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Full text of "Mechanics for beginners"

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 

1-2 




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 



__. 

a-x~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 two-thirds 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 (Oc-ca) + 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 
P-R 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 P-R 
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 ~ R-P 

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 self-evident 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 self-evident 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 self-evident 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 unrBB-f+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 



r-p 



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 Eb-EB 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 W-P: 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 
M-SB-IEZra- 

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. 

C-2 



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 BE-Eb 



q-r 
= - 



therefore 
therefore 



P (r - p} = Q (q-r) ; 



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 right-angled 
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 re-entrant 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 right-angled 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 obtuse-angled 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 crow-bar 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 wheel-barrow in use 
will also serve as an example. A pair of nut-crackers 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 fore-arm 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 
one-twelfth 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 one-sixth 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 6-h 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)-+TP-as 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 scale-pan 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(AG-AC)-, 
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) + ?i-2 .. 

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 
three-fifths 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 Power-wheel 
moment of W round N~ radius of Weight-wheel * 

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 watch-work and clock-work. 

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\ ^ = W-w 
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 left-hand figure, W=4P. The pressure at the 
fixed point K is P, and at the fixed point L is 4P. 

In the right-hand 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 left-hand 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 right-hand 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 right-angled 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 left-hand 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 
Power-arm. 

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 
Power-arm for radius. 

Let r be the radius of the cylinder, b the length of 
the Power-arm, 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 right-angled 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 Power-arm 
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 
Power-arm is n times the radius of the cylinder : find the 
mechanical advantage. 

8. The length of the Power-arm 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 Power-arm 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(rtana-r'tana / ); 

W b 

therefore -7T=-r - 77 - ,. 

P rtana-rtana 

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 th-9 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 2n-c. 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 2n-a 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 one-tenth 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 (/3-0) ' 

for the triangle AC a is similar to the triangle BCb. 

Now when 6 is made indefinitely small the right-hand 
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=KC-Kt. Now if KG and Kt coincided in direction 
we should have KC-Kt=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 Power-arm 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 
Power-arm, estimated on the horizontal plane in which the 
Power is supposed to act originally, is the circumference of 
a circle having the Power-arm 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 Power-arm 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 Power-arm 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), 

F-Ssina=0 (2). 

Take moments round D : thus 

W.DGcos 0=S.DEcos (a - 6\ 

that is TFcos0=2cos(a-0) (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 0-Fcos0=Q (1), 

Rcos0 + Fam0-P-W=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(a-rsin0) = 

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(/3-t) " 

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 
Power-arm, 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), 

ina-n(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 left-hand member of the last equation becomes 

tana. Also cos 0=1, and k~ e =l-6logk very approxi- 
mately, so that we have on the right-hand 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 Q-P 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 



-^ - 5-7-^ 
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=a-6, 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 ; 



13-2 



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 W-Z 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 right-angled 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 three-fifths 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 one-half 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. 

14-2 



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, ...... (ft-l)r, UT 

from starting, is, by Art. 20, respectively, 

/r, 2/r, 3/r, ...... (*l-l)/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 

(n-V}n fi* (n-l)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 air-pump. 

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 is-rfig; 
'2i 2t 

hence during the n ib second the space described is 
\n*g-\(n-V?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= + un-u 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 = 135-6. 



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 + (n-I)fr, u + nfr. 

Let s l denote the space which the body would describe 
if it moved during each interval r with the velocity whi-ch 
it has at the beginning of the interval ; and let s 2 denote 






(n-l)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=u-ft ........................ (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 ut--ft 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=u-2u=-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 AB-AT** 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 a-gt, 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 2a-8=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 (a-ft)_ 
#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 + - (/3-y). 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> . 
\9--l 

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. 

17-2 



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 two-thirds 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 rtig-T 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 

mg-T _T-m'g e 
m m' 

therefore 



m + m'' 
Hence the acceleration is 

2<7?ft' ,1 , . m-ni 

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 T-m'gsma, and the 

, T-m'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 

mgsina-T _ T-m'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 one-fourth 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 one-tenth of a 
second, while gravity in one-tenth 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 self-evident. 

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 

mu-E _ m'u' + E 
m m' 9 

r mm'(u-u') 
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 

_(m-m')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 

m-em' , 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 

18-2 



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'=u-u' -(- + 
\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(u-u'}. 

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, 



v-v'= -e(u-u 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- ^-7-5 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 r-p 

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 
(?i-l) 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 

k-e 

- -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 a-u' 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 1-e 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 re-ascend 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 A-p 

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 mg-kv\ 
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 = (n-l)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-=sin|cos0, 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 0-ff 
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 -- 



. 6-ff B-ff . . 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 (n-l)/3. 

But since sin m/3 = cos f ^ - wi/3 J = cos (n - m) /3, we have 

=l + sm 2 (rc-l)/3 + sm 2 (n-2)/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=n-SY.-.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 twenty-four 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 three-fourths 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 = -r-r ; 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 

-j-e, 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 foot-pound is used in some books in- 
stead of unit of work ; so that foot-pound may be con- 
sidered as an abbreviation for one pound weight raised 
vertically through one foot. 

T. ME. 22 



333 WOfiJt. 

194. The term Horse-Power 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 
water-wheel 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 water-wheel. 

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 one-third 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 
Horse-Power 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 Horse-Power 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 Horse-Power 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 Horse-Power 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 Horse-Power of a water-wheel, 
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 Horse-Power 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, 94-5, 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 Horse-Power 
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 saw-mill 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 fire-engine 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 
mill-pond which is 100 feet long, 50 feet broad, and 3 feet 
deep, and has a fall of 8 feet. 

24. In pile-driving 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 

23-2 



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 (TOX-LOX) = rsm (TOX-a) 
=r sin TO X cos a-r 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 co-ordinates 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 + (A-B} 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 -^(A-B) 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/Y|Y, 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 j-jr. 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 
St-atics 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^ + ^A-r-^ 

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 Q-C=(B-C] 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 co-ordinates are x and y ; it is plain that there 
is also a corresponding particle of mass m at the point of 
which the co-ordinates 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 7-2 

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 =X-r- , 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. 26|cwt. 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 
one-fifth of a side. 5. At the point at which the force 

of 8 Ibs. acts. 6. At the distance of the radius 

n-l 

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 one-sixth 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, 4|feet. 

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. 3|lbs. 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. 480n-lbs. 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 + S-2P, 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 

Msinc-g^ jo Q We must now take 

wcosa 
the lower sign of the preceding result : thus we get 

t= - r-r - ; 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 /? 

- H-TF > 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. f-g. 

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(msina-m'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 10-08. 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 

mg-Tj . ~ . 2T-m'(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=(l-e)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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