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Full text of "Mechanics for engineers"

MECHANICS FOR ENGINEERS 



- 



> 

MECHANICS FOR 
ENGINEERS 

A TEXT-BOOK OF INTERMEDIATE 
STANDARD 



BY 

ARTHUR MORLEY 

M.Sc., UNIVERSITY SCHOLAR (VICT.) 

SENIOR LECTURER IN ENGINEERING IN UNIVERSITY COLLEGE 
NOTTINGHAM 





WITH 200 DIAGRAMS AND NUMEROUS EXAMPLES 



LONGMANS, GREEN, AND CO. 

39 PATERNOSTER ROW, LONDON 
NEW YORK AND BOMBAY 



All rights reserved 



PREFACE 



ENGINEERING students constitute a fairly large proportion of 
those attending the Mechanics classes in technical colleges 
and schools, but their needs are not identical with those 
of the students of general science. It has recently become 
a common practice to provide separate classes in Mathematics, 
adapted to the special needs of engineering students, who are 
in most institutions sufficiently numerous to justify similar 
provision in Mechanics. The aim of this book is to provide 
a suitable course in the principles of Mechanics for engineering 
students. 

With this object in view, the gravitational system of 
units has been adopted in the English measures. A serious 
injustice is often done to this system in books on Mechanics 
by wrongly denning the pound unit of force as a variable 
quantity, thereby reducing the system to an irrational one. 
With proper premises the gravitational system is just as rational 
as that in which the "poundal" is adopted as the unit of 
force, whilst it may be pointed out that the use of the latter 
system is practically confined to certain text-books and exami- 
nation papers, and does not enter into any engineering work. 
Teachers of Engineering often find that students who are 
learning Mechanics by use of the " poundal " system, fail to 
apply the principles to engineering problems stated in the 
only units which are used in such cases the gravitational 



vi Preface 

units. The use of the dual system is certainly confusing 
to the student, and in addition necessitates much time being 
spent on the re-explanation of principles, which might other- 
wise be devoted to more technical work. 

Graphical methods of solving problems have in some 
cases been used, by drawing vectors to scale, and by esti- 
mating slopes and areas under curves. It is believed that such 
exercises, although often taking more time to work than the 
easy arithmetic ones which are specially framed to give exact 
numerical answers, compel the student to think of the relations 
between the quantities involved, instead of merely performing 
operations by fixed rules, and that the principles so illustrated 
are more deeply impressed. 

The aim has not been to treat a wide range of academic 
problems, but rather to select a course through which the 
student may work in a reasonable time say a year and 
the principles have been illustrated, so far as the exclusion 
of technical knowledge and terms would allow, by examples 
likely to be most useful to the engineer. 

In view of the applications of Mechanics to Engineering, 
more prominence than usual has been given to such parts 
of the subject as energy, work of forces and torques, power, 
and graphical statics, while some other parts have received 
less attention or have been omitted. 

It is usual, in books on Mechanics, to devote a chapter 
to the equilibrium of simple machines, the frictional forces 
in them being considered negligible : this assumption is so far 
from the truth in actual machines as to create a false impres- 
sion, and as the subject is very simple when treated experi- 
mentally, it is left for consideration in lectures on Applied 
Mechanics and in mechanical laboratories. 

The calculus has not been used in this book, but the 



Preface vii 

student is not advised to try to avoid it; if he learns the 
elements of Mechanics before the calculus, dynamical illus- 
trations of differentiation and integration are most helpful. 
It is assumed that the reader is acquainted with algebra to 
the progressions, the elements of trigonometry and curve 
plotting ; in many cases he will doubtless, also, though not 
necessarily, have some little previous knowledge of Mechanics. 

The ground covered is that required for the Intermediate 
(Engineering) Examination of the University of London in 
Mechanics, and this includes a portion of the work necessary 
for the Mechanics Examination for the Associateship of the 
Institution of Civil Engineers and for the Board of Education 
Examination in Applied Mechanics. 

I wish to thank Professor W. Robinson, M.E., and Pro- 
fessor J. Goodman for several valuable suggestions made 
with respect to the preparation and publication of this book; 
also Mr. G. A. Tomlinson, B.Sc., for much assistance in 
correcting proofs and checking examples; in spite of his 
careful corrections some errors may remain, and for any 
intimation of these I shall be obliged. 

ARTHUR MORLEY. 

NOTTINGHAM, 

June, 1905. 



CONTENTS 

CHAPTER I 

KINEMATICS 

PAGES 

Velocity ; acceleration ; curves of displacement, and velocity ; 
falling bodies ; areas under curves ; vectors ; applications to 
velocities ; relative velocity ; composition and resolution of 
acceleration ; angular displacement, velocity, and acceleration 1-26 

CHAPtER II 

THE LAWS OF MOTION 

First law ; inertia ; weight ; momentum ; second law ; engineers' 
units ; c.g.s. system ; triangle and polygon of forces ; impulse ; 
third law ; motion of connected bodies ; Atwood's machine 27-47 

CHAPTER III 
WORK, POWER, AND ENERGY 

Work ; units ; graphical method ; power ; moment of a force ; 

work of a torque ; energy potential, kinetic ; principle of work 48-67 

CHAPTER IV 
MOTION IN A CIRCLE: SIMPLE HARMONIC MOTION 

Uniform circular motion ; centripetal and centrifugal force ; 
curved track ; conical pendulum ; motion in vertical circle ; 
simple harmonic motion ; alternating vectors ; energy in 
S.H. motion ; simple pendulum 68-90 



x Contents 

CHAPTER V 

STA TICS CONCURRENT FORCES FRICTION 

PAGES 

Triangle and polygon of forces ; analytical methods ; friction ; 
angle of friction ; sliding friction ; action of brakes ; 
adhesion ; friction of screw 91-113 

CHAPTER VI 
STATICS OF RIGID BODIES 

Parallel forces ; moments ; moments of resultants ; principle of 
moments ; levers ; couples ; reduction of a coplanar system ; 
conditions of equilibrium ; smooth bodies ; method of 
sections ; equilibrium of three forces 114-139 

CHAPTER VII 

CENTRE OF INERTIA OR MASS CENTRE OF GRAVITY 

Centre of parallel forces ; centre of mass ; centre of gravity ; two 
bodies ; straight rod ; triangular plate ; rectilinear figures ; 
lamina with part removed ; cone ; distance of e.g. from 
lines and planes ; irregular figures ; circular arc, sector, 
segment ; spherical shell ; sector of sphere ; hemisphere . . 140-166 

CHAPTER VIII 

CENTRE OF GRAVITY PROPERTIES AND APPLICATIONS 

Properties of e.g. ; e.g. of distributed load; body resting on a 
plane ; stable, unstable, and neutral equilibrium ; work 
done in lifting a body ; theorems of Pappus 167-187 

CHAPTER IX 
MOMENTS OF INERTIA ROTATION 

Moments of inertia ; particles ; rigid body ; units ; radius of 
gyration ; various axes ; moment of inertia of an area ; 
circle ; hoop ; cylinder ; kinetic energy of rotation ; changes 
in energy and speed ; momentum ; compound pendulum ; 
laws of rotation ; torsional oscillation ; rolling bodies . . 188-222 



Contents xi 

CHAPTER X 

ELEMENTS OF GRAPHICAL STATICS 

PAGES 

Bows' notation ; funicular polygon ; conditions of equilibrium, 
choice of pole ; parallel forces ; bending moment and 
shearing force ; diagrams and scales ; jointed structures ; 
stress diagrams ; girders : roofs ; loaded strings and chains . 223-252 

APPENDIX 2 53~ 2 55 

ANSWERS TO EXAMPLES 256-259 

EXAMINATION QUESTIONS . 260-273 

MATHEMATICAL TABLES ,....; 274-278 

INDEX 279-282 



MECHANICS FOR ENGINEERS 







i. KINEMATICS deals with the motion of bodies without 
reference to the forces causing motion. 

MOTION IN A STRAIGHT LINE. 

Velocity. The velocity of a moving point is the rate of 
change of its position. 

Uniform Velocity. When a point passes over equal 
spaces in equal times, it is said to have a constant velocity ; the 
magnitude is then specified by the number of units of length 
traversed in unit time, e.g. if a stone moves 15 feet with a 
constant velocity in five seconds, its velocity is 3 feet per 
second. 

If s = units of space described with constant velocity v in 
t units of time, then, since v units are described in each second, 
(v x /) units will be described in / seconds, so that 

s = vt 
and v = - 

Fig. i shows graphically the relation between the space 
described and the time taken, for a constant velocity of 3 feet 

per second. Note that v=- = - or - or -, a constant 






Mechanics for Engineers 



velocity of 3 feet per second whatever interval of time is 
considered. 



1234- 

Time in- seconds 
FIG. i. Space curve for a uniform velocity of 3 feet per second. 

2. Mean Velocity. The mean or average velocity of a 
point in motion is the number of units of length described, 
divided by the number of units of time taken. 

3. Varying Velocity, The actual velocity of a moving 
point at any instant is the mean velocity during an indefinitely 
small interval of time including that instant. 

4. The Curve of Spaces or Displacements. Fig. 2 
shows graphically the relation between the space described and 




T N M 
in. seconds 

FIG. 2. Space curve for a varying velocity. 

the time taken for the case of a body moving with a varying 
velocity. At a time ON the displacement is represented by 



Kinematics 3 

PN, and after an interval NM it has increased by an amount 
QR, to QM. Therefore the mean velocity during the interval 

QR QR A 

NM is represented by 7 or --^ or by tan QPR, i.e. by the 



tangent of the angle which the chord PQ makes with a hori- 
zontal line. If the interval of time NM be reduced indefi- 
nitely, the chord PQ becomes the tangent line at P, and the 
mean velocity becomes the velocity at the time ON. Hence 
the velocity at any instant is represented by the gradient of the 
tangent line to the displacement curve at that instant. An upward 
slope will represent a velocity in one direction, and a down- 
ward slope a velocity in the opposite direction. 

5. If the curvature is not great, i.e. if the curve does not bend 
sharply, the best way to find the direction of the tangent line 
at any point P on a curve such as Fig. 2, is to take two ordi- 
nates, QM and ST, at short equal distances from PN, and join 

QV 
QS; then the slope of QS, viz. --, is approximately the same 

as that of the tangent at P. This is equivalent to taking the 
velocity at P, which corresponds to the middle of the interval 
TM, as equal to the mean velocity during the interval of 
time TM. 

6. Scale of the Diagram. Measure the slope as the 
gradient or ratio of the vertical height, say QV, to the hori- 
zontal SV or TM. Let the ratio QV : TM (both being 
measured in inches say) be x. Then to determine the velocity 
represented, note the velocity corresponding to a slope of 
T inch vertical to i inch horizontal, say y feet per second. 
Then the slope of QS denotes a velocity of xy feet per 
second. 

7. Acceleration. The acceleration of a moving body is 
the rate of change of its velocity. When the velocity is in- 
creasing the acceleration is reckoned as positive, and when 
decreasing as negative. A negative acceleration is also called 
a retardation. 

8. Uniform Acceleration. When the velocity of a point 
increases by equal amounts in equal times, the acceleration is 
said to be uniform or constant : the magnitude is then specified 



4 Mechanics for Engineers 

by the number of units of velocity per unit of time ; e.g. if a 
point has at a certain instant a velocity of 3 feet per second, 
and after an interval of eight seconds its velocity is 19 feet per 
second, and the acceleration has been uniform, its magnitude is 

increase of velocity TO 3 

.- , -. - = b = 2 feet per second in each of 
time taken to increase 8 

the eight seconds, i.e. 2 feet per second per second. At the end 
of the first, second, and third seconds its velocities would be 
(3 + 2 ): (3 + 4), and (3 + 6) feet per second respectively (see 
Fig. 3). 



K 10 



8 
8 7 

^ 
V) 5 



M 



1234-56 

Time- in, seconds 

FIG. 3. Uniform acceleration. 

9. Mean Acceleration. The acceleration from 3 feet 
per second to 19 feet per second in the last article was sup- 
posed uniform, 2 feet per second being added to the velocity 
in each second; but if the acceleration is variable, and the 
increase of velocity in different seconds is of different amounts, 
then the acceleration of 2 feet per second per second during 
the eight seconds is merely the mean acceleration during that 

increase of velocity 
time. The mean acceleration is equal to time taken for increase", 

and is in the direction of the change of velocity. 



Kinematics 



5 






The actual acceleration at any instant is the mean 
acceleration for an indefinitely small time including that 
instant. 

10. Fig. 3 shows the curve of velocity at every instant 
during the eight seconds, during which a point is uniformly 
accelerated from a velocity of 3 feet per second to one of 
19 feet per second. 

n. Calculations involving Uniform Acceleration. 
If u = velocity of a point at a particular instant, and^ = uni- 
form acceleration, i.e. f units of velocity are added every 
second 

then after i second the velocity will be ti +/ 
and 2 seconds u -f- 2/ 

jj 3 M ^ 3y 

/ v will be u +ff (i) 

e.g. in the case of the body uniformly accelerated 2 feet per 
second per second from a velocity of 3 feet per second to a 
velocity of 19 feet per second in eight seconds (as in Art. 8), 
the velocity after four seconds is 3 -f- (* X 4) = 1 1 feet per 
second. 

The space described (s) in t seconds may be found as 
follows : The initial velocity being u, and the final velocity 
being v, and the change being uniform, the mean or average 

i . u + v 
velocity is . 

u -f v u u -\- ft 
Mean velocity = SB -U = u 4- ^ft 

2 2 ' 2 

(which is represented by QM in Fig. 3. See also Art. 2). 
Hence u + \ft'= - 

and s=af + \fP ... (2) 
e.g. in the above numerical case the mean velocity would be 

=11 feet per second (QM in Fig. 3) 

and J=nx8 = 88 feet 
or s = 3 x 8 -f l x 2 x 8 2 = 24 + 64 = 88 feet 

It is sometimes convenient to find the final velocity in 



6 Mechanics for Engineers 

terms of the initial velocity, the acceleration, and the space 
described. We have-^ 

from (i) v = n -\-ft 

therefore V* = u 1 + zuft +/ 2 / 2 = 2 + /(/ + \ff) 
and substituting for (ut + |// 2 ) its value s from (2), we have 

. ****+2/S (3) 

The formulae (i), (2), and (3) are useful in the solution of 
numerical problems on uniformly accelerated motion. 

12. Acceleration of Falling- Bodies. It is found that 
bodies falling to the earth (through distances which are small 
compared to the radius of the earth), and entirely unresisted, 
increase their velocity by about 32-2 feet per second every 
second during their fall. The value of this acceleration varies 
a little at different parts of the earth's surface, being greater 
at places nearer to the centre of the earth, such as high lati- 
tudes, and less in equatorial regions. The value of the 
"acceleration of gravity" is generally denoted by the letter 
g. In foot and second units its value in London is about 
32' 1 9, and in centimetre and second units its value is about 
98 r units. 

13. Calculations on Vertical Motion. A body pro- 
jected vertically downwards with an initial velocity u will 
in / seconds attain a velocity u -)- gt, and describe a space 

Mf+'fa* 

In the case of a body projected vertically upward with a 
velocity //, the velocity after / seconds will be, */ gt, and will be 
upwards if gt is less than u, but downward if gt is greater than 
u. When t is of such a value that gt = u, the downward 
acceleration will have just overcome the upward velocity, and 
the body will be for an instant at rest : the value of / will then be 

u 

-. The space described upward after t seconds will be 

Sr-i/ft 

The time taken to rise h feet will be given by the equation 

h = ut - i/? 2 
This quadratic equation will generally have two roots, the 



Kinematics 



smaller being the time taken to pass through h feet upward, and 
the larger being the time taken until it passes the same point on 
its way downward under the influence of gravitation. 

The velocity z', after falling through " // " feet from the point 
of projection downwards with a velocity u, is given by the 
expression i? = u z + 2^, and if u = o, i.e. if the body be simply 
dropped from rest, v 2 = 2gh, and v = *J 2gh after falling h feet. 

14. Properties of the Curve of Velocities. Fig. 4 
shows the velocities at all times in a particular case of a body 




Time 
FIG. 4. Varying velocity. 

starting from rest and moving with a varying velocity, the 
acceleration not being uniform. 

(i) Slope of the Curve. At a time ON the velocity is 
PN, and after an interval NM it has increased by an amount 
QR to QM; therefore the mean acceleration during the 

OR OR 
interval NM is represented by ^rr or p^, i.e. by the tangent of 



the angle which the chord PQ makes with a horizontal line. 
If the interval of time NM be reduced indefinitely, the chord 
PQ becomes the tangent line to the curve at P, and the mean 
acceleration becomes the acceleration at the time ON. So that 
tJie acceleration at any instant is represented by the gradient of 
the tangent line at that instant. The slope will be upward if 
the velocity is increasing, downward if it is decreasing ; in the 
latter case the gradient is negative. The scale of accelerations 
is easily found by the acceleration represented by unit gradient. 
If the curve does not bend sharply, the direction of the 



8 



Mechanics for Engineers 



tangent may be found by the method of Art. 5, which is in this 
case equivalent to taking the acceleration at P as equal to the 
mean acceleration during a small interval of which PN is the 
velocity at the middle instant. 

(2) The Area under the Curve. If the velocity is 
constant and represented by PN (Fig. 5), then the distance 

described in an interval 
NM is PN.NM, and there- 
fore the area under PQ, 
viz. the rectangle PQMN, 
represents the space de- 
scribed in the interval 
NM. 

If the velocity is not 
constant, as in Fig. 6, sup- 
pose the interval NM 
divided up into a number 
of small parts such as 
CD. Then AC represents the velocity at the time represented 
by OC ; the velocity is increasing, and therefore in the interval 
CD the space described is greater than that represented by the 
rectangle AEDC, and less than that represented by the rect- 
angle FBDC. The total space described during the interval 



N 
Tirrve, 

FIG. 5. 



M 



I 






f 


f=l 


^-**" 




s 


A 


t 


' 




1 


/ 

















IN CD M 
Tinve- 


FIG. 6. Varying velocity. 



NM is similarly greater than that represented by a series of 
rectangles such as AEDC, and less than that represented by a 
series of rectangles such as FBDC. Now, if we consider 
the number of rectangles to be increased indefinitely, and 



Kinematics 



the width of each to be decreased indefinitely, the area 
PQMN under the curve PQ is the area which lies always 
between the sums of the 
areas of the two series of 
rectangles, however nearly 
equal they may be made by 
subdividing NM, and the 
area PQMN under the curve 
therefore represents the space 
described in the interval NM. 
The area under the curve 
is specially simple in the case 
of uniform acceleration, for 
which the curve of veloci- 
ties is a straight line (Fig. 7). F i G . 7 . 
Here the velocity PN being 

u, and NM being / units of time, and the final velocity being 
QM = v, the area under PQ is 




M 



PN + QM 



2 

u 4- v 



X NM = ST X NM 



X /(as in Art. u) 



v u 



QR 



And if / is the acceleration / = - (represented by -^ or 

OR 

fg, i.e. by tan QPR), 

.*.// = v u 

V = U -\-ft 

.u-\-v . u + u -\-ft 
and the space described - X / is - X /, which is 

2 2 

ut + |// 2 (as in Art. TI). 

15. Notes on Scales. If the scale of velocity is i inch 
to x feet per second, and the scale of time is i inch to y seconds, 
then the area under the curve will represent the distance 
described on such a scale that i square inch represents xy 
feet. 

1 6. In a similar way we may show that the area PQMN 



10 



Mechanics for Engineers 



(Fig. 8) under a curve of accelerations represents the total 
increase in velocity in the interval of time NM. 




If the scale of acceleration is i inch to z feet per second 
per second, and the scale of time is i inch to y seconds, then 
the scale of velocity is i square inch to yz feet per second. 

17. Solution of Problems. Where the motion is of a 
simple kind, such as a uniform velocity or uniform acceleration, 
direct calculation is usually the easiest and quickest mode of 
solution, but where (as is quite usual in practice) the motion is 
much more complex and does not admit of simple mathematical 
expression as a function of the time taken or distance covered, 
a graphical method is recommended. Squared paper saves 
much time in plotting curves for graphical solutions. 

Example i. A car starting from rest has velocities v feet per 
second after / seconds from starting, as given in the following 
table : 



V 


o 

o 


4 
iro 


9 

22'6 


17 

35-6 


24 
44'5 


30 
49-0 


32 
48-9 


40 


45 
33 7 


53 
26-8 


58 


62 

24'C 



Find the accelerations at all times during the first 60 seconds, and 
draw a curve showing the accelerations during this time. 

First plot the curve of velocities on squared paper from the 
given data, choosing suitable scales. This has been done in 



Kinematics 



II 



Fig. 9, curve I., the scales being i inch to 10 seconds and 
i inch to 20 feet per second. 

In the first 10 seconds RQ represents 24^2 feet per second 



f 50 

J 40 

^30 

S 20 



5 



*Al 



N I0 w 

-KffJI-Utr 



20 

-S4G6 



4-0 



50 



if 



Scale, of iTiches 



FIG. 9. 



gain of velocity, and OQ represents 10 seconds; therefore the 
acceleration at N 5 seconds from starting is approximately 

24*2 

, or 2 '4 2 feet per second per second. Or thus : unit 

gradient i inch vertical in i inch horizontal represents 

20 feet per second 
" 10 seconds = 2 feet P er second P er Sec nd 

RQ i '2 1 inch 

hence -^ = : = 1-21 

OQ i inch 

hence acceleration at N 1 

i c T'^T v \ = 2 '4 2 ' eet P er second per second 

Ib I 2 1 /% 2 

(see Art. 14) 

Similarly in the second 10 seconds, SV which is SM RQ, 
represents (39'8 24-2), or 15-6 feet per second gain of velocity ; 



12 



Mechanics for Engineers 



therefore the velocity at W 15 seconds from starting is approxi- 
mately -^ ,or 1*56 feet per second per second. 
10 

Continue in this way, finding the acceleration at say 5, 15, 
25, 35, 45, and 55 seconds from starting; and if greater ac- 
curacy is desired, at 10, 20, 30, 40, 50, and 60 seconds also. 
The simplest way is to read off from the curve I. velocities 
in tabular form, and by subtraction find the increase, say, in 
10 seconds, thus 



t 


O 


5 


10 


IS 


20 


2? 


30 


35 


40 


45 


50 


55 


60 


V ... 


o 

^_ 


13-5 


24'2 


32-8 

^/- 


.39/8:452 


49' o 47 '5 


40-6 

^- 


337 

^^ 


29-0 


25-1 


24/1 


Change in v 




24-2 




iS'6 


l 9-2 -8-4 




-11-6 




-5' 1 




for 10 sees. 






i 9' 3 




I2'6 




2'I 




'3*8 




-8-6 






Acceleration 




2-42 


i '93 


1-56 


1-26 




0'2I 


-0-84 


-1-38 


-1-16 


-0-86 


-0-51 





























From the last line in this table curve II., Fig. 9, has been 
plotted, and the acceleration at any instant can be read off 
from it. 

It will be found that the area under curve II. from the 
start to any vertical ordinate is proportional to the correspond- 
ing ordinate of curve I. (see Art. 16). The area, when below 
the time base-line, must be reckoned as negative. 



Example 2. Find the distance covered from the starting-point 
by the car in Example i at all times during the first 60 seconds, 
and the average velocity throughout this time. 

In the first 10 seconds the distance covered is found approxi- 
mately by multiplying the velocity after 5 seconds by the time, i.e. 
13*5 x 10 = 135 feet. This approximation is equivalent to taking 
13-5 feet per second as the mean velocity in the first 10 seconds. 

In the next 10 seconds the mean velocity being approximately 
32'8 feet per second (corresponding to / = 15 seconds), the distance 
covered is 32*8 x 10 = 328 feet, therefore the total distance covered 
in the first 20 seconds is 135 + 328 = 463 feet. Proceeding in this 
way, taking ro-second intervals throughout the 60 seconds, and 
using the tabulated results in Example i, we get the following 
results : 



Kinematics 





I 








t 


o 


IO 


20 


30 


40 


50 


60 


Space in 
















previous 


o 


135 


32 


454 


475 


.337 


251 


10 sees. 
















Total space 





135 


463 


917 


1392 


1729 


1980 



from which the curve of displacements, Fig. 10, has been plotted. 



2000 



1500 



1000 



500 



10 



2O 30 4-0 

Time in seconds 



50 



60 



Scale of Inches 



FIG. 10. 



Greater accuracy may be obtained by finding the space 
described every 5 instead of every 10 seconds. 

The average velocity = space described = 1980 = feet sec 
time taken 60 

Note that this would be represented on Fig. 9 by a height which is 
equal to the total area under curve I. divided by the length of 
base to 60 seconds. 



EXAMPLES I. 

I. A train attains a speed of 50 miles per hour in 4 minutes after starting 
from rest. Find the mean acceleration in foot and second units. 



14 Mechanics for Engineers 

2. A motor car, moving at 30 miles per hour, is subjected to a uniform 
retardation of 8 feet per second per second by the action of its brakes. 
How long will it take to come to rest, and how far will it travel during this 
time ? 

3. With what velocity must a stream of water be projected vertically 
upwards in order to reach a height of 80 feet ? 

4. How long will it take for a stone to drop to the bottom of a well 
150 feet deep? 

5. A stone is projected vertically upward with a velocity of 170 feet per 
second. How many feet will it pass over in the third second of its upward 
flight ? At what altitude will it be at the end of the fifth second, and also 
at the end of the sixth ? 

6. A stone is projected vertically upward with a velocity of 140 feet per 
second, and two seconds later another is projected on the same path with 
an upward velocity of 135 feet per second. When and where will they 
meet? 

7. A stone is dropped from the top of a tower 100 feet high, and at the 
same instant another is projected upward from the ground. If they meet 
halfway up the tower, find the velocity of projection of the second stone. 



The following Examples are to be worked graphically. 

8. A train starting from rest covers the distances s feet in the times 
seconds as follows : 



! 

r 




o 


5 

10 


ii 

54 


18 
170 


22 
260 


27 
390 


31 
450 


38 
504 


4 6 
55 


5 
570 



Find the mean velocity during the first 10 seconds, during the first 30 
seconds, and during the first 50 seconds. Also find approximately the 
actual velocity after 5, 15, 25, 35, and 45 seconds from starting-point, and 
plot a curve showing the velocities at all times. 

9. Using the curve of velocities from Example 8, find the acceleration 
every 5 seconds, and draw the curve of accelerations during the first 40 
seconds. 

10. A train travelling at 30 miles per hour has steam shut off and 
brakes applied ; its speed after / seconds is shown in the following table : 



* ... ... 

v, miles perj 
hour ...j 




30-0 


4 

26-0 


12 
2I"S 


2O 

167 


26 
i4'o 


35 
io'4 


42 

77 


5 
4-8 



Kinematics 1 5 

Find the retardation in foot and second units at 5-second intervals through- 
out the whole period, and show the retardation by means of a curve. Read 
off from the curve the retardation after 7 seconds and after 32 seconds. 
What distance does the train cover in the first 30 seconds after the brakes 
are applied ? 

II. A body is lifted vertically from rest, and is known to have the 
following accelerations / in feet per second per second after times t 
seconds : 



t ... 


o 

3-0 


0-8 
2-9 


1-9 
2-85 


3-0 

2'6o 


3 '9 

2'2O 


4-8 
175 


6-0 
1-36 


6-8 

I'2O 


8-0 
1*04 


8-8 
0-97 



Find its velocity after each second, and plot a curve showing its velocity at 
all times until it has been in motion 8 seconds. How far has it moved in 
the 8 seconds, and how long does it take to rise 12 feet ? 

VECTORS. 

1 8. Many physical quantities can be adequately expressed 
by a number denoting so many units, e.g. the weight of a 
body, its temperature, and its value. Such quantities are called 
scalar quantities. 

Other quantities cannot be fully represented by a number 
only, and further information is required, e. g. the velocity of a 
ship or the wind has a definite direction as well as numerical 
magnitude : quantities of this class are called vector quantities 
and are very conveniently represented by vectors. 

A Vector is a straight line having definite length and 
direction, but not definite position in space. 

19. Addition of Vectors. To find the sum of two vectors 




ab and cd (Fig. n), set out ab of proper length and direction, 
and from the end b set out be equal in length and parallel to 



i6 



Mechanics for Engineers 




FIG. 12. 



cd ; join ae. Then ae is the geometric or vector sum of ab and 
cd. We may write this 

ab -f be = ae 
or, since be is equal to cd 

ab -j- cd = ae 

20. Subtraction of Vectors. If the vector cd (Fig. 12) 

is to be subtracted from 
the vector ab, we simply 
find the sum ae as before, 
of a vector ab and second 
vector be, which is equal 
to cd in magnitude, but is 
of opposite sign or direc- 
tion ; then 

ae = ab + be = ab cd 

If we had required the difference, cd ab, the result would 
have been ea instead of ae. 

21. Applications: Displacements. A vector has the 
two characteristics of a displacement, viz. direction and magni- 
tude, and can, therefore, represent it completely. If a body 
receives a displacement ab (Fig. n), and then a further dis- 
placement completely represented by cd, the total displacement 
is evidently represented by ae in magnitude and direction. 

22. Relative Displacements. CASE I. Definition. If 

7 a body remains at 

rest, and a second 
body receives a dis- 
placement, the first 
body is said to re- 
ceive a displacement 
of equal amount but 
opposite direction re- 
lative to the second. 
CASE II. Where Two Bodies each receive a Displacement. 
If a body A receive a displacement represented by a vector ab 
(Fig- T 3)> an d a body B receive a displacement represented by 




FIG. 13. 





Kinematics \ 7 

cd, then the displacement of A relative to B is the vector 
difference, ab cd. For if B remained at rest, A would have 
a displacement ab relative to it. But on account of B's motion 
(cd), A has, relative to B, an additional displacement, dc (Case 
I.) ; therefore the total displacement of A relative to B is ab 4- dc 
(or, ab cd} = ab + be = ae (by Art. 20) ; where be is of equal 
ength and parallel to dc. 

23. A Velocity which is displacement per unit time can 
evidently be represented fully by a vector ; in direction by the 
clinure of the vector, and in magnitude by the number of units 
of length in the vector. 

24. Triangle and Polygon of Velocities. A velocity is 
said to be the resultant of two others, which are 

called components, when it is fully represented 

by a vector which is the geometrical sum of two 

other vectors representing the two components ; 

e.g. if a man walks at a rate of 3 miles per 

hour across the deck of a steamer going at 6 

miles per hour, the resultant velocity with which 

the man is moving over the sea is the vector 

sum of 3 and 6 miles per hour taken in the proper 

directions. If the steamer were heading due 

north, and the man walking due east, his actual velocity is 

shown by ac in Fig. 14; 

ab 6 be = 3 







ac = J6* + 3 2 = \/45 miles per hour 
= 671 miles per hour 

and the angle which ac makes with ab E. of N. is given by 
tan0=:=i = 26 35' 

Resultant velocities may be found by drawing vectors to 
scale or by the ordinary rules of trigonometry./ If the re- 
sultant velocity of more than two components (in the same 
plane) is required, two may be compounded, and then a third 
with their resultant, and so on, until all the components have 
been added. It will be seen (Fig. 15) that the result is repre- 
sented by the closing side of an open polygon the sides of 

c 





1 8 Mechanics for Engineers 

which are the component vectors. The order in which the 
sides are drawn is immaterial. It is not an essential condition 
that all the components should be in the same plane, but if 
not, the methods of solid geometry should be employed to 
draw the polygon. 





FIG. 15. 

Fig. 15 shows the resultant vector af of five co-planor 
vectors, ab, be, cd, de, and ef. 

If, geometrically, af = ab + be 
and ad = ac + cd 
then ad = ab + be -j- cd 

and similarly, adding de and ef- 

af = ab + be + cd + de -f ef 

In drawing this polygon it is unnecessary to put in the 
lines ac, ad, and ae. 

25. It is sometimes convenient to resolve a velocity into 
two components, i.e. into two other velocities in particular 
directions, and such that their vector sum is equal to that 
velocity. 

Rectangular Components. The most usual plan is to 
resolve velocities into components in two standard directions 
at right angles, and in the same plane as the original veloci- 
ties : thus, if OX and OY (Fig. 16) are the standard directions, 
and a vector ab represents a velocity v, then the component in 
the direction OX is represented by ac, which is equal to ab 
cos B, and represents v cos 0, and that in the direction OY is 
represented by cb, i.e. by ab sin 0, and is v sin 0. 



Kinematics 



This form of resolution of velocities provides an alternative 
method of finding the re- y 
sultant of several velocities. 
Each velocity may be re- 
solved in two standard 
directions, OX and OY, 
and then all the X com- 
ponents added algebraically 
and all the Y components 
added algebraically. This 
reduces the components to 
two at right angles, which 
may be replaced by a re- 

' FIG. 16. 

sultant R units, such that 

the squares of the numerical values of the rectangular com- 
ponents is equal to the square of R, e.g. to find the resultant 





FIG. 17. 



of three velocities Vj, V 2 , and V 3 , making angles a, /3, and y 
respectively with some fixed direction OX in their plane 
(Fig. 17). 

Resolving along OX, the total X component, say X, is 

X = Vj cos a + V 2 cos /3 + V :t cos y 
Resolving along OY 

Y = V, sin a + V 9 sin 4- V ;t sin y 
and R 2 = X 2 + Y 2 
or R = /X + Y 2 



and it makes with OX an angle 6 such that tan 6 = ^- 



2O 



Mechanics for Engineers 



Fig. 1 7 merely illustrates the process ; no actual drawing 
of vectors is required, the method being wholly one of calcu- 
lation. 

Exercise i. A steamer is going through the water 
at 10 knots per hour, and heading due north. The 
current runs north-west at 3 knots per hour. Find 
the true velocity of the steamer in magnitude and 
direction. 

(i) By drawing vectors (Fig. 18). 

Set off ab, representing 10 knots per hour, to scale 
due north. Then draw be inclined 45 to the direction 
ab, and representing 3 knots per hour to the same 
scale. Join ac. Then ac, which scales 12*6 knots per 
hour when drawn to a large scale, is the true velocity, 
and the angle cab E of N measures 10. 
! (2) Method by resolving N. and E. 

N. component =10 + 3 cos 45 = 10 + -4- knots per hour, 

V 2 



E. 



= 3 sin 45 = -7- knots per hour, or 2*12 

V 2 



Resultant velocity R = V( 12 ' 12 ) 2 + ( 2<I2 ) 2 = I2 '6 knots per hour 

And if is the) 3_ . /, i -/3\ 2 ' 12 

angle E. of N.) a ~ ^2 

:. = 9 55' 



RELATIVE VELOCITY. 

26. The velocity of a point A relative to a point B is the 
rate of change of position (or displacement per unit of time) 
of A with respect to B. 

Let v be the velocity of A, and u that of B. 

If A remained stationary, its displacement per unit time 
relative to B would be u (Art. 22). But as A has itself a 
velocity v, its total velocity relative to B is v + ( u) or v n, 
the subtraction to be performed geometrically (Art. 20). 

The velocity of B relative to A is of course u v, equal in 
magnitude, but opposite in direction. The subtraction of 
velocity z> u may be performed by drawing vectors to scale, 



Kinematics 



21 



by the trigonometrical rules for the solution of triangles, or by 
the method of Art. 25. 

Example. Two straight railway lines cross : on the first a 
train 10 miles away from the crossing, and due west of it, is ap- 
proaching at 50 miles per hour ; on the second a train 20 miles 
away, and 15 E. of N., is approaching at 40 miles per hour. 
How far from the crossing will each train be when they are nearest 
together, and how long after they occupied the above positions ? 

First set out the two lines at the proper angles, as in the left side 
of Fig. 19, and mark the positions A and B of the first and second 




FIG. 19. 



trains respectively. Now, since the second train B is coming 
from 1 5 E. of N., the first train A has, relative to the second, a 
component velocity of 40 miles per hour in a direction E. of N., 
in addition to a component 50 miles per hour due east. The 
relative velocity is therefore found by adding the vectors pq 50 
miles per hour east, and yr 40 miles per hour, giving the vector pr, 
which scales 72 miles per hour, and has a direction 57^ E. of N. 
Now draw from A a line AD parallel to pr. This gives the posi- 
tions of A relative to B (regarded as stationary). The nearest 
approach is evidently a distance BD, where BD is perpendicular 
to AD. The distance moved by A relative to B is then AD, which 
scales 23-2 miles (the trains being then a distance BD, which scales 
8' 1 2 miles apart). The time taken to travel relatively 23-2 miles 

2 3' 2 

at 72 miles per hour is - hours = 0^322 hour. 



22 



Mechanics for Engineers 




Hence A will have travelled 50 x o'322 or 16*1 miles 

and B ,, ,, 40 x 0-322 or 12-9 

A will then be 6'i miles past the crossing, and 
B ,, 7'i short of the crossing. 

27. Composition, Resolution, etc., of Accelerations. 

Acceleration being also a vector quantity, the methods of 
composition, resolution, etc., of velocities given in Arts. 23 
to 26 will also apply to acceleration, which is simply velocity 
added per unit of time. It should be noted 
that the acceleration of a moving point is not 
necessarily in the same direction as its velocity : 
this is only the case when a body moves in a 
straight line. 

If ab (Fig. 20) represents the velocity of a 
point at a certain instant, and after an interval t 
seconds its velocity is represented by ac, then 
the change in velocity in / seconds is be, for 
ab -\- be = ac (Art. 19), and be = ac ab (Art. 
20), representing the change in velocity. Then 
during the / seconds the mean acceleration is represented by 
be 4- 1, and is in the direction be. 

28. Motion down a Smooth Inclined Plane. Let 
be the angle of the plane to the horizontal, then the angle 
ABC (Fig. 21) to the vertical is (90 a). Then, since a 

body has a downward ver- 
tical acceleration g, its 
component along BA will 
be g cos CBA = g cos 
(90 a) = g sin a, pro- 
vided, of course, that there 
is nothing to cause a re- 
tardation in this direction, 
i.e. provided that the plane is perfectly smooth and free from 
obstruction. If BC = h feet, AB = h cosec a feet. The 
velocity of a body starting from rest at B and sliding down 
AB will be at A, \J 2 .^sin 6 X h cosec 6 = */ 2g/i, just as if it 
had fallen h feet vertically. 



FIG. 




FIG. 21. 



Kinematics 



Angular Displacement. If 
P 



FIG. 22. 



29. Angular Motion 

P (Fig. 22) be the posi- 
tion of a point, and Q 
a subsequent position 
which this point takes 
up, then the angle QOP 
is the angular displace- 
ment of the point about 
O. The angular displacement about any other point, such as 
O', will generally be a different amount. 

30. Angular Velocity. The angular velocity of a moving 
point about some fixed point is the rate of angular displacement 
(or rate of change of angular position) about the fixed point ; it 
is usually expressed in radians per second, and is commonly 
denoted by the letter w. As in the case of linear velocity, it 
may be uniform or varying. 

A point is said to have a uniform or constant angular 
velocity about a point O when it describes equal angles about O 
in equal times. The mean angular velocity of a moving point 
about a fixed point O is the angle described divided by the 
time taken. 

If the angular velocity is varying, the actual angular velocity 
at any instant is the mean angular velocity during an in- 
definitely small interval including that instant. 

31. Angular Acceleration is the rate of change of 
angular velocity ; it is usually measured in radians per second 
per second. 

32. The methods of Arts. 4 to n 
applicable to angular motion as well as 
to linear motion. 

33. To find the angular velocity about 
O of a point describing a circle of radius 
r about O as centre with constant speed. 

Let the path PP' (Fig. 23) be de- 
scribed by the moving point in / seconds. 
Let v be the velocity (which, although F|G - 2 3- 

constant in magnitude, changes direction). Then angular 



and 



1 6 are 




velocity about O is o> = -. 



But = 



Mechanics for Engineers 
arc PP 



and arc PP' = vt 



Vt 6 Tt V 

:. 6 = - and o> = -= / = - 
r t r r 

This will still be true if O is moving in a straight line with 
velocity v as in the case of a rolling wheel, provided that v 
is the velocity of P relative to O. 

If we consider / as an indefinitely small time, PP' will 
be indefinitely short, but the same will remain true, and we 

*V 

should have to = - whether the velocity remains constant in 

magnitude or varies. 

In words, the angular velocity is equal to the linear 
velocity divided by the radius, the units of length being the 
same in the linear velocity v and the radius /-. 

Example. The cranks of a bicycle are 61 inches long, and the 
bicycle is so geared that one complete rotation of the crank carries 
it through a distance equal to the circumference of a wheel 65 
inches diameter. When the bicycle is driven at 15 miles per hour, 
find the absolute velocity of the centre of a pedal (i) when 
vertically above the crank axle ; (2) when vertically below it ; 
(3) when above the axle and 30 forward of a vertical line 
through it. 

The pedal centre describes a circle of 13 inches diameter 
relative to the crank axle, i.e. 13* inches, while the bicycle travels 
6$v inches. Hence the velocity of the pedal centre relative to the 
crank axle is 1 that of the bicycle along the road, or 3 miles per 
hour. 

1 5 miles per hour = 22 feet per second 
3 > = 4'4 




FIG. 24. 



Kinematics 25 

(1) When vertically above the crank axle, the velocity of pedal 
is 22 + 4*4 = 26-4 feet per second. 

(2) When vertically below the crank axle, the velocity of pedal 
is 22 4'4 = 1 7 '6 feet per second. 

(3) Horizontal velocity X = 22 + 4-4 cos 30 = 22 + 2*2 ^3 feet 
per second. 

Vertical velocity downwards Y = 4-4 x sin 30 = 2 '2 feet per 
second. 

Resultant velocity being R 



R = 22^/(i + ^) + (~ J = 25-8 feet per second 



and its direction is at an angle below the horizontal, so that 
_ Y _ 2-2 i i 

and = 4-85 

EXAMPLES II. 

1. A point in the connecting rod of a steam engine moves forwards 
horizontally at 5 feet per second, and at the same time has a velocity of 3 
feet per second in the same vertical plane, but in a direction inclined 1 10 
to that of the horizontal motion. Find the magnitude and direction of its 
actual velocity. 

2. A stone is projected at an angle of 36 to the horizontal with a 
velocity of 500 feet per second. Find its horizontal and vertical velocities. 

3. In order to cross at right angles a straight river flowing uniformly at 
2 miles per hour, in what direction should a swimmer head if he can 
get through still water at 24 miles per hour, and how long will it take him 
if the river is loo yards wide ? 

4. A weather vane on a ship's mast points south-west when the ship is 
steaming due west at 16 miles per hour. If the velocity of the wind is 
20 miies per hour, what is its true direction ? 

5. Two ships leave a port at the same time, the first steams north-west 
at 15 knots per hour, and the second 30 south of west at 17 knots. 
What is the speed of the second relative to the first? After what time 
will they be IOO knots apart, and in what direction will the second lie from 
the first ? 

6. A ship steaming due east at 12 miles per hour crosses the track 
of another ship 20 miles away due south and going due north at 16 miles 
per hour. After what time will the two ships be a minimum distance apart, 
and how far will each have travelled in the interval. 



26 Mechanics for Engineers 

7. Part of a machine is moving east at 10 feet per second, and after ^ 
second it is moving south-east at 4 feet per second. What is the amount 
and direction of the average acceleration during the ^ second ? 

8. How long will it take a body to slide down a smooth plane the 
length of which is 20 feet, the upper end being 37 feet higher than the 
lower one. 

9. The minute-hand of a clock is 4 feet long, and the hour-hand is 
3 feet long. Find in inches per minute the velocity of the end of the 
minute-finger relative to the end of the hour-hand at 3 o'clock and at 
12 o'clock. 

10. A crank, CB, is I foot long and makes 300 turns clockwise per 
minute. When CB is inclined 60 to the line CA, A is moving along AC 




at a velocity of 32 feet per second. Find the velocity of the point B rela- 
tive to A. 

11. If a motor car is travelling at 30 miles per hour, and the wheels 
are 30 inches diameter, what is their angular velocity about their axes ? If 
the car comes to rest in 100 yards under a uniform retardation, find the 
angular retardation of the wheels. 

1 2. A flywheel is making 1 80 revolutions per minute, and after 20 seconds 
it is turning at 140 revolutions per minute. How many revolutions will it 
make, and what time will elapse before stopping, if the retardation is 
uniform ? 



CHAPTER II 

THE LAWS OF MOTION 

34. NEWTON'S Laws of Motion were first put in their present 
form by Sir Isaac Newton, although known before his time. 
They form the foundation of the whole subject of dynamics. 

35. First Law of Motion. Every body continues in its 
state of rest or uniform motion except in so far as it may be com- 
pelled by force to change that state. 

We know of no case of a body unacted upon by any force 
whatever, so that we have no direct experimental evidence of 
this law. In many cases the forces in a particular direction 
are small, and in such cases the change in that direction is 
small, e.g. a steel ball rolling on a horizontal steel plate. To 
such instances the second law is really applicable. 

From the first law we may define force as that which tends 
to change the motion of bodies either in magnitude or direction. 

36. Inertia. It is a matter of everyday experience that 
some bodies take up a given motion more quickly than others 
under the same conditions. For example, a small ball of iron 
is more easily set in rapid motion by a given push along a 
horizontal surface than is a large heavy one. In such a case 
the larger ball is said to have more inertia than the small one. 
Inertia is, then, the property of resisting the taking up of 
motion. 

37. Mass is the name given to inertia when expressed as 
a measurable quantity. The more matter there is in a body 
the greater its mass. The mass of a body depends upon its 
volume and its density being proportional to both. We may 
define density of a body as being its mass divided by its 
volume, or mass per unit volume in suitable units. 



28 Mechanics for Engineers 

If /;/ = the mass of a body, 

v = its volume, 
and p = its density, 

then P = 

v 

A common British unit of mass is one pound. This is 
often used in commerce, and also in one absolute system 
(British) of mechanical units ; but we shall find it more con- 
venient to use a unit about 32^2 times as large, f for reasons 
to be stated shortly. This unit has no particular name in 
general use. It is sometimes called the gravitational unit of 
mass, or the " engineer's unit of mass." 

In the c.g.s. (centimetre-gramme-second) absolute system, 

the unit mass is the gramme, which is about ? Ib. 

453-6 

38. The weight of a body is the force with which the 
earth attracts it. This is directly proportional to its mass, but 
is slightly different at different parts of the earth's surface. 

39. Momentum is sometimes called the quantity of 
motion of a body. If we consider a body moving with a 
certain velocity, it has only half as much motion as two 
exactly similar bodies would have when moving at that 
velocity, so that the quantity of motion is proportional to the 
quantity of matter, i.e. to the mass. Again, if we consider the 
body moving with a certain velocity, it has only half the quantity 
of motion which it would have if its velocity were doubled, so 
that the quantity of motion is proportional also to the velocity. 
The quantity of motion of a body is then proportional to 
the product (mass) X (velocity), and this quantity is given 
the name momentum. The unit of momentum is, then, that 
possessed by a body of unit mass moving with unit velocity. 
It is evidently a vector quantity, since it is a product of 
velocity, which is a vector quantity, and mass, which is a scalar 
quantity, and its direction is that of the velocity factor. It 
can be resolved and compounded in the same way as can 
velocity. 

40. Second Law of Motion. The rate of change of 



The Laws of Motion 29 

momentum is proportional to the force applied, and takes place in 
the direction of the straight line in which the force acts. This 
law states a simple relation between momentum and force, and, 
as we have seen how momentum is measured, we can proceed 
to the measurement of force. 

The second law states that if F represents force 

F oc rate of change of (m X v) 
where m = mass, v = velocity ; 

therefore Fa m X (rate of change v), if m remains constant 
or F cc m y, f 

where /= acceleration, 

w 

and/ cc 
m 

where F is the resultant force acting on the mass m ; 
hence F = m x f X a constant, 

and by a suitable choice of units we may make the constant 
unity, viz. by taking as unit force that which gives unit mass 
unit acceleration. We may then write 

force = (mass) x (acceleration) 
or F = m X / 

If we take i Ib. as unit mass, then the force which gives 
i Ib. an acceleration of i foot per second per second is called 
the poundal. This system of units is sometimes called the 
absolute system. 1 This unit of force is not in general use with 
engineers and others concerned in the measurement and calcu- 
lation of force and power, the general practice being to take 
the weight of i Ib. at a fixed place as the unit of force. We 
call this a force of i Ib., meaning a force equal to the weight 
of i Ib. As mentioned in Art. 38, the weight of i Ib. of 
matter varies slightly at different parts of the earth's surface, 
but the variation is not of great amount, and is usually negligible. 

1 The gravitational system is also really an absolute system, inasmuch as 
all derived units are connected to the fundamental ones by fixed physical 
relations. See Appendix. 



30 Mechanics for Engineers 

41. Gravitational or Engineer's Units. One pound 
of force acting on i Ib. mass of matter (viz. its own weight) 
in London 1 gives it a vertical acceleration of about 32^2 feet 

per second per second, and since acceleration = - C ce -, i Ib. of 

mass 

force will give an acceleration of i foot per second per second 
(i.e. 32*2 times less), if it acts on a mass of 32*2 Ibs. Hence, 
if we wish to have force denned by the relation 

force = rate of change of momentum, 
or force = (mass) X (acceleration) 
F = m X f 

we must adopt g Ibs. as our unit of mass, where g is the 
acceleration of gravity in feet per second per second in some 
fixed place; the number 32*2 is correct enough for most 
practical purposes for any latitude. This unit, as previously 

stated, is sometimes called the engineers' unit of mass. 

> 

Then a body of weight w Ibs. has a mass of W - units, 

g 

and the equation of Art. 40 becomes F = X f. 

Another plan is to merely adopt the relation, force = (mass) 
X (acceleration) x constant. The mass is then taken in 
pounds, and if the force is to be in pounds weight (and not in 
ponndals) the constant used is g (32*2). There is a strong 
liability to forget to insert the constant g in writing expressions 
for quantities involving force, so we shall adopt the former plan 
of using 32*2 Ibs. as the unit of mass. The unit of momentum 
is, then, that possessed by 32*2 Ibs. moving with a velocity of 
i foot per second, and the unit force the weight of i Ib. The 
number 32*2 will need slight adjustment for places other than 
London, if very great accuracy should be required. 

Defining unit force as the weight of i Ib. of matter, we 
may define the gravitational unit of mass as that mass which 
has unit acceleration under unit force. 

42. C.G.S. (centimetre-gramme-second) Units. In 
this absolute system the unit of mass is the gramme ; the 

1 The place chosen is sometimes quoted as sea-level at latitude 45. 



The Laivs of Motion 31 

unit of momentum that in i gramme moving at i centimetre 
per second; and the unit of force called the dyne is that 
necessary to accelerate i gramme by i centimetre per second 
per second. The weight of i gramme is a force of about 
981 dynes, since the acceleration of gravity is about 981 centi- 
metres per second per second (981 centimetres being equal to 
about 3 2 '2 feet). 

The weight of one kilogram (1000 grammes) is often used 
by Continental engineers as a unit of force. 

Example i. A man pushes a truck weighing 2*5 tons with a 
force of 40 Ibs., and the resistance of the track is equivalent to 
a constant force of 10 Ibs. How long will it take to attain a 
velocity of 10 miles per hour? The constant effective forward 
force is 40 10 = 30 Ibs., hence the acceleration is 

force 2*5 x 2240 

mass = 3 " ~~^T2~ ~ 0<I 7 2 5 foot per second per second 

10 miles per hour = * or - 4 / feet per second 

The time to generate this velocity at 0*1725 foot per second per 
second is then * -4- 0-1725 = 85 seconds, or i minute 25 seconds. 

Example 2. A steam-engine piston, weighing 75 Ibs., is at 
rest, and after 0*25 second it has attained a velocity of 10 feet per 
second. What is the average accelerating force acting on it 
during the 0^25 second ? 

Average acceleration = 10 * 0-25 = 40 feet per sec. 

per sec. 

hence average accelerating force is -~ x 40 = 93-2 Ibs. 

3 2 ' 2 

43. We have seen that by a suitable choice of units the force 
acting on a body is numerically equal to its rate of change of 
momentum ; the second law further states that the force and 
the change of momentum are in the same direction. Mo- 
mentum is a vector quantity, and therefore change of momentum 
must be estimated as a vector change having magnitude and 
direction. 

For example, if the momentum of a body is represented by 
ab (Fig. 25), and after t seconds it is represented by cd, then 
the change of momentum in / seconds is cd ab = eg (see 



32 Mechanics for Engineers 

Art. 20), where ef = cd and gf = ab. Then the average' rate 

of change of momentum in t seconds is represented by ^ in 

magnitude and direction, /.<?. the resultant force acting on the 
body during the / seconds was in the direction eg. Or Fig. 25 




FIG. 25. 

may be taken as a vector diagram of velocities, and eg as 

gcr 

representing change of velocity. Then represents accelera- 
tion, and multiplied by the mass of the body it represents the 
average force. 

Example. A piece of a machine weighing 20 Ibs. is at a certain 
instant moving due east at 10 feet per second, and after 1*25 seconds 
it is moving south-east at 5 feet per second. What was the average 
force acting on it in the interval ? 

The change of momentum per second may be found directly, 
or the change of velocity per second may be found, which, when 
multiplied by the (constant) mass, will give the force acting. 

Using the method of resolution of velocities, the 

final component of velocity E. = 5 cos 45 = 4- feet per second 

initial E. = 10 

hence gain of component 1/5 \ / $ \ 

, .. > = ( --, 10 ) east, or ( 10 -. west 

velocity / \ /v /2 ) \ Jil 

Again, the gain of velocity south is 5 sin 45 = / feet per second 

V 2 



The Laws of Motion 33 

If R = resultant change of velocity 



and R = \/54'3 = 7'37 f eet P er second in 1} seconds 
Hence acceleration = 7*37 -f- i'25 = 5^9 feet per second per second, 
and average force acting = ~ - x 5*9 = 3' 66 Ibs. in a direction 

south of west at an angle whose tangent is -4- -5- ( 10 -5- ) 

v 2 \ v 2 / 

or 0-546, which is an angle of about 28^ south of west (by table 
of tangents). 

44. Triangle, Polygon, etc., of Forces. It has been 
seen (Art. 27) that acceleration is a vector quantity having 
magnitude and direction, and that acceleration can be com- 
pounded and resolved by means of vectors. Also (Art. 40) 
that force is the product of acceleration and mass, the latter 
being a mere magnitude or scalar quantity ; hence force is a 
vector quantity, and concurrent forces can be compounded by 
vector triangles or polygons such as were used in Arts. 19 and 
24, and resolved into components as in Arts. 25 and 28. 

We are mainly concerned with uniplanar forces, but the 
methods of resolution, etc., are equally applicable to forces in 
different planes ; the graphical treatment would, however, in- 
volve the application of solid geometry. 

The particular case of bodies subject to the action of 
several forces having a resultant zero constitutes the subject of 
Statics. 

The second law of motion is true when the resultant force 
is considered or when the components are considered, i.e. the 
rate of change of momentum in any particular direction is pro- 
portional to the component force in that direction. 

45. Impulse. By the impulse of a constant force in any 
interval of time, we mean the product of the force and time. 
Thus, if a constant force of F pounds act for / seconds, the 
impulse of that force is F X A If this force F has during the 
interval / acted without resistance on a mass ;;/, causing its 
velocity to be accelerated from i\ to v, the change of momentum 



34 



Mechanics for Engineers 



during that time will have been from mv^ to mv. 2 , i.e. mr 2 mv^ 
or m(v^ V}). And the change of velocity in the interval / 
under the constant acceleration f is f x / (Art. 1 i), therefore 
z/ 2 v l =//, and t(v. 2 vj = m .f.t; but m X/= F, the 
accelerating force (by Art. 40), hence m(v 2 v^} = F/, or, in 
words, the change of momentum is equal to the impulse. The 
force, impulse, and change of momentum are all to be estimated 
in the same direction. 

The impulse may be represented graphically as in Fig. 26. 
If ON represents / seconds, and PN represents F Ibs. to scale, 



M 



Time 



FIG. 26. 



then the area MPNO under the curve MP of constant force 
represents F x /, the impulse, and therefore also the change of 
momentum. 

Impulse of a Variable Force. In the case of a 
variable force the interval of time is divided into a number of 
parts, and the impulse calculated during each as if the force 
were constant during each of the smaller intervals, and equal 
to some value which it actually has in the interval. The sum 
of these impulses is approximately the total impulse during the 
whole time. We can make the approximation as near as we 
please by taking a sufficiently large number of very small 
intervals. The graphical representation will illustrate this 
point. 

Fig. 27 shows the varying force F at all times during the 
interval NM. Suppose the interval NM divided up into a 



The Laws of Motion 



35 



A 



N C D M 

Tirr^e, 

FIG. 27. Impulse of a variable force. 



number of small parts such as CD. Then AC represents the 
force at the time OC ; the force is increasing, and therefore in 
the interval CD the impulse will be greater than that repre- 
sented by the rectangle AEDC, and less than that represented 
by the rectangle FBDC. 
The total impulse during 
the interval NM is simi- 
larly greater than that 
represented by a series 
of rectangles such as 
AEDC, and less than 
that represented by a 
series of rectangles such 
as FBDC. Now, if we 
consider the number of 
rectangles to be indefi- 
nitely increased, and the width of each rectangle to be decreased 
indefinitely, the area PQMN under the curve PQ is the area 
which lies always between the sums of the areas of the two 
series of rectangles however far the subdivision may be carried, 
and therefore it represents the total impulse in the time NM, 
and therefore also the gain of momentum in that time. 

It may be noticed that the above statement agrees exactly 
with that made in Art. 16. In Fig. 8 the vertical ordinates 
are similar to those in Fig. 27 divided by the mass, and the 
gain of velocity represented by the area under PQ in Fig. 8 is 
also similar to the gain of momentum divided by the mass. 

Note that the force represented by -, ~~i ir (^ e - by tne 

length NM 

average height of the PQMN) is the mean force or time- 
average of the force acting during the interval NM. This 

r u j c j total impulse 

time-average force may be defined as - =-- . 

total time 

The area representing the impulse of a negative or opposing 
force will lie below the line OM in a diagram such as Fig. 27. 
In case of a body such as part of a machine starting from rest 
and coming to rest again, the total change of momentum is 
zero ; then as much area of the force-time diagram lies below 



36 Mechanics for Engineers 

the time base line (OM) as above it. The reader should sketch 
out such a case, and the velocity-time or momentum-time curve 
to be derived from it, by the method of Art. 1 6, and carefully 
consider the meaning of all parts of the diagrams the slopes, 
areas, changes of sign, etc. 

The slope of a momentum-time curve represents accelerat- 
ing force just as that of a velocity-time represents acceleration 
(see Art. 14), the only difference in the case of momentum and 
force being that mass is a factor of each. 

It is to be noticed that the impulse or change of 
momentum in a given interval is a vector quantity having 
definite direction. It must be borne in mind that the change 
of momentum is in the same direction as the force and 
impulse. If the force varies in direction it may be split into 
components (Art. 44), and the change of momentum in two 
standard directions may be found, and 4he resultant of these 
would give the change of momentum in magnitude and 
direction. 

46. Impulsive Forces. Forces which act for a very 
short time and yet produce considerable change of momentum 
on the bodies on which they act are called impttlsive forces. 
The forces are large and the time is small. Instances occur in 
blows and collisions. 

47. The second law of motion has been stated, in Art. 40, 
in terms of the rate of change of momentum. It can now be 
stated in another form, viz. The change of momentum is equal 
to the impulse of the applied force, and is in the same direction. 

Or in symbols, for a mass m 

m(v 2 7'j) = F . / 

where # 2 and z> x are the final and initial velocities, the sub- 
traction being performed geometrically (Art. 20), and F is the 
mean force acting during the interval of time /. 

Example i. A body weighing W Ibs. is set in motion by 
a uniform net force P l Ibs., and in /j seconds it attains a velocity 
V feet per second. It then comes to rest in a further period of 
/ 2 seconds under the action of a uniform retarding force of P 2 Ibs. 
Find the relation between P,, P 2 , and V. 



The Laws of Motion 



37 



During the acceleration period the gain of momentum in the 

W 
direction of motion is .V units, and the impulse in that direction 

is Pj^, hence 

W 

P!/! = .V 
g 

During retardation the gain of momentum in the direction 

W 
of motion is -- .V units, and the impulse in that direction 



is P 2 . t z ', hence 

W 



and finally . V = P,/, = P,/ t = 



+ 



the last relation following algebraically from the two preceding 
ones. 

Example 2. If a locomotive exerts a constant draw-bar pull 
of 4 tons on a train weighing 200 tons up an incline of i in 120, 
and the resistance of the rails, etc., amounts to 10 Ibs. per ton, 
how long will it take to attain a velocity of 25 miles per hour from 
rest, and how far will it have moved ? 

The forces resisting acceleration are 

Ibs. 

(a) Gravity T |<j of 200 tons (see Art. 28) = - = 3733 

() Resistance at 10 Ibs. per ton, 200 x 10 = 2000 

Total ... 5733 

The draw-bar pull is 4 x 2240 = 8960 Ibs. ; hence the net 
accelerating force is 8960 5733 = 3227 Ibs. 

Let / be the required time in seconds ; then the impulse is 3227 
x / units. 

25 miles per hour = ^ x 88 feet per second (88 feet per second 
= 60 miles per hour) 

W 
so that the gain of momentum is . V 



therefore 



200 x 2240 5 

32'2 * 12 



32*2 12 

from which / = 1 59 seconds, or 2 minutes 39 seconds 



38 Mechanics for Engineers 

Since the acceleration has been uniform, the average speed 
is half the maximum (Art. 28), and the distance travelled will be in 
feet 

\ x ^Vj x 88 x 159 = 2915 feet 

Example 3. How long would it take the train in Ex. 2 to 
go i mile up the incline, starting from rest and coming to rest at 
the end without the use of brakes ? 

Let t l = time occupied in acceleration, 
/.j = time occupied in retardation. 

During the retardation period the retarding force will be as in 
Ex. 2, a total of 5733 Ibs. after acceleration ceases. The average 
velocity during both periods, and therefore during the whole time, 
will be half the maximum velocity attained. 

Average velocity = . feet per second 
*i + *a 

and maximum velocity = 2 x ~ T feet per second 

*1 T j 

200 V. 2240 . 5280 

.*. momentum generated = - - x 2 x / units 

32 2 l l -+- 1 2 

The impulse = 3227*1 = 5733^ 



i 3227 
and /, + /, = 



\3227 / 3227 



By the second law, change of momentum = impulse. 

200 x 2240 5280 

' - - x 2 x , . . = 5733/2 

32-2 (A + / 2 ) 

and substituting for / a the value found 



x 2240 5280 

* x 2 x 



/ 2 
agreeing with the last result in Ex. i. 

hence (^ + / 2 ) = 267 seconds = 4 minutes 27 seconds 

Example 4. A car weighing 12 tons starts from rest, and has 
a constant resistance of 500 Ibs. The tractive force, F, on the car 
after t seconds is as follows : 



The Laws of Motion 



39 



/ ... 

F ... 




1280 


2 
1270 


5 
1 220 


8 

1 1 10 


ii 
905 


13 
800 


16 
720 


19 
670 


20 
660 



Find the velocity of the car after 20 seconds from rest, and show 
how to find the velocity at any time after starting, and to find the 
distance covered up to any time. 

Plot the curve of F and /, as in Fig. 28, and read off the force, 



1200 
1000 
v; 600 
.g 600 
400 
200 




1 


> 


~*^ 






















^ 


s 






















V 
























^ 


^ . 


en 






























































24 6 8 10 12 14 16 18 2C 

. w& secoTtaLs 

FIG. 28. 



say every 4 seconds, starting from / = 2, and subtract the 500 Ibs. 
resistance from each as follows : 



/ 


2 


6 


10 


H 


18 


F Ibs. 


1270 


1190 


980 


760 


680 


F-SCO ... 


770 


690 


480 


260 


1 80 



The mean accelerating force in the first 4 seconds is approxi- 
mately 770 Ibs., and therefore the impulse is 770 x 4, which is also 
the gain of momentum, 



_, , , . 12 x 2240 _ 

The mass of the car is - = 835 units 

32*2 

. . .. , momentum 

The velocity after 4 seconds = 



770 x 4 



mass 835 

= 3*69 feet per second 



Mechanics for Engineers 



Similarly, finding the momentum and gain of velocity in each 
\ seconds, we have 



t 


o 


4 


8 


12 


16 


20 


Gain of momentum \ 
in 4 seconds . . . \ 


o 


3080 


2760 


1920 


1040 


720 


Momentum 





3080 


5840 


7760 


8800 


9520 


Velocity, feet per ? 
second \ 


o 


3-69 


7 'oo 


9'3i 


IQ'55 


11-41 



After 20 seconds the velocity is approximately ii'4i feet per 
second. The velocity after any time may be obtained approxi- 
mately by plotting a curve of velocities and times from the values 
obtained, and reading intermediate values. More points on the 
velocity-time curves may be found if greater accuracy be desired. 

The space described is represented by the area under the 
velocity-time curve, and may be found as in Art. 14. 

EXAMPLES III. 

1. The moving parts of a forging hammer weigh 2 tons, and are 
lifted vertically by steam pressure and then allowed to fall freely. What is 
the momentum of the hammer after falling 6 feet ? If the force of the 
blow is expended in 0^015 second, what is the average force of the blow ? 

2. A mass of 50 Ibs. acquires a velocity of 25 feet per second in 
10 seconds, and another of 20 Ibs. acquires a velocity of 32 feet per second 
in 6 seconds. Compare the forces acting on the two masses. 

3. A constant unresisted force of 7000 dynes acts on a mass of 20 kilo- 
grams for 8 seconds. Find the velocity attained in this time. 

4. A train weighing 200 tons has a resistance of 15 Ibs. per ton, sup- 
posed constant at any speed. What tractive force will be required to 
give it a velocity of 30 miles per hour in I '5 minutes ? 

5. A jet of water of circular cross-section and 1*5 inches diameter 
impinges on"a flat plate at a velocity of 20 feet per second, and flows off at 
right angles to its previous path. How much water reaches the plate per 
second ? What change of momentum takes place per second, and what force 
does the jet exert on the plate ? 

6. A train travelling at 40 miles per hour is brought to rest by a uniform 
resisting force in half a mile. How much is the total resisting force in 
pounds per ton ? 

7. A bullet weighing I oz. enters a block of wood with a velocity of 
1800 feet pet cecond, and penetrates it to a depth of 8 inches. What is 
the average resistance of the wood in pounds to the penetration of the 
bullet ? 

8. The horizontal thrust on a steam-engine crank-shaft bearing is 10 tons, 



The Laws of Motion 41 

and the dead weight it supports vertically is 3 tons. Find the magnitude 
and direction of the resultant force on the bearing. 

9. A bullet weighing i oz. leaves the barrel of a gun 3 feet long with a 
velocity of 1500 feet per second. What was the impulse of the force pro- 
duced by the discharge ? If the bullet took 0*004 second to traverse the 
barrel, what was the average force exerted on it ? 

10. A car weighing 10 tons starts from rest. During the first 25 
seconds the average drawing force on the car is 750 Ibs., and the average 
resistance is 40 Ibs. per ton. What is the total impulse of the effective force 
at the end of 25 seconds, and what is the speed of the car in miles per hour ? 

11. The reciprocating parts of a steam-engine weigh 483 Ibs., and 
during one stroke, which occupies 0-3 second, the velocities of these parts 
are as follows : 



Time 


O'O 




o-o 5 !o-075 


O'lOO 


0-125 


i 
0-I500-I75 


O'200 


0-225 


0-250 


0-275 


0-300 


Velocity 1 


























in feet > 


O'OO 


3'46 


6-558-91 


IO"22 


lO'OX) 


IO-48 


9-32 


7'75 


6'02 


4-14 


2-10 


O'OO 


per sec.) 
































' i 

















Find the force necessary to give the reciprocating parts this motion, and 
draw a curve showing its values on a time base throughout the stroke. 
Draw a second curve showing the distances described from rest, for every 
instant during the stroke. From these two curves a third may be drawn, 
showing the accelerating force on the reciprocating parts, on the distance 
traversed as a base. 

48. Third Law of Motion. To every action there is an 
equal and opposite reaction. By the word " action " here is meant 
the exertion of a force. We may state this in another way. 
If a body A exerts a certain force on a body B, then B exerts 
on A a force of exactly equal magnitude, but in the opposite 
direction. 

The medium which transmits the equal and opposite forces 
is said to be in a state of stress. (It will also be in a state of 
strain, but this term is limited to deformation which matter 
undergoes under the influence of stress.) 

Suppose A and B (Fig. 29) are connected by some means 
(such as a string) suitable to withstand tension, and A exerts 
a pull T on B. Then B exerts an equal tension T' on A. 
This will be true whether A moves B or not. Thus A may be 
a locomotive, and B a train, or A may be a ship moored to 



42 Mechanics for Engineers 

a fixed post, B. Whether A moves B or not depends upon 
what other forces may be acting on B. 

Again, if the connection between A and B can transmit a 




B 



FIG. 29. Connection in tension. 



thrust (Fig. 30), A may exert a push P on B. Then B exerts 
an equal push P' on A. As an example, A may be a gun, and 
B a projectile ; the gases between them are in compression. 





- P > 


< 
B 


A 









FIG. 30. Connection in compression. 

Or in a case where motion does not take place, A may be 
a block of stone resting on the ground B ; then A and B are 
in compression at the place of contact. 

49. An important consequence of the third law is that the 
total momentum of the two bodies is unaltered by any mutual 
action between them. For since the force exerted by A on 
B is the same as that exerted by B on A, the impulse during 
any interval given by A to B is of the same amount as that 
given by B to A and in the opposite direction. Hence, if B 
gains any momentum A loses exactly the same amount, and 
the total change of momentum is zero, and this is true for any 
and every direction. This is expressed by the statement that 
for any isolated system of bodies momentum is conservative. 
Thus when a projectile is fired from a cannon, the impulse or 
change of momentum of the shot due to the explosion is of 
equal amount to that of the recoiling cannon in the opposite 



The Laws of Motion 



43 



direction. The momentum of the recoil is transmitted to the 
earth, and so is that of the shot, the net momentum given to 
the earth being also zero. 

50. Motion of Two Connected Weights. Suppose 
two weights, Wj Ibs. and W. 2 Ibs., to be connected by a light 
inextensible string passing over a small and perfectly smooth 
pulley, as in Fig. 31. If W^ is greater 
than W a , with what acceleration (/) 
will they move (W 1 downwards and W 2 
upwards), and what will be the tension 
(T) of the string ? 

( W \ 

Consider Wj ( of mass ) : the 

downward force on it is W l (its weight), 
and the upward force is T, which is the 
same throughout the string by the 
" third law ; " hence the downward 
accelerating force is Wj T. FIG 

W 
Hence (by Art. 40) 1 ./= W, - T (i) 

o 

Similarly, on W 2 the upward accelerating force is T W 2 ; 

W 
hence 2 ./=T - W 2 . . . . (2) 




W, 



adding (i) and (2) 



and from (i) 



W 2 



W.-W, 



The acceleration / might have been stated from considering 
the two weights and string as one complete system. The 
accelerating force on which is W, W 2 , and the mass of 

\V -f- W 2 
which is : 



accelerating force _ W t 
hence / - = 



W 2 



total mass 



44 



Mechanics for Engineers 



As a further example, suppose W 2 instead of being suspended 
slides along a perfectly smooth horizontal table as in Fig. 32, 



W 2 



w 



FIG. 32. 



the accelerating force is W 15 and the mass in motion is 



hence the acceleration / = 



. 



accelerating force on W 2 T 
and since /also = - moeo nf ^ - = Wa -< 

we have T = ^ 



mass of 



W 2 



If the motion of W 2 were opposed by a horizontal force, F, 

W, F 

the acceleration would be r^tr-fr 
Wj + W a 

We have left out of account the weight of W 2 and the 
reaction of the table. These are equal and opposite, and 
neutralize each other. The reaction of the pulley on the 
string is normal to the direction of motion, and has therefore 
no accelerating effect. 

Atwood's Machine is an apparatus for illustrating the 
laws of motion under gravity. It consists essentially of a 
light, free pulley and two suspended weights (Fig. 31), which 
can be made to differ by known amounts, a scale of lengths, 
and clockwork to measure time. Quantitative measurements 
of acceleration of known masses under the action of known 
accelerating forces can be made. Various corrections are 



The Laws of Motion 45 

necessary, and this method is not the one adopted for measuring 
the acceleration g. 

Example I. A hammer weighing W Ibs. strikes a nail weigh- 
ing iv Ibs. with a velocity V feet per second and does not rebound. 
The nail is driven into a fixed block of wood which offers a 
uniform resistance of P Ibs. to the penetration of the nail. How 
far will the nail penetrate the fixed block ? 

Let V = initial velocity of nail after blow. 

W 

Momentum of hammer before impact = .V 

momentum of hammer and nail after impact = . V' 

g 

W + w , TI W Vr . , W Tr 

hence . V = V .'. V = ^r^ . V 

g g- <w + W 

Let / = time of penetration. 

W 

Impulse P/ = .V (the momentum overcome by P) 

/- 

During the penetration, average velocity = V (Arts. 1 1 and 14) 
hence distance moved by nail = $V x / 



' W 



W WV 

x 



+ 



Example 2. A cannon weighing 30 tons fires a icoo-lb. pro- 
jectile with a velocity of 1000 feet per second. With what initial 
velocity will the cannon recoil ? If the recoil is overcome by a 
(time) average force of 60 tons, how far will the cannon travel ? 
How long will it take ? 

Let V = initial velocity of cannon in feet per second. 

Momentum of projectile = - - x 1000 = momentum of cannon 

looo 30 x 2240 

or - x looo = x V 



= IOOOX roo 
Let t = time of recoil. 



and v = IOOOX roo = 14-87 feet per second 
30 x 2240 



46 MecJianics for Engineers 

Impulse of retarding force = 60 x 2240 x / = momentum of shot 

1000 x 1000 
60 x 2240 x t - 

32-2 

and hence / = 0*231 second 

14-87 x 0-231 
Distance moved = V x t - - - = 174 feet 

Example 3. Two weights are connected by a string passing 
over a light frictionless pulley. One is 12 Ibs. and the other ir Ibs. 
They are released from rest, and after 2 seconds 2 Ibs. are removed 
from the heavier weight. How soon will they be at rest again, 
and how far will they have moved between the instant of release 
and that of coming to rest again ? 

First period. 

. , . accelerating force 1211 g 

Acceleration = - - = - - x j* = 

total mass 12+11 23 

72*2 

velocity after 2 seconds = 2 x * = 2 '8 feet per second 
Second period. ^ 

Retardation = ; x g 

II + IO 21 

2 x - 

velocity 23 

time to come to rest = -3 r = = 2 x 4 = 1*825 sec. 

retardation g_ 

21 
average velocity throughout = | maximum velocity (Art. n) 

total time =2+1-825 seconds 
distance moved = \ x 2'8 x 3*825 = 5-35 feet 

EXAMPLES IV. 

1. A fireman holds a hose from which a jet of water i inch in diameter 
issues at a velocity of 80 feet per second. What thrust will the fireman 
have to exert in order to support the jet ? 

2. A machine-gun fires 300 bullets per minute, each bullet weighing 

1 oz. If the bullets have a horizontal velocity of 1800 feet per second, 
find the. average force exerted on the gun. 

3. A pile-driver weighing W Ibs. falls through h feet and drives a pile 
weighing w Ibs. a feet into the ground. Show that the average force of 

W 2 h 

the blow is ~ : Ibs. 

W + w a 

4. A weight of 5 cwt. falling freely, drives a pile weighing 600 Ibs 

2 inches into the earth against an average resistance of 25 tons. How far 
will the weight have to fall in order to do this ? 



The Laws of Motion 47 

5. A cannon weighing 40 tons projects a shot weighing 1500 Ibs. with 
a velocity of 1400 feet per second. With what initial velocity will the 
cannon recoil ? What average force will be required to bring it to rest in 
3 feet ? 

6. A cannon weighing 40 tons has its velocity of recoil destroyed in 
2 feet 9 inches by an average force of 70 tons. If the shot weighed 14 cwt., 
find its initial velocity. 

7. A lift has an upward acceleration of 3'22 feet per second per second. 
What pressure will a man weighing 140 Ibs. exert on the floor of the lift ? 
What pressure would he exert if the lift had an acceleration of 3-22 feet per 
second per second downward? What upward acceleration would cause 
his weight to exert a pressure of 170 Ibs. on the floor ? 

8. A pit cage weighs 10 cwt., and on approaching the bottom of the 
shaft it is brought to rest, the retardation being at the rate of 4 feet per 
second per second. Find the tension in the cable by which the cage is 
lowered . 

9. Two weights, one of 16 Ibs. and the other of 14 Ibs., hanging 
vertically, are connected by a light inextensible string passing over a 
perfectly smooth fixed pulley. If they are released from rest, find how far 
they will move in 3 seconds. What is the tension of the string ? 

10. A weight of 17 grammes and another of 20 grammes are connected 
by a fine thread passing over a light frictionless pulley in a vertical plane. 
Find what weight must be added to the smaller load 2 seconds after they 
are released from rest in order to bring them to rest again in 4 seconds. 
How many centimetres will the weights have moved altogether ? 

11. A weight of 5 Ibs. hangs vertically, and by means of a cord passing 
over a pulley it pulls a block of iron weighing 10 Ibs. horizontally along a 
table-top against a horizontal resistance of 2 Ibs. Find the acceleration of 
the block and tension of the string. 

12. What weight hanging vertically, as in the previous question, would 
give the lo-lb. block an acceleration of 3 feet per second per second on a 
perfectly smooth horizontal table ? 

13. A block of wood weighing 50 Ibs. is on a plane inclined 40 to 
the horizontal, and its upward motion along the plane is opposed by a 
force of 10 Ibs. parallel to the plane. A cord attached to the block, running 
parallel to the plane and over a pulley, carries a weight hanging vertically. 
What must this weight be if it is to haul the block IO feet upwards along 
the plane in 3 seconds from rest ? 






CHAPTER III 
WORK, POWER, AND ENERGY 

51. Work. When a force acts upon a body and causes motion, 

it is said to do work. 

In the case of constant forces, work is measured by the 

product of the force and the displacement, one being estimated 

by its component in the direction of the other. 

One of the commonest examples of a force doing work 

is that of a body being lifted against the force of gravity, its 

weight. The work is then 
measured by the product of 
the weight of the body, and 



the vertical height through 
which it is lifted. If we 
draw a diagram (Fig. 33) 
setting off the constant force 
F by a vertical ordinate, OM, 

(p N then the work done during 

Distance ,. , 

any displacement represented 

by ON is proportional to the 

area MPNO, and is represented by that area. If the scale of 
force is i inch =/ Ibs., and the scale of distance is i inch = q 
feet, then the scale of work is i square inch = pq foot-lbs. 

52. Units of Work. Work being measured by the 
product of force and length, the unit of work is taken as 
that done by a unit force acting through unit distance. In 
the British gravitational or engineer's system of units, this is 
the work done by a force of i Ib. acting through a distance 
of i foot. It is called the foot-pound of work. If a weight 



Work, Power, and Energy 



49 



W Ibs. be raised vertically through // feet, the work done is 
WA foot-lbs. 

Occasionally inch-pound units of work are employed, 
particularly when the displacements are small. 

In the C.G.S. system the unit of work is the erg. This is 
the work done by a force of one dyne during a displacement 
of i centimetre in its own direction (see Art. 42). 

53. Work of a Variable Force. If the force during 
any displacement varies, we may find the total work done 
approximately by splitting the displacement into a number 
of parts and finding the work done during each part, as if 
the force during the partial displacement were constant and 
equal to some value it has during that part, and taking the sum 
of all the work so calculated in the partial displacements. We 
can make the approximation as near as we please by taking 
a sufficiently large number of parts. We may define the work 
actually done by the variable force as the limit to which such 
a sum tends when the subdivisions of the displacement are 
made indefinitely small. 

54. Graphical Representation of Work of a Variable 
Force. Fig. 34 is a diagram showing by its vertical ordinates 



M 



C D 



FIG. 34- 



the force acting on a body, and by its horizontal ones the dis- 
placements. Thus, when the displacement is represented by 

E 



50 Mechanics for Engineers 

ON, the force acting on the body is represented by PN. 
Suppose the interval ON divided up into a number of small 
parts, such as CD. The force acting on the body is represented 
by AC when the displacement is that represented by OC. 
Since the force is increasing with increase of displacement 
the work done during the displacement CD is greater than 
that represented by the rectangle AEDC, and less than that 
represented by the rectangle FBDC. The total work done 
during the displacement will lie between that represented by 
the series of smaller rectangles, such as AEDC, and that 
represented by the series of larger rectangles, such as FBDC. 
The area MPNO under the curve MP will always lie between 
these total areas, and if we consider the number of subdivisions 
of ON to be carried higher indefinitely, the same remains true 
both of the total work done and the area under the curve MP. 
Hence the area MPNO under the curve MP represents the 
work done by the force during the displacement represented 
by ON. 

The Indicator Diagram, first introduced by Watt for 
use on the steam-engine, is a diagram of the same kind as 
Fig. 34. The vertical ordinates are proportional to the total 

force exerted by the steam on 
the. piston, and the horizontal 
ones are proportional to the dis- 
placement of the piston. The 
area of the figure is then pro- 
portional to the work done by 
the steam on the piston. 

In the case of a force vary- 



Space, N ing uniformly with the displace- 

FIG. 35. Force varying uniformly ment, the CUrVC MP is a Straight 
with space. .. /,-,. > , , 

line (rig- 35), and the area 



MPNO = - x ON, or if the initial force (OM) is F 3 

Ibs., and the final one (PN) is F 2 Ibs., and the displacement 

p i -p 
(ON) is d feet, the work done is - . d foot-lbs. 

In stretching an unstrained elastic body, such as a spring, 



Work, Power, and Energy 5 i 

the force starts from zero (or Y l = o). Then the total work 
done is ^F.y/, where F 2 is the greatest force exerted, and d is 
the amount of stretch. 

Average Force. The whole area MPNO (Figs. 34 and 
35) divided by the above ON gives the mean height of the 
area ; this represents the space-average of the force during the 
displacement ON. This will not necessarily be the same as 
the time-average (Art. 45). We may define the space-average 
of a varying force as the work done divided by the displacement. 

55. Power. Power is the rate of doing work, or the 
work done per unit of time. 

One foot-pound per second might be chosen as the unit 
of power. In practice a unit 550 times larger is used; it is 
called the horse-power. It is equal to a rate of 550 foot-lbs. 
per second, or 33,000 foot-lbs. per minute. In the C.G.S. system 
the unit of power is not usually taken as one erg per second, 
but a multiple of this small unit. This larger unit is called 
a watt, and it is equal to a rate of io 7 ergs per second. 
Engineers frequently use a larger unit, the kilowatt, which 
is 1000 watts. One horse-power is equal to 746 watts or 
0746 kilowatt. 

Example i. A train ascends a slope of i in 85 at a speed of 
20 miles per hour. The total weight of the train is 200 tons, and 
resistance of the rails, etc., amounts to 12 Ibs. per ton. Find the 
horse-power of the engine. 

The total force required to draw the load is 

N , 200 x 2240 
(200 x 12) -\ ~ ~ = 7670 Ibs. 

The number of feet moved through per minute is i\ x 88 x 60 
= 1760 feet; hence the work done per minute is 1760 x 7670 
= 13,500,000 foot-lbs., and since i horse-power = 33,000 foot-lbs. 
per minute, the H.P. is 13 : ${{$p Q = 409 horse -power. 

Example 2. A motor-car weighing 1 5 cwt. just runs freely at 
12 miles per hour down a slope of i in 30, the resistance at this 
speed just being sufficient to prevent any acceleration. What horse- 
power will it have to exert to run up the same slope at the same 
speed ? 

In running down the slope the propelling force is that of gravity, 
which is ;& of the weight of the car (Arts. 28 and 44) ; hence the 



52 Mechanics for Engineers 

resistance of the road is also (at 12 miles per hour) equivalent to 

15 x 112 

or 56 Ibs. 
30 

Up the slope the opposing force to be overcome is 56 Ibs. road 
resistance and 56 Ibs. gravity (parallel to the road), and the total 
112 Ibs. 

The distance travelled per minute at 12 miles per hour is \ 
mile = 5a 5 HO or 1056 feet ; hence the work done per minute is 

. 112 x 1056 foot-lbs., and the H.P. is U^ 2L- or y^ H p 

33000 

Example 3. The spring of a safety-valve is compressed from 
its natural length of 20 inches to a length of 17 inches. It then 
exerts a force of 960 Ibs. How much work will have to be done 
to compress it another inch, z>. to a length of 16 inches ? 

The force being proportional to the displacement, and being 
960 Ibs. for 3 inches, it is fi Q or 320 Ibs. per inch of compression. 

When 1 6 inches long the compression is 4 inches, hence the 
force is 4 x 320 or 1280 Ibs. ; hence the work done in compression 

is '- x i, or 1 120 inch-lbs f . (Art. 54, Fig. 35), or 93-3 

foot-lbs. 



EXAMPLES V. 

1. A locomotive draws a train weighing 150 tons along a level track 
at 40 miles per hour, the resistances amounting to 10 Ibs. per ton. What 
horse-power is it exerting ? Find also the horse-power necessary to draw 
the train at the same speed (a) up an incline of I in 250, (t>) down an incline 
of i in 250. 

2. If a locomotive exerts 700 horse-power when drawing a train of 
200 tons up an incline of I in 80 at 30 miles per hour, find the road 
resistances in pounds per ton. 

3. A motor-car engine can exert usefully on the wheels 8 horse-power. 
If the car weighs 16 cwt., and the road and air resistances be taken at 
20 Ibs. per ton, at what speed can this car ascend a gradient of I in 15 ? 

4. A winding engine draws from a coal-mine a cage which with the 
coal carried weighs 7 tons ; the cage is drawn up 380 yards in 35 seconds. 
Find the average horse-power required. If the highest speed attained is 
30 miles per hour, what is the horse-power exerted at that time ? 

5. A stream delivers 3000 cubic feet of water per minute to the highest 
point of a water-wheel 40 feet diameter. If 65 per cent, of the available 
work is usefully employed, what is the horse-power developed by the 
wheel ? 

6. A bicyclist rides up a gradient of I in 15 at 10 miles per hour. The 



Work, Power, and Energy 



53 



weight of rider and bicycle together is 180 Ibs, If the road and other 
resistances are equivalent to T J 5 of this weight, at what fraction of a horse- 
power is the cyclist working ? 

7. Within certain limits, the force required to stretch a spring is 
proportional to the amount of stretch. A spring requires a force of 
800 Ibs. to stretch it 5 inches : find the amount of work done in stretching 
it 3 inches. 

8. A chain 400 feet long and weighing 10 Ibs. per foot, hanging 
vertically, is wound up. Draw a diagram of the force required to draw 
it up when various amounts have been wound up from o to 400 feet. 
From this diagram calculate the work done in winding up (a) the first 
100 feet of the chain, (b) the whole chain. 

9. A pit cage weighing 1000 Ibs. is suspended by a cable 800 feet long 
weighing if Ibs. per foot length. How much work will be done in wind- 
ing the cage up to the surface by means of the cable, which is wound on a 
drum ? 

56. It frequently happens that the different parts of a body 
acted upon by several forces move through different distances 
in the same time ; an important instance is the case of the 
rotating parts of machines generating or transmitting power. It 
will be convenient to consider here the work done by forces 
which cause rotary motion of a body about a fixed axis. 

Moment of a Force. The moment of a force about a 
point is the measure of its turning effect or tendency, about 
that point. It is measured by the 
product of the force and the per- 
pendicular distance from the point 
to the line of action of the force. 
Thus in Fig. 36, if O is a point, and 
AB the line of action of a force F, 
both in the plane of the figure, and 
OP is the perpendicular from O on 
to AB measuring r units of length, 
the moment of F about O is F X r. 

The turning tendency of F about O will be in one direction, 
or the opposite, according as O lies to the right or left of AB 
looking in the direction of the force. If O lies to the right, the 
moment is said to be clockwise ; if to the left, contra-clockwise. 
In adding moments of forces about O, the clockwise and contra- 
clockwise moments must be taken as of opposite sign, and the 




IF) 



FIG. 36. 



54 Mechanics for Engineers 

algebraic sum found. Which of the two kinds of moments is 
considered positive and which negative is immaterial. If O 
lies in the line AB, the moment of F about O is zero. 1 

The common units for the measurement of moments are 
pound-feet. Thus, if a force of i Ib. has its line of action i 
foot from a fixed point, its moment about that point is one 
pound-foot. In Fig. 36, if the force is F Ibs., and OP represents 
r feet, the moment about O is F . r pound-feet. 

Moment of a Force about an Axis perpendicular 
to its Line of Action. If we consider a plane perpendicular 
to the axis and through the force, it will cut the axis in a point 
O ; then the moment of the force about the axis is that of the 
force about O, the point of section of the axis by the plane. 
The moment of the force about the axis may therefore be 
defined as the product of the force and its perpendicular distance 
from the axis. 

In considering the motion of a body about an axis, it is 
necessary to know the moments,, about that axis of all the 
forces acting on the body in planes perpendicular to the axis, 
whtether all the forces are in the same plane or not. The total 
moment is called the torque^ or twisting moment or turning 
moment about the axis. In finding the torque on a body 
about a particular axis, the moments must be added algebrai- 
cally. 

57. Work done by a Constant Torque or Twist- 
ing Moment. Suppose a force F Ibs. (Fig. 37) acts upon a 
body which turns about an axis, O, perpendicular to the line 
of action of F and distant r feet from it, so that the turning 

1 Note that the question whether a moment is clockwise or contra- 
clockwise depends upon the aspect of view. Fig. 36 shows a force (F) 
having a contra-clockwise moment about O, but this only holds for one 
aspect of the figure. If the force F in line AB and the point O be viewed 
from the other side of the plane of the figure, the moment would be called 
a clockwise one. This will appear clearly if the figure is held up to the 
light and viewed from the other side of the page. Similarly, the moment 
of a force about an axis will be clockwise or contra-clockwise according 
as the force is viewed from one end or the other of the axis. The motion 
of the hands of a clock appears contra-clockwise if viewed from the back 
through a transparent face. 



Work, Power, and Energy 55 

moment (M) about O is F.r Ib.-feet. Suppose that the 
force acts successively on different parts of the body all 
distant r from the axis O about which it rotates, or that the 
force acts always on the same 
point C, and changes its direc- 
tion as C describes its circular 
path about the centre O, so as 
to always remain tangential to 
this circular path ; in either case 
the force F is always in the same 
direction as the displacement it 
is producing, and therefore the 
work done is equal to the product 
of the force and the displacement 
(along the circumference of the 
circle CDE). Let the displace- 




FIG. 37. 



ment about the axis O be through an angle 6 radians correspond- 
ing to an arc CD of the circle CDE, so that 



CD 



= 0, or CD = r. 



(The angle is 2w, if a displacement of one complete cir- 
cuit be considered.) 

The work done is F X CD = F . rO foot-lbs. 

But M = F . r Ib.-feet 
therefore the work done = M X foot-lbs. 

The work done by each force is, then, the product of the 
turning moment and the angular displacement in radians. If 
the units of the turning moment are pound-feet, the work will 
be in foot-pounds ; if the moment is in pound-inches, the work 
will be in inch-pounds, and so on. The same method of calcu- 
lating the work done would apply to all the forces acting, and 
finally the total work done would be the product of the total 
torque or turning moment and the angular displacement in 
radians. 

Again, if w is the angular velocity in radians per second, 
the power or work per second is M . o foot-lbs., and the horse- 



56 Mechanics for Engineers 

power is , where M is the torque in Ib.-feet; and if N is 
the number of rotations per minute about the axis 

HP = 2 5.N^ 
33,000 

This method of estimating the work done or the power, 
is particularly useful when the turning forces act at different 
distances from the axis of rotation. 

We may, for purposes of calculation, look upon such a state 
of things as replaceable by a certain force at a certain radius, 
but the notion of a torque and an angular displacement seems 
rather less artificial, and is very useful. 

The work done by a variable turning moment during a 
given angular displacement may be found by the method of 
Arts. 53 and 54. If in Figs. 33, 34, and 35 force be replaced by 
turning moment and space by angular displacement, the areas 
under the curves still represent the work done. 

In twisting an elastic rod from its unstrained position the 
twisting moment is proportional to the angle of twist, hence 
the average twisting moment is half the maximum twisting 
moment ; then, if M = maximum twisting moment, and 6 = 
angle of twist in radians 

the work done = |M0 

Example i. A high-speed steam-turbine shaft has exerted on 
it by steam jets a torque of 2100 Ib.-feet. It runs at 750 rotations 
per minute. Find the horse-power. 
The work done per minute = (torque in Ib.-feet) x (angle turned 

through in radians) 
= 2100 x 750 x 2Tr foot-lbs. 

2100 X 750 X 2ir 

horse-power = = 300 H.P. 

33,000 

Example 2. An electro motor generates 5 horse-power, and 
runs at 750 revolutions per minute. Find the torque in pound-feet 
exerted on the motor spindle. 

Horse-power x 33,000 = torque in Ib.-feet x radians per minute 

horse-power x 33,000 
hence torque m Ib.-feet = - rad i an y^eFminute^ 

5 x 33.000 
= ?- - = 35 lb. feet 

750 X 27T " 



Work, Power, and Energy 57 

EXAMPLES VI. 

1. The average turning moment on a steam-engine crankshaft is 2000 
Ib.-feet, and its speed is 150 revolutions per minute. Find the horse-power 
it transmits. 

2. A shaft transmitting 50 H.P. runs at 80 revolutions per minute. 
Find the average twisting moment in pound-inches exerted on the shaft. 

3. A steam turbine develops 250 horse-power at a speed of 200 revolu- 
tions per minute. Find the torque exerted upon the shaft by the steam. 

4. How much work is required to twist a shaft through 10 if the 
stiffness is such that it requires a torque of 40,000 Ib. -inches per radian of 
twist? 

5. In winding up a large clock (spring) which has completely "run 
down," 8| complete turns of the key are required, and the torque applied 
at the finish is 200 Ib. -inches. Assuming the winding effort is always 
proportional to the amount of winding that has taken place, how much 
work has to be done in winding the clock ? How much is done in the last 
two turns ? 

6. A water-wheel is turned by a mean tangential force exerted by the 
water of half a ton at a radius of IO feet, and makes six turns per minute. 
What horse-power is developed ? 

58. Energy. When a body is capable of doing work, it is 
said to possess energy. It may possess energy for various 
reasons, such as its motion, position, temperature, chemical 
composition, etc. ; but we shall only consider two kinds of 
mechanical energy. 

59. A body is said to have potential energy when it is 
capable of doing work by virtue of its position. For example, 
when a weight is raised for a given vertical height above datum 
level (or zero position), it has work done upon it ; this work is 
said to be stored as potential energy. The weight, in returning 
to its datum level, is capable of doing work by exerting a force 
(equal to its own weight) through a distance equal to the 
vertical height through which it was lifted, the amount of 
work it is capable of doing being, of course, equal to the amount 
of work spent in lifting it. This amount is its potential energy 
in its raised position, e.g. suppose a weight W Ibs. is lifted h 
feet ; the work is W . // foot-lbs., and the potential energy of the 
W Ibs. is then said to be W . h foot-lbs. It is capable of doing 
an amount of work W . h foot-lbs. in falling. 



58 Mechanics for Engineers 

60. Kinetic Energy is the energy which a body has in 
virtue of its motion. 

We have seen (Art. 40) that the exertion of an unresisted 
force on a body gives it momentum equal to the impulse of the 
force. The force does work while the body is attaining the 
momentum, and the work so done is the measure of the kinetic 
energy of the body. By virtue of the momentum it possesses, 
the body can, in coming to rest, overcome a resisting force 
acting in opposition to its direction of motion, thereby doing 
work. The work so done is equal to the kinetic energy of the 
body, and therefore also to the work spent in giving the body 
its motion. 

Suppose, as in Ex. i, Art. 47, a body of weight W Ibs. 
is given a velocity V feet per second by the action of a 
uniform force F] Ibs. acting for ^ seconds, and then comes 
to rest under a uniform resisting force F 2 Ibs. in 4 seconds. We 
had, in Art. 47 

f W 
Impulse F^ = V = F 2 / 2 

<5 

But, the mean velocity being half the maximum under a 
uniform accelerating force, the distance d^ moved in accelerat- 
ing, is IVA feet, and that d. 2) moved in coming to rest, is ~Vt z 
hence the work done in accelerating is 

W W 

F 1 xiVA = -V X iV = ^V 2 

A A 

and work done in coming to rest is 

W W 

F 2 x !V/ 2 =-VxiV = i^V 2 

i> <!> 

W 

hence l-V a = F^ = F 2 

<r 
<S 

These two equalities are exactly the same as those of Ex. 

/ W \ 

i, Art. 47 f viz. V = F^ = F >2 / 2 1 , with each term multi- 

V 

plied by , and problems which were solved from considera- 
tions of changes of momentum might often have been (alter- 
natively) solved by considerations of change of kinetic energy. 



Work, Poiver, and Energy 59 

The amount of kinetic energy possessed by a body of 

W 

weight W Ibs. moving at V feet per second is therefore \ V' 2 

<S 

foot-lbs. 

Again, if the initial velocity had been it feet per second 
instead of zero, the change of momentum would have been 

W 

(v it), and we should have had 
A 

W 

Fj/! = (v ?/), v being final velocity 

o 

n + "v W. N n + v 

and the work done = F : X - X /" t = ( ) - 

W 

- 1- ( 7 ,2 _ 2\ 

J <r 

= change of kinetic energy 

Similarly, in overcoming resistance at the expense of its kinetic 
energy, the work done by a body is equal to the change of 
kinetic energy whether all or only part of it is lost. 

61. Principle of Work. If a body of weight W Ibs. be 
lifted through h feet, it has potential energy W/; foot-lbs. If 
it falls freely, its gain of kinetic energy at any instant is just 
equal to the loss of potential energy, so that the sum (potential 
energy) -f (kinetic energy) is constant ; e.g. suppose the weight 
has fallen freely x feet, its remaining potential energy is 
x} foot-lbs. It will have acquired a velocity V 2gx feet 



W 

per second (Art. 13), hence its kinetic energy ^ V 2 , will be 

& 

W W 

fh- X2gx=VJx foot-lbs., hence W(/$-*)-r-i V 2 = W^, which 

> & 

is independent of the value of x, and no energy has been lost. 

Note that for a particular system of bodies the sum of 
potential and kinetic energies is generally not constant. Thus, 
although momentum is conservative, mechanical energy is not. 
For example, when a body in motion is brought to rest by a 
resisting force of a frictional kind, mechanical energy is lost. 
The energy appears in other forms, chiefly that of heat. 

Principle of Work. Further, if certain forces act upon 



6o 



Mechanics for Engineers 



a body, doing work, and other forces, such as frictional ones, 
simultaneously resist the motion of the body, the excess of the 
work done by the urging forces over that done against the 
resistances gives the kinetic energy stored in the body. Or we 
may deduct the resisting forces from the urging forces at every 
instant, and say that the work done by the effective or net 
accelerating forces is equal to the kinetic energy stored. Thus 
in Fig. 38, representing the forces and work done graphically as 
in Art. 54, if the ordinates of the curve MP represent the 
forces urging the body forward, and the ordinates of M'P' re- 
present the resistances to the same scale, the area MPNO 
represents the work done ; the work lost against resistances is 
represented by the area M'P'NO, and the difference between 
these two areas, viz. the area MPP'M', represents the kinetic 
energy stored during the time that the distance ON has been 
traversed. If the body was at rest at position O, MPP'M' 
represents the total kinetic energy, and if not, its previous 
kinetic energy must be added toobtain the total stored at the 
position ON. From a diagram, such as Fig. 38, the velocity 



M 




FIG. 38. 

can be obtained, if the mass of the moving body is known, by 
the relation, kinetic energy = ^(mass) X (velocity) 2 . 

Fig. 39 illustrates the case of a body starting from rest and 
coming to rest again after a distance O^, such, for example, 
as an electric car between two stopping-places. The driving 
forces proportional to the ordinates of the curve abec cease 



Work, Power, and Energy 



61 



after a distance cc has been traversed, and (by brakes) the 
resisting forces proportional to the ordinates of the curve def 
increase. The area abed represents the kinetic energy of the 
car after a distance 0c, and the area efgc represents the work 




Distances 

FIG. 39. 

done by the excess of resisting force over driving force. When 
the latter area is equal to the former, the car will have come 
to rest. 

The kinetic energy which a body possesses in virtue of its 
rotation about an axis will be considered in a subsequent 
chapter. 

Example I. Find the work done by the charge on a projectile 
weighing 800 Ibs., which leaves the mouth of a cannon at a velocity 
of 1800 feet per second. What is the kinetic energy of the gun at 
the instant it begins to recoil if its weight is 25 tons? 

The work done is equal to the kinetic energy of the projectile 

I W I 8<DO , o \9 

K.E. = - x xV 2 = -x x (1800)* = 40,200,000 foot-lbs. 

The momentum of the gun being equal to that of the projectile, 
the velocity of the gun is 



1800 x 



800 



25 X 2240 

and the K.E. = - x 



= 2571 feet per second 

x (257 1 ) 2 = 577,ooo foot-lbs. 



It may be noticed that the kinetic energies of the projectile 



62 



Mechanics for Engineers 



and cannon are inversely proportional to their weights. The 

1 W i W 

K.E. is - x x V 2 , or - x - x V x V. which is i x momen- 

2 g 2 g 

turn x velocity. The momentum of the gun and that of the pro- 
jectile are the same (Art. 52), and therefore their velocities are 
inversely proportional to their weights ; and therefore the products 
of velocities and half this momentum are inversely proportional 
to their respective weights. 

Example 2. A bullet weighing i oz., and moving at a velocity 
of 1500 feet per second, overtakes a block of wood moving at 
40 feet per second and weighing 5 Ibs. The bullet becomes 
embedded in the wood without causing any rotation. Find the 
velocity of the wood after the impact, and how much kinetic energy 
has been lost. 

Let V = velocity of bullet and block after impact. 

Momentum of bullet = ? x -= = 

16 g g 



momentum of block = 

hence total momentum befo/e ) 
and after impact f 



5 200 

~ x 40 = 



Total momentum after impact = 



g 



29375 
xV = 



and therefore V = , = 58' i feet per second 
- 



5-0625 

x 1500 = 2183 foot- Ibs. 

= 124 
Total K.E. before impact = 2307 

Total K.E. after impact = I x 5 25 x 58-1 x 58-1 = 265 foot-lbs. 
2 32-2 



Kinetic energy of bullet = - x x x 

2 16 32-2 

Kinetic energy of block = - x. ^- x 40 x 40 



Loss of K.E. at impact = 2307 - 265 



= 2042 



Example 3. A car weighs i2'88 tons, and starts from rest ; 
the resistance of the rails may be taken as constant and equal to 
500 Ibs. After it has moved S feet from rest, the tractive force, 
F Ibs., exerted by the motors is as follows : 



s ... 

F ... 




1280 


20 
1270 


50 

1220 


80 

i no 


no 

905 


130 

800 


160 
720 


190 

670 


200 
660 



Work, Power, and Energy 63 

Find the velocity of the car after it has gone 200 feet from rest ; 
nlso find the velocity at various intermediate points, and plot a 
curve of velocity on a base of space described. 

Plot the curve of F and S as in Fig. 40, and read off the force 
every 20 feet, say, starting from S = 10, and subtract 500 Ibs. 
resistance from each, as follows : 



S ... 


10 30 


5 


70 


90 


IIO 


13 


1^0 


170 


190 


F ... 


1275 1260 


1 220 


1150 


1050 


905 


800 


740 


6QS 


670 


F-SOO 


775 76o 


720 


650 


55o 


405 


300 


240 


195 


170 



1200 
1000 

800 
1 

600 
) 

S 

; *oo 

200 







-o^ 


























!S 


k 






















X 


^fit- 




















^*-. 


" < 


>--^ 


^o-< 






























































20 40 60 80 100 120 140 160 ISO 2C 

S i-n feet 

FIG. 40. 



The mean accelerating force during the first 20 feet of motion is 
approximately equal to that at S = 10, viz. 775 Ibs. ; hence the 
work stored as kinetic energy (K.E.), i.e. the gross work done less 
that spent against resistance, is 

(1275 x. 20) - (500 x 20), or 775 x 20 foot-lbs. = 15,500 foot-lbs. 
Then, if V is the velocity after covering S feet, for S = 20 



K.E. = 
and W = 1 2 ' 



V*= 15,500 
x 2240 Ibs. 



W , . I2'88 x 2240 _ , 

therefore , the mass of the car is - - - : or 890 units, and 



64 



Mechanics for Engineers 
x V 2 = 15,500 



1 W W2 1 

- x V 2 = i x 

2 g 



V = 



= 5 '90 feet per second 



Similarly, finding the gain of kinetic energy in each 20 feet, the 
square of velocity (V 2 ), and the velocity V, we have from S = 20 
to S - 40 

gain of K.E. = 760 x 20 = 15,200 foot-lbs. 
/.total K.E. at S = 40 is 

15,500 + 15,200 = 30,700 foot-lbs. 
and so on, thus 



S 


o 


20 


4O 


60 


80 


IOO 


1 20 


I4.O 


1 60 


180 


200 


Gain of K.E. ) 
























in 20 feet, > 


o 


15500 


15200 


14400 


13000 


1 1000 


8100 


6000 


4800 


3900 


3400 


foot-lbs. ) 
























Total K.E., 1 
foot-lbs. / c 


15500 


30700 


45100 


58100 


69100 


77200 


83200 


88000 


91900 


95300 


V 2 or - ' 


o 


34-8 


68-5 


lOO'O 


129-4 


154-0 


172-1 


185-5 


196-2 


204-8 


212-5 


V ft. per sec. 







8-28 


10-03 


"'34 


12-40 


13-12 


13-62 


14-01 


14-30 


14-58 



These velocities have been plotted on a base of spaces in 
Fig. 41. 




20 



60 80 100 120 

S. in, feet 

FIG. 41. 



1*0 160 180 200 



Work, Power, and Energy 65 

Example 4. From the results of Example 3, find in what time 
the car travels the distance of 20 feet from S = 80 to S = 100, 
and draw a curve showing the space described up to any instant 
during the time in which it travels the first 200 feet. 

At S = 80, V = 11-34 feet per second 
at S = 100, V = 12*40 feet per second' 

hence the mean velocity for such a short interval may be taken 
as approximately 

11-34 + 12-40 

, or 1 1 -87 feet per second 

2 

Hence the time taken from S = 80 to S = 100 is approximately 
20 



11-87 



= 1-685 seconds 



Similarly, we may find the time taken to cover each 20 feet, and 
so find the total time occupied, by using the results of Ex. 3, 
as follows. The curve in Fig. 42 has been plotted from these 
numbers. 



200 



ISO 



100 



tt 



3 

* 



50 





10 15 

Time, in, seconds 

FIG. 42. 



20 



25 



66 



Mechanics, for Engineers 



s 


o 


20 


40 


60 


80 zoo 


I2O 140 


160 i 80 


2 


Mean velocity 






















for last 20 ft., 


o 


2'95 


7-09 


9'IS 


10-68 11-87 


I2-76 13-37 


1381 


14-15 


'4' 


feet per sec. 
























Time for last 
























20 feet, se- 


o 


6.780 


2-824 


2-188 


1-872 1-685 


I-S68 


1-496 


i '4*7 


i'4i3 


i'3 


conds 
























Total time,) 
/ seconds J 


o 


6780 


9-604 


11-792 


13-66415-349 


16-917 


18-413 


19-850 


21-263 


22'6 



EXAMPLES VII. 

1. Find in foot-pounds the kinetic energy of a projectile weighing 
800 Ibs. moving at 1000 feet per second. If it is brought to rest in 3 feet, 
find the space average of the resisting force. 

2. At what velocity must a body weighing 5 Ibs. be moving in order 
to have stored in it 60 foot-lbs. of energy ? 

3. What is the kinetic energy in inch-pounds of a bullet weighing I oz. 
travelling at 1800 feet per second? If it is fired directly into a suspended 
block of wood weighing 1-25 lb., how much kinetic energy is lost in the 
impact ? 

4. A machine-gun fires 300 bullets per minute, each bullet weighing 
I oz. and having a muzzle velocity of 1700 feet per second. At what 
average horse-power is the gun working ? 

5. A jet of water issues in a parallel stream at 90 feet per second from 
a round nozzle I inch in diameter. What is the horse-power of the jet ? 
One cubic foot of water weighs 62*5 Ibs. 

6. Steam to drive a steam impact turbine issues in a parallel stream 
from a jet J inch diameter at a velocity of 2717 feet per second, and the 
density of the steam is such that it occupies 26'5 cubic feet per pound. 
Find the horse-power of the jet. 

7. A car weighing 10 tons attains a speed of 15 miles per hour from 
rest in 24 seconds, during which it covers 100 yards. If the space-average 
of the resistances is 30 Ibs. per ton, find the average horse-power used to 
drive the car. 

8. How long will it take a car weighing 1 1 tons to accelerate from 
10 miles per hour to 15 miles per hour against a resistance of 25 Ibs. per 
ton, if the motors exert a uniform tractive force on the wheels and the 
horse-power is 25 at the beginning of this period ? 

9. A car weighing 12 tons is observed to have the following tractive 
forces F Ibs. exerted upon it after it has travelled S feet from rest : 



? '... 


o 

1440 


10 

1390 


30 

1250 


50 

1060 


65 

910 


80 

805 


94 
760 


100 

740 



Work, Power, and Energy 67 

The constant resistance of the road is equivalent to 600 Ibs. Find the 
velocity of the car after it has covered 100 feet. Plot a curve showing 
the velocity at all distances for loo feet from the starting-point. What is 
the space-average of the effective or accelerating force on the car ? 

10. From the results of the last question plot a curve showing the 
space described at any instant during the time taken to cover the first 
ico feet. How long does the car take to cover 100 feet ? 

11. A machine having all its parts in rigid connection has 70,000 foot- 
pounds of kinetic energy when its main spindle is making 49 rotations 
per minute. How much extra energy will it store in increasing its speed 
to 50 rotations per minute ? 

12. A machine stores 10,050 foot-lbs. of kinetic energy when the speed 
of its driving-pulley rises from ico to 101 revolutions per minute. How 
much kinetic energy would it have stored in it when its driving-pulley 
is making loo revolutions per minute? 



CHAPTER IV 



MOTION IN A CIRCLE: SIMPLE HARMONIC 
MOTION 

62. Uniform Circular Motion. Suppose a particle de- 
scribes about a centre O (Fig. 43), a circle of radius r feet 
with uniform angular velocity w radians per second. Then 
its velocity, ?', at any instant is of magnitude wr (Art. 33), and 
its direction is along the tangent to the circle from the point 

in the circumference 
which it occupies at that 
instant. Although its 
velocity is always of 
magnitude wr, its direc- 
tion changes. Consider 
T the change in velocity 
between two points, P 
and Q, on its path at 
an angular distance 6 
apart (Fig. 43)- Let 

the vector cb parallel to the tangent PT represent the linear 
velocity v at P, and let the vector ab, of equal length to cb and 
parallel to QT, the tangent at Q, represent the linear velocity 
v at Q. Then, to find the change of velocity between P and 
Q, we must subtract the velocity at P from that at Q; in 
vectors 

ab cb = ab + be = ac (Art. 27) 

Then the vector ac represents the change of velocity between 
the positions P and Q. Now, since abc = POQ = 0, length 




Motion in a Circle: Simple Harmonic Motion 69 

9 

ac= 2(ib . sin -, which represents 2V sin.-, and the time 

2 2 

Q 

taken between the positions P and Q is - seconds (Art. 33). 
Therefore the average change of velocity per second is 

. 6 

Sln 2 

2 v sin - -f- or <av . a 

2 0) U 

2 

which is the average acceleration. Now, suppose that Q is 
taken indefinitely close to P that is, that the angle is in- 


sin - 

2 

definitely reduced ; then the ratio & has a limiting value 

2 

unity, and the average change of velocity per second, or 
average acceleration during an indefinitely short interval is 

v* 
w, or w'V or -, since v = <ar. This average acceleration 

during an indefinitely reduced interval is what we have defined 
(Art. 9) as actual acceleration, so that the acceleration at P 

i? 
is wV or feet per second per second. And as the angle 

is diminished indefinitely and Q thereby approaches P, the 
vector al>, remaining of the same length, approaches cb (a and c 
being always equidistant from />), and the angle bai increases 
and approaches a right angle as 6 approaches zero. Ultimately 
the acceleration (o>V) is perpendicular to PT, the tangent at 
P, i.e. it is towards O. 

63. Centripetal and Centrifugal Force. In the 
previous article we have seen that if a small body is describing 
a circle of radius r feet about a centre O with angular velocity 
w radians per second, it must have an acceleration oV towards 
O; hence the force acting upon it must be directed towards 
the centre O and of magnitude equal to its (mass) X w'V or 

W 
- orr Ibs., where W is its weight in pounds This force causing 



7O Mechanics for Engineers 

the circular motion of the body is sometimes called the centri- 
petal force. There is (Art. 51), by the third law of motion, 
a reaction of equal magnitude upon the medium which exerts 
this centripetal force, and this reaction is called the centrifugal 
force. It is directed away from the centre O, and is exerted 

W 
upon the matter which impresses the equal force o>V upon 

0> 

the revolving body ; it is not to be reckoned as a force acting 
upon the body describing a circular path. 

A concrete example will make this clear. If a stone of 
weight W Ibs. attached to one end of a string r feet long 
describes a horizontal circle with constant angular velocity u> 
radians per second, and is supported in a vertical direction by 
a smooth table, so that the string remains horizontal, the force 

W 

which the string exerts upon the stone is o>V towards the 

d 

centre of the circle. The stone, on the other hand, exerts on 

W 
the string an outward pull urr away from the centre. In 

A 

other cases of circular motion the inward centripetal force 
may be supplied by a thrust instead of a tension ; e.g. in the 
case of a railway carriage going round a curved line, the centri- 
petal thrust is supplied by the rail, and the centrifugal force 
is exerted outward on the rail by the train. 

64. Motion on a Curved " Banked " Track. Suppose 
a body, P (Fig. 44), is moving with uniform velocity, v, round a 



CL 




FIG. 44 . 



smooth circular track of radius OP equal to r feet. At what 
angle to the horizontal plane shall the track be inclined or 



Motion in a Circle : Simple Harmonic Motion 7 [ 

"banked" in order that the body shall keep in its circular 
path? 

There are two forces acting on the body (r) its own weight, 
W ; (2) the reaction R of the track which is perpendicular to 

W v 2 
the smooth track. These two have a horizontal resultant 

g r 
towards the centre O of the horizontal circle in which the 

body moves. If we draw a vector, ab (Fig. 44), vertically, to 
represent W, then R is inclined at an angle a to it, where a 
is the angle of banking of the track. If a vector, be, be drawn 
from b inclined at an angle a to ab, to meet ac, the perpendicular 
to ab from a, then be represents R, and ac or (ab + be) represents 

W z; 2 

the resultant of W and R, viz. , and 

g r ' 

ac W V* V* 

tan a = -7 = -f- W = 

ab g r gr 

which gives the angle a required. 

65. Railway Curves. If the lines of a railway curve 
be laid at the same level, the centripetal thrust of the rails 
on the wheels of trains would act on the flanges of the wheels, 
and the centrifugal thrust of the wheel on the track would tend 
to push it sideways out of its place. In order to have the action 
and reaction normal to the track the outer rail is raised, and the 
track thereby inclined to the horizontal. The amount of this 
" superelevation " suitable to a given speed is easily calculated. 

Let G be the gauge in inches, say, v the velocity in feet 
per second, and r the 
radius of the curve in 
feet. Let AB (Fig. 45) 
represent G; then AC 
represents the height 
in inches (exaggerated) 
which B stands above 
A, and ABC is the angle 

of banking, as in Art. 64. Then AC = AB sin a = AB tan 
a nearly, since a. is always very small ; hence, by Art. 64, AC 

v 2 - 
represents G tan a, or G ff . inches. 




Mechanics for Engineers 




W 



FIG. 46. 



66. Conical Pendulum. This name is applied to a 
combination consisting of a small weight fastened to one end 

of a string, the other end of 
which is attached to a fixed 
point, when the weight keeping 
the string taut, describes a 
horizontal circle about a centre 
vertically under the fixed point. 
Fig. 46 represents a conical 
pendulum, where a particle, P, 
attached by a thread to a fixed 
point, O, describes the hori- 
zontal circle PQR with con- 
stant angular velocity about the centre N vertically under O. 
Let T = tension of the string OP in Ibs. ; 

w = angular velocity of P about N in radians per second ; 
W = weight of particle P in Ibs. ; 
r = radius NP of circle PQR in feet ; 
/ = length of string OP in feet ; . 
a = angle which OP makes with ON, viz. PON ; 
// = height ON in feet ; 
g- = acceleration of gravity in feet per second per 

second. 

At the position shown in Fig. 46 P is acted upon by two 
forces (i) its own weight, W; (2) the tension T of the string 
OP. These have a resultant in the line PN (towards N), 
the vector diagram being set off as in Art. 64, ab vertical, 
representing the weight W, of P, and be the tension T. Then 

W 

the vector ac = ab -f bc> and represents the resultant force - 

X wV along PN ; hence 



we- ., , TT 

tan a. = -, = w V W = 
ab g 



g 



Also ON or h = NP 4- tan a. = r 4- 



= ^ feet 



hence the height h of the conical pendulum is dependent only 
on the angular velocity about N, being inversely proportional 
to the square of that quantity. 



Motion in a Circle: Simple Harmonic Motion 73 



Since h or / cos a = -, a> 2 = , and w = 

t.\* ft 




FIG. 47. 



Also the time of one complete revolution of the pendulum is 

angle in a circle _ 27r _ /^ 

angular velocity ~~ w '\/ ^ 

the period of revolution being proportional to the square root 
of the height of the pendulum, and the number of revolutions 
per minute being therefore inversely proportional to the square 
root of the height. This principle is made use of in steam- 
engine governors, where a change 
in speed, altering the height of a 
modified conical pendulum, is 
made to regulate the steam 
supply. 

67. Motion in a Vertical 
Circle. Suppose a particle or 
small body to move, say, contra- 
clockwise in a vertical circle with 
centre O (Fig. 47). It may be 
kept in the circular path by a 
string attached to O, or by an inward pressure of a circular 
track. Taking the latter instance 

Let R = the normal inward pressure of the track ; 
W= the weight of the rotating body in pounds; 
v = its velocity in feet per second in any position P 
such that OP makes an angle to the vertical 
OA, A being the lowest point on the circum- 
ference ; 

?; A = the velocity at A ; 
r = the radius of the circle in feet. 

W 
Then the kinetic energy at A is \ z ; A 2 

d 

At P the potential energy is W X AN, and the kinetic energy 

W 
is r, - v*, and since there is no work done or lost between A 

*g 

and P, the total mechanical energy at P is equal to that at A 
(Art. 61). Therefore 



74 Mechanics for Engineers 

W W 

I .zr + W.AN = 1-V 
2 <sr V 

hence z> 2 4- 2^. AN = v^ . . . . (i) 

Neglecting gravity, the motion in a circle would be uniform, and 

W z' 2 
would cause a reaction from the track (Art. 63). And 

in addition the weight has a component W cos 6. in the 
direction OP, which increases the inward reaction of the track 
by that amount ; hence the total normal pressure 

W ?; 2 

R-.- 4-WCOS0 . . . . (2) 

g r 

The value of R at any given point can be found by sub- 
stituting for v from equation (i) provided V A is known. The 
least value of R will be at B, the highest point of the circle, 
where gravity diminishes it most. If V A is not sufficient to 
make R greater than zero for position B, the particle will 
not describe a complete circle. Examining such a case, the 
condition, in order that a complete revolution may be made 
without change in the sign of R, is 

R B > o 

W 7, 2 

i.e. B + W cos 180 > o . 

g r 

or since cos 180 = i 



g r 

or z/ B 2 > gr 

and since v s 2 = z> A 2 2g . AB = v? 4gr, substituting for t' B 2 , 
the condition is 

w . 2 A (ri- ^> (fr 



i.e. the velocity at A must be greater than that due to falling 
through a height fr, for which the velocity would be J $gr 
(Art. 28). For example, in a centrifugal railway (" looping the 
loop") the necessary velocity on entering the track at the 



Motion in a Circle: Simple Harmonic Motion 75 

lowest point, making no allowance for frictional resistances, 
may be obtained by running down an incline of height greater 
than two and a half times the radius of the circular track. 

If the centripetal force is capable of changing sign, as in the 
case of the pressure of a tubular track, or the force in a light 
stiff radius rod supporting the revolving weight, the condition 
that the body shall make complete revolutions is that Z' B shall 
be greater than zero, and since ' B 2 = Vj? 4gr, the condition is 



i.e. the velocity at A shall be greater than that due to falling 
through a height equal to the diameter of the circle. Similarly, 
the position at which the body will cease to describe a circular 
track (in a forward direction) if # A is too small for a complete 
circuit, when the force can change sign and when it can not, 
may be investigated by applying equations (i) and (2), which 
will also give the value of R for any position of the body. 

The pendulum bob, suspended by a thread, is of course 
limited to oscillation of less than a semicircle or to complete 
circles. 

Example i. At what speed will a locomotive, going round a 
curve of looo-feet radius, exert a horizontal thrust on the outside 
rail equal to T of its own weight ? 

Let W = the weight of loco, 

v = its velocity in feet per second. 

W Z/ 2 
Centrifugal thrust = . - - - = T J W 

jp" I OOO 

IOOO X 
.*. #2 _ - - 6. = 322 

100 

v = 1 7 '95 feet per second, equivalent 
to 12*22 miles per hour 

Example 2. A uniform disc rotates 250 times per minute 
about an axis through its centre and perpendicular to its plane. 
It has attached to it two weights, one of 5 Ibs. and the other of 7 
Ibs., at an angular distance of 90 apart, the first being I foot 
and the second 2 feet from the axis. Find the magnitude and 
direction of the resultant centrifugal force on the axis. Find, also, 



76 Mechanics for Engineers 

where a weight of 12 Ibs. must be placed on the disc to make the 
resultant centrifugal force zero. 

~, i 250 X 2ir 2?ir 

The angular velocity is , = -- radians per second 



The centrifugal pull Fj (Fig. 48) is 



32-2 \ 3 
and the centrifugal pull F 2 is 
7 



-5 Ibs. 






s x IT ' 

hence the resultant R of Fj and F 2 at right 
axis' i' " ou>3 - angles is 

FIG. 4 8. R = \fio6 z + 297 2 = 315 Ibs. 

at an angle tan" 1 = tan" 1 0*357 = ig'6 to the direction of F 2 

(Arts. 24 and 44) 

To neutralize this, a force of 31 5 Ibs. will be required in the opposite 
direction. 

Let x radius in feet of the i2-lbs. weight placed at 180 ig'6 
or i6o'4 contra-clockwise from F 2 . 

. - 



hence x 1-23 feet 

Example 3. Find in inches the change in height of a conical 
pendulum making 80 revolutions per minute when the speed 
increases two per cent. 

The increase in speed is T gg x 80 = r6 revolutions per minute 
to 8 1 '6 revolutions per minute. 

The height is 4r (Art. 66), where is the angular velocity in 

CO 

radians per second. 

At 80 revolutions per minute the angular velocity is 

2ir X 80 Sir ,. 

= radians per second 

60 3 

or ^2*2 X Q 

hence the height h^ = ^ = zr~a 

u) OA7T 

= 0-4585 foot 



Motion in a Circle: Simple Harmonic Motion 77 

At 8 1 -6 revolutions per minute the angular velocity is 
2T x 8r6 



ir 
r radians per second 



= 0-441 1 foot 



i = 0-6174 foot or 0*207 inch 



W 



and the height is ^ 81 . 6 = 

hence the decrease in height is 
0-4585 - 0-4411 

Example 4. A piece of lead is fastened to the end of a string 
2 feet long, the other end of which 
is attached to a fixed point. With 
what velocity must the lead be pro- 
jected in order to describe a hori- 
zontal circle of 2 feet diameter ? 

Let OP, Fig. 49, represent the 
string; then the horizontal line PN 
is to be i foot radius. 

In the vector triangle abc, ab N 1 
represents W, the weight of lead, FlG - 49- 

be the tension T of the string OP, and ac their resultant ; then 

NP _ ac _ W it _ v n ~ 

ON""^-^"^" 4 " J1TF 

where v = velocity in feet per second ; 





hence ^= 



= = 18-57 



and v = 4-309 feet per second 

Exercise 5. A stone weighing } Ib. is whirling in a vertical 
circle at the extremity of a string 3 feet long. Find the velocity of 
the stone and tension of the string (i) at the highest position, (2) 
at lowest, (3) midway between, if the velocity is the least possible 
for a complete circle to be described. 

If the velocity is the least possible, the string will just be slack 
when the stone is at the highest point of the circle. 

Let v be the velocity at the highest point, where the weight 
just supplies the centripetal force ; 



V* = 3 x 32*2 = 96-6 
and z'o = 9*83 feet per second. 

(2) At the lowest point let the velocity be v^ feet per second. 



78 Mechanics for Engineers 

Since there is no loss of mechanical energy, the gain of kinetic 
energy is \ x 6 foot-lbs., hence 

ill ill i 

- x - x v,* = - . - . -v? + - . 6 
24^ 2 4 <? - r 4 

and z>, 2 = r/ 2 + 2 .g . 6 

= 96-6 + 386-4 = 483 (or 5 x g x 3) 
v l = V483 = 22 feet per second (nearly) 

and the tension is \ , 

i i i 483 )>=- + - - = i'5lbs., or six times the weight 

~ -4- . 4 I2o'o 

4 4 32-2 3 J of the stone 

(3) When the string is horizontal, if v' = velocity in feet per 
second 

similarly i. - V 2 = - . - . -v 4-^-3 
y '2 4 g 2-4'^ ^4 J 

t-' 2 = r/ 2 + ?.g x 3 
= 96-6 + I93'2 

v' = .v/289'8 = 17 feet per second 
and the tension is 



i i 289/8 >= 075 lb., or three times the weight of the 
4 '32-2 X 3 J stone 

EXAMPLES VIII. 

1. How many circuits per minute must a stone weighing 4 ozs. 
make when whirled about in a horizontal circle at the extremity of a string 
5 feet long, in order to cause a tension of 2 Ibs. in the string ? 

2. At what speed will a locomotive produce a side thrust equal to g'g of 
its own weight on the outer rail of a level curved railway line, the radius of 
the curve being 750 feet ? 

3. What is the least radius of curve round which a truck may run on 
level lines at 20 miles per hour without producing a side thrust of more 
than T ig of its own weight ? 

4. How much must the outer rail of a line of 4 feet 8.J inches gauge be 
elevated on a curve of 800 feet radius in order that a train may exert a 
thrust normal to the track when travelling at 30 miles per hour ? 

5. The outer rail of a pair, of 4 feet 85 inches gauge, is elevated 25 inches, 
and a train running at 45 miles per hour has no thrust on the flanges of 
either set of wheels. What is the radius of the curve ? 

6. At what speed can a train run round a curve of 1000 feet radius 
without having any thrust on the wheel flanges when the outer rail is laid 
i '5 inches above the inner one, and the gauge is 4 feet 8^ inches ? 

7. To what angle should a circular cycle-track of 15 laps to the mile be 



Motion in a Circle: Simple Harmonic Motion 79 

banked for riding upon at a speed of 30 miles per hour, making no allow- 
ance for support from friction ? 

8. A string 3 feet long, fixed at one end, has attached to its other 
end a stone which describes a horizontal circle, making 40 circuits per 
minute. What is the inclination of the string to the vertical ? What is its 
tension ? 

9. What percentage change of angular speed in a conical pendulum 
will correspond to the decrease in height of 3 per cent. ? 

10. The revolving ball of a conical pendulum weighs 5 Ibs., and the 
height of the pendulum is 8 inches. What is its speed ? If the ball is 
acted upon by a vertical downward force of i lb., what is then its speed 
when its height is 8 inches ? Also what would be its speed in the case of 
a vertical upward force of I lb. acting on the ball ? 

1 1 . \Vhat will be the inclination to the vertical of a string carrying a 
weight suspended from the roof of a railway carriage of a train going 
round a curve of 1000 feet radius at 40 miles per hour ? 

12. A body weighing \ lb., attached to a string, is moving in a vertical 
circle of 6 feet diameter. If its velocity, when passing through the lowest 
point, is 40 feet per second, find its velocity and the tension of the string 
when it is 2 feet and when it is 5 feet above the lowest point. 

68. Simple Harmonic Motion. This is the simplest 
type of reciprocating motion. If a point Q (Fig. 50) describes 
a circle AQB with constant angular velocity, and P be the 
rectangular projection of Q on a fixed diameter AB of the 
circle, then the oscillation to and fro of P along AB is defined 
as Simple Harmonic Motion. 

Let the length OA of the radius be a feet, called the 
amplitude of oscillation. 

Let w be the angular velocity of Q in radians per second. 

Let 6 be the angle AOQ in radians, denoting any position 
ofQ. 

Suppose the motion of Q to be, say, contra-clockwise. 

A complete vibration or oscillation of P is reckoned in this 
country as the path described by P whilst Q describes a 
complete circle. 

Let T = the period in seconds of one complete vibration ; 
then, since this is the same as that for one complete circuit 
made by Q 

radians in one circle 2ir 

*~r* / - \ 

~~ radians described per second <> 



80 



Mechanics for Engineers 



Let x - distance OP of 
towards A, the 


/ v 






7" 




o , *~ 




1 r 


-: ' 


*L 




Trs 


1 


s, 


_ r "j ! 


S "" 


T^ 


^^ 


*~ \ 








7 S 


i ^ 


"^ ^ 


flL 








S / 




^^ ,' 




^ ~' 


^ 


J jf 


s. 


jt "^ . 


JJ ~ 


^ ' \ 


. 


~. 5 






j 5 "I ' 


^k 


d s *-* : 


^i " . / 


$ 


y 


'S-K 


c ' 


< * ^ 


j^* 




/ r ^ i 




\v / > 


7 


s \ 




^ ^ 


^ ~? 


cr "" 


\ s 


_ / 


. s 




^ . s 


j ' 


5s. -i 


lt- 


y % s 


e" ' 


~ >^ 


o- 8 - 


7 S 


u. ' - 


^ ^ 


L 


^ 


^ . 


^ ^ 


s ^_ 


"~ ^2T^iCiJK2L2Lj 
































c/ ^ - ^ j 


"~ ^ 




! . 


/ > 


"s 


~7 \ 


^ 


~7 




' \ 


. 


7 i 




L A 


J 






\ 


f\ 


^ 




. 


' 


r *** 


^ r 


jj ^ 




, - 




(/C 




^ 


_$ ^ 


,, 


\ j/ 


^ V. ~*~ 


- - -3r* 


^-~ . ! 


_ "^ 



of P from O in feet, reckoned positive 
then x = a cos ; 

and let v velocity of P in 
feet per second in position 6. 

Draw OS perpendicular 
to OQ to meet the circum- 
ference of the circle AQSB 
in S, and draw SM perpen- 
dicular to AB to meet it in M. 

Then for the position or 
phase shown in the figure, 
the velocity of Q is wa (Art. 
33) in the direction perpen- 
dicular to OQ, i.e. parallel to 
OS. Resolving this velocity 
along the diameter AB, OSM 
being a vector triangle, the 
component velocity of Q 

OM 
parallel to AB is -~c X wa, 

vJo 

or <aa sin 6, or w . OM. This 
is then the velocity of P 
towards O, the mid-path. 



Since sin 6 = 



OM 
OS 



v = <i>a sin 6 



which gives the velocity of 
P in terms of the amplitude 
and position. 

Or, if OS represents geo- 
metrically the velocity of Q, 
then OM represents that ot 
P to the same scale. 



Motion in a Circle: Simple Harmonic Motion 81 

Acceleration of P. The acceleration of Q is u?a along 
QO towards O (Art. 62). Resolving this acceleration, the 

PO 

component in direction AB is w 2 x pr or o> 2 0.cos % 0, or 

w 2 . x, towards O ; and it should be noted that at unit distance 
from O, when x = i foot, the acceleration of P is w 2 feet per 
second per second. 

The law of acceleration of a body having simple harmonic 
motion, then, is, that the acceleration is towards the mid-path 
and proportional to its distance from that point. When the 
body is at its mid-path, its acceleration is zero ; hence there is 
no force acting upon it, and this position is one of equilibrium 
if the body has not any store of kinetic energy. Conversely, 
if a body has an acceleration proportional to its distance from 
a fixed point, O, it will have a simple harmonic motion. If 
the acceleration at unit distance from O is /x, feet per second 
per second (corresponding to w 2 in the case just considered), 
by describing a circle with centre about its path as diameter, 
we can easily show that the body has simple harmonic motion, 
and by taking w = vV> !" corresponding to w 2 in the above 
case, we can state its velocity and acceleration at a distance 
x from its centre of motion O, and its period of vibration, viz, 
velocity v at x feet from O is vV V a * ~ x *> or V P-( a * ~ * 2 )- 
Acceleration at x feet from centre O is p . x, and the time 

27T 

of a complete vibration is == 
VJK- 

Alternating: Vectors. We have seen that, the displace- 
ment of P being OP, the acceleration is proportional also to 
OP, and the velocity to OM ; so that OP and OM are vectors 
representing in magnitude and direction the displacement and 
velocity of P. Such vectors, having a fixed end, O, and of 
length varying according to the position of a rotating vector, 
OQ or OS, are called " alternating vectors." It may be noted 
that the rate of change of an alternating vector, OP, of ampli- 
ade a is represented by another alternating vector, OM, of the 
same period, which is the projection of a uniformly rotating 
vector of length OS = w . OQ or <a (to a different scale), and 
one right angle in advance of the rotating vector OQ, of which 

G 



82 Mechanics for Engineers 

OP is the projection. A little consideration will show that the 
rate of change of the alternating vector OM follows the same 
law (rate of change of velocity being acceleration), viz. it is 
represented by a third alternating vector, ON, of the same 
period, which is the projection of a uniformly rotating vector 
of length OQ' = o> . OS or u?a (to a different scale), and one 
right angle in advance of the rotating vector OS, of which OM 
is the projection. 

The curves of displacement, velocity, and acceleration of 
P on a base of angles are shown to the right hand of Fig. 50. 
The base representing angles must also represent time, since 
the rotating vectors have uniform angular velocity <o. The 

fi ft 

time / = - seconds, since w = -. The properties of the curves 

0) / 

of spaces, velocities, and accelerations (Arts. 4, 14, and 16) 
are well illustrated by the curves in Fig. 50, which have been 
drawn to three scales of space, velocity, and acceleration by 
projecting points 90 ahead of Q, S, and Q' on the circle on 
the left. The acceleration of P, which is proportional to the 
displacement, may properly be considered to be of opposite 
sign to the displacement, since the acceleration is to the left 
from P to O when the displacement OP is to the right of O. 
The curves of displacement and acceleration are called " cosine 
curves," the ordinates being proportional to the cosines of angle 
POQ, or 6, or w/. Similarly, the curve of velocity is called a 
" sine curve." The relations between the three quantities may 
be expressed thus 

Displacement (x) : velocity (v) : acceleration 

= a cos wt : aw sin w/ : w 2 cos o>/ 

Curved Path. If the point P follows a curved path 
instead of the straight one AB, the curved path having the 
same length as the straight one, and if the acceleration of 
the point when distant x feet from its mid-path is tangential 
to the path and of the same magnitude as that of the point 
following the straight path AB when distant x feet from mid- 
path, then the velocity is of the same magnitude in each case. 
This is evident, for the points attain the same speeds in the 



Motion in a Circle: Simple Harmonic Motion 83 

same intervals of time, being, under the same acceleration, 
always directed in the line of motion in each case. Hence 

the periodic times will be the same in each case, viz. 

vV 

where ju, is the acceleration in feet per second per second 
along the curve or the straight line, as the case may be. 

69. There are numerous instances in which bodies have 
simple harmonic motion or an approximation to it, for in 
perfectly elastic bodies the straining force is proportional to 
the amount of displacement produced, and most substances 
are very nearly perfectly elastic over a limited range. 

A common case is that of a body hanging on a relatively 
light helical spring and vibrating vertically. The body is 
acted upon by an effective accelerating force proportional to 
its distance from its equilibrium position, and, since its mass 

f force \ 

does not change, it will have an acceleration I - - 1 also 

\ mass / 

proportional to its displacement from that point (Art. 40), and 
therefore it will vibrate with simple harmonic vibration. 
Let W = weight of vibrating body in pounds. 

e = force in pounds acting upon it at i foot from its 
equilibrium position, or per foot of displace- 
ment, the total displacement being perhaps 
less than i foot. This is sometimes called 
the stiffness of the spring. 

Then e . x = force in Ibs. x feet from the equilibrium position 

and if p = acceleration in feet per second per second i foot 
from the equilibrium position or per foot of displacement 

accelerating force _ . W _ eg 
^ mass g ~ W 

hence the period of vibration is - ^ or 2ir A / ; (Art. 68) 

Vj"- * 

The maximum force, which occurs when the extremities 
of the path are reached, is e.a, where a is the amplitude of 



84 Mechanics for Engineers 

the vibration or distance from equilibrium position to either 
extremity of path, in feet. 

The crank-pin of a steam engine describes a circle ABC 
(Fig. 51), of which the length of crank OC is the radius, with 




FIG. 51. 



fairly constant angular velocity. The piston P and other 
reciprocating parts are attached to the crank-pin by a con- 
necting-rod, DC, and usually move to and fro in a straight 
line, AP, with a diameter, AB, of the crank-pin circle. If the 
connecting-rod is very long compared to the crank-length, 
the motion is nearly the same as that of the projection N 
of the crank-pin on the diameter AB of the crank-pin circle, 
which is simple harmonic. If the connecting-rod is short, 
however, its greater obliquity modifies the piston-motion to 
a greater extent. 

70. Energy stored in Simple Harmonic Motion. 
If c = force in pounds at unit distance, acting on a body of 
weight W Ibs. having simple harmonic motion, the force at a 
distance x is ex, since it is proportional to the displacement. 
Therefore the work done in displacing the body from its equili- 
brium through x feet is \eo? (Art. 54 and Fig. 35). This 
energy, which is stored in some form other than kinetic energy 
when the body is displaced from its equilibrium position, 
reaches a maximum \ecf- when the extreme displacement a 
(the amplitude) has taken place, and the effective accelerating 
force acting on the body is ea. In the mid-position of the 
body (x = o), when its velocity is greatest and the force acting 
on it is nil, the energy is wholly kinetic, and in other inter- 
mediate positions the energy is partly kinetic and partly 
otherwise, the total being constant if there are no resistances. 



B 




Distance. A 



FIG. 52. 



Motion in a Circle: Simple Harmonic Motion 85 

Fig. 52 shows a diagram of work stored for various dis- 
placements of a body having simple harmonic motion. The 
amplitude OA = a, and 
therefore the force at A 
is ae, which is represented 
by AD, and the work done 
in moving from O to A is 
represented by the area 
AOD(Art.54andFig. 3 5). 
At P, distant x feet from 
O, the work done in motion 
from O is ^ex 1 , represented 
by the area OHP, and the 
kinetic energy at P is 
therefore represented by 
the area DAPH. 

71. Simple Pendulum. This name refers strictly to a 
particle of indefinitely small dimensions and yet having weight, 
suspended by a perfectly flexible weightless thread from a fixed 
point, about which, as a centre, it swings freely in a circular 
arc. In practice, a small piece of 
heavy metal, usually called a pendulum 
bob, suspended by a moderately long 
thin fibre, behaves very nearly indeed 
like the ideal pendulum defined above, 
the resistances, such as that of the 
atmosphere, being small. 

Let 0, Fig. 53, be the point of 
suspension of the particle P of a 
simple pendulum. 

Let OP, the length of thread, be 
/ feet. 

Let 6 = angle AOP in radians which OP makes with the 
vertical (OA) through O in any position P of the particle. 

Draw PT perpendicular to OP, i.e. tangent to the arc of 
motion to meet the vertical through O in T. 

The tension of the thread has no component along the 
direction of motion (PT) at P. The acceleration along PT is 




FIG. 53- 



86 Mechanics for Engineers 

then g sin 0, since FT is inclined 6 to the horizontal (Art. 28). 
If 6 is very small, sin 6 may be taken equal to 6 in radians. 
(If does not exceed 5, the greatest error in this approxi- 
mation is less than i part in 800.) Hence the acceleration 

along PT is g6 approximately. And 6 = T. ^^ ; therefore 

acceleration along PT =* j - , and the acceleration is 

proportional to the distance AP, along the arc, of P from A, 

g 

being -, per foot of arc. Hence the time of a complete oscilla- 
tion in seconds is 

^ Vf =2 x/^ (Art 68) 

and the velocity at any point may be found, as in Art. 68, for 
any position of the swinging particle. 

In an actual pendulum the pendulum bob has finite dimen- 
sions, and the length / will generally be somewhat greater than 
that of the fibre by which it is suspended. The ideal simple 
pendulum having the same period of swing as an actual pen- 
dulum of any form is called its simple equivalent pendiilum. 

For this ideal pendulum the relation /= 2tr\ /-holds, and 

V g 
Pg 
therefore / = -~g, from which its length in feet may be 

calculated for a given time, /, of vibration. 

The value of the acceleration of gravity, g, varies at different 
parts of the earth's surface, and the pendulum offers a direct 
means of measuring the value of this quantity g, viz. by 
accurate timing of the period of swing of a pendulum of known 
length. The length of an actual pendulum, i.e. of its simple 
equivalent pendulum, can be calculated from its dimensions. 

Example i. A weight rests freely on a scale-pan of a spring 
balance, which is given a vertical simple harmonic vibration of 
period o'5 second. What is the greatest amplitude the vibration 
may have in order that the weight may not leave the pan ? What 
is then the pressure of the weight on the pan in its lowest position ? 

Let a = greatest amplitude in feet. 



Motion in a Circle: Simple Harmonic Motion 87 

The greatest downward force on the body is its own weight, 
and therefore its greatest downward acceleration is g, occurring 
when the weight is in its highest position and the spring is about 
to return. Hence, if the scale-pan and weight do not separate, 
the downward acceleration of the pan must not exceed g, and 

therefore the acceleration must not exceed per foot of dis- 

a f 

placement. 

/2ir\ 2 

The acceleration per foot of displacement is f ) ; 
therefore 



a 
or a > 3p feet 

l6ir 2 

i.e. a ;j> o'2O4 feet or 2*448 inches 

If the balance has this amplitude of vibration, the pressure 
between the pan and weight at the lowest position will be equal to 
twice the weight, since there is an acceleration upwards which 
must be caused by an effective force equal to the weight acting 
upwards, or a gross pressure of twice the weight from which the 
downward gravitational force has to be subtracted. 

Example 2. Part of a machine has a reciprocating motion, 
which is simple harmonic in character, making 200 complete oscilla- 
tions in a minute ; it weighs 10 Ibs. Find (i) the accelerating force 
upon it in pounds and its velocity in feet per second, when it is 
3 inches from mid-stroke; (2) the maximum accelerating force; 
and (3) the maximum velocity if its total stroke is 9 inches, i.e. if 
its amplitude of vibration is 4^ inches. 

Time of i oscillation = = o'3 second 

200 

therefore the acceleration per foot) /2 w\ 2 4001^ 

> = I - ) = - feet per second 
distance from mid-stroke J \o'3/ 9 

per second 

and the accelerating force 0-25 foot from mid-stroke on 10 Ibs. is 

10 4OOT 2 

x 0-25 x * - = 34'oS Ibs. 
3* y 

and the maximum accelerating force 4^ inches from mid-stroke is 
i'5 times as much as at 3 inches, or 34 - o8 x 1*5 = 5i'i2 Ibs. 



88 Mechanics for Engineers 

The maximum velocity in feet per second occurring at mid-stroke 
= amplitude in feet x >J Acceleration per foot of displacement 

(Art. 68) 

= amplitude in feet x - 

period 

1 21T IjTT or 

= =j x = = 7*85 feet per second 
8 o'3 2 

Velocity at 3 inches) = V^LEI*' (Art . 68 ) 

from mid-stroke \ 4' 5 

= 7-85 x 'V " 25 = 5-85 feet per second 

Example 3. The crank of an engine makes 150 revolutions 
per minute, and is 1-3 feet long. It is driven by a piston and a very 
long connecting rod (Fig. 51), so that the motion of the piston may 
be taken as simple harmonic. Find the 
piston velocity and the force necessary 
to accelerate the piston and recipro- 
cating parts, weighing altogether 300 
Ibs., (i) when the crank has turned 
through 45 from its position (OB) in 
line with and nearest to the piston 
path ; (2) when the piston has moved 
forward o'65 foot from the end of its 
FIG. 54 . stroke. 

Let ABC (Fig. 54) be the circular 

path 1*3 feet radius of the crank-pin, CN the perpendicular from 
a point C on the diameter AB. 

The angular velocity of crank OC is 7 = 5*- radians per second 

(i) The motion of the piston being taken as that of N, the 
acceleration of piston when the crank-pin is at C is 

(57r)' 2 x i'3 x cos 45 (wVcos 6, Art. 68) 
and the accelerating force is 

300 / y, i 

X (Sir) 2 X I '3 X ->- = 2IIolbs. 

32-2 ,/a 

The velocity is 

5ir x i '3 x sin 45 = H'43 feet per second 




Motion in a Circle: Simple Harmonic Motion 89 

(2) When BN = 0*65 foot, ON = OB - BN = 1-3 - 0-65 = 0-65 

ON 
foot, and CON = cos" 1 _- = cos" 1 \ = 60. The accelerating 

force is then 

\ x (SO 2 x 1-3 x \ = 1493 Ibs. 

jz z 

and the velocity is 

5 IT x i '3 x sin 60 = 17*67 feet per second 

Example 4. A light helical spring is found to deflect 0*4 inch 
when an axial load of 4 Ibs. is hung on it. How many vibrations 
per minute will this spring make when carrying a weight of 
15 Ibs.? 

The force per foot of deflection is 4 * = 120 Ibs. 

1 2 



is 2w / _]> = o'3Q2 second 

V 32-2 x 120 



hence the time of vibration 

32'2 X 1 2O 

and the number of vibrations per minute is Tr;^ = I53' 2 

Example 5. Find the length of a clock pendulum which will 
make three beats per second. If the clock loses i second per 
hour, what change is required in the length of pendulum ? 

Let / = length of pendulum in feet. 
Time of vibration = ^ second 

' f Qxf* 32:2 fcet = inches 

4iH 36*-* 

The clock loses i second in 3600 seconds, i.e. it makes 3599 x 3 
beats instead of 3600 x 3. Since / oc / 2 oc y where n = number 

of beats per hour, therefore 

correct length _ 3599** _ ( . 

1-09 inches ~ 3600" ~ V 1 ** } 

i YgW approximately 

therefore shortening required = --^ inches = 0*000606 inch 



EXAMPLES IX. 

I. A point has a simple harmonic motion of amplitude 6 inches and 
period 1-5 seconds. Find its velocities and accelerations OT second, O'2 
second, and 0*5 second after it has left one extremity of its path. 



go Mechanics for Engineers 

2. A weight of 10 Ibs. hangs on a spring, which stretches O'lS inch 
per pound of load. It is set in vibration, and its greatest acceleration 
whilst in motion is i6 - i feet per second per second. What is the ampli- 
tude of vibration ? 

3. A point, A, in a machine describes a vertical circle of 3 feet diameter, 
making 90 rotations per minute. A portion of the machine weighing 400 
Ibs. moves in a horizontal straight line, and is always a fixed distance 
horizontally from A, so that it has a stroke of 3 feet. Find the accele- 
rating forces on this portion, ( I ) at the end of its stroke ; (2) 9 inches from 
the end ; and (3) 0^05 second after it has left the end of its stroke. 

4. A helical spring deflects \ of an inch per pound of load. How many 
vibrations per minute will it make if set in oscillation when carrying a load 
of 12 Ibs. ? 

5. A weight of 20 Ibs. has a simple harmonic vibration, the period of 
which is 2 seconds and the amplitude 1*5 feet. Draw diagrams to stated 
scales showing (i) the net force acting on the weight at all points in its 
path ; (2) the displacement at all times during the period ; (3) the velocity 
at all times during the period ; (4) the force acting at all times during the 
period. 

6. A light stiff beam deflects I'I4S inches under a load of I ton at 
the middle of the span. Find the period of vibration of the beam when so 
loaded. 

7. A point moves with simple harmonic motion ; when 0*75 foot from 
mid -path, its velocity is 1 1 feet per second ; and when 2 feet from the 
centre of its path, its velocity is 3 feet per second. Find its period and its 
greatest acceleration. 

8. How many complete oscillations per minute will be made by a 
pendulum 3 feet long? g= 32 '2. 

9. A pendulum makes 3000 beats per hour at the equator, and 3011 per 
hour near the pole. Compare the value of g at the two places. 



CHAPTER V 



STA TICS CONCURRENT FORCES FRICTION 

72. THE particular case of a body under the action of several 
forces having a resultant zero, so that the body remains at rest, 
is of very common occurrence, and is of sufficient importance 
to merit special consideration. The branch of mechanics which 
deals with bodies at rest is called Statics. 

We shall first consider the statics of a particle, i.e. a body 
having weight, yet of indefinitely small dimensions. Many of 
the conclusions reached will be applicable to small bodies in 
which all the forces acting may be taken without serious error 
as acting at the same point, or, in other words, being con- 
current forces. 

73. Resolution and Composition of Forces in One 
Plane. It will be necessary to recall some of the conclusions 
of Art. 44, viz. that any number of concurrent forces can be 
replaced by their geometric sum acting at the intersection of 
the lines of action of the forces, or by components in two 
standard directions, which are for convenience almost always 
taken at right angles to one another. 

Triangle and Polygon of Forces. If several forces, say four, 
as in Fig. 55, act on a particle, and ab, be, cd, de be drawn in 
succession to represent the forces of 7, 8, 6, and 10 Ibs. respec- 
tively, then ae, their geometric sum (Art. 44), represents a force 
which will produce exactly the same effect as the four forces, 
i*. ae represents the resultant of the four forces. If the final 
point e of the polygon abcde coincides with the point 0, then 
the resultant ae is nil, and the four forces are in equilibrium. 
This proposition is called the Polygon of Forces, and may be 



92 Mechanics for Engineers 

stated as follows : If several forces acting on a particle be 
represented in magnitude and direction by the sides of a closed 
polygon taken in order, they are in equilibrium. By a closed 
polygon is meant one the last side of which ends at the point 




FIG. 55. 



from which the first side started. The intersection of one side 
of the polygon with other sides is immaterial. 

The polygon of forces may be proved experimentally by 
means of a few pieces of string and weights suspended over 
almost frictionless pulleys, or by a number of spring balances 
and cords. 

This proposition enables us to find one force out of several 
keeping a body in equilibrium if the remainder are known, viz. 
by drawing to scale an open polygon of vectors corresponding 
to the known forces, and then a line joining its extremities is 
the vector representing in one direction the resultant of the 
other forces or in the other direction the remaining force neces- 
sary to maintain equilibrium, sometimes called the cqvilibrant. 

For example, if forces Q, R, S, and T (Fig. 56) of given 
magnitudes, and one other force keep a particle P in equili- 
brium, we can find the remaining one as follows. Set out vectors 
ab, be, cd, and de in succession to represent Q, R, S, and T 
respectively ; then ae represents their resultant in magnitude 
and direction, and ea represents in magnitude and direction the 
remaining force which would keep the particle P in equilibrium, 
or the equilibrant. 



Statics Concurrent Forces Friction 93 

Similarly, if all the forces keeping a body in equilibrium 
except two are known, and the directions of these two are 
known, their magnitudes may be found by completing the 





----'e 



FIG. 56. 



open vector polygon by two intersecting sides in the given 
directions. 

In the particular case of three forces keeping a body in 
equilibrium, the polygon is a triangle, which is called the 
Triangle of Forces. Any triangle having its sides respectively 
parallel to three forces which keep a particle in equilibrium 
represents by its sides the respective forces, for a three-sided 
closed vector polygon (i.e. a triangle) with its sides parallel 
and proportional to the forces can always be drawn as directed 
for the polygon of forces, and any other triangle with its sides 
parallel to those of this vector triangle has its sides also pro- 
portional to them, since all triangles with sides respectively 
parallel are similar. The corresponding proposition as to any 
polygon with sides parallel to the respective forces is not true 
for any number of forces but three. 

74. Lami's Theorem. If three forces keep a particle 
in equilibrium, each is proportional to the sine of the angle 
between the other two. 

Let P, Q, and R (Fig. 57) be the three forces in equilibrium 
acting at O in the lines OP, OQ, and OR respectively. Draw 
any three non-concurrent lines parallel respectively to OP, OQ, 
and OR, forming a triangle abc such that ab is parallel to OP, be 
to OQ, and ca to OR. Then angle abc = 180 - POQ, angle 



94 



Mechanics for Engineers 



bca = 1 80 QOR, and angle cab = 180 ROP, and there- 
fore 

A A. 

sin abc = sin POQ 
sin b'ca = sin QOR 
sin cab = sin ROP 

In the last article, it was shown that any triangle, such as 





P Q R 




ab be ca 
, ab be 


ca 


* . .^ . A _ 

sin oca sin cab 

ab be 


sin abc 
ca 



FIG. 57- 



abc, having sides respectively parallel to OP, OQ, and OR, has 
its sides proportional respectively to P, Q, and R, or 



(I) 



" sin QOR sin ROP sin POQ 
and multiplying equation (i) by equation (2) 

P Q R 

sin QOR "~ sin ROP ~ sin POQ 

that is, each of the forces P, Q, and R is proportional to the 
sine of the angle between the other two. 

This result is sometimes of use in solving problems in 
which three forces are in equilibrium. 

75. Analytical Methods. Resultant or equilibrant forces 
of a system, being representable by vectors, may be found by 
the rules used for resultant velocities, i.e. (i) by drawing 



Statics Concurrent Forces Friction 



95 



vectors to scale ; (2) by the rules of trigonometry for the solu- 
tions of triangles ; (3) by resolution into components in two 
standard directions and subsequent compounding as in Art. 25. 
We now proceed to the second and third methods. 

To compound two forces P and Q inclined at an angle 6 
to each other. 

Referring to the vector diagram abc of Fig. 58 (which need 



I (Q) 





not be drawn, and is used here for the purpose of illustration 
and explanation) by the rules of trigonometry for the solution 
of triangles 

(ac)* = (ab)* + (be)* - 2 ab.bccos abc 
= (ab)* + (be)* + 2 ab.be cos Q 

hence if ab and be represent P and Q respectively, and R is the 
value of their resultant 

R 2 = P 2 + Q 2 + 2PQ cos Q 

from which R may be found by extracting the square root, and 
its inclination to, say, the direction of Q may be found by 
considering the length of the perpendicular ce from c on ad 
produced 

Since ec = dc sin Q 
and de = dc cos & 

ec dc sin 6 P sin 6 

~ ae ~ ad -+ dc cos 6 ~ Q -f P cos 

which is the tangent of the angle between the line of action of 
ie resultant R and that of the force Q. 

When the resultant or equilibrant of more than two 
concurrent forces is to be found, the method of Art. 25 is 



g6 Mechanics for Engineers 

sometimes convenient. Suppose, say, three forces F 15 F 2) 
and F 3 make angles a, /3, and y respectively with some chosen 
fixed direction OX, say that of the line of action of Fj, so 
that a = o (Fig. 59). 





Resolve F 1} F 2 , and F 3 along OX and along OY perpen- 
dicular to OX. 

Let F x be the totalof the components along OX, 
and let F Y OY. 

Let R be the resultant force, and 6 its inclination to OX ; 
then 

F x = Fj + F 2 cos |8 + F 3 cos y 
F Y = o + F 2 sin /3 + F 3 sin y 

and compounding F x and F Y , two forces at right angles, R is 
proportional to the hypotenuse of a right-angled triangle, the 
other sides of which are' proportional to F x and F Y ; hence 

R2 = F x 2 + F Y 2 
and R = 



The direction of the resultant R is given by the relation 



If the forces of the system are in equilibrium, that is, if 
the resultant is nil 

R 2 = o 

or F x 2 + F Y 2 = o 
This is only possible if both F x = o and F Y = o. 






Statics Concurrent Forces Friction 



97 



The condition of equilibrium, then, is, that the components 
in each of two directions at right angles shall be zero. This 
corresponds to the former statement, that if the forces are in 
equilibrium, the vector polygon of forces shall be closed, as 
will be seen by projecting on any two fixed directions at right 
angles, the sides of the closed polygon, taking account of the 
signs of the projections. The converse statement is true, for 
if F x = o and F Y = o, then R = o ; therefore, if the com- 
ponents in each of two standard directions are zero, then the 
forces form a system in equilibrium, corresponding to the 
statement that if the vector polygon is a closed figure, 
the forces represented by its sides are in equilibrium. 

Example i. A pole rests vertically with its base on the ground, 
and is held in position by five ropes, all in the same horizontal 
plane and drawn tight. From the pole the first rope runs due 
north, the second 75 west of north, the third 15 south of west, and 
the fourth 30 east of south. The tensions of these four are 25 Ibs., 




FIG. 60. 

15 Ibs., 20 Ibs., and 30 Ibs. respectively. Find the direction of the 
fifth rope and its tension. 

The'directions of the rope have been set out in Fig. 60, which 

H 



98 Mechanics for Engineers 

represents a plan of the arrangement, the pole being at P. The 
vector polygon abate, representing the forces in the order given, 
has been set out from a and terminates at e. ae has been 
drawn, and measures to scale 18-9 Ibs., and the equilibrant ea is the 
pull in the fifth rope, and its direction is 7 north of east from the 
pole. 

Example 2. Two forces of 3 Ibs. and 5 Ibs. respectively act on 
a particle, and their lines of action are inclined to each other at an 
angle of 70. Find what third force will keep the particle in 
equilibrium. 

The resultant force R will be of magnitude given by the 
relation 

R2 = 32 + 52 + 2 . 3 . 5 C os 70 

= 9 + 25 + (30 x 03420) = 34 + 10*26 = 44*26 
R = -v/44'4 2 = 6*65 Ibs. 

And R is inclined to the force of 5 Ibs. at an angle the tangent of 
which is 

(,. 3 sin 70 3 x 0-9397 

5 + 3 cos 70 5 + (3 x 0-3420) 
2-8171 

= -; -r = 0*467 

6-026 

which is an angle 25. The equilibrant or 
third force required to maintain equilibrium 
is, therefore, one of 6*65 Ibs., and its line of 
action makes an angle of 180 25 or 155 
FIG. 6r. with the line of action of the force of 5 Ibs., 

as shown in Fig. 61. 

Example 3. Solve Example i Ly resolving the forces into 
components. Taking an axis PX due east (Fig. 60) and PY due 
north, component force along PX 

F x = - 15 cos 15 20 cos 15 + 30 cos 60 

= ( 35 x '9659) + (30 x 0-5) = -i8'8o6 Ibs. 

Component force along PY 

F Y = 25 + 15 cos 75 - 20 cos 75 - 30 cos 30 

= 25 (5 x 0-2588) 30 x o'866o = -2-274 Iks. 
hence R 2 = (i8-8i) + (2-27)2 = 359*3 
R = \/359'3 = 18-96 Ibs. 




Statics Concurrent Forces Friction 99 

R acts outwards from P in a direction south of west, being inclined 
to XP at an acute angle, the tangent of which is 

Fy 2'274 

F x = l8-8o6 = ' 121 

which is the tangent of 6 54' ; i.e. R acts in a line lying 6 54' 
south of west. The equilibrant is exactly opposite to this, hence 
the fifth rope runs outwards from the pole P in a direction 6 54' 
north of east, and has a tension of i8'g6 Ibs. 



EXAMPLES X. 

1. A weight of 20 Ibs. is supported by two strings inclined 30 and 45 
respectively to the horizontal. Find by graphical construction the tension 
in each cord. 

2. A small ring is situated at the centre of a hexagon, and is supported 
by six strings drawn tight, all in the same plane and radiating from the 
centre of the ring, and each fastened to a different angular point of the 
hexagon. The tensions in four consecutive strings are 2, 7> 9> and 6 Ibs. 
respectively. Find the tension in the two remaining strings. 

3. Five bars of a steel roof-frame, all in one plane, meet at a point ; 
one is a horizontal tie-bar carrying a tension of 40 tons ; the next is also a 
tie-bar inclined 60 to the horizontal and sustaining a pull of 30 tons ; the 
next (in continuous order) is vertical, and runs upward from the joint, and 
carries a thrust of 5 tons ; and the remaining two in the same order radiat-2 
at angles of 135 and 210 to the first bar. Find the stresses in the last 
two bars, and state whether they are in tension or compression, i.e. whether 
they pull or push at the common joint. 

4. A telegraph pole assumed to have no force bending it out of the 
vertical has four sets of horizontal wires radiating from it, viz. one due east, 
one north-east, one 30 north of west, and one other. The tensions of the 
first three sets amount to 400 Ibs., 500 Ibs., and 250 Ibs. respectively. Find, 
by resolving the forces north and east, the direction of the fourth set and 
the total tension in it. 

5. A wheel has five equally spaced radial spokes, all in tension. If the 
tensions of three consecutive spokes are 2000 Ibs., 2800 Ibs., and 2400 Ibs. 
respectively, find the tensions in the other two. 

6. Three ropes, all in the same vertical plane, meet at a point, and there 
support a block of stone. They are inclined at angles of 40, 120, and 
1 60 to a horizontal line in their common plane. The pulls in the first two 
ropes are 150 Ibs. and 120 Ibs. respectively. Find the weight of the block 
of stone and the tension in the third rope. 

76. Friction. Friction is the name given to that pro- 
perty of two bodies in contact, by virtue of which a resistance 



ioo Mechanics for Engineers 

is offered to any sliding motion between them. The resistance 
consists of a force tangential to the surface of each body at 
the place of contact, and it acts on each body in such a direction 
as to oppose relative motion. As many bodies in equilibrium 
are held in their positions partly by frictional forces, it will be 
convenient to consider here some of the laws of friction. 

77. The laws governing the friction of bodies at rest are 
found by experiment to be as follows : 

(1) The force of friction always acts in the direction opposite 
to that in which motion would take place if it were absent, and 
adjrists itself to the amount necessary to maintain equilibrium. 

There is, however, a limit to this adjustment and to the 
value which the frictional force can reach in any given case. 
This maximum value of the force of friction is called the 
limiting friction. It follows the second law, viz. 

(2) The limiting friction for a given pair of siirfaces defends 
upon the nature of the surfaces, is proportional to the normal 
pressure between them, and independent of the area of the sur- 
faces in contact. 

For a pair of surfaces of a given kind (i.e. particular sub- 
stances in a particular condition), the limiting friction F = p. . R, 
where R is the normal pressure between the surfaces, and p, is 
a constant called the coefficient of friction for the given surfaces. 
This second law, which is deduced from experiment, must be 
taken as only holding approximately. 

78. Friction during Sliding Motion. If the limiting 
friction between the bodies is too small to prevent motion, and 
sliding motion begins, the subsequent value of the frictional 
force is somewhat less than that of the statical friction. The 
laws of friction of motion, so far as they have been exactly 
investigated, are not simple. The friction is affected by other 
matter (such as air), which inevitably gets between the two 
surfaces. However, for very low velocities of sliding and 
moderate normal pressure, the same relations hold approxi- 
mately as have been stated for the limiting friction of rest, 
viz. 

F = ^R 

where F is the frictional force between the two bodies, and R 



Statics Concurrent Forces Friction 



101 



is the normal pressure between them, and ju, is a constant 
coefficient for a given pair of surfaces, and which is less than 
that for statical friction between the two bodies. The friction 
is also independent of the velocity of rubbing. 

79. Angle of Friction. Suppose a body A (Fig. 62) is 
in contact with a body B, and is being pulled, say, to the right, 
the pull increasing until the limiting amount of frictional re- 
sistance is reached, that is, until the force of friction reaches a 
limiting value F = ju,R, where R is the normal pressure between 




\S 



T & 

FIG 62. 

the two bodies, and p, is the coefficient of friction. If R and 
F, which are at right angles, are compounded, we get the 
resultant pressure, S, which B exerts on A. As the friction F 
increases with the pull, the inclination of the resultant S of 
F and R to the normal of the surface of contact, i.e. to the line 
of action of R, will become greater, since its tangent is always 

cb F 
equal to -7 or (Art. 75). 

Let the extreme inclination to the normal be X when the 
friction F has reached its limit, j,R. 



tanA = R = 



R 



= V- 



This extreme inclination, A, of the resultant force between 



IO2 



Mechanics for Engineers 



two bodies to the normal of the common surface in contact is 
called the angle of friction, and we have seen that it is the angle 
the tangent of which is equal to the coefficient of friction 

tan A = //,, or A = tan" 1 p, 

80. Equilibrium of a Body on an Inclined Plane. 

As a simple example of a frictional force, it will be instructive 
here to consider the equilibrium of a body resting on an 
inclined plane, supported wholly or in part by the friction 
between it and the inclined plane. 

Let |, be the coefficient of friction between the body of 
weight W and the inclined plane, and let a be the inclination 
of the plane to the horizontal plane. We shall in all cases 
draw the vector polygon of forces maintaining equilibrium, 
not necessarily correctly to scale, and deduce relations between 
the forces by the trigonometrical relations between the parts of 
the polygon, thus combining the advantages of vector illustra- 
tion with algebraic calculation, as in Art. 75. The normal 
to the plane is shown dotted in each diagram (Figs. 63-68 
inclusive). 

i. Body at rest on an inclined plane (Fig. 63). 



a 
W 
I 





w 



FIG. 63. 



If the body remains at rest unaided, there are only two 
forces acting on it, viz. its weight, W, and the reaction S of the 
plane; these must then be in a straight line, and therefore S 
must be vertical, i.e. inclined at an angle a 1o the normal to the 
plane. The greatest angle which S can make to the normal 



Statics Concurrent Forces Friction 



103 



is A, the angle of friction (Art. 79) ; therefore a cannot exceed 
A, the angle of friction, or the body would slide down the 
plane. Thus we might also define the angle of friction between 
a pair of bodies as the greatest incline on which one body 
would remain on the other without sliding. 

Proceeding to supported bodies, let an external force, P, 
which we will call the effort, act upon the body in stated 
directions. 

2. Horizontal effort necessary to start the body up the 
plane. Fig. 64 shows the forces acting, and a triangle of 
forces, abc. 



a, 




FIG. 64. 



When the limit of equilibrium is reached, and the body is 
about to slide up the plane, the angle dbc will be equal to X, 
the maximum angle which S can make with the normal to the 
plane; then 



or P = W tan (a. + A) 

which is the horizontal effort necessary to start the body up 
the plane. 

3. Horizontal effort necessary to start the body sliding 
down the plane (Fig. 65). 

When the body is about to move down the plane, the angle 
cbd will be equal to the angle of friction, A ; then 

P ca 

.-,. = , = tan (A a ) 

W ab 

or P = W tan (A - a) 



104 



Mechanics for Engineers 



If a is greater than A, this can only be negative, i.e. c falls 
to the left of a, and the horizontal force P is that necessary 




to just support the body on the steep incline on which it cannot 
rest unsupported. 

4. Effort required parallel to the plane to start the body up 
the plane (Fig. 66). 



angle acb = 90 X 
ab = W 

ca = P 





a, 



When the body is about to slide up the plane, the reaction 
S will make its maximum angle A (die) to the normal. 

P _ ac _ sin (A + a) 



rrii J- '* ' I 

Then = - = - 
W ab 

or P = W 



W ab sin (90 - A) 
sin (A + a) 



cos A 



which is the effort parallel to the plane necessary to start the 
body moving up the plane. 

5. Effort required parallel to the plane to start the body 
down the plane (Fig. 67). 



Statics Concurrent Forces Friction 



105 



When the body is just about to slide down the plane, ebd=- A. 

i _ sin (A - 

1) sin (90 

sin (A a) 



Then P -**- sin (A -a) 

J.11CI1,,, , / o _ * 

W ab sin (90 A) 



or P = W' 



cos A 



which is the least force parallel to the plane necessary to start 
the body moving down the plane. If a is greater than A, this 





W 



angle acb = 90 \ 
ab= W 
be- S 
ra = P 

FIG. 67. 

force, P, can only be negative, i.e. c falls between a and d, and 
the force is then that parallel to the plane necessary to just 
support the body from sliding down the steep incline. 

6. Least force necessary to start the body up the incline. 

Draw ab (Fig. 68) to represent W, and a vector, be, of 
indefinite length to represent S inclined A to the normal. 
Then the vector joining a to the line be is least when it 
is perpendicular to be. Then P is least when its line of action 
is perpendicular to that of S ; that is, when it is inclined 
90 A to the normal, or A to the plane ; and then 

P ca , 
w%7 =sm( a + A) 

Note that when a = o, 

P = W sin A 



io6 



Mechanics for Engineers 



which is the least force required to draw a body along the 
level. 



= S 




7. Similarly, the least force necessary to start the body 
down a plane inclined a to the horizontal is 

P = W sin (A - ) 

if A is greater than a. If a is greater than A, P is negative, and 
P is the least force which will support the body on the steep 
incline. In either case, P is inclined 90 A to the normal 
or A to the plane. 

8. Effort required in any assigned direction to start the 
body up the plane. 

Let be the assigned angle which the effort P makes with 
the horizontal (Fig. 69). 




angle bac = 90 6 

cub = 90 - a - \ + 

FIG. 69. 



Statics Concurrent Forces Friction 107 

TKr, P - ca s ' n ( A + <*) _ s i n (* + ) 

i nen . ; ^- . - 

w <* sin acb cos w (a + A)} 



or P - W 



cos [0 - (A + a)} 

which is the effort necessary to start the body up the plane in 
the given direction. 

9. The effort in any assigned direction necessary to pull 
the body down the plane may be similarly found, the resultant 
force S between the body and plane acting in this case at an 
angle A to the normal, but on the opposite side from that 
on which it acts in case 8. 

81. Action of Brake-blocks : Adhesion. A machine 
or vehicle is often brought to rest by opposing its motion by 
a frictional force at or near the circumference of a wheel or 
a drum attached to the wheel. A block is pressed against 
the rotating surface, and the frictional force tangential to the 
direction of rotation does work in opposing the motion. The 
amount of work done at the brake is equal to the diminution 
of kinetic energy, and this fact gives a convenient method of 
making calculations on the retarding force. The force is not 
generally confined to what would usually be called friction, as 
frequently considerable abrasion of the surface takes place, 
and the blocks wear away. It is usual to make the block of a 
material which will wear more rapidly than the wheel or drum 
on which it rubs, as it is much more easily renewed. If the 
brake is pressed with sufficient force, or the coefficient of 
"brake friction" between the block and the wheel is sufficiently 
high, the wheel of a vehicle may cease to rotate, and begin to 
slide or skid along the track. This limits the useful retarding 
force of a brake to that of the sliding friction between the 
wheels to which the brake is applied and the track, a quantity 
which may be increased by increasing the proportion of weight 
on the wheels to which brakes are applied. The coefficient of 
sliding friction between the wheels and the track is sometimes 
called the adhesion, or coefficient of adhesion. 

82. Work spent in Friction. If the motion of a body 
is opposed by a frictional force, the amount of work done 



io8 Mechanics for Engineers 

against friction in foot-pounds is equal to the force in pounds 
tangential to the direction of motion, multiplied by the distance 
in feet through which the body moves at the point of applica- 
tion of the force. 

If the frictional force is applied at the circumference of a 
cylinder, as in the case of a brake band or that of a shaft or 
journal revolving in a bearing, the force is not all in the same 
line of action, but is everywhere tangential to the rotating 
cylinder, and it is convenient to add the forces together arith- 
metically and consider them as one force acting tangentially to 
the cylinder in any position, opposing its motion. If the 
cylinder makes N rotations per minute, and is R feet radius, 
and the tangential frictional force at the circumference of the 
cylinder is F Ibs., then the work done in one rotation is 2?rR . F 
foot-lbs., and the work done per minute is 27rRF . N foot-lbs., 

2?rR . F . N , 

and the power absorbed is - horse-power (Art. 155). 

33,000 

In the case of a cylindrical journal bearing carrying a 
resultant load W Ibs., F = ^W, where p is the coefficient of 
friction between the cylinder and its bearing. 

83. Friction and Efficiency of a Screw. The screw 
is a simple application of the inclined plane, the thread on 




FIG. 70. 



either the screw or its socket (or nut) fulfilling the same functions 
as a plane of the same slope. For simplicity a square-threaded 
screw (Fig. 70) in a vertical position is considered, the diameter 



Statics Concurrent Forces Friction 109 

d inches being reckoned as twice the mean distance of the 
thread from the axis. 

Let/ = the pitch or axial distance, say in inches, from any 
point on the thread to the next corresponding point, so that 
when the screw is turned through one complete rotation in its 
fixed socket it rises / inches. Then the tangent of the angle 

of slope of the screw thread at its mean distance is ., which 

corresponds to tan a in Art. 80. Hence, if a tangential hori- 
zontal effort P Ibs. be applied to the screw at its mean diameter 
in order to raise a weight W Ibs. resting on the top of the 
screw 



where tan A = ju, (Art. 80 (2)) ; or, expanding tan ( + A) 

P _ tan a + tan A _ ird p + [tsird 

W i tan a tan A p.p ird p,p 

ird 

which has the value , or tan for a frictionless screw. 
Again, the work spent per turn of the screw is 
P x vd = W(tan a + A) . ird inch-lbs. 

The useful work done is W . p inch-lbs. ; therefore the work 
lost in friction is W tan (a + X)ird W/ foot-lbs., an expression 
which may be put in various forms by expansion and substitu- 
tion. The " efficiency " or proportion of useful work done 
to the total expenditure of work is 

Vffl tan a 



W tan (a + X)-rrd tan (a -f A) 



which may also be expressed in terms of /, d, and /x. The 

W 

quantity -p is called the mechanical advantage ; it is the ratio 

of the load to the effort exerted, and is a function of the 



IIO 



Mechanics for Engineers 



dimensions and the friction which usually differs with different 
loads. 

84. Friction of Machines. Friction is exerted at all 
parts of a machine at which there is relative tangential motion 
of the parts. It is found by experiment that its total effects 
are such that the relation between the load and the effort, 
between the load and the friction, and between the load and 
the efficiency generally follow remarkably simple laws between 
reasonable limits. The subject is too complex for wholly 
theoretical treatment, and is best treated experimentally. It 
is an important branch of practical mechanics. 

Example i, A block of wood weighing 12 Ibs. is just pulled 
along over a horizontal iron track by a horizontal force of 3^ Ibs. 
Find the coefficient of friction between the wood and the iron. How 
much force would be required to drag the block horizontally if the 
force be inclined upwards at an angle of 30 to the horizontal ? 

If /* = the coefficient of friction 

ju x 12 = 3^ Ibs. 

/x = = O'2QI 
12 

Let P = force required at 30 inclination ; 

S = resultant force between the block and the iron track. 




ttlbs 



12U>s 



FIG. 71. 



abc (Fig. 71) shows the triangle of forces when the block just 
reaches limiting equilibrium. In this triangle, cab = 60, since P is 
inclined 30 to the horizontal ; and 



Status Concurrent Forces Friction 



in 



hence sin 



tan abc = /* = 0-291 or -fa 
i 



cot 2 



7 

= = sin A 



and cos x = f * 
P__ <** . 

12 rt 



sin 
sin 



sin A. 



sin \ 



sin (A. + 60) 



i sin x + -- cos A 
2 



7x2 

-= = 0-289 

7 + 24^/3 



P = 1 2 x 0*289 = 3-46 Ibs. 
Or thus- 

Normal pressure between block > _ . 

and track / 

horizontal pull P cos 30 = /*(i2 P sin 30) 

p ( + &="*** 

\ 2 4o/ 

hence P = 3-46 Ibs. 

Example 2. A train, the weight of which, including locomotive, 
is 1 20 tons, is required to accelerate to 40 miles per hour from rest 
in 50 seconds. If the coefficient of adhesion is i, find the necessary 
weight on the driving wheels. In what time could the train be 
brought to rest from this speed, (i) with continuous brakes (i.e. 
on every wheel on the train) ; (2) with brakes on the driving-wheels 
only ? 

The acceleration is f X 88 X ^ = 1*173 ^ eet P er sec - P er sec - 

i **o 
The accelerating force is 1-173 x ~~r~ = 4*37 tons 

The greatest accelerating force obtainable without causing the 
driving-wheels to slip is \ of the weight on the wheels, therefore 
the minimum weight required on the driving-wheels is 7 x 4-37 
= 30-6 tons. 

(i) The greatest retarding force with continuous brakes is 120 x 1 
tons. Hence, if / = number of seconds necessary to bring the train 



to rest, the impulse 120 x 1 x / = x 
ton and second units. Hence 



x , the momentum in 






_ 7 x 



3 x 32-2 



= 12-75 seconds 



1 1 2 Mechanics for Engineei's 

(2) If the brakes are on the driving-wheels only, the retarding 
force will be restricted to \ of 30-6 tons, i,e. to 4*37 tons, which was 
the accelerating force, and consequently the time required to come 
to rest will be the same as that required to accelerate, i.e. 
50 seconds. 

Example 3. A square-threaded screw 2 inches mean diameter 
has two threads per inch of length, the coefficient of friction 
between the screw and nut being o - o2. Find the horizontal force 
applied at the circumference of the screw necessary to lift a weight 
of 3 tons. 

The pitch of the screw is \ inch. 

If a = angle of the screw, tan a = = 0*0794 

27T 

and if A. = angle of friction, tan A. = o'o2 
Let P = force necessary in tons. 

P _ , . _ tan o + tan A. _ 0*0794 + 0*02 
3 "~ I tan o tan A. i 0x3794 x 0*02 

= 



0- 
hence P = 0-2987 ton 



EXAMPLES XI. 



1. A block of iron weighing n Ibs. can be pulled along a horizontal 
wooden plank by a horizontal force of 1-7 Ibs. What is the coefficient of 
friction between the iron and the plank ? What is the greatest angle to the 
horizontal through which the plank can be tilted without the block of iron 
sliding off ? 

2. What is the least force required to drag a block of stone weighing 
20 Ibs. along a horizontal path, and what is its direction, the coefficient of 
friction between the stone and the path being 0*15 ? 

3. What horizontal force is required to start a body weighing 15 Ibs. 
moving up a plane inclined 30 to the horizontal, the coefficient of friction 
between the body and the plane being 0*25 ? 

4. Find the least force in magnitude and direction required to drag a 
log up a road inclined 1 5 to the horizontal if the coefficient of friction 
between the log and the road is 0*4. 

5. With a coefficient friction 0*2, what must be the inclination of a 
plane to the horizontal if the work done by the minimum force in dragging 
10 Ibs. a vertical distance of 3 feet up the plane is 60 foot Ibs. ? 

6. A shaft bearing 6 inches diameter carries a dead load of 3 tons, 
and the shaft makes 80 rotations per minute. The coefficient of friction 
between the shaft and bearing is o - oi2. Find the horse-power absorbed in 
friction in the bearing. 



Statics Concurrent Forces Friction 1 1 3 

7. If a brake shoe is pressed against the outside of a wheel with a force 
of 5 ton?, and the coefficient -of friction between the wheel and the brake is 
0-3, find the horse-power absorbed by the brake if the wheel is travelling 
at a uniform speed of 20 miles per hour. 

8. A stationary rope passes over part of the circumference of a rotating 
pulley, and acts as a brake upon it. The tension of the tight end of the 
rope is 120 Ibs., and that of the slack end 25 Ibs., the difference being due 
to the frictional force exerted tangentially to the pulley rim. If the pulley 
makes 170 rotations per minute, and is 2 feet 6 inches diameter, find the 
horse-power absorbed. 

9. A block of iron weighing 14 Ibs. is drawn along a horizontal 
wooden table by a weight of 4 Ibs. hanging vertically, and connected to 
the block of iron by a string passing over a light pulley. If the coefficient 
of friction between the iron and the table is O'i5, find the acceleration of 
the block and the tension of the string. 

10. A locomotive has a total weight of 30 tons on the driving wheels, 
and the coefficient of friction between the wheels and rails is O'I5. What 
is the greatest pull it can exert on a train 7 Assuming the engine to be 
sufficiently powerful to exert this pull, how long will it take the train to 
attain a speed of 20 miles per hour if the gross weight is 120 tons, and the 
resistances amount to 20 Ibs. per ton ") 

if. A square-threaded screw, 1*25 inches mean diameter, has five threads 
per inch of length. Find the force in the direction of the axis exerted 
by the screw when turned against a resistance, by a handle which exerts a 
force equivalent to 500 Ibs. at the circumference of the screw, the co- 
efficient of friction being o'o8. 



CHAPTER VI 
STATICS OF RIGID BODIES 

85. THE previous chapter dealt with bodies of very small 
dimensions, or with others under such conditions that all the 
forces acting upon them were concurrent. 

In general, however, the forces keeping a rigid body in 
equilibrium will not have lines of action all passing through 
one point. Before stating the conditions of equilibrium of a 
rigid body, it will be necessary to consider various systems of 
non-concurrent forces. We shall assume that two intersecting 
forces may be replaced by their geometric sum acting through 
the point of intersection of their lines of action ; also that a 
force may be considered to act at any point in its line of action. 
Its point of application makes no difference to the equilibrium 
of the body, although upon it will generally depend the dis- 
tribution of internal forces in the body. With the internal 
forces or stresses in the body we are not at present concerned. 

86. Composition of Parallel Forces. The following 
constructions are somewhat artificial, but we shall immediately 
from them find a simpler method of calculating the same 
results. 

To find the resultant and equilibrant of any two given like 
parallel forces, i.e. two acting in the same direction. Let P 
and Q (Fig. 7 2) be the forces of given magnitudes. Draw any 
line, AB, to meet the lines of action of P and Q in A and B 
respectively. At A and B introduce two equal and opposite 
forces, S, acting in the line AB, and applied one at A and the 
other at B. Compound S and P at A by adding the vectors 
Kd and de, which give a vector A^, representing Rj, the resultant 



Statics of Rigid Bodies 1 1 5 

of S and P. Similarly, compound S and Q at B by adding 
the vectors B/and/i, r , which give a vector sum Eg, representing 
R.,, the resultant of Q and S. Produce the lines of action of 
RI and R 2 to meet in O, and transfer both forces to O. Now 
resolve Rj and R 2 at O into their components again, and we 






Vector de represents P. 
Vector^' represents Q. 
Vectors Kd and B/" represent equal and opposite forces S. 

FIG. 72. 

have left two equal and opposite forces, S, which have a 
resultant nil, and a force P + Q acting in the same direction 
as P and Q along OC, a line parallel to the lines of action 
of P and Q. If a force P + Q acts in the line CO in the 
opposite direction to P and Q, it balances their resultant, and 
therefore it will balance P and Q, i.e. it is their equilibrant. 

Let the line of action of the resultant P + Q cut AB in C. 

Since AOC and Ked are similar triangles 



CA 
OC 



ae 



ii6 Mechanics for Engineers 

and since BOC and B?/are similar triangles 
CB B/ S 



and dividing equation (2) by equation (i) 

CB_ P 

CA~Q 

or the point C divides the line AB in the inverse ratio of the 
magnitude of the two forces ; and similarly the line of action 
OC of the resultant P + Q divides any line meeting the lines 
of action of P and Q in the inverse ratio of the forces. 

To find the resultant of any two given unlike parallel forces, 
i.e. two acting in opposite directions. 

Let one of the forces, P, be greater than the other, Q 
(Fig. 73). By introducing equal and opposite forces, S, at A 




Vector de represents P. 
Vector fg represents Q. 

Vectors A</ and B/~ represent equal and opposite forces S. 
FIG. 73. 

and B, and proceeding exactly as before, we get a force P Q 
acting at O, its line of action cutting AB produced in C. 
Since AOC and hfd are similar triangles 

_CA_A^_S M 

OC ~ de ~ P (3; 



Statics of Rigid Bodies 117 

and since BOC and Eg/ are similar triangles 
CB_B/_^ 

CQ-y^-Q ' 

Dividing equation (4) by equation (3) 

CB_ P 
CA~Q 

or the line of action of the resultant P Q divides the line 
AB (and any other line cutting the lines of action of P and Q) 
externally, in the inverse ratio of the two forces, cutting it 
beyond the line of the greater force. If a force of magnitude 
P Q acts in the line CO in the opposite direction to that of 
P (i.e. in the same direction as Q), it balances the resultant of 
P and Q, and therefore it will balance P and Q ; i.e. it is their 
equilibrant. 

This process fails if the two unlike forces are equal. The 
resultants Rj and R 2 are then also parallel, and the point of 
intersection O is non-existent. The two equal unlike parallel 
forces are not equivalent to, or replaceable by, any single force, 
but form what is called a " couple." 

More than two parallel forces might be compounded by 
successive applications of this method, first to one pair, then to 
the resultant and a third force, and so on. We shall, however, 
investigate later a simpler method of compounding several 
parallel forces. 

87. Resolution into Parallel Components. In the 
last article we replaced two 
parallel forces, P and Q, acting 
at points A and B, by a single 

force parallel to P and Q, acting A _/_ 

at a point C in AB, the posi- 
tion of C being such that it 
divides AB inversely as the mag- 
nitudes of the forces P and Q. 
Similarly, a single force may be *..* 

FIG. 74. Resolution into two like 

replaced by two parallel forces parallel components, 

acting through any two given points. Let F (Fig. 74) be the 
single force, and A and B be the two given points. Join AB 




1 1 8 Mechanics for Engineers 

and let C be the point in which AB cuts the line of action of F. 
If, as in Fig. 74, A and B are on opposite sides of F, then F 
may be replaced by parallel forces in the same direction as 
F, at A and B, the magnitudes of which have a sum F, and 
which are in the inverse ratio of their distances from C, viz. a 

CB AC 

force F X -^5 at A, and a force F X -TJ, at B. The parallel 

equilibrants or balancing forces of F acting at A and B are 

CB AC 

then forces F X -ps and F X -^ respectively, acting in the 

opposite direction to that of the force F. 

If A and B are on the same side of the line of action of the 
force F (Fig. 75), then F may be replaced by forces at A and B, 




'F 

FIG. 75. Resolution into two unlike parallel components. 

the magnitudes of which have a difference F, the larger force 
acting through the nearer point A, and in the same direction 
as the force F, the smaller force acting through the further 
point B, and in the opposite direction to the force F, and the 

magnitudes being in the inverse ratio of the distances of the 

/'p 
forces from C, viz. a force F X -y-n at A, in the direction of F, 

AC 

and an opposite force F X vg at B. 

CB 
The equilibrants of F at A and B will be F X -^g in the 

AC 

opposite direction to that of F, and F X -T-JT in the direction 

of F, respectively. 



Statics of Rigid Bodies 



119 



B 



As an example of the parallel equilibrants through two 
points, A and B, on either side of the line of action of a force, 
we may take the vertical up- 
ward reactions at the supports 
of a beam due to a load con- 
centrated at some place on 
the beam. 

Let W Ibs. (Fig. 76) be 
the load at a point C on a -^ 
beam of span /feet, C being 
x feet from A, the left-hand 
support, and therefore / x feet from the right-hand support, B. 

Let R A be the supporting force or reaction at A ; 
R B be the supporting force or reaction at B. 



k JC--*i 

i j : 



FIG. 76. 



Then R A = W x = 



Ibs. 



More complicated examples of the same kind where there 
is more than one load will generally be solved by a slightly 
different method. 

88. Moments. The moment of a force F Ibs. about a 
fixed point, O, was measured (Art. 56) by the product F X d 
Ib.-feet, where d was the perpen- 
dicular distance in feet from O 
to the line of action of F. Let 
ON (Fig. 77) be the perpen- 
dicular from O on to the line of 
action of a force F. 

Set off a vector ab on the 
line of action of F to represent 
F. Then the product ab . ON, 
which is twice the area of the 
triangle Oat>, is proportional to the moment of F about O. 
Some convention as to signs of clockwise and contra-clockwise 
moments (Art. 56) must be adopted. If the moment of F 
about O is contra-clockwise, i.e. if O lies to the left of the line 



/a, 




FIG. 77. 





1 20 Mechanics for Engineers 

of action of F viewed in the direction of the force, it is usual 
to reckon the moment and the area Oafr representing it as 
positive, and if clockwise to reckon them as negative. 

89. Moment of a Resultant Force. This, about any 
point in the plane of the resultant and its components, is equal 

to the algebraic sum of the 
moments of the components. 
Let O (Fig. 78) be any 
point in the plane of two 
forces, P and Q, the lines 
of action of which intersect 
at A. Draw Qd parallel to 
the force P, cutting the line 
of action of Q in c. Let 
FIG g the vector Ac represent the 

force Q, and set off Kb in 
the line of action of P to represent P on the same scale, 

P 
i.e. such that Kb = Kc X Q. 

Complete the parallelogram Kbdc. Then the vector Kd = 
Ac + cd Ac + Kb, and represents the resultant R, of P 
and Q. 

Now, the moment of P about O is represented by twice the 
area of triangle AO^ (Art. 88), and the moment of Q about 
O is represented by twice the area of triangle AO, and tlje 
moment of R about O is represented by twice the area of 
triangle AOrf. 

But the area KOd = area AcJ + area AO<r 
= area Kbd -f- area AOc 

Kbd and Acd being each half of the parallelogram Abdc \ 
hence area AO./ = area AOZ> + AO, since AO<5 and Abd are 
between the same parallels ; or 

twice area AQd = twice area AO + twice area AO^. 

and these three quantities represent respectively the moments 

of R, P, and Q about O. Hence the moment of R about O is 

equal to the sum of the moments of P and Q about that point. 

If O is to the right of one of the forces instead of tc the left 



Statics of Rigid Bodies 



121 



of both, as it is in Fig. 78, there will be a slight modification 
in sign ; e.g. if O is to the right of the line of action of Q and to 
the left of R and P, the area AO and the moment of Q about 
O will be negative, but the theorem will remain true for the 
algebraic sum of the moments. 

Next let the forces P and Q be parallel (Fig. 79). Draw 




a line AB from O perpendicular to the lines of action of 
P and Q, cutting them in A and B respectively. Then the 
resultant R, which is equal to P + Q, cuts AB in C such that 
BC _ P 
AC~Q' 

Then P . AC = Q . BC 

The sum of moments of P and Q about O is P . O A + Q . OB, 
and this is equal to P(OC - AC) + Q(OC + CB), which is 
squal to (P + Q)OC - P . AC + Q . CB = (P + QJOC, since 
P . AC = Q . CB. 

And (P + QJOC is the moment of the resultant R about O. 
Hence the moment of the resultant is equal to the sum of 
moments of the two component forces. The figure will need 
modification if the point O lies between the lines of action of 
P and Q, and their moments about O will be of opposite sign, 
but the moment of R will remain equal to the algebraic sum 
of those of P and Q. The same remark applies to the figure 
for two unlike parallel forces. 

The force equal and opposite to the resultant, i.e. the 
equilibrant, of the two forces (whether parallel or intersecting) 
has a moment of equal magnitude and opposite sign to that of 
the resultant (Art. 88), and therefore the equilibrant has a moment 



122 Mechanics for Engineers 

about any point in the plane of the forces, of equal magnitude and 
of opposite sign to the moments of the forces which it balances. In 
other words, the algebraic sum of the moments of any two forces 
and their equilibrant about any point in their plane is zero. 

90. Moment of Forces in Equilibrium. If several 
forces, all in the same plane, act upon a body, the resultant 
of any two has about any point O in the plane a moment equal 
to that of the two forces (Art. 89). Applying the same theorem 
to a third force and the resultant of the first two, the moment 
of their resultant (i.e. the resultant of the first three original 
forces) is equal to that of the three forces, and so on. By 
successive applications of the same theorem, it is obvious that 
the moment of the final resultant of all the forces about any 
point in their plane is equal to the sum of the moments of all 
the separate forces about that point, whether the forces be all 
parallel or inclined one to another. 

If the body is in equilibrium, the resultant force upon it in 
any plane is zero, and therefore the algebraic sum of the moments 
of all the separate forces about any point in the plane is zero. 
This fact gives a method of finding one or two unknown forces 
acting on a body in equilibrium, particularly when their lines 
of action are known. When more than one force is unknown, 
the clockwise and contra-clockwise moments about any point 
in the line of action of one of the unknown forces may most 
conveniently be dealt with, for the moment of a force about 
any point in its line of action is zero. 

The Principle of Moments, i.e. the principle of equation 
of the algebraic sum of moments of all forces in a plane acting 
on a body in equilibrium to zero, or equation of the clockwise 
to the contra-clockwise moments, will be most clearly under- 
stood from the three examples at the end of this article. 

Levers. A lever is a bar free to turn about one fixed 
point and capable of exerting some force due to the exertion 
of an effort on some other part of the bar. The bar may be of 
any shape, and the fixed point, which is called the fulcrum, 
may be in any position. When an effort applied to the lever 
is just sufficient to overcome some given opposing force, the 
lever has just passed a condition of equilibrium, and the relation 



Statics of Rigid Bodies 



123 



\4-tons 



between the effort, the force exerted by the lever, and the 
reaction at the fulcrum may be found by the principle of 
moments. 

Example I. A roof- frame is supported by two vertical walls 
20 feet apart at points A and B on the same level. The line of the 
resultant load of 4 tons on the 
frame cuts the line AB 8 feet 
from A, at an angle of 75 to the 
horizontal, as shown in Fig. 80. 
The supporting force at the point 
B is a vertical one. Find its 
amount. 

The supporting force through 
the point A is unknown, but its 
moment about A is" zero. Hence 
the clockwise moment of the 4-ton resultant must balance the 
contra-clockwise moment of the vertical supporting force RB at B. 

Equating the magnitudes of the moments 

4 x 8 sin 75 = 20 x R B (tons-feet) 
therefore R B = HJlILZ-L _ r g x 0^9659 = i'545 tons 

Example 2. A light horizontal beam of 12-feet span carries 
loads of 7 cwt., 6 cwt., and 9 cwt. at distances of i foot, 5 feet, and 
10 feet respectively from the left-hand end. Find the reactions of 
the supports of the beam. 

If we take moments about the left-hand end A (Fig. 81), the 




FIG. 80. 



AC = 
AD - 


17 'Curt/. \6curu 
\ 
C ID 


\f)curt-. 

TE B 


AE = 


10 feet 








AB = 


12 feet, i 


\ 










RA 







FIG. Si. 

vertical loads have a clockwise tendency, and the moment of the 
reaction R B at B is contra-clockwise ; hence 

R B x 12 = (7 x i) + (6 x 5) + (9 x 10) 
I2R B = 7 + 30 + 90 = 127 
RB = V/- = 10*583 cwt. 



/ 

/30 
/ 



B 



1 24 Mechanics for Engineers 

R A , the supporting force at A, may be found by an equation of 
moments about B. Or since 

RB + RA = 7 + 6 + 9 = 22 cwt. 

RA = 22 10-583 = 11-416 CWt. 

Example 3. An L-shaped lever, of which the long arm is 
1 8 inches long and the short one 10 inches, has its fulcrum at the 

right angle. The effort exerted on 
the end of the long arm is 20 Ibs., 
inclined 30 to the arm. The short 
arm is kept from moving by a cord 
attached to its end and perpendicular 
to its length. Find the tension of the 
chord. 

Let T be the tension of the string 
in pounds. 

Then, taking moments about B 
(Fig. 82), since the unknown reaction 
of the hinge or fulcrum has no moment 
FIG. 82. about that point 

AB sin 30 x 20 = BC x T 
18 x | x 20 = 10 x T 
T = 18 Ibs. 

EXAMPLES XII. 

1. A post 12 feet high stands vertically on the ground. Attached to 
the top is a rope, inclined downwards and making an angle of 25 with 
the horizontal. Find what horizontal force, applied to the post 5 feet 
above the ground, will be necessary to keep it upright when the rope 
is pulled with a force of 120 Ibs. 

2. Four forces of 5, 7, 3, and 4 Ibs. act along the respective directions 
AB, BC, DC, and AD of a square, ABCD. Two other forces act, 
one in CA, and the other through D. Find their amounts if the six forces 
keep a body in equilibrium. 

3. A beam of 15-feet span carries loads of 3 tons, \ ton, 5 tons, and 
I ton, at distances of 4, 6, 9 and 13 feet respectively from the left-hand end. 
Find the pressure on the supports at each end of the beam, which weighs 
| ton. 

4. A beam 20 feet long rests on two supports 1 6 feet apart, and over- 
hangs the left-hand support 3 feet, and the right-hand support by I foot. 
It carries a load of 5 tons at the left-hand end of the beam, and one of 
7 tons midway between the supports. The weight of the beam, which may 
be looked upon as a load at its centre, is I ton. Find the reactions at the 



Statics of Rigid Bodies 



125 



FIG. 83. 



supports, i.e. the supporting forces. What upward vertical force at the 
right-hand end of the beam would be necessary to tilt the beam ? 

5. A straight crowbar, AB, 40 inches long, rests on a fulcrum, C, near to 
A, and a force of 80 Ibs. applied at B lifts a weight of 3000 Ibs. at A. 
Find the distance AC. 

6. A beam 10 feet long rests upon supports at its ends, and carries 
a load of 7 cwt. 3 feet from one end. Where must a second load of 19 cwt. 
be placed in order that the pressures on the two supports may be equal ? 

91. Couples. In Art. 86 it was stated that two equal 
unlike parallel forces are not replace- 
able by a single resultant force ; they 

cannot then be balanced by a single 
force. Such a system is called a cotiple, 
and the perpendicular distance between 
the lines of action of the two forces is 
called the arm of the couple. Thus, 
in Fig. 83, if two equal and opposite 
forces F Ibs. act at A and B perpen- 
dicular to the line AB, they form a 
couple, and the length AB is called the arm of the couple. 

92. Moment of a Couple. This is the tendency to pro- 
duce rotation, and is measured by the product of one of the forces 
forming the couple and the arm of the couple ; e.g. if the two 
equal and opposite forces forming the couple are each forces 
of 5 Ibs., and the distance apart of their lines of action is 
3 feet, the moment of the couple is 5x3, or 15 Ib.-feet ; or 
in Fig. 83, the moment of the 

couple is F x AB in suitable 
units. 

The sum of the moments of 
the forces of a couple is the 
same about any point O in their 
plane. Let O (Fig. 84) be any 
point. Draw a line OAB per- 
pendicular to the lines of action of the forces and meeting 
them in A and B. Then the total (contra-clockwise) moment 
of the two forces about O is 

F . OB - F . OA = F(OB - OA) = F . AB 




FIG. 84. 



126 



Mechanics for Engineers 



This is the value, already stated, of the moment of the 
couple, and is independent of the position of O. 

A couple is either of clockwise or contra-clockwise ten- 
dency, and its moment about any point in its plane is of the 
same tendency (viewed from the same aspect) and of the same 
magnitude. 

93. Equivalent Couples. Any two couples in a plane 
having the same moment are equivalent if they are cf the same 
sign or turning tendency, i.e. either both clockwise or both 

contra-clockwise; or, if the 
couples are equal in magnitude 
and of opposite sign, they 
balance or neutralise one 
another. The latter form of 
the statement is very simply 
proved. Let the forces F, F 
(Fig. 85) constitute a contra- 
clockwise couple, and the forces 
F', F' constitute a clock-wise 
couple having a moment of the 
same magnitude. Let the lines 
of action of F, F and those of 
F', F' intersect in A, B, C, and D, and let AE be the perpen- 
dicular from A on BC, and CG the perpendicular from C on 
AB. Then, the moments of the two couples being equal 

F X AE = F' x GC 

F X AB sin ABC = F X CB sin ABC 

F x AB = F X CB 

F = C_B 

F' AB 

Hence CB and AB may, as vectors, fully represent F and F 
respectively, acting at B. And since ABCD is a parallelogram, 
CD = AB, and the resultant or vector sum of F and F is in 
the line DB, acting through B in the direction DB. 

Similarly, the forces F and F' acting at D have an equal 
and opposite resultant acting through D in the direction BD. 
These two equal and opposite forces in the line of B and D 
balance, hence the two couples balance. 




Statics of Rigid Bodies 



127 



F, 



B 



It has been assumed here that the lines of action of F and 
F' intersect ; if they do not, equal and opposite forces in the 
same straight line may, for the purpose of demonstration, be 
introduced and compounded with the forces of one couple 
without affecting the moment of that couple or the equilibrium 
of any system of which it forms a part. 

94. Addition of Couples. The resultant of several 
couples in the same plane and of given moments is a couple 
the moment of which is equal to the sum of the moments of 
the several couples. 

Any couple may be replaced by its equivalent couple 
having an arm of length AB (Fig. 93) and forces F^ F 1} pro- 
vided Fj X AB = moment of the 
couple. 

Similarly, a second couple may 
be replaced by a couple of arm AB 
and forces F 2 , F 2 , provided F 2 X AB 
is equal to the moment of this second 
couple. In this way clockwise 
couples must be replaced by clock- 
wise couples of arm AB, and contra- 
clockwise couples by contra-clock- 
wise couples of arm AB, until finally we have a couple of 
moment 

(F! + F 2 + F 3 + . . . etc.)AB = Fj x AB + F 2 X AB -f F 3 X 

AB + . . . etc. 

= algebraic sum of moments of 
the given couples 

the proper sign being given to the various forces. 

95. Reduction of a System of Co-planar Forces. 

A system of forces all in the same plane is equivalent to (i) a 
single resultant force, or (2) a couple, or (3) a system in equi- 
librium, which may be looked upon as a special case of (i), 
viz. a single resultant of magnitude zero. 

Any two forces of the system which intersect may be 
jplaced by a single force equal to their geometric sum acting 
irough the point of intersection. Continuing the same process 



FIG. 86. 



128 Mechanics for Engineers 

of compounding successive forces with the resultants of others 
as far as possible, the system reduces to either a single re- 
sultant, including the case of a zero resultant, or to a number 
of parallel forces. In the latter case the parallel forces may 
be compounded by applying the rules of Art. 86, and reduced 
to either a single resultant (including a zero resultant) or to a 
couple. Finally, then, the system must reduce to (i) a single 
resultant, or (2) a couple, or (3) the system is in equilibrium. 

96. Conditions of Equilibrium of a System of 
Forces in One Plane. If such a system of forces is in equi- 
librium, the geometric or vector sum of all the forces must be 
zero, or, in other words, the force polygon must be a closed 
one, for otherwise the resultant would be (Art. 95) a single 
force represented by the vector sum of the separate forces. 

Also, if the system is in equilibrium (i.e. has a zero re- 
sultant), the algebraic sum of all the moments of the forces 
about any point in their plane is zero (Art. 90). These are 
all the conditions which are necessary, as is evident from 
Art. 95, but they may be conveniently stated as three con- 
ditions, which are sufficient 

(i) and (2) The sum of the components in each of two 
directions must be zero (a single resultant has a zero component 
in one direction, viz. that perpendicular to its line of action). 

(3) The sum of the moments of all the forces about one 
point in the plane is zero. 

If conditions (i) and (2) are fulfilled the system cannot 
have a single resultant (Art. 75), and if condition (3) is ful- 
filled it cannot reduce to a couple (Art. 92), and therefore it 
must reduce to a zero resultant (Art. 95), i.e. the system must 
be in equilibrium. 

These three conditions are obviously necessary, and they 
have just been shown to be sufficient, but it should be remem- 
bered that the algebraic sum of the moments of all the forces 
about every point in the plane is zero. The above three con- 
ditions provide for three equations between the magnitudes of 
the forces of a system in equilibrium and their relative posi- 
tions, and from these equations three unknown quantities may 
be found if all other details of the system be known. 



Statics of Rigid Bodies 129 

97. Solution of Statical Problems. In finding the 
forces acting upon a system of rigid bodies in equilibrium, it 
should be remembered that each body is in itself in equi- 
librium, and therefore we can obtain three relations (Art. 96) 
between the forces acting upon it, viz. we can write three 
equations by stating in algebraic form the three conditions of 
equilibrium ; that is, we may resolve all the forces in two 
directions, preferably at right angles, and equate the com- 
ponents in opposite directions, or equate the algebraic sums to 
zero, and we may equate the clockwise and contra-clockwise 
moments about any point, or equate the algebraic sum of 
moments to zero. 

The moment about every point in the plane of a system of 
co-planar forces in equilibrium is zero, and sometimes it is 
more convenient to consider the moments about two points 
and only resolve the forces in one direction, or to take 
moments about three points and not resolve the forces. If 
more than three equations are formed by taking moments 
about other points, they will be found to be not independent 
and really a repetition of the relations expressed in the three 
equations formed. Some directions of resolution are more 
convenient than others, e.g. by resolving perpendicular to some 
unknown force, no component of that force enters into the 
equation so formed. Again, an unknown force may be elimi- 
nated in an equation of moments by taking the moments about 
some point in its line of action, about which it will have a zero 
moment. 

"Smooth " Bodies. An absolutely smooth body would 
be one the reaction of which, on any body pressing against it, 
would have no frictional component, i.e. would be normal to 
the surface of contact, the angle of friction (Art. 79) being 
zero. No actual body would fulfil such a condition, but it 
often happens that a body is so smooth that any frictional force 
it may exert upon a second body is so small in comparison 
with other forces acting upon that body as to be quite negli- 
gible, e.g. if a ladder with one end on a rough floor rest against 
a horizontal round steel shaft, such as is used to transmit power 
in workshops, the reaction of the shaft on the ladder might 

K 



130 



Mechanics for Engineers 



without serious error be considered perpendicular to the length 
of the ladder, i.e. normal to the cylindrical surface of the 
shaft. 

Example i. A horizontal rod 3 feet long has a hole in one 
end, A, through which a horizontal pin passes forming a hinge. 
The other end, B, rests on a smooth roller at the same level. Forces 
of 7, 9, and 5 Ibs. act upon the rod, their lines of action, which are 
in the same vertical plane, intersecting it at distances of 1 1, 16, and 
27 inches respectively from A, and making acute angles of 30, 75, 
and 45 respectively with AB, the first two sloping downwards 
towards A, and the third sloping downwards towards B, as shown 
in Fig. 87. Find the magnitude of the supporting forces on the 
rod at A and B. 



9 Ibs. 




FIG. 87. 

Since the end B rests on a smooth roller, the reaction RB at B is 
perpendicular to the rod (Art. 97). We can conveniently find this 
reaction at B by taking moments about A, to which the unknown 
supporting force at A contributes nothing. 

The total clockwise moment about A in Ib.-inches is 



7 * 



= 270^2 Ib.-inches 



The total contra-clockwise moment about A is R B x 36. Equating 
the moments of opposite sign 

RB x 36 = 270*2 Ib.-inches 



Statics of Rigid Bodies 131 

The remaining force R A through A may be found by drawing 
to scale an open vector polygon with sides representing the forces 
7, 9, 5, and 7-5 Ibs. (Rs); the closing side then represents R A . 

Or we may find R A by resolving all the forces, say, horizontally 
and vertically. Let H A be the horizontal component of R A 
estimated positively to the right, and V A its vertical component 
upwards. Then, by Art. 96, the total horizontal component of all 
the forces is zero ; hence 

H A 7 cos 30 9 cos 75 + 5 cos 45 = o 
HA = 7 x o'866 + 9 x 0*259 5 x 0707 = 4-85 Ibs. 

Also the total vertical component is zero, hence 

V A 7 sin 30 - 9 sin 75 - 5 sin 45 + 7-5 = 
VA = 7 x + 9 x 0-966 + 5 x 0-707 - 7*5 = 8-23 Ibs. 

Compounding these two rectangular components of RA 



RA = A/KfSs) 2 + (8'2 3 ) 2 } (Art. 75) 
RA = V9 1 ' 2 = 9'54 Ibs. 

Example 2. ABCD is a square, each side being 17-8 inches, 
and E is the middle point of AB. Forces of 7, 8, 12, 5, 9, and 
6 Ibs. act on a body in the lines and directions AB, EC, BC, BD, 



8 




FIG. 



CA, and DE respectively. Find the magnitude, and position 
with respect to ABCD, of the single force required to keep the body 
in equilibrium. 



132 Mechanics for Engineers 

Let F be the required force ; 

H A be the component of F in the direction AD ; 
V A be the component of F in the direction AB ; 
P be the perpendicular distance in inches of the force 
from A. 

Then, resolving in direction AD, the algebraic total component 
being zero 

H A + 8 cos OEC +12 + 5 cos 45 - 9 cos 45 \ 
- 6 cos EDA / ~ 

H A + 8 x -4j= + 12 - 4 x -7= - 6 x -= = o 

Vs V2 Vs 

H A + (2 x 0-895) + 12 - 4 x 0707 = o 

H A = 10-96 Ibs. 

Resolving in direction AB 

V A + 7 + 8 cos EEC - 5 cos 45 9 cos 45 \ _ 
+ 6 cos AED / = 

V A + 7 + 14 x -*= - 14 x -^ . = o 

V5 V 2 

V A = -7 - 6-26 + 9-90 = -3-36 



then F = J{( 10-96)2 + (3-36)2 } = 11-46 Ibs. 
and is inclined to AD at an angle the tangent of which is 
-3-36 



- 10-96 



= 0-3065 



i.e. at an angle 180 + 17 or 197. 

Its position remains to be found. We may take moments about 
any point, say A. Let p be reckoned positive if F has a contra- 
clockwise moment about A. 

1 1*46 x p + 6 x AD sin ADE - 5 x OA - 12 ) _ 
x AB - 8 x AE sin EEC / " 

89 , , 
- + 213-6 



v/5 



292-3 

p = i = 25-51 inches 
11-46 



This completes the specification of the force F, which makes 
an angle 197 with AD and passes 25-51 inches from A, so as to 
have a contra-clockwise moment about A. The position of F is 
shown in Fig. 89. 



Statics of Rigid Bodies 



133 



The force might be specified as making 197 with AD and 
cutting it at a distance 25'5i -4- sin 197 or 86'5 inches from A ; 
t\e. 86-5 inches to the left of A. 



197' 




FIG. 89. 

98. Method of Sections. The principles of the pre- 
ceding article may be applied to find the forces acting in the 
members of a structure consisting of separate pieces jointed 
together. If the structure be divided by an imaginary plane of 
section into two parts, 
either part may be looked 
upon as a body in equi- 
librium under certain 
forces, some of which are 
the forces exerted by 
members cut by the plane 
of section. 

For example, if a 
hinged frame such as 
ABCDE (Fig. 90) is in Fic - 

equilibrium under given forces at A, B, C, D, and E, and an 
imaginary plane of section XX' perpendicular to the plane of 
the structure be taken, then the portion kBzyw is in equilibrium 




134 



Mechanics for Engineers 



H 



12 



FIG. 91. 



under the forces at A and B, and the forces exerted upon it by 
the remaining part of -the structure, viz. the forces in the bars 
BD, BC, and AC. This method of sections is often the 
simplest way of finding the forces in the members of a jointed 
structure. 

Example. One end of a girder made up of bars jointed 
together is shown in Fig. 91. Vertical loads of 3 tons and 5 tons 
are carried at B and C respectively, and 
the vertical supporting force at H is 12 tons. 

;' The sloping bars are inclined at 60 to the 
horizontal. Find the forces in the bars 
CD, CE, and FE. 
The portion of the girder ACFH cut 
off by the vertical plane klm is in equili- 
brium under the action of the loads at B 
and C, the supporting force at H, and the 
forces exerted by the bars CD, CE, and 
FE on the joints at C and F. Resolving these forces vertically, 
the forces in CD and FE have no vertical component, hence the 
downward vertical component force exerted by CE on the left- 
hand end of the girder is equal to the excess upward force of the 
remaining three, i.e. 12 - 3 - 5 = 4 tons; hence 

Force in CE x cos 30 = 4 tons 

2 

or force in CE = 4 x -= = 4"6l tons 
V3 

This, being positive, acts downwards on the left-hand end, i.e. it 
acts towards E, or the bar C E pulls at the joint C, hence the bar 
CE is in tension to the amount of 4'6i tons. To find the force in 
bar FE, take a vertical section plane through C or indefinitely 
near to C, and just on the right hand of it. Then, taking moments 
about C and reckoning clockwise moments positive 

12 x AC - 3 x BC + A /3 x FE x (force in FE) = o 
12x2 3x1 + ^3 x (force in FE) = o 

2 1 

and force in FE = = = 12*12 tons 

V3 

The negative sign indicates that the force in FE acts on F in 
the opposite direction to that in which it would have a clockwise 
moment about C, i.e. the force pulls at the joint F ; hence the 
member is in tension to the extent of 12' 12 tons. 



Statics of Rigid Bodies 135 

Similarly, taking say clockwise moments about E, the force in 
CD is found to be a push of 14*42 tons towards C, i.e. CD has a 
compressive force of 14*42 tons in it, as follows : 

12x3-3x2-5x1 + ^3 (force in CD) = o 

force in CD = 14*42 

99. Rigid Body kept in Equilibrium by Three 
Forces. If three forces keep a body in equilibrium, they 
either all pass through one point (i.e. are concurrent) or are 
all parallel. For unless all three forces are parallel two must 
intersect, and these are replaceable by a single resultant acting 
through their point of intersection. This resultant cannot 
balance the third force unless they are equal and opposite and 
in the same straight line, in which case the third force passes 
through the intersection of the other two, and the three forces 
are concurrent. 

The fact of either parallelism or concurrence of the three 
forces simplifies problems on equilibrium under three forces by 
fixing the position of an unknown force, since its line of action 
intersects those of the other two forces at their intersection. 
The magnitude of the forces can be found by a triangle of 
forces, or by the method of resolution into rectangular com- 
ponents. 

Statical problems can generally be solved in various ways, 
some being best solved by one method, and others by different 
methods. In the following example four methods of solution 
are indicated, three of which depend directly upon the fact that 
the three forces are concurrent, which gives a simple method 
of determining the direction of the reaction of the rough 
ground. 

Example i. A ladder 18 feet long rests with its upper end 
against a smooth vertical wall, and its lower end on rough ground 
7 feet from the foot of the wall. The weight of the ladder is 40 Ibs., 
which may be looked upon as a vertical force halfway along the 
length of the ladder. Find the magnitude and direction of the 
forces exerted by the wall and the ground on the ladder. 

The weight of 40 Ibs. acts vertically through C (Fig. 92), and 
the reaction of the wall Fj is perpendicular to the wall (Art. 97). 
These two forces intersect at D. The only remaining force, F 2 , on 



136 



Mechanics for Engineers 



the ladder is the pressure which the ground exerts on it at B. This 
must act through D also (Art. 99), and therefore its line of action 

must be BD. F l may be found by 
an equation of the moments about B. 

Ft x AE = 40 x |BE. 




- 7 2 ) = 40 x | 



V(275) 
= 8-44 Ibs. 

And since F 2 balances the horizontal 
force of 8'44 Ibs. and a vertical force 
of 40 Ibs. 



F 2 = V{(8'44) 2 + 4Q 2 } = 40-8 Ibs. 
and is inclined to EB at an angle 

^ 

EBD, the tangent of which is 



AE 



_ 2 x ^(275) _ .._. 
~ - 



which is the tangent of 78-1. 

A second method of solving the problem consists in drawing a 
vector triangle, abc (Fig. 92), representing by its vector sides Fj, 
F 2 , and 40 Ibs. The 4o-lb. force ab being set off to scale, and be 
and ca being drawn parallel to F 2 and F l respectively, and the 
magnitudes then measured to the same scale. A third method 
consists (without drawing to scale) of solving the triangle abc 
trigonometrically, thus 

Fj : F 2 : 40 = ca : cb '. ab 

=HB:BD:HD 



from which Fj and F 2 may be easily calculated, viz. 

7 -^ 8-44 ibs. 



2 X 



F 2 = 40 x = 40-8 Ibs. 

\/275 

Fourthly, the problem might be solved very simply by resolving 
the forces F t and F 2 and 40 Ibs. horizontally and vertically, as in 



Statics of Rigid Bodies 



137 



this particular case the 4o-lb. weight has no component in the 
direction of F,, and must exactly equal in magnitude the vertical 
component of F 2 ; the horizontal component of F 2 must also be 
just equal to the magnitude of Fj. 

Example 2. A light bar, AB, 20 inches long, is hinged at A 
50 as to be free to move in a vertical plane. The end B is sup- 
ported by a cord, BC, so placed that the angle ABC is 145 and 
AB is horizontal. A weight of 7 Ibs. is hung on the bar at a 
point D in AB 13 inches from A. Find the tension in the cord 
ind the pressure of the rod on the hinge. 




FIG. 93. 

Let T be the tension in the cord, and P be the pressure on the 
linge. 

Taking moments about A, through which P passes (Fig. 93) 

T x AF = 7 x AD 
T x 20 sin 35 = 7 x 13 
U-47T = 91 

T = 7-94 Ibs. 

The remaining force on the bar is the reaction of the hinge, 
vhich is equal and opposite to the pressure P of the bar on the 
linge. 

The vertical upward component of this is 7 T sin 35 
= 2*45 Ibs., and the horizontal component is T cos 35 = 6-5 Ibs. 



Hence P = V(6'5) 2 + (2 45) 2 = 6-93 Ibs. 

The tangent of the angle DAE is -~ = 0^377, corresponding 

o an angle of 20 40'. 

The pressure of the bar on the hinge is then 6 93 Ibs. in a 



138 Mechanics for Engineers 

direction, AE, inclined downwards to the bar and making an angle 
20 40' with its length. 

EXAMPLES XIII. 

1. A trap door 3 feet square is held at an inclination of 30 to (and 
above) the horizontal plane through its hinges by a cord attached to the 
middle of the side opposite the hinges. The other end of the cord, which 
is 5 feet long, is attached to a hook vertically above the middle point of the 
hinged side of the door. Find the tension in the cord, and the direction 
and magnitude of the pressure between the door and its hinges, the weight 
of the door being 50 Ibs., which may be taken as acting at the centre of the 
door. 

2. A ladder 20 feet long rests on rough ground, leaning against a rough 
vertical wall, and makes an angle of 60 to the horizontal. The weight of 
the ladder is 60 Ibs. , and this may be taken as acting at a point 9 feet from 
the lower end. The coefficient of friction between the ladder and ground 
is o'25. If the ladder is just about to slip downwards, find the coefficient 
of friction between it and the wall. 

3. A ladder, the weight of which may be taken as acting at its centre, 
rests against a vertical wall with its lower end on the ground. The 
coefficient of friction between the ladder and the ground is , and that 
between the ladder and the wall \. What is the greatest angle to the 
vertical at which the ladder will rest ? 

4. A rod 3 feet long is hinged by a horizontal pin at one end, and 
supported on a horizontal roller at the other. A force of 20 Ibs. inclined 
45 to the rod acts upon it at a point 21 inches from the hinged end. Find 
the amount of the reactions on the rod at the hinge and at the free end. 

5. A triangular roof-frame ABC has a horizontal span AC of 40 feet, 
and the angle at the apex B is 120, AB and BC being of equal length. 
The roof is hinged at A, and simply supported on rollers at C. The loads 
it bears are as follow : (l) A force of 4000 Ibs. midway along and perpen- 
dicular to AB ; (2) a vertical load of 1500 Ibs. at B ; and (3) a vertical 
load of 1400 Ibs. midway between B and C. Find the reactions or 
supporting forces on the roof at A and C. 

6. Draw a 2-inch square ABCD, and find the middle point E of AB. 
Forces of 17, 10, 8, 7, and 20 Ibs. act in the directions CB, AB, EC, ED, 
and BD respectively. Find the magnitude, direction, and position of the 
force required to balance these. Where does it cut the line AD, and what 
angle does it make with the direction AD ? 

7. A triangular roof-frame ABC has a span AC of 30 feet. AB is 15 
feet, and BC is 24 feet. A force of 2 tons acts normally to AB at its 
middle point, and another force of I ton, perpendicular to AB, acts at B. 
There is also a vertical load of 5 tons acting downward at B. If the sup- 
porting force at A is a vertical one, find its magnitude and the magnitude 
and direction of the supporting force at C. 



Statics of Rigid Bodies 



139 



8. A jointed roof-frame, ABCDE, is shown in Fig. 94. AB and BC are 
inclined to the horizontal at 30, EB and DB are inclined at 45 to the 




FIG. 94. 

horizontal. The span AC is 40 feet, and B is 10 feet vertically above ED. 
Vertical downward loads of 2 tons each are carried at B, at E, and at D. 
Find by the method of sections the forces in the members AB, EB, 
and ED. 

9. A jointed structure, ACD . . . LMB (Fig. 95) is built up of bars all 




FIG. 95. 

of equal length, and carries loads of 7, 10, and 15 tons at D, F and L re- 
spectively. Find by the method of sections the forces on the bars EF, EG, 
and DF. 



CHAPTER VII 

CENTRE OF INERTIA OR MASS CENTRE OF 
GRA VITY 

100. Centre of a System of Parallel Forces. Let 

A, B, C, D, E, etc. (Fig. 96), be points at which parallel forces 
FI> F 2 , F 3 , F 4 , F 5 , etc., respectively act. The position of the 
resultant force may be found by applying successively the rule 




FIG. 96. 

of Art. 86. Thus Fj and F 2 may be replaced by a force 

AX F 

FJ + F 2 , at a point X in AB such that ^^ = =? /Art. 86). 

Ar> I 1 ! 

This force acting at X, and the force F 3 acting at C, may 
be replaced by a force F! + F 2 + F 3 at a point Y in CX such 
XY F 



Centre of Inertia or Mass Centre of Gravity 141 

Proceeding in this way to combine the resultant of several 
forces with one more force, the whole system may be replaced 
by a force equal to the algebraic sum of the several forces 
acting at some point G. It may be noticed that the positions 
of the points X, Y, Z, and G depend only upon the positions 
of the points of application A, B, C, D, and E of the several 
forces and the magnitude of the forces, and are independent of 
the directions of the forces provided they are parallel. The 
point of application G of the resultant is called the centre of the 
parallel forces F 1} F 2 , F 3 , F 4 , and F 5 acting through A, B, C, D, 
and E respectively, whatever direction those parallel forces may 
have. 

101. Centre of Mass. If every particle of matter in a 
body be acted upon by a force proportional to its mass, and 
all the forces be parallel, the centre of such a system of forces 
(Art. 100) is called the centre of mass or centre of inertia of 
the body. It is quite independent of the direction of the 
parallel forces, as we have seen in Art. 100. 

Centre of Gravity. The attraction which the earth 
exerts upon every particle of a body is directed towards the 
centre of the earth, and in bodies of sizes which are small 
compared to that of the earth, these forces may be looked 
upon as parallel forces. Hence these gravitational forces have 
a centre, and this is called the centre of gravity of the body ; it 
is, of course, the same point as the centre of mass. 

The resultant of the gravitational forces on all the particles 
of a body is called its weight, and in the case of rigid bodies it 
acts through the point G, the centre of gravity, whatever the 
position of the body. A change of position of the body is 
equivalent to a change in direction of the parallel gravitational 
forces on its parts, and we have seen (Art. 100) that the centre 
of such a system of forces is independent of their direction. 
We now proceed to find the centres of gravity in a number of 
special cases. 

1 02. Centre of gravity of two particles of given weights at 
a given distance apart, or of two bodies the centres of gravity 
and weights of which are given. 

Let A and B (Fig. 97) be the positions of the two particles 



142 Mechanics for Engineers 

(or centres of gravity of two bodies) of weights w^ and w 2 
A G B 

FIG. 97. 

respectively. The centre of gravity G is (Art. 86) in AB at 
such a point that 

GA w 



or GA = 



and GB = 



. AB 



.AB 



In the case of two equal weights, AG = GB = AB. 

A convenient method of finding the point G graphically 

may be noticed. Set off from A (Fig. 98) a line AC, making 

any angle with AB (preferably 
at right angles), and proportional 
to w z to any scale; from B set 
off a line BD parallel to AC 
on the opposite side of AB, and 
proportional to w l to the same 
scale that AC represents w z . 
Join CD. Then the intersection 
of CD with AB determines the 

point G. The proof follows simply from the similarity of the 

triangles ACG and BDG. 

103. Uniform Straight Thin Rod. Let AB (Fig. 99) 

be the uniform straight rod of length AB : it may be supposed 

to be divided into pairs of particles of equal weight situated at 

equal distances from the middle 
G b D point G of the rod, since there 

will be as many such particles 
between A and G as between G 




<B 



FIG. 99. 



and B. The e.g. (centre of gravity) of each pair, such as the 
particles at a and l>, is midway between them (Art. 102), viz. 
at the middle point of the rod, G, hence the e.g. of the whole 
rod is at its middle point, G. 



Centre of Inertia or Mass Centre of Gravity 143 




B 1 

FIG, 100. 



104. Uniform Triangular Plate or Lamina. The term 
centre of gravity of an area is often used to denote the e.g. 
of a thin lamina of uniform material cut in the shape of the 
particular area concerned. 

We may suppose the lamina ABC (Fig. 100) divided into 
an indefinitely large number of strips parallel to the base AC. 
The e.g. of each strip, such as PQ, 
is at its middle point (Art. 103), 
and every e.g. is therefore in the 
median BB,' i.e. the line joining B 
to the mid-point B' of the base 
AC. Hence the e.g. of the whole 
triangular lamina is in the median 
BB'. Similarly, the e.g. of the 
lamina is in the medians AA' and 
CC. Hence the e.g. of the triangle is at G, the intersection 
of the three medians, which are concurrent, meeting at a point 
distant from any vertex of the triangle by f of the median 
through it. The perpendicular distance of G from any side 
of the triangle is \ of the perpendicular distance of the oppo- 
site vertex from that side. 

Note that the e.g. of the triangular area ABC coincides 
with that of three equal particles placed at A, B, and C. For 
those at A and C are statically equivalent to two at B', and 
the e.g. of two at B' and one at B is at G, which divides BB' in 
the ratio 2 : i, or such at B'G = BB' (Art. 102). 

Uniform Parallelogram. If a lamina be cut in the 
shape of a parallelogram, 
ABCD (Fig. 101), the e.g. of 
the triangle ABC is in OB, 
and that of the triangle ADC 
is in OD, therefore the e.g. of 
the whole is in BD. Similarly 
it is in AC, and therefore it is 
at the intersection O. 

105. Rectilinear Figures in General. The e.g. of any 
lamina with straight sides may be found by dividing its area up 
into triangles, and finding the e.g. and area of each triangle. 




FIG. roi. 



144 



Mechanics for Engineers 



Thus, in Fig. 102, if G 1? G 2 , and G 3 are the centres of gravity 
of the triangles ABE, EBD, and 
DEC respectively, the e.g. of 
the area ABDE is at G 4 , which 
divides the length GiG 2 inversely 
as the weights of the triangles 
AEB and EDB, and therefore 
inversely as their areas. Simi- 
larly, the e.g. G of the whole 
figure ABCDE divides G 3 G 4 in- 
versely as the areas of the figures 
ABDE and BCD. The inverse 
division of the lines G : G 2 and 

of G 3 G 4 may in practice be performed by the graphical method 

of Art. 102. 

106. Symmetrical Figures. If a plane figure has an 

axis of symmetry, i.e. if a straight line can be drawn dividing it 




FIG. 102. 




FIG. 103. 



into two exactly similar halves, the e.g. of the area of the figure 
lies in the axis of symmetry. For the area can be divided into 
indefinitely narrow strips, the e.g. of each of which is in the axis 
of symmetry (see Fig. 103). If a figure has two or more axes of 




FIG. 104. 



symmetry, the e.g. must lie in each, hence it is at their intersection, 
e.g. the e.g. of a circular area is at its centre. Other examples, 
which sufficiently explain themselves, are shown in Fig. 104. 



Centre of Inertia or Mass Centre of Gravity 145 

107. Lamina or Solid from which a Part has been 
removed. Fig. 105 represents a lamina from which a piece, 
B, has been cut. The centre of gravity of the whole lamina, 
including the piece B, is 
at G, and the e.g. of the 
removed portion B is at g, 
The area of the remaining 
piece A is a units, and 
that of the piece B is b 
units. It is required to 
find the e.g. of the remain- 
ing piece A. 

Let G' be the required FIG. 105. 

e.g. ; then G is the e.g. of two bodies the centres of gravity of 
which are at G' and g, and which are proportional to a and b 
respectively. Hence G is in the line G'g, and is such that 
GG' :Gg:: b : a (Art. 102) 




That is, the e.g. G' of the piece A is in the same straight 
line gG as the two centres of gravity of the whole and the part 

b 
B, at times their distance apart beyond the e.g. of the whole 

lamina. The point G' divides the line G^ externally in the 
ratio ^ or G'G ; G'g:: b \ a + b. 

The same method is ap- 
plicable if A is part of a solid 
from which a part B has been 
removed, provided a repre- 
sents the weight of the part A, 
and b that of the part B. 

Graphical Construc- 
tion. The e.g. of the part A 
may be found as follows : 
from g draw a line gP (Fig. 
1 06) at any angle (preferably 

at right angles) to Gg and proportional to a -f b. From G 

L 




146 



Mechanics for Engineers 



draw GQ parallel to p and proportional to b. Join PQ, and 
produce to meet gG produced in G'. Then G' is the e.g. of 
the part A. 

1 08. Symmetrical Solids of Uniform Material. If a 

solid is symmetrical about one plane, i.e. if it can be divided 
by a plane into two exactly similar halves, the e.g. evidently 
lies in the plane, for the solid can be divided into lamina? the 




FIG. 107. 

e.g. of each of which is in the plane of symmetry. Similarly, 
if the solid has two planes of symmetry, the e.g. must lie in the 
intersection of the two planes, which is an axis of the solid, as 
in Fig. 107. 

If a solid has three planes of symmetry, the line of inter- 
section of any two of them meets the third in the e.g., which is 




FIG. 108. 



a point common to all three planes, e.g. the sphere, cylinder, 
etc. (see Fig. 108). 

109. Four Equal Particles not in the Same Plane. 

Let ABCD (Fig. 109) be the positions of the four equal 
particles. Join ABCD, forming a triangular pyramid or tetrahe- 
dron. The e.g. of the three particles at A, B, and C is at D', 
the e.g. of the triangle ABC (Art. 104). Hence the e.g. of the 
four particles is at G in DD', and is such that 

D'G : GD = i : 3 (Art. 102) 
or D'G = i DD' 



Centre of Inertia or Mass Centre of Gravity 147 




FIG. 109. 



Similarly, the e.g. of the four particles is in AA', BB', and CC', 
the lines (which are concurrent) 
joining A, B, and C to the centres 
of gravity of the triangles BCD, 
ACD, and ABD respectively. The 
distance of the e.g. from any face 
of the tetrahedron is \ of the per- 
pendicular distance of the opposite 
vertex from that face. 

no. Triangular Pyramid 
or Tetrahedron of Uniform 
Material. Let ABCD (Fig. no) be the triangular pyramid. 
Suppose the solid divided into indefinitely thin plates, such 
as abc, by planes parallel to the face ABC. Let D' be the 
e.g. of the area ABC. 
Then DD' will intersect 
the plate abc at its e.g., 
viz. at d, and the e.g. of 
every plate, and there- 
fore of the whole solid, 
will be in DD'. Simi- 
larly, it will be in AA', 
BB', and CC', where A', 
B', and C' are the centres 
of gravity of the triangles 
BCD, CDA, and DAB 
respectively. Hence the centre of gravity of the whole solid 
coincides with that of four equal particles placed at its vertices 
(Art. 109), and it is in DD', and distant \ DD' from D', in CC' 
and 5 CC' from C', and so on. It is, therefore, also distant 
from any face, \ of the perpendicular distance of the opposite 
vertex from that face. 

in. Uniform Pyramid or Cone on a Plane Base. 
If V (Fig. in) is the vertex of the cone, and V the e.g. of the 
base of the cone, the e.g. of any parallel section or lamina into 
which the solid may be divided by plates parallel to the base, will 
be in VV. Also if the base be divided into an indefinitely large 
number of indefinitely small triangles, the solid is made up of 




FIG. no. 



148 



Mechanics for Engineers 



an indefinitely large number of triangular pyramids having the 

triangles as bases and a common vertex, V. The e.g. of each 

small pyramid is distant from 
V f of the distance from its 
base to V. Hence the centres 
of gravity of all the pyramids 
lie in a plane parallel to the 
base, and distant from the 
vertex, f of the altitude of 
the cone. 

The e.g. of a right circular 
cone is therefore in its axis, 
which is the intersection of two 
planes of symmetry (Art. 108), 
and its distance from "the base 

is \ the height of the cone, or its distance from the vertex is f 

of the height of the cone. 

Example i. A solid consists of a right circular cylinder 3 feet 
long, and a right cone of altitude 2 feet, the base coinciding with 
one end of the cylinder. The cylinder and cone are made of the 
same uniform material. Find the e.g. of the solid. 

If r = radius of the cylinder in feet 

the volume of cylinder _ irr 2 x 3 _ 9 
volume of cone irr 2 x ^ x 2 2 

hence the weight of the cylinder is 4*5 times that of the cone. 

The e.g. of the cylinder is at A (Fig. 112), the mid-point of its 
axis (Art. 108), i.e. i'5 feet from the plane of the base of the cone. 




FIG. in. 




AC 




FIG. 112. 

The e.g. of the cone is at B, \ of the altitude from the base 
(Art. m), i.e. 0^5 foot from the common base of the cylinder and 
cone. Hence 

AB = AD + DB = 1-5 + 0-5=2 feet 

2 



And G is therefore in AB, at a distance 



2+9 



.AB from A 



(Art. 102), i.e. AG = ^ of 2 feet = j^ foot, or 4-36 inches. 



Centre of Inertia or Mass Centre of Gravity 149 

Example 2. A quadrilateral consists of two isosceles triangles 
on opposite sides of a base 8 inches 
long. The larger triangle has two 
equal sides each 7 inches long, and 
the smaller has its vertex 3 inches 
from the 8-inch base. Find the dis- 
tance of the e.g. of the quadrilateral 
from its 8-inch diagonal. 

Let ABCD (Fig. 113) be the 
quadrilateral, AC being the 8-inch 
diagonal, of which E is the mid- 
point ; then 

ED = 3 inches 
EB= 7^- : 4 2 




inches 



FIG. 113. 



The e.g. of the triangle ABC is in EB and \ EB from E ; or, 
if Gj is the e.g. 



EGj = 



= 1-915 inches 



Similarly, if G 2 is the e.g. of the triangle ADC 

EG 2 = i of 3 inches = I inch 
therefore GjG 2 = 1-915 + i = 2*915 inches 

This length is divided by G, the e.g. of the quadrilateral, so 
that 

G 2 G _ area of triangle ABC _ BE _ I'9i5 
GjG area of triangle ACD ED I 
G 2 G _ 1*915 _ 1*915 
G!^ I + 1-915 2-915 
G 2 G = 1-915 inches 
and EG = G 2 G G 2 E = 1*915 I = 0-915 inch 

which is the dista'nce of the e.g. from the 8-inch diagonal. 

Example 3. A pulley weighs 25 Ibs., and it is found that the 
e.g. is 0*024 inch from the centre of the pulley. The pulley is 
required to have its e.g. at the geometrical centre of the rim, and 
to correct the error in its position a hole is drilled in the pulley 
with its centre 6 inches from the pulley centre and in the same 
diameter as the wrongly placed e.g. How much metal should be 
removed by drilling ? 

Let x be the weight of metal to be removed, in pounds. 



150 Mechanics for Engineers 

Then, in Fig. 114, OA being 6 inches and OG 0*024 inch, the 
removed weight x Ibs. having its e.g. at A, and the remaining 



FIG. 114. 

25-.tr Ibs. having its e.g. at O, the e.g. G of the two together divides 
OA, so that 

r\r. 



OG _ x 
GA 25 -"x 
OG _ _^ 
)r OA 25 



25 x OG 0-024 

hence x - ^QT = 25 x ~ - o'i 



Ib. 



EXAMPLES XIV. 

1. A uniform beam weighing 180 Ibs. is 12 feet long. It carries a 
load of 1000 Ibs. uniformly spread over 7 feet of its length, beginning 
I foot from one end and extending to a point 4 feet from the other. Find 
at what part of the beam a single prop would be sufficient to support it. 

2. A lever 4 feet long, weighing 15 Ibs., but of varying cross-section, 
is kept in equilibrium on a knife-edge midway between its ends by the 
application of a downward force of I '3 Ibs. at its lighter end. How far is 
the e.g. of the lever from the knife-edge? 

3. The heavy lever of a testing machine weighs 2500 Ibs., and is poised 
horizontally on a knife-edge. It sustains a downward pull of 4 tons 
3 inches from the knife-edge, and carries a load of I ton on the same side 
of the knife-edge and 36 inches from it. How far is the e.g. of the lever 
from the knife-edge ? 

4. A table in the shape of an equilateral triangle, ABC, of 5 feet sides, 
has various articles placed upon its top, and the legs at A, B, and C then 
exert pressures of 30, 36, and 40 Ibs. respectively on the floor. Determine 
the position of the e.g. of the table loaded, and state its horizontal distances 
from the sides AB and BC . 

5. Weights of 7, 9, and 12 Ibs. are placed in the vertices A, B, and C 
respectively of a triangular plate of metal weighing 10 Ibs., the dimensions 
of which are, AB 16 inches, AC 16 inches, and BC n inches. Find the 
e.g. of the plate and weights, and state its distances from AB and BC. 

6. One-eighth of a board 2 feet square is removed by a straight saw-cut 
through the middle points of two adjacent sides. Determine the distance 
of the e.g. of the remaining portion from the saw-cut. If the whole board 
before part was removed weighed 16 Ibs., what vertical upward force 



Centre of Inertia or Mass Centre of Gravity 151 

applied at the corner diagonally opposite the saw cut would be sufficient to 
tilt the remaining \ of the board out of a horizontal position, if it turned 
about the line of the saw-cut as a hinge ? 

7. An isosceles triangle, ABC, having AB 10 inches, AC 10 inches, and 
base BC 4 inches long, has a triangular portion cut off by a line DE, 
parallel to the base BC, and 7 '5 inches from it, meeting AB and AC in D 
and E respectively. Find the e.g. of the trapezium BDEC, and state its 
distance from the base BC. 

8. The lever of a testing-machine is 15 feet long, and is poised on 
a knife-edge 5 feet from one end and 10 feet from the other, and in a 
horizontal line, above and below which the beam is symmetrical. The 
beam is 16 inches deep at the knife-edge, and tapers uniformly to depths 
of 9 inches at each end ; the width of the beam is the same throughout its 
length. Find the distance of the e.g. of the beam from the knife-edge. 

9. A retaining wall 5 feet high is vertical in front and 9 inches thick 
at the top. The back of the wall slopes uniformly, so that the thickness of 
the wall at the base is 2 feet 3 inches. Find the e.g. of the cross-section of 
the wall, and state its horizontal distance from the vertical face of the 
wall. 

10. What is the moment of the weight of the wall in Question 9 per 
foot length, about the back edge of the base, the weight of the material 
being 120 Ibs. per cubic foot ? What uniform horizontal pressure per 
square foot acting on the vertical face of the wall would be sufficient to 
turn it over bodily about the back edge of the base ? 

1 1 . The casting for a gas-engine piston may be taken approximately 
as a hollow cylinder of uniform thickness of shell and one flat end of uniform 
thickness. Find the e.g. of such a casting if the external diameter is 8 
inches, the thickness of shell \ inch, that of the end 3 inches, and the 
length over all 20 inches. State its distance from the open end. 

12. A solid circular cone stands on a base 14 inches diameter, and its 
altitude is 20 inches. From the top of this a cone is cut having a base 
3-5 inches diameter, by a plane parallel to the base. Find the distance of 
the e.g. of the remaining frustum of the cone from its base. 

13. Suppose that in the rough, the metal for making a gun consists 
of a frustum of a cone, 10 feet long, 8 inches diameter at one end, and 
6 inches at the other, through which there is a cylindrical hole 3 inches 
diameter, the axes of the barrel and cone being coincident. How far from 
the larger end must this piece of metal be slung on a crane in order to 
remain horizontal when lifted ? 

14. A pulley weighing 40 Ibs. has its e.g. 0-04 inch from its centre. 
This defect is to be rectified by drilling a hole on the heavy side of the 
pulley, with its centre 9 inches from the centre of the pulley and in 
the radial direction of the centre of gravity. What weight of metal should 
be drilled out ? 

15. A cast-iron pulley weighs 45 Ibs., and has its e.g. 0*035 mcn from 
its centre. In order to make the e.g. coincide with the centre of the 



152 



Mechanics for Engineers 



pulley, metal is added to the light side at a distance of 8 inches from the 
centre of the pulley and in line with the e.g. What additional weight 
is required in this position ? If the weight is added by drilling a hole in 
the pulley and then filling it up to the original surface with lead, how much 
iron should be removed, the specific gravity of lead being ii'35, and that of 
iron being 7-5 ? 

112. Distance from a Fixed Line of the Centre of 
Gravity of Two Particles, or Two Bodies, the Centres 

of Gravity of which are given. 

Let A (Fig. 1 15) be the position 
of a particle of weight w lt and let 
B be that of a particle of weight 
w. 2 , or, if the two bodies are of 
finite size, let A and B be the 
positions of their centres of gravity. 
Then the centre of gravity of the 




M 



N Q 

FIG. 115. 

two weights w l and w. 2 is at G in AB such that 
AG 



or AG = 



.AB 



and GB = 



AB 



Let the distances of A, B, and G from the line NM be 
x lt x 2 , and x respectively, the line NM being in a plane through 
the line AB. Then AN = x lt BM = x 2 , and GQ = x. 

GR AG w z 

Now ' ES = AB = ^+^ 



or GR = 



- . BS 



and 



GQ or x = RQ + GR = AN + *!* 

*&i ~r 



hence x = x l 



iu n 



, 




BS 



Distance of the e.g. from a Plane. If ^ 

the respective distances of A and B from any plane, then NM 



Centre of Inertia or Mass Centre of Gravity 153 

may be looked upon as the line joining the feet of perpen- 
diculars from A and B upon that plane. Then the distance x 
of G from that plane is 



_ 



This length x is also called the mean distance of the two 
bodies or particles from the plane. 

113. Distance of the e.g. of Several Bodies or of 
One Complex Body from a Plane. 

Let A, B, C, D, and E (Fig. 116) be the positions of 5 par- 
ticles weighing w lt w 2 , u> 3 , ze/ 4 , and w & respectively, or the 




FIG. 116. 

centres of gravity of five bodies (or parts of one body) of those 
weights. 

Let the distances of A, B, C, D, and E from some fixed 
plane be x lt x 2 , x 3 , x 4 , and x 5 respectively, and let the weights 
in those positions be w lt w. 2 , w 3 , w 4 , and w & respectively. It is 
required to find the distance x of the e.g. of these five weights 
from the plane. We may conveniently consider the plane to 



154 Mechanics for Engineers 

be a horizontal one, but this is not essential ; then x lt x%, x 3 , 
x 4 , and x 5 are the vertical heights of A, B, C, D, and E respec- 
tively above the plane. Let a, b, c, d, and e be the projections 
or feet of perpendiculars from A, B, C, D, and E respectively 
on the plane, so that Aa, B, O, ~Dd, and ~Ee are equal to x lt x^ 
x 3 , 0.4, and x s respectively. 

Let G! be the e.g. of w^ and > 2 , and let i be its projection 
by a vertical line on the plane ; then 

, . f .. 

(Art. 112. (i)) 
l 2 

Let G 2 be the e.g. of (u\ 4- w 2 ) and w 3 , and let g% be its 
projection by a vertical line on the plane ; then G 2 divides 
G X C so that 

1V 3 
GiG.) = 7 ; ^ ,~ GiC 

l 



Z/. 

and G^ = ~ ^~+T -* (Art. 112. (i) ) 

and substituting the above value of Gii 



Similarly, if G 3 is the e.g. of w lt w. lt w 3 , and ze/ 4 , and g 3 is its 
projection on the plane, then 



4- ss 

> and so on 



and finally 



Gg or x = - -* . . (2) 

Wi 4~ w.2 4~ ^'3 ~r w/4 4~ if 

which may be written 



where 2 stands for " the sum of all such terms as." If any 
of the points A, B, C, etc., are below the plane, their distances 
from the plane must be reckoned as negative. 



Centre of Inertia or Mass Centre of Gravity 155 



Plane -moments. The products u\x^ w^x.^ #ve 3 , etc., 
are sometimes called plane-moments of the weights of the 
bodies about the plane considered. The plane-moment of a 
body about any given plane is then the weight of the body 
multiplied by the distance of its e.g. from that plane. 

Then in words the relation (3) may be stated as follows : 
" The distance of the e.g. of several bodies (or of a body 
divided into parts) from any plane is equal to the algebraic 
sum of their several plane-moments about that plane, divided 
by the sum of their weights." 

And since by (3), x X 2(w) = 2(w), we may state that the 
plane-moment of a number of weights (or forces) is equal to 
the sum of their several plane-moments. 

This statement extends to plane-moments the statement 
in Art. 90, that the moment of the sum of several forces about 
any point is equal to the sum of the moments of the forces 
about that point. 

It should be remembered that a horizontal plane was chosen 
for convenience only, and that the formulae (2) and (3) hold 
good for distances from any plane. 

114. Distance of the e.g. of an Area or Lamina 
from a Line in its Plane. 

This is a particular case of the problem of the last article. 
Suppose the points A, B, C, D, and E in the last article and 
Fig. 116 all lie in one plane perpendicular to the horizontal 
plane, from which their distances are x lt x. 2 , x 3 , x 4 , and x s 
respectively. Then their projections a, l>, c, d, and e on the 
horizontal plane all lie in a straight line, which is the inter- 
section of the plane containing A, B, C, D, and E with the 
horizontal plane, viz. the line OM in Fig. 117. 

Thus, if x u x. u x 3) etc., be the distances of the centres of 
gravity of several bodies all in the same plane (or parts of 
a lamina) from a fixed line OM in this plane, then the 
distance of the e.g. of the bodies (or lamina?) from the line 
being x 



. . . , etc. _ ^(wx) 
>, 4- . . , etc. " So/ 



156 Mechanics for Engineers 

This formula may be used to find the position of the e.g. of 



G, 




a lamina or area by finding its distance from two non-parallel 
fixed lines in its plane. 

If the lamina is of irregular shape, as in Fig. 118, the dis- 
tance of its e.g. from a line OM in its plane may be found 

approximately by dividing 
it into a number of narrow 
strips of equal width by lines 
parallel to OM, and taking 
the e.g. of each strip as 
being midway between the 
parallel boundary-lines. The 
weight of any strip being 
denoted by w 

FIG. us. w = volume of strip X D 



M 



where D = weight of unit volume of the material of the lamina, 



or- 



w = area of strip X thickness of lamina X D 



If the weight of the first, second, third, and fourth strips be 
/!, w 2 , w 3 , and w 4 respectively, and so on, and their areas be 
a lt a 2) a 3 , and a respectively, the lamina consisting of a material 
of uniform thickness /, then u\ = aj. D, w z = a./. D, and 



Centre of Inertia or Mass Centre of Gravity 157 

so on. And if x is the distance of the e.g. of the area from 
OM, then by equation (4) 



. . . , etc. 



w 3 4- etc. 

h . . . i etc. 



(5) 



4- a 3 tD 4- j etc. 
or, dividing numerator and denominator by the factor /D 

x 4 4- ... , etc. 






83X3 



#1 + 2 + 3 4- <* 4 + . o . , etc. 

= SM or SM 



(6) 



where A = total area of the lamina, and 2 has the same 
meaning as in (3), Art. 113. 

Similarly, the distance of the e.g. of the area A from 
another straight line may 
be found, and then the 
position of the e.g. is 
completely determined. 

Thus in Fig. 119, if 
x is the distance of the 
e.g. of the lamina from 
OM, and y is its distance 
from ON, by drawing two 
lines, PR and QS, parallel 
to OM and ON and dis- 
tant "x and y from them 
respectively, the inter- 
section G of the two lines gives the e.g. of the lamina or 
area. 

Moment of an Area. The products a^ lt etc., may be 
called moments of the areas a lt etc. 

Regular Areas. If a lamina consists of several parts, the 
centres of gravity of which are known, the division into thin 
strips adopted as an approximate method for irregular figures 




FIG. 119. 



158 Mechanics for Engineers 

is unnecessary. The distance x of the e.g. from any line OM is 



^(product of each area and distance of its e.g. from OM) 
whole area 



or- 



x = 



S(plane mo. of each area about a plane perpend, to its own) 



whole area 



The product of an area and the distance of its e.g. from a 
line OM may be called the " line moment " of the area about 
OM, and we may write 

2(line moments of each part of an area) 
whole area 

For example, in Fig. 120 the area ABECD consists of a 
triangle, EEC, and a rectangle, ABCD, 
having a common side, BC. Let the 
height EF = h ; let AD = / and AB = d. 
Then the area ABCD = d x I, and the 
c area EEC = -j X / X h, and if G! is the 
e.g. of the triangle EEC, and G 2 that of 
the rectangle ABCD, the distance x of 
the e.g. of the area ABECD from AD is 
FIG. 120. found thus 




_ 



zd+h 

115. Lamina with Part removed. Suppose a lamina 
(Fig. 121) of area A has a portion of area a, removed. Let 
x = distance of e.g. G of A from a line OM in its plane ; let 
x l be the distance of the e.g. of the part a from OM; and 
let x z be the distance of the e.g. of the remainder (A a) from 
OM. 



Centre of Inertia or Mass Centre of Gravity 159 



- x^a -f- x. 2 (A. a} . 
Then* = - ^ -(Art. 114) 



x . A = 

and x 9 = 



a) 



A a 



In this way we can find the distance of the e.g. of the part 
A a from OM, and similarly we can find the distance from 




M 



FIG. 121. 

any other line in its plane, and so completely determine its 
position as in Art. 114. This method is applicable particularly 
to regular areas. 

1 1 6. Solid with Part removed. The method used in 
the last article to find the e.g. of part of a lamina is applicable 
to a solid of which part has 
been removed. 

If in Fig. 122 A is a 
solid of weight W, and a 
portion B weighing w is re- 
moved, the distance of the 
e.g. of the remainder (W w) 
from any plane is x. 2 where 

rW -x ?e> 

*A- V V ^^ i-vitc' 




M 



W w 



FIG. 122. 



by (i) Art. 112 and the method of Art. 115, where x = distance 
of e.g. of A from the plane, and x 1 = distance of e.g. of B from 
the plane. 



i6o 



Mechanics for Engineers 



117. Centre of Gravity of a Circular Arc. Let ABC 

(Fig. 123) be the arc, OA being the radius, equal to a units 
of length, and the length of arc ABC 
being / units. If B is the middle 
point of the arc, OB is an axis of 
symmetry, and the e.g. of the arc is 
in OB. Draw OM parallel to AC. 

Let the arc be divided into a 
B number of small portions, such as PQ, 
each of such small length as to be 
sensibly straight. Let the weight of 
the arc be w per unit length. The 
e.g. of a small portion PQ is at V, its 
mid-point. Draw VW parallel to 
OM, and join OV. Draw PR and QR 
parallel to OM and OB respectively. 
Then, if x = distance of e.g. of arc from the line OM, as in 
Art. 114 

_ 2(PQ X w X OW) _ S(PQ X OW) _ S(PQ . OW) 

'S(PQ) "~t~ 




Now, since OV, VW, and OW are respectively perpen- 
dicular to PQ, RQ, and PR, the triangles PQR and OVW are 
similar, and 

PQ_ RP 
"OV"" OW~ 

or PQ . OW = OV . RP = a . RP 
hence S(PQ . OW) = S( . RP) = S(RP) = a X AC 

and therefore 

S(PQ.OW) AC 

x = - ~ ~ 7 AC, or 7- X a 

LI & 

The e.g. of the arc then lies in OB at a point G such that 

OG = OB X - - or radius X - 

/ arc 

or, if angle AOC = 2, i.e. if angle AOB = a (radians) 

AC 2AD 2 . a sin a sin a 

OG = a X -j- = a X r = ax - - = a . 

I I a X 2tf a 



Centre of Inertia or Mass Centre of Gravity 161 



When the arc is very short, OG is very nearly equal to OB. 

118. Centre of Gravity of Circular Sector and 
Segment. Let the sector ABCO (Fig. 124) of a circle 
centred at O and of radius a, subtend 
an angle 20. at O. The sector may 
be divided into small parts, such as 
OPQ, by radial lines from O. Each 
such part is virtually triangular when 
PQ is so short as to be regarded as a 
straight line. The e.g. of the triangle 
OPQ is on the median OR, and # 
from O. Similarly, the centres of 
gravity of all the constituent triangles, 
such as PQO, lie on a concentric arc 
abc of radius and subtending an 
angle 20. at O. The e.g. of the sector coincides with 
the e.g. of the arc abc, and is therefore in OB and at a 




FIG. 124. 



distance 






- from O (Art. 117) ; e.g. the e.g. of a semi- 



circular area of radius " a " is at a distance \a - or from 

2 37T 

its straight boundary. 

The e.g. of the segment cut off by any chord AC (Fig. 124) 
may be found by the principles of Art. 115, regarding the 
segment as the remainder of the 
sector ABCO when the triangle 
AOC is removed. 

119. Centre of Gravity of 
a Zone of a Spherical Shell. 
Let ABCD (Fig. 125) be a zone 
of a spherical shell of radius a and 
thickness /, and of uniform material 
which weighs w per unit volume. 
Let the length of axis HF be /. 
Divide the zone into a number of 
equal smaller zones, such as abed, 
by planes perpendicular to the axis OE, so that each has an 
axial length //. Then the area of each small zone is the same, 

ftf 




1 62 Mechanics for Engineers 



viz. ZTraft, and the volume of each is then zirah . t, and each 
has its e.g. on the axis of symmetry OE, and midway between 
the bounding planes, such as a*/ and be, if h is indefinitely short. 
Hence the e.g. of the zone coincides with that of a large 
number of small bodies each of weight w . z-rrah . /, having their 
centres of gravity uniformly spread along the line FH. Hence 
the e.g. is at G, the mid-point of the axis FH of the zone, or 




e.g. the distance of the e.g. of a hemispherical shell from the 
plane of its rim is half the radius of the shell. 

120. Centre of Gravity of a Sector of a Sphere. Let 

OACB (Fig. 126) be a spherical sector of radius a. If the sector 

be divided into an indefinitely 
great number of equal small 
pyramids or cones having a 
common vertex O such that their 
bases together make up the base 
ACB of the sector, the c.g.'s of 
the equal pyramids will each be 
|<7 from O, and will therefore be 

evenly spread over a portion acb (similar to the surface ACB) 
of a spherical surface centred at O and of radius \a. The e.g. 
of the sector then coincides with that of a zone, acb, of a thin 
spherical shell of radius f a, and is midway between c and the 
plane of the boundary circle ab, i.e. midway between d and c. 

Solid Hemisphere. The hemisphere is a particular case 
of a spherical sector, and its e.g. will coincide with that of a 
hemispherical shell of f, where a is the radius of the solid 
hemisphere. This is a point on the axis of the solid hemisphere, 
and half of #, or \a from its base. 

Example i. The base of a frustum of a cone is 10 inches 
diameter, and the smaller end is 6 inches diameter, the height 
being 8 inches. A co-axial cylindrical hole, 4 inches diameter, 
is bored through the frustum. Find the distance of the e.g. of the 
remaining solid from the plane of its base. 

The solid of which the c g. is required is the remaining portion 



Centre of Inertia or Mass Centre of Gravity 163 




of a cone, ABC (Fig. 127), when the upper cone, DBE, and a 
cylinder, FGKH, have been removed. 

Since the cone diameter decreases 4 inches in a height of 
8 inches 

The height BM = 8 + 8xf = 2O inches 

and the e.g. of the conei . , .. 
AT ,~ . . , >= 5 inches from AC 

ABC is J x 20 inches j 

volume of cone ABC = n- . (5)2 . 2 .^- = TT . 5-Q& 
cubic inches 

distance from AC of e.g. 1 . , 

c i- j -c-rvu I = '* 4 inches 
of cylinder FGKH ) 

volume of cylinder) , . 

FGKH }=T.2.8 = 3 2T cublc 

inches 
volume of cone DBE = TT . 3 2 . * = 36^- cubic 

inches 
distance from AC of e.g. \ , 12 . , 

f 1^-0 T? ( = + 4 = JI WChCS 

of cone DBE 
then volume of remaining frustum is 

TT( A | Q 32 - 36) = TT . 2 fi cubic inches 

Let h = height of e.g. of this remainder from the base. 

Then equating the plane-moments about the base of the three 
solids, BDE, FGKH, and the remainder of frustum, to the plane- 
moment of the whole cone (Art. 1 13) (and leaving out of both sides 
of equation the common factor weight per unit volume) 

TT . S X 5 = rr{(32 X 4) + (36 X II) + (^ X //)} 

833'3 = 524 + -l~h 

h = -2 He x 39"3 = 3 -I 35 inches 

Example 2. An I-section of a girder is made up of three 
rectangles, viz. two flanges having their long sides horizontal, and 
one web connecting them having its long side vertical. The top 
flange section is 6 inches by i inch, and that of the bottom flange 
is 12 inches by 2 inches. The web section is 8 inches deep and 
i inch broad. Find the height of the e.g. of the area of cross-section 
from the bottom of the lower flange. 

Fig. 128 represents the section of the girder. 

Let x = height of the e.g. of the whole section. 

The height of the e.g. of BCDE is i inch above BE ; 

FGHKis 2 + f = 6 inches above BE ; 

LMNP is 2 + 8 + = 10-5 inches above BE. 



1 64 



Mechanics for Engineers 



Equating the sum of the moments of these three areas about 
A to the moment of the whole figure about A, we have 

(12 x 2)1 + (8 x i)6 + (6 x 1)10-5 = ~r{(i2X2) + (8xi) + (6xi)} 
24 + 48 + 63 = I<24 + 8 + 6) 

~* = W = 3' 5 5 inches 



;*- 



B A 

* / 2' 



FIG. 128. 



which is the distance of the c g. from the bottom of the lower 
flange. 

Example 3. Find the e.g. of a cast-iron eccentric consisting 
of a short cylinder 8 inches in 
diameter, having through it a cylin- 
drical hole 2-5 inches diameter, the 
axis of the hole being parallel to 
that of the eccentric and 2 inches 
from it. State the distance of the 
e.g. of the eccentric from its centre. 

This is equivalent to finding the 
e.g. of the area of a circular lamina 
with a circular hole through it. In 
Fig. 129 

AB = 8 inches CD = 2 inches 
EF = 2'5 inches 

Let the distance of the e.g. from A be x. 

If the hole were filled with the same material as the remainder 
of the solid, the e.g. of the \\ hole would be at C, its centre. 




S- 2 
FIG. 129. 



Centre of Inertia or Mass Centre of Gravity 165 

Equating moments of parts and the whole about A 

AC x (area of circle AB) = (AD x area of circle EF) 

+ (x x area of eccentric) 
4X 64 = 6x6 25+ -r(64 - 6-25) 

256 37 5 

x = :2i - 2 = 3783 

5775 

hence the distance of the e.g. from C is 4 3783 or 0*217 inch. 

Example 4. A hemispherical shell of uni- 
form material is 6 inches external radius and 
i '5 inches thick. Find its e.g. 

Let ABC (Fig. 130) be a solid hemisphere 
12 inches diameter, from which a concentric 
solid hemisphere abc, 9 inches diameter, has 
been cut, leaving a hemispherical shell ACB&rtf 
i '5 inches thick. 

Let x = distance of its e.g. (which is on the 
axis of symmetry OC) from O. 

Equating moments of volumes about O 
(i.e. omitting the factor of weight per unit 
volume) 

Volume of solid \ ., . r i-j L -m \ , / i r i. i 

ARr 3OT / = ( vomrne f solid acox.%Oc) + (volume of shell x.r) 

|7r6 3 X f X 6 = |TT X () 3 X X f + |7r{6 3 (|) 3 }^ 
from which x = 2*66 inches 

The e.g. of the shell is on the axis and 2'66 inches from the 
centre of the surfaces. 




H-C 



FIG. 130. 



EXAMPLES XV. 

1 . The front wheel of a bicycle is 30 inches diameter and weighs 4 Ibs. ; 
the back wheel is 28 inches diameter and weighs 7 Ibs. The remaining 
parts of the bicycle weigh 16 Ibs., and their e.g. is 18 inches forward of the 
back axle and 23 inches above the ground when the steering-wheel is 
locked in the plane of the back wheel. Find the e.g. of the whole bicycle ; 
state its height above the ground and its distance in front of the back axle 
when the machine stands upright on level ground. The wheel centres are 
42 inches horizontally apart. 

2. A projectile consists of a hollow cylinder 6 inches external and 3 
inches internal diameter, and a solid cone on a circular base 6 inches 
diameter, coinciding with one end of the cylinder. The axes of the cone 
and cylinder are in line ; the length of the cylinder is 12 inches, and the 



1 66 



Mechanics for Engineers 



Ct -40 



height of the cone is 8 inches. Find the distance of the e.g. of the 
projectile from its point. 

3. A solid of uniform material consists of a cylinder 4 inches diameter 
and 10 inches long, with a hemispherical end, the circular face of which 
coincides with one end of the cylinder. The other end of the cylinder is 
pierced by a cylindrical hole, 2 inches diameter, extending to a depth of 
7 inches along the cylinder and co-axial with it. Find the e.g. of the solid. 
How far is it from the flat end ? 

4. The profile of a crank (Fig. 131) consists of two semicircular ends, 
CED and AFB, of 8 inches and 12 inches radii respectively, centred at 

points P and O 3 feet apart, and joined by straight 
lines AC and BD. The crank is of uniform thick- 
ness, perpendicular to the figure, and is pierced 
by a hole 10 inches diameter, centred at O. Find 
the distance of the c.g of the crank from the axis O. 

5. Find the c.g. of a T girder section, the 
height over all being 8 inches, and the greatest 
width 6 inches, the metal being \ inch thick in the 
vertical web, and I inch thick in the horizontal 
flange. 

6. An I-section girder consists of a top flange 
6 inches by I inch, a bottom flange 10 inches by 
1*75 inches, connected by a web 10 inches by 1*15 
inches. Find the height of the c.g. of the section 
from the lowest edge. 

7. A circular lamina 4 inches diameter has two 
circular holes cut out of it, one I '5 inches and the 
other I inch diameter with their centres I inch and 
I '25 inches respectively from the centre of the 

lamina, and situated on diameters mutually perpendicular. Find the c.g. 
of the remainder of the lamina. 

8. A balance weight in the form of a segment of a circle fits inside the 
rim of a wheel, the internal diameter of which is 3 feet. If the segment 
subtends an angle of 60 at the centre of the wheel, find the distance of its 
c.g. from the axis. 

9. If two intersecting tangents are drawn from the extremities of a 
quadrant of a circle 4 feet diameter, find the distance of the c.g. of the 
area enclosed between the tangents and the arc, from either tangent. 

10. A balance weight of a crescent shape fits inside the rim of a wheel 
of 6 feet internal diameter, and subtends an angle of 60 at its centre. The 
inner surface of the weight is curved to twice the radius of the outer surface, 
i.e. the centre from which its profile is struck is on the circumference of the 
inside of the wheel. The weight being of uniform thickness perpendicular to 
the plane of the wheel, find the distance of its c.g. from the axis of the wheel. 

N.B. The profile is equivalent to the sector of a circle plus two 
triangles minus a sector of a larger circle. 




CHAPTER VIII 

CENTRE OF GRAVITY: PROPERTIES AND 
APPLICA TIONS 

I2i. Properties of the Centre of Gravity. Since the 
resultant force of gravity always acts through the centre of 
gravity, the weight of the various parts of a rigid body may 
be looked upon as statically equivalent to a single force equal 
to their arithmetic sum acting vertically through the centre of 
gravity of the body. Such a single force will produce the same 
reactions on the body from its supports ; will have the same 
moment about any point (Art. 90) ; may be replaced by the 
same statically equivalent forces or components ; and requires the 
same equilibrants, as the several forces which are the weights of 
the parts. Hence, if a body be supported by being suspended 
by a single thread or string, the e.g. of the body is in the same 
vertical line as that thread or string. If the same body is 
suspended again from a different point in itself, the e.g. is 
also in the second vertical line of suspension. If the two lines 
can be drawn on or in the body, the e.g., which must lie at 
their intersection, can thus be found experimentally. For 
example, the e.g. of a lamina may be found by suspending 
it from two different points in its perimeter, first from one and 
then from the other, so that its plane is in both cases vertical, 
and marking upon it two straight lindl which are continuations 
of the suspension thread in the two positions. 

Fig. 132 shows G, the e.g. of a lamina PQRS, lying in both 
the lines of suspension PR and QS from P and Q respectively. 
The tension of the cord acts vertically upwards on the lamina, 
and is equal in magnitude to the vertical downward force of 



1 68 



Mechanics for Engineers 



the weight of the lamina acting through G. The tension can 
only balance the weight if it acts through G, for in order that 
two forces may keep a body in equilibrium, they must be con- 




FIG. 132. 



current, equal, and opposite, and therefore in the same straight 
line. 

A " plumb line," consisting of a heavy weight hanging from 
a thin flexible string, serves as a convenient method of obtaining 
a vertical line. 

122. Centre of Gravity of a Distributed Load. If 
a load is uniformly distributed over the whole span of a beam, 
the centre of gravity of the load is at mid-span, and the 
reactions of the supports of the beam are the same as would 
I be produced by the whole load 

concentrated at the middle of 
the beam. Thus, if in Fig. 133 
a beam of 20-feet span carries a 
load of 3 tons per foot of span 
(including the weight of the beam) 
uniformly spread over its length, 
the reactions at the supports. A 
and B are each the same as would be produced by a load 
of 60 tons acting at C, the middle section of the beam, 
viz. 30 tons at each support. Next suppose the load on a 
beam is distributed, not evenly, but in some known manner. 
Suppose the load per foot of span at various points to be 



B 



FIG. 133. 



Centre of Gravity : Properties and Applications 1 69 



shown by the height of a curve ACDEB (Fig. 134). The 
load may be supposed to be piled on the beam, so that the 
curve ACDEB is its profile, and so that the space occupied is 
of constant thickness in a direction perpendicular to the plane 
of the figure. Then the e.g. of the load is at the e.g. G of 




FIG. 134. 

the area of a section such as ACDEB in Fig. 134, taken 
halfway through the constant thickness. The reactions of 
the supports are the same as if the whole load were concen- 
trated at the point G. The whole load is equal to the length 
of the beam multiplied by the mean load per unit length, 
which is represented by the mean ordinate of the curve ACDEB, 
i.e. a length equal to the area ACDEB divided by AB. 

Example. As a par- 
ticular case of a beam 
carrying a distributed load 
not evenly spread, take a 
beam of 2o-feet span carry- 
ing a load the intensity of 
which is 5 tons per foot 
run at one end, and varying 
uniformly to 3 tons per foot 
at the other. Fig. 135 
represents the distribu- 
tion of load. Find the 
reactions at A and B. 




FIG. 135. 



The total load = 20 x mean load per foot = 20 x - = 80 tons 

2 

Let I- be the distance of the e.g. of area ABCD from BD. 
jr(area ACFB + area CDF) = (loxarea ACFB) + (-Sfxarea CDF) 

X3 x 20 + 5 20 X 2 ) = ( 10 X 20 X 3) + * X 3p X 2 

600 + 133*3 ,. 

* = 8 -^ = 9-i6 feet 

and distance of e.g. from AC = 20 9' 1 6 = io'8.3 feet 



Mechanics for Engineers 

If R A and R B be the reactions at A and B respectively, equating 
opposite moments about B of all the forces on the beam 



R A x 20 = 80 x 9-16 

R A = 80 X * 
20 

R B = 80 36-6 = 43'3 tons 



R A = 80 x 2 = 36-6 tons 
20 



123. Body resting upon a Plane Surface. As in the 

case of a suspended body, the resultant of all the supporting 
forces must pass vertically through the e.g. of the body in 
order to balance the resultant gravitational forces in that 
straight line. The vertical line through the e.g. must then 
cut the surface, within the area of the extreme outer polygon 
or curved figure which can be formed by joining all the points 
of contact with the plane by straight lines. If the vertical 
line through the e.g. fall on the perimeter of this polygon 
the solid is on the point of overturning, and if it falls outside 
that area the solid will topple over unless supported in 
some other way. This is sometimes expressed by saying 



T G 

I 
I 

i 




FIG. 136. 

that a body can only remain at rest on a plane surface if 
the vertical line through the e.g. falls within the base. From 
what is stated above, the term " base " has a particular mean- 
ing, and does not signify only areas of actual contact; e.g. 
in Fig. 136 are two solids in equilibrium, with GN, the vertical 
line through G, the e.g., falling within the area of contact; 



Centre of Gravity : Properties and Applications 1 7 1 



but in Fig. 137 a solid is shown in which the vertical through 
the e.g. falls outside the area of contact when the solid rests 
upright with one end on a horizontal plane. If, however, 
it falls within the extreme area ABC, the solid can rest in 
equilibrium on a plane. 



Plan. 



FIG. 137. 

Two cases in which equilibrium is impossible are shown in 
Fig. 138, the condition stated above being violated. The first 
is that of a high cylinder on an inclined plane, and the second 




FIG. 138. 



that of a waggon-load of produce on the side of a high crowned 
road. It will be noticed that a body subjected to tilting will 
topple over with less inclination or more, according as its e.g. 
is high or low. 

Example. What is the greatest length which a right cylinder 
of 8 inches diameter may have in order that it may rest with one 
end on a plane inclined 20 to the horizontal ? 



172 Mechanics for Engineers 

The limiting height will be reached when the e.g. falls vertically 
over the circumference of the base, i.e. when G (Fig. 139) is 




FIG. 139. 

vertically above A. Then, G being the mid-point of the axis EF, 
the half-length of cylinder 

GE = AE cot AGE = AE cot ACD 
or GE = AE cot 20 = 4x2 7475 = 10-99 inches 

The length of cylinder is therefore 2 x 10^99 = 21 '98 inches. 

124. Stable, Unstable, and Neutral Equilibrium. 

A body is said to be in stable equilibrium when, if slightly 
disturbed from its position, the forces acting upon it tend 
to cause it to return to that position. 

If, on the other hand, the forces acting upon it after a 
slight displacement tend to make it go further from its former 
position, the equilibrium is said to be unstable. 

If, after a slight displacement, the forces acting upon the 
body form a system in equilibrium, the body tends neither 
to return to its former position nor to recede further from it, 
and the equilibrium is said to be neutral. ' 

A few cases of equilibrium of various kinds will now be con- 
sidered, and the conditions making for stability or otherwise. 

125. Solid Hemisphere resting on a Horizontal 
Plane. If a solid hemisphere, ABN (Fig. 140), rests on a 



Centre of Gravity: Properties and Applications 173 

horizontal plane, and receives a small tilt, say through an 
angle 0, the e.g., situated at G, of ON from O and in the 
radius ON, takes up the position shown on the right hand of 




FIG. 140. 

the figure. The forces acting instantaneously on the solid are 
then (i) the weight vertically through G, and (2) the reaction 
R in the line MO vertically through M (the new point of 
contact between hemisphere and plane) and normal to the 
curved surface. These two forces form a " righting couple," 
and evidently tend to rotate the solid into its original posi- 
tion. Hence the position shown on the left is one of stable 
equilibrium. Note that G lies below O. 

126. Solid with a Hemispherical End resting on 
a Horizontal Plane. Suppose a solid consisting of, say, 




FIG. 141. 



a cylinder with a hemispherical base, the whole being of 
homogeneous material, rests on a plane, and the e.g. G (Fig. 
741) falls within the cylinder, i.e. beyond the centre O of the 
hemispherical end reckoned from N, where the axis cuts the 



1/4 Mechanics for Engineers 

curved surface. On the left of Fig. 141 the solid is shown in 
a vertical position of equilibrium. Now suppose it to receive 
a slight angular displacement, as on the right side of the figure. 
The weight W, acting vertically downwards through G, along 
with the vertical reaction R of the plane, forms a system, the 
tendency of which is to move the body so that G moves, not 
towards its former position, but away from it. The weight 
acting vertically through G and the reaction of the plane acting 
vertically through O form an " upsetting couple " instead of a 
"righting couple." Hence the position on the left of Fig. 141 
is one of unstable equilibrium. Note that in this case G falls 
above O. If the upper part of the body were so small that G 
is below O, the equilibrium would be stable, as in the case of 
the hemisphere above (Art. 125). The lower G is, the greater 
is the righting couple (or the greater the stability) for a given 
angular disturbance of the body. While in the case of in- 
stability, the higher G is, the greater is the upsetting couple or 
the greater the instability, and we have seen that such a solid 
is stable or unstable according as G falls below or above O. 

127. Critical Case of Equilibrium neutral. If G 
coincides with the centre of the hemisphere (Art. 126), the 
equilibrium is neither stable nor unstable, but neutral. Suppose 
the cylinder is shortened so that G, the e.g. of the whole solid, 

falls on O, the centre of 

A^ /--^| R ^e hemisphere. Then if 

the solid receives a slight 
angular displacement, as 
in the right side of Fig. 
142, the reaction R of 
| w the plane acts vertically 

F, G . I42 . upwards through O, the 

centre of the hemisphere 

(being normal to the surface at the point of contact), and 
the resultant force of gravity acts vertically downward through 
the same point. In this case the two vertical forces balance, 
and there is no couple formed, and no tendency to rotate 
the body towards or away from its former position. Hence 
the equilibrium is neutral. 






FlG - 



Centre of Gravity : Properties and Applications 175 

In each of the above instances the equilibrium as regards 
angular displacements is the same whatever the direction of 
the displacement. As 
further examples of neu- 
tral equilibrium, a sphere 
or cylinder of uniform 
material resting on a 
horizontal plane may be 
taken. The sphere is 
in neutral equilibrium 
with regard to angular 
displacements in any direction, but the horizontal cylinder 
(Fig. 143) is only in neutral equilibrium as regards its rolling 
displacements ; in other directions its equilibrium is stable. 

Example. A cone and a hemisphere of the same homogeneous 
material have a circular face of i foot radius 
in common. Find for what height of the 
cone the equilibrium of the compound solid 
will be neutral when resting with the hemi- 
spherical surface on a horizontal plane. 

The equilibrium will be neutral when the 
e.g. of the solid is at the centre of the hemi- 
sphere, i.e. at the centre O (Fig. 144) of their 
common face. 

Let h be the height of the cone in feet. 
Then its e.g. G l is \h from O, and its volume 




is \h x - x 2 2 = \irh cubic feet. 
4 

The e.g. G 2 , of the hemisphere is at 
volume is TT cubic feet. Then 



FIG. 144. 

foot from O, and its 



G t O _ \h _ weight of hemisphere _ TT 
G 2 O weight of cone \-rrh 

and \h = 7 
h 



h = 



2 feet 



If h is greater than ^3 feet the equilibrium is unstable, and if it 
is less than ^3 f eet the equilibrium is stable. 



Mechanics for Engineers 

128. In the case of bodies resting on plane surfaces and 
having more than one point of contact, the equilibrium will 
be stable if the e.g. falls within the area of the base, giving 
the word the meaning attached to it in Art. 123 for small 
angular displacements in any direction. If the e.g. falls on 
the perimeter of the base, the equilibrium will be unstable for 
displacements which carry the e.g. outside the space vertically 
above the " base." 

The attraction of the earth tends to pull the e.g. of a body 
into the lowest possible position ; hence, speaking generally, 
the lower the e.g. of a body the greater is its stability, and 
the higher the e.g. the less stable is it. 

In the case of a body capable of turning freely about a 
horizontal axis, the only position of stable equilibrium will be 
that in which the e.g. is vertically below the axis. When it 




Unstable 



Stable, 

FIG. 145. 



is vertically above, the equilibrium is unstable, and unless the 
e.g. is in the axis there are only two positions of equilibrium. 
If the e.g. is in the axis, the body can rest in neutral equilibrium 
in any position. 

Fig. 145 represents a triangular plate mounted on a hori- 
zontal axis, C ; it is in unstable, stable, or neutral equilibrium 
according as the axis C is below, above, or through G, the e.g. 
of the plate. 

129. Work done in lifting a Body. When a body 
is lifted, it frequently happens that different parts of it are lifted 
through different distances, e.g. when a hanging chain is wound 
up, when a rigid body is tilted, or when water is raised from 
one vessel to a higher one. The total work done in lifting the 



Centre of Gravity: Properties and Applications 177 

body can be reckoned as follows : Let w^ w 2 , 7f 3 , w 4 , etc., be 
the weights of the various parts of the body, which is supposed 
divided into any number of parts, either large or small, but 
such that the whole of one part has exactly the same displace- 
ment (this condition will in many cases involve division into 
indefinitely small parts). Let the parts w^ w^ w 3 , etc., be at 
heights #1, x. 2 , x 3 , etc., respectively above some fixed horizontal 
plane ; if the parts are not indefinitely small, the distances x ly 
x. 2 , x 3 , etc., refer to the heights of their centres of gravity. 

2/K'jv) 
Then the distance x of the e.g. from the plane is - 

S(w) 

(Art. 113). After the body has been lifted, let .r/, x/, x 3 ', etc., 
be the respective heights above the fixed plane of the parts 
weighing w l} w z , w s , etc. Then the distance x' of the e.g. 

V/7/v'\ 

above the plane is -*/ r (Art. 113). 
Z(w) 

The work done in moving the part weighing w^ is equal to 
the weight w l multiplied by the distance (xj x^} through 
which it is lifted ; i.e. the work is w^Xi x^ units. 

Similarly, the work done in lifting the part weighing ?<" 2 is 
w. 2 (x. 2 x. 2 ). Hence the total work done is 

WT(XI - x^ + w z (x. 2 - x. 2 ) + w 3 (xs - x 3 ) +, etc. 
which is equal to 

(w^Xi + Wix. 2 + WaXs' -r, etc.) (w^x-i + W 2 x. 2 + Wi +, etc.) 

or 1(wx) ^(wx) 
But 2,(wx') =~x'2(w) and 2(w,r) =~x'$(w) 

therefore tne work done = x"^(w} x$(w) 
= (x 1 - x)1(w) 

The first factor, x ' x, is the distance through which the 
e.g. of the several weights has been raised, and the second 
factor, 2(w), is the total weight of all the parts. Hence the 
total work done in lifting a body is equal to the weight of the 
body multiplied by the vertical distance through which its e.g. 
has been raised. 

N 



178 



Mechanics for Engineers 



Example i. A rectangular tank, 3 feet long, 2 feet wide, and 
i '5 feet deep, is filled from a cylindrical tank of 24 square feet 
horizontal cross-sectional area. The level of water, before filling 

begins, stands 20 feet below 
the bottom of the rectangular 
tank. How much work is re- 
quired to fill the tank, the 
weight of i cubic foot of water 
being 62-5 Ibs. ? 

The water to be lifted is 
3 x 2 x 1-5 or 9 cubic feet, 
hence the level in the lower 
tank will be lowered by ^ or 
i of a foot, i.e. by a length BC 
on Fig. 146. The 9 cubic feet 
of water lifted occupies first 
the position ABCD, and then 
fills the tank EFGH. In the 
former position its e.g. is -|BC 
or j 3 ^ foot below the level AB, 
and in the latter position its 
e.g. is |GH or | foot above 
the level EH. Hence the 
e.g. is lifted (^ 6 + 20 + f ) feet, 
i.e. 2o}f feet, or 20*9375 feet, 
feet of water lifted is 9 x 62*5 




FIG. 146. 



The weight of the 9 cubic 
= 562-5 Ibs. 

Hence the work done is 562-5 x 20-9375 = 11,770 foot-lbs. 

Example 2. Find the work in foot-pounds necessary to upset 

a solid right circular cylinder 
3 feet diameter and 7 feet high, 
weighing half a ton, which is 
resting on one end on a hori- 
zontal plane. 

Suppose the cylinder (Fig. 
147) to turn about a point A on 
the circumference of the base. 
Then G, the e.g. of the cylinder, 
which was formerly 3-5 feet 
above the level of the hori- 




FIG. 147. 



zontal plane, is raised to a 
position G', i.e. to a height A'G' 
above the horizontal plane before the cylinder is overthrown. 



Centre of Gravity : Properties and Applications \ 79 



The distance the e.g. is lifted is then A'G' EG 
A'G' = 




+ EG2 ) = V0'5 2 + 3'5 2 ) = 3 - 8o7 feet 
The e.g. is lifted 3-807 3-5 = 0-307 foot 
and the work done is 1120 x 0*307 = 344 foot-lbs. 

Example 3. A chain 600 feet long hangs vertically ; its weight 
at the top end is 12 Ibs. per foot, and at the bottom end 9 Ibs. per 
foot, the weight per foot varying uniformly 
from top to bottom. Find the work necessary 
to wind up the chain. 

It is first necessary to find the total weight 
of the chain and the position of its e.g. The 
material of the chain may be considered to be 
spread laterally into a sheet of uniform thick- 
ness, the length remaining unchanged. The 
width of the sheet will then be proportional 
to the weight per foot of length ; the total 
weight, and the height of the e.g. of the chain, 
will not be altered in such a case. 

The depth of the e.g. below the highest 
point (A) of the chain (Fig. 148) will be the 
same as that of a figure made up of a rect- 
angle, ACDB, 600 feet long and 9 (feet or other units) broad, 
and a right-angled triangle, CED, having sides about the right 
angle at C of (CD) 600 feet and (CE) 3 units. 

The depth will be 

(600 x 9 x 300) + (\ x 600 x 3 x a 8 a ) , A , 

-. S-2 (Art. 114) 

(600 x 9) + (f x 600 x 3) 

which is equal to 2857 feet. 

The total weight of the chain will be the same as if it were 

600 feet long and of uniform weight - - or 10-5 Ibs. per foot, 

viz. 600 x 10-5 = 6300 Ibs. 

Hence the work done in raising the chain all to the level A is 

6300 x 2857 = 1,800,000 foot-lbs. 

130. Force acting on a Rigid Body rotating uni- 
formly about a Fixed Axis. 

Let Fig. 149 represent a cross-section of a rigid body of 
weight W rotating about a fixed axis, O, perpendicular to the 
figure. For simplicity the body will be supposed symmetrical 



i8o 



Mechanics for Engineers 




about the plane of the figure, which therefore contains G, the 
e.g. of the body. In the position shown, let w^ be the weight 

of a very small portion of the 
body (cut parallel to the axis) 
situated at a distance r from 
O. Let W be the uniform 
angular velocity of the body 
about the axis O. Then the 
force acting upon the small 
portion of weight w^ in order 
to make it rotate about O is 

-wV, directed towards O 

(Art. 63), and it evidently 
acts at the middle of the 
length of the portion, i.e. in the plane of the figure. Resolving 
this force in any two perpendicular directions, XO and YO, 

the components in these two directions are m*r cos and 

wV sin 6 respectively, where 6 is the angle which AO 
& 
makes with OX. 

These may be written - . w 2 . x and 'w 2 . y respectively, 

<$ O 

where x represents r cos and y represents r sin 0, the 
projections of r on OX and OY respectively. 

Adding the components in the direction XO of the centri- 
petal forces acting in the plane of the figure upon all such 
portions making up the entire solid, the total component 



F x = S =te - -l(u,x) = =-*SM = ~ . to' . * 



*' / g S g 

and the total component force in the direction YO is 



W 2 

= . or 



where x and y are the distances of G, the e.g. of the solid 
(which is in the plane of the figure), from OY and OX re- 
spectively. 



Centre of Gravity : Properties and Applications 1 8 1 

Hence the resultant force P acting on the solid towards 
O is 



= . w . (+ = w2 R 



vvhere R = V x' z + / 2 , the distance of the e.g. from the axis O. 
Hence the resultant force acting on the body is of the same 

/W \ 
magnitude as the centripetal force ( -^w 2 R ) which must act 

on a weight W concentrated at a radius R from O in order 

that it may rotate uniformly at an angular velocity w. Further, 

~if 

the tangent of the angle which P makes with XO is ^r- 



(Art. 75), which is equal to - , or where GN is perpen- 



y GN 

. nr 
X 

dicular to OX. Hence the force P acts in the line GO, and 
therefore the resultant force P acting on the rotating body is 
in all respects identical with that which would be required to 
make an equal weight, W, rotate with the same angular velocity 
about O if that weight were concentrated (as a particle) at G, 
the e.g. of the body. 

It immediately follows, from the third law of motionj that 
the centrifugal force exerted by the rotating body on its con- 
straints is also of this same magnitude and of opposite direction 
in the same straight line. 

Example. Find the force exerted on the axis by a thin 
uniform rod 5 feet long and weighing 9 Ibs., making 
30 revolutions per minute about an axis perpen- < 
dicular to its length. 

The distance from the axis O to G, the eg. of 
the rod (Fig. 150), is 2*5 feet, the e.g. being midway 
between the ends. The angular velocity of the 

, . 3O X 27T ,. 

rod is "* > = TT radians per second. The cen- 
60 

trifugal pull on O is the same as that of a weight of 

9 Ibs. concentrated at 2-5 feet from the axis and describing 

about O, TT radians per second, which is 

- x Ti- 2 x 2-5 = 6-89 Ibs. 
32-2 



1 82 Mechanics for Engineers 

131. Theorems of (iu Id in us or Pappus. (a) The 

area of the surface of revolution swept out by any plane curve 
revolving about a given axis in its plane is equal to the length 
of the curve multiplied by the length of the path of its e.g. 
in describing a circle about the axis. Suppose the curve 
ABC (Fig. 151) revolves about the 
axis OO', thereby generating a surface 
of revolution of which OO' is the axis. 
Let S be the length of the curve, and 
suppose it to be divided into a large 
number of small parts, s v s 2 , s 3) etc., 
each of such short length that if drawn 
straight the shape of the curve is not 
appreciably altered. Let the distances 
of the parts s 1} s 2 , s 3 , etc., from the 
axis be x lt x. 2) x 3 , etc. ; and let G, the 
e.g. of the curve which is in the plane of the figure, i.e. the plane 
of the curve, be distant x from the axis OO'. The portion s l 
generates a surface the length of which is 2irx l and the breadth 
^i ; hence the area is 2Trx- i s l . Similarly, the portion s 2 gene- 
rates an area 2irx. 2 . s 2 , and the whole area is the sum 




, etc., or 2ir2,xs 

If the portions s lt s 2 , s 3 , etc., are of finite length, this result is 
only an approximation ; but if we understand 2(xs) to represent 
the limiting value of such a sum, when the length of each part 
is reduced indefinitely, the result is not a mere approximation. 

Now, since 2<(xs) = x X S(j') = x X S, the whole area of 
the surface of revolution is 2irx . S, of which zirx is the length 
of the path of the e.g. of the curve in describing a circle about 
OO', and S is the length of the curve. 

(b) The volume of a solid of revolution generated by the 
revolution of a plane area about an axis in its plane is equal 
to the enclosed revolving area multiplied by the length of the 
path of the e.g. of that area in describing a complete circle 
about the axis. 

Suppose that the area ABC (Fig. 152) revolves about the 
axis OO', thereby generating a solid of revolution of which 



Centre of Gravity : Properties and Applications 183 

OO' is an axis (and which is enclosed by the surface generated 
by the perimeter ABC). 

Let the area of the plane figure ABC be denoted by A, 
and let it be divided into a large , 
number of indefinitely small parts 
#1, a 2 , a 3 , etc., situated at distances 
jc 1} x 2 , x 3 , etc., from the axis 
OO'. 

The area a- 1} in revolving about 
OO', generates a solid ring which 
has a cross- section a^ and a length 
27TJCJ, and therefore its volume is 
2irx 1 a l . Similarly, the volume swept 
out by the area a z is 2irx. 2 a. 2 , and 
so on. The whole volume swept 
out by the area A is the limiting value of the sum of the small 
quantities 



, etc., 
, etc.,) or 2ir'2 l (ax) 



And since 2(ax) = x^(a) = x . A (Art. 114 (6)), the whole 
volume is 2irx . A, of which 2irx is the length of the path of 
the e.g. of the area in describing a circle about the axis OO', 
and A is the area. 




FIG. 152. 



or 2ir(a l x l + a%x z + a 3 x 



B 



Example. A groove of semicircular section 1*25 inches 
radius is cut in a cylinder 8 inches diameter. Find (a) the area of 
the curved surface of the groove, 
and (b) the volume of material 
removed. 

(a) The distance of the e.g. of 
the semicircular arc ABC (Fig. 

153) from AB is ( 1*25 x - ) or 

\ It / 7T 

inches. Therefore the distance of 
the e.g. of the arc from the axis 

OO' is (4 --5 ) inches. The 



0' 



FIG. 153. 
length of path of this point in making one complete circuit about 



1 84 



Mechanics for Engineers 



OO' is 271/4 - ^ = (877 - 5) inches. The length of arc ABC 

is i'257T inches, hence the area of the surface of the semicircular 
groove is 

r257r(87r 5) square inches = icw 2 6 - 257r 

= 987 i9'6 
= 79' I square inches 
(6) The distance of the e.g. of the area ABC from AB is 

x I '25 = 0*530 inch, and therefore the distance of the e.g. from 

OO' is 4 0-53 = 3-47 inches. 

The length of path of this point in making one complete circuit 
about OO' is 2rr x 3*47 = 21*8 inches. The area of the semicircle 
isi(i*25)' 2 7r = 2'453 square inches, hence the volume of the material 
removed from the groove is 

2r8 x 2-453 = 53-5 cubic inches 

132. Height of the e.g. of a Symmetrical Body, 
such as a Carriage, Bicycle, or Locomotive. It was stated 
in Art. 121 that the e.g. of some bodies might conveniently 
be found experimentally by suspending the bodies from two 
different points in them alternately. This is not always con- 
venient, and a method suitable for some other bodies will now 
be explained by reference to a particular instance. The c.g. 
of a bicycle (which is generally nearly symmetrical about a 




FIG. 154. 



vertical plane through both wheels) may be determined by first 
finding the vertical downward pressure exerted by each wheel 
on the level ground, and then by finding the vertical pressures 
when one wheel stands at a measured height above the other one. 
Suppose that the wheels are the same diameter, and that 
the centre of each wheel-axle, A and B (Fig. 154), stands 



Centre of Gravity : Properties and Applications 185 



at the same height above a level floor, the wheels being locked 
in the same vertical plane. 

When standing level, let W A = weight exerted by the front 
wheel on a weighing machine table ; let W B = weight exerted 
by the back wheel on a weighing machine table ; then 

W A + W B = weight of bicycle 

Let AB, the horizontal distance apart of the axle centres, 
be d inches. If the vertical line through the e.g. G cuts AB in 
C, then 

W A 
BC = m rfw- </(Art. 87) 

Next, let the weight exerted by the front wheel, when A 
stands a distance " h " inches (vertically) above B, be W ; and 
let CG, the distance of the e.g. of the bicycle above AB, be H. 




FIG. 155. 

Then, since ABE and DGC (Fig. 155) are similar triangles 

GC _ BJ _ V"(<* a - yfe 2 ) 
CD ~ AE ~ h 

W W 

andCD = BC-BD = TT // ~ 



hence GC or H = - 



WA-W, 

W A +~W B 



In an experiment on a certain bicycle the quantities were 
d = 44 inches, // = 6 inches, weight of bicycle = 32'QO Ibs., 
pressure (W A ) exerted by the front wheel when the back wheel 



1 86 Mechanics for Engineers 

was on the same level = 14-50 Ibs., pressure (W a ) exerted 
by the front wheel when the back wheel was 6 inches lower 
= 1 3 '84 Ibs. 



TT 44 ^ 14x0 ir84 v 

Hence H = L X J ^ * X 44 

6 32-90 

= 6-54 inches 

or the height of the e.g. above the ground is 6-54 inches plus 
the radius of the wheels. The distance BC of the e.g. horizon- 

tally in front of the back axle is ^ v 44, or 10-4 inches. 

32-90 

A similar method may be applied to motor cars or locomotives. 
In the latter case, all the wheels on one side rest on a raised 
rail on a weighing machine, thus tilting the locomotive sideways. 

EXAMPLES XVI. 

1. A beam rests on two supports at the same level and 12 feet apart. 
It carries a distributed load which has an intensity of 4 tons per foot-run 
at the right-hand support, and decreases uniformly to zero at the left-hand 
support. Find the pressures on the supports at the ends. 

2. The span of a simply supported horizontal beam is 24 feet, and 
along three-quarters of this distance there is a uniformly spread load of 
2 tons per foot run, which extends to one end of the beam : the weight of 
the beam is 5 tons. Find the vertical supporting forces at the ends. 

3. A beam is supported at the two ends 15 feet apart. Reckoning 
from the left-hand end, the first 4 feet carry a uniformly spread load of 
I ton per foot run ; the first 3 feet starting from the right-hand end carry 
a load of 6 tons per foot run evenly distributed, and in the intermediate 
portion the intensity of loading varies uniformly from that at the right- 
hand end to that at the left-hand end. Find the reaction of the supports. 

4. The altitude of a cone of homogeneous material is 18 inches, and 
the diameter of its base is 12 inches. What is the greatest inclination on 
which it may stand in equilibrium on its base ? 

5. A cylinder is to be made to contain 250 cubic inches of material. 
What is the greatest height it may have in order to rest with one end on a 
plane inclined at 15 to the horizontal, and what is then the diameter of the 
base? 

6. A solid consists of a hemisphere and a cylinder, each 10 inches 
diameter, the centre of the base of the hemisphere being at one end of the 
axis of the cylinder. What is the greatest length of cylinder consistent 
with stability of equilibrium when the solid is resting with its curved end 
on a horizontal plane ? 



Centre of Gravity : Properties and Applications \ 87 

7. A solid is made up of a hemisphere of iron of 3 inches radius, and 
a cylinder of aluminium 6 inches diameter, one end of which coincides 
with the plane circular face of the hemisphere. The density of iron being 
three times that of aluminium, what must be the length of the cylinder if 
the solid is to rest on a horizontal plane with any point of the hemispherical 
surface, in contact? 

8. A uniform chain, 40 feet long and weighing 10 Ibs. per foot, hangs 
vertically. How much work is necessary to wind it up ? 

9. A chain weighing 12 Ibs. per foot and 70 feet long hangs over a 
(frictionless) pulley with one end 20 feet above the other. How much 
work is necessary to bring the lower end to within 2 feet of the level of 
the higher one ? 

10. A chain hanging vertically consists of two parts : the upper portion 
is 100 feet long and weighs 16 Ibs. per foot, the lower portion is 80 feet 
long and weighs 12 Ibs. per foot. Find the work done in winding up 
(a) the first 70 feet of the chain, (b) the remainder. 

11. A hollow cylindrical boiler shell, 7 feet internal diameter and 
25 feet long, is fixed with its axis horizontal. It has to be half filled with 
water from a reservoir, the level of which remains constantly 4 feet below 
the axis of the boiler. Find how much work is required to lift the water, 
its weight being 62*5 Ibs. per cubic foot. 

12. A cubical block of stone of 3-feet edge rests with one face on the 
ground : the material weighs 150 Ibs. per cubic foot. How much work is 
required to tilt the block into a position of unstable equilibrium resting on 
one edge? 

13. A cone of altitude 2 feet rotates about a diameter of its base at a 
uniform speed of 180 revolutions per minute. If the weight of the cone 
is 20 Ibs., what centrifugal pull does it exert on the axis about which it 
rotates ? 

14. A shaft making 150 rotations per minute has attached to it a pulley 
weighing 80 Ibs., the e.g. of which is o'l inch from the axis of the shaft. 
Find the outward pull which the pulley exerts on the shaft. 

15. The arc of a circle of 8 inches radius subtends an angle of 60 at 
the centre. Find the area of the surface generated when this arc revolves 
about its chord ; find also the volume of the solid generated by the revolu- 
tion of the segment about the chord. 

16. A groove of V-shaped section, i'5 inches wide and i inch deep, is 
cut in a cylinder 4 inches in diameter. Find the volume of the material 
removed. 

17. A symmetrical rectangular table, the top of which measures 8 feet 
by 3 feet, weighs 150 Ibs., and is supported by castors at the foot of each 
leg, each castor resting in contact with a level floor exactly under a corner 
of the table top. Two of the legs 3 feet apart are raised 10 inches on to the 
plate of a weighing machine, and the pressure exerted by them is 66" 5 Ibs. 
Find the height of the e.g. of the table above the floor when the table 
stands level. 



CHAPTER IX 



MOMENTS OF INERTIA ROTATION 



133. Moments of Inertia. 

(i) Of a Particle. If a particle P (Fig. 156), of weight w 



w 
and mass , is situated at a distance r from an axis OO', then 

o 

p its moment of inertia about that 

axis is denned as the quantity 

ant 

. r 2 , or (mass of P) X (distance 

from OO') 2 . 
0' (2) Of Several Particles. If 

several particles, P, Q, R, and 
S, etc., of weights a' lf a/ 2 > w z> '4> etc., be situated at distances 
r \, f'zt r& and r^ etc., respectively from an axis OO' (Fig. 157), 



FIG. 156. 





r 


S 






Q 


^X 





o 1 / 

_x^*7 
R 4C 3 



FIG. 157. 



End view of axis OO'. 



then the total moment of inertia of the several particles about 
that axis is denned as 



Moments of Inertia Rotation 189 

7f'j o , W. 2 o . W., o , WA 

_Jr* -f ~ V 2 2 + V 3 - + V 4 2 -f , etc. 
" 



or 2{(mass of each particle) x (its distance from OO') 2 } 

(3) Rigid Bodies. If we regard a rigid body as divisible 
into a very large number of parts, each so small as to be 
regarded as a particle, then the moment of inertia of the rigid 
body about any axis is equal to the moment of inertia of such 
a system of particles about that axis. Otherwise, suppose a 
body is divided into a large but finite number of parts, and the 
mass of each is multiplied by the square of the distance of 
some point in it from a line OO' ; the sum of these products 
will be an approximation to the moment of inertia of the whole 
body. The approximation will be closer the larger the number 
of parts into which the body is divided ; as the number of parts 
is indefinitely increased, and the mass of each correspondingly 
decreased, the sum of the products tends towards a fixed 
limiting value, which it does not exceed however far the 
subdivision be carried. This limiting sum is the moment of 

/ ' IV \ 

inertia of the body, which may be written ^(wr 2 ) or 2( ?" 2 ) 

Units. The units in which a moment of inertia is stated 
depend upon the units of mass and length adopted. No 
special names are given to such units. The " engineer's unit " 
or gravitational unit is the moment of inertia about an axis of 
unit mass (yz Ibs.) at a distance of i foot from the axis. 

134. Radius of Gyration. The radius of gyration of a 
body about a given axis is that radius at which, if an equal 
mass were concentrated, it would have the same moment of 
inertia. 

(7/' \ 
-f 1 ) of a body about some 
<S ' 

axis be denoted by I, and let its total weight 2(/) be W, and 

(w\ W 
- } = 
g) S 



Mechanics for Engineers 



Let k be its radius of gyration about the same axis. Then, 
from the above definition 



I 



W 7 o ^fiv 2 \ 
= ft* = 2 -r 2 ) 

\ ' 



w 



135' Moments of Inertia 
Y 



W 

of a Lamina about an 
Axis perpendicular to its 
Plane. 

Let the distances of any 
particle, P (Fig. 158), of a 
lamina from two perpen- 
dicular axes, OY and OX, 
in its plane be x l and ji re- 
spectively, and let w t be its 
weight, and i\ its distance 
from O, so that r? = x^ + y^. 

Then, if I x and I Y denote the moments of inertia of the 
lamina made up of such particles, about OX and OY re- 
spectively 




FIG. 158. 



and adding 



I 



g 



, etc. 



or 



g g g 

, which may be denoted by I . 
Then I = I x 4- IY 



(0 



This quantity I is by definition the moment of inertia 
about an axis OO' perpendicular to the plane of the lamina, 
and through O the point of intersection of OX and OY. 



Moments of Inertia Rotation 



191 



Hence the sum of the moments of inertia of a lamina about 
any two mutually perpendicular axes in its plane, is equal to the 
moment of inertia about an axis through the intersection of the 
other two axes and perpendicular to the plane of the lamina. 

Also, if < x > ^Y> an d k Q be the radii of gyration about 
OX, OY, and OO' respectively, OO' being perpendicular to the 

fui\ W 
plane of Fig. 158, and if 21 ) = , the mass of the whole 

^ o 

lamina 

. W 



W 



W 

and I Y = > Y 2 . 



and therefore, since I x + Iy = 



W 



by (i) 



(2) 



Or, in words, the sum of the squares of the radii of gyration of 
a lamina about two mutually perpendicular axes in its plane, 
is equal to the square of its radius of gyration about an axis 
through the intersection of the other two axes and perpendicular 
to the plane of the lamina. 

136. Moments of Inertia of a Lamina about 
Parallel Axes in its Plane. Let P, Fig. 159, be a 




0' 



constituent particle of weight 

ii\ of a lamina, distant x-^ from 

an axis ZZ' in the plane of 

the lamina and through G, the 

e.g. of the lamina, the distances 

being reckoned positive to the 

right and negative to the left 

of ZZ'. Let OO' be an axis 

in the plane of the lamina 

parallel to ZZ' and distant d 

from it. Then the distance 

of P from OO' is d x^ whether P is to the right or left 

of ZZ'. 



FIG. 159. 



I9 2 Mechanics for Engineers 

Let I be the moment of inertia of the lamina about OO' ; 
and let I z ZZ'. 

Then 

I = |V - Xl )* + V - xtf + W (d - xtf +,etcj 



0/2*2 + a'3#3 +, etc.) 



The sum 7# 1 ;c 1 4- ^2-^2 + ^3X3 +, etc., is, by Art 114, equal 
to 

(z^! + 7> 2 -r- 7/ 3 +, etc.) X (distance of eg. from ZZ') 

which is zero, since the second factor is zero. Hence 



<S ^ <b 6 



w 

orI =-rf a + I z ............. (i) 

o 

where W is the total weight of the lamina. And dividing each 

W 
term of this equation by 



(2) 



where and k z are the radii of gyration about OO' and ZZ' 
respectively. 

I37. 1 Extension of the Two Previous Articles to 
Solid Bodies. (a) Let ZX and ZY (Fig. 160) represent (by 
their traces) two planes perpendicular to the plane of the paper 
and to each other, both passing through the e.g. of a solid 
body. 

Let P be a typical particle of the body, its weight being w^ 

1 This article may be omitted on first reading. The student acquainted 
with the integral calculus will readily apply the second theorem to simple 
solids. 



Moments of Inertia Rotation 



193 




and its distances from the planes ZY and ZX being x and y l 
respectively. Then, if r^ is the distance of P from an axis ZZ', 
which is the intersection 
of the planes XZ and 
YZ, and passes through 
the c.g. r? - xf + y?. 

Let I z be the moment 
of inertia of the body 
about ZZ', and I that 
about a parallel axis 
OO'. Let OO' be distant 
d from ZZ', and distant 
p and q from planes ZY and ZX respectively. 

Let other constituent particles of the body of weights 
7i'. 2 , ' 3 , iv I, etc., be at distances x. ta x s , x 4 , etc., from the 
plane ZY, and distances y^ y*, y-> etc., from the plane ZX 
respectively, the x distances being reckoned positive to the 
right and negative to the left of ZY, and the y distances being 
reckoned positive above and negative below ZX. Let ;:,, r. A , r 4 , 
etc., be the distances of the particles from ZZj. Let w l + w 2 
-f- / :i + , etc. = 2(w) or W, the total weight of the body. 

By definition 



FIG. 160. 



therefore I = ~{ 



+ it>i(f - Ji) 2 + 



, etc.} 



etc. - <z 

7v. 2 y 2 -f- 



, etc.) - 2 



+, etc.)} 



I = / 

<*> 



' 3 ;- 3 2 +, etc.) 



*94 Mechanics for Engineers 



since the planes XZ and YZ pass through the e.g. of the body 
(Art. 113). 

Hence \ = -d* 4- I z ..... (,) 

i 

W 

and dividing both sides of (i) by 

<b 



where / = radius of gyration about OO', and k z = radius of 
gyration about ZZ'. 
(b) Also 

1t\ o i '< > i MS o 

Z== I l 7" r " / ; ' 3 +>etc ' 

= T(*i a + J'l 2 ) + ?<*-' +^ + ^ + ^> +' etc " 

o o o 

- -(W + w^ 2 4- ^s^s 2 +, etc.) + - (w-j? + w 2 y 2 2 + 

b O 

v>s}'3* +, etc.) 

W i i 

-kj=-*w*)+j(wf) ......... (3) 

d ^> ^> 

, 2 _ 5(o'*5 , a 

^ = 



which may be written 

...... (4) 



where j; 2 and _y 2 are the /<?^ squares of the distances of the 
body from the planes YZ and XZ respectively. The two 
quantities x 1 and _j^ are in many solids easily calculated. 

138. Moment of Inertia of an Area. The moment 
of inertia I of a lamina about a given axis OO' in its plane 

/ *iet \ 

is 2( - *' 2 ) (Art. 133), where w is the weight of a constituent 



Moments of Inertia Rotation 195 

particle, and r its distance from the axis OO'. This quantity 

W 
is equal to . & (Art. 134), where k is the radius of gyration 

^> 

about this axis OO', and W is the total weight of the lamina, 
so that 




g 

In a thin lamina of uniform thickness /, the area a (Fig. 
161) occupied by a particle of 
weight w is proportional to w, for 
w = a . t . D, where D is the weight 
per unit volume of the material ; 

hence ^wr 1 




and similarly, W = A . t . D, where 

A is the total area of the lamina ; FIG. 161. 






A/D 



Thus the thickness and density of a lamina need not be 
known in order to find its radius of gyration, and an area may 
properly be said to have a radius of gyration about a given 
axis. 

The quantity 2(ar*) is also spoken of as the moment of 
ineiiia of the area of the lamina about the axis OO' from which 
a portion a is distant r. 

The double use of this term " moment of inertia " is un- 
fortunate. The "moment of inertia of an area" ^(ar*) or 
2 . A is not a true moment of inertia in the sense commonly 
used in mechanics, viz. that of Art. 133 ; it must be multiplied 
by the factor " mass per unit area " to make it a true moment of 
inertia. As before mentioned, the area has, however, a radius 
of gyration about an axis OO' in its plane defined by the 
equation 



ig6 Mechanics for Engineers 

Units. The units of the geometrical quantity 2(r 2 ), 
called moment of inertia of an area, depend only upon the units 
of length employed. If the units of length are inches, a 
moment of inertia of an area is written (inches) 4 . 

139. Moment of Inertia of Rectangular Area about 
Various Axes. Let ABCD (Fig. 162) be a rectangle, AB = */, 
~ BC =.b. The moment of inertia of the 



B C 

area ABCD about the axis OO' in the side 



U* 



AD may be found as follows. Suppose AB 
^ divided into a large number , of equal 
parts, and the area ABCD divided into n 

equal narrow strips, each of width -. The 



Fi whole of any one strip EFGH is practically 

at a distance, say, FA from AD, and if 

EFGH is the/th strip from AD, FA => X -. 

n 

Multiplying the area EFGH, viz. b x -, by the square of 
its distance from AD, we have 

(area EFGH) x FA 2 = b X - X 

n 

There are n such strips, and therefore the sum of the 
products of the areas multiplied by the squares of their distances 
from OO', which may be denoted by ^(ar 2 ), is 



or J* = - X 



n 3 6 6 



2 I 

When n is indefinitely great, = o, and - 2 = o, and the sum 

// ft- 

^(ar 2 ) becomes ^ X 2 or This is the " moment of inertia 

6 3 

of the area " about OO' ; or, the radius of gyration of the area 

about OO' being k 



Moments of Inertia Rotation 



197 



If ABCD is a lamina of uniform thickness of weight w, its 

w w 

true moment of inertia about OO' is k z - \ . d 2 . 

S ' g 

The radius of gyration of the same area ABCD about an 
axis PQ (Fig. 163) in the plane of the 
figure and parallel to OO' and distant 

- from it, dividing the rectangle into 



halves, can be found from the formula (2), 
Art. 136, viz. 

2 



-Q 



where k f = radius of gyration about PQ ; 
whence / P 2 = (i 

The sum ^(ar 1 



O C iR D 0' 

FIG. 163. 



about PQ is then 2(<*) X k\ = I'd X = ^ 



12 



Similarly, if s is the radius of gyration of the rectangle 
about RS 



B 



and therefore, if k G = radius of gyration about an axis through 
G (the e.g.) and perpendicular to the figure 

W = ^ + k* or T V( 2 + d*) (Art. 135 (2)) 
which is also equal to yVB(7 or JGB 2 . 

Example. A plane figure consists of a rectangle 8 inches by 
4 inches, with a rectangular hole 6 inches 
by 3 inches, cut so that the diagonals of 
the two rectangles are in the same straight 
lines. Find the geometrical moment of 
inertia of this figure, and its radius of 
gyration, about one of the short outer 
sides. 

Let IA be the moment of inertia of the 
figure about AD (Fig. 164), and k be its A 
radius of gyration about AD. 

Moment of inertia of abed '\ 

ahnnt AD / = T2( area abcd ^ X ( Slde *) + ( area abcd ) 




FIG. i 4 . 



x (JAB)* (Arts. 139 and 136) 
Moment of inertia of "I _ t 3 

ABCD about AD / ~ :J X 4 X 



198 Mechanics for Engineers 

Hence I.\ = moment of inertia of ABCD moment of inertia of abed 
= i . 4 8 3 - (^ x 6 3 x 3 + 6 x 3 x 4*) 
= * a - (54 + 288) = 340-6 (inches) 4 

The area of the figure is 

8x4-6x3 = 14 square inches 



therefore k z = _ = 24*33 (inches) 2 

14 
and k = 4'93 inches 



140. Moment of Inertia of a Circular Area about 
Various Axes. (i) About an axis OO' through O, its centre, 
and perpendicular to its plane. 

Let the radius OS of the circle (Fig. 165) be equal to R. 
Suppose the area divided into a large 
number , of circular or ring-shaped 

"D 

strips such as PQ, each of width 
Then the distance of the /th strip from 

TJ 

O is approximately / X -, and its 

n 

area is approximately 

FIG. 165. 

R R R 2 

27T X radius X width = 2?r x /-- = 2irp^ 

n n n 

The moment of inertia of this strip of area about OO' is 
then 

R 4 . 




R 2 ^ //R\ 2 

-2 X (' ) = 

& \ n / 



27T- 





and adding the sum of all such quantities for all the n strips 

R4 

. , ov ** / o i 8 I *? I 3 I J\& I I T \\ 

71 

+ 2n 3 + n- 



R4 (( + i)) 2 R 4 ;/ 4 

= 27 vr~Y- r =27r -/7-- 



2 I 

When is indefinitely great, - = o and -^ = o, and the 



Moments of Inertia- -Rotation 



199 



rR 4 



becomes , which is the " moment of inertia of 



sum 

the circular area " about OO'. 






And since 2(ar 2 ) about OO' = , if we divide each side 



of the equation by the area 



of the circle 



R 2 



where k is the radius of gyration of the circular area about 
an axis OO' through its centre and perpendicular to its 
plane. 

(2) About a diameter. 

Again, if A and k c are the radii of gyration of the same 
area about the axes AB and CD 
respectively (Fig. 166) 



hence k* = k<? = 



R 2 




R 2 





from which the relations between 

the moments of inertia about AB, 

DC, and OO' may be found by 

multiplying each term by TrR 2 . 

That is, the moment of inertia of 

the circular area about a diameter is half that about an axis 

through O and perpendicular to its plane. 

Example. Find the radius of gyration of a ring-shaped 
area, bounded outside by a circle of radius a, and inside by a 
concentric circle of radius b, about a diameter of the outer 
circle. 

The moment of inertia of the area bounded by the outer circle, 

about AB (Fig. 167) is ; that of the inner circular area about 
4 



200 



Mechanics for Engineers 



the same line is - - ; hence that of the ring-shaped area is ^(rt 4 -^ 4 ). 

The area is ^(d 1 $} ; hence, if k is the radius of gyration of 
the ring-shaped area about AB 




7- IT (a 2 l>-) 



+ 



Note that & = * 



"+* 






= ("-+.} 

\ 2 / 



Y 

/' 



so that when a and b 



FIG. i6 7 . 



are nearly equal, i.e. when rt b 
is a small quantity, the radius of 



of the inner and outer radii. 



gyration / , about the axis O, approaches the arithmetic mean 
a_+ b 

2 

141. Moment of Inertia of a Thin Uniform Rod. The 

radius of gyration of a thin rod d units long and of uniform 
material, about an axis through one end and perpendicular to 
the length of the rod, will evidently be the same as that of a 
narrow rectangle d units long, which, by Art. 139, is given by 
the relation h* = % d 2 , where k is the required radius of gyration. 
Hence, if the weight of the rod is W Ibs., its moment of inertia 

, . W,_ W J* 
about one end is k* or 
g S 3 
Similarly, its moment of inertia about an axis through the 

W d" 2 - 
middle point and perpendicular to the length is 

142. Moment of Inertia of a Thin Circular Hoop. 

(i) The radius of the hoop being R, all the matter in it is 
at a distance R from the centre of the hoop. Hence the 
radius of gyration about an axis through O, the centre of the 
hoop, and perpendicular to its plane, is R, and the moment 

w 
of inertia about this axis is . R 2 , where W is the weight of 

A 

the hoop. 



Moments of Inertia Rotation 



201 



(2) The radius of gyration about diameters OX and OY 
(Fig. 1 68) being k x and k u respectively 



hence k% = k^ = 



R 2 




FIG. 



and the moment of inertia about 
any diameter of the hoop is 
W R 2 

143. Moment of Inertia of 
Uniform Solid Cylinder. (i) 

About the axis OO' of the cylinder. 

The cylinder may be looked upon as divided into a large 

number of circular discs (Fig. 169) by planes perpendicular 

to the axis of the cylinder. 

The radius of gyration of each 

disc about the axis of the cylinder 

R 2 

is given by the relation k z = > 

where k is radius of gyration of 

the disc, and R the outside radius 

of the cylinder and discs. If the 

weight of any one disc is /, and 

that of the whole cylinder is W, the moment of inertia of one 

disc is 

w R 2 




and that of the whole cylinder is 
R 2 \ R 2 



va/ R a \ R* 

l = _ 

\g 2 / 2g 



W R 2 



R 2 

and the square of the radius of gyration of the cylinder is . 

2 

(2) About an Axis perpendicular to that of the 
Cylinder and through the Centre of One End. Let OX 

(Fig. 170) be the axis about which the moment of inertia 



202 



Mechanics for Engineers 



of the cylinder is required. Let R be the radius of the 
cylinder, and / its length. 

Let o? = the mean square of the distance of the constituent 

particles from the plane YQO'Y' ; 
f = the mean square of the distance of the constituent 

particles from the plane OXX'O' ; 
= the radius of gyration of the cylinder about OO'. 

Then k* = j? + j? by Art. 137 (4) 
and from the symmetry of the solid, o? = y' 

R 2 
hence k* or = 2oc = z* 



and x 1 = = y* 

The cylinder being supposed divided into thin parallel rods 
all parallel to the axis and / units long, the mean square of the 




FIG. 170. 

distance of the particles forming the rod from the plane VOX 
of one end, is the same as the square of the radius of gyration 
of a rod of length / about an axis perpendicular to its length 

/2 

and through one end, viz. - (Art. 141). The axis OX is the 

<J 

intersection of the planes XOO'X' and VOX, the end plane ; 
hence, if x is the radius of gyration about OX 



TJ2 



. 137(2)) 



Moments of Inertia Rotation 203 

(3) Also, if k G is the radius of gyration about a parallel axis 
through G, the e.g. of the cylinder 






P R 2 / 2 

or k<? = k K -- = -- 
4 4 12 

The moments of inertia of the cylinder about these various 
axes are to be found by multiplying the square of the radius of 

w 
gyration about that axis by the mass , where u> is the weight 

cS 

of the cylinder, in accordance with the general relation I 
t. 134). 



Example. A solid disc flywheel of cast iron is 10 inches in 
diameter and 2 inches thick. If the weight of cast iron is o'26 Ib. 
per cubic inch, find the moment of inertia of the wheel about its 
axis in engineers' units. 

The volume of the flywheel is IT x 5 2 x 2 = SOTT cubic inches 
the weight is then 0*26 x 50^- = 40^9 Ibs. 

and the mass is ~ i'27 units 
32-2 

The square of the radius of gyration is i^o) 2 (feet)-. Therefore 
the moment of inertia is 

1*27 x oVs 0*1104 ur >it 

EXAMPLES XVII, 

1. A girder of I-shaped cross-section has two horizontal flanges 5 inches 
broad and I inch thick, connected by a vertical web 9 inches high and I 
inch thick. Find the " moment of inertia of the area " of the section about 
a horizontal axis through its e.g. 

2. Fig. 171 represents the cross-section of a cast-iron girder. AB is 4 
inches, BC I inch, EF I inch ; EH is 6 inches, KL is 8 inches, and KN is 
I '5 inches. Find the moment of inertia and radius of gyration of the area 
of the section about the line NM. 

3. Find, from the results of Ex. 2, the moment of inertia and radius of 
gyration of the area of section about an axis through the e.g. of the section 
and parallel to XM. 



204 



Mechanics for Engineers 



B 



i 
c 

( 




1 


F C 
G L 


1 E 
H 







M 



4. Find the moment of inertia of the area enclosed between two con- 
centric circles of 10 inches and 8 inches diameter respectively, about a 
diameter of the circles. 

5. Find the radius of gyration of the area bounded on the outside by a 

circle 12 inches diameter, and on the 
inside by a concentric circle of 10 
inches diameter, about an axis through 
the centre of the figure and perpen- 
dicular to its plane. 

6. The pendulum of a clock con- 
sists of a straight uniform rod, 3 feet 
long and weighing 2 Ibs., attached 
to which is a disc o - 5 foot in diameter 
and weighing 4 Ibs. , so that the centre 
of the disc is at the end of the rod. 
Find the moment of inertia of the 
pendulum about an axis perpendicular 
to the rod and to the central plane 
of the disc, passing through the rod 2*5 feet from the centre of the disc. 

7. Find the radius of gyration of a hollow cylinder of outer radius a and 
inner radius b about the axis of the cylinder. 

8. Find the radius of gyration of a flywheel rim 3 feet in external 
diameter and 4 inches thick, about its axis. If the rim is 6 inches broad, and 
of cast-iron, what is its moment of inertia about its axis? Cast iron 
weighs O'26 Ib. per cubic inch. 

144. Kinetic Energy of Rotation. If a particle of a 
body weighs w^ Ibs., and is rotating with angular velocity w 
about a fixed axis 1\ feet from it, its speed is o>i\ feet 
per second (Art. 33), and its kinetic energy is therefore 

1 1U 

l . (<ai\Y foot-lbs. (Art. 60). Similarly, another particle of 

2 ,s" 

the same rigid body situated r z feet from the fixed axis of 
rotation, and weighing w 2 Ibs., will have kinetic energy equal to 

-.-. (o>r 2 ) 2 ; and if the whole body is made up of particles 

weighing w^ w 2 , w 3 , ze> 4 , etc., Ibs., situated at r 1} r 2 , r 3 , r^ etc., 
feet respectively from the axis of rotation, the total kinetic 
energy of the body will be 



etc 



or W( -V, 8 + -V + -V 3 2 -K etc.) foot- 

- \^ g g J 



lbs. 



Moments of Inertia Rotation 205 

The quantity (*V + ^V + % +, etc.) or 

O <b u ' 

has been defined (Art. 133) as the moment of inertia I, of the 
body about the axis. Hence the kinetic energy of the body is 

W W 

Iw 2 , or \ . K 2 or, or \ V 2 foot-lbs., where K = radius of 

<^ S 

gyration of the body in feet about the axis of rotation, and 
V = velocity of the body in feet per second at that radius of 
gyration. This is the same as the kinetic energy ^MV 2 or 

W W 

- -V 2 of a mass M or , all moving with a linear velocity V. 

The kinetic energy of a body moving at a given linear 
velocity is proportional to its mass ; that of a body moving 
about a fixed axis with given angular velocity is proportional 
to its moment of inertia. We look upon the moment of 
inertia of a body as its rotational inertia, i.e. the measure of 
its inertia with respect to angular motion (see Art. 36). 

145. Changes in Energy and Speed. If a body of 
moment of inertia I, is rotating about its axis with an angular 
velocity w n and has a net amount of work E done upon it, 
thereby raising its velocity to w.>; then, by the Principle of 
Work (Art. 61) 

il(o> 2 2 - Wl 2 ) = E 

W 

or i-K 2 ( W3 2 - Wl 2 ) = E 

& 

W 

or i-(V 2 2 - V, 2 ) = E 



where K = radius of gyration about the axis of rotation, and 
V 2 and Vj are the final and initial velocities respectively at 
a radius K from the axis. 

Hence the change of energy is equal to that of an equal 
weight moving with the same final and initial velocities as a 
point distant from the axis by the radius of gyration of the 
body. If the body rotating with angular velocity o> 2 about 
the axis is opposed by a tangential force, and does work of 
amount E in overcoming this force, its velocity will be reduced 



2o6 Mechanics for Engineers 

to MU the loss of kinetic energy being equal to the amount of 
work done (Art. 61). 

146. Constant resisting Force. Suppose a body, such 
as a wheel, has a moment of inertia I, and is rotating at an 
angular velocity w 2 about an axis, and this rotation is opposed 
by a constant tangential force F at a radius r from the axis 
of rotation, which passes through the centre of gravity of the 
body. Then the resultant centripetal force on the body is 
zero (Art. 130). The particles of the body situated at a 
distance r from the centre are acted on by a resultant or 
effective force always in the same straight line with, and in 
opposite direction to, their own velocity, and therefore have 
a constant retardation in their instantaneous directions of 
motion (Art. 40). Hence the particles at a radius r have 
their linear velocity, and therefore also their angular velocity, 
decreased at a constant rate ; and since, in a rigid body, the 
angular velocity of rotation about a fixed axis of every point 
is the same, the whole body suffers uniform angular retardation. 

Suppose the velocity changes from o> 2 to wj in / seconds, 
during which the body turns about the axis through an angle 

radians. The uniform angular retardation a is - 2 - \ 

Also the work done on the wheel is Fr x 6 (Art. 57), 
hence 

F . r. 6 = I(w 2 2 w a 2 ) = loss of kinetic energy . (i) 
The angle turned through during the retardation period is 



Note that F . r is the moment of the resisting force or the 
resisting torque. 

Again, o> 2 2 c^ 2 = (o> 2 + wj)^ wj 
and w 2 o>i = a/ 

and o)j + u> 2 = twice the average angular velocity 
during the retardation 



Moments of Inertia Rotation 207 

Hence the relation 

F.r.0 = ll(<o 2 2 - V) 
may be written 

YrO = 11. a./. ~ 
or F . r = I . a . . , . . . . (2) 

i.e. the moment of the resisting force about the axis of rotation 
is equal to the moment of inertia of the body multiplied by its 
angular retardation. 

Similarly, if F is a driving instead of a resisting force, the 
same relations would hold with regard to the rate of increase 
of angular velocity, viz. the moment of the accelerating force 
is equal to the moment of inertia of the body multiplied by the 
angular acceleration produced. Compare these results with 
those of Art. 40 for linear motion. 

We next examine rather more generally the relation 
between the angular velocity, acceleration, and inertia of a 
rigid body. 

147. Laws of Rotation of a Rigid Body about an 
Axis through its Centre of Gravity. Let w be the weight 
of a constituent particle of the 
body situated at P (Fig. 172), 
distant r from the axis of rota- 
tion O ; let a> be the angular 
velocity of the body about O. 
Then the velocity v of P is tar. 

Adding the vectors repre- 
senting the momenta of all F ' G - 172- 
such particles, we have the total momentum estimated in any 
particular direction, such as OX (Fig. 172), viz. 

^./w \ w.. 

2( - v cos 6), or -2(wr cos 6) 

But 2 (wr cos 0) is zero when estimated in any direction if 
r cos 6 is measured from a plane through the e.g. Hence the 
total linear momentum resolved in any given direction is zero. 

Moment of Momentum, or Angular Momentum of 




2o8 Mechanics for Engineers 

a Rigid Body rotating about a Fixed Axis. This is 
defined as the sum of the products of the momenta of all the 
particles multiplied by their respective distances from the axis, 



^ 
or 2 .v 

r / 

Yw \ ^(w \ 

But 3(- .v.r)= ^- . cor . r) = 



or the angular momentum is equal to the moment of inertia 
(or angular inertia) multiplied by the angular velocity. 

Suppose the velocity of P increases from i^ to z> a , the 
angular velocity increasing from ^ to w. 2 , the change of 
angular momentum is 



If the change occupies a time / seconds, the mean rate of 
change of angular momentum of the whole body is 



where/" is the average acceleration of P during the time /, and 

w 

f or F is the average effective accelerating force on the 

o> 

particle at P, acting always in its direction of motion, i.e. 

acting always tangentially to the circular path of P (see Art. 40). 

Also 3 (F . r) is the average total moment of the effective 
or net forces acting on the various particles of the body or the 
average effective torque on the body. 

If these average accelerations and forces be estimated over 
indefinitely small intervals of time, the same relations are true, 
and ultimately the rate of change of angular momentum is 
equal to the moment of the forces producing the change, so 
that 

rate of change of Iw = 2(Fr) = M 

= total algebraic moment of effective 
forces, or effective torque 



Moments of Inertia Rotation 209 

Also 

rate of change of Iw = I X rate of change of o> 

or I . a, where a is the angular acceleration or rate of change of 
angular velocity. Hence 

= M = la 



a result otherwise obtained for the special case of uniform 
acceleration in (2), Art. 146. 

Problems can often be solved alternately from equation 
(i) or equation (2) (Art. 146), just as in the case of linear 
motion the equation of energy (Art. 60) or that of force (Art. 
51) can be used (Art. 60). 

Example i. A flywheel weighing 200 Ibs. is carried on a 
spindle 2*5 inches diameter. A string is wrapped round the spindle, 
to which one end is loosely attached. The other end of the string 
carries a weight of 40 Ibs., 4 Ibs. of which is necessary to overcome 
the friction (assumed constant) between the spindle and its 
bearings. Starting from rest, the weight, pulling the flywheel 
round, falls vertically through 3 feet in 7 seconds. Find the 
moment of inertia and radius of gyration of the flywheel. 

The average velocity of the falling weight is f foot per second, 
and since under a uniform force the acceleration is uniform, the 
maximum velocity is 2 x 2 or f foot per second. 

The net work done by the falling weight, i.e. the whole work 
done minus that spent in overcoming friction, is 

(40 - 4)3 foot-lbs. = 1 08 foot-lbs. 
The kinetic energy of the falling weight is 

*.^ 2 .(?) 2 = 0-456 foot-lb. 

If I = moment of inertia of the flywheel, and a = its angular 
velocity in radians per second. By the principle of work (Art. 61) 

I 2 4- 0-456 = 108 foot-lbs. 

ilw 2 = 108 - 0*456 = 107-544 foot-lbs. 

The maximum angular velocity <a is equal to the maximum 
linear velocity of the string in feet per second divided by the radius 
of the spindle in feet, or 

P 



2io Mechanics for Engineers 



12 / i-25 875 

= 8*22 radians per second 
therefore |I x (8 - 22) 2 = 107-544 

T 107-544 x 2 215*1 
1 = ,o \o -- = -TT-T = 3*io units 
(o 22) 15 67-0 

And if k = radius of gyration in feet, since the wheel weighs 
200 Ibs. 

32-2' 

2 = 0-5l8 (foot) 2 

k 0716 foot or 8'6 inches 

Example a. An engine in starting exerts on the crank-shaft 
for one minute a constant turning moment of 1000 Ib.-feet, and 
there is a uniform moment resisting motion, of 800 Ib.-feet. The 
flywheel has a radius of gyration of 5 feet and weighs 2000 Ibs. 
Neglecting the inertia of all parts except the flywheel, what speed 
will .the engine attain during one minute ? 

(1) Considering the rate of change of angular momentum 

The effective turning moment is loco 800 = 200 Ib.-feet 

The moment of inertia of the flywheel is x 5 2 = 1552 units 

32-3 

Hence if a = angular acceleration in radians per second per second 
200 = 15520 (Art. 146 (2)) 

o = - - = 0-1287 radian per second per second 

And the angular velocity attained in one minute is 

60 x 0*1287 = 774 radians per second 

774 x 60 
or - - = 74 revolutions per minute 

27T 

(2) Alternatively from considerations of.energy. 
If u = angular velocity acquired 

CO 

- = mean angular velocity 

Total angle turned through ) , u> ,. 

. t = DO x - = 3Ow radians 

in one minute 2 

Net work done in one minute = 200 x 300- foot-lbs. 
200 x 3oco = JIw 2 

6ooow = J. 1552. to 2 

w = I200 = 7*74 radians per second 
X 55 2 as before 



Moments of Inertia Rotation 



211 



Example 3. A thin straight rod of uniform material, 4-5 feet 
long, is hinged at one end so that it can turn in a vertical plane. 
It is placed in a horizontal position, and Q ~ 

then released. Find the velocity of the 
free end (i) when it has described an 
angle of 30, (2) when it is vertical. 

(i) After describing 30 the centre 
of gravity G (Fig. 173), which is then 



at G 
ON. 



has fallen a vertical distance 



ON = OGj cos 60 = AOG, = I x 2-25 
= 1-125 feet 




FIG. 173. 



If W is the weight of the rod in pounds, the work done by 
gravitation is 

W x i 'i 25 foot-lbs. 

/W \ 
The moment of inertia of the rod ( ^*J is 

W (4-5)2 _ 27 W 



If ta l is the angular velocity of the rod, since the kinetic energy 
of the rod must be ri25\V foot-lbs. 

Q ^ - . , 

' g 

w i 2 = I x 2 8 T x 3 2 '2 = 1073 

= 3'28 radians per second 
the velocity of A in position A l is then 

3'28 x 4'5 = 1474 feet per second 

(2) In describing 90 G falls 2 25 feet, and the kinetic energy is 
then 2'25\V foot-lbs. 

And if <a. 2 is the angular velocity of the rod 

J . 23-. _ a, 2 _ ->">r\y 

s ' 

wf = % x ./- x 32'2 = 2 1 '47 
o) 2 = 4-63 radians per second 

and the velocity of A in the position A 2 is 

4^63 x 4'5 = 20*82 feet per second 




212 Meclianics for Engineers 

148. Compound Pendulum. In Art. 71 the motion of 
a " simple pendulum " was investigated, and it was stated that 
such a pendulum was only approximated to by any actual 
pendulum. We now proceed to find the 
simple pendulum equivalent (in period) 
to an actual pendulum. 

Let a body be suspended by means 
of a horizontal axis O (Fig. 174) perpen- 
dicular to the figure and passing through 
the body. Let G be the e.g. of the body 
in any position, and let OG make any 
angle 6 with the vertical plane (OA) 
through O. 

FlG - *- Suppose that the body has been raised 

to such a position that G was at B, and then released. Let 
the angle AOB be <, and OG = OB = OA = //. 

The body oscillating about the horizontal axis O constitutes 
a pendulum. 

Let / = length of the simple equivalent pendulum (Art. 71) ; 
I = the moment of inertia of the pendulum about the 

axis O ; 

= radius of gyration about O ; 

k G radius of gyration about a parallel axis through G. 
Let W be the weight of the pendulum, and let M and N 
be the points in which horizontal lines through B and G 
respectively cut OA. 

When G has fallen from B to G, the work done is 

W x MN = W(ON - OM) = W(A cos 0-A cos <) 
= W/*(cos - cos <) 

Let the angular velocity of the pendulum in this position 
be at, then its kinetic energy is --Ito 2 (Art. 144), and by the 
principle of work (Art. 61), if there are no resistances to 
motion the kinetic energy is equal to the work done, or 



s - cos 
and therefore 



(cos 6 cos 0) . . . . (i) 



Moments of Inertia Rotation 



213 



Similarly, if a particle (Fig. 175) be attached to a point 
O' by a flexible thread of length /, and be released from a 
position B' such that B'6'A = <, O'A 
being vertical, its velocity v when passing 
G' such that G'O'A = 6 is given by 

V* = 2 g , M'N' = 2,/(cOS 6 COS <) 

and its angular velocity o> about O f 




being 



FIG. 175. 



= -(cos cos 



The angular velocity of a particle (or of a simple pendulum) 
given by equation (2) is the same as that of G (Fig. 174) 
given by equation (i), provided 



i.e. provided 



/ = 



This length r is then the length of the simple pendulum 

equivalent to that in Fig. 174, for since the velocity is the same 
at any angular position for the simple pendulum of length / 
and the actual pendulum, their times of oscillation must be the 
same. Also, since 

k? = k<? -f h 2 (Art. 137 (2)) 



r 
The point C (Fig. 174), distant -f- + h from O, and in the 

line OG is called the " centre of oscillation" The expression 

k ' l k(? 

-j- + h shows that it is at a distance -r- beyond G from O. 

A particle placed at C would oscillate in the same period 
about O as does the compound pendulum of Fig. 174. 



214 Mechanics for Engineers 

Example. A flywheel having a radius of gyration of 3*25 feet 
is balanced upon a knife-edge parallel to the axis of the wheel and 
inside the rim at a distance of 3 feet from the axis of the wheel. 
If the wheel is slightly displaced in its own plane, find its period of 
oscillation about the knife-edge. 

The length of the simple equivalent pendulum is 

ffacV 

3 + ^ *' = 3 + 3-5208 = 6-5208 feet 

Hence the period is 2H-* / = 276 seconds 
V 32*2 

149. The laws of rotation of a body about an axis may be 
stated in the same way as Newton's laws of motion as follows : 

Law T. A rigid body constrained to rotate about an axis 
continues to rotate about that axis with constant angular 
velocity- except in so far as it may be compelled to change 
that motion by forces having a moment about that axis. 

Law 2. The rate of change of angular momentum is pro- 
portional to the moment of the applied forces, or torque about 
the axis. With a suitable choice of units, the rate of change 
of angular momentum is eqtial to the moment of the applied 
forces, or torque about the axis. 

Law 3. If a body A exerts a twisting moment or torque 
about a given axis on a body B, then B exerts an equal and 
opposite moment or torque about that axis on the body A. 

150. Torsional Simple Harmonic Motion. If a rigid 
body receives an angular displacement about an axis, and the 
moment of the forces acting on it tending to restore equilibrium 
is proportional to the angular displacement, then the body 
executes a rotary vibration of a simple harmonic kind. Such 
a restoring moment is exerted when a body which is suspended 
by an elastic wire or rod receives an angular displacement 
about the axis of suspension not exceeding a certain limit. 

Let M = restoring moment or torque in Ib.-feet per radian 

of twist ; 
I = moment of inertia of the body about the axis of 

suspension in engineer's units ; 
p. = angular acceleration of the body in radians per 

second per second per radian of twist. 



Moments of Inertia Rotation 



215 



Then M = I . //. (Art. 147) 
M 



or //. = 



Then, following exactly the same method as in Art. 68, if 
Q (Fig. 176) rotates uniformly with angular velocity \f p. in a 
circle centred at O and of radius OA, which represents to scale 
the greatest angular displacement of 
the body, and P is the projection 
of Q on OA, then P moves in the 
same way as a point distant from O 
by a length representing the angular 
displacement 0, at any instant to the 
same scale that OA represents the 
extreme displacement. The whole 
argument of Art. 68 need not be 
repeated here, but the results are FlG '7& 

Angular velocity for an angular displacement 0, represented 
by OM, is 




Angular acceleration for an angular displacement 0, repre- 
sented by PO, is vV 0. 






T = time of complete vibration = = seconds 



or, since 



M 



Example. A metal disc is 10 inches diameter and weighs 
6 Ibs. It is suspended from its centre by a vertical wire so that 
its plane is horizontal, and then twisted. When released, how many 
oscillations will it make per minute if the rigidity of the suspension 
wire is such that a twisting moment of i Ib.-foot causes an angular 
deflection of 10 ? 



The twisting moment per radian twist is 



i 80 



573 lb.-feet 



The square of radius of gyration is ICi 5 ;-) 2 = 0*0862 (foot) 2 



2l6 



Mechanics for Engineers 



The moment of inertia is ----- x o'o862 = o'oi6i5 unit 
Hence the time of vibration 



ion is 277. / = 2n- / O ' OI " I 5 
V M ~ -'V 573 



573 
second 



The number of vibrations per minute is 

60 \ = 178 



then 



Q'337 



151. It is evident, from Articles 144 to 150, that the rotation 
of a rigid body about an axis bears a close analogy to the 
linear motion of a body considered in Chapters I. to IV. 

Some comparisons are tabulated below. 

Angular or Rotational. 

Moment of inertia, I. 

Angular displacement, 0. 
Angular velocity, w. 
Angular acceleration, a. 
Moment of force, or torque, M. 

Angular momentum, I . w. 


Average angular velocity, - 

Average angular acceleration, 



Linear. 
W 

Mass or inertia, - or m. 

g 

Length, /. 
Velocity, v. 
Acceleration,/. 
Force, F. 

W 

Momentum, . v or mv. 

Average velocity, - . 
Average acceleration, - t 



w (z\ z> 2 ) 
Average force, - - - - 

o 



Work of constant force, F . /. 

f UO 

Kinetic energy, \ v* or \ mi z . 

* 
Period of simple vibration, 



Im / 

* / or 27TA / 
\f e \/ 



m w , 

or 27TA -.where 
e 

e = force per unit displace- 
ment. 



Average moment or torque, 

!((! fa) 2 ) 

/ 

Work of constant torque, M . 0. 

Kinetic energy, |Iw 2 . 

Period of simple vibration, 
27r \/ ;r?, where M = torque 
per radian displacement. 



Moments of Inertia Rotation 



217 



The quantities stated as average values have similar mean- 
ings when the averages are reckoned over indefinitely small 
intervals of time, or, in other words, they have corresponding 
limiting values. 

152. Kinetic Energy of a Rolling Body. We shall 
limit ourselves to the case of a solid of revolution rolling along 
a plane. The e.g. of the 
solid will then be in the 
axis of revolution about 
which the solid will rotate 
as it rolls. Let R be 
the extreme radius of the 
body at which rolling 
contact with the plane 
takes place (Fig. 177); 
let the centre O be moving 
parallel to the plane with 

a velocity V. Then any point P on the outside circumference 
of the body is moving with a velocity V relative to O, the 
angular velocity of P and of the whole body about O being 

TJ, or say w radians per second. 

Consider the kinetic energy of a particle weighing w Ibs. 
at Q, distant OQ or r from the axis of the body. Let OQ 
make an angle QUA = with OA, the direction of motion of 
O. Then the velocity v of Q is the resultant of a velocity V 
parallel to OA, and a velocity wr perpendicular to OQ, and 
is such that 

v 1 = (w;-) 2 + V 2 + 2 cor . V . cos (90 + 6) 
Hence the kinetic energy of the particle is 




1 W 

|-(V a + V 2 - 

o 



sin 0) 



The total kinetic energy of the body is then 

V 2 ) = S| (o>V 2 + V 2 - 2torV sin 0)} 
2 J la ') 



V 2 , 

*s' 



*s 



. r sin 0) 



21 8 Mechanics for Engineers 

Now, 2(*w. sin (9) = o (Art. 113 (3)) 

(W 2\ 

and 2{~ r ) = I, the moment of inertia of the solid 



g 

w 



I TV 

hence 21 - 

\2j 






about the axis O 
W 



= kinetic energy of rotation about O + 
kinetic energy of an equal weight 
moving with the linear velocity of 
the axis. 



This may also be written 

iW/ 79 . ^,o 
i-( 2 + R 



or 



+ 



^ 2 \ 
J 



where k is the radius of gyration about the axis 0. The 

W / / 2 \ 
kinetic energy V 2 ( i +^ 2 ) ' s tnen tne same as that of a 

weight w( i + ITS) moving with a velocity V of pure trans- 

lation, i.e. without rotation. 

In the case of a body rolling down a plane inclined to 

the horizontal (Fig. 178), 
using the same notation as 
in the previous case, the 
component force of gravity 
through O and parallel to 
the direction of motion down 

W 
the plane is . sin 0. In 

g 

rolling a distance s down 
the plane, the work done 
is W sin . s. Hence the 




Fie. 178. 

kinetic energy stored after the distance s is 
,W T 



+ jp) = W sin 6 . s (Art. 61) 

a R2 
or V 2 = zsg sin "nirT~p 

This is the velocity which a body would attain in moving 



Moments of Inertia Rotation 219 

without rotation a distance s from rest under an acceleration 

R 2 
g sin ^irrnrw 1 Hence the effect of rolling instead of sliding 

down the-plane is to decrease the linear acceleration and linear 
velocity attained by the axis in a given time in the ratio 

R2-+Ta < See Art ' 28 >' 

We may alternatively obtain this result as follows ; 
Resolving the reaction of the (rough) plane on the body at T 
into components N and F, normal to the plane and along 
it respectively, the net force acting down the plane on the 
body is W sin 6 F ; and if a = angular acceleration of the 
body about O, and/= linear acceleration down the plane 



. 

But la = FR (Art. 146 (2)) 
F being the only force which has any moment about O ; 

la I/ 
hence F = R = ^2 

and the force acting down the plane is W sin 6 ^2- 

force acting down the plane /,.. I/\ . W 

Hence /=- f , ,- - = I W sin 0--^ 2 } - 

mass of body \ R 2 / g 

3 



R 2 
or/= gsm X 



R 2 + * 

Example. A solid disc rolls down a plane inclined 30 to the 
horizontal. How far will it move down the plane in 20 seconds 
from rest ? What is then the velocity of its centre, and if it weighs 
10 Ibs., how much kinetic energy has it ? 

The acceleration of the disc will be 

R 2 
32*2 x sin 30 x 3 = 32*2 x i x if 

R 2 + - 

2 

= 1073 feet per second per second 



22O Mechanics for Engineers 

In 20 seconds it will acquire a velocity of 

20 x 1073 = 214-6 feet per second 
Its average velocity throughout this time will be 

214-6 

^ = 107-3 feet per second 

It will then move 

107-3 x 20 = 2146 feet 

corresponding to a vertical fall of 2146 sin 30 or 1073 feet. 

The kinetic energy will be equal to the work done on it in 
falling 1073 f eet j *> e ' I0 73 x 10 = 10,730 foot-lbs. 

EXAMPLES XVIII. 

1. What is the moment of inertia in engineer's units of a flywheel 
which stores 200,000 foot-lbs. of kinetic energy when rotating 100 times 
per minute ? 

2. A flywheel requires 20,000 foot-lbs. of work to be done upon it 
to increase its velocity from 68 to 70 rotations per minute. What is its 
moment of inertia in engineer's units ? 

3. A flywheel, the weight of which is 2000 Ibs., has a radius of 
gyration of 3-22 feet. It is carried on a shaft 3 inches diameter, at the 
circumference of which a constant tangential force of 50 Ibs. opposes the 
rotation of the wheel. If the wheel is rotating 60 times per minute, how 
long will it take to come to rest, and how many rotations will it make in 
doing so? 

4. A wheel 6 feet diameter has a moment of inertia of 600 units, and 
is turning at a rate of 50 rotations per minute. What opposing force 
applied tangentially at the rim of the wheel will bring it to rest in one 
minute ? 

5. A flywheel weighing I -5 tons has a radius of gyration of 4 feet. 
If it attains a speed of 80 rotations per minute in 40 seconds, find the mean 
effective torque exerted upon it in pound-feet ? 

6. A weight of 40 Ibs. attached to a cord which is wrapped round the 
2-inch spindle of a flywheel descends, and thereby causes the wheel to 
rotate. If the weight descends 6 feet in 10 seconds, and the friction of the 
bearing is equivalent to a force of 3 Ibs. at the circumference of the spindle, 
find the moment of inertia of the flywheel. If it weighs 212 Ibs., what is 
its radius of gyration ? 

7. If the weight in Question 6, after descending 6 feet, is suddenly 
released, how many rotations will the wheel make before coming to rest ? 

8. A flywheel weighing 250 Ibs. is mounted on a spindle 2'5 inches 



Moments of Inertia Rotation 221 

diameter, and is caused to rotate by a falling weight of 50 Ibs. attached to 
a string wrapped round the spindle. After falling 5 feet in 8 seconds, the 
weight is detached, and the wheel subsequently makes 100 rotations before 
coming to rest. Assuming the tangential frictional resisting force at the 
circumference of the axle to be constant throughout the accelerating and 
stopping periods, find the radius of gyration of the wheel. 

9. A rod is hinged at one end so that it can turn in a vertical plane 
about the hinge. The rod is turned into a position of unstable equilibrium 
vertically above the hinge and then released. Find the velocity of the 
end of the rod (i) when it is horizontal; (2) when passing through its 
lowest position, if the rod is 5 feet long and of uniform small section 
throughout. 

10. A circular cylinder, 3 feet long and 9 inches diameter, is hinged 
about an axis which coincides with the diameter of one of the circular' ends. 
The axis of the cylinder is turned into a horizontal position, and then the 
cylinder is released. Find the velocity of the free end of the axis (i) after 
it has described an angle of 50! (2) when the axis is passing through its 
vertical position. 

n. A flywheel weighs 5 tons, and the internal diameter of its rim is 
6 feet. When the inside of the rim is supported upon a knife-edge passing 
through the spokes and parallel to its axis, the whole makes, if disturbed, 
21 complete oscillations per minute. Find the radius of gyration of the 
wheel about its axis, and the moment of inertia about that axis. 

12. A cylindrical bar, 18 inches long and 3 inches diameter, is suspended 
from an axis through a diameter of one end. If slightly disturbed from 
its position of stable equilibrium, how many oscillations per minute will it 
make ? 

13. A piece of metal is suspended by a vertical wire which passes 
through the centre of gravity of the metal. A twist of 8'5 is produced 
per pound-foot of twisting moment applied to the wire, and when the 
metal is released after giving it a small twist, it makes 150 complete 
oscillations a minute. Find the moment of inertia of the piece of metal 
in engineer's or gravitational units. 

14. A flywheel weighing 3 tons is fastened to one end of a shaft, the 
other end of which is fixed, and the torsional rigidity of which is such that 
it twists o'4 per ton-foot of twisting moment applied to the flywheel. If 
the radius of gyration of the flywheel and shaft combined is 3 feet, find the 
number of torsional vibrations per minute which the wheel would make if 
slightly twisted and then released. 

15. The weight of a waggon is 2 tons, of which the wheels weigh \ ton. 
The diameter of the wheels is 2 feet, and the radius of gyration 0*9 foot. 
Find the total kinetic energy of the waggon when travelling at 40 miles 
per hour, in foot-tons. 

16. A cylinder is placed on a plane inclined 15 to the horizontal, and 
is allowed to roll down with its axis horizontal. Find its velocity after 
it has traversed 25 feet. 



222 Mechanics for Engineers 

17. A solid sphere rolls down a plane inclined a to the horizontal. 
Find its acceleration. (NOTE. The square of the radius of gyration of a 
sphere of radius R is R 2 .) 

1 8. A motor car weighs W Ibs., including four wheels, each of which 
weigh w Ibs. The radius of each wheel is a feet, and the radius of 
gyration about the axis is k feet. Find the total kinetic energy of the 
car when moving at v feet per second. 



CHAPTER X 

ELEMENTS OF GRAPHICAL STATICS 

1 53- I N Chapter VI. we considered and stated the condi- 
tions of equilibrium of rigid bodies, limiting ourselves to 
those subject to forces in one plane only. In the case of 
systems of concurrent forces in equilibrium (Chapter V.), we 
solved problems alternatively by analytical methods of resolu- 
tion along two rectangular axes, or by means of drawing vector 
polygons of forces to scale. We now proceed to apply the 
vector methods to a few simple systems of non-concurrent 
forces, such as were considered from the analytical point of 
view in Chapter VI., and to deduce the vector conditions of 
equilibrium. 

When statical problems are solved by graphical methods, it is 
usually necessary to first draw out a diagram showing correctly 
the inclinations of the lines of action of the various known 
forces to one another, and, to some scale, their relative posi- 
tions. Such a diagram is called a diagram of positions, or 
space diagram ; this is not to be confused with the vector 
diagram of forces, which gives magnitudes and directions, but 
not positions of forces. 

154. Bows' Notation. In this notation the lines of 
action of each force in the space diagram are denoted by 
two letters placed one on each side of its line of action. Thus 
the spaces rather than the lines or intersections have letters 
assigned to them, but the limits of a space having a particular 
letter to denote it may be different for different forces. 

The corresponding force in the vector diagram has the same 
two letters at its ends as are given to the spaces separated by 



224 



Mechanics for Engineers 



its line of action in the space diagram. We shall use capital 
letters in the space diagram, and the corresponding small letters 
to indicate a force in the vector diagram. The notation will 
be best understood by reference to an example. It is shown 
in Fig. 179, applied to a space diagram and vector polygon for 



Space Diagram 




FIG. 179. 

five concurrent forces in equilibrium (see Chapter V.). The 
four forces, AB, BC, CD, DE, of 5 Ibs., 6 Ibs., s| Ibs., and 
6-g- Ibs. respectively, being given, the vectors ab, be, cd, de are 
drawn in succession, of lengths representing to scale these 
magnitudes and parallel to the lines AB, BC, CD, and DE 
respectively, the vector ea, which scales 57 Ibs., represents the 
equilibrant of the four forces, and its position in the space 
diagram is shown by drawing a line EA parallel to ea from the 
common intersection of AB, BC, CD, and DE. (This is ex- 
plained in Chapter V., and is given here as an example of the 
system of lettering only.) 

155. The Funicular or Link Polygon. To find 
graphically the single resultant or equilibrant of any system of 
non-concurrent coplanar forces. Let the four forces AB, BC, 
CD, and DE (Fig. 180) be given completely, i.e. their lines of 
action (directions and positions) and also their magnitudes. 
First draw a vector ab parallel to AB, and representing by its 
length the given magnitude of the force AB ; from b draw be 
parallel to the line BC, and representing the force BC com- 
pletely. Continuing in this way, as in Art. 73, draw the open 



Elements of Graphical Statics 



225 



vector or force polygon abcde; then, as in the case of con- 
current forces, Art. 73, the vector ae represents the resultant 
(or ea, the equilibrant) in magnitude and direction. The 
problem is not yet complete, for the position of the resultant 
is unknown. In Chapter VI. its position was determined by 
finding what moment it must have about some fixed point. 
The graphical method is as follows (the reader is advised to 







FIG. 180. 

draw the figure on a sheet of paper as he reads) : Choose any 
convenient point o (called a pole) in or about the vector 
polygon, and join each vertex a, b, c, d, and e of the polygon 
to o; then in the space diagram, selecting a point P on the 
line AB, draw a line PT (which may be called AO) parallel to 
ao across the space A. From P across the space B draw a 
line BO parallel to bo to meet the line BC in Q. From Q 
draw a line CO parallel to co to meet the line CD in R. From 
R draw a line DO parallel to do to meet the line DE in S, and, 
finally, from S draw a line EO parallel to eo to meet the line 
AO (or PT) in T. Then T, the intersection of AO and EO, 
is a point in the line of action of EA, the equilibrant, the magni- 
tude and inclination of which were found from the vector ea. 

Q 



226 Mechanics for Engineers 

Hence the equilibrant EA or the resultant AE is completely 
determined. The closed polygon PQRST, having its vertices 
on the lines of action of the forces, is called a funicular or link 
polygon. That T must be a point on the line of action of the 
resultant is evident from the following considerations. Any 
force may be resolved into two components along any two 
lines which intersect on its line of action, for it is only neces- 
sary for the force to be the geometric sum of the components. 
(Art. 75). Let each force, AB, EC, CD, and DE, be resolved 
along the two sides of the funicular polygon which meet on 
its line of action, viz. AB along TP and QP, BC along PQ 
and RQ, and so on. The magnitude of the two components 
is given by the corresponding sides of the triangle of forces 
in the vector diagram, e.g. AB may be replaced by components 
in the lines AO and BO (or TP and QP), represented in magni- 
tude by the lengths of the vectors ao and ob respectively, for 
in vector addition 

ao + ob = ab (Art. 19) 

Similarly, CD is replaced by components in the lines CO and 
OD represented by co and od respectively. When this process 
is complete, all the forces AB, BC, CD, and DE are replaced 
by components, the lines of action of which are the sides TP, 
PQ, QR, etc., of the funicular polygon. Of these component 
forces, those in the line PQ or BO are represented by the 
vectors ob and bo, and therefore have a resultant nil. Similarly, 
all the other components balance in pairs, being equal and 
opposite in the same straight line, except those in the lines TP 
and TS, represented by ao and oe respectively. These two 
have a resultant represented by ae (since in vector addition 
ao + oe = ae), which acts through the point of intersection T 
of their lines of action. Hence finally the resultant of the 
whole system acts through T, and is represented in magnitude 
and direction by the line ae; the equilibrant is equal and 
opposite in the same straight line. 

156. Conditions of Equilibrium. If we include the 
equilibrant EA (Fig. 180, Art. 155) with the other four forces, 
we have five forces in equilibrium, and (i) the force or vector 



Elements of Graphical Statics 



227 



polygon abcdc is closed ; and (2) the funicular polygon 
PQRST is a closed figure. Further, if the force polygon is 
not closed, the system reduces to a single resultant, which may 
be found by the method just described (Art. 155). 

It may happen that the force polygon is a closed figure, 
and that the funicular polygon is not. Take, for example, a 
diagram (Fig. 181) similar to the previous one, and let the 




FIG. 181. 



forces of the system be AB, BC, CD, DE, and EA, the force 
EA not passing through the point T found in Fig. 180, but 
through a point V (Fig. 181), in the line TS. If we draw a 
line, VW, parallel to oa through V, it will not intersect the line 
TP parallel to ao, for TP and VW are then parallel. Re- 
placing the original forces by components, the lines of action 
of which are in the sides of the funicular polygon, we are left 
with two parallel unbalanced components represented by ao 
and oa in the lines TP and VW respectively. These form a 
couple (Art. 91), and such a system is not in equilibrium nor 
reducible to a single resultant. The magnitude of the couple 
is equal to the component represented by oa multiplied by the 
length represented by the perpendicular distance between the 
lines TP and VW. A little consideration will show that it is 
also equal to the force EA represented by ea, multiplied by 
the distance represented by the perpendicular from T on the 



228 Mechanics for Engineers 

line VX. Or the resultant of the forces in the lines AB, BC, 
CD, and DE is a force represented by ae acting through the 
point T; this with the force through V, and represented by 
ra, forms a couple. 

Hence, for equilibrium it is essential that (i) the polygon 
of forces is a closed figure ; (2) that the funicular polygon is a 
closed figure. 

Compare these with the equivalent statements of the 
analytical conditions in Art. 96. 

Choice of Pole. In drawing the funicular polygon, the 
pole o (Figs. 1 80 and 181) was chosen in any arbitrary posi- 
tion, and the first side of the funicular polygon was drawn 
from any point P in the line AB. If the side AO had been 
drawn from any point in AB other than P, the funicular 
polygon would have been a similar and similarly situated figure 
to PQRST. 

The choice of a different pole would give a different 
shaped funicular polygon, but the points in the line of action 
of the unknown equilibrant obtained from the use of different 
poles would all lie in a straight line. This may be best appre- 
ciated by trial. 

Note that in any polygon the sides are each parallel to a 
line radiating from the corresponding pole. 

157. Funicular Polygon for Parallel Forces. To 
find the resultant of several parallel forces, we proceed exactly 
as in the previous case, but the force polygon has its sides all 
in the same straight line ; it is " closed " if, after drawing the 
various vectors, the last terminates at the starting-point of the 
first. The vector polygon does not enclose a space, but may 
be looked upon as a polygon with overlapping sides. 

Let the parallel forces (Fig. 182) be AB, BC, CD, and DE 
of given magnitudes. Set off the vector ab in the vector 
polygon parallel to the line AB, and representing by its length 
the magnitude of the force in the line AB. And from b set 
off be parallel to the line BC, and representing by its length 
the magnitude of the force in the line BC. Then be is evi- 
dently in the same straight line as ab, since AB and BC are 
parallel. Similarly the vectors cd t de, and the resultant ae of 



Elements of Graphical Statics 



229 



the polygon are all in the same straight line. Choose any 
pole o, and join a, b, t, d, and e to o. Then proceed to put 
in the funicular polygon in the space diagram as explained in 

a 






FIG. 182. 

Art. 155. The two extreme sides AO and EO intersect in 
T, and the resultant AE, given in magnitude by the vector ae, 
acts through this point, and is therefore completely deter- 
mined. 

158. To find Two Equilibrants in Assigned Lines 
of Action to a System of Parallel Forces. 

As a simple example, we may tatae the vertical reactions 

a 




FIG. 183. 

at the ends of a horizontal beam carrying a number of vertical 
loads. 

Let AB, BC, CD, and DE (Fig. 183) be the lines of action 



230 



Mechanics for Engineers 



of the forces of given magnitudes, being concentrated loads on 
a beam, xy, supported by vertical forces, EF and FA, at y and 
x respectively. Choose a pole, o, as before (Arts. 156 and 
157), and draw in the funicular polygon with sides AO, BO, 
CO, DO, and EO respectively parallel to ao, bo, co, do, and eo 
in the vector diagram. Let AO meet the line FA (i.e. the 
vertical through x) in /, and let q be the point in which EO 
meets the line EF (i.e. the vertical through y). Join pq, and 
from o draw a parallel line of to meet the line abcde in/. The 
magnitude of the upward reaction or supporting force in the 
line EF is represented by ef, and the other reaction in the line 
FA is represented by the vector fa. This may be proved in 
the same way as the proposition in Art. 155. 

of and fe represent the downward pressure of the beam at 
x and y respectively, while fa and ef represent the upward 
forces exerted by the supports at these points. 

159. In the case of non-parallel forces two equilibrants 
can be found one to have a given line of action, and the 
other to pass through a given point, i.e. to fulfil altogether three 
conditions (Art. 96). 




FIG. 184. 



Let AB, BC, and CD (Fig. 184) be the lines of action of 
given forces represented in magnitude by ab, be, and cd respec- 
tively in the vector polygon. Let ED be the line of action of 



Elements of Graphical Statics 231 

one equilibrant, and p a point in the line of action of the 
second. Draw a line, dx, of indefinite length parallel to DE. 
Choose a pole, o, and draw in the funicular polygon corre- 
sponding to it, but drawing the side AO through the given 
point p. Let the last side DO cut ED in q. Then, since the 
complete funicular polygon is to be a closed figure, ]<ympq. 
Then the vector oe is found by drawing a line, oe, through o 
parallel to pq to meet dx in e. The magnitude of the equili- 
brating force in the line DE is represented by the length de, 
and the magnitude and direction of the equilibrant EA through 
/ is given by the length and direction of ea. 

1 60. Bending Moment and Shearing Force. In con- 
sidering the equilibrium of a rigid body (Chapter VI.), we have 
hitherto generally only considered the body as a whole. The 
same conditions of equilibrium must evidently apply to any 
part of the body we may consider (see Method of Sections, 
Art. 98). For example, if a beam (Fig. 185) carrying loads 
Wj, W 2 , W 3 , W 4 , and W 5 , as shown, be ideally divided into two 



vy, Y 2 vy, % vy : 

I t> I f I 



:A : 



Y///W//////// 






FIG. 185. 

parts, A and B, by a plane of section at X, perpendicular to 
the length of the beam, each part, A and B, may be looked 
upon as a rigid body in equilibrium under the action of forces. 
The forces acting on the portion A, say, fulfil the conditions of 
equilibrium (Art. 96), provided we include in them the forces 
which the portion B exerts on the portion A. 

Note that the reaction of A on B is equal and opposite to 
the action of B on A, so that these internal forces in the beam 
make no contribution to the net forces or moment acting on 
the beam as a whole. 

For convenience of expression, we shall speak of the beam 



232 Mechanics for Engineers 

as horizontal and the loads and reactions as vertical forces. 
Let R A and R B be the reactions of the supports on the por- 
tions A and B respectively. 

Considering the equilibrium of the portion A, since the 
algebraic sum of the vertical forces on A is zero, B must exert 
on A an upward vertical force Wj + W. 2 R A . This force is 
called the shearing force at the section X, and may be denoted 
by F x . Then 

F x = W> + W 2 - R A , or W, + W 2 - R A - F x = o 

If the sum Wj + W 2 is numerically less than R A , F x is 
negative, i.e. acts downwards on A. 

The shearing force at any section of this horizontal beam is 
then numerically equal to the algebraic sum of all the vertical 
forces acting on either side of the section. 

Secondly, since the algebraic sum of all the horizontal forces 
on A is zero, the resultant horizontal force exerted by B on 
A must be zero, there being no other horizontal force on A. 
Again, if x lt d v and </ 2 are the horizontal distances of R A , W 15 
and W 2 respectively from the section X, since W ls W. 2) and R A 
exert on A a clockwise moment in the plane of the figure 
about any point in the section X, of magnitude 

R*.* 1 --W a .4-W v .4 

B must exert on A forces which have a contra- clockwise 
moment M x , say, numerically equal to R A . x l W^ W.^, 
for the algebraic sum of the moments of all the forces on A is 
zero, i.e. 



X = o 
Or M x = R A . #, - W,*/, - 

This moment cannot be exerted by the force F x , which has 
zero moment in the plane of the figure about any point in the 
plane X. Hence, since the horizontal forces exerted by B 
on A have a resultant zero, they must form a couple of 
contra-clockwise moment, M x , i.e. any pull exerted by B must 
be accompanied by a push of equal magnitude. This couple 
M x is called the moment of resistance of the beam at the 
section X, and it is numerically equal to the algebraic sum of 



Elements of Graphical Statics 



233 



moments about that section, of all the forces acting to either 
side of the section. This algebraic sum of the moments about 
the section, of all the forces acting to either side of the section 
X, is called the bending moment at the section X. 

161. Determination of Bending Moments and Shear- 
ing Forces from a Funicular Polygon. Confining our- 
selves again to the horizontal beam supported by vertical 
forces at each end and carrying vertical loads, it is easy to 
show that the vertical height of the funicular polygon at any 
distance along the beam is proportional to the bending moment 



W, W 2 




FIG. i 



at the corresponding section of the beam, and therefore repre- 
sents it to scale, e.g. that xl (Fig. 186) represents the bending 
moment at the section X. 

Let the funicular polygon for any pole o, starting say from 
z, be drawn as directed in Arts. 155 and 157, eg being drawn 
parallel to zp or GO, the closing line of the funicular, so that 
R 15 the left-hand reaction, is represented by the vector ga and 
R 2 by fg, while the loads W lf W 2 , W 3 , \V 4 , and W 5 are repre- 
sented by the vectors at>, be, cd, de, and ^respectively. Con- 
sider any vertical section, X, of the beam at which the height of 
bending-moment diagram is xl. Produce xl and the side zw 
to meet in y. Also produce the side win of the funicular 



234 Mechanics for Engineers 

polygon to meet xy in n, and let the next side mq of the 
funicular meet xy in /. The sides zw, wm, and mq (or AO, BO, 
and CO) are parallel to ao, bo, and co respectively. Draw a 
horizontal line, zk, through z to meet xy in k, a horizontal line 
through w to meet xy in r, and a horizontal 0H through o in 
the vector polygon to meet the line abcdef in H. Then in the 
two triangles xyz and gao there are three sides in either parallel 
respectively to three sides in the other, hence the triangles are 
similar, and 



ag ao 
Also the triangles zky and oHa are similar, and therefore 

zy _ zk 
ao 0H 



Hence from (i) and (2) 

xy zk ag . zk 

= -^, or xy.oH = ag X zk, or xy = g- 

Therefore, since ag is proportional to Rj, and zk is equal or 
proportional to the distance of the line of action of R! from X, 
ag . zk is proportional to the moment of Rj about X, and 0H 
being an arbitrarily fixed constant, xy is proportional to the 
moment of Rj about X. 
Similarly 

ab . wr 

1777 ?~~ _ 

~ 



and therefore yn represents the moment of Wj about X to the 
same scale that xy represents the moment of Rj about X. 
Similarly, again, nl represents the moment of W 2 about X to 
the same scale. 

Finally, the length xl or (xy ny In) represents the 
algebraic sum of the moments of all the forces to the left of 
the section X, and therefore represents the bending moment at 
the section X (Art. 160). 



Elements of Graphical Statics 235 

Scales. If the scale of forces in the vector diagram is 

i inch to/ Ibs. 

and the scale of distance in the space diagram is 
i inch to q feet ; 

and if 0H is made h inches long, the scale on which xl repre- 
sents the bending moment at X is 

i inch top.q.h. foot-lbs. 
A diagram (Fig. 187) showing the shearing force along the 



AJB 




FIG. 187. 

length of the beam may be drawn by using a base line, sf, of 
the same length as the beam in the space diagram, and in the 
horizontal line through g in the force diagram. The shearing 
force between the end of the beam s and the line AB is con- 
stant and equal to Rj, i.e. proportional to ga. The height ga 
may be projected from a by a horizontal line across the space 
A. A horizontal line drawn through b gives by its height above 
g the shearing force at all sections of the beam in the space B. 
Similarly projecting horizontal lines through c, d, e, and f we 
get a stepped diagram, the height of which from the base line 
st gives, to the same scale as the vector diagram, the shearing 
force at every section of the beam. 



236 Mechanics for Engineers 



EXAMPLES XIX. 

1. Draw a square lettered continuously FQRS, each side 2 inches 
long. Forces of 9, 7, and 5 Ibs. act in the directions RP, SQ, and QR 
respectively. Find by means of a funicular polygon the resultant of these 
three forces. State its magnitude in pounds, its perpendicular distance 
from P, and its inclination to the direction PQ. 

2. Add to the three forces in question I a force of 6 Ibs. in the direction 
PQ, and find the resultant as before. Specify it by its magnitude, its 
distance from P, and its inclination to PQ. 

3. A horizontal beam, 15 feet long, resting on supports at its ends, carries 
concentrated vertical loads of 7, 9, 5, and 8 tons at distances of 3, 8, 12, 
and 14 feet respectively from the left-hand support. Find graphically the 
reactions at the two supports. 

4. A horizontal rod AB, 13 feet long, is supported by a horizontal hinge 
perpendicular to AB at A, and by a vertical upward force at B. Four 
forces of 8, 5, 12, and 17 Ibs. act upon the rod, their lines of action cutting 
AB at I, 4, 8, and 12 feet respectively from A, their lines of action making 
angles of 70, 90, 120, and 135 respectively with the direction AB, each 
estimated in a clockwise direction. Find the pressure exerted on the 
hinge, state its magnitude, and its inclination to AB. 

5. A simply supported beam rests on supports 17 feet apart, and carries 
loads of 7, 4, 2, and 5 tons at distances of 3, 8, 12, and 14 feet respectively 
from the left-hand end. Calculate the bending moment at 4, 9, and 1 1 feet 
from the left-hand end. 

6. Draw a diagram to show the bending moments at all parts of the 
beam in question 5. State the scales of the diagram, and measure from it 
the bending moment at 9, II, 13, and 14 feet from the left-hand support. 

7. Calculate the shearing force on a section of the beam in Question 5 
at a point 10 feet from the left-hand support ; draw a diagram showing the 
shearing force at every transverse section of the beam, and measure from it 
the shearing force at 4 and at 13 feet from the left-hand support. 

8. A beam of 2O-feet span carries a load of 10 tons evenly spread over 
the length of the beam. Find the bending moment and shearing force at 
the mid-section and at a section midway between the middle and one end. 

162. Equilibrium of Jointed Structures. 

Frames. The name frame is given to a structure consist- 
ing of a number of bars fastened together by hinged joints; 
the separate bars are called members of the frame. Such 
structures are designed to carry loads which are applied mainly 
at the joints. We shall only consider frames which have just 
a sufficient number of members to prevent deformation or 
collapse under the applied loads. Frames having more 



Elements of Graphical Statics 



237 



members than this requirement are treated in books on 
Graphical Statics and Theory of Structures. We shall further 
limit ourselves mainly to frames all the members of which are 
approximately in the same plane and acted upon by forces all 
in this same plane and applied at the hinges. 

Such a frame is a rigid body, and the forces exerted upon 
it when in equilibrium must fulfil the conditions stated in Art. 
96 and in Art. 156. These "external" forces acting on the 
frame consist of applied loads and reactions of supports ; they 
can be represented in magnitude and direction by the sides of 
a closed vector polygon ; also their positions are such that an 
indefinite number of closed funicular polygons can be drawn 
having their vertices on the lines of action of the external 
forces. From these two considerations the complete system 
of external forces can be determined from sufficient data, as in 
Arts. 155 and 159. The "internal" forces, i.e. the forces 
exerted by the members on the joints, may be determined from 
the following principle. The pin of each hinged joint is in 
equilibrium under the action of several forces which are 
practically coplanar and concurrent. These forces are : the 
stresses in the members (or the " internal " forces) meeting at 
the particular joint, and the " external " forces, i.e. loads and 
reactions, if any, which are applied there. 

If all the forces, except two internal ones, acting at a given 
joint are known, then the two which have their lines of action 
in the two bars can be found by completing an open polygon 
of forces by lines parallel to those two bars. 

If a closed polygon of forces be drawn for each joint in the 
structure, the stress in every bar will be determined. In order to 
draw such a polygon for any particular joint, all the concurrent 
forces acting upon it, except two, must be known, and therefore 
a start must be made by drawing a polygon for a joint at which 
some external force, previously determined, acts. Remembering 
that the forces which any bar exerts on the joints at its two 
ends are equal and in opposite directions, the drawing of a 
complete polygon for one joint supplies a means of starting 
the force polygon for a neighbouring joint for which at least 
one side is then known. An example of the determination of 



238 



Mechanics Jor Engineers 



the stresses in the members of a simple frame will make this 
more easily understood. 

Fig. 1 88 shows the principles of the graphical method of 
finding the stresses or internal forces in the members of a 
simple frame consisting of five bars, the joints of which have 
been denoted at (a) by i, 2, 3, and 4. The frame stands in 



(b) 




For joint I For join I Z 



the vertical plane, and carries a known vertical load, W, at the 
joint 3 ; it rests on supports on the same level at i and 4. 
The force W is denoted in Bow's notation by the letters PQ. 
The reactions at i and 4, named RP and QR respectively, have 
been found by a funicular polygon corresponding to the vector 
diagram at (b), as described in Art. 158. 



Elements of Graphical Statics 239 

Letters S and T have been used for the two remaining 
spaces. When the upward vertical force RP at the joint i is 
known, the triangle of forces rps at (c) can be drawn by making 
rp proportional to RP as in (b), and completing the triangle by 
sides parallel to PS and SR (i.e. to the bars 12 and 14) 
respectively. After this triangle has been drawn, one of the 
three forces acting at the joint 2 is known, viz. SP acting in 
the bar 12, being equal and opposite to PS in (c). Hence the 
triangle of forces spt at (</), for the joint 2 can be drawn. Next 
the triangle tpq at (e) for joint 3 can be drawn, tp and pq being 
known; the line joining qt will be found parallel to the bar 
QT if the previous drawing has been correct ; this is a check 
on the accuracy of the results. Finally, the polygon qrst at (/) 
for joint 4 may be drawn, for all four sides are known in 
magnitude and direction, from the previous polygons. The 
fact that when drawn to their previously found lengths and 
directions they form a closed polygon, constitutes a check to 
the correct setting out of the force polygons. The arrow-heads 
on the sides of the polygons denote the directions of the forces 
on the particular joint to which the polygon refers. 

163. Stress Diagrams. It is to be noticed in Fig. 188 
that in the polygons (), (c), (d), (e), and (/), drawn for the 
external forces on the frame and the forces at the various joints, 
each side, whether representing an external or internal force, 
has a line of equal length and the same inclination in some 
other polygon. 

For example, sr in (c} corresponding to rs in (f), and // 
in (d) with tp in (<?). The drawing of entirely separate polygons 
for the forces at each joint is unnecessary ; they may all be 
included in a single figure, such as (g), which may be regarded 
as the previous five polygons superposed, with corresponding 
sides coinciding. Such a figure is called a stress diagram for 
the given frame under the given system of external loading. It 
contains- (i) a closed vector polygon for the system of external 
forces in the frame, (2) closed vector polygons for the (con- 
current) forces at each joint of the structure. 

As each vector representing the internal force in a member 
of the frame represents two equal and opposite forces, 



240 Mechanics for Engineers 

arrow-heads on the vectors are useless or misleading, and are 
omitted. 

Distinction between Tension and Compression 
Members of a Frame. A member which is in tension is 
called a " tie," and is subjected by the joints at its ends to a 
pull tending to lengthen it. The forces which the member 
exerts on the joints at its ends are equal and opposite pulls 
tending to bring the joints closer together. 

A member which is in compression is called a "strut;" 
it has exerted upon it by the joints at its ends two equal and 
opposite pushes or thrusts tending to shorten it. The member 
exerts on the joints at its ends equal and opposite " outward " 
thrusts tending to force the joints apart. 

The question whether a particular member is a " tie " or a 
" strut " may be decided by finding whether it pulls or thrusts at 
a joint at either end. This is easily discovered if the direction of 
any of the forces at that joint is known, since the vector polygon 
is a closed figure with the last side terminating at the point from 
which the first was started. E.g. to find the kind of stress in 
the bar 24, or ST (Fig. 188). At joint 4 QR is an upward 
force ; hence the forces in the polygon qrst must act in the 

->--> -> 

directions qr, rs, sf, and tq ; hence the force ST in bar 24 acts 
at joint 4 in the direction s to /, i.e. the bar pulls at joint 4, 
or the force in ST is a tension. Similarly, the force in bar 23, 
or PT, acts at joint 3 in a direction ty, i.e. it pushes at joint 
3, or the force in bar 23 is a compressive one. 

Another method. Knowing the direction of the force rp at 
joint i (Fig. 1 88), we know that the forces at joint i act in the 
directions rp, ps, and sr, or the vertices of the vector polygon 
rps lie in the order r p s. 

The corresponding lines RP, PS, and SR in the space 
diagram are in clockwise order round the point i . This order, 
clockwise or contra-clockwise (but in this instance clockwise) 
is the same for every joint in the frame. If it is clockwise for 
joint i, it is also clockwise for joint 2. Then the vertices of 
the vector polygon for joint 2 are to be taken in the cyclic 
order s p /, since the lines SP, PT, and TS lie in clockwise 



Elements of Graphical Statics 



241 



order round the joint 2, e.g. the force in bar 23, or PT, is in 

-> 
the direction//, i.e. it thrusts at joint 2. 

This characteristic order of space letters round the joints is a 
very convenient method of picking out the kind of stress in one 
member of a complicated frame. Note that it is the character- 
istic order of space letters round a joint that is constant not 
the direction of vectors round the various polygons constituting 
the stress diagram. 

164. Warren Girder. A second example of a simple 
stress diagram is shown in Fig. 189, viz. that of a common type 




FIG. 189. 

of frame called the Warren girder, consisting of a number of 
bars jointed together as shown, all members generally being 
of the same lengths, some horizontal, and others inclined 60 to 
the horizontal. 

Two equal loads, AB and BC, have been supposed to act 
at the joints i and 2, and the frame is supported by vertical 
reactions at 3 and 4, which are found by a funicular polygon. 
The remaining forces in the bars are found by completing the 
stress diagram abc . . . Mm. 

Note that the force AB at joint i is downward, />. in the 
direction ab in the vector diagram corresponding to a contra- 
clockwise order, A to B, round joint i. This is, then, the 
characteristic order (contra-clockwise) for all the joints, e.g. to 
find the nature of the stress in KL, the order of letters for 
joint 5 is K to L (contra-clockwise), and referring to the vector 



242 



Mechanics for Engineers 



diagram, the direction k to / represents a thrust of the bar KL 
on joint 5 ; the bar KL is therefore in compression. 

165. Simple Roof=frame. Fig. 190 shows a simple 
roof-frame and its stress diagram when carrying three equal 
vertical loads on three joints and supported at the extremities 
of the span. 




FIG. 190. 

The reactions DE and EA at the supports are each obvi- 
ously equal to half the total load, i.e. e falls midway between a 
and d in the stress diagram. The correct characteristic order 
of the letters round the joints (Art. 163) is, with the lettering 
here adopted, clockwise. 

1 66. Loaded Strings and Chains. Although not 
coming within the general meaning of the word " frame," stress 






Elements of Graphical Statics 



243 



diagrams can be drawn for a structure consisting partly of 
perfectly flexible chains or ropes, provided the loads are such 
as will cause only tension in flexible members. 

Consider a flexible cord or chain, Xi23Y (Fig. 191), sus- 
pended from points X and Y, and having vertical loads of 




W, 



Wj, W 2 , and W 3 suspended from points i, z, and 3 respectively. 
Denoting the spaces according to Bow's notation by the letters 
A, B, C, D, and O, as shown above, the tensions in the strings 
Xi or AO and i 2 or BO must have a resultant at i equal 
to Wj vertically upward, to balance the load at i. If triangles 
of forces, abo, bco, and cdo, be drawn for the points 1,2, and 3 
respectively, the sides bo and co appear in two of them, and, as 
in Art. 163, the three vector triangles maybe included in a 
single vector diagram, as shown at the right-hand by the 
figure abcdo. 

The lines ao, bo, co, and do represent the tensions in the 
string crossing the spaces A, B, C, and D respectively. If a 
horizontal line, 0H, be drawn from o to meet the line abed in 
H, the length of this line represents the horizontal component of 
the tensions in the strings, which is evidently constant through- 
out the whole. (The tension changes only from one space to 
the neighbouring one by the vector addition of the intermediate 
vertical load.) The pull on the support X is represented by 



244 Mechanics for Engineers 

ao, the vertical component of which is aU. ; the pull on Y 
is represented by od, the vertical component of which is Hd. 

A comparison with Art. 157 will show that the various 
sections of the string Xi23Y are in the same lines as the sides 
of a funicular polygon for the vertical forces W ls W 2 , and W 3 , 
corresponding to the pole o . If different lengths of string are 
attached to X and Y and carry the same loads, W 1( W 2 , and W 3 , 
in the lines AB, BC, and CD respectively, they will have 
different configurations ; the longer the string the steeper will 
be its various slopes corresponding to shorter pole distances, 
H0, i.e. to smaller horizontal tensions throughout. A short 
string will involve a great distance of the pole o from the line 
abed, i.e. a great horizontal tension, with smaller inclinations of 
the various sections of the string. The reader should sketch for 
himself the shape of a string connecting X to Y, with various 
values of the horizontal tension H0, the vertical loads remain- 
ing unaltered, in order to appreciate fully how great are the 
tensions in a very short string. 

A chain with hinged links, carrying vertical loads at the 
joints, will occupy the same shape as a string of the same 
length carrying the same loads. Such chains are used in sus- 
pension bridges. 

The shape of the string or chain to carry given loads in 
assigned vertical lines of action can readily be found for any 
given horizontal tension, H<?, by drawing the various sections 
parallel to the corresponding lines radiating from o, e.g. AO 
or Xi parallel to do (Fig. 191). 

Example i. A string hangs from two points, X and Y, 5 feet 
apart, X being 3 feet above Y. Loads of 5, 3, and 4 Ibs. are 
attached to the string so that their lines of action are i, 2, and 
3 feet respectively from X. If the horizontal tension of the string 
is 6 Ibs., draw its shape. 

The horizontal distance ZY (Fig. 192) of X from Y is 

\/5 2 - 3" = 4 feet 

so that the three loads divide the horizontal span into four equal 
parts. 

Let Vx and V Y be the vertical components of the tension of the 
string at X and Y respectively. 



Elements of Graphical Statics 



245 



The horizontal tension is constant, and equal to 6 Ibs. Taking 
moments about Y (Fig. 192) 

Clockwise. Contra-clockwise. 

V X x 4 = (4 x i) 4- (3 x 2) + (5 x 3) + (6 x 3) Ib.-feet 
4V X = 4 + 6+ 15 + 18 = 43 
V x = -^ =1075 Ibs. 

Since the vertical and horizontal components of the tension of 
the string at X are known, its direction is known. The direction 
of each section of string might similarly be found. Set out the 




vector polygon abed, and draw the horizontal line Ho to represent 
6 Ibs. horizontal tension from H, aH being measured along abed of 
such a length as to represent the vertical component 1075 Ibs. of 
the string at X. Join o to , l>, c, and d. Starting from X or Y, 
draw in the lines across spaces A, B, C, and D parallel respectively 
to ao, bo, co, and do (as in Art. 157). The funicular polygon so 
drawn is the shape of the string. 

Example 2. A chain is attached to two points, X and Y, 
X being I foot above Y and 7 feet horizontally from it. Weights 
of 20, 27, and 22 Ibs. are to be hung on the chain at horizontal 
distances of 2, 4, and 6 feet from X. The chain is to pass through 
a point P in the vertical plane of X and Y, 4 feet below, and 3 feet 



246 



Mechanics for Engineers 



horizontally from X. Find the shape of the chain and the tensions 
at its ends. 

Let V x and V Y be the vertical components of the tension at X 
and Y respectively, and let H be the constant horizontal tension 
throughout. 





Taking moments about Y (Fig. 193) 

Clockwise. Contra-clockwise. 

V x x 7 = (H x i) + (20 x 5) + (27 x 3) + (22 x i) 
7V X = H + 203 Ibs.-feet (i) 

Taking moments about P of the forces on the chain between 
X and P 

Clockwise. Contra-clockwise. 

V x x 3 = H x 4 + (20 x i) 

3V X = 4H + 20 (2) 

and 28V X = 4H + 812 from (i) 
hence 2$V X = 792 

V x = 3i'681bs. 
H = 7V X - 203 = 22176 - 203 = 1876 Ibs. 



Draw the open polygon of forces, abed (a straight line), and set 



Elements of Graphical Statics 247 

off am from a to the same scale, 31*68 Ibs. downwards. From in 
set off mo to represent 1876 Ibs. horizontally to the right of m. 

Then the vector ao am + mo = tension in the string XZ, 
which pulls at X in the direction XZ. By drawing XZ parallel 
to ao the direction of the first section of the chain is obtained, and 
by drawing from Z a line parallel to bo to meet the line of action 
BC, the second section is outlined. Similarly, by continuing the 
polygon by lines parallel to co and do the complete shape of the 
chain between X and Y is obtained. 

The tension ao at X scales 37 Ibs., and the tension od at Y 
scales 44 Ibs. 



167. Distributed Load. If the number of points at 
which the same total load is attached to the string (Fig. 191) 
be increased, the funicular polygon corresponding to its shape 
will have a larger number of shorter sides, approximating, if 
the number of loads be increased indefinitely, to a smooth 
curve. This case corresponds to that of a heavy chain or 
string hanging between two points with no vertical load but its 
own weight. If the dip of the chain from the straight line join- 
ing the points of the attachment is small, the load per unit of 
horizontal span is nearly uniform provided the weight of chain 
per unit length is uniform. In this case an approximation to the 
shape of the chain may be found by dividing the span into a 
number of sections of equal length and taking the load on each 
portion as concentrated at the mid-point of that section. The 
funicular polygon for such a system of loads will have one 
side more than the number into which the span has been 
divided ; the approximation may be made closer by taking 
more parts. The true curve has all the sides of all such poly- 
gons as tangents, or is the curve inscribed in such a polygon. 

The polygons obtained by dividing a span into one, two, 
and four equal parts, and the approximate true curve for a 
uniform string stretched with a moderate tension, are shown in 
Fig. 194. 

Note that the dip QP would be less if the tensions OH, OA ? 
etc., were increased. 

1 68. The relations between the dip, weight, and tension 
of a stretched string or chain, assuming perfect flexibility, can 



248 



Mechanics for Engineers 



more conveniently be found by ordinary calculation than by 
graphical methods. 

Assuming that the dip is small and the load per horizontal 

Q 





FIG. 194. 

foot of span is uniform throughout, the equilibrium of a portion 
AP (Fig. 195) of horizontal lengths, measured from the lowest 
point A, may be considered. 

l 




FIG. 195. 

Let w = weight per unit horizontal length of cord or chain ; 
y = vertical height of P above A, viz. PQ (Fig. 195) ; 
T = the tension (which is horizontal) at A ; 
T = the tension at P acting in a line tangential to the 
curve at P. 



Elements of Graphical Statics 249 

The weight of portion AP is then wx, and the line of action 
of the resultant weight is midway between AB and PQ, i.e. at 



OC 

a distance - from either. 
2 

Taking moments about the point P 
T x PQ = wx X - 

2 

wy? 
orTX7 = 

WX' 

S TT 

This relation shows that the curve of the string is a 
parabola. 

If d = the total dip AB, and / = the span of the string or 
chain, taking moments about N of the forces on the portion 

AN 





B 



which gives the relation between the dip, the span, and the 
horizontal tension. 

Returning to the portion AP, if the vector triangle rst be 
drawn for the forces acting upon it, the angle 6 which the 
tangent to the curve at P makes with the horizontal is given 
by the relation 

xw st 

- Y = - = tan6 

Also the tension T' at P is T sec 6, or 
T' = Ti+tan 2 = 



and at the ends where x = - 

2 



T' = 



250 



Mechanics for Engineers 



And since T = -7, the tension at N or M is 

wl I / 2 , wl' 1 

~2\i i6d 2 f 1 

wl* d 

which does not greatly exceed -x-j (or T), if -= is small. 

Example. A copper trolley-wire weighs \ Ib. per foot length ; 
it is stretched between two poles 50 feet apart, and has a horizontal 
tension of 2000 Ibs. Find the dip in the middle of the span. 

Let d the dip in feet. 

The weight of the wire in the half-span BC (Fig. 196) is 
25 x \ = 12-5 Ibs. 




The distance of the e.g. of the wire BC from B is practically 
I2'5 feet horizontally. 

Taking moments about B of the forces on the portion BC 

2OOO X d = I2'5 X I2'5 

d = 0^078 1 2 foot = 0-938 inch 

EXAMPLES XX. 

i. A roof principal, shown in Fig. 197, carries loads of 4, 7, and 5 tons 
in the positions shown. It is simply supported at the extremities of a span 




FIG. 197. 
of 40 feet. The total rise of the roof is 14 feet, and the distances PQ and 



Elements of Graphical Statics 



251 



RS are each 5'4 feet. Draw the stress diagram and find the stress in each 
member of the frame. 

2. A Warren girder (Fig. 198), made up of bars of equal lengths, carries 
a single load of 5 tons as shown. Draw the stress diagram and scale off 




FIG. 198. 

the forces in each member ; check the results by the method of sections 
(Art. 98). 

3. Draw the stress diagram for the roof-frame in Fig. 199 under the 



4lons 



JOfeet 




given loads. The main rafters are inclined at 30 to the horizontal, and 
are each divided by the joints into three equal lengths. 

4. A chain connects two points on the same level and 10 feet apart ; 
it has suspended from it four loads, each of 50 Ibs., at equal horizontal 
intervals along the span. If the tension in the middle section is 90 Ibs., 
draw the shape of the chain, measure the inclination to the horizontal, and 
the tension of the end section. 

5. Find the shape of a string connecting two points 8 feet horizontally 
apart, one being I foot above the other, when it has suspended from it 
weights of 5, 7, and 4 Ibs. at horizontal distances of 2, 5, and 6 feet 
respectively from the higher end, the horizontal tension of the string being 
6 Ibs. 

6. A light chain connects two points, X and Y, 12 feet horizontally apart, 
X being 2 feet above Y. Loads of 15, 20, and 25 Ibs. are suspended from 
the chain at horizontal distances of 3, 5, and 8 feet respectively from X. 
The chain passes through a point 7 feet horizontally from X and 4 feet 



252 Mechanics for Engineers 

below it. Draw the shape of the chain. How far is the point of suspension 
of the is-lb. load from X ? 

7. A wire is stretched horizontally, with a tension of 50 Ibs., between 
two posts 60 feet apart. If the wire weighs 0^03 Ib. per foot, find the sag 
of the wire in inches. 

8. A wire weighing o'oi Ib. per foot is stretched between posts 40 feet 
apart. What must be the tension in the wire in order to reduce the sag to 
2 inches ? 

9. A wire which must not be stretched with a tension exceeding 70 Ibs. 
is to be carried on supporting poles, and the sag between two poles is not 
to exceed i'5 inches. If the weight of the wire is 0^025 Ib. per foot, find 
the greatest distance the poles may be placed apart. 



APPENDIX 

UNITS AND THEIR DIMENSIONS 

Units. To express the magnitude of any physical quantity it has 
to be stated in terms of a unit of its own kind. Thus by stating 
that a stick is 275 feet long, we are using the foot as the unit of 
length. 

Fundamental and Derived Units. We have seen that the 
different quantities in common use in the science of mechanics 
have certain relations to one another. If the units of certain 
selected quantities are arbitrarily fixed, it is possible to determine 
the units of other quantities by means of their relations to the 
selected ones. The units arbitrarily fixed are spoken of as 
fundamental units, and those depending upon them as derived 
units. 

Fundamental Units. There are two systems of units in 
general use in this country. In the C.G.S. system (Art. 42), which 
is commonly used in physical science, the units chosen as funda- 
mental and arbitrarily fixed are those of length, mass, and time, 
viz. the centimetre, gramme, and second. 

In the British gravitational system the fundamental units chosen 
are those of length, force, and time, viz. the foot, the pound, i.e. the 
weight of i Ib. of matter at some standard place, and the mean 
solar second. 

The latter system of units has every claim to the name 
" absolute," for three units are fixed, and the other mechanical 
units are derived from them by fixed relations. 

The weight of a body of given mass varies at different parts of 
the earth's surface in whatever units its mass is measured. The 
value of i Ib. force, however, does not vary, since it has been 
defined as the weight of a fixed mass at a fixed place. 

Dimensions of Derived Units. 

(a) Length Mass Time Systems. In any such system other 
than, say, the C.G.S. system, let the. unit of length be L centimetres, 
the unit of mass M grammes, and the unit of time be T seconds. 



254 Mechanics for Engineers 

Then the unit of area will be L x L or L 2 square centimetres, 
i.e. it varies as the square of the magnitude of the unit of length. 
Similarly, we may derive the other important mechanical units as 
follows : 

Unit volume L x L x L or L 3 cubic centimetres, or unit 
volume varies as L 3 . 

Unit velocity is L centimetres in T seconds = ~ centimetres 

per second, or LT -1 centimetres per second. 
Unit acceleration is = centimetres per second in T seconds 

= ^2 centimetres per second, or LT~ 2 centimetres per 
second. 
Unit momentum is that of M grammes moving ~ centimetres per 

ML 
second, i.e. '-^- C.G.S. units of momentum, or MLT~ : 

C.G.S. units. 

MLT -1 
Unit force is unit change of momentum in T seconds, or ^ 

units in one second, or MLT~ 2 dynes (C.G.S. units offeree). 
Unit impulse is given by unit force (MLT~ 2 C.G.S. units) acting 

for unit time, T seconds generating a change of momentum 

(or impulse) MLT" 1 C.G.S. units. 
Unit work is that done by unit force (MLT~ 2 dynes) acting 

through L centimetres, i.e. ML 2 T~ 2 centimetre-dynes or 

ergs. 
Unit kinetic energy is that possessed by unit mass, M grammes 

moving with unit velocity (LT- 1 ),*.*. |M(LT- 1 ) 2 =1ML 2 T- 2 

C.G.S. units. 

ML 2 T~ 2 
Unit power is unit work in unit time T seconds, or , or 

ML 2 T~ 3 ergs per second. 

x , . , L units of arc ,. , . . , 

Note that unit angle T ; j-. = i radian, and is inde- 

L units radius 

pendent of the units of length, mass, or time. 

Unit angular velocity is unit velocity ^ divided by unit radius 
L centimetres, or LT -1 -i- L = T 



Appendix 



255 



Unit moment of momentum or angular momentum is unit 
momentum MLT" 1 at unit perpendicular distance L, or 
ML 2 *!- 1 C.G.S. units. 
Unit moment of force is unit force MLT~ 2 at unit distance L 

centimetres, or ML 2 T~ 2 C.G.S. units. 
Unit rate of change of angular momentum is ML 2 T -1 C.G.S. 

units in unit time T seconds = ML 2 T~ 2 C.G.S. units. 
Unit moment of inertia is that of unit mass M grammes at unit 

distance L centimetres, which is ML 2 C.G.S. units. 
Thus each derived unit depends on certain powers of the 
magnitudes of the fundamental units, or has certain dimensions of 
those units. 

(U) The dimensions of the same quantities in terms of the three 
fundamental units of length, force, and time may be similarly 
written as follows : 



Quantity. 

Length. 
Force. 
Time. 
Velocity. 

Acceleration. 
Mass. 

Momentum. 
Impulse. 
Work. 

Kinetic energy. 
Power. 

Angular momentum. 
Moment of force. 
Rate of change of angular 
momentum. 



Dimensions. 



L. 
F. 
T. 



or LT- 1 . 

-:- T or LT- 2 . 
Force 



T- 
Acceleration 

FT. 
FT. 
FL. 



or FL-'T 2 . 



FLT- 1 . 
FLT. 
FL. 
FL. ' 



Symbolical formulae and equations may be checked by testing 
if the dimensions of the terms are correct. Each term on either 
side of an algebraic equation having a physical meaning must 
necessarily be of the same dimensions. 



ANSWERS TO EXAMPLES 



EXAMPLES I. 

(i) 0-305 foot per second per second. (2) 5*5 seconds; 121 feet. 
(3) 7i'77 feet per second. (4) 3-053 seconds. 

(5) 89-5 feet ; 447-5 feet ; 440 '4 feet. 

(6) 5 '63 seconds after the first projection ; 278 feet. 

(7) 567 feet per second. (8) 4-5, 14-6, and 11*4 feet per second. 
(10) o'57 and 0-393 foot per second per second ; 880 feet. 

(n) 77 '3 feet ; 2-9 seconds. 

EXAMPLES If. 

(1) 4-88 feet per second ; 35 23' to the horizontal velocity. 

(2) 405 feet per second ; 294 feet per second. 

(3) 53 up-stream ; 2 minutes 16-4 seconds. (4) 10 6 south of west. 

(5) 19*54 knots per hour ; 5 hours 7'2 minutes ; 12 8' west of south. 

(6) 48 minutes ; 9*6 miles ; I2'8 miles. 

(7) 154-2 feet per second per second ; 2i"5 south of west. 

(8) 2-59 seconds. (9) 5-04; 4716. (ic) 16-83 feet per second, 
(n) 35*2 radians per second ; 2*581 radians per second per second. 

(12) 135 revolutions and 1-5 minutes from full speed. 

EXAMPLES III. 

(l) 2735 units; 182,333 Ibs. or 8i"4 tons. (2) $ or 1-172 to I. 

(3) 2*8 centimetres per second. (4) 9802 Ibs. 

(5) I 5'33 Iks. 5 9'53 units per second in direction of jet ; 9*53 Ibs. 

(6) 45-3. (7) 4720 Ibs. 

(8) 10-43 tons inclined downwards at 16 40' to horizontal. 

(9) 2-91 units ; 727-5 Ibs. (10) 8750 units ; 8-57 miles per hour. 

EXAMPLES IV. 

(1)67-8 Ibs. (2) 17-48 Ibs. (4) 34-54 feet. 

(5) 23-44 feet per second ; 255,000 Ibs. (6) 1005 feet per second. 

(7) 154 Ibs. ; 126 Ibs. ; 6*9 feet per second per second. 

(8) ii'243cwt. (9) 9-66 feet; 14-93 Ibs. 
(10) 4*69 grammes ; 477 centimetres. 

(n) 6-44 feet per second per second ; 4 Ibs. 
(12) i -027 Ibs. (13) 48-9 Ibs. 



Answers to Examples 



257 



EXAMPLES V. 

(1) 1 60 horse-power ; 303 '36 horse-power ; 1 6 '64 horse-power. 

(2) 15*75 Ibs. per ton. (3) 22' 15 miles per hour. 
(4) 929; 1253. (5) 147-5 horse-power. 
(6) 0*347 horse-power. (7) 60 foot-lbs. 

(8) 350,000 foot-lbs. 800,000 foot-lbs. (9) 1,360,000 foot-lbs. 



EXAMPLES VI. 



(i) 57' i horse-power. 

(3) 6570 Ib.-feet. 

(5) 534O inch-lbs. ; 2220 inch-lbs. 



(2) 39,390 Ib.-feet. 

(4) 609 inch-lbs. 

(6) i2'8 horse-power. 



EXAMPLES VII. 

(i) 12,420,000 foot-lbs. ; 4,140,000 Ibs. 
(3) 37>74 inch-lbs. ; 35,940 inch-lbs. 
(5) 7'O2 horse-power. 
(7) 19-6 horse-power. 
(9) 10*5 feet per second ; 467 Ibs. 
(ii) 2886 foot-lbs. 

EXAMPLES VIII. 



(2) 27-8 feet per second. 

(4) 25-5 horse-power. 

(6) 7-25 horse-power. 

(8) 8'65 seconds. 
(10) 15-3 seconds. 
(12) 500,000 foot-lbs. 



(i) 68-5 (2) 1 1-85 miles per hour. 

(4) 4-25 inches. (5) 3052 feet. 

(7) 47 to horizontal. 

(8) 52'5 ; i '64 times the weight of the stone. 

(9) !5 P er cent, increase. 

(10) 66'4 ; 72-7 ; 59-3 revolutions per minute. 
(12) 38'33 ; 35'68 feet per second, 7*79 ; 6'28 Ibs. 

EXAMPLES IX. 



(3) 2672 feet. 

(6) 20 miles per hour. 



(ii) 



(1) o 855, 1*56, i'8i feet per second ; 8-05, 5-96, 4-4 feet per second per 

second. 

(2) f inch. (3) 1654, 827, 1474 Ibs. 
(4) X 53'3- (6) 0*342 second. 

(7) 1*103 second ; 67*3 feet per second per second. 

(8) 31-23. (9) i to 1-0073. 



EXAMPLES X. 



(i) 14*65 Ibs. ; 17-9 Ibs. 

(3) 9-6 tons tension ; 55*6 tons tension. 

(5) 2250 Ibs. ; 2890 Ibs. 



(2) 3 Ibs. ; 13 Ibs. 

(4) 4i c "7 south of west ; 720 Ibs. 

(6) 220 Ibs. ; 58*5 Ibs. 



258 Mechanics for Engineers 



EXAMPLES XI. 

(i) 0-154; 8 '8 (2) 2-97 Ibs. ; 8 -5 to horizontal. (3) 14-51 Ibs. 

(4) O'6 times the weight of log ; 36'8 to horizontal. (5) io 0> 4 

(6) 0-3066 horse-power. (7) 179 horse-power. 

(8) 3*84 horse-power. 

(9) 3 '4 feet per second per second ; 3-57 Ibs. 
(10) 4-5 tons; 31-9 seconds. (n) 3820 Ibs. 



EXAMPLES XI [. 

(I) 261 Ibs. (2) 16-97 Ibs. ; 4-12 Ibs. 

(3) Left, 5*242 tons ; right, 5*008 tons. 

(4) Left, 10 tons ; right, 3 tons ; end, 2*824 ions. 

(5) I '039 inches. (6) 5-737 feet from end. 



EXAMPLES XIII. 

(1) Tension, 2i'68 Ibs. ; pressure, 33-4 Ibs. ; ig"j to vertical 

(2) 0-1264. (3) 36. 

(4) 15*3 Ibs. at hinge ; 8*25 Ibs. at free end. 

(5) 395 Ibs. at A ; 2954 Ibs. at C. 

(6) n-2 Ibs. cutting AD 2'i inches from A, inclined I9'3 to DA. 

(7) 4*3 tons ; 3-46 tons ; 46-7 to horizontal. 

(8) 8*2 tons compression ; 4-39 tons tension ; 4 tons tension. 

(9) 8*78 tons tension ; 25-6 tons compression ; 2P22 tons tension. 



, EXAMPLES XIV. 

(i) 1-27 feet from middle. (2) 2 '08 inches. 

(3) 43 inches. (4)' 1*633 f eet > I<22 5 feet. 

(5) 4-18 inches ; 4*08 inches. (6) lo'i inches ; 5-5 Ibs. 

(7) 2-98 inches. (8) 27-2 inches. 

(9) 975 inches. (10) 1293 Ib.-feet ; 103-5 Ibs. per square foot, 

(n) 11-91 inches. (12) 4-82 inches. 

(13) 4 feet 5-1 inches. (14) 0-17" Ib. (15) 0-197 Ib. ; 0-384 Ib. 



EXAMPLES XV. 

(i) 19*48 inches ; 16-98 inches. (2) I2'i6 inches. 

(3) 6"o8 inches. (4) 15-4 inches. 

(5) 2-52 inches from outside of flange. (6) 4-76 inches. 

(7) O'2O2 inch from centre. (8) 16*6 inches. 

(9) 5'36 inches. (10) 33-99 inches. 



Answers to Examples 259 

EXAMPLES XVI 

(i) 16 and 8 tons. (2) 25 and 16 tons. 

(3) Left, 16-5 tons ; right, 33-4 tons. (4) 53 10'. 

(5) 16-43 inches; 4-41 inches. (6) 3-53 inches. 

(7) 3-67 inches. (8) 8000 foot-lbs. 

(9) 1 188 foot-lbs. (10) 140,000 ; 74,400 foot-lbs. 

(1.1) 75,600 foot-lbs. (12) 2514 foot-lbs. 

(13) 1 10-3 Ibs. (14) 5-1 1 Ibs. 

(15) 37'6 square inches. (16) 7-85 cubic inches. 
(17) 4 feet 3-9 inches. 

EXAMPLES XVII. 

(i) 312 (inches) 4 . (2) 405 (inches) 4 ; 4-29 inches. 

(3) 1 9S (inches) 4 ; 2^98 inches. (4) 290 (inches) 4 . 

(5) 5'5 2 3 inches. (6) 0-887 gravitational units. 

.(7) 



(8) 16*1 inches ; 35*15 gravitational units. 

EXAMPLES XVIII. 

(0 3 6 47 gravitational units. (2) 13,215 gravitational units. 

(3) 10 minutes 46 seconds ; 323. (4) 17-48 Ibs. 

(5) 35olb.-feet. (6) 2-134 gravitational units ; 6-83 inches. 

(7) 141 '3- (8) 7-71 inches. 

(9) 22 feet per second ; 31*06 feet per second. 

(10) 14*85 feet per second ; 16-94 f eet per second. 

(11) 3-314 feet ; 3819 gravitational units. (12) 53-7. 
(13) 0-0274 units. (14) 125-5. 

(15) 117-5 foot-tons. (16) 167 feet per second. 

(17) 23 sin a feet per second per second. 



( 



EXAMPLES XIX. 

(1) 6-47 Ibs. ; 0-016 inch ; IO2'6. 

(2) 7-8 Ibs. ; 0-013 inch ; 36. (3) 17-7 right ; II'3 left. 

(4) 21 -6 Ibs. ; 134 measured clockwise. 

(5) 3'4 38*i5 34'85 tons-feet. 

(6) 38-15. 34'85, *9'6, 25-95 tons-feet. 

(7) 1-65 tons; 2-35 tons; 3-65 tons. 

(8) 25 tons-feet ; nil ; 18-75 tons-feet ; 2-5 tons. 

EXAMPLES XX. 

(4) 48 ; 134-5 Ibs. (6) 4-06 feet. (7) 3*24 inches. 

(8) 12 Ibs. (9) 53 feet. 



EXAMINATION QUESTIONS 

Questions selected from the Mechanics Examinations 
Intermediate (Engineering) Science of London 
University. 

r. What is implied in the rule : the product of the diameter of 
a wheel in feet, and of the revolutions per minute, divided by 28, is 
the speed in miles an hour ? 

Also in the rule : three times the number of telegraph posts per 
minute is the speed in miles an hour ? (i93-) 

2. Continuous breaks are now capable of reducing the speed of 
a train 3^ miles an hour every second, and take 2 seconds to be 
applied. Show in a tabular form the length of an emergency stop 
at speeds of 3f, 7^, 15, 30, 45, and 60 miles an hour. 

Compare the retardation with gravity ; express the resisting 
force in pounds per ton ; calculate the coefficient of adhesion of the 
break shoe and rail with the wheel ; and sketch the arrangement. 

093-) 

3. Prove that the horse-power required to overcome a resistance 
of R Ibs. at a speed of S miles an hour is RS -i- 375. Calculate 
the horse-power of a locomotive drawing a train of 200 tons up an 
incline of i in 200 at 50 miles an hour, taking the road and air 
resistance at this speed at 28 Ibs. a ton. ( I 93-) 

4. If W tons is transported from rest to rest a distance s feet 
in / seconds, being accelerated for a distance jj and time t l by a 
force Pj tons up to velocity v feet per second, and then brought to 
rest by a force P 2 tons acting for / 2 seconds through s 2 feet 



_ p 
g 

Wz/ 2 



S 9 S 

= 2 = 2 7 = 2 7 
* 



Examination Questions 261 

A train of 100 tons gross, fitted with continuous breaks, is to 
be run on a level line between stations one-third of a mile apart, 
at an average speed of 12 miles an hour, including two-thirds of a 
minute stop at each station. Prove that the weight on the driving 
wheels must exceed 22| tons, with an adhesion of one-sixth, 
neglecting road-resistance and delay in application of the breaks. 

(1903-) 

5. Give a graphical representation of the relative motion of a 
piston and crank, when the connecting rod is long enough for its 
obliquity to be neglected ; and prove that at R revolutions a 

o 

minute, the piston velocity is times the geometric mean of the 

distance from the two ends of the stroke. 

Prove that if the piston weighs W Ibs., the force in pounds 
which gives its acceleration is 

W ir 2 R 2 

(distance in feet from mid-point of stroke). 
g 900 

(1903-) 

6. Write down the formula for the time of swing of a simple 
pendulum, and calculate the percentage of its change due to i per 
cent, change in length or gravity, or both. 

Prove that the line in Question 4 could be worked principally 
by gravity if the road is curved downward between the stations to 
a radius of about 11,740 feet, implying a dip of 33 feet between 
the stations, a gradient at the stations of i in 13, and a maximum 
running velocity of 31 miles an hour. ( I 93-) 

7. Prove that if a hammer weighing W tons falling h feet 
drives a pile weighing w tons a feet into the ground, the average 
resistance of the ground in tons is 

W 2 h 
W + w a 

Prove that the energy dissipated at the impact is diminished by 
increasing 

W 

(1903.) 
w 

8. Prove that the total kinetic energy stored up in a train of 
railway carriages, weighing W tons gross, when moving at v feet 
per second is 



where W t denotes the weight of the wheels in tons, a their radius, 
and / their radius of gyration. 



262 Mechanics for Engineers 

Prove that W in the equations of Question 4 must be increased 

W ? 
by V- to allow for the rotary inertia of the wheels. (1903.) 

9. Determine graphically, by the funicular polygon, the reaction 
of the supports of a horizontal beam, loaded with given weights at 
two given points. 

Prove that the bending moment at any point of the beam is 
represented by the vertical depth of the funicular polygon. 

(1903-) 

10. A wheel is making 200 revolutions per minute, and after 
10 seconds its speed has fallen to 150 revolutions per minute. If 
the angular retardation be constant, how many more revolutions 
will it make before coming to rest? (!9O4-) 

11. A piston is connected to a flywheel by a crank and con- 
necting rod in the usual manner. If the angular velocity of the 
flywheel be constant, show that if the connecting rod be 
sufficiently long the motion of the piston will be approximately 
simple-harmonic ; and find the velocity of the piston in any 
position. 

If A, B, C, D, E be five equidistant positions of the piston, 
A and E being the ends of its stroke, prove that the piston takes 
twice as long to move from A to B as it does from B to C. 

(1904.) 

12. A train whose weight is 250 tons runs at a uniform speed 
down an incline of i in 200, the steam being shut off and the 
brakes not applied, and on reaching the foot of the incline it runs 
800 yards on the level before coming to rest. What was its original 
speed in miles per hour ? 

[The fractional resistance is supposed to be the same in each 
case.] (1904.) 

13. A weight A hangs by a string and makes small lateral 
oscillations like a pendulum ; another weight, B, is suspended by 
a spiral spring, and makes vertical oscillations. Explain why an 
addition to B alters the period of its oscillations, whilst an addition 
to A does not. Also find exactly how the period of B varies with 
the weight. (1904.) 

14. A steel disc of thickness t and outer radius a is keyed on to 
a cylindrical steel shaft of radius b and length /, and the centre of 
the disc is at a distance c from one end of the shaft. Find the 
distance from this end of the shaft of the mass-centre of the whole. 

(1904.) 

15. A uniform bar 6 feet long can turn freely in a vertical plane 
about a horizontal axis through one end. If it be just started from 



Examination Questions 263 

the position of unstable equilibrium, find (in feet per second) the 
velocity of the free end at the instant of passing through its lowest 
position. (1904-) 

1 6. Explain why, as a man ascends a ladder, the tendency of 
its foot to slip increases. 

A man weighing 13 stone stands on the top of a ladder 20 feet 
long, its foot being 6 feet from the wall. How much is the 
horizontal pressure of the foot on the ground increased by his 
presence, the pressure on the wall being assumed to be horizontal ? 

(1904.) 

17. A horizontal beam 20 feet long, supported at the ends, 
carries loads of 3, 2, 5, 4 cwts. at distances of 3, 7, 12, 15 feet 
respectively from one end. Find by means of a funicular polygon 
(drawn to scale) the pressures on the two ends, and test the 
accuracy of your drawing by numerical computation. (1904.) 



Questions selected from the Associate Members' 
Examinations of the Institution of Civil Engineers. 

1. A beam 20 feet long is supported on two supports 3 feet from 
each end of the beam ; weights of lolbs. and 20 Ibs. are suspended 
from the two ends of the beam. Draw, to scale, the bending- 
moment and shearing-force diagrams ; and, in particular, estimate 
their values at the central section of the beam. 

(I.C.E., February, 1905.) 

2. The speed of a motor car is determined by observing the 
times of passing a number of marks placed 500 feet apart. The 
time of traversing the distance between the first and second posts 
was 20 seconds, and between the second and third 19 seconds. If 
the acceleration of the car is constant, find its magnitude in feet 
per second per second, and also the velocity in miles per hour at 
the instant it passes the first post. (I.C.E., February, 1905.) 

3. In a bicycle, the length of the cranks is 7 inches, the diameter 
of the back wheel is 28 inches, and the gearing is such that the 
wheel rotates -2\ times as fast as the pedals. If the weight of the 
cyclist and machine together is 160 Ibs., estimate the force which 
will have to be applied to the pedal to increase the speed uniformly 
from 4 to 12 miles an hour in 20 seconds, frictional losses being 
neglected. (I.C.E., February, 1905.) 

4. A thin circular disc, 12 inches radius, has a projecting axle 
\ inch diameter on either side. The ends of this axle rest on two 
parallel inclined straight edges inclined at a slope of i in 40, the 



264 Mechanics for Engineers 

lower part of the disc hanging between the two. The disc rolls 
from rest through I foot in 53^ seconds. Neglecting the weight 
of the axle and frictional resistances, find the value of g. 

(I.C.E., February, 1905.) 

5. A gate 6 feet high and 4 feet wide, weighing 100 Ibs., hangs 
from a rail by 2 wheels at its upper corners. The left-hand wheel 
having seized, skids along the rail with a coefficient of friction of 
\. The other wheel is frictionless. Find the horizontal force that 
will push the gate steadily along from left to right if applied 2 feet 
below the rail. You may solve either analytically or by the force- 
and-link polygon. (I.C.E., October, '1904.) 

6. A ladder, whose centre of gravity is at the middle of its 
length, rests on the ground and against a vertical wall ; the co- 
efficients of friction of the ladder against both being \. Find the 
ladder's inclination to the ground when just on the point of slipping. 

(I.C.E., October, 1904.) 

7. The faceplate of a lathe has a rectangular slab of cast iron 
bolted to it, and rotates at 480 revolutions per minute. The slab 
is 8 inches by 12 inches by 30 inches (the length being radial). 
Its outside is flush with the edge of the faceplate, which is 48 inches 
diameter. Find the centrifugal force. (Cast iron weighs \ Ib. per 
cubic inch.) Where must a circular weight of 300 Ibs. be placed 
to balance the slab? (I.C.E., October, 1904.) 

8. A boom 30 feet long, weighing 2 tons, is hinged at one end, 
and is being lowered by a rope at the other. When just horizontal 
the rope snaps. Find the reaction on the hinge. 

(I.C.E., October, 1904.) 

9. A beam ABCD, whose length, AD, is 50 feet, is supported 
at each end, and carries a weight of 2 tons at B, 10 feet from A, 
and a weight of 2 -5 tons at C, 20 feet from D. Calculate the 
shearing force at the centre of the span, and sketch the diagram of 
shearing forces. (I.C.E., October, 1904.) 

10. Referring to the loaded beam described in the last question, 
how much additional load would have to be put on at the point B 
in order to reduce the shearing force at the centre of the span to 
zero? (I.C.E., October, 1904.) 

11. Estimate the super-elevation which ought to be given to the 
outer rail when a train moves round a curve of 2000 feet radius at 
a speed of 60 miles an hour, the gauge being 4 feet 8| inches. 

(I.C.E., February, 1904.) 

12. Show that in simple-harmonic motion the acceleration is 
proportional to the displacement from the mid-point of the path, 
and that the time of a small oscillation of a simple pendulum of 



Examination Questions 265 

length / is 2ir. /-. Deduce the length of the simple pendulum 

\i *> 

which has the same time of oscillation as a uniform rod of length 
L suspended at one end. (I.C.E., February, 1904.) 

13. To a passenger in a train moving at the rate of 40 miles 
an hour, the rain appears to be rushing downwards and towards 
him at an angle of 20 with the horizontal. If the rain is actually 
falling in a vertical direction, find the velocity of the rain-drops in 
feet per second. (I.C.E., February, 1904.) 

14. If it take 600 useful horse-power to draw a train of 335 tons 
up a gradient of i in 264 at a uniform speed of 40 miles an hour, 
estimate the resistance per ton other than that due to ascending 
against gravity, and deduce the uniform speed on the level when 
developing the above power. (I.C.E., February, 1904.) 

15. In a steam-hammer the diameter of the piston is 36 inches, 
the total weight of the hammer and piston is 20 tons, and the 
effective steam pressure is 40 Ibs. per square inch. Find the 
acceleration with which the hammer descends, and its velocity 
after descending through a distance of 4 feet. If the hammer then 
come in contact with the iron, and compress it through a distance 
of i inch, find the mean force of compression. 

(I.C.E., February, 1904.) 

16. A uniform circular plate, i foot in diameter and weighing 
4 Ibs., is hung in a horizontal plane by three fine parallel cords 
from the ceiling, and when set in small torsional oscillations about 
a vertical axis is found to have a period of 3 seconds. A body 
whose moment of inertia is required is laid diametrically across it, 
and the period is found to be 5 seconds, the weight being 6 Ibs. 
Find the moment of inertia of the body about the axis of oscillation. 

(I.C.E., February, 1904.) 

17. The acceleration of a train running on the level is found 
by hanging a short pendulum from the roof of a carriage and 
noticing the angle which the pendulum makes with the vertical. 
In one experiment the angle of inclination was 5 : estimate the 
acceleration of the train in feet per second per second and in miles 
per hour per hour. (I.C.E., February, 1904.) 

1 8. Two weights, one of 2 Ibs. and the other of i lb., are con- 
nected by a massless string which passes over a smooth peg. Find 
the tension in the string and the distance moved through by either 
weight, from rest, in 2 seconds. (I.C.E., February, 1904.) 

19. A solid circular cast-iron disc, 20 inches in diameter and 
2 inches thick (weighing 0*25 lb. per cubic inch), is mounted on 
ball bearings. A weight of 10 Ibs. is suspended by means of a 



266 Mechanics for Engineers 

string wound round the axle, which is 3 inches in diameter, and 
the weight is released and disconnected after falling 10 feet. 
Neglecting friction, find the kinetic energy stored in the wheel, 
and the revolutions per minute the wheel is making when the 
weight is disconnected, and also the time it would continue to run 
against a tangential resistance of \ Ib. applied at the circumference 
of the axle. (I.C.E., February, 1904.) 

20. A girder 20 feet long carries a distributed load of i ton per 
lineal foot over 6 feet of its length, the load commencing at 3 feet 
from the left-hand abutment. Sketch the shearing-force and 
bending-moment diagrams, and find, independently, the magni- 
tude of the maximum bending moment and the section at which 
it occurs. (I.C.E., February, 1904.) 

21. Explain what is meant by centripetal acceleration, and find 
its value when a particle describes a circle of radius r feet with a 
velocity of v feet per second. (I.C.E., October, 1903.) 

22. In an electric railway the average distance between the 
stations is \ mile, the running time from start to stop ITT minutes, 
and the constant speed between the end of acceleration and 
beginning of retardation 25 miles an hour. If the acceleration 
and retardation be taken as uniform and numerically equal, find 
their values ; and if the weight of the train be 1 50 tons and the 
frictional resistance 1 1 Ibs. per ton, find the tractive force necessary 
to start on the level. (I.C.E., October, 1903.) 

23. The mass of a flywheel may be assumed concentrated in 
the rim. If the diameter is 7 feet and the weight 2\ tons, estimate 
its kinetic energy when running at 250 revolutions per minute. 
Moreover, if the shaft be 6 inches in diameter and the coefficient 
of friction of the shaft in the bearings be 0-09, estimate the number 
of revolutions the flywheel will make before coming to rest. 

(I.C.E., October, 1903.) 

24. A plane inclined at 20 to the horizontal carries a load of 
1000 Ibs., and the angle of friction betweeen the load and the plane 
is 10. Obtain the least force in magnitude and direction which 
is necessary to pull the load up the plane. 

(I.C.E., October, 1903.) 

25. State the second law of motion. A cage weighing 1000 Ibs. 
is being lowered down a mine by a cable. Find the tension in the 
cable, (i) when the speed is increasing at the rate of 5 feet per 
second per second ; (2) when the speed is uniform ; (3) when the 
speed is diminishing at the rate of 5 feet per second per second. 
The weight of the cable itself may be neglected. 

(I.C.E., October, 1903.) 



Examination Questions 267 

26. Show that when a helical spring vibrates freely under the 
action of a weight, its periodic time is the same as that of a simple 
pendulum having a length equal to the static extension of the 
spring when carrying the weight, the mass of the spring itself 
being neglected. (I.C.E., October, 1903.) 

27. A flywheel, supported on an axle 2 inches in diameter, 
is pulled round by a cord wound round the axle and carrying 
a weight. It is found that a weight of 4 Ibs. is just sufficient to 
overcome friction. A further weight of 16 Ibs., making 20 Ibs. in 
all, is applied, and two seconds after starting from rest it is found 
that the weight has descended a distance of 4 feet. Estimate the 
moment of inertia of the wheel about the axis of rotation in 
gravitational units. (I.C.E., October, 1903.) 

28. With an automatic vacuum brake a train weighing 170 tons 
and going at 60 miles an hour on a down gradient of I in 100 was 
pulled up in a distance of 596 yards. Estimate the total resistance 
in pounds per ton ; and if the retardation is uniform, find the time 
taken to bring the train to rest. (I.C.E., October, 1903.) 

29. A string, ABCD, hangs in a vertical plane, the ends A 
and D being fixed. A weight of 10 Ibs. is hung from the point 
B, and an unknown weight from the point C. The middle-portion 
BC is horizontal, and the portions AB and CD are inclined at 30 
and 45 to the horizontal respectively. Determine the unknown 
weight and the tensions in the three portions of the string. 

(I.C.E , October, 1903.) 

30. Two masses, of 10 Ibs. and 20 Ibs. respectively, are attached 
to a balanced disc at an angular distance apart of 90 and at radii 
2 feet and 3 feet respectively. Find the resultant force on the axis 
when the disc is making 200 turns a minute ; and determine the 
angular position and magnitude of a mass placed at 2'5 feet radius 
which will make the force on the axis zero at all speeds. 

(I.C.E., October, 1903.) 

31. A crane has a vertical crane-post, AB, 8 feet long, and a 
horizontal tie, BC, 6 feet long, AC being the jib. It turns in 
bearings at A and B, and the chain supporting the load passes 
over pulleys at C and A, and is then led away at 30 to AB. Find 
the stresses in the bars and thrusts in the bearings when lifting 
i ton at a uniform rate. (I.C.E., February, 1903.) 

32. A man ascends a ladder resting on a rough horizontal floor 
against a smooth vertical wall. Determine, graphically or other- 
wise, the direction of the action between the foot of the ladder and 
the floor. (I.C.E., February, 1903.) 

33. A train on a horizontal line of rails is accelerating the speed 



268 Mechanics for Engineers 

uniformly so that a velocity of 60 miles an hour is being acquired 
in 176 seconds. A heavy weight is suspended freely from the roof 
of a carriage by a string. Calculate, or determine graphically, the 
inclination of the string to the vertical. 

(I.C.E., February, 1903.) 

34. Explain the meaning of the term " centrifugal force." With 
what speed must a locomotive be running on level railway lines, 
forming a curve of 968 feet radius, if it produce a horizontal thrust 
on the outer rail equal to fa of its weight ? 

(I.C.E., February, 1903.) 

35. Explain how to determine the relative velocity of two 
bodies. A is travelling due north at constant speed. When B 
is due west of A, and at a distance of 21 miles from it, B starts 
travelling north-east with the same constant speed as A. Determine 
graphically, or otherwise, the least distance which B attains from A. 

(I.C.E., February, 1903.) 

36. Two men put a rail way- waggon weighing 5 tons into motion 
by exerting on it a force of 80 Ibs. The resistance of the waggon 
is 10 Ibs. per ton, or altogether 50 Ibs. How far will the waggon 
have moved in one minute ? Calculate at what fraction of a horse- 
power the men are working at 60 seconds after starting. 

(I.C.E., February, 1903.) 

37. State and explain fully Newton's third law of motion. A 
loo-lb. shot leaves a gun horizontally with a muzzle velocity of 
2000 feet per second. The gun and attachments, which recoil, 
weigh 4 tons. Find what the resistance must be that the recoil may 
be taken up in 4 feet, and compare the energy of recoil with the 
energy of translation of the shot. (I.C.E., February, 1903.) 

38. An elastic string is used to lift a weight of 20 Ibs. How 
much energy must be exerted in raising it 3 feet, supposing the 
string to stretch i inch under a tension of i Ib. ? Represent it 
graphically. If the work of stretching the string is lost, what is the 
efficiency of this method of lifting ? (I.C.E., February, 1903.) 

39. Explain how to determine graphically the relative velocity 
of two points the magnitudes and directions of whose velocities are 
known. Find the true course and velocity of a steamer steering 
due north by compass at 1 2 knots, through a 4-knot current setting 
south-west, and determine the alteration of direction by compass in 
order that the steamer should make a true northerly course. 

(I.C.E., October, 1902.) 

40. Find, by graphic construction, the centre of gravity of a 
section of an I beam, top flange 4 inches by i inch ; web, between 
flanges, 14 inches by i| inches ; bottom flange 9 inches by 2 inches. 

(I.C.E., October, 1902.) 



Examination Questions 



269 



41. A crankshaft, diameter \i\ inches, weighs 12 tons, and it 
is pressed against the bearings by a force of 36 tons horizontally. 
Find the horse-power lost in friction at 90 revolutions per minute 
(coefficient of friction = o'o6.) (I.C.E., October, 1902.) 



Questions selected from the Board of Education 
Examinations in Applied Mechanics. 

(Reprinted by permission of the Controller of His Majesty's 
Stationery Office.) 

1. A truck, weighing 5 tons without its wheels, rests on 4 
wheels, which are circular discs. 40 inches in diameter, each 
weighing | ton, and moves down an incline of i in 60. Find the 
velocity of the truck in feet per second after moving 100 feet from 
rest, if the resistance due to friction is I per cent, of the weight. 
What percentage of the original potential energy has been wasted 
in friction ? (Stage 3, 1905.) 

2. A flywheel is supported on an axle aj inches in diameter, and 
is rotated by a cord, which is wound round the axle and carries a 
weight. It is found by experiment that a weight of 5 Ibs. on the 
cord is just sufficient to overcome the friction and maintain steady 
motion. A load of 25 Ibs. is attached to the cord, and 3 seconds 
after starting from rest it is found that the weight has descended 
5 feet. Find the moment of inertia of this wheel in engineers' 
units. 

If the wheel is a circular disc 3 feet in diameter, what is its 
weight ? (The thickness of the cord may be neglected.) 

(Stage 3, 1905.) 

3. The angular position D of a rocking shaft at any time / is 
measured from a fixed position. Successive positions at intervals 
at g\j second have been determined as follows : 



Time t, se- ) 
conds j 
Position D, | 
radians ) 


O'O 

o'lo6 


O'O2 
0'208 


0-04 
Q'337 


o'o6 
0-487 


0-08 
0-651 


O'lO 
O'SlQ 


O'I2 
0-978 


O-I4 

mi 


0*16 

I '201 


0-18 

I '222 



Find the change of angular position during the first interval 
from / = o to / = 0*02. Calculate the mean angular velocity during 
this interval in radians per second, and, on a time base, set this up 
as an ordinate at the middle of the interval. Repeat this for the 



2/0 



Mechanics for Engineers 



other intervals, tabulating the results and drawing the curve show- 
ing approximately angular velocity and time. 

In the same way, find a curve showing angular acceleration and 
time. Read off the angular acceleration in radians per second per 
second, when / = 0x175 second. (Stage 2, 1905.) 

4. A motor car moves in a horizontal circle of 300 feet radius. 
The track makes sideways an angle of 10 with the horizontal 
plane. A plumb-line on the car makes an angle of 12 with what 
would be a vertical line on the car if it were at rest on a horizontal 
plane. What is the speed of the car? If the car is just not side- 
slipping, what is the coefficient of friction ? (Stage 3, 1904.) 

5. A body whose weight is 350 Ibs. is being acted upon by a 
variable lifting force F Ibs. when it is at the height x feet from 
its position of rest. The mechanism is such that F depends upon 
x in the following way ; but the body will stop rising before the 
greatest x of the table is reached. Where will it stop ? 




















i 




X ... 


o 


15 


*j 


50 


70 


100 


125 


150 


180 


2IO 


F 


530 


525 


56 


490 


425 


300 


210 


160 


no 


90 



Where does its velocity cease to increase and begin to diminish ? 

(Stage 3, 1904.) 

6. Part of a machine weighing i ton is moving northwards at 
60 feet per second. At the end of 0-05 second it is found to be 
moving to the east at 20 feet per second. What is the average 
force (find magnitude and direction) acting upon it during the 
interval ox>5 second ? What is meant by " average " in such a 
case ? What is meant by force by people who have to make exact 
calculations? (Stage 3, 1904.) 

7. A flywheel and its shaft weigh 24,000 Ibs. ; its bearings, which 
are slack, are 9 inches diameter. If the coefficient of friction is 
ox>7, how many foot-pounds of work are wasted in overcoming 
friction in one revolution ? 

If the mean radius (or rather the radius of gyration) is 10 feet, 
what is the kinetic energy when the speed is 75 revolutions per 
minute ? If it is suddenly disconnected from its engine at this 
speed, in how many revolutions will it come to rest ? What is its 
average speed in coming to rest? In how many minutes will it 
come to rest ? (Stage 2, 1904.) 

8. A train, weighing 250 tons, is moving at 40 miles per hour, 
and it is stopped in ten seconds. What is the average force during 



Examination Questions 271 

these ten seconds causing this stoppage ? Define what is meant by 
force by people who have to make exact calculations. 

(Stage 2, 1904.) 

9. A tram-car, weighing 15 tons, suddenly had the electric 
current cut off. At that instant its velocity was 16 miles per hour. 
Reckoning time from that instant, the following velocities, V, and 
times, /, were noted : 

V, miles per hour ... ... 16 14 12 10 

/, seconds o 9^3 21 35 

Calculate the average value of the retarding force, and find the 
average value of the velocity from / = o to / = 35. Also find the 
distance travelled between these times. (Advanced, 1903.) 

10. A projectile has kinetic energy = 1,670,000 foot-lbs. at a 
velocity of 3000 feet per second. Later on its velocity is only 
2000 feet per second. How much kinetic energy has it lost? 
What is the cause of this loss of energy? Calculate the kinetic 
energy of rotation of the projectile if its weight is 12 Ibs., and its 
radius of gyration is 075 inch, and its speed of rotation is 500 
revolutions per second. (Advanced, 1903.) 

1 1. A weight of 10 Ibs. is hung from a spring, and thereby causes 
the spring to elongate to the extent of 0^42 foot. If the weight is 
made to oscillate vertically, find the time of a complete vibration 
(neglect the mass of the spring itself). (Advanced, 1903.) 

12. A flywheel weighs 5 tons and has a radius of gyration of 
6 feet. What is its moment of inertia? It is at the end of a shaft 
10 feet long, the other end of which is fixed. It is found that a 
torque of 200,000 Ib.-feet is sufficient to turn the wheel through i. 
The wheel is twisted slightly and then released : find the time of a 
complete vibration. How many vibrations per minute would it 
make ? (Honours, Part I., 1903.) 

13. A flywheel of a shearing machine has 150,000 foot-lbs. of 
kinetic energy stored in it when its speed is 250 revolutions per 
minute. What energy does it part with during a reduction of speed 
to 200 revolutions per minute ? 

If 82 per cent, of this energy given out is imparted to the shears 
during a stroke of 2 inches, what is the average force due to this on 
the blade of the shears ? (Advanced, 1902.) 

14. A weight of 5 Ibs. is supported by a spring. The stiffness of 
the spring is such that putting on or taking off a weight of i Ib. 
produces a downward or upward motion of 0^04 foot. What is the 
time of a complete oscillation, neglecting the mass of the spring ? 

(Advanced, 1902.) 



272 



Mechanics for Engineers 



15. A car weighs 10 tons : what is its mass in engineers' units? 
It is drawn by the pull P Ibs., varying in the following way,/ being 
seconds from the time of starting : 



P ... 
t 


IO20 



980 
2 


882 

5 


720 
8 


702 

10 


650 

13 


713- 

16 


722 
19 


805 

22 



The retarding force of friction is constant and equal to 410 Ibs. 
Plot P 410 and the time /, and find the time average of this 
excess force. What does this represent when it is multiplied by 22 
seconds ? What is the speed of the car at the time 22 seconds 
from rest? (Advanced, 1902.) 

16. A body weighing 1610 Ibs. is lifted vertically by a rope, 
there being a damped spring balance to indicate the pulling force 
F Ibs. of the rope. When the body had been lifted x feet from its 
position of rest, the pulling force was automatically recorded as 
follows : 



X 

F 


o 
4010 


ii 
3915 


20 
3763 


34 
3532 


45 
3366 


55 
3208 


66 
3100 


76 
3007 



Using squared paper, find the velocity v feet per second for 
values of x of 10, 30, 50, 70, and draw a curve showing the probable 
values of v for all values of x up to 80. In what time does the 
body get from x = 45 to x = 55 ? (Honours, Part I., 1901.) 

17. A machine is found to have 300,000 foot-lbs. stored in it as 
kinetic energy when its main shaft makes 100 revolutions per 
minute. If the speed changes to 98 revolutions per minute, how 
much kinetic energy has it lost ? A similar machine (that is, made 
to the same drawings, but on a different scale) is made of the same 
material, but with all its dimensions 20 per cent, greater. What 
will be its store of energy at 70 revolutions per minute ? What 
energy will it store in changing from 70 to 71 revolutions per 
minute ? (Honours, Part I., 1901.) 

1 8. A body of 60 Ibs. has a simple vibration, the total length of 
a swing being 3 feet ; there are 200 complete vibrations (or double 
swings) per minute. Calculate the forces which act on the body at 
the ends of a swing, and show on a diagram to scale what force acts 
upon the body in every position. (Advanced, 1901.) 

19. An electric tramcar, loaded with 52 passengers, weighs 
altogether 10 tons. On a level road it is travelling at a certain 



Examination Questions 273 

speed. For the purpose of finding the tractive force, the electricity 
is suddenly turned off, and an instrument shows that there is a 
retardation in speed. How much will this be if the tractive force is 
315 Ibs. ? If the tractive force is found on several trials to be, on 
the average 

342 Ibs. when the speed is 12 miles per hour 
3'5 10 j) 

2 94 8 

what is the probable tractive force at 9 miles per hour ? 

(Advanced, 1901.) 



274 



LOGARITHMS. 








1 


2 


3 


4 


5 


6 


7 


8 


9 


123 4 


5 

21 
20 


6789 


10 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


4 9 13 17 
4 8 12 16 


25 30 34 38 
24 28 32 37 


11 

12 


0414 
0792 


0153 
0828 


0492 
0864 


0531 
C899 


0569 
0934 


0607 
0969 


0645 
1004 


0682 
1038 


0719 
1072 


0755 
1106 


4 8 12 15 
4 7 11 IS 
3 7 11 14 
3 7 10 14 


19 
19 
18 
17 


2327 31 35 
2J 26 30 33 
21 25 28 32 
20 24 27 31 


13 
14 


1139 
1461 


1173 
1492 


1206 
1523 


1239 
1553 


1271 
1581 


1303 
1614 


1335 
1641 


1367 
1673 


1399 
1703 


1430 
1732 


3 7 10 13 
3 7 10 12 
3 6 9 12 
3 6 9 12 


16 
16 
15 
15 


20 23 26 30 
19 22 2i 29 
18 21 24 28 
17 20 23 26 


15 


1761 


1790 


IfclS 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


3 6 9 11 
3 5 8 11 


14 
14 


17 20 23 26 
16 19 22 25 


16 
17 


2041 
2304 


2068 
2330 


2095 
2355 


2122 
2380 


2148 
2405 


2175 
2430 


2201 
2455 


2227 
2480 


2253 
2504 


2279 
2529 


3 5 8 11 
3 5 8 10 
3 5 8 10 
2 5 7 10 


14 

13 
13 
12 


16 19 22 24 
15 18 21 23 
15 18 20 23 

15 17 19 22 


18 
19 


2553 

2788 


2577 
2810 


2601 
2833 


2625 
2856 


2648 

2878 


2672 
2900 


2695 
2923 


2718 
2945 


2742 
2967 


2765 
2989 


2579 
2 5 7 9 
2479 
2468 


12 
11 
11 
11 

11 


14 16 19 21 
14 16 18 21 
13 16 18 20 
13 15 17 la 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


2468 


13 15 17 19 


21 
22 
23 
24 


3222 
3424 
3617 
3802 


3243 
3444 
3636 
3820 


3263 
3461 
3655 
3838 


3284 
3483 
3674 
3356 


3304 
3502 
3692 
3874 


3324 
3522 
3711 
3892 


3345 
3541 
3729 
3909 


3365 
3560 
3747 
3927 


3385 
3579 
3766 
3945 


3104 
3598 
3784 
3962 


2468 
2468 
2467 
2457 


10 

Id 
9 
9 


12 14 16 18 
12 14 15 17 
11 13 15 17 
11 12 14 16 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


2367 


9 


10 12 14 15 


26 
27 
28 
29 


4150 
4314 
4472 
4624 


4166 
4330 

4487 
4639 


4183 
4316 
4502 
4654 


4200 
4362 
4518 
4669 


4216 
4378 
4533 
4683 


4232 
435)3 
4548 
4698 


4249 
4409 
4564 
4713 


4265 
4125 
4579 
4728 


4281 
4440 
4594 
4742 


4298 
4456 
4609 
4757 


2357 
2356 
2356 
1346 


8 
8 

8 
7 


10 11 13 15 
9 11 13 14 
911 12 14 
9 10 12 13 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


1346 


7 


9 10 11 13 


31 
32 
33 
34 


4914 
5051 
5185 
5315 


4928 
5065 
5198 
6328 


4942 
5079 
5211 
5340 


4955 
5092 
5224 
5353 


4969 
5105 
5237 
5366 


4983 
5119 
5250 
5378 


4997 
5132 
5263 
5391 


5011 
6145 
5276 
5103 


5024 
5159 
5289 
5416 


5038 
5172 
5302 
5428 


1346 
1345 
1345 
1345 


7 
7 
6 
6 


8 10 11 12 
X 9 11 12 
8 9 10 12 
8 9 10 11 


35 


5441 


5453 


5465 


5478 


5490 


5502 


5514 


5527 


6539 


5551 


1245 


6 


7 9 10 11 


36 
37 
38 
39 


5563 
5682 
579b 
5911 


5575 
5694 
6809 
5922 


5587 
6705 
5821 
5933 


5599 
5717 
6832 
5944 


5611 
5729 
5843 
5955 


5623 
5740 
5855 
5906 


5635 
5752 
5866 
5977 


5647 
5763 

5877 
5988 


5658 
5775 

5888 
5999 


5670 
5786 
5S99 
6010 


124 5 
1235 
1235 
1234 


6 
6 
6 

5 


7 8 10 11 
78 1) 10 
7 8 9 JO 
7 8 9 10 


40 


6021 


6031 


6042 


6053 


6064 


6U75 


6085 


6096 


6107 


6117 


1234 


6 8 9 10 


41 
42 
43 
44 


6128 
6232 
6335 
6435 


6138 
6243 
6345 
6444 


6149 
6253 
6355 
6454 


6160 
6263 
6365 
6464 


6170 
6274 
6375 
6474 


6180 
6284 
6385 
6484 


6191 
6294 
6395 
6493 


6201 
6301 
6405 
503 


6212 
6314 
6415 
6513 


6222 
6325 
6425 
6522 


1234 
1234 
1234 
1234 


5 

5 
5 


6789 
6789 
6789 

6789 


45 


6532 


6542 


6551 


6561 


6571 


05-0 


6590 


6599 


6609 


6618 


1234 


5 


6789 


46 
47 
48 
49 

50 


6628 
6721 
6812 
6902 


6637 
6730 
6821 

(-'.II 1 


6646 
6739 
6830 
6920 


6656 
6749 
6839 
6928 


6665 
6758 
6848 
6937 


6675 
6767 
6857 
6946 


6684 
6776 
6566 
6955 


6693 

6785 
6S75 
6964 


6702 
6794 
6884 
6972 


6712 
6803 
6893 
6*81 


1234 
1234 
1234 
1234 


5 
5 
4 

4 

4 


6778 
5 6 7 H 
5678 
5678 

5678 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


1233 



LOGARITHMS. 



275 








1 


2 


3 


4 


5 


6 


7 


8 


9 


123 4 


5 


6789 


51 
52 
53 
54 


7076 
7160 
7243 
7324 


7084 
7168 
7251 
7332 


7093 
7177 
7*59 
7340 


7101 
7185 
7267 
7348 


7110 
7193 
7275 
7356 


7118 
7202 
7284 
7364 


7126 
7210 
7292 
7372 


7135 

7218 
7300 
7380 


7143 

7226 
7308 

7388 


7152 
7235 


1233 
1223 


4 

4 


5678 
5677 


73M6 


1223 


4 


5667 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


1223 


4 


5567 


56 
57 
58 
59 


7482 
7559 
7631 
7709 


7490 
7566 
7642 
7716 


7497 
7574 
7649 
7723 


7505 

7582 
7657 
7731 


7513 

7589 
7664 
7738 


7520 
7597 
7672 
7745 


7528 
7604 
7679 
7762 


7536 
7612 
7686 
7760 


7543 
7619 
7694 
7767 


7551 
7b27 
7701 

7774 








122 3 
1123 
1123 


4 
4 
4 


556V 
4567 
4 5 G 7 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


1123 


4 


4566 


61 
62 
63 
64 


7853 

7924 
7993 
8062 


7860 
7931 
8000 
8U69 


7868 
7938 
8007 
8075 


7875 
7945 
8014 
8082 


7882 
7952 
8021 
8089 


7889 
7959 
80^8 
8096 


7896 
7966 
8035 
8102 


7903 
7973 
8041 
8109 


7910 
7980 
8048 
8116 


7917 
7987 
8055 
8122 


1123 
1123 
1123 
1123 


4 
3 
3 
3 


4566 
4566 
4556 
4556 


65 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


1123 


3 


4556 


66 
67 
68 
69 

70 


8195 
8261 
8325 
8383 


8202 
8267 
8331 
8395 


8209 
8274 
8338 
8401 


8215 
8280 
344 
84U7 


8222 
8287 
8351 
8414 


8228 
8293 
8357 
8420 


8235 
8299 
8363 
8426 


8241 
8306 
837u 
8432 


8248 
8312 
8376 

813!) 


8254 
8319 
8382 
8445 


1123 
1123 
1123 
112 2 


3 
3 
3 
3 


4556 
4556 
4456 
4456 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


1122 


3 


4456 


71 
72 
73 
74 


8513 
8573 
8633 
8692 


8519 
8579 
8639 
869s 


8525 
8585 
8645 
8704 


8531 
8591 
8651 
8710 


8537 
8597 
8657 
8716 


8543 
8603 
8663 
8722 


8549 
8609 
8669 
8727 


8555 
8615 
8675 
8733 


8561 
8621 
3681 
3739 


8567 
8627 
8686 
8745 


11 1 2 2 
1122 
1 1 li -.2 
112 2 


3 
3 
3 
3 


4465 
4465 
4455 
4455 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


1122 


3 


3455 


76 
77 
78 
79 


8808 
8865 
8921 
8976 


8814 
8871 
8927 
8982 


8820 
8876 
8932 
8987 


8825 
8882 
8938 
8993 


8831 
8887 
8943 
8998 


8837 
8893 
8949 
9004 


8842 
8899 
8954 
9009 


8848 
8904 
8960 
9015 


8854 
8910 
8965 
9020 


8859 
8915 
8971 
9025 


1122 
1122 
1 1 2, 2 
112 2 


3 
3 
3 
3 


3455 
3 4 1 f) 
3446 
3445 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


1122 


3 


3445 


81 
82 
83 
84 

85 


9085 
9138 
9191 
9243 


9090 
9143 
9196 
9248 


9096 
9149 
9201 
9253 


9101 
9154 
9206 
9258 


9106 
9159 
9212 
9263 


9112 
9165 
9217 
9269 


9117 
9170 
9222 
9274 


9122 
9175 
9227 
9279 


9128 
9180 
9232 
9284 


9133 
9186 
9238 
9289 


1122 
112 2 
1122 
112 2 


3 
3 
3 
3 


3445 
3445 
3445 
3445 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9336 


9340 


1122 


3 


3445 


86 
87 
88 
89 


9345 
9395 
9445 
9494 


9350 
9400 
9450 
9499 


9355 
9405 
9455 
9504 


9360 
9410 
9460 
9509 


9365 
9415 
9465 
9513 


9370 
9420 
9469 
9518 


9375 
9425 
9474 
9523 


9380 
9430 
9479 
9528 


9385 
9435 
9484 
9533 


9390 
9440 
9489 
9538 


1122 
0112 
0112 
0112 


3 
2 
2 
2 


3445 
3344 
3344 
3344 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 


0112 


2 


3344 


91 
92 
93 
94 


9590 
9638 
9685 
9731 


9595 
9643 
9689 
9736 


9600 
9617 
9694 
9741 


9605 
9652 
9699 
9745 


9609 
9657 
9703 
9750 


9614 
9661 
9708 
9754 


9619 
9666 
8713 
9759 


9624 
9671 
9717 
9763 


9628 
9675 
9722 
9768 


9633 
9680 
9727 
9773 


0112 
0112 
II 1 1 2 
0112 


2 
2 
2 
2 


3344 
3344 
3344 
3344 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


0112 


2 


3344 


96 
97 
98 
99 


9823 
9868 
9912 
9956 


9827 
9872 
9917 
9961 


9832 
9877 
9921 
9965 


9836 
9881 
9926 
9969 


9841 
9886 
9930 
9974 


9845 
9890 
9934 
9978 


9850 
9894 
9939 
9983 


9864 
9899 
9943 

'J9*7 


9859 
9903 
9948 
9991 


9863 
9908 
9952 
9996 


0112 
0112 
0112 
0112 


2 
2 
2 
I 


3344 
3344 
3344 
3334 



2/6 



ANTILOGARITHMS. 








1 


2 


3 


4 


5 


6 


7 


8 


9 


1234 


5 


6789 


00 


1000 


1002 


1005 


1007 


1009 


1012 


1014 


1016 


1019 


1021 


0011 


i 


1222 


01 


1023 


1026 


1028 


1030 


1033 


1035 


1038 


1040 


1042 


1045 


0011 


i 


1222 


03 
04 


107^ 
1096 


1074 
1099 


1076 

no; 


1079 
1104 


1081 
1107 


1084 
1109 


1086 
1112 


1089 
1114 


1091 
1117 


1094 
1119 


0011 
0111 


i 

i 


1222 
2222 


05 


1122 


1125 


1127 


1130 


1132 


1135 


1138 


1140 


1143 


1146 


0111 


i 


2222 


06 
07 
08 
09 


1148 
1175 
1202 
1230 


1151 
1178 
1205 
1233 


1153 
1180 
1208 
1236 


1156 
1183 
1211 
1239 


1159 
1186 
1213 
1242 


1161 

1189 
1216 
1245 


1164 
1191 
1219 
1247 


1167 
1194 
1222 
1250 


1169 
1197 
1225 
1253 


1172 
1199 
1227 
1256 


0111 
0111 
0111 
0111 


i 
i 
i 
i 


2222 
2222 
2223 
2223 


10 


125!) 


1262 


1265 


1268 


1271 


1274 


1276 


1279 


1282 


1285 


0111 


i 


2223 


11 
12 
13 
14 


1288 
1318 
1349 
1380 


1291 
1321 
1352 
1384 


1294 
1324 
1355 
1387 


1297 
1327 
1358 
1390 


1300 
1330 
1361 
1393 


1303 
1334 
1365 
1396 


1306 
1337 
1368 
1400 


1309 
1340 
1371 
1403 


1312 
1343 
1374 
1406 


1315 
1346 

1377 
1403 


0111 
0111 
0111 
0111 


2 
2 
2 
2 


2223 
2223 
2233 
2233 


15 


1413 


1416 


1419 


1422 


1426 


1423 


1432 


1435 


1439 


1442 


0111 


2 


2233 


16 
17 
18 
19 


1445 
1479 
1514 
1549 


1449 
1483 
1517 
1552 


1452 
1486 
1521 
1556 


1455 
1489 
1524 
1560 


1459 
1493 
1528 
1563 


1462 
1496 
1531 
1567 


1466 
1500 
1535 
1570 


1469 
1503 
1538 
1574 


1472 
1507 
1542 
1578 


1476 
1510 
1545 
1581 


0111 
0111 
0111 
0111 


2 
2 
2 
2 


2233 
2233 
2233 
2333 


20 


1585 


1589 


1592 


1596 


1600 


1603 


1607 


1611 


1614 


1618 


0111 


2 


2333 


21 
22 
23 
24 


1622 
1660 
1698 
1738 


1626 
1663 
1704 
1742 


1629 
1667 
1706 
1746 


1633 
1671 
171U 
1750 


1637 
1675 
1714 
1754 


1641 
1679 
1718 
1758 


1644 
1683 
1722 
1762 


1648 
1687 
1726 
1766 


1652 
1690 
173(1 
1770 


1656 
1694 
1734 
1774 


0112 
0112 
0112 
0112 


2 
2 
2 
2 


2333 
2333 
2334 
2334 


25 


1778 


1782 


1786 


1791 


1795 


1799 


1803 


1807 


1811 


1816 


0112 


2 


2334 


26 
27 
28 
29 


1820 
1862 
1905 
1950 


1824 
1866 
1910 
1954 


18^8 
1871 
1914 
1959 


1832 
1875 
1919 
1963 


1837 
1879 
1923 
1968 


1841 
1884 
1928 
1972 


1845 
1888 
1932 
1977 


1849 
1892 
1936 
1982 


1854 
1897 
1941 
1986 


1858 
1901 
1945 
1991 


0112 
0112 
0112 
0112 


2 
2 
2 
2 


3334 
3334 
3 3 4 4 
3344 


30 


1995 


2000 


2004 


2009 


2014 


2018 


2023 


2028 


2032 


2037 


0112 


2 


3344 


31 
32 
33 
34 


2012 
2089 
2138 

2188 


20461 
2094 
2143 
2193 


2051 
2099 

2148 
2198 


2056 
2104 
2153 
2203 


2061 
2109 
2158 
2208 


2065 
2113 
2163 
2213 


2070 
2118 
2168 
2218 


2075 
2123 
2173 
2223 


2080 
2128 
2178 
2228 


2084 
2133 
2183 
2234 


0112 
0112 
0112 
1122 


2 
2 
2 
3 


3344 
3344 
3344 
3445 


35 


2239 


2244 


2249 


2254 


2259 


2265 


2270 


2275 


2280 


2286 


1122 


3 


3445 


36 
37 
38 
39 


2291 
2344 
2399 
2155 


2296 
2350 
2404 
2460 


2301 
2355 
2410 
2466 


2307 
2360 
2415 
2472 


2312 
2366 
2421 
2477 


2317 
2371 
2427 
2483 


2323 
2377 
2432 
2489 


2328 
2382 
2438 
2495 


2333 
2388 
2443 
2500 


2339 
2393 
2449 
2506 


1122 
1122 
1122 
1122 


3 
3 
3 
3 


3445 
3445 
3445 
3455 


40 


2512 


2518 


2523 


2529 


2535 


2541 


2547 


2553 


2559 


2564 


1122 


3 


4455 


41 

42 
43 
44 


2570 
2630 
2692 
2754 


2576 
2636 
2698 
2761 


2582 
2642 
2704 

2767 


2588 
2649 
2710 
2773 


2594 
2655 
2716 
2780 


2600 
2661 
2723 
2786 


2606 
2667 
2729 
2793 


2612 
2673 
2735 
2799 


2618 
2679 
2742 
2805 


2624 
2685 
2748 
2812 


1122 
1122 
1123 
1123 


3 
3 
3 
3 


4455 

4456 
4450 
4456 


45 


2818 


2825 


2831 


2838 


2844 


2851 


2858 


2864 


2871 


2877 


1123 


3 


4556 


46 
47 
48 
49 


2881 
2951 
3020 
3090 


2891 
2958 
3027 
3097 


2897 
2965 
3034 
31U5 


2904 
2972 
3041 
3112 


2911 
2979 
3048 
3119 


2917 
2985 
3055 
3126 


2924 
2992 
3062 
3133 


2931 
2999 
3069 
3141 


2938 
3006 
3076 
3148 


2944 
3013 
3083 
3155 


1123 
1123 
1123 
1123 


3 
3 
4 
4 


4556 
4556 
4566 
4566 



ANTILOGARITHMS. 



277 








1 


2 


3 


4 


5 


6 


7 


8 


9 


1 234 


5 


6789 


50 


3162 


3170 


3177 


3184 


3192 


3199 


3206 


3214 


3221 


3228 


1123 


4 


4567 


51 
52 
53 
54 


3236 
3311 
3388 
3467 


3243 
3319 
3396 
3475 


3251 
3327 
3404 
3483 


3258 
3334 
3412 
3491 


3266 
3342 
3420 
3499 


3273 
3350 
3428 
3508 


3281 
3357 
3436 
3516 


3289 
3365 
3443 
3524 


3296 
3373 
3451 
3532 


3304 
3381 
3459 
3540 


1223 
1223 
1223 
1223 


4 
4 
4 
4 


5567 
5567 
5667 
b 6 6 7 


55 


3548 


3556 


3565 


3573 


3581 


3589 


3597 


3606 


3614 


3622 


1223 


4 


5677 


56 
57 
58 
59 


3631 
3715 
3802 
3890 


3639 
3724 
3811 
3899 


3648 
3733 
3819 
3908 


3656 
3741 
3828 
3917 


3664 
3750 
3837 
3926 


3673 
3758 
3846 
3936 


3681 
3767 
3855 
3945 


3690 
3776 
3864 
3954 


3698 
3784 
3873 
3963 


3707 
3793 
3882 
3972 


1233 
1233 
1234 
1234 


4 
4 
4 
5 


5678 
5678 
5678 
5678 


60 


3981 


3990 


3999 


4009 


4018 


4027 


4036 


4046 


4055 


4064 


1234 


5 


6678 


61 
62 
63 
64 


4074 
4169 
4266 
4365 


4083 
4178 
4276 
4375 


4093 
4188 
4285 
4385 


4102 
4198 
4295 
4395 


4111 
4207 
4305 
4406 


4121 
4217 
4315 
4416 


4130 
4227 
4325 
4426 


4140 
4236 
4335 
4436 


4150 
4246 
4345 
4446 


4159 
4256 
4355 
4457 


1234 
1234 
1234 
1234 


5 
5 
5 
5 


6789 
6 7 8 !i 
6789 
6789 


65 

66 
67 
68 
69 


4467 


4477 


4487 


4498 


4508 


4519 


4529 


4539 


4550 


4560 


1234 


5 


6789 


4571 

4677 
4786 
4898 


4581 
4688 
4797 
4909 


4592 
4699 
4808 
4920 


4603 
4710 

4819 
4932 


4613 

4721 
4831 
4943 


4624 
4732 
4842 
4955 


4634 
4742 
4853 
4966 


4645 
4753 
4864 
4977 


4656 
4764 
4875 
4989 


4667 
4775 
4887 
5000 


1234 
1234 
1234 
1235 


5 
5 
6 
6 


6 7 9 10 
7 8 9 10 
7 8 9 10 
7 8 9 10 


70 


5012 


5023 


5035 


5047 


5058 


5070 


5082 


593 


5105 


5117 


1245 


6 

6 
6 
6 
6 


7 8 9 11 


71 
72 
73 
74 


5129 
5248 
5370 
5495 


5140 
5260 
5383 
5508 


5152 
5272 
5395 
5521 


5164 
5284 
5408 
6534 


5176 
5297 
5420 
5546 


5188 
5309 
5433 
5559 


5200 
5321 
5445 
5572 


5212 
5333 
5458 
5585 


5224 
5346 
5470 
5598 


5236 
5358 
5483 
5610 


1245 
1245 
1345 
1345 


7 8 10 11 
7 9 10 11 
8 9 10 11 
8 9 10 12 


75 


6623 


5636 


5649 


5662 


5675 


5689 


5702 


5715 


5728 


5741 


1345 


7 


8 9 10 12 


76 
77 
78 
79 


5754 
5888 
6026 
6166 


5768 
5902 
6039 
6180 


5781 
5916 
6053 
6194 


5794 
5929 
6067 
6209 


5808 
5943 
6081 
6223 


5821 
5957 
6095 
6237 


5834 
5970 
6109 
3252 


5848 
5984 
6124 
6266 


5861 
5998 
6138 
6281 


5875 
6012 
6152 
6295 


1345 
1345 
1346 
1346 


7 
7 
7 
7 


8 9 11 12 
8 10 11 12 
8 10 11 13 
9 10 11 13 


80 


6310 


6324 


6339 


6353 


6368 


6383 


6397 


6412 


6427 


6442 


1346 


7 


9 10 12 13 


81 
82 
83 
84 


6457 
6607 
6761 
6918 


6471 
6622 
6776 
6934 


6486 
6637 
6792 
6950 


6501 
6653 
6808 
6966 


6516 
6668 
6823 
6982 


6531 
6683 
6839 
6998 


6546 
6699 
6855 
7015 


6561 
6714 
6871 
7031 


6577 
6730 
6887 
7047 


6592 
6745 
6902 
7063 


2356 
2356 
2356 
2356 


8 
8 
8 
8 


9 11 12 14 
9 11 12 14 
9 11 13 14 
10 11 13 15 


85 


7079 


7096 


7112 


7129 


7145 


7161 


7178 


7194 


7211 


7228 


2357 


8 


10 12 13 15 


86 
87 
88 
89 


7244 
7413 
7586 
7762 


7261 
7430 
7603 
7780 


7278 
7447 
7621 
7798 


7295 
7464 
7638 
7816 


7311 
7482 
7656 
7834 


7328 
7499 
7674 
7852 


7345 
7516 
7691 
7870 


7362 
7534 
7709 
7889 


7379 
7551 
7727 
7907 


7396 
7568 
7745 
7925 


2357 
2357 
2457 
2457 


8 
9 
9 
9 

9 


10 12 13 15 
10 12 14 16 
11 12 14 16 
11 13 14 16 


90 


7943 


7962 


7980 


7998 


8017 


8035 


8054 


8072 


8091 


8110 


2467 


11 13 15 17 


91 
92 
93 
94 


8128 
8318 
8511 
8710 


8147 
8337 
8631 
8730 


8166 
8356 
8551 
8750 


8185 
8375 
8570 
8770 


8204 
8395 
8590 
8790 


8222 
8414 
8610 
8810 


8241 
8433 
8630 
8831 


8260 
8453 
8650 
8851 


8279 
8472 
8670 
8872 


8299 
8492 
86SIO 
8892 


2468 
2468 
2468 
2468 


9 
10 
10 
10 


11 13 15 17 
12 14 15 17 
12 14 16 18 
12 14 16 18 


95 


8913 


8933 


8954 


8974 


8995 


9016 


9036 


9057 


9078 


9099 


2468 


10 


12 15 17 19 


96 
97 
98 
99 


9120 
9333 
9550 
8772 


9141 
9354 
9572 
9795 


9162 
9376 
9594 
9817 


9183 
9397 
9616 
9840 


9204 
9419 
9638 
9863 


9226 
9441 
9661 

9886 


9247 
9462 
9683 
9908 


9268 
9484 
9705 
9931 


9290 
9506 
9727 
9954 


9311 
9528 
9750 
9977 


2468 
2479 
2479 
2579 


11 
11 
11 
11 


13 15 17 19 
13 15 17 20 
13 16 18 20 
14 16 18 20 



2/8 



Angle. 


Chord. 


Bine. 


Tangent. 


Co- 

tangent. 


Cosine 








De- 
grees. 


Radians. 








000 








8 


1 


1-414 


1-5708 


90 

89 
88 
87 
86 


1 
2 
3 

4 


0175 
0349 
0524 
0698 


017 
035 
052 
070 


0175 
0349 
0523 
0698 


0175 
0349 
0524 
0699 


57-2900 
28-6363 
19-0811 
14-3007 


9998 
9994 
9986 
9976 


1-402 
1-389 
1-377 
1-364 


1-5533 
1-5359 
1-5184 
1-5010 


5 

6 

7 
8 
9 


0873 

1047 
1222 
1396 
1571 


087 


0872 


0875 


11-4301 


9962 


1-351 


1-4835 


85 


105 

122 
140 
157 


1045 
1219 
1392 
1564 


1051 

1228 
1405 
1584 


9-5144 
8-1443 
7-1154 
6-3138 


9945 
9925 
9903 
9877 


1-338 
1-325 
1-312 
1-299 


1-4661 
1-4486 
1-4312 
1-4137 


84 
83 
82 
81 


10 


1745 


174 


1736 


1763 


5-6713 


9848 


1-286 


1 -3963 


80 

79 

78 
77 
76 


11 

12 
13 
14 


1920 
2094 
2269 
2443 


192 
209 
226 
244 


1908 
2079 
2250 
2419 


1944 
2126 
2309 
2493 


5-1446 
4-7046 
4-3315 
4-0108 


9816 

9781 
9744 
9703 


1-272 
1-259 
1-245 
1-231 


l-a788 
1 3614 
1-3439 
1-3265 


15 


2618 


261 


2588 


2679 


3-7321 


9669 


1-218 


1 '3090 


75 


16 
17 
18 

19 


2793 
2967 
3142 
3316 


278 
296 
313 
330 


2756 
2924 
3090 
3256 


2867 
3057 
3249 
3443 

3640 


3-4874 
3-2709 
3-0777 
2-9012 


9613 
9563 
9511 
9455 


1-204 
1-190 
1-176 
1-161 


1-2915 
1-2741 
1-2566 
1-2392 


74 
73 
72 
71 


20 


3491 


347 

364 
382 
399 
416 


3420 


2-7475 


9397 


1-147 


1-2217 


70 


21 
22 
23 

24 


3665 
3840 
4014 
4189 


3584 
3746 
3907 
4067 


3839 
4040 
4245 
4452 


2-6051 
2-4751 
2-3559 
2-2460 


9336 
9272 
9205 
9135 


1-133 
1-118 
1-104 
1-089 


1-2043 
1-1868 
1-1694 
1-1519 


69 
68 
67 
66 


25 


4363 


433 


4226 


4663 


2-1445 


9063 


1-075 


1-1345 


65 

64 
63 
62 
61 


26 
27 
28 
29 


4538 
4712 
4887 
5061 


450 
467 
484 
501 


4384 
4540 
4695 
4848 


4877 
5095 
5317 
5543 


2-0503 
1-9626 
1 8807 
1-8040 


8988 
8910 
8829 
8746 


1-060 
1-045 
1-030 
1-015 


1-1170 
1-0996 
1-0821 
1-0647 


30 


5236 


518 


5000 


5774 


1-7321 


8660 


1-000 


1-0472 


60 


31 
32 
33 
34 


5411 
5585 
5760 
5934 


534 
551 

568 
585 


5150 
5299 
5446 
5592 


6009 
6249 
6494 
6745 


1-6643 
1-6003 
1-5399 
1-4826 


8572 
8480 
8387 
8290 


986 
970 
954 
939 


1-0297 
1-0123 
9948 
9774 


59 
58 
57 
56 


35 


6109 


601 

618 
635 
651 
668 


5736 


7002 


1-4281 


8192 


923 


9599 


55 

54 
53 
52 
51 


36 
37 

38 
39 


6283 
6458 
6632 
6807 


6878 
6018 
6157 
6293 


7265 
7536 
7813 
8098 


1-3764 
1-3270 
1-2799 
1-2349 


8090 
7986 
7880 
7771 


908 
892 
877 
861 


9425 
9250 
9076 
8901 


40 

41 
42 
43 
44 


6981 


684 


6428 


8391 


1-1918 


7660 


845 

829 
813 
797 
781 


8727 


50 


7156 
7330 
7505 
7679 


700 
717 
733 
749 


6561 
6691 
6820 
6947 


8693 
9004 
9325 
9667 


1-1504 
1-1106 
1-6724 
1-0355 


7547 
7431 
7314 
7193 


8552 
8378 
8203 
8029 


49 

48 
47 
46 


45 


7854 


765 


7071 


1-0000 


1-0000 


7071 


765 


7864 


45 








Cosine. 


Co- 
tangent. 


Tangent. 


Sine. 


Chord. 


Radians. 


De- 

greesw 


Angle. 



INDEX 



( The numbers refer to pages) 



Acceleration, 3 
Adhesion, 107 
Alternating vectors, 81 
Amplitude, 79 
Angular acceleration, 23 

momentum, 207 

motion, 23 

velocity, 23 

Atwood's machine, 44 
Average force (space), 51 
force (time), 35 

B 

Bending moment, 231 

diagram, 233 

Bicycle, centre of gravity, 184 
Bows' notation, 223 
Brakes, 107 



Centre of gravity, 141-165 

of mass, 141 

of parallel forces, 140 

Centrifugal force, 69, 181 
Centripetal force, 69 
C.g.s. units, 30 
Chains, loaded, 243 



Circular arc, 160 

motion, 68 

sector and segment, 161 

Coefficient of adhesion, 107 

of friction, 100 

Compound pendulum, 212 
Conditions of equilibrium, 97, 128, 

226 

Conical pendulum, 72 
Couple, 125 
Curve, motion on, 70, 71 



D 

Density, 27 
Derived units, 253 
Displacement curve, 2 

, relative, 16 

Distributed load, 168, 247 



E 

Efficiency of machines, 1 10 

of screw, 108 

Energy, 57 

in harmonic motion, 84 

, kinetic, 58 

Equilibrant, 92 

Equilibrium, conditions of, 97, 128, 

226 
, stability of, 172 



280 



Mechanics for Engineers 



(The numbers refer to pages) 



First law of motion, 27 
Force, 27, 29 
Forces, coplanar, 127 

, parallel, 114 
, resolution and composition 

of, 91 
-, triangle and polygon of, 33, 91 



Frames, 236 
Friction, 99 

, angle of, 101 

, coefficient of, 100 

, laws of, 100 

of machines, I IO 

of screw, 1 08 

, sliding, 100 

-, work spent in, 107 



Fundamental units, 253 
Funicular polygon, 224, 228, 233, 
243 



Gravitational units, 30, 253 
Gravity, acceleration of, 6 
Guldinus, 182 



H 

Harmonic motion, 79 
Hemisphere, 162, 172 
Horse-power, 51 



Impulse, 33 
Impulsive force, 36 
Inclined plane, 102 

, smooth, 22 

Indicator diagram, 50 
Inertia, 27 



Inertia, moment of, 188 
, (areas), 194 

K 

Kinematics, Chapter I. 
Kinetic energy, 58 

of rotation, 204 

of rolling body, 2 



Lami's theorem, 93 

Laws of motion, Chapter II. 

Levers, 122 

Lifting, work in, 176 

Limiting friction, 100 

Load, distributed, 168 

Locomotive, centre of gravity, 186 

M 

Machines, HO 
Mass, 27 
Mechanical advantages of screw, 

109 

Method of sections, 133 
Moment, 53, 119-122 

of an area, 157 

of inertia, 1 88 

of areas, 194 

of momentum, 207 

Momentum, 28 
Motion, first law of, 27 

of connected weights, 43 

, second law of, 28 

, simple harmonic, 79 

, third law of, 41 

Motor-car, centre of gravity, 1 86 

N 

Neutral equilibrium, 172, 174 
Newton's laws of motion, 27 



Index 



281 



( 77/<? numbers refer to pages) 



Pappus, 182 

Parallel axes, moment of inertia 

about, 191, 192 

forces, 114 
Pendulum, compound, 212 

, conical, 7 2 

, simple, 85 

, simple equivalent, 86, 213 

Plane-moments, 155 
Polygon offerees, 33, 91 

, funicular or link, 224 

of velocities, 17 

Pound, unit of force, 29, 253 

Poundal, 29 

Power, 51 

Principle of moments, 122 

of work, 59 



R 

Radius of gyration, 189 
Railway curve, 71 
Reduction of forces, 127 
Relative displacement, 1 6 

velocity, 20 

Resolution of accelerations, 22 
of forces, 9 1 

of velocity, 18 

Rolling body, 217 

Roof, 242 

Rotation about axis, 179, 204, 207 



Screw friction, 108 
Second law of motion, 28 
Sections, method of, 133 
Sector of circle, 161 

of sphere, 162 

Segment of circle, 161 
Shearing force, 231 



Shearing-force diagram, 235 
Simple equivalent pendulum, 
213 

harmonic motion, 79 

, torsional, 214 



86, 



pendulum, 85 
Smooth body, 129 
Space-average force, 51 



curve, 2 

diagram, 223 
Spherical shell, 161 
Spring, vibrating, 83 
Stable equilibrium, 172 
Statics, 91 
Stress diagram, 239 

, tensile and compressive, 240 

String, loaded, 243 

polygon, 242 

Strut, 240 

T 

Theorem of Guldinas or Pappus, 

182 

Third law of motion, 41 
Tie, 240 

Time-average force, 35, 36 
Torque, 54 

Torsional oscillation, 214 
Triangle of forces, 33, 91 

of velocities, 17 

Twisting moment, 54 



U 

Uniform circular motion, 68 

Units, 253 

Unstable equilibrium, 172 

V 

Vector diagram, 223 
Vectors, 15 
Velocity, I 



282 



Mechanics for Engineers 
( The numbers refer to pages) 



Velocity, angular, 23 

, component, 18 

curves, 7 

, polygon of, 17 

, relative, 20 
Vertical circle, motion in, 73 

motion, 6 

Vibration of spring, 83 



W 



Warren girder, 133, 241 
Weight, 28 
Work, 48 

in lifting, 176 

of a torque, 54 

, principle of, 59 



THE END 



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