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Full text of "Science Pertaining To Civil Engineering"

International Library of Technology 
398 



Sciences Pertaining to 
Civi. Engineering 



167 ILLUSTRATIONS 



By 

EDITORIAL STAFF 

INTERNATIONAL CORRESPONDENCE SCHOOLS 



FUNDAMENTAL PRINCIPLES OF MECHANICS 

ANALYTIC STATICS 

KINEMATICS AND KINETICS 

HYDROSTATICS 

PNEUMATICS 
RUDIMENTS OF ANALYTIC GEOMETRY 



Published by 
INTERNATIONAL TEXTBOOK COMPANY 

SCRANTON, PA. 
1927 



GJ? 



Fundamental Principles of Mechanics: Copyright, 1906, by INTERNATIONAL TEXT- 
BOOK COMPANY. Entered at 'Stationers' Hall, London. 

Analytic Statics, Parts 1 and 2: Copyright, 1906, by INTERNATIONAL TEXTBOOK 
COMPANY. Entered at Stationers' Hall, London. 

Kinematics and Kinetics: Copyright, 1906, by INTERNATIONAL TEXTBOOK COMPANY. 
Entered at Stationers' Hall, London. 

Hydrostatics: Copyright, 1906, by INTERNATIONAL TEXTBOOK COMPANY. Entered at 
Stationers' Hall, London. 

Pneumatics: Copyright, 190.6, 1901, by INTERNATIONAL TEXTBOOK COMPANY. 
Copyright, 1901, 1897, 1895, 1893, by THE COLLIERY ENGINEER COMPANY. 

Rudiments of Analytic Geometry: Copyright, 1906, by INTERNATIONAL TEXTBOOK 
COMPANY. Entered at Stationers' Hall, London. 

All rights reserved 
Printed in U. S. A. 



^BB^g, 




PRESS OF 

INTERNATIONAL TEXTBOQJK COMPANY 
SCRANTON, FA, 

398 



92109 



PREFACE 

The volumes of the International Library of Technology are 
made up of Instruction Papers, or Sections, comprising the 
various courses of instruction for students of the International 
Correspondence Schools. The original manuscripts are pre- 
pared by persons thoroughly qualified both technically and by 
experience to write with authority, and in many cases they are 
regularly employed elsewhere in practical work as experts. 
The manuscripts are then carefully edited to make them suit- 
able for correspondence instruction. The Instruction Papers 
are written clearly and in the simplest language possible, so as 
to make them readily understood by all students. Necessary 
technical expressions are clearly explained' when introduced. 

The great majority of our students wish to prepare them- 
selves for advancement in their vocations or to qualify for 
more congenial occupations. Usually they are employed and 
able to devote only a few hours a day to study. Therefore 
every effort must be made to give them practical and accurate 
information in clear and concise form and to make this infor- 
mation include all of the essentials but none of the non- 
essentials. To make the text clear, illustrations are used 
freely. These Illustrations are especially made by our own 
Illustrating Department in order to adapt them fully to the 
requirements of the text 

In the table of contents that immediately follows are given 
the titles of the Sections included in this volume, and under 
each title are listed the main tppics discussed. 

INTERNATIONAL TEXTBOOK COMPANY 
B 



CONTENTS 



NOTE. Tills volume is made up of a number of separate Sections, the page 
numbeis of which usually begin with 1. To enable the reader to distinguish 
between the different Sections, each one is designated by a number preceded by a 
Section mark (), which appears at the top of each page, opposite the page number. 
In this list of contents, the Section number is given following the title of the Section, 
and under each title appears a full synopsis of the subjects treated. This table of 
contents will enable the reader to find readily any topic covered. 

FUNDAMENTAL PRINCIPLES OF MECHANICS, 27 

Pages 

Motion 1-10 

Introductory Definitions 1-4 

Matter; Body, substance, and material; Motion and rest; 
Path; Initial and final position; Displacement. 

Velocity 5-10 

Velocity in uniform motion; Formulas for uniform 
motion ; Velocity in variable motion ; Average or mean 
velocity; Instantaneous velocity; Direction and magni- 
tude of velocity. 

Force and Mass 1 1-52 

Force ' 11-14 

Definition of force; Balanced and unbalanced forces; 
Equilibrium; Weight; Measure of force; Magnitude of 
a force; Direction of a force; Mechanics; Dynamics; 
Kinematics; Statics. 

Fundamental Laws of Dynamics 15-35 

Law of inertia ; Facts on which the law is founded ; 
Galileo's law of the independent effect of forces; 
Acceleration; Formulas for uniformly accelerated mo- 
tion; Initial and final velocities; Mass; Determination 
of the mass of a body; Law of action and reaction; 
Stress ; Tension, compression, and 'pressure. 

Immediate Consequences of the Preceding Laws 36-52 

Important formulas; Space passed over in uniformly 
accelerated motion; Modification of the formulas 
when the body has an initial velocity; Retardation; 
Composition and resolution of forces. 

:v 



vi CONTENTS 

ANALYTIC STATICS, 28, 29 

28 Pages 

Concurrent Coplanar Forces 1-33 

Definitions 1-3 

Coplanar and non-coplanar forces; Analytic statics; 
Graphic statics ; System of forces ; Equilibrants ; Inter- 
nal and external forces. 

Fundamental Principles and Formulas 4-10 

Moments 11-13 

Conditions of Equilibrium 14-23 

General statement of the conditions of equilbrium; Equi- 
librium of three forces ; Selection of axes. 

Stresses in Framed Structures 24-33 

Structure ; Machine ; Frames ; Trusses ; Supports ; Reac- 
tions ; Struts and ties ; Determination of stresses. 

Parallel Forces 34-49 

Coplanar Forces 34-45 

Two forces ; Theorem of moments ; Definition of a couple ; 
Any number of forces. 

Non-Coplanar Parallel Forces 46-49 

29 

Center of Gravity 1-28 

Definitions and General Properties 1-6 

Important Cases .' 7-28 

Symmetrical figures ; Determination of the center of grav- 
ity by addition and subtraction; Center of gravity of 
polygons; Center of gravity of areas bounded by cir- 
cular arcs ; Center of gravity of a plane area bounded 
by an irregular curve; Center of gravity of solids. 

Coplanar and Non-Concurrent Forces 29-43 

Couples 29-35 

Effect of a circle; Equivalance and equilibrium of coax- 
ial couples. 

Equivalence and Equilibrium of Coplanar Non-Concur- 
rent Forces ". 36-43 

Friction 44-65 

Sliding Friction 44-54 

Definitions and general principles; Angle and coefficient 
of friction. 

Resistance to Rolling 1 55-57 

The Inclined Plane 58-65 



CONTENTS vii 

KINEMATICS AND KINETICS, 30 Pages 

Composition and Resolution of Velocities 1-7 

Uniform Motion in a Circle 8-22 

Angular Velocity 8-12 

Centripetal and Centrifugal Forces 13-17 

Motion of a Train on a Curved Track 18-22 

Work and Energy 23-34 

Work 23-28 

Energy 29-34 

HYDROSTATICS, 31 

Properties of Liquids : 1-2 

Liquid Pressure 3-24 

Pascal's Law and Its Applications 3-6 

General Theory of Liquid Pressure 7-14 

Pressure on an Immersed Surface 15-24 

Buoyant Effort of Liquids 25-33 

Immersion and Flotation 25-28 

Specific Gravity 29-33 

PNEUMATICS, 32 

Properties of Air and Other Gases 1-16 

Vacuum; Barometer; Tension of gases; Mariotte's law; 
Manometers and gauges ; Gay-Lussac's law. 

The Mixing of Gases. 14-16 

Pneumatic Machines 17-41 

The Air Pump 17-22 

Air Compressors 23-25 

The Siphon 26-27 

The Locomotive Blast 28 

Pumps 29-41 

Suction pump; Lifting pump; Force pump; Plunger 
pump; Double-acting pumps; Steam pump; Centrif- 
ugal pumps; Hydraulic ram; Power necessary to work 
a pump. 

RUDIMENTS OF ANALYTIC GEOMETRY, 33 

Graphs of Equations 1-14 

Equations of Lines 15-31 



viii CONTENTS 

RUDIMENTS OF ANALYTIC GEOMETRY 

(Continued') Pages 

The Straight Line 16-22 

Equation of the straight line, Graph of any equation of 
the first degree, Applications, Pressure on the back of 
the dam 

The Parabola 23-31 

Focus; Directrix, Vertex, Axes, Parameter, Applica- 
tions , Jet of water issuing from a small orifice , Moment 
in a beam. 



FUNDAMENTAL PRINCIPLES OF 
MECHANICS 

MOTION 

INTRODUCTORY DEFINITIONS 



MATTER-BODY 

1. Matter may be defined as either- (a) whatever is 
known, or can be supposed capable of being known, through 
the sense of touch; or (b) that which occupies space 

2. A phenomenon is whatevef happens in nature. The 
motions of the heavenly bodies, the rolling of a ball on the 
ground, the generation of steam from water, the formation 
of rust on the surface of iron, the circulation of the blood, 
the beating of the heart, thinking, speaking, walking ali 
these are phenomena It will be seen from the definition 
and examples just given that the word phenomenon, as 
understood and used in science, does not mean "something 
extraordinary or wonderful." For science, there is nothing 
extraordinary, and a phenomenon is simply a fact anything 
that happens. 

3. Body, Substance, Material. A body is any 

limited portion of matter, as a block of wood, a coin, a 
stone, a piece of flesh, an animal. Another meaning of the 
word will be given presently. 

copvmoHTio BY INTRHNATIONAU TEXTBOOK COMPANY ALL RIHT MMMVBD 

27 

I I. T 8-3 



2 FUNDAMENTAL PRINCIPLES 27 

4. All bodies do not possess the same properties, or do 
not possess them in the same degree. This fact makes it 
necessary to distinguish different kinds of matter, to which 
different names are given 1 as, iron, water, air, flesh. 

The word substance is often employed in the same sense 
as the word matter, but in its more common use, it refers 
to matter of a special kind. Thus, water, steel, iron, are 
substances, or kinds of matter. 

The word body is also used, especially in physics and 
chemistry, to denote matter of a particular kind. In this 
sense, water, air, hydrogen, gold, are called bodies. In 
mechanics, however, the term is usually taken in the sense 
given to it in Art. 3. 

5. Those substances that are used for works of art are 
called materials: steel, iron, brick, stone are examples. 

6. Particle. A particle is a body so small that its 
dimensions may be disregarded. A particle is accoidmgly 
treated as if it were a geometrical point, except that it is 
sometimes necessary to consider some of its physical 
properties, such as weight. A particle is often called a 
material point. All bodies are treated as being com- 
posed of an indefinitely large number of particles. 

7. It will be noticed in what follows that, in many cases, 
properties and principles that have been established and 
stated as relating to material points are extended to whole 
bodies. Thus, after investigating and explaining the laws 
governing the motion of a material point under certain 
circumstances, these laws are illustrated by taking as 
examples the motions of wheels, cars, engines, etc. The 
reason is that in cases of this kind the properties considered 
do not depend on the size of the bodies. For example, the 
speed of a car moving in a straight track is measured by the 
speed of any of its points, and what applies to the speed of 
a point applies to the speed of the car. 

So, too, the force requj v to move in a :traight track a 
train of a certain weigh the same as the force necessary 
to move a particle of the same weight; and, therefore, the 



27 OF MECHANICS 3 

laws obtained m the case of a particle may be applied to the 
motion of a tram or any othei body, bo long as the only 
thing to be taken into account is the weight of the body. 

8. Deformation. When a body is pulled, pressed, or 
struck m any direction, its shape is more or less changed, 
either permanently or temporarily. A change of this kind 
is called deformation, and the body undergoing it is said 
to be deformed. 

9. A rigid body is a body that is not deformable, that 
is, a body whose foim and dimensions remain the same, to 
whatever action it may be subjected 

There are no absolutely rigid bodies m nature; but it often 
simplifies matters to consider bodies at first as being per- 
fectly rigid, which is equivalent to neglecting their deform- 
ability, the latter being afterwards taken into account. 
Besides, in many cases, deformability plays so small a part 
that, for practical purposes, it may be neglected entirely. 



MOTION AND REST 

10. Relative Position. The relative position of a 

point with respect to another is determined by the length 
and direction of the straight line between the two points 
When either the length or the direction, or both the length 
and direction, of this line change, the relative position of the 
two points is said to change 

The relative position of two bodies with respect to each 
other is determined by the relative position of the points of 
one body with respect to those of the other. When the rela- 
tive position of one or more points m one of the bodies with 
respect to the points of the other body changes, the relative 
position of the two bodies, with respect to each other, is also 
said to change. 

11. Motion is a change iiT-rtie relative position of two 
bodies, and is said to be post* 1 ' M by either body with 
respect to the other. 



4 FUNDAMENTAL PRINCIPLES 27 

12. Rest is the condition of two bodies not in motion 
with respect to each other, and is said to belong to either 
body with respect to the other. Each body is said to be 
fixed with respect to the other body. 

13. When the motion or rest of a body is referred to, 
without specifying 1 any other body, it is generally understood 
that the other body is the earth or its surface. Thus, the 
motion of a train, of a steamer, of a horse, as usually spoken 
of, means the change of position of the train, steamer, or 
horse with respect to the ground that is, to the surface of 
the earth, or to objects on that surface. 

14. A body may be in motion with respect to another 
body, and at rest with respect to a third body. For example, 
the smokestack of a locomotive is at rest with respect to the 
boiler, since their relative position does not change; yet, both 
may be moving with respect to objects on the ground. Like- 
wise, a man standing on the deck of a moving vessel is at 
rest with respect to the vessel, but in motion with respect to 
the water, the shore, etc. 

15. Path. The path, or trajectory, of a moving point 
is the line over which the point moves, or, in geometrical 
language, the line generated by the moving point. The path 
of a point may be straight or curved, or a combination of 
straight and curved lines. 

16. Initial and Final Position. When the motion 
of a point during a certain time or over a certain portion of 

its path is considered, tha 
position occupied by the 
point at the beginning of 
the time is called the Ini- 
tial position of the point, 
Flt< ' and the position at the end 

of the time is called the final position. Suppose that a 
particle starts from A, Fig. 1, and moves so that it describes, 
in a certain time, the curve A M B\ this curve is the path of 
the particle. With respect to the time considered, A is the 
initial and B the final position of the particle,. , , , 




27 OF MECHANICS 5 

17. Displacement. The length of the straight line 
joining the initial and the final position of a moving point 
is called the displacement of the point while the point 
passes from the former to the latter position. In Fig 1, 
in which the point moves from A to B m the curved path 
AM B, its displacement during this motion is the length of 
the straight line AB 

18. The space passed over, or described, m a certain 
time by a moving point is the length of the part of the path 
over which the body passes during that time. In Fig 1, the 
space described by the particle between A and B is the length 
of the curve A MB 

19. Direction of Motion In a Curve. By the direc- 
tion of a curve at any point is meant the direction of its 
tangent at that point So, too, when a point is moving in a 
curved path, the direction of motion at any moment is the 
direction of the tangent to the path at the point occupied at 
that moment by the moving point. In Fig 1, the line PT 
just touches the curve AMD at P, it is, therefore, a tangent 
to the curve at P, and shows the direction in which the 
particle is moving when it reaches the position P 

VELOCITY 



VELOCITY IN UNIFORM MOTION 

20. Uniform Motion. A point is said to move with 
uniform motion when it passes over equal distances in 
any and every two equal intervals of time The fact, for 
instance, that a point moves over 10 feet during the first 
second of its motion, and over an equal space during the 
fiftieth second, is not enough to define the motion as uni- 
form: that the motion may be uniform, the point must 
describe the same space (10 feet m this case) in every 
second, whatever instant is chosen in which to begin to 
count the time; thus, the space described by the point 
between the middle of the third second and the middle of 



6 FUNDAMENTAL PRINCIPLES 27 

the fourth must be the same as the space described in the 
twentieth second, or during the second in which the time of 
motion changes from 27 15 to 28.15 seconds, or, in short, 
during any and every interval of 1 second. 

21. Definition of "Velocity. It is a familiar fact that 
bodies move more or less rapidly, more or less slowly By 
this is meant that they require more or less time to move 
from one place to another, or that they move over longer or 
shorter distances in a given time. In ordinary language, 
the term speed is used to denote the degree of quickness or 
rapidity of motion. The measure of the speed, or of the 
degree of quickness, of a motion is called velocity. How 
this measure is obtained will now be explained. 

22. Uniform. Velocity. Let a point M> Fig. 2, move 
along the path A B, which may have any form, either curved 

or straight; and suppose the 
motion to be such that equal 
lfIG 2 ""* spaces are passed over in equal 

times; that is, suppose the motion to be uniform. Also, let 
the number of units of length passed over in a unit of time 
(as the number of feet passed over in 1 second) be denoted 
by v. It is evident that the magnitude of this number v 
depends on how fast the point is moving. Thus, if the point 
passes over 10 feet in 1 second, it is moving twice as fast as 
if it passed over only 5 feet in 1 second Hence, the num- 
ber v may be used to measure the rapidity of the motion, in 
which case v is the velocity of the moving point. In general 
terms, then, the velocity of a point moving with uniform 
motion is the space passed over by the point in a unit of time. 

When, as here assumed, the motion is uniform, the veloc- 
ity also is said to be uniform, or constant. What is meant 
by variable motion and variable velocity will be explained 
presently. 

Velocity is expressed in units of length per unit of time. 
For example, if a body moving uniformly passes over 7 feet 
in 1 second, its velocity is 7 feet per second. In 1 minute, 
the body passes over a space of 7 X 60 = 420 feet; hence, 




27 OF MECHANICS 7 

its velocity can be expressed also as 420 feet per minute. A 
train that moves uniformly and travels 45 miles an hour has 
a velocity of 45 miles per hour. 

23. Formulas for Uniform Motion. Let a point or 
body moving uniformly pass over a space j during any time t. 
Then, it is obvious that the space described in a unit of time 
is j t. Hence, 

v = S - t (1) 

For example, if the point moves over a space of 10 feet in 
2 seconds, v = 10 2 = 5 feet per second 

If the velocity and the time are known, the preceding 
formula gives the space passed over. 
s = vt (2) 

Similarly, if the velocity and the space are given, the 
same formula, or formula 2, gives the time. 

t= s - (3) 

v 

EXAMPLE 1 The wheels of a carnage moving with uniform velocity 
make 3,760 revolutions between two stations A and B The diameter 
of each wheel is 4 feet, and the time employed by the carnage to 
travel from one station to the other is 7 6 hours It is required to find 
the velocity of the carnage, m feet per second (By the velocity of 
the carriage is meant the velocity of any of its points ) 

SOLUTION During each revolution, the carriage passes over a 
distance equal to the circumference of the wheel, or 4 X 3 1416 ft. 
The total distance traveled, or space passed over, m feet, is s = 3,760 
X 4 X 3.1416 The time required to travel this distance is 7 6 br. 
As the velocity is required in feet per second, this time must be 
reduced to seconds, which gives / = 75X60X60 Substituting In 
formula 1 the values of s and / just found, 



EXAMPLE 2. The distance between two ports is 400 miles What 
must be the velocity, in knots, of a vessel that will cover the distance 
in 18 hours? (A knot is a velocity of 6,080 feet per hour.) 

SOLUTION To find the velocity m feet per hour, we have, 
i - 400 mi. = (400 X 5,280) ft , t - 18 hr Therefore, by formula 1, 

, = *Lxpo f , per hr . . x M knote . 19 . 298 lnot , 



FUNDAMENTAL PRINCIPLES 27 



VELOCITY IK VARIABLE MOTION 

24. Variable motion is that motion in which the 
moving point, or body, passes over unequal spaces m equal 
times, or over equal spaces in unequal times 

25. Average or Mean Telocity. It will be remem- 
bered that the distinguishing characteristic of uniform 
motion is the equality of all distances passed ovei m equal 
intervals of time, and that the quotient obtained by dividing 
any space 5 by the number / of units in the conesponding 
interval of time is a constant quantity (v, formula 1 of 
Art 23). Unless all these quotients are equal, the motion, 
and therefore the velocity, is not uniform 

Suppose, for example, that a tram moves over a distance 
of 60 miles in 45 minutes. If it is known beforehand that 
the motion is uniform, the velocity may be found by 

dividing 60 miles by 46, which gives v = - miles 

45 3 

per minute. In this case, the space described in any num- 
ber t of minutes would be, by formula 2 of Art. 23, 
v t = &/ miles. 

Suppose, on the contrary, that it is not known whether 
the motion has been uniform during the 45 minutes, but 
that the distance passed over during the first 5 minutes is 
known to be 5.5 miles, and that the distance passed over 
between the end of the thirtieth and the end of the fortieth 
minute is 13.98 miles Then, the quotients obtained by 
dividing the spaces by the numbers measuring the corre- 
sponding times are, respectively, 

5-5 _ T in 13.98 _ , OQQ 60 _ 4 
T " ' ~W ~ lf6W] 46 " 8 

Since these quotients are not equal, there is no uniform 
motion, and therefore no uniform velocity. 

If, however, another train is imagined moving with a uni- 
form velocity of $ miles per minute, this tram will pass over 
the distance of 60 miles in 45 minutes. For this reason, 
\ miles per minute is called mean, or average, velocity 
of the actual train considered; it is not a velocity that 



27 OF MECHANICS 9 

the train really possesses, but rather the equivalent velocity 
with which a body moving uniformly would move over 
the same space in the same time Likewise, if the motion 
of the tram during; the first 5 minutes is considered, its 

5 5 

mean velocity, foi that interval, is - = 1 10 miles per 

5 

minute. In general, if a body moves over a space j in the 
time /, its mean velocity v m , during that time, is defined 
mathematically by the formula 



26. Variable Velocity Instantaneous Velocity. 

When a point is moving with variable motion, its velocity 
at any instant is the velocity that the point would have if, 
at that instant, its speed ceased to change, that is, if, after 
that instant, the point moved neither faster nor more slowly 
than it is moving. Since the motion is variable, the velocity 
at any instant is not any velocity with which the body actually 
moves over any part of its path, but simply the velocity 
with which it would move if its speed became invariable. 

The velocity of a moving point or body at any given 
moment is called the instantaneous velocity at that 
moment. In variable motion, the velocity is said to be 
variable, because it changes or varies from instant to 
instant. 

27. To illustrate the character of variable motion and 
velocity, imagine a train A to start from rest and leave a 
station at the same time that another train B is passing the 
station with a uniform velocity of 60 miles per hour, and 
suppose the two trains to run on two parallel tracks. At 
first, B will move ahead of A, since A' starts from rest. 
vSuppose, however, that A moves faster and faster, so that 
to a person in B the motion of B with respect to A will 
appear to become slower and slower; there may then be a 
time when B will be losing instead of gaining speed, with 
respect to A, or when A will appear to be moving past B. 
Thus, during the first minute, B will move over 1 mile, and 



10 FUNDAMENTAL PRINCIPLES 27 

A may move over only 4 mile; but, during: the following 
minute, A may move over li miles, and then, it will be 
i mile ahead of JS. Since A is at first losing space with 
respect to B and then is gaming, there must be an instant 
at which A is neither losing nor gaining; at that instant the 
two trains are evidently at rest with respect to each other, 
and if the speed of A ceased to change, A would continue 
to move with the same velocity as , or 60 miles per hour. 
The velocity of A at that instant is, therefore, 60 miles per 
hour. 

It must be understood that, as already stated, the tram A 
does not actually move with the velocity of 60 miles per 
hour during any interval of time; but this is the velocity with 
which A would move if, after the instant considered, its 
motion underwent no further change. 

28. Direction and Magnitude of Velocity. By the 
direction of the velocity of a moving point is meant the 
direction in which the point is moving. 

29. By the magnitude of a velocity is meant the 
numerical value of the velocity, expressed in units of length 
per unit of time, as feet per second or miles per hour. 

EXAMPLE A float moves down a stream from a point A to o. 
point D, a distance of 800 feet, in 2-J minutes The float is observed 
at two intermediate points B and C, whose distances from A are, 
respectively, 350 and 575 feet, and it Is found that it moves from A 
to B in 45 seconds, and from B to Cm 70 seconds. Required, the 
mean velocity, m feet per second, of the surface of the stream 
(a) between A and D, (6) between A and B t (c) between B and C; 
(rf) between C and D 

SOLUTION (a) Here s = 800 ft , / = 2| min = 150 sec ; and the 
formula of Art 25 gives 

v m = r = r^r = 5.33 ft. per sec. Ans. 
(6) Here s = 350 ft , t = 45 sec., and, therefore, 

v m = 7 = ~r=- = 7 78 ft per sec. Ans. 
/45 

(c) Here s = 575 ft - 360 ft = 225 ft., t = 70 sec , and, therefore, 
225 

t ** 70 



27 OF MECHANICS 11 

(<f) Here s = 800 ft - 575 ft = 225 ft , t = 150 sec - (45 + 70) sec. 
= 35 sec , and, therefore, 

v m = - = -T- = 6 429 ft per sec. Ans. 



EXAMPLES FOB PRACTICE 

1 A tram moves with uniform velocity between two stations 
375 miles apart, the distance is traveled in 7i hours. What is the 
velocity of the train, in feet per second? Ans 73 333 ft per sec 

2 A train moves from station A to station , a distance of 
90 miles, in 1$ hours, from station B to station C, a distance of 
17 miles, m % hour; and from station C to station D, a distance 
of 64 miles, in 2 hours Find its mean velocity, m feet per second 
(a) between stations A and >, (b) between stations B and C\ 
(c) between stations B and D. ( (a) 59 012 ft per sec. 

Ans { (d) 49 867 ft per sec 
[(c) 4752ft. per sec. 

FORCE AND MASS 



FORCE 



DEFINITIONS RELATING TO FORCE MEASURE OF FORCE 

30. Definition of Force. It is known from experience 
that, when a body is at rest, it can be set in motion by the 
action of another body. Thus, if a block of iron is lying on 
the ground, it can be set m motion by pulling it with a rope, 
by pushing it with the hand, or by placing a strong magnet 
near it. Furthermore, the block may be pushed by one per- 
son in one direction, while another person pushes it m the 
opposite direction; or, the magnet may be placed on one 
side of the block, at the same time that the block is pulled 
by a string from the opposite side. In cases of this kind, 
there may be no motion, on account of the neutralizing 
effects of two or more actions on the same body. We say, 
however, that there is a tendency to motion; for the 
moment one of the opposing actions is removed, the other 
causes the body to move 



12 FUNDAMENTAL PRINCIPLES 27 

31. This action of one body on another, producing, 01 
tending to produce, motion in the latter body, is called force, 
and the former body is said to exert force on the latter 

The characteristic of the action known as force is, then, 
that, when exerted on a body originally at rest, and not 
influenced by other bodies, it results in the motion of the 
first-mentioned body. The nature of the ensuing motion, 
and the result of the action in question, when the body 
acted on is already in motion or under the action of other 
bodies, are complicated phenomena to be either actually deter- 
mined by experiment or inferred from experimental data. 

32. Balanced and Unbalanced Forces. If two or 

more forces act on a body in such a manner that they cause 
no motion, owing to the neutralizing effects they tend to pro- 
duce, each force is said to be balanced by the combined 
action of the others, and is referred to as a balanced force. 

When a force acts on a body, and there is no opposing 
force preventing the motion of the body, the force is called 
an unbalanced force. 

These definitions apply to moving bodies as well as to 
bodies at rest If two or more forces are balanced when 
exerted on a body at rest, they are also balanced when the 
body is in motion; that is, they do not affect the motion of 
the body. 

33. Equilibrium. A body is in equilibrium when it 
is under the action of balanced forces. The balanced forces 
themselves are also said to be in equilibrium. 

It should be noticed that equilibrium and rest are not 
equivalent terms A body may move uniformly while acted 
on by balanced forces, in which case it is in equilibrium, but 
not at rest In this case, however, the motion of the body 
is not due to the balanced forces acting on it. This subject 
will be better understood after the law of inertia, presently 
to be explained, has been studied 

34. Weight. Experience teaches that, when a body is 
unsupported, it falls to the ground. This fact is ascribed to a 
force exerted by the earth on all bodies, which force is known 



27 OF MECHANICS 13 

by the general name of force of attraction, or simply 
attraction, and also force of gravity, or simply gravity. 
The atti action of the earth on any particular body is called 
the weight of the body The methods of comparing the 
weights of bodies are well known. 

35. Measure of Force. For engineering purposes, 
force is expressed in units of weight, such as pounds, kilo- 
grams, etc. The reason for this will be more apparent when 
it is considered that every force can be replaced by a weight. 
Thus, suppose a body P, Fig. 3, to be suspended from the 
extremity A of a per- 
fectly symmetrical A Aj B 
beam AB resting at 

its center on a knife -^ 

edge F. If we wish / f\ \,Q 

to prevent P from 

falling, we may pull downwaids on the string at Q, 01 
attach at Q a. weight equal to P, or tie a small piece of 
iron at Q and place a magnet underneath. In all these cases, 
the weight of P balances, and is therefore equivalent to, the 
force at Q, by whatever means the latter may be produced. 
As the two forces are equivalent, the one may be meas- 
ured by the other, and we may say that the force acting at Q 
is 20 pounds, kilograms, tons, etc., according as P weighs 
20 pounds, kilograms, tons, etc. Suppose, for instance, that 
the weight of P is 10 pounds, and that we pull at Q with 
sufficient force to keep P from falling, but without moving it. 
Then, the force with which we are pulling is 10 pounds. If 
P falls, the pull is less than 10 pounds; and if it rises, the 
pull is greater than 10 pounds. 

36. Magnitude of a Force. By the magnitude of a 

force is meant the numerical value of the force, expressed 
in units of weight. If, for example, a force is equivalent to 
a weight of 12 pounds, its magnitude is 12 pounds. 

37. Direction of a Force. The direction of a force 
is the direction in which the force moves, or tends to move, 
the body on which it acts. 



14 FUNDAMENTAL PRINCIPLES 27 

DEFINITION AND DIVISIONS OF MECHANICS 

38. Mechanics is the science of force and motion. This 
science is divided into two general branches, (a) dynamics, 
(b) kinematics or phoronomics. 

39. Dynamics is the science of force and its effects. 
In this science, the forces applied to bodies are given, and 
the resulting effects of these forces are determined, or the 
conditions of motion are given, and the forces necessary to 
produce that motion determined. Rest is considered as a 
special case of motion in which the velocity is zero. 

40. Kinematics treats of motion alone, without refer- 
ence to either force or the physical or mechanical propeities 
of bodies. Suppose, for instance, that a body is moving in 
a circle with uniform velocity, and that it is required to deter- 
mine what force is necessary to preserve this kind of motion, 
this is a problem in dynamics, for it relates to force Suppose 
that it is required to determine what space the body travels 
in a certain time; this is a problem in kinematics, for it can 
be solved without having regard to anything but the velocity 
and path of the body. 

41. Subdivisions of Dynamics. Dynamics is sub- 
divided into two branches: (a) kinetics, (b] statics. 

42. Kinetics treats of unbalanced forces and their effects. 

43. Statics treats of the equivalence and equilibrium of 
forces. 

NOTE The preceding divisions and definitions are of very recent 
origin Formerly, the term dynamics was used in the sense m which 
kinetics 13 now used, the latter terra being then unknown Even 
today, the old meaning is frequently given to the term dynamics, 
the best modern writers, however, use it in the sense just defined 
that is, to denote the science of force in general, whether it pro- 
duces motion or not 

44. Applied mechanics is the application of mechan- 
ical principles to the works of human art. 



27 OF MECHANICS 16 



FUNDAMENTAL LAWS OF DYNAMICS 



I-iAW OF INKltTIA 

45. Statement of l ho IJHW. In giving the definition 
of force (Art. 30), icference was made to the fuel, known 
from experience, that a body may act on anothei body, 
originally at rest, in such a manner as to pioduce motion 
in the latter body. This action was defined as foice That 
definition relates only to the effect of force on a body origi- 
nally at lest. The law presently to be stated expresses the 
general effect of force under any eiieumsUnces; that is, 
whether the body on which it is excited is at icst or in 
motion. Tins law, known as the lir\v of lm>rltn, or the 
first law of motion, may be formulated as follows: 

// a body is not acted on by fom\ the body, m> u tvhol<\ /.s 
either at test ot inovmg uniformly in a sh (tight faith. 

This is not as definite a statement of the law of ineilia as 
can be given. But the complete statement requires, in order 
to be understood, some familiarity with mechanical piinciples 
and conceptions of a high order, with which the beginner 
cannot be supposed to be acquainted 

By saying that a body as a whole is in motion, it is meant 
that all the points of the body are in motion. Each of the 
wheels of a locomotive, for instance, has a motion us a 
whole, and, if the track is straight, the wheel, as a whole, is 
said to be moving in a straight path, whose direction is that 
of the track. Any one would understand what was meant 
by saying that a billiard ball was rolling in a diiection par- 
allel to one of the cushions of the billiard table. 

In the simple cases of motion just given, the direction of 
the motion of the body as a whole is plainly seen to be flu* 
direction in which its center of figure is moving. In the OIKO 
of unsymmetrical bodies, the direction of their motion is the 
direction in which a certain point in them, called the oonlor 
of fifravlty, is moving, For the present, however, the 
student may think of the motion of symmetrical bodies 



16 FUNDAMENTAL PRINCIPLES 27 

only, such as balls, disks, cubes, prisms, etc., all of which 
have a center of figure. 

46. Facts on Wlilcli the Law is Founded. That a 
body does not start to move by itself is a fact so familiar 
that it is scarcely necessary to state it That a body, once 
set in motion, does not come to rest unless acted on by force 
is not so obvious. Bodies often come to rest apparently by 
themselves, without any assignable cause, and this gave 
rise to the belief, prevalent among the ancients, and still 
entertained by some, that rest is the natural condition of 
all bodies, and that all moving bodies have a tendency to 
come to rest. A little consideration will show that this is 
an erroneous view of the nature of motion. If a block is 
placed on a stone pavement and struck sidewise, it will 
slide for a short distance and soon come to rest; if the 
experiment is tried on an asphalt pavement, the block will 
move farther and come to rest after a longer time; it will 
slide longer and farther on a piece of marble, and longei and 
faither still on a sheet of ice. These facts plainly indicate 
that the tendency to come to rest is not inherent in the block 
itself, but depends on the action of other bodies namely, the 
bodies on the surfaces of which the block slides, and we 
naturally infer that, were it not for this action (or frlc- 
tlonal resistance, as it is called), the block would not 
come to rest at all. 

Another common obstacle to motion is the resistance of 
the medium, by which is meant the action of the substance 
through which the body moves. Thus, if a block is placed 
on a horizontal suiface, immersed in mercury, and then 
struck sidewise, it will move through a very short distance, 
If the experiment is repeated m water, the block will describe 
a longer space before coming to rest. The space will be 
longer in air, and longer still in a space (called a vacuum) 
from which the air has been removed by means of an air 
pump. Here, as before, the inference is that, did the medium 
offer no resistance, or, more properly, were there no medium, 
the body would, move forever in a straight path. 



27 OF MECHANICS 17 

47. In the preceding examples, it will be observed that, 
whether, the body is brought to rest by fnctional resistance 
or by the resistance of the medium, the diminution of its 
velocity is continuous, in other words, the body comes to rest 
gradually, and its velocity constantly diminishes while the 
resisting force acts 

An example of the action of force in increasing the velocity 
of a moving body is afforded by the familiar phenomenon of 
a falling body Here, the body is constantly acted on by the 
attraction of the earth, and its velocity is observed to increase 
very lapidly As another example, suppose that a person 
pushes a car on a horizontal surface. The car moves at first 
with almost no velocity; but, as the person continues to push, 
the velocity keeps growing greater and greater. 

48. The necessity of applying force to a moving body in 
order to change the direction of its motion is also a familiar 
fact. If a prismatic block is sliding on a surface in the diiec- 
tion of its axis, no change m the direction of its motion will 
be observed unless the prism is struck, or pulled, or pushed, 
or in some other way interfered with by some other body. 
Of course, all foices do not change the direction of motion, 
but only those exeited by bodies that are, so to speak, outside 
the path of the moving body. Thus, in the example just 
given, a pull exerted by a rope in the direction of the axis of 
the prism would cause no change in the direction of motion; 
but, if the lope were pulled at right angles to the axis of the 
prism, the direction of motion would evidently change. 

49. Summing up, it may be said that, whenever the motion 
of a body has been observed to change \ cither in direction or in 
velocity , or in both, it has always been possible to trace the change 
to the influence of other bodies ikat ZA, of force. Furthermore ', 
bv diminishing' those influences, the changes referred to are corre- 
spondingly diminished; and the conclusion has been reached thai, 
could these influences be entirely eliminated, the said changes 
would not take place at all. 

If, then, a body is at rest, and no unbalanced force acts on 
it, it will continue at rest; if it is moving 1 in a straight line 

ILT398 3 



18 FUNDAMENTAL PRINCIPLES 27 

with uniform velocity, it will continue to move m the same 
straight line with the same velocity If a body moves under 
the action of a force for a certain time, its velocity will 
constantly change during that time; but, if the force ceases 
to act, the body will continue to move uniformly in the 
direction and with the velocity it had at the instant the foice 
was withdrawn. It is not necessary, however, in order that 
the body may continue to move uniformly, to withdraw the 
force or forces acting on it. the same result will be obtained 
if other forces are introduced equal and opposed to those 
already acting. An instance of this is afforded by the 
motion of a train- at first, the tractive force of the engine is 
greater than the combined resistance of friction and the air, 
and the velocity of the train constantly increases, but, as will 
be explained elsewhere, the resistance increases with the 
velocity, so that there is a moment when the traction of the 
engine is balanced by the resistance, and from that moment 
on, the train moves uniformly. 

50. Conversely, if either the velocity or the direction of 
motion of a body changes^ the body must be under the action of 
unbalanced forces, for, according to the law of inertia, 
rectilinear motion with uniform velocity (in which rest must 
be included as a special case) is the only possible motion of 
a body not acted on by unbalanced forces. Thus, if a body 
moves in a curve, it may be concluded that, whatever its 
velocity may be, some unbalanced force or forces must be 
constantly acting on the body. The moment these forces 
cease to act, the body will continue to move uniformly in 
the direction of the tangent to the path at the point occupied 
by the body at that moment. For example, if a stone is tied 
to a string and swung around, the pull of the string will keep 
the stone moving in a circle; but the moment the string 
breaks, the stone will fly off on a tangent to this circle. A 
train is kept on a curve by the resistance of the rails acting 
against the flanges of the wheels; but if this resistance is not 
great enough, the wheels get off the rails, and the train, in 
leaving the curve, moves straight ahead along the tangent 




27 OF MECHANICS 19 

51. Inertia of Bodies. The fact that the motions of 
all bodies follow the first law of motion is often expiess>ed 
by saying that all bodies possess the property of Inez-tin. 
In this sense, it is very common in mechanics to refer to 
what the motion of a body 
is or would be, "by virtue 
of the inertia of the body", 
that is, according to 
the law of inertia. Sup- 
pose, for instance, that a -* 
particle is moving along the path A B, Fig. 4 It follows 
fiom the first law of motion that the particle must be under 
the action of unbalanced forces, since the path is not straight. 
Suppose, also, that, by some means, it is found that the veloc- 
ity of the particle, when at /*, is 6 feet per second. Then, 
we say that, if the forces were suddenly withdrawn (or bal- 
anced), the particle, "by virtue of its inertia," would move 
uniformly along the tangent P T, describing a space of 6 feet 
in every second 



GALILEO'S LAW OF THE INDEPENDENT EFFECT OF 
FORCES 

52. Preliminary Explanation. Let A, Fig. 5, be a 
body acted on by a force Suppose the force to be such 
that, if it acts on the body for a certain time /, and the body 

is at rest when the force 

A r~~ 7- 1 *" begins to act, the body 

/ / moves over the path A A' 

I /in that timej so that its 

/ / final position is A 1 . Sup- 

/ / pose, now, that the body, 

/ / instead of being at rest 

/, /, when the force is applied, 

A PlGlG Al is moving along the path 

A A lt in such a manner that, if the force did not act, the body 

would move to A l in the time /. It is found from experiment 

that if, while the body has this motion, the force referred to 

above constantly acts on it during the time *, the direction of 



20 FUNDAMENTAL PRINCIPLES 27 

the force remaining; unchanged, the final position of the body 
after this time is a point A,', such that the line A^AJ is equal 
and parallel to A A 1 . In other words, the final position of 
the body, with respect to the position the body would have 
occupied if the force had not acted, is the same whether the 
force starts the body from rest or acts on the body while 
the latter is in motion and acted on by other foiceb Had 
the body been originally at rest, the effect of the force would 
have been to cause the displacement A A', in the direction of 
the force, in this case, the final position of the body, had not 
the force acted, would have been A. If the force had not 
acted and the body had been moving along A A t , its final 
position would have been As, the effect of the force is to 
cause the displacement A^AJ of the final position, which 
displacement is equal and parallel to A A'. 

53. Statement of Galileo's Law. From the pre- 
ceding facts, and from others of a similar character, is 
derived the following general law, which is Galileo's law 
of the Independent effects of forces: 

The effect of a force on a body is the same whether the body js 
at rest or in motion, and, if several forces act simultaneously on 
a body, each force produces its effect independently of the othct 
forces. 

The meaning of this proposition is that each force produces 
the same amount of displacement, parallel to its direction, as 
if it acted alone and moved the body from a state of rest 

Galileo's law may be otherwise stated as follows: 

When a force acts on a body, the position of the body at any 
time relatively to the Position the body would have if the force 
had not acted, is independent of the motion Produced m the body 
by any other forces. 

54. Experimental Verification. The following 
experiment affords an easy verification of Galileo's law In 
Fig. 6, a ball e is supported in a cup, the bottom of which is 
attached to the lever o in such a manner that a movement of o 
will swing the bottom horizontally and allow the ball to drop. 
Another ball b rests in a horizontal groove that is provided 



27 



OF MECHANICS 



21 



with a slit in the bottom. A swinging arm is actuated by 
the spring d m such a manner that, when drawn back, as 
shown, and then released, it will strike the lever o and the 
ball b at the same time 
This gives b an impulse 
in a horizontal direction 
and swings o so as to 
allow c to fall 

On trying the experi- 
ment, it is found that b 
follows a path shown by 
the curved dotted line, 
and reaches the floor at 
the same instant as <?, 
which drops vertically 
This shows that the 
force that gave the first 
ball its horizontal move- 
ment had no effect on 
the vertical force that 
compelled both balls to 
fall to the floor, the 
vertical force producing 
the same effect as if the 
horizontal force had not 
acted. 

ACCELERATION 

55. Change of 
Velocity Caused by 
an Unbalanced Uni- 
form Force. A uni- 
form force is a force 
whose magnitude and 
dnection do not change. Both from the law of inertia and 
fiom the definition of force, it follows that the effect of an 
unbalanced uniform force on a body originally at rest is to set 
the body m motion and impart to it a continually increasing 




22 FUNDAMENTAL PRINCIPLES 27 

velocity Galileo's law affords the means to determine 
exactly how that increase takes place. Suppose that a uni- 
form force acts on a body originally at rest, imparting to it, 
in 1 second, a velocity of 10 feet per second This means 
that, if the force acts during 1 second and then ceases to act, 
and the body is not interfered with by other forces, the body 
will, by the law of inertia, continue to move uniformly with a 
velocity of 10 feet per second If the force, instead of being 
withdrawn at the end of the first second, acts for 1 second 
longer, its effect will be independent of the motion already 
acquired by the body; that is, during the second second, the 
force will impart to the body a velocity of 10 feet per second, 
rn addition to the velocity the body had at the end of the first 
second, so that, at the end of the second second, the body 
will have a velocity of 20 feet per second Likewise, the vel- 
ocity at the end of the third second will be 30 feet per second; 
and so on. 

A similar law of change applies whenever a body is acted 
on by an unbalanced uniform force, that is, the velocity of 
the body increases by a fixed amount during every unit of 
time; or, in other terms, the velocity increases at a constant 
rate. If, for example, the change of velocity per second is a 
feet, the change m t seconds will be at feet per second. Of 
course, it is not necessary to use the second as a unit of time, 
nor the foot as a unit of length; any other units may be used. 

56. Acceleration. The acceleration of a body 
moving in a straight path under the action of an unbalanced 
force is the amount by which the velocity of the body in the 
direction o the force increases in a unit of time. 

57. When a body moves under the action of a uniform 
force, its change of velocity is the same for any two equal 
intervals of time, as has just been explained. In this case, 
the acceleration is said to be uniform, and the motion is 
said to be uniformly accelerated. 

58. In the example given in Art. 55, the acceleration is 
10 feet per second per second The import of this appar- 
ently confusing expression is this: J!f at any moment the 



I V J 



27 OF MECHANICS 23 

felocity of the body is v feet per second, this means that, if 
at that moment the force ceased to act, the body would con- 
tinue to move uniformly with a velocity of v feet per second. 
If, however, the uniform force continues to act for 1 second 
longer and then ceases to act, the velocity will have increased 
to v + 10 feet per second; that is, the body, if left to itself, 
will continue to move uniformly with that velocity, descri- 
bing v + 10 feet every second, or 10 feet more than before. 
Instead of saying that the acceleration is 10 feet per second 
per second, it is customary and sufficient to say simply that 
the acceleration is 10 feet per second, it being understood 
that the velocity is also measured in feet per second. If 
the acceleration of a body was said to be 50 miles per 
hour, the velocity would be understood to be measured in 
miles per hour, and to be, at the end of any one hour, 
50 miles per hour greater than at the beginning of that hour. 

59. Formulas for Uniformly Accelerated Motion. 

Let a body start from rest and move under the action of a 
uniform force for t seconds. If the acceleration due to this 
force is a, the velocity at the end of 1 second will be a, at 
the end of 2 seconds, 2 a; and at the end of / seconds, ta, or 
a t. Denoting by v the velocity at the end of / seconds, we 
have, then, 

v = at (1) 

"i (2) 

If, at any instant, the body has a velocity z> , the veloc- 
ity v t t seconds after that instant, is given by the formula 

v z/ + at (3) 
From this formula follows 

"V 1)a I A\ 

a - - * (4) 

The velocity v t) taken with respect to the interval of time 
considered, is called the initial velocity of the body; and v 
is called the final velocity. It is immaterial how the 
body has acquired the velocity V Q ; in any case, the effect of 
the applied force is to produce an increase of velocity equal 



24 FUNDAMENTAL PRINCIPLES 27 

to at, or v TV feet per second, in t seconds The change 

1) -~ i -' 7/ 

of velocity produced in 1 second is, therefore, - - , as 



expressed by formula 4, and this agrees with the general 
definition of acceleration In all this, it is assumed that the 
force acts in the direction of motion. 

If the body starts from rest, v = 0, and formulas 3 and 
4 become identical with formulas 1 and 2, respectively. 



MASS 

60. Force Proportional to Acceleration. Accord- 
ing to Galileo's law, if several forces act simultaneously on 
a body, each force produces its effect independently of the 
others Therefore, if the forces act m the same direction, 
they will produce an acceleration equal to the sum of the 
accelerations that the forces would produce if each force 
acted alone. Furthermore, if the forces are equal, the accel- 
eration will be equal to their number multiplied by the 
acceleration due to any one of them. It is also evident that 
any force may be supposed to be equivalent to any number 
of forces acting in the same direction as the given foice, and 
whose sum is equal to the given force Thus, a weight of 
5 pounds is equivalent to a weight of 3 pounds and one 
of 2 pounds, or to a weight of 4 pounds and one of 1 pound, 
or to five weights of 1 pound each, etc. 

Let a force of 3 pounds act on a body, imparting to it 
an acceleration of 5 feet per second. Then, a force of 
12 pounds, which is equivalent to four forces each equal to 
3 pounds, will produce an acceleration four times as great, 
or 4 X 5 = 20 feet per second. Any other force will produce 
an acceleration that will be as many times 5 feet per second 
as the force contains 3 pounds. For instance, a force of 

49 

49 pounds will produce an acceleration of -- X 5 = 81.67 feet 

3 

per second, nearly. In general, let a force F l impart the 
acceleration a^ to a certain body. Let F n be another force 
imparting the acceleration a n to the same body. If F n is 



27 OF MECHANICS 25 

twice F lt then will a n be twice ,; and, generally, if F n = nF l 
where n is any number, fractional or integral, then a n n ,. 
Now, 



= = ; s = - = ; 

7*i ^*i ! tfl 

that is, ~ = 



Therefore, /or /// &z#z body, any two forces are to each other 
as the accelerations they impart to the body. 
From the last equation follows 

^ = 

a n ai 

61. If several forces F lt F a) F,, etc are capable of pro- 
ducing, respectively, the accelerations a lt a a , a,, etc. in the 
same body, then, 

5 = 5 = 5, etc. 
fti a, a, 

Each of these quotients may be considered to represent 
the force necessary to give the body an acceleration of 1 foot 
per second (or, in general, 1 unit of length per unit of time). 
If the value of this force, or the common value of the pre- 
ceding quotients, is denoted by m, then 

w = = = 5, etc.; 
a a, #o 

and hence, 

F! = ma^F, = ma,, F a = ma a , etc. 

In general, if Fis any force producing in a body the accel- 
eration a, we have 

m = (1) 
a 

and F = ma (2) 

62. Definition of Mass. Experience teaches that it 
requires different forces to produce the same acceleration 
m different bodies. Therefore, the value of m is different for 
different bodies. When two equal forces produce the same 
acceleration in two bodies, the two bodies are said to have 
the same mass. If one body requires a greater force than 



26 FUNDAMENTAL PRINCIPLES 27 

another in order to receive a certain acceleration, the former 
body is said to have a greater mass than the latter. 

Mass, then, is that property of bodies on which alone the 
acceleration they receive when under the action of given 
forces depends. The acceleration, of course, depends on 
the applied force also, what the definition means is that, so 
long as the force remains the same, the acceleration varies 
with only that property of bodies called mass 

According to this definiton, the mass of a body may be 
measured by the force necessary to impart to the body an 
acceleration of a unit of length per unit of time For this 
reason, the factor m of formulas 1 and 2 of Art. 61 is called 
a measure of the mass of the body, or, for shortness, the mass of 
the body. It should be remembered that m has different 
values for different bodies. 

63. Determination of the Mass of a Body Accel- 
eration Due to Gravity. The mass of a body has to be 
determined experimentally. A known force is applied to 
the body for a certain time, and the velocity at the end of the 
time is ascertained by some method Then the acceleration 
is found by dividing the velocity by the time (formula 2 of 
Art. 59), and the mass by dividing the applied force by the 
acceleration (formula 1 of Art. 61). The mass is, of 
course, expressed m the same units as the force, and its 
numerical value further depends on the unit of time and on 
the unit of length, these units being involved in acceleration. 
In this work, mass will, unless otherwise stated, be referred 
to the pound, the foot, and the second as units. 

The most convenient force to use for determining the mass 
of a body is the force of gravity that is, the weight of the 
body. It has been ascertained by actual experiment that a 
body falling freely in vacua follows the laws of uniformly 
accelerated motion. This might have been anticipated; for, 
while the body falls, it is under the action of a uniform force 
(gravity), which is measured by the weight of the body. 

The velocity of a falling body has been found to increase 
at the constant rate of about 32.16 feet per second per second; 



27 OF MECHANICS 27 

in other words, when acting on a falling body, gravity pro- 
duces an acceleration of about 32.16 feet per second. As this 
acceleration is due to the weight of the body, we have, deno- 
ting this weight by W (from formula 1 of Art. 61), 

= W 
m 32.16 

64. The acceleration produced in a body by the force of 
gravity is called the acceleration due to gravity, or the 
acceleration of gravity. The magnitude of this accelera- 
tion decreases from the pole, where it is about 32.26 feet per 
second, toward the equator, where it is about 32.09 feet 
per second For the United States, its average value is 
32.16 feet per second. This value will be used here, unless 
a different value is especially stated. 

65. The acceleration due to gravity is usually denoted 
by g. If the weight of a body i& W, we have, 

W i-t \ 
m = (1) 

and W = mg (2) 

The weight vanes proportionately to g t so that the value 
of m, as determined from formula 1, is always the same, as 
it should be. It is, of course, understood that the weight W 
is the weight of the body at the place where the acceleration 
of gravity is jr.* 

Substituting m formula 2 of Art. 61 the value just found 
for m t 

r-i K (I ( o \ 

F (3) 

g 

which gives the force necessary to produce a given acceler- 
ation a in a body having a given weight W. 

Both in formula 3 of this article and in formula 2 of 
Art. 61, the force F is the force acting in the direction of 

*The exact value of g; at any particular place, is determined by 
careful observations with a pendulum, to which formulas derived m 
advanced mechanics are applied. The following formula gives a 
fairly approximate value of ff (feet per second) at any point whose 
latitude is L and whose elevation above sea lex el is h (feet): 
g = 32.173 - 084 cos L - .000002 h 



28 FUNDAMENTAL PRINCIPLES 27 

motion. If there is another force acting m an opposite 
direction, F must be understood to be the difference 
between the two. How forces acting on the same body 
or particle are combined will be explained further on. 

EXAMPLE 1 A block of any material is placed on a horizontal 
smooth surface (that is, a suiface supposed not to offer any factional 
resistance, or whose fnctional resistance may be neglected), and Is 
pulled by a string for 4 seconds, a registering appaiatus (called a 
dynamometer] placed between two portions of the string shows that 
the pull is 3 pounds The weight of the block being 20 pounds, it is 
required to find (a) the mass of the block, (6) the acceleration of its 
motion; (c) its velocity at the end of the fourth second 

SOLUTION (a} The mass of the block is found by formula 1. 
Taking g equal to 32 16, 

JQ 

m = sire = 622 Ans - 

(6) By formula 2 of Art 61, 

F 3 

a = ~ = -jjKo = 4 823 ft per sec. Ans. 
in ,\j&i 

(c) By formula 1 of Art 59, 

v = 4 823 X 4 = 19 292 ft per sec Ans 

EXAMPLE 2 A uniform force acts on a body originally at rest for 
10 seconds, at the end of which the velocity of the body is 40 feet per 
second, the force then ceases to act, and a force of 15 pounds is 
applied for 5 seconds, at the end of which the velocity of the body is 
76 feet per second To find (a} the weight of the body, (/>) the force 
that acted during the first 10 seconds 

SOLUTION (a) During the second interval (5 seconds), the velocity 
has changed from 40 to 75 ft per sec Therefore, the acceleration, by 

formula 4 of Art 59, is = = 7 ft. per sec. (= a). The mass of 
the body is = -=-, and the weight is 

d I 

15 X 32.16 _. _, . . 
mg = - ~ = 68.914 Ib. Ans. 

(d) The acceleration due to the fust force is (formula 2 of 
Art. 59), 40 T- 10 = 4 ft per sec By formula 2 of Art. 01, the 
force necessarv to produce this acceleration is 

m X 4 = x 4 - 8.5714 Ib Ans. 



27 OF MECHANICS 29 



EXAMPLES FOB PRACTICE 

1. Find the force necessary to start a weight of 15 pounds from 
rest and give it a velocity of 80 feet per second in 4 seconds 

Ans F = 9 3284 Ib. 

2 An engine and train of cars having an aggregate weight of 
160 tons start from rest and move for 1 minute (60 seconds) over 
a level track Assuming the traction of the engine to exceed the 
resistances by 2 50 tons, find the velocity of the tram at the end 
of the minute Ans v = 32 16 ft per sec 

3 A body has an initial velocity of 20 feet per second. A force of 
40 pounds is applied, and after 12 seconds the velocity is found to be 
85 feet per second Find (a) the acceleration due to the force, 
(b) the weight of the body &/() a = 5 4167 ft per sec. 

Ans I (d) W = 237.49 Ib 

4 A body weighing 980 pounds moves for a certain time under the 
action of a uniform force of 50 pounds. At the end of that time, 
another force of 50 pounds is applied in the opposite direction After 
the application of the second force (the first not being removed) , the 
body moves over 300 feet in 15 seconds Find the time during which 

the first force acted alone . , v 300-15 , , , 

Ans / == - = - w = 12. 189 sec. 
a f 



LAW OF ACTION AND REACTION 

66. Action and Reaction. Force has been defined as 
the action of a body on another, producing, or tending to 
produce, a change in the motion of the latter body. When 
the weight of a body is spoken of as the action of the earth 
on the body, we are concerned only with the effects of this 
action on the body in question, and for this reason disregard 
whatever action the body may, in its turn, exert on the earth. 
In the same manner, we say that a magnet attracts a piece of 
iron; this is sufficient for us to know, if the only thing being 
dealt with is the condition and motion of the piece of iron. 
But it must not be concluded from these common and con- 
venient forms of expression that a body can act on another 
without itself being acted on by the other body. All action 
between bodies is mutual. Take, for instance, a magnet, 
hold it in the hand, and bring it near a small piece of iron- 
the iron will immediately move toward the magnet and 



607 6100 



SO FUNDAMENTAL PRINCIPLES 27 

adhere to it. Now, place the magnet on the table and 
hold the piece of iron in the hand, bringing it near the 
magnet: the magnet will move toward the piece of iion 
and adhere to it. If both the magnet and the piece of 
iron are placed on corks and let float on the .surface of water, 
they will both move, each toward the other, and adhere to 
each other. Furthermore, in this motion, the acceleiation of 
the magnet and that of the iron are found to be in the inverse 
ratio of the masses of the two bodies. This shows that the 
two bodies are acted on by the same force (Art. (52), 01 that 
the action of the magnet on the iron is equal to the action of 
the iron on the magnet. 

When a body is at rest, it is evident that the rorces rtCting 
on it must balance among themselves (law of inertia) Take 
a body weighing 4 pounds lying on a flat surface If the 
weight of the body were not balanced, the body would move; 
the surface, therefore, must exert an upward force equal to 
4 pounds, in order to keep the body from falling If we 
press against the wall with a force of 10 pounds, the wall 
presses on the hand with the same force, but in an opposite 
direction. A weight of 20 pounds hanging from a string 
will evidently pull the string downwards with a force of 
20 pounds; but the string must pull the weight upwards 
with the same force, as otherwise the weight would fall. 

In general, whenever one body acts on another, the latter 
body acts on the former with the same force, but in the 
opposite direction. 

In considering the mutual action of two bodies, we are 
usually concerned with the condition of only one of them; the 
force acting on it is called the action, while the force exerted 
by it on the other body is called the reaction. 

67. Statements of the iJaw of Action and Reaction. 

From the facts stated in the preceding article, the following 
general law has been derived; it is known as the law of 
action and reaction, and was first stated by Newton: 

To every action^ there is always an equal and opposite 
reaction. 



27 OF MECHANICS 31 r -> 

68. As the accelerations produced in two bodies by the 
same force are inversely proportional to the masses of the 
two bodies, the law of action and reaction may be stated in 
the following terms 

Whenever two bodies aft on each other, they produce > or tend to 
produce, in each other accelerations ni opposite directions, -and 
these accelerations are inversely propoi tional to the masses of the 
two bodies. 

This happens whether the two bodies act on each other by 
means of a string stretched between the two, or by means of 
a rod connecting them, or through some unknown medium, 
as in the case of a magnet and a piece of iron. In every case 
where a body moves another, the latter moves, or tends to 
move, the former 

69. Sti-ess. When only the action of one body on 
another is considered, this action is called force. If the 
mutual action of two bodies, or two parts of a body, is 

Jp 

lOOlb.. I .20026. .lOOlb. \.100lb. 

'' 


FIG 7 

considered, that action is called stress. A stress includes a 
pair of equal and opposite forces, an action and a reaction, 
a force and a cotmterforce. A stress is measured by the 
magnitude of either of the forces of which it consists. 

In applied mechanics, the term stress is generally restricted 
to the forces acting between two parts of a body Thus, 
when a string is pulled at both ends with a force of 10 pounds, 
every portion of the string is under a stress of 10 pounds; for, 
if the two portions of the string are considered as being 
situated on the two sides of a plane perpendicular to its 
direction, each portion will be pulled away from the other 
with a force of 10 pounds. Therefore, to keep the two por- 
tions from separating, each must pull the other with a force 
of 10 pounds. Again, consider the bar A B, Fig. 7, pushed 
from each end with a force of 100 pounds. Imagine the bar 



32 FUNDAMENTAL PRINCIPLES 27 

lo be divided by a plane PQ- when the portion AM presses 
on B M with a force of 100 pounds toward the right, the por- 
tion B M must press on AM with a force of 100 pounds 
toward the left. 

In these examples, the string is under a stress of 10 pounds, 
and the bar under a stress of 100 pounds. 

70. Tension, Compression, Pressure. When, as in 
the case of the string just considered, the forces tend to pull 
each portion of a body from the other, or to lengthen the 
body on which they act, the stress is called tension. When, 
as in the case of the bar, the forces tend to move each por- 
tion toward the other, or to shorten the body on which they 
act, the stress is called pressure or compression. The 
term pressure, however, is more commonly applied to the 
stress between two contiguous bodies, and the term com- 
pression to the stress between two contiguous parts of the 
same body. In the example of the bar, the stress is ordi- 
narily called compression. In the case of a heavy body 
resting on a table, the force exerted by the body on the 
table and by the table on the body is a pressure. 

EXAMPLE 1 Two bodies MI and M,, Fig 8, of masses OTI and m t> 
respectively, rest on a smooth horizontal surface, and are connected by 

a string S a A force of /'' 

jf a j * a [ jf t | Sl - pounds is applied to Mi 

' ' ' by means of a string 5, 

FlG 8 m the direction shown. 

To find the acceleration of the resulting motion, and also the tension 
in each string. 

SOLUTION In all problems of this kind, the strings are supposed 
to be rigid and of no mass or weight, if the mass and stretching of 
each string were considered, the problem would be more compli- 
cated. In almost all cases that occur in practice, the ma<tf of the 
string, rope, chain, etc may be neglected without any sensible error; 
and the same is true of the extensibility, or stretching capacity, of 
the connection For example, the mass of the coupler between two 
railroad cars need not be considered in determining the pull between 
one car and the other; nor is it necessary to take into account the 
amount of stretching m the coupler 

Since the two bodies ^/"i and M, are rigidly 1 connected, they must 
move with the same acceleration Let this acceleration be a, and let 



27 OF MECHANICS 33 

the tensions m the strings Si and S a be 7i and T a , respectively Since 
the body MI is pulled by the string S\ with a force equal to F, it must 
pull the string toward the left with the same force, so that 7~i is 
evidently equal to F Also, Af,. acts on M, through the string S, and 
imparts to it an acceleration a towaid the right, therefore, the force 
exerted by MI on Af a is (from foimula 2 of Art 61) in a a By the 
law of action and reaction, Jlf t exerts on Jlf t the hame force, m,a, but 
directed toward the left The tension m S, is, therefore, T, = m 9 a 
The net force acting on M^ is the difference between F and m a a, 
or Fm a a. Consequently, since the resulting acceleration is a, we 
have, by formula 2 of Art 01, F w, a = m t a; whence, 

a = (1) 

7i + Wa 

Also, T, = m 3 a = -^-^ (2) 

&! + m, v ' 

The same results may be obtained by assuming at the outset that 
the two masses m l and m, are equivalent to one single mass (m l + tti) 
acted on by the force F From this, equation (1) follows at once as a 
special case of formula 2 of Art 6 1 . Knowing the acceleration of Af t , 
which is a, the unbalanced force T a acting on Hf, is found by for- 
mula 2 of Art 61: 

~, itt t F 

T, = m a a = 7 

j + m, 

which is the same as before. 

EXAMPLE 2 A body Af lt Fig 0, whose weight is W pounds, is 
attached to a string, the latter 
is passed over a smooth peg P, 
and then tied to a second body 
7l/ a having a weight of 
pounds and free to move along 
a smooth horizontal surface 
To find the pull m the string 
and the resulting acceleration 
of the two bodies. Pro. 9 

SOLUTION This problem does not diffei from the preceding 1 in 
principle, and the method of solving it is identically $he same, but it 
has been given in order to caution against a common mistake made 
by beginners The object of the peg is simply to change the direc- 
tion of the force, it has no effect on the magnitude of the pull, which 
is the same in the vertical as In the horizontal portion of the string 
It is to be carefully borne in mind, however, that the pull on the 
string is not the weight of M^ and that therefore the force pulling M t 
is not Wi pounds, as might appear at first sight. The weight W* is 
a force acting on the bodies Jtfi and Jtf t , and, like the force F m the 
preceding example, has to move these two bodies. If there were 
I LT39S-4 




34 FUNDAMENTAL PRINCIPLES 27 

no body M t , the unbalanced force acting on AF* would he W tt 
but, in the present case, this force is diminished by the leaction 
of M, transmitted through the string to M^ If the pull in the string 
is denoted by T and the common acceleration of the two bodies 
is denoted by a, the unbalanced force acting on Jlfi is, as before, 
W l - T, and the unbalanced force acting on M, is simply T The 

mass of Mi is - (formula 1 of Art. 65), and the mass of Jlf a is . 

ff 

Substituting in the formulas of the preceding solution, we obtain 

a = Wj. w,. "* Wi + w m M 

g g 

Tff 

* or Jjf jy 

TXf W = Ef _1_ TV \*J 



EXAMPLE 3 A body weighing W pounds lies on the surface of a 
horizontal table. Both the table and the body are moving upwards 
with an acceleration of a feet per second To find the pressure of 
the body on the table. 

SOLUTION The pressure of the body on the table is equal to the 
pressure of the table on the body, which is the total upward force 
acting on the body, call this force P The downward force acting 
on the body is W. Therefore, the unbalanced force producing the 

W a. 
acceleration a is P W, and, as P W => am ^~ t we find, for 

the downward pressure exerted by the body, 

/.-ZS+Wr-^ + f) (1) 

ff g 

This shows that the weight of the body is increased by the amount 

- . Ifthetableismovingdownwards.ais negative, andjP Ji!.lj!rfv t 

If in this case a, ff, then, P = 0, or there is no pressure on the table. 
If a is negative (downward) and greater than g t 8 ay a g + a', then, 



g 

This shows that, if the body remains in contact with the surface of 



the table, it will exert an upward pressure equal to -^-A; but this can 

only be done by placing the body under the table. In this case, it is 
evident that, since the table tends to move faster than the body, the 
latter must offer a resistance, or exert an upward pressure on the table. 
This problem should be carefully studied, as it is a good illustration 
of how the results of mathematical formulas {Ire to be Interpreted. 



27 



OF MECHANICS 



35 



EXAMPLES FOB PRACTICE 

1. In example 1 of Art 70, the force .F is 500 pounds, and M* and M t 
weigh 8 tons and 3 tons, respectively Find (a) the resulting acceler- 
ation, in feet per second, (b) the tension on the rope, in pounds (1 ton 
= 2,000 pounds) . A ,/()= 731ft per sec 

Ans \(t>) T= 13636 Ib. 

2 With the weights of M t and Af, as in example 1, find: (a) the 
force (tons) necessary to give the two bodies an acceleration of 10 feet 
per second, (b) the tension in the rope, m tons 

F = 3 4204 tons 
T = .9328 ton 



Ans 



3. In Fig. 10, the three bodies M lt M,, M a , weighing 18 tons, 
16 tons, and 20 tons, respectively, are connected by the ropes S L 




FIG 10 

and S a A force F of 15 tons is, applied to M l as shown. Find, 
(a) the resulting acceleration a, m feet per second, (b) the tension T, 
in the rope S tt (c) the tension T, in the rope S 3 

f a) a = 8 9333 ft per sec. 
Ans { b) T a = 5 5556 tons 
[ c) T a = 10 000 tons 

4 A body that on the surface of the earth weighs 10 pounds, is 
weighed m an ascending balloon by means of a spring balance, and 
found to weigh 15 pounds With what acceleration is the balloon 
ascending? [See equation (1} m example 3 of Art. 70.] 

Ans a = 18 08 ft per sec. 



5. An engine pulls a tram of 
The total force of the locomotive i 
train (the engine not included) 
resistance of 10 pounds per ton of 
during the first minute, find (a) 
feet per second, (b) the velocity v 
minute, in miles per hour, (c) the 
the second car to the first. 



seven cars weighing 80 tons each 
that is, the total force acting on the 
-is 8 tons. Assuming a constant 
tram (engine and tender excluded) 
the acceleration a of the train, in 
of the tram at the end of the first 
tension T m the coupler connecting 

a = 1 0643 ft per sec 
Ans. 4 b) v = 43.541 mi per hr. 
T = 6.8571 tons 



86 FUNDAMENTAL PRINCIPLES 2? 



CMMEDIATE CONSEQUENCES OF THE PRE- 
CEDING 1LAW8 



IMPORTANT FORMULAS 

71. Space Passed Over In Uniformly Accelerated 
Motion. Let a body start from rest and move during: t 
seconds under the constant action of a force acting in the 
direction of motion. Let a be the acceleration of the body, 
and v the velocity at the end of the time t. As explained 
in Art. 59, v is equal to a t. It can be shown by the use of 
advanced mathematics that the space j desciibed by the 
body in the time t is the same as if the body had moved din- 
ing that time with a constant velocity equal to one-half the 

actual final velocity, or |; that is, 

2i 

.-\* (i) 

If, in the second member of this formula, v is replaced 

by its equivalent a t, the result is ~ t = a f. Therefoie, 

2 

j = i/ a (2) 

This is one of the most important formulas in mechanics 
and should be memorized. 

72. If, in formula 1 of Art. 71, t = 1, then, s = $ a; 
or, the space described during the first second is numerically 
equal to one-half the acceleration. 

73. From the general formula v = at, we get t - . 

a 

This value substituted in formula 1 of Art. 71 gives 

t 

s = - , whence, v* = 2 as, and 
a 

v = V2 as 
which gives the final velocity at the end of a given space. 

74. To find the space and the velocity in terms of the mov- 
ing force and the mass or the weight of the moving body, we 



27 OF MECHANICS 37 

have, by formula 2 of Art. 61, and formula 1 of Art. 65, 

Zp 

a = = g. These values in formula 2 of Art. 71 give 
m W 

s - * = -* *r/ a (1) 

*~ 2 2 X W*' UJ 

The same values in the formula of Art. 73 give 

(2) 

75. Modification of the Formulas "WTien the Body 
Has an Initial Velocity. If, before the force begins to 
act, the bodv has an initial velocity v , or, what is the same 
thing, if only an interval of time during which the velocity 
changes from v a to v is considered, the preceding formulas 
must be modified as follows: 

As the force has no effect on the motion already acquired 
by the body, the latter, having a velocity z/ , will describe, 
by virtue of its inertia alone, during the time /, a space equal 
to v 1-, to this must be added the space described by the body 
in virtue of the action of the force, as given by formula 2 of 
Art. 71. Therefore, the total space is given by the formula 

J = v,t + \af (1) 

or, putting a = v ~ v (formula 4 of Art. 59), 



j =.* + *(-.)* = ifa + w.)/ (2) 
which is the mean of the initial and the final velocity multi- 
plied by the time. 

76. The equation a = v ~ v gives / = "L=J!* f Substi- 

/ a 

tuting this value in formula 2 of Art. 75, 



i W 

s = + . x - 

\ a- 
whence, v' v*' => 2 a s, and 

v = Vz>o" + 2 a s 

This gives the final velocity in terms of the initial velocity, 
the space, and the acceleration. 

To find s and v in terms of force and mass, it is only 

p 

necessary to substitute for a. 
m 



38 FUNDAMENTAL PRINCIPLES 27 

77. Falling Bodies. It has been explained (Art. 63) 
that, when a body falls freely under the action of gravity, its 
acceleration is g feet per second, the value of g being neaily 
constant and equal to 32 16. The preceding formulas apply 
to falling bodies, for a falling body is simply a body mov- 
ing under the action of a force equal to its own weight, with 
a uniform acceleration equal to g. It is, therefoie, not 
strictly necessary to write new formulas for the motion 
of falling bodies. As, however, it is customary to use 
the symbol g for the acceleration of a falling body, and h 
for the space described, or height fallen through, it is con- 
venient to have the general formulas expressed in terms of 
these symbols. By putting h for s and g for a in formula 1 
of Art. 59, in formulas 2 and 1 of Art. 71, and in the for- 
mula given in Art. 73, the following formulas are obtained 
for a body falling freely from rest during / seconds: 

v=gt (1) 

k = iff (2) 

h = %vt (3) 

v = *hgh (4) 

For a body falling for t seconds, after having acquired a 
velocity z , formula 3 of Art. 59, formulas 1 and 2 of 
Art. 75, and the formula of Art. 76, give 

v = v +gt (5) 

h = v.t + lgf = (v. + ir/)* (6) 

*-*(. + )/ (7) 

(8) 



78. It should be remembered that, theoretically, these 
formulas are exactly correct only for bodies falling m vacua. 
When bodies fall through the air, the latter offers a resist- 
ance that varies with the surface of the body; hence, all 
bodies do not fall in air with exactly the same acceleration^ 
but in a vacuum they do. This has been verified by the fol- 
lowing experiment: A feather and a ball of lead are sup- 
ported at the upper (inside) part of a long tube from which 
the air has been removed; by a convenient arrangement the 
support is quickly turned from the outside, so that both 



27 OF MECHANICS 39 

bodies will begin to fall at exactly the same time; and it is 
found that they reach the bottom of the tube simultaneously; 
nor does the ball get ahead of the feather during any portion 
of the time of falling. Again, if a paper disk is placed on 
top of a dollar, and the latter let fall, the paper will remain 
constantly on the dollar and fall in the same time. The 
reason is that in this case the paper does not encounter any 
more resistance than does the coin. 

EXAMPLE 1 What is the velocity at the end of 30 seconds of a 
body moving with an acceleration of 8 feet per second? 

SOLUTION Substituting the given values in formula 1 of Art 59, 
v = 8 X 30 = 240 ft per sec. Ans 

EXAMPLE 2 A force of 10 pounds starts a body from rest and 
causes it to move over 65 feet If the weight of the body is 40 pounds, 
what is the final velocity? 

SOLUTION. By formula 3 of Art 65, 

Fff 10 X 32 16 . ., .. 
a = -^ = 4Q - = 8 04 ft. per sec. 

Substituting the known values in the formula of Art 73, 

v = V2 as = -N/2 X 8 04 X 65 = 32 33 ft per sec Ans. 
EXAMPLE 3. A ball is thrown downwards with a velocity of 5 feet 
per second from a tower 200 feet high. What will be its velocity when 
it reaches the ground? 

SOLUTION Here, the initial velocity v a ~ 6 ft per sec , and the 
height h = 200 ft., are given. Substituting these values in formula 8 
of Art 77, 

v - V6 a + 2 X 32 16 X 200 = 113 53 ft per sec. Ans 
EXAMPLE 4 If a ball dropped from a bridge reaches the water in 
2 seconds, at what height is the bridge above the surface of the water? 
SOLUTION From formula 2 of Art. 77, the height through which 
the ball drops is found to be 

iX 32.16X2" 6432ft Ans. 



EXAMPLES FOR PRACTICE 

1. A body starts from a state of rest and moves with an acceleration 
of 20 feet per second for 30 seconds. Over what distance has the body 
passed? Ans. 9,000 ft. 

2. A body weighing 160.8 pounds is started from rest by a force of 
40 pounds. Over what space has the body passed if the final velocity is 
16 feet per second? Ans. 16 ft. 



40 FUNDAMENTAL PRINCIPLES 27 

3 A body having an initial velocity of 25 feet per second moves, 
with an acceleration of 8 feet pei second, over a distance of 29 feet 
Find the final velocity of the body. Ans 33 ft per sec 

4 A body falls freely for 2 3 seconds If the initial velocity of the 
body is 5 feet per second and its final velocity is 79 feet per second, 
through what distance has the body fallen? Ans 96 6 ft 

5 A body having an initial velocity of 10 feet per second falls 
freely for 12 seconds Through what distance does the body pass? 

Ans 2,435 5 ft 



RETARDATION 

79. Uniformly Retarded Motion. When a body is 
moving in a certain direction and a force acts in the opposite 
direction, the effect of the force is to diminish the velocity 
by equal amounts in equal times, and the motion is said to 
be uniformly retai-ded. The amount by which the veloc- 
ity is decreased per unit of time is called the retardation 
due to the force. 

According to Galileo's law, if a force acting 1 on a. body at 
rest produces in it an acceleration a, the same force acting 
on the same body, when the latter is m motion, will pro- 
duce the same acceleration a and the velocity v = a t, in 
its own dnection, whatever the previous direction of the 
motion may have been Therefoie, if the force acts in a 
direction opposite to that in which the body is moving, and if 
the latter has an initial velocity v , the velocity after t seconds 
will be v, a t. Retardation, then, is nothing but negative 
acceleration, and all formulas relating to uniformly acceler- 
ated motion apply to uniformly retarded motion, by simply 
changing the sign of a. For this reason, the term accelera- 
tion, when taken in its most general sense, includes letarda- 
tion as a special case, and the latter term is seldom used in 
mechanics. 

80. Law of Retardation. Let a. body be moving in a 
straight line, and let its velocity at a certain moment be TV 
If at that moment a force is applied in the opposite direc- 
tion, and allowed to act during / seconds, the final velocity v 



27 OF MECHANICS 41 

and the space s described are given by the following formulas 
(See Arts 59 and 75): 

v = v at (1) 

j = v t t %at* (2) 

To find the time in which the body will be brought to rest, 

v must be equal to zero; that is, v a t = 0, or / = . 

a 

Substituting this value in equation (2), and denoting by s 1 
the space described during the time /, we have, 

/ . &' - J a ! = '- 

a a 2 a 

If the velocity v a is supposed to have been generated by a 
force equal to the retarding force, the time required for the 

force to produce this velocity must evidently have been , 

a 

which is the same as the time required for the retarding force 
to destroy the velocity z/ The space described during the 
action of the force, up to the time the velocity v, is attained, 
is given by the foimula 

j = $af=l;a^=^ = S > 
a 2 a 

It follows, therefore, that, if a uniform force acts on a 
body during a certain time and carries it through a certain dis- 
tance, an equal but opposite foice, it applied after the first force, 
will destroy the velocity generated by the first force in the same 
time and in tlie same space (that is, after the body has described 
the same space] required by the first force to generate the velocity 
m question. 

SOJLDT1ON OF PROBLEMS 

81. Fundamental Equations of Motion. The fol- 
lowing are the three fundamental equations of motion: 

F=ma (1) 

v-v a = at 1 ,v 



They form two groups, (1) and (2), one consisting of one 
equation and the other of two. In group (1), there are three 
quantities; and as there is only one equation, two of the 



42 FUNDAMENTAL PRINCIPLES 27 

quantities must be known before the other can be determined. 
If /'is to be determined, m and a must be known, if m is to be 
determined, F and a must be known, etc. In group (2) , there 
are five quantities, and, as there are only two equations, three 
of the quantities must be known in order to determine the other 
two; the process of finding these two is simply the process 
of solving two equations with two unknown quantities. 

If the two groups are taken together, it will be observed 
that a is common to both If two of the quantities m group 
(1) are given, a becomes known, and to solve group (2), 
only two more quantities of that group have to be given 
Or, if three of the quantities of group (2) are given, a becomes 
known, and only one of the quantities of gioup (1) (either 
/'or m) is required to find the other. The process of elimina- 
tion is exceedingly simple. When there is no initial velocity, 
v = 0. 

82. General Solution of Some Important Problems. 

As an illustration of the use of the fundamental equations, let 

it be required to find F, a, and /, when ?;/, v, v ot and are given. 

In group (2), the quantities v, v ot and s are known. The 

unknown quantities are a and /. From the first equation of 

the group is obtained a = ^-~r^'> and fr m th e second, 



a = ^ s ~~ Vt -. 



Equating these two values of a, transpo- 

sing, and canceling the common factor /, 
(v v ) / = 2 (s v t}; 

whence / = -- - 

v + v a 

which is the same value that would have been obtained from 
formula 2 of Art 75. This value of t substituted in the 
equation a = ^-^-^ gives 

a _ - Q _ (*LrJ?o) ( v + v ) - ?. "J!s! 

~" ~ " ~ 



Finally, F - ma * 

2s 



27 OF MECHANICS 43 

83. Again, let m, v t v,, and t be given, and let it be 
required to find F, s, and a. 

The value of a follows at once from the relation v z/ 

= at, which gives a = ^^. Substituting this value in 

the second equation of group (2), formula 2 of Art. 75 is 
again found. From group (1) is obtained 

F=ma= (-.) 
t 

The values for F given in this and the preceding article are 
very important, and should be memorized. 

EXAMPLE 1 A body weighing 250 pounds is moving with a velocitj' 
of 25 feet per second, a force acting in the same direction through a 
space of 75 feet changes the velocity to 60 feet per second To find 
the magnitude of the force 

SOLUTION Here, m = SHTB (formula 1 of Art 85), v = 26, 
Aft ID 

v 50, and s = 75 Substituting these values in the formula given 
in Art 82, 

-(50 -26") 
m(v* - zO 32 16 ^ ; 260 X (2,600 - 626) 

2s = 2X75 " " 8216X2X76 

= 97.17 Ib. Ans 

EXAMPLE 2 A body whose weight is 192 96 pounds moves with a 
velocity of 15 feet per second. To find the magnitude of a force neces- 
sary to change the velocity to 45 feet per second, that force acting during 
15 seconds in the direction of motion 

192 96 
SOLUTION Here, tn = - pn -' - =6, v 15, v = 45, and / <a 15. 

oZ 10 

Substituting these values m the formula given in this article, 



f = , = - . tt 



EXAMPLES FOR PRACTICE 

1 A body weighing 100 pounds moves for 6 seconds along a smooth 
horizontal surface under the action of a force of 10 pounds Find 
the space passed over. Ans. s = 57.888 ft. 

2 A body moving with uniformly accelerated motion passes over 
100 feet in 12 seconds. What is the acceleration? 

Ans. a 1.3880 ft. per sec, 



44 FUNDAMENTAL PRINCIPLES 27 

3 What force is necessary to give a body weighing 6 tons a velocity 
of 10 feet per second in 75 feet? 

W-31* 

Ans F = ~- = 12438 T = 248 76 Ib 



4 A body starts fiom rest and moves with an acceleration of 
40 feet per second Find the velocity after the body has passed ovei 
75 feet Ant. v = 77 46 ft per sec 

5 Find the time in which the body in the preceding example has 
described the space of 75 feet Ans t = 1 9365 sec. 

6 A body is thrown vertically downwards from a height of 10,000 
feet and reaches the ground m 6 seconds Find the velocity with 
which the body was projected (Use formula 6 of Ait 77.) 

Ans z/o = 1 ,570 2 ft per sec 



COMPOSITION AND RESOLUTION OF FORCES 

84. Point of ApplI cation, Line of Action, and 
Direction of a Force. When a force is applied to a particle 
or material point, the particle or point is called the point 
of. application of the force. 

The line of action of a force is the straight line along 
which the force tends to move its point of application The 
direction of the force is the direction of this line (see 
Art. 37). 

85. Graphic Representation of Forces. As a great 
many mechanical problems are solved by means of geometry, 
c ^ A it is convenient, and almost 

FIC, 11 necessaiy, to represent forces 

graphically that is, by means of lines. A force is repre- 
sented graphically by drawing a line parallel to the line 
of action of the force, and of a length proportional to the 
magnitude of the force, the direction of the latter being 
indicated by an arrowhead marked on the line. In Fig. 11, 
O is the point of application of a force, and the force is repre- 
sented by O A, whose length is as many units of length as 
there are pounds in the force. The arrow indicates that the 
force tends to move O from O toward A along the line OA, 
If the force is 100 pounds, OA may be made A | A inch, or 
mch, or 100 millimeters, etc.; the unit of length used is 



27 OF MECHANICS 45 

immaterial, provided that the same unit is used for all forces 
in the solution of any one problem 

It is not necessary, in ordei to represent a foice by a line, 
to draw the line through the point of application, or any 
other special point. Any line parallel to the line of action of 
the force may be used for the purpose 

86. Vectors. Any quantity that, like force and velocity, 
has magnitude and direction is called a vector qumitlty; 
and the line, as OA, Fig 11, by which the quantity is repre- 
sented graphically is called a vector. A vector is, then, 
simply a line having a length proportional to the magnitude 
of a vector quantity, and an airowhead to indicate the direc- 
tion of that quantity Such quantities as mass and volume, 
which have no direction, cannot be represented by vectors. 

87. That extremity of a vector to which the arrowhead 
points is called the end of the vector; the other extremity 
is the origin. In Fig. 11, O is q 

the origin and A the end of the 
vector OA. 




88. Two vectors having the 
same origin or the same end will 
here be referred to as being in 11011- Pl 12 B 

cyclic order. If the end of one coincides with the origin 
of the other, they will be described as being in cyclic oi-der. 
In Fig 12, the vectors OA and O B have the common origin O, 




PIG 18 




and are, therefore, in non-cyclic order; so are the vectors OA 
and J3A, Fig. 13, which have the common end A. In 
Fig. 14, the end of the vector O A coincides with the origin 
of the vector A B, these two vectors are in cyclic order. 



46 FUNDAMENTAL PRINCIPLES 27 

89. Resultant and Components. When a body is 
acted on by several forces, there is usually one single force 
that, if acting alone, would produce the same effect as the 
several forces combined This single force is called the 
resultant of the other forces. With respect to the result- 
ant, the combined forces are called components. 

90. Composition and Resolution of Forces. The 

process of finding the resultant when the components are 
known is called the composition of foi-ces. The process 
of finding the components when the resultant is known is 
called the resolution of forces. 

Both the resolution and the composition of forces are of 
the utmost importance in dynamics, especially in statics, and 
will be explained more fully further on. But here the funda- 
mental principles must be stated and explained. 

91. Composition of Colllnear Forces. Forces hav- 
ing the same line of action are called colllnear forces. 

The resultant of several colhnear fotces is equal to thein alge- 
braic sum. 

For example, if a force of 15 pounds pulls a body upwards, 
and a force of 10 pounds pulls the body downwards, along 
the same line, the effect will be the same as if the body were 
pulled upwards only, with a force of 15 10 = 5 pounds. 
If the upward direction is treated as positive, the downward 
direction will be negative, and the downward pull of 10 
pounds will be represented by ( 10). The resultant is, 
therefore, 15 + ( 10) = +5. The same principle applies 
to any number of forces. Thus, the resultant of the forces 
15 pounds, 37 pounds, 39 pounds, 13 pounds, and 78 
pounds is 15 + 37 - 39 + 13 - 78 = 65 - 117 * -52 pounds. 
The negative sign indicates that the resultant is a downward 
pull of 52 pounds. This principle, which is sufficiently 
obvious, has already been applied in this Section. 

92. Concurrent and Non-Concurrent Forces. Two 

or more forces not having the same line of action, but whose 
lines of action meet at a point, are called concurrent 




27 OF MECHANICS 47 

forces. If the lines of action of several forces do not meet 
at one point, the forces are said to be no n- concurrent. 

93. Resultant of Two Concurrent Forces: Paral- 
lelogram of Forces. Let O t Fig 15, be a particle acted 
on by two concurrent forces F l and F, in the directions O A 
and OB, respectively. These lines are vectors representing 
the two forces. Com- 
plete the parallelogram 
OA CB, by drawing 
AC parallel to OB, 
and BC parallel to OA. 
Draw the vector C, 
forming the diagonal 
of the parallelogram. 
It can be proved, both Fl<< Jr> 

mathematically and by experiment, that this vector represents 
the resultant of the forces jf\ and F a ; in other words, that, if 
the forces fl and F; are represented in magnitude and direc- 
tion by the sides OA and OB of a parallelogram, the result- 
ant will be represented in magnitude and direction by the 
diagonal C. This principle is called the law of the par- 
allelogram of forces, and can be stated in general terms 
as follows: 

// two concurrent forces are represented in magnitude and 
direction by two vectors taken in non-cyclic order, their resultant 
will be represented^ in magnitude and direction, by another 
vector forming the diagonal of the parallelogram constntcted 
on the vectors representing the two forces. 

94. Formulas Derived From the Parallelogram of 
Forces. When the magnitudes of the forces F,. and F at 
Fig. 15, and the angle between their lines of action are 
given, the magnitude and direction of the resultant R are 
very readily computed by the principles of trigonometry. 
In the triangle OA C, 

OC = 4O A* + AO~*TOA. A~Cl~wTA 
But OC = R, OA = F lt A C * OB = F tt and angle 
A = 380 . Substituting these values, and writing 



48 FUNDAMENTAL PRINCIPLES U7 

cos L for cos (180 Z,), the preceding equation 
becomes 

R = V/T,' + /,' + 2 /s/; cos Z, (1) 

The angle Mi that R makes with F^ is found from the 

same triangle by the principle of smes, which gives sm 

sin A 

A C 
= - -; or, replacing sin A by sin L (for A = 180 L), 

and writing F a for A C, and R for C, 
sin Mi _ F. m 
sin Z, R ' 

z? 

whence, sin Mi = -j sin Z, (2) 

y? 
r 

Similarly, sin M, = -~ sin Z, (3) 

R 

The values of M and yfl/ a may be checked by the i ela- 
tion Mi + M a = Z. 

EXAMPLE Two concurrent forces having the magnitudes 450 pounds 
(= F,.) and 675 pounds (= F,} make an angle of 60 Required 
(a) the magnitude jR of their resultant, (fl) the angles MI and M, 
made by the line of action of R with the lines of action of FI and f lt 
respectively 

SOLUTION. (a) Here, /? t = 450 Ib ,/?". = 075 Ib , Z = B0, and 
formula 1 gives 

A = -\I450 1 + 676 s + 2 X 450 X 675 X cos 60 = 980 75 Ib Ans 

(d) Substituting m formulas 2 and 3 the values of F tt F lt A", 
and L (angles are given to the nearest 10 seconds), 

fl7^ 

sin Mi = ~Y & X sin 60, whence, M* = 86 SB' 10". Ans 

sm M t = 935-75 X sin 60, whence, Jlf a = 23 24' 50". Ans 
Checking, M v + Jlf, = 36 SB'- 10" + 23 24' 50" = H0. 

95. Rectangular Components. When the two forces 
are perpendicular to each other, they are called rectangular- 
components. In this case, angle L = 90, sin L = 1, 
cos L = 0, and the preceding formulas become 

= V/^'+T^ (1) 

sin M, = cos M, = 5 = ^^__ (Q) 

* V/v + /; 



27 OF MECHANICS 49 

For this condition, however, the angles MI and M* are 
more easilv determined by their tangents or cotangents. 
This special case is illustrated in Fig. 16, which is similar to 
Fig. 15, except that the 

angle A OB is a right ^ ^ f 2 * 

angle The parallelo- ^ % V^ fa 
gram A CB is a rect- 
angle, and the right ^\ 
triangle O A C gives R ' 



= OC = 

= ^F\* + F t ', as found A 

before Also, Fl0 - 16 



Similarly, tan M t = = (4) 

96. Resolution of a Force Into Two Components. 

The next thing to consider is the converse problem of find- 
ing the components of a foice when the force is given. This 
is called resolving a force into components. We shall 
o -Ja B v here confine ourselves 

p 7 >- -7 Xz 

fa 7 i to the particular prob- 

/ lem of resolving a force 

/ into two components 

whose directions are 
given. 

Let O C = R, Fig. 17, 
be a force acting at O. 
17 It is required to find its 

components along two given directions. To do this, draw 
through O two indefinite lines Xi and OX ti parallel to the 
given directions. From C draw a parallel to OX,, meeting 
OXt at A, and a parallel to OX lt meeting OX, at B. Then, 
OA and OB are the required components Ft. and F a . 

To find Ft. and F t by calculation. Since the directions 
of Ft. and F, are given, M* and M* are known, and also 
L = M, + Af n . Then, from formula 2 of Art. 94, 

ILT398 5 




FUNDAMENTAL PRINCIPLES 27 

_ R sin Mi _ R sin M^ / \ 

sm L sin (M l -f- 



Similarly, f.- . 
sin 

If one of the components is to make an angle M l with the 
>rce R, and the other an angle of 90 yl/i, that is, if the 
70 components are to be perpendicular to each other, for- 
ula 2 of Art. 95 gives 

F. = R sin M, (2) 

Similarly, F t = R sin M a = -ff cos M l (3) 

The student should again consider Fig. 16, and see how 
iese formulas are derived. If the figure is kept in mind 
id the geometrical relations of the quantities involved are 
imembered, no difficulty in deriving, remembering, and 
Dplying the formulas- will be experienced 

97. By the component of a force in a certain direction is 
sually understood one of two rectangular components. The 
>rce is supposed to be resolved into two components at 
ght angles to each other, one of which has the direction in 
uestion. Such a component is also called the resolute and 
le resolved port of the force in the given direction. 



EXAMPLES JFOB PRACTICE 

1. If, in Fig 17, C represents a force of 40 pounds acting; on a 
idy at 0, and the angles that the components Fi and F, make with 
ie given force are 46 and 20, respectively, what are the values of 
,ese components? . f/^ 15.10 Ib. 

Ans 't/?; - 31.21 Ib. 

2 A rectangular component F 9 of 60 pounds makes an angle of 40 
ith the resultant J?, Find the other component FI and the resultant. 

Ana /^ " 60 ' 34 lb - 
An8 'lj? . 78.82 lb, 

98. The Triangle of Forces, In the construction 
lown in Fig. 17, the complete parallelogram OACB has 
2en drawn, in order that the principle on which the oonatruc- 
on is founded may be understood. But, in practice, it is not 
scessary to draw more than one-half of the 







27 OF MECHANICS 51 

that is, one of the equal triangles OAC, O B C. For, since 
OB = AC, the line CA drawn through C parallel to OX, in 
order to deteimme the point A, gives at once the length of 
the line OB representing the component F a . Also, the 
angle M a between F a o 

and R is equal to the 
angle OCA. Hence 
the following construc- 
tion (Fig. 18). 

When a force R is 
given, and it is required 
to hnd its components in 
two given directions, ^1 RIO is 

draw from one extremity, as O, of the vector representing the 
given force, a line parallel to one of the given directions, and 
from the other extremity, as C, a line parallel to the other 
given direction The two sides O A and AC of the triangle 
thus formed ate the requited components. The vectors repre- 
senting the two components aie in cyclic order with each other, 
but in non-cyclic order with the vector representing their 
resultant R. 

This construction is called the triangle of forces. 

Notice that, although A C gives the magnitude and direc- 
tion of F t , the force F* must be supposed to be really acting 
at O, as shown by the dotted line OB. 

The line C B might have been drawn through C, to meet 
OX, at B In this case, however, the triangle would have 
been as shown by the light dotted lines, but the results 
would have been the same as before. 



EXAMPLES FOR PRACTICE 

1. Two forces, F,. = 1,000 pounds and F, = 8,000 pounds, act 
simultaneously on a particle, the lines of action of the two forces 
make with each other an angle of 30. Find (a.) their resultant jR; 
(t>) the inclination of the resultant to the two components (angles to 
nearest 10"). f(a) J? - 3,898.2 Ib. 

Ans \ IK f^/t = 22 37' 60" 
I (0 > Utf. = 7 22' 10" 



52 FUNDAMENTAL PRINCIPLES 27 

2, Given a force R = 20 tons, to resolve it into two components /", 
and F t making with R angles M* = 30, Jlf a = 45 

Aus 1 F * = 14 41 tons 
Aus l/> 3 = 10 SHU tons 

3 Find- (a) the resultant of two rectangular forces FI = 40 pounds 
and F, = 80 pounds, (d) the Inclination of the resultant to the com- 
ponents f (a) R = 50 Ib 

An s 1 ih\ JM* = JW r >-' 10" 
L w l^/, = f)8 7'fiO" 

4 A force of 100 tons is inclined to the vertical at an angle of <H) 
Resolve it into a horizontal component H and a vertical component / 

AriB Iff => Sfi(K)3 tons. 
Ans I V = 50 tons. 



ANALYTIC STATICS 

(PART 1) 
CONCURRENT COPLANAR FORCES 



DEFINITIONS 

1. Coplanar and Non-Coplanar Forces. When the 
lines of action of several forces he m the same plane, the 
forces are said to be coplanar. If the lines of action are 
not in the same plane, the forces are non-coplanar. 

2. Analytic Statics and Graphic Statics. So far as 

the mathematical treatment of statics is concerned, this 
science may be considered as being mainly a branch of 
applied geometry. And, as geometrical problems may be 
solved either analytically or graphically that is, either 
by computation or by construction so, too, statics may be 
either analytic or graphic, according to the method of solu- 
tion used. 

Analytic statics treats of the equilibrium and equivalence 
of forces by means of the arithmetical and algebraic relations 
existing among the forces, which relations, in the case of 
forces represented by vectors, are the same as the relations 
existing among the vectors, and depend on the geometric 
properties of the figures formed by combinations of such 
vectors. In analytic statics, all results are found by 
calculation. 

Graphic statics treats of the equilibrium and equivalence 
of forces by means of geometric figures. In graphic statics, 
all results are found by measurement. 

COPYNIIIHTBD MY INTBRNATIONAI. TEXTBOOK COMPANY. ALL HlaHTI MMBRVU 

28 



2 ANALYTIC STATICS 28 

3. System of Forces. The aggregate of all the forces 
acting on a body or on a group of bodies is called a system 
of forces. The simplest sybtem of forces is obviously that 
in which there is only one force. 

4. Equivalent Systems. Two or more systems of 
forces are equivalent when they produce the same effect, 
or, what is the same thing, when they have the same 
resultant The resultant of any numbei of forces is itself 
equivalent to the system formed by the components. 

5. Equillbrants. When a system of forces is balanced 
by a single force, the latter i& called the equill brant of the 

system Conversely, if several 
forces balance one single force, 
they are termed the equilibrants of 
that force. 

The equilibrant of several forces 
is evidently equal in magnitude, 
but opposite in direction, to the re- 
sultant of those forces. Hence, it 
is also called the antl -resultant. 
In Fig. 1, Ft and F a are two 
forces acting at O. Their resultant 
R is found by the principle of the 
parallelogram of forces. The force Q, equal to jR, but acting 
in the opposite direction, will evidently balance R, or the sys- 
tem of forces F l and F t to which R is equivalent. Therefore, 
Q is the equilibrant of F,. and F t . Conversely, Fi and F, are 
the equilibrants of Q. 

It is also evident that, if the forces F l} F tt and Q are in 
equilibrium, any of them may be considered as the equili- 
brant of the two others. The same principle applies to any 
number of forces. 

6. The Kesultaiit Not an Actual Force. It should 
be understood that, when several forces act on a body, the 
resultant is not an actual force acting on the body, but an 
imaginary force whose effect would be the same as the com- 
bined effects of the components. Similarly, when a single 




28 ANALYTIC STATICS 3 

force acts on a body, its components are imaginary forces 
that would produce the same effect as the single force. 
The resultant may replace the components and the compo- 
nents may replace the resultant, but in no case are the com- 
ponents and the resultant supposed to act simultaneously. 

The equilibrant, on the contrary, is a real force applied in 
order to balance a system of other forces; or, if the system 
is balanced, any one of the forces is the equilibrant of the 
others. 

7. Internal and External Forces. With reference to 
a body or system of bodies, a force is said to be external 
when it is exerted by a body outside of the system. The 
forces exerted by parts of a body, or of a system of bodies, 
on one another are called internal forces. 

8. The terms "internal" and "external" are relative; and 
it is evident that a $ p 
force may be external _ | 

with respect to a body %\ I 

and internal with re- A 1, W 

72 r 

spect to another | | 

body, or to a system >, | j Jj, 

of which the body in A fb) Q 

question forms a part. Fl 2 

For instance, the forces F t and F, shown in Fig. 2 (a) are 
external to the body A B CD on which they act. If the body 
is in equilibrium, any part of it, as AB PQ, to the left of a 
section PQ, must push on the other part with a force equal 
to F lt while the other part pushes on it with a force F tt = F t . 
This mutual push, or stress, between AB PQ and CDQP 
consists of two forces, both of which are internal with regard 
to the whole body AB CD. But if the condition of only the 
part AB PQ is under consideration, the push of CDQP on it 
may be regarded as an external force. The portion CDQP 
may be removed, as shown in Fig. 2 (), and the portion 
AB PQ considered as an independent body acted on by the 
external forces Fi and F. 



i I 



4 ANALYTIC STATICS 28 

FUNDAMENTAL PRINCIPLES AND FORMULAS 

9. Free Body: Principle of Separate Equilibrium. 

When a body is cousideied by itself as a whole disconnected 
from other bodies, it is called a free body. 

Any pait of a system or body may be treated as a free 
body, provided that it is assumed to be under the action of 
external foices equal to the internal forces that keep it con- 
nected with the rest of the body or system of which it forms 
a part. Thus, in Fig. 2 (/>), the portion ABPQ may bfe 
treated as a free body, after introducing the force F t) equal 
O to the force exerted by the rest of the body A B CD, 
that is, by CD QP, on the portion ABPQ. 

As another example, take the system formed by a 
string OP, Fig 3, carrying a weight W. If it is 
desired to study the condition of a part of the string, 
as O A, the part A P and the weight W may be sup- 
. posed to be removed; but, in order that the condition 
of O A may remain unaltered, it is necessary to intro- 
duce at A an external force equal to the sum of the 
weight Wand the weight of the part AP of the 
string. 

In general, when a body or a system of bodies is 
in equilibrium, every part of it must be in eqtiihb- 



Flt * - H rium and may be treated as a separate body by 
itself, provided that external forces are introduced to replace 
tht; internal forces exerted by the rest of the body or system 
on the pait removed. This principle is called the principle 
of separate equilibrium; its application, called the free- 
body method, is of great value in the solution of static 
problems. 

10. TruiiHfetiiblllty of the Point of Application. 

Concurrent forces have already been defined as those whose 
lines of action intersect at a point, They may or may not 
be actually applied at their point of intersection. Their 
point of application is not essential, provided that the forces 
act on a rigid body. This Is expressed by the following 



28 



ANALYTIC STATICS 



5 



principle, which is usually known as the principle of the 
transmissibdity or transference of force, but will here be called 
the principle of the transferability of the point of 
application: 

When a force acts on a rigid body, the force may be supposed 
to be applied at any point on its line of action^ provided that 
this point and the body are rigidly connected. 

Let a force f, Fig. 4, act on the body M ' N m the direction 




Kio 4. 

OX. A string may be tied at A, or at^?,, and a pull exerted 
from X with the force /''and along the line OX; or it may 
be imagined that the string 
is tied to a point A, and that 
this point is the end of a 
rigid rod connected to the 
body at A l} A t or to any 
other point along OX; no 
matter where the string is 
tied, and no matter from 
what point on OX a pull is 
exerted, the effect will be 
the same. Again, it may 
be imagined that a rigid 
strut ,S is attached to the 
body and that the force is 
applied at A, either by pulling from X or by pushing from O. 
It may thus happen that several concurrent forces acting 
on a body do not intersect within the body. For example, 
in Fig fi, the points of application of the forces F^ F t) and/ 7 , 
actually applied at the points M^ M lt and M, of the body 
PQ t may be transferred to the common point of inter- 
section O of their lines of action, assuming this point to be 




Fio 5 



6 



ANALYTIC STATICS 



28 



rigidly connected to the body. Thus, the force F, may be 
imagined to be a pull at the end O of a string tied to M,, 
and the force F a as a push at the end O of a strut between O 
and M 3 . 

11. Resultant of Any Number of Concurrent 

Forces. The resultant of two concurrent foices can be 

easily determined by either the parallelogram or the triangle 
of forces, as explained in Fundamental Principles of Mechanics. 
The resultant of more than two forces can be found by 
successive applications of either of these principles. 
Let four concurrent forces F lt F t , F a , F t be represented, 

respectively, by the 
vectors O A, OB, O C, 
OD, Fig. 6. The re- 
sultant R? of F l and F, 
is found from the par- 
allelogram OAEB 
constructed on the vec- 
tors O A and OB. This 
resultant is now com- 
bined with another of 
the given forces, say 
F t , by constructing the 
parallelogram OEGC, 
which gives the result- 
ant R" of R' and F tt 
and, therefore, of F lt F tt 
and TV Finally, the parallelogram G KD gives the result- 
ant R of R" and F tl and, therefore, of F lt F t , F and F 4 . 

When analytic methods are used, this process is exceed- 
ingly laborious, as it involves the calculation of the magni- 
tude and direction of each resultant, A much simpler process 
is afforded by the resolution of each force Into two rect- 
angular components, as explained in tha next article. If the 
problem is solved graphically, the triac&te of forces is used 
instead of the parallelogram: A E ia fleawn equal and parallel 
to OB or F t ; then, EG equal and pafalllipj to /i; then, GK 




Fu 6 



I \ 



28 



ANALYTIC STATICS 



equal and parallel to F 4 , the vector O If, between O and K, 
gives the resultant This method will be further explained 
and illustrated in connection with graphic statics. At present, 
only analytic methods are under consideration. 

12. Method by Rectangular Components. Let fi, 

F,, F a , Ft, Fig. 7, be four concurrent forces represented, 



r 



Of 



I 

1 

f* 


1 

1 




FIG 



respectively, by the vectors A,, OA a , OA a , OA 4 . Take any 
two lines X f X, Y' Y through O and at right angles to each 
other. These lines are used as lines of reference; they are 
called coordinate axes, and may be taken in any convenient 
position. The axis X' X is usually referred to as the x axis, 
or axis of x, and the axis Y' Y t as the y axis, or axis of y. 
The angles made by the given forces with the axis of x are 
denoted by #,,-#., etc., as shown. 



8 ANALYTIC STATICS 28 

As explained in Fundamental Principles of Mechanics, the 
force /*, can be icsolved into two components in the direc- 
tions OX and Y, by drawing from A t a line .-/, C paral- 
lel to OX, and a line A,B^ parallel to OY. These lines 
determine a parallelogram OB^A^ C, in which OB, and O C, 
represent the components of F, in the directions OX and O Y, 
respectively. The component OB, is called the x compo- 
nent of F,; and O C, the y component. These components 
will be denoted by A' and K M respectively. Instead of con- 
structing the parallelogram OBiA^C,., it suffices to draw 
from A t the line A/?i perpendicular to OX] this determines 
the two components of F lt since B^A, = O C, = ]',. The 
components of F,, F , F< are similarly found; they will 
be denoted by X t and Y 3 , A' a and Y a , etc. It should be 
understood that, although K,, for instance, may be repre- 
sented by the vector B^A^ that component really ads 
through 0, its position being O C. The given forces aic 
now reduced to the system formed by the A components ( ) /?,, 
OB t , Z? D , #>9,, and the y components O C, OC,, O C,, OC 4 , 
and the resultant of the given forces is the same as thai of 
these components. Since the x components act in the same 
line, their resultant is equal to their algebraic sum (sou 
Fundamental Pnucifilfs of Mechanics}. The same principle 
applies to the resultant of the y components. Taking forces 
acting toward the right or upwards as positive, and those 
acting toward the left or downwards as negative, we luivc, 
X, <= OB X t = fl/?., X a = - Off,, A' t = - Oh\ 
Y, = 0C, Y, = OC, Y, = r)C, K 4 =r - 0(\ 
Denoting the resultant of the x components ant] that of 
the y components by X r and Y ft respectively, we hnve, 
X f = A' t + A", + X tt + A". = OB, + OB, - OB, - OJi, [a] 
Y,= Y,+ Y.+ K.+ K 4 = OC + (9C + OC - ^f, U) 
These resultants are represented in the figure by Oft, 
and OC n respectively. The given system of forces has, 
therefore, been reduced to the two forces X? O J7 f and K- 
=s C r ; and their resultant /?, found by the parallelogram or 
by the triangle of forces, is the resultant of the given forces, 
In other words, the forces X r and Y r are the rectangular 



28 ANALYTIC STATICS 9 

components, in the directions OX and O Y, of the required 
resultant. 

13. The calculation of the magnitude and direction of R 
is accomplished as follows. The light tuangle O .?, A* gives: 

OB, = OA, cos Jf lt B,A, = OA t sin // 
or, since OB^ = X^ OA r = F lt and B l A l = Y lt 
X, = F t cos ff lt Y! = F l sm //i 

The values of X, and Y a , X e and K n , and X t and Y<. are found 
in the same manner It may be well to remembei the pnn- 
ciple that the component or resolute of a force hi any direction is 
equal to the force multiplied by the cosine of the acute angle 
that the line of action of the force makes with that direction. 
Thus, the x component of F, is F, cos H a For the other 
component, the sine should be used instead of the cosine 

Substituting m equations (a) and (b] of the preceding 
article the values of A',, K,, etc., the following expressions 
are obtained: 

X r = F, cos H, + F, cos H a F* cos H 3 F< cos H,. 
Y r = Fi sin H, + F* sin H, + F, sin H, - F. sm H* 

Having determined X r and Y r , the right triangle OB r A r 
gives the magnitude of the resultant R, and the inclination H f 
of this resultant to OX- 

R = OA r = ^OB r 3 + B r AS = *&' + Y r " (1) 



tan H - - - ; (2) 

14. As explained in Plane Trigonometry, Part 2, the sym- 
bol of summation -V, read sigma, is often used to indicate 
the algebraic addition of several quantities denoted by the 
one letter affected by subscripts or accents. If, for example, 
several quantities are denoted by X^X^X^X^ etc., their 
algebraic sum X l + A r , + A' + A' 4 ... is denoted by 
the expression - A', read signia x. 

With this notation, the algebraic sum of the x components 
A', Xv, etc. of any number of forces may be denoted by - X, 
and the algebraic sum of the y components by 2 Y. Also, 
the sum of the expressions F t cos ff lt F, cos H t , etc. may be 
denoted by -i'^cos If, If the components of the resultant 



10 



ANALYTIC STATICS 



are, as before, denoted by A' r and )',., the following equations 

apply to any number of forces 

X r = 2X = .i'/'cob // (1) 
Y r = 2Y = ZF sin // (2) 

Having computed X r and K,, the magnitude and dnection 
of the resultant R are determined by formulas 1 and 2 of 
Art. 13. 

EXAMPLE It is required to hml tlie resultant of the three c<>n- 




the 



l-ll. S 

current foices J'\, /'\, J'\ represented in Fitf H, their magnitudes and 
relative positions being as shown 

SOLUTION Take A" A' inclined nt W to the line of m-lion of /,. 
The Inclinations of the other two forces to A'' V aio u-iulily determined 
from the given angluH. Thus, 

//, = 45", //. = A' ; Os1, - 1HO" - (7fi n H ifi") ** IMI" 

//n A'' O A* B HO" - (!()" ^ ^0" 

For the JT componontH, with the usual convention iw to 
following eqimtioliK uru obtained; 

A', = 1(K) )H4ri 11 * 70.711 Ih, 
A' =-- - JK) ros UO" -- ~ ir>.(MH) Hi, 
A; ~lo() cos L>(] - LlO.IUi lh. 
and for they corupononts: 

Y, KH)siu4f 11 *. 70.711 lh. 

r, m IH) Niu HO" 77,iu n>. 
K M * ~ifiHina -fii.:ttiij, 

Therefore, 

X, i 1 ^ - 70.711 - 4fi.OOO - 14JI.WV - - 115.24 Ib, 
K- - A 1 X 70.711 + 77.48 - ftl ,OS 7 ,ST>1 Ib. 



28 ANALYTIC STATICS 11 

The negative sign of X e indicates that this component of the 
resultant R is directed toward the left, as indicated by the vector O B r 
Formula 1 of Art 13 now gives, 

R = <\/115.24' + 97~3oT n -= 150 86 Ib. 
and formula 2 of the same article gives, 

tan H r = ?Tj^, Hr = 40 11' 20" 
llo U4 

In finding the value of tan H r , the negative sign of X r is dis- 
regarded, as the only thing that is required is the numerical value of 
the angle The signs of X r and Y r show in what quadrant the result- 
ant is In the present case, X r (= O B r } is negative and Y f (B,A,} 
is positive. This at once shows that R lies in the angle X 1 O Y t and 
that H r is, therefore, the angle that R makes with O X' 

NOTE In this, as in many other problems In this Course, angle s aie jdyen to the 
nearest 10" Thus, an angle of 86 10' 37" is called Bfi 10' 40", an angle of 17 47' 28" is 
called 17 47' 20" 



MOMENTS 

15. Definitions. The moment of a force about u 
point, or witli respect to a point, is the product obtained 
by multiplying the magnitude of the 

force by the perpendicular distance from 
the point to the line of action of the 
force. In Fig. 9, the moment of F 
about the point C is Fp\ about the 
point C, it is Fpv 

16. The point to which a moment 
is referred, or about which a moment is 
taken, is called the center of moments, 
or origin of moments. The perpen- 
dicular p or p l from the origin on the 
line of action of the force is called the 

lever arm, or simply the arm, of the force with respect 
to the origin. 

17. A moment is expressed in foot-pounds, inch-tons, 
etc., according to the units to which the force and its arm 
are referred. If, for instance, the force is 10 pounds and the 
arm is 60 feet, the moment is 10 X 60, or 600, foot-pounds; 
if the force is 8 tons and the arm is 6 inches, the moment i& 




12 



ANALYTIC STATICS 



8 X 6, or 48, inch-tons. The term foot-pound is used in 
kinetics m another sense; but the two meanings are so 
different that there is no danger of confusion 

18. Sign ami Direction of u Moment. When, look- 
ing in the direction of the anowhead of a vector represent- 
ing a force, the center of moments lies on the right of the 
line of sight, as C and C, Fig. 10, the moment is considered 
positive; if the center is at the left, as C v , the moment is 
considered negative This may be stated otheiwise thus* 
Imagine the level aim p to be levnlvmg about C in the 

direction indicated by 
../>. the arrowhead. In this 
, c , case, / will t evolve from 
' left to nght, 01 clock- 
wise, and the moment is 
considered positive. In 
the case of /> on the 
contraiy, if this line re- 
volvc'd about C, follow- 
ing the direction of the 
ai low-head, its motion 
would be counter-clock- 
wise; the moment is 
then considered nega- 
tive. The direction of the motion just refcned to is called 
the direction of th iiioniciil, and is supposed to have 
the same sign as the moment itself. 

If M l M^ and Jlf, are the moments of /''about C, (',, and C" fl 
respectively, then, 

M - 4- Fj> - /> 
//, - + /'/, jffr 
/!/ * - />, 

19, Itcin'cHentutlon of ti Momont by an Ai-ou, If 

the lines CO and CA, Kitf. 10, ure drawn, the area of the 
trianffle CO A is 

10 Ax CD m |/-> * \M 
whence, M * 2(i OA X CD) 2 X area OA C 




i'io 10 



28 ANALYTIC STATICS 13 

The magnitude of the moment M is, therefoie, numeri- 
cally equal to twice the aiea of the triangle whose base is 
the vector representing; the force F and whose opposite 
vertex is the origin of moments. 

20. Moment of ti Force In Terms of the Moments 
of Its Components. The moment of a force about any point 
in //\ plane is equal to the algebraic sum of the moments of 
r/<; (omponcnts in any diiections. 

Let the foice R, Fig. 11, be resolved into the two com- 
ponents Ft and F, in any directions, and let C be the origin 
a 



* V<*, . 




PIG. 11 

of moments. The arms of , F l} and F a are, respectively, 
p r , Ai and p at as shown. The moments of these three forces 
about C will be denoted by M r , M^ and M t , respectively. 
Draw CO, and <9 A" perpend ictilai to CO. Denote the angles 
made by the lines of action of the three forces with OX "by 
If,, #i, .//., as shown, and the resolutes of the forces in the 
direction OA'by X,,X lt and JC. Then (Art. 12), 

X r = X,. + X 9 ', 

that is (Ait. 13), 

R cos H r = F! cob HI + F, cos Hj> 

whence, multiplying both members of this equation by CO, 
ft X CO cos H r = F l X CO cos ff^ + F^X CO cos ff, (a) 

1 LT398- 



14 ANALYTIC STATICS 2* 

But H f = 90 COE, and, therefore, cos /7, = sin CO& 

and 

CO cos ff, = CO sin (TO^ 

= C^( triangle C6>>) =A 
Also, CO cos // t = C(9 sin C<9 A = A 
CO cos // = CO sin COD, = A 
Substituting these values in (a), 



that is, jWr = ^ + ^/,- 

In general, if the algebraic sum of the moments of any 
number of concurrent forces is denoted by 2 Af t and the 
moment of the resultant by Af rt as above, then, 

Mr = XM 



CONDITIONS OP EQUILIBRIUM 

21. The Absolute Condition. When any number of 
concurrent forces are in equilibrium, it is evident that they 
can have no rebultant; that is, if their resultant is R, we 
must have R 0. This is the general condition, sometimes 
called the absolute condition, of equilibrium Expressed 
in the form of R 0, however, this condition is of no value 
in the solution of problems. It is necessary to express R in 
terms of other quantities, or to derive some other relations 
by means of which unknown quantities can be detei mined. 

22, Condition of lUtsoluton. Since the resultant of 
several forces in equilibrium is zero, itb resolutes m any two 
directions mUvSt be zero; and, as each resolute is equal to the 
algebraic sum of the corresponding resolutes of the given 
forces, it follows that the sum of the tesolutes of the forces in 
any direction must be zero* 

The resolute of a force in a direction perpendicular to 
itself is zero. There is, then, one direction (at right angles 
to the resultant) ftlongr which the sum of the resolutes of any 
number of unbalanced concurrent forces is zero ; but if the 
sum of the resolutea is zero for any two directions, the forces 
are in equilibrium; for both directions cannot be perpen- 
dicular to the line of action of the resultant. 



28 ANALYTIC STATICS 15 

23. Condition of Moments. Since the resultant is 0, 
its moment about any point must be zero; and, as the 
moment of the resultant is equal to the sum of the moments 
of the components, we must have, 

M + M, + M',+ . . . =0, 

where, M lt M,, Jlf a , etc. are the moments of the components 
about any point in their plane. 

It is necessary to specify that the sum of the moments 
must be zero, when taken with respect to any point, or every 
point in their plane, for, m a system of forces not in equilib- 
rium, the sum of their moments about a point on the 
resultant is also equal to zero, because in this case the lever 
arm of the resultant is zero, and its moment, therefore, is 
zero. But, if the forces have a resultant, the sum of their 
moments about a point outside of the resultant cannot be 
zero. Conversely, if the sum of the moments of the forces 
about three points not in the same straight line is zero, the 
forces have no resultant; for these three points cannot all lie 
on the line of action of the resultant, and the only case m 
which the sum of the moments of a system of unbalanced 
forces can be zero is that in which the origin of moments is 
a point m the line of action of the resultant. 

24. Genei-al Statement of the Conditions of Equi- 
librium. Summing up: Any balanced system of coplanar 
concurrent forces must satisfy the following conditions, each 
of which is necessary and sufficient for equilibrium, and 
involves the other two: 

1. The resultant of the forces must be sero. 

2. T/ie sum of the resolutes of the forces in each of any two 
directions must be zero. 

3. The sum of the moments of the forces about three points in 
their plane \ not in the same straight line^ must be zero. 

Condition 1 is expressed algebraically by the equation R 0. 
Condition 2 may be expressed thus (see Art. 14): 
EX = S 



Condition 3 may be expressed thus (see Art. 20): 



16 



ANALYTIC STATICS 



1 28 



If the lever arms of the forces F n F* t etc are denoted 
by Pi t A, etc., the algebraic sum of their moments, or 2 Af, 
jt> is equal to F 1 p^ 4- F a /> a + F 3 p* + , etc 

Denoting this sum by 2 Fp, formula 2 
may be written 
*** vpp = (3) 

25. Equilibrium of Three 
Forces. Problems i elating to the 
equilibrium of three forces are usu- 
ally most conveniently solved by 
means of the triangle of foices. It 
is to be observed that the eqiuhbrant 
of any number of forces is numer- 
ically equal to their resultant, but 
acts in the oppobite direction. Let 
the forces F lt F yi Fig. 12(tf), act 
through O. By di awing through O f , 
us shown at (), the vectors (Y AS and 
\-l Aj ) representing the given forces, and completing the 

triangle O 1 Aj AJ> the re- 
sultant R of /'I and F, is 
obtained. In this triangle 
of forces, the two compo- 




FIG lli 




nents are in cyclic order with each other, but in non-cyclic 
order with the resultant (see Fundamental Principles of 
Mechanics}. 



28 ANALYTIC STATICS 17 

In Fig. 13 (a), the three forces F^F,, and Q, acting at O, 
are supposed to be m equilibrium. The force Q is the equi- 
librant of F, and F t ; it is numerically equal to their result- 
ant R, but acts in the opposite direction. Consequently, 
a triangle O 1 AS A* 1 , Fig. 13 (b), equal to the ti tangle giving 
the resultant /?, can be formed with the vectors lepresent- 
ing /-" F yt and O\ but the anowheads on Q and 7? must 
point in opposite directions, which means that Q must be 
taken in cyclic order with F l and F tt . 

26. Selection of Axes. When there are more than 
three forces in equilibimm, the most convenient method for 
determining any of them, when the others aie known, con- 
sists m finding expiessions foi the resolutes of all the forces 
m two rectangular dnections, and using formulas 1 and 2, 
Art 24. A similat method applies to the determination of 
the resultant, as aheady explained Theoretically these 
directions are entirely aibitrary; but, practically, the direc- 
tion of one of the foices, often one whose magnitude is not 
known, is almost always a very convenient one to use, as in 
this case one of the resolutes of that force is equal to zero, 
and the other is equal to the force itself. When this is done, 
either direction along the line of action of the force may be 
taken as positive and the other as negative; but this does 
not necessarily mean that the direction taken as positive is the 
direction of the force acting along that line. For example, 
forces acting at a point O may be resolved into components 
parallel and perpendicular to the line of action, say OK, of 
one of the forces; and, for the purpose of this resolution, the 
direction OK may be treated as positive and the opposite 
direction as negative. This simply means that, if the resolutes 
parallel to OK are to be added algebraically, the arithmetical 
difference must be taken between those resolutes whose direc- 
tion is OK and those whose direction is opposite; if the dif- 
ference is positive, the direction of the resultant resolute is 
from O toward K\ if negative, from A" toward O. 

Similarly, in taking moments, any point can be used as an 
origin; but it is convenient to take the origin on the line or 



18 



ANALYTIC STATICS 



28 



action of one of the forces, as in this case the moment of 
that force is zero, and the force is thereby eliminated from 
the equation of moments. 

These methods will be better understood by a study of the 
following problems: 

EXAMPLE 1. A weight of 500 pounds (see Fig. 14) hangs by three 




j t 



/ I \ 



W-BOOlb. 



FIG. 14 

ropes tied to a ring at O, the inclinations of the two slanting ropes O JS t 
and O E, to each other and to the horizontal being as shown. It Is 
required to find the tensions in the ropes OEi and O E 9 . 

SOLUTION BY TRIANGLE OF FORCES. The tensions F,. and F n in 
the ropes are represented by the vectors A v and A tt and the weight 
by the vector W. These three forces form a balanced system, each 
force being the equilibrant of the other two. Considering W an the 



28 ANALYTIC STATICS 19 

equilibrant of F l and F a , the triangle of forces OBC is obtained by 
drawing through B a line parallel to OA lt and through O a line 
parallel to O A* In the present case, CO is in the prolongation of 
O At The triangle might have been constructed anywhere else; It is 
not necessary to draw it m the position here shown Nor is it neces- 
sary to draw it accurately to scale, as its only object is to serve as a 
guide in the calculation 
In this triangle, 

K= AiOC = 180 - Ai OA, = 180 - 75; 
and, therefore, 

sin K = sin (180 - 76) = sm 76 
Also, Y being vertical, 

Li = 90 - ff m = 00 - 60 = 30 
L, = A, O Y = 90 - Hi = 90 - 46 = 46 
The triangle O B C now gives 

sm L, = -^U sin 30 = 258 82 Ib. Ans. 




- , 

sin A sm 7o 

F t = -^-sin L 3 = -^osm 45 = 366 03 Ib. Ans. 
9 ^ sin 75 



SOLUTION BY RESO LUTES Let the vertical line Y and the 
horizontal hue OX be taken as axes of coordinates As usual, the 
resolutes of Fl will be denoted by A' and YI, and those of F, by X* 
and K,, as shown in the figure. They are here drawn for purposes of 
illustration, but it is not necessary to draw them in order to apply the 
general formulas. The vertical resolute of the weight is W\ the 
horizontal resolute is zero. 

Placing the sum of the horizontal resolutes equal to zero, we have, 
noticing that X l is negative, 

- Ft cos Hi + F t cos H t + = 0; 
that is, - Fi cos 45 + F a cos 60 - (a) 

Similarly, for the vertical resolutes, 

Fi sin Hi + F, sm H, - W = 0; 
or, Fi sm 45 + F t sin 60 - W = (*) 

Prom (a), cos 46 



which, substituted in (b), gives 

etn ftn 

Ft sin 45 + F t cos 45 ^L^_ _ 
whence, cos w 

W _ 

" sm 46 + cos 45 sm ^ " 8in 46 cos ^ + cos 45 sln 
cos 60 



W OAO ?f . OAO 

sin (45 + 60-) = STW Sm ^ = iSTTS 5 sin 80 
as found before. 
The value of F t may now be found from (c) . 



20 



ANALYTIC STATICS 



28 



The horizontal and vertical resolutes have been used simply to 
illustrate the general method, but it will be seen that, by resolving 
the forces m these directions, the solution is long and tedious If 
however, resolutes perpendicular to the lines of action of the unknown 
forces FI and F a are taken, the solution will be very much simplified 
First, let the resolutes be taken perpendicular to O J5 y Then, we shall 
have, considering the direction from O toward E 3 as positive, 
resolute of F a = 
resolute of F t = f,. sm 75 
resolute of W = -W^smL,= - W sm 30 
Therefore (formula 1, Art 24), 

/r lS m75- W sin 30 = 



whence, 



-=a sm 30 
sin 75 




PIG 1C 



i&oo zfc. 



In a similar manner, , may 
be found by taking the resolutes 
perpendicular to FI 

This method should be care- 
fully studied, as its simplicity 
makes it of very great practical 
importance. 

EXAMPLE 2. A weight W of 
1,500 pounds is hung from the 
extremity O, Fig 15, of a hori- 
zontal bar projecting out of a 
vertical wall and held by a 
rope ON, Inclined to the vertical 
at an angle of 40 Required 
the tension T in the rope and 
the compression C in the bar 

SOLUTION UY TRIANGLE OK 
FORCES. It is obvious that W 
is the equilibrunt of C and T. 
The actual lines of action of W 
and T are shown by the vectors 
O B and O A. Draw B D per- 
pendicular to O //, to meet A' O 
produced at D. Then, in the 



triangle O B D, the vectors D O and B D represent T and C } respec- 
tively. Also, BOD = 40. 

C - W tan 40 = 1,500 tan 40 - 1,258.0 Ib. Ans 

T - 1 -^so = 1,958.1 Ib Ans. 

cos 40 cos 40 

SOLUTION BY MOMENTS. If moments are taken about N, the 
moment of T will evidently be zero, since //lies on the line of action 



28 



ANALYTIC STATICS 



21 



of T, the moment of W will be WxNS=lf>WXEO The 
moment of C whose line of action is OE, will be CX EN There- 
fore (Art 2O) , 

CXEN = i,mxo, 

whence, C = 1,600 J^ = 1,600 tan 40, 

as found before 

If moments aie taken about E, the moment of C, the resultant, 
will be 7cio, that of H' will be the same as befoie, 01 1,500 X E O, and 
that of ywill be TX E A negative, because the origin E is on the 




A* 



FIG 10 

left of O A (Art. 18) Equating the sum of the moments of the com- 
ponents to the moment of the resultant, 

1,500X^0- TxEK~ 0, 

v. T -i n EO 1.600 L500 

whence, y . 3 ,600 ^. - -^y - ^ 

as found before 

The method of moments is sometimes very convenient. The center 
of moments should, if possible, be taken on the line of action of one 
of the unknown foices, as that force is thereby eliminated 

EXAMPLE 3 To find the equilibrant of the five forces represented 
by the full-line vectors in Fig. 10. 

SOLUTION Let OA lt the line of action of F,, be taken as the 
x direction, and Y, the perpendicular to it at 0, as the y direction of 



22 ANALYTIC STATICS 28 

resolutes. The angles of F lt F a , F 3> etc , with O X will be denoted 
by HI, // a , &3, as in Art 12, and the resolutes of the forces by 
X lt X a , y it y,, etc. Here we have H* = 0, H, = 50, ff, = A n O X 1 
= 180 - (//, + 90) = 180 - 140 = 40, H< = 60 - H* = 60 - 40 
= 20, H, = 30. 

Therefoie, JT X - F = 20 

X 3 = F, cos 50 = 35 cos 50 

X, = - F, cos 40 = - 25 cos 40 

X< = - F t cos 20 = - 40 cos 20 

X, - F t cos 30 = 45 cos 30 
Denoting the x resolute of the equihbrant by X gt we must have 

(Art 24), 

X f + X,+ X, + X, + Xi + X, = 
whence, 

X 9 = -(Xi+X, + X, + X< + X*) 

= _ 20 - 35 cos 50 + 26 cos 40 + 40 cos 20 - 45 cos 30 

- - 24.730 Ib 
Similarly, K x - 

Y, = - F t sin 50 = - 35 sin 50 
Y, = - F 3 sin 40 = - 25 sin 40 
K = F* sin 20 40 sin 20 
y t = F, sm 30 = 45 sm 30 

and y f s -(y 1 + y.+ y.+ y 4 + y 1 ) 

= 35 sm 50 + 26 sm 40 - 40 sin 20 - 45 sin 30 

- 6.700 Ib 

The magnitude of the equihbrant Q may be found from the rela- 
tion Q = VAV + y,', but it is better to determine first the angle H q 
between Q and OX. For this purpose, it is not necessary to take the 
sign of X g into account (Art 14) 



and, therefore, 

Y a 6 700 



EXAMPLES FOR PRACTICE 

1 Two concurrent forces Fj. = SO tons and F t = 25 tons act at an 
angle of 170. Find (a) the magnitude Q of their equilibrant ; (*) the 
angle L made by Q with F,. Ano / (a) Q - 6.0132 tons 

Ans \(t>)L - 48 64' 

2. Resolve a force /? 100 pounds into two equal components Fi 
- J" 1 , making an angle of 70. Ans. F l F, 61 039 Ib. 

3. Find a general relation between a force F and either of two 
equal components F^ making an angle If with each other. 

Ans. F" 2-F, costtf 



28 



ANALYTIC STATICS 



23 



4 Given the equilibrant Q = 10 tons of the two forces F and F t , 
Fig 17, find the magnitudes of those two forces, the angles being as 
shown (The vectors F l and F, in the figure are drawn of arbitrary 
lengths, their purpose being to indicate direc- 
tions, not magnitudes, the latter being as yet 
unknown ) Ans IF,. = 11 154 T. 

Ans \F, - 2 9886 T. 

6 Find the equilibrant Q of the forces 
represented in Fig 18, and the angle H, it 
makes with the line of action O X of the 20-ton 
force (Find resolutes parallel and perpen- 



dicular to O X ) 



Anc 
Ans 



W 
\Jf f 



10 735 tons 
5 is/ 50" 



6 A weight of 15 tons is supported by two 
ropes, one horizontal and the other making an 
angle of 45 with the horizon (135 with the 
former rope) Find the tensions Ti and T t in 
the two ropes, 7i being tension In horizontal 

r P e< S 

21.213 tons 




Q-lOTon* 



FIG 17 




PIG 18 



24 ANALYTIC STATICS 



STRESSES IN FRAMED STRUCTURES 



DEFINITIONS 

NOIE The theoiy of eoncuirent forces finds sunie of its principal 
applications in the determination of stiesset, in framed structuies 
Although a complete treatment of such a subject would be foreign to 
the scope of this Section, a. few simple examples will be given, in order 
that the utility of these principles may be seen Before giving those 
examples, it will be necessary to define some of the teims employed 

27. Structures. A (structure is a statical combina- 
tion of parts designed for the transmission of foice. By 
saying that a structure is a statical combination, it is meant 
that the forces transmitted by the parts are supposed to be 
all balanced and produce no motion. Bridges and buildings 
are examples 

28. MacliluoB. A machine is a combination of paits 
designed for the transmission of motion In any special case 
in which a machine produces no motion, its vanous parts 
being at rest, the machine is to be treated as a stiuctiue. 

29. Frames*. The term frame is applied to any combi- 
nation of bats, strings, ropes, or other straight paits con- 
nected together so that their center lines form a polygon or 
a pait of a polygon. Eveiy one of the straight parts so con- 
nected is called a member of the frame, and the connection 
of two or more members is called a Joint. 

Although in some cases the center lines of the various 
members connecting at a joint do not meet exactly in a point, 
they are considered so to meet, and the forces acting along 
the members are treated as concurrent forces meeting at the 
center of the joint. 

30. Trusses. A trubs is a rigid frame contesting; of 
triangles, such as occur in bridges and buildings, and in some 
branches of carpentry. 

31. Supports. The place or places on which a stuiuture 
rests, and to which, therefore, the forces acting on the stiiic 1 - 
ture considered as a whole are transmitted, are called 



28 ANALYTIC STATICS 25 

siipports, because they support, or sustain, the structuie. 
Such are the piei s of a bridge and the foundation of an engine. 
When a stiucture, as a bridge, rests on diffeient supports, 
the point ot application of the resultant pressure on any sup- 
port is called <\ point of support. 

32. Reactions. At every joint of a frame there are 
seveial forces acting, namely: 

1 The external forces, or forces applied to the frame 
from the outside 

2 The mutual reactions, or the forces exerted by the 
various members on one another at the joint. 

At some joints, there maybe no external foices diiectly 
applied, but the membeis may act on one another on account 
of external foices acting at other joints. 

As to every action there is an equal and opposite reaction, 
eveiy suppoit of a stiuctuie exerts on the btiucture a force 
equal and opposite to the foice transmitted by the structure to 
the support. This foice excited by the support on the struc- 
ture is called the reaction at the support, and is to be con- 
sidered as one of the external forces acting on the structure. 

33. Struts ami Ties. The two kinds of stress ten- 
sion and compression that may occur in the members of a 
frame are defined in Fundamental Principles of Mechanics. A 
tension is often called a pull, and a compression a thrust. 

Those members that are designed to resist compression 
are called struts; those that are designed to resist tension 
are called tie*?. 

1>KTKRM I NATION OF STRKS8ES 

34. Introductory Explanations. Here those frames 
only will be dealt with in which the forces acting on every 
member aie applied al its extremities, and m which the con- 
ditions are such that the mutual reactions at the joints may 
be taken to have lines of action meeting at the common 
intersection of the center lines of the membeis. 

This being understood, lei A^A, A, A, A,A t Fig. 19, be 
three members of a frame, A being the point of intersection 



26 ANALYTIC STATICS 28 

of the center lines of the members, and A lt A t ,A a , the joints 
connecting the three members with other members. It will 
be first assumed that there is no external force acting at A. 
Let F^P\,F t be the resultants of the forces acting on 
A,. A, A, A, and A, A at A^A a , and A tt respectively. By the 
principle of separate equilibrium (Art. 9), every member, 
considered as a free body, must be in equilibrium; therefore, 
the resultant of the forces acting at A on A^A must bal- 
ance F^ and that resultant and F must be collinear and act 
along A As, so that the resultant at A is a force Fi equal 




FIG. 19 

in magnitude to F lt but acting in an opposite direction. 
Likewise, AA t is kept in equilibrium by the forces F, 
and F,, both having the direction of the member, and 
F, being the resultant of the forces exerted by the other 
members on A, A at A. Similarly for A, A, The vector 
representing F, is drawn on one side of the member, in 
order to avoid confusion, although F, really acts along 
the member A A,. Now, transfer the forces Fi,F, t and F t 
to A (Art 10), where they are represented by accented 
letters. For convenience, as well as for the sake of uni- 
formity, the forces thus transferred are represented by veer 
FtftF, 1 , having their common origin at A that is, 



28 ANALYTIC STATICS 27 

with the arrowheads pointing away iromA. It is evident 
that the force F lt being the resultant of the foices acting 
directly on A* A at A, is the resultant of F, f and FJ, and, 
since F,.' is the equihbrant of F lt it is likewise the equili- 
brant of F,' and F a f ; in other words, the forces FS, FJ ', and 
Fa' (or their equivalents F^F t , F 3 ) form a system of concur- 
rent forces in equilibiium 

Substantially the same explanations would apply in case 
there weie external forces acting at A. These forces, with 
the forces Fi f , F,' t and F a f , would form a system of balanced 
concurrent forces. 

The problem of determining the forces acting on the 
members is thus reduced to the general problem of the equi- 
librium of concurrent forces. It is now necessary to ascer- 
tain what effect those forces have on the members of the 
frame that is, what members are in tension and what 
members are in compression. 

35. Character of the Stress In a Member. Considui 
the condition of A l A It has been explained that this mem- 
ber is held in equilibrium by the two forces F l and F[', this 
pair of forces constitutes the stress in the member, and the 
magnitude of either force is a measure of the stress (see 
Fundamental Principles of Mechanics) . It will be observed 
that the tendency of the forces 7*1 and FI is to stretch the 
member, and that, therefore, the member is m tension. The 
stress in A,. A is, therefore, a tension of /i pounds, tons, etc., 
as the case may be. Similarly, the stress in A t A is a com- 
pression equal to F, (by which is meant that each of the two 
forces constituting the stress has a magnitude equal to F a ), 
and the stress m A a A is a tension equal to F*. 

36. In determining the character of the stress in a 
member, however, it is not necessary to consider the two 
forces acting at the extremities of the member. Referring 
again to the joint A, where the forces FS, F t f , and F<J (or 
their equals F lt F F a ) form a balanced system, it is seen 
that FJ and F, f have their arrowheads pointing away from 
the ioint; these forces may be said to be pulling at the joint 



28 



ANALYTIC vSTATIC\S 



28 



in the direction of the membeis /, / and /,./, icspectively, 
and these two members are in tension. The force /*/, on 
the contrary, when supposed, as it should be, to be acting 
on the joint along ./M, as shown by the dotted vectoi /.y, 
pushes against the joint in the direction of the member . / /; 
hence, this member is in compulsion. 

37. Summing up T/iefonfathuX'il(>ni>aHit'iHbe> Hica^itcs 
tension ot compiesvon atconiing "S ? '//<"'/ sttflfiw/ to mt on the 
joint through the membo > it pttlh at ot frtewA on the joint. 

It has been seen, for instance, that /'V, when supposed to 
be acting on the joint thiough the membei /,./ (as shown 
by F t ") presses on that joint, and that / . / is in compression. 

38. The forces F,', /<"/, Fl (which, for shoitness, are 
called the stresses in the membeis) have been represented 
by vectors having their common unj>i at the joint. This is 

convenient in cornpli- 
1 ezited problems in 

which the method of 
resolutes is used 
^ Hut, in almost all 
cases, the vectors 
may be drawn along 
the lines represent- 
ing the members, and 
then the arrowhead 
on any force shows at 
once whether the 
force is (that is, 
measures) a pull or a 
thrust. 

39, Noot'MHary 
Data For Deter- 
mining? Htrosaea, 
In determining the 
stresses in the members meeting at a joint, the external foroas 
or the stresses in some of the other members must be known. 
Suppose, for instance, that the stress in the member 





28 ANALYTIC STATICS 29 

Fig. 20, is known to be a compiession equal to F lt repre- 
sented by the vectoi A N, and that it is required to find the 
stresses F t and F, in the members A, A and A a A Since the 
forces F t ,F 3 , and F-, ai e in equilibrium, a vectoi triangle can be 
constiucted with the vectoi s lepresenting them Through N 
draw A T < I/ parallel to A* A, meeting A, A at Jlf, and mark the 
arrowheads on N Jlf and M 'A so that the vectors AN, NAf, 
and /If. 1 will be in cyclic order Then, N .If = F 3 ; MA = F n . 
As F, pi esses on the joint A, the member A\A is m com- 
piession. Similaily, F a) although given in direction by the 
vectoi JVM, must be supposed to be acting along A a A, and 
it is seen that, when thus applied, it presses on the joint A\ 
therefoie, 1 3 A also is in compression 

40. Opposite Arrowheads 011 a Member. By refer- 
ring to Fig. 21, it will be seen that the member connecting 

the two joints M and O has two 
arrowheads pointing in opposite 
directions The student should 
not fall into the en or of think- 
ing that each arrowhead indi- 
cates the direction in which the 
part of the member on which it 
is marked acts on the other part. 
Thus, if the member is cut by 
a plane X K, the arrowhead F 
does not indicate the direction 
in which the part Z acts on 
the part MZ\ if that were the 
case, the member would be in compression, whereas it is in 
tension. The arrowhead F indicates the direction in which 
MO acts on O, and, as this action is transmitted from MZ 
to Z, the arrowhead F really indicates the direction in which 
M Z acts on O Z\ which shows that MZ pulls on OZ, and 
that, therefore, OZi& in tension. Likewise, the arrowhead F' 
indicates the direction in which the member acts on the 
joint M, or the direction in which, a^y part Q the member, 
as Z, acts on the part Jtf pelow it. - The forces acting in 

ILTJ9S-7 f ' 




30 



ANALYTIC vSTATICS 



28 



the direction of /'"and JP lire, of course, equal, as each measures 
the stress in the member M O\ either may be considered as the 
action, and the other as the equal and opposite reaction. 

EXAMPI.B. A tiuss consisting of tlnue horizontal membeis PC, 
CQ t and B f), Fig liii, and four equal inclined members/*./?, 12 C t CB t 




and B Q, carries a weight of ft' pounds hung from the cent IT C. The 
truss rests ou two piers /' untl )\ its length, lietwutni thu joints at /* 
and Q, IB 2/, and itn height is h. Neglecting thu weight of thts Htrwt:- 
ture, it IB required to find the Htreiw in eavh momlier und thts rtruc-tiniiH 
at the supports. 

SOLUTION BY TKIANOI.R OF FORCRH.- On iiwcmnt of the Hymmetry 
of the figure, the angle* PCJ) and Q CH MO eqtwl; tliwy ure both 
denoted by //. Also, /; C IH parallel to Ji Q t und D CJt f'/tQ - L, 
say. The anglea // and A, however, are nttt ntipptihcd to bo known 
and have to be exprowed in terms of / nnd h. The figure 

/;t '-~. A <> 

(i\ 



Niu // - 



tan // - 



m 180 



In - aln 2 /^ - 



W 



28 ANALYTIC STATICS 31 

Of the forces that balance at C, it is evident that, owing to the sym- 
metry of the tiuss, the foices along CQanA CP are equal and opposite, 
and form, therefore, by themselves a balanced system. Hence, the 
weight Wmust be balanced by the foices F and f, acting; along the 
members CD and C B. The stresses in these two members are 
evidently pulls, and it is obvious that /<\ = F, 

In the triangle CMN, the vector CM represents the weight W, 
i\I ' N is parallel to CB and CN Is in the prolongation of D C 
Therefore, MN FI and NC F a . Since F^ => F tt then, also, A", 
= A\ 

Now, #1 = 180 - (90 + H) = 00 - H 

N = 180 - (Ki 4- A.) - 180 - 2 A', - 180 - 2 (90 - H) = 2ff 

W W 

Then, F t -F m - ^sin A', = -sm (90 - H) 



W rr 

, --- ?- r cos H = 



S sm H cos H " 2 sin// 

Coming now to the joint B t It is known that the force /*i acts on 
It along the member CB, hence, the two forces F a and F^ acting 
along DB and QB can be determined Take B M 1 = F v and 
draw .fl/'.A" parallel to DB, meeting BQ at A r/ Then, .rt/'JV' 
represents the force F* acting along D B t and N 1 B the force F+ acting 
along QB The stresses in these members are evidently thrusts 
(Art 37) As M> N' is parallel to C Q, the angles B M' N' and B JV' M f 
are both equal to B C Q, or H. The triangle B M 1 N 1 being isosceles, 
we have F* = FI . 

For F, t we have, _ 

F a = -^.smZ 

sin /f 
jq/ 
Putting Fi = - jr> as found above, and sin L = 2 sin H cos .#, 

^. = ~ X 2 B 



The forces acting at Q are the thrust /?! of the member B Q, the 
horizontal force F t along CQ, and the reaction of the pier, which, 
as will be shown hereafter, acts vertically upwards Take Q M" to 
represent F t (notice that it is not necessary to measure the distance 
from Q in the direction of the arrowhead), draw M" N" horizontal 
and QN 1 ' vertical, and, starting with F*, whose direction is known, 
mark the arrowheads on the sides of the triangle M" Q N" in cyclic 
order Then, QN" will represent the reaction /? and N" M" the 
force F t in Q C. The latter force acts in the direction Q C, thus 
showing the stress in Q C to be a pull. The triangle M" QN" gives 

= ZT rr p rr W COS If W t Wl 

F.-F.cMjr-FtixHir- 2sln/r - T cotjsr- ^ 

For the reaction, , , 

JZ - /^ tan // ~ cot /f tan /f = ~ 
* 



32 ANALYTIC STATICS 28 

The reaction being equal and opposed to the pressure exerted by 
the truss on the pier, or to the part of the load transmitted by the truss 
to the pier, it is seen that one half of the load is transmitted to Q, and, 
as the truss is symmetrical, the other half is transmitted to the other 
pier P This subject will be more fully explained in connection with 
the theory of parallel forces By this theoiy, the value of the reaction 
can be determined first and the calculation begun at the joint through 
which the reaction acts, this is the method used in piactice. 

On account of the symmetry of the truss, the stress in DP \^ 
obviously the same as the stress in B O 

SOLUTION BY MOMENTS The solution of this pioblem by the 
method of moments is as follows 

For joint C, momenta are taken about a point on the line of action 
of one of the unknown forces Fj. or F, The equal and opposite forces F t 
need not be taken into account If the joint 13 is taken as the origin 
of moments, the lever arm of W is fi U = 1 /, and that of F a is 



BG = J&Csm L = DC shi L = ~ sin /. = ~* ~ 

cos H 2 COB H 

The moment of W\& negative, and that of F 9 is positive (Art. 18). 
Therefore (Art. 24), 

- WXBU+F,XBG = 0, 
or, writing the values of B U and B G just found, 
Wl Ft I sin L 

*M 



whence, 



2 cos// 
W cof, H 



sin L 

This was the value found for F l (which is equal to /",) by using the ' 

triangle of forces, and may be transformed in the same mannei t 

Passing now to joint B, F t is determined by taking moments about D t \ 

in the line of action of F a . The lever arms of F t is DJ, and the lever f 

arm of F Is the perpendicular from t) on Q R produced, but this per- \ 

pendtcular is equal to B G 1 , because QB and CD are parallel. 

Therefore, F L X DJ- F t X B G = 0, i 



whence, F* = F* X 



because the triangle B CD is isosceles, arid DJ*=BG. 
*To find F a , take moments about Q, in the line of F t . The lever 
arm of F 9 is Q S = k, that of F^ is Q Z = / sin H. Then, 
Fi A - F l I sin H = 



' h ~> 

W 
or, because FI - ^ g (see the solution by the triangle of forces 

given above) , 



28 



ANALYTIC STATICS 



33 



Passing now to joint Q, to find F t , take moments about S 1 in the line 
of R The lever arm of F t is S/ = S sm H = ~ sin H, that of F, 



is (PS = /; Then, 



whence, 



t I sm // 



= 0, 

sin H 



t . , . \ F t I sin H . , 
but (see above) -j = F at therefore, 

F = f = Wj, 

To hnd R, take moments about C^m the line of F, The lever arm 
of R is Q C = /, that of F t la C T = ~Q Z = / sm JV. 
Then, Fl s\n H - I? I = 0, whence, 

J? = FI sm /f = /?, sm // = 2~^-X sm // = -^ 

In practice, it is not necessary to draw the perpendiculars C U, B G, 
C T, etc , as their values can be at once written down by means of the 
fundamental trigonometric relations among the elements of a right 
triangle 



EXAMPLES FOR PRACTICE 

1. A weight of 2 tons is suspended from a derrick, Fig. 23; the 
length of the boom A B is 40 feet; the guy rope CB is fastened at a 
point C, 30 feet from A, and the boom is 10 feet out of vertical. Find 




FIG 28 

the tension FI in the guy rope and the thrust F in the boom. 

= .96840 tons 
, 2 7542 tons 



a 



34 



ANALYTIC STATICS 



28 



2 A trapezoidal frame having the dimensions shown in Pig 24 
rests on two piers and carries two weights of 5 tons each at the 
joints A and B Find the stresses F tt F,, and F 3 in the members A t 




A C, and CD, respectively, and the reaction /? ut either pier The 
members A B and CD are horizontal, and the inclinations of A C 
and B D to CD are equal. [F^ 2 fi tons, thrust 

Ans \ F ' = 5 - 5002 tons > 
Ans 



F a 



2 fi tons, pull 
5 tons 



PARALLEL FORCES 



COPLANAR FORCES 



TWO FORCES 

41. Resultant or Two Parallel Forces Having the 
Same Direction. Let two parallel forces /*i and F, Fig. 26, 
act on a rigid body A BCD. The lines of action of the two 
forces are K^L^ and K*L*, respectively, and the points of 
application Ei and , are any two points on those lines (see 
Art 10). Draw E^E t . The effect of the forces will not be 
changed if any two equal and opposite forces R^ T^ and fi t 7 1 ,, 
acting 1 along 1 the line ./?, /?,, are introduced, for these two 
forces will evidently balance each other. The two forces F l 
and F t may, therefore, be replaced by the four forces f t and 
E, 71, F, and /?. T.. The resultant of F, and t T, is -fi 1 , ,S lf 
the diagonal of the parallelogram E t A^ 5, 7*,. Similarly, 
the resultant of F and E* T, is ^, 5 1 ,. The lines of action 



28 



ANALYTIC STATICS 



35 



of ^ ,5*! and E, S, meet at a point E 1 . By the principle of the 
transferability of the point of application (Art. 10), the 
points of application of E^ S,. and E* S, may be transferred 
to ', provided that the point Ef is supposed to be rigidly 
connected to the body A CD. In this new position, the 
forces are represented by E 1 SJ and E 1 S, f . The force E 1 57 
may now be resolved again into its components E' 77, equal 
and parallel to j\ 7\, and E 1 'A7, equal and parallel to F^\ 




PIG 26 



and E' S, f into its components E 1 T,', equal and parallel to 
E t T, t and & W, equal and parallel to /".. As E' TJ and 
E 1 T, f balance each other, only two forces, acting along the 
line K L, now remain, whose resultant R is equal to F v -f- F*. 

The point of application of this force may be taken any- 
where, as at E } on the line K L, 

In the similar triangles E f E* and NJJS'SJ we have 
EEi E'E 
WSJ ~ & 

E'E X W 5' 



whence 



36 ANALYTIC STATICS 28 

In the same manner, we get, from the triangles E E' E* 



E 1 N,' = E 1 E X N,' S,' 
or, because NJ S a f = E 1 T. f = E' 77 = AV SJ t 

EE,XE' N.' = E 1 E X A r / S/ 
This, compared with (), gives 

- t X EN! = ^/T 3 X /;' AY, 
that is, 

", X Fi = ; X /=;, /^ X Vi\ = 1*\ X EE* (b) 



, JC> AI j'a f \ 

whence, - = (c) 

i\ 



From this we obtain, according to the laws of piopoition 
(see Geometry), 



_ 

EE, , ' 

that is, 

r = , and F, X E,E* = R X E E, (d} 



Likewise, 

- 5 = ' and ^ X ^ ^ = /v> X 7 ' : ^ 



The preceding results may be stated as follows 

1. 77w resultant of two paiallcl font's having the same 
direction is equal to their turn > its line of attioti ts parallel to 
the lines of action of the two foitt's, and its d nation is the same 
as the common direct ton of the fwo fanes. 

2. The line of action of the tfsidtant is so silimtt'tl that it 
divides any line mteicepted between the components into two seg- 
ments niveisely proportional to those components [equation (r)]. 

3. The resultant and the lomponcnts aie so t elated and 
situated that, if the points of applitation of thf fhiee are taken 
on any straight line (as JK^E^ Fig, 25) tutfi netting then lines 
of action^ the produtts of any two of thf three forces by the dis- 
tances of their respective points of application fnwi (he point of 
application of the third force are equal [equations {) , (</) , (<)]. 

The last statement applies to distances taken on any line 
between the lines of action of the two forces; fur it will be 
remembered that E l and , were taken arbitrarily on K* t 



28 ANALYTIC STATICS 37 

and K a L*. Any other points, as <7, and G a , might have 
been taken, and the same reasoning would have led to the 
same conclusions 

Let /?!/?* = /, REi = l^EE* /. Then, equations (6), 
(rf), and (<?) may be written 

K /, = /:/. i 

F.I = X. 
FJ = /?/J 

42. Equilibrium of Three Parallel Forces. By 

leversing the direction of A*, Fig 25, we have the equili- 
brant Q of the forces F t and /'", Arithmetically, the value 
of Q is F! -\- Fj. But, if we consider forces acting in one 
direction ass positive and those acting in the opposite direc- 
tion ab negative, we have 

whence, Q + Ft + F a = 

This, with any one of equations (/) of Art. 41, gives the 
conditions of equilibuum of three parallel forces. The 
eqmhbrant must always lie between the other two forces. 

43. Theorem of Moments. Let O, Fig. 25, be any 
point on the plane of the forces, and draw O P, perpendicular 
to their lines of action. Let // M^ Kf r be the moments 
of F lt F 9) and A' about O Then, 

Mi = FiXOJ\ (a} 

M* - F t X ( O P, + A /\) (b) 

Afr^RxloPi+PtP) 
Adding (a) and ((>) , 



= tf X 

or, because /% X A A = JtxPiP (see Ait. 41 ) , 

W + Af.^/tXOPi + JtxPiP^XX (OP, + I\P] = ^/ r 

Therefore, the moment of the ) exultant of /wo parallel fotces 
about any point in their plane equals the algebiait. sum of the 
moments of the components. 

The student may take the point O' and verify this prin- 
ciple, paying due attention to the signs of the moments, 
according: to the convention explained in Art. 18. 



38 ANALYTIC STATICS 28 

If, instead of the resultant, the equilibrant Q is taken, its 
moment M q is equal to M r , and 

M, = - Mr = -Mi-M t 
whence, M, + M + ^ = 

44. Two Pai*allel Forces Acting in Opposite Direc- 

tions. When the two component forces act in opposite 

directions, as /'land /*" Fig. 26, 
the resultant is found as fol- 
lows. If a force Q = /*! F a 
is applied at a point E in , E^ 
produced, such that /*",/= Q / 
the three forces F lt 1?,, and Q 
will be in equilibrium, accord- 
ing to the principles stated in 
Arts. 42 and 43. Since Q 
balances F l and F tt , it must be 
FlG ^ equal and opposite to their 

resultant R. Therefore, in this case, the resultant is equal to 
the difference of the components, and the line E t E l produced is 
divided by the three forces F lt F*, and R m the manner indi- 
cated by equations (/) of Art. 41. The law of moments applies 
in this case also, but due attention must be paid to signs 

45. The principle of moments just stated for the case of 
two parallel forces and their resultant (or of three balanced 
parallel forces) is known as the principle of the lever, or the 
law of the lever, as it was first discovered by Archimedes 
in the determination of the conditions of equilibrium of a lever. 

46. Definition of a Conple. There is an apparent 
exception to the foregoing conclusions that must be particu- 
larly noticed. If F^ = F,, Fig. 26, the formulas would give 
R = F l F, - 0, and, fiom equations (/) of Art. 41, 




So, although the resultant is 0, the distance of its point of 
application from the points of application of the other two 

F I F I 
forces cannot be determined, since the fractions ~- and ~~ 



28 ANALYTIC STATICS 89 

do not represent any numbers. This simply means that, in 
the case here considered, it is impossible to replace the two 
given forces by a single force. 

A system of two equal parallel forces acting in opposite 
directions but not in the same line, is called a couple. 

The theory of couples is veiy important, and will be more 
fully treated further on. But here it may be stated, from 
what has just been explained, that a couple cannot be either 
replaced or balanced by a single force. 

47. Resolution of a Force Into Two Parallel Com- 
ponents. Equations (/) of Ait. 41 afford a means to 
resolve any force into two parallel components passing 
through any two given points situated on a line intersecting 
the line of action of the given force. For, if ./?, / and /, are 
given, we have: 



whence, F, = l 

* i T *a 

Similarly, ft = ' 

*i T * 

If R and one of the distances, as / lt from the line of action 
of J? to the line of action of one of the components are given, 
and also the magnitude Ft of 
this component, the other 
component is found from the 
relation 

Its distance /, from the line 
of action of R is determined 
as follows: 

F.I. = FJ 
whence, l* Fi 

The distances A and /, may be either perpendicular or 
oblique. 

EXAMPLE 1. It is required to find the magnitude and line of action 
of the resultant R of the parallel forces ^i(* 500 pounds) and 
F t (= 260 pounds), the distance / being 4 feet, as shown in Fig. 27. 



w +0* 




40 



ANALYTIC STATICS 



828 



SOLUTION According to statement 1 of Art 41, the rebitlUnt is 
equal to the sura of Fj. and F a , that is, /t" = 500 + 250 = 750 Ib Ans 

The distance / t from the hue of action of FI to the line of action of 
>?, along the hue /, is given by equations (/) of Art 41 : 



l t , fiom which, A = ~- 

A 



F,l = 
From the above, F, = 250, / = -i, and A 1 = 7oO, therefore, 



750 



500 X 4 
"750~ 



Ans 



EXAMPLE 2 In Fig 28, the forces F^= 200 pounds) and /'(= r>0 
pounds) act in opposite directions; It is required to find the resultant A' 
when the distance I is 6 feet. 



_rf_ / -.--< 

r^ 'I ' 




~~~--l- 

'/a- 

FIG 28 




R 



Kio '."I 



SOLUTION The resultant is equal to the difference between the 
forces (Art. 44); that is, R A 1 , - A, = 200 Ib. - 50 Ib = 150 Ib. 

Ans. 
From equations (/) of Art 4 1 , 

Fii _ 200 X 



/. 



160 



8 ft 



COX 6 _ 0f 

TBO" " w t< 



Ans. 



EXAMPLE 3 In Fig 29, resolve the force V\'( 450 pounds) into 
two parallel forces FI and F a , when A = 10 feet and /, = 8 feet 

SOLUTION. From equations (/) of Art 41, 

AV. , ., A^/! 
/f,-- 7 ',and^- -/ 

In this example, A = 450 Ib , / = 10 ft., /, 8 ft., and //, + / 
= 10 + 8 - 18 ft. Therefore, 



460 X 10 



Ans, 



260 Ib. 



28 



ANALYTIC STATECS 



41 



ANY NUMBER OF FOHCPS 

48. Magnitude of the Remiltant. Let /*,, /^ 7s, /%, 

Fig 30, be parallel coplanai foices acting through points 
A i, A,, etc of a rigid body, their lines of action being, 
respectively, K* Z,,, K t L lt etc. By the pimciple of the trans- 
ferabihty of the point of application, each force may be sup- 




i. 



posed to be applied at any point on its line of action; but, 
for reasons presently to be explained, it will be assumed that 
the points of application are the fixed points /,, .,/,, etc. 

In the first place, the i exultant R /.s equal to the algebra it 
sum of the components ', and //A I me of action Is parallel to the 
lines of action of the components 

For, according to Arts. 41 and 44, the icsultant A" of 
/'I and F t is parallel to I<\ and /" and equal to /^ + J<\\ 
the resultant R" of R' and /". is parallel to A" and /;, and 
equal to A" + F* /", + / + /'" (the negative sign being 
implied); the resultant A? of R" and F t is parallel to R n and 
fl, and equal to.-^" + -ft = A + /?", + F, + /;. 



42 ANALYTIC STATICS 28 

The same reasoning applies to any number of forces; so 
that, if their algebraic sum is denoted by 2' F, then R = 2 F. 

49. Line of Action of the Resultant. To locate the 
line of action K L of the resultant, let N be any point m the 
plane of the forces, and NP<. a. perpendicular to the common 
direction of the forces, intersecting JK' 1 L l at /*,, A' 3 L a at /> 
etc.; and KL, the still unknown line of action of the 
resultant, at P Let N P* - p NP, = A, etc., and NP = p r 

If moments are taken about N, the following equations 
obtain (Art 43). ' 

Moment of JR' = F^ + F.p* 
Moment of R" = moment of R' + F,p, 



Moment of R - Rp = moment of R" + F t p< 

= Ft pi + F t p t + F a A + F t A 

In general, if the algebraic sum of the moments of any 
number of coplanar parallel forces is denoted by ~ Fp, and 
the lever arm of the resultant R by p r , then 
Rp, = SFfr 

whence, p r = ^ = ^ (1) 

This locates the line of action of R with respect to the 
point N. 

If, instead of the perpendicular distances p t) Ai etc. of 
the lines of action of the forces from A'', the oblique dis- 
tances NTi. = a,, NT, = a, } etc., along any line NT* (as 
when the forces are applied at several points of the same 
straight line), are given, the following equations are 
obtained: 

A = NPi = -AT 71 cos H = a, cos H 
p a = NP t = N T, cos H = a, cos H, etc. 
A = Af/* = NT cos ff = a r cos .#" 
Substituting these values of A, A. etc. in the equation of 
moments, 
Ra r cos H = /!! cos ff -\- F t a, cos #+ . . . etc.; 

whence, * r = *<" + *+ = ( 2 ) 



28 ANALYTIC STATICS 43 

If N is taken on the line of action of one of the forces, 
that force is eliminated from formulas 1 and 2, since in this 
case both its p and its a are equal to zero Thus, for two 
forces, the piecedmg equations become identical with equa- 
tions (/) of Art. 41. 

It should be carefully borne in mind that, in using the 
inclined distances a lt a t , etc., the same rules for signs are to 
be observed as for the perpendicular distances A)A> etc - 
Thus, F^a,. is to be taken as positive and F a t as negative. 

50. Conditions of Equilibrium. When the forces are 
in equilibrium, we must have 2 F = 0, and S(Fp) = 0; that 
is, the algebraic sum of the forces must be zero, a?id the algebraic 
sum of their moments about any point in their plane must be aero 

Both of these conditions are necessary for equilibrium 
For, in the first place, it is evident that if S F is not zero, 
there is a resultant, 
and no equilibrium is 
possible. In the sec- 
ond place, if the 

resultant of all the h 8 ' 

forces acting in one-^1 ^ J* 

direction is equal I JL 



w 



100U> 



FIG 81 






to the resultant of all 

the forces acting in 

the opposite direction, the condition -F = obtains, but, 

if those two resultants have not the same line of action [in 

which case 2(Fp) is not zero] , they form a couple (Art. 46), 

and the system cannot be in equilibrium 

It should be remembered that the arithmetical meaning of 
the expression 2F = is that the arithmetical mm of all the 
forces acting in one direction is equal to the arithmetical sum of 
all the forces acting in the opposite direction. 

EXAMPLE 1 A weight of 100 pounds is hung from the extremity A 
of a straight lever A JB } Fig. 31. The lever is suspended by a rope passed 
around a smooth peg and carrying a weight W\ and from the extremity 
B is hung another weight W*. The dimensions being as shown, it is 
required to find the magnitudes of W* and W, that the lever may remain 
in equilibrium. (The weight of the lever is neglected.) 



44 ANALYTIC STATICS 28 

SOLUTION Since JTis the eqmlibrant of H 7 , and 100 lb , the sum 
of the moments of the three forces about any point in their plane-must 
be zero If moments are taken about 5 1 , which is in the line of action 
of W (for W acts upwards at .S 1 ), the foice W will be eliminated, its 
moment about 5 being zero The moment of /f, being positive, and 
that of the 100-lb weight negative, we have, 
6 Wi - 100 X 8 = 0, 

Q/V) 

whence, W, = = 133 S3 lb Ans 

To find W, we have, 

100 + Wi - W=Q 

W = 100 + /Ft = 100 + 133 33 = 233 33 lb Ans. 
Notice that it Is not necessary that the lever should be horizontal 
From the mathematical conditions of equilibrium (Art. 5OJ, it follows 
that the three weights given above will hold the lever in equlhbnuiu 
in any position 
EXAMPLE 2. A rigid bar A, Fig 1 32, rests on two supports Si and S lt 



OOOlb. 

^~~^ son. 




G) 



O 



la 
I V* I 

*. I ^ Jt f _ - I - //7^ _ J 

I ^" " V ~~'^|^ ' ' /{/ *^ \ 



and carries weights of 600, 300, 250, and 50 pounds placed as shown. 
The weight of the bai .being neglected, it is required to find the 
reactions J?i and A* a at .5\ and S y , respectively 

SOLUTION Here we have a system of balanced parallel forces, 
consisting of the given weights, which act downwards, and the two 
reactions, which act upwards. Taking moments about >S" U , and 
observing the rule of signs, we have, 

- 600 X -10 + ^ X 28 - 300 X 23 - 250 X H + 50 X 10 =0; 

whence, j?, = --.^ = l,ir>7.1 lb. 

To find J! a , we have: 

J?i + #, - 800 + 300 + 250 + SO = 1,200, 
whence, J? t = ],200 - JK V = 42.9 lb 

In order to check these results, the value of Jf may be found by 
taking moments about 5^ The moments of the weights at the right 



28 



ANALYTIC STATICS 



45 



of Si are positive, the moment of /? and the moment of the weight 
at the left of Si are negative. Therefore, 

- 600 X 12 + 300 X 5 + ,250 X 20 - ./?, X 28 + 50 X 38 = 0, 



whence, 



1 200 
= -W" = 42 -9 Ib i approximate to tenths. 




EXAMPLES FOR PRACTICE 

NOTB -lu these examples the weights of bars and ropes are neglected 

1 A straight bar is supported at two points S 1 , and 5, 25 feet apart, 

and carries five loads between the supports as follows 100 pounds, 

placed 8 feet from 5,, 150 

pounds, placed 8 feet Vj^ 

from Sn 200 pounds, | 

placed in the middle of 

the bar, 300 pounds, 

placed 15 feet from J?,, 

and 450 pounds, placed FlG M 

20 feet from 5\ Find the reactions fa and fa. at Si and S a 

Ans. fa = 500 Ib.; fa, = 700 Ib 
2. A man carries a weight of 30 pounds hung from a stick resting on 

his shoulder, the distance from the weight to the shoulder is 2 5 feet. Find 

the pressure P on the shoulder- (a,) when the man holds the stick at a 

distance of 1 foot from the shoulder, (b] when he holds the stick at 

a distance of 1.5 feet from the shoulder. a / (a) P = 105 Ib 

Ans -\(,5) />= 80 Ib 

ii and A,, Fig 33, is to 
will crush under a pres- 
sure exceeding 275 
pounds, but it is desired 
to have the weight as far 
from A t as possible 
Find the maximum dis- 
tance x at which the 
weight can be placed 
from A, 

Ans x 13 75 ft 
4 Five coplanar par- 
allel forces act on a body 
Aff, Fig 34; magnitudes 
and direction of forces, 
Find the magnitude and 

direction of their resultant fa and the distance a of its line of action 
from 0, measured along the hue X. 

Ans. R - 350 Ib., acting upwards; a <* 4 875 ft to the right of O. 
NOTB First take moments about 0, using the inclined distances instead of the 
lever armi then, as a oheok, take moments about P. 

I LT 398-8 



3 A rigid bar, supported on two piers 
carry a weight of 500 pounds. The pier A 




FlO 34 

and distances along OX, are as shown 



46 



ANALYTIC STATICS 



28 




E 



5 In example 4, find two forces Ft and F a parallel to the given 
forces and equivalent to them, F l to act at a distance of 10 feet to the 
right of 0, and F, at a distance of 14 feet to the right of O 

A (F t = 798 44 Ib , acting upwards 
\F 3 = 448 44 Ib , acting downwards 

6 A bar A 1 A 3 , Fig 35, is to carry two equal weights W, sus- 
B! ^B 2 pended as shown The 

I s bar is supported by two 

^Aa ropes AI BI and A* 

J I of which AI J3i cannot 

r' -j- s' ] be subjected to a tension 

greater than 1,000 
pounds, and A, B* can- 
not be subjected to a ten- 
FIG 85 sion greater than 600 

pounds. What is the greatest value that W can have, and what are 
the tensions F* and F t in the two ropes when the bar carries its 
greatest load? f W - 720 Ib 

Ans IF* - 840 Ib 
[F, = 000 Ib. 

NOTE Assume first Fi 1,000, and find Wby taking momenta about As; tlion, 
Fa - 2 W Fi As this is greater than 600, start by ausumint ft = 000 pounds, and 
proceed to find W and ^"i 

NON-COPLANAR PARALLEL FORCES 
51. Magnitude of the Resultant of Any Number of 
^, A B Parallel Forces. 

*"""""" T ^f 1 /** /7* /P 77* 

,.-'^- : ~^ -^ a ..._.^ 4 Fl ^- 36, be parallel 

N^I--"'" forces acting on a 

\ \ ^ 4 body at the points^,, 

Ai, etc. The forces 
may be either copla- 
nar or not. In either 
case, they may be 
combined in pairs, as 
in Art. 48. Hence, 
the resultant is equal 
to the algebraic sum 
of the components. 

52. Center of 
*io Any System of Par- 

allel Forces. According to the principles stated in Art. 41, 



R 



28 ANALYTIC STATICS 47 

the resultant R 1 of F l and F,, Fig. 36, passes through a point A' 

P 

in the line A t A a , whose distance from A l is A^ A,X-~ f = A t A* 

p 

X - Since this distance does not depend on the 
F l + F 3 

inclination of the forces to A^A^ and since the fraction 

p 

, "--rr remains the same when for F, and F, are substituted 
A + F n 

any forces proportional to the latter, such as n /\ and n F* 
(n being any number, integral or fractional), it follows that, 
so long as the points A v and A* remain fixed and the forces 
remain parallel and their relative magnitudes and directions 
unchanged, the lines of action of the forces may be revolved 
about A,, and A, and made parallel to any direction whatever 
m space, without changing the position of the point A f , 
through which the resultant R 1 must constantly pass. The 
same principle may be proved of the point A" through which 
the resultant R" of R' and F a must pass, and of the point A, 
traversed by the total resultant R. It follows that, for 
every system of parallel forces applied at or, more properly ', 
passing 1 through any points whatever^ there zs a fixed point 
through which the resultant must pass, whatever the (common} 
direction of the forces may be. Furthermore^ that point is the 
same for all systems of parallel forces in which the relative 
magnitudes of the forces are the same, whatever their absolute 
magnitudes may be. 

Such a point, as A in Fig. 36, is called the center of the 
system of forces considered. 

53. Center of Gravity Defined. In the particular 
case in which the points of application A^ A,, etc. are the 
particles of a tyody, and the forces acting are the weights of 
those particlea, the .center of the system of parallel forces 
thus formed Is called the center of gravity of the body. 
The weight of tihe body, which is the resultant of the 
weights of Us $rarttcls> must therefore be treated as a force 
whose line of fldifbli 'passes through the center of gravity of 
the body. TN |K|^tI^^s for determining the position of the 
center of 




48 ANALYTIC STATICS 28 

54. Coordinates of Center of Parallel Forces. 

When all the points of application of the forces lie in one 
plane, the center of the forces is easily found as follows: 

Let the parallel forces F lt F,, F a act at the points A,, A,, A,, 
Fig. 37, and let <9,Yand Y be any two mutually perpen- 
dicular lines drawn in the plane containing A 1} A a , A a These 
reference lines are called coordinate axes, and the distances 
of any point from them are called the coordinate's of that 
point Distances perpendicular to O K, or parallel to OX, 
are usually denoted by the letter x, sometimes with an accent 
or subscript; distances perpendicular to OX, or parallel to 
OY, are denoted by the letter y. Distances above OXa.nd 

T 



JK 



/ ,fl| **<2*f J-f& 

Pro 87 

those to the right of O Y are treated as positive; distances 
below X, and those to the left of O Y, as negative. 

Let the coordinates of At be O HI K^ A l = x, and O A", 
= Jf 1 A l = y lt Similarly, let the coordinates of A, be j. a 
and ;/,; those of A a be x a and y a ; and those (still unknown) 
of the center A of the forces be x c and y e . 

Since the position of A is independent of the common direc- 
tion of the forces, the latter may be imagined as acting 
in the plane X Kin a direction parallel to OX. If, now, 
moments are taken about any point on OX, the lever 
arm of F l will evidently be equal to A^ //i = y lt the lever arm 
of Ft will be equal to y t , that of A a equal to y,, and that of R 
equal to y c . Then, by formula 1 of Art. 49, 

\-F t y t - 



28 ANALYTIC STATICS 49 

If, on the contrary, the forces are imagined as acting in a 
direction parallel to O Y t and moments are taken about any 
point on O K, the following formula is obtained, by pursuing 
the same method as before. 

- &*+ F >x> + F 'Xi 

ft + K + ft 

The coordinates x e and y e locate the center A. 
As the foregoing reasoning is evidently applicable to any 
number of parallel forces whose points of application all he 
in one plane, the following general formulas can be at once 
written 



(2) 



55. Moment About a Line. The product F l y l con- 
sidered in the last article is called the moment of the force F t 
about the lino OX, or with respect to OX. In general, 
when a force is considered as applied to a special point, the 
moment of the force about any line perpendicular to the line 
of action of (although not necessarily in the same plane with) 
the force is the product of the magnitude of the force and the 
perpendicular distance of the point of application of the force 
from that line. 



ANALYTIC STATICS 

(PART 2) 
CENTER OF GRAVITY 



DEFINITIONS AND GENERAL PROPERTIES 

1. The center of gravity of a. body has already been 
defined as the center of the parallel forces of gravity acting 
on the particles of the body, or the center of the weights of 
all the particles, each particle being taken as the point of 
application of its own weight (see Analytic Statzcs> Part 1). 
These forces are considered parallel because they are all 
directed toward the earth's center, whose distance from the 
surface is so great, compared with the dimensions of ordinary 
bodies, that the angle between the lines of action of the 
weights of any two particles of a body is practically zero. 
(Two terrestrial radii meeting the surface at two points 
distant 100 feet from each other make an angle at the center 
equal to about 1 second.) 

The abbreviation c. g. will here be used to signify center 
of gravity. The c. g. of a body is called also center of 
mass, center of inertia, and centrold. These terms, 
however, will not be used here. 

2. Immediate Consequences of the Definition. 

Three important consequences follow at once from the 
preceding definition, namely: 

1. As stated in Analytic Statics, Part 1, the weight of a body 
may be treated as a single force acting vertically through the 
o/ (fo body, , This is expressed by saying that the 

TBXTHOOK COMPANY, ALL MIHT MMBMVKD 

529 



ANALYTIC STATICS 



29 



weight of a body may be supposed concentrated at the 
c. g of the body 

2 The position of the c. g. of a homogeneous body (that is, a 
body whose substance is the same throughout, or whose particles 
have all the same weight] depends on the form of the body only, 
not on the material of which the body is made. For the centei 
of a system of parallel forces depends on the relative, not on 
the absolute, magnitudes of the forces, and on the relative 
positions of their points of application. 

3 // the c. g. of a body acted on by its own weight is sup- 
ported, the body will remain in equilibrium, whatever its Posi- 
tion. For the weight 
of a body is equivalent 
to a single force acting 
through its c. g., and, if 
this point is supported 
that is, if a force equal 
to the weight, but acting 
in the opposite direc- 
tion, is applied at this 
point the two forces 
will balance each other. 

Conversely, // a body 
acted on by its own weight and by other forces is in equilibrium, 
the resultant of the other forces must be vertical, act upwards, 
and pass through the c. g of the body. Thus, if the triangular 
plate ABC, Fig 1, hangs from three strings A O, BO, CO, 
and is in equilibrium, its weight W, acting through the center 
of gravity G, must be balanced by the resultant of the 
reactions of the strings. This resultant must, therefore, be 
vertical, act upwards, and pass through G. 

3. The Center of Gravity of a Body May be Outulde 
the Body. When several forces act on a body, the line of 
action of either their resultant or their equilibrant may be 
altogether outside the body. For example, in Fig. 2, the 
equilibrant Q of the forces F l and F t acting on the lever A^ A t 
passes through a point A, outside the lever. This means 




FIG 



ANALYTIC STATICS 



that the only single force that can balance Ft. and F t is the 
force Q acting through A 9 ', but, as the point A,, is outside the 
lever, that point must be imagined to be rigidly connected to 
the lever, in order that Q may act on the latter. 

In the same manner, it often happens that the c. g. of a 
body is outside the body. Thus, the c. g. of the curved 



Q 



FIG. 2 




FIG 8 



rod A /?, Fig. 3, is a point G outside the rod If it is desired 
to suspend the rod so that it will be in equilibrium in all 
positions, the point G must be connected to the jod by some 
means, as by another rod G H. The additional weight of the 
rod GH alters the position of the c. g. of the whole system; 
but the change may be made small, and its amount easily 
calculated, as will be explained presently. 

4. Center of Gravity or a Line. Properly speaking, 
geometrical lines and surfaces have no c. g , since they have 
no weight. By an extension of 
the definition, however, every 
line and surface is said to have 
a c. g., the expression being 
taken in the sense now to be 
explained. 

Consider a body A fi, Fig, 4, 
of uny form, and a section, as 
PQ, containing the centers of a 
number of particles of the body. 
The resultant of the weights of 
these particles will pass through a fixed point 0, which is the 
center of the system of parallel forces constituted by those 
weights. Likewise, 0> is the center of the system of parallel 




FIG 4 



4 ANALYTIC STATICS 29 

forces constituted by the weights of the particles in the sec- 
tion P'ff. If a sufficient number of sections is taken to 
include all the particles of the body, and the centers thus 
found are joined, a line A O 'B will be obtained, which will 
be intersected by the resultants of the various groups of 
particles considered The center of the system formed by 
these resultants is the c. g of the body, and is said to be 
also the c g of the line ADO'S. 

If the body is homogeneous and of uniform cross-section, 
all the resultants referred to will be equal, and the number 
of them acting on any portion of the line A O' B will be 
proportional to the length of that portion. Now, in deter- 
mining the center of any system of parallel forces, the 
forces may be replaced by any quantities proportional to 
them For instance, if the x coordinates of the points of 
application of two forces F l and F* are # and jc tt the x 
coordinate x t of their center will be given by the formula 
(see Analytic Statics, Part 1) 

r* i z? Xi -)r -=;Xi 

F^XI-\- F, x, _ F 1 



If F a and /i are proportional to two given lengths / B and / 

F I 

that is, if ' = -', we have 
A l\ 

,/. 
*i + 7 x, 

l lQ it * 
' 



Therefore, .-+/* = L#* +.** 
F> +F. /, + /. 

So, in finding the center of a system of parallel forces 
whose points of application lie in a line, and whose dis- 
tribution along any part of the line is proportional to the 
length of that part, that length may be used instead of the 
corresponding forces. 

5. Center of Gravity of a Plane Area. Take now a 
homogeneous prism ACGE, Fig. 5. If all the particles 



29 



ANALYTIC STATICS 



5 




FIG. 5 



enclosed by a cylindrical surface PQ are considered, the 
resultant of their weights will be a vertical force whose line 
of action will meet the upper surface of the prism at some 
point O. Similarly, other points 
may be found where the resultants 
of the weights of vertical rows of 
pai tides meet the surface ABCD, 
thus obtaining 1 a system of parallel 
forces whose points of application 
may be taken on the plane AB CD, 
and whose center is called the c. g. 
of the surface A B CD. This point 
is evidently the same as the point 
where the vertical line through 
the c g. of the prism pierces the 
plane A B CD. As the forces act- 
ing on any such space as P are proportional to the area of 
this space, areas may be substituted for forces when dealing 1 
with the c. g of a pl?ne surface. 

6. General .Definition of the Center of Gravity of 
Jjlnes and Surfaces. A line is said to be homogeneous, 
or of uniform weight, when parallel forces are distributed 
along the line in such a manner that the resultant of the 
forces acting on any part of the line is proportional to the 
length of that part. The center of the parallel forces thus 
acting is the c. g. of the Hue; and their resultant is called 
the weight of the line. 

The same terms apply to a surface when the parallel forces 
acting through it (that is, whose lines of action pierce it) are 
such that the resultant of the forces acting through any 
part of the surface is proportional to the area of that part. 
The center of the parallel forces thus acting is the c. g, of 
the surface; and their resultant is called the weight 
of the surface. 

7. Static Moment. Let Wb& the weight, or any num- 
ber (length, area) proportional to the weight, of any figure 
line, surface, or solid; and let x be the distance of the center 



6 ANALYTIC STATICS 29 

of gravity of the figure from any point or line. Then, the 
product Wx is called the static moment of the figure with 
respect to that point or line 

According to the theory of parallel forces, the static 
moment of any figure or system of figures about any point or 
line is equal to the algebraic sum of the moments of the parts 
of which the figure or system is composed If the distances 
of the centers of gravity of several areas A^A^ etc from 
a given point or line are denoted, respectively, by AH .a,., etc , 
the static moments of these areas about the point or line 
are XiA lt x,A a , etc. If the distance of the c g of the system 
formed by the aggregate of these areas from the given point 
or line is denoted by x c) then, 

x e (Ai + A,+ . . . ) = x^Ai + x*A> + 

In general, denoting the sum of the areas by 2 A, and 
the algebraic sum of their static moments by SxA, the 
following equation may be written: 

x t 2 A = IxA\ 

V y. A 

whence, x c = =~^ 

A similar formula applies to lines and volumes. The 
formula shows that 

1. The algebraic sum of the static moments of the patts 
of any figure or system of figures about the c. g. of the figure 
or system of figures, or about a line containing that center, is 
equal to zero. For, in this case, x e = 0, and, therefore, 
SxA (= x t !A) = Conversely, 

2. // the algebraic sum of the moments of the parts of a 
figure or system of figures about a point is zero, that point is 
the c. g. of the figure or system. 



29 ANALYTIC STATICS 



IMPORTANT CASES 
SYMMETRICAL FIGURES 

8. Definitions. A figure is symmetrical with respect 
to a point O when every straight line passing through that 
point meets the figure in pairs of points equidistant from 
the point O, the two points of each pair being; on opposite 
sides of the point O. 

The circle, the ellipse, the sphere, the circular ring, are 
each symmetrical with respect to its center. This center is 
called the geometric centei', center of figure, or cen- 
ter of symmetry of the figure in question. A parallel- 
ogram is symmetrical with lespect to the point of intersection 
of its diagonals. 

9. A figure is symmetrical with respect to a straight 
line, called an axis of symmetry, if every straight line 
perpendicular to the first-mentioned line meets the figure in 
pairs of points equidistant from said line or axis, the two 
points of each pair being on opposite sides of the axis 
Thus, a rectangle is symmetrical with respect to either of the 
lines joining the middle points of two opposite sides. An 
ellipse is symmetrical with respect to either of its principal 
axes, and n, ciicle with respect to any diameter. The axis of 
a right circular cylinder is an axis of symmetry, and so is the 
axis of a right circular cone. 

10. A solid is symmetrical with respect to a plane, 
called a piano of symmetry, when every straight line 
perpendicular to the plane meets the solid in pairs of points 
equidistant from the plane. 

11. Center of Gravity of Symmetrical Figures. It 
is evident that, if a figuie is symmetrical with >espect to a point, 
a line, or a plane, the c. g. of t/ie figure coincides with the point, 
or lies in the line or plane, of symmetry, as the case may be. 

Thus, the c. g. of a circle coincides with the center of the 
circle; the c. g. of a parallelogram coincides with the inter- 
section of the diagonals; the c, g. of an isosceles triangle 



8 ANALYTIC STATICS 29 

lies in the perpendicular from the vertex to the base (at what 
distance will be seen further on), and the c g of a icgular 
pyramid lies in a plane through the vertex peipendicular to 
the plane of the base 

12. Again, if a figure has two axes of symmetry, the c. jg- m 
will be at the intersection of those axes, and 

7? a solid has three planes of symmetry meeting at a 
that point is the c. g. of the solid. 



DETERMINATION OF THE CENTER OF GRAVITY BY 
ADDITION AND SUBTRACTION 

13. Addition Method. If a figure (by which is meant 
either a line, a surface, or a solid) or system of figures can 
be divided into parts whose centers of gravity are known, 
the c. g. of the whole figure or system is easily found by the 
principles explained in Analytic Statics, Part 1. 

For example, let it be required to find the c. g of a system 
consisting of two homogeneous spheres C and C,, Fig. 6, 

whose weights are Z-J^, 




and whose distance 
FlQ fl apart, measured be- 

tween centers, is a. This is equivalent to finding 1 the center 
of two parallel forces W t and W t acting through d and C,. 
Let this center be G; then, 



whence, G C, = - ^ a 

W, + H7. 

14. Subtraction Method. If, on the contrary, the 
c. g. of a system is known, and also the c. g. of all its parts 
but one, the c. g. of the remaining part is found by the same 
general principles just referred to, the only difference being 
that subtraction is used instead of addition. 

As an illustration, let it be required to find the c. g-. of a 
figure obtained by cutting from a rectangle B CDR, Fig. 7, 
a rectangular corner EFHI and a circular piece 0, the 



29 



ANALYTIC STATICS 



9 



dimensions being as shown Take EDior the axis of x and 
EB for the axis of y\ and let x,' and y' be the coordinates 
EH' and H 1 G' of the c. g of the rectangle B CDE\ x" and 
y, the coordinates EH" and //" O of the c. g. of the circle, 
x'" and y'", the coordinates //'" and H'" G 1 " of the c g. 
of the rectangle EFH '/; and ^ and j/, the coordinates of G, 
the required c. g. According to Art. 5, areas may, when 
dealing with surfaces, be substituted for forces in the equa- 
tions of moments and of centers of parallel forces. In this 



a j{j 




sense, the moment of CD with respect to ED is the 
same as if the whole area were concentrated at &; that is, 
the moment of B CD E with respect to ED is a by'. 

The area A of the figure whose center of gravity is required 
is a b cii bi TT r*. The c. g. of BCDE is at the inter- 
section of the two diagonals, or at the middle point of CE, 

and, therefore, xf = EH' = ? andj/' = H' G 1 = . 

2 2 



Likewise, 
x" 



EH" = c, y" 



and 



H" O = b - r, 
~ ~~2 



The formula of Art. 7 now gives 

+ aibt x' 1 ' 
A-\- ctibi 



10 
whence, 



x e = 



ANALYTIC STATICS 29 

(A + a, bi + n r')^ - a, b, x"' - x r> x" 






a b ! 61 TT r* 



Similarly, 
whence, 



_ 

y 



_ Ay e 



ff, b.y" 1 

, 

ab 



, 

= a by 1 - a, b,y'" - TT r'y" = 1 \ab* - a, bS -_2_7r r*(b--_r) "I 
A 2L ab-atbi-'ni* \ 



15. The subtraction method, just illustrated, is very 

much simplified when the 
c. g. of a figure consisting 
of two parts and the c. g. 
of one of the parts are 
known, and it is required 
to find the c g. of the other 
part. This special case is 
illustrated in Fig. 8, where 
C, the c. g. of the figure 
BCDJS, and the c. g. 
of the part C E D, are 
known, and it is required 
to find G,, the c. g. of the 




. .X 

PIG 8 



part BCE. 

Let total area = A\ and area CDE = A,. Then (Art. 13) , 
Ai X G l G, A x G G: 



whence, 



GG,= 



A 



EXAMPLE 1 -To find the c g. of the channel section represented 
in Fig 9 L 

SorurioN.-Owmg to the symmetry of the figure with respect to 
the center line OX, its center of gravity G t must be on that line. To 
find the distance OG e . *-, of G e from the back of the channel, notice 

la " er i8 the dlfference be tween the rectangles 
Q D>, whose centers of gravity G and G' are at the 



29 



ANALYTIC STATICS 



11 



distances O G = 5 inches, and O G> 



- + 5 
J 8 



i + 5 = ?j 
2 8 ID 



inches from A D Denoting the area of the section by A, and talcing 
moments about A /?, 

A x t = (AJtXAD)XOG- (A> JB> X A' D 1 ) X O G> 



= 150 X 8 X 16 -36 X 11 X 46 
X16 



" 



I G 



I 



1 *- 
1_ ! * 

S \JLZ 



i<if. i) 

The area of the section is 

A = A B X A D - A 1 B> X A 1 W = (M) - ^ X 11 
Hence, 

.v c t= OG e ** ] 
1,870 



X 8 X W - x n x45 



Km Id 

B x_eo_ r _3B_x 11 

8 

XJJO-36X 11 
flT" 



in 



EXAMPLE 2. To find the c. g. of the angle section represented in 
Fig. 10, and having dimensions as shown. 
ILT3W-9 



12 



ANALYTIC STATICS 



29 



SOLUTION Produce CD to meet HF at / The section is now 
divided into two rectangles, BCIH and DEFI The c g of a 
rectangle being at the point of intersection of the diagonals, itp 
distance from either of two parallel sides is equal to one-half of either 
of the other two parallel sides Taking moments about H B, and 
denoting the distances of the c g of the section from H B and ffF 
by x e and y t , respectively, and the aiea of the section by A, 



whence, 



Xc = HP = 



9 11X13 
8" 1 " 16 



r 16 ' 
9 , 11 XJ8 

T-CTT - 1.3879" 



Moments about H ' F\ 



whence, 




= 2 8879 In Aus. 



EXAMPLES FOB PRACTICE 

1 Find the center of gravity G 1 of the area obtained by taking the 
rectangles ^and D H, Fig 11, from the rectangle A BCD The 




FIG 11 
dimensions are as shown 



.in., nearly 
In., nearly 



ANALYTIC STATICS 



13 



2 Find the c g of the area obtained by cutting off the circle A BI C lt 
Fig 12, whose radius O l A is 3 inches, from the circle ABC, whose 
radius OA is 10 Inches Ans O G = 69231 in = H in., nearly 

3 Find the c g of the angle section 
represented in Fig 13 

= 92500 m. = ff m , nearly 
= 2 1750 m = 2Vfc In., nearly 



r 



L 




FIG 13 

4 Find the c g of the Z section represented in Fig 14 

NOTE In talcing moments about QM, remember tlmt the moments of areas on 
opposite sides have opposite sljrns 



Ario fOP = .41667 in = 
ADS \ PG = l 4167 in _ 



in , nearly 
i( nearly 



CENTER OF GRAVITY OF POLYGONS 

NOTE In some of the articles that follow, formulas and rules are 
given without explaining how they nre obtained This is done when- 
ever the processes Involved require the use of advanced mathematics, 
or, being elementary, are too long and complicated to be given in 
connection with this instruction 

16. A Fundamental Principle. // a straight line 
divides a plane figute in such a manner that eveiy hue paral- 
lel to a hxed direction 
meets the perimeter of 
the figure at one point 
on each side of the first- 
mentioned line and is 
bisected by said hne> 
the c. g. of the figure 
lies on that line. 

Let OX, Fig. 15, 
be a line dividing the 




Pro. 10 



figure OQXQi into two parts, in such a manner that all lines, 



14 



ANALYTIC STATICS 



29 



as QQRR etc., parallel to the fixed line /P"are bisected 
at their point of intersection with OX. Then, according to 
the principle just stated, the c. g. of the area OQXQ^ lies m 
the line OX. 

17. Triangle. Let ABC, Fig. 16, be any triangle. 
DtawAA' from A to the middle point A' of the opposite 
side. Any line, as RS, parallel to B C is bisected at its 
intersection J with A A'. Therefore, according to the 
proposition of the preceding article, the c g. of the triangle 
lies on A A'. For a similar reason, the c. g. must lie 
on B B 1 , joining the vertex B and the middle point /}' of 
the opposite side A C Therefore, the c g of the triangle 

is at the intersection 
A of the lines A A 1 and 

B B't or of either of 
them with the line 
CO from C to the 
middle of An. It is 
shown in geometry 
that the distance of 
the point G, where 
A A 1 , /?/?', and CO 
meet, from any of the 
vertexes is equal to 
two-thirds the length 
of the line joining that vertex to the middle point of the 
opposite side, or A G = \AA' t BG = Inn 1 , CG - \ CC f . 
The lines A A 1 , BB>, CO are called the modlan linen.' 
Therefore, 

The c. g. of a triangle lies at the intersection of the median 
lines, and its distance from any vertex is egual to two- 
thirds the length of the median line from that vertex to the 
opposite side. 

18. The perpendicular distance from the e.g. of a triangle 
to any of the sides is equal to one-third the altitude of the tri- 
angle, when that side is taken as the base. 




29 



ANALYTIC STATICS 



15 



Foi, drawing 1 A ff and < /if perpendicular to B C, Fig. 16, 
two similar triangles AA'H and GA'K are formed, which 

A A' GA 1 



whence, bearing in mind that GA 1 = \AA' or 

A A' 
GA' 

A A 1 

19. Sometimes, the distances of the vertexes of a tri- 
angle from an axis or line of reference are given, and it is 
required to find the distance of the c. g. from the same 
line. In Fig. 17, let .1,, 

the distances of the 
vertexes of the tri- 
angle A^AtA* from 
the line A' A 7 be y lt 
y,, y a . The c. g. of 
the triangle is on the 
median A* M, and, as 
already explained, x ^ 
GM = \AiM. Let 
the distance G P of 
the c. g. from X' X be denoted by y e . Draw MN perpen- 
dicular and MB parallel to A 7 A'. The figure gives 

y e = GP= DP + GD = MN + GD (1) 

In the trapezoid A,P t P a A tt the line M N joins the middle 
point of A a A* and P t P tt therefore, 

In the similar triangles A^M B and GM D, 

1:1? YB' that is - -/J- fi ! 

whence, 

GD = ^AijB = ^(AiPi MN} = ijj/, i(y, + j/ a )] (3) 
Substituting in equation (1) the values of MN and GD 
given by equations (2) and (3), and reducing, 

This formula is perfectly general, provided that due atten- 
tion is paid to the signs of the coordinates. That is, distances 




16 



ANALYTIC STATICS 



29 



measured on one side of the reference line should be consid- 
ered positive, and those measured on the opposite side, neg- 
ative. Thus, if the vertex A, lies below A'' A", and y t and j/, 
are considered positive, y, should be considered negative. 

20. Trapezoid. Let B CD , Fig 18, be any trapezoid, 
having the bases BE = ,, CD = b,; altitude DPI = /i, 
median line Mi M, = m The median line /*/, Af, (that is, a 
line through the middle points of the bases) bisects every 
line parallel to the bases; therefore, G, the c. gf of the trape- 
zoid, must lie on that line (Art 16). It is necessary, there- 
fore, only to find the distance, as GP^ of G from either base. 




Fin 18 

Draw B D; this line divides the trapezoid into two triangles 
having the same altitude //, and d l and b, for their bases. 
The distance of the c. g. oiBCD from CD is \ h (Art. 18), 
and its distance from BE is h h = % h. The distance 
of the c. g. of JS D E from B E is h. Denoting the area of 
the trapezoid by A, and taking moments about /?/j, 
A X GP, = area B CD X I h + area BDE X \ h 



whence, 



= \ b, h X 



6 



X i h => (b, + 2 b,) 



6' 



m 



A similar value may be found for /> by simply Inter- 
changing fa and b* in this formula. 



29 ANALYTIC STATICS 17 

Draw M t K* perpendicular to the bases. The similar 
triangles M, Mi A" t and GM^P* give 
MnM* = MjJCi m = _ h 
GM, ~GP\~* GM, 



whence, GM, = ^-~- (2) 

3 \ 0i + b* / 

which gives the distance from the middle point of z to 
the c. g., measured along the median line. A similar 
expression may be found for GM^ by simply interchanging 
&i and b t . 

The distance B P l is found by the following formula, in 
which N is the angle between CB and CJ, the latter line 
being perpendicular to the bases: 

+*, + (bl + 2 ^ a) k ^ N ~ ***} (3) 

*i + o, J 

21. There is another formula that is often useful for the 
determination of the c g. of a trapezoid. Let the non- 
parallel sides EB and D C, Fig. 19, meet at V. As before, 
Mi M, is the line through the 
middle points of the bases. 

From geometric princi- 
ples, it is known that this 
line passes through V. Let 
d be the c. g. of the triangle 
DVE, G, the c g. otCVB, 
and G that of the trapezoid 
BCDE. Also, let V M, 

VM, = a t) VG a. As explained in Art. 17, 
= $ a,, and V G t = s . Taking moments about , we 
have, 

area DVEXG^G = area CVBX G, G\ 
whence, 

t g area C E^g VMf 




VMS 

since the areas of two similar triangles are proportional to 
the squares of their homologous sides. In terms of the 



.it U ^ 



18 ANALYTIC STATICS 29 

quantities a lt a,, and a, the preceding proportion may be 
written: 

a 3 dj. _ af 
a "f a, 1" 

Solving this equation for a, the following result is finally 
obtained: 

_ 2 a," - / 
a 3 ,' - a,' 

22. Trapezoid: Graphic Solution. Produce CD to E'> 
Fig. 38, making D E' = d lt draw -fi 7 (7, and produce it to its 
intersection d with EB produced. The similar triangles 
GM,E' and GMi d give- 

fad M.Ef. 
GM,' 



whence, M 1 d = ~XfaE' (1) 

GM, 



The value of C^/, is given by formula 2 of Art. 20, and 
the value of GM t is given by the same formula, by inter- 
changing bt and b,. Therefore, 
m J. + 2J. 



x a , , 

3 ^ + ^ 

By construction, fa E' = M t D + D E 1 - i b t + <5 X . Sub- 
stituting these values in (1), 

fa d = I- 6 / + * X (i*. + ^) = U + ^ a = faj5 + * 

t ^a + 0i 

therefoie, B d = b, 

Hence, the following construction for finding the c. g. of 
any trapezoid. 

Join the middle points of the bases. Produce the bases in 
opposite directions, making the prolonged part of each base equal 
to the other base Draw a line joining the extremities of the 
prolonged segments The point where thts line intersects the 
median line is the required c. g. 

23. Any Quadrilateral. Let B CDE, Fig, 20, be any 
Quadrilateral. Draw the diagonals BD and EC, and find the 



29 



ANALYTIC STATICS 



19 




middle point M of one of them, in this case EC. Take 
on D B the distance B N D /, and draw MN. The c. g. 
of the quadrilateral is a point G 
on MN, obtained by taking on 
M 'N a distance AfG = i M ' N. 



EXAMPLK 1 To find the c g. of the 
channel section represented in Fig 21, 
the dimensions being as shown 

NOTE The section being symmetrical, the 
dimensions of the lower part are the same an 
the corresponding: dimensions of the upper 
part, and need not be given 

SOLUTION The c, g lies on the line 
of symmetry O Q drawn through the 
middle point of, and perpendicular to, 
SJ The only thing to be determined 
is the distance O G = x e 





FIG 20 

In complicated cases like this, it is often convenient to hnd a general 
formula first, and then substitute the numerical values. In order to 
do this, the dimensions will be represented by the letters written beside 
the figures on the diagram Since the channel section to the differ- 
ence between the rectangle B CIJ and the trapezoid DE FH> then, 
denoting the area of the rectangle by 1?, the area of the trapezoid 
by T, that of the section by A, and their centers of gravity by G>, Gt, 
and G, respectively, 

x t - G, - G G f - \b - G G r (1) 

Taking moments about a line YZ drawn parallel to BJ through 
the center G r of the rectangle B CIJ, we have, since the moment of 
R about G r is 0, 

G,_G t 
~A 



GG, 



(2) 



20 



ANALYTIC STATICS 



29 



Now, 
T = \ (EF+DH) X PQ = i[(A - 2 j - 2/) + (A - 20] (* - ** 

= (A - $-20 (b -*,) 
and (formula 1 of Art 2O) , 



> 
X 



- 2 j - 20 



, 
* 



iT 



b- t, 3A-4s-6t 
3 X 2(/;-J-20 



^1 




S 

l\ *' 

/.__!p__\/__^ 

i-v U>- * T 

( 1125' I 

U*f J 

* 



&S> ZZlf-^'lZS'i 

JS J 



T G 



Ljff. 



A 



U 




w 



y tf ^ 



FIG 22 
Substituting the values of T'tind <7 r G 1 / in (2), 






- j - 2 - (* - O (8 A - 4 j - 0] 



29 



ANALYTIC STATICS 



21 



Substituting the values of O G, and G G r m (1), 
r = J 6 _ <* -M[M*-J- 2 

The dimensions, expiessed in sixteenths of an inch, are 6 = 54; 
// = 20S, / = (, s = 8, f t = 7 Therefore, 

A = b !i - (t> - t,) (// - s - 2 1) = 54 X 208 - (54 - 7) (208 - 8 - 12) ; 
and 

_ 54 _ ^T)4 - 7) [7(208 - 8 - 12) + $(54 - 7|] 
Xf ~ 2 2[fi4 X 208 - (54 - 7) (208 - 8 - 12)] 



= 27 - 



47 (7 X 188 + 



X47\ 
3 / 



27 - 14 187 = 12 HHS sixte. nths 



2 X 2,396 
= 804 in , or, approximately, x e = f in. Ans. 

EXAMPLE 2 To find the c. g of the plane figure represented in 
Fig 22, the dimensions being as shown 

SOLUTION The figure may be divided into the two rectangle^ 
B CL M and EFIJ, the trapezoid C 'D K L, and the isosceles tri- 
angle FH1 Momenta will be taken about BX and B K The 
altitude of the trapezoid is DN ' - CJVtan 60 = (CL - DJf)ta.a 60 
= 3 4641 ft The operations are given m the following table, which 
needs no explanation. 





Area 


Lever Arm With Respect to 


Moment About 


Figure 


Square 










Feet 


BX BY 


BX BY 


BCLM 


ao oooo 


i oooo 5.0000 


200000 100 OOOO 




f 


DN CL+aDK 




CDhL 


\ 


2 3 CL-\-DK 50000 


99.4353 138.5640 






= 3 5877 




El'IJ 


10 1250 < 


4 s ' f a 25 
-+DJV+CJi J + ED+CX 

= 7 7141 L -=5fi350 


78 1053 56 PS3I 


FIff 


i 2656 I 


3 5 6350 


130853 7iigo 




I 


= 10 3391 




59 1034 


210 6l57 303 636l 



Having the area and the resultant moments, 
,r c = BP 



302.6361 _ 10rtl . _ 4 . 
,or 5.1205 ft. Aus. 



PG 



59 1034 
210.6167 
59.1034 



3.5635ft. Ans. 



22 



ANALYTIC STATICS 



29 



EXAMPLE FOR PRACTICE 

Find the area and the c. g of the Irregular T section represented in 

Fig. 23 

3 3242 sq in 




Ans 



\A : 
\OP: 

\PG-- 



2 0695 in 
7826 in 



CENTER OF GRAVITY 
OF AREAS BOUNDED 
BY CIRCULAR ARCS 

24. Circular Sec- 
tor. Let O PS V, 

Fig 24, be a circular sec- 
tor, having the radius 
O P = r, and the central 
angle P O V = L. As 
the sector is symmetrical with respect to the line O S> bisect- 
ing the angle L t the c. g. lies on that line, and it is only 
necessary to determine the distance OG of the c. g. from 
the center of the sector This # 

distance is given by the for- 
mula p < ^ 
240 r sin 

G 



= OG = 



In the denominator of this 
formula, the angle L is ex- 
pressed in degrees and deci- 
mals of a degree. 

25. For the sectors whose 
arcs are, repectively, a semi- 
circumference, a quadrant, and a sextant, 
formula takes the following special forms: 

4 r 
Semicircle, y e = - = .4244 r 




preceding 



(I; 



Quadrant, y e = 8 ^ in45 -! = .8002 r (2) 



Sextant, y c 



^ Bin80 --.6866r (3) 

7T 




29 



ANALYTIC STATICS 



23 



26. Circular Segment. Let BCD, Fig. 26, be a 
circular segment, having c 

the radius OB = r, and 
the central angle BOD 
= L, expressed m de- s \ 
grees and decimals of a 
degree. Here, G, the c. g. 
of the segment, lies on 
the line of symmetry O C. 
The distance y e = <O G is 
found by the following 

formula: 

_ Q G = 240 r sin 3 
rcL 180 si 

If the length of the choid B D is denoted by c and the area 
of the segment B CD by A, the preceding formula can be 
reduced to 

(2) 




y. - 



12 A 



27. CircularTrapezoid, or Flat Ring. In Fig. 26, let 
B CD EH I be a circular trapezoid, or flat ring, having the 

radii rt. and r tt and the 
central angle L, expressed 
in degrees and decimals 
of a degree. The c. g. 
lies on the line of sym- 
metry O C, and its dis- 
tance y e from the center O 
of the ring is given by the 
formula: 

= 240 sin Z, r t ' -*,' 
Fie,. -jo nL r**-r,* 




EXAMPLE To find the c. g. of the area represented In Fig 27 

SOLUTION Take the center line O y, bisecting the two circular 
arcs, for the axis of y, and the perpendicular OX, through the center 
of the arcs, for the axis of A' As usual, x e and y e are the coordinates 
of G, the required c g , and A is the area of the figure The diagram 
gives A = rectangle B FH T+ segment CD E segment KJ I. The 



24 



ANALYTIC STATICS 



82ft 



moment of A with respect to OX\& equal to the algebraic sum of the 
moments of these three areas To find the area of the rectangle, 
KI must first be found, since TK and IH are given To find the 



IS? 




JF 
S 



Pis 27 

areas of the segments and their lever arms, the angles Z, and Z, must 
be determined The figure gives, expressing angles to the nearest 
minute, 

K I = 1KQ = 2^lr a a O Q* = 2VJ^ fl ~(r^ 

= 2Vr 1 " - (r, -~4p = W 967 ft 
sin i Z, = , whence Z, = 112 3( 



sin 



, 
-^-, whence Z t = 



23' 



Having all the required elements, the coordinates .r, nnd y r ure 
found by the usual methods, with which the student is now supposed 
to be perfectly familiar. In finding x ct however, the operations will 
be much shortened by observing that, if a distance K ' H' = / H is 
taken, and the line H' F< drawn perpendicular to TK, the figure at 
the right of F' H> is symmetrical with respect to O y, and its moment 
about Fis consequently zero Therefore, the moment of the whole 
area about Xis simply the moment of the rectangle B F 1 H 1 T, and 
hence, ' ' 



A 
The results, which should be verified by the student, are- 

A = 118 35 sq. ft. Ans. 
x t = OP= 18741ft. Ans 
y c - PG = 7 7607 ft. Ans. 



29 



ANALYTIC STATICS 



25 



EXAMPLES FOR PRACTICE 

1. Find the c g of a circular sector whose radius is 10 feet and 
whose central angle is 45 y 

Ans y c = 6 497 ft j 



2 Find the c g of a segment whose 
chord is 8 inches and whose radius is B 
5 inches Ans. y e = 3 815 m. 

3 Find the c g of a circular trapezoid 
whose radii aie (i inches and 3 inches, and 
whose shorter chord (as E /, Fig, 26) is 
3 inches. Ans. y e = 4 466 m. 

4. Find the c g of the half circular seg- 
ment BCD (Fig 28). 

x t = 2.839 m 
y e = 13.798 in. 




PIQ 28 



CENTER OF GRAVITY OP A PLANE AREA BOUNDED BY 
AN IRREGULAR CURVE 

28. Approximate Analytic Method. To determine 
the c. g. of a figure, as /? CD E, Fig. 29,, having- an irregular 
contour, proceed as follows: 

Draw two lines of reference OX and O Y perpendicular to 
each other in any con- Y 
venient positions; 
preferably, one of the 
lines, as OX, should 
as nearly as pos- 
sible bisect the area. 
Divide OX into a 
sufficient number of 
parts, so that, by 
erecting perpendic- 
ulars at the points of. 
division, the par 01 $he perimeter of the curve intercepted by 
two conseGUt$TO;c$rtoates, as HiH^ and /;/ may be treated 




Pro. 29 



as a straight 
a trapezoid., 




oorceflgondlng strip .fifr.ff'././i, as 
.#/,//, etc., and the 



I i t 



26 ANALYTIC STATICS 

ordinates ff^H^/if^ etc. Find the distances of the centers 
of gravity of the strips, considered as trapezoids, from the 
lines OX and. Y. Treat the area of the whole figure BCDE 
as equal to the sum of the areas of the trapezoids, and apply 
the method of moments, as usual That is, if the sum of the 
areas of the trapezoids is denoted by 2 T, and the sum of 
their moments about OX and O Y by - Ty_ and 2 Tx t 
respectively, the coordinates x t and y e of the c. g. of the 
whole figure are given by the formulas 

STx 



_ _ 

*'~ 



Tf 



t v ~, 

Although this method is only an approximation, it is 
sufficiently close for almost all practical purposes. The c. g. 
of each trapezoid may usually be taken at the middle point 
of its median line 

29. Experimental Methods. A very convenient 
method of finding the c. g. of an irregular figure (or of any 
other figure) consists in drawing the figure to scale on a 
piece of cardboard of uniform thickness, then cutting off the 
remaining part of the cardboard and balancing the part thus 
left on a knife edge in. two positions. 
In Fig. 30, let B B' CO be the piece cut out of the card- 
board, having the form of the figure 
whose c. g. is lequired; let B C and 
B 1 C 1 be two positions of the knife 
edge, for each of which the piece of 
cardboard remains in equilibrium. 
Then, their intersection G is evi- 
FlG M dently the required c. g. 

This method may be used not only for irregular figures, 
but also for those figures the determination of whose center 
of gravity leads to complicated formulas. The method, 
however, is not very accurate, and, as a check, the piece of 
cardboard should be balanced in more than two positions 
say in six or eight. This will give an idea of the degree of 
accuracy attained. 




29 ANALYTIC STATICS 27 

30. The method just described is also very convenient 
for determining the distance of the c g. of a symmetrical 
body from a plane perpendicular to a plane or axis of sym- 
metry. A locomotive, for instance, is symmetrical with 
respect to a. plane midway between and parallel to, the 
axis of the cylinders. The distance of the c. g from either 
end of the locomotive (that is, from a plane through either 
end perpendicular to the axes of the cylinders) is found by 
balancing the locomotive on a horizontal rod perpendicular 
to the plane of symmetry, passed through ungs attached to 
the locomotive and symmetrically located with respect to 
that plane. The position of the rod and rings is changed 
until the locomotive is found to remain balanced (that is, 
with its two ends at the same level) when suspended. The 
distance of the rod, when the locomotive is thus balanced, 
from either end of the locomotive, is, approximately, the 
distance required. 

31. Still another method for finding the e.g. of a plane 
figure forming: the faces of a thin plate of uniform thickness 
consists in suspending the plate by a string and marking on 
either face a vertical line that is the prolongation of the 
direction of the string; then suspending the plate in a differ- 
ent position (that is, tying the string to a different point In 
the plate) and marking a line similar to the one marked 
before. The intersection of the two lines is the required c. g. 

This method can be conveniently used for determining 
the c. g. of any body when that center is outside the body 
(as an angle section, a bent rod, etc.). Here, when the body 
is suspended in one position by a string, a line containing 
the c. g. is obtained by making two marks on the body: one 
at the point where the string is fastened and another directly 
under it, By tyio$ the string to another point of the body, 
another lin is( determined, whose intersection with the first 
gives the requfrf&jS 




28 



ANALYTIC STATICS 



29 



CENTER OF GRAVITY OF SOLIDS 

32. Right Cylinder or Prism. The c g. of a homo- 
geneous right cylinder or prism evidently coincides with the 
middle point of the line joining the centers of the bases. 

33. Right Cone or Regular Pyramid. The c. g. of 

any right cone or regular pyramid lies 
on the perpendicular from the vertex to 
the base (line joining vertex with cen- 
ter of base) at a distance from the ver- 
tex equal to three-quarters of the length 
of that perpendicular. 

34. Conical Frustum. The c. g. 

of a conical frustum, Fig. 31, lies on 
the line joining the centers of its bases. 
Its distance from the lower base is given 
by the formula 




PIG 81 



v - o G-X r* 

y e LA LJ- 7 PV. 7- v 

4 (r, + rj 

in which r t = radius of lower base; 

r, = radius of upper base; 
A = altitude of frustum. 



29 ANALYTIC STATICS 29 



COPLANAR NON-CONCURRENT FORCES 

COUPLES 

DEFINITIONS-EFFECT OF A COUPLE 

35. A statical couple, or simply a couple, has already 
been defined as a system of two equal non-collmear parallel 
forces having opposite direc- / 

tions. 

In Fig. 32, the forces F,. and n 
F, constitute a couple acting on 
the body BCD. Although F, 
and F, aie equal in magnitude, 
they are denoted by different 
letters for convenience in refer- 
ring to their lines of action. 

36. The lever arm, or 

simply the arm, of a couple is FlQ 32 

the perpendicular distance (A/ 5 . = p) between the lines of 

action of the two forces constituting the couple. 

37. The plane of a couple is the plane determined by 
the lines of action of the two forces constituting the couple. 

38. The axis of a couple is any line perpendicular to 
the plane of the couple. Such will be the meaning given to 
the term here, although some writers use it in a different 
sense. It follows from this definition that all couples whose 
planes are either coincident or parallel have the same axis. 

39. Coaxial couples are couples having the same axis. 
Couples not having the same axis are called non-coaxial. 

40. Resultant Moment of the Forces of a Couple. 

The resultant moment of the two forces of a couple about any 




30 ANALYTIC STATICS 29 

point in their plane is constant and equal to the moment of either 
fotce about any point on the line of action of the other fotcc. 

This is easily shown Paying due attention to bigns, the 
resultant moment of the two foices A and 1*\ about any 
point 0, Fig 32, is 

-F a Xff*0 + F l Xtf l O = -F*(ff t O- //, 0} = 

- F. X Hi H* = - F*p = - I\p 

For a point <7 between the lines of action of the forces, 
the resultant moment is 

- F, X ff. 0> - F, X Hi O 1 = -F, (// (V + //, 0} = 
-F t p = -Ftp 

41. The constant resultant moment of the foices of a 
couple about any point m their plane is called the moinout 
of the couple, and is numerically equal to the pioduct of 
either force by the arm of the couple. 

42. Notation. A couple is expressed cither by its 
moment (for all coaxial couples having the same moment 
are equivalent, as will be shown piesently), or by writing 
its two forces in parenthesis with the arm between; thus, 
(F^fiiFt). The latter is a more convenient form of expres- 
sion for some purposes, and will be often used here. 

43. Effect of a Couple. The effect of a couple acting on 
a rigid body not acted on by other forces is to turn the body about 
an axis passing through its c g. 

A full demonstration of this proposition cannot be given 
here, as it would be necessary to make use of some kinetic 

principles that have not yet 
been explained. Moreover, it 
is to be observed that, for the 

A\ Q. \ IB purposes of statics, it is not 
i necessary to know what the 

* jf* ~n 

V.F, effect of a couple is, for all 

the theorems relating to the 

FlG 83 equilibrium and equivalence of 

couples can be stated and proved without any reference to 

what the effect would be if the couples were unbalanced, 

The foregoing principle, however, has been stated here, in 



29 ANALYTIC STATICS 31 

order to caution the student against a common error preva- 
lent among beginners. If the couple (F lt p t F a ), Fig 33, 
acts on the body A B, its tendency is to turn the body about 
an axis passing through its center of gravity G, not about an 
axis or point situated between the two forces. 

44. Direction tind Sign of a Cotiple. The sign of a 
couple will here be treated as positive or negative according 
as the moment of either force about a point in the line of 
action of the other is positive or negative (see Analytic 
Statics, Part 1). In both Fig. 32 and Fig. 33, the moment 
of Fi about any point in the line of action of F, t or of F, 
about any point in the line of action of F lt is negative, and 
the moment of the couple is considered negative. 

The motion that the couple (F 1} ^,F a ), Fig 32, tends to 
produce is evidently the reverse in direction of the motion 
of the hands of a watch with its face placed upwards and its 
center at H,. or H t . This motion is said to be counter-clock- 
wise, or left-handed Motion similar to that of the hands of 
a watch is said to be clockwise, or right-handed. Clockwise 
motion is said to have a right-handed direction; counter- 
clockwise motion, a left-handed direction. 

It will be noticed that the sign of a couple is positive 
when the couple tends to produce clockwise motion; other- 
wise, however, the couple is negative. This distinction is 
of value only when couples are to be combined by algebraic 
addition. 

The direction and sign of a couple can be very readily 
determined by imagining the arm of the couple to be a line 
rotating about its center so that each extremity follows the 
direction of the force applied at it. Thus, in Fig. 32, if we 
conceive Hi ff* to begin to rotate about its center so that H 
will follow the direction of F l} and H a the direction of F a , it 
will be seen that the rotation will be counter-clockwise. 
The couple is, therefore, a left-handed couple, and its sign 
is negative. 



32 ANALYTIC STATICS 29 



EQUIVALENCE AND EQUILIBRIUM OF COAXIAL COUPLES 

45. Equilibrant and Equivalent Couples. Since a 
couple cannot be replaced by a single force, it follows that 
no single force can balance a couple In order to balance a 
couple, another couple must be opposed to it. Either couple 
is called the equilibi-ant of the other. 

Two couples are equivalent when they can each be 
balanced by one and the same couple that is, when they 
have the same equihbrant. 

46. Equilibrium of Two Coaxial Couples. Two 

coaxial couples balance each other if they have equal but opposite 
moments that is, if their moments are numerically equal but 
have opposite directions (or signs} 

DEMONSTRATION The demousttation of this principle is given 
below Although it is not essential that it be learned or even read, it 
affords a very useful and interesting exercise in the composition and 
resolution of forces It will be necessary to distinguish three cases, as 
follows 

Case I. When the two couples ate in the same plane and the fotces 
of one are parallel to the forces of the other 

Let the couples be (F lt p t F s ) and (F,',p', F a '), Pig. 34 It Is 
assumed that, with the usual notation as to signs, F^p = F^f p', or 
F,.p + Fj.'p' = On the other hand, the algebraic sum of the four 

forces Fi, F*, /V, FJ is equal to 

F i zero, or 2F = Therefore, the 

ji A a four forces, or the two couples, 

a satisfy the two necessary con- 

jp * o tf , ditions of equilibrium of parallel 

3 ^--^.7 j,. -,Q| forces (Analytic Statics, Part 1). 

and so form a balanced system. 

\\ f uy It follows from this that a 

1 ' couple may be replaced by an- 

other couple in the same plane, 
IG> M If the two couples have the same 

moment and their forces are all parallel, and that the effect of a couple 
is not altered by moving the couple In Its plane parallel to itself. 

Case 11. When the two couples are in the same plane and the lines 
of action of the forces of one intersect the hnes of action of the forces of 
the other. 



29 



ANALYTIC STATICS 



Let the couples be (F lt p, F,) and (F^p 1 , F,') , Fig 36. According to 
the preceding demonstration, the couple (F^f, F,') may be replaced 
by another couple with one of its forces acting through 0,, provided 
that the direction of the forces is not changed, and, moreover, the 
magnitude of the forces may be changed, provided that the arm is so 
changed as to keep the moment constant Thus, if Pi is made 
parallel to FJ and F t ', and equal to F lt and if, at the end of the 
arm t 0," = p, the force P t is applied, equal in magnitude to />,, 
but opposite in direction, then the two couples (F l ',p l ,F, 1 ) and 
(/* <9i0,", A) will be equivalent, for, by hypothesis, FJ & is 
numerically equal to F t p t and therefore to P t X 0i O," By trans- 
ferring the point of application of P, to the intersection / of the lines 



A 



"*- A. 



/ X 



I/ 



A 

f i 



o a 



^ i 

yi 




FIG 86 

of action of P t and F,, and the point of application of F t to the same 
point, the two couples (F lt p, F m ] and (FJ.f, F,') are replaced by the 
forces P t and F t acting at 1( and P, and F, acting at 7 Since ^ 
and PI are equal, their resultant must act along the bisector of the 
angle />, d F lt which evidently is the same as the bisector of O, O-, O," 
Similarly, the resultant of P, and F, must act along the bisector 
of O,fO t , aud, as P a and F t are equal and parallel, respectively, 
to PI and FI, the resultants of the two pair of forces are numerically 
equal Now, owing to the equality of O^ O a and t O," t the right 
triangles (not fully shown) X O a f and t 0,"7 are equal, and the 
line 70 t (not shown) is the bisector of both O, 0i 0," and 0,70,". 
The two resultants have, therefore, the same line of action, and, as 
their algebraic sum is zero, they balance each other. 

It follows that a couple may be replaced by any other couple acting 
in the same plane, provided that the two have the same moment 



34 



ANALYTIC STATICS 



29 



Case III. When the planes of the couples do not coincide, but are 
parallel (as they must be, since the couples are supposed to be 
coaxial) . 

Let PQ and P' Q, Fig 36, be the planes of the two couples, and 
let (F^pi.F,) be the couple acting in the plane PQ Since the 
moment of the couple in the plane P' Q is F t p lt that couple may be 
replaced by another couple (F l ' l pS, F a ') having the same force and arm 
as (F lt p lt F a ), and the lines of action of whose forces are parallel to 
the lines of action of F l and F 3 This follows from the two cases pre- 
viously considered, for the couple (/V.A', FJ) is equivalent to any 
other couple acting m the plane P' Q and having the same moment. 
Draw 0, OJ and O OJ, meeting at / We may now compound F l 




Q 



Fit, 30 

with FJ, and F t with FJ. Since d O, is equal and parallel to OJ OJ, 
the point /is the middle point of O l OJ and O, OS The resultant of 
jFi and FJ Is R = F, + F,' = 2 F,, acting through 7 The resultant 
of F, and FJ is ' = F t + FJ = 2 F, = 2 F lt acting through /. As 
these two resultants act In opposite directions along the same line (for 
they are both parallel to the forces of the couples) and have the same 
magnitude, they balance each other. Therefore, the two couples to 
which the two resultants are equivalent balance each other. 

47. It follows from the preceding principle that the only 

couple are its moment and its axis. 

ver, that the axis is not any 

ion. Any line perpendicular 



29 ANALYTIC STATICS 35 

to the plane of a couple may be taken as its axis; and, con- 
versely, if the axis of a couple is given, the couple may be 
supposed to act in any plane perpendicular to that axis. The 
force and the arm of the couple may be changed atpleasuie, 
provided that their product, which is the moment of the 
couple, remains unchanged. 

48. Resultant and Etinlllbrant of Any Number of 
Coaxial Couples. 77/6' resultant of any numhet of coaxial 
confilcs is a single couple having the same axis as the component 
couples and whose moment is the algebraic sum of the moments of 
the component couples. This principle is a consequence of the 
one stated in the last article, as can be shown by a process of 
mathematical reasoning that it is not necessary to give here. 

In general, let M^ M^ M a , etc. be the moments of any 
number of coaxial couples, M r the moment of their result- 
ant, and M q the moment of their equilibrant Then, 
Mr = 2M> and M t = - IM 

Also, if several coaxial couples are in equilibrium, the 
following equation must obtain: 

Mr = SM = 



EXAMPLES FOB PRACTICE 

1 Find the resultant of the following couples, in which the 
sign before each parenthesis indicates the direction of the couple 
(10 Ib , 2 ft , 10 Ib ), - (7 Ib , 6 ft , 7 lb.), (25 lb., 12 ft , 25 lb.); 
- (8 Ib , 40 ft., 8 lb ) Aus M r = - 42 ft -lb. 

2 (fl) Find the lever arm/ of a couple with a foice of 100 pounds 
that will balance the following couples: (30 lb , G f t , 30 lb.), (20 lb , 
6ft.,201b.), - (125 lb ,10ft , 1251b ); (70 lb ,4 ft , 70 lb ), - (80 Ib., 
3 ft., 80 lb.). () What is the sign of the balancing couple? 

Ana /()/ = 9 3 
Ana \ (b) Couple positive 

3. One of the forces of a couple acts through a certain point O, 
and is equal to 300 pounds, the moment of the couple is 2,574 foot- 
pounds. How far from O is the line of action of the other force? 

Ans R 58 ft. 

4. The moment of a couple is 1,500 foot-pounds. Express it 
(a) as a couple having a force of 75 pounds; () as a couple having 
an arm of 12 feet. ._ f (a) (75 lb , 20 f t , 75 Jh ^ 

Ans V v ;, V I25 lb, 12 it., us& It./ 



36 ANALYTIC STATICS 29 



EQUIVALENCE AND EQUILIBRIUM OF COPLANAR 
NON-CONCURRENT FORCES 

49. To Make the Line of Action of a Force Pass 
Through a Given Point. Let F, Fig. 37, be a force act- 
ing on a body, and O, a point in the body or rigidly connected 
with the body. Let the perpendicular distance OP of O 

from the line of action of the force 

* be denoted by x. It is obvious that, 

if the two forces P and F f , equal 

Jijl | to each other and to F, are applied 

jfx at 0, they will have no effect on the 

I j condition of the body, since they will 

2>j- so !o balance each other. The single force 

I | Fis, therefore, equivalent to the sys- 

I i tern of forces F, F', F', applied as 

\-Jf' shown. Now, the two forces F and 

1 F' form a couple whose moment is 

Fx Therefore, the force F acting 

along LK is equivalent to an equal and parallel force F 1 
acting through O, together with a couple whose moment is 
equal to the moment of F about O. In general, 

The line of action of a force may be shifted parallel to itself > 
so that it -will pass through any chosen point, provided that a 
couple is introduced having a moment equal to the moment of the 
force about that point. 

50. Resultant of a Couple and a Force In the Plane 
of the Couple. Let the force 

F, Fig. 38, and a couple (F tt p, 
F t ), whose plane contains the 
line of action of the force, act 
on a rigid body. The couple 
may be replaced (Art. 46) by 
another couple (F' t x t F')> 
whose forces are each numer- 
ically equal to F, provided that 
the lever arm x is such that F 1 x 




29 ANALYTIC STATICS 37 

Also, the couple may be so turned and shifted that one of its 
forces F 1 will act along the line of action of F, but m an 
opposite direction, as shown. The two forces F and F 1 
balance each other, so that the system is reduced to the 
single force F 1 equal and parallel to F. Therefore, 

The resultant of a couple and a single force in the plane of the 
couple is a single force equal and parallel to the given force, 
acting along a line whose distance from the line of action of the 
given force is equal to the moment of the couple divided by the 
magnitude of the given force. 

The direction in which the distance x, or O P, should be 
measured is indicated by the character of the couple. The 
moment F' x of the resultant force about any point in the line 
of action of the given force must have the same sign as the 
moment of the couple. In the figure, the couple is right- 
handed, or positive The moment of F 1 about O must, there- 
fore, be rig-ht-handed, which indicates that F 1 is on the left of F. 

51. Resultant of Any Number of Coplanar Non- 
Concurrent Forces. Let F^F^F^F^ Fig. 39, be four 
non-concurrent forces having their lines of action in one 
plane, and let O be any point in the plane whose perpen- 
dicular distances from the lines of action of the forces are, 

respectively, A, A, A, A- 
As explained m Art. 49, 



, 
I 

* 




kjp' S 

\ ' X^a 1 ' /"* F l m&y be replaced by a 

V I S 

">X / force F-!, equal and par- 

\ \ allel to F lt acting through 

/^i O, combined with a couple 
>. whose moment is /^A- 
^ The other forces may be 
similarly replaced. The 

whole system is thus replaced by the four concurrent forces, 
/^ F a ',F t f ,F t ', acting through O and equal and parallel, 
respectively to F it F,,F tt , /;, together with the four couples 
Fip^Fipt, F a p a ,F t p t . The resultant R of the concurrent 
forces is found as explained elsewhere; the resultant M r of 
the couples is a single couple whose moment is the algebraic 



38 ANALYTIC STATICS 29 

sum of the moments of the forces about O. Finally, the 
resultant of R and M r is found as explained in Ait 50. 

If the given forces are resolved into components in two 
directions perpendicular to each other, then, with the usual 
notation ( see Analytic Statics, Part 1 ) , 

X r =2X=ZFcosff (1) 
Y r = IY = 2Fsrnff (2) 



R = Vx r +">7 (3) 

\r V v" 

tan H r = ^ = ^J-- (4) 

A r - sL 

And also, M r = 2 Fp (5) 

52. Conditions of Equilibrium. When the forces 
are in equilibrium, both R and M f must be zero. The reso- 
lutes of R must, therefore, be zero, and the algebraic condi- 
tions of equilibrium are 

2X = 2Fcostf = Q (1) 
2Y = 2Fsrnff = Q (2) 

M r =ZM=!Fp = (3) 

These conditions may be stated in words thus: 
1 The algebraic sum of the rcsolutes of the fat CM in each 

of any two directions at right 
angles to each other must be 
sero. 

i B 2. The algebraic sum of the 

\ j / moments of the forces about any 

\ \ / and every point in their plane 

NO must be scio 

I 53. Equilibrium of 

I Three Forces. If three 

forces F t , F at F , Fig. 40, are 
in equilibrium, tmy one of 

}fi\ them is the eqtiilibrant of the 

other two, and, therefore, 
equal and opposite to their 

I lesultant. Thus, /'I must be 

FlQl 40 equal and opposite to the re- 

sultant R of F, and F,. As R passes through the point of 





29 ANALYTIC STATICS 39 

intersection of the lines of action of F, and F t and ft must 
balance V?, the line of action of F, must pass through O. In 
general, 

// three coplanar foices are in equilibrium, they must be con- 
curtent (unless they aie parallel), and if the point of nitersection 
of the lines of action of two of the forces is known, the line of 
action of the other must pass through that point. 

EXAMPLE 1 Four forces, F l = 100 pounds, F a = 200 pounds, 
F 3 = 125 pounds, F t = 150 pounds, Fig 41, act on a horizontal 




lever O B The inclinations of their lines of action to the horizontal 
are as shown, and O A-, - 4 feet, O A a = 10 feet, O A = 15 feet, 
O At. = 20 feet Required the magnitude, direction, and line of 
action of the resultant A J 

SOLUTION Using formulas 1 and 2 of Art 51, we have, since 
F, = 100, F tt - 200, F t = 125, F< = ISO, and H, = 45, H* = 50, 
77, = 80, 77* = (50, 
X r - S F cos H - - F, cos 77 t - F, cos H, + F, cos H 3 + F*. cos ff t 

- - 100 cos 4f) - 200 cos 50 + 125 cos 80 + 150 cos 60 = - 102 56 
Y r 2 F sin H = 7^ sin ff^. F, sm H? F, sm 7/ a + F t sin ff t 

- 100 sin 45 - 200 sin 50 - 125 sin 80 + 150 sin 60 - - 75.694 
By formula 3 of Art 51, 

A' = VX r + YS = Vl02~5H a + 757694" = 127 47 Ib Ans 
By formula 4 of Art 51, 

tan H r = % - Jh.rsn; whence H, = 36 25' 40". Ans. 

J\f !VM OU 

Since p, = O A, sin 45 = 4 sin 45 - 2.8284; />, - O A, sin 50 
= 10 sin 50 = 7.6604, p, = OA> sin 80 = 15 sm 80 14.772, and 



40 ANALYTIC STATICS 29 

p* = OAi sin 60 = 20 sin 60 - 17 321, formula 5 of Art 51 gives 

M r = IFp = - ^i A + F, p, + F, p* - F. p. 
= - 100 X 2 8284 + 200 X 7 6604 + 125 X 14 772 - 150 X 17 321 

= 497 59 ft -lb 

The value of Y r shows that the force R acts downwards, its line of 
action must, therefore, be on the right of O, since its moment about 
this point is positive Hence, 

Mr 49769 
Mr = Rpr, whence p r = -g 127 47 = 



EXAMPLE 2 Two forces, F, = 2,000 pounds and F, = 800 pounds, 
Fig 42, act on a horizontal beam 25 feet long resting on two supports 




PIQ 42 



at its extremities O and B The distances O AI and O A t are, respect- 
ively, 8 and 16 feet. The inclinations of FI and F, are as shown. 
It is known that the reaction R" at B is vertical Required the magni- 
tudes of the two reactions R" and R 1 1 and the inclination //' of R 1 to 
the horizontal 

SOLUTION Since the moment of R' about O is zero, R" may be 
found by taking moments about that point anil using formula 3 of 
Art 62, which gives, 

Fipi + F* X OA, - R" X OB - 0; 
that is, 2,000 X 8 sin 30 + 800 X 16 - A 1 " X 2T> 0; 

, _.. 2,000 X 8 sin 30 + 800 X 16 M01K .. 

whence, R" OK ~ - = 832 lb. Ans. 

it) 

By formula 1 of Ai. 52, IF cos H = 0; that is, 

R' cos H' - F, cos 30 = 0; 
whence, 

R 1 cos H 1 = F, cos 30 - 2,000 cos 30 1,732 (1) 

Similarly, by formula 2 of Art. 52, 2 F a\n // - 0; that is, 

Ri sm If' - F, sin 30 - F, + R" - 0; 
whence, R' sin H 1 = ^ sm 80 + F, - R" 

2.000 sin 30 + BOO - 832 - 968 (3) 



29 



ANALYTIC STATICS 



Dividing (2) by (1), 

R< sin H 1 _ . .,, _ 968 
X'cosff' ~ taU ** -l7732 ; 
whence, H' = 29 12', nearly Ana. 



From (2), 



ff 



1,984 Ib. Ans. 



APPLICATION'S 

54. Mutual Reactions. In Fig. 43 (a) are represented 
two bars hinged at A,, and A at and to each other at B The 
bars are acted on by forces F t and F,. Through the joint B t 
the bar A^. B exerts \ 

on A,B a force R a f , 
which is equal and 
opposite to the force 
Rt f exerted by A, B on 
A* B. This force RJ 
R,' is the mutual 
reaction between 
the two bars at the 
joint B. How it is 
determined will be 
explained presently. 
Since the system 
formed by the two 
bars is m equilibrium 
under the action of 
the external forces F lt 
F t) R lt R tt these forces 
form a balanced sys- 
tem to which the 
general equations of 
equilibrium can be 
applied. From these Pl i 4S 

equations, the reactions Ri. and R, can be found when all 
other conditions, such as distances, etc., are known. 

The mutual reaction at B is determined by applying the 
principle of separate equilibrium (see Analytical Statics, 




42 



ANALYTIC STATICS 



29 



Part 1). The part A,B may be removed, and A^B 
treated as a free or separate body, provided that a foice is 
introduced at B equal to the force exerted by A. B on A* B t 
that is, a force equal to RJ. This condition is represented 
in Fig '43 (b), where A,B is shown as a free body acted on 
by the external forces R i} -ft, and RJ. By applying to the 
system constituted by these forces the general equations of 
equilibrium, RJ may be determined. 

55. Method of Sections. The free-body principle 
finds a very useful application in the determination of 
stresses in framed structures by the method or sections, 
illustrated in Fig. 44. The truss A, B* B* A, rests on piers A, 
and A n and is loaded at the joints d, C, and C 91 as shown. 




TT, 



W 

PIG. 44 



Let the reactions R* and R, first be determined. Consider- 
ing the truss as a whole, the external forces acting on it are 
the weights W lt W, and W, and the reactions JR l and R tt 
These forces form a balanced system, and, therefore, the 
algebraic sum of their moments about any point mubt be 
zero (Art. 52). Taking moments about A^ in order to 
eliminate R lt 

W. XA 1 C*+ WX A, C+ W t XAtCt-R.X A,A t 0, 
from which R, can be found. To find ^?i, moments may be 
taken about A a ; or, knowing ./?, R l may be found from the 
equation J 1 Y = 0, which in this case gives 

Ri + R> - W> - W* - W 

The stresses in the members may be found by the method 
explained in Analytic Statics, Part 1, proceeding joint by joint, 



29 ANALYTIC STATICS 43 



beginning either atA^ or at/ a , since now R* and R* are known. 
Or the method of sections, referred to above, may be used, 
as follows: 

Let it be required to find the stresses in the member CC a . 
Imagine the truss to be cut in two by a plane PQ inter- 
secting CC a and the members CB, and B^B*, which meet 
at JS 3 . 

The part of the truss at the right of PQ may be supposed 
to be removed by introducing at the points of separa- 
tion E, F, ff, forces 5, T, U, equal to the actions of B* E on 
EB^BiF on FC> and C t H on H C, respectively, which are 
the measures of the stresses m the three members inter- 
sected by the plane The part A^B^EFH may now be 
treated as a free body acted on by the external forces R^ W lt 
W, S, T, and 7, of which only the last three are unknown. 
If moments are taken about B t (the point of intersection 
of 6" and T), the forces S and T will be eliminated, and an 
equation will be obtained in which the only unknown quantity 
will be U. 

The same result would have been obtained if the mem- 
bers CC at BI G, and B t A n had been cut by the plane PQ. 

To find S, moments are taken about C. In every case, 
several members should be cut, of which all but one meet 
at a joint, and this joint should be, taken as the origin of 
moments. If, however, the stresses in some of the mem- 
bers cut are known, it is immaterial whether they meet at a 
joint or not: the only thing necessary is that, of the mem- 
bers whose stresses are unknown, all but one should be 
concurrent. 



ILT3D8 M 



44 



ANALYTIC STATICS 29 



FRICTION 

SLIDING FRICTION 
DEFINITIONS AND GENERAL PRINCIPLES 

66. Definition of Sliding Friction. Let a block 
CDE, Fig. 45, rest on a horizontal surface AB capable of 
resisting or balancing the weight W of the block. The 
block will then be in equilibrium under the action of the 
weight W and the reaction R, the latter being equal and 
opposite to W. If, now, a horizontal force F( which, for 

convenience, will be 

1 supposed to act along 

ir ^ a line passing through 

the c. g of the block) 
is applied, and it is 
assumed that no other 
force than W, R, and 
F acts on the block, 
the latter being under 
the action of an un- 
Fl 45 balanced force, will 

move in the direction of F with an acceleration equal to 

g (see Fundamental Principles of Mechanics}. So long as 
Vv 

there is no other force acting on the block, motion will 
ensue, however small /'may be. 

Experience, however, shows that a small force, whether 
horizontal or not, often produces no effect on a body resting 
on a surface, and sometimes a very great force is required 
before the equilibrium of the body is disturbed. Thus, to 
drag a trunk or a box over the floor may require the efforts 
of several strong men. We also know that the resistance is 




29 ANALYTIC STATICS 46 

greater the rougher the surfaces in contact: a rough box Is 
not so easily dragged over the sidewalk as a polished stone 
block over a smooth wooden floor. 

There is, then, a force brought into action whenever there 
is a tendency of a body to slide on another. This force, 
whose effect is to prevent, or which tends to prevent, motion, 
is called sliding- friction, or simply friction. It is 
obviously caused by the roughness of the surfaces m con- 
tact. No matter how smooth a surface may appear, it 
always has small projections or elevations separated by 
small depressions. When the surfaces of two bodies are m 
contact, the projections of one surface go into the hollows 
of the other, the two surfaces thus become more or less 
interlocked and cannot slide freely on each other. In order 
to cause sliding, a force is necessary, whose magnitude 
depends on the roughness of the two surfaces. As this 
roughness varies with different bodies, it may be anticipated 
(and this is known from experience to be the case) that 
friction must be a very variable force, depending both on 
the nature and on the conditions of the bodies m contact. 

57. Limiting Equilibrium. Given the block CDE, 
Fig. 45, resting on the surface A B, the force F may 
either move it or leave its equilibrium undisturbed. In 
the former case, F must be greater than the friction; in 
the latter case, .Fmust be less than the friction Let F m be 
the greatest force that can be. applied to the block without 
moving it. Then, a force equal and opposite to F m will 
represent the maximum friction that can exist between the 
block and the surface AB, any force greater than F m will 
cause motion, and any force less than F m will be balanced by 
the friction. But it is not to be supposed that the friction is 
constantly equal to F m : so long as the applied force is less 
than F nn the friction is just equal and opposite to the 
applied force; the friction grows with the applied force up 
to the value F m , beyond which equilibrium ceases to exist 
and the body moves under the action of the difference 
between the applied force and the maximum friction F m . 



46 ANALYTIC STATICS $29 

When the block is acted on by a force equal to /v l} it is 
said to be in a condition of limiting equilibrium, or on 
the point of moving, for the least mciease in the applied 
force is sufficient to produce motion In (Ins <MSU, there 
exists between the two bodies A />' and C HH tin 1 greatest 
possible friction that under the given cucuinstancus can 
exist between them. This maximum force of fiietion is 
called limiting friction, and will heieaftcr be designated 
by POT. Numerically, P m = J ? m . 

58. Passive Forces. Friction, like many other resist- 
ances, is a passive force that is, a force preventing 
motion, but not producing it. The reason foi this is that 
friction is brought into action by the application of other 
forces to which it is opposed; and, as it can never exceed 
those forces, it can never pioduce motion in the direction of 
its own line of action. The .same is true of this reactions of 
supports. A pier may be capable of exerting n reaction 
of 1,000 tons; but, if a stone weighing 1 pound is placed 
on it, the pier will exert on the stone an upward pressure 
of only 1 pound, or just enough to balance the weight of 
the stone. 

Forces producing, or tending to produce, motion, are 
called active forces. 

Passive forces are balancing forces and never acquire 
greater magnitudes than the active forces they oppose. 
Usually, as in the ca.se of friction and reactions, n passive 
force cannot exceed a certain limit; but between Kero and 
that limit it can have any value, and so long as the active 
force to which the passive force is opposed does not exceed 
that limit there will be equilibrium. 

59. Shifting of tlio Ijlno of Action of the tteaotlou. 

The force of friction is a tangential force, by which is 
meant a force acting along the surface of contact of the two 
bodies between which it is exerted, In Fig, 46, the force of 
friction is not directly opposed to /; but acts along the 
surface CE, in the direction EC, How, then, can this force 
balance F, not being in line with it? 



29 



ANALYTIC STATICS 



47 



The reason is that, on the application of F, the reaction R 
is shifted so that it no longer passes through the point where 
the vertical through the c g. of the body meets the sup" 
porting surface. This is illustrated in Fig. 46. The line of 
action of the applied force F meets the vertical line 
through G at 0. The resultant of W and F, found in the 
usual manner, is F r , whose line of action meets AB at O 1 . 
Transferring F r to O 1 (where, for convenience, it is repre- 
sented by Fr'), and again resolving it into its components 
F 1 F and W = W t it is seen that, in order that there 
may be equilibrium, the reaction R must be equal and 




PIG. 40 

opposite to F r ', and its components Y and P must be, 
respectively, equal and opposite to W'(= W} &ndF / (=s F). 
The vertical component Y is the resistance of A B to direct 
or normal pressure; and at present it will be assumed that 
the surfaces in contact are capable of offering this resistance, 
whatever the value of Y or W may be. The horizontal 
component P is the force of friction acting along the surface 
of contact. In the case illustrated in the figure, F is sup- 
posed less than F m that is, less than the maximum force 
that can be opposed by the friction. Therefore, P, which 
can take any value between and F m , will, in this case, be 
just equal to F 1 , or F, and balance the latter force. 



48 



ANALYTIC STATICS 



29 



It thus appears that the effect of the friction is to shift the 
point of application of the reaction from K to O', and to 
change the line of action of that reaction from the vertical 
direction to the direction O 1 L. 

The body CD E may be considered as being acted on by 
the two equal and opposite forces R and F r , or by the two 
couples - (W,KO f , Y) and (F, OK, P). Let the student 
show that the moments of these two couples are numerically 
equal. 

60. Case In Wnlch F Is Greater Than /*,. We 

shall now consider the case in which F is greater than F m , 
or P m In Fig. 47, the block CDE is acted on by its own 




Fin 47 



weight W and the horizontal force F. As before, F r ' F r is 
the resultant of W and F, transferred to the point (V on the 
surface of contact; its components are F 1 = F and W W< 
As, m this case, /MS greater than F mt the force of friction, 
which has its maximum value P, H = F m , is not sufficient to 
balance F 1 , and there will be sliding of the block on A B. 
The reaction R of the latter surface is the resultant of 
Y = W = H^and P m . This reaction is evidently less than 
F r ', and its line of action makes with the normal O M to AB 
an angle Z less than the angle / made by the line of action 
of F r with that normal, 



29 



ANALYTIC STATICS 



49 



If, Y (that is, W or W) remaining constant, F is made 
less than P m , say equal to O' S, the resultant will be O 1 T, 
and the reaction will be O 1 U, whose components are Y = W 
and MU equal to the friction />, which in this case, will be 
equal to OS. 

The conditions of equilibrium may, therefore, be stated 
by saying; either that /"must not be greater than P m , or that 
/ must not be greater than Z. 



ANGLE AND COEFFICIENT OF FRICTION 

61. Maximum Resistance. The results of the fore- 
going discussion may now be generalized. Let N, Fig. 48, be 
the normal force between two bodies whose surface of con- 
tact is A B. The force N is the normal component of the 
resultant force acting on CD E (or on the other body), when 
that resultant, after its point of application has been trans- 




"jr 



FIG. 48 

ferred to the intersection of its line of action with the surface 
of contact, is resolved into two components, one parallel, and 
one normal, to that surface. Given, besides the normal 
component N (usually called the normal pressure), all 
other conditions such as the nature of the two bodies, 
the extent of their surface of contact, etc, the maximum 



50 ANALYTIC STATICS 29 

friction P m that can exist between them is determined by 
actual experiment Once P m is known, the maximum reac- 
tion R, which we shall call the maximum resistance that 
can exist between the two surfaces, is 



The angle made by the line of action of the reaction with 
the normal OS to the surface of contact is given by the, 
familiar expression, 

Li S fm -fft 



If, the normal pressure and all other conditions remaining 
constant, a force F greater than R is applied to CDE (in 
which case N is the normal component of F) , it is obvious 
that there cannot be equilibrium, since the component of F 
parallel to A B, which component is equal to M S, is greater 
than P m . If, on the contrary, F is less than R t or equal 
to R that is, if its line of action falls within the angle Z or 
coincides with L equilibrium will obtain. 

62. Angle of Friction: Condition of Equilibrium. 

The angle Z, Fig 48, is called the angle of friction, and 

may be defined as the angle between the line of action of the 
maximum resistance and the normal to the surface of contact. 

The condition of equilibrium explained in the preceding 
articles may be thus stated: 

In order that there may not be sliding between two bodies in 
contact, it is necessary and sufficient that the resultant of the 
applied forces (as F r , Fig. 47} shall not make wtth the normal 
to the surface of contact an angle greater than the angle of friction. 

63. Coefficient of Friction. Until recently, it was 
thought that, for any two given substances, the maximum 
friction P m was directly proportional to the normal pres- 
sure .Wand independent of all other circumstances. Accord- 
ing to this view, if P m ', Pa/'t /V", etc. are the maximum 

frictions corresponding to the normal pressures N' % N", N" f , 

pi p tf p tff " 
etc., then ^ = = ; = ^. The common value of these 

ratios will be denoted by c. 



29 ANALYTIC STATICS 61 

Having determined the ratio c of the friction to the nor- 
mal pressure for any particular case, the friction in any other 

D 

case could be at once found from the relation = c, 

N 

whence, P m = c N. Also, since 

tan Z = ^2 = c 
N 

it would follow that the angle of friction was constant for 
every two substances sliding on each other. 

That there is generally some dependence of the force of 
friction on the pressure is a familiar fact. Thus, referring 
again to Fig. 45, daily experience shows that, if a pressure Q 
is applied to the block, the effort required to drag the block 
will increase as the pressure Q increases. The relation 
between pressure and friction, however, is not always so 
simple as stated above. 

The law that friction is proportional to pressure is approxi- 
mately true only m some cases, or under certain conditions. 
What these conditions are, and how friction varies under 
different conditions, are problems to be solved by direct 
experiment. 

Although the ratio of P m to N is usually variable, it is 
customary and convenient to express P m as a fraction of the 
normal pressure, and write, 

P m = fN (1) 

The factor /, or the ratio of the maximum friction to the 
normal pressure, is called the coefficient of fi-Iction, or 
friction factor. It has been determined experimentally 
for various substances under various circumstances; its 
values have been tabulated, and the approximate laws of its 
variations stated. Here, the theory only of friction will be 
considered. In order, however, that a general idea of the 
values of / may be obtained, it will be remarked that, in the 
majority of cases coming within the practice of the mechanical 
engineer, / is usually much less than .5. Thus, for unlubri- 
cated metals sliding on one another, the average value of / 
is about .18; for dry and smooth wood sliding on the same, 
/ averages about .38; and for wood sliding on metal, both 



52 ANALYTIC STATICS 

dry and smooth, about A. Tn civil-cntfinerring wink, how- 
ever, values of /often oceui that e\eeeil ..", ,is in the i ,IHI nf 
brick sliding on hiick or on stone, masoniy on hiKk\\otk, civ. 



Since tan X ~ / "" 1 , which is the same .is the v.ihut *f / 

from foimuhi 1, \vo may write 

tan /. ~- f (2) 

The angle of friction m.iy, therefore, he defined .is .in 
angle whose trigoiuuneti tu tangent is uinuil ti the encHu ieut 
of friction, it being understood th.it one of the sides if tins 
angle is the normal to the surf.iee of contuet nf llic tw> 
bodies whose fiiction is considered. 



64. titatlo Frletlon and Kliiftlc l^rlctloiu In 

to set a hody in motion over another hody, u fojve is neecs- 
sary whose component parallel to tlic snrf.iee of roni;<el i^ 
greater than P m . Once the body is in motion, it seem** th.it 
it should continue in motion, by virtue of its inert ut, if n 
force just equal to />, were constantly uiiplied to if; fur this 
force would ho sufficient to overcome the frietion. KXJHTI- 
ence, however, shows that the force nceessary t Ktiut n 
body sliding on another is almost always greater th;ui the 
force necessary to keep it sliding, once motion has Itegmi. 
In other words, the resistance of friction in greater when 
motion is to bo produced, than when it iff to IMJ nminlaiitct). 
In the former case, tjie friction IH culled fHetloii of n-nt, 
or static fi'totlon; in the latter case, fi'teflon of auttflon, 
clyiminle ri-It-tlon, or Idnettc frietion. In cither ease, 
the maximum frietion will bo here designated by A,, niul 
the coefficient of friction by /, it being undorwtood that. IM 
a rale, P m and /have different values for the two 



65. In practice, it is necessary to bear in inlmt what the 
function of friction is when any particular problem in tp lie 
solved. In designing a nmcliine, where niotitm han to 1 
produced and maintained, sufficient force should JMJ allowed 
to overcome the friction of rest, an othnrwive the machine 
could not be started. But, in calculating the c<rriolt*m<y 
(A term to be defined elsewhere) of the machine while ia 



29 



ANALYTIC STATICS 



53 



motion, the friction of motion should be used. In providing 
for the equilibrium of a structure, however, where friction is 
a favorable force, the friction of motion should be used; for, 
although under ordinary circumstances the structure, being 
in a state of rest, will offer a frictional resistance equal to 
the maximum friction of rest, the least shock is often suf- 
ficient to produce a disturbance of equilibrium; the structure 
is, so to speak, started, and then the only resistance prevent- 
ing it from continuing to move will be the friction of motion. 
This is especially the case m structures subjected to shocks, 
such as bridges and engine foundations. 




FIG. 49 

EXAMPLE A block CD E, Fig 49, whose weight is W, rests on a. 
horizontal surface A B The coefficient of friction between the block 
and the surface is /. A force F, acting in a. vertical plane containing 
the c g of the block, IB applied at an angle H to the horizontal. 
Required the magnitude of F, that the block may be on the point 
of sliding along A JB. 

SOLUTION All problems similar to this may be solved in two 
manners, and the result found directly m terms of either / or Z, Of 



64 ANALYTIC STATICS 29 

course, one result can be transformed algebraically into the other 
from the relation tan Z = f 

1 Let O be the intersection of the line of action of F with the vertical 
through the c g of the block The latter is held in equilibrium by the 
forces F, W, and the total reaction J? of A B t whose line of action 
O>O must pass through O (Art 53) As explained in Art 62, this, 
line of action must make with the normal to A 2?, or with the vertical, 
an angle equal to Z Therefore, F r being equal to and colllnear with 
R t angle OJT= WOJ = Z The triangle O JT gives 
TJ W 



W 

- WOJ) snZ 
W W 

= sin (90 + H - Z) sm Z = cos (H - Z) sin Z (1) 

2 The line of action of the resultant F r of the forces F and W 
meets the surface A B at 0'. Let F and JP be transferred to this 
point, as shown. Resolving F into its horizontal and vertical com- 
ponents F cos H and F sin H, the resultant normal pressure acting at 

N= W-FsinH 
Therefore, the resistance of friction is: 

P m = Nf = (W-FtinH)f 

As this force must balance the horizontal component of F, there 
results- ( W - F sin H )f=F cos H\ 

W f 

whence, F = __ , , rr (2) 

1 cos /f+ /sin If ^ f 

To reduce equation (2) to equation (1), we have 

Wf ^ w tan Z 

cos ff+fsinH cos /f + tan Z sm ^T 

sin Z 
*,* cosZ _, sin Z 



smZ 
cos(H-Z) 



EXAMPLES FOR PRACTICE 



1. Find the angle of friction, to the nearest minute, corresponding 
to each of the following coefficients of friction: (a) /=*.16; (d) f** .25; 



(c) f - .50, (d) f = .65. 



Ans. 



8 32' 
2T 14 2' 
2T 26 84- 
Z - 33 1' 



29 ANALYTIC STATICS 55 

2. Find the coefficient of friction corresponding to each of the fol- 
lowing angles of friction (a) Z => 12 15', (6) Z = 30, (c) Z = 8 35'; 
(d) Z = 3 17'. f (a) f = .217 

A O J(*) f = 577 

Ans lW /= .151 

[(d) /= 057 

3 A block of marble weighing 100 pounds is kept sliding with 
uniform velocity on a horizontal pine floor by a force inclined to the 
horizontal at an angle H = 30. If the coefficient of kinetic friction 
between marble and pine is 45, what must the magnitude of the force 
be (a) if H is an angle of depression? (6} if H is an angle of eleva- 
tion? & IW F= 70221b. 

Ans \(b) F= 41251b. 

4. A force of 15 pounds, inclined to the horizontal at an angle of 
elevation of 30, is just enough to keep a block of cast iron, weighing 
100 pounds, sliding uniformly on a horizontal cast-iron plate Find 
the coefficient and the angle of friction . f / = 14 

Ans \Z = 7 68' 

5. Taking the angle of friction of rest for brick sliding on brick 
as 35 30' , with what normal pressure N must a brick weighing 
6 pounds be pressed against a vertical brick wall that the brick may 
not slide down? (Use only two decimal places for /.) 

Ans N = 8 45 Ib. 



RESISTANCE TO ROLLING 

66. Cause of Resistance to Rolling. Let a homo- 
geneous cylinder A D, Fig. 50, of radius r, rest on a hori- 
zontal surface X' X. If a horizontal force F is gradually 
applied to the cylinder along any line NL perpendicular to 
the axis of the cylinder, it will be found that no motion can 
be produced before the force F exceeds a certain limit. 
Now, the line of action of the weight W of the cylinder 
meets the supporting surface at A, directly under the 
center O. Did the cylinder touch the surface X' X only at 
A, it is obvious that any horizontal force, however small, 
would cause motion; for, as the resultant of that force and 
the weight could not be vertical, its line of action could not 
pass through A, and, therefore, such a resultant could not be 
balanced by the reaction of the supporting surface. And, 
since experiment shows that it is possible to apply a horizon- 
tal force to the cylinder without causing motion, it follows 



56 



ANALYTIC STATICS 



29 



that not only the point A, but a part AB of the cylinder 

must be in contact with X' X; in which case it is easy to 

understand how the 
resultant F r of F and 
W may be counter- 
acted by the icaction 
R of the suiface X 1 X. 
That contact cannot 
take place at the point 
it A only is otherwise 
evident from the fact 
that all substances 
are more or less 
compressible, so that, 
while the weight of 
the cylmdei causes a 
small depression in 
the supporting sur- 
face, the reaction of 
the latter causes a 

small flattening of the cylindrical surface, as shown in the 

figure. 

67. Coefficient of Boiling: Friction. Suppose that 
the cylinder is in a state of limiting equilibrium with respect 
to rolling; that is, that any increase of the force F will cause 
the cylinder to roll. Let the line of action of the resultant 
of Wand -Fmeet the surface of contact at Af t at a horizontal 
distance c from the theoretical point of contact A. Resolving 
the reaction R into its components F and W, it will be 
seen that the cylinder is in equilibrium under the action of 
two couples; namely, (F, Afff, -F), and -(W,NH,-W). 
Therefore, 




Fio 60 



But, as the deformation of the surfaces of contact is 
very small, we may write, with sufficient approximation, 
MH = A N = h; and, therefore, Fh = Wc\ whence 

F = W- 
h 



29 ANALYTIC STATICS 57 

It has been ascertained by experiment that the distance c 
is independent of the dimensions and weight of the cylinder 
and depends only on the materials of the two surfaces in 
contact The particular value of c for any two materials 
rolling on each other (it is not necessary that one of the 
surfaces should be a plane, as X 1 X} is called the coefficient 
of rolling friction for those two materials This' coeffi- 
cient is not an abstract number, but a length, and its value, 
of course, depends on the unit of length used. The follow- 
ing are approximate values of the coefficient c. 

For elm rolling on oak, c = .032 inch 

For iron on iron and steel on steel, c .02 inch. 

It appears from the foregoing formula that, for any two 
materials, the magnitude of the force F producing limiting 
equilibrium depends on its lever arm h. What is necessary 
and sufficient in order to produce limiting equilibrium is that 
the moment Fh of the applied force should balance the con- 
stant moment We The resistance to rolling may, there- 
fore, be said to be expressed by a couple We rather than by 
a single force. 

This constant couple, whose moment is obtained by multi- 
plying the normal pressure acting between the two surfaces by the 
coefficient of rolling friction , is called a friction couple. 

68. To Determine Whether Equilibrium Will Be 
Broken by Sliding or by Rolling. Referring again to 
Fig. 50, it must be noticed that rolling about M (practically 
about A} cannot occur if the cylinder slides before Fh reaches 
the limit We (here Wdenotes the sum of all the normal forces 
acting between the two surfaces) If / is the coefficient of 

sliding friction, ^must not be greater than Wf, or W - must 

not be greater than Wf\ therefore, - must not be greater 

/i 

than /, or c must not be greater than fh. 

If c is greater than fh, sliding will begin before rolling can 
take place. If c fh, sliding and rolling will begin 
simultaneously. 



58 



ANALYTIC STATICS 



29 



TUB INCTJNKl) 




/ I 
/ I 



/V 
JL 




Fifi.6i 



69. Definitions. An hiulliuMl piano is, as its 

implies, a plane surface inclined to the hon/.on. In Fitf. ftl, 
the plane PQRS, making with the horizontal plane /' 7V/.V 

(or any other hori- 
zontal plane) an angle 
' T, is an inclined 
plane. 

70. Any vertical 
plane perpendicular to 
an inclined plane is 
called a principal plane, and its intersection with the 
inclined plane is called a lino of Uot'llvlty. A line of 
declivity may alho be defined as a line lyintf in the inclined 
plane and perpendicular to the intersection of the ItitU-r plane 
with any horizontal plane. In Fig, fil, PQ /", . I />' ( ', A' A 1 V t 
being vertical planes perpendicular to /'j^A'A*. aie principal 
planes, and the lines Ql\ J1 A } RS are lines of declivity. 

71. The angle between an inclined plane tind the horizon- 
tal is called the aii^lo of the inclined plane, and is the same us 
the angle that any line of declivity imilcea with the horizontal. 
Thus, in Fig, 51, the angle of the plane is the common value 
of the angles QPT, BAC } .flSf/tnade by the lines of 
declivity QP, BA, RS with the horizontal. 

72. In practice, an inclined plane is always the surface 
of some body, as a plank, an inclined rail, the side of a 
hill, etc. If, for any special purposes, it is desirable to take 
into account a definite extent of this surface, as /'? RS in 
Fig. 61, the line A n (or any other parallel to it and included 
between PS and QR) is called the length of the inclined 
plane; JB C is the height, and A C the base. 

73. Equilibrium of n Body on an Inclined Plane, 

Let J t MJ t ', Fig. 62, be a body resting on an inclined 
plane AB. The view here represented IB a section made 
by a principal plane through the c. g. of the body. All 



29 



ANALYTIC STATICS 



59 



forces are supposed to lie in that plane. The angle of the 
inclined plane is B A C = H. If only the portion AB were 
considered, AB would be the length, BC the height, and 
A C the base. But these dimensions are not needed for the 
purpose of the present discussion. The line W is a 
vertical through the c. g. of the body, W is the weight of 

'\ 




Fib 62 

the body, and Fa. force whose line of action meets W 
at O. Through O draw YY 1 and XX 1 , the former per- 
pendicular, the latter parallel, to A B. Let the line of action 
of F make an angle K with Y Y 1 . Denote the equilibrant 
of F and W by ?, their resultant by F rt and the angle that 
Q and F, make with Y Y' by L. The triangle O T U gives 

UT _ sin f/0 r 

0tf sin C>r/ 

Now, UT= F-,OU= W\*\\\UOT = sin ( + #); and 
sin 7Y7 sin TOF - sin (180 QOF) sin _ 
= sin (L + A"). Substituting in the preceding equation, 
_^ _ sin ( + #). 



whence, 

I L T 398-U 



sn 



+ 



60 ANALYTIC STATICS 29 

The equihbrant Q of /*" and W is the rctiction A 1 oi the 
inclined plane, acting through the point / whcic the line ot 
action of the resultant 7v meets the plane. When the body 
is in a position of limiting 1 equilibiium with respect to sliding, 
R makes with the normal J N an angle equal to the angle 
of faction Z (Ait. <>2). As JN is parallel to )')'', it 
follows that, in this case, L = xf, and the foiou /*' is the 
maximum foiec acting; along the line OF that the body can 
resist without sliding:. It is often said that this is the foico 
that is just enough to start the body moving ovt i r the plane; 
but this is not coirect. If the body is aheady moving, this 
force, being just enough to balance the friction, will keep the 
body moving with constant velocity. If the tootlv is at rest, 
the force will simply keep it in a condition of limiting equilib- 
rium; the body will not move, but the least increase in the 
force will be sufficient to produce motion. 

74. To find the value of /"for which the body is in a 
condition of limiting equilibrium (or will move with con- 
stant velocity, once staited), it is necessary to distinguish 
two cases. 

CUHO I. The body is on the point of moving up tht plane. 

In this case, ATmuat be on the right (in the figure) of Y Y' t 
but it may be acute, right, or obtuse. If, however, A' is 
acute, it must be greater than //; that is, /''must act on the 
right of V V\ for, if /* lay on the left of V l\ the re.sult- 
ant F r would act either in the angle VOX' or in the 
angle V OX 1 : in the former case, the equilibrant would be 
some force directed like Q' t and, in order that the plane 
might furnish this equilibrant by its resistance, the reaction JR 1 
should have a normal component acting downwards. This 
cannot take place, as the plane is not supposed to be capable 
of exerting any downward reaction. If the resultant fell in 
the angle VOX 1 , the equilibrant would be a force like Q"; 
this equilibrant might be furnished by the reaction R" of 
the plane; but in this case the friction, being: the component 
1 of R" parallel to the plane, wouid act upwards, and the body 
could not be on the point of moving upwards, 



29 



ANALYTIC STATICS 



61 



It being, then, understood that, in the case under con- 
sideration, K must be greater than H, the force j^may be 
found from the preceding equation by writing Z instead of L\ 

l (1) 

r l ^ 



p _ 



This value may be expressed in terms of the coefficient of 
friction / as follows: 

Sin(Z + ff) _ ^y sin Z cos H + cos Z sm H 
sm(Z + K} sm Z cos K + cos Z sin K 




PIG fiB 



Dividing both terms of the fraction by cos 2", and writing / 

instead of ^4 ( = tan Z, Art. 63), 
cos Z 



/ cos K -f sm K 
Leaving: W, ff, and / unchanged, it is seen from for- 
mula 1 that, if K is changed, F will have its least value 



62 ANALYTIC STATICS 29 

when the denominator is the greatest possiblethat is, when 
sin (Z+K] = 1; whence, 

Z + K = 90, and 90 - K = Z, or FOX = Z 
This very important result is expressed by saying that 
the best angle of t> action up an inclined plane is the angle of 
fnctton. 

Case 11. The body is on the point of moving down the 
plane (Fig. 53). 

In this case, the friction, which is the component parallel 
to the plane of the total reaction R, acts upwards. The 
line of action of R, and therefore of F ft must lie on the 
right (in the figuie) of the normal JN. By a process of 
reasoning similar to that employed in the preceding case, 
the following formulas are obtained for the present condi- 
tions: 



F = W !-^L~ f cos *L (4) 
sin K f cos K 

75. Discussion of Formula 3 of Art. 74 Angle of 

epose. So long as H is greater than Z, formula 3 will 
give a positive value for F. In this case, F, being the 
equilibrant of Q and W> must act outside the angle WOQ, 
that is, K must be greater than, Z. The positive value of F 
indicates that, if the body is left to itself, it will slide down 
the plane, and that, therefore, a force is necessary to keep it 
from sliding. 

If H= Z, then sin (H - Z) = 0, and, therefore, F=Q. 
This means that the body, if left to itself, will rest on the 
plane in a state of limiting equilibrium; no force is necessary 
to keep the body from sliding; but if the angle of the plane 
is increased and no force applied, the body will slide. For 
this reason, the angle of friction is often called the angle of 
repose, and defined as the greatest angle with the horizon 
that the surface of contact of the two bodies to whose friction 
it refers can make without the bodies sliding on each other 
This relation is made use of in the determination of the 



29 



ANALYTIC STATICS 



63 



coefficient of friction. Suppose the plane AB t Fig. 54, to be 
the tipper surface of a plank hinged at A, and that it is desired 
to find the coefficient of friction between cast iron and marble. 
The plank is lined with a plate of cast iron and turned about 
the hinge until it is nearly horizontal. A block of marble is 
then laid on the plank, and the latter turned upwards until 
the block begins to slide. The angle of inclination of the 




PIG. 64 

plank to the horizontal at which this takes place is the angle 
of friction, and its tangent is the coefficient of friction. 

When H is less than Z (see Fig 54), the component 
of W along the plane is not sufficient to overcome the fric- 
tion. Therefore, in order that the body may be on the point 
of sliding down, the force F must have a component acting 
down the plane; that is, K must be on the left of YY' 
In this case, 

. / T rr\ 

(1) 



or 



sin (Z + JK) 

= W -~ c ~^ H ~~- S i5-^. 
/ cos K + sm K 



(2) 



EXAMPLE 1 A wooden box 10 feet long, 6 feet wide, and 4 feet 
deep is used for carrying coal up and down an inclined steel-rail track, 
the grade of the track being 10 in 100 (which means that the track 
rises 10 feet for every 100 feet of length, measured horizontally) . The 
weight of coal will be taken as 54 pounds per cubic foot, and the 



64 



ANALYTIC STATICS 



coefficient of kinetic friction between wood and steel as .4. The box 
being full, required- (a) the magnitude and inclination of the least 
force that will keep the box moving up the plane with constant veloc- 
ity; (d) the magnitude of a force parallel to the rails necessary to 
produce the same effect; (c) the magnitude of a force parallel to the 
rails that will keep the box moving downwards with constant velocity. 

(d) If the available force 
acting upwards, parallel 
to the track, is 4,000 
pounds, to what depth 
can the box be filled? 

SOLUTION (a) For 
the weight W we have, 
neglecting the weight of 
the box, W = 10 X X 4 
X 54 = 12,960 Ib The 
angle Z of friction, to 
the nearest minute, is the 
angle whose tangeut is 
4, that is, Z = 21 48'. 
For the inclination H of 
the track we have tan H 
= Vnr, whence H = 5 
43' Foi the least force 
for which the box will 
be on the point of mov- 
ing upwards, or that will 
keep the box, after the 

latter has been started, moving upwards with constant velocity, we 

must have (Art. 74) K = 90 - Z = 68 12'. These values in formula 

1 of Art. 74 give 

f = 12,960 sin (5 43' + 21 48') = 6,988 Ib. Ans. 
(5) In this case, K = 90, and formula 1 of Art. 74 gives, noticing 

that sin (90 + Z] = cos Z, 




FIG. 66 



(c) As here H is less than Z, formula 1 of Art. 75 is used, making 
(see Fig 54) K = Y O X' = 90, and sin (Z+ K] = cos Z. 

'-^"^.-^-.^U. An, 

(d) Let x be the height of the coal above the bottom of the box; 
then, W '= 10 X 6 X .r X 54 Formula 1 of Art 74 gives, noticing 
that here F= 4,000 and K = 90, 



W 



8ln (21 



A nrw 
4>00a 



coa 



sin 27 81? 



29 ANALYTIC STATICS 65 

or, substituting the value of ZTjust given, 

10X6X^X54 = 4,000 ^ ^ gp! 
whence, 

^ = ^X^|^|~ = 2481ft =2ft 5fm, nearly. Ans. 

EXAMPLE 2 A piece of rock ABC, Pig 55, lying on the floor of 
a mine Is kept from sliding down by a prop M N Weight of A B C 
is 3 tons; inclination of CA to horizontal, 60; inclination of NM, 
45; coefficient of friction, .75 Required the pressure on the prop. 

SOLUTION The pressure on the prop is equal and opposite to 
the reaction F of the prop The inclination of NM to the horizontal 
being 45, its inclination VO F to the vertical is likewise 45. Here, 
H = 60, theiefore, 
K => YOF = 60 + 45 = 105, sin K = cos 15, cos K = - sin 15 

Formula 4 of Ait 74 gives 

sin 60 -3X cos 60 _ 866 - -f X 5 



EXAMPLES FOB PRACTICE 

1. What force will keep an iron block weighing 4 tons, placed on 
an iron plate inclined at 45 to the horizontal, in a condition of limit- 
ing equilibrium with respect to upward motion: (a) if the force is 
parallel to the plate? (b) if the force makes with the plate an angle 
of 30? Take / - 20. 

NOTE In this case, it is more convenient to use formulas Involving /, rather 

=i 8394T. 
= 3.613 T. 

2 If, in the preceding example, the force is parallel to the plate 
and acts upwards, what must its magnitude be, that the block may 
be on the point of moving downwards? Ans F = 2 263 T. 

3. Taking the coefficient of kinetic friction between steel and pine 
as .16, what is the least force that can keep a steel block having a 
weight of 2 tons moving with constant velocity up a pine plank 
inclined at 20 to the horizontal? Ans F - 1,944 Ib 

4. A block of marble weighing 1,000 pounds is to be kept moving 
with constant velocity up an inclined white-pine plank by a force of 
600 pounds; what must be the inclination of the plank, assuming that 
the best angle of traction is used, and that / = .45? 

Ans, H - 12 SSf 



KINEMATICS AND KINETICS 



COMPOSITION AND RESOLUTION OF 
VELOCITIES 

1. Graphic Representation of Velocity. Velocity, 
like force, is a vector quantity that is, a quantity having 
both magnitude and direction and, like all vector quantities, 
can be represented by a straight line, called a vector (see 
Fundamental Principles of Mechanics] . The vector is drawn 
parallel to the direction of the velocity represented; its 
length is made, to any convenient scale, equal to the mag- 
nitude of that velocity, and 
the arrowhead on the vector 
is placed so that it will 
point in the direction of 
the motion under consid- 
eration. 

In Fig l.let AB be the 
path of a moving point, 
and v the velocity the point 
has when it occupies the 
position P on its path. 
The direction of the motion at the instant considered is that 
of the tangent PTto the curve A B. The velocity v may be 
represented by the vector OM drawn parallel to P T through 
any convenient point. If v is expressed in feet per second, 
and a scale of 5 feet per second to the inch is adopted, the 

Similarly for any 
o 

other scale. 




length of the vector O M should be \ . 

5 



eOPYRIOHTRD Y INTERNATIONAL TIX^BOOK COMPANY, ALL RIHT RBHRVBD 



30 



2 



KINEMATICS AND KINETICS 



30 



2. Parallelogram ol Velocities. Let a body or par- 
ticle O, Fig. 2, be moving in the direction OX, on a flat 
surface A B, with a velocity v, relative to that surface, that 
is, in such a manner that, if the line OX, is fixed on the sur- 
face A B, the particle O will move in that line describing v, 
units of length per unit of time. Let P, be the position of 
the moving particle after the time /. Then, 
OP, = v>t (1) 

While the particle has this motion, let the surface A B move 
in the direction O A' with uniform velocity z/ l( and let A' B' be 






FIG 2 

the position of the surface after the time t, O / being the cor- 
responding position of 0, and O'X a f , parallel to OX,, the 
corresponding position of OX t > Then, 
O0 f = v l t (2) 

The moving particle will now be at PJ t the distance O'PJ 
being equal to P a , or v, t. The figure O O' PJ P. is evidently 
a parallelogram. 

In the same manner, it can be shown that, after a time /', 
in which the moving particle has described the distance 
OQ t along the line OX t) while the surface has moved so 



30 KINEMATICS AND KINETICS 3 

that O" is the position of the starting point O, the final 
position Q t ' of the particle will be the end of the diagonal of 
the parallelogram O O" Q, 1 Q t) in which 

0"Q,' = OQ a = Vm f (3) 

O 0" = Q a Q, f = v, f (4) 

Dividing (2) by (1) gives 



v t OP, a PS 
and dividing (4) by (3) gives 

2i = " 
v, 0QJ 

Equating these two values of , 

v, 

oa = o o" 

OPJ O"Q,' 

According to the theory of similar triangles, the last 
equation shows that O, Q, f , and /V are in the same straight 
line. It follows, therefore, that, at every instant, the mov- 
ing particle is on the straight line O P a f or, what is the 
same thing, that the particle moves in that line. The spaces 
OP,', OQJ described by the particle in times t and /', 
respectively, are to each other as O O r is to O O" that is, 
as z/,/ is to Vt ^, or as t is to t'. Therefore, the motion of 
the particle along OP,' is a uniform motion. If O O" repre- 
sents v lt and OQ a represents v, t the diagonal OQJ will 
evidently represent the space described by the particle in a 
unit of time that is, the velocity v of the particle along its 
path O /V . Hence, the following construction: 

From any point P, draw two vectors PMi and PJbf,, 
representing, to any convenient scale, the velocities z/i and v,, 
respectively. Construct a parallelogram PM l MM* on those 
two vectors. Then will the diagonal PM represent, to the 
scale adopted, the velocity of the moving particle in its path. 

3. It will be observed that the velocity v is determined 
by the same general method used for finding the resultant 
of two concurrent forces. With respect to the velocities v t 
and zr., the velocity v, which is the actual velocity of the 



KINEMATICS AND KINETICS 



30 



particle, is called the resultant velocity, and v t and v t are 
called the components of v in the directions OX t and OX,, 
respectively. Since the particle has the two velocities z^ 
and v t at the same time, these velocities are said to be 
simultaneous . 

The principle of the pavallelogfram of velocities, which 
is similar to that of the parallelogram of forces, may be 
stated as follows: 

// two simultaneous velocities of a particle are represented in 
magnitude and direction by two vectors drawn from the same 
origin, and a parallelogram is constructed on these two vectors, 
the resultant velocity is tepresented in magnitude and direction 
by that diagonal of the parallelogram that passes through the 
common origin of the two vectors, this diagonal being treated a? 
a vector having tJie same origin 

4. The process of finding the resultant of two or more 

simultaneous velocities is called composition of velocities. 

As in the case of forces, any velocity may be considered 

to be the resultant of two velocities in any given directions. 

Thus, the velocity v t 
Fig. 3, may be consid- 
ered as the resultant 
of the simultaneous 
velocities v l and v t in 
the direction OX, and 
OX,, respectively; or 
as the resultant of the 
simultaneous veloci- 
ties vl and v tt ' in the 
direction O AV and 
OX,', respectively, 
etc. To resolve a 

PIG 8 . .^ . . 

velocity into its com- 
ponents in given directions is to find the values of these 
components, or to replace the given velocity with these com- 
ponents, the process whereby this is accomplished is called 
resolution of velocities. 




V*. 



30 



KINEMATICS AND KINETICS 



5. Since velocities are combined and resolved in the 
same manner as forces, all that has been said relating to the 
composition and resolution of forces applies to the composi- 
tion and resolution of velocities. Thus, instead of the par- 
allelogram of velocities, the triangle of velocities is often 
used. In Fig. 2, for example, the resultant v may be deter- 
mined by drawing PM^ to represent z/,, and then MM to 
represent v,, and drawing PM. The magnitude of PMcan 
be ascertained either graphically (by constructing the par- 
allelogram or the triangle accurately to scale), or analytically 
(by applying the principles 
of trigonometry), as in the 
case of forces. 

EXAMPLE 1 A ship is pro- 
pelled by its screw in a north- 
east direction at the rate of 15 
knots, while the current carries 
it due south at the rate of 5 
knots Find the resultant 
motion of the ship (A knot is 
a velocity of 1 nautical mile, or 
6,080 feet, per hour ) 

GRAPHIC SOLUTION Draw 
the north-and-south line NS, 
Fig 4 From any point A and 
to any convenient scale, draw 
A B = 15, making an angle of 
45 with A N. This will represent the velocity of 15 knots toward the 
northeast From B draw B C due south, that is, parallel to N S, and 
equal to 5, using the same scale as before Join A C, and mark the 
arrowhead so that it will be in non-cyclic order with A and B C. 
The length of A C, measured to the scale used for i and v,, will give 
the magnitude of the resultant velocity. The direction is determined 
by measuring the angle NA C with a protractor. Ans. 




FIG 4 



ANALYTIC SOLUTION The tuaugle ABC gives 

i) = A C = V15" + fi" - 2 X 5 X 15 cos 45 = 12 knots 

Also, sin M 



. ain 45 5 sin 45 



v 12 

The angle M, taken to the nearest minute, may be either 17 8' or 
180 - 17 8' = 162 52' As B C, opposite M t is the shortest side 
of the triangle, the value 17 & must be taken. Then, NA C 



6 



KINEMATICS AND KINETICS 



830 



= 45 + 17 ' = (JU H' The ship, therefore, is moving in .1 dm c luui 
N (i2 K' E, at the rate of li! knots, nearly Ans. 

EXAMPLB 2 A river 4 miles wide h.is a turiunt velmilj of .'I milts 
per houi A boat whose paddle wheels can cany it thiough <"> miles 
per houi in still water is to cross the rivei fiom a point ,/, Kig. f>, so 

as to land at a point />' dneitly 
opposite the starting point A, It 
is leqmied to hud: (a) the ilutv- 
turn in which the hoal must lu: 
headed, (ft) the time lequiic-d for 
a trip across the i ivei 

KOIUTION () Snue I he veloc- 
ity of the cm rent and that im- 
parted to the boat by the paddle 
wheels are both uniform, the boat 
must move along the lim; .,-/ // 
with a uniform velocity v (to hi 
determined). Let /'be th posi- 
tion of the boat at any instant, 
and I* Q the diieetum in whieh it 
is headed, and let i\ he thevelonty 
in]]>arted to it in that direotion by 
the paddle wheel. In addition to 
,,the boat has a velocity v t down the stieam, t-qiuil to the vdoeity of 
the current. The velocity v is the resultant of the. velocities f, and v t , 
The triangle PNN lt being right-angled at N, gives 

v = PN = V/ 5 7/; r - //, A/ a V, - r,' i V' - .'! 
= fi.1110 mi. per lir. 
:/, :i 1 , 




Also, sm M - 

(&) The time required is 
AR 4_ 
"" ** 5ilOO "" 



(5 



AIIH. 



nearly. Ant. 



EXAMPLES JTOB PKACTICK 

1. A point has two aimultaneous velocities, nun of 100 fct't pur 
second and one of 200 feet per second. The vectors represent ing 
these two velocities make an angle of 4fi with cueh other. Kind: 
(a] the resultant velocity v ; (b) its inclination M to iho velocity of 
200 feet. (Angles are given to the nearest 10 



2. A balloon moves upwards with a velocity of fiO feet par necund, 
and at the same time the wind carries it la a horizontal direction at the 



30 KINEMATICS AND KINETICS 7 

rate of 20 feet per second Find (a) the resultant velocity z/, (6) its 
inclination M to the vertical . f (a) v = &3 852 ft per sec. 

Ans \ (A) M = 21 48' 1C?' 



6. Absolute and Kelatlve Velocity. Velocity, like 
motion, is always relative, it represents the rate of motion 
of a body with respect to another, and the same velocity can 
have different values according to the condition of the objects 
to which it is referred. Thus, when a locomotive is running, 
the velocity of the piston with respect to the cylinder in 
which it moves is the space that the piston describes in the 
cylinder per unit of time, the velocity of the piston with 
respect to the ground is equal to the velocity of the piston 
with respect to the cylinder added to or subtracted from the 
velocity that, in common with the whole engine, the cylinder 
has with reference to the ground added, if piston and engine 
are moving in the same direction; otherwise, subtracted. 

7. In nearly all practical questions, it is customary to 
refer velocities to the surface of the earth. When the 
velocity of a body is given without any qualification, it is 
usually understood to be the velocity of the body relative 
to the ground. This velocity is customarily, although not 
properly, called absolute velocity; while the term relative 
velocity is restricted to indicate velocity with respect to 
objects that are themselves in motion with respect to the 
ground. Thus, in the example of the preceding article, the 
relative velocity of the piston is its velocity with respect to 
the cylinder; while the absolute velocity of the piston is its 
velocity with respect to the ground. In the case represented 
m Fig. 2, the velocity of the moving particle, relative to the 
surface A J3, is v,; the absolute velocity of the particle is v\ 
the absolute velocity of AB is /. It is obvious that, if a 
body moves on another with a certain relative velocity, 
while the latter is in motion, the absolute velocity of the 
former body is the resultant of its relative velocity and the 
absolute velocity of the other body. 



KINEMATICS AND KINETICS 30 



UNIFORM MOTION IN A CIRCLE 



ANGUXAR VELOCITY 

8. Angular Displacement. Let a point P, Fig. 6, be 
moving uniformly in a circular path of radius r. The center 
T of the circle is C, and the velocity of the 
moving point is v. If the point moves 
from a position P to a position />, in the 
time /, the angle H swept over by the 
radius in passing from the position CP 
to the position CP,. is called the angular 
displacement of P with respect to C. 
Angular displacement may be measured 
PlQi6 in degrees, but is usually measured 

in radians. As explained in Geometry, Part 2, the radian 

measure of the angle H is arc , or 

r 

H (in radians) = arc PP - 
r 

If the length of the arc PP l is denoted by s t 




r r 

Also, s = rff (2) 

9. Angular Velocity. Since the velocity v is uniform, 

s = vt\ that is, rH = vt\ whence 

tf-H/ (1) 
r 

The angular displacement is, then, proportional to the 
time, and - is evidently the angular displacement per unit of 

time, since, when / is made equal to 1 in the preceding 
formula, H becomes equal to -. This displacement per unit 



30 KINEMATICS AND KINETICS 9 

of time is called the angular velocity of the moving 1 point. 
The angular velocity may be defined also as the angle 
described by the moving point in a unit of time. As already 
stated, it is customary to express angular velocity in radians 
per unit of time. It is also customary to represent this 
quantity by the Greek letter to (o-md-ga). Therefore (see 
formula 1), 

= ~ (2) 

10. Halations Between Angular and Linear Teloc- 
ity. The velocity of a moving point in the direction of the 
tangent to its path that is, what in previous articles has 
been called the velocity of the point is often called linear 
or tangential velocity, in order to distinguish it from 
angular velocity. When the word velocity is used without 
any qualification, linear velocity is meant The velocity in 
the direction of the path, especially when the latter is a circle, 
is also called circumferential velocity. 

From Art. 9, the following' obvious and very important 
relation between linear and angular velocity is obtained: 

-* (1) 

r 

v = 1 to (2) 

If two points move in two circles of radii r^ and r, with 
the same angular velocity to, and linear velocities v and v,, 
then, v l = r, to, and v, = r, CD; whence, by division, 

= (3) 

V, r, 

that is, the linear velocities of two points moving in different 
circles with the same angular velocity are directly proportional 
to the radii of the respective paths. 

11. If the linear velocities are the same, and the angular 
velocities are to,, and to,, we must have v = r w, = r a to,; 
whence, 

^i = TJL 
to, r* 

that is, for the same linear velocity \ the angular velocities of two 
moving points are inversely as the corresponding radii of the paths. 

I LT 398-13 



10 KINEMATICS AND KINETICS 80 

12. If, in Fig. 6, the radius CP is imagined to move 
with P t all the points in CP will evidently have the same 
angular velocity, since they will all describe the same angle 
in the same time. If the linear velocity of the point P', 
situated at unit's distance from the center, is v r , then, 

v f = 1 X a) = of 

The angular velocity of P may, therefore, be defined as 
being the linear velocity of, or the length of the arc 
described in a unit of time by, a point on the radius CP sit- 
uated at unit's distance from the center, when the radius is 
considered as moving with the point P. 

13. Angular Velocity In Terms of Number of Revo- 
lutions Per Unit of Time. In practical engineering prob- 
lems, it is customary to state the velocity of circular motion 
in terms of the number of revolutions per unit of time 
(usually per minute); that is, the number of times that the 
moving point passes over an entire circumference in a \init 
of time. If this number is denoted by n, and the radius of 
the circle by r, the space passed over by the point in a unit 
of time is 2 n r X n; hence, for the linear velocity of the 
point, we have 

v = 2nrn 
and for its angular velocity (formula 2 of Art. 10), 

CD = - = 2;rw 
r 

It n is the number of revolutions per minute, the veloc- 
ities v and oj, referred to the second, are: 



Conversely, if v or CD, referred to the second, is given, 
the number of revolutions per minute is given by the 
formula 

n = ^ = 9.5493 ^ = 9.5493 o> (3) 

nr r 

14. An$rular Velocity of a Rotating Body. When- 
ever any body, as a wheel, revolves about a fixed axis, every 



30 KINEMATICS AND KINETICS 11 

point in the body describes a circle whose radius is equal to 
the distance of the point from the axis. Since all points 
revolve through the same angle in the same time, they all 
have the same angular velocity. This common angular 
velocity is called the angular velocity of the rotating body. 

EXAMPLB 1 A drum W, Fig. 7, whose radius J? is 1.5 feet is fixed 
to a revolving shaft 5". The motion of W is transmitted to another 
shaft s by means of a 
belt B B passing around 
the drum W and a drum 
w fixed on the shaft s. 
If the velocity of the belt 
is 10 feet per second and 
the drum w makes 100 
revolutions per minute, 
it is required to deter- 
mine the number N of 
revolutions that S or W makes per minute, the radius r of the drum w, 
and the angular velocity of the two drums in radians per second. It 
is assumed that the circumferential velocity of each drum is the same 
as the velocity of the belt 

SOLUTION The circumferential velocities of the two drums, in feet 
per second, are, respectively, g^ and "on- (Art 13). Since each 
of these velocities is equal to the velocity of the belt, 




From (1) we get (see also formula 3 of Art. 13), replacing Jf by 
its value 1.5, 

N - 9.6493 X ~ = 63 602 A ^ s - 
1.5 

And from (2) (see the same formula) , 

r - 9.5493 X -^ - .95493 ft. Ans. 

For the angular velocity Wj (radians per second) of W, formula 1 of 
Art. 1O gives 

Wt *" 3? * n " 6 6667< AnSi 

And for the angular velocity <a, of w (formula 2 of Art. 13). 
ta t - 2 - 10.472. Ans. 



12 



KINEMATICS AND KINETICS 



30 



EXAJMPLB 2 Two drums D and ZX, Fig 8, of radii r and r 1 feet, 
are fixed on a shaft 55 A rope M N L Q is wound around the drums 
in opposite directions, so that when MN rises and winds around D t the 





FIG 8 

portion descends by unwinding from ZX. The rope passes around 
a pulley P carrying a weight W. It is required to find the motion of 
the weight W when the shaft 5 5 revolves at the rate of revolutions 
per minute 

NOTE This arrangement is known by the various named of Alfferoutlal 
windlass, differential -wheel and axle, and Chinese wheel and axle, 

and is used to obtain a slow motion of IV, which, under some conditions, is n 
mechanical advantage 

SOLUTION. It is assumed that every turn of the rope around the 
drums is a circle, and that M N and Q L remain parallel. This is near 
enough to the actual conditions occurring in practice 

Let the shaft turn in the direction indicated by the curved arrows. 
Then N M will rise and Q L will descend For the angular velocity 
of the shaft and the drums (formula 2 of Art 13), we have 
ta = 10472 n radians per sec. Let v and v 1 be the velocities otNfif 
and QL t or of M and Q, respectively, m feet per second. Then (for- 
mula 2 of Art. 10), 

v sa r ta = 10472 n r 

v> s* #-/ at .10472 nr 1 



30 KINEMATICS AND KINETICS 13 

During any time f, the center O of the pulley, and therefore the 
weight W, moves through a distance OO lt which is easily determined. 
The length of the rope between M and Q is reduced by an amount 
equal to NNi + L L t = 2 OOi But, as during this time a portion v / 
of the rope has wound about D, and a portion v' / unwound from -Z7, 

2 O O l = v t - v' f, 

whence O Oi = i( v')t 

Therefore the point O and the load W move uniformly with a 
velocity z/a, equal to \(v v 1 } Substituting the values of v and j/ 
found above, 

Vw = \(r -r')ta = t( 10472 n r - .10472 n r') 
- 05236 n(r - r 1 ) Ans 



EXAMPLES FOR PRACTICE 

1. A flywheel 6 feet in diameter makes 76 revolutions per minute 
Find 1 (a) its angular velocity ru, in radians per second, (b) the veloc- 
ity v of a point on the rim, In feet per second A _ / (a) u> = 7 854 

Ana X (d) v - 23.562 

2. A point revolves in a circle 8 feet in diameter with an angular 
velocity of 4 5 .radians per second Find: (a) its linear velocity v, 
(6) the number n of revolutions it makes per minute. 

v ~ 18 ft< P er sec> 
ft - 42972 

3. Two drums A and .Z? on parallel shafts are driven by a belt 
traveling at the rate of 1,000 feet per minute; the radius of A is 5 feet, 
and B makes 300 revolutions per minute. Find, (a) the number n of 
revolutions that A makes per minute; (d) the angular velocities io a 
and <y* of the two drums, m radians per second, (c) the radius r of B 

\ (a) n = 31.831 

Ana J W/ w " 333S 
Ans ' d H<w, = 31.416 
(c} r - .531 ft. 



CENTRIPETAL AND CENTRIFUGAL FORCE 

15. Restatement ol the Law of Inertia. A provi- 
sional statement of the law of inertia was made in Funda- 
mental Principles of Mechanics. Now that the theory of the 
center of gravity has been explained, the law can be expressed 
in a more complete and definite manner as follows: 

// a body is under the action of no force t its center of gravity 
is either at rest or moving uniformly in a straight line, 



14 



KINEMATICS AND KINETICS 



30 



16. Centripetal Force. Let a body be- moving in such 
a manner that its center of gravity G, Fig. 9, travels uni- 
formly in a circle A B of radius r. As examples of this kind 
of motion may be mentioned a car moving m a curved track, 
and the balls of a ball governor. According to the law of 
inertia, such motion cannot take place unless the body is 
acted on by unbalanced forces, for, were the body under the 

action of no force, 
the path of its center 
of gravity would be a 
straight line instead 
of a circle It can be 
shown by the use of 
advanced mathemat- 
ics that, in order to 
preserve this motion, 
the body must be con- 
stantly acted on by a 
force P whose line 
of action is directed 
toward the center of 
the circle and passes 

through the center of gravity of the body. This force, 
which is exerted on the revolving body, is called centrip- 
etal force. 

If the mass of the body is denoted by m, and the linear 
velocity of its center of gravity by v, the magnitude of the 
centripetal force P is given by the formula 




FIG 9 



P = 



m v 



W 



Since m = , where W denotes the weight of the body, 
and g the acceleration of gravity, we have also, 

P-Z& (1) 
gr 

In applying this formula, v should be expressed in feet 
per second, and r in feet, since g is referred to the foot and 
second as units. 



30 KINEMATICS AND KINETICS 16 

If, instead of v t the angular velocity a) is given, we have, 
from Art. 10, v = rto, v* = r* a)'. Substituting this value in 
formula 1, and reducing, 



To express P m terms of the number n of revolutions per 
inute, we have (Art 

stituting in formula 2, 



minute, we have (Art. 13), <a ^ , and, therefore, sub 

30 



If the revolving body is acted on by any system of forces, 
their resultant must pass through the center of gravity of the 
body, be directed toward the center of the circle, and have 
the magnitude given by any of the preceding formulas. 

17. Illustrations of Centripetal Force. A familiar 
example of centripetal force is afforded by the circular motion 
of a ball tied to a string and swung around, the other end of 
the string being held in the hand. The ball constantly tends 
to move along the tangent, or "fly off on the tangent," and 
would do so, if it were not kept in the circular path by the 
pull of the string. 

It should be noticed that, while the centripetal force is 
always directed toward the center of the circle, it is not 
necessarily exerted from that center; in other words, the 
body to the action of which the centripetal force is due does 
not need to be at the center. In the case of the ball given 
above, the hand, which exerts the centripetal force, is placed 
at the center of the circle. In the case of a train moving on 
a curved track, on the contrary, the centripetal force is] the 
resultant of the weight of the car and the pressure of the 
rails on the wheels; and this force, although directed toward 
the center of the curve, is exerted at the circumference. 

18. Centrifugal Force. By the law of action and 
reaction, a particle (or body) moving in a circle must exert 
on the body to whose action the centripetal force is due a 
reaction equal and opposite to that force. This reaction is 



16 KINEMATICS AND KINETICS 30 

called centrifugal force. Thus, in the case of the ball 
considered in the preceding article, the pull exerted on the 
hand through the string is the centrifugal force exerted by 
the ball on the hand. 

Suppose the source of the centripetal force P, Fig. 9, to 
be at the center C of the circle. The force P that acts 
on the body G may be represented by the vector G M, 
while the centrifugal force Q, which is exerted by the 
body G, may be represented by the equal and opposite 
vector C N. 

19* Caution Against Some Common Misconcep- 
tions Regarding Centrifugal Force. The following 
facts must be clearly understood, as this subject is one 
about which students (and even teachers) often fall into 
very gross errors: 

In studying' the motion of a body in a circular Path, the 
centripetal, not the centrifugal, force must be considered as 
the only force acting on the body. In Fig. 9, the body G is 
under the action of the force P, not under the action of the 
force Q. 

A body moving in a circular path has no tendency to move 
along the radius, its tendency being to move along the tangent. 
In Fig. 9, the body G, if left to itself, would move along G T, 
not along G C'. 

A body moving in a circle is at every instant under the action 
of an unbalanced force (the centripetal force}; as, otherwise, 
the center of gravity of the body would move in a straight 
line. 

Although the centripetal and the centrifugal force are 
collinear, equal in magnitude and opposite in direction, they 
do not act on the same body, and cannot, therefore, balance 
each other. It should be kept in mind that, when it is said 
that two or more forces balance, the meaning of the state- 
ment is that they produce no motion on the body on which they 
all act; and, if several forces act on different independent 
bodies, it cannot be said that they form either a balanced or 
an unbalanced system, as, under such circumstances, the 



30 



KINEMATICS AND KINETICS 



17 



expressions balanced system and unbalanced system have 
absolutely no meaning. 

EXAMPLE A ball J3 t Pig 10, whose weight is W, is suspended by 
a string of length / from a point O, and made to revolve uniformly in 
a horizontal circle MN, at " 

the rate of n revolutions per 
minute It is required to 
determine (a) the radius r 
of the circle M N, (b) the 
centripetal force P acting on 
the ball, (c) the tension T 
m the string 

NOTE This contrivance is 
called a conical pendulum, 

and its operation Is very similar 
to tbat of the ball erovomor of an 
engine 

SOLUTION (a) The two 
forces acting on the ball are 
the weight Wai the ball and 
the tension 7"of the string. 
According to the principles 
stated in the preceding 
articles, the resultant P of 
these two forces must be 
directed toward the center C 

of the circle, and have the magnitude given by formula 3 of Art 
that is, 



In the triangle BED, 




900 jr 
P - W tan V, 



or. since tan V = tan B O C 



(2) 



Equating the second members of (1) and (2), 
Wjjt'v^ = _Wr_ . 
~900> " <>lP-~r*' 

whence V/^ 

Squaring and solving for r, 






18 KINEMATICS AND KINETICS 30 

(6) Substituting in equation (2) the values ]ust found for V/ 1 r* 
and r, 



D __ 

90Q.g- ~ 900 



, . 

~ * Ans 



In the triangle BDE, 

T= ED 



EXAMPLES FOR PRACTICE 

1 A ball weighing 50 pounds revolves uniformly on a smooth 
horizontal surface, about an axis to which the ball is connected by a 
steel rod (a) If the rod is 10 feet long and the ball makes 00 revolu- 
tions per minute, what is the tension in the rod? (&) If the greatest 
tension that the rod can stand is 2,500 pounds, what is the greatest 
number of revolutions per minute that the ball can make without 
breaking the rod? / t a \ 1,381 Ib. 

Ans I W 121 

2 The height h (= CO, Pig 10) of a conical pendulum Is 2 feet. 
How many revolutions per minute can the pendulum make? (Observe 
that the number of revolutions is independent of the weight of the 
pendulum and of the length of the suspending string.) Ans. 38.3 

3 If, in example 2, the weight of the pendulum is 10 pounds and 
the length of the string 16 feet, find (a) the centripetal force P\ 

(5) the tension T in the string. . / (a) 70 4 Ib 

Ans.< ) 801b< 



MOTION OF A TRAIN ON A CURVED TRACK 
20. General Theory. In Fig. 11 are represented the 
rails Ri and R, of a curved track, and the flanges F^ and F t of 
two directly opposite wheels of a car. The inclination H, or 
^i R t N, of the track to the horizontal is shown very much 
exaggerated for the sake of clearness. The weight W/that 
comes on the two wheels acts through G, at a height a above 



30 



KINEMATICS AND KINETICS 



19 



the track equal to the height of the center of gravity of the 
car. The difference NR^. in elevation between the outer rail 
Rt and the inner rail R t is called the superelevation of the 
outer rail When the car is moving uniformly in the curve, 
G moves m a horizontal circle whose radius is practically the 
same as the radius of the center line of the track. According 
to the theory of centripetal force, the resultant P of all the 
forces acting on that part of the car whose weight is carried 
by the two wheels here considered must pass through G 




and be directed toward the center of the circle in which G 
moves; that is, P must be horizontal. For convenience 
all forces will be resolved into components parallel and 
perpendicular to R^R t . The components of P in these 
directions are denoted by X and Y t as shown in the tri- 
angle G DE. 

The forces acting on the car (or on the part here con- 
sidered) are W, which is resolved into the component X, and 
y., and the pressures of the rails. The components of these 



20 KINEMA ^ICS AN ) KINETICS 30 

pressures perpendiculs r 'to R l R, e re denoted by Y l and K,; 
the component paralle to or alon^ RI R, is denoted by X'. 
If, as shown m the figure, X 1 acts toward R a , the pressure is 
exerted by the rail R! on the flange ft; if X' acts toward R lt 
the pressure is> exerted by R, on the flange F,. In the gen- 
eral derivation of the formulas, X' will be treated as acting 
toward R a ; if, in any particular case, the value of X 1 is 
found to be negative, this means that X 1 acts in the oppo- 
site direction, and that it is, therefore, exerted by the lower 
rail R,. 

Let v = linear velocity of car, in feet per second, 
e = superelevation of outer rail, in feet; 
b = R^R, = distance between centers of rails, m feet; 
r = radius of center line of track, in feet. 
According to the principles of statics, we have, since P 
is the resultant of the forces acting on the car, 

X' + X = X (1) 

Y* + y> - Y, = Y (2) 
From (1), X' = X-X, (3) 

Now, 



X = P cos H = cos H (Art. 16) = X 

gr gr 



gr b 



R l R a b 
These values in (3) give 

X ' = ~(^F-e*-e} (D 

b \gr ] 

To find Y lt moments are taken about R a . Since the 
moments of X' and Y, about R, are each zero, the algebraic 
sum of the moments of Y 1 and W must be equal to the 
moment of /*; that is, 

Y*b- WXKR, = P^GK (4) 

The figure gives, MI being horizontal and MS vertical, 
KR* = JR t cos H = (MR, - MJ) cos H 

= (i6- MG tan H) cos H = i b cos H- a sin H 



30 KINEMATICS AND KINETICS 21 



Substituting these values in (4), writing -^-^- for P, and 

gr 
solving for Y 1} 

(2) 



The value of Y, is found from equation (2) by substi- 
tuting the expressions for the values of Y, Y^ and Y a . The 
result is 



Y, = ** - *- cos H+ iA + a sin tf (3) 

b L\ "/ \ gr I J 

21. Car Moving Without Exerting Lateral Pres- 
sure. The pressure A'' exerted between the rail and the 
flange increases with the velocity v, as is evident from 
formula 1 of the preceding article. As this pressure is very 
injurious both to the rail and to the wheels, the track is so 
constructed as to eliminate it, that is, the superelevation e is 
so made that, for the greatest velocity of trams moving on the 
track, the pressure X' shall be zero. Putting the second 
member of formula 1 of the last article equal to zero, 



whence, squaring and solving for e, 

b^, 



(1) 



Usually, (---) is a very small fraction. If this fraction is 
\ffr/ 

neglected, the following approximate formula is obtained: 

e = b ?*- (2) 

gr 

22. When A'' is zero, the algebraic sum of the moments 
of Yi and Y, about G is zero, since the moments of both 
W and P are zero; and, as the lever arms of Ki and Y, are 
equal, it follows that Y l == F a . Also, since the resultant P of 
Yi, K B> and W is horizontal, the sum of the vertical compo- 
nents of yi and X, must be numerically equal to W, and 



22 KINEMATICS AND KINETICS 30 

the sum of their horizontal components must be equal to 1 
If, therefore, a horizontal line A Q is drawn through A, the 
vector Q G will represent Xi + Y,, and the vector O si 
will represent the horizontal component of Y l + y a , this 
component being equal to P It is thus seen that, in this 
case, the centripetal force P is obtained by resolving the 
weight W into two components one, GQ, peipendicular 
to the track, and one, GJS, horizontal: GQ simply bal- 
ances the pressure Y l + Y, of the rails, while GE is the 
centripetal force. 

23. Car on the Point of Upsetting About the 
Outer Wheel. If the velocity increases sufficiently, the 
pressure X' will become so great that the car will upset by 
turning about the rail R-,.. When the car is on the point of 
upsetting, the wheel F, is on, the point of being lifted from 
the rail J? at and there is, consequently, no pressure between 
the wheel and the rail at J? 3 . The relation that exists 
between e and v for this condition is obtained by making 
Y a = in formula 3 of Art. 20, expressing cos // and 
sm H in terms of e and b, and solving for v or e. The pioe- 
ess is comparatively complicated, and will not be given 
here. There is, however, an approximate solution that is 
sufficiently close for practical purposes. In railroad work, 
the angle H is always very small, and its cosine can, with- 
out any appreciable error, be taken equal to 1. Putting the 
second member of formula 3 of Art. 20, equal to zero, and 

writing 1 for cos H and 7 for sin H (see Fig, 11), 

b 



and 



. 
gr \ gr I b 



d(2a 



EXAMPLE 1. A curved track is to be constructed for a maximum 
train velocity of 45 miles per hour. If the radius of the curve is 
2,600 feet, and the distance between centers of rails 6 feet, what must 



30 KINEMATICS AND .KINETICS 23 

be the superelevation of the outer rail, that there may be no lateral 
pressure? 

SOLUTION Formula 2 of Art. 21 will be used Here, b 5; 

r = 2,500, g 32 16, and v = ' = 66 ft per sec. Therefore, 

66 a 

e = 5 X Q4> , A _ Knn = .271 ft = 3i in , nearly Ans . 
32 16 X 2,500 (_ 

EXAMPLE 2 With the track constructed as m example 1, what is 
the maximum velocity a train can have without upsetting, the height 
of the center of gravity of each car above the track being 6 feet? 

SOLUTION The formula m Art. 23 gives 

.-f *" + **-** persec . 

= 134 mi. per hr., nearly. Ans. 



EXAMPLES FOR PRACTICE 

1. A curved track of 1,600 feet radius is to be built for a maximum 
tram velocity of 40 miles per hour, the distance between rails is 5 feet 
What must the superelevation be, that there may be no lateral 
pressure? Ans 535 ft 

2 The radius of a curved track being 1,500 feet, tne superelevation 
3| inches, and the distance between, the centers of rails 5 feet, what is 
the maximum velocity a tram can Jiav> without exerting any lateral 
pressure? Ans. 36.17 mi. per hr. 

WORK AND ENERGY 

WORK 

24, Definition of Work. Let a force F, supposed to 
remain constant in magnitude and direction, act on a body 
while the point of application of the force undergoes a 
certain displacement. This displacement does not neces- 
sarily have the direction of the force, nor is the force F 
necessarily the only force acting on the body. Let the pro- 
jection of the displacement on a line parallel to the direc- 
tion of the force be denoted by s. Then, the product Fs is 
called the work of the force F for the given displacement. 

Thus, in Pig. 12, if the point of application of the force F 
moves from A to A f , its displacement is A A'\ the projection 
of this displacement on a line parallel to the line of action 




24 KINEMATICS AND KINETICS 30 

of the force is AS, A' S being perpendicular to the direc- 
tion of the force. In this case, then, s = A S, and the 

t work of the force, while 

B / the point of application 

/^X S moves from A to A 1 , 

*\^ fr, is FX AS. The pro- 

/ jection A S is called the 

MT component of the dis- 

f ,--*'* placement A A' parallel 

to the line of action of 
the force When, as in 
Fig. 12, the projection 

Pl - 12 of A' on A S moves in 

the same direction as the force, the force is said to do work 
on the body on which it acts When, as in Fig 13, the pro- 
jection of A 1 on A S moves in a direction opposite to that of 
the force, the body is said to do work against the force, or the 
force is said to do the negative work FX AS on the body. 

If the line of action of the s ^ 

force F, Figs. 12 and 13, coincide / """"--^^ 

with the path A A' of the point / "~"^'7^ 

of application, then s = A A'. If / /' \ 

the point of application moves / 

in the direction of the force, the / 

work of the force is F X A A 1 } / 

otherwise, Fx A A'. / > 

If the line of action of F is / /' 
perpendicular to A A f , the projec- 
tion A S reduces to the point A, 
and F X A S = F X = 0; that 
is, the work done by the force 
is zero. This shows that, when ' Fr0tW 

the point of application of a force moves in a direction per- 
pendicular to that of the force, the force does no work, 

25. Formulas for Work. Denoting the work of the 
force F, Fig. 12, by U, we have 

U=FXAS (1) 



30 KINEMATICS AND KINETICS 25 

If Fis resolved into two components Tand N, the former 
along A A', and the latter perpendicular to A A', then, 

T 

T = /''cos M, and F - - 



This value in (1) gives 



cos y// cos M 

This formula shows that the work of a force can be obtained 
by multiplying the component of the force in the direction of the 
displacement of its point of application, by the length of that 
displacement. 

26. Work of the Resultant of Several Forces. 

When several forces act on a body, the algebraic sum of their 
works is equal to the woik of their resultant. This is 
an immediate consequence of the principle stated in the 
preceding article. Let F he the resultant of several forces, 
say F^ J 7 ,, F a . Let the body on which these forces act move 
in any direction through a distance x, and let the correspond- 
ing work of the forces be U, /i, /, /. Denoting the com- 
ponents of the forces in the direction of the displacement by 
X, X lt A' a , X 3 , we have, according to the preceding article, 
U = Xx, U, = X t x, U* = A',*, U, = A'.*. Therefore, 
#++/. = (X t + Xn + X.)x 

According 1 to the principles of statics, Ji 4- X, + X a = X\ 
therefore, 7 X + /, + U, = Xx = U, as stated^ It should 
be remembered that the resultant F'\& a force that can replace 
the forces F^F*, F,, not a force acting simultaneously with 
them. 

Since, when seveial forces are in equilibrium, their result- 
ant is zero, it follows that, if a body is moving under the 
action of balanced forces (in which case the body must be 
moving with uniform velocity), the algebraic sum of the 
works of the forces for any displacement of the body is equal 
to zero. 

27. Effort and Resistances. Usually, forces are 
applied to bodies in order to produce or to prevent motion. 
The forces thus applied are called efforts, and those that 

I L T 398-14 



26 KINEMATICS AND KINETICS 30 

oppose them are called resistances. Thus, when a train is 
moved by a locomotive, the pull of the locomotive is the 
effort; the friction of the wheels and the resistance of the 
air are the resistances. Generally, when the work done on 
a body is referred to, the work done by the effort is meant, 
although, in some cases, its value is determined by calcula- 
ting the work of the resistance. Thus, if a weight W is 
raised through a height h, the work of the resistance W 
is numerically Wh> which, if the body comes to rest at the 
end of the distance h } is also the work of the effort. 

28. Unit of "Work. As work is measured by the prod- 
uct of a force and a distance, its numerical value depends 
on the units of force and length employed. The unit of 
work may be defined as the work done by a force equal to 
the unit of force acting through a distance equal to the unit 
of length. If the unit of force is the pound, and the unit of 
length is the foot, work is expressed in foot-pounds; if the 
unit of force is the ton, and the unit of length is the inch, 
work is expressed in inch-tons; etc. 

29. Work Done In Raising a System of Bodies. 

It can be shown by the use of advanced mathematics that 
the work performed in raising a system of bodies is equal to the 
aggregate weight of all the bodies multiplied by the height 
through which their center of gravity is raised. Thus, if stones 
piled in any manner at the bottom of a tunnel shaft are 
raised and formed into a pile on the surface, the work per- 
formed is equal to the combined weight of all the stones 
multiplied by the vertical distance between the centers of 
gravity of the two piles. The stones may be raised one or 
more at a time and deposited on the surface in any manner 
whatever; the work will always be the same. 

EXAMPLE 1. The average pull of a locomotive on the train being 
7 5 tons, what is the work done by the locomotive in hauling the 
train 1 mile? 

SOLUTION. Here F = 7 5 T., J = 1 mi. = 5,280 ft Therefore, 
V - 7.6 x 6,280 = 39,600 ft. -tons. Ana. 



30 KINEMATICS AND KINETICS 27 

EXAMI'LB ->. fiiven an inclined plane CD, Pig 14, 500 feet long 
and inclined to the horizontal at an tingle //of .SO", it is reqtmed to 
determine the work necessary tn just pull up from C to D a block 
whose weight W is SiO pounds The coefficient of friction / between 
the block and the plane 
is 2fi 

SOIUTION The re- 
sistances opposing the 
motion of the block are 
the weight W and the 
friction P The work U 
of the effort is numeri- 
cally equal to the sum 
of the works of those re- 
sistances Resolving Jf 
into two components A r and T, the foimer noinial to the plane and the 
lattei along the plane, \ve have (.see Analytic Statics}, T = Wsln H, 
N = H r cos //, /' = fN = fW cos //. The work done against I\ 
in foot-pounds, is /' X CD fiOO / J , and that done against W is 
7'X C/) - . r >(H) T. Therefore, 
U -= f)(K) T + r>(H) /' = fiOO (T + P) = 5(M) ( W sin // + / W cos //) 

= fi(K) W (sin U + /cos //) = fi()0 X SX) (iln + ,2T> X cos M ) 
= HH.rtM ft -Ib. Ans. 

EXAMPI.K U. A cylmdnctil cistern 10 feet in diameter is filled with 
water to u distance of fiO feet above the bottom. What work must be 
done in pumping all the water into a tank l> r > feet in diameter, whose 
bottom is 1(X) foot above the bottom of the cistern? The weight of 
water is taken ocjual to (12.5 pounds per cubic foot. 

SorimoN. The- volume F"o the water to be raised iu 




(* x 4 10 'x>) 



cu. ft. 



I'or the wuSght of thiw volume wo have 

IT V 1(1" 

W = l r X (JB.fi , X 50 X 02.5 

i 

If y is the height of the water In thu tank after pumping;, 

r , it X ir> 
V*> 4 y 

4 X * X J ' X nO 
4 ^ * x 4 X W 10" X 50 

whence * - x 1B . - ff x 1B - ' - lfi 

Tne rtiHtancu of the center of gravity of the water In the clBtern 
from the bottom of the cistern is 1'5 ft. The distance of the center 

y 
>f gravity of the water in the tauk from the bottom of the tank ia ~. 



28 KINEMATICS AND KINETICS '30 

*or the distance h between the two centers of gravity we have, 
therefore, 



Then (Art 29), 

U=Wh = 1L ^~ X 60 X 62 5 (?5 + ^^) = 21,135,000 ft -Ib 
4 N J X 15 I Ans. 

NOTE In practical problems, it Is seldom necessary to use moie than five 
significant figures 



POWER 

30. Work, as defined in Art. 24, has no reference to 
time. The work performed in raising a given weight through 
a given height is the same whether the time occupied is a 
second or an hour. However, when the capacities of diffei- 
ent agents for doing work are to be compared, time must be 
considered; and to indicate the amount of work performed in 
a given time, or the rate of doing work, the term power is 
used. If an engine does in an hour twice as much work as 
another engine, it is said to have twice the power of the 
second engine. 

31. Horsepower. Power may be expressed in tenns 
of any convenient units, as foot-pounds per second, foot- 
tons per minute, etc. Thus, if an engine woiking uniformly 
raises 40,000 pounds through a distance of 10 feet in 1 min- 
ute, the work it performs per minute is 400,000 foot-pounds, 
and its power may be stated as 400,000 foot-pounds per min- 
ute. The foot-pound per minute is too small a unit for many 
purposes, and a larger unit is more generally used by engi- 
neers. This unit is called the horsepower, and is equal to 
33,000 foot-pounds per minute, or 550 foot-pounds per 
second. The abbreviation for horsepower is H. P. 

Let F = effort (or resistance), in pounds; 

s = distance, in feet, that point of application is 

moved; 

/ = time occupied, in minutes; 
t, = time occupied, in seconds; 
ff = horsepower. 



30 KINEMATICS AND KINETICS 29 

TT PS Fs 



8 

EXAMPLE 1 What horsepower is required to pull a tram weighing 
400 tons at a speed of a mile per minute, if the resistance at this speed 
is 12 pounds per ton? 

SOLUTION The force F necessary to overcome the resistance is 
400 X 12 = 4,800 Ib s = 6,280 ft Substituting In the foimula, 
4,800X5,280 _ 
H= 33,000 X~r = 768H P ' AnS 

EXAMPLE 2 A crane hoists a load of 5 tons a height of 22 feet m 
20 seconds What horsepower is developed? 

SOLUTION The resistance F is 6 X T= 10,000 Ib.; 20 sec. = i mm 
Substituting in the formula, 

10,000 X_22 H 
H ~ 33,000 X * ** H P 
10,000 X 22 _ 

H = ~~- = 2 H p 



ENERGY 

32. Kinetic Energy. Let a body of mass m and 
weight W be moving 1 with a velocity v a . If a force F is 
applied to the body in a direction opposite to the direc- 
tion of motion, that force will bring the body to rest in a 
distance s, such that 

F s = a 1 m V* ( 1 ) 

This formula, which should be committed to memory, is 
obtained from the following, derived in Fundamental Prin- 
ciples of Mechanics 

F^'Jii^-J^l (2) 

2s 

In the present case, v = 0, and, therefore, 



F = ~~, Fs = - i m v? 
2s 

The negative sign simply indicates that F and v have 
opposite directions. When only numerical values are con- 
sidered, the sign is disregarded, and the equation written as 
in formula 1 . 

Notice that Fs is the work done by the body against the 
force in the space j. Whatever the force may be, this work 



30 KINEMATICS AND KINETICS 30 

is always the same, since it is always equal to i m v a ' It 
follows, then, that the body can, on account of its velocity 
and mass, perform an amount of work numerically equal 
to a" m v,". 

33. The capacity that an agent has for performing work 
is called enei-gy. The energy that a moving body has, on 
account of its velocity and mass, is called the kinetic 
energy of the body; and, as ]ust explained, is measured by 
the product -3 m v a *. Since this quantity is equivalent to 
work, it is expressed in units of work, such as foot-pounds 
or foot-tons 

Denoting the kinetic energy of a moving body by jR" t and 

replacing m by (see Fundamental Principles of Mechanics] , 

g 
formula 1 of Art. 32 may be written, 



Also, Fs = ^p- (2) 

EXAMPLE Find the kinetic energy, m foot-tons, of a car weighing 
25,000 pounds, and moving at the rate of 30 miles an hour. 

SOLUTION Here, W = 25,000 Ib = -^ tons. The velocity of the 
tram, in feet per second, is 

30X5,280 , . 

60 X 60 ^ ^ V ' 
Therefore (formula 1), 

K = \ X |^- X 44* 376 24 ft -tons. Ans. 

NOTU The velocity has been reduced to feet per necond, because jf In expressed 
in feet per second We might also find v In feet per hour and reduce jf to feet per 
hour In all cases, it is necessary to refer v andjr to the same units, 

34. Equation of Energy. From formula 2 of Art. 32, 

we have, leplacing m by , 



Fs = (z>* zJo") (1) 

2.P" 

This formula gives the work performed by an unbalanced 
force in changing the velocity of a body from z> to v, or its 



30 KINEMATICS AND KINETICS 31 

kinetic energy from v a " to v". If the force F is the 
difference between an effort P and a resistance R, then, 

(P R} s = (v* v ") 
*f 

whence Ps = R s + -^ ( - v, a ) (2) 

2.T 

This formula is called the equation of energy, and may 
be stated thus: 

The work done by the effort zs equal to the work done against 
the resistance plus the increase of kinetic energy. 

EXAMPLE In hoisting coal from a mine, the load to be hoisted, 
including cage and car, is 12,000 pounds, the load starts from rest, 
and when it is 50 feet from the bottom, it is moving with a velocity of 
30 feet per second What is the pull in the hoisting rope? 

SOLUTION. The resistance R is the weight, 12,000 Ib. The effort P 
is the unknown pull in the rope, v, = 0, v = 30 ft per sec., and 
s = 50 ft Substituting in the formula, 

PS = tfj + |(- -.-)_ 12,000 x so + 2 x 32i6 (30> " QI) 

= 767,910 ft.-lb. 
767,910 



50 



15,358 2 Ib. Ans. 



NOTE In the solution of this example It is assumed that, when the load has 
reached a height of 60 feet, it is still being accelerated at the average rate of accelera- 
tion necessary to fflve it a velocity of 30 feot per second in a space of 60 feet When 
the velocity becomes uniform, the pull in the rope is that due to the load only, since 
there la then no Increase in the kinetic energy. 

35. Potential Energy. As has been explained, a body 
may have a capacity for doing 1 work by reason of its velocity. 
There are, however, other states or conditions than that of 
motion that give a body a capacity for work: (1) Suppose 
that a body of weight W is raised through a height h\ in 
thus raising the body, the work Wh is dons against the 
force W. If, now, the body is permitted to descend through 
the same vertical distance h, the weight W becomes the 
acting force, and the work Wh is done by it. The body in 
its highest position has the work Wh stored in it, and thus 
possesses a stock of energy equal also to W h. (2) Suppose 
that a spring is extended or compressed; work is done 
against the resistance of the spring to change of lengtH. 



32 KINEMATICS AND KINETICS 30 

If the spring is released, it will return to its original form, 
and in so doing can do precisely the amount of woik that 
was expended on it. Thus, in the extended or compressed 
state, there is an amount of work stored in the spimg, and 
the spring possesses energy because of its sti etched 01 
compressed condition. Similarly, compressed air possesses 
energy merely because it is compressed and can do work in 
returning to its original state. 

The energy that a body possesses by reason of its position, 
state, or condition is called potential energy. 

36. It is to be remarked that the potential energy that a 
body has, due to its position, is relative to a certain plane 
below the body. Thus, if the body weighs W and its 
height above the ground is h, its potential energy with 
respect to the ground is Wh With lespect to a plane 
whose distance below the body is h lt the potential energy 
is 



37. Conservation of Enerjyy. The law of the con- 
servation of eneigy asserts that energy cannot be destroyed. 
When energy apparently disappears, it is found that an 
equal amount appears somewhere, though perhaps m another 
form. Frequently, potential energy changes to kinetic 
energy or vice versa. To illustrate, take the case of a 
steam hammer or a pile driver. The ram is raised to a 
height h above the pile, and, on being released, strikes the 
head of the pile and comes to rest. In its highest position, 
the ram has the potential energy Wh In falling, it attains 
a velocity v that, just at the instant of striking, has the 
magnitude ^%ffh, hence, at this instant, the ram has lost 

all its potential energy, but h'as a kinetic energy -- = IV h. 

** 

The loss of one kind is therefore just balanced by the gain 
of the other kind. After the blow, the ram comes to rest 
and has neither potential nor kinetic energy. Apparently 
there is a loss of energy, but really the energy has for the 
most part been expended in doing the work of driving the 
pile a certain distance into the earth. A small part has 



30 KINEMATICS AND KINETICS 33 

been expended in heating the ram and head of the pile, 
and another part in producing sound. 

Heat is a form of energy due to the rapid vibration of the 
molecules of bodies. In many cases where energy appar- 
ently disappears, it reappears in the foim of heat Thus, the 
work clone against frictional resistances appears as heat, 
which is usually dissipated into the air as fast as produced 

Conversely, there aie examples of heat energy transformed 
into other forms. A pound of coal has potential energy, 
which, when the coal is burned, is liberated and changed 
into heat energy. The heat applied to water produces steam, 
and the steam has potential energy. Finally, the steam m 
giving up its energy does work in a steam engine, and this 
woik is expended in overcoming the friction of shafts, belts, 
and machine parts, and ultimately reappears m the form of 
heat. However, in all these transformations, the total 
amount of energy the sum of the kinetic and potential 
energy remains always the same. Such is the law of the 
conservation of energy. 



EFFICIENCY 

38. In every machine, there is an effort or driving force 
that produces motion in the machine, and a resistance against 
which the machine does work. Take, for example, the rais- 
ing of a load by a crane. The effoit is exerted by the work- 
man on the crank of the windlass, and the resistance is the 
weight of the load lifted. In the case of a machine tool for 
cutting metal, the effort is the pull of the belt on the driving 
pulley, and the resistance is the resistance of the metal to 
the cutting tool. In a given time, say 1 minute, a quantity 
of work is done by the effort and another quantity of work is 
done against the resistance. Were it not for frictional resist- 
ances, these two works would be just equal. In actual cases, 
however, only a part of the woik of the effort is usefully 
expended in doing work against the resistance. The remain- 
der is used in overcoming the friction between the various 
sliding 1 surfaces, journals and bearings, etc. The work thus 



34 KINEMATICS AND KINETICS 30 

expended appears in the form of heat energy, and serves no 
useful purpose. 

The ratio ful work Qr useful work 

total work of effort total energy supplied 
the efficiency of the machine. 

Representing the efficiency by E, the useful work by /, 
and the total work (= energy supplied) by /, we have, 



EXAMPLE A water motor receives 150 cubic feet of water per 
second, with a velocity of 45 feet per second If the maximum power 
that can be obtained from the motor is 459 H P , what is its efficiency? 

SOLUTION The weight of the water delivered per second is 
150 X 62 5 Ib , the kmet:c energy of which (Art 33) is 

150X626X45' .. , , n 

2 ft -Ib per sec ( = U) 

Since 1 H P = 650 ft -Ib per sec , the useful work per second done 
by the motor is (459 X 550) ft -Ib ( = /) . Therefore, 



150 X65X 45' 



- 8513 ' r ffi 5 per Centl Ans ' 



HYDROSTATICS 



DEFINITIONS PROPERTIES OF LIQUIDS 

1. Hydromechanics is that branch of mechanics that 
deals with liquids, their properties, and their applications 
to engineering. It may be divided into two chief branches: 
hydrostatics and hydraulics. 

2. Hydrostatics treats of liquids in a state of rest. 

3. Hydraulics treats of liquids in motion. 

4. In the first of these divisions are included such prob- 
lems as the pressure of water on enclosing vessels and sub- 
merged surfaces; in the second are included problems relating 
to the flow of liquids through orifices, in pipes, and m channels. 

5. !Llquid Bodies. A liquid body, or simply a liquid, 
is a body whose molecules change their relative positions 
easily, being, however, held in such a state of aggregation 
that, although the body can freely change its shape, it retains 
a definite and invariable volume, provided the pressure and 
temperature are not changed. Water and alcohol are exam- 
ples of liquid bodies. 

6. A perfect liquid is a liquid without internal friction; 
that is, one whose particles can move on one another with 
absolute freedom. On account of this characteristic property, 
a perfect liquid offers no resistance to a change of form. 

7. A viscous liquid is a liquid that offers resistance to 
rapid change of form on account of internal friction, or vis- 
cosity. Tar, molasses, and glycerine are examples of 
viscous liquids. 

COPYRIaHTBD BY INTIBNATIONAL TEXTBOOK COMPANY ALU KIOHTB 

31 



2 HYDROSTATICS 31 

All liquids are more or less viscous. For the purposes of 
hydrostatics, however, water, which is the liquid mainly dealt 
with in this work, may be treated as a perfect liquid, its vis- 
cosity at ordinary temperatures being too small to be taken 
into account. 

8. Compressibility. All liquids offer great resistance 
to change m volume; that is, they can be compressed but 
little. Under the pressure of one atmosphere (about 14 7 
pounds per squaie inch), water is compressed about 20067 
of its original volume. Foi engineering purposes, it may 
be assumed that water is practically incompressible. 

9. Density. The density of a homogeneous body is 
the mass of the body per unit of volume, and may be obtained 
by dividing the total mass of the body by its volume. 

If V, W, ;, and D are, respectively, the volume, weight, 
mass, and density of a body, then, 



or, since m (see Fundamental Principles of Mechanics), 



Both from the definition and from this formula, it follows 
that the density of any one body varies inversely as the 
volume of the body; in other words, when the mass of the 
body remains the same, the density is greater the less 
the volume of the body, or the space that the body occupies. 
This agrees with the ordinary use of the words density and 
dense, which are employed to denote the degree of com- 
pactness, a body being said to be more or less dense accord- 
ing as its molecules are supposed to be more or less close 
to one another. 

10. Weiglit of "Water. The weight of 1 cubic foot of 
water varies with the temperature: at 39.2 F., which is the 
temperature of maximum density, it is 62.425 pounds. For 
nearly all engineering purposes, 62.5 pounds is used as a 



31 HYDROSTATICS 3 

convenient and sufficiently approximate value. This value 
will be used throughout this work 

Since a. column of water 1 square inch in cross-section 
and 1 foot high is -j4r cubic foot, its weight is 62 5 -f- 144 
= 434 pound. This value is of very frequent application, 
and should be memorized. 



LIQUID PRESSURE 

PASCAL'S LAW AND ITS APPLICATIONS 

11. Statement of Pascal's Law. A perfect liquid 
transmit* pressure equally in all directions This principle is 
Pusc-aPs lu\v, and follows di- 

7 A 

rectly fiom the definition of a per- "* 

feet liquid. i 

The difference between a liquid H 
and n solid as regards transmis- \ 



sion of pressure may be illus- 
trated as follows: In Fig. 1 are 
shown two cylindrical vessels of 
the same size. The vessel a is 
fitted loosely with a wooden block 
of the same size as the vessel. 
The vessel b is filled with water, 
whose depth is the same as the 
length of the wooden block in a. m * l a 
Both vessels are fitted with air- FlGl 

tight pistons P. For convenience, let the weight of the block 
and that of the water be neglected, and suppose that a force 
of 100 pounds is applied to each piston. Assume the piston 
area to be 10 square inches; then the pressure per square inch 
is 100 -T- 10 <= 10 pounds. In the vessel a, this pressure will 
be transmitted, undiminished, to the bottom of the vessel; 
it is easy to see that there will be no pressure on the sides. 
In the vessel b, an entirely different result is obtained. The 
pressure on the bottom will be the same as in the other 




HYDROSTATICS 



31 



case, that is, 10 pounds per square inch; but, owing to the 
fact that the molecules of water are perfectly free to move, 
this pressure of 10 pounds per square inch is transmitted in 
every direction with the same intensity; that is to say, the 
pressure at all points, such as c, d, e, f,g,h, due to the external 
force of 100 pounds, is exactly the same, namely, 10 pounds 
per square inch. 

12. Verification of Pascal's Law. An experimental 
proof of Pascal's law may be effected with the apparatus 
shown in Fig. 2 The vessel is filled with water, and is fitted 

with pistons of different 
diameters. Let a force be 
applied to the small piston c , 
as a result, the water in con- 
tact with c is subjected to a 
certain pressure According 
to Pascal's law, this pressure 
must be transmitted with 
undimimshed intensity in 
all directions. For the sake 
16 of convenience, let the area 
of the piston c be 1 square 
inch, and let a force of 6 
pounds be applied to it. 
The pressure exerted by the 
piston on the water is, there- 
fore, 5 pounds per square inch. If the area of the piston a 
is 40 square inches, and Pascal's law is true, the water must 
exert on the face of a a total pressure of 40 X 5 = 200 pounds. 
It is found by experiment that a force of 200 pounds must be 
applied to the piston a to prevent it from moving outwards. 
This shows that the fluid pressure against a is 200 pounds, or 
5 pounds per square inch, which is the same as that against c. 
Similar results are obtained with the pistons b, d, <?, and /. If 
their areas are, respectively, 7, 6, 8, and 4 square inches, the 
pressures exerted on them are found to be, respectively, 35, 
30, 40, and 20 pounds. 




31 HYDROSTATICS 6 

Pascal's law may be formally stated as follows: The pres- 
sure i>er unit of area exerted anywhere on a mass of liquid is 
transmitted undimimshed in all directions, and any surface in 
contact with the liquid will be subjected to this pressure in a 
direction at right angles to the surface. 

13. Application of Pascal's Law. In Fig. 3, let the 
area of the piston a be 1 square inch, and that of b, 40 square 
inches. According to Pascal's law, 1 pound placed on a 
will balance 40 pounds placed on b. If a moves down- 
wards 10 inches, then 10 cubic inches of 

water will be forced into "the tube b. 

This will be distributed in the tube b in 

the form of a cylinder whose cubical 

contents must be 10 cubic inches, whose 

base has an area of 40 square inches, 

and whose altitude must therefore be 

10 - 40 = } inch. This shows that a 

movement of 10 inches of the piston a 

causes a movement of i inch of the 

piston b. This is an example of the 

familiar principle of work that the force, PlQ 3 

or effort, multiplied by the distance through which it moves, is 

equal to the resistance multiplied by the distance through which 

it moves. 

14. The Hydraulic Press. The principle just stated 
finds an important application in the hydraulic press, which 
is shown in Fig. 4. As the lever a is depressed, the piston b 
is forced down on the water in the cylinder c. The water is 
forced through the bent tube d into a cylinder fitted with 
a large piston e> and causes that piston to rise, the plat- 
form / is thus lifted, and the bales placed between it and 
an upper fixed platform are compressed. Assume the area 
of the piston b to be 1 square inch, and that of the piston e 
to be 100 square inches. Also, assume the length of the 
lever between the hand and the fulcrum to be ten times the 
length between the fulcrum and the piston rod b. If the end 
of the lever is depressed, say 10 inches, the piston b is 




6 



HYDROSTATICS 



31 



depressed one-tenth of 10 inches, or 1 inch, and the piston e 
is raised rihr inch, since 1 cubic inch of water is displaced. 
If P denotes the force applied and Q the piessure on the 
platform /, then P X 10 inches = Q X Toir inch, hence, Q 
= 1,000 P, or P = ToW Q. A force of 40 pounds applied 




PIG 



by hand thus produces a pressure of 40 X 1,000 = 40,000 
pounds. But, if the average movement of the hand per 
stroke is 10 inches, it will require 1,000 -r 10 = 100 strokes 
to raise the platform 1 inch, which again shows that what is 
gained in power is lost in speed. 



31 



HYDROSTATICS 



GENERAL THEORY OF LIQUID PRESSURE 

15. Downward Pressure. The weight of a mass of 
liquid is a force, and will produce a piessure on any surface 
in contact with the liquid independently of the pressure pro- 
duced by other forces. Thus, in the case of the vessel shown 
in Fig. 2, if the vessel is vertical so that the piston d is at 
the bottom, d is subjected, in addition to the pressure of 
5 pounds per squaie inch transmitted from the piston c, to an 
additional pressure due to the weight of the water above it. 
Evidently, the piston e will be under an additional pressure 
due to the weight of the water, but the pistons a, b, and /, 
being above the liquid, will be 
subjected only to transmitted 
pressuie. 

To arrive at the laws govern- 
ing the pressure due to the 
weight of a liquid, let us con- 
sider first the pressure on the 
bottom of a vessel containing 
the liquid. In Fig. 5 are shown 
two vessels of different shape 
but with the water level in each 
at the same height 24 feet. 
Assume the area of the bottom 
of a to be 500 square inches, or $M square feet; then the vol- 
ume of the water contained in a is -fH X 24 = 83i cubic feet, 
and the weight is 83i X 62.5 = 6,208.3 pounds. The pressure 
per square inch is 5,208.3 -5- 500 = 10.42 pounds, nearly. 

An easier solution, howevei, is effected as follows: The 
weight of a column of water 1 square inch in cross-section 
and 1 foot long is .484 pound (Art. 10). Now, above 
each square inch of the bottom there is a column of water 
24 feet high, whose weight is, therefore, 24 X .434 = 10.42 
pounds, and this weight is the pressure per square inch on 
the bottom of the vessel, 

In vessel 6, assume the cross-section of the lower part 
to be 600 square inches and that of the upper part to be 

ILT39&-1S 





PIG 5 



8 



HYDROSTATICS 31 



100 square inches. It is evident that the volume of water 
contained is smaller than before, and it might be concluded 
that the pressure on the bottom is, consequently, smaller; 
but this conclusion would not be true, as will appear from 
the following reasoning: Let water be poured into the 
empty vessel until the lower part is filled; the pressure on 
the bottom is, according to the method of calculation just 
given, 10 X .434 = 4.34 pounds per square inch. Imagine, 
now, a thin piston or disk fitted tightly into the smaller part 
and resting on the surfa'ce of the water below, and then let 
water be poured in until the smaller part is full. This 
piston acts as the bottom of a vessel 14 feet high and with 
a uniform cross-section of 100 square inches. Hence, the 
pressure on this piston, due to the weight of the water above, 
is 14 X .434 = 6.08 pounds per square inch. According to 
Pascal's law, this pressure is transmitted in all directions, 
and therefore acts on the bottom of the vessel; hence, the 
total pressure on the bottom is 4.34 + 6.08 = 10.42 pounds 
per square inch, the same as in the first case. It is evident 
that the result will be the same if the piston is left out. 

16. From the reasoning used in the preceding paragraph, 
it appears that the pressure per unit of area on the bottom of 
a vessel depends only on the vertical distance between the 
bottom and the liquid surface, and not at all on the shape of 
the vessel. This principle is one of great importance. 

17. Intensity of Pressure. When a surface is sub- 
jected to a uniformly distributed pressure, the pressure per 
unit of area is called the intensity of pressure, or unit 
pressure. Usually, the unit of force is the pound, and that 
of area is either the square inch or the square foot; and so 
the intensity of pressure is expressed either in pounds per 
square inch or in pounds per square foot. 

In general, if P is the total pressure uniformly distributed 
over a surface, and A is the area of the surface, the intensity 
of pressure p is given by the formula 



31 HYDROSTATICS 9 

18. Head. The distance from any horizontal layer of a 
liquid body to the surface of the liquid is termed the head 
on that layer. 

Let h = head, in feet, on any horizontal layer; 

p = pressure per square inch on the layer, in pounds; 

w = weight of a column of liquid 1 foot long and 

1 square inch in cross-section. 
Then, p = wh (1) 

For water, w = .434, and, therefore, 
p = .4347* (2) 

Also, h => -f~-r; that is, 
.4o4 

h = 2.304^ (3) 

EXAMPLE 1 The depth of water in a stand pipe is 80 feet 
(a) What is the pressure per square inch on the bottom ? (b} What 
is the pressure per square inch on a horizontal layer 65 feet from the 
surface? 

SOLUTION. (a) Substituting the value of h in formula 2, 
p = 434 X 80 = 34 72 Ib per sq in. Ans 

(b) Here // = 65, hence, substituting in formula 2, 
p = 434 X 65 = 28 21 Ib. per sq. in. Ans 

EXAMPLE 2. What must be the height of water in a stand pipe to 
give a pressure of 80 pounds per square inch on the bottom? 

SOLUTION. Substituting in formula 3, 

h - 2.304 X 80 = 184 32 ft. Ans. 

19. Upward and Lateral Pressure. So far, only 
downward pressure has been discussed. It is necessary now 
to consider upward pressure, as well as lateral (sidewise) 
pressure on vertical surfaces. Let the vessel shown in 
Fig. 6 (a) be filled with a liquid to the level a. The part 
of the liquid in a b rests on the layer at 6, and produces over 
that surface an intensity of pressure of w// t pounds per 
square inch, where A x is the head, in feet, on the layer at b. 
According to Pascal's law, this intensity of pressure is trans- 
mitted to all the bounding surfaces below the level b\ hence, 
there is a pressure of wh^ pounds per square inch, due to the 
liquid in ad, exerted at c,d,e,f t g, and k } at right angles to 



10 



HYDROSTATICS 



31 



the surfaces. At any point below the level b there is, how- 
ever, additional pressure due to the weight of the liquid 
between the point and the level b. At c, which is at the 
same level as b, the total upward pressure per unit of area 
is w h lt the same as the downward pressure on the layer at b. 
Consider now the layer at de at a distance fi a below the 
surface a. The pressure on this layer, due to the weight of 
the liquid above, is wh t pounds per square inch, and by 
Pascal's law this pressure is transmitted to all parts of the 
bounding surface below the level de, just as if the layer dc 
were a solid piston. The liquid below de can exert no pres- 
sure at the points d and e\ hence, at these points th2 pressure 
per unit of area is the same as the downward pressure on the 







to 



layer de, namely, w h, pounds per square inch. The same 
reasoning shows that the lateral pressure per unit of area at 
the points / and g is w A 3 , where h, is the head on the layer fg. 

The following important law, which is a direct consequence 
of that of Pascal, may now be stated: 

The intensity of pressure at any point of a surface enclosing a 
liquid is normal to that surface; it depends only on the depth of 
the point below the surface of the liquid^ and is equal to the 
intensity of downward pressure on a horizontal layer of the 
liquid having' the same head as the point in question. 

The formula p = w h is, therefore, general, and expresses 
the intensity of pressure at any point whose distance from 
the surface of the liquid is A. 



31 HYDROSTATICS 11 

EXAMPLE In Pig 6, suppose the depth of the various layers below 
the level a to be as follows: depth of t>, 10 feet, of de, 17 feet, of fg t 
26 feet, and of k, 30 feet. The liquid being water, what are the 
pressures per square inch at the points c, d, <?, /, g , kt 

SOLUTION Using formula 2 of Art 18 in each case, the pressures 
at the various points are found to be as follows 

c = 434 X 10 = 4 34 Ib per sq m. Ans. 
d = .434 X 17 = 7.38 Ib. per sq. in Ans. 

Pressure at, * = <4S4 X 1? = 7 88 lb P er ** in " AnS 
Pressure at, f _ 434 x 25 10 85 lb per sq in. Ans. 

g = .434 X 25 = 10 85 lb. per sq in. Ans 
k = .434 X 30 = 13 02 lb. per sq in Ans. 

20. Graphic Determination of Pressure. The 

varying intensity of pressure for different points below the 
surface of a liquid may be determined graphically as follows: 
Let OM, Fig. 6 (), represent, to any convenient scale, the 
head h t on the bottom k of the vessel. Draw MN perpen- 
dicular to MO, and representing, to any convenient scale, 
the intensity of pressure w h* on the bottom of the vessel. 
It should be observed that OM represents a distance and 
MN a pressure, and that there is no necessary relation 
between the two scales. For instance, OMma.j represent h* 
to a scale of 2 feet to the inch; and MN } the pressure w 7i 4 
to a scale of tsV inch to the pound. Draw ON. Then, any 
horizontal line limited by OM and ON will represent the 
intensity of pressure at any depth equal to the vertical 
distance of that line below O. Thus, CJ3 t whose distance 
below is A,, represents w k^ Likewise, D 2? represents w h t . 
If the pressure scale is -sV inch to the pound, it will be 
found that the line FG measures about H, which represents 
11 pounds per square inch. For very accurate work, the scale 
should be as large as possible. It should be observed that 
the tangent of the angle MONis numerically equal to w, 
since 



MO hi 

21. Pressure Due to Matternal Load, If the surface 
of a liquid is subjected to a pressure, this pressure, accord- 
ing to Pascal's lav?, is transmitted undiminished to all parts 



2 HYDROSTATICS 81 

f the enclosing vessel, and must be added to the pressure 
ue to the weight of the liquid. Assume the surface a of 
'ig. 6 to be subjected to a pressure of 30 pounds per square 
ich. In the example of Art. 19, the pressure at c, due to the 
ead of water, is 4.34 pounds per square inch. Adding the 
D pounds per square inch due to the external force applied 
t a, the result is 4.34 + 30 = 34 34 pounds per square inch 
.ikewise, the pressure at d and e is 7.38 + 30 = 37 38 pounds 
er square inch; the pressure at / and g is 10 85 + 30 = 40.85 
ounds per square inch; and the pressure at k is 13.02 + 30 
= 43.02 pounds per square inch. 
Let G = total load on surface of liquid, in pounds; 

A = area of surface loaded, in square inches; 

/^ 
pt = = intensity of pressure on surface, in pounds 

A 

per square inch; 

P = intensity of pressure, in pounds per square inch, 
at a point h feet below liquid surface. 

Then, p = wh+p* = wh + (1) 



When the liquid is water, 



p = .434^+A = .434>% + (2) 

A 

EXAMPLE A vessel filled with salt water weighing 1 03 times as 
uch as fresh water has a circular bottom 13 inches in diameter The 
p of the vessel is fitted with a piston 3 inches in diameter, on which 
laid a weight of 75 pounds. What is the intensity of pressure on the 
ttom, if the depth of the vessel is 18 inches? 

SOLUTION In this case, w = 1 03 X 434 = .447 lb.; h = 18 in. 
1 6 ft , and A = 3 1 X 7864 sq. m By formula 1, 

p = .447 X 1 5 + 3 , x ?6 7854 - 11.28 lb per sq. in. Ans, 

22. Atmospheric Pressure. When a liquid is exposed 
the air, as in the cases represented in Figs. 5 and 6, there 
an external pressure, due to the weight of the air, on the 
rface of the liquid. This pressure is called atmospheric 
essure. It varies according to locality and atmospheri 
nditions, but, for nearly all practical purposes, it may 
taken as 14.7 pounds per square inch. This pressure is 



31 



HYDROSTATICS 



13 



transmitted by the liquid and should be treated as the pres- 
sure p n considered in the preceding article. It is customary, 
however, in hydrostatic and hydraulic calculations to take the 
atmospheric pressure as the zero of reference. When the 
pressure of a liquid is referred to, the pressure in excess of 
the atmospheric pressure is meant. This is called gauge 
pi-essure. When the atmospheric pressure is added to the 
gauge pressure, the resultant pressure, which is really the 
total pressure on the liquid or surface considered, is called 
absolute pressure. Thus, a gauge pressure of 20 pounds 
per square inch is equivalent to an absolute pressure of 
20 + 14.7 = 34.7 pounds per square inch. 

23. Equilibrium of Liquids at Best. Since the 
pressure on a horizontal layer due to the weight of a liquid 




PIG 7 

is dependent only on the height of the liquid, and not on the 
shape of the vessel, it follows that, if a vessel has a number 
of radiating tubes, as in Fig. 7, the water in each tube will 
be at the same level, no matter what may be the shape of 
the tubes. For, if the water is higher in one tube than in 
the others, the downward pressure at the level , due to the 
height of water m this tube, is greater than that due to the 
height of the water in the other tubes. This excess of pres- 
sure will cause a flow toward the other tubes, which will 
continue until there is no further excess that is, until the 



14 HYDROSTATICS 31 

free surfaces are at the same level. Then the liquid will 
come to rest, and will be in equilibrium. The principle 
here stated is embodied in the familial saying, Wafer seeks 
tts level. This principle explains why city reservoirs are 
located on high elevations, and why water, when leaving 
the hose nozzles, spouts so high. If there were no resist- 
ance by friction and air, the water would rise to a height 
equal to the level of the reservoir If a long pipe with a 
length equal to the vertical distance between the nozzle and 
the level of the water in the reservoir were attached to the 
nozzle and held vertically, the water would just reach the end 
of the pipe. If the pipe were lowered slightly, the water 
would trickle out. Fountains, canal locks, and artesian wells 
are examples of the application of this principle. 

EXAMPLE 1 The water in a city reservoir is 150 feet above a hydrant 
What is the pressure per square inch at the hydrant? 

SOLUTION By formula 2, Art 18, 

p = 434 X 150 = 65 1 Ib per sq in Ans 

EXAMPLE 2 The pressure on a water mam, when the water is not 
flowing, is shown by a gauge to be 72 pounds per square inch What 
is the elevation of the reservoir above the mam? 

SOLUTION. By formula 3, Art 18, 

h = 2.304 X 72 = 165.89 ft Ans. 



EXAMPUJ8 FOR PRACTICE 

1. What is the intensity of pressure on the bottom of a stand pipe 
90 feet high? Ans 39 06 Ib, per sq In. 

2. A cylindrical vessel 15 inches m diameter is filled with water. 
The top of the vessel is fitted with a piston on which is laid a weight 
of 300 pounds. The depth of the vessel being 24 inches, determine 
the intensity of pressure on the bottom, of the vessel 

Ans 2 566 Ib per sq. in. 

3. If the intensity of pressure, due to an external load, on a vessel 
filled with water is 30 pounds per square inch, what is the intensity of 
pressure at a point 12 inches below the surface of the water? 

Ans. 30.434 Ib. per sq. in 



31 



HYDROSTATICS 



15 



PRESSURE ON AN IMMERSED SURFACE 
24. Total Pressure on a Flat Immersed Surface. 

Let M N, Fig. 8, be a plate immersed in water. It is pro- 
posed to determine the total pressure acting on the upper 
surface of this plate. In the first place, the total piessure />, 
being the resultant of the pressures acting at all the points 
of M N, is perpendicular to MN. The magnitude of P is 
determined by means of the following principle, which is 
derived by advanced mathematics: 

The total pressure on any plane surface immersed in a liquid 
is equal to the product of the following quantities ( 1 ) the area 
of the surface, (2) the dis- 
tance of the center of gravity 
of the surface ftom the level 
of the liquid, (5) the 
weight of the liquid per unit 
of -volume. 

In applying this princi- 
ple, length, area, and vol- 
ume should be referred to 
the same unit. Thus, if dis- 
tances are expressed in 
feet, areas should be ex- 
pressed in square feet and volumes in cubic feet. The princi- 
ple obviously applies to the pressure on the bounding sur- 
faces of the vessel or receptacle containing the liquid as well 
as to the pressure on the immersed surface. 

Let A be the area of the surface M N> Fig 8; G, the center 
of gravity of that surface, its distance from the level of the 
liquid being h a \ and W, the weight of the liquid per unit of 
volume. Then, the foregoing principle may be stated in 
symbols thus: 

P = A h e W 

It should be particularly noticed that the total pressure P 
does not act through the center of gravity G. The deter- 
mination of the point of application of P will be dealt with 
farther on. 




Pro. 8 



16 



HYDROSTATICS 



31 



EXAMPLB To find the total pressure on the side ML N K oi the 
rectangular tank shown in Fig 9, the dimensions being as indicated. 

SOLUTION Since ML N K 'is a rectangle, its center of gravity G is 
at a distance from the upper side L N equal to 5, or 6 25 ft 2 

LI 

Also, A = MK X KN = 3 2 X 6.25, 



= 3 125 ft.; hence, h r = 3 125 ft 
and W = 62 5 Ib 

Substituting these values in the formula for P, 

P = 3 2 X 6.25 X 3 125 X 62 5 = 3,906 Ib , nearly 



Ans. 



25. Center ol Pressure. The point where the line 
of action of the total or resultant pressure acting on an 

immersed surface meets that surface is 
called the center of pressure. This 
point does not coincide with the center 
of gravity of the surface unless the 
latter is horizontal In all other cases, 
the center of pressure lies below the 
center of gravity. Thus, in Fig. 8, 
the center of pressure C lies below the 
center of gravity G. A general for- 
mula for determining the position of 
the center of pressure cannot be either 
derived or applied without the use of 
advanced mathematics. Special for- 
mulas applying to some important cases are given in the fol- 
lowing articles. The distance of the center of pressure from 
the level of the liquid will in all cases be denoted by A as 
shown in Fig. 8. 

26. Center of Pressure ol a Rectangular Surface. 

In Fig. 10 is represented a vessel V containing a liquid. 
The bounding surface RSTU is shown inclined to the ver- 
tical; but the formulas derived in this and the following two 
articles apply both to an inclined and to a vertical surface. 
Let K L M N be a rectangular part of the inclined surface, 
the sides K L and M N being horizontal The line EFYis 
an axis passing through the middle points of K L and M N. 
The heads on the points E and F are denoted by hi and Jt t , 
respectively. It should be observed that the formulas given 




31 



HYDROSTATICS 



17 



below apply whether the rectangular surface considered forms 
part of one of the bounding surfaces of the vessel or whether 
it is the surface of any immersed body. 

The center of pressure Cis on the line E F, and its depth h c 
below the level of the liquid is given by the formula 

CD 



To find the distance FC, draw FHQ perpendicular to the 

JZ 




FlQ 10 



verticals through F t C t and ., and denote the length EF, or 

f* T 

NK, by 5. The similar triangles FCH and FEQ give ^ 

/ y jD 

~^r] whence, the following equa- 



= -\ that is, 
EQ b 

tion obtains: 



PC 



' X 



(2) 



If the value of h e found by formula 1 is substituted in 
formula 2, the result, after several transformations have 
been made, is 

X* (3) 



18 



HYDROSTATICS 



31 



27. If the edge K ' L is flush with the surface of the 
water, h* = 0, and formulas 1 and 3 of the preceding article 
become, respectively, 

A. -iXv- 1 ! = *A, (1) 



FC= 



' 



(2) 



28. Center of Pressure of Triangular Surface. 

Only the case of an isosceles triangle with its base horizontal 
will be treated in this Section. And first, a triangle MFN t 




FIG 11 

Fig. 11, with its vertex F above its base will be considered 
The line EFYis perpendicular to the base MN The alti- 
tude FE is denoted by a The rest of the notation is 
similar to that in Fig. 10. The center of pressure is on the 
line FE, and its depth below the surface of the liquid is 
given by the formula 

h t = 3/ '' a +-' (1) 

The distance PC is determined in the same way as for the 
rectangle treated in Art 26, the result being 



FC = 



fit. A, 



(2) 



31 



HYDROSTATICS 



19 



29. If the vertex is flush with the surface of the liquid, 
//, = 0, and the two formulas of the preceding article become 

7 3 til n i / t X 

h c - = i /it ( 1 ) 

4 fli 

30. Let, now, the vertex be below the base, as shown 
in Fig 12. With the same notation as before, 

_ R 

.. i- , -.~ T ~" ?s " rm /\ 

.J'..i _A_____^__ f __/.__\ 

1 i ' 
1 i ' 

! i '/ , 

* ^ / 
! /rJ y 




S 




PIG. 12 



3 A. 1 



, i 
2 



i h t 



(1) 
(2) 



31. If the base is on the surface of the liquid, h t 0, and 
the two formulas of the preceding article become 

h e = %hi (1) 

B *xi (2) 



32. Center of Pressure of Circular Surface. Let 

M ' N, Fig. 13, be a circular surface. The line EFY con- 
tains the lowest point E and the highest point F of the circle. 



S30 



HYDROSTATICS 



31 



The diameter of the circle is denoted by d. The rest of the 
notation is the same as in previous articles. 

The center of pressure is on the diameter EF, and the 
distances k e and FC are given by the formulas 

7B 

'. T-T (D 



(2) 



FC = ' ~ ' X 

kt h. 



/js'^T^r,^-'^" -^ ,.""' 

A 1 ,*: V . " .' x T-.F- -/ -\ 

r> * .* ,-. i i / / 

' - i / / 

i \\ !/ / )tf 

I H/fc, / 




/'l^ 



PlQ 18 



33. If the circle just touches the surface of the liquid, 
h t = 0, and formulas 1 and 2 of the preceding article 
become 

+& (1) 



h c = 



EXAMPLE. The vertical circular plate a, Fig. 14, 8 inches In diam- 
eter, covers an opening of equal size. The center G is 24 inches 
below the surface of the -water, and 20 inches below the hinge b, 
What horizontal force applied at c is sufficient to move the plate ? 

SOLUTION. Area of plate is 50 266 sq. in Head above center of 
gravity is 2 ft. Then (Art. 24), the total pressure on the plate is 

X 62.5 X 2 = 43.63 Ib. 



144 



31 



HYDROSTATICS 



21 



Here, A, = 24 + f = 28 in , and A, = 24 - f = 20 m The depth 

of the center of pressure below the surface is, by formula 1, Art 32, 

h e = 24 167 in , nearly The center of pressure is 167 in below the 

center of gravity G, and, therefore, 20 167 m. below the hinge b The 

' ;_ _y moment of P, the total pressure, about 

r the hinge must be equal to the moment 

of .P 1 about the hinge, hence, 

FX 40 = P X 20 167 = 43 63 

X 20.167 = 879 89 in -Ib ; 
whence, F = 879 89 -*- 40 = 22 
Ib , nearly Ans. 

5 

$ 34. Plates 01- Gates With 

Liquid Press xi i-e on Both 

Sides. The plate A B, Fig. 15 
is subjected to water pressure 
with the level a' on one side and 
a!' on the other. The total pres- 




_ _ B ~ -^ 

FIG 14 FIG. 15 

sure on one side is /", acting through the center of pres- 
sure C'\ the total pressure on the other side is P' r , acting 
through the center of pressure C". The heads on the centers 
of pressures O and C" are denoted by h e ' and h e ", respect- 
ively; and the difference between the levels a 1 and a" is 
denoted by e. The two horizontal forces P' and P" have a 
resultant P whose magnitude is P' P". Let h t be the 
depth, below a f , of the point of application C of this result- 
ant. Taking moments about the point E> we have, 

Ph e = /"^'-/>"U," + *); 

P' h f P" ( h it 4- e\ 

p /l < ( + e 



whence, 



h t = 



22 



HYDROSTATICS 



31 



EXAMPLE Let the gate in Fig 15 be rectangular and 6 feet wide; 
let the depth BE be 9 feet and the difference *, 2 4 feet. Required 
the pressures /*, P", P, and the distance h e 

SOLUTION From formula of Art 24, the total pressure on the 
right side is 

/ J/ = 0X6x625Xo = 15,188 Ib. Ans 
The total pressure on the left side is 

P" = 6X66x625X-fj- = 8,167.5 Ib Ans. 

The resultant force is 

P = 15,188 - 8,167 5 = 7,020 5 Ib 

From formula 1 of Art 27, h c ' - % X 9 = 6, and h = X 6 6 

= 44, also, e = 9 - 6.6 = 2 4. Substituting in formula of Art 34, 

. 15,188 X 6 - 8,167 5 X (4 4 + 2 4) ? 

7,020.5 

35. Assume the plate B, Fig 16, to be entirely below 

the water levels a and b, 
and, therefore, subjected 
to pressure on both sides. 
Let the triangle OM ' N be 
drawn to represent the 
variation in pressure due 
to the weight of the water 
on the left-hand side (see 
Art. 20), and let the tri- 
angle 0' M f N' be drawn 
to represent the variation 
of pressure due to the weight of the water on the right-hand 
side Since the angles M O N and M' O 1 N' are equal, the 
tangent of each being w (Art. 20), it follows that O 1 N' and 
ON are parallel. Take M0 n = M 1 1 , and through O" draw 
O"N" parallel to O ' N'\ then, the triangles O 1 M' N 1 and 
O" MN" are equal. Now, the intensity of pressure on the 
left of the gate, at any level, as x t below the surface, is given 
by the intercept K ' E, and that on the right by the intercept 
DE = D'E'. The resultant intensity of pressure at that 
point is the difference KE DE KD. Since ON and 
O" N" are parallel, the horizontal intercepts between them 
are equal, which means that the intensity of pressure is the 



_ _ 







N' 



M 



FIG 16 



31 HYDROSTATICS 23 

same at all points of the plate B It follows that the center 
of pressure for such a submerged plate coincides with its 
center of gravity. 

The total resultant pressure P on the plate is found as 
follows: The intercept SO" represents the intensity of pres- 
sure on the left side at the level /?, that is, for the head e, 
which is the difference between the levels a and b. Hence, 
since the intensity of pressure on B is everywhere equal 
to KD, or SO", or we, it follows that, if the area of B 
is denoted by A, the total pies&ure is 

P = Awe 

EXAMPLE In the example of Art 33 (see Fig 14) , suppose there 
is water on the right-hand side of the partition If the level of this 
water surface is 8 inches lower than that of the level d, find the total 
resultant pressure on the plate, and also the force F 

SOLUTION The head e = 8 in = f ft The total pressure is 

A w e - 50 206 X 434 X ij = 14 54 Ib Ans 

This pressure acts through the center of gravity, 20 ill below the 
hinge, hence, taking moments about the hinge, 
F X 40 = P X 20, 

whence, F=\IP=\P= ^~ = 7.27 Ib Ans. 

36. Pressure on Surfaces That Are Not Plane. If 

the surface sustaining liquid pressure is not plane, but is 






(b) 

Fio 17 

curved or irregular, the total pressure on it in any direction, 
neglecting the pressure due to the weight of the liquid, is the 
same as the total pressure would be on the projection of the surface 
on a plane at light angles to the given direction. To illustrate 
this statement, consider the three pistons in Fig. 17. At (a) is 
shown a piston with a curved end, at (b), a piston with 
an irregular end; and at (f), a piston with a flat end. In 
each case, the projection of the surface sustaining pressure, 

I L T 398-W 



24 HYDROSTATICS 31 

on a plane perpendicular to the piston rod, is the circular 
cross-section of the cylinder, and if the pressures per unit 
of area in the cylinders are the same, the pressure on the 
face of the piston is the same in each case. It is assumed 
that the pressure per unit of area is the same at all points 
of the surface; that is, the change of pressure due to the 
varying depth of the liquid is neglected With this restnc- 
tion, the law applies to all fluids, both liquid and gaseous. 

If the curved or irregular surface forms a part of the wall 
of a vessel, or is submerged, so that the pressure on it is 
due to the weight of the liquid, the law of Art. 24 is not 
true except for a projection on a vertical plane. 

37. The whole subject of pressure in any direction on a 
curved or irregular surface may be summarized as follows: 

1. If the external pressure is so great that the pressure 
due merely to the weight of the liquid may be neglected, the 
total pressure in any direction is equal .to the product of the 
projection of the surface on a plane perpendicular to that 
direction, and the pressure per unit of area. 

2. When the pressure is due wholly, or in part, to the 
weight of the liquid, as m the case of submerged surfaces, 
the total pressure in any direction cannot, in general, be 
determined except for regularly curved surfaces, such as 
spheres, cylinders, cones, etc., and for these the computa- 
tions are difficult and must be made by advanced mathematics. 

3. In one direction, however the horizontal the total 
pressure on a surface is easily found. It is precisely the 
same as the horizontal pressure on the projection of the 
given surface on a vertical plane. 



EXAMPLES FOR PRACTICE 

\ What is the pressure on a layer of water 33 feet below tne 
surface? Ana 14.32 Ib. per sq. In. 

2 What elevation of a reservoir is required to make the pressure 
in a. water main 40 pounds per square inch? Ans. 92.16 ft 

3. A vertical triangular plate forms the side of a vessel that con- 
tains water The base of the tnangle is 4 feet long and lies at the 



31 HYDROSTATICS 25 

water level; the vertex is 4^ feet below the water level (a) What is 
the total pressure on the triangle? (0) What is the distance of the 
center of pressure below the water surface? A \ (a) 843 75 Ib 

S U*) 2J ft 

4. A circular plate 2 feet in diameter is held vertically so that its 
upper edge is 5 feet below the liquid surface Find the depth of the 
center of pressure Ans ft ^ in 

5 In the gate shown in Fig 15, the depth of the head-water is 
7 feet, and that of the tail-water is 6 feet (a) What is the resultant 
pressure on the gate per foot of length? (f>) How far above the 
bottom B is the line of action of the resultant? 



An8 /() 760 Ib 
Ans \ (6) 3 03 ft 



6 The diameter of the plunger of a hydraulic press used m an 
engineering establishment is 12 inches. Water is forced into the 
cylinder of the press by means of a small pump having a plunger 
whose diameter is f inch What pressure is exerted by the large 
plunger when the force acting on the small plunger is 125 pounds? 

Ans 32,000 Ib. 

BUOYANT EFFORT OF LIQUIDS 



IMMERSION AND FLOTATION 

38. Principle of Archimedes. In a mass of liquid at 
rest, suppose a part M N, Fig. 18, of the liquid to become 
solid without changing its form or density. 
This solid part, having the same density as 
before, will be held in equilibrium, or 
remain at rest. Let B denote the weight 
of the solid part; D, the total downward 
pressure; and U, the total upward pressure 
on this part. The oidinary static condi- 
tions of equilibrium require that the 
upward force U should balance the down- 
ward forces D and B\ that is, - 

U=D + B\ F <o-w 

whence, /- D = B (1) 

The difference U L) between the upward and the down- 
ward pressure is called the buoyant effort of the body MN. 
Suppose, now, that the imaginary solid part of the water 
is replaced by an actual solid body having precisely the same 




26 HYDROSTATICS 31 

shape and volume, and let the weight of this body equal W. 
The solid will be subjected to the same vertical pressures 
D and /as was the part M N of the liquid. Consider, now, 
the vertical forces acting on it. The weight \V and the 
pressure D are downwards, and tend to cause the body to 
sink; the pressure U is upwards, and tends to cause the 
body to rise. The resultant downward force is 

D+ W- U = W-(U-D) = W-B (2) 

The weight B is the weight of a mass of liquid whose 
volume is equal to that of the solid, or, what is the same 
thing, B is the weight of the liquid displaced by the solid. 
The following principle may therefore be stated- 

When a solid body is immersed in a liquid, a buoyant effort 
equal to the weight of the hqmd displaced acts upwanh and 
opposes the action of gravity. The weight of a body, as shown 
by a scale, is decreased by an amount equal to the buoyant effot t, 
that is, by an amount equal to the -weight of liquid displaicd. 
This principle is called the principle of Archimedes, from 
the name of its discoverer. 

EXAMPLE In Fig 18 is shown a cube immersed in water Ihe 
edge is 6 inches, the sides are vertical, and the lateral pressures me 
balanced On the upper face, there is a downward pressure I) due 
to the head of water of 15 inches, and on the lower face there is an 
upward pressure U due to the head of 21 inches. To fancl the buoyant 
effort J 

SOLUTION. The downward pressure due to the head of 1C in. or 
H ft , is '' 

6' X 434 X It = 19 53 Ib 

The upward pressure on the lower face due to a head of 21 in., or 
1-f ft , is 

6 X 434 X 1* = 27 34 Ib. 
The buoyant effort of the body is, therefore, 

27 34 - 19 53 = 7 81 Ib. Ans. 

Also, the buoyant effort is equal to the weight of the water dis- 
placed by the body The volume of the cube is (j-) or } cu. ft The 
weight of the water displaced is 

i X 62 5 7 81 Ib. Ans 

39. The principle of Archimedes may be experimentally 
verified with a beam balance, as shown in Fig. 19. From 
one scale pan, suspend a hollow metallic cylinder l> and 



31 



HYDROSTATICS 



27 



below that a solid cylinder a of the same size as the hollow 
part of the upper cylinder. Put weights m the other scale 
pan until they exactly balance the two cylinders If a is 
immersed in water, the scale pan containing the weights will 
descend, showing that a has lost some of its weight. Now 
fill / with water; the volume of water that can be poured 
into t is obviously equal to that displaced by a. The scale 
pan containing the 
weights will rise gradu- 
ally until / is filled, when 
the scales will balance 
again 

40. If a body im- 
mersed in a liquid has 
the same weight as the 
liquid it displaces, then 
IV = B, the resultant 
vertical force IV B is 
zero, and the body will 
remain at rest at any 
depth below the surface. 
If the body is heavier 

than the liquid it displaces, W is greater than .5, and the 
resultant vertical force W B is downwards, hence, the body 
will sink to the bottom. If the body is lighter than the 
liquid it displaces, B is greater than W, the resultant vertical 
force B IV is upwards, and the body will rise to the surface. 

41. An interesting experiment in confirmation of the 
conclusions just derived may be performed as follows: Drop 
an egg into a glass jar filled with fresh water. The mean 
density of the egg being a little greater than that of fresh 
water, the egg will fall to the bottom. Now dissolve salt 
in the water, stirring the mixture; as soon as the salt water 
becomes denser than the egg the latter begins to rise. If 
fresh water is then ponied in until the egg and water have 
the same density, the egg will remain stationary in any 
position it may be placed below the surface of the water. 




PIG 19 



28 HYDROSTATICS ' 31 

42. Floating Bodies. A body lighter than a liquid, 
bulk for bulk, rises to the surface, when immersed, and 
floats. For equilibrium, the buoyant effect B must be just 
equal to the weight W of the body. But, since B is the 
weight of liquid displaced, it follows that the weight of the 
liquid displaced by a floating body is equal to the weight of 
the body. 

The depth of a floating body m a liquid depends on the 
relative weights of equal volumes of the body and the liquid. 
If the body is nearly as heavy as the liquid, it will sink until 
it displaces nearly its whole volume; if very light, compared 
with the liquid, the larger part of the body will be above the 
liquid surface. For example, the density of ice being about 
nine-tenths that of water, about one-tenth of an iceberg 
appears above the surface and nine-tenths is submerged. 
The density of pine is about one-half that of water; hence, 
about one-half of a floating pine log is submerged, and 
one-half is above water. 

EXAMPLE 1. Water-tight canvas air bags are used for raising 
sunken ships These bags are sunk when collapsed, attached to the 
ship by divers and then filled with air from the pumps above, (a) If 
the capacity of a bag is 200 cubic feet, what Is the buoyant effort? 
(b) How many bags will be required to lift 600 tons? 

SOLUTION. (a) The weight of the bag and the enclosed air may 
be neglected. The buoyant effort is the weight of the water displaced 
by the full bag, that is, 

200 X 62.5 - 12,500 Ib. Ans. 
(b) To raise 600 tons, 

600 X 2,000 __ . 

12 500 = ^ are necessarv ' Ans. 

EXAMPLE 2 A cast-iron cylinder 14 inches long and 8 inches in 
diameter is closed at the ends, and the metal is i inch thick through- 
out. Will the cylinder float or sink in water? 

SOLUTION. The volume of the entire cylinder is 
7854 X 8 1 X 14 = 703.72 cu. in. 

The hollow portion has a length of 18 in. and a diameter of 7 in.; 
tts volume is, therefore, 

.7854 X 7* X 13 - 600.80 cu. !n. 
The volume of metal is 

703.72 - 600.30 = 203.42 cu. in. 



31 HYDROSTATICS 29 

Taking the weight of cast iron as 450 Ib per cu. ft., the weight of 
the cylinder is 

450 X ^|| = 52.97 Ib 

If Immersed, the cylinder displaces 703.72 cu. in of water, which 

weighs 

703 T> 
62 5 X j~~ = 25.45 Ib 

The buoyant effort being less than the weight, the cylinder will 
sink Ans. 



SPECIFIC GRAVITY 

43. Definition. The specific gravity of a solid or 
of a liquid substance is the ratio of the weight of any volume 
of that substance to the weight of an equal volume of water. 
This ratio varies slightly with temperature, and the values 
given by physicists are exactly correct only for a temperature 
of about 39 F , at which water has its maximum density. 
For practical purposes, however, it is not necessary to take 
changes of temperature into account. The abbreviation 
Sp. Gr is often used for specific gravity. 

44. Specific Gravity of Solids Not Soluble in 
Water. The principle of Archimedes affords a very easy 
manner of determining 1 the specific gravity of a solid not 
soluble in water. The body is first weighed in air; it is then 
attached to a scale pan and weighed in water. The difference 
between the two weights will be the weight of an equal volume of 
water. The ratio of the weight in air to the difference thus found 
will be the specific gravity. 

Let W be the weight of the body in air and W the weight 
in water; then W W is the weight of a volume of water 
equal to the volume of the solid, and 

Sp. Gr. 
** w- W 

EXAMPLE. A body weighs in air 364- ounces and in water 30 ounces. 
What is its specific gravity? 

SOLUTION. Here W 36}, and W 30. Substituting In the 
formula, 

Sp. *. - ** - f - 6.8. AM. 



30 HYDROSTATICS 3J 

45. If the body is lighter than water, a piece of iron or 
other heavy substance must be attached to it, sufficiently 
heavy to sink it. Then the two bodies are weighed together 
both in air and in water, both are weighed separately in air, 
and the heavier body in water. Subtracting the combined 
weight of the bodies in water from then combined weight 
in air, the result will be the weight of a volume of water 
equal to the volume of the two bodies. The difference 
between the weight of the heavy body m air and in water 
gives the weight of a volume of water equal to the volume 
of the heavy body. Subtracting this last result from the 
former, the result will be the weight of a volume of water 
equal to the volume of the light body. The weight of the 
light body in air divided by the weight of an equal volume 
of water, as just determined, is the specific gravity of the 
light body. 

Let W = weight of both bodies in air; 

W = weight of both bodies in water; 

w = weight of light body in air; 
Wi = weight of heavy body in air; 
W, = weight of heavy body in water. 

Then, the specific gravity of the light body is given by the 
formula 



Sp. Gr = 



w 



(W- W}-(W l - W,) 

EXAMPLE. A piece of cork weighs, in air, 4 8 ounces To it is 
attached a piece of cast iron weighing 36 ounces in air and 31 ounces 
m water The weight of the iron and cork together, in water, is 
15 8 ounces, what is the specific gravity (a) of cork? (<J) of cast iron? 

SOLUTION. (a) Here w = 4 8, W = 40 8, W = 15 8; W v - 30, 
W t = 81 Substituting in the formula, 

Spec,fic gravity is (4Q g _ ^*_ (M __ _ ^ ^ 
(b) To apply formula of Art 44, W = 36, W - 31. 

no 

Specific gravity of cast iron is Q . ., => 7 2. Ans. 

oo SI 

46. Specific Gravity of a Liquid. To determine the 
specific gravity of a liquid, proceed as follows: Weigh an 
empty flask; fill it with water, then weigh it and find the 



31 HYDROSTATICS 31 

difference between the two results; this will be equal to 
the weight of the water Then weigh the flask filled with 
the liquid, and subtract the weight of the flask; the result 
is the weight of a volume of the liquid equal to the volume 
of the water. The weight of the liquid divided by the 
weight of the water is the specific gravity of the liquid. 
Let w = weight of empty flask, 

W = weight of flask when filled with the liquid; 
W 1 = weight of flask when filled with water 

Then, Sp.Gr. = *-' 

W 1 w 

EXAMPLE A flask when empty weighs 8 ounces; when filled with 
water, 33 ounces, and when filled with alcohol, 28 ounces. What is 
the specific gravity of the alcohol? 

SOLUTION Here W = 28, w = 8, W = 33 Substituting in the 
formula, 

no __ Q 

Sp Gi. = ^__| = 8 Ans. 

47. Nicholson's Hydrometer. Instruments called 
hydrometers are in general use for detei- 
mining quickly and accurately the specific 
gravity of liquids and of some solids. One of 
the principal forms is Nicholson's hydrometer, 
which is shown in Fig. 20. It consists of a 
hollow cylmdei, carrying at its lower end a 
basket //, heavy enough to keep the apparatus 
upright when placed in water. At the top of 
the cylinder is a veitical rod, to which is 
attached a shallow pan a The cylinder is made 
of such size and weight that the apparatus is 
somewhat lighter than water, and a certain 
weight W must be placed in the pan to sink Fl 
it to a given point c on the rod. The body whose specific 
gravity it is desired to find must weigh less than W. It is 
placed in the pan a, and enough weight w is added to sink 
the point c to the water level. It is evident that the weight 
In air of the given body is W w. The body is now 
removed from the pan a and placed in the basket d, an 




32 HYDROSTATICS 31 

additional weight being added to sink the point e to the 
water level. Represent the weight now m the pan by W 1 . 
The difference W w is the weight of a volume of water 
equal to the volume of the body. Hence, 



EXAMPLE The weight necessary to sink a hydrometer to the 
point c is 16 ounces, the weight necessary when the body is In the 
pan a is 7 3 ounces; and when the body is in the basket d, 10 ounces. 
What is the specific gravity of the body? 

SOLUTION Here W = 16, w = 7 3, W = 10 Substituting in the 
formula, 

I Q IT Q 

Specific gravity = ^ ~ = 3.22 Ans. 

1U " i o 

48. Volume of an Irregular Solid. The principle of 
Archimedes affords a very easy and accurate method of find- 
ing the volume of an irregularly shaped body. Let the 
weight of the body in air be W, and in water W. The 
difference W W is, according to the principle of Archi- 
medes, the weight of a volume of water equal to the volume of 
the body. If this volume is denoted by F, and the weight of 
water per unit of volume is denoted by W , the weight 
of the volume V is W, V, and, therefore, 
W V = W- W'\ 

whence, F= W ~ W> (1) 

W, 

If the volume is expressed in cubic feet, W,is 62.5 pounds, 
and, therefore, 

V= W ^^~ = -016(^- W) (2) 

If the volume is expressed in cubic inches, W* = .03617 
pound, and, therefore, 

v = "~ = 27<647( w " w ' } 



EXAMPLE The weight of a body in air is 96 pounds, and m water, 
48 8 pounds What is the volume of the body? 

SOLUTION To apply formula 2, we have W = 96 and W * 48.8. 
Substituting in the formula, 

V .016(98 - 48.8) - .76 cu. ft. Ans. 



31 HYDROSTATICS 33 

49. If the specific gravity of a body is known, the cubical 
contents of the body can be found by dividing its weight by 
its specific gravity, and then dividing again by either .08617 
or 62 5, according as the volume is desired in cubic inches 
or in cubic feet. 

EXAMPLE A certain body has a specific gravity of 4 38 and weighs 
76 pounds What is the volume of the body m cubic Inches? 

76 
SOLUTION.- __-_ . 479.7 en. In. Ans. 



EXAMPLES FOB PRACTICE 

1 If a certain quantity of red lead weighs 5 pounds in air and 
4 441 pounds in water, what is its specific gravity? Ans. 8.94 

2. A piece of iron weighing 1 pound in air and 861 pound in water 
is attached to a piece of wood weighing 1 pound in air. When both 
bodies are placed in water they weigh 2 pound. What is the specific 

gravity: (a) of the iron? (6) of the wood? Aric /() 7 194 

Ans> \() 602 

3. An empty flask weighs 13 ounces; when filled with water, it 
weighs 22 ounces, and when filled with sulphuric acid, 29.56 ounces. 
What is the specific gravity of the acid? Ans. 1 84 

4. How many cubic feet of brick having a specific gravity of 1.9 
are required to make a total weight of 260 pounds? Ans. 2.39 cu. ft. 



PNEUMATICS 



PROPERTIES OF AIR AND OTHER 

GASES 

1. Pneumatics is that branch of science that treats of 
the mechanical properties of gases. 

2. The distinguishing property of a gas is that, no matter 
how small the quantity may be, the gas -will always till the 
vessel or vessels that contain it. If 

a bladder is partly filled with 
air and placed under a glass jar 
(called a receiver) from which 
the air has been exhausted, the 
bladder will immediately 
expand, as shown in Fig. 1. The 
force that a gas exerts when con- 
fined in a limited space is called 
tension. In this case, the 
word tension means pressure, 
and is only used in this sense 
when referring to gases. Fm. i 

3. As water is the most common type of fluids, so air is 
the most common type of gases. It was supposed by the 
ancients that .air was "imponderable," by which was meant 
that it weighed nothing, and it was not until about the year 
1650 that it was proved that air really has weight. A cubic 
inch of air, under ordinary conditions, weighs about .31 gram. 
The ratio of the weight of air to that of water is about 1 . 774; 
that is, air is only T^T as heavy as water. 

OOPYMiaHTID BY INTERNATIONAL TKXTMQOK COMPANY. AU. RIOHTft RMKHVtO 

532 




2 PNEUMATICS 32 

It was shown in Hydrostatics that if a body is immersed 
in water, and weighs less than the volume of water it dis- 
places, the body will rise and project partly out of the water. 
The same principle, which is the principle of Archimedes, 
applies to gases. If a vessel made of light material is filled 
with a gas lighter than air, so that the total weight of the 
'. vessel and gas is less than the weight 

i of the volume of air they displace, the 

\ vessel will nse. It is on this pnnci- 

! pie that balloons are made. 

! 4. Since air has weight, it is evi- 

, dent that the enormous quantity of 

air that constitutes the atmosphere 
must exert a considerable pressure 
on the earth. This is easily proved 
by taking a long glass tube, closed at 
one end, and filling it with mercury. 
If the finger is placed over the open 
end, so as to keep the mercury from 
running out, and the tube is inverted 
and placed in a glass partly filled with 
the same liquid, as shown in Fig. 2, 
the mercury in the tube will fall, then 
rise, and after a few oscillations will 
come to rest at a height above the top 
of the mercury in the glass equal to 
about 30 inches This height will 
always be practically the same under 
PlQ 2 the same atmospheric conditions 

Now, since the atmosphere has weight, it presses on the 
upper surface of the mercury in the glass with equal force on 
every square unit, except on that part of the surface occupied 
by the tube. According to Pascal's law (see Hydrostatics), 
this pressure is transmitted in all directions. There being 
nothing in the tube, except the mercury, to counterbalance 
the upward pressure of the air, the mercury falls in the tube 
until it exerts an upward pressure on the upoer surface of 




32 PNEUMATICS 3 

the mercury in the glass sufficiently great to counterbalance 
the downward pressure produced by the atmosphere. In 
order that there may be equilibrium, the pie&sure of the air 
per unit of area on the upper surface of the mercury in the 
glass must equal the pressure (weight) exerted per unit of 
area by the mercury inside of the tube. Suppose that the 
area of the inside of the tube is 1 square inch, then, since 
mercury is 13 6 times as heavy as water, and 1 cubic inch ot 
water weighs about 03617 pound, the weight of the mer- 
curial column is .03617 X 13.6 X 30 = 14.7574 pounds The 
actual height of the mercury is a little less than 30 inches, 
and the actual weight of a cubic inch of distilled water is a 
little less than .03617 pound. When these considerations 
are taken into account, the average weight of the mercmial 
column at the level of the sea is 14.696 pounds, or, as it is 
usually expressed, 14.7 pounds. Since this weight, exerted 
on 1 square inch of the liquid in the glass, just produces 
equilibrium, it is plain that the pressure of the outside air 
is 14 7 pounds on every square inch of suiface. This pres- 
sure is often referred to as one atmosphere. A pressure of 
two atmospheres is a pressure of 2 X 14.7 = 29.4 pounds 
per square inch, a pressure of three atmospheres is a pres- 
sure of 3 X 14.7 = 44.1 pounds per square inch; etc. 

5. Vacuum. Referring to Fig. 2, the space between 
the upper end of the tube and the upper surface of the mercury 
in the tube is called a vacuum, or empty space. If this 
space contained a gas of some kind, no matter how small the 
quantity might be, the gas would expand and fill the space, 
and its tension would, according to the amount present, 
cause the column of mercury to fall and become shorter; the 
space would then be called a partial vacuum. If the mer- 
cury fell 1 inch, so that the column was only 29 inches high, 
this would be expressed by saying that there were 29 inches 
of vacuum; a fall of 8 inches would be referred to as 22 inches 
of vacuum; a fall of 16 inches, as 14 inches of vacuum, etc. 
Hence, when the vacuum gauge of a condensing engine 
shows 26 inches of vacuum, there i$ enough air in the 



PNEUMATICS 



32 



condenser to produce a pressure of 



_ no 

X 14.7 = 



30 



X 14 7 




= 1.96 pounds per square inch. In all cases where the mer- 
^ __ cunal column is used to measure a vacuum, the 

height of the column, in inches, gives the number 
of inches of vacuum Thus, if the column were 5 
inches high, or the vacuum gauge showed 5 inches, 
the vacuum would be 5 inches 

If the tube had been filled with water instead 
of mercury, the height of the column of water to 
balance the pressure of the atmosphere would 
have been about 30 X 13.6 = 408 inches = 34 feet 
This means that if a tube is filled with water, and 
is inverted and placed in a dish of water in a 
manner similar to that shown m Fig 2, the result- 
ing height of the column of water will be about 
34 feet 

6. The barometer is an instrument used for 
measuring the pressure of the atmosphere There 
are two kinds in general use the mercurial 
barometer and the aneroid barometer. The latter 
was described in Leveling The mercurinl 
barometer is shown in Fig 3. The principle is 
the same as in the case of the inverted tube shown 
in Fig 2 The tube and cup at the bottom are 
protected by a brass or an iron casing At the 
top of the tube is a graduated scale that can be 
read to rsW inch by means of a vernier. Attached 
to the casing is an accurate thermometer for deter- 
mining the temperature of the outside an at the 
time the barometric observation is taken. This 
is necessary, since mercury expands when the 
temperature rises, and contracts when the temper- 
ature falls; for this reason a standard temperature 
is assumed, and all barometer readings are reduced to this 
temperature. This standard temperature is usually taken as 
32 F., at which temperature the height of the mercurial 




PlG 3 



32 PNEUMATICS 5 

column is 30 inches. Another correction is made for the alti- 
tude of the place above sea level, and a third correction for 
the effects of capillary attraction. It is not necessary here 
to go into details regarding these corrections. 

7. The pressure of the atmosphere varies with the alti- 
tude above sea level, being greater in low than in high 
places. At the level of the sea, the height of the mercurial 
column is about 30 inches, at 5,000 feet above the sea, it is 
24.7 inches; at 10,000 feet above the sea, it is 20 5 inches; 
at 15,000 feet above the sea, it is 16.9 inches; at 3 miles, 
it is 16 4 inches; and at 6 miles above the sea level, it is 8 9 
inches. 

The density also varies with the altitude; that is, a cubic 
foot of air at an elevation of 5,000 feet above the sea level 
does not weigh as much as a cubic foot at sea level. This 
is proved conclusively by the fact that at a height of 83" miles 
the mercurial column measures but 15 inches, indicating 
that half the weight of the entire atmosphere is below that 
height It is known that the height of the earth's atmos- 
phere is at least 50 miles; hence, the air just before reach- 
ing the limit must be in an exceedingly rarefied state. It 
is by means of barometers that great heights are measured. 
The aneroid barometer has the heights marked on the dial, 
so that it can be read directly. With the mercurial barom- 
eter, the heights must be calculated from the reading. (See 
Leveling.) 

8. The atmospheric pressure is everywhere present, 
and presses all objects in all directions with equal force. If 
a book is laid on the table, the air presses on it in every 
direction with an equal average force of about 14.7 pounds 
per square inch. It would seem as though it would take 
considerable force to raise a book from the table, since, if 
the size of the book were 8 inches by 5 inches, the pressure 
on it would be 8 X 6 X 14 7 = 588 pounds But there is an 
equal pressure beneath the book to counteract the pressure 
on the top. It would now seem, as though it would require 
a great force to open the book, since there are two pressures 

I LT 398-17 



6 PNEUMATICS 32 

of 588 pounds each, acting in opposite directions, and tending 
to crush the book, and so it would but for the fact that there 
is a layer of air between each two leaves acting upwards and 
downwards with a pressure of 14.7 pounds per square inch. 
If two metal plates are made as perfectly smooth and flat as 
it is possible to make them, and the edge of one is laid on 
the edge of the other, so that one may be slid on the othei, 
and the air thus excluded, it will take an immense force, 
compared with the weight of the plates, to separate them. 
This is because the full pressure of 14 7 pounds per square 
inch is then exerted on each plate with no counteracting 
equal pressure between the plates 

If a piece of flat glass is laid on a flat surface that has been 
previously moistened with water, it will require considerable 
force to lift it off the surface. This is due to the fact that 
the water helps to fill up the pores m the flat surface and 
glass, and thus creates a partial vacuum between the glass 
and the surface, thereby reducing the counterpressure beneath 
the glass. 

9. Tension of Gases. In Art. 5, it was said that the 
space above the column of mercury in Fig. 2 was a vacuum, 
and that if any gas or air were present it would expand, its 
tension forcing the column of mercury downwards. If 
enough gas is admitted to cause the mercury to stand at 
15 inches, the tension of the gas is evidently 14,7 2 
= 7 35 pounds per square inch, since the pressure of the out- 
side air of 14.7 pounds per square inch only balances 15 inches, 
instead of 30 inches, of mercury, that is, it balances only 
half as much as it would if there were no gas in the tube; 
hence, the pressure (tension) of the gas in the tube is 7.35 
pounds. If more gas is admitted until the top of the mercurial 
column is just level with the mercury in the cup, the gas m 
the tube has then a tension equal to the outside pressure of 
the atmosphere. Suppose that the bottom of the tube is 
fitted with a piston, and that the total length of the inside of 
the tube is 36 inches. If the piston is shoved upwards so 
that the space occupied by the gas is 18 inches long, instead 



32 PNEUMATICS 7 

of 36 inches, the temperature remaining the same as before, 
it will be found that the tension of the gas within the tube is 
29.4 pounds per square inch. It will be noticed that the 
volume occupied by the gas is only half that in the tube 
before the piston was moved, while the pressure is twice as 
great, since 14.7 X 2 = 29.4 pounds. If the piston is shoved 
up so that the space occupied by the gas is only 9 inches, 
instead of 18 inches, the temperature still remaining the 
same, the pressure will be found to be 58.8 pounds per square 
inch. The volume has again been reduced one-half, and the 
pressure increased two times, since 29.4 X 2 = 58.8 pounds. 
The space now occupied by the gas is 9 inches long, whereas, 
before the piston was moved, it was 36 inches long, as the 
tube was assumed to be of uniform diameter throughout its 
length, the volume is now TUT = i of its original volume, and 

CO O 

its pressure is -"- = 4 times its original pressure. More- 
over, if the temperature of the confined gas remains the 
same, the pressure and volume will always vary in a similar 
way. The law that states these effects is called Marwtte*s 
law, and is as follows: 

10. Marietta's Law. The temperature remaining the 
same, the volume of a. given quantity of gas varies inversely as 
the pressure. 

The meaning of this is: If the volume of the gas is dimin- 
ished to one-half, one-third, one-fifth, etc. of its former 
volume, the tension will be increased two, three, five, etc. 
times; or, if the outside pressure is increased two, three, five, 
etc. times, the volume of the gas will be diminished to one- 
hnlf, one-third, one-fifth, etc. of its original volume, the tem- 
perature remaining constant. 

Suppose 3 cubic feet of air to be under a pressure of 
60 pounds per square inch in a cylinder fitted with a mov- 
able piston; then the product of the volume and pressure is 
3 X 60 = 180. Let the volume be increased to 6 cubic feet; 
then the pressure will be 80 pounds per square inch, and 
30 X 6 SB 180, as before, Let the volume be increased to 



8 PNEUMATICS 32 

24 cubic feet; it is then 24 3 = 8 times the original volume, 
and the pressure is one-eighth of the original pressure, or 
60 X i = 71 pounds, and 24 X 7i = 180, as in the two pre- 
ceding cases. It will now be noticed that, if a gas is allowed 
to expand without change of temperature, the pioduct of any 
Pressure and the corresponding volume is the same as for any other 
pressure and the corresponding volume If the air were com- 
pressed, the same result would be obtained. 

Let p = pressure corresponding to volume v; 
pi = pressure corresponding to volume z^. 

Then, p v = p v 

Knowing the volume and the pressure for any position of 
the piston, and the volume for any other position, the pres- 
sure may be calculated; or, if the pressure is known for any 
other position, the volume may be calculated. 

EXAMPLE 1 If 1 875 cubic feet of air is under a pressure of 
72 pounds per square inch, what will be the pressure when the volume 
is increased (a) to 2 cubic feet? (b) to 3 cubic feet? (c) to 9 cubic feet? 

SOLUTION Solving the last equation for/,, the unknown pressure 
gives 

. . . p v 72 X 1 875 _, ., 

(a) p! = r -- = - = --- = 67 Ib per sq m Ans. 

v\ & 

... . 72 X 1 875 ._ ,. . 

(b) pi = = 45 Ib per sq in Ans. 

, . . 72 X 1 875 .,_ ,. 

(c) pi = g =15 Ib. per sq. in Ans 

EXAMPLE 2 Ten cubic feet of air has a tension of 5.6 pounds per 
square inch, what is the volume when the tension is (a) 4 pounds? 
(b) 8 pounds? (c) 25 pounds? (d) 100 pounds? 

SOLUTION Solving the same equation for v,. gives 

(a) , l =^ = 56 * 10 = 14cu ft Ans 

pi 4 

7 cu ft Ans 



, 

(c) Vl = 5 6 * 10 = 2 24 cu ft Ans 

(<0 i = ^ = 56 cu ft Ans. 



11. For the same quantity of gas, the weight per unit 
of volume varies inversely as the volume. For example, if 



32 PNEUMATICS 9 

1 pound of gas occupies n volume of 4 cubic feet, its weight 
per cubic foot will be i pound. If it occupies a volume 
twice as large, or 8 cubic feet, its weight per cubic foot 
will be i pound, or only one-half of what it was before. In 
general, it, the temperature of a fixed quantity of gas 
remaining the same, the weights per unit of volume when 
the gas occupies the volumes v and z>,, respectively, are 
denoted by w and w lt then, 

w _ v\ 

Wj. V 

and wv z0, v, (1) 

Also, since = 

v i 



and w p l = iv ip (2) 

EXAMPLE 1 The weight of 1 cubic foot of air at a temperature of 
60 F , and under a pressure of 1 atmosphere (14.7 pounds per 
square inch), is 0763 pound, what would be the weight per cubic foot 
if the volume were compressed until the tension was 5 atmospheres, 
the temperature still being 60 F ? 

SOLUTION. Applying formula 2, 1 X w\ = 6 X .0763. Hence, 
a/, = .3815 Ib. per cu. ft. Ans. 

EXAMPLE 2. If in the last example the air had expanded until the 
tension was 5 pounds per square inch, what would have been its 
weight per cubic foot? 

SOLUTION. Here, p 14 7, p l = 5, and w = 0703 Hence, apply- 

ing formula 2, 14.7 X w, = 5 X .0763, whence a/, = ~~y = 02695 Ib. 
per cu. ft. Ans. 

EXAMPLE 3 If 6.75 cubic feet of air, at a temperature of 60 F. 
and a pressure of 1 atmosphere, is compressed to 2 25 cubic feet 
(the temperature still remaining 60 P.), what is the weight of a cubic 
foot of the compressed air? 

SOLUTION Applying formula 1, 6.75 X .0763 = 2 25 X w t . Hence, 



w, m - 22 2289 Ib. per cu. ft. An 8 . 

12. Manometers and Gauges, There are two ways 
of measuring the pressure of a gas: by means -of an instru- 
ment called a manometer, and by means of a gauge. 



10 PNEUMATICS 32 

The manometer generally used is practically the same as a 
mercurial barometer, except that the tube is much longer, so 
that pressures equal to several atmospheres may be meas- 
ured, and is enlarged and bent into a U shape at the lower 
end, both the lower and the upper ends are open, the lower 
end being connected to the vessel containing the gas whose 
pressure it is desired to measure. The gauge is so com- 
mon that no description of it will be given here. With both 
the manometer and the gauge, the pressure recorded is the 
amount by which the piessure being measured exceeds the 
atmospheric pressure, and is called gauge pressure. To 
find the total pressure, or absolute pressure, the atmos- 
pheric must be added to the gauge pressure. In all for- 
mulas in which the pressure of a gas or steam occurs, the 
absolute pressure must be used, unless the gauge pressure is 
distinctly specified as the proper one to use. For conve- 
nience, all pressures given in this Section and in the ques-- 
tions referring to it will be absolute pressures, and the word 
"absolute" will be omitted, to avoid repetition. 

13. In all that has been said, it has been stated that the 
temperature was constant; the reason for this will now be 
explained. Suppose 5 cubic feet of air to be confined in a 
cylinder placed in a vacuum, so that there will be no pressure 
due to the atmosphere, and suppose the cylinder to be fitted 
with a piston weighing, say, 100 pounds, and having an 
area of 10 square inches. The tension of the air will be 
1 r q u a = 10 pounds per square inch. Suppose, now, that the 
air, originally at a temperature of 32 F., is heated until its 
temperature is 33 F that is, until its temperature is raised 
1. It will be found that the piston has risen a certain 
amount, and, consequently, the volume has increased, while 
the pressure remains the same as befoie, 01 10 pounds per 
square inch. If more heat is applied, until the temperature 
of the gas is 34 F , it will be found that the piston has again 
risen, and the volume again increased, while the pressure 
still remains the same. It will be found that for every 
increase in temperature there will be a corresponding increase 



32 PNEUMATICS 11 

of volume. The law that expresses this change is called 
Gay-Lussac 's law^ and is as follows: 

14. Gay-IJussac's Liaw. // the pressure remains con- 
stant, every increase tn temperature of 1 F. produces in a given 
quantity of gas an expansion 0/To of its volume at 32 f. 

If the pressure remains constant, it will also be found that 
every decrease of temperature of 1 F. will cause a decrease 
of TU of the volume at 32 F. 

Let z/ volume of any quantity of gas at 32 F.; 
v = volume of gas at temperature t; 
z^ =* volume of gas at temperature /,. 

It is assumed that the pressure of the gas remains 
unchanged. Then, m passing from the temperature 32 to 
the temperature /, the volume of the gas will be increased 

J _ QO 

algebraically by the amount TUYZ/O (t 32) = jnn v*. 

492 

Therefore, 



or, reducing, v = flo T~- (D 

492 



Likewise, v, = v, --' (a) 

492 

Dividing formula 1 by equation (a), 
v _ 460 + t_ 
v\ 460 + t, 

whence v - V *M (2) 



EXAMPLE If 5 cubic feet of air at a temperature of 45 is heated 
under constant pressure up to 177, what is the final volume of the air? 

SOLUTION To apply formula 2, we have v^ = 5; , 46; t 177. 

Therefore, 

460 + 177 

v * 5 x 07 cu ' ft AnSi 



15. Suppose that a certain volume of gas is confined 
in a vessel so that it cannot expand; in other words, sup- 
pose that the piston of the cylinder before mentioned is 
fastened so that it cannot move. Let a gauge be placed on 



ihl 



12 PNEUMATICS 32 

the cylinder so that the tension of the confined gas can be 
registered If the gas is heated, it will be found that, for 
every increase of temperature of 1 F , there will be a cor- 
responding increase of TUT of the tension That is, the 
volume remaining: constant, the tension increases Tn of the 
original tension for eveiy degree rise of temperature. 

Let p tension of gas at temperature /, 

pi = tension of gas at temperature ti. 

Then, as in the preceding article, 



= 
y ^460 + ^ 

EXAMPLE If a certain quantity of air is heated under constant 
volume from 45 to 177, what is the resulting tension, the original 
tension being 14 7 pounds per square inch? 

SOLUTION Applying the formula, 

p = 14 7 X t' 7 - 18 542 lb per sq in Ans 



16. According to the modern and now generally accepted 
theory of heat, the atoms and molecules of all bodies are in 
an incessant state of vibration. The vibratory movement 
in liquids is faster than in solids, and in gases faster than 
in either of the other two Any increase of heat increases 
the vibrations, and a decrease of heat decreases them From 
experiments and advanced mathematical investigations, it has 
been concluded that at 460 below zero, on the Fahrenheit 
scale, all these vibrations cease. This point is called the 
absolute zero, and all temperatures reckoned from this 
point are called absolute temperatures. The point of 
absolute zero has never been reached, nevertheless, it 
has a meaning, and is used in many formulas, absolute tem- 
peratuies being usually denoted by T. Ordinary temperatures 
are denoted by /. When the word temperature alone is used, 
it refers to the ordinary way of measuring temperatures, but 
when absolute temperature is specified 460 F. must be added 
to the ordinary temperature The absolute temperature cor- 
responding to 212 F. is 460 -f 212 = 672 F, If the abso- 
lute temperature is given, the ordinary temperature may be 
found by subtracting 460 from the absolute temperature. 



32 PNEUMATICS 13 

Thus, if the absolute temperature is 520 P., the ordinary 
tempeiature is 520 - 460 = 60. 

17. Let p = pressure, in pounds per square inch, of 

W pounds of air; 
V = volume of air, in cubic feet, 
T absolute tempeiature of air. 

It is shown in advanced works on the theory of heat that 
these quantities are related by the following- equation: 
pV = .37 WT 

EXAMPLE 1 The pressure on 9 cubic feet of air weighing 1 pound 
is 20 pounds per squaie inch, what is the ordinary temperature of the 
air? 

SOLUTION Here, W= 1. Substituting the other values in the 

180 

formula gives 20 X 9 = 37 T, hence, T = ~ = 486.6, nearly. 

.37 

40 5 - 4(50 = 26 6, the ordinary temperature. Ans 

EXAMPLE 2 What is the volume of 1 pound of air whose tempera- 
ture is 00 F under a pressure of 1 atmosphere? 

SOLUTION Here, W - 1, and T = 460 -f 60 - 520, therefore, 14 7 

07 \f Knn 

X V = .87 X 520, whence V = ~^~ = 18 088 cu. ft Ans. 

EXAMPLE 3 If 3 cubic feet of air weighing 35 pound Is under a 
pressure of 48 pounds per square inch, what Is the ordinary tem- 
perature of the air? 

SOLUTION Applying the formula, 48 X 8 - .37 X .35 X T, 

JO v ^ 

whence, T= --%. - 1,112 Then, 1,112 - 460 - 652. 

A I X <50 

Ans. 

EXAMPLK 4. What is the weight of 1 cubic foot of air at a tempera- 
ture of 32, and under a pressure of 1 atmosphere? 

SOLUTION Here T - 400 + 32 = 492; V = 1; and p = 14 7 
(Art. 4). Substituting these values in the formula gives 14 7 X 1 
- .37 X 492 X W, whence, 

Ans - 



If the pressure is taken as 14.696 Ib. per sq. in , the weight of 1 cu. ft. 
of air, at 32 and atmospheric pressure, is found to be 

- - 08078 lb 



14 PNEUMATICS 32 



EXAMPLES FOR PRACTICE 

1 A vessel contains 25 cubic feet of gas at a pressure of 18 pounds 
per square inch, if 125 cubic feet of gas having the same pressure Is 
forced into the vessel, what will be the resulting pressure? 

Ans 108 Ib. per sq. in. 

2 A pound of air has a temperature of 126, and a pressure of 1 
atmosphere, what volume does it occupy? Ans. 14 75 cu. ft. 

3 A certain quantity of air has a volume of 26 7 cubic feet, a pres- 
sure of 19.3 pounds per square inch, and a temperature of 42; what 
is its weight? Ans. 2 77 Ib 

4 A receiver contains 180 cubic feet of gas at a pressure of 20 
pounds per square inch, if a vessel holding 12 cubic feet is filled from 
the receiver until its pressure is 20 pounds per square Inch, what will 
be the pressure m the receiver? Ans 18$ Ib per sq in. 

5 Ten cubic feet of air, having a pressure of 22 pounds per square 
inch and a temperature of 75, is heated until the temperature is 300, 
the volume remaining the same, what is the new pressure? 

Ans 31 25 Ib per sq in. 

THE MIXING OF GASES 

18. If two liquids that do not act chemically on each 
other are mixed together and allowed to stand, it will be 
found that after a time the two liquids have separated, and 
that the heavier has fallen to the bottom. If two equal ves- 
sels containing: gases of different densities are put in com- 
munication with each other, it will be found that after a short 
time the gases have become mixed in equal proportions. 
If one vessel is higher than the other, and the heavier 
gas is in the lower vessel, the same result will occur. The 
greater the difference of the densities of the two gases, 
the faster they will mix. It is assumed that no chemical 
action takes place between the two gases. When the two 
gases have the same temperature and pressure, the pressure 
of the mixture will be the same. This is evident, since the 
total volume has not been changed, and, unless the volume 
or temperature changes, the pressure cannot change This 
property of the mixing of gases is a very valuable one, since, 
if gases acted like liquids, carbonic-acid gas (the result 



32 PNEUMATICS 15 

of combustion), which is li times as heavy as air, would 
remain next to the earth, instead of dispersing into the 
atmosphere, the result being that no animal life could exist. 



19. Mixture of Equal Volumes of Gases 
Unequal Pressures. // two gases having equal volumes and 
temperatures, but different Pressures, ate mixed in a vessel vu hose 
volume equals one of the equal volumes of the gases, the pressure 
of the mixture will be equal to the sum of the two pressures, 
provided that the temperature remains the same as before. 

20. Mixture of Two Gases Having; Unequal Vol- 
umes and Pressures. Let v and p be the volume and 
pressure, respectively, of one of the gases. 

Let v 1 and PI be the volume and pressure, respectively, of 
the other gas. 

Let f-'and P be the volume and pressure, respectively, of 
the mixture. Then, if the temperature remains the same, 

yp= vfi + v^ 

That is, if the temperature is constant, the volume after mix- 
ture, multiplied by the resulting pressure, equals the volume of 
one gas before mixture multiplied by its pressure, plus the volume 
of the other gas multiplied by its pressure* 

EXAMPJ.B Two gases at the same temperature, having volumes of 
7 cubic feet and 4J cubic feet, and pressures of 27 pounds and 
18 pounds per square inch, respectively, are mixed together In o vessel 
whose volume is 10 cubic feet. What is the resulting pressure? 

SOLUTION. Applying the preceding formula, J V p v + p^ v tl or 
/>X 10 - 27 X 7 + 4t X 18. Hence, P - i?-il - 27 Ib. per sq. in. 

Aim. 

21. Mixture of Two Volumes of Air Having 
Unequal Pressures, Volumes, and Temperatures. If 

a body of air having a temperature /, a pressure p^ and a vol- 
ume v l is mixed with a volume of air having 1 a temperature * 
a pressure /> and a volume z>,, to form a volume f having a 
pressure /* and a temperature t, then, either the new temper- 
ature /, the new volume V, or the new pressure P may be 
found, if the other two quantities are known, by the following 



16 PNEUMATICS 32 

formula, in which T lt T at and Tare the absolute temperatures 
corresponding to / / a , and /: 

PV= /A^i + A 
\Ti T 

EXAMPLE Five cubic feet of air having a tension of 30 pounds per 
square inch and a temperature of 80 F are compressed, together 
with 11 cubic feet of air having a tension of 21 pounds per square inch 
and a temperature of 45 F , in a vessel whose cubical contents are 
8 cubic feet The new pressure is required to be 45 pounds per square 
inch What must be the temperature of the mixture? 



SOLUTION Substituting in the formula, 45 X 8 = 

Qflft 

X T, or 360 = .7352 T Hence, T = ~^ = 489 66, nearly, and 
/ = 29 66 Ans 

EXAMPLES FOB PRACTICE 

1 Two vessels contain air at pressures of 60 and 83 pounds per 
square inch The volume of each vessel is 8 47 cubic feet If all the 
air in both vessels is removed to another vessel, and the new pressure 
is 100 pounds per square inch, what is the volume of the vessel, the 
temperature remaining unchanged? Ans. 12 11 cu ft 

2 A vessel contains 11 83 cubic feet of air at a pressure of 
33 3 pounds per square inch It is desired to increase the pressure to 
40 pounds per square inch by supplying air from a second vessel that 
contains 19 6 cubic feet of air at a pressure of 60 pounds per square 
inch What will be the pressure in the second vessel after the pressure 
in the first has been raised to 40 pounds per square inch? 

Ans. 55.96 Ib. per sq m. 

3 If 4 8 cubic feet of air having a tension of 52 pounds per square 
inch and a temperature of 170 is mixed with 13 cubic feet having a 
tension of 78 pounds per square inch and a temperature of 265, what 
must be the volume of the vessel containing the mixture in older that 
the tension of the mixture may be 30 pounds per square inch and the 
temperature 80? Ans. 32 31 cu ft. 



PNEUMATICS 



17 



PNEUMATIC MACHINES 



THE AIR PUMP 

22. The air pump is an instrument for removing air 
from an enclosed space. A section of the principal parts is 
shown in Fig. 4, and a view of the complete instrument is 
given in Fig. 5. The closed vessel R is called the receiver, 





FIG 4 

and the space it encloses is that from which it is desired to 
remove the air. The receiver is usually made of glass, and 
the edges are ground so as to be perfectly air-tight. When 
made in the form shown, it is called a boll-Jar receiver, 
The receiver rests on a horizontal plate, in the center of 
which is an opening communicating with the pump cylin- 
der C by means of a bent tube /. The pump piston fits the 
cylinder accurately, and has a valve V 1 opening upwards. 
At the junction of the tube with the cylinder is another 
valve V, also opening upwards, When the piston is raised, 



1 



PNEUMATICS 



the valve V closes, and, sinct 1 no .111 can m't into tin- i \liu 
dcr from above, the piston leave's a \aamm brliind it. Tin 
pressure on top of rbemjj now u-ninvi'd, tin 1 ti'iismu il tli 

,ur in llu- rm-iMx IM A' 

iMllsi's / * tn MSI-, lh' .111 

in tlu- ii'i'i'ivvi thru t'\ 
pands and uivupu-s thr 
sp.u't* U'tt rmpty by 
the pistnn, as wi-ll as 
tin- spaiv in the- tube / 
and tin* IT or i VIM A*. 
Tbtj piston is iiuw 
pnslu-tl down, tin 1 vahc- 
/"i-losfs, tlk 1 \,dvt* / ' 
opens, and ihr an nt (' 
csoapcs. Tilt* 
valv /" is snni 
suppiirli'tl, as shown in 
Kitf. 1, by a nu-tal nxl 
passing tlimuuh llu- 
piston and itttinu it 
somewhat tightly. 

When the piston is raised or lowered, this rod moves with it. 

A button near the upper end of the rod confines its motion 

to within very narrow limits, the piston sliding on the rod 

during the greater part of the journey. 

S3. JUetfi'ooH und JLlmltH of Kxlvauwtlon, Suppose 
that the volume o /? and / together is four times that of ( \ 
and that there are, ny, 200 grains of nir in 'A* and /, and 
50 grains in C, when the piston is at the top of the i-ylimU-i. 
At the end of the first stroke, when the piston is a^ain at 
the top, 60 grains of air in the cylinder ( ' will have been 
removed, and the 200 grains in /? and / will omipy the 
spaces jR t /, and C. The ratio between the sum of the 

spaces .# and / and the total space jV + / f Tis ^; hence, 

j 
200 X - 160 grains - the weight of air in / and t after 




32 



PNEUMATICS 



19 




the first stroke. After the second stroke, the weight of the 
air in R and / will be 200 X X = 200 X (*)' = 200 X iif 
= 128 grams. At the end of the third stroke, the weight 
will be [200 X ()'] X = 200 X (*)' = 200 X rW = 102 4 

grains. At the end of n 
strokes, the weight will be 
200 X (&)" It is evident 
that it Zi impossible by this 
method to remove all the air 
contained in R and t. It 
requires an exceedingly 
good air pump to reduce 
the tension of the air in R 
to -gV inch of mercury. 
When the an has become 
so rarefied as this, the 
valve V will not lift, and, 
consequently, no more air 
can he exhausted 

24. Spreiigel's Air 
Pump. In Fig 6, cd is a 
glass tube longer than 30 
inches, open at both ends, 
and connected by means of 
India-rubber tubing with a 
funnel A filled with mer- 
cury and supported by a 
stand. Mercury is allowed 
to fall into this tube at a 
rate regulated by a clamp 
at c. The lower end of the 
tube c d fits in the flask ./?, 
which has a spout at the 
side a little higher than the lower end of c d\ the upper part 
has a branch at x to which a receiver R can be tightly fixed. 
When the clamp a.t c is opened, the first portions of the mer- 
cury that run out close the tube and prevent air from enterirg 




Fia. 6 



20 



PNEUMATICS 



32 



from below. These drops of mercury act like little pistons, 
carrying the air in front of them and forcing it out through 
the bottom of the tube. The air in R expands to fill the tube 
every time that a drop of mercury falls, thus creating a partial 
vacuum in R, which becomes more nearly complete as the 
process goes on. The escaping mercury falls into the dish H, 
from which it can be poured back into the funnel from time 
to time. As the exhaustion from R goes on, the meicury 
rises in the tubecd until, when the exhaustion is complete, it 
forms a continuous column 30 inches high, in other words, 
it is a barometer whose vacuum is the receiver R. This 
instrument necessarily requires a great deal of time for its 
operation, but the results are very complete, a vacuum of 
^aiUa inch of mercury being sometimes obtained. By use of 
chemicals in addition to the above, a vacuum of TBTMMTO inch 
of mercury has been obtained. 

25. Magdeburg: Hemispheres. The pressure of the 
atmosphere can be made manifest by means of two hollow 
hemispheres, such as are shown in Fig. 7. 
This contrivance was devised by Otto Von 
Guencke, of Magdeburg, and is known as 
the Magdeburg hemispheres. One of 
the hemispheres is provided with a stop- 
cock, by which it can be screwed on to an 
air pump. The edges fit accurately and 
are well greased, so as to be air-tight. 
When the hemispheres contain air, they 
can be separated easily; when the air is 
pumped out by an air pump, they can 
be separated only with great difficulty. 
The force required to separate them will 
be equal to the area of the largest circle 
of the hemisphere (projected area) m 
square inches, multiplied by 14.7 pounds. 
This force will be the same in whatever 
position the hemispheres may be held, .which proves that 
the pressure of air on them is the same m all directions. 




FIG 7 



32 



PNEUMATICS 



21 



26. The Weight Liifter. The pressure of the atmos 
phere is shown by means of the apparatus illustrated in 
Fig 8. Here, a cylinder fitted with a piston is held in 
suspension by a chain. At the top of the cylinder is a 
plug a, which can be taken out. This plug is removed and 
the piston is pushed up until it touches the 
cylinder head. If the plug is then screwed 
in, the piston will remain at the top until 
a weight has been hung on the rod equal 
to the area of the piston multiplied by 14.7 
pounds, less the weight of the pibton and 
rod. If a force is applied to the rod suffi- 
ciently great to push the piston down- 
wards, the piston will, on the removal of 
the force, raise to the top of the cylinder 
any weight that is less than the one men- 
tioned. Suppose the weight to be removed, 
and the piston to be supported midway 
between the top and bottom of the cylinder. 
Let the plug be removed, air admitted 
above the piston, and the plug screwed 
back into its place; if the piston is shoved 
upwards, the farther up it goes, the greater 
will be the force necessary to push it, on 
account of the compression of the air. If 
the piston is of large diameter, it will also 
require a great force to pull it out of the 
cylinder, as a little consideration will show. 
For example, let the diameter of the piston 
be 20 inches, the length of the cylinder 86 
inches, plus the thickness of the piston, and 
the weight of the piston and rod 100 pounds. If the piston 
is in the middle of the cylinder, there will be 18 inches of 
space above it, and 18 inches of space below it. The area 
of the piston is 20' X .7864 =* 314.16 square inches, and the 
atmospheric pressure on it is 814.16 X 14.7 4,618 pounds, 
nearly. In order to shove the piston upwards 9 inches, the 
pressure on it must be twice as great, or 9,286 pounds, and 

I LT 398-18 





Pro 8 



22 



PNEUMATICS 



32 



to this must be added 100 pounds, the weight of the piston 
and rod, which gives 9,236 + 100 = 9,336 pounds. The 
force necessary to cause the piston to move upwards 9 inches 
will then be 9,336 4,618 = 4,718 pounds. Now, suppose 
the piston to be moved downwards until it is just on the point 
of being pulled out of the cylinder The volume above it will 
then be twice as great as before, and the pressure one-half as 
great, or 4,618 2 = 2,309 pounds. The total upward pies- 
sure will be the pressure of the atmosphere less the weight 
of the piston and rod, or 4,618 100 = 4,518 pounds, and 
the force necessary to pull it downwards to this point will 
be 4,518 - 2,309 = 2,209 pounds. 

27. Tlie Baroscope. The buoyant effect of air is very 
clearly shown by means of an instrument called the baro- 
scope, shown in Fig. 9. It con- 
sists of a scale beam, from one 
extremity of which is suspended 
a small weight, and from the 
other a hollow copper sphere. 
In air, they exactly balance each 
other, but when they are placed 
under the receiver of an air 
pump and the air is exhausted 
the sphere sinks, showing that 
it is really heavier than the small 
weight. Before the air is 
exhausted, each body is buoyed 
up by the weight of the air it 
displaces, and, since the sphere displaces more air, it loses 
more weight by reason of this displacement than the small 
weight. Suppose that the volume of the sphere exceeds that 
of the weight by 10 cubic inches; the weight of this volume 
of air is 3.1 grains. If this weight is added to the small 
weight, it will overbalance the sphere in air, but will exactly 
balance it m a vacuum. 




FIG 9 



32 PNEUMATICS 23 



AIR COMPRESSORS 

28. For many purposes, compressed air is preferable to 
steam or other gases for use as a motive power; m such 
cases, air compressors are used to compiess the air. 
These are made in many forms, but the most common one 
consists of a cylinder, called the air cylinder, placed in front 
of the crosshead of a steam engine, so that the piston of the 
air cylinder can be driven by attaching its piston lod to the 
crosshead, in a manner similar to a steam pump. A cross- 
section of the air cylinder of a compressor of this kind is 
shown in Fig 10, in which a is the piston and b is the piston 
rod, driven by the crosshead of a steam engine not shown 
in the figure. Both ends of the lower half of the cylinder 
are fitted with inlet valves d and d', which allow the air to 
enter the cylinder, and both ends of the upper half aie fitted 
with discharge valves / and f, which allow the air to escape 
fiom the cylinder after it has been compressed to the 
required pressure. 

Suppose the piston a to be moving in the direction of the 
arrow; then the inlet valves d in the left-hand end of the 
cylinder from which the piston is moving will be forced 
inwards by the pressure of the atmosphere, which over- 
comes the resistance of the light spring: c t thus allowing 
the air to flow in and fill the cylinder. On the other side of 
the piston, the air is being compressed, and, consequently, 
it acts with the springs ^ to force the inlet valves d' in 
the right-hand end of the cylinder to their seats. In the 
right-hand end of the cylinder, the discharge valves f are 
opened when the pressure of the air in the cylinder is great 
enough to overcome the resistance of the light springs f 
and the tension of the air in the passages leading to the 
discharge pipe h, and the discharge valves / are pressed 
against their seats by the springs c and the tension of the air 
in the passages. Suppose it is desired to compress the air 
to 59 pounds per square inch, and to find at what point of the 
stroke the discharge valves will open. Now, a pressure of 59 
pounds per square inch equals a pressure of 4 atmospheres, 



24 



PNEUMATICS 



32 



very nearly, hence, when the pressure in the cylinder becomes 
great enough to force air out through the discharge valves, 
the volume must be one-quarter of the volume at atmos- 
pheric pressure, or the valves will open when the piston has 
traveled three-quarters of its stroke, provided that the air is 
compressed at constant temperature. 

The air, after being discharged from the cylinder, passes 
out through the delivery pipe h, and from there is con- 




FIG 10 

veyed to its destination. It has been shown that when air or 
any other gas is compressed its temperature is increased. 
For high pressures, this increase of temperature becomes a 
serious consideration, for two reasons: (1) When the air 
is discharged at a high temperature, the pressure falls con- 
siderably when the air has cooled down to its normal temper- 
ature, and this represents a serious loss in the economical 
working of the machine. (2) The alternate heating and 



32 



PNEUMATICS 



25 




cooling of the compressor cylinder by the hot and cold air 
is very destructive to it, and increases the wear to a great 
extent. To prevent the air from heating, cooling devices 
are lesorted to, the most common one being 1 the so-called 
water-Jacket. This is effected in the following manner 
The cylinder walls are hollow, as shown in Fifi 10; the cold 
water enters this hollow space 
in the cylinder wall through the 
pipe k k, and flows around the 
cylinder, finally passing out 
through the discharge pipe /. 
The water keeps cold the cyl- 
inder walls, which cool the air 
as it is compressed. 

29. Hei-o's Fountain. 
Hero's fountain derives its 
name from its inventor, Hero, 
who lived at Alexandria about 
120 B. C. This fountain, which 
is shown in Fig. 11, consists of 
a brass dish A and two glass 
globes B and C, and depends for 
its operation on the elastic prop- 
erties of air The dish com- 
municates with the lower part 
of the globe C by a long tube D, 
and another tube E connects 
the two globes. A third tube 
passes through the dish A to ^ 
the lower part of the globe B. _ 
This last tube being taken out, 
the globe B is partially filled with water; the tube is then 
replaced and water is poured into the dish. The water flows 
through the tube D into the lower globe, and expels the air, 
which is forced into the upper globe. The air thus com- 
pressed acts on the water and makes it jet out through the 
shortest tube, as represented in the figure. Were it not for 



33 




Fie. 11 



26 



PNEUMATICS 



.12 



the resistance offered by the atmosphere and by friction, 
the issuing water would rise to a height above the water 
in the dish equal to the difference of the level of the water m 
the two globes. 

THE SIPHON 

30. The action of the siphon illustrates the effect of 
atmospheric pressure. A siphon is simply a bent tube with 
unequal branches, open at both ends, and is used to convey a. 

liquid from a higher point to a 
lower, over an intermediate point 
higher than either of the other two. 
In Fig 12, a and b are two vessels, b 
being lower than a, and a c b is the 
bent tube or siphon. Suppose this 
tube to be filled with water and 
placed in the vessels, as shown, 
with the short branch a c in the 
vessel a. The watei will flow fiom 
the vessel a into , so long as the 
level of the water in b is below 
the level of the water in a and the 
level of the water in a is above 
the lower end of the tube ac. Tne 
atmospheric pressure on the surfaces of a and b tends to 
force the water up the tubes ac and be. When the siphon 
is filled with water, each of these pressures is counteracted 
in part by the pressure of the water in that branch of the 
siphon that is immersed in the water on which the pres- 
sure is exerted. The atmospheric pressure opposed to the 
weight of the longer column of water will, therefore, be 
more resisted than that opposed to the weight of the 
shorter column; consequently, the pressure exerted on 
the shorter column will be greater than that on the longer 
column, and this excess of pressure will produce motion, 

The action of the siphon is of great importance, and 
should be thoroughly understood. The following considera- 
tions and computations will make the subject clear: 




PNEUMATICS 



27 



Let a = area of tube, in square inches; 

// = dc = vertical distance, in inches, between the 
surface of water in b and highest point 
of the center line of tube, 

//, = e c = distance, in inches, between the surface of 
water in. a and highest point of center 
line of tube. 

The weight of the water in the short column is .03617 a hi, 
and the resultant atmospheric pressuie, tending to force 
the water up the short column, is 14 7 X a .03617 a //,. 
The weight of the water in the long column is 03617 a/i t 
and the icsultant atmospheric pressure, tending to force the 
water up the long column, is 14.7 a .03617 a k. The differ- 
ence between these two is (14.7 a .03617 ah^} (14.7 a 
- 03617 ah} = .03617 a (A - k,) But h - //, = ed = dif- 
ference between the levels of the water in the two vessels. 

It will be noticed that the short column must not be higher 
than 34 feet for water, or the siphon will not work, since the 



,r\ 




PIG 18 

pressure of the atmosphere will not support a column of 
water that is higher than 84 feet; 28 feet is considered to be 
the greatest height for which a siphon will work well. 

31. Intermittent Springs.- Sometimes a spring is 
observed to flow for a time and then cease; then, after an 



28 



PNEUMATICS 



32 



interval, to flow again for a time The generally accepted 
explanation of this is that there is an underground reservoir 
fed with water through fissures in the earth, as shown in 
Fig. 13 The outlet for the water is shaped like a siphon, 
as shown. When the water in the reservoir reaches the same 
height as the highest point of. outlet, it flows out until the 
level of the water in the reservoir falls below the mouth of 
the siphon, if the flow of water is greater than the supply to 
the reservoir, in which case the flow ceases until the water 
in the reservoir again reaches the level of the highest point 
of the siphon. 

THE LOCOMOTIVE BLAST 

32. Fig. 14 shows the front end of a locomotive: B is 
the exhaust pipe, the center of which is directly in line with 




PIG 14 



the center of the smokestack S, T, T are the tubes through 
which the hot furnace gases are discharged. The exhaust 



32 



PNEUMATICS 



29 



steam has a pressure of about 2 pounds above the atmos- 
phere, and rushes through the exhaust pipe E and up the 
smokestack 5 1 with a very high velocity, taking the air out 
with it, and producing 1 a partial vacuum in the space in 
front of the tubes. No air can get in this space except 
through the grates of the firebox, consequently, the partial 
vacuum created in front of the tubes causes an influx of air 
through the grate, and produces the forced draft, or blast. 
The faster the engine runs, the greater is the quantity of 
air drawn through the grate. 



PUMPS 

33. The Suction Pump. A section of an ordinary 
suction pump is shown in Fig. 15. Suppose the piston 
to be at the bottom of 
the cylinder and to be 
just on the point of 
moving upwards in 
the direction of the 
arrow. As the piston 
rises, it leaves a 
vacuum behind it. 
The air in P then 
raises the valve P, 
and expands in the 
cylinder .#, whereby 
its pressure is dimin- 
ished below that of 
the atmosphere. The 
atmospheric pressure 
on the surface of the 
water in the well 
causes the water to rise in the pipe P. When the piston 
descends, the air in B escapes through the valves . After 
a few strokes, the water fills completely the space under 
the piston in cylinder JB, so that, when the piston reaches the 
end of its stroke, the water entirely fills the space between the 




30 PNEUMATICS 32 

bottom of the piston and the bottom of the cylinder and also 
the pipe P. The instant that the piston begins its down 
stroke, the water in the chamber B tends to fall back into 
the well, and its weight forces the valve Fto its seat, thus 
preventing any downward flow of the water. The piston 
now tends to compress the water in the chamber B, but this 
is prevented through the opening of the valves u, u in the 
piston. When the piston has reached the end of its down- 
ward stroke, the weight of the water above closes the valves 
u, u. All the water resting on the top of the piston is then 
lifted with the piston on its upward stroke, and discharged 
through the spout A, the valve V again opening, and the 
water filling the space below the piston as before 

It is evident that the distance between the valve V and 
the surface of the water m the well must not exceed 34 feet, 
the highest column of water that the pressure of the atmos- 
phere will sustain, since otherwise the water in the pipe 
3 would not reach to the height of the 

valve V In practice, this distance 
should not exceed 28 feet This is 
due to the fact that there is a little 
air left between the bottom of the 
piston and the bottom of the cylinder, 
a little air leaks through the valves, 
which are not perfectly air-tight, and 
a pressure is needed to raise the valve 
against its weight, which, of course, 
acts down wards. There are many vari- 
eties of the suction pump, differing 
principally in the valves and piston, 
but the principle is the same in all. 

34. The Lifting Pump. A 
section of a lifting pump is shown 
in Fig 16 These pumps are used 
FIG IB when water is to be raised to greater 

heights than can be done with the ordinary suction pump. 
As will be perceived, it is essentially the same as the suction 




32 



PNEUMATICS 



31 






PlO. 17 



pump, except that the spout is fitted with a 
cock and has a pipe attached to it, leading to 
the point of discharge. If it is desired to dis- 
charge the water at the spout, the cock may 
be opened, otherwise, the cock is closed, and 
the water is lifted by the piston up through 
the pipe P' to the point of discharge, the 
valve c preventing it from falling back into 
the pump, and the valve V preventing the 
water in the pump from falling back into the 
well. It is not necessary that there should 
be a second pipe P f , as shown in the figure, 
for the pipe P may be continued straight 
upwards, as shown in Fig 17. This figure 
shows a section of a lifting pump for raising 
water from great depths, as from the bottom 
of mines to the surface. The pump consists 
of a series of pipes connected together, of 
which the lower end only is shown in the 
figure. That part of the pipe included 
between the letters A and B forms the pump 
cylinder, in which the piston P works. That 
part of the pipe above the highest point of 
the piston travel, thiough which the water is 
discharged, is called the delivery pipe, and 
the part below the lowest point of the piston 
travel is called the suction pipe. The 
lower end of the suction pipe is expanded, 
and has a number of small holes m it, to 
keep out solid matter. C is a plate covering 
an opening, and may be removed to allow the 
suction valve to be repaired. D is a plate 
covering a similar opening, through which 
the piston and piston valves may be repaired. 
The pibton rod, or rather the piston stem, is 
made of wrought iron, inserted with wood, 
and connected with the piston. The only 
limit to the height to which a pump of this 



32 PNEUMATICS 32 

kind can raise water is the strength of the piston rod Lifting 
pumps of this kmd are used to raise water from great depths 
to the earth's surface; hence, a very long piston rod is neces- 
sary. In the lifting pump shown in Fig. 16, the water is raised 
from a point a few feet below the earth's surface to a point 
considerably higher. This requires the piston rod to move 
through a stuffingbox, as shown at 5 1 , and also necessitates 
the rod being round, in order that the water may not leak out. 

35. Force Pumps. The force pump differs from the 
lifting pump in several important particulars, but chiefly in 
the fact that the piston is solid, that is, it has no valves A 
section of a suction and force pump is shown in Fig. 18. The 

water is drawn up the 
suction pipe as be- 

H - -T ^ F fore, when the piston 

| 5 rises, but when the 

piston reverses the 
pressure on the water 
caused by the descent 
of the piston opens 
the valve V and 
forces the water up 
the delivery pipe P'. 
When the piston 
again begins its up- 
ward movement, the 
valve V is closed by 
Fro I8 the pressure of the 

water above it, and the valve V is opened by the pressure of 
the atmosphere on the water below it, as in the previous 
cases. For an arrangement of this kind, it is not necessary 
to have a stuffingbox. The water may be forced to almost 
any desired height. The force pump differs again from the 
lifting pump in respect to its piston rod, which should not be 
longer than is absolutely necessary in order to prevent it 
from buckling, while, in the lifting pump, the length of the 
piston rod is a matter of indifference. 




34 



PNEUMATICS 



.".2 



pressure below the plunger being less than the pressure of 
the atmosphere above, the air would rush in instead of being 
expelled. 

37. Double-Acting- Pumps. In the pumps previously 
described, the discharge was intermittent, that is, the pump 
could only discharge when the piston was moving in one 
direction. In some cases, it is necessary that there should 




PIG 20 

U a continuous discharge; in all cases, it takes more power 
to run the pump with an intermittent discharge, as a little 
consideration will show. If the height that the water is to 
be raised is considerable, its weight will be very great, and 
the entire mass must be put in motion during one stroke 
of the piston. 

In order to obtain the advantage of a more continuous 
discharge, double-acting pumps are used. Fig 20 shows a 
part sectional view of such a pump. Two pistons a and 6 



32 PNEUMATICS 35 

are used, which are operated by one handle c in the manner 
shown. The pump has one suction pipe s and one discharge 
pipe d The cylinders e and / are separated by a diaphragm , 
so that they cannot communicate with each other above the 
pistons In the figure, the handle c is moving to the right, 
the piston a upwards, and the piston b downwaids In 
moving upwards, the piston a lifts the water above it, causing 
it to flow through the delivery valve h into the discharge 
pipe d. This upward movement of the piston creates a. 
partial vacuum below it m the cylinder <?, and causes the 
water to rush up the suction pipe j into the cylinder, as 
shown by the arrows In the cylinder /, the downwaid 
movement of the piston b raises the piston valve v, and the 
weight of the water on the suction valve z keeps it closed. 
When the handle c has completed its movement to the right 
and begins its return, all the valves on the right-hand side 
open except v, and those on the left-hand side close except /; 
watei is then discharged into the delivery pipe by the cylin- 
der /, and only at the instant of reversal is the flow into the 
delivery pipe d stopped. 

38. Air Chambers. In order to obtain a continuous 
flow of water in the delivery pipe, with as nearly a unifoim 
velocity as possible, an air cluimber is usually placed on 
the delivery pipe of force pumps as near to the pump cylinder 
as the construction of the machine will allow. The air 
chambeis aie usually pear-shaped, with the small end con- 
nected to the pipe. They are filled with air, which the water 
compresses during the discharge. During the suction, the 
air thus compressed expands and acts as an accelerating force 
on the moving column of water, a force that diminishes with 
the expansion of the air, and helps to keep the velocity of 
the moving column more nearly uniform. An air chamber 
IH sometimes placed on the suction pipe. These air chambers 
not only tend to promote a uniform discharge, but they also 
equalize the stresses on the pump, and prevent shocks due 
to the incompressibility of water. They serve the same pur- 
pose in pumps that flywheels do m steam engines. Unless 



36 



PNEUMATICS 



32 



the pump moves very slowly, it is absolutely necessary to 
have an air chamber on the delivery pipe. 

39. Steam Pumps. Steam pumps are force pumps 
operated by steam acting on the piston of a steam engine, 
directly connected to the pump, and in many cases cast with 
the pump. A section of a double-acting steam pump showing 1 
the steam and water cylinders, with other details, is illustrated 
in Fig. 21. Here G is a steam piston, and R the piston rod, 
which is secured at its other end to the plunger P. Fis a 
partition cast with the cylinder, which prevents the water in 




FIG 21 

the left-hand half from communicating with that m the right- 
hand half of the cylinder. Suppose the piston to be moving 
in the direction of the arrow The volume of the left-hand 
half of the pump cylinder will be increased by an amount 
equal to the area of the circumference of the plunger multi- 
plied by the length of the stroke, and the volume of the right- 
hand half of the cylinder will be diminished by a like amount 
In consequence of this, a volume of water in the right-hand 
half of the cylinder equal to the volume displaced by the 
plunger m its forward motion will be forced through the 
valves V. V into the air chamber A, through the orifice D, 



32 PNEUMATICS 37 

and then dischaiged through the delivery pipe H. By reason 
of the partial vacuum in the left-hand half of the pump cylin- 
der, owing to this movement of the plunger, the water will 
be drawn from the reservoir through the suction pipe C into 
the chamber K, A", lifting the valves S', S', and filling the 
space displaced by the plunger. During the return stroke, 
the watei will be drawn through the valves 5, 5 into the 
right-hand half of the pump cylinder, and discharged 
through the valves V, V in the left-hand half. Each of the 
four suction and four discharge valves is kept to its seat, 
when not working, by light springs, as shown. 

There ate many varieties and makes of steam pumps, the 
majority of which are double-acting. In many cases, two 
steam pumps are placed side by side, having a common 
dehveiy pipe. This arrangement is called a duplex pump. 
It is usual so to set the steam pistons of duplex pumps that 
when one is completing the stroke the other is in the middle 
of its stroke A double-acting duplex pump made to run in 
this manner, and having an air chamber of sufficient size, 
will deliver water with a nearly uniform velocity 

In mine pumps for forcing water to great heights, the 
plungers are made solid, and in most cases are extended 
through the pump cylinder. In many steam pumps, pistons 
are used instead of plungeis, but when very heavy duty is 
icquired plungers are preferred. 

40. Centrifugal Pumps. Next to the direct-acting 
steam pump, the centrifugal pump is the most valuable 
instrument for raising water to great heights. As the name 
implies, the effects produced by centrifugal force are made 
use of. Fig. 22 represents a centrifugal pump with half of 
the casing removed. The hub 5* is hollow, and is connected 
directly to the suction pipe. The curved arms a, called vanes 
or wliiffs, are revolved with a high velocity in the direction 
of the arrow, and the air enclosed between them is driven out 
through the discharge passage and delivery pipe DD* This 
creates a partial vacuum in the casing 1 and suction pipe, and 
causes the water to flow in through S, This water is also 

I LT 398-19 



58 



PNEUMATICS 



82 



nade to revolve with the vanes, and, of course, with the same 
velocity. The centrifugal force of the revolving water causes 
t to fly outwards toward the end of the vanes, and becomes 
jreater the farther away the water gets from the center This 
:auses the water to leave the vanes, and finally to leave 
he pump by means of the discharge passage and delivery 
npe D D. The height to which the water can be forced 

depends on the velocity 
of the revolving vanes. 
In the construction of 
a centrifugal pump, 
particular care is re- 
quired in giving the 
correct form to the 
vanes, for the effi- 
ciency of the machine 
depends greatly on this 
feature. What is 
required is to raise the 
water, and the energy 
used to drive the pump 
hould be devoted as much as possible to this one purpose. 
"*he water, when it is raised, should be delivered with as 
ittle velocity as possible, for any velocity that the water 
hen possesses has been secured at the expense of the energy 
sed to drive the pump. The form of the vanes is such 
nat the water is delivered at the desired height with the 
33 st expenditure of energy. 

The number of vanes depends on the size and capacity of 
lie pump. It will be noticed that, in the pump shown m 
lie figure, the vanes have sharp edges near the hub. The 
bject of this is to provide for a free ingress of the water, 
nd also to cut any foreign substance that may enter the 
ump and prevent it from working properly. 
Almost any liquid can be raised with these pumps, but, 
rhen they are intended for pumping chemicals, the casing 
nd vanes should be made of materials that will not be acted 
n by the chemicals. 




FIQ 22 



32 



PNEUMATICS 



39 



41. Tlie Hydraulic Ram. The construction of a 
hydraulic ram is shown in Fig. 23. This machine is used 
for raising water from a point below the level of the water 
in a spring or reservoir to a point considerably higher, with 
no power other than that afforded by the inertia of a moving 
column of water. In the figure, a is a pipe called the drive 
pipe, connecting the ram with the reservoir; the valve b slides 
freely in a guide, and is provided with locknuts to legulate 
the distance that it can fall below its seat. When the water 
is first turned on by opening the valve , the valve b is 
already opened, and the water flows out through c, as shown. 




PlO. 28 

As the discharge continues, the velocity of the water in the 
drive pipe will increase until the upward pressure against 
the valve b is sufficient to force the valve to its seat. The 
actual closing of the valve takes place very suddenly, and 
the momentum of the column of water, which was moving 
with an increasing velocity through the drive pipe a, will 
very rapidly force some water through the valve d into the 
air chamber /. Immediately after this, a rebound takes 
place, and for a short interval of time the water flows back 
up the drive pipe a and tends to form a vacuum under the 
air-chamber valve d\ this opens the snifter valve g and admits 
a little air, which accumulates under the valve d and is forced 
into the air chamber with the next shock. This air keeps the 



40 PNEUMATICS 32 

air chamber constantly charged, otherwise, the water, being 
under a greater pressure in the air chamber than in the reser- 
voir, would soon absorb the air in the chamber and the ram 
would cease to work until the chamber was recharged with 
air The rebound also takes the pressure off the under side 
of the valve b and causes it to drop, and the above-described 
operations are repeated. The delivery pipe is shown at c; a 
steady flow of water is maintained through it by the pressure 
of the air in the chamber /; this air also acts as a cushion 
when valve b suddenly closes, and prevents undue shock to 
the parts of the ram. 

The height to which water can be raised by the hydraulic 
ram depends on the weight of the valve b and the velocity 
of the water in a. 

42. Power Necessary to "Work a Pump. 

Principle I. In all pumps, whether lifting, force> steam 
single- or double-acting, or centrifugal, the number of foot- 
Pounds of power needed to work the pump is equal to the weight 
of the water in pounds, multiplied by the vertical distance, in 
feet, between the level of the water in the well, or sowcc, and the 
point of discharge, plus the work necessary to ovej come the fric- 
tion and other resistances. 

Principle II. The work done in one stroke of a pump zs 
equal to the weight of a volume of water equal to the volume 
displaced by the piston during the stroke, multiplied by the total 
vertical distance, in feet, through which the water is to be raised, 
plus the work necessary to overcome the resistances. 

A little consideration will make Principle II evident. 
Suppose that the height of the suction is 25 feet; that the 
vertical distance between the suction valve and the point 
of discharge is 100 feet; that the stroke of the piston is 
15 inches, and that its diameter is 10 inches. Let the diam- 
eters of the suction pipe and delivery pipe be 4 inches each. 
The volume displaced by the pump piston or plunger in one 

stroke equals 10 X ; 7 _ 8 5 4 X *- = .68177 cubic foot. The 
1,728 

weight of an equal volume of water is .68177 X 62.6 



32 PNEUMATICS 41 

= 42 611 pounds Now, in order to discharge this water, 
all the water in the suction and delivery pipes has to be 
moved through a certain distance, in feet, equal to 68177 
divided by the area of the pipes, in square feet. 

4 inches = foot, (i)" X .7854 = ^f- = .0812$ square 

y 

foot. .68177 -^ .08723- = 7 8126 feet. 

The weight of water in the delivery pipe is (^) a X .7854 
X 100 X 62.5 = 545 42 pounds. 

The weight of water in the suction pipe is (i) X .7854 
X 25 X 62 5 = 136.35 pounds. 

545 42 + 136 35 = 681 77 pounds, which is the total weight 
of water moved in one stroke. The distance that the water 
is moved m one stroke is 7.8125 feet; hence, the number of 
foot-pounds necessary for one stroke is 681.77 X 7.8125 
= 5,326 3 foot-pounds. Had this result been obtained by 
Principle II, the process would have been as follows: The 
weight of the water displaced by the piston in one stroke 
was found to be 42.611 pounds. 42.611 X 125 = 5,326.4 
pounds, which is practically the same as the result obtained 
by the previous method, and is a great deal shorter. The 
slight difference between the two results is due to neglected 
decimals. 

EXAMPLE. What must be the necessary horsepower of a, double- 
acting steam pump if the vertical distance between the point of dis- 
charge and the point of suction is 96 feet? The diameter of the 
pump cylinder is 8 inches, the stroke is 10 inches, and the number of 
strokes per minute is 120. Allow 25 per cent, for friction and other 
resistances. 

SOLUTION. Since the pump IB double-acting, it raises ft quantity 
of water equal to the volume displaced by the plunger at every stroke. 
The weight of the volume of water displaced at one stroke la (A)* 
X .7854 X itf X 02.5 = 18 18 lb., nearly. 

18.18 X 90 X 120 - 209,430 ft.-lb. per minute. 

Since 25 per cent. Is to be allowed for friction, the actual number 
of foot-pounds per minute is 209,430 + .75 - 279,240. 1 H. P. 

= 33,000 ft.-lb. per min,; hence, 8.462 H. P., nearly. Ans. 



2 RUDIMENTS OF ANALYTIC GEOMETRY 33 

If the acceleration a is supposed to have a fixed value, as 
8 feet per second, and different values are assigned to /, 
different values will be obtained for s. Here, too, / and .v 
are variables /, which is varied at pleasure, is the independent 
variable; and s, whose values depend on those of t, is the 
dependent variable, or a function of /. The acceleration a, 
although represented by a letter, is assumed to be fixed or 
invariable, and is therefore a constant. 

2. In general, when, for any particular purpose, some of 
the quantities represented by letters in an equation are made 
to take (or are considered as being such that they can take) 
different values, they are called variables. Those to which 
values are assigned arbitrarily are called Independent 
variables; those whose values depend on the values of 
the independent variables are called dependent variables, 
or functions of the independent variables. Those quantities 
that are supposed to remain fixed are called constants. 

A function may also be defined as a quantity whose value 
depends on the value or values of one or more other quan- 
tities; the very word "depends" indicates that the function 
can have different values (that is, can vaiy) according to the 
values assigned to other quantities. Thus, the area of a 
triangle is a function of the base and altitude; if the base 
is assumed to be fixed, it becomes a constant, and the area 
is a function of only the altitude. The velocity of a body 
moving under the action of an unbalanced force is a function 
of the magnitude of the force and the mass of the body; if 
the mass is assumed to be fixed, it becomes a constant, and 
the velocity is a function of only the force. 

3. Graph of an Equation. It was stated in Art. 1 
that from the equation A = xi*, a table can be made 
giving the values of the function A corresponding to different 
values of the independent variable r. Instead of a table, a 
diagram may be constructed to represent fte relation between 
the values of A and r. Such a diagram; w&ick is the graphic 
representation of the equation just givei^, is palled the graph 
of that equation, and is constructed as foj 




I \ 



4 RUDIMENTS OF ANALYTIC GEOMETRY $33 

value of ns laid off along OX from O to, say, A/, a per- 
pendicular is erected at M, intersecting the graph at P. 
Then MP will represent the value of A corresponding to 
the value OM of r. 

4. The perpendicular distances of any point of the graph 
from the axes are called the coordinates of that point. 
Thus, the coordinates of P are MP (= OM') and M 1 P 





(= OM}. The horizontal coordinate OM, or, more gen- 
erally, the coordinate representing the independent variable, 
is usually called the abscissa, and the other coordinate 
OM, the ordinate. It is customary to reckon the abscissa 
along the axjs OX, called the axis of x, and the ordinate 
on a line parallel to O Y t through the foot of the abscissa. 
Thus, the coordinates of P are stated as OM and MP 
instead of M'P and MP. 



6 RUDIMENTS OF ANALYTIC GEOMETRY 33 

connects with the crank OJfby the connecting-! od ff K It 
is shown m mechanics that when the crank has described an 
angle X from the position OA, which is in line with the axis 
of the cylinder, the velocity u of the piston is given approxi- 
mately by the formula 

/ . v , a sm 2 X\ 

u = v ( sm X -\ - 1 

\ 2/ / 

in which v = linear velocity of crankpm K; 
I = length of connecting-rod; 
a = length of crank. 

If the ratio - is represented by <:, the formula may be 

If 

written, 

- = sinAr+sin2AT (1) 
v 2 

When c is given, the graph of equation (1) can be con- 
structed by taking X as the independent variable, and u as 

v 



L 

r 




PlQ 3 



the function Then, the value of - for any value of X can 

v 

be found from the graph, and, when v is given, u can be 

determined by multiplying by v the value found for --. 

v 

EXAMPLE 1 To construct the graph of equation (1) when c = $, 
for values of X varying from to 180, that Is, for one-half a revolu- 
tion of the crank, from the position OA to the position O A', Fig. 3. 

SOLUTION Writing, for shortness, y for -, and substituting the 

given value of c, equation (1) becomes 

y = em X + -rV sm 2 X 

By giving to lvalues from to 180, at intervals of 6, the follow- 
ing table is obtained: 



RUDIMENTS OF ANALYTIC GEOMETRY 



EXAMPLES FOR PRACTICE 

1 Plat the following equations for values of x varying from 
3 to 8 (a) y = X j* - 7 x* + 4 x - 6 (6) v = t x 9 - 1<> x + 4 



a) The general form of the guiph is shown in Fig f) 
6) The general form of the graph Is shown in Pig 
c) The general form of the graph is shown in Fig 7 







PIG 5 



10 RUDIMENTS OF ANALYTIC GEOMETRY 33 

have been proposed to determine the relation between these 
two quantities, no exact formula has yet been found. Sup- 
pose that a series of experiments on wrought-iron columns 
gives the following results: 

RATIOS OF LENGTH BREAKING LOAD 

TO DIAMETER POUNDS PER SQUARE INCH 

6 51,200 

7 47,400 

8 44,600 

9 42,200 

10 40,200 

11 38,700 

12 37,900 

13 37,100 

14 36,900 

These results may be represented graphically, as shown 
in Fig 9. Having drawn two coordinate axes OX and O Y, 

r\ 

V, 

5 



Batio of Length to Diamter 
JUT | 

FIG. 



the values of the ratio of length to diameter, which is the 
independent variable, are laid off from O, along OX, to 
any convenient scale. Thus, OM* represents the ratio 6; 
OJIf represents 7; OM, t 8; etc. The corresponding values 



12 RUDIMENTS OF ANALYTIC GEOMETRY 33 

values written on the respective projections. Thus, the 
point P, is projected at MJ, on which is written 31,500,000, 
the population in 1860. 

The probable population m any intermediate year can be 
readily obtained from the curve. For instance, if it is 
desired to find the population in 1878, it will be observed 
that 1878 lies between 1870 and 1880, and that the interval 
between 1870 and 1878 is .8 of the interval between 1870 
and 1880. Therefore, the required population is found by 
laying off M t M equal to .8 of M*M t , and drawing the 

ordmate MP, which 

JT' 

P-r represents the approx- 

/ imate population in 

Pa/ 1878. 



P B / 8. In some cases, a 

*?' diagram is constructed 

m yf | 

y rather for the purpose 

_3isoopoo_ J:*/ of presenting to the 

o -"a p *s e y e > m a striking man- 

Pl ^ ner, the variations ol 

certain quantities, than 

~ ^r f or the purpose of de- 



1840 Bo70 

Year values, the variations 

Pl 10 m the latter being too 

irregular. In the examples so far given, the curves are fairly 
regular, which shows that the variations in the functions 
(ordinates) are not too abrupt, and that the curves may be 
depended on to give tolerably approximate values of the 
function corresponding to intermediate values of the inde- 
pendent variable Example 1 of the following Examples for 
Practice is a case in which the curve, on account of its too 
great irregularity, could not be depended on to give inter- 
mediate values. In such cases, the extremities of the ordi- 
nates are joined by straight lines, instead of by a curve. 



33 RUDIMENTS OF ANALYTIC GEOMETRY 13 



EXAMPLES FOR PRACTICE 

1 Draw a. graph representing the average daily water consumption 
in the city of Brooklyn between January, 1H71, and December, 1873, 
from the following data. 



YE.R 



MO.TH 





January 


21,000,000 


s. 


April . . . 


17,500,000 


oo 

rH 


July . . ... 


. . 19,600,000 




, October . 


. . 18,500,000 






23,600,000 


2 


April . 


20,500,000 


SB' 

I-H 


July 


. 22,500,000 




October 


. . . 23,000,000 




January . . 


. . . 29,000,000 


CO 


April . . . 


22,500,000 


t- , 

00 


July ... 


26,500,000 


rH 


October ... 


. 22,500,000 




December ... 


24,500,000 




Ans. 


The curve is shown in Pig. 11 




o 



PIG. 11 



2. (a) Taking the hours as abscissas, and the discharges as ordi- 
nates, construct a graph for the water discharged by a pipe from the 
following observed values for 1 day, (b] determine, by means of the 
curve, the probable discharge at 11:30 A. M.; (c] determine the prob- 
able discharge at 5:80 F. M. 



. The scale of ordlnatoi should be chosen sufficiently large to 
show the differences In discharge say 4 Inches to 1 cable loot. 



RUDIMENTS OF ANALYTIC GEOMETRY 



HOUR 

A. M. 


DISCHARGE 
CUBIC FEBT PER 
SKCOND 


HOUR 

A M 


DISCHARGE 
CUBIC FEET PER 
SECOND 


1 


2.45 


11 


2.54 


o 


2 67 


M. 




3 


2 69 


12 


233 


4 


2.62 


f M. 




5 


2.36 


1 


264 





2.30 


2 


215 


7 


2.23 


3 


1 19 


8 


2.19 


4 


119 





2 2O 


5 


226 


10 


2.OO 


8 


237 




Aus. 



FIG 12 

The general form of the graph is shown in Fig. 12 
2 43 cu. ft. per sec. 
2 31 cu. ft per sec. 



33 RUDIMENTS OF ANALYTIC GEOMETRY 15 



EQUATIONS OF LINES 



INTRODUCTION 

9. Equation of a Line. As already explained, a 
graph is constructed from an equation expressing a relation 
between two variables, or giving the value of one variable as 
a function of the other. In 

V 

the graph, these variables 
are the coordinates of 
points on a line, usually 
curved. Sometimes, on 
the contrary, a line is given, 
and it is required to find the 
equation of which the line 
is the graph. That equa- 
tion is called the equation 
of the given line, and is a 
general expression of the 
relation between the two 
coordinates of any point of 
the line, with reference to two coordinate axes conveniently 
chosen. 

Take, for instance, a circle of radius r, Fig. 13, and two 
rectangular axes OX, OY, passing through its center. Let 
P be any point on the circumference, and x and y its coordi- 
nates, as shown. The right triangle OMP gives 

OM* + MP* = OP 1 -, 
that is, x'+y* - r' (1) 

This is the equation of the circle referred to two rectangu- 
lar axes through the center. In that equation, x and. y 
are the coordinates of any point on the circumference. No 




16 RUDIMENTS OF ANALYTIC GEOMETRY 33 

matter where the point is located, its coordinates satisfy 
equation (1). Thus, for the point P lt 

x = -OM lt y = -M 1 P 1 , 
and 




= OP,' = S 

10. The equation of a line is useful in the study of the 
geometric properties of the line, and it often serves to 
recognize the form of a graph corresponding to a given 
equation If, for example, it is found in the solution of a 

\f mechanical problem that 

the path AB, Fig. 14, of a 
moving point is such that 
the coordinates x and y of 
any of its points, with re- 
gard to the axes OX and 
v O y, are related by the 

equation 

x" + y = a' 

o = - ' -- x tne quantity a being con- 

1 Pl 14 stant, it can be at once 

concluded that the path of the point is a circle whose center 
is and whose radius is a. 

11. Analytic geometry is that branch of mathematics 
m which geometric figures are studied by means of their 
equations. Surfaces, as well as plane lines, have equations; 
but in this Course only a few plane lines will be treated. 



THE STRAIGHT LINE 

12. Equation of the Straight Line. Let X' X and 

y y, Fig. 15, be two axes of coordinates, and A B a straight 
line making with X'X an angle H, and intersecting Y' Y 
at 7, the distance b (= 01} being known. It should be 
understood that the angle H is always measured from the 
axis X> X upwards. Thus, if the line were 4iB the 
angled would be XJ^B,. Also, b, like the ordinates, is 



53 RUDIMENTS OF ANALYTIC GEOMETRY 17 



positive upwards and negative downwards. Thus, for the 
line A t Bi, the value of b is Of, This being understood, 
the equation now to be derived is entirely general, and, with 
the symbols interpreted as just explained, applies to all 
cases. 

The right triangle PMJ gives 

PM = JM tan H\ 
y = (x 4- JO] tan H. 

L j. r T 



that is, 
Now, 



therefore, y =[x 



- , 

tan H 

-- -} tan H = x tan H + b 
tan HI 



This is the equa- 
tion of the line A B. 
It is customary to 
denote tan H by a, 
and write the equa- 
tion in the form 
y = ax + b 

The distances 01 
and J are called the 
Intercepts of AB 
on the axes of y and*, 
respectively . 1 1 i s 
evident that, at 7, the 
abscissa x is 0, and 
the ordinate is 01] 
and at J the ordi- 
nate y is 0, and the 
abscissa is OJ.t 



13. Graph of Any Equation of the First Degree. 

Any equation of the first degree between two variables can 
be represented by a straight line; in other words, the graph 
of any equation of the first degree between two variables is 
a straight line, 

Let mx+ ny = P (1) 

be any equation of the first degree between the variables 




18 RUDIMENTS OF ANALYTIC GEOMETRY 38 

x and y, the other quantities m,n,p being constants, either 
positive or negative. Solving the equation for y, 



n n 

Since the tangents of the angles between and 180 con- 
tain all possible numbers, both positive and negative, it is 
always possible to find an angle whose tangent is . Let 

that angle be denoted by H, and denote * by b. Then, 

n 

= tan H, and equation (2) becomes 
n 

y = jirtan/f + b (3) 

If is positive, 
n 

H is less than 90; if 
negative, H is greater 
than 90. 

If on the y axis the 
distance b is laid off 
from O t upwards if b 
x is positive, down- 
wards if b is negative, 
and from the extrem- 
ity of b a line is drawn 
making with the x 
axis an angle equal 
to H, that line is the 
graph of equation (3) , 
r and, therefore, also 

FIG 16 of equation (1). If b 

and tan H are positive, the line will have such a position 
as A,B^ Fig 16, in which O A = b, and XJ,B, = H. If & 
is positive and tan H negative, the graph will have a position 
like A t B t > in which OI t = b, and XJ,B t = H. If b is neg- 
ative and tan H positive, the graph will be like A, JB,, in 
which OI> = b, and XJ,B* = H. If both b and tan H are 
negative, the graph will have such a position as A^ B,.. 




33 RUDIMENTS OF ANALYTIC GEOMETRY 19 



14. Because any equation of the first degree between 
two variables can be represented by a straight line, formulas 
in which the value of a quantity is given in terms of the first 
power of another are called straight-line formulas. For 
example, the formula 

p = 10,000 - 45 - 
r 

which expresses the intensity p of pressure that a column can 

stand, m terms of the ratio - of length to radius, is a straight- 

r 

line formula. If y is 
written instead of p, 

and x instead of-, the 
r 

formula becomes 

y = 10,000-45* 

= _45*+ 10,000 
which is the standard 
form of the equation 
of a straight line. 

15. The simplest 
way to draw the 
straight line corre- 
sponding to an equa- 
tion of the first de- 
gree is as follows: 
Let the equation be 
mx -\-ny-\-p = 0. 

Select the axes in any convenient position, as X 1 X and V K, 
Fig. 17. Making x = in the equation, and solving for y, 
the intercept (say <?/,) on the y axis is obtained, and the point 
7, where the line intersects that axis is determined. Making 
y = 0, and solving for x, the intercept 07, and the point ,/i are 
obtained. The line is then drawn through the points 7, and /,. 

EXAMPLE 1. To draw the graph of the equation 8# l&y + 20-0. 

SOLUTION^ Draw the axes X'X< Y> Y, Fig. 17 Making x In 
the equation, we have -16^ + 20 = 0; whence 
y - 0/ 1.26 




20 RUDIMENTS OF ANALYTIC GEOMETRY 33 



Making y = in the equation, we have 8x + 20 = 0, whence 

x = OJ= -26 

Laying off Of = 1 25, and OJ =25, the line A B drawn through 1 
and J is the required graph Ans 

EXAMPLE 2 To draw the graph of the equation 10 x + 8y + 40 = J. 
SOLUTION Making x 0, and solving for y, Pig. 17, 

= 01 = - = -5 
Making y = 0, and solving for x, 

* - n T 40 _ 4 

* ~ l = ~ 10 ~ ~ 4 

Laying off OJ* = 4, 07; = 5, and joining/! and J lt the required 
graph A! B* is obtained Ans. 

EXAMPLE 3 To draw the graph of the equation 4j/ 8x = 0. 

SOLUTION It will be noticed that this equation has no term 

independent of x and y, 
This shows that the line 
passes through the origin 
of coordinates, or that its 
intercepts on the two axes 
are zero This follows at 
once from Art 13, for 
it was there shown that 




>-, and, as 
n 



in this 



case,/ = 0, it follows that 
b = In this case, the 
graph cannot be con- 
structed as in the two pre- 
ceding examples. Since 
the line passes through 
the origin O, Pig. 17, it 
is only necessary to de- 
termine another point. 
This is done by assu- 
ming any convenient value for x and solving for y Making*- = |, the 
equation becomes 4j/ - 4 = 0, whence y = 1 Laying off OM^ i, 
and the ordmate MP = 1, the line A, t , drawn through and JP, 
is the required graph 



FOR PRACTICE 

Draw the graphs of the equations (a) 8 

(c) 5x+4y = -20. rf I2y - 10 . 



Ans 



The 



are shown, respectively, at 
tt A, B, s AtJBt in Pig. 18 



33 RUDIMENTS OF ANALYTIC GEOMETRY 21 



2 Find the angle that the graph of the equation x + &y 4 = 
makes with the x axis. Ans. 141 20> 20" 

3 Given the equation x-\- 4y = 20, find (a) the intercept of the 
graph on the y axis, (b) the angle that the graph makes with the 
x axis (Give seconds in angle to nearest multiple of 10.) 

165 57' 60" 



Ans 



. 
{(?) 



APPLICATIONS 

16. Reactions on a Beam. Let a beam A B t Fig. 19, 
resting on the supports A and B> carry a movable load W. 
Let the distance of the load from the left support at any 
instant be x, and let the length of the beam be denoted 
by /. For this position of the load, the reaction Ri at A is 
obtained by taking moments about B\ thus, 
RJ- W(l-x) = 0; 



whence 



I 



(1) 



Since the load W moves, x is a variable, and so is 
Equation ( 1 ) gives R l 
as a function of x. As 
that equation is of the 
tirst degree, it can be 
represented by a 
straight line, Ri being 
used instead of the y 
used in previous arti- 
cles. A convenient 
and usual way of 
drawing the graph is 
as follows: The ori- 
gin is taken direct- 
ly under A, and the 
axis OX is drawn 
parallel to A B\ the axis O Y is drawn through O. Equa- 
tion (1) shows that the y intercept is W. Therefore, laying 
off, to any convenient scale, along O Y the distance Of to 
represent W, one point / of the graph is obtained. The 
equation also shows that the x intercept (fotind by making 




Pio, 19 



22 RUDIMENTS OF ANALYTIC GEOMETRY 33 

J?, = 0) is /. Therefore, projecting the point B on OX at /, 
the point J of the graph is found, and the graph is the 
straight line IJ. The ordmate M P represents the left leac- 
tion when the load is at W\ the ordmate M' P' represents 
the left reaction when the load is at W; etc. If 1 ' O is 
drawn parallel to OX, the ordinates M,P, AfJP', etc. will 
represent the corresponding values of the right reaction, 
since jR, + R* = W= Of. If, for example, the load is 
2,000 pounds, and a scale of 1,000 pounds to the inch is 
used, Of should be made 2 inches; and, if the ordinate M'P' 
is found to measure li inches, the reaction R^ when the 
load is at W>, is 1,000 X 1* = 1,500 pounds. 



17. Pressure on the Back of a Dam. Let 

Fig. 20, be the back or inner face, supposed to be vertical, 
o of a dam, the water reach- 

ing to the top From 
hydrostatics it is known 
that the pressure p per 
square foot, at any point M 
whose depth below the sur- 
face is x feet, is given by 
the formula 

p = wx 

-* m which w is the weight 

of 1 cubic foot of water. 
Fl M As this is an equation of 

the first degree, it can be represented by a straight line. 
Taking O M,. as the axis of x and as the origin, the latter 
point is a point in the graph, since there is no intercept. 
Making* = h, the formula gives p = w h t which is the pres- 
sure at Mr Laying off, horizontally, M r P t = wh, the line 
O P l is the required graph. The intensity of pressure at 
any point M is given by the ordinate M P. 

If, for instance, the height k is 24 feet, and w is taken 
equal to 62.5 pounds, wh = 62 5 X 24 = 1,600 pounds per 
square foot. If a scale of 500 pounds per square foot to the 
inch is used, M^P^ should be made 1,500 -T- 500 = 3 inches; 




33 RUDIMENTS OP ANALYTIC GEOMETRY 23 



and, if M ' P measures li inches, the pressure at M is 

500 X H = 625 pounds per square foot. 
It will be observed that here p is used instead of y, that 
the x axis is vertical; and that positive values of x are 
counted downwards, and positive values of p toward the 
left It is often necessary to make such changes in notation, 
so as to adapt the construction to given conditions. The 
general principles and methods, however, remain the same. 



THK PARABOLA 

18. Definitions. A parabola, Fig. 21, is a curve such 
that, if any point P on it is taken, the distance P F of that 
point from a fixed point F is s r 
equal to its distance PN 
fiom a fixed line DE. The 
fixed point F is called the 
focn&j the fixed line DE, 
the directrix. The line N, X 
passing through the focus 
and perpendiculai to the di- 
rectrix is called the axis of 
the curve, and the point O 
where the curve crosses the 
axis is called the vertex. 
Twice the distance FN<> 
from the focus to the di- 
rectrix is called the par- 
ameter, and is denoted by 
2/>; so that 2/ = 2 N.F, 
and p = N,F. 

The curve is symmetrical with respect to the axis; that is, 
to every point P on one side of the axis there corresponds 
another point P 1 on the other side, at the same distance from, 
and on the same perpendicular to, the axis. Any line, as JPP', 
perpendicular to the axis and bounded at its two ends by 
the curve is called a double ordluate. 




FIG. 21 



24 RUDIMENTS OF ANALYTIC GEOMETRY 33 

19. Equation of the Parabola Referred to the Axis 
and Vertex. Let P, Fig. 21, be any point on the curve, 
and O M = x and M ' P y its coordinates, the axes being 
the axis OX of the parabola and the perpendicular Y at O 
Let the parameter be 2/>. Then, the distance FN from the 
focus F to the directrix DE is equal to p, and OF = vfl, 
since ON a OF, the point O of the curve being, according 
to the definition of a parabola, equally distant from the 
directrix DE and the focus F. According to the same 
definition, we have 

PF= PN 
and, therefore, 

PF' = PN' (1) 

Now, 

PF' = PM' + FM' = y 9 + (OM- OF)' 



Also, 

= N M' = (OM+N t O)' = (x + p)* = A 

-\ 

Substituting in equation (1) these values of PF' and PN', 



4 4 

whence y' = %px 

which is the required equation. 

20. Let y^ and y tt be two ordinates corresponding to the 
abscissas x* and x, Then, since the equation applies to all 
points, y* = 2px lt y' = ZpXi, whence 



That is, in any parabola, any two coordinates parallel to the 
axis of the curve are to each other as the squares of the corre- 
sponding coordinates perpendicular to the axis. 

21. It is important that the equation of the parabola 
should be so mastered that it can be applied to parabolas in 
different positions, and when the notation is different from 
the one here used. It should be borne in mind that in the 



S33 RUDIMENTS OF ANALYTIC GEOMETRY 25 

general equation, the coordinate to be squared is that per- 
.pendicular to the axis of the parabola. In Fig. 22 is repre- 
sented a parabola with the axis vertical, coinciding with the 
y axis. In this case, the equation should be written 



Fig. 23 represents a vertical parabola with its axib down- 

T x t o 




Fio. 22 Pro 23 

wards The coordinate axes are denoted by O S and O T, 
as shown, and the coordinates of any point P by s and /, 
5 being positive downwards, and t positive toward the left 
Under such conditions, the equation of the curve should be 
written 



22. Problem I. To find the parameter and equation of 
a parabola, when a double ordinate and the corresponding abscissa 
are given. 

This is the usual way in which the parabola occurs in 
practice. Thus, in road construction, in which the cross- 




section of a road is often made parabolic, the width a of the 
road, Fig. 24, and the height h of the center above the 
ends A and B are given. In order to determine the fall of 
the cross-section at different distances from the center, it is 
necessary, or at least advisable* to determine the equation 
and parameter of the curve A OB. 



26 RUDIMENTS OF ANALYTIC GEOMETRY 33 



As usual, the parameter will be denoted by 2 p. At the 

point M , for which x = -, and y = A, we have, since here 

L 

the x coordinate is perpendicular to the axis, 



whence e lp = ^~ 

4:h 

The general equation of the parabola is, therefore, 

* 



In practice, points on the curve are determined by assu- 
ming values for x, and computing the corresponding values 
of y. The equation may, therefore, be more conveniently 
written in the form 

a' \i aj 

Let OM t be divided into any convenient number of equal 

Jf, 

>F 

FIG 25 

parts, say n, and give to x successive values correspond- 
ing to the points of division; thus, x = ", x = 2 -^-^, 

n n 

x = 3 -, etc.; or x = , x - 2 , # = 3 , etc. Then, 

n n n n 

the corresponding values of y beccme 




Thus, if OM, is divided into ten equal parts, and OM con- 
tains six of those parts, 

-Axr-ixH 



33 RUDIMENTS OF ANALYTIC GEOMETRY 27 



EXAMPLE Given the double ordinate A B = 40 feet, Fig 25, and 
the corresponding abscissa OK = 15 feet, it is desired to hnd points 
on the curve at intervals of 5 feet on each side of O 

SOLUTION Draw O M t parallel to A B, and B MI parallel to KO 
Since the points are to be located every 6 ft from O, and O M t 
= 20 ft , the latter line should be divided into four equal parts O MI, 
Jlf, Mt, etc Here, i0=jx*0 = 20, = ^=4, and h = 1 5. 



Therefore, - = ^ = 



094, nearly, and the values of y are 



SLtM lt Mi PI = 094 X 1 

at M., M, /> = 004 X 2 

at Af a , M a /> = .094 X 3 1 

at Jtf t , Jlf t n = 1 5 ft 



09 ft , nearly, 
094 X 4 = 38 ft , nearly, 
.094 X9 = .85 ft., nearly, 



23. Problem II. To construct a parabola when a double 
ordinate and the con espotidmg abscissa are given. 

In practice, what is usually required is to locate points 
of the curve on the % ft 

ground, as in the case 
of road and street 
construction, and in 
railro ad curves. 
Under such circum- 
stances, the ordinates 
are calculated as ex- 
plained in Art. 22; if 
desirable, they may 
be platted, and the 
curve drawn through 
the points thus deter- 
mined. A purely 
graphic method of 
constructing the 
curve is explained in 
Geometrical Drawing* 
Another method, 
which is often convenient in the drafting room, is as follows: 

Let AB> Fig. 26, be the given double ordinate, and OK 
the corresponding abscissa. Bisect ATX at Cand draw CI 
parallel to KO and OI perpendicular to K O. Draw /A", 

ILT39&-21 




Fro 26 



28 RUDIMENTS OF ANALYTIC GEOMETRY 33 

and IN a perpendicular to it, meeting the axis produced at N . 
The line D E, drawn through N perpendicular to OK pro- 
duced, is the directrix It is shown here for the purpose 
of explanation, but it is not necessary to draw it. Lay off 
OF = ON.. The point/' is the focus. From 7V , lay off 
along the axis any convenient distances, as N M, N M^ etc , 
and at the points M, M 1} etc. draw indefinite perpendiculars 
8Q f > QiQS> to the axis. From F as a center, and with 
a radius equal to NM, describe an arc, cutting QQ' at 
P and P'\ these are points of the curve. Likewise, the points 
P! and P/ are the intersections of ?, Q l / with an arc described 
from Fwiih a radius equal to N M* Other points may be 
determined in a similar manner, and the curve drawn through 
them. 

It is usually more convenient to find the points N 9 and F 
by calculation, instead of by the geometrical construction 
described above. The parameter 2^ is computed as explained 
in Art. 22, and then N and OF are laid off each equal 



NOTE The correctness of the preceding construction will now be 
shown It is not necessary for the student to study the following 
demonstration, but he is advised to read it carefully, as it is a good 

exercise As explained In Art 22, 2/ = ~ In Fig 26, fl = 2 A K 
and A = OK. Therefore, 

P 
AK* 



In the right triangle N, IK, the perpendicular Of, which is equal 
to \A K t is a mean proportional between O K and O N a . that is, 
(IAK)*= OJV.XOJf, 

whence ON, = ^frjj = | [by equation (1)] 

Therefore, O N a Is the distance from the vertex to the directrix 
(Art 18) 

The point P was so determined that FP = N a M = PN. That 
point is, therefore, at the same distance from the focus as from the 
directrix, and, according to the definition of a parabola, must be a 
point of the curve The same reasoning applies to the points P 1 , A, etc. 



33 RUDIMENTS OF ANALYTIC GEOMETRY 29 



APPLICATIONS 

24. Projectiles. In the most general sense of the 
term, a projectile is any body thrown into the air The 
velocity with which a projectile is thrown is called the initial 
velocity, or velocity of projection. Here, only pro- 
jectiles thrown horizontally will be considered, and the 
resistance of the air will be neglected. 

Let a projectile be thrown horizontally in the direc- 
tion OX from a point O, Fig 27, with a velocity v. It is 
required to determine the path OA of the projectile. The 
horizontal line OX and the vertical line O Y will be taken 
as axes of coordinates, y being positive downwards. Were 
the projectile not acted on by gravity, it would describe, in 
any time /, a space OM = vt, which o 
will be denoted by x. If it were not 
thrown at all, and were allowed to fall 
freely from O during the time t, it would 
fall through a distance O G = %gf, 
denoting, as usual, by g the accelera- 
tion due to gravity. This distance O G 
will be denoted by y. When the pro- PIG. 27 

jectile, after being thrown along OX with the velocity v, is 
acted on by giavity, its position P, at the end of the time t, 
will be such that M P will be equal and parallel to O G. We 
have, therefore, 

x = vt 
and y = i g t*. 

yjt 

From the first of these two equations is found /* = -. 

v 

Substituting this value in the second equation, there results 




2 w" 
whence x' = y, 

# 

which is the equation of a parabola with its vertex at O and 
axis vertical. This parabola is, therefore, the path followed 

2 if 
by the projectile. Its parameter is . 







30 



RUDIMENTS OF ANALYTIC GEOMETRY 33 



25. Jet of Water Issuing 1 From & Small Orifice. 

One of the most important applications of the theory of pro- 
jectiles is to the determination of the velocity of water 
flowing from a small orifice. This determination is of much 
importance in hydraulics. In Fig. 28 is represented a tank T 
from which water flows through a small orifice O. The 
water issues horizontally with a velocity v (to be deter- 
mined), and each particle, being under the same conditions 
as a projectile thrown horizontally with the same velocity, 

describes a parabola. 
As the jet is narrow, 
it may be treated as a 
whole as a parabolic 
arc OJ. A horizon- 
tal string OX is 
stretched from O, and 
any distance OM^ is 



measured. From M* 
Pl another string carry- 

ing a heavy weight is suspended, and the distance M l P t is 
measured. Let OM l = x lt Af l P l = y lm Then, 




whence 



EXAMPLE. What was the velocity of a jet for which the measured 
distances x^. and jy, were, respectively, 6 and 5 feet? 

SOLUTION Substituting m the formula the given values of .*-, 
and y lt and 32 16 for^-, 

x g - = V3 216 X 36 = 10 76 ft. per sec Ans 

26. Moment In a Beam. Let a beam A B, Fig. 29, of 

length /, carry a movable load W. When the load is at a 

distance x from the center of the beam, the left reaction R 1 

is found by taking moments about B\ thus, 

RJ = 

whence R^ = 



33 RUDIMENTS OF ANALYTIC GEOMETRY 31 

The moment of .#, about I^is called the bending moment 

at W. Denoting it by M, we have 



The bending moment at the center, which will be denoted 
by M^ is obtained by making x = 0, which gives 

". - \ L 



Equation (1) may p 1 



be written 
M M 
whence 



If M - M is de- 
noted by y, equation 
(1) may be written 
/ 



x 




which is the equa- 
tion of a parabola 
whose axis is perpen- 
dicular to A B, and whose parameter is 



The curve 



is shown at A 1 OB 1 . It may be constructed by points, 
assuming values of x and finding the corresponding values 
of M and M t M\ or by the method of Art. 23, noticing 
that to the double ordinate A' #', or /, corresponds a value 
of y (= fCO) equal to M^ since, when x = J/, M 0, and 
y = M, M M,. Having drawn the curve, the bending 
moment, when the weight is at any point W, is found by 
projecting W on A' /?', and measuring the distance QP, 
which is equal to the moment at W> to the scale by which 
KO represents M,, For -tfie ordinate NP represents y, or 
M, M, and, therefore, 

M.-M= NP, M = M.-NP = Qfi-NP= QP