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AN ELEMENTARY TREATISE ON

THEORETICAL MECHANICS

J. H. JEANS, M.A., F.RS.

FELLOW OF TRINITY COLLEGE, CAMBRIDGE (ENGLAND)

AND PROFESSOR OF APPLIED MATHEMATICS

IN PRINCETON UNIVERSITY

GINN AND COMPANY

BOSTON • NEW YORK - CHICAGO • LONDON
ATLANTA • DALLAS • COLUMBUS • SAN FRANCISCO

ENTERED AT STATIONERS' HALL

atftenaum

GINN AND COMPANY • PRO-
PRIETORS • BOSTON • U.S.A.

The primary aim of the present book is to supply for students
beginning the study of Theoretical Mechanics a course of such a
nature as shall emphasize the fundamental physical principles of
the subject. Different students will of course approach the study
of mechanics with different interests, different aims, and different
amounts of mathematical equipment, so that it may not be possible
to produce a single book which shall exactly fit the requirements
of every class of student. But I believe that all students of me-
chanics, no matter what their aims and intentions may be, will
be in the same position in one respect, namely that they will best
begin the study of the subject by trying to acquire a firm grasp
of the physical principles, leaving aside at first all mathematical
developments and all practical applications, except in so far as these
contribute to the elucidation of the fundamental physical principles.

I am aware that this belief is not held by all teachers of
mechanics, some of whom regard the laws of mechanics simply as
working rules to be acquired as rapidly as possible for their utili-
tarian value, while others appear to regard them in the same light
as the rules of a game, the game consisting in the solution of
mathematical puzzles, most of which have no conceivable refer-
ence to the facts of nature. I find it hard to believe that there
can be any considerable class of students for whom either of these
points of view is the best. As regards the former, I feel that a
student who cannot get, or does not wish to get, a clear under-
standing of mechanical principles would be well advised not to
enter a profession in which his work will consist in the handling
of mechanical problems ; and as regards the latter, that a student
who wishes merely to obtain material for puzzle solving would do
better to turn his attention to chess or double acrostics.

iii

iv PKEFACE

If I have taken some space to express my private convictions, it
is because the method I have embodied in the present book arises
directly out of these convictions. Mathematical analysis is, of
course, not excluded from the book, because without mathematics
there can be no serious study of mechanics, but I have tried to
reduce the amount of mathematics to a minimum, and I have
regarded it (hi the present book) as the servant and not as the
master. Again, practical applications of mechanics have not been
excluded, — on the contrary, these have been introduced wherever
possible as illustrations of principles or results, — but I have tried
to place principles first and applications second. And problems
have not been excluded : I have inserted a great number, because
the solution of problems seems to me to be the one and indis-
pensable way of emphasizing a group of abstract principles and of
fixing them in the mind of the student. But I have regarded the
problems as an adjunct to the study of the principles, and not the
principles as a framework round which to build problems.

Besides explaining the method and objects of a book, a preface
may be expected to explain where the book starts and where it
ends. The present book is intended to start from the very begin-
.ning of its subject, assuming no previous knowledge of mechanics
on the part of the student. The question of how much knowledge
of mathematics ought to be assumed has been a more difficult one
to settle. I finally decided to rely as little as possible on the stu-
dent's knowledge of trigonometry, and to employ the calculus as
little as possible in the earlier chapters, but felt that the subjects
of the later chapters could not be advantageously treated without
a very considerable use of the calculus. Until the later chapters
the use of the calculus is confined almost exclusively to unimpor-
tant branches and extensions of the subject, and to the working of
illustrative examples. Thus a student who has no knowledge at
all of the calculus will, I hope, be able to omit the sections of the
book in which it is used, while at the same time acquiring a con-
siderable and continuous knowledge of the essentials of theoretical
mechanics.

PREFACE V

The point at which the book ought to close seemed in the pres-
ent instance to be determined by the method of the book itself.
If, as I believe, a study of physical principles ought to be the com-
mon preliminary to the study of every branch and every appli-
cation of mechanical science, then the book might clearly try to
cover all this common ground, and ought to stop at the point at
which detailed specialization becomes feasible and profitable. It
ought, in fact, to cover the range which will be covered by all stu-
dents, and stop short of subjects which will be of interest or impor-
tance only to a few. Judged by this criterion the book will perhaps
be thought by some to be open to the criticism of covering too much
ground; it may be thought that the final chapter on generalized
coordinates can hardly be regarded as essential to the student
whose study of mechanics is a preliminary to his entering the pro-
fession of, say, engineering. I am nevertheless convinced that, even
if the study of generalized coordinates is not absolutely indispen-
sable to such students, it is of extreme value and ought not to be
neglected by a student, possessed of the requisite ability, who can
possibly find time for it. The student who omits it shuts himself
off from a point of view which sums up and illuminates the whole
of dynamical theory; at the same time he denies himself the
opportunity of studying, or at least of fully understanding, the
theory of electricity and magnetism. And as regards the student
who intends to continue his studies in the direction of theoretical
physics, the theory of generalized coordinates formS so essential
a preliminary to the study of most branches of physics that the
advantages of including a short treatment of this subject in the
preliminary mechanics course will hardly be disputed.

PRINCETON J- H«

November, 1906

CONTENTS

PAGE
CHAPTER I. REST AND MOTION 1

Introduction. Motion of a point. Velocity. Acceleration. Vec-
tors.

CHAPTER II. FORCE AND THE LAWS OF MOTION 26

Newton's laws. Frame of reference. Laws applicable only to the
motion of a particle.

CHAPTER III. FORCES ACTING ON A SINGLE PARTICLE .... 37

Composition and resolution of forces. Particle in equilibrium.
Types of forces, — weight of a particle, tension of a string, reac-
tion between two bodies. Friction.

CHAPTER IV. STATICS OF SYSTEMS OF PARTICLES 59

Moments. System of particles in equilibrium. Forces in one plane.
Strings, — the suspension bridge, the catenary.

CHAPTER V. STATICS OF RIGID BODIES 90

Rigidity. Conditions of equilibrium for a rigid body. Transmissi-
bility of force. Composition of forces acting in a plane. Parallel
forces. Couples. Forces in space.

CHAPTER VI. CENTER OF GRAVITY 117

Center of gravity of a lamina. Center of gravity obtained by inte-
gration. Center of gravity of areas and volumes.

CHAPTER VII. WORK 145

Measurement and units. Work done against a variable force.
Work done in stretching an elastic string. Work represented by an
area. The principle of virtual work. Potential energy. Kinetic
energy. Conservation of energy. Stable and unstable equilibrium.

vii

viii CONTENTS

PAGK
CHAPTER VIII. MOTION OF A PARTICLE UNDER CONSTANT FORCES 188

Body falling under gravity. Motion on an inclined plane. At-
wood's machine. Motion referred to a moving frame of reference.
Frictional reactions between moving bodies. Flight of projectiles.

CHAPTER IX. MOTION OF SYSTEMS OF PARTICLES 220

Equations of motion. Conservation of momentum. Motion of
center of gravity. Kinetic energy. Impulsive forces. Elasticity.

CHAPTER X. MOTION OF A PARTICLE UNDER A VARIABLE FORCE 254

Equations of motion. The simple pendulum. Simple harmonic
motion. The cycloidal pendulum. Motion of a particle about a
center of force, — force proportional to the distance. General theory
of motion about a center of force. The law of the inverse square.

CHAPTER XI. MOTION OF RIGID BODIES 286

Angular velocity. Kinetic energy. Radii of gyration. Moment of
momentum. General theory of moments of inertia. General equa-
tions of motion of a rigid body. Euler's equations. Rotation of a
planet. Motion of a top.

CHAPTER XII. GENERALIZED COORDINATES 320

Hamilton's principle. Principle of least action. Lagrange's equa-
tions. Small oscillations. Stability and instability of equilibrium.
Forced oscillations. The canonical equations.

INDEX 361

THEORETICAL MECHANICS

CHAPTEE I
REST AND MOTION

INTRODUCTION

1. Uniformity of nature. If we place a stone in water, it will
sink to the bottom ; if we place a cork in water, it will rise to
the top. These two statements will be admitted to be true not
only of stones and corks which have been seen to sink or rise in
water but of all stones and corks. Given a piece of stone which
has never been placed in water, we feel confident that if we place
it in water it will sink. What justification have we for supposing
that this new and untried piece of stone will sink in water ? We
know that millions of pieces of stone have at different times been
placed in water ; we know that not a single one of these has ever
been known to do anything but sink. From this we infer that
nature treats all pieces of stone alike when they are placed in
water, and so feel confident that a new and untried piece of stone
will be treated by the forces of nature in the same way as the
innumerable pieces of stone of which the behavior has been
tested, and hence that it will sink in water. This principle is
known as that of the uniformity of nature ; what the forces of
nature have been found to do once, they will, under similar condi-
tions, do again.

2. Laws of nature. The principle just stated amounts to say-
ing that the action of the forces of nature is governed by certain
laws ; these we speak of as laws of nature. For instance, if it
has been found that every stone which has ever been placed in

2 REST AND MOTION

water has sunk to the bottom, then, as has already been said, the
principle of uniformity of nature leads us to suppose that every
stone which at any future time is placed in water will sink to the
bottom ; and we can then announce, as a law of nature, that any
stone, placed in water, will sink to the bottom.

That part of science which deals with the laws of nature is
called natural science. Natural science is divided into two parts,
experimental and theoretical. Experimental science tries to dis-
cover laws of nature by observing the action of the forces of
nature time after time. Theoretical science takes as its material
the laws of nature discovered by experimental science, and aims at
reducing them, if possible, to simpler forms, and then discovering
how to predict from these laws what the action of the forces of
nature will be in cases which have not actually been subjected to
the test of experiment. For example, experimental science dis-
covers that a stone sinks, that a cork floats, and a number of sim-
ilar laws. From these theoretical physics arrives at the simple laws
of nature which govern all phenomena of sinking or floating, and,
going further, shows how these laws enable us to predict, before
the experiment has been actually tried, whether a given body will
sink or float. For instance, experimental science cannot discover
whether a 50,000-ton ship will float or sink, because no 50,000-
ton ship exists with which to experiment. The naval architect,
relying on the uniformity of nature, on the laws of nature deter-
mined by experimental science, and on the method of handling
these laws taught by theoretical science, may build a 50,000-ton
ship with every confidence that it will behave in the way pre-
dicted by theoretical science.

3. The science of mechanics. The branch of science known
as mechanics deals with the motion of bodies in space, and with
the forces of nature which cause or tend to cause this motion.
The laws of nature which govern the action of these forces and the
motion of bodies have long been known, and were reduced to their
simplest form by Newton. Thus we may say that experimental
mechanics is a completed branch of science.

MOTION OF A POINT 3

The present book deals with theoretical mechanics. We start
from the laws supplied by experimental mechanics, and have to
discuss how these laws can be used to predict the motion of bodies,
— for instance, the falling of bodies to the ground, the firing of
projectiles, the motion of the earth and the planets round the sun.
An important class of problems which we shall have to discuss
will be those in which no motion takes place, the forces of nature
which tend to cause motion being so evenly balanced that no
motion occurs. Such problems are known as statical.

MOTION OF A POINT

4. State of rest. Before we can reason about the motion of a
body we have to determine what is meant by a body being at
rest. In ordinary language we say that a train is at rest when the
cars are not moving over the rails. We know, however, that the
train, in common with the rest of the earth, is not actually at rest,
but moving round the sun with a great velocity. Again, a fly
crawling on the wall of a railway car might in one sense be said to
be at rest, if it remained standing on the same spot of the wall.
The fly, however, would not actually be at rest ; it would share in
the motion of the train over the country, the country would share
in the motion of the earth round the sun, and the sun would
share in the motion of the whole solar system through space.

These instances will show the necessity of attaching a clear
and exact meaning to the conceptions of rest and motion. Obvi-
ously our statements would have been exact enough if we had
said that in the first case the train was at rest relatively to the
earth, and that in the second case the fly was at rest relatively to
the car.

5. Frame of reference. Thus we find it necessary, before dis-
cussing rest and motion, to introduce the conception of a frame of
reference. The earth supplied a frame of reference for the motion
of the train, and when a train is not moving over the rails we
may say that it is at rest, the earth being taken as frame of

4 BEST AND MOTION

reference. So also we could say that the fly was at rest) the car
being taken as frame of reference. Obviously any framework, real
or imaginary, or any material body, may be taken as a frame of ref-
erence, provided that it is rigid, i.e. that it is not itself changing
its shape or size.

We may accordingly say that a point is at rest relatively to any
frame of reference when the distance of the point from each point
of the frame of reference remains unaltered.

6. Motion relative to frame of reference. Having specified a
frame of reference, we can discuss not only rest but also motion
relative to the frame of reference. When the train has moved
a mile over the tracks we say that it has moved a mile rela-
tively to its frame of reference, the earth. When the fly has
crawled from floor to ceiling of the car we say that it has moved,
say, eight feet relatively to its frame of reference, the car.

In fixing the distance traveled by the fly 'relatively to the train in an
interval between two instants tl, t%, we notice that the actual point from
which the fly started is, say, a mile behind the present position of the
train ; but the point from which we measure is the point which occupies
the same position in the par at time t2 as this point did at time tr So, in
general, to fix the distance moved relatively to a given frame of reference
in the interval between times tl and £2, we first find the point A which stands
in the same position relative to the frame of reference at time t2 as did
the point from which the moving point started at time t^. The distance
from this point A to the point B, which is occupied by the moving point at
instant t2, is the distance moved relatively to the moving frame of reference.

By the motion of a particle B relative to a particle A, is meant
the motion of B relative to a frame of reference moving with A.

7. Composition of motions. Suppose that in a given time the
moving point moves a certain distance relatively to its frame of
reference, while this frame of reference itself moves some other
distance relatively to a second frame of reference, — as will, for
instance, occur if a fly climbs up the side of a car while the car
moves relatively to the earth.

Let us suppose that there is a frame of reference moving in
the plane of the paper on which fig. 1 is drawn, and that the

MOTION OF A POINT

FIG. 1

paper itself supplies a second frame of reference. Suppose that
the moving point starts at A, and that during the motion that
point of the first frame of reference which originally coincided
with the moving point has moved
from A to B, while the point itself
has moved to C. Then the line AB
represents the motion of frame 1
relative to frame 2, while BC repre- --'"'
sents the motion of the moving point
relative to frame 1. The whole mo-
tion of the point relative to frame 2 is represented by AC. The
motion AC is said to be compounded of the two motions AB, BC,
or is said to be the resultant of the two motions. Thus :

If a point moves a distance BC relatively to frame 1, while
frame 1 moves a distance AB relatively to frame 2, the resultant
motion of the point relative to frame 2 will ~be the distance AC,
obtained by taking the two distances AB, BC and placing them in
position in such a way that the point B at which the one ends is
also the point at which the other begins.

There is a second way of compounding two motions. Let x, y
represent the two motions. The rule already obtained directs us to
construct a triangle ABC, to have x, y for the sides AB, BC, and
then AC will be the motion required. Having constructed such a

triangle ABC, let us
D O complete the paral-
lelogram ABCD by
drawing AD, CD
parallel to the side
of the triangle.
Then AD, being

equal to BC, will also represent the motion y, so that we may say
that the two edges of the parallelogram which meet in A represent
the two motions to be compounded, while the diagonal A C through
A has already been, seen to represent the resultant motion. Thus
we have the following rule for compounding two motions x, y :

6 BEST AND MOTION

Construct a parallelogram AB CD such that the two sides AB, AD
which meet in A represent the two motions x, y to be compounded,
as regards both magnitude and direction; then the diagonal AC
which passes through A will represent the resultant obtained by
compounding these two motions.

VELOCITY

8. Uniform and variable velocity. Velocity means simply
rate bf motion. It may be either uniform or variable. If a point
moves in such a way that a feet are described in each second of
its motion, no matter which second we select, we say that the
velocity of the pgint is a uniform velocity of a feet per second.
If, however, the point moves 'a feet in one second, b feet in another,
c feet in a third, and so on, we cannot say that any one of the
quantities a, b, or c measures the velocity. The velocity is now
said to be variable : it is different at different stages of the motion.
To define the velocity at any instant, we take an infinitesimal in-
terval of time dt and measure the distance ds described in this

tune. We then define the ratio — to be the velocity at the instant

at which the interval dt is taken. If the velocity is uniform, —

is the space described in unit time, and so the present definition
of velocity becomes the same as that already given.

Average velocity. If a point moves with variable velocity, and

describes a distance of a feet in t seconds, we speak of - as the

" average velocity " of the moving point during the time t. This
average velocity is the velocity which would have to be possessed
by an imaginary point moving with uniform velocity, if it were to
cover the same distance in time t as the actual point moving with
variable velocity.

Units. In measuring a velocity we need to speak in terms of a
unit of length and of a unit of tune ; for instance, in saying that
a point has a velocity of a feet per second we have selected the foot

VELOCITY 7

as unit of length and the second as unit of time. We can find the
amount of this same velocity in other units by a simple proportion.

Thus suppose it is required to express a velocity of a feet per second
in terms of miles and hours.

The point moves a feet in one second, and therefore a x 60 x 60 feet in
one hour, and therefore

a x 60 x 60 15 a

3 x 1760 22

miles

in one hour. Thus the velocity is one of miles per hour.

EXAMPLES

1. A railway train travels a distance of 918 miles in 18 hours. What is its
average velocity in feet per second ?

2. Compare the velocities of a train and an automobile which move uni-
formly, the former covering 100 feet a second and the latter 1500 yards a minute.

3. A man runs 100 yards in 9| seconds. What is his average speed in miles
per hour ?

4. The two hands of a town clock are 10 and 7 feet long. Find the velocities
of their extremities.

5. Taking the diameter of the earth as 7927 miles, what is the velocity in
foot-second units of a man standing at the equator (in consequence of the daily
revolution of the earth about its axis) ?

6. Two trains 230 and 440 feet long respectively pass each other on parallel
tracks, the former moving with twice the speed of the latter. A passenger in
the shorter train observes that it takes the longer train three seconds to pass
him. Find the velocities of both trains.

9. Composition of velocities. All motion, as we have seen, must
be measured relatively to a frame of reference. Thus velocity,
or rate of motion, must also be measured relatively to a frame
of reference. A point may have a certain velocity relative to a
frame of reference, while the frame of reference itself has another
velocity relative to a second frame. It may be necessary to find
the velocity of the moving point with reference to the second
frame, in other words, to compound the two velocities.

To do this we consider the motions which take place during
an infinitesimal interval of time dt. Let the moving point have
a velocity v^ in a direction AB relative to the first frame, while

8 BEST AND MOTION

the frame has a velocity v2 in a direction AC relative to the
second frame. Then in time dt the moving point describes a dis-
tance v^dt, say the distance AD, along AB relative to the first
frame, while the frame itself describes a distance v2dt, say AE,
along AC relative to the second frame. Let AF be the diagonal

of the parallelogram of which AD, AE
are two edges; then AF will be the
resultant motion of the point in time
dt relative to the second frame. Since
the moving point describes a distance
AF in time dt, the resultant velocity

•11 i AF
will be — - — •

Let us now agree that velocities are

to be represented by straight lines, the direction of the line being
parallel to that of the velocity and its length being proportional ,
to the amount of the velocity, the lengths being drawn according
to any scale we please ; for example, we might agree that every
inch of length is to represent a velocity of one foot per second, in
which case a velocity of three feet a second will be represented by a
line three inches long drawn parallel to the direction of motion.

In fig. 3 let Ap, Aq represent the velocities vv vl drawn on any
scale we please. Since the scale is the same for both, we have

Ap : Aq = v2 : vr
Now AE = vzdt, AD = v^dt, so that

and hence Ap : Aq = AE : AD.

If we complete the parallelogram Aprq, the diagonal Ar will pass
through F, and we shall have

If V is the resultant velocity, it has already been seen that

r-^.

so that

VELOCITY

= Vdt : vzdt

and hence

Ar : Ap =

Thus Ar represents the magnitude of the velocity V on the
same scale as that on which Ap represents the velocity v2. Also
since Ar is in the direction of AF, the resultant motion, we see
that Ar represents the velocity Fboth in magnitude and direction.
We have accordingly proved the following theorem :

THEOREM. If two velocities are represented in magnitude and
direction l>y the two sides of a parallelogram which start from any
point A, then their resultant is represented in magnitude and direc-
tion on the same scale "by the diagonal of the parallelogram which
starts from A.

This theorem is known as the parallelogram of velocities. We
may illustrate its meaning by two simple examples.

1. Suppose that a carriage is moving on a level road with velocity F.
As a first frame of reference let us take the body of the carriage; as
a second frame take the road itself. The velocity of frame 1 relative to
frame 2 is then F. Relatively to frame 1 , the center of any wheel P is
fixed, so that any point
on the rim describes
a circle about P. Rela-
tively to frame 1 the
road is moving backward
with velocity F, so that
if there is to be no slip-
ping between the rim and
the road, the velocity of
any point on the rim, rel-
ative to the first frame
(the carriage), must be F.
Thus the velocity of any
point Q on the rim rela-
tive to frame 1 will Jbe a velocity F along the tangent QT. Representing
this by the line QT, the velocity of the carriage relative to the road is
represented by an equal line QH parallel to the road. Thus the resultant
velocity of the point Q is represented by the diagonal QS of the parallelo-
gram QHST. Clearly its direction bisects the angle HQT. Let L be the

FIG. 4

10 BEST AND MOTION

lowest point of the wheel, and let X complete the parallelogram QPLX.
Obviously this parallelogram is similar to the parallelogram QTSH, corre-
sponding lines in the two parallelograms being at right angles. Thus

QS : QT = QL : QP.

So that on a scale in which the velocity of the carriage is represented in
magnitude by QP, the radius of the wheel, the velocity of the point Q will
be represented by QL. Thus the velocities of the different points on the
rim are proportional to their distances from L, their directions being in
each case perpendicular to the line joining the point to L.

2. A battle ship is steaming at 18 knots, and its guns can fire projectiles
with velocities of 2000 feet per second relative to the ship. How must
the guns be pointed to hit an object the direction of
which from the ship is perpendicular to that of the
ship's motion?

Let AB be the direction of the ship's motion, and
let us suppose the gun pointed in a direction AC.
Then the velocity of the shot relative to the ship
can be represented by a line Ap along A C, while that
of the ship relative to the sea can be represented by
a line Aq along AB. Completing the parallelogram
Aprq, we find that the diagonal Ar will represent the

velocity of the shot relative to the sea in magnitude and direction. Hence
Ar must, from the data of the question, be at right angles to AB. If 6 is
the angle pAr through which the gun must be turned after sighting the
object to be hit, we have

velocity of ship
- — - — *

sin 6 =

Ap velocity of firing of shot

The velocity of the ship is 18 knots, or 18 nautical miles per hour.
Now 1 nautical mile = 1.1515 ordinary miles = 6080 feet, so that a
velocity of 18 knots is equal to 109,440 feet per hour, or 30.4 feet per

30 4
second. Thus sin 0 = -~- = .0152, whence we find that 0 = 0° 52' 16".

Triangle of Velocities

10. We can also compound velocities by a rule known as the
triangle of velocities. In fig. 3 the two velocities were represented
by Ap, Aq, and their resultant by Ar. The two velocities, how-
ever, might equally well have been represented by Ap, pr, and
their resultant by Ar, from which we obtain the following rule :

VELOCITY 11

If two velocities are represented ly the two sides of a triangle
taken in order, their resultant will be represented hy the third side,
taken in the direction from the, first side to the second side.

For example, let OP^ OP2 be two lines drawn through O to
represent, on any scale, the velocities of a mov-
ing point at instants tlf t2. Then P^PZ will, on
the same scale, represent the additional velocity
acquired by the point in this interval.

For we can imagine a frame moving with
the uniform velocity OPl of the particle at
instant tr The velocity OPZ at instant tz may
be supposed compounded of the velocity OPX
of the frame and a velocity P±PZ relative
to the frame. Obviously this latter is the increase of velocity.

EXAMPLES

1. A car is running at 14 miles an hour, and a man jumps from it with a
velocity of 8 feet per second in a direction making an angle of 30° with the
direction of the car's motion. What is his velocity relative to the ground ?

2. A railway train, moving at the rate of 60 miles an hour, is struck by a
bullet, which is fired horizontally and at right angles to the train with a velocity
of 440 feet a second. Find the magnitude and direction of the velocity with
which the bullet appears to meet the train to a person inside.

3. A ship whose head points northeast is steaming at the rate of 12 knots in
a current which flows southeast at the rate of 6 knots. How far will the ship
have gone in 2|- hours ?

4. A train is traveling at the rate of 30 miles an hour, and rain falls with
a velocity of 22 feet per second at an angle of 30° with the vertical in the same
direction as the motion of the train. Find the direction of the splashes made
on the windows by the raindrops.

5. A steamer's course is due south, and its speed is 20 knots ; the wind is
from the west, but the line of smoke from the steamer is observed to point in a
direction 30° east of north. What is the velocity of the wind ?

6. A man rows across a stream a mile wide, pointing his boat upstream at
an angle of 30° with the bank. How long does he take to cross, if he rows with
a velocity of 4 miles an hour and if the current has an equal velocity ?

7. A stream has a current velocity a, and a man can row his boat with a
velocity 6.. In what direction must he row, if he is to land at a point exactly
opposite his starting point ? And in what direction must he row so as to cross
in the shortest time ?

12 KEST AND MOTION

8. A ship whose head is pointing due south is steaming across a current run-
ning due west ; at the end of two hours it is found that the ship has gone 36 miles
in the direction 15° west of south. Find the velocities of the ship and current.

9. A person traveling eastward at the rate of 3 miles an hour finds that the
wind seems to blow directly from the north ; on doubling his speed it appears
to come from the northeast. Find the direction of the wind and its velocity.

ACCELERATION

11. Acceleration is rate of increase of velocity. If we find that
the velocity of a moving point increases by an amount / in a sec-
ond, no matter which second is selected, we say that the motion
of the point has a uniform acceleration / per second. For instance,
a stone or other body falling under gravity is found to increase
its velocity by a certain constant velocity f per second, where f
denotes a velocity of about 32 feet per second. Thus we say that
a falling stone has a uniform acceleration of / per second, or of
about 32 feet per second per second.

Generally, however, an acceleration will not be uniform ; the
rate of increase of velocity will be different at different stages of
the journey. To find the acceleration at any instant, we observe
the change in velocity during an infinitesimal interval dt of time.

If dv is the increase of velocity, we say that — is the acceleration

Cut

at the instant at which dt is taken. An acceleration will of course
have sign as well as magnitude, for the velocity may be either
increasing or decreasing. When the velocity is decreasing, the
acceleration is reckoned with a negative sign. A negative accelera-
tion is spoken of as a retardation. Thus a retardation / means
that the velocity is diminished by an amount / per unit of time.

EXAMPLES

1. A workman fell from the top of a building and struck the ground in
4 seconds. With what velocity did he strike the ground, the acceleration due to
gravity being 32 feet per second per second ?

2. A train has at a given instant a velocity of 30 miles an hour, and moves
with an acceleration of 1 foot per second per second. Find its velocity after
20 seconds.

ACCELERATION

3. A train comes to rest after the brakes have been applied for ten seconds.
If the retardation was 8 feet per second per second, what was the velocity of
the train when the brakes were first drawn ?

4. How long does it take a body starting with a velocity of 22 feet per second
and moving with an acceleration of 6 feet per second per second, to acquire a
velocity of 60 miles an hour ?

5. Two bodies start at the same instant with velocities u and v respectively;
the motion of the first undergoes a retardation of / feet per second per second,
while that of the second is uniform. How far will the second have gone by the
time that the first comes to rest ?

6. A body starting from rest moves for 4 seconds with a uniform accelera-
tion of 8 feet per second per second. If the acceleration then ceases, how far
will the body move in the next 5 seconds ?

7. A train has its speed reduced from 40 miles an hour to 30 miles an hour
in 5 seconds. If the retardation be uniform, for how much longer will it travel
before coming to rest ?

8. A body falling under gravity has an acceleration of 32.2 feet per second
per second. Express this acceleration when the units are (a) centimeter,
second ; (6) mile, hour.

12. Parallelogram of accelerations. THEOREM. Let the velocity
of a point be compounded of two velocities v^, vz along given direc-
tions, and let these velocities be variable, their accelerations being
fi> /2- Then if two lines be drawn in the direction of the velocities,
to represent flt fz on any scale, the resultant acceleration will be
represented on the same scale ~by the diagonal of the parallelogram
of which these ,

lines are edges.
To prove the
theorem, we
consider the
motion dur-
ing any small
interval dt
at which the
component ac-
celerations are

/!, /2. In fig. 7 let AB, AC represent the two velocities vlt v2 at the
beginning of this interval. Let BBf, CCf represent, on the same
scale, the infinitesimal increments in velocity in the interval dt,

FlQ. 1

14 BEST AND MOTION

namely f^dt, fzdt. Then A B', A C' will represent the velocities at
the end of the interval dt.

In the figure the lines BDF, B'ED', CDE, C'FD' are drawn
parallel to AB and A C. Thus AD represents the resultant velocity
at the beginning of the interval dt, and AD1 that at the end of the
interval. The velocity AD1 can be regarded as compounded of the
two velocities AD, DD', and, as in § 10, DD' represents the incre-
ment in velocity in time dt. Thus, if F is the resultant acceleration,
the line DD' will represent a velocity Fdt. On the same scale DE,
DF represent velocities f^dt, f2dt, and DED'F is a parallelogram.

If OFl} OFZ (fig. 8) represent the accelerations/!,/^ on any scale,
and if OG is the diagonal of the completed parallelogram, we
clearly have OFl : OF2 = f^ :/2 = DE : DF,
so that the parallelograms OF^GFZ (in
fig. 8) and DED'F (in fig. 7) will be simi-
lar and similarly situated. Thus
OG : OFt = DD1 : DE = Fdt :f^dt = F:fv
so that OG represents the acceleration F
on the same scale as that on which OFlf
OF2 represent /p/2 ; and OG, being parallel

to DD' , will also represent the direction of F, proving the theorem.
Clearly the acceleration at any instant need not be in the same
direction as the velocity. In fig. 7 the directions AD, AD' repre-
sent velocities at the beginning and end of the interval dt. When
in the limit we take dt = 0, these lines coincide, and the direction
of the velocity at the instant at which dt is taken is that of AD.
The direction of the acceleration at this instant is, however, DD'.

As an illustration of this, let us consider the motion of a particle mov-
ing uniformly in a circle ; e.g. a point on the rim of a wheel, turning with
uniform velocity V about its center.

Let A, B (fig. 9) be the positions of the point at two instants, let the
tangents at A , B meet in C, and let D complete the parallelogram A CBD.

The velocity at the first instant is a velocity V along A C. Let us agree
to represent this by the line A C itself. At the second instant the velocity
is a velocity V along CB ; this may, on the same scale, be represented by
the line CB, or more conveniently by AD. Since AC, AD represent the

ACCELERATION 15

velocities at the two instants, the line CD will represent the change in
velocity between these two instants.

Now let the two instants differ only by an infinitesimal interval dt, so
that the points A , B coincide except for an infinitesimal arc Vdt. In the
figure, CD passes through
P wherever A , B are on the
circle, so that when B is
made to coincide with A,
CD coincides with the ra-
dius through A. But if F
is the acceleration of the
moving point, the change P ^

in velocity produced in

time dt must be Fdt. Thus CD represents the change of velocity Fdt in
direction and magnitude, so that the change of velocity, and hence the
acceleration at A , is along the radius at A .

Here, then, we have a case in which the acceleration is at right angles
to the velocity.

To find the magnitude of the acceleration, we notice that CD = 2 CE,
and that, by similar triangles,

EC : CB = BE : BP.

Now EC, or I CD, represents the velocity £ Fdt, while CB on the same
scale represents the velocity V.

Thus i Fdt : V = BE : BP.

In the limit when BA is very small, BE, or %BA, becomes identical
with half of the arc BA of the circle, and therefore with i Vdt. Thus, if a
is the radius of the circle,

iFdt: V=

giving F = — as the amount of the acceleration.
a

EXAMPLES

1. A windmill has sails 20 feet in length, and turns once in ten seconds.
Find the acceleration of a point at the end of a sail.

2. A wheel of radius 3 feet spins at the rate of 10 revolutions a second and
is at the same time falling freely with an acceleration of 32 feet per second
per second due to gravity. Find the resultant accelerations of the different
points on the rim of the wheel.

3. Taking the earth to have an equatorial diameter of 7927 miles, find the
acceleration towards the earth's center of (a) a point at rest, relative to the
earth's surface, on the equator; (6) a body falling under gravity at the equator,

16 BEST AND MOTION

with an acceleration, relative to the earth's surface, of 32.09 feet per second
per second.

4. Supposing that the moon describes a circle of radius 240,000 miles round
the earth in 29£ days, find its acceleration towards the earth.

5. Assuming that the planets describe circles round the sun with different
periodic times, such that the squares of the periodic times are proportional to
the cubes of the radii of the circles, show that the accelerations of the planets
are inversely proportional to the squares of their distances from the sun.

VECTORS

13. We have found three kinds of quantities, — motion, velocity,
and acceleration, — all of which can be compounded according to
the parallelogram law.

Quantities which can be compounded according to the parallelo-
gram law are called vectors. A vector must have magnitude and
direction, and hence must be capable of representation, on an
assigned scale, by a straight line. We have seen that motion,
velocity, and acceleration are all vectors.

Composition and Resolution of Vectors in a Plane

14. By definition of a vector, two vectors can be compounded
into one, by application of the parallelogram law. It also fol-
lows from the definition that any one vector may be regarded as
equivalent to two, these two being represented by the edges of a
parallelogram constructed so as to have the original vector repre-
sented by the diagonal ; or, as we shall say,
any vector can be resolved into two others.

In particular, if we construct a rectangu-
lar parallelogram so as to have a line which
represents a vector R as its diagonal, we find
that the vector R can be resolved into two
vectors R cos e and R sin e, at right angles to one another, and in

7T"

directions such that R makes angles e, — — e with them.

If we take two fixed rectangular axes Ox, Oy in a plane, we see
that any vector R can be resolved into two components R cos e,

VECTOES IN A PLANE 17

R sin e parallel to these axes, where e is the angle which R
makes with Ox. The components R cos e, R sin e are spoken of
as the components of R along Ox and Oy.

There are two ways of compounding a number of vectors Rl}
R2, ••-, Rn. In the first place, we can construct a polygon
ABODE - • • N, such that the sides AB, BC, CD, •••, MN repre-
sent the vectors RI} Rz, Rz, •••:, Rn. Then AN will represent the
resultant. For RI} Rz can first be compounded into a vector R1
represented by A C. Combining R9
with this vector, we obtain a vec-
tor represented by AD, and so on
until finally AN is reached.

As a second way, we can resolve
each vector,, such as Rs, into its
two components

Rs cos es, Rssin.€s,

along rectangular axes Ox, Oy.

The n vectors are now resolved

into 2 n vectors, of which n are parallel to Ox and n are parallel

to Oy. The first set of n can be compounded into a single vector

X = Rl cos et -f R2 cos e2 -f • • •

parallel to Ox, while the second set can be compounded into a
single vector Y = R^ sin €l + R2 sin e2 + • • •

parallel to Oy. We now have two vectors X, Y parallel to Ox, Oy.
If their resultant is a vector R making an angle e with Ox, we

have ^cose = X=^1cos€1 + ^2cos€2 + .... (1)

R sin e = Y = R1 sin e1 + R2 sin e2 + • • • . (2)

To find the numerical value of R, we square and add (1) and
(2) and obtain
R2 = X2 + Y2

= (Rl cos el 4- R2 cos e2 H )2 + (Rt sin Cj -f- Rz sin e2 -| )2

= R\ + Rl-\ h 2^1^2(cose1cose2 + sin ex sin ea) H

= JBJ + R\ + ... + 2 R^R2 cos (€l - e2) + ....

18 BEST AND MOTION

To find the direction of the resultant, we divide the correspond-
ing sides of (1) and (2) and obtain

tane = ^ = ^8iDe' + ^sin *' + •••.
X RI cos e1 + Rz cos e2 + • • •

If we nave only two vectors Rlf Rz, making an angle 0 with
one another, we may put e1 — e2 = 0, and obtain

Since R is obviously the diagonal of *a parallelogram having two
edges of lengths Blt J22) meeting at the angle 0, this result can
be obtained' directly from the geometry of the triangle ADC, in
D which the angle at C is evidently TT — 0.
_ , Thus

^2>

"^^ ' 7?2 T>2 i 7?2 O T> T> I u\

Jt = Ji1 T -ti2 — .Z JK^Jt^ cos (TT — i/j,

which is clearly identical with the above
expression.

We may take two examples to illustrate
the method of resolving vectors into rectangular components in a plane.

1. In Ex. 2, p. 10, suppose that the direction of the ship (AB in fig. 5)
is taken for axis Ox, and that that in which the shot is to travel is taken
for axis Oy. Let the shot be fired with velocity V, making an angle e with
Ox, the velocity of the ship being v. The resultant velocity is to be along
Oy, so that the velocity along Ox, say X, is to be nil. We have, however,

X = v + Fcos0,

so that we must have cos 6 = > giving

the result already obtained.

2. To find the acceleration of a point
moving with uniform velocity F in a circle
of radius a. Let A be the position of the
particle at time t = 0, and take OA for
axis of x. After time t the particle has
described a length Vt of arc, so that if
B is its position after time t, the angle

BOA is — in circular measure. The
a

direction of velocity at B, namely the tangent at B, will accordingly

VECTORS IN SPACE 19

7T Vt

make an angle - H with Ox, so that the components of the velocity

along Ox, Oy, say vly vz, will be

/IT Vt\ Vt

v* = V sin ( - H ) = V cos

\2 a I a

The acceleration along Ox is — -> which, on differentiating vl with
respect to t, is found to be

F2 F«

cos — ,

a a

while that along Oy is similarly found to be — > or

F2 . F*

sin — •

a a

Compounding th^ese, we obviously obtain an acceleration — along BO,

the result already obtained on page 15.

Composition and Resolution of Vectors in Space

15. It may be that the vectors to be compounded are not all in
one plane. However, the method of determining the resultant is
essentially the same. Thus we can con-
struct a polygon in space ABCD -> N
such that the sides AB, BC, •••, MN rep-
resent the vectors R^ • • -, Rn. As in the
preceding case, it is readily shown that
AN is the resultant.

It is usually more convenient to resolve
each vector into three components par- /y FIQ

allel to rectangular axes in space. Given

a vector AB, we draw through A, and likewise through B, three
planes parallel to the coordinate planes. They inclose a rectangular
parallelepiped of which AB is a diagonal. The edges AC, AD,
AE represent three vectors by which AB can be replaced ; they
are the components parallel to the axes of the vector AB.

Suppose there are n vectors, and that the direction angles of
the vector Rs are denoted by as, /3S) ys. As above, each vector Rs

20 KEST AND MOTION

can be replaced by three components parallel to the axes; these
vectors are of amount

Of the 3n vectors thus obtained, the n vectors parallel to the
aj-axis can be compounded into the single vector

X — RI cos o^ + Rz cos #2 + • • • + Rn cos an. (3)

The whole system of vectors can thus be replaced by this vector
and two others parallel to the y and z axes respectively, namely

Y=Itl cos ft 4- R, cos j32 H ----- h Rn cos /3n. (4)

Z = R^ cos 7X + J22 cos 72 H ----- h Rn cos 7B. (5)

Evidently the resultant of these three vectors, and consequently of
the original n vectors, is a diagonal of a rectangular parallelepiped
whose edges are of lengths X, F, Z. If the length of the resultant
be denoted by R, and the direction angles by a, /3, 7, we have

R2 = Z2 + F2 + ^2,

X. Y Z

and cos a, = — > cos ft = — > cos, 7 = — •

.K ^ a

Hence the resultant is completely determined in magnitude and
direction.

Centroids

16. Let a system of vectors be represented in direction by OAl}
OA2, • • •, OAn, and let their magnitudes be mf>Av • - •, mnOAn, where mlt
mz,--,mn are any quantities. Denote by xr, yr, zr the coordinates of
Ar with respect to axes through 0 ; by art j3r, jr the direction angles
of OAr with respect to these axes ; and by Rr the magnitude of the
vector mrOAr. The components of this vector along these axes are

y Rr cos ar = mrOAr cos ar =

JKr cos /3r = mrOAr cos /3r =
J2r cos 7r = mrOAr cos 7r =
Hence equations (3), (4), (5) can be written thus :

ryr, Z = m^r. (6)

THE CENTKOID

For the interpretation of this result, we make use of the idea of
the centroid of a system of points. By definition the centroid of a
system of points is the point such that its distance from any one
of three coordinate planes is the average of the distances of all the
points of the system from this plane, it being understood that each
distance is measured with its proper algebraic sign.

From this definition, it follows that the distance of the centroid
from any plane whatever is equal to the average of the distances
of the n points from this plane. For if xr, yr, zr are the coordinates
of the rth point, the coordinates of the centroid, say x, y, z, will be

and the perpendicular distance from the centroid to any plane
ax -f ly + cz + d = 0

+ b2 + c2

(ax + ly + cz -f d)

lyr + czr+ d

V a2 + b* -f c2

which proves the result.

Let us imagine that of the n points a number ma all coincide
at the point xa9 ya, za, a number mb at the point xb, yb) zb, and so on.
Then the centroid has coordinates (by equations (7)),

x = -

m x^ =

(8)

22 KEST AND MOTION

where the summation is now taken over the various points in space
at which the original points are accumulated. Calling these points
in space A, B, C,--, the point x, y, % is said to be the centroid of the
points A, B, C, • • • , corresponding to the multipliers ma, mb) mc>
By means of these results, equations (6) are reducible to

X=x--j.mr> r=y.V»r> Z=z-^.mr. (9)

1 1 1

Hence the resultant of the above set of vectors is directed along

the line 0(7, and its magnitude is 0(7- Vrar. As denned by equa-

i
tions (9), the multipliers mr can be any numbers whatever, positive

or negative, so that the sum Vmr may be positive, zero, or nega-
tive. In particular, when the vectors are represented in magnitude
as well as direction by 0 A^ • • • , OAn) the resultant is directed along
OCoy and its magnitude is n • OC0) where n is the number of vectors
and the point C0 is the centroid, as denned above. Thus we have
the theorem :

THEOREM. If vectors of magnitude m^OA^ m2OA2, ••• act along
the lines OAlt OA2, •••, then their resultant is of magnitude

(ml + ra2 H )OG, and acts along OG, where G is the centroid of

the points Av A2, •- for the multipliers ml} m2,

Obviously the parallelogram law is a particular case of this
theorem.

EXAMPLES

1. Find the resultant of two vectors of magnitudes 5P and 12 P which meet
at right angles.

2. A vector P is the resultant of two vectors which make angles of 30° and
45° with it on opposite sides. How large are the latter vectors ?

3. Show how to determine the directions of two vectors of given magnitude
so that their resultant shall be of given magnitude and direction. When is this
impossible ?

4. Show that if the angle at which two given vectors are inclined to each
other is increased, their resultant is diminished.

5. Under what conditions will the resultant of a system of vectors of magni-
tudes 7, 24, and 25 be equal to zero ?

EXAMPLES 23

6. Three vectors of lengths P, P, and P V2 meet in a point and are mutu-
ally at right angles. Determine the magnitude of the resultant and the angles
between its direction and that of each component.

7. Three vectors of lengths P, 2 P, 3 P meet in a point and are directed
along the diagonals of the three faces of a cube meeting at the point. Determine
the magnitude of their resultant.

8. Show that the resultant of three vectors represented by the diagonals of
three faces of a parallelepiped meeting in a vertex A is represented by twice
the diagonal of the parallelepiped drawn from A.

9 D is a point in the plane of the triangle ABC, and I is the center of its
inscribed circle. Show that the resultant of the vectors a • AD, b • BD, c • CD is
(a + b + c) ID, where a, 6, c are the lengths of the sides of the triangle.

10. ABCD, A'B'C'Dr are two parallelograms in the same plane. Find the
resultant of vectors drawn from a point proportional to and in the same direc-
tion as AA', B'B, (7(7, D'D.

11. If 0 is the center of the circumscribed circle of the triangle ABC and
P its orthocenter, show that OP is the resultant of the vectors OA, OB, and
OC ; also that 2 PO is the resultant of PA, PB, PC.

12. The chords AOB and COD of a circle intersect at right angles. Show
that the resultant of the vectors OA, OB, OC, OD is represented by twice the
vector OP, where P is the center of the circle.

GENERAL EXAMPLES

(In these examples take the acceleration produced by gravity to be 32 feet per second

per second)

1. A point possesses simultaneously velocities of 2,3,8 feet per second,
in the directions of a point describing the three sides of an equilateral tri-
angle in order. Find the magnitude of its velocity.

2. A point possesses simultaneously velocities, each equal to v, in the
directions of lines drawn from the center of a regular hexagon to five of
its angular points. Find the magnitude and direction of the resultant
velocity.

3. When a steamer is in motion it is found that an awning 8 feet above
the deck protects from rain the portion of the deck more than 4 feet behind
the vertical projection of the edge of the awning ; but when the steamer
comes to rest the line of separation of the wet and dry parts is 6 feet in
front of this projection. Find the velocity of the steamer, if that of the
rain be 20 feet per second.

4. A ship sailing along the equator from east to west finds that from
noon one day (local time) to noon the next day (local time) the distance
covered is 420 miles. What would be the day's run, if the ship were sail-
ing at the same rate from west to east ?

24 BEST AND MOTION

5. A railroad runs due east and west in latitude X. At what rate
must a train travel along the road to keep the sun always directly
south of it?

6. Determine the true course and velocity of a steamer going due north
by compass at 10 knots through a 4-knot current setting southeast; and
determine the alteration of direction by compass in order that the steamer
should make a true northerly course.

7. A bicyclist rides faster than the velocity of the wind, and makes the
error of judging the direction of the wind to be the direction in which it
appears to meet him when he is in motion. Show that the wind will
always appear to be against him, in whatever direction he rides.

8. One ship sailing east with a speed of 20 knots passes a light-
ship at 11 A.M. ; a second ship sailing south at the same rate passes the
same point at 1 P.M. At what time are they closest together, and what is
then the distance between them ?

9. Two particles move with velocities v and 2 v respectively in oppo-
site directions, in the circumference o,f a circle. In what positions is their
relative velocity greatest and least, and what values has it then?

10. Find the relative motion of two particles moving with the same
velocity v, one of which describes a circle of radius a while the other moves
along a diameter.

11. Two particles move uniformly in straight lines. At a given time
the distance between them is a and their relative velocity is F, the com-
ponents of the latter in the direction of a and perpendicular to it being u
and v. Show that, when they are nearest together, their distance is av/V,
and that they arrive at this position after the interval au/ V2.

12. Three horses in a field are at a certain moment at the vertices of
an equilateral triangle. Their motion relative to a person driving along a
road is in a direction round the sides of the triangle (in the same sense),
and in magnitude equal to the velocity of the carriage. Show that the
three horses are moving along concurrent lines.

13. Two points describe concentric circles, of radii a and Z>, with speeds
varying inversely as the radii. Show that the relative velocity is paral-
lel to the line joining the points when the angle between the radii to
these points is

cos

-i

2ab

14. A stone dropped from a balloon moving horizontally is observed to
be 4 seconds in the air, and to strike the earth in a direction making an
angle of 15° with the vertical. Find the velocity of the balloon.

EXAMPLES 25

15. A ball is thrown from the top of a building with a velocity of 64 feet
per second at an angle of 30° with the horizontal in an upward direction.
Find the directions of its motion at the end of the first and second seconds,
and also the velocities at these instants.

16. A ball is tossed into the air with a velocity of 20 feet a second, and
at the end of a second is seen to be moving in a line at right angles to the
direction of projection. What is its velocity at the instant?

17. If the velocity of a bullet is supposed to be a uniform horizontal
velocity equal to n times that of sound, show that the points at which the
sounds of the firing and of the bullet striking the target are heard simul-
taneously lie on a hyperbola of eccentricity n. Examine the case in which
n is very nearly equal to unity.

18. Assuming that the earth moves in a circular orbit about the sun
with a velocity 29.6 kilometers per second, and that the velocity of light
is 300,000 kilometers per second, 'find the apparent displacement of the sun
due to the earth's motion.

19. Assuming that the earth in a year describes a circle uniformly
about the sun as center, that the distance between the centers is 220 radii
of the sun, and that the radius of the sun is 108 times that of the earth,
find the velocity of the vertex of the earth's shadow, taking the sun's
radius as the unit of space and a year as the unit of time.

CHAPTEE II
FORCE AND THE LAWS OF MOTION

NEWTON'S LAWS

17. The laws of motion, as we have said, form the material sup-
plied by experimental mechanics for theoretical mechanics to work
with. These laws have been stated in compact form by Newton :

LAW I. Every body continues in its state of rest, or of uniform
motion in a straight line, except in so far as it is compelled by
impressed force to change that state.

LAW II. The rate of change of momentum is proportional to the
impressed force, and takes place in the direction of the straight line
in which the force acts.

LAW III. To every action there corresponds an equal and oppo-
site reaction.

18. These laws introduce several new terms, — "force," "mo-
mentum," " action," " reaction," — which must be explained before
the laws can be fully understood.

The first law involves the idea of motion, which has already
been discussed, and of force, which is new.

The .word " force " is in common use. It is associated in the
first instance with muscular effort ; for example, we exert force in
pushing against an obstacle. Scientifically, however, the word has
a wider use ; we say, for instance, that two railway trucks when
they collide exert force on one another, and that the earth exerts
force on all bodies, causing them to fall towards it unless they are
supported in such a way that they resist this force.

The first law of motion, in fact, explains what is to be understood
by force : it is that which tends to change the state of rest of a
body, or of uniform motion in a straight line.

THE FIRST LAW 27

Consider, for instance, a railway truck standing at rest on a level line
of rails. If a second truck runs into it the first truck will start into motion,
so that force has been applied.

The first law, however, tells us more than this. It tells us that
if a body is kept free from the action of forces, it will remain in
its state of rest or of uniform motion in a straight line. Thus the
normal state for a body to be in is one of rest or of uniform motion
in a straight line, i.e. motion with uniform velocity; it is only
the presence of force which can alter this normal state.

Consider again the case of the railway truck. Let us suppose it has
been set in motion by collision, and that it starts off with a velocity of,
say, 10 miles an hour. The first law tells us that unless forces act on the
truck, it will continue its motion with an unaltered velocity of 10 miles
an hour in the same straight line in which it started. When a truck is
actually started into motion by collision, it may be taken for granted that
it will not continue in uniform motion in a straight line, but will sooner
or later be brought to rest. Thus forces must be at work. Let us consider
the nature of these forces.

, In the first place there is a force known as the resistance of the air.
The air in front of the truck presses against it in such a way as to retard
its motion. The air therefore exerts force on the truck just as a man might
exert force on the truck by pressing against it with his hand. This force
alone would stop the truck in time.

Let us suppose that the brakes are applied, and that the wheels are
gripped so firmly as to be at rest relatively to the truck, so that they slide
on the rails. There is then a large force applied to the truck by the rails,
and this again tends to stop the motion of the truck. Even if the brakes
are not applied, so that the wheels are left free to turn, there will still be a
force applied by the rails, although this force will be smaller than before.

Suppose that the track is curved instead of straight. We can imagine
the motion continuing for some time, but it will be motion along the
curve, and not motion in a straight line, such as we are told by the first
law would take place if it were not for the action of force. Force has
therefore been applied, the force being that of the rails on the flanges of
the wheels, tending to turn the truck round the curve. If the flanges were
not present, this force could not act, so that the motion would continue in
a straight line — the truck would run off the rails.

As another illustration of the meaning of the first law, let us examine
the motion of a bullet fired from a gun. Here the forces which start the
motion are supplied by the pressure of the powder. After the bullet has
left the gun, the forces which act are small compared with those which

28 FOBCE AND THE LAWS OF MOTION

have started the motion, so that we get an approximation to uniform
motion in a straight line. The forces acting to alter this motion are
(a) the resistance of the air and (&) the weight of the bullet. The former,
as we have seen, tends to stop the motion by pressure on the ends and sides
of the bullet ; the latter tends to drag the bullet down to the earth, and
so causes it, instead of describing a straight line, to describe a path which
curves downward towards the earth.

19. The conception of uniform motion in a straight line, or of
rest (the particular case of uniform motion in which the velocity
is nil), as being the normal state of a body is due to Galileo (1564-
1642). An interesting account of the discovery of this law will be
found in Chapter II of Mach's Science of Mechanics? or in Chap-
ter IX of Cox's Mechanics?1 Before the time of Galileo it was
commonly supposed, on the authority of Aristotle, that every body
had a natural place, and that its normal state was one of rest in
this natural place. For instance, a stone was supposed to sink in
water, not because the force of gravity was acting on it and setting
it into downward motion, but because its natural place was at the
bottom of the water; a cork was supposed to rise because its
natural place was at the top. Thus Girard,3 in 1634, speaks of
" millions de matieres, qui sont disposees chacunes en leurs lieux,"
and defines gravity as " la force qu'une matiere demonstre a son
obstacle, pour retourner en son lieu." Thus the effect of force,
before Galileo, was supposed to be to keep a body out of its
natural place. Galileo perceived that bodies had no natural
places at all, but natural states, namely of rest or of uniform
motion in a straight line, and the effect of force is not to move a
body from its natural place but to disturb it from its natural
state, — i.e. to alter its speed. This discovery of Galileo is what is
expressed by Newton's first law of motion.

20. Having settled what is meant by the natural state of a
body and also what is meant by force, — namely that which tends

1 Ernest Mach, Science of Mechanics (Eng. trans, by McCormack).

2 J. Cox, Mechanics, Cambridge, University Press, 1904.

3 In the Elzevir edition of Stevin, Leyden, 1634. See Cox, Mechanics, loc. cit. J
ante.

THE SECOND LAW 29

to alter the natural state, — we next inquire as to what is the law
which governs the effect produced by force. Given a force, by
how much will this alter the natural state of uniform motion in a
straight line ? An answer to this is provided by the second law :

LAW II. The rate of change of momentum is proportional to the
impressed force, and takes place in the direction of the straight line
in which the force acts.

The force, then, produces change in a certain quantity, — the
momentum of the body on which the force acts, — and the force
is proportional to the rate of change of this momentum.

By momentum is meant the product of the velocity of the body
by a quantity known as the mass of the body. The mass meas-
ures simply the quantity of matter of which a body is composed,
and so does not depend on the motion of the body. Thus

rate of change of momentum = mass X rate of change of velocity

= mass X acceleration,

by the definition of acceleration. We therefore see that the force
is proportional to the product of two quantities, the mass of the
body and its acceleration.

21. Measurement of mass. If we drop a body from our hand,
it will, in general, be acted on by two forces, the resistance of the
air and its weight. If we suspend the body in a vacuum, with an
arrangement for letting it drop at any instant we please, we get
rid of the resistance of the air, and the only force acting on the
body will be its weight. Now if any two bodies are suspended
side by side in a vacuum, and are let fall at precisely the same
instant, it will be found that they remain side by side during the
whole time they are falling towards the earth. Thus at any
instant their accelerations are the same.

It follows from the second law of motion that the forces acting
are proportional to their masses. These forces, as we have seen, are
simply the weights of the bodies, so that, as the experimental
result is true whatever the two bodies may be, we have the gen-
eral law : The masses of bodies are proportional to their weights.

30 FOKCE AKD THE LAWS OF MOTION

This gives us a means of comparing the masses of any two
bodies. In every country a certain mass is taken as standard, and
the mass of any other body is then compared with this standard, or
with a copy of it, and in this way we get a knowledge of the actual
mass of a body. For instance, in saying that the mass of a body
is n pounds we mean that its mass (or weight) is equal to n times
the mass (or weight) of a certain standard body kept at London.

22. Measurement of force. The weight of a unit mass is a
force which may conveniently be taken to represent a unit force,
and if this is done all other forces may be compared with this
force. Thus a force of m pounds weight will mean a force m times
as great as the weight of the standard pound.

This unit of force, however, is convenient rather than scientific,
since it varies when the mass is moved about from place to place
on the earth's surface. A unit pound mass will weigh more at
London than at Washington ; for instance, it will be found to
extend or compress the spring of a spring balance more at London
than at Washington, so that if the pound weight is taken as unit
of force, we must remember that the unit of force is different at
different parts of the earth's surface, and that a force of m pounds
weight at London will be different from a force of m pounds
weight at Washington.

For this reason a second unit of force is generally used for
scientific purposes. This is called the absolute unit of force, and
is chosen so as to be independent of position on the earth's surface.
The second unit of force is defined to be one which produces unit
acceleration in unit mass, whereas the former unit produced an
acceleration equal to the value of gravity at the point. Thus, if g is
the value of gravity, i.e. the acceleration of a body falling freely in
a vacuum, the practical unit equals g times the absolute unit.

If unit force produces unit acceleration in unit mass, a force P

will produce in mass m an acceleration — • Hence, denoting the

acceleration by /, we have the fundamental equation

P = mf, (10)

THE THIED LAW . 31

which is the mathematical expression of Newton's second law.
Here the force P must be measured in absolute units.

23. LAW III. To every action there corresponds an equal and
opposite reaction.

It is a matter of common observation that a body A cannot
exert force on a second body B without B at the same time exert-
ing force on A. Thus an athlete trying to throw the hammer has
to be oh his guard that the hammer does not throw him ; the force
he exerts on the hammer is accompanied by the hammer exerting
force on him, and he must steady himself against the effects of
this force. So also when a gun fires a shot by exerting force on it,
the shot exerts force on the gun, which is shown in the recoil of
the gun. Thus all forces Qj2£llLJiL-Daiis, which may conveniently
be spoken of as action and reaction. The third law of motion tells
us thai; the two forces which constitute such a pair are equal in
magnitude and opposite in direction.

The meaning of the third law will be seen on examining the reaction
corresponding to the forces which we have already used for illustrative
purposes. The first illustration employed was that of a collision between
two railway trucks. Truck A runs into truck B, exerting force on it and
setting it in motion. The third law tells us that at the instant of collision
B must exert force on A , this force being equal in amount to that exerted
by A on B, and opposite in direction. The force of reaction will result in
a change of velocity of A , lasting during the instant of collision only, and
this may either merely check the motion of A , so that after the collision
A proceeds with diminished velocity, or it may reverse the motion of A,
so that truck A is seen to rebound from B and return in the direction in
which it came.

After B has been set in motion we have imagined it to be acted on by
three forces :

(a) the resistance of the air ;

(6) the friction of the rails ;

The reaction corresponding to the first force is a force exerted by the
truck on the air in front of it and near it, tending to set the air in motion
in the direction in which the car is moving ; it is, in fact, this force which
clears the air away from the space occupied at any instant by the truck.

32 FOKCE AND THE LAWS OF MOTION

The reaction corresponding to the second force is a force tending to
drag the rails along with the truck. The rails are, of course, fastened
down, so that this force cannot actually produce motion.

The reaction corresponding to the third force is a force exerted by the
flanges of the truck wheels on the outer rail of the curve. The rails press on
the flanges in a direction towards the center of the curve, so that the flanges
press the rails outwards and away from the center of the curve. If the rails
are not securely fixed, this pressure will cause them to move in the direction
just mentioned ; the rails will " spread" and the truck will run off the track.

In the illustration of the bullet we again had three forces operating on
the bullet :

(a) the pressure of the powder before the shot leaves the barrel ;

(6) the resistance of the air during the flight of the bullet ;

The reaction corresponding to the first force is the pressure of the shot
driving the powder back. This in turn is transmitted to the gun, producing
the " recoil " of the gun.

The reaction corresponding to the second force, just as with the truck,
sets the air in motion, carving out a path for the bullet and producing
the wind which accompanies its flight.

The reaction corresponding to the third force, the weight of the bullet,
is more interesting, because we can obtain no direct evidence as to its
existence. We merely infer from the principle of the uniformity of nature
that as, in every case which has ever been tested, an action is accompanied
by an equal and opposite reaction, therefore in this case, which is similar
except in that it cannot be tested, we may suppose the action to be accom-
panied by its equal and opposite reaction.

The force which we can observe is the weight of the bullet, dragging
it earthwards. This, we believe, represents a force exerted by the earth
itself on the bullet, — the force of gravitation. This force must be accom-
panied by its reaction, so that the bullet must act on the earth with
a force equal to the weight of the bullet, this force dragging the earth
upwards to meet the bullet. The force exerted by the bullet on the
earth is, by the third law, just as great as that exerted by the earth on the
bullet. The upward acceleration produced in the earth by the bullet is,
however, very much less than the downward acceleration produced in the
bullet by the earth ; for the force is jointly proportional to the mass and
acceleration of the body acted on, and as the mass of the earth is very
great compared with that of the bullet, its acceleration will be very small
in comparison with that of the bullet.

Although for these reasons the acceleration produced in the earth by
a bullet flying above it cannot be observed directly, yet in a very similar
case the acceleration can be observed directly.

FRAME OF REFERENCE 33

The moon, in describing a circle round the earth, is believed to be acted
upon by the earth's gravitation in just the same way as the bullet. If no
force acted on the moon, it would describe a straight line ; as it is, it is
continually dragged down towards the earth, as we believe, by the
same force of gravitation as the bullet. Just in the same way, then, the
earth ought to experience an acceleration towards the moon. This accelera*
tion is one which admits of astronomical observation.

24. In terms of ideas which have now been explained, the three
laws may be restated as follows :

I. The normal state of a body is one of no acceleration. De-
partures from this normal state are produced by the action of
force.

II. When a force acts so as to disturb the normal state of a
body, the force is proportional to the product of the mass of the
body by the acceleration produced.

III. Forces occur in pairs, every action being accompanied by a
reaction, and each pair of forces being equal and opposite.

FKAME OF REFERENCE

25. In stating the laws of motion we have spoken of the motion
of a body without specifying the frame of reference relatively to
which this motion is to be measured. In practice, motion is gen-
erally measured relatively to the surface of the earth, whereas
Newton believed it to be possible to imagine a frame of reference
actually fixed in space, and intended all motion to be measured
relatively to this frame. Thus Newton's laws of motion apply to
motion referred to axes fixed in space, whereas what we require
to know, for all problems except those of astronomy, are the laws
of motion referred to axes moving with the earth.

Let us first consider the effect of referring motion to a set of
axes moving with uniform velocity in a straight line through
space. A body under the action of no forces will have no accel-
eration in space, and, therefore, as the axes themselves have no
acceleration in space, will have no acceleration relatively to the
moving axes. Again, an acceleration has the same value whether

34 FORCE AND THE LAWS OF MOTION

referred to axes fixed in space or to the moving' axes ; for the
acceleration referred to the moving axes is obtained by compound-
ing the acceleration referred to axes fixed in space with that of the
moving axes, and this latter acceleration is nil.

Thus it appears that the laws of motion retain exactly the same
form when the motion is referred to axes which move in space,
provided that these axes move with no acceleration.

This condition of no acceleration is not satisfied by a set of
axes fixed in the earth's surface. A point on the earth's surface
describes, on account of the earth's rotation, a circle about the
earth's axis. If a is the radius of this circle, and v the velocity
with which it is described, the point will have, by § 12, an accelera-

tion — towards the earth's axis of rotation. Thus a set of axes
a

fixed in the earth's surface will have an acceleration of this
amount, and this has to be borne in mind in applying the laws of
motion. At a point on the equator v = 46,510 centimeters per
second and a = 637 X 106 centimeters, so that the acceleration is

— = 3.4 centimeters per second per second. A body dropped at the

equator will appear to have an acceleration of 978.1 centimeters per
second per second, if the motion is referred to axes fixed in the
earth ; but will have a true acceleration of amount

978.1 + 3.4 = 981.5,
if the acceleration is referred to axes fixed in space.

This explains part of the reason why the force of gravity appears
to vary from point to point at the earth's surface. The weight of
a mass of one kilogramme will produce a certain extension of the
spring of a spring balance at the North Pole. If taken to the
equator, part of the weight goes towards producing the acceleration
of tnVmass towards the earth's center, and it is only the remain-
der which extends the spring of the balance. The first part is the
weight of about 31 grammes ; the remainder is the weight of about
996J- grammes. Thus we may say that, owing to the acceleration
of the earth's surface towards its center, a mass of a kilogramme

FRAME OF REFERENCE 35

at the equator will appear to act on a spring balance with a force
equal only to the earth's attraction on 99 6 J grammes.

A second set of errors would be introduced by referring motion
to axes in the earth's surface, these being caused by the change in
the directions of the axes. For instance, if we use the laws of
motion as though they were true for motion referred to axes fixed
in the earth, and apply these laws to the fall of a stone, we shall
find that the stone ought to strike the ground at a point vertically
below that from which it is dropped. If we allow for the rotation
of the earth, we shall find that the point at which the stone actually
strikes must be somewhat to the east of the point vertically below
that from which it started.

The errors introduced by treating motion on the earth as though
it were motion with reference to axes fixed in space are, in gen-
eral, either extremely small or very easily corrected. We shall,
therefore, proceed at present by neglecting such errors altogether,
and shall apply the laws of motion to motion with reference to
the earth's surface.

LAWS APPLICABLE ONLY ;ro MOTION OF A PAKTICLE

26. There is a further limitation to the completeness of New-
ton's laws which ought to be noticed here. The second law would
lead us to suppose that from a knowledge of the force acting on a
body, and the mass of the body, we could deduce a definite accel-
eration of the body. But if the body is of finite size, the accelera-
tion will be different at different points of the body ; for example,
we have seen that, as a consequence of the earth's rotation, the
acceleration of a point at the equator of the earth is different from
that of a point at the North Pole. Which acceleration, then, is it
that is determined by the second law ?

The answer to this difficulty is that the second law must be
supposed to apply only to particles, i.e. to pieces of matter so
small that they may be regarded as points. A moving particle has
a single definite acceleration, just as a moving point has. From the

36 FORCE AND THE LAWS OF MOTION

law as applied to particles we shall be able to deduce laws which
shall apply to bodies of finite size. This problem will be treated
in a later chapter. Although, however, in strictness, the laws
ought to be applied only to particles, it is obvious that there may
be many problems in which we can treat bodies of finite size as
particles without introducing any appreciable error. Such a case
occurred when we discussed the flight of the bullet in § 18 : the
size of the bullet did not come into the question, as we could
imagine all the points of the bullet to have the same acceleration.
Many cases will occur in which a body of finite size may be treated
as though it were a particle. In the next chapter we shall con-
sider the application of forces to particles and to bodies which we
find it is permissible to treat as particles.

CHAPTER III
FORCES ACTING ON A SINGLE PARTICLE

COMPOSITION AND RESOLUTION OF FORCES

27. The second law of motion enables us to find the acceleration
produced when a particle of known mass is acted upon by a known
force. In nature, however, forces do not generally act singly.

Consider, for example, the flight of the rifle bullet, discussed in
§ 18. While the bullet is in the air it is acted on by its weight
and by the resistance of the air simultaneously. In addition to
these, there may be a cross wind blowing and acting on the bullet
with a horizontal pressure in a direction perpendicular to its
motion. The resistance of the air retards the motion of the bul-
let, i.e. produces an acceleration in a direction opposite to that of
the bullet's motion; the weight of the bullet drags it down, Le.
produces an acceleration towards the earth ; while the cross wind
will blow the bullet out of its course, i.e. will produce an accel-
eration in the direction in which the wind is blowing. Thus we
can regard the three forces as each producing its own acceleration.
The three accelerations can each be calculated from the second
law of motion, and on compounding these three accelerations we
shall have the resultant acceleration of the bullet. This resultant
acceleration could have been produced by the action of a certain
single force, so that we may say that this single force is equiva-
lent, as regards the acceleration produced, to the combination of
the three separate forces, or that the single force is the resultant
of the three separate forces.

We must now put these ideas into exact mathematical form.
As a preliminary, let us notice that a force has magnitude and
direction, so that it can be represented by a straight line. We shall
show that forces may be compounded according to the parallelogram

38 FORCES ACTING ON A SINGLE P ARTICLE

law. Having proved this, it will follow that forces are vectors,
and may be resolved and compounded according to the general
rules already given.

28. Parallelogram of forces. THEOREM. If two forces are
represented in magnitude and direction l>y the two sides of a par-
allelogram, their resultant will be represented ly the diagonal of
the parallelogram.

Let AB, AC represent the two forces, and let Ab, Ac represent
the accelerations they would produce if they acted on any particle

separately. Since, by the second law
of motion,- the acceleration is propor-
tional to the force, we must have

Construct the parallelograms Abdc,

ABDC. On account of the proportion
FIG. 15

just obtained, the two parallelograms
will be similar, so that AdD will be a straight line, and we shall have

d = AB: Ab.

But Ad, the diagonal of the parallelogram of edges Ab, Ac, repre-
sents the resultant acceleration. Since AB represents the force
necessary to produce acceleration Ab, it follows from the proportion
just obtained that AD will represent the force necessary to produce
acceleration Ad. In other words, the acceleration of the particle
is the same as if it were acted on by a single force represented by
AD. Thus AD represents the resultant of the forces AB, A C.

It now follows that force is a vector, so that forces can be
compounded according to the laws explained in §§ 14-16.

PARTICLE IN EQUILIBRIUM

29. In statics we are concerned only with particles, or systems
of particles, at rest. The resultant force on each particle must
accordingly be nil. It is therefore important to consider cases in
which the resultant of a system of forces is nil.

CONDITIONS FOE, EQUILIBRIUM 39

30. Polygon of forces. THEOREM. If forces acting on a particle
are represented by straight lines, the particle will be in equilibrium
if the polygon formed by taking all these straight lines as edges is
a closed polygon, i.e. if after putting all the lines end to end we
come back to the starting point.

Let AB, BC, CD, • • •, MN represent in magnitude and direction
any number of forces which act simultaneously on a particle.
Since force is a vector the forces represented by AB and BC are
equivalent to a single force repre-
sented by AC, and may therefore
be replaced by this force.

Thus the system of forces may
now be supposed to be forces repre-
sented by the lines A C, CD, •••, MN.
The first two of these may again
be replaced by a single force repre-
sented by AD, so that the system
is reduced to forces represented by
AD, DE, ••-, MN. We can proceed FlG< 16

in this way until we are left with only a single force represented
by AN. This therefore represents the resultant of all the forces.

If the polygon is a closed polygon, the points A and N coincide,
so that the resultant force represented by AN vanishes and the
particle is in equilibrium. Conversely, if the particle is in equilib-
rium, AN vanishes, so that the polygon is a closed polygon. Thus
the condition for equilibrium expressed by the theorem just proved
is necessary and sufficient, — necessary because the condition must
be satisfied if the particle is to be in equilibrium, and sufficient
because equilibrium is insured as soon as the condition is satisfied.

31. Triangle of forces. If there are only three forces, the theo-
rem reduces to a simpler theorem known as the triangle of forces.
This is as follows :

THEOREM. If a particle is acted on by three forces represented
by straight lines, the particle will be in equilibrium if these three
straight lines placed end to end form the sides of a triangle.

40 FORCES ACTING ON A SINGLE P ARTICLE

As this is a particular case of the polygon of forces no separate
proof is needed. As before, the converse is also true, so that the
condition is a necessary and sufficient condition for equilibrium.

When there are only three forces acting, the condition for equi-
librium can be expressed in a still simpler form :

32. LAMI'S THEOREM. When a particle is acted on by three
forces, the necessary and sufficient condition for equilibrium is that
the three forces shall be in one plane and that each force shall be
proportional to the sine of the angle between the other two.

Suppose that a particle is acted on by three forces P, Q, R. The
necessary and sufficient condition for equilibrium is that we can
form a triangle by placing end to end three
lines which represent the forces P, Q, R in

magnitude and direction.

Let us begin by taking AB to represent
P, and placing against it at B a line BC to
represent Q. Then CA must represent R, if
the conditions for equilibrium are to be satis-
fied. Thus the three forces must be in one
plane, namely, the plane parallel to ABC
through the point of action of the forces.

Assume that there is equilibrium, so that
the three forces are represented by the sides of
the triangle ABC. Let us denote the angles of the triangle as usual
by A, B, C, and its sides by a, b, c. Then, from a known property

of the triangle, 7

a o c .

sin A sin B sin C

By our construction, however, a, b, c are proportional to the
magnitudes of the forces : we have

c b a

Thus P « *

sin C sin A sin B

CONDITIONS FOR EQUILIBRIUM 41

If pq denote the angle between the lines of action of the forces
P and Q, we have pq = TT — B, so that sin B = sin pq, and hence

^- = ^-=^-. (ID

sin qr sin rp sin pq

The converse is true because, if the relation (11) is satisfied and
if the lines of action of the forces are in one plane, we can con-
struct a triangle of which the sides will represent the forces P, Q, R,
so that there is equilibrium.

33. Analytical conditions for equilibrium. Expressed in an ana-
lytical form, the condition for equilibrium is that the resultant of all
the forces acting shall be zero. If the individual forces are known,
the resultant force can be obtained at once from the rules for the
composition of vectors, which have already been given in §§ 14-16.

If the forces all act in a plane, let their magnitudes be Rv Rv • • •,
Rn) and let their lines of action make angles e^ e2, • • •, en with the
axis of x. Then the resultant has components X, Y, where (cf. § 14)

X= RI cos el + R2 cos e2 H ---- ,
Y = Rl sin €j -j- Rz sin e2 + • • -.

The magnitude of the resultant is Vj;2+ F2, and this vanishes
only if X and Y vanish separately. Thus the condition for equi-
librium is that the components along the two axes shall vanish
separately, i.e. that the sum of the components of the separate
forces acting shall vanish when resolved along each axis.

Similarly, if the forces do not all act in one plane, the condition
for equilibrium is that 'the sums of the components along three
axes in space shall vanish separately.

EXAMPLES

1. Forces of 12 and 8 pounds weight act in two directions which are at right
angles. Find the magnitude of their resultant.

2. Three forces each equal to F act along three rectangular axes. Find their
resultant.

3. The resultant of two forces PI and P2 acting at right angles is R • if P!
and P2 be each increased by 3 pounds, R is increased by 4 pounds and is now
equal to the sum of the original values of PI and P2. Find Px and P2.

42 FORCES ACTING ON A SINGLE PARTICLE

4. Forces acting at a point 0 are represented by OA, OB, OC, • • • , 07V. Show
that if they are in equilibrium, 0 is the centroid of the points A, B, C, • • • , N.

5. ABCDEF is a regular hexagon. Find the resultant of the forces repre-
sented by AB, AC, AD, AE, AF.

6. ABCDEF is a regular hexagon. Show that the resultant of forces repre-
sented by AB, 2 AC, 3 AD, 1AE, 5AF is represented by V351 • AB, and find
its direction.

7. ABC is a triangle, and P is any point in BC. If PQ represent the resultant
of the forces represented by AP, PB, BC, show that the locus of Q is a straight
line parallel to BC.

TYPES OF FORCES
Weight of a Particle

34. The weight of a particle acts always vertically down-
ward; for at a given place on the earth it is found that the
weights of all particles act in parallel directions, and this direc-
tion is called the vertical at the place in question. The weight is
the gravitational force with which the particle is attracted by the
earth, except for a small correction which has to be introduced
on account of the facjt that axes fixed in the earth do not move
without acceleration. This correction we shall not discuss here.
When the weight of a body is said to be W, it is meant that to
keep the body at rest relatively to the earth's .surface a force W is
required to act vertically upward.

Tension of a String

35. A string or rope supplies a convenient means of applying
force to a body, and this force is spoken of as the tension of the

pQrs string. Let ABCD • • • be the string,

P* \l B l G ' D ' E an(* ^et ^ be a particle tied to the

string at its end. Let the divisions A B,
BC, • • • of the string be so small that
each may be regarded as a particle.

There will be three forces acting on any particle such as BC:
first, its weight; second, a force exerted on BC by the particle CD
of the string ; and third, a force exerted on BC by the particle AB.

STEINGS 43

Generally the weight- of a string is very slight compared with
the other weights in the problem. It is therefore convenient to
regard a string as having no weight at all. In this case there are
only two forces acting on the particle AB, so that for equilibrium
these must be equal and opposite.

36. Flexibility. A string is said to be perfectly flexible when
the force exerted by one particle on the next is in the direction
. Thus, if the string now under discussion

is perfectly flexible and weightless, the forces acting on the parti-
cle BC are along the directions pq, qr. To hold BC in equilibrium
these must be equal in magnitude. Let T be taken as the magni-
tude of each. Also the two forces must be in opposite directions,
so that pqr must be a straight line.

Since action and reaction, by the third law, are equal and oppo-
site, the force exerted by BC on CD must also be Tin the direc-
tion qr. This must, for equilibrium, be equal and opposite to the
force exerted by DE on CD. This force must accordingly be of
amount T, and qrs must be a straight line.

We can continue in this way, and find that all the particles
must lie in a straight line pqrs • • -, and that each acts on the next
with the same force T. Also the particle A at the end of the string
acts on P with this same force T in the direction of the string.
The force T is called the tension. Thus we have the following :

The tension of a string at any point P is defined as the force
with which the particle of the string on the one side of P acts on
the particle on the other side of P.

The tension is the same in magnitude and direction at every point
of a perfectly flexible, weightless string acted on by no external forces.

Hence it follows that

A perfectly flexible, weightless string acted on ~by no external
forces must ~be in a straight line when in equilibrium.

If the tension vanishes, there is equilibrium whatever the direc-
tion of the elements of length pq, qr, • •• . When the tension van-
ishes the string is said to be unstretched. Clearly an unstretched
string can rest in equilibrium in any shape.

44 FOECES ACTING ON A SINGLE PAETICLE

It will be proved later that when a perfectly flexible, weightless
string passes over a smooth peg or pulley, the tension has the same
magnitude at all points of the string, and at points of contact with the
peg or pulley its direction is along the tangent to the peg or pulley.

37. If the string is not absolutely weightless, but is very light, any
particle such as q will be acted on by three forces, — its weight vertically
down, and the two forces from the adjacent particles acting along pq,

rq. By Lami 's theorem , each force
must be proportional to the sine
of the angle between the remain-
ing two forces. Since the weight
. 19 is small, sin pqr must be small ; i.e.

pqr must be very nearly a straight

line. The line cannot be perfectly straight, however, unless the string is
absolutely weightless ; thus in a real string there must always be a certain
" sag," due to the weight of the string, although this sag may be so slight
as to be imperceptible.

38. Extensible and inextensible strings. The tension, as will
have been seen, is a force acting at every point of the string, and
tending to stretch the string in the direction of its length. The
string either may or may not yield to this tendency to stretch. A
string which stretches under tension is called extensible ; a string
which does not stretch at all, or which stretches so little that the
amount of stretching is inappreciable, is called inextensible.

Thus an inextensible string remains of the same length what-
ever tension is applied to it, while the length of an extensible string
depends on its tension.

In 1660 Hooke discovered a law which expressed a relation
between the tension and the amount of stretching in a string : the
one is proportional to the other.

DEFINITION. The length of a string when the tension is zero is
called the " natural length " of the string.

DEFINITION. The amount by which the length of a stretched
string exceeds the natural length of the same string is called the
"extension" of the string.

HOOKE'S LAW. The tension of a string is proportional to the
extension.

STRINGS 45

Although Hooke discovered this law in 1660, he did not publish it until
1676, and then only in the form of the anagram ceiinosssttuv.

In 1678 he explained that the letters of the anagram were those of the
Latin words ut tensio sic vis, — "the power of any spring is in the same
proportion with the tension thereof." By tension (tensio} Hooke meant
the quantity which we have called the " extension "; by the power (vis)
he meant the force tending to stretch the spring, i.e. the tension.

39. Hooke's law only enables us to compare the extensions
produced by different tensions. To find the actual extension pro-
duced by a given tension we must know that produced by some
other tension for comparison.

DEFINITION. The force required to stretch a string to double its
natural length is called the modulus of elasticity of the string.

Thus if a is the natural length of a string, and X the modulus
of elasticity, we know that a tension X produces an extension a,
so that a tension T produces an extension Ta/\

When we say that a string is inextensible, we mean that X is so
large that the extension Ta/\ may be neglected.

Hooke's law only holds within certain limits. If we go on
increasing the tension in a string indefinitely, we find that, after a
certain limit is passed, Hooke's law ceases to be true, and when a
certain still greater tension is reached the string breaks in two
parts.

EXAMPLES

1. A weight W hangs by a string and is pushed aside by a horizontal force
until the string makes an angle of 45° with the vertical. Find the horizontal
force and the tension of the string.

2. A weight suspended by a string is pushed sideways by a horizontal force.
Show that as it is pushed farther from its position of rest, in which the string
is vertical, the tension continually increases.

3. A weight of 100 pounds is suspended by two strings which make angles
of 60° with the vertical. Find their tensions.

4. A weight of 30 pounds is tied to two extensible strings of natural length
2 feet, modulus of elasticity 100 pounds, and the other ends of the strings are
tied to two points at a horizontal distance 4 feet apart. Find the position in
which the weight can rest in equilibrium.

5. A weight W is suspended by three equal strings of length I from hooks
which are the vertices of a horizontal equilateral triangle of side a. Find the
tensions of the strings.

46 FORCES ACTING ON A SINGLE PARTICLE

Reaction between Two Bodies

40. A second way in which force can be applied to a particle is
by the pressure between the particle and the surface of a solid
body. Such a force is commonly spoken of as a reaction.

A body standing on the floor of a room is acted on by its
weight acting downwards, but is kept at rest by the action of a
second force acting upwards from the floor; this is the reaction
between the body and the floor. Clearly, in order that the body
may rest in equilibrium, the reaction in this case must be equal to
the weight of the body and must act vertically.

FRICTION

41. Imagine a small body standing on a plane of which the
slope can be varied, such as the lid of a desk. If the plane is held
horizontally, the body can stand at rest as already described. Let
the plane be gradually tilted, and it will be found that as soon
as the tilting reaches a certain angle the body will begin to slide
down the plane. The angle at which sliding first occurs is found
to be different for different pairs of substances ; thus for wood
sliding on wood it may vary from 10° to 25°, for iron on wood it
varies from 10° to 30°, while for iron sliding on iron it is only
about 10° or 15°.

When two substances are such that this angle is zero, — i.e. such
that one can only rest on the other when the surface of contact is
perfectly horizontal, — then the contact between them is said to be
perfectly smooth. The nearest approximation to a perfectly smooth
contact which we experience in actual life is probably that of steel
on ice, as in skating.

f" It is found that the angle to which a plane made of one sub-

j stance has to be tilted before a second substance begins to slide

< on it is independent both of the amount of the second substance

I and of the area in contact. 7 Thus the angle depends only on the

nature of the two substances in contact.

a ~

FKICTION

Further, when the two bodies are pressed together in any
way it is found that the directkwa^of the reaction can make any
angle up to a certain limiting angle with the normal to the
plane of separation without sliding taking place, but that as soon
as this angle is reached sliding takes place. This -angle is
known as the angle of friction. It is clearly the same as the
angle through which the plane before considered can be tilted,
for the angle between the normal to the plane and the direction
of the reaction (i.e. the vertical) is simply the slope of the plane.

42. In any case in which frictional forces act, let R denote the
normal component of the reaction, and let F denote the compo-
nent in the plane of the contact which
is caused by friction. When slipping is
just about to occur, the resultant must
make an angle e with the normal, where
e is the angle of friction. Thus, if S de-
notes the whole reaction, we must have

R = S cos e, F = S sin e,
and hence F = E tan e.

FIG. 20

The quantity tan e is called the coefficient of friction and is denoted
by the single letter p. Then, when slipping is just about to take
place, we have F= E

It must be clearly understood that this equation gives the true
value of the frictional force only when slipping is just about to
take place. It sets an upper limit to the value of the frictional
force, but does not give the actual value of this force unless we
know that the system is on the verge of sliding.

43. Consider, for instance, the experiment already discussed, in
which a particle is placed on a horizontal plane which is gradually
tilted up. When the plane is horizontal the particle is at rest,
acted on only by gravity and the reaction with the plane. Thus
the reaction is vertical, so that here F= 0. Consider next the
state of things when the plane makes an angle a with the horizon.

FOECES ACTING ON A SINGLE PAETICLE

IVcosa

If slipping does not take place, the particle is in equilibrium
under its weight W and the reaction between it and the plane.
Thus the reaction must consist of a vertical force W. We can

resolve this into components
W cos a and W sin a perpen-
dicular to and up the plane.
The former is the normal com-
ponent of the reaction, the
latter is the frictional com-
ponent. Thus in the notation
already used we have
R = W cos a,

FIG. 21

so that in this case we have
F = ft tan a.

As a increases, F and F/R both increase until, when a reaches
the value e, F/R reaches its limiting value //<, or tan e, and after
this slipping takes place.

EXAMPLES

1. A mass of 100 pounds placed on a rough horizontal plane is on the point
of starting into motion when acted on horizontally by a force equal to the
weight of 100 pounds. Find the arfgle of friction.

2. A body placed on an inclined plane which makes an angle of 30° with the
horizontal is just on the point of moving down the plane when acted on by a
horizontal force equal to the weight of the body. Find the coefficient of friction.

3. A man capable of exerting a pull of 200 pounds tries to drag a mass of
700 pounds over a horizontal road (coefficient of friction -J-). To help him, the
chain from a crane is attached to the mass, the chain hanging vertically. How
much tension must there be in the chain before the man can move the block ?

4. An insect tries to crawl up the inside of a hemispherical bowl of radius a.
How high can it get, if the coefficient of friction between its feet and the bowl is £ ?

5. A man trying to push a block of stone over ice pushes horizontally and
finds that just as soon as the stone begins to move his feet begin to slip. Show
that if he pushes upwards on the stone he will get it along without difficulty, but
that if he pushes downwards he cannot possibly move it.

6. A smooth pulley is placed at the edge of a horizontal plane. A string
passes over it, having at one end a weight w hanging freely, and at the other

ILLUSTRATIVE EXAMPLES

end a weight W resting on the plane. If the coefficient of friction /* is so large
that motion does not take place, find through what angle the plane must be
tilted before motion begins.

7. A tourist of mass M is roped to a guide of mass TO on the side of a moun-
tain, the side of which may be taken to be an inverted hemisphere. The length
of the rope subtends an angle a at the center of the mountain, and the rope is
not supposed to touch the mountain at any point. If the coefficient of friction
between either man and the mountain is yw, how far can the tourist venture
down the side of the mountain before both he and the guide fall to the bottom ?

ILLUSTRATIVE EXAMPLES

1. A heavy particle C rests on a smooth inclined plane, being supported by
two strings of lengths Zi, 1%, which are attached to two points A, B in the plane,
these points being in the same horizontal line and at a distance h apart. Find the
tensions of the strings and the reaction with the plane.

Let W be the weight of the particle and let a be the inclination of the plane to
the horizon. The particle is in equilibrium, being acted on by the following forces :

(a) Its weight W, which acts vertically downwards.

(6) The reaction between the particle and the plane. Since the plane is
smooth, this reaction acts at right angles to the plane. Let the amount of the
reaction be E.

Since these four forces pro-
duce equilibrium, the sum of
their resolved parts in any di-
rection must vanish. The two
tensions have no resolved parts
at right angles to the plane ;
hence, by resolving at right an-
gles to the plane, we shall get

FIG. 22

an equation in which only two of the forces are involved.

The resolved part of the weight at right angles to the plane is TFcos a. The
reaction is wholly at right angles to the plane ; hence the equation for which

we are in search is

E - Wcosa = 0.

This gives us the amount of the reaction at once.

Let us now consider the resolved parts of the forces in the inclined plane.
The only forces which have components in this plane are the following :

(a) The weight, of which the component isTFsina, down the line of greatest slope.

(6) The tensions of the strings, which are entirely in the plane and which
act along the strings CA, CB.

FORCES ACTING ON A SINGLE PAETICLE

The three forces TFsina, Zi, and T2 must be in equilibrium; hence, by
Lami's theorem, each must be proportional to the sine of the angle between
the other two.

In fig. 23, CD, CA, CB are the lines of action of these three forces. The
line CZ>, being the line of greatest slope through O, is at right angles to the
line AB, which we are told is horizontal. Thus if DC is pr6duced to meet AB

in P, the angle APC is a right angle. Hence

sin ACD = sin AC P = cos CAP,
and similarly

sin BCD = sin BCP = cos GBP.
By Lami's theorem we have

Wsina _ TI T2

sin AC B ~ sin BCD ~ sinACD

From the relations just obtained we can
obtain these ratios in terms of the angles of
the triangle ABC. We have

Wsina= T± T2

FIG. 23 sin C cos B cos A

We can now, if required, express cos J., cos B, and sin C in terms of the sides
Zi, ^2, and h of the triangle, by means of the ordinaiy formulae of trigonometry.

The student is advised to examine for himself the form assumed by the result
in the two special cases

(a) A = B — 0, in which ACB are in a straight line ;

(b) C = 0, A = B = — , in which the strings are parallel.

2. Find the direction and magnitude of the smallest force which will start a
body resting on an inclined plane into motion down the plane.

Let a be the angle of the plane, and fj, the coefficient of friction between it
and the body to be moved. Let W be the weight of the body, and let a force F
be applied in a direction making
an angle 0 with the line of greatest
slope down the plane, this force
being supposed to be just sufficient
to move the body.

The forces acting on the body
consist of

(a) its weight W ;

(b) the applied force F ;

FIG. 24

Let the last force be resolved into two components along and perpendicular
to the plane. Taking the latter to be E, the former will be /xR acting up the
plane, for, by hypothesis, the body is on the point of motion down the plane.

ILLUSTRATIVE EXAMPLES 51

The resultant of all these forces vanishes, so that the sum of their com-
ponents in any direction vanishes. Resolving normal to the plane, we obtain

R + .Fsin0 — Wcosa = 0,
and resolving along the plane,

F cos 0 + Wsma- fj.R = 0.

Eliminating the unknown reaction B, we obtain

F(n sin 0 + cos 6) — W(fj. cos a — sin a) = 0,

sothat F=H^coso--sina)_x

H sin 6 + cos 8

Replacing /* by tan e, we obtain

_ W(cos a tan e — sin a:) _ Wsin(e — a)
cos 6 + tan e sin 0 cos (0 — e)

The value of F is a minimum when cos (0 — e) is a maximum, and this occurs
when cos (0 — e) = 1 ; i.e. when 0 = e. In this case the value of F is

F= "FT sin (e- a).

Thus this is the smallest force by which motion can be produced, and it must
act so as to make an angle 0 with the plane equal to the angle of friction e.
Since, by hypothesis, the weight rests without slipping when no force is applied,
the angle e must be greater than a. Thus the direction of the force F must
always be inclined in an upward direction. The function performed by the
force F is twofold : it supports part of the weight of the body (through its
component normal to the plane), and so lessens the amount of friction to be
overcome ; and it also supplies (through its component in the inclined plane)
the motive power for overcoming the frictional resistance. When
these two parts of the force are balanced in the most advantageous
way the value of F is a minimum, and this, as we have proved,
occurs when 0 = e.

An interesting and instructive solution of this problem can
also be obtained geometrically. For equilibrium, the three forces
already enumerated must satisfy the condition of forming a tri-
angle of forces.

Let AB represent the weight and BC the reaction between the
mass and the plane, then CA must represent the applied force F.
If the body is on the point of motion, the reaction must make an
angle e with the normal to the plane, so that the angle ABC must
be e — a. Thus the line BC is fixed in direction, and the problem
is that of finding the direction and magnitude of AC, when the length AC is a
minimum. Obviously the minimum occurs when A C is perpendicular to BC, so
that A C must be in a direction making an angle e — a with the horizontal, as
already found, and since A C = AB sin A BC = AB sin (e — a), the magnitude of
the force required will be TF"sin(e — a).

FORCES ACTING ON A SINGLE PARTICLE

3. A particle is tied to an elastic string, the other end of which is fixed at a
point in a rough inclined plane. Find the region of the plane within which the
particle can rest.

The forces acting on the particle are

(a) its weight, say TF, vertically down ;

(6) the tension of the string ;

Let the natural length of the string be Z, the modulus of elasticity X ; then

when the actual length of the
string is r, where r is greater

than Z, the tension is — •

Let a be the inclination of
the plane, and let /* be the
coefficient of friction between
the plane and the particle. Let
the reaction with the plane be
resolved into a normal component
R, and a component F along the
plane. The condition that the
particle can remain at rest is
that F shall be less than pR.
Resolving at right angles to the plane, the only forces which have compo-
nents in this direction are found to be the weight of the particle and its reaction

with the plane. Thus

R — W cos a = 0.

Consider the equilibrium of the particle when at some point P, distant r ( > I)
from 0. The components in the inclined plane, of the forces acting on it, are
(a) W sin a down the line of greatest slope through P ;

(6) the tension — along PO ;

Let OP make an angle 6 with the line of greatest slope in the plane. Then,
since the resultant of the first two forces must be of magnitude F, we must have

2(r-

FIG. 26

W sin a cos 0,

giving the magnitude of the f rictional force required to maintain equilibrium.
If the particle is on the point of motion, F = /J.R = /tTFcos a, so that

= 0.

(a)

Since r, 6 are polar coordinates of the point P, equation (a) is the polar equa-
tion of the boundary of the region within which the particle can remain at rest.

ILLUSTRATIVE EXAMPLES

The equation is most easily interpreted by noticing that if r — I is replaced
by r, the equation becomes

•\ 2 \

TF2(sin2ar - ^2cos2or) + — r2 + 2 - Wsin a - r cos 0 = 0, (6)

which is the polar equation of a circle. Thus the original locus represented by
equation (a) can be drawn by first drawing the circle represented by equation
(6), and then producing each radius vector through the origin to a distance I
beyond the circumference of this circle.

FIG. 28

The same result can be obtained by a geometrical treatment of the problem.
The particle is acted on by only three forces in the plane on which it rests, so
that lines parallel and proportional to these forces must form a triangle of forces.

In fig. 29 let OP be the string, and let AP be a
length I measured off from P, so that AO is the
extension r — I of the string. The tension is. always
proportional to AO and acts along AO. Let us
then agree that in the triangle of forces the tension
shall be represented by the actual line AO. On
the same scale let the component of the weight,
W sin a, be represented by the line OG, the direc-
tion of this being, of course, down the line of greatest
slope through 0. Then AOG must be the triangle
of forces, so that GA must represent the f rictional
reaction between the particle and the plane. The
maximum value possible for this is /j.W cos or, so that
if slipping is just about to occur, GA will represent
a force ^W cos a. Thus corresponding to a position
of P in which slipping is just about to occur, the positions of A are such that GA
represents the constant force /j,W cos or, — in other words, the locus of A is a circle
of center G. This leads at once to the construction previously obtained.

FIG. 29

FORCES ACTING OK A SINGLE PARTICLE

The region in which equilibrium is possible assumes two different forms
according as the angle of the inclined plane a is less or greater than the angle
of friction e. In the former case the region of equilibrium is of the kind repre-
sented in fig. 27. On passing the value a. = e the circle used in the construction
passes through the point O, and for values of a greater than e the region of
equilibrium becomes an area of the kind drawn in fig. 28. On passing through
the value a = e a sudden change takes place in the shape of the region of
stability. For values of a which are greater, by however little, than e, a circle
of radius Z, center 0, is entirely outside the region of stability ; while for values
of a which are smaller, by however little, than e, this circle is inclosed within
the region of equilibrium. Clearly this circle maps out the region within which
the weight can rest with the string unstretched, and this will be one of equi-
librium or not according as a < or > e.

Thus this circle falls inside or outside the region of equilibrium in the way
predicted by analysis. At the same time we could not have been sure, without
a separate investigation, that the result given by analysis would be accurate as
regards the region within a distance I of O. For the analysis began by assuming
the string to be stretched, and so had no application except to the region at a
distance greater than I from 0.

4. Two weights w, wf rest on a smooth sphere, being supported by a string
which passes through a smooth ring at 0, a point vertically above the center of the
sphere. Find the configuration of equilibrium.

Let P, Q, in fig. 30, be the positions of the two weights
in a configuration of equilibrium. The weight w at P is
acted on by the following forces :

(a) its weight w vertically downwards ;

(6) the tension of the string along PO;

(c) the reaction between the sphere and the weight.
Since the sphere is supposed smooth, the direction of
this reaction is at right angles to the plane of contact
between the particle and the sphere ; i.e. along CP.

The three forces acting on the particle P are accord-
ingly parallel to the three sides of the triangle OPC.
Thus the triangle OPC may be regarded as a triangle of
forces for these forces, so that the magnitudes of the
forces must be proportional to the sides of this triangle.
Denoting the tension and reaction by T and J?, we obtain

Fia. 30

oc~ OP~ CP

(a)

In the same way the triangle OCQ may be regarded as a triangle of forces for
the particle Q. (This triangle does not represent force on the same scale as the
former triangle OCP ; for in the former case 0(7 represented a weight w, whereas
it now represents a weight w'.)

ILLUSTRATIVE EXAMPLES 55

. From this second triangle of forces we obtain

«1 =*!=*!, (b)

oc OQ CQ

where T', R' represent the tension and reaction acting on Q.

Since the ring at O is supposed to be smooth, the tension in the string POQ
is the same at all points. Thus T = T'. We now obtain, from equations (a)

and (6),

w • OP = w' • OQ, (c)

since each product is equal to T • OC. If Ms the whole length of the string,
we have

w w' w + w' , ,v

- =i - = - ) (en

OQ OP I

showing that the string arranges itself so that it is divided by the ring at 0
in the inverse ratio of the two weights. We notice also from equations (a)
and (b) that

B-^

w w'

for each ratio is that of the radius of the sphere to OC. Thus the reactions are
in the direct ratio of the weights.

If the string is inextensible, the length I is known, so that equations (d)
determine the lengths OP, OQ completely. Suppose, however, that the string is
an extensible string, say of natural length a and modulus X. Then, instead of I
being a known quantity, we have one additional equation between unknown
quantities, namely

Remembering that equation (c) gives two values for the quantity T> OC,
we have

A • C/U W • t

a
OP

JC W

_°Q_

1 1

__ -1

w w'

l~a\ oc l

a i

- +

so that

This equation determines the value of I, and having found this we proceed
as before.

5. A weight W is supported by strings of which the tensions are TI, T2, > - Tn.
The strings do not hang vertically, but the angles between the different pairs of
strings are known, being e12, ei3, etc. Find the weight W in terms of the tensions
and of these angles.

56 FOECES ACTING ON A SINGLE PARTICLE

Clearly the weight is equal to the resultant of the tensions TI, T2, • • • Tn.

Let us take any three rectangular axes in space. Let the direction cosines
of the first string be /i, mi, n\ ; let those of the second string be J2, ^2, n2 ; and so
on. Then, resolved along the axes, the components of the first tension will be

The components of the other tensions are similar expressions, so that if
X, y, Z denote the three components of the resultant, we have

X=liTi + 12T2 + • • • + lnTn,

Y = miTi + m2T2 + • • • + mnTn,
Z = niTi + n2T2 + . . . + nnTn.

Since the magnitude of the resultant is equal to W, we have

W2 - X* + F2 + Z2

12T2 + • • . + Znrn)2 + (miTi + m2T2 + • • •

mifOl'+ ml + *fi + ••• +

= Tf + T| + • • • + 2 TiT2 cose12 + • • •,
which gives the result required.

GENERAL EXAMPLES

1. ABC is a triangle, with a right angle at ^4; AD is the perpendicular

on BC. Prove that the resultant of forces, acting along AB and

1 AB AC

acting along AC, is — — acting along AD.
AD

2. At a point O there acts a force P, whose line of action is in the plane
determined by two lines OA, OB, meeting at 0. The resolved part of P in
the direction OA is represented in magnitude and direction by OX, that
in the direction OB by OF. Show that the force P is represented in mag-
nitude by the diameter of the circle OXY, and find its direction.

3. Forces P1? P2, • • •, PM acting in one plane at a point 0 are in equi-
librium. Any transversal cuts their lines of action in points Ll , L2, • • •, Ln ;
and a length OL{ is considered positive when the direction from 0 to Lt is
the same as to Pt. Prove that ^PJOL, = 0.

4. A body is sustained on a smooth inclined plane by two forces, each
equal to half the weight, the one acting horizontally, and the other along
the plane. Find the inclination of the plane.

5. The angle of a smooth inclined plane is 30°, and a force P acting
horizontally sustains a body. In what other direction can P act and sup-
port the body ? Compare the pressure upon the plane in the twro cases.

EXAMPLES 57

6. Two smooth planes, whose inclinations are a and /3, meet in a hori-
zontal line AB. At a point in A B is a small smooth ring through which
passes a string with a weight at either end, resting one on each of the
given planes, and in the same vertical plane with the ring. If the weights
are in the equilibrium, find the tension of the string and the ratio of the
weights.

7. Two smooth rings of weights W± and W2 are connected by a string
and rest in equilibrium on the convex side of a circular wire in the vertical
plane. Show that, if the string subtends the angle a at the center of the
circle, the angle of inclination 6 of the string to the vertical is given by

Wl + Wt a

8. Two weights rest on a rough inclined plane and are connected by a
string which passes over a smooth peg in the plane ; if the angle of inclina-
tion a is greater than the angle of friction c, show that the least ratio of
the less to the greater is sin (a: — e)/sin(a + e).

9. Two weights support one another on a rough double inclined plane,
by means of a fine string passing over the vertex, and both weights are
about to move. Show that if the plane be tilted un-til both weights are
again on the point of motion, the angle through which the plane will be
turned is twice the angle of friction.

10. Two weights P, Q of similar material, resting on a double inclined
plane, are connected by a fine string passing over the common vertex, and
Q is on the point of motion down the plane. Prove that the greatest weight
which can be added to P without disturbing the equilibrium is

P sin 2 e sin (a + /3)
sin (a — e) sin (/3 — e)

<r, /3 being the angles of inclination of the planes, and e the angle of
friction.

11. A body is supported on a rough inclined plane by a force acting
along it. If the least magnitude of the force, when the plane is inclined
at an angle a to the horizon, be equal to the greatest magnitude, when the
plane is inclined at an angle /3, show that the angle of friction is \(a.— /3).

12. Two equal rings of weight W are movable along a curtain pole, the
coefficient of friction being p. The rings are connected by a loose string
of length I, which supports by means of a smooth ring a weight Wv How
far apart must the rings be so that they will not come together ?

13. Two weights P, Q of different material are laid on a rough plane,
whose inclination is 0, and connected by a taut string inclined at 45° to the
intersection of the plane with the horizon. Both weights are on the point

58 FORCES ACTING ON A SINGLE PAETICLE

of motion. Determine the coefficients of friction of P and Q, it being
known that that of the upper weight is twice that of the lower.

14. A heavy ring is free to slide on a smooth elliptic wire of eccen-
tricity e in a vertical plane, the major axis of the ellipse making an angle a
with the horizontal, and a string fastened to the ring passes over a smooth
peg at the center of the ellipse and supports a body of equal weight. Show
that the angle <£ which the tangent to the wire at the ring makes with the
major axis is given by the equation

tan (0 + a) (sec2 0 - e2) = ez tan 0.

15. Two small smooth rings of weights W, W are connected by a string,
and slide on two fixed wires, the former of which is vertical and the latter
inclined at an angle a to the horizontal. A weight P is tied to the string,
and the two portions of it make angles 6, 0 with the vertical. Prove that

cot 6 : cot 0 : cot a = W : P + W : P + W + W.

16. Two particles of unequal mass are tied by fine inextensible strings
to a third particle. They lie on a rough inclined plane, with the strings
stretched and making angles a, /3 with the horizontal line in the plane.
Find the magnitude and direction of the least horizontal force which, on
being applied to the* third particle, will move all three.

17. A heavy particle is placed on a rough inclined plane of which the
inclination a is equal to the angle of friction. A thread is attached to
the particle, and passed through a hole in the plane which is lower than
the particle, but is not in the line of greatest slope through it. Show that
if the thread be gradually drawn through the hole, the particle will describe
a straight line and a semicircle in succession.

CHAPTEE IV

STATICS OF SYSTEMS OF PARTICLES

44. So far we have been considering the action of forces on a
single particle. A different class of problems arises in considering
the action of forces on a body composed of a great number of par-
ticles, to which forces are applied in such a way as to act on the
different particles of the body.

Consider what happens when a force F is applied to one parti-
cle A of a body which is composed of a great number of particles
A, B, C, D, • • -. If the particle A
were in no way influenced by the
other particles B, C, D, ••• the
particle A would start into motion
under the action of the applied
force, and would soon become sepa-
rated from the other particles B,
C, D,--. If, however, the particles

A, By (7, D, • - • constitute a single
continuous body, this does not
happen. What happens is that as

soon as the particle A begins to move relatively to the other parti-
cles, systems of actions and reactions come into play between the
particle A and the adjacent particles B, (7, X>, • • •. Speaking loosely,
we may say that the forces acting on A tend to check the motion
of A, while the corresponding reactions tend to impart motion to

B, C,D,--. When B, C,D,-- start into motion, further systems of
forces begin to operate on the particles next beyond B, C, D, -•,
and so on. Thus all the particles are set into motion, and instead
of the particle A moving singly the complete body moves as a whole.
We have now to discuss whether such a body, or system of bodies,

FIG. 31

60 STATICS OF SYSTEMS OF PAETICLES

will move or will remain at rest, when systems of forces act from
outside on its different particles. We shall have to remember
throughout that the forces applied from outside are not the only
forces acting, but that these are accompanied by actions and reac-
tions between the different particles.

45. One consequence of this last fact appears at once. Applying
a force to one particle A of a body is not the same thing as apply-
ing an exactly similar force to another particle B. For the systems
of internal actions and reactions will be different in the two cases.
Any simple example will show that the resulting motion will, in
general, also be different; e.g. a horizontal force applied to the
middle point of the back of a chair will probably cause the chair
to overturn. A similar force applied to one foot will drag it along
the ground and also cause it to turn about a vertical axis.

The position occupied by the particle to which a force is applied
is called the point of application of the force. The line drawn
through this point in the direction of the force is called the line
of action of the force.

Clearly, in order to have full data as to the action of a force,
we must know

(a) its magnitude;

(b) its point of application j

MOMENTS

46. DEFINITION. The moment of a force about a line at right
angles to the line of action of the force is defined to be the product
of the force and of the shortest distance between the two lines.

This moment, as we shall soon find, measures the tendency to turn around
the line about which the moment is measured ; e.g. if the arm of a balance
is of length I, a weight w at its end has a moment Iw about the pivot of the
balance, and we shall find that this measures the tendency of the arm to turn.

DEFINITION. The moment of a force about a line L which is not
at right angles to the force is defined to be the same as the moment
about L of the component of the force in a plane perpendicular to L.

MOMENTS 61

Resolving the force into two components, one parallel to L and one
perpendicular to Z, it is clear that the former will not give any tendency
to turn about L, so that the whole tendency to turn comes from the second
component.

The two definitions which have now been given suffice to deter-
mine the moment of any force F about any line L. It may be
noticed that the moment vanishes

(a) if the line of action of F is parallel to L ;

(b) if the line of action of F intersects L.

Obviously, in either of these cases, the tendency to turn about L is zero.

47. Let the line L be at right angles to the plane of the paper,
and intersect it in the point M. Let PA be the line of action of
a force F in the plane of the paper, acting on a particle at A, and
let MN be the perpendicular from M on
to PA. Then, by definition, the moment
of the force F about L is F x MN.

Let the angle PAS be drawn equal
to the angle NMA, say equal to 0, so
that AS is perpendicular to MA. Then
the moment of the force F about L

= Fx MN

= Fx AM cos 0

= AMx Fcos 6

= AM x resolved part of F along SA.

Instead of F being the actual force acting at A, suppose that F
is the resolved part, in the plane perpendicular to the line L, of
some other force R. Then the moment of R about L is, by defini-
tion, the same as the moment of F9 and the resolved part of R
along AS = Fcos 6. Hence what has just been proved may be
put in the form

moment about L of any force R acting at A

= AM x resolved part of R along SA,

and SA is now determined as the direction which is perpendicular
to L, and also to AM, the perpendicular. from A on to L.

62 STATICS OF SYSTEMS OF PARTICLES .

Thus we have a new definition of a moment, which is exactly
equivalent to that previously given, namely:

The moment about a line L of a force R acting at A is equal
to AM, the perpendicular from A on to L, multiplied by the com-
ponent of R in a direction perpendicular to AM and to L.

48. From this conception of a moment we have at once the
theorem :

The sum of the moments about any line L of any number of
forces acting at a point A is equal to the moment of their resultant
about L.

For, let Rlf Rz, • • • be the forces, and R their resultant. Let AM,
as in § 47, be the perpendicular from A on to L, and let AS be a
direction perpendicular to AM and to L. The theorem to be proved

is that

AM x component of R^ along AS

+ AM X component of R2 along AS -\- •••
— AM x component of R along AS.

On dividing through by AM the theorem to be proved is seen
to be simply that the component of R along AS is equal to the
sum of the components of R±, Rz, •• - along AS, which is known
to be true.

We can now see more clearly how it is that the moment of a
force, defined as we have defined it, gives a measure of the tendency
to turn. In fig. 32 we are taking moments about a line L which
is at right angles to the plane of the paper and meets this plane
at M. The force whose moment is being considered is a force R
acting at the point A. At A we have three directions mutually at
right angles, namely

AS, AM, and the direction of a line through A parallel to L.
The moment of R about L has been defined to be
AM x component of R along AS.

MOMENTS 63

Now the component of R along AM is a force of which the line
of action intersects L, and so can produce no tendency to turn a
body about L, while the component of R along the line through A
parallel to L can again produce no tendency to turn about L. Thus
R can be resolved into three components, of which only the first,
the component along AS, tends to set up rotation about L. We
have defined the moment of the whole force R in such a way that
it becomes identical with the moment of that one of its components
which tends to set up rotation.

It will be noticed that a moment has sign as well as magnitude.
In moving along the line of action of a force ft, we may turn in
either one direction or the other about a line L. We agree that
when the turning is in one direction the moment of R about L is
to be regarded as positive; when the turning is in the other
direction the moment is taken to be negative.

49. If a particle is in equilibrium under the action of any num-
ber of forces, the resultant of all these forces must be nil. The
sum of the moments of the separate forces, taken about any
line whatever, is equal to the moment of the resultant and is
therefore nil.

Hence we have the result :

When a particle is in equilibrium under the action of any forces,
the sum of the moments of these forces about any line whatever
must vanish.

SYSTEM OF PARTICLES IN EQUILIBRIUM

50. Consider a system of particles supposed to be in equilibrium
under the action of any number of forces. As we have seen, the
forces acting on any single particle will be of two kinds :

(a) external forces, forces applied to the particle from outside,
as for instance the weight of the particle ;

(b) internal forces, forces of interaction between the particle and
the remaining particles of the system.

64 STATICS OF SYSTEMS OF PARTICLES

Now if the whole system of particles is in equilibrium, it follows
that each particle separately must be in equilibrium. It follows
from § 33, that

(a) the sum of the components in any direction, of all the forces
acting on any single particle, must vanish;

and from the theorem just proved in § 48, that

(b) the sum of the moments about any line, of all the forces
acting on any single particle, must vanish.

If, however, the sum of the components of the forces acting on
each particle vanishes, it follows by addition that the sum of the
components of all the forces acting on all the particles must vanish.
The sum of the components of the internal forces, however, van-
ishes by itself, for the internal forces consist of pairs of actions
and reactions, and the two components in any direction of such a
pair of forces are equal and opposite.

Since the total sum vanishes, and the sum of the components
of internal forces vanishes, it follows that the sum of the com-
ponents of external forces vanishes.

A similar proposition is true of the moments of the external
forces. The sum of the moments about any line L of all the internal
forces is nil, for the moments of an action and reaction are equal
and opposite. The sum of the moments of all the forces, internal
and external, is zero, for each sum of the moments of the forces
acting on each particle is zero separately. Thus the sum of the
moments of the external forces is zero.

Thus we have proved the following theorems :

When a system of particles is in equilibrium under the action
of any system of external forces,

(a) the sum of the components of all these forces in any direction
is zero ;

(b) the sum of the moments of all these forces about any line is zero.
Speaking loosely, we may say that these theorems express that

there is no tendency to advance in any direction or to turn about
any line.

ILLUSTRATIVE EXAMPLE

Wheel and axle. The apparatus known as the "wheel and axle" consists of
a circular axle free to turn about its central axis, to which a circular wheel
is rigidly attached, so that its center is on the center of the axle. A rope or
string is wound round the axle, and has a weight attached to its end. A second
rope or string is wound round the circumference of the circle in the opposite
direction, and this again has a weight attached to its end. By a suitable choice
of the ratio of these two weights, the apparatus may be balanced so that there
is no tendency for it to turn about its axis.

Let us consider the equilibrium of the system consisting of the wheel and
axle and of those parts of the strings or ropes which are wound round them.
To simplify the problem, let us disregard altogether the weight of the system.
Then the externally applied forces are

(a) the tension of the rope wound round
the wheel ;

(6) the tension of the rope wound round
the axle ;

Let the weights be denoted by P and Q, so
that these are also the tensions of the strings,
and let the rad'ii of the wheel and axle be a,
b respectively. Let us express mathematically
that the sum of the moments of the externally
applied forces about the axis is nil.

The moment of the tension of the string
on the wheel is Pa, for P is the amount of
the tension, which acts at right angles to the
axis, and a is the shortest distance from the
axis to the line of action of this tension.

Similarly the moment of force (b) is — Q6,
the negative sign being taken because this tends to turn the system in the
direction opposite to that in which the first tension tends to turn it.

If we imagine the system to be supported by forces acting on the axis itself,
the moment of forces (c) vanishes, for the lines of action of these forces intersect
the line about which we are taking moments. Thus the required equation is

Pa - Qb = 0.

This equation simply expresses that

[tendency of P to turn system] — [tendency of Q to turn system] =0.

Thus when the system is balanced so as to remain at rest we must have

so that the weights must be inversely as the radii. Practical examples of the
principle of the wheel and axle are supplied by the windlass and capstan.

FIG. 33

66 STATICS OF SYSTEMS OF PARTICLES

EXAMPLES

1. Eight sailors, each pressing on the arm of a capstan with a horizontal force
of 100 Ibs., at a distance of 8 feet from its center, can just raise the anchor.
The radius of the axle of the capstan is 12 inches. Find the pull on the cable
which raises the anchor.

2. In the apparatus of fig. 33 the weight Pis disconnected, and the free end
of the string is tied to the same point on Q as the other string. Show that in
equilibrium this point is vertically below the axis.

3. A wheel is free to turn about a horizontal axis, and has fastened to it
two strings which are wound round its circumference in opposite directions.
The other ends are both tied to a small ring from which a weight is suspended.
Show that when the system is at rest the two strings will make equal angles
with the vertical.

4. A man finds that he can just move a lock gate against the pressure of the
water, by pressing with a horizontal force of 150 Ibs. at a distance of 8 feet
from the pivot. What force must he exert if he presses at a distance of 9 feet
from the pivot ?

5. A wheel capable of turning freely about a horizontal axis, has a weight
of 2 pounds fixed to the end of a spoke which makes an angle of 60° with the
horizontal. What weight must be attached to the end of a horizontal spoke to
prevent motion taking place ?

6. A drawbridge is raised by a chain attached to the end farthest removed
from the hinges. When the bridge is at rest in a horizontal position, the chain
makes an angle of 60° with the bridge, and the pull on the chain necessary to
move the bridge is equal to the weight of three tons. Find what additional
pull is required in the chain when a weight of one ton is placed at the middle
point of the bridge.

FOKCES IN ONE PLANE

51. The simplest problems in statics are always those in which
all the forces have their lines of action in one plane. In such a
problem it is obviously most convenient to take moments about
a line perpendicular to the plane in which the forces act. Let any
such line intersect the plane in a point P. Each force is entirely
perpendicular to the line about which moments are taken, so that
the moment is equal to the product of the force and the shortest
distance of the line of action of the force from P.

Taking moments about an axis which intersects the plane of
the forces at right angles in a point P is often spoken of as taking

CO-PLANAK FORCES 67

moments about the point P, and the perpendicular from P to the
line of action of a force is spoken of as the arm of the moment
of this force.

52. THEOREM. When three forces, acting in a plane, keep a body
or system of bodies in equilibrium, these three forces must meet in
a point.

For let P, Q, R be the forces, and let P, Q intersect in the
point A. Then the sum of the moments of P, Q, and R about A
must vanish, and those of P and Q are already known to vanish.
Thus the moment of R about A must vanish, — that is, R must
pass through the point A, or, what is the same thing, the three
forces must intersect in a single point.

An application of this principle is often sufficient in itself for
the solution of statical problems in which the applied forces can
be reduced to three.

ILLUSTRATIVE EXAMPLES

1. The seesaw. Two persons of weights Wi, TF2 stand on a plank which rests
on a rough support about which it is free to turn. Neglecting the weight of the
plank, find how the persons must place themselves in order that the plank may
balance.

The forces may be supposed all to act in one plane, namely the vertical
plane through the central line of the
plank. The forces are

(a) the weight Wi of the person at
one end ; r-

(b) the weight Wz of the person at W\ /\ ^2
the other end ; / \

K IG. OT:

and its support.

Let a, 6 be the distances of the persons from the support ; then, on taking
moments about the point of support, we have

Thus the two persons should stand at distances from the support which are
inversely proportional to their weights.

Notice that in this problem the system is acted on by three forces, which meet in
a point, the point being at infinity.

STATICS OF SYSTEMS OF PARTICLES

100 tf>s-

FIG. 35

2. The nutcracker. It is found that a weight of 100 pounds placed on top of a
nut will just crack it. How much force must be applied at the ends of the arms of
a nutcracker 6 inches long to crack the nut when it is placed | inch from the hinge ?

Let a force F applied at the extreme end of each arm be supposed just suffi-
cient to crack the nut. Then when a force F is applied at the end of the arm, the
pressure between the nut and the arm must be the weight of 100 pounds. Thus

the forces acting from outside on either arm
of the nutcracker will consist of

(a) the force F applied at the end of the arm ;
(6) the pressure of 100 pounds weight exerted
by the nut on the arm at a distance of i inch
from the hinge ;

Taking moments about the hinge, we obtain

6 x F = i x 100 pounds weight,
so that F = 8| pounds weight.

NOTE. When, as here, au unknown force neither enters in the data nor is required
in the answer, we can always obtain equations in which the force does not occur, by
taking moments about a point in its line of action. So again, if two such forces occur,
we can obtain an equation into which neither force enters, by taking moments about
the point of intersection of their two lines of action.

3. A ladder stands on a rough horizontal plane, laining against a rough ver-
tical wall, the contacts at the two ends of the

ladder being equally rough. Find how far a
man can ascend the ladder without its slipping,
it being supposed that the weight of the ladder
may be neglected.

The forces acting on the system composed
of the man and ladder are three in number :
(a) the reaction with the horizontal plane ;
(6) the reaction with the vertical wall ;
(c) the weight of the man.

These forces are all in one plane ; hence,
by the theorem of § 52, their lines of action
must meet in a point.

In the figure let AB be the ladder, C

the position of the man, and P the point in

which the three forces meet, so that PC is A

vertical, and AP, BP are the lines of action FlG- 36

of the reactions at A, B. When slipping is just about to begin, each of these

reactions must make with the normal an angle equal to the angle of friction.

ILLUSTRATIVE EXAMPLES

Let e be this angle of friction, and let a be the inclination of ladder to the
horizontal. Then, from the geometry of the triangle ACP, we have

AC _ AP
sine

and since APB is a right angle,

AP - AB cos

Thus

lit \

( e — a]-

\2 /

A C = AP sin e sec a = AB sin e sin (e + a) sec a.

Thus slipping will begin as soon as the man has climbed a height equal to
sin e sin (e + a) sec a times the whole height.

The condition that the man can reach the top without slipping is that
sin e sin (e + a) sec a shall be greater than unity, or that

sin e sin (e + a) > cos [(e + a) — e]

> sin e sin (e + a) + cos e cos (e + «)•

Thus for the condition to be satisfied cos e cos (e + a) must be negative ; i.e.
e + a must be greater than 90°. Thus the angle between the ladder and the
vertical must be less than the angle of friction. This is also clear from the
figure, for when the man reaches B, two of the forces, namely the reaction at B
and the weight of the man, both pass through B, so that the third force must
also pass through B ; i.e. the reaction at A must have AB for its line of action,
and if the ladder just slips here, the angle between AB and the vertical
must be e.

4. I/, in the last problem, the man has
ascended to some point C without the ladder
slipping, what are the reactions at A and B ?

Here it is not known what angles the
reactions make with the normals : all that
is known is that these angles are less than
the angle of friction.

Let us resolve the reaction at A into two
components NI, FI, and that at B into two
components N%, F%, these being horizontal
and vertical as in the figure. Then the
forces acting on the system composed of
the man and ladder are the five forces
NI, FI, N2, F2, and W.

Resolving vertically, W — NI — Fz = Q.

Resolving horizontally, FI — N% = 0.

Taking moments about A,

W- AC cos a - F2 • AB cos a - N2 • AB sin a = 0.

(a)

70 . STATICS OF SYSTEMS OF PARTICLES

There are four quantities which it is required to find. So far we have obtained
only three equations. We can, of course, obtain other equations by resolving
in other directions and by taking moments about other points, but it will be
found that the equations so obtained will not be new equations, but simply
equations of which the truth is already implied in the equations already
obtained. Thus we cannot, by resolving and taking moments, obtain more than
three independent equations, and these do not suffice to determine the four
unknown quantities.

Here we have illustrated a problem which cannot be solved by the methods
explained in this chapter, and which requires for its solution a consideration
of the systems of forces set up between the separate particles of the bodies acted
upon. It is important that the student should realize that such problems exist,
although he may not yet be able to solve them.

5. Force required to drag a car along. To simplify the problem as far as
possible, let us suppose that the car is mpunted on four equal wheels, each of
radius a and revolving round an axle of radius 6, the coefficient of friction
between wheel and axle being the same for each wheel. Let us suppose that a
force P applied horizontally is found to be just sufficient to start the car into
motion.

1*1

FIG. 38

Let us consider first the equilibrium of the whole car. We are most con-
veniently able to enumerate the forces acting on any system by taking a tour,
in imagination, over the whole surface of a cover made just to fit the system,
and noting the forces which act across this cover at its different points. These,
together with the weight of the whole system, will give the whole system of
forces. The forces acting on the car are in this way found to be

(a) its weight, say W;

(b) the horizontal applied force P ;

(c) the reactions between the wheels and the ground. Let us resolve each
reaction into a vertical component R and a horizontal component F', let us
denote the components of the reaction between the first wheel and the ground
by J?i, RI ; let the corresponding quantities for the second wheel be F2, R2 ', and
so on.

ILLUSTRATIVE EXAMPLES

(As regards the f rictional force acting between the wheel and the ground, we
notice that although motion is about to take place, this motion is not one of
slipping between the wheel and the ground, so that the ratio of F to B for any
wheel is not the coefficient of friction between the wheel and the ground.)

The forces just enumerated hold the car in equilibrium. Thus the sum of
their components in any direction must vanish and the sum of their moments
about any line must vanish. Resolving horizontally and vertically, we obtain

P = FI + F2 + F3 + *V (a)

W = B! + U2 -f Rs + £4- (b)

There is nothing to be gained by taking moments about any line ; as we do not
know the line of action of P, we cannot
know its moment.

Next let us consider the equilibrium of
a single wheel. The wheel touches the
ground and also touches the axle at some
point C. (We may think of the axle as a
circle of radius very slightly less than
that of the inside of the hub of the
wheel.) The forces acting on the wheel
are accordingly

(a) its reaction with the ground ;

(6) its reaction with the axle ;

Neglecting the third force, the two former forces must be equal and opposite.
The line of action of each is accordingly the line joining B, the point of contact
with the ground, to C the point of contact with the axle. Since slipping is just
about to take place at (7, the reaction at C will make an angle e, equal to the
angle of friction, with AC the normal at C. Thus the angle BCA is equal to e.

In the triangle A CB we have AC = 6, AB = a, and the angle ACS = e.

a b

Thus

sin c sin ABC

Since, however, a force along BC has components
pendicular to AB, we have

tan ABC = —•

I and FI along and per-

Thus

sin ABC

Vl_ sin2 ABC

b sin e

Va2-

72 STATICS OF SYSTEMS OF PAKTICLES

Now e, a, and 6 are supposed to be the same for each wheel, so that
6 sin e

by equations (a) and (6).

Thus P = , (c)

giving the horizontal pull required.

The value of 6, the radius of the axle, will generallyjDe small in comparison
with a, the radius of the wheel. Thus, without serious error, we may neglect
62 sin2 e in comparison with a2, and replace the denominator in equation (c) by a.
The equation now becomes

p_ Wb sin e
a

By making b/a very small, we see that the car can be made to run very
smoothly. We notice also that even if there is so much friction between wheel
and axle that the coefficient of friction may be regarded as infinite, we have
sin e — 1, and hence

so that the force required to drag the car along will still be small compared
with that required to drag the same weight over a fairly smooth surface.

This analysis has assumed that the wheels may be supposed to touch the
ground only at their lowest point. It applies pretty accurately to the case of
steel wheels rolling on steel rails, but does not apply to the problem of an
ordinary road carriage moving over a soft road, where the wheels are embedded
to a small extent in the road. In fact, if the analysis just given took account of
all the facts of the case, it is clear that the force required to haul a car would
be independent of the state of the road.

EXAMPLES

1. A weight of 250 pounds is suspended from a light rod which ;s placed
over the shoulders of two men and carried in a horizontal position. If the men
walk 10 feet apart and the weight is 4 feet from the nearer of them, find the
weight borne by each.

2. A weight is suspended from a light rod which passes over two fixed sup-
ports 6 feet apart. On moving the weight 6 inches nearer to one support, the
pressure on that support is increased by 10 pounds. What is the amount of the
weight ?

EXAMPLES 73

3. A balance has two pans, each of weight 8 ounces, suspended from a beam,
each at distance 7 inches from the pivot. A dishonest tradesman moves one
pan half an inch nearer to the pivot, adding weight to this same pan in order
that the two pans may still balance. Find how much weight he must add, and
by how much his profits will be increased by his dishonesty.

4. A balance has a weight of 20 ounces suspended from one end of the beam.
A string is tied to the other end of the beam and at equal distance from the
pivot, and this string makes an angle of 45° with the horizontal. With what force
must it be pulled to maintain the beam of the balance in a horizontal position ?

5. A crowbar 8 feet long is to be used to move a body, it being required to
apply a force of 500 pounds weight vertically upwards to this body to move it.
How near to the end of the crowbar must the fulcrum be placed in order that a
man of 140 pounds weight may be able to apply the required force by pressing
on the other end of the crowbar ?

6. A table of negligible weight has any number of legs. A heavy particle is
placed on the table. Show that the table will tilt over if the vertical through
the particle meets the floor under the table, in a point outside the polygon
formed by joining the points of contact of the feet with the floor.

7. A table of negligible weight has three legs, the feet forming an equilateral
triangle. A heavy particle is placed on the table in a position such that the table
does not tilt over. Find the proportion of weight which is carried by each foot.

8. A card is suspended in a horizontal position by three equal inextensible
strings fastened to three points A, B, C in the card which form an equilateral
triangle, and also to a point P above the card. A weight is placed on the card
at any point Q inside the triangle ABC. Find the tensions of the strings.

9. A card is suspended by four equal inextensible strings which pass through
the four points A, B, (7, D of a square in the card and are tied to four points
A', B', C', D' at equal heights h vertically above the points A, B, C, D. A
weight is placed on the card at any point P inside the square ABCD. Show
that the tensions in the strings cannot be determined without discussing the
internal stresses in these strings.

10. If in the last question the internal stresses in the strings stretch the
strings very slightly, so that Hooke's law is obeyed, show that the tensions can
be found, and find them.

11. A seesaw rolls on a rough circular log of radius a, fixed horizontally. **
Two persons stand at distances 6, c from the middle point of the seesaw, their
weights being such that the seesaw is just balanced horizontally with the
middle point resting on the log. The first person moves a distance d towards
the center of the log. Through what angle will the seesaw turn ? How far
can the person advance before the seesaw slips off the log altogether ?

12. A pair of wheels of radius a are connected by an axle of radius 6, and run
on horizontal rails. A string is wound round the axle and the end leaves the
axle making an angle 6 with the horizontal. If this string is pulled, show that
the wheels will run toward or away from the person pulling according as
cos 6 is greater or less than b/a. What happens when cos 6 — b/a ?

74 STATICS OF SYSTEMS OF PARTICLES

13. If the weight of the wheels and axles of a car is to, and if this may not
be neglected in comparison with W, the total weight of the car, show that
equation (c) of example 5, p. 72, must be replaced by

_ ( W — w) b sin e

14. A locomotive of weight 134 tons rests on a bogie, of which the wheels and
axles weigh 4 tons, and two pairs of driving wheels, of which the wheels and
axles weigh 10 tons. The weight taken on the axles of the bogie is 40 tons,
that taken on the axles of the driving wheels being 80 tons. The diameters of
the wheels are 2 feet 10 inches and 7 feet 1 inch respectively. Each axle, where
it passes through the axle box, is of radius 2£ inches, and the coefficient of fric-
tion is A. Find the horizontal force necessary to move the engine.

15. In question 14 the rails are greased so that the coefficient of friction
between them and the wheels is less than yi^; show that the engine cannot
be started without the wheels skidding on the greased rails, and explain the
dynamical processes by which the engine is set in motion in this case.

STRINGS

53. Strings, ropes, and chains frequently form part of the systems
of bodies with which statical problems are concerned, so that it is
important to discuss the equilibrium of a string (or rope or chain).
The first problem we shall consider is that of a string stretched
over a surface, — as for example a pulley wheel, — it being supposed
that the weight of the string may be neglected, and that the con-
tact between the string and the surface is equally rough at all
points. It will also be supposed that the string is all in one plane.

Let P, Q be two adjacent points of the string so near together
that the portion PQ of the string may be treated as a particle.

The forces acting on this particle will be

(a) Tpt the tension at P, acting along the tangent to the string
at P;

(b) TQ, the tension at Q, acting along the tangent to the string
atC;

By Lami's theorem, each force must be proportional to the sine
of the angle between the remaining two forces.

STRINGS

Let A be the point of the surface at which the string leaves it.
Let the normals to the surface be drawn at A, P, Q, and let the
normal at P make an angle 6 with the normal at A. If the points
A, P, Q come in this order, as in fig. 40, the normal at Q will make
with the normal at A an angle slightly greater than 6, — say 6 -f- dd,
— so that dd is the small angle between the normals at P and Q.

FIG. 40

With this notation the angle between the tensions Tp and TQ is
TT — dd. Let the angle between the reaction R and the tension Tp
be a, then the angle between the tension TQ and H is TT — a + dd.
Thus we have

sin(7r — dd} sin(7r — a + dO)
Since sin(?r — a + dd) = sin (a — dO), we have

sin a

sin (a — dd)
and by a known theorem in algebra, each fraction is equal to

sin a — sin (a — dd)

Now TQ—TP is the increase in T when 6 changes from 6 to 6 + dO,
and this, in the notation of the differential calculus, may be written

76 STATICS OF SYSTEMS OF PARTICLES

Also the denominator sin a — sin (a — d6) is the increase in
sin a when a changes from a — dO to a, and this in the same way

is equal to

d (sin a) a a

— *— - - du or cos a du.
da

Thus the original fraction is equal to

Te^ dT

- — or sec a — - -
cos a dd du

T0 dT

Thus — «- = sec a —r>

sin a du

fJT
or Tg = tana — (12)

When in the limit the particle PQ is supposed to become van-
ishingly small, Tp and TQ become indistinguishable. Let us denote
either by the single letter T, so that T is now simply the tension
at a point at which the normal makes an angle 6 with that at A.
If the string is just on the point of slipping in the direction APQ,
the angle between the reaction R and the normal of either Q or P
will be e, the angle of friction. Thus we shall have

• -•£-<,

so that tan a = cot e, and equation (12) becomes

54. If the contact between the surface and the string is per-

dT
fectly smooth, e = 0, so that — = 0. It follows that T is a con-

CLv

stant ; i.e. the tension is the same at all points of the string. Thus
the tension of a string is not altered by its passing over a smooth
surface, — the result already given in § 36.

STRINGS 77

55. In the more general use in which the contact is not perfectly
smooth, let fi be the coefficient of friction, so that p = tan e, then
equation (13) may be written

and integrating this, — = d

d (log T) = d (fJL0),

or log T = pd + a constant.

Let T0 be the tension at A, then we find, by putting 6 = 0, that
the constant must be equal to log Tot so that

or T= T0e»0. (14)

If the string leaves the surface again at some point B at which
the normal makes an angle ^r with the normal at A, we find for
the tension at B

so that the tension is multiplied by e^ on passing over the surface
from A to B.

If the string (or rope) is passed round and round a post or bollard,
the tension is increased in the ratio e2^ for each complete turn.
For a hemp rope on oak the coefficient of friction, according to
Morin,is/* = 0.53. Thus 2/A-Tr = 3.34 and e2l"r= 28.1. The tension
of a rope wound round an oak post is accordingly increased about
twenty-eight fold for each complete turn.

EXAMPLES

1. A weight is suspended by a rope which, after being wound round a hori-
zontal beam, leaves the beam horizontally, its end being controlled by a work-
man. If the rope makes 1| complete revolutions round the beam, what force
must be exerted by the man

(a) to keep the weight from slipping ?

(b) to raise the weight ?
(Take /* = £.)

78 STATICS OF SYSTEMS OF PARTICLES

2. A weight of 2 1 pounds stands on a rough table. A string tied to the
base of the weight hangs over the edge of the table and has attached to it
a second weight which hangs freely. If the coefficients of friction between
the weight and the table and the string and the table are ^ and £ respec-
tively, find how heavy the hanging weight must be to start the other weight
into motion.

3. A weight of 2500 pounds is to be raised from the hold of a ship. A rope
attached to the weight makes 3^ turns round a steam windlass, its other end
being held by a seaman. With what force must he pull his end of the rope so
as to raise the weight when the windlass is in motion ? (/* = |.)

4. In the last question, find what pull would have to be exerted on the rope
if the windlass were at rest.

5. It is found that two men can hold a weight on a rope by taking three
turns about a post, and that one of them can do it alone by taking one half
turn extra. If each can pull with a force of 220 pounds weight, find the weight
sustained.

6. In a tug of war the rope is observed to rub against a post at the critical
moment, in such a way that the two parts of the rope make an angle of 1° with
one another. If the coefficient of friction between the rope and the post is |,
show that this imposes a handicap on the winning side equal to about .0029
times its aggregate pull.

Suspension Bridge

56. An interesting problem is afforded by the kind of suspen-
sion bridge in which the weight of the bridge (supposed horizontal)
is taken by a suspension cable by means of vertical chains con-
necting the bridge with the cable.
Let us, for simplicity, agree to
neglect the weights of the chains
and cable, and suppose the weight
of the bridge to be distributed
evenly along its length.

Let 0 be the lowest point of
the cable, and let P be any other
point. Let o, p be the points of the bridge vertically below 0, P,
and let op = x. Let the tension at P be T, and let that at 0 be H.
Let the direction of the cable at P make an angle 6 with the
horizontal.

SUSPENSION BRIDGE 79

The forces acting on the piece OP of the cable consist of

(a) the tension at 0, of amount 77" acting horizontally;

(b) the tension at P, of amount T acting at an angle 6 with
the horizontal;

Resolving horizontally, we obtain

H-Tcos0 = Q. (15)

Resolving vertically, we obtain

Tsm6- S = 0,

where S is the sum of the tensions of all the chains which leave
the cable between 0 and P. These tensions support the portion
op of the bridge, and if the weight of the bridge is w per unit
length, the weight of op will be wx. Thus S = wxy and therefore

Tsm0 = wx. (16)

This and equation (15), T cos 0 = H, (17)

will give us all the information we require.

To find the shape which the cable must have in order that the
bridge may hang horizontally, we require to obtain a relation
between 6 and x. We accordingly eliminate Tfrom equations (16)
and (17), and obtain

tan 6 = — x.
H.

If y is the height of the cable above the bridge, xt y may be
regarded as the Cartesian coordinates of a point P on the cable,
and we have ,

tan0 = ^.
ax

Thus the coordinates x, y of P are related by

dy _ w
dx = HX>

"I f\ ij

giving on integration y = x2 4- O,

U -tL

where C is a constant of integration.

80 STATICS OF SYSTEMS OF PAETICLES

This is the Cartesian equation of the cable. It is easily seen to

2 H

represent a parabola of latus rectum --- Thus the cable must

hang in the form of a parabola. The greater the horizontal ten-
sion, the greater the latus rectum of the parabola, and therefore
the natter the curve of the cable. A perfectly straight cable is of
course an impossibility — this would require infinite tension.

57. To find the tension at any point of the cable, we square
equations (16) and (17), and add corresponding sides. Thus

giving the tension at a point distant x from the center. If the
bridge is of length 2 a, the tension at either pier must be

The Catenary

58. In the problem of the suspension bridge, we neglected the
weight of the cable. A second problem arises when the cable is
supposed to be acted on by no external forces except its own
weight. The problem here is simply that of
a string of which the two ends are fastened
to fixed points, and which hangs freely be-
tween these points.

As before, let 0 be the lowest point, and
let P be any other point. The forces acting
on the portion OP of the string are

(a) the tension at 0, of amount If acting
0 p horizontally ;

Fm 40 (b) the tension at P, of amount T acting

at an angle 0 with the horizontal ;

(c) the weight of OP. If we take the string to be of weight w
per unit length, and denote the distance OP by s, this weight is
ws acting vertically.

THE CATENARY 81

Resolving horizontally, we obtain

= 0. . (18)

Resolving vertically, we obtain

Tsm6-ws = Q. (19)

To find the shape of curve in which the string will hang, we
must obtain a relation between 6 and s. Eliminating T, we obtain

H tan 6 = ws,
or, if we replace H/w by the single constant c,

(20)

This is one form of the equation of the curve, s and 0 being taken as
coordinates. The equation in this form is known as the intrinsic equation
of the curve. We require, however, to deduce the equation in its Carte-
sian form.

59. If the point o in fig. 42 is taken as origin, the axes being
horizontal and vertical, we have at once the rela- >

fcl°ns dx : dy :ds = cos 6 : sin d : 1, (2 1) ds/

/ dy

for dx and dy are the horizontal and vertical pro-

jections of the small element ds of length of the
string. As a first step, let us use relations (21) to
change the variables of equation (20) from s and 6 to s and y.

We have //7 \2

c2 = s2 cot2 0 = s2 cosec2 0 - s2 = s2 1 — } - s2,

\dy]

,, ds

so that s — =

dy.

Thus dy =

and integrating this, we obtain

y = Vs2 + c2 + a constant. (22)

We can determine the constant of integration as soon as we
decide where the origin is to be taken — so far we have not fixed

82 STATICS OF SYSTEMS OF PARTICLES

the point o. Since s denotes the arc of the curve measured from
0, we have s = 0 at the point 0, and therefore the y coordinate
of 0 (putting s = 0 in equation (22)) is

y = c + a constant.

Let us agree that Oo is to be made equal to c, so that y — c at 0.
Then the unknown constant of integration must be zero. Thus
equation (22) will be y^^ + c2. ^>3)

The last step is to transform the variables from y and s to y and x.
The relation which enables us to do this is obtained by eliminating
6 from relations (21), and is

(ds)2 = (dy? + (dx)\ (24)

The equation already obtained is

s = V?/2 — c'2'

so that ds = / y y

^/f-c*

and on eliminating ds from this and relation (24), we obtain

From this, (dxf = (dy? \~^—z - 1~|

f-<?

Integrating this, we obtain

so that dx= .Cdy - (25)

/ o o \ »

-> (26)

c c

•

x - --

where cosh - = ^ (ec -f e c).

The student who is not familiar with the hyperbolic cosine (cosh) func-
tion will easily be able to verify equation (26) by differentiating it back
into equation (25).

THE CATENARY 83

Equation (26) is the Cartesian equation of the curve formed by
the string; this curve is known as the catenary.

From equation (23) we obtain the value of s in the form

= c2 sinh2 - >
c

s . , x

so that - = smh->

c c

•/P 5 _*

where sinh - = \ (ec — e c).

- -- x

60. Expanding the exponentials ec, e c, we obtain cosh - in

the form

tJC

So long as x is small, we may neglect all the terms of this series
beyond the second. Using the value obtained in this way, we
obtain instead of equation (26)

, a-2
'-'+TS

showing that so long as x is small the curve coincides very approxi-
mately with a parabola of latus rectum 2 c or 2 H/w.

This parabola, it will be noticed, is one which would be formed by the
cable of a suspension bridge of horizontal tension H, w being the weight
per unit length of the bridge itself. Indeed, it is clear that when the cable
is almost horizontal, it is a matter of indifference whether the cable
itself possess weight w per unit length of its arc, or whether a weight w
per unit length is hung from it so as to lie horizontally.

We can also obtain a simple approximation to the shape of the
catenary when x is large, i.e. at points far removed from the lowest

flfi

point. When x is very large, so that - is very large, the value of

* -* c

ec becomes very large, while that of e c becomes very small. Thus

STATICS OF SYSTEMS OF PAKTICLES

the value of cosh - becomes approximately ^ ec, and the equation

of the catenary (equation (26)) becomes

Thus for large values of x the catenary coincides with the
exponential curve.

Fig. 44 shows the form of the catenary. The thin curves are

(a) the parabola, to which the catenary approximates for small
values of x;

(b) the exponential curve, to which the catenary approximates
for large values of x.

THE CATENARY

61. Sag of a tightly stretched string. A string or wire stretched
so as to be nearly horizontal all along its length — as for instance
a telegraph wire — may, as we have seen, be supposed to form a
parabola to within good approximation. Thus let A, B be two poles
at equal height between
which a wire is stretched ;
let C be the middle point
of AB, and let D be the
point of the wire verti-
cally below C. Then,
from symmetry, D will be the lowest point of the wire and there-
fore will be the vertex of the parabola. Thus, from the equation
of the parabola, 2 jf

since, by § 60, its latus rectum is 2 H/w.

Thus if h = AB, the distance between the poles, the " dip " CD is
given by

FIG. 45

(27)

To obtain the length of the wire we have to introduce a higher
order of small quantities, and so are compelled to return to the
equation of the catenary.

X _X

We have s = l c (ec — e c)

laf

^ec2"1

The quantity we require is s — x, namely DB — CB in fig. 45.
When the string is tightly stretched c is very great, so that we
may neglect the terms in s beyond those written down in the
above equation, and obtain

s — x = > approximately,

6 G

\ w2x9
= 'E'W

86 STATICS OF SYSTEMS OF PARTICLES

Putting x = ^ h, we find as the total increase in a span of
length h, caused by sagging,

EXAMPLES

1. The entire load of a suspension bridge is 320 tons, the span is 640 feet,
and the height is 60 feet. Find the tension at the points of support, and also
the tension at the lowest point.

2. The weight of a freely suspended cable is 320 tons ; the distance between
the two points of support, which are in the same horizontal line, is 640 feet,
and the height of these points above the lowest point of the cable is 60 feet.
Find the tension at the points of support, and also the tension at the lowest point.

3. The wire fora telegraph line cannot sustain a weight of more than a mile
of its own length without breaking. If the wire is stretched on poles at equal
intervals of 88 yards, what is the least sag permissible ?

4. In the last question how much wire is required for a mile of the line ?

5. A telegraph line has to be built of a certain kind of wire, stretched over
evenly spaced posts. Show that if the number of posts is very large, the line
will be built most economically as regards the cost of wire and posts, if the
cost of the posts equals twice that of the additional length of wire required by
"sagging."

GENERAL EXAMPLES

1. A block of stone weighing \ ton is raised by means of a rope
which passes over a pulley vertically above it and. is wound upon a wind-
lass one foot in diameter. The windlass is worked by two men who turn
cranks of length 2 feet. What force must each man exert perpendicular
to the cranks ?

2. A man sitting in one scale of a balance presses with a force of 60
pounds against the beam in a vertical direction and at a point halfway
between the fulcrum and the end of the beam from which his scale is sup-
ported. If the beam is 5 feet long, find the additional weight which must
be put in the other scale to maintain equilibrium.

3. The scales of a false balance hang at unequal distances a, b from
the fulcrum, but balance when empty. A weight appears to have weights
P, Q, respectively when weighed in the two scales. Find its true weight
and prove that

EXAMPLES 87

4. A weightless string 24 inches in length is fastened to two points
which are in the same horizontal line, and at a distance of 16 inches apart.
Weights are fixed to two points at distances of 9 and 7 inches from the
ends of the string and hang in such a way that the portion of the string
between them is horizontal. Determine the ratio of the weights.

5. A light wire has a weight suspended from its middle point, and is
itself supported by a string fastened to its two ends and passing over a
smooth peg. Show that the wire can rest only in a horizontal or vertical
position.

6. Three smooth pegs A, B, C stuck in a wall are the vertices of an
equilateral triangle, A being the highest and the side BC horizontal ; a
light string passes once around the pegs and its ends are fastened to a
weight W which hangs in equilibrium below BC. Find the pressure on
each peg.

7. Two rings of weights P and Q respectively slide on a weightless
string whose ends are fastened to the extremities of a straight rod inclined
at an angle 0 to the horizontal. On this rod slides a light ring through
which the string passes, so that the heavy rings are on different sides of
the light ring. All contacts are smooth and, in equilibrium, <f> is the angle
between the rod and those parts of the string which are close to the light
ring. Prove that tan ^ p -Q

tan 0 ~ P + Q '

8. Two small heavy rings slide on a smooth wire, in the shape of a
parabola with axis horizontal ; they are connected by a light string which
passes over a smooth peg at the focus. Show that their depths below the
axis are proportional to their weights when they are in equilibrium.

9. Two equally heavy rings slide on a wire in the shape of an ellipse
whose major axis is vertical, and are connected by a string which passes
over a smooth peg at the upper focus. Show that there are an infinite
number of positions of equilibrium.

10. ABCD is a quadrilateral; forces act along the sides AB, BC, CD,
DA measured by a, /3, 7, 8 times those sides respectively. Show that if
these forces keep any system of particles in equilibrium, then

ay = /35.

11. A light rod rests wholly within a smooth hemispherical bowl of
radius r, and a weight W is clamped on to the rod at a point whose dis-
tances from the ends are a and &. Show that 0, the inclination of the rod
to the horizon in the position of equilibrium, is given by the equation

2 Vr2 - ab sin 0 = a - 6.

88 STATICS OF SYSTEMS OF PAKTICLES

12. A weightless rod, to which are fixed two rough beads of masses m
and m', lies on an inclined plane and is free to turn about an axis through
the rod perpendicular to the plane. If it be in a horizontal position, show
that it will not begin to slip round unless

ma ~ m'l) ,
/A < - tan a,
ma + m'b

where a is the angle of the plane, and a and b the distances of m and mf
respectively from the axis.

13. A bead of weight W, run on a smooth weightless string, rests on an
inclined plane of angle a, the coefficient of friction between the bead and
plane being /A. The ends of the string are tied to two points A , B in the
plane at the same height. Show how to find the positions of limiting
equilibrium for the bead, and show that in such a position P, the tension

of the string is

\W sec \ APR - (tan2 a - M2)*.

14. A uniform string is placed on a rough sphere so as to lie on a hori-
zontal small circle in altitude a. Prove that, if the string be on the point
of slipping along the meridians, the tension is constant and equal to W cot
(a + e), where W is the weight of a length of the string equal to the radius
of the circle, and e is the angle of friction.

15. A weightless string is suspended from two fixed points and at given
points on the string equal weights are attached. Prove that the tangents
of the inclinations to the horizon of different portions of the string form
an arithmetical progression.

16. A smooth semicircular tube is just filled with 2n equal smooth
beads, each of weight W, that just fit the tube, and stands in a vertical
plane with the two ends at equal height. If Rm is the pressure between
the rath and (m + l)th beads from the top, show that

17. In the last question, let the beads be indefinitely diminished in
size. Prove that the pressure between any two beads is proportional to
the depth below the top of the tube.

18. A heavy string hangs over two smooth pegs, at the same level and
distance a apart, the two ends of the strings hanging freely and the central
part hanging in a catenary. Show that for equilibrium to be possible, the
total length of the string must not be less than ae.

EXAMPLES 89

19. A string of weight W is suspended from two points at the same
level, and a weight W is attached to its lowest point. If a, j8 are now the
inclinations to the vertical of the tangents at the highest and lowest points,
prove that tang W

tan/3 W

20. A heavy string of length I is supported from two points, and at
these points the string makes angles a, ft with the vertical. Show that
the height of one point above the other is

I cos !(<* + 0) sec I (a - /S).

21. Prove that the direction of the least force required to draw a
carriage is inclined at an angle 6 to the ground, where asm 6 = bsine,
a and b being the radii of wheels and axles respectively, and e being the
angle of friction.

CHAPTEE V
STATICS OF RIGID BODIES

ElGIDITY

62. If we press a lump of wet clay or of soft putty with the
finger, we find that a dent is left in the clay or putty ; the force
applied to the substance by the finger has caused it to change its
shape. If we press a mass of jelly with the finger, we do not find
any dent left in the jelly, but we notice that so long as the force
is applied the shape of the jelly is altered, although it returns to
its original shape as soon as the pressure is removed.

On the other hand, if we press a lead bullet or an ivory billiard
ball with the finger, we do not notice any change of shape either
while the pressure is applied or after. In ordinary language we
say that the lead and ivory are harder than the clay and putty;
in scientific language we say that they are more rigid.

63. A perfectly rigid body would be one which showed no
change of shape under any force, no matter how great this force
might be. A bullet and a billiard ball are not perfectly rigid,
billiard ball is pressed out of shape during the interval of collisioi
with a second billiard ball, but regains its shape immediately,
while a lead bullet is pressed permanently out of shape by striking
a target. A perfectly rigid body does not exist in nature, althougl
such bodies as a billiard ball or a bullet may be regarded as per-
fectly rigid, so long as the forces which act upon them are n<
too great.

We can give a mathematical definition of a perfectly rigid bod]
as follows:

A body is perfectly rigid when the distance between any pail
of particles in it remains unaltered, no matter what forces act
the body.

RIGIDITY 91

64. A rigid body can move about in space without changing
the direction of any line in it — such a motion is called a motion
of translation. It can also turn about any point P without the
position of P altering — such a motion is called a motion of rota-
tion about P. It can again have a motion compounded of a motion
of translation and one of rotation, and this we shall show to be
the most general motion which it can possibly have.

65. First we must notice that a rigid body is fixed when any
three points in it are fixed, provided that these three points do not
lie in a straight line. For let A, B, C be the three points. If we
fix A and B, any motion which can take place must, since the
body is by hypothesis perfectly rigid, be one in which the dis-
tances of any other point P from A and B remain unaltered.
Thus P must describe a circle with AB as axis, and the motion
of the body must be one of rotation about the line AB. Thus
if A, B, C are not in a straight line, C must describe a circle
about AB. But if C also is fixed, this cannot happen; in other
words, no motion can take place, so that the body is fixed in
position.

Thus the position of a rigid body is determined when the posi-
tions of three non-collinear points in it are determined.

66. We can now prove the theorem :

TJie most general motion of a rigid ~body is compounded of a
motion of translation and one of rotation.

In fig. 46 let the figure on the left represent the body in its
original position, and let the heavy curve on the right represent
the body after it has been moved in any way. Let P be the
position of any particle of the body in its original position, and
let Q be the position of the same particle after the motion has
taken place.

Imagine that the body is first moved from its original position
in such a way that the point P. moves to Q, while all lines in the
body remain parallel to their original positions. This motion is
one of pure translation. After this motion has taken place, we
can turn the body about the point Q in such a way as to turn it

STATICS OF KIGID BODIES

into its final position. For let us take any two other particles jR, S
in the body (not in the same straight line with Q), and let R'} Sf be
their final positions. Since the body is supposed to be perfectly

FIG. 46

rigid, it follows that all distances between particles of the body
remain unaltered. Hence the distances QR, RS, SQ are respectively
equal to QR't fi'S', S'Q. Thus the triangles QRS, QR'S', being
equal in all respects, can be superposed, and the motion of super-
position of these triangles is the motion required and is a motion
of pure rotation about Q, since Q does not move.

Since the position of a rigid body is fixed when any three points
in it are fixed, it follows that the rigid body can only have one
position in which the three points Q, It, S have given positions. But
after the motion we have described, the three points Q, R, S are
placed in their final positions. Hence the whole body must be in
its final position, and this proves the theorem.

67. Axis of rotation.
In a motion of rotation,
let P be the point which
remains fixed. Take any
plane A through P, and
let B be the position of
the plane A after the
rotation has occurred.
These two planes both pass through P, and must therefore

FIG. 47

CONDITIONS OF EQUILIBEIUM 93

intersect in some line PQ passing through P. This line is called
the axis of rotation. The rotation can be imagined as a turning
about an imaginary pivot running along the axis of rotation.

CONDITIONS OF EQUILIBRIUM FOR A EIGID BODY

68. We have seen that a rigid body is fixed when three non-
collinear points in it are fixed. It follows that, whatever forces
act on a rigid body, we can always hold it at rest by applying
three suitably chosen forces at three points which are not in a
straight line.

"We can, however, select these forces in a special way.

Let us select any three points A, B, C, subject only to the con-
dition that they are not in a straight line. By a suitably chosen
force acting on the particle at A we shall always be able to keep
the point A at rest.

When A is fixed B may or may not tend to move. If B tends
to move, the direction of motion of B must be perpendicular to
BA, since A cannot move. Hence after A is fixed it must be
possible to fix B by applying at B a force perpendicular to BA.

When A and B are both fixed the only motion possible for the
third point C is one perpendicular to both AC and BC,Le. perpen-
dicular to the plane ABC. Thus C can be held at rest by a force
perpendicular to the plane ABC, and the whole body is now held
at rest. Thus it has been proved that a rigid body can be held at
rest, in opposition to the action of any system of forces, by the
application of the following forces at three arbitrarily chosen
points A, B, C not in a straight line:

(a) a force at A, direction unknown ;

(I) a force at B, direction perpendicular to the line AB ;

The condition that the original system of forces should hold
the body in equilibrium is of course that no additional forces are
required to fix" the body, and hence that the three forces introduced
at the points A} B, C should each vanish.

STATICS OF RIGID BODIES

TRANSMISSIBILITY OF FORCE

69. Consider a rigid body acted on by two forces WA) WB at
two points A, B, these forces being equal in magnitude but acting
in the opposite directions AB, BA.

Either the rigid body will be in equilibrium under the action of
these two forces, or else it can be held at rest by three forces PA,
PB, Pc at the points Ay B, and any third point C not in the line
AB, these forces being in the directions already mentioned, namely
Pc being perpendicular to ABC, and PB perpendicular to AB.

FIG. 48

Let these forces, if necessary, be put in so that the body is in
equilibrium under the action of the forces WA, WB, PA, PB, Pc.
The body being in equilibrium, the sum of the moments of these
forces about any line, or of their components in any direction, must
vanish, by § 50.

Let us consider what is the sum of the moments about the line
AB. The forces WA> WB, PA, PB all meet this line, so that the
moment of each of these forces vanishes. Thus the sum of the
moments about the line AB consists of the moment of the single
force Pc, and for the sum of these moments to vanish, the moment
of Pc must vanish. Now Pc is perpendicular to the line AB, and
does not intersect it, so that the moment of Pc can vanish only if
the force Pc is itself equal to zero. This means that no force is
required to keep the body from turning about AB as axis of
rotation.

TRANSMISSIBILITY OF FORCE 95

The body is now held at rest by the two forces PA, P£, and is
therefore in equilibrium under the forces PA) P£, WA, WB. Tak-
ing moments about a line through A perpendicular to ABy we find
that the moments of PA, WA, WB vanish, so that in order that the
sum of the moments of the four forces taken about this line may
vanish, we must have the moment of PB equal to zero, and there-
fore PB itself equal to zero. Thus the only force required to keep
the body at rest is the force PA at A.

A condition for equilibrium is now that the sum of the com-
ponents of WA, WB) and PA shall vanish in any direction. The
components of WA and WB are, however, equal and opposite, so
that the component of PA must vanish in every direction. That
is to say, we must have PA = 0.

It has now been proved that the rigid body is in equilibrium
under the action of the two forces WA, WB.

70. This establishes at once a principle known as the transmissi-
lility of force.

The effect of a force acting on a rigid ~body depends on its mag-
nitude and on the line along which it acts, but not on the particular
particle in this line to which it is applied.

For, let the same force be applied at any
two points Q, R of its line of action. An

equal and opposite force at R can neutralize either of the two
forces, which are therefore equivalent.

COMPOSITION OF FORCES ACTING IN A PLANE

71. Suppose that we have two forces P, Q acting at two points
A, B of a rigid body, it being supposed that the two lines of action
of these forces lie in one plane. Then the two lines of action,
produced if necessary beyond the points A, B, will meet in some
point C.

By the principle of the transmissibility of force it is imma-
terial whether the force P acts at A or at (7; let us suppose it to
act at C. In the same way let us suppose the force Q to act at C

STATICS OF KIGLD BODIES

instead of at B. Then we have the body acted on by two forces
P, Q which act on the same particle C. These may be compounded,
according to the rules explained in Chapter III, into a single force

n acting at C. Thus we can
compound forces of which
the lines of action intersect,
even though they do not act
on the same particle.

Having compounded two
forces into a single force, we
may compound this resultant
with any third force which
lies in the same plane as the
two original forces, and in
this way obtain a resultant
FIG. so of three forces, and so on.

Thus any number of forces which all lie in one plane may be
compounded into a single force. This force is called the resultant of
the original force.

72. An exception arises when we attempt
to compound two parallel forces, for their
lines of action do not meet. This difficulty,
however, is easily surmounted. Let P, Q
be the two forces to be compounded, and
let AB be any line cutting their lines of
action in A, B. Let us add to the system

P, Q two forces : • ,

(a) a force E acting along BA ;

(b) a force R acting along AB.

FIG. 51

These two forces being equal and opposite
can be introduced without producing any
effect. On compounding the first with P
we obtain a resultant P' acting at Ay and on compounding the
second with Q we obtain a resultant Q' acting at B. Thus the

COMPOSITION OF FORCES 97

original forces P, Q have been replaced by the new forces P', Qr.
The lines of action of these forces, however, will not in general be
parallel, so that they may be compounded into a single resultant
force acting through their point of intersection.

73. Let us suppose that the forces originally to be compounded
were Rlf Rz,--, and that these have been compounded into a
single resultant R. Let us take axes x, y in the plane in which
these forces act, and let the components of R1 along these axes be
X^ Ylt those of Rz being Xv Y2) and so on. Finally let the com-
ponents of R be X, Y.

The system of forces which consists of the original forces Rv
^2» " "> together with the resultant R reversed, constitutes a system
in equilibrium. Resolving parallel to the axes, we obtain

X\ + X, + XZ + ---- X=0,

Y!+ r2+ r3 + ---- r=o.

Thus the components of R are given by the equations
X=Xl + Xt + X9 + >-,

Y= *!+ ra+ y, + ....

The magnitude of R can be found from the equation

R2 = x2 + r2,

while the angle 6 which the line of action of R makes with the
axis of x can be found from the equation

To obtain the position of the line of action of R we use the fact
that the sum of the moments of

RI} Rz, RS) • • •, and — R

taken about any point in the plane must vanish. This gives us
the moment of R about any point, and hence, since we know the
magnitude and direction of R, we can find the position of its line
of action.

STATICS OF EIGID BODIES

ILLUSTRATIVE EXAMPLE

Forces P, Q, B act on a rigid body, all the forces being in one plane,_and their
lines of action forming a right-angled isosceles triangle of sides a, a, Vj? a. Find
their resultant,

Let ABC be the triangle, the forces P, Q, E acting along BC, CA, AB
respectively. Take C for origin and CB, CA for axes of x, y. Let the resultant
have components X, Y. Then, resolving along Cx, we obtain

and similarly resolving along Cy,

Thus the resultant is of magnitude R given by

>» Q'

R2 = X* + F2 = ( -

V2/
= P2 + Q2 + E2 - V2 £ (P + Q) .

The angle 0 which it makes with Cx is given by

P B

FIG. 52

To find the line of action, we express that the moment of R about C must be
to the sum of the moments of P, Q, and R. If p is the perpendicular
from C on to the line of action of the resultant R, this gives

Rp = E

VI'

so that

E a

and this determines the line of action of R.

EXAMPLES

(All forces are supposed to be acting on rigid bodies)

1. ABCD is a square, and along the sides AB, BC, CD there act forces of
1, 2, 3 pounds respectively. Determine the magnitude and line of action of the
resultant.

2. ABCD is a square and forces P, Q, E, S act along the sides A B, BC, CD,
DA respectively. What is the condition that their resultant shall pass through
the center of the square ?

3. In question 2, what is the condition

(a) that the resultant shall pass through A ?

(b) that it shall pass through B ?

PARALLEL FORCES 99

4. Forces P, Q, E act along the sides of a triangle ABC, and their
resultant passes through the centers of the inscribed and circumscribed
circles. Prove that

P = Q = R

cos B — cos C cos C — cos A cos A — cos B

5. If four forces acting along the sides of a quadrilateral are in equi-
librium, prove that the quadrilateral must be plane.

6. ABCD is a plane quadrilateral and forces represented by AB, CJ5, CD,
AD act along these sides, respectively, of the quadrilateral. Show that if there
is equilibrium, the quadrilateral must be a parallelogram.

7. If a quadrilateral can be inscribed in a circle, prove that forces acting
along the four sides and proportional to the opposite sides will keep it in equi-
librium. Show also that the converse is true, namely that for equilibrium, the
forces must be proportional to the opposite sides.

8. A quadrilateral is inscribed in a circle, and four forces act along the
sides, and are inversely proportional to the lengths of these sides. Show that
the resultant has for line of action the line through the intersections of pairs
of opposite sides.

9. Forces act along the four sides of a quadrilateral, equal respectively to
a, 6, c, and d times the lengths of those sides. Show that if there is equilibrium,

ac = bd,

and that the further conditions necessary to insure equilibrium are that the
ratios a : 6 and 6 : c shall be the ratios in which the diagonals are divided at
their points of intersection.

10. In the last question show that the perpendicular distances to the first
side, from the two points of the quadrilateral which are not on that side, are in

the ratio

a(c-6) :d(b-a).

PARALLEL FORCES

74. Let us use the method just explained, to determine the
resultant of two parallel forces P, Q.

Take any point 0 on the line of action of P as origin, and take
this line of action of P for axis Oy, as in fig. 53. Let the resultant
be R, components X, Y. Then resolving we obtain

X= 0,
F-P + ft

so that the resultant force is of magnitude P + Q and acts parallel

100

STATICS OF RIGID BODIES

P+Qi

,—•6.

"""""- a-

"•---"-"

FIG. 53

to Oy. Let us suppose that its distance from Oy is b, that of Q
being denoted by a. Then taking moments about 0 we have

(P+ Q)b = Qa,

, , . I a a — I

so that — = — = >

Q P+Q P

showing that the line of action
divides the distance between P and
Q in the ratio Q : P.

Thus we have shown that the
resultant of two parallel forces P,
Q is a force of magnitude P + Q,
parallel to these forces, of which
the line of action divides the dis-
tance between the lines of action of
P and Q, in the ratio Q : P.

75. Alternative treatment of parallel forces. We can prove directly from
§ 68 that parallel forces P, Q will be in equilibrium with a force

— (P + Q) parallel to them and acting along a line which divides the dis-
tance between them in the ratio Q : P.

Take two points A, B on the line of action of P, Q, and a third point
C not on the line AB. Then the body, acted on by the forces P, Q, and

— (P + Q), can be kept in equilibrium by the further application of
(a) a force JRcat C, perpendicular to ABC ;

(&) a force RB at B, perpendicular to AB ;

Thus the system of forces

P, Q, -(P + Q), Rc, RB, RA
will be in equilibrium.

Taking moments about the line AB, we find
that Rc = 0. Taking moments about A , we find
that RB=Q, or else that it acts along BA, in
which case it can be absorbed into RA. Resolv-
ing perpendicular to the plane of the forces P, Q,
we find that RA can have no component perpen-
dicular to the plane. Thus the four remaining
forces

P, Q,
are all in one plane.

Fia. 54

COUPLES

Next, resolving parallel and perpendicular to the line of action of P, we
find that both components of RA vanish, and hence that RA — 0. Thus the
original forces were in equilibrium.

76. Clearly these methods of compounding forces can be
extended, so that any number of parallel forces can be compounded
into a single resultant force. We see at once, on reversing the
resultant and resolving, that the resultant is parallel to the lines
of action of the original forces, while its magnitude is equal to
their algebraic sum.

This result is of importance in connection with the weights of
bodies. It shows that the effect of gravity on any rigid body —
i.e. the resultant of the weights of the individual particles of which
the body is composed — may be regarded as a single force acting
vertically along a single line.

In the next chapter it will be shown that, whatever position the
rigid body is in, this line always passes through a definite point,
fixed relatively to the body, known as its center of gravity.

77. Without assuming this, we can find the line of action in a
number of simple cases. Suppose, for instance, we are dealing with
a uniform rod. The weights of two equal particles equidistant from
the center may be compounded into a single force acting through
the middle point of the rod. Treating the weights of all the parti-
cles in this manner, we find that the weight of a uniform rod may
be supposed to act at its middle point.

In the same way we can see that the weight of a circular disk,
of a circular ring, or of a sphere may be supposed to act at its
center ; the weight of a cube or a parallelepiped at the intersection
of its diagonals, and so on.

COUPLES

78. If we try to compound two parallel forces which are equal
in magnitude but opposite in sign, we obtain as the resultant a
force of zero amount of which the line of action is at infinity.
Although such a force is of zero amount, its effect cannot be

•102 STATICS OF EIGID BODIES

neglected : its moment does not vanish, being equal to the sum of
the moments of the component forces. If, in fig. 55, AA', BB' are
the parallel lines of action of two opposite forces each equal to R,

and if PAB is a line at right angles to
their direction, then the sum of their
moments about a line through P, at
right angles to the plane in which the
forces act,

where d is the distance between the
line of action of the forces. A pair of
forces, equal in magnitude and opposite
in direction, but not acting in the same

line, is called a couple. Their moment about any point P in the
plane containing their lines of action is independent of the position
of the point P, and is spoken of as the moment of the couple.

CONDITION OF EQUILIBRIUM

79. Since the resultant of a system of forces in a plane may be
either a single force or a couple, the condition for there being no
resultant will be that the resultant single force shall be nil, and
that there shall be no couple. The component of the resultant
force in any direction vanishes if the sum of the components
vanishes. Thus, in order that the resultant force may vanish, it is
necessary that the resolved parts in two different directions should
vanish. If this condition is satisfied, there can be no resultant
except a couple, and since the moment of a couple is the same, no
matter about what point the moment is taken, it appears that there
can be no couple if the moment about any one point is zero. Thus,
as a necessary and sufficient condition of equilibrium for a system
of forces in a plane, we have found the following :

A system of forces in a plane will be in equilibrium if the sum
of the resolved parts in two directions each vanishes, and if the sum
of the moments about any point also vanishes.

CONDITION OF EQUILIBRIUM 103

We can express the condition for equilibrium in a different form :

A system of forces in a plane will be in equilibrium if the sums
of the moments about any three non-collinear points are each zero.

For, if the moment about any one point is zero, the resultant
cannot be a couple. It must, therefore, be a single force. If the
moments about each of two points A, B vanish, the line of action
of this force must in general be AB, but if the moment about
some third point C, not in the line ABy also vanishes, then the
force itself must vanish.

EXAMPLES

1. Parallel forces of 6, 12, and 7 pounds act at the two ends and middle
point, respectively, of a line 2 feet in length. Find the magnitude and line of
action of their resultant.

2. Find the resultant of the forces in the last question when their magnitudes
are respectively 5, — 12, and 7 pounds.

3. Find the resultant of three forces, each of amount P, acting along the
sides of an equilateral triangle, taken in order.

4. Prove that a system of forces acting along and represented by the sides
of a plane polygon taken in order, is equivalent to a couple, whose moment is
represented by twice the area of the polygon.

5. If the sums of the moments of any co-planar forces about three points
which are not in a straight line are equal, and not each zero, prove that the
system is equivalent to a couple. .

6. A uniform rod is of length 3 feet and weight 24 pounds. Weights of 16
and 18 pounds are clamped to its two ends. Find at what point the rod must be
supported so as just to balance.

7. A uniform beam weighing 20 pounds is suspended at its two ends,
and has a weight of 50 pounds suspended from a point distant 7 feet and
3 feet from the two ends. Find the pressures at the points of suspension of
the beam.

8. A uniform rod of weight 60 pounds and length 18 feet is carried on the
shoulders of two men who walk at distances of 2 feet and 3 feet respectively
from the two ends. A weight of 60 pounds is suspended from the middle point
of the beam. Find the total weight carried by each man.

9. A dumb-bell weighing 32 pounds is formed of two equal spheres, each of
radius 3 inches, connected by a bar of iron so that the centers of the spheres
are 16 inches apart. One of the spheres is now removed, and the remainder of
the dumb-bell is found to weigh 20 pounds. Find where this remaining part
must be supported in order that it may just balance.

104 STATICS OF EIGID -BODIES

COUPLES IN PARALLEL PLANES

80. The result of § 79 shows that two couples acting in the
same plane produce the same effect if their moments are equal.
For, on reversing one of them, all the conditions of equilibrium
are satisfied.

Thus we can determine the effect of a couple in any plane by
knowing its moment only. We shall now show that the actual
plane in which the couple acts is immaterial, the direction of this
plane alone being of importance. In other words :

Couples of equal moments acting in parallel planes produce the
same effect.

To prove this, we reverse one couple and show that the two
couples are then in equilibrium. Let the first couple consist of
I? v two forces each equal to R,

and let a common perpen-
dicular to their lines of
action meet the latter in
points A, B. Let A'B' be
a line equal and parallel
to AB in the plane of the
second couple, and let the
R second couple reversed be

represented by two forces

R, R at A'B'. We can regard this couple as representing the second
couple reversed, for its moment is equal and opposite to that of
the second couple, while it is acting in the same plane as the
second couple.

We have now to show that the four forces each equal to R
acting at A, B, A', B' are in equilibrium. By construction, ABB' A' is
a parallelogram, so that C, the point of intersection of its diagonals,
is also the middle point of each diagonal.

The two parallel forces R, R acting at A, B1 may be compounded
into a single force 2 R acting at (7, the middle point of AB', and
similarly the two forces R, R at B, A' may be compounded into

COMPOSITION OF COUPLES

105

a force 2 R at C, the middle point of A'B. These two forces 2 R,
2 R are equal and act in opposite directions at the same point C.
There is therefore equilibrium, proving that two couples are equiv-
alent if they have equal moments and if the planes in which they
act are parallel.

81. The direction perpendicular to the plane in which a couple
acts is called the axis of a couple. Thus :

Two couples having the same axis and the same moment are
equivalent.

COUPLES COMPOUNDED ACCORDING TO THE PARALLELOGRAM LAW

82. A couple, as we have just seen, is determined by a quantity
(its moment) and a direction (its axis). Thus it may be fully
represented by a straight line, the direction of this line being that
of the axis, and the
length representing,
on any scale we
please, the magni-
tude of the moment
of the couple.

We shall now
show that couples
can be compounded
according to the
parallelogram law.

THEOREM. Two couples, represented in magnitude and direction
ly two lines AB, A C, will have as resultant a couple represented in
magnitude and direction ~by AD, the diagonal of the parallelogram
of which AB, A C are edges.

Let AB, AC be lines representing by their direction and magni-
tude the axes and moments of two couples. Let 8 AS' be a line
perpendicular to the plane ABC, A being its middle point. Let
us draw planes through S, S' parallel to the plane ABC, and let
the couple AB be replaced by two forces PS, P' S' in these two

FIG. 57

106 STATICS OF RIGID BODIES

planes, the lines PS, P'S' being both perpendicular to AB. In the
same way let the couple AC be replaced by two forces QS, Q'S'
in these same two planes.

The two couples have now been replaced by the four forces
PS, QS, P'S', Q'S'.

Let us complete the parallelograms PSQR, BACD, P'S'Q'R'.
Obviously these parallelograms are all similar to one another, and
corresponding lines in the first and second parallelograms are at
right angles to one another. Thus a couple represented by AD
may be replaced by forces US, R'Sf. But these two forces are
exactly equivalent to the four forces PS, QS, P'S', Q'S' to which,
as we have seen, the couples AB, AC may be reduced, and this
proves the theorem.

FORCES IN SPACE

83. When the forces acting on a body are not all in one plane,
their resultant will not in general be a single force.

THEOREM. Any system of forces acting on a rigid body can be re-
placed ~by a force acting at an arbitrarily chosen point, and a couple.
Let G be the chosen point, and let R be any force of which the
line of action does not pass through G. At G let us introduce two
equal and opposite forces, each equal to R
and parallel to the line of action of R. By
combining one of these forces with the
original force R, we get a couple, so that
the original force R can be replaced by a
force parallel and equal to the original
force but acting at G, and a couple.

Treating all the forces of the system in this way, we find that
the original system of forces may be replaced by

(a) a number of forces acting at the chosen point G\

(b) a number of couples.

The forces acting at G can be combined into a single force at G,
and the couples into a single couple, proving the result.

FORCES IN SPACE

107

84. THEOREM. Any system of forces acting on a rigid body can
be replaced by a force and a couple of which the axis is parallel to
the line of action of the force.

By the theorem just proved, the system may first be replaced
by a force acting at any point 0, and a couple. Let the force be
of amount R, having OP as its line of action, and let the couple
be of moment 6r, having OQ for its axis. If the angle POQ is
denoted by 6, we may resolve the couple into two couples :

(a) a couple of moment G cos 0,
having OP for axis;

(b) a couple of moment G sin 0,
having its axis perpendicular to OP.

The second of these couples may
be replaced by any two forces pro-
vided these are chosen so as to be
equivalent to the couple. Let us
choose the first force to be a force
— R acting along OP, i.e. the force
which will exactly neutralize the
force R which we already have acting along OP. The second
force of the couple must then be a force R acting along a line
parallel to OP but at a distance from it equal to G sin 9/R.

The system has now been replaced by

(a) forces + R, — R acting along OP;

(b) a force R parallel to OP;

The two forces (a) neutralize and we are left with a force R and
a couple G cos 0 of which the axis is parallel to the line of action
of the force. This proves the theorem.

The line of action of the force, which is now also the axis of
the couple, is called the central axis of the system of forces.
A system of forces is most simply specified by a knowledge of the
magnitude of the force and couple, and of the position and direction
of the central axis. Such a system is called a " wrench."

FIG. 59

108

STATICS OF EIGID BODIES

ILLUSTRATIVE EXAMPLES

1. Two equal uniform planks are hinged at one end, and stand with their free
ends on a smooth horizontal plane, being prevented from slipping by a rope which
is tied to each plank at the same height up. Find the tension in this rope and
the action at the hinge.

In the figure let AB, AC represent the two planks, hinged at A, and let
PQ be the rope. The forces acting on the plank AB will consist of
A (a) the action at the hinge A •

(b) the tension of the rope acting along PQ;

(d) the weight.

Of these four forces, (a) and (b) are the forces which
it is required to find. Force (c) also is at present
unknown. • Force (d), as explained in § 77, can be
regarded as a single force W, the total weight of the
plank, and since we are told that the plank is uniform,
this must be supposed to act through its middle point.

FIG. 60

There is a simple way of finding force (c), the reaction at B. Since we are
told that the contact at B is smooth, the direction of the reaction must be
vertically upwards. Let its amount be R. From symmetry, there must be
an exactly similar reaction at the foot C of the second plank. Now consider
the equilibrium of the whole system which consists of the two planks and the
rope. The only external forces which act on this system consist of

(a) the weight ;

(b) the two reactions at B and C.

If we resolve vertically, we obtain, since this
system is in equilibrium,

2 W - 2 E = 0,

so that E = W ; each reaction is Just equal to the
weight of one plank, as we might have anticipated.

Of the four forces acting on the plank AB,
the last two are now known, while the first two
remain unknown. If we take moments about A,
we shall get an equation between forces (&), (c),
and (d), and this will enable us to find the un-
known force (b), the tension.

If we denote the tension by T, and the angle
BA C by 2 6, ^ the equation obtained on taking
moments about A is /

FIG. 61

R • AB sin 6 - W - \ AB sin d - T- AP cos 6 = 0,
so that, remembering that R = W, we have

ZAP

ILLUSTKATIVE EXAMPLES

109

Also on resolving horizontally and vertically, it is evident that the action
at A must consist of a horizontal force of amount T and of direction opposite
to that of T.

2. A ring (e.g. a dinner napkin ring) stands on a table, and a gradually
increasing pressure is applied by a finger to one point on the ring. Having given
the coefficients of friction at the two contacts, examine how equilibrium will first
be broken.

Let A be the point of contact of the ring and table, and let B be the point
of contact of the ring and ringer. Let e, e' be the angles of friction at A and B
respectively. Let the line BA make an angle a with the vertical.

The forces applied to the ring from outside are

(a) the reaction at A ;

(b) the reaction at B ;

' Regarding the latter as a single force W acting along the vertical diameter
CA of the ring, we see that so long as the ring remains at rest it is in equilibrium
under the action of three forces.

Hence, by the theorem of § 52, the lines of action of the three forces must
meet in a point.

The line of action of the weight is already known to be the vertical CA,
and the line of action of the reaction at A must pass through A. Hence either

(a) the point in which the three lines of action meet must be A ; or

(/3) the reaction at A must act along CA, so that the point in which the
three lines of action meet will be some point in CA, other than A.

The second alternative may be dismissed at once. For if the reaction at A
acts along CA, this and the weight may be combined into a single force, and
there must now be equilibrium under this force and the reaction at B. This
requires that each force should vanish, i.e. there must be no pressure at B,
and the reaction at A must be just equal to the weight of the ring. This

110 STATICS OF EIGID BODIES

obviously gives a state of equilibrium — the ring is standing at rest on the table,
acted on solely by its weight — but this state of equilibrium is not the one with
which we are concerned in the present problem.

Let us now consider .the meaning of alternative («). If the three lines of
action meet in A, the reaction at B must act along BA, and this must be true
no matter how great the pressure at B. Hence the reaction at B will always
make an angle a with the normal.

If a is less than e', the angle of friction at B, this will be a possible line of
action for the reaction, and no slipping can take place at B, no matter how
great the pressure applied at B may be.

On the other hand, if a is greater than e', equilibrium is impossible, no matter
how small the pressure applied at B may be. Thus, when there is equilibrium,
the pressure at B must vanish, and we are led to the same
state of equilibrium as was reached from case (j3). As soon
as the pressure at B becomes appreciable, equilibrium is
obviously broken by slipping taking place at .B, since for
equilibrium to be maintained at B, the reaction would
have to act at an angle greater than the actual angle of
friction.

Thus the solution resolves itself into two different cases :
CASE I. If a > e', as soon as pressure is applied at Z?,
motion takes place. The ring slips at B, and consequently
rolls at A.

CASE II. If a < e', we have seen that no matter how
great a pressure is applied at JB, there can never be
FIG 63 slipping at B. It remains to examine whether there can

be slipping at A.

To settle this question, we have to determine whether the reaction at A can
ever be made to act at an angle with the vertical which is as great as the angle
of friction at A, namely e. Now the ring is acted on by three forces, the
reactions at A and .B, say EA and RB, and its weight W. The lines of action
of these forces meet in the point A , and from Lami's theorem we can connect
the magnitudes of the forces with the angles between them.

The lines of action of the three forces are represented in fig. 63. The angle
between W and BB is, as we have seen, always equal to a. Let the angle
between RA and the vertical be supposed to be 0. Then, by Lami's theorem,

EA = ^B_ = w
sin a sin 6 sin (a — 6)

The value of EA is not given, but on equating the last two fractions we have

W sin (a - 6)

— = — 5 '- = sin a cot 6 — cos a,

EB sin 6

(W\
cos a H ) cosec a.
EB/

EXAMPLES 111

This equation enables us to trace the changes in the value of the angle 6 as
EB is gradually increased. We find that when RB = 0 the value of 0 is 6 = 0,
and that as RB increases 0 increases continually, but never exceeds the value
0 = a, which is reached when Rj$ = oo. .

If e < a, the value of 6 will pass through the value e when Eg reaches a
certain value, namely

Wsine

RB = - — ; - -»
sin (a — e)

and slipping at A will take place at this point.

If c > a, the value of 8 will never reach the value e, so that slipping at A can
never occur. Thus equilibrium is never broken, and the harder we press at B
the more firmly the ring is held between the finger and the table.

We can now summarize the results which have been obtained, as follows :

If a > e', the ring rolls along the table as soon as we begin to press at B.
If a < e', there are two cases :

(a) a > e, the ring will slip at A as soon as sufficient pressure is applied ;
(6) a. < e, the ring cannot be made to move under any amount of pressure.

To make the ring shoot out from under the finger by slipping at A (in which
case it returns to the hand, as in the well-known trick), it is necessary to press
at a point on the ring at which a is greater than e, while being less than e'.
We notice that if c is greater than e', it is impossible to project the ring in this
way ; this can only be done if the contact with the finger is rougher than the
contact with the table.

GENERAL EXAMPLES

1. A pair of steps is formed by two uniform ladders each of length 12
feet and weight 20 pounds, jointed at the top, and having their points at
distances 5 feet from the ground connected by a rope. The steps stand on
a smooth horizontal plane, and a man of weight 160 pounds ascends to a
height of 9 feet on one side. Find the tension in the rope.

2. A heavy uniform rod is supported by two strings of lengths a, b.
The upper ends of the strings are tied to the same point, the lower ends
being tied to the two ends of the rod. Show that the tensions of the strings
are proportional to a and b respectively.

3. Two small fixed pegs are in a line inclined at an angle 8 to the
horizon. A rough thin rod passes under the lower and rests on the higher,
this latter being lower than the center of gravity of the rod. The distances
of the center of gravity from the two pegs are a and b respectively, and the
coefficient of friction is /*. Show that if the rod is on the point of motion,

b + a

112 STATICS OF EIGID BODIES

4. Two heavy uniform rods have their ends connected by two light
strings, and the whole system is suspended by the middle point of one rod.
Prove that in equilibrium either the rods or the strings are parallel.

5. Two rods AB, CD lying on a smooth table are connected by stretched
strings AC, BD. If the system is kept in equilibrium, by forces acting at
the middle points of the rods, prove that if the strings are not parallel

(a) the rods must be parallel ;

(&) the tensions must be proportional to the strings.

6. ABCD is a parallelogram and E is the intersection of the diagonals
AC, BD. Show that parallel forces 7, 5, 16, 4 at A, B, C, D respectively
are equivalent to other parallel forces, 8 at the middle point of CD, 10 at
the middle point of BC, and 14 at E.

7. A solid cube is placed on a rough inclined plane of angle a with two
edges of its base along lines of greatest slope. The angle of friction is e.
Prove that if a > 45° it will at once topple over, while if e < a < 45° it will
slide down the plane. If a is less than either e or 45°, find the friction
brought into action.

8. A uniform rod of length 2/ and weight W rests over a smooth peg
at distance h (< Z) from a smooth vertical wall at an angle 6 with the hori-
zontal, its lower end pressing against the wall, and its upper being held by a
vertical string. Find the tension of the string and show that it vanishes if

9. Two equal uniform spheres, each of weight W and radius a, rest in
a smooth hemispherical bowl of radius b. Find the pressure between the
two spheres and also the pressure of each on the bowl.

10. A uniform rod rests with its two ends on smooth inclined planes,
inclined to the horizontal at angles a and p. Find the inclination of the
rod to the horizontal.

11. In the last question a weight equal to that of the rod is clamped to it.
At what point must it be clamped in order that the rod may rest horizontally?

12. A uniform circular ring of weight W has a bead of weight w fixed
on it and hangs on a rough peg. Show that if sin e > , then the ring

can rest without slipping, whatever point of it rests on the peg, e being
the angle of friction.

13. A pentagon ABCDE, formed of equal uniform heavy rods connected
by smooth joints at their ends, is supported symmetrically in a vertical
plane with A uppermost, and AB, AE in contact with two smooth pegs
in the same horizontal line. Prove that if the pentagon is regular, the
pegs must divide AB and AE each in the ratio

1 + sin j1^ TT : 3 sin T^ TT.

EXAMPLES 113

14. A uniform beam of length / leans against the horizontal rim of a
hemispherical bowl of radius a, with its lower end resting upon the smooth
concave surface. Find its inclination to the vertical.

15. A bowl in the shape of a paraboloid of revolution is placed with its
axis vertical. A uniform rod rests on a peg at the focus and has its lower
end resting on the inner surface. Both contacts are perfectly smooth.
Find the inclination of the rod to the vertical.

16. A uniform beam of weight W rests against a vertical wall and a
horizontal plane with which it makes the angle a. Both contacts are
perfectly smooth. The lower end of the beam is attached by a string to
the foot of the wall. Find the tension of the string.

17. One end of a straight uniform heavy rod rests on a rough horizontal
plane, the other end being connected with a fixed point by a string. If
0, 0, \f/ be the inclinations of the string, the rod, and the total reaction of
the horizontal plane respectively to the vertical, show that

cot 6 ± 2 cot 0 - cot ^ = 0.

18. Two uniform rods AB, BC of the same material but of different
lengths are jointed freely at B and fixed to a vertical wall at A and C. Show
that the direction of the reaction at B bisects the angle ABC.

19. A uniform regular-hexagonal board ABCDEF of given weight W
is supported in a horizontal position on three pegs, placed at the corners
A , B and the middle point of DE. Find the pressures on the pegs.

20. Two spheres of radii a, b and weights W, W respectively are
suspended freely by strings of lengths /, /' respectively from the same hook
in the ceiling. If V > I + 2 a, show that the angle which the first string
makes with the vertical is

. Wa

21. A uniform rod hangs by two strings of lengths I, I' fastened to its
ends and to two hooks in the same horizontal line at the distance a. If
the strings cross one another and make the respective angles a, a', with
the horizontal, show that when the rod is in equilibrium

sin (a + a') (I' cos a' — I cos a) = a sin (a — a').

22. A uniform plank of length 2 b rests with one end on a rough hori-
zontal plane, touches a smooth fixed cylinder of radius a lying on the plane,
and makes an angle 2 a with the plane, the angle of friction being e.
Show that equilibrium is possible if

a sin e > b tan a cos 2 a sin (2 a + e).

23. Two equal and similar isosceles wedges, each of weight W and
vertical angle 2 a, are placed side by side with their bases on a rough

114 STATICS OF KIGID BODIES

horizontal table so as to be just in contact along an edge. A smooth
sphere of weight w and radius r is supported between them, being in con-
tact with a face of each. Prove that for equilibrium it is necessary that

wcota I W\

> r < 2 a sin a tan a [ H -- ,
\ wl

where /* denotes the coefficient of friction and 2 a is the length of either
base.

24. A seesaw consists of a plank of weight w laid across a fixed rough
log whose shape is that of a horizontal circular cylinder. The inclination
to the horizontal at which it balances is increased to a when loads W, W
are placed at the lower and higher ends respectively; and the inclination
is reduced to /9 when these loads are interchanged. Show that the inclina-
tion of the plane when unloaded is

w) (W'a -

w' being the weight which, placed at the higher end, would balance the
plank horizontally.

25. A chain is formed of 2 n exactly similar links, the contacts between
consecutive links being perfectly smooth. The two end links can slide on
a horizontal wire, the contact here being rough, coefficient of friction fj..
Show that in the limiting position of equilibrium, the inclination of either
of the upper links to the vertical is

tan-1 ^- •

, 2n -1

26. Two equal circular disks of radius r, with smooth edges, are placed
on their flat sides in the corner between two smooth vertical planes inclined
at an angle 2 a, and touch each other in the line bisecting the angle.
Prove that the smallest disk which can be pressed between them without
causing them to separate is one of radius r(sec a — 1).

27. How is the result of the last question modified if all the contacts
are rough, the angle of friction at each being e?

28. Two uniform ladders are jointed at one end and stand with their
other ends on a rough horizontal plane. A man whose weight is equal to
that of one of the ladders ascends one of them. Prove that the other will
slip first.

If it begins to slip when he has ascended a distance x, prove that the

coefficient of friction is — — — tan a. a being the length of each ladder
2a + x

and a the angle each makes with the vertical.

29. A weightless ladder rests against a smooth cube of weight Wt
standing on smooth ground, with the foot of the ladder tied to the middle

EXAMPLES 115

point of one of the lowest edges of the cube ; a man of weight w ascends
the ladder. Prove that, if the ladder projects above the top of the cube,
the cube will tilt before he reaches the top of the cube unless

W>2w cos a (sin a — cos a),
where a is the inclination of the ladder to the horizontal.

30. Four equal spheres rest in contact at the bottom of a smooth
spherical bowl, their centers being in a horizontal plane. Show that if
another equal sphere be placed upon them the lower spheres will separate
if the radius of the bowl be greater than (2 Vl3 + 1) times the radius of
a sphere.

31. Three equal spheres rest in contact on a smooth horizontal plane,
so that their centers form an equilateral triangle, and are bound together
by a fine string passing around them on the level with their centers. If
another equal sphere be placed symmetrically on them, show that the

tension of the string is increased by — — to, where w is the weight of the
upper sphere. ° "

32. A right circular cone of vertical angle 2 a rests with its base on a
rough horizontal plane. A string is attached to the vertex, and is pulled
in a horizontal direction with a gradually increasing force. Find in what
way equilibrium will first be broken.

33. A heavy particle is placed on a rough inclined plane whose inclina-
tion is exactly equal to the angle of friction. A thread is attached to the
particle and is passed through a hole in the plane which is lower than the
particle, but is not in the line of greatest slope through it. Show that if
the thread be gradually drawn through the hole, the particle will describe
a straight line and a semicircle in succession.

34. A uniform cubical block of weight W rests with one edge horizontal
on a rough inclined plane, and against the block rests a rough sphere of
weight W of radius less than an edge of the cube. The inclination of the
plane is gradually increased. Examine the different ways in which equi-
librium may be broken, and determine which will actually occur in a
given case.

35. A rough uniform rod is placed on a horizontal plane and is acted
on, at one of the points of trisection of its length, by a horizontal force in
a direction perpendicular to its length. Find about what point the rod
will begin to turn.

36. A heavy bar AB is suspended by two equal strings of length /,
which are originally parallel. Find the couple which must be applied to
the bar to keep it at rest after it has been twisted through an angle 0 in
the horizontal plane.

116 STATICS OF EIGID BODIES

37. The line of hinges of a door is inclined at an angle a to the vertical.
Show that the couple required to keep it in a position inclined at an angle
/3 to that of equilibrium is proportional to sin a sin ft".

38. Show that any system of forces acting on a rigid body can be
reduced to two equal forces equally inclined to the central axis.

39. Prove that the central axis of two forces P and Q intersects the
shortest distance c between their lines of action and divides it in the ratio

Q(Q + Pcostf) :P(P + Qcos0),
B being the angle between their directions. Also prove that the moment

of the principal couple is

. cPQ sin $

VP2 + Q2 + O PQ COS 6

40. Prove that the axis of the resultant of two given wrenches (721} HJ
and (^2' ^2)' ^^e axes °f which are inclined to each other at an angle 0,
intersects the shortest distance, 2c, between their axes at a point the dis-
tance of which from the middle point is

(R? - Rf) c +
R* +

CHAPTER VI

CENTER OF GRAVITY

85. As we have seen, the action of gravity on a system of
masses may be represented by a system of parallel forces, these
forces consisting of a force acting on each particle equal to the
weight of the particle, its direction being vertically downwards.
By the rules explained in the last chapter, these forces may be
compounded into a single force. The magnitude of this force is
the sum of all the component

forces, and is therefore the total
weight of the body, while the
direction of the force, being par-
allel to the component forces,
is itself vertically downwards.
The problem discussed in the
present chapter is that of de-
termining the position of the
line of action of this force.

86. Let the particles be of

masses ml} m2, ••-. Let rectangular axes be taken, the axis of z
being vertical, and let the coordinates of the first particle be
xi> 2/i' zi> the coordinates of the second be #2, yz, z2, and so on.

The weight of the first particle is m^g, and its line of action
cuts the plane Oxy in a point of which the coordinates are
xlt y13 0. Hence the moment of the force about the axis Oy is
mlgxr Let the line of action of the resultant cut the plane Oxy
in the point x, y, 0. Then the moment of the resultant about the
axis Oy is C^m^\gx, where Vmx is the sum of the masses of all
the particles.

117

FIG. 64

118 CENTER OF GRAVITY

Since the moment of the resultant is equal to the sum of the
moments of the separate forces, we must have

so that x =

Similarly y =

These equations determine the coordinates x, y of the point in
which the line of action of the resultant meets the plane Oxy.

We have, however, seen that the coordinates of the centroid o
masses ml at xlt ylt zv ra2 at x2, yz) z2, etc., are

so that the point in which a vertical through the centroid wil
meet the plane Oxy must be

> m:v.

i.e. the point must be the point x, y, 0 in which the line of action
of the resultant force meets the plane Oxy. Thus

The line of action of the resultant force of gravity is the ver-
tical line through the centroid of the particles.

For this reason the centroid of a number of points, weighted
according to the masses of the particles which occupy these points,
is cal^d the center of gravity of the particles. The effect of
gravity acting on a rigid body is, as we have now seen, repre-
sented by a single force acting vertically downwards through the
center of gravity of the body, the amount of the force being equal
to the total weight of the body. The action of gravity is, accord-
ingly, the same as if the whole mass of the body were concentrat
in a single particle placed at the center of gravity.

SYSTEM OF MASSES 119

87. It is clear that if we suspend a rigid body or system of
bodies by a string, the center of gravity must be vertically below
the string. For all the forces acting on the system reduce to two,
— the tension of the string and the weight acting at the center
of gravity, — and in equilibrium these two must act along the
same line.

In the same way it will be seen that if a body is placed on a
point in such a way as to balance in equilibrium on this point,
then the center of gravity must be vertically above the point.

88. A few simple instances of the position of the center of
gravity have been mentioned in § 77. These were as follows :

(a) the center of gravity of a uniform rod is at its middle point ;

(b) the center of gravity of a uniform circular disk, circular
ring, or sphere is at the center;

89. It is easy to find the center of gravity of a system of bodies
when the center of gravity of each of the component parts is known.
For, regarding the weight of each of the bodies as a single force
acting through its center of gravity, we have a number of parallel
forces in action, and on compounding these according to the rules
already explained, the line of action of the resultant determines
the line along which the total weight will act. Thus the center
of gravity of the whole system of bodies will be the centroid of
the centers of gravity of the separate bodies, weighted
according to the masses of these bodies.

90. For instance, let us suppose we require to find the
center of gravity of a pendulum which consists of a wire,
of length / and weight w, to which is affixed a circular bob
of weight W, the center of the circle being at a distance a
from the end of the wire. Let AB be the wire, C the center
of the bob, and D the middle point of the wire. The center
of gravity of the wire will be at D and that of the bob at C, ]

so that the center of gravity of the whole will be at the centroid

120 CENTER OF GRAVITY

of the points D and C, these being weighted in the ratio w : W.
Denoting this center of gravity by G, we have, from the formula

on treating the line ADCB as the axis of x and taking A as origin,

_ W-AC + w-AD

W + w
_W(l-a) + \ wl

W+w

EXAMPLES

1. Weights of 3 pounds are placed at each of three corners of a square, and
a weight of 5 pounds at the fourth corner. Find their center of gravity.

2. From one corner of a cardboard square of edge 6 inches, a square of edge
3 inches is cut. Find the center of gravity of the remainder.

3. A thin rod of weight 6 ounces and length 6 inches is nailed on to a circle
of cardboard of weight 6 ounces and radius 6 inches so that its two ends are on
the circumference of the circle. Find the center of gravity of the whole.

4. A bicycle wheel of diameter 26 inches weighs 3 pounds. Each spoke is of
length 11 inches, and starts from the hub at a distance of i inch from the cen-
tral axis of the wheel. If one spoke is taken out, find the center of gravity of
the wheel.

5. A hammer has for handle a wooden cylinder, length 8 inches, radius
| inch, weight 8 ounces, and for head an iron cylinder, from which a hollow is
cut into which the handle exactly fits, the handle coming through so as to be
exactly flush with the iron. The head is of length 3 inches, radius 1| inch, and
weight 3 pounds. Find the approximate position of the center of gravity.

6. A box, without lid, is made of 1-inch wood so as to have internal dimen-
sions 12 x 12 x 12 inches. Find the position of its center of gravity.

7. A uniform thin rod 28 inches in length is bent so that the two parts, of
lengths 12 and 16 inches, are at right angles to one another. Find the center of
gravity.

8. A uniform wire is bent into the form of a triangle. Show that the center
of gravity of the wire coincides with the center of the circle inscribed in the
triangle formed by joining the middle points of the sides.

9. A T-square is made of cedar of uniform density, the crosspiece being of
dimensions 6 x 2 x \ inches, and the arm being of dimensions 8 x 1| x } inches.
The crosspiece is cut away so that the under surface of the instrument is
plane. Find the position of the center of gravity of the whole.

TRIANGLE 121

10. Three beads of weights TFa, TF&, Wc are placed on a circular wire, and
when the beads are at the points A, B, C on the circle, the center of gravity
of the whole is found to coincide with 0, the center of the circle. Show that

Wg = Wb WC

sinBOC sinCOA sin A OB

CENTER OF GRAVITY OF A LAMINA

91. It is often of importance to be able to find the position
of the center of gravity of a lamina, i.e. of a thin, plane shell of
uniform thickness and density, such, for instance, as is obtained
by cutting a figure out of a sheet of cardboard.

92. Center of gravity of a triangle. Let ABC represent a
triangular lamina of which it is required to find the position of
the center of gravity. Let us imagine the triangle divided by
lines parallel to the base B C into a very

great number of infinitely narrow strips.
Let pq be any single strip. Since, by
hypothesis, we may regard this strip as
of vanishingly small width and thick-
ness, we may treat it as a thin uniform
rod. The center of gravity of a thin B
uniform rod is at its middle point, so
that the weight of the strip pq may be supposed to act at r, the
middle point of pq.

The weights of the other strips may be treated in the same
way, so that the weight of the whole triangle may be replaced by
the weights of a system of particles situated at the middle points
of these strips.

Now if D is the middle point of the base BC, the middle points
of all the strips lie in the line AD. Thus the weight of the tri-
angle is replaced by the weights of a number of particles, all of
which are situated in the line AD. It follows that the center of
gravity of the whole triangle must lie in the line AD.

We might equally well have supposed the triangle divided into
strips parallel to the side AC. We should then have found that

122 CENTER OF GRAVITY

the center of gravity must lie in the line BE, where E is the
middle point of AC.

These two results fully fix the position of the center of gravity ;
it must be at the intersection of the lines AD, BE.

Join DE. Then the triangles DCE, BCA are two similar
triangles, the former being just half the size of the latter. Thus
DE must be parallel to AB, and of half the length of AB.

It now follows that DGE, AGB are similar triangles, of which

the former is half the size of the latter.
Hence GD is half of AG.

Thus G divides AD in the ratio 2 : 1.
If we join C to F, the middle point of
AB, we can in the same way show that
CF must divide AD in the ratio 2:1.
Thus CF must also pass through G.
FIG 67 The three lines AD, BE, CF, which

join the vertices of the triangle to the

middle points of the opposite sides, are called the medians of the
triangle. We have shown that the three medians meet in the same
point G, and that this point is the center of gravity of the triangle.
We have also shown that the center of gravity divides any median
in the ratio 2:1, i.e. that it is one third of the way up the median,
starting from the base.

93. Center of gravity of any polygon. The center of gravity
of any rectilinear polygon can be found by dividing it up into
triangles and replacing each triangle by a particle at its center of
gravity.

EXAMPLES

1. Show that the center of gravity of a triangle coincides with that of three
equal particles placed at its angular points.

2. Prove that if the center of gravity of a triangle coincides with its ortho-
center the triangle is equilateral.

3. A cardboard square is bent along a diagonal until the two parts are at
right angles. Find the position of its center of gravity.

4. A quarter of a triangular lamina is cut off by a line parallel to its base.
Where is the center of gravity of the remainder?

EOD OF VARYING DENSITY 123

5. A right-angled isosceles triangle is cut out from a lamina in the shape of
an equilateral triangle so as to have the same base as the original triangle. Find
the center of gravity of the V-shaped piece left over.

6. The center of gravity of a quadrilateral lies on one of its diagonals. Show
that this diagonal bisects the other diagonal.

CENTERS OF GRAVITY OBTAINED BY INTEGRATION

94. Center of gravity of a rod of varying density. Let AB be

a rod of which the weight per unit length varies from point to
point, and let p denote the weight per unit length at any point.

Let P, Q be two adjacent points, the distances of P, Q from the
point A being x and x + dx respectively. Then the length PQ is
dx, and its mass is pdx, where p is the PQ

mass per unit length at this point. ' '

When dx is made vanishingly small, FlG- 68

the distance of the center of gravity of PQ from A may be taken
to be x. Hence if x denotes the distance of the center of gravity
of the whole rod from A,

X =

where m is the mass of any element such as PQ, and the summa-
tion is taken over all the particles of which the rod is formed.
Putting m = pdx, this becomes

x —

or, in the notation of the integral calculus,

I pxdx

x =

fpdx

(28)

where the integration extends in each case over the whole rod.
The variable p will be a function of x, and the integrations cannot
be performed until the exact form of this function is known.

124 CENTER OF GRAVITY

95. To take a definite instance, let us suppose that the density
increases uniformly from one end to the other. Let the density at
A be 0, and that at B be k. If the rod is of length a, the density

at a distance x from A will be k I - \ Thus we must put

in formula (28), and so obtain

x
kl- \xdx

(H-

J \«

where the integration is from x = 0 to x = a. Dividing numerator

k
and denominator by -> we obtain

. jf

xdx

showing that the center of gravity is two thirds of the way along
the rod.

96. We can use this result to obtain the center of gravity of a
triangle. As in § 92, we divide the triangle into parallel strips,
and replace each strip by a particle at its middle point. The mass
of each particle must be that of the strip which it replaces, and
this is jointly proportional to the width and the length of the
strip. If x is the distance of any particle from A measured along
the median AD, the width of a strip is proportional to dx, the
length intercepted on the median, while the length of the strip is
proportional to x, the distance from a. Thus p dx must be simply

CIECULAR AEG

125

proportional to x dx, and as we have just found, this at once leads

to the result _ 9

x = I a,

where a is the length of the median. This is exactly the result
previously obtained.

97. Center of gravity of a circular arc. The same method can
be used to find the center of gravity of a wire bent into the form
of a circular arc PQ. Let 0 be the center of the circle, and A the
middle point of the arc, and let the whole arc subtend an angle
2 a; at the center. Consider a small element cd of the half PA of
the wire. Let the angle dOA be 6,
and cOA be 6 + dO, so that the
element subtends the angle dO at
the center. If a is the radius of
the circle, the length of this ele-
ment is a d6, so that if w is the
mass of the wire per unit length,
the mass of the element will be
wa d6. This and the similar ele-
ment c'd' in the half AQ of the
wire form a pair of equal particles
equidistant from the central line
OA. They may be replaced by a
single particle of mass 2 wa dO at their center of gravity. This
center of gravity is in OA, at the point at which the line joining
the two elements meets OA, and hence at a distance a cos 0. from 0.
Denoting this by x, and the mass 2 wa d6 by ra, we have, for the
distance x of the center of gravity of the whole wire from 0,

x =

IwadB

126 CENTER OF GRAVITY

where the integration is from 6 = 0 to 6 — a. Simplifying, we find

•r

i/0 =

f>6= a

Je = o

a sn a
a

(29)

giving the position of the center of gravity.

When a is very small, sin a and a become equal, so that for
very small values of a, formula (29) reduces to x = a, as it ought.
This simply expresses that as the curvature of the arc decreases,
the center of gravity approximates more and more closely to the
middle point of the arc. Finally, when a = 0, the arc becomes a
straight rod, and the center of gravity is, of course, found to be
exactly at the middle point.

For an arc bent into a semicircle, we take a = — > and obtain

x =

= — = .6366 a.

98. The center of gravity of a cir-
cular arc PQ can be found in an in-
teresting manner, without the use of
the integral calculus.

From symmetry it is clear that the
center of gravity of the arc AP must
lie in the radius which bisects the
angle A OP. Let p be this center of
gravity, and let q be the center of
gravity of the arc AQ. Then the
center of gravity or the whole arc PQ
must be N, the middle point of pq.

Now since the angle pON = \ a,
we have

ON = Op cos \ a .

FIG. 70

CIRCULAR ARC 127

This relation shows that

(the distance of c. g. of arc 2 a from center)

= cos — x (the distance of e.g. of arc a from center).
Similarly

(the distance of e.g. of arc a from center)

= cos — x (the distance of c. g. of arc — from center),
and so on. Continuing in this way, and substituting, we obtain
(the distance of c. g. of arc 2 a from center)

= cos — • cos — • cos — • • • cos — ^— -
2 4 8 2n + 1

x (the distance of c. g. of arc -^ from center).

If we make n very great, the value of — becomes zero. Thus the dis-
tance of the c. g. of an arc 7^ from the center becomes equal to a, the radius
of the circle. Making n infinite, we have

(the distance of c. g. of arc 2 a from center)

= a cos ^ cos j cos ^ ... to infinity.

a sin or
Now cos — = - ,

2sinf

...
so that

Making n infinite, the value of sin -^ becomes identical with -2- , so
that 2n sin ^ becomes identical with or, and we have

a a a • ^ • , sin a

cos — cos — cos — • • • to infinity = - >
248 a

so that the distance of the c. g. of the arc 2 a from the center is found to

, sin a

be a - > as before.

128

CENTER OF GRAVITY

99. Center of gravity of a segment of a circle. Suppose next
that we require to find the center of gravity of a segment PAQN

of a circle, cut off by a chord PNQ,

which subtends an angle 2 <z at
the center 0 of the circle. Let us
divide the whole segment into thin
strips parallel to the chord, and
let cc'dd' in fig. 71 be a typical
strip bounded by chords cc1 ', ddf.
Let the angle cOA be 0, and let
dOA be 0 + dO. Then the width
of the strip is cd sin 0 or a sin 0 e£0,
while its length is 2 c% or 2 a ski 0.
Thus the area is 2 a sin2 0 e£0. Its
mass may be supposed to be all
F ^ concentrated at n, of which the

distance from 0 is a cos 0.

Thus if # is the distance from 0 of the center of gravity of the
whole segment, we shall have

h'
fd'

f (a cos 0) (2 a sin2

f(2asi

and the integration has to be taken from 6 = 0 to 0 = a. Simpli-
fying, we have

f"sin2

J 0

0 cos 0^0

f sin2
Jo

l- (a — sin a cos a)

3 a — sin a cos a

SECTOR OF A CIRCLE

129

We find, on putting a = ~» that the center of gravity of a semi-

4
circle is at a distance - — a from the center.

O 7T

100. Center of gravity of a sector of a circle. The center of
gravity of a sector of a circle can be found by regarding the sector
as made up of a triangle and a segment. The center of gravity of
the triangle and of the seg-
ment both being known, it
is easy to find the center of
gravity of the whole figure.
A simpler way is the
following: We can divide
the sector by a series of
radii into a great number o<
of very narrow triangles.
The weight of each triangle
may be replaced by the
weight of a particle placed
at its center of gravity.
Now, in the limit, when the
triangles become of infini-
tesimal width, the center of
gravity of each is on its median at a distance from the center of
the circle equal to f a, where a is the radius of the circle. Thus
all the particles lie on a circle of radius f a.

The weight of any particle must be equal to the weight of the
triangle OPQ which it replaces. It must, therefore, be propor-
tional to the base PQ of the triangle, and this again is proportional

P to pq, the piece of the circle of
radius J a which is inclosed by
the triangle. Thus the weight of
the particle which has to be
placed in the small element pq of this circle is proportional to the
length pq. On passing to the limit, and making the number of
triangles infinite, we find that the string of particles may be

FIG. 72

FIG. 73

130

CENTER OF GRAVITY

replaced by a wire of uniform density. The center of gravity of
this wire has already been determined. If 2 a is the angle of the
wire, the center of gravity lies on the radius to the middle point

of the wire at a distance —a— - from the center.

o a

Thus the center of gravity of the original sector of a circle of
radius a and angle 2 a is found to lie on the central radius of the

sector at a distance - a from the center.

3 a

101. Center of gravity of a spherical cap. The piece cut off

from a spherical shell by a plane is called a spherical cap.

The center of gravity of a spherical cap
cut from a uniform shell can easily be
found by the methods already explained.
Let PQ be the spherical cap, 0 being
the center of the sphere from which it
is cut. Let OE be the radius perpen-
dicular to the plane PQ by which the
cap is bounded, and let a denote the
radius of the sphere.
Any plane parallel to PQ will cut the sphere in a circle of

which the center will lie on OE. Hence by taking a great num-
ber of planes parallel to PQ, we can

divide' the spherical cap into a number

of narrow circular rings, each having its

center on the line OE. Let us consider a

single circular ring cut off by the planes

A a A', BIB1. Let the angles AOE, BOE

be equal to 6 and 0 + dO respectively, so

that the ring itself subtends an angle dd

at the center. The width AB of the ring is

a d6. Its circumference may, in the limit,

be supposed equal to the circumference

of the circle AaA'. Since Aa = a sin 6,

this circumference is 2 TTO, sin 6. Thus the ring under consideration

FIG. 74

SPHERICAL CAP AXD BELT

131

may be regarded as a narrow strip of length 2 IT a sin 6 and of
width a d6. Its area is accordingly 2 TTO? sin 0 dO.

When dO is made very small, the arc BA may be regarded as a

straight line of length a dd, making an angle — — 6 with OE.

Thus

the length of la, the projection of BA on OE, is add cos •- — 6
or a sin 6 d6. The area of the ring BA is now seen to be

= 2 Tra2 sin
— 2 Tra • 5a.

0 dd

Thus the mass of the ring is the same as the mass of the ele-
ment ba of a rod OE, if this rod is of uniform density such that
its mass per unit length is that of an area 2 Tra of the shell. The
center of gravity of the ring we have been considering clearly lies
on the axis OE, so that in finding the center of gravity of the
spherical cap this ring may obviously be replaced by the element
la of this rod.

In the same way each small ring may be replaced by the cor-
responding element of the rod. Thus the whole cap may be
replaced by the length rE of the rod (fig. 74) which is inter-
cepted between the boundary-plane PQ and the sphere. Since
the rod is uniform, the center of gravity of the portion rE of
the rod is at its middle point. This point is therefore the center
of gravity of the spherical cap.

102. Center of gravity of a belt cut from a
spherical shell by two parallel planes. In the
same way we can find the center of gravity of
the belt cut off from a uniform spherical shell
by two parallel planes. In fig. 76 let PQ, P'Q'
be the two planes. Then we can divide the
belt into narrow rings by planes parallel to
PQ. Each ring, as before, may be replaced by
the corresponding element of a uniform rod
along the axis OE, so that the whole belt may
be replaced by -the portion rr' of this rod, the
portion intercepted between the two planes PQ,
Pr Qr. The center of gravity is now seen to be the middle point of rrf.

FIG. 76

132

CENTER OF GRAVITY

CENTEK OF GRAVITY OF A SOLID

103. Center of gravity of a pyramid on a plane base. Let a

pyramid be formed having any plane figure OPQR as base and

any point A as vertex. We can find the center of gravity of a

homogeneous pyramid by dividing it into thin layers parallel to

its base, by a series of parallel planes.

Let opqr be any such layer, this layer being regarded as an

infinitely thin lamina. Let G be the center of gravity of a uni-
form lamina coinciding with
the base OPQR, and let the
line AG meet the lamina opqr
in g. Then, from the geometry
of similar figures, it is clear
that g occupies a position in
the lamina opqr which corre-
sponds exactly with that occu-
pied by the point G in the
lamina OPQR. Thus g will be
the center of gravity of the
lamina opqr. The mass of this
lamina may, accordingly, be

replaced by the mass of a single particle at g.

In the same way each of the laminas into which we are sup-
posing the pyramid to be divided may be

replaced by a single particle at the point

at which the lamina intersects the line AG.

Thus the whole pyramid may be supposed

replaced by a series of particles lying along

AG. These form a rod of varying density,

and the center of gravity of the pyramid

will coincide with that of this rod.

The center of gravity of the rod may be

found by the method already explained in

§ 94. Consider the lamina which lies between two adjacent

FIG. 77

FIG. 78

PYEAMID 133

parallel planes meeting AG in g, g' respectively. Let Ag = x and
Agf = x + dx, so that the lamina intercepts a length dx on ^4 G.

Let 0 be the angle between AG and the perpendicular from A
on to the base of the lamina. Then the thickness of the lamina

= gg1 cos 6 = dx cos 9.

If S is the area of the base of the pyramid, the area of the
lamina under discussion is

for the areas of the different laminas are proportional to the squares
of their linear dimensions. Thus the volume of the lamina we are
considering 2

= - - S dx cos 6.

If this is to be replaced by a particle occupying the length dx
of the rod AG, the density of the rod must be

AG2

Thus the rod AG must be of a density which varies as the
square of the distance (x) from the end (A).

The distance of the center of gravity of this rod from A is now,
by the formula of § 94,

C°

I pxdx

x =

pdx
A

/"* "

I x3dx
JA

134

CENTER OF GRAVITY

Thus the center of gravity of the pyramid is in the line AG,
three quarters of the way down from A.

104. Center of gravity of the sector of a sphere. We can now
find the center of gravity of the sector of a sphere, — the volume
cut out of a solid sphere by a right circular cone having its ver-
tex at the center of the sphere. To do this we divide the base
PQ of the sector into a number of small elements of area, and
then divide the volume of the sector into a number of pyra-
mids of small cross section by taking these elements of area as
bases and joining them to the common vertex 0. These pyra-
mids are all of the same
height, so that their masses
are proportional to their
bases. The center of grav-
ity of each pyramid is three
quarters of the distance
down from 0 to its base,
and is, therefore, at a dis-
tance from 0 equal to three
quarters of the radius of
the sphere. Thus, if we con-
struct a second sphere hav-
ing 0 as its center and of
radius equal to three quar-
ters of the radius of the original sphere, the center of gravity
of each small pyramid will lie on this new sphere. Each pyra-
mid may be replaced by a particle at its center of gravity, so
that the whole spherical sector may be replaced by a series of
particles lying on this sphere and forming the spherical cap peq
(fig- 79). '

The mass of each pyramid is proportional to the base, and this
again is proportional to the part of the spherical shell peq which
is intercepted by the pyramid. Thus the spherical shell peq which
is to replace the original volume must be supposed to be of uni-
form density.

FIG. 79

SECTOE OF A SPHERE

135

The sector of a sphere OPQ has now been replaced by the
uniform spherical shell pq, and the center of gravity of this shell
is known to be G, the middle point of re in fig. 79. This point G
is, accordingly, the center of gravity required.

If the semivertical angle of the cone by which the sector is
bounded is a, and if a is the radius of the sphere, we have

Oe = J a, Or = J a cos a,
so that OG = f a (1 + cos a).

In particular, if a = — > the sector becomes a hemisphere, and

Thus the center of gravity of a hemisphere is three eighths of
the way along the radius which is perpendicular to its base.

CENTEK OF GRAVITY OF AREAS AND VOLUMES OBTAINED BY
DIRECT INTEGRATION

105. Center of gravity of a lamina. To find the center of
gravity of a lamina of any shape by integration, we take any con-
venient set of axes Ox, Oy in the plane of the lamina, and imagine
the lamina divided into small elements
by two series of lines, one parallel to the
axis Ox, and the other parallel to the
axis Oy.

Consider the small rectangular element
for which the values of x for the two
edges parallel to Oy are x and x -f- dx,
and the values of y for the two other
edges are y and y + dy. The area of
this element is dxdy, so that if p is the mass of the lamina per unit
area at this point, the mass of the element will be p dxdy. More-
over, when dx, dy are made vanishingly small in the limit, the
mass may be treated as a particle. Thus the whole mass of the
lamina may be regarded as the masses of a number of particles.

FIG. 80

136 CENTER OF GRAVITY

In § 86 we obtained for the center of gravity of a number of
particles the formulae

x — ~\ ' y —

(30)

In the present instance these become

I I px dxdy I I py dxdy

x = — - — y = — - ->
j (p dxdy \\p dxdy

the sign of summation being replaced by an integration which is
to extend over the whole area of the lamina.

If the lamina is uniform, the value of p is constant, so that

/ / px dxdy = p I lx dxdy,
and so on, and on dividing throughout by p, the formulae reduce to

I lx dxdy I \y dxdy

-X — JA. _ , y^lLl __

I I dxdy I I dxdy

106. Center of gravity of a solid. To find the center of gravity
of a solid we divide it into small solid elements by three systems
of planes parallel to the three coordinate planes. The volume of
any small element is then dxdydz, and its mass is p dxdydz. The
formulas of § 86 now give the coordinates of the center of gravity
in the form

I I I px dxdydz 1 1 1 ?& dxdydz

_ = JJJ - _? y = JJJ -, etc. (31)

I I ip dxdydz I / I p dxdydz

INTEGRATION FORMULA 137

If the solid is homogeneous, p is constant, and the formulae
become

z
etc.

I I \xdxdydz I I \ydxdyd

•% = JJJ _ , y = JJJ

i I \dxdydz I I \dxdydz

107. Use of polar coordinates. Any other system of coordinates
can, of course, be used for finding a center of gravity by integra-
tion. The only coordinates besides Cartesians which are of much
use for this purpose are polar coordinates.

We can find the center of gravity of a lamina in polar coordinates
by supposing the Cartesian coordinates xy y connected with the
polar coordinates r, 6 by the usual transformation

x = r cos 6, y = r sin 6.
Formulae (31) then become

rcosfl

r sin0 =

CCp (r cos 6) (r drd0) CCpr2 cos 0 drdd
CCp(rdrdO) CCprdrdO

CCp (r sin 6} (r drdO) CCpr* sin 0 drd0
CCp (r drdd) CCpr drdd

in which r, 0 are the polar coordinates of the center of gravity.
On dividing corresponding sides of these equations, we can obtain
an equation giving the 0 coordinate alone, namely

CCpr*s
JJ

sm0drd0

cos 0 drdd

138 CENTER OF GRAVITY

Similarly we can find the center of gravity of a solid in three-
dimensional polars by supposing the polar coordinates r, 0, <f> con-
nected with xy y, z by the usual transformation

x = r sin 6 cos $, y = r sin 6 sin <£, z = r cos 6.
Using this transformation, the first of formulae (31) becomes

CCCp (r sin 0 cos 0) (r2 sin 6 drd0d<f>)
r sin 6 cos (f> =

CCCp(r2sm0drd0d<t>)
I I I prs sin2 0 cos $ di

(32)

pr2 sin 0 drd0d(f>
while similarly we have, from the remaining two formulae,

fffpr* sin2 0 sin <£ drdOdj
rsni<9sin0=^^_ _,

/ / Ipr2 sin 6> drdOdj*

sin ^ cos 0 drd0d<f>

Pr2s

(33)

(34)

108. An exactly similar method will lead to formulae giving
the position of the center of gravity in any system of coordinates.

The methods which have already been employed, or a combina-
tion of them, will suffice to determine any center of gravity. As
illustrations of the use and combination of these methods, we
shall find the center of gravity of the same solid figure in three
different ways.

ILLUSTEATIVE EXAMPLE

139

ILLUSTRATIVE EXAMPLE

A right circular cone OPQ is scooped out of a solid homogeneous sphere, the
vertex of the cone 0 being on the surface of the sphere, and its axis being a diameter
of the sphere. It is required to find the center of gravity of the remainder.

METHOD I. Polar coordinates. First let us use polar coordinates, taking the
vertex 0 of the cone as origin, and the axis of the cone as initial line. If a is
the semivertical angle of the cone, the equation of the cone is 6 = a. If a
is the radius of the sphere, the equation of the sphere is r = 2 a cos 6. The
center of gravity must from symmetry lie on the axis 0 = 0, so that 6 = 0, and
equation (34) becomes

r =

fffpr* sin 6 cos 0 drd0d<j>
CCCPr*sm0drd0d<t>

The solid is supposed to be homo-
geneous, so that p is a constant, and
may, therefore, be taken outside the
sign of integration in both numerator
and denominator. The limits of inte-
gration for <f> are from <f> = 0 to 0 = 2 TT,
so that this integration may be per-
formed in each case. Doing this, and dividing out by 2 irp, we are left with

FIG. 81

r =

r3 sin e cos 0 drd0

r2 sin 0 drd0

We may next integrate with respect to r, the limits being r= 0 to r = 2acos0,
and obtain

C\ (2 a cos 0)* sin 0 cos 0 d0

r =

f}(2acos0)3sin0d0
I cos5 0 sin 0 d0

cos30sin0d0

The limits of integration for 0 are obviously from 0 = a (the cone) to
0 = - (the tangent plane to the sphere). We have

TT *

f 2 cos50 sin 0 dd = - $ [cos6 0]2 = fc cos6 a,

. " J a

v «;

cos3 0 sin 0 de = - \ [cos4 0]2 = \ cos* a.

140

CENTER OF GRAVITY

Substituting these values, we find

r =

cos6 a.

= a cos2 a.

Thus the center of gravity is on the axis of the cone at a distance a cos2 a
from the vertex.

METHOD II. Cartesian coordinates. We may next employ Cartesian coordi-
nates, taking 0 as origin and the axis of the cone as axis of x. The equation of

the cone is now

while that of the sphere is

x2 + y2 + z2 - 2 ax = 0.
From § 106, we have

« =

In each integral we may inte-
grate first with respect to y and z
together. We have to evaluate
the same integral in both cases, namely \ Cdydz, the limits being given by

FIG. 82

and y2 + z2 = 2 ax — x2.

The problem is the same as that of finding the area of a circular ring of
inner and outer radii xtano: and V2 ax — x2 respectively. (This ring is, of
course, the intercept of the solid on the plane parallel to the yz plane.) The
area of the ring is

IT (2 ax — x2) - TT (a;2 tan2 a) = TT (2 ax - x2 sec2 a),

and on substituting this value for J \ dydz, the formula becomes

TTX (2 ax — x2 sec2 a) dx

AT (2 ax - x2 sec2 a) dx

The limits of integration are now from x = 0, the origin, to x = 2 a cos2 cr,
the value of x on the plane PQ. Evaluating the integrals, and substituting
these limits, we obtain

__ 2 a-rr | (2 a cos2 a)8 - IT sec2 a j- (2 a cos2 or)4
~~ 2 a?r i (2 a cos2 a)2 - TT sec2 a $ (2 a cos2 a)8
= a cos2 a,
giving the same result as before.

ILLUSTRATIVE EXAMPLE

141

METHOD III. Geometrical Method. The center of gravity can also be found
by regarding the given volume as the sums
and differences of simpler volumes of which
the center of gravity is already known.

The volume is obtained by taking the
complete sphere OPsQ and subtracting from
it the cone OPrQ and the spherical segment
PrQs. The center of gravity of the sphere
and cone are known, — that of the segment
PrQs is most easily found by regarding it as
the difference between the sector CPsQ and
the cone CPrQ. Thus we regard the original
figure as made up of -pIG

(sphere OPsQ) - (cone OPrQ) - (sector CPsQ) + (cone CPrQ).

The volumes of these, and the distances of their centers of gravity from 0
measured along OC, are as follows :

FIGURE

VOLUME

+ sphere
— cone OPr Q
- sector CPsQ
-f cone CPrQ

f Tra3

— $ (2 a cos2 a) (Tra2 sin2 2a)

- f Tra3 (1 - cos 2a)

£ (a cos 2a) (Tra2 sin2 2a)

DISTANCE OF C.G. FROM O

f(2 a cos2 a)

a + | a (1 + cos 2a)

a + f a cos 2a

In this table the negative sign denotes that a figure is to be removed, so that
its volume must be reckoned as of negative sign.

Denoting the distance of any center of gravity from O by x, and using
the formula

of § 86, we obtain as the distance of the center of gravity of the whole figure
from O

-|(2acos2ar)2(Tra2sin22ar)-|Tra4(l-cos2a:){l+|(l + cos2a:)}
| Tra3 - 1 (2 a cos2 a) (Tra2 sin22a) - f Tra3(l - cos2a) -f i(acos2a)(Tra2sin22a)

+ |(a cos 2a) (Tra2 sin2 2a:)a(l + f a cos 2 a)
| Tra3- |(2a cos2 a)(7ra2 sin22a) - f Tra3(l- cos2a)+|(a cos 2 a) (Tra2 sin2 2 a) '

which, after reduction, gives

x = a cos2 or,
the same result as before.

142 CENTER OF GRAVITY

GENERAL EXAMPLES

1. A plane quadrilateral A BCD is bisected by the diagonal AC, and
this diagonal is divided in the ratio a : b by the diagonal BD. Prove
that the center of gravity of the quadrilateral lies in AC and divides it
into two parts in the ratio 2 a + & : 2b + a.

2. A uniform wire is bent into the form of a circular arc and the two
bounding radii, and the center of gravity of the whole is found to be
at the center. Show that the angle subtended by the arc at the center is
tan-i(-f).

3. The three feet of a circular table are vertically below the rim and
form an equilateral triangle. Prove that a weight less than that of the
complete table cannot upset it.

4. A triangular table is supported by three legs at the middle points
of its sides, and a weight W is placed on it in any position. It is found
that the table will just be upset if a weight P is placed at one angular
corner. The corresponding weights needed to upset it at the other corners
are Q, E. Prove that P + Q + R is independent of the position of the
weight W.

5. Weights are nailed to the three corners of a triangular lamina, each
proportional to the length of the opposite side of the triangle, and of com-
bined weight equal to the original weight of the lamina. Show that the
center of gravity of the triangle is at the center of the nine-point circle.

6. A uniform triangular lamina of weight W and sides a, b, c is sus-
pended from a fixed point by strings of lengths Z1? /2, ls attached to its
angular points. Show that the tensions of the strings are

WUlt WMZ, Wkls,
where k = [3 (tf + ij + 1$) - a2 - &2 - c»]~*.

7. Explain how a clock hand on a smooth pivot can be made to show
the time by means of watchwork, carrying a weight round, concealed in
the clock hand.

8. A spindle-shaped solid of uniform material is bounded by two right
circular cones of altitudes 6 and 2 inches with a common circular base of
radius 1 inch. It is suspended by a string attached to a point on the rim
of the circular base. Find the inclination of the axis of the spindle to the
vertical when it is hanging freely.

9. A pack of cards is laid on a table, and each projects beyond the one
below it in the direction of the length of the pack to such a distance that
each card is on the point of tumbling, independently of those below it.
Prove that the distances between the extremities of successive cards will
form a harmonic progression.

EXAMPLES 143

10. Prove that the center of gravity of any portion PQ of a uniform
heavy string hanging freely is vertically above the intersection of the
tangents at P, Q.

11. A hemispherical shell has inner and outer radii a, b. Show that
the distance of its center of gravity from its geometrical center is

3 (a + 6)(a2 + 62)
8 a2 + ab + b2

12. An anchor ring is cut in two equal parts by a plane through
its center which passes through its axis. Find the center of gravity of
either half.

13. Prove that the pull exerted by a man in a tug of war is - of his

weight, where a is the horizontal projection of a line joining his heels to
his center of gravity, and b is the height of the rope above the ground.

14. Prove that a horse weighing W pounds can exert a horizontal pull
of Wa/h pounds at a height h above the ground by advancing his center
of gravity a distance a in front of its position when he is standing upright
on his legs.

15. A rod of varying density and material is supported by a man's two
forefingers, across which it rests in a horizontal position. The man moves
his fingers toward one another, keeping them in the same horizontal plane,
and allowing the rod to slip over one or both of his fingers. Show that
when his fingers touch, the center of gravity of the rod will be between
the points of contact of his fingers with the rod.

16. A semicircular disk rests in a vertical plane with its curved edge on
a rough horizontal and an equally rough vertical plane, the coefficient of
friction being /*. Show that the greatest angle that the bounding diameter
can make with the vertical is

17. A hemisphere of radius a and weight W is placed with its curved
surface on a smooth table, and a string of length l(l<d) is attached to a
point on its rim and to a point on the table. Prove that the tension of the
string is 3 a _ i

8 V2 al - I2

18. A triangular lamina of weight W is supported by three vertical
strings attached to its angular points so that the plane of the triangle is
horizontal ; a particle of weight W is placed at the orthocenter of the
triangle. Prove that the tensions of the strings are given by

1 + 3 cot.B cot C ~ 1 + 3 cot C cot A 1 + 3 cot A cotB 2

144 CENTER OF GRAVITY

19. Find the center of gravity of a lamina bounded by a parabola and
a line perpendicular to its axis.

20. Find the center of gravity of the volume cut from a solid parabo-
loid by a plane perpendicular to its axis.

21. Find the center of gravity. of the area inclosed by two radii of
an ellipse.

22. Find the center of gravity of the volume cut off from a solid ellipsoid
by a plane through the center.

23. Find the center of gravity of half of an ellipsoidal shell, this being
bounded by two similar concentric and coaxial ellipsoids, and a plane
through the center.

24. A right circular cone whose base is of radius r is divided into two
equal parts by a plane through the axis. Prove that the distance of the

center of gravity of either half from the axis is — •

25. Find the center of gravity of a lamina bounded by the semicubical
parabola a;3 = ay2, the axis of x, and the ordinate x = a.

26. Find the center of gravity of a single loop of the curve

r = a sin 30.

27. Find the center of gravity of an octant of a sphere.

28. A cylindrical hole of radius a is drilled through a hemisphere of
radius b so that the radius perpendicular to the base of the hemisphere is
also the central line of the hole. Find the center of gravity of the figure.

29. Find the center of gravity of the area inclosed between the two
circles a? + yi = fl2 . xi + yi = 2 ab.

30. Find the center of gravity of a lens made of homogeneous glass,
having spherical surfaces of radii r, s, and of which the thickness is t at
the center and zero at the edge.

CHAPTER VII
WORK

109. Measurement of work. There are various kinds of work,
but in mechanics we are concerned only with the work done in
moving bodies which are acted on by forces. Such work is
described as mechanical work. We say that mechanical work is
done whenever a body is moved in opposition to the forces acting
on it, as, for instance, in raising a weight, in dragging a heavy
body over a rough surface, or . in stretching an elastic string. In
the first case work is performed against the force of gravity, in
the second case against the frictional force exerted on the moving
body by the rough surface, and in the third case against the
tension of the string.

Obviously in estimating the amount of work done, two factors
have to be taken into account, namely the amount of the force
acting on the body and the distance through which the body is
moved in opposition to this force. The amount of work will clearly
be directly proportional to the force, — in raising a weight of 200
pounds through a given distance we do twice as much work as
in raising a weight of 100 pounds through the same distance. It
will also be proportional to the distance moved, — hi raising a
weight through two feet we do twice as much work as in raising
the same weight through one foot. Thus the amount of work done
varies as the product of the force and the distance.

The amount of work done in raising a weight of one pound
through a height of one foot is called one foot pound.

From what has been said, it is clear that the work done in
raising a weight of w pounds through a height of h feet is
wh foot pounds. Also, the work done in moving a body a distance
of s feet in opposition to a force of F pounds weight is Fs foot

145

146 WORK

pounds. Thus we may say that the work done in moving a body
through any distance against a uniform force is the product of the
distance and the force.

Suppose, for instance, that it is found that the force required to drag a
railway train along a level track is equal to the weight of 10,000 pounds,
then the work done in hauling this train a distance of 100 miles

= 100 x 5280 x 10,000 foot pounds.

110. Rate of performing work. Work frequently has to be done
within a given time, so that it is often necessary to measure the
rate at which work is being done. The rate of doing work in
which 33,000 foot pounds are done per minute is called one horse
power (I H. P.).

This unit was introduced by Watt, and was supposed to measure the
rate of working of an ordinary horse. It is found, however, that very few
horses are capable of working continuously at one horse power for any
length of time.

As an example of the calculation of horse powrer, let us find the horse
power required of an engine to haul a train at 30 miles an hour, the fric-
tional resistance being equal to the weight of 10,000 pounds. A velocity
of 30 miles an hour = 44 feet per second, so that the work done per second
= 44 x 10,000 foot pounds. Since one horse power = 550 foot pounds per
second, we see that the horse power required

44 x 10,000

550

= 800 horse power.

This gives the horse power required to haul the train at a steady speed
of 30 miles per hour. We shall find that if the speed is not constant the
horse power will be different, part of the work being used up in producing
the acceleration of the motion. For the present, however, we confine our
attention to motion with uniform velocity.

ABSOLUTE UNIT OF WORK

111. We have already seen that besides the practical unit of
force, which is the weight of a unit mass, there is also a second
unit of force, known as the absolute unit, which is denned as being
a force capable of producing unit acceleration in unit mass. As
the practical unit produces acceleration g in unit mass, where g is

MEASUKEMENT AND UNITS 147

the acceleration due to gravity, it follows that the practical unit
is g times the absolute unit.

In practical British units, the unit force is the pound weight.
In absolute units, the corresponding unit is known as the poundal;
it is the force which will produce unit acceleration in a mass of
one pound.

The practical unit of work, as we have said, is the work done in
raising a mass of one pound through one foot, i.e. in moving
through one foot the point of application of one pound weight.
There is also an absolute unit of work, namely the work done in
moving through one foot the point of application of one poundal.
This unit is called the foot poundal. Since one pound weight is
equal to g poundals, we obviously have the relation

1 foot pound = g foot poundals.

EXAMPLES

1. At what speed can a horse of 1 horse power draw a cart weighing 1 ton,
friction being supposed to cause a horizontal force equal to one fortieth of the
weight of the cart ?

2. A body resisted by a force of P poundals is moved against this resistance
with a velocity v. What horse power is required ?

3. At what rate can a steam roller of 7 horse power and weight 1 ton roll a
path, the resistance due to friction being equal to the weight of the roller ?

4. A snail weighing -j- ounce climbs a wall 6 feet in height in 4 hours. At
what horse power is he working ?

5. A load of bricks weighing 5 tons has to be raised to the top of a house
50 feet in height by 10 laborers, each of whom works at an average rate of
^ horse power. How long ought the job to take ?

6. The piston of an engine has an area of a square feet and a stroke of
I feet, and the engine makes > re volutions per minute. If the pressure per unit
area acting on the piston is p pounds weight per square foot, prove that the
horse power at which the engine is working is

plan
33,000 '

7. A locomotive has a circular piston of diameter 17 inches, and stroke 26
inches. It makes 250 revolutions per minute, the pressure being 225 pounds
weight per square inch. Find its horse power.

148 WORK

8. If 200 horse power is required to drive a steamer 150 feet long at a speed
of 9 knots, prove that 25,600 horse power will be required to drive a similar
steamer, 600 feet long, similarly immersed, at 18 knots, assuming that the
resistance is proportional to the wetted surface and the square of the velocity
through the water. Prove also that the cost of coal per ton of cargo will be
the same in the two steamers.

9. Fifty horse power is transmitted from one shaft to another by means of
a belt moving over two wheels on the shafts with a linear velocity of 250 feet
per minute. Find the difference of tensions on the two sides of the belt.

10. A locomotive consumes 1£ pounds of coal per horse-power-hour. How
much coal is required to haul a train of total weight 1000 tons over 50 miles of
level road on which the resistance to friction is 12 pounds weight per ton ?

11. A liner of 22,000 horse power makes a run of 3300 miles in six days.
Find the resistance to the ship's motion.

WORK DONE AGAINST A VARIABLE FORCE

112. If a body is moved in opposition to a force which is not
of constant intensity but varies from point to point on the path
of the moving body, we can no longer use the formula JFs for the
amount of work performed.

To calculate the amount of work done, we divide up the whole
range over which motion takes place into an infinite number of
infinitesimally small ranges, each of these ranges being so small
that the force opposing the motion may be regarded as of constant
magnitude during the motion through any one of them.

If ds is any small range at a distance s from the starting point,
and if F is the intensity of the force opposing the motion while
the body moves through the small range ds, then the work done in
moving through this range is Fds. The total work done, the sum
of the amounts of work done in all the ranges, is, accordingly,

Fds.
WORK DONE IN STRETCHING AN ELASTIC STRING

113. As an example of the use of this formula, let us find th<
work done in stretching an elastic string. Let the natural lengtl
of the string be /, and let X denote its modulus of elasticity.

WOEK OF STEETCHING A STEING 149

When the length of the string is x9 its tension T, by the
formula of § 39, is given by

In stretching the string through a further distance dx, — i.e. from
length x to length x + dx, — the work done

= Tdx

= — (x — / ) dx.

By integration, we find that the work done in stretching a string
from length a to length b

= C^(x~l)dx

The distance stretched is I — a, while — (b 4- a — 2 /) is the

-^ /

tension when half of the stretching has been completed, i.e. when
x = \(a 4- I).

Thus we have found that

The work done in stretching an elastic string from any length a,
greater than the natural length of the string, to a length b, is equal
to the tension at length ^(a + b) multiplied by (b — a).

Obviously, if the tension is measured in pounds weight and the
extension (b — a) in feet, the product will give the amount of work
measured in foot pounds. If the tension is measured in poundals,
and (b — a) in feet, the product will give the amount of work in
foot poundals.

150

WOKK

s r
FIG. 84

WORK REPRESENTED BY AN AREA

114. Let PQ represent the path described by a moving body,
and let us draw ordinates at each point in PQ to represent, on
any scale we please, the force opposing the motion of the body at

that point. Let s, r be two adja-
cent points, and let ssr, rrf be
the ordinates at these points.

Then the area of the small
strip ss'rr' may, in the limit, be
supposed equal to sr multiplied
by ss1. On the scale on which
we are representing forces, this
product will represent the dis-
tance sr multiplied by the force opposing the motion of the body
from s to r. In other words, the small area ss'rr' will represent
the work done in moving the body from s to r.

By addition of such small areas, we find that the complete
PP'QQ' represents the work done in moving from P to Q.

115. This method gives a simple way of investigating the work done ii
stretching an elastic string, already calculated in § 113. Let OP be the
natural length. For the sake of definiteness suppose that the end 0
held fast, and that as the string is stretched the point P moves along
line OP. Let it be required to find the work done in stretching the string
from a length OA to a length OB.

Let Q be any point of the line
OPAB, and let QQ' be drawn to
represent the tension when the
length of the string is OQ.

For different positions of Q, the

ordinate QQ' will be of different °
heights. Since, by Hooke's law,

the tension is proportional to the extension, the height of the ordinal
QQ' (representing the tension) will always be in the same ratio to PQ (tl
extension). Thus Q? is always on a certain straight line through P. If A A'
BB' are the ordinates which represent the tensions at A , B, this line will,
of course, pass through the points A', B'. The work done in stretchii
the string through the range AB is now, in accordance with § 114, repi
sented by the area AA'B'JB, the area which is shaded in fig. 85.

FIG. 85

GRAPHICAL REPRESENTATION OF WORK 151

The area of this figure is clearly equal to AB multiplied by the ordinate
at the middle point of AB. This ordinate represents the tension of the
string when its length is equal to \(OA + OB), so that we again obtain the
result of § 113, namely

(work done) = (range of stretching,

x (tension at halfway stage of stretching).

116. The indicator diagram. The graphical representation of
work explained in § 114 is made use of in practical engineering.
Suppose that 00' is the distance traveled by a piston inside a
cylinder. When the piston is in any position P, let the pressure
acting on the piston be measured,
and let a line PPf be drawn at
right angles to 00' to represent it
on any assigned scale. As the
piston moves along the range 00r
and then back along the range
O'O, the point Pr will describe a
closed curve AP'BP"A, which is
called the indicator diagram of the
motion of the piston.

The work done by the steam on
the piston in its forward motion is,
as we have seen, represented by the area AP'BO'POA inclosed be-
tween the curve AP'B and the axis 00r. This work is expended in
moving the piston forward in opposition to the thrust in the piston
rod. Similarly the work done by the steam on the piston in its back-
ward motion is represented by the curve BO'POAPnB inclosed be-
tween BP"A and the axis O0'} this area being taken negatively, since
the piston is now moving in opposition to the pressure at work on it.

Thus the whole work done on the piston is represented by the
difference of these two areas, and this is easily seen to be the area
AP'BP"A of the indicator diagram itself. Hence, to find the rate
at which an engine is performing work, it is only necessary to
measure the area of its indicator diagram and the number of
revolutions per unit time.

FIG. 86

152 WORK

WORK DONE AGAINST FORCE OBLIQUE TO DIRECTION OF MOTION

117. So far we have only considered cases in which the force
acts in a direction exactly opposite to that in which the particle
moves. We may, however, have to calculate the work when the
motion makes any angle with the direction of the force.

When a body is moved at right angles to the force acting on it,
the work done will clearly be -nil ; e.g. in moving a weight about
on a horizontal surface no work is done against gravity.

We can now find the amount of work done when a body is moved
in a direction making any angle with the force acting on it. Let a
body be moved from P to Q, a small distance ds of its path, while
acted on by a force R, of which the line of
action makes an angle <£ with QP. Eesolve
, R into two components, R cos </> along QP

and R sin <£ perpendicular to QP. The work
done against the force R is the same as the

work which would be done if these two forces R cos <£, R sin <f> were
acting on the body simultaneously. The work done against th(
former force would be R cos <f> ds ; that against the latter would
be nil. Thus the whole amount of work done is R ds cos c/>.

118. Let R have components X, Y, Z, and let the element oi
path PQ have direction cosines I, m, n. The direction cosines of the
line of action of R are

X Y Z

— > — » — >
R R R

and since this makes an angle TT — <£ with PQ, we must have

~X~ V 7

COS(TT - <£) = I— + m- + n->

Hence R ds cos </> = — ds (IX + mY + nZ)

= - (Xdx + Ydy + Zdz),

where dx, dy, dz are the projections of ds on the axes. This gives
an analytical expression for the work done in a small displace-
ment. By integration, we can find the work done in any motion.

WORK OF RAISING BODIES AGAINST GRAVITY 153

119. Work of raising a system of bodies against gravity. If a

particle of mass m is moved a distance ds along a path making an
angle <£ with the vertical (upwards), the work done is mg cos <£ ds.
Since the distance through which the particle is raised is ds cos <£,
we may say that the work done is equal to the weight of the
body (mg) multiplied by the distance through which the particle
is raised.

By taking the particle along any path, and adding together the
amounts of work done on the successive elements of the path, we
find that the total work done against gravity is equal to the weight
of the particle multiplied by the total vertical distance through
which the body has been raised.

120. Let us suppose that we move a number of particles of
masses ml} m2, • • •. Let their heights above the ground before the
motion be hlt h2, • - •, and let their heights at the end of the motion
be h[,h'2,--. The work done against gravity on the first particle
is ffi^gil^ — 7^); by addition of such quantities, the total work
done against gravity

(36)

Now let M be the total mass of the particles, and let H, H'
denote the heights of the center of gravity of all the particles
above the ground before and after the motion respectively. Then,
by the formula of § 86, we have

so that

and, similarly, ^m^J = MH'.

Thus the total work, as given by expression (35), becomes
g(MHJ — MH) = Mg(Hr — H).

154 WOKK

Thus the total work done against gravity is equal to the total
weight of the particles multiplied by the vertical height through
which the center of gravity of the particles has been raised.

WORK PERFORMED AGAINST A COUPLE

121. THEOREM. If a rigid body acted on by a system of forces
be given any small rotation through an angle e about any axis, the
work done is Ge, where G is the moment about this axis of the
forces opposing the motion.

Let the axis of rotation be supposed to be a line perpendicular
to the plane of the paper, meeting it in the point L. Let a typical

force be a force F acting on the particle
A of the body.

As the result of the rotation, let A
move to a position A1, so that the angle
ALA1 is equal to e, the angle through
which the body has been turned.
Then, during the rotation, the point

FIG. 88

of application of the force F moves
from A to A', and, therefore, the work done

= F-AAf-cos<l>,
where <£ is the angle between F and AAr,

= AA' x component of F along AAf

= e X LA x component of F along AA1

= e x moment of F about the axis of rotation.

If the rigid body is acted on by a number of forces applied to
its different particles, we find, on summation, that the total work
done

= e x sum of the moments of all these forces

about the axis of rotation
= Ge, where G is the moment about the axis of
rotation of all the forces.

VIBTUAL WORK 155

EXAMPLES

1. A man who weighs 140 pounds walks up a mountain path at a slope of
30 degrees to the horizon at the rate of 1 mile per hour. Find his rate of work-
ing in raising his own weight in horse power.

2. At what horse power is an engine working which hauls a train of 1000
tons up an incline of 1 in 200 at 12 miles an hour, the resistance due to friction
being ^ of the weight of the train ?

3. An automobile weighing 1 ton can run up a hill of 1 in 60 at 8 miles an
hour. Taking the resistance due to friction as ^ of the weight of the car,
find at what rate it could run down the same hill, assuming the horse power
developed by the engine to remain the same.

4. A cargo of stone weighing 18 tons is unloaded from a barge on to a quay
30 feet above the barge by cranes worked by an engine. If the unloading takes
three hours, find the average horse power at which the engine has been working.

5. Assuming that a man in walking raises his center of gravity through a
vertical height of one inch at every step, find at what horse power a man
is working in walking at 4 miles an hour, his stride being 33 inches, and his
weight 168 pounds.

6. A cyclist and his machine weigh 200 pounds, and he rides up an incline
of 1 in 80 at 16 miles an hour. His bicycle is geared to 72 inches, and the
length of his cranks is 7 inches. Find the average vertical pressure of his foot
on the pedal, assuming this pressure to exist only during the downward motion
of the pedal.

7. A single-screw ship has engines of 5000 horse power, and, when working
at full power, the engines make 75 revolutions per minute. Find the couple
transmitted by the shaft.

8. When one body rolls on another, there is found to be a couple opposing
the motion, equal to that produced by the normal reaction at the end of an
arm of length Z, where / is called the coefficient of rolling friction.

If a railroad truck runs on wheel of radius a, show that the resistance to its
motion produced by rolling friction is I/a times its weight.

THE PRINCIPLE OF VIRTUAL WORK

122. By a small displacement is meant for the present a motion
in which each particle of a system is displaced from its original
position through a distance which is so small that it may be treated
as an infinitesimal quantity of which the square may be neglected.
If the system is under the action of forces, work will be done in
performing any small displacement. Since the displacement is
supposed to be a small quantity, the work performed will also be
a small quantity.

156 WORK

If any particle is in equilibrium, the resultant force acting
on it vanishes, so that the work done in any small displacement
of the particle vanishes to a higher order than the displacement.
If a rigid body, or system of rigid bodies or particles, is in equilib-
rium, and any small displacement is given to it, the work done
on each particle is nil, so that the aggregate work done is nil.

123. The forces acting on the particles of the system may, as
in § 50, be divided into two classes :

(a) forces applied to the bodies from outside ;

(b) pairs of actions and reactions acting between the particles
of the bodies, or between two bodies in contact.

In calculating the work done in a small displacement, we must
take account of the work done against all the forces of both classes,
but shall find that a great number of the terms arising from the
forces of the second class cut one another out.

124. Let us first consider the pair of forces which constitute
the action and reaction between two particles P, Q of a rigid body.
Let the amount of each force be R, its direction being QP or P(

according as it acts on P or Q. Let
the effect of a small displacement
to move P, Q to Pr, Q' respectively, and
let Prp, Qfq be perpendiculars drawn
from Pf, Q' to PQ. The work done

against the force R acting on P is R X Pp, while that done against
the force R acting on Q is — R X Qq. Thus the total work performed

= X(Pp-Qq)

= R(PQ — projection of P'Q' on PQ).
Since the body is rigid, the length P'Q' is equal to the length
PQ, and since the displacement is, by hypothesis, small, the angl<
between P'Q' and PQ is small. Thus the projection of P'Q1 on P(
= P'Q', except for small quantities of order higher than the first

so that the work performed vanishes,

VIRTUAL WORK 157

125. Again, the work performed against the pair of forces which
constitute action and reaction between two smooth surfaces can
be seen to vanish.

First consider the case in which one body is held at rest while
the second is made to slide over its surface. In such a displace-
ment the work performed, if any, is performed against the reaction
which acts on the moving body. Since the
force acts along the normal, while its point of
application necessarily moves in the tangent
plane, — i.e. at right angles to the normal,—
we see that the work done is nil.

The most general motion possible for the two
surfaces is compounded of a motion of the kind
just described and a motion in which the two
surfaces move as a rigid body. The work done
in the first part of the displacement has just been seen to be zero,
the work done in the second part of the displacement vanishes by
§ 124; hence the total work vanishes, proving the result required.

126. The results just proved are not true if the contact between the
surfaces is rough. The work done in such a case depends on the magni-
tude of the frictional forces, and as it is generally as difficult to determine
the amount of these forces as to solve the whole problem, the method of
virtual work is not of any value in such cases.

127. We have now seen that a large number of forces may be
left out of account altogether in calculating the work done in a
small displacement, and the principle of virtual work, which states
that when a system is in equilibrium the work done in any small
displacement is zero, requires us only to calculate the work per-
formed against external forces, and not that performed against the
internal actions and reactions of rigid bodies.

128. Systems of pulleys. An important application of the prin-
ciple of virtual work is the following : Let us suppose that we have
any arrangement of pulleys and inextensible ropes, the ropes hav-
ing two free ends, — to one of which the weight to be raised is
attached, and to the other of which the power is applied. Let these

158

WOKK

two free ends of rope be called the weight end and the power end
respectively, and let us suppose that the arrangement is such that,
in order to move the weight end through 1 inch, the power end
must be moved through n inches. Let a weight W be attached to
the weight end, and let us suppose that it is found that a force P
must be applied to the power end to maintain equilibrium.

We now have forces P and F in equilibrium. To find the rela-
tion between them, let us give the system a small displacement.
Let us move the weight W a distance ds, then, if the rope is not to
be stretched, we must suppose the power P moved through a dis-
tance nds. The work done by external force consists solely of
the work performed on the power end of the rope, namely P n ds,
and the work performed in moving the weight against gravity,
namely W ds. These are of opposite signs, — if we raise the weight,
W ds must be taken positively and P nds negatively, and vice versa.
If the system was initially in equilibrium, the total work performed
by external forces in this small displacement must vanish, so that
the equation of equilibrium is seen to be

K Wds-Pnds = Q,

so that P = — >

giving the relation between power and weight.
This investigation assumes that friction,
etc., may be neglected, and also neglects the
weight of the moving ropes and pulleys.

As an instance of a system of pulleys, let us
consider the arrangement shown in fig. 91.
FlG 91 There are two blocks of pulleys, A and B.

The former is fixed, while the latter is free to

move, and has the weight W suspended from it. The rope, starting
from the power end, passes first round a pulley of block A, then round
one of block B, then round one of block A, and so on any number of
times, until finally its end is fastened to block B. To find the relation
between P and W, we need only find the number n. Let us suppose that
in addition to the free power end of the rope the number of vertical
ropes is s. Then, if we pull the power end until the weight end is raised

VIRTUAL WOEK

159

1 inch, we shall shorten each of these s ropes by 1 inch, and so lengthen

the power end by s inches. Thus n = s, so that, in this case, P =

For instance, with two pulleys in the lower block and three in the
upper block the value of n will be 5, so that each pound of power will
support 5 pounds of weight, — a man pulling with a vertical pull of
100 pounds could support a weight of 500 pounds, and as soon as his pull
exceeds 100 pounds, he will raise the weight of 500 pounds.

ILLUSTRATIVE EXAMPLES

1. As a first example of the principle of virtual work, let us suppose that we
have an endless elastic string of natural length a, modulus X, placed over a
sphere of radius 6, and allowed to stretch under gravity. We might, of course,
find the amount of stretching in the equilibrium
position by resolving forces, but we can get it
more readily by the method of virtual work. Let
us suppose that, when in equilibrium, the string
lies on a small circle of angular radius 6. Let a
small displacement be given, this consisting of
each element of the string being displaced down
the surface of the sphere, so that the string forms
a new circle of angular radius 6 + d&. The length
of the string when forming a circle of angle 0 was
2 Trb sin 0 ; the increase in this when 6 is changed

to e + dd is d6 — (2 Trb sin 0) or 2 Trb cos 0 d0. The

work done in stretching the string by this amount
is T • 2 -n-b cos 0 dd, where T is the tension. Work

is also done against (or, in this particular case, with) the force of gravity. The
height of the center of gravity of the string when forming a circle of angle 0 is
b cos 0 ; on increasing 0 to 0 + dd, this increases by — b sin 0 dd, so that the work
done against gravity is - wb sin 0 d0. We have now calculated all the work
performed in the small displacement; by the principle of virtual work, the
total amount of this work must be nil, so that

FIG. 92

Thus,

- wb sin 0 de + T • 2 Trb cos 0 d0 = 0.

T= — tan*,

2ir

and the length of the string corresponding to tension T is, as we have seen,

Hence

an equation giving 0.

all + -^-tan0) =

\ 27T\ /

2 -rrb sin 0,

160

WORK

2. Gearing of a bicycle. As a second example, let us apply the principle of
virtual work to the mechanism of a bicycle. Let the length of the crank be a,
and let the bicycle be geared to 6 inches, so that each revolution of the pedals
causes the machine to move as far forward as it would in one revolution of a
wheel of 6 inches diameter. Let us find what pressure must be exerted on the
pedal by a rider in order that the machine may move forward against an
opposing frictional force of w pounds weight.

Let us give the machine a small displacement, the cranks being supposed to
turn through an infinitesimal angle e, and the wheels and machine moving
forward accordingly. Since the gearing is to 6 inches, the distance moved by
the machine as a whole will be 1 be inches, while the distance moved by the pedal,
taking the machine itself as frame of reference, will be ae. Let W pounds weight
be the force exerted on the pedal when the machine is just on the point of motion,
so that the machine is in equilibrium under this force acting on the pedal, and
the backward pull of w pounds due to friction. The equation of virtual work is

so that the required force is

nr=±*.

.Thus the force is directly proportional to the gearing of the machine, but
inversely proportional to the length of the cranks.

3. Four rods of equal weight w and length a are freely jointed so as to form a
rhombus ABCD. The framework stands on a horizontal table so that CA is vertical,
and the whole is prevented from collapsing by a weightless inextensible string of

length I which connects the points B, D. It is
required to find the tension in this string.

To find the tension by the principle of
virtual work, we must of course find a small
displacement such that work is done in oppo-
sition to the tension, or otherwise the tension
would not enter into the equations at all.
Since the string is inextensible, it is not
possible in actual fact to stretch it and so
perform work against its tension. We can
however imagine it to be stretched in spite of its actual inextensibility, or,
what comes to the same thing, we can imagine it replaced by an extensible
string of the same length and having the same tension. It is now easy to
arrange a displacement of the kind required.

Let us imagine that the framework is displaced in such a way that A moves
vertically downwards towards (7, while C remains at rest. Let the displacement
be such that the angle D A C is increased from e to 6 + d0. The length I of the
string which corresponds to the angle 6 is given by

I — 2 a sin 8,
from which, by differentiation, we obtain

dl = 2 a cos e d8,

FIG. 93

ILLUSTRATIVE EXAMPLES

161

giving the relation between the increments dl, dd in I and 0. The work done
against the tension of the string (T) in this displacement is Tdl. The height
of the center of gravity of the whole figure above C is initially £ A C, or a cos 0,
so that, as in § 120, the work done against gravity is

4wd(acos6).
Thus the total work performed by external forces in the displacement is

4wd(acos6) + Tdl,
or, on substituting the values of dl and d(a cos 6),

- 4 wa sin 0 cZ0 -f T-2acos0d0.
For equilibrium this must vanish. We must therefore have

T = 2 w tan 0,
giving the required tension.

4. A rod of length I and weight w is suspended by its two ends from two points
at the same height and distant I apart, by two strings each of length a. Find the
couple required to hold the rod in a position
in which it makes an angle 6 with its equi-
librium position.

In equilibrium the strings are vertical,
the two ends A, B of the rod lying exactly
underneath the two points of suspension

P, Q-

As the rod is turned from its equilibrium
position, we can imagine its middle point to
rise gradually along the vertical line through
the original position of this middle point.
When the rod has been turned through any
angle 6, let the height through which this
point has risen be x.

Then the projection of the length PA'
on a vertical line will be a — x, while its projection on a horizontal plane,

being equal to the horizontal projection of A A', will clearly be I sin - .

Thus, expressing that the length of the displaced string PA' remains equal
to its original value a, we have

a2 = (a-z)2 + Z2sin2-. (a)

To find the couple required to hold the rod ai an angle 6, let us suppose that
the rod is held in equilibrium in this position by a couple G, and that a small
displacement occurs in which 0 is increased to 0 + dd. The work done against
the couple is equal, by § 121, to — V(?d0, the negative sign being taken, since
the couple aids, instead of opposing, the motion. The work done against gravity
is equal to w dx. Thus the equation of equilibrium is

- G d 0 + w dx = 0,

162 WORK

To obtain the relation between d0 and dx we differentiate equation (a),
obtaining

- 2(a - x)dx + I* sin - cos -cZ0 = 0.
2 2

Thus G = w —

. e e

wl2 sin - cos -
_ 22

2(a -x)

wl2 sin 0

4 A /a2- Z2 sin2 -
giving the couple required.

EXAMPLES

1. A square ABCD is formed by joining four equal rods by freely moving
hinges. The points A, C are joined by an elastic string of natural length equal
to a diagonal of the square, and of modulus X. What forces must be applied
to the points 5, D to stretch the string to 1£ times its length ?

2. Three spheres each of radius a and weight w are tied to a point P by
strings of natural length I and modulus X, and hang freely, touching each other.
Find the depth of their centers below P.

3. The mechanism by which a Japanese umbrella is opened is such that each
rib turns through an angle of 5° for every inch that the sliding piece is moved
up the stick. If there are 18 ribs, each of weight i ounce, and having their
centers of gravity 10 inches from their pivots, find with what force the sliding
piece must be pushed up the stick to open the umbrella, when the stick is held
vertically, and the ribs are inclined at an angle of 30° to it.

4. The hands of a clock are balanced with counterpoises, so as to be in
equilibrium in any position. When the time indicated by the clock is 5.10, a
bird of weight w suspends itself from a point on the minute hand which is six
feet from the pivot. How large a vertical thrust must be applied to the hour
hand, also at a point six feet from the pivot, to restore equilibrium ?

5. A clock is wound by raising a weight of 20 pounds through a distance of
3 feet, this enabling the clock to run for 30 hours. The pendulum and escape-
ment are removed, so that the hands will "race" unless held fast. How large
a couple must be applied to the minute hand to prevent this occurring ?

6. The coupling between two English railway carriages consists of a rod with
a right-handed and a left-handed screw cut at its opposite ends and turning
in nuts attached to the carriages. If the pitch of each screw is one inch, and
the rod is turned by a force of 56 pounds acting at best advantage at the
end of a lever 15 inches long, find the force by which the carriages are
drawn together.

POTENTIAL ENERGY 163

POTENTIAL ENERGY

129. It will have been noticed that we are concerned with two
different kinds of work. The first is typified by the work done in
raising a weight in opposition to gravity, the second by the work
done against friction in hauling a train along a level road. The
essential difference between the two kinds is that work of the first
kind can be recovered from the system of bodies by making these
bodies themselves perform mechanical work, whereas work of the
second kind, when once expended, can never be regained. In rais-
ing a weight we may be said to be storing up work rather than
spending it, for the weight can at any time be made to yield back
all the work devoted to raising it. If we raise a weight w through
a distance h, the work done on the Weight is wh ; on letting the
weight descend to its original position, the work done for us by
the weight is wh, so that the total work performed on the weight
is nil.

On the other hand, in hauling a mass a distance s against
a frictional force F the work performed is Fs. To bring the
mass back to its original position, we have to expend an addi-
tional amount of work Fs, so that the total work performed is
2 Fs. This brings out the essential difference between the two
types of work and between the two systems of forces against
which the work is performed.

130. DEFINITION. When the forces acting on a system of bodies
are of such a nature that the algebraic total work done in perform-
ing any series of displacements which bring the system back to its
original configuration is nil, the system of forces is said to be a
conservative system.

The algebraic work being nil, the work done on the system in taking
it to any configuration is equal and opposite in sign to the work done on
the system in allowing it to .resume its former configuration, so that all
the work spent can be regained. The work is accordingly stored up, or
conserved.

164 WOKK

A small amount of reflection will show that a system of forces
is conservative if the only forces which come into play are some
or all of the following :

(a) gravity;

(b) reactions in which the contact is perfectly smooth ;

On the other hand, if any one or more forces of the following
types come into play (so that work is performed against them),
the system of forces is non-conservative :

(a) reactions in which the contact is rough ;

(b) resistance of the air.

131. THEOKEM. The work done on a system of bodies acted on
by conservative forces, in moving from one configuration P to a
Q second configuration Q, is independent of the series of

configurations through which the system moves in
passing from P to Q.

To prove this, let us denote the work done in pass-
ing from P to Q through one series of configurations
by Wl , that done in passing through any second series
by Wz, and that done in returning from Q to P by any
P third series of configurations by W3. If we pass from

FIG. 95 p t0 Q by the first series and back from Q to P by
the third, the total work done is nil, so that

So, also, if we pass from P to Q by series 2 and back from Q to
P by series 3,

Thus Wl=Wz, which proves the theorem.

132. DEFINITION. Taking any configuration P as standard, the
work done in moving a system of bodies from the configuration
P to the configuration Q is spoken of as the potential energy of
configuration Q.

POTENTIAL ENERGY 165

The potential energy, accordingly, measures the work which
has been stored up in placing the system in configuration Q.

THEOREM. The work done in moving a system from a config-
uration (1) to a second configuration (£) against conservative forces
is Trs —Wlt where Wl} Wz are respectively the potential energies in
configurations (1) and (2).

For if P is the standard configuration, the work from P to (1)
is W1 ; the work from P to (1) plus that from (1) to (2) is Wz, so
that the work from (1) to (2) is WZ-W^.

133. THEOREM. If a system of bodies is in a configuration of
potential energy W, and if x, y, z are the coordinates of any
particle, the resultant force acting on the particle has components

dW dW dW

-- - — , -- - — , -- - —
ex cy cz

To prove this, let us imagine that we give the system a small
displacement, which consists in moving the single particle at x, y, z
a distance dx parallel to the axis of x. If X, Y, Z are the compo-
nents of the force acting on it, the work we do in the displace-
ment is, as in § 118, equal to — X dx. This work is also equal to

the increase in the potential energy, namely — • dx, so that we have

— Xdx = —-dx.

Thus X— -- -—> and- similarly we may prove that

mm
Y = -- — , Z = -- - —

dy cz

134. THEOREM. If a system of bodies is in a configuration of
potential energy W, and if 6 is an angle giving the orientation of
a rigid body of the system about any line, the moment about this line
of the forces acting on the rigid body (reckoned positive if tending
to rotate it in the direction of 0 increasing) is

166 WORK

For, let us give the system a small displacement, which consists
in turning the body in question through a further angle dO about
the selected line, so that 6 becomes changed into 0 -f- d6. The

increase in potential energy is — r- d6, while the work performed

is, by the theorem of § 121, equal to — G d6, where G is the moment
about the axis of all the forces acting on the rigid body.

so that e = -^'

the result required.

135. THEOREM. In a position of equilibrium of a system of
bodies, the potential energy W is either a maximum or a minimum

The potential energy is a function of all the coordinates of al
the particles of which the system of bodies is composed, say

x if z ' x ii z ' etc

If the position is one of equilibrium, each particle is in equilib-
rium, so that the components of the forces acting on each particle
vanish separately by § 33. By § 133 the condition for this is

dW A

— = 0, etc.

dxz

But these are exactly the conditions that W shall be a maximum
or a minimum.

136. The converse of this theorem is also true.

THEOREM. If the potential energy of a system of bodies is either
a maximum or a minimum in any configuration, then the con-
figuration is one of equilibrium.

For, with the notation of the previous section, if W is a maxi-
mum or a minimum, it follows that

n SW~Q ?*-_
", — — — \J) — — — v.

dyl dzl

POTENTIAL ENERGY 167

dW dW dW

Since > , are the components of the force

dx, 0^ dz,

acting on particle (1), these equations indicate that particle (1)
is in equilibrium. Similarly, it follows that the other particles
are in equilibrium, giving the result.

137. An important special case of these theorems arises when
the only forces which do any work in a displacement are the
weights of the bodies of which the system is composed. If M is
the mass of the whole system, and if h is the height of its center
of gravity above any standard horizontal plane, the potential
energy is, by § 120, Mgh, and this is a maximum or a minimum
when h is a maximum or a minimum. Thus we have the theorem :

In a system of bodies in which the only forces which perform
work in a, displacement are those of gravity, the configurations of
equilibrium are those in which the height of the center of gravity
is a maximum or a minimum.

EXAMPLES

1. Two uniform rods, each of length Z, are freely jointed at their extremities
and placed over a smooth cylinder of radius a of which the axis is horizontal.
Find the angle which the rods make with the horizontal when in equilibrium.

2. An elliptic disk is weighted so that its center of gravity is halfway
between its center and one extremity of its major axis. Show that if its

eccentricity is greater than — — there will be four positions of equilibrium in

V2
which the disk stands vertical on a horizontal plane, but otherwise only two.

3. A horizontal rod of weight W has its center pierced by a fixed vertical
screw on which it turns, one revolution raising or lowering it by \ inch. If
there is no friction, find the couple required to hold it at rest.

4. A plug of weight W is made in the shape of a pyramid of square cross
section. It is placed with its axis vertical in a square hole of side c, the depth
of its vertex in this position being d below the plane of contact. Find the couple
required to hold it turned through an angle 0 and still having its axis vertical.

5. A smooth parabolic wire is placed with its axis vertical. Two beads are
strung on it, and are connected by a string which passes through a smooth ring
at the focus. Show that there are an infinite number of positions of equilibrium.

6. A smooth bowl in the shape of an ellipsoid of semi-axes a, 6, c has one axis
vertical. Find the couple required to hold a rod of length I in a horizontal posi-
tion in the bowl, making an angle 0 with a position of equilibrium.

168 WORK

KINETIC ENERGY

138. Suppose that a moving particle is acted on by a force
of which the direction is opposite to that of the motion of the
particle. The effect of the force, according to the second law of
motion, is to produce a retardation in the velocity of the particle.
The velocity of the particle decreases so long as the force acts, so
that if the force continues to act for a sufficient time, the particle
must ultimately be reduced to rest.

Consider, for example, a hammer striking a nail. The reaction between
the hammer and nail^ is a force in the direction opposite to that of the
motion of the hammer, and this ultimately brings the hammer to rest.
Again, when a particle is projected vertically upwards, its weight after a
time reduces it to rest, after which of course it falls back to the ground.

By the time that the moving body has been reduced to rest
the point of application of the force, which has moved with the
moving body, has moved through a certain distance. Thus a cer-
tain amount of work has been done by the moving body. We are
thus led to the conception of the motion of a body possessing a
capacity for doing work.

For instance, in the previous examples, the motion of the hammer has
driven the nail into position, and the motion of the particle projected into
the air has raised it to a certain height above the earth's surface.

139. Let us suppose that a particle moving with velocity v is
opposed by a force P (in absolute units) acting in the direction
opposite to that of the motion of the particle. Let the particle
describe a distance ds in opposition to this force in time dt, an(
let its velocity change from v to v — dv in this time. The particle

then has a retardation — in the direction of its motion, or, what is
dt ,

the same thing, an acceleration — in the direction in which

ctt

is acting, so that by the second law of motion

p = m —
dt

KINETIC ENEBGY 169

The work done by the particle in moving the distance ds in
opposition to the force P is

Pds = m — ds,

or, since — is the same as the velocity v of the particle,
at

Pds = mv dv.

Integrating, we find that the whole work done by the particle
before being reduced to rest is

(36)

Since P has to be measured in absolute units (cf. § 22), it
follows (§ 111) that the work ^mv2 will also be measured in
absolute units.

Thus whatever the magnitude of the force opposing the motion
of a particle, the work performed by the particle before being
reduced to rest is the same, namely J- mv2 absolute units of work.

The quantity \mv2 (measured in absolute units) is called the kinetic
energy of a moving particle. It is equal to the amount of work which
can be performed by the particle before being reduced to rest.

Suppose, for instance, that the resistance offered by a nail to being
driven into a board is equal to the weight of 5000 pounds, i.e. that it would
require a weight of 5000 pounds to press it into the board. Suppose that
it is driven into the board by being struck with a hammer, of which the
head weighs 10 pounds, and hits the nail with a velocity of 50 feet per
second. Let s be the distance the nail is driven in at each stroke measured
in feet, then the work done by the hammer at each stroke is that of moving
a force of 5000 pounds weight — or 5000 x g poundals — through a distance
of s feet. It is therefore equal to 5000 gs foot poundals. The kinetic energy
of the hammer in striking the nail is

±mv2 = I • 10 • 502 = 12,500

in absolute foot-pound-second units. Thus from the relation (36) we have

the equation

5000 gs = 12,500,

in which, since the units are foot-pound-second units, we may take g = 32,

and so obtain

s = JU. feet = 44 inches.

170 WORK

140. THEOREM. During the motion of a particle under any sys-
tem of forces, the increase in kinetic energy is equal to the total
work done on the particle by external agencies.

Let us consider motion of a particle from one position P to a
second position Q, and let the velocities of the particle at these
two points be vp, VQ respectively.

Let us examine the motion over any element ds of the path,
and let the velocities at the beginning and end of this element be
v and v + dv. Let P be the force, or component of force along ds,
which acts on the particle while it describes the element ds of its
path. If dt is the time taken to describe this element of path, the

ci/ij

acceleration is — > and since the force acting in the direction of
at

motion is P, we have, by the second law of motion,

P = m —
dt

SjftJ

Hence, as in § 139, Pds = m ~— ds

da'

= m — dv
dt

= mv dv.

Integrating over the whole path from P to Q, we obtain

/»Q /»<?

I Pds = m I vdv
Jp JP

= $mv*-$mv* (37)

= increase in kinetic energy.

The left-hand side of this equation represents the work done on
the particle, proving the result required.

141. The work performed on the particle by external forces may
be regarded also as equal to minus the work performed by the parti-
cle on external agencies. For if P is the force acting on the particle
along ds, it follows from the equality of action and reaction that
the force acting on the external agencies from the particle is — P,

CONSERVATION OF ENERGY 171

so that the total work performed by the particle is — P ds. Thus
the theorem can be stated in the following alternative form :

During the motion of a particle under any system of forces, the
decrease in kinetic energy is equal to the total work done by the
particle against external agencies.

142. If the system of forces acting on the particle is a conserv-

CQ
ative system, the value of — / Pds, the total work performed by

Jp
the particle on external agencies, is equal, by § 132, to WQ — WP.

Thus equation (37) becomes

or again WQ + \mv%=WP+ ^mvp> (38)

so that the sum of the potential and kinetic energies is the same
at Q as at P, proving the theorem.

The sum of the potential and kinetic energies is called the total
energy of the particle.

CONSERVATION OF ENERGY

143. The kinetic energy of a system of bodies is obviously equal
to the sum of the kinetic energies of the separate particles. The
potential energy of the system, as has been seen, is the sum of
the potential energies of its particles.

Thus the total energy of a system is equal to the sum of the
total energies of the separate particles. Since the total energy of
each particle remains constant, it follows that the total energy of
the system remains constant.

The fact that the total energy remains constant is spoken of as
the Conservation of Energy. An equation expressing that the total
energy at one instant is equal to that at any other instant is
spoken of as an equation of energy.

144. As an illustration, let us consider the firing of a stone from a
catapult.

Work is performed in the first place in stretching the elastic of the
catapult, and the work is stored as potential energy of the stretched elastic.
As soon as the catapult is released, the stone is acted on by the tension of

172

WORK

the elastic ; the stone moves under the accelerating influence of this tension,
and the tension of the elastic slackens. While this is in progress the stone
is acquiring kinetic energy, while the stretched elastic is losing potential
energy. By the theorem just proved, the kinetic energy gained by the
stone must be just equal to the potential energy lost by the elastic.

When the Stone escapes from the^catapult, most of the potential energy
of the elastic will have disappeared, having been transformed into the
kinetic energy of the stone. After this a further transformation of energy
may take place while the stone is in motion. If the stone moves upwards,
its potential energy will increase, so that there must be a corresponding
decrease in its kinetic energy — its speed must slacken. On the other
hand, if the stone moves downwards, the potential energy will decrease,
so that its kinetic energy will increase — it will gain in velocity.

145. A very important deduction from the principle of the con-
servation of energy is the following :

THEOREM. If a particle slide along any smooth curve, being acted
on by no forces except gravity and the reaction with the curve, and
if u, v be the velocities at two points P, Q of its path, then

v* = u*+2gh, (39)

where h is the vertical distance of Q below P, — i.e. is the vertical
projection of the path PQ described by the particle.

Let hp, hQ denote the heights of P and Q above any horizontal
plane — - for instance, the earth'a surface.
Then when the particle is at P its kinetic
energy is ^ mu*, and its potential energy
is mghp. Thus its total energy is

1- mu2 + mghp.

Similarly at Q its total energy is
J- wy2 + mghQ.

Since the system of forces acting is a
conservative system, the total energy remains unaltered. Thus

1 mil? + mghp = ^ mv* + mghQ,

so that v2 — if = 2 g (hp — hQ) = 2 gh,

proving the theorem.

CONSERVATION OF ENERGY 173

146. The theorem of § 145 is clearly true when the particle is
ascending, in which case li is negative — or if the particle ascends
during part of its path and descends during the remainder. More-
over, the particle may move under any conservative system of
forces, provided only that the whole potential energy arises from
the weight of the particle, and the theorem remains true.

It is true, for instance, of a particle tied to an inextensible
string, or of a particle moving 'freely in a vacuum.

To illustrate the use of the theorem, let us suppose that a bicyclist,
riding with a velocity of 15 miles an hour, comes to the top of a hill of
height 60 feet, down which he coasts. Let us find his velocity at the
bottom, on the supposition that friction, air resistance, etc., may be
neglected.

Taking the top and bottom of the hill to be the points P, Q respectively
of the theorem just proved, we have, from the data of the problem,

h = 60 feet,

u = 15 miles per hour = 22 feet per second.

Thus, using foot-second units, we have

y2 = w2 + 2 gh = 222 + 2-32-60 = 4324,
so that v = 66 feet per second, approximately,

= 45 miles per hour.

Thus the velocity of the bicycle, if unchecked by friction or air resist-
ance, would be one of about 45 miles per hour.

EXAMPLES

1. An automobile running 40 miles an hour comes to the foot of a steep hill,
and at the same instant the engine is shut off. To what height up the hill will
the automobile go before coming to rest (neglect friction, etc.)?

2. A laborer has to send bricks to a bricklayer at a height of 10 feet. He
throws them up so that they reach the bricklayer with a velocity of 10 feet per
second. What proportion of his work could he save if he threw them so as only
just to reach the bricklayer ?

3. A gun carriage of mass 3 tons recoils on a horizontal plane with a velocity
of 10 feet per second. Find the steady pressure that must be applied to it to
reduce it to rest in a distance of 3 feet.

4. A ship of 2000 tons moving at 30 feet a minute is brought to rest by a
hawser in a distance of 2 feet. Find in tons what pull the hawser has to sustain.

174 WORK

5. A bicycle and rider weigh 200 pounds, and, when riding along a level
road at 26 miles an hour, the rider suddenly applies a brake which presses
on the tire with a force equal to the weight of 60 pounds. If the coefficient of
friction between the brake and the tire is |, find how far the machine will go
before coming to rest.

6. In the last question, how far will the machine go if, instead of the road
being level, it is down an incline of 1 in 20 ?

7. A bullet fired with a velocity of 1000 feet per second penetrates a block
of wood to a depth of twelve inches. Prove that if it were fired through a board
of the same wood, two inches thick, its velocity on emergence would be about
913 feet per second. (Assume the resistance of the wood to the bullet to be
constant.)

8. Two equal weights P and P are supported by a string passing over two

small smooth pulleys A and B in the same horizontal line, and a weight

o
W = — P is attached to the middle point of the string between A and B.

Prove that W will continue to descend until WAB forms an equilateral triangle,
and examine what will happen after this.

9. A string of natural length I and modulus X is suspended between two
points A, B in the same horizontal line and at a distance h apart, and has a
weight W attached to its middle point. The weight W is held at rest midway
between the points J., 5, and is suddenly set free. Find how far it will fall
before being brought to rest by the strings.

10. A heavy particle hangs by a string of natural length Z, which it stretches
to a length l\ the other end of the string being fixed. The particle is pulled
down to a length 2 1' below the point of support, and is then set free. How high
will it rise V

11. Determine the horse hower which could be obtained from the kinetic
energy of a river at a place where the width is 100 feet, the mean depth 20 feet,
and the mean velocity 4£ miles per hour. (A cubic foot of water weighs 62.6
pounds.)

12. The river of the last question ends in a waterfall of which the bottom
is 60 feet below the river bed. Find the horse power which could be obtained
from the water.

13. A locomotive burns l\ pounds of coal per horse-power-hour. How much
coal must be burned, beyond that consumed in overcoming gravity, friction, etc. ,
in giving to a train of 300 tons a velocity of 65 miles an hour ?

STABLE AND UNSTABLE EQUILIBRIUM

147. Let us consider a system at rest in a position of equilibrium,
and capable of moving from this position by only one path, over
which it may, of course, move in either direction. As an illustra-
tion of a system of this kind we may take a locomotive standing

STABILITY AND INSTABILITY 175

on a pair of rails, a door turning about a hinge, or a bead sliding
on a wire. The system is supposed to be acted on by any number
of conservative forces, but to be in a position of equilibrium under
these forces.

Let P denote the position of equilibrium, and let Wp be the
potential energy when the system is in configuration P. Let x
denote any coordinate which measures how far the configuration
of the system has moved from P — for instance, returning to our
former illustrations, x might denote the distance the locomotive had
moved along the track, the angle through which the door had
turned about its hinges, or the distance the bead had moved along
the wire. The value of x will of course be considered positive if the
system moves in one direction, and negative if it moves in the other.

As the system moves away from its equilibrium configuration
P, the value of x will change. The value of W, the potential
energy, will also change, and as it depends only on the value of x if
the forces are conservative, we may say that W is & function of x.

By a well-known theorem, we can expand W in powers of x in

in which the subscript P denotes (as it has already been supposed
to denote in the case of WP] that the quantity is to be evaluated
in the configuration P. Since the configuration P is supposed to
be one of equilibrium, we have by the theorem of § 135,

— o

ox '

so that equation (40) becomes

_ l 2 /32W\ * .. v

\ /-P
For configurations near to P, a; is small, so that the term

I ("a^rl m ®°luatio11 (41)' although itself small, is yet very

\ ^^ /P
large compared with the terms in xs, x*, etc., which follow it.

176 WORK

Thus, for configurations near to P, we may neglect these latter
terms altogether, and write the equation in the form

(C&TT/~\
— - ) may be either positive or negative.
^ IP
If it is positive, then W — WP is positive whatever the value of

x, so that the potential energy W in every configuration near to
P is greater than that in configuration P. In other words, W7 is a
minimum at P.

So also if / -— - J is negative, W — WP is negative for all small

\ °x /P
values of x, and we find that W is a maximum at P.

148. Suppose now that the system is placed at rest in some con-
figuration near to P. This configuration is not one of equilibrium,
so that the system cannot remain at rest. To determine the direc-
tion in which it begins to move, we^need only notice that as the
system moves it acquires kinetic energy, and as this must, by
§ 143, be acquired at the expense of its potential energy, we see
that the system will begin to move in such a direction that its
potential energy will be diminished.

A glance at equation (42) will show whether this direction is

towards or away from P. We see that if ( — — ) is positive, the

\ °x IP
value of a? must decrease, so that the motion will be towards P,

/32W\
whatever the value of x. Similarly, if ( — — - J is negative, the

\ d* IP
value of xz must increase, so that the motion will be always away

from P.

We have now seen that if the system is placed in a con-
figuration adjacent to P, the question of whether the motion
which ensues is towards or away from P does not depend on the
configuration in which the system is placed, but depends on the

* t&W\

Slgn of

STABILITY AND INSTABILITY 177

We have seen that if P is a configuration of equilibrium, and
if the system is slightly displaced from P to a neighboring con-
figuration, then

/O2TT7-\
v~7 ^ . — — ) is positive, the system, when set free, will return

\ d* IP
to its original position of equilibrium ;

(b) if / — — ) is negative, the system when set free will move
\ &zr /p

farther away from its original position of equilibrium.

Equilibrium of the first kind is called stable equilibrium ; equi-
librium of the second kind is called unstable equilibrium.
We can summarize the results as follows :

~. „ nf t&w\

Potential Energy W

Equilibrium

81911 °f ( ex* )P

minimum
maximum

stable
unstable

149. THEOREM. Positions of stable and unstable equilibrium
occur alternately.

We can assume that we are dealing only with finite forces, so
that the function W will always be finite : it can never pass
through the values W = ± oo . It must be continuous, for, by
hypothesis, the work done in placing the system in any configura-
tion must have a definite value, so that the potential energy can
have only one value for a given configuration. Also the differential
coefficients of the potential energy must be finite, for these measure
the forces (§ 133) which can have only finite values in any given
configuration.

Thus if the graph of the function W is drawn, we see that it
must consist of portions in which W is alternately increasing and
decreasing. On -passing from a portion in which W increases to
one in which it decreases, we pass through a point at which W is
a maximum, while in passing from a region in which W decreases

178

WORK

to one in which it increases, we pass through a minimum. Thus
maximum and minimum values of W must occur alternately, or,
what is the same thing, configurations of stable and unstable
equilibrium must occur alternately.

150. Examples of these two kinds of equilibrium can be found
in the illustrations already employed.

1. Locomotive moving on a pair of rails. Let h be the height of the center
of gravity in any position, let x denote distances measured horizontally
along the track, and let M be the mass of the locomotive. The potential

energy is then Mgh. The con-
dition for equilibrium in the
configuration x — 0 is

h maximum
EquiLunstable

h minimum
Equil. stable

FIG. 97

or — =0, expressing that the
dx

value of h must be either a maxi-
mum or a minimum. The table
on page 177 shows that if h is a
minimum, — i.e. if the center of
gravity is at its lowest point, —
the equilibrium will be stable.
Thus, if the locomotive is moved slightly from this position, it will roll
back to it again. If h is a maximum, — i.e. if the center of gravity is at
its highest point, — the equilibrium will be unstable. The locomotive is
now at the summit of a hill, and if displaced to either side of the summit,
will continue rolling down the hill.

NOTE. If the moving parts of the engine are not " balanced " properly, the center
of gravity may not always be at the same height above the rails, so that the maxima
and minima of h do not necessarily occur at points where the height of the track is a
maximum or a minimum. For instance, a position of equilibrium might occur where
the track was not level, or again a position of stable equilibrium might occur at a
point at which the track was at its highest point, the height of the center of gravity
above the rails being of course a minimum at this point. Thus if the engine were dis-
placed to a point slightly lower on the track, and set free, it would return of itself to
the highest point. The principle here is the same as that of mechanical toys which,
on being placed at rest at the foot of an inclined plane, start to roll up the plane as
soon as set free.

We notice that positions of stable and unstable equilibrium must occur
alternately, as already proved in § 149.

2. Door turning on hinges. Here again the potential energy is Mgh, where
h is the height of the center of gravity of the door above any standard

STABILITY AND INSTABILITY

179

level. As the door turns on its hinges, its center of gravity describes a

circle about the line of hinges. If this line is perfectly vertical, the circle

described by the center of gravity

lies entirely in a horizontal plane, so

that every position is one of equilib-

rium, and the question of stability or

instability does not arise. If, how-

ever, the line of hinges is not per-

fectly vertical, the circle will lie in

an inclined plane. The points at

which the height above the standard

horizontal plane is a maximum or

minimum are two in number :

P, the highest point of the circle, FIG. 98

at which equilibrium is unstable ;

Q, the lowest point of the circle, at which equilibrium is stable.

3. Bead sliding on wire. To obtain a definite problem, let us suppose
that the bead P slides on an elliptic wire placed so that its major axis A A'
is vertical, and let it be acted on by its weight, and
also by the tension of a stretched elastic string of
which the other end is tied to the center of the
ellipse. Let a, b be the semi-axes of the ellipse, and
let I, X be the natural length and modulus of the
string, I being greater than a, so that the string is
always stretched. Let w be the weight of the bead.
The first step is to calculate the potential energy
in any configuration. Let the configuration be
specified by the eccentric angle <f> of the point on
the ellipse occupied by the bead. The height of
the bead above the center of the ellipse is then
acos^, so that that part of the potential energy
which arises from gravitational forces is wa cos <f>.
The length of the string r is given by

r2 = a2 cos2 0 -f &2 sin2 0, (a)

and the work done in stretching the string from length I to length r is (§ 113)

This may be taken to be the part of the potential energy which arises
from the stretching of the string. Thus the total potential energy will be

W = wa cos

+ ^(r-02-

180

WORK

dW
The positions of equilibrium are now given by - = 0, or

-Z) — =0,
d<p

or, substituting for r from equation (a),

X, X (a2 — bz~) sin 0 cos 0 _

wa sm0 + -(a2 — &2) sm0 cos0 -- ^ } — 0.

* Va2COS20 _|_ J2sin20

Rationalizing, we find that roots are given by sin 0 = 0, and also by
[wa + -(a2 - 62) cos 01 2(a2 cos2 0 + Z>2 sin2 0) - X2 (a2 - &2)~ cos2 0 = 0,

which reduces to

[wa + - (a2 - 62) cos 01 2 f(a2 - 62) cos2 0 + &21 - X2 (a2 - &2)2 cos2 0 = 0, (c)

an equation of the fourth degree in cos 0.

The roots of sin 0 = 0 are 0 = 0, TT, so that there are always two

positions of equilibrium at A, A', the ends of the major axis. Equation

(c), being of the fourth degree, may have
0, 2, or 4 real roots in cos 0. The equa-
tion as it stands has been obtained by
squaring both sides of the equation to
be satisfied, and in doing this we have
doubled the number of roots of the true
equation. Thus the true equation will
only be satisfied by 0, 1, or 2 real roots
in cos 0. In other words, between A and
A', on either side of the wire, there can
be at most two positions of equilibrium.
It would be a tedious piece of work to find the actual values of the roots

d2W
for cos0, and then determine the signs of the values of -- — correspond -

«02

ing to these roots. The question is, however, very much simplified by
using the general theory of stable and unstable configurations.

If we put X = 0 in expression (&), we obtain as the potential energy in
the case in which X is vanishingly small in comparison with w,

W — ^

0=0

of which the graph is shown in fig. 100. Here there are only two positions
of equilibrium, namely 0 = 0 and 0 = TT, the former being unstable (£7)
and the latter stable (S).

STABILITY AND INSTABILITY

181

Again, if we put w = 0 in expression (&), we obtain as the potential
energy in the case in which X is infinitely great in comparison with w,

and the graph of W in this case is shown in fig. 101. There are four posi-

tions of equilibrium, TT 3 TT

0 = 0, -, TT, — ,

which are respectively unstable, stable, unstable, and stable.

The general case in which X stands in a finite ratio to w is intermediate
between the two extreme cases which have been considered. The graph
for W in the general case can be obtained by compounding the two graphs
already drawn. To obtain the ordinate corresponding to any value of 0,
we multiply the corresponding ordinates
in the graphs already obtained by the
appropriate constants, and add. The two
ordinates give the two terms of expres-
sion (5) separately : their sum gives the
total value of W as required.

From this geometrical construction it
is clear that 0 = 0 remains a configura-

FIG. 101

Between these two

0=0

tion of unstable equilibrium. The con-
figuration 0 = TT is also a configuration of
equilibrium, but may be either stable or unstable,
configurations there may be one other configuration of equilibrium, as in
fig. 101 ; or there may be none, as in fig. 100. Since, by § 149, stable and
unstable configurations occur alternately, it is clear that if the configuration
0 = TT is stable, there can be no other configuration of equilibrium between
this and 0 = 0, while if 0 = TT is unstable, there must be one configuration
of equilibrium between 0 = TT and 0 = 0, and this must be stable.

The stability or instability of the configuration 0 = TT accordingly deter-
mines the nature of the solution for a given value of X. This stability or

instability is in turn determined by the sign of — — at 0 = TT. To deter-
mine this, let us write TT — 0 = 6 near to 0 = TT, and neglect terms smaller
than 02. We have, to this approximation,

r2 = a2cos20 + &2sin20

— 7)2^ #2

so that by equation (&),

W = wa cos0 -\ (r —

182 WORK

d*W X(a2-

Thus — = wa -- ^

It follows that equilibrium at 0 = TT is stable or unstable according as

wan

X< or >

(a2 -

To sum up, there are two cases :

7*1/727

I. X < — The only positions of equilibrium are 0 = 0

(a2-62)(a-0
and 0 = TP, which are respectively unstable and stable.

II. X > There are positions of equilibrium 0 = 0 and

0 — TT, both unstable, and also an intermediate position which is stable.
This last position is determined by equation (c).

Critical and Neutral Equilibrium

d2W
151. If the value of — - at a position of equilibrium is zero,

the equilibrium is called critical. So far, we have not discovered
what happens when a system is slightly displaced from a posi-
tion of critical equilibrium.

In general, the value of W in the neighborhood of any position
of equilibrium can be expanded in the form (cf. equation (41))

-r-r-r- -i •/"»'.! -I Q / " '' \ . 1 A * V - r r 1 . / J O \

-^ + t^(-«3-|+l^«-s3-) + A^(-5Lr) + -- (43)

If -— - vanishes at P, the most important term in the value of

dx*

W— WP is that in #8, so that we have approximately

Here W—WP changes sign on passing through x = 0, the con-
figuration of equilibrium, so that the graph of W is as shown in

STABILITY AND INSTABILITY 183

fig. 102, having a horizontal tangent and point of inflection at P.
On one side the potential energy is less than at P, on the other
side it is greater.

Let Q, Q' be two adjacent configurations on these two sides
of P. If the system is placed at Q, it must move so that its poten-
tial energy decreases, and therefore moves away from P. If it is
placed at Qr, for the same reason it must move
at first towards P, but it will move beyond P
and will then continue to move away from P,
— for it cannot come to rest until its potential
energy is again equal to that at Q', and this - "
cannot happen in the neighborhood of P. Thus
if the system starts from any configuration in the neighborhood
of P, it will ultimately be moving away from P. In other words,
the equilibrium is unstable.

Thus if — — = 0 at P, the equilibrium is, in general, unstable.

.
An exception has to be made when — - = 0; for then we have

This case may be treated as in § 148, and we find that the

/ &W\
equilibrium is stable or -unstable according as / — — I is positive

, . \ 0X1 /ft

or negative. x

152. Higher degrees of singularity may be treated in the same
way, and we easily obtain the following general rules :

If the first differential coefficient which does not vanish is of odd
order, the equilibrium is unstable.

If the first differential coefficient which does not vanish is of even
order, the equilibrium is stable or unstable according as this differ-
ential coefficient is positive or negative.

It is possible for all the differential coefficients to vanish, in
which case the problem is best treated by other methods.

WOKK

For instance, if the potential energy is of the form

it will be found that all the differential coefficients vanish in the
configuration given by x = 0. On drawing a graph of the function
W it appears that- the equilibrium is stable.

It may be that all the differential coefficients of W vanish
because W is a constant throughout the whole of a range sur-
rounding the configuration under consideration. If this is so,
the system may be displaced, and there will be no force tending
to move it from its new configuration — every configuration is
one of equilibrium. Equilibrium of this kind is called neutral
equilibrium.

A case of neutral equilibrium has already occurred in Ex. 2, p. 179, — a
door free to swing about a vertical line of hinges. A second case is that
of a sphere rolling on a horizontal plane.

Systems possessing Several Degrees of Freedom

153. So far we have considered only systems which are limited
to moving through a single series of configurations — systems
with only a single degree of freedom. The determination of the
stability or instability of a system having more than one degree
of freedom is a more complex problem.

If the potential energy is absolutely a minimum in a position
of equilibrium, so that every possible motion involves an increase
of potential energy, then the equilibrium is stable. This obviously
can be proved by the same argument as has served when there is
only one degree of freedom.

If the potential energy is not an absolute minimum, — that
to say, if displacements are possible in which the potential ener^
decreases while moving away from the position of equilibrium,
then the configuration is one of unstable equilibrium. This will
proved later. It cannot be proved by the methods used in this chaj
ter, and so we defer the question until later (Chapter XII).

EXAMPLES 185

GENERAL EXAMPLES

1. Prove that the horse power of an engine which overcomes a resist-
ance of R pounds at a speed of S miles an hour is

RS -4- 375.

2. A train weighing, with the locomotive, 500 tons is kept moving at
the uniform rate of 30 miles an hour on the level, the resistance of air,
friction, etc. , being 40 pounds per ton. Find the horse power of the engine.

By how much must this horse power be increased if the rate is to be
maintained while water is taken up from a trough between the rails to the
amount of 20 pounds per foot passed over, the height to which the water is
raised above the trough being 10 feet, and the kinetic energy imparted to
the water in the trough, as well as that of the motion of the water taken up,
relatively to the tank, being neglected ?

3. The sides of a conical hill are of such a shape that a given mass will
just rest on them without slipping. A man wishes to move this mass from
a point at the base of the hill to a second point diametrically opposite to
the first. Show that the work of dragging it over the hill is less than the
work of dragging it round the base of the hill, in the ratio 2 : TT.

4. Show that the work a man does in dragging a weight up a hill from
a given point A to the summit B depends only on the positions of A and
B, and is independent of the shape of the hill, provided he keeps always
in the vertical plane through A and B.

5. A catapult is made by tying the two ends of a piece of elastic, natural
length a, modulus of elasticity X, to the two prongs of a forked piece of
wood, distant I apart, I being greater than a. A stone of mass m is placed
at the middle point of the catapult, and is drawn back until the string is
stretched to double its natural length. If it is then set free, find the
velocity with which it will leave the catapult.

6. If , in the last question, the stone is projected vertically upwards from
the catapult, find the height to which it will rise before coming to rest.

7. A necklace of mass m is made of beads threaded on a light string of
modulus X. It is held in a horizontal plane, with the string unstretched,
resting on the surface of a smooth right circular cone of semivertical angle
a, of which the axis is vertical. If the necklace is let go, how far will it
slip down the cone before coming to rest ?

8. A fly wheel is of radius 2 feet 6 inches, and the weight of the spokes,
etc., may be neglected in comparison with that of the rim. It is rotating
at the rate of 250 revolutions per minute about a fixed axle, which is
3 inches in diameter, the coefficient of friction between the wheel and axle
being ^. If it is left to itself, find how many revolutions it will make
before stopping.

186 WORK

9. A spider hangs from the ceiling by a thread of modulus of elasticity
equal to its weight. Show that it can climb to the ceiling with an expend-
iture of work equal to only three quarters of what would be required if
the thread were inelastic.

10. A fine thread having two masses each equal to P suspended at its
ends is hung over two smooth pegs in the same horizontal line, distant
2 a apart. A mass Q is then attached to the middle point of the portion of
the string between the pegs and allowed to descend under gravity. Show
that its velocity after falling a depth x will be

l{2g(x2 + a2)(Qx + 2 Pa - 2Pa:2 + a2)>[
\t Q(x2 + a*) + 2Px* JT

11. Assuming that the attraction of the earth on a body outside the
earth falls off inversely as the square of the distance of the body from the
earth's center, find with what velocity a shot would have to be fired ver-
tically upwards from the earth's surface so as never to return to the earth
at all.

12. A steam hammer of weight 30 tons is pressed down partly by its
weight and partly by the pressure of steam in a vertical cylinder acting on
a piston which moves with the hammer. The area of the piston is 4 square
feet, and the steam pressure is 225 pounds to the inch. If the hammer is
raised a height of 2 feet above its block, and set free, find the velocity with
which it will strike the block.

13. The ends of a uniform rod of length / are connected by a string of
length a which is placed over a smooth peg. Show that the rod can only
hang in a horizontal or vertical position, and examine the stability or
instability of these positions.

14. Two equal uniform rods are rigidly jointed in the shape of the let-
ter L, and placed astride a smooth circular cylinder of radius a. Find
the smallest length of the rods consistent with stability of equilibrium, the
rods being constrained to remain in a vertical plane perpendicular to the
axis of the cylinder.

15. A cube of stone of edge a rests symmetrically and with its base
horizontal on a rough circular log of diameter b. Show that the equilib-
rium is stable or unstable according as b > or < a.

16. A rocking stone rests on a fixed stone, the contact being rough, and
the common normal at the point of contact being vertical. If p, p' be
the radii of curvature of the surfaces of the two stones at the point of
contact, and if h be the height of the center of gravity of the movable stone,
show that the equilibrium of the rocking stone will be stable or unstable
according as 111

- > or < - + — •
h p p

EXAMPLES 187

17. A ladder of length h and weight w stands in a vertical position on
a rough floor, an elastic string being tied to its topmost point and to a
point in the ceiling at a height b above the floor, its tension being T. Show
that the equilibrium is stable or unstable according as

b -

18. If the tension in question 17 is equal to w(b — h)/2b, determine
whether the equilibrium is stable or unstable.

19. If in question 14 the rods are of the critical length which separates
stability from instability, show that the equilibrium is neutral, so that the
rods can, within certain limits, rest in any position in the plane perpen-
dicular to the axis of the cylinder.

20. Show that the rods in the last question are in stable equilibrium
as regards displacements in which the plane of the rods rotates about a
vertical axis, and find the couple required to hold the rods in a position in
which the plane makes any given angle 6 with the axis of the cylinder.

21. Find the smallest length of the rods in the last question, in order
that the equilibrium may be stable for all possible displacements.

22. The radii of curvature at the blunt and pointed ends of a hard-boiled
egg are a and b respectively, and the egg can just be made to balance on
its blunt end when stood on a rough horizontal surface. Show that it can
be made to balance on its pointed end if stood inside a hemispherical basin
of radius less than

where c is the longest axis of the egg. If the radius of the basin is just
equal to a(c — b)/(a + b — c), would the equilibrium be stable or unstable?

23. A, B, C are three equidistant smooth pegs in the same horizontal
line, and a heavy uniform string has its ends tied to A, C, and is looped
over B. Show that there may or may not be a position of equilibrium in
which the two catenaries AB, EC are unequal, and that if there is such a
position it will be stable.

Show also that the position of equilibrium in which the middle point of
the string is at B is unstable or stable according as an unsymmetrical
position of equilibrium does or does not exist.

CHAPTEE VIII
MOTION OF A PARTICLE UNDER CONSTANT FORCES

154. The simplest case of motion of a single particle occurs
when the particle is acted upon only by constant forces and moves
in a straight line.

If P is the component force in the direction of the motion of
the particle, there will, by the second law of motion, be an

acceleration / given by

P = mf,

where m is the mass of the particle. Since the forces are, by
hypothesis, constant, the acceleration / is also constant.

Let the particle start with a velocity u, and move with a con-
stant acceleration /. In time t the increase in velocity is ft, so
that, after any time t, the whole velocity is u + ft. Denoting this

velocity by v, we have

v = u+ft. (44)

By definition, v is equal to — > where s is the space described

Cbv

from the beginning of the motion. We accordingly have

an equation giving the rate of increase of s at any instant. Integrat-

ing, we obtain

s = ut + \ft\ (45)

no constant of integration being needed, because the distance
described at time t = 0 has to be 0, from the definition of s,
By equation (44), u = v — ft, so that equation (45) can be written

. = vt - \ft\ (46)

188

MOTION UNDER CONSTANT FORCES 189

This gives the distance described in time t, when we know the
velocity v with which the particle arrives at the end of its
journey.

Combining equations (45) and (46), we have

s = $(u + v)t, (47)

showing that the space described is the arithmetic mean of ut and
vt : the former is the space that would be described if the particle
maintained its original velocity u through the whole time t ; the
latter is that which would be described if the particle had its final
velocity throughout the whole time.

Combining equation (47) with equation (44), which can be

written in the form

ft = (v- u),

we obtain, on eliminating t,

2fs = v2 - u2, (48)

an equation connecting the space described with the initial and
final velocities.

This last equation may also be deduced from the equation of
energy. Since the work done on the particle is equal to the change
in its kinetic energy, we have

Ps = 1- mv2 — |- mu*,
and since P = mf, equation (48) follows at once.

BODY FALLING UNDER GRAVITY

155. The simplest application of these equations is to the
motion of a body which is allowed to fall freely under the influ-
ence of gravity, so that the acceleration is g.

If the body starts from rest, we put u = 0, and measure s
vertically downwards. We find from equation (45) that after
time t the body has fallen a distance \gt2, while its velocity

190

MOTION UNDER CONSTANT FORCES

time

is ft. After falling a distance h its velocity is, by equation (48),
equal to ^J'lgh. This is frequently spoken of as the "velocity

due to a height h"

We notice that the distance
fallen varies as the square of the
time during which the body has
been falling. In fig. 103 the
time is measured horizontally,
while the distance fallen is
measured vertically. The thick
curve gives a graphical repre-
sentation of the distance fallen.
Denoting the horizontal dis-
tance by x and the vertical by
y, we have x = t,y = ^gt*, so that

distance
fallen

FIG. 103

156. This is the equation of a parabola, so that the curve is a
parabola. The graph can be obtained experimentally by a method
known as Morin's method. A weight
P is free to fall vertically in a slot
formed in a rod AB, and is arranged
so that, as it falls, a pencil attached to
it makes a mark on a drum CD which
is covered with paper. The drum is
made to rotate uniformly. On unroll-
ing the paper from the drum we obtain
the graph of fig. 103, — for the hori-
zontal distance is proportional to the
time, while the vertical is the distance
fallen through. The fact that the curve
obtained in this way is accurately a
parabola gives experimental confirma-
tion of the fact that motion under gravity is motion with uni-
form acceleration.

FIG. 104

BODY FALLING UNDER GRAVITY 191

157. If the body is projected vertically upwards with velocity
u, we may measure the distance s vertically upwards, and the
acceleration in this direction will be — g. Thus we have

v = u — gt,
2 gs — u2 — vz,

where s is the distance upwards described after time t, and v is
the upward velocity. From the first equation we see that s = 0

not only when t = 0, but also when t = --- Thus the particle

returns to its original position after time 2 u/g. When s = 0 the
third equation shows that u2 = v*. Thus when the particle returns,
its velocity is the same as when it started. Clearly this must be
S0j for the potential energy is the same, and therefore the kinetic
energy also is the same.

EXAMPLES

1. If an express train is doubled, the first half being given 5 minutes' start,
and attaining its maximum booked speed of 48 miles an hour after moving with
constant acceleration for a mile, prove that the two halves will run about 4
miles apart, but the first half will have gone 3 miles before the start of the
second half.

\>/2. A train passes another on a parallel track, the former having a velocity
of 45 miles an hour and an acceleration of 1 foot per second per second, the
latter a velocity of 30 miles an hour and an acceleration of 2 feet per second
per second. How soon will the second be abreast of the first again, and how
far will the trains have moved in the meantime ?

3. A body is dropped from a balloon at a height of 70 feet from the ground.
Find its velocity on reaching the ground, if the balloon is (a) rising, (&) falling,
with a velocity of 30 feet a second.

(y/4. A stone is dropped into a well, and the sound of the splash reaches the
top after 9 seconds. Find the depth of the well, the velocity of sound being
1100 feet per second.

1 5. An elevator descends with an acceleration of 5 feet per second per second
until its velocity is 20 feet per second, after which its velocity remains uniform.
After it has been in motion for 6 seconds, a stone is dropped on to it from the
point at which the elevator started. How soon will it strike the elevator ?

6. A juggler keeps three balls going with one hand, so that at any instant
two are in the air and one in his hand. If each ball rises to a height of 4 feet,
show that the time during which a ball stays in his hand is £ second.

192 MOTION UNDEE CONSTANT FOECES

7. A body was observed to take t seconds in falling past a hatchway to the
bottom of a hold h feet deep. Prove that it fell

and struck with velocity

- 4- - gt feet per second.
t 2

8. A chain 12 feet long hangs from its upper end. If this be released, find
the time the chain will take in passing a point 60 feet below the initial position
of the highest point.

9. A body whose mass is 5 pounds, moving with a speed of 160 feet per
second, suddenly encounters a constant resistance equal to the weight of
^ pound, which lasts until the speed is reduced to 96 feet per second. For
what time and through what distance has the resistance acted ?

10. Two wagons, coupled together, are pulled along a horizontal track by
a steady force, and move over 100 feet in the first ten seconds from rest. The
rear wagon is then uncoupled, and it is found that at the end of the next ten
seconds the interval between the two wagons is 150 feet. Compare the masses
of the two wagons, all resistance being neglected.

11. A balloon of weight W is rising with acceleration /. If a weight w of
sand be emptied out of the car, find the increase in the acceleration of the
balloon, neglecting the resistance of the air and the buoyancy of the sand.

MOTION ON AN INCLINED PLANE

158. Suppose we allow a particle to slide down an inclined
plane, the contact between the two being supposed perfectly

smooth. If m is the mass of the par-
ticle, the forces acting on it are its
weight mg, and the reaction R normal
to the plane. The component down the
plane is mg sin a, so that the particle
moves with uniform acceleration g sin a.
We can obtain the distance described

FIG. 105

in the time t from the usual formula.

If the particle starts from rest, the distance described in time t
is ^ g sin a • t2.

159. Suppose that through a point 0 we have a great number of
smooth wires on which smooth beads are free to slide. Let these"

MOTION ON AN INCLINED PLANE

193

wires make all possible angles with the vertical, one of them, 00',

being vertical. Let us imagine that the beads are all collected at

0 and are set free simultaneously.

After time t, let the bead which is

falling vertically be at P, and let the

bead which is falling along a wire

inclined at an angle {3 to the vertical

be at Q. This latter bead moves with

acceleration g cos 0. Thus OP = \ gf,

while OQ = \g cos /3 • f. Hence

OQ=OPcos{}, and therefore OQP

is a right angle. It follows that Q

is on the sphere constructed on OP

as diameter, and obviously the same

will be true of every other bead.

Thus at any instant all the beads

will be on a sphere of which 0 is the highest point, and of which

the lowest point is at a distance \gt* below 0. Hence as the

motion proceeds the beads will appear to form a sphere which

continually swells out in size, the highest point appearing to

remain fixed at 0, while the lowest point appears to fall freely

under gravity.

160. This imaginary experiment
indicates a way of solving a prac-
tical problem. Suppose we wish
to place a smooth plane or wire in
such a position that a particle will
pass down it from a fixed point 0
to a given fixed surface in the least
time possible. Let us suppose that
we fix the apparatus of wires and
beads at 0, that we set the beads
free simultaneously, and watch the
increase in size of the sphere which
FIG. 107 they form. As soon as the sphere

194

MOTION UNDER CONSTANT FOECES

reaches such a size that it touches the fixed surface at some point
P, one of the beads has arrived at this surface and, moreover, has
arrived in shorter time than any of the others. Thus it has found
the quickest path from 0 to the surface. This path is OP, and
we can now fix the path without performing the experiment, from
the knowledge that a sphere drawn so as to have 0 as its highest
point, and to pass through P, must touch the surface at P.

In the same way, if we wish to
find the quickest time from a surface
to a fixed point 0 below it, we have
to find a sphere which touches the
surface at some point P and has 0 for
its lowest point. Then PO will be the
path required. For it is seen at once
that the time down all the chords of
this sphere which pass through 0 is
the same, so that the time down PO
is equal to the time down any other
chord QO, and therefore less than the time down the complete
path Q'O from the surface to 0, of which the chord QO is a part.

ILLUSTRATIVE EXAMPLE

A ship stands some distance from its pier,
and it is required to place a chute at some point
of the ship's side, so that the time of sliding
down the chute on to the pier may be as short

as _

Clearly the lower end of the chute must just
rest on the nearest point O of the pier, and the
problem reduces to that of drawing a sphere to
have 0 as its lowest point and to touch the ship's
side. Assuming the ship's side to be vertical,
the tangents to this circle at the ends of the
chute must be horizontal and vertical ; whence it
is easily seen that the chute must be placed so )
that it makes an angle of 45° with the vertical.

FIG. 109

ATWOOD'S MACHINE 195

EXAMPLES

1. A body is projected with a velocity of 20 feet per second up an inclined
plane of angle 45°. Find how high up the plane it will go, and how long it will
take in going up.

2. Two particles slide down the two faces of a double inclined plane, the
angles being a and /3. Compare the times they take to reach the bottom, and
the velocities they acquire.

3. A body is projected down an inclined plane of length I and height &, from
the summit, at the same instant as another is let fall vertically from the same
point. Prove that if they strike the base at the same time, the velocity of
projection of the first must be

[7
\2h'

4. Give a construction for finding the line of quickest descent from a fixed
point to a circle in the same vertical plane.

5. Particles are sliding down a number of wires which meet in a point, all
having started from rest simultaneously at this point. Prove that at any instant
their velocities are in the same ratio as the distances they have described.

^/6. A railway carriage is observed to run with a uniform velocity of 10 miles
an hour down an incline of 1 in 250, and on reaching the foot of the incline
runs on the level. Find how many yards it will run before coming to rest, assum-
ing the resistance to be constant and the same in each stage of the motion.

7. Prove that if a motor car going at 100 kilometers an hour can be stopped
in 200 meters, the brakes can hold the car on an incline of about 1 in 6 ; and
determine the time required to stop the car.

8. A carriage weighing 12 tons becomes uncoupled from a train which is
running down an incline of 1 in 250 at a rate of 40 miles per hour. The f ric-
tional resistance is 14 pounds weight per ton. Find how far the carriage will
go before coming to rest.

9. The pull of the locomotive exceeds the ordinary resistances to the motion
of a train by -fa of its whole weight ; and when the brakes are full on there is a
total resistance of T^ of its whole weight. Find the least time in which the
train could travel between two stopping stations on the level 3 miles apart.

10. In the last question, find the time if the track is down a gradient of
1 in 100.

ATWOOD'S MACHINE

161. It is difficult to measure the acceleration produced by
gravity from direct observations on a body falling freely, because
either the distance fallen must be very great or else the time
of falling very small. These difficulties are to some extent obviated
in a machine designed by Atwood. •

196

MOTION UNDER CONSTANT FORCES

If a string having two equal weights attached to its ends is placed
over a smooth vertical pulley, so that the weights hang freely, it is
obvious that there will be equilibrium. If the weights are unequal,
equilibrium cannot exist. In Atwood's machine the difference
between the weights is made small, so that the
motion is slow and is therefore easily measured.
Let mlt m2 be the masses of the weights,
of which the former will be supposed to be the
greater. When set free, let us suppose that
the former descends with an acceleration /.
Regarding the string as inextensible, the second
mass must ascend with an acceleration /.

The string will be treated as weightless, so
that the mass- of any element of it may be dis-
regarded. The second law of motion accordingly shows that the
resultant force acting on any element must vanish. Thus the forces
acting on the string must be in equilibrium (even although the
string is not at rest), and it follows, as in § 54, that the tension
must be the same at all points, say T.

. The forces acting on either mass consist of its weight acting
downwards and the tension of the string acting upwards. Thus
the resultant downward forces on the two masses are respectively
m^g — T and mzg — T. The equations of motion for the two masses
are accordingly m^g _ T = mj^

m*9 — T = — m2f.

If we eliminate T, we obtain

mi-m ^

/~ '

giving the acceleration. On eliminating /, we obtain as the value

of the tension o m

(50)

Clearly, if ml is nearly equal to ra2, the acceleration will be small. For
instance, if the weights are 100 and 101 grammes, we find that
/= ^jg =® feet per second per second.

MOVING FRAME OF REFERENCE 197

An acceleration of this smallness could easily be measured ; for instance,
the heavier mass would only descend eight feet in ten seconds. In prac-
tice, the difficulty arises that if the difference of the weights is made too
small, the forces acting on the pulley are so evenly balanced that their
difference is not sufficient to overcome the friction of the bearings, etc.

MOTION REFERRED TO A MOVING FRAME OF REFERENCE

162. It has already been seen (§25) that the second law of
motion remains true when the motion is measured relative to a
frame of reference which is not at rest but is moving with a uni-
form velocity. It is easy to find how the statement of this law
must be modified when the frame of reference moves with a known
acceleration.

Let a be the acceleration of the frame of reference, let / be the
component of the acceleration of a moving particle in the direction
of the acceleration a, and let P be the component in this direction
of the force acting on the particle. By the second law of motion

P = mf, (51)

where /' is the component acceleration referred to a frame of refer-
ence at rest. The acceleration/' may, however, be regarded as com-
pounded of the acceleration / of the particle relative to the moving
frame of reference, together with the acceleration a of this frame
relative to one at rest. Since these accelerations are all in the
same direction, we have ff=f-}-a,so that equation (51) becomes

P = m(f+a).

We can also write this in the form

P — ma = mf, (52)

showing that the motion is the same as if the frame were at rest,
provided we imagine the force P diminished ~by an amount ma.

This result can easily be interpreted physically. Of the force P,
a part equal to ma is used up in causing the particle to keep pace
with the moving frame of reference. It is only the remaining part,
P — ma, which is available for producing accelerations relative to
the moving frame.

198

MOTION UNDER CONSTANT FORCES

163. Frame moving with vertical acceleration. If the frame
of reference moves with an acceleration a vertically downwards
we see that before measuring accelerations relative to this frame
we must suppose the vertical component of force on each particle
of mass m to be diminished by ma. Whatever forces are acting
there will be amongst them the weights of the particles mg, etc
We can conveniently suppose the diminution ma to be taken
from these, so that the weight of a particle, instead of being
taken to be mg, will be taken to be m (g — a).

Thus the acceleration of the frame of reference may be allowec
for by supposing the acceleration due to gravity to be diminishec
from g to g — a.

For example, if an Atwood's machine is placed in an elevator, then at
the instant at which the elevator has an upward acceleration a, the accel
eration of the masses relative to the machine will be (cf . equation (49))

while the tension of the string will be (cf . equation (50))

164. Effect of earth's rotation on the value of g. As we have
seen (§ 25), a frame of reference which is fixed relatively to the

earth's surface possesses an acceleration
in consequence of the rotation of the
earth about its axis.

Let 00' be the earth's axis, and let
P be any point on the earth's surface
in latitude X. Regarding the earth as a
sphere of radius a, the point P will de-
scribe a circle of radius a cos X having
its center N on the earth's axis. If v is
the velocity with which P describes the
circle, the acceleration of P is, by § 12

a cosX

towards the center of the circle, — i.e. along PN.

MOVING FRAME OF REFERENCE 199

Let co be the angular velocity of the earth, — i.e. let it turn through
ft> radians per unit time. Then the time of P describing a com-
plete circle is the same as the time required for the earth to perform

, , , . , 2 7T , . . , 27rttCOSX

a complete revolution, namely This time is also

TT 1_ ® ^

Hence we have

i) = aco cos X.

The acceleration of the frame of reference is now seen to be

— = co2a cos X
a cosX

along PN. The motion of any particle referred to a frame moving
with P may accordingly be calculated as though with reference to
a fixed frame, provided the component of force in the direction
PN is diminished by mcaPa cos X.

Thus the total force acting may be supposed to consist of the
forces which actually do act, combined with a force mca?a cos X
along NP. Compounding this last force with the earth's attrac-
tion, we obtain a force which may be called the apparent force of
gravity at P. Thus the motion of the frame of reference may be
allowed for by using the apparent force of gravity in place of the
true attraction of the earth. It is this apparent gravity which is
determined experimentally, and which is always meant in speaking
of the weight of a particle at any point.

To find the apparent weight of a
body at the point P, we have to com-
pound its true weight, say mG acting
along PC, with a force mco2a cos X along

NP. Let the latter force be resolved

C
into its components FIG. 112

— mco^a cos2 X, m&>2a cos X sin X
along PC, PT respectively, PT being the tangent at P.

Compounding with the force mG along PC, we find for the com-
ponents X, Y of the apparent weight along PC, PT respectively,

X = m (G - a>2a cos2 X), (53)

Y = rao>2 a cos X sin X. (54)

200 MOTION UNDER CONSTANT FORCES

Squaring and adding, and denoting the apparent weight, as usual,
by mg, we obtain

m*g* = X2+Y2 = ra2(<92- 2<A G cos2X-f o>Vcos2X). (55)

Taking the diameter of the earth to be 7927 miles, and the
value of G (the acceleration due to gravity at the North Pole)
to be 32.25, we easily find that

~ = 290'

The square of this is so small that to a first approximation it
may be neglected, and equation (55) may be written in the form
g — G — a?a cos2 X.

Thus the apparent weight in latitude X is less than the true
weight by mcoza cos2 X, or about -%%-Q cos2 X of the whole weight.

The apparent weight does not act along the radius CP. If we
suppose it to act at an angle 6 with this radius, we obtain, from
equations (53) and (54),

a _ Y _ ca?a cos X sin X
~~

cos X sin X, approximately,

giving the deviation of the plumb line from the earth's radius at
any point.

FRICTIONAL REACTIONS BETWEEN MOVING BODIES
165. It is found experimentally that the relation

(in which F, R are the tangential and normal components of the
reaction between two bodies) remains very approximately true
when the bodies are sliding past one another. The value of ,
the coefficient of friction, is not quite the same as when the
bodies are at rest, the latter being always somewhat larger.

Friction between two bodies which are sliding past one another
is called dynamical friction, that between two bodies at rest being
called statical friction.

ILLUSTEATIVE EXAMPLES 201

ILLUSTRATIVE EXAMPLES

1. Two particles of masses mi, m2 are placed on two inclined planes of angles
a, /3, placed back to back, and are connected by a string which passes over a smooth
pulley at the top of the planes. If the coefficients of friction between the particles
and the planes are /*i , /xa , find the resulting motion.

If motion occurs at all, one particle, say mi, must move down its plane, while
the other, mg, will move up. Since the string is inextensible, the acceleration
of each will be the same, say / in the direction in which motion is taking place.

The forces acting on the
first particle are

(a) its weight mig ver-
tically down ;

(6) the tension of the
string, say T, up the plane ;

fj.E normal to and up the

, FIG. llo

plane.

Since the particle mi has no acceleration normal to the plane, the component
of the resultant force in this direction must be zero. Resolving in this direction

we obtain

R — mig cos a = 0.

Resolving down the plane,

mig sin a — /J.R — T = mi/,

and if we eliminate the unknown reaction 22, we obtain

•

migr (sin a — p. cos a) — T = mi/. (a)

A similar equation can be obtained for the motion of the second particle,

namely

mzg (sin p + /j. cosjS) — T = — m2/. (6)

Solving equations (a) and (6) for/, we obtain

_ mj(sinor — /j, cos a) — mgKsinff -t- /* cos/3)

mi +. m2
giving the acceleration.

If this value of / comes out negative, we see that the acceleration cannot be
in the direction in which motion has been assumed to take place.

If the system starts from rest, motion in the direction assumed is found to
be impossible, and we must proceed to examine whether motion in the opposite
direction is possible. If this also is found to be impossible, the system will
remain at rest.

202 MOTION UNDEK CONSTANT FORCES

If, however, the system is known to have been started in motion in the direc-
tion assumed, then the acceleration given by equation (c) will be in operation,
increasing the velocity if positive, and decreasing it if negative. In the latter
case the system will in time be reduced to rest, and we must then examine
whether or not it will start into motion in the reverse direction.

2. To one end of the string of an Atwood's machine a weight of mass mi is
attached. To the other end a smooth pulley of mass m% is attached, over which
passes a string with masses m^, m± hanging at its ends. Find the motion.

Let the mass mi be supposed to have an acceleration /, measured downwards.
Then m2 must have an acceleration / upwards. The masses m3, m4 will them-
selves form an Atwood's machine, the whole of which
moves upwards with an acceleration/. Thus the ten-
sion in the string of this machine, say TI, is (cf . § 163)

(a)

m2 If we denote the tension in the string connecting mi
and m2 by T2, we have as the equation of motion of m2,

while the equation of motion of mi is

FIG. 114

Eliminating 2\ and T2 from equations (a), (6), and (c), we obtain as the
value of the acceleration /,

mi — m2 —

The accelerations of the masses

, m* relative to m2 are known, by § 163,

3. At equal intervals on a horizontal circle n small smooth rings are fixed,
and an endless string passes through them in order. If the loops of the string
between each consecutive pair of rings support n pulleys of masses P, Q, R, • • •
respectively, the portions of string not in contact with the pulleys being vertical,
show that the pulley P will descend with acceleration

ILLUSTRATIVE EXAMPLES

203

The tension of the string must be the same throughout, say T. If the accel-
erations of the pulleys are /p, /Q, • • • , all measured down, we have equations
of motion of the type

Pg-2T=PfP,

(a)

one equation for each pulley. The unknown
quantity T enters these equations, as well as
the n unknown quantities /p, /Q, • • • . Thus
there are n + 1 unknown quantities, and so
far only n equations connecting them. An-
other equation is therefore required, and this
is obtained by noticing that the accelerations
/p, /Q, • • • cannot be independent, for the
length of the string must remain unaltered.

Let us denote the depths of P, Q, • • • below the horizontal ring by SP,

Then SP + SQ -\ ----

must be constant throughout the motion. It follows that

/P+/Q+ •••=().

Substituting the values of /p,

from equation (a), we obtain

so that

-4- |
i^

and on substituting this value for 2 T in equation (a), we obtain the required
value of /p.

EXAMPLES

1. Show that the tension of the string in an Atwood's machine is intermediate
between the weights of the two masses. Show also that it is nearer to the
smaller than to the larger of these weights.

2. Two weights 16 and 14 ounces respectively are connected by a light inex-
tensible string which passes over a smooth pulley. The weights hang with the
strings vertical and the string is clamped so that no motion can take place. If
the string is suddenly undamped find the change in the pressure exerted on the
pulley.

3. A string passing across a smooth table at right angles to two opposite
edges has attached to it at the ends two masses P, Q which hang vertically.
Prove that, if a mass M be attached to the portion of the string which is on the
table, the acceleration of the system when left to itself will be

P-Q

204 MOTION UNDEK CONSTANT FORCES

4. Two masses ra, m' are tied to the two ends of a string which is slung over
a peg, as in an Atwood's machine. The peg is not smooth, the angle of friction
between it and the string being e. Find the motion.

5. In question 3, let the coefficient of friction between the table and the
weight M be /A, that between the table and string being /*'. Find the motion.

,, 6. A rope hangs over a smooth pulley. Find the uniform acceleration
with which a man weighing 10 stone must pull himself up one end of the
rope, for the rope to be. kept at rest by a weight of 12 stone hanging from
the other end. »' Of"

7. A monkey is tied to one end of the string of an Atwood's machine, the
other end having attached to it a weight exactly equal to that of the monkey,
and at just the same depth below the pulley. The monkey suddenly starts to
climb up his string. Which will rise the faster, the monkey or the weight on
the other string ?

8. Weights of 10 pounds and 2 pounds, hanging by vertical strings, balance
on a wheel and axle. If a mass of 1 pound be added to the smaller weight, find
the acceleration with which it will begin to descend, and the tension of each
rope. (The inertia of the wheel and axle is to be neglected.)

9. A mass of 5 pounds resting on a smooth plane inclined at 30 degrees to
the horizon is connected by a fine thread, which passes over a pulley at the
summit of the plane, with a mass of 3 pounds hanging vertically. Compare the
pull in the thread when the mass on the plane is "held fixed and when it is let
go. If the thread is severed or burnt 8 seconds after this mass has been let go,
find how far it will rise on the plane before falling back.

10. A light thread passes over two fixed pulleys A and B, and carries
between them a movable pulley block O, under which it passes. A mass M is
attached to each end of the thread, and a mass m to the movable block. The
masses of the pullers- are negligible, and the pulleys are so arranged that all
the segments of the thread are vertical. Show that, when the system is let go,
the tension in the thread is mM/(M + ^ m) pounds, and find the acceleration
with which the mass m falls.

11. An elastic band of mass m, natural length a, and modulus X is placed
round a rough horizontal wheel of circumference b (> a). How fast must the
wheel be made to rotate for the band to leave the wheel ?

12. The elastic band of question 11 is placed on a smooth sphere of circum-
ference b rotating with angular velocity w. Find the position of rest.

13. If the earth rotates faster and faster until ultimately bodies fly off from
its equator, show that by the time this stage is reached the plumb line at any
point will be parallel to the earth's axis.

14. A body is placed on a spring balance when in a ship which is sailing
along the equator with velocity v. Show that if the balance weighs accurately
when the ship is at rest, its reading when the ship is in motion will show an

error of times the weight of the body (approximately), where w is the

9
angular velocity of the earth.

PROJECTILES 205

FLIGHT OF PROJECTILES

166. By a projectile here is meant any body which is small
enough to be regarded as a particle, and which is projected in such
a way that it describes a path under the influence of gravity.

A projectile will, in general, be influenced by the resistance of
the air as well as by gravity, but we shall suppose the resistance
of the air to be negligible, so that gravity will be the only force
which need be taken into account.

To take the simplest case first, let us imagine that the projectile
is projected horizontally from the point 0 (fig. 116), with velocity u.
The only force acting is gravity, which o
has no horizontal component, so that
the horizontal velocity remains equal to
u throughout the motion. The initial
vertical component of velocity is nil,
but there is a downward acceleration g.
Thus after time t, the horizontal dis-

tance described is ut, while the vertical

• i o -r^ • FIG. 116

distance fallen is \gt. Denoting the

horizontal distance described by x, and the vertical distance fallen
by y, we have x = uf

The equation of the path described is obtained by eliminating t
from these equations, and it is found to be

2 u*
This is a parabola, of which the latus rectum is --

Clearly the problem of determining the curve is essentially the same as
in § 156. There we have a body falling freely, and tracing its path on a
paper which moves post it with a uniform horizontal velocity. Here we have
a body falling freely, and can imagine it to trace its path on a paper past
which it moves with a uniform horizontal velocity. The relative motion is
the same in the two cases, so that the curves are necessarily the same.

206

MOTION UNDER CONSTANT FORCES

167. At 0 the velocity of the particle is u, which is the velocity

due to a height This is equal to a quarter of the latus rectum,

and is therefore equal to the depth of 0, the vertex of the parabola,
below the directrix XM. Thus the total energy of the projectile

when at 0 is equal to that of the same
projectile at rest at Xy or, of course, at
any other point of the directrix, since
this is horizontal.

Since the total energy remains con-
stant, we see that when the particle is
at any point P of its path, its kinetic
energy is that due to a fall through PM,
the distance from P to the directrix.

FIG. 117

This is expressed by saying that

The velocity of a projectile at any point is that due to a fall
from the directrix.

168. Instead of supposing that the particle is projected horizon-
tally at 0, the vertex of the parabola, we can suppose that it has
arrived at 0 in its flight through the air, having been previously
projected from some point A. The same reasoning which shows
'that the part of the path described
after passing 0 is parabolic, will
show that the path described be-
fore reaching 0 is parabolic also.
Thus the path of a particle pro-
jected from any point in any
manner is a parabola.

Suppose that a particle is pro-
jected from A with velocity v, in

FIG. 118

a direction which makes an angle

a with the horizontal. Let 0 be the vertex of its path, and let us
suppose that when the projectile passes through 0, its velocity is
u, this velocity being of course horizontal

PROJECTILES 207

There is no horizontal force acting on the particle, so that its
horizontal velocity remains unaltered throughout its flight. Thus

u = v cos a.

The latus rectum of the parabola is accordingly
2u2 2 v* cos2 a

9 9

The velocity v is that due to a fall from the directrix to A, so
that if NX is the directrix in fig. 118,

The time of flight from A to 0 is the tune required for gravity

to destroy a vertical velocity vsina; it is therefore -- In

9
this time the horizontal distance AM is described with a uniform

horizontal velocity u, so that

v sin a v2 sin a; cos a
AM— - u = - •

The vertical distance described, OM, is by equation (47) equal
to half of the time multiplied by the initial vertical velocity.

2 9

The total range on a horizontal plane, AAf, is twice A M, so that

2 v2 sin a cos a ^2 sin 2 a

A A' = = = —

9 9

169. If the value of v is fixed (as, for instance, it would be if
we were firing a shot with a given charge of powder), while the

angle a can be varied, then the range AA1 can never exceed — > for

the factor sin 2 a can never exceed unity. Thus the greatest range
attainable on a horizontal plane with a given velocity of projection

208 MOTION UNDER CONSTANT FORCES

v2
v will be — j and to obtain this range we make sin 2 a: = 1, or

a = 45 degrees. Thus, to send a projectile as far as possible on
a horizontal plane we project it at an angle of 45 degrees.

170. These results can also be obtained analytically. Let us
take the point of projection for origin, and the plane in which
the flight takes place as plane of xy, the axes of
x and y being respectively horizontal and vertical.
The ^-coordinate of the point reached by the
particle after time t is equal to the horizontal
distance described in time t with uniform hori-
~x zontal velocity v cos a. Thus

FIG. 119

x = vcosa-t. (56)

Similarly the ^-coordinate of this point is the distance described
in time t, starting with initial velocity v sin a, and with retarda-
tion a. Thus . , ,2 /cr?x
y = v sin a; • t — \g&. (57)

If we eliminate t between equations (56) and (57), we obtain
the equation of the path. It is found to be

/~r, (58)

2 v2 cos2 a
This can be expressed in the form

1 v2 sin2 a g I v2 sin a cos a\

which is clearly the equation of a parabola, of which the vertex is

at the point .

vz sin a cos a: 1 tr sin2 a /Km

(59)

9 2 9

and of which the latus rectum is of length

2 v2 cos2 a

PROJECTILES

209

To obtain the range on a horizontal plane, we have to find the
point in which the parabola intersects the line y = 0. Putting
y = 0 in equation (58), we obtain at once

2 v2 cos2 a v* sin 2 a

x = — — tan a = >

9 9

agreeing with the value obtained in § 168.

Range on an Inclined Plane

171. Suppose, next, that the projectile is fired so as to strike
an inclined plane through 0, the point of projection. Let ft be
the inclination of this plane to the
horizon, and let r be the range of the y
projectile on this plane. Then the co-
ordinate of the point at which the
projectile meets the plane must be

x = r cos ft, y = rsinft.

This point is a point on the parab-
ola, so that its coordinates must sat- °
isfy equation (58). Substituting these
coordinates, we obtain

n or2 cos2 ft

r sin p = r tan a cos p — - — — >

2 v2 cos2 a

giving as the value of the range r,

_ 2v2 cosasm(a — ft)
g ' cos2/3

Since 2 cos a sin (a — ft) = sin (2 a — ft) — sin/3, (61)

it is clear that if a alone is allowed to vary, the range r will be a
maximum when sin (2 a — ft) is a maximum, i.e. when it is equal
to unity. To obtain this value, we make

FIG. 120

210

MOTION UNDER CONSTANT FORCES

Thus, to get the maximum range, we project in the direction which
bisects the angle between the inclined plane and the vertical.

When projection takes place in this direction, the maximum
range R is given by putting sin (2 a — ft) = 1 in the value for r
given by equation (60). Thus we have

_ v2 2 cos a sin (a — ft)
~~^~ cos2 ft
_ v2 sin (2 a - ft) - sin ft
g cos2 ft

v*l- sin ft

=
g

cos ft

172. This equation enables us to find the greatest distance
which can be reached in any direction by a projectile fired with

velocity v. Let us replace ft by — — 6, so that 6 is the angle

which the direction makes with the vertical. Then the relation
between R and 6 is v2

r*r (63)

Regarded as an equation in polar coordinates R, 0, this is clearly
the equation of a curve such that we can hit any point inside it

with a projectile fired with
velocity v, but cannot reach
any point outside it. The
polar equation of a parabola
of latus rectum I, referred
0 ^ **• \ to its focus and axis, is

known to be

FIG. 121 1 + cos 0

Comparing this with equation (63), we see that this equation
represents a parabola, of which the point of projection is the
focus, the axis is vertical, and the semi-latus rectum is v*/g.

PROJECTILES

211

of Paths

173. If we imagine all the parabolas drawn, which can be
described by projectiles fired from the point 0 with a given
velocity v, we shall obtain a figure similar to fig. 122. The out-
side curve obviously separates the points which can be reached

FIG. 122

from those which cannot be reached. Thus this is the parabola of
which the equation is given in equation (63). A study of fig. 122
will now show that this curve is the envelope of the system of
parabolas which correspond to the different directions of firing.

174. The envelope of the system of parabolas can be found more
directly by analytical methods. If we write m for tan a in equa-
tion (58), we obtain the equation of a parabola of the system in

the form

qx*
y = mx-j—^(l + m2),

and the whole system is obtained by giving different values to m.
The condition for this equation to have equal roots in m is that

or, in reduced form,

= _ _

(64)

If x, y satisfy this relation, two parabolas which only differ
infinitesimally pass through x, y, and therefore x, y is a point on

212

MOTION UNDER CONSTANT FORCES

the envelope. Thus equation (64) is the equation of the envelope,
and is easily seen to give the same parabolic envelope as has already
been obtained.

175. There is also a very simple geometrical way of deter-
mining the envelope of the system of parabolas. We notice first
that as the projectiles are all fired from the same point A with
the same velocity v, their paths must all have the same directrix
NM (fig. 123).

Let any two parabolas of the system intersect in P, and let

S, Sf be the foci of these parabolas. Let ANy PM be the perpen-

N M diculars from A and P to

-\ the directrix.

Then AS = AS', since
each is equal to AN, and
PS = PS', for each is
equal to PM. Thus S, S'
^ are the two points of in-
tersection of two circles of
which the centers are A, P.
If the two parabolas are

FIG. 123

supposed to be adjacent,

their foci S, S' are adjacent points, and therefore the two circles
touch, and ASP is, in the limit, a straight line. We now have

AP = AS + SP
=AN+PM

= the perpendicular from P on to a fixed horizontal
line at a distance AN above MN.

The point P, then, satisfies the condition that its distance from
this fixed line is equal to its distance from the fixed point A. It
therefore is always on a certain parabola of focus A. But also it
always a point on the envelope, this being the locus of the point
of intersection of adjacent pairs of the parabolas of the systei
Thus the envelope is the parabola just obtained, of which the foci
is A, and this is the same parabola as was obtained before.

ILLUSTKATiVE EXAMPLES

213

ILLUSTRATIVE EXAMPLES

1 . A carriage runs along a level road with velocity V, throwing off particles
of mud from the rims of its wheels. Find the greatest height to which any of them
will rise.

Let a be the radius of the wheel, then we
have seen (p. 9) that any point, such as Q,
moves with a velocity F • QL/a in the direc-
tion QM at right angles to QL. This will be
the velocity of inud projected from Q.

If the angle QLP is 0, the height above the
ground at which the mud starts is

LN=LP + PN = a(l+ cos 20),
while the vertical component of its velocity is
F • (QL/a) sin 6 = 2V sin 6 cos 6 = V sin 2 6.
The mud projected with this vertical velocity will attain a further vertical

height

(F sin 20)2

so that the total height attained is

a + a cos 2 6 -\ -- sin2 2 6.

This may be written as a quadratic function of cos 2 0 in the form
f — 1 cos2 2 0 4- a cos2 6

CLQ

The maximum value of this expression, as 6 varies, occurs when cos 2 0 = — ,

if it is possible for cos 2 6 to have this value • — i.e. if F2 > ag. In this case the
maximum height attained is

20 2F2 20F2
measured from the ground.

If, however, F2 < ag, we cannot make cos 20 -- —I vanish. We accord-

ingly make it as small as possible, so that we take cos 20 = 1. Thus the mud
which carries to the highest point is that which starts at the top point M of the
wheel, and' obviously this never gets higher than its starting point.

214

MOTION UNDER CONSTANT FORCES

2. Find what area of a vertical wall can be covered by afire-hose projecting
water with velocity v at a distance h from the wall

Let S be the nozzle of the fire-hose, and le
us regard it as capable of projecting particl
of water in any direction we please with
velocity v. The points which can be read
will, by § 172, be all the points which lie insic
a paraboloid of revolution having its axis v<

tical, S for focus, and latus rectum If

^ take S as origin, and the vertical through
|j for axis of z, the equation of this paraboloi
|j will be

Fia. 1*5

The curve in which this cuts the vertical wall, of which the equation m£
be taken to be y = A, will be

g \2g 2i)2 Z)'

2 I)2

This is a parabola, of latus rectum — - , having its axis vertical, and it
vertex at a height

2g 2v2

above S. All the points inside this parabola will be within range of the jet of
water. The points on the wall which are outside this parabola will b
inaccessible.

EXAMPLES

1. A revolver is fired horizontally from the top of a tower 100 feet high, th
bullet leaving the muzzle with a velocity of 600 feet per second. Where will th
bullet strike the ground ?

2. A rifle bullet, fired horizontally at a height of 10 feet above the surface o
a lake, strikes the water at a distance of 600 yards. Find its velocity in fee
per second, the resistance of the air being supposed negligible.

3. Prove that the claim for a rifle, that the bullet does not rise more thai
one inch in a range of 100 yards, implies that the velocity must be greater thai
2078 feet per second.

4. Find the greatest range on a horizontal plane of a cricket ball throwi
with a velocity of 100 feet per second.

EXAMPLES 215

5. A shot fired from a gun whose muzzle is close to the ground just clears a
man 6 feet high standing 10 yards away, and embeds itself in the ground a
quarter of a mile off. Show that the shot rises to a height above the ground
which is certainly greater than 22 yards.

6. A projectile has a maximum horizontal range of 256 feet ; what is its
velocity of projection ?

If it be projected with this velocity from a point on the floor of a long
corridor 24 feet high, what will be its greatest range, if it is not to strike the
ceiling ?

7. Prove that the velocity required for a range of 20 miles will be at least
1840 feet per second, with a time of flight of 81.3 seconds.

8. Determine the charge of powder required for the range of 20 miles in the
last question, supposing the shot to weigh a ton, and the strength of the powder
capable of realizing 100 foot-tons per pound of powder.

9. Show that the range E of a projectile fired from a height h above a level
plane with velocity v at an angle a is given by

2 u2 (h + R tan a) = gE2 sec2 a.

10. Show that the area of a level plane swept by a gun at a height h above
the plane increases proportionally with ft, being equal to

where A is the area commanded when the gun is at the level of the plane.

11. A projectile can be fired with a velocity of 1720 feet per second from a
fort at a height of 300 feet above a horizontal plane. Find what area of the
plane is covered by the gun.

12. A particle is projected so as just to graze the four upper corners of a
regular hexagon of side a, placed vertical with one edge resting on a horizontal
table Find the highest point in the flight of the particle, and show that the range
on the table is a y?.

13. A machine-gun is placed on an armored train which runs along a hori-
zontal line of rails with velocity v. The muzzle velocity of shots fired from the
gun is V. Find the greatest range

(a) in front of the train ;
(6) behind the train.

GENERAL EXAMPLES

1. A train is going at 60 miles an hour, when it comes to a curve hav-
ing a radius of f of a mile. There is a perfectly smooth horizontal shelf
in the train, its edge being parallel to the rails and on the side of the shelf
away from the center of the curve. A small object stands on the shelf at
a distance of 8 inches from the edge. Show that the object will fall off
the shelf after the car containing the object has described about 24 yards
of the curve, and find what its horizontal velocity will be when, it leaves
the shelf.

216 MOTION UNDEK CONSTANT FORCES

2. A balloon is moving upwards with a speed which is increasing at the
rate of 4 feet per second in each second. Find how much the weight of a
body of 10 pounds, as tested by a spring balance on it, would differ from
its weight under ordinary circumstances.

<^3. An Atwood's machine is placed, with the string clamped, on one
scale of a weighing machine. Show that as soon as the string is undamped
the apparent weight of the machine is diminished by

where m, mf are the suspended weights.

4. A uniform chain of length / and weight W passes over a smooth
peg, hanging vertically on each side. If the chain be running freely, prove
that when the length on one side is x, the pressure on the peg is

5. A jet of water falls vertically from a hose to the ground, starting with
a velocity which is negligible. Show that the center of gravity of the
water which is in the air at any instant is two thirds of the way up from
the ground to the hose.

6. A heavy uniform chain of weight w is tied to a string which is pulled
up with tension T. Find the tension in the chain at any point.

7. Prove that the shortest time from rest to rest, in which a chain,
which can bear a steady load of P tons, can lift or lower a weight of W
tons through a vertical distance of h feet is

seconds.

8. In a system of pulleys with one fixed and one movable block, in
which the cord is attached to the axis of the movable block, then passes
over the fixed one, then under the movable one, and then over the fixed
one, find the weight P which, when attached to the cord, will support a
given weight W hung from the movable block. (The blocks are so small
that all the straight portions of the cord may be considered parallel.)

Show that, if the weights do not balance, the downward acceleration of
W-, when let go, will be w — ^ P

W + 9P9'

the weight of the cord being neglected, and that of the movable block
being included in W.

9. A pulley carrying a total load W is hung in a loop of a cord which
passes over two fixed pulleys, and has weights P and Q freely suspended
from its ends, each segment of the cord being vertical; show that, when

EXAMPLES 217

the system is let go, W will remain at rest or move with uniform velocity,
provided -- 1 -- = — and there is no friction anywhere.

If this relation does not hold, find the acceleration of W.

10. A particle, falling under gravity, describes 100 feet in a certain
second. How long will it take to describe the next 100 feet, the resistance
of the air being neglected ?

If, owing to resistance, it takes .9 second, find the ratio of the resist-
ance (assumed to be constant) to the weight of the particle.

sll. Prove that the line of quickest descent from any curve to any other
curve in the same vertical plane makes equal angles with the normals to
the two curves at the points at which it meets them.

12, Find the position of a point on the circumference of a vertical
circle, in order that the time of rectilinear descent from it to the center
may be the same as that to the lowest point.

13. Find the line of quickest descent from the focus to a parabola of
which the axis is vertical and vertex upwards, and show that its length is
equal to the latus rectum.

^14. An ellipse is suspended with its major axis vertical. Find the diam-
etefdown which a particle can fall in the least time. What is the least value
of the eccentricity in order that this diameter may not be the major axis ?
15. A bullet is to be fired at a vertical target so as to hit it exactly at
right angles. If v is the velocity of the bullet and a the distance of the
target from the point of firing, show that the angle of elevation of the

shot must be \ sin-1/ — ^M, and show that the point of the target which is

hit will be at half the height of the point aimed at.

16. A bullet is fired at a vertical target. Show that the projection of
the bullet on the target, as seen by the firer of the shot, appears to move
with uniform velocity.

17. A gun fires two shots, one with velocity v at elevation a, and the
second with velocity v' at a smaller elevation a', in the same vertical plane.
Show that the shots will collide if the interval between firing is

2 vv' sin (a — a'}
g v cos a + v' cos a'

18. A, B, C are three points in order in a horizontal line, AB being
640 feet ; a particle is projected from A with a velocity of 390 feet per
second in a direction making an angle tan~ 1T5J with A C ; at the same instant
another particle is projected from B with a velocity of 250 feet per sec-
ond in a direction making an angle tan-1f with BC ; show that these par-
ticles will collide, and find when and where.

218 MOTION UNDER CONSTANT FORCES

19. A howitzer gun has a muzzle velocity of 400 feet per second. What
distance immediately behind the top of a hill 200 feet high on a level plane
is safe, if the gun is distant 1000 yards from the point in the plane verti-
cally below the top of the hill ?

20. The sights of a gun are inaccurately marked, the gun carrying
always 3 per cent farther than the distance indicated on the sights. A
marksman, not aware of the error, aims at a target distant 1000 yards. If
the velocity of the shot is 1200 feet per second, show that he will hit the
target about one yard too high.

21. The sighting of a rifle is accurate, and to aim at a point on a ver-
tical target distant a feet the rifle ought to be elevated to an angle a.
Owing to the unsteadiness of the marksman's hand, the rifle is pointed in
directions which lie anywhere within a small angle 6 of the true direction.
Show that if a succession of shots is fired, the points at which they hit
the target will all lie within a small ellipse of semi-axes ad cos a and
a0(l — tan2 a) respectively.

When a = —) the minor axis of this ellipse vanishes, so that the shots
ought all to lie in a straight line. Interpret this result.

22. A particle slides down the outer surface of a smooth sphere, start-
ing from rest at the highest point. It leaves the sphere at a point P and
describes a parabola in space. Prove that the circle of curvature of the
parabola at P will touch the directrix.

23. A particle is projected horizontally from the lowest point of the
interior of the surface of a smooth sphere. It leaves the surface of the
sphere at P, and after describing a parabola strikes the sphere again at Q.
Show that PQ and the tangent at P make equal angles with the vertical.

24. Show that the whole area commanded by a gun planted on a hill-
side, supposed plane, is an ellipse, whose focus is at the gun, eccentricity
the sine of the inclination of the hill, and semi-latus rectum equal to twice
the height to which the muzzle velocity of the shot is due.

25. Show that the area of a plane hillside of inclination a, which is
commanded by a gun placed on a fort of height H above the hill, is

4 irh (h + H cos2 a) sec3 a,
where V2 gh is the muzzle velocity of the shot.

26. A spherical shell of mass m explodes when moving with negligible
velocity at a height of h feet above the ground. The shell is divided into
very small particles, each of which moves, after the explosion, away from
the center of the shell with velocity vt and ultimately falls to the ground.
Find the total mass of the fragments which will be found per unit area a1
any specified distance from the point vertically underneath the shell.

EXAMPLES 219

27. A shell bursts while in the air, all the fragments receiving equal
velocities from the explosion. Show that the fragments at any instant lie
on a sphere, that the foci of the paths they describe also lie on a sphere,
and that the vertices lie on a spheroid.

28. A particle slides down a rough inclined plane AB, starting at rest
from A and describing a parabola freely after leaving the plane at B. If

F is the focus of the parabola described, show that the angle AFB = - -f e,
where c is the angle of friction.

29. From a fort a buoy was observed at a depression i below the horizon ;
a gun was fired at it at an elevation a, but the shot was observed to strike
the water at a point whose depression was i'. Show that, in order to strike
the buoy, the gun must be fired at an elevation B, where

cos 6 sin (0 -f f) _ cos2 i sin i'
cos a sin (a + i') cos2 i' sin i

30. Show that the least energy which will project a particle over a wall
which is at distance a from the point of projection is

1 + tan £ a

\ mqa — ,

1 - tan i a

where a is the elevation of the top of the wall at the point of projection.

31. A mill wheel of radius a revolves so that its rim has a velocity F,
and drops of water are thrown off from the rim of the wheel. Show that
the envelope of their paths is a parabola whose axis is vertical and whose

focus is at a distance — -j- vertically above the center of the wheel.

CHAPTEK IX
MOTION OF SYSTEMS OF PARTICLES

EQUATIONS OF MOTION

176. The present chapter will deal with the motion of systems
of particles, taking account of the actions and reactions which may
be set up between the different pairs of particles. As a prelimi-
nary to this, it will be convenient to recapitulate the results which
have been obtained for a single particle, stating these results in a
more analytical form than before.

The whole system of forces which act on a particle must,
since they act at a point, have a single force as resultant. Let
us call this resultant P, and denote its components along three
rectangular axes by X, Y, Z.

Also the particle, being regarded as a point, must have a definite
acceleration /, and, since / is a vector, this acceleration may be
supposed to be compounded of three components fx,fy,fe along the
three coordinate axes.

The second law of motion supplies the relation

P = mf. (65)

We are, however, told more than this by the second law of
motion : we are told that the directions of P and of / are the same.
Let X, ft, v, be the direction cosines of this single direction, then
we have X=XP, Y=pP, Z=vP,

and also fx = X/, /, = pf, /, = vf.

From these relations, combined with relation (65), we clearly
have =

Y=mf

(66)

220

EQUATIONS OF MOTION 221

These are the equations of motion of a particle in analytical
form. They simply express the second law of motion in mathe-
matical language.

177. Let xt y, z be the coordinates of the particle at any instant,
and let u, v, w be the three components of its velocity. The com-
ponent u is the velocity, along the axis of x, of the projection of
the moving point on the axis of x, and the distance of this point
from the origin at any instant is simply <c. Thus, by the defini-
tion of velocity, we have

dx
u = -, (67)

and similarly, of course, v = -p> •

w = —
dt

The rate at which the ^/-component of velocity increases is — >

but it has also been supposed to be/x, for this is the ^-component
of the acceleration. Thus we have

L=Tt'

—,

, . ., ,
and similarly y

Using the values just found for u, v, w, these equations become

ffx
fx *'

222 MOTION OF SYSTEMS OF PARTICLES

Substituting these expressions for the components of accelera-
tion into the equations of motion (66), we obtain these equations in
the new form

„

Z = m

*• <68)

178. Suppose we have a system of particles, — ml at xlt y^ z^
m2 at x2, y2, z2; etc., — and let the components of force acting on
them be Xlt Ylf Z^ X2, F2, ^2; etc.

Then, from the equation just obtained,

so that, by addition,

»
2==m*~d/' '

where denotes summation over all the particles of the system.
The left-hand member 2jX is the sum of the components along
Ox of all the forces acting on all the particles of the system. As
in § 50, these forces may be divided into two classes :

(a) external forces — forces applied to the particles from outside
the system ;

(b) internal forces — forces of interaction between pairs of par-
ticles of the system.

As in § 50, we find that the contribution to 5}-3T from the
second class of forces is nil. For all these forces fall into pairs,
each pair consisting of an action and reaction, of which the com-
ponents are equal and opposite.

Thus hi calculating ^X we need only take account of exter-
nal forces.

CONSERVATION OF LINEAR MOMENTUM 223

d?x
The right-haud term of equation (69), namely Vm ~T^> can also be

dx d?x

modified. Since, by equation (67), we have u = — , the value of -

. du dt dt

is — , so that #x

By the momentum of a particle, as we have already seen (§20),
is meant the product of its mass and velocity. The momentum of
a particle is therefore a vector of components mu, mv, mw, and mu
may be spoken of as the ^-component of the momentum. Each
particle of the system will have momentum, and the sum of the
^-components will be C^mu\ the quantity which appears on the
right hand of equation (70).

We may now replace equation (69) by

where VX denotes the sum of the ^-components of the external
forces, and ^mu is the sum of the ^-components of momentum.

CONSERVATION OF LINEAR MOMENTUM
179. When there are no external forces acting, 2/^T = 0, so that

— (2)ra^) = 0. (72)

Similarly we have — (^mv\ =0, (73)

dt\^ '

/ ^r^ \ __ f\ /r7/i\

*
These equations express that the quantities

do not vary with the time. That is to say, the components of the
total momentum are constant, so that the total momentum,

224 MOTION OF SYSTEMS OF PARTICLES

regarded as a vector, is constant. This is known as the principle
of the conservation of momentum. Stated in words it is as follows :

When any system of particles moves without being acted on by
external forces, the total momentum of the system remains constant
in magnitude and direction.

MOTION OF CENTER OF GRAVITY OF SYSTEM

180. Let us now return to the consideration of the general
equations (71),

'» - (75)

Let x, y, z denote the coordinates of the center of gravity of the
particles of the system at any instant, and let the components of
the velocity of this point be denoted by u, v} w. Then we have

* = f,etc. . ' (76)

The value of x is, by equation (8),

so that if M is the total mass, > of all the particles, we have

MOTION OF CENTER OF GRAVITY 225

Equation (75) now becomes

***.**•* dt9

and similarly we have the equations

:. (79)

^Y=M% (78)

Remembering that

du dv dw
dt dt dt

are the components of acceleration of the center of gravity, we see
that the motion of the center of gravity is the same as it would
be if it were replaced by a particle of mass M, acted upon by a
force of components 2/X, 2/F, ^Z. This force again is simply
the force which would be the resultant of all the external forces,
if they were all applied to the imaginary particle which we are
supposing to move with the center of gravity.

181. In the particular case in which there are no external forces,
the center of gravity moves as if it were a particle acted on by no
forces, so that its motion will be a motion of uniform velocity in a
straight line.

182. The motion of the center of gravity in this particular case,
and in the more general case in which external forces act, may
accordingly be supposed to be governed by the two following laws :

LAW I. The center of gravity of every system of particles con-
tinues in a state of rest, or of uniform motion in a straight line,
except in so far as the action of external forces on the system com-
pels 'it to change that state.

LAW II. When external forces act on the system, the motion of
the center of gravity is the same as it would le if all the masses of
the particles were concentrated in a single particle which moved
with the center of gravity, and all the external forces were applied
to this particle.

226 MOTION OF SYSTEMS OF PARTICLES

These laws may be regarded as the extensions of Newton's Laws
I and II to the motion of a system of particles. We can see now
why it is often legitimate to apply Newton's second law to the
motion of bodies of finite size, as though they were particles
(cf. § 26).

The principle of conservation of momentum is often sufficient
in itself to supply the solution of a dynamical problem in which
only two bodies are in motion.

ILLUSTRATIVE EXAMPLE

A shot of mass m is fired from a gun of mass M, which is free to run back on
a pair of horizontal rails. Find the velocity of recoil of the gun, and examine the
influence of the recoil on the motion of the shot.

Let us suppose that, before firing, the gun stands pointing at an angle a to the
horizon, and let the muzzle velocity of the shot — i.e. the velocity relative to
the gun with which the shot emerges — be V.

Let us suppose that the velocity of the shot relative to the earth has compo-
nents w, v horizontal and vertical, and let the velocity of recoil of the gun be U,
measured in the horizontal direction opposite to that in which the gun is pointing.

The system consisting of the gun, powder, and shot is not free from the
action of external forces, but these forces, namely the weight of the system and
its reaction with the earth, have no horizontal component. Thus the horizontal
momentum of the system must remain unaltered by the explosion. This hori-
zontal momentum was zero initially : it is therefore zero when the shot leaves
the gun. Thus we have, neglecting the weight of the powder,

M U -mu = Q. (a)

The velocity of the shot relative to the gun has components

u + U, v.

This velocity must, however, be a velocity V making an angle a with the
horizontal. We therefore have

u + U = V cos a, (6)

v = V sin a. (c)

From equations (a) and (6) we find

u _ U _ V cos a
M~ m ~ M -f m

Thus the velocity of recoil is

m
— T^coscr.

ILLUSTRATIVE EXAMPLE 227

The components of actual velocity of the shot are

u — — — Fcosa,
M + m

v — F sin a.
Thus the actual velocity of the shot is

= r m(23f+m) f

(M+m)*

while the angle of elevation, 0, is given by

v M +

u M

^
tana.

EXAMPLES

./I. An empty railway truck weighing 8 tons, originally at rest, is run into by
a similar truck carrying a load of 24 tons and moving at the rate of a mile an
hour, and the two trucks then move on together. Find their common velocity.
- 2. A gun of mass M fires a shot of mass m horizontally. Show that of the

work done by the powder, a fraction is wasted in producing the recoil

of the gun. M+m

^3. A particle of mass m slides down a smooth inclined plane of angle or, the
plane itself (mass M ) being free to slide on a smooth table. Find the acceleration
of the particle and the plane.

4. A shell is observed to explode when at the highest point of its path. It
is divided into two equal parts of which one is seen to fall vertically. Prove
that the other will describe a parabola of which the latus rectum will be four
times the latus rectum of the original parabola.

5. A shot of \ ounce weight strikes a bird of weight 6 pounds while in the
air. At the moment of striking, the shot has a horizontal velocity of 1000 feet
a second and the bird is flying horizontally in the same direction at a height of
64 feet above the ground, with a velocity of 20 feet a second. Show that the
bird will fall at a distance of about 62.2 feet beyond the place where it was
struck by the shot.

6. A ship of 5000 tons steaming at 20 knots suddenly runs into a whale
whose weight is 12 tons, asleep on the surface of the water. By how much is
the ship's speed reduced ? (Neglect motion of water.)

7. A mail package weighing 2 hundredweight is thrown out from a train
going at 60 miles an hour, with a horizontal velocity relative to the train of 11
feet per second at right angles to the track. It falls into a handcart of weight
3 hundredweight, which is free to move on a level platform, its wheels being
set so that its motion will make an angle of 30 degrees with the track of the
train. With what velocity will the cart start into motion ?

228 MOTION OF SYSTEMS OF PARTICLES

8. A mass of 8 pounds moving north at a speed of 10 feet per second is struck
by a mass of 6 pounds moving east at 14 feet per second, and its direction of
motion is thereby deflected through 30 degrees, while its speed is increased by
1 foot per second ; show that the velocity of the other is diminished by 7.3 feet
per second, approximately, and find its new direction of motion.

9. Two ice yachts, each of mass M, stand at rest on perfectly smooth ice,
with their keels in the same direction. A man of mass m jumps from the first
to the second, and then immediately back again on to the first. Show that the
final velocities of the yachts are in the ratio of M + m : M.

KINETIC ENERGY

183. We may best begin the study of the kinetic energy of a
system of particles by drawing attention to a difficulty which has
not so far been encountered in the present book. This difficulty
will be best illustrated by a particular example.

Suppose that a ship is moving through the water with a velocity
of 20 feet per second, and that a person on deck throws a ball of
mass m forward with a velocity of 30 feet per second relative to
the ship. If the person were fixed in space, we might say that the
work he did was equal to the final kinetic energy of the ball, and
was therefore Jw(30)2, or 450 m.

. On board ship, however, the ball originally had a velocity of 20
and the thrower increases this velocity to 50. The change in the
kinetic energy of 'the ball is accordingly

lm(50)2-lm(20)2,

or 1050 m. If this represents the work done by the thrower, then
we are driven to suppose that it would be more than twice as
hard to throw the ball on board ship as on land. This would
clearly be erroneous.

184. The error lies in this, that the thrower not only imparts a
velocity to the ball but also to the ship. If he throws the ball
forward he must, from the principle of conservation of momentum,
impart a backward velocity to the ship, of momentum equal and
opposite to the forward momentum of the ball. The total work per-
formed is equal to the change produced in the total kinetic energy
of the ship and the ball.

KINETIC ENERGY 229

Since, by the third law of motion, no force can act singly, it fol-
lows that, in every case of calculation of work from kinetic energy,
it will be necessary to consider the kinetic energy of more than one
body. For instance, a man throwing a ball on land will not only
jerk the ball forward, but will also jerk the whole earth backward,
and the energy of both must be taken into account, or we shall get
erroneous results.

185. A second difficulty, closely connected with the first, suggests
itself at once. Suppose we have a ball thrown with a velocity v
along the deck of a ship moving with velocity V. We have seen
that we must not suppose the kinetic energy of the ball to be
\ mv*, but is it any more legitimate to suppose it to be \ m (v + F)2 ?
For the sea in which the ship sails will have a further velocity V'
in consequence of the earth's rotation, so that the energy might
equally well be taken to be

and so we might go on indefinitely. Knowing of no frame of
reference which is absolutely at rest, it would seem to be impos-
sible to find the true value of the kinetic energy. Moreover,
it ought to be noticed that the expressions for the kinetic energy
referred to different frames of reference differ by more than mere
constants. For instance, the difference between the two expres-
sions we have found for kinetic energy relative to the sea and
kinetic energy relative to the earth's center is

Jm(v+F+F/)2-lm(

This difference not only depends on m and V' but also on v and V.
It is not a constant difference, and so does not disappear when we
calculate the increase in kinetic energy resulting from the action
of forces.

The theorems which follow serve the purpose of showing a way
through these and similar difficulties.

230 MOTION OF SYSTEMS OF PAETICLES

186. THEOREM. The kinetic energy of any system of moving par-
ticles is equal to the kinetic energy of motion relative to the center
of gravity of the particles, plus the kinetic energy of a single par-
ticle of mass equal to the aggregate mass of the system, moving
with the center of gravity.

Let the particles be ml at xly yl} zlt etc., and let the coordinates
be measured with the center of gravity taken as origin. Let the
velocities be denoted by ulf vlt wlt etc., and let these also be meas-
ured relative to a frame moving with the center of gravity, so that

dx.
u, = ^, etc.

Let the velocity of the center of gravity referred to any frame
of reference, moving or fixed (provided only that the directions of
the axes do not turn), have components u, v, w. Then the velocity
of the particle m^ is compounded of the velocity of the particle, rel-
ative to the center of gravity of the system, of components ult vlt
wv together with the velocity of the center of gravity, of com-
ponents u, v, w. Thus the whole velocity of the particle ml has

components

u + ult v -h vlt w + wr

The kinetic energy of the first particle is accordingly

1 ml [(u + ^)2 + (v + vtf + (w + wtf\,
so that the kinetic energy of the system is

[(u + u)2 + (v + v)2 + (w + wf}>
or, on expanding squares,

Since, when the center of gravity is taken as origin, the coor-
dinates of the particles are xl)yl)zl) etc., we have, by equations (8),

KINETIC ENERGY 231

0 = £L_, etc.,
2jnn

so that V mx = 0. It follows that V m — = 0, or Yraw = 0.
^ ^ at ^

Similarly ^mv = 0 and ^?mw = 0. The whole of the second
line of expression (80) is now seen to disappear, so that we find
for the kinetic energy the expression

which proves the theorem.

187. Next, suppose that the coordinates of the center of gravity
at any instant, referred to an imaginary set of fixed axes, are
x, y, z, the velocity of the center of gravity having, as before, com-
ponents Uy V, W.

We have supposed that, relative to the center of gravity, the
particle mx has coordinates xlt yl} zlf and components of velocity
uv vv wr Thus, referred to the imaginary fixed axes, the coordi-
nates of the particle m^ will be

while its components of velocity, as before, are
u -f- ulf v + vlt w + wr

Let the force applied to the particle mjhave components XlfTv Z{.
As in § 141, the work done on this particle by the external forces
is equal to minus the work performed by the particle against these
forces. Thus the work done on the particle while it moves over
any small element of its path is, by § 118, equal to

The total work done on all the particles in any small displace-
ment is therefore

x,) + Y,d(y + y]) + Z,d(z +
and this can be separated into two parts as follows:

232 MOTION OF SYSTEMS OF PARTICLES

The first part may be taken to be

^X^dx+^Y^dy +2^i, (82)

while the second is

^Xldxl 4-_2)ri%i +2X^i- (83)

By equation (77), we have

v d/H

^ = MM'

where M is the total mass of the system, expressing that the
center of gravity moves as though it were a particle of mass M

acted on by a force of components 2^fi, 2)-^i> 2/^i- ^ ^s a^ once
clear that expression (82) represents the work done in the motion
of this imaginary particle, and this we know must be equal to the
increase in its kinetic energy.

The total work done is the sum of expressions (82) and (83).
This total work is equal to the increase in the total kinetic energy
of the system (by § 140), and this again (by § 186) is equal to the
increase in the kinetic energy of motion relative to the center of
gravity of the particles, plus the increase in the kinetic energy of
the imaginary particle of mass M moving with the center of gravity.

This latter increase, as we have just seen, is represented by
expression (82), so that the former must be represented by
expression (83).

Thus the increase in kinetic energy relative to the center of
gravity is

and is therefore equal to the work done by the forces, calculated
as though the center of gravity were at rest.

188. Thus we see that, in the theorem that the increase in
kinetic energy is equal to the work done, it is legitimate to calcu-
late both the kinetic energy and the work done by considering
motion relative to the center of gravity only ; i.e. the system may
be treated as though its center of gravity remained at rest.

IMPULSIVE FORCES 233

As an illustration, consider the problem of firing a shot on board a
moving ship. The mass of the shot being small compared with that of the
ship, we may suppose the center of gravity of shot and ship to have
exactly the motion of the ship. The velocity of the shot relative to this
center of gravity may accordingly be taken simply to be that relative to
the deck of the ship. The work done by the powder in ejecting the shot
from the barrel is the same as though the ship were at rest, so that the
velocity of the shot relative to the ship will be the same as though the
ship were at rest.

EXAMPLES

1. A cart is moving with velocity V and a man on the cart throws out sand
horizontally from the back of the cart at the rate of m pounds per minute, the
sand having a velocity v relative to the road. At what rate is the man working ?

2. A gun capable of firing a shot vertically upwards to a height h is placed
on an armored train running with velocity V. What is the greatest range to
which a shot can reach (a) behind the train, (6) in front of the train ?

3. In the last question find the nearest point to the track, which is out of
range of the gun.

4. A shell of mass M is moving with velocity V. An internal explosion gen-
erates an amount E of energy, and thereby breaks the shell into masses of which
one is k times as great as the other. Show that if the fragments continue to
move in the original line of motion of the shell, their velocities will be

V + V2 kE/ M , V - V2 E/kM .

5. Two men, each of mass M, stand on two inelastic platforms each of mass
m, hanging over a smooth pulley. One of the men, leaping from the ground,
could raise his center of gravity through a height h. Show that if he leaps with
the same energy from the platform, his center of gravity will rise a height

2(M+m)

IMPULSIVE FORCES

189. There are many instances in dynamical problems in which
the action of a force begins and terminates within so short an
interval of time that the action may be regarded as instantaneous.
Such forces are called impulsive forces. As instances of impulsive
forces we may take the forces brought into play by the jerking of
an inextensible string, or by the collision between two hard bodies.

The change of momentum produced by the action of an impul-
sive force is, in general, of finite amount. As the force only acts for

234 MOTION OF SYSTEMS OF PARTICLES

an infinitesimal time, the rate of change of momentum must be in-
finite. By the second law of motion, the rate of change of momen-
tum is equal to the force, so that the force itself, while it lasts,
must be infinite. Thus an impulsive force may be regarded as an
infinite force acting for an infinitesimal time.

190. At the outset of our study of impulsive forces, it will be
well to notice one physical peculiarity of these forces. A perfectly
rigid body was defined as one which kept its shape under the
action of any forces, no matter how great. At the same time it
was mentioned that no perfectly rigid bodies exist in nature.
Under the action of infinite, or very great, forces such as occur in
impulses, no body may be treated as perfectly rigid.

The consequence of this is that when any impulsive forces are
brought into action, relative motion is set up between the different
small particles of which continuous bodies are composed. This
relative motion possesses energy of a kind which cannot be regain*
from the system by mechanical processes ; in fact, the relative
motion of these particles simply represents the he"at of the body.
Inasmuch as this energy cannot be recovered from the system as
mechanical work, we see that the impulsive forces which do work
in producing this energy cannot be treated as conservative forces.
Thus we see that

The sum of the potential and kinetic energies of a system does
not remain constant through the action of impulsive forces.

For clearly part of the total energy is left, after the impulses, in the
form of heat.

Consider, for instance, a lead bullet striking a steel target. Suppose
that, before striking the target, the bullet is moving horizontally with
velocity v at a height h. Its kinetic energy is ^ raw2, its potential energy
being mgh. After striking, we may suppose the bullet to have no horizontal
velocity, but to fall to the bottom of the target. At the instant at which
this fall begins, the kinetic energy is nil, while the potential energy is mgh,
as it was before the impact. Thus an amount of energy \ mv2 has disap-
peared from the total energy. This has been used up in producing motions
of the particles of the bullet and target relative to one another ; tliese show
themselves in the form of heat, and also, perhaps, partly in permanent
changes of shape, — a dent in the target, or a flattening of the bullet.

IMPULSIVE FORCES 235

Measure of an Impulse

191. The change of momentum produced by an impulsive force
is called the impulse of the force. Thus if an impulse / acts on
a mass m, changing its velocity (or component of velocity in the
direction of the impulse) from u to v, we have

I=m(v-u). (84)

The force acting at any instant is, by the second law, equal
to the rate of change of momentum of the particle (or body) on
which it acts. If the force is of constant amount, the whole
change of momentum is equal to the product of the force by the
time over which it acts. If the force is of variable amount, the
change of momentum will be equal to the integral of the force
with respect to the time over which it acts. Thus if P is the
value of the force acting at any instant of the whole time t, we
see that the impulse

= Pt, if the force is of constant amount,

•I-

Pdt, if the force is of variable amount.

Work done ly an Impulse

192. The work done by an impulse / in changing the velocity
of a mass m from u to v will be

the increase in the kinetic energy of the mass. Since / = m (v — u),
we may write the expression for the work done in the form

^ m (v — u) (v + u)
+ u\

Thus the work done ly an impulse is equal to the impulse multi-
plied by the mean of the initial and final velocities of the mass
acted upon.

236 MOTION OF SYSTEMS OF PARTICLES

If the mass is not moving in the same direction as the line of
action of the impulse, the foregoing result will obviously be true
if u, v are taken to be the components of the velocities along the
line of action.

ILLUSTRATIVE EXAMPLES

1. A shot of 14 pounds is fired into a target of mass 200 pounds which is
suspended by chains so that it is free to start into motion horizontally. If the
shot, before impact, was moving with a horizontal velocity of 1000 feet a second,
and afterwards remains embedded in the target, find the loss of energy caused
by the impact.

Let V denote the horizontal velocity, measured in feet per second, with
which the target and bullet together start into motion after the impact. Then,
by the conservation of momentum, equating the momentum before the impact
to that after,

1000 x 14 = V x 214,

so that F

The kinetic energy before impact was | • 14 • (1000)2 ; that afterwards is
\ - 214 • F2. Thus the loss of energy is

| (14,000,000 - 214 F2) = 6,540,000 foot poundals, approximately.

2. A heavy chain, of length I and mass m per unit length, is held with a length
h hanging over the edge of a table, and the remainder coiled up at the extreme edge
of the table. If the chain is set free, find the velocity at any stage of the motion.

Suppose that at any stage of the motion a length x is hanging vertically, so
that a length I — x is coiled up on the table. After an infinitesimal time dt let
x be supposed to have increased from x to x + dx. Then if v is the downward
velocity of the chain, we clearly have

At the beginning of the interval dt, the downward momentum of the chain
was that of a mass mx moving with a velocity v. It was accordingly mvx. At
the end of the interval, the momentum is that of a mass m (x + dx) moving with
a velocity which may be denoted by (v + dv). Thus the gain in momentum is

m (x + dx) (v + dv) — mxv,

or, neglecting the small quantity of the second order dv dx, the gain is

m(xdv + vdx).

ILLUSTRATIVE EXAMPLES 237

The gain of momentum per unit time is, however, by equation (71), equal
to the total force acting, and this is mgx at the beginning of the interval dt and
mg (x -f dx) at the end. Neglecting the small quantity of the second order dx dt,
we find that the total gain of momentum in the interval dt must be mg x dt.

Thus we have

m (x dv -f v dx) = mgxdt

dx
= mgx — ->

or, simplifying, vx h u2 = gx.

To integrate this equation, we multiply by 2 x, and then we obtain
v*x2 = f gxB + a constant.

To determine the constant, we note that v = 0 when x = h, so that the value
of the constant must be — f ghs. Thus we have

giving the velocity when a length x is off the table. When the last particle of
the chain is pulled off, the value of x is Z, so that at this instant

I3 -h*
V* = ZQ _, .

We notice that this value of vz is not the value which would be obtained from
the equation of energy. Clearly this equation must not be employed, since
impulses are in action all the time, jerking new particles of the chain into
motion.

EXAMPLES

-' 1. An empty car of 10 tons weight runs into a similar car loaded with 50
tons of coal, and the two run on together with a velocity of 5 feet per second.
What was the velocity of the first car originally, and what was the amount of
the impulse between the cars ?

.-• 2. A stone of weight | ounce is dropped on to soft ground from a height of
5 feet. Find the impulse exerted before the stone is brought to rest.
• 3. A mass of 1 ton falls from a height of 16 feet on the end of a vertical
pile, and drives it half an inch deeper into the ground. Assuming the driving
force of the mass on the pile to be constant while it lasts, find its amount and
the duration of its action.

4. A body of mass 10 grammes is moving with a speed of 8 centimeters
a second. Suddenly it receives a blow which causes it to double its speed, and
to change its direction through half a right angle. Determine the direction of
the blow, and the velocity with which the body would have moved off, had it
been at rest.

238 MOTION OF SYSTEMS OF PARTICLES

5. The string of an Atwood's machine has masses mi, m2 attached to its
ends, m-i being the heavier. After it has been in motion for 1 second mi strikes
the floor. Find (a) for how long w2 will continue to ascend, (6) with what
velocity mi will start into motion again when the string becomes tight.
v 6. On a certain day one inch of rain fell in 10 hours, the drops falling with
a velocity of 20 feet per second. Find the average pressure per square foot on
the canvas roof of a tent, supposed horizontal, produced by the impact of the
raindrops. (One cubic foot of water weighs 62 1 pounds.)

7. The earth, moving in its orbit with velocity "P", runs into a swarm of small
meteorites, of density one kilogramme to the cubic mile, moving with a velocity
v in a direction exactly opposite to that of the earth. Find the rate of decrease
of the earth's speed in consequence of its bombardment by the meteorites,
and find also the increase in the height of the barometer at different points on
the earth's surface, it being assumed that all the meteorites are dissipated into
dust before they reach the ground. (The earth's mass is 6 x 1027 grammes, its
diameter is 7927 miles.)

w 8. A uniform chain is coiled in a heap on a horizontal plane, and a man
takes hold of one end and raises it uniformly with a velocity v. Show that
when his hand is at a height x from the plane, the pressure on his hand is equal

v2
to the weight of a length x -\ — of the chain.

ELASTICITY

193. It is a matter of common experience that if we drop a
ball of steel on to a hard floor it rebounds to a considerable height,
while a ball of wood will rebound to a much smaller height, and a
ball of wool, paper, or putty will hardly rebound at all.

When the contact between the surfaces of two bodies is of such
a nature that they do not rebound at all after impact, it is said to
be perfectly inelastic, while if the bodies rebound, the contact is
said to be elastic. Obviously there are varying degrees of elasticity.

Moment of Greatest Compression

194. Probably the collision of two billiard balls supplies the
most familiar instance of an impact with a high degree of elasticity.
We shall discuss this impact best by referring the motion of the
second ball to a frame of reference moving with the first. Before
impact the center of the second ball is approaching that of the first,

ELASTICITY 239

after impact it is receding. Hence at some instant during the
impact, its motion must have changed from one of approaching to
one of receding; at this instant the distance between the two
centers was a minimum.

Suppose that, before the experiment, we had chalked the two
faces of the balls on which the collision takes place. On examin-
ing the balls after impact it will be found that the chalk has been
disturbed, not only at a single point, but all over a circle of con-
siderable size, — perhaps of diameter half an inch for billiard balls
moving with a fair velocity. This shows that at the moment at
which the centers of the balls were closest to one another, their
distance was less than if they had been placed in contact and at
rest, — the balls were compressed.

The instant at which the two centers were nearest is called the
moment of greatest compression.

In general, for any two surfaces in collision, the instant at which
the relative velocity along the common normal vanishes is called
the moment of greatest compression. Obviously this is the instant
at which the motion of the two surfaces changes from one of
approach to one of recession.

195. By the time the moment of greatest compression is reached,
the velocities of both bodies will, in general, have been changed,
so that forces must have been at work to produce this change. The
whole time of action of these forces, the time from the instant at
which the bodies first touch to the instant of greatest compression,
is so small that these forces may be treated as impulsive. The
impulses acting on the two bodies, being action and reaction, must
be equal and opposite. If the surfaces are smooth, the direction of
these impulses will be along the common normal. If the surfaces
are rough, we cannot specify the direction until we know the direc-
tion of sliding, if any, of the surfaces over one another. In either
case, let us denote the component of the impulse along the com-
mon normal by /. The quantity / is called the impulse of com-
pression. Clearly it is the forces of which this impulse is composed
which reduce the relative normal velocity to zero.

240 MOTION OF SYSTEMS OF PARTICLES

After the moment of greatest compression, a second system of
forces must come into play to set up the velocities with which the
bodies separate from one another. In fact, at the instant of great-
est compression, the compressed parts of the bodies act like a com-
pressed spring, and we can suppose the velocities of separation
produced by the action of this imaginary spring. The forces which
separate the bodies may again be treated as impulsive, and the
component of this impulse along the common normal will be
denoted by I'. The impulse /' is called the impulse of restitution.

196. When the motion of the bodies before impact is known, we
can calculate the velocities at the instant of greatest compression
by an application of the principle of conservation of momentum.
It is therefore possible to calculate the impulse J, the impulse
of compression.

The amount of the impulse /', on the other hand, depends on
the nature of the contact between the two bodies ; for instance,
if the bodies are perfectly inelastic, there is no separation at all
after impact, so that /' = 0. In general, it is found as a matter
of experiment that the impulse I1 is connected with the impulse

/ by the simple law

I'=el,

where e is a quantity which depends only on the nature of the
contact between the two surfaces, and not on the amount of the
impulse I. The quantity e is called the coefficient of elasticity for
the two bodies.

It is important to understand that this coefficient of elasticity is a quan-
tity entirely different from the coefficients or elastic constants which occur
in the theory of elastic solids. Indeed, the term coefficient of elasticity is
somewhat unfortunate as a description of the quantity e ; what is measured
is resilience rather than elasticity, and doubtless coefficient of resilience would
be a better description than coefficient of elasticity. The term coefficient of
elasticity has, however, been generally adopted.

197. The value of e, as we have seen, is zero for perfectly in-
elastic bodies. For iron impinging on lead, the value of e is about
.14, for iron on iron about .66, and for lead on lead about .20. We

PARTICLE IMPINGING ON A FIXED SURFACE 241

notice that resilience depends on the nature of the contact between
two bodies, being in this respect similar to the coefficient of fric-
tion. The resilience does not arise partly from one body and partly
from the other, for if it did the value of e for iron impinging on
lead would be intermediate between the values for iron on iron
and for lead on lead.

As examples of bodies for which the coefficient of elasticity is
large, it is found that the value of e for the impact of two ivory
billiard balls is about .81, while for glass impinging on glass it is
.94. The most perfect elasticity conceivable is that of two bodies
for which e = 1, in which case the impulse of restitution is equal
to the impulse of compression. The bodies in this case are spoken
of as perfectly elastic. The peculiarity of perfectly elastic bodies is
that no energy is lost on impact. It is clear that the value of e can-
not exceed unity, for if the value of e were greater than unity, the
kinetic energy set up by the impulse of restitution would be greater
than that absorbed by the impulse of compression, so that the total
energy would be increased.

We shall now apply these principles to some important cases
of impact.

PARTICLE IMPINGING ON A FIXED SURFACE
Direct Impact

198. Suppose first that the impact is direct — i.e. that, at the
instant of collision, the particle is moving along the normal to the
surface at the point at which it strikes. Let m be its mass, and v
its velocity before impact. At the moment of greatest compression,
the particle will be at rest relatively to the plane, so that its
momentum is reduced by the impulse of compression from mv
to 0. Thus we must have

/= mv.

If e is the coefficient of elasticity,

/' = e I = emv.

242 MOTION OF SYSTEMS OF PARTICLES

Thus there is a normal impulse of amount emv, and this gener-
ates a velocity ev in the particle. There is no tangential impulse,
for there is no sliding of the surfaces past one another. Thus the
velocity of rebound is a velocity ev normal to the surface.

Oblique Impact : Smooth Contact

199. If the impact is oblique, let us suppose that the components
of velocity along the tangent plane and along the normal before
impact are u, v. As before, w"e find

/= mv, I1 = emv,

so that the normal velocity after impact, say v', is

v' = ev.

If the contact is supposed smooth, there can be no force in the
tangent plane, so that the momentum in the tangent plane remains
unaltered. Thus the velocity in the tangent plane remains equal
to u, and the velocity after impact will be one of components u, ev.
Let 6 be the angle which the velocity before
impact makes with the normal, and let </> be the
corresponding angle after impact. Then

tan 6 = — >

u
tan (b — —y

so that tan 0 — e tan <f>.

If the bodies are perfectly elastic, e = 1, so that
0 = (f> • i.e. the particle rebounds at an angle equal
to the angle of incidence. Its reflexion obeys the same law as
that of a ray of light.

If the bodies are imperfectly elastic, 0 is less than </>, so that
the path is bent away from the normal.

If the bodies are perfectly inelastic, e = 0, so that <f> = — ; the

particle simply slides along the plane, as of course it obviously
must since /'= 0.

PARTICLE IMPINGING ON A FIXED SURFACE 243

The kinetic energy before impact is
%m(u2+v2),
that after impact is ^ m (u2 + v'2).

Thus there is a loss of kinetic energy of amount

lm(^__^2)
or ^mv2(l — e2).

This vanishes if e = 1, i.e. if the bodies are perfectly elastic.
In all other cases there is a loss of energy. We again see that e
cannot be greater than unity, or it would be possible to gain
energy by causing bodies to impinge on one another.

Oblique Impact : Rough Contact

200. As in the case of a smooth contact, we obtain the relation
v' = ev connecting the components of velocity along the normal.
The reaction, however, no longer acts entirely along the normal, so
that it is not now true that the tangential component of velocity
remains unaltered.

Let us consider the case in which the surface of the particle
slides in the same direction over the fixed surface during the
whole time that the two surfaces are in contact. Then at every
instant of the impact there will be a tangential force equal to
//. times the normal force, so that the total tangential impulse
must be /z times the total normal impulse, and therefore equal
to /*(/+/').

Thus if uf is the tangential velocity after the impact, we have

m(u—u') = /*(/+/')
«/*(!+«)!

= ft(l -\-e)mv,
so that u' = u — ( 1 + e) pv.

244 MOTION OF SYSTEMS OF PABTICLES

If, as before, we suppose that 6, <f> are the angles which the path
makes with the normal before and after the impact (see fig. 126),
we have

tan 9 = — - =

vr ev

so that e tan (/> = tan 6 — (1 -f- e) p.

The value of (1 4- e) //, is always positive, so that <f> is less than it
would be if the plane were smooth ; in other words, the roughness
of the plane causes the particle to rebound nearer to the normal.

This equation, however, is only true within certain limits, for
we have assumed that there is sliding during the whole time of
impact. It may be tha^p. a certain stage of the motion sliding
will give place to rolling, and if so the equation we have obtained
is no longer valid.

IMPACT OF Two MOVING BODIES

201. Suppose now that two bodies A, B of masses m, m' im-
pinge at the point C, the common normal to C being the line CP.
Let it be supposed that the centers of
gravity of the two bodies both He in the
line CP at the moment of impact, and let
the components along CP of the velocities
of the centers of gravity of the masses
A, B respectively be

uy uf before impact,

V, V at the instant of greatest

compression,
127 v, vr after impact.

Then if we denote the impulse of compression by 7, and the
impulse of restitution by 7', we have

-r , / -rr\ rm1 InJ V\ (?t 5^

7'= m(V- v) = - m'(V- vf). (86)

IMPACT OF TWO MOVING BODIES 245

From the first line u = V H >

i-V- —

so that u—u'=I( 1 -\>

\m m'/

an equation connecting / with the relative velocity before collision.
Similarly, from equations (86),

The experimental relation I' = el is now seen to be exactly
equivalent to the relation

v — vr= — e(u — ur), (87)

or, in words : The normal component of relative velocity of the
centers of gravity after collision is equal to e times the relative
velocity before collision, and is in the opposite direction.

This law is known as Newton's experimental law ; it expresses
the same property of matter as the relation /' = el.

A second relation, connecting velocities before impact with
velocities after, is given by the conservation of momentum; we

nave mv + m'v' = mu + m'u'.

Combining this with equation (87), we can determine the
velocities v, vr after collision in terms of the velocities u, uf
before collision.

Solving, we find that

mu + raV — em'(u — ur) oox

V = - r-5 - -> (00)

m + m'

f _

ra + ra'

giving the normal velocities.

If the bodies are rough, we find the tangential velocities in
the same way as in § 200 ; while if the bodies are smooth, the
velocities in directions perpendicular to CP remain unaltered.

246 MOTION OF SYSTEMS OF PARTICLES

If the center of gravity of the two bodies is at rest, — or, what
comes to the same thing, if we measure all velocities relatively to
the center of gravity, — we have

mu + m'u' == 0,

,, m'(u — ur)

so that "' — — a — ^—

m + m'

m(u — u')
vr = e — — — -/•
m -f m'

Using the relation mu = — m'u', these become

v = — eu,
v'= — eu',

so that the bodies rebound from one another as though they had
impinged on a fixed plane of elasticity e.

The kinetic energy, either before or after the collision, is equal
to the kinetic energy of a single particle moving with the center of
gravity, together with that of the system relative to the center of
gravity. The former remains unchanged by collision, so that the
loss in the total kinetic energy produced by collision is equal to
the loss in the kinetic energy relative to the center of gravity.

If the bodies are smooth, this loss of kinetic energy

= ^ (mu2 + m'u'2 — mv2 — .raV2)
= ^(mu2+m'u'2)(l-e2).

Thus the loss of kinetic energy is (1— e2) times the original kinetic
energy relative to the center of gravity. If the bodies are perfectly
elastic, e = 1, so that there is no loss of energy ; while if e = 0, the
original energy relative to the center of gravity is all lost.

Impact of Two Smooth Spheres

202. Let us apply the principles just explained to determining
the motion, after impact, of two smooth spheres.

At the moment of impact let A, B be the centers of the two
spheres, so that the line AB is the common normal to the surfaces
at the point of impact C.

IMPACT OF TWO MOVING BODIES 247

As before, let the velocities along AB before impact be u, uf, I
these both being measured in the direction AB, and let the |
velocities in the same direction after impact be v, v'. Then we
have, by the conservation of momentum along AB,

mu •+• m'ur = mv + mfvf,
and, by Newton's law, v — v' = — e (u — u'}.

From these equations (88) and (89) follow as before.

FIG. 128

If the velocities of A before impact make angles a, a' with AB
as marked in the figure, the tangential velocities of A before and
after impact are

so that, since the tangential velocities remain unaltered, we
must have ^ tana'= - » tana;

while similarly, from the motion of B,

Thus equations (88) and (89) become

. mu + m'yJ — em'(u — u')

cot a' = -- — — * - - cot a,

(m + m)u

cot/3'= - *"* + «'«'+ «»(«-»'

(m -\- m!}u'

giving a', @f in terms of the initial motion.

248 MOTION OF SYSTEMS OF PARTICLES

If, as in the game of billiards, the spheres are of equal mass and
the second sphere is originally at rest, we take m = m', u' = 0, and

obtain cot a' = - 1 (1 - e) cot a, 0' = 0.

Thus B starts into motion along the line of centers, as it obviously
must since the forces which set it in motion act along this line.

Since e is always less than unity and a is necessarily acute, cot a'
must be negative, so that af will be obtuse. If e. = 1, then a1 = 90°.
Thus if the spheres were perfectly smooth and perfectly elastic, A
would move at right angles to the line of centers after impact ; its
motion would be the same as if it had impinged on a perfectly
smooth and inelastic plane.

ILLUSTRATIVE EXAMPLE •

A row of similar coins is placed on a rough table, the coins being at equal dis-
tances apart and in a straight line. The first coin is projected along this line so as
to impinge directly on the second. Find the resulting motion.

Let e be the coefficient of elasticity for an impact between the two coins, and
/a the coefficient of friction between the coins and the table. Let m be the mass
of each coin, and d the distance between the nearest points of two adjacent coins.

The normal reaction between a coin and the table is mg, so that the f rictional
force opposing the coin's motion is /iragr, and the retardation produced is ng.
Thus if a coin is started from its original position with a velocity F, its velocity
on reaching the next coin is reduced to M, where

F2 - M2 = 2 ngd. (a)

We now have two coins of equal mass impinging with velocities w, 0. Their
velocities after impact, say u, w', are given by the equations
v — v'= — eu (Newton's law),
v + u'= u (conservation of momentum).

Thus «=-i-w(l-e),

v'=iu(l + e).

After impact the coin originally in motion has a velocity u, and is retarded
by a f rictional retardation ^g. It accordingly comes to rest, if it does not collide
again in the meantime, after a distance s given by

,= -t (6)

2flr Sg

while the coin which has been started into motion sets off with a velocity

(c)

ILLUSTRATIVE EXAMPLE 249

The coin before this started into motion with a velocity V given by equa-
tion (a). Eliminating u from equations (a) and (c), we find as the relation

between successive velocities of starting

47/2

a difference equation with constant coefficients.

If e = 1, we notice from equation (6) that s = 0, so that each coin remains
absolutely at rest after striking the coin next in front of it — it transmits the
whole of its momentum to this coin. Also from equation (a),

After the momentum has been transmitted over the space between n coins,
the value of the square of the velocity is reduced by 2 n/tgrd. Thus at any point
the velocity of a moving coin is that which would be possessed by a coin which
had been started with velocity F, and made to move over a distance equal to all
the intervals between the coins over which the motion has been transmitted.

If d = 0, so that the coins were originally in contact, we have

V-

1 + e
Thus, if there are n coins, the nth coin will start in motion with a velocity

EXAMPLES

1. Hailstones are observed to strike the surface of a frozen lake in a direction
making an angle of 30 degrees with the vertical, and to rebound at an angle of
60 degrees. Assuming the contact to be smooth, find the coefficient of elasticity.

2. If the hailstones of the last question rise after impact to a height of 2 feet,
find the velocity with which they originally struck the ground.

3. In the last question find the height to which the hailstones will rise in
their second rebound from the ice.

4. A ball is dropped on to a horizontal floor and, after rebounding twice,
reaches a height equal to half that from which it was dropped. Find the
coefficient of elasticity.

"5. A bullet strikes a rough target at 45 degrees, and rebounds at the same
angle. Show that I — e

"=T+~e'

6. A shot fired from a distance a strikes a target at right angles, and
rebounds. Show that it will fall at a distance ae from the target (neglecting
the resistance of the air).

^ 7. A sphere of mass m collides with a sphere of mass ra' at rest, the contact
between them being smooth, and their paths after collision are observed to be
at right angles. Prove that m — em'.

250 MOTION OF SYSTEMS OF PARTICLES

8. Two billiard balls stand in contact, and a third ball is made to strike
them simultaneously, and is observed to remain at rest after the impact.
Show that e = f .

9. A particle is projected from a point on a smooth horizontal plane, with
velocity V at an elevation a, and after striking the plane rebounds time after

time. Show that its total time of flight is - , and that its total range
. F2sin2a g^~ e^

3 0(1- «) '

10. A player stands at a horizontal distance d from a wall, and throws a
ball towards the wall at an inclination a to the horizontal. Show that if it is
to return to him after bouncing, he must throw it with a velocity V given by

2 e cos a (sin a — /x, cos a)

where e, /* are coefficients of elasticity and friction.

11. In the last question consider the cases of (a) e = 0, (6) /i = tana,
(c) M > tan a.

GENERAL EXAMPLES

1. A particle is placed on the face of a smooth wedge which can slide
on a horizontal table ; find how the wedge must be moved in order that
the particle may neither ascend nor descend. Also find the pressure between
the particle and the wedge.

2. It is required to run trains of 100 tons on a level electric railway,
with stations half a mile apart, at an average speed of 12 miles an hour,
including half a minute stop at each station. Prove that the electric loco-
motives must weigh at least an additional 8 tons, taking a coefficient of
friction of £, and supposing the trains fitted with continuous brakes.
(Neglect passive resistances.)

Prove that the railway can be worked by gravity, if the line is curved
downward between the stations to a radius of about 46,000 feet; and
that the dip between the stations will be about 20 feet, the inclines
at the stations about 1 in 33, and the maximum velocity about 23£ miles
an hour.

3. A cylinder of height h and diameter d stands on the floor of a rail-
way car, which suddenly begins to move with acceleration f. Show that
the cylinder will only remain at rest relative to the car if fis less than
both /j.g and dg/li.

4. If a circular hoop is projected, spinning steadily without wobbling,
prove that the center describes a parabola, and that the tension of the rim
is the weight of a length vz/g of the rim, where v denotes the rim velocity
relative to the center of the hoop.

EXAMPLES 251

5. A uniform chain 6 feet long, having a mass of 2 pounds per foot, is
laid in a straight line along a rough horizontal table, for which the coeffi-
cient of friction is |, a portion hanging over the edge of the table so that
slipping is just about to occur. If a slight disturbance sets the chain slip-
ping, find the tension at the edge of the table when x feet have slipped off.

6. Two equal balls A, B, each of mass m, are at a distance a apart.
An impulse / acts on A in the direction AB, and a constant force F acts
on B in the same direction. Show that A will not overtake B if

72 < 2 aFm.

7. A bullet weighing one ounce is fired with a velocity of 1200 feet per
second at an elevation of 1 degree so as to hit a bird weighing 2\ pounds
when the bullet is at the highest point of its path. Supposing the bird to
have been at rest when hit and afterwards to fall with the bullet embedded
in it, find how far from the point of firing the bird will fall.

8. If a bullet weighing w pounds is fired with velocity v at a body
weighing W pounds, advancing with velocity F, prove that the body will
retain the velocity

WV-wv Tr w ,

___, or ! -_(„-„),

according as the bullet is embedded, or perforates and retains a velocity u.
Calculate the energy liberated, and thence infer the average resistance of
the body from the length perforated by the bullet.

9. A pile is being driven in by repeated impacts of a falling weight.
How does the extent to which the pile is driven in by each blow depend
(a) on the magnitude of the weight, and (&) on the height to which it is
raised before being released ?

If the weight be 1 ton, and the height from which it falls be 10 feet,
and the pile be driven in a tenth of an inch, find the resistance in tons.

10. An inelastic pile, of mass m pounds, is driven vertically into the
ground a distance of a feet at each blow of a hammer of mass M pounds,
which falls vertically through h feet. Show that the weight which would
have to be placed on the top of the pile to clrive it slowly into the ground
would be

i/r • M ll A

M H -- pounds.
(M + m)a*

11. A hammer head of W pounds, moving with a velocity of v feet a
second, strikes an inelastic nail of w pounds fixed in a block of M pounds
which is free to move. Prove that if the mean resistance of the block to
penetration by the nail is a force of R pounds, then the nail will penetrate
each blow a distance, in feet,

MW* v*

(M + W + w)(W + w) 2gR

252 MOTION OF SYSTEMS OF PARTICLES

12. In the system of pulleys described in § 128, show that if P is a weight
which is not equal to W/n, the acceleration produced in the weight W will be

nP-W
n*P+W9'

13. Two masses m, m' connected by an elastic string are placed on a
smooth horizontal table, the masses being at rest and the string unstretched.
A blow of impulse P is given to the first mass, in the direction away from
the second. Show that when the string is again unstretched, the velocity
of the second mass is

m + m'

14. Three equal particles are tied at the ends and middle point of an
inextensible string, which is placed, fully extended, on a smooth table.
The middle particle is jerked into motion in the direction towards and
perpendicular to the line joining the other two. Find the loss of energy
when the other particles are jerked into motion.

15. A coal train consists of a number of similar trucks hauled by an
engine whose weight is just equal to that of three trucks. The train is at
rest on a level track, the couplings, which are of equal length, being al]
equally slack. The engine then begins to move with a constant tractive
force, and each truck is jerked into motion as its coupling tightens. Show
that the speed of the engine will be greatest just before the tenth jerk
occurs.

16. Snow is evenly spread over a roof. If a mass commences to slide,
clearing away a path of uniform breadth as it goes, prove that its acceler-
ation is constant, and equal to a third of that of a mass sliding freely
down the roof.

17. A heavy, perfectly flexible uniform string hanging vertically with
its lowest point at a height h above an inelastic horizontal plane is sud-
denly allowed to fall on to the plane. Show that the pressure on the table
when a length x of the string has fallen on to the table is

(3 x + 2 7i) mg.

18. Show that if two equal balls impinge directly with velocities
e V and — V, the former will be reduced to rest.

1 — €

19. Show that the mass m of a sphere which must be interposed between
a sphere of mass M at rest and one of mass M' moving directly on to it
with velocity V, in order that the former may acquire the greatest possible
velocity from the impact, will be VjOf7, and that the velocity acquired

e M'V(\+ef

M+M'+2m

EXAMPLES 253

20. Prove that an elastic ball, let fall vertically from a height of h feet
on a hard pavement, and rebounding each time vertically with e times the
striking velocity, will have described

l+e /2ft

- - \l —
1 — e \

k feet, in - - \l — seconds

I— e2
before the rebounds cease.

Work this out for h = 1, e = |.

21. A ball is dropped from the top of a tower, height A, and at the same
time another ball of equal weight is projected upwards with the velocity
•\/2gh from the base of the tower and collides directly with the falling ball.
If the coefficient of restitution be e, prove that the falling ball will, in

the rebound, rise to a height short of the top of the tower by - (1 — e2).

22. A boy standing on a railway bridge lets a ball fall on the horizontal
roof of a car passing under the bridge at 15 miles an hour. If /* = 1, e = *
between the roof and the ball, find the least height of the boy's hand above
the roof in order that the second rebound of the ball may be from the
same, point "of the roof as the first.

If the boy's hand is at a greater height than this, what will happen?

23. A perfectly elastic particle is projected so as to strike the inside of
a surface of revolution of which the axis is a given vertical line. Show
that the vertices of all the parabolas described after successive rebounds
lie on a surface of which the shape is independent of that of the surface
of revolution.

24. Prove that in order to produce the greatest possible deviation in the
direction of motion of a smooth billiard ball of diameter a by impact on
another equal ball at rest, the former must be projected in a direction
making an angle

with the line, of length c, joining the two centers.

25. A pendulum hangs with its bob just in contact with a smooth verti-
cal plane. The bob is drawn aside until it is 5 inches higher than it was,
and is then released so as to strike the plane normally ; and on the first
rebound it rises vertically through 4 inches. What would have been the
vertical rise on rebound if the pendulum had been drawn aside through
the same angle, but so that the bob strikes at an angle of 60 degrees with
the normal?

CHAPTEE X
MOTION OF A PARTICLE UNDER A VARIABLE FORCE

203. In almost all the cases of motion of a particle which
have so far been considered, the forces acting on the particle
have remained constant throughout the whole of the path, so
that the acceleration of the particle has been constant. We pro-
ceed now to consider the motion of a particle which is acted
upon by forces which vary from point to point of the path of
the particle.

These problems fall into two classes, according to whether the
path described by the particle is or is not given as one of the data
of the problem. The former class is the simplest and is considered
first. It includes such cases as the motion of a pendulum, in which
the " bob " of the pendulum is constrained to describe a circle by
the mechanism of suspension of the pendulum, as also that of the
motion of a bead on a wire, in which the bead is compelled to
describe the path marked out for it by the wire.

EQUATIONS OF MOTION

204. Let s denote the distance described by the particle along
its path at any instant t, this distance being measured from any

fixed point 0 on the path. The velocity along

ds
the path is then — • Calling this v, the accelera-

u/t

. dv d?s
tion is — or — •

We can also obtain a value for the accelera-
tion from a knowledge of the forces acting. To find the acceler-
ation, we must resolve all the forces which act on the particle in
the direction of the path. If S is the component of force in this

254

EQUATIONS OF MOTION 255

direction, the equation of motion of the particle, by the second
law of motion, will be

ft W

S = m-> (90)

We shall suppose the field of force to be permanent, so that the
quantity S may be supposed to depend only on the position occu-
pied by the particle on its path, and not on the instant at which
it arrives there. In other words, S is a function of s but not of t.
Equation (91) is a differential equation connecting s and t\ if we
can solve this equation, we shall have a full knowledge of the
motion of the particle provided its path is known.

The equation is a differential equation of the second order, but
can easily be transformed into one of the first order. For

dzs _ dv dv ds _ dv
dt2 dt ds dt ds

so that the equation can be written

S = mv — •
ds

Since S is a function of s, this equation can be integrated with
respect to s, so that we obtain

(92)

where C is a constant of integration.

Since v is equal to — > this equation can be written in the form
at

<93)

which is an equation of the first degree. If this can be solved, the
solution of the problem is complete.

256 MOTION UNDER A VARIABLE FORCE

We notice that the right hand of equation (92) is the kinetic
energy of the particle. Also, since the force opposing the motion

of the particle along its path is — S, its potential energy is — I S ds.

Thus equation (92) expresses that the sum of the kinetic and
potential energies remains constant — it is the equation of energy
for the motion of the particle. From a knowledge of the total
energy at any instant of the particle's motion, we can determine
the constant (7, and can then proceed to integrate equation (93),
if possible.

ILLUSTRATIVE EXAMPLE

Assuming that the value of gravity falls off inversely as the square of the dis-
tance from the earth's center, determine the motion of a projectile fired vertically
into the air, the diminution of gravity being taken into account.

Let a be the radius of the earth, and g the value of gravity at the surface.
Then, at a distance r from the earth's center, the value of gravity will be — •

Since the particle moves along a radius drawn through the center of the
earth, we may measure all distances from the earth's center, and the distance
r from the earth's center may replace the coordinate s of § 204. The value of

the force S resolved along the path is — — , so that the equation of motion is

mga? _ d2r

The equation of energy, as in equation (92), is

rmga'2 ,

C — I — - — dr — | mv2,

or G H — - — = \ mv2. (a)

Let us suppose that the particle was projected from the earth's surface with
velocity V. Putting r = a in equation (a), the value of v must be V, so that
we must have

and this equation determines the value of C. Eliminating C from equations (a)
and (6), we obtain

ILLUSTRATIVE EXAMPLE 257

giving the velocity at any point in the form

Since v = — , this equation becomes
dt

(d)

so that t = I dr (e)

On performing the integration, we can find the time required to describe any
portion of the path. Let us first consider the different types of solution.
We see from equation (c) that v vanishes when

so that if F2 < 2 gra, there is a positive value of r, intermediate between + a
and + oo , for which the velocity vanishes. Thus if F2 < 2 gra, the projectile

goes to the point at distance — » and then falls back on to the earth. If

F2 > 2 gra, we find that there is no positive value of r for which v vanishes, so
that the particle goes to infinity : it escapes from the earth altogether.

When F2 = 2 gra, the velocity vanishes at infinity ; thus the particle just
escapes from the earth's attraction, but is left with zero velocity. Its kinetic
energy of projection is just used up in overcoming the earth's attraction.

Let us consider first the special case in which F2 = 2 ga. We find that equa-
tion (e) reduces to

where C' is a new constant of integration.

If we measure time from the instant of projection, we must have t = 0 when
r = a, so that

and on eliminating (7',

258 MOTION UNDER A VARIABLE FORCE

In the case in which V2 > 2 ga, we obtain after integration of equation (e),

(9)

where C' is a new constant of integration. If the time is to be measured from
the instant at which the particle is projected, we must have t = 0 when r — a,
so that

and on eliminating (7', we can again obtain the time required to describe any
portion of the path. The case in which V2 < 2 ga can be treated in a similar
manner. This is left as an example for the student.

EXAMPLES

1. In the foregoing illustrative example, suppose that V2 <2gra, and find
(a) the greatest height reached ;

(&) the time of flight of the particle.

2. A meteorite falls on to the earth. Assuming it to start from infinity with
zero velocity, and to fall directly on to the earth, find the velocity with which
it reaches the earth's surface, and the time taken to fall to the earth's surface
from a point distant r from the earth's center.

3. A particle falls from distance a into a center of force which attracts
according to the law /A/r2. Show that the average velocity on the first half of
the path is to the average velocity on the second half in the ratio

7T - 2 = 7T + 2.

4. Find the time of falling to a center of force which attracts according
to the law pr~*.

5. A particle moves in a straight line from a distance a to a center of attrac-

tion towards which the force is — • Show that the time required to reach the
center is

6. A particle begins to move from a distance a towards _a fixed center which
repels according to the law /wr. If its initial velocity is V/^a, show that it will
continually approach the fixed center, but will never reach it.

THE SIMPLE PENDULUM

259

THE SIMPLE PENDULUM

205. One of the most important cases of a variable force arises
in the motion of a simple pendulum. To obtain a first approxima-
tion, we can suppose that the whole weight of the pendulum is
concentrated in the bob, which may be

treated as a particle, and that this is sus-
pended from a fixed point by a weight-
less string or rod so that it is constrained
to move in a vertical circle.

Let s denote the distance along this
circle described by the particle, this dis-
tance being measured from the lowest
point 0. Let the angle PCO between
the string and the vertical be denoted
by 0, so that s = ad. The forces acting
on the particle consist of its weight and the tension of the
string. The latter has no component in the direction in which
the particle moves. The former has a component — mg sin 6.

Thus the equation of motion is

— ^ = — #sin0, (94)

where 6 = s/a.

206. This equation cannot be solved by elementary mathe-
matics, except in the simple case in which the angle 0 is small,

— i.e. the case in which the pendulum never swings through
more than a small angle from the vertical. Confining our atten-
tion to this case, we may replace sin 6 by 0, and 9 by s/a, and
write the equation of motion in the form

Thus the acceleration of the bob of the pendulum is proportional
to its distance from 0, and is towafds O.

260 MOTION UNDER A VARIABLE FORCE

Writing the equation in the form

v-r = -
ds \a

and integrating with respect to s, we obtain

(95)

Clearly the constant C must be positive, and the velocity will
vanish as soon as s reaches a value such that

Let us denote the two values of s which satisfy this equation by
.± s0 , then the motion of the bob is clearly confined within two
points at distances s0 from the point 0 on opposite sides. Thus
we may call s0 the amplitude of the swing.

Replacing C by ( -js2, equation (95) becomes

*2=!(*o2-*2), (96)

so that

and the integral of this equation is

t=, r ds

J AJ^o-

where e is a constant of integration.
This equation gives

so that s = s0 cos

THE SIMPLE PENDULUM 261

This equation contains the complete solution of the problem.
We notice that the values of s continually repeat at intervals of
time t0 for which

Thus the motion of the pendulum repeats itself indefinitely.
The interval between two instants at which the pendulum is in
the same position, namely t0> given by

is called the period.

207. Seconds pendulum. To construct a pendulum which is to
beat seconds, we choose a so that t0 shall be equal to two seconds,
for a seconds pendulum is one which takes one second to move
from left to right, and then one second more to move from right
to left. Thus we must have

*r,- =

In foot-second units we may take g — 32.19 for London, and

so obtain

a = 39.14 inches,

as the length of the seconds pendulum at London.

We notice that the period of a pendulum varies as the square
root of its length. Thus, for a pendulum to beat half-seconds, its
length would have to be only a quarter of that of the seconds
pendulum, and therefore 9.78 inches at London.

Since g varies from point to point on the earth's surface, the
length of the seconds pendulum will also vary. If we observe
the length of a pendulum and also measure its period with a
chronometer, we shall be able to calculate the value of g at
the place at which the experiment is performed; in fact, this
method affords the easiest and most accurate way of obtaining
the value of g at any point of the earth's surface.

262 MOTION UNDER, A VARIABLE FORCE

ILLUSTRATIVE EXAMPLE

A pendulum which beats seconds accurately at New York is found to gain 3
seconds a day when taken to Philadelphia. Compare the values of g at Philadel-
phia and New York.

At Philadelphia the pendulum makes 24 x (60)2 + 2 beats in 24 x (60)2
seconds. Thus the time of a beat is

24 x (60)2

seconds,

24 x (60)2 + 2
and this is equal to »A/-i where a is the length of the pendulum and g is the

value of gravity at Philadelphia. If go denote the value of gravity at New York,
we have

24 x (60)2

so that

24 x (60)2/

= 9o (1 + HEISTS) approximately.
Thus gravity is greater at Philadelphia than at New York by about one part in

21,600.

EXAMPLES

1. Calculate the length of a pendulum to beat time to a march of 100 paces
a minute.

2. A pendulum which beats seconds in London requires to be shortened by
one thousandth of its length if it is to keep time in New York. Compare the
values of gravity at London and New York.

3. The value of g at a point on the earth's surface in latitude X is

g = g0 (1 - .00257 cos 2 X),

where g0 = 32.17 is the value of g in latitude 45 degrees. Show that the lati-
tude in which a short journey of given length will produce the greatest error
in the rate of a pendulum clock is latitude 45 degrees, and find the error per
mile in this latitude. (One minute of latitude = 6075 feet.)

4. In a building, at height h feet above the ground, the value of gravity is

g0-. 000003 fc,

where g0 is the value of gravity at the foot of the building. In New York,
go = 32.14. Find the error in the rate of a pendulum clock, produced by taking
it from the ground to the top of a building 300 feet high.

SIMPLE HARMONIC MOTION 263

5. The length of a pendulum which makes 2 n beats per day is changed
from I to I + L. Show that the pendulum will lose — beats per day approxi-
mately.

6. A balloon ascends with constant acceleration, and reaches a height of
3600 feet in two minutes. Show that during the ascent a pendulum clock will
have gained about one second.

7. A pendulum of length I is adjusted by moving a small part only of the
bob of the pendulum, this being of mass equal to one nth of the complete bob.
How far must this be moved to correct an error of p seconds a day ?

SIMPLE HARMONIC MOTION

208. We have seen that throughout the motion of a pendulum
which moves so that its maximum inclination to the vertical
is small, the acceleration is proportional to the distance from
the middle point of its path, and is directed towards that point.
A point which moves in this way is said to move with simple
harmonic motion. Thus if s is the distance from a fixed point,
of a point which moves with simple harmonic motion, we have
an equation of the form

d?s

d?=-^s'

where k is a constant.

Integrating, we obtain, as before in the case of the pendulum
(cf. equation (96)),

4,2 _ Z>2/C2 2\

v — fa (so — s )y

and from this again

s=sQcosk(t — €). (97)

The constant k is known as the frequency of the motion. Thus

the frequency of a simple pendulum is \\ —

^ff

209. A simple geometrical interpretation can be given of simple

harmonic motion, and this enables us to obtain a complete knowl-
edge of the motion without any use of the integral calculus or of
the theory of differential equations. In fig. 131 let the arm OP
rotate about 0 with uniform angular velocity Jc, so that P describes
a circle of radius a with uniform velocity ka. Let a perpendicular

264

MOTION UNDER A VARIABLE FORCE

PN be drawn from P to a fixed diameter AA'. Then we shall find
that the point N moves backwards and forwards on the line A A '

with simple harmonic motion.

The acceleration of P is, by § 12,
an acceleration k*a along PO. This
can be regarded as compounded of
the acceleration of P relative to Nt
which must be along NP, and the
acceleration of N relative to 0,
which must be along ON. Thus
the acceleration of N is that com-
Fia- 131 ponent of the acceleration of P

which is in the direction AA1. This, however, is known to be
kza cos 0, or k2 • ON, along NO. Putting ON = s, we have an
acceleration — J<?s in the direction in which s is measured, namely
ON. Thus the point N moves with simple harmonic motion.

This geometrical interpretation of simple harmonic motion
enables us to obtain expressions for v and s directly. The value of
s is ON, or a cos 0. Let t = e be an instant at which the point P
was passing through the point A! in its motion round the circle,
then, at any subsequent instant t, the tune since P was at A' will
be t — e, so that the angle described by OP will be 6 = k(t — e).

Thus we have

s = ON= a cosk(t - e). (98)

This is the same result as is contained in equation (97). We
notice that the amplitude s0 of the motion is the same as the
radius a of the circle, and that the frequency k is identical with
the angular velocity. On differentiating equation (98), we obtain
at once for the velocity

v = — = — ka sin k(t — e)
at

This result can also be obtained by resolving the velocity ka of
the moving point P into two components, along and perpendicular

THE CYCL01DAL PENDULUM 265

to AAf. The former is obviously the velocity of JV along AAf, and
it is at once seen to be of amount — ka sin 6, or

v = — ka sin Jc (t — e)
= k Va2 — s2, as before.

In this motion, as in the motion of the simple pendulum, the

quantity a is called the amplitude, while the time after which

the motion repeats itself is called the period.

EXAMPLES

1. A point moves with simple harmonic motion of period 12 seconds, and has
an amplitude of 6 feet. Find its maximum velocity, and find its position and
velocity one second after an instant at which its velocity is a maximum.

2. A particle moving with simple harmonic motion of period t is observed
to have a velocity v when at a distance a from its mean position. Find its
amplitude.

3. A particle is free to move along a line AB and is acted on by an attract-
ive force directly proportional to its distance from a point P in AB, and con-
sequently moves with simple harmonic motion. Prove that its average kinetic
energy is equal to its average potential energy.

4. A point moving with simple harmonic motion is observed to have
velocities of 3 and 4 feet per second when at distances of 4 and 3 feet respec-
tively from its mean position. Find its amplitude and period.

5. A point moves with simple harmonic motion relative to one frame, and
the frame itself moves with simple harmonic motion relative to a second frame,
the directions of the two motions being parallel, and their periods the same.
Show that the motion of the moving point relative to the second frame is simple
harmonic motion, of the same direction and period as that of the frame.

6. A weight w is tied to an elastic string of natural length a and modulus X,
and is allowed to hang vertically in equilibrium. The weight is now pulled
down vertically through a further distance h. Show that on being set free it
will describe simple harmonic motion of amplitude A, provided this does not
involve the string ever becoming unstretched. Find the period of the motion.

THE CYCLOID AL PENDULUM

210. We have seen that the motion of a simple pendulum is
simple harmonic motion only so long as the amplitude of the
motion is small It is, however, possible to constrain a particle to
move under gravity in such a way that its motion shall be simple
harmonic motion no matter how great the amplitude.

266

MOTION UNDER A VARIABLE FORCE

To find the curve in which the particle must be constrained to
move, let us go back to equation (94), namely

which is the equation of motion of a particle constrained to move
in any curve, provided 0 is the angle which the tangent to the
curve at a distance s along it makes with the horizontal. For this
equation to represent simple harmonic motion, the acceleration
d2s

dt2

must be equal to — k2s. Thus we must have

(99)

g sin 0 = tfs,

so that sin 6 must be proportional to s.

211. This relation expresses a property of the cycloid, — i.e. of
the curve described in space by a point on the rim of a circle which
rolls along a straight line. For, in fig. 132, let P be a point on
a cycloid which is formed by a circle rolling along the line EF.
When the point on the rim of the moving circle is at P, let A be

the point of the circle which is in contact with the line EF, and
let AB be the diameter of the circle which passes through A.

At the instant considered, we know that the motion of the point
P on the rim of the circle is perpendicular to the line AP (see
example 1 on p. 9). Thus since APB is a right angle, the motion
must be along BP. Thus BP is the tangent to the cycloid.

If EF is supposed horizontal, the angle 6 between the tan-
gent at P and the horizontal is equal to the angle PAB, so that

THE CYCLOIDAL PENDULUM 267

the radius of the circle through P will make an angle 2 6 with
the vertical.

Suppose that the circle rolls along EF until the tangent to the
cycloid at P makes an angle 6 + dO with the horizontal. The
radius at P must now make an angle 2 (9 -f- d6) with the vertical,
so that the circle must have rotated through an angle 2 dO. Since
the motion of P may be regarded as one of rotation about A, the
small element of path ds described by P will be given by

ds = AP • 2 d0.

Now AP — AB cos 6 = D cos 6, where D is the diameter of the
rolling circle. Thus

giving, on integration, s = 2 D sin 6.

No constant of integration is required if we agree to measure s
from the point at which 0 = 0, i.e. the lowest point of the cycloid.

The property of the cycloid is now proved, and we see that
equation (99) is true throughout the motion of a point which
describes a cycloid, this being generated by the rolling of a circle
of diameter D given by

212. If the cycloid is given, the frequency k of the simple har-
monic motion will be k = AO"' an^ ^ne Period ^s ~T~J or

Thus the motion is of the same period as that of a simple
pendulum of length 2Z>.

213. The importance of cycloidal motion is as follows. It has
been seen that the motion of a simple pendulum is only strictly
simple harmonic when the amplitude is so small that it may be
treated as infinitesimal. For finite amplitudes the motion is not

268 MOTION UNDER A VARIABLE FORCE

simple harmonic, and consequently the period is different from that
of the simple harmonic motion described when the amplitude is
very small. Thus the period depends on the amplitude, so that a
clock which beats true seconds when the pendulum swings through
one angle will gain or lose as .soon as the pendulum is made to
swing through any different angle. Variations of amplitude must
always occur during the motion of any pendulum, and these cause
irregularities in the timekeeping of the clock.

We have, however, seen that if a particle describes a cycloid,
the period is independent of the amplitude, so that variations of

amplitude cannot affect the
timekeeping powers of a par-
ticle moving in a cycloid.

The simplest way of caus-
ing a particle to move in a
cycloid is, in practice, to sus-
pend it from a fixed point by
a string, in such a way that
FIG 133 during its motion the string

wraps and unwraps itself

about two vertical cheeks. If the curve of these cheeks is rightly
chosen, the particle can be made to describe a cycloid, and it can
easily be shown that the curves o£ the cheeks must be portions of
two cycloids each equal to the cycloid which is to be described by
the particle.

EXAMPLES

1. In cycloidal motion prove that the vertical component of the velocity of
the particle is greatest when it has described half of its vertical descent.

2. A particle oscillates in a cycloid under gravity, the amplitude of the
motion being 6 and the period being r. Show that its velocity at a time t meas-

. 2irb . 2-n-t
ured from a position of rest is sin

T T.

3. A particle of mass ra slides on a smooth cycloid, starting from the cusp.
Show that the pressure at any point is 2 mg cos ^, where \f/ is the inclination to
the horizontal of the direction of the particle's motion.

MOTION ABOUT A CENTER OF FORCE 269

MOTION OF A PARTICLE ABOUT A CENTER OF FORCE
Force Proportional to the Distance

214. Let us suppose that a particle moves under no forces ex-
cept an attraction to a fixed point 0, the force of attraction being
directly proportional to its distance from 0.

Taking 0 as origin, let the coordinates of the point Py the posi-
tion of the particle at any instant, be denoted by x, y, z. Let the
force acting on the particle be /* • OP directed along PO, where /z.
is a constant. The components of this force along the three
coordinate axes are _^ _^ _^

The components of acceleration are, as in § 177,

d*x tfy cPz
~dt^ W d?'

Thus the equations of motion of the particle are

/72r

m — = -^x, (100)

TO = -^> (101)

(102)

These three equations are all of the same type, namely the type
of equation which denotes simple harmonic motion. Thus the foot
of the perpendicular from the moving point on to each of the
coordinate axes moves with simple harmonic motion.

The solution of equation (100) has already been seen to be

x=Acosp(t — e),
where p* = fi/m. This can be written

x = A cos pe cos pt + A sin pe sin pt,
or again x — C cos pt + D sin pt,

270 MOTION UNDER A VARIABLE FORCE

if we introduce new constants C, D to replace the constants
Acospe and Asmpe. The other two equations have similar
solutions, so that we can take the complete solution to be

x = C cos pt + D sin pt, (103)

y = cf cos pt + D1 sin pt, (104)

z = C" cos pt + D" sin pt. (105)
We can always solve the equations

f<7 + rC"+s<7"=0, (106)

\D+rD'+sD"=Q, (107)

and so obtain values of r and s for which these relations are true.
Let us multiply equations (104) and (105) by these values of r and
s, and add corresponding sides to the corresponding sides of equa-
tion (103). We obtain
(x + ry + sz) = ( C + r Cf + sC") cospt + (D + rD' + sD") sinpt

= 0, (108)

since equations (106) and (107) are satisfied. The meaning of
equation (108) is that for all values of t we have the relation
x 4- vy + sz = 0, and, therefore, that throughout its motion the
particle remains in the plane of which this is the equation.

The axes of coordinates have been supposed to be chosen arbi-
trarily. We can always choose the axes so that the plane in which
the whole motion takes place is that of xy. The motion is then
given by two equations of the form

x = C cos pt + D sin pt,
y = cr cos pt + Df sin pt.
Solving for sinpt and cospt we obtain

C'x—Cy

D'x —Dy
- -

so that on squaring and adding, we obtain

( C'x - CyY + (D'x -Dy)* = ( C'D - CD')*.

MOTION ABOUT A CENTER OF FORCE 271

This is the equation of an ellipse.

Thus the most general motion possible for the particle consists
in describing the same ellipse over and over again. The period is
2 TT/P, this being the time required for cos pt and sin pt both to
repeat their values.

215. The axes of x, y are still undetermined ; let us imagine
them to be the principal axes of the ellipse.

Then if we suppose the time measured from one of the instants
at which the particle is at one of the extremities of the major
axis, we shall have equations of the form

x = A cos pt^
y = B sin pt.

Thus pt is the eccentric angle of the ellipse described by the
particle, so that the eccentric angle increases with uniform angular

velocity p or A— • The motion repeats itself as soon as p increases

f/A Ira

by 2 TT. Thus the frequency is p or -vl — > while the period is 2 TT A| — •

216. This motion is realized experimentally in the motion of
a pendulum which is not constrained to move in one vertical
plane, but of which the deviations from the vertical

/*t

remain small.

Let the pendulum be of length a, and let its bob be
displaced from its equilibrium position 0 to some near
point P, such that the angle PCO may be treated as
small. Calling this angle 6, the weight of the bob may
be resolved into mg cos 0 along CP, which is exactly neu-
tralized by the tension of the string, and a force mg sin 0 -Q
along PO. If 6 is small, ski 6 may be put. equal to 0, and FIG. 134

this in turn to Thus the bob may be supposed to experience

a force- — OP along OP. The motion is therefore of the kind
a/

which has been described, the value of //, being — > and the value

I — d

of p therefore being A) Thus we see that a hanging weight

272 MOTION UNDER A VARIABLE FORCE

drawn from its position of equilibrium and projected in any way
will always describe an ellipse in the horizontal plane in which it
is free to move, having the point immediately below its point of
suspension as center.

An arrangement may sometimes be seen at village fairs in England, in
which the showman ingeniously takes advantage of this result. A weight
is suspended by a string, and a skittle is placed on the floor exactly under
the point of suspension of the weight. Passers-by are invited to pay an
entrance fee and compete for a prize which is awarded to any one who can
project the weight so that on its return it knocks the skittle over. The
problem is, of course, as impossible as that of describing an ellipse which
shall pass through its own center.

217. Another way in which motion under a force proportional
to the direct distance may be realized, is as follows: An elastic
string of natural length I has one end fastened to a small particle
which is free to move on a smooth horizontal table ; the other end,
after passing through a small hole in the table, is fastened to a
fixed point at a distance I from the hole. If the particle is pulled
away from the hole to a point distant r from it, the total length

of the string is I + r, so that its tension is - X, where X is its

modulus of elasticity. The force acting on the particle, namely
the tension of the string, is therefore proportional to the distance
of the particle from a fixed point — namely the hole in the table
— and its direction is toward the hole. Thus the particle will
move in elliptic motion on the table.

EXAMPLES

1. The point P is describing an ellipse under an attractive force to the center,
and p is the corresponding point on the auxiliary circle. Show that p moves
round the auxiliary circle with uniform velocity.

2. A particle describes an ellipse about a center of force, the attraction
being that of the direct distance. Show that the radius vector from the center
of the ellipse to the particle sweeps out equal areas in equal times.

3. A particle is describing an ellipse under a force proportional to the dis-
tance, when it receives a blow in a direction parallel to the major axis of the
ellipse. Show that the minor axis of the new orbit is the same as that of the
old, and show how to find the change produced in the major axis.

MOTION ABOUT A CENTER OF FORCE 273

4. A particle is acted on by attractions to a number of centers of force, each
Deing proportional to the distance. Show that it describes an ellipse.

How could a mechanical model be constructed to illustrate this motion ?

5. A particle is acted on by a repulsion proportional to its distance from a
ter of force. Show that it describes a hyperbola.

6. Show that in the last question the radius vector joining the particle to
center of force sweeps out equal areas in equal times.

GENERAL THEORY OF MOTION ABOUT A CENTER OF FORCE

218. Suppose that we have a particle acted on only by a force
directed towards a fixed center of force, the magnitude of this force
being any function of the distance from the center.

Let 0 be the center of force, P the position of the particle at
any instant, and PP' the direction of the velocity of the particle
at this instant. Then
the plane OPP' con-
tains the velocity of the

particle, which is along ( o«

PP'y and also the accel-
eration, which is along
•PO. Hence, after any

, , . , , , , ' FIG. 135

short interval the veloc-
ity of the particle will still be in the plane OPP1. The particle is
still in this plane, say at P1, so that the acceleration which is
along P'O is also in this plane.

Hence we can show that, after a further small interval, the posi-
tion, velocity, and acceleration of the particle are all in the plane
OPP', and so we can proceed indefinitely.

It follows that the particle will never leave the plane OPP1.
We accordingly have the theorem :

The orbit described by a particle about a fixed center of force lies
entirely in one plane.

This theorem has already been ^emplified in § 214 by the
case of the orbit described under an attraction proportional to the
distance from the center of force.

274 MOTION UNDER A VARIABLE FORCE

Moment of a Velocity

219. The velocity of a point is a vector, and the line of action
of this vector may be supposed to be the line through the moving
point in the direction of its velocity. We can define the moment
of a velocity in just the same way as the moment of a force has
been defined. Moreover, all the properties of the moments of a
force followed from the fact that forces could be compounded

according to the parallelogram law, so
that the same properties will be true of
the moments of velocities, because veloci-
ties also can be compounded according to
the parallelogram law.

Let P be a particle describing an orbit
about 0, and let OQ be a perpendicular
from 0 on to the line through P in the

direction of the particle's velocity. Then the moment of the par-
ticle's velocity about 0 is OQ x (velocity of particle).

After a short interval dt, let the particle be at P'. Its velocity
at Pr is compounded of its velocity at P together with dt times its
acceleration at P. Hence

(moment about 0 of velocity at Pr)

= (moment about 0 of velocity at P)

4- (moment about 0 of [dt x acceleration at P] ).

The acceleration at P being along PO, the last term of tl
equation is zero, so that we see that the moments about 0 of tl
velocities at P and at P' are equal.

We can extend this step by step as in the former theorem,
obtain finally that

The moment about 0 of the velocity of a particle describing
orbit about 0 is constant.

220. We have supposed that the particle moves from P to

in time dt, so that if v is its velocity at P, then PP' = v dt. As
the particle describes its orbit, the line OP may be regarded as

MOTION ABOUT A CENTER OF FORCE 275

sweeping out an area in the plane of the orbit. The area described
in time dt is the small triangle OPP1. We now have

area described in time dt
= area OPP'

= \OQ • vdt

= l- dt X moment of velocity about 0.

Thus the area described per unit time is half the moment of the
velocity about 0, and this by the theorem of the last section is a
constant. Thus we have the theorem :

Equal areas are described in equal times.

Differential Equation of Orbit

221. The theorem just proved, in combination with the theo-
rem of the . conservation of energy, enables us to determine the
equation of the orbit in which a parti-
cle will move. This equation is most
conveniently expressed in polar coordi-
nates, the center of force being taken
as origin.

If r, 0 are the polar coordinates of the
particle, the velocity may be regarded FlG- 137

7 7/1

as compounded of a velocity — along OP, and a velocity r —
at right angles to OP.

The velocity is therefore given by

The moment of the velocity about 0 is equal to the moment of
the second component, for the moment of the first component

vanishes. Thus the moment of the velocity about 0 is r X r — >
and since this has a constant value, say h, we have

r*—=h. (109)

276 MOTION UNDER A VARIABLE FORCE

If m is the mass of the particle, and if /(r) is the attraction per
unit mass when at a distance r from 0, we find that the potential
energy of the particle is ~r

m I f(r)dr.

*Jao

The kinetic energy is |- mv2, or

[/dr\2 2 /deVl

Ht)+rU)l ]

Expressing that the total energy is constant, we have

(SMSy+'Jpw*-* <"»>

where E is a constant.

Equations (109) and (110) lead to the differential equation of
the orbit. We have, since r and 6 are both functions of tt

dr_dr<M
di~d0di'

so that equation (110) may be expressed in the form

7/1

and on eliminating — from this and equation (109), we have

Cut

the differential equation of the orbit.

LAW OF INVERSE SQUARE

222. Let us now suppose that the attraction follows the law of
the inverse square of the distance, so that

where JJL is a constant. Then

(»•)&• = -£, (li:

LAW OF INVERSE SQUARE 277

and equation (111) becomes

, , .7/3 hdr

whence we obtaui du -

*L-»
r h
giving, on integration, 6 = sin"1 — + e,

where e is a constant of integration.
Simplifying, this becomes

'-sin(0-e),
and if we compare with the equation

1 = e cos 0,

we see that equation (113) represents a conic, having the origin
as focus, and being of semi-latus rectum I — — and eccentricity
e = -xjl H — • In order that the line 0 = 0 may coincide with

the major axis of the conic, the value of e must be

223. We notice that if

E is positive, then e > 1, and the orbit is a hyperbola ;
E is zero, then e = 1, and the orbit is a parabola ;
E is negative, then e < 1, and the orbit is an ellipse.

Thus the class of conic described depends solely on the value
of E, and not. on that of h. And it should be noticed that, if we
are given the point of projection of a particle, and also its velocity
of projection, the value of E is determined, for by equation (110)

278 MOTION UNDER A VARIABLE FORCE

Thus the class of conic described depends only on the velocity
of projection, and not on the direction : the conic is a hyperbola,
parabola, or ellipse, according as

v2 > = or < —
r

The actual eccentricity depends on both E and h, for if e is the
eccentricity, we have ,2

224. In order that the particle may describe a circle we must
have e*= 0, and therefore * ^,2

Putting E = v2 -- — and h = pv (so that p is the perpendicular

from the center of force on to the direction of projection), this
reduces to 2 o „

t:_*ff>V*-o,

Since p is necessarily less than r, neither term in this equation
can be negative. Thus, in order that the equation may be satisfied,
both terms must vanish, and we must have

p = r and v* = — •
r

The first equation expresses that the projection must be at right
angles to the line joining the particle to the center - of force. The
second equation, which can be written

72==7'

shows that the attractive force must just produce the acceleration
appropriate to motion in a circle of radius r.

LAW OF INVERSE SQUARE 279

225. For an elliptic orbit, the periodic time is that required to
sweep out an area Trab, where a, b are the semi-axes of the ellipse.
Since the area is swept out at a rate J- h per unit time, the periodic
time T will be

b*
The semi latus-rectum / is equal to — > and has also been seen

h2 a

to be equal to — > so that

— 2 Trab 2 TTO /IIA\

whence T = — - — = — -=- - (114)

Since this does not depend on the eccentricity, it is clear that
the periodic time of any orbit is the same as that in a circle of
radius equal to the semi major-axis.

226. The law of force of the inverse square is that of gravitation :
the law which we have been investigating is therefore that which
governs the motions of the planets in their orbits round the sun,
as well as the motions of comets and meteorites. For reasons
which cannot be explained here, the conies described by the planets
are all of them ellipses of small eccentricity. A wider range is
found in the orbits of comets. These bodies generally come from
far outside the solar system. To a close approximation many of
them may be treated as coming from infinity, and as starting xwith
relatively small velocity. In this case the orbit is approximately
parabolic.

Kepler's Laws

227. Long before the theory of the planetary orbits had been
worked out mathematically by Newton, three of the principal
laws governing the motion of the planets had been discovered
empirically by Kepler. Kepler's three laws are as follows :

LAW I. Every planet describes an ellipse having the sun in one
of its foci.

280 MOTION UNDER A VARIABLE FORCE

LAW II. The areas described by the radii drawn from the planet
to the sun are, in the same orbit, proportional to the times of
describing them.

LAW III. The squares of the periodic times of the various orbits
are proportional to the cubes of their major axes.

From the first of these laws Newton proved that the law of
force between the planets and the sun must be the law of the
inverse square. The third law is seen to express the same fact as
equation (114).

Motion of Two Particles about One Another

228. A pair of objects known as a double star is of common
occurrence in the sky. This consists of two stars describing orbits
about one another, neither star being fixed.

By the theorems proved in Chapter IX, the center of gravity of
the two stars must either remain at rest, or else must move with
uniform velocity in a straight line, in which case it may, as we
have seen, be treated as fixed, provided all motion is measured
relative to a frame of reference moving with it.

Let A, B be the positions of the two stars at any instant, and let
G be their center of gravity. Let the masses of the stars be m, m',
and let a, b denote their distances from G. Then

m mf m + m'

The complete law of gravitation is expressed by the law

where m, mf are the two masses, r the distance between them, 7 a
constant whose value can be found by experiment, and F is the
force of attraction between the two masses. Thus the force acting

on the star B is f

//Lin

F— ry - ,

LAW OF INVERSE SQUARE 281

acting along BA. This force can always be regarded as acting from
the fixed point G, for its line of action is always BG. Moreover,
its magnitude per unit mass of star B is

or, from relations (115),

This is a force ^ acting towards G if we take

(ra + ra')2

Thus the two stars each describe a conic about the center of
gravity of the two. It is possible astronomically to observe the
values of the periodic time T and the major axes of the orbits of
these conies. From these quantities we can determine the values
of /*, so that we know the values of

(m + ra')2 ' (m + ra')2 '

and these at once lead to the values of ra, ra'. In this way it has
been found possible to determine the masses of some of the stars.

EXAMPLES

(Take the gravitation constant to be -y = C6.6 x 1(T9 in centimeter-gramme-second units.)

1. Given that the earth attracts as though its mass were concentrated at its
center, and that the value of g at the equator, distant 6.378 x 108 centimeters from
the earth's center, is 978. 1 centimeters per second per second, find the mass of the
earth.

2. Taking the masses of the earth and moon as 6.14 x 1027 and 7.94 x 1025
grammes respectively, and assuming their distance apart to be always 3.84 x 1010
centimeters, find the periodic time of the moon.

3. Taking the sun's mass to be 2 x 1038 grammes, and the year to be 365.24
days, find the semi major-axis of the earth's orbit, regarding the sun as a fixed
center of force.

4. If the sun's mass is 324,000 times that of the earth, by how much must
the result of question 3 be altered when the sun's motion is taken into account?

282 MOTION UNDER A VARIABLE FORCE

5. Taking the mass of Jupiter to be TJ?7 of the mass of the sun, and its
greatest distance from the sun to be 498| million miles, show that, on account of
Jupiter's attraction, the sun will describe an ellipse of semi major-axis equal to
about 461,000 miles, and find the length of Jupiter's year.

6. The maximum velocity attained by the earth in its orbit is 3,000,000
centimeters per second, and the minimum velocity is 2,920,000 centimeters per
second. Find the eccentricity of the earth's orbit.

GENERAL EXAMPLES

1. A particle, attached by a string to a point, has just sufficient energy
to make complete revolutions in a vertical circle. Show that the tension
of the string is zero and six times the weight of the particle respectively,
when the particle is at the highest and lowest points of its path.

2. A particle moves under gravity in a vertical circle, sliding down the
convex side of a smooth circular arc. If its velocity is that due to a height
h above the center, show that it will fly off the circle when at a height f h
above the center.

3. If the angle a through which a simple pendulum swings on each
side of the vertical is small, but not infinitesimal, show that to a first
approximation the time of oscillation is

Deduce that a pendulum, which beats seconds accurately when performing
infinitesimal oscillations, would lose about 40 seconds a day when attached
to a clock which caused it to oscillate to 5 degrees on each side of the
vertical.

4. A train is moving uniformly round a curve at 60 miles an hour, and
in one of the carriages a seconds pendulum is found to beat 121 times
in two minutes. Show that the radius of the curve is about a quarter
of a mile.

5. One end of an elastic string, natural length a, modulus X, is tied to
a fixed point on a smooth horizontal table, and the other end is tied to a
particle of mass m which rests on the table. If the mass is pulled to a
distance 2 a from the other end of the string, and is then let go, show that

it will return to its original position at regular intervals 2(?r + 2)-i/— •

\ A

6. Two balls weighing W1 and W2 pounds are connected by a thread a
feet long ; and Wl is held in the hand while W2 is whirled round. Deter-
mine the motion which ensues if Wl is released from rest when W2

EXAMPLES 283

is moving with velocity F at inclination a ; and prove that in the air
the tension of the thread is

— pounds.
W2 ga r

7. Two masses, ml and m2 , are connected by a weightless spring of such
strength that when ml is held fixed, m2 performs n vibrations a second.
Show that if ?n2 be held fixed, ml will perform n "^m2/ml vibrations a sec-
ond, while if both masses are free, they will perform n-^l— ^ vibra-
tions per second, the vibrations in every case being in the line of the
spring.

8. A particle of mass m, moving in a smooth curved tube of any shape,
is in equilibrium under the tensions of two elastic strings in the tube, of
natural lengths I, I' and moduli of elasticity X, X', of which the other ends
are attached to fixed points of the tube. If the particle makes oscillations,
large or small, in the tube, show that the time of oscillation is

9. A string passes through a small hole in a smooth horizontal table,
and has equal particles attached to its ends, one hanging vertically and the
other lying on the table at a distance a from the hole. The latter is pro-
jected with a velocity ^fga perpendicular to the string. Show that the
hanging particle will remain at rest, and that if it be slightly disturbed,

\/— •
\3<

the time of a small oscillation will be 2 T

10. A particle moves in a circular groove, under an attraction — to a

point P which is in the plane of the circle and distant b from its center.
The particle is projected with velocity Ffrom the point of the circle nearest
to P. Show that for the particle to perform complete revolutions, the value

4 id)

of F2 must not be less than --

a2 — bz

11. A smooth ellipse, semi-axes a and &, is placed with its major axis
vertical, and a particle is projected along the concave side of the arc, with
velocity due to a height h above the center. Find the point at which the
particle will leave the arc, and show that it will pass through the center of

the ellipse if

, 2 2

h =

284 MOTION UNDEE A VARIABLE FOECE

12. A particle is constrained to move in a circle of radius a, under an
attraction pr per unit mass to a point inside the circle distant c from its
center. If the particle be placed at its greatest distance from this point,
and started with an infinitesimal velocity, prove that it will pass over the
second quadrant of the circle in a time

13. A particle describes an ellipse about a center of force in one focus.
Show that the velocity at the end of the minor axis is a mean proportional
between the velocities at the ends of any diameter.

14. A comet describes a parabola. Show that its velocity perpendicular
to the axis of its orbit varies inversely as the radius vector from the sun.

15. A comet of mass m, describing a parabola about the sun, collides
with an equal mass m at rest, and the masses move on together. Show that
their center of gravity will describe a circle about the sun as center.

16. Assuming that a projectile, after allowing for variations in gravity,
describes an ellipse about the earth's center as focus, show that the maxi-
mum range on a horizontal plane through the point of projection, for a
given velocity v, is

where R is the distance from the earth's center to the point of projection.

17. When the earth is at the end of the major axis of its orbit, a small
meteor, of mass one rath of that of the sun, suddenly falls into the sun.

Show that the length of the year will be diminished by — of itself.

18. A planet P moving about the sun S picks up a small meteor, and
consequently has its velocity reduced by one nth of its former amount,
although unaltered in direction. Treating n as small, show that the eccen-
tricity of the planet's orbit will be reduced by 2 n(e + cos 0), where 6 is the
angle between SP and the major axis of the orbit.

Show also that the new major axis will make an angle - with
the old axes.

19. A particle describes an ellipse about the focus. Show that the
greatest and least angular velocities occur at the ends of the major axis, and
also that if a, j8 be these angular velocities, the mean angular velocity is

c* + Vp

EXAMPLES 285

20. A comet describes a parabola about the sun, its nearest distance
from the sun being one third of the radius of the earth's orbit, supposed
circular. For how many days will the comet remain within the earth's
orbit ?

21. If the attraction on a particle varies as the inverse square of the
distance from a center of force 0, show that there are two directions in
which a particle can be projected from a given point P so that its orbit
may have a given major axis. If OP — c, and if o^, a2 are the angles
which the two directions of projection make with OP, show that

where a is the semi major-axis.

22. A particle is projected from a point P under a force to a fixed point
S at a distance R from P, so as to describe a circle passing through S. The
initial velocity is F, and the moment of the velocity about S is h. Show
that the particle will describe a semicircle in time

JZ*

4 h*

23. A block of mass M, whose upper and lower faces are smooth horizon-
tal planes, is free to move along a groove in a parallel plane, and a particle
of mass m is attached to a fixed point in the upper face by an elastic string
whose natural length is a and modulus X. If the system starts from rest
with the particle on the upper face, and the string stretched parallel to
the groove to 1-f n times its natural length, prove that the block will per-
form oscillations of amplitude

(n + 1) am

M+m
and period

aMm

CHAPTEE XI
MOTION OF RIGID BODIES

229. The present chapter is devoted to a discussion of the
motion of rigid bodies, when the motion is such that the bodies
may not be treated as particles.

It has already been proved in § 66 that the most general motion
possible for a rigid body is one compounded of a motion of trans-
lation and a motion of rotation. As a preliminary to discussing
the general motion of a rigid body under the action of forces of
any description, we shall examine in greater detail than has so far
been done the properties of a motion of rotation.

ANGULAR VELOCITY

230. We have seen (§67) that for every motion of a rigid body
in which a point P remains fixed, there is an axis of rotation,
which is a line passing through P, of which every point remains
fixed. If a rigid body is moving continuously we may analyze its
motion in the following way. We select a definite particle P of
the rigid body, and we refer the motion to a frame of reference
having P as origin, and moving so as always to remain parallel
to its original position. Eelative to this frame, the motion of the
body between any two instants is a motion of rotation about P.

Now let the two instants be taken very close to one another,
the interval between them being dt. Let us find the axis of rota-
tion of the motion which takes jplace in the interval dt, and call it '
PQ. Then PQ is called the axis of rotation at the instant at which
the interval dt is taken.

Let us suppose that during the interval dt the rotation of the
body about its axis of rotation PQ is found to be a rotation

286

ANGULAR VELOCITY 287

through an angle d6. Then the limit, when dt is made to vanish,

of the rate — is called the angular velocity of the body, — it
dt

measures the angle turned through per unit time.

Thus to have a full knowledge of the motion of a rigid body at
any instant we must know

(a) the direction and magnitude of the velocity of the point P
which has been selected to give a frame of reference ;

(5) the direction of the axis of rotation through P ;

231. The angular velocity has associated with it a direction —
the axis of rotation — and a magnitude. Thus it may be repre-
sented by a line. We shall now prove that it is a vector, i.e.
that angular velocities may be compounded according to the
parallelogram law.

Let a rigid body have a ro-
tation about P compounded
of (a) a rotation of angular
velocity o> about an axis PQ,
and (6) a rotation of angular
velocity o>' about a second
axis PQ'. Let the lengths
PQ, PQ' be taken proportional to co, co', so that the lines PQ, PQf
will represent the directions and magnitudes of the angular
velocities on the same scale.

Let the parallelogram PQRQ* be completed, and let L be any
point on the diagonal PR. Let LNy LN1 be drawn perpendicular
to PQ,PQ' respectively.

In time dt there is, from the first angular velocity, a rotation of
the rigid body through an angle CD dt about PQ. The effect of this
rotation is to move the particle of the body which originally coin-
cided with L through a distance LN- codt at right angles to the
plane PLN. Similarly the effect of the rotation about PQ' is to
move the same particle through a distance LN' • co'dt at right

288

MOTION OF EIGID BODIES

angles to the plane but in the direction opposite to that of the
former motion. Thus the total displacement of the particle is

LNcodt-LN'oo'dt. (116)

Since L is on the diagonal of the parallelogram, we see that the
area of the triangle PLQ is equal to that of the triangle PLQf, so

thafc

Again, since PQ, PQ' are in the ratio of GO : &/, this equation may
be written in the form

and on comparing with expression (116) we see that the displace-

ment of the particle L vanishes.

Thus the resultant of the two angular velocities is a motion such

that the points P and L both remain at rest. It is therefore an

angular velocity having PR, the diagonal of the parallelogram, as

axis of rotation.

We must next find the magnitude of this angular velocity.

Let us denote it by fl. From Q draw perpendiculars QX, Q Y to
Q p PQ'au&PR. The displace-

ment of the particle Q in
time dt will be QY-£ldt
at right angles to the plane.
This displacement, however,
can also be obtained by com-
pounding the displacements
produced by the two angular

velocities a>, ft/. That produced by the former is nil, since Q is on

the axis of rotation ; that produced by the latter is QXco' dt. Thus

QY-Cldt = QX- ft/ dt. (117)

We have QY- Pit = QX- PQ',

each being equal to the area of the parallelogram, and on combin-
ing this relation with (117), we find

FIG. 139

ANGULAB VELOCITY 289

Thus if a/ is- represented by PQ', then H will, on the same scale,
be represented by PR.

We have now proved the following :

The resultant of two angular velocities represented by the edges
PQ, PQ' of a parallelogram is an angular velocity represented by
the diagonal PR of the parallelogram.

Thus angular velocity is a vector, and possesses the properties
which have been proved to be true of all vectors.

232. It follows that an angular velocity fl about an axis of
rotation of which the direction cosines are I, ra, n may be replaced
by three angular velocities mlf a>2, o>3 about the axes of coordinates,
such that

o>3 = nQ. (118)

Squaring and adding, we find that

ft2 = cof + o>22 + o>32. (119)

We now see that the motion of a rigid body is given when we know

(a) u, v, w, the components of velocity of the point P;

(b) (olf o>2, ew3, the components of angular velocity.

KINETIC ENERGY OF KOTATION

233. Suppose that at any instant a rigid body is rotating about
an axis of rotation PQ with
angular velocity O.

Let L be any particle of
the body, its mass being m,
and let LN, the perpendicu-
lar distance from L to PQ,
be denoted by p. Then the
velocity of the particle L is
p£l, and its kinetic energy p^ 140

is Jra/n2.

On summation, the kinetic energy of the whole body is seen to be

290 MOTION OF RIGID BODIES

. The quantity V/7i£>2 is called the moment of inertia about the
axis PQ.

If we introduce a quantity k, denned by

so that k* is the mean value of p2 averaged over all the particles of
the body, then k is called the radius of gyration about the axis PQ.
The kinetic energy can now be written in the form

so that the energy is the same as if the whole mass were concen-
trated in a single particle at a distance k from the axis of rotation.

KINETIC ENERGY OF A EIGID BODY

234. The point P is at our disposal : let us suppose it to be the
center of gravity of the body. Then the most general motion may
be compounded of a motion of translation, this being identical with
that of the center of gravity, and a motion of rotation about an
axis through the center of gravity.

Let V be the velocity of the center of gravity, let H be the
angular velocity, and let k be the radius of gyration about the
axis of rotation through the center of gravity. Let M be the total
mass, Vm, of the body.

By the theorem of § 186, the total kinetic energy of the body is
the sum of two parts : ,

(a) the kinetic energy of a single particle of mass M moving
with the center of gravity of the body ;

(b) the kinetic energy of motion relative to the center of gravity.

The value of part (a) is J. M Vz ; that of part (b) is \
Hence we have for the total kinetic energy

This expression is of extreme importance in itself, but is also of
interest because it enables us to prove the following theorem.

KINETIC ENERGY OF A RIGID BODY 291

235. THEOREM. Let k be the radius of gyration about any axis
through the center of gravity, and let k1 he the radius of gyration
about a parallel axis distant a from the former, then

Let PQ be any axis through the center of gravity G, and let
P'Q' be any parallel axis distant a from the former. Suppose that
the rigid body has a motion of rotation
about P'Q', the angular velocity being fl.

Then the velocity of G is aft, and the
motion may be regarded as compounded
of a motion of translation of velocity aft
together with a rotation ft about the
axis PQ. By formula (120), the kinetic
energy is

It is also ^ Jf&'2ft2, where kf is the radius of gyration about P'Q'.
Hence we have

2 = $ M (a2ft2+

and the result follows on dividing through by \ Jfft2.

236. Alternative proof. This theorem may also be proved geometrically.

Let L be any particle of the body, and let the plane of fig. 142 be
supposed to be the plane through L at right angles to the two axes of
, rotation, these axes cutting the plane in

the points A, A' respectively. Let LA =p,
LA' = p', and let LN be drawn perpendicu-
lar to A A'. Then Mk2 = ^mp2, and also

Mk'2 = %mp'2

__^_M = &"& +
142

- 2p • AA' cos 0)

Now ^N is the projection of the line from L to the center of gravity,
upon the line A A'. Hence ^ra • AN = 0, and we have

giving the result to be proved, after division by M.

292 MOTION OF KIGID BODIES

237. From the theorem just proved, it follows that the radius
of gyration about any axis can be found as soon as we know that
about a parallel axis through the center of gravity, and vice versa.
We now give some examples of the calculation of radii of gyration.

CALCULATION OF EADII OF GYRATION

. 238. Uniform thin rod. Let the rod AB be of length 2 a, and
let k be its radius of gyration about an axis through A perpendic-
ular to its length. Let r be its mass per unit
1 1 ' length, and let x be a coordinate which meas-
ures distances from A. The element which
extends from x to x -f dx is of mass r dxt and its perpendicular
distance from the axis of rotation is x. Hence

7,2 _ ^ _ __ JO ¥ '"" __ 1 ^2

?a , " 2ra 3
rax

so that the radius of gyration is —= •

V o

About the center of gravity, which is distant a from A, the
radius of gyration is given by

*•=$««-«•= | ,

so that the radius of gyration about the center of gravity is — -= -

239. Rectangular lamina. Let us suppose the lamina to be of
edges 2 a, 2 6, and let us find its radius of gyration about an axis
through its center at right angles to its plane. Let us take axes as
in fig. 144, and let cr denote the mass per unit area. Then

CALCULATION OF RADII OF GYKATION

293

The integration is over the lamina, and therefore between the limits
x — a to x = — a and y = b to ?/

y = — b. On integrating, we find

On taking 6 = 0, the lamina
becomes a thin rod, and the re- FlG- 144

suit agrees with that obtained in the last section.

240. Homogeneous solid ellipsoid. Let the semi-axes of the
ellipsoid be a, b, c, and let us find the radius of gyration about the
marjor axis. Taking the principal axes of the ellipsoid as axes of
coordinates, and denoting the density of the ellipsoid by p, we have

fj J

where the integration is over the whole volume of the ellipsoid.
On performing the integrations, we obtain

EXAMPLES

1. Find the radius of gyration of a rod 12 inches lo"ng ab6ut a point distant
4 inches from one end.
is 2. Find the radius of gyration of a circular disk,

(a) about an axis through its center perpendicular to its plane ;

(6) about a diameter.

3. Show that the radius of gyration of a sphere, radius a, about any diameter
is f a2, and about any tangent line is | a2.
V 4. What is the radius of gyration of a cube about an edge ?

5. What is the radius of gyration of a square lamina about a diagonal ?

6. Find the radius of gyration of a solid circular cylinder,
(a) about an axis ;

(6) about a generator ;

7. Prove that the radius of gyration of a solid conical spindle about its axis
is V^~ a, where a is the radius of its base.

294 MOTION OF RIGID BODIES

KOUTH'S RULE

241. The following convenient rule, given by Dr. Routh, (Rigid
Dynamics, § 8), provides an easy way of remembering the values
of several radii of gyration. The rule applies to linear, plane, and
solid bodies which are

(a) rectangular (rod, lamina, or parallelepiped) ;

(b) elliptical or circular (disk or lamina) ;

and states that the radius of gyration about an axis of symmetry
through the center of gravity is given by

,2 _ sum of squares of perpendicular semi-axes

3, 4, or 5

where the denominator is 3, 4, or 5 according as the body comes
under headings (a), (6), or (c) of the above classification.

ILLUSTRATIVE EXAMPLE

A coin rolls down an inclined plane. Find its velocity after any distance and
also its acceleration.

Let the coin be treated as a uniform circular disk, and let a be its radius.
When its velocity down the plane is F, its angular velocity will be V/a. The

axis of rotation is perpendicular to the plane of
the coin. Its semi-axes of symmetry, regarding it
as a lamina, will be a, a. The radius of gyration
about the axis of rotation through its center is,
by Routh's rule,

so that the kinetic energy is

F,o.U5

After rolling a distance s down the plane, the center of gravity of the coin
has fallen a distance s sin a, so that from the conservation of energy

and therefore the velocity is given by

F2 = | sg sin a.

Comparing with the formula (48), V2 - 2/s, for motion under uniform accel-
eration, we see that the disk rolls down the plane with a uniform acceleration
fflr sin a.

MOMENT OF MOMENTUM

EXAMPLES

1. Show that the acceleration of a hoop rolling down a hill of inclination a
is \ g sin a.

2. Find the acceleration of a pair of locomotive wheels running down a gra-
dient of 1 in 50, each wheel consisting of a rim and of spokes of uniform thick-
ness, the weight of the rim being twice that of the spokes and the weight of the
axle being half of that of a wheel. (Neglect the thickness of the axle.)

3. Two bicyclists, riding exactly similar machines, coast down a hill, start-
ing with equal velocities at the top. Neglecting the forces of friction and the
resistance of the air, show that the heavier rider will reach the bottom first.

4. The pulley of an Atwood's machine is a uniform disk of mass M. When
masses mi , w2 are attached to the ends of the string, show that the acceleration

of mi is im - m2

gr.

mi + m2 + £ M

5. Two spheres, one a hollow shell and the other a homogeneous solid, roll
down hill together, starting simultaneously from rest at the top. Show that
their times over any part of the path are in the ratio 5 : V21.

6. If the masses of the wheels of a carriage are supposed to be all collected
at the rim, show that the energy of the carriage when moving with velocity V
is \ M F2, where M is the weight of the complete carriage plus the weights of
the wheels.

7. A straight piece of uniform wire is stood vertically on end and allowed to
fall over. With what velocity does it strike the ground ?

8. A homogeneous solid cigar-shaped spheroid, semi-axes a and 6, is stood
on its point on a horizontal plane and is allowed to roll over. Find its angular
velocity when the end of its minor axis is in contact with the plane, and find
the pressure on the plane at this instant.

MOMENT OF MOMENTUM

242. Let x, y, z be the coordinates of any particle of mass m.
Let the components of the total resultant force acting on the par-
ticle be X, Y, Z. Then the equations of motion are

-3-'-

296 MOTION OF BIGID BODIES

The moment about the axis of x of the force acting on the
particle is yZ~ zY, and from the foregoing equations we have

The velocity of the particle has components — - > -j- > — > so that

at (Jut Cut

the moment of this velocity about the axis of x, as denned in
§ 219> is d, dy

nj __ 2 •

" dt dt

The momentum of the particle is m times its velocity, so that
the moment of momentum about the axis of as is m times the
moment of the velocity, and therefore

dz dy

*#-**

On differentiating, we have
d f / dz t

Kty<te.<^\_/dz<fy + z ^y\l
dt dt df/ \dt dt dt2/]

m ( y -j-2 — z -
= yZ-zY, (122)

by equation (121). Thus we have proved that

The rate of change of the moment of momentum of a particle
about any axis is equal to the moment, about the same axis, of the
forces acting on the particle.

243. Equation (122) is true for each particle of any system of
bodies. Let us sum the equation over all particles, then we obtain

- IV™ (y — -* — ^1 =^\(yZ - ssY). (123)

dt [^ \ dt dt/\ ^

The right-hand side of this equation is the sum of the moments
of the external forces acting on the body or system of bodies,

MOMENT OP MOMENTUM 297

for the internal forces occur in equal and opposite pairs which
contribute nothing.

The term ^m / y — — z — \ > which is the sum of the moments

\ Ol/V Ctl/ I

of momentum of the separate particles, is called the moment of
momentum of the system.

Thus equation (123) expresses that

The rate of change of the moment of momentum of any system
about any axis is equal to the sum of the moments of the external
forces about this axis.

244. Several important consequences of this theorem follow
at once.

I. If a system of bodies is acted on by no external forces, the
moment of momentum about every axis remains constant.

This expresses the principle known as the conservation of angu-
lar momentum.

The sun affords an instance of a body which may practically be sup-
posed to be acted on by no external forces. It is generally supposed that
the sun is gradually shrinking in size ; if this is so, we see that its velocity
of rotation about its axis must continually increase, in order that its
moment of momentum may remain constant.

II. If all the forces acting on a system are either parallel to a
given line, or else intersect this line, then the moment of momentum
of the system about this line must remain constant.

A peg top is acted on only by the reaction at the peg and gravity. The
moment of the latter about a vertical line through the peg vanishes, and
the moment of the former may be supposed to vanish to a close approxi-
mation. Hence the moment of momentum about a vertical through the
peg will remain constant, to a close approximation.

III. If a rigid body is free to rotate about a fixed axis, and if
ft> is its angular velocity at any instant, then

where Ml? is the moment of inertia about the fixed axis, and L is
the sum of the moments about this axis, of all the external forces.

298

MOTION OF RIGID BODIES

To see the truth of this, it is only necessary to notice that a par-
ticle of mass m at distance p from the axis has momentum mpw,
so that the moment of momentum of the whole system will be

and since M and kz do not vary with the time, the rate of change
of angular momentum will be M J<? — -

Oscillation of a Pendulum

245. An important application of the last theorem enables us
to find the time of oscillation of a pendulum of any description.
Let 0 be the pivot about which the pendulum turns, let G be
its center of gravity, let OG = h, and let the line OG
make an angle 0 with the vertical at any instant, so

= — is the angular velocity of the pendulum

(Jut

FIG. 146

that

about its axis.

Let M be the mass, and Jc the radius of gyration
about its axis, of the whole pendulum. Then the
equation of motion is

whicn <» = — The value of L is equal to the

moment of the weight about the axis through 0, and is therefore
Mgh sin 0.

Thus the equation becomes

or,

MOMENT OF MOMENTUM 299

The equation of motion for a simple pendulum of length I is

so that we see on comparison that the motion is the same as that
of a simple pendulum of length I = tf/h.

For instance, the complete period of small oscillations is

ILLUSTRATIVE EXAMPLE

A ring (e.g. a dinner napkin ring] stands vertically on a table, and a gradually
increasing pressure is applied by a finger to one point of the ring in such a way
that equilibrium is broken by the point of contact with the table slipping along the
table. Find the subsequent motion of the ring.

We have seen in example 2, p. 109, that it is
possible to apply pressure in the manner described.

Let us suppose that when the ring leaves the
finger it is observed to be moving with a velocity
V forward and a rotation 12 in the direction oppo-
site to that in which it would rotate if it were
rolling without sliding. Let r, w be the values of —
the velocity and rotation at any instant, measured
in the same directions as V and Q.

Let a be the radius of the ring and m its mass. The forces acting on it are

(a) its weight mg ;

(b) the 'vertical component of the reaction with the table, which is equal to
mg since the center of gravity of the ring has no vertical acceleration ;

By the theorem of § 180 we have

FlQ

We can obtain a second equation from the theorem of § 243. Let us take as
axis the axis of the ring at instant t. The moment of inertia at this instant is
ma2. To obtain the moment of momentum we regard the whole motion as com-
pounded of a motion of translation of the center of gravity (velocity u), and a

300 , MOTION OF RIGID BODIES

motion of rotation about an axis through the center of gravity (velocity w).
The former contributes nothing to the moment of momentum, so that the whole
moment of momentum is

At the end of a small interval dt the ring will have moved forward a distance
vdt, so that we are now considering the moment of inertia about an axis which
is distant v dt from the center of gravity of the ring. The moment of inertia
after an interval dt is, accordingly, by § 235,

We may, however, neglect the small quantity of the second order (dt)2 and
treat the moment of inertia as though it remained constant and equal to ma2.

The rate of increase of the moment of momentum is, accordingly, ma2 — •

The moment of the external forces, measured about the same axis and in the
same direction, is

— mg jta,

du
so that we have the equation ma2 — = — mg /tta, (6)

or, simplified, a -^ = - w, (c)

while equation (a) reduces to — = — p.g. (d)

These relations give the rates of decrease of v and u so long as sliding is
taking place. Sliding clearly ceases as soon as we have v + wa = 0, for v + ua
is the forward velocity of the lowest point of the ring. From equations (c) and
(d) we have

— (7) + wa) = -2/tgr,

and initially the value of v + wa is V+ fla. The time required to reduce v + wa
to zero is, accordingly,

T+fla

After this interval sliding ceases. The velocity of the ring at this instant is
given by

so that the motion may be either forwards or backwards according as we had
initially V > or < fla. After sliding has once ceased there is no force tending
to start it afresh, so that the ring simply rolls on with uniform velocity v. If
V > fla, it rolls farther from its point of projection ; while if F< fia, it will
return to the point of projection.

GENERAL THEOEY OF MOMENTS OF IKEKTIA 301

EXAMPLES

1. The line of hinges of a door makes an angle a. with the vertical, and the
door swings about its position of equilibrium. Show that its motion is the same
as that of a certain simple pendulum, and find the length of this pendulum.

2. A target consists of a square plate of metal of edge a and of mass 3f,
hinged about its highest edge, which is horizontal. When at rest it is struck by
an inelastic bullet of small mass ra moving with velocity v, at a point at depth h
below the line of hinges. Find the subsequent motion of the target.

3. A homogeneous sphere is projected without rotation up a rough inclined
plane of inclination a and coefficient of friction /*. Show that the time during
which the sphere ascends the plane is the same as if the plane were smooth, and
that the time during which the sphere slides stands to the time during which it
rolls in the ratio 2 tana : 7/x.

4. A sphere of radius a is held at rest at a point on the concave surface of
a spherical bowl of radius 6. It is suddenly set free and allowed to roll down
the surface. Show that the line joining the centers of the two spheres swings
in the same way as a simple pendulum of length |(6 — a).

^.--5. A sphere of radius a is held at rest at the highest point of the rough con-
vex surface of a sphere of radius 6. It is then set free and allowed to roll down
this sphere. Show that the spheres will separate when the line joining their
centers makes an angle cos-1^ with the vertical. Examine the case of 6 = 0.

6. A circular hoop, which is free to move on a smooth horizontal plane, has
sliding on it a small ring of I/nth its mass, the coefficient of friction between
the two being /*. Initially the hoop is at rest, and the ring has an angular
velocity w round the hoop. Show that the ring comes to rest relative to the

hoop after a time .

ft*

GENERAL THEORY OF MOMENTS OF INERTIA
Coefficients of Inertia

246. Suppose that a rigid body is rotating about an axis of
rotation of which the direction cosines, referred to any three fixed
coordinate axes, are I, ra, n. Let us
take any point O on the axis of rota-
tion for origin, and let L be any par-
ticle of mass mlt distant p from the
axis of rotation. Let the coordinates
of L be x, y, z, and let LN(=p) be

the perpendicular from L on to the

axis of rotation. "y FIG. 148

302 MOTION OF KIGID BODIES

We have OL2= x2 + y2 + z2,

ON2 = (Ix + my + nz}2,
so that p2=OL2-ON2

= x2+y2+z2-(lx + my + nz)2

= x2 (m2 + n2) + y2 (n2 + I'2) + z2 (I2 + m2)

— 2 wm • yz — 2nl • zx — 2 Im • xy
= I2(y2+z2) + m«(aa+ ai8) + n

— 2 raw -yz—Znl-zx —

Hence the moment of inertia, say J, is given by

— 2 mn

= I2 A + m2^ + 7i2(7 - 2 miiD -ZnlE-1 ImF, (124)

where ^4 = 2^ (y2 + z2), etc.,

D = mz etc.

The quantities -4, J?, (7 are seen to be the moments of inertia
about the axes of x, y, z respectively. The quantities Z>, Ey F are
called products of inertia.

By giving different values to I, m, n in equation (124), we can
find the moment of inertia about any line through 0, as soon as
we know the values of the six coefficients A,. By C, D, E, F.

Ellipsoid of Inertia
247. The equation

Ax*+ Bf+ Cz2- 2 Dyz - 2 Ezx - 2 Fxy = K9

where K is any constant, being of the second degree, represents a
conicoid. If r is the radius vector of direction cosines I, m, n, we have

r2(Al* + Bm2+ Cn2— 2 Dmn — 2 Enl — 2 Flm) = K,
or, from equation (124), r2 = — • (125)

GENERAL THEORY OF MOMENTS OF INERTIA 303

Since / is positive for all values of I, m, n, it follows that r2 is
positive for all directions of the radius vector. Thus the conicoid
is seen to be an ellipsoid.

This ellipsoid is called the ellipsoid of inertia of the point 0.

Equation (125) may be written

and now expresses that the moment of inertia about any axis
through 0 is inversely proportional to the square of the parallel
radius vector of the ellipsoid of inertia.

Principal Axes of Inertia

248. This physical property of the ellipsoid shows that the
ellipsoid itself remains the same, no matter what axes of coor-
dinates are chosen. The ellipsoid has three principal axes, which
are mutually at right angles. The directions of these axes are called
the principal axes of inertia, at the point 0.

If the principal axes of inertia at the point 0 are taken as axes
of coordinates, then the coefficients of yz, zx, xy in the equation of
the ellipsoid must disappear. Thus we must have

D = E = F = 0.

Taking the principal axes of inertia at 0 as axes of coordinates,
equation (124) assumes the form

The kinetic energy of a rotation of angular velocity II is

(126)

where a>v a>2, o>3 are the components of H (see § 232).

304

MOTION OF KIGID BODIES

i- tr

GENERAL EQUATIONS OF MOTION OF A KIGID BODY

249. Let 0 be any point of a rigid body, and let Ox, Oy, Oz be
a set of axes moving so that the point 0 maintains its position in
the rigid body, while the axes remain parallel to their original

position.

Let the velocity of 0 have compo-
nents u, v, w along these axes. The
motion of the rigid body relative to
these axes will be a motion of rotation
about some axis OP which passes
through 0. Let us regard this as
compounded of rotations cox, coy) &>2
about the three axes.
Let x, y, z be the coordinates of any point of the rigid body rela-
tive to these axes. The velocity of this point relative to the frame
supplied by the axes moving with 0 has components

L/r

FIG. 149

_
dt

dz
dt

while the velocity of this frame in space has components

Thus the whole velocity of the point x, y, z has components

37
dt

At any instant let L, M, N denote the sums of the moments of
the external forces about the axes of x, y, z respectively, so that, as
in § 243,

etc.

The moment of momentum of a particle of mass m, at xy y, zy
about the axis of x is

dy\\
*)\

GENERAL EQUATIONS OF MOTION 305

Hence, by the theorem of § 243,

and there are similar equations for the other axes.

250. Relative to the moving axes of coordinates the particle m
has coordinates x, y, z, so that a rotation cox about Ox gives the
particle a velocity of components

0, - o>^,

Similarly the rotations a>y, (oz give velocities respectively of

components n

(oyZj U, — coyx,

and — o)^, &sx, 0.

Compounding these velocities, we obtain as the components of
the resultant velocity, relative to the axes,

2,77 w*^.- - -

&=«*-«,*

-Tt = a>*y- <»vx'

and on differentiation of this equation with respect to t, we obtain
as the value of part of the left-hand member of equation (127)

X/ 2 , 2\ a X^ w X^ s

m (?/ + 2T) — - — > m xy — u — > m xz — =
; dt *4 ' fa £4 dt

— 2m yz(<o* - o>/) +m(^ - ^)o)y«2

dt dt ' dt

306 MOTION OF RIGID BODIES

251. Let x, y, z be the coordinates of the center of gravity of
the rigid body, and let M be its total mass. Then

etc.

As the value of the 'remaining part of the left-hand member of
equation (127) we now have

— Mzv)
dt d>t

= M-(yw-zv).
Thus equation (127) now assumes the form

a)x = L. (128)

If 2)x, 2)F, ^.Z denote the total components along the axes,
we have, by § 180, the further equations

*«+=2 <129>

Equations (128) and (129) and the two other pairs of equations
corresponding to the two other axes are the equations of motion
for a rigid body moving under any forces.

o y — *

EULER'S EQUATIONS

252. Let us now suppose that we have a second set of axes,
which we shall denote by 1, 2, 3. Let these axes move so as
always to retain the same position in the rigid body, the point 0
(which we have already supposed always to retain the same posi-
tion in the rigid body) being the origin. Let the axes 1, 2, 3 coin-
cide with the axes x, y, z at the instant under consideration. Then
the values of the coefficients of inertia referred to axes 1, 2, 3 are
the same as those referred to axes' x, y, z, namely A, B, C, D, E, F.

EULER'S EQUATIONS 307

Moreover, all velocities referred to axes 1, 2, 3 have the same val-
ues as they would have if referred to axes x, y, z. Let us denote
the rotations about the axes 1, 2, 3 by o^, o>2, <w3, then at the
instant under consideration we shall have

This is not necessarily true at any instant except the instant at
which the axes coincide, so that it is not permissible to differen-
tiate these equations with respect to the time and deduce that

do). dcox
— *= — -» etc.
dt dt

Nevertheless, it can be shown that this last result is true at the
instant under consideration. Let OQ denote any line through 0,
let cos a, cos ft cos 7 be its direction cosines relative to axes 1, 2, 3,
and let £lq be the component of angular velocity about OQ. If the
resultant angular velocity is one of amount H about an axis OP
of which the direction cosines referred to axes 1, 2, 3 are I, m, n,
then we have -

C0-&9' —

= fl (/ cos a + m cos fi + n cos 7)
= o^ cos a -f- fc>2 cos /3 + o>8 cos 7.

Whatever line OQ may be, this equation is always true ; hence we
may legitimately differentiate it with respect to the time, and so
obtain

d£lq _ d^

2 Q s

cosa + -jrcosp + —f cos 7
dt dt dt dt

da . Qd$ dy

— o^sintf— G>2sinp— w3 sin 7— -• (130)

dt dt dt

Now let the line OQ be supposed to coincide with Ox, so that
Hg = cox. At the instant under consideration, /3=-7 = — > a = 0.
Moreover, — is the rate at which the angle between Ox and axis

1 increases, and clearly this is ft>8. Similarly, -p = — o)2 and -^ = 0.

a^ dt

308 MOTION OF EIGID BODIES

Making all these substitutions, we find that at the moment under
consideration, at which the two sets of axes coincide, equation (130)
assumes the form

dco1
~

Thus at the instant at which the two sets of axes coincide, we
have the relations

°>x = <°v etc->

da) da>,

and also — * = - — •

dt dt

Let us introduce the further simplification of supposing that the
origin is either a fixed point or the center of gravity of the body-
In the former case we have

u = v = w = 0, always ;
in the latter case

x = y = z = 0, always.

Let us further suppose that the axes are chosen to be the prin-
cipal axes of inertia through the origin, so that

Introducing all these simplifications into equation (128) and the
two similar equations, we find that these assume the form

A^-(B-C)a>^=L, (131)

^ ~ (£-^)*V»i =&, (132)

C^ - (A-B)**^ =N. (133)

(JUv

These equations are known as Euler's equations.

DOTATION OF A PLANET 309

KOTATION OF A PLANET

253. As a first example of the use of these equations, let us
examine the motion of a rigid body, symmetrical about an axis,
acted on by forces all of which pass through the center of gravity.
These conditions approximately represent those which obtain when
a planet moves in its orbit, or a star in space.

Let us take the center of gravity as origin and the axis of sym-
metry as axis 1. Let the moments of inertia be A, By B. Then the
equations of motion are

' ^ = °' If <134>

*^=*(S-4>«V»1, (135)

B8 = -(U-4)«y»,. (136)

The first equation gives at once that ®1 is constant, say equal
to ft. If we write

equations (135) and (136) become

— ** , <138>

-* =

of which the solution is o>2=JE cos (Jet -+- e) ;
and equation (137) now leads at once to

&)3 = — E sin (Jet + e).

Thus the components of angular velocity at the instant t are
ft, E cos (Jet + e), E sin (Jet + e),

310 MOTION OF RIGID BODIES

and we see that the axis of rotation describes a cone in the solid,

. , 27T 27T B

with period — - or — — — •

K &L -D — A.

If B is very nearly equal to A, the period may be very great,
and the motion consequently very slow. This happens in the case
of the earth : the motion of the axis of rotation gives rise to the
phenomenon known as the variation of latitude, of which . the

period is about 428 days. Since a period — — - represents roughly

B—A

one day, we conclude that for the earth is of the order of -^J-g-.

-The true value of this quantity is .00328, the discrepancy resulting from
the imperfect rigidity of the earth.

. MOTION OF A TOP

254. As a second example of the methods of this chapter, let us
consider the motion of a spinning top. This we shall suppose to
be a solid of revolution spinning on a peg of which the end will be

treated as a point, the contact between
the peg and the surface on which it
rests being assumed rough enough to
prevent slipping. The point of contact
is now a fixed point 0. Let us take
axes Ox, Oy, Oz fixed in space, the axis
of z being vertical, and also axes 1, 2,
fixed in the body, and coinciding witl

the principal axes of inertia through 0
FIG. 150

Let axis 1 be the axis of symmetry o

the top, and let the moments of inertia about axes 1, 2, 3 be A, B, B
The first of Euler's equations becomes

A d(°l - 0

JL — — - — U,

dt
since B = C and L = 0. Thus c^ is a constant, say H.

Let the axis of the top cut a unit sphere about 0 at a point whos(
polar coordinates are 1, 0, <£, the axis of Oz being taken for pole, so
that 0 is the angle between the vertical and the axis of the top.

MOTION OF A TOP 311

The kinetic energy of the top is, by § 248,

while the potential energy is Mgh cos 6, where h is the distance
of the center of gravity of the top from 0. Thus the equation of
energy is

+ B (o>; + a>l) + 2 Mgh cos 6 = E, (139)

where E is a constant. This may be put into a different form.
For o>2 + °>l ig the square of the angular velocity of the axis of
the top : it is therefore the square of the actual velocity of the
point 1, 6, <f> on the unit sphere, and hence we have

The equation of energy now assumes the form

AW + B —%• sin2 0 /Yl + 2 Mgh cos 6=E. (140)

We can obtain a third equation from the fact that the angular
momentum about Oz, the vertical, is constant. The angular momen-
tum may be regarded as compounded of

(a) the momentum due to the rotation fl about axis 3 ;

(&) the momentum due to the motion of the axis of the top.

The rotation fl about axis 3 may be further decomposed into
rotations H cos 6, fl sin 6 about the horizontal and vertical, giving
moments of momenta A£l cos 9, A£l sin 6 about the horizontal and
vertical. Thus the moment of momentum contributed by part (a)
is A£l cos 6.

The motion of the axis of the top may be resolved into a rota-

tion of angular velocity sin 6 — about an axis making an angle

d()

— — 6 with the vertical, and one of angular velocity — about a

312 MOTION OF EIGID BODIES

horizontal axis. The former may be replaced by sin2 6 — about

dd>

the vertical, and sin 9 cos 6 -~ about^a horizontal axis. Thus the

clt

moment of momentum about the vertical contributed by part (6)
of the motion is j j

***•$

and since the moment of momentum about the vertical has a con-
stant value, say G, we have

AD, cos 0 + B sin2 0(Q = G. (141)

cut

If we eliminate — — from this equation and equation (140), we
obtain

B sin20 \AW + B(— Y+ 2 Mgh cos 6-E\

L\*/ J

+ (G — AQ, cos 0)2 = 0, (142)

a differential equation giving the variations in the value of 6, and
therefore allowing us to trace the changes in the inclination of the
axis of the top to the vertical. _

The maxima and minima of 6 are given by putting — = 0, and
are therefore the roots of

B(l — cos2 0) [AW + 2 Mgh cos 6 — E]

+ (G—AQ, cos 0 )2 = 0. (143)

Let us call the left hand of this equation /(cos 6). Since/ is a
function of degree three, there will be three roots for cos 6. Let
us suppose that the top is started at an angle 6 = 00, and with the

value of — equal to ( — )• Then, from equation (142),
dt \«Vi

B sin2 00 AQ* + B l^- V + 2 Mgh cos 00 - E\

so that /(cos 0Q) =B sin2(90[^n2+ 2 Mgh cos 0e—E]

+ (G - A£l cos 60)2

MOTION OF A TOP

313

so that /(cos 00) is negative. We easily find, from equation (143),
that

so that /(I) is positive.

Again, f(-

so that /(— 1) is positive, and

/(+ oo) = — 2 Mgh B (+ oo)8,
which is negative. Thus we have seen that

when cos 0 = + oo, /(cos 0) is — ;

when cos 0=1, /(cos 0) is + ;

when cos 0= cos 00, /(cos 0) is — ;

when cos 0= — 1, /(cos 0) is +.

Thus the three roots of the cubic /(cos 0) = 0 lie as follows :

a root 0 = 0l between cos 0 = 1 and cos 0 = cos 00 ;
a root 0 = 02 between cos 0 = cos 00 and cos 0 = — 1 ;
a root for which cos 0 is numerically greater than unity, giving
no real value for 0.

We see, therefore, that
the only points at which

— can vanish are 0=0j,

CLL

and 0 = 02. Moreover, at

these points - - vanishes,
at

and as there are not co-
incident roots at either

. . d0 ,

point — - changes sign on
at

reaching these points, so that 0 can range only between the values
Bl and 02.

Thus the axis of the top oscillates between the two cones 0 = 0l
and 0=02.

314 MOTION OF E1GID BODIES

255. Let us find what is the least angular momentum which the
top must have so as to spin without falling over. To do this, we
may assume that falling will occur if ever 6 exceeds a certain limit
03, either through the peg slipping or through its side touching
the ground. The condition that the top shall not fall over is that
#2 must be less than 03, and hence that /(cos 03) must be positive.
Thus the values of E, G, and fl must be such that

B sin2 08 (AW + 2 Mgh cos 03 - E) + ( G -An cos 08)2

is positive.

Suppose that the top is started at an inclination 00 to the ver-
tical, having no motion except one of rotation H about its axis.
We then have, from equations (140) and (141),

E = AW + 2 Mgh cos 00,

G=A£l cos 00.
Thus

/(cos 08) = B sin2 03 (AW + 2 Mgh cos 08 - E) + ( G - Aft cos 03)2
= B sin2 03 • 2 Mgh (cos 03 - cos 00) + ^2ft2 (cos 03 - cos 00)2
= (cos 03 - cos 00) [2 MghB sin2 03 + ^2H2 (cos 03 - cos 00)].

(144)

Since the top is necessarily started in a position in which it can
spin, the value of cos 03 — cos 00 is necessarily negative. Thus in
order that /(cos 03) may be positive, we must have

A*W (cos 00 - cos 03) - 2 MghB sin2 03 (145)

2 MghB sin2 03

positive, or W > y -^-. (146)

A2 (cos 00 — cos 03)

We notice that if A is very small, the value of fl required to
keep the top from falling is very large. It is therefore very hard
to spin a top of small cross section, such as a lead pencil or a
pointed wire*.

If we can choose the angle at which we start the top, cos 00 is
at our disposal. We see that the necessary value for H is least

MOTION OF A TOP 315

when cos 00 is a maximum, i.e. when the top is started vertical.
In this case the top will spin if

2 MgJiB sin2 03
> .42(l-cos03)'

or if Qa>

256. In general, for a top started vertically and with no velocity
except one of pure rotation about its axis, we find, on putting
cos 00 = 1 in equation (144),

/ (cos 0) = (1 — cos 0)2 [A*W - 2 MghB (I + cos (9)].

The roots of the equation /(cos 6) = 0 are

22
cos0=+l, +1,

iMghB
Let us write il; = - ^ —

2 MghB

then when H2 = fl2, the roots are

cos0 = + l, +1, +1.

When H2 > H2 the third root is greater than unity, and when
fl2 < fl2 the third root is less than unity, say cos 6 = cos ®,
where ® is a real angle, given by

y-X -O- \M — —(! ^"^ »*

1=1-- • <147>

Thus, as long as H2 > H2 the oscillations are confined within
the coincident limits 0 = 0 and 6 = 0, so that the top remains
vertical, but as soon as we have H2 < H2 the oscillations are
between the limits 0 = 0 and 0 = ®.

Suppose we start a top with angular velocity fl greater than H0,
so that at first its axis is vertical and the only motion of the top
is one of rotation about its axis. Then the real roots for 0 are
0, 0 ; there is therefore no range of oscillations, and the axis of

316 MOTION OF EIGID BODIES

the top remains strictly vertical, — in common language, the top
is " asleep."

If the conditions were the ideal conditions supposed, this motion
would continue forever, but in nature such ideal conditions can-
not exist. The region of contact between the peg and the surface
on which it spins is not strictly a point, but a small circle or
ellipse, on account of the small compression which takes place at
the point of contact. By making the peg of hard steel and spin-
ning on a hard surface, this region is very small, but is still of
finite dimensions. The consequence is that the reactions on the
peg do not all meet the axis. There is a small frictional couple
resisting the rotation of the top, and O gradually decreases.

When H has so far decreased as to be less than H0, the ranges
of oscillation are 6 — 0 and 0 = ©. The top is no longer asleep,
but is now wobbling through an angle ®. As fl continues to
decrease, ® continually increases, as is clear from equation (147),
and finally ® reaches so large a value that the top rolls against
the ground and so falls over.

257. The interest of these results will perhaps be enhanced by exam-
ining the form they assume when the top is of a very simple kind.
Let us suppose that a top is formed by running
a pin through the center of a uniform disk of
mass M, radius a. Let the length of the pin
which protrudes through the disk on its lower
side be h, and let the mass of t*he pin be
neglected in comparison with that of the disk.

The h is the same as the h of our previous analysis. The values of
A and B are

When spinning at the critical velocity O0 at which wobbling sets in, the
velocity of a point on the rim is O0a or 2 V^A. Thus wobbling begins when
the velocity of a point on the rim is reduced to 2 V^A, a velocity which
depends only on the height of the disk and not on its radius. We see that
the lower the disk is, the slower it can spin without wobbling. If we take
h = 2 inches, we find that wobbling begins when the rim velocity is about
4.7 feet a second.

EXAMPLES 317

The rim will touch the ground when the range of wobbling is given by

a
a = 6 inches and h = 2 inches, as before, this gives tan® = |, so that

o o 2 _ O2 1 Q

Thus such a top will " sleep " until its rim velocity is reduced to 4.7 feet
a second. After this it will wobble, and as soon as its rim velocity is
reduced to about 4.5 feet a second the top will begin to roll on the ground.
For an ordinary small, pear-shaped peg top we may take roughly h = 1^
inches, and the radii of gyration about axes through the point of contact
as | inch and 2 inches. Thus in inches

A = ^ M, B = 4;M,

9 _ 4 Mgh B _ 2048
~~A*~ ''^Tg'

Taking g — 386 inches per second per second, this gives O0 = 170 revolu-
tions per second. If thrown from a string of which the end is coiled round
the top in circles of radius 1 inch, the string must be withdrawn with a
velocity of about 60 miles an hour relative to the top to set up the required
angular velocity.

GENERAL EXAMPLES

1. A fly wheel whose moment of inertia is / has a string wound round
its axle of radius b. A tension equal to a weight w is applied to the string
for 1 second. What is the angular velocity of the fly wheel at the end of
1 second? I^^\J*2^

2. A fleet of total displacement 200,000 tons steams from east to west
along the equator, covering 20 minutes of longitude per hour. Regarding
the earth as a homogeneous sphere of mass 6 x 1021 tons, find the change
in the earth's angular velocity produced by the motion of the fleet. Show
that the day is lengthened by, roughly, 16 x 10~14 seconds.

3. The earth's mass is 6 x 1021 tons, and icebergs and melted snow
weighing 1010 tons move from the north pole to latitude 45 degrees. Find
the change in the length of the day.

4. A train of mass m runs due north at 60 miles an hour. Show that
there must be a pressure between the eastern rail and the flanges of the
wheel in consequence of the earth's rotation, and find the amount of
the pressure.

5. A thin layer of dust, thickness h feet, is formed on the earth by
'the fall of meteors, reaching the earth from all directions. Show that the

318 MOTION OF RIGID BODIES

change in the length of a day is about — -^ of a day, where a is the radius

of the earth in feet and D, p are the density of the earth and the meteoric
dust respectively.

6. Two masses M and m suspended from a wheel and axle of radii
a, b do not balance. Show that the acceleration of M is

Ma — mb

where / is the moment of inertia of the machine about its axis.

7. A uniform cylinder has coiled round its central section a light, per-
fectly flexible, inextensible string. One end of the string is attached to a
fixed point, and the cylinder is allowed to fall. Show that it will fall with
acceleration f g.

8. Two equal uniform rods of length 2 a, loosely joined at one extremity,

are placed symmetrically upon a fixed sphere of radius — - and raised

into a horizontal position so that the hinge is touching the sphere. They
are then allowed to descend. Show that when they are first at rest they
are inclined at an angle cos-1^ to the horizontal, that the pressure on the
sphere at each point of contact is one quarter of the weight of a rod, and
that there is no strain at the joint.

9. A rod rests with one extremity on a smooth horizontal plane and the
.other on a smooth vertical wall, the rod being inclined at an angle a to

the horizon. If it is allowed to slip down, show that it will separate from
the wall when its inclination to the horizontal is sin-1(| sin a).

10. If the sun gradually contracts in such a way as always to remain
similar to itself in constitution and form, show that when every radius has
contracted an nth part of its length, where n is large, the angular velocity

will have increased to 1 1 + — J times its former value. Examine the change
in the kinetic energy of rotation.

11. An elastic band of natural length 2?ra, mass m, modulus X, 'rests
against a rough wheel of radius a in a horizontal plane. The string is held
against the circumference of the wheel, which is made to rotate with angu-
lar velocity O. If the string is left to itself, show that it will expand, and

that when its radius is r its angular velocity will be — • , and that its radial
velocity will be

. 2»X(,-art

Tfld J

EXAMPLES 319

v 12. A uniform triangular disk ABC is so supported that it can oscillate
in its own plane, which is vertical, about A. Show that the length of the
simple equivalent pendulum is

1 3 62 + c2 - a2

13. A spherical hollow of radius a is made in a cube of glass of mass M,
and a particle of mass m is placed inside. The cube is then projected with
velocity V on a smooth horizontal plane. If the particle just gets round
the sphere, remaining in contact with it all the way, show that

14. Three equal particles are attached to the ends and middle point of a
rod of negligible mass, and one of the end particles is struck by a blow at
right angles to the rod. Show that the velocities of the particles at starting
are in the ratio KOI

O . <£ . J..

15. A rough horizontal cylinder of mass M and radius a is free to turn
about its axis. Round it is coiled a string, to the free extremity of which
is attached a chain of mass m and length /. The chain is gathered close
up and then let go. Show that if 0 is the angle through which the cylinder
has turned after a time t before the chain is fully stretched, then

16. A uniform flat circular disk is projected on a rough horizontal table,
the friction on any element moving with velocity V being cF3 x (mass of
element), in a direction opposite to that of V. Find the path of the center
of the disk.

CHAPTEE XII
GENERALIZED COORDINATES

258. So far we have dealt with the mechanics (dynamics and
statics) of material bodies on the supposition that these bodies
consist of innumerable small particles which, in the case of a rigid
body, are held firmly in position and serve the purpose of trans-
mitting force from one part of the body to another.

259. Even when dealing with rigid bodies this conception of the
structure of matter has not led to entirely consistent results. Tor
instance, we have found that after an impact between two im-
perfectly elastic bodies, or after sliding between two imperfectly
smooth bodies, a certain amount of energy disappears from view,
and we have had to suppose that this energy is used in starting
small motions, relative to one another, of the ultimate particles of
which the bodies are composed. In other words, after an impact
or sliding has taken place, a rigid body can no longer be supposed
to satisfy the conditions postulated for a rigid body.

When dealing with bodies which are obviously not rigid the case
is worse. Here the conceptions which we have introduced into the
study of rigid bodies do not help at all, and very little progress is
possible without introducing some other conceptions to replace these.

260. There are two ways of proceeding at this stage. We may
introduce new conceptions which seem plausible, and in this way
try to form a picture of the structure of the matter with which we
are dealing. We cannot be certain that the results obtained in this
way will be true, for we can never be sure that our conceptions of
the nature of the ultimate structure of matter are accurate. But
it may be worth trying what results are obtained by introducing a
set of provisional conceptions as to the structure of matter. If
these results are in agreement with the phenomena observed in

320

GENERALIZED COORDINATES 321

nature, the probability that our provisional conceptions are near to
the truth is strengthened. If, on the contrary, the results obtained
are not found to agree with what is observed in nature, the pro-
visional conceptions from which these results have been deduced
must be either modified or withdrawn.

Different sets of conceptions as to the structure of the matter
dealt with will lead to different branches of mathematical physics.
As in stances of such branches of mathematical physics may be
mentioned the theory of elastic solids which is based upon certain
provisional conceptions as to the behavior of the particles of
which solid bodies are composed, and the kinetic theory of gases
which is based upon certain provisional conceptions as to the
behavior of the particles of a gas. The tracing out of the conse-
quences of different sets of provisional conceptions as to the struc-
ture of matter cannot, however, be regarded as coming within the
scope of a book such as the present one.

261. There is, however, an alternative way of proceeding. We
have taken Newton's laws of motion as the material supplied by
experimental science for theoretical science to work upon. The^,
truth of these laws as applied to the ultimate particles of the
material universe is by no means certain, because we cannot obtain
the ultimate particles to experiment upon. Suppose, however, that
we examine whether any further progress can be made in the
study of mechanics without introducing any hypothesis beyond
the single one (admittedly uncertain) that Newton's laws apply to
the ultimate particles. If we can make progress in this direction,
the results obtained will of course apply to all further extensions
of mechanics, whether or not additional hypotheses are introduced
as to the nature and arrangement of the ultimate particles.

262. The standpoint from which we are regarding the matter
can, perhaps, be explained by an analogy, first suggested by Pro-
fessor Clerk Maxwell. Suppose that we have a complicated ma-
chine in a closed room, and that the only connection between
this machine and the outer world is by means of a number of
ropes which hang through holes in the floor into the room beneath.

322 GENERALIZED COORDINATES

A man introduced into the room beneath will have no opportunity
of inspecting the machinery above, but he can manipulate it to a
certain extent by pulling the different ropes. If on pulling one
rope he finds that the others are set into motion, he will under-
stand that the different ropes must be connected above by some
kind of mechanism, but will not be able to discover the exact
nature of the mechanism.

This concealed mechanism may be supposed to represent those
parts of the mechanism of the universe which are hidden from our
view, while the ropes represent those parts which we can manipu-
late. In nature, there are certain acts which we can perform, cor-
responding to the pulling of the ropes in our analogy, and we see
that these are followed by certain consequences, analogous to the
motion of the other ropes ; but the ultimate mechanism by which
the cause produces the effect remains entirely unknown to us.
For instance, if we press the key of an electric circuit, we may
find that the needle of a distant galvanometer is moved, but the
mechanical processes which transmit the action through the wires
of the circuit and through the ether surrounding the galvanometer
needle remain unknown.

263. Now suppose that the imaginary man is at liberty to handle
the ropes and that he wishes to study the connection between them.
He may begin by conjecturing that the connecting mechanism in
the room above consists of arrangements of, say, levers, pulleys, and
cogwheels, and he may work out for himself the manner in which
the ropes ought to move if his conjectures are correct. This proced-
ure would be analogous to that we have described in § 260; it is
not the procedure we are going to follow here.

On the other hand, without any conjecture at all as to the
nature of the mechanism above, the man will know that certain
laws will govern any manipulation of the ropes, if the ropes are
connected by mechanism of any kind whatever, such that each
particle obeys Newton's laws of motion.

To explain this, let us take the simplest case, and suppose that
there are two ropes only and that when A is pulled down ijH^ an

HAMILTON'S PRINCIPLE 323

inch, then B invariably rises through two inches. The mechanism
may be a lever, an arrangement of pulleys, or clockwork. But
whether it is any one of these, or something entirely different from
any of them, it will be known that the motion of rope A down-
wards can be restrained by exerting on rope B a force equal to half
of that applied to A. This fact follows from the principle of virtual
work, quite apart from any conjecture as to the nature of the
hidden mechanism. Now the question before us is as follows:
Can we, without any knowledge of the hidden mechanism, discover
what motion of the ropes will ensue, if they are started in any
given way. And the answer is that we can, provided we know
the amount of energy involved in a motion of any kind, — i.e. pro-
vided we know the kinetic energy of every motion, and also the
potential energy of every configuration.

So also, to pass from analogies to realities, we can, without any
knowledge of the ultimate mechanism of the universe, discover
what motion will ensue from any initial conditions, provided that
we know the kinetic and potential energies of all configurations
of the portion of the universe with which we are dealing.

HAMILTON'S PRINCIPLE

264. Let us suppose that any single particle of a material sys-
tem has at any instant coordinates xlt yltzlt its mass being ra^and
that it is acted upon by forces of which the resultant has compo-
nents Xlt Ylt Zr Let the velocity of this particle have components
ult vlt wlt so that dx

u. = — - > etc.
dt

Then, if the motion of this particle is governed by Newton's laws,
we shall have ,

(149)

324 GENERALIZED COORDINATES

Let us compare this motion with a slightly different motion in
which Newton's laws are not obeyed. In this second motion let the
coordinates of mlf at the instant at which they are xlt y^ z1 in the
actual motion, be supposed to be x[, y{, z[, and let the components
of velocity at this instant be u(, v[, w(, so that

, dx[

<*5»*a

Let us agree that the modified motion is to differ so slightly
from the actual, that any quantity such as x[ — xl} u[ — ult which
measures part of this difference, may be treated as a small quan-
tity. Let us denote x{ — xl by Bxv and use a similar notation for
the other differences.

Multiply equations (148), (149), (150), which are true at every
instant, by Sxv Syl} Szl} and add. We obtain

du^ ~ dvl r, dwl £

1 dt 1 l dt l l dt 1

= X18xl-}-YlSi/1-}- Z-fz^. (1^1)

Now — - Sx* = — (

= -
~~ dt

d .

= dt(U^~
Hence

du, ~ dv. ^ dw,

m^ — * oxl -\- ml — • oy^ -\- TYI^ — -

dt dt dt

= ™l \Tt (U^X* + V

= JT1&B1+ri8y1 + Z1&1, (152)

by equation (151).

HAMILTON'S PRINCIPLE 325

An equation of this kind is true for each particle of the system
and at every instant of the motion. It is moreover true whatever
the displaced motion may be. On summing this equation for all
particles we obtain

1 di ^Xl + v^y^ + w£z^ ~ (ufa + vfa + wfiwd

=£ (Xfa+Y^y, + ZA)- (153)

Now let T denote the kinetic energy of the motion, so that

Then ST= 1

Now u[z — u\ = fa + §u^f — u\ = lufiu^,

if we neglect the small quantity of the second order (fo^)2, so that

we have

oT = m,i (ufiui + v1ovl + wlowl).

265. Assuming for the moment that the system of forces is
conservative, let W denote the potential energy of the system at the
instant under consideration, and W1 that of the imaginary system
in the slightly displaced configuration. Then, by § 118, we have

= (work done in moving system from actual

to displaced configuration)
= - V (xfa+Yfa +Z1Szl). (154)

Substituting into equation (153) for the expressions which have
been found to be equal to ST and &W, we find that this equation
reduces to the simpler form

or again,

326 GENERALIZED COORDINATES

This equation is true at every instant of the motion. Let us
integrate it between any two instants of the motion, say from
t = t: to t = £2. We obtain

(ufa + vjft + wfa)~ '* = f *S (T - W) dt. (155)

The displaced motion has so far been subject to no restrictions
except that the difference between it and the actual motion must
always remain small. Let us now introduce the further restriction
that at times ^ and tz the configurations in the displaced motion
are to be identical with those in the actual motion. The displaced
motion is now one in which the imaginary system starts in the
same configuration as the actual system at time t — tl} swerves
from the course of the actual system from time ^ to time t2
(because the actual system obeys Newton's laws, while the imagi-
nary system does not), and ultimately ends in the same position
as the actual system at time t2.

In consequence of this restriction on the motion of the imagi-
nary system, we have at times tt and t2,

and similar relations for the other particles. Thus

and equation (155) reduces to

C*S(T-W)dt = Q. (156)

Here we have an equation which depends only on the amounts
of the kinetic and potential energies of the system, and not on the
mechanism of the system. We shall find that from this single equa-
tion we can determine the motion of all the known parts of the
system as soon as T and W are known, without any knowledge of
the mechanism of the unknown parts.

PBINCIPLE OF LEAST ACTION 327

266. Before proving this, however, we may attempt to interpret
equation (156). Let us denote T — Why L. Then

C*S(T-W)dt= C &Ld$

Jti Jt!

= f \L' - L) dt

Jt!

X<2 /»«2

L'dt- I Ldt
Jti

-«(£***).

/*<2

If we denote / Ldt

by S, the equation becomes SS = 0, or

S' = S.

Thus the value of the function S for the actual motion is the
same, except for small quantities of the second and higher orders,
as the corresponding function Sr for any slightly different motion,
which begins and ends with the same configuration at the same
instants. In other words, the function S is either a maximum or a
minimum when the series of configurations is that which actually
occurs in nature.

PRINCIPLE OF LEAST ACTION

267. The total energy will, by the theorem of § 143, remain
constant during the actual motion, say equal to E, so that at every
instant we shall have

In the slightly varied series of configurations it is not true that
the total energy remains constant throughout the motion, but out
of the infinite number of slightly varied series of configurations.
there will still be an infinite number for which the conditions
already postulated are satisfied, together with the condition that
the total energy at every instant shall have the value E. For such

328 GENERALIZED COORDINATES

a series we have

Thus we have L = 2T — E, Lf = 2T' — E,

/"* ^2

so that S = I Ldt

•At

= f\2T-E)dt

-f

Jtl

Thus if S is a maximum or a minimum, it follows that

2Tdt

is a maximum or a minimum. This integral is called the action
of the motion. We now see that of all possible series of configura-
tions which bring the system from one configuration to another in
a given time, and in such a way that the total energy has always
a specified constant value, that one which can be described by a
natural system is the one on which the action is a maximum or
a minimum. Since the action is in general a minimum, this prin-
ciple is known as the principle of least action.

The statement of this principle was first given by Maupertius
(1690-1759), who did not deduce it by mathematical reasoning, but
believed it could be proved by theological arguments that all changes in
the universe must take place so as to make the action a minimum (Essai
deCosmologie, 1751).

NON-CONSERVATIVE FORCES

268. If the forces are non-conservative, we may no longer, as in
equation (154), replace

by — §W, and consequently, instead of equation (156), we shall have

dt = °- (167)

LAGKANGE'S EQUATIONS 329

LAGRANGE'S EQUATIONS

269. If the coordinates xlf yly zlt etc., of every particle of
the system are known, we know not only the configuration of
the system but also the mechanism by which the different parts
of the system are connected. It may, however, be that we can
determine the configuration of the system, by knowing a smaller
number of quantities which do not give us a knowledge of the
mechanism.

For instance, in our former illustration we imagined two ropes to hang
from an unknown machine, the ropes being connected in such a way that
a motion of one inch in the one invariably produced a motion of two inches
in the other. In this case the configuration is fully determined when we
know the single coordinate which measures the position of the end of the
first rope, but a knowledge of this coordinate does not imply a knowledge
of the mechanism connecting the ropes.

Again, the position of a rigid body is, as we have seen (§ 65), determined
by the values of sufficient quantities (six) to fix the positions in space of
three non-collinear particles of the body, but a knowledge of these quanti-
ties does not give us information as to the arrangement of the particles of
which the body is formed.

Let 6V 02, • • • , 6n be a set of quantities such that when their
value is known, the configuration of a system of bodies is fully
determined. Then the quantities 0V 02, • • • , 6n are called general-
ized coordinates of the system.

270. Let Xy y, z be the, coordinates of any particle of the system.
Then x is fully determined by the values of 0V 02, • • • , 0n, so that
it is a function of these quantities, say

If the system is in motion, all the quantities which enter in
equation (158) are functions of the time. We have, on differ-
entiation with respect to the time,

dx^df^dO^ df^d^ • df_ dj^

dt ~ d6l dt + W2 dt " ddn dt '

330 GENERALIZED COORDINATES

To abbreviate, let us denote — > — - > • • • by x, 0,, - • •. Then the

at at

equation just obtained may be written

Q/3 1 ' y}/3 ^ ^^ £)/•) ""* \ /

so that x is a linear function of 0^ 02, • • • , 0B, the coefficients being
functions of 01? 02, • • -, 0n.
The kinetic energy,

is now seen to be a quadratic function of 0V 02) • • •, 0n) the coeffi-
cients being functions of 0lt 02, • • • , ^n.

The potential energy W depends only on the configuration of
the system, so that W is a function of 0lt 62, • • •, 6n only.

Thus the function L, or T — W, is a function of

L = lift, 62, - . ., 0n, 619 6V . - -, tfj. (160)

say

The corresponding function L' in the displaced motion is the
same function of

9, + W,, 02 + S02, ..-, etc.,
so that

By Taylor's theorem, we may expand L' in the form

or, from equation (160),

LAGBANGE'S EQUATIONS 331

Equation (156), namely

may be written in the form

and this, we now see, may be replaced by

?*^+?*'^*-a <162>

Now we have

sothat

and, integrated by parts, this becomes

Since the disturbed configuration, by hypothesis, coincides with
the actual configuration, we have 80l = 0 at times tr and tz. Thus
the first term in expression (163) vanishes, and leaves

Equation (162) now assumes the form

The limits tl and tz are entirely at our disposal ; the equation is
true whatever values we assign to them. In other words, the sum
of a number of small differentials vanishes, no matter how many

332 GENERALIZED COORDINATES

of them are included in the sum. It follows that each term of the
sum must vanish. Thus we must have

rS®]-

at every instant.

271. At this point we have to consider two alternatives. It may
be that whatever values are assigned to B0l} 802) - • • , B0n> the new
configuration, specified by coordinates

will be a possible configuration ; that is to say, will be one which
the system can assume without violating the constraints imposed
by the mechanism of the system. In this case the system is said
to have n degrees of freedom.

If the system has n degrees of freedom, equation (165) is true
for all values of $0l} S02, • • • , B0n. For instance, it is true if we take

£/3 £/3 _ Sk/D 5"/3 _ f\

°Vi = e, °V2 = °V9 = • • = t>Vn = U,
where e is any small quantity. In this case we must have

cL d I uL \ I

/ I I — n

oTT ~T, I T i I — v»

, .. , d /dL\ dL .

and therefore — ( -7- — -r = 0.

\
J

dt \d0

A similar equation will, of course, hold for each of the coordinates
#i> 02> ' ' • i @n' These equations are known as Lagrange's equa-
tions. There are n equations between the n unknown quantities
#i> 02> ' • ' > &n an(i their differential coefficients with respect to the
time. Thus they enable1 us to find the way in which 0lt 02, - • •, 0n
change with the time. To use the equations we require a knowl-
edge only of the function L, and therefore only of the kinetic and
potential energies of the system ; we do not need a knowledge of
the internal mechanism of the system. Thus the problem proposed
in § 263 is solved, if we can solve Lagrange's equations.

LAGKANGE'S EQUATIONS

333

ILLUSTRATIVE EXAMPLE

Common pendulum. As a simple example of the use of Lagrange's equations,
let us consider the problem of the motion of the common pendulum. A rigid
body is constrained to move so that one point 0 remains fixed, while the line
06? joining 0 to the center of gravity moves in a vertical plane. Let 0 be the
inclination of OG to the vertical; then the position of the system is entirely
fixed as soon as the value of 0 is known. The kinetic and potential energies are,
in the notation of § 245,

so that

T =
L =

W = Mgh (I - cos 0) ,
- Mgh (1 - cos 0).

Thus — = MWd. and Lagrange's equation,
50

^ /5IA _ dL

~ '

becomes

Mk2 — = - Mgh sine,
dt2

FIG. 153

the same equation as was obtained in § 246, and from this the
motion can be deduced.

We notice, however, that Lagrange's method shows that the motion is inde-
pendent of the method of suspension of the pendulum, provided only that it is
constrained to move in the way described. For instance, the result is true if
there is no pivot at all at 0, the constraints being imposed by a suspension of
strings.

272. Let us now consider the second alternative to that exam-
ined in § 271. It may be that if we assign arbitrary values to
86 j, 8#2, • • • , $0n, the new configuration obtained is not in every
case a possible one. It may be that there are certain relations
which must be satisfied, in order that the constraints imposed by
the mechanism may not be violated.

For instance? in the illustration already employed, let there be two
ropes hanging from a ceiling of a room, such that on pulling one down one
inch the mechanism compels the second to rise two inches. Let 0lt 02
denote the lengths of ropes below the ceiling. Then a displacement in
which 50J = -j^ inch, 502 = -£$ inch, is not a possible displacement ; such
a displacement is not permitted by the mechanism above. We must always
have 50l5 502 connected by the relation

501 + \ 502 = 0.

334 GENERALIZED COORDINATES

In general let us suppose that we have certain relations imposed
by the mechanism, these being of the form

a^ + azW,+ . . . + anWn= 0, (166)

b^ + b^ + . - - + bnS0n = ' 0, . . .. (167) - - •

Then equation (165), namely

.=°.

is true only if 80V S02, - • •, $0n satisfy relations (166), (167),

For a possible displacement, however, S0lt'80z, - • •, 80n will be
such that equations (166), (167) • • •, and (168) are all true. Let
us multiply by X, //., • • • and unity, and add, X, JJL, • • • being quan-
tities as yet undetermined — undetermined multipliers, we may call
them. Then we have the equation

The quantities W^ S02, • • . , Wn are not at our disposal. If, how-
ever, the relations of the type (166) are m in number, we may say
that of the quantities 80 lf S02, • . ., 80 n all except m are at our dis-
posal, and after arbitrary values have been assigned to n — m of
these quantities, the remaining m quantities must be obtained by
solving equations (166), (167) • • • . The configuration obtained in
this way must necessarily be a possible one.

Let us assign arbitrary values to

LAGRANGE'S EQUATIONS 335

and then find the values of &0l} S02, • • -, 80m from equations (166),
(167) • • •. Let us, moreover, choose the m undetermined multi-
pliers X, ft, • • • so that they satisfy the m equations

4 (^-}~ IF +*«i + /*&i+ • • • = 0, (170)

dt\xAj fa

the suffixes ranging from 1 to m. Then equation (169) reduces to

Inasmuch as S#TO+1, S0m+2, • • • , Wn are all arbitrary, we may take

and obtain

and similarly we may obtain the same equation for all suffixes from
m + 1 to n. The equation has, however, already been supposed
true for suffixes I to m [cf. equations (170) • • •, (171)].
Thus we have the complete system of equations

in which the suffixes range from 1 to n. On eliminating the
m multipliers X, //., • • • from these n equations, we are left with
n — m equations, which enable us to determine the changes in
the coordinates.

336

GENEKALIZED COOKDINATES

ILLUSTRATIVE EXAMPLES

1. A homogeneous sphere of radius a rolls down the outer surface of a fixed
sphere of radius b without sliding. Find the motion.

At any instant let the line of centers make an angle 0 with the vertical,
and let the angular velocity of the rolling sphere be 0. The velocity of the
center of the rolling sphere is (a + 6)-0, so that

The potential energy is

W — mg (a + b) cos 0,
so that L = T - W

FIG. 154

— mg (a + b) cos 0. (a)

The variations in 0 and 0 are not
capable of having any values we
please, for the velocity of the center
of the moving sphere is (a + b) 0,
and also must be a0', since the sphere
is rolling with angular velocity 0*
without sliding. Thus

ad = (a + b) 0. (6)

This is true at every instant of every possible motion, so that we must have,
on integrating with respect to the time,

a0=;(a + &)0 + a constant,

and hence we must suppose changes in the coordinates 0, 0 to be connected by
the relation a 50 = (a + 6) 50.

Thus Lagrange's equations are

Eliminating X, we obtain

Substituting from equation (a), this becomes

(a + 6)ri/?ma«Al + af£(m(a.+ &)20) - mg (a + b) sin 0 J =

ILLUSTEATIVE EXAMPLES

After replacing a6 by (a + b) <j> from equation (6), we have

- ma (a + 6)2 — = mga (a + b) sin 0,
5 dt2

(a + 6)-^ = -

337

showing that the center of the moving sphere moves with five sevenths of the
acceleration of a smooth particle sliding down a sphere of radius a -\- b.

The same result could have been obtained by eliminating 0 from equations
(a) and (6), and then regarding 0 as a single Lagrangian coordinate.

2, A flywheel is connected by a crank and rod to a piston moving in a hori-
zontal cylinder. When there is no steam in the engine the flywheel rests in its
position of equilibrium. Find its motion if displaced.

Let a, 6 denote the length of the crank and rod, and let 0, 0 be the angles
they make with the horizontal in any position of the flywheel. Then the posi-
tion of the engine is known fully when 6 and 0 are known. Not only do the

FIG. 155

values of 6 and 0 suffice to determine the position of the engine, but if we
assign arbitrary values to 6 and 0, we do not necessarily obtain a possible
position for the engine.

The velocity of rotation of flywheel, axle, and crank is 0, so that the kinetic
energy of this motion is \ Id2, where J is the moment of inertia of this part of
the engine about the axis of the flywheel. The coordinates of the center of
gravity of the rod, which we shall assume to be its middle point, measured
from the axis of the flywheel, are :

horizontal : a cos 0 + £ 6 cos 0,
vertical : ^ 6 sin 0.

Thus the velocity of its center of gravity has components
— (a sin 0 - 0 + i 6 sin 0 . 0)

338 GENEBALIZED COOBDINATES

horizontally, and % b cos 0 • 0 vertically. The whole velocity v of the center of
gravity of the rod is therefore given by

•U2 = (a sin 6 • 0 + 1 6 sin 0 • 0)2 + (| 6 cos 0 • 0)2
= a2 sin2 0 • 02 + ab sin 0 sin 0 • 00 + £ 6202.

The angular velocity of the rod is 0, and its radius of gyration k is given by

H6H-

Thus, if m is the mass of the rod, the kinetic energy of the rod is

= | m (a2 sin2 0 02 + a6 sin 6 sin 0 00' + | &202).

Lastly, the horizontal distance of the end of the piston rod from the center
of the flywheel is a cos 0 + 6 cos 0, so that the velocity of the piston and piston

rod is

— a sin 6-6 — 1) sin 0 • 0.

If J\f is the mass of the piston and piston rod, the kinetic energy of this part
of the engine is Wa sin 0 • 0 + 6 sin 0 • 0)2.

We now have, for the whole kinetic energy T,

2 T = Id* + m (a2 sin2 6 • 02 + ab sin 6 sin 0 • 00' + i 6202)

+ M (a sin 0 • 0 + 6 sin 0 • 0)2. (a)

The potential energy TF, measured from a standard configuration in which

(6)

Here 3fC is the total mass of flywheel and crank, and A, e are the polar coor-
dinates of its center of gravity when 0=0.

The changes in 0 and 0 are not independent. A glance at the figure shows
that we must always have

a sin 0 = 6 sin 0, (c)

and on differentiating this, we can see that if 0 and 0 are taken as generalized
coordinates, we must suppose them connected by

a cos 0 50 — 6 cos 0 50 = 0.
Thus Lagrange's equations will be

0, (d)

d /dL\ dL

—( — )-- -- X6cos0 = 0.

LAGRANGE'S EQUATIONS 339

The elimination of X from these equations gives

\~d/dL\ dL
— I — ) ---

\_dt\ce) de

dL~]

6cos0 — I — ) --- + acos0 — (— 1 --- =0,

a0J

and on substituting for L from equations (a) and (&), this equation becomes an
equation between 0, 0, and their differential coefficients with respect to the time.
From this and the geometrical relation (c),

a sin B = b sin 0, (/)

we can proceed to determine 0 and 0 in terms of the time.
Using equation (/), we can transform equation (6) into

W = — £%Cgh sin (0 + e) — \ inga sin 0.

It will be possible to arrange counterpoises on the flywheel in such a way as
tomake

and if this is done, the center of gravity will always be at the same height.
This is called balancing the engine.

If we suppose the engine balanced in this way, we have W = 0 and there-
fore L = T. We can, however, determine the motion much more simply than
by using Lagrange's equations, for we know that T must remain constant
throughout the motion ; and by differentiation of equation (/) we have

a cos 0 0 = b cos 0 0,
so that we can replace equation (a) by

2 T = I02 + m(a2 sin20 02 + a2 sin 0 cos 0 tan 0 02 + 1 a2 cos20 sec20 e2)

+ M (a sin 0 • 0 + a cos 0 tan0 • 0)2
= 02 [I + wia2 sin 0 sin (0 + 0) sec 0 + $ ma2 cos2 0 sec2 0
+ Ma2 sin2 (0 + 0) sec2 0].

This is constant throughout the motion, but we see that it does not follow
that 0 is constant. Thus, although the engine is balanced so as to remain at rest
in any position, it will not necessarily run evenly if started into motion.

Lagrange's Equations for Non- Conservative Systems

273. For non-conservative systems it has been shown (§ 268)
that equation (156), namely

f\L^
J

, (172)

must be replaced by

YSy+ZSzfldt = 0. (173)

340 GENERALIZED COORDINATES

Now since, as in equation (158),

we must have

Sx = x' — x

neglecting small quantities of the second order.
Thus

where ©^ ©2, •••,©„ depend on the configuration of the system,
and therefore are functions of 0lf 02, • • ., 6n only.
Equation (173) now becomes

Just as in § 270 we found that

•£••

SLdt
could be transformed into

so the first term of equation (174) can now be transformed into

» Substituting this, the equation becomes

LAGRANGE'S EQUATIONS 341

Since this is true for all possible ranges of time, we must have

at every instant.

If the 0's can vary independently, each coefficient must vanish,
and the system of equations will be

etc., while if B61) 502, • • • are connected by the constraints implied
in equations (166), (167), • • •, we find, as in § 272, that the sys-
tem of equations must be replaced by

(177)

274. These systems of equations reduce to those previously
obtained in the special case in which the forces are conservative.
For in this case consider the work done in a slight displacement in
which 0l alone varies, 0l being increased by 80r It is O^^, and

is also — — S0lt so that]

L oeingjncreasi

FES

rm, 3T _ dT dW dL

Thus _ + 0l = ____=__,

cu_iu.

since TTdoes not contain 0r Thus equation (175) reduces to

d_ fd_L\ _ dL_ _
dt \30J 00! ~

as before, and in the same way, equations (177), ••• can be trans-
formed into the equations obtained in § 272.

342 GENERALIZED COORDINATES

Lagrange's Equations ~by Direct Transformation

275. Instead of deducing Lagrange's equations from equation
(156), they may be obtained directly by transformation of the
equations of motion.

We have, as before,

so that, on differentiation,

dx^ df_ d6± df_d0s

dt ~ 0^ dt 862 dt '

*-'' + '•+•••• -•' <178>

Thus x is a linear function of 0V 02, • • •, and

— - — - '

^ ~ ^

We have T = £ m (x* + f + z2),

(179)

so that T, as before, is a quadratic function of 6^ 02, • • •, which
also involves 0lt 02) • - -. By differentiation,

dT /. dx

dx . dy . dz\
~r + y -7- + * -T- I

a^ ^x a^/

or, by equation (179),

Thus

. . d

+ + (180)

LAGRANGE'S EQUATIONS 343

Since —r- is a function of 0lt 02, • • • , we have

_ = _, _,
dt(ddj del dt 20^ dt

while, by differentiation of equation (178),
dx &x * d*x

The right-hand members of equations (181) and (182) are seen
to be identical, so that

and the last line of equation (180) transforms into

/ . dx . dy . dz
I ™ l_ ni iL _i_ 2

of which the value is

dT
or —

Equation (180) now becomes

CC/tX/C™ "^"^7 ~^7 | *•" ™ - ™-

Cc-6
|

From the equations of motion,

X=W^7"V^
so that this again becomes

d/d_T\_dT_^/vix^v<ly^l7te
dt\x

344 GENERALIZED COORDINATES

If we give the system a small displacement, in which 0t is
increased to 01 + &0l> 02 to 02 + 802, etc., we have, on equating
two different expressions for the work done :

On substituting the value of &x obtained on page 340, this

and the equation reduces to

This is the same equation as equation (175), and the different
forms of Lagrange's equations can be deduced as before.

Lagrange's Equations for Impulsive Forces

276. Let a system of impulses act during the short interval
from t = tt to t = t2. Let 0lt 02, • • • , 0n now be supposed to be
independent coordinates, so that Lagrange's equations are

If we multiply by dt, and integrate from t = tltot = tt) we have

, e0 ,,

The value of the first term is

LAGKANGE'S EQUATIONS 345

and when the interval from ^ to t2 is made vanishingly small, this

measures simply the change in — produced by the impulse.

B01

C** dT dT

In the second term / •— - dt, the integrand -^r- is finite, so that

when the interval of tune is supposed to vanish, this term will
vanish with it. Thus the equation becomes

dT Ctz

change in — = | 6X dt. (183)

d6l Jti

277. If F is an ordinary force acting impulsively through the

/>*2
interval ^ to £,, we call I Fdt the impulse. By analogy we call

Jti

the generalized impulse, corresponding to the generalized coordi-
nate #!. Thus we have equation (183) in the form

change in - — = generalized impulse.
00,

From analogy with the relation,
change in momentum of a particle = impulse on particle,

we call — T- the generalized momentum corresponding to the coor-

dinate Or Thus with these meanings attached to the terms
"impulse" and "momentum," the relation

change of momentum = impulse
is true in generalized coordinates.

When our coordinates are x, y, z, the coordinates in space of a moving
particle, the generalized momenta will of course become identical with the
ordinary components of momentum. We have

so that — = mx, etc.

346

GENERALIZED COORDINATES

EULER'S EQUATIONS FOR A RIGID BODY

278. Euler's equations (§ 252) can be derived from those of
Lagrange.

Let the moments of a rigid body about its principal axes of
inertia at a point 0, which is fixed in the body and is also
either fixed in space or is the center of gravity of the body, be
A, B, C. Then if o^, &)2, o>3 are the components of rotation about
these axes, we have, as in § 248,

T = %(Aa)\ +B(D\ + Ca)l). (184)

As Lagrangian coordinates, let us take 6, <f> the spherical polar
coordinates of the third axis OC of the body, and ty a third coor-
dinate which measures the angle between the first axis OA of the

rigid body and the plane
through OC and the axis
6 = 0, say the plane COz.
We have first to find
<ol9 G>2 and <»3 in terms of
6, <£ and ty, so as to ex-
press 2T as a function of
these coordinates. The
motion of the body is com-
pounded of the motion
relative to the plane COz,
together with the motion
of the plane COz relative
to fixed axes. The former
motion consists of a rotation ijr about OC, and this, resolved along
the axes OA, OB, OC, has components

0, 0, i^.

The motion of the plane COz is compounded of

(a) a rotation 6 about an axis at right angles to its plane ;

(b) a rotation (j> about the axis 0 = 0.

FIG. 156

EULER'S EQUATIONS FOR A RIGID BODY 347

Resolved along the axes OA, OB, OC, the first part has compo-
nents

6 sin i/r, 6 cos ty, 0,

while the second has components

— <j> sin 6 cos T/T, <j> sin 6 sin -v/r, <£ cos 0.

Compounding these motions, we obtain

tw1 = 6 sin i/r — <£ sin 0 cos i|r 1
G>2 = 6 cos A/T -f- $ sin d sin T/T L.
co3 = ijr -f <£ cos 0

Let the work done in a small displacement be

®80 + <&ty + ¥ty;
then Lagrange's equation for the coordinate M* is

on substituting from equations (185). Also

dT da*. a®2

— - — i — «

We have, by differentiation of equation (184),

j (0 cos T/T + <^> sin 6 sin i/r)

+ ^o>2 (— 6 sin ^r + 0 sin 0 cos

Finally ^Sijr is the work done by external forces in a small
rotation &»/r, and therefore, by § 121, "9 is equal to N, the sum of
the moments of these forces about the axis OC.

348 GENERALIZED COORDINATES

Making all these substitutions, equation (186) becomes

which is Euler's third equation, and the other two equations
follow from symmetry.

SMALL OSCILLATIONS

279. Let 019 02) - • • , 6n be generalized coordinates of any system,
and let it be supposed that these coordinates are all ^independent,
so that any set of values of 0lt 02, ••-,#„ gives a possible configura-
tion of the system.

Suppose that the configuration

f\ f\ f\ /i f\ __ /j / -| orrv

11> 22' > n n \/

is known to be a configuration of equilibrium. Then, if

the quantities <j>lt (f>2, - • • , <f>n may be taken to be generalized coordi-
nates of the system, and will possess the property of all vanishing
in the position of equilibrium.

Let WQ denote the value of the potential energy in the configu-
ration of equilibrium. The potential energy in any other configu-
ration may, by Taylor's theorem, be expanded in the form

where all the differential coefficients are evaluated in the position
of equilibrium. In this position of equilibrium, however, we have,
by the theorem of § 135,

dfF=dJF =dW

90, ~ M2= = dOn '' U'

SMALL OSCILLATIONS 349

so that we can write the value of W in the form

W=W0 + «u# + 2 altfafa + - • - + «„,,<#, (188)

in which rjbwers of fa, fa, • • • higher than the second are left out
of account, because we are going to confine our attention to motions
in which fa, fa, • • • are all small quantities.

The kinetic energy, as before (§ 270), is a quadratic function of
<£i>4> •••><£«• Letus'say

^ =6U# + 2 &ufc4 + .-. + &..&. (189)

The coefficients Jn, 6la, • • • , ~bnn are, strictly speaking, functions of
^i » 02> ' ' '» 0n> but we mav regard their values as being equal to
the values in the configuration of equilibrium, and so may treat
them as constants.

280. Now consider two quadratic functions of n variables xv
xv ' ' ') xn> defined by

f(xlf x2, -.., xn) = aux\ + 2
FX x • • - aj = bx* + 2

Since the function T defined by equation (189) is necessarily
positive, it follows that F(xlt x2, • • -, xn) is positive for all values
of xlt x2) - • -, xn. Hence, by a known theorem in algebra, we can
find a transformation of the type

\ /

in which the coefficients /cn, etc., are real, which is such that / and
F transform into expressions of the type

f(xlf x2, •

^(^, *2, •

and all the coefficients yS^ /S2, • • • , @n will be positive.

350 GENEBALIZED COORDINATES

Algebraic proofs of this theorem will be found in treatises on analysis,
or in Salmon's Higher Algebra, Lesson VI. The theorem will be readily
understood on considering a geometrical interpretation in the case in which
the number of variables is three. Calling the variables x, y, and z, the
equations •

/(*,y,*) = l, F(z,y,z) = l (191)

will be the equations of concentric quadrics ; and since F is positive for all
values of x, y, z, the second quadric will be an ellipsoid. It is known that
two concentric quadrics, of which one is an ellipsoid, always have one real
set of mutually conjugate diameters in common. A transformation of the
type expressed by equation (194) enables us to transform to these axes as
axes of coordinates, and the equations of the quadrics are then of the
required forms

«i{i + «s6+a8g = l, fcg + &g + /386 = l. (192)

[Simple reasoning will show the truth of the geometrical theorem that
an ellipsoid and a second quadric always have one common set of real
mutually conjugate diameters. For a real linear transformation will trans-
form the ellipsoid into a sphere, and the second quadric into a new, but
still real, quadric. The principal axes of this real quadric are now real
mutually conjugate diameters for the sphere and the quadric, and on trans-
forming back, real mutually conjugate diameters remain real mutually
conjugate diameters.]

The algebraic proof that equations (191) could be transformed into
equations (192) would, however, clearly not be limited to the case of three
variables, so that the theorem must be true for any number of variables.

281. This theorem proves that we can find new coordinates
^i> ^2* * ' * > ^n connected with ^, <£2, ••-,<£„ by relations of the
type

*i = M"i + "12^2 + ' • - + *!>„, (194)

such that, expressed in terms of these coordinates, the potential
and kinetic energies assume the forms

W- W0 +

T = P +& + ...+ /8.J. (196)

The coordinates tyv ^2, • • •, ^rn are called the principal coordi-
nates of the system, or, by some writers, the normal coordinates.

SMALL OSCILLATIONS 351

Lagrange's equations, in terms of these coordinates, are

d /dT\ dT dW

- 7TT- = — TTT- ' etc->

which become ft — = — a^lt etc. (197)

Stable Equilibrium

282. If ^ is positive, let us put — - = &J, so that ^ will be
real. The equation is now

of which the solution is

•^1 = -41cos(^ — €j, (198)

as in § 208. Thus the motion is a simple harmonic motion of
frequency kr If all the coefficients alt av • • •, an are positive, the
complete solution of the equations will be of the form

— €2), etc.,

and the coordinate x of any particle, of which the value in the
equilibrium position is a?0, will be

where ^, 52, • • • are new constants.

Thus the motion of any single particle will be a motion com-
pounded of a number of simple harmonic motions.

352 GENERALIZED COORDINATES

283. The potential energy corresponding to any principal coor-
dinate ^ is tfi^p or, if we suppose tyl given by equation (198), is

Similarly, the kinetic energy corresponding to this principal
vibration is &41}, or

Averaged over a very long time, the average values of cos2 (kj — ex)
and of sin2 (kj — ej are each ^, so that the average potential and
kinetic energies are respectively

and these are equal since kl = -±- - Thus in any vibration the aver-
age kinetic and potential energies are equal.

Unstable ^Equilibrium
284. Suppose now that any one of the coefficients in equation

(195) is negative, say ar Let us put — - =— %%, so that \ will be

PI
real. Equation (197) now assumes the form

and this has as solution

showing that -^ increases indefinitely with the time, and does not
oscillate about the value -^ = 0. Thus the motion is unstable, and
we now see that the motion can only be stable provided all the
coefficients ccv azt • • •, an are positive. In other words,

For stable equilibrium the potential energy in the configuration
of equilibrium must be an absolute minimum.

This is the result which has already been stated without proof
in § 153.

FORCED OSCILLATIONS 353

FORCED OSCILLATIONS

285. The oscillations which have so far been considered are of
the type known as free vibrations, — that is to say, the forces act-
ing arise entirely from the potential energy of the system itself.

A second type of oscillation occurs when the system is acted on
by forces from outside, in addition to those arising from its own
potential energy. These oscillations are known as forced oscillations.

Let us suppose that the potential and kinetic energies of the
system are given by equations (195) and (196), and that the sys-
tem of external forces acting at any instant is such that the work
done in a small displacement is

Then Lagrange's equations for this system are

1W_ — - _^ + ^
dt \9%J 0^ a^

which becomes 2 & — ^ = - 2 a^ + ¥„ (199)

U/t

in which Wlt it must be remembered, is now a function of the
time. This equation can be solved according to the rules given
in any treatise on differential equations. If, as before, we take

^ = -p the general solution is found to be

<'= t
^ = A, cos (\t - 6,) + ~= f W<= ,sin k, (t - t') dtr,

the lower limit of integration being either t1 = — oo, or the instant
of which the external forces first came into operation.

286. A case of extreme importance occurs when ^ is simply
periodic with respect to the time, say

354 GENERALIZED COORDINATES

The solution is then found to be

^ = Al cos (k^t - €l) + — - — — cos (Plt - 7l),
A #! — A plp1

or, since al = ftJQ

~n<

^ = A! cos (kj - €X) + - -, - -5T- cos (^ - 7l).

Thus the variation hi ^ is now compounded of a simple
harmonic motion of frequency klt and also one of frequency plt
the frequency of the impressed force.

We notice that if pt is very nearly equal to k19 then the second
vibration is of very large amplitude. In the limiting case in which
Pi = ^i> the amplitude of the second vibration becomes infinite,
but now the two vibrations are of the same period, so that they
may be compounded, and we cannot say that the resultant vibration
is one of infinite amplitude, because we do not know the values of 4i
and €1, and these may just be such as to destroy the infinite ampli-
tude of the second term. The result we have obtained may be
enunciated in the following form :

When a system is acted on ~by a periodic force, of frequency
very nearly equal to that of one of the principal vibrations of the
system, then the forced oscillations will be of very great amplitude.

This is known as the principle of resonance.

The principle is one of which many applications appear in nature. For
instance, a bridge, not being absolutely rigid, maybe regarded as a system
having a number of free vibrations. A body of men marching over the
bridge in regular step will apply a periodic force, and if the period of their
step happens to nearly coincide with one of the free periods of the bridge,
the amplitude of the vibrations forced in the bridge may be so large as to
endanger the bridge. For this reason troops are ordered to " break step "
when crossing a bridge.

Again, a ship is not perfectly rigid, and so will possess a number of free
vibrations. The motion of its engines will apply a periodic force of period
equal to that of its revolution, and if this coincides with that of one of the
vibrations of the ship, large pulsations will be set up. This can be reme-
died by altering the speed of the engine until it no longer is in resonance
with the free vibrations of the ship.

THE CANONICAL EQUATIONS 355

As a last example, it may be noticed that a ship will have a free period
of rolling about its vertical position. If it is in a rolling sea, the waves
which meet it will apply external forces which may be regarded as approxi-
mately periodic. If the period of the waves happens to coincide with that
of the ship, the ship will roll heavily even though the waves may be com-
paratively small. This danger can be remedied by altering the course of
the ship and so causing it to meet the waves at a different interval. Another
way is to give the ship a list by spreading canvas, and so causing it to
oscillate about a different position of equilibrium, about which the periods
of free vibrations are different.

THE CANONICAL EQUATIONS

287. If 0lt 02, • • • are Lagrangian coordinates of any system, the
kinetic energy T is a quadratic function of 6lt 62, 03, • • •. Let the
corresponding momenta be ult u2, • • • , un, these being given by

u^ = ^?> etc. (200)

30i

Now let us introduce a function T1, defined by
T' = U& 4- ujz + ---- T,

so that T1 is a function of ult u2, - • •, 01} 02) • • •, 6V 02, • • • ; and
ult uz, • • - are of course functions of 6lt 02, • • •, 0lt 0Z,

On differentiation of Tf we have

dT' = u + ud0 + • • •

and this, by equation (200), reduces to

dT' = e.dn, + 0idu2+...-Sddl--d02 ---- . (201)

356 GENERALIZED COORDINATES

Since the differentials ddl} d&2, - • . do not occur, it appears that T'
can be expressed as a function of uv u2, - • •, 6lt 02, . . . only. We
can easily find its value; we have

= 2 T, since T is a homogeneous quadratic func-

tion of 015 02, • • • + 0n.
Thus T' = 2 T — T = I7,

showing that T' is equal to T, but is expressed as a function of
**!, tta, ••-, 0J, 02, .... Thus

To illustrate, let

so that M! = 2 (a^i + h62) , w2 = 2 (7^i +

Then, by definition,

T = tii*! + tiai + ---- ^

+ 2 &, (A^ + 6

2
J,/ ^From equation (201), we have

In Lagrange's equations

_^ /^A _ ^ _ 0

^WJ ^~

ai d(T-W) dT
we have — = — i - '- = — =

THE CANONICAL EQUATIONS 357

Thus Lagrange's equations may be written as

du. cL 3(T — W) d (T1 + W)

-er

while, by equation (202), — - * = — — •

If we write H = T' + W, these equations assume the symmet-
rical form

d0, m

^L-± — _l±,
dt du,

du, m

-j7 = -W^c' I4W

288. This is known as the canonical form of the dynamical
equations. The function H is called the Hainiltonian function, and
since H = T' + W, we notice that H is the total energy expressed
as a function of the coordinates 0X, 02, • • •, 6n and of the momenta
ul9 u2, •-., un.

The canonical form is the simplest and most perfect form in
which the generalized dynamical equations can be expressed. For
this reason the canonical system of equations forms the starting
point of a great many investigations in higher dynamics, mathe-
matical physics, and mathematical astronomy.

289. We may appropriately terminate the present book by giv-
ing illustrations of the use of generalized coordinates from two
branches of mathematical physics.

Illustration from hydrodynamics. Let a solid of any shape be in a stream of
water flowing with uniform velocity V. If the solid is at a sufficient depth from
the surface, its presence will not disturb the flow at the surface, and the only
disturbance in the flow of the water will be in the neighborhood of the solid.
It can be proved, from elementary hydrodynamical principles, that there is
only one way in which the water can flow past the solid. Hence it follows
that the kinetic energy of the flow of the water is given by

358 GENERALIZED COORDINATES

where T0 is the value which the kinetic energy would have if the solid were
removed. Suppose that the solid is acted on by external forces, besides the pres-
sure of the water. Let the sum of the moments of these forces about any axis
be ®, and let 6 be a coordinate which measures the angle turned through about
this axis. Then Lagrange's equation corresponding to the coordinate 6 is

dt\s0

If the external forces just suffice to hold the body at rest in the liquid, we
have — ( — - } = 0, so that
™'

Hence the sum of the moments of the liquid pressure must be — ®, or

We can calculate a from the shape of the solid, and so can obtain a knowledge
of the couples acting on the solid.

Illustration from electromagnetism. The energy required to establish the flow
of two steady currents of electricity of strengths i, i' in two given closed circuits
is known to be of the form

where L and N depend on the shape of the first and second circuits respectively,
while M depends on the shape of both circuits, and also on their positions
relative to one another.

Suppose that the second circuit is free to move along any line towards the
first circuit. Let x be a coordinate measured along this line, and let the force
required to hold the second circuit at rest be JT in the direction in which x is
measured.

Let L denote the usual function T — W, and let the second circuit be acted
on by an externally applied force X. Then Lagrange's equation for the coor-
dinate x is

1/?A_£^ = x
dt\dx) dx

•0 that, since there is no acceleration,

X.-S5.

dx
As a matter of experiment, it is found that

EXAMPLES 359

If the energy of the two currents were potential energy, we should have

_<5L_ oW _<W__^

dx ~ dx' dx dz '

so that the force X would be exactly opposite to that observed.
On the other hand, if the energy is kinetic energy, we have

dL _ dT _ dE

dx dx dx

so that the value of X agrees with that observed.

Hence we conclude that the energy of an electric current is wholly kinetic.

GENERAL EXAMPLES

1. The friction of an engine is such that one horse power- can run it at
250 revolutions per second when it is doing no external work. The inertia
of its moving parts is such that when running at 125 revolutions per sec-
ond, and acted on by one horse power, its speed is accelerated at the rate
of 10 revolutions per second. If the engine is left to itself when running
at its full speed of 250 revolutions per second, find how many revolutions
it will make before coming to rest.

2. A square is moving freely about a diagonal with angular velocity o>,
when suddenly one of the angular points not in that diagonal becomes fixed.
Determine the impulsive pressure on the fixed point, and show that the
new angular velocity will be \ u.

3. Four equal rods, each of length 2 a and mass m, are freely jointed so
as to form a rhombus. The system falls from rest with one diagonal ver-
tical, and strikes a fixed horizontal inelastic plane. Find the impulse and
the subsequent motion.

4. Two particles connected by a rigid rod move on a smooth vertical
circle. Find the time of a small oscillation.

5. A uniform rod of length I has the two points at distance c from
its middle point connected by equal strings of length L to two fixed points
at distances 2 c apart in the same horizontal line.

Find the principal coordinates and the corresponding periods of
vibration.

6. If the rod of the last question receives a horizontal blow of impulse
I at one extremity and at right angles to its length, find the subsequent
motion.

7. A rough uniform cylinder of radius a has an inextensible string coiled
round its central section. One end of the string is fastened to a fixed point
P, and the cylinder is rolled up the string until it is touching P, with the
tangent to the cylinder at P vertical. The cylinder is then let go. Find the
motion.

360 GENERALIZED COORDINATES

8. In the last question, find the motion if the tangent at P is perpen-
dicular to the axis of the cylinder, but is not quite vertical.

9. In spherical polar coordinates prove that the kinetic energy of a
moving particle of unit mass is given by

T =

Hence, prove that the acceleration of the particle has components in the
direction of r, 6, <f> increasing, of amounts

1 [".I I?I\ _ #T~| 1 d_ /8T

r \_dt\w) dO J' rsin6dt\^

dt\dr dr r
Show that the actual values of these accelerations are

— - — rff2 — r sin20 02, -- (V20) — r sin 0 cos 6 d>2, -- (r2 sin20 0).
dt2 r dt r sin 0 dt

10. The velocity of a particle in its orbit is found to vary in the inverse
square of its distance from a fixed point. Apply the principle of least
action to find the orbit, and thence the law of attraction.

Deduce the same results from the law of conservation of energy.

11. Suppose that all forces are annihilated in the universe, and that
there is a concealed mechanism capable of possessing kinetic energy. Sup-
pose that the amount of this kinetic energy depends only on the positions
of the material bodies in the universe, being equal in magnitude except for
a constant, and opposite in sign, to the potential energy which the system
would have if the forces had not been annihilated.

Show that the dynamical phenomena of a universe of this kind will be
identical with those of a universe in which both forces and kinetic energy
exist, the changes in the latter being determined by Newton's laws of
motion.

12. A number of spheres without mass, of radii a, b, c, . . ., move in a
straight line through an infinite ocean of density p0 , the distances apart of
their centers being r^, r6c, etc., and their velocities va, vb, vc, . * .. When
a, b, c, . . . are small compared to r^, etc., the kinetic energy of the motion
of the ocean is given by

n^W

2T= f irpa«»» + • • - + 2irp^-vavb + - . ..

rab

Show that to an observer who is unconscious of the presence of the ocean,
the spheres will appear to move as though having masses f 7r/>a3, f 7r/o&3,
etc., and as though forces of attraction acted between every pair of spheres,
proportional to the product of the masses, to the product of their veloci-
ties, and to the inverse fourth power of the distance between them.

INDEX

(The numbers refer to pages.)

Absolute units, of force, 30 ; of work,
146.

Acceleration, 12 ; parallelogram of, 13 ;
in circular motion, 14, 18.

Action, 328 ; principle of least, 328.

Amplitude, of a pendulum, 261; of sim-
ple harmonic motion, 265.

Angle of friction, 47.

Angular momentum, 297; conservation
of, 297.

Angular velocity, 286 ; composition of,
287.

Arc, center of gravity of circular, 125.

Atwood's machine, 195.

Average velocity, 6.

Axes of inertia, 303.

Axis of rotation, 92.

Balancing, of an engine, 337.

Belt, center of gravity of spherical, 130.

Canonical equations, 355.

Catenary, 80.

Center of force, motion of point about,
269.

Center of gravity, 117 ; of a lamina,
121, 135 ; of a solid, 132, 135 ; of a
triangle, 121 ; of a pyramid, 132 ; of
a circular arc, 126 ; of a segment and
sector of a circle, 128, 129; of a
spherical belt and cap, 130 ; of a sec-
tor of a sphere, 130 ; motion of, of a
system, 224.

Central axis, of a system of forces, 107.

Centroid, 20.

Circular arc, center of gravity of, 125.

Coefficient, of friction, 47; of elas-
ticity, 240.

Coefficients of inertia, 301.

Composition, of motions, 4; of veloci
ties, 7 ; of accelerations, 13 ; of forces
acting on a particle, 37; of forces
acting in a plane, 95; of parallel
forces, 99 ; of couples, 105 ; of rota-
tions, 286.

Compression, moment of greatest, 238.

Conical pendulum, 271.

Conservation, of energy, 171 ; of linear
momentum, 223 ; of angular momen-
tum, 297.

Conservative system, of forces, 163.

Coordinates, generalized, 320, 329; nor-
mal or principal, 350.

Couples, 101; in parallel planes, 104;
composition of, 105 ; work performed
against, 154.

Cycloidal pendulum, 265.

Degrees of freedom, number of, 184,

332.

Descent, line of quickest, 193.
Diagram, indicator, 151.
Differential equations, of orbits, 275.
Double stars, 280.

Earth's rotation, 198, 310.
Elasticity, of a string, 45; modulus

of, 45 ; of a solid, 238 ; coefficient

of, 240.
Ellipsoid of inertia, 302.

361

362

INDEX

Energy, potential, 163; kinetic, 168,
228 ; total, 171; conservation of , 171;
of motion of a system, 230 ; of a rigid
body, 290.

Envelope of paths, of projectiles, 211.

Equation, of energy, 171, 256; of mo-
tion of a particle, 254 ; of orbit of
a particle, 275 ; of a rigid body, 304.

Equations, Euler's, 306, 346 ; La-
grange's, 329, 342 ; canonical (Ham-
ilton's), 355.

Equilibrium, of a particle, 38, 41 ; of
a system of particles, 63 ; of a rigid
body, 93; stability and instability
of, 174, 351.

Euler's equations, 306, 346.

Extensibility, of strings, 44.

Flexibility, of strings, 43.

Force, 26 ; measurement of, 30 ; trans-
missibility of, 94.

Forced oscillations, 353.

Forces, composition and resolution of,
37-39 ; in one plane, 66, 95 ; parallel,
96, 99 ; in space, 106 ; impulsive,
233.

Frame of reference, 3, 33 ; motion re-
ferred to moving, 197 ; kinetic energy
referred to moving, 228.

Frequency, of a vibration, 263.

Friction, 46 ; coefficient of, 47. Reac-
tion between moving rough bodies,
200.

Generalized coordinates, 320, 329 ; im-
pulse, 345; momentum, 345.

Gravitation, law of, 279.

Gravity, work performed against, 153 ;
motion of body falling under, 189 ;
variation with latitude, 200.

Gyration, radii of, 290.

Hamilton's Principle, 323.
Harmonic motion, simple, 261.
Hooke's law, 44.
Horse power, 146.

Impact,- 238 ; of particle on fixed sur-
face, 241 ; of any two moving bodies,
244 ; of two smooth spheres, 246.

Impulse, 233 ; of compression, 239 ; of
restitution, 240 ; generalized, 345.

Impulsive forces, 233, 345.

Inclined plane, motion of particle on,
192.

Indicator diagram, 151.

Inertia, moment of, 290; coefficients
and products of, 301, 302 ; ellipsoid
of, 302 ; principal axes of, 303.

Inverse square, law of, 276.

Kepler's laws, 279.

Kinetic energy, 168 ; of system of par-
ticles, 228; of rotation, 289; of a
rigid body, 290.

Lagrange's equations, 329 ; for impul- /
sive forces, 344; for non-conserva-
tive systems, 339.

Lamina, center of gravity of, 121, 135.

Latitude, variation of gravity with,
200; variation of terrestrial, 310.

Laws, of nature, 1 ; of motion, 26.

Least action, 327.

Line of action, of a force, 60.

Mass, and measurement of mass, 29.

Measurement, of velocity, 6 ; of accel-
eration, 12 ; of mass, 29; of force, 30 ;
of work, 145 ; of acceleration due to
gravity, 195; of an impulse, 235.

Modulus of elasticity, of a string, 45.

Moment, of a force, 60 ; of a velocity,
274 ; of inertia, 290 ; of momentum,
295.

Moment of greatest compression, 238.

Moments, principal, of inertia, 301.

Momentum, 29; conservation of linear,
223 ; moment of, 295 ; conservation
of angular, 297 ; generalized, 345.

Motion, referred to frame of reference,
3 ; of a rigid body, 91, 286 ; referred
to moving frame of reference, 197 ;

INDEX

363

of system of particles, 220 ; of cen-
ter of gravity of any system, 224;
simple harmonic, 261; of particle
about a center of force, 26ft ; of par-
ticle under law of inverse square,
276.

Neutral equilibrium, 182.
Newton's law, of elasticity, 245.
Newton's laws, of motion, 26.
Normal coordinates, 350.

Orbit, general theory, 273 ; differential
equation of, 275.

Orbit of a particle, law of direct dis-
tance, 269 ; law of inverse square of
distance, 276.

Oscillations, of a pendulum, 298;
small, of a general dynamical sys-
tem, 348 ; forced, 363.

Parallel forces, 96, 99.

Parallelogram law, velocities, 9; ac-
celerations, 13 ; forces, 38 ; couples,
105 ; angular velocity, 286.

Pendulum, simple, 259 ; seconds, 261 ;
cycloidal, 265 ; general motion of,
298.

Period, of vibration, 261; of simple
harmonic motion, 265.

Plane, composition of forces in one, 95 ;
orbit about a center of force confined
to one, 273.

Planet, rotation of a, 309.

Point of application, of a force, 60,
95.

Potential energy, 163.

Principal axes of inertia, 303.

Principal coordinates, 350.

Principle, of least action, 327 ; Ham-
ilton's, 323.

Products of inertia, 302.

Projectiles, 205 ; range on a horizontal
plane, 209 ; range on an inclined
plane, 209; envelope of paths with
given initial velocity, 211..

Pulleys, systems of, 157.
Pyramid, center of gravity of, 132.

Quickest descent, line of, 193.

Radius of gyration, 290.

Range, of a projectile, 209.

Reaction, 31 ; frictional, between
bodies at rest, 46; frictional, be-
tween bodies in motion, 200.

Reference, frame of, 3, 33; motion of
frame of, 197.

Relative motion, 4.

Resonance, principle of, 354.

Rest, 3.

Restitution, impulse of, 240.

Retardation, 12.

Rigidity, 90.

Rotation, axis of, 92 ; of earth, 198 ;
of a rigid body, kinetic energy of,
289 ; of a planet, 309.

Sag, of a string, 85.

Sector, of a circle, center of gravity of,
129; of a sphere, center of gravity
of, 134.

Segment, of a circle, center of gravity
of, 128.

Simple harmonic motion, 261.

Spherical cap, center of gravity of, 130.

Spinning top, motion of, 310.

Stability and instability of equilib-
rium, 174, 351.

Stars, double, orbits of, 280.

Strings, tension of, 42 ; flexibility of,
43 ; extensibility of, 44 ; on surface,
74 ; sag of a stretched, 85 ; work of
stretching, 150.

Suspension bridge, 78.

System, of pulleys, 157 ; conservative,
of forces, 163.

System of particles, statics of, 69 ; mo-
tion of, 220 ; kinetic energy of, 230.

Tension, of a string, 42, 74.
Top, motion of, 310.

364

INDEX

Transmissibility of force, 94.
Triangle, of velocities, 10.
Triangular lamina, center of gravity
of, 121.

Uniformity, of nature, 1.
Unit, of velocity, 6 ; of force, 30 ; of
work, 145.

Variation, of value of gr, 200 ; of ter-
restrial latitude, 310.

Vectors, 16 ; in one plane, 16 ; in space,
19.

Velocity, uniform and variable, 6 ;
average, 6 ; composition of, 7 ; mo-
ment of, 274 ; angular, 286.

Vibrations, 348, 353.

Virtual work, principle of, 155.

Weight, of a particle, 42 ; of a system
of particles, 118.

Wheel and axle, 65.

Work, measurement of, 145 ; against a
variable force, 148 ; of stretching a
string, 148 ; represented by an area,
150 ; against an oblique force, 152 ;
performed against gravity, 153 ; per-
formed by a couple, 154 ; principle
of virtual, 166; performed by an
impulse, 235.

Wrench, 107.

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