjects we discuss later in this Chapter, These techniques can be applied in a 
wide range of fields. Architects and engineers must be able to calculate the 
forces on the structural components of buildings, bridges, machines, vehicles, 



226 



and other structures, since any material will buckle or break if too much force is 
applied (Fig. 9-1). In the human body a knowledge of the forces in muscles and 
joints is of great value for doctors physical therapists and athletes. 



The Conditions for Equilibrium 



Objects in daily life have at least one force acting on them (gravity). If they arc 
at rest, then there must be other forces acting on them as well so that the net 
force is zero. A book at rest on a table, for example, has two forces acting on it, 
the downward force of gravity and the normal force the table exerts upward on 
it (Fig, 9-2). Since the net force on the book is zero, the upward force exerted by 
the table on the book must be equal in magnitude to the force of gravity acting 
downward on the book. Such an object is said to be in equilibrium (Latin Tor 
"equal forces" or "balance") under the action of these two forces. 

Do not confuse the two forces in Fig. 9-2 with the equal and opposite forces 
of Newton's third law, which act on different objects. Here, both forces act on the 
same object. 



EXAMPLE 9-1 



Straightening teeth. The wire band shown in Fig. *)-3a 
has a tension F T of 2.0 N along it. It therefore exerts forces of 2,0 N on the 
highlighted tooth (to which it is attached) in the two directions shown. Calcu- 
late the resultant force on the tooth due to the wire, F^. 

APPROACH Since the two forces t\- are equal, their sum will be directed 
along the line that bisects the angle between them, which we have chosen to 
be (he v axis. The x components of the two forees add up to zero. 

SOLUTION The y component of each force is (2.0 N)(cos70°) = 0,684 N: 
adding the two together, we get a resultant force F R = I.37N as shown in 
Fig. 9-3b. We assume that the looth is in equilibrium because (he gums exert 
an equal and opposite force, Actually that is not quite so since the objective is 
to move the tooth ever so slowly. 

NOTE If the wire is firmly attached to the tooth, the tension to the right, say, 
can be made larger than that to the left, and the resultant force would corre- 
spondingly be directed more toward the right. 




FIG URE 9-1 Elevated walkway 
collapse in a Kansas Cily hotel in 1981, 

How a simple physics calculation could 
have prevented the tragic loss, of over 
100 lives is considered in F.xample 9-12. 

(j S 1 PHYSICS APPLIED 

Orthodontia 



FIGURE 9-2 The book is in equi- 
librium; the net force on il is zero. 



Normal force 



" ~1 










Gravity 























FIGURE 9-3 Forees on a tuolh. 
Example 9-1. 



The First Condition for Equilibrium 

For an object to be at rest, Newton's second law tells us that the sum of the 
forces acting on it must add up to zero. Since force is a vector, the components 
of the net force must each be zero. Hence, a condition for equilibrium is that 

2F, = 0, LF y = 0, XF, = 0. (9-1 > 

We will mainly be dealing with forces that act in a plane, so we usually need 
only the x and y components. We must remember that if a particular force 
component points along the negative x or y axis, it must have a negative sign. 
Equations 9-1 are called the first condition for equilibrium. 

SECTION 9-1 The Conditions for Equilibrium 



First condition for equilibrium: 
the sum of ail forces is zero 



111 




W 



200 kg 



•'■■ 



(b) fa 

FIGURE 9-4 Example 9-2- 



FIGURE 9-5 Although the net 
force on it is zero, the ruler will move 

(rotate). A pair of equal forces acting 
in opposite directions but at different 
points on an object (as shown heTe) 
is referred to as a couple. 



EXAMPLE 3-2 



Chandelier cord tension. Calculate the tensions P A and F B 
in the two cords that are connected lo the vertical coto" supporting the 200-kg 
chandelier in Fig. 9-4. 



APPROACH We need a free-body diagram, but for which object? Tf we 
choose the chandelier, the cord supporting il must exert a force equal lo Ihe 
chandeliers weight mg = (200 kg)(9.8 m/s 2 ) = 1%0N. But the forces F A 
and F B don't get involved- Instead, let us choose as out object the point where 
the three cords join (it could be a knot). The free-body diagram is then as 
shown in Fig, 9-4a. The three forces — F A ,F B , and tire tension in the vertical 
cord equal to the weight of the 200-kg chandelier — act at this point where the 
three cords join. For this junction point we write iLF^ = and XF V = 0, 
since the problem is laid out in two dimensions. The directions of F A and F B 
are known, since tension in a rope can only be along the rope — any other 
direction would cause the rope to bend, as already pointed out in Chapter 4. 
Thus, our unknowns are the magnitudes F A and F R . 

SOLUTION We first resolve F A into its horizontal (x) and vertical (y) compo- 
nents. Although we don't know the value of F A , we can write (see Fig. 9-4b) 
F Al = -F A cos60 u and F Ay = F A sin60' J , F u has only an x component. In the 
vertical direction, we have the downward force exerted by the vertical cord 
equal to the weight of the chandelieT = (200kg)(#), and the vertical compo- 
nent of F A upward. Since XF,, = 0, we have 



2F>, = F A sin 6Q J 



(200kg)(£)-0 



M> 



'■', 



(200 kg) ff (200 kg)* 



sinG0 u 
In the horizontal direction, 



0.866 



(23lkg}g - 2260 N. 



Thus 



SF, = F B - F A cos60° = 0. 

F„ = F A cosfi0° = (231 kg) (g) (0.500) = (115 kg)* = 1130N. 



The magnitudes of F A and Fu determine the strength of cord or wire that must 
be used. In this case, the wire must be able to hold more than 230 kg. 

NOTE We didn't insert the value ol ' g, the acceleration due to gravity, until 
the end. Tn this way we found the magnitude of the force in terms of g times the 
number of kilograms (which may be a more familiar quantity than newtons). 



EXERCISE A 
Why? 



In Example 9-2, F A has to be greater than the ehandelier's weight, mg. 




Second condition for equilibrium 

lilt' sum Of all torques is zero 



The Second Condition for Equilibrium 

Although Eqs, 9-1 are a necessary condition for an object to be in equilibrium, 
they are not always a sufficient condition, Figure 9-5 shows an object on which 
the net force is zero. Although the two forces labeled F add up to give zero net 
force on the object, they do give rise to a net torque that will rotate the object. 
Referring to Eq, 8-14, Xt = /«, we see that if an object is to remain at rest, 
the net torque applied to it (calculated about any axis) must be zero. Thus we 
have the second condition for equilibrium: that the sum of the torques acting on 
an object, as calculated about any axis, must be zero: 

St = 0, (9-2j 

This condition will ensure that the angular acceleration, «, about any axis 



228 CHAPTER 9 Static Equilibrium; Elasticity and Fracture 



will be zero. If the object is not rotating initially (w - 0), it will not start 
rotating, Equations 9-1 and 9-2 are the only requirements for an object to 
be in equilibrium. 

We will mainly consider cases in which the forces all act in a plane (we call 
it the xy plane), Tn such cases the torque is calculated about an axis that is 
perpendicular to the xy plane. The choice of this axis is arbitrary. If the object is 
at rest, then St - about any axis whatever. Therefore we can choose any 
axis that makes our calculation easier. Once the axis is chosen, all torques must 
be calculated about that axis. 



<^ 



CAUTION 



Axil choice for Ef = U is arbitrary. 
All torques must be calculated 
about the same axis, 



CONCEPTUAL EXAMPLE 9-3 I A lever. The bar in Fig. 9-6 is being used { ^physics applied 



as a lever to pry up a large rock. The small rock acts as a fulcrum (pivot point). 
The force F ? required at the long end of the bar can be quite a bit smaller than 
the rock's weight mg, since it is the torques that balance in the rotation about 
the fulcrum, If, however, the leverage isn't sufficient, and the large rock isn't 
budged, what are two ways to increase the leverage? 

RESPONSE One way is to increase the lever arm of the force /> by slipping a 
pipe over the end of the bar and thereby pushing with a longer lever arm. A 
second way is to move the fulcrum closer to the large rock. This may change 
the long lever arm R only a little, but it changes the short leveT arm r by a 
substantial fraction and therefore changes the ratio of R/r dramatically. In 
order to pry the rock, the torque due to F v must at least balance the torque 
due to mg, so mgr - F^R and 






my 



With /■ smaller, the weight mg can be balanced with less force F ? . The ratio 
R/r is the mechanical advantage of the system. A lever is a "simple machine."' 
We discussed another simple machine, the pulley, in Chapter 4, Example 4-14. 



The lever 




FIGURE 9-6 Example 9-3. A lever 
can "multiply" your foTce, 



EXERCISE B For simplicity, we wrote the equation in Example 9-3 as if the lever were 
perpendicular to the forces. Would thy equation he valid even for a lever at an angle as 
shown in Fig. 9-67 



Solving Statics Problems 



This subject of statics is important because it allows us to calculate certain 
forces on (or within) a structure when some of the forces on it are already 
known. We will mainly consider situations in which all the forces act in a plane, 
so we can have two force equations (.v and v components) and one torque equa- 
tion, for a total of three equations. Of course, you do not have to use all three 
equations if they are not needed. When using a torque equation, a torque that 
tends to rotate the object counterclockwise is usually considered positive, 
whereas a torque that tends to rotate it clockwise is considered negative. (But 
the opposite convention would not be wrong.) 

One of the forces that acts on objects is the force of gravity. Our analysis in 
this Chapter is greatly simplified if we use the concept of center of gravity (en) 
or center of mass (cm), which for practical purposes are the same point. As we 
discussed in Section 7-8, we can consider the foree of gravity on the object as 
acting at its c:n. For uniform symmetrically shaped objects, the CO is at the 
geometric center. For more complicated objects, the oo can be determined as 
discussed in Section 7-8. 

There is no single technique for attacking statics problems, but the following 
procedure may be helpful. 



*• PROBLEM SOLVING 
i > counterclockwise 
t < i) clockwise 



SECTION 9-2 Solving Statics Problems 229 



PROBLEM SOLVING 



Statics 



Choose one object at a time for consideration. 

Make a careful free-body diagram by showing all 

the forces acting on that object and the points at 

which these forces act. If you aren't sure of the 

direction of a force, choose a direction; if the actual 

direction is opposite, your eventual calculation will 

give a result with a minus sign. 

Choose a convenient coordinate system, and resolve 

the forces into their components. 

Using letters to represent unknowns, write down 

the equilibrium equations for the forces: 



SF, = and £/\ 



0. 



assuming all the forces act in a plane. 
For the torque equation. 

St = 0, 
choose any axis perpendicular to the xy plane that 



might make the calculation easier, (For example, 
you can reduce the number of unknowns in the 
resulting equation by choosing the axis so that 
one of the unknown forces acts through that 
axis; then this force will have zero lever arm 
and produce zero torque, and so won't appear 
in the equation,) Pay careful attention to deter- 
mining the levei" arm for each force correcfly. 
Give each torque a + or -- sign to indicate 
torque direction. For example, if torques tending 
to rotate the object counterclockwise are posi- 
tive, then those tending to rotate it clockwise are 
negative. 
5, Solve these equations for the unknowns. Three 
equations allow a maximum of three unknowns to 
be solved for. They can be forces, distances, or 
even angles. 



PHYSICS APPLIED 

Balancing a seesaw 



EXAMPLE 9-4 



Balancing a seesaw. A board of mass M = 2.0 kg serves 
as a seesaw for two children, as shown in Fig, 9-7a, Child A has a mass of 30 kg 
and sits 2.5 m from the pivot point, P (his center of gravity is 2.5m from the 
pivot). At what distance x from the pivot must child B,of mass 25 kg, place herself 
to balance the seesaw? Assume the board is uniform and centered oveT the pivot. 

APPROACH We follow the steps of the Problem Solving Box explicitly. 
SOLUTION 

1. Free-bttdy diagram. We choose the board as our object, and assume it is 
horizontal. Its free-body diagram is shown in Fig. 9-7b. The forces acting 
on the board are the forces exerted downward on it by each child, F A and 
F u , the upward force exerted by the pivot F N , and the force of gravity on 
the board (= Mg) which acts at the center of the uniform board. 

2. Coordinate system. We choose y to be vertical, with positive upward, and. v 
hon/onta! to the right, with origin at the pivot. 

y. Forte equation. All the forces are in the y (vertical) direction, so 



2R = 



f\ 



m A g - m b g - Mg = 0, 

and F n = m a g because each child is in equilibrium 
when the seesaw is balanced. 



where F A = »» A g 



FIGURE 9-7 (a) Two children 
on a seesaw. Example 9-4. 

(b) Free-body diagram 

of the board. , 

+ Torque- [ c 
(a) 



m A = 30 kg 



> 



-2.5 m- 



A 



m h = 25 kg 

B 

-*■•' 



W 



■ Tonquc 



-2.5 m- 



±F. 



I 1 



*/g = (2.[')kg>g 



(b) *F A ='n A g 

230 CHAPTER 9 Static Equilibrium; Elasticity and Fracture 



F B = w B gi 



4, Torque equation. LeL us calculate Lhe kuque about an axis through the board 
at the pivot point, P. Then the lever arms for F N and for the weight of the 
board are zero, and they will contribute zero torque (about point P) in our 
torque equation, Thus the tOTque equation will involve only the forces F A and 
F[j , which are equal to the weights of the children. The toique exerted by 
each child will be >ng times the appropriate lever arm, which here is the 
distance of each child from the pivot point. Hence the torque equation is 

St -0 
m A g(2.5m) - m B gx + M^(0m) + F N (0m) = 
or 

m A g(2,5 m) — m^BX = 0, 

where two ternis were dropped because their lever aims were zero. 

5. Solve. We solve the torque equation for x and find 

x = — (2.5 m) = : -— - (2.5 m) = 3.0 m. 

To balance the seesaw, child B must sit so that her cm is 3.0 m from the 
pivot point. This makes sense: since she is lighter, she must sit farther from 
the pivot than the heavier child. 



EXERCISE C We did not need to use the force equation to solve Example 9-4 because 
of our choice of the axis, Use the force equation to find the force exerted by the pivot. 



EXAMPLE 9-5 



Forces on a beam and supports. A uniform 1500-kg beam, 
20,0 m long, supports a 1 5,000-kg printing press 5.0 m from the right support 
column (Fig. 9-8). Calculate the force on each of the vertical support columns. 

APPROACH We analyze the forces on the beam (the force the beam exerts 
on each column is equal and opposite to the force exerted by the column on 
the beam). We label these forces P A and P u in Fig. 9-8. The weight, of the 
beam itself acts at its center of gravity, 10.0 m from either end. We choose a 
convenient axis for writing the torque equation: the point of application of F A 
(labeled P), so P A will not enter the equation (its lever arm will be zero) and 
we will have an equation in only one unknown, F u . 

SOLUTION The torque equation, Xt = 0, with the counterclockwise direc- 
tion as positive gives 

St = -(IC.0m)(l500kg}# - ( 1 5.0 m)(l 5,000 kg)# + (20,0m)Fu = 0. 

Solving for 1%, we find F u - ( 12,000 kg)g - 1 18,000 N. To find F A , we use 
2F^ =0, with + v upward: 



SF. = F 4 



(l500kg) A > - (1 5,000 kg) A > + F E = 0. 



Putting in F„ = ( 12,000 kg)g, wc find that F A = (4500 kg}# = 44,100 N. 



Figure 9-9 shows a uniform beam that extends beyond its support like a 
diving board. Such a beam is called a cantilever. The forces acting on the beam 
in Fig. 9-9 arc those due to the supports, F A and F B , and the force of gravity 
which acts at the CO, 5.0 m to the right of the right-hand support. If you follow 
the procedure of the last Example and calculate F A and F B , assuming they point 
upward as shown in Fig. 9-9, you will find that F A comes out negative. If the 
beam has a mass of 1200 kg and a weight mg - 12,000 N, then F u - 15,000 N 
and F A — — 3000 N (see Problem 10). Whenever an unknown force comes out 
negative, it merely means that the force actually points in the opposite direction 
from what you assumed. Thus in Fig. 9-9, P A actually points downward. With a 
little reflection it should become clear that the left-hand support must indeed 
pull downward on the beam (by means of bolts, screws, fasteners and/or glue) if 
the beam is to be in equilibrium; otherwise the sum of the torques about the CG 
(or about the point where P B acts) could not be zero, 

SECTION 9-2 



CG 



Q 



(1500 kg)g 
■ 1 0.0 m .4-5.0 m 



■5.0 m- 



<l5,00Gkg)g 

FIG U RE 9 -8 A 1 SOU- k g beam 
supports a 1 5,000-kg machine. 
Example 9-5. 



PHYSICS APPLIED 
Cantilever 

*» PROBLEM SOLVING 
If a three comes out negative 

FIGURE 9-9 A cantilever. 

Fa P B 

-20.0 mJ .10.0 in -| 



t— 20.0 m— t 



Z2E 



Solving Statics Problems 231 



Hinges and earth 




FIGURE 9-10 F.xamplc 9-6. 



Our next Example involves a beam lhal is attached to a wall by a hinge and 
is supported by a cable or cord (Fig- 9-10), It is important to remember that a 
flexible cable can support a force only along its length. (If there were a compo- 
nent of force perpendicular to the cable, it would bend because it is flexible,) 
But for a rigid device, such as the hinge in Fig, 9- 1 0, the force can be in any 
direction and we can know the direction only after solving the problem. The 
hinge is assumed small and smooth, so it can exen no internal torque (about its 
center) on the beam. 



EXAMPLE 9-6 



Hinged beam and cable. A uniform beam, 2.20m long 
with mass m - 25.0 kg, is mounted by a hinge on a wall as shown in Fig. 9-10, 
The beam is held in a horizontal position by a cable that makes an angle 
8 - 30.0^ as shown. The beam supports a sign of mass M - 28.0 kg suspended 
from its end. Determine the components of the force P H that the hinge exerts 
on the beam, and the tension F T in the supporting cable, 

APPROACH Figure 9-10 is the free-body diagram for the beam, showing all 
the forces acting on the beam, It also shows the components of F H and F T . Wc 
have three unknowns, F Hx , F Hv , and F T (wc are given 8), so wc will need all 
three equations, 2/^ = 0, %F y — 0, St = 0. 

SOLUTION The sum of the forces in the vertical (y) direction is 



Fxi., + F-, 



| :■ 



i'. 



mg 



EF, = 
Mg = 0. 



In the horizontal (jc) direction, the sum of the forces is 
ZF X = 
Fm ~ Fix - 0, 



til 



liil 



For the torque equation, wc choose the axis at the point where F T and A/g act 
(so our equation then contains only one unknown, F Hv ). Wc choose torques that 
tend to rotate the beam counterclockwise as positive. The weight mg of the 
(uniform) beam acts at its center, so we have 

Sr = 
-(F H ,)(2.20m) + mg( 1.10 m) = 0. 

We solve for F Hy : 



(■):, 



l.lDrn 
2.20 rn 



mg = (G.500)(25,0kg)(9.80m/s : ) = 123 N. (iii) 



Next, since the tension F T in the cable acts along the cable (8 = 30.0"), we see 
from Fig. 9-10 that tanfl = F Ty /F Tx , or 

F l} - /v t tantf - F Ti (tan 30.0") - 0.577 F ix . <iv) 

Equation (i) above gives 

F fy = (m + M)g - F Hv = (53.0kg)(9.S0m/s 2 ) - 123 N = 396 N; 

Equations (iv) and (ii) give 

F Tl = F T ,/0.577 = 687 N; 
F lis -F ls -687N, 

The components of Pn are Fj u . - 123 N and F, u - 687 N. The tension in the 
wire is F r = \/F\ x + F\ y = 793 N. 

Alternate Sofution Let us see the effect of choosing a different axis 
for calculating torques, such as an axis through the hinge. Then the lever arm 



232 CHAPTER 9 Static Equilibrium; Elasticity and Fracture 



for F lt is zero, and the torque equation (St = 0) becomes 
-mK(l.lOm) - Mg(220m) + F ty (2.20m) = 0. 
We solve this for F Ty and find 

Fry = -^8 + M S = ('2.5 kg + 28.0kg)(9.80 m/s 2 ) = 397 N. 

We get the same result, within the precision of our significant figures, 

NOTE It doesn't matter which axis we ehoose for St = f). Using a second 

axis can serve as a check. 



Additional Example — The Ladder 



EXAMPLE 9-7 



Ladder. A 5,0-m-long ladder leans against a wall at a point 
4.0 m above a cement flour as shown in Fig, 9- 1 1 .The ladder is uniform and has 
mass m = 12.0 kg. Assuming the wall is frictionless (but the floor is not), 
determine the forces exerted on the ladder by the floor and by the wall. 

APPROACH Figure 9- 1 1 is the free-body diagram for the ladder, showing 
all the forces acting on the ladder The wall, since it is frictionless, can 
exert a force only perpendicular to the wall, and we label that force P w - 
The cement flooT exeTts a force F c which has both horizontal and vertical 
force components: F Cx is frictional and F C| . is the normal force. Finally, 
gravity exerts a force ing = (12.0 kg}(9,80m/s 2 ) = I IS N on the ladder at 
its midpoint, since the ladder is uniform, 

SOLUTION Again we use the equilibrium conditions, 2F, - 0, SF V - 0, 
Sr - 0. We will need all three since there are three unknowns: F w , tjt. an£ l 
F Cl ., The y component of the force equation is 

^ F v = F cy - mg = 0, 
so immediately we have 

F €y = mg= 118N. 
The x component of the force equation is 



2.F r — Ft 






f., - n. 



To determine both F tl and F w , we need a torque equation. If we choose to 
calculate torques about an axis through the point wheje the ladder touches 
the cement floor, then F c , which acts at this point, will have a lever arm of 
zero an d so won't enter the equation. The ladder touches the floor a distance 
x c = \/(5.0m) 2 - (4.0m) 2 = 3.0 m from the wall. The lever arm Cot mg is 
half this, or 1,5 m, and the lever arm for F w is 4.0 m. Fig, 9-11. We get 

St = (4.0m)F w - (l.5m)mg = 0. 
Thus 

(1.5 m )( 12.0 kg)(9.8m/s : ) 

F w = — = 44 N. 

4.0 m 

Then, from the x component of the force equation, 

F Cx = F w =44N. 

Since the components of F c are F Ci . = 44 N and F Cv = I IS N, then 

F c - V"(44N) 2 + (II8N) : - 126 N « 130 N 
(rounded off to two significant figures), and it acts at an angle to the floor of 
= tan-'(118N/44N) = W. 

NOTE The force f c does no{ have to act along the ladder's direction because 
the ladder is rigid and not flexible like a cord or cable, 



EXERCISE D Why is il reasonable lo ignore friction along the wall, bill no! reasonable 
to ignore it along the floor? 




4.0 m 



FIGURE 9-11 A ladder leaning 
against a wall. Hxample 9-7. 



SECTION 9-2 Solving Statics Problems 233 



Tendon'; 



Trieeps 

muscle 

(extensor! 




FIGURE 9-1 2 The biceps (flexor) 
and triceps (extensor) muscles in 
(he human arm. 



PHYSICS APPLIED 



Forces in muscles ivut joints 



Applications to Muscles and Joints 

The techniques we have been discussing for calculating forces on objects in 
equilibrium can readily be applied to the human (or animal) body, and can be of 
great use in studying the forces on muscles, bones, and joints for organisms in 
motion or at rest. Generally a muscle is attached, via tendons, to two different 
bones, as in Fig. 9-12. The points of attachment are called insertions, Two hones 
arc flexibly connected at a joint, such as those at the elbow, knee, and hip. A 
muscle exerts a pull when its fibers contract under stimulation by a nerve, but a 
muscle cannot exert a push. Muscles that tend to bring two limbs closer 
together, such as the biceps muscle in the upper arm (Fig. 9-12) arc called 
flexor?, those that act to extend a limb outward, such as the triceps muscle in 
Fig. 9-12, arc called extensors. You use the flexor muscle in the upper arm when 
lifting an object in your hand; you use the extensor muscle when throwing a ball, 



FIGURE 9-13 F.xample 9-8. 




EXAMPLE 9-8 



Force exerted by biceps muscle. How much force must 
the biceps muscle exert when a 5.0-kg mass is held in the hand (a) with the 
arm horizontal as in Fig. 9-1 3a, and (b) when the arm is at a 45" angle as in 
Fig. 9- 13b? Assume that the mass of forearm and hand together is 2.0 kg and 
their cg is as shown. 

APPROACH The forces acting on the forearm are shown in Fig. 9-13 and 
include the weights of the arm and ball, the upward force F M exerted by the 
muscle, and a force Fj exerted at the joint by the bone in the upper arm (all 
assumed to act vertically). We wish to find the magnitude of F M , which is done 
most easily by using the torque equation and by choosing our axis through the 
joint so that F, contributes zero torque. 

SOLUTION {a) We calculate torques about the point where Fj acts in 
Fig. 9- 13a, The St - equation gives 

(0.050 m)F M - (0.15m)(2.0kg)g - (0.35 m)(5.0kg) ff = 0. 



Wc solve for F hi : 



Fu - 



(u.l5m)(2.0kg) g + (0.35m)(5.0fcg)g 

0.050 m 



- (41 kg)# -400N. 



(h) The lever arm, as calculated about the joint, is reduced by the factor 
sin 45 ° for all three forces. Our torque equation will look like the one just 
above, except that each term will have its lever arm redueed by the same 
factor, which will cancel out. The same result is obtained, F M - 400 N. 

NOTE The force required of the muscle (400 N) is quite large compared to 
the weight of the object lifted (49 N). Indeed, the muscles and joints of the 
body arc generally subjected to quite large forces. 



PHYSICS APPLIED 

Muscle insertion and 
lever arm 



J P H Y S I C S APPLIED 

Forces on the spine, and back pain 



The point of insertion of a muscle vanes from person to person, A slight 
increase in the distance of the joint to the point of insertion of the biceps 
muscle from 5.0 em to 5.5 em can be a considerable advantage for lifting and 
throwing. Champion athletes are often found to have muscle insertions farther 
from the joint than the average person, and if this applies to one muscle, it 
usually applies to all. 

As another example of the large forces acting within the human body, we 
consider the muscles used to support the trunk when a person bends forward 
(Fig. 9- 14a). The lowest vertebra on the spinal column (fifth lumbar vertebra) 
acts as a fulcrum for this bending position. The "erector spinae" muscles in the 
back that support the trunk act at an effective angle of about 12 ' to the axis of 
the spine, Figure 9- 14b is a simplified schematic drawing showing the forces on 
the upper body. We assume the trunk makes an angle of 30" with the horizontal, 



Z3fl CHAPTER 9 Static Equilibrium; Elasticity and Fracture 



The farce exerted by the back muscles is represented by F M , the force exerted 
on the base of the spine at the lowest vertebra is F v , and w H , w A , and w r repre- 
sent the weights of the head, freely hanging arms, and tTunk, respectively. The 
values shown are approximations taken from Table 7-1, Tie distances (in cm) 
refer to a person 180 cm tall, but are approximately in the same ratio of 1:2:3 for 
an average person of any height, and the result in the following Example is then 
independent of the height of the person. 

1 3If.il fi I a ^ agigl Forces on your back. Calculate the magnitude and 
direction of the force F v acting on the fifth lumbar vertebra for the example 
shown in Fig. 9-1 4b. 

APPROACH Wc use the model of the upper body described above and shown 
in Fig. 9- 14b. Wc can calculate F M using the torque equation if wc take the 
axis at the base of the spine (point S);with this choice, tire other unknown, F v , doesn't 
appear in the equation because its lever arm is zero. To figure the lever arms, 
we need to use trigonometric functions. 

SOLUTION For F M , the lever arm (perpendicular distance from axis to line of 
action of the force) will be the real distance to where the force acts (48 cm) 
multiplied by sin 12', as shown in Fig. 9- 14c. The lever arms for w H , w A , and w T 
can be seen from Fig. 9- 14b to be their respective distances from S times sin 60 '. 
F M tends to rotate the trunk counterclockwise, which we take to be positive. 
Then w H , w A , w T will contribute negative torques. Thus 2t = gives 

(0.48m)(sinl2*)(F M ) - (0.72 m) (sin 6(F)(™ H ) 

- (0,48 m) (sin 60° )(w A ) - (0.36 m)(sin 60°)(tt>r) = 0. 
Solving for F M and putting in the values for w^ , , w; A , w r given in Fig 9- 14b, we find 
(0.72 m) (0.07m;) + (0.48 m) (0.1 2w) + (0.36 m) (0.46m;) 

F " = (0.48m)(sinl2<>) (sin ^ 

= 2.37m; ss 2Auh 

where w is the total weight of the body. To get the components of F v we 
use the x and y components of the force equation (noting that 30° - 12° - I8 U ): 



EF V = F, 



so 



and 



so 



SF r 



F M sinl8" 

1.38hj ~ 1.4m;, 



Wh - w A - iv, = 



!■■,, 



F M cos 18" =0 



F Vx = 2.25w « 2,3m?, 



where we keep 3 significant figures for calculating, but round off to 2 for giving 
the answer. Then 



F v = yFvx + F vy = 2,6h;. 
The angle 8 that F v makes with the horizontal is given by tan# - F Vi ./F Vli 
- 0,6 L so 6 - 32°. 

NOTE The force on the lowest vertebra is over 2t times the total body 
weight! This force is exerted by the "sacral" bone at the base of the spine, 
through the fluid-filled and somewhat flexible intervertebral disc. The discs at 
the base of the spine are clearly being compressed under very large forces. [If 
the body was less bent over (say, the 30° angle in Fig. 9- 14b becomes 40" or 
50°), then the stress on the lower back will be less (see Problem 35). | 



Tf the person in Fig. 9-14 has a mass of 90 kg and is holding 20 kg in his 
hands (this increases m* a to 0.34m;), then F v is increased to almost four times the 
person's weight (3.7m;). For this 200-lb person, the force on the disc would be 
over 700 lb! With such strong forces acting, it is little wonder that so many 
people suffer from low back pain at one time or another. 



Fifth 

lumbar 
vertebra 



Erector spinae 
muscles 




CO 




IC H = 0.07-jj; 
[head) 

W A = 0.12W 

(2 arm*) . 

hi = Toial wcighi 
Wj=OA(vtl! of person 

(irunkj 

(b) 



I^ver arm 
for F u 

Axis for Xt 
calculation 




{--i 



FIGURE 9-14 (a) A person 
bending over, (h) Forces on the 
back exerted by the buck muscles 
(F M ) and by the vertebrae (F v ) 
when a person bends over. 
(c) Finding the lever arm for P M . 



*SECTION 9-3 Applications to Muscles and Joints 235 



Stable and 
unstable equilibria 



Stability and Balance 



An object in static equilibrium, if left undisturhed, will undergo no tra relational 
or rotational acceleration since the sum of all the forces and the sum of all the 
torques acting on it are zero. However, if the object is displaced slightly, three 
outcomes are possible: (I) the object returns to its original position, in which 
case it is said to be in stable equilihrium; (2) the object moves even farther from 
its original position, and it is said to be in unstable equilibrium; or (3) the object 
remains in its new position, and it is said to be in neutral equilibrium. 

Consider the following examples. A ball suspended freely from a string is in 
stable equilibrium, for if it is displaced to one side, it will return to its original 
position (Fig. 9- 1 5a) due to the net force and torque exerted on it. On the other 
hand, a pencil standing on its point is in unstable equilibrium. If its center of gravity 
is directly over its tip (Fig. 9- 15b), the net force and net torque on it will be zero. 
But if it is displaced ever so slightly as shown — say, by a slight vibration or tiny air 
current — there will be a torque on it, and this toTque acts to make the pencil 
continue to fall in the direction of the original displacement. Finally, an example of an 
object in neutral equilibrium is a sphere resting on a horizontal tabletop, Tf it is placed 
slightly to one side, it will remain in its new position — no net torque acts on it. 



FIGURE 9-15 (a) Stable equilibrium, and 
(b) unstable equilibrium. 




FIGURE 9-16 Equilibrium of a 

refrigerator resting on a flat floor. 

FIGURE 9-17 Humans adjust 

their posture to achieve stability 
when carrying loads. 




PHYSICS APPLIED 

Humurts and balam < 



I 




i_fc 



& 

i 



(■) (b) 

In most situations, such as in the design of structures and in working with the 
human body, we arc interested in maintaining stable equilibrium, or balance, as we 
sometimes say, Tn general, an object whose center of gravity (CG) is below its point 
of support, such as a ball on a string, will be in stable equilibrium. If the cg is 
above the base of support, we have a more complicated situation. Consider a 
standing refrigerator (Fig. 9-l6a). If it is tipped slightly, it will return to its original 
position due to the torque on it as shown in Fig, 9- 16b. But if it is tipped too far, 
Fig. 9-l6c, it will fall over. The critical point is reached when the cg shifts from 
one side of the pivot point to the other. When the co is on one side, the torque 
pulls the object back onto its original base of support. Fig, 9- 1 6b. If the object is 
tipped further, the cg goes past the pivot point and the torque causes the object to 
topple. Fig, 9- 1 frc. In general, cm object whose center of gravity is above its base of 
support will be stable if a vertical line projected downward from the cg falls within 
the base of support- Thin is because the normal force upward on the object (which 
balances out gravity) can be exerted only within the area of contact, so if the force 
of gravity acts beyond this area, a net to ■ que will act to topple the object. 

Stability, then, can be relative. A brick lying on its widest face is more stable 
than a brick standing on its end, for it will lake more of an effort to lip il over, In 
the extreme case of the pencil in Fig. 9- 15b, the base is practically a point and 
the slightest disturbance will topple it. In general, the larger the base and the 
lower the cg, the more stable the object. 

In this sense, humans are much less stable than four-legged mammals, which 
not only have a larger base of support because of their four legs,, but also have a 
lower center of gravity. When walking and performing other kinds of movement, 
a person continually shifts the body so that its en is over the feet, although in the 
normal adult this requires no conscious thought. Even as simple a movement as 
bending over requires moving the hips backward so that the en remains over the 
feet, and you do this repositioning without thinking about it, To sec this, position 
yourself with your heels and back to a wall and try to touch your toes, You won't 
be able to do it without falling. Persons carrying heavy loads automatically adjust 
their posture so that the cg of the total mass is over their feet, Fig. 9-17. 



236 CHAPTER 9 Static Equilibrium; Elasticity and Fracture 



Elasticity; Stress and Strain 



In the first part of this Chapter we studied how to calculate the forces on 
objects in equilibrium. In this Section we study the effects of these forces: any 
object changes shape under the action of applied forces. If the forces are great 
enough, the object will break, or fracture, as we will discuss in Section 9-6. 

Elasticity and Hooke's Law 

If a force is exerted on an object, such as the vertically suspended metal rod 
shown in Fig. 9-18, the length of the object changes. If the amount of elonga- 
tion, AL, is small compared to the length of the object, experiment shows that 
AL is proportional to the force exerted on the object. This proportionality, as we 
saw in Section 6-4, can be written as an equation: 



F = k AL. 



Here F represents the force pulling on the object, AL is the change in length, 
and k is a proportionality constant. Equation 9-3, which is sometimes called 
Hooke's law 1 after Robert Hooke (1635-1703), who first noted it, is found to be 
valid for almost any solid material from iron to bone — but it is valid only up to 
a point. For if the force is too great, the object stretches excessively and eventu- 
ally breaks. 

Figure 9-19 shows a typical graph of applied force versus elongation. Up to 
a point called the proportional limit, Cq. 9-3 is a good approximation for many 
common materials, and the curve is a straight line. Beyond this point, the graph 
deviates from a straight line, and no simple relationship exists between F and 
AL. Nonetheless, up to a point farther along the curve called the elastic limit, 
the object will return to its original length if the applied force is removed. The 
region from the origin to the clastic limit is called the elastic region. If the object 
is stretched beyond the elastic limit, it enters the plastic region: it does not 
return to the original length upon removal of the external force, but remains 
permanently deformed (such as a bent paperclip). The maximum elongation is 
reached at the breaking point. The maximum force that can be applied without 
breaking is called the ultimate strength of the material (actually, force per unit 
area as we discuss in Section 9-6). 



H: 



"V\ 



FIGURE 9-18 Hooke's law: 
AL oc applied force. 



(9-3) t/< >< ik e 's in w (again) 



Ultimate strength 
\l Proportional limit Ll n. 




FIGURE 9-19 Applied force vs. elongation fur 
a typical metal under tension, 



Elongation, &L 



Young's Modulus 

The amount of elongation of an object, such as the rod shown in Fig. 9- 1 8, 
depends not only on the force applied to it. but also on the material of which it 
is made and on its dimensions. That is, the constant k in Eq. 9-3 can be written 
in terms of these factors. 



'The rerm "law" applied to this relation is not really appropriate, since first of all. it is only an 

approximation, and secondly, it refers only to a limited set of phenomena, Most physicists prefer to 
reserve tlie word "law" for those relations That are deeper and mure cneompassinj and precise, such 
as Newton's lawi of motion or the law of conservation of energy. 

^SECTION 9-5 Elasticity; Stress and Strain 237 



TABLE 9-1 


Elastic Moduli 










Material 






Young's Modulus, 
E (N/m 2 ) 


Shear Modulus, 
O (N/m 2 ) 


]'. ill. Modulus. 
B (lN/in 2 ) 


Solids 


Fran, cast 






100 X 10* 


40 X 10* 


90 X 10* 


Sleel 






200 X 10* 


80 x 10* 


[40 x 10' 


Brass 






ioo x io* 


35 X 10* 


80 X 10* 


Aluminum 






70 X 10' 


25 X 10* 


70 X Id* 


Curi cretc 






20 X 10 9 




Brick 






14 X 10* 






Marble 






50 X 10* 




70 X 10* 


Granite 






45 X 10* 




45 X 10* 


Wood (pine 


) (parallel to grain; 
(perpendicular to 


grain) 


10 X 10" 
1 X 10* 






Nylon 






5 X 10" 




Bone (limb 






15 X 10* 


SO x 10* 




Liquids 


Water 










2.0 X 10* 


Alcohol (ethyl) 








1.0 X 10" 


Mercury 










2.5 X 10* 


Gases' 


Air, H,, He 


co 2 








1.01 x 10 s 


' At normal atmospheric pressure; no variation 


ti tempcraiurc during process. 





Young's modulus 



If we compare rods made of the same material but of different lengths and 
cross-section a 1 areas, it is found thiit far the same applied force, the amount of 
stretch (again assumed small compared to the total length) is proportional to 
the original length and inversely proportional to the cross-sectional area. That 
is, the lungeT the object, the more it elongates for a given force; and the thicker 
it is, the less it elongates. These findings can be combined with Eq. 9-3 to yield 

where L Q is the original length of the object, A is the cross-sectional area, 
and A£. is the change in length due to the applied force F. E is a constant of 
proportionality' known as the elastic modulus, or Young's modulus: its value 
depends only on the material. The value of Young's modulus for various mate- 
rials is given in Table 9-1 (the shear modulus and bulk modulus in this Table 
are discussed later in this Section). Because £ is a property only of the material 
and is independent of the object's size or shape, Eq. 9-4 is far more useful for 
practical calculation than Eq. 9-3. 



EXAMPLE 9-10 



Tension in piano wire. A I.GO-m-long steel piano wire 
has a diameter of 0,20cm. How great is the tension in the wire if it stretches 
0.25 cm when tightened? 

APPROACH We assume Hooke's law holds, and use it in the form of Eq. 9-4, 
finding E for steel in Table 9-1. 

SOLUTION We solve for F in Eq. 9-4 and note that the area of the wire is 
A = nr 2 = (3.l4)(0XI010m} 2 = 3J4 X 10~"m\ Then 

0.0025 m \ 



1 ■: 



(2.0 X 10"N/m z )( „ )(3.14 x 10" ft m 2 ) = 980 N. 
\ 1 .60 m / 



1.60 m 
The large tension in all the wires in a piano must be supported by a strong frame 

''llic fact that E is in the denominator, so l/'E is On; actual proportionality coos tan t. is merely a 
convention. When we rewrite Lq, 9-4 to get Lq. 9-5, E is found in the numerator. 



238 CHAPTER 9 Static Equilibrium; Elasticity and Fracture 



* Stress and Strain 

Let us return to solid objects. From Eq. 9-4, we see that the change in length 
of an object is directly proportional to the product of the object's length /.„ 
and the force per unit area F/A applied to it. It is general practice to define 
the force per unit area as the stress; 

force F 
area A 



stress — 



strain = 



which has ST units of N/rrr. Also, the strain is defined to be the ratio of the 
change in length to the original length: 

change in length i£, 

original length L n 

and is dimensionlcss (no units). Strain is thus the fractional change in length of 
the object, and is a measure of how much the rod has been deformed. Stress is 
applied to the material by external agents, whereas strain is the material's 
response to the stress, Equation 9-4 can be rewritten as 



Stress defined 



Strain defined 



— = E- 



A£. 



or 



I., 
F/A 



<»-S> 



stress 



Young's modulus 



A /.//.() strain 

Thus we see that the strain is directly proportional to the stress, in the linear 
(elastic) region of Fig. 9-19. 

* Tension, Compression, and Shear Stress 

The Tod shown in Fig. 9-20a is said to be undeT tension or tensile stress. Not only Tension and compression 

is there a force pulling down on the rod at its lower end, but since the Tod is in 

equilibrium we know that the support at the top is exerting an equal' upward p 

force on the rod at its upper end. Fig, 9-20a, In fact, this tensile stress exists 

throughout the material. Consider, for example, the loweT half of a suspended 

rod as shown in Fig, 9-20b. This lower half is in equilibrium, so there must be an 

Upward force on it to balance the downward force at its lower end. What exerts 

this upward force? It must be the upper part of the rod. Thus we see that 

external forces applied to an object give rise to internal forces, or stress, within 

the material itself. (Recall also the discussion of tension in a cord, page S6.) 

Strain or deformation due to tensile stress is but one type of stress to which 
materials can be subjected. There are two other common types of stress: compres- 
sive and shear. Compressive stress is the exact opposite of tensile stress. Instead of 
being stretched, the material is compressed: the forces act inwardly on the object. 
Columns that support a weight, such as the columns of a Greek temple (Fig. 9-21), 
arc subjected to compressive stress, Equations 9-4 and 9-5 apply equally well to 
compression and tension, and the values for the modulus E are usually the same. 

; If the weight of The rod can be ignored compared, to F. 



V 



T t 



l 
(a) 

FIGURE 9-20 
the ma tonal - 



F 
(b) 

Stress; exists -within 




FIGURE 9-21 This Greek" temple, in Agrigento, 
Sicily, built 2500 years ago. shows the- post-and- 
heam construction. The columns are under 
compression. 



*SECTION 9-5 Elasticity; Stress and Strain 239 



FIGURE 9-22 The three types of 
stress for rigid ohjects. 



She 



Mii-itr Mtiiiiilttx 



FIGURE 9-23 

The fatter book (a) shifts 

more than the thinner book (t>) 

with the same applied shear force. 



Bulk modulus defined 






-f t 


-Al 


+ 




\ 




1 

1 

.' 

1 

1 

1 


/I 

', ^ 

s 

i 



Compression 

00 



Shear 
(c) 



Figure 9-22 compares tensile and compressive stresses as well as the third type, 
shear stress. An object under shear slress has equal and opposite forces applied 
across its opposite faces. A simple example is a book or brick firmly attached to a 
tabletop, on which a force is exerted parallel to the top surface. The table exerts an 
equal and opposite force along the bottom surface Although the dimensions of the 
object do not change significantly, the shape of the object does change, Fig. 9-22c 
An equation similar to Eq. 9-4 can be applied to calculate shear strain: 

1 F 



AL = 



G A 

but. &L, /.(), aud A must be reinterpreted as indicated in Fig. 9- 22c. Note that A 
is the area of the surface parallel to the applied force (and not perpendicular as 
foT tension and compression), and A J. is perpendicular to /. (1 . The constant of 
proportionality G is called the shear modulus and is generally one-half to one- 
third the value of Young's modulus E (see Table 9-1). Figure 9-23 illustrates 
why AL oc L n : the fatter book shifts more for the same shearing force. 




(a) I 



Volu me Change — Bulk Modulus 

If an object is subjected to inward forces from all sides, its volume will decrease. A 
common situation is an object submerged in a fluid; in this case, the fluid exerts a 
pressure on the object in all directions, as we shall see in ChapteT 10. Pressure is 
defined as force per unit area, and thus is the equivalent of stress. For this situation 
the change in volume, AF, is proportional to the original volume, Fi , and to the 
change in the pressure, AP. We thus obtain a relation of the same form as Eq. 9-4 
but with a proportionality constant called the bulk modulus ft: 

AV I 

— ~-AP (9-7) 



C| 



B = 



b 



AP 



AV7K, 
The minus sign means the volume decreases with an increase in pressure. 



240 CHAPTER 9 Static Equilibrium; Elasticity and Fracture 



Values for the bulk modulus are given in Table 9-1. Since liquids and gases 
do not have a fixed shape, only the bulk modulus (not the Young's or shear 
moduli) applies to them. 



Fracture 



It" the stress on a solid object is too great, the object fractures, or breaks 
(Pig. 9-24). Table 9-2 lists the ultimate strengths for tension, compression, and 
shear for a variety of materials. These values give the maximum force per unit 
area, or stress, that an object can withstand under each of these three types of 
stress foT various types of material. They are, howeveT, representative values 
only, and the actual value for a given specimen can differ considerably, Tt is 
therefore necessary to maintain a safety factor of from 3 to perhaps 10 or 
more — that is, the actual stresses on a structure should not exceed one-tenth to 
one-third of the values given in the Table. You may encounter tables of "allow- 
able stresses"' in which appropriate safety factors have already been included. 



Tensilon 



f t 




Shear 



Compression 



FIGURE 9-24 Fracture as a result 
of the three types of stress. 



TABLE 9-2 


Ultimate Strengths 


of Materials (force/area 1 




Material 






Tensile Strength 

(N/m 2 > 


Compressive 

Strength 


Shear Strength 

(N/m 2 ) 


Iron, east 






170 x 10 6 


55(i x I0 tt 


170 x 10* 


Steel 






500 x ltf 5 


500 X 10 6 


250 x 10* 


Brass 






250 x 10 6 


250 x tfl 6 


200 x 10* 


Aluminum 






200 x 10" 


200 X 10 h 


200 x 10" 


Concrete 






2 x LO 6 


20 x in 6 


2 x 10* 


Brick 








35 X 10" 




Marble 








80 X I0 A 




Granite 








170 X 10* 




Wood (pine) (paralle 
(perpen 


to grain) 
dicular to grain) 


40 X ltf 5 


35 X I0 6 

io x io h 


5 x 10* 


Nylon 






500 x 10 6 






Bone (limb) 






130 X 10" 


170 X ID 6 





EXAMPLE 9-11 



Breaking the piano wire. The steel piano wire we 
discussed in Example 9- If) was l.fifJm long with a diameter of 0.20 em. Approx- 
imately what tension force would break it? 

APPROACH We set the tensile stress F/A equal to the tensile strength of 
steel given in Table 9-2, 

SOLUTION The area of the wire is A = irr 2 , where r = 0.10 cm = 
1.0 x 10 'm.Then 

F 

— = 500 x l0 6 N/m- 
A 

so the wire would likely break if the force exceeded 

F = (500 x 10 A N/m 2 )(7r)(l.O X lO'-'m) 3 = 1600 N. 



As can be seen in Table 9-2, concrete (like stone and brick) is reasonably 
strung under compression but extremely weak under tension. Thus concrete 
can be used as vertical columns placed under compression, but is of little 
value as a beam because it cannot withstand the tensile forces that result 
from the inevitable sagging of the lower edge of a beam (see Fig. 9-25). 



FIGURE 9-25 A beam sags, at least 

a little (but is exaggerated here), even 
under its own weigh!. The beam thus 
changes shapes the upper edge is 
compressed, and the lower edge is 
under tension (elongated), Shearing 
stress also occurs within the beam. 

Compression 



TfiiMoil 



'SECTION 9-6 Fracture 241 



FIGURE 9-26 Steel rods around 
which concrete will he poured to 
form a new highway. 




f 



■m. 



v-./v 



PHYSICS APPLIED 

Reinforced concrete 
and 

prestressed concrete 



PHYSICS APPLIED 

A tragic collapse 



Reinforced concrete, in which iron rods arc embedded in the concrete 
(Fig. ^-26), is much stronger. But the concrete on the lower edge of a loaded 
beam still tends to crack because it is weak under tension. This problem is 
solved with prestressed concrete, which also contains iron rods or a wire mesh, 
but during the pouring of the concrete, the rods or wire are held under tension. 
After the concrete dries, the tension on the iron is released, putting the 
concrete under compression. The amount of compressive stress is carefully 
predetermined so that when loads arc applied to the beam, they reduce the 
compression on the lower edge, but never put the concrete into tension. 



CONCEPTUAL EXAMPLE 9-12 | A tragic substitution. Two walkways. 



one above the other, arc suspended from vertical rods attached to the ceiling 
of a high hotel lobby. Fig. 9-27a.The original design called for single rods I4m 
long, but when Such long rods proved to be unwieldy to install, it was decided 
to replace each long rod with two shorter ones as shown schematically in 
Fig, 9-27b. Determine the net force exerted by the rods on the supporting pin 
A (assumed to be the same size) for each design. Assume each vertical rod 
supports a mass m of each bridge. 



FIGURE 9-27 Example 9-12. 






r 

■ 



(a) 



(hi 



mg 



(c) Force on pin A exerted 
bv vertical rod 



■■■":; 

(d) Forces on pins at A exerted 
by vertical rods 



242 CHAPTER 9 Static Equilibrium; Elasticity and Fracture 



RESPONSE The single long vertical rod in Fig. 9 -27a exerts an upward force 
equal to mg on pin A to support the mass m of the upper bridge- Why? Recause 
the pin is in equilibrium, and the other force that balances this is the downward 
force mg exerted on it by the upper bridge (Fig, 9-27c). There is thus a shear 
stress on the pin because the rod pulls up on one end of the pin, and the bridge 
pulls down on the olheT end- The situation when two shoTteT rods support the 
bridges (Fig. 9-27b) is shown in Fig. 9-27d, in which only the connections at 
the upper bridge are shown, The loweT rod exerts a force of mg downward on 
the loweT of the two pins because it supports the loweT bridge. The upper rod 
exerts a force of 2mg on the upper pin (pin A) because the upper rod supports 
both bridges. Thus we see that when the builders substituted two shorter rods 
for each single long one, the stress in the supporting pin A was doubled. What 
perhaps seemed like a simple substitution did, in fact, lead to a tragic collapse 
in 1981 with a loss of life of over 100 people (see Fig. 9-1). Having a feel for 
physics, and heing able to make simple calculations based on physics, can have 
a great effect, literally, on people's lives. 



Spanning a Space: Arches and Domes 

There arc a great many areas where the arts and humanities overlap the 
sciences, and this is especially clear in architecture, where the forces in the mate- 
rials that make up a structure need to be understood to avoid excessive 
deformation and collapse, Many of the features we admire in the architecture of 
the past were introduced not simply tor their decorative effect, but for technical 
reasons, One example is the development of methods to span a space, from the 
simple beam to arches and domes, 

The first important architectural invention was the post-and-beam (or post- 
and-lintel) construction, in which two upright posts support a horizontal beam. 
Before steel was introduced in the nineteenth century, the length of a beam was 
quite limited because the strongest building materials were then stone and 
brick. Hence the width of a span was limited by the size of available stones. 
Equally important, stone and brick, though strong under compression — are 
very weak under tension and shear; all three types of stress occur in a beam 
(see Fig. 9-25), The minimal space that could be spanned using stone is shown 
by the closely spaced columns of the great Greek temples (Fig. 9-21), 

The introduction of the semicircular arch by the Romans (Fig. 9-28), 
aside from its aesthetic appeal, was a tremendous technological innovation. 



Physics APPLIED 



Architecture: 

and domes 



. arcke 




FIGURE 9-28 (a) 2000 year-old round 
arches in Rome. The one in the background is 
the Arch of Titus, (b) A modem arch used to 

span a chasm on the California coast, 



'SECTION 9-7 Spanning a Space: Arches and Domes 243 




FIGURE 9-29 Stones in a round 

(or "true") arch art mainly under 
compression. 

FIGURE 9-30 Flying buttresses 
(on the cathedral of Notre Dame, in 

Paris). 




The advantage of the "true" or round (semicircular) arch is that, if well designed, 
its wedge-shaped Stones experience stress which is mainly compressive (Fig, 9-29) 
even when supporting a large load such as the wall and roof of a cathedral. 
A round arch consisting of many well -shaped stones could span a very wide 
space. However, considerable buttressing on the sides was needed to support 
the horizontal components of the forces, which we discuss shortly. 

The pointed arch came into use about a.d. 1 100 and became the hallmark 
of the great Gothic cathedrals. Tt too was an important technical innovation, and 
was first used to support heavy loads such as the tower of a cathedral, and as 
the central arch. Because of the steepness of the pointed arch, the forces due to 
the weight above could be brought down more nearly vertically, so less hori- 
zontal buttressing would be needed. The pointed arch reduced the load on the 
walls, so there could be more openness and light. The smaller buttressing 
needed was provided on the outside by graceful flying buttresses (Fig. 9-30). 

The technical innovation of the pointed arch was achieved not through 
calculation but through experience and intuition; it was not until much later that 
detailed calculations, such as those presented earlier in this Chapter, came into 
use. To make an accurate analysis of a stone arch is quite difficult in practice, 
But if wc make some simplifying assumptions, wc can show why the horizontal 
component of the force at the base is less for a pointed arch than for a round 
one. Figure 9-31 shows a round arch and a pointed arch, each with an 8.0-rn 
span. The height of the round arch is thus 4,0 m, whereas that of the pointed 
arch is larger and has been chosen to be 8.0 tn. Each arch supports a weight of 
12.0 X 10 4 N (a 12,000 kg X g), which, for simplicity, wc have divided into two 
parts (each 6.0 x 10 4 N) acting on the two halves of each arch as shown. For the 
arch to be in equilibrium, each of the supports must exert an upward force of 
6,0 x 10 4 N, Each support also exerts a horizontal force, F H , at the base of the 
arch, and it is this we want to calculate. We focus only on the right half of each 
arch. We set equal to zero the total torque calculated about the apex of the arch 
due to the forces exerted on that half arch, as if there weTe a hinge at the apex. 
For the round arch, the torque equation (St - 0) is 

(4,0mp x I0 4 N) - (2.0m)(6.0 x 10 4 N) - (4,0 m)(f„) - 0, 

Thus F H = 3.0 x I0 4 N for the round arch. For the pointed arch, the torque 
equation is 

(4.0m}(6.0 X I0 4 N) - (2.0m)(6.0 X 10 4 N) - (S,0m)(F H ) = 0. 

Solving, wc find that F H - 1. 5 x 10 4 N — only half as much as for the round 
arch! From this calculation we can see that the horizontal buttressing force 
required for a pointed arch is less because the arch is higher, and there is there- 
fore a longer lever arm for this force. Indeed, the steeper the arch, the less the 
horizontal component of the force needs to be, and hence the more nearly 
vertical is the force exerted at the base of the arch. 



2.0 m 



FIGU RE 9-31 Forces in (a) a 
round arch, compared with those 
in (b) a pointed arch. 



6.0 X 10 4 N 



2.0 m 

CI 



6,0 X IQ 4 N 



/ / 4,0 m \\ 



6.0 x 10 4 N 



f v = fi.flx I0 4 NJ 



I 8.0 in 1 

(a) 




6.0 x 10 4 N 



■* ^u 



24fl CHAPTER 9 Static Equilibrium; Elasticity and Fracture 



Whereas an arch spans a two-dimensional space, a dome — which is 
basically an arch Totaled aboul a vertical axis — spans a three-dimensional 
space. The Romans buill the first large domes. Their shape was hemispherical 
and some still stand, such as that of the Pantheon in Rome (Fig, 9-32), built 
2000 years ago. 

Fourteen centuries later, a new cathedral was being built in Florence. Tt was 
to have a dome 43 m in diameter to rival that of the Pantheon, whose construc- 
tion has remained a mystery, The new dome was to rest on a "drum" with no 
external abutments. Filippo Brunelleschi (1377-1446) designed a pointed dome 
(Fig. 9-33), since a pointed dome, like a pointed arch, exerts a smaller side thrust 
against its base, A dome, like an arch, is not stable until all the stones are in place. 
To support smaller domes during construction, wooden frameworks were used. 
But no trees big enough or strong enough could be found to span the 43-m space 
required. Brunei leschi decided to try to build the dome in horizontal layers, each 
bonded to the previous one, holding it in place until the last stone of the circle 
was placed. Each closed ring was then strong enough to support the next layer. Tt 
was an amazing feat. Only in the twentieth century were larger domes huilt, the 
largest being that of the Supcrdomc in New Orleans, completed in 1975. 



FIGURE 9-33 The skyline of 
Florence, showing Brunelleschi 's 
dome on the cathedral, 





ws* ? 




FIGU RE 9-32 Interior of the 
Pantheon in Rome, buill in Ihe first 
century, This view, showing the great 
dome and its central opening for light, 
was painted about 1740 by Panini, 
Photographs do not capture its 
grandeur as well as Ihis painting does, 



EXAMPLE 9-13 



A modern dome. The 1.2 x |(rkg dome of the Small 
Sports Palace in Rome (Fig, 9-34a) is supported by 3b buttresses positioned at 
a 38 a angle so that they connect smoothly with the dome. Calculate the 
components of the force, F H and /y , that each buttress exerts on the dome so 
that the force acts purely in compression — that is, at a 38° angle (Fig. 9-34b). 

APPROACH We can find Ihe vertical component F v exerted upward by each 
buflTess because each supports jg of the domes weight. We find F H knowing thai 

the buttress needs to he under compression so F = F v + F H acts at a 38 <J angle. 

SOLUTION The vertical load on each buttress is j^ of the total weight. Thus 
mg (1.2 X lO^kgXQ.gm/s 2 ) 
36 " 



!-, 



36 



= 330,000 N. 



The force must act at a 38° angle at the base of the dome in order to be purely 
compressive. Thus 



tan 38° = 



F 



Fn 



/■,: 



F v 



tan 38" 



330,000 N 

tan 38' 



42.0.000 N, 



NOTE For each buttress to exert this 420,000-N horizontal force, a 

prestressed-concrete tension ring surrounds the base of the buttresses beneath 
the ground (see Problem 56 and Fig. 9-70). 



FIGURE 9-34 Example 9-13. 

(a) The dome of the Small Sports 
Palate in Rome, built for the i960 
Olympics, (b) The force components 
each buttress exerts on the dome. 




in) 




(b) 



*SECTION 9-7 Spanning a Space: Arches and Domes 245 



I Summary 



An object at rest is said to he in equilibrium. The subject 
concerned with the determination of the forces within a struc- 
ture at rest is called statics. 

The two necessary conditions for an object to he in equi- 
librium are that ( I) the vector sum of all the forces on it must 
be zero, and (2) the sum of all the torques (calculated about 
any arbitrary axis) must also be 7.ero: 

EF, = 0, £F V = 0, 2r = 0. (9-1,9-2) 

It is important when doing statics problems to apply the equi- 
librium conditions to only one object at a time. 

[*An object in static equilibrium is said to be in 
U'l) si ah le. (ft) unstable, or (t) neutral equilibrium, depending 
on whether a slight displacement leads to (a) a return to the 
original position, (ft) further movement away from the orig- 
inal position, or (c) Test in the new position. An object in 
stable equilibrium is also said to be in balance. | 

|* Hookers law applies to many elastic solids and states 
that the change in length of an object is proportional to the 



applied force: 



F = k A/.. 



(9 1) 



If the force is too great, the object will exceed its elastic limit, 
which means it will no longer return to its original shape 
when the distorting force is removed, If the force is even 
greater, the ultimate strength of the material can he exceeded, 
and the object will fracture. The force per unit area acting on 
an object is called the stress, and the resulting fractional 
change in length is called the strain. The stress on an ohject 
is present within the object and can be of three types: 
compression, tension, or shear. The ratio of stress to strain is 
called the elastic modulus of the material. Young's modulus 
applies for compression and tension, and the shear modulus 
for shear: hulk modulus applies to an object whose volume 
changes as a result of pressure on all sides. All three moduli 
are constants for a given material when distorted within the 
elastic region.] 



| Questions 



1. Describe several situations in which an object is not in 
equilibrium, even though the net force on it is zero. 

2. A bungee jumper momentarily comes to rest at the 
bottom of the dive before he springs back upward. At that 
moment, is the bungee jumper in equilibrium? Explain. 

i. You can find the center of gravity of a meter stick by 
resting it horizontally on your two index fingers, and then 
slowly drawing your fingers together. First the meter stick 
will slip on one finger, and then on the other, but eventu- 
ally the fingers meet at the <x\. Why does this work? 

4. Your doctor's scale has arms on 
which weights slide to counter your 
weight. Fig. 9-35. These weights are 
much lighter than you are. How 
does this work? 



FIGURE 9-35 
Question 4. 



A ground retaining wall is shown in Fig. 9-36a,The ground, 
particularly when wet. can exert a significant force F on the 
wall, (a) What force produces the torque to keep the wall 
Upright? (b) Explain why the retaining wall in Fig. 9-36b 
would be much less likely to overturn lhan that in Kg, 9-36a. 



/% Weights 




6. Explain why touching your toes while you are seated on 
the floor with outstretched legs produces less stress on 
the lower spinal column than when touching your toes 
from a standing position, Use a diagram, 

7. A ladder, leaning against a wall, makes a 60" angle with 
the ground. When is it more likely to slip: when a person 
stands on the ladder near the top or near the bottom? 
Explain, 

S. A uniform meter stick supported at the 25-cm mark is in 
equilibrium when a 1-kg rock is suspended at the 0-cm 
end (as shown in Fig. 9-37). Is the mass of the meter stick 
gTeater than, equal to, or less than the mass of the rock? 
Explain your reasoning. 



— 




^ 



FIGURE 9-37 Question «. 



9. Can the sum of the torques on an object be zero while the 
net force on the object is nonzero? Explain. 

10, Figure 9-38 shows a cone. Explain how to lay it on a flat 
table so that it is in [a) stable equilibrium, (ft) unstable 
equilibrium, (c) neutral equilibrium. 



r^ 



(a) (b) 

FIGURE 9-36 Question 5. 



FIGURE 9-38 Question HI. 



246 CHAPTER 9 Static Equilibrium; Elasticity and Fracture 



II, Which of the configurations of brick, («) or (b) of 
Fig. 9-39, is the more likely to be stable? Why'' 



t 




(a) 



!b) 



FIGURE 9-39 Question 1 1 . The dots indicate the 






CG of each brick. The fractions I and { indicate what 
portion of each brick is hanging beyond its support. 

12. Why do you tend to lean backward when carrying a heavy 
load in your arras? 

13. Place yourself facing the edge of an open door, Position 
your feet astride the door with your nose and abdomen 
touching the door's edge. Try to rise on your tiptoes. Why 
can't this he done? 

14. Why is it not possible to sit upright in a chair and rise to 
your feet without first leaning forward? 



15, Why is it more difficult to do sit-ups when your knees are 
bent than when your legs aTe stretched out? 

16, Name the type of equilibrium for each position of the ball 
in Fig. 9-40. 



A 




FIGURE 9-40 
Question 16. 



* 17, Is the Young's modulus for a bungee cord smaller or 

larger than that for an ordinary rope? 

* 18, Examine how a pair of scissors or shears cuts through a 

piece of cardboard- Is the name "shears" justified? Fx plain. 

* 19. Materials such as ordinary concrete and stone are very 

weak undeT tension or shear. Would it be wise to use such 
a material for either of the supports of the cantilever 
shown in Fig. 9-9? If so. which one(s)? Fxplain. 



Problems 



9-1 and 9-2 Equilibrium 

1. 0) "lliree forces aTe applied lo a uree sapling, as shown in 
Fig. 9-41, to stahilize it. If F A = 310 N and F B = 42.5 N. 
find F c in magnitude and 



direction. 



FIGURE 9-41 
Problem 1, 




2. (I) Calculate the torque about the front support post (R) 

of a diving board, Fig. 9-42, exerted by a 58- kg 



person 3.0 m from that post. 



FIGURE 9-42 

Problems 2, 4. and 6, 



A B 



' 



.1.0 



h -t- 

I.Om 



3. fl) Calculate the mass m needed in order to suspend the 
leg shown in Fig. 9-43. Assume the leg (with cast) has a 
mass of 15.0 kg. and its crj is 35.0cm from the hip joint: 
the sling is 80.5 cm from the hip joint. 



FIGURE 9- 

Problem 3- 



43 



1 Hip join) 



• CG 




4. (I) How faT out on the diving board (Fig. 9-42) would a 
5S-kg diver have to be to exert a torque of 1100 m-N on 
[he board, relative to the left (A) support post? 

5. (II) Two cords support a chandelier in the manner shown 
in Fig. 9-4 except that the upper wire makes an angle of 
45 n with the ceiling. If the cords can sustain a force of 
1550 N without breaking, what is the maximum chandelier 
weight that can be supported? 

f>. (II) Calculate the forces F A and F n that the supports 
exert on the diving boaTd of Fig. 9-42 when a 58-kg 
person stands at its tip, (a) Ignore the weight of the 
board, (b) Take into account the board's mass of 35 kg. 
Assume the board's cc is at its center, 

7. (II) A uniform steel beam has a mass of 940kg, On it is 
resting half of an identical beam, as shown in Fig, 9-44. 
What is the vertical support force at each end? 



iW 



=1 



*M 



FIGURE 9-44 Problem 7. 

8. (II) A 140-kg horizontal beam is supported at each end. A 
320-kg piano rests a quarter of the way from one end. 
What is the vertical force on each of the supports? 

M. (II) A 75-kg adult sits at one end of a 9.0-m-long board. 
His 25-kg child sits on the other end, (a) Where should 
the pivot be placed so that the board is balanced, 
ignoring the board's mass? (b) Find the pivot point if the 
board is uniform and has a mass of 15 kg, 

111. (II) Calculate F A and F & for the uniform cantilever shown 
in Fig. 9-9 whose mass is 1200 kg. 



Problems 247 



11. (IT) Find the tension in 
Fig. 9-45. Neglect the mass 
of the cords, and assume 
that the angle 6 is 33" and 
the mass m is 170 kg. 

FIGURE 9-45 

Problem I I 

12. (II) Kind the tension in 
the two wires supporting 
the traffic light shown in 
Eg. 9-46. 



FIGURE 9-46 
Problem 12. 



the two cords shown in 



\h. 



14. 




13. (II) How close to the edge of the 20.0-kg table shown in 
Fig, 4-47 can a 66,0-kg person sit without tipping it Liver? 



2 20 !, 



on ill 




0.50 m 

FIGURE 9-47 Problem 13. 

(II) A 0.60-kg sheet hangs from a massless clothesline as 
shown in Fig, 9-48. 'lhe clothesline on either side of the 
sheet makes an angle of 3.5° with the horizontal. Calculate 
the tension in the clothesline on either side of the sheet. Why 
is the tension so much greater than the weight of the sheet? 




FIGURE 9-48 Problem 14. 

15. (II) Calculate F A and F^ for the beam shown in Fig, 9-49. 
The downward forces represent the weights of machinery 
on the beam. Assume tho beam is uniform and has a mass 
of 250 kg. 

F A 4300 N 31(H)N 220(1 \ F n 



FIGURE 9-49 

Problem 1.5. 



I i I II 



ff 



-4" — 4.0 m- 



2.0 m 



-3.0 m -~-|>| 
1.0 m 



17, 



(II) Three children are trying to balance on a seesaw, 
which consists of a fulcrum rock, acting as a pivot at the 
center, and a very light board 3.6 m long (Fig. 9-50). Two 
playmates are already on either end- Boy A has a mass of 50 kg. 
and girl B a mass of 35 kg, Where should girl C, whose 
mass is 25 kg, place herself so as to balance the seesaw? 



* 



iji = 50 kg 



~~ 



m = 35 kg 



m = 25 kg 

FIGURE 9-50 Problem 16. 

(II) Figure 9-51 shows a pair of foTceps used to hold a 
thin plastic rod firmly If each finger squeezes with a force 

fy = Fr = 11.0 N. what force do the forceps jaws exert 
on the plastic rod? 



Rod 




>.70 l-iil FIGURE 9-51 
PToblem 17. 



IS. (II) Calculate (a) the tension F T in the wire that supports 
the 27-kg beam shown in Fig, 9-52, and (ft) the force F w 
exerted by the wall on the beam (give magnitude and 

direction). 




FIGURE 9-52 

Problem 18. 



19. (II) A 172-cm-tall person lies on a light (massless) board 
which is supported by two scales, one under the top of heT 
head and one heneath the bottom of her feet (Fig. 9-53). 
The two scales read, respectively, 35,1 and 31.6 kg. What 
distance is the center of gravity of this person from the 
bottom of her feet? 




[mrajkiB 



FIGURE 9-53 Problem 19. 



248 CHAPTER 9 Static Equilibrium; Elasticity and Fracture 



2tt. (II) A shop sign weighing 245 N is supported by a 
uniform 155-N beam as shown in Fig. 9-54, Find the 
tension in the guy wire and the horizontal and vertical 
fortes exerted by the hinge on the beam. 



FIGURE 9-54 
Problem 20. 




ZL 



(II) A traffic light hangs from a pole as shown in 
Fig. 9-55 The uniform aluminum pule AR is 7.50 m long 
and has a mass of 12.0 kg, The mass of the traffic light is 
21.5 kg, Determine (a) the tension in the horizontal mass- 
less cable CD. and (b) the verticil I and horizontal compo- 
nents of the force everted by the pivot A on the 
aluminum pole. 



380 m 




FIGURE 9-55 Problem 21. 

21. (II) The 72-kg-man 1 s hands in Fig. 9-56 are 36 cm apart- 
His CG is located IS^'A-. of the distance from his right hand 
toward his left. Find the force on each hand due 
to the ground. 



23. (II) A uniform meter stick with a mass of 180 g is supported 
horizontally by two vertical strings, one at the 0-cm mark 
and the other at the 90-cm mark (Fig. 9-57). What is the 
tension in the string (a) at U cm? (b) al 9U cm'? 



T-Tn E 1 X""""«"" , T" , "i;"T''*'P 



FIGURE 9-57 
Problem 23. 



24. (II) The two trees in Fig. 9-58 are 7.6 m apart. A back- 
packer is trying to lift his pack out of the reach of bears. 
Calculate the magnitude of the force F that he must exert 
downward to hold a 1 9-kg backpack so that the rope sags 
at its midpoint by (a) 1 .5 m, (b) 0.15 m. 




FIGURE 9-58 Problem 24. 

25. (Ill) A door 2.30 m high and 1.30 m wide has a mass of 
13.0 kg. A hinge t).4IJm from the top and another hinge 
0,40 m from the bottom each support hal( the door's 
weight (Fig. 9-59). Assume that the center of gravity is 
al the geometrical center of the door, and determine 
the horizontal and 
vertical force compo- 
nents exerted by 
each hinge on the 
door, 



FIGURE 9-59 
Problem 25, 



in . m 





26. (Ill) A uniform ladder of mass m and length / leans at an 
angle # against a frictionless 
wall. Fig. 9-C0. If the coeffi- 
cient of static friction between 
the ladder and the ground is fi, 
determine a formula for the 
minimum angle at which the 
ladder will not slip. 



FIGURE 9-60 
Problem 26, 




FIGURE 9-56 Problem 22. 



Problems 249 



27. 



(Ill) Consider a ladder with a painter climbing up it 
(Fig. 9-61), If the mass of the ladder is 12.0 kg, the mass of 
the painter is 55.0 kg. and the ladder begins to slip at its 
base when her feet are 7Q% of the way up the length of 
the laddeT, what is the coefficient of static friction between 
the ladder and the floor? Assume the wall is Motionless. 



FIGURE 9-61 

Problem 27. 




- 2.1 m ~ 

28. (Ill) A person wants to push a lamp (mass 7.2kg) across 
the floor, foT which the coefficient of friction is 0,20, 
Calculate the maximum height x above the floor at which 
the person can push the lamp so that it slides rather than 
tips (Fig. 9-62). 



FIGURE 9-62 

Problem 28. 




29. (Ill) Two wires run from the top of a pole 2.6 m tail that 
supports a volleyball net. 'I"he two wires are anchored to 
the ground 2,0 in apart, and each is 2,0 m from the pole 
(Fig, 9-63). The tension in each wire is 95 N. What is the 

tension in the net, assumed horizontal and attached at the 
top of the pole? 




:; 9-3 Muscles and Joints 

*■ 30. (I) Suppose the point of insertion of the biceps muscle into 
the lower arm shown in Fig. 9- 1 3a (Example 9-8) is 6.0 cm 

instead of 5.0 cm: how much mass could the person hold 
with a muscle exertion of 450 IN ? 
* 31. (I) Approximately what magnitude force, Fm- must the 
extensor muscle in the upper arm exert on the lower ami to 
hold a 7.3-kg shot put (Fig. 9-64)? Assume the lower arm has 
a mass of 2.8 kg and its CG is 12 cm from the elbow-joint pivot. 




30.0 cm- 






FIGURE 9-64 

Problem;! I. 



* 32. (II) (a) Calculate the force, Fy , required of the "deltoid" 

muscle to hold up the outstretched arm shown in 
Fig, 9-65, The total mass of the arm is 3,3kg, 
(b) Calculate the magnitude of the force F } exeTted by the 
shoulder joint on the upper arm. 




24 on 



FIGURE 9-65 

Problems 32 and 33. 



* 33, 



34. 



(II) Suppose the hand in Problem 32 holds a 15-kg mass. 
What force, F M , is required of the deltoid muscle, 
assuming the mass is 52 cm from the shoulder joint? 
(II) The Achilles tendon is attached to the rear of the foot 
as shown in Fig. 9-66. When a person elevates himself 
just barely off the floor on the "ball of one foot." estimate 
the tension F v in the Achilles tendon (pulling upward). 
and the (downward) force F B exerted by the lower leg bone 
on the foot. Assume the person has a mass of 72 kg and D 
is twice as long as d. 



Achilles 
lerkkm 



i 



■ Lea hone 



I, 



Ball of fool 



FIGURE 9-66 

Problem 34. 



U-d- 



1> 



FIGURE 9-63 Problem 29. 



* 35. (II) Redo Example 9-9, assuming now that the person is 
less bent over so that the 30° in Fig. 9-1 4b is instead 45". 
What will be the magnitude of Fv on the vertebra? 

9-4 Stability and Balance 

36. (II) The Leaning Tower of Pisa is 55 m tall and about 
7.0 m in diameter. The top is 4.5 m off center. Is the tower 
in stable equilibrium? If so. how much larther can it lean 
before it becomes unstable' Assume the tower is of 
uniform composition. 



250 CHAPTER 9 Static Equilibrium; Elasticity and Fracture 



37. (Ill) Four bricks arc to be stacked at the edge of a table, 
each brick overhanging the one below it, so that the top 
brick extends, as far as possible beyond the edge of the 
tabic- (a) To achieve (his, show that successive bricks must 
extend no more than (starting at the top) 5,3.6, ar| d k oi 
their length beyond the one below (Fig. 9-67a). (ft) Is the 
lop brick completely beyond the base? (c) Determine a 
general formula for the maximum total distance spanned 
by n bricks if [hey are to remain stable, (d) A builder 
wants to construct a corbeled arch (Fig, 9-67b) based on 
the principle of stability discussed in («) and (c) abovc. 
What minimum number of bricks, each 0,30 m long, is 
needed if the arch is to span 1.0 m? 





FIGURE 9-67 Problem 37. 

9-5 Elasticity; Stress and Strain 

* 38, (I) A nylon string on a tennis racket is under a tension of 

275 N. If its diameter is 1.00 mm, by how much is it length- 
ened from its untensioned length of 30.0cm? 

*3*>. (I) A marble column of cross-seelional area 1.2 m" 
supports a mass of 25.OO0 kg, (a) What is the stress within 
the column? (b) What is the strain'? 

* 4tt. (I) By how much is the column in Problem 3s> shortened 

if it is 9.6 m high? 

* 41. (1) A sign (mass 2100 kg) hangs from the end of a vertical 

steel girder with a cross-sectional area of 0.15 m 2 . (a) What 
is the stress within the girder? (ft) What is the strain on the 
girder? (t) If the girder is 9.50 m long, how much is it 
lengthened? (Ignore the mass of the girder itself.) 

s 42, (II) One liter of alcohol ( 1000 cm -1 ) in a flexible container 
is carried to the hottom of the sea, where the pressure is 

2.6 x lO^N/m-. What will be its volume there? 

*43. (II) A 15-cm-long tendon was found to stretch 3.7 mm by 
a force of 13.4 N. The tendon was approximately round 
with an average diameter of 8-5 mm. Calculate the 
Young's modulus of this tendon. 

* 44. (11) How much pressure is needed to compress the 

volume of an iron block by 0,1091? Express your answer 
in N/m , and compare it to atmospheric pressure 
(1.0 X 1 IF N/m 2 ). 

* 45. (II) At depths of 2000 m in the sea, the pressure is about 

200 times atmospheric pressure (I atm = 1,0 x 10' N/m 2 ). 
By what percentage does the interior space of an iron bathy- 
sphere's volume change at this depth? 
*46. (Ill) A scallop forces open its shell with an elastic mate- 
rial called abductin, whose Young's modulus is about 
20 X Kl h N/m 2 . If this piece of abductin is 3.0mm thick 
and has a cross-sectional area of 0.50 em - , how much 
potential energy does it store when compressed 1.0 mm? 



47. (Ill) A pole projects horizontally from the front wall of a 
shop, A 5.1-kg sign hangs from the pole at a point 2.2 m 
from the wall (Fig, 9-68). (a) What is the torque due to 
this sign calculated about the point where the pole meets 
the wall? (ft) If the pole Is not to fall off, there must be 
another torque exerted to balance it. What exerts this 
torque? Use a diagram to show how this torque must act. 
(c) Discuss whether compres- 
sion, tension, and/or shear play 
a role in part (ft). 

H — 2.2m h 



The 



FIGURE 9-6B 
Problem 47. 



: 9-6 Fracture 

* 48. (I) The femur bone in the human leg has a minimum 

effective cross section of about 3.0 cm 1 (=3.0 X 10 *m 2 }. 
How much compressive force can it withstand before 
breaking? 

* 49. (II) (a) What is the maximum tension possible In a 

1 .00 -mm -diameter nylon tennis racket string? (h) If you 
want tighter strings, what do you do to prevent breakage: 
use thinner or thicker strings? Why? What causes strings 
to break when they are hit by the ball? 

* 50. (II) If a compressive force of 3.6 x 10^ N is exerted on 

the end of a 22-cm-long hone of cross-sectional area 
3.6 cm 2 , (w) will the bone break, and (ft) if not. by how- 
much does it shorten? 

* 51. (II) (a) What is the minimum cross-sectional area 

required of a vertical steel cahle from which is suspended 
a 320-kg chandelier? Assume a safety factor of 7.0 (ft) If 
the cable is 7.5 m long, how much does it elongate? 

* 52, (II) Assume the supports of the uniform cantilever shown 

in big. 9-69 (mass = 2600 kg) are made of wood. Calcu- 
late the minimum cross-sectional area required of each, 
assuming a safety factor of S.5. 



!•— 20.0 m— X> 



-30.0 m- 



,u; 



FIGURE 9-69 
Problem 52, 



* 53. (II) An iron bolt is used to connect two iron plates 

together. The bolt must withstand shear forces up to 

about 3200 N, Calculate the minimum diameter for the 
bolt, based on a safety factor of 6,0. 

* 54. (Ill) A steel cable is to support an elevator whose total 

(loaded) mass is not to exceed 3100 kg. If the maximum 
acceleration of the elevator is 1.2 m/s 2 . calculate the 
diameter of cable required. Assume a safety factOT of 7-0. 

: 9-7 Arches and Domes 

:: 55. (II) How high must a pointed arch be if it is to Span a 
space S.Om wide and exert one-third the horizontal force 
at its base that a round arch would? 



Problems 251 



BH. (II) The subterranean tension ring that exerts the 
balancing horizontal force on the abutments for the dome 
in Fig. 9-34 is 36-sided, so each segment makes a 10° 
angle with the adjacent one (big. 9-70). Calculate the 
tension F that must exist in each segment so that the 
required foree of 4.2 x 10 5 N can be exerted at each 
comer (Example 9—13). 




420,000 N 



FIGURE 9-70 

Problem 56. 



| General Problems 



57. The mobile in Fig. 9-71 is in equilibrium. Object R has 
mass of 0.885 kg. Determine the masses of objects A, C 
and D. (Neglect the weights of the crossbars.) 




FIGURE 9-71 Problem57. 



5X. A tightly stretched "high wire" is 46 m long It sags 2.2 m 
when a 60,0-kg tightrope walker stands at its center. What 
is the tension in the wire? Ts it possible to increase the 
tension in the wire so that there is no sag? 

59. What minimum hori7.ontal force F is needed to pull a 
wheel of radius R and mass M over a step of height It as 
shown in Fig. 9-72 (R > h}7 (a) Assume the force is 
applied at the top edge as shown, (h) Assume the force is 
applied instead at the wheel's center. 



*- F (in a) 




FIGURE 9-72 Problem 59. 



nil. 



A 25-kg round table is supported by thTee legs equal 
distances apart on the edge. What minimum mass, placed 
on the table's edge, will cause the table to overturn? 

When a wood shelf of mass 5-0 kg is fastened inside a slot 
in a vertical support as shown in Fig. 9-73, the support 
exerts a torque on the shelf, (a) Draw a free -body 
diagram for the shelf, assuming three vertical forces (two 
exerted by the support slot — explain why). Then calculate 
(b) the magnitudes of the three forces and (<.-) the torque 
exerted by the support (about the left end of the shelf). 



(.2. 



32.0 em - 



3.0 cm 



H —2,0 cm I 

FIGURE 9-73 Problem 61. 

A 50-story building is being planned. It is to be 200.0 m high 
with a base 40.0m by 70.0m. Its total mass will be about 
L.8 x ]0 7 kg. and its weight therefore about 1.8 x 10 s N. 
Suppose a 200-km/h wind exerts a force of 950 N/m~ over 
the 70.0-m-wide face (Fig. 9-74). Calculate the torque about 
the potential pivot point, the rear edge of the building 
(where F H acts in Eg. 9-74), and determine whether the 
building will topple, Assume the total force o( the wind acts 
at the midpoint of the building's face, and that the building 
is not anchored in bedrock, [flint: Fk i fl F'8- 9-74 represents 
the force that the Earth would exert on the building in the 
case where the huilding would just lx;gin to tip] 



en e^ 

en en 
|= rr 
ei e - 
en c - ; 
en n 

*rn ci 
cr. r-. 
ee CT 



en n en m 



□ nnn 



□nanun 



en ce 
ee c^ . 
a crt I 
en en i 
u n [ 



□ nun 



u n u n 



. nun 

I en m en m 
I en m en m 

. nun 

I en m en m 
innnn 
nnnn 
I en n tn en 




FIGURE 9-74 b'orces 
on a building subjected 
to wind (F A ). gravity (wig), 
and the (orce F^ on 
the building due to the 
Farth if the building 
were just about to tip, 
Problem 62. 



252 CHAPTER 9 Static Equilibrium; Elasticity and Fracture 



63. The center of gravity of a loaded truck depends on how 
the truck is packed. If it is 4,0 m high and 2,4 m wide, and 
its CG is 2.2 m 
above- [he- ground, 
how steep a slope 

be parted on 
without. tipping 
over (Fig. 9-75)'? 




FIGURE 9-75 
Problem 63. 



64. In Fig. 9-76. consider the right-hand (northernmost) 
section of the Golden Gate Bridge, which has a length 
d\ = 343 m. Assume the CG of this spun is halfway 
between the tower and anchor. Determine F M and F V2 
(which act on the northernmost cable) in terms of mg, the 
weigh! of the northernmost span, and calculate (he lower 
height h needed for equilibrium. Assume the roadway is 
supported only by the suspension cables, and neglect (he 
mass of the cables and vertical wires. \Hint: Fp does not 
act on this section.] 




FIGURE 9-76 Problem 61 

65. When a mass of 25 kg is hung from the middle of a fixed 
straight aluminum wire, the- wire sags to make an angle of 
12" with the horizontal as shown in Fig. 9-77. Determine 

the radius of the wire. 



12 



FIGURE 9-77 
Problem 65. 



I 



25 k 



fci 



66. The forces acting on a 67,000-kg aircraft flying at constant 
velocity are shown in Fig. 9-78. The engine thrust. 
F T = 5.0 X 10 s N, acts on a line 1.6 m below the CM. 
Determine the drag force Fn and the distance above the 
CM that it acts. Assume Fp and F-|- are horizontal. 



FIGURE 9-73 
Problem 66. 




67. A uniform flexible Steel cable of weight mg is suspended 
between two points at the same elevation as shown in 
Fig. 9-79, where B = 60°. Determine the tension in the 
cable (a) at its lowest point, and (b) al (he points of 
attachment, (e) What is the direction of the tension force 
in each case? 



;>- 



a 



FIGURE 9-79 
Problem 67 

68. A 20-U-m-long uniform beam weighing 55U N rests on walls 
A and B, as shown in Fig. 9-80, (a) Find the maximum 
weight of a person who can walk to the extreme end D 
without tipping the beam. Find the forces (hat the walls A 
and B exert on the beam when the person is standing: 
(h) at D: (c) at a point 2.0m to the right of B; (rf) 2.0m to 
the right of A. 



- 

I- 



20.0 in 



B 



D 



h 3.0 m - 



12.0 m 



69. 



FIGURE 9-80 Problem 68. 

A cube of side- I rests on a rough floor. It is subjected lo a 
steady horizontal pull F, exerted a distance h above (lie floor 
as shown in Fig 9-S1, As F is increased, the block will either 
begin to slide, or begin to tip over. Determine (he coefficient 

of static friction m so that {a) the block begins to slide rather 



than tip; (b) the block begins to 
tip, {Hint. Where will the normal 
force on the block act if it tips?] 

FIGURE 9-31 
Problem 69. 



I 



70. 



A 60.0-kg painter is on a uniform 25-kg scaffold 
supported front above by ropes (Fig. 9-S2). There is a 
4,0-kg pail of paint to 
one side, as shown, Can 
the painter walk safely to 
both ends of the scaf- 
fold? If not. which end(s) 
is dangerous, and how 
close to the end can he 
approach safely? 

FIGURE 9-82 
Problem 70. 




1.0 m 



4.0 m 



■T ' 



1.0m 



71. A woman holds a 2.0-m-long uniform ll).0-kg pole as shown 
in Fig, 9-83. (a) Determine the forces she must exert with 
each hand (magnitude and direc- 
tion) To what position should 
she move her left hand so that 
neither hand has to exert a force 
greater than (b) 150N? (<:) 85 V? 






FIGURE 9-83 
Problem 71 . 



30cm H 



General Problems 253 



72. 



A man doing push-ups pauses in the position shown in 
Fig. 9-84. His mass m = 75 kg. Determine the normal 
force exerted by the floor (a) on each hand: (/>) on each foot. 




r*-40em-M— 



95 cm 



Fl G U Ft E 9- 84 Probl em 72 . 

73. A 20-kg sphere rests between two smooth planes as 
shown in Fig. 9-85- Determine the magnitude of the foTce 
acting on the sphere exerted by each plane. 




FIGURE 9-85 Problem 73. 



* 75, Parachutists whose chutes have failed to open have been 

known to survive if they land in deep snow, Assume that a 
75-kg parachutist hits the ground with an area of impact 
of 0.30 m 2 at a velocity of 60m/s, and that the ultimate 
strength of body tissue is 5 X 10 s N/nr, Assume that the 
person is brought to rest in 1.0 m of snow. Show that the 
person may escape serious injury. 

* 76, A steel wire 2.0 mm in diameter stretches by (J.03tr% 

when a mass is suspended from it. How laTge is the mass? 

* 77. In Example 7-6 in Chapter 7, we calculated the impulse 

and average force on the leg of a person who jumps 
3.0 m down to the ground. If the legs aTe not bene upon 
landing, so that the body moves a distance d of only 
1 ,0 cm during collision, determine (a) the stress in the 
tibia (a lower leg bone of area = 3.0 X 10 _4 m 2 ), and 
(t>) whether or nol the bone will break- (c) Repeal for a 
bent-knees landing (ti = 50.0 cm). 

* 78. The roof over a 7.0-m x I0,0-m room in a school has a 

total mass of 12.600 kg. The roof is to be supported by 
vertical "2 X 4s" (actually alxiut 4.0 cm x 9.0 cm) along 
the 100-m sides. How many supports are required on 
each side, and how far apart must they he? Consider only 
compression, and assume a safety factor of 12. 

* 79. A 25-kg object is being lifted by pulling on the ends of a 

1 .UO-mm-diameter nylon string that goes over two 
3,00-m-high poles that are 4.0 m apart, as shown in Fig. 9-87, 
How high above the floor will the object be when the 
String breaks? 



74. A 2200-kg trailer is attached to a stationary truck at 
point B. Fig. 9-86, Determine the normal force exerted by 
the road on Che rear tires at A, and the vertical force exerted 
on the trailer by the support B. 





FIGURE 9-86 Problem 74, 



FIGURE 9-87 Problem 79 

* 80. There is a maximum height of a uniform vertical 
column made of any material that can support itself 
without buckling, and il is independent of the cross- 
sectional area (why?). Calculate this height for 
(<i) steel (density = mass/ volume = 7,8 x I U J kg/m 3 ). and 
(b) granite (density = 2.7 X 10 3 kg/m 3 ). 



Answers to Exercises 



A: F A also has a component to balance the sideways 

force F B . 
B: Yes: sin (t appears on both sides and cancels out- 
er F s = m A g + m h g + Mg = 560 N. 



D: Static friction at the cement floor (= F Cl ) is crucial, or else the 

ladder would slip. At the top. the I udder can move and 
adjust, so we wouldn't expect a strong static friction force 
there. 



25fl CHAPTER 9 Static Equilibrium; Elasticity and Fracture 




Scuba divers and sea creatures 
under water experience a buoyant 
force (F B ) that almost exactly 
balances their weight »ig. The 
buoyant force is equal to the weight 
of the volume of fluid displaced 
(Archimedes' principle) and arises 
because the pressure increases with 
depth in the fluid. Sea creatures 
have a density very close to that of 
water, so their weight very nearly 
equals the buoyant force Humans 
have a density slightly less than 
water, so they can float. 

When fluids flow, interesting effects 
occur hecause the pressure in the fluid 
is lower where the fluid velocity is 
higher (Bernoulli's principle). 



CHAPTER 



10 



Fluids 



In previous Chapters we considered objects that were solid and assumed to 
maintain their shape except for a small amount of elastic deformation, We 
sometimes treated objects as point panicles. Now we are going to shift our 
attention to materials that aTe very deform able and can flow. Such "fluids 1 ' 
include liquids and gases. We will examine fluids both at rest (fluid statics) and 
in motion (fluid dynamics). 



I Phases of Matter 



The three common phases, or states, of matter are solid, liquid, and gas. We can 
distinguish these three phases as follows. A solid maintains a fixed shape and a 
fixed size; even if a large force is applied to a solid, it does not readily change in 
shape or volume, A liquid does not maintain a fixed shape — if takes on the 
shape of its container — but like a solid if is not Teadily compressible, and its 
volume can be changed significantly only by a very large force. A gas has 
neither a fixed shape nor a fixed volume — it will expand to fill its container. For 
example, when air is pumped into an automobile tire, the air does not all run to 
the bottom of the tire as a liquid would; it spreads out to fill the whole volume 
of the tire. Since liquids and gases da not maintain a fixed shape, they both have 
the ability to flow; they are thus often referred to collectively as fluids. 



Phases of matter 



255 



The division of matter into three phases is not always simple. How, Tot 
example, should butter be classified? Furthermore, a fouHh phase of mailer car 
be distinguished, the plasma phase, which occurs only at very high temperatures 
and consists of ioni/ed atoms (electrons separated from the nuclei). Some scien- 
tists believe that so-called colloids (suspensions of tiny particles in a liquid) 
should also be considered a separate phase of matter. Liquid crystals, which aTe 
used in laptop computer screens, calculators, digital watches, and so on can be 
considered a phase of matter intermediate between solids and liquids. However, 
foT our present purposes we will mainly be interested in the three ordinary 
phases of matter. 



10-2 



Density and Specific Gravity 



Density defined 



TABLE 10-1 

Densities of Substances 1 


Substance 


Density, 
t> (kgM 


Solids 


Aluminum 


2,70 X I0 3 


Iron and steel 


7,8 x 10 3 


Copper 


8,9 x 10 3 


I.ead 


11.3 x in- 1 


Gold 


19.3 x 1Q 3 


Concrete- 


2,3 x I0 3 


Granite 


2,7 x 10 3 


Wood (typical) 


0,3- (J, 9 x in 3 


Glass, common 


2.4-2.8 X It) 1 


lce(H 2 0) 


0.917 X I0 3 


Bone 


1.7-2.0 X 10 3 


Liquids 


Water (4 "Q 


1.00 x It) 1 


Blood, plasma 


1.03 X 10 3 


Blood, whole 


1.05 x in- 1 


Sea water 


1.025 X 10 3 


Mercury 


13.6 X I0 3 


Alcohol, ethyl 


0.79 x in 3 


Gasoline 


0.68 x I0 3 


Cases 


Air 


1.29 


Helium 


0.179 


CaTbon dioxide 


1.98 


Water (steam) 
(lOCrC) 


0.598 


' Densities are given ai = C and I atm 

prL'ssun; unless mherwise specifiuri. 



It is sometimes said that iron is "heavier" than wood. This cannot really be true 
since a large log clearly weighs more than an iron nail. What we should say is 
that iron is more dense than wood. 

The density, p, of a substance (p is the lowercase Greek letter rho) is 
defined as its mass per unit volume: 



P = 



m 

17' 



(10-1) 



where m is the mass of a sample of the substance and V its volume. Density is a 
characteristic property of any pure substance. Objects made of a particular pure 
suhstanee, such as pure gold, can have any size or mass, but the density will he the 
same for each. (We will sometimes use the coneept of density, Eq. 10-1, to write 
the mass of an object as m = pV, and the weight of an abject, mg, as pVg,) 

The ST unit for density is kg/m*. Sometimes densities are given in g/em 3 . 
Note that since lkg/m 3 = 1000 g/(l00 em) 3 = I0 :! g/10"cm 3 = I0"*g/em\ then 
a density given in g/cm 3 must be multiplied by 1000 to give the result in kg/m 3 , 
Thus the density of aluminum is p = 2.70 g/cm 3 , which is equal to 2700 kg/m 3 . 
The densities of a variety of substances arc given in Table 10-1. The Table spec- 
ifics temperature and atmospheric pressure because they affect the density of 
substances (although the effect is slight for liquids and solids). 



EXAMPLE 10-1 



Mass. given volume and density. What is the mass of a 
solid iron wrecking ball of radius 18 cm? 

APPROACH First we use the standard formula V = firr 1 (see inside rear 
etiver) to obtain the volume of the sphere. Then Eq, 1 0-1 and Table 1 0-1 give 
us the mass m. 

SOLUTION The volume of the sphere is 

V = \ nr" = ^(3. 14) (0.1 8 m) 3 = 0.024 m 3 . 
From Table 10-1, the density of iron is p = 7800 kg/m 3 , so Eq. 10-1 gives 
m = pV = (7800 kg/m 3 )( 0,024 m 3 ) = 190 kg. 



The specific gravity of a substance is defined as the ratio of the density of 
that substance to the density of water at 4,0"C. Because specific gravity (abbre- 
viated SG) is a ratio, it is a simple number without dimensions or units. The 
density of water is 1.00 g/cm 3 = 1.00 x 10'kg/nr 1 , so the specific gravity of 
any substance will be equal numerically to its density specified in g/cm 3 , or I0~ 3 
times its density specified in kg/m 3 . For example (sec Table 1 0-1), the specific 
gravity of lead is 1 1.3, and that of alcohol is 0.79. 

The concepts of density and specific gravity are especially helpful in the 
study of fluids because we are not always dealing with a fixed volume or mass. 



256 CHAPTER 10 Fluids 



1 Pressure in Fluids 



Pressure is defined as force per unit area, where the force Fis understood to he 
the magnitude of the force acting perpendicular to the surface area A: 

F 
pressure = P = —- (10-2) 

Although foTce is a vector, pressure is a scalar. Pressure has magnitude only. 
The ST unit of pressure is N/rrr, This unit has the official name pascal (Pa), in 
honor of Blaise Pascal (see Section 10-5); that is, I Pa = 1 N/m 2 . However, 
for simplicity, we will often use N/m 2 . Other units sometimes used are 

dynes/cm 2 , and lb/in. 3 (abbreviated "psi"). Several other units for pressure 
are discussed, along with conversions between them, in Section 10-6 (see also 
the Table inside the front cover). 



EXAMPLE 10-2 



Calculating pressure. The two feet of a 60-kg person 
cover an area of 500 cm 3 , (a) Determine the pressure exerted by the two feet 
on the ground, (b) If the person stands on one foot, what will the pressure be 
under that foot? 

APPROACH Assume the person is at rest. Then the ground pushes up on 
her with a foree equal to her weight trig, and she exerts a force mg on the 
ground where her feet (or foot) contact it. Because I em 2 = ( 1 -2 m ) 2 = 1 -4 m 2 , 
then 500 cm 2 = 0.050 m 2 . 
SOLUTION (a) The pressure on the ground exeTted by the two feet is 

F_ _ >ng_ (60kg)(9.Sm/s 3 } 

A ~ A 



P = 



= 12 X t0 3 N/m 2 . 



(0,050 m 2 ) 

(b) If the person stands on one foot, the force is still equal to the person's 
weight, but the area will be half as much, so the pressure will be twice as much: 

24 x 10- 5 N/m 2 . 



Pressure is particularly useful for dealing with fluids. It is an experimental 
observation that a fluid can exert a pressure in any direction. This is well known 
to swimmers and divers who feci the water pressure on all parts of their bodies. 
At any point in a fluid at rest, the pressure is the same in all directions at a 
given depth. This is illustrated in Fig. 10-1. Consider a tiny cube of the fluid 
which is so small that we can ignore the force of gravity on it. The pressure on 
one side of it must equal the pressure on the opposite side. If this weren't true, 
there would be a net force on the cube and it would start moving. If the fluid is 
not flowing, then the pressures must be equal. 

Another important property of a fluid at rest is that the force due to fluid 
pressure always acts perpendicular to any solid surface it is in contact with. If 
there were a component of the force parallel to the surface, as shown in 
Fig. 10-2, then according to Newton's third law the solid surface would exert a 
force back on the fluid that also would have a component parallel to the 
surface. Such a component would cause the fluid to flow, in contradiction to our 
assumption that the fluid is at rest. Thus the force due to the pressure in a fluid 
at rest is always perpendicular to the surface. 



Pressure defined 



<$L 



CAUTION 



Pressure is a scalar, not a rector 
The pascal (unit) 



Raids exert 
pressure in all 
directions 




FIGURE 10-1 Pressure is the 

same in every direction in a fluid at 
a given depth: if it weren't, [he fluid 
would l>e in motion. 




FIGURE 10-2 If there were a component of force 
parallel to the solid surface of the container, the liquid 

would move in response to it. For a liquid at rest. Pj = (J 



SECTION 10-3 Pressure in Fluids 257 











K—P- 


h 

i 








A / 





FIGURE 10-3 Calculating the 

pressure at a depth h in a liquid. 

Pressure variation with depth 




u 



\1) 



FIGURE 10-4 Forces on a thin 
slab of fluid (shown as a liquid, but 
il could instead be a gas), 



Change in pressure with 
change in depth in a fluid 



(g j PHYSICS APPLIED 

Water supply 
FIGURE 10-5 Rsamplc 10-3. 



"I 



A/i = 

30 m 



/H 



m 






Let us now calculate quantitatively how the pressure in a liquid of uniform 
density varies with depth. Consider a point at a depth h below the surface of the 
liquid (that is, the surface is a height h above this point), as shown in Fig. 10-3. 
The pressure due to the liquid at this depth h is due to the weight of the column 
of liquid above it. Thus the force due to the weight of liquid acting on the 
area A is F = mg = (pV)g = pAhg, where Ah is the volume of the column of 
liquid, p is the density of the liquid (assumed to be constant), and g is the 
acceleration of gravity. Tie pressure P due to the weight of liquid is then 



P = 



pAhg 



A 



P = pgh. 



[liquid] (10-3a> 



Note thai the area A doesn't affect the pressure at a given depth. The fluid pres- 
sure is directly proportional to the density of the liquid and to the depth within the 
liquid, In general, the pressure at equal depths within a uniform liquid is the same, 

Equation 10— 3a is extremely useful. It is valid for fluids whose density is 
constant and does not change with depth — that is, if the fluid is incompyessible. 
This is usually a good approximation for liquids (although at great depths in the 
ocean, the density of water is increased substantially by compression due to the 
great weight of water above). 

Gases, on the other hand, are very compressible, and density can vary signif- 
icantly with depth. For this more general case, in which p may vary, Eq. IO-3a 
may not be useful. So let us consider a thin slab of liquid of volume V - A Ah 
as shown in Fig. 10-4. Wc choose Ah thin enough so that p doesn't vary signifi- 
cantly over the small thickness Ah. Let P be the pressure exerted downward on 
the top surface, and let P + AP be the pressure upward on the bottom surface, 
The forces acting on our thin slab of fluid, as shown in Fig. 10-4, are 
(F + AP)A upward and PA downward, and the downward weight of the slab, 
mg = {pV)g - pA Ah g. We assume the fluid is at rest, so the net force on the 
slab is zero. Then 

(P + AP)A - PA - pA Ah g = 0. 

The area A cancels from each term, and when we solve for AP we obtain 

AP — pgAh. \p « constant over Ah\ (10-3I>J 

Equation IO-3b tells us how the pressure changes over a small change in 
depth (Ah) within a fluid, even if compressible, 



EXAMPLE 10-3 



Pressure at a faucet. The surface of the water in a 
storage tank is 30 m above a water faucet in the kitchen of a house. Fig. 10-5, 
Calculate the difference in wafer pressure between the faucet and the surface 
of the water in the tank. 

APPROACH WateT is practically incompressible, so p is constant even for a 
Ah - 30 m when used in Eq. 10— 3b. Only Ah matters; we can ignore the 
"route" of the pipe and its bends. 

SOLUTION The same atmospheric pressure acts both at the surface of the 
water in the storage tank and on the water leaving the faucet. So, the water 
pressure difference between the faucet and the surface of the water in the tank is 

AP = pg Ah = (1.0 X 10 3 kg/m i )(9.8m/s 2 )(3Qm) 
= 2,9 x l0 5 N/m 2 . 

NOTE The height h is sometimes called the pressure head. In this Example, 
the head of wafer is 30 m at the faucet, Tie very different diameters of the 
tank and faucet don't affect the result — only pressure does. 



EXERCISE A A dam holds back a lake that is 85 m deep at the dam. If the lake is 20 km 
long, how much thicker should the dam be than if the lake were smaller, only 1.0km long? 



258 CHAPTER 10 Fluids 



10-4 



Atmospheric Pressure and Gauge Pressure 



Atmospheric Pressure 

The pressure of the Earth's atmosphere, as in any fluid, changes with depth. But 
the Earth's atmosphere is somewhat complicated: not only docs the density of 
air vary greatly with altitude but there is no distinct top surface to the atmo- 
sphere from which h (in Eq. 10-3a) could be measured, We can, however, calculate 
the approximate difference in pressure between two altitudes using Eq. IO-3b. 

The pressure of the air at a given place varies slightly according to the 
weather, At sea level, the pressure of the atmosphere on average is 
1.013 x ItTN/m 2 (or 14, 7 lb/in."). This value lets us define a commonly used 
unit of pressure, the atmosphere (abbreviated atm): 

1 atm = 1,013 X lu^N/m 2 = 101.3 kPa. 

Another unit of pressure sometimes used (in meteorology and on weather 
maps) is the bar, which is defined as 

1 bar = 1.00 x lO'N/nr. 

Thus standard atmospheric pressure is slightly more than 1 bar. 

The pressure due to the weight of the atmosphere is exerted on all ohjects 
immersed in this great sea of air, including our bodies, How does a human body 
withstand the enormous pressure on its surface? The answer is that living cells 
maintain an internal pressure that closely equals the external pressure, just as 
the pressure inside a balloon closely matches the outside pressure of the 
atmosphere. An automobile tire, because of its rigidity, can maintain internal 
pressures much greater than the external pressure, 



One atmosphere {unit of pressure) 



The bar (unit of pressure) 



PHYSICS APPLIED 
Pressure on living i ells 



CONCEPTUAL EXAMPLE 10-41 Finger holds water in a straw. You 
insert a straw of length L into a tall glass of water. You place your finger over the 
top of the straw, capturing some air above the water but preventing any additional 
air from getting in or out, and then you lift the straw from the water. You find that 
the straw retains most of the water. (See Fig. 10 -6a.) Does the air in the space 
between your finger and the top of the water have a pressure P that is greater 
than, equal to, or less than, the atmospheric pressure P A outside the straw? 

RESPONSE Consider the forces on the column of water (Fig. 10— fib). Atmospheric 
pressure outside the straw pushes upward on the water at the bottom of the straw, 
gravity pulls the water downward, and the air pressure inside the top of the 
straw pushes downward on the water. Since the water is in equilibrium, the 
upward force due to atmospheric pressure must balance the two downward 
forces. The only way this is possible is for the air pressure inside the straw to 
be less than the atmosphere pressure outside the straw. (When you initially 
remove the straw, a little water may leave the bottom of the straw, thus 
increasing the volume of trapped air and reducing its density and pressure.) 



P=? 



t 



I PA 






pgAh 



(a) 
FIGURE 10 -G Example 10-4. 



Gauge Pressure 

It is important to note that tiro gauges, and most other pressure gauges, register the 
pressure above and beyond atmospheric pressure, This is called gauge pressure. 
Thus, to get the absolute pressure, P, we must add the atmospheric pressure, P A , to 
the gauge pressure, P G : 

P = ?* + Pq- 

If a tire gauge registers 220 kPa, the absolute pressure within the tire is 
220kPa+ 101 kPa = 32l kPa, equivalent to about 3.2 atm (2,2atm gauge 
pressure). 



Gauge pressure 



Absoluts' pressure = 

atmospheric pressun + gnH^t pn-^m-i 



SECTION 10-4 Atmospheric Pressure and Gauge Pressure 258 



Pascal's Principle 



Pascal's principles 



PHYSICS APPLIED 



Hydraulic lift 



The Earth's atmosphere exerts a pressure on all objects with which it is in 
contact, including other fluids. External pressure acting on a fluid is transmitted 
throughout thai fluid. For instance, according to Eq. 10-3a, the pressure due to 
the water at a depth of 100 m below the surface of a lake is P = pg&H = 
(l000kg/m v )(9.8m/s 2 )(100m) = 9.S x 10 5 N/m 2 , or 9.7 atm. However, the 
total pressure at this point is due to the pressure of water plus the pressure of 
the air above it. Hence the total pressure (if the lake is near sea level) is 
9,7 atm + 1.0 atm = 10.7 atm. This is just one example of a general principle 
attributed to the French philosopher and scientist Blaise Pascal (1623-1662), 
Pascal's principle states that if an external pressure is applied to a confined fluid, 
the pressure at every point within the fluid increases by that amount. 

A number of practical devices make use of Pascals principle. One example 
is the hydraulic lift, illustrated in Fig. IO-7a, in which a small input force is used to 
exert a large output force by making the area of the output piston larger than the area 
of the input piston. To see how this works, we assume the input and output 
pistons are at the same height (at least approximately). Then the external input 
force F ia , by Pascals principle, increases the pressure equally throughout. 
Therefore, at the same level (see Fig. 10-7a), 

P = P 

' (ml ' in 

where the input quantities are represented by the subscript "in" and the output 
by "out." Since P - F/A, we write the above equality as 



Mechanical advantage 



or 



"out 

*L1UL 

F 

' in 



■1. 



The quantity F aul /F in is called the mechanical advantage of the hydraulic lift, and 
it is equal to the ratio of the areas. For example, if the area of the output piston 
is 20 times that of the input cylinder, the force is multiplied hy a factor of 20: 
thus a force of 200 lb could lift a 4000-lb car. 



FIGURE 10-7 Applications of 
Pascal's principle: (a) hydraulic lift: 
(b) hydraulic brakes in a car. 



(a.) 



£ 




Master 
. cylinder 




Disk,- 

attached lo wheel 



PHYSICS APPLIED 
Brake 



Figure 1 0-7b illustrates the brake system of a car. When the driver presses 
the brake pedal, the pressure in the master cylinder increases. This pressure 
increase occurs throughout the brake fluid, thus pushing the brake pads against 
the disk attached to the car's wheel. 



Manometer 



260 CHAPTER 10 Fluids 



10-6 



Measurement of Pressure; 
Gauges and the Barometer 



Many devices have been invented to measure pressure, some of which arc 
shown in Fig, 10-8. The simplest is the open-tube manometer (Fig. IO-8a), 
which is a U-shaped tube partially filled with a liquid, usually mercury or water. 
The pressure P being measured is related (by Eq. 10— 3b) to the difference in 



Atmospheric 
pressure 






I" 



r. 



% 



(Picsaire being 
measured! 



Flexible 
dvambef 



(a) Open-lube manometer 



(h) Anerukt gauge »jsi\I m.iirik 
for air pressure, and ihen 
called an aneroid barometer \ 



Pressure of J 

air in tire ■* 



Scale 

reading 



S prins; 



(c) lire gauge 



FIGURE 10-8 Pressure gauges: (a) open-tube manometer, (b) aneroid gauge, and (c) common tire 
pressure gauge. 



height Ah of the two levels of the liquid by the relation 

P = P + pg Ak, (10-3*) 

where P it is atmospheric pressure (acting on the top of the liquid in the left-hand 
tube), and p is the density of the liquid. Note that the quantity pg Ah is the gauge 
pressure — the amount by which P exceeds atmospheric pressure P t) . Tf the liquid 
in the left-hand column were lower than that in the right-hand column, P would 
have to be less than atmospheric pressure (and Ah would be negative). 

Instead of calculating the product pg Ah, sometimes only the change in height 
Ah is specified. In fact, pressures are sometimes specified as so many "millimeters of 
mercury" (mra-Hg) or "mm of water"' (mm-H 2 0). The unit mm-Hg is equivalent to 
a pressure of 133 N/m 2 , since pg Ah for 1 mm - 1.0 X I0~' m of mercury gives 

pghh = (13.6 X 10 3 kg/m a )(9.80m/s 2 )(l.00 X l(r 3 m) = 1.33 X 10 2 N/m 2 . 

The unit mm-Hg is also culled the torr in honor of Evangelisla Torricelli 
(1 60S- 1647), a student of Galileos who invented the barometer (see below). 
Conversion factors among the various units of pressure (an incredible nuisance!) 
are given in Table 10-2. Tt is important that only N/m 2 = Pa, the proper SI 
unit, be used in calculations involving other quantities specified in ST units. 

Another type of pressure gauge is the aneroid gauge (Pig. 10-8b) in which 
the pointer is linked to the flexible ends of an evacuated thin metal chamber. In 
an electronic gauge, the pressure may he applied to a thin metal diaphragm 
whose resulting distortion is translated into an electrical signal by a transducer. 
How a common tire gauge is constructed is shown in Fig, 1 0-8c. 



TABLE 10-2 Conversion Factors Between Different Units of Pressure 


In Terror of 1 Pa = 1 N/m 2 


1 atm in Different Units 


lalm = 1.013 X It) 5 N/m 2 


latm = 1.013 X 10' N/m 2 


= 1.013 X Iff Pa = lOUfcPa 


Ibar = 1.000 X 10 s N/m 2 


latm = 1.013 bar 


1 dyne/cm 3 = 0.1 N/m 2 


1 atm = 1.013 X 10* dyne/cm 2 


lib/in. 2 = 6.90 X KTN/m 2 


1 atm = 14.7 lb/in. 2 


I lb/ft 2 = 47.9 N/m 2 


latm =2.12 x 10' lb/ft 2 


lcm-Hg = 1.33 X 10 1 N/m 1 


1 atm = 76cm-Hg 


1 mm-Hg = l33N/m ; 


1 atm = 760 mm-Hg 


1 torr = 133 N/m 2 


1 aim = 760 torr 


1 mm-H 2 (4°C) = 9.81 N/m 2 


1 atm = 1.03 x lO^mm-H^O (i'C) 



Pressure beneath surface of 
liquid open to the atmosphere 



The torr (unit of pressure) 



*» PROBLEM SOLVING 
Use SI unit in calatlntkms: 

: Pa - i N/m 2 



SECTION 10-6 Measurement of Pressure; Gauges and the Barometer 261 



P = 



76 cm 



P = I ami 




FIGURE 10-9 A mercury 
barometer, invented by Torricelli, is 

shown here when the air pressure 
is standard atmospheric, 76 cm-Hg. 



Barometer 




FIGURE 10-10 A water barometer: 
a full tube of water is inserted into a 
tub of water, beeping the spigot at the 
top closed. When the bottom end of 
the tube is uncovered, some water 
flows out of the tube into the tub, 
leaving a vacuum between the water's 
upper surface and the spigot. Why"? 
Because air pressure can nol support a 
column of wateT more than 10 m high, 



Atmospheric pressure can be measured hy a modified kind of mercury 
manometer with one end closed, called a mercury barometer (Fig. 10-9). The 
glass tube is completely filled with mercury and then inverted into the bowl of 
mercury. If the tube is king enough, the level of the mercury will drop, leaving a 
vacuum at the top of the tube, since atmospheric pressure can support a column 
of mercury only about 76 cm high (exactly 76.0 cm at standard atmospheric 
pressure). That is, a column of mercury 76 cm high exerts the same pressure as 
the atmosphere 1 : 

P = pg Ah 

= (13,6 x 10*kg/m , )(9,80m/s 2 )(0.7fi0m) = 1.013 X 10 5 N/rrr = 1.00 atm. 

Household barometers are usually of the aneroid type, either mechanical 
(Fig. 10— Sb) or electronic. 

A calculation similar to that above will show that atmospheric pressure 
can maintain a column of watei 10.3 m high in a tube whose top is under 
vacuum (Fig. 10-10). No matter how good a vacuum pump is, it cannot lift 
water moTe than about 10 m. To pump wateT out of deep mine shafts with a 
vacuum pump requires multiple stages for depths greater lhan 10m. Galileo 
studied this problem, and his student Torricelli was the first to explain it. 
The point is that a pump does not really suck water up a tube — it merely 
reduces the pressure at the top of the tube. Atmospheric air pressure pushes 
the water up the tube if the top end is at low pressure (under a vacuum), 
just as it is air pressure that pushes (or maintains) the mercury 76cm high in 
a barometer. 



CONCEPTUAL EXAMPLE 1 Q-5~| Suction. You sit in a meeting where a 



novice NASA engineer proposes suction cup shoes for Space Shuttle astro- 
nauts working on the exterior of the spacecraft. Having just studied this 
Chapter, you gently remind him of the fallacy of this plan. What is it? 

RESPONSE Suction cups work by pushing out the air underneath the cup. 
What holds the cup in place is the air pressure outside the cup. (This can 
be a substantial force when on Earth. For example, a lO-cm-diameteT 
cup has an area of 7.9 X 10~ 3 nr. The force of the atmosphere on it is 
(7.9 X IO~ 3 m 2 )(l.O X 10 5 N/m 2 ) ^ SOON, about ISO lbs!). But in outer space, 
there is no aiT pressure to hold the suction cup onto the spacecraft. 

We sometimes mistakenly think of suction as something we actively do. For 
example, we intuitively think that we pull the soda up through a stTaw. Instead, 
what we do is lower the pressure at the top of the straw, and the atmosphere 
pushes the soda up the straw. 

'This talculaliun confirms the entry in Table 10-2, I aim = 76 cm-Hg. 



262 CHAPTER 10 Fluids 



Buoyancy and Archimedes' Principle 

Objects submerged in a fluid appear to weigh less than they do when outside 
the fluid. For example, a large rock that you would have difficulty lifting off 
the ground can often be easily lifted from the bottom of a stream. When the 
rock breaks through the surface of the water, it suddenly seems to be much 
heavier. Many objects, such as wood, float on the surface of water. These are 
two examples of buoyancy. In each example, the force of gravity is acting 
downward, But in addition, an upward buoyant force is excited by the liquid. 
The buoyant force on fish and underwater divers (as in the chapter-opening 
photo) almost exactly balances the force of gravity downward, and allows 
them to "hover" in equilibrium. 

The buoyant force occurs because the pressure in a fluid increases with 
depth. Thus the upward pressure on the bottom surface of a submerged object is 
greater than the downward pressure on its top surface. To sec this effect, 
consider a cylinder of height i/j whose top and bottom ends have an area A 
and which is completely submerged in a fluid of density p v , as shown in 
Fig. 10-11. The fluid exerts a pressure /' p t gh l at the top surface of the 



Rocks seem to weigh less 
under water 



Wood floats 





1 

''1 
1 "T 

T 



FIGURE 10-11 Determination 

of the buoyant force, 



cylinder (Eq. IO-3a). The force due to this pressure on top of the cylinder is 
F, = P, A = p ¥ gh { A, and it is directed downward. Similarly, the fluid exerts an 
upward force on the bottom of the cylinder equal to F 2 = P 2 A = p^ghjA. 
The net force on the cylinder exerted by the fluid pressure, which is the buoyant 
force, F b , acts upward and has the magnitude 

F B = F 2 - F, = pygA(h 2 - /;,) 
= prgALh 
- foVg 
= w't#. 

where V = A Ah is the volume of the cylinder, the product p^V is its mass, 
and p?Vg = wi h g is the weight of fluid which takes up a volume equal to the 
volume of the cylinder. Thus the buoyant force on the cylinder is equal to 
the weight of fluid displaced by the cylinder. This result is valid no matter what 
the shape of the object- Tls discovery is credited to Archimedes (2£7?-2l2 n.c), 
and it is called Archimedes' principle: the buoyant force on an object immersed 
in a fluid is equal to the weight of the fluid displaced by that object - 

By "fluid displaced; 1 we mean a volume of fluid equal to the volume of the 
submerged object, or that part of the object submerged if it floats or is only partly 
submerged (the fluid that used to be where the object is). If the object is placed 
in a glass or tub initially filled to the brim with water, the water that flows over 
the top represents the water displaced by the object. 



A rchimede s ' principle 



SECTION 10-7 Buoyancy and Archimedes' Principle 263 



FIGURE 10-12 Archimedes' principle 





We can derive Archimedes' principle in general by the following simple but 
elegant argument. The irregularly shaped object D shown in Fig, 10- 12a is acted on 
by the force of gravity (its weight, wg, downward) and the buoyant force. F B , 
upward, We wish to determine F B , To do so, we next consider a body (D' in 
Fig, 10- 12b), this time made of the fluid itself, with the same shape and size as the 
original object, and located a! the same depth. You might think of this body of fluid 
as being separated from the rest of the fluid by an imaginary membrane. The 
buoyant force F B on this body of fluid will be exactly the same as that on the orig- 
inal object since the surrounding fluid, which exerts F a , is in exactly the same 
configuration, This body of fluid D' is in equilibrium (the fluid as a whole is at rest). 
Therefore, F u = m'g, where m'g is the weight of the body of fluid. I lencc the 
buoyant force F a is equal to the weight of the body of fluid whose volume equals 
the volume of the original submerged object, which is Archimedes' principle, 

Archimedes' discovery was made by experiment. What we have done in the 
last two paragraphs is show that Archimedes' principle can be derived from 
Newtons' laws. 



CONCEPTUAL EXAMPLE 10-6 | Two pails of water. Consider two iden- 
tical pails of water filled to the brim. One pail contains only water, the otheT 
has a piece of wood floating in it. Which pail has the greater weight? 

RESPONSE Rolh pails weigh the same. Recall Archimedes' principle: the wood 
displaces a volume of water with weight equal to the weight of the wood, Some 
water will overflow the pail, but Archimedes' principle tells us the spilled wateT 

has weight equal to that of the wood; so the pails have the same weight, 



FIGURE 10-13 Rxample 10-7. 
The force needed to lift the statue 

i- F 




EXAMPLE 10-7 



Recovering a submerged statue. A 70- k« ancient 
statue lies at the bottom of the sea. Its volume is 3.0 X I0 4 cm\ How much 
force is needed to lift it? 

APPROACH The force F needed to lift the statue is equal to the statue's 
weight mg minus the buoyant force F a . Figure 10-13 is the free-body diagram. 

SOLUTION The buoyant force on the statue due to the water is equal 
to the weight of 3,0 x lO^cm 3 = 3.0 x 10 _2 m 3 of water (for seawater, 
p = 1.025 X lO'kg/m'): 

^b = >"h,o£ = Ph,o^S 

- (1.025 X 10' kg/m 3 )(3.0 X 10 'm'^.Sm/s 2 ) 

- 3.0 x I0 2 N. 

The weight of the statue is ira# = (70kg)(9.8 m/s ! ) = 6.9 X I0 2 N. Hence the 
force F needed to lift it is 690 N - 300 N = 390 N. ft is as if the statue had a 
mass of only (390N)/(9.8m/s : ) = 40 kg. 

NOTE Here F - 390 N is the force needed to lift the statue without acceler- 
ation when it is under water. As the statue comes out of the water, the force F 
increases, reaching 690 N when the statue is fully out of the water. 



264 CHAPTER 10 Fluids 



Archimedes is said to have discovered his principle in his bath while thinking 
how he might determine whether the king's new crown was pure gold or a take- 
Gold has a specific gravity of 1 9.3, somewhat higher than that of most metals, but 
a determination of specific gravity or density is not readily done directly because, 
even if the mass is known, the volume of an irregularly shaped object is not easily 
calculated, However, if the object is weighed in air (=«;) and also "weighed" 
while it is under water (=w'), the density can be determined using Archimedes' 
principle, as the following Example shows, The quantity w' is called the apparent 
■weight in water, and is what a scale reads when the object is submerged in water 
(see Fig, 10-14): w/ equals the true weight {w = mg) minus the buoyant force. 



(.«) 




(F T - -mg) 



CM 




FIGURE 10-14 (a) A scale reads [he mass of an object 
in air — in this case the crown of Example 10-8. All 
objects are at rest, so the tension F v in the connecting 
cord equals the weight w of the object: F T = mg. We 
show the free-body diagram of the crown, and Fj is what 
causes the scale reading (it's equal to the net downward 
force on the scale, by Newton's third law). 
(b) Submerged, the object has an additional force on it, 
the buoyant force f B - The net force is zero, so 
{•'j + F h = mg (= w). The scale now reads 
in' = 13,4 kg. where m' is related to the effective weight 
by w' = m'g. Thus F' r = w' = u> - F n . 



EXAMPLE 10-8 



Archimedes: Is the crown gold? When a crown of 
mass 14.7 kg is submerged in water, an accurate scale reads only 13.4 kg. Is the 
crown made of gold? 

APPROACH Tf the crown is gold, its density and specific gravity must be very high, 
SG = 19.3 (see Section 10-2 and Table 10-1). We determine the specific gravity 
using Archimedes' principle and the two free-body diagrams shown in Fig. 10-14. 
SOLUTION The apparent weight of the submerged object (the crown) is w', 
and it equals F r in Fig. 1 0-1 4b. The sum of the forces on the object is zero, so 
w' equals the actual weight w (- mg) minus the buoyant force Fy,: 

if' - F-'[- - w — F a 
so 

it 1 — w' - F u . 
Let V be the volume of the completely submerged object and p its density 
(so p V is its mass), and let p, be the density of the fluid (water). Then 
{pi.V)g is the weight of fluid displaced (- F B ). Now we can write 
w = mg = p Vg 

w - w' = F B = ppVg. 
We divide these two equations and obtain 
W _ p Q Vg _ po_ 

Pv 



w - til 
We sec that w/(w 



Pv Vg 



w') is equal to the specific gravity of the object if the 
fluid in which it is submerged is water (p F = 1.00 X 10' kg/m v ). Thus 

(14.7 kg)?. 14.7 kg 



Pi :■ 



If 



- 11.3. 



PHfi w - w' ( 14.7 kg - 13.4 kg}# 1 .3 kg 
This corresponds to a density of 1 1 ,300 kg/rrr\ The crown seems to be made of 
lead (see Table 10-1)! 



Recall m = pV ('(Tc/. ID-!/ 



SECTION 10-7 Buoyancy and Archimedes' Principle 265 



FIGURE 10-15 (a) The fully submerged lug 
accelerates upward because F n > mg. It comes 
to equilibrium (b) when SF = 0, so 
F B = mg = (\2QQkg)g. Thus 1200 kg, or 
1.2 m \ of water is displaced. 



F B ={2000fcg) 5 



»!q = 1 200 ka 
r»2.0ro 3 



(a) 



/' B = (1200kgte 




[bj 



rictdiii}- 





' B = flKi 


spl & 
















»'tf = A#i 


F 



FIGURE 10-16 An object I'loiuins; 
in equilibrium: F B = nig. 

Fraction of floating object 
submerged in water = its SG 



FIGURE 10-17 A hydrometer. 
Example 1 0-9, 



25.0 
cm 



I.0O0 ; 



U 



Archimedes' principle applies equally well to objects that float, such as wood. 
Tn general, an object floats on a fluid if its density is less than that of the fluid. This 
is readily seen from Fig. 10- 1: 5a, where a submerged object will experience a net 
upward force and float to the surface if F b > mg: that is, if p\Vg > p tJ Vg 
or p ¥ > p . At equilibrium — that is, when floating — the buoyant force on an 
object has magnitude equal to the weight of the object- For example, a log whose 
specific gravity is 0.60 and whose volume is 2.0m 3 has a mass m - p V ' = 
(0.(50 x 10 3 kg/m 3 )(2.0m 3 ) = 1200kg. Tf the log is fully submerged, it will 
displace a mass of water m ¥ = p ¥ V = (1000 kg/m 3 )(2.0 m :i ) = 2000 kg. Hence 
the buoyant force on the log will he greater than its weight, and it will float 
upward to the surface (Fig. 10-15). The log will come to equilibrium when it 
displaces 1200 kg of water, which means that 1.2 m 3 of its volume will be 
submerged. This 1.2 m 3 corresponds to 60% of the volume of the log 
(1.2/2.0 = 0.60), so 60% of the log is submerged. In general when an object 
floats, we have F K = trig, which we can write as (see Fig. 10-16) 

where V is the full volume of the object and V d j sp ] is the volume of fluid it 
displaces (- volume submerged). Thus 



Khspl 



Po i 

Pi 



That is, the fraction of the object submerged is given by the ratio of the object's 
density to that of the fluid. Tf the fluid is water, this fraction equals the specific 
gravity of the object. 



EXAMPLE 10-9 



Hydrometer calibration. A hydrometer is a simple 
instrument used to measure the specific gravity of a liquid by indicating how 
deeply the instrument sinks in the liquid. A particular hydrometer (Fig, 10-17) 
consists of a glass tube, weighted at the bottom, which is 25,0 cm long and 
2.00 cm ? in cross-sectional area, and has a mass of 45.0 g. How far from the end 
should the 1 .000 mark be placed? 

APPROACH The hydrometer will float in water if its density p is less than 
pv, - l.000g/cm 3 , the density of water. The fraction of the hydrometer 
submerged (Kuspiaced/^otai) is equal to the density ratio p/p v , 
SOLUTION The hydrometer has an overall density 

m_ '15.0 g 

V ~ (2.00em : )(25.0em) 



P 77 



- 0.900 g/cm-\ 



Thus, when placed in water, it will come to equilibrium when 0.900 of 
its volume is submerged. Since it is of uniform cross section, 
(0.900) (25,0 cm) -22.5 cm of its length will be submerged. The specific 
gravity of water is defined to be 1 .000, so the mark should be placed 22.5 cm 
from the end. 



266 CHAPTER 10 Fluids 



I EXERCISE B On the hydrometer of Fxample 10-9, will the mark's above Che 1.000 
| mark represent higher or lower values of density of the liquid in which it is submerged? 

Archimedes principle is also useful in geology. According to the theories of 
plate tectonics and continental drift, the continents float on a fluid "sea" of slightly 
deform able Toek (mantle rock). Some interesting calculations can be done using 
very simple models, which we consider in the Problems at the end of the Chapter. 

Air is a fluid, and it too exerts a buoyant force. Ordinary objects weigh less 
in air than they do if weighed in a vacuum. Because the density of air is so 
small, the effect for ordinary solids is slight. There are objects, however, that 
float in air — helium -filled balloons, for example, because the density of helium 
is less than the density of air. 



PHYSICS APPLIED 
( 'ontinental drift— plate tectonics 



Weight affected by 
buoyancy of air 



F R 



A 



">H e g 



i '"Joadl 
FIGURE 10-13 Fxample 10-10. 



EXAMPLE 10-10 



Helium balloon. What volume Voi helium is needed if a 
balloon is to lift a load of 180 kg (including the weight of the empty balloon)? 

APPROACH The buoyant force on the helium balloon, F^, which is equal to 
the weight of displaced aiT, must be at least equal to the weight of the helium 
plus the weight of the balloon and load (Fig, 10-18). Table 10-1 gives the 

density of helium as 0.179 kg/nri. 

SOLUTION The buoyant force must have a minimum value of 

F n = (m m + 180kg)g. 
This equation can be written in temis of density using Archimedes' principle: 

PairVg = (p„ e y + I80kg)g. 
Solving now for V, we find 

ISO kg 



PHc 



I SO kg 



(1.29kg/nr s - 0.179 kg/m 3 ) 



I60m\ 



NOTE This is the minimum volume needed near the Earth's surface, where 
p air = 1.29 kg/ m\ To reach a high altitude, a greater volume would be needed 
since the density of air decreases with altitude. 



SECTION 10-7 Buoyancy and Archimedes' Principle 267 



FIGURE 10-19 

fa) Streamline-, ot laminar, flow; 
(b) turbulent flow. 




h^'H 




FIGURE 10-20 Fluid flow 

through a pipe of varying diameter. 



Etjttalitm of continuity 
(general) 



10-8 



Fluids in Motion; 

Flow Rate and the Equation of Continuity 

We now turn to the subject of fluids in motion, which is called fluid dynamics, or 
(especially if the fluid is water) hydrodynamics. Many aspects of fluid motion 
are still being studied (for example, turbulence as a manifestation of chaos is a 
"hot" topic today). Nonetheless, with certain simplifying assumptions, we can 
understand a lot about this subject. 

We can distinguish two main types of fluid flow, Tf the flow is smooth, such that 
neighboring layers of the fluid slide by each other smoothly, the flow is said to be 
streamline pr laminar flow. In streamline flow, each particle of the fluid follows a 
smooth path, called a streamline, and these paths do not cross one anotheT 
(Fig, I0-I9a), Above a certain speed, the flow becomes turbulent. Turbulent flow is 
characterized by erratic, small, whirlpool-like circles called eddy currents or eddies 
(Fig, I0-I9b). Eddies absorb a gTeat deal of energy, and although a certain amount 
of internal friction called viscosity is present even during streamline flow, it is much 
greater when the flow is turbulent. A few tiny drops of ink or food coloring dropped 
into a moving liquid can quickly reveal whether the flow is streamline or turbulent. 

Let us consider the steady laminar flow of a fluid through an enclosed 
tube or pipe as shown in Fig. 1 0-20. First we determine how the speed 
of the fluid changes when the size of the tube changes. The mass flow rate 
is defined as the mass Am of fluid that passes a given point per unit time Ar: 

Am 
mass flow Tate = — — ■ 
if 

Tn Fig. 10-20, the volume of fluid passing point I (that is, through area A { ) in a 

time Af is A\ A?, , where A/, is the distance the fluid moves in time A(. Since the 

velocity 1 of fluid passing point I is V\ = A/ L /Ar, the mass flow Tale Awi/Af 

through area A , is 

Aw i PiAVl pj/l|A/| 

where Af', = A i All is the volume of mass A/«, , and p t is the fluid density. Simi- 
larly, at point 2 (through area A 2 )> the flow rate is piA 2 v 2 . Since no fluid flows in 
or out the sides, the flow rates through A t and A 2 must be equal. Thus, since 

A/H, Am, 



M 



A; 



then 



PjAyVi - p 1 A 2 v 1 . 
This is called the equation of continuity. 



(10-4a) 



268 CHAPTER 10 



'The word laminar means "in layers." 

*ff there were no viscosity, the velocity would be the same across a cross section of the tube. Real 
fluids have viscosity, and this internal friction causes different layers of the fluid to flow at different 
speeds. In this case e, and u ; represent the average speeds at each cross section. 



Tf the fluid is incompressible (p doesn't change with press utc), which is an 
excellent approximation for liquids under most circumstances (and sometimes 
for gases as well), then p, = p 2 , and the equation of continuity becomes 

A^Vj - A 2 v 2 . \p - constant! (10-4b) 

The product Am represents the volume rale of flow (volume of fluid passing a 
given point per second), since AV/Af - A A//Af = Av, which in SI units 
is m'/s. Equation ID— 4b tells us that where the cross-sectional area is large, the 
velocity is small, and where the area is small, the velocity is large. That this is 
reasonable can be seen by looking at a river. A river flows slowly through a 
meadow where it is broad, but speeds up to torrential speed when passing 
through a narrow gorge. 



ESTIMATE I Blood flow. In humans, blood flows from 



EXAMPLE 10-11 



the heart into the aorta, from which it passes into the major arteries. These 
branch into the small arteries (arterioles), which in turn branch into myriads of 
tiny capillaries. Fig. 10-21. The blood returns to the heart via the veins. The 
radius of the aorta is about 1,2 cm, and the blood passing through it has a 
speed of about 40cm/s. A typical capillary has a radius of about 4X10 'cm, 
and blood flows through it at a spe«d of about 5 x 10 4 m/s, Estimate the 
number of capillaries that are in the body. 

APPROACH We assume the density of blood doc;sn't vary significantly from 
the aorta to the capillaries, By the equation of continuity, the volume flow rate 
in the aorta must equal the volume flow rate through alt the capillaries. The 
total area of all the capillaries is given by the area of one capillary multiplied 
by the total number N of capillaries. 

SOLUTION Let .4, be the area of the aorta and A 2 be the area of till 
the capillaries through which blood flows. Then A 2 - iVjrr^ p , where 
lap ~ 4 X I0 _4 cm is the estimated average radius of one capillary. From the 
equation of continuity (Eq. 10-4), we have 

jj 2 A 2 - i>] A 1 
»i derm I 0.40 m/s Vl2xl0 2 m 



Si I 



V 



■ 'T' 



5 X IO" 4 m/s/\ 4 X 10 



7 x 10", 



or on the order of 10 billion capillaries. 



EXAMPLE 10-12 



Heating duct to a room. What area must a heating 
duct have if air moving 3.0 m/s along it can replenish the air every 15 minutes 
in a room of volume 300 m l 7 Assume the air's density remains constant, 

APPROACH We apply the equation of continuity at constant density, 
Eq. 10-4, to the air that flows through the duet (point I in Fig. 10-22) and 
then into the room (point 2). The volume flow rate in the room equals the 
volume of the room divided by the 15-minute replenishing time. 
SOLUTION Consider the room as a large section of the duct. Fig, 10-22, and 
think of air equal to the volume of the room as passing by point 2 in 
f = 15 minutes = 900 s. Reasoning in the same way we did to obtain Eq. 10-4a 
(changing Af to f), we write v 2 = l 2 /t so A 2 v 2 = Ajhjt = V 2 /t, where V 2 
is the volume of the room. Then the equation of continuity becomes 
yli'U| = jA >i v i — F^A an d 

V^ 300 m 3 

<■ 1 



'i 1 



(3.0m/s)(900s) 

If the duct is square, then each side has length / 
rectangular duct 20 cm x 55 cm will also do. 



- 0.1 1 m\ 



0-33 m, or 33 cm. A 



Equation of continuity 
(p = constant) 



PHYSICS APPLIED 



Blood flow 



llud 




V = valve* 
c = capillaries 

FIGURE 10-21 

Human circulatory system. 



PHYSICS APPLIED 



//t'ci'iV'jL;' LtttC! 



FIGURE 10-22 Example 10-12. 



Poinl I A 1 




SECTION 10-8 Fluids in Motion; Flow Rate and the Equation of Continuity 269 



Bernoulli's principle 



~A1. h 




FIGURE 10-20 (repeated) 
Fluid flow through a pipe of 
varying diameter. 



Bernoulli's Equation 



Have you ever wondered why an airplane can fly, or how a sailboat can move 
against the wind? These are examples of a principle worked out by Daniel 
Bernoulli (1700-1782) conce