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http://www.archive.org/details/cu31924004606756
STATICS AND DYNAMICS
'h^'^1
X) (o
STATICS AND DYNAMICS
BY
C. GELDARD, M.A.
FORMERLY SCHOLAR OF TRINITY COLLEGE, CAMBRIDGE
MATHEMATICAL LECTURER UNDER THE NON-COLLEGIATE STUDENTS BOARD
CAMBRIDGE
LONDON
LONGMANS, GREEN, AND CO.
AND NEW YORK : 15 EAST le"- STREET
1893
K
A II eights reserved
PREFACE
The object of this Elementary Text-book is to
present both Statics and Kinetics in a combined form
suitable for beginners. I have given the proof of the
Parallelogram of Forces which is based on Newton's
Laws of Motion, since that is the form which appears
to me to be most intelligible to the student, and to
follow most naturally from the Definition of Force ; at
the same time, Duchayla's proof is so interesting and
ingenious, that it is entitled to a place in any book on
dynamics ; I have therefore added it in an Appendix.
Several chapters have been introduced dealing with
subjects often omitted in elementary text-books — e.g.
Couples, the General Conditions of Equilibrium, Virtual
Work, and Normal Acceleration. I have endeavoured,
when possible, to familiarise the student with more
advanced methods of work, by introducing the idea of
limits in the definitions of Variable Velocity and of
Variable Acceleration which I use in the chapter on
Normal Acceleration and the Hodograph.
A knowledge of the trigonometrical relations of one
angle, such as is provided in my ' Introduction to
Elementary Trigonometry,' is assumed ; articles requiring
vi Statics and Dynamics
a more advanced knowledge of trigonometry, and
chapters and articles which present difficulties to the
beginner, are usually marked with an asterisk. Articles
and paragraphs printed in smaller type are subsidiary
to the more important matter printed in the ordinary
type. Explanatory examples are worked out in each
chapter.
The book is designed to meet the requirements of
students preparing for the London University Matricu-
lation, the Further Examinations for admission into
Sandhurst and Woolwich, the first and second stages
of the Examination in Theoretical Mechanics of the
Science and Art Department, and the Oxford Local
Examination for Senior Students. In the case of the
following examinations, students may find it advisable
to omit the parts noted : —
Cambridge Local Examination (Seniors) : Chapters XV.
and XVI.
Cambridge Previous Examination (Additional Subjects) :
Articles 74 to "jg, 91 to 93, 97 to 102 ; Chapters IX.,
X. ; Articles 181 to 186 ; Chapters XV. to the end.
Cambridge General Examination : Articles 74 to 79, 91
to 9l< 97 to 102 ; Chapters IX. to the end.
My thanks are due (i) to Messrs. A. and C. Black,
the publishers of the 'Encyclopaedia Britannica,' for
permission to make quotations from the articles on
' Mechanics,' ' Applied Mechanics,' and ' Gunnery,' and
also to the authors of those articles ; but especially
to Professor Tait, to whose teaching I am indebted for
Preface vii
much of what is best in my own ; (2) to the Publishing
Conimittee of the Society for Promoting Christian
Knowledge, for permission to quote from the late
Professor Clerk Maxwell's ' Matter and Motion ' — a
book which every student of dynamics should read.
I am much indebted to Mr. PI. C. Knott, of Selwyn
College, for correcting the proof sheets, and for many
valuable suggestions ; also to Mr. J. M. Dodds, of
St. Peter's College, for much valuable criticism in the
Introductory Chapter.
C. GELDARD.
Cambridge:
February 22, 1893.
CONTENTS
[Chapters and Articles marked with an asterisk *
may be omitted by beginners.'\
CHAPTER I.
PAGE
Definitions, Units, La\vs o? Motion i
Examples on Chapter 1 15
CHAPTER H.
Parallelogram of Velocities 18
Parallelogram of Accelerations 19
Parallelogram of Forces 20
STATICS.
CHAPTER HI.
Magnitude of the Resultant of two Forces . . . 24
Triangle of Forces 27
Lami's Theorem 29
Polygon of Forces 30
Resolution of Forces 32
Resultant of any number of Forces meeting at a Point . 33
Conditions of Equilibrium 34
Examples on Chapter III. 37
CHAPTER IV.
Resultant of Parallel Forces 43
Centre of Parallel Forces . ' 4S
Resultant of any Coplanar Forces . .... 52
Examples on Chapter IV 53
X Statics and Dynamics
CHAPTER V.
PACE
Moments 5"
Theorems relating to Moments 60
Couples* -64
Theorems relating to Couples* 65
Conditions OF EQuiLiBRiUiM OF ANY CopLANAR Forces* . . 71
Examples on Chapter V 7'
CHAPTER VI.
Centre of Gravity 74
Centre of Gravity of a Parallelogram — Triangle — 78
Pyramid 81
Centre of Gravity of a System of Particles ■ . . 83
Examples on Chapter VI 89
CHAPTER VII.
General Conditions of Equilibrium 94
Theorem on the Equilibrium of any Coplanar Forces . 95
Nature of Pressures and Tensions 98
Examples on Chapter VII 107
CHAPTER VIII.
Machines no
Levers. Steelyards. Balance. Wheel and Axle. Pulleys. 1 1 1
Inclined Plane. Screw. Wedge 136
Examples on Chapter VIII 144
CHAPTER IX.
Friction. Laws of Friction 153
Rough Inclined Plane. Rough Screw. Rough Wedge 159
Examples on Chapter IX 168
CHAPTER X.*
Virtual Work 170
Principle of Virtual Work applied to various Machines 173
Theorem relating to Virtual Work 170
Contents xi
CHAPTER XI.
KINEMATICS.
PAGE
Change of Velocity 184
Relative Velocity 186
Acceleration 189
Formulae connecting Space, Time, Velocity, and Accele- 192
RATION 193
Theorems relating to Accelerated Motion . . . 197
Examples on Chapter XI 201
CHAPTER XII.
Motion due to Gravity . 206
Motion on an Inclined Plane 21c
Motion down the Chords of a Vertical Circle . .211
Lines of Quickest Descent 212
Examples on Chapter XII. . . ... 216
CHAPTER XIII.
Projectiles. Parabolic Path ... ... 220
Time of Flight. Range in vacuo 223
Range on an Inclined Plane 227
Elements of the Path . 230
Geometrical Treatment of certain Problems . . . 232
Equation to the Path 234
Empirical Formula for the Range in Air . . . 236
Examples on Chapter XIII. .... 239
KINETICS.
CHAPTER XIV.
Laws of Motion considered more fully 243
Second Law, '¥=m . u 248
Acceleration, of a System of two Masses connected by a 249
String 251
Pressure on a Plane moving with uniform acceleration 253
Atwood's Machine . . 254
Examples on Chapter XIV. , . ... 260
xii Statics and Dynamics
CHAPTER XV.*
PAGE
Normal Acceleration ... . ■ • ^°4-
HoDOGRAPH . . • • 265
Uniform Circular Motion .... • 266
Conical Pendulum . . ... 268
Change of Velocity on a Curve . ... 269
Motion in a Vertical Circle ... ... 270
Examples on Chapter XV. 272
CHAPTER XVI.*
Impulse. Impact 274
Nature of an Impact . . , 275
Impact of a Sphere on a Plane . . . . 277
Direct Impact of two Spheres 280
Oblique Impact of two Spheres 281
Impact of a Sphere on a Rough Plane .... 284
Motion of the Centre of Mass of a System . . . . 284
Examples on Chapter XVI 286
CHAPTER XVII.
Work. Energy 289
Examples of Work done . 290
Horse-power
291
Work done = Change of Kinetic Energy .... 292
Kinetic Energy lost by Impact* 294
Conservation of Energy 296
Examples on Chapter XVII. 297
APPENDIX.
Duchayla's Proof of the Parallelogram of Forces . . 299
Answers to Examples 303
DYNAM ICS
CHAPTER I
DEFINITIONS, UNITS, LAWS OF MOTION
r. Dynamics is the Brancli of Natural Philosophy which
treats of the action of Force upon Matter, and of the laws
by which Motion is governed.
It may be divided into three parts— Statics, Kinematics,
Kinetics.
Statics relates to forces which so act upon bodies as to
keep them at rest.
Kinematics is the theory of motion considered without
reference to the mutual action between bodies ; and when
this mutual action is taken into account it is called Kinetics.
2. In the above definitions the word body is used to denote
any portion of matter of finite size. The conception of
matter must be familiar to every intelligent person ; it refers to
that which is concrete as opposed to that which is abstract.
A portion of matter indefinitely small in all its dimensions
is called a Particle.
A particle is then a material point of such small linear
dimensions that it presents no difference in aspect however it
be turned about itself ; and is such, that when it is acted on
by a force, the effect will be that of translation only ; that is to
say, it will begin to move in a straight line, and its rotation
about itself may be neglected.
Bodies of finite size may be supposed to consist of an
infinite number of particles.
V
2 Dynamics
3. Force is any action between two bodies which changes
or tends to change the physical relation between them as
regards rest or uniform motion in a straight line.
4. Our first impression of force is the result of a muscular
effort in the act of pushing or pulling a body ; but when any
other agent produces the same or a similar effect on a body,
it is said to exert force. This effect is always reciprocal-
equal and opposite ; the force which A exerts on B is equal
and opposite to that which B exerts on A ; e.g. a book is rest-
ing on a table ; the pressure of the table on the book is equal
and opposite to the pressure of the book on the table. Again,
a horse pulls a cart ; the pull of the horse on the cart is equal
and opposite to the pull of the cart on the horse ; the former
makes the cart move, the latter is counterbalanced by the
muscular exertion of the horse. Thus force is always the
action between two bodies ; it is, however, often convenient to
consider only its effect upon one of them, without considering
the agency by which it is produced ; it may then be considered
as any cause which changes or tends to change a body's
existing state of rest or uniform motion in a straight line.
5. The effect of a force in changing a body's shape is called
a strain. The consideration of this effect of force is of too
difficult a nature to be attempted in an elementary work.
The effect of a force in changing a body's motion may be
of two kinds— a translation and a rotation. When a force
acts upon a particle, the effect is that of translation only (see
definition of Particle in Art. 2) ; but when it acts upon an
extended body, in addition to the motion of translation, there
is also usually a rotation about an axis within the body.
6. There are various ways in which Force acts, for which we
have different names. Push, or Pressure, is the name given to
the force exerted at the point or points of contact when one
body presses against another. Pull, or Tension, is the name
given to the force exerted by one body on another with which
it is connected by means of a string or rod. The nature of a
Force, however, is always the same, whatever be the name which
is given to it.
7. There is a third kind of force— which is apparently
Definitions, Units, Lcxws of Motion 3
neither pressure nor tension — which is exerted by one body on
another in no way connected with it, and which we call Attrac-
tion. The most familiar example of this force is the attraction
which the earth exerts upon all matter, called the Force of
Gravity. The measure of the force of gravity on any body is
called the body's Weight.
8. There are three elements of a force which must be known
before the effect of the force on a body can be determined ;
they are as follows : — •
(i) The direction of the force ; that is, the direction in
which a particle would begin to move under the action of the
given force only.
(2) A point in the Hne of action of the force, usually called
ihs. point of application of the force.
(3) The magnitude of the force.
9. It is immaterial what point in the line of action of the
force is given ; the effect of the force on the body will be the
same if we suppose it to act at one point in its line of action as
at another, provided that they are rigidly connected with the
body. This is called the Principle of the Transmissibility of
force.
The following illustration will assist in explaining it : Suppose
a weight attached to the hand by means of a thread, whose weight
may be neglected. The muscular effort required to support the
weight is always the same, at whatever point of the thread we
suppose the weight to be attached.
It will be found that the motion of translation of a. body
will not be changed by changing the line of action of the force,
provided the direction of the force remain the same ; but when
we consider the rotation of the body, it is necessary that we
should know the line of action as well as the direction of
the force.
Geometrical Representation of Forces.
10. Forces may be completely represented by straight
lines. For a straight line is completely ascertained when we
know :
(i) The point from which it is drawn.
B 2
4 Dynamics
(2) The direction in which it is drawn.
(3) Its length.
Thus a straight line may be drawn
(i) From any point which may represent a point m the line
of action of the force.
(2) In the direction of the force.
(3) Of such a length as to contain as many units of length
as the force contains units of force.
It will then represent the force completely.
11. In Art. I we used the terms 'rest' and 'motion.'
These terms are relative : the state of absolute rest is un-
known. A body is said to be at rest when it is relatively at
rest ; that is, when it continually occupies the same position
with reference to other objects ; and it is said to be in motion
when its position is changing ; that is, when it occupies dif-
ferent positions at different times among the objects to which it
is referred.
Consider, for example, the state of a stone apparently at
rest on the surface of the earth. Its actual motion in space is
very complicated ; for the earth is rotating on its own axis
once a day ; it is moving in its orbit round the sun once a
year ; and, lasdy, the whole solar system is moving towards a
certain point of the heavens.
Hence the actual motion in space of any point on the
earth's surface is a combination of at least three motions,
although it is at rest with reference to the earth.
Again, consider the case of a stone resting on a table in a
railway carriage which is moving.
The stone is at rest with reference to the table and the
carriage, but the stone and the table and the carriage are all
moving in the same manner with reference to the earth. It is
possible, again, for the stone to be moving on the table and the
table in the carriage in such a manner that the position of the
stone in the carriage remains the same ; that is, it will still be
at rest with reference to the carriage.
12. Before proceeding further with the definitions of the
physical quantities which we employ in the Science of
Dynamics, it will be well to consider briefly the method of
Definitions, Units, Laws of Motion g
measuring quantities and the choice of a unit of measure-
ment.
In order to measure any quantity we choose some unit of
the sa7!ie kind as the quantity under consideration, and then
compare the given quantity with the unit chosen ; the measure
of the quantity is the number of times it contains the unit
chosen ; the greater the unit is, the smaller will be the
measure. Thus, if m, ni be the measures of the same quantity
in terms of the units U, U' respectively,
m\5 ^m' U'
hence the measure of a quantity varies inversely as the unit
chosen.
e.g. 60 pence = 5 (12 pence) = 5 shilHngs
= -^ (20 shillings) = \ of j£i
and 6 feet = - (3 feet) = 2 yards
3
or = 6 (12 inches) = 72 inches
thusA = ^ = I.in_£h_
72 12 I foot
The measurement of all dynamical quantities depends upon
the three primary or fundamental conceptions of space, time,
and matter, for the measurement of which we have three
fundamental units which are perfectly arbitrary, and which
vary in different countries.
The unit of length — or linear space — employed in England
is one foot, which is equal to one-third of the length of a piece
of platinum which, as well as various duplicates, is carefully
preserved in London, and is called the Standard Yard.
The unit of area is the area of a square surface whose sides
are each one foot long, and is called a ' square foot.'
The unit of volume is the space contained in a cubical box
whose edges are each one foot long, and is called a ' cubic foot.
In France the unit of length is one centimetre, which is
the one-hundredth part of a metre— a metre being an arbitrary
g Dynamics
length of platinum which is kept in Paris, chosen originally as
the length of the seconds pendulum.
One metre = 39-37 inches approximately,
= 1-09 yard,
One centimetre = -3937 inches.
The unit of time is the same in both countries : it is one
second.
One second = 7^ (a minute) = ^~Er (^"^ ^o""")
60 60 X 00
= ^ (a day),
60 X 60 X 24
and a day is the length of time it takes the earth to make a
complete revolution on its axis. Hence a second is the time
during which the earth revolves through aa angle of 15".
The English unit of matter is the quantity of matter con-
tained in a lump of platinum which is kept in London, and
which is called a 'pound.'
The French unit of matter is a different quantity, and is
called a ' gramme.'
All dynamical quantities will be measured in terms of these
three fundamental units : in England, in terms of a foot, a
second, and a pound ; in France, in terms of a centimetre, a
second, and a gramme.
The latter system of measurement is called the C.G.S.
system, and is largely employed in scientific investigations in
England as well as in France.
13. When a body is in motion it is said to have a certain
velocity, which measures the rate at which it is moving in a
straight line.
If the motion is not in a straight line the rate of motion is
sometimes called speed.
- This velocity may be constant or variable.
A body is said to move with constant velocity when it
passes over equal distances in equal intervals of time ; the
velocity is then measured by the number of units of length
which the body passes over in one unit of time.
Definitions, Units, Laws of Motion 7
Thus, if a body passes over j units of length in / units of
time, it will pass- over ^ units of length in one unit of time ; if
we call this velocity v,
s
t
The unit in which v is measured is called the 'unit of
velocity,' and is the velocity of a body which passes over one
unit of length in one unit of time ;
thus » = - (units of length) per unit of time.
The measure - is always the same, however great ^ and /
may be, provided that the velocity remains constant.
The unit of velocity in English units is the velocity of a
body which passes over one foot in one second ; thus, if a body
pass over s feet in / seconds,
its velocity = - feet per §econd.
The unit of velocity in French units is the velocity of one
centimetre per second.
14. A body is said to move with variable velocity when it
passes over unequal distances in equal intervals of time.
The velocity of the body may be different at every point of
its course, but its average velocity during any time will be
equal to the velocity of a body which moves over the same
distance in the same time with constant velocity.
15. We have noticed that when a body is moving with
constant velocity the measure of that velocity \v ■= -\ is the
same, however large t may be ; and this formula is always true
of the average velocity of a body during any time. Now,
when a body is moving with variable velocity, the smaller we
.takeV the more nearly does the average velocity during that
time approximate to the actual velocity at any instant during
■ 8 Dynamics
that time ; and when t is indefinitely small, the actual velocity
and the average velocity are equal.
Hence the actual velocity of a body, whether it is moving
uniformly or not, may be measured at any instant by the
indefinitely small space described in a small portion of tmie
including the instant, divided by the indefinitely small time
required to describe it.
1 6. When a body is moving with variable velocity in a
straight line, the rate at whicli the velocity changes is called
Acceleration. Acceleration may, like velocity, be constant or
variable.
A body is said to be moving with constant acceleration
when it receives equal increments 'of velocity in equal
intervals of time ; its acceleration is said to be variable when
it receives unequal increments of velocity in equal intervals
of time.
When acceleration is opposite to the direction of motion it is
sometimes called a ' retardation.'
17. When a body is moving with constant acceleration,
this acceleration is measured by the number of units of
velocity gained in a unit of time.
Hence, if a body gain v units of velocity in t units of time,
it will gain - units of velocity in one unit of time ;
if we call this acceleration a,
a = - units of velocity per unit of time.
The unit of acceleration is therefore an increment in
velocity of one unit of velocity in one unit of time.
In English units the unit of acceleration is an increment in
velocity of one unit of velocity per second ; that is, of one foot
per second, per second.
Thus the above equation would be read :
a = - feet per second, per second.
Similarly the unit of acceleration in French units is an
acceleration of one centimetre per second, per second.
Definitions, Units, Laws of Motion 9
When the acceleration of a body is variable, it is measured
at any instant by the indefinitely small change in velocity in a
small portion of time including the instant, divided by the
indefinitely small time during which the change takes place.
18. Hitherto we have confined our attention to explaining
the meaning of certain terms used in the Definitions of the
Subject. We shall now consider the relation between force
and matter more fully, dealing only with those very small
portions of matter which we have called ' particles.'
It must be borne in mind that these particles contain
different quantities of matter, but their size and shape need
not be considered.
19. We are already prepared by our definition of Force for
Newton's First Law of Motion. This may be stated as
follows : —
'Every body remains in its state of rest or of uniform
motion in a straight line, unless it is compelled by external
forces to change that state.'
This property of Matter, in virtue of which it requires force
to change its state of rest or uniform motion, is called its
' Inertia,' and the first law of motion is sometimes called the
Law of Inertia.
20. This property of Inertia is familiar to all of us — eg. :
We are sitting in a railway carriage at rest, with our faces
towards the engine ; the carriage receives an impulse from the
engine forwards, we feel impelled against the back of the seat
behind us. We were at rest and tended to remain at rest after
the carriage has begun to move.
Again, let us watch a pendulum, consisting of a bullet tied
to a thread and suspended from the roof of a railway carriage
going at full speed, the pendulum being at rest with reference
to the carriage ; directly the brakes are put on, and the train
commences to slacken speed, the pendulum begins to oscillate
in the direction of the train's motion, and then back.
Again, suppose we roll a ball along a road : the ball is
quickly brought to rest on account of its friction with the
ground and the resistance of the air. If we now take a very
smooth round ivory ball and roll it on very smooth ice, it will
lo Dynamics
continue to move for a very much longer time than on the
road ; and we argue that if the friction between the ball and
the surface, and the resistance of the air, could be got rid of
altogether, the ball would go on moving for ever with uniform
velocity in a straight line.
21. Let us now consider another example. Suppose we
jump out of a carriage which is in motion : we remain in this
state of motion until our feet touch the ground. We then feel
a sudden impulse in the direction of motion, because our feet
are brought to rest by contact with the ground, while our body
still retains its original state of motion.
A heavy man will feel this impulse much more severely
than a light man ; the latter will often be able to keep on his
feet when the former falls prostrate on the ground.
We see in this case that the inertia of a body depends
upon and is a measure of the quantity of matter it contains, or,
as it is usually called, its ' mass.'
As a further illustration of this, consider the following
case : — A croquet ball and a cannon ball of the same size are
at rest on the ground, and we give a kick of equal strength to
each of them : the croquet ball will begin to move with a much
greater velocity than the cannon ball, although the impulse given
to each is the same. Again, in catching or stopping a cricket
ball, we exert a much greater force than in stopping a woollen
ball of the same size which reaches us with the same velocity.
2 2. From these examples we see that the quantity of
motion in a moving body depends upon two things : (i) the
velocity, (2) the mass.
23. The mass of a body depends upon two things— its
volume or cubical content, and the material of which it is com-
posed. The quality of the material is measured by a number
which we call its ' density.'
If 7n be the mass of a body, v its volume, and p its density,
w := w . p.
If V = I, ;« = p ; thus the density of a substance is the
mass of the unit volume of the substance.
24. As already stated in Art. 12, the British Standard unit
Definitions, Units, Laws of Motion \ i
of mass is one pound avoirdupois, and is a certain mass of
platinum kept in the Standards Office of the Board of Trade.
N.B. — The mass one pound must be carefully distinguished
from the weight of one pound ; the former is merely a lump
of matter, the latter measures the force with which such a
lump of matter is attracted by the earth.
The French unit of mass is one gramme ; this is the one-
thousandth part of a mass of platinum kept in the French
archives. The gramme was originally chosen as being the
mass of a cubic centimetre of pure water at 4°C.
One pound is equivalent to 453"S93 grammes.
25. Bodies are said to be of equal mass when equal
impulses generate in them equal velocities ; or when equal
forces acting upon them for the same time generate in them
equal velocities.
For example : place three balls made of wood, stone, and
lead on a smooth table and let the mass of each be i lb. ; now
apply simultaneously to each an equal impulse by means of a
spring ; it will be found that they reach the other side of the
table at the same time ; that is, they move with equal velocities.
Again, place three balls on the table as before, and let their
masses be i lb., 2 lbs., 3 lbs. ; now apply an equal impulse to
each ; it will be found that the times they take to get to the
other side of the table will be in the ratio of i : 2 : 3, which
means that their velocities are in the ratio of 3 : 2 : i. This
will not be exactly true, on account of the friction of the table.
Hence we say that quantity of motion is measured by the
product mass x velocity.
This is usually called Momentum.
Hence, if M stand for the momentum of a body of mass
m moving with a velocity v,
M = m . V.
The unit of momentum is the momentum of unit mass
moving with the unit velocity. In English units it is the
momentum of a mass of one pound moving with a velocity of
one foot per second.
26. We find, then, that when the same force acts for the
1 2 Dynamics
same length of time upon particles of different mass, different
velocities are generated, but in every case the momentum
generated is the same.
Hence, force may be measured by the momentum
generated in a unit of time.
Thus, if V, v' be the velocities in the same straight Ime of
a particle of mass m at the beginning and end of a time /,
mv = momentum at the beginning of the trnie
nn/ = „ „ end „ »
mi/ - mv = change of momentum in / seconds
nwl^-jm ^ ^^ ^^ in J second.
If we call the moving force in this direction F,
1^ mv' — mv
^= -t
m{v' — v)
t
= m .a. (Art. 17),
where a is the acceleration produced by the force.
27. The equation F = ma, states in an algebraical form
Newton's Second Law of Motion, which is as follows : —
Change of momentum is proportional to the impressed
force, and takes place in the direction of the straight line in
which the force acts ; provided that the unit of force is taken
to be that force which produces the unit of acceleration in the
unit of mass.
28. The British Absolute unit of Force is that force
which, when acting on a mass of one pound, communicates to
it an acceleration of one foot per second, per second.
This unit of Force is called a Poundal.
29. We shall see later that when a body is allowed to fall
freely in vacuo it moves with a constant acceleration which we
will denote by g. This acceleration is due to the lorce of
gravity ; that is, it is the acceleration caused by the constant
force of the body's weight acting on its mass.
Definitions, Units, Laws of Motion 13
The value of ^varies at different points on the earth's surface.
Its value in England in English units is 32-2 approximately.
Hence the weight of one pound acting on a mass of one
pound produces an acceleration of 32-2 feet per second, per
second.
But the unit force is that which acting on one pound pro-
duces an acceleration of one foot per second, per second,
therefore the unit force = x (weight of one pound)
32-2
= the weight of half an ounce approximately.
Or, calling the acceleration due to gravity g, we may say that
g . Poundals ^ weight of one pound.
In Statics it is customary to consider the weight of one
pound as the unit of Force. This unit of force varies at
different points on the earth's surface, whereas the Poundal is
always the same at every point. Hence the Poundal is called
the Absolute unit of Force.
30. In the French system of measurement, where the units
of length, mass, and time are one centimetre, one gramme,
and one second respectively, the unit of Force is called a Dyne.
Hence a Dyne is the force which, acting on a mass of one
gramme, produces an acceleration of one centimetre per
second, per second.
Now 32 '2 feet = 980 centimetres approximately,
so that 980 is the acceleration due to the action of the weight
of one gramme on the mass of one gramme.
Hence 980 Dynes = weight of one gramme.
If we compare the units of Force in the two systems we
shall find that
one Poundal = 13825 Dynes.
For I lb. = 4i5^"'5Q'5 grammes, and i foot = cms.
approximately.
/. one poundal = 15J:5"-3 dynes.
°'3937
31.* The equation F = ot . a is true of all finite forces
acting for an appreciable time.
14 Dynamics
It is also true for blows or impulses which produce a
finite change of motion in an indefinitely short time.
Thus, if F be the force acting on mass m, u the acceleration
generated, t the small time during which F acts, and v and v
the velocities before and after the blow, we have
v' — v = a- . T and Y — m . a.
.; F T = m . u. . T
=. m {2/ — v).
Thus, the product F . t measures the change in momentum
caused by the blow, that is, the impulse, just as F . / measures
the effect of the force F during a finite time A
It must, of course, be borne in mind that t is often so small
as to be inappreciable, but F may be so large that the product
F . T is finite and is measured by the change of momentum
which is produced by the blow.
32. Newton's Third Law of Motion is as follovre : —
' To every action there is always an equal and opposite
reaction.'
The truth of this law was illustrated when we considered
the meaning of Force (Art. 4). We then saw that there could
be no force without a reaction.
Take, for example, the recoil of a gun when it is fired off.
Momentum generated 1 j Momentum of the
in the bullet J 1 gun's recoil.
Mass of projectile x vel. of projection = mass of gun
X vel. of recoil.
This law embodies the Principle of the Conservation of
Energy.
33. For the sake of impressing the three Laws of Motion
on the memory of the student we will again state them.
Law I. Every body remains in its state of rest or of
uniform motion in a straight line, unless it is compelled by
external forces to change that state.
Law II. Change of momentum is proportional to the
impressed force, and takes place in the direction of the
straight line in which the force acts.
Definitions, Units, Laws of Motion 15
Law III. To every action there is always an equal and
opposite reaction.
34. Of these three laws, the second is at present by far the
most important.
It involves the following principles : — •
If there is no force, there is no change of momentum .
The effect of a force is exactly the same whether the body
-be at rest or in motion— it is always measured by the momen-
tum generated, and this change of momentum is in the
direction of the force.
If two or more forces act upon a body at the same time,
the change of motion due to each is independent of the others,
and may be considered without reference to them.
Ex, I. — Find the force which acting on a mass of 4 lbs. gene-
rates a velocity of 20 feet per second in 10 seconds.
Here the acceleration = — = 2
10
.". F = ?« a = 4 X 2 = 8 poundals.
Ex. 2. — Find the velocity generated by a force of 7 poundals
in a mass of i stone after it has acted for I minute.
Y = in a
7 = 14.0
I.
. , a = —
Z
Velocity generated in one minute
= u . ^ = — X 60
2
<= 30 feet per second.
Examples.
1. Express a velocity of 60 miles an hour in feet per second.
2. A man walks 30 miles in 8 hours. What is his velocity in
feet per second ?
3. Assuming an inch is 2 '5 centimetres, express a velocity of 4
miles an hour in terms of centimetres and seconds.
4. Express in centimetres per second a velocity of 4 kilometres
per hour.
1 6 Dynamics
5. A man travels a feet in b seconds. Find the distance he will
travel in a minute, and the time he will take to travel 100 yards.
6. What is the velocity of a point on the earth's equator, the
radius of the earth being 4,000 miles ?
7. A man 6 feet high walks at the rate of 4 miles an hour away
from a lamp 10 feet above the ground. Find the velocity of the
end of his shadow.
8. A body starts from rest and is subject to a uniform accelera-
tion. It acquires a velocity of 20 feet per second at the end of the
fourth second. What is the acceleration ?
9. A body moves over i, 3, 5, 7 feet during the 1st, 2nd, 3rd,
and 4th seconds respectively. Find the average velocity.
10. A body moving with uniform acceleration has a velocity la
A minute later its velocity is 40. What is the acceleration ? Have
the units employed any effect on the answer ?
11. Compare the velocities 66 feet per second and 40 miles per
hour.
12. What is the momentum of a body whose mass is 5 lbs.
moving with a velocity of 2 feet per second ? What force is re-
quired to produce this momentum in one second ? What force
is required to produce this momentum in the hundredth part of a
second ? How many pounds weight is this latter force ?
13. Compare the momentum of a mass of 100 lbs., moving with
a velocity of 1,000 feet a second, with that of a mass of I lb.
moving with a velocity of 30 feet a minute.
14. What force is required to reverse the motion of a mass of
I lb. movipg with a speed of 100 feet per second, supposing the
whole change to take place in one-tenth of a second ?
15. A foice of 20 pbundals acts on a mass of 4 lbs. What is
the acceleration ? Determine the velocity at the end of one
minute if it starts from rest.
16. A force of i poundal acts on a mass of i ounce. Determine
the velocity acquired in 5 seconds.
17. A force acts on a mass of i cwt. and communicates a
velocity of 20 feet per second in 4 seconds. Find the force.
18. A force of 56,000 poundals acts on a truck on smooth rails
and communicates to it a. velocity of 300 feet per second in I
minute. Find the mass of the truck in tons.
17
CHAPTER II
PARALLELOGRAM OF VELOCITIES AND FORCES
35. We shall now consider how to represent the velocity of
a point and the composition of two or more velocities.
The velocity of a point is known when its direction and
magnitude are known. We can then completely represent the
velocity of a point O by a straight line O A.
For let V be the velocity of a point O moving in a direction
making an angle 6 with
a fixed straight line ^
OX. Now if A be
drawn, making the
same angle 9 with O X,
and containing as many
units of length as the
point has units of
velocity, O A com-
pletely represents the velocity of the point O.
The unit of length in which O A is measured may be any
unit that is convenient, provided always we keep our unit
unchanged during the consideration of any mechanical
theorem. If the velocity of a point in one direction be called
V, an equal velocity in the opposite direction will be — v.
Thus, if O A represent the velocity v, O A' equal and
opposite to O A will represent a velocity — v.
36. A body may have at the same instant velocities in two
or more different directions quite independent of each other.
For example, if a person walks from one side of a railway
carriage in motion to the other, his apparent motion in the
carriage is the same as if it were at rest, but his actual motion
referred to the surface of the earth will be compounded of the
two motions — that of the carriage and his own motion in
c
Fig. I.
1 8 Dytiamics
the carriage ; and if these two motions are uniform, his actual
motion must also be uniform.
Bef. — The velocity which is equivalent to two or more
velocities is called their Resultant, and with reference to this
resultant each of the single velocities is called a Component.
We shall now show how the resultant of two velocities may
be found.
Parallelogram of Velocities.
37. If the two simultaneous velocities of a moving point
be represented in magnitude and directioa by two straight
lines drawn from a point, and if the parallelogram having
these two lines for adjacent sides be completed, then the
diagonal of the parallelogram which passes through the
point will represent the resultant velocity in magnitude and
direction.
Let u and v be the two simultaneous velocities of the point,
and let them be represented in magnitude and direction by the
straight lines O A, O B.
Complete the parallelogram O A C B, O C shall represent
the resultant velocity.
We may imagine the point to move from O to A in a unit
of time, while the page of the book
moves parallel to O B with a velo-
city repres.ented by O B. Thus,
while the point moves from O to A
the line O A will move into the posi-
tion B C ; and owing to these two
simultaneous velocities the moving
point will arrive at C in the unit of time.
Now, since the two simultaneous velocities are constant in
magnitude and direction, the actual velocity of the point from
O to C must be constant in magnitude and direction ; that is,
its path from O to C must be in a straight line. Hence O C
represents in direction and magnitude the path described by O
in the unit of time.
.-. O C represents the resultant of the two simultaneous
velocities represented by O A, O B.
Parallelogram of Velocities and Forces ' 1 9
This proposition holds of any two simultaneous velocities
which a point may have, whether it has other independent
velocities or not. It is also true of two variable velocities by
the extended definition of velocity in Art. 15.
38. The acceleration of a point is the rate at which its
velocity is changing, and is measured by the number of units
of velocity the point gains in a unit of time (Art. 17).
Like velocity, it may be represented completely in magni-
tude and direction by a straight line, and may be positive or
negative according to the direction in which the straight line is •
drawn.
Thus, if a positive acceleration, that is, an acceleration in
the direction in which the point or particle is moving, be
represented by a straight line in one direction, an acceleration
equal and opposite to this, that is, a retardation, will be repre-
sented by an equal straight line, in the opposite direction.
Acceleration may likewise be compounded and resolved in
the same way as velocities.
We shall thus have the Parallelogram of Accelerations,
which may be stated as follows : —
39. If a point have simultaneously two accelerations
wMch are represented in magnitude and direction by two
straight lines drawn from the point, and if the parallelo-
gram having these two liaes for adjacent sides be com-
pleted, the diagonal of the parallelogram drawn through
the point wiU represent the resultant of the two accelera-
tions in magnitude and direction.
For let O A, O B represent the two accelerations of the
point.
Complete the parallelogram
O A C B, and join O C.
Then O A, O B will represent in
magnitude and direction the veloci-
ties due to each acceleration which
are added in a unit of time. '^"^ 3-
Therefore, by the Parallelogram of Velocities, O C represents
in magnitude and direction the resultant velocity added in that
unit of time.
c 2
20 . Dynamics
Hence O C represents in magnitude and direction the
resultant acceleration of the two simultaneous accelerations
represented by O A, O B.
40. We have seen (Art. 26) that Forces are measured by
the momenta which they produce in a unit of time ; that is,
they are measured by the product, mass x acceleration. Hence,
if two or more forces act upon particles of equal mass, they
will be proportional to the accelerations which they produce in
them, since the mass moved is the same for each force.
The Second Law of Motion further states (Art. 34) that, if
two or m.ore forces act upon a body at the same time, the
change of momentum due to each is independent of the others,
and is the same as if that force acted alone.
We may therefore say that, if several forces act simul-
taneously on the same particle, each force will be proportional
to the acceleration which it produces in it.
Definition. — The resultant of two or more forces is the
single force whose effect is equivalent to the combined effect
of the forces, and with reference to this resultant each of the
original forces is called a Component.
Forces also may be represented completely by straight
lines (Art. 10).
We will now state and prove the Parallelogram of Forces.
41. If two forces acting on a point be represented in
magnitude and direction by two straight lines drawn from
tlie point, and if the parallelogram having these two lines
for adjacent sides be completed, the diagonal of the
parallelogram drawn through the point will represent the
resultant of the two forces in magnitude and direction.
Let O A, O B represent in magnitude and direction the
By ^7C two forces which act on a par-
/ ^^^'•^^ / Complete the parallelogram
iL......^^^^^ I O A C B, and join O C.
Z^.^''''^/ / Then since O A, O B repre-
o a * J[ sent the forces acting on O,
^'t^- 4- they will also be proportional
to the accelerations which those forces produce in O (Art. 40).
Parallelogram of Velocities and Forces 21
Hence \l Q)a,Q)b represent the accelerations produced by
the forces represented by O A, O B,
OA : Oa :: OB : Ob,
and if the parallelogram O acb be completed, the point c must
lie on the diagonal O C (Euclid VI. 4). And by the parallelo-
gram of accelerations, O c represents the resultant of the tvvo
accelerations represented by O «, O b.
Now, since the mass acted upon remains the same, the
resultant force must have the same ratio to the resultant
acceleration as either of the component forces bears to the
acceleration which it produces.
But OC : 0, or it can be
easily proved geometrically
Forces at a Point 25
Example.
Find the magnitude of the resultant of two forces, 9 poundals
and 7 poundals, at an angle of 120°.
R2 = p^ + Q^ + 2 P . Q cos 5
= 9' + 7' + 2.9.7(-^)
= 130 - 63
= 67
.•. Resultant force = '/Uj poundals.
To find the magnitude and direction of the resultant of two
forces 5 and 4 acting at an angle of 45°.
Resultant force = a/ 5^ + 4^ +2. 5. 4. — :
V ./i
^2
'/25 + 16+20 V2
= -/4I + 20 v/2
= 8'3 approximately.
If ^ be the angle it makes with the force 5,
tan F, H K > O A, and C, C fall on opposite sides of O. We
still have tvi'o parallelograms formed, but only one of them satisfes
the required conditions ; for if the angle C O A = 5, the angle
C O A must be = 180'' - e.
Examples.
1. Resolve a given force into two components, making given
angles with it. Find the ratio of these components if the angles
are 30° and 45°- Ans. I : V2.
2. Resolve a given force into two others whose magnitudes are
given. What conditions are necessary as to the given magnitudes
of the components ?
3. Resolve a force of 12 poundals into two others, one of which
is 6 poundals and the other makes an angle of 30° with the force 12.
4. In Art. 53, Ex., if P = F sin 6, show that the other compo-
nent = v'F^ — P^.
5. If a force 4 be resolved into forces 3 and 2, find the cosines
of the angles which the components make with the given force.
Ans. I and \\.
6. Resolve a given force into two components, one of which is
given in magnitude and direction. What is the magnitude of the
other component ? Ans. Q' = F'^ + P^ — 2 F . P . cos 9.
7. If a force equal to the weight of 5 lbs. be resolved into two
others, one of which is equal to the weight of 4 lbs., making an
angle of 60° with it, find the magnitude of the other component.
If (^ be the angle which it makes with the given force, show that
, 2
sm =
54. To find the resolved part oj a force P in a direction
making an angle 6 with the given force.
Since a force can produce no change of position in a direc-
tion at right angles to itself, we must resolve the given force
Forces at a Point
33
into two components — one in the given direction, and the
other at right angles to it.
Let O C represent the given
force Pj and D E the given
direction.
Draw O X parallel to D E,
making an angle Q with O C^
and draw O Y at right angles
to OX.
Through C draw C A, C B,
at right angles to O X, OY 5 E
respectively. ■^'°' '''
Then O A C B is a parallelogram, and therefore O A, B
represent two components of P.
Further, since O B is _!_■■ to O X, it produces no displace-
ment in the direction O X ; therefore O A represents the
total force in the direction O X.
Now O A = O C cos 61 = P cos 61,
and O B =0 C sin 61 = P sin Q.
Hence the resolved part of a force P in a direction making
an angle Q with it is equal to P cos B.
The other component \j to this one is equal to P sin 6.
55. To find the resultant of any number of forces acting at
a point in the same plane, their directions and magnitudes being
given.
This can be done by applying the parallelogram of forces
as many times as is required.
Thus, first find the resultant of two of the forces, then find
the resultant of this force and a third, and so on.
This process will be found extremely tedious, except in the
most simple cases.
Again, the resultant might be found by the method of
Art. 52.
This method is well adapted to geometrical measurements
depending on accurate drawing, but the determination of the
length and position of the line A E by trigonometrical calcula-
tion is very troublesome.
D
34 Dynamics
The following is the method which we almost invariably
adopt : —
Resolve all the forces along two lines at right angles to
each other.
A\'e shall thus get two equivalent forces at right angles to
each other, one of which is the sum of all the components in
one direction, and the other the sum of all the components in
a direction at right angles to it. We then find the magnitude
and direction of the resultant of these two forces.
Thus, let Pi, P2, P3, &c., be the forces acting on a particle
at O, making angles 6^, 6^, 63, &c., respectively with a given
line O X.
Draw Oy perpendicular to O x.
Resolve P, along O x and Oj'.
The resolved part of Pi along O jc = P, cos 6, (Art. 54)
Oji' = Pi sin(9i.
Proceed in a similar manner with all the other forces.
Let X be the algebraical sum of all the components along Ox.
ji ^ )) j> )) » ^y*
X = P, cos ^1 + P2 cos 62 + P3 cos ^3 +
Y = P, sin ^1 + P2 sin 63 + P3 sin ^3 +
The directions of X and Y will be given by their signs.
Hence, if R be the re-
sultant force and <^ the
angle it makes with O x,
R2 = X2 + Y2
tan -^ = -j^-
Corollary. — If the re-
sultant force be zero,
R = o
.-. X2 + Y2 = O
.•. X = o and Y = o.
Fig. i6.
We thus get the following condition that any number of
forces acting at a point should be in ecLuilibrium,
Forces at a Point
35
The sum of the components of the forces resolved in any two
directions at right angles to each other must each be zero.
Example. — Find the magnitude and direction of the resultant
of forces i, 2, 3, 4, 5,
6, 7, each making an
angle of 30° with the next
one.
Let O jr be the direction
of force I.
Draw Oy J.' to O x.
Let X and Y be the sum
of all the components in
the directions O x, Oy
respectively, R their resultant.
X = I + 2 cos 30° + 3 cos 60° + 4 cos 90" + 5 cos 120°
+ 6 cos 150° + 7 cos 180°
= x.2.f,3.i,.(_i),.(_^),7(_,)
2 2
7 - 2 -/3
Y = 2 sin 30° + 3 sin 60° + 4 sin 90° + 5 sin 120° + 6 sin 150°
= 2.1 + 3.^+4 + 5.^3 + 6,
22 2
. I + 4 + 3 + 3 ^^ + 5^3
= 8 + Wz
R' = (-7-2V3f + (8 +4V3f
= 173 + 92V'3
R = V173 + 92 V3
= 1 8 '2 approx.
02
36 Dynamics
If 6 be the angle this resultant makes with O x,
tan 6 = — y-^^-'.
7 + 2V^
32 + I2v/3
37
= - 1-427.
This lies between - i and - v^, and therefore 6 lies between
120° and 135°.
Its actual value is about 125°.
Example. — To explain by means of the resolution offerees how
a sailing vessel can be made to go from one point to another
against the wind.
This is done by the pro-
cess called tacking; that is,
taking a zigzag course in
directions making acute
angles with the direction of
the wind. In such a case
the sail is placed so that its
plane lies between the di-
rection of the wind and the
ship's course.
Suppose that F is the
force of the wind on a square
inch of sail. Resolve F into
two components, P and Q
respectively parallel and
normal to the plane of the
sail ; of these, P has no
effect on the sail, therefore
Fif.. 18. Q represents the total avail-
able effect of the wind on each square inch of sail.
Resolve Q into its two components R and S respectively
parallel and normal to the proposed course of the ship ; of these
R is entirely available to move the ship in the required direction.
Although the component S is very often larger than R, it only has
a small effect on the motion of the ship, since the shape of all
ships is such that they move easily in the direction of their length
Forces at a Point 37
and only with difficulty in a direction perpendicular to their
length.
The effect of S is called leeway, and is largely counteracted by
the effect of the rudder, though there will always be a little leeway.
Thus we see how the total effect of the wind on the whole sail
can be so resolved as to make the ship move in the direction of
its length at an acute angle to the direction of the wind.
So, in the case of a kite, we have to consider the equilibrium
of the weight of the kite, the tension of the string, and the force of
the wind perpendicular to the plane of the kite.
56. Since velocities and accelerations are compounded
by the Parallelogram Law in the same manner as Forces are,
and since the results of Arts. 46-55 have all been derived
from the Parallelogram of Forces, it follows that we shall have
analogous propositions for velocities and accelerations. In
fact, all the propositions contained in the above articles will
hold mutatis mutandis for velocities and accelerations.
We shall thus have the Triangle and Polygon of Velocities
and Accelerations, and we shall resolve and compound them
in the same manner as we have resolved and compounded
forces.
Examples.
1. What forces will just balance a system of four forces of 6, 7,
8, and 9, two of which act in one direction and the other two in
the same straight line in the opposite direction ?
2. A heavy uniform chain has 10 lbs. and 16 lbs. attached to its
ends, and hangs in equilibrium over a smooth peg. If the greatest
tension of the chain be 20 lbs. weight, find the weight of the chain.
3. In the above question, if the lengths of the chain on either
side of the peg be as 3 : 2, find the weight of the chain.
4. Forces 3, 4, and 5 act in the same straight line. What is the
greatest and least resultant they can have ? How must they act
if they are in equilibrium ?
5. Forces 99 and 20 act on a point at right angles to each
other. Determine the magnitude of their resultant.
6. Forces i and 2 act on a particle at an angle of 135°. Find
the resultant.
38 Dynamics
7. Forces 5 and 4 act on a particle at an angle cos-'(^^j.
Find the resultant.
8. Forces 3 and 4 act at an angle of 60°. Find the magnitude
of the resultant and the sines of the angles it makes with the com-
ponents.
9. Find the resultant of forces 13 and 14 acting at an obtuse
angle whose sine is — .
13
10. Forces i, 2, and \/^ acting at a point are in equilibrium.
What are the angles between their directions ?
11. A lo-lb. weight is supported by two forces, one of which
acts horizontally and the other at an angle of 30° with the horizon.
Find the magnitude of these forces.
12. Forces P and 2 P act at a point at an angle of 60° with
each other. Find the magnitude and direction of the resultant.
13. Forces 7 and 8 act at an angle of 60°. Determine their
resultants.
14. If the angle at which two forces act be increased, show that
their resultant will be diminished.
15. Forces of 2i, 6, 6J act on a particle and keep it at rest.
Show that two of them are at right angles, and find the cosine of
the angle between the greatest and least.
16. R and R' are the smallest and greatest forces which along
with P and Q can keep a particle at rest. Show that if P, Q,
-v/RR' are in equilibrium, two of these forces are perpendicular to
each other.
17. A picture is suspended by a cord attached to its upper
corners, and passing round a nail. Show that the longer the
string the smaller will the tension be.
18. Two pictures of equal size and weight are suspended like
the one in the preceding question. The strings are inclined at
angles of 45° and 60° to the top of the pictures. Compare the
tensions.
19. The resultant of two forces acting at a point is double of
one of the forces P and at right angles to it. What must be the
magnitude and direction of the other force ?
20. A mass of 10 lbs. is suspended by a string and is acted on
Forces at a Point 39
by a horizontal force so that the string is inclined at an ang'e ut
30° to the vertical. Find this force and the tension of the string.
21. Forces of 5 and 6 acting at a point have a resultant of 7.
Find the angle between them.
22. Forces of 7 and 8 acting at a point have a resultant of 6.
Find the angle between them.
23. Forces of 4 and 5 have a resultant of 7. Find the angle
between them.
24. Forces of 5 and 6 have a resultant of 8. Find the angle
between them.
25. Forces 2, P, and Q are in equilibrium ; the angle between
2 and P is 135° and the angle between 2 and Q is 90°. Find
P and Q.
26. Two forces act at an angle of 135°. Find the ratio between
them in order that the resultant may be equal to the smaller.
27. P, and Q are two forces in the ratio of 2 : ^/3 ; R is the
resultant of P and Q when at right angles to each other ; S is the
resultant of R and Q. Find the magnitude and direction of S.
28. R is the resultant of two forces P and Q, S the resultant
)f R and P, and T the resultant of R and Q. Find the resultant
of S and T.
29. Forces P + Q and P — Q act on a point at an angle whose
1 P' + 2 O-
cosine is - --^-^ ^ . Find the force which will keep them in
2 Q2 _ pi
equilibrium. Show that P must be greater than 2 Q.
30. ABC is a triangle ; D the middle point of B C. Show
that the resultant of forces represented by A B, A C is represented
by 2 A D.
31. Forces represented by lines drawn from any point O to the
angular points of any triangle are equivalent to forces represented
by the lines drawn from O to the middle points of the sides.
32. If forces act at the angular points of a triangle in directions
towards and perpendicular to the opposite sides, and are pro-
portional to those sides in magnitude, prove that they are in
equilibrium.
33. Forces acting at the angular points of a triangle are repre-
sented in magnitude and direction by lines drawn from the angular
40 Dynamics
points to the middle points of the opposite sides. Prove that they
are in equiHbrium.
34. Three equal forces (P) act at a point parallel to the sides
of a triangle, these sides being in the ratio of 3 : 4 : 5- Find the
magnitude of their resultant.
35. Two forces are represented in magnitude and direction by
two chords A P, A Q of a circle, which are at right angles to each
other. Find the magnitude and direction of their resultant.
36. Forces are represented by lines O A, O B, O C, O D drawn
from any point O to the fixed points A, B, C, and D. Show that
their resultant is 4 O G, G being a fixed point.
37. R is the resultant of two forces P and Q acting at a point
at right angles to each other ; S is the resultant of R and P : if
Q = 2 P prove that S = 2 Pv'2.
38. In the above question if Q = 3 P prove that S = Pv'^-
39. Two forces acting at a point are represented by a side and
diagonal of a square, the length of the side being a. Show that
their resultant is av^ and find its direction.
40. Forces P, Q, and R act at a point parallel to the sides of
an equilateral triangle taken in order. Show that their resultant is
-v/P^'TiyT'R^^'^njTr^^R'P^^P Q.
41. Forces P, Q, R, and S act at a point parallel to sides of a
square taken in order. Find the magnitude of the resultant.
42. A B C D is a rectangle, E is the middle point of A B ; forces
represented by A D, D E, E C, C B act at a point. Find their
resultant.
43. A B C D is a parallelogram ; E and F are the middle points
of A B and B C ; forces acting at a point are represented by A F,
F B, A E, E D. Determine their resultant.
44. ABCDEF.is a regular hexagon and forces act at the
point A in the directions A C, A F, D A, and are in equilibrium.
The force along A F is 2 lbs. wt. ; find the other two.
45. Forces 1, 2, 3 act at a point parallel to the sides of an
equilateral triangle taken in order. Show that their resultant is -Ji.
46. Forces i, 2, 3, 4, 5, 6 act from the centre of a regular hexagon
to its corners. Find their resultant.
Forces at a Point 41
47- Forces 5, 13 and 20 act on a particle and are equally
inclined to each other. Find their resultant.
48. The sides of a triangle taken in order are 3, 4, and 5 inches.
Forces of 12, 16, 19 act along them. Find the resultant.
49. The sides of a quadrilateral are i, 2, 9, 7 inches. Forces
of 2, 4, 8, 14 act along them respectively. Find the resultant.
50. Forces 3, 5, and 6 act on a particle ; can they keep it at
rest ? If so, what is the angle between 6 and 5 ?
51. Forces 2 P, v'S? have a resultant P. Find the angle
between them.
52. AB C D is a quadrilateral, and forces acting at a point are
represented in magnitude and direction by AB, BC, DC, AD.
Find their resultant.
53. If the resultant of two forces be perpendicular to one of the
components and make an angle of 30° with the other, find the ratio
of the two components.
54. Resolve a force of 7 into two components, making angles
of 30° and 60° with it.
55. Two forces acting at a point at right angles to each other
are kept in equilibrium by a third force, making an angle of 30°
with one of them. The greatest force is 3 lb. wt. ; what must the
other two be ?
56. A weight of 10 lbs. is suspended by a string 25 inches long.
The weight is drawn aside until its vertical distance below the
point is 20 inches. Find the least force which will keep it in this
position.
57. A B C D E F is a regular hexajron, and forces are repre-
sented by A B, A C, A D, A E, A F. Find their resultant.
58. Forces i, 2, 3 act at the angular points of an isosceles right-
angled triangle towards and perpendicular to the opposite sides,
the last force being perpendicular to the hypotenuse. Find the
resultant
59. A quadrant is trisected, and forces i, 2, 4, 8 act along the
four radii to the points of section. Find the magnitude and
direction of the resultant.
60. Forces l and 2 are at right angles to each other, and a
force 3 »/2 bisects the angle between them. Find the resultant.
42 Dynamics
6i. Forces lo and 3 are at right angles, and a force 2 ^/2
bisects this angle. Show that the resultant of the three is equal to
the sum of the two first.
62. If any number of forces in one plane act at a point O, and
a fixed line O A be drawn through that point, find an expression
for the tangent of the angle which their resultant makes with O A.
63. If a given force be resolved into two equal forces, prove
that the extremities of the lines representing these equal forces
always lie on a fixed straight Ime.
64. If a given force be resolved into two components at right
angles to each other, prove that the extremities of the lines repre-
senting those components always lie on the circumference of a
fixed circle.
65. If a force 8 P be resolved into two forces, each equal to
5 P, find the sine of the angle between the equal components.
66. Give a geometrical construction for resolving a force into
two others inclined at a given angle, one of which is to be of given
magnitude. Under what conditions would you get two values for
the other component ?
67. Resolve the force represented by the diagonal of a square
into three equal forces, each being represented in magnitude by a
side of the square, and one of them being coincident with a side
of the square.
68. Explain the equilibrium of a kite in the air.
69. A string is attached to a kite at a point P, whose distance
from the centre of gravity G is 5 feet. The resultant effective
pressure of the air is perpendicular to the kite's surface and acts at
■a point R in P G such that P R. equals 18 inches. The plane of
the kite is inclined at an angle of 45° to the horizon. Find the
tension of the string if the kite be 4 lb. wt.
70. A B C D is a parallelogram, P any point in its plane ;
through P lines are drawn parallel to the sides, cutting A B, C D
in L, N respectively, and B C, A D in M, O respectively. Show
that forces represented by M L, N O, A C if acting at a point
would be in equilibrium.
43
CHAPTER IV
PARALLEL FORCES
57. Parallel forces are said to be like when they act in the
same direction, and unlike when they act in opposite directions.
To find the magnitude and position of the resultant of two
like parallel forces acting on a body.
Let P and Q be two Hke parallel forces acting on a body
at points A and B respectively. Join A B and produce it
both ways. At A and B apply two equal and opposite forces F
in the line A B. This will not affect the action of P and Q
(Art. 44). The resultant of P and F acting at A can be found
by the parallelogram of forces ; let it be Rj ; similarly, let the
resultant of Q and F acting at B be R2 ; let the directions of
Ri, R2 meet at O ; this they must do, since the angles between
Ri, F and Rj, F are together less than two right angles.
44 Dynamics
Through O draw O C parallel to the direction of P and Q,
and also a line parallel to A B the direction of F.
Suppose R, the resultant of P and F to act at O (Art. 9),
and at O replace Ri by its original components P and F
acting parallel to their former directions ;
Similarly let R2 acting at O be replaced by its original
components Q and F acting parallel to their former direc-
tions.
We then have four forces acting at O, of which P and Q
act along O C, and the two equal and opposite forces F act
parallel to A B. These two forces F may be removed without
disturbing the effect of the system ; and we are left with two
forces P and Q acting along O C, and producing exactly the
same effect as P and Q acting at A and B respectively.
Therefore if R be the resultant of P and Q,
R=P+Q;
and it. acts parallel to the direction of P and Q. Since C is a
point in its line of action we may suppose R to act at C.
To find the position of C.
Since P, F, R, are parallel to O C, C A, OA respectively,
they are proportional to them. (Art. 50.)
. P oc
••f~ca
0)
SoalsoQ-OC
F CB
(2)
Dividing (i) by (2) we have - = ^— ,
or P . C A = Q . C B.
58. To find the magnitude and position of the resultant of
two unlike parallel forces acting on a body.
The proof of this is almost identical with that of the pre-
Parallel Forces
45
ceding article ; we will, however, go through it in detail on
account of its importance.
Let P and Q be two un-
like parallel forces acting on
a body at points A and B
respectively.
Fig.
Suppose Q greater than P. Join A B and produce it both
ways. At A and B apply two equal and opposite forces F in
the line A B. This will not disturb the action of P and Q.
The resultant of P and F acting at A can be found by the
parallelogram offerees : let it be R,; similarly let the resultant
of Q and F be R2. Since the angle between P and F is equal
to the angle between Q and F, but Q is greater than P, there-
fore the angle between Rj and F is greater than the angle
between R, and F (Art. 46, Cor.) ;
.•. the directions of Ri and Rj meet above AB (Euclid,
Axiom 12).
Let them meet at O ;
through O draw O C parallel to the direction of P and Q, and
also a line parallel to A B, the direction of F.
Suppose Ri the resultant of P and F, to act at O, and at
O replace Ri by its original components P and F acting
parallel to their former directions ; similarly let Rj acting at O
be replaced by its components Q and F acting parallel to their
former directions. We now have four forces acting at O, of
46 Dynamics
which P and Q act along O C in opposite directions, and the
two equal and opposite forces F act parallel to A B.
These two forces F may be removed without disturbing the
effect of the system, and we are left with the two forces P and Q
acting along O C in opposite directions and producing exactly
the same effect as P and Q acting at A and B.
Therefore if R be the resultant of P and Q,
R = Q - P in the direction C O,
= P — Q in the direction O C,
and it acts parallel to the direction of P and Q. Since C is a
point in its line of action we may suppose R to act at C.
To find the Position of C.
Since P, F, Ri are parallel to O C, C A, D A respectively,
they are also proportional to them (Art. 50),
••r CA^'''
So also Q = ^^^ (2).
P BC
Dividing (i) by (2) we have -^ = -^,
or P . C A = Q . C B.
Note I. — The result of this article is identical with that of
the previous article.
The resultant of the unlike parallel forces P and Q is the
algebraical sum of P and Q, and the distances of C from A
and B have the same ratio in both cases.
Note 2. — We have supposed Q > P.
If P had been > Q, O, the point where R, and R2 meet,
would have been below A B and to the left of the line of
action of P.
Hence, C would have been in A B to the left of A, such
that P . C A = Q . C B.
59. The relation P . C A = Q . C B enables us to find the
distance of C from either A or B when the distance A B is
Parallel Forces 47
known and the magnitudes of the forces P and Q are also
known.
Thus with the figure of Art. 58 put C B = C A — AB in
the equation.
P. CA = Q. CB
.-. P . C A = Q (C A - A B)
/. CA(Q-P)=Q.AB
Similarly, C B = %r^
When P = Q
R=Q-P=o
CA
_Q. AB_Q. AB
Q-P o
.-. C A = 00
So also C B = 00.
Hence the point C is at an infinite distance from A and B.
The reason for this is apparent. For if P = Q in the
figure of Art. 58 the directions of Rj and R2 are parallel, and
consequently the point O is removed to an infinite distance.
60. Definition. — If two equal parallel forces act on a body
in opposite directions, but not in the same straight line, the
system is called a Couple.
From the results at the end of the last article, i.e. R = O
and A C = B C =00, we see that when a couple acts on a body
it cannot be replaced by a single force at a finite distance. It
may be looked upon as an infinitely small force acting at an
infinitely great distance from the body.
The result of such a force cannot partake of the nature of
a translation at all ; its effect on the body is that of a twist or
spin only, about a point in the body.
We shall see in the next chapter how this effect may be
measured.
Fig. 21.
48 Dynamics
61. To find the resultant of any number of parallel forces
acting on a body in the
I same plane.
Let P, Q, R, &c. be
parallel forces acting on
C a body at points A, B, C,
&c. respectively.
The resultant of the
parallel forces P and Q
is a parallel force P + Q
which acts at a point E
suchthatP.AE=Q.BE.
P and Q may now be
replaced by this resultant acting at E. Join E C.
Then the resultant of the parallel forces P + Q acting at E
and R acting at C is the parallel force P + Q + R acting at a
point F such that (P + Q) E F = R . C F, and so on for any
number of parallel forces.
Hence the resultant of any number of parallel forces is the
algebraical sum of the forces, and acts at a point F, whose
position depends only on the magnitude of the forces and the
position of their points of application.
The position of F will not be altered if we suppose the
lines of action of all the forces to be turned through the same
angle at their points of application, provided that they remain
parallel to each other.
On this account F, the point at which the resultant acts, is
called the Centre of the Parallel Forces.
62. To find the position of the centre of a system of parallel
forces whose points of application lie in a straight line.
Oh
/ / /
Fig. 22,
Let Pj, P2, P3; &c. be parallel forces acting at points
Parallel Forces 49
A, B, C, &c. in a straight line, and let the distances of A, B, C
&c. from a fixed point O in the line be «,, x^, x^, &c.
Let the resultant of Pi and P2 act at D.
Then P, . AD^Pj-BD
P, (O D - X,) = Ps (x, - O D),
.•.OD(P, + P2) = P,*1 + P2.*2. .
Similarly if the resultant of P^ + Pj at D and P3 at C act
at E,
(P, +P2)DE = P3.CE
(P, +P2)(OE-OD) = P3(OC-OE),
.-. OE(Pi + P2 + P3) = OD(P, +P2) + P3.0C
^ Tl Xi + P2 ^2 "T P3 -^Sj
and so on.
Hence if x be the distance of the point at which the resul-
tant R acts from O, we have
x.R = FiXi + ^2X2 + ¥3X3 + .
62a. To find the position of the centre of a system of
parallel forces .
Let Pj, P2, P3, &c. be parallel forces acting at points
A, B, C, &c. in the same plane.
Let O X, O Y be any two lines at right angles to each other
in the plane in which the forces act.
We will find the distances of the centre of the parallel
forces from O X and O Y in terms of the distances of the
points A, B, C from these lines and the magnitudes of the
forces Pi, Pa, P3, &c.
Let D be the centre of the parallel forces P, and P2, arid
let E be the centre of the parallel forces P,, P2, P3.
Draw A a, B (5, Qc, D d, Ee perpendicular to O X.
Draw B H, D K perpendicular to D d, A a respectively,
the triangles A D K, D B H are similar.
Then, since D is the centre of the parallel forces Pi, P2,
Pi'. A b = P2 . B D
50
Dynamics
"Pa AD AK Ka-Y>d
.'. Pi (Aa-D^)=P2 (D "" 5
feet respectively.
Hence, if "x be the distance of their centre from O we have
- 2(P;r)
2(P)
_ I ■ 5 + (- 2).3 + 3. I + (-4)(- I) + 5- (- 3) +(-6) (-5)
1-2+3-4+5-6
- 3
that is, the resultant force is 3 in the direction of the forces 2, 4
and 6 acting at a point 7 feet to the left of O.
63. To find the resultant of any forces acting on a body in
the same plane.
We first take any two of the forces, which do not form a
couple, and find their resultant by the method of Arts. 46 or 57.
This reduces the number of forces by one.
We then combine this resultant with a third force, which
does not form a couple with it, and find their resultant ; and
so on.
Proceeding in this way we shall eventually be left with two
forces only.
These two forces may be,
(i) equal and opposite and collinear,
or (2) unHke, parallel and equal, but not collinear,
or (3) unlike, parallel and unequal,
or (4) like, parallel and equal, or unequal,
or (5) they may meet in a point, not being equal, and collinear.
In Case (i) the resultant is zero and the forces are in
equilibrium ;
in Case (2) the resultant is a couple ;
in Cases (3), (4), and (5) the resultant is a single force.
Hence, if any number of forces act on a body in one plane,
and are not in equilibrium, their resultant must be a single
force or a couple.
Parallel Forces 5 3
Examples.
1. Like patallel forces 3 and 7 act at points A and B 20 inches
apart. Where does the resultant act ?
2. Parallel forces 4 and 9 act in opposite directions at points
15 inches apart. Where does the resultant act ?
3. The resultant of like parallel forces 7 and 5 acts at a point
2j ft. from the lesser one. Find the distance between the forces.
4. The resultant of unlike parallel forces 9 and 7 acts at a
distance of 4^ ft. from the lesser one. Find the distance between
them.
5. The resultant of two like parallel forces is 12 J and acts at a
point 2 feet from one force and 3 feet from the other. Find the
forces.
6. Parallel forces 3 and 5 act in opposite directions at A and B,
and their resultant acts 3 feet from A. Find A B.
7. The resultant of two unlike parallel forces is 2 and acts at a
point 6 inches from one and 8 inches from the other. Find the
forces.
8. A line is divided into four parts in the ratio of i : 3 : 5 : 7, and
five equal parallel forces act in the same direction at the extremi-
ties of the line and at the points of section. Show that the point at
which the resultant acts will divide the line in the ratio of 3 : 5.
9. Masses of i lb., 7 lb., 5 lb. and 3 lb. are placed i foot apart
on a rod without weight. Find the point on which the rod will
balance.
10. Particles of 2, 3, 4, and 5 lbs. are placed along a weightless
rod at distances 2, 3, 4, and 5 feet from one end. Find the dis-
tance from that end of the point about which the rod will balance.
It. A beam 15 feet long and 300 lbs. weight is supported in a
horizontal position by props at the ends. It will balance on a
point 7 feet from one end. Find the pressure on the prop at that
end.
12. A uniform rod O A, 12 inches long, is suspended by two
vertical strings at its ends, which break when the tension is more
54 Dynamics
than 7 lbs. wt. Weights of 2 lbs. and 7 lbs. are attached to the
rod at distances of i inch from O and 2 inches from A respectively.
Find the greatest weight the rod can have.
13. Two like parallel forces of 10 lbs. wt. and 11 lbs. wt. act on
a body, the distance between their Unes of action being i foot. A
force of ;ir lbs. wt. is added to the former and taken from the latter,
in consequence of which the resultant is moved 3 inches. Fmd x.
14. If the forces in the above question be unlike, and x lbs. be
added to the 11 lbs., and taken from the 10 lbs., in consequence of
which the resultant is moved 3 feet. Find x.
15. A tricycle (5 st. 4 lbs.) has the small wheel placed sym-
metrically 3 feet behind the two large wheels, which are 3 feet
apart. The centre of gravity of the machine is 9 in. behind the
front wheels and that of the rider (9 st.) 3 in. behind. Find the
pressures on the ground of the different wheels.
16. Weights of I lb., 4 lb., 3 lb., and 2 lb. are fastened to a
straight rod without weight at distances from one end of 1,2, 3,
and 6 inches respectively. Find the point about which the rod
will balance.
17. Two like parallel forces, P and Q, have a resultant R. If
Q be changed to -— show that the new resultant will occupy the
same position as if the forces had been interchanged.
18. A weightless rod, A B, i foot in length, is kept horizontal
by a vertical string, C D, attached to a point, C, in the rod, 4 in.
from A, and also by two equal vertical strings attached to the ex-
tremities A and B, and to two weights of 5 lbs. and 3 lbs. on a
horizontal plane below the rod. What is the greatest force that
can be applied along C D without disturbing equilibrium ?
19. Four forces in the ratio of i : 2 : 4 : 3 act along the sides
A B, B C, D C, A D respectively, whose side is one inih. Prove
that their resultant is parallel to a diagonal of the square and find
where it cuts A B.
20. Like parallel forces act at the vertices of a triangle, each
force being proportional to the opposite side. Show that the centre
of the parallel forces is at the centre of the circle inscribed in the
triangle.
21. A B is a uniform rod of length n inches, and weight in + i)
Parallel Forces 55
W. To the rod, weights W, 2 W, 3 W, . . . . N W, are attached
at distances i, 2, 3, .... « inches respectively from A. Find
the distance from A of the point on which the rod will balance.
22. A B C is a triangle, and D is the middle point of A B. Find
the position of the resultant of forces represented by two parallel
lines D E, B F, which trisect A C at E, and F.
23. Forces 3, 4, and 5 act towards the same parts through
points A, B, and C. Show that if their resultant bisects A C in
D, the line of action of the force through B will bisect A D.
24. A B C is a triangle right angled at A, whose sides are «, b,
c ; A D is the perpendicular form A on B C ; like parallel forces
P, Q, and R acting through A, B, C respectively have their centre
at the middle point of A D. Find the ratio of the forces.
56 Dynamics
CHAPTER V
MOMENTS, COUPLES
64. Hitherto we have only considered the traiislaiing
power of a force on a body ; we must now consider the
rotating power also (Art. 5).
There are three elements which determine a force : —
(1) Its direction,
(2) Its line of action or position,
(3) Its magnitude.
The translating power of the force depends on (i) and (3)
only, but the rotating power depends on (2) also.
This rotating power is called the Moment of the Force.
Definition. — The moment of a force about a point is the
power of the force to produce rotation about the point.
The measure of this power will obviously depend on two
things : —
(i) The magnitude of the force,
(2) The distance of its line of action from the fixed point.
Let us consider one or two examples.
In opening or shutting a door which rotates on hinges,
we apply force by pulling or pushing near to the handle, since
the further it is applied from the hinge the greater is the
rotating power.
Again, in turning the capstan, the sailor is provided with a
long pole which he fits into a hole in the capstan ; he then
applies his pressure at the extremity of this arm, and thereby
the rotating power of his push is largely increased.
Moments, Couples
57
In the figure, suppose that xy z is the cross section of a
capstan movable about an axis
perpendicular to the plane of
the paper and cutting this plane
inO.
Let F be the force which
a man can exert, and let it be
applied at the extremity of the
arm A. Fig. 24.
The further A is from the axis through O the greater is the
rotating power of F about that axis.
Suppose that O B = O A, and O C = 2 O A, the rotating
power of F acting at C is exactly double of its rotating power
when acting at A or B, and we find that F acting at C will
exactly balance, in rotating power, F acting at A, and F acting
at B, or 2 F acting at A or B.
We thus see that if we double the distance of the line of
action of the force from the fixed point we double the rotating
power or moment about that point, or if we double the force
and keep the distance the same, we shall also double the
moment : if we double both force and distance the moment
will be quadrupled.
The distance of the line of action of the force from the
point or axis of rotation must always be taken the shortest
distance ; i.e. it is the length of the perpendicular from the
point on the line of action of the force or between the axis of
rotation and the line of action of the force.
Thus, if F be the force, D the perpendicular distance of
its line of action from the point O or axis of
rotation.
Moment of the force is proportional to F
when D is constant, and is proportional to D
when F is constant.
It is therefore proportional to the product
F. D, and may be taken equal to F . D, pro-
vided we take our unit moment to be the rotating power of the
unit force acting at unit distance ; because
Fig. 25.
58 Dynamics
Moment of force F acting at distance D F . D
Moment of force i acting at distance i i . i
Thus, if we take i lb. wt. as our unit force and i foot as
our unit distance, then the moment of a force of 4 lbs. wt.
acting at a distance of 3 feet from a point or axis of rotation is
12 (units of moment).
The measure of the moment of a force F about a point or
axis is F . D, where D is the perpendicular distance of the
line of action of the force from the point or axis of rotation.
The conception of the moment of a force and the method of
measuring moments are of the greatest importance ; the use of
them in solving questions of equiUbrium in Statics will occur over
and over again. The student is advised to read again the previous
article before passing on to the next.
65. Geometrical representation of the moment of a force
about a point.
Let F be a force acting on a body at the point A.
^^O Let O be any point, and D the
^y''^/ \ length of the perpendicular O C,
^^^^^^ / I drawn from O to A X, the line of
^.^•^"^ / I action of F ; then the moment of
A B C X F about O is F . D.
^"=- =^- Let A B represent the force F,
so that A B contains F units of length ; join O A, O B.
Then the area of the triangle O A B
= i X base x altitude. (Euclid I. 41.)
=- AB . OC
2
= - F . D. units of area.
2
Hence the measure of the moment of F about O is
numerically equal to twice the area of the triangle formed by
jommg O to the extremities of the line which represents the
force F.
Moments, Couples 59
Note. — A B may be taken anywhere in the line of action
of F without altering the area of the triangle O A B, because
triangles on equal bases and of the same altitude are equal in
area.
Corollary. — If the point O is in the line of action of F,
O C {i.e. D) becomes = o, hence the value of the moment of
F about O becomes = zero ; that is, a force has no tendency
to produce rotation about a point in its own line of action,
or a force has no moment about a point in its own line of
action.
In this case the area of the triangle O A B obviously
becomes zero also.
Conversely : If the moment of a force about any point is
zero, the point is in the line of action of the force.
66. This method of representing a moment geometrically
only gives us a measure of the value of the moment ; it does
not represent the moment completely, for it does not give us
the direction of rotation.
If the direction of F were reversed, or if O were the same
distance on the other side of A X, the moment of F would be
reversed, but the area of the triangle O A B would remain the
same. We therefore make the same convention with reference
to direction of rotation as we make in trigonometry with
■ reference to the direction of revolution and the measurement
of angles. We consider moments which tend to turn the body
in one direction as positive moments, and those which tend to
turn the body in the opposite direction as negative moments.
It is usual to take the direction opposite to that of the
hands of a watch as the positive direction of rotation, the
reverse direction being negative. This convention is not
necessary ; the student may take either direction of rotation
as the positive one, provided that the reverse direction is taken
as negative.
If a number of forces act on a body and the sum of all the
moments about a point in one direction is equal to the sum of
all the moments about the same point in the opposite direc-
tion, the body has no tendency to rotate about that point, and
the algebraical sum of the moments about that point is zero.
6o Dynamics
67. We have defined the Resultant of two or more forces
to be the single force whose effect is equivalent to the
combined effect of the forces, ^^'e have shown how to find
the Resultant of Forces in one plane as far as the translating
power of the forces, is concerned ; it is now necessary for us to
prove that the force which is the resultant of other forces with
reference to their translating power will also be their resultant
with reference to their rotating power. This we shall first prove
in the case of two forces which meet at a point or are parallel.
68. Jf two forces meeting at a point act on a body, the alge-
braical sum of their moments about a foi?tt O in their plane is
equal to the moment of their resultant about the same point.
Case I. When the lines of action of the forces are on the
Q same side of O.
f R Case 2. "When the lines of
SJ. .yfC action of the forces are on oppo-
V >// site sides of O.
A y/^ / Let P, Q be two forces meeting
\/^ / at A, R their resultant ;
/^ \ / Let O be a point in the plane
\/ , p so that P and Q act on the same
A B side of O.
^"'■^''- . Through O draw OCD parallel
to the line of action of P, cutting the lines of action of Q and
R in C and D ;
Through D draw D B || to A C.
Then A B D C is a parallelogram whose diagonal is A D,
and A B, A C, A D are in the directions of P, Q, R respectively.
Therefore AB, AC, AD represent the forces P, Q, R
completely.
Join O A, O B.
Then aOAB=aDAB=aCAD,
.■. the algebraical sum of the moments P and Q about O
= 2 aOAB + 2 aOAC
=2ACAD+2A0AC
= 2 A O A D
= moment of R about O.
Fig. 28.
Moments, Couples 61
Case 2. Let O lie between the line of action of P and Q.
With the same construc-
tion as in Case i,
the algebraical sum of the
moments of P and Q about O
= 2 aOAB - 2 aOAC
=2 aCAD— 2 aOAC
= 2 A O A D
= moment of R about O.
69. If two parallel forces act on a body, the algebraical sum
of their moments about a point O in
their plane is equal to the moment of .
their resultant about the same point.
Case I. When the parallel forces
are in the same direction,
Draw O A C B perpendicular to
the lines of action of P, Q, R, cutting
them in A, B, C respectively.
Then P . A C = Q . B C.
Sum of the moments of P and Q about O
= P.OA+ Q.OB
= P (O C - A C) + Q (O C + C B)
= (P + Q)OC-P.AC + Q.BC
= R . O C since P . A C = Q . B C.
Note. — AVe have here taken the positive direction of rota-
tion the same as that of the hands of a watch.
Case 2. When the parallel forces act in opposite direc-
tions.
Suppose Q > P so
that R = Q-P in the
direction of Q.
Fig. 30.
62
Dynamics
Sum of the moments of P and Q about O.
= P . OA-Q. OB
= P (O C + C A) - Q (O C + C B)
= P.OC-Q.OC + P.CA-Q.CB
= (P - Q) O C since P . C A = Q . C B
= -R. OC
= moment of R about O.
Note. — We have taken the positive direction of rotation
opposite to that of the hands of a watch.
70. The sum of the moments of any two forces in one plane
about a point in tiie line of action of their resultant is zero.
This follows at once from the result of the two previous
articles and the Corollary of Art. 65.
It can be proved directly in any given case.
Conversely : Jf tlie algebraical sum of the moments of any
two forces about a point in
their plane be zero, that point
must lie in the line of action
of their resultant.
With the usual notation.
Since the sum of the
moments of P and Q about O
is zero,
.-. AOAB = AOAC;
.-, since they are on the same base O A, they are of equal
altitude.
i.e. B and C are equally distant from A O.
.-. A O is the direction of the diagonal of the parallelo-
gram A B D C.
.•. O lies on the line of action of R.
71. The results of the preceding articles may now be
extended to any number of forces in a plane which have a
single force for their resultant.
For we have proved the theorem to be true for the resultant
of two forces whether ihey meet in a point or are parallel. By
Fig. 31.
Moments, Couples 63
taking 'this resultant with a third force, as in Art. 63, we can
prove the theorem true for three forces ; and so on for any
number of forces which have a single force for their resultant.
We may therefore say : —
If any number of forces, wMch have a single force for
their resultant, act on a body, the algebraical sum of their
moments about a point in their plane is equal to the moment
of their resultant about that p.oint.
Whence it follows that
The algebraical sum of the moments of any number of
forces in a plane about a point in the line of action of their
resultant is zero.
And conversely :
If the sum of the moments of any number of forces in a
plane which are not in equilibrium about a point in their
plane be zero, that point must lie in the line of action of
their resultant.
Note. — If the forces are in equilibrium their resultant is
zero ; consequently the algebraical sum of their moments
about any point in their plane must be zero ; since a system of
forces in equilibrium is equivalent to no force at all.
To make this clearer, suppose that R is the resultant of any
number of forces in one plane, and let/ be the length of the
perpendicular drawn from any point O in the plane on the line
of action of R.
Then, sum of the moments of all the forces about O
= moment of R about O
= R./.
If R . / = o,
R = o, i'r / = o ;
if R = o, the forces are in equilibrium,
if / = o, the point O lies on the line of action of R.
72. In the preceding article we have supposed that the
forces had a single force as their resultant.
We have seen in Art. 59 that in certain cases the resul-
64
Dynamics
tant of a number of coplanar forces is not a single force acting
at a finite distance, but a couple ; that is, a system of two
equal and parallel forces, acting in opposite directions, and not
in the same straight line. The translating power of such a
system is zero, because R = o (Art. 60) ; but its rotating
power is never zero ; in fact, as we shall show in the next
article, the rotating power of such a system about any point in
its plane is always the same.
73. The algebraical sum of the moments of the two forces
of a couple about any point in their plane is constant.
Let P, P be the forces of the couple acting at points
A and B.
O any point in
their plane.
Draw O ab I' to
the lines of action of
P, P.
The algebraical sum
of the moments of P, P
about O
= P. 0<5 - P. 0<2
= P(0,5-0fl)
= P . a ^.
Similarly, if O were between the lines of action of the
forces, as at O',
The algebraical sum of the moments about O'
= P. O'^ + P. O'a
= P (O' <^ + O' a)
= 'P .ab.
The length of a b, the perpendicular distance between the
lines of action of the two forces, is the same at every point.
Let a 1^ =/, / is called the Arm of the couple.
P . / is called the Moment of the couple. It is the pro-
duct of either force into the arm, and is constant.
Note. — The result of this article is not inconsistent with the
result of Art. 71, though at first sight it might be thought that
since R = o, the moment of R about a point in the plane should
Fig. 32.
Moments, Couples 65
also be = o. It must, however, be remembered that though R = o,
yet its line of action is removed to an infinite distance (Art. 59),
consequently its moment about any point O — which is = R x (its
J."" distance from O) — becomes = o x co ; and since the expression
o X CO may have any finite value, there is no reason why it should
not be equal to P . ^.
74. Since the translating power of every couple is zero, and
the rotating power about any point in its plane is measured by
its moment, it follows that the effect of a couple on a rigid body
depends only upon the magnitude of its moment.
That is to say, the effect of a couple is not altered by
moving the points of application of the equal forces or by
turning the arm of the couple through any angle, provided
we do not alter the forces or the length of the perpendicular
distance between them.
Further, any couple can be replaced by another couple of
equal moment in its own plane.
Thus, if it be required to find the arm of a couple Q, Q
which will have the same effect as the couple P, P with an
arm /, the only necessary condition is
Moment of couple Q, Q = moment of couple P, P.
Hence if q be the arm of the new couple,
q_.q = Y .p
^ -P
Similarly, if the arm be given and the magnitude of the
forces be required,
1
75. A couple may be transferred to ajiy plane parallel to its
own without altering its effect on a rigid body.
This also is obviously true since the effect of a couple is
solely a rotatory effect about some axis perpendicular to its
plane ; but the proof of the above by means of Art. 57 is of
sufficient interest to authorise its insertion here.
Suppose the couple P, P to act with an arm A B, and to
F
66 Dynamics
be^noved out of the plane of the paper into a parallel plane,
keeping the arm A B parallel to itself.
Let abh& the new position of A B.
p^
42 P
2P'I' \
Ar
^J.p
Fig. 33.
At both a and b
apply the two equal and
opposite forces P, P
^P parallel to the forces of
the couple. This will
have no effect on the
equilibrium of the sys-
tem.
Join Kb,^a. Then
since A B, a ,5 are equal and parallel, Kb, B a bisect each
other at O ; therefore, the resultant of the equal parallel forces
P and P acting downwards at A and /5 is 2 P acting
downwards through O ; and the resultant of the forces P, P
acting upwards at B and a is 2 P acting upwards through O.
Now the equal and opposite forces 2 P and 2 P at the
point O may be removed without disturbing the equilibrium
of the system, and we are left with the couple P, P acting with
the arm ab, producing exactly the same effect as it did with
the arm A B.
Hence we see that a couple may be replaced by any other
couple of equal moment acting in any position in its own
plane, or in a plane parallel to its own, provided the direction
of rotation remain the same.
76. To find the resultant of any number of couples acting in
parallel planes.
Let P, Q, R . . . &c. be the forces of the couples
/, q, r, &c. their arms,
and let the planes in which the couples act be all parallel to
the plane of the paper.
First, all the couples may be transferred to the plane of *
the paper without altering their effect.
moments, Couples
67
They may then all be replaced by couples whose arms are
equal to a. Thus, if P' Q' R' be the forces of the new
couples,
P' = — ^; (y .
a
Q-^^R' = R
, &c.
They may then be moved about in the plane until their
arms are coincident with any line A B of length = a-
Then, if X = P' + Q' + R' + . . . we have as the resul-
tant of all the couples, a single couple = X . a
= (P' + Q' + R' + . . . . ) X «
= P'a + Q'a + R'a + . . . ,
= P/ + Q^ + Rr + ....
That is, the resultant of all the couples is a couple whose
moment is equal to the algebraical sum of the moments of the
original couples.
Note that in estimating the algebraical sum of the moments
of the couples we choose either direction of rotation as the positive
one, and then call those moments which are in the opposite
direction ' negative.'
Thus, if the moments P . / and Q . q have rotations in opposite
directions, they are of opposite signs. So also in the finding of
the value of X, if P . / and Q . §■ be of opposite sign, P' and Q' at
A will be in opposite directions, and likewise at B.
77. To find the resultant of a couple and a force in the same
plane.
Let the moment of
the couple be P . / and
let the force be Q.
Let Q meet the direc-
tion of one of the forces
of the couple P . / in A.
Let R be the resul-
tant of P and Q, produce
R to cut the direction of
the other force of the
couple in B.
F i
P>^
68
Dynamics
-^
Suppose R to act at B and replace it by its original conv-
ponents P and Q acting parallel to their original directions.
Then the two equal forces P, P acting at B are in equi-
librium and may be removed, and we are left with the single
force Q acting at B parallel to the original direction of Q acting
at A.
Hence the resultant of a single force and a couple is a
single force equal and parallel to the original single force.
Or thus : replace the couple P . / by the couple Q . ^ of
equal moments, and move the arm of this new couple until one
end of it is at A, a point in the line of action of Q, and the
arm is at right angles to A Q.
Then the equal and opposite forces Q, Q at the point A
may be removed without dis-
turbing the equilibrium of the
system, and we have left the
single force Q acting at C
parallel to its original direction
and producing the same effect
as the couple P . /, and the force Q acting at A.
From this it is obvious that a single force and a couple can
never produce equilibrium.
78. As an example of the results contained in this chapter
let us find the resultant of a number of parallel forces acting
on a rigid body in one plane.
Let the plane be the plane ot the paper. Let O be a fixed
point in the plane and
draw O x perpendicular
to the direction of the
parallel forces.
Let P„ P2, P3, &c.
be the parallel forces, and
let O X cut their lines of
«s, &c..
Fig. 35.
->Q
action at a.
Fig. 36.
O a^, 0^3, &c. be x^, X2, Xg, &c.
At O apply two equal and
parallel to P,.
'3!
and let the distances Oa,,
opposite forces equal and
Moments, Couples
eg
Then Pi acting at A; may be replaced by Pi acting at O,
and the couple whose moment is Pi . %,.
Similarly all the other forces may be replaced by an equal
force acting through O, and a couple.
LetPi+Po + P3+ =2(P)
andP,:<;i +P2a;2 + P3a;3+ . . = S (P x).
Then the system of parallel forces P,, Pj, Pj, &c. may be
replaced by S (P) acting through O parallel to the directions
of the forces, and the couple S (P x).
To find the conditions of equilibrium of such a system.
It is obvious that there cannot be equilibrium if neither the
single force S (P) nor the couple S (P . ^) vanish (Art. 77) ;
neither will there be equiUbrium if one of these vanishes and
not the other ; therefore they must both vanish ;
/?. S(P) = o S(P..a:) = o
i.e. the algebraical sum of the forces must be equal to zero,
and the algebraical sum of the moments of these forces about
any point in their plane must also be zero.
Note. — This single force % (P) and the couple 2 (P x) can
always be reduced to the single force 2 (P) acting at a distance
from O equal to _ /p/ by the last article.
This result agrees with that of Art. 62.
79.* Similarly we may find the resultant of any number
of forces in one plane.
Take any point O in
the plane and draw any
two lines Ox, O y at
right angles to each other.
Let Pi be one of these
forces acting at a point A,
whose coordinates referred
to O^ and Oj' are (je,, j'l). ■<-
Let the components of '
Pi parallel to Ox and Oy
be Xi, Yi respectively, and
let a, be the angle Pi makes with O x,,
then Xi = Pi cos a, and Yj = Pi sin a^.
Fig. 37.
70 Dynamics
At O apply two equal and opposite forces Xi parallel to Ox
and „ „ Y, „ Oy
Thus P| may be replaced by the forces Xi Y, acting at O,
the couple Y, . x^ from the axis of x to that of y in one
direction and the couple X, . j', from the axis of y to that
of X in the opposite direction, — that is, the couple Y, . x^
— Xi . ji from the axis of x to the axis of j.
Proceeding in this manner with all the forces, proper
attention being paid to the signs of the coordinates of the
points of application of the forces and to the measurements
of the angles a,, aj, a,, &c. in the same direction from the
line O X, we shall have as the final resultant, with the usual
notation,
5 (X) = 5 (P cos tt) II to O X
2 (Y) = 2 (P sin a) II to O^
and the couple 2 (Y x — X7) from the axis of x to that of j'.
If R be the resultant of the single forces acting through O
we have
R2= [2(X)}2+ {2(Y)}2
and if be the angle it makes with O x
As in the previous articles, if neither R = o nor 2 (Y x
— X j) = o, the force and the couple may be reduced to a
single force R parallel to this R and at a distance from O
equal to ^ — ^'^^ — ^ .
R
For the equilibrium of the system we must have, as in the
previous article,
R = o and 2 (Ya- - Xjy) = o
or 2 (X) = o, 2 (Y) = o and 2 (Yx - X^') = o.
We have seen in Arts. 68, 69 that the moment of the
resultant P, about any point O is equal to the moments of its
components X,, Y, about the same point.
Hence if/i be the perpendicular distance of P, from O
2(Yx-Xj') = 2(P.i«);
Mmnaits, Gn^ks
71
nmnber of
hence our conditiaDS of eqpiHIiri— . for any
forces acting on a rigid body in one plane are :
The SMms «f Hk remkied parts «f the fortes faraild to any
Ug0 strvti^t lines at right ai^es to each ether vanish, axd the
sum tf the mcmeii/s «f these forces about a fmnt in the pbme
must abo -•.imsh.
Exsaatple. — If the moments of two co(i{des be equal in magni-
tude but (^losite in AiectiaD, prove that the foor forces trhich
make op the om^es aie in eqoiUlHinm.
Let the conges be P, P at an aim^
and Q, Q „ ^-
Let (Hke lorce of each coo^e P and Q meet in A, and let the
odier fince of each coo^ meet
inB^
Diaw ACAD perpeadicalar
to the diiectians of die fiuces ^y>^ \ J^
P acd Q, nhich meet at B.
Then since P .jl=Q . q
P.AC=Q.AD
that IS, the momoits of P and Q
acting at B about A aie equal ;
.*. A is a point in the lesnhant
rfPandQ;
.*. the lesohant of P and Q
acting at B B akmg B A.
Sinulady the lesnhant vS
P and Q actii^ at A is almg A B.
But die tno resultants are obnoDsly equal in magnitude,
because the ai^le b^veen P and Q at A is the same as it is at R
HoKe diey are bodi eqcial in magnitude and (qgposite in diiec-
tian,andaie diaefoie in eqnilibrinm.
Extai^les.
I. K^tplain vfay a slid^ vhich win suj^ixt a laige kmgitDdinal
tension, can be broken by a much smaller force at ri^it angles to
itste^th.
Z- Cbmpaie die moments about a point of a fince of 12 as. trt.
acting in a fine vbose £stance fiom the pinnt is 2 ft. 6 in^ and of
amdiercf 5 oz. «t. acting at a distance of 5 ft.
J'a Dynamics
3. A triangular lamina has a point within it fixed, and is acted
on by forces completely represented by the sides of the triangle
taken in order. Show that the moment of the force required to
maintain equilibrium is represented by twice the area of the
triangle.
4. Extend the above question to the case of a polygon.
5. What is the result if the fixed point be outside the lamina ?
6. Masses of 7 lb., i lb., 3 lb., and 5 lb. are placed on a rod with-
out weight I foot apart. Find the point on which the rod will
balance.
7. The sides A B, B C of a rectangle A B C D are respectively
3 ft. and 4 ft., and forces of 6, 8, 10 act along A B, B C, and C A
respectively. Find the algebraical sum of their moments about A
and D respectively.
8. The sides of a triangle are 9, 1 2^ and 1 5 ft. , and a force P acts
along each of its sides. Find the moments about each of the
angular points.
9. A straight rod A B without weight hinged at A is in equili-
brium under the action of two forces P and Q in one plane ; P acts
at right angles to the rod at a point 2 in. from A ; Q acts at
an angle 30° with the rod at a point i inch from A. Find the ratio
of P to Q that the rod may have no tendency to turn about A.
10. A B, A D are two straight lines, and O a point outside the
angle A B D ; B M is drawn parallel to O A, meeting A D in M.
Prove that the moments about O of forces represented by A B,
A M are equal.
11. A B C is a triangle, and O A is parallel to B C. Prove that
the moments of forces represented by A B, A C about O are equal.
12. Show that four forces, represented in all respects by the
sides of a quadrilateral, taken in order, are equivalent to two equal
and unlike parallel forces, through the extremities of either dia-
gonal, each being represented by a line equal and parallel to the
other diagonal.
13. A circular disc suspended by its centre has strings from the
centre hanging down over the edge, and supporting weights of
20 oz., 21 oz., and 29 oz. ; the disc rests horizontally. Find the
inclination to each other of the strings supporting the lighter
weights.
Moments, Couples 73
14. A plank 5 ft. long and weighing 2 cwt. is supported at each
end by a prop. Find at what point of the plank a weight of 2i cwt.
must be placed in order that one prop may then support twice as
much of the whole weight as the other.
15. A bar 16 in. long is balanced on a fulcrum at its middle.
On the right arm are suspended 4 oz. and 3 oz. at distances of
5 in. and 7 in. respectively from the middle, and on the left arm
5 oz. at a distance of 5 in. from the middle and w at the end.
Determine w.
16. A rod 6 in. long and I lb. wt. is supported by two vertical
strings at its ends : a 3 lb. wt. is attached to the rod at a distance
of I in. from one end. At what distance from the other end must a
4lb. wt. be attached in order that the tensions of the two strings
may be equal ?
1 7. A heavy beam A B of known length and weight is movable
in a vertical plane about a hinge at A, and is supported by a cord
P Q of given length, P being fastened to the beam and Q to a
point vertically above A. A P = / and A Q = ^. Prove that the
tension of the cord will be the same when A P = ^ and A Q = /.
18. AB C D is a square whose side is a ; forces of magnitudes
2, 4, 2, 4 act along A B, B C, C D, D A respectively. Find their
resultant.
74 Dynamics
CHAPTER VI
CENTRE OF GRAVITY
80. Before proceeding to consider the general conditions
of equilibrium of forces which keep a body at rest, we shall in
the present chapter consider that force which is the most
important of all forces, since it acts upon all bodies with which
we are concerned ; i.e. the force which we call ' gravity '
(Art. 7).
This force is due to the attraction of the earth for all
matter within a finite distance of its surface, and the measure
of this force acting on any particular body is called the ' weight
of the body.'
The measure of the force of attraction between any two
particles is proportional to — ,j- where m, m' are the masses of the
particles and r the distance between them ; and the Hne of action
of this force is along the line joining the particles.
We shall assume that the attraction due to the whole mass of
the earth is the same as that due to a particle of equal mass placed
at the earth's centre. Hence the line of action of the force of
gravity on any particular particle is the line joining the particle to
the centre of the earth.
This is called the vertical line at the point.
Since the earth is not a perfect sphere, the force of gravity
varies from point to point of the earth's surface, being least at the
equator and greatest at the poles ; because the distances of points
on the equator from the earth's centre are greater than the
distances of other points on the earth's surface from its centre.
Since the mass of the earth is constant, it follows that the
weight of any particle at any particular spot on the surface of the
earth varies as the mass of the particle only.
Centre of Gravity 75
Hence, at the same place, equal masses have equal weights ;
and at different places, the same mass has different weights.
Thus if w be the weight, m the mass of any particle,
the ratio — = g
m
ox w = m . g
where g is always the same at the same place, but varies at
different places on the earth's surface.
8i. Now any body may be considered as an agglomeration
of particles, and the weight of any body is the resultant of the
weights of all the particles of which the body is composed.
All these weights act in vertical lines, that is, in lines joining
the various particles to the centre of the earth ; but since the
centre of the earth is about 4,000 miles away from points on
its surface, we may consider that all the lines drawn to the
earth's centre from various points of a body of finite size on
the earth's surface are parallel lines ; and we can therefore find
the weight of any body or system of particles in the same way
as we find the resultant of a system of parallel forces. The
point at which this resultant acts — that is, the centre of these
parallel forces — is called the centra of gravity of the body or
system of particles.
Hence the weights of all the particles of which a body is
composed may be replaced by a single weight, equal to their
sum, and acting through the centre of gravity of the body.
We therefore have the following definition :
The Centre of Gravity of a Body is the point at which
the whole weight of the body may be supposed to be
collected.
When we consider the mass of the body only, this point is
called the centre of mass.
Since the centre of gravity is the point at which the result-
ant weight of a body may be supposed to act, it follows that
the line of action of the resultant weight of a body must pass
through its centre of gravity in whatever position the body is
placed.
Now a number of lines cannot meet together in more than
76 Dynamics
one point ; hence no body can have more than one centre of
gravity. If this point be supported, the body or system will
rest in any position.
82. If a heavy body be suspended by a single string, or by *
any number of strings, to a fixed point, the centre of gravity
must be vertically beneath this point.
Foi all the forces which act on the body may be com-
pounded into two — the resultant of all the tensions which
must pass through the fixed point, and the resultant weight of
the body acting vertically downwards through the centre of
gravity. If there is equilibrium these two must be equal and
opposite, therefore the point of suspension and the centre of
gravity must be in the same vertical line.
This gives us a useful practical method of determining the
centre of gravity of a body. Suspend the body by a string,
produce the direction of the string through the body ; this
must also be the direction of the weight, therefore it must pass
through the centre of gravity. Now suspend it from any other
point and do the same thing. The point where these two
lines meet must be the centre of gravity.
83. If a body be placed upon a horizontal surface it will
stand or fall, according as the vertical through its centre of
gravity falls within or ivithout its base.
All the forces which
act on the body may be
compounded into two —
the resultant weight act-
ing vertically downwards
through its centre of gra-
vity, and the resultant
pressure of the surface on
Fig. 39. the body.
Of these two the latter must act within the base of the
body, and, if there is equilibrium, it must be equal and oppo-
site to the former ; that is, the vertical through the centre of
gravity must fall within the base.
If this vertical line fall outside the base, the body must
vw
Centre of Gravity y^
fall over towards the point where this vertical line cuts the
surface.
If this line fall on the boundary of the base, the body will
just stand, but will be on the point of falling over.
84. Equilibrium may be of three kinds —
Stable, Unstable, or Neutral.
Suppose a body to be in such a position that the forces
acting upon it are in equilibrium, and that the body is then
slightly displaced :
(i) If the forces now acting tend to make the body return
to its original position, it is said to be in stable equilibrium.
(2) If these forces tend to make the body recede from
its original position, it is said to be in unstable equilibrium.
(3) And if the body after displacement tends neither to
return nor to recede, it is said to be in neutral equilibrium.
Examples : —
(i) A body supported at a point above its centre of gravity, e.g.
a pendulum.
Bodies resting on a surface such that the vertical line through
their centres of gravity falls well within the base.
(2) All bodies resting on very small bases, so that when
displaced the weight acts in a line falling outside the base ; e.g.
an e^g balanced on its end, and all bodies which are top-heavy.
(3) All bodies which are supported at their centre of gravity,
and which, therefore, rest in any position ; or bodies which, when
displaced, still have their centre of gravity vertically above the
point of support, e.g. a sphere on a horizontal table.
We will now proceed to find the centre of gravity of some
bodies of simple shape and of uniform or homogeneous
material.
]\lote. — Matter is said to be uniform or homogeneous when
it is of the same density throughout — that is, when equal
volumes of it are of equal mass and therefore of equal weight.
78 Dynamics
85. To find the centre of gravity of a uniform material
straight line or of a uniform rod.
A 711 G rn: B
I-H i i— I
Fig. 40.
Let A B be the rod, G its middle point. The rod may be
supposed to consist of a number of particles of equal weight.
The resultant of the weights of any two equal particles m and
m' equidistant from G must act through G. Hence the re-
sultant weight of the rod must act through G, which must
therefore be the centre of gravity.
86. To find the centre of gravity of a parallelogram.
Let A B C D be the parallelo-
gram ; E, K, F, H the middle points
of the sides.
The parallelogram may be di-
vided up into strips of material
parallel to A B and D C.
The centre of gravity of each of
these strips is at its middle point ;
but E F bisects all such strips ;
therefore the centre of gravity of
each strip lies in the line E F ;
hence the centre of gravity of the
parallelogram lies in E F.
Similarly, by dividing the parallelogram into strips parallel
to A D, the centre of gravity can be proved to lie in the line
H K ; therefore it must lie in G, the point where E F and
H K intersect. It can easily be proved by geometry that this
point is also the point in which the diagonals intersect.
Similarly the centre of gravity of a circle must be its centre ;
for it can be divided into parallel strips, of which any pair
equally distant from the centre are of equal length and of
equal mass.
So, also, the centre of gravity of a sphere must be its centre.
87. To find the centre of gravity of a uniform material
triangle.
Centre of- Gravity
79
Let A B C be the triangle, D the middle point of B C.
The triangle may be di- p^
vided into a number of
thin material strips parallel
to BC.
Let M N be one of these
strips ; then, since M N is
parallel to B C, it can be
proved that A D bisects
M N ; therefore the centre
of gravity of M N lies in
AD.
Fig. 42.
Hence the centre of gravity of every strip parallel toB C lies
in A D ; therefore the centre of gravity of the triangle lies in A D.
Similarly, if E be the middle point of A C, it can be proved
that the centre of gravity of the triangle must lie in B E.
Therefore the centre of gravity must lie at G, the point of
intersection of A D and B E.'
Note. — The straight lines joining the angular points of a
triangle to the middle points of the opposite sides all pass
through one point.
88. We have shown that the centre of gravity of a triangle
lies at the point of intersection of certain lines in the figure ;
it will be often more convenient to know its exact position in
one of these lines.
Join D E.
Then, since D E bisects
C A and C B, therefore D E is
parallel to AB and is equal
to i A B.
Again, since E D is parallel
to AB, the triangles GDE,
GAB are similar ;
. G D ^ G A
■■ DE AB'
But DE = iAB; .-. GD = iGA;
.-. G D = i A D.
' A thin sheet or plate of material is often called a lamina.
8o
Dynamics
Similarly, also, G E = ^ B E.
Hence, to find the centre of gravity of a triangular lamina,
join any angular point A to the middle point of the opposite
side D, and measure off a distance D G equal to ^ A D.
89. The ce7itre of gravity of a triangle coincides with the
centre of gravity of three equal particles placed at its angular
points.
Let A, B, C be the angular points of a triangle, and let
particles of equal weight
w be placed there. Bi-
sect AC at D. The
resultant of the equal
weights w at the points
A and C is 2 w, acting
at the point D.
The resultant of the
weight w at B and 2 w
at D is 3 W acting at a
C point O, such that
w. OB = 2W. OD;
i.e. O B = 2 O D.
Hence O coincides in position with the centre of gravity of
the triangular lamina.
We can therefore suppose that the weight w of a uniform
triangular table may be replaced by three weights, each equal
to \ w, at its angular points.
Hence we have the following result :
If a uniform horizontal triangular table rest on legs at its
angular points, the pressure on each
leg is the same.
90.* To find the centre of gravity
of a uniform wire bent into the shape
of a triangle.
Let ABC be the triangular
wire ; D, E, F the middle points
of the sides, and let the lengths of
Centre of Gravity
the sides be a, b, c. Then the lengths of E F, F D, D E
are k a, \b,\c respectively.
Let w be the weight of a unit of length of the wire ;
then the weight of B C = ?« a acting at D, its middle point.
Hence the weight of the wire may be replaced by the three
weights wa, wb, wc, acting at the points D, E, F respectively.
Now the centre of gravity of the weights wb at F, and w c
at F is at a point H in the line E F such that
z£/i5xEH=:z£Ji:xFH;
• EH ^ ■w_^c _ £_ED.
^■'^' F H w .b ' b F D '
therefore H D bisects the angle E D F (Euclid VI. 3), and the
centre of gravity of the wire must lie in this line.
Similarly it can be shown that it must lie in the line E K
which bisects the angle D E F, or in the bisector of the
angle DEE.
Therefore these three lines all pass through . one point,
which is the centre of gravity of the wire. This point is the
centre of the circle inscribed in the triangle D E F.
(Euclid IV. 4.)
91.* To find the centre of gravity oj a pyramid on a
triangular base}
Let A B C D be the pyra-
mid ; it may be considered as
made up of a large number
of triangular laminae parallel
to the base BCD.
Let b cdh^ one of these.
Bisect CD in E. Join
A E, cutting cd m. e ; e is the
D middle point oi cd.
Join B E and cut off E F
equal to |^ B E ; F is the centre
of gravity of the lamina BCD.
]o\nbe. The plane ABE
contains the line b e.
Join A F, cutting be'mf.
' Such a pyramid is usually called a ' tetrahedron.
S2 Dynamics
Then, by similar triangles,
-?/ = A/ _ /^
E F A F F B'
EF = iFB; .•.ef=\fb;
.: /is the centre of gravity of the lamina bed ;
.', the centre of gravity of the pyramid lies in the line A F.
Similarly, if H be the centre of gravity of the face A C D,
it can be proved that the centre of gravity of the pyramid lies
in the line B H. Now the plane ABE contains both the
lines A F and B H. Therefore the centre of gravity of the
pyramid lies at the point G, where A F and B H intersect.
Again, since E H = i E A, and E F = ^ E B,
.'. F H is parallel to B A and is = ^ B A ;
.•. the triangles G F H, GAB are similar ;
. GF_GA
■■ FH- AB"
But FH = AAB; .•.GF = ^GA;
.-. FG = i AF.
Similarly, H G = J H B.
Hence, to find the centre of gravity of a pyramid, join the
vertex A to F, the centre of gravity of the base, and measure a
distance F G = J F A.
As in the case of the triangle, we can prove that the centre
of gravity of a pyramid coincides with the centre of gravity
of four particles of equal mass placed at the angular points.
92.* We can now find the centre of gravity of a pyramid
on any base, polygonal or curvilinear, as follows :
Take any point within the base ; if the base is polygonal,
we can divide the base into as many triangles as it has
sides, by joining this point to the angular points of the
base ; if the base is curvilinear, we can imagine the curve to
Centre of Gravity 83
be made up of an infinite number of very short straight lines;
and we can then divide the base up into a very large number
of triangles with very small bases.
By joining the vertex of the pyramid to the point within
the base, and also to the other angular points of all these
triangles, we shall split up the pyramid into a number of
pyramids, all having a common vertex and triangular bases in
the same plane.
If we now draw a plane parallel to the base, and at a
distance from the base equal to one-fourth of the distance of
the vertex from the base, this plane will contain the centres
of gravity of each of these pyramids on triangular bases.
Hence the centre of gravity of the whole pyramid must lie
in this plane.
But by dividing the pyramid into thin laminae all parallel
to the base it can be shown, as before, that the centre of
gravity of each lamina lies in the line joining the vertex to the
centre of gravity of the base. Hence it must lie at the point
where this line cuts the plane which contains the centre
of gravity.
Therefore, to find the centre of gravity of a pyramid on
any base, or of a cone on a 'circular base or on any curvilinear
base, join the vertex to .the centre of gravity of the base, and
measure off from the base end a distance equal to one-fourth
of the length of this line.
93. We v/ill now apply the results of Article 62 to find the
centre of gravity of a system of particles in a straight line or in
the same plane.
(i) In a straight line.
Let zc,, zc;.,, z^j, &c., be the weights of the particles ; x-^, x^,
Xi, &c., their distances from a fixed point O in the line.
X the distance of their centre of gravity from O.
Then, ^ = ^(^-,4
(2) In the same plane.
G 2
84
Dynamics
y
B*
X\
Take any point O in the plane, and through O draw two
lines xO x', y O y' at right
angles to each other.
Then if w,, Wj, W3, &c.,
be the weights of the parti-
cles at the points A, B, C, Src,
x^, y^ the co-ordinates of A,
Xj, yi the co-ordinates of B,
and so on, and x, y the
co-ordinates of the centre of
gravity.
y
Fig. 47.
X :—
^{w)
y-
2H
These results might also be derived from the principle that
the moment of the resultant of any number of forces about a
point or about a straight line is equal to the sum of the
moments of its components.
Example I. — Masses of i lb., 2 lbs., 3 lbs., 4 lbs., and 5 lbs. are
placed in a straight line at distances i foot apart. Find their
centre of mass.
Take the position of the I lb. mass as the fixed point. Then,
with the usual notation,
+ 2 • 1+3 • 2 + 4 • 3 + 5 ■ 4
- I
x= -
n-2 + 3 + 4 + 5
2
IS 3 "%'
i.e. the centre of gravity is 2 ft. 8 ins. from O.
Example 2. — To find the distance of the centre of gravity of a
triangle from a point in its plane, the
co-ordinates of the angular points with
reference to rectangular axes through
the fixed point being (i, 2), (2, i),
(2, 2).
If W be the weight of the triangle,
it can be replaced by three weights,
, W
each = — , at its angular points.
Fig. 48.
Centre of Gravity 85
Then, with the usual notation,
W ^ W ^ W = W = 3"
+ + '^
3 3 3
So, also, y = 1 ;
"^ 3
therefore, if r be the distance from O,
S - 5 -
.-. r = - a/2 = g (2 a/2)
= 5. o C.
6
Example 3. — Masses 01 I lb., 2 lbs., 3 lbs., 4 lbs. are placed at
the angular points of a square. Find the distance of their centre
of mass from the I lb.
Let AB C D be the square, with the masses i, 2, 3, 4 at the
points A, B, C, D respectively.
Let a be the side of the square.
Take A as origin, A B as axis of ;ir, A D as axis of y.
Then the co-ordinates of A, B, C, D are {o, o), {a, 0), (a, a),
{o, a) respectively.
Then, \i x y be the coordinates of G, the centre of mass.
I
X = -
. + 2 .
I +
a + 3
2 + 3
. a +
+ 4"
A^ =
10
I
- I .
. (7 + 2 .
+ 3
. a +
4. a _
I? =
T a
I +
2 + 3
+ 4
10
10
r
be the distance of G from A
r' =
x' +J^
100
a' +
49^^ =
100
= 74„,.
100
'. r =
VTa
. a
94. If a body be made up of two parts whose centres of
gravity are known, the centre of gravity of the whole can be
found as follows :
86
Dynamics
Let G„ G2 be the centres of gravity of the two parts whose
weights are w^, w^ respectively.
Then, if G be the centre of gravity of the whole body, it is
the point at which the resultant of z£/, and w^ acts ; therefore,
by Art. 57, it must lie in the line joining G, G2 in such a
position that
w^ X G G, = ze'2 >^ G G2,
from which equation G G, or G G2 can be found.
If a body whose weight and centre of gravity are known has
a known portion removed from it, to find the centre of gravity of
the remainder.
Let W be the weight of the body, G its centre of gravity.
Let w be the weight and G]
the centre of gravity of the
part which is taken away ;
then G.;, the centre of
gravity of the remainder, must
lie in Gj G produced, at such
a distance from G that
Fig. 49.
(W-ze/)xGG2 = wxGG, ;
xGGi.
GG, = ..^-
\N-w
Example l. — A T-shaped figure is made up of two rectangular
pieces of cardboard, 5 x 2 sq. ins. and
4x3 sq. ins., as shown in the figure.
Find its centre of gravity.
Call the pieces A and B.
Let G and H be their centres of
gravity respectively.
Then G H bisects their coincident
sides in K.
Let O be the centre of gravity of
the figure.
Then weight of A x G O = weight of B x H O.
Now weight of A : weight of B : : area of A : area of B ;
:: 10 : 12
.-. 10 G O = 12 H O ; .-. 5 G O = 6 H O ;
.-. 500= 6(HG-G0); .-. iiGO= 6 H G.
5
A
f "^
K
\
^H
4
B
3
Fig. 50.
Centre of Gravity
87
Now G K = I inch and K H = 3 inches ; .'. H G = 3 inches ;
.'. II G = 18 ins. ;
G O = i/i in. ;
Example 2.— To find the centre of gravity of a quadrilateral of
which two sides are parallel and their lengths in a given ratio.
.e,t| = .
Join B D. Then
Fig. 51.
the areas of triangles of the same
altitude are proportional to their
bases ;
. AAB D ^ A B ^
■ ■ A B D C DC
Let sw = weight of the triangle
BDC.
Then ^rw = weight of the triangle A B D.
But the weight of the triangle can be replaced by three equal
weights at its angular points.
Hence, BDC can be replaced by iv at the points B, D, C,
and A B D „ rw „ A, B, D.
Therefore the weight of the quadrilateral can be replaced by
the weights,
rw at A, w + rw at B and at D, and w at C.
Hence the position of the centre of gravity can be found by
means of Article 93.
Example 3. — A lamina is made up of a square whose side is
2 inches and a triangle, on the same base but on opposite sides of
it ; find the height of
Dl ik ^^^^^^^^ the triangle in order
-Q ^1 ■ — • — \,^y^ 'hat the figure should
balance when this
common base is sup-
ported.
Let A B C D be
the square, ABE the
triangle whose alti-
tude E N is required.
Call it h inches.
^'=- 5^- Let G, H be the
centres of gravity of the square and the triangle respectively.
8S
Dynamics
The centre of gravity of the figure must lie in G H, and by the
question it must He in A B ; therefore it must He at the point O
where G H cuts A B ;
.•. weight of triangle x H O = weight of square x G O.
Now the weights of the triangle and the square are proportional
to their areas :
the area of the square = (A B)'^ = 4 sq. ins.,
and the area of the triangle = iAB.EN = ^.2.^=^sq. ins. ;
.-. y^ X H O = 4 X G O.
HO ^ HK .
GO G F'
•. /4xHK = 4xGF (i)'
HK ^ HF ^ i .
EN E F ^ '
H K = i E N = ^ y^, and G F = i in. ;
h
Now
Now
.-. h
= 4
/^■' = 3 X 4
:. h = 2 V3 ins
= \/s (the side of the square).
JVofe. — The equation (i) might have been obtained at once
from the consideration that, since the whole figure balanced about
A B, the moments of the weights of the two parts about A B must
be equal in magnitude.
Hence, weight of triangle x HK = weight of square x G F.
Example 4. — A square A B C D, whose centre is O, has the
portion A O B cut away. Find the centre of gravity of the re-
mainder.
Through O draw E F bisecting the sides
AB, CD.
Then the centre of gravity of A O B lies
at G, such that O G = f O E, and the centre
of gravity of the square lies at O ;
.•. the centre of gravity of remainder lies
at H in G O produced, such that
weight of part A O B x O G
= weight of remainder x O H.
A e: B
\
G /
\
"o
H
F
Fig. 53-
Centre of Gravity 89
Now the triangle A O B = J- (the square) ;
.•. the remainder = 3 (the triangle A O B)
.-. O G = 3 O H ;
.-. O H = i O G
= i(|OE)=|OE
= J (side of the square).
Examples.
1. Weights of 1 lb. and 5 lbs. are placed 8 ins. apart ; where is
their centre of gravity ?
2. Weights of W and 8 W are placed 5 ins. apart ; where is
their centre of gravity ?
3. Masses of i, i, i and 2 lbs. are placed at the angular points
of a square ; find their centre of mass.
4. Weights of 2, 2, 2 and i lb. are placed at the angular points
of a square ; find their centre of gravity.
5. Five equal weights are placed in a straight line at distances
apart of i, 2, 3, 4 ins. respectively. Show that the centre of gravity
lies at a point whose distances from the outside weights are in the
ratio 2 : 3.
6. Masses of 3 lbs., 5 lbs., 7 lbs., 9 lbs. are in a straight line
at intervals of a foot ; find their centre of mass.
7. If there are a number of heavy particles situated at fixed
points, and another heavy particle move along a given straight
line, prove that the centre of gravity of the whole system will also
move along a certain straight line.
8. A ■ triangular lamina is suspended from one of its angular
points, and rests with the opposite side horizontal ; show that the
triangle is isosceles.
9. O is the centre of gravity of a triangle ABC; forces are
represented by O A, O B, O C ; show that they are in equilibrium.
10. The sides A B, A C of a triangle ABC, right-angled at A,
are respectively 18 and 12 inches long. Find the distance of the
centre of gravity from C.
11. If the base of a triangle be fixed, and its vertex always lie
in a straight line, prove that the centre of gravity will also lie in
a certain straight line.
90 Dynamics
12. D, E, F are the middle points of the sides of a triangle
ABC; find the centre of gravity of the triangle D E F.
13. A B C D is a square plate, E and F being the middle points
of the sides A B and B C ; the plate is bent along E F, so that the
triangle E B F lies flat on the other part of the plate. Find the
centre of gravity.
14. A circular board has two circular holes cut in it, the centres
of these holes being at the middle points of two radii of the board
at right angles to each other. Find the centre of gravity of the
remainder if each of the pieces cut out be equal to one-ninth of the
board.
1 5. A circular board of radius a has a hole of radius b cut out of
it. Show that the centre of gravity of the remainder must lie
within a circle whose radius is ,.
a -v b
16. Two equal rods are fixed on a circular board so as to coin-
cide with the chords of two adjacent quadrants. If the weight of
each rod be equal to the weight of the board, find the centre of
gravity of the whole.
17. A triangular table of 3 lbs. weight, whose sides are 3, 4, and
5 ins., is supported by legs of equal length at its angular points.
Find the pressure on each leg.
18. If the centre of gravity of three equal particles placed on the
circumference of a circle is at the centre of the circle, prove that
the particles are equally distant from each other.
19. A B C is a uniform triangular plate of weight 3 w, weights of
5 7^, w and w are placed at A, B, C respectively. Find the point
about which the triangle will balance.
20. An isosceles triangle is suspended (i) from the vertex, (2)
from one of the equal angles. The angle between the two positions
of the base is 60°. Find the angles of the triangle.
21. A right-angled isosceles triangle is suspended (i) from the
right angle, (2) from one of the equal angles. Find the tangent of
the angle between two positions of any one of its sides.
22. A lamina in the shape of a right-angled triangle, such that
one of the sides containing the right angle is three times the other,
is suspended freely from the right angle ; prove that in equilibrium
the hypothenuse makes an angle sin"' - with the vertical.
Centre of Gravity 91
23. A rectangle is divided into four parts by lines joining the
middle points of opposite sides ; one of them is cut away ; find the
centre of gravity of the remainder.
24. A B C D is a square, O its centre, E, F the middle points of
A B, A D ; A E F is cut away ; find G, the centre of gravity of the
remainder.
25. An equilateral triangle and a square are described on oppo-
site sides of the same base. Find the centre of gravity of the
figure so formed.
26. Two squares, of which one is four times the other, are
placed so that the sides about an angular point of the one are
co-linear with those about an angular point of the other. Find the
centre of gravity of the figure so formed.
27. A quarter of a square is cut away by lines drawn from the
centre to two adjacent corners. Find the centre of gravity of the
remainder.
28. In a quadrilateral A B C D the sides A B, A D are 15 ins.,
and B C . C D are 20 ins. If B D = 24 ins., find the distance of
the centre of gravity of the quadrilateral from A.
29. A figure is made up of a square and an isosceles triangle on
opposite sides of the same base. Find the relation between the
altitude and base of the triangle in order that the centre of gravity
of the figure should lie in the common base.
30. Masses of i lb., 2 lbs., and 3 lbs. are placed at the angular
points A, B, C of a triangle. Find their centre of gravity, and show
that it lies on the line joining the middle points of C A, C B.
31. Masses of 2, 6, 4, 5, 3, 7 lbs. are placed at the angular
points of a hexagon taken in order. Show that their centre of
gravity is at the centre of the hexagon.
32. Particles of 2 lbs., i lb., 2 lbs,, 3 lbs. are placed at A, B, C, D
respectively, the angular points of a square. Find the distance
of the centre of gravity from the centre O.
33. Squares are described on the three sides of an isosceles
right-angled triangle, outside the triangle. Find the centre of
gravity of the figure so formed.
34. A B C is a triangle and A' a point inside it. Find the posi-
tion of A' when it is the centre of gravity of the area which lies
between the two triangles A B C, A' B C.
92 Dynamics
35. Where must a circular hole of i ft. radius be punched out of
a circular disc of 3 ft. radius, so that the centre of gravity of the
remainder may be 2 ins. from the centre of the disc ?
36. How would you place a brick, whose length, breadth, and
thickness were all different, on a rough inclined plane so that it
would be least likely to tumble over ?
37. A brick whose dimensions are 8x4x3 ins. rests on a
rough plane in such a way that it cannot slip, and the plane is
tilted about a line parallel to one edge of the brick. Find the
greatest and least angles of inclination for which the brick will just
not upset.
38. Masses of 3 lbs., 5 lbs., 7 lbs., 9 lbs. are at the angular
points of a square A B C D. Find their centre of mass.
39. A square lamina A B C D is suspended from A ; weights of
4 lbs., 5 lbs., 16 lbs. are suspended -from B, C, D respectively, and
it hangs so that A is vertically above the middle point of C D.
Find the weight of the square.
40. A uniform rod A B, 8 lbs. weight and 20 ins. long, is sus-
pended horizontally by two vertical strings attached to the points
B and C, C being distant 4 ins. from A. Find the greatest weight
which can be attached to the end A without disturbing the equi-
librium of the rod.
41. G is the centre of gravity of a triangle A B C ; a hne is
drawn though G parallel to B C cutting A B, AC in P and Q.
Show that the centre of gravity of P B C Q divides G D in the ratio
of 8 : 7, D being the middle point of B C.
42. The sides of a quadrilateral lamina taken in order are 3, 5,
4, 10 respectively ; the side 5 is parallel to 10. Find the distance
of the centre of gravity of the quadrilateral from each of the sides
. 3 and 4.
43. A heavy body is suspended by a number of strings, all
attached to a fixed point. Show that the centre of gravity must be
vertically beneath this point.
44. A flat triangular board ABC, right-angled at A, stands with
its plane vertical and its side A C on a hori;^ontal plane ; D is the
middle point of A C. If the portion B A D be cut away, will' the
remainder stand or fall ?
45. The wheels of a hay-cart are lo feet apart, and the centre of
gravity of the cart and load is 12 feet above the ground and mid-
Centre of Gravity 93
way between the wheels. How much could either wheel be raised
without the cart falling over ?
46. A B C D is a square standing on C D as base in a vertical
plane, E is a point in C D, and the triangle A C E is cut away.
Find the least length of D E in order that the remainder should
not fall.
47. The radius of the base of a cone is to the altitude as
2:15, the cone is placed on its base on a smooth inclined plane,
and is kept from slipping by a string fastened to a point in the
plane and to the rim of the base. Find the greatest ratio of the
height of the plane to its base which is consistent with equilibrium.
48. A brick is laid with a quarter of its length projecting over
the edge of a wall ; a brick and a quarter brick are laid on the
first with a quarter of its length over the edge of the first brick ;
a brick and a half laid on this, and so on. Prove that four such
courses can be laid in the above manner, but that if a fifth course
be added the mass will topple over.
49. A B C D is a rectangle, E the middle point of C D ; the
triangle A D E is cut away. Find the centre of gravity of the re-
mainder.
50. The lengths of the sides of an isosceles triangle are 2 a, 2 a,
2 iJ ; on each side an equilateral triangle is described. Find the
distance of the centre of gravity of the whole figure from the side
zb.
Verify your result by applying it to the case of an equilateral
triangle.
94 Dynamics
CHAPTER VII
GENERAL CONDITIONS OF EQUILIBRIUM
95. Two forces cannot be in equilibrium unless they are
equal and opposite in the same straight line. Otherwise, if
they act in the same plane, they have a single resultant or form
a couple.
If they do not act in the same plane they cannot be in
equilibrium, nor can they be replaced by a single force or a
couple.
Consequently, if three forces acting on a body are in equi-
librium, they must all act in the same plane, for otherwise it
■would be impossible for the third force to balance the resultant
of the other two.
96. If three forces acting on a body in one plane are in
equilibrium, they must all be parallel or they must all pass
through one poiat,
For consider any two of them— their lines of action must
be parallel or they must meet in a point.
If they are parallel, they must have a resultant which is
parallel to them ; and, since there is equilibrium, the third
force must be equal and opposite to this resultant — that is, it
must be parallel to them.
If the two are not parallel, produce their lines of action to
meet at a point ; then they have a resultant which passes
through this point ; and, since there is equilibrium, the third
force must be equal and opposite to this resultant ; that is, it
must pass through the point.
Note. — If there is equilibrium, no two of the three forces
can form a couple.
General Conditions of Equilibrium 95
97. To find the conditions oj equilibrium of any number of
forces which act on a body at one point.
The algebraical sum of the resolved parts of the forces in
any two directions through the point must each be zero,
(^^t. 55.)
It is not necessary that these directions should be at right
angles to each other, but it is generally more convenient to
take them so.
This condition is sufficient for such a system of forces. For
since all the forces pass through one point they cannot reduce
to a couple ; hence, if the resultant vanishes, the forces must
be in equilibrium.
98. To find the conditions of equilibrium of any number of
co-planar forces which act on a rigid body.
(i) If the forces are parallel.
The algebraical sum of the forces must be zero, and the
algebraical sum of the moments of the forces about any point
in the plane must also be zero. (Art. 78.)
(2) If the forces are not parallel.
The algebraical sum of the resolved parts of the forces in
two directions at right angles to each other must each be
zero, and the algebraical sum of the moments of the forces
about any point in the plane must also be zero. (Art. 79.)
99. The results of the previous articles might also have been
obtained from Article 63.
We then proved that, if there is not equilibrium, the system
reduces to a single force or a couple.
Now in part (2) of the last article, if the first condition
holds, the resultant vanishes ; that is, the system cannot reduce'
to a single force.
If the second condition holds, there is no rotation ; that
is, the system cannot reduce to a couple. Therefore, if both
conditions hold, the system must be in equilibrium.
100. A system of coplanar forces will be in equilibrium if the
algebraical sum of the moments of the forces about any three
points not in the same straight line vanishes in each case.
Suppose A, B, C to be the three points.
g6 Dynamics
We have seen that if the sum of the moments of any
number of co-planar forces about any point in their plane
vanishes, the forces must be in equilibrium, or the point must
be in the line of action of their resultant. (Art. 71.)
Hence, in the above case, if the forces are not in equili-
brium, A must be a point in the direction of their resultant.
Similarly, B and C must be points in the direction of the
resultant.
This is impossible unless A, B, C are in the same straight
line. Hence, if they are not in the same straight line, the
system of forces must be in equilibrium.
loi. If one point of a rigid body be fixed, the forces will
be in equilibrium if the algebraical sum of the moments of the
forces about the fixed point vanishes.
This is often found to be an extremely useful condition,
since in many cases the reaction of the fixed point is unknown,
and if we consider the rotation about that point we shall get a
relation between the other forces excluding this unknown
reaction.
In solving problems it will be found that when three forces
acting on a body are in equilibrium the position of equilibrium
will be determined by Art. 96.
When more than three forces act on a body, and are in
equilibrium, we apply the conditions laid down in Art. 98 ; of
these sometimes one set of conditions is sufficient to solve the
problem without the other.
The latter condition is usually the most useful of the two,
provided that sufficient care is taken in choosing the points
about which we take moments.
For example, if four forces are in equilibrium and we take
moments about the point where two of them intersect, we shall
get a simple relation between the other two.
Note.—T\A expression take moments about a point 1% a phrase
meaning, express the fact that the algebraical sum of the moments
of the forces about that point is zero.
102. In certain cases of equilibrium we shall often find it
Pt
General Conditions of Equilibrium 97
convenient to consider the rotation of the body about a certain
fixed hne in it.
For example, if two points of a body be fixed, the body
can only have one motion— namely, a rotation about the line
joining these two points.
This line is then called the axis of rotation.
We shall only consider cases in which the forces which act
upon the body are perpendicular to this fixed axis ; in which
cases the turning powers of the forces about the axis are
measured by the product of the force and the perpendicular
distance between its line of action and the fixed axis.
It is easy to see that this moment represents the total
effect of a force on a body which is fixed in this manner. For
let A B be the fixed axis, O C the
perpendicular distance between
A B and the line of action of P ;
at C apply two equal and opposite
forces P, P parallel to the direction
of P. Then the original force P
acting at O may be replaced by a O
single force P acting at C and a ^'°- 54-
couple whose moment is P . O C ; of these the single force
has no effect on the body since it acts through the fixed axis,
and we have left the couple only. Hence, the only effect of P
is a rotation which is measured by the product P . O C.
103. Before proceeding to work examples of forces in equi-
librium, it will be useful here to consider in what manner
various forces act.
In Article 6 we mentioned that the forces which affect a
body's state of equilibrium must arise from some external
cause or agency which produces a force of the nature of a
push or a pull or an attraction. In all cases a force is one part
of a mutual or reciprocal action, to which we give the name
oi stress. A push or pressure may be due (i) to the action
of a rod on the body, (2) to the action of another body in
contact with it, or (3) to the action of a hinge about which the
body can turn.
In all cases of smooth bodies this mutual reaction must be
H
98
Dynamics
along the common normal to the surfaces ; for if there were
any tangential action there could not be equilibrium, since
there is no tangential force to counteract it.
In the case of a rod pushing a body, the pressure must act
along the normal to the surface at the point of contact.
In the case of two surfaces in contact, the mutual pressure
will be along the common normal at the point of contact of
the two surfaces.
Fig. 55.
Fia. 50.
The direction of the reaction of a hinge will often be
apparent from the conditions of the question ; but if it is not
we must assume the magnitude and direction
•fY of this unknown reaction, or assume the mag-
nitude of its components in two given directions
_^ at right angles, and find the value of these
X unknown components from the conditions
given in the problem.
Then, with the usual notation, R^ = X" + Y^,
Y
X
and tan 6 ■
A tension may be due to the action of a string or rod on
the body.
In this case every particle aboi the string or rod is kept in
equilibrium by two equal and opposite forces /, t, which are in
equilibrium, and are the
Jb a, hi; halves of stresses whose
"^ M-H- H^-« > other halves act on the
^"'- 58. next particle, and so on.
General Conditions of Equilibrium
99
Fig. 59;
Hence the tension of the string is the same throughout
its length.
Further, the tension of a string is not changed by passing
the string over any smooth surface, such as a peg or pulley.
For if we consider any small particle of the string, it is kept in
equilibrium by the tensions at its ends and the reaction of the
surface along the normal.
Since the two tensions at the ex-
tremities of the particle make equal
angles with the normal, we see, by
resolving them along and J^' to this
normal, that they must be equal.
The force of attraction between two bodies always acts
along the hne joining their centres of mass. The particular
case of gravity we have fully considered in the previous
chapter.
Note. — The student has several times been warned against
confounding mass with weight, and he has been told that i lb.
is a mass and not a weight ; yet for the sake of brevity we do
use the expressions weight of 6 lbs., or 6 lb. weight ; but
we always mean by such expressions, a weight equal to the
weight of the mass 6 lbs., or the weight of the mass 6 lbs.
Example I. — A uniform rod whose weight is W is movable
about a hinge at one end ; it is kept in equilibrium in a position
making an angle of 30° with the horizontal by a force making an
angle of 30° with the rod at its other end.
Determine the reaction of the
hinge and the direction of its line of
action.
Let A B be the rod, hinged at
B ; its weight W acts vertically down-
wards through the middle point C.
Let F be the force at A, R the
reaction of the hinge.
Let F and W meet at O ; then ,
smce there is equilibrium between
the forces F, W, and R, R must also
pass through O.
H 2
lOO
Dynamics
It is obvious from the geometry of tlie figure that since
OAC = 30'andCAH = 30° .-. alsoCOA= 3o°andOCB
Also CO = CA=CB; .-. COB = CBO = 6o°;
.". R makes an angle of 60° with the rod.
To find R, we have by Lami's theorem, Art. 51,
_R_ ^ F^ ^ JW_ .
sin 150° sin 120° sin 90° '
. R W
60°.
.-. R = A W.
This result could also be obtained by taking moments about A.
Example 2. — A ladder is placed with one end on a smooth hori-
zontal plane and its other end against a smooth vertical wall ;
show that it must slip down. If the ladder be uniform and of
weight W, and is kept at rest at an angle 6 with the wall by a man
putting his foot against the lowest rung so as to apply a horizontal
pressure at that point, find this pressure.
Let A B be the ladder. It is acted on by three forces — its
weight W, and the pressures of the
ground and the wall ^ to the ground
and wall respectively.
Call these pressures P and R.
Now P, R, and W cannot be in
equilibrium because P and W are
parallel and R is x^ to them ; there-
fore the ladder must slip down.
Let F be the horizontal force
applied by the man.
Resolve the forces horizontally and
vertically, then we have
P = W;
F = R.
Take moments about B, and let 2 / = length of the ladder :
R . 2 / cos ^ = W . / sin 5 ;
W . tang . . Y = WjanJ
2 ' • • 2 ■
Note I. — If R and W meet in O the resultant of P and F must
pass through O.
R =
General Conditions of Equilibrium
lOI
Note 2. — The four forces P, W, R and F form two couples which
are equal in magnitude and opposite in direction.
Example 3. — A sphere of weight W rests in contact with two
inclined planes whose inclinations to
the horizon are a,'/3. Find the pres-
sures on the planes.
Let R, R' be the pressures at A
and B the points of contact of the
sphere with the planes.
Then R, R' must be J.' to the
planes, and make angles a, S with the
vertical, and the three forces R, R'
and W all pass through the centre of
the sphere.
.•. by Lami's theorem,
Fig. 62.
R : R' : W : : sin /3 : sin a : sin (a + /3) ;
sin S __ J -n/ sin a
R
W sin (a + /3)
and R' =
W sin (a + /3)
Example 4. — A uniform beam of weight W is placed with one
end on a smooth horizontal plane, and the other end rests on
another plane inclined at an angle a to the horizon ; to this end a
string is fastened which passes over a pulley at the top of the
plane and supports a
weigh TV hanging freely.
Show that the beam
will rest in any position
if2w=Wsina. "^^
Let A B be the beam,
■R, R' the pressures of
the planes on the ends
A and B respectively.
Resolve vertically and
horizontally p.,^^ ^^
W = R + R' cos a + w sin a . . . . (l)
R' sin a = w cos a . ' (2)
Take moments about B :
W = 2 R •'• (3)
>l'W
I02
Dynamics
Eliminate R and R' by substituting their values from equations
(2) and (3) in equation (i) ;
W = — + w . ^^i-^ . cos a + w sin a :
W
- . — + sm a 1 = -; — ;
\ sm a / sm a
2 •je' = W sin a.
Example 5. — A uniform rod A B, 18 feet long and of 20 lbs.
weight, is hinged at A, and has a weight of 5 lbs. suspended from B.
It is kept at rest by a string 13 ft. long, one end of which is
attached to a point D on the rod 13 ft. from A, and the other end
to a point O 10 ft. vertically above A. Find the magnitude and
direction of the reaction cf the hinge.
~^^^^^:::^:;-^^^
E
Let 6 be the angle
the rod makes with
the horizontal.
Draw DE ±r to
OA.
20"!
f
. A
i
Then, since D
"^^ = D A, D E bisects
the angle D A ;
Fig. 64.
.-. ODE = EDA = ^.
If A F is Xr to
OD, OAF = e.
Let X and Y be the horizontal and vertical component.'; of the
reaction of the hinge at A, and T the tension of the string.
Resolve vertically and horizontally
T sin 5 + Y = 20 + 5 (i)
T cos 5 + X = O (2)
Take moments about A :
T . A F = 20. A C cos 5 + 5 . A B cos (9.
But A F = O A cos fl ;
.-. T . O A = 20. A C + 5 . A B ;
.'. 10 T = 20 X 9 + 5 X 18
= 270;
/. T = 27 lbs. wt.
sin 5 =
OE 5
OD 12
.-.
cos e =
12
13'
Y
= 25-
-Tsinfl
= 25-
-A'
= 14- lb
13
X
= -Tcos5
— ;-3
lbs. wt,
General Conditions of Equilibrium 103
Now
From (i)
Y = 25-Tsinfl
^ o
lbs. wt.
13 13
From (2)
which shows that X acts to the left, and not to the right, as
drawn ;
.'. if R be the reaction of the hinge,
R^ = X'HY2;
.-. R=375:i =28-9 lbs. wt.
13
If ^ be the angle R makes with the axis of x,
tan,^=Y^_^o^_
X 324
We find from the tables that ^ lies between 149° and 150° ;
that is, it makes an angle of about 60° to the left of A O.
Example 6*. — A smooth uniform rod whose length is 2 / and
weight W is at rest in a smooth hemispherical bowl whose radius
is r, which is fixed with its axis vertical, one end of the rod project-
ing over the edge of the bowl. Find the conditions of equilibrium
and the pressures of the bowl on the rod.
Since there are only three forces acting on the rod, they must
all act in one plane ; hence the rod must rest in the vertical plane
through the centre O of the bowl.
Let the figure represent the section of the bowl and rod by this
plane. Let A B be the rod resting against the bowl at the points
A and D.
Let R, R' be the pressures at A and B respectively.
Then, since the pressures of smooth surfaces must be normal to
the surfaces, R is perpendicular to the tangent at A, and must
therefore pass through the centre O ; R' is perpendicular to the
rod A B, ...
I04
Dynamics
Let R, R' meet in E ; then, since E D A is a right angle, A E
must be a diameter of the circle ; therefore E hes on the circum-
ference of the completed circle. Since R, R' and W arc in cqif
librium they must all pass through one point ; therefore C,
centre of gravity of the rod, must be vertically below E.
the
Fig. 65.
Y W
The above considerations give us the position of equilibrium.
Let 5 = the angle of inclination of the rod to the horizon.
From the geometry of the figure we get the following :
since O D A = ^, O A D = 5, and .-. E O D = 2 ^.
If AF be drawn perpendicular to the line of action of W,
F must lie in the circle A E D, since A F E is a right angle ; and
since A F is || to O D, D A F = i9.
Resolve the forces along the rod and at right angles to the rod.
R cos i9 = W sin fl (i)
R' + R sin 5 = W cos ^ (2)
Taking moments about A, wc have
WxAF = R'xAD;
.'. W X / cos 5 = R' X 2 r cos ;
.-. W./=2R'.r (3)
General Conditions of Equilibrium 105
From (i)
R = Wtan5 (4)
substitute for R in (2) ;
.-. R' = Wcosfl-Wtan5. sin5 ... (5)
R' ^„ a sin'^ cos'(9-sin^
~ cos a —
But from (3)
W cos 6 cos 6
W 2/'
I cos'' 5 -sin- 5 2 cos'' 5-
(6)
" 2r cos fl cos 6
This is a quadratic for cos 6, from which we have
l±j/y2rf+V
cos fl =
8 r
Since 5 is less than 90°
.". cos 6 is positive ;
.". the upper sign only of the root need be taken ;
■•■COSg = ^^^32^'W^,
8r
This equation gives us the position of equilibrium of the rod
when / and r are known.
The angle E O D cannot be greater than 90° ;
.•. 6 cannot be greater than 45° ;
.■. 6 must lie between 0° and 45°.
R and R' can be found from equations (4) and (5).
Case I. If 6 = 0,
R = Wtan5 = o,
and R' = W S^^^'I^ = w,
cos e '
and l=ir.
Hence C and D coincide; that is, the rod balances on the point D.
Case 2. It /= ^ -, COS ^ = ^ ;
■v/3 2
.-. ^ = 30°;
.". R = — and R = — .
V3 -v/3
io6 Dynamics
Case 3. The least length the rod can have s when the extremity
B just rests on the edge of the bowl.
- In this case
a AD 2/ /
cos5=-r-^ = — = ~.
A E ■zr r
Equating this with the value already obtained, we have
/_V2
when
r ^
In this position cos 6 = a /-•
This gives the greatest value that 6 can have, and in this case
6 Ues between 35° and 36°.
If the rod be any shorter B will fall within the bowl, and the rod
will slip down into a horizontal position at the bottom of the bowl.
Hence, provided the length of the rod is less than /\,r and
2 A/2
greater than — ^^ . r, there is always one position, and one posi-
a/3
tion only, in which the rod will rest, which position is given by
the equation,
^ a / + -v/32 r- + P
cos 6 = —-^ .
8r
Note. — By Lami's theorem we get
R : R' : W : : sin (180° - 6) : sin (90° + 26): sin (90° - 6)
: : sin 5 : cos 2 6 : cos 6 ;
which relation includes equations (4) and (5). Equation (5) might
have been obtained from the geometry of the figure instead of from
equations (3) and (5).
„, AC _ / _ si n A E C _ cos 2 6
A E ~ 2 r ~ sin A C E ~ cos 5 ■
Case 3 could be obtained by putting I = r cos 6, since A B can-
not be less than A D, in which case equation (6) becomes
r cos 6 _ 2 cos''^ — I
2 r cos 6.
:. cos"^ 5 = 4 cos' 6-7.
or cos' 6 = 1-
General Conditions of Equilibrium \0^
Examples.
1. A uniform rod is suspended from a peg by two strings
attached to each end, and of such a length that the angles between
them and the rod are 30° and 60°. Find the tensions of the strings,
the rod being i lb. weight.
2. A lamina in the shape of a rectangle, whose length is double
its breadth, is suspended freely from a corner. Prove that in the
position of equilibrium the diagonal makes with the vertical an
angle sin - .
3. Two rods O A, O B of equal weight, whose lengths are a
and b, are rigidly connected at O, so that the angle A O B is a right
angle. The point O is attached to a string ; if 5 be the inclination
of O A to the horizon, find sin 6.
4. In the preceding question if the rods be of equal length but
their weights in the ratio in '. n, show that tan 6 = - .
n
5. A uniform wire is bent into two arms inclined at an angle of
120°, one arm being twice the length of the other. The wire is
suspended from its angular point ; find the position of equilibrium.
6. A uniform rod of length 2 / is in equilibrium in a vertical
plane. One end of the rod rests against a vertical wall, and the
rod also rests on a prop whose distance from the wall is a. Find
the angle the rod makes with the vertical.
7. A heavy uniform beam of weight W, resting on a smooth
horizontal plane and against a smooth vertical wall, is held by a
horizontal cord one-fourth the length of the beam attached to its
middle point and to the wall. Find the tension of the cord and
the pressures on the planes.
8. A ring of weight W slides freely on a string of length a ^2
whose ends are fastened to two points at a distance a apart in a
line making an angle of 45° with the horizon. Find the tension of
the string in the position of equilibrium.
9. A uniform rod has a 5 lb. weight fixed at one end, and balances
about a fulcrum 7 ins. from its middle point. If the 5 lb. weight be
replaced by a 10 lb. weight the fulcrum must be moved i inch.
Find the length of the rod and its weight.
10. A string is tied to two points. A ring weight W can slip
freely along the string, and is pulled by a horizontal force P. If
io8 Dynamics
the parts of the string when in equilibrium are inclined at 90° and
45° to the horizon, find the value of P.
ri. A smooth horizontal bar is placed at a distance of r foot
from a smooth vertical wall and parallel to it. A uniform rod is
supported by the bar and the wall and is inclined at an angle of
30° to the vertical ; show that its length is 16 feet.
12. A uniform rod i foot long and 3 lbs. weight has one end
on a smooth table ; a string is attached to a point 3 ins. from the
other end, and is fastened to the ceiling so that the rod makes
an angle with the table. Find the tension of the string.
13. One end of a uniform ladder 84 lbs. weight rests against a
smooth vertical wall at a height 12 feet above the ground, and
other end rests on the ground at a distance 10 feet from the wall.
Find the pressure on the ground.
14. A uniform beam of mass i ton is suspended in a hori-
zontal position by two ropes attached to its ends ; one of the ropes,
of the same length as the beam, is attached to a peg ; the other
rope passes over a pulley and is attached to a weight W ; the
pulley is fixed in the same horizontal line as the peg, and at a
distance from it equal to twice the length of the beam. Find W.
15. The two legs of a light step ladder are connected by a
smooth joint at the top and a cord at the bottom. The ladder
stands on a smooth floor with one leg, which is 3 feet long,
vertical. A man of 1 1 st. weight stands on the other leg at a height
of 2 feet above the ground. Find the pressure on the vertical
leg. What is the tension of the cord?
16. A rod 5 feet long is inclined to the vertical at an angle of
45°, and stands on the ground resting against a smooth rail with
I foot of its length projecting over the rail. Show by a figure the
direction of the ground's action, and find the magnitude of this
action if the mass of the rod be 6 lbs.
17. A smooth sphere of radius a and weight W is supported on
a smooth plane at an angle of 30° to the horizon by a string, one
end of which is fastened to a point in the plane and the other end
to the surface of the sphere ; in the position of equilibrium the
string is horizontal. Find the length of the string and the pressure
on the plane.
18. A heavy ladder is sustained at an inclination 6 to the hori-
zon by means of a force applied at the lowest round of the ladder
General Conditions of Equilibrium 109
and by another force perpendicular to the ladder at a point half-
way between the lowest round and the C.G. of the ladder. Find
the direction of the first force, and compare the forces with the
weight.
19. A heavy uniform beam P Q of length 2 / has one end
connected by means of a smooth ring with a vertical rod P A C, and
rests on a vertical circle whose centre is C and whose radius
C A = a. Find an equation for 6, the inclination of P Q to the
horizon in the position of equilibrium.
no Dynamics
CHAPTER VIII
MACHINES
04. A Machine is an instrument by means of which a
force, which is applied at one point, is made available to do
work or overcome resistance at some other point.
The simplest forms of machines are called the Mechanical
Powers, and may be classified as follows : —
1. The Lever, which includes
id) The Steelyards,
yb) The Common Balance,
\c) The Wheel and Axle.
2. The Pulley and Systems of Pulleys.
3. The Inclined Plane, which includes
(a) The Screw,
\b) The Wedge.
All machines consist of one or other of the above mechani-
cal powers, or of combinations of them.
In the present chapter we shall consider only the simple
machines, and find the relation between the forces which act
upon them when they are at rest ; we shall also suppose that
the various parts of the machines are perfectly smooth.
The primary object of a machine is to make a force over-
come a resistance greater than itself.
The external force which we apply to the machine is called
the Power, and is denoted by the letter P ; the resistance
which this force can be made to balance is called the Weight,
and is denoted by the letter W.
Machines
I II
The ratio -^ is called the modulus or working power of
the machine, and is the measure of its efS.ciency.
W
If -p- IS > unity— that is, if W is > P— the machine is
said to have mechanical advantage ; but if — is < unity, the
machine is said to have mechanical disadvantage.
It will be our object now to determine the efficiency of each
of the simple machines when the power and the weight are in
equilibrium.
The Lever.
105. The lever is a rigid rod capable of turnmg aoout a
fixed point called the fulcrum. A lever may be straight cr
bent.
The power and the weight may act on opposite sides of
the fulcrum or on the same side. Those portions of the lever
between the fulcrum and the points of application of the
power and the weight are called the arms of the lever.
Levers are divided into three classes, according to the
relative positions of the fulcrum and the points of application
of the power and the weight.
In levers of the First Class the power and the weight act
on opposite sides of the fulcrum.
In the Second Class the power and the weight act on the
same side of the fulcrum, but the power is further from the
fulcrum than the weight.
In the Third Class the power and the weight are on the
same side of the fulcrum, but the power is nearer to the
fulcrum than the weight.
We will first consider the case of a straight lever without
weight, in which the power and the weight are parallel and at
right angles to the lever.
112
Dynamics
Let A B be the lever, F the fulcrum, P the power, W the
■weight, R the reaction of the fulcrum.
w
Fig. 66.
_F
■a
w
w
Fig. 67.
Fig. 68.
Fig. 66 represents the^fw/ class of levers, fig. 67 the second
class, and fig. 68 the third class. In each case P and W tend to
turn the lever in opposite directions round the fulcrum C, and
the condition of equilibrium is obtained from the fact' that the
algebraical sum of the moments of P and W about F must
vanish.
Hence P x F A = AV x F B ;
W^FA
■ P F B'
efficiency :
In fig. 66 R = P + W,
„ 6.7 R = W-P,
„ 68 R = P-W.
The lever has mechanical advantage or disadvantage,
according as F A is > or < F B.
In the first class there will be mechanical advantage or
disadvantage according as the fulcrum F is nearer to the
weight or to the power.
Machines
ir3
In the second class there will always be a mechanical
advantage.
In the third class there will never be a mechanical
advantage.
Examples of the first class of lever are the common balance,
the steelyard, a poker, a see-saw.
A pair of scissors is a double lever of this class.
Examples of the second class area wheelbarrow, an oar (the
blade of the oar in the water being the fulcrum), a tobacco-cutter.
A pair of nutcrackers is a double lever of this class.
Examples of the third class, which is rarely made use of,
are : a catapult, the forearm of a man with a weight in his
hand (the elbow being the fulcrum).
A pair of tongs is a double lever of this class.
If the weight of the lever be w acting at the centre of
gravity G, we must also take into account the moment
a' X FG.
Thus, in fig. 66, if G is to the left of F,
P . F A = W . F B - ze/ . F G,
R = P + WfZ£/.
In figs. 67 and 68
P.FA = W.FB+ze;.FG,
R = (W + ze')~P.
1 06. If the lever be bent and the forces P and W are not
parallel nor at right angles to the
arms, the condition ofequihbrium
is still obtained by taking mo-
ments about the fulcrum.
Draw F A, F B perpendicular
to the directions of P and ^V
respectively. Then
P . F A = W . F B.
In this case it is obvious that
R, the reaction of the fulcrum,
must pass through O, the point
Fig. 69. where the directions of P and W
meet, since R, P, and W are in equilibrium.
I
114
Dynamics
P^^
107, If a heavy straight lever with weights attached to its
extremities be in equilibrium in any one position, it will also be
in equilibrium in any
other.
Let 6 be the angle
which A B makes with
the horizontal.
Through C draw the
common perpendicular
to the directions of the
forces, cutting their lines of action in a, b and g.
Then, taking moments about C,
P.Cfl + w.C^=W.C^;
.-. P . C A cos e + z£^ C G cos e = W . C B cos e;
.-. P. CA + w. CG = W. CB,
which is independent of Q, and is therefore true for all values
oid.
Hence, the lever will be in equilibrium in any position
Example. — A uniform rod of 10 lbs. weight is hinged at one end
and has a lolb. weight suspended from the other end. Itis supported
in a horizontal position
by a string making an
angle of 60° witlrthe
rod. Find the point to
which the string must
be attached when the
tension of the string is
= 40 lbs. weight.
Let A B be the rod
hinged at A, C its middle
point ; let D be the
point to which the string is attached. Draw AE ±"- Id the
direction of the string, and take moments about A :
40 X A E = 20 X A B + 20 X A C,
40 X A D sin 60" = 20. A B + 10 A B,
40
2
A D = 30 A B ;
Machines
IIS
AD =
30
20v/3
a/3
AB
AB.
The principle of the lever is the principle which we use in
the various machines for weighing heavy substances, such as the
Common or Roman Steelyard, the Danish Steelyard, and the
Common Balance.
The Common or Roman Steelyard.
108. This steelyard consists of a straight lever of the first
class. One end of the lever is usually considerably heavier
than the other, and the fulcrum or point of suspension is nearer
to this heavy end.
The substance whose weight is required is attached to the
extremity of the short arm by means of a hook, or by placing
it upon a platform suspended from this end. Its weight is
ascertained by sliding a movable weight along the longer arm
until there is equilibrium, and noting the distance of the
movable weight from the fulcrum when this is the case.
We will now proceed to graduate the steelyard, so that, by
noting the graduation at which the movable weight rests, we
should at once know the weight which is being supported.
Let A B be the steelyard, C the point of suspension. Let
W be the weight to be weighed, which is attached to the end
A ; P the movable weight attached to the ring at D. Let
Q be the weight of the whole beam acting through G the
centre of gravity.
/
vD
S
ly
Fig. 72.
Now express the fact that the algebraical sum of the
moments of the forces about C must be zero —
.-. W.AC = P.DC + Q.GC.
12
Ii5 Dynamics
If G coincided with C, the moment Q . G C would vanish, and
the equation of equiUbrium would become
W. A C = P . D C,
W ^ DC
P AC
Hence W could be found in terms of P by measuring C D in
terms of C A.
But if G does not coincide with C, mark off a distance C O
along the rod such that the moment Q . G C = the moment
P. CO.
This can always be done, because P, Q, and C G are known
quantities in any particular steelyard, and hence the distance C O
must be known also.
Then the equation of equilibrium will become
W. AC = P.DC + P.CO
= P (D C + C O)
= P . D O.
TT W DO
Hence ^ = ^^
IfDO = AC, W = P;
„D0 = 2AC, W = 2P;
„ D O = ;^ . A C, W = « P.
Hence O is taken as the zero point of graduation, and dis-
tances are marked off from O equal to A C, 2 A C, 3 A C, &c. ;
then, as P coincides with the ist, 2nd, 3rd, &c., of these gradua-
tions, W = P, 2 P, or 3 P, &c.
Note. — The student may find some difficulty in the algebraical
treatment of the mechanical equation of equilibrium, where we
change the sum of the moments P . D C + P . C O into the moment
P. DO.
He may be helped by looking at the change from the follow-
ing point of view :
We wish to get a relation between W and P only ; now
W.AC = P.DC + Q.GC.
We then do away with the moment Q . G C by supposing that
P's momenf is taken about a point O, further from D than C is by
Machines IiJ'
a distance C O, such that the additional moment P . C O = the
moment Q . C G.
If G lies on the right of C, Q will help W, and O the zero of
graduation will be on the left of C.
If on account of the wearing away of the steelyard by constant
use Q becomies diminished, the old graduations will no longer give
the true weight
Example. — In using a common steelyard whose weight has di-
minished by wear, and whose centre of gravity lies on the same
side of the fulcrum as the weight to be weighed, a tradesman uses
the old graduations ; does he defraud his customers or himself?
Using the figure of the last article we have
W.AC + Q.GC = P.DC;
hence W. A C = P . D C-Q . G C
= P . D O, when P . C O = Q . G C, and O
lies to the left of C.
If Q is decreased while G remains in the same position, C O
must also be decreased ; that is, O ought to be moved to the right.
Hence eveiy graduation ought to be moved to the right by the same
distances ; that is, the customer gets a little more than the true
weight denoted by the graduation.
109. In most steelyards the weighing of heavy weights is
accomplished by suspending small weights from the extremity
of the long arm, and so avoiding the necessity of having a very
long steelyard. Thus, in making a steelyard to weigh
substances in stones, pounds, and fractions of a pound with
a movable weight of 2 lbs., we might graduate the rod up to
7 A C, calling the consecutive graduations 2, 4 ... 14, the last
graduation denoting 7x2 lbs., that is, 14 lbs. or i stone.
Bisecting each of the graduations we should get the marks
corresponding to i, 3, 5 . . .13 lbs.
We can then subdivide these graduations into whatever
fractions of the whole would be convenient — say, quarters.
At the last graduation B a small platform is suspended on
which we place multiples of 2 lbs. ; each multiple of 2 lbs.
which we place at B denoting an additional weight of i stone
on the platform at A.
By means of this steelyard we can then weigh a substance
in stones, pounds, and quarters of a pound.
II!
Dynamics
The Danish Steelyard.
no. In this steelyard, which consists of a straight lever
heavily weighted at one end, the substance to be weighed is
suspended from the other end, the weight of the steelyard
acting as \}!\q power.
The weight of the substance is determined by moving the
point from which the steelyard is suspended to one side or the
other until the steelyard rests in a horizontal position ; we then
compare the lengths of the arms.
To graduate it.
^
o
f
W
Fig. 73-
Take moments about C, the point of suspension
P . O C = W . A C,
P(A O - A C) = W . A C,
AC(P + W) = P. AO,
P
(See note.)
\l\N = n P.
AC =
AC =
Now, put « = I, 2, 3, &c..
P + W
P
P + «P
AO.
. A0 =
AO
AC = — ^
2 ' 3 '
AO
&c.,
which are the distances of the successive graduations from A.
Hence, to graduate this steelyard we measure distances
from A equal to i A O, ^ A O, A A O, &c., and mark them i,
2, 3, &c. ; and when the fulcrum coincides with the ist, 2nd,
3rd, &c., of these graduations W = P, 2 P, 3 P, &c.
Machines 1 19
Note. — Tliis equation might have been obtained at once by
taking moments about A, since the reaction of the fulcrum at
C = P + W.
This steelyard is obviously not adapted for weighing heavy
weights, for it is very tedious to move the steelyard one way or the
other until it balances ; and when the weight is a considerable
multiple of P, the successive graduations approach very close to
each other; thus the distance between the graduations 9 and 10 is
only .
no
Again, in subdividing the space between successive graduations,
each subdivision has to be calculated, since equal differences in
weight do not produce equal distances between the corresponding
graduations.
Thus, if C, C, C be the graduations for 7 P, ^\ P, 8 P,
AC = 3 AO, AC' = — AO,AC" = -A0;
8 17 9
.-. C C = ^ . A O, and C C" = — . A O.
136 153
Example. — A Danish steelyard weighs i^ lb. Graduate it to
weigh ounces.
Let W = « oz., P = 24 oz.,
« . A C = 24 O C,
«(A0-0 C) = 24 O C,
(24 + «)0 C = «. AO;
.-. 0C = ^— AO.
24 + «
Let«=i, 0C = — AO;
25
« = 2, OC=^AO;
26
27
oc=Aao,
and so on.
These give us the successive graduations measured from O.
120
Dynamics
Tlie Common Balance.
III. The balance consists of a lever A B suspended from
C, such that its arms are perfectly symmetrical about the line
C D, which bisects A B at right angles. From the ends
A and B are suspended two scale-pans of equal size and
weight, so that when the pans are empty the beam should be
horizontal.
The substance to be weighed is placed in one of the pans,
and known weights are placed in the other pan until the beam
rests in a horizontal position.
Fig. 74.
To determine the horizontality of the beam, a long needle
is attached at right angles to the beam at D, which oscillates in
front of a scale attached to the stand of the balance.
When the beam is in a horizontal position the weight W of
the substance in the one pan ought to be equal to the weight P
in the other.
If P and W differ by a httle the beam would come to rest
in a position inclined at an angle to the horizontal ; we will
now proceed to determine this angle 6 for a given difference
of P and W.
Suppose P > W.
Let w be the weight of the beam and scale-pans acting
through the centre of gravity G ; the point G is always below
C in the line C D, or C D produced.
Let A B = 2 fl CT> = h,CCj = k.
To determine the position of equilibrium we shall take
Machines 1 2 r
moments about C ; we shall therefore require the perpendicular
distances of C from the lines of action of P, W and w.
The perpendicular from D on the direction of P and W
= a cos 6 ; the perpendicular from C on the vertical through
D = ^ sin ^ ;
.*, the perpendicular from C on the direction of P
= a cos Q — h€v[\Q;
.: the perpendicular from C on the direction of W
= ccos 6+A sin ;
and the perpendicular from C on the direction of w = /^ sin 5 ;
.•. P(a cos Q—h sin fl) = W(i2 cos 6+,^ sin G)-\-w . ,^ sin ^ ;
.-. sin 6'(W . /^ + P . h^w . k) = cos BC? . a-W . a) ;
• ta /)_ (P-W )a
•■ ^^ {V + -W)h + w.k'
The quantities a, h, w, and k are always the same for the
same balance ; the above equation therefore gives us the value
of 6 when P and W are known.
112. The requisites for a good balance are sensibility or
sensitiveness, stability and truth.
By sensible we mean that when P and W differ by a very
small amount the beam should show a distinct deflection from
the horizontal ; that is, for a given difference of P and W, B
should be large.
Now tan (9 = ,^ ^^^"^ . (P _ W) ;
.*. Tz; TTTTT , is a measure of the sensibility.
(P + W) /4 + ze) . /4 ■'
Consequently a should be large in comparison with h and
k, and w should be small.
It is not convenient to have the arms of the balance very
long, but h and k can be made as small as we please, and for
very accurate weighing w is made as small as possible, due
regard being paid to the strength of the balance.
By stable we mean that when the balance is slightly dis-
placed from its position of equilibrium it should tend to return
quickly.
122 Dynamics
For this it is necessary that h and'/4 should be as large as
possible ; hence the requirements for stability are at variance
with the requirements for sensibility.
Thus in making a balance we have to take into account the
uses to which it will be put in the future ; whether great accuracy
or speed in weighing is most essential.
Thus in making a chemical balance for use in the laboratory
we pay special attention to the sensibility without neglecting
stability ; while for a balance required for heavy goods, where
extreme accuracy is not of great importance, stability is the chief
requisite.
By true we mean that the weights should be equal when
the beam is horizontal.
This is a necessary requisite for all balances, and is easily
tested by interchanging the weights P and W ; if the balance
does not remain in a horizontal position, P and W are not
equal and the balance is false.
This is probably due to the fact that one arm is longer than
the other, but so weighted that the balance is horizontal when
the pans are empty ; but when weights are placed in the pans,
the larger weight suspended from the shorter arm will be
balanced by the smaller weight suspended from the longer arm.
113. Proposition. To determine the true weight of a
body in a false balance.
Let W be the true weight of the substance which is required.
Weigh the substance first in one pan and then in the other.
Let P and Q be the apparent weights ; let x, y be the lengths
of the arms.
Take moments about C, the point of suspension:
y^ .x='S .y; W.j = Q.^.
Eliminate x and y by multiplying the equations together
and dividing both sides of the resulting equation by the pro-
duct xy.
Then W^ = P . Q ;
.-. W = .v/Trij.
Hence the true weight is the geometric mean between
the apparent weights.
Machines 1 23
If we eliminate W by division, we get
x_ P . J
"y n/'q'
which gives us the ratio of the arms.
114. If the falseness of a balance is not due to the unequal
lengths of the arms, it may be due to the fact that one of the
pans is loaded,\n which case the balance would not be horizontal
when the pans are empty.
In this case, if we weighed the substance W in both pans,
and if the apparent weights were P and Q, and if w were the
weight of the loading given to one pan, we should have
W + w — . P
W = Q -(- a/.
Eliminate w,
2
So, if we eliminate W we shall get the amount of the
loading,
P- Q
2
Example. — A grocer who knows that his balance is false weighs
his customer \ lb. of tea out of one pan and | lb. out of the other.
Does the customer get more or less than i lb. ?
Let X and y be the lengths of the anns, P and Q the weights
he actually gets.
Then V . x = \ . y, and Q . J' = \ ■ x ',
:. P = '^lb., Q =— lb.;
2x 2y
... P + Q = X+ JL = 2iiJ^lb.
2x 2y 2xy
Now ^r' + /' > 2 xy^ because x" + y'^ - 2 xy = {x — jif, which
is positive ;
.-. P + Q > I lb.
This result could also have been obtained at once from the
result of Article 113, since the arithmetic mean between two
quantities is greater than their geometric mean.
124
Dynamics
The "Wheel and Axle.
115. This machine, which is a simple adaptation of the
principle of the lever, consists of a long cylinder — the Axle —
whose axis ends in two pivots resting in fixed sockets.
Rigidly attached to this cylinder is a larger cylinder of no
great thickness — the Wheel— \\zy\x\g the same axis.
The power is applied to a rope
which is wrapped round the cir-
cumference of the wheel, and the
weight is attached to a rope which
is wrapped in the opposite direction
round the axle.
Thus, when we apply the power
we turn the wheel, which also turns
the axle, and thereby raises the
weight.
This machine is often used for raising a bucket of water
from a well, where the power is applied by the hand to a handle
which takes the place of the wheel.
The capstan for raising the anchor on board ship is another
example of the same machine, where the power is applied at
the ends of long spokes fitted into the axle.
116. To determine the condition of equilibrium. — Let the
figure represent a cross section of the
machine by a vertical plane perpendicular
to the common axis, and cutting it at C.
Let A and B be the points where the
ropes leave axle and wheel respectively,
their directions here being tangential to the
wheel and axle, so that A C B is perpen-
dicular to P and W.
Take moments about the common axis
— that is, about C.
P . C B = W . C A ;
. (radius of wheel) = W x (radius of axle).
Efficiency = ?dius of wheel^
radius of axle
Machines
I2S
117. A greater mechanical advantage may be obtained by
means of the Differential Wheel and Axle, in which the weight is
attached to a pulley which hangs in the loop of a string, one end
of which is wrapped round the axle as before, and the other end is
wrapped in the opposite
direction round an axle of c
smaller radius projecting .[ vj.
out of the main axle.
Let T be the tension
of the string round the
pulley.
Then 2 T = W. (See
Art. 119.)
Let the radii of the
power arm and the two
axles be a, b, c respec-
tively.
Taking moments about
the common axis we have ;
P.a = T.i5-
V .a = T{b-
Fig. 76a.
•T.c,
W
P ''
la
Now b-c may be made as small as we please by making the
radii of the axles nearly equal ; hence the mechanical advantage
may be increased indefinitely.
The Toothed or Cogged Wheel.
118*. — A chapter on machines would be incomplete if we
omitted to mention the cogged wheel, which is constantly employed
in such things as clocks, watches, and bicycles, and frequently in
combination with the wheel and axle, as in cranes.
Two wheels of different radii are made with teeth or cogs of
equal size cut on the outside of their circumferences. They are
then placed in the machine so that the tooth of one wheel fits
between two consecutive teeth of the other, the two wheels revolving
in the same plane in opposite directions.
126
Dynamics
If P and W act at right angles to the radii of the two wheels
there is no mechanical advantage, but if W is attached to an axle
fixed to the small wheel then, if a, b, c be the radii of the large
Fig. 77.
wheel, small wheel, and axle respectively, and if F be the mutual
action between the cogs, we have :
P . « = F . a,
W. f=F.^;
• P =^
■■ W ^'
The chief use made of the cogged wheel is due to the difference
in the velocities of revolution of the two wheels owing to their
difference in radius, for their velocities of revolution vary inversely
as the number of cogs on their
circumference — that is, inversely as
their radii.
A modification may be seen in
the apparatus for raising the gates
of sluices, in which case the cogs of
the wheel fit into the spaces of a
cogged vertical bar which carries the
gate.
This is called the Rack and
Pinion.
For the condit'ons of equilibrium
take moments about C :
Wxr=Px/,
Fig. 78.
Machines
127
where r is the radius of the cogged wheel, and / the length of the
arm at which P acts.
The Pulley.
119. Thepulky is a small circular disc of wood or metal,
movable about an axis through its centre, the ends of which
axis rest in sockets in the framework of the pulley-block.
A groove is cut in the outer edge of this disc, so as to keep
in position the rope or cord which is passed round it. A hook
is fastened to one end of the pulley-block for the purpose of
supporting the pulley or attaching a weight to it.
The power is applied at one end of the string, which passes
round a portion of the pulley, the other end being attached to
the weight or to a fixed support, or sometimes (in combinations
of pulleys) to another pulley.
c
Fig. 79.
If the pulley-block is attached to a fixed support the pulley
is said to be fixed, as in fig. 79 ; otherwise it is said to be
movable, as in fig. 81. In fig. 80 one pulley is fixed, the other
movable.
We suppose that the pulley is perfectly smooth, so that the
tension of the string is the same on one side of the pulley as
it is on the other.
A fixed pulley or smooth peg is often used to change the
direction of a force, without altering its effect on the weight, as
in fig. 79.
Here the tension of the string is the same throughout,
,-. P = W.
128 Dynamics
In fig. 80 we have a single movable pulley to which the
weight is attached ; the pulley is supported in the loop of a
string, at one end of which the power is applied, the other end
being attached to a fixed support • the two portions of the
string A D, B C being parallel.
The tension all along the string is = P.
Hence W acting vertically downwards is balanced by two
tensions, each = P, acting vertically upwards ;
.-. W = 2 P.
If the weight of the pulley be w,
W + w = 2 P.
The fixed pulley on the left merely changes the final direc-
tion of P.
In fig. 81 we also have a single movable pulley, but the
portions of the string on either side of it are not parallel.
Since the tension of each portion is = P, they, -must make
equal angles with the direction of the third force W, i.e. with
the vertical. Calling this angle &, and resolving the forces
vertically, we have,
W + w = P cos ^ + P cos
= 2 P cos 6.
Systems of Pulleys.
120. There are three so-called systems or combinations of
pulleys usually given in statics.
The First System is that in which each pulley hangs in the
loop of a separate string, one end of which is attached to a
fixed beam or support, and the other end to the pulley next
above it, and so on, the power acting at the other end of the
string which passes round the highest pulley.
The Second System consists of two blocks of pulleys, in
which the pulleys are placed side by side and movable about
a common axis, or else they are placed one underneath the
other.
The upper block is attached to a fixed support, the weight
being attached to the lower block ; the same string passes
round each pulley, beginning with the upper block, when
Machines
129
the power acts downwards. The power acts at one end of
this string, the other end is finally attached to the upper or
lower block. If the power acts upwards, the string must first
pass round one of the pulleys in the lower block.
The Third System is that in which a string passes over each
pulley, one end being attached to the weight and the other to
the pulley next below it, and so en, the power acting at the
other end of the string which passes over the lowest pulley ;
the top pulley being attached to a fixed support.
(See the figures on the following pages.)
Note. — In considering the conditions of equilibrium, we
shall suppose that all the strings are parallel and vertical.
121. Condition of equilibrium in the first system of
pulleys.
This is the system in which each pulley han^s in the loop
of a separate string. _ RUW
Let A, B, C, &c., be the pulleys, and
let ^1, tc, /j, &c., denote the tensions of
the strings which pass round the pulleys
C, B, A, &c., respectively.
The pulley on the right merely alters
the direction of the last string.
(i) When the weights of the pulleys
are neglected.
For the equilibrium of the pulley C we
W
have
W = 2 A ;
A =
t, =: 2 t„
t — -i
t,- ^
For the equilibrium of the pulley B we have
A W
4 '
For the equilibrium of the pulley A we have
t -2t ■ • ,- - '^ - W
2 6
I'hus, if there are only three movable pulleys,
we have
but
h = ^;
W:
P.
130
Dynamics
Similarly, if there were four movable pulleys, W = 2^ . P,
and so on ; and if there were n movable pulleys, W = 2" . P.
(2) When the weights of the pulleys are taken into con-
sideration.
Let z£/|, 14^2) z^*.? • • • w„ be the weights of the n pulleys C,
B, A, &c.
Then, considering the equilibrium of each pulley separately,
as in the accompanying figure, and beginning
'f with the lowest pulley, we have the following
equations :
vy
«f
Fig. 823.
2 ^1 = W + Z£*i
2 t.^ = ^i + W^
2 4 = ^2 + ze/3
and so on ;
2 4 = 4-1 + K>„
and 4 = P.
If we now multiply the equations (2), (3)
2, 2^, 2' . . . 2"~' we get thefoUowing :
2 /[ := W + Z£/,,
(i)
(2)
(3)
, («)
(«) by
2^4
24 + 2 w.
2)
and so on ;
2=* 4 = 2^ . ^2 + 2^ ze^j,
2" P = 2"-' . 4_, + 2"
' w„,
Now add the corresponding sides of the equations together,
and cancel out similar terms on both sides of the resulting
equation, and we get
■ W + Wi + 2 w^ + 2^ . w^ + ... +2"
w„
or
W
P= — +
2" 2
W2
^ + -^ +
Ul,
+
, w„
2
If the weights of all the pulleys are equal, and each = w,
we have,
P=W + W+2W + 2^ZW+ . . . +2"
' w
= W + zf (2" — i) by geometrical progression.
Machines
131
'Note. — The resultant force on the beam (omitting the right-
hand pulley)
= 2'i + ^2 + 4 + . . .
If we call this force R,
R + P = \V + ze/i + a^j + . . , +w^.
122. Condition of equilibrium in the second system of
pulleys.
This is the system in which there are two blocks and the
same string passes round each pulley.
If the string is fastened to the upper block, the number of
pulleys in each block will be the same ; but if the string is
fastened to the lower block, there must be one more pulley
in the upper block than in the lower block.
K2
132
Dynamics
^^'hen P acts downwards, the two blocks are connected by
the various portions of the string just as many times as there
are pulleys. Now the tension of the string is the same
throughout and = P ; hence, if there be n pulleys, the resul-
tant upward tension on the lower block = nY ; hence, if W
be the weight supported, and w the weight of the lower block,
;»Z P = W + Z£f.
Note. — The resultant force on the beam
= W + P + the weights of the two blocks.
If P were to act upwards, the top pulley in the above
figures would be unnecessary, and the conditions would be
slightly altered.
:j^ A modification of the differential wheel
ii and axle, called the Differential Pulley o^
the American Block, is of very great practical
value.
The upper block consists of two pulleys
of different radii movable about a common
axis ; the lower block consists of a single
movable pulley, to which the weight is
attached.
An endless chain passes over these two
upper pulleys and under the lower one ; this
chain cannot slip over the upper pulleys on
account of their being made with projections
in their rims which fit the links of the chain ;
consequently, W hangs in equilibrium without
any power being applied if the upper pulleys
are nearly equal.
To raise W, P is applied to the left-hand
portion of the chain.
Let T be the tension of the portion of
chain between the bottom pulley and the
upper block, and let b and c be the radii of
the pulleys in the upper block ; then, neglect-
ing the weights of the bottom pulley and the
chain, we have
Fig. 84.
and
2 T = \Y, for the equilibrium of the bottom pulley,
P . (5 + T . (T = T . (5, for the equilibrium of the upper block;
Machines
133
^^'
,-. P.* = T(^-f) = - ifi-c).
w
b-c
2b'
123. Condition of equilibrium in the third system of
pulleys.
This is the system in which each string is attached to the
weight.
Let A, B, C, &c., be the n pulleys,
and let /j, f^, ^3 ... 4 denote the
tensions of the strings which pass
round them respectively.
(i) AVhen we neglect the weights
of the puUeys we have the following
equations :
For the equilibrium of A, t^^= 2 /,
= 2P,
J) j> i*) '3 ^ 2 ^i
= 22P,
and so on, p,^.
/^i-'A
4
W
For the equilibrium of the }i^ pulley 4^2 4-i = 2"'^ . P.
Now W is supported by all the tensions ;
.-. W = /, + /j + 4 + • • • + 4
= P + 2P + 22p+...+ 2"-i P
= P(2''-l).
{2) When the weights of the pulleys are taken into con-
sideration.
Let the weights of A, B, C, &c., be 7v^, w^_, w-^ . . . !»„_„
the weight of the «"* pulley not being taken into account since
it is attached to a fixed beam and is not supported by any of
the strings.
134
Dynamics
We then get the following equations :
/, =P,
/,, = 2 /, + ze;, = 2 P + Je'i,
/■j = 2 A + W2 = 2M' + 2 7e', + Tf/j.
and so on;
tn = 2 /„_, + M'„_,
= 2"-' P + 2"-'"W| + 2"-''W2 + • ■ • w,,_-^ ;
/. W = /, + ;^2 + 4 + • • • ''.
= p
+ 2 P + Zy,
+ 2^ P + 2 7£;| + ze'2
+ 2' P + 2^ Ze', + 2 2^2 + ?«'3
+ &c.
Fig. 85n.
+ 2"-lP + 2"^=Wi + 2"-''lV.i + • . . +Z«'„-1
= P(2"-l) + W, (2"-'-!) + Z«2(2"-''- l)
+ . .. + (2-1) W„ ,.
If the weights of the pulleys are all equal and each = re
we have :
W = P (2"- i) + w[2"-' + 2"-'' + . . . + 2~{n-i)]
= P(2"-l) + Z£'(2(2"-' -l) — ?t + 1}
= P(2"— l) + w[2"-2- n -
= P(2"-l) + w{2" — n-i].
1}
In this system when the weight is suspended from a bar it
must be attached to the point in the bar at which the resultant
of all the tensions /,, /21 ^3 • • • ^cis, otherwise the bar would
not be in equilibrium, and would tilt up to one side or the
other.
In general, the weight should be suspended from a point in
the bar between the ends of the two last strings.
NoU. — In the third system the weights of the pulleys assist
P in supporting the weight, whereas in the first and second
systems P has to support them as well as W. Theoretically,
then, the third system would be the most advantageous to use ;
Machines 135
practically, we rarely see either the first or third in use with
more than two movable pulleys, since the pulleys soon come
in contact with the beam unless the strings are made of very
great length.
The second system is the one commonly found in use.
Example I. — In the first system of pulleys there are four mov-
able pulleys each weighing I lb. weight ; what weight can a power
of 2 lbs. weight support ?
With the usual notation let /„ t.^, /j, t^ be the tensions of the
strings, commencing with the lowest pulley.
Then t^ = 2,
^3 + I =2/4, ,'. /3 = 3 J
4 + I = 2 ^37 •'. 4=5;
^1 + 1 = 2 /,, .-. ifi = 9 ;
W + I = 2/„ .-. W = 17 ;
.'. W = 17 lbs. weight.
Example 2. — In the first system a power of 3 lbs. weight
balances a weight of 10 lbs. when there are three movable pulleys ;
find the weight of each.
Let w = the weight of each pulley.
Then, with the usual notation
2/,=
10 + tu.
.-. A =
■ 5
.f,
2/,=
t^ + -d)
=
5 + If,
.-. 4 =
5
*':-■
24 =
4 + w
=
5 ^ 7'^,
2 4
.-. 4 =
5
4
*'7-
h =
P = 3;
•■• 3 =
5
4
^'-f--
.'. TW =
24 -
10 = 14;
.'. iv =
2 lbs.
weight.
But
No/e. — In the first system of pulleys, if Wis given and P is re-
quired, begin with the lowest pulley and work up ; but if P is given
and W is required, begin with the top pulley and work down.
1 36 Dynamics
In all examples where the weights of the pulleys are given the
student is advised never to use the formulae which have been ob-
tained in the previous articles.
Example 3. — In the second system a power of 5 lbs. weight can
support a weight of 23 lbs., the weight of the lower block being
7 lbs. What power can support a weight of 41 lbs. 1
Let n = the number of strings.
Then
« P = W + w,
«.5 = 23 + 7
= 30;
.-. « = 6.
When V/ =
= 41,
6P = 41 + 7
= 48;
.-. P = 8 lbs. weight.
Example 4. — In the third system, a power of 3 lbs. weight can
support a weight of 67 lbs. when there are four pulleys of equal
weight. Find the weight of each pulley.
Let w = the weight of each pulley.
Then, with the notation adopted in Art. 123,
A = P = 3,
4 = 2 /j + Tf/ = 6 + ■Z£/,
/^ = 2 ^2 ^■ W = 1 2 + 3 w,
/,, = 2 ^3 + ray = 24 + 7 w,
67 = l^ + l, + I, + l^ = 4S + 11 w ;
.'. llw = 22 ;
.'. W = 2.
The Inclined Plane.
1 24. The Inclined Plane is usually a slanting rigid board
inclined to the horizon at an angle less than a right angle,
which angle is called the angle of the plane. It may frequently
be seen in use in getting barrels out of cellars, or in trans-
ferring luggage between a wharf and the deck of a ship.
Machines 137
Since the inclined plane itself supports a portion of the
weight, the power required to keep the weight in equilibrium
will usually be less than the weight.
To find the condition of Equilibrinm.
(i) When the power acts up the plane and parallel to it.
Let the figure represent a section of
the inclined plane by a vertical plane
through the weight and perpendicular
to the edge of the plane.
Let P be the power, W the weight,
R the pressure of the plane which is
perpendicular to the plane, and let i ^'°' ^'
be the angle of the plane.
Then, by Lami's theorem,
P : W : R :: sin (W, R) : sin (R, P) : sin (P, W)
:: sin (180°—/) : sin (90°) : sin (90° + /)
: : sin i : \ : cos /' ;
.•. P = W sin ?', and R = W cos /.
These results might also have been obtained by resolving
the forces along A B and X^ to A B, or by applying the tri-
angle of forces directly.
Since P=W sin /, and sin / is always < i, P must be v w, ift^ weights, the angles of the planes
Fig. 89. being i, z".
Let T be the tension of the string.
Then
^ = sinz, -^^„ = sin/';
.-. Wsinz=W'sin2';
W
smz
sin i '
sinz = , sm z' -■
If /, /' be the lengths of the planes, and A the common altitude,
h
■ ^^ J' = 1
■■ W h /'■
/
Example 2. — On a certain inclined plane the power required to
support a weight when acting parallel to the plane is half that
v.'hich will support the same weight when it acts horizontally. Find
the angle of the plane.
Machines
139
Let a = the angle of the plane,
P the power acting up the plane which will support W,
2P „ „ horizontally „ „
P 2 P
Then — - = sin o, — - = tan a ;
W W
.
. 2 sin n = tan a,
or
sin a
2 sina = ,
cos a
or
2 sin a cos a = sin a.
Now
sin a
cannot
= 0;
therefore, divid
2 cos a= I,
ng
by
sin a,
we
have
or
cos a = 1 ;
.'. a = 60°.
The Screw.
125. The Screw in its simplest form consists of a cylinder
of wood or metal with a projecting spiral thread running round
its circumference, which thread always makes the same angle
with all lines parallel to the axis of the cylinder.
We may imagine it to be constructed in the following
manner :
Let Aa dD he a rectangle whose base A a is equal to the
circumference of the cylinder. In the sides AD,ad mark off
A B, B C, CD, a d, be, cd, all equal to the vertical distance
between two consecutive revolutions of the spiral thread.
D
— <
t
C
= •■^_^
J
V
B
. — -^
X
A
. — -j:;:*^
Fig. 90
Fig. 91.
Join A 1^, Be, Cd, &c. Now apply the rectangle to the
surface of the cylinder, so that A a coincides with the base of
140 Dynamics-
the cylinder, the points a, b, c, d, &c., coinciding with the
points A, B, C, D, &c., respectively ; the hnes Ab, Be, Cd
will now form a continuous spiral line on the surface of the
cylinder, their middle points x, y, z becoming the points x, y, z
on the cylinder. If we now suppose this line to project from
the surface of the cylinder \ve shall have the thread of the
screw.
When the cylinder is vertical the angle which the thread
makes with the horizontal is constant and is called the pitch of
the screw ; it is, of course, the complement of the constant
angle which the thread makes with lines parallel to the axis of
the cylinder. It is the angle a Kb.
If we call this angle a,
. ab vertical distance between two threads
tan a ^ ^ .
A a circumference of the cylmder
Let a be the radius of the cylinder, then circumference of
the cylinder = 2 ir a ;
.•. vertical distance between two threads = 2 Tra . tan a.
This cylinder with its projecting thread works inside a
hollow cylinder of equal radius, in which there is a spiral
groove of the same pitch and the same shape as the thread on
the cylinder.
This hollow cylinder and groove are cut in a block, which is
usually fixed. The solid cylinder and the hollow cylinder into
which it fits are called companion screws. When in working,
the weight or resistance which has to be overcome is at the end
of the screw ; the power is applied in a plane J.' to the .axis,
at right angles to an arm which is also \y to the axis ; since
the screw can only move through its companion in the fixed
block in the direction of the axis, this is the motion which the
power produces, and so the resistance at the end of the screw
is overcome.
Some idea of the mode of action will be conveyed by the
following figure.
Machines
141
126. The thread of the screw and the groove of the
companion screw may be of any shape — in considering the
Fig. 92
condition of equihbrium we shall take it to be rectangular — the
mutual pressure between the surfaces of the thread of the
screw and the groove being perpendicular to these surfaces ;
thus the portion of the thread which at any time is within the
groove will be of the nature of a continuous inclined plane
whose inclination to the horizon is u.
Thus, if O be any point of the thread of the screw inside
the groove, w that portion of the weight which is supported at
O, / the horizontal force at O r 1
which is due to the force P at A,
and r the mutual pressure at O xO-'
between the screw and its com-
panion \j to the surface at O,
we have the second case of
the inclined plane (Art. 1 24),
p : w : r :: sin a : cos u, : i.
Thus/ = w . tan a at every point of the thread ;
,*. 2 (/) = 2 (w tan a) = tan a . S (w), since a is constant
— tan U..W (i)
14^ Dynamics
Further, the sum of the moments of all the forces p about
the axis is = the moment of P ; therefore if (5 = the distance
of A from the axis,
= a . S (/), since a is constant
= a tan a . W, by equation (i) ;
. P a tan a
2 77 a tan a
2 irb
__ vertical distance between two consecutive revolutions of the thread
circumference of the circle whose radius is b
Theoretically there is no limit to the mechanical advantage
which can be gained by the screw, since P's arm can be made
as large as we please, and the distance between two threads
can be made as small as we please ; yet, in practice, it is not
convenient to have P's arm very long, nor is it advisable to
have consecutive turns of the thread very close together, since
the strength of the screw is thereby considerably impaired.
It is largely made use of in cases where large weights have
to be moved slowly forward, as in the case of raising the roof
of a building.
The Wedge.
127. The Wedge is a solid triangular prism which is used
for the purpose of splitting a body into two parts.
The edge of the wedge is introduced between the particles
of the substance at the required place, and is then thrust
forward by the application of some force, as the blow of a
hammer ; by such means a very great pressure is produced,
which is capable of tearing asunder the particles of substances
which adhere to each other most closely.
We will suppose the wedge isosceles, so that it presents the
appearance of two equal inclined planes placed base to base.
Machines
143
In the figure we may suppose the wedge to be used for
splitting the trunk of a tree.
Let R, R' be the pressures at
the points of contact of the
wedge with portions of the trunk ;''
let 2a be the angle of the wedge,
so that each face makes an
angle a with the vertical ; and
let P be the force applied at the
top.
Resolving the forces verti-
cally and horizontally, we have
Fig. 94.
P = Rsina + R'sina . . . . (i)
R cos a = R' cos a (2)
From (2) R = R' ;
.". P — 2R sin a.
It is obvious that, if P is constant, R increases as « de-
creases ; and when P is great and u, is small R can be made
very great indeed.
Besides the wedge in its ordinary well-known form most
of our cutting tools are modifications of it ; thus knives, axes,
and chisels are all modifications of the wedge.
128. It is scarcely necessary to remind the student that in
the practical use of machinery the power employed has not
only to counterbalance the resistance, but to overcome it —
that is to say, work has to be done. Hence in all cases the
power required to work the machine must be greater than that
obtained in the investigations of this chapter ; and if the power
applied be sufficiently great to overcome the resistance of the
weight and the friction of the various parts of the machine,
and to set the mass of the machinery in motion, the machine
will continue to work at an ever-increasing rate of speed (by
the second law of motion) ; and as soon as the required rate
has been attained the power may then be decreased so as just
144 Dynamics
to counterbalance or be in equilibrium with these various
forces, when the machine will continue to work with uniform
speed so long as this power is kept up.
Examples.
The Lever.
1. 3 lbs. and 7 lbs. balance each other on a lever ol the first-
class whose length is i foot. Where is the fulcrum ?
2. 5 lbs. and 8 lbs. balance each other on a lever whose short
arm is 7 ins. What is the length of the lever ?
3. 3 lbs. and 4 lbs. balance on a lever whose long arm is g ins.
What is the length of the other arm?
4. A uniform rod of 4 lbs. weight and i foot long has 6 lbs. and
8 lbs. suspended from its ends. About what point will it balance ?
5. Find the length of a lever of the second class in order that
a force of 5 lbs. weight may balance a weight of 10 lbs., their points
of application being i foot apart.
6. A uniform beam 10 feet long of 140 lbs. weight turns freely
about a point 4 feet from one end, and at that end a weight of
41 lbs. is hung. What weight must be hung from the other end
to keep it in equilibrium ?
7.. A uniform beam 8 feet long of 24 lbs. weight turns freely
about one end and is kept in a horizontal position by a force of
24 lbs. weight at the other end. At what angle must this force
act ?
8. Masses of 7 lbs. and 20 lbs. are hung at the ends of a straight
rod whose length is 2 ft. 8 ins. and whose weight is 25 lbs. About
what point will the rod balance ?
9. The arms of a straight lever are in the ratio i : 2 ; a weight
W is suspended from the extremity of the shorter arm. What
force acting at the other end at an angle of 30° with the lever will
support it?
10. A straight uniform lever of lo lbs. weight and length 5 feet
has its fulcrum at one end ; weights of 3 lbs. and 6 lbs. are attached
to it at distances of i foot and 3 feet from the fulcrum, and it is
kept horizontal by a vertical force at the other end. Find the
pressure on the fulcrum.
Machine's. 145
11. Forces of 13 and 14 act on a bent lever at an angle whose
cosine is - J.. Find the pressure on the fulcrum.
12. Parallel forces balance each other on a straight lever, and
the pressure on the fulcrum is ten times the difference between the
forces. Find the ratio of the arms.
13. A B C is a piece of uniform wire bent at B into two straight
arms such that the angle A B C is 135°. It is suspended from B,
and rests with A B horizontal. Find the ratio of A B : B C.
14. A O B is a bent lever whose arms O A, O B are equal anil
straight. When weights P and Q are suspended from A and B it
is in equilibrium with O A horizontal. What weight must be sus-
pended from B so that the lever should be in equilibrium with O B
horizontal ?
15. A uniform lever of weight W is acted on at the ends A and
B by weights P and Q. Show that if C be the fulcrum A C : C B
: : 2 Q + W : 2 P + W.
16. A uniform heavy rod having a 10 lb. weight at one end
balances about a pwint 4 feet from that end, and if the lo lb. weight
is replaced by a 5 lb. weight the fulcrum has to be moved I foot.
Find the length and weight of the rod.
17. A O B is a bent lever whose arms A O, O B are at right-
angles ; O is the fulcrum. It is in equilibrium when weights P and
Q are suspended from A and B, O A being inchned at an angle of
5o° to the horizon. What force at B will keep P in equilibrium
with O A inclined at an angle of 30° to the horizon ?
Soman Steelyard.
18. A B is a Roman steelyard 10 feet long ; its centre of gravity
is 9 ins. from A, and the fulcrum 6 ins. from A. If the weight of
the steelyard be 8 lbs., and that of the movable weight i lb., how
far from B will the graduation 12 be ? What is the greatest weight
that can be weighed ?
19. If a common steelyard lose one-tenth of its weight, the
centre of gravity remaining unaltered, how would you correct the
graduations ?
20. In the common steelyard the distance of the centre of
gravity from the fulcrum is 2 ins., and the movable weight 4 oz.,
■and the weight of the beam 2 lbs. Where is the zero of gradua-
L
146 Dynamics
tion, and what is the least weight that can be weighed if the ful-
crum is 3 ins. from the heavy end ?
21. In a Roman steelyard the fulcrum is 8 ins. from the heavy
end and 5 ins. from the centre of gravity, and Hes between them ;
the weight of the steelyard is 4 lbs. and the movable weight 3 lbs.
Where is the first graduation corresponding to 3 lbs. ? What is
the smallest weight that can be weighed ? Where is the gradua-
tion for a weight of 15 lbs. ?
22. A common steelyard is made of a uniform bar 18 ins. long
and of 3 lbs. weight. It is suspended 3 ins. from one end, and the
movable weight is 2 lbs. What is the greatest weight that can be
weighed ?
23. A common steelyard is formed of a uniform rod i foot long,
the fulcrum being l inch from the end ; the sliding weight and the
weight of the rod are each i lb. If the sliding weight be changed
into one of 2 lbs. find what error will be made by using the old
graduations.
24. A uniform rod is divided into 20 equal parts, and the fulcrum
is placed at the first graduation. The movable weight can move
between the second division and the end, and substances from
2 lbs. up to 20 lbs. can be weighed. Show that the weight of the
rod is ^ lb. weight.
25. The beam of a steelyard is ■^'i 'ns. long, the body to be
weighed being attached to the end A ; the fulcrum is distant 4 ins.
and the centre of gravity 5 J ins. from A. The weight of the beam
is I lb. weight, and the greatest weight that can be weighed 24 lbs*
weight. Find the movable weight.
Danish Steelyard.
26. The distance between the zero of graduations and the end
of the steelyard is divided into 20 equal parts, and the greatest
weight that can be weighed is 3 lbs. 9 oz. Find the weight of the
steelyard, and the weight which can be weighed when the fulcrum
is at the first gi'aduation.
27. A weight of 4 oz. is in equilibrium on a Danish steelyard
when the fulcrum is 6 ins. from the end to which the weight is
attached, and a weight of 8 oz. is in equilibrium when the fulcrum
is 4 ins. from that end. Find the centre of gravity and the weight
of steelyard.
Machines 147
28. The weight Of a Danish steelyard is i lb., and the nearest
distance of the fulcrum from the end to which the weight is hung
is I inch, and the centre of gravity is 3 feet from that end. What
is the greatest weight that can be weighed ?
29. If the bar rests with the fulcrum half- way between the first
and second graduations, show that the weight to be weighed is
seven-fifths of the weight of the bar.
30. Find the length from the end of a Danish steelyard (i lb.
weight) to the zero of graduation when the distance between the
graduations for 4 lbs. weight and 5 lbs. weight is i inch.
Balance.
31. The arms of a balance are 8 J and 9 ins. respectively. What
is the real weight of a substance vv-hich, when suspended from the
longer arm, apparently weighs 27 lbs. ?
32. A tradesman's balance has arms whose lengths are 1 1 and
12 ins., and it rests horizontally when the scales are empty. If he
sell to each of two customers i lb. of tea at is. gd. per lb., put-
ting his weights into different scale-pans for the two transactions,
how much does he gain or lose owing to his balance being in-
correct ?
33. The beam of a false balance is uniform and heavy ; show
that the arms are proportional to the differences between the true
and the apparent weights.
34. A false balance has its beam horizontal when no weights
are in the pans, but one arm is longer than the other by one-ninth.
The seller puts the substance to be weighed as often into one scale
as the other ; show that he loses f per cent, on his transactions.
35. The arms of a balance are 10 ins. and 11 ins. respectively.
A body apparently weighs 10 lbs. when weighed from the shorter
arm ; what will it appear to weigh when weighed from the other
arm?
36. If one of the pans of a common balance be loaded, and the
apparent weights of a body are iS oz. and 20 oz., find the amount
of the loading.
37. The arms of a balance are unec|ual. What is the true
weight of a body which apparently 'weighs 4 lbs. and 3 lbs. I oz. ?
Find also the ratio of the arms.
L 2
148 Dynamics
Wheel and Axle.
38. If the radius Of the axle is 4 ins., and that of the wheel
3 feet, find what power will support a weight of 1 50 lbs.
39. Four sailors, each capable of raising i cwt., can just raise
an anchor by means of a capstan whose radius is I ft. 2 ins., and
whose spokes are 7 feet long (measured from the axis). Find the
weight of the anchor.
40. The radii of a wheel and axle are 2 ft. and 2 ins. respec-
tively, and the strings which hang from them support the two
ends of a rod 2 ft. 2 ins. long and 10 lbs. weight. What weight
must also hang from one of thejtrings that the rod may rest in a
horizontal position ?
Pulleys.
41. In the case of a single mov-able pulley whose weight is
J lb., and in which the two portions of the string are at right angles
to each other, find the power required t-o support a weight of
10 lbs.
42. If the inclination of one branch of the string to the vertical
be 30°, find the power necessary to support a weight W.
First System of Pulleys.
43. If the weight be 2 cwt. and the power the weight of i st.,
the weights of the pulleys being neglected, how many pulleys are
there ?
44. With three movable pulleys .whose weights, beginning
with the lowest, are 5 lbs. ,4 lbs., and 3 -lbs., and the weight 7 lbs.,
find the power.
45. If there be four movable pulleys, the weight of each being
71.', what power is required to support the pulleys only without any
weight being attached ?
46. In a system of three pulleys, each of which is 8 oz. weight,
what power will support a weight of 5 lbs. attached to the lowest ?
47. What weight can be supported by a power of 6 lbs. weight
in a system of four pulleys each of i lb. weight ?
48. With three movable pulleys whose weights, beginning
with the lowest, are 4 lbs., 2. lbs., i lb. respectively, find what
power will support a weight of 12 lbs.
Machines 149
49. In a system of four movable pulleys, each of which weighs
1 oz., find the weight which will be supported by a power of i lb.
weight.
50. In a system of three movable pulleys, of weights 4 lbs.,
5 lbs., and 6 lbs., find what weight a power of 1 cwt. will support.
51. Find the power required to support a 10 lb. weight when,
there are four movable pulleys each of i oz. weight.
52. What power will support a 54 lb. weight in a system of
four movable pulleys whose masses, beginning, from the lowest, are
4, 3, 2, I lbs. ?
53. Show that if P and W be in equilibrium, and one of them be
raised or lowered, the position of the centre of gravity remains ths
same.
54. By means of a system of three movable pulleys, each of
2 lbs. weight, a man of 12 st. weight raises himself in a basket
attached to the lowest pulley by the exertion of a force of 22 lbs.
weight. Find the weight of the basket.
55. How many movable pulleys, each weighing I lb., must be
used in order that a power of 2 lbs. weight may support a weight
of 65 lbs. ? If the weight be raised 3 ins. how many feet of string^
must pass round the highest pulley ?
Second System of Pulleys,
56. The two blocks each contain twto pulleys, and the movable
block weighs 8 lbs., what weight will i cwt. support .'
57. There are nine pulleys in the two blocks, the lower of which
Weighs 8 lbs., what force will support a weight of 100 lbs. ?
58. With eight pulleys, the weight of the lower block being
I lb., what weight can a force of 2 lbs. weight raise.'
59. Draw the arrangement of pulleys in this system so that
five men, each capable of exerting a force equal to the Weight of
I cwt., may be able to raise ij ton.
60. In this system of pulleys it is found that forces of 5 lbs
vifeight and 6 lbs. weight can just support weights of 18 lbs. and
22 lbs. attached to the lower block. Find the number of strings
and the weight of the lower block.
1 50 Dynamics
Third System of Pulleys.
61. If the weight be 2 cwt. i lb,, and the power i St. I lb., the
weights of the pulleys being neglected, how many pulleys are
there ?
62. With two movable pulleys of weights I lb. and 2 lbs.
what weight could be supported by a power of 3 lbs. ?
63. With two pulleys, of which the lower weighs 5 lbs. weight,
what weight will a power of 3 lbs. weight suppoit ?
64. If there be three pulleys, each weighing 4 oz., find the
power necessary to support a weight of 8 lbs.
65. In a system of four pulleys, each weighing 2 oz., find the
weight which will be supported by a power of I lb. weight.
66. In a system of four movable pulleys, the weight of each
being iv^ what weight will be supported without the application of
any power ?
67. Find the power required to support a weight of 1 1 lbs.
when there are four pulleys (three movable) each of i oz. weight.
68. If there be four movable pulleys, each of i Hi. weight, and
the weight attached to the lowest be 3 cwt., find the power.
69. With three movable pulleys, of weights 71:',, 10., w^, a
force P balances W ; if the first and third pulleys be interchanged
Pj balances W. Show that P, - P = f {w^-w^).
70. If there be three movable pulleys whose masses, beginning
with the lowest, are l, 2 and 3 lbs., what power is required to sup^
port a weight of 91 lbs. ?
71. If this system be modified by making the string which
passes over each pulley pass under a small pulley attached to the
weight and then attaching it to the former pulley, show that for JZ
pulleys, whose weights may be neglected, the mechanical advantage
will be 3" - 1.
Inclined Plane.
72. A weight of 12 lbs. is supported on a plane of an angle of
30° by a force acting up the plane. Find this force.
73. What force acting horizontally will support a weight of
12 lbs. on a plane whose angle is 60° ?
74. What is the inclination of the plane on which a power
parallel to the plane will support twice its own weight ?
Machines i S I
75. Find the inclination of the plane on which a horizontal force
will support a weight equal to itself.
76. What power acting up the plane will support a weight of
52 lbs. on a plane whose height is to its base as 5 is to 12 ?
77. A particle weighi'ng 9 lbs. is kept at rest by a horizontal
force of 3 v/3 lbs. weight. Find the angle of the plane and the
pressure upon it.
78. If the weight, power, and pressure on the plane are in the
ratio \/3 : I : 1, find the inclination of the plane and the direction
of the force.
79. A weight W is supported on a plane whose angle is 45° by
a force P such that W = ^ 3 YmA the direction of P.
80. The angle of the plane is cos~' J, and there is equilibrium
when the force acts up the plane. If the same force now act
horizontally, show that an equal force applied in the proper direc-
tion will still preserve equilibrium, and find the direction of this
force.
8 1. The angle of a plane is 30°, and a force P, acting horizontally,
keeps the weight in equilibrium. If P now act at an angle of 30°
with the plane, show that there will still be equilibrium, and that
the pressure on the plane will be reduced one-half.
82. A weight W is supported on a plane whose angle is 30° by
a force P such that W = P a/2. Find the direction of P.
83. Find the direction in which the power must act in order
that the pressure on the plane may be double the pressure when
the power is the least possible.
84. Two planes, whose angles are 60° and 45° respectively, have
the same altitude and are placed back to back ; two weights con-
nected by a string passing over their common vertex are in equi-
librium upon them. Compare these weights.
85. A rope inclined to the vertical at an angle of 30° is just
strong enough to support 180 lbs. weight on a plane whose angle
is 30°. Find approximately the greatest weight it would support
if hanging vertically.
86. A "weight 2 P is kept in equilibrium on a plane by a hori-
zontal force P and an equal force acting up the plane. Find the
ratio of the length, base, and height of the plane, and the pressure
of the plane.
152 Dynamics
87. A weight W is kept in equilibrium by two forces P P acting
parallel to the plane and horizontally. The angle of the plane is «,
such that sin o = A ; show that W = 5 P.
13
88.* If the pressure on the plane be an arithmetic mean
between the weight and the power, and the inclination of the power
to the horizon be double that of the plane, prove that the sine of
the angle of inclination of the power to the horizon will be |.
89. W is supported on an inclined plane (l) by a power acting
up the plane, (2) by a power acting horizontally. Show that W =
Vr Rj, where R, Rj are the pressures on the plane in the two cases.
go. If P be the power required to support a weight W on an
inclined plane, and the plane be then placed so that its vertical
side is now the base and Pj is the power requix-ed, show that W =
91. A power of 3 lbs. weight cart support 5 lbs. weight on an
inclined plane ; the plane is no^ turned on to the other side as
base. Show that the same power can only support a weight of
3f lbs.
92. Find the direction in which a force of 8 lbs. weight must
act in order that it should just support a weight of 12 lbs. on a
plane whose angle is 30°.
93. If the inclined plane is the upper surface of a smooth wedge
v/hose under surface rests on a smooth horizontal table, find the
horizontal force necessary to keep the wedge from sliding on the
table.
94. A weight of 7 cwt. is supported on a smooth plane inclined
at an angle of 30° to the horizon by means of a rope which, after
passing over a smooth pulley at the top of the plane, is wound
round the axle of a windlass below, the handle describing a circle,
of 16 ins. radius and the radius of the axle being 4 ins. Find the
pressure on the handle.
Screw.
95. A screw having its threads \ inch apart is turned by an arm
7 ins. long, and moves in a smooth nut with its axis vertical. What
force must be applied at the end of the arm to support a body of
1 1 St. ?
153
CHAPTER IX
FRICTION
129. In the preceding chapter we have supposed that -when
two surfaces are in contact the mutual reaction between them
is always along their common normal ; in other words, we 'have
supposed that the surfaces of the bodies in contact were per-
fectly smooth. In nature this is only true when the surfaces
in contact have no tendency to move the one against the
other, as in the case of a body resting on a horizontal table.
Xn this case the weight of the body acting vertically downwards
can only be supported by the pressure of the table, which must
therefore act vertically upwards — that is, perpendicular to the
surface of the table.
Now suppose the table to be tilted, so that it is no longer
horizontal ; the resultant
pressure of the table must
still act vertically upwards
in order to support the
weight, and consequently
cannot be perpendicular to
the table. It may in fact
be resolved into two forces
R and F, of which R acts
normal to the surface, and
F acts along the surface nig
up the line of greatest slope. Fig. 95.
and consequently in a direction opposite to that in which the
body would move if it could.
In this case R is called the normal pressure, and F is called
i}a& force of friction.
This force of friction is due to the roughness of the two
Note. — In. fig. 95, mg = W (Art. 80).
154 Dynamics
surfaces, and is always called into action whenever a body has
a tendency to move upon the surface with which it is in con-
tact ; but only so much is called into action as is required to
prevent motion. Hence we say that friction is a passive
force and not an active force— it prevents motion, but never
causes it.
The amount of friction that can be called into action in
any particular case is necessarily limited.
Thus, in the case of the body on the slanting table, as we
increase the angle of inclination, the body has a greater ten-
dency to slip, and the force of friction increases so as always
to prevent motion up to a certain point ; after which, if we
increase the angle of inclination, F is no longer able to resist
the tendency to motion, and the body begins to slide.
The amount of friction which is called into action when the
body is just on the point of sliding is called the limiting frictioti.
If we resolve these forces R, F, W perpendicular to and
along the plane, we must have
R = W cos a,
F = AV sin a,
and W = v/ R^^TF^
a being the angle of inchnation of the plane to the horizon
Hence so long as there is equilibrium F = W sin a ; thus F
increases as a increases, until the limiting position of equilibrium
is reached, when F can increase no longer. If a is still in-
creased, the body will begin to move with an acceleration
, m being the mass of the body. (See Art. 27.)
ni
The amount of limiting friction which can be called into action
between two substances varies with the nature of the substances ;
some materials, being muth rougher than others, are consequently
able to exert a much greater limiting friction. In nature no sub-
stances are so smooth as to be incapable of exerting a small
resistance to motion, though it is possible so to polish the surfaces
in contact as to make this resistance very small indeed.
In speaking of a smooth surface we do not merely mean a sur-
face which is not very rough, but a surface which is perfectly smooth
- -that is, which offers no resistance to motion. We shall consider
Friction 1^5
s.uch surfaces in future examples for the sake of illustration, though,
like much of the work in the earlier stages of a physical subject,
they must be regarded as purely hypothetical.
130. We will now proceed to describe briefly how the
laws of limiting friction were experimentally determined by
Coulomb in the last century.
Take a number of blocks of the same materia! but of
various shapes (not spherical), and place them upon a table
with a uniform surface ; now gradually raise one end of the
table ; it will be found that when the limiting position of
equilibrium is reached for one block, it is reached for all — that
is to say, they will all begin sliding together. If we now
change the materials, we find that blocks of different material
begin to slide at different inclinations of the table.
We therefore infer that the limiting friction does not
depend upon the size or the shape of the surfaces in contact,
provided that one cannot roll on the other. Further, since
F" F
the ratio ^ or -— depends only on a in the preceding example,
R W
and since u. remams the same for the same substances, we
infer that the limiting friction varies as the normal pressure.
131. We may now state the laws of Statical Friction :
Law I. — The direction of the force of friction is opposite
to the direction in which motion would take place if there
were no friction.
Law IL — In all eases of equilibrium the magnitude of
the force of friction is equal to the least force required to
prevent motion.
Laws of Limiting Friction.
Law III. — So long as the substances in contact remain
the same the limiting friction varies as the normal pressure.
Law IV. — So long as the normal pressure remains the
same the limiting friction is independent of the size and
shape of the surfaces in contact.
Definition. — The constant ratio of the hmiting friction to
the normal pressure is called the coefficient of friction.
156
Dynamics
Thus, if /*, = coefficient of friction,
1 = ^; .-. F = /.R.
Coulomb also discovered that Laws III. and IV. of statical
friction were also true when the body was in motion, but that
in this case the friction exerted opposite to the direction of
motion was less than the limiting statical friction. Hence the
coefficient of dynamical friction is less than the coefficient of
statical friction.
He also found that the dynamical friction was independent
of the velocity of the moving body. Though this is not
rigidly accurate, it is sufficiently so to be embodied in our
law of dynamical friction.
Law V. — The coefficient of dynamical friction is less than
the coefficient of statical friction, but is independent of the
velocity with which the body is moving.
The value of ^ varies greatly for different substances, being
much greater for substances like wood and stone than for
metal, glass, or marble. It can usually be considerably lessened
by smearing the surfaces in contact with oil or soap.
Rolling friction is much less than sliding friction.
132. The actual pressure between two substances is the
resultant of the normal pressure and the force of friction.
Definition. — The angle between the resultant pressure in
the limiting position of equilibrium and the normal is called
the angle of friction.
. Thus, if (^ be the angle of friction, O A the direction of
the resultant of the normal
pressure R and the limiting
friction F, represented by
OB, O C respectively, then,
completing the rectangular
parallelogram BOCA, we
have
tan<^
_/^
R
= tr-=/*-
^
Fig. s«.
BA _
BO R ^ R
Hence ^ = tan~y,
Friction
157
Since the resultant reaction cannot be inclined to the normal
at a greater angle than , it cannot act outside the angle A O A'
when acting in the plane of the paper.
Hence it cannot act outside the cone whose axis is the normal
O B and vertical angle 2 1^. This cone is called the cone of
friction.
133. To find the coefficient of friction between any two
substances experimentally.
We take a plane board of the one substance hinged along
one edge, so that it can revolve about that edge ; on it we
place a portion of the- other substance with a plane base, and
turn the first board about its fixed edge until the other sub-
stance is in the Hmiting posi-
tion of equiUbrium.
Thus, let W be the weight
of the substance, R the normal
pressure, ya R the limiting fric-
tion, so that fx. is the coefficient
of friction between the two
substances. F'g. 97.
Let 6 be the angle of inclination of the plane to the
horizon in the limiting position of equilibrium. Resolving the
forces parallel and perpendicular to the plane we have
R = w cos e,
/tR = Wsin^i;
W sin 61
/^R
R '
W cos e '
/, /x = tan p.
6 is carefully measured, and thus /a may be determined.
Note. — Since /i = tan ^, where ,4).is the angle of friction, it follows
that fl = (^.
Hence the angle of inclination of the plane in the limiting
position of equilibrium is equal to the anglp of friction. For this
reason the angle tan~'^ is often called the angle of repose.
This result might also have been obtained from the following
considerations : — -
158
Dynamics
(f> is the angle between the resultant pressure and the normal
■pressure — that is, between the direction of R and the vertical ;
6 is the angle between the plane and the horizontal ; the lines
v>'hich contain the angle i (W cos a — P sin 6),
P(cos^ + /xsine) = W(sina + /*cosa),
p _ W sin g + /^ co s a
~" ■ cos ^ + /n sin ^'
So also if the body were on the point of moving down the
plane, ^k, R would act up the plane, and we should have
p ^ sin g — //. co s g
' cos ^ — ;«, sin 6
134B.* To find the magnitude and direction of the least
force which will move a body up a rough inclined plane.
Using the first result of the previous article, we have, when
W is on the point of moving up the plane,
p T^ sin g + /t cos g
' cos ^ + /A sin Q
If <^ be the angle of friction, /a = tan <^,
p yfj si n g + /It cos g
' cos ^ -+ /A sin 5
-^ s in g + tan <^ . cos g
cos h + tan <^ sin B
y, sin g cos <^ + cos g^in^
' cos Q cos (^ + sin ^ sin <^
=:W sin (g + ) 1
■cos(e-<^y
P is a minimum when cos (^ — 0) is a maximum ;
i.e. when cos {Q — ^) = i,
when ^ — ^ = o,
when = 4>,
and then P = W sin (g + <^).
Hence the least force which will just move a weight W up
a rough inclined plane must act at an angle with the plane
equal to the angle of friction ((/>), and be just greater than
W sin (g + ).
' This result might have been obtained at once by resolving P and W
perpendicular to the direction of the total resistance, thus P cos (9 ± (j>)
= W sin (a T respectively. In
the first case the effect of the limiting friction is the same as if the
angle of the plane were increased by an angle equal to the angle
of friction ; and in the second case the effect is the same as if the
angle of the plane were diminished by an angle equal to the angle
of friction.
135. Equilibrium on a rough screw.
With the notation used for the smooth screw in Article 126,
suppose that the screw is on the
point of rising, so that the friction
/A r acts down the tangent at the
point O :
cos a-= w sin a + /A '',
r ■=■ w cos a -f / sin a ;
Fig. 100,
•. p cos a-:= w sin a -\- ij.(w cos a + / sin a),
p (cos a — fx. sin a) = W (sin a + yii cos a),
sin a + yu. cos a 2
p ■= W ~
COS a — fji sm a
V COS a — /J, sm a/
-y^ sin a + {JL COS a
COS a — jjL sin a
' See note on previous page.
2 This result might have been obtained from the result of Art. 134A
by putting 9 = a.
1 62 Dynamics
Again, P. :
P a sin a + tan <^ cos a
W b ' cos a — tan (^ sin a
a sin (a + <^)
b ' cos (a + ^^
= 1 . tan(a + <^)-
which is the same result that we should have obtained with a
smooth screw whose pitch was a + <^.
Similarly, if the screw were on the point of moving down,
we should get
|r = f •tan(a-<^).
Rougli Wedge.
136. Using the description and notation employed in
Article 127, and supposing that the wedge
is on the point of moving down, we have
P = (R + R') sin a + /i (R + R') cos a.
R cos a—ji. Rsin a = R' cos a — /x R' sin a;
:. R = R'.
Fig. ioi.
Hence, P = 2Rsina+2/AR cos a
= 2 R (sin a + /A cos a).
Similarly, it the wedge were on the point of moving
upwards,
P = 2 R (sin a — jU cos a).
Friction
163
It is obvious that in this case it is possible for the wedge to
be in equilibrium under the action of the pressures alone
without any force being applied at the top.
Thus, putting P = o, we have
sin li — ft, cos a = o ; /. tan a ^ /j..
Hence the wedge will be in equilibrium without any force
being applied if its vertical angle is equal to or less than
twice the angle of friction.
Note. — In this case the resultant pressures at A and A'
must be horizontal.
Example i. — A uniform ladder rests against a vertical wall with
its lower end on a horizontal pavement. If the coefficients of
friction between the ladder and the
pavement and between the ladder and
the wall be ^ and ^' respectively,
determine the greatest inclination the
ladder can make with the vertical.
Let A B be the ladder, 2 / its
length, 6 the angle it makes with the
vertical. Rand R' the normal pressures
at A and B respectively, /x R, /i' R'
the limiting friction, W the weight of
the ladder acting through C the middle
point.
Resolving the forces vertically and
horizontally,
R + /R'=W . . . (i)
R' = fi R . . . (2)
Take moments about B :
R . 2 i sm 6 — n R . 2I cos 6 ■
Fig. 102.
W . / sin fl = o
(3)
From (i) and (2) R (i + ,i /) = W.
Substitute for W in (3)
R . 2 sin 5 - /x R . 2 cos 5 - R (I + iiii') sin 6
sin 0(i-ii fi') = 2fi cos 6,
tan 6 = -, .
Note. — The res Itant pressures at A and B and the direction of
the weight must all pass through one point O.
M 2
164 Dynamics
Corollary. — If the wall be smooth /x' = o, then tan 5 = 2 ^.
(See Example 2 at the end of Chapter vii.).
If the ground is smooth \x. = o; then tan 6 = 0,
Hence equilibrium is impossible. This is also obvious by Art. 96.
* If there is a man of weight w on the ladder, let D be his posi-
tion, and let the ratio — —- = r. so that our equations must be
B A
modified as follows :
R + /x' R' = W + w ,
R' = H R,
R . 2 / sin 5' - /i R . 2 / cos 6/' - W . / sin 5' - w . 2 / r sin 5' = o ;
since B D = 2 / . r.
Here R = — ;
I + fi/
.-. ^^^i^ . 2 sin 5' - W . sin 5' - 2 r -^ sin 5' = t^^^^ . ?cos 5';
I + nil' 1 + fifi'
tan 6' = 2j^ W + w)
2 (W + w) — (W + 2 rzu) (i +n/i')
_ 2 fX (W + w)
W(i — fi. fi') + 2ZU — 2wr(i i-/i/x')
- ^-^ (1 + ^)
_ I - ^^'\ W/
W \ I -/x^i' /
Now — - = tan fl, where 5 is the inclination without the man;
I - fj.li
2 W I —r(l +M|a')
. tan d _ W ' I -fifi
tanfl' T +-^
W
Hence -tan fi is > or < tan 6',
according as 2 . — — ^- t^J^ > or < i ;
I -/I /I
i.e. as 2 -2 r (i +/i/i') > or < i—fiji.',
a.s 2 r (i i- fj. fi') < or > I f fi/,
as 2 r is < or > i,
as r < or > ^.
Hence, ^ is <6' \f ris > J ; that is, if the man is on the lower
half of the ladder ; so that if the ladder is on the point of slipping
Friclioh
i6s
down when a man gets on it, it will be less inclined to slip so long
as he is on the lower half of it, but as soon as he gets beyond the
middle it will slip. (Encyc. Brit. ' Mechanics.')
Example 2.— A brick is placed on its end on a rough plane
and the plane is tilted.
Find the proportion of its
dimensions in order that
it should slide before it
tumbles over.
Let A B C D be the
brick standing on the base
A B, and let u, b be the
length and depth of the
brick respectively ; its
breadth perpendicular to ''"'°- "s-
the plane .of.the paper will not affect the question. Let G be the
centre of gravity of the brick, and let C A D = a.
Let 6 be the inclination of the plane to the horizon ; then 6 will
also be the inclination of D A to the vertical. The brick will be
on the point of sliding when 6 is equal to (^ the angle of friction.
The brick will be on the point of falling over when the vertical
through the centre of gravity passes through A the extremity of
the base ; that is, when 6 = a.
When 5>0 the brick will sHde ;
when 6>a the brick will fall over ;
.■. \i^iia.
This gives the ratio of the depth of the brick to its length.
Example 3. — A weight W is attached by a string to a point O
on a rough inclined plane whose
inclination a is greater than the
angle of friction, so that the
body can only describe the arc
of a circle whose centre is O.
What is the greatest angle that
the string can make with the
hne of greatest slope ? Fig. 104.
1 66 Dynamics
Let 6 be the angle the string makes with the line of greatest
slope in the limiting position of equilibrium. ; A A' the arc of the
circle the body can describe.
Then the forces which act on the body are
W vertically downwards,
R X' to the plane,
T along the string A O, ) These two are
ft R along the tangent to the circle at A j in the plane.
W may be resolved into W cos a x' to the plane, and W sin a
along the line of greatest slope, of which W cos a is in equilibrium
with R, and W sin a is in equilibrium with T and \i. R ;
.-. R = Wcosa (i)
Now, resolving W sin a parallel and perpendicular to A O in the
plane, we have
T = Wsin a . cos 5 ,.,... (2)
ju R = W sin a . sin 5 (3)
Substituting for R from (i in (3)
\i. W . cos n = W sin a . sin 5 ;
.'. sin d = fj. cot a.
This gives the greatest value 6 can have. T can then be found by
substituting for 8 in equation (2).
Example 4. — A rough vertical screw is in equilibrium without
any power being applied. Find the least number of revolutions
the thread can be allowed to make in a given length / of the screw.
There will be the least number of revolutions of the thread
when the pitch of the screw is greatest. In the limiting position
of equilibrium, when the screw is on the point of moving down-
wards, the pitch of the screw is equal to the angle of friction.
Hence, if a be the greatest pitch of the screw,
tana =/i.
Now if a be the radius of the cylinder of the screw,
2 TT a . tan a = vertical distance between two consecutive revolutions ;
.•. 2'n- a. 11 = greatest distance between two revolutions.
Hence, if n be the least number of revolutions in a given length
/ of the screw,
fi . 2% afi = l ;
/
. . n = .
2t! jxa
Friction
167
Example 5. — A picture is hung from a peg in a rough wall by a
string attached to the peg and the middle poinl; of the upper side
of the picture, so that the length of the string (/) is equal to the
height of the picture. Determine the angle 6 which the picture
makes with the wall in the limiting position of equilibrium, and
the tension of the string when the picture is in that position.
Let A B represent the cross section of the
picture, W its weight acting through C, O A the
string, so that OA, AB' make equal angles
vvith the wall.
Let R be the normal pressure at B, m R
the limiting friction acting up the wall ;
T the tension of the string.
Then, resolving the forces vertically and
horizontally,
W = Tcos5 + ^R,
R = T sin 5 ;
.-. W = T (cos e + /i sin ff). Fig. 105.
Take moments about B.
W. BCsin^ = T. B O sin ^,
W . - = T . B O,
2
T(cos5 + fisin6)- = T. 2 /cos 5;
.•. cos 5 + fi sin ^ = 4 cos 6 ;
.•. fi sin 5 = 3 cos 6y
tan (9 = 3.
Hence sin 6 =
W = T
and cos 6 ■■
V9 + H-'
~'^W9+H-y'
T = W.
1 68 Dynamics
Examples.
1. Find the greatest and least forces required tosuport a lo lb.
weight on a rough plane inclined at an angle tan-' \ to the horizon,
the coefificient of friction being \, the forces acting parallel to the
plane.
2. A 5 lb. weight can just be supported on a rough inclined
plane by a 2 lb. weight, or can just support a 4 lb. weight sus-
pended by a string passing over a smooth pulley at the top of the
plane. Find the coefficient of friction and the inclination of the
plane to the horizon.
3. A straight uniform rod A B of weight W rests with its lower
end A on a smooth horizontal plane, and its upper end B supported
by a rough peg ( /i = — 7-). Show that the inclination of the rod
to the horizon is 30°, and find the pressure on the plane at A.
4. A uniform heavy beam rests with one end on a rough hori-
zontal plane and the other against a rough vertical wall, and is
inclined at an angle of 30° to the horizon. Find /i, supposed to be
the same for both surfaces.
5. A body of weight W can just rest on a rough inclined plane,
inclined at an angle of 30° to the horizon. What is the least force
acting (i) along the plane, (2) horizontally, which will make it begin
to move up the plane ? Determine the pressure in the two cases.
6. Find the tangent of the angle at which a heavy uniform
beam will rest against a smooth vertical wall, with its lower end on
a horizontal rough plane on which the resistance to sliding is two-
thirds of the weight of the beam.
7. A weight rests on a rough inclined plane, and is just pre-
vented from sliding down the plane by a given force P. Show that
the weight supported will be the greatest when P is inclined to
the plane at an angle equal to the angle of friction measured below
the plane.
8. A heavy particle is in equilibrium on a rough plane inclined
to the horizon at twice the angle of friction, being just supported
by a force whose direction bisects the angle between the plane and
the horizon. Show that the resolved part of the weight in the
direction of the force just balances the force, and find the pressure
on the plane.
JFrtcfioM 169
9. A ladder 13 fiset long rests with one end on a rough pave-
ment smd the other agiunst the smooth vertical wall of a house at
a height of 12 feet bma the gionnd. A man ascends it Show
that the frictional resistance of the g^round teust increase as he
ascends, and if the coefiSdent ot friction be less than ^, the ladder
(whose w^ght may be n^ected) will slip before he reaches the
tqp.
10. A heavy aniform straight rod rests with its extremities on
two fixed roo^ similar rodSj each inclined at an angle of 45° to
the horizon, all three bdng in the same vertical plane. Find the
greatest angle at ^diich the movable rod can be inclined to the
horizon.
11. On a rough horizontal floor is placed a uniform equilateral
triangle ABC with its base B C verticaL The vertex A isin con-
tact with a smooth slojvng wall, which is perpendicular to the
plane of the trian^e and makes an angle of iao° with the floor. If
the triangle just rests in diis position, find the coefficient of firiction.
IX A wtight is supported on a rough inclined plane by a force
acting parallel to the plane, whose inclination is a and coefGdent
of fiiction fL If the force which would just pull it up the plane is
doable the force which just prevents it boat shpping down, prove
that tan a = 3 |t.
13. A and B are two small equal heavy rings which slide on a
rough horizontal rod, the coeffident of fiiction bang 3— I. Another
equal heavy ring C slides on a smooth weightless string connecting
A and 6, and hangs below the rod in a position of limiting equilib-
rium. Show that A, B, C are the comers <^ an equilateral triangle.
14.* Two particles, whose wdgfats are equal, are connected by a
light string, and rest on a rough vertical drde, the coeffident of
friction bdng tan c. Show that when there is no fiiction between
the string and the circle, and both wd^ts are on the point of
motion, the radios of the cirde through the middle point of the
string wiQ make an angle e with the vertical
15.* A uniform rod rests at an aqgle a to the horizon, with one
end on a rough horizontal plane, and on the point of slij^ing, and
the other end supported by a string inclined at an angle 9 to
the horizon. Prove tiiat the coeffident of limiting fiiction is
I
tan # - ^ tan a*
I/O Dynamics
CHAPTER X*
VIRTUAL WORK
. 137. It has been stated at the end of Chapter VIII. that
the use of a machine is to transfer work. Now, when is a
force said to do work ?
A force is said to do work when its point of application
is displaced in the direction in which the force acts.
Thus, let A be the point of appHcation of the force F ;
suppose that, under the action of F, A is displaced to A', and
draw A' a perpendicular to the line of action of F ; then A a
is the projection of the dis-
A' placement of A on the line of
^^n action of F, and F . Aa is the
^ ^ y^ work done by F during the
displacement. If a falls on
the side of A towards which
F acts, the force is said to do work, and the work is said to be
positive ; if a falls on the other side, work is said to be done
against the force, and is negative.
Thus F . 5 is always the measure of the work done by or
against F during any displacement ; s being the projection of
the displacement on the direction of the force, which will be
positive or negative according to the magnitude of the angle
which A A' makes with F.
Energy is the capacity of doing work. All forces have
energy, and when a force does work some of its energy dis-
appears ; but it is not lost, it is merely transferred to some
other part of the system.
This is the Principle of the Conservation of Energy, which
may be stated as follows :
Fig. 106
Virtual Work 171
The total energy of any material system is a quantity
wMcli can neither be increased nor diminished by any action
between the parts of the system, though it may be trans-
formed into any of the forms of which energy is susceptible.
(Clerk Maxwell, ' Matter and Motion,' p. 60.)
This great principle is really embodied in Newton's Third
Law of Motion —
Action and Eeaction are equal and opposite.
No work can be done without the expenditure of an equal
quantity of work. There can be no increase of energy in any
system without the loss of at least an equal quantity of energy
in some other system. If, in any conservative system, one
form of energy is increased, it must be done at the expense of
some other form of energy in the system.
The different forms of energy are various ; thus, mechanical
energy may be transformed into heat, light, sound, elec-
tricity, &c. ; though in the transformation from one form of
energy to another there is always an apparent loss of energy
due to various causes, which is in fact a deterioration and
not an actual loss.
138. AVe will consider this subject more fully later On ; our
immediate object is to apply the principle to the machines
considered in Chapter VIII.
In all the cases which we have there considered we have
found the conditions of equilibrium, this being the primary
object of statics. When there is equilibrium there is no
motion, and consequently no work is done. We shall now
imagine that there is a small displacement ; this is called a
virtual displacement, since it is imaginary and cannot- actually
take place, and the work which would be done during such a
displacement is called virtual work.
By the principle of the conservation of energy the total
work done by a machine in motion is zero — that is, the work
done by P is equal to the work done against W ; or, since
the one is positive and the other negative, the algebraical
sum of the work done is zero ; so in all machines which are at
rest, the algebraical sum of the virtual work done by the
172 Dynamics
various forces during any virtual displacement must be
zero.
We may now assume this principle, and from it derive the
conditions of equilibrium ; or we may, by "using the conditions
of equilibrium already obtained in Chapter VIII., prove the
principle in any particular case.
Note. — If the virtual displacement be perpendicular to the
direction of any force, the virtual work done by the force
is zero.
Hence, in solving problems by the principle of virtual
work, if there is a force in the system the value of which we
do not wish to ascertain, we shall suppose the displacement to
take place in a direction 1.^ to the line of action of this force,
so that it will not then enter into the equation of virtual
work.
Again, in certain cases we may have to consider more than
one virtual displacement, in order that we may get several
equations between the forces of the system, since one equation
is often insufficient to completely determine the magnitude of
each force.
139. Sometimes h.a — the projection of the displacement
of the point of application on the line of action of F — is called
the virtual velocity of the force F, and the product of F . A a
is called the virtual moment of the force.
Thus the principle stated at the end of the last article is
called the principle of virtual velocities, and is stated thus : the
algebraical sum of the virtual moments of the various forces of a
system in equilibrium is zero.
140. To prove tlie principle of virtual work in a lever of the
second class.
Let ABC be a lever of the second class, movable about
the fulcrum C.
Then P . A C = W . B C.
.Suppose that the points of application of A, B to be
slightly displaced by the rotation of the lever through a small
angle a.
Virtual Work 173
Then A A' and B B' are the small arcs of circles whose
centre is C ; but when u. is indefinitely small they may be
looked upon as straight lines perpendicular to A C B, and
b'
ABC
Fig. 107.
consequently represent the displacements of A and B in the
lines of action of P and W.
Now AA' = CA. aandBB' = CB . u.
Hence the algebraical sum of the virtual work of the forces
= P. AA'-W . BB'
= P.CAa - W. CB .c.
= (P. CA- W. CB)a
= 0, sinceP. AC = W. BC,
which proves the principle.
Note. — The virtual work of the third force — namely, the
reaction of the fulcrum — is zero, because there is no virtual
displacement of the fulcrum.
It will be noticed that in the preceding example, as in
those which follow, if work is supposed to be done by P it
must be done against W, and vice versa.
Since Displacement of P _ CA _ W
Displacement of W C B P
the principle used to be stated as follows :
What is gained in power is lost in distance.
141. Assu7ning the principle of virtual work to be true in the
case of a straight lever, when the forces P and W are not parallel,
find the condition of equilibrium.
Let A C B be the lever, movable about the fulcrum C ;
suppose A to receive a small displacement to A', in conse-
1/4 Dynamics
quence of which B is moved to B', the lever being turned
through a small angle a.
Draw A'fl, CM perpendicular to the direction of P
and draw '&' b, C N „ „ „ W.
Fig. io8.
Then A A' and B B' are small arcs of circles whose radii
are C A, C B respectively ; hence
AA' = CA. a, BB' = CB. a.
Since a is very small, A A', B B' may be considered to be
straight lines, and the angles C A A', C A' A, C B B', C B'B tg
be right angles.
Hence the triangles A A' a, CAM are similar ;
. Aa ^CM
"AA' CA"
Now A A' = C A. a; .-. A« = ^ X C A . a = C M . a.
Similarly B^ = ^-^ X C B . a = C N . a.
Now, by the principle of virtual work,
P. Aa-A¥. B^ = 0;
.-. P. Afl = W. B^,
P.CM.a = W.CN.a;
.-. P . C M = W . C N ;
that is, the sum of the moments of P and W about C is zero.
Virtual Work
175
142. Apply the principle of virtual work to find the con^
ditions of equilibrium on tlie wheel and axle.
Suppose that there is a small
displacement of the machine, in
consequence of which it turns
about its axis through a small
angle a.
Aa- W. D^= O;
P ^D^
" W Ka
OD . a
Then P
•^p
O A. a
_OD
~OA'
P X radius of wheel = W x radius of axle.
Fig, 109.
143. Prove the principle of virtual work for the first system
of pulleys when the weights of the pulleys are taken into account.
Using the figure and notation of Article 121, let us suppose
that W is raised through a small distance x, then A must
also be raised through x ; to do this B must be raised through
a distance 2 x-, because the other end of the string round A is
fixed ; similarly C must be raised through a distance 2 x 2 x,
that is 2^ . jc, and so on. The «th pulley will have to be
raised through a distance 2""^ . x ; consequently, P's point of
application must be moved through a distance 2" . x.
Hence the sum of the virtual work
= 'S y, 2"- X — "W X — Wx . x — w^ . 2x — w^ . 2'x — . . . .
— w^ . 2'^~^X,
= X (P . 2" — W — W, — 2 W2 — 2'^ W3 — . . . . — 2''~^W„)
i= X X o, by the equation of equilibrium,
= o.
144. Apply the principle of virtual work to find the con-
dition of equilibrium in the third system of pulleys.
Using the figure and notation of Article 123, let us
suppose that W is raised through a small distance x, then C
176
Dynamics
must descend through a distance x ; B will descend through
a distance 2 x, on account of the descent of C, and also
through a distance x on account of the ascent of W ; that is,
B will descend through a distance 2 x + x.
So A will descend through a distance 2 (2 x + x) + x,
that is, through (2^ + 2 + i)x; and if there are n pulley*
the point of apphcation of P will descend through a distance
that is,
(2"-^ + 2"-
2"— I
2 — I
■"+....+ 2 + i)x;
. X, or (2" — i)x.
Now the algebraical sum of the virtual work is zero ;
P . (2" — l)x+ Wi(2"~^ — l)x + . . . . + W,^_i{2 — i)x
-\Y . x = o;_
.: P (2" - i) + w, (2"-' — i) + .
which is the equation of equihbrium.
+ Z£'„_i = W,
145. Prove the principle of virtual work in the case of a
smooth inclined plane when the power acts along the plane.
Suppose the point
of application O to
be displaced to O' ;
draw 0'« 1.'' to the
vertical through O.
Then, since O O' is
Fig. ho. J.r to the direction
of R, the virtual work done by R during this displacement
is zero.
Hence the algebraical sum of the virtual work
= P . O O' - w . o «
= P . O O' - W . O O' sin O O' «
= P . O O' - W . O O' sin a
= O O' (P - W sin a)
= o.
We might have imagined a small displacement x in the
Virtual Work
177
line of action of W, in which case the sum of the virtual
work
= W . * — P . jjc sin a — R x cos a
= X (W — P sin a — R cos a)
= a: . o = o.
146. Use the principle of virtual work to find the condi-
tion of equilibrium on a
rough inclined plane
when the body is on the
point of moving down
the plane, the power
acting up the plane.
Suppose a small dis-
placement O O' up the
plane. Fig. m.
Sum of the virtual work
= W.O«-(P-|-yu.R)0 0' = o;
.-. W . O O' sin <. = (P + JU.R) O O' ;
.'. Wsina = P + /xR (i)
Suppose now a small displacement O O" ±^ to the plane.
Sum of the virtual work
= R . O O" - W . O OT = o ;
.-. R . O O" = W (O O" cos a) ;
.-. R = W cos a . . . . (2)
Substitute in (i)
P + ju, W cos a = \V sin a ;
.•. P = W (sin a — /4 cos a). Fig. 112.
147. The preceding case may be solved more quickly
thus :
I^et S be the resultant pressure of the plane making an
angle ^ with the normal to the plane, so that tan <^ = /i.
Now suppose the virtual displacement to take place ly
to the direction of S, as O O'.
N
178 Dynamics
Draw O' n, 0' m !>• to the directions of P and W respec-
tively. Then, since sum of virtual work is zero,
-S ^
' W .On-V .Om—o.
^/oi.
i i w
^P
Now 0' makes the
same angle with the hori-
zontal that S makes with
the vertical, that is
1 .
ti — ip ,
Fig. 113
.'. the angle O O' « = a — ■
the angle O' O »2 == ^ ;
O « = O O' sin (a - .^) ;
, O ;« = O O' cos <^ ;
.-. P . O O' cos ^ = W . O O' sin (a - )_
cos (j)
148. To find the condition of equilibrium on a smooth screw
by the principle of virtual work.
(See figure at the end of Article 125.)
Suppose that A were to move completely round, then W
would be raised through a distance equal to the vertical
distance between two turns of the thread.
Hence, if displacement of A = - (complete revolution of A)
the displacement of W = ~ (
n \
vert. dist. between two turns of
the thread
)
Now P (P_'s displacement) = W (W's displacement) ;
. _P _ W's displacement
' * W P's displacement
_ vertical distance between two turns of thread
circumference of circle described by A
Virtual Work 179
149. If any number of forces act at a point, the algebraical
sum of the virtual work of all the forces = the virtual work of
their resultant.
Let O be the point, O O' the displace-
ment, P any one of the forces making an
angle Q with O O'.
Draw O'a ±'' to the direction of P,
and suppose that R, the resultant of all the
forces, makes an angle with O O'.
The sum of the virtual work of all the
forces
Fig. 114.
= 2(P,
. Ofl)
= 2(P
. 0' cos 6)
= 00'.
S (P cos e)
= 0'
. R cos (^
= R . 0' cos .^
= virtual work of the
resultant.
(Art. 55.)
Hence, if the forces are in equilibrium, their resultant is
gero, and consequently the sum of the virtual work of all the
forces is zero.
For examples in this chapter the student may work the
converse of any of the propositions in Articles 140 to 148.
N 2
Part II
KINEMATICS AND KINETICS
183
CHAPTER XI
KINEMATICS
The student is here advised to read again the first seven •
teen articles of Chapter I.
We shall in the present chapter consider abstract motion
without reference to the mass of the moving body, or the force
which produces change of motion.
150. The rate at which a point moves is called its velocity,
and, when uniform, is measured by -, where s is the space
passed over in / seconds (Article 12).
Thus V = - feet per second ;
.'. s = V . t
151. To find iJie measure of a velocity v when the units of
space and time a7-e altered.
Suppose o- feet and t seconds to be the new units of space
and time.
The body describes v feet in one second,
or - (o- feet) in one second,
or — (tr feet) in t seconds ;
O"
hence the new measure = — .
184 Dynamics
152. In Article 37 we have proved the parallelogram of
velocities.
Hence all the results of Chapter III. which have been
derived from the parallelogram of forces will hold mutatis
mutandis for velocities.
Thus we shall have the following :
If V be the resultant velocity of two simultaneous velocities
u and V, inclined to each other at an angle Q,
Y"^ = u^ + v'^ + 2 uv cos 0.
If <^ be the angle this resultant makes with u,
V sin d
tan <^ ;
u -\- V cos S
Triangle of Velocities.
If a point has simultaneously three velocities which can be
represented in magnitude and direction by the sides of a
triangle, taken in order, it will be at rest.
The converse is also true.
So also we have the Polygon of Velocities.
Similarly, the resolved part of a velocity z; in a direction
making an angle Q with the direction of z; is z^ cos 0.
Again, the resultant of any number of simultaneous
velocities may be found by resolving them all along two
directions at right angles to each other, and finding the resul-
tant of the algebraical sums of these two sets of components.
Thus V2 = {% {v cos e)} 2 + {2 (w sin 6)} K
Change of Velocity.
153. When the velocity of a point changes in either magni-
tude or direction, or both, it is necessary for us to know what
the change of velocity is.
Given therefore the velocity both before and after a certain
change, we have to ask ourselves the following question :
' What velocity, if compounded with the original velocity, will
Kinematics
185
have a resultant velocity equal to the velocity after the
change ? '
Let OA represent the ve-
locity of the point before the
change, OB the velocity after.
Join BA and complete the
parallelogram O A B C.
Then OA, OC represent
two velocities whose resultant
is represented by OB; that is,
O C represents the velocity which must be communicated to a
point in order to change its velocity from a velocity represented
by O A to that represented by O B.
Hence O C or A B represents the change in velocity.
If O A = «, O B = », and the angle AO B = 61,
A B^ = «2 + »2 — 2 « z; cos ^.
If
= 2^^ -^ v"^ — 2UV cos 6.
This gives the magnitude.
If ^ be the angle C O A
tan d> = — '^^—^ (as in Article 11;^.)
The preceding result may also be conveniently arrived at
in the following manner :
Let O B in the figure represent v, the velocity of B, as
before. Resolve O B into two components, one of which is
equal and parallel to u, the velocity of A, by completing the
parallelogram B A O C.
O A, O C will therefore represent the whole velocity of B ;
but the velocity O A is A's velocity and produces no change in
the distance between A and B ;
.". O C represents the velocity of B relative to A.
Example i. — Two points have equal velocities inclined to each
other at an angle of 60°. Find the relative velocity.
1 88 Dynamics
Let the common velocity be v.
(Relative velocity) ^ = w' + ?/' - 2 w ' cos 60°
= 7/^ + ■z/'^ - 2 w '^ (^)
= •z/"- ; _
.'. relative velocity = v.
Hence the relative velocity is an equal velocity, and makes an
angle of 120° with either velocity.
This result could also have been obtained at once from the
geometry of the figure.
Example 2. — Two points A and B are distant a feet apart at a
certain time. Their relative velocity is v, making an angle 6 with
A B. How near to each other can they get ? Find also the time
that elapses before they arrive at their nearest distance.
Suppose that A is brought to
rest by giving each point an
additional velocity equal and op-
posite to A's velocity ; then if
A B C = 5, B C represents the
direction of the velocity of B re-
lative to A. If AC be X' to BC,
A C is the shortest distance be-
tween them.
A C = «z sin 5.
If t be the time,
. _ B C _ a cos 6
V V
156. When a point is moving with variable velocity in a
straight line, the rate at which the velocity changes is called its
Acceleration, and is in the direction of motion.
It is then said to be uniform when there are equal incre-
ments of velocity in equal intervals of time. Uniform accele-
ration in the direction of motion is measured by -, where v is
the velocity gained by the point in t seconds. (See Article 17.)
Thus, if a be the acceleration,
a = - units of velocity per second.
When we say that the acceleration of a point in any
particular direction is a, we mean that its velocity in that
Kinematics 1 89
direction is increased at the rate of a feet per second, per
second.
Acceleration need not necessarily be in the direction of
motion. Thus a point may have a uniform velocity in one
direction and a uniform acceleration in another. The path ot
the point will then be a curve, and its velocity will be the
resultant of two velocities — namely, the uniform velocity in
the one direction, and the velocity due to the acceleration in
the other. The rate of motion along\}as. curve is not uniformly
accelerated, but the rate of change of velocity is the given uni-
form acceleration.
In this case we consider the two motions separately, and
the acceleration is measured by the change of velocity in the
particular direction in which the acceleration takes place.
But so long as the acceleration remains uniform and in the
same direction, it is measured by - ; where v is the change of
velocity in that direction in / seconds.
Hence v = at,
where v stands for the velocity gained in time / in the direction
in which the acceleration a takes place.
Velocity may change in magnitude only, as in the case of
an acceleration in the direction of motion ; or it may change
both in msgnitude and direction, as in the case of an accele-
ration which is not in the direction of motion ; or it may
change in direction without changing in magnitude, as in the
case of a ball tied to a string and made to revolve in a
horizontal circle with uniform velocity.
At present we shall only deal with uniform acceleration in
the direction of motion.
157. To find the measure of an acceleration a when the units
of space and time are altered.
Suppose cr feet and t seconds to be the new units of space
and time.
The velocity of the point is accelerated by a feet per second
in I second.
190 t)y7iamics
or - (o- feet) per second in i second,
or — (o- feet) per t seconds in i second,
or — ^ (o- feet) per t seconds in t seconds.
rr
Hence the new measure of the acceleration is — .
or
158. To find the unit of acceleration when the units of space
and time are altered.
Let o- feet and t seconds be the new units of space and
time.
The new unit of acceleration
= a velocity of a- feet per x seconds in t seconds
= anu so on ;
( „ U + ta,
and these form a series in arithmetical progression.
Hence, if v be the velocity at the end of time t,
v= u + ai (i)
rr^i 1 -^ u + V u + u + at
The average velocity = — ' — = — ■
3 2
= U -\ .a
2
= velocity at the middle of the time.
161. It must be borne in mind that the velocity is not constant
during any second, but is ahvays changing uniformly. This does
not affect the result of the last article.
For we may suppose that each second is divided up into n
very small equal intervals of time, during each of which we may
suppose that the velocity is constant, and at the end of each of
which we may suppose the velocity to be increased by -.
n
Thus during each second the velocities at the end of each of
these small intervals will form an arithmetical series whose con-
stant difference is -.
?i
Thus the average velocity during the first second
_ « + M + a
= « + i a
= velocity at the middle of the second,
and so on.
The a\ erage velocity duinng the ^th second
^ U+{t— l)a-¥U-¥ia
2
= U + it — \)a
= velocity at the middle of the /th second.
192 Dynamics
Hence the average velocity during the whole time
= the average of all the averages
2
t
= M + - . u.
2
162. To find the space passed over by a point in t seconds,
which starts with an initial velocity u, and has a constant
acceleration a in the -direction of motion.
Let s = space passed over.
Since the point is moving with constant acceleration, the
average velocity during the time is equal to the velocity at the
middle of the time ; and hence the space passed over in time
t is the 'same as vpould be described by a point moving with a
uniform velocity equal to the velocity at the middle of the
time ;
.'. i' = (average velocity) x t
= (u +- .a)t
= ut + \af^ (2)
To find the relation between v, a, and s, eliminate t. The
above equation may also be written in the form
.^C-t-"),,.
But v=u + at (j) (Art. 160); :.t='"- ^.
He„c. , = (^«)(^«) =
W — u'
« ■* = 2 a J-,
or
V^ = U^ -\- ■zas (3)
Equation (3) may also be obtained by substituting the value of
/ from (i; in (2).
Thus s = u{^-=-^)^\a{^^".,
,\ 2 a s = 2 u (v — u) -v (v — uf
= v'' ~ u''.
A:'!c-r'liltU-S 195
163. We now have the three equations —
T =11 + ot (i)
s=iit + iats . . . . {:)
V* = U* + J as (3)
\\"hen » and o are giren
{i) gives us the velocity at the end of a given time ;
{::) gives us the space described in a given time :
^3^ gi\es us the velocity af:er describing a given space.
It' .'« ^ o. the body starts from rest, and we have —
:■ = u /,
If a is opposite to the direction of motion — that is, a
retardation — put — a for « in the above equations, and we
b.aN e —
f ^ « — a .■',
.«- = »/ — I o /*,
r^ = .'*- — .: a.i-.
-^'i'.V. — If a ^ c the eo^uations become
P ^ «f, y = a /, T' ^ .'«--
164. It will be interesting to obtain equation (3) of the pre\ ious
article by the ^i," "/ -'''i' method.
The jpace described by a point moving: uniformly is = velo-
citv \ t;me. If, therefore, we represent time by a straight line in
one direction, and \-elocity by a straight line at right angles to it,
the space described may be represented by a rectangle whose
base represents the titr.e and whose he;i;ht represents the velocity.
Thas, if A B C D ;s 3. rectange whose base A B conUiins r
units of lenjith. and height ,\ D a units of lengih, the area = a/,
,\na represent; the space described by a point in time ."' movuig
V. .-.h uniform velocity ;..
Now snp|.vse the point to have an acceleration a in addition
to the uniform velocity ».
Prcduce B C to E, makng C E =/i!, so that B £ = ;. +/a = the
\-elocity at the end of time i. Join D E,
To find the veloci:> at any time represented by A P.
Draw P/> j_" ,A. B, cutting D E in /. Pj~ shall represent the
vekKity after time A F,
o
194
Dynamics
Let AV ==n seconds, where n may be integral or fractional, and
let P;>cut D C in/'.
«*
p a
Fig. ii8.
Then, by similar triangles,
p p' ^V) p' ^KV ^n ,
EC DC AB / '
•'• PP' = ~ {ai) = na.
Hence P/ = « + i^ a = velocity after n seconds. Similarly, the
velocity at any other time will be given by the ordinate drawn
from the point in A B which represents the time, to cut the
line D E.
Let Q represent a time very near to P, so that P Q represents
a small interval of time during the body's motion ; the ordinate
Q q represents the velocity at Q, and during the time P Q the
velocity increases uniformly from V p to Q g.
Draw /.r parallel to P Q to cut Qq m x, and qy parallel to
P Q to cut P/ my.
Then, if the velocity during the time P Q were constant and
equal to the velocity at P, the space described during the time P Q
would be represented by the rectangle P Q xp ; so also, if the
velocity during the time P Q were constant and equal to the
velocity at Q, the space described would be represented by the
rectangle P Q qy.
But the velocity is not constant, but changes from V p to Q y !
hence the actual space described is represented by an area which
lies between the rectangles P Q xp and P Q qy.
But in the limit where the time P Q is very small, the areas of
these two rectangles are very nearly equal to each other and also
to the area P Q q p which lies between them ; and if P Q be taken
small enough, these three areas are all equal. Hence the actual
Kinematics
195
space described d-uring the time P Q is represented by the area
It follows therefore that the space described during the time t
or A B is represented by the area ABED;
.-. j = area ABED
= area of rectangle A C + area of triangle D C E
= AB. BC + ^DC. CE
= ut-^\ . t . at
= ut + ^ at^.
165. In the previous article A B is called the time-line and D E
the curve of velocity. In the case we have considered D E is a
straight line, because the \'elocity is uniformly accelerated in the
direction of motion ; if this were not the case D E would no longer
be a straight line, but an irregular line or a curve. This curve is
of such a nature that if A T
be taken to represent any
time on the time-line, and
and the ordinate T/ be
drawn to cut the curve in /,
then T/ represents the
velocity at the time repre-
sented by A T.
Then, whether the ac-
celeration be uniform or
not, the space described
during any time represented
by A B on the time-line will be represented by the area ABED,
which is inclosed between the time-line, the curve of velocity, and
the ordinates at the beginning and end of the time-line.
The method used in the preceding article is of the very greatf st
importance in the more advanced parts of the subject.
166. To find the space described by a point whose velocity is
uniformly accelerated in any particular second.
Let u = initial velocity, ui = the acceleration, then
s = ut ->r\o.t\
Fig. 119.
1 96 Dynamics
The space described in the /th second
= space described in / seconds — space described in {t— i) sees.
= ut-\-\a.t'^ -u{t -T.)-\a.(t -if
= u{t—t+ i) + \a{t'^ - t"^ + 2t— i)
^U + \a {2 t — 1)
= u + a{t—\).
This might have been obtained more quickly by the
following consideration :
Space described in the ^th second
= space described in i second by a point moving with uniform
velocity equal to the velocity at the middle of the rth
second
= .K + {i - i) a.
Putting / = I, 2, 3, Src, we have —
space described during ist second = u + \a,
„ „ 2nd „ = « + 1 a,
„ „ 3rd „ =U + ^a.
Hence the spaces described in consecutive seconds form a
series in arithmetical progression whose constant difference
is a.
If a body starts from rest subject to a constant acceleration,
the velocity at any time varies as the time, and the space
passed over varies as the square of the time.
For V = a f, and s = \a t'^.
Now a is constant ;
.•. w a /, and j' oc /"^
167. A body is projected with a velocity u, and is subject to a
uniform retardation a opposite to the direction of motion. To
prove that the time that elapses before it comes to rest is equal to
the time it takes to return to the point of projection, and that the
Telocity with which it passes the point of projection on its return
is equal to the velocity with which it was projected.
Kinematics 197
The equations which determine the motion are —
» = M — at (i)
j- = a/ — ^a/^ (2)
v"^ =^u'^ — 2aS (3)
To find the time that has elapsed when it comes to rest,
put o = o in equation (r),
o =t^ — at ; . . t ^=-—.
a
To find the time that has elapsed when it is again at the
starting point, put ^ = o in equation (2),
o =s «/ — ^at^ ;
.: i (u — ^ a t) = o ; .', t ^ o, or — ;
a
that is- to say, it is at the starting point when it starts, and
again after a time — , which is twice the time that elapsed
a
before it came to rest ; hence, time taken to go out = time
taken to return.
To find the velocity with which it returns to the starting
point, put J- = o in equation (3),
z;^ = «2 .
.•. » = ± K ;
that is to say, at the starting point the velocity is u when it
starts and — u, or ti in the opposite direction, when it returns.
Similarly, the velocity with which it passes any point on its
way back is equal to the velocity with which it passed it on its
way out. For, put s — a in equation (3), then
ifi ==■ u^ — 2a. a ;
.". » ^ it ^/a* — 2aa.
The positive value denotes the velocity on the way out ; the
negative value is the velocity at the same place on the way
bacL
198
Dynamics
i68. If a body has an acceleration which is not in the
direction of motion, its path is a curve on which we can find
' points in the following manner.
Suppose u the uniform velo-
city with which it starts from O
in the direction O x, and a the
uniform acceleration in the direc-
tion Qy.
Let O A represent u ; then
if there were no acceleration the
body would be at the points
A, B, C, &c., at the end of
consecutive seconds.
But in consequence of the
acceleration a in direction Oy
it also describes distances pa-
rallel to O y, which distances
vary as the square of the time
which has elapsed.
Fig.
Thus at the end of i second O a
2 seconds O (5 = ^ a .
„ „ 3 " O c = ^ a . 9,
and so on.
These two motions are quite independent of each other
and may be considered separately. (Art. 34.)
Thus at the end of consecutive seconds the body will be
at the points a', b', c', d', &c., and its path will be the curved
line O a' b' c' d'. The path of the body is in fact the same as
if it were projected with uniform velocity u along a smooth
tube O X while the tube moved with uniform acceleration a in
the direction Qy.
To find the velocity after any time f.
Let 6 be the angle between O x and Oy,
the velocity in direction O x = ti,
„ „ Oy = a^;
,'. (velocity)' = u'^ + (aty + 2 uat cos 0.
The direction of this resultant velocity can be found as in
Kinematics 1 99
Article 46. This direction is the actual direction in which
the body is moving at the time, and is a tangent to the curve
at the point which represents the body's position at that time.
This curve is a parabola.
Note. — It is found that the path of the body is a curve
when the two motions are of different kinds. If the body
started from rest at O with two uniform accelerations, its path
would be a straight line by the parallelogram of accelerations.
Example I. — The measure of an acceleration in the foot-second
units is 3. The units of space and time being altered, the measure
of the acceleration is 8. If the new units of space be 1 yards,
what is the new unit of time ?
Let t seconds be the new unit of time.
The acceleration
= 8 (2 yards) per t seconds in t seconds
= 48 feet per / seconds in / seconds
= ~- feet per second in t seconds
= 4^ feet per second in i second;
..3 ^,,
or 3 ^'-48,
or /^=i6;
.-. ^ = 4-
Hence the new unit of time is 4 seconds.
Example 2. — A velocity of ^ and an acceleration of ^ in the
foot-second units become 10 and 120 when the units of space and
time are altered. Find the new units of space and time.
Let s feet and t seconds be the new units.
A velocity of \ foot per second
= a velocity of — (j feet) per t seconds ;
.-. — =10 (I)
2J
An acceleration of ^ foot per second per second
= an acceleration of is feet) per t seconds per / seconds ;
\o s
f
:. — = 120 (2)
10 J-
200 Dynamics
Substitute from (i) in (2)
(20 j)^= 1200 s ;
.■. ■f=3
and hence < = 60 ;
/. the new units of space and time are i yard and i minute.
Example 3. — A point moving with uniformly accelerated velocity
has a velocity of 3 feet per second ; 5 seconds later its velocity
is 10. Find the space passed over in the time, and the accelera-
tion.
Space described in the time = space tliat would be described by
a body moving with uniform velocity equal to the velocity at the
middle of the time
= 3iLo ,
= ^ = 32ifeet.
2 '
Acceleration = change in velocity
time
= Z feet per second per second.
To work it by our formulae :
K = 3, j/=io, /=5;
substitute in the equation v = u + at,
whence a = -.
5
To find s, substitute in either of the equations
i^)
or 5 = u i + ^ af.
Example 4. — A body has an initial velocity 20, and moves with
constant acceleration 3 opposite to the direction of motion. Find
(l) its velocity after 5 seconds,
' (2^ space passed over in 5 seconds,
(3) velocity when it is 56 feet away,
(4) time which has elapsed when it is 56 feet away,
(5) the distance traversed before it stops,
(6) the time which elapses before it stops,
(7) its position after 14 seconds.
TCinematics 201
Here
tt = 20, a.= - l-
(I)
V = U ■¥ at
= 20 - 3 . 5 = 5.
(2)
S = ut -V iaf
= 20. 5 -f ,25)
= lOO - "^5 = 62i.
(3)
:■' = k' + 2aj-,
1-' = (20)» - 6 . 56
= 400 - 336 = 64 ;
.-. :■ = i S.
(4)
s = ut -^ ^at'-,
56 = 20/- §/-,
3 /• - \Ot + 112 =
.-. / = 4. or 9I.
(5)
t'- = k' + 2 a J ;
put '■ =
o,
= (20^- — 6s ;
.-. J = ±ja = 66§.
6)
V ^ u + af;
put :• =
o.
O = 20 — , / ;
.-. / = 4"^ = 6f .
This result must ob\-iously lie half-way between the two
answers to (|4\
7) s = ui ^ iat-
= 20 "X 14 — f I4"-
= 280 - 3 X 98
= - 14;
that is to say, it ^v-iU have returned past the place where it started
and have readied a place 14 feet on the other side.
JExamp/es.
I. Find the actual velocity of a particle which has two inde-
pendent velocities of 5 and 3 in directions making an angle of 60°
with each other.
For other excunphs on the p.j>-a!!-:'cg-ra;n of --dodtics iak; any
of the c.fjmpics 5 to 9, 12 to 15. 21 to 24 a/ tr'-.c ;r.J of Ouipter
III.^ subsliiu,ting ■ i.'CU\'ity for -jor:^:
202 Dynamics
2. A distance of i,ioo yards is described by a point moving
with a uniform velocity of a mile an hour in the time 2|-. What is
the unit of time?
3. Express a velocity of i mile an hour in terms of a unit
velocity of 10 feet a second.
4. Express the acceleration 32 feet per second per second in
terms of the units 1 yard and I minute.
5. If I inch, I second, and I ounce be the units of length,
time, and mass respectively, find the measure of the force equal to
the weight of i pound.
6. A stone is projected horizontally with a velocity of 16 feet
per second from a steamer which is moving at the rate of 12 miles
per hour in a direction perpendicular to that in which the stone is
projected. Find the initial velocity of the stone. Find also the
velocity at the end of I second if the stone also acquires a vertical
velocity of 32 feet per second in i second.
7. A vessel sailing due north at the rate of 7 miles an hour is
under the influence of an easterly current running 3 feet a second.
What is the actual velocity of the vessel, and how far will she be
carried out of her course in 1 1 hours ?
8. A train is moving 40 miles an hour when a ball is thrown
horizontally out of the window in a direction at right angles to that
of the train with a velocity relative to the train of 44 feet per
second. What is the actual velocity of the ball at the moment of
projection ?
g. The unit of length is 5 feet, and the unit of time 6 seconds.
How many units of velocity are there in the speed of a train going
10 miles an hour?
10. If a mile per minute be the unit of velocity, and 32 feet per
second per second the unit of acceleration, find the units of space
and time.
11. If the acceleration of a particle be 32 in terms of the units
I foot and i second, what will be the measure of the acceleration
when a mile is the unit of space and a minute the unit of time ?
12. The unit of length is 6 feet, and the unit of time 5 seconds.
What number will represent the acceleration of a train which
acquires a velocity of 10 miles an hour in one minute from the
start, the acceleration being assumed constant ?
13. Find the velocity which would be acquired from rest by a
Kinematics 203
particle moving with a constant acceleration of 3 feet per second
per second after traversing 22 miles.
14. A bucket is ascending the shaft of a mine at a uniform rate
of 10 feet per second. When it is 50 feet from the top the tension
of the cord is diminished, so that it now ascends with a velocity
which is uniformly retarded, and comes to rest at the top. What
is the retardation ?
15. If the spaces described by a particle moving with uniform
acceleration be 168 feet and 288 feet in the first three and the first
four seconds respectively, find the initial velocity and the accele-
ration.
16. A point moving from rest with uniform acceleration ac-
quires in 3 seconds a velocity of 1,000 yards a minute. In what
time will it acquire a velocity of 100 miles an hour?
17. A body which was moving at the rate of 140 yards per
minute increased its velocity uniformly, so that at the end of a.
quarter of a minute it was moving at the rate of 1 5 miles an hour.
What was its acceleration ?
18. A particle acquires a velocity of 1,000 miles an hour in 22
minutes. Compare its acceleration with that of gravity.
ig. A particle moving with uniform acceleration a passes over
a space s in time r. Prove that its velocity at the beginning of
time T is equal to — .
J- 2
20. A train moving from rest with uniform acceleration attains
a velocity of 20 miles an hour in i minute. Find its acceleration,
adopting a foot and a second as the units.
21. A train moving from rest with uniform acceleration acquires
after three minutes a velocity of 30 miles an hour. Find the
distance travelled.
22. In what time would a body acquire a velocity of 225 miles
an hour if the acceleration were 33 feet per second per second ?
■23. Find the mean acceleration of a train which increases its
speed from 40 to 49 miles per hour while descending an incline
in 4^ minutes.
24. In the above question, if the acceleration due to gravity be
32, determine the slope of the incline.
25. A train is travelling at the rate of 44 feet per second on
204 Dynamics
level rails, and it is observed that when steam is shut off the train
stops after travelling a quarter of a mile further ; to what retar-
dation is it subject ? If the mass of the train be i ton, find the
resistance in poundals. If this resistance be all due to the friction
of the rails, find the coefficient of friction.
26. A body is moving with uniformly accelerated velocity.
Compare the spaces passed over in the «th and {n + i )th seconds
after the motion commenced.
27. If a ball with a velocity of 900 feet per second enters a
block of wood to the depth of 9 inches, what velocity will produce
a penetration of 16 inches, the resistance of the wood being
supposed uniform ?
28. Using only the formute lor accelerated motion, determine
with what velocity a stone will strike the ground which falls"
through a height of 257-6 feet in 4 seconds.
29. A body observed three times at intervals of a second is
found to have velocities 36, 60, 84. Find the spaces passed over
in the intervals, the acceleration being uniform.
30. An engine-driver on passing through a station reduces
speed from 60 miles an hour to 40 miles an hour in the space of
400 yards. What is the retardation ?
31. A train moving with a constant retardation travels 87 feet
in the first second and 85 in the next. How far will it go before it
comes to rest ?
32. A train starting from rest travels i foot in the first second,
3 feet in the next, 5 feet in the next, and so on. How long will it
be before it attains a velocity of a mile a minute ?
33. A point moving with uniform acceleration describes 52 feet
in the half second which elapses after the third second of its
motion. Compare its acceleration with that of a falling body.
34. A uniformly accelerated body passed two points 6 feet
apart in one third of a second ; four seconds after reaching the
first of these points the body had a velocity of no feet per second.
Find the distance travelled in one second after reaching the
second point.
35. A passenger in a train moving with uniform acceleration
observed that in one minute and a quarter the train travelled a
mile. He waited a minute and then found that the train took
Kmematies 205
another minute to travel the next mile. What was the velocity at
the end of the last minute and what was the acceleration ?
36. It is observed that as a vessel is steaming 10 knots an hour
due N., while the wind is blowing from E.N.E., the direction of
the smoke track is from N.E. to S.W. Show by a diagram how
you could ascertain the velocity of the wind.
yj. A ship is steaming due N. with a speed of 15 miles per
hour, and another due E. with a speed of 20 miles per hour. The
second passes the point of intersection of the two tracks i hour
before the first reaches it. Find the least distance between the
ships.
2o6 Dynamics
CHAPTER XII
MOTION DUE TO THE FORCE OF GRAVITY
169. The measure of the force of gravity on a body is
called the body's weight (Article 7).
On account of this force, we find that when a body is
allowed to fall freely in vacuo it moves vertically downwards
with a constant acceleration, which we denote by g.
Since the earth is not a perfect sphere, the force of gravity
varies somewhat at different points on the earth's surface ;
hence the value of g varies slightly. Its value at the Equator
is about 32-08, and at the Poles 32'2S when measured in feet
and seconds.
In England its value is approximately 32'2, or. more
accurately, 32' 18.
In working numerical examples, its value may be taken to
be 32, unless it is otherwise stated.
Thus g= a.n acceleration of velocity vertically downwards
of 32 feet per second per second.
This value of g only applies to the motion in vacuo.
In nature the resistance of the air tends to retard the
velocity. This retardation will not be taken into account.
170. The formulae of the preceding chapter may now be
used for vertical motion under the action of gravity by putting
a = ^ when the motion is measured vertically downwards.
Thus v = u + gi,
V'^ — u"^ + 2gS.
Motion aue to uravity 207
If the body starts from rest u = o, and we have
v=gt,
r;2 := 2gS.
If the body is projected vertically upwards with a velocity
u we put a = — g, and we have for the upward motion
v = u- gt,
s=ut-\gf,
V^ -^U^ — 2gS.
If we use these latter equations in solving numerical ex-
amples, a negative answer for v means that the body is on its
way down, and a negative answer for j means that the body is
below the point from which its motion is considered.
171. A body is projected vertically upwards with a velocity
u. To find the greatest height to which it will go, and the time
it takes to reach this height.
When it arrives at its greatest height its velocity will be
zero ; hence we put » = o in the equation,
v^ ^ u^ — 2 gs,
=
u"-
-■2gs;
.*. ^ =
«2
2g
To find the time.
put v =
■ in
the equation,
V =
: u —
gt.
=
- u -
-gt;
.-, i =
u
g
Also, as in Article 167 of the last chapter, the time it takes to
go up is equal to the time it takes to come down ; the velocity
with which it returns to the earth is equal to the velocity with
which it was projected ; the velocity with which it passes any
point on its way down is equal to the velocity with which it
passed the same point on its way up; the time between any
two points on the way up is equal to the time between the same
two points on the way down.
2o8 Dynamics
Example I. — A man stands at the bottom of a cliff loo feet
high ; with what velocity must he throw a stone up in order that
it may just reach the top ?
put Z/ = O, J = ICO,
o = M^ - 64 X 100 ;
.•. a' = 6400,
or « = 80 feet per second.
The time taken by the stone to get there = 2i seconds.
Example 2. — A stone is projected vertically upwards, and on
its way up passes a point 640 feet above the ground with a velocity
of 96. Find the time which elapses before it reaches the ground.
Here we will consider that j is measured upwards from the
point 640 feet above the ground, and substitute in the equation
s = ut — ^gf,
— 640 = 96 ^ - 16 /',
^^ - 6 i" - 40 = o
(/- 10) (if + 4) =0;
.'. / = 10, or — 4.
Hence the stone will reach the ground 10 seconds after it
passes the point 640 feet above the ground.
The negative answer means that the stone started from the
ground 4 seconds before it reached the above point.
The answer of 10 seconds is thus made up as follows : —
3 seconds from the given point to the highest point,
3 seconds back to the given point,
4 seconds from the given point to the ground.
Example 3. —A stone is dropped down a perpendicular clift h
feet high. At the same time another stone is thrown up from the
bottom with a velocity which would just enable it to reach the top.
When and where will they meet ?
Let / be the time after which they will meet, x the distance
from the top, u the velocity with which the stone is projected from
the bottom.
Then • u' = 2g ^ h,
h - X = ut - ^gf;
Motion due to Gravity 209
adding, we get h = u i ;
n^\l-
h
V2g/i V 2g
' A /— = ~ (time required tcJ fall h feet);
2 V
.'. V = u . y-J^ = u (sec a — tan a).
cos j3
i86.* Sometimes in solving problems we shall find it
advantageous to use the equation to the path referred to the
horizontal and vertical lines through the point of projection as
axes.
Definition.-^Th& equation to a curve referred to any two
lines as rectangular axes expresses the relation which must
exist between the distances of any point on the curve from
these two lines.
These distances are called the co-ordinates of the point.,
(See Trigonometry, Art. 4.)
Thus, in Article 178, if Ox, Or are the rectangular axes
through the point of projection a-, y the co-ordinates of P after
any time t,
X := O N = z< cos a . /,
J)/ = P N = asina . t—\gt'^-
Eliminate t.
y = uw\a\ \ — \g\ J I
\U cos a. J \u cos a.)
y =^ X tan a —
Projectiles 235
nte^
whose latus rectum is
The curve represented by this equation must be a parabola
2 u^ cos^ a
S
Thus, to find the range on an incHned plane, let yS be the
inclination of the plane and R = the range ; then
jc = R cos /3, and jv = R sin j8 ;
.-. R sin S = R cos S . tan ^ - S^\ '^°^^ ^ ■
2 u^ COS^ a
/. ^R^ cos^ 13 = R (cos /? tan a — sin yS) x 2 u^ cos'^ a ;
. Tj _ 2 it^ cos^ g / sin (g — /3) ^
^ cos^ y8 V cos a y
2j^ cos g sin (g — /S)
g ' cos^/8
187. Many experiments have been made with the object of
determining' a practical formula for the horizontal range of a
projectile when the resistance of the air and the size and weight of
the projectile were taken into account.
Up to a certain point the resistance of the air increases with
the velocity. A large number of experiments carried on in i85r
led Professor Helie to the conclusion that the resistance of the air
at practical velocities was more nearly proportional to the cube of
the velocity than to any other working expression.
He constructed a formula for the trajectory by empirically
modifying the formula given above for the path of the projectile in
vacuo thus —
^ = ;.tana--^:^(i+^Y
2 cos'' a \u- U /
where k = 0-0000000458 — ,
w
d = diameter of projectile in inches,
w = weight of projectile in pounds.
The above value of k is for ogival heads which suffer least
retardation from the air ; with other shaped heads a different
coefficient of resistance would be required.
Putting J/ = o in the above equation, and dividing the equation
by X and simplifying, we get the following quadratic equation for
X the horizontal range —
236 Dynamics
gx (i -^ kux) = ic' %\rv2a = 'K . g,
where R is the range of a particle in vacuo, which gives
an expression for the range which at low elevations for initial
velocities between 800 £s. and 1,400 f.s. is very fairly accurate, and
which furnishes the simplest method yet devised of roughly con-
structing a range-table. (Encyc. Brit. 9th Edit. vol. xi. p. 302.)
If we expand the surd in the above expression by the binomial
theorem, and neglect terms which contain higher powers of k than
the second, we get
zkuy 1 o J
= -4— (2 kuK- i k" u' R\
iku^ '
= R{i - iuR],
Example i. — A bullet is projected with a velocity of 200 feet
per second at an angle of 30° with the horizon. Find the horizontal
range and the time of flight ; find also the maximum range for this
velocity of projection.
Horizontal velocity = 200 cos 30° = 100 a/j.
Vertical velocity =200 sin 30°= loa
Time of flight = seconds.
Range = 100 ^/3 x
20000 a/3
= — = 1,020 feet.
o
Maximum range =— = 4oooo ^ j , cq feet.
Example 2. — A particle is projected from the top of a per-
pendicular clifif A feet high with a horizontal velocity u. Find the
distance from the cliff of the point when it strikes the ground.
The time it is in the air = time it would take to fall freely
through h feet. If / be the time, h = \ gt\
• /_ l^h
/2h
"Vg-
Projectiles 237
The horizontal velocity u is constant ;
.'. the distance of the particle from the diff when it strikes the
ground =ut = u 4/ —
= ?^^feet.
4
Example 3. — A man can just throw a cricket ball 108 yards.
With what velocity does the ball leave his hand, how high will
it rise, and how long will it be in the air ?
If w be the initial velocity, the maximum range = — ;
g
- = 3.4;
.-. ^' = 324^,^
= 102 feet per second nearly.
The direction of projection for maximum range = 45° ;
.■. vertical velocity = 72 feet per second = horizontal velocity.
Greatest height = i^ = 8 1 feet.
Time of flight = — ^'— = 4^ seconds.
g
Example 4. — A number of particles are projected at the same
instant with different velocities from the same point in the same
direction and in the same vertical plane. Prove that at any subse-
quent time they all lie in a straight line.
Let O A B be the common direction of projection, and suppose
that A and B would be the positions of any two of the particles after
a time t if gravity did not act ; now, in consequence of the action of
gravity, they are at A' and B' vertically beneath A and B, such
that A A' = B B' = \gt'^ ; :. A' B' is parallel to A B ; hence after
any time all the particles lie on a line parallel to the direction of
projection.
238
Dynamics
Example 5-— A particle is projected from a point O with
velocity u at an angle a ■
A p is its position at any
subsequent time. If A
^1 X be the highest point, P N
the perpendicular from
P on the vertical through
^ A, prove (by dynamical
Fig. 134. considerations) that
PN- = AN X constant.
Draw P M l' to OC. Then
and
AC = '
^ r^ zt sin a
O C = . z^cos a;
g
PN=OC-OM
ic^ sin a cos a
g
tc cos g
g
— U cos a . /
(u sin a-gt) \
AN=AC-NC
= AC-P]VI
u' sin' a
PN- =
■^g
U^ cos' a
ti' cos' Q
■2 k' cos'
- (z^ sin a.t-^gf) ;
{u sin a-gff
{ii'^ sin' a — 2u sin a.gt+g" f-)
g
1 u' cos' a
a /a' sin' a . J 1 jo\
- I u s\na.t+^gt-\
\ 2 g /
^g
.AN;
.-. P N' = constant x A N.
This result proves that P is on a parabola whose axis is A C,
and whose latus rectum is
2 u' cos' a
Example 6. — An army is encamped on a hill h feet high. An
opposing force is on the level ground at such a distance from the
hill that it subtends an angle ^. What is the least velocity of
jt^ reject ties 239
■projection that will enable a cannon ball to reach the top of the
hill?
The distance of the top of the hill from the opposing force is
-. - . This must be the maximum ranee on a plane whose in-
sm/i or
chnation is for the least velocity of projection.
Hence if u be the least velocity of projection,
u' h
^(i + sinj3) sin/3
„2_^'^(i + sin/:
(Art. 182.)
sin/3
= gh (cosec/3+ i).
Examples.
I. If the greatest height attained by a projectile be 36 feet,
prove that the vertical velocity is 48.
2. A bullet is projected at an angle cos-' f to the horizon with
a velocity of 1,200. Find the greatest height and the horizontal
range.
3. If the velocity of projection be 660 feet per second, find the
maximum range in miles.
4. What is the direction of projection when the horizontal range
is four times the greatest height ?
5. Determine the direction of projection when the greatest
height acquired is equal to the horizontal range.
6. If the horizontal range of a cannon ball be 12 miles when
the direction of projection is sin-' f , find the velocity at the highest
point of the path.
7. A stone is projected at an angle 30° with the horizon.
Find the velocity of projection in order that the greatest height
may be 9 feet above the horizontal plane through the point of pro-
jection, and find the range on this plane.
8. A bullet leaves a gun with a horizontal velocity of 1,200 feet
per second and a vertical velocity of 20 feet per second. Find
the greatest height and the horizontal range.
9. If V be the velocity and a the angle of projection, and if v
be the velocity and ^ the angle which the direction of v makes
with the horizontal at any subsequent time, v = N cos a . sec ^.
(r;-«>
240 Dynamics
10. A body is projected with a velocity of 160 feet per second
at an angle a with the horizon, determine the velocity with which
it strikes an object 100 feet above the ground.
11. If H be the greatest height of a projectile, R the horizontal
range, V the velocity of projection, a the angle of projection, prove
that tan a = 4Ji^ and that R' = 16 H
R
12. Prove that the maximum horizontal range of a projectile
is equal to the latus rectum of the parabolic path.
13. Prove that the maximum range is equal to twice the height
to which the velocity of projection is due.
14. A man can just throw a stone vertically upwards to a height
of 100 feet ; how far could he send it in the same horizontal plane ?
15. The maximum range of a projectile is 1,600 yards. Find
the time of flight.
16. A stone is thrown at an angle of elevation of 60°, so as just
to strike the top of an object 54 feet high, whose base is on the
same level as the point of projection, and which subtends an angle
of 30° at that point. Find what the velocity of projection must be,
and determine the time of flight.
17. If a particle be projected with velocity u at an angle a from
a point in an inclined plane of inclination (3, then its inclination to
the horizon at the point where it strikes the plane is given by the
equation, tan 6=1 tan ^ — tan a.
18* If a projectile strike an inclined plane (/3) at right angles,
the range on the plane is
2 u"- sin/3
^ ■ I + 3 sin^ (3
19. A smooth square board, whose side is 51 feet, is placed with
two of its sides horizontal and is inclined at an angle 45° to the
horizon. With what velocity must a particle be projected horizon-
tally in the plane of the square from one of the top corners in order
that it may strike the opposite corner? (Take a/t = i^^-,.)
20. A body is projected from a given point with velocity v at
an angle a to the horizon. When it reaches the highest point
another body is projected from the same point with velocity v' at
an angle a' so as to strike it. Prove that
v_ _ cos a' __ 2 sin a'
1/ COS a sin a
Projectiles 241
21. If at two instants during the motion of a projectile the
direction of motion make angles a, 8 with the horizontal, then the
time between the two observations is (tan a — tan j3) _, where u is
the horizontal component of the velocity of the projectile.
22.* A shell is projected with velocity V at an angle a with the
horizon, and when at its highest point it explodes into two equal
parts, one of which rises vertically to a height equal to the height
at the moment of explosion ; where will the other part strike the
ground ?
23. If two particles be projected from the same point in the
same vertical plane with equal velocities (z/) in different directions,
so as to have the same horizontal range (R), find the difference
between their times of flight.
24. If the velocity of projection be that due to the vertical
height a, and h be the greatest height reached by the projectile,
then the horizontal range = 4 '^ah — K'.
25. Prove that the expression for the range in the preceding
question is a maximum when h = ^ a.
26.* A particle, after sliding down an inclined plane, moves in a
parabola of given latus rectum. Find the inclination and length
of the plane in order that the focus of the parabola may be verti-
cally below the point from which the particle started.
27. Prove that the velocity at any point of the path of a pro-
jectile varies as the perpendicular let fall from the focus on its
direction at that point.
28. Prove that the line joining the two foci of the paths which
have the same range on an inclined plane is perpendicular to the
plane.
29.* Two particles are projected with the same velocity from the
same point, but with different inclinations, in the same plane ; prove
that when their directions are parallel they are moving at right
angles to the line which bisects the angle between their directions
of projection.
30. If a be the direction in order that a particle with a given
velocity of projection should pass through a given point, find an
equation for tan a. What is the condition that the roots of this
equation should be equal ? Hence find an expression for the maxi-
mum range.
R
242 Dynamics
31.' A particle is projected at an elevation of 45° from the ex-
tremity of the horizontal diameter of a vertical circle with such a
velocity that, if it had been projected vertically upwards, it vi'ould
have risen to a height of one-eighth of the vertical diameter.
Find where it will strike the circle.
32.* The maximum horizontal range for a particle with a given
velocity being R, determine the elevation necessary for it to strike
an object distant 2 R sec a (tan o)* on a plane through the point of
projection depressed a° below the horizon.
243
CHAPTER XIV
FURTHER CONSIDERATIONS OF THE LAWS OF MOTION
1 88. The three Laws of Motion as given by Newton are
as follows : —
Lex L — Corpus omne perseverare in statu suo qidescendi vel
movendi uiiiformiter in directum, nisi quatenus a viribus
impressis cogitur statum suum mutare.
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 forces to change that state.
Lex II. — Mutationem motus proportionalem esse vi motrici
impresscE et fieri seauidum lineam reciam qua vis ilia iinprimitur.
Change of motion is proportional to the impressed force,
and takes place in the straight line in which that force is
impressed.
Lex III. — Actioni contrariam semper et cequalem esse re-
actionem : sive, corporum duorum actiones in se mutuo se??iper
esse cequales et in partes contrarias dirigi.
To every action there is always an equal and opposite
reaction; that is to say, the mutuil actions of any two
bodies are always equal and in opposite directions.
There is no possible a priori proof of these laws. They
have been deduced in the first instance from observation and
experiment, and from them the whole science of dynamics
and astronomy has been built up ; if rigorous proof be looked
for, it will be found in the fact that the ' Nautical Almanac ' for
each year is published some years in advance, and that the
E 2
244 Dynamics
predictions contained in it concerning the places of the
heavenly bodies from day to day are found to agree most
exactly with our observations. These predictions depend on
the theory of astronomy, which is based on the laws of
motion ; we cannot, therefore, but believe in the truth of these
laws, since the results which are deduced from them are so
universally corroborated. Any deviation between deduction
and observation would not make us doubt the truth of our
laws, but would lead us to the discovery of the disturbing
cause ; and, in fact, such deviations in the past have led to the
discovery of a planet whose existence was not previously
known.
189. Though called Newton's laws of motion, it cannot be said
that they were discovered by him. Previous to the time of Galileo
the popular ideas about force and motion were in a very confused
state ; Galileo was the first to put Laws I and 2 into a satisfactory
form, and it is probable that his death (1641) alone prevented him
from arriving at a satisfactory statement of Law 3. About 1669
papers were sent to the Royal Society of London by Wren, Wallis,
and Huyghens, which show that people were gradually coming to
apprehend the third law of motion in its most general sense.
These papers stated that in the mutual impact of two bodies ' the
quantity of motion remains unaltered,' or ' the momentum gained
by the one body is equal to the momentum lost by the other.'
Newton was the first to collect the knowledge acquired by his
predecessors — much of which he extended and greatly improved
upon — and to present the whole subject in a systematic and
intelligible treatise. -(Whe well's 'History of the Inductive
Sciences,' Book VI.)
190. As we have already seen in Chapter I., the first law
states the property of matter called Inertia, by virtue of which
ilmaintains its state of rest or uniform motion in a straight
line ; it also gives us our definition of Force, as that which
changes the inertia of a body. If a body changes its state of
inertia, it must be on account of the action of some force or
forces ; if a body's state remain the same, no force acts upon
it. Newton extends this property also to the state of a body
which is rotating about an axis, as in the case of a hoop ; so a
Laws of JMotion 245
wheel rotating about an axis through its centre will maintain
this state of rotation if it is not acted upon by forces such as
the resistance of the air and the friction of the axis in the
sockets.
In no case can we prove this law by experiment, since it is
not possible in nature to entirely eliminate the forces which
resist motion ; but the more nearly we eliminate these forces,
the more nearly does the motion of a body, unacted on by
force, approach to being uniforiTL
The nearest approach to uniform motion which we can
observe in nature is the rotation of the earth on its axis. This
motion is scarcely altered at all by external forces, and is so
nearly uniform that we use it as our measurement of time, and
say that equal intervals of time are the times during which the
earth turns through equal angles.
Thus, if 24 hours be the time which the earth takes to rotate
through an angle of 360°, it rotates through 15° per hour, or 15'
per minute, or 15" per second.
191. The second law not only contains the first law by
implication, but makes the definition of force, which is con-
tained in it, more complete by supplying us with a measure of
force.
The word motus in the original stands for quantitas motus,
or quantity of motion, to which we always give the name
momentum. The second law says ' that change of momentum
is proportional to the force which produces it,' implying that if
there is no change pi momentum there is no force, and vice
versa ; and further adds that if there is a change of momentum
in any direction it is due entirely to the force which acts in
that direction, which force is proportional to, and may there-
fore be measured by, that change of momentum. We therefore
choose our unit force to be the force which produces in one
second a change in momentum of one unit of momentum, and
we measure a force of any magnitude by the change of
momentum which it produces in one second in the direction
in which it acts. Thus, F = ma ; where F is the force, 7H the
mass of the body it acts upon, a the acceleration produced
246 Dynamics
in the direction of the force. (See Chap. I., and especially
Art. 26.)
Further, the second law implies that the change of
momentum due to any force which acts on a body is the same
whether that body be at rest or in motion, the same whether
other forces act on the body or not, the same as if it alone
acted on the body at rest. This law enabled us in Chapter II.
to deduce the Parallelogram of Forces from the Parallelogram
of Velocities, and from the Parallelogram of Forces we have
worked through the whole subject of statics. For this reason
it was stated in Article 34 that of the three laws the second
was, for present purposes, by far the most important.
Not only does the second law enable us to measure force
but it also enables us to measure mass. For if F = m a,
711 = ; consequently, if the same force act on bodies of dif-
u
ferent mass, these masses must be inversely proportional to the
accelerations which are communicated to them by the force.
In the present chapter the equation F = m a, which states
the second law in an algebraical form, will be applied to the
working out of various questions depending on the relations
which exist between the force which acts, the mass of the
particle it acts upon, and the change of velocity which is
thereby produced.
192. Thus we shall find that the second law is sufficient to
determine the motion of a single particle when it is acted upon
by any forces, as it has also been sufficient to deduce the
conditions of equilibrium of a number of forces in statics ; but
when we come to consider the motion of a system consisting
of two or more particles we shall find that this knowledge is
not sufficient for us. The additional knowledge required is
then supplied to us by the third law, which tells us that the
effect of A on B is always equal and opposite to the effect of
B on A.
It has been pointed out in the first chapter (Art. 4) that
force is always of the nature of a stress — that is to say, there
can be no force without an equal and apposite reaction.
La%vs of Motion 247
Now since force is measured by the change of momentum
which it can produce, we see that if the momentum of any
part of a system be changed without the action of a force
external to the system there must be an equal and opposite
change of momentum communicated to the rest of the system.
This condition must hold all through the system. It therefore
follows that the sum of the momenta of the various parts of a
system, parallel to any given direction, must remain constant
so long as there is no external action, whatever be the mutual
action between the various parts. This is called the principle
of Conservation of Momentum. Hence it follows that the
centre of mass of a system of mutually influencing particles
obeys the first law of motion — that is to say, it moves uni-
formly unless acted on by some force external to the system.
193. This law is of the very greatest importance, although
i'ls full value cannot be appreciated by the student until he has
advanced beyond the limits of this book. To the beginner
the second law (F = m a) seems of paramount importance, for
it enables him to deduce the conditions of equilibrium of a
system of forces, and it enables him to determine the motion
of a particle acted on by any forces. But in nature we more
often have to deal with motion than rest, and we never meet
with particles ; consequently the results achieved by means of
the second law are of small practical importance until they are
developed by means of the third law.
As examples the student may consider the following : —
When a gun is fired off, the kick on the shoulder is
measured by the momentum generated in the shot.
When two moving bodies meet, the momentum gained by
one is equal to the momentum lost by the other — i.e. there is
no change of momentum.
If two bodies, connected by a string, be projected into
space, gravity being the only force, the motion of their centre
of mass is a parabola, though the paths of the bodies are very
various.
The attraction of the sun for the earth is equal and
opposite to the attraction of the earth for the sun, and similarly
248 Dynamics
for the mutual action between any two of the planets of the
solar system ; hence, although the motion of the centre of
mass of the solar system is uniform, the motion of each planet
is very complicated.
194 To deduce the equation F = 7« a from the second
law of motion.
Although ti7ne is not mentioned in the law, it is obvious
that in estimating the effect of the force we must take into
consideration the time during which it acts.
Let u, V be the velocities of the mass m at the beginning
and end of time /, F the force. Then, change of momentum
oc force,
inv — mu'x.Y . t,
m(z>-u) ^-p
t
We now choose F = 1, when ot ^ i and a = ij that is, we
take our unit force to be the force which, acting on the
unit of mass, produces in it the unit acceleration ;
^, m . a. F
then =— :
I.I I
.*. ma = F.
195. Taking the unit force to be the force which can
produce the unit of momentum in the unit of time, deduce
the equation F = w a.
Let M = number of units of momentum produced in the
mass m in time t ; then
M. = mv —,m u.
Unit force produces i unit momentum in i unit of time ;
.•. F produces F units of momentum in i unit of time ;
.-. F
))
F . ^
t „
!ut F
»
M
.-. F . /■ = M
^ mv — mu;
F = — ^— ' = m a.
/unitsoftime;
F ^ m I
249
196. The equation W =^ 'ug is a particular form of the
equation F = ma.
W is the measure of the force of gravity on a body of
mass m, g is the acceleration it produces in it.
Moving force ^ mass moved x acceleration produced,
W = m X g.
Hence, taking the ft. lb. sec. units, weight of i lb. = ^
poundals ; and, generally, the measure of a force in poundals
^ g times the measure of the force in lb. wt. units {g = 32 'a).
So also, taking the C. G. S. units, the measure of a force in
dynes = g times the measure of the force in gramme wt. units
Cg-=98o).
197. Two masses m and ui' are connected hy a light string
passing over a smooth pulley. Find the acceleration of the
system and the tension of t/ie string.
Let T be the tension of the string, a the acceleration of
the system.
Suppose m greater than m'. Then m moves
downwards with the same acceleration as m' moves
upwards.
Moving force acting downwards on m = m g
— T; moving force acting upwards on m' = T
— fi'g; .: by the second law of motion
mg — T = ma .... (i)
T — m'g = m a . . . . {2}
Adding, we get mg — nig = {ni + m') a ;
. „ — (^ - m ')i
(3)
T,>
Tl(^
Fig. 135.
(4)
m + 7n
Substituting for a in equations (i) or (2) we get
ry 2 m m' g
m + m'
Note. — Equation (3) might have been obtained at once by
considering the two masses as one mass {?n + ;«') acted on
by a force equal to the difference between the two weights
{mg-m' g).
250 Dynamics
T the tension is greater than m' g and less than mg.
If the masses be equal there is no acceleration, and
9
rp _ 2 m g _ ^^ _ jjjg weight of either ; that is, the tension
2 m
is the same as if the system had been at rest.
This example embodies the principle of Atwood's machine,
which we will consider at the end of the chapter.
It will be noticed that we have neglected
(i) the mass of the string ;
(2) the inertia of the pulley ;
(3) the friction of the pulley.
Of these (i) is inconsiderable if iine silk thread be used,
but in practical work (2) and (3) must be taken into account.
Example. — Masses of 7 lbs. and 9 lbs. are connected by a
string passing over a smooth peg ; find the acceleration of the
system and the tension of the string. With the same notation
9^-T = 9a,
T-7.g' = 7a;
.-. 2^= 16 a,
or a = ^g = 6, feet per second per second.
T = 9^-9a = 9Cg--^^) = ^|.^
= 252 poundals.
198. A mass m hanging freely is connected by a light string
with a 7>iass xsx' on a smooth table ; to find the acceleration a.
In considering the motion of m', we notice that T is the
„;^ only force which acts on it in the direction
:^a of motion ;
.". T = ;«' a . , . (i)
For the motion of m vertically down-
wards we have
e
HI
Fig. 136. rp / X
mg —T: = ma .... (2)
Adding (i) and (2) mg= m' a-\- ma.;
m' + m
F=-mcL 251
Hence it is obvious that, however small m may be, and
however large ni' may be, motion must take place provided
that the table be perfectly smooth.
199. If two masses m' and m be comiected as in the preceding
article, and it is observed that m' moves over s feet in t seconds,
starting from rest, to determine the coefficient of friction between
m' and the table.
If F = force of friction on ;«',
R = pressure of the table on ni ,
a. = acceleration of the system,
we have the following equations : — •
T — F = ot' a for the motion along the table,
R — m' g^o for the motion U to the table,
F = /.R;
,'. T — jx m' g^ m' a.
For the motion of m we have
mg — T ^ ma ;
.'. mg — fji m'g = (m' + m) a;
m f , m\ a
m \ m'J g
and
2 s
hence /x, can be found.
It is obvious that ;« must be greater than fxm', or there
will be no motion.
200. To find the motion of a particle rafree to move on a
rough inclined plane of inclination &.
Let a = the acceleration,
R = the pressure of the plane on the particle,
F = the force of friction.
Resultant force U to the plane = R — w^ cos 0.
Resultant force down the plane = ot^ sin ^ — F j
.•. R — mg cos 6=0,
252
and
and
Dynamics
tng sin 6 ■
F = ma,
,*. mg sin 6 — ft.mg cos 6 = m a;
.'. a = (sin 6 — fi. cos 6) g.
We note that tan 6 must be greater than ft..
Similarly, if the particle were projected up the plane with
any velocity u, the retardation down the plane would be
(sin 6 + /x, cos 6) g,
and it would come to rest after passing over a distance
^ u^
2 g (sin ^ + /i cos 6)'
201. A mass m' resting on a smooth inclined plane of in-
clination & is connected witli a mass xo. by a light string passing
over a pulley at the top of the
' -|- plane so that m hangs freely ; to
find the acceleration a.
(i) If the plane be smooth,
considering the motion of m',
since there is no acceleration
perpendicular to the plane, the
forces perpendicular to the
plane must be in equilibrium ; /. R ^ m' g cos 6'.
The resultant force up the plane =^T — m' g sin 6' ;
.'. T — m' g sin 6' = m' u. (t)
Moving force on m = mg — T ^ ma (2)
.". mg — m' g sin & ^ (m -r ni) a
m — m' sin 6'
a = . g.
m + m'
If m be less than m' sin 6', the motion will be down the
plan?.
(2) If the plane be rough.
Let /i = the coefficient of friction,
F = the force of friction opposite to the direction of
motion.
+ 7710
Fig. 137.
F= ma
25 J
Then F = /^.R = ju, m' g cos 6'.
Hence equation (i) becomes
T — m' g sin Q' — )xm' g cos B' = m' a ;
m — wYsin ^' + «. cos &\
• • a = ^^ ; r~ • S-
If the acceleration were down the plane F would act in the
same direction as T, and the equations of motion would be — •
tn' g sin & — F — T = ni a,
T — m z = ma\
m' g sin ff — F — m g
{m + vi') a ;
m'ism 0' — u cos 6') — m
• • '^ = — ^^ ^—-1 • S-
m + m
Hence, for motion on the plane either up or down, m
must not lie between
w/(sin & — fi. cos 6') and 7« (sin $' + fi cos 6').
This result coincides with the result obtained in Article 134.
If both m and m! be on rough planes whose inclinations
are 6 and 6', the coefficient of friction being /a, the student
will easily obtain the following expression for a when m is
moving down the plane
m (sin & — u, cos
a = — ^
— OT'(sin 6' + fx cos 6)
m + m'
202. To find the pressure of a mass m ivhich is at rest on a
ho7-izontal plane which itself has a vertical acceleration a.
Let R be the pressure between the body
and the plane, then B. — jn g is the resultant
moving force upwards.
Hence, if a = the vertical acceleration,
'R — m g^ 7na ;
.-. R = m(g+ a).
Similarly, if the motion be downwards,
mg — R = OT a ;
.-. R = m{g— a).
-nR
'"3
Fig. 138.
254 Dynamics
If W be weight of the body, W = mg;
.•.R = ^(f±.)
according as the motion is up or down.
Atwood's Machine.
203. This is a machine constructed for the purpose of
illustrating the first two laws of motion ; it can also be used for
making a rough calculation of the value of ^.
Referring to equation (3) of Article 197 we have,
m g — m! g = {in + w') a ;
m + m! may be kept constant, while m — m' may be made to
take any value less than m + m'. We may thus determine the
accelerations produced by different forces on the same mass.
Force oc acceleration.
Again, m — m' may be kept constant, while m + m may
be made to have any value we please. We may thus compare
the effects of the same force on different masses,
Acceleration oc .
mass
Again, the equation may be written
m + m'
* 7n — m'
We can choose m and m' so that a may be made so small
as to enable us to determine it by observation ; we can thus
determine the value of ^.
204. In using the machine we take two equal weights, each
of mass i?i, and fasten them to the ends of a fine thread
passing over a pulley in which the friction is reduced to a
minimum by letting the extremities of its axis rest between
friction wheels. (See Deschanel's 'Natural Philosophy.')
To one of the equal masses m a rider of mass m' is
Atwood's Machine
255
attached ; this at once makes the system begin to move with
an acceleration
2 m + m'
The distances traversed by the ascend-
ing or descending mass in any time can
be observed by means of a graduated
scale placed in close proximity to their
path.
The rider can be detached at any
time by making the descending mass pass
through a ring R, through which the rider
cannot pass. After the rider is detached
the acceleration vanishes and the velocity
becomes uniform.
The distance traversed with this uni-
form velocity can now be observed for
any given time, and hence the velocity can
be calculated.
Thus, suppose the descending mass to
start from rest at A, to move with uniform
acceleration a to R, when the rider is de-
tached, and then to move with uniform
velocity v to S. We can make the follow-
ing observations : —
Distance A R = « ;
„ R S = ^ ;
Time from A to R = ^ ;
„ R to S = t'.
By varying t we can prove that a oc ("2,
and that v^t.
Dfl-OR
iri
^■--O'^
We have also from our formulae iz = ^ a . t^
v^ at .
v'' = 2 a . a
b — v.t'
256 Dynamics
2 a
Theoretically a can be calculated from (i), a = — ; or
from the equation, b =■ a . t . f . Practically it is easier to
measure f than t ; we therefore eliminate v between equations
(3) and (4) ; thus
la . a \
. . a
i =
2 m + iri
; — • "■
m
2 m + m! l/^
m 2 at
'2'
an equation from which we may determine g.
For a description of the apparatus used for measuring the time,
see Deschanel's ' Natural Philosophy.'
For the allowances to be made for the friction and inertia of
the pulley, see also Deschanel, or Glazebrook and Shaw's ' Practical
Physics,' p. 137.
It is shown in the latter place that the above equation must be
written
_ 2 in + W + m' — F b"-
^ ~ m' - F ■ 2 at'-'
where W and F can be determined by experiment.
205.* With the most delicate instruments that can be made
the value of ^ thus obtained is only a rough approximation to its
actual value, owing to many unavoidable sources of error.
The most accurate method of calculating the value of g at any
particular spot on the earth's surface is derived from observations
of the time of oscillation of a simple pendulum, which consists of a
small heavy particle suspended from a fixed point by a fine thread.
Thus, if / be the length of the thread and t the time of a com-
plete oscillation,
IT
27r
^\ ^')
•■S ^2 '
or if ;z be the number of complete oscillations in i second,
t = - ; : . g = 4 ir'^ n"- 1,
n
Laivs of Motion 257
Both n and / are capable of the most exact measurement.
It is necessary for the student to defer the proof of equation (i)
until he has advanced a good deal further in the subject.
Example i. — A particle whose mass is 15 lbs. is at the centre
of a smooth circular table whose radius is 4 feet, and is connected
by a fine thread with a mass of i lb. hanging just over the edge.
If the height of the table be 3 feet, how long will the heavy
particle take to get off the table ?
Let T be the tension of the string, a the acceleration, then
T=I5a,
I.^-T = a;
.-. \6a=g;
.'. a = 2 feet per second per second.
Let t be the time taken, v the velocity acquired in passing over
3 feet ; then
3 = ^ a /-, and •z/'^ = 2 a X 3 ;
.'. /= ^73, and^' = 2^/3.
The string now becomes slack, and the particle moves over the
remaining distance of i foot with uniform velocity v ;
:. whole time taken = v^ + -^ = ^-7^ seconds
= 2 seconds approximately.
Example 2.— A sand bag of 50 lbs. is on the floor of a balloon,
which is ascending with an acceleration of 8 feet per second per
second. Determine its pressure on the floor.
Let R be the mutual pressure between the bag and the floor.
Then the moving force = mass moved x acceleration.
R- 50^= 50x8,
R = So(^+8)
= 50 X 40
= 2,000 poundals,
or 62 1 lbs. wt.
Example 3. — Two masses of 5 lbs. each are connected by a
string and placed on a rough plane inclined to the horizon at an
angle sin"^|, so that one just hangs over the top when the other is
S
258
Dynamics
at the bottom. If the coefBcient of friction be \, and the height of
the plane 12 feet, find the time taken by the freely hanging mass
to reach the base of the plane, and
determine how near the top the
other mass will get.
Let A B C be the plane,
angle B A C = 5 = sin-if ;
.•. sin 5 = f, and cos 5 = ^.
Since B C= 12 feet,
F>=- 14°- A B = f X 1 2 feet = 20 feet.
Calling the particles P and Q, when Q is at C, P will be at D,
a point 8 feet from B.
If F be the force of friction, and R the pressure between P and
the plane,
R = 5^cos^=5^x ^ = 4_^poundals ;
.". F=;aR = J X 4^=^poundals.
Let T be the tension of the string, a the acceleration, then the
moving force on P up the plane
= T-5^sin5-F
.-. T-4^=5a.
For the motion ofQ S^-T = 5a;
.'. g=ioa.
Let V be the velocity acquired, and t the time taken in passing
over 12 feet from rest with an acceleration — .
10
Then I2 = i^ . /^;
^ 10
f =
12 X 20
7-5,
i=2-j seconds.
After this time Q reaches the ground, and P passes the point D.
»' = 2 . *_ .
10
10 '
12
Laivs of Motion
259
As P passes D the string becomes slack, consequently P con
tinues the ascent with an initial velocity v, vvhich is subject to a re
.F
m
tardation due to the friction of the plane, which is = '' =-g, and
also to the iretardation of gravity resolved down the plane.
Let J = distance travelled by P above D before it comes to rest,
then o = z/^ — 2 . Vl\^°\s;
9,g Sgxjo ^
.•. P gets within 6 J feet from the top of the plane.
Example 4. — P, Q and R are three masses, of which P and Q
are connected by a light string passing over a smooth pulley, and
R is attached to Q by another light string of length / ; P is > Q
and < Q + R. Motion is allowed to commence when Q is on the
point of leaving the plane on which R rests ; to find the time that
will elapse between the time that R leaves the plane and strikes it
again.
P-Q
Acceleration of P and Q =
P + Q
r'^
Let V = the velocity acquired in passing through
the distance /,
P-Q ,
"" = ^pTq-^^-
Let z/' = velocity with which R leaves the plane ;
then, since the momentum immediately after the
addition of R = momentum before.
-y (P + Q + R) = (P + Q) V,
V
Fig. 141.
P + Q
— v.
P + Q + R
Acceleration of the system is now opposite to the direction of
, Q+R-P ^ ,
motion and = ^-p^-pp .g=a.
Time in which v' is destroyed by a retardation a'= - ;
a
:. time of R's flight = ^.
s 2
26o
,'. time =
Dynamics
. -+Q
'P + Q + R
Q + R
(Q + R-
+ P
2(P + Q)
V
Q+R-P ■
S
2(P + Q)
Q+R-P'
^hl=,
Q
g
2^/2^/P^
-Q^ 1'.
Q + R-P 'V^
1. What is the force required to communicate to a mass of
36 lbs. a velocity of 10 yards a minute in a quarter of an hour ?
2. Of two forces, one acts on a mass of 5 lbs. and produces in
it a velocity of 5 feet per second in L second, and the other acts
on a mass of 625 lbs., and produces in it a velocity of 18 miles per
hour in one minute. Compare the two forces.
3. A particle of 10 lbs. is placed on a plane which is made to
descend with a uniform acceleration of 10 feet per second per
second. Find the pressure on the plane.
4. A particle of mass m is moving with a constant velocity v
along one of the sides of a square, and when it arrives at the end
of that side it receives a blow which makes it move along the adja-
cent side of the square with the same velocity v. Find the mo-
mentum imparted by the blow.
5. A mass of i lb is laid on a smooth table and connected by a
string with a mass of i oz., which hangs freely. Determine the
acceleration.
6. A mass of 112 lbs. is placed on a smooth horizontal table
with its centre of gravity one foot from the edge of the table ; to it
is attached a string which is also attached to a mass of 14 oz.,
hanging over the edge of the table. How long will it be before the
latter pulls the larger mass off the table ?
7. A mass of 10 lbs. rests on a rough horizontal table, and is
connected by a string passing over a pulley at the edge of the table
to a mass of 3 lbs. hanging freely. If the coefficient of friction
= i, find the acceleration and the tension of the string.
8. Masses of 23^ lbs. and 24 J lbs. are attached to the ends of a
String which passes over a smooth pulley. Find (i) the accelera-
Lau's of Motion 261
tion of the system ; (2) the space described in four seconds from
rest ; (3) the space described in the fourth second.
9. In Atwood's machine the equal masses are each i lb. and
the small mass i oz. Find the space described in 2j seconds after
the motion commences ; and if the small mass be then taken off,
find the space described by the others in the next second. Com-
pare the tensions of the string before and after the i oz. is taken off.
10. Masses of i lb. and i:^lb. are connected by a string passing
over a smooth pulley. Find the tension of the string and the
distance moved from rest in three seconds.
11. If the heavier of two masses suspended at the ends of a
string which passes over a pulley descends \2^ feet in 2i seconds,
and if when I oz. is added to each it only descends 10 feet in the
same time, find the masses.
12. The equal masses in an Atwood's machine are each
12^ oz. ; the mass of the rider is if oz. Find the acceleration.
13. Two equal bodies A and B, each of mass M, ar£ connected
by a string passing over a smooth pulley. To A a body C of mass
m is attached. A and C are started upwards with velocity v, B
starting downwards with the same velocity. How much string will
pass over the pulley before the system comes to rest ?
14. Two equal masses, the one of iron, the other of brass, are
connected by a string over a frictionless pulley. A magnet is
placed below the iron so as to attract it. Show that the attractive
force is measured by the mass of either multiplied by the space
descended by the iron in two seconds, the force of the magnet
being supposed constant during the motion.
1 5. Masses M and ;« hang at the ends of a string which passes
over a smooth pulley. IfM strike the ground one second after
motion commences, how much longer will it be before in first
comes to rest, and what will its distance then be from the ground
if the two masses were originally at the same level .'
16. A mass M is connected with a mass 2 M by means of a
string which passes over a smooth pulley, the strings being vertical.
The first mass falls freely for one second and then begins to
lift the second mass from rest. Find how high the latter will
rise, and the time that elapses before it returns to its original
position.
17. Two weights are connected by a string passing over a
262 Dynamics
smooth pulley. In the fourth second of the motion seven feet of
string pass over the pulley. Find the ratio of the weights.
i8. Find the force of friction acting upon a train whose mass is
l,ooo tons which, starting up an incline of i in 50 with a velocity
of 30 miles per hour, is brought to rest in 300 yards.
19. Ahorse continually exerts a pull equal to the vi eight of
150 lbs. upon a truck whose mass is 5 tons placed on smooth level
rails. The maximum speed of the horse is 10 miles an hour ; how
long is it before this speed is attained ?
20. A force equal to the weight of 10 lbs. is required to make a
mass of 10 lbs. move with uniform velocity up a plane board whose
inclination to the horizon is 45". Find the coefficient of friction.
21. In the prece'Hing question, if the force be equal tt) the weight
of II lbs., determine the acceleration.
23. A train is travelling at the rate of 45 miles an hour on level
rails when steam is shut off. Assuming that g = 32, and that the
resistance of friction and the air to the train's motion is 1 1 lbs.
weight per ton, find how far it will run, and for what time, before it
comes to rest.
23. A mass m resting on a smooth plane of inclination 30° is
connected by a light string passing over a pulley at the top to an
equal mass hanging freely. Find the acceleration and the tension
of the string.
24. What must be the masses attached to the ends of the string
of an Atwood's machine, and the mass of, the rider, in order that
the action of a force of 10,000 dynes upon amass of 1,000 grammes
may be investigated. (Acceleration due to gravity is 981 cm. per
sec. per sec.)
25. Two masses of 5 oz. and 4 qz. are connected by a string
over a frictionless pulley. A rider of l oz. is supported by a ring
one foot above the initial position of the smaller weight so that it
is lifted off as the weight passes the ring. If the masses start
from rest, find the uniform velocity of the system after the rider is
lifted off.
26. Masses of 7 lbs. and 9 lbs. are connected as in Atwood's
machine. The system starts from rest ; after moving through
2 feet I lb. is detached from the 9 lbs. Find the whole distance
travelled by the masses in three seconds from their position of
rest.
Lazvs of Motion 263
27. In the above question, if 3 lbs. were detached from the
9 lbs., find the time at which the masses would be at rest again.
28. Find the space through which two masses of 7 and 8 lbs.,
connected as in Atwood's machine, would move in four seconds,
starting from rest, if, after the system had moved through 5 inches,
I lb. was taken from the larger weight.
29. If the weights in Atwood's machine were equalised by the
smaller mass picking up a rider, instead of the larger one losing a
rider, how would you modify your result for the velocity after the
rider had been picked up ?
In the preceding question, find the velocity of the system if,
after motion through 5 inches, the 7 lb. mass picks up a rider of
I lb.
30. How v/ould a man, ascending in a balloon, determine his
upward acceleration by means of a spring balance ?
31.* A weight W hangs over a pulley. A monkey takes hold
of the other end, and when W is at rest commences to climb, and
climbs up a height h in / seconds without disturbing W. Find his
weight. If at the end of / seconds he ceases to climb, how much
further will he ascend in the next / seconds ?
32.* A light string has masses P and Q attached to its ends,
and is put over two fixed pulleys, the portion between them sup-
porting a movable pulley of mass R. Write down the equations of
motion for P, Q, and R, and find the fourth equation necessary to
determine the motions.
264 Dynamics
CHAPTER XV
HODOGRAPH— NORMAL ACCELERATION — MOTION IN A CIRCLE
206. In Chapters XI., XII., and XIV. we have considered
the case of a change of velocity in the direction of motion and
the force which produces such change, the motion being in a
straight line.
In Chapter XIII. we have considered the case of a
constant force always acting in the same direction but not in
the direction of motion, and we have seen that the path of the
body so acted upon is a parabola. In this case there is a
change of velocity both in magnitude and direction.
Generally, when a body is moving under the action of any
force which does not act in the direction of motion, the force
may be resolved into two components at right angles to each
other — one along the tangent to the path, and the other along
the normal ; the former is in the direction of motion, and pro-
duces a change in the magnitude of the velocity only ; the latter
is perpendicular to the direction of motion, and produces a
change only in the direction of the velocity.
If the latter vanishes there is no change in the direction,
and the path is a straight line ; if the former vanishes, the
path is curved, but the speed does not alter.
In the present chapter we shall first consider the case of a
particle moving with uniform speed in a circle ; the force pro-
ducing the change of motion will always be normal to the
path— that is, towards the centre.
In order to determine the normal acceleration in such a
case, we will first define and consider the properties of a curve
called the Hodograph.
HodograpJi
265
Definition. — If a point P be moving in any manner along a
curve and if from a fixed point O a straight line O Q ie drawn
representing the velocity of P in tnagniiude and direction, the
locus of Q is called the hodograph of the path ofV.
207. We will now prove that the velocity ofQ in the hodo-
graph represents in magnitude and direction the acceleration of P
at the corresponding point of the path
Fig. 142.
Fig. 143.
Let P, P', P" be consecutive points in the path of a particle
close to each other, at which the velocities are v, v', v" respec-
tively; from O draw O Q, O Q', O Q" parallel and numerically
equal to v, v', v" respectively ; then Q, Q', Q" are the points
on the hodograph corresponding to P, P', P".
Then, if t be the indefinitely small time required to describe
the arc P P', when P P' is indefinitely small,
PP'
» = -
(See Definition, Art. 15.)^
Now O Q =: velocity of P at beginning of time t,
and O Q' = velocity of P at end of time t ;
therefore Q Q' = change of velocity in time r. (Art. 153.)
Hence velocity of Q in the hodograph
_ Q Q' change of P's velocity in time t
= acceleration of P.
(Art. 17.)
266 Dynamics
Again, just as the limiting position of the chord P P', when
P P' is indefinitely small, becomes the tangent at the point P,
and is the direction of the velocity at P, so the limiting position
of Q Q' is the tangent to the hodograph at the point Q, and
consequently represents the direction of the acceleration of P.
Hence the velocity of the point Q in the hodograph
represents the acceleration of P in magnitude and direction.
208. To determine tlu normal acceleration a of a particle
describing a circle of radius r with uniform speed v.
Fig. 144. F"=- ^45.
Let P, P' be contiguous positions of the particle in its
circular path ; since P's rate of motion is uniform, its hodo-
graph is a circle whose radius is v. Let O be the centre of
this circle, Q, Q' points on it corresponding to the points
P and P' on the path.
Then P P', the tangent at P, is the direction of P's velocity ;
and Q Q', the tangent at Q, is the direction of Q's velocity—
that is, of I"s acceleration ; but Q Q' is perpendicular to O Q,
and o'q is parallel to P P' ; therefore, Q Q' is perpendicular to
P P' ■ hence, Q Q' is parallel to P A— that is, the acceleration
of P 'is inwards towards the centre of the circle.
Again, since P describes a circle with uniform speed v, Q will
describe the hodograph with uniform speed a in the same time.
Hence, the speeds are proportional to the radii ;
,: a : '0 ::v \ r;
r '
Circular Motion 267
Hence, when a particle describes a circle of radius ;- with
uniform speed v, there is always a uniform acceleration towards
7,2
the centre of the circle equal to — .
r
If the mass of the particle be m, the force necessary to
2
produce this acceleration is ''— .
r
209. The following are examples of this force : the tension
of the string attached to a particle which is being whirled
round in a circle ; the pressure from the outside of a circular
groove on a marble which is travelling round it ; the pressure
from the inside surface of the outside rail of a railway track
on the flanges of the wheels of a train travelling round a
curve. It is a force which is familiar to everybody ; when a
boy is running quickly round a sharp curve, he knows that he
has to incline his body towards the centre of the Qurve, and put
his feet on the ground in such a way that a great amount of
lateral friction may be brought into action. We will now cal-
culate the magnitude of the force exerted in such a case.
Example. — A boy of 9 stone is running round a curved running-
path whose radius is 121 feet at a rate of 15 miles an hour; find
the normal force necessary to keep him in his circular path.
Normal force = = ^-^ — \=^ poundals
r 121
= 504 poundals
= weight of i6| lbs. nearly.
It, is this force which causes the moon to describe its
curvilinear orbit round the earth ; in this case the force is due
to the attractive or gravitative action between the two bodies.
Formerly this force received the name of ' centrifugal
force,' as though a body desired to flee from the centre ; later
this was changed by some writers to ' centripetal force,' as
though a body wished to reach the centre. Both terms are
misleading, since a body has no wish to do the one or the
other ; if left to itself it would continue to move with uniform
velocity in a straight line, and, therefore, if its path is curvi-
linear it must be on account of the action of some force along
268
Dynamics
the normal. We may call it the normal force, or the force
required to produce the normal acceleration ; its measure is
always .
r
If the path of the body is not circular, but is curvilinear,
the normal acceleration is — , where p is the radius of curvature
P
at any point — that is, the radius of the circle passing through
three contiguous points on the curve. Hence, whatever be the
path described by a body, we may look upon its motion at any
point as consisting of two parts — an acceleration along the
tangent equal to the limit of , when t is indefinitely
small, and an acceleration along the normal equal to — , where p
is the radius of curvature at the point.
Note. — If the path be a straight line, p = inSnity, and the
expression for the normal acceleration vanishes.
2 lo. A particle of mass m attached to a fixed point by a
string of length 1 is projected so that it describes a circle in a
horizontal plane with uniform speed v. To find T, the tension
of the string, and its inclination 6 to the vertical.
Let P be the fixed point, A the position of the particle at
any time, O the centre of the horizontal
circle which it is describing.
Let O A = ;-.
The only forces which act upon m
are its weight m g vertically downwards,
and T the tension of the string ; but the
normal force necessary to make ;« de-
scribe the circle whose radius is r is
^^ ; therefore ^ — is the resultant of T
Fig. 146.
and mg;
T sin (9 =
;«»■'
and
T cos B — m g,
Circular Motion
269
now
sino =
-. T =
From these equations we can find T and B in terms of /
and V.
The results may be given in a more convenient form if we
take n as the number of revolutions per second. In this case
& = 2 Trr . n ;
. rp ml{ziTr . n)"^
and
:= 4 ;« /tt^ . n^ poundals ;
r 4 / TT^ 71^
U f^= time of a complete revolution, /= - ;
n
. ,2 4 TT^ /cos 6 .
... ,.„^^.
The above is known as the Conical Pendulum.
211. By an extension of Article 172 of Chapter XII. we
can prove that tke velocity acquired by a particle in sliding down .
a smooth curve is exactly the same as that which would be
acquired by a particle in falling through the same vertical height.
Thus, let P, P' be contiguous points on the curve, and let
PP' the tangent at P make an angle B
with the horizontal ; let », v' be the velo-
cities at P and P' respectively : then for
an indefinitely short space of time we
may consider that the particle is moving
through the space P P' with an accelera-
tion ^ sin 0;
.-. z;'2 = ^2 + 2 ^ sin (9 . P P'
= &2 + 2^. PN,
Fig. 147.
270
Dynamics
and so on all the way down the curve. Hence, if v be the
final velocity, u the initial velocity, and h the vertical height,
v^
U^ + 2 .
Similarly, if a particle be projected up a smooth curve,
ip- ■=. u'- — 2 g h.
These results are exactly the same if the motion of the
particle is constrained by means of a string instead of a
smooth surface ; in all such cases v''- =^ u^ + 2gh, according-
as the motion is with or against gravity.
The results of this Article can also be obtained directly by
the method of Chapter XVII.
212. A particle m is attached to a string of length 1, and is
whirled round in a vertical circle ; to find the tension of the
string at any point, and the condition that m shall just describe
a complete circle.
We will suppose that it passes the lowest point A with a
velocity u; let P be its position at any subsequent time, zrits
velocity, h the vertical height above its lowest point, Q the
angle which the string makes with the vertical. Then,
- - "'^ — "^ — 2 gh . , .
l-h
I ' ' '
cos Q ■■
(I)
(2)
The normal force at P required
to make the particle move in a
V circle
mv'
along P O; but the
only forces which act on the par-
ticle are T along P O and m g verti-
•i'7a be the angle its direction
makes with the line of impact
after it strikes.
In this case let v be the velocity
with which the ball leaves the
surface ; then, since the velocity
perpendicular to the line of impact
remains unaltered,
!'"=■ '51- z; sin <^ = « sin 6.
By Newton's law v cos (j>-= eu cos 0.
Squaring and adding, v^ = u^ (sin^ + e^ cos^ 6) ; also
^ , tan
tan (f> = .
e
The measure of the impulse = mu cos ^ (i + e).
Note. — If the ball and the plane be inelastic e = o.
The vertical velocity is destroyed, but the horizontal
velocity remains the same ; that is to say, the ball after impact
will move along the plane with a velocity u sin 0.
Example. — A ball falls from a height of h feet on to a hori-
zontal plane, the coefficient of elasticity being \ \ find the height to
which it will rise after impact.
Let u = the velocity with which it strikes the plane ;
V = „ „ leaves „
Then u' = 2gh, v = eu = \u.
Let X = the height to which it will rise ; then
o = t/' — 2gx,
_ v^_it^ _ 2g/l _ h,
" ^g~^g~ 8^ ~ 4
Example. — A ball of elasticity j is dropped
from a height of loo feet on to a plane whose
inclination to the horizon is 30° ; find the
velocity and the direction in which it leaves
the plane.
Let u = the velocity with which it reaches
Fig. JS2. the plane, v = the velocity with which it
leaves it, making an angle ^ with the line of impact.
Impact 279
Then «' = 2^. 100 = 6,400 ;
.•. « = 8o;
v%\n^ = u sin 30° = 80 X J = 40,
. o I o -4/3 40
f cos (f> = eu cos 30 = - X 80 X -!-j2 = — r ;
3 2 -v/3
.". z'-sin'<^ + z/'cos'(^ = (4o)^(l +i) ;
. s (80)'
3
Soa/J
tan<> = ^=,/3-;
a/3
.-. ^ = 60°;
.', its direction immediately after impact is horizontal.
Example. — What is the condition that the direction of an elastic
ball after striking a horizontal plane should be perpendicular to its
direction before impact ?
With the notation of the preceding article, put 6+(p = 90° ;
.1 . , tan 5
then tan = ;
e
. a tan 6
:. cot fl = — ,
e
or tan' 6 = e;
:. tan 6 = n/1.
This gives the direction of projection required. The necessary
condition is independent of the velocity.
217. Two inelastic spheres of mass m, m', moving with
velocities u, u', impinge directly ; to find the common velocity
after impact.
Let V be the common velocity.
By the third law of motion, sum of the momenta after
impact = sum of the momenta before impact ;
.•. mv + m' V ^ mu + m' u' ;
mu + m' t^
.". v = ,-.
m + 7Ji
28o Dynamics
If I be the measure of the mutual impulse,
I = m u — ;« w = Til! V — m' u'
VI m' ( u — u')
m + 111'
by substituting for v the value just obtained.
218. Two spheres, of elasticity e and mass m, m', moving with
velocities u, u', impinge directly. To find their velocities after
impact.
Let V, v' be their velocities after impact.
Then, by Newton's experimental law,
V — v' = — e {u — u') (i)
and by the third law of motion
mv + m' v' ^=mu + m' u' (2)
Multiply (i) by ;«', and add the equation thus formed to (2) :
(in + ;;/') v = {m — e fn') u + (m' + e ?«') u\
Multiply (i) by m, and subtract the equation thus formed
from (2) :
(in + ;«') v' = (tn + eni) u + (ni' — em') u'. ]
These equations give v and v'.
If u' were in the opposite direction to ti, we should write
— u' for u' all through the above equations.
219. When two elastic spheres impinge directly on each other,
to compare the impulse which causes the compression with the
impulse due to the restitution of the particles.
It has already been pointed out that the first part of the
impact — the compression — goes on until the two bodies are
moving with a common velocity, at which instant the force of
restitution commences.
Let V be this common velocity, I the impulse of com-
pression, I' the impulse of restitution.
Then \ = m u — mY ^ m' V — m' u' ;
whence V = ^''^ + ^','^' , and I = ^^ ^' ('^ -,"' ),
m + m m + m
(as in Article 216).
Impact 281
Similarly, V = niY — mv= m' v' — m'Y ;
whence V = ^^^^il^.and I' = '"^'i^'-^) .
m + m! m + m
Substituting the values obtained for v and v' in the last
article, we have
I' = - — ? [e mu •{■ etn' u — e in u' — e m' u'\
(ni + niy
mm
X e {m + ;«') (u — a')
{m + m'y
e m m! (u — u') t .
m + m
that is,
impulse of restitution _
impulse of compression '
hence the name ' coefficient of restitution.'
The measure of the entire impulse
= i + r
= (I + ^) I
/ , s, mm' (u — u')
m + m'
This could also have been obtained from either of the
expressions m (u — v), or m (»' — u'), by substituting the values
of V and v' just obtained.
220. Two spheres A and A', of elasticity e and mass m and
m', impinge obliquely, with velocities u and u', in directions
making angles a and a' respectively with the line of impact ; to
determine the velocity and direction after impact.
A
Fig. ,53.
Let V and v' be the velocities after impact, ^ and /8' the
angles which their directions make with the line of impact ;
282 Dynamics
and suppose that a, a! and ji, /3' are measured as in the
accompanying figure — a and a' above the line of impact from
left to right, y8 and §! below the line of impact from right to
left.
Then we have four unknown quantities, v, v', P, /S', to deter-
mine which we must have four equations, which are obtained
as follows : —
(i) Velocity of ot_l'" to the line of impact remains unchanged.
\'^) J) ^^ 3J J) » »>
(3) Newton's law for the relative velocity in the line of impact.
(4) Total momentum in the line of impact remains unchanged.
Thus,
V sin 13 :=u sin a (i)
v' sin /3' = 2^' sin a' . . . ^ . . (2)
V C05 p — w'-cos yS' = — e{ucosa — u' cos a') (3)
m V cos y8 + m' v' cos /3' = ;« u cos a + m' u' cos a' (4)
From equations (3) and (4), as in Article 217, we get
V cos p and v' cos yS'; thus :
(;« + m') » cos ;S = {m — em')u cos a + m' (i + e) u' cos a'(s)
(»« + m')v' cos ;S' = OT (i + «) « cos a-\- (m! - em) u' cos a' (6)
From equations (i) and (5) we obtain v sin ,8 and & cos y8 ;
squaring and adding we obtain »^ ; dividing one by the other
we obtain tan ;8. Similarly we obtain z/' ^ and tan |8'.
The results of the last article hold also in this case, when
u cos a, u' cos a', V cos yS, p' cos y8' are substituted for w, «',
», »' respectively.
Thus, the whole impulse due to the impact
/ , \ ;« m' (u cos a — u' cos a')
ir;i'a;«/&^Spherical masses of 2 lbs. and 3 lbs., moving re-
spectively with velocities of 2 feet per second and 3 feet per second
in opposite directions, impinge directly. If the coefficient of elas-
ticity be ^, find the velocities after impact.
Let V and v' be the velocities after impact, measured in the
direction of the 2 lbs.
Impact 283
Then, v-v' = - \{p.-^i\
2^ + 3 w' =2x2-3x3;
.". 2 57 — 2 Z/'= — 5,
27/+3'z/' 5 ;
.•. z/' = o,
and z'=-f.
Hence the larger mass is reduced to rest, and the velocity of
the 2 lbs. is i\ feet per second in the opposite direction to that in
which it was moving before impact.
Example. — Two balls are moving with equal momenta in
parallel but opposite directions, making angles of 30° with the line
of impact, the mass of one being 3 lbs. and the velocity 2, and the
mass of the other being 2 lbs. with velocity 3. If the elasticity be
— i^, find the velocity and direction of each after impact.
-^3
Let V, v' be the velocities whose directions make angles 0, 0'
with the line of impact, the measurements being taken as indicated
by the figure.
I/'
^IG. 154.
Then 7/ sin ^ =2 sin 30°=! (i)
v' sin 0' = 3 sin 30° = | (2)
V cos (j) + v' cos ({>' = £ {2 cos 30° + 3 COS 30°)
-^(= f )-i «
3 V COS (j) — 2v' COS (f)'= - 3 X 2 COS 30° + 2 X 3 COS 30°
= ° (4)
Multiply equation (3) by 2 and add it to (4), and we get
' ■ V COS 0=1 (5)
.-. ■z/'cos0' = | (6)
284 Dynamics
From (i) and (5) 7/= ^2", = 45°;
and from (2) and (6) v'=^, ^' = 45°
221.* To find the motion of an elastic ball after impact
with a rough plane.
In this case there is a frictional impulse perpendicular ta
the line of impact which diminishes the momentum of the
ball perpendicular to the line of impact ; and this frictional
impulse ^ /x x (total impulse in the line of impact), where
/i = coefficient of friction.
Using the notation of Article 215, we have the frictional
impulse := /x »z « cos 6(1+ e),
.•, m vsin cj> = mu sin — /J, mu cos ^ (i + e);
.'. V sin = u {sin 6 — ij.{i + e) cos 0} ;
and V cos =^ eu cos 6;
.". tan <^ = - {tan — /jl^i +«)}.
■ 222.* To find the motion of the centre of mass of a ftumber
of masses moving with different velocities in a plane.
Let ;;/, ot,, m^, &c., be the masses.
Let O X, Oy be two lines at right angles to each other
through a fixed point O in the plane.
Let the co-ordinates of m, w,, &c., referred to 0.x, 0^ as
axes, be {x,y), («,, y^\ &c., at the beginning of time t, and
(.%-', y'), (jc'i, jc']), &c., at the end of time t.
Let u, «|, z<2, &c., be their velocities parallel to Oa-;
z', W|, z/j) &c., be their velocities parallel to Oy ; and let
.V, y ; x', y' ; u, v denote the same quantities for the centre of
mass.
Then, considering only the motion parallel to O x, we
have — . . . • .
X 2 — ^2 ^ ^2 • h
and so on ;
lastly, ^x' ~ x=-u . t
Impact 285
Further, by Article 93,
- _^(mx) - 1{mx')
. y _ - _ 2 (mx') - 2 (m x)
2 {m)
2 ( OT (y —x)]
2 (m) '
%{ mui\ t .%{m u)
since t is the same for all ;
- 2 (»« It)
2,(tn)
Similarly, » =
- 2 {mv)
"'- 2H"
By the third law of motion 2 (w u) and 2 {m v) are not
altered by any forces within the system — that is to say, some
of the particles may be connected by strings, others may im-
pinge against each other, others may explode, yet there is no
change in 2 (m u) or 2 {m v), consequently u and v remain un^
changed. Hence the statement, that the motion of the centre of
mass of a system of moving particles remains unifor7n so long as
no force acts which is external to the system.
An exactly similar proposition holds with reference to the
acceleration in any given direction of the centre of mass of a
number of accelerated particles ; thus
-_2(wa)
" - 2 (m) •
Example. — Masses m and 7n' are connected by a light string
passing over a smooth pulley ; to determine the acceleration of the
centre of mass.
Let n, a! be the accelerations of m and m' respectively ; then
measuring positive accelerations downwards, we have
, (m — m')e'
m + m
286 Dynamics
— ;« a + m' a'
• _ _
" ~ m + 111'
_ m {m — m') - m'{m - m')
[in + m'Y
\in + m'}
This is positive, and therefore downwards.
Examples.
r. A person of lo stone weight is skimming along smooth ice
with a stone of 3^ lbs. weight in his hand ; he throws it in front of
him, so that it leaves him with a relative velocity of 20 feet per
second ; by how much will his own velocity be diminished ?
2. A ball drops from a height of 50 feet, and after striking the
ground rebounds. Determine the height to which it \V\\\ rise, the
coefficient of elasticity being e.
3. A glass ball, whose coefficient of restitution is f, is dropped
from a height of 4 feet on a stone pavement, and rebounds. How
long will it be before it again strikes the ground ?
4. A perfectly elastic body whose mass is 10 lbs., moving g
yards per second, comes into direct collision with another whose
mass is 1 1 lbs., moving 8 yards per second in the opposite direction.
Find the initial velocities of each after the collision.
5. Two perfectly elastic balls of given mass are moving with
given velocities in opposite directions such that their momenta are
equal. Find the velocities after impact.
6. A ball of elasticity e is projected vertically upwards with a
velocity V ; it falls and rebounds. Find the whole space it will
have travelled when it again reaches the ground, and also when
it finally comes to rest.
7. A ball moving with velocity k impinges directly upon another
ball at rest whose mass is n times that of the first ball. If the
first ball is reduced to rest, find the coefficient of elasticity and the
rate at which the second ball will begin to move.
8. Give a geometrical construction to show the direction in
which a billiard ball must be propelled in order that, after striking
a perfectly elastic cushion, it may strike a ball at another part of
the table.
Impact 287
9. A ball is projected directly against another ball lying at rest
with such a velocity that, after the impact, it returns with a velocity
double that imparted to the other ball. If « = g, find the ratio of
the masses of the balls.
10. A smooth ball of mass ?« moving on a horizontal plane
strikes another ball of mass m' at rest, and is itself reduced to rest.
Find the coefficient of elasticity.
11. Two perfectly elastic balls of 4 oz. and 6 oz., moving with
velocities of 6 feet and 4 feet per second, impinge directly. Find
the momentum of each ball after impact.
12. A sphere of mass m impinges obliquely on a sphere of mass
in' which is at rest. Show that if the modulus of elasticity be — ,
711
they will separate in directions which are at right angles to each
other.
13. Find the angle at which an imperfectly elastic ball must
strike a plane in order that its direction of motion after impact
may be at right angles to its direction before impact.
14. A ball of mass m is projected with a velocity V = \^~^
from a point A up a smooth plane A B inclined to the horizon at
an angle of 30°, and at the same moment a second ball of mass m'
begins to move from rest from B towards A. If A B = a, and the
elasticity of the balls = , , find the velocity of each after
impact.
1 5. An elastic particle drops from a height /4 on a smooth plane
of inclination a ; if « be the coefficient of elasticity, find the dis-
tance between the first and second points where it strikes the
plane.
16. A perfectly elastic particle is let fall from a height of 50 feet
above the ground, and when half way down strikes a smooth sur-
face inclined at an angle of 45° to the horizontal, and rebounds. At
what distance from the original line of descent will the particle
strike the ground ?
17. A body of elasticity e is projected with velocity ic from a
point in a horizontal plane in a direction making an angle a with
the plane ; find the time that will elapse before it ceases to re-
bound.
18. A set of five balls of elasticity |, whose masses form a series
288 Dynamics
in geometrical progression whose common ratio is 2, is arranged
in a straight line. The smallest impinges directly on the first with
velocity u. Find the velocity of the last ball after the fourth
impact.
19. A ball (e = J) is thrown against a smooth vertical wall from
a point whose distance from the wall is u. Find the relation which
must exist between the velocity of projection and the angle of pro-
jection in order that it may return to the starting point.
20. Two equal imperfectly elastic balls, moving in opposite di-
rections along the lines B A, C A, the velocity of one being double
that of the other, impinge at the same moment on a third equal ball
at rest at A. Find the subsequent motion of the ball at A.
21. A ball of mass ni and velocity u strikes two other balls of
equal mass in contact with each other at the same instant. Show
that if £ = I the first ball will be brought to rest, and find the
initial velocity of the other two.
22. A smooth sphere slides down between a vertical wall and
an inclined plane (45°) which can move along a smooth horizontal
plane. If the masses of the sphere and incUned plane be equal,
show that their accelerations are each equal to half that due to
gravity.
289
CHAPTER XVII
WORK— ENERGY
223. AVoRK has already been defined in Chapter X.
A force is said to do ivork when its point of application is
displaced in the direction in which the force acts ; and the work
done by a force is measured by the product of the force and the
projection of the displacejnent of its point of applicat on in the
direction of the force.
Thus, if F be the force, j the displacement measured in the
direction of F, then F. s is the measure of the work done by
the force F.
If the displacement be in the opposite direction to that of
F, j- is negative, and the work done is negative ; work is then
said to be done againstthe force. (Article 137.)
E.^'. a body of mass m falls to the ground through a height h ;
then m.g\% the measure of the force of gravity which acts on the
body, h is the distance through which the force acts ; therefore
tn gh is the measure of the work done by gravity on the body. If
the body had been raised from the ground to the height h, the dis-
placement would be in the opposite direction to that of the force of
gravity, and 7ngh would be a measure of the work done against
gravity.
224. To measure work we take as our unit the work done
by the unit force acting through the unit distance.
The unit of force usually adopted by engineers in this
country is the gravitation unit one pound weight ; and the unit
of work is the work done by a force of i lb. wt. in acting
through a distance of i foot.
This is called a foot-pound ; thus the work done by an
agent which raises a mass of i lb. through j foot in opposition
to gravity is one foot-pound.
u
290 Dynamics
This is the gravitation unit of work, and depends on the
value of g at the place where it is adopted.
The absolute unit of work is a foot-poundal — that is, the
work done by a poundal in acting through a distance of one
foot ; thus, I foot-pound = ^ foot-poundals.
The unit of work adopted in France is the work done by a force
of I dyne in acting through a distance of i cm. This is called an
erg.
Since a poundal = 453'59 x 30'4797 dynes,
and a foot = 30'4797 cm.,
I foot-poundal = 421390 ergs approx.
225. To find the work done in pulling a body along a rough
horizontal plane.
If F be the limiting friction, s the distance through which
the body is moved, the work done = F . j-.
Now F = yu, R, if R be the pressure on the plane,
= /x W, if W be the weight of the body ;
/. the work done =: /a W 5.
26. To find the work done in pulling a body up a rough
inclined plane of length I.
Let h be the height, b the base of the plane.
By Article 134 the force which is required to move it up
must be just greater than W (sin a+yn cos a).
Hence the work done
:= W (sin a +yti cos a) X /
= W . /sina + /x W . /cos a
= W. /%+/^W. b
= work done in lifting W up a vertical distance h
+ work done in dragging W along the rough horizontal
surface /'.
227. Definition. — The rate at which a machine or agent
can do work is called its power.
In estimating the capabilities of different machines of doing
work we have to compare the rate at which they can do work
with some standard rate of doing work. The standard adopted
by engineers in England is a horse-power.
Work — Energy 2gi
Definition. — A horse-power is the power of doing work at
the rate of 33,000 foot-poujids per minute, or 550 foot-pounds
per second.
This is rather greater than the rate at which the average
horse can do work.
Example. — A man of 1 1 st. weight can climb a vertical height
of 3,000 feet in ij hour. If he is put to the treadmill, compare his
rate of doing work with a horse-power.
He can raise 1 1 x 14 lbs. through 3,000 ft. in 90 minutes;
/. he does — ^ — 4jl3i foot-pounds of work per minute ;
90
. the man's power _ 11 x 14 x 3,000 _ T_
a horse-power 90 x 33,000 45'
228. Energy is the capacity of doing work.
When work is spent on a body there is an increase of
energy in the body and a corresponding loss of energy in the
agent ; and when an agent does work it loses some of its
energy by parting with it to the body on which it is working.
Energy may be made up of two kinds — kinetic, 01 potential.
Kinetic energy is the energy which a body has by virtue
ef its motion.
Potential energy is the energy which a body has by
virtae of its position.
E.g. a body of mass wz, moving with a velocity v, has
kinetic energy, because in being brought to rest it is capable
of doing work ; and the work which it can do before it is
brought to rest is a measure of the energy which it has by
virtue of its motion.
The kinetic energy of a moving body is measured by
\ (momentum x velocity), or one-half of the product of the
mass into the square of the velocity — i.e. \ m v'-.
The reason for this will appear in the next article. Kinetic
energy is sometimes called vis viva.
As examples of the energy due to the position of a body
we may take a stone lifted to a certain height above the surface
of the earth, — the weight of a clock which has to be wound up
to keep it going, — a reservoir of water which can be made to
u 2
292 Dynamics
work machinery producing mechanical power or heat energy
or electric energy.
Let us consider the simplest case — a stone of mass m
lifted by a man to a height h above the earth.
The work done on the stone against gravity is measured by
High. This work is done by the man, who consequently
expends the energy which the stone gains as potential energy.
So long as the stone remains in this position it retains this
potential energy, but when it begins to descend it can be made
to do work. The quantity of work which it can do cannot be
greater than mgh ; if it descends freely it does no work in its
descent, but, although it loses potential energy, it gains in
kinetic energy, and the total kinetic energy which it acquires
just before striking the earth is a measure of the work which it
can then do, or of the potential energy which it had before it
commenced to fall, or of the work which was originally done
upon it by the man.
229. To find an expression for the work done on a body in
terms of the mass of the body, and its velocity at the beginning
and end of the action of the force.
Work done = Y . s.
Let m be the mass of the body, ti and v its velocity at the
beginning and end of the action of F.
Then F = ot a ; 2 as ^v^ — «^ ;
.•. work done = mas
= \m {v^ — tt''-)
= \mv^ — \ m u^
= change of kinetic energy.
Note.— In this and the following articles F is measured in
absolute units.
If V is greater than u, the work done by F on m is measured
by the increase of kinetic energy ; if » is less than i/, the work
done by m against the force is measured by the loss of kinetic
energy ; if n = o, the work done on pi is measured by the
kinetic energy which it acquires ; if » = o, the work done by
IVork — Energy 293
m is measured by the kinetic energy which it had before the
force began to bring it to rest.
Example. — A train travelling with velocity u on level rails
suddenly shuts off steam, and comes to rest after traversing a
distance j. If the resistance to motion be entirely due to friction,
find the coefficient of friction.
Let
F = force of friction,
m = mass of the train ;
then the work done = the kinetic energy destroyed.
Y .s = \mu^;
but
Y = fimg;
.: iimgs^^mu'';
eg-
« =
= 60 miles an hour = 88 ft. per second.
s =
= 1 mile = 5,280 ft.
88x88
2 X 32 X 1760 X 3
II
480'
230. When a body of mass m is raised from the ground to
a height h, the work done against gravity is mgh.
This is the measure of the potential energy which m now
possesses. That is to say, by virtue of its position it is
capable of doing mgh units of work at some future time.
Suppose, now, that m be allowed to fall freely from the
height h to the ground ; let v be the velocity with which it
reaches the ground ; then its kinetic energy
=-\mv'''
= \m. 2gh
= mgh;
that is, the energy which it now has by virtue of its motion is
equal to the energy which it had before by virtue of its
position.
294 Dynamics
231. A body of mass m falls to the ground from a height h /
to show that the sum of its potential and kinetic energies is
constant throughout the motion.
Let us consider its energy at a height x from the ground,
when its velocity is v ; then its potential energy = m g x, and
its kinetic energy
■=.\m . 2g{h—x) ;
.*. the sum of its potential and kinetic energy
= mgx + mg{h—x)
■=.m gh.
This is a simple example of the principle of the conservation
of energy.
After m has been reduced to . rest by contact with the
ground, it has neither potential energy nor kinetic energy.
What, then, has become of its energy ? It has been used up
in molecular change, in compressing the particles when it
struck the earth, and in an addition of heat-energy to itself and
the immediate neighbourhood. In this "form its energy is not
available for practical purposes ; it is, therefore, said to have
deteriorated.
Similarly, in Article 225, the work done on W is
this ought all to be present in the form of an increase of
energy, but the only visible increase of energy is the potential
energy W . ^ ; the remainder has been converted by friction
into heat energy, partly in W and partly in the surface which
it has traversed.
232.* When two imperfectly elastic balls impinge directly,
the kinetic energy after impact is less than the kinetic energy
before impact.
Let E, E' be the kinetic energy of the two balls before and
after impact respectively; then, with the notation of Article 218,
E = I OT «2 4. ^ ;«' u' 2,
E' = iwz/2 + |w'»'2;
Work— Energy 29 S
. y—^ < »iu + m'u'—em'(u—u' )^ m'{ pm + m'u' + em(u — u' )
2 1 m + m' ] 2 I m + ?>i'
= —. — --{(mu + m'u'Y(m + m') + e^mm' (u — u'y(m + m')}
2 {tn + m'Y
(mu + m! u'Y + g^ >ti m' (u -- u'Y
2 (m + m')
_ (mu + m! u'Y + m tn' {u — u' Y — (i — e"-) m m' (u — u') '^
2 {m + m')
m"^ ur' + m m! u^ + m' ^ u ''^ + m m' u'^ {i—e''')mm'{u — u' Y
2 (m + m') 2 (ot + m')
{m u^ + w! u'^ ) (m + m) (i — e'^) tn ml ( u — u'Y
2 (;« + m') 2 (m + m')
= E - -7^^^ (I - e^) {u - u'^).
Hence E = E' + /" '"' ,^ U-e'^\{u- u')K
2 (in + m')
Therefore —r'^'^M - e^) (u - uj
2 (in + tn'y ' ^ '
is the kinetic energy lost by the impact; this has been trans-
formed into some other form of energy in which it is not so
apparent.
If «=i — that is, if the balls are perfectly elastic— this expres-
sion vanishes, and E = E'; hence there is no deterioration of
energy when perfectly elastic balls impinge.
Examjile. — To show that in Atwood's machine the work done
by gravity in any time is equal to the kinetic energy of the system.
With the notation of Article 197, let x be the distance through
which the weights have moved in time t from rest, v the velocity
they have acquired, then
x=^ a . /^, v = at,
, (m — m')
and a = i ^.
m + tn
Work done on tn by gravity = »z^;ir,
„ „ m' „ =-m'gx.
296 Dynamics
Whole work done by gravity = ;«^.r-OT'^.r
/ /N / /N 1 j2 (m - m'Y g- t'^
- (m — ni) erx= im — ;« ) e y-i at' = -, — -^-y^. — •
^ '^ ^ ; i 2 2 (;« + m')
Sum of the kinetic energies of m and m'
= I wz w'^ + I m' v'' = ^ {in + m') v^
= \ {m + m') a' f
233. In Chapter X. we have given the general statement of
the principle of the conservation of energy :
' Tke total energy of a?iy material system is a quantity which
can neither be increased nor diminished by any action between the
parts of the system, though it may be transformed into any of the
forms of which energy is susceptible' (Maxwell's 'Matter and
Motion,' chap, v.)
We have seen that this is the case in a system consisting of
the earth and a stone at a distance h from it.
It is also the case when one system does work upon
another system ; the quantity of energy lost by the one system
is equal to the quantity of energy gained by the other ; con-
sequently, if the two be looked upon as one system, the total
quantity of energy remains unchanged.
Newton's third law of motion says : — 'Action and Reaction
are equal and opposite ' ; but in a scholium to this law he
estimates action and reaction by 'the product of the force
into the velocity of its point of application ' ; thus, the action,
of an agent is ' the rate at which it does work ' — e.g. the power
of a steam engine ; thus the third law becomes ' The energy
of a conservative system is constant,' or.
Work done on the system = change of energy in the system.
234. As an example of the transformation of energy we
may consider the transformation of mechanical energy into
heat energy.
Thus Dr. Joule, of Manchester, by a series of careful
experiments, determined that in order to raise the temperature
of I lb. of water from 39° F. to 40° F. it was necessary to
expend 772 foot-pounds of work.
IVork — Energy 2^7
The above is called the unit of heat, or a thermal unit.
Hence i unit of heat = 772 foot-pounds of work
= 24,858 foot-poundals.
Thus, in Article 232, if the whole loss of kinetic energy is
converted into heat energy, the quantity of heat gained by the
impact
2 (m + m) X ^14,858
Examples.
1. Find the work done by gravity upon a stone having a mass
of \ lb. during the tenth second of its fall from rest.
2. What is the work done against gravity by a man of 10 stone
in climbing a mountain half a mile high ?
3. How many foot-pounds of work can be obtained by bringing
to rest a body with a mass of i lb. moving with a velocity of 40 feet
per second.''
4. A shot of ^ ton is travelling at the rate of 1,600 feet per
second ; find its energy in foot-pounds, and determine for how long
a time this energy, if stored, would yield two horse-power.
5. A IQ lb. mass falls 100 feet ; what is the time of its fall, and
what is its kinetic energy at the moment it touches the ground ?
6. A lump of clay of 10 lbs. is thrown with a velocity of 50 feet
per second against an equal lump at rest. If the two travel toge-
ther with a velocity of 25 feet per second, find the loss in energy,
estimated in foot-pounds.
7. Two bodies have the same momentum ; the mass of the first
is double that of the second ; show that the kinetic energy of the
second is double that of the first.
8. A mass of i lb. is dropped from a height of 120 feet ; what
will be its velocity when its potential energy is 56 foot-pounds ?
9. A horse capable of working at one horse-power is put to
grind corn by making it pull the long arm of an axle ; he walks
round in a circle at the rate of 4 miles an hour. What is the force
(in poundals) exerted by the horse ?
10. A ball whose mass is i lb., moving at the rate of 12 feet a
second, impinges directly on an equal ball at rest. Find the velo-
298 Dynamics
cities of the bnlls after impact and the amount of kinetic energy
lost, the coefficient of elasticity being ^.
11. A ball of mass A, moving with velocity z^, impinges directly
on another ball B whose velocity is v, whch in its turn impinges
on a ball of mass at rest. Show that the sum of the kinetic ener-
gies of the three balls is unchanged by the two impacts, the balls
being perfectly elastic.
12. Show that if equal forces produce the same energy (hey
must act for times which vary as the square roots of the masses
acted on.
13. A man of 12 stone weight walks up a hill 3,000 feet high in
3| hours. What is the average power which he exerts as compared
with a horse-power.
14. A blow with an inelastic hammer-head of mass ^ lb. drives
a nail to a depth of lA inch into a board, the hammer-head being
reduced to rest. If the velocity of the hammer-head when it first
touched the nail was 1 5 feet per second, find the average resistance
of the board to the nail's motion.
15. A shot of mass half a ton is discharged from a gun of mass
1 10 tons ; the gun's backward motion is checked by a constant
pressure equal to the weight of 10 tons ; the recoil is observed to
be 6^5 feet. Show that the muzzle velocity of the shot is 1,320 feet
a second [g = 32).
299
APPENDIX
ruchayla's Proof of the Parallelogram of Forces.
For the purposes of this proof we make the following assump-
tions : —
(i) That two forces, whose directions meet, have a resultant.
(2) That this resultant may be substituted for the two com-
ponents, and vice versd.
(3) The principle of the transmissibility of force. (Art. 9.)
We shall divide the proof of the proposition into four parts, and
prove that it is true —
(i) For the direction of the resultant of two equal forces.
(2) For the direction of the resultant of two commensurable *
forces.
(3) For the direction of the resultant of two incomtnensurable
forces.
(4) For the magnitude of the resultant of all forces.
(1) For the direction of the resultant of two equal forces,
Let A B, AC represent two
equal forces acting through the
point A.
Complete the parallelogram
A B D C ; join A D.
Then A B D C is a rhombus,
since A B = A C, and therefore A D '''
bisects the angle B A C ; but the
resultant of two equal forces must bisect the angle between the
' Commensurable forces are forces which can be expressed as multiples
of the same unit of force ; if they cannot be so expressed they are
incommensurable.
Thus, 2^ and 3^ are commensurable, but i and Vz are incom-
mensurable.
300 Dynamics
forces, because there is no reason why it should act nearer to one
component than the other ; therefore the direction of the resultant
of two equal forces coincides with the diagonal of the parallelogram.
(2) For the direction of the resultant of commensurable forces.
We will assume that the proposition is true for two sets of
1^ P c ff f forces, P, Q and P, R, and
we will prove that it is
true for the forces P and
Q + R.
Let A B, A C repre-
sent the two forces P and
Q, and let C E, measured
"^/? along A C produced, re-
p\ ' f\ present R, so that A E
Fig 156. represents Q + R.
Complete the parallelograms ABDC, CDFE; then since
A C E is a straight line, B D F is also a straight line, and therefore
A B F E is a parallelogram.
Let the resultant of P and Q acting at A be S acting along
A D, and let the resultant of P and R acting at C be T acting
"along C F.
Replace P and Q by their resultant S, suppose S to act at the
point D along A D produced, and then replace it by its components
P and Q acting parallel to their original directions — that is, along
C D produced and along D F respectively ; suppose P to act at C
and Q to act at F.
We now have two forces, P and R, acting at C ; replace P and
R by their resultant T along C F ; suppose T to act at F, and
then replace T by its components P and R acting parallel to their
old directions — that is, along E F produced and along D F pro-
duced respectively ; we have thus replaced P and Q + R acting
at A by P and O + R acting at F without altering their effect.
Therefore F must be a point in the line of action of the re-
sultant of P and Q + R ; therefore A F must be the direction of
this resultant. Hence, if the assumption is true of the forces
P and Q, and also of the forces P and R, it is true of the forces
P and Q + R.
But by (i) it is true of P and Q, and also of P and R, if Q and
R are each equal to P.
Therefore it is true of P and 2 P, and of P and 3 P, and so on ;
therefore it is true of P and m P.
Appendix-
301
Again, since it is true of wz P and P, and also of w P and P,
therefore it is true of m P and 2 P, and therefore of m P and 3 P,
and so on ; therefore it is true of m P and n P — that is, it is tiue
for the direction of the resultant of commensurable forces.
(3) For the direction of the resultant of incommensurable
forces.
r--5^
Fig. 157-
Let A B, A C represent two incommensurable forces P and Q
acting at the point A. Complete the parallelogram A B D C ;
then A D shall represent the direction of the resultant of P and
Q ; for if this resultant does not act in the direction A D, let us
suppose that it acts in any other direction as A V, and show the
absurdity of such a supposition.
Let A V cut C D in V.
Divide A C into a number of equal parts, each less than V D ;
from C D cut off parts equal to these commencing from the end C ;
the last point of division must fall between V and D, since each
part is less than V D, and it cannot fall on D since A C and C D
are incommensurable lines ; let it fall at E. Complete the parallelo-
gram A C E F, and join A E.
Now the force P is represented by A B — that is, by A F, F B, of
which A F is commensurable with A C ; therefore the resultant of
P and Q is the same as the resultant of forces represented by
AC, A F, and F B ; now the resultant of the commensurable
forces represented by A C, A F acts along A E by (2), and the
force represented by F B may be supposed to act at A in the
direction A B ; therefore the resultant of P and Q is in the same
direction as the resultant of these two forces acting along A E and
A B ; hence it 7mcst act within the angle E A B ; therefore the
supposition that it acts along A V is absurd.
Similarly it can be shown that it is absurd to suppose that the
302 Dynamics
resultant of P and Q is in any other direction than A D ; therefore
A D must be the direction of the resultant.
We have thus proved that the proposition is true for the
direction of the resultant of all forces ; (4) we will now prove that
it is also true of the magnitude of the resultant.
Let A B, A C represent any two forces P and Q ; complete the
parallelogram A B D C, join A D ; then A D represents the direc-
tion of the resultant of P and Q ; it shall also represent the
resultant in magnitude.
Produce D A through A to E, making A E of such a length as
to represent the magnitude of the resultant of P and Q ; complete
the parallelogram E A C F, and join A F.
Then, since A E has been drawn equal in magnitude and
opposite in direction to the resultant of P and Q, it follows that the
forces represented by A B, A C, and AE must be in equilibrium,
and either of them must be equal and opposite to the resultant of
the other two ; but A F represents the direction of the resultant of
forces represented by AC, A E by (3), therefore A F must be in the
same straight line as A B ; but A B is parallel to C D ; therefore
F A is parallel to C D, and F C is parallel to E A D ; therefore
F A D C is a parallelogram ; therefore A D = F C ; but F C = A E ;
therefore AD = AE; but AE represents the magnitude of the
resultant of P and Q ; therefore A D also represents the magnitude
of the resultant of P and Q.
Hence, if any two forces acting at a point be represented in
magnitude and direction by two lines drawn from that point, and
if the parallelogram having these two lines for adjacent sides be
completed, the diagonal of this parallelogram, which passes through
the point, represents the resultant of the two forces both in
magnitude and direction.
303
ANSWERS
CHAPTER I.
1. 88. 2. 5i. 3. 28'i6. 4. ml. 5. ^, 3££i
p a •
6. 1,536 ft. per sec. 7. 10 miles an hour. 8. 5 ft. per. sec. per sec.
9. 4 ft. per sec. 10. \ the unit of velocity is added every second.
11. 9'8. 12. 10; 10 poundals ; 1,000 poundals ; . 13. 200,000: 1.
g
14. 2,000. 15. 5 ; 300. 16. 80. 17. 560. 18. Acceleration = 5 ;
mass = 5 tons.
CHAPTER III.
1. 4, 2, or o. 2. 14 lb. weight. 3. 30 lb. vifeight. 4. 12, 2 ;
3 and 4 at right angles. 5. loi. 6. -^^5 — 2 a/2. 7. 7.
8. Jyi. 9. 15. 10. 120°, 150°, 90°. 11. 10 a/J, 20.
12. P Vt"; sin— \~ir) with P. 13. 13. 18. V3 : v'i.
19. P VJ; tan-' J with resultant. 20. — ~, — -—■
21. Cos-"(- 1). 22. Cos-'(- U). 23. Cos-' i. 24. Cos"' (i).
25. P = 2 a/2, Q = 2. 26. ^2 : I. 27. S = P and makes an
angle of 60° with it. 28. 3 R. 29. P. 34. i^
35. The diameter through A. 39. Sin"' — p with the side.
V5
41. a/(P - Rf + (Q - S)^ 42. A B or D C. 43. A C.
44. 4, 2 V3. 46. 6 in the direction of the force 5. 47. 13.
48. I opposite to the direction of 19. 49. 10 opposite to thp direction
of 8. 50. Cos-' (!§). 51. 150°. 52. 2 AC. 53. 1:2.
304 Dynamics
54. |, ^3. 55_ |_ 3^3. 5e_ g n,. ^^eight. 57. 5 A ^
along AD. 58. V29 - 18 ^2. 59. -^^105 + 42 Vs at an angle
tan-' y Y^) ^'^h the force 8. 60. v'4'i. 64. ^1 65. "When
the given angle 9 is obtuse and the given component is greater than the
given force but less than P cosec fl. 68. ?J'^.
3
CHAPTER IV.
1. 6 ins. from 7. 2. 12 ins. from 9. 3. 4 ft. 4. i ft.
5. 5, 7|. 6. \\ ft. 7. 3i, ij. 9. l| ft. from I lb. 10. 3§ ft.
•H. 160 lb. wt. 12. 2 lb. wt. 13. 5^. 14. \. 15. 85I, 85I, 58.
16. At the 3 lb. wt. 18. 7^ lb. wt. 19. f in. from A.
21. ?i^±_2) ins. 22. A O = 5 A B. 24. P : Q ; R : : a' : *= : f=.
3 6 ^
CHAPTER V.
2. 2:1. 5. The same. 6. 1} ft. from 7. 7. Each is 24.
8. 12 P, 9 P, « P. 9. I : 4. 13. 90". 14. I ft. from one end.
15. 2 oz. 16. 1 1 in. 18. A couple whose moment is 6 a.
CHAPTER VI.
1. \\ in. from the S lbs. 2. % in. from 8 W. 3. f (diagonal)
from the 2 lbs. 4. f (diagonal) from the i lb. 6. llj ft. from 3 lbs.
10. 10 ins. 12. It coincides with that of the triangle ABC.
13. II B D from D. 14. If O be the centre of the board, G the centre
of gravity required, O G = -^ radius. 16. \ (radius) from the centre.
17. I lb. wt. 19. It will balance at O if A O = | A G. 20. 60°.
21. 3. 23. It divides the diagonal in the ratio 11 : 13.
24. O G = — O C. 25. 5^-^ from the base. 26. It divides
21 8 + :i V3
the diagonal of the larger square in the ratio 7 : 13. 27. It divides the
perpendicular from the centre in the ratio 2 : 7. 28. iii ins.
29. a/3 : I. 32. OG = JOn. 33. It divides the perpendicular
from the right angle upon the hypothenuse in the ratio 26 ." i. 34. It
bisects the line joining A to the middle point of B C. 35. 16 ins. from
Answers 305
the centre. 37. Tan"' a and tan"' |. 38. \a from AD,
\a from KM. 39. 6 lb. wt. 40. 12 lb. wt. 42. The sides 3
and 4 are at right angles to each other ; 2.\, 2|. 44. It will be on the
point of falling. 45. 311 ft. 46. ^3 - ' (side of square).
47. 8:15. 48. II < J < l|. 49. Its distances from A D and A B
are 11 A B and * AD. 50. 7 'i' l> - 4 ^' + .1 ^/3 '^^ ^^ 4ZZE ; if
3 (2 a/3 a= + a'3 i5= + ^ a'4 fl= - ^2)
b = a this = — ^.
V3
CHAPTER VII.
— . Cl
1' li v'3- 3- Sin 9 = — . 5. The longer arm is inclined
\'a- -V b-
at an angle tan-' ^-5^3 to the horizon. 6. Sin-' » / "^^ 7. ^ ^3
3 V / 3 ■
W
8. — -. 9. l8|ft., iflb. wt. 10. W (,/2 - I). 12. 2 lb. wt.
V3
v'r!
13. 91 lb. wt. 14. -r* ton. 15. |= st. wt. ; o. 16. ■/146.
17. 2 3, 2 W. 18. First force = W and makes an angle 90° — 2 8
below the horizon ; the other force = 2 W cos 9. 10. Tan fl = - cos- 8.
CHAPTER VIII.
Lever.
1. ;„- ft. from the 3 lbs, 2. 181 ins. 3. 6J ins. 4. 4 ft. from
the 8 lbs. 5. if ft. 6. 4 lb. wt. 7. 30°. 8. 12 ins. from
the 20 lbs. 9. W. 10. gj lb. wt. 11. 15. 12. fj. 13. 1:2*-
14. ^'. 16. i3Ht., IS lb. wt. 17. 3Q.
The -Common Steelyard.
18. 5I ft. ; 23 lbs. 20. 18 ins., § lb. 21. % in. from C ; 2\ lbs. ;
4I5 ins. from A. 22. 16 lb. wt. 23. 5 lbs. excess. 25. 3j lb. wt.
The Danish Steelyard.
26. 3 oz., \ oz. 27. 12 ins. from the end ; 4 ot, wt. 28. 35 lbs.
30. 2ift.
Balance.
31. 261^ lbs. 324a;. 35. 8ifjlbs. 36. I oz. 37. 3^ lb, wt.; 7 ; 8.
X
3o6 Dynamics
Wheel and Axle.
38. l6| lb. wt. 30. 24 cwt. 40. 55 lb. wt.
Pulleys.
W
41. 84 \/2 oz. 42. — ^ First System. 43. 4. 44. 4 lb. wt.
v'3-
45. ifw. 46. 17 oz. wt. 47. 81 lb. wt. 48. 3 lb. wt. 49. Wt. of
15 lb. I oz. 50. The middle pulley being 5 lbs. —864 or 858 lb. wt.
51. loif oz. wt. 52. 5 lb. wt. 54. 16 lb. wt. 55. 7 ; 32 ft. Second
System. 56. 4401b. wt. 57. I2lb. wt. 58. 15 lb. wt. 59. 5 pulleys.
60. 4; 2 lb. wt. Third System. 61. 4. 62. 26 or 28 lb. wt.
63. 14 lb. wt. 64. I lb. wt. 65. 16 lbs. 6 oz. 66. 15 w.
67. II oz. wt. 68. 10 lb. wt. 70. 5 lb. wt.
Inclined Plane.
72, 6 lb. wt. 73. 12 v'3 lb. wt. 74. 30°. 75. 45°.
76. 20 lb. wt. 77. 30°, 6 v'3 lb. wt. 78. 30°, 30° with the plane.
79. 30°. 80. At an angle cos-' f with the plane. 82. 45°.
83. P makes an angle with R equal to the angle of the plane.
84. v'l : v'2. 85. 103-92 lb. wt. 86. 5 : 3 : 4 ; 2 P.
92. Cos-' f with the plane. 93. W cos a . sin a. 94. 7 St. wt.
CHAPTER IX.
1. 8 lb. wt., 4 lb. wt. 2. \; sin-' |.
5. P = (I)W, (2) V3W; R= (I)W, (2)2W.
are at right angles. 10. Twice the angle of friction,
CHAPTER XI.
1. 7. 2. i hour. 3. \\. 4. 38,400. 5. 192^.
6. 24, 40 nearly. 7. About 7fj miles per hour ; 22^ miles.
8. 50 miles an hour. 9. W. 10. 242 ft., 2f sec. 11. =Ao.
12. If. 13. 264 yfa 14. I. 15. 8, 32. 16. 8|. 17. i.
18. ii?. 20. §. 21. J mile. 22. 10 sec. 23. JL ;„ ft.-
144
sec. units ; 120 in mile-hour units. 24. i in 654. 25. — ■ _"
•5 I5.S-"
w
2 •
6. |.
8. P and R
Dn.
11. ^3.
5
Answers 307
26. '" ~ ' 27. 1,200. 28. 128-8. 29. 48, 72. 30. 4,440 miles
2 K + I ■
per hour per hour. 31. 1,936 ft. 32. 44 sees. 33. Acceleration = 32.
34. 34 ft. 35. Velocity = ij% miles per min. ; acceleration = /j miles
per min. per min. 37. 12 miles.
CHAPTER XII.
5 V2 _
1. 2537. 2. — -,40-/2. 3. 32-18. 4. 2a 5. 72.
6. 576 ft. 7. 320 ft. 8. Tan-' 'Li^_ g, 720 ft., 21 sec.
10. 4-65 sec. 11. 64 ft. 12. 6i sec, 3121 ft. 13. Vl2^.
14. 5 sec. 15. 24 + 8 -/"S ft. 16. 9 sec. 17. 64 ft. 19. IS9'04-
20. 676 ft. 21. 17s ft. ^ sec. 22. 30°- 23. | sec. 25. a sec ",
CHAPTER XIII.
2. 19,600 ; 58,800. 3. 2|| miles. 4. 45°. 5. Tan-" 4.
6. 1 162-6. 7. 48; 36^3. 8. 61ft.; 1,500 ft. 10. 80 ^/3.
14. 200 ft, 15. 10 ~/l sec. 16. 72 ; 2| sec. 19. 24.
22. (2 v'i — i) (V^ sin u cos a) j g from the point of projection.
2 L
23. - Vv"- - g'R. 26. — =. 31. At the lowest point. 32. Tan S
= (tan a)i or (cot o)J — (tan a)l.
CHAPTER XIV.
^- ^ poundal. 2. They are equal. 3. 220 poundals.
4. mv ^2 along the diagonal. 5. — . 6. About 2-84 sees.
7- ^ ; ^ poundals. 8. (i) |; (2) 5I ; (3) 2^. 9. 3f;2f;||.
10. ^ poundals; 16 ft. 11. 4J oz., 3I oz. 12. 2.
«^ 2M + OT ,g M - OT . (3 M + m) (M - ot) g fj
^''' 2^' « ■ ■ M + »i ■' 2(M + w)''
16. 5^ ft. ; 2 sees. 17. 17 : 15. 18. ° '" poundals, where /« = mass
225
_ g
of train in lbs. 19. 34g sees. 20. -/2 - I. 21. - ■
3o8 Dynamics
22. 2f miles ; 7 mins. 23. ^ ; Vll poundals. 24. ■ajj"f" and 12222.
4 4
25. 2-4. 26. I4|. 27. 2| sees. 28. 4 ft. 11 ins. 29. J ft. per sec.
31. w . — SJl — . i J*' + 4 /' _ ^^ 32. Acceleration of R
gf' + 2h 2 {gt- + -^)
__ 4 P Q - R (P + Q ) ^
4PQ + R(P + Q) ' *■
CHAPTER XV.
1,1
6. - =^cos9- 10. 31 '92.
9
CHAPTER XVI.
1. \. 2. 50 e". 3. I sec. 4. 8|f and 85^ yards per sec. in
the opposite directions. 5. Same as before, but in opposite directions.
VS .". ,, V^ / I \ „ I «
e. Y!(x^..);r(_L_y 7.1," 9.5:28.
10.
^ . .. .
11- TO> To- 13- Tan-' a/c. 14. They meet with equal velocities - ;
VI is at rest after impact, and vi' starts up the plane with velocity e V.
15. 4«(i +e)/;sina. 16. Soft. 17. ^ " ^'" ° . ^J— . 18. — .^■
^ I — ^ 10
19. »'i sin 2 a = 3 ag 20. « -(i_±_£l. 21. ?^3.
3 3
CHAPTER XVII.
1. 152 foot-pounds. 3. 25. 4. n-3hrs 5. 2| sees. ; 10a mg.
6. 6,250. 8. 64. 9. 3,oco poundals. 10. 3, ii ; 27. 13. ^.
14. 450 poundals.
PRINTED BY
SPOTTISWOODE AND CO., NEW-STREET SQt/ARE
LONDON