MECHANICS FOE BEGINNEKS
MECHANICS FOR BEGINNERS
WITH NUMEROUS EXAMPLES
BY
I. TODHUNTER, D.Sc., F.R.S.
LATE HONORARY FELLOW OF ST JOHN'S COLLEGE, CAMBRIDGE.
Honfcon :
MACMILLAN AND CO.
AND NEW YORK.
1887
[The Eight of Translation is reserved.]
Printed by C. J. CLAY, 1867.
Reprinted 1870, 1874.
Stereotyped 1878. Reprinted 1880, 1882, 1884, 1887.
PEEFACE.
THE present work is constructed on the same plan as
the author's Algebra for Beginners and Trigonometry
for Beginners; and is intended as a companion to them.
It is divided into short Chapters, and a collection of
Examples follows each Chapter. Some of these examples
are original, and others have been selected from College
and University Examination papers.
The work forms an elementary treatise on demon
strative mechanics. It may be true that this part of
mixed mathematics has been sometimes made too abstract
and speculative; but it can hardly be doubted that a
knowledge of the elements at least of the theory of the
subject is extremely valuable even for those who are
mainly concerned with practical results. The author has
accordingly endeavoured to provide a suitable intro
duction to the study of applied as well as of theoretical
Mechanics.
The demonstrations will, it is hoped, be found simple
and convincing. Great care has been taken to arrange
them so as to assume the smallest possible knowledge
of pure mathematics, and to furnish the clearest illus
tration of mechanical principles. At the same time there
has been no sacrifice of exactness; so that the beginner
may here obtain a solid foundation for his future studies :
afterwards he will only have to increase his knowledge
without rejecting what he originally acquired. The ex
vi PREFACE.
perience of teachers shews that it is especially neces
sary to guard against the introduction of erroneous
notions at the commencement of the study of Mechanics.
The work consists of two parts, namely, Statics and
Dynamics. It will be found to contain all that is usually
comprised in elementary treatises on Mechanics, together
with some additions. Thus, for example, an investigation
has been given of the time of oscillation of a simple
pendulum. The more important cases of central forces
are also discussed; partly because they are explicitly
required in some examinations, and partly because by
the mode of discussion which is adopted they supply
valuable exemplifications of fundamental mechanical theo
rems. It would be easy to treat in the same manner
the other cases of central forces which are contained in
the first three sections of Newton's Principia. Some notice
has been taken of D'Alembert's Principle and of Moment
of Inertia, with the view of introducing the reader to the
subject of Rigid Dynamics.
As the Chapters of the work are to a great extent
independent of each other, it will be possible to vary the
order of study at the discretion of the teacher. The
Dynamics may with advantage be commenced before the
whole of the Statics has been mastered.
I. TODHUNTER.
CAMBEIDGE,
April, 1878.
CONTENTS.
STATICS.
PAGE
I. Introduction i
II. Parallelogram of Forces 9
in. Forces in one Plane acting on a particle 24
IV. Resultant of two Parallel Forces 32
V. Moments 40
VI. Forces in one Plane 45
VII. Constrained Body 55
VIII. Centre of Parallel Forces 63
IX. Centre of Gravity 72
X. Properties of the Centre of Gravity 91
XI. The Lever 100
XII. Balances 109
XIII. The Wheel and Axle. The Toothed Wheel ... 118
XIV. ThePully 124
XV. The Inclined Plane 135
XVI. The Wedge. The Screw 142
XVII. Compound Machines 147
XVIH. Virtual Velocities 154
XIX. Friction 166
XX. Miscellaneous Propositions 1/9
XXI. Problems 191
Miscellaneous Examples in Statics 198
viii CONTENTS.
DYNAMICS.
I. Velocity 209
II. The First and Second Laws of Motion 212
III. Motion in a straight line under the influence of
a uniform force 216
IV. Motion in a straight line under the influence of a
uniform force, with given initial velocity 230
V. Second Law of Motion. Motion, under the in
fluence of a uniform force in a fixed direc
tion, but not in a straight line. Projectiles.. 237
VI. Projectiles continued 250
VII. Mass 261
VIII. Third Law of Motion 266
IX. The Direct Collision of Bodies 271
X. The Oblique CoUision of Bodies 281
XI. Motion of the Centre of Gravity of two or more
bodies 293
XII. Laws of Motion. General Eemarks 297
XIII. Motion down a Smooth Curve 306
XIV. Uniform motion in a Circle 314
XV. Motion in a Conic Section round a focus 326
XVI. Motion in an Ellipse round the centre 333
XVII. Work 337
XVin. D'Alembert's Principle 350
XIX. Moment of Inertia 355
XX. Motion round a fixed axis 370
XXI. Miscellaneous Theorems 376
Miscellaneous Examples in Dynamics 384
Answers 388
STATICS.
I. Introduction.
1. WE shall commence this work with some prelimi
explanations and definitions.
We shall assume that the idea of matter is familiar to
student, being suggested to us by everything which we
can touch.
2. A body is a portion of matter bounded in every
direction. A maternal particle is a body which is indefi
nitely small in every direction: we shall speak of it for
shortness as a particle.
3. Force is that which moves or tends to move a body,
or which changes or tends to change the motion of a body.
4. When forces act on a body simultaneously it may
happen that they neutralise each other, so as to leave the
body at rest. When a body remains at rest, although acted
on by forces, it is said to be in equilibrium.
5. Mechanics is the science which treats of the equili
brium and motion of bodies. Statics treats of the equili
brium of bodies, and Dynamics of the motion of bodies.
6. Mechanics may be studied to a certain extent as a
purely experimental subject ; but the knowledge gained in
this way will be neither extensive nor sound. To increase
the range and to add to the security of our investigations
we require the aid of the sciences of space and number.
T. ME. 1
2 INTRODUCTION.
These sciences form Pure Mathematics: Arithmetic and
Algebra relate to number, and Geometry to space : Trigo
nometry is formed by the combination of Arithmetic and
Algebra with Geometry. When Mechanics is studied with
the aid of Pure Mathematics it is often called Demonstra
tive Mechanics, or Rational Mechanics, to indicate more
distinctly that the results are deduced from principles by
exact reasoning. Demonstrative Mechanics constitutes a
portion of Mixed Mathematics.
7. In the present work we shall give the elements of
Demonstrative Mechanics. We shall assume that the
student is acquainted with Geometry as contained in
Euclid, and with Algebra and Trigonometry so far as they
are carried in the Treatises for Beginners.
8. There are three things to consider in a force acting
on a particle : the point of application of the force, that
is, the position of the particle on which the force acts ; the
direction of the force, that is, the direction in which it
tends to make the particle move; and the magnitude of
the force.
9. The position of a particle may be determined in
the same way as the position of a point in Geometry. The
direction of a force may be determined in the same way
as the direction of a straight line in Geometry. We shall
proceed to explain how we measure the magnitude of a
force.
10. Forces may be measured by taking some force as
the unit, and expressing by numbers the ratios which other
forces bear to the unit. Two forces are equal when on
being applied in opposite directions to a particle they keep
it in equilibrium. If we take two equal forces and apply
them to a particle in the same direction we obtain a force
double of either; if we combine three equal forces we
obtain a triple force ; and so on.
11. Thus we may very conveniently represent forces by
straight lines. For we may draw a straight line from the
point of application of the force in the direction of the force,
and of a length proportional to the magnitude of the force.
INTRODUCTION.
3
Thus, suppose a particle acted on by three forces
P, Q, R in three different directions. We may take straight
lines to represent these forces.
For let denote the position *
of a particle. Draw straight
lines OA, OB, OC in the di
rections of the forces P, Q, R
respectively ; and take the
lengths of these straight lines
proportional to the forces:
that is, take
OB Q , OG_R
In saying that OA represents the force P, we suppose
that this force acts from towards A ; if the force acted
from A towards we should say that AO represents it.
This distinction is indeed sometimes neglected, but it may
be observed with great advantage.
It would be convenient to distinguish between the
phrases line of action and direction: thus, if we say that
OA is the line of action of a force P, it may be under
stood that the force merely acts in this straight line,
either from to A, or from A to 0; bub if we say that OA
is the direction of a force P it may be understood that the
force acts from towards A.
12. Sometimes an arrow
head is used in a figure to
indicate which end of the
straight line representing the
force is that towards which
the force tends. Sometimes
the letters, as P, Q, tf, which
denote the magnitudes of the
forces, are inserted in the
figure.
13. We find by experiment that if a body be set free
it will fall downwards in a certain direction ; if it start
12
4 INTRODUCTION.
again from the same point as before it will reach the
ground at the same point as before. This direction is
called the vertical direction, and a plane perpendicular
to it is called a horizontal plane. The cause of this
effect is assumed to be a certain power in the earth which
is called attraction, or sometimes gravity.
If the body be prevented from falling by the interposi
tion of a hand or a table, the body exerts a pressure on
the hand or table.
"Weight is the name given to the pressure which the
attraction of the earth causes a body to exert on another
with which it is in contact.
14. In Statics forces are usually measured by the
weights which they can support. Thus, if we denote the
force which can support one pound by 1, we denote the
force which can support five pounds by 5.
15. Force may be exerted in various ways ; but only
three will come under our notice :
We may push a body by another ; for example, by the
hand or by a rod : force so exerted may be called pressure.
We may pull a body by means of a string or of a rod :
force exerted by means of a string, or by means of a rod
used like a string, may be called tension.
The attraction of the earth is exerted without the in
tervention of any visible instrument: it is the only force
of the kind with which we shall be concerned.
16. A solid body is conceived to be an aggregation of
material particles. A rigid body is one in which the par
ticles retain invariable positions with respect to each other,
No body in nature is perfectly rigid; every body yields
more or less to the forces which act on it. But investiga
tions as to the way in which the particles of a solid body
are held together, and as to the deviation of bodies from
perfect rigidity, belong to a very abstruse part of Mechanics.
It is found sufficient in elementary works to assume that
the rigidity is practically perfect.
INTRODUCTION. 5
17. Wo shall now enunciate an important principle of
Statics which the student must take as an axiom : when a
force acts on a body the effect of the force will be un
changed at whatever point of its direction it be applied,
provided this point be a point of the body or be rigidly
connected with the body.
Thus, suppose a body kept in equilibrium by a system
of forces, one of which is the force P applied at the point
A. Let B be any point on the straight line which coincides
with the direction of P. Then the axiom asserts that P
may be applied at B instead of at A, and that the effect of
the force will remain the same.
This principle is called the transmissibility of a force
to any point in its line of action. It may be verified to
some extent by direct experiment ; but the best evidence
we have of its truth is of an indirect character. We as
sume that the principle is true ; and we construct on this
basis the whole theory of Statics. Then we can compare
many of the results which we obtain with observation and
experiment ; and we find the agreement so close that we
may fairly infer that the principle on which the theory rests
is true. But we cannot appeal to evidence of this kind at
the beginning of the subject, and we therefore take the
principle as an axiom the truth of which is to be assumed.
18. The following remarks will give a good idea of the
amount of assumption involved in the axiom of the preced
ing Article.
6 INTRODUCTION.
At the point E suppose we apply two forces Q and R,
each equal to P, the former in the direction of P, and the
latter in the opposite direction. Then we may readily
admit that we have made no change in the action of P.
Now P at A and R at B are equal forces acting in
opposite directions ; let us assume that they neutralise each
other : then these two forces may be removed without dis
turbing the equilibrium of the body, and there will remain
the force Q at , that is, a force equal to P and applied
at B instead of at A.
19. When we find it useful to change the point of
application of a force, we shall for shortness not always state
that the new point is rigidly connected with the old point ;
but this must be always understood.
20. "We shall have occasion hereafter to assume what
may be called the converse of the principle of the trans
missibility of force, namely, that if a force can be trans
ferred from its point of application to a second point with
out altering its effect, then the second point must be in the
line of action of the force.
21. We shall frequently have to refer to an important
property of a string considered as an instrument for exert
ing force, which we will now explain.
Let AB represent a string pulled at one end by a force
P, and at the other end by a force Q, in opposite directions.
It is clear that if the string is in equilibrium the forces
must be equal.
INTRODUCTION. 7
If the force Q be applied at any intermediate point (7,
instead of at B, still for equilibrium we must have Q equal
to P. This is sometimes expressed by saying that force
transmitted directly by a string is transmitted without
change.
Again, let a string Q
ACB be stretched round ^0 >~ Q
a smooth peg C\ then we /'
may admit that if the /
string be in equilibrium /^
the forces P and Q at its / , /
ends must be equal. This if
is sometimes expressed p
by saying that for ce trans
mitted by a string round
a smooth peg is transmitted without change.
Or in both cases we may say briefly, that the tension of
is the same throughout.
We suppose that the weight of the string itself may be
left out of consideration.
22. Experiment shews that the weight of a certain
volume of one substance is not necessarily the same as the
weight of an equal volume of another substance. Thus,
7 cubic inches of iron weigh about as much as 5 cubic
inches of lead. We say then that lead is denser than iron ;
and we adopt the following definitions :
When the weight of any portion of a body is propor
tional to the volume of that portion the body is said to be
of uniform density. And the densities of two bodies of
uniform density are proportional to the weights of equal
volumes of the bodies. Thus we may take any body of uni
form density as the standard and call its density 1, and then
the density of any other body will be expressed by a num
ber. For example, suppose we take water as the standard
substance ; then since a cubic inch of copper weighs about
as much as 9 cubic inches of water, the density of copper
will be expressed by the number 9.
EXAMPLES. I.
EXAMPLES. I.
1. If a force which can just sustain a weight of 5 Ibs.
be represented by a straight line whose length is 1 foot 3
inches, what force will be represented by a straight line
2 feet long ?
2. How would a force of a ton be represented if a
straight line an inch long were the representation of a force
of seventy pounds ?
3. If a force of P Ibs. be represented by a straight
line a inches long, what force will be represented by a
straight line b inches long?
4. If a force of P Ibs. be represented by a straight line
a inches long, by what straight line will a force of Q Ibs. be
represented?
5. A string suspended from a ceiling supports a weight
of 3 Ibs. at its extremity, and a weight of 6 Ibs. at its middle
point : find the tensions of the two parts of the string. If
the tension of the upper part be represented by a straight
line 3 inches long, what must be the length of the straight
line which will represent the tension of the lower portion?
6. If 3 Ibs. of brass are as large as 4 Ibs. of lead, com
pare the densities of brass and lead.
7. Compare the densities of two substances A and B
when the weight of 3 cubic inches of A is equal to the
weight of 4 cubic inches of B.
8. A cubic foot of a substance weighs 4 cwt. : what
bulk of another substance five times as dense will weigh
7 cwt.?
9. Two bodies whose volumes are as 3 is to 4 are in
weight as 4 is to 3 : compare their densities.
10. If the weight of a cubic inches of one substance
and of b cubic inches of another substance be in the ratio
of m to n, compare the densities of the substances.
PARALLELOGRAM OF FORGES. 9
II. Parallelogram of Forces.
23. When two forces act on a particle and do not
keep it in equilibrium, the particle will begin to move in
some definite direction. It is clear then that a single
force may be found such that if it acted in the direction
opposite to that in which the motion would take place,
this force would prevent the motion, and consequently
would be in equilibrium with the other forces which act
on the particle. If then we were to remove the original
forces, and replace them by a single force equal in magni
tude to that just considered, but acting in the opposite
direction, the particle would still be in equilibrium.
Hence we are naturally led to adopt the following
definitions :
A force which is equivalent in effect to two or more
forces is called their resultant; and these forces are called
components of the resultant.
24. We have then to consider the composition of
forces, that is, the method of finding the resultant of two
or more forces. The present Chapter will be devoted to
the case of two forces acting on a particle.
25. When two forces act on a particle in the same
direction their resultant is equal to their sum and acts
in the same direction.
This is obvious. For example, if a force of 5 Ibs. and a
force of 3 Ibs. act on a particle in the same direction their
resultant is a force of 8 Ibs., acting in the same direction.
26. When two forces act on a particle in opposite
directions their resultant is equal to their difference, and
acts in the direction of the greater force.
This is obvious. For example, if a force of 5 Ibs. and a
force of 3 Ibs. act on a particle in opposite directions their
resultant is a force of 2 Ibs., acting in the same direction
as the force of 5 Ibs.
27. We must now proceed to the case in which two
forces act on a particle in directions which do not lie in
the same straight line; the resultant is then determined
by the following proposition :
10 PARALLELOGRAM OF FORCES.
If two forces acting on a particle be represented in
magnitude and direction by straight lines dra/ivn from
a point, and a parallelogram be constructed having these
straight lines as adjacent sides, then the resultant of the two
forces is represented in magnitude and direction by that
diagonal of the parallelogram which passes through the
point.
This proposition is the most important in the science of
Statics ; it is called briefly the Parallelogram of Forces.
We shall first shew how the proposition may be verified
experimentally ; we shall next point out various interesting
results to which it leads ; and finally demonstrate it.
28. To verify the Parallelogram of Forces experi
mentally.
Let A and B be smooth horizontal pegs fixed in a
vertical wall. Let three strings be knotted together; let
represent the knot. Let one string pass over the peg
A and have a weight P attached to its end; let another
string pass over the peg B and have a weight Q attached
to its end; let the other string hang from and have a
weight R attached to its end : any two of these three weights
must be greater than the third. Let the system be allowed
to adjust itself so as to be at rest. By Art. 21 the pegs do
not change the effects of the weights P and Q as to
magnitude.
We have three forces acting on the knot at 0, and
PARALLELOGRAM OF FORCES.
11
g it in equilibrium; so that the effect of P along
)A and of Q along OB is just balanced by the effect of R
ting vertically downwards at 0. Therefore the result.
of P along OA and of $ along OB must be a force
nial to .#, acting vertically upwards at 0.
Now on OA take 6>/> to contain as many inches as the
sight P contains pounds ; and on B take Oq to contain
as many inches as the weight Q contains pounds; and
complete the parallelogram Oqrp. Then it will be found
by trial that Or contains as many inches as the weight R
contains pounds ; and that Or is a vertical line.
"We may change the positions of the pegs, and the
magnitudes of the weights employed, in order to give due
variety to the experiment ; and the general results afford
sufficient evidence of the truth of the Parallelogram of
Forces. The strings should be fine and very flexible in
order to promote the success of the experiment ; and it is
found practically that small pullies serve better than fixed
pegs for changing the directions of action of the weights P
and Q without changing the amounts of action.
We proceed now to the numerical calculation of the
resultant of two forces.
29. The case in which two forces act on a particle in
directions which include a right angle deserves especial
notice.
Let AB represent a force P, and AC a force Q] and
let BAG be a right angle. Complete the rectangle ABDC :
then AD represents the resultant of P and Q; we will
denote this resultant by R.
Now by Euclid I. 47,
so that *
For example, let P be 15 Ibs.,
and Q be 8 Ibs. : then
therefore ^=17.
Thus the resultant force is 17 Ibs,
12 PARALLELOGRAM OF FORCES.
30. We will now give the general expression for the
resultant of two forces which act on a particle whatever be
the angle between their directions.
Let AB represent a force P, and AC a force Q\ and
let a denote the angle BAG. Complete the parallelogram
ABDC: then AD represents the resultant of P and Q ;
we will denote the resultant by R.
Now by Trigonometry,
so that R 2 =P 2 + Q* + 2PQ cos a.
For example, let Q be equal to P } and let a be 60, then
cos 60=^, so that
therefore R=P*J3.
Whatever be the angle a if P= Q we have
cos a=2P 2 (1 + cos a);
but 1 + cos a= 2 cos 2  ;
2
therefore IP = 4P 2 cos 2 , and R = 2P cos ? .
'
31. Let .4J5 and 4(7 represent two forces; and AD
their resultant. Draw CB the other diagonal of the pa
rallelogram.
PARALLELOGRAM OF FORCES.
13
Then since the diagonals of a
parallelogram bisect each other,
CE=EB, and AD=ZAE. Hence
the resultant of the two forces
may be determined thus : join CB
and bisect it at E; then AE is
the direction of the resultant, and
the magnitude of the resultant is
twice AE. This mode of determin
ing the resultant is often useful.
32. Suppose three equal forces to act on a particle,
and the direction of each to make an angle of 120 with
the directions of the two others : see the diagram in Art. 12.
Then it is obvious that the particle will be in equilibrium,
for there is no reason why it should move in one direction
rather than in another.
This result is in accordance with the Parallelogram of
Forces.
For let OA and OB be equal
straight lines; and let the angle
AOB be 120. Complete the
parallelogram OACB.
Then AC=OB=OA\ and the
angle OAG is 60; therefore the
angle A CO = the angle A OC= 60.
Hence the triangle OAG is equi
lateral, so that 00=0 A.
Thus the resultant of two equal
forces, the directions of which
include an angle of 120, is equal
to either of the forces, and bisects
the angle between them.
33. Suppose that three forces act on a particle and
keep it in equilibrium, and that two of the forces are
equal : then the third force must be equally inclined to the
directions of the other two.
Hence, if the resultant of two forces is equal in magni
tude to one of the forces, the other force is at right angles
to the straight line which bisects the angle between the
resultant and the force to which it is equal.
14 PARALLELOGRAM OF FORGES.
34. y three forces acting on a particle be represented
in magnitude and way of action by the sides of a triangle
taken in order they will keep the particle in equilibrium.
Let ABC be a triangle; let P, Q, R be three forces
proportional to the sides BC, CA, AB\ let these forces
act on a particle, P parallel to BC, Q parallel to CA, and
R parallel to AB : then the particle will be in equilibrium.
For draw AD parallel to BC, and CD parallel to BA.
Forces represented by AB and AD in magnitude and
direction will have a resultant represented by AC in mag
nitude and direction. Therefore forces represented by
AB, AD, and CA in magnitude and direction will be in
equilibrium; and AD is equal and parallel to BC. Thus
the proposition is true.
35. The preceding proposition is usually called the
Triangle of Forces. The student should pay careful at
tention to the enunciation, in order to understand distinctly
what is here asserted. The words taken in order must bo
noticed. If one force is represented by AB the others
must be represented by BC and CA, not by BC and AC,
nor by CB and CA, nor by CB and AC. Also, it is to be
observed that the forces are supposed to act on a particle,
that is, to have a common point of application. Thus the
directions of the forces are not represented strictly by AB,
BC, and CA, but by straight lines parallel to these drawn
from a common point. Beginners frequently make a mis
take in this respect; they imagine that forces, actually
represented in magnitude and situation by AB, BC, and
PARALLELOGRAM OF FORCES. 15
CA, would keep a body in equilibrium : that is, they forget
the limitation that the forces must act on a particle. In
order to direct the attention of the student to this impor
tant limitation we have employed the words way of action
in the enunciation, instead of the usual word direction,
which is liable to be confounded with situation.
36. If three forces acting on a particle keep it in
equilibrium, and a triangle be drawn having its sides
parallel to the lines of action of the forces, the sides of the
triangle will be proportional to the forces respectively
parallel to them.
p
Let forces P, Q, R acting on a particle at keep it in
equilibrium. In the direction of P take any point p, and
in the direction of Q take a point q, such that 7T = p'
Complete the parallelogram Oprq. Then Or represents the
resultant of P and Q in magnitude and direction. Hence
r must be on the straight line RO produced. Therefore
any triangle having its sides parallel to the directions of
the forces will be similar to the triangle Opr, and will
therefore have its sides proportional to the forces in
magnitude.
37. If three forces acting on a particle keep it in
equilibrium, and a triangle be drawn having its sides, or its
sides produced, perpendicular to the lines of action of the
forces, the sides of the triangle will be proportional to the
forces respectively perpendicular to them.
For suppose ABC to denote any triangle. Then straight
16 PARALLELOGRAM OF FORCES.
lines perpendicular respectively to AB and A C will include
an angle equal to A ; and so on. Thus the triangle which
has its sides perpendicular to those of ABC will be equi
angular to ABC, and therefore similar to it. The side
which is perpendicular to BC will be opposite to an angle
equal to A ; and so on. Now by Art. 36 the forces will be
represented by the sides of a triangle parallel to the lines
of action ; and therefore they will also be represented by
the sides of a triangle perpendicular to the lines of action.
38. If three forces acting on a particle keep it in
equilibrium, each force is proportional to the sine of the
angle between the directions of the other two.
Let forces P, Q, R acting on a particle at keep it in
equilibrium. Then, as in Art. 36, we have
P : Q : R=0p : pr : rO.
But, by Trigonometry,
Op \ pr : rO=sinprO : &mpOr : sin rpO
=sin QOR : sin ROP : sin POQ.
39. Conversely, if three forces act on a particle, and
each force is proportional to the sine of the angle between
the other two, the forces will keep the particle in equilibrium
provided a certain condition be fulfilled. This condition
corresponds to that expressed by the words taken in order
of Arts. 34 and 35. Thus if Op and Oq in the diagram of
Art. 36 represent the directions of two of the forces,
the third force must be in the direction rO, and not
in the direction Or. We may express this by saying that
the direction of the third force must lie outside the angle
formed by the directions of the other two forces ; the word
direction being taken strictly as in Art. 11.
Similarly the converse of Art. 37 is true, provided this
condition holds.
40. If two forces act on a particle their effect is equi
valent to that of a third force of suitable magnitude and
direction ; if the two forces act on a body at a point, we may
take it as obvious that their effect is also equivalent to that of
PARALLELOGRAM OF FORCES. 17
the same third force acting on the body at the point. Hence
suits obtained with respect to forces acting on a particle
lay be applied with respect to forces acting on a body at
a point: and in stating such results we often, for brevity,
omit the words on a body and speak of forces acting at
a point. Thus the important enunciation of Art. 27 may
be modified by changing the words on a particle to at a
rint.
If three forces act at a point we may find their
mltant thus : form the resultant of two of them, then
the resultant of this and the third force. Thus we
have the resultant of the three forces. This process may
be extended to more than three forces ; we shall consider
it more fully in the next Chapter.
41. If three forces acting in one plane maintain a
rigid body in equilibrium their lines of action either all meet
at a point or are all parallel.
For suppose the lines of action of two of the forces to
meet at a point ; these forces may be supposed to act at
this point, and may be replaced by their resultant. This
resultant and the third force must then be equal and oppo
site in order to maintain the body in equilibrium : so that
the line of action of the third force must pass through the
intersection of the lines of action of the other two. Then
the forces must satisfy the conditions of Art. 36.
The conditions which must hold among three parallel
forces which keep a rigid body in equilibrium will be given
in Chapter IV.
42. As we can compound two forces into one, so on
the other hand we can resolve one force into two others.
For let AD represent a force ; .2 I?
draw any parallelogram ABDG
having AD as a diagonal ; then the
force represented by AD is equi
valent to two forces represented
by AB and AC respectively.
Thus we can resolve any force into two components
which shall have assigned directions.
T. ME. 2
18
PARALLELOGRAM OF FORCES.
The case in which we resolve a force into two others at
right angles deserves special notice.
Let BAG be a right angle; and
let a denote the angle DAB. Then
AB=AD cos a, AC=BD=AD sin a.
Thus any force P may be resolved
into two others, P cos a and P sin a,
which are at right angles : the direc
tion of the component P cos a making
an angle a with the direction of P.
43. "We see from the former part of the preceding
Article that a given force may be resolved into two others
in an infinite number of ways. In future when we speak
of the resolved part of a force in a given direction we shall
always suppose, unless the contrary is expressed, that the
force is resolved into two forces, one in the given direction,
and the other in the direction at right angles to the given
direction; and the former component we shall call the
resolved force in the given direction.
44. The resolved part in any direction of the resultant
of two forces a'cting at a point is equal to the sum of the
resolved parts of the components in that direction.
Let AB and AC denote two forces, and AD their
resultant.
Let AK bo any straight line through A. Draw BE
and DF perpendicular to AK, and BG parallel to AK.
Then AF represents the resolved part of the resultant
along AK; and AF=AE+EF. Now AE is the resolved
PARALLELOGRAM OF FORCES. 19
part of the component AB along AK ; and EF is equal to
BG, that is to the resolved part along AK of the compo
nent AC; for BD is equal and parallel to AC.
45. We shall now give the demonstration of the truth
of the Parallelogram of Forces which is most suitable for
an elementary work ; it is called by the name of Duchayla,
to whom it is due.
The demonstration is divided into three parts : in the
first part the proposition is established so far as relates to
the direction of the resultant, the forces being commen
surable ; in the second part this result is extended to in
commensurable forces ; in the third part the proposition is
established with respect to the magnitude of the resultant.
46. We have first to notice a preliminary assumption.
We assume that if two equal forces act on a particle the
direction of the resultant will be in the same plane as
the directions of the forces and will bisect the angle be
tween them. This seems obvious, for there is no reason
why the resultant should lie on one side of the plane of the
forces rather than on the other ; and there is no reason why
it should be nearer to one force than to the other.
If a parallelogram be constructed on two equal straight
lines meeting at a point as adjacent sides, the diagonal
which passes through that point bisects the angle between
the sides which meet at that point.
Hence the parallelogram of forces is true, so far as the
direction is concerned, when the forces are equal.
47. To demonstrate the Parallelogram of Forces so
far as relates to the direction of the resultant, the forces
being commensurable.
Suppose that the proposition is true for two forces P
and $, inclined at any angle ; and also for two forces P
and R inclined at the same angle : we shall shew that it
will be true for two forces P and Q + R inclined at tho
same angle.
22
20 PARALLELOGRAM OF FORGES.
Let A be the point of ap c l4 ,
plication of the forces; let
the force P act along AB and
the force Q + R along AE.
Let AB and AC represent
P and $ in magnitude, and
let CE represent R in mag ^ JjK~ G\
nitude. Complete the paral 1 X \
lelogram ACDB, and the pa L
rallelogram CEOD.
By Art. 17 the force R may be supposed to act at C
instead of at A ; and thus may be denoted in magnitude
and situation by CE.
Now by hypothesis the resultant of P and Q acts along
AD; let P and Q be replaced by their resultant, and
let this resultant be supposed to act at D instead of
at A. Resolve this force at D into two components, one
along CD and the other along DO : these two components
will be respectively P and Q, the former may be supposed
to act at (7, and the latter to act at G.
Again, the resultant of P along CD and R along CE
acts by hypothesis along CG ; let P and R be replaced by
their resultant, and let this resultant be supposed to act
at G.
Thus by this process we have transferred the forces
which acted at A to G, without altering their effect.
Hence, by Art. 20, we infer that G is a point in the
direction of the resultant of the forces P and Q + R at A ;
that is, the resultant of P and Q + R acts in the direction
of the diagonal AG; thus, if the proposition hold for
P and Q and for P and R, it holds for P and Q + R.
But the proposition holds for equal forces P and P, there
fore it holds for P and 2P, and therefore for P and 3P,
and so on ; hence it holds for P and nP> where n is any
whole number.
And since it holds for P and nP, it holds for 2P
and nP, and therefore for 3P and nP, and so on ; hence
it holds for mP and nP, where m is any whole number.
PARALLELOGRAM OF FORCES. 21
Thus the proposition holds for any two commensurable
forces.
48. To demonstrate the Parallelogram of Forces so
far as relates to the direction of the resultant, the forces
being incommensurable.
The result in this case may be inferred from the fact
that when two magnitudes are incommensurable, so that
the ratio of the one to the other cannot be expressed
exactly by means of numbers, we can find numbers which
shall represent the ratio within any assigned degree of
closeness.
The result may also be established indirectly thus :
A C
Let AB, AC represent
two incommensurable forces.
Complete the parallelogram
BC. Then if their resultant
do not act along AD sup
pose it to act along AE : \\j_r..
draw EF parallel to CA. c /~ N/K
35 J>
Divide A C into a number of equal portions, each less
than DE\ mark off on CD portions equal to these, and
let K be the last division : then K evidently falls between
D and E. Draw KG parallel to CA.
Then two forces represented by AC and AG have a
resultant in the direction AK, because the forces are
commensurable. Therefore the forces AC and AB are
equivalent to a force along A K, together with a force equal
to GB applied at A along AB. And we may assume
as obvious that the resultant of these forces must lie
between AK and AB; but by hypothesis the resultant
is AE, which is not between AK and AB: this is
absurd.
In the same way we may shew that the resultant cannot
act along any straight line except AD.
Thus the Parallelogram of Forces is demonstrated so
far as relates to the direction of the resultant whether
the forces are commensurable or incommensurable.
22
EXAMPLES. II.
49. To demonstrate that the Parallelogram of Forces
holds also with respect to the magnitude of the resultant.
Let AS, AC be the directions of
the given forces, AD that of their re
sultant. Take AE opposite to AD,
and of such a length as to represent
the magnitude of the resultant.
Then the forces represented by
AS, AC, AE balance each other. On
AE and AS as adjacent sides con
struct the parallelogram ABFE: then
the diagonal ^l^is the direction of the
resultant of AE and AS.
Hence AC^ must be in the same
straight line with A F\ therefore A FED
is a parallelogram; therefore A D =BF.
But BF=AE. Therefore AE=AD: hence the resultant
is represented in magnitude as well as in direction by the
diagonal AD.
Thus the Parallelogram of Forces is completely de
monstrated.
EXAMPLES. II.
1. Two forces act on a particle, and their greatest and
least resultants are 72lbs. and 561bs. : find the forces.
2. Find the resultant of two forces of 121bs. and 35lbs.
respectively which act at right angles on a particle.
3. Two forces whose magnitudes are as 3 is to 4, act
ing on a particle at right angles to each other, produce
a resultant of 15lbs. : find the forces.
4. Two forces, one of which is double of the other, act
on a particle, and are such that if Gibs, be added to the
larger, and the smaller be doubled, the direction of the
resultant is unchanged : find the forces.
5. Shew that if the angle at which two given forces are
inclined to each other is increased their resultant is di
minished.
L
EXAMPLES. II. 23
6. If one of two forces acting on a particle is 5 Ibs.,
and the resultant is also 5 Ibs., and at right angles to the
known force, find the magnitude and the direction of the
other force.
7. If the resultant of two forces is at right angles to
one of the forces, shew that it is less than the other force.
8. If the resultant of two forces is at right angles to
one force and equal to half the other, compare the forces.
9. If forces of 3 Ibs. and 4 Ibs. have a resultant of 5 Ibs.,
at what angle do they act?
/10. If two forces acting at right angles to each other
be in the proportion of 1 to A/ 3 > an( i their resultant be
10 Ibs., find the forces.
V/ll. How can forces of 43 and 65 Ibs. be applied to a
particle so that their resultant may be 22 Ibs. ?
y!2. Find at what angle the forces P and 2P must act
at a point in order that the direction of their resultant may
be at right angles to the direction of one of the forces.
v^!3. Two strings, the lengths of which are 6 and 8 inches
respectively, have their ends fastened at two points the
distance between which is 10 inches ; their other ends
are fastened together, and they are strained tight by a
force equivalent to 5 Ibs. at the knot acting perpendicu
larly to the straight line joining the two points : find the
tension of each string.
^14. AB and AC are adjacent sides of a parallelogram,
and AD is a diagonal; AB is bisected at E: shew that the
resultant of the forces represented by AD and AC is
double the resultant of the forces represented by AE
a,u<LAC.
^15. Two forces are represented in magnitude and direc
tion by two chords of a circle, drawn from a point on the
circumference at right angles to each other : shew that the
resultant is represented in magnitude and direction by the
diameter which passes through the point.
t/16. A and B are fixed points; at a point M forces of
given magnitude act along MA and MB : if their resultant
is of constant magnitude, shew that M lies on one or other
of two equal arcs described on AB as chord.
24 FORGES IN ONE PLANE
III. Forces in one plane acting on a particle.
50. We shall now shew how to determine the resultant
of any number of forces in one plane acting on a particle ;
we have already briefly noticed this subject : see Art 40.
51. To find_ the resultant of a given number of forces
acting on a particle in the same straight line.
When several forces act on a particle in the same
direction their resultant is equal to their sum.
When some forces act in one direction, and other forces
act in the opposite direction, the whole force in each direc
tion is the sum of the forces in that direction. Hence the
resultant of all the forces is equal to the difference of
those two sums, and acts in the direction of the greater
sum.
If the forces acting in one direction are reckoned posi
tive, and those acting in the other direction negative, then
the resultant of all the forces is equal to their algebraical
sum ; and the sign of this sum determines the direction in
which the resultant acts.
If the algebraical sum is zero the forces are in equi
librium ; and conversely, if the forces are in equilibrium their
algebraical sum is zero.
52. To determine geometrically the resultant of any
number of forces acting on a particle.
Let forces P, Q, R, S act on a particle : it is required
to determine their resultant.
ACTING ON A PARTICLE. 25
Take any point A and draw the straight line AB to
represent the force P in magnitude and way of action;
from B draw BC to represent Q in magnitude and way of
action ; from C draw CD to represent R in magnitude and
way of action; and from D draw DE to represent S in
magnitude and way of action. Join AE; then AE will
represent the resultant in magnitude and way of action.
This is obvious from the preceding Chapter. For the
resultant of P and Q would be represented by the straight
line AC; and then the resultant of the forces represented
by AC and CD would be represented by AD; that is, AD
would represent the resultant of P, Q, and R ; and so on.
The method is applicable whatever be the number of
forces acting on a particle.
Hence we easily see that Art. 44 may be extended to
the case of any number of forces.
53. If any number of forces acting on a particle
be represented in magnitude and way of action by the
sides of a polygon taken in order, they will keep tlie par
tide in equilibrium.
Take for example a polygon of five sides. Let forces
represented in magnitude and way
of action by AB, BC, CD, DE, EA
act on a particle : they will keep the
particle in equilibrium.
For, by the preceding Article,
the forces represented by AB, BC,
OD, DE have a resultant which may
be represented by AE: and two forces represented by AE
and EA respectively will balance each other.
54. The preceding proposition is usually called the
Polygon of Forces. The remarks made in Art. 35 respect
ing the Triangle of Forces are applicable here also.
The converse of the Triangle of Forces is true, as is
shewn in Art. 36; but the converse of the Polygon of
Forces is not true ; that is, if four or more forces acting on
a particle keep it in equilibrium, we cannot assert that the
26 FORCES IN ONE PLANE
forces are proportional to the sides of any polygon which
has its sides parallel to the lines of action of the forces.
For one polygon may be equiangular to another without
being similar to it. If in the figure of the preceding Article
we draw any straight line parallel to one of the sides, as
AE for example, we can form a second polygon, which like
the first has its sides parallel to the lines of action of the
forces; but the sides of the one polygon are not in the
same relative proportion as the sides of the other.
55. It will be seen that the geometrical process of
Art. 52 is applicable when the forces do not all He in one
plane. Also in Art. 53 the polygon need not be restricted
to be a plane polygon. But the method which we shall
now give for calculating by the aid of Trigonometry the
resultant of any number of forces acting at a point assumes
that the forces are all in one plane.
56. Forces act on a particle in one plane: required
the magnitude and the direction of their resultant.
Let P, Q, R,... denote the forces; let a, ft, y,... denote
the angles which their directions respectively make with a
fixed straight line drawn through the particle.
By Art. 42 the force P can be resolved into P cos a
and Psina along the fixed straight line and at right
angles to it respectively ; similarly Q can be resolved into
Q cos /3 and Q sin /3 ; and R can be resolved into R cos y
and R sin y ; and so on.
Then, by algebraical addition of the components which
act in the same straight line, we obtain
P cos a + Q cos /3 + R cos y + . . .
along the fixed straight line, and
P sin a + Q sin /3 + R sin y + ...
at right angles to the fixed straight line.
We shall denote the former sum by JT, and the latter
by Y; hence the given forces are equivalent to the two
forces X and Y in directions which are at right angles to
each other. Let K be their resultant^ and 6 the angle
ACTING ON A PARTICLE. 27
which the direction of K makes with that of X. Then, by
Art. 29,
Thus the magnitude and the direction of the resultant
are determined.
57. To find the conditions of equilibrium when any
number of forces act on a particle in one plane.
The forces will be in equilibrium if their resultant
vanishes; that is, by the preceding Article, if jST=0; and
this will be the case if X=Q and Y=0. These conditions
then are sufficient for equilibrium; we may express them
in words thus :
A system offerees acting in one plane on a particle will
be in equilibrium if^ the sum of the resolved parts of the
forces along two straight lines at right angles to each other
vanishes.
Conversely, if the forces are in equilibrium K=0 and
then JT=0 and 7=0; and X denotes the sum of the re
solved parts of the forces along any straight line. Hence
if forces acting in one plane on a particle are in equili
brium, the sum of the resolved parts of the forces along any
straight line will vanish.
58. We have hitherto spoken of forces acting at a
point; but we can often investigate the resultant of forces
in one plane which act at various points by repeated use of
the principles of transference and composition.
For let there be any number of such forces ; take two of
them, produce their directions to meet, and suppose the
two forces to act at this point; determine their resultant
and replace the two forces by this resultant. Then produce
the direction of this resultant to meet the direction of one
of the remaining forces; and replace the two by their
resultant. Proceeding in this way we may represent the
resultant of all the forces by carefully drawing the suc
cessive diagrams; and we may if we please calculate the
numerical value of the resultant.
28
FORCES IN ONE PLANE
The only case which could present any difficulty is that
in which we should have finally forces in parallel direc
tions ; and this will be considered in the next Chapter.
"We will now give two examples.
(1) Let ABCD be a square; suppose a force of 1 Ib.
to act along AD, a force of 2 Ibs. along AB, and a
force of 3 Ibs. along CB : required the resultant of the
forces.
The forces along AB and AD will have their resultant
along the straight line AE which is so situated that EB
is to A B as 1 Ib. is to 2 Ibs. ; that is, E is the middle point
of EG. Suppose this resultant applied at E; and resolve it
again into its components parallel to AB and AD. Thus
we have a force of 1 Ib. along EC, and a force of 2 Ibs.
acting at E parallel to AB. There is also a force of 3 Ibs.
along CB. Thus on the whole we have a force of 2 Ibs.
along JEB, and also a force of 2 Ibs. at E parallel to AB.
Hence the resultant of all the forces is V(^ + 4) Ibs.,
that is 2 v/2 Ibs. ; and its direction passes through E, and
makes an angle of 45 with EB.
(2) ABCD is a parallelogram; forces represented in
magnitude and situation by AB, J3C, and CD act on a
body : required the resultant of the forces.
The forces AB and BC may be supposed to act at B.
Produce AB to E, so that BE=AB. Then the forces
AB and BC may also be represented by BE and BC;
and their resultant will be represented by BF, the
diagonal of the parallelogram BEFC.
ACTING ON A PARTICLE. 29
The force represented by CD may also be represented
by FC which is equal to CD. Thus the three given forces
are reduced to the two BF and FC, which may be sup
posed to act at F.
Produce BF to (7, so that FG=BF. Then we require
the resultant of forces represented by FG and FC.
Produce EF to 77, so that FH=EF; join HG and CH.
Then, by Geometry, CFGH is a parallelogram; and FH
represents the resultant of FG and FC.
Thus finally the resultant of the forces AB, BC, and
CD is represented in magnitude and in situation by FH.
This example deserves the attention of a beginner. We
shewed that BF is the resultant of AB and BC. Begin
ners are very apt to say that BD is the resultant of AB
and BC; this is wrong : BD is the resultant of BA and
BC, but not of AB and BC. Or beginners sometimes say
that AC is the resultant of AB and BC; this is wrong.
AC is the resultant of AB and .4Z>, but not of AB and 5(7.
Again, beginners sometimes say that the forces AB and
CD being equal and opposite balance each other ; this is
wrong : the forces would balance if they acted in the same
straight line, but they do not so act.
"We may observe that BFHC is a parallelogram ; and
that FH is equal to BC.
30 EXAMPLES. III.
EXAMPLES. III.
1. If three forces represented by the numbers 1, 2, 3
acting in one plane keep a particle at rest, shew that they
must all act in the same straight line.
2. Three forces represented by the numbers 1, 2, 3 act
on a particle in directions parallel to the sides of an equi
lateral triangle taken in order : determine their resultant.
3. Can a particle be kept at rest by three forces whose
magnitudes are as the numbers 3, 4, and 7 1
4. Three forces act at a point parallel to the sides of a
triangle, taken in order, and are inversely as the perpen
diculars from the angular points of the triangle, on the
sides parallel to which the forces act respectively: shew
that the forces are in equilibrium.
5. A weight of 25 Ibs. hangs at rest, attached to the
ends of two strings, the lengths of which are 3 and 4 feet,
and the other ends of the strings are fastened at two points
in a horizontal line distant 5 feet from each other : find the
tension of each string.
6. Three forces acting at a point are in equilibrium ;
the greatest force is 5 Ibs., and the least force is 3 Ibs., and
the angle between two of the forces is a right angle : find
the other force.
7. Two equal forces act at a certain angle on a par
ticle, and have a certain resultant; also if the direction
of one of the forces be reversed, and its magnitude be
doubled, the resultant is of the same magnitude as before :
shew that the resultant of these two resultants is equal to
each of them.
8. Two equal forces act at a certain angle on a particle,
and have a certain resultant ; also if the direction of one of
the forces be reversed, and its magnitude be doubled, the
resultant is of the same magnitude as before : shew that
the two equal forces are inclined at an angle of 60.
EXAMPLES. III. 31
9. AOB and COD are chords of a circle which in
tersect at right angles at ; forces are represented in
magnitude and direction by OA, OB, OC, OD: shew that
their resultant is represented in direction by the straight
line which joins to the centre of the circle, and in magni
tude by twice this straight line.
10. Eight points are taken on the circumference of a
circle at equal distances, and from one of the points straight
lines are drawn to the rest : if these straight lines represent
forces acting at the point, shew that the direction of the
resultant coincides with the diameter through the point,
and that its magnitude is four times that diameter.
11. The circumference of a circle is divided into a
given even number of equal parts, and from one of the
points of division straight lines are drawn to the rest:
shew that the direction of the resultant coincides with the
diameter through the point.
12. Forces of 3, 4, 5, 61bs. respectively act along the
straight lines drawn from the centre of a square to the
angular points taken in order : find their resultant.
13. Perpendiculars are drawn from any point on the
four sides of a rectangle : find the magnitude and the di
rection of the resultant of the forces represented by the
perpendiculars.
14. The circumference of a circle is divided into any
number of equal parts; forces are represented in magni
tude and direction by straight lines drawn from the centre
to the points of division: shew that these forces are in
equilibrium.
15. Shew that the result in Example 11 is true also
when the number of equal parts is odd. Find also the
magnitude of the resultant.
16. A BCD is a parallelogram; forces represented in
magnitude and situation by AB, BC, and DC act on a
body : required the resultant of the forces.
RESULTANT OF
IV. Resultant of two Parallel Forces.
59. Forces which have their lines of action parallel
are called parallel forces; if they tend in the same way
they may be called like, and if they tend in opposite ways
they may be called unlike.
60. To find the magnitude and the direction of the re
sultant of two like parallel forces acting on a rigid body.
Let P and Q be the forces acting at A and B re
spectively.
The effect of the forces will not be altered if we apply
two forces equal in magnitude, and acting in opposite di
rections along the straight line A B. Let S denote each of
these forces, and suppose one to act at A along BA pro
duced, and the other at B along AB produced.
Then P and S acting at A are equivalent to a single
force X acting in a direction between those of S and P ;
and Q and S acting at B are equivalent to a single force
Y acting in a direction between those of S and Q.
TWO PARALLEL FORCES. 33
Produce the directions of X and T to meet ; let them
meet at (7, and draw CD parallel to the directions of P and
Q, meeting AB at D.
Transfer X and T to (7, and resolve them along CD
and a straight line through C parallel to AB ; each of the
latter components will be equal to S, and they will act in
ite directions, and balance each other: the sum of
former components is P + Q.
Hence the resultant of the like parallel forces P and Q
is P + Q, and it acts parallel to the directions of P and Q
in a straight line which cuts AB at D ; so that it may be
supposed to act at D.
"Wo shall now shew how to determine the point D.
The sides of the triangle ADC are parallel to the direc
tions of the forces , P, X\ hence, by Art. 36,
DC
Similarly g*
Therefore
Thus the point D divides AB into segments which are
inversely as the forces at A and B respectively.
Let AB=aj and AD=x; then
__.
ax~P'
therefore Px=Q(a #)
therefore x = .
T. ME.
34
RESULTANT OF
61. To find the magnitude and the direction of the re
sultant of two unlike parallel forces acting on a rigid body.
Suppose Q the greater of the two forces. By the same
method as in the preceding Article we shall arrive at the
following conclusion :
The resultant of the unlike parallel forces P and Q is
Q  P, and it acts parallel to the directions of P and Q
in a straight line which cuts AB produced at a point D
such that
AD_Q
BD P'
Thus D divides AB produced through B into segments
which are inversely as the forces at A and B respectively.
Let AB= a, and ADx\ then
TWO PARALLEL FORCES. 35
therefore Px = Q (x  a) ;
therefore x = Q_p
It will be observed that the results of this Article may
be deduced from those of the preceding Article by changing
P into  P.
62. If three parallel forces keep a rigid body in equili
brium one must be equal and opposite to the resultant of
the other two. Hence they must all act in one plane ; one
of them must be unlike the other two ; and its line of action
must lie between theirs, dividing the distance between
them in the inverse ratio of the two forces.
63. "We may find the resultant of any number of paral
lel forces by repeated application of the process of Arts. 60
and 61. First find the resultant of two of the forces ; then
find the resultant of this and the third force ; and so on.
64. There is one case in which we are unable to find a
single resultant for two parallel forces, namely, when the
forces are equal and unlike. The process of Art. 61 will
not apply to this case ; for the lines of action of the forces
X and Y are then parallel, so that the points C and D do
not exist.
Two such forces are usually called a couple. The theory
of couples includes some important propositions ; it will be
sufficient for our purpose to demonstrate one of these :
some preliminary definitions and remarks will be required.
65. A couple consists of two parallel forces which are
equal and uiitike.
The arm of a couple is the perpendicular distance be
tween the lines of action of its forces.
The moment of a couple is the product of one of the
equal forces into the arm ; that is, the number whiqh ex
32
36 RESULTANT OF
presses the force must be multiplied by the number which
expresses the arm to produce the moment.
66. The student will readily conceive that the tendency
of a couple which acts on a free rigid body is to make the
body turn round ; and it is shewn in works on the higher
parts of mechanics that such is the case : the rotation begins
to take place round a straight line which always passes
through a certain point in the body called its centre of
gravity, but is not necessarily perpendicular to the piano
of the couple.
67. Two couples in the same plane may differ as to the
direction in which they tend to turn the body round on
which they act.
Take for example the couple denoted in the figure of
the next Article by the two equal forces P, and that de
noted by the two equal forces Q. Suppose a board ABCD
capable of turning in its own plane round a pivot fixed
at any point within ABCD. The couple Q, Q tends to
turn the board in the same way as the hands of a watch
revolve, and the couple P, P tends to turn the board in
the opposite way.
Couples which tend to turn a body round in the same
way may be called like, and couples which tend to turn a
body round in opposite ways may be called unlike.
68. Two unlike couples in the same plane will balance
each other if their moments are equal.
Let each force of one couple be P, and each force of
the other couple Q. Let ABCD be the parallelogram
formed by the lines of action of the forces.
Draw AM perpendicular to CD, and AN perpendicular
to CB.
By hypothesis the moments of the couples are equal,
that is,
TWO PARALLEL FORCES.
Q
37
therefore
Hence, by
P and Q at A
Similarly,
direction CA,
tude ; so that
Hence the
Since two
plane balance
equal moment
A 7?
, by Euclid vi. 4.
the parallelogram of forces, the resultant of
acts in the direction AC.
the resultant of P and Q at C acts in the
and is equal to the former resultant in magni
the two resultants balance each other.
two couples balance each other.
unlike couples of equal moment in the same
each other, it follows that two like couples of
in tlie same plane produce equal effects.
69. The demonstration of the preceding Article as
sumes that the forces of one couple are not parallel to
those of the other. When the four forces are all parallel,
the theorem may be shewn to be true by the aid of Arts.
60 and 61. Or we may proceed thus. Suppose the couples
unlike, and all the forces parallel; take a couple in the
38 TWO PARALLEL FORCES.
same plane, and of equal moment, with its forces not parallel
to the four forces. Then by Art. 68 this new couple
balances one of the two couples and is equivalent to the
other. Therefore the two couples balance each other.
70. Hence we see that two like couples of equal mo
ments in the same plane are equivalent to a single like
couple in that plane of double moment. And any number
of like couples in the same plane are equivalent to a like
couple in that plane with a moment equal to the sum of
the moments of these couples.
Hence finally, if any number of couples act in one plane,
some of one kind and some of the other, they are equi
valent to a single couple in that plane, having a moment
equal to the difference of the sums of the moments of the
two kinds, and being of the same kind as the couples which
have the greater sum of moments.
71. A single force and a couple in one plane are
equivalent to the same single force acting in a direction
parallel to its original direction.
Let P denote the single force, Q each force of the
couple.
If the directions of all the forces are parallel, P com
bined with the like force of the couple will produce a re
sultant P+Q also parallel to the former force. Then
P + Q combined with the other force of the couple will
produce a resultant P parallel and like to the original P.
If the directions of all the forces are not parallel, let A
denote the point at which the line of action of P crosses
that of one force Q of the couple. Form the resultant of
P and this Q, and denote the resultant by R Let B de
note the point at which the line of action of jR crosses that
of the other force Q of the couple ; and suppose R to act
at this point. Eesolve E into its components P and Q.
The two forces Q balance each other, leaving the force P
acting at B parallel to its original direction.
EXAMPLES. IV. 39
EXAMPLES. IV.
1. ABCD is a square. A force of 3 Ibs. acts from A
to B, a force of 4 Ibs. from B to C, a force of 6 Ibs. from Z)
to C, and a force of 5 Ibs. from A to D : find the magnitude
and the direction of the resultant force.
2. Two men carry a weight of 152 Ibs. between them
on a pole, resting on one shoulder of each ; the weight is
three times as far from one man as from the other : find
how much weight each supports, the weight of the pole
being disregarded.
3. A man supports two weights slung on the ends of a
stick 40 inches long placed across his shoulder : if one
weight be twothirds of the other, find the point of support,
the weight of the stick being disregarded.
4. ABC is a triangle, and any point within it ; like
parallel forces P and Q act at A and B such that P is to
Q as the area of BOG is to the area of AOC: shew that the
resultant acts at the point where CO produced meets AB.
5. If the point be outside the triangle and the forces
P and Q in the same proportion as in Example 4, shew that
the result is still true, provided P and Q be like or unlike
according as the intersection of CO and AB is on A B or
on AB produced.
6. ABC is a triangle, and any point within it ; like
parallel forces act at A, B, and C which are proportional
to the areas HOC, CO A, and AOB respectively : shew that
the resultant acts at 0.
7. If the point be outside the triangle, and the
forces in the same proportion as in Example 6, but a cer
tain one of the three be unlike the other two, shew that
the resultant acts at 0.
8. P and Q are like parallel forces ; an unlike parallel
force P + Q acts in the same plane at perpendicular dis
tances a and 6 respectively from the two former, and
between them : determine the moment of the couple which
results from the three forces.
9. Like parallel forces, each equal to P, act at three
of the corners of a square perpendicular to the square ; at
the other corner such a force acts that the whole system is
a couple : determine the moment of the couple.
40 MOMENTS.
V. Moments.
72. The product of a force into the perpendicular
drawn on its line of action from any point is called the
moment of the force with respect to that point.
Thus suppose AB the line of action of a force, any
point, and OM the perpendicular from
on AB. Then if P denote the force A M R
the moment with respect to is P x OM.
Hence the moment of a force never
vanishes except when the point about
which the moment is taken is on the
line of action of the force.
73. Suppose that OM were a rod which could turn
round in the plane of the paper ; if a force were to act at
M in the direction AB it would tend to make the rod turn
round in the same direction as the hands of a watch
revolve; if the force were to act in the direction BA it
would tend to make the rod turn round in the opposite
direction. It is found very convenient to distinguish be
tween these two kinds of moment ; and thus moments of
one kind are called positive, and moments of the other
kind negative. It is indifferent in any investigation which
kind of moment we consider positive, and which negative ;
but when we have made a choice we must keep to it
during that investigation.
74. Since the area of any triangle is equal to half the
product of the base into the altitude, the moment of a
force may be geometrically represented by twice the area
of a triangle having for its base the straight line which
represents the force, and for its vertex the point about
which the moments are taken.
We shall now give a very important proposition re
specting moments.
75. The algebraical sum of the moments of two forces
round any point in the plane containing the two forces is
equal to the moment of their resultant.
Let AB, AC represent two forces; complete the paral
lelogram ABCD and draw the diagonal AD, which will
represent the resultant force. Let be the point round
which the moments are to be taken; join OA, OB, OC, OD.
MOMENTS.
41
I. Let fall without the angle BAG and that which
is vertically opposite to it.
The area of the
triangle AOC is equal
to half the product of
the base AC into the
perpendicular on it
from ; and this per
pendicular is equal
to the sum of the
perpendiculars from
A and from on ED.
Now ED is equal to AC] so that the product of AC into
the perpendicular on it from is equal to the product of
ED into the sum of the perpendiculars on it from A and
from 0.
Thus triangle AOC= triangle A BD + triangle OBD
= triangle A OD  triangle A OB;
therefore triangle A OC+ triangle AOB= triangle AOD.
Hence moment of AC+ moment of AB= moment of AD.
II. Let fall within the angle BAG or that which is
vertically opposite to it.
The area of the triangle AOC is equal to half the
product of the base .4 (7 into the perpendicular on it from ;
and this perpendicular is
equal to the difference of
the perpendiculars from A
and from on ED. Now
ED is equal to AC; so
that the product of AC
into the perpendicular on
it from is equal to the C
product of ED into the difference of the perpendiculars on
it from A and from 0.
Thus triangle AOC= triangle ABD  triangle OBD
= triangle AOB + triangle AOD ;
therefore triangle AOC triangle AOB= triangle AOD.
Hence moment of AC moment of A B= moment of AD.
The moments of AC and AB round are now of oppo
site kinds: thus the moment of the resultant is always
42 MOMENTS.
equal to the algebraical sum of the moments of the com
ponents.
76. The preceding Article applies to the case in which
the lines of action of the forces meet: we have still to shew
that the proposition holds for parallel forces.
Let P and Q be like parallel forces, acting at A and B ;
let R be their resultant, C the
point where the direction of
the resultant cuts AB. Take
any point 0, not between the
lines of action of the forces,
and draw a perpendicular
from on the lines of action
of P, Q, and It, meeting them
at a, b, and c respectively.
Now ^ = ~ , by Art. 60, = , by Euclid vi. 2 ;
(# 0.4 CCi
therefore Pxca=Qxcb.
And PxOa + QxOb=P (Occa) + Q (Oc + cb)
Thus the required result is obtained in this case.
Similarly, it may be shewn that when falls between
the lines of action of P and Q the moment of R is equal
to the arithmetical difference of the moments of P and Q.
Thus for two like parallel forces the moment of the
resultant is always equal to the algebraical sum of the
moments of the components.
In a similar manner the proposition may be established
for the case of unlike parallel forces.
77. The algebraical sum of the moments of the two
forces which form a couple is constant round any point in
the plane of the couple.
If the point is between the lines of action of the forces
the moments of the two forces are of the same kind, and
their sum is equal to the product of one of the forces into
the perpendicular distance between the lines of action.
If the point is not between the lines of action of the
forces the moments of the two forces are of opposite kinds,
MOMENTS. 43
and their arithmetical difference is equal to the product of
one of the forces into the perpendicular distance between
the lines of action.
Thus in every case the algebraical sum of the moments
of the forces of a couple round a point in the plane of the
couple is equal to the product of one force into the perpen
dicular distance between the lines of action ; that is, to the
moment of the couple. See Art. 65.
78. When two forces act in one plane one of three
cases must hold. Either the forces balance each other so
that their resultant is zero, or they have a single resultant,
or they form a couple. The algebraical sum of the mo
ments of the forces about a point in the plane always
vanishes in the first case, vanishes in the second case only
when the point is on the line of action of the resultant, and
oever vanishes in the third case.
79. The result of Art. 75 can be extended to the case
of any number of forces acting in one plane at a point.
Take two of the forces ; the algebraical sum of their mo
ments round any point is equal to the moment of their
resultant. Replace the two forces by their resultant ; then
apply Art. 75 again to this resultant and the third force.
And so on.
80. Similarly, by the aid of Art. 63 we may extend
Art. 76 to the case of any number of parallel forces acting in
one plane.
81. If any number of forces in one plane act on a parti
cle either they are in equilibrium or they have a single re
sultant. The algebraical sum of the moments round a point
in the plane always vanishes in the former case, and in the
latter case vanishes only when the point is on the line of
action of the resultant.
82. Hence we may use the following instead of Art. 57 :
Forces acting in one plane on a particle will be in
equilibrium if the algebraical sum of the moments vanishes
round any two points in the plane which are not situated
on a straight line passing through the particle.
Conversely, if the forces are in equilibrium the alge
braical sum of the moments will vanish round any point in
the plane.
44 EXAMPLES. V.
EXAMPLES. V.
1. P and Q are fixed points on the circumference of
a circle ; QA and QB are any two chords at right angles to
each other, on opposite sides of QP : if Q A and QB denote
forces, shew that the difference of their moments with
respect to P is constant.
2. If two or more forces act in one plane on a particle,
shew that the algebraical sum of their moments round a
point in the plane remains unchanged when the point
moves parallel to a certain straight line.
3. If the algebraical sum of the moments of forces
acting in one plane on a particle has the same value for
two points in the plane, then the straight line which joins
these two points is parallel to 4he resultant force.
4. ABC is a triangle ; D, E, F are the middle points
of the sides opposite to A, B, C respectively : shew by
taking moments round A, B, and C, that forces denoted by
AD, BE, and CF are in equilibrium.
5. Forces are denoted by the perpendiculars drawn
from the angular points of a triangle on the opposite sides :
shew by taking moments round the angular points that the
forces will not be in equilibrium unless the triangle is equi
lateral.
6. Forces act at the angular points of a triangle along
the perpendiculars drawn from the angular points on the
respectively opposite sides, each force being proportional
to the side to which it is perpendicular: shew by taking
moments round the angular points that the forces are in
equilibrium.
7. ABC is a triangle, any point within it; AO,
BO, and CO produced cut the respectively opposite sides
at //, 7, and K : shew that forces denoted by AH, BI, and
CK, cannot be in equilibrium unless ZT, /, and K are the
middle points of the respective sides.
 8. ABC is a triangle, any point within it ; straight
lines are drawn through parallel to the sides and termi
nated by the sides: shew that forces denoted by these
straight lines cannot be in equilibrium unless each straight
line is bisected at 0.
FORCES IN ONE PLANE. 45
VI. Forces in one Plane.
83. In the present chapter we shall investigate the
conditions of equilibrium of a system of forces acting in
one plane on a rigid body.
84. A system of forces acting in one plane on a rigid
body, if not in equilibrium, must be equivalent to a single
resultant or to a couple.
For take any two of the forces of the system, and de
termine their resultant ; then combine this resultant with
another force of the system ; and so on. By proceeding in
this way we must obtain finally a single resultant or a
couple ; unless one force of the system is equal and oppo
site to the resultant of all the others, so that the whole
system is in equilibrium.
85. If a system of forces acting in one plane on a
rigid body is equivalent to a single resultant, the moment
of the resultant round any point in the plane is equal to
the algebraical sum of the moments of the forces.
This proposition is established by repeated applications
of Arts. 75 and 76. Take any two of the forces ; then the
moment of their resultant round any point in the plane is
equal to the algebraical sum of the moments of the two
forces. Combine the resultant of these two forces with
another of the forces ; then the moment of then* resultant
is equal to the algebraical sum of the moments of the com
ponents, that is, to the algebraical sum of the moments of
the three forces of the system. And so on.
Hence the algebraical sum of the moments of the forces
round any point in the plane will not vanish unless the
point about which the moments are taken is on the line of
action of the resultant.
86. If a system of forces acting in one plane on a rigid
body is in equilibrium, the algebraical sum of the moments
round any point in the plane vanishes.
For when the system of forces is in equilibrium, one of
the forces is equal and opposite to the resultant of all the
46 FORCES IN ONE PLANE
others. Hence the moment of the one force round any
point in the plane is equal and contrary to the moment of
the resultant of all the other forces ; see Art. 78. Thus,
by Art. 85, the moment of the one force is equal and con
trary to the algebraical sum of the moments of all the
other forces. Therefore the algebraical sum of the mo
ments of all the forces vanishes.
87. If a system of forces acting in one plane on a rigid
body is equivalent to a couple, the algebraical sum of the
moments of the forces round any point in the plane u
equal to some constant which is not zero.
For proceeding as in Art. 85 we have the algebraical
sum of the moments of the forces equal to the moment of
the couple ; and the moment of the couple is constant for
all points of the plane, namely, equal to the product of one
force of the couple into the perpendicular distance between
the two forces.
Conversely, if tlie algebraical sum of the moments of the
forces round three points in the plane not in the same straight
line is constant, and not zero, the system of forces is equivalent
to a couple.
For if the system is not equivalent to a couple it must
be in equilibrium or reduce to a single force ; by Art. 84.
If it were in equilibrium the sum of the moments would
be zero ; by Art. 86. If it reduced to a single force the
sum of the moments could not have a constant value except
for points in a straight line parallel to the direction of the
single force ; by Art. 85.
88. A system of forces acting in one plane on a rigid
body will be in equilibrium if the algebraical sum of the
moments of tJie forces vanishes round three points in the plane
which are not in a straight line.
For if the system of forces be not in equilibrium, it is
equivalent to a single resultant or to a couple.
In the present case the system of forces cannot be equi
valent to a couple; for then the algebraical sum of the
moments would not vanish for any point in the plane.
Nor can the system of forces be equivalent to a single
resultant; for then the algebraical sum of the moments
FORCES IN ONE PLANE. 47
would vanish only for points on the line of action of the
resultant.
Hence the system of forces must be in equilibrium.
89. The preceding Article gives the conditions of
equilibrium of a system of forces acting in one plane on a
rigid body ; if these conditions are satisfied the body is in
equilibrium, or in other words these conditions are suffi
cient for equilibrium. And these conditions are necessary
for equilibrium; because we have shewn in Art. 86 that
they must hold if there be equilibrium.
We shall give a proposition in the next Article which
will enable us to put the conditions of equilibrium in
another form.
90. If a system of forces acting in one plane on a
rigid body is in equilibrium the algebraical sum of the
forces resolved parallel to any fixed straight line vanishes.
For when the system of forces is in equilibrium, one of
the forces is equal and opposite to the resultant of all the
others. Hence the resolved part of the one force parallel
to any fixed straight line is equal and opposite to the alge
braical sum of the resolved parts parallel to the fixed
straight line of all the other forces ; see Arts. 44 and 52.
Therefore the algebraical sum of the forces resolved pa
rallel to any fixed straight line vanishes.
91. A system of forces acting in one plane on a rigid
body will be in equilibrium if the algebraical sum of tJie
moments round two points in the plane vanishes, and the
algebraical sum of the forces resolved parallel to the straight
line which joins these points also vanishes.
For if the system of forces be not in equilibrium it is
equivalent to a single resultant or to a couple.
In the present case the system of forces cannot be
equivalent to a couple ; for then the algebraical sum of the
moments would hot vanish for any point in the plane.
Nor can the system of forces be equivalent to a single
resultant; for then the algebraical sum of the moments
would vanish only for points on the line of action of the
48 FORCES IN ONE PLANE.
resultant, so that the resultant would act along the straight
line joining the two points : but the algebraical sum of the
forces resolved parallel to this straight line vanishes by
supposition ; therefore there cannot be a resultant acting
along this straight line : see Arts. 44 and 52.
Hence the system of forces must be in equilibrium.
92. In the proposition of the preceding Article we
speak of forces resolved parallel to the straight line which
joins two points ; this means of course that every force is
resolved into two components, one parallel to this straight
line, and the other parallel to another fixed straight line :
the second straight line is not necessarily at right angles
to the first, although we may for convenience generally
take it so.
We shew in the preceding Article that the conditions
of equilibrium there stated are sufficient; and we see by
Arts. 86 and 90 that they are necessary.
There is still another form in which the conditions of
equilibrium of a system of forces acting in one plane on a
rigid body may be put : and this we shall now give.
93. A system of forces acting in one plane on a rigid
body will be in equilibrium if the algebraical sum of the
forces resolved parallel to two fixed straight lines in the
plane vanishes, and the algebraical sum of the moments
round any point in the plane also vanishes.
For if the system of forces be not in equilibrium it is
equivalent to a single resultant or to a couple.
In the present case the system of forces cannot be
equivalent to a couple ; for then the algebraical sum of the
moments would not vanish round any point in the plane.
Nor can the system of forces be equivalent to a single
resultant; for, by supposition, the algebraical sum of the
forces resolved parallel to two fixed straight lines in the
plane vanishes, and therefore the resolved part of the
single resultant parallel to these two straight lines would
vanish : so that the resultant would vanish. See Arts. 44
and 52.
Hence the system of forces must be in equilibrium.
FORCES IN ONE PLANE. 49
94. In the proposition of the preceding Article the
two fixed straight hues are not necessarily at right angles,
although it may generally be convenient to take them so.
"We shew in the preceding Article that the conditions
of equilibrium there stated are sufficient; and we see by
Arts. 86 and 90 that they are necessary.
We have thus presented in three forms the conditions
of equilibrium of a system of forces acting in one plane on
a rigid body; see Arts. 88, 91, and 93: the last form is
generally the most convenient to use.
95. The preceding Articles relate to any system of
forces acting in one plane on a rigid body ; the particular
case in which the forces are parallel deserves separate
notice ; for some of the general propositions may then be
simplified. We have the following results :
A system of parallel forces acting in one plane on a
rigid body, if not in equilibrium, must be equivalent to a
single resultant parallel to the forces or to a couple. See
Arts. 63 and 84.
A system of parallel forces acting in one plane on a
rigid body will be in equilibrium if the algebraical sum
of the moments of the forces vanishes round two points in
the plane, which are not in a straight line parallel to the
direction of the forces. See Art. 88.
A system of parallel forces acting in one plane on a
rigid body will be in equilibrium if the algebraical sum
of the forces vanisJies, and also the algebraical sum of
the moments of the forces round any point in the plane.
See Art. 91 or 93.
96. Many of the results of the present Chapter depend
on the theorem of Art. 84, and although the simple reason
ing of that Article is quite satisfactory, it may be desirable
to give another investigation. Accordingly we shall now
demonstrate a new theorem which includes that of Art. 84.
97. A system of forces acting in one plane on a rigid
body can in general be reduced to a couple and a single
force acting at an arbitrary point in the plane.
T. II K. 4
50 FORCES IN ONE PLANE.
Let P acting at A be one of the forces of the system.
At any arbitrary point in the plane apply two forces
each equal and parallel to P, in opposite directions. This
will not alter the action of the system of forces. Thus, in
stead of P at A we have P at and a couple formed by
P at A and P at 0.
Let Q acting at B be another of the forces of the
system. Then, as before, we may replace Q at B by Q at
and a couple formed by Q at ^ and Q at 0.
Proceeding in this way we can replace the given system
of forces by the following :
(1) A system of forces at which are respectively
parallel and equal to the original forces ; this system may
be combined into a single force at 0.
(2) A set of couples in the plane which may be com
bined into a single couple by Art. 70. As the moment
round of the force at P is equal to the moment of the
couple consisting of P at A and P at 0, it follows that the
moment of the single couple thus obtained is equal to the
sum of the moments of the forces.
Thus in general a system of forces acting in one plane
on a rigid body may be reduced to a couple and a single
force ; the latter acting through any arbitrary point in the
plane, and being equal in magnitude and parallel in direc
tion to what would be the resultant of all the forces if they
acted at one point parallel respectively to their original
directions.

FORCES IN ONE PLANE. 51
It may happen that the couple vanishes and then the
system can be reduced to a single force ; or the single force
may vanish and then the system reduces to a couple ; or
both the single force and the couple may vanish and then
the system is in equilibrium.
If neither the couple nor the single force vanishes they
be reduced to a single force by Art. 71.
98. We will now give some examples which illustrate
the subject of the present Chapter.
Jl) If a system of forces is represented in magnitude
position by the sides of a plane polygon taken in order
the system is equivalent to a couple.
The sum of the moments of the forces is constant round
any point in the plane. For if the point be taken within
the polygon the moments are all of the same kind, and
their sum is represented by twice the area of the polygon.
And if the point be taken without the polygon the mo
ments are not all of the same kind, but their algebraical
sum is constant, being still represented by twice the area
of the polygon.
Since the sum of the moments is equal to a constant
which is not zero, the system of forces is equivalent to a
couple ; see Art. 87.
(2) ABC is a triangle; a force P acts from A to B,
a force Q from B to C, and a force R from A to C : re
quired the relation between the forces in order that they
may reduce to a single resultant passing through the centre
of the circle inscribed in the triangle.
The algebraical sum of the
moments round the centre
of the inscribed circle must
vanish ; see Art. 85.
Let r denote the radius
of the inscribed circle, then
Pr + Qr = Rr, therefore
R=P + Q. This is the neces A~
sary and sufficient condition.
42
52 FORCES IN ONE PLANE.
(3) Forces P, Q, R, S act in order round the sides of a
parallelogram : required the direction of the resultant.
D R *? o F
The resultant of P and R will be a force equal to
PR acting parallel to AB, through the point E on CB
produced through B, such that .
BE_R
CE~P'
The resultant of Q and S will be a force equal to Q  S
acting parallel to BC, through the point F on DC produced
through (7, such that
CF_S
DF Q'
Draw EO parallel to AB, and FG parallel to CB.
Then the resultant of all the four forces passes through G.
Produce EO to // so that GH may be to GF as PR
is to Q  S. Then the straight line joining G to the middle
point of FH is the direction of the resultant.
We have supposed P greater than R and Q greater
than S; it is easy to make the necessary modifications for
any other case.
Or we may proceed thus :
Let AB=h, AD=k; from
any point draw OM pa
rallel to CB; let
AM=x, OM=y.
We can now express
the moments of the forces
round 0.
EXAMPLES. VI. 53
The moment of P=P x 0J/sin OMB=Py sin A,
the moment of Q= Q x MB sinA = Q(h x) sin A,
the moment of R=E x (.2(7 ay>in A =R (Jc y) sin A,
the moment of S=Sx A Jfsin .4 =# sin A.
Now if be a point on the line of action of the result
ant the algebraical sum of the moments of the forces round
vanishes, that is,
This gives the relation which must hold between AM and
3W, when is a point on the line of action of the resultant.
For example, suppose to be on the straight line
AD; then # =0 : thus we get
y ~ RP
which determines the point where the direction of the re
sultant cuts AD. Similarly we may determine the point
where this direction cuts AB.
EXAMPLES. VI.
1. A BCD is a square. A force of 3 Ibs. acts from A
to B, a force of 4 Ibs. from B to C, and a force of 5 Ibs.
from C to D : find the single force which will preserve
equilibrium.
2. A man carries a bundle at the end of a stick over
his shoulder: as the portion of the stick between his
shoulder and his hand is shortened, shew that the pressure
on his shoulder is increased. Does this change alter his
pressure on the ground?
3. If forces in one plane reduce to a couple, shew that
if they were made to act on a particle, retaining their mu
tual inclinations, they would keep the particle at rest.
54 EXAMPLES. VI.
4. ABC is a triangle ; H, /, and K are points in the
sides BC, CA, and AB respectively, such that
BH_ C^_AK
HC~ IA~ KB :
shew by taking moments round A, B, C that forces denoted
by AH, SI, and CK are equivalent to a couple, except
when H, I, and K are the middle points of the sides, and
then the forces are in equilibrium.
5. A and B are fixed points; at any point C, in the
arc of a circle described on AB as a chord, two forces act,
namely, P along CA and Q along CB : shew that their re
sultant passes through a fixed point on the other arc which
makes up the complete circle.
6. ABCD is a quadrilateral inscribed in a circle: if
forces P, Q, E act in directions AB, AD, CA so that
P : Q : R as CD : BC : BD, shew that they are in
equilibrium.
7. Two forces are denoted by MA and MB, and two
others by NC and ND : shew that the four forces cannot
be in equilibrium unless MN bisects both AB and CD.
8. Find a point within a quadrilateral such that if
forces be represented by straight lines drawn from it to
the angular points of the quadrilateral the forces will be
in equilibrium.
9. Forces proportional to 1, *J3, and 2 act at a point
and are in equilibrium: find the angles between their
lines of action.
10. If two equal forces P and P acting at an angle of
60 have the same resultant as two equal forces Q and Q
acting at right angles, shew that P is to Q as v/2 is to J3.
11. C and B are fixed points; CA and CB represent
two forces : if A move along any straight line shew that
the extremity of the straight line which represents the re
sultant moves along a parallel straight line.
12. Forces denoted by the sides of a polygon, except
one side, act in order : shew that they are equivalent to a
single resultant which is parallel to the omitted side.
CONSTRAINED BODY. 55
VII. Constrained Body.
99. A body is said to be constrained when the manner
in which it can move is restricted. A simple example is
that in which a body can only turn round a fixed axis,
that is, can receive no other motion. In such cases forces
may act on the body besides the restraints which restrict
the motion, and we may require to know the conditions
which must hold among these forces in order to ensure the
equilibrium of the body.
100. When a body can only turn round a fixed axis
and is acted on by a system of forces in a plane perpen
dicular to the axis, such that the algebraical sum of the
moments of the forces round the point where the axis meets
the plane vanishes, the body will be in equilibrium.
If the system of forces be not in equilibrium it is
equivalent to a single resultant or a couple.
In the present case the system of forces cannot be
equivalent to a couple; for then the algebraical sum of
the moments would not vanish for any point in the plane.
Suppose that the system of forces is equivalent to a
single resultant. Since the algebraical sum of the mo
ments of the forces vanishes round the point where the
axis meets the plane, the line of action of the resultant
must pass through the point. Therefore the resultant
has no tendency to turn the body round the axis ; and the
body is therefore in equilibrium.
101. The investigation of the preceding Article shews
that the condition there stated is sufficient for equilibrium.
The condition is also necessary for equilibrium ; for if the
condition does not hold, the system of forces is equivalent
either to a couple or to a single resultant which does not
pass through the axis, and in either case the body would
be set in motion round the axis.
102. The most simple case of the preceding two
Articles is that of the lever. A lever is a rigid body
which is moveable in one plane about a point which is
called the fulcrum, and is acted on by forces which tend
56 CONSTRAINED BODY.
to turn it round the fulcrum. In order that the lever
may be in equilibrium the moments of the two forces round
the fulcrum must be equal and contrary, by Art. 101.
Hence the condition of equilibrium stated in Art. 100 is
often called the Principle of the Lever.
103. A body which is not constrained is called a free
body. From considering the equilibrium of a constrained
body we may render our conception of the equilibrium of
a free body more distinct. Any condition which is neces
sary for the equilibrium of a constrained body will also be
necessary for the equilibrium of a free body; although a
condition which may be sufficient in the former case will
not generally be sufficient in the latter case.
For example, in Art. 86 a certain principle is established
with respect to the equilibrium of a free rigid body, and
the investigation of Art. 100 shews us the interpretation
of the principle. Suppose a body in equilibrium under
the action of a system of forces in one plane. Imagine
two points in the body, which lie in a straight line perpen
dicular to the plane, to become fixed. This cannot disturb
the equilibrium, for we do not communicate any motion
to the body by fixing two points in it; we merely restrict
to some extent its possible motion. The body has still
the power of turning round the straight line which joins
the fixed points; and, by Art. 101, the body will not be in
equilibrium unless the algebraical sum of the moments of
the forces round the point where the straight line cuts the
plane vanishes.
104. Suppose a body can only turn round a fixed axis,
and that it is acted on by forces which are not all in one
plane perpendicular to the axis; a strict demonstration
of the condition of equilibrium is rather beyond our
present range, but by assuming some principles which
are nearly selfevident we shall be able to give a sufficient
investigation.
First suppose the forces to consist of various systems
in planes which are all perpendicular to the axis. It may
be assumed as nearly selfevident that the tendency of the
systems to set the body in motion will not be altered if
all the other planes are made to coincide with one of
CONSTRAINED BODY. 57
them ; and then the forces reduce to a system in one plane
perpendicular to the axis, and Arts. 100 and 101 apply.
Next suppose the forces to be any whatever. Kesolve
each force into two components at right angles to each
other ; one component being parallel to the fixed axis. It
may be assumed as nearly selfevident that the compo
nents parallel to the axis have no tendency to set the
body in motion round the axis; and they may accordingly
be left out of consideration.
The other components form various systems of forces
in planes which are perpendicular to the axis; and, as in
the first case, they may be supposed all to act in one
plane, and Arts. 100 and 101 apply.
105. Suppose a body capable of moving only in such
a manner that all points of the body describe parallel
straight lines. For example, two fixed rigid parallel
straight rods may pass through the body, and so the body
be only capable of sliding along the rods. Suppose also
that a system of forces acts on the body. Resolve each
force into two components at right angles to each other,
one component being parallel to the fixed rods. Then the
necessary and sufficient condition of equilibrium is that the
sum of the components parallel to the fixed rods, that is to
the direction of possible motion, should vanish.
If there were only one rigid straight rod passing through
it the body could both slide and turn round ; in such a case,
besides the condition just obtained, that of Art. 104 must
also hold for equilibrium.
We see from these cases the interpretation of the con
dition in Art. 90 relative to the equilibrium of a free rigid
body.
106. When three forces maintain a body in equi
librium their lines of action must lie in the same plane.
Suppose a body in equilibrium under the action of three
forces. Imagine two points in the body, one on the line
of action of one force, and the other on the line of action of
another force, to become fixed, the points being so taken
that the straight line which joins them is not parallel
58 CONSTRAINED BODY.
to the line of action of the third force. This cannot
disturb the equilibrium. The body has still the power of
turning round the straight line which joins the fixed
points, as an axis, and it will not be in equilibrium unless
the line of action of the third force pass through the axis.
Thus any straight line which meets the lines of action
of two of the forces, and is not parallel to the line of
action of the third force, must meet that line of action;
and therefore all the three lines of action must lie in one
plane.
By combining this result with those in Arts. 41 and 62
we have a complete account of the conditions of equilibrium
of a rigid body when acted on by three forces.
107. In the present work on Mechanics we have begun
with the Parallelogram of Forces and have deduced the
Principle of the Lever; this is the course which is now
generally adopted. Formerly it was usual to begin with the
Principle of the Lever and to deduce the Parallelogram
of Forces. We will briefly indicate the principal steps of
the process.
Various axioms are laid down: for example the fol
lowing : Equal forces acting at right angles at the extremi
ties of equal arms of a lever, exert equal efforts to turn the
lever round.
From these axioms certain propositions are deduced;
for example the following : A horizontal rod or cylinder
of uniform density will produce the same effect by its
weight as if it were collected at its middle point.
In this way the Principle of the Lever is established,
first for a straight lever, and then for a lever of any form.
We will give one proposition as an example of this method
in the next Article, and in the following Article we will
shew how the Parallelogram of Forces is deduced.
108. Two forces acting at right angles on a straight
lever on opposite sides of the fulcrum will balance each
other if they are inversely proportional to their distances
from the fulcrum and tend to turn the lever round in
contrary directions.
CONSTRAINED BODY. 59
Let the forces
P and Q which act
at M and N at
right angles to a
straight lever on
opposite sides of P
the fulcrum C be
such that
P^_CN
Q~CM'
and let them be like forces, so that they tend to turn the
lever in contrary directions : they will balance each other.
If P=Qj the proposition is true by the axiom stated
in Art. 107.
If P be not equal to Q, suppose P the greater.
On NM take ND=MC '; then NC=MD. Make
MA=MD, and NB=ND.
Let the forces P and Q be measured by the weights
which they would support ; and let AB be a uniform rod
of weight equal to P + Q.
Now CA
therefore AB is bisected at C.
AD 2MD 2NC P
And
therefore unrBBf+V
But the weight of AB is P + Q; and therefore the
weight of the portion AD is P ; and therefore the weight
of the portion DB is Q.
Since C is the middle point of AB the rod AB will
balance about C; and by Art. 107 if the part AD be
attached at its middle point M to the lever MN, and the
part BD at its middle point JV, the effect will be the same
as before. Therefore in this case also the weights balance ;
that is P at M and Q at N balance.
60
CONSTRAINED BODY.
109. To deduce the Parallelogram of Forces from the
Principle of the Lever.
Let a force P act
along Op and a force
Q along Oqi let Or
be the direction of
their resultant.
From any point C
in the direction of the
resultant draw CA
parallel to Oq and
CB parallel to Op ;
also CM perpendicular
to Op and CN perpen
dicular to Oq.
If a force equal to the resultant of P and Q act along
rO } it will with the forces P and Q keep a particle at in
equilibrium. Suppose these forces applied by means of
rods, connected at ; these rods will then be in equili
brium. Imagine the point on the rod Or to become
fixed ; this will not disturb the equilibrium. The system
can still turn round C, and it will do so unless the moments
round C are equal and contrary. Thus if there is equi
librium we must have, by the Principle of the Lever,
Therefore
P CN
jgp by Euclid vi. 4, = ^.
Thus the forces P and $ are proportional to the sides
A and OB of the parallelogram OA CB, and the diagonal
00 represents the direction of iheir resultant.
This demonstrates the Parallelogram df Forces so far
as relates to the direction of .the respHfiiit ; then as in
Art. 49 we can demonstrate it also for the magnitude of
the resultant.
EXAMPLES. VII. Cl
EXAMPLES. VII.
1. A BCD is a square ; a force of 3 Ibs. acts from A to
?, a force of 4 Ibs. from B to C, and a force of 5 Ibs. from
C to D : if the centre of the square be fixed, find the
force which acting along AD will keep the square in equi
librium.
2. The length of a horizontal lever is 12 feet, and
the balancing weights at its ends are 3 Ibs. and 6 Ibs.
respectively : if each weight be moved 2 feet from the end
of the lever, find how far the fulcrum must be moved for
equilibrium.
3. If the forces at the ends of the arms of a horizontal
lever be 8 Ibs. and 7 Ibs., and the arms 8 inches and 9
inches respectively, find at what point a force of 1 Ib. must
be applied at right angles to the lever to keep it at rest.
4. The arms of a lever are inclined to each other :
shew that the lever will be in equilibrium with equal
weights suspended from its extremities, if the point mid
way between the extremities be vertically above or verti
cally below the fulcrum.
5. A weight P suspended from one end of a lever
without weight is balanced by a weight of 1 Ib. at the other
end of the lever ; when the fulcrum is removed through
half the length of the lever it requires 10 Ibs. to balance P :
determine the weight of P.
G. A rod capable of turning round one end, which is
fixed, is kept at rest by two forces acting at right angles
to the rod ; the greater force is 6 Ibs. and the distance
between the points of application of the forces is half the
distance of the greater force from the fixed end : find the
smaller force. Shew that if any force be added to the
smaller force, a force half as large again must be added
to the greater force in order to preserve equilibrium.
. 7. ABC is a triangle without weight, having a right
angle at (7, and CA is to CB as 4 is to 3 ; the triangle is
fixed at C, and two forces P and Q acting at A and B in
directions at right angles to CA and CB keep it at rest:
find the ratio of P to Q.
62 EXAMPLES. VII.
8. In Example 7 if the force P act at A at right
angles to A (7, and the force Q act at B at right angles to
JBA, find the ratio of P to Q in order that the triangle
may be at rest.
9. The lower end B of a rigid rod without weight 10
feet long is hinged to an upright post, and its other end A
is fastened by a string 8 feet long to a point C vertically
above , so that A CB is a right angle. If a weight of one
ton be suspended from A find the tension of the string.
10. ABC is a bent lever without weight of which B is
the fulcrum ; weights P, Q suspended from A, (7 respective
ly are in equilibrium when BG is horizontal ; weights P,
2Q similarly suspended are in equilibrium when AB is hori
zontal : shew that the angle ABC=\Z&.
11. A triangle can turn in its own plane round a point
which coincides with the centre of the inscribed circle ;
forces acting along the sides keep the triangle in equili
brium : shew that one of the forces is equal to the sum of
the other two.
12. A string passes through a small heavy ring, and
the ends of the string are attached to the ends of a lever
without weight : shew that when the system is in equi
librium the ring is vertically under the fulcrum.
13. Forces P, Q, E, S act at the middle points of a
rhombus, outwards, in the plane of the rhombus, at right
angles to the sides : find the condition of equilibrium when
the rhombus can only move in its own plane round the
point of intersection of its diagonals : find also the condi
tions of equilibrium when the rhombus is free.
14. Two forces P and Q acting at a point have a re
sultant R ; any straight line meets the directions of P, Q, E
at A, B, C respectively : shew that
P Q R
OA.BC OB. AC OC.AB'
15. A rod A B without weight 1 4 inches long is suspend
ed by two strings from a peg 6 y ; the string AC is 15 inches
long, and the string BG is 13 inches long ; 130 Ibs. is sus
pended from A, and 52 Ibs. from B : when the whole is in
equilibrium find the tensions of the strings.
CENTRE OF PARALLEL FORCES. 03
VIII. Centre of Parallel Forces.
110. Suppose we have two parallel forces acting re
spectively at two points ; we know that their resultant is
equal to the algebraical sum of the forces, and is parallel
to them, and that it may be supposed to act at a definite
point on the straight line which passes through the two
points. See Arts. 60 and 61. Moreover this definite point
remains the same however the direction of the two forces
be changed, so long as they remain parallel. This point is
called the centre of the two parallel forces. Hence we
adopt the following definition :
The centre of a system of parallel forces is the point
at which the resultant of the system may be supposed to act,
whatever may be the direction of the parallel forces.
111. To find the resultant and tJie centre of any sys
tem of parallel forces.
Let the parallel
forces be P, Q, It, S,
acting at the points
A, B, C, D, respec
tively.
Join AB, and divide it at Z, so that AL may be to
LB as Q is to P ; then the resultant of P at A and Q at
B is P + Q parallel to them, at L.
Join LC, and divide it at J/, so that LM may be to
MC as R is to P + Q, then the resultant of P + Q at L
and R&tCisP + Q + R parallel to them, at M.
Join J/Z), and divide it at JV, so that MN may be to
ND as S is to P + Q + R ; then the resultant of P + Q + R
at M and S at D is P + Q + R + S parallel to them, at N.
Thus we have found the resultant and the centre of four
parallel forces ; and in the same way we may proceed
whatever be the number of the forces.
C4 CENTRE OF PARALLEL FORCES.
112. In the diagram and language of the preceding
Article we have implied that the forces are all like : it is
easy to make the slight modifications which are required
when this is not the case. We may, if we please, form two
groups, each consisting of like parallel forces, and obtain
the resultant and the centre of each group ; and then by
Art. 61 deduce the resultant and the centre of the whole
system of parallel forces.
We shall always obtain finally a single resultant and a
definite centre, except in the case where the algebraical
sum of all the forces is zero ; and then either the forces
are in equilibrium or they form a couple : see Art. 95.
Suppose that a system of parallel forces is formed into
two groups in the manner just indicated ; then if the re
sultant of one group is equal to the resultant of the other,
and the centres of the two groups coincide, the whole
system is in equilibrium. And conversely, if the whole
system is in equilibrium the resultant of one group must be
equal to the resultant of the other, and the centres of the
two groups must coincide.
113. We have thus shewn how to determine geometri
cally the position of the centre of a system of parallel
forces : we shall now shew how we may attain the same
end by the aid of algebraical formula.
114. TJie distances of the points of application of two
parallel forces from a straight line being given, to deter
mine the distance of the centre of the parallel forces from
that straight line; the straight line and the points being
all in one plane.
First let A and B be the points
of application of two like parallel
forces, P and Q ; their resultant is
P+ Q, parallel to them, and it may
be supposed to act at the point (7,
which is such that
P_^CB
Q ~~ CA '
CENTRE OF PARALLEL FORCES.
65
Let AD, BE, CF be perpendiculars from A, B, C on
any straight line which is in a plane containing A and B.
Let AD=p, BE=q, CF=r : then we have to find the
value of r, supposing the values of p and q to be known.
Through C draw aCb parallel to DFE meeting AD
and BE at a and b respectively. Then
CB
CA
~ , by Euclid VI. 4 ;
thus
therefore
therefore
thus
rp
P + Q
Next let A and B be the points of application of two
unlike parallel forces P and Q.
Suppose Q the greater. Then
using the same construction and
notation as before, we have
P == CB = b
Q ~ CA ~ Aa ;
P EbEB r
therefore
thus
It will be observed that the result in the second case
can be deduced from that in the first case by changing P
into P.
T. MB. 5
66
CENTRE OF PARALLEL FORCES.
115. The distances of the points of application of
any number of parallel forces from a straight line being
given, to determine the distance of the centre of the paral
lel forces from that straight line, the straight line and the
points being all in one plane.
Let the parallel forces be P, Q, R, S acting at the
points A, B, C, D respectively. Let p, q, r, s be the dis
tances of A, B, C, D respectively from a straight line Ox
in the same plane as the points.
Join AB and divide it at L, so that AL may be to LB
as Q is to P ; then L is the centre of P at A and Q at B,
and these forces are equivalent to P + Q at L : let I denote
the distance of L from Ox, then, by Art. 114,
l== P + Q '
Join LC and divide it at Jf, so that LM may be to MO
as R is to P + Q ; then M is the centre of P + Q at L and
R at (7, and these forces are equivalent to P + Q + R at
M: let m denote the distance of M from Ox. then, by
Art. 114,
P+Q+R P+Q+R '
Join MD and divide it at N so that MN may be to
ND as S is to P + Q + R, then N is the centre of P + Q + R
at M and S at D, and these forces are equivalent to
P + Q + R + S &i N: let n denote the distance of N from
Ox, then, by Art. 114,
P+Q+R+S
P+Q+R+S
CENTRE OF PARALLEL FORCES. 67
Thus we have determined the distance from Ox of the
centre of four parallel forces ; and in the same manner we
may proceed whatever be the number of the forces.
The symmetrical form of the expression for n should be
noticed. We see that we shall obtain the same result in
whatever order we combine the given forces, as we might
have expected.
116. In the same way if the distances of A, B, 0, and
D from a second straight line, as Oy, in the plane be given,
we can deduce the distance of the centre of the parallel
forces from the same straight line.
And when we know the distances of the centre from two
straight lines in the plane we can determine the position of
the centre ; for the centre will be the point of intersection
of straight lines parallel to Ox and Oy, and at the respec
tive distances from them which have been found.
117. In the figure and language of Art. 115 we have
implied that the forces are all like; it is easy to make the
slight modifications which are required when this is not
the case. We may, if we please, form two groups, each
consisting of like parallel forces, and obtain the centre of
each group; and then by Art. 114 deduce the centre of the
whole system of parallel forces.
The final result will be like that of Art. 115, the sign of
those forces which act in one way being positive, and the
sign of those which act in the other being negative.
118. We supposed in Art. 114 that the straight lines
AD, BE, and CF were all perpendicular to DFE. But
this is not necessary; it is sufficient that these straight
lines should be all parallel. And so also in Art. 115 the
distances denoted by p, q, r, and s need not necessarily be
measured perpendicularly to the straight line Ox\ it is
sufficient that they should all be measured in parallel direc
tions.
119. It is easy to extend our investigation to the case
in which the points of application of the parallel forces are
not all in one plane.
52
CS CENTRE OF PARALLEL FORCES.
In the fundamental investigation of Art. 114 we may
suppose A D, CF, and BE to be the distances of A, B, and
C, not from a given straight line but from a given plane;
either perpendicular distances or distances measured paral
lel to a given straight line. Then, as in Arts. 114 and 115,
if we know the distances of the points of application of the
parallel forces from a given plane, we can obtain the dis
tance of the centre of the parallel forces from that plane.
120. The weight of a body may be considered to be
the aggregate of the weights of the particles which com
pose the body. The weights of these particles form a
system of like parallel forces, and such a system always has
a centre; see Art. 111. The centre of the parallel forces
which consist of the weights of the particles of a body
is called the centre of gravity of the body.
Thus the centre of gravity is a particular case of the
centre of parallel forces ; but it is found convenient to give
especial attention to this particular case, and accordingly
we shall consider it in the next Chapter. It will be ob
served that the theory of the centre of gravity is rather
simpler than the general theory of the centre of parallel
forces, because the weights of the particles of a body are
all like forces, and thus we shall not have to consider the
second case of Art. 114.
121. The following examples contain an interesting
result.
(1) ABC is a triangle ; parallel forces act at B and C,
each proportional to the opposite side of the triangle :
determine the position of the centre of 't/ie parallel forces.
First let the forces be
like. In EC take D so that
BD is to DC as the force at
C is to the force at fi, that
is as AB is to AC; then D
is the centre of the parallel
forces.
Hence, by Euclid, vi. 3, the point D is such that AD
bisects the angle BAG.
EXAMPLES. V1IL 69
Next let the forces be unlike. Suppose AB greater
than AC. Then, proceeding as before, we find that the
centre of the parallel forces is at E on BC produced, such
that AE bisects the angle between AC and BA produced.
See Euclid, vi. A.
(2) Parallel forces act at the angular points of a
triangle, each force being proportional to the opposite side
of the triangle: determine the position of the centre of the
parallel forces.
First let the forces be all like. By the preceding
example D is the centre of the parallel forces at B and C\
hence the centre of all the three parallel forces lies on the
straight line AD which bisects the angle BAG. Similarly
the centre lies on the straight line which bisects the angle
ABC, and on the straight line which bisects the angle BCA.
Therefore the centre of all the parallel forces must coin
cide with the centre of the circle inscribed in the triangle
ABC.
Next let the forces be not all like. Suppose that the
forces at B and C are unlike, and the forces at A and C
like. By example (1) the centre of all the three parallel
forces must lie on the straight line which bisects the angle
between CA and BA produced, and also on the straight
line which bisects the angle ABC, and on the straight
line which bisects the angle between AC and BC produced.
Therefore the centre of all the parallel forces must coincide
with the centre of the circle which touches AC, and BA
and BC produced. See Notes on Euclid, Book iv.
EXAMPLES. VIII.
1. A body is acted on by two parallel forces 2P and
5P, applied in opposite directions, their lines of action
being 6 inches apart: determine the magnitude and the
line of action of a third force which will be such as to keep
the body at rest.
2. Parallel forces P and Q act at two adjacent
corners of a parallelogram : determine the forces parallel
to these which must act at the other corners, so that tho
70 EXAMPLES. VIII.
centre of the four parallel forces may be at the intersection
of the diagonals of the parallelogram.
3. A rod without weight is a foot long ; at one end a
force of 2 Ibs. acts, at the other end a force of 4 Ibs., and at
the middle point a force of 6 Ibs., and these forces are
all parallel and like : find the magnitude and the point of
application of the single additional force which will keep
the rod at rest.
4. Equal like parallel forces act at five of the angular
points of a regular hexagon : determine the centre of the
parallel forces.
5. Find the centre of like parallel forces of 7, 2, 8, 4,
6 Ibs. which act in order at equal distances apart along
a straight line.
6. The circumference of a circle is divided into n equal
parts, and equal like parallel forces act at all the points of
division except one : find their centre.
7. Like parallel forces of 1, 2, and 3 Ibs. act on a bar
at distances 4, 6, and 7 inches respectively from one end :
find their centre.
8. ABC is a triangle ; parallel forces Q and R act at
B and C such that Q is to R as tan B is to tan C : shew
that their centre is at the foot of the perpendicular from A
onBC.
9. Parallel forces act at the angular points A, B, C of
a triangle, proportional to tan A, tan B, tan C respectively :
shew that their centre is at the intersection of the perpen
diculars drawn from the angles of the triangle on the oppo
site sides.
10. Parallel forces P, Q, R act at the angular points
A, B, C of & triangle : shew that the perpendicular dis
tance of their centre from the side BG is
P 2 area of triangle
PT#TIs x BC
11. Parallel forces P, Q, R act at the angular points
EXAMPLES. VIII. 71
A, B, C of a triangle: shew that the distance of their
centre from BC measured parallel to AB is
PxAB
P+Q+E'
12. Parallel forces P, Q, E act at the angular points
A) B, C of a triangle : determine the parallel forces which
must act at the middle points of BC, CA, AB, so that the
second system may have the same centre and the same
resultant as the first system.
13. Like parallel forces of 3, 5, 7, 5 Ibs. act at A, B, C,
D, which are the angular points of a quadrilateral figure,
taken in order : shew that the centre and the resultant will
remain unchanged if instead of these forces we have acting
at the middle points of AB, BC, CD, DA respectively
P, 10  P, 4 + P, 6  P Ibs., where P may have any value.
14. Parallel forces P, Q, E act at the angular points
A, B, C of a triangle ; and their centre is at : shew that
P = Q E
area of BOO area of CO A area of AOB '
15. Parallel forces act at the angular points A, B, C of
a triangle, proportional to a cos A, bcosB, ccos C respec
tively : shew that then centre coincides with the centre of
the circumscribed circle.
16. Parallel forces P, Q, E, S act at A, B, C, D; and
P = Q = E = S
axea.ofBCD area of CD A area, of DAB area, of ABC '
shew that their centre is at the intersection of AC
and BD.
17. Find the centre of equal like parallel forces acting
at seven of the angular points of a cube.
18. Parallel forces P, Q, E, S act at the angular
points A, B, C, D of a, triangular pyramid : shew that the
perpendicular distance of their centre from the face BCD is
P 3 volume of pyramid
P+Q+E+S* area, of BCD '
72 CENTRE OF GRAVITY.
IX. Centre of Gravity.
122. We begin with the following definition :
The centre of gravity of a body or system of bodies
is a point on wfiick the body or system will balance in all
positions, supposing t/ie point to be supported, the body or
system to be acted on only by gravity, and the parts of the
body or system to be rigidly connected with the point.
123. To find the centre of gravity of two heavy par
ticles.
Let A and B be the positions of the two
particles whose weights are P and Q respec
tively.
Join AB and divide it at L, so that AL
may be to LB as Q is to P; then L is the
centre of gravity.
For, by Art. 60, the resultant of the weights
P and Q acts through L\ and therefore if A and B are
connected by a rigid rod without weight the system will
balance in every position when L is supported.
As the resultant of P and Q is P + Q the pressure on
the point of support will be P + Q.
124. To find the centre of gravity of any number of
heavy particles.
Let A, B, C, D
be the positions of par
ticles whose weights
are P, Q, R, S, respec
tively.
CENTRE OF GRAVITY. 73
Join AB and divide it at Z, so that AL may be to LB
as Q is to P : then L is the centre of gravity of P at .4
and Q at 5; arid these weights produce the same effect
as P + Q at L. See Art. 123.
Join Z(7, and divide it at J/, so that .Z/J/ may be to
MC as R is to P + Q : then J/ is the centre of gravity of
P + Q at L and R at (7; and these weights produce the
same effect as P + Q + R at M.
Join 3/Z), and divide it at JT, so that JO" may be to
ND as is to P+ Q + R : then N is the centre of gravity
of P + Q + R at J/ and at Z> ; and these weights produce
the same effect as P + Q + R + S at N.
Then N is the centre of gravity of the system ; for the
resultant of the weights passes through N, and therefore
if the particles are connected with N by rigid rods without
weight the system will balance in every position when N
is supported.
Thus we have found the centre of gravity of four
heavy particles; and in the same way we may proceed
whatever be the number of the particles.
125. The investigation of the preceding Article shews
that every system of heavy particles has a centre of gravity ;
for the construction there given is always possible.
We see that the resultant weight of a system of heavy
particles always acts through the centre of gravity; so
that the effect of the weight of the system is the same as
if the whole weight were collected at the centre of gravity :
this result might have been anticipated from the definition
of the centre of gravity.
When we speak of a body or system balancing about
its centre of gravity we shall not always explicitly say that
the parts of the body or system are supposed to be rigidly
connected with the centre of gravity ; but this must always
be understood.
74 CENTRE OF GRAVITY.
126. A body or a system of bodies cannot have more
than one centre of gravity.
For, if possible, suppose that the body or system of
bodies has two centres of gravity, G and H; and let G and
H be brought into the same horizontal plane. Then when
G is supported the body or system balances, and therefore
the vertical line in which the resultant weight of the body
or system acts passes through G. Similarly the resultant
weight acts through H. Thus a vertical line passes through
two points which are in a horizontal plane; but this is
absurd.
127. If a body or a system of bodies balances itself on
a straight line in every position, the centre of gravity of
the body or system lies in that straight line.
Let AB be the straight
line on which the body or
system of bodies will balance
in every position.
Suppose, if possible, that
the centre of gravity is not
mAB\ let it be at G.
Place the body or system so that AB is horizontal, and
G not in the vertical plane through AB. Suppose two
points of the body or system situated on the straight
line AB to become fixed; this cannot disturb the equi
librium. The body or system has still the power of turning
round AB as an axis, and since it is in equilibrium the
resultant weight of the system must pass through AB,
by Art. 101. But this is impossible, because G is neither
vertically above nor vertically below AB.
Hence, the centre of gravity cannot be out of the
straight line AB.
128. We shall now give two propositions which are
almost immediately obvious, but which it is convenient to
enunciate formally ; and then we shall determine the posi
tion of the centre of gravity for some bodies of simple
forms.
CENTRE OF GRAVITY. 75
129. Given the centres of gravity of two parts which
compose a body or system of bodies, to find the centre of
gravity of the whole body or system of bodies.
Let A and B be the centres
of gravity of the two parts ; P
and Q the respective weights
of the parts.
Join AB and divide it at
C, so that AC may be to CB
as Q is to P : then is the
centre of gravity of the whole body or system.
130. Given the centre of gravity of part of a body or
system of bodies, and also the centre of gravity of the
whole body or system, to find the centre of gravity of the
remainder.
Let A be the centre of gravity of the part, C the centre
of gravity of the whole ; let P be the weight of the part,
and W the weight of the whole.
Join AC and produce it to B, so that CB may be to
CA as P is to WP: then B is the centre of gravity of
the remainder.
131. To find the centre of gravity of a straight line.
By a straight line here we mean a uniform material
straight line, that is, a fine straight wire or rod, the
breadth and thickness of which are constant and indefinitely
small.
The centre of gravity of a uniform material straight
line is at its middle point. For we may suppose the straight
line to be made up of an indefinitely large number of equal
particles. Take two of these which are equidistant from
the middle point of the straight line ; their centre of gravity
is at the middle point. And since this is true for every
such pair of particles the centre of gravity of the whole
straight line is at the middle point of the straight line,
VG
CENTRE OF GRAVITY.
132. Tofi'iid the centre of gravity of a parallelogram.
By a parallelogram here we mean a uniform material
parallelogram; that is, a thin
slice or lamina of matter, the
thickness of which is constant
and indefinitely small.
Let ABCD be the paral
lelogram. Bisect AB at E,
and CD at F\ join EF. Draw
any straight line aeb parallel to
AEB, meeting AD, EF, BC, at a, e, b respectively. Then
DFea and FCbe are parallelograms, from which it will
follow that ae=eb.
Suppose the parallelogram to be made up of indefinitely
thin strips parallel to AB; the centre of gravity of each
strip will be at its middle point by Art. 131 ; and will
therefore be on the straight line EF. Hence the centre
of gravity of the parallelogram is on the straight line EF.
Similarly the centre of gravity of the parallelogram is
on the straight line which joins the middle points of AD
and EC.
Hence the centre of gravity of a parallelogram is at
the intersection of the straight lines which join the middle
points of opposite sides. This point coincides with the
intersection of the diagonals of the parallelogram.
133. To find the centre of gravity of a triangle.
The meaning of the word M
triangle here is similar to
that of the word parallelo
gram in the preceding Arti
cle.
Let ABC be the triangle ;
bisect BC at E\ join AE.
Draw any straight line bee
parallel to BEC, meeting
AB, AE, AC at b, e, c respec
tively.
CENTRE OF GRAVITY. 77
Then ^=2^' by Euclid VL 4 ;
similarly ^h'~ J~fi'
be ce be BE
therefore Wp~ CF'' therefore ~~ = TTZ^ 
But BE=CE; therefore be=ce.
Hence e is the middle point of be.
Suppose the triangle made up of indefinitely thin strips
parallel to BC ; the centre of gravity of every strip will be
ut its middle point by Art. 131, and will therefore be on
the straight line AE. Hence the centre of gravity of the
triangle is on the straight line AE.
In the same way if AC be bisected at F the centre of
gravity of the triangle is on BF.
Hence the centre of gravity of the triangle must be at
G, the point of intersection of AE and BF.
Join EF; then .E^is parallel to AB, by Euclid, vi. 2 ;
,, EG AG , ^
therefore ~WF = ~~AR ' ^ Euc ^ K ^ VL 4 >
Thus AG is twice EG, and therefore AE is three times
EG; that is, EG is one third of EA.
Hence the centre of gravity of a triangle is determined
by the following rule : Join any angular point with the
middle point of the opposite side ; the centre of gravity is
on this straight line at one third of its length from the
side.
78
CENTRE OF GRAVITY,
The following statement will be very obvious, but it is
useful to draw attention to it : Join an angle A of a triangle
wjth any point L in the opposite side EG or EC produced ;
take M in LA so that LM is one third of L A ; and through
M draw a straight line parallel to EC : then the centre of
gravity of the triangle is in this straight line.
134. The centre of '_ gravity of a triangle coincides
with the centre of gravity of three equal heavy particles
placed at the angular points of the triangle.
Suppose equal heavy particles placed at the angular
points of a triangle ABC.
U JK J}
The centre of gravity of equal heavy particles at E
and C is at E, the middle point of EC. Join AE and
divide it at G, so that AG may be to GE as 2 is to 1 :
then G is the centre of gravity of equal heavy particles at
A, , and C. And G coincides with the point which was
found in the preceding Article to be the centre of gravity
of the triangle ABC.
135. The centre of gravity of any plane rectilineal
figure may be obtained in the following way : divide the
figure into triangles, find the centre of gravity of each
triangle, and then by successive applications of Art. 129
determine the centre of gravity of the proposed figure.
For example, suppose ABCD to be any quadrilateral
figure. Draw a diagonal DB and bisect it at E\ join
EA and EC.
CENTRE OF GRAVITY. 79
Take EH=\EA y and EK=\EC. Then H is the
o o
centre of gravity of the triangle ABD, and K is the centre
of gravity of the triangle BCD.
Join .071 and divide it at G, so that EG may be to KG
as the triangle CBD is to the triangle ABD : then G is
the centre of gravity of the quadrilateral figure.
Draw the diagonal AC, and let be the point of
intersection of the two diagonals; let HK meet BD at
L. Then the triangle CBD is to the triangle ABD as
CO is to A0\ thus
KG~ AO
KL
triangles ;
.,  HG KL EG KL
therefore S QT^Q = rr , rrr > tliat 777? = m I
therefore IIG=KL.
This gives a simple mode of determining G.
The case in which two sides of the quadrilateral are
parallel may be specially noticed ; to this we shall now pro
ceed.
80 CENTRE OF GRAVITY.
136. To find the centre of gravity of a quadrilateral
figure which has two sides parallel.
Let ABCD be a quadrilateral figure, having AB paral
lel to CD : it is required to find the centre of gravity of
the figure. Produce AD and BG to meet at ; let E be the
middle point of AB ; join OE meeting CD at F. Then, as
in Art. 133, we can shew that DF=FC, and that the centre
of gravity of the quadrilateral is on EF.
The centre of gravity of the triangle AOB is on OE,
o
at a distance  OE from ; and the centre of gravity of
o
o
the triangle DOC is on OF, at a distance = OF from 0.
o
Let G denote the centre of gravity of the quadrilateral
ABCD.
therefore
OG\OF
o
area of DOC
area, of AOB'
CENTRE OF GRAVITY. 81
Now, by Euclid, vi. 19 and vi. 2,
area of DOC = OP 2 = OF 2 m
areaof^OB OA* OE*'
therefore
OE+OF
2 OE 2 + OE. OF+ OF 2  ...
therefore FG=    OF
 OF] (20E+ OF} = FE 20E+ OF
3(OE+OF) = 3 ' OE+OF'
OF CD 20E+OF 2AB + CD
Thus 7^(7 is expressed in terms of the lengths of the
two parallel sides and the distance of their middle points.
Or we may proceed thus. By drawing DE and CE the
quadrilateral is divided into three triangles ADE, BCE,
CED. Since these triangles are between the same paral
lels their areas will be in the proportion of their bases AE,
JBE, CD. The distances of the centres of gravity of these
three triangles from AB, measured parallel to EF } will be
respectively \ EF, \ EF,  EF. Thus, by Art. 115,
o o o
EG (AE+ BE+ CD] = EF(AE+ BE} + EF. CD ;
therefore EO *.
T. ME.
82
CENTRE OF GRAVITY.
From this we get for FG, that is for EF EG, the same
value as before.
137. To find the centre of gravity of a triangular
pyramid.
Let ABC be the base, D the vertex; bisect AC at E\
join BE and DE. Take F on EB so that EF=\EB,
a
then ^ is the centre of gravity of the triangle ABO.
Join FD. From any point b in Z).# draw ba and 6c
parallel to BA and J5(7 respectively.
Let DF meet the plane afrc at/; join &/, and produce
it to meet DE at e.
Then, by similar triangles, ae=ec.
Also
W 3L'
JJF~ DF
CENTRE OF GRAVITY. 83
therefore =^.
But BF is twice EF\ therefore bf is twice ef\ and
therefore /is the centre of gravity of the triangle abc.
Suppose the pyramid made up of indefinitely thin slices
parallel to ABC; then, as we have just seen, the centre
of gravity of every slice will be on the straight line DF.
Hence the centre of gravity of the pyramid is on the
straight line DF.
Again, take // on ED so that EH=ED, and join
o
BH. Then, as before, the centre of gravity of the pyramid
is on BH.
Hence the centre of gravity of the pyramid must be at
G, the point of intersection of DF and BH.
Join FH\ then FH is parallel to BD by Euclid, vi. 2.
HG BG .
TlF = JD ' y EucM > VL 4 >
HG II F EF 1
MSBIEZra
Thus BG is three times 7/<7, and therefore BH is four
times HG } that is, IIG is one fourth of BH.
Hence the centre of gravity of a triangular pyramid is
determined by the following rule : Join any angular point
with the centre of gravity of the opposite face ; the centre
of gravity of the pyramid is on this straight line at one
fourth of its length from the face.
The following statement will be obvious : Join a vertex
D of a triangular pyramid with any point L in the plane of
the opposite face ABG\ take M in LD so that LM is one
fourth of LD ; and through M draw a plane parallel to
ABC: then the centre of gravity of the triangular pyramid
is in this plane.
C2
84
CENTRE OF GRAVITY.
jOTVvrO t// (y/ wi/bt/tj \jj \Jv t// V\A/IV\J u/t/w/ ~mrij i umi+w*
coincides with the centre of gravity of four equal heavy
at the angular points of the pyramid.
138. The centre of gravity of a triangular pyramid
with t
particles placed
Suppose equal heavy par
ticles placed at the angular
points of a pyramid ABCD.
The centre of gravity of
equal heavy particles at A, B,
and C, is at a point F which co
incides with the centre of gra
vity of the triangle ABC; see
Art. 134. The effect of equal
weights at A, JB, and C is the
same as that of a triple weight
at F. Join DF, and divide it at
G so that DG may be to GF
as 3 is to 1 : then G is the centre of gravity of equal heavy
particles at A, J3, C, and D. And G coincides with the
point which was found in the preceding Article to be the
centre of gravity of the pyramid ABCD.
139. To find the centre of gravity of any pyramid
having a plane rectilineal polygon for its base.
The pyramid may be divided into triangular pyramids
determined by drawing straight lines from any point on
the base to all the angular points of the base, and to the
vertex. The centre of gravity of every one of these trian
gular pyramids is in a plane which is parallel to the base,
and at one fourth of its distance from the vertex. Hence
the centre of gravity of the whole pyramid is in this plane.
Again, suppose the pyramid made up of indefinitely thin
slices parallel to the base. It may be shewn, as in Art. 137,
that the centre of gravity of every slice is on the straight
line which joins the centre of gravity of the base of the
whole pyramid with the vertex. Hence the centre of
gravity of the whole pyramid is on this straight line.
Therefore the centre of gravity of the whole pyramid is
on the straight line which joins the vertex with the centre
CENTRE OF GRAVITY. 85
of gravity of the base, at one fourth of the length of this
straight line from the base.
140. To find the centre of gravity of a cone.
A cone may be considered as a pyramid which has for
its base a polygon with an indefinitely large number of
sides. Hence the result obtained for a pyramid in Art. 139
holds for a cone. Therefore the centre of gravity of a cone
is on the straight line which joins the vertex with the
centre of gravity of the base, at one fourth of the length of
this straight line from the base.
141. The principle of symmetry will often aid us in
finding the position of the centre of gravity of a body.
A body is said to be symmetrical with respect to a
plane when the body may be supposed to be made up of
pairs of particles of equal size and weight, the two which
form a pan* being on opposite sides of the plane, equi
distant from it and on the same perpendicular to it.
If a body be symmetrical with respect to a plane,
that plane contains the centre of gravity of the body.
For the weights of the two portions into which the
plane divides the body are equal, and their centres of
gravity are at equal distances from the plane on opposite
sides of it : therefore the centre of gravity of the whole
body is in the plane. See Art. 129.
142. If a body be symmetrical with respect to each
of two planes, the centre of gravity will be in each of the
planes, and therefore in the straight line in which they
intersect. If a body be symmetrical also with respect to
a third plane, the centre of gravity is in that plane ; if the
three planes have not a common line of intersection they
will meet at a point, and this point will therefore be the
centre of gravity of the body.
Take, for example, a sphere. Any plane passing through
the centre of the sphere divides the sphere symmetrically,
and so contains the centre of gravity : therefore the centre
of the sphere is its centre of gravity.
86
CENTRE OF GEA VITT.
143. The propositions respecting the centre of parallel
forces, given in Arts. 11 4... 11 9, are applicable to the centre
of gravity, with the simplification which arises from the
fact that the weights of particles are like parallel forces.
It will be convenient to repeat these propositions.
144. TJie distances of two heavy particles from a
straight line being given, to determine the distance of ' the
centre of gravity of the particles from that straight line ;
the straight line and the particles being all in one plane.
Let A and B be the posi
tions of the particles ; P and Q
their respective weights.
Join AB, and divide it at C,
so that AC may be to CB as Q
is to P : then C is the centre of
gravity of the particles.
Let AD, BE, CF be perpen
diculars from A, B, C on any straight line which is in a
plane containing A and B. Let AD=p, BE=q, CF=r:
then we have to find the value of r, supposing the values
of p and q to be known.
Through C draw aCb parallel to DFE, meeting AD
and BE at a and b respectively.
Then
^ = , by Euclid, vi. 4 ;
P Bb BEEb
qr
= 
therefore
therefore
P (r  p} = Q (qr) ;
CENTRE OF GRA V1TY.
87
145. The distances of any number of heavy particles
in one plane from a straight Line in the plane being given,
to determine the distance of the centre of gravity of the
system from that straight line.
Let A, B, C, D be the positions of particles whose
weights are P, Q, R, S. Let p, q, r, s be the distances of
A, B t C, D respectively from a straight line Ox in the
same plane.
Join AS, and divide it at L, so that AL may be to LB
as Q is to P ; then L is the centre of gravity of P at A
and Q at B, and these weights produce the same effect as
P + Q at L. Let I denote the distance of L from Ox, then,
by Art. 144,
Join LC, and divide it at J/, so that LM may be to
MG as R is to P + Q ; then Jf is the centre of gravity of
P + Q at L and R at (7, and these weights produce the
same effect as P + Q + R at M. Let m denote the distance
of M from Ox, then, by Art. 144,
_(P + Q)l + Rr _Pp + Qq + Rr
P+Q+R P+Q+R '
Join MD, and divide it at N, so that MN may be to
ND as S is to P + Q + R; then JV" is the centre of gravity
of P + Q + R at M and S at D, and these weights produce
the same effect as P + Q + R + S a& N. Let n denote the
distance of N from Ox, then, by Art. 144,
P+Q+R+S
P+Q+R+S '
88 EXAMPLES. IX.
Thus we have determined the distance from Ox of the
centre of gravity of four heavy particles ; and in the same
manner we may proceed whatever be the number of heavy
particles.
146. In the same way if the distances of A, 13, C, and D
from a second straight line, as Oy, in the same plane be
given, we can deduce the distance of the centre of gravity
of the system from the same straight line.
And when we know the distance of the centre of gravity
from two straight lines in the plane we can determine the
position of the centre of gravity ; for it will be at the point
of intersection of straight lines parallel to Ox and Oy and
at the respective distances from them which have been
found.
It is easy to extend our investigation to the case in
which the heavy particles are not all in one plane; see
Art. 119. Thus if p, q, r, s denote the distances from any
fixed plane of particles whose weights are respectively
P, Qj R, S, the value of n in Art. 145 gives the distance
of the centre of gravity of the particles from the same
fixed plane. The distances may be either perpendicular
distances, or distances measured parallel to any given
straight line.
EXAMPLES. IX.
1. If two triangles are on the same base, shew that
the straight line which joins their centres of gravity is
parallel to the straight line which joins their vertices.
2. A rod 3 feet long and weighing 4 Ibs. has a weight
of 2 Ibs. placed at one end : find the centre of gravity of the
system.
3. A quarter of a triangle is cut off by a straight line
drawn parallel to one of the sides : find the centre of
gravity of the remaining piece.
EXAMPLES. IX. 89
. 4. Find the centre of gravity of a uniform circular disc
out of which another circular disc has been cut, the latter
being described on a radius of the former as diameter.
5. If three men support a heavy triangular board at
its three corners, compare the force exerted by each man.
6. Shew that the centre of gravity of a wire bent into
a triangular shape coincides with the centre of the circle
inscribed in the triangle formed by joining the middle
points of the sides of the original triangle.
7. If the centre of gravity of a triangle be equidistant
from two angular points of the triangle, the triangle must
be isosceles.
8. If a straight line drawn from an angular point
through the centre of gravity of a triangle be perpendicular
to the opposite side, the triangle must be isosceles.
9. A triangle AEG has the sides AB and BC equal ;
a portion APG is removed such that AP and PC are
equal : compare the distances of P and B from AC in
order that the centre of gravity of the remainder may be
at P.
10. A heavy bar 14 feet long is bent into a right angle
so that the lengths of the portions which meet at the angle
are 8 feet and 6 feet respectively : shew that the distance
of the centre of gravity of the bar so bent from the point
of the bar which was the centre of gravity when the bar
Q /2
was straight, is ~ feet.
11. If the centre of gravity of three heavy particles
placed at the angular points of a triangle coincides with
the centre of gravity of the triangle, the particles must be
of equal weight.
12. Two equal uniform chains are suspended from the
extremities of a straight rod without weight, which can
turn about its middle point : find the position of the centre
of gravity of the system, and shew that it is independent of
the inclination of the rod to the horizon.
90 EXAMPLES. IX.
13. The middle points of two adjacent sides of a
square are joined and the triangle formed by this straight
line and the edges is cut off : find the centre of gravity of
the remainder of the square.
14. If n equal weights are to be suspended from a
horizontal straight line by separate strings, and a given
length I of string is to be used, determine the distance of
the centre of gravity of the weights from the straight line.
15. If the sides of a triangle be 3, 4, and 5 feet, find
the distance of the centre of gravity from each side.
16. A piece of uniform wire is bent into the shape of
an isosceles triangle ; each of the equal sides is 5 feet long,
and the other side is 8 feet long : find the centre of gravity.
17. Find the centre of gravity of the figure consisting
of an equilateral triangle and a square, the base of the tri
angle coinciding with one of the sides of the square.
18. Two straight rods without weight each four feet
long, are loaded with weights 1 lb., 3 Ibs., 5 Ibs., 7 Ibs., 9 Ibs.
placed in order a foot apart : shew how to place one of the
rods across the Bother, so that both may balance about a
fulcrum at the middle point of the other.
19. A rod of uniform thickness is made up of equal
lengths of three substances, the densities of which taken in
order are in the proportion of 1, 2, and 3 : find the position
of the centre of gravity of the rod.
20. A table whose top is in the form of a rightangled
isosceles triangle, the equal sides of which are three feet in
length, is supported by three vertical legs placed at the
corners ; a weight of 20 Ibs. is placed on the table at a point
distant fifteen inches from each of the equal sides : find
the resultant pressure on each leg.
21. In the diagram of Art. 136 shew that if AC and
BD be joined, intersecting at , then S is on FE : shew
also that SG is equal to two thirds of the difference between
SE and SF.
PROPERTIES OF CENTRE OF GRA VITY. 91
X. Properties of the Centre of Gravity.
147. WJien a body is suspended from a point round
which it can move freely it will not rest unless its centre
of gravity be in the vertical line passing through the point
of suspension.
For the body is acted on by two forces, namely its own
weight in a vertical direction through the centre of gravity,
and the force arising from the fixed point. The body will
not rest unless these two forces are equal and opposite.
Therefore the centre of gravity must be in the vertical line
which passes through the point of suspension.
148. The preceding Article suggests an experimental
method of determining the centre of gravity of a body
which may sometimes be employed. Let a body be sus
pended from a point about which it can turn freely, and let
the direction of the vertical line through the point of sus
pension be determined. Again, let the body be suspended
from another point so as to hang in a different position,
and let the direction of the vertical line through the point
of suspension be determined. The centre of gravity is in
each of the two determined straight lines, and is therefore
at their point of intersection.
149. When a body can turn freely round an axis
which is not vertical, it will not rest unless the centre of
gravity be in the vertical plane passing through the axis.
The weight of the body may be supposed to act at the
centre of gravity. Kesolve it into two components at right
angles to each other, one component being parallel to the
axis. The component parallel to the axis will not produce
nor prevent motion round the axis ; but the other compo
nent will set the body in motion round the axis, unless the
centre of gravity be in the vertical plane passing through
the axis.
150. A body which is suspended from a fixed point by
means of a string will not rest unless its centre of gravity
be below the fixed point to which the string is fastened.
But a body which can turn freely round a fixed point
rigidly connected with it may rest with its centre of
gravity either vertically above or vertically below the fixed
92
PROPERTIES OF THE
point. And in like manner when a body can turn freely
round a fixed axis which is not vertical it may rest with its
centre of gravity either above or below the axis. There is
an important difference between the two positions of equi
librium, which is shewn by the following proposition.
151. ^ When a body ivhich can turn freely round a
fixed point is in equilibrium, if it be slightly displaced it
will tend to return to its position of equilibrium or to re
cede from it according as the centre of gravity is below or
above the fixed point.
This may be taken as an experimental fact ; or it may
be established thus :
Let be the fixed point, G the centre of gravity of the
body. Draw GH vertically downwards. The weight of
the body acts along Gil ; resolve it into two components at
right angles to each other, one along the straight line
which joins the centre of gravity with the fixed point : let
GK be the direction of the other component.
When G is nearly below the former component acts
along OG ; and thus the latter obviously tends to move the
body towards the position in which G is vertically below 0.
When G is nearly above the former component acts
along GO ; and thus the latter obviously tends to move
the body away from the position in which G is vertically
above 0.
In the former case the body is said to be in stable
equilibrium, and in the latter case in unstable equilibrium.
CENTRE OF GRA VIT7. 93
152. When a body is placed on a horizontal plane it
stand or fall according as the vertical line drawn
through its centre of gravity passes within or without the
base.
Let Q be the centre of gravity of a body. Let the
. vertical line through G cut the horizontal plane on which
the body stands at H. Let any horizontal straight line be
drawn through H, and let AB be that portion of it which
is within the base of the body.
First suppose II to be between A and B.
No motion can take place round A. For the weight of
the body acts vertically downwards at G ; and it may be
resolved into two components, one along GA, and the
other at right angles to GA. The former component has
no tendency to produce motion round A. The latter com
ponent tends to turn G round A in the direction GK ; now
if this motion could take place such a point as B would
turn round A in a like direction, but this is prevented by
the resistance of the plane at B : therefore G cannot move
round A.
Similarly no motion can take place round B ; therefore
the body cannot fall over either at A or at B.
Next suppose H not to be between A and B : let it be
on AB produced through B.
Then, as before, no motion will take place round A.
But motion will take place round B ; for the tendency of
the component of the weight at right angles to GB is
to move G round B in the direction GK ; and there is
nothing to prevent this motion.
94 PROPERTIES OF THE
153. The sense in which the word base is used in the
preceding proposition may require some explanation. The
portions of surface common to the body and the hori
zontal plane may form one undivided area, or may consist
of various separate areas ; a smooth brick placed on a
smooth horizontal plane will exemplify the former case,
and a chair will exemplify the latter case. Moreover
these areas may be indefinitely small, that is, may be mere
points.
The boundary of the base for the purposes of the pre
ceding proposition must be determined thus : let a polygon
be formed by straight lines joining the points of contact, in
such a manner as to include all the points of contact, and
to have no reentrant angle. See Notes on Euclid, I. 32.
154. Forces are represented in direction by the straight
lines drawn from any point to a system of heavy par
tides, each force being equal to the product of the length of
the straight line into the weight of the corresponding
^article: to shew that the resultant force is represented
in direction by the straight line drawn from the point to
the centre of gravity of the particles, and is equal to the
product of the length of this straight line into the sum of
the weights.
First, let there be two heavy par
ticles. Suppose P and Q their weights ;
A and B their respective positions.
Let L be their centre of gravity.
Let be any point ; and suppose
there are two forces, namely P x OA A
along OA, and Q x OB along OB.
The force PxOA along OA may be resolved into
P x OL along OL, and P x LA parallel to LA. The force
Q x OB along OB may be resolved into QxOL along OL
and QxLB parallel to LB.
The two forces P x LA and Q x LB are thus equal and
opposite, and therefore balance each other. Hence the re
sultant is (P + Q) OL along OL.
CENTRE OF ORA VITT. 95
Next, let there be three heavy particles. Suppose
P, Q, R their weights; A, B, C
their respective positions. Let
L be the centre of gravity of P
and Q ; and M the centre of gra
vity of P, Q, and R.
Let be any point, and
suppose there are three forces,
namely, P x OA along OA,
QxOB along OB, and jRxtftf
along OC.
By what has been already shewn, these forces are equi
valent to (P+ Q) OL along OL, and RxOC along OC: and
these again are equivalent to (P + Q + R) OM along OM.
If there be a fourth heavy particle, of weight S, at a
point D, there are four forces which, by what has been
shewn, are equivalent to (P + Q + R) Olf along OM, and
S x 02) along OD : and these again are equivalent to
(P + Q + R + S) ON along ON, where N is the centre of
gravity of the four heavy particles.
In this manner the proposition may be established,
whatever be the number of heavy particles.
If the point at which the directions of the forces meet
coincides with the centre of gravity of the system of heavy
particles, the resultant is zero ; that is, the forces are then
in equilibrium.
155. Forces are represented in magnitude and direc
tion by straight lines drawn from any point: to shew
that the resultant force is represented in direction by tJie
straight line draim from this point to the centre of gra
vity of a system of equal particles situated at the other
extremities of tlie straight lines, and is equal to the product
of this straight line into the number of particles.
This is a particular case of the proposition of the pre
ceding Article, obtained by supposing all the heavy par
ticles there to be of equal weight.
If the point at which the directions of the forces meet
coincides with the centre of gravity of the system of equal
96 PROPERTIES OF THE
particles, the resultant is zero ; that is, the forces are then
in equilibrium.
156. The sum of the products of the weight of each
particle of a system of heavy particles into tlie square of
its distance from any point exceeds the sum of the products
of the weight of each particle into the square of its dis
tance from the centre of gravity by the product of the sum
of the weights into the square of the distance between the
point and the centre of gravity.
First, let there be two heavy par
ticles. Suppose P and Q their U
weights; A and B their respective
positions. Let L be their centre of
gravity.
Let be any point : then shall
Let OH be the perpendicular from on AB.
By Euclid, n. 12, 13,
OB 2 = BL* + OL*  2BL . Z77 ;
therefore
for P x AL Q x BL, by the nature of the centre of gravity.
In the figure 77 falls between L and B ; the demon
stration is essentially the same for
every modification of the figure.
Next, let there be three heavy
particles. Suppose P, Q, R their
weights ; A, B, C their respective
positions. Let L be the centre
of gravity of P and Q, and M
the centre of gravity of P, Q t
and R.
Then we have, by three applications of the result already
obtained,
CENTRE OF GRAVITY. 97
Similarly, by three applications of results already ob
tained, we can shew that the proposition is true when
there are four heavy particles : and so on universally.
157. If the weight of each of a system of heavy par
ticles be multiplied into the square of the distance of the
particle from a given point, the sum of the products is
least when the given point is the centre of gravity of the
system.
This follows immediately from the proposition of the
preceding Article.
158. Examples may be proposed respecting the centre
of gravity which do not involve any new mechanical con
ception, but are merely geometrical deductions.
For example, required the distances of the centre of
gravity of a triangle from the three angular points in
terms of the sides of the triangle.
Let ABC denote a triangle, D the middle point of BC,
G the centre of gravity. Then G is in AD, and AG= AD.
Now, by the Appendix to Eudid, Art. 1,
AB 2 + AC*=2 (AD* + BD~) ;
therefore AD^=AB Z + AC 2 BC 2 \ And AGP
therefore AG Z =
Similar expressions may be found for EG 2 and CG 2 .
159. The theory of the centre of gravity will furnish
us with indirect demonstrations of geometrical theorems,
We will give an example.
T. MB. 7
98 PROPERTIES OF CENTRE OF GRAVITY.
Let A, B, C, D be four points,
which need not be all in the same
plane ; and let equal heavy particles
be placed at these points. Let E be
the middle point of AB, and F the
middle point of CD. Then E is the
centre of gravity of the particles at A
and B, and Fis the centre of gravity of the particles at C and
D. Therefore the centre of gravity of the system is at the
middle point of EF. In the same way the centre of gravity
of the system is at the middle point of the straight line
which joins the middle points of AD and BO. But there
is only one centre of gravity of the system ; and hence we
obtain the following result : The straight lines which join
the middle points of the opposite sides of any quadri
lateral bisect each other.
Similarly, from the process for finding the centre of
gravity of a triangle, we might infer that the straight lines
which join the angular points of a triangle with the
middle points of the opposite sides meet at a point.
EXAMPLES. X.
1. A square stands on a horizontal plane : if equal
portions be removed from two opposite corners by straight
lines parallel to a diagonal, find the least portion which
can be left so as not to topple over.
2. Find the locus of the centres of gravity of all tri
angles on the same base and between the same parallels.
3. A portion of the surface of a heavy body is spheri
cal, and the body is in equilibrium when any point of this
portion is in contact with a horizontal plane : find the posi
tion of the centre of gravity of the body.
4. Given the base and the height of a triangle, con
struct it so that it may just rest in equilibrium with its
base on a horizontal plane.
5. A quadrilateral lamina which has all its sides equal
will be in equilibrium if its plane be vertical and any one
of its sides on a horizontal plane.
EXAMPLES. X. 99
6. Two weights W and 2 W are connected by a rod
without weight, and also by a loose string which is slung
over a smooth peg : compare the lengths of the string on
each side of the peg when the weights have assumed their
position of equilibrium.
7. If a number of rightangled triangles be described
on the same straight line as hypotenuse, their centres of
gravity all lie on a circled
8. If the sides of a triangle be bisected, and the tri
angle formed by joining these points be removed, shew
that the centre of gravity of the remainder will coincide
with that of the whole triangle.
9. A round table stands on three legs placed on the
circumference at equal distances : shew that a body whose
weight is not greater than that of the table may be placed
on any point of it without upsetting it.
10. A BCD is a parallelogram having the angle
AJ3C=60, and the base EC six inches in length : deter
mine the greatest possible length of AB if the figure is
to stand on EG.
11. A heavy triangle is to be suspended by a string
passing through a point on one side : determine the posi
tion of the point so that the triangle may rest with one
side vertical
12. A triangle obtuseangled at B is placed with its
side CB resting on a horizontal plane ; a vertical straight
line from A meets the plane at D : shew that the triangle
will stand or fall according as BD is less or greater
than EG.
13. The sides of a heavy triangle are 3, 4, 5 respec
tively : if it be suspended from the centre of the inscribed
circle shew that it will rest with the shortest side hori
zontal.
14. The altitude of a right cone is h, and a diameter
of the base is b ; a string is fastened to the vertex and to
a point on the circumference of the circular base, and is
then put over a smooth peg : shew that if the cone rests
with its axis horizontal the length of tho string is V(/t a + 6 2 ).
72
100 THE LEVER.
XI. Tlie Lever.
160. Machines are instruments used for communi
cating motion to bodies, for changing the motion of bodies,
or for preventing the motion of bodies.
The most simple machines are called Mechanical
Powers; by combining these, all machines, however com
plicated, are constructed. These simple Machines or Me
chanical Powers are usually considered to be seven in
number; namely the Lever, the Wheel and Axle, the
Toothed Wheel, the Fully, the Inclined Plane, the Wedge,
and the Screw.
We shall investigate the conditions of equilibrium of
the Mechanical Powers ; that is, we shall suppose these
simple machines employed to prevent motion. We shall
in every case have two forces which balance each other by
means of a machine ; one force for the sake of distinction
is called the Power, and the other the Weight: we shall
find that in every case for equilibrium the Power must
bear to the Weight a certain ratio which depends on the
nature of the machine.
We shall assume, unless the contrary is expressly stated,
that the parts of the machine are smooth and without
weight.
In the present Chapter we shall consider the Lever.
161. The Lever is an inflexible rod moveable, in one
plane, about a point in the rod which is called the fulcrum.
The parts of the Lever between the fulcrum and the points
of application of the Power and the Weight are called the
arms of the Lever. When the arms are in the same
straight line the lever is called a straight Lever; in other
cases it is called a bent Lever. The plane in which the
Lever can move may be called the plane of the Lever.
The forces which act on the Lever are supposed to act in
the plane of the Lever.
THE LEVER. 101
162. Levers are sometimes divided into three classes,
according to the positions of the points of application of
the Power and the Weight with respect to the fulcrum.
In 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, the Weight being the nearer
to the fulcrum.
In the third class the Power and the Weight act on the
same side of the fulcrum, the Power being the nearer to
the fulcrum.
Thus we may say briefly that the three classes have
respectively the Fulcrum, the Weight, and the Power in
the middle position.
163. The following are examples of Levers of the first
class: a crowbar used to raise a heavy weight, a poker
used to raise coals in a grate, the brake of a pump. In
scissors, shears, nippers, and pincers we have examples of
a double Lever of the first class.
The oar of a boat furnishes an example of a Lever of
the second class. The fulcrum is at the blade of the oar
in the water; the Power is applied by the hand; the
Weight is applied at the rowlock. A wheelbarrow in use
will also serve as an example. A pair of nutcrackers is a
double Lever of the second class.
A pair of tongs used to hold a coal is a double Lever of
the third class. The fulcrum is the pivot on which the two
parts of the instrument turn; the Power is the pressure
applied by the hand ; the Weight is the resistance of the
coal at the end of the tongs. An example of the third
class of Lever is seen in the human forearm employed to
raise an object taken in the hand. The fulcrum is at the
elbow; the Power is exerted by a muscle which comes
from the upper part of the arm, and is inserted in the fore
arm near the elbow; the Weight is the object raised in the
hand.
102 THE LEVER.
164. The necessary and sufficient condition for the
equilibrium of two forces on the Lever is that their moments
round tho fulcrum should be equal in magnitude but of
opposite kinds. This has been already demonstrated ; see
Art. 102. But on account of the importance of the prin
ciple of the Lever we shall give a separate investigation.
165. Wlien there is equilibrium on the Lever the
Power is to the Weight as the length of the perpendi
cular from the fulcrum on the direction of the Weight is
to the length of the perpendicular from the fulcrum on the
direction of the Power.
Let ACE or ABC be a Lever, C boing the fulcrum. Let
forces P and W act at A and B respectively and keep
the Lever in equilibrium.
Let the directions of P and W meet at 0. Then
the resultant of P and W will be some force which may
be supposed to act at 0\ and this resultant must pass
through C, since the Lever is in equilibrium. Hence 00
is the direction of the resultant of P and W.
Draw Ca parallel to OA, and Cb parallel to OB, to
meet OB and OA respectively; and draw CM perpen
dicular to OA, and CN perpendicular to OB.
Then, by the Parallelogram of Forces,
P _Ca
W ~ Cb '
THE LEVER. 103
and by the similar triangles ONa and (7J/6,
Cb = CM'
. f P ON
therefore TF = tfJT
T) (7/V
Conversely, if ^= 77,,, and P and W tend to turn
rr LM
the Lever in opposite directions, they will keep it in equi
librium.
For with the same construction we have
'W = CM = Cb'
and therefore 00 is the direction of the resultant of P
and W] and since the resultant passes through the Lever
will be kept in equilibrium.
166. There is no substantial difference in theory be
tween the second and the third class of Levers considered
in Art. 162 ; but there is considerable practical difference.
For it follows from the condition of equilibrium of the
Lever that in the second class the Power is less than the
Weight, and in the third class the Power is greater than
the Weight. Thus it is said that a mechanical advantage is
gained by a Lever of the second class, and lost by a Lever
of the third class.
The word advantage is used in a popular sense in the
remark just made ; more strictly the advantage of a machine
may be denned as the ratio of the Weight to the Power
when there is equilibrium.
167. In the investigation of Article 165 we assume that
the directions of P and W will meet if produced ; but the
point of intersection may be at any distance from the ful
crum, so that we may readily admit that the result will
hold even when the directions of P and Q are parallel.
But it may be useful to give an investigation of this case.
104 TEE LEVER.
Let ACE or AEG be a Lever, C being the fulcrum. Let
parallel forces P and TF act at A and .Z? respectively 'and
keep the Lever in equilibrium.
r
p
Through C draw a straight line perpendicular to the
directions of the forces meeting them at M and N respec
tively.
Now by Arts. 60 and 61 the resultant of P and W is a
force parallel to them at distances from them which are in
versely proportional to them. But since the Lever is in
equilibrium the resultant must pass through (7; and there
fore
P__GN
W~CM'
~P C*W
Conversely, if r?r=^T>, and P and T
the Lever in opposite directions, they will keep the Lever in
equilibrium.
n /^/V"
For since "TK^TYT/} the resultant of P and W passes
through (7, and therefore the Lever will be kept in equi
librium.
168. It appears from the foregoing Articles that equi
librium is maintained on the Lever by the aid of the ful
crum which supplies a force equal and opposite to the re
sultant of P and W. Thus we see that the pressure on
the fulcrum will be equal to the resultant of P and TF;
if P and W are parallel this resultant is equal to their
algebraical sum, in other cases it may be determined by
the Parallelogram of Forces.
THE LEVER.
105
169. If two weights balance each other on a straight
Lever in any one position inclined to the vertical, they will
balance each other in any other position of the Lever.
Let AB be the position of the Lever when the weights
P and TF balance each other; let C be the fulcrum.
Let ab be any other position of the lever in the same
vertical plane.
Through C draw a horizontal line, meeting the vertical
lines which represent the lines of action of the weights
at M and N and m and n respectively.
Now since P and W balance in the position AB of the
Lever,
P _Ctf
W~CM'
And by similar triangles,
CN _ CB _ Cb _ Cn_
CM~CA~ Ca~ Cm'
therefore
^_
W~ Cm'
Hence P and TF will balance each other in the posi*
tion ab of the Lever.
10G EXAMPLES. XL
EXAMPLES. XI.
1. A weight of 5 Ibs. hung from one extremity of a
straight Lever balances a weight of 15 Ibs. hung from the
other : find the ratio of the arms.
2. Two weights of 3 Ibs. and 4 Ibs. are hung at the
ends of a straight Lever whose length is 92 inches : find
where the fulcrum must be for equilibrium.
3. Two weights which together weigh 6 Ibs. are hung
at the ends of a straight Lever and balance: if the fulcrum
is four times as far from one end as from the other find
each weight.
4. A Lever 7 feet long is supported in a horizontal
position by props placed at its extremities : find where a
weight of 28 Ibs. must be placed so that the pressure on
one of the props may be 8 Ibs.
5. Two weights of 12 Ibs. and 8 Ibs. respectively at the
ends of a horizontal Lever 10 feet long balance : find how
far the fulcrum ought to be moved for the weights to
balance when each is increased by 2 Ibs.
6. If the pressure on the fulcrum be equivalent to a
weight of 15 Ibs., and the difference of the forces to a weight
of 3 Ibs., find the forces and the ratio of the arms at
which they act.
7. A Lever is in equilibrium under the action of the
forces P and Q, and is also in equilibrium when P is
trebled and Q is increased by 6 Ibs. : find the magnitude
of Q.
8. The pressure on the fulcrum is 12 Ibs., and the dis
tance of the fulcrum from the middle point of the Lever is
onetwelfth of the whole length of the Lever : find the
forces which acting on opposite sides of the fulcrum will
produce equilibrium.
9. One force is four times as great as the other, and
the forces are on the same side of the fulcrum, and the
pressure is 9 Ibs. on it : find the position and the magnitude
of the forces.
EXAMPLES. XT. 107
10. AEG is a straight weightless rod 9 inches long,
placed between two pegs A and B which are 4 inches
apart, so as to be kept horizontal by means of them and a
weight of lOlbs. hanging at C: find the pressures on the
pegs.
11. A Lever bent at right angles, with the angle for
fulcrum and having one arm double the other, has two
weights hanging from its ends : if in the position of equi
librium the arms are equally inclined to the horizon com
pare the weights.
12. The pressure on the fulcrum is 3 Ibs., and the sum
of the forces 10 Ibs. : find the distance of each from the
fulcrum, if their distance apart be 2 feet.
13. If the pressure on the fulcrum be 5 Ibs., and one of
the weights be distant from the fulcrum onesixth of the
whole length of the Lever, find the weights, supposing them
on opposite sides of the fulcrum.
14. If the arm of a cork compressor be 18 inches, and
a cork be placed at a distance of one inch and a half from
the fulcrum, find the pressure produced by a weight of
twelve stone suspended from the handle.
15. If the fulcrum be between the two forces, and its
distance from one of them be a third of the whole length of
the Lever, shew that when the direction of either of the
forces is reversed, the fulcrum must then be placed at three
times its former distance from the same force.
16. Two forces of 2 Ibs. and 4 Ibs. act at the same point
of a straight Lever on opposite sides of it, and keep it at
rest, the less force being perpendicular to the Lever : deter
mine the direction of the greater force, and the pressure
on the fulcrum.
17. A weight of Plbs. hangs from the end of a Lever
2 feet long, at the other end of which is a fulcrum, and the
Lever is kept in equilibrium by such a force Q that the
fulcrum bears  P Ibs. : determine the magnitude of Q t
and the point of its application.
108 EXAMPLES. XI.
18. If three weights P, Q, S hang from the points
J, J3, C of a straight Lever which balances about a fulcrum
I), shew that
19. ABC is a straight Lever ; the length of AB is
7 inches, that of BG is 3 inches ; weights of 6 Ibs. and 10 Ibs.
hang at A and J5, and an upward pressure of 6 Ibs. acts at C:
find the position of a fulcrum about which the Lever so acted
on would balance, and determine the pressure on the ful
crum.
20. Weights of Gibs, and 4 Ibs. hang at distances
2 inches and 6 inches respectively from the fulcrum of a
Lever on the same side of it : find where a single force
of 9 Ibs. must be applied to support them so as to leave
the least possible pressure on the fulcrum.
21. Shew that the proposition of Art. 169 holds when
there are more than two weights if they are applied at
points of one straight line passing through C.
22. ACS is a bent Lever; the arms CA, CB are
straight, and inclined to one another at an angle of 135.
When CA is horizontal a weight P at A just sustains a
weight W at .5; and when CB is horizontal the weight
W at B requires a weight Q at A to balance it : find the
ratio of Q to P.
23. A Lever ACB is bent at (7, the fulcrum, and from
B a weight Q is hung ; when P is hung at A the Lever
rests with AC horizontal; but when S is hung at A then
CB becomes horizontal : shew that
CA : CB :: Q : J(P.S).
24. A Lever is 5 feet long, and from its ends a weight is
supported by two strings 3 feet and 4 feet long respectively :
shew that the fulcrum must divide the Lever into two parts
the ratio of which is that of 9 to 16, if there be equilibrium
when the Lever is horizontal.
BALANCES. 10!)
XII. Balances.
170. The Lever is employed to determine the weights of
substances ; and under this character it is called a Balance :
we shall now describe various forms of the Balance.
In the preceding Chapter we considered a Lever to be a
rod without weight; but in practice a rod always has
weight, and we shall accordingly attend to this fact in our
investigations.
We shall first consider the Common Balance,
171. The Common Balance. The Common Balance
consists of a beam with a scale suspended from each extre
mity ; the beam can turn about a fulcrum which is above
the centre of gravity of the beam, and therefore above the
centre of gravity of the system formed by the beam, the
scales, and the tnings which may be put in the scales. The
arms of the beam must be of equal length, and the system
should be in equilibrium with the beam horizontal when
the scales are empty : if these conditions hold, the Balance
is said to be trite, if they do not hold, the Balance is said
to be false.
The substance to be weighed is placed in one scale,
and weights in the other until the beam remains at rest in
the horizontal position. In this case, if the Balance be true,
the weight of the substance is indicated by the weights
which balance it. "We may test whether the Balance is true,
by observing whether the beam still remains at rest in the
horizontal position when the contents of the scales are
interchanged.
If the Balance be true and the contents of the two
scales be made of unequal weight, the beam will not remain
in the horizontal position, but after oscillating for a time
will finally rest in some position inclined to the horizon.
110 . BALANCES.
172. In the construction of a Balance, the following
requisites should be satisfied :
(1) When loaded with equal weights the beam should
be perfectly horizontal : that is, the Balance should be
true.
(2) When the contents of the two scales differ in
weight, even by a small quantity, the Balance should detect
this difference : that is, the Balance should be sensible.
(3) When the Balance is disturbed, it should readily
return to its state of rest : that is, the Balance should
be stable.
173. To find how the requisites of a good Balance may
be satisfied.
Let AB be the beam, C the fulcrum; let AB=2a,
and let h be the distance of C from AB. Let P and Q
be the weights of the contents of the two scales. Let
W be the weight of the beam ; let k be the distance from
G of the centre of gravity of the beam, this centre of
gravity being supposed to lie on the perpendicular from C
on AB. Let S be the weight of each scale, so that
P and S act vertically through A, and Q and S act verti
cally through B. Let 6 be the angle which the beam
makes with the horizon when there is equilibrium.
BALANCES. Ill
The sum of the moments of the weights round C will
be zero when there is equilibrium, by Art. 86. Now the
length of the perpendicular from C on the line of action of
P and S is a cos 6 h sin 6 ; the length of the perpendicular
from C on the line of action of Q and S is a cos 6 + h sin 6 ;
the length of the perpendicular from C on the line of
action of W is Jc sin 6. Therefore
(Q + 8) (a cos + Asin 6)(P + &) (a cos 6h sin 6}
therefore
This determines the position of equilibrium.
(1) When P=Q we have tan 0=0 ; thus the Balance
is true : so that by making the arms equal and having the
centre of gravity of the beam on the perpendicular from
the fulcrum on the beam the first requisite is satisfied.
(2) For a given difference of P and Q the sensibility
is obviously greater the greater tan 6 is ; and for a given
value of tan 6 the sensibility is greater the smaller the
difference of P and Q is. Thus we may consider that the
sensibility varies as tan 6 when P Q is constant ; also
that it varies inversely as P  Q when tan 6 is constant ;
and so when both tan 6 and P  Q vary the sensibility will
be measured by ^ : see Algebra for Beginners, Art 389.
Therefore the second requisite will be satisfied by making
(P + # + 2/Sf)+TPas small as possible.
(3) The stability is greater the greater the moment of
the forces which tend to restore the equilibrium when it
has been destroyed. Now this moment is
or supposing P and Q equal, it is
Hence to satisfy the third requisite this should be mado
as large as possible. This is, in part, at variance with the
112 BALANCES.
second requisite. The two requisites may however both be
satisfied by making (P + Q + 2$) h + Wk large, and a large
also ; that is, by increasing the distances of the fulcrum
from the beam and from the centre of gravity of the beam,
and by lengthening the arms.
174. It may be observed, that the sensibility of a
Balance is in general of more importance than the stability,
since the eye can judge pretty accurately whether the beam
makes equal oscillations on each side of the horizontal line ;
that is, whether the position of rest would be horizontal ;
if this be not the case, then the weights must be altered
until the oscillations are nearly equal. Accordingly in
practice the sensibility is secured at the expense of the
stability ; Tc is made small ; and h very small. In fact h
is usually zero, or is extremely small ; and thus the whole
of this investigation might be simplified.
175. Another kind of Balance is that in which the arms
are unequal, and the same weight is used to weigh different
substances, by varying its distance from the fulcrum. The
common Steelyard is of this kind.
176. To graduate the common Steelyard.
Let AB be the beam of the Steelyard, C the fulcrum.
Let A be the fixed point from which the substance to be
weighed is suspended. Let Q be the weight of the beam
together with the hook or scalepan at A ; let G be the
centre of gravity of these. Let P be a weight which can
be placed at any distance from the fulcrum. Suppose that
BALANCES. 113
the machine is in equilibrium with the beam horizontal
when P is suspended at N, and a substance of weight W
is suspended at A. Then, taking moments round C, we
have by Art. 86,
W. AC Q. CG P. CN=Q,
therefore TF=   P.
AC
On GC produced through C, take the point D such that
00=2(70; then
CN+CD
AC ~ AC
Now, let DB be graduated by taking on it from D
distances equal respectively to AC, 2 AC, 3 AC, 4AC,...:
and let the figures 1, 2, 3, 4,... be placed over the points of
graduation: these distances may also be subdivided if
necessary. Then, by observing the graduation at N, we
know the ratio of IF to P; and P being a given weight,
we know IF.
In this manner any substance may be weighed.
177. The sensibility of the common Steelyard is greater
the greater the distance between the two points at which
P must be suspended in order to balance two weights
of given difference. Hence it will follow that the sensi
bility is increased by increasing CA, or by diminishing P.
For suppose that N' denotes the point of suspension of the
moveable weight when the weight at A is W. Thus
P.Dir=W.AC,
and P.DN=W.AC;
therefore P.NN'=(W W} AC;
, (W'W)AC
therefore NN'=  ~^ .
T. ME. 8
114 BALANCES.
Now W  W is supposed to be given ; therefore NN'
A C 1
varies as p, and is increased by increasing AC, or by
diminishing P.
Since the sensibility varies as p it is independent of
the weight of the beam ; it is also independent of the posi
tion of JV, that is, a given Steelyard is equally sensible
whatever be the weight which is to be determined.
178. Another kind of Balance is called the Danish
Steelyard. This consists of a heavy beam which termi
nates in a knob at one end, and tho substance to be
weighed is placed at the other end j the fulcrum is move
able.
179. To graduate the Danish Steelyard.
Let AB be the beam; let
P be its weight, G its centre
of gravity.
Suppose that the machine is
in equilibrium with the beam
horizontal when the fulcrum is at (7, and a substance of
weight W is suspended at A. Taking moments round (7,
we have, by Art. 86,
W.AC=P.CG=P(AGAC),
therefore AC
Hence, making W=P, 2P, 3P, 4P,... successively, we
can mark on the beam the corresponding positions of the
fulcrum. If intermediate graduations are required they
must be determined by giving to W intermediate values,
as for example, g P, ^P, j^V.
It will be seen that if the successive values of W form
an Arithmetical Progression, the distances from A of the cor
responding graduations will form an Harmonical Progression,
EXAMPLES. XII. 115
EXAMPLES. XII.
1. If a substance be weighed in a Balance having un
equal arms, and in one scale appear to weigh a Ibs. and in the
other scale 6 Ibs., shew that its true weight is >J(db} Ibs.
2. A body, the weight of which is one lb., when placed
in one scale of a false Balance appears to weigh 14 ounces :
find its weight when placed in the other scale.
3. The arms of a Balance are in the ratio of 19 to 20 ;
the pan in which the" weights are placed is suspended from
the longer arm : find the real weight of a body which appa
rently weighs 38 Ibs.
4. If a Balance be false, having its arms in the ratio of
15 to 16, find how much per lb. a customer really pays for*
tea which is sold to him from the longer arm at 3s. 9c?.
per lb.
The next six questions relate to the common Steelyard :
5. The moveable weight for which the Steelyard is
constructed is one lb., and a tradesman substitutes a weight
of two Ibs., using the same graduations, thus giving his
customers a weight which he says is the same number of
times the moveable weight as before : shew that he defrauds
his customers if the centre of gravity of the Steelyard be
in the longer arm, and himself if it be hi the shorter arm.
6. The moveable weight is one lb., and the weight of
the beam is one lb.; the distance of the point of suspension
from the body weighed is 2^ inches, and the distance of
the centre of gravity of the beam from the body weighed is
3 inches : find where the moveable weight must be placed
when a body of 3 Ibs. is weighed.
7. If the fulcrum divide the beam, supposed uniform,
in the ratio of 3 to 1, and the weight of the beam be equal
to the moveable weight, shew that the greatest weight
which can be weighed is four tunes the moveable weight.
8. If the beam be uniform and its weight of the
M
moveable weight, and the fulcrum be  of the length of the
82
116 EXAMPLES. XII.
beam from one end, shew that the greatest weight which can
. . . . 2m(n l) + ?i2 ..
be weighed is  =* times the moveable weight.
2iin
9. Find what effect is produced on the graduations by
increasing the moveable weight.
10. Find what effect is produced on the graduations
by increasing the density of the material of the beam.
11. A straight uniform Lever whose weight is 50 Ibs.
and length 6 feet, rests in equilibrium on a fulcrum when a
weight of 10 Ibs. is suspended from one extremity : find the
position of the fulcrum and the pressure on it.
12. Two weights P and Q hang at the ends of a straight
heavy Lever whose fulcrum is at the middle point : if the
arms are both uniform, but not of the same weight, and the
system be in equilibrium, shew that the difference between
the weights of the arms equals twice the difference between
P and Q.
13. A uniform heavy rod AB, seven feet long, is sup
ported in a horizontal position between two pegs C and J) t
two feet apart, of which C is half a foot from the end A :
find the pressures on the pegs. If a force act upwards at a
distance of half a foot from the end JB, sufficient to remove
all pressure from the peg (7, shew that the pressure on the
peg D will be half of what it was before.
14. A bar of iron of uniform section and 12 feet long
is supported by two men, one of whom is placed at one end:
find where the other must be placed so that he may sustain
threefifths of the whole weight.
1 5. Two weights of 2 Ibs. and 5 Ibs. balance on a uniform
heavy Lever, the arms being in the ratio of 2 to 1 : find the
weight of the Lever.
16. If a heavy uniform rod c inches long be supported
on two props at distances a and b inches from the ends,
compare the pressures on the props.
17. A uniform heavy bar ten feet long and of given
weight W is laid over two props in the same horizontal
line, so that one foot of its length projects over one of the
EXAMPLES. XII. 117
props. Find the distance between the props so that the
pressure on the one may be double that on the other. Also
find the pressures.
18. A straight Lever weighing 20 Ibs. is moveable about
a fulcrum at a distance from one extremity equal to one
fourth of its length : find what weight must be suspended
from that extremity in order that the Lever may remain at
rest in all positions.
19. A bent Lever is composed of two straight uniform
rods of the same length, inclined to each other at 120, and
the fulcrum is at the point of intersection : if the weight of
one rod be double that of the other, shew that the Lever
will remain at rest with the lighter arm horizontal.
20. Two men carry a uniform beam 6 feet in length
and weighing 2 cwt., and at 2 feet from one end a weight of
1 cwt. is placed : if one man have this end resting on his
shoulder, find where the other man must support the beam
in order that they may share the whole weight equally.
21. A cylindrical bar of lead a foot in length and 8 Ibs.
in weight is joined in the same straight line with a similar
bar of iron 15 inches long and 6 Ibs. in weight : find the
point on which they will balance horizontally.
22. A uniform rod 10 feet long and 48 Ibs. in weight is
supported by a prop at one end : find the force which must
act vertically upwards at a distance of 2 feet from the other
end to keep the rod horizontal.
23. A straight uniform rod is suspended by one end :
determine the position in which it will rest when acted on
by a given horizontal force at the other end.
24. Two weights acting perpendicularly on a straight
uniform Lever at its ends on opposite sides of the fulcrum
balance : if one weight be double the other, and the weight
of the Lever equal to their sum, find where the fulcrum
must be.
25. If the common Steelyard be correctly constructed
for a moveable weight P, shew that it may be made a cor
rectly constructed instrument for a moveable weight nP
by suspending at the centre of gravity of the Steelyard a
weight equal to n  1 times the weight of the Steelyard.
118 TEE WHEEL AND AXLE.
XIII. The Wheel and Axle. The Toothed Wheel.
180. The present Chapter will be devoted to the Wheel
and Axle, and the Toothed Wheel. It will be seen that
these two Mechanical Powers are only modifications of the
Lever.
181. The Wheel and Axle. This machine consists
of two cylinders which have a common axis ; the larger
cylinder is called the Wheel, and the smaller the Axle.
The two cylinders are rigidly connected with the common
axis, which is supported in a horizontal position so that the
machine can turn round it. The Weight acts by a string
which is fastened to the Axle and coiled round it ; the
Power acts by a string which is fastened to the Wheel and
coiled round it. The Weight and the Power tend to turn
the machine round the axis in opposite directions.
182. When there is equilibrium on the Wheel and
Axle, the Power is to the Weight as the radius of the
Axle is to the radius of the WJieel.
Let two circles having the common centre C represent
sections of the Wheel and A^le respectively, made by planes
perpendicular to the axis of the cylinder.
THE WHEEL AND AXLE.
119
It may be assumed, that
the effects of the Power and
the Weight will not be
altered if we suppose them
both to act in the same
plane perpendicular to the
axis. Let the string by
which the power, P, acts
leave the "Wheel at A, and
the string by which the
weight, TF, acts leave the
Axle at B. Then CA and
CB will be perpendicular
to the lines of action of P
and W. We may regard
ACB as a Lever of which C is the fulcrum, and hence, by
Art. 165, the necessary and sufficient condition for equi
librium is
P_Z?
W~CA'
183. If we wish to take into account the thickness of
the strings by which P and W act, we may consider that
the line of action of each of these forces coincides with the
middle of the respective strings. Thus, in the condition of
equilibrium, CA will denote the radius of the Wheel in
creased by half the thickness of the string by which P
acts, and CB will denote the radius of the Axle increased
by half the thickness of the string by which W acts.
184. We have supposed that the Power in the Wheel
and Axle acts by means of a string ; but the Power may
act by means of the hand, as in the familiar example of the
machine used to draw up a bucket of water from a well.
A windlass and a capstan may also be considered as
cases of the Wheel and Axle.
The windlass scarcely differs from the machine used to
draw up water from a well : the windlass however has
more than one fixed handle for the convenience of working
it ; or it may have a moveable handle which can be shifted
from one place to another.
120 THE WHEEL AND AXLE.
In the capstan the fixed axis of the machine is vertical ;
the hand which supplies the Power describes a circle in a
horizontal plane, and the rope attached to what we call
the Weight leaves the Axle in a horizontal direction.
185. In the Wheel and Axle, as described in Art. 182,
the whole pressure on the fixed supports is equal to the
sum of the Weight and the Power ; for the machine re
sembles a Lever with parallel and like forces. If the
Power be directed vertically upwards, the Power and the
Weight being then on the same side of the axis of the
machine, the whole pressure on the fixed support is equal
to the difference of the Weight and the Power. In prac
tice however with the windlass and the capstan, although
P and W act at right angles to their respective radii,
they do not necessarily act in parallel directions : in such
cases the pressure on the fixed supports must be found
by the Parallelogram of Forces. See Art. 168.
186. The Toothed Wheel. Let two circles of wood or
metal have their circumferences cut into equal teeth at
equal distances. Let the circles be moveable about axes
perpendicular to their planes, and let them be placed with
their axes parallel, so that their edges touch, one tooth of
one circumference lying between two teeth of the other
circumference. If one of the wheels of this pair be turned
round its axis by any means, the other wheel will also be
made to turn round its axis. Or a force which tends to
turn one wheel round may be balanced by a suitable force
which tends to turn the other wheel round.
187. When there is equilibrium on a pair of Toothed
Wheels, the moments of the Power and the Weight about
the centres of their respective wheels are as the perpendi
culars from the centres of the wheels on the direction of
the pressure between the teeth in contact.
Let J/ and N be the fixed centres of the wheels.
Suppose the Power, P, and the Weight, W, to act by
strings which are attached to axles concentric with the
wheels. Let these strings leave the axles at A and B
respectively. Then MA and NB will be perpendicular to
the lines of action of P and W.
THE WHEEL AND AXLE.
121
Let Q denote the mutual pressure at the point of con
tact of the teeth j so that a force Q acts at the point of
contact in opposite directions on the two wheels. Draw
perpendiculars from M and N on the line of action of Q t
meeting it at m and n respectively.
Then, since the wheel which can turn round M is in
equilibrium, the moments round M must be equal; that is,
Similarly, since the wheel which can turn round ^Vis in
equilibrium,
,,  PxAM QxMm Mm
Iheretore =57 ^_ = ^ s= ;
W x BN Q x Nn An
this establishes the proposition.
Draw the straight line MN meeting mn at : then,
by similar triangles,
_
Nn~ NO*
If the teeth are very small compared with the radii of
the wheels, will nearly coincide with the point of contact
122 EXAMPLES. XIII.
of the teeth, and NO and NO will be nearly the radii of
the wheels. Thus we have very nearly
moment of P round M _ radius of Powerwheel
moment of W round N~ radius of Weightwheel *
188. In practice the machine is used to transmit mo
tion ; and then it is necessary to pay great attention to the
form of the teeth, in order to secure uniform action in the
machine, and to prevent the grinding away of the surfaces :
on this subject however the student must consult works
which treat specially of mechanism.
Toothed Wheels are extensively applied in all ma
chinery, as in cranes and steam engines, and especially
in watchwork and clockwork.
189. Wheels are sometimes turned by means of straps
passing over their circumferences : in such cases the minute
protuberances of the surfaces prevent the sliding of the
straps. The strap passing partly round a wheel exerts a
force on the wheel at both points where it leaves the wheel :
the effect at each point would be measured by the moment
of the tension of the strap round the centre of the wheel.
If it were not for friction and the weight and stiffness of
the strap the tension would be the same throughout ; and
so the action at one point of the wheel would balance the
action at the other point. The subject of friction will be
considered in Chapter XIX.
EXAMPLES. XIII.
1. Find the radius of the Wheel to enable a Power of
llbs. to support a Weight of 281bs., the diameter of the
Axle being 6 inches.
2. Find what Weight suspended from the Axle can be
supported by 3 Ibs. suspended from the Wheel, if the radius
of the Axle is l feet, and the radius of the Wheel is 3
feet.
3. A man whose weight is 12 stone has to balance by
his weight 15 cwt. : shew how to construct a Wheel and
Axle which will enable him to do this.
EXAMPLES. XIII. 123
4. A Weight of 14 ounces is supported by a certain
Power on a Wheel and Axle, the radii being 28 inches and
16 inches respectively : if the radii were each shortened by
4 inches, find what Weight would be supported by the same
Power.
5. If the radius of the Wheel be to the radius of the
Axle as 8 is to 3, and two weights of 61bs. and 15lbs.
respectively be suspended from the circumferences of the
Wheel and Axle, find which weight will descend.. Suppos
ing that the weight which tends to descend is supported
by a prop, find the pressure on the prop and on the fixed
supports of the Wheel and Axle.
6. The radius of the Axle of a capstan is 2 feet, and
six men push each with a force of one cwt. on spokes
5 feet long : find the tension they will be able to produce
in the rope which leaves the Axle.
7. The difference of the diameters of a Wheel and
Axle is 2 feet 6 inches ; and the Weight is equal to six
times the Power : find the radii of the Wheel and the
Axle.
8. The radius of the Wheel being three times that of
the Axle, and the string on the Wheel being only strong
enough to support a tension of 361bs., find the greatest
Weight which can be raised.
9. If the string to which the Weight is attached be
coiled in the usual manner round the Axle, but the string
by which the Power is applied be nailed to a point in the
rim of the Wheel, find the position of equilibrium, the
Power and the Weight being equal.
10. In the Wheel and Axle if the two ropes were coiled
each on itself, and their thickness not neglected, find
whether the ratio of the Power to the Weight would be
increased or diminished as the Weight was raised, sup
posing the ropes of the same thickness.
124
THE FULLY.
XIV. The Pully.
190. The Pully consists of a small circular plate or
wheel which can turn round an axis passing through the
centres of its faces, and having its ends supported by a
framework which is called the Block. The circular plate
has a groove cut in its edge to prevent a string from slip
ping off when it is put round the Pully.
191. Let A denote a Pully the
Block of which is fixed ; and suppose
a Weight attached to the end of a
string passing round the Pully. If
the string be pulled at the other end
by a Power equal to the Weight there
will be equilibrium.
Thus a fixed Pully is an instru
ment by which we change the direc
tion of a force without changing its
magnitude. We have already ad
verted to this in Art. 28.
As we proceed with the present Chapter it will be seen
that by the use of a moveable Pully we can gam mechan
ical advantage.
Theoretically the fact that the Pully can turn round
its axis is not important ; but practically it is very im
portant. When the Pully can turn round it is found that
the tension of the string is almost exactly the same on
both sides of the Pully in the condition of equilibrium.
But when the Pully cannot turn round it is found that
there may be considerable difference between the tensions
of the two parts of the string : this is owing to Friction,
which we shall consider hereafter.
In all that follows we shall assume that the tension of
a string is not changed when the string passes round a
Pully. We shall always neglect the weight of the strings ;
and also the weight of the Pullies unless the contrary be
stated.
THE FULLY.
125
192. In a single moveable Pully with the strings
parallel when tJiere is equilibrium the Weight is twice
the Power.
Let a string pass round the Fully
A, have one end fixed, as at A",
and be pulled vertically upwards by a
Power, P, at the other end.
Let a Weight, W, be attached to
the Block of the Fully.
The tension of the string is the
same throughout. Hence we may
regard the Pully as acted on by two
parallel forces, each equal to P, up
wards, and by the force W downwards. Therefore W=2P.
It may be observed that the line of action of W must
be equally distant from the two parts of the string ; that
is, it must pass through the centre of the Pully.
The pressure on the fixed point K is equal to P, that
is, to \ W.
The string by which P acts sometimes for convenience
passes over a fixed Pully, and P acts downwards.
193. The preceding Article will probably present no
difficulty to the student ; but perhaps the following remarks
should be made. The Wheel and the Block of the Pully
are really two distinct bodies; but when there is equi
librium we shall not disturb it by rigidly connecting the
two bodies : thus we obtain one rigid body, and the condi
tion of equilibrium follows by Art. 62. Sometimes the
principle of the Lever is employed in obtaining this condi
tion of equilibrium ; the strict mode of employing the
principle is as follows : The Wheel of the Pully is capable
of turning round its axis, and for equilibrium the moments
of the forces round this axis must be equal ; this condition
is satisfied if the axis be equidistant from the two parts of
the string. The pressure on the axis is equal to the sum
of the two forces ; and this pressure is supported by the
Block. Thus the Block is acted on by 2P upwards, and
by IF downwards. Therefore W=2P
126 THE FULLY.
We may add that the action of the string in the pre
ceding Article is best explained, for elementary purposes,
by supposing all that part which is in contact with the
Fully to become rigidly connected with it ; thus we are left
with a rigid body of Weight W supported by the tensions
of two strings acting at the points where the string leaves
the Fully.
194. To find the ratio of the Power to the Weight
in the single moveable Fully with the parts of the string
not parallel.
Let a string pass round
the Fully, A, have one end
fixed, as at K, and be pulled
by a Power, P, at the other \
end. Let a weight, W, be
attached to the block of
the Fully.
The tension of the string
which passes round the
Fully is the same through
out. Hence we may regard
the Fully as acted on by two forces, each equal to P, and
a force W. Therefore the line of action of W must bisect
the angle formed by the lines of action of the two forces
Pi that is, the two parts of the string must be equally
inclined to the vertical. Suppose them each to make an
angle a with the vertical. Then W is equal and opposite
to the resultant of two equal forces P, which are inclined
at an angle 2a. Thus the ratio of P to W is known by the
Parallelogram of Forces. By Art. 30, we have
TF=2Pcosa.
195. We now pass on to investigate the conditions of
equilibrium of various combinations of Pullies.
196. In the system of Pullies in which each Fully
hangs by a separate string and all the strings are parallel^
when there is equilibrium the Weight is equal to the Power
multiplied by 2", where n is the number of Pullies.
In this system the string which passes round any Fully
except the highest has one end attached to a fixed point,
TEE PULLY.
127
and the other end attached to the block of the next higher
Pully ; the string which passes round the highest Pully has
one end attached to a fixed point, and the other end sup
ported by the Power.
Suppose there are four moveable Pullies. Let W de
note the weight, which is suspended from the block of the
lowest Pully ; and P the Power which
acts vertically upwards at the end
of the string which passes under the
highest Pully.
By the principle of the single
moveable Pully, the tension of the
string which passes under the lowest
W
Pully is ; the tension of the
string which passes under the next
W
Pully is half of this, that is ^ j the
tension of the string which passes
under the next Pully is half of this,
W
that is xj ; the tension of the string which passes under
W
the next Pully is half of this, that is ^. This last tension
must be equal to the Power which acts at the end of the
W
string. Therefore P=^ t or Tf=2 4 P.
Similarly, if there be any number of moveable Pullies
and n denote this number, W=Z"P.
This system of Pullies is sometimes called the First
System of Pullies.
197. Let K, L, N, N denote the points at which the
ends of the strings are fixed in the system of Pullies con
sidered in the preceding Article. Then the pressure at
W W W
K is , the pressure at L is ^g, the pressure at M is ^j,
v/
TF
the pressure at N is ^.
Hence the sum of these pres
128
THE PULLY.
sures is TF^ + ^ + i + 1 J ; b 7 summing the Geo
metrical Progression, we find that this is TFM. A
Thus the sum of these pressures together with the Power is
equal to the whole Weight.
198. In the system of Putties in which the same string
passes round all the Putties and the parts of it between
the Putties are parallel, when there is equilibrium the
Weight is equal to the Power multiplied by the number
of parts of the string at the lower block.
Suppose there are four parts of
the string at the lower block.
Let W denote the weight which
is suspended from the lower block ;
and P the power which acts verti
cally downwards at one end of the
string. The tension of the string is
the same throughout and is equal to
P ; thus we may regard the lower
block as acted on by four parallel
forces each equal to P upwards, and
by the force IF downwards. There
fore
TF=4P.
Similarly, if there be any number of parts of the string
at the lower block, and n denote this number, W=nP.
This system of Pullies is sometimes called the Second
System of Pullies.
199. In the figure one end of the string is fastened to
the upper block, and the number of parts of the string at
the lower block is even ; if one end of the string is fastened
to the lower block the number of parts of the string at the
lower block will be odd.
In the figure there are five parts of the string at the
upper block, so that the pressure at the fixed point K is
5P, that is JF+P.
THE FULLY.
129
200. In the system of Putties in which each string is
attached to the Weight, and all the strings are parallel,
when there is equilibrium the Weight is equal to the
Power multiplied by 2  1, where n is the number of
Putties.
In this system the string which passes round any Fully
except the lowest has one end attached to the Weight, and
the other end attached to the block of the next lower Polly ;
the string which passes round the lowest Fully has one end
attached to the Weight, and the other end supported by
the Power. The highest Pully is fixed; the others are
moveable.
Suppose there are four Pullies.
Let W denote the Weight to which
all the strings are attached ; and P
the Power which acts vertically
downwards at the end of the string
which passes over the lowest Pully.
The tension of the string which
passes over the lowest Pully is P ;
the tension of the string which passes
over the next Pully is 2P ; the ten
sion of the string which passes over
the next Pully is twice this, that is
2 2 P; the tension of the string which
passes over the next Pully is twice
this, that is 2 3 P.
The Weight is equal to the sum of these tensions by
Art. 111. Thus
W=P + 2P + 2 2 P + 21P=P (1 + 2 + 2 2 + 2 3 ) = P (2 4  1).
Similarly, if there be any number of Pullies, and n
denote this number, W= P (2"  1).
This system of Pullies is sometimes called the Third
System of Pullies.
T. ME. 9
130 THE FULLY.
201. The pressure at the fixed point K is 2 x 2 3 P, that
is2 4 P, that is W+P.
202. We have hitherto neglected the weights of the
Pullies ; but it is easy to take account of them, and we
shall now do so.
203. In Art. 192 let w denote the weight of the Fully ;
we have only to put W+w instead of W in the condition
of equilibrium. Thus W+w=2P.
Similarly, in Art. 194, TF"+w=2Pcosa.
In Art. 198 let w denote the whole weight of the Pullies
at the lower block ; then W+w=4P.
204. In Art. 196 let the weights of the four Pullies,
beginning with the lowest, be w, x, y, z respectively.
Then, the tension of the string which passes round the
lowest Pully is  ( W+ w) \ the tension of the next string
2i
is ^( W+ W) + JT x ; the tension of the next string is
+  2 # + y ; the tension of the next string is
Thus P + J +  +  + .
In fact, here P supports simultaneously W and w by
the aid of four Pullies, x by the aid of three Pullies, y by
the aid of two Pullies, and z by the aid of one Pully.
If the weight of each Pully is the same, and equal to w,
we have
THE FULLY. 131
Similarly, if n denote the number of Pullies, and the
weight of each Pully be w,
TF /. 1\ ^ = Ww
2* '
205. In Art. 200 let w denote the weight of the low
est Fully, x that of the next, y that of the next. Then
the tension of the string which passes over the lowest
Fully is P ; the tension of the next string is 2P + w ; the
tension of the next string is 2 2 P + 2w + x ; the tension of
the next string is 2 3 P + 2 2 w + 2x+y.
Thus Tr=P
=P (2*  1) + w (2 3  1) + x (2 2  1) +y.
In fact here TF is supported by the simultaneous
action of P with the aid of four Pullies, w with the aid of
three, x with the aid of two, and y with the aid of one.
If the weight of each Fully is the same and equal to
w, we have
Similarly, if n denote the number of Pullies, and the
weight of each Fully is w,
W= P (2"  1) + w (2  n  1).
206. If we take the weights of the Pullies into account
the expressions for the pressures on the fixed points which
have been given in preceding Articles will require some
alteration. There will be no difficulty in making these
alterations ; we will take Art. 197 as an example.
With the notation of Art. 204, the pressure at K is
2 ( TF+ 20) ; the pressure at L is ^ ( ^+ w ) + o 5 ^ ne pres
1 xv
sure at M is ^ ( TF+ w) + ^ + 1 > *^ e pressure at N is
92
132 THE PULLY.
Hence the sum of these pressures
207. There are two systems of Pullies which are usu
ally called Spanish Bartons : they will be understood from
the annexed figures.
First, neglect the weights of the Pullies.
In the lefthand figure, W=4P. The pressure at the
fixed point K is P, and at the fixed point L is 4P.
In the righthand figure, denoting by 2a the angle be
tween the parts of the string round the upper moveable
Fully, W= 5P cos a. The pressure at the fixed point K is
2P cos a, and at the fixed point L is 4P cos a.
Next, let w denote the weight of the lower moveable
Fully, and x that of the upper moveable Fully.
In the lefthand figure W+w=4P + x. The pressure
at the fixed point K is P, and at the fixed point L is
4P + 2z together with the weight of the fixed Fully.
EXAMPLES. XIV. 133
In the righthand figure W+w=5F cos a + Zx. The
pressure at the fixed point K is 2P cos a + x, and at the
fixed point L is 4P cos a + 2.T together with the weight of
the fixed Fully.
208. It will be seen that in every system of Pullies we
find the condition of equilibrium by beginning at one end
of the system and determining in order the tensions of all
the strings of the system. We have always begun with
the Power end, except in Art. 196; and in that Article we
might also have begun with the Power end.
EXAMPLES. XIV.
1. In a single moveable Pully if the weight of the
Pully be 2 Ibs., find the force required to raise a Weight of
4lbs.
2. If there be two strings at right angles to each other
and a single moveable Pully, find the force which will sup
port a Weight of ^2 Ibs.
3. A man stands in a scale attached to a moveable
Pully, and a rope having one end fixed passes under the
Pully, and then over a fixed Pully : find with what force the
man must hold down the free end in order to support him
self, the strings being parallel.
4. If on a Wheel and Axle the mechanical advantage
be six times as great as on a single moveable Pully, com
pare the radii of the Wheel and the Axle.
The following ten Examples relate to the First System
of Pullies; see Art. 196:
5. If n=6, and P=28 Ibs., find W.
6. If TF=4 Ibs., and P=\ ounce, find n.
7. If w=3, find the consequence of adding one ounce
to P, and ten ounces to W.
8. If a man support a Weight equal to his own, and
there are three Pullies, find his pressure on the floor on
which he stands.
134 EXAMPLES. XIV.
9. If there are three Pullies, each weighing one lb.,
find the Power which will support a Weight of 17 Ibs.
10. If there are three Pullies of equal weight, find the
weight of each in order that a Weight of 56 Ibs. attached
to the lowest Pully may be supported by a Power of
7 Ibs. 14 ounces.
11. If the weight of each Pully is P, find W and the
tension of each string.
12. If there are three Pullies each of weight w, and
W= P, find W.
13. If there are two Pullies each of weight 4^0, and the
Power be 3w, shew that no Weight can be supported by
the system.
14. In a system of three Pullies if a weight of 5 Ibs.
is attached to the lowest, 4 Ibs. to the next, and 3 Ibs. to
the next, find the Power required for equilibrium.
The following six Examples relate to the Second System
of Pullies; see Art. 198:
15. Find the number of parts of the string at the lower
block in orcjer that a Power of 4 ounces may support a
Weight of 4 Ibs.
. 16. Find the number of Pullies at the lower block if
P=12 stone and W=IS cwt.
17. If there are four parts of the string at the lower
block, find the consequence of adding one ounce to P, and
three ounces to W.
18. If there are six parts of the string at the lower
block, find the greatest Weight which a man weighing 10
stone can possibly raise.
19. A man supports a Weight equal to half his own
weight ; if there are seven parts of the string at the lower
block, find his pressure on the floor on which he stands.
20. Find what Weight can be supported if there are
three Pullies at the lower block, the string being fastened
to the upper block, and the weight of the lower block being
three times the Power.
THE INCLINED PLANE. 135
XV. The Inclined Plane.
209. An Inclined Plane in Mechanics is a rigid plane
inclined to the horizon.
When an Inclined Plane is used as a Mechanical Power
the straight lines in which the Power and the Weight act
are supposed to lie in a vertical Plane perpendicular to
the intersection of the Plane with the
horizon. Thus the Inclined Plane is
represented by a rightangled tri
angle, such as AGB\ the horizontal
side Ada called the base; the verti
cal side CB is called the height ; and
the hypotenuse AB is called the
length. The angle BAG is the incli
nation of the Plane to the horizon.
The Plane is supposed perfectly rigid, and, unless the
contrary be stated, it is supposed to be perfectly smooth;
so that the Plane is assumed to be capable of supporting
any amount of pressure which is exerted against it in a
perpendicular direction.
210. If a Weight be supported on an Inclined Plane
at a point L, and LM be drawn in the direction of the
Power, and LN at right angles to the Plane, so as to meet
a vertical line at M and N, the Power is to the Weight as
LM is to MN.
Let BAG be the Inclined Plane.
Let a heavy body whose weight is W
be placed on it at any point L, and
be kept at rest by a Power, P, acting
in the direction LM. Let MN be
drawn vertically downwards, and LN
at right angles to the Plane.
The body at L is acted on by
three forces ; the Power P in the
direction LM, its own Weight in a di
rection parallel to MN, and the resist
ance of the Plane in the direction lYL.
136 THE INCLINED PLANE.
Hence, by Art. 36, since there is equilibrium, the sides
of the triangle LMN are respectively proportional to the
forces. Therefore
P _LM
W MN'
Let R denote the resistance of the Plane ; then
R L _NL
W~MN'
We may write these results thus :
P : W : R :: LM : MN : NL.
It is usual to consider separately two special cases of
the general proposition; and this we shall do in the next
two Articles.
211. When there is equilibrium on the Inclined
Plane, and the Power acts along the Plane, the Power is
to the Weight as the height of the Plane is to the length.
Let W denote the Weight of
a heavy body, and P the Power.
From any point L in the Plane, draw
LN&i right angles to the Plane, meet
ing the base at N; and draw NM
vertical, meeting the Plane at N.
Then the sides of the triangle LMN are parallel to the
directions of the forces which keep the heavy body at rest ;
therefore, by Art. 36,
P _LM
W~MN'
But the triangle LMN is equiangular to the triangle
CBA ; for the angle LMN is equal to the angle A J5C, by
Euclid, i. 29 ; the right angle NLM is equal to the right
angle ACB; and therefore the third angle MNL is equal
to the third angle BAG.
THE INCLINED PLANE. 137
Hence, by Euclid, vi. 4,
LM CB
MN~BA'
^ * P OB
W = BA'
Let jR denote the resistance of the Plane ; then
W~ AB'
We may write these results thus :
P : W :E r.CB : BA .AC.
212. When there is equilibrium on the Inclined Plane,
and the Power acts horizontally, the Power is to the Weight
as tJie height of the Plane is to the base.
Let W denote the Weight of
a heavy body, and P the Power.
From any point L in the Plane draw
LN at right angles to the Plane,
meeting the base at N; and draw
NM vertical meeting at M the hori
zontal straight line drawn through L.
Then the sides of the triangle LMN are parallel to the
directions of the forces which keep the heavy body in
equilibrium ; therefore, by Art. 36,
P__LM
W~ MN'
But the triangle LMN is equiangular to the triangle
BAG. For the angle BLN, being a right angle, is equal
to the sum of the two angles BAG and ABC', and BLM
is equal to BAC, by Euclid, I. 29 : therefore MLN is equal
to ABC. And the right angles LMN and BCA are
equal. Therefore the third angle LNM is equal to the
third angle BA C.
138 THE INCLINED PLANE.
Hence, by Euclid, vi. 4,
LM _BC
MN~CA'
therefore W^&L'
Let E denote the resistance of the Plane ; then
n_AE_
W CA'
We may write these results thus :
P : W:E ::BC :CA : AS.
213. We may obtain convenient expressions for the
proportionate values of P, FT, and E by the aid of Trigo
nometry.
Let the angle JBAC=a, and the
angle MLB = ft therefore the angle
J/Z,:
Then
P : W : E :: LM : MN : NL
:: sin LNM : sin MLN : sin NML
:: sin a : sin (90 + /3) : sin (90  a  0)
:: sin a : cos /3 : cos (a + /3).
214. In the figure of the preceding Article the re
sistance of the Plane is represented by the straight line
NL ; that is, the resistance acts from N towards L. Thus
if the body be placed on the Plane, and be in equili
brium, the straight line LN must be below the Plane ; that
is, the sum of the angles MLB and BAG must be less than
a right angle.
215. The results of Art. 213 may also be obtained by
the method of resolving the forces given in Arts. 56, 57 ; and
thus we obtain a good example of the method, and assis
tance in remembering the results.
EXAMPLES. XV. 139
Resolve the forces along the Plane : this gives
Pcos/3=TFsin a.
Resolve the forces at right angles to the Plane : this
gives
Hence we deduce
E W cos a Waiu a sin = Wcoa ( a + ^
cos cos j3
The general formulae of course include the particular
cases of Arts. 211 and 212 :
When the Power acts along the Plane =0 ; then
P=TFsina, JR=JFcosa.
When the Power acts horizontally = a ; then
P=JFtana, J2=IFseca.
216. Perhaps it may seem that an Inclined Plane can
scarcely be called a Machine; it is not obvious that it can
be usefully employed like the other Mechanical Powers.
But we may observe that if we have to raise a body we
may draw it up an Inclined Plane by means of a Power
which is less than the Weight of the body.
EXAMPLES. XV.
In the following twelve Examples the Power is sup
posed to act along the Plane :
1. If the Weight be represented by the height of the
Plane, shew what straight line represents the pressure on
the Plane.
2. If 17= 12 Ibs., and the height of the Plane be to
its base as 3 is to 4, find P.
3. If W= 10 Ibs. and P= 6 Ibs., find E.
4. If P=R, find the inclination of the Plane, and tho
ratio of P to IF.
5. If P is to R as 3 is to 4, express each of them
in terms of IF.
140 EXAMPLES. XV.
6. If P=9 Ibs., find W when the height of the Plane
is 3 inches, and the base 4 inches.
7. An Inclined Plane rises 3 feet 6 inches for every
5 feet of length : if JF=200, find P.
8. If the length of the Plane be 32 inches, and the
height 8 inches, find the mechanical advantage.
9. When a certain Inclined Plane ABC, whose length
is ABj is placed on AC as base, a Power of 3 Ibs. can
support on it a Weight of 5 Ibs. : find the Weight which the
same Power could support if the Plane were placed on BG
as base, so that AC is then the height of the Plane.
10. A railway train weighing 30 tons is drawn up an
Inclined Plane of 1 foot in 60 by means of a rope and
a stationary engine ; find what number of Ibs. at least the
rope should be able to support.
11. A Weight of 20 Ibs. is supported by a string
fastened to a point in an Inclined Plane, and the string is
only just strong enough to support a Weight of 10 Ibs. :
the inclination of the Plane to the horizon being gradually
increased, find when the string will break.
12. If it takes twice the Power to support a given
Weight on an Inclined Plane ABC when placed on the
side AC, that it does when the Plane is placed on the side
BC, find the greatest Weight which a Power of one Ib. can
support on the Plane.
In the following four Examples the Power is supposed
horizontal :
13. If TF=12 Ibs., and the base be to the length as
4 is to 5, find P.
14. If TF=48 Ibs., and the base be to the height as
24 is to 7,,find P and R.
15. If R=2 Ibs., and P=l Ib., find TF, and the incli
nation of the Plane.
16. If W= 12 Ibs., and P= 9 Ibs., find R.
EXAMPLES. XV. 141
17. If the Power which will support a Weight when
acting along the Plane be half that which will do so acting
horizontally, find the Inclination of the Plane.
18. If E be the pressure on the Plane when the
Power acts horizontally, and 8 when it acts parallel to the
Plane, shew that BS= IF 2 .
19. A Power P acting along a Plane can support W t
and acting horizontally can support x : shew that
P*=W*x 2 .
20. A Weight W would be supported by a Power P
acting horizontally, or by a Power Q acting parallel to the
Plane : shew that
. 21. Give a geometrical construction for determining
the direction in which the Power must act when it is equal
to the Weight, but does not act vertically upwards ; and
shew that if S be the pressure on the Plane in this case,
and R the pressure when the Power acts along the Plane,
S=2B.
22. The length of an Inclined Plane is 5 feet, and the
height is 3 feet. Find into what two parts a Weight of
104 Ibs. must be divided, so that one part hanging over the
top of the Plane may balance the other part resting on
the Plane.
23. The inclination of a Plane is 30; a particle is
E laced at the middle point of the Plane, and is kept at rest
y a string passing through a groove in the Plane, and
attached to the opposite extremity of the base : shew that
the tension of the string is equal to the Weight of the
particle.
24. A Weight of 20 Ibs. is supported by a Power of
12 Ibs. acting along the Plane : shew that if it were required
to support the same Weight on the same Plane by a Power
acting horizontally, the Power must be increased in the
ratio of 5 to 4, while the pressure on the Plane will be
increased in the ratio of 25 to 16.
142 THE WEDGE. THE SCREW.
XVI. The Wedge. The Screw.
217. The Wedge. The Wedge is a solid body in the
form of a prism ; see Euclid, Book xi. Definitions. In
the Wedge two parallel faces are equal and similar tri
angles, and there are three other faces which are rect
The Wedge may be employed to
separate bodies.
We may suppose the Wedge
urged forward by a force P acting
on one of the rectangular faces,
and urged backwards by two re
sistances Q and R on the other rect
angular faces arising from the bodies
which the Wedge is employed to se
parate. These forces may be supposed to act in one plane
perpendicular to the rectangular faces ; and we shall assume
that the Wedge and the bodies are smooth, so that the force
acting on each face is perpendicular to that face.
Let the triangle ABC represent a section of the Wedge
made by a plane perpendicular to its rectangular faces ;
and suppose the Wedge kept in equilibrium by the forces
P, Q, R perpendicular to AB y BC, CA respectively : then
by Art. 37
P :Q:Rr.AB:BC :CA.
If AC=BC the Wedge is called an isosceles Wedge ; in this
case Q=R, and P : R :: AB : CA.
Let the angle ACE be denoted by 2a, then when the Wedge
is isosceles AB= 2 A C sin a, and
P : R :: 2sina : 1, so that P=2sino.
218. There is very little value or interest in the pre
ceding investigation, because the circumstances there sup
posed scarcely ever occur in practice. A nail is sometimes
THE WEDGE. THE SCREW.
143
given as an example of the Wedge, but when the nail is at
rest the resistances on its sides are counterbalanced by
friction and not by a Power on the head. The nail is
indeed driven into its place by blows on the head ; but it
does not belong to Statics to investigate the effect of blows
in producing motion.
219. The Screw. The Screw consists of a right cir
cular cylinder AB with a
uniform projecting thread
abed,.. traced round its sur
face, making a constant an
gle with straight lines pa
rallel to the axis of the
cylinder. This cylinder fits
into a block C pierced with
an equal cylindrical aper
ture, on the inner surface
of which is cut a groove the
exact counterpart of the
projecting thread abed...
Thus when the block is
fixed and the cylinder is in
troduced into it, the only manner in which the cylinder can
move is backwards or forwards by turning round its axis.
220. In practice the forms of the threads of Screws
may vary, as we see exemplified in the accompanying two
figures.
The lefthand figure
most nearly resembles that
which is taken for investi
gation in elementary books ;
it is usual to disregard the
thickness and the breadth
of the projecting thread,
that is to consider both of
these as practically very
small. We may form a good
idea of the figure of the
thread in the following geometrical manner ;
144
THE WEDGE. THE SCREW.
Let ABNM be any rectangle.
Take any point C in BN, and make
CD, DE, EF,... all equal to BC.
Join CA and through D, E, F,... draw
straight lines parallel to CA, meeting
AM at c, d, e,... Then if we conceive
ABNM to be formed into the convex
surface of a right cylinder, the straight
lines AC, cD, dE, eF,... will form the
curve which determines the figure of the
Screw.
Let the angle CAB be denoted by
a; then CB = AB tan a; if r be the A B
radius of the right circular cylinder and ?r express as usual
the ratio of the circumference of a circle to its diameter,
A=2irr; thus CJ3=27rriana. CB is the distance be
tween two consecutive threads of the Screw measured
parallel to the axis. The angle a may be called the angle
of the Screw.
221. Suppose the axis of the cylinder to be vertical;
and let a Weight IF be placed on the Screw. Then the
Screw would descend unless prevented by some Power, P.
This Power we shall suppose to act at the end of a
horizontal arm firmly attached to the cylinder ; the direc
tion of the Power being horizontal and at right angles to
the arm : the length between the axis of the cylinder and
the point of application of the Power we shall call the
Powerarm.
222. When there is equilibrium on the Screw the
Power is to the Weight as the distance between two adja
cent threads is to the circumference of a circle having the
Powerarm for radius.
Let r be the radius of the cylinder, b the length of
the Powerarm, a the angle of the Screw. The Screw is
acted on by the Weight IF, the Power P, and resistances
exerted by the block. These resistances act at various
points of the block, but as the thread is supposed smooth
they all act at right angles to the thread ; thus their direc
THE WEDGE. THE SCREW. ]45
tions all make an angle a with vertical straight lines.
Denote these resistances by R, S, T,... Resolve each re
sistance into two components, one vertical and the other
horizontal. Thus the vertical components are R cos a,
/S'cosa, 7* cos a,...; and the horizontal components are
R sin a, Ssin a, Tsin a,...
By reasoning as in Arts. 104 and 105 we find that there
are two conditions which must hold when the machine is
in equilibrium, namely :
The components parallel to the axis must balance each
other, thus
and the moments of the forces round the axis must
balance each other, thus
Hence, by division,
Pb _ r sin a t
~W~~ cos a '
, i  P r sin a 2?rr tan a
therefore ^=7  = ;n
W b cos a 27T&
distance between two consecutive threads
circumference of circle of radius b
223. The most common use of a Screw is not to sup
port a Weight, but to exert a pressure. Thus suppose a
fixed bar above the body denoted by W in the figure of
Art. 219; then, by turning the Screw, the body will be com
pressed between the head of the Screw and the fixed bar.
A bookbinder's press is an example of this mode of using
a Screw. The theory of the machine will be the same as in
Art. 222 ; W will now denote the pressure exerted parallel
to the axis of the Screw by the body which is compressed.
T. ME. 10
146 EXAMPLES. XVI.
EXAMPLES. XVI.
1. A Wedge is rightangled and isosceles, and a force
of 50 Ibs. acts opposite to the right angle : determine the
other two forces.
2. A Wedge is in the form of an equilateral triangle,
and two of the forces are 40 Ibs. each : find the third force.
3. Find the vertical angle of an isosceles Wedge when
the pressure on the face opposite this angle is equal to
half the sum of the two resistances.
4. The tangent of the angle of a Screw is , the radius
of the cylinder 4 inches, and the length of the Powerarm
2 feet : find the ratio of W to P.
5. The circumference of the circle corresponding to the
point of application of P is 6 feet : find how many turns
the Screw must make on a cylinder 2 feet long, in order
that TFmay be equal to 144 P.
6. The distance between two consecutive threads of a
Screw is a quarter of an inch, and the length of the Power
arm is 5 feet: find what Weight will be sustained by a
Power of 1 Ib.
7. The angle of a Screw is 30, and the length of the
Powerarm is n times the radius of the cylinder : find the
mechanical advantage.
8. The length of the Powerarm is 15 inches : find the
distance between two consecutive threads of the Screw,
that the mechanical advantage may be 30.
9. A Screw having a head 12 inches in circumference
is so formed that its head advances a quarter of an inch
at each revolution : find what force must be applied at the
circumference of the head that the Screw may produce a
pressure of 96 Ibs.
10. If a Screw makes m turns in a cylinder a foot long,
and the length of the Powerarm is n feet, find the me
chanical advantage.
COMPOUND MACHINES. 1 47
XVII. Compound Machines.
224. We have already spoken of the mechanical
advantage of a machine, and have denned it to be the ratio
of the Weight to the Power when the machine is in equi
librium; see Art. 166. Now we might theoretically obtain
any amount of advantage by the use of any mechanical
power. For example, in the Wheel and Axle the advantage
is expressed by the ratio of the radius of the Wheel to the
radius of the Axle ; and this ratio can be made as great as
we please : but practically if the radius of the Axle be too
small the machine is not strong enough for use, and if the
radius of the Wheel be too great the machine becomes of
an inconvenient size.
Hence it is found advisable to employ various compound
machines, by which great mechanical advantage may be
obtained combined with due strength and convenient size.
We will now consider a few of these compound machines.
225. Combination of Levers.
K B L O M
T
Let AB, BC, CD be three Levers, having fiilcrums at
K) L, M respectively. Suppose all the Levers to be hori
zontal, and let the middle Lever have each end in contact
with an end of one of the other Levers. Suppose the system
in equilibrium with a Power, P, acting downwards at A,
and a Weight, TF, acting downwards at D.
Let Q be the pressure at B between the two Levers
which are in contact there, and R the pressure at C between
the two Levers which are in contact there ; these pressures
may be supposed to act vertically.
102
148 COMPOUND MACHINES.
v B L
If
D
Since the Lever AKB is in equilibrium  = wirl
wir
7? 7? T
since the Lever BLC is in equilibrium ^ = ^ ;
W CM
and since the Lever CMD is in equilibrium  =
Hence, by multiplication, = x x
Hence the mechanical advantage of the combination of
Levers is equal to the product of the mechanical advantages
of the component Levers.
The result holds even if Q and jR do not act vertically.
Suppose for example that Q does not act vertically, but in
some other direction ; let k and I denote the lengths of the
perpendiculars from K and L on this direction. Then
Q AK E I n , r en
we have p = T 75 = 777 But by similar triangles
T = r ; and thus the value of ^ is the same as before.
Ic JJh. r
226. Combinations of Wheels and Axles are frequently
used. The Wheel of one component is made to act on the
Axle of the next component by means of teeth, or of a
strap. It may be shewn that the mechanical advantage of
COMPOUND MACHINES. 149
the combination is the product of the mechanical advan
tages of the components ; that is, we shall have
W _ Product of the radii of the Wheels
P ~ Product of the radii of the Axles *
227. The Differential Axle, or Chinese Wheel.
This machine may L. . ,i
be considered as a
LL
combination of the
Wheel and Axle with
a single moveable
Fully.
Two cylinders of
unequal radii have a
common axis with
which they are rigidly
connected ; the axis
is supported in a hori
zontal position so that
the machine can turn
round. A string has one end fastened to the larger cy
linder, is coiled several times round this cylinder, then
leaves it, passes under a moveable Fully, and is coiled
round the smaller cylinder, to which the other end is
fastened. The string is coiled in opposite ways round the
two cylinders, so that when it winds off one it winds on the
other. A weight W is suspended from the moveable
Fully. The equilibrium is maintained by a Power, P,
applied at the end of a handle attached to the axis.
Let a denote the radius of the larger cylinder, b the
radius of the smaller cylinder, c the length of the arm at
which the Power acts.
Suppose the machine in equilibrium, and the parts of
the string on both sides of the Pully to be vertical.
The tension of the string is the same throughout; and
is equal to i W by Art. 192.
150
COMPOUND MACHINES.
At the point where the string leaves the larger cy
linder the tension tends to turn the machine round in one
direction, and at the point where the string leaves the
smaller cylinder the tension tends to turn the machine
round in the opposite direction. Hence, taking moments
round the axis, we have by Art. 100,
Wb=
therefore
TF
Thus we see that by taking a and 6 very nearly equal
we can obtain any amount of mechanical advantage.
228. Hunter's Screw, or the Differential Screw.
AB is a right circular
cylinder, having a Screw
traced on its surface ; this
fits into a corresponding
groove cut in the block
CE, which forms part of
the rigid framework
CDFE.
AB is hollow, and has
a thread cut in its inner
surface, so that a second
Screw GH can work in it.
The second Screw does
not turn round, for it has
a cross bar KL the ends
of which are constrained by smooth grooves, so that the
piece GIIKL can only move up and down. The machine is
used to produce a great pressure on any substance placed
between KL and the fixed base on which the framework
CDFE stands.
Let P denote the Power applied by a handle at the top
of the outer Screw. Let W denote the pressure exerted
below KL in the state of equilibrium.
COMPOUND MACHINES. 151
Let a denote the angle of the outer Screw, r the radius
of the cylinder; let a denote the angle of the inner Screw,
/ the radius of the cylinder. Let b be the length of the
arm at which the Power acts. We shall now proceed as in
Art. 222.
Let the resistances which act between the outer Screw
and the block be denoted by R, S, T,..., and those between
the two Screws by R, S' t T 1 ,...
Then, as in Art. 222, since the outer Screw is in equi
librium,
and since the inner Screw is in equilibrium,
W=(R' + S'+T + ...)cosa'.
From the first and third of these equations
and then from the second equation
Pb= IF(rtanar'tana / );
W b
therefore 7T=r  77  ,.
P rtanartana
Thus we see that by making r tan a and / tan d very
nearly equal we can obtain any amount of mechanical
advantage.
229. We see by what has been said respecting the
mechanical powers and compound machines that we can
obtain any amount of mechanical advantage. But it must
be observed that we have hitherto considered machines in
the state of equilibrium, that is as used to support weights ;
practically however machines are more commonly used to
move weights. Now it is found that although with the aid
of a machine we can move a Weight by a Power much
smaller than the Weight, yet in order to make the Weight
152 EXAMPLES. XVII.
move through a line of any length the Power must describe
a much longer line. Let us take as a simple example the
single moveable Fully described in Art. 192 ; suppose the
Power somewhat greater than half the Weight, so that
instead of equilibrium we have the Power prevailing over
the Weight. If we have to raise the Weight through one
foot, the vertical part of the string which ends at the fixed
point K must be shortened one foot, and this requires the
end at which the Power P acts to move through two feet,
in order to keep the string stretched. Thus the length of
the line which P describes is to the length of the line
which W describes as W is to P.
The principle is popularly enunciated thus : what is
gained in power is lost in speed. We will give another
illustration of it.
We will take th9 case of the Differential Axle; see
Art. 227. Suppose the cylinders to turn once completely
round so as to raise the Weight; then the point of appli
cation of the Power P moves round the circumference of a
circle of radius c, that is describes a length 2nc. The depth
of the centre of the Pully below the axis is half the sum of
the lengths of the two parts of the string. Now in turning
once round the length 2na is wound on one cylinder, and
the length 2nb is wound off the other. Thus the Weight
is raised through the height  (2/ra  27r6), that is through
TT (a  6). Therefore
length described by P 2?rc 2c _ W
length described by W *" IT (a  b) ~~ a^6 ~ P '
EXAMPLES. XVII.
1. Three horizontal Levers AKB, BLO, CMD without
weight, whose fulcrums are K, L, J/, act on one another
at B and C respectively, and are kept in equilibrium by
Weights of lib. at A and 24lbs. at D: if AK, KB, BC,
CM, MD are equal to 2, 1, 4, 4, 1 feet respectively, find
the position of L and the pressure on it.
EXAMPLES. XV11. 153
2. In a combination of Wheels and Axles each of the
radii of the Wheels is five times the radius of the cor
responding Axle : if there be three Wheels and Axles de
termine what Power will balance a Weight of 375 Ibs.
3. A rope, the ends of which are held by two men
A and B, passes over a fixed Pully L, under a moveable
Fully M, and over another fixed Pully N. A Weight of
120 Ibs. is suspended from M. Supposing the different
parts of the rope to be parallel find with what force A and
B must pull to support the Weight.
4. In the preceding Example if B fastens his end of
the rope to the Weight find whether any change takes
place in the force which A must exert.
5. A is a fixed Pully ; B and C are moveable Pullies,
A string is put over A ; one end of it passes under C and
is fastened to the centre of B ; the other end passes under
B and is fastened to the centre of A. Compare the
Weights of B and C that the system may be in equilibrium,
the strings being parallel
6. Two unequal Weights connected by a fine string
are placed on two smooth Inclined Planes which have a
common height, the string passing over the intersection of
the Planes : find the ratio between the Weights when there
is equilibrium.
7. A Weight W is supported on an Inclined Plane
by a string along the Plane. The string passes over a fixed
Pully, and then under a moveable Pully to which a weight
W is attached, and having the parts of the string on each
side of it parallel ; the end of the string is attached to a
fixed point : shew that in order that the system may be in
equilibrium the height of the Plane must be half its length.
8. In an Inclined Plane if the Power P be the tension
of a fine string which passes along the Plane and over a
email fixed Pully and is attached to a Weight hanging
freely, shew that if P be pulled down through a given
length the height of the centre of gravity of P and W will
remain unchanged.
154 VIRTUAL VELOCITIES.
X.VII1. Virtual Velocities.
230. We have already drawn attention to a very re
markable fact with respect to machines, which is popularly
expressed in these words : what is gained in power is lost
in speed. This fact is included in the general Principle
of Virtual Velocities, which we will now consider.
231. Suppose that A is the a
point of application of a force P; /\
conceive the point A to be moved A^ ' : p
in any direction to a new position
a at a very slight distance, and from a draw a perpen
dicular ap on the line of action of the force P : then Ap is
called the virtual velocity of the point A with respect to
the force P; and the complete phrase is abbreviated,
sometimes into the virtual velocity of the point A, and
sometimes into the virtual velocity of the force P.
The virtual velocity is considered to be positive or
negative according as p falls on the direction of P or
on the opposite direction. Thus in the figure the virtual
velocity is positive.
We see that Ap= Aa cos aAp.
232. Now it is found that the following remarkable
proposition is true : suppose a system of forces in equi
librium, and imagine the points of application of the forces
to undergo very slight displacements, then the algebraical
sum of the products of each force into its virtual velocity
vanishes; and conversely if this sum vanishes for all
possible displacements ilie system of forces is in equili
brium.
This proposition is called the Principle of Virtual
Velocities.
233. We shall not attempt to demonstrate this im
portant principle, or even to explain it fully; the student
VIRTUAL VELOCITIES. 155
may hereafter consult the larger work on Statics. We may
however notice two points.
The displacements which the principle contemplates
are such as do not destroy the connexion of the points of
application of the forces with each other. Thus any rigid
body must be conceived to be moved as a whole, without
separation into parts; also any rods or strings which
transmit forces must be conceived to remain unbroken.
The word virtual is used to ultimate that the dis
placements are not really made but only supposed to be
made. The word velocities is used because we may con
ceive all the points of application of the forces to move
into their new positions in the same time, and then the
lengths of the paths described are proportional to the
velocities in the ordinary meaning of this word. But
there is no necessity for introducing this conception, and
it would probably be advantageous for beginners if the
term virtual velocity could be changed into virtual dis
placement.
234. In the present work we shall follow the usual
course of elementary writers, and shew that the Principle
of Virtual Velocities holds for all the Mechanical Powers,
by special examination of each case.
Thus in every case which we shall examine there will
be two forces, the Power, P, and the Weight, IF; and we
shall have to establish the result
P x its virtual velocity + JFx its virtual velocity =0.
We shall find that in every case the virtual velocities
of P and W will have opposite signs ; but as there are only
two forces we shall not fall into any confusion by dropping
the distinction between positive and negative virtual
velocities. We shall accordingly shew that in every case
we have numerically
P x its virtual velocity = Wx its virtual velocity.
Suppose for example that in any case of equilibrium P
is onetenth of IF; then the virtual velocity of P must be
ten times that of W. Thus the reader will see after care
fully studying this Chapter that the Principle of Virtual
Velocities includes the fact stated in Art. 230.
156 VIRTUAL VELOCITIES.
235. To demonstrate the Principle for the Lever.
Let ACB ;be
a Lever, having
its fulcrum at
C; and kept in
equilibrium by
forces P and W
which act at A
and B respec
tively.
Suppose the
Lever to be turned round C so as to come into the position
aCb. Join A a and Bb.
The angle ACa=fhQ angle BCb\ denote it by 20.
Let CM and CN be perpendiculars from C on the lines
of action of P and W respectively. Let the angle MAC=a,
and the angle NBC=&.
Then CAa= 90  6, and CBb = 90  6.
The displacement of A resolved along AM
= Aa cos MAaAa cos (90  a  6}= Aa sin (a + 6}.
The displacement of B resolved along NB
=Bb cos (180  bBC CBN)
=Bb cos (90  /3 + 6) =Bb sin (/3  0).
Resolved displacement of A
Resolved displacement of B
_ Aa sin (a + 6) _ CA sin (a + 6)
~ 6 sin (0) ~ UB sin (/30) '
for the triangle AC a is similar to the triangle BCb.
Now when 6 is made indefinitely small the righthand
CA sin a CM
Therefore =
side of this equation becomes
which is
VIRTUAL VELOCITIES. 157
W
equal to p by the principle of the Lever.
Hence ultimately, P multiplied by the resolved dis
placement of its point of application is equal to W multi
plied by the resolved displacement of its point of applica
tion.
236. To demonstrate the Principle for the Wheel and
Axle.
Let two circles having
the common centre C re
present sections of the
Wheel and Axle respec
tively.
Let the machine be
in equilibrium with the
Power P acting down
wards at A, and the
Weight W acting down
wards at B.
Suppose the machine
to be turned round its
axis so that A comes to
a, and B comes to b ; then aCb is a straight line.
The displacement of A resolved along the line of action
of P is Ca sin ACa ; the displacement of B resolved along
the line of action of W is Cb sin BCb.
_ , Kesolved displacement of A
iherefore = . .. .  ; 7r ~. n
Kesolved displacement of B
Ca sin ACa Ca CA W
Hence P multiplied by the resolved displacement of its
point of application is equal to W multiplied by the resolved
displacement of its point of application.
158
VIRTUAL VELOCITIES.
237. To demonstrate the Principle for the single
moveable Putty with parallel strings.
Suppose the Weight to be raised
through any height s ; then the
part KG of the string between the
fixed end and the Fully must be
shortened by s : and to keep the
string stretched the end at which P
acts must be raised through the
height 2s. Therefore the point of
the string which is on the line of
action of P, and in contact with the
Pully, is raised through the height
2s. Denote this point of the string,
at which P may be supposed to act,
by A ; and denote the centre of
the Pully, at which W may be supposed to act, by B.
Then
Displacement of A _ 2s _ 2 _ W
Displacement of B s ,1 P '
Hence P multiplied by the displacement of its point of
application is equal to W multiplied by the displacement
of its point of application.
238. To demonstrate the Principle for the single
moveable Pully with strings not parallel.
Let the system be displaced so that the strings are
still inclined at the same angle as before* the part of the
string with the fixed end being kept in its original direc
tion. Let 2a be the angle between the parts of the string.
Let A denote the point of the string where the string
leaves the Pully, at which we may suppose the Power to
act. Let A be displaced to a ; draw ap perpendicular to
the line of action of P : then Ap is the resolved displace
ment of A.
Let B, the centre of the Pully, be displaced to 6 : then
Bb cos a is the displacement of B resolved in the direction
of the Weight. Draw am parallel to Bb meeting the line
VIRTUAL VELOCITIES. 159
of action of P at m. Let C and c denote the points at
which the part of the string with the fixed end K leaves
the Pully in its two positions.
Now Ap
But since the length of the part KG A of the string is
equal to the length of the part Kca we have Am=Cc=ljb.
Also it may be shewn that am is equal to Bb.
Thus Ap =Bb(l + cos 2a) = 2JBb cos 2 a.
Therefore
Resolved displacement of A ZJBb cos 2 a _ _ _ TF
Resolved displacement of B ~ Bb cos a " ~~ P '
Hence P multiplied by the resolved displacement of
its point of application is equal to W multiplied by the
resolved displacement of its point of application.
239. In the preceding Article we considered such a
displacement as left the two parts of the string inclined at
the same angle as they were originally : it is however usual
to consider another displacement, in which this condition
is not fulfilled. We will now give the investigation ; but it
involves so many approximations, instead of exaot equalities,
that it is difficult for a becrinner.
160
VIRTUAL VELOCITIES.
Let K be the fixed end of the string. Suppose that
part of the string to which the Power is applied to pass
over a small fixed peg or Fully at Z, such that K and L are
in the same horizontal line. Let 2a be the angle between
the parts of the string. Let the system be displaced so
that the point of application of the Weight rises vertically
through the height Bb. Let A denote the point of the
string where the string leaves the Fully, at which we may
suppose the Power to act ; and suppose A displaced to a.
Draw ap perpendicular to AL\ then Ap is the resolved
displacement of A.
Let AC be the part of the string in contact with the
Pully in the original position, and st the part of the string
in contact with the Pully in the second position.
Draw am vertical meeting A L at m : then the angle
amp = a, and mp = am cos a.
Now when the displacement is very small we may con
sider that am=Bb. For if sa were parallel to Am then
am would be parallel and equal to Bb ; and since the angle
between sa and Am is supposed very small we may treat
these straight lines as if they were parallel.
Also when the displacement is very small we may con
sider Am=Bbcosa. For the length of the part KG A of
the string is exactly equal to the length of the part Kta ;
the parts in contact with the Pully, AC and st, wilJ be
VIRTUAL VELOCITIES.
161
very nearly equal: and therefore we may consider that
as=KCKt. Now if KG and Kt coincided in direction
we should have KCKt=Bbcosa.
And we regard Am as parallel and equal to sa, so that
we take Am=as=Bb cos a.
Thus Ap = Am + mp= 2Bb cos a.
Therefore, when the displacement is very small,
Resolved displacement of A 2Bbcosa_ _ W
Kesolved displacement of B Eb P'
Hence P multiplied by the resolved displacement of its
point of application is equal to W multiplied by the re
solved displacement of its point of application.
240. To demonstrate the Principle for the First System
ofPullies.
Let there be four Pullies.
Suppose the Weight raised
through a height s. Then the
lowest Fully is raised through
a height s, the next Fully is
raised through a height 2s, the
next Fully is raised through a
height 4, and the highest Fully
is raised through a height 8s.
And as the highest Fully is
raised through a height 8s the
point at which the Power acts
is raised through a height 16s :
see Art. 237.
Therefore
Displacement of the point at which P acts _ 16s _ 2_ l _ W
Displacement of the point at which W acts' s ~~ 1 ~~ P '
Hence P multiplied by the displacement of its point
of application is equal to W multiplied by the displace
ment of its point of application.
In a similar manner the result may be established
whatever be the number of Pullies,
v/
T. ME.
11
162
VIRTUAL VELOCITIES.
241. To demonstrate the Principle for the Second
System of Pullies.
Let there be four parts of the
string at the lower block.
Suppose the Weight raised
through a height s, then each of
the four parts of the string at the
lower block is shortened by s ; and
therefore the point at which the
Power acts must descend through 4s.
Therefore
Displacement of the point at which P acts 4s 4 _ W
Displacement of the point at which TFacts 5 1 ~~ P '
Hence P multiplied by the displacement of its point of
application is equal to W multiplied by the displacement
of its point of application.
In a similar manner the result may be established
whatever be the number of parts of the string at the lower
block.
242. In this system of Pullies it is easily seen that
when the Weight is raised through a height * a length s of
string passes round the highest Pully of the lower block, a
length 2s passes round the lowest Pully of the upper block,
a length 3s passes round the next Pully of the lower block,
a length 4s passes round the next Pully of the upper block ;
and so on if there are more Pullies. Hence it follows that
if the radii of the Pullies at the lower block are proportional
to 1, 3, 5,... the Pullies will turn through equal angles in
equal times when the machine is used to raise a Weight.
Thus all these Pullies may be connected together so as to
turn on one common axis. Also if the radii of the Pullies
at the upper block are proportional to 2, 4, 6,... these
Pullies may be connected so as to turn on one common
axis. This arrangement was invented by James White,
and is called White's Pully.
VIRTUAL VELOCITIES.
163
243. To demonstrate the Principle for the Third
System of Putties.
Let there be four Pullies.
Suppose the "Weight raised
through a height s. Then the
highest moveable Pully descends
through a depth s. The next
Pully descends through 2s in
consequence of the descent of the
Pully above it, and through s be
sides in consequence of the ascent
of the Weight ; thus it descends
through (2 + 1)5 on the whole.
The next Pully descends through
twice this depth in consequence
of the descent of the Pully imme
diately above it, and through
s besides in consequence of the ascent of the Weight ; thus
it descends through 2(2 + l)s + s on the whole, that is
through (2 2 + 2 + l)s. Similarly the end of the string at
which the Power acts descends through (2 3 + 2 2 + 2 + 1) s.
Therefore
Displacement of the point at which P acts
Displacement of the point at which W acts
W
Hence P multiplied by the displacement of its point of
application is equal to W multiplied by the displacement
of its point of application.
In a similar manner the result may be established
whatever be the number of Pullies.
244. To demonstrate tJie Principle for the Inclined
Plane.
Let a Weight W be supported at L on an Inclined
Plane by a Power, P, the direction of which makes an
112
164 VIRTUAL VELOCITIES.
angle /3 with the Plane. Suppose
the Weight displaced along the
Plane to a point K. Then the
displacement of L resolved along
the direction of P is LK cos /3 ;
and the displacement of L re
solved along the direction of W is
: see Art. 231.
Therefore
Eesolved displacement of the point at which P acts
Resolved displacement of the point at which W acts
= ZJrcosff = cosft = W
LK&in a sin a P *
Hence P multiplied by the resolved displacement of its
point of application is equal to W multiplied by the resolved
displacement of its point of application.
245. To demonstrate the Principle for the Screw.
Suppose the Powerarm to make a complete revolution,
then the Weight would rise through a height equal to the
distance between two consecutive threads measured parallel
to the axis of the Screw. The path of the end of the
Powerarm, estimated on the horizontal plane in which the
Power is supposed to act originally, is the circumference of
a circle having the Powerarm for radius. If the Power
arm instead of making a complete revolution makes only a
small part of a revolution, the ratio between the displace
ments of the end of the Powerarm and of the Weight re
mains the same. Therefore
Resolved displacement of the point of application of P
Displacement of W
_ circumference of circle with Powerarm for radius _ W
distance between two consecutive threads P *
Hence P multiplied by the resolved displacement of
its point of application is equal to W multiplied by the dis
placement of its point of application.
EXAMPLES. XVIII. 165
246. The student will not find any difficulty in shewing
that the Principle is true with respect to the various com
pound machines described in Chapter xvn.
EXAMPLES. XVIII.
1. On a Wheel and Axle a Power of 7 Ibs. balances
1 cwt., and in one revolution of the Wheel the point of
application of the Power moves through 32 inches: find
through what height the Weight is raised.
2. The radius of the Wheel is 15 times that of the
Axle, and when the Weight is raised through a certain
height it is found that the point of application of the Power
has moved over 7 feet more than the Weight: find the
height through which the Weight was raised.
3. In the First System of Pullies find how much string
passes through the hand in raising the Weight through one
inch, there being four Pullies.
4. In the First System of Pullies it is found that 5 feet
4 inches of string must pass through the hand in order to
raise the Weight 2 inches : find how many Pullies are
employed.
5. In the First System of Pullies the distance of the
highest Pully from the fixed end of the string which passes
round it is 16 feet : find the greatest height through which
the Weight can be raised, there being four Pullies.
6. In the Second System of Pullies if P descends
through 12 feet while W rises through 2 feet, find the
number of parts of the string at the lower block.
7. In the Second System of Pullies if there be five
parts of the string at the lower block and W rise through
6 inches, find how much P descends.
8. In the Second System of Pullies if P descend
through 1 foot, W will rise through  inches, where n is
the number of Pullies in the lower block.
166 FRICTION.
XIX. Friction.
247. We have hitherto supposed that all bodies are
smooth; practically this is not the case, and we must now
examine the effect of the roughness of bodies.
248. The ordinary meaning of the words smooth and
rough is well known ; and a little explanation will indicate
the sense in which these words are used in Mechanics.
Let there be a fixed plane horizontal surface formed of
polished marble; place on this another piece of marble
having a plane polished surface for its base. If we attempt
to move this piece by a horizontal force we find that there
is some resistance to be overcome : the resistance may be
very small, but it always exists. We say then that the
surfaces are not absolutely smooth, or we say that they are
to some extent rough.
Thus surfaces are called smooth when no resistance
is caused by them to the motion of one over the other;
they are called rough when such a resistance is caused by
them; and the resistance is called friction.
249. The following is another mode of defining the
meaning of the words smooth and rough in Mechanics.
When the mutual action between two surfaces in contact
is perpendicular to them they are called smooth; when it
is not perpendicular they are called rough.
When two surfaces in contact are both plane surfaces
the definition is immediately applicable ; when one or each
of the surfaces is a curved surface some explanation is re
quired. When one surface is curved and the other plane,
the perpendicular to the plane surface at the point of con
tact is to be taken for the common perpendicular. When
FRICTION. 167
each surface is curved, a plane must be supposed to touch
each surface at the point of contact, and the perpendicular
to this plane at the point of contact is to be taken for the
common perpendicular.
250. The following laws relating to the extreme
amount of friction which can be brought into action be
tween plane surfaces have been established by experi
ment.
(1) The friction varies as the normal pressure
when the materials of the surfaces in contact remain
the same.
(2) The friction is independent of the extent of the
surfaces in contact so long as the normal pressure re
mains the same.
These two laws are true not only when motion is just
about to take place, but when there is sliding motion.
But in sliding motion the friction is not always the same as
in the state bordering on motion ; when there is a difference
the friction is greater in the state bordering on motion than
in actual motion.
(3) The friction is independent of the velocity when
there is sliding motion.
251. Coefficient of friction. Let P denote the force
perpendicular to the surfaces in contact by which two
bodies are pressed together; and let .F denote the extreme
friction, that is, let F be equal to the force parallel to the
surfaces in contact which \ajust sufficient to move one body
along the other ; then the ratio of F to P is called the
coefficient of friction. It follows from the experimental
laws stated in the preceding Article that the coefficient of
friction is a constant quantity so long as we keep to the
same pair of substances.
The following experimental results are given by Pro
fessor Rankine ; they apply to actual motion :
168 FRICTION.
The coefficient of friction for iron on stone varies be
tween '3 and '7, for timber on timber varies between '2
and '5, for timber on metals varies between '2 and '6, for
metals on metals varies between '15 and 25. Friction acts
in the opposite direction to that in which motion actually
takes place, or is about to take place.
252. Angle of friction. Let a body be placed on an
Inclined Plane ; if the Plane were perfectly smooth the body
would not remain in equilibrium, but would slide or roll
down the Plane: but practically owing to friction it is
quite possible for the body to remain in equilibrium.
R
Let W denote the weight of
the body, which acts vertically
downwards. Let R denote the
resistance of the Plane which
acts at right angles to the Plane.
Let F denote the friction which
acts along the Plane.
Suppose the body to be in
equilibrium; then we have, as in Art. 215, by resolving the
forces along the Plane and at right angles to it,
F T7sina=0,
R TFcosa=0.
Hence by division, ^ = tan a.
This result is true so long as the body is in equilibrium,
whatever be the inclination of the Plane to the horizon.
Now suppose the Plane to be nearly horizontal at first and
let the inclination be gradually increased until the body
is just about to slide down the Plane; the value of ~
in this state is by our definition the coefficient of friction:
so that the coefficient of friction is equal to the tangent
of the inclination of the Plane when tJie body is just about
to slide.
In this way we may experimentally determine the value
of the coefficient of friction for* any proposed pair of sur
FRICTION. 169
faces. The inclination of the Plane when the body is just
about to slide is called the angle of friction.
253. The only case of friction besides that of plane
surfaces which is practically important in Statics, is that of
a hollow cylinder which can turn round a fixed cylindrical
axis, and that of a solid cylinder which can turn within a
fixed hollow cylinder or on a fixed plane. It is found by
experiment that in these cases the extreme friction is very
nearly proportional to the pressure ; but the coefficient of
friction is much less than it would be for plane surfaces
of the same material kept in contact by the same pressure.
254. Statical Problems respecting friction involve two
points, namely, the determination of the extreme or limit
ing position or positions at which equilibrium is possible,
and the determination of the range of positions within
which equilibrium is possible. Thus in Art. 252 the
limiting position is that at which the tangent of the in
clination of the Plane is equal to the coefficient of fric
tion, and equilibrium will subsist as long as the inclination
is less than the value thus determined. We may describe
the process of solving statical problems involving friction
thus : let F denote a friction and R the corresponding pres
sure ; put nR for F in the conditions of equilibrium ; then
the limiting position of equilibrium is found by making p.
equal to the coefficient of friction for the substances in
consideration; and the range of positions within which
equilibrium is possible is found by ascribing to \i values
less than the coefficient of friction. If there are various
frictions, and the pairs of surfaces in contact not all of the
same material, we shall require different symbols to denote
the ratio of the friction to the pressure in each case.
This general description will be illustrated by the re
maining Articles of the present Chapter.
255. We will now solve the following problem :
One end of a uniform beam is on a smooth inclined
plane, and the other end ' on a rough horizontal plane :
determine the limiting positions of equilibrium.
170
FRICTION.
Let DE be the beam, BA the inclined plane, DAG
the horizontal plane. Let
a denote the angle BAG,
and 6 the angle EDA.
The forces acting on
the beam may be denoted
as follows: a vertical force,
M, and a horizontal force,
F, arising from the action
of the rough horizontal
plane at D ; a force, S, at
right angles to the smooth inclined piano at E\ and the
weight of the beam, TF, which acts vertically downwards
through G, the centre of gravity of the beam.
We apply the conditions of equilibrium of Art. 93.
Resolve the forces vertically and horizontally : thus
R + Scosa TT=0 (1),
FSsina=0 (2).
Take moments round D : thus
W.DGcos 0=S.DEcos (a  6\
that is TFcos0=2cos(a0) (3).
Put pR for F; then from (1) and (2)
H ( W Scos a) =tSsin a,
therefore 8= . ^ . . . (4) ;
Sin a + fj. cos a
and then from (1) R= . TFsinct ...(5).
sm a + /* cos a
Substitute the value of S in (3) : thus
2 " 003 ^... ...(0).
sm a + p. cos a
FRICTION. 171
The limiting position of equilibrium is assigned by the
value of 6 found from this equation, p. being put equal to
the coefficient of friction which is supposed known. We
proceed to investigate the range of positions of equilibrium.
From (6) we obtain
_ cos & sin a
^~ 2 cos (a  6}  cos a cos Q
cos 6 sin a tan a .*
cosacos0 + 2sin#sina 1 + 2 tan tana '"^ '"
Now it is obvious that 6 must lie between and a.
Hence we deduce the following results :
I. If the coefficient of friction is not less than tan a
every position is one of equilibrium.
II. If the coefficient of friction lies between tan a and
 , then 6 may have any value between a and the
limiting position assigned by (6) or (7).
III. If the coefficient of friction is less than
l + 2tan 2 a
there is no position of equilibrium.
In cases II. and III. suppose the beam in equilibrium
when 6 has an assigned value ; and let p. be determined from
(7) : then (4) and (5) will determine the values of the forces
jR and S, and (2) will determine the value of F.
The resultant action of the rough horizontal plane on
the beam is the resultant of the forces R and P', and is
therefore equal to V(^ 2 + ^ 2 ), that is to ^(1 +f 2 ) R in the
limiting position. This remark is applicable also in all
similar cases.
256. The beginner may perhaps object to the investi
gation of the preceding Article that it is impossible to
obtain a perfectly smooth inclined plane, so that the pro
172
FRICTION.
blem cannot correspond with any real experience. We may
reply, that although it is impossible to obtain a perfectly
smooth inclined plane, yet there is no difficulty in imagining
such a plane ; and that the problem illustrates the principles
of the subject very well even although the conditions which
it supposes cannot be experimentally secured : and we may
add, that it would be practically possible to make the inclined
plane so very much smoother than the horizontal plane, that
the friction arising from the former might be neglected in
comparison with the friction arising from the latter.
We proceed to consider the effect of friction on some of
the Mechanical Powers.
257. The Lever with friction.
Suppose a solid bar pierced with a cylindrical hole,
through which passes a solid fixed cylindrical axis. Let
the outer circle in the figure represent a section of the
cylindrical hole made by a plane perpendicular to its axis ;
and let the inner circle represent the corresponding section
of the solid axis. In the plane of this section we sup
pose two forces, P and TF, to act on the bar at right
angles to the bar. Also at (?, the point of contact of the
bar and the axis, there will be the action of the rough axis
on the bar ; denote this by a force E along the common
radius and a force F along the common tangent. Suppose
that these four forces keep the bar in equilibrium in a hori
zontal position.
FRICTION. 173
Let r denote the radius of the outer circle, a and b
the lengths of the perpendiculars from the centre of this
circle on the lines of action of P and W respectively. Let
6 be the inclination of CR to the vertical.
"We apply the conditions of equilibrium of Art. 93. Re
solve the forces parallel to the bar and at right angles to
it: thus
R sin 0Fcos0=Q (1),
Rcos0 + Fam0PW=0 (2)
Take moments round C: thus
(3).
Put pR for F; then in the limiting position of equili
brium fi is equal to the coefficient of friction which is sup
posed known : see Art. 253.
Thus from (1) sin0/icos0=0; therefore tan#=/i.
This determines 0; and then (3) determines the ratio of
P and W. From (1) and (2) we can find R and F.
If the point of contact C be supposed to fall on the
other side of the vertical through the centre of the outer
circle, we should have instead of (3),
P(arsin0) =
From tan 0=p we deduce sin
and (4) we infer that the bar will be in equilibrium pro
vided the ratio of ^ lies between
rr
b
where /z is the known coefficient of friction
174 FRICTION.
258. The Inclined Plane with friction.
Let a be the inclina
tion of the Plane. Sup
pose a body of Weight
W placed on the Plane,
and acted on by a Power
P in the direction which
makes an angle /3 with
the Plane.
First suppose the body
just on the point of mov
ing down the Plane. Let
R denote the resistance which acts at right angles to the
Plane, and p.R the friction which acts along the Plane
upwards, p being the coefficient of friction. Kesolve the
forces along the Plane, and at right angles to it, as in
Art. 215 : thus
 Tfsina=0 .................. (1),
 T7cosa=0 .................. (2).
Substitute in (1) the value of R from (2) ; thus
p = TFsina^TFcosa ^
cos /3  p. sin /3 "
Next suppose the body just on the point of moving up
the Plane. The friction now acts down the Plane; and
proceeding as before we obtain
TTsin a + p TFcos a .
cos/3 + /zsin/3
It will be seen that this result can be deduced from the
former by changing the sign of p.
The equations (3) and (4) give the ratio of P to W in
the limiting states of equilibrium ; and we may infer that
p
the body will be in equilibrium if ^ lies between
sin a  p cos a , sin a + \i cos a
cos  /i sin j8 cos ^3 + /x sin 8 '
FRICTION. 175
A particular case of the general result may be noticed.
Suppose 3=0 so that the force acts along the Plane; then
(3) and (4) give respectively
P= W (sin a  p. cos a), P= TF(sin a + p cos a).
Instead of an Inclined Plane suppose a body of weight
W placed on a rough horizontal Plane, and acted on by a
force P inclined at an angle /3 to the horizon. Then we
shaU find that
It will be seen that this result is the same as we should
obtain by putting a=0 in (4).
259. Let denote the angle of friction; see Art. 252.
Then tan f =/x. The first value of P of the preceding
Article
_ TTsin a  tan c TFcos a __ TFsin (a  e)
cos /3  tan e sin " cos (/3 + e) *
The second value of P
Wsm a + tan TFcos a _ TFsin (a + *)
cos /3 + tan sin /3 cos(/3t) "
Suppose we require to know the least Power which will
suffice to prevent the body from moving down the Plane,
the Power being allowed to act at the inclination to the
Plane which we find most suitable.
Consider then that j3 may vary : the first value of P will
be least when cos ()3 + e) = l, that is when j3 + e=0, so that
/3= f. Hence the Power must be supposed to act at an
inclination to the Plane equal to the angle of friction,
measured below the Plane. And the value of P is
IF sin (a e).
Again, suppose we require to know the least Power
which will suffice to move the body up the Plane. The
second value of P is least when cos (/3  e) = l, that is when
/3 = f ; and the value is then TFsin (a + c) : this Power would
keep the body on the point of motion up the Plane, and any
greater Power would move the body.
176 FRICTION.
As a particular case of the last result suppose a=0, so
that instead of an Inclined Plane we have a horizontal
Plane ; thus we find that in order to move a given Weight
along a rough horizontal Plane with the least Power we
should make the Power act at an inclination to the Plane
equal to the angle of friction ; and then the body will be on
the point of motion when the Power is equal to the Weight
multiplied by the sine of the angle of friction.
260. The Screw with friction. See Art. 222.
Let r be the radius of the cylinder, b the length of the
Powerarm, a the angle of the Screw. Suppose that the
Weight is about to prevail over the Power. Let /* be the
coefficient of friction. The Screw is acted on by the
following forces: the Weight W; the Power P; the re
sistances R, /S, T)... at various points of the surfaces in
contact, at right angles to the surfaces, and so all making an
angle a with the vertical; and frictions pR, pS, pT,...
which will all make an angle 90  a with the vertical
Then using the same conditions of equilibrium as in
Art. 222, we have
......... (1),
inan(R + S+T+...)r
From (1) we obtain
COS a + n Sin a
then, substituting in equation (2)
p, _ r sin a pr cos a ^r
cosa + /isina
P r(sina J
therefore
If we suppose P about to prevail over W we obtain
P _ r (sin a + /i cos a) ...
W~ b (cos a  p. sin a) ' "
EXAMPLES. XIX. 177
The equations (3) and (4) give the ratio of P to W in
the limiting states of equilibrium ; and we may infer that
p
there will be equilibrium if rp lies between
r (sin a  /A cos a) , r (sin a + /* cos a)
b (cos a + p sin a) c b (cos a /* sin a) *
If we put tan e for p the expressions become
r tan (a  ) and r tan (a + e).
261. If a body be placed on an Inclined Plane, and
the friction be great enough to prevent sliding down the
Plane, the body will stand or fall according as the vertical
line drawn through the centre of gravity of the body falls
within or without the base. This may be established in
the manner of Art. 152.
EXAMPLES. XIX.
1. Find the coefficient of friction if a "Weight just
rests on a rough Plane inclined at 45 to the horizon.
2. A weight of 10 Ibs. rests on a rough Plane inclined
at an angle of 30 to the horizon : find the pressure at
right angles to the Plane and the force of friction exerted.
3. A force of 3 Ibs. can, when acting along a rough
Inclined Plane, just support a weight of 10 Ibs., while a force
of 6 Ibs. would be requisite if the Plane were smooth : find
the resultant pressure of the rough Plane on the Weight.
4. A body is just kept by friction from sliding down
a rough Plane inclined at an angle of 30 to the horizon :
shew that no force acting along the Plane would pull the
body upwards unless it exceeded the Weight of the body.
5. A body placed on a horizontal Plane requires a
horizontal force equal to its own weight to overcome the
friction: supposing the Plane gradually tilted, find at what
angle the body will begin to slide.
T. MB. 12
178 EXAMPLES. XIX.
6. In the preceding Example if additional support be
given by means of a string fastened to the body and to a
point in the Plane, so that the string may be parallel to
the Plane, find at what inclination of the Plane the string
would break, supposing the string would break on a
smooth Inclined Plane at an inclination of 45.
7. If the height of a rough Inclined Plane be to the
length as 3 is to 5, and a Weight of 15 Ibs. be supported by
friction alone, find the force of friction in Ibs.
8. If the height of a rough Inclined Plane be to the
length as 3 is to 5, and a Weight of 10 Ibs. can just be
supported by friction alone, shew that it will be just on the
point of being drawn up the Plane by a force of 12 Ibs.
along the Plane.
9. Find the force along the Plane required to draw a
weight of 25 tons up a rough Inclined Plane, the coefficient
of friction being , and the inclination of the Plane being
128
such that 7 tons acting along the Plane would support the
Weight if the Plane were smooth.
10. Find the force in the preceding Examj
it to act at the most advantageous inclination to the
11. A ladder inclined at an angle of 60 to the horizon
rests between a rough pavement and the smooth wall of a
house. Shew that if the ladder begin to slide when a
man has ascended so that his centre of gravity is half way
up, then the coefficient of friction between the foot of the
ladder and the pavement is ^'
12. A heavy beam rests with one end on the ground,
and the other end in contact with a vertical wall. Having
given the coefficients of friction for the wall and the ground,
and the distances of the centre of gravity of the beam
from the ends, determine the limiting inclination of the
beam to the horizon.
MISCELLANEOUS PROPOSITIONS. 179
XX. Miscellaneous Propositions.
262. In the present Chapter we shall give some miscel
laneous propositions of Statics.
263. We have hitherto confined ourselves almost en
tirely to the equilibrium of forces which act in a plane : the
following Articles will contain some propositions explicitly
relating to forces which are not all in one plane.
264. To find the resultant of three forces which act on
a particle and are not all in one plane.
Let OA, OB, OC repre
sent three forces in magnitude
and direction which act on a
particle at 0. Let a parallel
epiped be formed having these
straight lines as edges; then
the diagonal OD which passes
through will represent the
resultant in magnitude and
direction.
For OE, the diagonal passing through of the parallelo
gram OAEB, represents hi magnitude and direction the
resultant of the forces represented by OA and OB. Also
OEDC is a parallelogram, and OD, the diagonal passing
through 0, represents in magnitude and direction the
resultant of the forces represented by OE and OC, that is,
the resultant of the forces represented by OAj OB, and OC.
265. The preceding investigation is only a particular
case of the general process given in Art. 52, but on account
of its importance it deserves special notice. As we can
thus compound three forces into one, so on the other
hand we can resolve a single force into three others which
act in assigned directions. Most frequently when we have
thus to resolve a force the assigned directions are mutually
at right angles; that is with the figure of Art. 264, the
angles AOB, BOG, CO A are right angles. The angle
OCD is then a right angle, so that OC=OD cos COD:
thus when the three components are mutually at right
122
180 MISCELLANEOUS PROPOSITIONS.
angles a component is equal to the product of the resultant
into the cosine of the angle between them.
Also by Euclid, I. 47, we have
OD*= 00* + OD 2 = OC 2 + OE*= OC 2 + 01P + OA 2 :
thus when the three components are mutually at right
angles the square of the resultant is equal to the sum of
the squares of the three components.
266. The process of Art. 52 for determining the re
sultant of any number of forces acting on a particle is
applicable whether the forces are all in one plane or not;
the process in Art. 56 assumes that the forces are all in
one plane: we shall now extend the latter process to the
case of forces which are not all in one plane.
267. Forces act on a particle in any directions:
required to find the magnitude and the direction of their
resultant.
Let denote the position of the particle; let Op, Oq,
Or, Os,... denote the directions of the forces; let P, Q,
R, $,... denote the magnitudes of the forces which act
along these directions respectively. Draw through three
straight lines mutually at right angles; denote them by
Ox, Oy, Oz : and resolve each force into three components
along these straight lines, by Art. 265. Thus P may be
replaced by the following three components: PcospOx
along Ox, PcospOy along Oy, and PcospOz along Oz.
Similarly Q may be replaced by QcosqOx along Ox,
Q cos qOy along Oy, and Q cos qOz along Oz. And so on.
Let X denote the algebraical sum of the components
along Ox, so that
X=P cospOx + Q cos qOx + JR cos rOx + cos sOx + ..'.
Similarly let Y and ^denote the algebraical sums of the
components along Oy and Oz respectively.
Thus the given system of forces is equivalent to the
three forces X, Y, Z which act along three straight lines
mutually at right angles.
MISCELLANEOUS PROPOSITIONS. 181
Let V denote the resultant of the given system of forces,
and Ov its direction; then, by Art. 265,
Thus the magnitude and the direction of the resultant
are determined.
268. Since Fcos vOx=X, the resolved part of the
resultant in any direction is equal to the sum of the
resolved parts of the components in that direction: see
Arts. 44 and 52.
269. Suppose that any force Q acts at a point in any
direction OF; let OD be any other direction: then the
resolved part of Q along OD is Q cos FOD.
Now let OF make angles a, ft y respectively with three
straight lines OA, OB, OC mutually at right angles; and
let OD make angles a', ft, y respectively with OA, OB,
OC. Then the force Q may be resolved into the three
forces Q cos a, Q cos ft Q cos v respectively along OA, OB,
OC. Resolve each of these three components along OD;
thus we obtain Q cos a cos a, Q cos ft cos ft, Q cos y cos y
respectively. And as we may admit that the sum of these
must be equal to the resolved part of Q along OD, we
have
Q cos FOD = $ cos a cos a + Q cos pcosfi+Q cos y cos y ;
thus cos FOD = cos a cos a' + cos /3 cos ft + cos y cos y.
Thus by the aid of statical considerations we arrive at the
preceding formula which expresses the cosine of the angle
between two straight lines in terms of the cosines of the
angles which these straight lines make with three others
mutually at right angles.
A particular case of the formula is obtained by sup
posing OD to coincide with OF; then a'=a, ft=$, y =y :
and we obtain
1 = cos 2 a + cos 2 (3 + cos 2 y.
182 MISCELLANEOUS PROPOSITIONS.
270. In Art. 39 we have given the conditions under
which three forces acting on a particle will maintain it in
equilibrium; we will now present these conditions in a
slightly different form, and then demonstrate a correspond
ing result for the case of four forces which are not all in
one plane.
271. OA, OB, 00 are three straight lines of equal
length in one plane, and they are not all on the same side
of any straight line in the plane passing through O/
forces P, Q, R respectively act along these straight lines
sudi that
P = Q _R
area of OBC area of OCA area of OAB '
these forces will maintain a particle at in equilibrium.
For the area of a triangle is half the product of two
sides into the sine of the included angle ; hence each force
is proportional to the sine of the angle between the direc
tions of the other two : and the proposition follows imme
diately from Art. 39.
272. OA, OB, 00, OD are four straight lines of
length, no three of them being in the same plane,
and they are not all on the same side of any plane passing
through 0; forces P, Q, R, S respectively act along these
straight lines such that
P Q R S
vol. OBCD ~ vol. OCDA ~ vol. ODAB ~ vol. OABC ''
these forces will maintain a particle at in equilibrium.
Let p denote the length of the perpendicular from on
the plane BCD. Resolve each force into three along direc
tions mutually at right angles, one direction being perpen
dicular to the plane BCD. The sum of the components of
Q, R, and S in the direction perpendicular to the plane
BCD
Let h denote the length of the perpendicular from A on
the plane BCD. The component of P perpendicular to
MISCELLANEOUS PROPOSITIONS. 183
7j _ fQ
the plane BCD is P TTJ . Now the direction of the com
ponent of P is opposite to the direction of the sum of the
components of Q, R, and S by reason of the condition that
the straight lines OA, OS, 00, OD are not all on the
same side of any plane through 0. Moreover by reason of
the given ratios we have
P _ vol. of BCD _
P + Q + R + S sum of vols of OBCD, OCDA, ODAB, OABC
_vol. of OBCD _p
~vol.ofABCL>~h'
therefore Ph = (P + Q + R + S) p,
and
Thus the algebraical sum of the components perpen
dicular to the plane BCD vanishes.
Similarly the algebraical sum of the components esti
mated perpendicular to CDA, DAB, and ABC vanishes.
Hence the resultant of the four forces vanishes ; for if it
did not the component estimated in all the four assigned
directions could not vanish. See Art. 268.
273. Conversely, if four forces acting on a particle
maintain it in equilibrium, and no three of the forces are
in the same plane, the forces must be in the proportion
specified in the preceding Article.
For take OA, OB, OC, OD all equal on the directions
of the forces; then, resolving perpendicular to the plane
BCD, we have as a necessary condition of equilibrium
P p P P vol. of OBCD
' ^
vo\. of ABCD'
and similar expressions hold for the ratios of Q, of R, and
of S, ioP + Q + R + S.
274. We will now give some additions to our account
of the theory of couples.
275. Two unlike couples in parallel planes will balance
if their moments are equal.
184 MISCELLANEOUS PROPOSITIONS.
Let there be two unlike couples in parallel planes of
equal moments. By Art. 68 we may replace a couple by
any like couple in the same plane which has an equal
moment. Hence we may suppose the forces of one couple to
be equal and parallel to the forces of the other couple : then
as the moments are equal the arms of the couples will also
be equal. Let P and p denote the forces of one couple ; and
Q and q the forces of the other ; where P, p, Q, and q are
all numerically equal. Suppose P and Q to be like forces,
and therefore p and q to be like. The resultant of P and
Q will be 2P, parallel to the direction of P and Q, and
midway between them. The resultant of p and q will be
2p, parallel to the direction of p and q, and midway between
them.
Suppose any plane to cut
the lines of action of P and
p at A and a respectively,
and of Q and q at B and b
respectively. Then since the
couples are in parallel planes
Aa is parallel to Eb\ and
since the arms are equal Aa
is equal to Bb. Thus AaBb
is a parallelogram. And since
the couples are unlike, A and B are at the opposite ends of
a diagonal. The resultant of P and Q acts at the middle
point of AB, and the resultant of p and q at the middle
point of ab ; so that they act at the same point. And as the
two resultants are equal but unlike they balance each other.
276. Hence a couple is equivalent to another like couple
of equal moment in any plane parallel to its own.
277. We will briefly consider the case of two couples
in two planes which are not parallel.
By Art. 68 we may transform each couple in its own
plane until the two couples have a common arm situated in
the straight line which is the intersection of the two planes.
Let Cc denote this common arm. Let P and p be the
forces which form one couple, and Q and q the forces which
MISCELLANEOUS PROPOSITIONS. 185
form the other; and suppose that P and Q act at C', and
p and q at c. Then P and may be replaced by a single
resultant jR, and jp and q by a single resultant r ; also .R
and r will be equal and parallel but unlike. Thus the two
couples are compounded into a single couple.
278. An example of Art. 97 which is of some interest
may be noticed.
Let A, JB, C denote the positions of heavy particles
in a plane; suppose at each of these points a force to act in
the plane proportional to the product of the weight of
the particle into its distance from a fixed point 0, and at
right angles to the distance ; and suppose these forces all
tend to turn the same way round : it is required to
replace the system of forces by a single force at and a
couple.
First with respect to the single force. Suppose each
force of the system to be p times the product of the weight
of the particle into the corresponding distance. If the
direction of each force were along the corresponding dis
tance instead of at right angles to it, the direction of the
resultant would be along the straight line from to the
centre of gravity of the particles, and the magnitude of the
resultant would be p. times the product of the distance of
the centre of gravity from into the sum of the weights of
the particles : see Art. 154. Then in the actual case the
magnitude of each force, supposed transferred to 0, is the
same as in the other case, but the direction is at right
angles. Hence finally, in the actual case the direction of
the single force is at right angles to the straight line drawn
from to the centre of gravity ; and the magnitude of the
single force is //, times the product of the length of this
straight line into the sum of the weights of the particles.
Next with respect to the couple. Let P denote the
weight of the particle at A\ then the force at A is
p.P x OA ; and the moment of this force round is
pPxOAxOA, that is p.Px.OA 2 . Hence finally, the
moment of the couple is p times the sum of the product of
the weight of each particle into the square of the corre
sponding distance from 0.
186 MISCELLANEOUS PROPOSITIONS.
279. We have not discussed the case of a system of
forces acting at any points and in any directions on a rigid
body; but the investigations which have been given will
enable the student to discuss such a case.
For suppose forces P, Q, JR, #,... to act at various points
of a rigid body, and in various directions. Since a force
may be supposed to act at any point in its line of action
we may suppose these forces to be applied at points which
are all in one plane. Then resolve each force into two
components, one in the plane and one at right angles to
the plane : thus we obtain two systems of forces, one in the
plane and one at right angles to the plane. The former
system if not in equilibrium will reduce to a single force or
a couple : see Art. 84. The latter system if not in equili
brium will also reduce to a single force or a couple : see
Art. 112. Hence in any particular case we shall be able to
find the resultant of all the forces.
It is plain that the original system of forces will not be
in equilibrium unless each of the two systems into which we
have resolved it is separately in equilibrium ; for one of the
two cannot balance the other. Hence from considering the
system at right angles to the plane we arrive at the fol
lowing result : if a system of forces is in equilibrium the
algebraical sum of the forces resolved parallel to any straight
line must vanish. We say that this is necessary to equili
brium ; we do not say that it is sufficient.
280. When we have spoken of a string passing round
a peg or a pully we have hitherto assumed the peg or pully
to be smooth. But in practice there may be a sensible
amount of roughness; and every one must have observed
that if a rope be coiled two or three times round a post, it
is possible for a force at one end to balance a much larger
force at the other end. This is owing to the friction be
tween the rope and the post ; and we shall now give some
investigations relating to this subject.
281. A string is stretcJied round a rough right cir
cular cylinder in a plane perpendicular to the axis: to shew
that as the portion of string in contact with the cylinder
increases in Arithmetical Progression the mechanical advan
tage increases in Geometrical Progression.
MISCELLANEOUS PROPOSITIONS. 187
Suppose AB the por p
tion of the string in con
tact with the cylinder;
and let C be the centre of
the circle of which AB is
an arc. Let P and Q be
the tensions of the string
at A and B respectively ;
and suppose Q the greater.
Suppose the string to be
in the limiting condition
of equilibrium, so that it is just about to move from A
towards B.
I. The relation between P and Q will be of the
form p=K ) where K is some quantity which does not
depend on the forces.
For suppose that without changing the angle ACJB, or
the radius AC, or the coefficient of friction, we double P;
then equilibrium will still hold if we also double Q. For
the result is the same as if we had two strings in contact
with the same cylinder, over equal arcs, and each acted
on by a force P at one end, and a force Q at the other.
Similarly, if P be changed to 3P, and Q to 3$, equili
brium will still hold; and so on. Thus if the angle, the
radius, and the coefficient of friction remain the same, Q
varies as P.
II. Let the angle ACB be denoted by 6 : then K must
be of the form & , where k is some quantity which does not
depend on the forces, nor on 6.
For take any arbitrary angle a, and suppose that 6=na.
Imagine AB to be divided into n equal parts; and let
Qv> 2J Qs)" be the tensions at the end of the first, second,
third,... of these parts, beginning from A. Then by what
has been already shewn, we have
188 MISCELLANEOUS PROPOSITIONS.
where H is some quantity which does not depend on the
forces, nor on 6. Hence by multiplication
j =//=//=/,
i
where k=H a ; and Tc does not depend on the forces, nor
on 6.
If the length of string in contact with the cylinder
increases in Arithmetical Progression, then 6 increases in
Arithmetical Progression ; and thus the ratio of Q to P in
creases in Geometrical Progression.
This result explains the very great mechanical advan
tage which is gained by coiling a rope two or three times
round a post. Suppose, for example, that when a rope is
coiled once round a post we have 73=8; then when the
rope is coiled twice round the post ^=3 2 ; when the rope
is coiled three times round s=3 3 ; and so on.
282. We will now determine the value of &, as the
process is instructive, although it requires more knowledge
of mathematics than we have hitherto assumed.
The forces which act on the portion AB of the string
are the following : P along the tangent at A, Q along the
tangent at J3, and a resistance and a friction on every
indefinitely small element of the string AB.
The resistance on every element is a force the direction
of which passes through C : the corresponding friction is p
times this resistance and its direction is at right angles to
that of the resistance. Suppose jR to denote the resultant
of all the resistances, and to denote the angle its direction
makes with CA ; then fiR will denote the resultant of all
the frictions, and its direction will make an angle < with
the tangent at A.
Suppose the string to be in equilibrium ; if it were to
become rigid, equilibrium would still subsist; the forces
therefore must satisfy the conditions of Art. 93. Hence
MISCELLANEOUS PROPOSITIONS. 189
resolving parallel to the tangent and to the radius at A we
have
......... (1),
(2).
From (2) we have
substitute in (1) ; thus
^
cos + p sin
Put tan a for /* ; thus
(sin 6 cos a  cos d> sin a) .
P= Q cos 6 + =^ f 2L.  i sm 0.
cos 9 cos a + sm sin a
therefore P = (cos 6  tan (a  d>) sin 0} $.
P 1
But ^ = = =0, so that 0=cos  sin 6 tan (a 
therefore tan (a  6) ==
This is an exact equation which is true for all values
of 0, and is therefore true when 6 is indefinitely small : from
this consideration we shall deduce the value of k. The
value of k will depend on the unit of angular measure
which we adopt : we will take the unit to be that of circu
lar measure. Now when 6 is indefinitely small, so also is
<, and the lefthand member of the last equation becomes
tana. Also cos 0=1, and k~ e =l6logk very approxi
mately, so that we have on the righthand side of the equa
tion . /. ; and this by Trigonometry is equal to log,
sin u
when 6 is indefinitely small.
Thus log &=tana=/i, therefore k=e*.
283. It will be seen that in the preceding Article we
only had occasion to employ two out of the three equations
of equilibrium of Art. 93. To form the third equation we
will take moments round. C: thus we find that QP is
equal to the sum of all the frictions exerted.
190 EXAMPLES. XX.
EXAMPLES. XX.
1. Three forces of 11, 10, and 2 Ibs. respectively acton
a particle in directions mutually at right angles : determine
the magnitude of the resultant.
2. Three forces P, P, and P*/2 act on a particle in
directions mutually at right angles : determine the magni
tude of the resultant, and the angles between the direction
of the resultant and that of each component.
3. Three forces each equal to P act on a particle, and
the angle between the directions of any two forces is 2a ; if
jR denote the resultant, and 6 the angle between the direc
tion of the resultant and that of each component, shew that
sin0= sin a, R 2 =P 2 (9  12sin 2 al
\M
4. A particle is placed at the corner of a cube, and is
acted on by forces P, Q, R along the diagonals of the faces
of the cube which meet at the particle : determine the re
sultant force.
5. Two couples act in planes which are at right angles
to each other; each force of one couple is 3 Ibs., and the
arm is one foot; each force of the other couple is 2 Ibs.,
and the arm is two feet: determine the moment of the
resultant couple.
6. D is the vertex of a pyramid on a triangular base
A BG; forces P, Q, R act at the centres of gravity of the
faces DBG, DCA, DAB, at right angles to these faces
respectively, and such that
P = Q = R .
area of DBG area of DCA area of DAB '
shew, by resolving P, Q, and R, parallel and perpendicular
to the base ABC, that their resultant is perpendicular to
ABC, and passes through the centre of gravity of ABC;
and that the resultant bears the same ratio to the area of
ABCasP bears to the area of DBG.
PROBLEMS. 191
XXI. Problems.
284. We will close the part of the work relating to
Statics with some observations on the solution of Mechani
cal Problems.
285. Problems may be proposed which have been
formed by combining some definition or principle in Me
chanics with some theorem of Pure Mathematics, and
which cannot be solved briefly and simply without the aid
of that theorem. The results given in Art. 121 exemplify
this remark ; they are obtained by combining Euclid vi. 3
and vi. A with an elementary principle respecting the centre
of parallel forces. It is obvious that in order to solve pro
blems of this kind the student requires a knowledge of the
most important theorems in Pure Mathematics, together
with a readiness in selecting the appropriate theorem,
which can only be acquired by practice.
286. On the other hand problems may be proposed
which do not depend so much on a knowledge of Pure
Mathematics as on a correct use of mechanical principles ;
and respecting this class of problems we may make a few
general remarks.
If forces act in one plane on a rigid body three condi
tions must be satisfied for equilibrium. These conditions
may be expressed in various forms, as we have shewn in
Chapter vi. The most interesting problems, and at the
same time the most difficult, are such as relate to a system
of two or more bodies which are in contact or connected
by hinges or strings. The beginner should pay great at
tention to the following statements :
When a system of bodies is in equilibrium each body of
the system must be in equilibrium ; and so the forces which
act on each body must satisfy the conditions of equilibrium.
When two bodies are in contact some letter should be
used to denote the mutual action between them ; and the
conditions of equilibrium will enable us to find the magni
192 PROBLEMS.
tude of the force. With respect to the direction of the
mutual action see Art. 249.
"We assume that if two bodies A and B are in contact
the force which A exerts on B is equal and opposite to
that which B exerts on A : this principle is called the
equality of action and reaction, and it may be admitted
as an axiom.
If two bodies are connected by a string some letter
should be used to denote the tension, and the value of the
tension found from the conditions of equilibrium.
Beginners frequently make mistakes by assuming in
correct values for the action which takes place between
two bodies, or for the tension of a string, instead of deter
mining the values of these forces by the conditions of equi
librium.
When a body is in equilibrium under the action of forces
in one plane three conditions of equilibrium must be satis
fied, yet it may happen that we do not require to express
all these conditions. For example, in the case of a Lever
we may require only the one equation which is obtained by
taking moments round the fulcrum ; the other two equa
tions would serve to determine the magnitude and the
direction of the resistance of the fulcrum, and need not be
formed if we do not wish to know these.
We shall illustrate these remarks by solving some pro
blems.
287. Four beams without weight are connected by
smooth joints so as to form a parallelogram ; the opposite
corners are connected by strings in tension : compare the
tensions of the strings.
Let ABCD represent the
parallelogram. Let P be the D
tension of the string AC, and /\
Q the tension of the string / \
BD. Let be the intersec
tion of AC and BD.
The simplest mode of form A
ing a joint is to pass a smooth
PROBLEMS. 193
peg or pivot through the beams which are to be connected.
Thus in the present case we have four beams and four
pegs, and each of these must be in equilibrium. The
strings may be supposed to join the pegs, and so not to be
immediately connected with the beams.
Thus the beam AB is acted on by only two forces, one
from the peg at A, and the other from the peg at B. The
two forces must therefore be equal and opposite, so that
their line of action must coincide with AB. Denote each
force by R.
Similarly AD must be acted on by two equal and oppo
site forces, the line of action of which must coincide with
AD. Denote each force by S.
The rod AB exerts on the peg at A a force R equal
and opposite to that which the peg exerts on the rod ; simi
larly the rod AD exerts on the peg at A a force S equal
and opposite to that which the peg exerts on the rod.
Thus the peg at A is in equilibrium under the action of
forces P, jR, S along AC, BA t and DA respectively.
Therefore, by Art. 38,
P _sinDAB
8 ~~ sin BAG'
In the same way by considering the equilibrium of the
peg at D we obtain
Q _ sin ADC
S~sinBDC'
P sin BDC
Therefore
^  57^
Q Bin B AC
smDBAAO AC
Thus the tensions of the strings are as the lengths of
the diagonals along which they act
T. ME. 13
194 PROBLEMS.
288. It will be instructive to treat the preceding
problem also in another manner.
The joint may be made by a peg or pivot which is
rigidly attached to one beam and passes through the other.
Suppose for example that the pegs are rigidly attached to
the beams AB and CD. We have then only four bodies
to consider, namely the four beams. The strings may be
supposed attached to the beams AB and CD.
The beam AD is acted on by forces at A and D arising
from the other beams. The two forces must be equal and
opposite, so that their line of action must coincide with
AD. Denote each force by 8.
Similarly the beam EG is acted on by two forces which
are equal and opposite, having EC for their line of action.
Denote each force by T.
The beam AB is acted on by four forces ; namely P
along AC, Q along BD, S having AD for its line of action,
and T having EG for its line of action.
We shall apply the conditions of equilibrium of Art. 88.
Take moments round E. Thus
S. AE sin BAD=P. AB sin BAO \
so that sin BAD=P sin BAO :
and S must act on the beam AB in the direction DA.
Take moments round A. Thus
T. A B sin ABC= Q.ABsinABO',
so that T sin ABC= Q sin ABO :
and T must act on the beam AE in the direction CB.
Take moments round 0. Thus T=S.
sin BAD P sin BAO
xience = . TJ ~ = ^ = . r)r . :
sm ABC Q sin ABO
P sin ABO AO
therefore  = =.
PROBLEMS.
195
289. A heavy rod rests with its ends on two given
smooth inclined planes : required the position of equi
librium.
Let AB be the rod,
AOM and BON the in
clined planes ; MON being
a horizontal line.
The forces acting on
the rod are the resistance
of the plane at A, at right
angles to OA, the resist
ance of the plane at B, at
right angles to OB, and
the weight vertically down
wards through the centre
of gravity of the rod. Let
AC and BG be the lines of action of the resistances of the
planes, and G the centre of gravity of the rod ; then the
directions of the three forces must meet at a point by
Art. 41, so that O must be vertically under C. Join CQ.
Let AG=a, BG=b, the angle AOM=a, and the angle
; and let 6 be the inclination of AB to the hori
zon, that is, the angle between AB and MN produced.
Since CA and CG are perpendicular to OA and OM
respectively, the angle GCA=th& angle AOJtf=a. Simi
larly the angle
The angle OAB=a6, and the angle ABO=ft + 0, by
Euclid, i. 32.
NOW ^r;
sin GAG
'smGCA'
sin GBG
GG
GA
GG
GB~s>mGCB
therefore, by division,
b _ sin ft cos (a  6} m
a ~ sin a cos (ft + 6) '
cos (a  ff) m
sin a
cos (ft + 0)
sin ;
132
196
therefore
PROBLEMS.
b sin a _ cos (a  ff} _ cos a cos 6 + sin a sin 6
a siii /3 ~~ cos (p + 6} ~ cos ft cos, 6 sin /3 sin
cos a + sin a tan
therefore tan#=
cos /3  sin j3 tan 6 '
Z> sin a cos /3  a sin /3 cos a
(a + 6) sin a sin /3
6 cot /3  a cot a
a + 6 "
290. As another example we will explain the Balance
called RobervaVs Balance.
AB and CD are
equal beams which
can turn in a ver
tical plane round
fixed points E and
F in the same ver
tical line ; AE being
equal to OF.
AC and BD are
equal beams connected with the former beams by pivots
at A, B, C, and J). HK is a beam rigidly attached to
AC, and ZJ/is a beam rigidly attached to BD ; the angles
A HK and BLM being right angles. A weight P is hung
at K and a weight W is hung at M : it is required to
find the ratio of P to W when there is equilibrium, neg
lecting the weights of the beams.
Since EF is vertical so are AC and BD ; and KU
and LM are horizontal.
The piece formed of KII and AC is acted on by the
weight P, by a force at A arising from the beam AB, and
by a force at C arising from the beam CD. Resolve the
force at A into two components, one vertical and the other
in the straight line A B. Resolve the force at C into two com
ponents, one vertical arid the other in the straight line CD.
PROBLEMS. 107
The components in the straight lines AB and CD must
be equal and unlike; for if they were not the sum of the
horizontal components of the forces on the piece would not
be zero.
Let Y denote the vertical force on the piece at A, sup
posed upwards : then the vertical force on the piece at C
must be P Fup wards: for if it were not the sum of the
vertical components of the forces on the piece would not
be zero.
Thus AB is acted on at A by some force in the straight
line AB, and by a vertical force T downwards. And CD is
acted on at C by some force in the straight line CD, and by
a vertical force P  Y downwards. Similarly AB is acted
on at B by some force in the straight line AB, and by a
vertical force downwards which we may denote by Z. And
CD is acted on at D by some force in the straight line CD t
and by a vertical force WZ downwards.
Also AB is acted on by some force at E, and CD by
some force at F.
Take moments round E for AB : thus
Yx EA sin EAC=Zx EB sin EBD ;
therefore Yx EA =Zx EB.
Take moments round F for CD ; thus
therefore (P  Y} EA = ( W  Z] EB.
Hence, by addition,
PxEA=WxEB.
Thus the ratio of P to TF is independent of the lengths
of HK and LM; and if EA=EB then P= W.
In practice CA and DB are produced vertically upwards,
and have pans rigidly attached to them, in which P and
Q are placed, instead of being hung at the ends of Kll
and LM.
198 MISCELLANEOUS EXAMPLES
Miscellaneous Examples in Statics.
1. The magnitudes of two bodies are as 3 is to 2, and
their weights are as 2 is to 1 : compare their densities.
2. Two forces act on a particle in directions at right
angles to each other ; they are in the ratio of 5 to 12, and
their resultant is equal to 65 Ibs. : find the forces.
3. Three forces represented by 24, 25, and 7 are in
equilibrium when acting on a particle : shew that two of
them are at right angles.
4. The resultant of two forces which act at right
angles on a particle is 51 Ibs.; one of the components
is 24 Ibs. : find the other component.
5. Two forces acting in opposite directions to one
another on a particle have a resultant of 28 Ibs.; and if
they acted at right angles they would have a resultant of
52 Ibs. : find the forces.
6. ABC is a triangle and D is the middle point of EC\
three forces represented in magnitude and direction by
AJ3, AC, DA act on a particle at A : find the magnitude
and the direction of the resultant.
7. Three forces 3, 4, 5 act on a particle in the centre
of a square in directions towards three of the angles of the
square : find the magnitude and the direction of the force
which will keep the particle at rest.
8. Forces */3 + 1, V?  1, and *J6 act on a particle : find
the angles between their respective directions that there
may be equilibrium.
9. Three forces, represented by those diagonals of
three adjacent faces of a cube which meet, act at a point :
shew that the resultant is equal to twice the diagonal of
the cube.
10. A string passing round a smooth peg is pulled at
each end by a force of 10 Ibs., and the angle between the
parts of the string on opposite sides of the peg is 120 : find
the pressure on the peg, and the direction in which it acts.
IN STATICS. 199
11. Three smooth pegs are fastened in a vertical piano
so as to form an isosceles triangle with the base horizontal
and the vertex downwards, and the vertical angle is equal
to 120. A fine string with a weight W attached to each
end is passed under the lower peg and over the other two
pegs. Find the pressure on each peg. Find also the
vertical pressure on each peg.
12. Find a point within an equilateral triangle at which
if a particle be placed it will be kept in equilibrium by
three forces represented by the straight lines joining the
point with the angular points of the triangle.
13. Forces represented in magnitude and direction by
the diagonals of a parallelogram act at one of the angles :
find the single force which will counteract them.
14. If R be the resultant of two forces P and Q
acting on a particle, and S the resultant of P and R,
shew that the resultant of S and Q will be 2.R.
15. Three equal forces act at a point, in directions
parallel to three consecutive sides of a regular hexagon:
find the magnitude and the direction of the resultant.
16. Shew that if one of two forces acting on a particle
be given in magnitude and position, and also the direction
of their resultant, the locus of the extremity of the straight
line representing the other force will be a straight line.
17. A weight is supported by two strings which are
attached to it, and to two points in a horizontal straight line :
if the strings are of unequal length, shew that the tension
of the shorter string is greater than that of the longer.
18. Two weights of 3 Ibs. and 4 Ibs. respectively are
connected by a string which is passed over two smooth pegs
in the same horizontal straight line : find what weight must
be attached to the string between the pegs in order that
when the weights have assumed their position of equilibrium
the string may be bent at right angles.
19. A weight is supported by two strings equally in
clined to the vertical : shew that if instead of one of them
we substitute a string pulling horizontally so as not to
disturb the position of the other, the tension of the latter
will be doubled.
200 MISCELLANEOUS EXAMPLES
. 20. Three forces act on a particle; the forces are
1 lb., 4 Ibs., and 6 Ibs. respectively, and the force of 4 Ibs.
is inclined at an angle of 60 to each of the other forces :
find the magnitude and the direction of their resultant.
21. Two couples acting along the sides of a parallelo
gram are in equilibrium : find the ratio of the forces.
22. A straight rod two feet in length rests in a hori
zontal position between two fixed pegs placed at a distance
of three inches apart, one of the pegs being at the end of
the rod ; a weight of 5 Ibs. is suspended at the other end of
the rod : find the pressure on each of the pegs.
23. A bent Lever has equal arms making an angle of
120: find the ratio of the weights at the ends of the arms
when the Lever is in equilibrium with one arm horizontal.
24. A heavy bent Lever of which one arm is twice the
length of the other, and of which the arms form a right
angle, is suspended by its angle, the point of suspension
being two feet above a horizontal table; the extremity of
the longer arm is just close to the table when the Lever is
in equilibrium by its own weight : find the height above
the table of the extremity of the shorter arm.
25. The ends of a uniform rod are connected by strings
with the ends of another uniform rod which is moveable
about its middle point : shew that when the system is in
equilibrium either the rods or the strings are parallel.
26. Two cylinders of the same diameter whose lengths
are 1 foot and 7 feet respectively, and whose weights are
in the ratio of 5 to 3 are joined together so as to form one
cylinder : find the position of the fulcrum about which the
whole will balance.
27. A uniform bar 1 1 feet in length and 4 Ibs. in weight
rests in a horizontal position upon a fulcrum 3 inches
distant from one end : find what weight acting at this end
will keep the rod at rest. Find also the pressure on the
fulcrum.
28. Find the centre of gravity of four weights 1 lb.,
2 Ibs., 3 Ibs., 4 Ibs., placed at the angular points of a square.
AV STATICS. 201
29. If a quadrilateral be such that one of its diagonals
divides it into two equal triangles, the centre of gravity of
the quadrilateral is in that diagonal.
30. Having given the positions of three particles A)B,C,
and the positions of the centre of gravity of A and 23, and
of the centre of gravity of A and C, find the position of the
centre of gravity of B and (7.
31. A heavy rightangled triangle is suspended by its
right angle, and the inclination of the hypotenuse to the
horizon is 40: find the acute angles of the triangle.
32. Two scale pans are suspended from the two ends
of a straight Lever whose arms are as 3 is to 4, and an iron
bar of 20 Ibs. weight is laid on the scale pans, and will just
reach from the one to the other : find what weight must be
put into one scale to preserve equilibrium.
33. A uniform rod 3 feet long and weighing 6 ounces is
held horizontally in the hand, being supported by means of
a finger below the rod two inches from the end, and the
thumb over the rod at the end : find the pressures exerted
by the finger and thumb respectively.
34. On a uniform straight Lever weighing 5 Ibs. and
5 feet in length, weights of 1, 2, 3, 4 Ibs. are hung at the
distances 1, 2, 3, 4 feet respectively from one end : find the
position of the fulcrum on which the whole will rest.
35. A uniform stick 6 feet long lies on a table with one
end projecting beyond the edge of the table to the extent
of two feet; the greatest weight that can be suspended
from the end of the projecting portion without destroying
the equilibrium is 1 Ib. : find the weight of the stick.
36. Two equal particles are placed on two opposite
sides of a parallelogram : shew that their centre of gravity
will remain in the same position, if they move along the
sides through equal lengths in opposite directions.
37. A beam capable of moving about one end is kept
in a position inclined to the horizon at an angle of 60
by a string attached to the other end; the string is in
clined to the horizon at an angle of 60 in an opposite
direction : compare the tension of the string with the
weight of the beam.
202 MISCELLANEOUS EXAMPLES
38. Two strings have each one of their ends fixed to a
peg, and the other to the ends of a uniform rod : when the
rod is hanging in equilibrium, shew that the tensions of
the strings are proportional to their lengths.
39. A sugar loaf whose height is equal to twice the dia
meter of its base stands on a table, rough enough to prevent
sliding, one end of which is gently raised until the sugar
loaf is on the verge of falling over : when this is the case
find the inclination of the table to the horizon.
40. A beam ten stone in weight and ten feet long
rests on two points distant four feet from each end : find
the greatest weight which is unable to turn it over, on
whatever point of the beam it be placed.
41. A heavy uniform beam of weight W is supported
in a horizontal position by two men, one at each end ; and
a weight Q is placed at threefifths of the beam from one
end : find the weight supported by each man.
42. A heavy beam is made up of two uniform cylin
ders whose lengths are as 3 is to 2, and weights as 3 is to
5 : determine the position of the centre of gravity.
43. Three weights of 2 Ibs., 3 Ibs., and 4 Ibs. respec
tively, are suspended from the extremities and the middle
point of a rod without weight: determine the point in
the rod about which the three weights will balance. If
the three weights be interchanged in all possible ways,
find how many such points there will be.
44. Four weights of 3 ounces, 2 ounces, 4 ounces, and
7 ounces respectively are at equal intervals of 8 inches on a
Lever without weight, two feet in length : find where the
fulcrum must be in order that they may balance. If the
Lever is uniform and weighs 8 ounces, find the position to
which it would be necessary to shift the fulcrum.
45. A rod 8 feet long balances about its middle point
with a weight of 5 Ibs. at one end, and a weight of 4 Ibs. at
the other end. If the weight of 5 Ibs. be removed it is
found that the rod will then balance about a point 1 foot 8
inches nearer the other end. Find the weight of the rod.
IN STATICS. 203
46. A rod 11 inches long has a weight of 7 ounces at
one end, and a weight of 8 ounces at the other end, and is
found to be in equilibrium when balancing on a fulcrum
5 inches from the heavier weight. If the weights are
interchanged the fulcrum must be shifted of an inch.
Find the weight of the rod, and the position of its centre
of gravity.
47. If any triangle be suspended from the middle
point of its base, and likewise a plumb line from the same
point, shew that the plumb line will pass through the
vertex of the triangle. If now we place a weight equal to
one third of the weight of the triangle at either angle of
the base, shew that the triangle will assume a position
such that all the angles will have their perpendicular
distances from the plumb line equal.
48. A heavy triangle is hung up by the angle A, and
the opposite side is inclined at an angle 6 to the vertical :
if B be the smaller of the other two angles of the triangle
shew that 2 cot Q =cot B  cot C.
49. Find the centre of gravity of a uniform wire 16
inches long, bent so as to form three sides of a rectangle,
the middle length being 6 inches. If the ends be brought
together so as to form a triangle, shew that the centre
of gravity will be of an inch nearer to the base.
50. A uniform plank 20 feet long and weighing 42 Ibs.
is pkced over a rail ; two boys, weighing respectively 75 Ibs.
and 99 Ibs., stand on the plank, each one foot from the end:
find the position of the rail for equilibrium. If the two
boys change places, find where a third boy weighing 72 Ibs.
must stand so as to maintain equilibrium without shifting
the plank on the rail.
51. Find the centre of gravity of a cube from one
corner of which a cube whose edge is onehalf the edge of
the first has been removed.
204 MISCELLANEOUS EXAMPLES
52. A pyramid is cut from a cube by a plane which
passes through the extremities of three edges that meet at
a point : find the distance of the centre of gravity of the
remainder of the cube from the centre of the cube.
53. Two forces of 6 and 8 Ibs. respectively act at the
ends of a rigid rod without weight 10 feet long; the forces
are inclined respectively at angles of 30 and 60 to the
rod : find the force which will keep the rod at rest, and the
point at which its direction crosses the rod.
54. A Wheel and Axle have radii respectively 2 feet
4 inches, and 5 inches. Find the Power which will balance
a Weight of 3 cwt.
55. In the Wheel and Axle, supposing the rope which
supports the Power to pass over a fixed pully so as to be
horizontal on leaving the Wheel, find what difference would
be made in the pressures on the fixed supports of the
machine.
56. Find the magnitude of the Weight in the second
system of Pullies if it exceed the Power by 40 Ibs., and
there are 6 parts of the string at the lower Block.
57. In the single moveable Pully with parallel strings
a weight of 100 Ibs. is suspended from the block, and the
end of the string in which the Power acts is fastened at
the distance of 2 feet from the fulcrum to a straight hori
zontal Lever 5 feet long, the fulcrum being at one end : find
the force which must be applied at the other end of the
Lever to preserve equilibrium.
58. If the weights of the Pullies in the first system,
commencing with the highest, be 1, 2, 5, 6 Ibs. respectively,
find what Power will sustain a Weight of 24 Ibs.
59. A capstan has four spokes, each projecting 8 feet
from the axis. The cylinder round which the rope is
wound has a diameter of 7 inches, and the rope itself is
half an inch thick. If four men exert a force of 60 Ibs.
each at the ends of the spokes, find the tension of the rope.
60. A weight of 56 Ibs. rests on a rough Plane inclined
at an angle of 45 to the horizon : find the pressure at right
angles to the plane.
IN STATICS. 205
61. A body whose weight is >/ 2 Ibs. is placed on a
rough Plane inclined to the horizon at an angle of 45. The
coefficient of friction being rr, find in what direction a
vj
force of (V3  1) Ibs. must act on the body in order just to
support it.
62. A uniform pole leans against a smooth wall at an
angle of 45, the lower end being on a rough horizontal
plane : shew that the amount of friction required to prevent
sliding is half the weight of the pole.
63. A rough Plane is inclined to the horizon at an angle
of 60 : find the magnitude and the direction of the least
force which will prevent a body weighing 100 Ibs. from
sliding down the Plane, the coefficient of friction being ^.
V3
"64. A triangular plate is suspended by three parallel
strings attached to the three corners ; one of the strings
can bear a weight of 2 Ibs. without breaking, and each of
the other two can bear a weight of 1 Ib. without breaking :
find the point of the triangular plate on which a weight of
4 Ibs. may be placed without breaking any of the strings.
65. ABCD is a triangular pyramid, is a point within
it ; like parallel forces act at A, B, C, D proportional re
spectively to the volumes of the triangular pyramids
OBCD, OCDA, ODAB, OABC : shew that the centre of
the parallel forces is at 0.
66. Parallel forces act at the angular points of a tri
angular pyramid, each force being proportional to the area
of the opposite face ; shew that the centre of the parallel
forces is either at the centre of the inscribed sphere, or at
the centre of one of the escribed spheres.
67. Two equal spheres are strung on a thread, which
is then suspended by its extremities so that its upper por
tions are parallel : find the pressure between the spheres.
68. Two uniform rods AB, BG of similar material are
connected by a smooth hinge at B, and have smooth rings
at their other ends which slide upon a fixed horizontal
wire : shew that in equilibrium the smaller rod is vertical.
206 MISCELLANEOUS EXAMPLES
69. A rod AB is fixed at an inclination of 60 to a
vertical wall ; and a heavy ring of weight W slides along
it. The ring is supported by a tight string attached to
the wall. Shew that the tensions of this string, when the
ring is respectively pulled up and pulled down the rod by a
W
force acting along the rod, are as 1 is to 3.
70. Parallel forces P, Q, R, S act at the angular points
of a tetrahedron : determine the parallel forces which must
act at the centres of gravity of the faces of the tetrahedron,
so that the second system may have the same centre and
the same resultant as the first.
71. Perpendiculars are drawn from the angles of a
triangle on the opposite sides ; and at the feet of these
perpendiculars act parallel forces proportional to sin 2 A,
sin 2.5, sin 2(7 : shew that their centre coincides with the
centre of parallel forces proportional to tan A, tan B } tan C
at the angular points.
72. Two equal heavy rods of weight W are joined by a
hinge at one end, and connected at the other ends by a
thread on which a weight w is capable of sliding freely :
the system is then placed with the hinge resting on a
horizontal plane, so that the two rods are in a vertical
plane : shew that in the position of equilibrium the hanging
weight cuts the vertical between the hinge and the hori
zontal straight line through the extremities of the rods in
the ratio of W to w.
73. Three equal rods AB, BC, CD without weight,
connected by hinges at B and C, are moveable about
hinges at A and D, the distance AD being twice the length
of each rod. A force P acts at the middle point of each
of the rods AB and CD, and at right angles to them :
shew that the pressure on each of the hinges A and
D will be jz , and that its direction will make an angle
of 60 with *AB.
74. Two weights support each other on a rough double
Inclined Plane by means of a fine string passing over the
vertex, and no friction is called into operation : shew that
IN STATICS. 207
the Plane may be tilted about either extremity of the base
through an angle 2e without disturbing the equilibrium,
e being the angle of friction, and both angles of the Plane
being less than 90  1 .
75. A Lever without weight is c feet in length, and
from its ends a weight is supported by two strings in
length a and b feet respectively : shew that the fulcrum
must divide the Lever into two parts, the ratio of which is
that of a 2 t c 2  6 2 to 6 2 + c 2  a 2 , if there be equilibrium
when the Lever is horizontal.
76. A uniform rod rests with one extremity against a
rough vertical wall, the other extremity being supported
by a string three times the length of the rod, attached to a
point in the wall ; the coefficient of friction is  : shew that
the tangent of the angle which the string makes with the
wall in the limiting position of equilibrium is f r or .
77. If when two particles are placed on a rough double
Inclined Plane, and connected by a string passing over a
smooth peg at the vertex, they are on the point of motion,
and when their positions are interchanged, no friction is
called into play, shew that the angle of friction is equal to
the difference of the inclinations of the two Planes.
78. A plane equilateral pentagon is formed of five
equal uniform rods AJ3, BC, CD, DE, EA loosely jointed
together. The angular points B, D of the pentagon are
capable of sliding on a smooth horizontal rod, and the
plane of the pentagon is vertical, the point C being upper
most. Shew that if 0, be the respective inclinations of
the rods A B, BC to the horizon in the position of equili
brium, 2 tan $ = tan 6.
79. A uniform wire is formed into a triangle ABC, the
lengths of the sides of which are a, 6, c respectively : shew
that if x, y, z be the respective distances of the centre of
gravity of the wire from the middle points of its sides,
80. If a particle be in equilibrium under the action of
four equal forces, tending to the angular points of a tetra
hedron, prove that the three straight lines passing through
208 MISCELLANEOUS EXAMPLES.
the point, and through each pair of opposite edges of tho
tetrahedron are at right angles to each other.
81. Two weights are connected by a fine inextensible
string which passes over a Fully ; and one rests on a rough
Inclined Plane, while the other hangs freely ; if the string
make angles 6 lt 2 with the Plane in the highest and
lowest positions of equilibrium of the free weight, and 6
when no friction is called into play, shew that
cos d 2 + cos 6 l  2 cos 6 =p, (sin 2  sin QJ,
where \L is the coefficient of friction.
82. A cylinder open at the top, stands on a horizontal
plane, and a uniform rod rests partly within the cylinder,
and in contact with it at its upper and lower edges in a
vertical plane containing the axis of the cylinder : sup
posing the weight of the cylinder to be n times that of the
rod, find the length of the rod when the cylinder is on the
point of tumbling.
83. Two equal rough balls lie in contact on a rough
horizontal table ; another ball is placed upon them so that
the centres of the three are in a vertical plane : find the
least coefficient of friction between the upper and lower
balls and between the lower balls and the table, in order
that the system may be in equilibrium.
84. Two uniform beams of equal weight but of unequal
length, are placed with their lower ends in contact on a
smooth horizontal plane, and their upper ends against
smooth vertical planes : shew that in the position of equi
librium the two beams are equally inclined to the horizon.
85. A bowl is formed from a hollow sphere of radius a ;
it is so fixed that the radius of the sphere drawn to each
point in the rim makes an angle a with the vertical, and the
radius drawn to a point A of the bowl makes an angle /3
with the vertical : if a smooth uniform rod remains at rest
when placed with one extremity at A, and with a point in
its length on the rim of the bowl, shew that the length of
the rod is 4a sin B sec  (a  /3).
DYNAMICS.
I. Velocity.
1. DYNAMICS treats of force producing or changing
the motion of bodies.
Before we consider the influence of force on the motion
of bodies we shall make some remarks on motion itself:
we confine ourselves to the case of motion in a straight
line.
2. The velocity of a point in motion at any instant is
the degree of quickness of the motion of the point at
that instant.
3. If a point in motion describe equal lengths of path
in equal times the velocity is called uniform or constant.
Velocity which is not uniform is called variable.
4. Uniform velocity is measured by the length of path
described in the unit of time. "We may take any unit of
time we please ; and a second is usually chosen. We may
also take any unit of length we please : and a foot is usu
ally chosen. Thus by the velocity 16 we mean the velocity
of a point which moves uniformly in such a manner that
the length of path described in one second is sixteen feet.
The word space is used as an abbreviation of the term
length of path: thus in the example just given it would
be said that the space described in one second is sixteen
feet.
T. MB. 14
210 VELOCITY.
5. If a point moving with tJie uniform velocity v describe ^
the space s in the time t, then s=vt.
For in one unit of time v units of space are described,
and therefore in t units of time vt units of space are de
scribed ; therefore s=vt.
6. Variable velocity is measured at any instant by the
space which would be described in a unit of time, if the
velocity were to continue during that unit of time the
same as it is at the instant considered.
Hence, as in Art. 5, if v denote the measure of a vari
able velocity at any instant, a point moving for the time
t with this velocity would describe the space vt.
7. The mode of measuring variable velocity is one
with which we are familiar in practice. Thus a railway
train may be moving with variable velocity, and yet we
may say that at a certain instant it is moving at the rate of
30 miles an hour ; we mean that if the train were to con
tinue to move for one hour with just the same speed as at
the instant considered it would pass over 30 miles.
8. The illustration just employed suggests that a velo
city may be given expressed in any units of time and
space ; it is easy to express the velocity in terms of the
standard units.
For example, suppose that a body is moving at the rate
of 30 miles an hour. The body here is moving at the rate
of 30 x 5280 feet in an hour, that is, in 60 x 60 seconds :
, 30 x 5280 , ,
hence it is moving at the rate of ^ ^ feet in one
bO x bO
second, that is, at the rate of 44 feet in one second. Hence
44 denotes the velocity when expressed in the standard
units.
In like manner we may pass from the standard units to
any other units.
For example, if v denote a velocity when a second is
taken as the unit of time, the same velocity will be denoted
by 60v when a minute is taken as the unit of time. For to
EXAMPLES. I. 211
say that a body is moving at the rate of v feet per second
is equivalent to saying that it is moving at the rate of 60 v
feet per minute.
In like manner if we wish to take a yard for the unit of
length instead of a foot, as well as a minute for the unit of
time instead of a second, the velocity denoted by v with
the standard units will now be denoted by 5.
o
Generally, let v denote a velocity when a second is the
unit of time, and a foot is the unit of length ; then if we
take m seconds as the unit of time, and n feet as the
unit of length, the same velocity will be denoted by .
EXAMPLES. I.
1. Compare the velocities of two points which move
uniformly, one through 5 feet in half a second, and the
other through 100 yards in a minute.
2. Compare the velocities of two points which move
uniformly, one through 720 feet in one minute, and the
other through 3 yards in three quarters of a second.
3. Two points move uniformly with such velocities that
when they move in the same direction the distance between
them increases at the rate of 5 feet per second, and when
they move in opposite directions the distance between them
increases at the rate of 25 feet per second : find the velo
city of each.
4. A railway train travels over 100 miles in 2 hours :
find the average velocity referred to feet and seconds.
5. One point moves uniformly round the circumference
of a circle, while another point moves uniformly along the
diameter : compare their velocities.
6. One point describes the circumference of a circle of
a feet radius in b minutes ; and another point describes the
circumference of a circle of b feet radius in a minutes :
compare their velocities.
142
212 THE FIRST LAW OF MOTION.
II. The First and Second Laics of Motion.
9. The science of Dynamics rests on certain principles
which are called Laws of Motion. Newton presented
them in the form of three laws ; and we shall follow him.
It is not to be expected that a beginner will obtain a
clear and correct idea of these laws on reading them for the
first time ; but as he proceeds with the subject and ob
serves the applications of the laws he will gradually discover
their full import. In like manner a beginner of geometry
rarely comprehends at first all that is meant by the defini
tions, postulates, and axioms ; but the imperfect notions
with which he starts are corrected and extended as he
studies the propositions.
In the present Chapter we shall chiefly discuss the First
Law of Motion.
10. First Law of Motion. Every body continues in
a state of rest or of uniform motion in a straight line,
except in so far as it may be compelled to change that state
by force acting on it.
It is necessary to limit the meaning of the word motion
in the First Law. By the motion of a body is here meant
that kind of motion in which every point of the body
describes a straight line ; in other words, there is to be no
rotation. The rotation of bodies is discussed in works
which treat of the highest branches of dynamics, and many
important results are demonstrated : for example, it is
shewn that if a free sphere of uniform density be rotating
about a diameter at any instant, it will continue to rotate
about that diameter if no force act on it.
In order to exclude all notion of rotation, some writers
use the word particle instead of body in enunciating the
First Law of Motion.
We must now proceed to consider the grounds on
which we rest our belief in the truth of the First Law of
Motion.
THE FIRST LAW OF MOTION. 213
11. Little direct experimental evidence can be brought
forward in favour of the truth of the Law. It is in fact
impossible to preserve a body which is in motion from the
action of external forces ; and so it is impossible to obtain
that perseverance in uniform motion of which the Law
speaks. If we start a stone to slide along the ground we
find that the stone is soon reduced to rest ; but we have
no difficulty in perceiving that the destruction of motion
is due mainly to the friction of the ground. Accordingly
we find that if the same stone is started with the same
velocity to slide on a smooth sheet of ice, it will proceed
much farther before it is reduced to rest. And we may
easily imagine that if all such external forces as friction of
the ground and resistance of the ah 1 were removed the
motion would continue permanently unchanged.
In this illustration we suppose the stone to slide along
the ground ; we do not suppose the stone to roll, for the
reason which is assigned in Art. 10.
12. But although the direct experimental evidence of
the truth of this and of the other Laws of Motion is weak,
the indirect evidence is very strong. For on these laws
as a foundation the whole science of dynamics rests ; the
theory of astronomy forms a part of dynamics, and it is a
matter of every day experience that the calculations and
predictions of astronomy are most closely verified by ob
servation. It seems in the highest degree improbable that
numerous and intricate results, deduced from untrue laws,
should be uniformly true ; and accordingly we say that the
agreement of theory and observation in astronomy justifies
us in accepting the Laws of Motion.
13. From the First Law of Motion then we see that a
body has no power to put itself in motion, or to change its
motion ; but a commencement or change of motion must
be ascribed to the action of some external force.
14. It will be readily conjectured from common expe
rience, that the effect of a given force in communicating
or changing motion may depend partly on the size and the
214 THE SECOND LAW OF MOTION.
kind of the body to which the force is applied ; and this
point will be discussed hereafter, so that we shall be able
to compare the effect of a force on one body with the effect
of the same force on another body. But at present we
confine ourselves to the case in which a given force acts on
a given body, so that we have only to consider the influence
of the force on the velocity of the body.
15. Second Law of Motion. Change of motion is pro
portional to the acting force, and takes place in the direction
of the straight line in which the force acts.
This law will require to be developed in order to place
before the student all which its concise statement includes ;
but this development we shall reserve, as at present we
only require a part of the law. We suppose a body in
motion in a straight line, and acted on by a force in the
direction of that straight line. Then we require so much of
the Second Law of Motion as to enable us to assume that a
given force communicates the same velocity in a given
time, whatever be the velocity which the body already has.
This is in fact included in the first clause of the Law:
change of motion is proportional to the acting force. The
whole meaning of this clause will be exhibited hereafter :
see Art. 84.
It was scarcely necessary to introduce here even this
brief notice of the Second Law of Motion ; but without it
the definition of uniform force which is given in the next
Chapter might appear arbitrary and unnatural.
Although the student must not consider that he has
mastered the subject until he understands the Laws of
Motion, yet it is by no means necessary to weary himself by
trying to understand these Laws fully before he passes on
to any results deduced from them. He will learn more by
examining the way in which these Laws are applied than
by confining himself to the Laws themselves.
EXAMPLES. IT. 215
EXAMPLES. II.
1. Two bodies start together from the same point and
move uniformly along the same straight line in the same
direction ; one body moves at the rate of 15 miles per hour,
and the other body at the rate of 18 feet per second : deter
mine the distance between them at the end of a minute.
2. If the bodies move with the velocities of the pre
ceding Example but in opposite directions, find when they
will be 200 feet apart.
3. A body starts from a point and moves uniformly
along a straight line at the rate of 30 miles per hour. At
the end of half a minute another body starts from the
same point after the former body, and moves uniformly at
the rate of 55 feet per second. Find when and where the
second body overtakes the first.
4. Two bodies start together from the same point and
move uniformly in directions at right angles to each other;
one body moves at the rate of 4 feet per second, and the
other body at the rate of 3 feet per second : determine the
distance between them at the end of n seconds.
5. Supposing the earth to be a sphere 25000 miles in
circumference, and turning round once in a day, determine
the velocity of a point at the equator.
6. A mill sail is 7 yards long, and is observed to go
round uniformly ten times in a minute : find the velocity of
the extremity of the sail
7. Two bodies start from the same point and move
uniformly with the same velocity along straight lines in
clined at an angle of 60: find their distance apart at the
end of a given time.
8. Two bodies start from the same point and move
uniformly along straight lines inclined at an angle a : if the
velocity of one body be u and the velocity of the other body
v, find their distance apart at the end of n seconds.
216 NOTION IN A STRAIGHT LINE
III. Motion in a straight line under the influence of a
uniform force.
16. We confine ourselves to the case of motion in a
straight line, and the direction of the force is supposed to
be in the same straight line as that of the motion ; and we
consider only the effect of the force on the velocity without
regard to the size and the kind of the body moved.
17. If a force acting on a body adds equal velocities
in equal times, the force is called uniform or constant.
Force which is not uniform is called variable.
18. Uniform force acting on a given body is measured
by the velocity which is added in each successive unit of
time. Variable force acting on a given body is measured
at any instant by the velocity which would be added in a
unit of time, if the force were to continue during that unit
the same as it is at the instant considered.
19. We are now about to give some propositions re
specting uniform force acting on a given body. The term
acceleration is used as an abbreviation for the velocity
added in a unit of time; so that when we speak of an
acceleration /, we mean that by the action of a given force
on a given body the velocity /is added in a unit of time.
20. A uniform force acts on a body in a fixed direction
during the time t : iffbe the acceleration, and v tJie velocity
generated, then v=ft.
For by the definition of uniform force, in each unit of
time the velocity / is communicated to the body ; and
therefore in t units of time the velocity ft is communicated.
21. A body starting from rest is acted on by a uniform
force in a fixed direction: if f be the acceleration, and s the
('pace described in the time t, then s= ft 2 .
UNDER A UNIFORM FORCE. 217
Let the whole time t be divided into n equal intervals ;
denote each interval by r, so that nr=t.
Then the velocity of the body at the end of the times
T, 2r, 3r, ...... (ftl)r, UT
from starting, is, by Art. 20, respectively,
/r, 2/r, 3/r, ...... (*ll)/r, tt/r.
Let s 1 denote the space which the body would describe
if it moved during each interval r with the velocity which
it has at the beginning of the interval ; and let s 2 denote
the space which the body would describe if it moved
during each interval T with the velocity which it has at the
end of the interval. Then
*2=/r.T+2/r.r+3/r.r
that is,
Hence, by the theory of Arithmetical Progression in
Algebra, we have
(nV}n fi* (nl)n ffif \
~
1
Now *, the space actually described, must lie between
s l and 2 ; but by making n large enough we can make 
as small as we please ; so that we can make s t and s 2 differ
from nft* by less than any assigned quantity. Hence
218 MOTION IN A STRAIGHT LINE
22. Thus if a body start from rest and be acted on for
the time t in a fixed direction by a uniform force of which
the acceleration is/, we have the following values, of v the
velocity acquired, and s the space described,
*=f* (1),
*=2^ 2 (2).
From (1) and (2) by eliminating t we have
* 2 =2/s (3);
this gives the velocity in terms of the acceleration and
the space described.
We may of course modify the forms of these expres
sions by common Algebra ; for example, (2) may be written
thus:
f"
From (1) and (2) we may deduce
this shews that the space actually described is half that
which would be described by a body moving for the time t
with a uniform velocity equal to v.
These formulas are very important, and will often be
applied.
23. Falling bodies. When bodies are allowed to fall
freely to the surface of the earth from heights above it,
we find that different bodies fall through equal spaces
from rest in a given time, and that the space fallen through
in any time from rest varies as the square of the time.
These laws at least hold approximately, and the resistance
of the air appears to be the reason of such deviations from
exact conformity with these laws as may be observed. For
example, a sovereign and a feather do not fall to the
ground in the same time if the experiment be tried in the
UNDER A UNIFORM FORCE. 219
open air; but they do if the experiment be tried in the
exhausted receiver of an airpump.
From these observed facts, compared with the results
given in the preceding Article, we infer that the Earth
exerts a focce in the vertical direction on all bodies, that
this force is a uniform force, and that it produces the same
acceleration in all bodies. This force is called gravity.
24. The letter g is invariably used to denote the acce
leration produced by gravity.
It is found that the value of g increases slightly as we
pass from the equator towards the poles. At London
<7= 32.19 feet nearly, the unit of time being one second.
That is, when a body falls freely in the latitude of London
a velocity of 32.2 feet nearly is communicated to it every
second.
Moreover, the value of g is not the same at different
heights above the same point of the Earth's surface ; the
force which the Earth exerts on a given body varies very
nearly inversely as the square of its distance from the cen
tre of the Earth. But as any heights to which we can
ascend are very small compared with the radius of the
Earth, the change thus produced in the force of gravity will
also be very small.
The direction of the force of gravity is perpendicular
to the horizontal plane at every place, and so really varies
from point to point on the Earth's surface. But this
variation will be scarcely sensible so long as we do not
move more than a few miles from an assigned spot.
Thus on the whole we may practically, in the vicinity of
any assigned spot, regard the direction of the force of
gravity as parallel to the same straight line, and the value
of g as constant.
25. The laws respecting the variation of the force of
gravity which we have stated in the preceding Article are
suggested by observation and experiment; their exact
truth is established by shewing that results deduced by
calculation from these assumed laws are verified in nume
rous cases; see Art. 12.
220 MOTION IN A STRAIGHT LINE
26. Thus, by Art. 22, when bodies fall from rest the
following formula) apply :
v=gt...(I); =i^...(2); * 2 =2<7*...(3).
For example, take the second formula, and put for t in
succession 1, 2, 3, 4,...; thus we obtain the following re
sults: in 1, 2, 3, 4,... seconds respectively from rest the
1 4 9 16
spaces described are g t g t g t g.... Hence by sub
tracting each of these numbers from that which follows it,
we find that the spaces described in the second, third,
fourth, . . . seconds respectively are  g,  g,  g, ....
And generally in n 1 seconds the space described
is ^(n Yfg; in n seconds the space described isrfig;
'2i 2t
hence during the n ib second the space described is
\n*g\(nV?g 9 that is, \(2n\}g.
The velocity acquired by a body falling from rest
through the space s is sometimes called the velocity due to
the space s under the action of gravity; and the space
through which a body must fall from rest to acquire a
velocity v is sometimes called the space due to the velocity
v under the action of gravity.
27. Motion down an Inclined Plane. We may now
consider the motion of a body sliding down a smooth In
clined Plane.
Suppose there is a smooth Plane inclined at an an
gle a to the horizon. The force which the earth exerts
on a body acts in the vertical direction; we may resolve
this force into two components, one along the Plane, and
the other perpendicular to the Plane. The component
along the Plane is obtained by multiplying the whole force
by sin a, and so we may naturally assume that the accele
ration due to this component is obtained by multiplying
the whole acceleration by sin a : thus, this acceleration is
g sin a. The force perpendicular to the Plane has no influ
ence on the motion down the Plane ; it is counteracted by
the resistance of the Plane.
UNDER A UNIFORM FORCE.
221
Hence we conclude that the motion of a body sliding
down a smooth Inclined Plane is similar to that of a body
falling freely ; the only difference is that g sin a must be
put instead of g in the formulae of Art. 26, so that the
motion of the sliding body is slower than that of the body
falling freely.
"We shall in Chapter vn. consider the reason which
justifies the assumption of the present Article.
28. Thus if I be the length of an Inclined Plane, and
v the velocity acquired by a body in sliding down it from
rest, we have v z =2gl sin a by equation (3) of Art. 26. Let
h be the height of the Plane; then A=?sin a; thus v 2 =2gh.
Hence the velocity acquired in sliding down a smooth
Inclined Plane is the same as would be acquired in falling
freely through a vertical space equal to the height of the
Plane.
29. The time of falling from rest down a chord of a
vertical circle drawn from the highest point is constant.
Let A be the highest point of
a vertical circle, AB a diameter,
AC any chord. Let a be the
inclination of AC to the horizon;
then the angle BAG '=90 a, and
therefore the angle ABC=a.
Let t be the time of falling
down AC; then by Art. 27
And AC=ABama; so that AB sm
therefore
That is, t is equal to the time of falling freely down the
vertical diameter AB. This establishes the proposition.
222
NOTION IN A STRAIGHT LINE
In the same manner we may shew that the time of
falling from rest down a chord passing through the low
est point is constant.
30. If two circles touch each other at their highest or
lowest point, and a straight line be drawn through this
point, the time of falling fi'om rest down a straight line
intercepted between the circumferences is constant.
Let two circles touch
each other at their high
est point A. Through
A draw any straight line
ADE, cutting the cir
cumferences at D and
E. Let the vertical
straight line through A
cut the circumferences
at B and C. On BG
as diameter describe a
circle ; join EC, cutting
the circumference of
this circle at F. Join
The angles at D, E, and F are right angles. Therefore
BF is parallel and equal to DE.
Hence the time from rest down DE is the same as the
time from rest down BF', and is therefore equal to the
time from rest down BG, by Art. 29. Thus the time is
the same for every such straight line as DE.
Similarly the proposition may be established when the
circles touch at their lowest point.
31. The two preceding results will enable us to solve
various problems with respect to straight lines of quickest
descent. We will give some examples : we suppose in
every example that the entire figure is in one vertical
UNDER A UNIFORM FORCE. 223
plane, and the moving body is supposed to start from rest.
The first six Examples depend on Art. 29. the rest on
Art. 30.
Required the straight line of quickest descent :
(1) From a given point to a given straight line.
Let A be the given point, R A
BG the given straight line,
AB a horizontal straight
line through A. Draw a
circle touching AB at A and
also touching BC ; let D be
the point of contact with
BC: then AD is the re
quired straight line. For
draw through A any chord of the circle AE, and produce
it to meet the straight line BO at F. Then the time down
AD is equal to the time down AE, and is therefore less
than the time down AF.
Since the two tangents BA and BD are equal, the
point D is determined simply by taking BD down BC,
equal to BA.
The demonstration of this will give sufficient aid for the
next five cases.
(2) From a given straight line to a given point.
Let A denote the given point ; let a horizontal straight
line through A meet the given straight line at B ; take
BD up the given straight line =.#.4 : then DA is the
required straight line.
(3) From a given point without a given circle to a
given circle.
Join the given point with the lowest poiut of the given
circle : the part of the joining straight line which is out
side the given circle is the straight line required.
For the geometrical part of this and the next three
cases see Appendix to Euclid, No. 9.
224 MOTION IN A STRAIGHT LINE
(4) From a given circle to a given point without it.
Join the given point with the highest point of the given
circle : the part of the joining straight line which is outside
the given circle is the straight line required.
(5) From a given point within a given circle to the
Join the given point with the higJiest point of the given
circle : the part of the joining straight line produced which
is between the point and the circle is the straight line
required.
(6) From a given circle to a given point within it.
Join the given point with the lowest point of the given
circle : the part of the straight line produced which is
between the circle and the point is the straight line re
quired.
(7) From a given straight line without a given circle
to the circle.
Through A the lowest point of the circle draw a straight
line touching the circle, and meeting the given straight
line at B; take BG up the given straight 5ne>2Ll, and
join AC meeting the circle at D : then CD is the required
straight line.
For it follows from (3) that whatever be the point on
the straight line, the straight line produced must pass
through the lowest point of the given circle. And then,
by Art. 30, the point on the straight line must be the point
of contact of a circle drawn to touch this straight line and
also to touch the given circle at its lowest point.
(8) From a given circle to a given straight line without
the circle.
Through A the highest point of the circle draw a
straight line touching the circle, and meeting the given
straight line at B ; take BC down the given straight line
=J3A, and join A C meeting the circle at D\ then DC is
the required straight line.
The demonstration is like that in (Y).
UNDER A UNIFORM FORCE. 225
(9) From a given circle to another given circle
without it.
Join the highest point of the first circle with the lowest
point of the second circle; the part of this straight line
which is between the two circles is the straight line
required.
For it follows from (4) that whatever be the point on
the second circle the straight line produced must pass
through the highest point of the first circle. And then,
by Art. 30, the point on the second circle must be the
point of contact of a circle drawn to touch the first circle
at its highest point, and also to touch the second circle.
The demonstration of the next two cases is similar
to this.
(10) From a given circle to another given circle
within it.
. Join the lowest point of the first circle with the lowest
point of the second circle : the part of the joining straight
line which ia between the two circles is the straight fine
required.
(11) From a given circle within another given circle
to the outer circle.
Join the highest point of the first circle with the
highest point of the second circle : the part of the joining
straight line which is between the two circles is the straight
line required.
32. In the preceding three Articles we have supposed
for simplicity that the motion takes place in a vertical
plane : but similar results will hold if the motion takes
place down a fixed smooth Inclined Plane. If /3 be the
inclination of such a Plane to the horizon, then we shall
merely have to put g sin /3 instead of g in the investigation
of Arts. 29 and 30. And in Arts. 29 and 30 we may put
sp/iere instead of circle.
33. The following problem furnishes an interesting
application of the formulae of the present Chapter. A
T. ME. 15
26 MOTION IN A STRAIGHT LINE
person drops a stone into a well and after n seconds
hears it strike the water : find the depth of the surface of
the water.
We neglect the resistance of the air. It appears from
experiments that the velocity of sound is uniform and
equal to about 1130 feet per second : we will denote this
number by u.
Let x be the number of feet in the depth of the sur
face ; then the number of seconds taken by the stone to
fall to the surface of the water is / . by Art. 26 ; and
v' 9
the number of seconds taken by the passage of the sound
is  : therefore
u
**x
By solving this quadratic equation we obtain
9 V \ff J
The upper sign must be taken because *Jx is by sup
position a positive quantity. By squaring we obtain
~ 9 9 U V W 2
u 2 //u 2 2un\
therefore x= + unu 4 / , H
9 V \9 2 9 I
It will be found that =35 nearly : thus we have very
approximately
# =w{35 + 71 V(1225 + 70?i)}.
For example, if n=3 ; then
#tt{38  Vl435}=w(38 ~ 3 7'88) nearly
=wx 12 = 1356.
UNDER A UNIFORM FORCE. 227
34. In Art. 8 we have explained the change made
in the expression of a given velocity by changing the units
of time and space : we must now consider the change made
in the expression of a given acceleration.
Let / denote an acceleration when a second is taken
as the unit of time, then the same acceleration will be
denoted by (60) 2 / when a minute is taken as the unit of
time. For an acceleration is measured by the velocity
communicated in a unit of time. In the present case/ is
communicated in one second, therefore 2/ in two seconds,
....and 60/ in 60 seconds. But by Art. 8 a velocity which
is denoted by 60/ when a second is the unit of time will be
denoted by 60 x 60/ when a minute is the unit of time.
Hence (60) 2 / is the measure of the acceleration referred to
a minute as the unit of time.
In like manner if we wish to take a yard for the
unit of length instead of a foot, as well as a minute for the
unit of time instead of a second, the acceleration denoted
by/ with the standard units will now be denoted by
Generally, let / denote an acceleration when a second
is the unit of time and a foot is the unit of length ; then if
we take ra seconds as the unit of time, and n feet as the
unit of length, the same acceleration will be denoted by
2
n J
EXAMPLES. III.
The following examples all relate to uniformly acce
lerated motion :
1. A body has described 50 feet from rest hi 2 seconds :
find the velocity acquired.
2. A body has described 50 feet from rest in 2 seconds :
find the time it will take to move over the next 150 feet.
3. A body moves over 63 feet in the fourth second,
find the acceleration.
228 EXAMPLES. III.
4. A body describes 72 feet while its velocity increases
from 16 to 20 feet per second : find the whole time of
motion, and the acceleration.
5. A body in passing over 9 feet has its velocity
increased from 4 to 5 feet per second : find the whole
space described from rest, and the acceleration.
6. Two bodies uniformly accelerated in passing over
the same space have their velocities increased from a to b,
and from u to v respectively : compare the accelerations.
7. Find the numerical value of the acceleration when
in half a second a velocity is produced which would carry
a body over four feet in every quarter of a second.
8. A body moving from rest is observed to move
over 80 feet and 112 feet respectively in two consecutive
seconds : find the acceleration, and the time from rest.
9. A body moving from rest is observed to move
over a feet and b feet respectively in two consecutive
seconds : shew that the acceleration is b  a, and find the
time from rest.
10. A body uniformly accelerated is found to be
moving at the end of 10 seconds with a velocity which
would carry it through 45 miles in the next hour : find
the acceleration.
11. A body moving with uniform acceleration describes
20 feet in the half second which follows the first second
of its motion : find the acceleration.
12. Two bodies are let fall from the same point at
an interval of one second : find how many feet they will
be apart at the end of five seconds from the fall of the
first.
13. Two particles are let fall from two given heights :
find the interval between then: starting if they reach the
ground at the same instant.
14. A body is let fall : find how many inches it moves
over in the first half second of its motion : if it were
EXAMPLES. III. 229
made to move uniformly during the next half second with
the velocity then acquired, find over what space it would
move.
15. A body slides down a smooth Inclined Plane of
given height : shew that the time of its descent varies
as the secant of the inclination of the Plane to the vertical.
1 fi
16. A body falls to the ground; it describes of the
whole space during the last second of the motion : find the
whole time.
17. Find the position of a point on the circumference
of a circle so that the time of descent down an Inclined
Plane to the centre of the circle may be equal to the
time of descent down an Inclined Plane to the lowest point
of the circle.
18. Find a point in a vertical circle such that the
time down a tangent at that point terminating in the
vertical diameter produced may be equal to the time down
the vertical diameter.
19. Find the measure of the force of gravity when
half a second is taken as the unit of time.
20. Also when the unit of space is a metre, that is,
about 3'28 feet.
21. Also when the unit of time is ten seconds, and the
unit of space is a yard.
22. Also when the unit of time is a quarter of a
second, and the unit of space is half a yard.
23. If / be the measure of an acceleration when m
seconds is the unit of time, and n feet the unit of length,
find the measure of acceleration when a second and a foot
are the units.
24. If / be the measure of an acceleration when m
seconds is the unit of time, and n feet the unit of length ;
find the measure of the acceleration when p. seconds is the
unit of time, and v feet the unit of length.
230 NOTION IN A STRAIGHT LINE
IV. Motion in a straight line under the influence of a
uniform force, with given initial velocity.
35. In the preceding Chapter we confined ourselves to
the case in which the body was supposed to have no velo
city before the force began to operate ; this supposition is
usually expressed by saying that the body has no initial
velocity. We shall now suppose that the body has an
initial velocity, the direction of which coincides with the
straight line in which the force acts.
36. A body starts with the velocity u, and is acted on by
a uniform force in the direction of this velocity during the time
t:iffbe tJie acceleration, and v the velocity of the body at the
time t, then v=u + ft.
For, by the definition of uniform force, in each unit of
time the velocity / is communicated to the body ; and
therefore in t units of time the velocity ft is communi
cated : therefore at the end of the time t the velocity is
u+ft.
37. A body starts with the velocity u, and is acted on by
a uniform force in the direction of the velocity during the time
t : if t be the acceleration, and s the space described in the
time t, then s=ut +  ft 2 .
Let the whole time t be divided into n equal intervals ;
denote each interval by T, so that nr=t. Then the velo
city of the body at the end of the times
T, 2r, 3r, (nl)r, nr
from starting is, by Art. 36, respectively
u+fr, w + 2/r, w + 3/r u + (nI)fr, u + nfr.
Let s l denote the space which the body would describe
if it moved during each interval r with the velocity which
it has at the beginning of the interval ; and let s 2 denote
(nl)fr}T,
+ {u + 2/r} r + ...
WITH GIVEN INITIAL VELOCITY. 231
the space which the body would describe if it moved
during each interval r with the velocity which it has at the
end of the interval. Then
that is,
Hence, by the theory of Arithmetical Progression in
Algebra, we have
Now s, the space actually described, must lie between
s l and s 2 ; but by making n large enough we can make 
as small as we please; so that we can make s l and s z
differ from ut + ~ft z by less than any assigned quantity.
Hence s=*tf + /<! 2 .
38. The result just obtained has been deduced by an
independent investigation founded on first principles ; if we
are allowed to assume the result obtained in Art. 21 we
may put the investigation more briefly as follows :
If the body at a certain instant is moving with a cer
tain velocity, its subsequent motion will be the same, how
ever we suppose that velocity to have been acquired. Let
us suppose that the velocity u was generated by the action
of the force, of which the acceleration is /, during the
232 MOTION IN A STRAIGHT LINE
time tf; and let the body have moved from rest through
the space s' during this time. Then we have, by Art. 21,
therefore s=flt + ft' i =ut + ft\
39. The result of Art. 37 is sometimes obtained in
the following way :
If no force acted on the body the space described in
the time t would be ut, by Art. 5. If there were no initial
velocity the space described in the time t under the in
fluence of the force would be ^ft 2  Now if the body start
with the velocity u, and be also acted on by the force, the
space actually described must be the sum of these two
spaces; because by the nature of uniform force the velo
city at any instant is exactly the sum of what it would be in
the two supposed cases.
40. Hence we have the following results when a body
starts with a given velocity and is acted on by a uniform
force in the direction of this velocity :
Let f be the acceleration, u the initial velocity, v the
velocity at the end of the time Z, and s the space de
scribed; then
v=u+ft ........................ (1).
s=ut + \ft* ..................... (2).
From (1) and (2) we have
thus v 2 =u z + 2fs .................. (3).
WITII GIVEN INITIAL VELOCITY. 233
41. The student must observe that during the motion
which we consider in Art. 37 the only force acting is that
of which the acceleration is /. The body starts with the
velocity u, and this must have been generated by some
force, which may have been sudden, as a blow or an ex
plosion is usually considered to be, or may have been
gradual like the force of gravity. But we are only con
cerned with what takes place after this velocity u has
been generated, and so during the motion which we con
sider no force acts except that of which the acceleration
is/.
42. Hitherto we have supposed the direction of the
force to be the same as that of the initial velocity ; we
will now consider the case in which the direction of the
force is opposite to that of the initial velocity. It will be
sufficient to state the results, which can be obtained as in
Arts. 36, 37, 38, and 40.
Let / be the acceleration, u the initial velocity, v the
velocity at the end of the time t, and * the space de
scribed, the force and the initial velocity being in opposite
directions; then
v=uft ........................ (1),
(2),
* 2 =w 2 2/s ..................... (3),
These formulae will present some interesting conse
quences ; the student will obtain an illustration of the in
terpretation ascribed in Algebra to the negative sign.
As long as ft is less than u we see from (1) that v is
positive, so that the body is moving in the direction in
which it started. When ft  u = 0, that is when *=^,
we have v=0, so that the body is for an instant at rest.
When t is greater than ^ the value of v is negative;
that is, the body is moving in the direction opposite to that
in which it started. Thus we see that the body continues
to move in the direction in which it started, until by the
234 MOTION IN A STRAIGHT LINE.
operation of the force, which acts in the opposite direction,
all its velocity is destroyed ; after this the force generates
a new velocity in the body in the direction of the force,
that is, in the direction opposite to that of the original
velocity.
u w 2 1 w 2 n?
From (2) when => we have s=^  ? = TT>; this
"
> 
/ J " J *
des
.
gives the whole space described by the body while moving
in the direction in which it started. This value of s may
also be obtained from (3) by putting i>=0; for then we have
u 2  2/5=0.
From (2) we have s=0 when utft 2 =(), that is when
2w
=0 and when t=f The value =0 corresponds to the
instant of starting ; the other value of t must correspond
to the instant when the body in its backward course
reaches the starting point again. Thus the time taken in
moving backwards from the turning point to the starting
point is f~ft or ^> which is equal to the time taken
in moving forwards from the starting point to the turning
2w
point. Put t=f in (1), then we get v=u2u=u; so
that at this instant the velocity of the body is the same
numerically as it was at starting, but in the opposite
direction. Equation (3) shews that the velocity at any
point of the forward course is numerically the same as at
the same point of the backward course. When t is greater
2M
than j the value of s becomes negative, indicating that the
body is now on the side of the starting point opposite to
2l4
that on which it was while t changed from to ?.
It will be useful to remember these two results : the
original velocity u is destroyed in the time >, and tho
4*2
space described in that time is <r>.
EXAMPLES. IV. 235
43. Tlie most important application of the preceding
Article is to the case of gravity. If a body be projected
vertically upwards with a velocity u it rises for a time  ,
u* 9
reaches the height , falls to the ground in the same
time as it took to rise, and strikes the ground with the
velocity u downwards.
EXAMPLES. IV.
1. A stone is thrown vertically upwards with a velo
city 3<7 : find at what times its height will be 4#, and find
its velocity at these times.
2. A body is projected vertically upwards with a
velocity which will carry it to a height 2g : find after
what interval the body will be descending with the velo
city g.
3. A body moves over 20 feet in the first second of
time during which it is observed, over 84 feet in the third
second, and over 148 feet in the fifth second : determine
whether this is consistent with the supposition of uniform
acceleration.
4. A particle uniformly accelerated describes 108 feet
and 140 feet in the fifth and seventh seconds of its motion
respectively : find the initial velocity and the numerical
measure of the acceleration.
5. A body starts with a certain velocity and is uni
formly accelerated : shew that the space described in any
time is equal to that which would be described in the
same time with a uniform velocity equal to half the sum
of the velocities at the beginning and at the end of the
time.
6. A bullet shot upwards from a gun passes a certain
point at the rate of 400 feet per second : find when tho
bullet will be at a point 1600 feet higher.
236 EXAMPLES. IV.
7. A. body is dropped from a given height and at
the same instant another body is started upwards, and they
meet half way : find the initial velocity of the latter body.
8. At the same instant one body is dropped from a
given height, and another body is started vertically up
wards from the ground with just sufficient velocity to attain
that height : compare the time they take before they meet
with the time in which the first would have fallen to the
ground.
9. A smooth Plane is inclined at an angle of 30 to
the horizon; a body is started up the Plane with the
velocity g : find the time it takes to describe a space g.
10. A smooth Plane is inclined at an angle of 30 to
the horizon ; a body is started up the Plane with the
velocity 5g : find when it is distant Qg from the starting
point.
11. A body is thrown vertically upwards, and the time
between its leaving a given point and returning to it again
is observed : find the initial velocity.
12. A particle is moving under the action of a uni
form force, the acceleration of which is fi if p be the
arithmetic mean of the first and last velocities in passing
over any portion * of the path, and q the velocity gene
rated, shew that pq=fs.
13. Two small heavy rings capable of sliding along
a smooth straight wire of given length inclined to the
horizon are started from the two extremities of the wire
each with the velocity due to their vertical distance :
find the time after which they will meet, and shew that
the space described by each is independent of the inclina
tion of the wire.
14. A body begins to move with the velocity u, and
at equal intervals of time an additional velocity v is com
municated to it in the same direction : find the space
described in n such intervals. Hence deduce the space
described from rest under the action of a force constant in
magnitude and direction.
SECOND LA W OF MOTION. 237
V. Second Law of Motion. Motion under the influence
of a uniform force in a fixed direction, but not in a
straight line. Projectiles.
44. "We are still confining ourselves to the case of a
uniform force in a fixed direction ; but the body will
now be supposed to start with a velocity which is not
in the same direction as the force : it will appear that
a body under such circumstances will not describe a straight
line but a certain curve called & parabola.
It is necessary at this stage to introduce the Second
Law of Motion.
45. Second Law of Motion. Change of motion is
proportional to tJie acting force, and takes place in the
direction of the straight line in which the force acts.
So long as we keep to the same force and the same
body change of motion is measured by change of velocity ;
the law then asserts that any force will communicate
velocity in the direction in which the force acts : and it is
implied that the amount of the velocity so communicated
does not depend on the amount or the direction of the
velocity which may have been already communicated to
the body. It will appear hereafter that the law contains
more than this : see Art. 84.
For the reason explained in Art. 10 we ought to sup
pose the Second Law to relate to the motion of a particle.
46. In confirmation of the truth of the Second Law of
Motion it is usual to adduce the following experiment : if
a stone be dropped from the top of the mast of a ship in
motion the stone will fall at the foot of the mast notwith
standing the motion of the ship. The stone does not fall
in a straight line ; it starts with a certain horizontal velo
city, namely, the same as that of the ship, and gravity
acts on it in a vertical direction. The fact that tlte stone
falls at the foot of the mast shews that the vertical force
238 SECOND LAW OF MOTION.
of gravity makes no change in the horizontal velocity
with which the stone started ; so that the vertical force
can only have communicated a vertical velocity. If the
time of the descent of the stone were observed, and found
to be the same as of a stone falling from rest through the
same space, the confirmation of the truth of the Second
Law of Motion would be much more decisive.
However it is obvious that few persons can perform this
experiment : but most persons can observe the fact that the
motion of a steamer or of a railway train will not affect the
circumstances of the fall of bodies through small heights.
As we have already indicated in Art. 12, the best evi
dence of the truth of the Laws of Motion is the agree
ment of results deduced from these Laws with observed
phenomena, especially those furnished by Astronomy.
47. Newton gives the following as one of the Corol
laries to his Laws of Motion :
A body acted on by two forces will describe the
diagonal of a parallelogram in the time in which it
would describe the sides under the influence of the forces
singly.
The following is the substance of Newton's exposition
of this statement.
Suppose that a body, in a given
time, under the influence of a single
force J/, which acted at A, would
move with uniform velocity from A
to B ; and suppose that the body
in the same time under the influ
ence of another single force N 9
which acted at A, would move with uniform velocity from
A to C ; complete the parallelogram ABCD : then if both
forces act simultaneously at A the body will move uni
formly in the given time from A to D.
For since the force N acts along the straight line AC,
which is parallel to BD, this force, by the Second Law of
Motion, will not change the velocity of approach towards the
straight line BD, which is produced by the other force.
SECOND LAW OF MOTION. 239
Thus the body will reach the straight line BD in the
same time whether the force N act or not : and so at
the end of the given time will be found somewhere in
the straight line BD. By the same reasoning it follows
that the body at the end of the given time will be found
somewhere in the straight line CD. Therefore the body
will be at D.
The body must move in a straight line from A to D, by
the First Law of Motion.
48. Thus it appears that, according to Newton's view,
the Second Law of Motion tells us that when forces act
simultaneously on a body each force communicates in a
given tune the same velocity as if it acted singly on
the body originally at rest ; and then by the Corollary we
learn how to compound the velocities thus generated into
a single velocity.
It will be seen that Newton supposes in his expo
sition that the two forces act instantaneously ; that is,
they are of the kind which we naturally suppose a blow
to be, and communicate velocity by sudden action, not by
continuous action.
49. The principle contained in Art. 47 is called the
Parallelogram of Velocities, and is usually enunciated
thus : if a body have communicated to it simultaneously
two velocities which are represented in magnitude and
direction by two straight lines drawn from a point, then
the resultant velocity will be represented in magnitude and
direction by the diagonal, drawn from that point, of the
parallelogram constructed on the two straight lines as adjacent
sides.
This principle gives rise to applications similar to that
deduced from the Parallelogram of Forces in Statics.
We may use the principle either to compound two velo
cities into one, or to resolve one velocity into two.
50. Thus if velocities u and v be simultaneously commu
nicated to a body in directions which include an angle a,
the resultant velocity is */(w 2 + v 2 + 2uv cos a). Let /3 be
the angle between the direction of the velocity u and that of
240
PROJECTILES.
the resultant velocity, and y the angle between the direc
tion of the velocity v and that of the resultant velocity j
then /3 + 7=0, and
sin j3 v
In the special case in which a =90, the resultant velo
city is J(u? + *> 2 ) 5 a 180
v u
Bin j3= // o o\ siny= 77 s <
V(^ + v ) V(** + v )
See ^aw;s, Art. 30.
V/ 51. A body projected in any direction not vertical and
acted on by gravity will describe a parabola.
Let a body be projected from
the point A in any direction
which is not vertical ; let AT be
the space which would be de
scribed by the body in the time t
if the force of gravity did not
act. Draw AM vertically down
wards, equal to the space through
which a body would fall from
rest, in the time , under the
action of gravity. Complete the
parallelogram ATPM. Then P,
the corner opposite to A, will be
the place of the body at the end
of the time t.
For, by the Second Law of Motion, gravity will com
municate the same vertical velocity to the body as it would
if the body had not received any other velocity. Thus at
any instant there will be the same vertical velocity as if
there had been no velocity parallel to AT, and the same
velocity parallel to AT as if there had been no vertical
velocity. Therefore the spaces described parallel to AT
and A M respectively will be the same as if each alone had
been describea. Thus P will be the place of the body at
the end of the time t.
PROJECTILES. 241
Let u be the velocity with which the body is projected
at A\ then AT or PM=ut; also AM=gt\ therefore
9
Thus PJ/ 2 bears a constant ratio to AM, and there
fore by Conic Sections the path of the body is a parabola,
U*
having its axis vertical and AT for a tangent. And
is the distance of A from the focus of the parabola, and
also from the directrix.
52. Produce TP to Q, so that PQ=PT, and produce
AM to N so that MN=AM. Then
AT ut u
Now at the end of the time t the body has its original
velocity parallel to AT, and also the vertical velocity gt
downwards which has been communicated to it by gravity ;
these velocities are proportional to AT and AN, and in
these directions. Hence the resultant velocity at P is
parallel to AQ, by Art. 49. Thus if this direction be
drawn at P, and be produced, it will meet the vertical
straight line through A at a point whose distance from A
is equal to PQ, that is to TP. The direction thus deter
mined for the velocity at P is, by Conic Sections, that of
the tangent at P, which might have been anticipated.
53. A body projected in any manner and acted on by
gravity is called a Projectile; thus we have shewn that the
path of a Projectile is in general a parabola, and is a
straight line in the particular case in which the body is
projected vertically upwards or downwards. The student
must observe that during the motion which we consider in
Art. 51 the only force acting is that of gravity; see Art. 41.
54. Tlie velocity of a projectile at any point of its
path is tliat which would be acquired in falling from the
directrix to the point.
T. ME. 16
242 PROJECTILES.
For we have seen in Art. 51 that the distance of the
directrix from the point of projection is , where u is
u z
the velocity of projection ; and is equal to the vertical
space through which a body must fall from rest under the
action of gravity in order to acquire the velocity u. Now
any point of the parabolic path may be regarded as the
point of projection, and the velocity at that point as the
velocity of projection. Thus the required result is obtained.
55. The preceding result may also be obtained thus :
Suppose the direction of projection to make an angle a
with the horizon ; resolve the velocity of projection u into
u cos a in the horizontal direction and u sin a in the verti
cal direction. Then at the end of the time t the hori
zontal velocity is still u cos a, and the vertical velocity is
u sin a  gt.
Let v denote the resultant velocity; then, by Art. 50,
v 2 = (u cos a) 2 + (u sin a  gff
u 2  2gtu sin a +g 2 fi= u 2  2g (tu sina 
Now tu&iaa^gt 2 is the vertical height of the body
21
above the horizontal plane through the starting point by
Art. 43 ; we will denote this by y : thus v 2 =u 2  2gy.
Let h denote the distance of the directrix from the
u 2
starting point, so that h= \ thus v 2 =2g (h y).
This shews that the velocity is that which would be
acquired in falling from rest through the space h y, that
is, in falling from the directrix to the point considered.
56. To determine the position of the focus of the para
bola described by a projectile.
Let u be the velocity of projection, and a the angle
which the direction of projection makes with the horizon.
PROJECTILES. 243
The distance of the focus from the point of projection
is ~ by Art. 51. By the nature of the parabola the tan
gent at any point makes equal angles with the focal dis
tance of that point and the diameter at the point. Hence
the straight line from the point of projection to the focus
makes an angle 2 (90  a) with the vertical, and therefore
an angle 2a  90 with the horizon. Thus the situation of
the focus is determined.
The height of the focus above the horizontal plane
through the point of projection is ^ sin (2a  90), that is
*y2
 g cos 2a. Thus the focus is below the horizontal plane
through the point of projection if 2a is less than 90, and
above it if 2a is greater than 90.
If a perpendicular be drawn from the focus on the
horizontal plane through the point of projection, the dis
tance of the foot of the perpendicular from the point of
projection is cos (2a  90), that is sin 2a.
57. To find the time in which a projectile reaches its
greatest height, and the greatest height.
Let u be the velocity of projection, a the angle which
the direction of projection makes with the horizon ; then at
the end of the time t the vertical velocity is u sin a  gt.
Now at the instant of reaching the greatest height the
vertical velocity vanishes, so that we have ttsina<jtf=0 ;
, ,  , u sin a
therefore t= .
9
' By Art. 55 the height of the projectile at the time t
above the horizontal plane through the point of projection is
tu sin a  jr3tf 2 ; substitute the value of t just found : thus
the greatest height is U  u sm a , that is
Compare Art. 43.
16 2
244 PROJECTILES.
The position of the highest point may be easily deter
mined in the manner of Art. 56.
58. To determine the Latus Rectum of the parabola
described by a projectile.
Let u be the velocity of projection, a the angle which
the direction of projection makes with the horizon.
At the highest point the velocity is entirely horizontal,
so that it is parallel to the directrix ; and thus the highest
point is the vertex of the parabola. The velocity at the
highest point is u cos a. By Art. 54 this velocity would be
acquired in falling from the directrix; therefore the dis
tance of the vertex from the directrix is . The
latus rectum is equal to four times this distance, so that it
. 2M 2 cos 2 a
9
59. The interval between the projection of a projectile
and its return to the horizontal plane through the point
of projection is called the time of flight. The distance
from the point of projection of the point at which the body
meets the horizontal plane is called the range on the
horizontal plane through the point of projection.
60. To find the time of flight of a projectile.
Let u be the velocity of projection, a the angle which
the direction of projection makes with the horizon.
The height of the body at the time t is tu sin a  ^ gt' 2 .
This vanishes when =0 and when t= . The value
g
=0 corresponds to the instant of starting; the other value
of t must correspond to the instant when the body again
reaches the horizontal plane through the point of projec
tion. Thus by Art. 57 we see that the time which "the
projectile takes in descending from the highest point to
the horizontal plane through the point of projection is
2u sin a u sin a . , , . u sin a . , . , ,
. that is ; so that the time of
99 9
descent is equal to the time of ascent.
PROJECTILES. 245
61. On reaching the ground the vertical velocity ia
, u sin a , , , . . . , ...
u Bin a  2g , that is  u sin a ; thus it is numeri
cally the same as at starting, but in the opposite direction.
The horizontal velocity is the same at the two points.
And generally at the two points which are in the same
horizontal plane the whole velocities are the same by
Art. 54 ; and the horizontal velocities are the same : hence
the vertical velocities are numerically the same, but must
be in opposite directions.
62. To find the range on the horizontal plane through
the point of projection.
It is shewn in Art. 60 that the time of flight is " ;
and the horizontal velocity is u cos a : hence the horizontal
, ., , . 2w sin a mi . ,
space described is x u cos a. This may be put in
U 2 . 9
the form sin 2a.
9 2
It is often useful to observe that it is  u sin a. u cos a,
2 g
that is  x vertical velocity at starting x horizontal velocity.
y
63. The time of flight and the range may also be in
vestigated thus :
Let A be the point of projection,
A T the direction of projection, AB the
range on the horizontal plane through
A, and TB vertical. Let u be the
velocity of projection, a the angle TAB,
t the time of flight.
ight.
Then AT=ut, TB=^gP; hence
0*
TB 2 y gt .. . , 2Msina
8ma= 2r = = 4 ; therefore^
And ABAT** a=tu cos a= 2u * sin a COS a
246 PROJECTILES.
64. To determine the inclination to the horizon of the
direction of motion of a projectile at any instant.
By Art. 55 the vertical and horizontal velocities at the
end of any time t are respectively
u sin a gt and u cos a.
Thus if v be the resultant velocity, and < the angle
which its direction makes with the horizon,
v sin <=w sin agt, v cos <f>=u cos a;
u sin a at
therefore tan d>= .
u cos a
65. An inclined plane passes through the point of pro
jection of a projectile and is at right angles to the plane of
motion: to fond the time of 'flight ', the greatest distance from
the plane, and the range on the plane.
Let AP, the inclined plane, make
an angle with the horizon AN ; let
AT, the direction of projection, make
an angle a with the horizon ; let u be
the velocity of projection.
Kesolve the initial velocity along
the plane and at right angles to it;
the latter part is u sin (a  ). Resolve the acceleration g
parallel to the plane and at right angles to it ; the latter
part is g cos /3.
The motion in the direction at right angles to the plane
is independent of the motion parallel to the plane. Hence
as in Arts. 43 and 57 the body reaches its greatest distance
from the plane at the end of the time ^ sm ( q ~/ 3 ) . an( j j t
g cos p
takes the same time to move from this point to the plane,
so that the time of flight is 2 " sin (/3) .
#cos
And, as in Arts. 43 and 57, the greatest distance from
., . . u 2 sin 2 (a /3)
the plane is = v ^' .
2g cos 3
PROJECTILES.
247
Let P be the point where the body meets the inclined
plane : draw PN perpendicular to the horizon. Thus
AN=AP cos 0.
But AN is the horizontal space described in the time
^ cos
g cos 2
The preceding result may also be obtained thus
Let A be the point of projection,
P the point where the body meets the
inclined plane. Draw a vertical line
meeting the direction of projection at
T and the horizon at N.
Then, with the same notation as be
fore,
AT=ut, TP=gt z ;
TP sin TAP sin TAP
also lr = iSrZFF
thus
L*
*L
ut
Sn ("
cos /3
, whence t= ^
Also AN=AT cos TAN=tu cos a
CG. The theory of projectiles which has now been
given is of no practical use, because it is found by experi
ment that the resistance of the air exercises a veiy power
ful influence on the motion of a body, especially when the
velocity is large. On this account the actual path of a
cannon ball is not a parabola, and the range and time of
are quite different from the values determined
flight
248 EXAMPLES. V.
above. The following is an example: a ball of certain
size and weight being projected at an angle of 45, with
a velocity of 1000 feet per second, it is found that on
taking the resistance of the air into account the range
is about 5000 feet, instead of being S292..
aSt
The discussion of the motion of a projectile, taking
into account the resistance of the air, is however far too
difficult to find a place in this book.
EXAMPLES. V.
1. Velocities of 5 feet and 12 feet per second in direc
tions at right angles to each other are simultaneously com
municated to a body : determine the resultant velocity.
2. A body is projected with the velocity 3# at an incli
nation of 75 to the horizon : determine the range.
3. If at the highest point of the path of a projectile
the velocity be altered without altering the direction of
motion, will the time of reaching the horizontal plane
which passes through the point of projection be altered?
4. From the highest point of the path of a projectile
another body is projected horizontally with a velocity equal
to the original vertical velocity of the first body: shew
that the focus of the path described by the second body is
in the horizontal plane which passes through the point of
projection of the first body.
5. A ship is moving with a velocity u, a cannon ball is
shot from a cannon which makes an angle a with the hori
zon, with powder which would give a velocity v to the ball
if the cannon were at rest : find the range supposing the
ship and the ball to move in the same vertical plane.
6. Two bodies are projected simultaneously from the
same point, with different velocities and in different direc
tions in the same plane : find their distance apart at the
end of a given time.
EXAMPLES. V. 249
7. Determine how long a particle takes in moving
from the point of projection to the further end of the
latus rectum.
8. A body slides down a smooth Inclined Plane : shew
that the distance between the foot of the Inclined Plane
and the focus of the parabola which the particle de
scribes after leaving the Plane is equal to the height of
the Plane.
9. Two parabolic paths have a common focus and their
axes in the same straight line : shew that if tangents be
drawn to the two paths from any point in their common
axis the velocities at the points of contact are equal.
10. If two projectiles have the same initial velocity
and the same horizontal range, the foci of their paths are
at equal distances from the horizontal plane, which passes
through the point of projection.
11. A heavy particle is projected from a point with
a given velocity, and in a given direction : find its distance
from the point of projection at the end of a given time.
12. A number of particles are projected simultaneously
from a fixed point in one plane, so that their least velocity
is constant : shew that all of them will be found at any the
same instant on the same vertical line.
13. A body is projected with a given velocity and in
a given direction : determine the velocity with which
another must be projected vertically so that the two may
reach the ground at the same instant.
14. A ball fired with velocity u at an inclination a to
the horizon just clears a vertical wall which subtends an
angle /3 at the point of projection : determine the instant
at which the ball just clears the wall.
15. In the preceding Example determine the horizon
tal distance between the foot of the wall and the point
where the ball strikes the ground.
16. If one body fall down an Inclined Plane, and
another be projected from the starting point horizontally
along the Plane, find the distance between the two bodies
when the first has descended through a given space.
250 PROJECTILES.
VI. Projectiles continued.
67. Although, as we have stated at the end of the
preceding Chapter, the theory of projectiles is of no use
in practice, yet it deserves careful study on account of the
valuable illustration which it affords of the principles of
Dynamics; and a thorough knowledge of the elementary
principles is the true foundation for those higher investiga
tions which apply to the phenomena actually presented by
nature. A very large number of deductions and problems
may be given which serve to impress the methods and
results of the preceding Chapter on the memory : some of
these Examples we will now discuss.
68. We have seen in Art. 62 that the range on the
horizontal plane through the point of projection is sin2a.
Hence we deduce the following results :
The greatest range for a given velocity of projection is
found by supposing 2a=90, that is a =45: this greatest
. 2
range is .
Suppose the range to be given; denote it by c: then
sin 2a=c, thus if either a or u is also given we may find
the other.
Since v?= . ^ , the least value of u is when sin 2a is
sm 2a '
greatest, that is when a =45.
Thus when a =45 we have the greatest range corre
sponding to a given velocity, and also the least velocity cor
responding to a given range.
Again, suppose c and u given, and a to be found ; we
have sin 2a=^: it is known by Trigonometry that if eg
PROJECTILES. 251
is less than u z there are two values of 2a between and
180 which satisfy this equation, and one value is the sup
plement of the other. Hence there are two values of a
(between and 90, and one value is the complement of the
other.
69. We have seen in Art. 65 that the range on a plane
inclined at an angle 8 to the horizon which passes through
the point of projection and is at right angles to the plane
. 2w 2 cos a sin (a  8) m ,
of motion is  . We shall now investigate
g cos 2 8
for what angle of projection this range is greatest, the velo
city being given.
We have to investigate for what value of a the expres
sion cos a sin (a  8) has its greatest value. Now we know
by Trigonometry that
2 cos a sin (a  8) =sin (2a  8)  sin 8 ;
hence the greatest value is when 2a8=90, that is when
a=^(8 + 9QO);
the greatest range is
</2 f\ Qin ff\ /2
A', that is
Suppose the range to be given ; denote it by c : then
2 M 2 cos a sin (aft)_
#cos 2 /3
Hence the least value of u is when cos a sin (a  /3) is
greatest, that is, as before, when a=^ (/3 + 90).
Thus, for this value of a we have the greatest range
corresponding to a given velocity, and also the least velo
city corresponding to a given range.
252 PROJECTILES.
Suppose the range and the velocity of projection given,
and that we have to find the angle of projection; then
snce
we have sin (2a/3)=sin p +
Hence we have in general two values of 2a  ft between
and 180, and one value is the supplement of the other.
Suppose one of these values is y, then the other is 180  y;
from the former we obtain a.=($ + y\ and from the
latter a=90 +  (/3y). The sum of these values of a is
90 + /3, that is twice the angle of projection which gives
the greatest range corresponding to a given velocity : hence
the two directions which correspond to a given range are
equally inclined to that which corresponds to the greatest
range, but on opposite sides of it.
70. To determine the direction in which a body must be
projected from a given point with, a given velocity so as to hit
a given point.
Let A denote the point of projection, B the other given
point. The velocity at A is known; and therefore the
distance of A from the directrix is known by Art. 54, so
that the position of the directrix is known. Then since B
is a given point, the distance of B from the directrix is
also known.
Now the distance of any point in the parabola from
the focus is equal to the distance of that point from the
directrix : hence the distances of A and B from the focus
are known.
Describe a circle with A as a centre, and radius equal
to the known distance of the focus from A ; describe an
other circle with B as centre, and radius equal to the
known distance of the focus from B. The focus of the
PROJECTILES. 253
parabola will be at the intersection of the circles ; and as
the directrix is also known, the parabola is determined.
If the circles do not meet the problem has no solution ;
if they touch there is one solution ; if they cut, since either
point of intersection may be taken, there are two solutions.
71. Let u be the velocity of projection, a the angle
which the direction of projection makes with the horizon;
and let y be the height of the projectile at the time t
above the horizontal plane through the point of projection.
Then as in Art. 55,
^ 2 (1).
Suppose a perpendicular to be drawn from the projectile
at the end of the time t on the horizontal plane through
the point of projection; and let x be the distance of the
foot of the perpendicular from the starting point ; then
tf = ft*COSa (2).
From (2) we have t= ; substitute in (1), thus
These equations are often useful in solving problems
respecting projectiles.
72. We may apply equation (3) of the preceding Arti
cle to give another mode of solving the problem in Art. 70.
For since the point to be hit is given, the values of x
and y will be known ; we may then by solving the quadra
tic equation determine tan a. For we have
or tan 2 a
g*
254 PROJECTILES.
Hence, by solving the quadratic equation,
u 2 // u* 2uw
tana= . /(s  1 ^
## V Vr 4
Hence we see that tan a has two values, or one, or none,
according as the quantity under the radical sign is positive,
zero, or negative, that is, according as 2 is greater than,
/
equal to, or less than x 2 + & .
a
73. If particles are projected from the same point, at
the same instant, with the same velocity, in different direc
tions, they will all at any future instant be on the surface
of a sphere.
Let u be the velocity of projection; then with the
figure of Art. 51 we have at the end of the time t
MP=AT=ut.
This shews that whatever be the direction of projection
all the particles at the end of the time t are on the surface
of a sphere of which the radius is ut, and the centre is M\
so that the centre is at the distance  gt 2 below the point
of projection.
74. If two particles are projected from the same point
at the same instant, with different velocities, and in different
directions, the straight line which joins them will always
move parallel to itself.
Let A be the point of pro
jection; suppose one body pro
jected along AP, and the other
along AQ.
First suppose the force of
gravity not to exist. Then each
body would move uniformly in a
straight line. Suppose one body
PROJECTILES. 255
to be at P at the end of the time T, and at p at the end
of the time t\ and suppose that the other body is at Q
at the end of the time T, and at q at the end of the time t.
AP T AQ
therefore PQ is parallel to pq, by Euclid, VL 2. Now
suppose the force of gravity to act; then in the time t each
body would be drawn down through the vertical space
2 gt\ Thus take pR and qS vertically downwards, and
each equal to ^gt 2 ; then R and S are the positions of the
bodies at the end of the time t; and RS is parallel to PQ
by Euclid, XL 9.
75. If three particles are projected from the same point,
at the same instant, with different velocities, and in different
directions, the plane which passes through them always moves
parallel to itself.
This is demonstrated in the same manner as the preced
ing proposition] Euclid, XL 15 will be required.
76. Let v be the velocity at any point P of the parabola
described by a projectile ; let S be the focus : it is shewn
in Art. 54 that v z =2gSP. Let p be the perpendicular
from S on the tangent to the parabola at P; then it is
known from Conic Sections that SP varies as p 2 . Hence
v varies as p. And the perpendicular from S on the tan
gent at P is at right angles to the tangent, that is at right
angles to the direction of the velocity at P.
Since then the perpendicular varies as the velocity and
is alwavs at right angles to the direction of the velocity,
it may be conveniently used to furnish a representation of
the velocity at any point of the parabola described by a
projectile.
77. It mav be shewn that the path of a projectile is
a parabola in the following way :
256 PROJECTILES.
Let u be the velocity of projection, a the angle which
the direction of projection makes with the horizon. Then
the vertical height of the body at the end of the time t is
tu sin. a  ^gt z , and therefore the distance of the body from
a straight line parallel to the horizon and at the height
above the point of projection is  tu sin a + 5 gt 2 .
Now
{u 2 1 u 2 1 2
(sin 2 a  cos 2 a)  tu sin a + 5 gfi + cos 2 ah
(u? 1 I 2
5 (sin 2 a  cos 2 a)  tu sin a + <7^r
2w 2 cos 2 a fw 2 . 1 9 1 /W 2 co
+  \ty( sma  cos a )  tusma + 2^7 + \
{ u z 1 I 2
(sin 2 a  cos 2 a)  tu sin a + 5 ^ 2 [
2iiPt ifl
+ t z W 2 cos 2 a  cos 2 a sin a + s sin 2 a cos 2 a
# 2
(M 2 1 ) 2
5 (sin 2 a  cos 2 a)  tu sin a + 5 gt z [
+ J<w cos a  sin a cos a> .
\9l
Now the horizontal distance which the body moves
through in the time t is Zwcosa; and so the expression
just given is the square of the distance of the body at the
time t from a certain fixed point, namely the point which is
EXAMPLES. VI. 257
u z
at the height (sin 2 a  cos 2 a) above the horizontal plane
through the point of projection and at the horizontal dis
tance sin a cos a from this point.
Thus we see that the distance of the body from a
certain fixed straight line is always equal to its distance
from a certain fixed point. Therefore from the definition
of a parabola the path of a projectile must be a parabola.
Hence it follows that we could thus by the aid of
mechanical principles demonstrate that the property em
ployed in Art. 51 belongs to a parabola, without assuming
it from geometry.
EXAMPLES. VI.
1. Find the velocity and the direction of projection in
order that a projectile may pass horizontally through a
given point.
2. Find the velocity with which a body must be pro
jected in a given direction from the top of a tower so as to
strike the ground at a given point.
3. A body is projected with a given velocity at an
inclination a to the horizon ; a plane inclined at an angle /3
to the horizon passes through the point of projection : find
the condition in order that the body when it strikes the
plane may be at the highest point of its path.
4. Two bodies are simultaneously projected in the
same vertical plane with velocities u and v at inclinations
a and 8 to the horizon. Shew that their directions are
parallel after the time 
g (v cos /3  u cos a)
5. Bodies are projected from the same point in the
same vertical plane and in such a manner that the para
bolas have a latus rectum of given length : shew that the
locus of the vertices of these parabolas is a parabola with a
latus rectum of the same length.
T. ME. 17
258 EXAMPLES. VI.
6. Bodies are projected from the same point in the
same vertical plane so as to describe parabolas having a
latus rectum of given length : shew that the locus of the
foci is a parabola with a latus rectum of the same length,
having its vertex downwards, and its focus at the point of
projection.
7. Bodies are projected simultaneously from the same
point, and strike the horizontal plane through that point
simultaneously : shew that the latera recta of the paths
vary as the squares of the horizontal ranges.
8. A body is projected at an inclination a to the
horizon : determine when the motion is perpendicular to a
plane which is inclined at an angle /3 to the horizon.
9. A body is projected at an inclination a to the
horizon : determine the condition in order that the body
may strike at right angles the plane which passes through
the point of projection and makes an angle /3 with the
horizon.
10. A body is projected with the velocity u and strikes
at right angles a plane which passes through the point of
projection and is inclined at an angle (3 to the horizon :
shew that the height of the point struck above the hori
zontal plane through the point of projection is
2w2 sin 2 ft
g
11. Shew by mechanical considerations that any dia
meter of a parabola bisects the chords which are paraUel to
the tangent at the extremity of the diameter.
12. Shew that the time of describing any arc of a
parabola by a projectile is equal to the time of moving
uniformly over the chord with the velocity which the pro
jectile has when it is moving parallel to the chord.
13. The time of describing any arc of a parabola by a
projectile is equal to twice the time of falling vertically
from rest from the curve to the middle point of the
chord.
EXAMPLES. VI. 259
14. Two bodies are projected from two given points
in the same vertical line in parallel directions and with
equal velocities : shew that tangents drawn to the path of
the lower will cut off from the path of the upper arcs
described in equal times.
15. A smooth plane of length I is inclined at an angle
a to the horizon ; a body is projected up the plane with
the velocity u, and after leaving the plane describes a
parabola : shew that the greatest vertical height reached
u z
above the point of projection is I sin a cos 2 a + =p sin 2 a.
16. A heavy body is projected from a given point in a
given vertical plane with a given velocity so as to pass
through another given point : shew that the locus of the
second point in order that there may be only one parabolic
path is a parabola having the given point as focus.
17. A ball is shot from a cannon with velocity v, at an
inclination a to the horizon ; the cannon is moving horizon
tally with velocity u in a direction inclined at an angle /3 to
the vertical plane which is parallel to the cannon : find the
range of the ball on the horizontal plane.
18. A stone is thrown in such a manner that it would
just hit a bird on the top of a tree, and afterwards reach a
height double that of the tree ; if at the moment of throw
ing the stone the bird flies away horizontally, shew that the
stone will notwithstanding hit the bird if the horizontal
velocity of the stone be to that of the bird as 1 + JZ
is to 2.
19. Find the time in which a projectile would reach a
plane inclined to the horizon at an angle equal to the angle
of projection, and bisecting the range on the horizontal
plane.
20. A particle is projected from the top of a tower
at an inclination a to the horizon, with the velocity which
would be acquired in falling down n times the height of
the tower. Obtain a quadratic equation for determining
the range on the horizontal plane through the foot of tho
tower.
172
260 EXAMPLES. VI.
21. If h be the height of the tower in the preced
ing Example, shew that the greatest possible range is
+ n), and that the tangent of the corresponding
angle of projection
/ n
18 V 77+1
22. A particle being projected with velocity u, at an
inclination a, just clears a cube of which the edge is c,
which stands on the horizontal plane : find the relation
between u, a, and c.
23. From the result of the preceding Example form
a quadratic equation for finding tan a ; and thence shew
that the least possible value of u 2
24. In the last Example shew that when u is least
tan a = A/ 5 j an< i * na ^ the point of projection is at the
distance C  (</5  1) from the cube.
25. If t be the time in which a projectile describes an
arc of a parabola,, and v the velocity which a particle
would acquire in falling from the intersection of tangents
at the extremities of the arc to the chord of the arc,
shew that
26. Shew that the greatest range up an inclined plane
of 30 is twothirds of the greatest range on a horizontal
plane, the initial velocity being the same in the two cases.
27. A number of heavy particles are projected simul
taneously from a point ; if tangents be drawn to their paths
from any point in the vertical straight line through the
point of projection, prove that the points of contact will be
simultaneous positions of the particles.
28. Two projectiles start from the same point at the
same instant with any velocities : prove that they will
both be moving in a common tangent to their paths at the
same instant after an interval of time which is the Arith
metical mean between the times in the two paths from the
point of projection to the point where the paths meet
again.
MASS. 261
VII. Mass.
78. The word matter is in common use ; and it is not
easy to define it so as to give a notion of it to any person
who does not already possess the notion. The following
definitions have been proposed :
Body or matter is any thing extended, and possessing the
power of resisting the action of force.
Matter is the Substance, Material, or Stuff, of which all
bodies are composed that are capable of having forces applied
to them.
79. The word mass is used as an abbreviation for
quantity of matter.
80. We assume that at the same place on the Earth's
surface, the masses of bodies are proportional to their
weights. We will explain the grounds of this assumption.
If we take a cubic inch of lead, we find by experiment
that it produces the same effect by its weight as another
cubic inch of lead ; and thus two cubic inches of lead
produce by their weight twice the effect which one cubic
inch of lead produces by its weight. Now it is a very
natural supposition that so long as we keep to one kind of
substance the mass is proportional to the volume; and
therefore, so long as we keep to the same kind of substance
the mass is proportional to the weight. We assume then
that this will also be true when we compare bodies which
are not of the same kind of substance.
81. Now suppose we have two bodies containing equal
volumes of the same kind of substance. If a certain force
acting for a certain time on one of these bodies generates a
certain velocity, an equal force acting for an equal time
on the other body will generate an equal velocity. Imagine
that the two bodies are united into one body, and that
the two forces are made to act on the united body : it is
most natural to conclude that a velocity equal to the for
mer will still be generated in an equal time. We are thus
262 MASS.
led to suppose that when bodies of different masses, but
composed of substance of the same kind, are similarly acted
on by forces proportional to the masses, the velocities gene
rated in equal times will be equal
Thus, as long as we keep to the same kind of substance,
we see that in order to generate a certain velocity in a
certain time, the force must vary as the mass. We assume
that this is also true for bodies which are not of the same
kind of substance.
We have already seen that for the same body the force
varies as the velocity generated in a given time ; and we
now see that for the same velocity the force varies as the
mass. Hence, by Algebra, when both the velocity and the
mass vary the force varies as their product ; or in other
words, when a force acts on a body, the product of the mass
moved into the velocity generated in a given time is propor
tional to the force.
82. We see then that the velocity generated in a given
time by a given force, varies inversely as the mass. This
fact, that the greater the mass the less the effect which a
given force produces, is sometimes expressed by saying
that matter is inert, or that inertia is a property of matter.
The words inert and inertia however are sometimes used
in reference to the fact involved in the First Law of
Motion, namely, that a body cannot change its own state of
rest or motion.
83. The word momentum is used as an abbreviation of
the product of the mass moved into the velocity.
84. We now repeat the Second Law of Motion. Change
of motion is proportional to the acting force, and takes place
in the direction of the straight line in which the force acts.
By motion here we are to understand motion as mea
sured by momentum; so that we can now remove the
restriction of having only one body and one force, which
we have hitherto regarded, and may proceed to those more
complex cases in which different bodies and different forces
occur.
MASS. 263
85. Oue case of the general principle of Art. 82 will
be as follows ; the weight of a body at a given place is
proportional to the product of the mass moved into the
velocity generated in a given time. Let the given time
be one second, and the unit of length one foot ; then the
velocity generated is denoted by g. Let J/ be the mass
of a body, and W its weight ; then W varies as Mg, so
that by Algebra TF= CMg, where C is some constant.
It is convenient to have this constant equal to unity ;
this we can secure by making a suitable connexion between
the units of mass and of weight which have not yet been
fixed: then W=Mg.
Suppose, for example, we resolve to have one Ib. as the
unit of weight : required to determine the unit of mass.
Let M=l ; then we obtain W=g, that is 32 2 ; so that
the unit of mass is so much mass as weighs 32 '2 Ibs.
Again, suppose, for example, we resolve to have the
mass of one cubic foot of water as the unit of mass,
required to determine the unit of weight. Let W= 1 ;
then we obtain M=^^: so that the unit of weight is such
''
a weight that its mass is ^rx, that is, the mass of the unit
''
of weight is ;r r of the mass of a cubic foot of water. Now
it is known by experiment that a cubic foot of water
1000
weighs 1000 ounces, so that the unit of weight is ^.
o'2t'2>
ounces.
86. We may illustrate the preceding remarks by
discussing the motion of a body sliding on a rough In
clined Plane.
Suppose a Plane inclined at an angle a to the horizon ;
let a body be placed on the Plane. Let M denote the
mass of the body, and therefore Mg its weight The
resolved force of gravity down the Plane is Mg sin a. The
pressure on the Plane is Mg cos a. If ft denote the co
efficient of friction, the friction will be pMg cos a.
264 EXAMPLES. VII.
If the body is moving down the Plane, the friction acts
up the Plane. Hence the resultant force down the Plane
is Mg (sin a  p cos a). Now when a body is acted on by
its own weight, the velocity generated in a unit of time
is g\ that is, the force Mg generates in a body of mass M
the velocity g in a unit of time : therefore, by Art. 81,
the force Mg (sin a  /* cos a) will generate the velocity
g (sin a p. cos a) in a unit of time.
Thus the motion of a body sliding down a rough In
clined Plane is similar to that of a body sliding down a
smooth Inclined Plane, or to that of a body falling freely :
the acceleration is g (sin a  \JL cos a) for the rough Plane,
#sina for the smooth Plane, and g for the body falling
freely.
In the same manner it may be shewn that if a body is
sliding up a rough Inclined Plane the acceleration is
g (sin a + /x cos a) downwards.
87. We have then the following important general result :
if a force F act on a body of mass M the acceleration is
jj. This result follows from Art. 82 by making as in
Art. 85 a suitable connexion between the unit of force and
the unit of mass.
EXAMPLES. VII.
1. A body weighing nibs, is moved by a constant
force which generates in the body in one second a velocity
of a feet per second : find the weight which the force could
support.
2. Find in what time a force which would support a
weight of 4 Ibs., would move a weight of 9 Ibs. through
49 feet along a smooth horizontal plane : and find the
velocity acquired.
3. Find how far a force which would support a weight
of n Ibs., would move a weight of m Ibs. in t seconds :
and find the velocity acquired.
EXAMPLES. VII. 265
4. Find the number of inches through which a force of
one ounce constantly exerted will move a mass weighing
one Ib. in half a second.
5. Two bodies urged from rest by the same uniform
force describe the same space, the one in half the time the
other does: compare their final velocities and their mo
menta.
6. If a weight of 8 Ibs. be placed on a plane which
is made to descend vertically with an acceleration of 12
feet per second, find the pressure on the plane.
7. If a weight of n Ibs. bs placed on a plane which is
made to ascend vertically with an acceleration /, find the
pressure on the plane.
8. Find the unit of time when the unit of space is two
feet, and the unit of weight is the weight of a unit of mass ;
assuming the equation W=Mg.
9. A body is projected up a rough Inclined Plane, with
the velocity which would be acquired in falling freely
through 12 feet, and just reaches the top of the Plane; the
inclination of the Plane to the horizon is 60, and the
coefficient of friction is equal to tan 30 : find the height of
the Plane.
10. A body is projected up a rough Inclined Plane
with the velocity 2^; the inclination of the Plane to the
horizon is 30, and the coefficient of friction is equal to
tan 15: find the distance along the Plane which the body
will describe.
11. A body is projected up a rough Inclined Plane ; the
inclination of the Plane to the horizon is a, and the coeffi
cient of friction is tan : if m be the time of ascending,
and n the time of descending, shew that (^Y =
12. Find the locus of points in a given vertical plane
from which the times of descent down equally rough In
clined Planes to a fixed point in the vertical plane vary as
the lengths of the Planes.
266 THIRD LA W OF MOTION.
VIII. Third Law of Motion.
88. Newton's Third Law of Motion is thus enunciated :
To every action there is always an equal and contrary
reaction: or the mutual actions of any two bodies are
always equal and oppositely directed in the same straight
line.
Newton gives three illustrations of this Law :
If any one presses a stone with his finger, his finger is
also pressed by the stone.
If a horse draws a stone fastened to a rope, the
horse is drawn backwards, so to speak, equally towards
the stone.
If one body impinges on another and changes the
motion of the other body, its own motion experiences an
equal change in the opposite direction. Motion here is to
be understood in the sense explained in Art. 84.
The first of Newton's illustrations relates to forces in
Statics; and the law of the equality of action and reaction
in the sense of this illustration has been already assumed
in this work ; see Statics, Art. 286. The second illustration
applies to a class of cases of motion which we shall consider
in the present Chapter. The third illustration applies to
what are called impulsive forces, which we shall consider
in the next Chapter.
89. Two heavy bodies are connected by a string ichich
passes over a Jixed smooth Fully: required to determine the
nnotion.
Let m be the mass of the heavier body, and m' the
=mass of the other. Let T be the tension of the string,
which is the same throughout by the Third Law of Motion,
the weight of the string being neglected as usual
The forces which act on each body are its weight and
the tension of the string ; and these forces act in opposite
THIRD LAW OF MOTION. 2G7
directions. Thus the resultant force on the
heavier body is rtigT downwards, and on
the lighter body T m'g upwards. Therefore
the acceleration on the heavier body is ,
and on the lighter body T ~ v f3 > (^ 37 )
Now as the string is supposed to be in
extensible, the two bodies have at every instant
equal velocities; and therefore the accelera
tions must be equal. Thus
mgT _Tm'g e
m m'
therefore
m + m''
Hence the acceleration is
2<7?ft' ,1 , . mni
q ; , that is ; q.
y m + m"  /y
This is a constant quantity. Hence the motion of the
descending body is like that of a body falling freely, but
is not so rapid : for instead of q we have now
If m=m' there is no acceleration; and so if there is
any motion it is a uniform motion.
90. In the investigation of the preceding Article no
notice is taken of the motion of the Fully : thus the result
is not absolutely true. But it may be readily supposed
that if the mass of the Fully be small compared with that
of the two bodies, the error is very slight ; and the supposi
tion is shewn to be correct in the higher parts of Dynamics.
Theoretically instead of a Fully, we might have a smooth
peg for the string to pass round, but practically it is found
that owing to friction this arrangement is not so suitable :
see Statics, Arts. 191 and 281.
91. The system of two bodies considered in Art. 89
forms the essential part of a machine devised by Atwood,
for testing experimentally the results obtained with respect
268 THIRD LAW OF MOTION.
to rectilinear motion under the action of uniform forces.
Atwood's machine contains some contrivances for diminish
ing friction, and some for assisting in the arrangement
and observation of the experiments ; but the principle is
not affected by these contrivances.
The chief advantage secured by Atwood's machine is
that by taking two bodies of nearly equal weight we can
make  . q as small as we please, and thus render the
' y
motion slow enough to be observed without difficulty.
The results of experiments with Atwood's machine are
found to agree with those assigned by the investigations
already given; and thus they confirm the two important
statements that g is constant at the same place, and that
its value at London is about 32*2.
92. Two bodies are connected by a string, which passes
over a small smooth Pully fixed at the top of two Inclined
Planes having a common height: required to determine the
motion, supposing one body placed on each Plane.
Let m and m' be the masses of the two bodies ; a and a'
the inclinations of the Planes on which they are respectively
placed. Let T denote the tension of the string.
Suppose the body of mass m to be descending. The
weight of this body is mg ; the resolved part of the weight
along the Plane is mg sin a ; hence the resultant force
down the plane is mg sin a  T, and therefore the accelera
. mg sin a  T
tion is .
m
Similarly, for the other body, the resultant force up
the Plane on which it moves is Tm'gsma, and the
, Tm'gsiua
acceleration is , .
in
EXAMPLES. VIII. 269
Now as the string is supposed to be inextensible, the
two bodies have at every instant equal velocities : and
therefore the accelerations must be equal. Thus
mgsinaT _ Tm'g ski a'
therefore y
Hence the acceleration is
m'g (sin a + sin a') , , . . m sin a  m' sin a
g sm a  " . that is  .  g.
' m + m 1
Thus we see, that in order that this may be positive,
and so the body of mass m be acquiring downward velocity,
we must have m sin a greater than m' sin a.
If m sin a =m f sin a' there is no acceleration; and so
if there is any motion it is a uniform motion.
EXAMPLES. VIII.
1. If the two weights in Art. 89 are 15 ounces and 17
ounces respectively, find the space described and the
velocity acquired in five seconds from rest.
2. If the string in Art. 89 were cut at the instant
when the velocity of each body is , find the distance be
tween the two bodies after a time t.
3. In the system of Art. 89 shew that if the sum of
the weights be given, the tension is greater the less the
acceleration is.
4. A weight P is drawn along a smooth horizontal
table by a weight Q which descends vertically, the weights
being connected by a string passing over a smooth Fully
at the edge of the table: determine the acceleration.
5. A weight P is drawn up a smooth plane inclined at
an angle of 30 to the horizon, by means of a weight Q
which descends vertically, the weights being connected
270 EXAMPLES. V1IL
by a string passing over a small Fully at the top of the
plane : if the acceleration be onefourth of that of a body
falling freely, find the ratio of Q to P.
6. Two weights P and Q are connected by a string ;
and Q hanging over the top of a smooth plane inclined
at 30 to the horizon, can draw P up the length of the
plane in just half the time that P would take to draw
up Q : shew that Q is half as heavy again as P.
7. Four equal weights are fastened to a string: find
how they must be arranged so that when the string is laid
over a fixed smooth Fully, the motion may be the same as
that produced when two of the weights are drawn over
a smooth horizontal table by the weight of the other two
hanging over the edge of the table.
8. Two weights of 5 Ibs. and 4 Ibs. together pull one of
7 Ibs. over a smooth fixed Fully, by means of a connecting
string; and after descending through a given space the
4 Ibs. weight is detached and taken away without in
terrupting the motion : find through what space the re
maining 5 Ibs. weight will descend.
9. Two weights are attached to the extremities of
a string which is hung over a smooth Fully, and the
weights are observed to move through 6 '4 feet in one
second ; the motion is then stopped, and a weight of 5 Ibs.
is added to the smaller weight, which then descends
through the same space as it ascended before in the same
time : determine the original weights.
10. Find what weight must be added to the smaller
weight in Art. 89, so that the acceleration of the system
may have the same numerical value as before, but may be
in the opposite direction.
11. Solve the problem in Art. 92, supposing the In
clined Flanes rough.
12. If the Fully in Art. 89 can bear only half the sum
of the weights of the two bodies, shew that the weight of
the heavier body must not be less than (3 f 2 N /2) times the
weight of the lighter body.
DIRECT COLLISION OF BODIES. 271
IX. The Direct Collision of Bodies.
93. "We have hitherto spoken of force as measured
by the momentum which it generates in a given time ;
and the force with which we are most familiar is that of
gravity, which takes an appreciable time to generate in any
body a moderate velocity. There are however examples
of forces which generate or destroy a large velocity in a
time which is too brief to be appreciated. For example,
when a cricket ball is driven back by a blow from a bat,
the original velocity of the ball is destroyed, and a new
velocity generated; and the whole time of the action
of the bat on the ball is extremely brief. Similarly when a
bullet is discharged from a gun, a large velocity is gene
rated in an extremely brief time. Forces which produce
such effects as these are called impulsive forces, and the
following is the usual definition : An impulsive force is a
force which produces a finite change of motion in an indefi
nitely brief time.
94. Thus an impulsive force does not differ in Tdnd
from other forces, but only in degree : an impulsive force is
a force which acts with great intensity during a very brief
time.
As the Laws of Motion may be taken to be true what
ever may be the intensity of the forces which produce or
change the motion, we can apply these laws to impulsive
forces. But since the duration of the action of an im
pulsive force is too brief to be appreciated, we cannot
measure the force by the momentum generated in any
given time : it is usual to state that an impulsive force is
measured by the whole momentum which it generates.
95. We shall not have to consider the simultaneous
operation of ordinary forces and impulsive forces for the
following reason : the impulsive forces are so much more
intense than the ordinary forces, that during the brief
time of simultaneous operation, an ordinary force docs not
272 DIRECT COLLISION OF BODIES.
produce an effect comparable in amount with that pro
duced by an impulsive force. Thus, to make a supposition
which is not extravagant, an impulsive force might gene
rate a velocity of 1000 in less time than onetenth of a
second, while gravity in onetenth of a second would gene
rate a velocity of about 3.
96. The student might perhaps anticipate that diffi
culties would arise in the discussion of questions relating to
impulsive forces, but it will appear as we proceed that the
cases which we have to consider are sufficiently simple.
We may observe that the words impact and impulse
are often used as abbreviations for impulsive action.
97. We are about to solve some problems relating to
the collision of two bodies ; the bodies may be considered
to be small spheres of uniform density, and, as before, we
take no account of any possible rotation: see Art. 10.
The collision of spheres is called direct when at the instant
of contact the centres of the spheres are moving in the
straight line in which the impulse takes place ; the collision
of spheres is called oblique when this condition is not
fulfilled.
98. When one body impinges directly on another, the
following is considered to be the nature of the mutual
action. The whole duration of the impact is divided into
two parts. During the first part a certain impulsive force
acts in opposite directions on the two bodies, of such an
amount as to render the velocities equal. During the
second part another impulsive force acts on each body in
the same direction respectively as before, and the magni
tude of this second impulsive force bears to that of the
former a ratio which is constant for any given pah of
substances. This ratio lies between the limits zero and
unity, both inclusive. When the ratio is unity the bodies
are called perfectly elastic; when the ratio is greater than
zero and less than unity the bodies are called imperfectly
elastic; and when the ratio is zero the bodies are called
inelastic. The ratio is called the coefficient of elasticity, or
the index of elasticity.
DIRECT COLLISION OF BODIES. 273
99. There are three assumptions involved in the pre
ceding Article.
We assume that there is an epoch at which the velo
cities of the two bodies are equal ; this will probably be
admitted as nearly selfevident.
We assume that during each of the two parts into which
the whole duration of the impact is divided by this epoch,
the action on one bo^.y is equal and opposite to the action
on the other; this is justified by the Third Law of Motion.
We assume that the action on each body after the
epoch is in the same direction as before, and bears a
certain constant ratio to it ; this assumption may be taken
for the present as an hypothesis, which is to be established
by comparing the results to which it leads with observa
tion and experiment. See Art. 104.
100. We have still to explain why the words elastic
and inelastic are used in Art. 98. It appears from experi
ment that bodies are compressible in various degrees, and
recover more or less their original forms after the com
pression has been withdrawn : this property is termed elas
ticity. When one body impinges on anpther, we may
naturally suppose that the surfaces near the point of
contact are compressed during the first part of the impact,
and that they recover more or less their original forms
during the second part of the impact.
101. A body impinges directly on another : required to
determine the velocities after impact, the elasticity being im
perfect.
Let a body whose mass is m, moving with a velocity u t
impinge directly on another body whose mass is m\ moving
with a velocity u'. Let .B denote the impulsive force
which during the first part of the impact acts on each
body in opposite directions. Then at the end of the first
part of the impact, the momentum of the body of mass m
is mu R) and therefore its velocity is : and the
T. ME. 18
274 DIRECT COLLISION OF BODIES.
momentum of the body of mass m' is m'u' + R, and there
fore its velocity is , . These velocities are equal
by hypothesis, that is
muE _ m'u' + E
m m' 9
r mm'(uu')
therefore E= m + m > 
Let e denote the index of elasticity ; then during the
second part of the impact an impulsive force eR acts on
each body in the same direction respectively as before.
Let v denote the final velocity of the body of mass m,
and 1/ that of the body of mass m' ; then
mu  (1 + e) E (1 + e}mf (u  ')
v= a ' =u S 
m m + m
_ mu + m'u'  em' (u  u')
m + m'
=ll ,
m' m + m'
_ mu + m'u' + em (u w')
m + m'
102. From the general formulae of the preceding
Article many particular results may be deduced ; we will
give some examples.
If the bodies axe perfectly elastic, e=l; then we have
_(mm')M + 2mV ,_ 2mu (m m')u r
m + m' m + m'
If the bodies are inelastic, e=0; then we have
_ f _mu + m'u'
m + m'
.
DIRECT COLLISION OF BODIES. 275
Again, suppose w'=0, so that a body of mass m, moving
with a velocity u, impinges on a body of mass m! at rest;
then we have
mem' , m(l + e)
t'= T u. v = u.
m + m m + m
Thus the body which is struck goes onwards, and the
striking body goes onwards, or stops, or goes backwards,
according as m is greater than, equal to, or less than em'.
If m'=em, then v=(l  e) u, and v'=u.
103. The formulae of Art. 101 supply two important
inferences. Multiply the value of v by m, and the value of
v' by ra', and add ; thus we obtain
mv + m'v' =mu + m'u'.
This is usually expressed by saying that the^ momentum
of the system is the same after impact as before. It will
be seen that by the momentum of the system, we mean the
result obtained by the algebraical addition of the momen
tum of each body.
Again, subtract the value of v' from that of v\ thus we
obtain
v v'= e(u u').
This is usually expressed by saying that the relative
velocity after impact is e times the relative velocity before
impact. It will be seen that by the relative velocity, we mean
the algebraical excess of the velocity of the one body over
that of the other.
104. The results of the preceding Article have been
deduced from the principles assumed in Art. 98 : if these
results were contradicted by observation and experiment
we should infer that the principles are partly or entirely
inadmissible. On the other hand, assuming these results
to be confirmed by observation and experiment, we may
proceed to examine what support is thus furnished to the
principles.
The first result in the preceding Article may be put in
the form m (u  v} =m! (i/  u'}. This furnishes a corrobo
ration of the truth of the Third Law of Motion ; for it
182
276 DIRECT COLLISION OF BODIES.
shews that the whole force which has acted on one body is
equal and opposite to that on the other.
Wo now pass to the second result. Let R denote, as
before, the impulsive action between the two bodies during
the first part of the impact, and R f that during the second
part of the impact : we shall shew that it will follow that
R' bears a constant ratio to R.
For since v and if are the respective velocities at the
end of the impact we have
m'u'
v'=
therefore v v'=uu' ( +
\m m
Now let us suppose that experi
e pah 1 of substances v v' is alw
Then ()(!+)= (I .,. i,
Now let us suppose that experience shews that for the
same pah 1 of substances v v' is always equal to e (u u').
m + m m+m
This is the required result.
105. A body impinges directly on another: required
to determine the conditions in order that the bodies should
interchange velocities.
Using the same notation as before, we require that
v=u', and v'=u. Hence, by Art 103,
mu' + m'u = mu + m'u')
and u'u= e(uu'}.
The first of these conditions maybe written in this form :
(m m'} (u u'}=Q. Hence we must have m=m'. The
second condition shews that we must have e = l. Thus the
bodies must be of equal mass and perfectly elastic.
106. In Art. 101 we supposed the collision to be
caused by one body overtaking the other. If the bodies
DIRECT COLLISION OF BODIES. 277
move originally in opposite directions, the collision will be
caused by one body meeting the other ; the investigation
in this case will be similar to that already given, and the
results will coincide with those which would be obtained
by changing the sign of u' in the formulae for v and i/.
107. The product of the mass of a body into the
square of its velocity is called the vis viva of the body.
The vis viva of a system of bodies is the sum of the vis viva
of every body of the system. In the higher parts of
Dynamics the consideration of vis viva frequently occurs ;
and it is usual in the elementary parts to demonstrate one
proposition respecting vis viva: this will form the next
Article.
108. By the direct collision of two imperfectly elastic
bodies the vis viva of the system is diminished.
Let u and u' be the velocities before impact of two
bodies whose masses are m and m' respectively, and v and
t/ their velocities after impact. Then by Art. 103,
vv'= e(uu r ).
Therefore (mv + m'v') 2 = (mu + m'u'}\
mm'(v  tf) 2 =mm'e 2 (u  u') 2
=mm'(u  u') 2  mm' (I  e 2 ) (u  u'} 2 ;
therefore, by addition, (m + m'} (mv 2 + m'v' 2 }
= (m + m'} (mu 2 + mV 2 )  mm' (I  e 2 ) (u  u'} 2 ;
therefore mv 2 + m'v' 2 =mu 2 + m'u' 2 ^^ (1  e 2 ) (u  u'} 2 .
m + m ^
Now e cannot be greater than unity, so that 1  e 2 cannot be
negative ; hence mv 2 + m'v' 2 is always less than mu 2 + m'u'*
except when e=l, and then the two expressions are equal.
The expression for the vis viva after impact shews that
during compression vis viva to the amount of t (u u'}~
is lost; and then during restitution e 2 times this amount
is regained.
278 DIRECT COLLISION OF BODIES.
109. It is usual to give the following example of the
subject of the present Chapter : Let A, B, C denote the
masses of three bodies, such that the first and third are
formed of the same substance; let e be the index of
elasticity for the first and second bodies, and therefore
also for the second and third. Suppose the first body to
impinge directly with the velocity u on the second at rest ;
then the second acquires the velocity ^ ^. Suppose
that the second body now impinges directly with this
velocity on the third at rest ; then the third acquires the
A(l + e) ,. ..
2 thatis
supposing every quantity given except B, required to deter
mine B 50 that the velocity communicated to C may be the
n
"We have to make j ^75 TA as great as possible.
B
B A c + AC'
B
We must therefore make the denominator of the last
fraction as small as possible.
But B + A + C+^ = (jB /^^\ +(JA + >JCY,
so that the least value is th'at when *JB ~ \/ ~fi~ van i snes >
that is when B=*J(AC}.
Hence the velocity communicated to the third body is
greatest when the mass of the second body is a mean
proportional between the masses of the first and third.
110. The theory of the collision of bodies appears to
be chiefly due to Newton, who made some experiments
and recorded the results: see the Scholium to the Laws
of Motion in Book I. of the Principia. In Newton's experi
ments however the two bodies seem always to have been
formed of the same substance. He found that the value
EXAMPLES. IX. 279
of e for balls of worsted was about , for balls of steel
about the same, for balls of cork a little less, for balls of
Q  p
ivory , for balls of glass rp
An extensive series of experiments was made by Mr
Hodgkinson, and the results are recorded in the Report
of the British Association for 1834. These experiments
shew that the theory may be received as satisfactory, with
the exception that the value of e, instead of being quite
constant, diminishes when the velocities are made very
large.
EXAMPLES. IX.
1. An inelastic body impinges on another of twice its
mass at rest: shew that the impinging body loses two
thirds of its velocity by the impact.
2. A body weighing 5 Ibs. moving with a velocity of
14 feet per second, impinges on a body weighing 3 Ibs., and
moving with a velocity of 8 feet per second: find the
velocities after impact supposing 6=5.
3. Two bodies are moving in the same direction with
the velocities 7 and 5; and after impact their velocities
are 5 and Q : find the index of elasticity, and the ratio of
the masses.
4. Two bodies of unequal masses moving in opposite
directions with momenta numerically equal meet : shew
that the momenta are numerically equal after impact.
5. A body weighing two Ibs. impinges on a body
weighing one Ib. ; the index of elasticity is ^ : shew that
, and that v'=u.
280 EXAMPLES. IX.
6. The result of. an impact between two bodies moving
with numerically equal velocities in opposite directions is
that one of them turns back with its original velocity, and
the other follows it with half that velocity : shew that
one body is four times as heavy as the other, and that
7. Find the necessary and sufficient condition in order
that v' may be equal to u.
8. A strikes J3, which is at rest, and after impact the
velocities are numerically equal : if r be the ratio of JB's
mass to A'a mass, shew that the index of elasticity is
a
 , and that J3's mass is at least three times A J a mass.
9. A, B and C are the masses of three bodies, which
are formed of the same substance ; the first impinges on
the second at rest, and then the second impinges on the
third at rest : determine the index of elasticity in order
that the velocity communicated to C may be the same as if
A impinged directly on C.
10. A body impinges on an equal body at rest : shew
that the vis viva before impact cannot be greater than
twice the vis viva of the system after impact.
11. A series of perfectly elastic bodies are arranged in
the same straight line ; one of them impinges on the next,
then this on the next, and so on : shew that if their masses
form a Geometrical Progression of which the common
ratio is r, their velocities after impact form a Geometrical
o
Progression of which the common ratio is  .
12. A number of bodies A,B,C,... formed of the same
substance, are placed in a straight line at rest. A is then
projected with a given velocity so as to impinge on B\
then B impinges on C; and so on. Find the masses of the
bodies J3, C,... so that each of the bodies A, B, (7,... may
be at rest after impinging on the next; and find the
velocity of the n* ball just after it has been struck by the
(?il) th baJL
OBLIQUE COLLISION OF BODIES. 281
X. The Oblique Collision of Bodies.
111. In the present Chapter we shall consider the
oblique collision of bodies; see Art. 97. It will be found
that the problems discussed involve only a more extensive
application of principles already explained. We shall
confine ourselves to cases in which the line of impact
and the directions of the motions of the bodies are in one
plane.
112. A body impinges obUqmly on another: required
to determine tJie velocities after impact, the elasticity being
imperfect.
Let a body whose mass is m, moving with a velocity u,
impinge on another whose mass is m', moving with a velo
city u'. Let the direction of the first velocity make an
angle a with the line of impact, and that of the second an
angle a'. After impact let the velocities be denoted by v
and i/, and the angles which their directions make with
the line of impact by ft and ft'.
Eesolve all the velocities along the line of impact and
at right angles to it. No impulsive force acts on the
bodies in the direction at right angles to the line of im
pact, and so the velocities at right angles to the line of
impact remain unchanged. Hence
vsmft=usiua (1),
v'sin/3 / =w'sina' (2).
The velocities along the line of impact are affected just
as they would be if the velocities in the other direction did
not exist. Hence, preceding as in Art. 101, we obtain
o_>u cos a + m'u' cos a  em' (u cos a  u' cos a')
08 P m + m' ( 3 ;>
, a, mu cos a + m'u' cos a' + em (u cos a  u' cos a')
V COS a = ^ ....(4).
m + m' ^ '
If we divide (1) by (3) we obtain the value of tan ft ; this
determines the direction of the velocity of the impinging
282 OBLIQUE COLLISION OF BODIES.
body after impact. If we square (1) and (3) and add, we
obtain the value of v 2 ; this determines the magnitude of
the velocity. Similarly from (2) and (4) we can determine
the direction and the magnitude of the velocity of the
other body after impact.
113. In the accom
panying figure C repre
sents the centre of the
body which we call the
impinging body, consi
dered to be a sphere, at
the instant of impact; and
C' the centre of the other
body. CC' is the line of
impact. The directions of
the velocity of the impinging body before and after impact
are represented by CA and CB ; and those of the other body
by C'A' and C'B'. Thus if D be a point on CO' produced,
angle ACD=a, angle A'C'D=a,
angle BCD = /3. angle B'C'D = $.
This figure may serve to illustrate the problem; it
will however be easily perceived that the general formulae
admit of application to a large number of special cases,
and that the figure would have to be modified in order to
apply accurately to such special cases. For instance, we
have supposed CA and C'A' to fall on the same side of
CD, but it is of course possible that they should fall on
different sides. It will be found on careful investigation
that u and u' may always be considered to be positive
quantities; and all the cases which can occur will be
included in the general formulas, where the angles have
values lying between and 180, positive or negative.
The student should notice the particular results which
may be deduced from the general formulas as in Art 102.
114. Multiply equation (3) of Art. 112 by m, and equa
tion (4) by m' and add ; thus we obtain
mv cos Q + m'v' cos /3' = mu cos a + m'u' cos a' ;
this shews that the momentum of the system resolved in
OBLIQUE COLLISION OF BODIES. 283
tlic direction of tJie line of impact is the same after impact
as before.
The momentum of each body resolved in the direction
at right angles to the line of impact is the same after im
pact as before, and therefore so also is the momentum of
the system resolved in this direction.
Subtract equation (4) of Art. 112 from equation (3);
thus we obtain
v cos $  v' cos ' =  e (u cos a  u' cos a').
This result may be expressed in words thus : the rela
tive velocity, resolved along the line of impact, after impact
is e times its value before impact.
115. We have hitherto treated of the collision of two
bodies each of which is capable of motion ; in the next
Article we shall apply the principles already explained to
a case of collision in which one body is fixed.
116. A body impinges obliquely on a fixed smooth
plane : required to determine the velocity after impact, the
elasticity being imperfect.
Let m be the mass of the body.
Let AC represent the direction
of the velocity before impact,
meeting the plane at C, and CB
the direction after impact. Draw
CD at right angles to the plane ;
then, since the plane is smooth,
CD represents the line of impact.
Let u denote the velocity before impact, and v that
after impact: let a denote the angle ACD* and 8 the
angle BCD.
Resolve the velocities along the line of impact and at
right angles to it. No impulsive force acts on the body
at right angles to the line of impact, and so the velocity
at right angles to the line of impact remains unchanged.
Hence
(1).
284 OBLIQUE COLLISION OF BODIES.
Let R denote the impulsive force which acts on the
body during the first part of the impact. Then at the end
of the first part of the impact the velocity of the body
. mu cos a R ., . .
resolved along the line of impact is  ; this is
wi
zero by hypothesis, therefore R=mucosa. Let e denote
the index of elasticity ; then during the second part of
the impact an impulsive force eR acts on the body ; and
therefore the final velocity along the line of impact
mucosa(I + e}R eR
=  *   =  =  eu cos a.
m m
Thus
vcos/3= eucosa .................. (2).
From (1) and (2) we obtain, by division,
cot/3= ecota .................. (3).
The negative sign indicates that CB and CA are on
opposite sides of CD, as represented in the figure : the
velocity after impact along the line of impact, that is at
right angles to the plane, is numerically e times its value
before impact. From (1) and (2) we have by squaring and
adding
(4).
Thus (3) determines the direction of the velocity after
impact, and (4) determines its magnitude.
The angle ACD is called the angle of incidence) and
the angle BCD the angle of reflexion. Thus from (3) we
see that the cotangent of the angle of reflexion is always
numerically equal to e times the cotangent of the angle of
incidence.
117. Some particular results of interest may be de
duced from the preceding Article.
Suppose e=l; then cot ft = cot a, and v 2 =w 2 . Thus
if the elasticity be perfect the angles of incidence and
reflexion are numerically equal, and the velocities before
and after impact are equal.
OBLIQUE COLLISION OF BODIES. 285
Suppose e=0; then /3 is a right angle. Thus if there
be no elasticity, the body after impact moves along the
plane with the velocity u sin cu
Suppose a=0, so that the impact is direct. Then after
impact the body rebounds along its former course with Q
times its former velocity.
Suppose a=0 and e=0. Then the body is brought to
rest by the impact.
118. In the equations of Art. 112 suppose w'=0;
then the equations become
v sin j3=wsin a .............................. (1),
v'sinj3'= .............................. (2),
_ in em /0 .
vcosj3=  r wcosa .................. (3).
, ,. N
v' cos/3'=   r^cosa .................. (4).
Let ,=, so that m=km r ; then the last two equa
tions become
ke
 r',
Thus if k be very small indeed we .have very nearly
v cos /3 =  eu cos a ..................... (5),
v'cos/3'= ..................... (6).
Now the results (1) and (5) agree with those denoted
by (1) and (2) in Art. 116. Thus we see that the case
of a body impinging on a fixed plane is practically the
same as that of a body impinging on another body of very
much larger mass which is at rest. The comparison we
have here made between the two cases is an example of a
kind of exercise which is very valuable for students.
286 OBLIQUE COLLISION OF BODIES.
119. The theory of the collision of bodies gives the
opportunity of forming a large number of illustrative pro
blems ; we will now solve some as examples.
120. A body is to start from one given point, and after
reflexion at a given fixed smooth plane it is to pass through
another given point : required to determine the direction of
incidence, the index of elasticity being supposed known.
Let A be the point from A
which the body is to start, B the
point through which the body is
to pass after reflexion at the plane.
Draw BC perpendicular to
the plane, meeting it at C ; pro
duce BC to D so that CD may be
equal to  BC, where e is the
index of elasticity. Join AD, cutting the plane at E;
then AE is the required direction of incidence, and EB
is the direction of reflexion.
For the cotangent of the angle of incidence at E is
the tangent of CED, that is 7^; and the cotangent of
GJ&
the anglo of reflexion at E is the tangent of BEG, that is
TIC 1
%. Therefore
cotangent of the angle of reflexion _ BC _
cotangent of the angle of incidence ~~ CD ~
Hence, by Art. 116, a body impinging on the plane in
the direction AE will be reflected in the direction EB.
121. A body is reflected in succession by two fixed smooth
planes, of the same substance, which are at right angles to
each other, the body moving in a plane at right angles to the
intersection of the fixed planes : required to shew that the
directions of motion before the first reflexion and after the
second reflexion are parallel.
OBLIQUE COLLISION OF BODIES. 287
Let PQES be the course of
the body, the first reflexion being
at Q and the second at R. Let e
be the index of elasticity.
Suppose the velocity before
reflexion at Q to consist of u per
pendicular to A /?, and v parallel
to AB. After reflexion at Q the
velocity will consist of  eu per
pendicular to AB, and v parallel
to AB. After reflexion at jR the velocity will consist of
eu perpendicular to AB, and  ev parallel to AB.
Hence the value of each component velocity after re
flexion at jR is  e times its value before reflexion at Q.
This shews that ES is parallel to QP.
And the whole velocity after reflexion at R is numeri
cally equal to e times the whole velocity before reflexion
at.
We here assume that no force acts on the body during
its motion except the impulses at Q and R ; so that we
must suppose that gravity does not exist, or that it is prac
tically neutralized by the motion taking place on a fixed
smooth horizontal table.
122. If a body be projected in a direction inclined to
the horizon it describes a parabolic arc ; on reaching the
ground it will in general rebound and describe another
parabolic arc : we shall now investigate the connexion be
tween these two arcs.
Let u be the velocity of projection, and a the inclina
tion of the direction of projection to the horizon. Thus
at starting the vertical velocity is u sin a, and the hori
zontal velocity is wcosa. The horizontal velocity is not
changed during the motion. When the body reaches the
ground its vertical velocity is the same as at starting ; and
accordingly it rebounds with a vertical velocity ewsina,
if the ground be a smooth hard plane, and the index of
elasticity be e.
288 OBLIQUE COLLISION OF BODIES.
Hence, on starting for the second parabolic arc, the
body has the horizontal velocity MCOSO, and the vertical
velocity eu sin a ; and all the circumstances of the motion
can be determined; see Arts. 57, 58, 60, 62. Thus:
The latus rectum = : , which is the same as that
of the first arc.
2eu sin a
The time of flight:
9
e 2 M 2 sin 2 a
The greatest height reached
*&
2 2w 2
The range =  eu sin aw cos a =  e sin a cos a.
9
After describing the second parabolic arc the body will
rebound and describe a third parabolic arc; and so on.
The following results are easily seen to hold :
All these parabolic arcs have the same latus rectum.
The times of flight form a Geometrical Progression of
which the common ratio is e. The greatest heights form a
Geometrical Progression of which the common ratio is e 2 .
The ranges form a Geometrical Progression of which the
common ratio is e.
123. In like manner if a projectile describe succes
sive arcs by rebounding from an inclined plane which
passes through the point of projection, it will be found that
the times of flight form a Geometrical Progression of which
the common ratio is e, and that the greatest distances
from the inclined plane form a Geometrical Progression of
which the common ratio is e 2 .
124. By the oblique collision of two imperfectly elastic
bodies the vis viva of t/ie system is diminished.
Let m and m' be the masses of the bodies, u and u'
their respective velocities before impact, v and i/ their
velocities after impact ; let a and a' be the angles which
I
EXAMPLES. X. 289
their directions of motion make with the line of impact
before impact, /3 and /3' the corresponding angles after
impact.
tThen, by Art. 114,
(mv cos ft + m'v' cos @') 2 =(mu cos a + m'u' cos a') 2 ,
' (v cos /3  1/ cos ft} 2 =mm' e 2 (u cos a  u' cos a') 2
vm! (u cos a  u' cos a') 2  mm' (1  e 2 ) (w cos au' cos a') 2 .
Hence by addition, and by division by m + m' y
mv 2 cos 2 p + m'v' 2 co$ 2 ff
_ , mm' (1  e 2 ) , , A9
= mu? cos 2 a + TO u ~ cos 2 a   , ' (u cos a  u cos a) *.
Also mv 2 sin 2 /3 = mw 2 sin 2 a,
and m'v' 2 sin 2 # = m'u' 2 sin 2 a'.
Therefore, by addition,
u 2 + m'u' 2  mm '( l ~f) ( u cos a u f cos a') 2 ;
and as 1e 2 cannot be negative the required result is
obtained.
If the elasticity is perfect the vis viva of the system is
the same after the collision as before.
EXAMPLES. X.
[The elasticity is to be supposed imperfect unless tho
contrary is stated.]
1. A ball impinges on an equal ball at rest, the
elasticity being perfect; if the original direction of the
striking ball is inclined at an angle of 45 to the straight
line joining the centres, determine the angle between the
directions of motion of the striking ball before and after
impact.
11 MS, 19
290 EXAMPLES. X.
2. A ball falls from a height h on a horizontal plane,
and then rebounds : find the height to which it rises in its
ascent.
3. A ball falls from a height h on a horizontal plane,
and then rebounds, falls and rebounds again ; and so on ;
find the sum of the spaces described.
4. A ball of mass m impinges on a ball of mass m'
at rest : shew that the tangent of the angle between the
old and new directions of motion of the impinging body is
l + e mf sin 2a
2 m + m' (sin 2 a  e cos 2 a) *
5. A ball of mass m impinges on a ball of mass m' at
rest : find the condition which must hold in order that the
directions of motion of the impinging ball before and after
impact may be at right angles.
6. A ball impinges on an equal ball at rest, the
elasticity being perfect; the angle between the old and
new directions of motion of the impinging body is 60 : find
the velocity after impact.
7. A ball impinges on an equal ball at rest, the
elasticity being perfect : find the condition under which
the velocities will be equal after impact.
8. A body is projected at an inclination a to the
horizon ; and by continually rebounding from the horizon
tal plane describes a series of parabolas : find the tangent
of the angle of projection at the ?i th rebound.
9. % A body is projected with the velocity u, at the
inclination a to the horizon, and by continually rebounding
from the horizontal plane describes a series of parabolas :
find the sum of the ranges.
10. In the preceding Example find the time which
elapses before the body ceases to rebound.
11. A ball is projected from a point in a smooth
horizontal billiard table, and after striking the four sides
in order returns to the starting point : shew that the
sides of the parallelogram described are parallel to the
diagonals of the table, the elasticity being perfect.
EXAMPLES. X. 291
12. A ball is projected from the middle point of one
side of a billiard table, so as to strike first an adjacent side,
and then the middle point of the side opposite to that from
which it started : determine the direction of projection.
13. Two balls moving in parallel directions with equal
momenta impinge : shew that if their directions of motion
be opposite they will move after impact in parallel direc
tions with equal momenta.
14. In the preceding Example find the condition in
order that the direction after impact may be at right
angles to the original direction.
15. A and B are given positions on a smooth hori
zontal table : and AC, BD are perpendiculars on a hard
plane at right angles to the table. If a ball struck from A
rebounds to B after an impact at the middle point of
CD, shew that when the ball is sent back from B to A,
the point of impact on CD will divide it into parts whose
ratio is that of e 2 to 1.
16. A BCD is an ordinary rectangular billiard table
perfectly smooth ; E is a ball in a given position : it is
required to select the proper position for another ball F
in all respects like the first, so that the player, striking E on
F, may cause F to run into the corner pocket A, and E to
run into D with equal velocities, the elasticity being perfect.
It is assumed that the radius of each ball may be
neglected in comparison with the dimensions of the billiard
table.
17. A ball is projected from a point between two
vertical planes, the plane of motion being perpendicular to
both : shew that the latera recta of the parabolic arcs
described form a Geometrical Progression having the
common ratio e 2 .
18. A body slides down a smooth Inclined Plane of
given height; at the bottom of the Inclined Plane the
particle rebounds from a hard horizontal plane: find the
range on the latter plane.
192
292 EXAMPLES. X.
19. A ball is projected from a given point at a given
inclination towards a vertical wall : determine the velocity
of projection so that after striking the wall the ball may
return to the point of projection.
20. Two equal balls start at the same instant with
equal velocities along the diagonals of a square from the
ends of a side, and when they meet, the line of impact
is parallel to that side : determine the angle which the
direction of motion of each ball after impact makes with
the line of impact.
21. A perfectly elastic ball is projected with a given
velocity from a point between two parallel walls, and
returns to the point of projection after being once reflected
at each wall : find the angle of projection.
22. An imperfectly elastic ball is thrown from a given
point against a vertical wall : find the direction in which
it must be projected with the least velocity, so as to return
to the point of projection.
23. There are two parallel walls whose distance apart
is equal to their height, and from the top of one wall a
perfectly elastic ball is thrown horizontally so as to fall at
the foot of the same wall after rebounding from the other :
determine the position of the focus of the first path.
24. Bodies of different elasticities slide down a smooth
Inclined Plane through the same vertical height, and
impinge on a horizontal plane at its foot : shew that all the
parabolas which are afterwards described have the same
latus rectum.
25. A ball is projected in a given direction within a
fixed horizontal hoop, so as to go on rebounding from the
surface of the hoop : if the velocity at the end of every
impact be resolved along the tangent and the normal to
the hoop at the point, shew that the former component is
constant, and that the latter component diminishes in
Geometrical Progression.
26. Shew how to determine the direction of projection
of a ball lying at a given point on a smooth billiard table,
so that after striking all the sides in succession the ball
may hit a given point.
MOTION OF THE CENTRE OF GRAVITY. 293
XI. Motion of the Centre of Gravity of two or
more bodies.
125. We have explained in the Statics what is meant
by the centre of gravity of a body or a system of bodies ;
and have shewn that for a given body or system there
is only one centre of gravity. If a change takes place
in the position of any body of the system, there is a
corresponding change in the position of the centre of
gravity of the system ; and thus we are led to consider the
motion of the centre of gravity of two or more bodies.
126. Having given the velocities of two bodies estimated
in any direction, required the velocity of their centre of
gravity estimated in the same direction.
Suppose m and ml the masses of the bodies ; let their
distances from a fixed plane at a certain instant be a and
a' respectively; then the distance of the centre of gravity
from the fixed plane is ; see Statics. Arts. 119
m + m
and 146.
Let the velocities of the two bodies estimated at right
angles to the plane be b and 6'; then at the end of a
time t the distances of the bodies from the fixed plane are
a + bt and a' + b't respectively. Therefore the distance
of the centre of gravity from the fixed plane
_ m (a + bt] + m' (a' + b't] _ma + m'a' mb + m'b'
m + m' m + m' m + m'
This shews that the distance of the centre of gravity
from the fixed plane increases uniformly with the time;
and that the velocity of the centre of gravity at right angles
. . . mb + m'b'
to the fixed plane is r .
m + m
127. In the preceding Article we have assumed that
the two bodies have uniform velocities in the assigned
294 MOTION OF THE CENTRE OF GRAVITY.
direction; but the result may be easily extended to the
case in which the velocities are not uniform. For the
time t may be as short as we please ; and if the velocities
of the bodies are really variable in the assigned direction,
no error will ultimately arise from regarding them as
uniform for an indefinitely short time. Thus we have
the following general result : the velocity of the centre of
gravity of two bodies estimated in any direction at any
instant is found by dividing the momentum of the system
estimated in that direction at that instant by tlw sum of
the masses.
128. The result just enunciated for the case of two
bodies is true for any number of bodies; the mode of
demonstration is the same as that given for two bodies.
129. The motion of the centre of gravity of two bodies is
not affected by the collision of the bodies.
First suppose the collision to be direct.
Let m and m' be the masses of the bodies, u and u'
their velocities before impact, v and v' their velocities after
impact. The velocity of the centre of gravity, by Art. 126,
. mu + m'u' , f , mv + m'tf .,,
is ; before impact, and j after impact;
m + m m + m
and these are equal by Art. 103.
Next suppose the collision to be oblique.
Let m and m' be the masses of the bodies, u and u'
their velocities before impact, a and a' the angles which
the directions of motion make with the line of impact;
let v and v 1 be the corresponding velocities, and /3 and ft
the corresponding angles after impact.
The velocity of the centre of gravity, estimated in
the direction of the line of impact, by Art. 126, is
mu cos a + m'u' cos a'
m + m'
before impact, and is
mv cos ft + m'v' cos ft
m + m'
after impact; and these are equal by Art. 114.
MOTION OF THE CENTRE OF GRAVITY. 295
The velocity of the centre of gravity estimated in the di
rection at right angles to the line of impact, by Art. 126, is
mu sin a + m'u' sin a'
m + m'
before impact, and is
mv sin /3 + m'v' sin /3'
m + m'
after impact; and these are equal by Art. 114.
Thus the component velocity of the centre of gravity
in two directions is the same after impact as before ; and
therefore the resultant velocity is the same in magnitude
and direction after impact as before.
130. It follows from the investigation of Art. 126, that
if two bodies move in straight lines, each with uniform
velocity, then their centre of gravity moves also in some
straight line, with uniform velocity. Hence we may esta
blish the following proposition : the centre of gravity of two
projectiles, which are moving simultaneously ', describes a
parabola. For suppose at any instant that gravity ceased
to act; then each body would move in a straight line
with uniform velocity, and so would also the centre of
gravity. The effect of gravity in a given time is to draw
each body down a vertical space which is the same for each
body, and which varies as the square of the time: and the
centre of gravity is drawn down through the same vertical
space. Hence, by reasoning as in Art. 51, we find that the
path of the centre of gravity is a parabola.
And this result may be extended to the case of any
number of bodies : see Art. 128.
131. By the method of Arts. 126 and 127, we may
establish the following result : If f and f be the accelera
tions, estimated in any direction, of two moving bodies,
whose masses are m and m' respectively, the acceleration of
the centre of gravity of the two bodies estimated in the same
,. . . mf+m'f
direction is  r .
And this result may be extended to the case of any
number of bodies : see Art. 128.
296 EXAMPLES. XL
EXAMPLES. XI.
1. A body weighing 4 Ibs. and another weighing 8 Ibs.
are moving in the same direction, the former with the
velocity of 8 feet per second, and the latter with the
velocity of 2 feet per second : determine the velocity of the
centre of gravity.
2. Equal bodies start from the same point in directions
at right angles to each other, one with the velocity of 8 feet
per second, and the other with the velocity of 6 feet per
second : determine the velocity of the centre of gravity.
3. In the system of Art. 89 supposing the initial
velocity zero, find the velocity of the centre of gravity at
the end of a given time.
4. A heavy body hanging vertically draws another
along a smooth horizontal plane ; supposing the initial
velocity zero, find the horizontal and the vertical velocity
of the centre of gravity at any instant.
5. Shew that the centre of gravity in the preceding
Example describes a straight line with uniform acceleration.
6. In the system of Art. 92 supposing the initial velo
city zero, find the velocity of the centre of gravity at the end
of a given time resolved parallel to the two planes.
7. Shew that the centre of gravity in the preceding
Example describes a straight line with uniform acceleration.
8. Two balls are dropped from two points not in the
same vertical line, and strike against a horizontal plane,
the elasticity being perfect : shew that the centre of
gravity of the balls will never reascend to its original
height, unless the initial heights of the balls are in the
ratio of two square numbers.
9. Three equal particles are projected, each from one
angular point of a triangle along the sides taken in order,
with velocities proportional to the sides along which they
move: shew that the velocity of the centre of gravity
estimated paraUel to each side is zero ; and hence that the
centre of gravity remains at rest.
10. P, Q, R are points in the sides EC, CA, AB re
Ttp riQ Ap
spectively of the triangle ABC, such that ~p = ^~ = ^ :
shew that the centre of gravity of the triangle PQR
coincides with that of the triangle ABC.
LAWS OF MOTION. 297
XII. Laws of Motion. General Eemarks.
132. We propose in the present Chapter to make some
general remarks concerning the Laws of Motion. It is not
necessary that a student should devote much attention to
this Chapter on his first reading of the subject. He should
notice the points which are here considered, and when in
his subsequent course he finds any difficulty as to these
points he can examine the remarks which bear upon the
difficulty.
133. We will here repeat the Laws of Motion.
I. Every body continues in a state of rest or of uni
form motion in a straight line, except in so far as it may be
compelled to change that state by force acting on it.
II. Change of motion is proportional to the acting
force, and takes place in the direction of the straight line
in which the force acts.
III. To every action there is always an equal and
contrary reaction : or the mutual actions of any two bodies
are always equal and oppositely directed in the same
straight line.
It is manifest that instead of Laws of Motion it would
be more accurate to call these statements, Laws relating
to the connexion of force with motion.
134. We have already observed that the motion of a
body here considered is of that kind in which all the. points
of the body describe curves identical in form, though vary
ing in position. For example, when we speak of the motion
of a falling body we mean such a motion that every point
of the body describes a straight line. The motion which
is here considered is called motion of translation, to dis
tinguish it from motion of rotation, which we do not con
sider.
135. We have also stated, in connexion with the
distinction just explained, that the Laws of Motion ought
to be enunciated with reference to particles rather than to
bodies. It might appear to a beginner that there can be
298 LAWS OF MOTION.
little advantage in studying the theory of the motion of
particles, because in practice we are always concerned with
bodies of finite size. But it is not difficult to shew the
importance and value of a sound theory of the motion of
particles. For it is easy to conceive that a solid body is
made up of particles, and that the forces acting may be
such as to render the motion of one particle exactly the
same as the motion of another ; and so the motion of the
body is known when that of one particle is known. The
case of a falling body illustrates this remark ; see also
Art. 81. Again, it is shewn in the higher parts of Me
chanics that the motion of the centre of gravity of a rigid
body is exactly the same as the motion of a particle having
a mass equal to the mass of the rigid body, and acted on by
forces equal and parallel to those which act on the rigid
body. Although the student could not at the present stage
follow the reasoning by which this remarkable result is
obtained, nor even fully apprehend the result itself, yet he
may readily perceive that great interest is thus attached to
the theory of the motion of particles.
136. The terms relative velocity and relative motion are
sometimes used in Mechanics ; their meaning will be ob
vious from an illustration. Suppose two bodies, A and B
moving along the same straight line, A with the velocity
of 4 feet per second, and B foremost with the velocity of
6 feet per second. Then the distance between A and B
increases at the rate of 2 feet per second ; this rate is called
the relative velocity. We may say that the motion of one
body with respect to the other, or the relative motion, is
the same as if A were brought to rest, and B moved for
wards with the velocity of 2 feet per second ; or, it is the
same as if B were brought to rest, and A moved backwards
with the velocity of 2 feet per second. See also Art. 103.
137. Up to the end of the sixth Chapter we considered
the effect which a force produces on the velocity of a body
without regard to the mass of the body moved. It is
usual to apply the name accelerating force to force so con
sidered ; and hence the two following definitions are used :
Force considered only with respect to the velocity
generated is called accelerating force.
LA WS OF MOTION. 299
Force considered with respect to the mass to which
velocity is communicated as well as to the velocity generated
is called moving force.
The terms tend to confuse a beginner, because they lead
him to suppose that there are two kinds of force. There is
really only one kind of force, namely that which is called
moving force in the foregoing definitions ; for when force
acts it always acts on some body. It is not necessary to
make any use of the term accelerating force: when the
beginner hears or reads of an accelerating force / he must
remember that this means a force which produces tha
acceleration f in the motion of the body which is con
sidered.
138. We have followed Newton in our enunciation of
the Laws of Motion ; but it is necessary to observe that
this course is not universally adopted. Many writers in
effect divide Newton's Second Law into two, which they
term the Second and Third Laws, presenting them thus :
Second Law. When forces act on a body in motion
each force communicates the same velocity to the body as
if it acted singly on the body at rest.
Third Law. When force acts on a body the momentum
generated in a unit of time is proportional to the force.
Then Newton's Third Law is presented as another
principle which must be admitted to be true, although
apparently not difficult enough or not important enough to
be ranked formally with the Laws of Motion.
We have followed Newton for two reasons. In the first
place, his mode of stating the Laws of Motion seems, to say
the least, as good as any other which has been proposed :
and in the second place, there is very great advantage in a
uniformity among teachers and students as to the first
principles of the subject, and this uniformity is more likely
to be secured under the authority of Newton than under
that of inferior names.
139. We have given in Art. 47 Newton's form of the
parallelogram of velocities ; some writers omit this, and
supply its place by a purely geometrical proposition, which
is substantially as follows :
300 LAWS OF MOTION.
Let OACB be a par
allelogram : from a point
P within it draw PM
parallel to OB, meeting
OA at M, and PN par
allel to OA, meeting OB
at N : then if a point
moves in such a manner
that ~p]v = ~Q J A always, P must move along the diago
nal OC.
PM OB
Since ir^j i* follows by Euclid, vi. 26, that the
parallelograms OMPN and 0^Z? are about the same
diagonal. Thus P must be on the straight line OC.
Also P arrives at C when M is at A and JV is at B ;
and if J/ and JV move uniformly along OA and &Z? respec
tively, then P moves uniformly along OC.
Thus we have demonstrated the result, without any
reference to the notion of force, chiefly by the .aid of
Euclid, vi. 26. Students are sometimes perplexed by find
ing that while the theorem is asserted to be purely geo
metrical the enunciation and demonstration are expressed
by the aid of language borrowed from Mechanics. In
Newton's mode we arrive at the result as a deduction from
the First and Second Laws of Motion. We have already
seen, by an example given in Art. 77, that it is possible to
obtain geometrical truths indirectly by the aid of Mechanics ;
and such a process is both interesting and valuable ; but
when we wish to draw special attention to the fact that
a certain result is purely geometrical, it is advisable to
restrict ourselves to geometrical language in the enunciation
and investigation.
140. We have already stated that the direct experi
mental evidence for the truth of the Laws of Motion is not
very strong ; strictly speaking we might assert that there
is no direct experimental evidence : for the Laws of Motion
ought to be enunciated with respect to particles, and we
LAWS OF MOTION. 301
cannot make the requisite experiments on particles. In
fact the Laws of Motion should be assumed in the outset as
hypotheses, and their truth verified by the agreement of
results deduced from them with accurate observations. We
are enabled to institute some such comparisons by the aid
of Atwood's machine ; but, as we have said, it is from the
close agreement of theory with observation in Astronomy
that we derive the most convincing evidence of the truth of
the Laws of Motion. The history of the progress of Me
chanics confirms the statement that the Laws of Motion
cannot be regarded as obviously true or even as readily
admissible when enunciated. The Greeks excelled in
Geometry, and were not ignorant of Statics ; but even men
so illustrious as Aristotle and Archimedes completely failed
in their attempts at Dynamics ; and the honour of laying
the foundations of this subject was reserved for Galileo.
141. In Art. 51 we have devoted a few lines to shewing
that P is the position of the body at the end of the time t.
It is usually considered sufficient to make the following
statement : AT is the space which a body moving with the
velocity u would describe in the time t, and TP is the
space through which the body would be drawn by gravity
in the time t ; and therefore by the Second Law of Motion
P is the position of the body at the end of the time t. This
statement implies that the result is an immediate deduction
from the Second Law of Motion. But the Second Law of
Motion does not give us any immediate information about
the position of a body when forces act on it ; the Law is
directly concerned only with the velocity of the body, and
when we have determined the velocity of a body at any
instant we have a further investigation to make in order
to find the position of the body at any instant.
The point is perhaps of small importance in this case ;
but a beginner might easily be led into error on other
occasions, if his attention had never been drawn to the
exact meaning of the Second Law of Motion.
142. It will be interesting to give a brief account of
that part of Newton's Principia which is devoted to the
302 LA WS OF MOTION.
Laws of Motion ; the student will thus have his attention
drawn to some important principles, which will be of
service to him as he proceeds, although he may be unable
at present to master them completely.
After enunciating and briefly illustrating the Laws
of Motion, Newton adds a series of Corollaries ; these we
shall now state, omitting the commentary by which he
supports them.
I. The proposition which is now called the Parallelo
gram of Velocities : see Art. 47.
II. The statical proposition which is now called the
Parallelogram of Forces. Newton deduces this from his
first Corollary, and points out some applications to the
theory of machines.
III. The momentum of a system estimated in any
direction is unaffected by the mutual actions of the bodies
which compose the system. Newton considers principally
the case of the collision of two bodies : see Art. 114.
IV. The position of the centre of gravity of two or
more bodies is not changed by the mutual actions of the
bodies; so that the centre of gravity of bodies acting
on each other, and subject to no external forces, either
remains at rest or moves uniformly in a straight line :
see Art. 129.
V. The relative motions of bodies comprised within a
given space are the same whether that space is at rest,
or moving uniformly in a straight line. This is illustrated
by the fact that motions take place in a ship in the same
way whether the ship is at rest or moving uniformly in a
straight line.
VI. If bodies be in motion in any manner their rela
tive motions will not be changed if they are all acted
on by forces producing equal accelerations in parallel
directions. Arts. 74 and 75 illustrate this statement.
After the Corollaries Newton gives in a Scholium an ac
count of experiments on the collision of bodies, and some
additional remarks on the Third Law of Motion.
LA WS OF MOTION. 303
143. We have already stated that in our investi
gations respecting falling bodies, we leave out of considera
tion the resistance of the air; and that in consequence our
results may in some cases deviate considerably from
practical exactness. We will make some remarks on the
nature of the resisting force exerted by the air.
Let us take for example the case of a falling body.
It appears from experiments that the resistance of the
air varies as the square of the velocity of the body;
or at least this is very approximately the case. Let v
denote the velocity of the body; then the resistance of
the air may be denoted by kv 2 , where k is some constant.
Let m be the mass of the body, and mg the weight of the
body; then the downward force on the body is mgkv\
and so the acceleration, at the instant the velocity is v, is
m y~ . Thus we see that the acceleration is not con
m
stant. In order to determine the motion of the body
under this acceleration, more mathematical knowledge
would be required than the student is at present supposed
to possess; but two interesting results will be readily
understood.
If there are two bodies of the same external form and
substance, experiment shews that the coefficient k is the
fa.2
same for both. Now the acceleration is a ; and
m
therefore for a given value of &, the effect of the resistance
is smaller the larger m is. For example, suppose we have
a solid sphere and a hollow sphere, made of the same sub
stance, and having the same external radius ; then the re
sistance of the air has less influence on the motion of the
solid sphere than on the motion of the hollow sphere.
Thus we are able to understand why the resistance of the
air produces less effect on the motion of dense bodies, than
on the motion of light bodies, other circumstances being
the same.
If kv*=mg the acceleration is zero, and the nearer the
value of v is to / ~;r the smaller the acceleration be
comes. This expression is called the terminal velocity.
304 LAWS OF MOTION.
If the body falls from rest its velocity will never exceed
this value, but will approach indefinitely near to this value
if the motion can continue long enough. If the body
be projected downwards with a velocity greater than this
expression the velocity will always exceed this value, but
will approach indefinitely near to this value if the motion
can continue long enough. Thus in each case the motion
tends to become uniform.
144. The following example will illustrate the effect of
the resistance of the air on falling bodies. In the fortress
of Konigstein in Saxony water is raised from a great
depth below the surface of the ground. For the amuse
ment of visitors a man draws up a bucket of water, and
then pours the water back into the well. The depth is
known to be about 640 feet, so that if there were no resist
ance from the air the sound of the splash should reach
the ear in about 7 seconds ; practically the time is about
15 seconds.
] 45. We have seen in Art. 43, that, if we neglect the
resistance of the air, a body projected vertically upwards
will take the same time in its descent as in its ascent, and
will reach the ground with a velocity numerically the
same as that at starting. These results will not hold when
we take into account the resistance of the air; the time
of ascent is then less than the time of descent, and the
velocity on reaching the ground is less than the velocity at
starting. The demonstration of these results will furnish
a valuable exercise, and we will therefore give it.
The velocity on reaching the ground must be less than
that at starting for two reasons : In the first place, in
consequence of the resistance of the air, the body will not
rise to so great a height as if there were no resistance ;
and therefore it falls down through a space less than that
in which gravity, if unopposed, would generate a velocity
equal to that in starting : and so if there were no resistance
during the motion downwards the velocity on reaching the
ground would be less than at starting. In the next place,
while the body falls the resistance of the air opposes tho
action of gravity, and thus the velocity generated while the
LA WS OF MOTION. 305
body falls through any small space is less than that which
would have been produced by the action of gravity alone,
while the body falls through the same space : see equation
(3) of Art. 40. Thus for both reasons the velocity on
reaching the ground is less than the velocity at starting.
Again, in the same way it follows that the velocity at
any point in the descent is less than it was at the same
point in the ascent : and thus each indefinitely small part
of the straight line described is moved over in less time
in the ascent than in the descent ; and therefore the
whole time of ascent is less than the whole time of
descent.
EXAMPLES. XII.
1. If the velocities and directions of motion of two
bodies moving in the same plane be known, find the direc
tion and the magnitude of the velocity of one body relative
to the other.
2. If a be the distance at a given instant between two
bodies which are moving uniformly, V their relative velocity
and u, v the resolved parts of V in, and at right angles
to, the direction of a respectively, shew that the distance
of the bodies when they are nearest to each other is y ,
and find the time of arriving at this nearest distance.
3. Two bodies move with constant accelerations/and/"'
in given straight lines ; they start with velocities u and u' :
find the relative velocity at the end of the time t estimated
along the straight line which makes angles a and a' with the
directions of motion.
4. A ball is thrown up vertically with a velocity u and
meets with a uniform resistance equal to half the force of
gravity both in the ascent and descent : if it reach the
ground again with the velocity v, shew that u=v/j3.
5. Two straight lines of railway cross each other at a
given angle, and a train on each of them is approaching the
junction, each with a given velocity : find geometrically or
otherwise the velocity with which the trains are approaching
each other.
T. MB. 20
306 MOTION DOWN A SMOOTH CURVE.
XIII. Motion down a Smooth Curve.
146. We shall now proceed to consider cases of motion
in which the force acting is not constant in magnitude and
direction ; in the present Chapter we shall suppose a body
to be acted on by two forces, namely, gravity and the
resistance of a smooth fixed curve. The student may
imagine a fine tube in the form of a curve, and a body in
the shape of an indefinitely small sphere moving in the tube.
We shall not attempt to determine the motion completely,
for the mathematical difficulties would be too great for the
student at present, but we shall demonstrate some import
ant results.
147. When a body descends down a smooth curve in a
vertical plane the velocity acquired at any point is the same
as if the body had fallen freely down the same vertical
height.
We shall consider the motion down a curve as the
limiting case of the motion down an indefinitely great
number of successive inclined planes.
Let AB, BC, CD,... represent sue A
cessive inclined planes. Let h^ be the
vertical height of A above B, h 2
the vertical height of B above C,
h z the vertical height of C above Z>,
and so on.
Suppose a body to slide down this
series of planes. Let v v v 2 , ^...de
note the velocities at B, C, D,... respectively. Then if
no velocity were destroyed in passing from plane to plane
we should have the following equations by Arts. 28 and 40,
and so on.
Hence, by addition, supposing there are n planes,
where h denotes the whole vertical height.
MOTION DOWN A SMOOTH CURVE. 307
We must now consider what velocity is lost in passing
from plane to plane. We assume that the body is in
elastic.
Let us suppose the angle between any plane and the
next plane produced to be the same ; denote it by a.
Resolve the velocity at B along BC and at right angles
to BC, the former component is ^cosa, and the latter is
v l sin a : the former will be the velocity at the beginning
of the motion down BC, for the latter is destroyed by the
plane BC. See Art. 117. Hence instead of v<?=vf + 2yh^
we have v^v^ cos 2 a + 2^A 2 .
Similarly we obtain v=v cos 2 a + 2^ 3 , and so on.
Hence, by addition,
i a + V + v s 2 + + v = ( v i + v z + v * + .+^ 2 i)cos 2 a + 2#A;
therefore vJ^Sgh  2, where 2 stands for
sin 2 a (v? + v 2 2 + v 3 2 + . . . + v\.^.
We shall now shew that 2 vanishes when the number of
planes is made indefinitely great.
It is obvious that 2 is less than (n  1) v 2 n ^ sin 2 a. Let
/3 be the angle between the first plane and the last plane
produced; then p = (nl)a. Hence 2 is less than
v 2 ,_i sin 2 a, that is less than sin a. ^_? tf Now we
a a
know from Trigonometry that f is less than unity, so
a
that 2 is less than /3 sin a tf n \> Hence by making a small
enough we can make 2 less than any assigned quantity.
Thus ultimately 2=0.
Hence y w 2 =2^A, which was to be shewn.
148. In a similar way we may obtain the following
result : if a body start with the velocity u and move in
contact with a smooth curve in a vertical plane the velocity
when the body has risen through the vertical height h is
V(u 2 2gh).
202
308 MOTION DOWN A SMOOTH CURVE.
149. The demonstration in Art. 147 is that which is
usually given in elementary works ; when the student has
sufficient mathematical knowledge to read more elaborate
treatises on Dynamics, he will find that the result can be
obtained in a more satisfactory manner, and without assum
ing that the body is inelastic.
150. Let one end of a fine string be fastened to a fixed
point ; and let a heavy particle be attached to the other
end. In the position of equilibrium the string will be ver
tical. Let the particle be displaced from this position, the
string being kept stretched, and then allowed to move. The
particle wiU oscillate backwards and forwards describing
an arc of a circle ; the arc continually diminishes owing
to the resistance of the air, until the particle comes to rest.
The system is called a simple pendulum.
Now it is a matter of great interest to determine the
time in which the particle describes an arc, or rather the
time in which it would describe an arc neglecting the re
sistance of the air. This we shall consider in the next
Article. The investigation is somewhat complex; but it
deserves attention because the student will find hereafter
when he has the Differential and Integral Calculus at his
command, that although some of the steps may be abbre
viated, yet the process cannot be essentially improved.
We assume as obvious that the motion is exactly the
same whether the particle is compelled to describe an arc
of a circle by means of a string or a fine straight wire in
the manner just explained, or whether the particle moves
in a fine tube in the manner of Art. 147.
151. To find the time of descent of a particle moving in
Q, circle under the action of gravity.
Let APE be an arc of a circle of which C is the cen
tre, and B vertically under (7; let a particle start from A
and move along the curve to B : required the time of
the motion.
Let P be any point in the arc; let the radius CPr;
let the angle BCA = ^ and the angle BCP=6. Let TT de
note as usual the ratio of the circumference of a circle to its
diameter.
MOTION DOWN A SMOOTH CURVE. 309
When the particle is at P the square of its velocity, by
Art. 147,
=2^ (r cos 6  r cos a)=tyr ( sin 2 1  sin 2  J .
a 1
^/?=2rsin, PJ3
Now
Assume sin=sincos0, so that cos$ denotes the ratio
of BP to BA, and thus as the particle moves from A to
B the angle <f> changes from to ^ .
Describe a quarter of a circle o^6 with radius r ; let c
be the centre, and let the angle acp=(f). Then we have in
fact to find the time in which the point p describes the arc
ctb. We must first determine the velocity of p.
Suppose P to move to a new position P 7 , such that the
angle BCP = & ; and let p move to a corresponding new
position jo', such that the angle acp'=<fi. Then the velo
city of P is to the velocity of p as the chord PP is to the
chord pp' when these chords are indefinitely small. But
we have
. 6 . a
sin =sin o cos
mm
B &
therefore
. & . a
sm ;r=sin  cos
a
sin   sin = sin  (cos  cos <'),
*u i 0ff
that is, sin cos
. a .
sm sin
.
sin
310 NOTION DOWN A SMOOTH CURVE.
BID. 9 COS
therefore
. 6& . a .
sin  sin  sin
n
4 22
. >
. , 2r sm y n
>T chord p 2
Now
+ ff
cos 
. 6ff Bff . . a .
2 sin 3 cos  2 sin  sm y . y cos
44 224
Hence when the chords are indefinitely small this ratio
6
COS 2
becomes .
2 sin  sin <f>
The velocity of P
therefore the velocity of p
2 sin  sin
If we assume a to be very small, sin 2  is extremely small,
MOTION DOWN A SMOOTH CURVE. 311
and thus the velocity of p is very nearly */gr. Henco
/r
the required time is very nearly  , that is,  / .
\'gr * V 9
152. The particle will take the same time in rising from
the lowest point of the arc to a height equal to that from
which it descended : see Art. 148. Thus the whole time of
moving from the extreme position on one side of the vertical
to the extreme position on the other side is IT I  : this
extent of motion is called an oscillation.
If we wish to find the length of a simple pendulum
which will oscillate once in a second we put n /=!;
v^ y
thus r=^>. Thus taking <7=32 feet, the length of the
32
seconds' pendulum is about /ee; this will be found
^o 14)
to be about 39 inches. The British standard of length
is connected with the length of the seconds' pendulum
by an Act of Parliament, which defines the inch to be such
that the length of a simple pendulum which oscillates in a
second in the latitude of London shall be 391393 inches.
153. It will be seen that the investigation in Art. 151
is exact up to the point at which we find that the velocity
of p is *Jgr * /(l  sin 2 1 cos 2 J , and then we take an
approximate value of this expression instead of the exact
value. It is not difficult to make a closer approximation,
assuming still that a is small.
Suppose n a large number, and let n/3= 5 so that /3 is
SB
a very small angle. Let 0=7/2/3, and assume that while the
angle acp changes from mp to (m + l)fi we may consider
the velocity of p to be always *fgr I ( 1  sin 2  cos 2 ^ J .
312 MOTION DOWN A SMOOTH CURVE.
Then the time of describing this portion of the quadrant
__Pvf / ! , s in 2 1 cos 2 m/3 J nearly, by the Binomial Theorem.
Then we have to find the sum of the values of this ex
pression for all values of m from to n 1 inclusive. Thus
the time required is ^4 ( n/3 + ^ S sin 2 1 j , where S stands
for 1 + cos 2 /3 + cos 2 2/3 + ...... + cos 2 (nl)/3.
But since sin m/3 = cos f ^  wi/3 J = cos (n  m) /3, we have
=l + sm 2 (rcl)/3 + sm 2 (n2)/3 + ...... + sin 2 /3,
and also
...... + cos 2 /3 + l;
thus by addition 2= n + I. Also np =  .
Therefore the required time
n + I .
Let n increase indefinitely : then we obtain finally for
the required time
1 .
154. Thus in Art. 151 we have found an approximate
value of the time of motion ; and in Art. 153 a still closer
approximation : the smaller the value of a is the less will
be the error in taking these approximations for the exact
time. By the aid of the higher parts of mathematics we
can find an expression, in the form of a series, which will
determine the time, as nearly as we please, whatever be
the value of a.
EXAMPLES. XIII. 313
EXAMPLES. XIII.
1. A particle slides down an arc of a circle to the
lowest point : find the velocity at the lowest point if the
angle described round the centre is 60.
2. If the length of the seconds' pendulum be 39'1393
inches find the value of g to three places of decimals.
3. A pendulum which oscillates in a second at one
place is carried to another place where it makes 120 more
oscillations in a day : compare the force of gravity at the
latter place with that at the former.
4. Suppose that I is the length of the seconds' pendu
lum, and that the lengths of two other pendulums are
I c and I + c respectively, where c is very small : shew that
the sum of the number of oscillations of these two pendu
/ 3c 2 \
lums in a day is very nearly 2 x 24 x 60 x 60 ( 1 + ^ J .
5. A pendulum is found to make p oscillations at one
place in the same time as it makes q oscillations at another.
Shew that if a string hanging vertically can just support
n cubic inches of a given substance at the former place it
will just support ^~ cubic inches at the latter place.
6. A seconds' pendulum hangs against the smooth face
of an inclined wall and swings in its plane : find the time
of a small oscillation.
7. A seconds' pendulum is carried to the top of a
mountain ra miles high : assuming that the force of gra
vity varies inversely as the square of the distance from
the centre of the earth, find the time of a small oscillation.
8. Shew that the length of a pendulum which will
make a small oscillation in one second at the top of a moun
tain m miles high is (TT^T ~) ^ where I is the length
of the seconds' pendulum at the surface of the earth.
314 UNIFORM MOTION IN A CIRCLE.
XIV. Uniform motion in a Circle.
155. If the direction of a force always passes through
a fixed point the force is called a central force; and the
fixed point is called the centre of force.
In the present Chapter and the next two Chapters we
shall be occupied with cases of central forces : we begin
with some propositions due to Newton which are contained
in the next five Articles.
156. When a body moves under tJie action of a central
force the areas described by the radius drawn to the centre of
force are in one plane and are proportional to the times of
describing them.
Let S be the centre
of force ; and suppose
a body acted on by no
force to describe the
straight line AB, with
uniform velocity, in a
given interval of time.
In another equal inter
val, if no force acted,
the body would de
scribe Be equal to AB,
in AB produced, so
that the equal areas
ASB and BSc would
be described by the radius drawn to S in equal intervals.
But when the body arrives at B let a force tending to
S act on it by an impulse, and cause it to proceed in the
direction BC instead of Be ; then if C be the position of
the body at the end of the second interval Cc is parallel to
BS\ see Art. 47. Join SC \ then the triangle BSC is
equal to the triangle BSc, by Euclid, I. 37 ; therefore the
triangle BSC is equal to the triangle ASB, and the two
triangles are in the same plane.
UNIFORM MOTION IN A CIRCLE. 315
In like manner if impulses tending to S act on the body
at C, D, J?,... causing the body to describe in successive
equal intervals the straight lines CD, DE,..., the triangles
CSD, DSE,... are all equal to the triangle ASB, and
are in the same plane with it.
Thus equal areas are described in equal intervals, and
the sum of any number of areas is proportional to the time
of description.
Now let the number of triangles be indefinitely in
creased, and the base of each indefinitely diminished ; then
the boundary ABCDE. . . will ultimately become a curve,
and the series of impulses will become a continuous central
force by the action of which the body is made to describe
the curve. And the areas described being always propor
tional to the times will be so also in this case.
157. The proposition of the preceding Article is true
also if S be a point which instead of being fixed moves
uniformly in a straight line. For by the fifth Corollary
in Art. 142 the relative motion is the same whether the
plane in which the curve is described be at rest or be
moving with the body and the curve and the point S uni
formly in a straight h'ne.
158. If v be the velocity of the body at any point A,
and p the perpendicular from S on the tangent at that
point, the area described in the time t=^ P^ y>
Draw SY perpendicular to AE. Let t be divided
into n equal intervals, and let AB be the space described
in the first interval, the force at S being supposed to act
by impulses at the end of each interval.
Then the polygonal area which is described in the time t
=n times the triangle SAB=nSY..v=^ST.t.v.
In the limit the straight line AB, which is the direction
of the velocity at A, becomes the tangent to the curve
at A ; and the curvilinear area described in the time
t = \ptv.
316 UNIFORM MOTION IN A CIRCLE.
Thus the area described in a unit of time is pv, it is
usual to denote twice the area described in a unit of
time by h : therefore h=pv, and v= .
159. If a body move in one plane so that the areas
described by the radius drawn to a fixed point are pro
portional to the times of describing them the body is acted
on by a force tending to that point.
Let S be the fixed point about which areas proportional
to the times are described, and suppose a body acted on by
no force to describe the straight line AB with uniform
velocity in a given interval of time. In another equal in
terval if no force acted the body would describe Be equal
to AB, in AB produced : so that the triangles ASB and
BSc would be equal. But when the body arrives at B let
a force act on it by an impulse which causes it to describe
BC in the second interval, such that the triangle SBC is
equal to the triangle ASB, and in the same plane.
Then the triangle BSC is equal to the triangle BSc,
and therefore Cc is parallel to SB, by Euclid, i. 39 : there
fore the impulse at B is in the direction BS : see Art. 47.
In like manner if impulses act on the body at C, D, E,...
causing the body to describe in successive equal intervals
the straight lines CD, DE,...so that the triangles CSD,
DSE,...3XQ all equal to the triangle ASB, and are in the
same plane with it, then all the impulses tend to S.
Hence if any polygonal areas be described proportional
to the times of describing them, the impulses at the an
gular points all tend to S.
Now let the number of triangles be indefinitely in
creased, and the base of each indefinitely diminished ;
then the boundary ABCDE ... will ultimately become a
curve, and the series of impulses will become a continuous
force by the action of which the body is made to describe
the curve : and the force always tends to S.
UNIFORM MOTION IN A CIRCLE. 317
160. The proposition of the preceding Article is true
also if S be a point which instead of being fixed moves
uniformly in a straight line ; see Art. 157.
161. We have already observed in Art. 49 that the
principle called the Parallelogram of Velocities gives rise
to applications similar to those deduced from the Paral
lelogram of Forces in Statics ; some illustrations of this
remark will occur as we proceed, one of great interest
being given in the next Article.
162. The direction of the resultant action of a central
force on a body while it describes an arc of a curve is
the straight line which joins the intersection of the tangents
at the extremities of the arc with the centre of force.
Let PQ be an arc of a curve
described by a body under the
action of a centre of force at S.
Let PT, QT be the tangents
at P and Q respectively. Sup
pose the body to move from P
to Q. Produce PT to any
point p.
The resultant action of tho
central force during the motion
changes the direction of the
velocity from Tp to TQ, and thus the direction of the
resultant action must pass through T. But the direction
of the action of the central force passes through S at
every instant, and therefore the direction of the resultant
action must pass through S. Thus TS must be the direc
tion of the resultant action.
This proposition has been given on account of its
simplicity and interest ; but it is not absolutely necessary
for the purposes of the present work, for it will be found
that so much of the result as we may hereafter require will
present itself naturally in the course of our investigations.
See Arts. 163 and 175.
318 UNIFORM MOTION IN A CIRCLE.
163. If a body describes a circle of radius r, with
uniform velocity v, the body is acted on by a force tending
v 2
to the centre of the circle, the acceleration of which is .
Since the body moves with uniform velocity the arc
described in any time is proportional to the time. Hence,
by Euclid, vi. 33, the area described in any time by the
radius drawn to the centre is proportional to the time.
Therefore, by Art. 159, the body is acted on by a force
always tending to the centre of the circle.
Let PQ be an arc of the circle,
S the centre, P^and QT the tan
gents at P and Q respectively.
Let the angle PSQ be ex
pressed in Circular Measure and
denoted by 2<, and let u denote
the velocity communicated to the
body by the action of the central
force while the body moves from
P to Q. Then the velocity v
along TQ is the resultant of v along PT and of u com
municated by the central force. Hence as in Art. 33 of
the Statics the direction of u makes eaual angles with
TP and TQ ; and must in fact coincide with TS. Hence,
as in Art. 38 of the Statics,
u sin PTQ _ sin PSQ _ sin 2ft
v sin PTS~~ cos PST~ cos
Let t denote the time in which the body moves from P to $,
and let / denote the accelerating effect of the central force.
Then if we suppose Q very near to P we have u=ft, be
cause during a very small change of position of the body
the force may be considered as constant in magnitude and
in direction. Hence ft=2vsm<j>. But since the velocity
is uniform we have 2r<=ztf, as in Art. 3 ; for 2r< is tho
length of the arc PQ. From these two equations we obtain
_
2;0
UNIFORM MOTION IN A CIRCLE. 319
But when is indefinitely small we have by Trigonometry,
.
r
164. The preceding investigation of the value of the
central force should be carefully studied.
In the first part of the investigation we obtain the
exact direction and amount of the velocity u communicated
by the central force while the body describes a given arc.
In the second part we have to use the method of limits,
that is, we write down equations which are true in the
limit, namely when the arc described is supposed indefi
nitely small.
The reasoning in the second part of the investigation
might.be given more fully in the following manner. Let/j
be the greatest value of the acceleration, and / 2 the least,
while the body describes the arc PQ. Then u cannot be
so large as frf, and cannot be so small as fy cos 20 : for f^t
would be the velocity generated if the force always acted
in the same direction, and had its greatest possible value ;
and f z t would be the velocity generated if the force always
acted in the same direction and had its least possible value ;
and as 20 is the angle between the extreme directions of
the force, if the force always had its least value the velocity
generated would be greater than f 2 t cos 20. Hence u lies
between fj and f 2 t cos 20 ; that is 2v sin lies between
fit and / 2 2 cos 20; therefore  7^ lies between / x and
/ 2 cos 20. Since this is always true it is true at the limit.
" sin <f> . v 2
Now the limit of is ; and the limit of / t and of
/ 2 cos 20 is/. Thus  =/.
165. It will be seen that we demonstrate that the ac
celeration has the same value at every point of the circle :
this might have been anticipated but we did not assume it.
In the manner thus exemplified we may in similar
cases develop the reasoning, so as to render it more rigor
ous hi form : the student will have no difficulty in supply
ing such a development for himself on other occasions if
required.
320 UNIFORM MOTION IN A CIRCLE.
166. We have thus demonstrated the following result :
if a body of mass m describes a circle of radius r, with
uniform velocity v, then whatever be the forces acting on
the body their resultant tends to the centre of the circle,
mv 2
and is equal to . No single fact in the whole range
of Dynamics is of greater importance than this, and the
student should regard it with earnest attention.
167. For example, suppose a body of mass m fastened
to one end of a string, and the other end of the string fast
ened to a fixed point in a smooth horizontal table. Let
the body be started in such a manner as to describe a
circle with uniform velocity, v, on the table round the fixed
point, the string forming the radius, r, of the circle. The
forces acting on the body are its weight, the resistance of
the table, and the tension of the string. The weight and
the resistance act vertically and balance each other. The
tension of the string acts horizontally, and its value must
," , . mv*
be equal to .
This may be verified experimentally. Instead of fast
ening the string to a fixed point in the plane, let the
string be prolonged and pass through a hole in the position
of the fixed point, and have a body fastened to its end.
Let m'g denote the weight of this body ; then if it remains
at rest we shall find that m'g = .
168. For another example we will take the conical
pendulum. One end of a fine string is fixed ; to the other
end a particle is fastened. The particle is set in motion in
such a manner as to describe a horizontal circle with
uniform velocity : thus the string traces out the surface
of a right cone, from which the name conical pendulum
is derived.
Let mg be the weight of the particle, v its velocity, T
the tension of the string, I its length, a the inclination of
the string to the vertical.
The vertical forces acting on the particle are its weight
and the resolved part of the tension ; these must be in
UNIFORM NOTION IN A CIRCLE. 321
equilibrium, so that
The only horizontal force is the resolved part of the ten
sion; therefore by Art. 166
r
also r=sino.'
v z v 2 sin 2 a r 2
Hence tan a= = , : , or ==.
rg Igsma' cos a Ig
This relation then must hold between v, , and a, in order
that the supposed motion may take place.
169. For another example, we will take the case of the
moon moving round the earth ; this example is of special
interest as being that by which Newton teste*d his law of
gravitation.
It appears from observation that the moon moves
nearly in a circle, with uniform velocity, round the earth as
centre. Let v denote the moon's velocity and r the dis
tanee of the moon from the earth's centre. Hence the
acceleration on the moon is .
r
Now Newton conjectured that this acceleration was
owing to the earth's attraction, that the fall of heavy
bodies to the earth's surface was due to the same cause,
and that the force of the earth's attraction varied inversely
as the square of the distance.
Let a denote the earth's radius ; then, since g denotes
the acceleration produced by the earth's attraction at the
surface of the earth, the acceleration produced at the
distance of the moon will be ^ . Hence, if Newton's
conjecture be true, we must have
2 ,,2
r
It is said that when Newton first tried the calculation the
result was not satisfactory ; the value of a not being known
at that time with sufficient accuracy; but at a subsequent
T. ME. 21
322 UNIFORM MOTION IN A CIRCLE.
period, having obtained a more accurate value of a, he re
turned to the calculation and obtained the desired agreement.
But there does not appear to be decisive authority for
the statement : see Rigaud's Historical Essay on the first
publication of Sir Isaac Newtoris Principia, page 6.
The student can easily verify the result approximately,
taking the following facts as given by observation :
a =4000 miles =4000 x 5280 feet,
_
~~ Time of moon's revolution ~~ 27 j x 24 x 60 x 60 *
The preceding investigation is sufficient to give a
general idea of the circumstances of the motion of the
moon; but many additional considerations enter into an
exact discussion of the subject. The moon does not move
accurately in a circle round the earth, nor with quite uni
form velocity: and the sun exerts an influence on the
motion : but the investigation of these points is altogether
beyond the student at present.
170. Suppose a body to describe a circle of radius r
with uniform velocity v. Then as the circumference is
Snr, the body moves once round in the time . This is
called the periodic time; and generally when a body
describes any closed curve the time of going once round is
called the periodic time.
171. "When a body is describing a circle with uniform
velocity the straight line drawn from the body to the
centre describes equal angles in equal times. The rate at
which angles are described is called the angular velocity
of the radius. Thus with the notation of the preceding
Article the periodic time is ; and in this time the angle
2?r is traced out : thus the angular velocity is 2/r r '
that is .
r
UNIFORM MOTION IN A CIRCLE. 323
7\
The velocity v is called the linear velocity when it is
necessary to distinguish it from the angular velocity,
which is equal to the linear velocity divided by the radius.
172. To find the pressure * R
which a body on the surface of
the earth at any point exerts
on the earth, supposing the earth
to be a sphere of uniform den
sity.
Let NCS represent the axis
on which the earth turns ; sup
pose a body at P resting on the
earth and turning round with it.
Let C be the centre of the earth,
and suppose PC to make an angle Q with the plane of the
equator, so that 6 is the latitude of the body, and PCN is
the complement of 0.
Let m be the mass of the body. The body is acted on
by the following forces : the attraction of the earth towards
its centre, which we will denote by mf, the resistance of
the earth along the radius CP, which we will denote by R,
the friction in the plane NCP along the tangent at P,
which we will denote by F. Draw PM perpendicular to
CN\ then as P revolves with the earth it describes a circle
of radius PM with uniform velocity. Let PM=r, and
let v denote the velocity of P, and a the earth's radius.
Then the forces which act on the body at P must have their
resultant along PM, and equal to ^ . Therefore
(mf JR) .cos 6 + Fsm 6 = , (mf R) sin 6  Fcos 6 = 0.
Multiply the first of these equations by cos, and the
second by sin 6, and add ; thus
f j? mvZ
~ r
that is
n f >V
li=mf cos 0.
212
324 UNIFORM MOTION IN A CIRCLE.
Let co denote the angular velocity of the earth, and there
fore of P; then v=ra> : also r=a cos 6.
Thus R = mf ma&> 2 cos 2 0.
Similarly we find F=ma<o' 2 sin 6 cos 6.
The resultant pressure on the earth is equal and oppo
site to the resultant of R and F. Denote it by O. Then
the direction of O makes with the raolius CP an angle
, , , . , . F ., , . mac* 2 sin 6 cos 6
the tangent of which is 55 , that is ?  s   .
R ' mf maw 2 cos 2 6
And C7 2 = m 2 {(/  aa> 2 cos 2 0) 2 + aV sin 2 cos 2 6}
= m?{f 2  2a/a> 2 cos 2 6 + a V cos 2 6}
 a 2 a> 4 ) cos 2 ^}.
The quantity O is what we have hitherto denoted by mg ;
we see that it is not the same for all places on the earth's
surface. We shall proceed to some numerical estimate.
The earth revolves once in twentyfour hours; thus
<0= 24x 60x60 : als a=4000x5280  Here we t^ 6 ^
usual a second as the unit of time, and a foot as the unit
of length.
2
We shall find that ao> 2 = nearly; the square of this,
that is a 2 4 , we shall neglect in comparison with 2a/o> 2 .
Thus approximately # 2 =/ 2 U ~^~ cos 2 A and therefore
approximately g=f(^^j cos 2 J . We know that /= 32
nearly, and thus y 2 =^ nearly; so that <7=/(l  C 
nearly.
EXAMPLES. XIV. 325
"We have assumed that the earth is a sphere, and that
the attraction which it exerts on a body placed at any
point on the surface is directed towards the centre; but
these assumptions are not strictly accurate, so that the
result must not be considered absolutely true.
EXAMPLES. XIV.
1. Find the force towards the centre required to make
a body move uniformly in a circle whose radius is 5 feet,
with such a velocity as to complete a revolution in 5
seconds.
2. A stone of one Ib. weight is whirled round horizon
tally by a string two yards long having one end fixed : find
the time of revolution when the tension of the string
is 3 Ibs.
3. A body weighing P Ibs. is at one end of a string,
and a body weighing Q Ibs. at the other end ; the system is
in motion on a smooth horizontal table, P and Q describing
circles with uniform velocities : determine the position of
the point in the string which does not move.
4. A string I feet long can just support a weight of
P Ibs. without breaking ; one end of the string is fixed to
a point on a smooth horizontal table ; a weight of Q Ibs. is
fastened to the other end and describes a circle with uni
form velocity round the fixed point as centre : determine
the greatest velocity which can be given to the weight of
Q Ibs. so as not to break the string.
5. One end of a string is fixed; to the other end a
particle is attached which describes a horizontal circle
with uniform velocity so that the string is always inclined
at an angle of 60 to the vertical : shew that the velocity of
the particle is that which would be acquired in falling freely
from rest through a space equal to threefourths of the
length of the string.
326 MOTION IN A CONIC SECTION
XV. Motion in a Conic Section round a focus.
173. The cases of motion which we shall discuss in the
present Chapter are of great interest on account of the
application of them to the earth and the planets which de
scribe ellipses round the sun in a focus.
In the present Chapter and the next Chapter we shall
consider the action of a force on a given body, so that
we shall be occupied only with the influence of the force
on the velocity of the body : see Arts. 14, 45.
174. If a body describes an ellipse under the action
of a force in a focus, the velocity at any point can be
resolved into two components, both constant in magnitude,
one perpendicular to the major axis of the ellipse, and the
other at right angles to the radius drawn from the body to
the focus.
Let S be the focus which
is the centre of force, H the
other focus, P any point on the
ellipse, SY and HZ perpen
diculars from S and H on the
tangent at P. Let C be the
centre of the ellipse, A one end
of the major axis.
By Art. 158 the velocity at P varies inversely as SY,
and therefore directly as HZ; for SYx HZ is constant, by
a property of the ellipse, being equal to the square on half
the minor axis of the ellipse. Thus HZ may be taken to
represent the velocity in magnitude, and it is at right
angles to the velocity in direction. Now a velocity repre
sented by HZ may be resolved into two represented by
HC and CZ. And by the nature of the ellipse CZ is
parallel to JSP and equal to CA.
Hence a velocity represented by HZ in magnitude,
and at right angles to HZ in direction, may be resolved
into two, one represented by CA in magnitude and at right
ROUND A FOCUS. 327
angles to SP in direction, and the other represented by
HC in magnitude and perpendicular to IIS in direction.
It is convenient to have expressions for the magnitudes
of these component velocities. Let CA=a, let b denote
half the minor axis, and let e be the excentricity of the
ellipse. Let h represent twice the area described by
the radius SP in a unit of time; then the velocity at
P= p! Ji7 = ~~7z Therefore the component at right
angles to SP is ',. ,"" , that is r^ ; and the component
TT ~ . ll . CII ., , . JlO.e
perpendicular to HS is r a > that is p .
175. A body describes an ellipse under the action of a
force in a focus: find the law of force.
Let 8 be the focus which is
the centre of force ; let P and Q
be any two points on the ellipse ;
and suppose the body to move
from P to Q.
Resolve the velocity at P into
two components one at right
angles to &P, and the other perpendicular to IIS; denote
these by t', and v% respectively. When the body arrives at
Q its velocity is composed of v l and v 2 parallel to their
directions at P, and the velocity generated by the action
of the central force during the motion, which we will
denote by u. But by Art. 174 the velocity at Q can be
resolved into v at right angles to SQ } and v 2 perpendicular
to SH.
Hence it follows that v l at right angles to SP together
with u in its own direction have for their resultant v, at
right angles to SQ. Hence, as in Art. 33 of the Statics,
the direction of u makes equal angles with the straight
lines at right angles to SP and SQ, and therefore with
SP and JSQ. And, by Art. 38 of the Statics,
328 MOTION IN A CONIC SECTION
Let SP=r, and PSQ=2cj). Let t denote the time in
which the body moves from P to Q ; and let / denote the
accelerating effect of the force. Then if we suppose Q
very near to P so that t is very small, we have u=ft ; hence
/if=2z; 1 sin$. The area described in passing from P to
Q=^ht by Art. 158; and this area may be taken to be
= g r 2 sin 2$, for it may be considered ultimately as a tri
anle.
Therefore fr 2 sin 20 = 2/w 1 sin ;
i ^ 2/iv* Avi ,.. . ,
therefore /""si L  r= r ultimately.
2r* cos i^
when is made indefinitely small.
This shews that the force varies inversely as the square
of the distance.
It is usual to denote the constant hv L by /*; thus
^i=Axrf = y^, The quantity /z is called the absolute
force.
176. In the preceding investigation it was shewn that
the direction of the velocity u communicated by the cen
tral force while the body moves from P to Q bisects the
angle PSQ. But we know by Art. 162 that this direction
is that of the straight line which joins S with the intersec
tion of the tangents at P and Q. Thus our dynamical in
vestigation suggests that in an ellipse the two tangents
from an external point subtend equal angles at a focus;
and this is a known property of the ellipse.
177. A body describes an ellipse under the action of
a force in a focus: required to determine the periodic
time.
ROUND A FOCUS. 329
Let a and b denote the semiaxes of the ellipse ; and h
twice the area described by the radius in the unit of time.
By Art. 156 the periodic time
_ twice the area of the ellipse
~~~
Now it is known that the area of the ellipse is nab, and
by Art. 175 we have h = *. Hence the periodic time
178. We can now apply the results obtained to the
motions of the earth and the planets round the sun..
There are certain facts connected with these motions
which were discovered in the seventeenth century by the
diligence of Kepler, a famous German astronomer, and
which are justly called Kepler's Laws. These laws are the
following :
(1) The planets describe ellipses round the sun in a
focus.
(2) The radius drawn from a planet to the sun de
scribes in any time an area proportional to the time.
(3) The squares of the periodic times are proportional
to the cubes of the major axes of the orbits.
From the second law it follows, by Art. 159, that each
planet is acted on by a force tending to the sun.
From the first law it follows, by Art. 175, that the force
on each planet varies inversely as the square of the dis
tance.
From the third law an important inference can be
drawn, as we will now shew. Let a be the semiaxis major
of the ellipse described by one planet, /u the absolute force,
T the periodic time ; let a', //, T' denote similar quantities
for another planet : then, by Art. 177,
m ,
T= j , T = rr ; therefore ^, = ~
Jn x 1*
330 MOTION IN A CONIC SECTION
T 2 a 3 u!
But by Kepler's third law ^ = 73; therefore = 1,
so that //=/*. This shews that the constant which denotes
the absolute force is the same for all the planets; so that
the acceleration produced by the sun depends solely on
the distance from the sun, and not on the nature of any
particular planet.
1V9. By investigations similar to those in Arts. 174
and 175 it may be shewn that if a body describes an
hyperbola under the action of a force in a focus, the force
varies inversely as the square of the distance. If the body
describes the branch which is the nearer to the focus, the
force is attractive as in the case of the ellipse. But if the
body describes the branch which is the more remote from
the focus the force is repulsive; the body at any instant
instead of moving along the tangent as it would if there
were no central force, or deviating from the tangent on the
side towards the centre of force as it would do if the force
were attractive, deviates from the tangent on the side
remote from the centre of force.
"We proceed to consider the case of motion in a para
bola round a force in the focus.
180. If a body describes a parabola under the action
of a force in the focus, the velocity at any point can be
resolved into two equal constant velocities, one perpendicular
to the axis of the parabola, and the other at right angles
to the radius drawn from the body to the focus.
Let S be the focus, P any point on
the parabola, A the vertex, ST the per
pendicular from S on the tangent at P.
Let AS=a\ and let h denote twice
the area described by the radius SP in a
unit of time.
ST bisects the angle ASP', therefore
the resultant of two velocities, each equal
to , one along 8A, and the other along
ROUND A FOCUS. 331
2A ST ., . A /ST 2
that
by the nature of the curve, that is, ~ But ~. is the
magnitude of the velocity at P, and its direction is at
right angles to /ST". Hence the velocity at P can be re
solved into two velocities, each equal to , one perpen
dicular to AS, and the other at right angles to SP.
Hence it may be shewn, as in Art. 175, that if a body
describes a parabola, under the action of a force in the
focus, the force vanes inversely as the square of the dis
tance. And if /i denote the absolute force, we have fi= .
181. In the figure of Art. 174 we have the velocity at
P=~y. Now by a property of the ellipse we might ex
press SY in terms of SP and the major axis of the ellipse ;
and thus obtain another formula for the velocity at P.
But instead of appealing to a property of the ellipse we
can arrive at the result by the aid of mechanical principles,
as we will now shew.
Let v denote the whole velocity of the body at P ; and
let v 1 and v 2 have the same meanings as in Art. 175. Let
a denote the angle SPY, and j3 the angle between TZ and
AH produced
Suppose we resolve v t and Vo along the tangent at P,
and at right angles to it ; then the algebraical sum of the
former two components must be v, and the algebraical sum
of the latter two components must be zero : that is,
v l sin a  v 2 sin =v, v x cos a  v 2 cos /3=0.
From the second equation we have cosfl= yiCOSa ; substi
v
tute in the first equation ; thus v= v sin a  J(v  V* cos 2 a)
=v l sin a  V( V 2 2 ~ v \ + v i sin 2 a).
332 EXAMPLES. XV.
SY ,, ,
sin a= ; therefore v f v
Now
Y ,, ,
; theref
A 2
Hence sT
Using the values of ^ and v 2 which were found in
Art. 174 we obtain
_ 2H W (1  e 2 ) _ 2^ A 2
"/^ " 6 4 ~ &P 6 2
je, by Art. 175,
^
In the same way we shall find that when a body moves
in an hyperbola the square of the velocity => +
EXAMPLES. XV.
1. If a planet revolved round the sun in an orbit with
a major axis four times that of the earth's orbit, deter
mine the periodic time of the planet.
2. If a satellite revolved round the earth close to its
surface, determine the periodic time of the satellite.
3. A body describes an ellipse under the action of a
force in a focus : compare the velocity when it is nearest the
focus with its velocity when it is furthest from the focus.
4. A body describes an ellipse under the action of a
force to the focus S ; if II be the other focus shew that the
velocity at any point P may be resolved into two veloci
ties, respectively at right angles to SP and HP, and each
varying as HP.
5. In the diagram of Art. 174 if ASP =6 shew that
the velocity of the revolving body at P may be resolved
into T (e + cos 6) perpendicular to AC and ^nr
parallel to AC.
MOTION IN AN ELLIPSE. 333
XVI. Motion in an ellipse round the centre.
182. "We shall give in this Chapter investigations
respecting the motion of a body in an ellipse round the
centre. The results have not the practical interest which
those in the preceding Chapter derive from their appli
cation to Astronomy ; but the investigations will furnish
valuable illustrations of mechanical principles.
183. It will be necessary to return to a result already
established.
Suppose that a body describes a circle of radius r with
uniform velocity v. We have shewn that the body is acted
on by a force to the centre of the circle of which the
accelerating effect is . Let P be a point on the circum
ference of the circle, C the centre, CA a radius. Draw
PM perpendicular to CA. Then a force of constant mag
nitude, acting along PC, may be represented by PC ; and
so may be resolved into two represented by PM and MO
respectively. Thus we may say that if a body describes a
circle with uniform velocity tha forces acting on it may
be represented by PM and MC respectively.
334
NOTION IN AN ELLIPSE
tllQ
184. A body describes an ellipse round a force
centre: required to find the law of force.
Let AC A' be the major axis of the ellipse, P any point
on the ellipse ; draw PM perpendicular to the major axis.
Produce HP to meet the circle described on AA' as dia
meter at p.
Now by Art. 156 the elliptic area ACP varies as
the time of 'moving from A to P ; and by a property of
the ellipse the circular area ACp bears a constant ratio to
the elliptic area ACP. Hence, as P moves, a body always
occupying the position p would describe the circle uni
formly, and would therefore be acted on by a constant
force along pC ; or by Art. 183 it would be acted on by
forces which we may represent by pM and MO. Now
PM bears to pM a constant ratio, by a property of the
ellipse, so that the velocity of P estimated parallel to
CB always bears a constant ratio to that of p estimated in
the same direction. Therefore the force parallel to CB on P
bears to the force on p in the same direction a constant
ratio, namely that of PM to pM ; so that the force on P
in this direction may be represented by PM.
The force on P parallel to CA is the same as that on
p, and so may be represented by MG.
Hence P is acted on by two forces which may be
denoted by PM and MO respectively ; and the resultant
of these will be a single force which may be denoted in
magnitude and direction by PC.
ROUND THE CENTRE. 335
Therefore the force required varies as the distance.
Since the force varies as the distance CP we may denote
it by pCP, where p. is a constant ; p. is usually called the
absolute force.
185. Let A denote twice the area described by CP in
a unit of time ; and let a and b be the semiaxes of the
ellipse : then will A 2 =/ia 2 6 2 . For the force at B is /i6, and
therefore the force at E is pb x T , that is /*. Now the
velocity of P when at B is the same as that of p when at
E\ denote it by v: then, by Art. 163, ^=H a > so that
c3=pa 2 . But, by Art. 158, v=  = T ; therefore p =/i 2 , so
that k 2 =pa 2 b 2 .
186. J. body describes an ellipse under the action of a
force in the centre: required to determine the periodic time.
Let a and b denote the semiaxes of the ellipse ; and
h twice the area described by the radius in a unit of time.
By Art. 156 the periodic time
twice the area of the ellipse
But h=ab*Jp, by Art. 185 ; therefore the periodic time
187. It will be seen that the result in the preceding
Article is independent of the size of the ellipse ; it will
therefore hold even if we suppose b indefinitely small, that
is it will hold when the body moves in a straight line,
oscillating backwards and forwards through the centre of
force.
188. It is easy to give a geometrical representation of
the velocity of a body moving in an ellipse under the action
of a force in the centre.
336
EXAMPLES. XVI.
Draw Cq at right an
gles to Cp, meeting the
circle at q ; and draw qN
perpendicular to A A', cut
ting the ellipse at Q.
The velocity of p is
constant in magnitude, and
its direction is at right
angles to Cp, so that it may
be represented by Cq. The
velocity represented by Cq
may be resolved into two
represented by CN and
Nq respectively. The velocity of P parallel to CA is
to that of jo, and the velocity of P parallel to CB is to
that of p as 6 is to a: see Art. 184. Hence the velocity of
P parallel to CA may be represented by CN, and that
paraUel to CB by NQ,
~ Nq. Thus the velocity
of P may be resolved into two components, denoted by
CN and NQ respectively; so that the resultant velocity
may be represented by CQ : that is, the resultant velocity
of P is parallel to CQ in direction, and is proportional
to CQ in magnitude.
Since the velocity is proportional to CQ in magnitude
it will be equal to the product of CQ into some constant;
and by Art. 185, we see that this constant is <Jp.
EXAMPLES. XVI.
1. A body describes an ellipse under the action of
n force in the centre : if the greatest velocity is three times
the least find the excentricity of the ellipse.
2. A body describes an ellipse under the action of a
force in the centre : if the major axis is 20 feet and the
greatest velocity 20 feet per second, find the periodic time.
WORK. 337
XVII. Work.
189. In many modem treatises on practical mechanics
the term work is employed in a peculiar sense ; and various
useful facts and rules are conveniently stated by the aid of
the term in this sense. We propose accordingly to give
some explanations and illustrations which will enable the
student to understand and apply such facts and rules.
190. The labour of men and animals, and the power
furnished by nature in wind, water, and steam, are employ
ed in performing operations of various kinds, such as
drawing loads, raising weights, pumping water, sawing
wood, and driving nails. In these and similar operations
wo may perceive one common quality which has been
adopted as characteristic of work, and suggests the follow
ing definition : work is the production of motion against
resistance.
191. This definition will not be fully appreciated at
once ; the beginner may be inclined to think that it will
scarcely include every thing to which the term work is
popularly applied : he will however find as he proceeds
that the definition is wide enough for practical purposes.
According to this definition a man who merely supports
a load without moving it does not work ; for here there is
resistance without motion. Also while a free body moves
uniformly no work is performed ; for here there is motion
without resistance.
192. Work, like every other measurable thing, is mea
sured by a unit of its own kind which we may choose at
pleasure. The unit of work adopted in England is the
work which is sufficient to overcome the resistance of a
force of one pound through the space of one foot : or we
may say practically that the unit of work is the work dona
in raising a pound weight vertically through one foot.
193. The term footpound is used in some books in
stead of unit of work ; so that footpound may be con
sidered as an abbreviation for one pound weight raised
vertically through one foot.
T. ME. 22
333 WOfiJt.
194. The term HorsePower is used in measuring the
performances of steam engines. Boulton and Watt esti
mated that a horse could raise 33000 Ibs. vertically through
one foot in one minute ; this estimate is probably too high,
on the average, but it is still retained : so that a Horse
Power means a power which can perform 33000 units of
work in a minute.
195. The term duty is also used with respect to steam
engines ; it means the quantity of work which is obtained
by burning a given quantity of fuel. In good ordinary
engines the duty of one pound weight of coal varies between
200000 and 500000.
196. Observations have been made of the amount of
work which can be performed by men and by animals
labouring in various ways ; and the results are given in
treatises on practical mechanics. The following table is an
example : the first column states the kind of labour, the
second column the number of hours in a day's labour, the
third column the number of units of work performed in a
minute.
Man raising his own weight on a ladder
Man raising a weight with a cord and a Fully
Man turning a windlass
Man lifting earth with spade to the height
of five feet
6
10
4230
1560
2600
470
197. Many examples can be given which involve no
thing more than the application of the rules of Arithmetic
to the measurement of work. We proceed now to some
new propositions which arise from the combination of the
definition of work with the known principles of mechanics.
198. When weights are raised vertically through various
heights the whole work is the same as that of raising a
weight equal to the sum of the weights vertically from, the
Jirst position of the centre of gravity of the system to the last.
Suppose for example that there are three weights. Let
P, Q, It denote the weights ; p, q, r their respective heights
above a fixed horizontal plane in the first position of the
^YORK. 339
system. Then by Arts. 119 and 146 of the Statics the
distance of the centre of gravity of the system above the
same fixed horizontal plane is
Pp + Qq + Rr
~
Now suppose the weights raised vertically through the
heights a, b, c respectively. Then the distance of the
centre of gravity of the system in the new position above
the same fixed horizontal plane is
Thus the vertical distance between the two positions of
the centre of gravity of the system is
Pa + Qb + Rc
P + Q + R '
Now the work of raising a weight equal to the sum of
the weights vertically through this space is the product of
this space into the sum of the weights, that is into
P + Q + R ; hence this work is equal to Pa + Qb + Re, that
is to the sum of the work of raising P, Q, R vertically
through the heights a, b, c respectively. In the same
manner the proposition may be demonstrated whatever
may be the number of the weights.
199. The work done in raising a heavy body along a
smooth inclined plane is equal to the work done in raising
the same body through the corresponding vertical space.
Let a be the inclination of the plane to the horizon, T7
the weight moved, s the distance along the plane through
which the weight is moved. The weight W may be re
solved into two components, TFcosa at right angles to the
plane and TFsin a down the plane ; the latter is the com
ponent which resists motion along the plane. Hence the
work done = W sin a x s= Wx. s sin a ; and s sin a is the
vertical space corresponding to the space 5 on the plane.
This establishes the proposition.
222
340 WORK.
200. If a body of weight "W be dragged along a rough
horizontal Plane through a space s, and p. be the coefficient
of friction for motion, the work done is /uWs.
For the weight being W the force which resists the
horizontal motion is p. W; and therefore the work done
is p, Ws.
201. If a body of weight W be dragged up a rough
Plane inclined to the horizon at an angle a through a space
s, and p, be the coefficient of friction for motion, the work
done is W (sin a + p. cos a) s.
For the weight W may be resolved into two compo
nents, W cos a at right angles to the Plane and W sin a
along the Plane. The work done consists of two parts,
namely, raising the weight along the Plane, and overcom
ing the resistance along the Plane ; the former part is
W sin a x s, and the latter part is p. W cos a x s. Hence
the whole work is W (sin a + p. cos a) s.
This may also be obtained in another way. By Art. 86
the resolved force down the Plane is TF(sin a + p. cos a), and
therefore by the definition of work W (sin a + p. cos a) s is
the work done in dragging the body up the Plane.
Since s sin a represents the vertical height through which
the weight is raised, and s cos a the horizontal space through
which it is drawn, we may say that the work consists of
two parts, one being that which would be required to raise
the weight through the vertical height passed over, and
the other that which would be required to overcome the
friction supposing the Plane to be horizontal. In most
cases which occur in practice a is so small that cos a may
without any important error be taken as equal to unity,
and the expression for the work becomes Ws sin a + p, Ws.
Similarly if a body be dragged through a space s down
an Inclined Plane which is too rough for the body to slide
down by itself, the work done is W (p. cos a  sin a) *.
202. The preceding Article may be applied to the case
of carriages drawn along a common road or a railroad ; in
this there is indeed a rotatory motion of the wheels which
WORK. 341
is not contemplated in the preceding Article ; but the
weight of the wheels will in general be small compared
with the weight of the whole mass which is moved, and we
will assume that no important error will arise from neg
lecting the rotatory motion.
The numerical value of p will depend on various cir
cumstances. Take the case of a cart on a common road ;
then observation indicates that the value of p. depends on
the size of the wheels and on the velocity of motion as well
as on the nature of the road. For a cart having wheels
four feet in diameter drawn with a velocity of six miles an
hour along a good road, the value of /z may lie between
and . Again, consider a train drawn along a railroad ;
then observation indicates that the value of p depends on
the velocity. For a velocity of 30 miles an hour the value
1 f\
will be about *kat * s ^ e friction is about 16 Ibs.
per ton, estimated on the whole weight of the engine and
the load. There is however besides this the resistance of
the air, which depends on the square of the velocity and
the area of the frontage of the train.
203. There is one mode in which the labour of men and
of animals is employed which is not directly comparable with
the application of a force to raise a weight, namely, that of
carrying burdens along a horizontal road. It seems that a
portion of the labour is spent in merely supporting the
burden, and this portion does no work in the sense in
which the term is used here : the remainder of the labour
does the work of carrying the burden. By observation we
can find the amount of useful effect which can be produced
by this mode of labour ; thus, for example, it is said that a
porter walking with a burden on his back through a day
of seven hours long can carry a weight of 90 Ibs. through
145 feet in a minute. Here the useful effect is equivalent
to 13050 units of work per minute. But this must not be
taken to measure the labour of the porter ; for, as we have
said, some labour is spent in merely supporting the bur
342 WORK.
den : moreover some labour is also spent by the porter in
carrying the weight of his own body.
See Young's Lectures on Natural Philosophy, Lecture
XIL, and Whewell's Mechanics of Engineering, page 178.
204. In the Statics we investigated the conditions of
equilibrium of the simple machines and of some compound
machines. In practice however machines are generally
used not to maintain equilibrium but to assist in doing
work. By the aid of machines the labour of men and of ani
mals and the powers furnished by nature are transmitted
and applied in various ways. Now it is a general prin
ciple that if we set aside friction and the weights of the
parts of a machine, then the work applied to the machine
is equal to the work done by the machine: this principle
is called by some writers the principle ofworL
205. It would be impossible in an elementary treatise
to demonstrate the principle just stated, or even fully to
explain its meaning : we will however give two simple
illustrations.
Take the case of the Wheel and Axle ; see Art. 236 of
the Statics. Suppose P moving uniformly downwards, and
therefore W moving uniformly upwards. Then the rela
tion of P to W is found to be the same as in the state of
equilibrium. Hence it follows that the product of P into
the vertical space which it describes is equal to the product
of W into the vertical space which it describes ; that is in
this case the work applied to the machine is equal to the
work done by the machine.
Again, suppose a body of weight TF drawn up an In
clined Plane by means of a string which passes over a
smooth Pully fixed at the top of the Piano and has a weight
P attached to it which hangs vertically.
Let a be the inclination of the Plane. It may be shewn
that if the motion is uniform we shall have the same rela
tion between P and W as in the state of equilibrium : see
Statics, Arts. 211, 244, and Dynamics, Art. 92. Hence it
will follow that the product of P into the vertical space
which it describes is equal to the product of W into the
WORK. 343
space which it describes resolved vertically. Thus the work
applied to the machine is equal to the work done by the
machine.
206. But in practice, owing to friction and the weights
of the parts of a machine, the principle of Art. 204 must
be modified. The whole work performed by a machine is
distinguished into two parts, namely, the useful part and
the lost part : the useful work is that which the machine is
designed to produce ; the lost work is that which is not
wanted but which is unavoidably produced, such for ex
ample as the wearing away by friction of the machine itself.
It is still true that the work applied to a machine is equal
to the whole work, useful and lost, done by the machine ;
and consequently the useful work done by the machine is
always less than the work applied to the machine.
207. The ratio of the useful work done by a machine to
the work applied to the machine is called the efficiency of
the machine, or sometimes the modulus of the machine.
The efficiency is thus a fraction, and it is of course the ob
ject of inventors and improvers to bring this fraction as
iiear to unity as possible.
208. Accumulated Work. If a body is moving it is
said to have work accumulated in it. In fact if a body
possesses any velocity it can be made to do work by parting
with that velocity. For example, a cannon ball in motion
can penetrate a resisting body; water flowing against a
waterwheel will turn the wheel.
We will now shew how the amount of work accumulated
in a body may be conveniently estimated.
209. Let a body of mass J/ be moving with a velocity
v ; let a constant force F acting on the body through a
space s bring it to rest ; then we shall take Fs as the mea
sure of the work accumulated in the body.
F
We know by Art. 87 that the acceleration is ^ ; there
2F Mo*
fore v 2 =jrfS, by Art. 42 : thus Fs= ~ . Hence we may
say that the work accumulated in a moving body is
344 WORK,
measured by half the Vis Viva of the body: see Arts. 85
and 107.
Let h be the height through which the body must fall
to acquire the velocity v, and TF the weight of the body.
Then v*=2gh; and W=Mg\ therefore Wh=Fs. Hence
we may say that the work accumulated in a moving body
is measured by the product of the weight of the body into the
height through which it must fall to acquire the velocity.
We know from Art. 124 that by the collision of bodies
there is always a loss of Vis Viva if the elasticity be
imperfect; so we may say in such a case that there is
always a loss of accumulated work.
210. Suppose a body of weight TF to be moving in a
straight line, and urged on by a force in that straight line ;
if in moving through a space s the velocity changes from u
to v the work done on the body as it moves through that
TF
space is (v 2  w 2 ) : if there be one force urging the body
on, and another force resisting the body, this expression
gives the excess of the work done by the former force over
the work done by the latter force.
211. We have seen that a body in motion may be said
to have work accumulated in it, because it may be made to
do work by parting with its velocity ; in like manner, if a
body be in such a position that it can fall and thus acquire
velocity, it may be considered to possess a store of work
ready to be used. Thus suppose that a weight has been
drawn up to a certain height ; it may then be allowed to
fall, and do work in various ways, as for instance in driving
piles. So men may employ their labour in ascending to a
high point, and then descending as weights in a machine.
Water which can fall from a certain height contains a store
of work the amount of which is measured by the product of
the weight of the w r ater into the vertical descent : this
store of work can be used to turn a waterwheel.
212. Hitherto we have supposed that the force which
performs work is a constant force ; we will now make some
remarks concerning the work performed by a variable
WORK. 345
force, that is a force which is not constant : but the com
plete discussion of this subject is beyond the range of an
elementary treatise.
213. Suppose we have a c 3>
force of 41bs., which acts
through 2 feet ; then let the
force be changed into a force
of 5 Ibs., and act through 2
feet, and then be changed
into a force of 7 Ibs., and act
through 3 feet. Then the
whole work done is A JB C JD
4x2 + 5x2 + 7x3, that is 39.
Now there is a convenient mode of representing this
calculation. Take on a straight line successive lengths to
represent the spaces through which the forces act : thus let
AB represent two feet, EG two feet, and CD three feet.
Draw at A, , and C straight lines at right angles to AEG
to represent the forces which act respectively through the
distances represented by AB, EG, and CD : thus let Aa
represent the force of 4 Ibs,, Bb the force of 5 Ibs., and
Cc the force of 7 Ibs. Complete the rectangles Ea, C6, and
DC; then the area of each rectangle represents the 'work
done by the corresponding force. This is obvious, because
the area of a rectangle is equal to the product of its base
into its altitude.
Hence the sum of all the areas represents the whole work.
214. Now let us suppose that the changes in the value
of the force are more gradual than in the preceding
Example; that is, suppose the changes to be more fre
quent, but each change to be of less amount. For instance,
suppose a force of 5 Ibs. to act through 6 inches, then a
force of 5 Ibs. to act through 10 inches, then a force of
5 Ibs. to act through 8 inches, then let the force change to
6 Ibs. ; and so on. Still the above geometrical mode of
representing the calculation may be conveniently applied.
Instead of the rectangle Ba we should now have three
rectangles, the breadths representing respectively , f , and
 of a foot, and the heights 5, 5, and 5 Ibs.
346
WORK.
0/1
01) E t
215. By proceeding in this way we can obtain the con
ception of a force which is always changing its value, so
that it does not remain con
stant even for a very small
space. Let a straight line
AG represent the whole
space through which a vari
able force acts ; conceive that
at every point of A O straight
lines are drawn at right an
gles to AG, so that any
straight line Del represents
the amount of the force at
the point D. Then the other
extremities of these straight
lines will form a curve adg :
and ike whole area bounded
by the curve and the straight lines will represent the whole
work done.
216. No exact Rule can be given for finding the area
of such a figure as that of Art. 215, which will apply what
ever may be the form of the curve : but a Rule called
Simpson's Rule will furnish results which are sufficient
approximations to the truth for many practical purposes.
The straight lines Aa, Bb,...Gg are called ordinates ;
the values of an odd number of them, drawn at equal dis
tances, are supposed to be known ; then the Rule for finding
the area is this : Add together the first ordinate, the last
ordinate, twice the sum of all the other odd ordinates, and
four times the sum of all the even ordinates ; multiply the
result by onethird of the common distance between two ad
jacent ordinates.
See Mensuration for Beginners, Chapter xvur.
217. An important application of the preceding
method occurs in the Steam Indicator. While the piston
of a steam engine is making a stroke the pressure of the
steam on the piston varies. By a suitable contrivance the
amount of the pressure in any position is registered : in fact
a curve is drawn corresponding to the curve adg in Art. 215.
Then by the Rule of Art. 216 the whole work is calculated.
EXAMPLES. XVII. 347
218. We may remark that the process of Art. 215 is
applicable to other investigations as well as to that of the
work done by a variable force.
For example, consider the velocity generated in a given
time in a particle by a variable force. Here let the
straight line AG represent the whole time during which
the force acts ; and let the straight lines at right angles to
this represent the force at the corresponding instants.
Then the area will represent the whole velocity generated
in the given time.
Again, consider the space described in a given time by
a particle moving with variable velocity. Here let the
straight line AG represent the whole time of motion; and
let the straight lines at right angles to this represent the
velocity at the corresponding instants. Then the area will
represent the whole space described in the given time.
EXAMPLES. XVII.
1. Find how many units of work are performed in
raising 2 cwt. of coal from a pit 50 fathoms deep.
2. Find how many cubic feet of water an engine of 40
HorsePower will raise in an hour from a mine 80 fathoms
deep, supposing a cubic foot of water to weigh 1000
ounces.
3. Find how many bricks a labourer could raise in a
day of 6 hours to the height of 20 feet by the aid of a
cord and a Fully, supposing a brick to weigh 8 Ibs. See
Art. 196.
4. Find the HorsePower of an engine which is to
move at the rate of 30 miles an hour, the weight of the
engine and load being 50 tons, and the resistance from
friction 16 Ibs. per ton.
5. Find the HorsePower of an engine which is to
move at the rate of 20 miles per hour up an incline
which rises 1 foot in 100, the weight of the engine and load
being 60 tons, and the resistance from friction 12 Ibs, per ton,
348 EXAMPLES. XVII.
6. A well is to be made 20 feet deep, and 4 feet in
diameter : find the work in raising the material, supposing
that a cubic foot of it weighs 140 Ibs.
7. Supposing that a man by turning a windlass can
perform 2600 units of work per minute, find how many
cubic feet of water he can raise to the height of 24 feet in
8 hours ; the efficiency of the machine being '6.
8. If an engine of 50 HorsePower raise 2860 cubic
feet of water per hour from a mine 60 fathoms deep, find
the efficiency of the engine.
9. Find the work accumulated in a body which weighs
300 Ibs. and has a velocity of 64 feet per second.
10. Find the useful HorsePower of a waterwheel,
supposing the stream to be 5 feet broad and 2 feet deep
and to flow with a velocity of 30 feet per minute; the
height of the fall being 14 feet, and the efficiency of the
machine "65.
11. Determine the HorsePower of a steam engine
which will raise 30 cubic feet of water per minute from a
mine 440 feet deep.
12. It is found that a man with a capstan can do 3120
units of work per minute for 8 hours a day, but if he carries
weights up a ladder he can do only 1120 units of work per
minute : find in each case how long he will take in raising
36 tons from a depth of 130 feet.
13. A steam engine is required to raise 70 cubic feet
of water per minute from a depth of 800 feet : find how
many tons of coal will be required per day of 24 hours,
supposing the duty of the engine to be 250000 for a Ib. of
coal.
14. A variable force has acted through 3 feet; the
value of the force taken at seven successive equidistant
points, including the first and the last, is in Ibs. 189, 15 T2,
126, 108, 945, 84, 75'6 : find the whole work done.
15. A railway truck weighs m tons; a horse draws
it along horizontally, the resistance being n Ibs. per ton ; in
passing over a space s the velocity changes from u to v :
find the work done by the horse in this space.
EXAMPLES. XVII. 349
16. A stream is a feet broad, and b feet deep, and
flows at the rate of c feet per hour ; there is a fall of h
feet ; the water turns a machine of which the efficiency is
e : find the number of bushels of corn which the machine
can grind in an hour, supposing that it requires m units of
work per minute for one hour to grind a bushel.
17. A shaft a feet in depth is full of water : find the
depth of the surface of the water when one quarter of the
work required to empty the shaft has been done.
18. Find at what rate an engine of 30 HorsePower
could draw a train weighing 50 tons up an incline of 1 in
280, the resistance from friction being 7 Ibs. per ton.
19. The French unit of work is the work done in rais
ing a kilogramme vertically through a metre : find how
many English units of work are equivalent to a French unit,
taking the metre at 39 '37 inches, and the kilogramme at
15432 grains.
20. According to Navier it requires 43333 French
units of work to saw through a square metre of green oak :
find how many English units of work are required to saw
through a square foot of green oak.
21. A sawmill was supplied with 111500 units of
work per minute : in eleven minutes 13 square feet of green
oak were sawn by the mill : find the ratio of the lost work
to the useful work. See the preceding Example.
22. Find the useful work done by a fireengine per
second which discharges every second 13 Ibs. of water with
a velocity of 50 feet per second.
23. Find how many units of work are stored up in a
millpond which is 100 feet long, 50 feet broad, and 3 feet
deep, and has a fall of 8 feet.
24. In piledriving 38 men raised a rammer 12 times
in an hour; the weight of the rammer was 12 cwt., and the
height through which it was raised 140 feet : find the work
done by one man in a minute.
350 PALEMBERTS PRINCIPLE.
XYIII. PAlemlerfs Principle.
219. The three Laws of Motion which have been
already given are found to be sufficient for the discussion of
all problems connected with the motion of particles. The
mathematical investigations may be in some cases extremely
long and difficult, but no new mechanical principles are
required. In the discussion of problems connected with
the motion of bodies however another Law is required;
this was first explicitly stated by D'Alembert, and is
called D'Alemberfs Principle: we shall now give some
account of it.
220. The particles of a solid body are acted on by
various forces which bind the particles together, and
which are usually called molecular forces. If we knew the
direction and the amount of the molecular forces which
act on an assigned particle as well as the other forces,
like gravity, which act on it, we could determine the
motion of the particle. But the nature of the molecular
forces is unknown, and it is the object of D'Alembert's
Principle to obtain results respecting the motion of a body
without considering the molecular forces which act on the
particles of the body.
221v It is necessary to explain the distinction between
impressed force and effective force. By the impressed force
on* a particle is meant the external force which acts on it,
that is all the force which acts on it except the molecular
force. By the effective force is meant the force which is
just sufficient to produce the actual motion of the particle.
Thus the effective force is equivalent to the resultant of
the impressed force and the molecular force. By reversed
effective force is meant a force equal in magnitude and
opposite in direction to the effective force. These defini
tions have to be given before we state D'Alembert's
D'ALEXBERrS PRINCIPLE. 351
Principle, but they will probably not be fully understood
until the Principle and some of its applications have been
considered.
222. D'Alembert's Principle may be stated thus : the
impressed forces together with the reversed effective forces in
a body or system of bodies satisfy tJte conditions of equi
librium for the body or system of bodies.
223. It must be observed that the utility of the Prin
ciple depends on the fact that we can find expressions for
the effective forces in terms of the motion produced.
224. We will give some illustrations of the Principle
from cases already considered : these illustrations will
serve to explain the meaning of the terms which are used,
although they will supply little notion of the importance
and the extent of the Principle.
225. Suppose a particle of mass m to describe a circle
of radius r with uniform velocity v, then, as we have
shewn in Art. 166, the force acting on the particle must be
equal to , and must tend towards the centre of the
circle. Thus if F denote the force, F = 0. Now
this equation may be considered an example of the Prin
ciple ; F is the impressed force, and is the reversed
effective force : thus the equation expresses the condition
for the equilibrium of the impressed force and the reversed
effective force.
226. If a particle is moving in a straight line with
uniform velocity the effective force is zero ; for no force is
required to maintain such a motion. Suppose a particle
of mass m to be moving in a straight line, the velocity not
being uniform ; let/ denote the acceleration at any instant,
that is the velocity which is added in a unit of time if the
force be uniform, or which would be added in a unit of
time if the force were to continue uniform during that
unit : them mfis the effective force at that instant.
352 UALEMBERTS PRINCIPLE.
227. We shall now investigate the value of the
effective force when a particle describes a circle, the
velocity not being uniform. Take the diagram of Art. 163;
let v denote the velocity at P, and v^ the velocity at Q.
Suppose that the force acting at P is resolved into two
components, one in the direction PS, and the other in the
direction PT. Let the angle PSQ be denoted by 20.
At Q the velocity is along TQ produced ; and it may be
resolved into v t cos 20 along a straight line parallel to P T,
and v sin 20 along a straight line parallel to PS. Thus in
passing from P to Q the change of velocity is v l cos 20  v
in the direction PT, and v 1 sin 20 in the direction PS.
Let t denote the time in which the particle moves from P
to Q. Then if we assume the force to continue uniform we
have as the measure of the force along PT\
i 1 1 ' i L v \ v 20, sin 2 . T
and this expression is equal to * L *. Now, as
t t
in Art. 163, we have 2r<j>=Vjt when t is made small enough,
..... , . . , , V.V 2v 1 2 sin 2
so that the above expression is equal to * * ^ .
When is made indefinitely small the second term
vanishes, and the first term remains alone : this is the
same as if the motion at P were along the tangent there
instead of along the arc of the circle. Thus if m be the
mass of the particle, and / the acceleration at P considering
the motion to be along the tangent PT, then the effective
force in the direction P T is mf. The effective force in the
direction PS is  ; this is obtained in the manner of
Art. 163: the only difference is that the equation 2r0=v^
which is used is now not absolutely true, but only true
when t is taken small enough.
228. We can now give some notion of the way in which
the important result stated in Art. 135 is obtained. Sup
pose m v ra 2 , ra 3 ,... to denote the masses of the particles of a
rigid body; and /^/2,/g,... the respective accelerations of
the particles estimated parallel to an assigned direction.
Let F denote the sum of the impressed forces acting
UALEMEEETS PRINCIPLE. 353
parallel to this direction; then by Statics, Art. 90, and
D'Alembert's Principle
F m lfl ~ m 2/2 ~ 3/3  = 0.
But if/ be the acceleration of the centre of gravity we have,
by Art. 131,
(?! + m 2 + m z + . . .)/= %/j + m 2 f 2 + m 3 / 3 + . . .
Hence F(m 1 + m 2 + m 3 + ...)/=0:
that is the motion of the centre of gravity parallel to the
assigned direction is the same as that of a particle having a
mass equal to the mass of the rigid body, and acted on by
forces parallel to those which act on .the rigid body. The
Article 90 of the Statics to which we appeal contemplates
forces in only one plane, though the result can be shewn to
be true universally: but even with this restriction we
obtain a good notion of the way in which Art. 135 is
established.
229. Suppose that F=0 in the preceding Article, then
/=0; but when/=0 the centre of gravity is either at rest,
or moving uniformly in a straight line, so far as regards the
assigned direction : hence if the forces which act on a body
in an assigned direction vanish, the centre of gravity either
has no motion in that direction or moves uniformly in it.
Therefore if no forces act on a rigid body the centre of
gravity either remains at rest or moves uniformly in some
straight line. Thus it appears that by the aid of D'Alem
bert's Principle we can make a statement with regard to a
rigid body analogous to that which is made in the First Law
of Motion with respect to a particle.
230. With regard to the truth of D'Alembert's Prin
ciple remarks may be made similar to those which have
been already applied to the Laws of Motion : see Arts. 12
and 140. But although the Principle may be confirmed
in an indirect manner yet the evidence for it is probably
somewhat short of that which establishes the truth of the
Laws of Motion. Strictly speaking it is only by means of
D'Alembert's Principle that we are justified in treating the
heavenly bodies as particles in our theories; and thus it
T. ME. 23
354 EXAMPLES. XVIII.
may be said that Astronomy verifies the truth of D'Alem
bert's Principle as well as that of the Laws of Motion : but
owing to the small dimensions of these bodies compared
with the enormous distances which separate them, and to
their nearly spherical forms, we may calculate the motions
as if these bodies were particles, by means of special con
siderations of a reasonable character, without appealing to
the Principle. There is indeed one very important problem
in Astronomy, namely that of Precession and Nutation,
which distinctly involves the Principle ; but then various
assumptions have to be made in order to arrive at numeri
cal results, so that the testimony to the truth of the Prin
ciple is not decisive. If we regard the evidence of Astronomy
as inconclusive we must turn to other cases in which theory
involving D'Alembert's Principle may be compared with
observation and experiment ; probably the most satisfactory
instances of this kind are those which depend on the motion
of pendulums, which will be considered in Chapter XX.
231. Perhaps however on due reflection the Principle
will appear to involve only what may be readily accepted.
The impressed forces and the molecular forces, together
with the reversed effective forces, are in equilibrium by
definition; thus D'Alembert's Principle amounts to the
assertion that the molecular forces will be in equilibrium
taken alone. It is plain from observation that this is practi
cally the case when the body is not in motion ; thus we
merely assume that the molecular forces when the body is
in motion satisfy the same conditions as they do when the
body is not in motion.
EXAMPLES. XVIII.
1. Point out the impressed forces and the effective
force in the case of Art. 92.
2. Point out the impressed forces and the effective
force in the case of Art. 1 68.
MOMENT OF INEETIA. 355
XIX. Moment of Inertia.
232. Let the mass of every particle of a body be
multiplied into the square of its distance from an assigned
straight line ; the sum of these products is called the
moment of inertia of the body about that straight line.
The straight hue is often called an axis. In the discussion
of problems respecting the motion of rigid bodies the
moment of inertia occurs very frequently, so that it be
comes necessary to consider this subject in detail: we
shall demonstrate some general results, and shall calculate
the value of the moment of inertia in various special cases.
233. The moment of inertia of any body about an
assigned axis is equal to the moment of inertia of the body
about a parallel axis through the centre of gravity of the
body, increased by the product of the mass of the body into
the square of the distance between the axes.
Let m be the mass of one
particle of the body ; let this
particle be at A. Suppose a
plane through A, at right
angles to the assigned axis,
to meet that axis at 0, and
to meet the parallel
<Z1
through the centre of gravity at G. From A draw a
ight line AM, perpendic
Let GM=x, where x is a positive or a negative quantity
.
straight line AM, perpendicular to OG or to OG produced.
according as M is to the right or left of G. By Euclid II.
12 and 13 we have OA 2 = OG' 2 + GA 2 + WG . x ; therefore
m. OA 2 =m. OG 2 + m. GA 2 + 20G.m. x.
A similar result holds with respect to every particle of
the body. Hence we see that the moment of inertia with
respect to the assigned axis is composed of three parts :
namely, first the sum of such terms as m . OG' 2 , and this
will be equal to the product of the mass of the body into
OG 2 ; secondly the sum of such terms as m . GA 2 , and this
will be the moment of inertia of the body about the axis
through G; and thirdly the sum of such terms as WG.m.x,
which is zero by Art. 146 of the Statics. Hence the moment
232
356 MOMENT OF INERTIA.
of inertia about the assigned axis has the value stated in
the proposition.
234. For the sake of abbreviation the symbol 2 is
used to indicate the sum of terms of any specified kind;
thus 2m. 0.4 2 means the sum of such terms as m.OA 2 .
The context will always make it evident what collection of
terms we are considering. In future unless the restriction
is expressly removed we shall confine ourselves to plane
figures; that is the bodies we consider are to be supposed
of uniform thickness, but that thickness is to be infini
tesimal. Also the straight lines we have to use will be in
the plane of the figure unless the contrary is stated.
235. The moment of inertia of any plane figure about
any straight line at right angles to its plane is equal to the
sum of the moments of inertia about any two straight lines at
rig/it angles to each other in the plane which intersect at the
first straight line.
Let denote any point in the plane ; through draw
any two straight lines at right angles to each other in the
plane. Suppose one particle of the body to be at the
point T of the plane; let m be the mass of the particle.
Denote by x and y respectively the perpendiculars from T
on the two straight lines in the plane. Then oP + y 2 de
notes the square of the distance of T from the straight
line drawn through at right angles to the plane ; the
moment of inertia of the body about this straight line
then is 2m(^ 2 +y 2 ), that is 2mx' 2 + 2my 2 . But 2m^ 2 is the
moment of inertia about one of the straight lines in the
plane, and 2 my 2 is the moment of inertia about the other.
Thus the proposition is established.
236. Hence when the moments of inertia of a plane
figure about two straight lines at right angles to each
other in the plane of the figure are known, the moment of
inertia about the straight line at right angles to the
plane through the common point is known immediately.
The moment of inertia about any straight line in the plane
through the common point is also connected with the
moments of inertia about the first two straight lines, but
MOMENT OF INERTIA. 357
the relation is not so simple as in the case of the straight
line at right angles to the plane. To this we now proceed.
237. Through any point in the plane of a plane
figure, draw any two straight lines OX and Y at right
angles to each other in the plane ; also through draw
any straight line OL in the plane. It is required to
connect the moment of inertia about OL with the moments
of inertia about OX and OY.
Let a denote the angle LOX. Suppose one particle of
the body to be at the point T of the plane ; let m be the
mass of the particle. Let OT be denoted by r; and let
the perpendicular from T on OX be denoted by y, and the
perpendicular from T on OY by x. Let A denote the
moment of inertia of the body round OX, and B the
moment of inertia round OY.
The perpendicular from T on OL=r sin TOL
=r sin (TOXLOX) = rsm (TOXa)
=r sin TO X cos ar cos TOX sin a
=y cos a  x sin a.
The moment of inertia round OL = ~S,m (y cos a  x sin a) 3
=2m (y 2 cos 2 a + x z sin 2 a  %xy sin a cos a)
= cos 2 a 2wiy 2 + sin 2 a 2m# 2  2 sin a cos a 'S.mxy
= A cos 2 a + B sin 2 a  2 sin a cos a 'S.mxy.
Thus we see that we cannot determine the moment of
inertia about OL from the three quantities A, B, and a ;
because the preceding expression involves another quantity,
namely
238. The quantities x and y of the preceding Article
are called the coordinates of T with respect to the axes OX
and OF; the point is called the origin; the quantity
'Smxy is called the product of inertia of the body for the
axes OX and OY. If at a point in a plane figure two
straight lines can be drawn at right angles to each other
in the plane such that the corresponding product of inertia
vanishes, the two straight lines are called principal axes of
the plane figure at the point.
358 MOMENT OF INERTIA.
239. We shall now shew that at every point in the
plane of a plane figure principal axes exist. The moment
of inertia of an assigned body about an assigned axis is by
definition a finite positive quantity. Thus among all the
axes which pass through a given point there must be one
for which the moment of inertia is greater than it is for all
other axes, or at least for which it is as great as it is for any
other axis. Let denote the point considered, and sup
pose that OX is an axis for which the moment of inertia
is as great as it is for any axis through 0. Let Y be at
right angles to OX, and OL any other straight line through
; and denote the angle LOX by B. Let the moment of
inertia about OX be denoted by J, that about OY by B,
and that about OL by Q. Also let P denote the product
of inertia with respect to OX and OY. Then by Art. 237,
Now by hypothesis A  Q cannot be negative ; but
A  Q=A  A cos 2  B sin 2 + 2P sin 6 cos Q
so that the sign of this expression is the same as the sign
of A  B + 2P cot 6. But if P have any value different from
zero we can make 2P cot 6 numerically as great as we please
by taking 6 small enough : thus if P were positive, by taking
6 very small and negative we should make A  B + 2P cot 6
negative; and if P were negative we should attain the
same object by taking 6 very small and positive. Hence
it follows that P must be zero, for otherwise A Q could
be rendered negative; and since P is zero, OX and OY
constitute principal axes.
Thus Q=A cos 2 + B sin 2 B\ and as this can be put in
the form Q=B + (AB} cos 2 6 it follows that Q cannot be
less than B. Thus principal axes at a point have this
property with respect to all axes through the point : no
moment of inertia can be greater than that for one of these
axes, and no moment of inertia can be less than that for
the other axis. If A = B we have Q=A; and then the
moment of inertia is the same for all axes through the
point.
MOMENT OF INERTIA. 359
240. We may express this result with sufficient clear
ness by saying that the axes of greatest and least moments
of inertia for an assigned point are at right angles to each
other; and we see that this is consistent with Art. 235.
For that Article shews that the sum of the moments of
inertia about two axes at right angles to each through an
assigned point is constant, namely equal to the moment of
inertia about the axis through the point at right angles to
the plane : hence when one of the two has the greatest
possible value the other must have the least possible value.
241. The position of the axes for which the product of
inertia vanishes, that is the position of principal axes,
is often sufficiently obvious. Suppose, for example, we
require the position of principal axes for a rectangle
when the given point is the centre. Draw through the
centre straight lines parallel to the sides of the rectangle ;
then these will be the principal axes. For consider a
particle of mass w, at a point of which the coordinates are
x and y ; it is plain that thero is also a corresponding
particle of mass m at the point of which the coordinates
are x and y. Hence we see that 'Zmxy consists of terms
which may bs arranged in pairs, so that the two terms in
a pair are numerically equal but of opposite signs ; and
therefore 'S.mxy vanishes.
242. The value of the product of inertia at any point
may be made to depend on the value of the product of
inertia for parallel axes through the centre of gravity.
Let x and y be the coordinates of a particle of mass m
referred to axes through any assigned point ; and let x 1
and ;/ be the coordinates of the same particle referred to
parallel axes through the centre of gravity ; also let h and
k be the coordinates of the centre of gravity referred to the
first pair of axes. Then
therefore 2 mxy = 2m (tf + h} ( y' + k]
But by Art. 146 of the Statics
2m/ = 0, 2w.r' = 0;
therefore 2 mxy = ?,mx'y' + hk'S.m.
360 MOMENT OF INERTIA.
243. The result of the preceding Article will often
enable us to calculate the value of the product of inertia
for an assigned origin and axes. Suppose, for example,
that we require the product of inertia in the case of a
rectangle, when the origin is at a corner, and the axes are
the edges which meet at that corner. Then by Art. 241
we have 2?n#y=0; and therefore ^mxyhJc^m\ and h
and k are known, being half the lengths of the edges of the
rectangle to which they are respectively parallel.
244. If the moments of inertia about three axes pass
ing through a point are known, the position of the principal
axes at that point and the moments of inertia about them
can be determined. Let Q, It, S denote the three known
moments of inertia ; suppose /3 the angle between the axes
corresponding to Q and It, and y the angle between the
axes corresponding to Q and S. Let 6 denote the unknown
angle between one of the principal axes, and the axis
corresponding to Q ; let A denote the moment of inertia
about this principal axis, and B the moment of inertia
about the other principal axis. Then, by Art. 239,
It=A cos 2 (6 + &+B sin 2 (6 + /3),
= A cos 2 (6 + y) + B sin 2 (6 + y).
These may be written
(4 ) (cos 20 cos 2/3  sin 26 sin 2/3),
= (A + B] + (A  B} (cos 20 cos 2 y  sin 20 sin 2y).
These are three simple equations for finding  (A + B\
MOMENT OF INERTIA. 361
 (A  B) cos 20, and ^(AB) sin 26 ; thus we can deduce
the value of tan 20, and thence the value of 6 ; and then
we shall know A  B and A + J3, and finally A and B.
When Q, E, S are not all equal we shall obtain one
system of principal axes ; when Q, B, S are all equal, the
values of tan 20 will be indeterminate, and any pair of
lines at right angles to each other will be principal axes.
245. Hence if two different plane figures have the same
moment of inertia about three axes through a common
point, they will have the same principal axes at the point,
and the same moment of inertia for every axis through the
point. And if the plane figures have also the same mass
and the same centre of gravity they will have the same
moment of inertia about any straight line in the plane or
at right angles to it: see Arts. 235 and 233. The term
plane figure here includes any collection of particles which
are all in one plane.
We shall now determine the value of the moment of
inertia for various special cases.
246. To determine the moment of inertia of a rectangle
about an edge.
Let ABCD be a rectangle ; it
is required to determine the
moment of inertia of the rect
angle round the edge AB. Let
BC=b\ divide BC into n equal
parts, and suppose straight lines
drawn through the points of
division parallel to AB. Thus
ABCD will be divided into n
equal strips ; let PQ represent
the r th strip counting from AB.
If M denote the mass of the rectangle will denote the
362 MOMENT OF INERTIA.
mass of PQ ; and if n is very great the distance of every
point in the strip from AS may be taken to be  b ; so
that the moment of inertia of the strip about AB will be
x T TT. Hence the moment of inertia of the whole
n n*
rectangle will be equal to the value when n is made in
definitely great of the expression
But, by Algebra,
thus the above expression is equal to
and when n is made indefinitely great this becomes
247. The method of the preceding Article is sub
stantially the same as that which is used in the higher
parts of the subject for finding moments of inertia, but by
the aid of the notation and the principles of the Integral
Calculus the process is rendered shorter. It would be
inconsistent with the plan of the present work to give any
great attention to such investigations ; we will however
notice an indirect method of obtaining the result of the
preceding Article which may admit of application in other
cases.
248. Suppose two rectangles having the common edge
AB ; let M denote the mass of one, and b the length at
right angles to AB ; let M' and b' denote the corresponding
quantities for the other. Then we shall shew that about
the common axis AB
Moment of inertia of first rectangle _ 3fb z
Moment of inertia of second rectangle M'b'* '
For, divide the second rectangle into the same number
of very slender strips as the first, in the manner of
Art. 246. Let p denote the mass of a strip of the first
MOMENT OF INERTIA. 363
rectangle, and // the mass of the corresponding strip of
the second ; let x denote the distance of the strip of the
first rectangle from AS, and xf the distance of the cor
responding strip of the second : then it is obvious that
a M . X* b*
= = 
Therefore
Since this relation holds for every pair of correspond
ing strips we obtain the result which had to be established.
From the nature of the result it is easily understood and
remembered ; and the same remark will apply to various
similar cases which will occur as we proceed. We shall
now apply this to the problem of Art. 246.
249. To find the moment of inertia of a rectangle about
It follows from Art. 248 that the moment of inertia
varies as the product of the mass into the square of the
length of the rectangle. Let M denote the mass, and b the
length; then the moment of inertia will be \Jtfb' 2 , where X
denotes some quantity which does not change when 6
changes or when J/ changes. Thus if there be another
rectangle of mass M' and length 6', having the same edge
AB as the former, then the moment of inertia of this
rectangle about AB will be XJ/'6' 2 . Hence the moment of
inertia of the difference of the two rectangles about AB will
be XJT6' 2  X J/6 2 : we will denote this by Q. Put a for AB,
and let r denote the infinitesimal thickness of the rect
angles. We suppose both of the rectangles to be of the
same substance, so that the masses may be represented by
the volumes ; thus we may put
M ' abr, M' = ab'r;
therefore
_ 6 3 ) =Xra (b'b} (6'
where p denotes the difference of the masses of the two
3G4
MOMENT OF INERTIA.
rectangles, that is the mass of the body which has Q for its
moment of inertia round AB. This result is true whatever
may be the values of b' and b. But when the difference
between b' and b is infinitesimal the body which has Q for
its moment of inertia becomes a strip, every particle of
which may be considered to be at the distance b from AB,
so that the moment of inertia is /z6 2 . Hence when b' =b
we must have X (b' 2 + b'b + 6 2 ) = 6 2 j therefore X = \ .
o
250. To find the moment of inertia of a rectangle about
a diagonal.
Let ABCD be the rectangle; let AB= a, and BC=b\
and let J/ denote the mass of the rectangle. Let the
diagonals AC and BD intersect
at 0. Draw OX parallel to AB,
and OY parallel to BC: let 6
denote the angle COX. The
moment of inertia of the rect
angle about OX by Arts. 233
and 246 is If ?  J/YY, that
is M r^ . Similarly the moment
of inertia of the rectangle about _ .
Y is M^ . Now OX and Y
are principal axes of the rect
angle at by Art. 241 ; therefore, by Art. 239, the moment
of inertia of the rectangle about OC is
But
moment of inertia is = . ? ^ .
6 a 1 + 6 2
and sm 2 6= 2 ^; hence the required
MOMENT OF INERTIA. 3G5
251. Let A denote the length of the perpendicular from
D on AC, then h x AC AD x DC, for each expresses twice
the area of the triangle ADC: thus h 7 5 jjr. There
fore the moment of inertia of a rectangle about a diagonal
A 2
is M , where M denotes the mass, and h the perpendicular
on the diagonal from an opposite angle.
252. The moments of inertia of ABC and ADC about
AC are equal; for these two triangles are equal and are
symmetrically situated with respect to AC. Thus the
moment of inertia of each of them about AC is half the
moment of inertia of the rectangle A BCD; and therefore
if J/" now denote the mass of ACD the moment of inertia of
A CD about AC is J/.
253. If M denote the mass of any triangle and h the
perpendicular from the vertex on the base the moment of
7i 2
inertia of the triangle about the base is M . This is
shewn in the preceding Article for the case in which the
angle opposite to the base is a right angle, and it may be
readily extended to the case in w r hich this condition does
not hold. For suppose S to denote any point in AC or AC
produced, and join DS, thus forming a triangle ADS.
Then by drawing straight lines parallel to AC we can
divide the triangle ADC and also the triangle ADS into the
same number of strips parallel to AC; the particles in each
strip may be considered to be at the same distance from
AC; and thus the moments of inertia of the strips will be in
the same proportion as the masses of the strips, that is as
the lengths of the strips, that is as AC is to A3. Since this
is true for each pair of strips it will be true for the whole
triangles ; and thus the required result is obtained.
254. The moment of inertia of any indefinitely thin
wire bent into the form of a circle, about a straight line
through the centre of the circle at right angles to its
366 MOMENT OF INERTIA.
plane is Mr 2 ', where M denotes the mass and r the radius
of the circle. For every particle of the ring thus formed is
at the same distance r from the axis about which the
moment of inertia is required. Hence the moment of
.r2
inertia of such a ring about a diameter is M ; for the
moment of inertia must be the same about any diameter,
and the sum of the moments of inertia about two diameters
at right angles to each other is equal to Mr 2 by Art. 235.
255. To find the moment of inertia of a circle about an
axis through its centre at right angles to its plane.
It may be shewn as in Art. 248 that the moment of
inertia must vary as the product of the mass into the
square of the radius. Let M denote the mass, and r the
radius ; then the moment of inertia will be \Mr* where X is
some quantity which does not change when r changes or
when M changes. Thus if there be a concentric circle
of mass M' and radius /, its moment of inertia about
the axis will be XJ/V 2 . Hence the moment of inertia
of the difference of the two circles about the axis will be
X J/V 2  X Mr 2 : we will denote this by Q. Let r denote the
infinitesimal thickness of the circles ; then representing the
masses by the volumes, as in Art. 249, we may put
M' = 7r/ 2 r ;
therefore Q=\TTT (/ 4  r 4 ) =\nr (r' 2  r 2 ) (r' 2 + r 2 } =X/* (r' 2 + r 2 ),
where p, denotes the difference of the masses of the two
circles, that is the mass of the body which has Q for its
moment of inertia round the axis considered. This result
is true whatever may be the values of / and r. But when
the difference between / and r is infinitesimal the body
becomes a ring every particle of which may be considered
to be at the distance r from the axis, so that the moment
of inertia is p.r 2 . Hence when r'=r we must have
r 2 ; therefore X = Q: thus the required moment
r 2
of inertia is M.
MOMENT OF INERTIA. 3fi7
256. The moment of inertia of a circle about a
r 2
diameter is J/ . For the moment of inertia must be the
4
same about any diameter ; and the sum of the moments of
inertia about two diameters at right angles to each other
r 2
is equal to J/ by Art. 235. The moment of inertia of
a circle about any chord at the distance of h from the centre
will be M + ^ 2 b Art. 233.
257. If a and b denote the lengths of the edges of a
rectangle, and M the mass, the moment of inertia about
72 2
the former edge is M , and about the latter N . Hence,
o o
by Art. 235 the moment of inertia about an axis through a
corner of the rectangle at right angles to its plane is
258. If the moment of inertia of a body of mass M
about any axis is equal to Mk 2 , then k is called the radius
of gyration of the body about the assigned axis. Thus,
for example, the moment of inertia of a circle of radius r
about an axis through its centre at right angles to its
plane is M ; so that 2 =o , and therefore the radius of
gyration about this axis is ^ .
259. Hitherto we have restricted ourselves as we stated
in Art. 234, to the case of plane figures; we shall now
shew by two examples that the results which we have thus
obtained may be used to determine the moments of inertia
of bodies which are not indefinitely thin. Let there be a
rectangular parallelepiped of mass J/; and let a, b, c be the
lengths of three edges which meet at a point : then the
moment of inertia about the edge of length c is (a 2 + b 2 ).
o
For suppose the body cut up into an indefinitely large
number of indefinitely thin slices by planes at right angles
to the axis; then, by Art. 257, the moment of inertia of
3G8 EXAMPLES. XIX.
each slice is the product of its mass into  (a 2 + 6 2 ) : hence
o
the moment of inertia of the whole mass is the product of
that mass into  (a 2 + 6 2 ). Next let there be a right cir
o
cular cylinder of mass J/, and let r be the radius of the
cylinder : then the moment of inertia of the cylinder about
*M
its axis of figure is J/ . For we may suppose the cylinder
cut up into slices as in the preceding Example ; and then
the result follows by the aid of Art. 255.
EXAMPLES. XIX.
1. Shew that the moment of inertia of a triangle
about the straight line which is drawn from an angle to
the middle point of the opposite side is J/^; where M is
the mass of the triangle, and p the perpendicular drawn
to this straight line from either end of the side.
2. Let ABC be a triangle; let D, E, F be the middle
points of the sides opposite to A, B, C respectively ; and
denote by J/ the mass of the triangle : shew that the
moment of inertia of the triangle about AD is the same as
that of two particles each of the mass ^ placed at J3 and C
respectively : or of two particles each of the mass placed
o
at E and F respectively.
3. Shew that the moment of inertia of a triangle of
mass M about any axis in its plane is the same as that of
three particles each of mass at the middle points of the
sides.
EXAMPLES. XIX. 3G9
4. Shew that the moment of inertia of a triangle of
mass M about any axis in its plane is the same as that of
three particles each of mass at the angular points, and
3J/
a particle of mass ^ at the centre of gravity.
5. Shew that the moment of inertia of a triangle of
mass M about any axis at right angles to its plane is the
same as that of three particles each of mass at the
a
middle points of the sides.
6. Shew that the moment of inertia of a triangle of
mass M about any axis at right angles to its plane is the
same as that of three particles each of mass at the
3M
angular points, and a particle of mass at the centre of
gravity.
7. Find the moment of inertia of a triangle about an
axis through its centre of gravity at right angles to its
plane.
8. Find the moment of inertia of a rectangular paral
lelepiped about an axis through the centre of gravity
parallel to an edge.
9. A given mass is to be formed into a plane figure so
as to have the least moment of inertia about an axis
passing through a given point, at right angles to the
plane of the figure : shew that the figure must be a circle
having its centre at the given point.
10. If at a given point of a plane figure the moments
of inertia are equal about two straight lines unequally
inclined to the principal axes, the moments of inertia
about the two principal axes through that point must be
equal.
T. ME. 24
370 MOTION ROUND
XX. Motion round a fixed axis.
260. We have already introduced the term angular
velocity in the particular case in which this velocity is
constant ; see Art. 171 : we shall now have to consider it
more generally. When a particle describes a circle the
straight line drawn from the particle to the centre describes
an angle. The rate at which the angle is described is
called angular velocity, and this may be uniform or variable.
When the angular velocity is variable it may be considered
as uniform for an indefinitely short time ; and thus, as in
Art. 171, we have a relation between the angular velocity
of the describing straight line and the linear velocity of its
extremity at any instant, namely this : the linear velocity
is the product of the angular velocity into the radius.
261. Corresponding to the term velocity we have the
term acceleration; this denotes the amount of velocity
which is added in a unit of time if the velocity is increased
uniformly, or the velocity which would be added in a
unit of time if the rate of increase were to continue for a
unit of time what it is at the instant : see Arts. 18 and 19.
In like manner corresponding to the term angular velocity
we have the term angular acceleration; this denotes the
amount of angular velocity which is added in a unit of
time if the angular velocity is increased uniformly, or
the angular velocity which would be added in a unit of
time if the rate of increase were to continue for a unit of
time what it is at the instant. And from the conclusion
of the last Article it follows that the linear acceleration in
the direction of the tangent is the product of the angular
acceleration into the radius.
262. When a body turns round a fixed axis every
point of it describes a circle or an arc of a circle in a plane
at right angles to the axis. The angle which the radius
belonging to one point describes is equal to the angle which
the radius belonging to any other point describes in the
same time. Thus the angular velocity is the same for
every point of the body, and the angular acceleration is
A FIXED AXIS, 371
the same for every point of the body. It is the object
of the present Chapter to shew how the angular accelera
tion is to be determined ; and when this is known the
angular velocity may be sought, and finally the angle through*
which the body has turned from any assigned position.
263. To investigate the angular acceleration of a body
'which can turn round a fixed axis and is acted on by given
forces.
Let m^ wi 2 , ??i 3 ,... denote the masses of the particles of
the body; and let r w r 2 , r 3 ,... denote the radii of the
corresponding circles which they can describe. Let o>
denote the angular velocity at any instant, and x the
angular acceleration. Then the effective forces correspond
ing to m l are m^co 2 along the radius towards the centre
of the circle which this particle can describe, and ffi^x
along the tangent : see Art. 227. Suppose the impressed
forces to consist of P l acting at an arm p lt P 2 acting at an
arm jo 2 , P 3 acting at an arm jo 3 , and so on ; each force
being supposed to be in some plane at right angles to the
axis. By D'Alembert's Principle the impressed forces and
the effective forces reversed must satisfy the condition of
statical equilibrium ; hence by Arts. 100 and 104 of the
Statics the sum of the moments round the axis must
vanish. Now the moment of such a force as w^eo 2 is
zero, because the direction of the force passes through the
axis ; and the moment of such a force as ??V'iX * s m i r iX x r i
because the direction of the force is at right angles to the
radius r t . Therefore
This equation may be written
that is, according to a mode of abbreviation already used,
2G4. In Art. 87 we obtained a result which may be
expressed verbally thus : when force acts on a free body
212
372 MOTION ROUND
the acceleration is equal to the quotient of the force by the
mass. If we use the word inertia as synonymous with mass,
which some of the old writers did, we may say that the
acceleration is equal to the quotient of the force by the
inertia. In the formula just obtained for % we have in
the numerator the moment of the forces, and hence the
name moment of inertia might be given by analogy to the
denominator, so that we may have this result : when a
body can turn round a fixed axis the angular acceleration
is the quotient of the moment of the forces by the moment
of inertia about the axis. It is obvious that the term
moment of inertia now has precisely the same sense as it
had in Art. 232.
265. In the Chapters on Dynamics which we have
given we have mainly confined ourselves to cases in which
the acceleration in the direction of the motion is constant;
the reason for this is that the mathematical difficulties in
other cases are greater than the student at this stage can
be assumed able to overcome. The example of Art. 151
shews how complex the investigation may be of even simple
problems which do not fall within the restriction we have
named.
266. As an illustration of Art. 263 we will now discuss
the motion of a system of the nature of the Wheel and
Axle : see the figure in Art. 180 of the Statics. Suppose
there to be a body of mass ^ t and therefore of weight p^,
which acts at an arm a l ; let this be rising. Suppose there
to be a body of mass /z 2 and therefore of weight /z^, which
acts at an arm a. 2 ; let this be descending. Let the mass
of the body which forms the Wheel and Axle be J/, and
let its moment of inertia about the axis be J/fc 2 . Let 7\
be the tension of the string attached to the mass p l9 and
T 2 the tension of the string attached to the mass /^ 2 ;
then if % denote the angular acceleration of the body which
forms the Wheel and Axle we have, by Art. 263,
J/F X = 7V* 2 7X (1).
The axis is supposed to pass through the centre of
gravity of the body which forms the Wheel and Axle, so
that the moment of the weight of the body round the
.1 FIXED AXIS. 373
axis vanishes. If the system were in equilibrium the
values of T l and T 2 would be known ; namely the former
would be equal to fi^, and the latter to p 2 g : but these
will not be the values when there is motion, and we shall
proceed to find two other equations by the aid of which we
can eliminate T l and T 2 . The velocity of the descending
weight is at any instant the same as that of the circum
ference of the circle to which its string is attached, so that
the acceleration is a lX > therefore, by Art. 87,
Ti~toff .................. (2).
Similarly from considering the descending weight we have
F 2 2X=M 2 5 f ^2 .................. (3).
Substitute in (1) the values of T t and T 2 from (2) and (3) ;
thus Mtf x = 2 (ptf  fjL 2 a 2 x)
therefore (J/ 2 + ^ + M2 2 2 ) *
Thus the angular acceleration is constant, namely equal to
Denote this by X : then by the same processes as we have
used in Arts. 36 and 37, we see that if the body which
forms the Wheel and Axle starts with an angular velocity
y, the angular velocity at the end of the time t will be
y + \t ; and the angle which the body has turned through
in the time t will be t + \t 2 .
267. To investigate the motion of a heavy body which
can turn round a fixed horizontal axis.
Let M denote the mass of the body, h the length of
the perpendicular from the centre of gravity of the body
on the axis, 6 the angle which this straight line makes at
any instant with the horizontal plane through the axis.
Then the weight of the body is Mg, and the moment of
this round the axis at the instant considered is Mgh cos 6.
Therefore by Art. 263 the angular acceleration at the instant
_ Mgh cos 6
S/nr* *
374 MOTION ROUND A FIXED AXIS.
Now let J/P denote the moment of inertia of the body
about a straight line through the centre of gravity parallel
to the axis ; then, by Art. 233,2mr 2 =J/(/t 2 + F) ; therefore
the angular acceleration =^ 2 7.2
Now suppose that instead of the heavy "body we had
a heavy particle attached to the axis by a rigid rod without
weight of the length I, thus forming what is called a simple
pendulum; we shall find, on investigating the motion by
the principles applied to the case of the heavy body, that
the angular acceleration when the rod is inclined at an
angle 6 to the horizon is ^^ , that is ^ S . Hence
t" (/
by comparing the two cases we arrive at the following
important result : the motion of the heavy body is the
same as that of a simple pendulum of length I where
h 2 + & 2
1= T . Therefore the results already obtained with
respect to the simple pendulum in Art. 151 will hold for
the heavy body. Thus if o> be the angular velocity acquired
from rest while 6 changes from 90  a to 90  /3 we shall
have
cos /3  cos a) ;
and the time of a small oscillation will be TT / .
V 9
268. Thus we see that the motion of a heavy body
round a fixed horizontal axis is exactly the same as that of
a certain simple pendulum; this pendulum is caUed the
equivalent simple pendulum. Let a straight line be drawn
at right angles to the axis so as to pass through the centre
of gravity of the heavy body, and let it be produced till
its length measured from the axis is equal to the length of
the equivalent simple pendulum; then the extremity of
this straight line is called the centre of oscillation of the
body.
269. The most important property connected with the
centre of oscillation is that which is briefly stated by saying
that the centres of oscillation and suspension are convertible:
EXAMPLES. XX. 375
the meaning of this statement will appear from the inves
tigation to which we now proceed. Let the distance of the
centre of gravity from the centre of oscillation be denoted
by A'. Suppose the body to be put in motion round an
axis through the centre of oscillation parallel to the original
axis, instead of round the original axis ; then the length of
the equivalent simple pendulum in this case will be
A^and^A^ + ^Ar^
and this is the length of the original equivalent simple
pendulum. Thus the motion round the new axis will be
precisely the same as the motion round the original axis :
for instance, the time of a small oscillation in either case is
. / ,
V 9
where I stands for
270. We have  =   V^ + 2. Now suppose
the body to be put in motion in succession round various
axes which are all parallel; then Jc remains the same in all
these cases while h may vary : and we see that the least
value of the length of the equivalent simple pendulum is
/*
when jj Jh vanishes, that is when h=k .
V"
EXAMPLES. XX.
1. Find the length of the equivalent simple pendulum
when a rectangle turns round an edge which is horizontal
2. Also when a cube turns round an edge which is
horizontal.
3. Also when an equilateral triangle turns round a
horizontal axis through an angular point at right angles to
its plane.
4. A circular arc turns round a horizontal axis through
its middle point at right angles to the plane of the arc :
shew that the length of the equivalent simple pendulum is
equal to the diameter of the circle.
376 MISCELLANEOUS THEOREMS.
XXI. Miscellaneous Theorems.
271. In Chapter XIX. we confined ourselves to the
case of plane figures, and easy deductions from this case ;
we now proceed to the more general problem, namely
that in which the moment of inertia of any body is con
sidered. Through any point draw three straight lines
X) Y, OZ mutually at right angles, like the edges of a
cube which meet at a common point. Let A, B, 7 denote
the moments of inertia of an assigned body about OX, OY,
OZ respectively. Through draw any straight line OL ;
let Q denote the moment of inertia of the body about OL :
we shall now connect Q with A, B, C. Let a, ft y denote
the angles which OL makes with OX, OY, OZ respectively.
Suppose one particle of the body of mass m to be at a point
T\ let x, y, z denote the perpendiculars from T on the
planes YOZ, ZOX, XO Y respectively ; and let r denote OT.
The perpendicular from Ton OL=rsin TOL; the square
of this =r 2 sin 2 TOL=r 2  r 2 cos 2 TOL : and, by Art. 269 of
the Statics, this
=r 2  r 2 (cos mTcos LOX+ cos TOYcos LOT
=r 2  (# cos a +y cos /3 + 2 cos y) 2
= (# 2 + y 2 + z 2 ) (cos 2 a + cos 2 /3 + cos 2 y)
 (x cos a +y cos /3 + z cos y) 2
= (y 2 + 2 2 ) cos 2 a + (s 2 + tf 2 ) cos 2 /3 + (^ 2 + /) cos 2 y
tyz cos /3 cos y  2zx cos y cos a  2.ry cos a cos /3.
Then Q=A cos 2 a + B cos 2 /3 + C cos 2 y
 2 cos j3 cos y ^myz  2 cos y cos a 2wiz#  2 cos a cos /3 'Zmxy.
272. The quantity 'S.myz is called the product of inertia
of the body for the axes Y and OZ\ and similar names are
given to 'S.mzx and 'S.mxy. If at a point in a body three
straight lines can be drawn at right angles to each other,
MISCELLANEOUS THEOREMS. 377
such that the three products of inertia of the body all
vanish, the three straight lines are called principal axes of
the body at the point.
273. We shall now shew that at every point of a body
principal axes exist. The moment of inertia of an assigned
body about an assigned axis is by definition a finite positive
quantity. Thus among all the axes which pass through a
given point there must be one for which the moment of
inertia is greater than it is for all other axes, or at least for
which it is as great as it is for any other axis. Let denote
the point considered, and suppose that OX is an axis for
which the moment of inertia is as great as it is for any axis
through 0. Then, with the notation of Art. 271, it follows
that A  Q cannot be negative. Now
A  Q=A (cos 2 a + cos 2 j8 + cos 2 y)Q
2
f 2 cos /3 cos y 'S.myz + 2 cos y cos a 'S.mzx + 2 cos a cos /3 'Smxy.
Suppose now that OL is in the plane XOY, so that y=90,
and therefore cos y=0; then
A  Q = (A  B} cos 2 /3 + 2 cos a cos /3 ^mxy ;
and since OL is in the plane XO Y we have a + /3=90; thus
A  $=sin 2 a (A  B + 2 cot a ~2mxy\
Then proceeding as in Art. 239 we see that 'Zmxy must be
zero. In precisely the same way, by supposing OL to be
in the plane XOZ we see that 'S.mxz must be zero. Hence
by taking for OX what we may call the axis of greatest
moment of inertia it follows that the two products of inertia
^mxy and 'S.mxz vanish.
We have not yet fixed the position of the axes OY and
OZ in the plane in which they must lie. Let then OZ be
such that C is the least moment of inertia for all axes in
this plane, or at least as small as any. Suppose OL to be
in the plane YOZ, so that cos a=0, and /3 + y=90.
Then Q = B cos 2 + C cos 2 y  2 cos /3 cos y ?.myz,
and QC=(BC] cos 2 /3  2 cos ft cos
= sin 2 y (B  C  2 cot y 'S.myz}.
378 MISCELLANEOUS THEOREMS.
Tliis expression cannot be negative, and hence in the
manner of Art. 239 it follows that 'S.myz must be zero.
Thus the existence of principal axes at any point
is established. And if the moments of inertia about these
are known the moment of inertia about any other axis is
known from the formula
Q=A cos 2 a + B cos 2 /3 + C cos 2 y.
We may observe that the least of the three A, B, G will
also be the least moment of inertia about all axes through 0.
For
Q  C=Aco$ 2 a + Bcos 2 j3 + (7cos 2 y  (7(cos 2 a + cos 2 /3 + cos 2 y)
= (A  C} cos 2 a + (B  C) cos 2 /3 ;
and this cannot be negative if G be not greater than either
A or B.
274. The position of axes for which the three pro
ducts of inertia vanish, that is the position of principal
axes, is often sufficiently obvious. Suppose, for example,
that we require the position of principal axes for a
right circular cone at the vertex: take the axis of figure
and any two straight lines at right angles to each other
and to this, and they will constitute principal axes. For
suppose OX to coincide with the axis of figure ; then *S,mxy
vanishes. For consider a particle of mass m at a point
of which the coordinates are x and y ; it is plain that there
is also a corresponding particle of mass m at the point of
which the coordinates are x and y. Hence we see that
"Smxy consists of terms which may be arranged in pairs, so
that the two terms in a pair are numerically equal but of
opposite signs; and therefore *S,mxy vanishes. Similarly
^mxz and 'Zmyz vanish.
275. The values of the products of inertia at any point
may be made to depend on the values of the products of
inertia for parallel axes through the centre of gravity. For
as in Art. 242 we can shew that
'S.mxy = 'S.mx'y' + hJc 2m ;
and that similar equations hold for the other two products
of inertia.
MISCELLANEOUS THEOREMS. 379
276. The sum of the moments of inertia of a given body
about any three axes at right angles to each ot/fcr passing
through a given point is constant.
Let denote the given point; and OX, OY, OZ three
straight lines at right angles to each other. Let m, x, y, z
have the same meaning as in Art. 271. Then the moment
of inertia about OX is 2?n(y 2 + 2 2 ), that about OY is
2m (z 2 + x 2 \ and that about OZ is 2m (x* + y 2 }. The sum of
these three expressions is 22/?i(# 2 + ?/ 2 + z 2 ), that is 22mr 2 ,
where r is the distance of the particle of mass m from 0.
And 2mr 2 has obviously the same value whatever may be
the position of OX, OY, OZ.
277. To find the moment of inertia of an indefinitely
thin spherical shell about a diameter of the sphere.
Let M denote the mass of the shell, a the radius of the
sphere, Q the required moment of inertia. It is plain that
the value of Q is the same for any diameter; hence, by
9 72
Art. 276, we have 3=22mr 2 =2J/r 2 : therefore Q=M .
3
278. To find the moment of inertia of a sphere about a
diameter.
It may be shewn as in Art. 248 that the moment of
inertia must vary as the product of the mass into the
square of the radius. Let M denote the mass, and r the
radius, then the moment of inertia will be \Mr 2 , where X
is some quantity which does not change when r changes
or when M changes. Thus if there be a concentric sphere
of mass M' and radius r 1 , its moment of inertia about the
diameter will be X J/V 2 . Hence the moment of inertia of
the difference of the two spheres about the diameter will
be X (M'r 2  Mr 2 }. Let p denote the mass of the difference
of the spheres, and pJc 2 its moment of inertia about the
diameter ; so that
Now it is known that the volume of a sphere is the
3SO MISCELLANEOUS THEOREMS.
product of into the cube of the radius ; therefore repre
5
senting the masses by the volumes, as in Art. 249, we
have
,, 4?r , ^,, 4?r 4?r , ~
M= y r 3 , M' = y r' 3 , /i = (r' 3  r 3 ).
Hence ~ (r' 3  r 3 ) F = ^ X (/ 5  r 5 ),
X (
so that 2 =
Now, this, being always true, holds when the difference
between / and r is infinitesimal; and then by Art. 277 we
, 79 2r 2 ,, 2> 2 . 5r 2 , , . . 2
have F= r : thus =Xr , so that X= .
o o 3 O
279. We may propose questions relative to the effect
of impulsive forces on a body which can turn round a fixed
axis; the discussion of such questions will resemble that
which we have given in Chapters IX. and X. with respect
to the subject of the Collision of Bodies.
280. To determine the change of motion produced by
impulsive forces acting on a body which can turn round a
Jixed axis.
Let m v m 2 , m 3 ,... denote the masses of the particles of
the body ; let r 1? 7*25 r v~ denote the respective distances of
the particles from the fixed axis ; and suppose that o> is
the angular velocity with which the body is turning round
the axis just before the impulses. Let Q^ Q 2 , Q&... denote
impulsive forces which act simultaneously on the body,
at distances q v q 2 , q^,... respectively from the axis, each
force being supposed to be in some plane at right angles
to the axis. Suppose that the angular velocity is thus sud
denly changed to o>'. Then the effective impulsive force on
m^ is m,r l (a/  CD) in the direction at right angles to that of
r^ ; ana similar expressions hold for the other particles.
By D'Alembert's Principle and Arts. 100 and 104 of the
Statics we have
MISCELLANEOUS THEOREMS. 381
+ Q&z +  ~ (%i 2 + m 2 r 2 2 + m 3 r 3 2 + ...) ('  ) =O.
"We may write this result thus,
that is the numerator of the fraction which gives a/  < is
the sum of the moments of the impulsive forces round the
fixed axis, and the denominator is the moment of inertia of
the body about that axis.
281. In such a case as the preceding there are forces
which act on the body at the points where the axis is fixed ;
but these do not enter into the equation which gives o>'  o>,
because they have no moment round the axis. If the
direction of any one of the impulsive forces is not at right
angles to the fixed axis, this force must be resolved into
two, one parallel to the axis, and the other at right angles
to it : the former component will have no effect on the
angular velocity, the latter alone will occur in the equa
tion.
282. Suppose a body at rest but capable of turning
round a fixed axis : we may enquire whether it is possible
to put the body in motion by a blow without producing
any impulsive pressure on the axis. We shall confine
ourselves to the case in which the blow is given in a plane
which is at right angles to the axis and which divides the
body symmetrically, so that we may assume that there can
be no impulsive pressure on the axis out of this plane.
Let Q denote the blow, q the arm at which it acts, '
the angular velocity given to the body. As there is to be
no impulsive pressure on the axis the force Q with the
effective forces reversed must satisfy the conditions of
equilibrium. Now, as in Art. 280, the effective force on
the particle of mass m 1 is ra^ta', the direction being at
right angles to that of r t ; and similarly for the other
particles. Let G denote the centre of gravity of the body,
and let the perpendicular from G on the axis meet it at 0.
382 EXAMPLES. XXL
As in Art. 97 of the Statics the system of effective forces
reduces to a certain single force at and a couple. Then, as
in Art. 278 of the Statics, the direction of the single force is
at right angles to OG, and its magnitude is w'OGSm ; and the
moment of the couple is n'Smr*. Hence this force and
couple reversed must be in equilibrium with Q. Moreover
Q may be replaced by a force Q at parallel to its original
direction, and a couple of which the moment is Qq. Hence
for the equilibrium required we must have
(1) The direction of Q must be at right angles to Off.
(2) Q must be equal to u'
(3) Qq must be equal
From (2) and (3) by division we deduce
Hence finally q must have the value determined by
(4), and Q must have the direction determined by (1) :
these are the necessary and sufficient conditions in order
that the blow may produce no impulsive pressure on the
axis.
283. It will be observed that the preceding investi
gation does not determine the point at which the blow
must be given, but only the value of the perpendicular
from the axis on the direction of the blow. This might
have been anticipated from the fact that the effect of
a force will be the same at whatever point of its line of
action we suppose it applied. The point at which the
direction of the blow meets OG produced is called the
centre of percussion. By comparing the value of q with
that given for the length of the equivalent simple pendulum,
according to Art. 269, we see that the centre of percussion
and the centre of oscillation relative to the same axis
coincide.
EXAMPLES. XXI.
1. Shew that the two systems mentioned in Example 3
of Chapter XIX. have the same moment of inertia for any
axis whatever. Shew also that a similar statement may be
made with respect to Examples 4, 5, and 6.
EXAMPLES. XXL 383
2. Shew that the moment of inertia of a rectangle of
mass M about any axis is the same as that of four particles
each of mass at the angular points, and a particle of
mass ^ at the centre of gravity.
o
3. Find the position of principal axes at the centre
of gravity of a rectangular parallelepiped, and the moment
of inertia about each of them.
4. Shew that the moment of inertia of a uniform
straight rod of infinitesimal thickness about an axis
through one end at right angles to the rod is the same
as that of a particle at the other end equal in mass to one
third of the rod. Shew also that this is true for an axis
through the end inclined at any angle to the rod.
5. Shew that the moment of inertia of a uniform
straight rod of infinitesimal thickness of mass M about any
axis through one end is equal to that of a particle of
mass at the other end, and a particle of mass ~ at the
o o
middle point.
6. Shew that the moment of inertia of a uniform
straight rod of infinitesimal thickness of mass M about
any axis is equal to that of a particle of mass at each
O IT
end, and a particle of mass at the middle point.
3
384 MISCELLANEOUS EXAMPLES.
MISCELLANEOUS EXAMPLES.
1. A stone falls down 100 feet, determine the time of
motion.
2. A stone falls down a well and is heard to strike the
water after 3 seconds : find the depth of the well, supposing
sound to be transmitted instantaneously.
3. If the space described in falling for 11 seconds from
rest be 556'6 feet, find the acceleration.
4. A body, starting from a given point, moves vertically
downwards at the rate of 32*2 feet per second. After four
seconds a heavy body falls from the same point under the
action of gravity. Shew that it will overtake the first body
at a distance of 257 '6 feet from the starting point.
5. Any number of smooth fixed straight rods, not in
the same plane, pass through a given point; and a heavy
particle slides down each rod, the particles starting simul
taneously from the given point. If the rods be so situated
that the particles are at one instant of their motion in the
same plane, prove that they will be so throughout it, and
that a circle can be described passing through them.
6. AB is the vertical diameter of a sphere ; a chord is
drawn from A meeting the surface at P, arid the tangent
plane at B at Q : shew that the time down PQ varies as
JBQ, and that the velocity acquired varies as BP.
7. Find a point at a given distance from the centre of
a given vertical circle, such that the time of falling from it
to the centre is less than the time of falling to any point
on the circumference except one, and equal to the time of
falling to this point.
8. Find the locus of points in a given vertical plane
from which the times of descent down smooth Inclined
Planes to a fixed point in the vertical plane vary as the
length of the Planes.
9. A body is projected along a smooth horizontal table
with a velocity g : find the length to which the table must
be prolonged in the direction of the body's motion, so that
the body after leaving the table may strike a point whose
distances measured horizontally and vertically from the
point of projection are 3# and 2^ respectively.
10. A heavy particle is projected from a given point in
a given direction so as to touch a given straight line : give
IN DYNAMICS. 385
a geometrical construction for determining the point of
contact and the elements of the path described. If the
direction of projection be not fixed, find the path so that
the velocity of projection may be the least possible.
11. A chord is drawn joining any point on the circum
ference of a vertical circle with the lowest point : shew
that if a heavy body slide down this chord the parabola
which it describes on leaving the chord has its directrix
passing through the upper end of the chord.
12. Chords are drawn joining any point on the circum
ference of a vertical circle with the highest and lowest
points ; a heavy body slides down the lower chord : shew
that the parabola which it will describe after leaving the
chord is touched by the other chord, and that the locus of
the points of contact is a circle.
13. A heavy body is projected from one fixed point so
as to pass through another which is not in the same hori
zontal line with it : shew that the locus of the focus of its
path is an hyperbola.
14. A force acting uniformly during one tenth of a
second produces in a given body the velocity of one mile
per minute : compare the force with the weight of the body.
15. One end of a string is fastened to a weight P ; the
string passes over a fixed Fully, and under a moyeable
Pully, and has its other end attached to a fixed point ; a
weight Q is attached to the moveable Pully : determine the
motion, supposing the three parts of the string all parallel.
16. In the formulae of Art. 101 shew that if the veloci
ties u and u' are each increased by the same quantity, so
are the velocities v and v'.
17. From the formula) of Art. 101 determine the values
of v and v' if m=em' ; also if m' =em. 
18. A body of given mass is moving in a given direc
tion : determine the magnitude and the direction of a blow
which will cause it to move with the same velocity in a
direction at right angles to the former.
19. A projectile at the instant it is moving with the
velocity v at an inclination a to the horizon impinges on a
vertical plane which makes an angle /3 with the plane of
motion of the projectile : find the velocity after impact.
20. Small equal spherical balls of perfect elasticity are
T. ME. 25
386 MISCELLANEOUS EXAMPLES
placed at the corners of a regular hexagon ; one of them is
projected with the velocity u, so as to strike all the others
in succession and to pass through its original position : find
the velocity with which it returns.
21. In the preceding Example shew that each of the
five balls starts at right angles to an adjacent side of the
hexagon ; and find the velocity with which each starts.
22. Two perfectly elastic balls of equal mass impinge :
shew that if the directions of motion after impact are paral
lel, the cosine of the angle between their original directions
is equal to the ratio of the product of the velocities after
impact to the product before impact.
23. Of two equal and perfectly elastic balls one is pro
jected so as to describe a parabola, and the other is drop
ped from the directrix so as just to faU upon the first when
at its highest point : determine the position of the vertex
of the new parabola.
24. A mark in a vertical wall appears elevated at an
angle /3 at a certain point in a horizontal plane ; from this
point a ball is projected at the mark and after striking it
returns to the point of projection : shew that if a be the
angle of projection tan a= (1 + e) tan /3.
25. A plane is inclined at an angle /3 to the horizon ; a
particle is projected from a point in the plane at an inclina
tion a to the horizon, with the velocity u, and the particle
rebounds from the plane: find the time of describing n
parabolic arcs.
26. In the preceding Example find the condition which
must hold in order that after describing n parabolic arcs
the particle should be again at the starting point.
27. A particle is projected with a given velocity at a
given inclination to the horizon from a point in an inclined
plane : find the whole time which elapses before the par
ticle ceases to hop.
28. In the preceding Example find the condition which
must hold in order that the particle may cease to hop just
as it is again at the starting point.
29. In Example 25 find the cotangent of the inclination
to the plane of the direction of motion of the particle at
the beginning of the (n + l) th arc.
30. In Example 25 if the elasticity be perfect find the
IN DYNAMICS. 387
condition which must hold in order that the particle may
rise vertically at the n ih rebound.
31. Shew that the time of descent to the lowest point
of a very small circular arc is to the time of descent down
its chord as the circumference of a circle is to four times
its diameter.
32. If the resistance on similar steamers moving uni
formly is proportional to the product of the transverse
section and the square of the velocity, while their Horse
power is proportional to the tonnage, find how the velocity
varies according to the tonnage.
33. A particle descends down a smooth circular tube
of very small bore, and impinges on an equal particle at
rest at the lowest point of the tube : if Ji denote the vertical
height through which the particle descends, determine the
vertical height to which each particle will rise after impact.
34. If the weight attached to the free end of the string
in a system of Pullies in which the same string passes round
each of the Pullies be m times that which is necessary
to maintain equilibrium, shew that the acceleration of the
ascending weight is 4^. where n is the number of
mn + 1
parts of the string at the lower block, and the grooves of
the Pullies are supposed perfectly smooth. Compare the
tension of the string with the ascending weight.
35. Two particles move with constant accelerations in
given straight lines. If at any instant their relative velo
cities in any two directions are as their relative accelera
tions in the same directions, shew that the velocity of one
particle always bears a constant ratio to the velocity of the
other.
36. Two equal perfectly elastic balls are let fell at the
same instant, one from the height ^ and the other from the
height jj above a horizontal table ; shew that at the end of
6n  1 seconds the velocity of the centre of gravity changes
suddenly from to g t and at the end of 6n + 1 seconds the
velocity of the centre of gravity changes suddenly from
</toO.
252
STATICS. ANSWERS.
I. 1.8 Ibs. 2. 32 inches. 3. Ibs. 4. ^ a inches.
a . P
5. 9 Ibs. and 3 Ibs. : 1 inch. 6. As 3 is to 4. 7. As 4 is to 3.
8. of a cubic foot. 9. As 16 is to 9. 10. As is to ^ .
20 a b
II. 1. 64, 8. 2. 37. 3. 9, 12. 4. 3, 6. 6. 5^2 Ibs.,
at an angle of 45 with the resultant. 8. As /v/3 is to 2.
9. A right angle. . 10. 5, 5</3. 11. In a straight line.
12. 120. 13. The tension of the shorter string is 4 Ibs.,
and of the longer string 3 Ibs.
III. 2. By Art. 34, forces 1, 1, 1 are in equilibrium
and may be omitted ; thus the resultant is equivalent to
that of forces 1 and 2 at an angle of 120. 4. See Art. 34.
5. 15 Ibs., 20 Ibs. 6. 4 Ibs. 7. Let OA and OB denote
the equal forces, OD their resultant; produce A to C so
that OC= 2 OA ; and let OE be the resultant of OB and
OC : then it is given that OE=OD. The resultant of OE
and OD is equivalent to that of twice OB and half OC,
and is therefore equal to OE. 8. It follows from Example 7
that the angle EOD=IZCP. 12. The resultant is 2^2 Ibs.,
and it is parallel to a side of the square. 13. The re
sultant coincides in direction with the straight line from
the point to the intersection of the diagonals of the
rectangle, and is equal to twice that straight line. 14. Use
the polygon of forces. 15. Use Ex. 14 : if n be the number
of equal parts the resultant is represented by n times the
radius. 16. The straight line joining AF in the second
diagram of Art. 58.
IV. 1. The resultant is 9^2 Ibs. : it is parallel to the
a
diagonal AC, and it crosses AD at the distance  AD
y
from A. 2. 38 Ibs. and 114 Ibs. 3. 16 inches from
the heavier weight. 8. Pa~Qb. 9. Pa2j2, where
a is the side of the square.
V. 7. Take moments round A : thus we find that KI
is parallel to BC. 8. Take moments round an end of
one force : thus we find it must be bisected at 0.
STATICS. ANSWERS. 389
VI. 1. V(?0) Ibs. : see Art. 58. 5. The angle ACS
is given ; and since P, Q, and R are given, the angles which
the direction of R makes with AC and CB are given.
6. See Art. 39, and Euclid, in. 21, 22. 8. The point
must be at the intersection of the straight lines which join
the middle points of opposite sides. 9. The forces 1
and V3 are at right angles ; the forces 2 and 1 at 120.
11. Let CD be the resultant of CA and CB. Let A come
to a. Take Dd equal and parallel to Aa\ then ad is equal
and parallel to CB. Thus Cd is the resultant of Ca
and CB.
VII. 1. 12 Ibs. 2. 8 inches. 3. One inch from
the fulcrum. 5. 2 Ibs. or 5 Ibs. 6. 4 Ibs. 7. 3 to 4.
8. 9 to 20. 9. 26cwt. 13. P + R=Q + S',
P=Q=R=S. 14. See Art. 38. 15. It may be
shewn that the point in the rod at which the resultant of
the two weights acts is 13 inches from C. Then use Ex
ample 14 : the tensions will be found to be 150 Ibs. and
52 Ibs.
VIII. 2. P and Q. 3. A force of 12 Ibs. at
5 inches from the end at which the force of 4 Ibs. acts.
4. At a distance from the centre of the hexagon equal to
onefifth of a side. 5. At the point at which the force
of 8 Ibs. acts. 6. At the distance of the radius
nl
from the centre. 7. 6J inches from the end. 12. The
force at the middle point of BC must be Q + R P ; and
so on. 17. On the diagonal through the point where no
4
force acts, at = of the diagonal from this point.
IX. 2. One foot from the end. 3. Suppose the straight
line parallel to BC; let D be the middle point of BC: the
centre of gravity is on AD at the distance ^ AD from A.
u
4. At a distance from the centre of the larger circle
equal to onesixth of the radius. 5. Equal forces,
9. The ratio must be . 13. At a distance from the
390 STATICS. ANSWERS.
centre of the square equal to of the diagonal of the
I 44
square. 14. . 15. , 1, feet. 16. At a dis
71 3 O
tance from the base of the triangle equal to of the
altitude of the triangle. 17. At a distance from the
o
base of the triangle equal to  of the base.
o + 2i/>J o
18. Put the rods so that the points in contact may be  of
a foot from the middle of each, towards the 1 Ib. of the
lower rod, and towards the 9 Ibs. of the upper rod.
19. At of the whole length from the end of the densest
part. 20. 8, 8, 3 Jibs.
X. 1. Three quarters of the square. 2. A straight
line parallel to the base. 3. The centre of the spherical
surface. 6. One is double the other. 10. Twelve
inches. 11. The distance of the point from one end of
the side must be twice its distance from the other end.
XI. 1. Ito3. 2. 52 inches from the end. 3. lAlbs.,
5 ^ Ibs. 4. Two feet from the end. 5. Two inches.
6. 9 Ibs., 6 Ibs. : ratio that of 2 to 3. 7. 3 Ibs. 8. 5 Ibs.,
7 Ibs. 9. The forces are 3 Ibs. and 12 Ibs. 10. 12ilbs.,
22i Ibs. 11. One is double the other. 12. 2J, 4feet.
13.  Ib., 4 Jibs. 14. 144 stone. 16. 30 with Lever ;
v/(12) Ibs. 17.  P at a distance 1 feet from the fulcrum.
o
19. One inch from A ; 10 Ibs. 20. 4 inches from the
fulcrum. 22. #=2P.
XII. 2. 18 ounces. 3. 40 Ibs. 4. He gets 15
ounces for 35. 9d. ; which is at the rate of 4s. per Ib.
6. Seven inches from the point of suspension. 9. The
point D, from which the graduations begin, is brought
nearer to the point of suspension C. 10. The point D
is taken further from C. 11. 2 J feet from the end at
STATICS. ANSWERS. 391
which 10 Ibs. is suspended ; 60 Ibs. 13. Pressure on C is
o
half the weight of the rod, on D is  of the weight of the
rod. 14. Two feet from the other end. 15. Two Ibs.
16. As 6 is to C  a. 17. 6 feet; \ W, f W.
2i 2i 30
18. 20 Ibs. 20. 8 inches from the other end. 21. At
O
of an inch from the end of the lead bar. 22. 30 Ibs.
14
24. At a distance of of the Lever from the end where
the greater force acts.
XIII. 1. 56 inches. 2. 6 Ibs. 3. The radius of
the Wheel must be 10 times the radius of the Axle.
4. 16 ounces. 5. The weight of 6 Ibs. The prop must
O
support  lb., leaving 5f Ibs. on the Wheel to balance the
15 Ibs. on the Axle. The pressure on the fixed supports is
20^ Ibs. 6. 15cwt. 7. 18 inches ; 3 inches. 8. 108 Ibs.
9. The string which is nailed to the Wheel hangs vertically
so that its direction just touches the Axle. 10. In
creased.
XIV. 1. 3 Ibs. 2. One lb. 3. A force equal to a
third of his weight. 4. As 12 is to 1. 5. 16cwt. 6. 6.
Q
7. The Weight will overcome the Power. 8. 5 of his
o
weight, supposing him to pull upwards, as in Art. 196 ; but
*7
 of his weight if the Power end of the string passes over
o
a fixed Pully so that he pulls downwards. 9. 3 Ibs.
10. Onelb. 11. W=P. 12. W=w. 14. 3* Ibs.
15. 16. 16. 6. 17. The Power will overcome the
13
Weight. 18. 7 cwt. 19. of his own weight.
20. Three times the Power.
XV. 1. The perpendicular from the right angle on the
length. 2. 7i Ibs. 3. 8 Ibs. 4. 45 ; as 1 is
392 STATICS. ANSWERS.
5. P=j?TF, E =\ w  6  15lbs  ? 140 Ibs. 8. 4.
9. 3lbs. 10. 1120. 11. At an inclination of 30.
12. V^lbs. 13. 9 Ibs. 14. 14 Ibs.; 50 Ibs.
15. v/3lbs., 30. 16. 15 Ibs. 17. 60. 22. 39 Ibs.
to hang over.
XVI. 1. 25V2lbs. 2. 40 Ibs. 3. 60. 4. As 24
istol. 5. 48. 6. 480nlbs. 7. nj& 8. TT inches.
9. 2 Ibs. 10.
XVII. 1. #=3 ft. ; 8 Ibs. 2. 3 Ibs. 3. 60 Ibs.
4. .4 must now exert a force of 40 Ibs. 5. The weight
of C is twice the weight of B. 6. The weights are as
the lengths of the Planes on which they are placed.
XVIII. 1. 2 inches. 2. 6 inches. 3. 16 inches.
4. 5. 5. 2 feet. 6. 6. 7. 30 inches.
XIX. 1. 1. 2. Pressure 5^/3 Ibs. ; friction 5 Ibs.
3. V(8 2 + 3 2 ) Ibs. ; that is JT3 Ibs. 5. 45. 6. 75. 7. 9 Ibs.
9. Any force greater than 17 tons. 10. 15 T % tons.
12. tan Q=j ~ , where p. and p.' are the coefficients of
friction for the ground and wall, a and b the distances of
the centre of gravity from the lower and upper ends; 6
the inclination to the horizon.
XX. 1. 15 Ibs. 2. 2P; 60, 60, 45.
4. P^+^ + ^ + PQ + QE + RP. 5. 5.
MISCELLANEOUS EXAMPLES. 1. As 4 is to 3. 2. 25 Ibs.,
60 Ibs. 4. 45 Ibs. 5. 48 Ibs., 20 Ibs. 6. It is
represented by AD. 7. It is equal to the resultant of
2 and 4 acting at right angles. 8. 75, 165, 120.
10. 10 Ibs. ; bisecting the angle formed by the parts of the
string. 11. On the lower peg the resultant pressure is
W in a vertical direction ; on each of the other pegs the
resultant pressure is WJ'3, and the vertical component is
O TIT"
 . 18. 5 Ibs. 20. 5^/3 Ibs. inclined at an angle of
30 to the 4 Ibs. component. 21. That of the sides.
22. 35 Ibs., 40 Ibs. 23. As 2 is to 1. 24. 21 inches.
STATICS. ANSWERS. 393
26. 2 feet from end. 27. 8 Ibs., 12 Ibs. 31. 25, 65.
32. 3 Jibs. 33. 54oz., 48oz. 34. 1 feet from the
4 Ibs. weight. 35. 2 Ibs. 37. Weight = 2 */3 x Tension.
W 9O W *3<0
39.450. 40. 2 stone. 41.^ + ^,^ + ^. 42. lit
Z O X
the shorter cylinder at a point which divides it in the ratio
of 1 to 31. 43. The point divides the rod in the ratio of
5 to 4 : six points. 44. 15 inches from one end ; shifted
1 finches. 45. 8 Ibs. 46. 2 ounces ; 4 inches from one end.
50. 9 feet from the end near the heavier boy ; 6 feet from
the rail. 52. ^> where a is an edge of the cube.
40
53. 10 Ibs. ; at a point j feet from the 6 Ibs. end.
4 + nj o
54. 60 Ibs. 55. The pressures would now be a hori
zontal force equal to the Power, and a vertical force equal
to the Weight. 56. 48 Ibs. 57. 20 Ibs. 58. 3^ Ibs.
59. 6144 Ibs. 60. 28^/2 Ibs. 61. At an angle of
30 to the plane. 63. 50 Ibs. inclined at 30 to the
plane. 64. At the centre of gravity of the weights
2 Ibs., 1 lb., and 1 Ib. at the angular points of the tri
angle. 67. It is equal to the weight of a sphere.
70. The force on the face opposite to P must be
Q + R + S2P, and so on. 71. The centre is the point
of intersection of the perpendiculars from the angles of the
triangle on the opposite sides. 82. Let r be the radius of
the cylinder, a the inclination of the rod to the horizon ;
then the extreme length of the rod is 2 (n + 2) r sec a.
83. Let w be the weight of the upper ball, w that of each
lower ball, a the inclination to the vertical of the straight
line joining the centre of the upper ball with the
centre of one of the lower balls ; then the least coefficient
of friction between the upper and lower ball is tan  ; and
between the lower balls and the table is ^f= tan ? ,
2W+w 2
394
DYNAMICS. ANSWERS.
I. I. As 2 is to 1. 2. As 6 is to 7. 3. 15,10.
4. 73. 5. As TT is to 1. 6. As a 2 is to 6 2 .
II. 1. 240 feet. 2. At the end of 5 seconds.
3. At the end of two minutes ; at the distance 6600 feet
from the starting point. 4. 5n feet. 5. 1527 feet
per second. 6. 7ir feet per second. 7. The distance
between them is equal to the distance each has described.
8. n *J(u 2 + v 2  2uv cos a).
III. 1. 50. 2. 2 seconds. 3. 18. 4. 20; 1.
5. 25; i. 6. As 6 2  a 2 is to v 2  u 2 . 7. 32. 8. 32;
the first second is the third from rest. 9. The first
second is the T^th from rest. 10. 6. 11. 32.
12. ff. 13. /  / ; where h and h' are the
5& \/ g \l g
heights. 14. 48 inches ; 8 feet. 16. 2 seconds.
17. The radius to the point is inclined at 60 to the
radius which is vertically upwards. 18. The radius to
the point is inclined at 60 to the radius which is vertically
downwards. 24. f .  . ^ .
J v m?
IV. 1. 2 or 4 seconds; respective velocities g and
g. 2. 3 seconds. 3. Yes,/=32. 4. 36; 16.
6. 5 or 20 seconds. 7. >J(gh\ where h is the given
height. 8. It is half the time. 9. 2 seconds.
10. 2 or 18 seconds. 11. Let u be the initial velocity,
h the height of the given point, n the number of seconds
between passing this point and coming to it again ; then
The time is 1^ .where Z
g
is the length of the wire, and a its inclination to the
horizon. One ring describes y^ , and the other .
14. Let T denote each interval^ then the space is
DYNAMICS. ANSWERS. 395
nur+ ?r  ; put nr=t, and nv=f; eliminate T and
2i
v, and finally suppose n infinite.
Y. 1. 13 feet per second. 2. j . 3. No.
2i
2
5.  v sin a (u + v cos a). 6. Let u and v be the velo
y
cities of projection, and a and ft the angles of pro
jection. The square of the distance at the time t
is (u sin a  v sin /3) 2 ^ 2 + (u cos a  v cos /3) 2 * 2 , that is
Zuv cos (a  /3)} 2 . 7. + . 11. The
Cj CJ
f 1 \ 2
square of the distance is t 2 u? cos 2 a + (tu sin a  ^gt 2 J , that
is fiu z  gtu sin a + j , 13. Suppose the first body pro
jected with the velocity u at an inch" nation a ; then the
second body must be projected vertically with the velocity
u sin a. 14. ^ =tan 8. 15. cos 2 a tan ft.
w cos a g
16. Let w be the velocity with which the body is projected
horizontally ; then the distance at the end of the time /
is ut ', and t= / ^ s , where s is the given space, and a
'V g sin a
the inclination of the plane to the horizon.
VI. 1. The point must be the vertex of the parabolic
,, w 2 sin 2a , w 2 sin 2 a ,
path, so that the values of  ^ and   are known ;
2 # ty
see Arts. 56 and 57. 2. See Art. 71. 3. From Arts. 57
, . , u sin a 2% sin (a  3)
and 65 we have  = v Q or tan a=2 tan ft.
g gcoft
Msincg^ jo Q We must now take
wcosa
the lower sign of the preceding result : thus we get
t=  rr  ; and another value of t is found from
9 s^ ft
396 DYNAMICS. ANSWERS.
Art. 65. Hence we get 2 tan (a/3)=cot; this gives
tan a=  ,f m n . 10. Let x be the horizontal space
sin /3 cos /3
described, and y the vertical space. Then x=\
2u 2 sin 8 cos a sin (a  8}
l=tatan/3cosa=
2w 2 sin 8 cos 2 a ,. . ~ s m , . ,
=  ^TT  (tan a cos 3  sm /3). This reduces to
2w 2 cos 2 a , .,. 2w 2 sin 2 /?
 HTF > and this to r . 7o\ 12. This amounts
gcos 2 fi ' (1 + 3 sin 2 /?)
to the fact that the time of describing a space I with a
velocity V is the same as the time of describing a space
I cos )3 with a velocity F cos j8. 16. See Art. 70.
17.  v sin a */(^ 2 + *> 2 cos 2 a + Swvcosacos /3). 19. ^ A/2.
20. x? 4nxh sin a cos a 4n^ 2 cos 2 a = 0. 21. From the
preceding result obtain a quadratic in tan a ; solve it, and
examine the expression under the radical sign. 22. Let t
be the time between just passing the cube and reaching
the highest point, or between reaching the highest point,
and just passing the cube again ; then
1 u 2 sin 2 a
therefore kifi sin 2 a cos 2 a 8cgu? cos 2 a  c 2 <7 2 = 0.
23. cy tan 4 a + tan 2 a (Qcgu? + 2c 2 # 2  4w 4 ) + c 2 # 2 + 8cgu 2 = 0.
Solving the quadratic for tan 2 a, we find that under the
radical sign we have the expression w 4  4cgu 2 + 3c 2 <7 2 , that is
(w 2 3 eg} (u 2 eg}. From this we infer that u 2 must be
greater than 3^7, for it cannot be less than eg, as we see
from the first formula in Ex. 22. 25. See Ex. 13.
vii. Lib, * ** 3..
m'tf =Zmv. 6. Let the pressure be p Ibs. : then ~ = .
n
7. Let the pressure be p Ibs. : then ^ = . 8. Take
the unit of mass, then J/=l and W=l ; thusr=l. Let
DYNAMICS. ANSWERS. 397
the unit of time be t seconds; then as^^ is the space
through which a body falls in a unit of time, we must take
the unit of time such that a body should fall through 1 foot
during it. Let t be the number of seconds, then . 32= 1 ;
. 9. 9 feet. 10. ^3 + 1). 12. If r
denote the length of a plane inclined at an angle 6 to the
horizon, we find that r (sin 6  p cos &} must be constant ;
that is rsin($f) must be constant, where /n=tane.
Thus the starting point must be at a constant distance
from a straight line drawn through the origin which makes
an angle e with the horizon. Two such straight lines can
be drawn ; and the required locus is two straight lines
parallel to these respectively.
VIII. 1. s = 25, v=W. 2. 2ut added to the
distance at the time of cutting. 4. p Q* 5 * QP>
7. Three on one side of the Fully, and one on the other
3 9
side. 8. Through  of the given space. 9. ^Ibs.,
21., w 2 m' 2
_lbs. 10. fg.
IX. 2. 11, 13. 3. ^; m'=2m. 7. m'=em.
9 ' n~TT^\ 12  jB'smass= times A' a mass; and so
Jj \jo. + C) 6
on ; e n ~ l u where u is the original velocity of A.
an 2
2 ^ sin cos
X. 1. 45 2. e*h. 3. h + ^. 5. tan 2 a=^^.
1  e 2 " m +m
6. u sin 30. 7. a =45. 8. e"tana. 9.
9 V ~ e )
10. 2 ^ 1 S  P . 12. Let b be the length of the adjacent side ;
the ball must hit this side at the distance from the
398 DYNAMICS. ANSWERS.
end nearest to the opposite side. 14. tan 2 a=e. 16. The
angle AFD must be 90, and the angle DFE must be 135.
18. 4eh sin a cos a where h is the height of the plane,
and a its inclination to the horizon. 19. Let c be the
distance of the wall from the point of projection ; then the
time of motion = and = H ; therefore
g v cos a ev cos a
v 2 sin 2a= qc( 1 +  \ 23. At the foot of the first wall.
XL 1. 4 feet per second. 2. 5 feet per second.
3. (  7 ) at. 4. Let m be the mass of the body hang
\m + mj *
ing over the plane, m' that of the other : then at the end of
the time t, the vertical velocity of the centre of gravity is
ni 2 qt i ,1 i . , i i ., mm' at
.  sZ  7 r . and the horizontal velocity is .  ~r.
(m + m') 2 ' J (m + ra') 2
6. The velocity of the centre of gravity is composed of
m(msinam'sina')^ n i . .1 ,
!  .  7715  ^ parallel to the plane on which is the
(m + m) 2
, , 7 7 , m' (m sin a  m' sin a') at
body of mass m, downwards, and  1  .  ^5  *
(m + m ) 2
parallel to the other plane upwards.
XII. 2. The square of the distance at the time t will
be found to be F>  2uta + a 2 , that is
hence the distance is least at the end of the time .
3. U cos a  u' COS a' + (/cos a f cos a') t.
where a is the inclination of the plane to the horizon;
I _ 4000 + m
XI V. 1. Acceleration . 2. 2n A/  seconds.
DYNAMICS. ANSWERS. 399
3. The point is the centre of gravity of the P Ibs. and the
* *
XV. 1. 8 years. 2.  5 of the moon's period.
(80JP
3. As 1 + e is to 1  e t where e is the excentricity.
XVI. 1. s. 2. TT seconds.
o
XVII. 1. 67200. 2. 2640. 3. 3510. 4. 64.
5. 1 1008. 6. 1 12000 x TT ; for the centre of gravity of the
part removed is at the depth of 10 feet. 7. 499*2.
8. 65. 9. 19200. 10. 5'2 nearly. 11. 25.
12. 7 days, 19 days. 13. 9. 14. 346 4.
. 7nx20xll2. ,. _ lOOOabche
15.  _  (v 2 u?) + mns. 16. ^ ^ .
2ff 16 x 60m
17.  . 18. 1320 feet per minute. 19. 7 '233 nearly.
20. 29120 nearly. 21. 2'3 nearly. 22. 508 nearly.
23. 7500000. 24. About 990.
XIX. 7. ( 2 + 6 2 + c 2 ). 8. (a 2 + 6 2 ) about the
ot> LL.
axis parallel to the edge c.
XX. 1. . , , 3.
MISCELLANEOUS. 1. 2J seconds. 2. 144 feet. 3. 9 '2.
8. A horizontal straight line. 9. g. 12. This may
be deduced from the geometrical fact that the two
tangents to a parabola from any point in the directrix are
at right angles. 13. See Art. 70 ; the difference of the
distances of the focus from the two fixed points is constant.
14. 27. 15. Let T be the tension of the string, ra the
mass of Pj and m' the mass of Q ; the acceleration of P is
mgTj . ~ . 2Tm'(7 .
 downwards, and that of Q is  7^ upwards ;
and at any instant P is moving downwards with twice
400 DYNAMICS. ANSWERS.
the velocity with which Q is moving upwards : thus
<7 11 rrt QmtflfCt T .
; thus T  ,. 17. Ifm=em
m m
then v=u', and v'=ew + (l e)w'; if m'=em then v'=u,
and tf=(le)w + ett'. 18. The blow must communicate
a momentum *J2, times that which the body has, in a
direction making an angle of 135 with that of the original
motion. 1 9. v V(sin 2 a + cos 2 a cos 2 + e 2 cos 2 a sin 2 /3).
20. 5) where u is the original velocity. 21. ^ ,  ,
, /o .. /o . /o
23. In the old directrix.
8 ' 16 ' 32 '
24. We get two expressions for the w r hole time, namely
(!+) (u sin a  u cos a tan /3) and ' : equate them.
g cos 8 ' 1  e ' cos /2 " 1  e ~~ sin/3
27. ^f^'fj. 28.
29< cot(q^) 2(1^
30. cot(a/8) = (2n +
32. The velocity varies as the sixth root of the tonnage.
THE END.
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