B 448096
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7 a Hyett
ЭКИН
ARTES
1837
VERITAS
LIBRARY
SCIENTIA
OF THE
UNIVERSITY OF MICHIGAN
E PLUATOUR US
TUEDOR
SI QUERIS PENINSULAM AMINAM
CIRCUMSPICE
ИНИЦИИ!
PROF.
THE GIFT OF
ALEXANDER ZIWET
1

2.
QA
807
789
1859

7а
Hjelt
Tonbridge Cartle
April 29- 18.63
5286
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AN
1.3.2.2
ELEMENTARY TREATISE
ON
MECHANICS
FOR THE USE OF JUNIOR UNIVERSITY STUDENTS.
BY
RICHARD POTTER, A.M., F.C.P.S.
LATE FELLOW OF QUEENS' COLLEGE, CAMBRIDGE; LICENTIATE OF THE ROYAL COLLEGE
OF PHYSICIANS, LONDON; HONORARY MEMBER OF THE LITERARY AND
PHILOSOPHICAL SOCIETY OF ST. ANDREW'S;
PROFESSOR OF NATURAL PHILOSOPHY AND ASTRONOMY
IN UNIVERSITY COLLEGE, LONDON.
FOURTH EDITION, REVISED.
LONDON:
WALTON AND MABERLY,
UPPER GOWER STREET; AND IVY LANE, PATERNOSTER ROW.
DEIGHTON AND CO., CAMBRIDGE; PARKER, OXFORD.
1859.
L
PRINTED BY TAYLOR AND FRANCIS,
RED LION COURT, FLEET STREET.
Pry. Alex Jewet
1-31-1923
រ
!
PREFACE.
THE present treatise was undertaken to supply a book for the
course of instruction of the Junior Mathematical Class of Natural
Philosophy in University College, London. Although there was
no scarcity of treatises within nearly the same limits as the present,
yet the author had to regret that the students who went forward
into his Senior Mathematical Class had to re-learn the subject in
an entirely different manner; so that their previous reading of it
was in a great measure lost to them.
It has been the author's wish to supply a work which, whilst it
presented to the less advanced student the more modern method
of treating Mechanics, and taught him a general analytical method
of solving the new problems he met with, as far as his mathematical
attainments would reach, should at the same time be an advanta-
geous foundation on which the superstructure of a more advanced
study might be reared.
Some experience in Professorial teaching leads the author to
believe that he has succeeded to some extent in the object which
he had in view; and he concludes that the book which will supply
a desideratum in the Natural Philosophy course in his own lectures
will be also acceptable to other teachers similarly situated.
42-91-01 fi
416653
}
CONTENTS.
INTRODUCTION
•
Definitions-mass-matter, 1; forces, how measured, 2; what is re-
quired that a force may be fully known-tension in cords-reac-
tions in rods-friction, 3: rigid bodies-elastic bodies, 4
•
STATIC S.
CHAPTER I.
ON FORCES WHICH ACT AT THE SAME POINT
Forces which act in the same line, 5; forces which act in different
directions the parallelogram of forces, 7; the triangle of forces,
9; the parallelopiped of forces, 10; the polygon of forces, 11;
graphical problem, 12; the moment of the resultant equals the
sum of the moments of the components, 13; examples, 14.
•
CHAPTER II.
ON FORCES WHOSE DIRECTIONS ARE PARALLEL
The resultant of two parallel forces, 16; any number of parallel
forces, 18; the moment of the resultant equals the sum of the
moments of the components, 20; examples, 22
CHAPTER III.
ON THE THEORY OF COUPLES
A couple may be turned round in its plane, 23; may be removed to
a parallel plane, 24; may be changed to another couple of equal
moment, 25; axis of a couple, 25; composition of couples in one
plane, 26; in different planes, 27
PAGE
1
1
5
15
16
22
23
28
vi
CONTENTS.
CHAPTER IV.
ON ANALYTICAL STATICS OF TWO DIMENSIONS
Any number of forces acting at one point, 29; at different points, 30;
the principle of virtual velocities, 33.
PAGE
29
36
ON THE CENTER OF GRAVITY
CHAPTER V.
37
•
Definition-properties of the center of gravity, 37; stable, unstable,
and neutral equilibrium, 38; nature of the equilibrium when one
body rests on another, 39; conditions that a body may stand or
fall, 40; the center of gravity of any number of particles, 41; of a
straight line—of a triangle, 43; of a parallelogram, 44; of a polygon
—of a triangular pyramid, 45; of any pyramid and cone, 47; of a
frustum of a pyramid, 48; of the perimeter of a triangle, 48; of a
triangular plate, 49; of the surface of a cone, 49.
Examples on the preceding Chapters, 50
· •
•
•
•
•
50
62
CHAPTER VI.
ON THE MECHANICAL POWERS
•
Levers, 63; wheel and axle, 68; toothed wheels, 69; pullies, 71;
inclined plane, 77; wedge, 80; screw, 81...
63
82
CHAPTER VII.
ON COMBINATIONS OF THE MECHANICAL POWERS AND BALANCES.... 83
The mechanical advantages of the simple machines recapitulated, 83;
of combinations of machines, 84; endless screw, 84; a combina-
tion of levers, 85; the knee, 85; properties of the common balance,
87; steelyard balance, 89; bent lever balance, 91; Roberval's
balance, 92...
93
CHAPTER VIII.
APPLICATION OF THE PRINCIPLE OF VIRTUAL VELOCITIES TO THE
MECHANICAL POWERS..
·
Wheel and axle, 94; toothed wheels, 95; pullies, 96; inclined plane,
98; wedge, 99; screw, 99; lever, 100; moveable pulley with
inclined cords, 101
Examples on Chapters VI. VII. and VIII., 102
•
• •
94
102
4
103
CONTENTS.
vii
CHAPTER IX.
ON THE EQUILIBRIUM OF A SYSTEM OF WEIGHTS SUSTAINED BY
CORDS OR BEAMS
Tensions in two cords which sustain a weight, 104; the funicular
polygon, 105; reactions in beams sustaining a weight, 107; the
equilibrium of a framework of beams with weights, 108
•
PAGE
104
109
CHAPTER X.
ON FRICTION
110
Measure of force of friction, 110; coefficient of friction, 111; inclined
plane with friction, 111; screw with friction, 112; Problems,
113
114
DYNAMICS.
CHAPTER I.
ON DEFINITIONS AND THE LAWS OF MOTION
115
Definition, 115; Laws of motion, 118; Parallelogram of velocities,
120; composition and resolution of velocities, 121
122
CHAPTER II.
ON THE IMPACTS OF BODIES
123
Measure of elasticity, 123; table of moduli of elasticity, 125; im-
pact of non-elastic bodies, 126; impact of imperfectly elastic
bodies, 127; motion of the center of gravity before and after
impact, 128; impacts on planes, 130; examples in impacts, 132.. 133
CHAPTER III.
ON UNIFORM ACCELERATING FORCES AND GRAVITY.
The relations of time, space, and force, 134; measure of gravity,
136; bodies projected in the direction of the force, 137; time of
motion, 138; examples on the direct action of gravity, 139......
134
142
ON PROJECTILES
CHAPTER IV.
Form of a path of a projectile, 143; equation of path, 144; time of
flight-range, 145; greatest height, 146; velocity is what would
be acquired in falling from the directrix, 147; range on an in-
clined plane, 147; examples on projectiles, 148.
143
149
viii
CONTENTS.
CHAPTER V.
ON CONSTRAINED MOTION
Motion on an inclined plane, 150; motion down the chords of a
circle, 151; motion of two bodies hanging from the ends of a cord
over a pulley, 152; body falling down a curve, 153; velocity
down a circular arc, 154; time of falling down an arc of a cycloid,
154; centripetal force equals the centrifugal force in a circle, 157;
conical pendulum, 159; curve in a railway, 160; examples on
constrained motion, 160.
•
PAGE
150
162
•
•
ELEMENTARY MECHANICS.
INTRODUCTION.
MECHANICAL Science is that in which the laws of forces, and
the effects they produce on bodies, are investigated.
It is subdivided into four Sciences Statics, Dynamics,
Hydrostatics, and Hydrodynamics.
In Statics the effects of forces on solid bodies at rest are
examined.
In Dynamics the effects when motion is produced.
In Hydrostatics the effects of forces on fluid bodies at rest
are considered.
In Hydrodynamics the effects on fluid bodies when motion
ensues.
In the present treatise, Elementary Statics and Dynamics
only are treated of.
The mass of a body is the quantity of matter which it con-
tains; and matter is defined to be whatever possesses bulk and
affords resistance to the occupation of the same portion of space
by other matter.
We are ignorant of the ultimate nature of matter, but we
know that dense matter consists of atoms*, which have each their
* See Dr. Daubeny on the Atomic Theory.
B
->
INTRODUCTION.
peculiar masses constant and unchangeable by any mechanical
or chemical means within our reach.
The term subtile matter has been applied to the agents which
cause the phenomena of electricity, heat, &c.
Though evidently closely connected with the development of
forces, we as yet only know some of the properties and laws
of the effects of these agents upon dense matter. Whenever
the term matter is used in Mechanics, it is understood to mean
what is called above dense matter. The quantity of matter in a
body is measured by its inaptitude to receive motion (inertia)
when acted on by a given force; and is proportional to the
weight at the same place on the Earth's surface. So that a
body of two, three, &c. pounds weight contains twice, thrice, &c.
respectively the matter that a body of one pound does.
We define force to be whatever causes or tends to cause
motion, or change of motion, in bodies. We see force acting
continually around us, and developed by various means, though
we cannot trace it to its ultimate origin. We measure forces
by their effects; and in Statics they are often called pressures:
being compared with the pressures produced by known weights,
they can thus be expressed numerically. We speak, for in-
stance, of a pressure of twelve ounces, of thirty pounds, of two
tons, &c. when the unit of measure is an ounce, a pound, or a
ton respectively.
In Dynamics, forces are measured in two different manners,
according to the nature of the problem,—namely, in some cases
by the velocity generated in a unit of time, and in other cases
by the momentum (or velocity multiplied into the mass of the
body moved) generated in a unit of time.
In Statics, we continually represent forces by lines of definite
lengths. A unit of length being taken to represent the unit
INTRODUCTION.
3
of pressure, the length of the line represents the magnitude of
the force; its direction represents the direction of the force; and
the commencement or first extreme of the line, the point at
which the force acts, or its point of application. Lines which
are parallel are said to have the same direction.
In order that a statical force may be known, its magnitude,
direction, and point of application must be given.
The weight of bodies, being their gravitation vertically down-
wards, arises from the attraction of the Earth upon them, ac-
cording to the laws of universal gravitation. These laws we
shall have to consider in the science of Dynamics.
In statical problems we frequently have other forces arising
from the effects of the original forces, which have to be consi-
dered in the same manner,-as the tension in cords, and the
reaction in rods. Unless the contrary is stated, the cord is sup-
posed without weight, perfectly flexible, and to pass perfectly
freely round any object which changes its direction: the force
applied at one extremity must then be transmitted without loss
along the whole length of the cord, or the tension in the cord is
the same in every part. Unless the contrary is stated, the rods
are supposed to be inflexible and without weight; so that a
straight rod transmits a force applied at one extremity in the
direction of its length, to the other extremity unchanged: thus
the reaction of the rod equals the original force, and may be
supposed to act at any point in it.
In other statical problems there arise forces of a different
nature to the original forces, and which therefore have to be
considered differently,-as the friction, which arises from the
roughness of the surfaces of bodies in contact, and the adhesion,
which arises when one of the surfaces, at least, is of an adhesive
nature. The laws of friction have been ascertained, and will be
treated of in a distinct chapter. The properties of adhesive sur-
B 2
4
INTRODUCTION.
faces are of less theoretical importance, and are seldom discussed
in treatises on Mechanics.
By a particle of a body, we mean a portion of it whose di-
mensions are smaller than any possible means of measurement.
By a rigid body, we mean one in which the relative positions
of its particles remain unchanged.
In some statical problems the
elastic bodies have to be considered.
properties of flexible and
In these the relative posi-
tions of the particles change by the action of the forces.
STATICS.
CHAPTER I.
ON FORCES WHICH ACT AT THE SAME POINT.
1. A force acting at any point is balanced by an equal force
acting at the same point in an opposite direction. It is clear
that this must be true; for whatever tendency to motion the
point might receive from one of the forces, it would receive an
equal and opposite tendency from the other; and these would
neutralize each other.
2. If two forces be in equilibrium at a point, they must be
equal in magnitude and opposite in direction. If possible, let
the forces P and Q', acting in the Q'
directions of the arrows, as in the
figure, keep the point A at rest.

Let Q be a force equal and oppo- Q
A
P
site to P; P and Q will balance, and therefore Q produces the
same tendency to motion that Q', a different force, does, which is
absurd.
DEFINITION. The resultant of two or more forces is the
single force which produces the same mechanical effect as the
forces themselves, which are called the component forces.
3. When any number of forces act at a point in the same
straight line, the resultant equals the algebraic sum of the com-
ponent forces.
First, let all the forces act in the same direction, as in the
figure. Let the line AB
represent the force P₁.
If the point B be rigidly
C
D P₁
P2
P 3
A
B
connected with A, we may suppose the force P₂ to act at B, and
2
it will produce the same effect as if it acted directly at A. Let
6
ELEMENTARY MECHANICS.
1
2
BC represent the magnitude of P. The line AC represents
a force equivalent to P, and P₂ acting at A; or, their resultant
equals their sum. If we had a third force, P3, represented by
CD, we should have the resultant of P₁, P2, and P, a force
represented by AD, their sum as before; and so onwards for
any number of forces.
When some of the forces act in the contrary direction, we
must take the lines which represent their magnitudes on the
contrary side of A, and subtract them from the sum of the other
forces. Let P be repre-
sented by the line AB,
and P', a force acting in
P/
A
+
b'
B
P
the opposite direction, by Ab; we may suppose P' to act at B,
and measuring Bb' equal to Ab, we have A b', a line representing
the resultant of P and P'. So that, having taken the sum of
the forces acting in one direction and the sum of those acting
in the contrary direction, the difference of these sums is the re-
sultant force, and it acts in the direction of the forces which
form the larger sum. This is equivalent to calling the forces in
one direction positive, and those in the contrary direction nega-
tive; and then their algebraic sum gives the magnitude of the
resultant, and its sign determines its direction. When the
resultant = 0, the forces balance; and the condition of equili-
brium of forces acting in the same straight line is, that their
algebraic sum = 0.
Hence we may add equal and opposite forces at any point
without affecting a system of statical forces; this is called the
superposition of equilibrium. In the same way we may remove
from any point in a system those forces which are equal and
opposite. In finding the resultant of given forces, we are said
to compound them; and when we find the components of a given
force, we are said to resolve it into its components.
THE PARALLELOGRAM OF FORCES.
4. PROP. When two forces act at a point, if the parallelogram
upon the lines representing them be completed, the diagonal from
that point represents their resultant in magnitude and direction.
First, to prove that the diagonal is the direction of the result-
STATICS.
7
ant. If the forces are equal, as P, P in the
figure, this is evidently true; for no reason
can be assigned why it should incline more
to one force than the other, and the diagonal
bisects the angle A in which the forces meet.
Let us assume that this holds good for
forces Ρ and q, and Ρ and r, we show that it
must also be true for a force p and a force
q+r.
Ρ
Let and
q act at A, A
as in the figure, in the
directions AD and AB,
and be represented in
magnitude by these lines.
P
A

R
4
B
7"
C
respectively. We may
consider the forcer to be D
E
P

F
applied at B, a point in its direction; let BC represent its
magnitude. Complete the parallelogram ADFC, and draw BE
parallel to AD. Draw the diagonals A E, AF, B F.
Then, p at A in AD, and q at A in AB, are equivalent to
a force in AE which we may suppose to act at the point E.
Resolving it again, parallel to the original directions, we have a
force p acting at E, in the direction BE, and a force q acting at
E in EF. We may suppose them to act at the points B and F
in their directions; but p at B in BE, and r at B in BC, are
equivalent to a force in BF. Therefore we may suppose all the
forces to act at the point F, parallel to their original directions,
and F must be a point in their resultant; or force p at A in
AD, and forces q and r in 4 C, have their resultant in the direc-
tion of the diagonal AF.
Now, our assumption is true when q and are each equal to
p, therefore the proposition for the direction of the resultant is
true for forces p and 2p: again, putting q=2p, r=p, it is
true for forces p and 3p, and so for p and up; also, putting up
P, and
q=p, r=p, it is true for np and 2p, and so onwards
for
ELEMENTARY MECHANICS.
for all commensurable forces. We
see also in the annexed figure,
that the resultant lies nearer to
the greater than the weaker force;
and if one of the two forces be increased, the resultant lies still
nearer to the increased force.

G
B
E
F
When the forces are incommensurable, the proposition still
holds good.
Let the lines AB, AC A
represent the incommensurable forces:
complete the parallelogram, and draw
the diagonal AD; then AD represents
the direction of the resultant. If not,
let some other line, as AE, be its di-
rection. Take a quantity which di-
vides A B without remainder, and being
applied to A C, leaves a remainder G C
less than ED. Complete the parallelogram ABFG, and draw
the diagonal AF. Now AB, AG represent commensurable
forces, and therefore their resultant is in the direction AF; but
AE, the resultant of AB and greater force A C, is nearer AB
than the resultant of A B and the less force A G, which is impos-
sible. Similarly it can be proved that no other direction than
AD can be that of the resultant.

Р
C
D
E
D
Secondly. To prove that the diagonal of the parallelogram
represents the magni-
tude of the resultant
also, when the sides
respectively represent
the component forces.
Let P, Q, R, acting in
the directions of the

C
arrows, as in the figure, a
keep the point A at
B
R
A
F
rest. Let the lines AD, AB, AF represent respectively the
forces. Complete the parallelograms A C, AE, and draw the
diagonals. The resultant of any two of the forces must be
equal in magnitude, and opposite in direction, to the third
force, since there is equilibrium: therefore CAF, BAE are
STATICS.
9
straight lines. AE is parallel to CD, and AC is parallel to
DE; and ACDE is a parallelogram, of which the side AC is
equal to the side DE. But DE by construction is equal to
AF; therefore, AD and AB representing the forces P and Q,
A C represents a force equal in magnitude to the third force R,
and opposite in direction, and thus represents the resultant of
P and Q in magnitude as well as direction.
5. PROP. If three forces acting at a point keep it in equili-
brium, and a triangle be formed by three lines drawn in their di-
rections, the sides of the triangle, taken in order, will represent the
forces. Conversely, if three forces which act at a point be repre-
sented by the sides of a triangle, taken in order, they will be in
equilibrium. This proposition is called the triangle of forces.
P
If the three forces, P, Q, R, in equilibrium, act at the
point A, as in the figure,
the triangle ABC, which
is half the parallelogram
formed on the lines re-
presenting P and Q, will
have its side BA repre-
senting the force R, by
taking the sides in order,

as shown by the direction
A
B
R
Q
of the arrows; for AB, when taken in the opposite direction,
represents the resultant of P and Q.
Conversely, If the three forces were represented by the sides
of the triangle ABC taken in order, we might form a parallelo-
gram with any one of the sides,
B
D

as BC, for its diagonal, and
the resultant of the other two
forces, represented by CA and
CD (or AB), would be repre-
A
C
sented by this diagonal taken in the opposite direction, which
would make equilibrium with the third force.
By means of this proposition we can resolve a given force into
two others which are equivalent to it, in any given directions.
10
ELEMENTARY MECHANICS.
6. PROP. If three forces acting at a point are in equilibrium,
they are proportional each to the sine of the angle contained
between the other two.
In the figure the sides of the
triangle ABC represent the forces
P, Q, R, respectively; or,
P:Q:R::AC: CB:BA
:: sin ABC: sin BAC: sin ACB
:: sin BAQ: sin BAC: sin PAQ
:: sin QAR : sin PAR: sin PAQ R

Since the sines of the angles equal
the sines of their supplements.
A
C
7. PROP. If P and Q be two forces which act at a point, the
angle between their directions being 0, then if R equals their re-
sultant, we have R2P2+Q2+2. P. Q cosine 0.
By trigonometry we have in triangle ABC, art. 6,
A B² = A C² + BC2-2A C. BC. cos ACB
and
cos ACB= cos BCP=— cos 0.
... R² = P² + Q²+2PQ cos 0.
8. PROP. If three forces acting at a point in different planes
be represented in direction and magnitude by the three edges of
a parallelopiped, then the diagonal will represent their resultant
in direction and magnitude; and reciprocally, if the diagonal re-
presents a force, it is equivalent to three forces represented by the
edges of the parallelopiped.
B
F
Let the three edges AB, AC, AD of the parallelopiped in
the figure represent the three
forces. Then AE, the diagonal
of the face ACED, represents
the resultant of the forces AC
and AD. Compounding this with A
the third force represented by 4B,
we have AF, the diagonal of the
D
C
E
P
B

STATICS.
11
parallelogram AEFB, representing the resultant of A E and A B,
or of the forces A C, AD, A B.
Reciprocally, the force AF is equivalent to the components
A B, AE, or to the component forces, A B, A C, and A D.
9. PROP. If any number of forces are represented in direc-
tion and magnitude by the sides of a polygon taken in order, they
will, when applied at one point, produce equilibrium.

Let P₁, P2, P3,
P4 P5, be forces
acting at the point
A, as in the figure.
P
A
B
P₁
·P.
C
D
Pi
P₂
E
Let P₁ be represented by A B.
ور
رو
P 2
>>
JJ
the parallel line B C.
C D.
DE.
E A.
""
P3
4
P₁
P₂
>>
ور
وو
Or, let the forces be represented in magnitude and direction by
the sides of the polygon ABCDE, taken in order. If we com-
plete the parallelograms AC, AD, AE, and draw the diagonals
A C, AD, we see that AC represents the resultant of forces P₁
and P₂; compounding this resultant with the force Pg, we see
that AD represents their resultant, or the resultant of P1, P2
and P3; compounding this last resultant with the force P, we
see that their resultant is represented by the line A E, acting in
the direction from A to E, which would consequently balance
the last force P5, represented by the last side E A of the poly-
gon, and acting in the direction from E to A. It will be seen
from the proof that it is not necessary the forces should lie all in
one plane.
This proposition is called the polygon of forces.
12
ELEMENTARY MECHANICS.
When three forces act in equilibrium at a point, any three
lines taken parallel to their directions will form a triangle, the
sides of which respectively will represent the forces; but when
there are four or more forces, this will not hold, because the
relation which subsists between the sides and angles of triangles
does not hold in four-sided figures or polygons.
For instance, in the figures, if CB and cb be parallel, and
also FG and
fg, the triangles
g

G
II

ABC, Abc, being
C
similar and si-
milarly situated,
the sides being
A
D
B
b
F
E
respectively proportional, would represent the same three forces;
but, although the sides of the polygon DEFGH might repre-
sent a system of five forces in direction and magnitude, yet they
could not be represented by the sides of the polygon DEfgH in
magnitude also.
10. When any number of given forces act at given points
a plane, we may find, graphically, the magnitude and direction
of the resultant of the system.
Let P1, P2, P3, in the figure, have their points of application
A, B, C. Producing the directions
of the forces P₁ and P₂, until they
meet at a point a, we form the
parallelogram ab upon the lines re-
presenting them, and the diagonal
is their resultant R, in the figure.
Producing R₁, until it meets at c
the direction of the force P3, and
drawing the parallelogram cd on
the lines representing R, and P3,
PI
C

C
P3
d
B
R 2
P.
R₁
we have the diagonal representing R₂ their resultant, or the
resultant of P₁, P2, and Pg. By pursuing the same method we
may find the resultant for any number of forces.
DEFINITION. The moment of a force about any point is
the product of the force into the perpendicular let fall from the
STATICS.
13
point upon the direction of the force. The moment, as we shall
see in the next chapter, measures the tendency of the force to
produce rotatory motion about the fixed point.
11. PROP. The moment of the resultant about any point in
the plane of the forces equals the sum of the moments of the
forces.
A
Let the forces P and Q, act-
ing at A, be represented by the
lines AB, A C, and their result-
ant R by AD, the diagonal of
the parallelogram drawn upon o
AB, AC. Let O be the point
about which the moments are
taken; join O A, and draw AEFG
E
ni
n
B
F
G

H
R
D
P
perpendicular to OA; draw 01, 0m, On, respectively perpendi-
cular to AB, AC, AD; and BE, CF, DG, perpendiculars to
AEFG; and CH parallel to that line.
Now the triangles OlA, OmA, OnA are respectively similar
to the triangles AEB, AFC, AGD.
Whence
AE οι
ΟΙ
A E =
AB OA
AB.Ol
O A
AF Om
AC.Om
or
AF=
ACTO A
O A
AG On
AD. On
or
AG=
AD O A
OA
AB.O1+AC. Om
AD. On;
but
or
or
AECH FG, . AF+AE= AG;
P.Ol+Q.Om R. On.
Or the sum of the moments of the components equal the mo-
ment of the resultant about any point in their plane, or about
an axis perpendicular to their plane.
If the point O fell within the angle formed by the forces, we
should have the moment of one of the forces tending to cause ro-
14
ELEMENTARY MECHANICS.
tation in the opposite
direction to the other,
and it must then be
considered as negative
B
E
A
711
if the other were po-
sitive, and the alge-
braic sum of the mo-
ments of the compo-
P
G
F

LI
C
Q
D
nents still equals that
R
of the resultant. In the annexed figure we have the proof the
same as before, except that A G = AF AE,
and
or
AC. Om AB. 01 AD. On,
=
Q.Om-P.01 R.On.
=
By compounding R with another force acting at A, we should
obtain a like result: or the moment of the resultant of three
forces acting at A cquals the algebraic sum of the moments of
the forces. The proposition may, in the same manner, be ex-
tended to any number of forces acting at a point.
EXAMPLES.
1. Show that if be the angle between two forces of given
magnitudes, their resultant is the greatest when 0, least
when =π, and intermediate for intermediate values of 0. If
the component forces be P and Q, what is the magnitude of the
resultant when 0, and also when 0 = π? Ans. (P+Q)
and (P-Q).
2. If two equal forces (P) meet at an angle of 60°, show that
their resultant = P3.
3. If two equal forces meet at an angle of 135°, show that
their resultant =P (2 — √/2)*.
4. If three forces, whose magnitudes are 3m, 4m, and 5m, act
at one point and are in equilibrium, show that the forces 3m
and 4m are at right angles to each other.
STATICS.
15
5. If two equal forces are inclined to each other at an angle
of 120°, show that their resultant is equal to either of them.
6. If the magnitudes of two forces are 6 and 11, and the
angle between their directions 30°, show that the magnitude of
their resultant is 16.47 nearly.
7. Show that, in the last question, the resultant makes with
the force 6 the angle whose sine is 3339, and with the force 11
the angle whose sine is 1821, which are the sines of 19° 30′ and
10° 30′ nearly.
8. Apply the proof of the polygon of forces to the case of five
equal forces represented by the sides of a regular pentagon taken
in order.
9. Enunciate all the propositions requisite to prove that the
resultant is in every respect mechanically equivalent to the com-
ponent forces.
10. A cord PAQ is tied round a pin at the fixed point d,
and its two ends are drawn in different directions by the forces
P and Q show that the angle between these directions is found
3 (P² + Q²) - 2 PQ
:
from the expression cos 0=—
when the
8 P Q
pressure on the pin is equal to
P+Q
2
11. A cord whose length is 27 is tied at the points A and B
in the same horizontal line, whose distance is 2 a: a smooth
ring upon the cord sustains a weight w: show that the force of
tension in the cord =
น
1
12
CHAPTER II.
ON FORCES WHOSE DIRECTIONS ARE PARALLEL.
THOUGH the propositions in the last chapter will not apply at
once to forces acting at different points, of which the directions
are parallel, yet we can reduce the proof of the method of find-
ing their resultant to that of two forces acting at one point in
different directions.
12. PROP. If two parallel forces P and Q act at points A
and B respectively, then their resultant equals their algebraic
sum in magnitude, and acts at a point C in the same straight
line with A and B, such that P× AC=Q × B C.
Let the forces P
and Q act as in the
figures at A and B ;
at these points apply
equal and opposite
forces, Sat A and S'
at B; they will not
affect the system. P
and S at A will have
a resultant in
the direction
DA; Q and
S' at B will
have a result-
ant in the di- s
rection DB;
and in the
S
A
D

S
S
B
S'
C
P
R=P+Q
Q
R=Q-P

D
S
S'
A
S'
B
lower figure,
where
the
P
forces act in opposite directions, we suppose Q greater than P,
so that the resultant of Q and S will lie nearer Q than the re-
STATICS.
17
sultant of P and S does to P; and therefore the directions of
the resultants will meet at some point, as D, in both figures.
We may suppose the whole of the forces to act at the point D.
Resolving the forces parallel to the original directions, we shall
have forces S and S' parallel to AB, which will destroy each
other; and forces P and Q acting parallel to their original direc-
tions, giving a resultant R=P+Q in the upper figure, and
R=Q-P in the lower figure. To find the point C in the line
AB, or AB produced, where R acts, we have,
from triangle ACD,
P:S:: CD: A C,
from triangle BCD,
S': Q :: BC: CD.
Compounding these ratios, we have
or
P: Q:: BC: AC,
PxAC=Q x B C.
13. If AB be perpendi-
cular to the direction of the
forces, A C and B C are called
the arms of the forces, and
the products P.AC, Q.BC
are the moments of the forces,
about the point C. If the
forces are inclined at an an-
gle a to AB, which is angle
A
M

Q
C
N
PAC or QBC in the figure, then the perpendicular line MCN
being drawn through C on the lines of action of the forces, we
have CMA C. sin a, and CN=BC. sin a, and the moments
are P× CM and Q× CN, or P. A C. sin a and Q . BC . sin a.
α
14. If C be a fixed point, its resistance will destroy the
effect of the resultant force; so that P and Q will be in equili-
brium about such a point, when their moments, tending to cause
rotation opposite ways about it, are equal to each other.
15. To find A C in terms of A B, we have
P × A C= Q (A B— A C)
in the upper figure;
AB; and similarly
and similarly
BC=
P
P+Q
AB,
in the lower figure;
Q
P+Q
P× AC=Q(A C— A B)
ΟΙ
A C=
and
C
18
ELEMENTARY MECHANICS.
or
Q
P
A C= AB; and similarly BC= A B.
Q-P
Q-P
The point C is determined in both cases; unless in the latter
P=Q, when A C=infinity; but then the resultant R=Q—P=0.
This is a peculiar case, the effect of two equal and parallel
forces which do not act at the same point being to produce
rotatory motion only. Such forces constitute what is called a
statical couple; and all tendency to rotatory motion can be re-
ferred to forces forming such couples. If the forces are inclined
at an angle ɑ to A B, then P. AB. sin a is the moment of the
couple.
16. PROP. To find the resultant of any number of parallel
forces which act at any points in one plane.
First, let the parallel forces
P1, P2, P3, &c. .
Pm, have
n
their points of application A,
B, D, &c. and act towards the
same part. Take any two lines,
Ox, Oy, at right angles to each
other; join AB, and let C be
the point of application of the
resultant, R₁, of P, and P2.
Then
Ri
P3

P₁ R2
У
Pi
D
B
E
C
A
C
R₁=P₁+P2,
and
P₁ × A C=P₂ × B C.
O F
H
G
L
K
Draw the lines AF, BG, CH, DK parallel to Oy; and Ac,
Cb parallel to Ox. The triangles ACc, CBb are similar, and
or
1
or
AC BC
Cc Bbi
.. P₁ × Cc=P₂× Bb ;
2
P₁× (CH-AF) = P₂ × (B G—CH)
(P₁+P₂) × CH=P₁ × A F+ P₂× B G ;
2
R₁× CH=P₁× AF+ P₂x BG.
2
Taking another force, Pg, and compounding with R₁ acting
at C, we find the second resultant, R½=R₁+P3=P₁+P₂+P3,
and R₂ × EL=R₁ × CH+P¸ × D K
3
2
=P₁ × A F+ P₂ × B G+P¸× DK,
2
and so onwards for any number of forces.
2
STATICS.
19
If we put
AF=y, BG=Y₂, DK=y, &c.,
OF=x, OG=x2,
or if x₁y1, x2Y2, X3Y3, &c.
OK=x3, &c.,
Xnyn are the co-ordinates of
the points A, B, D, &c., the points of application of the n
forces, the above formula becomes
R₂ × EL=P₁·y₁ + P₂• Y₂+P3 • Y8;
1
2
and if x, ÿ were the co-ordinates of the point of application of
R, the resultant of the n forces, we should have
and
R=P₁+P₂+P3+ &c. . . . +Pn
R.ÿ=P₁.Y₁+P₂· Y₂+ P3.Y3+ &c. . . .
1 1
2 2
+Pn⋅yn
If we had drawn lines parallel to Ox, we should have found,
in the same way,
R.ã=P₁.x₁+P₂.≈2+P3.X3+&c. ... +Pr.x,
n.
These formulæ are often written more concisely by using
the Greek letter E as the sign of summation; and P being any
one of the forces, x, y being the co-ordinates of its point of
application, then
R.x=(P.x),
R.y=Σ(P.y),
R=Σ(P).
The point whose co-ordinates are x, y, is called the center of
parallel forces. It depends on the magnitude of the forces and
their points of application; but is independent of the angle
which their direction makes with any given line.
Secondly. When some of the forces act in opposite direc-
tions, they must be taken negative; and so also, when the co-
ordinates of the points of application are negative, they must be
applied with their proper signs; and then the above formulæ
will apply to all cases.
If R₁, the resultant of P, and "
P₂, as found in the previous case,
be compounded with a force P3
acting as in the figure at D, by
joining C D, and producing it in
the direction of the greater force,
say R₁, we have R, the second
resultant R₁,- P3; and E being
its point of application,
=
R₁ × E C=P¸× ED.
A
Pi
P
D

R 1
R₂
B
P3
d
સ
E
O F
L
H G
K
21
د
}
J
c 2
20
ELEMENTARY MECHANICS.
Drawing Ce, Ed parallel to Ox and meeting DK in d, EL
produced in e, the triangles ECe, EDd are similar, and
or
or
or
EC DE
Ee Dd ;
.. R₁× Ee=P¸× D d,
3
R₁(CH—EL)=P¸(DK—E L),
(R₁— P¸). EL=R₁ × CH−P¸× DK,
3
R₁₂ × E L=P₁× A F+ P₂ × BG-P¸× DK;
and so for any other forces act-
ing in the opposite direction to
P, and P₂, &c.
1
Again. Let y₂, the ordinate
of the point of application of
P₂, be negative=B G in figure;
draw A a, Cb parallel to Ox,
P₁× AC=P₂ × CB,
1
2
or P₁ × Ca=P₂x Bb by similar
triangles,
2
ย
3

P₁
RI
A
a
b
G
P
H
Cz
P₁(A F—CH) = P₂(CH+B G)
(P₁+P₂)CH=P₁ × A F— P₂× B G,
2'
2
B
P2
8
or R₁.y=Py₁-P2.2, as we should have found by putting
Y2 with its proper sign in the general formula, which thus ap-
plies generally to all cases of parallel forces.
When some of the forces are negative and others positive,
we may have the sum Σ(P)=0,
or
R=0,
and the system of forces may be equivalent to a couple.
17. PROP. The algebraic sum of the moments of any number
of parallel forces which act in one plane about any point in the
plane, equals the moment of their resultant about that point.
Let P1, P2, P3 be parallel forces acting at the points A, B,
D respectively; R₁ and R₂ the resultants, as before; G the
2
STATICS.
21
point about which
the moments are
taken, and called
the center of mo-
ments.
Draw Gdef and
Acb perpendicu-
lars to the direc-
tion of the forces
in both figures. In
the first figure both
1
2
P₁ and P₂ will tend to
cause rotation the same
way round G; but in the
other figure they tend op-
posite ways, and must
therefore be taken with
different signs.
In the first figure, the
sum of the moments of P₁
2
A
A
d
P₁
and P₂ about G=P₁× Gd+P₂× Gƒ
=P₁(Ge-de) +P₂(Ge+ef)
e
R₂

P2
R₁
P₂
C
B
с
G
e
03
R₁
b
C
B
b
E
D

R₁₂
P₂
E
D
Ps
(P₁+P₂)Ge+P₂ × cb-P₁ Ac... since c bef,de=Ac
=R₁ × Ge+ (P₂ × C B− P₁ × A C) cos BAb
1
=R₁x Ge=moment of the resultant about G,
since P₁× A C=P₂× C B when C is the point of application of
the resultant.
1
2
In the second figure, the algebraic sum of the moments about G
=P₂× Gf-P₁× Gd
2
=P₂(Ge+ef)— P₁(de— Ge)
= (P₁+P₂) Ge+ P₂ xef-P₁x de
2
=R₁× Ge . . . as before, since P₂ xef=P₁ × de.
If we take another force, Pg, we find, in the same way, the
moment of the second resultant, R₂, equals the algebraic sum of
the moments of R, and P3, or of P1, P2, and P, and so for any
number of forces.
22
ELEMENTARY MECHANICS.
EXAMPLES.
1. Two parallel forces acting in the same direction have
their magnitudes 5 and 13, and their points of application 6 feet
distant; show that their resultant acts at a point 43 feet from the
point of application of the force 5, and 12 feet from that of the
force 13. What is the magnitude of the resultant?
2. If the forces in the last question act in opposite directions,
show that the point of application of their resultant is distant
33 feet from the point of application of force 13, and 93 feet
from that of force 5. What is the magnitude of the resultant?
3. If two parallel forces, P and Q, act in the same direction
at the points A and B, and make an angle with the line AB,
show that the moment of each of them about the point of appli-
cation of their resultant =
P.Q
P+ Q
A B sin 0.
4. If three forces which act at a point be represented in
direction and magnitude by the sides of a triangle taken in
order, they will make equilibrium; show that if, instead of act-
ing at one point, they each act in the line which is the side of
the triangle representing it, they are equivalent to a statical
couple.
5. If three equal parallel forces act at the three angles of
an equilateral triangle, show that their center, or, the point of appli-
cation of their resultant, is in the line drawn from any angle to
the middle of the opposite side, and at the length of the line
measured from the angle,-being independent of the angle which
the forces make with the plane of the triangle.
R
6. In question 4, if a, b, c be the sides of the triangle op-
posite to the angles A, B, C respectively, then the moment of
the couple equals ab sin C-a c sin B-be sin A.
CHAPTER III.
ON THE THEORY OF COUPLES.
WE saw in the last chapter that two equal and parallel forces
acting in opposite directions and at different points of a body
had no single resultant, but constituted a statical couple, tend-
ing to produce rotatory motion. This tendency can be balanced
only by an equal and opposite tendency produced by an oppo-
site couple. Statical couples have peculiar properties, which we
will now discuss and chiefly by employing the superposition of
equilibrium. See page 6.
18. PROP. A couple may be turned round in any manner in
its own plane without altering its statical effect.
2
5
=
6
Ps
P2
P
b
E
P
B
a
D
Let P₁ABP, be the ori-
ginal couple; take ab AB,
and turned round any point C;
apply equal and opposite forces,
PP perpendicular to ab at
a; and similarly P, and P at
b; these will not affect the
system, being in equilibrium
amongst themselves: let each
of them equal P, or P2. Then
P₁ at A, and P4 at a, are equi-
valent to a force bisecting the angle P,EP, between them, or a
force in CE; similarly, P, and P are equivalent to an equal
force in CD. These forces being equal and opposite may be
removed; that is, we may remove from the system the forces
P1, P2, P4, P6, and we have remaining the forces P, and P, at
a and b, forming the couple Pab B5, which is the same as if we
had turned the original couple round the point C until its arm
came to the position a b.

1
2
Ps
5
24
ELEMENTARY MECHANICS.
19. PROP. A couple may be removed to any other part of its
own plane, its arm remaining parallel to the original direction;
and it may be removed to any other plane, in the body on which
it acts, parallel to its own plane, its arm being parallel to the
original arm.
2
First. Let the arm AB of the
original couple P₁ABP₂ be re-
moved in its own plane to the
parallel position ab; let forces
P3, P4 P5 P6, each equal to the
original forces, be applied per-
pendicularly to ab at the extre-
mities a and b in opposite pairs,
as in the figure.
Join Ab and a B, these lines
will bisect each other in C; and
P₁ at A, and P¸ at b, are equiva-

2
5
1
4
a
P4
VP₁
A
PIV
CA
P
6
b
P₂
B
2
lent to a force =2P₁ at C, parallel to the original direction;
similarly, P₂ at B, and P₁ at a, are equivalent to a force =2P₂
at C, opposite to the former; these will consequently balance
each other, and may be removed, or the forces P1, P2 P4 P5
may be removed, and we have remaining the couple Pab Po
equivalent to the original couple removed parallel to itself in
its own plane.
1
D
A
B
P4
C
F
PG
Secondly. Let the arm A B of the original couple be re-
moved from its own plane
DE, parallel to itself, to a b
in the parallel plane FG;
let equal forces, each equal
to P₁ or P2, be applied at a
and b, as in the former case.
Join Ab and Ba; these lines
will bisect each other in C.
The forces P, and Ps will be
equivalent to a force =2P, at
C, parallel to the original di-

1
5
Pi
a
E
4
G
P
STATICS.
25
4
rection, and P2 and P will be equivalent to an equal and oppo-
site force at the same point; these equal and opposite forces at
C may be removed, and there remains the couple Pab P6, equi-
valent to the original couple removed to the plane FG.
20. PROP. All statical couples are equivalent to each other,
whose planes are parallel and moments equal.
DEFINITION. The moment of a couple is the product of one
of the forces into the arm, or P. AB in the foregoing proposi-
tions.
2
Let P, ABP, be the original couple,
whose moment is P. A B.
Produce
A B to any point C, and apply there
equal and opposite forces Q, and Q₂,
such that Q.A C=P.A B. We may
suppose P₁ at A to be the sum of
two forces, P' and Q₁; now
P₁AB=Q₁. A C=Q₁(A B+B C) ;
.. (P₁-Q₁)A B=Q₁. BC
1
=P'. AB;
P₂

A
B
C
P₁ = P'+Q,
and forces P' at A, and Q, at C, have a resultant parallel to
their directions and equal to their sum at B. This force,
P'+Q=P₁, will balance P₂, and therefore P' at A, Q, at C,
and P₂ at B may be removed; and there remains the couple
Q. A C, which is therefore equivalent to the original couple.
By the previous propositions this couple, Q. AC, may be
removed into any plane parallel to its own, and turned round in
any manner in that plane.
DEFINITION. The axis of a couple is a line perpendicular
to the plane of the couple; and its length being taken propor-
tional to the moment of the couple, represents it in magnitude.
The tendency to rotatory motion being round the axis, and the
length of the axis representing this tendency as measured by
the moment, the axis represents completely the couple. The
effect of the previous propositions is consequently this,-that the
26
ELEMENTARY MECHANICS.
axis, of fixed length, may be removed any where within the body
acted on, parallel to itself.
When any number of couples act upon the body, they can
be compounded into one resultant couple.
21. PROP. When any number of couples act upon a body in
parallel planes, the moment of the resultant couple equals the
sum of the moments of the component couples.
Let P, Q, R, &c. be the forces; a
b, c, &c. their arms respectively; the
couples can be removed all into one
plane, turned round, and moved in that
plane, and their arms changed to a
common arm, whilst their moments re-
main unchanged. Let m be the com- P'
mon arm AB; P', Q', R', &c. the
forces; so that P'.m=P. a, Q'.m=Q.b, q'\
R'.m=R. c, &c.; but the forces P', Q', RV
R', &c. at A are equivalent to a force

P.a Q.b
P'+Q'+R'+ &c.= +
m
A
B
R.c
+
+ &c.
m
m
And similarly, the forces at B are equal to
the same sum; and the moment of the re-
sultant couple (P' + Q'+R'+ &c.) AB
=
=(P'+Q'+R'+ &c.) m
= Pa+Qb+Rc+&c.
R'
P/
O
K
L
M
P
This is the same thing as taking the algebraic sum of the
axes, as OK, KL, LM, &c., for the resultant axis, when the
component axes are parallel. If any of the couples tend to
cause rotation the contrary way round,
we must take them with contrary
signs, or their axes must have been
measured in the opposite direction
from 0. An axis is therefore ba-
lanced by an equal and opposite axis,
or a couple by an equal and oppo-
site couple. If PABP, QDEQ were
couples whose moments were equal and

P
A
B
E
STATICS.
27
opposite, or P.AB=-Q.DE, they would evidently make
equilibrium with each other.
22. PROP. If two sides of a parallelogram represent the axes
of two component couples, the diagonal represents the axis of the
resultant couple.
Let OK, OL be the axes
of the component couples,
then OM, the diagonal of
the parallelogram formed on
them, represents the axis of
a couple equivalent to them.
The planes of the couples
being perpendicular to their
axes respectively, will be in-
clined at the same angle to
P
R

A
B
K
R
L
M
each other as the axes themselves are. The couples can be
moved and turned round, each in their own plane, until their
arms are in the intersection of their planes; and their moments
being kept the same, they can be brought to have the same arm.
Let A B be this common arm in the intersection of the planes;
PABP, QABQ the couples. Completing the parallelograms
on the lines representing P and Q respectively, the diagonal
represents their resultants R, at A and B, and the two couples
are equivalent to a couple RABR.
If be the angle between P and Q, or between OK and OL,
R²= P²+ Q²+2PQ cos 0... from the triangle of forces;
:. R.AB=AB.√ P²+Q² +2 P Q cos 0
√ P² . A B² + Q² . A B² + 2 P.AB× Q.A B cos 0
= √OK²+O L² + 2 OK. OL cos 0
=0M.
And OK, OL being respectively perpendicular to the planes of
the couples PABP, QABQ, we have O M perpendicular to
the plane of the resultant couple RABR; therefore OM repre-
sents the axis of the resultant couple. Let L and M be the
28
ELEMENTARY MECHANICS.
moments of the component couples, G the moment of their re-
sultant; then
G²=L2+ M²+2L. M. cos 0.
COR. If L, M, N were the axes of three component couples,
we could show, as in article 8, that G, the axis of the resultant
couple, would be represented in direction and magnitude by the
diagonal of the parallelopiped formed upon them.
If L, M, N were at right angles each to the plane passing
through the other two, the parallelopiped would be rectangular,
and we should have
G²=L²+M² + N².
CHAPTER IV.
ON ANALYTICAL STATICS IN TWO DIMENSIONS.
In this chapter we refer the points of application and the
directions of the forces to co-ordinate axes Ox, Oy, at right angles
to each other.
23. PROP. Required the magnitude and direction of the re-
sultant of any number of forces in one plane acting at one point,
and the conditions in order that there may be equilibrium.
Let P1, P2, P3, &c. . .. Pn be the n
forces, and let the point at which they act
be taken for the origin of co-ordinates.
Let P₁ make the angle a, with Ox

Ра
P3
&c.
"}
az
"
y
da
P
"}
A
N
>>
""
&c.
""
))
an
""
"}
O
M
P n
"}
Let OA represent the force P₁; com-
plete the right-angled parallelogram
OMAN, then OM represents the resolved part of P₁ in Ox,
ON that in Oy.
Let X1, X2, X, &c. . . . X be the resolved parts of the
forces respectively in Ox,
Y₁, Y₂, Y's, &c. . . . Y
we have 0 M=0 A cos a, or X₁ = P₁ cos a₁,
or
and similarly, X₂= P₂ cos α2,
2
X3 = P3 cos α3,
&c.
ვა
&c.
X₂= P₂ cos αn,
1
. in Oy,
ON=0 A sin a₁,
Y₁ = P₁ sin α₁;
2
Y₂= P₂ sin d₂,
Y₁=P3 sin ag
&c.
&c.
Y₁₁ = P₁₂ sin a,.
Y₁₂=Pn
But the components in Ox are equivalent to a single force,
=X₁+X2+X3+&c.
•
+ X₂.
30
ELEMENTARY MECHANICS.
This sum we often write more concisely by employing the
Greek letter Σ as the sign of summation;
or
2
•
X₁+X₂+ X3+ &c.
+Xn=Σ(X).
y we have
In the same way, in the axis of
or we have
Y₁+Y₂+ Y₂+&c. ... + Y₂=Σ(Y);
2
Σ(X)=Σ(P cos a),
Σ(Y)=Σ(P sin a).
Let R be the resultant required, & the angle its direction makes
with Ox. The resolved parts of R in the axes must equal the
resolved parts of the forces in the same directions; or
R cos 0=E(X),
R sin 0=Σ(Y),
•• tan @_E(Y)
E(X)'
(1)
=(E(X))²+(E(Y))².
(2)
R² = R²(sin²0+cos²0)
These equations (1) and (2) give the magnitude and direction of
the resultant.
If the forces are in equilibrium, their resultant =0, and
R=0 gives
0=(E(X))² + (Σ(Y))².
But square quantities being essentially positive, this equation
cannot be true unless we have
Σ(X)=0,
Σ(Y)=0.
These are the two necessary and sufficient equations for equili-
brium, when any number of forces in one plane act at one point.
24. PROP. To investigate the expressions for the resultant
force and resultant couple, and to find the conditions of equili-
brium, when any number of forces act at various points of a rigid
body in one plane.
Let Ox, Oy be the co-ordinate axes, and the plane passing
through them be that in which the forces P1, P2, &c. . . . P₂ act.
an be the angles they make with Ox
Let α1, d2,
respectively.
&c. .
STATICS.
31
Let x₁, y₁ be the co-ordinates
O M₁, A, M₁ of A, the point of appli-

1 1
1
cation of the force P₁, and let x2, y2
&c. . Xn, Yn, be the co-cordinates
of the points of application of the
others respectively.
1
Resolving P, in the directions of
Ox and Oy, we shall have compo-
nents
X₁ = P₁ cos α1,
at the point A₁.
५)
Y₁
αι
A₁
-X₁ O X1
M
Y₁ = P₁ sin a₁
P₂
Apply at O, in Ox, equal and opposite forces, each equal to
X₁, which will not affect the system; and X₁, at A₁, with -X, at
O, form a couple, with its arm A, M₁.
19
1
Then X, at A₁ is equivalent to a force X, at O, and a couple
X₁ × A₁ M₁=X₁ Y₁•
1
1
Similarly, applying at O in Oy forces equal and opposite,
and each equal to Y₁, we have Y₁ at A, equivalent to Y, at O,
and a couple
1
— Y₁× OM₁ = — Y₁.≈₁·
-
I
1'
This last couple will be negative if the former be taken positive,
as tending to cause rotation the contrary way. We consider
those couples positive which tend to cause rotation in the direc-
tion of the hands of a watch.
Proceeding in the same way with all the other forces, we
shall have a sum of forces
at O in Ox;
at O in Oy;
X₁ + X₂+ X3+&c. . • +Xn=E(X)
1
2
Y₁+Y₂+Y₂+&c. . . . + Y₂=Σ(Y).
1
2
and couples X₁₁+X2Y2+X3Y3+&c.
- Y₁₁ — Y₂x½ — Yzz―&c.
2
+X,3 = (Xỳ)
Yn~n=Σ(— Yx).
The couples being in the same plane, we have their resultant
axis G equal to the algebraic sum of the component axes; or
32
ELEMENTARY MECHANICS.
G=E(Xy)—Σ(Yx)
= Σ(Xy — Yx).
(1)
If R be the resultant force acting at 0, 0 the angle it makes
with Ox,
R cos 0=Σ(X),
R sin 0=(Y),
Σ(Υ)
tan 0=
;
(2)
Σ(Χ)
(3)
and
R²=(Σ(X))²+(Σ(Y))².
The
The equations (1), (2), and (3) determine G and R; they
can, however, be simplified when neither G nor R = 0.
moment of the couple remaining the

same, let its forces be each made
A
***
R
y
B
R
equal to R; then let it be moved and
turned round until one of its forces
acts at O in an opposite direction to
the resultant force; AO being the
arm. These two forces balancing,
may be removed, and there remains.
the other force of the couple acting
in ABR in the figure. This final re-
sultant being parallel to R, makes the angle @ with the axis of x.
R
To find the equation of the line ABR, we have, since G≈R. AO,
the arm of the couple, O A=
or
G
O A
G
; and O B=
R
cos
R cos '
G
0 B={(X)'
and equation of the line ABR is
or
y=tan 0.x+OB;
Σ(Υ)
G
y=Σ(X)
Σ(X)⋅ x + 2(X)⋅
When there is equilibrium, we must have both the resultant
force 0 and the resultant couple 0. These conditions give us
=0.
Σ(X)=0,
Σ(Υ)=0,
Σ(Xy— Ya)=0.
STATICS.
33.
These are the three necessary and sufficient equations of equili-
brium, when any forces act on a free body in one plane.
25. If there were a fixed point in the plane of the forces, we
might take it for the origin of co-ordinates O, and its resistance
would destroy the effect of the resultant force R, and we should
have the condition of equilibrium only G=0;
or
Σ(Xy—Yx)=0;
or there must be no tendency to rotation around the fixed point.
26. PROP. To prove the principle of virtual velocities for
forces acting in one plane on a point, and on a rigid body at dif-
ferent points.
DEFINITION. If any forces as P₁ and P₂
act at a point as A in the figure, and this
point is displaced through an indefinitely
small space A a, and we draw perpendiculars
P₁ a and p½ a from a upon the directions of
the forces, then the distances Ap₁ and A

1
2
Ap2
P₂
P
P、
a
P₁
A
are called the virtual velocities of the forces P₁ and P₂; and
Ap₁ being measured in the direction of force P₁ is called positive,
Ap, being measured in the direction of force P₂ produced is called
negative.
The principle of virtual velocities is thus enunciated: If any
number of forces be in equilibrium at one or more points of a
rigid body, then if this body receive an indefinitely small disturb-
ance, the algebraic sum of the products of each force into its
virtual velocity is equal to zero.
This principle is true when the forces in equilibrium act at
any points and in any planes on a rigid body; but we shall in
this treatise only prove the case when the forces act either at
one point or at different points in one plane, because the ge-
neral case requires a knowledge of analytical geometry of three
dimensions.
First. To prove the principle when the forces act all at one
point.
Let A be the point at which the forces P1, P2, &c. . . P₂ act.
D
34
ELEMENTARY MECHANICS.
Let α1, d2, dg, &c.
an be the y

angles they make respectively with Ox.
Let be the angle which the displace-
ment A a makes with Ox.
Let v1, v2, v3, &c.
V3
.
vn be the virtual
velocities of the forces respectively; then in
figure v₁=Ap₁= A a cos a Ap₁
Ар 1
= Aa cos (a,—0)
=Aa (cos a, cos
✪
+ sin a, sin 0),
P1
a
A
01
P
and
e
P₁.v₁=P₁ A a (cos a, cos + sin a, sin )
1
= Aa (cos 0. P₁ cos a₁ + sin 0. P, sin a₁).
Similarly, for P, we have
1
P2. v₂ = A a (cos 0. P₂ cos a₂+ sin 8. P₂ sin α),
2
and so for the other forces, therefore, we have
2
P₁. v₁+P₂. v₂+Pg. Vy+&c. . . . Pn. vn=Σ(P. v); say
1 1
2
=A a {cos 0 (P₁ cos a₁ + P₂ cos a₂+ &c.
2
+ Pn cos an)
+ sin (P, sin a₁ + P₂ sin a₂+ &c. . . . + P sin an)}.
Ꮎ
But when there is equilibrium at a point,
2
1
P₁ cos α, + P₂ cos α₂+ &c.
2
+Pn cos an=0,
P₁ sin a₁+ P₂ sin a₂+ &c. . . . + P₂ sin α=0,
2
n
N
... we have Σ(P.v)=0; or the principle is true when the forces
act all at one point.
Second. Let the forces act at different points or particles of
the body in one plane. We have now to consider these points
connected together by rigid lines or rods without weight, which
transmit the reactions of the particles upon each other. These
reactions must be considered together with the other forces.
Let A1, A2, A3, &c.
Let ra, az
Am be the particles.
be the reaction of the particle A, upon the particle A₁,
رو
A₁
1
разаг
"
A3
>>
Гараз
>>
A₁
"
razar
&c.
&c.
"
""
دو
A
A₁
Ag,
&c.
Let vaja,ɔ Vazar, Varaz, Vaza₁, &c. &c. be the corresponding virtual
velocities;
STATICS.
35
&c. &c. from the nature of reactions.
- Vaza,
then r
=)
аза, я до а
α193
raja,"
Also v
บ
บ
Ալջ
a1a3
азая
&c., which we must show.
B
A
P
7
b
Let A and B be
the particles dis-
placed to a and b.
C
Draw the perpendiculars ap, bq.
a
Then, if the line ab is parallel
to AB, Ap=Bq, and the point to be proved is evidently true.
When ab is not parallel to A B, let them meet when produced, if
necessary, in some point C. Since the displacements are indefi-
nitely small, the perpendiculars ap, bq coincide with circular arcs
having C for center, and
but
Ca=Cp,
Cb=Cq;
Ap=Cp-CA-Ca-CA,
Bq=Cb−CB=(Ca+ab) — (C A+ A B)
= Ca-CA. . . since A B=ab,
=Ap, but measured in the opposite direction to
the reaction of B upon A, and is therefore negative.
Let the sum of the products of all the external forces into their
virtual velocities, acting on particle A, be
an
(Pa,va),
ar
those on
وو
A₂ be
those on
""
Ag be
а-3
(Pa, Va),
Σ(P₁₂-va)
υ
&c.
&c.
those on
A be Σ(Pumam).
772
Since each particle is in equilibrium from the action of the forces
upon it, we have from the first case,
•
0=E(Pa₁• Va₁)+r บ +r
α
Ալջ
аз
• Vajaz
Va₁₂+ &c.
•
0=E(Pa₂• Va₂) +ˆ˜a27, • V2, +10273 • Vazz + &c.
04
UN
Vázlı
аза
0=E(Paz•Vas) +raza,•Vaza, +" Vaz?z
&c.
&c.
+ &c.
0=E(Pam•V&m)+rama,• Vama, +ramaz • Vamas + &c.
In taking the sum of the products for all the particles, the
products of the reactions into their virtual velocities will disap-
pear, being in pairs equal in magnitude with contrary signs;
therefore we have
Σ(Pa₁•va,+E(Pa₂•vas) +Σ(Pas•Vas)+
'a₁ • Va₁ + E (Pa₂ • Va₂) + Σ(Pas• Vas) + &c. . . . +Σ(Pom• Vam)=0;
૧
D 2
36
ELEMENTARY MECHANICS.
or generally, when there is equilibrium,
Σ(P.v)=0.
27. CONVERSELY. If the sum of the products of the forces into
their virtual velocities be equal to zero, or Σ(P.v)=0, then there
will be equilibrium.
For if the forces be not in equilibrium, they will be equivalent
either to a single force or a single couple. (Art. 24.)
In the first case, let R be the single resultant force, then a
force equal and opposite to R will reduce the system to equili-
brium; let u be its virtual velocity for any displacement. Since
there is now equilibrium, we have, by the preceding article,
Σ(P.v)+R.u=0.
But by hypothesis Σ(P.v)=0, .. R.u=0; which being true for
all small displacements of the body, we must have R=0, or the
body was in equilibrium from the action of the original forces.
In the second case, if the forces were equivalent to a resultant
couple, it would be balanced by an equal and opposite couple.
Let the forces of this opposite couple be Q and Q', and their
virtual velocities for any displacement be q and q' respectively.
Since they will reduce the system to equilibrium, we have, by
the preceding article,
Σ(P.v)+Q.q+Q'.q'=0;
but Σ(P.v)=0, .. Q.q+Q'.q'=0 for all displacements, which
is impossible unless Q and Q' each =0, since they are equal and
parallel forces, and act at different points.
CHAPTER V.
ON THE CENTER OF GRAVITY.
28. The center of gravity of a body is that point at which the
whole weight of a body may be considered to act, and would pro-
duce the same mechanical effect as the weight of the body actually
does.
The weights of all the particles of a body, acting vertically
downwards, are parallel forces, so that the center of gravity co-
incides with the center of parallel forces for such weights.
From the definition it arises that, if the center of gravity of
a body be a fixed point, the body will balance about that point
in all positions. This property of the center of gravity often
furnishes the means of determining its position practically. In
regular and symmetrical figures, as cubes, spheres, cylinders,
thin plates which are circular, elliptic, or regular polygons, &c.,
it is evidently the center of the body, or point about which it is
symmetrical.
29. PROP. If a body be in equilibrium, suspended from any
point, or resting with one point of contact upon another body,
then the center of gravity lies in the vertical line through that
point of suspension or contact respectively.
Let A in the figures be
the points of suspension and
contact respectively; draw
the vertical lines A w. If the
whole weight of the body act
in these vertical lines, it will
be supported by the reac-
tions of the fixed points A,
or when the centers of gra-
A

IL
772
771
A
10
38
ELEMENTARY MECHANICS.
vity g are in these lines. If the centers of gravity were not in
these lines, but at some points as g'; drawing the vertical lines
g'm through g', and the horizontal lines Am, then w, the weight
of the body acting at g', would have a moment w× Am, which
being unbalanced, the body could not be in equilibrium, con-
trary to the supposition.
DEFINITIONS. A body is said to rest in stable equilibrium
when, after receiving a slight disturbance, it returns to its first
position.
It is said to rest in unstable equilibrium when, after receiving a
slight disturbance, it moves from its position of equilibrium.
It is said to rest in neutral equilibrium when, after being dis-
turbed slightly, it still rests in equilibrium.
30. PROP. When a body rests in stable equilibrium, its center of
gravity is in the lowest position it can take; and when in unstable
equilibrium, it is in the highest position it can take.
In the first figure of the last article, the body rests in stable
equilibrium, and the center of gravity g would, on disturbance,
describe a circular arc about the point of suspension A, and
therefore would rise on the body being disturbed. In the
second figure also, if the equilibrium be stable, the center of
gravity will rise on disturbance, from the change of the point
of contact from A to a neighbouring point. In the annexed
figures, whilst the vertical lines through
the centers of gravity g pass through the
points of suspension or contact A, the
body will rest in equilibrium; but on a
small disturbance being given to the
bodies, the centers of gravity, falling out
of the vertical lines through the points
of suspension and contact, will come to
a lower position than at first, and the
weight will have a moment turning the

9
น
A
A
g
น
body further from its position of equilibrium, which therefore in
this case is unstable.
STATICS.
39
31. PROP. To find the conditions that the equilibrium may be
stable, unstable, or neutral, when the spherical surface of a body
rests upon another spherical surface.
C'
C
A'
B
A
Let A be the point of contact of
the spherical surfaces, C the center
of the upper surface, O that of the
lower. Let CA=r, OA=r. Let
the body receive a small disturbance,
so that the point C comes to C' in
the plane of the figure, and the point
of contact is now B, A' being the
new position of A. Join O and C',
then OC'=r+r. Draw Bb, a ver-
tical line meeting C'A' in b. Then
if the center of gravity of the body
falls between A' and b, as at g in the figure, the equilibrium will
be stable, for the moment of the weight (w) of the body will
bring the body back to its first position. If the center of gra-
vity falls beyond b from A', as at g', the vertical line through
g' will fall beyond B, and the moment of the weight will cause
the body to move further from its first position, and the equili-
brium will be unstable.

w'
10
If the vertical line through the center of gravity passes
through b, the body will be still in equilibrium, which is there-
fore neutral.
When the displacement is indefinitely small, A' will be inde-
finitely near A, and, by similar triangles, we have
or
A'b: A'C':: OB: OC',
A' b : r : : r': r+r",
or
r₁
A'b=
r+pl
consequently, when the equilibrium is
stable,
rel
Ag is less than
unstable,
Ag' is greater than
за дов
neutral,
Ag is equal to
•
r+x
40
ELEMENTARY MECHANICS.
When the lower surface is a horizontal plane, the radius of the
body at the point of contact is always vertical, and the equili-
brium is
stable, when Ag is less than r;
unstable, when Ag' is greater than r;
neutral, when Ag is equal to r.
32. PROP. To find the condition that a body placed on a plane
surface may stand or fall.
Let figures 1 and 2
represent the sections of
bodies by vertical planes
through their centers of
gravity (g), which rest on
a horizontal plane.
Let figures 3 and 4 re-
present, similarly, bodies
resting on an inclined
plane, down which they
are prevented from slip-
ping by friction.
Drawing
vertical
lines through the cen-
ters of gravity of the
bodies, they will be the
A
?
Fig. 1.
g
V
พ
B
C
D
Fig. 2.

Fig. 4.
w
Fig. 3.

༡
H
E
G
F
3
directions in which the
พ
weight of each acts, as g w in the figures.
Now in figures 1 and 4 the weight cannot cause the body
to turn about either A or B in figure 1, or G or H in figure 4,
because its effect is destroyed by the resistance of the plane.
But in figures 2 and 3 the weight will have a moment about D in
figure 2, and about E in figure 3, which is not neutralized by
the resistance of the plane, and the bodies consequently will fall
over.
The condition, therefore, that a body placed on a plane shall
stand or fall, is that the vertical line through its center of gravity
falls within or without the base respectively.
STATICS.
41
TO FIND THE POSITION OF THE CENTER OF GRAVITY IN SYSTEMS
OF PARTICLES AND IN RIGID BODIES.
33. PROP. To find the center of gravity of any number of
heavy particles whose places are given.
Let A, B, C, D, E, &c. be the
particles whose weights w,, w, wz,
&c. act in the vertical direction A
indicated by the arrows from A,
B, C, &c., and therefore consti-
tute a system of parallel forces; w
w₁ and we will have a resultant
=w₁+w, acting at a point a, such
that w₁ A a=w, Ba. (Art. 12.)
α
wi+w2
D
WA
B

1າ ກ
E.
C
wi+wa+w3
พ
3
2
Compounding the weight w₁+w, at a with another weight
w, acting at C, they will have a resultant =w₁+wa+w, acting
at a point b, such that
(w₁+w₂) × a b=w3× Cb.
Compounding the weight w₁+wa+w, acting at b, with another
weight w acting at D, we should find the point at which the
resultant w₁+wa+w+w, acted, and so onwards for any number
of particles whose positions were given.
The position of the point at which the final resultant weight
acts is thus determined, and is the same point whatever be the
order in which we compound the weights; so that a system of
particles or a rigid body can never have more than one center of
gravity.
The positions of the points a, b, &c. depend on the weights
w, w, ws, &c., and on the positions of the points A, B, C, &c.
with respect to each other, but not at all on the directions of the
arrows with regard to the lines Ab, a C, &c.; so that gravity
acting vertically downwards, we may turn the whole system into
any new position, the weights and relative positions of the par-
ticles remaining the same, and shall find the center of gravity in
the same point as before.
42
ELEMENTARY MECHANICS.
COR. If the center of gravity of a system of particles rigidly
connected, or of a rigid body, be supported, the whole system
will be supported, and the system or body will balance about
the center of gravity in all positions.
34. If the positions of the particles be given with respect to
a fixed origin and co-ordinate axes, we should find the position
of the center of gravity by the same process as in article 16,—the
weights of the particles being the forces, and the whole weight
of the body being the resultant, or the sum of the weights of
all the particles.
If m = the mass of any particle, and therefore proportional
to its weight, x, y its co-ordinates, and x, y the co-ordinates
of the center of gravity of the system, we have, by article 16,
x =
Σ(m.x)
Σ(m)
m₁ x₁ + m₂ x ₂+MgXg+ &c.
or =
or =
m₁ + m₂ + m² + &c.
m₁ YĮ + mq Y₂+M3Y3+&c.
y
Σ(m.y)
Σ(m)
M₁ + M₂ + Mg + &c.
where x, y, are the co-ordinates of the particle m,; 2, y2 those
of m₂, &c. &c. If the particles are all in the axis of x, y₁=0,
y=0, &c. &c. and y=0.
35. The formulæ of the last article are applicable to the so-
lution of problems where the centers of gravity of the parts are
given to find the center of gravity of the whole body, and where
the centers of gravity of the whole body and some of the parts
are given to find the center of gravity of the remaining part;
for the weight of each part being taken at its center of gravity,
we treat the problem as if a heavy particle of that weight were
placed there. Thus, let M equal the whole mass of a body; a, y
the co-ordinates of its center of gravity; M₁, M, X₁, ÿ₁, X, Y₂
the corresponding quantities for its two parts; we have
M₁x₁+M₂x2
M
I
Ꮖ
M₁ ÿ₁ + M₂ Y 2
y=
1 1
If M, M, and x, y, z₁, y₁ were given, we have
x1, Y1
1
Mx-M₁x₁
X2
M2
M₂=M—M₁,
2
M
2
My - M₁ y₁
Y 2
Y₂ =
M2
STATICS.
43
36. Ex. 1. To find the center of gravity of a uniform physical
straight line.
C
B
If AB be the uniform straight A
line, it will balance on a fulcrum or
fixed point at C, its middle point, which will be its center of
gravity by art. 28; for we may consider the line as made up of a
series of equal particles in pairs, at equal distances on opposite
sides of C, and the weights of each pair would have their result-
ant weight acting at C, the middle point between them, or the
resultant weight of the whole line would act at C; and this
weight being supported by the reaction of the fulcrum, the line
will be supported, and its center of gravity will be at C, its
middle point.
Ex. 2. To find the center of gravity of a triangular plate of
uniform thickness and density.
a
E
d
G
C
Let ABC be the triangular
plate of which the thickness is
inconsiderable. Draw from C the
line CD bisecting A B in D, and
from B the line BE bisecting
AC in E. Let G be the inter-
section of CD and B E, then G is
the center of gravity of the triangle. For we may consider the
triangle to be made up of physical lines, each parallel to AB;
let adb be any one of these lines meeting CD in d, which will
be its middle point, because by similar triangles we have

A
Cd: da:: CD: DA
:: CD:D B
:: Cd: db.
D
The line ab has therefore its center of gravity at d.
B
Similarly it is shown that every line parallel to AB will be
bisected by CD, and have its center of gravity in that line; there-
fore the center of gravity of the triangle will be in this line.
In the same way it is shown that the center of gravity will
44
ELEMENTARY MECHANICS.
be in the line BE; consequently it must be at the intersection
of these lines, or at the point G in the figure.
Join the points D and E. The line DE is parallel to BC,
because the sides AC, AB are cut proportionally in D and E,
and DEB C.
Also the triangles EDG, BCG are similar;
or
So also
.. ED: BC:: DG: CG
::1:2;
DG=1.CG=. CD.
EG=.BE.
Therefore to find the center of gravity of a triangular plate,
we draw a line from any angle to the middle of the opposite
side, and measure along this line its length from the angle, or
from the bisection of the side, and the point so found is the
center of gravity required.
3
Ex. 3.
To find the center of gravity of a parallelogram of
which the density is the same at every point, and the thickness
uniform but very small.
Let ABCD be the parallelogram ;
bisect the sides AD, BC in E and F,
and join EF; also bisect A B and CD
in H and K, and join HK; let G be
the intersection of EF and HK, then
G is the center of gravity of the paral- A
lelogram.
E
D d K
C

a H
B
F
For the parallelogram may be considered made up of physical
lines, as a d parallel to AD; each of these will be bisected by the
line EF, and therefore the center of gravity of the parallelogram
will be in this line. Similarly, the center of gravity will be in
the line HK, and is therefore the point G at the intersection of
these lines.
It is also evidently the intersection of the diagonals of the
parallelogram.
STATICS.
45
Ex. 4. To find the center of gravity of a polygonal plate of
uniform density and thickness.
Let ABCDEF be the poly-
gon. Draw the lines A E, A D,
A C, dividing it into triangles.
When the polygon is given,
these triangles will be known,
and their centers of gravity will
be found by Ex. 2. Let 91, 92
93 94 be these centers of gra-
vity respectively. We may con-
sider the mass of each triangle
B
A
F

E
G'
G"
9a
G
94
93
D
C
as a heavy particle at its center of gravity. Compounding the
masses at g₁ and g, by art. 33, their center of gravity will be at
91
some point, as G', in the line joining g₁ and 92. Compounding
the mass of the two triangles at G with the mass of the next
triangle at g, we shall have the center of gravity of the three
triangles at some point as G", and so onwards; the last point so
found will be the center of gravity G of the whole figure.
Ex. 5. To find the center of gravity of a triangular pyramid
of uniform density.
Let ABCD be the triangular
pyramid. Bisect the edge BC in
E, and draw the lines AE, DE;
in AE take Af=AE, in DE
take De=DE, then e and ƒ are
the centers of gravity of the tri-
angular faces of the pyramid DCB,
ABC respectively. Join Df, Ae;
these lines intersect in a point G,
which is the center of gravity of
the pyramid.
D

b
E
B
made up of triangular
Let abc be such a plate.
For we may consider the pyramid
plates parallel to any one of its faces.
parallel to the face ABC. The parallel planes meet the plane
DCB in cb, CB, therefore these lines are parallel, and cb is bi-
sected by the line DhE in h; for
46
ELEMENTARY MECHANICS.
Dh: ch:: DE: CE
:: DE: EB
therefore
:: Dhhb;
ch=hb.
Similarly, the lines ah and
AE are parallel, and Df, a line
in the plane of the triangle AED,
cuts them proportionally.
Let
g be the intersection of Df and
ah;
then
and
Compounding,
A
Dgag: Df: Aƒ,
gh: Dg::fE: Dƒ.
gh: agfE: Af
"
G
D

b
E
::1:2;
therefore g is the center of gravity of the triangle abc. Simi-
larly it may be shown that the center of gravity of every section
parallel to ABC is in the line Df. In the same way it may be
shown that the center of gravity of every section parallel to the
face BCD is in the line Ae. The center of gravity of the whole
pyramid must therefore be in each of these lines, which are both
in the plane AED. Let G be the intersection of Ae and Df, or
the center of gravity of the pyramid; join fe. Since AE, DE
are cut proportionally in e and f, ef is parallel to A D, and
ef: AD::ƒE: AE
::1:3.
Also the triangles AGD, fGe are similar, and
or
Similarly
fe: AD::fG: GD
::1:3;
ƒ G= } G D= { Dƒ, and DG= { Dƒ.
e G=Ae, and A G=ž A e.
Or, to find the center of gravity of a triangular pyramid, we
must draw a line from any one of the solid angles to the center
of gravity of the opposite face, and measure of that line from
the angle for the point required.
Ex. 6. To find the center of gravity of a pyramid whose
base in any polygon.
Let BCDEF be the polygon which is the base of the pyra-
B
STATICS.
47
B
b
G
K
d
mid whose vertex is A. Joining
CF and DF, we divide the poly-
gon into triangles; and planes
passing through CF, DF, and the
vertex A will divide the pyramid
into triangular pyramids. Draw-
ing a line from A to the center of
gravity of any one of the triangular
bases, and measuring of that line
from A, we shall have the center
of gravity of the triangular pyra-
mid on that base. If we take a
plane bcdef through this point parallel to the base, it will cut all
lines drawn from A to the polygonal base in the same proportion;
and therefore the centers of gravity of all the triangular pyramids
will be in this plane, and consequently the center of gravity of
the whole pyramid also, because the mass of each pyramid may
be considered a heavy particle at its center of gravity.

C
D
E
Again. If g be the center of gravity of the polygon, and we
join Ag, it can be shown that the center of gravity of every sec-
tion parallel to the base will be in the line Ag, and therefore the
center of gravity of the whole pyramid will be in this line; and
since it is also in the plane bcdef, it is at the point G where Ag
meets the plane.
Hence to find the center of gravity of any pyramid on a
polygonal base, we must draw a line from the vertex to the
center of gravity of the polygon, and measure of it from the
vertex, or from the base.
COR. The above rule holds good whatever may be the
number of sides of the polygon, and is therefore true when the
number becomes indefinitely great, or when the base becomes a
continued closed curve, as a circle, an ellipse, oval, &c. Or the
center of gravity of a cone, right or oblique, and on any base, is
found by drawing a line from the vertex of the cone to the
center of gravity of the base, and measuring of the line from
the vertex, or from the base.
48
ELEMENTARY MECHANICS.
Ex. 7. To find the center of gravity of a frustum of a cone
or pyramid cut off by a plane parallel to the base.

Let the length of a line drawn from the
vertex of the cone, when complete, to the
center of gravity of the base A C, in the
figure, a.
=α. Let the length of the same line
to where it meets the smaller end of the
frustum, Ac, =d'. Let A G₁₁, A G₂=πv
A G=x, G being the center of gravity of the
whole cone, G₁ that of the part cut off, G₂
that of the frustum. Using the formula of
article 35, we have
А
G
C
G+
G
G₂t
Mx-M₁x₁
1
X2=
x=
M₂
where x=α, x₁= {a'.
Also, similar solids have their volumes proportional to the
cubes of their lines similarly situated, and the part of the cone
or pyramid cut off by a plane parallel to the base is similar to
the whole cone or pyramid, therefore we have
and
and
1
M₁ a's
M
M₁ =M_M₁ =M(1–3)
2
18
11
13
M} a − M (~~) } a
+ficu
13
a
M(1-2)
a4 — a¹4
a³ — a¹³
(a + a') (a² + a¹²)
a²+aa+a¹2
which gives the distance of the center of gravity from the vertex
of the cone or pyramid; and the distance from the center of
gravity of the base along the same line is
a-32
12
(a + a') (a² + a¹²)
3 a'3
13
a
12
4
4(a² + a a' + a¹²) *
a² + a a' + a¹²
Ex. 8. To find the center of gravity of the perimeter of a given
triangle in terms of the co-ordinates of its angular points.
STATICS.
49
We suppose the perimeter of the triangle to be three uni-
form physical lines whose weights are proportional to their
lengths. Let a, b, c be the sides respect-
ively opposite to the angles A, B, C.
The centers of gravity will be each at
the middle point of the side, as at 91, 92
9, in the figure.
Let x¡y₁ be co-ordinates of A to origin 0,
xi yı
X
2Y2
XzY3
رو
دو
رو
"2
B,
C ;
92
C

93
A
B
then the co-ordinates of the middle point of each line being the
arithmetic means of those of its extremities, the co-ordinates of
9₁ are 1(x2+x3), † (Y2+Y3),
2
92 are 1(x1+x3), 1 (Y₁+Y3),
2 1
93 are 1(x1+x2), 1 (Y₁+Y2),
2
and a, 7 being the co-ordinates required, the formulæ of article
34 give us
ac=
y=
a{x2+x3)+b(x1+x3)+c(x1+x2)
2(a+b+c)
a(y₂+Y3)+b(Y1+Y3)+c(Y₁+Yg)
2(a+b+c)
Ex. 9. To find the co-ordinates of the center of gravity of a
triangular plate.
Let x1 y1 x2 y2 y3 be the co- y
ordinates of the points A, B, C re-
spectively.
Let the line AD bisect B C in D;
the center of gravity being G, we
have A G=AD. The co-ordinates
of D are (x2+Xy), ¦(Y2+Y3); and
if x=ON, y=GN be the co-ordi-
nates of G, we have

or
Similarly
A
OL
ON=OL+(0M-0 L),
GN=AL+3(DM—AL);
x = x₁ + } { } ( X 2 +X3) −xX1}
1
= } (X1+X2+X3).
ÿ=} (Y₁+Y₂+Ys).
C
D
G
B
N M
E
50
ELEMENTARY MECHANICS.
Ex. 10. To find the center of gravity of the surface of
right cone.
We consider the surface of the cone as a sheet of matter
equally dense at every point; and as it is symmetrical with.
respect to the axis of the cone, its center of
gravity must be in that line.
B
100
A

G
Again. If we draw the straight lines Ab,
Ac from the vertex to the circumference of
the base, so that bc is an indefinitely small
arc, the center of gravity (g) of the triangle
Abc is at a distance Ag from A=Ab. This
is true for every such elementary triangle
which can be formed on the surface of the
cone; or their centers of gravity are in a circle whose center is
G in the axis AD of the cone, such that A G=AD; and G is
the center of gravity of the surface of the cone required.
b c
D
C
EXAMPLES ON THE PRECEDING CHAPTERS.
Ex. 1. Two beams, connected together at a given angle,
turn about a horizontal axis at their point of meeting; find the
position of equilibrium which they will take by the action of
their own weights.
Let AC, BC be the beams
suspended from C, and inclined to
each other at an angle a. Since
C is a fixed point, the only con-
dition of equilibrium is, that the
moments of the weights about C
may balance.
A
M
C
N

92
B
91
W1
W2
Let 91, 92 be the centers of gravity of the beams, and 91 C=a,
92 C=b. Let w
Let w,
BC acting at 92.
weight of beam A C acting at g₁; w₂=that of
Draw MCN, a horizontal line, meeting the
vertical lines in which the weights of the beams act in M and N.
STATICS.
51
In equilibrium we have w₁× CM=w₂ × CN. Let 0=angle
BCN, which is to be found; we have
or
whence
w₁ × Cg₁ cos ACM=w, x Cg, cos BCN,
1
w₁ a cos (180—a +0)=w₂ b cos 0;
tan 0=
w₂ b+w₁ a cos a
w₁ a sin a
which gives the position of the beam as required.
Ex. 2. When a given weight (W) is hung from the end of
one of the beams (A), as in the last question, show that
tan 0: =
wą b + (W. A C+w,a) cos a
(W.AC+w, a) sin a
Ex. 3. Two beams, as in Ex. 1, are suspended from (B)
one end; show that if O be the angle which the upper one makes
with a horizontal line, we have in equilibrium,
tan 0=
=
(w₁+w2)B C—(w₂b+w₁a cos a)
w₁ a sin α
Ex. 4. Two spheres of unequal
radii, but of the same material, are
placed in a hemispherical bowl; find
the position they take when in equili-
brium.
M
с
N

8
A
B
W₁
W2
N.B.—This, and similar problems
of bodies resting in equilibrium in a
hemispherical bowl, can be reduced to problems like the pre-
ceding. For if the center of the hemisphere C in the figure
were a fixed point, and connected by rigid rods A C, B C, without
weight, to the centers of the spheres, we might suppose the hemi-
sphere removed without changing the conditions of equilibrium.
1
Let w, and w₂ be the weights of the spheres A and B, whose
radii are r₁ and r₂ respectively. Let R be the radius of the
bowl. Then if the angle ACB=α in triangle ABC, we have
AB=r₁+r₂, AC=R-r₁, BC=R—r, and
1
COS α=
A C²+B C2-A B2
2A C. BC
(R—r,)² + (R—r₂)² —(1+r)², which give a.
2(R—r₁) (R—r2)
E 2
52
ELEMENTARY MECHANICS.
The position of the spheres will be known if the angle BCN
be known; let it=0. The weights are proportional to the
volumes of the spheres, or to the cubes of the radii,
ΟΙ
or
3
W]
and w₁ x CM=w₂ × CN;
W X
W2 r2
w₁(R—r₁ cos 180—a+0=w₂(R−r₂) cos 0 ;
W
whence
tan 02
r₂³(R—r₂)+r,³(R—r₁) cos a
3
r₁³(R—r₁) sin a
Ex. 5. A heavy beam in a given position has one end resting
against a smooth wall, and the other tied to a cord which is
fastened at a point directly above the point where the beam rests;
find the forces which keep the beam in equilibrium.
Let CB be the beam in the figure, AB the cord;
A and C being points on the wall. The weight of
the beam (w), the distance (Cg=a) of the center of
gravity from the end against the wall, the length of P
the beam (7), the length of the cord (c), and the di-
stance () of the points A and C must be given. The
angles A, B, C will be known.
Let t=the tension in the cord.
A

R
V
B
w
The beam will press at C against the wall, and we
may resolve this pressure into a vertical and horizon-
tal part; the latter, perpendicular to the wall, will
be destroyed by its reaction (R); but since the wall is
smooth, the vertical component can be balanced only by an
opposite force P.
If we take into account all the forces which act on the beam,
we may treat it as a free body in equilibrium from their action,
and apply the conditions of equilibrium investigated in Chapter IV.,
namely,-
Σ(X)=0,
Σ(Y)=0,
Σ(Xy-Yx)=0.
Therefore, resolving the forces vertically and horizontally, we
have
STATICS.
53
P+t cos A-w=0,
R-t sin A-0.
(1)
(2)
Since the origin of co-ordinates may be taken any where, we
may fix it at a point where we avoid moments of the unknown
forces; therefore, fixing it at C, we have, for the equation of
moments,
or
w × sin Cx Cg-tx sin B× CB=0,
t=w
a. sin C
1. sin B
а с
=w
which gives the tension in the cord.
Substituting in the equations (1) and (2), we have
R=w
a c
sin A, which gives the pressure against the wall;
I h
a c
P=w(1–72, cos 4), which gives P, as required.
h
When P=0, substituting the value of cos A in the triangle
ABC, we find
h=
Vale
'a (c² — 1²)
21-a
Ex. 6. A heavy beam rests upon a peg, with one end against
a smooth vertical wall: find the position of equilibrium.
Let ACB be the beam resting at E
A against the vertical wall ADE, and
upon the peg C.
The center of gravity, when there
is equilibrium, will be evidently at
some point, as g, beyond C from A.
Let w=the weight of the beam ;
R=reaction of the wall perpendicular

=
D
B
R'
K
C
8
A
R
心
​to itself at A; R' reaction of the peg perpendicular to the
beam at C. These three forces keep the beam in equilibrium
when making some angle with the horizontal direction, which
is to be found.
Let Ag=a, and CD=b=perpendicular distance of the peg
from the wall, which must be given.
54
ELEMENTARY MECHANICS.
Resolving the forces horizontally and vertically, we have
R-R' sin 0=0,
w-R' cos 0=0.
(1)
(2)
w.Cg cos 0-R.CD. tan 0=0,
(3)
Taking the moments about C, we have
Οι
w(a-b sccant 0) cos 0-R b. tan 0=0.
Multiply (1) by cos 0, (2) by sin 0, and subtract, and we have
R cos 0-w sin 0=0;
.. Rw tan 0.
w(a-b sec 0) cos 0-w b tan20=0,
Substituting,
or
3
whence
cos 0=/
a cos 0 — b(1+tan²0) =0 ;
and b must be less than a.
Ex. 7. Prove that in the last question the same result is
obtained if we resolve parallel and perpendicular to the beam,
and take the moments about either A or g in place of C.
Ex. 8. A heavy beam lies partly in a smooth hemispherical
bowl and partly over one edge: find the position of equilibrium.
Let ABC be the beam
resting on the surface of the
hemispherical bowl at A, and
on the edge at B.
Let O be the center of the
bowl DAB, whose radius=r,
and BOD horizontal.
D
A
R
R'

9
B.
C
The center of gravity of the
W
beam will be at some point g within the bowl. Let Ag=a.
The beam is supported by the reaction of the bowl at A per-
pendicular to the surface, or in AO, let it R; by the reaction
of the edge at B perpendicular to the beam, let it R'; and by
the weight of the beam (w) acting at g.
STATICS.
55
Let 0= angle ABO= angle BAO, which is to be found.
By the artifice of resolving in the direction of the beam and
taking moments about B, we avoid expressions involving the
unknown reaction R', and have, parallel to A B,
or
For moments about B,
or
R cos 0-w sin 0=0,
R=wtan 0.
R. AB sin 0-w. Bg cos 0—0,
R.2 r cos 0. sin 0-w(2 r cos 0— a) cos 0=0.
Substituting for R, and omitting the common factors,
2 r tan 0.sin 0 - 2 r cos 0 + a=0,
whence
or
Ө
2r-4r cos² 0 + a cos 0=0 ;
a + √ 32 r² + a²
cos >=
at
8 r
in which the sign only is admissible.
Ex. 9. Solve the last example by following the method of
article 24 in every step; taking A for the origin of co-ordinates,
and A B for the axis of x.
Ex. 10. Find the horizontal strain on the hinges of a given
door, and show that the vertical pressures are indeterminate.
Let the figure annexed represent
the door, of which the hinges are A
and B.
Let g be the center of gravity at
which the weight (w) of the door acts.
The door is in equilibrium from its
weight w at g, and the reactions of the
hinges represented by the oblique ar-
rows at A and B.
Let A be the origin of co-ordinates
as in the figure; Ax the axis of a; Ay
B
R₁

Y
RA
A
9
10
56
ELEMENTARY MECHANICS.
the axis of y; and let x=a, y=b be the co-ordinates of g,
x=0, y=h those of the hinge B.
Let the resolved parts of the reactions at B be Q, horizontally,
and R₁ vertically, and let Q2, R₂ be those at A respectively, as
in figure.
1
Then
2
Σ(X)=0=Q₂— Q₁, or Q₁ = Q₂
Σ(Y)=0=w-R₁ — R₂,
Σ(Xy—Yx)=0=w.a—Q₂.h.
From (3) and (1) we have
1
Q₁ = 2a = Q2
h
(1)
(2)
(3)
which gives the horizontal strain; and it is the same at each
hinge in magnitude, but opposite in direction.
2
Again. From (2) we have R₁+R₂=w; but we have no
other relation to enable us to determine the separate values of
R₁ and R₂, which are therefore indeterminate.
Ex. 11. Two given smooth spheres rest in contact on two
smooth planes inclined at given angles to the horizon; to find
their position of equilibrium.
Let AB, AC be
R₁

the planes, making the
anglesa and ẞrespect-
ively with the horizon-
tal line through A.
Let 0, 0, be the D
02
centers of the spheres
at which their weights
w, and w₂ respectively act.
→
8
R₁
C
O₂
B
S
β
A
W₁
w2
Let R₁ and R₂ be the reactions of the planes at the points of
contact, perpendicular to themselves, and therefore passing through
the centers of the spheres to which they are tangents.
STATICS.
57
Let S equal the mutual pressure of the spheres at their point
of contact, acting in the line passing through their centers; let
this line 02, 0₁, D make the angle with the horizontal line
AD. It is required to find 0.
Each sphere is in equilibrium from its own weight, the
reaction of the plane against which it rests, and the pressure of
the other sphere. By the artifice of resolving in the directions
of each plane for the equilibrium of each sphere, we avoid equa-
tions involving the unknown reactions R, and R, and have, in
the direction of A B,
wą sin α- Scos (x — 0) = 0,
in direction of A C,
w, sinẞ-S cos (B+0)=0.
Eliminating S, we have
whence
tan 0=
W2
(w₁+w₂) sin a. sin B
(1)
(2)
w₂ sin a.cos (B+0)=w, sin ẞ cos (α—0) ·
sin a.cosß—w, sin B. cosa
1
w₂ cot ẞ-w, cot a
2
w₁ + w z
Ex. 12. A sphere is sustained upon an inclined plane by
the pressure of a beam moveable about the lowest point of the
inclined plane; given the position of the beam, required that of
the plane.
Let AgB be the beam
moveable about A.
Let w weight of the
=
beam acting at its center of
gravity g; B the point of con-
tact with the sphere, whose
center is C; let w' weight A
of the sphere.
0
R

B
C
D
9
W
w!
P
E
The sphere is in equilibrium, from the reaction (R) of the
plane at the point of contact, from the pressure (P) of the beam
at B, and from its own weight; these three forces all act through
the center C.
58
ELEMENTARY MECHANICS.
Let Ag=a, AB=b, angle BAD, which the beam makes
with the plane, =a, these are given; or, in place of either one of
the two latter, we may have the radius of the sphere given.
Let the angle DAE=0 the elevation of the inclined plane,
which is to be found, when there is equilibrium.
For the condition of equilibrium of the beam, taking moments
about A,
or
P× AB=w.Ag cos a +0,
α
w= cos a + 0.
P=w7
For the condition of equilibrium of the sphere, resolving the
forces in the direction of AD, we have
or
whence
w' sin 0-P sin a=0,
a
w'sin 0-w sin a. cos a +0=0;
tan 0=
b
w a cos a. sin a
α
w a sin²a+w'b
which gives, the elevation of the plane, as required.
Ex. 13. A heavy beam turns about a hinge at the lower end,
with the other end pressing on an inclined plane, a part of the
surface of a body which rests on a smooth horizontal plane passing
through the hinge; find the horizontal force necessary to keep
the body from moving.
Let BCD be the body
resting on the smooth hori-
zontal plane ACD. Let
AB be the beam turning
about the hinge at A; let g A
be the center of gravity of
the beam at which its weight
(w) acts.
The body is to be in equi-
librium from the pressure of
the beam upon it at B, and a
horizontal force (F) acting at
Fig. 1.
R

9
R'
B
F
K
P
C
D
w

Fig. 2.
F
B
C
m n
D
STATICS.
59
some point K; and the beam is to be in equilibrium from the
reaction (R) of the inclined plane upon it at B, and its own
weight acting at g.
The point B is in equilibrium, from the reaction (R) per-
pendicular to the inclined plane, the reaction (R') of the beam
in the direction of its length, and a force (P) acting perpendicu-
larly to AB, arising from the moment of its weight (w) about A.
Taking the moments about A for the equilibrium of the
beam, we avoid expressions involving R', and have
w.Ag cos gAC-R.A B sin ABR=0.
Let the angle BAC=ß, the angle of the inclined plane with
CD=a, Ag=a, A B=1; then
R=w
a cos B
Icos (a-B)*
(1)
To find R in terms of F we must consider the conditions of
equilibrium of a right-angled triangular wedge sliding along a
smooth plane, as CD, figure 2, from the action of a force (F)
acting parallel to CD, a pressure (R) perpendicular to CB at B,
and the reaction of the plane CD. If Bn be perpendicular to
CB, and Bm to CD, these three forces will be proportional to
the sides of the triangle Bmn respectively.
Resolving parallel to CD, we have
R sin BCD-F=0;
or in figure 1,
Substituting in (1), we have
F
R=
sin α
a sina.cos B
F=w7* cos(a—B)
which gives the horizontal force required.
'
Ex. 14. Solve the last example by taking the conditions of
equilibrium at the point B; and show that the whole pressure
a
on the hinge A=w(sin B+cos B. tan (a-B)}.
60
ELEMENTARY MECHANICS.
Ex. 15. A beam turning about a hinge is supported in
equilibrium by the tension in a cord tied to its lower end: the
cord passes over a pulley in the same horizontal line with the
hinge, and sustains a given weight; find the position of equili-
brium of the beam.
In the previous examples we obtained
the solution from the equations for equi-
librium only; but many statical problems
require, for the determination of all the
unknown quantities, equations to be formed
from geometrical conditions also, of which
this simple problem is an example.
Let A be the hinge, C the pulley, and
A C=c.
C

P
B
g
A
Let A B be the beam, whose length =1, g its center of gravity
at which its weight (w) acts, and Ag=a.
Let P be the weight hung from the cord, which is equal to the
tension (t) in the cord.
Let 0-angle CBA, p-angle CAB; these are both unknown
quantities.
or
Taking moments about A, we have
t.AB sin 0=w. Ag cos &,
w a
sin 0= cos $.
ΡΙ
From the geometrical data we have
sin
AC
sin ACB™ AB'
(1)
or
sin 0=7. sin (0+$).
(2)
The equations (1) and (2) suffice to determine 0 and .
Ex. 16. A uniform beam rests with its lower end in a smooth
hemispherical bowl, and its upper end against a smooth vertical
plane; find the position of equilibrium.
STATICS.
61
Let AB be the beam resting
against the vertical plane at A, and
upon the bowl at B.
Let C be the center of the bowl;
ECF a horizontal diameter which,
being produced, meets the vertical
plane at D.
Let the radius of the bowl = r;

A
R
Ꭱ
E
C
D
F
9
m
B
A B=length of the beam=1; Ag=2, since the beam is uniform ;
w=its weight: let also CD=d; these must be given.
Let angle BCE=0, and let angle of the beam with the
horizon; these have to be determined.
The beam is supported by its weight (w) acting at g, the
reaction (R') of the plane perpendicular to itself at A, and the
reaction (R) in the radius C B.
Resolving vertically, we have
or
R sin 0-w=0,
R=
sin 0
W
Taking the moments about A, we have
therefore
R.A B.sin (0-6)-w.Ag.cos &=0;
sin (0-6) cos
cos $=0.
sin
2
(1)
This equation containing two unknown quantities, we require still
a geometrical relation between them.
Let Cm be a vertical line meeting A B in m;
sin BCm cos 0 Bm AB-Am
sin BmCcos &¯BC
2°
l-d.seco
;
2
...cos =
0
l.cos &— d
(2)
2°
62
ELEMENTARY MECHANICS.
From (1) we have cos -cot 0.sin — ½ cos &=0,
or
tan 0=2 tan p.
These two equations suffice to determine & and as required.
Ex. 17. A weight (w) hangs from one end of a cord, of
which the other end is fastened to a vertical wall: the cord is
pushed from the wall by a rod tied to it, which is perpendicular to
the wall. Show that if the cord, where it is fixed to the wall,
makes an angle & with it, and R be the pressure of the rod on the
wall, then
α
R=w.tan a.
Ex. 18. A heavy beam lies with its upper end against a
smooth vertical plane, and its lower end on a smooth horizontal
one. Show that if the beam makes an angle a with horizontal
direction, its length being 1, and weight w, and the distance of
its center of gravity from the lower end being a; then the force
required to be applied horizontally at its lower end to maintain
the equilibrium being F, we have F=wcot a = the pressure
against the vertical wall. Show also that the pressure on the
horizontal plane =w.
a
Ex. 19. A body is suspended by a cord of given length from
a point in a horizontal plane, and is thrust out of its vertical
position by a rod, without weight, acting from another point in
the plane; show that if t = the tension in the cord, w = the
weight of the body, 7 the length of the cord, d= the distance
of the two points, and be the angle which the rod makes with
the horizon,
t=w=cot 0.
Ex. 20. A triangular plate of uniform thickness and density
is supported horizontally by a prop at each angle; show, by
drawing perpendiculars on the sides respectively, from the oppo-
site angles and the centre of gravity, that the pressure on each
prop = the weight of the plate.
CHAPTER VI.
ON THE ELEMENTARY MACHINES, OR MECHANICAL powers.
THE effects of forces in practical mechanics are continually
modified through the agency of instruments which we call
machines.
The simplest of these instruments are Cords and Rods, which,
with hard planes, may be considered as forming, by their combi-
nations and recombinations, all other machines, however com-
plicated.
Cords are considered in the first instance as without weight,
and perfectly flexible. A cord transfers the action of a pulling
force, applied at one extremity, to any other point in it, un-
changed in magnitude, as long as it is in a straight line to that
other point, or only passes over smooth obstacles without friction.
The force which is thus transmitted along the cord we call the
tension in the cord.
Rods are considered in the first instance as without weight,
and inflexible or rigid. They transmit the action of either a
pulling or a pushing force in the line joining their extremities
unchanged in magnitude. The force which is transmitted along
this line we call the reaction of the rod.
The machines which are next in simplicity to simple cords and
rods are called the Mechanical Powers. They comprise the Lever,
the Wheel and Axle, Toothed Wheels, the Pulley, the Inclined
Plane, the Wedge, and the Screw.
ON THE LEVER.
The simple lever is a straight rod, having a fixed point some-
where in its length, and supposed without weight. The fixed
point about which the lever may freely turn is called its fulcrum.
64
ELEMENTARY MECHANICS.
The conditions of equilibrium of any heavy lever may be reduced
to those of a lever without weight, by taking the weight of the
lever itself, acting at its center of gravity, with the other forces
producing equilibrium.
The arms of a lever are the portions of it on each side of the
fulcrum. When the arms are not in the same straight line, it is
called a bent lever.
The mechanical powers being
most familiar to us as employed
to support or raise heavy bodies
or weights, it is usual to call one
of the forces the Power, and the
other the Weight.
Levers have been divided into
three kinds, according to the rela-
tive positions of the Power, the
Weight, and the Fulcrum.
Figure 1 is an example of a
lever of the first kind, AB being
the lever, C its fulcrum; the power
(P) and weight (1) acting on op-
posite sides of the fulcrum.
Figure 2 is an example of a
lever of the second kind, A C being
the lever, C the fulcrum; the
power (P) and the weight (IV)
acting on the same side of the ful-
crum, but W nearer to it.
Figure 3 is an example of a
lever of the third kind, B C being
the lever, C the fulcrum; the
power (P) and the weight (W)
acting on the same side of the ful-
crum, but the power nearer to it.
P
Fig. 1.
A
C
B

P
Fig. 2.
B
A
W
Fig. 3.
P
W


B
A
C
W
STATICS.
65
A crow-bar, according to the way in which it is used, is a
lever of the first or second kind. Scissors and carpenter's pincers
are examples of double levers of the first kind. An oar is an
example of a lever of the second kind, the fulcrum being a point
in the blade of the oar which rests for an instant stationary in
the water. Nut-crackers are double levers of the second kind.
Tongs, shears, &c., are double levers of the third kind. The
bones of the arm act as levers of the third kind.
37. PROP. To find the condition of equilibrium when two
parallel forces act upon a straight lever, and to find the pressure
on the fulcrum.
Since the fulcrum is a fixed point, the only effect of either
of the forces tends to turn the lever round this point; and in
equilibrium, the tendency to turn it one way round must be
balanced by an equal tendency to turn it the other way round;
or the moments of the forces about the fulcrum must be equal
and opposite.
If a be the angle which they make
with it, we must have, in both figures,
P. AC sin a=Q. B C sin a,
or
P.AC=Q.BC,
P BC
Q=AC
which being independent of a, there
will be equilibrium in every inclina-
tion of the lever to the forces if there
be equilibrium in any one; and the
forces are inversely as the distances
from the fulcrum at which they act.
We may solve this proposition by
going through all the steps of Article
12, Chap. II.; for in equilibrium
the resultant of the two parallel forces
must pass through the fulcrum, and
P
A
C
R


Р
B
( '
F
66
ELEMENTARY MECHANICS.
be destroyed by its reaction; therefore the pressure on the ful-
crum is always equal to the algebraic sum of the parallel forces,
and acts in the direction of the greater force when they are op-
posite.
38. PROP. To find the condition of equilibrium when any
two forces in the same plane act upon a straight lever, and the
pressure on the fulcrum.
Let the forces P and Q
make the angles a and ẞ re-
spectively with the lever, as in
the figures, and let their direc-
tions when produced, if neces-
sary, meet in D. Since their
moments about C'must be equal
and opposite when there is equi-
librium, we must have
P. A C sin α=Q. B C sin ß.
For the pressure on the fulcrum
and its direction we must find
the magnitude (R) of the result-
ant of P and Q, and the angle
it makes with AB; since in equi-
librium it must be destroyed by
the reaction of the fulcrum.
P
P
By article 7 we have, in figure 1,
and
A
α
Fig. 1. D

C
Fig. 2.
R
A
B
R²= P² + Q²+2 P Q cos ADB,
ADB=180°-a-B,
... R² = P² + Q² −2 P Q cos (a+B).
R
B

D
In figure 2, ADB=B-a, and ADQ is the angle between
the forces;
··. R² = P² + Q² — 2P Q cos (B—α).
To find the inclination (0) of R to the lever in figure 1. By
resolving parallel and perpendicular to the lever, taking R the
reaction of the fulcrum opposite to the resultant of the forces,
we have
STATICS.
67
P cos a-Q cos B+ R cos 0=0,
P sin a+Q sin B-R sin 0=0.
(1)
(2)
From these equations (1) and (2), we have
tan 0=
P sin a+ Q sin B
which gives 0.
lower figure.
39. PROP.
Q cos B-P cosa
A similar expression may be found for the
To find the condition of equilibrium when two
forces in the same plane act in any manner on a lever of any
form.
If P and Q, acting as in the
figures at A and B respectively,
be in equilibrium about the ful-
crum C, and we draw perpendicu-
lars CM, CN upon their direc-
tions, we have, by the equality of
moments about C,
Οι
PxCM=Q× C N
P CN
QCM³
or in equilibrium the forces are
inversely as the perpendiculars
upon their directions from the
fulcrum.
When the directions of P and
P
P
M

M
C
B
A
B

Q, with respect to any given straight line through C, are known,
the magnitude and direction of the pressure on the fulcrum can
be found, as in the last proposition.
40. PROP. To find the condition of equilibrium and the pres-
sure on the fulcrum when any number of forces act in any manner,
in one plane, on a lever of any form.
Let P₁, P2, P3, &c. be the forces acting in the plane of the
F 2
68
ELEMENTARY MECHANICS.
figure at the arms CA, CB, CD, &c.
respectively, or at the points A, B,
D, &c. in a plane turning about an
axis through C perpendicular to it.
Take any two lines perpendicular
to each other through C, as Cx, Cy,
for the axes of co-ordinates.
Let P be any one of the forces
which makes the angle a with Cx,

P3
P₁
E
P₁
P₁
F
A
B
D
P₂
8
and let x, y be the co-ordinates of its point of application. Then
proceeding as in article 24, we shall have
at C, in Ca, a force
at C, in Cy,
""
(P cos a),
=Σ(P sin a),
and a resultant couple whose moment must=0 when there is
equilibrium, or we must have
(P cosa.y-P sin a . x)=0.
The resultant force at C will be destroyed by the reaction of
the fulcrum, and we have the pressure (R) upon the fulcrum
from the equation
R=√✓ {Σ(P cos a)}² + {E(P sin a)}².
If R makes an angle 0 with
Cx, we have
tan 0=
Σ(P sin a)
Σ(P cos a)
ON THE WHEEL AND AXLE.
41. This machine consists of a
wheel, A B, firmly fixed to a cylinder
or axle, CDEF, with a common axis,
GH. The pivots or extremities Go
and H of the axis generally turn in
steps which support the whole; and
the forces act by cords which are
wrapped round the wheel and the
axle.
B

C
F
H
D
E
W
A
P
STATICS.
69

The annexed figure being an end
view, we see that by drawing a hori
zontal line through the center of the
axis C, we have the power (P) and
the weight (W) acting at A and B,
the extremities of a straight lever
ACB; and in equilibrium we must
have their moments about C equal
and opposite; or
or'
P× AC=W × BC,
W AC
P˜B C
W/
B
C
A
P
W
P
The same relation will exist if the wheel and axle be turned
round either way; so that the machine may be called a perpetual
lever. Also P and W being forces equal respectively to P
and W, there will be equilibrium at whatever points the cords
leave the wheel and axle respectively, since the moments about
C remain the same as before.
ON TOOTHED WHEELS.
42. Toothed Οι Cogged
Wheels are thin cylinders, on
the circumference of which are
projections called teeth or cogs,
as in the figure. If two such
wheels have their cogs set at
equal distances, they will work
together, and if one be set in
motion, it will communicate
motion to the other through
the mutual pressure of the
cogs which are at any instant
in contact.
A
C
Av

B
C
P
W
t
Let S be this mutual pressure in the figure, which acts in
the line S'm'mS, a common normal to the cogs at their point of
70
ELEMENTARY MECHANICS.
contact; Cm, C'm' being perpendiculars from the centers C, C'
of the wheels, upon that line.
Let the power (P) act by a weight from a cord wrapped
round an axle with center C' and radius C'A, and let the
weight (W) act similarly from an axle with center C and radius
C B.
Taking the moments about C', we have, in equilibrium,
PxC'A=Sx C'm',
and about C,
WXCB=SxCm.
Dividing the latter by the former, we have
W CA
Cm
X
PC B C'm"
Now if the axles from which P and I act are of equal radii, the
effect of the combination will depend on the cog-wheels only, and
we have then
Cm
W
P¯¯¯C'm'
Let S'm'mS meet the line joining the centers C, C' in o, then
the triangles C'm'o, Cmo will be similar, and
Cm
Co W
C'm'¯¯C' o¯¯P'
Let the dotted circles be described with radii Co, C'o, these
are called the pitch-lines of the wheels, which roll uniformly
upon each other, when the cogs are made of a proper form. In
planning wheel-work these pitch-lines are first laid down, and
the places of the cogs marked at the same equal distances upon
each; their radii are to be so taken that there may be the re-
quisite number of cogs on each circumference; and as they are
at equal distances, their number will be in proportion to the
circumference;
W Co
circumference to radius Co
P¯Co¯¯¯circumference to radius C'o
number of teeth in wheel of H
number of teeth in wheel of P
STATICS.
71
The form of the teeth most used in machinery is, that the
inner part is formed by radii from the center of the wheel, and
the outer part by epicycloidal curves,-the bottoms of the spaces
being rounded to give strength to the teeth. See Willis's Prin-
ciples of Mechanism.
There are several forms of toothed wheels, which are all sub-
ject to the above rule. When the teeth project from the flat
face of the wheel instead of the edge, they form the crown wheel.
When the wheel contains very few teeth, it is called a pinion,
and its teeth or cogs are then called leaves. These forms may
be seen in most watches.
In the preceding instances the axes of the wheels working
together were either parallel, as
in the first form, or at right angles
to each other, as in the crown
wheel and pinion. But wheels
are in continual use in which the
axes form any given angles; and
the cogs are then placed on thin
frustums of cones, as in the figure,
in place of cylinders; and the wheels are called bevelled wheels.

ON THE PULLEY AND SYSTEMS OF PULLIES.
43. The pulley is a small wheel, with a groove on its edge to
admit a cord which passes over it; it turns round an axis or pivot
through its center, which is fastened to the frame-work, called
the block, in which the pulley moves.
It is called a moveable or a
fixed pulley, according as the block is moveable or fixed.

A fixed pulley serves merely to change
the direction of the force in the cord passing
over it; for, neglecting the friction of the
pulley, the tension of the cord must be the
same in every part.
P
44. PROP. To find the relation of the power to the weight in the
single moveable pulley with the cords parallel.
72
ELEMENTARY MECHANICS.
The annexed system consists of a move-
able pulley, from which the weight (W) is
suspended, and a fixed pulley over which the
cord sustaining the power (P) passes and has
the other end fastened at A.
Since the cord is supposed to pass freely
over the pullies, the tension will be the same
at every point, and the two vertical cords
from the moveable pulley sustain W;
or, if t= the tension,
2 t=W=2 P;
W
... P = 2
A

о
Р
W
If we take into account the weight of the moveable pulley,
we may either add this weight (w) to W, or we may suppose
the pulley counterpoised by a weight suspended with P, but not
reckoned with it.
Taking W+w= the whole weight sustained by the tension of
the two vertical cords, we have
2t=W+w,
W+w
ΟΙ
P=
2
45. PROP. To find the relation of the power to the weight in
the single moveable pulley with the cords inclined.
Let t the tension in
=
cord, which, being fastened
at A, passes freely over the
moveable pulley sustaining
the weight (W), and the
fixed pulley, and is there-
fore the same at every part
of the cord, and equals the
power (P).
Let w the weight of the
A

O
W
P
moveable pulley; a= the angle which the inclined cords make
STATICS.
73
with the vertical direction, and which is the same for both cords,
since the resultant of the two equal tensions must be a vertical
force sustaining W and the weight of the pulley.
Therefore, resolving in the vertical direction, we have
2t. cos a=W+w
or
W+w
P=
2 cos a
If w be neglected, or counterpoised independently, we have
P=
W
2 cos a
We note here, that the two parts of the cord can never be
drawn into one straight line, or a become 90°, whilst P and W
remain finite, however great P may be, and however small W
may be.
Amongst the various ways in which pullies can be combined,
there are three, which have been called the first, second, and
third systems of pullies.
46. PROP. To find the relation of the power to the weight in a
combination of pullies, with the cords parallel, where the power at
each pulley acts as weight to the next above
it. This is called the first system of pullies.
The figure represents this system with
three moveable pullies, a,, a, and ag. Let
W be the weight supported at the block of
the pulley a₁, and P the power acting at the
last cord after passing over the fixed pulley.
Let w₁, wą, wз be the weights of the
pullies a₁, 2, as respectively. Let t₁ be the
tension in the cord passing round the pulley
a₁, to that in the cord round a2, tg that in
the cord round ag.

2
Then, for the equilibrium of a,, we have
2t₁ = W+w₁
IT
a
aɔ
4-3
P
74
ELEMENTARY MECHANICS.
or
Ww₁
1₁ = 2 + 2/2.
For the equilibrium of a₂, we have
2t₂ = t₂+w2
or
W2
2
W2
WW₁
=22 + 22 + 2²
For the equilibrium of a,, we have
or
2tq=tq+ wg
ts=P= +213 +
22 + 23,
W
23 23 22
which gives the relation of P to W, as required.
If the system contains n moveable pullies, we have similarly
WW1
P= + +
102
+
W3
2n 2n 2n-1 2n−2+ &c. .
Wn
+
2
2
If the pullies are all equal, and their weights each equal to
w₁, we have
P= + (1+2+22+ &c. ...+2"-1)
2n
W
2n
w₁
+
(2n − 1)
Qn
n
W
+ Wi
2n
2n
In this system a part of the power is expended in sustaining
the pullies. If the pullies be counterpoised independently, we
have
W
=2n.
P
47. PROP. To find the relation of the power to the weight in
a combination of pullies in two blocks where the same cord passes
round all the pullies. This is called the second system of
pullies.
STATICS.
75
Fig. 1.
Fig. 2.
W
P
•
P
This combination of pullies is
the one in most common use. The
pullies are arranged in the blocks in
a variety of ways, but figure 1 repre-
sents the most usual. White's Pulley
is a modification of the system, in
which all the pullies of each block
were formed upon one piece of wood
or metal, with their circumferences
each proportional to the length of
cord which would pass over them
when put into use to raise weights,
so that all those in each block would
have the same angular motion, and
they might be then combined into
one wheel, by which a great propor-
tion of the friction in the common
form would be saved. This plan, which pro-
mised so well in a theory constructed upon im-
perfect data, has been found useless in practice,
from that fertile source of error, the omission of
essential natural properties, namely, here, the
elasticity of all cords, which is continually chang-
ing with the wetness or dryness of the atmosphere,
and the friction of the cords upon the pullies,
as well as the changing thickness of the cord in use.
the friction, we shall have the tension the same at every part of
the cord, and counting the number of cords at the lower block, we
shall have the number of times the tension which supports the
weight (II) and the weight of the block (w).


Therefore, if n=number of cords at the lower block,
nt = IT + w
W
Neglecting
but
t=P
IIT
W
··. P =
+
n
n
If, as in the figures, the end of the cord be tied to the
upper block, the number of pullies in each block will be the
76
ELEMENTARY MECHANICS.
same, =m say; and the number of cords at the lower block will
be 2m, and then
P=
W
+
พ
2m¹ 2m®
If the weight of the pullies be counterpoised or neglected,
we have
W
P
=n.
48. PROP. To find the relation of the power to the weight in
a combination of pullies when the cords are parallel, and each
attached to the weight. This is called the third system of
pullies.

In this system the uppermost pulley is fixed,
and its weight is supported by the beam to
which they are attached. Let w₁ be the weight
of the pulley a₁; w₂, w3 those of the pullies a2,
ay respectively. Let t,, to, tg, to be the tensions.
in the cords respectively, and
and
t₁=P,
t2=2t₁+w₁=2P+w₁,
4
t3=2t₂+w₂=2²P+2w₁+wq,
ts=2tz+w3=2³P +2²w₁+2w₂+w3,
W = t₁ + t₂ + ts + ts
1
2
3
= P(1 + 2 + 2² +2³) + w₁(1+2+2²) +
w₂(1+2)+wg
= P(2ª — 1) + w₁ (2³ — 1)+w₂(2² —1) +
w3(2—1).
If the system contains n moveable pullies,
we shall have similarly,
W
143
ua
W=P(2n+1—1)+w₁(2″-1)+w₂(2n-1-1)+ &c. ... +wn(2−1).
If the n pullies were of the same weight w₁, we should have
W= P(2n+¹ −1)+w₂ (2n+2n−1 +22-2+ &c. ... +2)-nw、
= P(2n+1-1)+w₁(2n+1—2—n)
=(P+w₁) (2n+1 — 1) — (n+1)w,.
STATICS.
77
In this system the weights of the pullies assist the power.
they are balanced independently, or neglected, we have
W
p=(2n+1-1).
P
ON THE INCLINED PLANE.
If
When a particle or body is in contact with any hard surface,
plane or curved, and the forces press it against the surface, the
normal reaction of the surface is one of the forces concurring to
produce equilibrium, and must be considered with the other
forces.
The inclined plane, as a mechanical power, is supposed per-
fectly hard and smooth (unless the friction is considered); and
having some angle of elevation above the horizontal plane, has a
heavy particle or body resting on it in equilibrium, by the action
of one or more forces.
49. PROP. To find the conditions of equilibrium when a body
rests on an inclined plane by the action of a force acting up the
plane.
Let the figure represent a section
A B, of the inclined plane by a verti-
cal plane through the body at a.
AC be a horizontal line, and the
angle BAC=a.
Let
Let P be the power acting up the
plane, R the reaction perpendicular
A
४
OL
R

P B
a
W
C
to the plane at a, W the weight of the body acting vertically
downwards.
These three forces keep the body in equilibrium; and two
out of the three quantities W, P, and a, must be given when
the other and R are required. Using the method of article 23,
taking the axis of x in the plane, the axis of y perpendicular to
it, with a for origin of co-ordinates, we have the equations of
equilibrium,
78
ELEMENTARY MECHANICS.
Σ(X)=0,
Σ(Y)=0;
or here
P-Wsin a=0,
(1)
R-W cosα=0.
(2)
From any point B in the plane, draw BC vertical; AB is
called the length of the plane, BC its height, and A C its base.
W
1
AB
From (1),
P
sin a
BC
length of the plane
height of the plane'
W
length of the plane
From (2),
R
base of the plane
50. PROP. To find the conditions of equilibrium when a body is
supported on an inclined plane by a force whose direction makes an
angle e with the plane.
P, W, and R being the forces in
equilibrium at a, in the figure, where
angle PaB=e, angle BAC-a.
solving parallel and perpendicular to
the plane, we have, in equilibrium,
P cos e- W sin a=0,
Re-
P

R
B
C
or
or
(1) A
R+P sin e
W cos a=0;
(2)
W
COS €
P
sin a'
R=W cos a—- P sin e
α
sin a.
sin e
- W cos a
W
COS €
=W
cos (a + €)
;
W
COS €
COS €
R
cos (a + e)'
W
If a +e=90, P=W, and R=0.
If e>90°-a, R is negative, and the body must be on the
under side of the plane.
STATICS.
79
51. PROP. To find the conditions of equilibrium when a body is
supported on an inclined plane by a horizontal pushing force.
The body being supported on the
plane at a by the action of the horizon-
tal force P in Pa; resolving parallel
and perpendicular to the plane, we have,
in equilibrium,
P cos α- W sin α=0,
a
R− P sin α — Wcos a=0.
a.
R

P
a
8
(1)
A
(2)
From (1),
TI
W
AC
base of the plane
or
=cota=
Р
BC
PW tan a;
B
C
From (2),
height of the plane
R=W cosa + P sin a
IV
COS α
W
AC
or
COS α=
Ꭱ
AB
base of the plane
length of the plane'
52. PROP. To find the relation between the weights of two
bodies which rest on two inclined planes having a common summit,
the bodies being connected by a cord passing over a pulley at the
summit, when they are in equilibrium.
Let a and a' be the bodies whose
weights are II and II. Let AB,
BC be the two planes.
Let a=
angle BAC, B= angle BCA, and
BD a perpendicular on A C.
If t be the tension in the cord
aBa', we have, for equilibrium on
the plane AB,
A
TI
B

D
W
and on plane B C,
or
t=W' sin a,
t = W'sin ß;
... I sin a= II' sin ß,
Ᏼ Ꭰ
И
A B
Ᏼ Ꭰ
FIZI
;
BC
80
ELEMENTARY MECHANICS.
or
W AB
WBC
or the weights are proportional to the lengths of the planes on
which they rest respectively.
ON THE WEDGE.
This mechanical power, not less simple than any of the
others, has been discussed very differently by different writers,
and those who have given the most elaborate solutions have
treated it the most erroneously. The student who wishes to
understand the mode of action of the wedge should consider
attentively the example 13, page 58.
53. PROP. To find the conditions of equilibrium on the isosceles
wedge.

By the wedge, we mean a triangular
prism whose perpendicular section is an
isosceles triangle, as in the figure, where
A is the section of the edge of the prism;
A B, AC sections of the sides; and C B
of the back. The prism is considered
hard, and perfectly smooth, if we do not
introduce the friction as one of the forces
involved.
R
P P
C
B
D
a
४
A
R
α
Draw AD bisecting the angle of the wedge, and let
BAD=CAD=α. Let 2 P be the power applied at the back
of the wedge, which is in equilibrium with the pressures (R) on
its two sides, which must be equal, and act through the same
point a in the power's direction, and also perpendicularly to
the sides of the wedge, since they are supposed perfectly
smooth.
Resolving in the vertical direction, we have
2 R cos RaA-2 P=0;
or
P
R=
sin a
STATICS.
81
ΟΥ
P
back of the wedge
sin a=
R
side of the wedge
which gives the force exerted by the wedge perpendicular to its
sides.
If there be more than one force acting on each side of the
wedge, their resultant in equilibrium must be R, as thus found;
and we require such additional data, in order to determine the
components, as will enable us to solve the triangle of forces when
one of the forces (R) is known.
Hatchets, knives, carpenters' chisels, &c., are examples of
different forms of the wedge.
ON THE SCREW.
The Screw consists of a projecting rib or thread passing round a
cylinder at the same angle with its axis everywhere. This screw
works in a hollow screw to which it fits.
54. PROP. To find the conditions of equilibrium on the
screw.
If we suppose one revolution of the thread to be unwrapped,
it forms an inclined plane, of which the height equals the di-
stance of two threads, and the base the circumference of the
cylinder.
Let ABC represent
the inclined plane formed
by the unwrapping of one
revolution of a thread, and
B C the distance of two
contiguous threads, A C—
circumference of the cy-
linder.
Let aangle BAC of
the plane, r-radius of
the cylinder.
P
F
A
α
E D

R
B
9
a
V &c.
獻
​Q
82
ELEMENTARY MECHANICS.
The power P acts perpendicularly at the extremity F of a lever,
which turning the screw round, produces a pressure in the direc-
tion of the axis of the cylinder of the screw. In equilibrium let
this pressure be balanced by a weight W.
Let the arm of the lever of P=FD=a. Let Q be the equi-
valent force to P at the circumference of the cylinder, so that
Q=P
FD
ED
=P
pa
α
r
We may suppose a small part (w) of W to be supported at each
point of AB; let q be the corresponding part of Q, and let R be
the perpendicular reaction of the thread at a.
Proceeding as in article 51, we have
BC
q=w tan a=w
AC
w.distance of two threads
circumference of the cylinder
and the same holds for all other points on the plane; therefore
we have
Q=W-
distance of two threads
circumference of the cylinder
a
=P.
jo
πα
=P.
2π r
or
P-W
distance of two threads
circumference described by P in one revolution
W_circumference described by P
or
P
distance of two threads
We note here, that the relation of P to W is independent of
the radius of the cylinder on which the thread is wrapped.
CHAPTER VII.
ON COMBINATIONS OF THE MECHANICAL POWERS AND
BALANCES.
By the mechanical advantage of any machine, we mean the
number of times the weight contains the power, or the value of
W
P
the fraction as used in the preceding propositions.
Recapitulating, we have for the elementary machines the
mechanical advantage,
In the lever,
In the wheel and axle,=
the arm of the power
the arm of the weight'
the radius of the wheel
the radius of the axle
the number of teeth in the wheel of I
the number of teeth in the wheel of P
In toothed wheels,
In the single moveable pulley, =2.
In the first system of pullies, =2",
where n = number of move-
able pullies.
In the second system of pullies,=n, where n = number of cords
at the lower block.
In the third system of pullies, =2"+¹-1, where n = number of
moveable pullies.
the length of the plane
In the inclined plane,
the height of the plane'
In the wedge,
In the screw,
the side of the wedge
half the back of the wedge'
the circumference described by the power
the distance between two contiguous threads'
G 2
84
ELEMENTARY MECHANICS.
55. PROP. To find the mechanical advantage of a combination of
any number of elementary machines.
We may suppose the machines to be connected together by
cords or rigid rods, and the tension or reaction in any cord or
rod will be the weight to the machine above, and the power to
the machine below. Let the number of machines be n; and let
t₁, të tâ, &c. . . . t- be the tensions or reactions in the con-
necting cords or rods.
1
Let P-the power for the whole combination.
W = the weight
"
""
رو
""
if
The mechanical advantage of the combination=
W tn-1 tn-2
tr tn-2 tr-3
1
— ·
•
n
An An−1 · An-2
•
ɑ1, ɑ2, ɑz, &c.
the separate machines.
W
P
to t₁
&c. .
t₁
P
•
an be the mechanical advantages of each of
Or, the mechanical advantage of a combination of machines
equals the product of the mechanical advantages of the separate
machines.
56. PROP. To find the mechanical advantage in the endless
screw.

This machine is a combination of the
screw and the wheel and axle. The wheel
has projections or teeth on its circumfer-
ence, set obliquely so as to fit the thread
of the screw, the power being applied at
the handle of a winch, by which the screw
presses against the teeth of the wheel and
supports a weight hanging from the axle.
W
ч
The mechanical advantage of the endless screw=
the circumference described by the power
X
P
radius of the wheel
the distance of two threads of the screw radius of the axle
STATICS.
85
EXAMPLE. Let the winch be 10 inches, the distance of con-
tiguous threads of the screw inch, the radius of the wheel 10
inches, and that of the axle 1 inch.
The mechanical advantage=
5
2π × 10 10
X
X
•2
=3141
or a power of one pound will sustain more than three thousand
pounds by such a machine.
57. PROP. To find the mechanical advantage of a combination of
levers.
C B
Let C, C', C, C be the
fulcrums of the levers. Let the
power (P) act at A, and the weight
(W) at B"; the last lever being "B"
of the first kind, and the other
three of the second kind. Let
B A', B' A", B" A"" be rigid rods
connecting the levers.
The mechanical advantage of
the combination=

B'
C'
A'
P
A”
C'
A"
B"
CA× C'A' × C" A" × C'"" {"
C B × C' B' × C" B" × Cꞌꞌꞌ B'II ·
W
Ex. Let the mechanical advantage of each lever10; the
mechanical advantage of the combination=10410000; or a
power of one pound will sustain a weight of ten thousand pounds
by such a combination of levers.
The weighing machine for carts and waggons is a combination
of levers; the first being a pair of framework levers, which support
the platform, on which the cart or waggon is placed when
weighed.
58. PROP. To find the relation of the power to the pressure
produced in the combination of levers called the knee.
86
ELEMENTARY MECHANICS.

A combination of levers like the
annexed figure is used with advan-
tage in cases where a very great
pressure is required to act through
only a very small space, as in coining
money, in punching holes through
plates of iron, in the printing-press,
&c. The lever, AB, turns about
a firmly-fixed pivot at A, and is
connected by another pivot at B to
the rod BC, whose extremity, C,
produces the pressure on the ob-
stacle, as E in the figure, being
retained in its proper motion by
R
A
M
B
N
P
F
L
C
D
E
some contrivance producing a similar action to the lever DC in
the figure.
Let the power (P) act horizontally at some point F, in the
lever A B. Let ANE be a vertical line meeting the direction of
P in N, and DE a horizontal plane with the substance subject to
pressure at E.
Let R= the reaction of the rod BC in the direction of its
length, and AM, DL perpendiculars upon its direction from A
and D. Let W be the vertical pressure of the substance at E.
Taking the moments about A and D, in equilibrium, we have
P.AN=R.AM
W.DE=R.DL
or
or
AN DL
W=P
DE AM
W AN DL
PAM DE
When BC comes nearly to the vertical direction, DL comes
nearly equal to DE, and A N becomes A F nearly, whilst A M is
very small.
So that we have
W AF
Р AM nearly, which is then very great.
STATICS.
87
ON THE COMMON BALANCE.
The common balance, as ordinarily constructed, is a bent
lever, in which we have to take into consideration the weight of
the lever itself.

In the figure, A and B
are the points from which
the scale-pans and weights
are suspended; C is the
fulcrum, being the lower
edge of a prismatic rod of
steel projecting on each
side of the beam; when
the balance is in use, these
A
B
C
edges on each side of the
beam, as at C, rest on hard
M
Cad
9
1
D
-
B
E
N
surfaces, so that the beam turns freely about C as fulcrum. In the
lower figure, let C, A, and
B be as before; AB being
the line joining the points
of suspension. Draw CD
a perpendicular on AB;
when the beam is symme-
trical on each side of C, its
center of gravity will be at
some point as g in CD.

The requisites of a good
balance are:
A
P
20
R
1st. That the beam rests in a horizontal position when loaded
with equal weights.
2nd. That the balance possesses great sensibility.
3rd. That it possesses great stability.
For the first condition, it is necessary that the arms are of
equal lengths, and that the beam is symmetrical on each side of
88
ELEMENTARY MECHANICS.
C, with the points of suspension and the center of gravity below
that point. When this is the case, we shall have the perpendi-
cular CD bisecting AB in D.
59. PROP. To investigate the conditions that a balance may
possess great sensibility and great stability.
A balance possesses great sensibility, when, for a small differ-
ence between the weights P and Q with which it is loaded, the
line A B is considerably deflected from the horizontal position in
which it rests when loaded with equal weights. Let MCEN
be a horizontal line through C, meeting AB in E, and let the
angle MEA=0. The sensibility depends on the magnitude of
O compared with the difference P-Q of the weights.
Let w
equal the weight of the beam. Let the horizontal line through
C meet the vertical lines through A and B in M and N respect-
ively; and those through g and D in a and d respectively. Then
M N is bisected in d.
or
ㅎㅎ
​ΟΙ
Taking the moments about C, we have, in equilibrium,
P.CM-Q.CN-w.Ca=0
P(Md-Cd)-Q(Nd+Cd)-w.Ca=0
(P–Q) AD cos 0— (P+Q) C D sin 0-w. Cg sin 0=0.
Let the length AD or BD of the arms =a, CD=d, and
Cg=h; then
(P−Q) a− { (P+Q) d+w.h}tan 0=0
.. tan 0=
(P–Q).a
(P+Q) d+w.h
We see that 0, and therefore the sensibility, will be increased
for given valucs of P and Q by increasing a, and by diminishing
w, d, and h; or, by increasing the lengths of the arms, by dimi-
nishing the weight of the beam, and by diminishing the di-
stances of the fulcrum from the center of gravity of the beam,
and from the line joining the points of suspension of the scale-
pans.
A balance possesses great stability, when, being loaded with
STATICS.
89
equal weights, on being disturbed it returns quickly towards its
position of equilibrium. The stability is therefore greater as
the moment bringing the beam towards its horizontal position is
greater.
as
Or, since P=Q,
or as
P.CN-P.CM+w.Ca
P(Nd+Cd-Md-Cd)+w.Ca
=(2 P.d+w.h) sin
is greater;
is greater.
This is greater for given values of P and e, as d, w, and h are
increased.
We see that the sensibility of the balance is diminished as
we increase the stability, but that we may increase the sensibility
without injuring the stability by increasing the length of the
arms.
We see also that balances must be adapted to the uses they
are to be applied to. The fine delicate balance of the chemical
laboratory must possess great sensibility, but we must not ex-
pect great stability. For weighing coarse wares, it is of more
consequence that the balance possesses great stability than that
it shows very small differences of weight, where the material
weighed is not of great value.
ON THE STEELYARD BALANCE.
The steelyard has a longer and a shorter arm, as in the
figure; the substance to be weighed being hung from the point.
A in the shorter arm, its weight is found from the distance to
which the constant weight P must be moved on the longer arm
in order to balance it.
60. PROP. To show that the divisions on the longer arm of the
steelyard, which correspond to equal additions of weight in the body
weighed, must be in a succession of equal distances.
Let the projecting knife-edge at be the fulcrum, and A
90
ELEMENTARY MECHANICS.
the point from which the
weight W is suspended.
Let P be the moveable
weight. When no weight
is suspended from A, the
longer arm will prepon-
derate; let a be the point
from which P must be
A
W
a
B

P
suspended to produce equilibrium. Take Cb Ca, then P at b
would balance P at a, if the lever was without weight, conse-
quently we may consider it without weight if we suppose a
weight equal to P to be hung from b. Let the steelyard be in
equilibrium from the weight of Wat A, of P at B and the effect
of the weight of the lever.
From the equality of the moments about C we have
W.CA-P.CB-P.Cb=0
or
W
CB+Cb
P
AC
a B
AC
If
W=P then a BAC
if
W=2 P
a B=2 AC
""
if
if
W-3 P
W=n.P
a B=3A C
a B=n.AC
or for every additional weight P, by which W is increased, the
moveable weight will have to be moved a distance equal to AC
further along the arm to balance it; and for multiples of P the
divisions will be in a succession of distances equal to A C, counted
from a.
This same rule holds for increments of W corresponding to
any fractional part of P. Let it be required to graduate the
steelyard for a succession of weights increasing by
P
m
From the expression
W a B
PAC
STATICS.
91
we have
W
P˜AC
a B
m
m
where N is any integer,
P
Let W=N.
กน
a B
N=
AC
M
AC
or
a B=N.
m
Giving N a succession of integral values, we shall have for
AC
a B a succession of distances counted from a, increasing by m
It is scarcely necessary to remark that the method supposed
to be followed above would never be the practical method of
graduating the steelyard; but the discussion is of use, to show
that the divisions must be at equal distances.
ON THE BENT LEVER BALANCE.
61. This balance is similar to the
figure, where BCD is the bent lever
turning about a pivot at C. A scale
(A) hangs from B; and at D an index.
points to some division on the graduated

arc.
D
10
m
B
W
A
Let g be the center of gravity of the
beam at which its weight w acts. The weight of the scale (4)
and the weight (IV) of a body placed in it will act vertically
through B. If we draw a horizontal line through C, meeting
the vertical lines through g and B respectively in m and b, we
must have, in equilibrium,
w.Cm-(A+W)Cb=0.
Now, as greater weights are put into the scale A, the point B
comes more nearly to the vertical line through C, from the bent
form of the beam; and the distance C'm increases, so that the
arc may be graduated from the positions of the index at D for
a succession of weights put into A. When the arc has been
-
92
ELEMENTARY MECHANICS.
thus graduated experimentally, the weight of a body placed in
the scale is told very quickly by the division on the arc to which
the index rises. When it is required to sort a great number of
bodies into classes of different weights, but where extreme nicety
is not essential, this balance is the most convenient for the
purpose.
ON ROBERVAL'S BALANCE.
62. This balance is of greater interest from its paradoxical
appearance than from its use as a machine for weighing bodies.
It consists of an upright stem upon a heavy base, A, with equal

do
F
B
a
C
}
k
O
b
A
P
crossbeams turning about pivots at a and b. These crossbeams.
are connected by pivots at c, d, e, f with other equal pieces in
the form of the letter T; the weights are hung from the horizon-
tal arms of the latter pieces; and the peculiar property of the
balance is that equal weights balance at all distances from the
upright stem; thus two equal weights (P), as in the figure, ba-
lance, although one may hang at B much nearer the upright
stem than C, the point from which the other hangs.
This property is easily
P2
P3

proved from the theory of
couples. Let the letters Q
in the annexed figure in-
dicate the same parts as
in the former.
Let equal and oppo-
site forces P, and P2, as
in the figure, act in ec,
B
d
R2
C
k
b
Qi
R1
P
P
P
P₁
A
STATICS.
93
each equal to P; and similarly let P, and P4 act in df. These
forces, P1, P2, P3, P4, will not affect the equilibrium; and P at
B is equivalent to P, at e and the couple P, Bh, P. Similarly,
P at C' is equivalent to P₁ at f, and the couple P, Ck, P3. The
forces P₁ at e and P₁ at ƒ will evidently balance.
1
1
2
4
4
The couple P, Bh, P, is equivalent to a couple Q₁, e c, Q₂ in
its own plane, of equal moment, in which we may take the
forces Q₁ and Q₂ acting in the directions of the crossbeams eb,
ac, which always remain parallel to each other as they turn on
the pivots. These forces Q, and Q, are destroyed by the resist-
ance of the pivots b and a; and similarly, if R₁, fd, R₂ were the
corresponding couple on the other side, R, and R₂ would be de-
stroyed by the resistance of b and a. These couples therefore
would not affect the equilibrium, and the original forces P at B
and P at C must be in equilibrium.
If the beams cad, ebf be moved round the pivots into any
oblique position, the same reasoning holds good, and the equili-
brium still subsists.
It is easy to see that unequal weights, as P and Q, could not
balance when hanging from any points.
CHAPTER VIII.
APPLICATION OF THE PRINCIPLE OF VIRTUAL VELOCITIES TO THE
MECHANICAL POWERS.
WE saw in articles 26 and 27 that the principle of virtual velo-
cities holds good for all cases of the equilibrium of a free body
under the action of any number of external forces.
We may consider the effective parts of the mechanical powers
as free bodies if the reactions of the parts which support them
be taken with the other external forces; and the internal re-
actions and tensions do not enter the fundamental expression
E(P. v)=0. We shall also generally have the virtual velocities
of the reactions of the supporting parts equal to zero for the
possible displacements of the system.
In some of the mechanical powers we have the principle ap-
plying to all possible displacements whether great or small, since
they are always in the direction of the forces,-as in the wheel
and axle, toothed wheels, the pullies with parallel cords, the
inclined plane, the wedge, and the screw. In the lever and
the pullies with inclined cords we must take the displacement
indefinitely small.
63. PROP. To show that the principle
of virtual velocities holds good for the wheel
and axle in equilibrium.

The forces which act on the wheel and
axle are the power P, the weight W, and
the reaction R of each of the steps which
support each end, C, of the pivot about
which it turns. When the wheel and axle
receives a displacement turning about C,
the virtual velocity of R equals 0.
A'
A
R
B
B'
P
W
P'
W
STATICS.
95
Let A and B be the points at which the cords left the wheel
and the axle respectively before displacement; A', B' afterwards.
Then W ascends through the space W W' = arc B B', and P de-
scends through PP' arc AA. WW' is a negative virtual
velocity if P P' be positive.
or
or
or
By the principle of virtual velocities,
P× PP'-W× WW'=0
Px arc A A'-W× arc B B'=0
PxACx angle A'CA-Wx BC x angle BCB'=0
PXAC-W × BC=0
W AC
PBC
X
the condition of equilibrium as found in article 41.
64. PROP. To show that the principle of virtual velocities holds
good in a pair of toothed wheels.
Let the circles in the figures
represent the pitch-lines of the
wheels which roll on each other
without slipping, and let 0,, 02
be the points which were in
contact in the line C C before
disturbance. Then the other
letters being as in article 42,
we have
arc 0 0₁ = arc 0 02,

C
A
ó
O
B
P
V P
W
W
P's displacement,
=P P'
=AC'x angle O₁C'O
=AC' ×
arc 0,0
C'O
W's displacement,
= WIV!
= C B x
arc 0, 0
CO
By the principle of virtual velocities,
PxPP-W× WIF'=0
96
ELEMENTARY MECHANICS.
or
P× AC'×
arc 0, 0
W× CBX
C'O
arc (2
CO
=0
A C
CB
P.
W =0
C'O
CO
W AC
CO
or
when A CB C
P¯¯¯ B C
C O
CO
C'O
as found for the condition of equilibrium in article 42.
65. PROP. To show that the principle of virtual velocities holds
good in the single moveable pulley with the cords parallel.
In the figure, if the pulley A be raised to A', we
shall have A A=WWP P', since each of the
cords passing round the pulley A must be shortened
by a length = WW'. And WW' is a negative virtual
velocity;

gives
.. P × P P' — W × WW' =0
P× PP-W × ≥P P' = 0
PP'
IA'
P
A
W
P'
or
p=2
W
the condition of equilibrium as found in article 44.
66. PROP. To show that the principle of virtual velocities holds
good in the first system of pullies.
Taking the same figure as in article 46,
we see that if P descended through a space
PP',
the pulley a, would be raised a space PP

Az
P
О
the pulley an-1"
""
&c.
the pulley as
وو
ܙܕ
the pulley 2
دو
•
P P
""
22
&c.
P P
دو
2n-2
PP'
2n-1
PP
"
2n
W
دو
وو
the weight Wor the pulley a₁
and the equation of virtual velocities
E
ar
STATICS.
97
or
PXP P-W× WW'=0 becomes
P P
PX PP-W×
=0
2n
W
=2n
P
the condition of equilibrium when the weights of the pullies are
counterbalanced or neglected.
67. PROP. To show that the principle of virtual velocities holds
good in the second system of pullies.
Referring to the figure in article 47, we see that if the weight
be raised through a space WW', each of the n cords at the lower
block must be shortened the same quantity, or P must descend
through a space nx WW". The equation of virtual velocities
is
PxPP-W× WW'=0
which becomes Pxn.WW'—W × WW'=0
or
W
12.
P

the condition of equilibrium as in article 47.
68. PROP. To show that the principle of
virtual velocities holds good in the third system of
pullies.
Taking the same figure as in article 48, we
see that if W be raised a space = WW', each cord
will be shortened in consequence a space equal to
it. The highest moveable pulley, an, will de-
scend through the same space, WW'. The next
pulley, an-1, will descend through a space 2. IV IV',
in consequence of the descent of a,, and WW' in
consequence of the elevation of IV, or will de-
scend on the whole (2+1) W IV'.
Similarly, the pulley an-2 will descend through
{ 2 (2 + 1) + 1 } W W' = (2² + 2 + 1) II'I''.
H
W
163
13
P
98
ELEMENTARY MECHANICS.
Proceeding in the same way, we find that pulley ag, or
ɑn—(n−3), will descend through the space
(2n−3
-3+2n−4+&c.
3
+2+1)WW!
and a through the space (2"-2+2n−³ +&c. . . . +2+1)WW'
(2n-1+2n−2+ &c. ..+2+1) WW'.
a₁
>>
""
P descends through twice the last found space in consequence
of the descent of the pulley a,, and through the space WW', in
consequence of the elevation of the weight;
or
1
P P'=WW' { 2 (2n−¹ +2n−2+ &c.
= WW' (2n+¹ — 1).
The equation of virtual velocities is
. +2+1)+1}
PXP P-W×WW'=0
which becomes
P× WW' (2n+1-1) — Wx WW'=0
or
W
P
2n+1-1
the condition of equilibrium when there are n moveable pullies, as
in article 48.
69. PROP. To show that the equation of virtual velocities holds
in the inclined plane.
Taking the most general case,
when the force (P) makes any angle
e with the plane. Let a=<BAC;
a the first position of the body whose
weight is W; a' the position of it
after a disturbance.
Drawing the perpendiculars av, A
a'u, we have av, the virtual velo-
city of P=—a a' cos e, and a u, the
virtual velocity of W=a d' sin a.

or
By the equation of virtual velocities,
Px av-Wx au=0
છે
P × a a' cos e — W × a a' sin α = 0
a
16
2
TI
P
ལ
B
C
or
W
P
COS €
sin a
as found for the condition of equilibrium in article 50.
STATICS.
99
70. PROP. To show that the principle of virtual velocities holds
for the wedge.
Let 2 P be the whole power, R and
R the pressures perpendicular to the
smooth sides of the wedge ABC, which
produce equilibrium.

Let the wedge be displaced to the
position A'B'C'. The displacement of
the point of application of P is a a' in
the figure, = A A'; that of b, the point
of application of R, is bb'Am, a per-
pendicular from A on A B'; and
Am A A' sin
BAC
2
The equation of virtual velocities is
P
P
G
B
B
R
h
25
C
la'
C
R
or
Pxaa-Rx bb' = 0
BAC
PxAA-Rx A4' sin
=0
2
P
·R=
BAC
sin
2
as found in article 53 for
the condition of equili-
brium of the wedge.
71. PROP. To show that
the principle of virtual velo-V P
cities holds for the screw.
Suppose the power P
to act by means of a cord
passing over a pulley upon
a wheel, as perpetual lever,
fixed to the cylinder of the
screw.
Let A be the point
where the cord left the
P'
A

A
C
I'
V W
H 2
100
ELEMENTARY MECHANICS.
wheel when the power was at P and the weight at W; and let
A', P', W' be their positions after a disturbance,
or
P P'arc A A'A Cx angle ACA
WW' distance of two threads ×
By the equation of virtual velocities,
PxPP-W× WW'=0
angle ACA
2π
P× AC× angle ACA' — W × distance of two threads x
angle ACA
or
=0,
27
W
2п АС
P
distance of two threads
the condition of equilibrium as found in article 54.
72. PROP.
To show that the principle of virtual velocities holds
in a lever of any form.
Let ACB be the lever
before displacement, A'CB'
its position after.
From A draw Av per-
pendicular to AP, and from
B', B' u perpendicular to B Q
produced. Av is the virtual
velocity of P, Bu that of Q,
and negative. Now, when
the displacement is indefi-
nitely small, the circular arcs
A
P
p/
A
A A', B B' become straight lines, and
and
C

B
Q
Q
A v=A A' × cos A' Av=A C× angle ACA' × cos (PAC—90°)
=A C.sin PAC. angle ACA
Bu=B C.sin QBC. angle BCB'
angle ACA'=angle BCB'.
The equation of virtual velocities is
Px Av-Q× Bu=0
B'
STATICS.
101
or
or
P.A C. sin PAC-Q.B C.sin QBC=0
P BC.sin QBC
QA C.sin PAC
as found in article 39 for the condition of equilibrium.
73. PROP. To show that the principle of virtual velocities holds
in the single moveable pulley with the cords inclined.
Let A be the point where the
cords produced would meet at
the first position of the pulley,
when P and W are the positions
of the power and the weight.
Let P be displaced to P'
when the weight is raised to W',
or the point of meeting of the
B
C

小
​P
m
A
P/
W
B
cords to A'. Draw the circular arcs, A'm, A'n, with centers B
and C. When the displacement is indefinitely small, the arcs
A'm, A'n become straight lines, and
Am-A A' cos BAA'—An
BAC
PP'=Am+An=2AA' cos
2
WW = A A'.
The equation of virtual velocities is
PxPP-W× WW'=0
and becomes P × 2AA' cos
BAC
2
W× A A' =0
W
BAC
or
=2 cos
P
2
as found for the condition of equilibrium in article 45.
or
In the preceding propositions, the expression
PxPP'W×WW¹
WW P
PPW
explains the principle that, "in using any machine, what we
102
ELEMENTARY MECHANICS.
gain in power we lose in time." For, in order that W
may be
raised through any given space, we must have the space moved
through by P increased in the same ratio that the magnitude of
P is diminished.
EXAMPLES ON CHAPTERS VI. VII. AND VIII.
1
Ex. 1. A lever 30 feet long balances itself upon a prop
of its length from the thicker end; but when a weight of 10
pounds is suspended at the other end, the prop must be moved
2 feet towards it to maintain the equilibrium; show that the
weight of the beam is 90 pounds.
Ex. 2. The forces P and Q act at arms a and b respectively
of a straight lever which rests upon a fixed point to which it is
not attached. When P and Q make angles a and B with the
lever, show that the conditions of equilibrium are
P cos a + Q cos ẞ=0
Pa sin a-Q b sin ß=0.
Ex. 3. The beam of a false balance being of uniform density
and thickness, it is required to show that the lengths of the arms
(a and b) are respectively proportioned to the differences between
the true () and apparent weights (P and Q). Or to show that
the weight of the beam being considered, we have
Ex. 4. Two given
α
P-WV
b¯¯¯ W — Q
>>
weights hanging vertically from two
given points in the rim of a wheel, find the position in which
the greatest weight will be sustained on the axle.
Let P and Q be the weights, a the angle contained between
the radii drawn to the points of suspension, which are given.
Let be the angle which the lower radius (that of P) makes
with the vertical direction. It is shown by means of a subsidiary
angle, and without the differential calculus, that
P+Q cos a
tan 0=
Q sin a
STATICS.
103
Ex. 5. If I be the mechanical advantage of a lever, and s
the mechanical advantage of a screw, show that the mechanical
advantage of the common vice is l. s.
Ex. 6. Apply the principle of virtual velo-
cities to find the relation of the power to the
weight in the endless screw, as found in article
56.

Ex. 7. Show that, in the annexed system of
pullies, W=5 P. Apply the principle of virtual
velocities to find the same result.
Ex. 8. In the annexed system of three move-
able pullies, with the cords at each pulley inclined
60° to each other, show that
W=P(3)*.
Obtain the same result by
means of the principle of
virtual velocities.
Ex. 9. Show that in
every combination of the
mechanical powers, the re-
lation of the power to the
weight may be found by the
principle of virtual velo-
cities.
to
W
W.
P

i
CHAPTER IX.
ON THE EQUILIBRIUM OF A SYSTEM OF WEIGHTS SUSTAINED BY
CORDS OR BEAMS.
74. If a weight W be hung from a knot C at which the cords
A C, BC (supposed without weight) meet, and the lengths of the
cords and the position of A and B are given; to find the tensions
in the cords.
The geometrical data give the angles.
which AC and B C make with the ver-
tical direction; let them be a and ẞ re-
spectively.
2
If t, and to be the tensions in the
two cords AC and B C respectively, we
may find their values by the properties
of the parallelogram of forces or other-
wise analytically as in article 23, as follows :
Resolving horizontally and vertically,
t, sin a-t, sin ß=0
2
t₁ cos a + t₂ cos ß— W=0.
B
12
Τ
W
C
Multiply (1) by cos ß, (2) by sin ß, and add, we find
t, (sin a cos ẞ+ cos a. sin B) - W. sin ß=0
(1)
(2)
A

or
Similarly,
W.sin B
t₁ =
sin (a+B)*
W sin a
tq=
sin (a+B)*
We should find these results at once from article 6; but the
equation (1) shows us a property in addition, which we shall find
STATICS.
105
to hold for systems of any number of cords and beams, namely,
that the resolved parts of the tensions horizontally are the same
in each.
75. PROP. To investigate the form and properties of the
funicular polygon, or a system of cords connected by knots from
which given weights are hung.
Let
Let ABCDEF be the A
polygon formed by cords.
connected by knots at the
points B, C, D, E.
AF be a horizontal dis-
tance which is given, as
well as 1, 2, 3, &c. the
lengths of the cords AB,
BC, CD, &c., which we
suppose without weight.
Let W₁, W., W₂, &c. be
the weights suspended
from B, C, D, &c. re-
spectively.
The above being given,
we have to determine the
3.
α11
1
1
E
33
B
WI
βι
aq
W
D
W
C
02
W
G
W3/
F

tensions in the cords, and the angles they make with the vertical
direction.
Let t₁ = tension in AB, t=tension in BC, t=tension in
CD, &c.
Let a₁, B₁ be angles of the cords with the vertical direction at B.
az, B₂
""
"}
C.
dg, Bz
&c.
D.
>>
""
&c.
If n be the number of cords, we have n tensions, and 2(n-1)
angles to determine, from the following equations:
106
ELEMENTARY MECHANICS.
AF-1, sin a-l sin a₂-&c. . . . -In-1 sin an—1—
I, sin ßn−1=0
(a)
2
1½ cos a₁ + 1½ cos a₂+lz cos a3+&c.
•
··
+In-1 cos an−1+
In cos (π-B₂-1)=0.
(b)
1
(1)
Resolving horizontally and vertically at each knot, we have
at B, t₁ sin a₁—t₂ sin ß,=0,
at C, to sin α-t。 sin ẞ2=0,
&c.
tn-1sinan-1-tnsinẞn-1=0,
t₁ cos a₁ + t₂ cos ẞ₁-W₁=0
2
1
t₂ cos a₂+tз cos ẞ2-W₂=0 (2)
2
&c.
tn- cosan-1+tn cosẞn-1-
Wn-1=0.
(n-1)
We have also ẞ₁+α₂=π, ß₂+αz=π, &c. . . Bn-3+an-1=π,
which are n-2 equations.
We have, consequently, the equations (a) and (b), 2(n−1)
and n-2; or, in all, 3n-2 equations, to find the n tensions.
and 2 n-2 angles as required.
=
Since B₁+α=π, we have sin ẞ, sin a, and so onwards,
sin ẞ₂=sin a¸, &c.
sin Bn_=sin an; therefore we see
2
2
1
that the first of each pair of the equations (1), (2), (n − 1)
shows the horizontal component of the tension in each cord to
be the same; for we have
2
t₁ sin a₁ = t₂ sin ß₁=t₂ sin aq=t, sin ẞ₂=&c. . . . =t₂ sin ß₂-1
but
t₁ sin a₁ =t, cos A=tn sin ẞn-1=t₂ cos F
α
1
or the resolved parts of the extreme tensions in the line AF are
equal and opposite.
From the equations (1), (2), . . . (n-1), we find by elimi-
nating alternately one of the tensions in each pair,
W
tą
t₁
sin ß,¯sin (α, +B₁)
sin
a
t2
W 2
#8
sin B₂sin (2+ẞ2) ¯¯¯ sin
2
α
sin a2
&c.
&c.
&c.
STATICS.
107
tn-1
Wn-1
tn
sin Bn_1 sin (an - +Bn-i) sin d
From these we find the horizontal tensions to be
W₁
W₂
2
cot a₁+cot B₁ cot a₂+cot B₂
Wn-1
= &c.
&c. . .
1
cot an+ cot bai
1
When the weight of a chain supplying the place of the cords
of the polygon is taken into account, the problem becomes one of
considerable difficulty, and is connected with the construction of
chain bridges.
76. PROP. Three uniform beams connected together form a
triangle, ABC, with AB horizontal, and have a weight, W, hung
from C; to find the reaction in each of the beams, and to show that
the horizontal part is the same in each beam.
Since the triangle is given,
the angles which the sides.
AC, BC make with the
vertical direction will be
known let them be a and ß
respectively.
Let R₁ be the reaction in
2
A
C

a
R₁
R
II
B
the beam A C, R₂ that in BC; and let b₁ be the weight of A C,
b₂ that of B C.
Since the beams are uniform, the center of gravity of each
will be at its middle point, and we may consider half the weight
of each to be at each end, so that the whole weight hanging
from C will be I++ b 2
1 =W!
W say.
2
Resolving horizontally and vertically at C, we have
R₁ sin a-Ro sin ß=0
R₁ cos a + R₂ cos B-I=0.
(1)
(2)
The equation (1) shows that the horizontal parts of the reactions
arc equal.
108
ELEMENTARY MECHANICS.
Eliminating alternately between the equations, we have
R₁
WI
R2
sin ẞsin (a+B) sin a
and the horizontal part of R, or R2, or, as it is called, the horizontal
thrust when the beams form a roof, is equal to
WI
cot a + cot ß
which is the reaction in the horizontal beam A B, called the tie-
beam.
If the side A B were taken away, and its place supplied by a
cord, the last expression would be the tension in the cord. If the
points A and B rested on a smooth horizontal plane, we should
have the pressure on the plane
at A=R₁ cos a +
and at B=R2 cosß+
b₁
20/09 201
77. PROP. A number (n) of beams connected by hinges form a
frame-work in a vertical plane, having weights hung from the hinges;
to find the position of equilibrium, and to show that the horizontal
reaction at each hinge is the same.
Let ABCDEF be the
frame-work, of which the ex-
tremities A and F are fixed
points. Let AH and FH,
a horizontal and vertical line,
be given, as well as the lengths R,
b₁, bq, bз, &c. of the beams.
The beams being given,
B
βι
Rs.
аз
R₂
B2
02
D

E
F
W.
W
3
VWA
WI
H
A
1
we may suppose the weight to be applied at the ends, and W,
W₂, Wз, &c. to be the whole weights acting at the hinges B, C,
D, &c. respectively.
Let a₁, B₁ be the angles which the beams AB, BC make
STATICS.
109
with the vertical line at B; a, Ba, ag, B3, &c. the corresponding
angles at C, D, &c.
Let R₁, R2, R3, &c.
R₂ be the reactions in the beams
commencing at A.
We shall have to find n reactions, and 2(n-1) angles. From
the geometrical relations we have
A H−b, sin a, b, sin a,— bg sin ag—&e. ...−b, sin Bn-1=0 (a)
H-b₁ α
FH-b₁ cosa,-b₂ cos a₂-b₂ cos ag- &c. . . . —b₂ x
bn
ɑ2—b3
b¸
3
cos (π-B-1)=0.
(b)
Resolving horizontally and vertically at each hinge, we have, in
equilibrium,
2
at B, R₁ sin α, — R₂ sinß₁ =0
at C, R₂sina-R¸ sinß₂ = 0
=0
&c.
R₂-1 sin an-1-R, sin ẞ-1=0
2
R₁cosα, +R₂cosß₁-W₁ =0 (1)
1
R₂cosα₂+R₂cos B₂-IF₂=0
2
&c.
2
(2)
Rn-1 cos an-1+ R₂ cos ẞn-1-
W-1=0.
(n − 1)
We have also B₁+α₂ =πT, B₂+αg=πT, &c. ... ẞn-2+αn-1=π,
or n-2 equations.
These equations being 3n-2 in number, suffice to find the n
tensions and 2 n-2 angles.
The expressions would have been identical with those for the
funicular polygon if we had taken Fin the horizontal line through
A; and they lead to the same consequences.
The first of each pair of equations (1), (2), &c. shows that the
horizontal component of the reaction at each hinge is the same;
and by the same process as in article 75, we show that it is the
same at every hinge, and equal to
W .
cot a,+cot B₁cot a₂+ cotß,
W 2
= &c.
&c...
2
W
n-1
cot a₂-1+ cot ẞr - 1°
CHAPTER X.
ON FRICTION.
WE have hitherto considered the surfaces on which bodies
pressed to be perfectly smooth, so that they offered no resistance
to motion parallel to themselves, their only reaction being per-
pendicular.
When rough surfaces are in contact, the motion, or tendency
to motion, parallel to the surfaces, is affected by the roughness,
and we call the effect friction.
Experiments have been made to determine the laws of friction,
which we may subdivide into rubbing friction, when one body rubs
on the other, and rolling friction, when one rough surface rolls.
upon another: the former only will be considered here, under the
term friction or statical friction.
m
F
تا
78. If a body rest, as at A, upon a rough plane, BC, it is
found that a force, within certain
limits, may act upon it parallel
to the plane without motion en-
suing, as would be the case if the
plane were smooth. The greatest
force which can be so applied,

B
II
A
C
without the body moving, measures the friction. If I be a
weight acting by a cord passing over a pulley, as in the figure,
on the body A, when motion is about to take place, F being the
opposing force of friction which balances W, we have F=W.
It is found that if we put various weights, as m, upon the body
A, then W or F is proportional to the weight of A and m, or is
proportional to the pressure perpendicular to the surfaces in
contact. It is also found that it is independent of the magni-
tude of the surfaces in contact, the friction being the same when
the pressure is the same, whether the surface of the body be
increased or diminished; except in extreme cases, where the
STATICS.
111
pressure is exceedingly great compared with the surfaces in con-
tact. So that if R be the pressure on the plane which is equal
and opposite to its normal reaction, we have
F=mR
where μ is a constant depending on the nature of the surfaces in
contact, and called the coefficient of friction.
We may determine the coefficient of
friction by placing the body on a plane
of which we can increase the inclination
to the horizon until the body begins to
slide down.
79. PROP. To show that the coeffi-
cient of friction between two given sub-
α
R
a
IT'
μR

stances is the tangent of the inclination of the plane formed of
one of the substances, when the body formed of the other is about to
slide down it.
The body a in the above figure is in equilibrium from the
normal reaction of the plane R, the friction μR acting up the
plane, and its weight acting vertically. Resolving parallel and
perpendicular to the plane, we have
or
μ R-IV sin a=0
Eliminating R, we have
R-IF cos x=0.
µ cos a — sin a=0
μ=tan a.
80. PROP. To find the limits of the
ratio of P to W on an inclined plane,
when friction acts up or down the plane.
Let the power P, as in the figure,
make an angle e with the plane whose
inclination to the horizon is a. Let I
be the weight of the body.
R
α
P

R
W
u R
Friction being considered as an inert force resisting the ten-
112
ELEMENTARY MECHANICS.
dency to motion, will act up or down the plane as the body is on
the point of moving down or up respectively.
Resolving parallel and perpendicular to the plane, we have
P cose +μR-Wsin a=0
P sin e+R— W cos a= 0.
Multiply (2) by μ and subtract and add, we have
P (cos eμ sin e)-W (sin aμ cos a)=0,
(1)
(2)
or
W
cos e μ sin e
P
sin au cos a
where the upper sign is to be taken when friction acts up, and the
lower when friction acts down the plane.
No notion will occur whilst the relation of P to W lies between
the two values.
81. PROP. To find the limits of the ratio of P to Win the screw
when friction acts assisting the power or the weight.
Proceeding as in article 54,
let ABC be the inclined plane
formed by the unwrapping of
one revolution of the thread;
the angle BAC=a.
Let W be the whole weight
sustained by the screw; w be
the part of it supported at a;
Q the whole force acting at
the circumference of the cy-
linder, whose radius E D=r,
P
F
A
9
α
E
D

KR
FD=a the lever at which the power P acts, and
. . P = = Q
グ
​R
W
and let q be the part of Q which supports w at a.
MR
B
C
The forces
which are in equilibrium at a are the weight w, the reaction R,
the friction acting up or down the plane μ R, and the horizontal
pressure q.
STATICS.
113
or
and
Resolving parallel and perpendicular to the plane, we have
q
cos a ±µ R—w sin a=0
sin « — R+w cos a=0.
q a
Multiply (2) by μ, and add and subtract, we have
9 (cos a±µ sin a) — w (sin a Fµ cos α) =0
W
q
W
W cos a+μ sin a
Qsin a μ cos a
a cos atμ sin a
α
Pr sin aµ cos a
The two values of this expression give the limits required.
(1)
(2)
82. PROB. A uniform beam rests on a cylinder of given
radius (a); to find the weight (W) which may be hung from one
end, so that the beam may be just about to slide off when friction
acts.
Let g be the center of gravity of the
beam B C, whose length is 2B; and Bg=b,
since the beam is uniform. Let w be its
weight. Before the weight W was hung
at B, the point g must have been at A,
the highest point of the cylinder. Let A'
be the point of contact when the beam is
on the point of sliding off the cylinder, в
and let a be the angle which the beam
then makes with the horizontal direction.
IP
R
1
tu
Resolving parallel and perpendicular to the beam, we have
A
0
μR

μ
R-w sin a- W sin a=0
R-
—w cos a — Icos a=0
whence μtan α.
or
Taking the moments about A', we have
but
W × BA' cos a-w × A' g cos a=0
W(Bg-A'g)-w.A'g=0
A' g=arc A A'= radius × angle AOA'
= α a
I
(1)
(2)
114
ELEMENTARY MECHANICS.
or
..W (b-a α)—w.aα=0
w.aa
W=
b-aa
Ιμ
w a tan
tan-¹μ
b- a tan -1
the weight required.
83. PROB. A ladder rests with its
foot on a horizontal plane, and its upper
extremity against a vertical wall; having
given its length (1), the place of its center
of gravity, and the ratios of the friction
to the pressure both on the plane and on
the wall; find its position when in a state
bordering on motion.
R
M'R'

R'
B
8
C
A
μη
W
If AB be the ladder in the figure,
whose length A B=l, and weight acting.
at the center of gravity g=W, Bg=b,
'the coefficient of friction against the
vertical wall, μ against the horizontal plane AC, and angle
BAC=0; we find
tan 0=
l—b (1 + µ µ')
μι
DYNAMICS.
CHAPTER I.
ON DEFINITIONS AND THE LAWS OF MOTION.
WHEN forces produce motion, or change of motion, in bodies,
their effect being different to that in statical problems, we require
other methods of measuring them, in addition to the statical
measures.
The motion of a body may be considered only with respect
to its change of place, either absolutely, or relatively to some
other body which is itself in motion. It may also be considered
with respect to the power the body acquires of overcoming ob-
stacles; and then the magnitude of the body itself has to be
considered.
The velocity of a body is its rate of motion, and it is measured,
when uniform or constant, by the space passed over in a unit of
time, or by the space in any time divided by the time. The
units of time and space must be known. Thus we say, he
travelled at the rate of thirty miles per day; he travelled at the
speed or velocity of eight miles per hour; the bullet was fired
from the gun with a velocity of 1000 feet per second.
When no
mention is made of different units, we shall take a foot for the
unit of space, and a second for the unit of time.
Let v be put for velocity, s for space in feet, t for time in
seconds, we have for a uniform velocity,
v = space described in one second
space described in t seconds
t seconds
= 1/1
or s=vt.
I 2
116
ELEMENTARY MECHANICS.
When the velocity is continually changing, it is measured by the
space passed over in an indefinitely short space of time divided by
the time.
Let s be the space described with a variable velocity in t seconds,
s' that in t' seconds indefinitely near the former time; then, using
the symbol & to signify difference,
s' — s
ds
ย
t'
t-t
St
When pressures such as we have considered in Statics are not
balanced, the body on which they act will be put in motion. Such
pressures may act on the body during a definite time, or act
through a definite space, and then cease to act.
When pressures act only for a very short space of time, they
are called impulsive forces,-as, for instance, the mutual pressure
of bodies which impinge, the force of the string exerted upon an
arrow shot from a bow, &c.
We call a force an accelerating force whilst it continually
increases the velocity of a body; and a retarding force whilst
it continually diminishes it. A uniform or constant accelerating
force is one that increases the velocity of the body uniformly,
or adds the same amount of velocity to the previous velocity of
the body in every successive equal interval of time. A uniform
retarding force is one that diminishes the velocity of a body
according to the same law. Variable accelerating or retarding
forces are those whose effects on the velocities of bodies are con-
tinually changing.
By the moving force acting on a body, we mean the mass moved
multiplied by the accelerating force, which we shall shortly see,
by the third law of motion, is proportional to the pressure exerted
on the body, by whatever means that pressure arises, whether by
the unbending of a spring, by the attraction of other bodies upon
it, by the explosion of gunpowder, &c.
By the momentum of a moving body, we mean the mass of the
body multiplied by its velocity. The mass multiplied into the
square of the velocity is called the vis viva of a moving body.
DYNAMICS.
117
As forces are measured by the effects they produce, the
dynamical measure of an accelerating force must be the velocity
which it generates in the body in a given time, when uniform or
constant. Let f represent the force, and let the given time be
taken as the unit, we have f = the velocity generated in a unit of
time. Since the force is constant, the same velocity is added in
every unit of time; therefore, if v be the velocity acquired from
rest by the action of the force in t units of time, we have v=ft.
From the above definitions we see that moving force is measured
by the momentum generated in a unit of time.
From the expression v=ft, we have
f=
ย velocity generated in any time.
time
When the force is variable, and changing from one instant to
another, we must take the time indefinitely small; so that, for a
variable force, we have
f=
velocity generated in an indefinitely small time
time
.
v-v
Sv
t' — t d t
if v = velocity of the body at the time t, and v' the velocity at the
time t' indefinitely near to t.
In the above, it will be shortly seen that we have anticipated
the first law of motion.
By the path of a body we mean the line, straight or curved,
which it describes in passing from one point to another in space.
The pressure produced by the weight of a body depends
upon the attraction of gravitation towards the earth, which is
sensibly different at different parts of the earth, and upon the
mass of the body; and varies directly as the force of gravity
when the mass is the same, and directly as the mass when the
force of gravity is the same; therefore, by the rules of algebra,
when both vary, the weight varies as their product. Let m be
the mass, w the weight of the body, and g the force of gravity: we
118
ELEMENTARY MECHANICS.
have wamg; and taking our units of measure of w, g, and m
พ
accordingly, we may put w=mg, or m=—.
g
The relative velocity of two bodies is the velocity with which
they approach each other, or separate from each other.
A body is said to move freely when its path depends on the
action of the impressed forces only. Its motion is said to be
constrained when its path is limited to be in a given line,
straight or curved, or to be upon a given surface. A stone
thrown in any direction from the hand is an example of the
former; and a pendulum swinging, or a body rolling down a hill,
are examples of the latter.
ON THE THREE LAWS OF MOTION.
The first law of motion. When a body in motion is not acted on
by any external force, it will move in a straight line, and with a
uniform velocity.
We may satisfy ourselves of the truth of the first part of
the law, that the body not acted on by any external force will
describe a straight line, from the consideration that whatever
reason could be alleged for its deviating to one side, as good a
reason could be given for its deviating to the opposite; and since
it could not move in two directions at the same time, the reasons
could not be valid, and therefore it would move in a straight
line only.
The second part of the law we conclude to be true by in-
duction from the results of our experience. If a body be thrown
along a rough surface, it deviates from a straight path, and soon
loses all its velocity; if it be thrown along a smoother level
surface, it moves in a path nearer a straight line, and with a
velocity more slowly diminished; if it be thrown along a sheet
of ice, it moves very nearly in a straight line, and retains its
motion for a considerable time. If a heavy ball be suspended
DYNAMICS.
119
by a very fine thread in a vessel from which the air has been
withdrawn by the air-pump, and it is set oscillating, it remains
in motion for a very great length of time, although it still
experiences resistance from the remaining air in the vessel, and
the want of perfect flexibility in the thread. We therefore con-
clude that if we could remove all the resistance to a body in
motion, it would retain its velocity unchanged.
The second law of motion. If several forces act upon a body
at the same time, each produces its full effect in the direction of
its action, whether the body be at rest or in motion.
We see this to be true when a ball is rolled along the deck
of a vessel moving uniformly in smooth water: the ball takes
the same course along the deck as it would if the vessel were at
rest; and if it struck any body in its motion, the forces called
into play in the collision would produce the same effects in the
two cases. If a body be let fall from the top of the mast, it
falls to the foot of the mast, whether the vessel be at rest or in
motion. A more complete proof is afforded by experiments
with the pendulum. Whatever be the vertical plane in which it
oscillates, at the same place on the earth's surface, whether north
and south, east and west, or in any other azimuth, the time of
oscillation is the same; showing that the effect of gravity on the
pendulum is unaffected by the rotation of the earth on its axis,
and by its motion in its orbit.
A
The third law of motion. When a pressure acting on a body
puts it in motion, the moving force, measured by the
momentum generated in a unit of time, is proportional
to the pressure.
This law is proved experimentally by Atwood's
machine, which consists of a pulley A, having its axle
resting on two friction-wheels at each end, as repre-
sented by the dotted circles. These friction-wheels
have their pivots, or axles, accurately and delicately
supported, so that the pulley A may be subject to as little a
effect of friction as possible. A clock which goes for
a few minutes, with a second's pendulum and dead-

P
120
ELEMENTARY MECHANICS.
beat escapement, is connected with the frame-work supporting
the machine.
Let P and Q be the weights of the bodies suspended from
the ends of the cord passing over the pulley A. If P-Q, they
will balance; but if one (P) be greater than the other (Q), it
will descend and draw the other upwards, with a force-P-Q,
which is the pressure setting in motion the masses of P, Q, and
P Q
of the wheel-work A. The masses of P and Q are and
g
respectively. Let I represent the inertia or mass of the wheel-
work, acting upon the cord connecting P and Q.
P+Q
g
g
+I, and the pressure
The whole mass set in motion
producing motion=P-Q. Now it is found that when these
two quantities are kept in the same proportion to each other,
the velocity acquired in a unit of time is the same; or the
pressure varies as the mass moved, when the velocity is constant.
If part of Q be taken successively away and added to P, so
that the mass moved remains the same, it is found that the
velocity acquired in a unit of time is proportional to P-Q; or
the pressure varies as the velocity acquired in a unit of time,
when the mass moved is constant.
When both the mass and velocity change, by the rules of
variation, we have generally,
pressure varies as mass moved multiplied by the velocity
generated in a unit of time,
varies as mass multiplied into accelerating force;
-varies as moving force.
ON THE PARALLELOGRAM OF VELOCITIES.
1. PROP. If two velocities be impressed upon a body at the
same instant, the actual velocity of the body will be represented in
direction and magnitude by the diagonal of the parallelogram
formed upon the lines representing the impressed velocities.
DYNAMICS.
121
B
ხი
63
D
Let a body at A have a velocity
impressed upon it which would carry
it with a uniform motion from A to
B in a given time, and another velo-
city at the same instant which would
carry it similarly from A to C in the
same time. If we complete the pa-
rallelogram ABDC, the actual path of the body will be the dia-
gonal AD, described in the same given time.

A
C1
C3 C3
C
By the second law of motion each velocity is impressed in-
dependently of the other, and if the body had passed over the
spaces Ac₁, A c₂, Ac3 in any times in the direction of A C, it
would have passed over the spaces c₁ b₁, cq b₂, cg by parallel to A B,
in the same times respectively; so that the latter spaces bear to
the former respectively the same ratio that AB bears to AC,
and therefore the points b₁, ba, bg will be all in the straight line
AD, which is the diagonal of the parallelogram. The body
will also have arrived at D in the same time as it would have
passed over either of the spaces A B or A C.
3
2. PROP. Having given a velocity, to find the component
velocities in any directions at right angles to each other; and
having given two component velocities at right angles to erch
other, to find the resultant velocity of a body.
Let Ax, Ay be the directions, at y

>
right angles, in which the components
are required. Let AD represent the
velocity (v) of the body in direction
and magnitude, a the angle which
AD makes with Ax.. If we com-
plete the right-angled parallelogram
ABDC, the side A B will represent
the velocity (a) in Ax, and AC (b)
that in Ay.
Then
C
D
A
B
AB-AD cos a
a=v cos a
A CAD sin a
b=v sin a.
01'
Again, if the components a and b in Aa, Ay are given to
122
ELEMENTARY MECHANICS.
find the magnitude and direction of the resultant velocity v, we
have
v²=a²+b²
b
and
tan a=- as required.
a
We have seen in Statics that if two bodies are in equili
brium from their mutual pressures, their actions upon each
other must be equal and opposite; so also with two bodies in
which motion takes place, the mutual pressures or actions are
equal and opposite. By the third law of motion, the moving
force is proportional to the pressure; and hence when two bodies
move by the effect of their mutual actions, the moving force pro-
duced in each of the bodies is the same in magnitude, but oppo-
site in direction.
This consequence of the third law of motion is called the
principle, that action and reaction are equal and opposite, each
being measured by the momentum generated in a given time,
the effect being considered uniform during that time.
CHAPTER II.
ON THE IMPACT OR COLLISION OF BODIES.
WHEN two bodies in motion impinge, they exert a mutual but
varying pressure, during an interval of time which is generally
very short. The forces called into play are subject to the prin-
ciple that action and reaction are equal and opposite; and since
this is true at each instant of the mutual pressures, the whole
effects of the impulsive forces will be subject to the same principle.
When it is the final and completed result that we require,
we have only to consider the mutual pressures in the collision to
have produced their full effect and to have ceased; we can, how-
ever, find the circumstances of the motion of the bodies during
the short interval of mutual pressure, by employing a higher
analysis than can be admitted in this treatise. For example,
if the mutual pressures of

the bodies A and B in the
figure acted by a spiral
spring of known elasticity,
A
GOOOOOOOOD
0 0 0 0 0 0
B
the circumstances of the motion of each body during the com-
pression of the spring are easily determined.
When natural bodies impinge, we have a similar case to the
one just taken for example.
If two surfaces of india-rubber be pressed against each other,
we see them flattened by the pressure, but recover their former
shape as the pressure is removed. Although not so perceptible,
the same takes place in other elastic bodies.
During the impact of two elastic bodies, the force urging
them towards each other is called the force of compression; and
the opposing force, causing them to separate again, is called the
force of restitution. The ratio which the force of restitution
bears to the force of compression in any bodies of the same ma-
terial or substance, is nearly the same for all degrees of compres-
124
ELEMENTARY MECHANICS.
sion, and measures the elasticity of the substance. The value of
this ratio is called the modulus of elasticity.
The whole force of compression is measured by the momen-
tum destroyed during the approach, and the whole force of
restitution by that generated during rebounding; and the mo-
mentum destroyed or generated in one of the bodies equals that
destroyed or generated in the other, because action and reaction
are equal and opposite.
Let e be the modulus of elasticity of two bodies whose masses
are A and B, which, moving with their centers of gravity in the
same straight line, meet directly with the opposite velocities a
and b, and rebound with the opposite velocities a' and b' respect-
ively: let v be the velocity supposed in the direction of a, common
to both bodies at the instant when compression ends and rebound-
ing commences; we have
whole force of restitution
modulus of elasticity=whole force of compression
momentum generated in either A or B in rebounding
momentum destroyed in either A or B during compression
which gives us
€=
A (a' + v)
A(a-v)
or
€ a — € v = a' to
B(b'-v)
and
€ =
or
eb+ev=bv.
B(b+v)
Adding, we eliminate v, and have
€ =
a' + b'
a+b
velocity of separation
velocity of approach
so that when experiments are tried in which the velocities of
rebounding can be determined for any given velocities of impact,
the above ratio gives the modulus of elasticity.
Bodies suspended by fine cords, and allowed to oscillate in
circular arcs of given radius about a fixed point as center,
acquire, as will be shown in the chapter on constrained motion,
velocities at the lowest point which are proportional to the chords.
of the arcs fallen through, and similarly they ascend through arcs
DYNAMICS.
125
of which the chords are proportional to the velocities impressed
upon them at the lowest point of the arcs. They are also shown
to fall down all small arcs of the same circle in the same time.
If two bodies be suspended in this manner, so as to impinge
at the lowest points of the arcs they describe, and be let fall
simultaneously from given points, we know the velocities with
which they impinge, and by observing the arcs through which
they ascend after rebounding, we know the velocities of re-
bounding, and so have sufficient data to determine the modulus
of elasticity.
All known solid bodies are imperfectly elastic; that is, in all,
the force of restitution is less than the force of compression; but
we have none without some force of restitution, or which are per-
fectly non-elastic. Experiments to prove the mathematical results
for the impact of non-elastic bodies require to be tried with balls
of soft clay recently worked up, so that they adhere together after
impact, or balls of wood with metallic spikes fixed in one of them,
which entering the other ball prevent them separating again after
impact.
We are indebted to the excellent experiments of Mr. Eaton
Hodgkinson for our more accurate knowledge of the properties of
impinging bodies. The following Table of Moduli, and the rule
for bodies of different hardnesses, are from his results:-

Modulus of
Elasticity.
Perfect elasticity
Glass ..
Hard-baked clay
•
Ivory
Limestone
1
Steel (hardened)
Cast iron
Steel (soft).
•
•
•
4
•
1.00
·
·94
•
.89
• •
.81
•79
·
*79
+ •
73
•67
Bell-metal
• ·
Cork
• • •
Brass
Lead
·
+
• •
1
•
•
* 晶
​•67
'65
Elm-wood, across the fibres
•60
•
•
-
•41
•
•20
Clay, just malleable by the hand
·17
126
ELEMENTARY MECHANICS.
In the impacts between bodies whose hardness differs in any
degree, the resulting elasticity is made up of the elasticities of
both, according to the following formula:
modulus of elasticity from both =
h'e + he
h+h'
where h and h' are the relative hardnesses, and e and e' the moduli
of elasticity respectively of the bodies. Or putting the compres-
1
h
1
sibility =c=1, and d=1, then the modulus of elasticity
cε+ c'é
c+d'
From this rule, it results that if one of the bodies is much
harder than the other, the effective elasticity is that of the softer
body nearly.
3. PROP. To find the common velocity of two non-elastic spherical
bodies after direct impact.
Let A and B be the masses
of the bodies as in the figure,
moving in the line joining

A
B
their centers as indicated by the arrows, so that they impinge
directly, and not obliquely, when A overtakes B.
Let a be the velocity of A, b of B, and a greater than b. Let v
be the velocity after impact, which is to be found. The velocity
lost by A is (a-v), and that gained by B is (v—b).
Since action and reaction are equal and opposite, we have the
momentum lost by A equal to the momentum gained by B, or
A(a—v)=B(v-b)
whence
v=
Aa+Bb
A+B
When the bodies are moving in opposite directions, we must
take the velocities with contrary signs. Let B's velocity be -b,
we have
V
Aa-Bb
A+ B
If A a=Bb, then v=0, or the bodies remain at rest after
DYNAMICS.
127
impact. If Bb is greater than Aa, then v is negative, or in the
direction of B's motion.
4. PROP. The velocities of two imperfectly elastic spherical
bodies of the same substance which impinge directly being given, it
is required to find their velocities after impact.
Referring to the figure of the last Prop., let A and B be the
masses of the impinging bodies A and B; let a, b be their velo-
cities respectively before impact; a', b' those after impact, which
are to be found.
Then, ife be the modulus of elasticity of the bodies, we
have
and
€ =
velocity of separation
velocity of approach
b' — a'
a_b
momentum lost by A-A (a-a′)
momentum gained by B=B(b-b).
(1)
Since action and reaction are equal and opposite, we have
A(a-a') B(b'—b).
=
(2)
Substituting in (2) the values of b' and a' successively from (1),
we have
Aa+Bb Be(a—b)
a':
A+B
A+ B
Aa+Bb, Ae(a—b)
b'
+
A+B
A+B
If B be moving in the opposite direction to A, and we con-
sequently take its velocity negative and equal to -b, the expres-
sions become
Aa-Bb Be(a+b)
a' =
=
A+B
A+ B
A a—Bb, A e(a+b),
b' =
+
A+ B
A+B
128
ELEMENTARY MECHANICS.
If B were at rest, or b=0, the expressions become
a'_a(A-B €)
=
A+B
b'
A a(1 + €)
A+B
We see that b' can never in this case =0, but a' will =0 if
A
B=
€.
If in the preceding expressions we put e=1, the results will be
those for perfectly elastic bodies.
If we take A=B, and e=1, when the bodies meet, moving in
opposite directions with any velocities a and b respectively, we have
the expressions becoming
a = (a - b) -
(a + b) = — b
b' = (a—b) +
-b
(a + b) = a
or in direct impact the balls exchange velocities.
If we have a series of n perfectly elastic and equal balls ar-
ranged in one straight line, and the ball at one extremity be pro-
jected against the
next, each ball will

Ө
Өөөөөе
remain at rest after striking the succeeding one until we come to
the last, which will fly off, having the velocity with which the
first was projected. It is immaterial at what distances the balls
be placed; and if in contact, the impinging of the first ball ap-
pears to produce no visible effect but causing the last one to fly
off with its velocity of impact, the others remaining stationary.
The same holds good for imperfectly elastic balls, if each bears
to the one it strikes the ratio e:1, as we see above in the case of
A
B
=e, when a'=0 whatever a may be, when b=0; also b'=a e, or
the velocity of each ball is less than that of the preceding one.
5. PROP. If any two spherical bodies of the same substance
impinge directly, the motion of their center of gravity after impact is
the same as before impact.
Since the bodies impinge directly, their common center of
DYNAMICS.
129
gravity, both before and after impact, will be always in the line
joining their separate centers of gravity; or its motion will be
always in this line.
Let u be the velocity of the common center of gravity of the
bodies A and B, moving in the same direction with velocities a and
b respectively, before impact.
Let x1, x2, ≈ be the distances of the centers of A and B, and of
their center of gravity, from any fixed point in their line of motion,
at any instant; x', ', ' the same quantities after an interval of
time t, so that
X x₁ = x₁+at
1
1
x 2 = x 2 + b t
π =x+ut.
By the properties of the center of gravity (articles 34 and 16 in
Statics), we have
(A+B)x=Ax₁ + Bx2
(A+B)π'=Ax₂'+Bx2'
(1)
or substituting for ', x, x' the values above, the latter equation
becomes
(A+B)(x+ut) = A(x₁+ a t) +В(x₂+bt).
Subtracting (1) from this, we have
2=
A a+ Bb
A+B
which is the same as the velocity of the center of gravity of non-
elastic bodies after impact.
Let u' be the velocity of the center of gravity of the bodies after
impact, when imperfectly elastic; we have similarly,
A d'+ B b'
u!
A+ B
Substituting the values of a' and b' from article 4, we have
u =
A Aa+Bb_Be(ab)) +
4B (
A+B A+B A+B
Aa+Bb
B Aa+Bb Ae(a — b)
A+B A+B A+ B
+
A+B
=u
or the velocity of the center of gravity is unchanged by impact.
K
130
ELEMENTARY MECHANICS.
When a body impinges against a fixed plane surface, we can
determine its motion after impact, if the direction of the motion
and velocity before impact be given, together with the modulus of
elasticity between it and the plane.
The angle which the direction of the motion before impact
makes with the perpendicular to the surface at the point of impact
is called the angle of incidence; and the angle which the direction
of the motion after impact makes with the same line is called the
angle of rebounding or of reflexion.
6. PROP. If a smooth non-elastic body impinge on a smooth, non-
elastic, hard, and fixed plane, it will, after impact, slide along the
plane.
Let the body A impinge upon the
plane BC at B, making an angle a with
BN, the perpendicular to the plane at
B; let the velocity of A be v.
The component of v parallel to the
plane v sin a; this will not be changed

A
α
B
N
C
by the impact, since the ball and the plane are smooth. The
velocity perpendicular to the plane after impact will be nothing,
since there is no elasticity; therefore the body after impact will
slide along the plane with the velocity v sin a.
7. PROP. To determine the velocity and direction of motion after
impact, when an imperfectly elastic spherical body impinges obliquely
on a smooth, hard, and fixed plane.
Let the body A in the figure impinge
on the plane BC at B, making the
angle of incidence ABN=<; let a'=
the angle of reflexion NBD.

Ρ
N
D
M
B
C
Let the line PB represent v the
velocity of the body before impact;
draw PN parallel to BC, then PN represents the velocity of
the body parallel to the plane =v sin a, and BN represents the
DYNAMICS.
131
velocity perpendicular to the plane v cosa. If e be the modulus
of elasticity between the body and the plane, the velocity perpen-
dicular to the plane after impact will be ev cos a=M B, if we
take
M B
=¤.
NB
From M measure MQ-NP, the parallel velocity, which is
unchanged in impact, since the plane and body are smooth; BQD
is the direction of the motion after impact, and the line B Q repre-
sents the velocity. Now
B Q²=M Q²+B M²
=v² sin²a + €²v² cos²α.
Or if u be the velocity after impact, we have
u=v v sin² a + €² cos²
α.
Also
MQ
PN
tan a'=
MBC.NB
tan a
€
which give the direction of the motion and the velocity after
impact.
If the elasticity be perfect, or e=1, we have
a²= a
and u=v.
8. PROP. To find the path of a body which, having passed
through one given point, after rebounding from a given plane passes
through another given point.
Let MBE be the given plane ;
A the point through which the body
is projected; D the point through
which it passes after rebounding from
the plane.

N
A
D
M
B
E
C
Draw a perpendicular AMC to
the plane through A, and if e be the modulus of elasticity be-
MC
tween the body and the plane, take
AM
= €. Draw through the
:
K 2
132
ELEMENTARY MECHANICS.
points C and D the line CBD, cutting the plane in B. Join AB ;
then AB, BD is the path required, or the body projected from A
in the direction AB will rebound in the direction B D.
For the angle of reflexion DBN= angle BCM, and angle of
incidence ABN=angle BAM,
.. tan DBN=tan BCM
BM
MC
BM
€. MA
tan ABN
€
or the angles ABN and DBN have the required relation as found
in the last proposition.
EXAMPLES IN IMPACTS.
Ex. 1. Two bodies A and B, whose elasticity is m, moving in
opposite directions with velocities a and b, impinge directly upon
each other; show that their distance at a time t after impact is
tm(a+b).
Ex. 2. If two perfectly elastic balls, whose masses are in the
ratio 1:3, meet directly with equal velocities; show that the
larger one remains at rest after impact.
Ex. 3. When two perfectly elastic bodies impinge directly, show
that the sum of the vires vivæ after impact equals their sum before
impact; or prove the expression
A a² + Bb²= A a¹² + B b¹².
Ex. 4. When two perfectly inelastic bodies, A and B, moving
in the same direction, impinge, show that
A+B : A :: relative velocity before impact: velocity gained by B.
DYNAMICS.
133
Ex. 5. An imperfectly elastic ball is projected from a point in
the circumference of a circle, and after twice rebounding from the
circle returns to the same point again; show that the direction of
projection makes an angle a with the radius drawn to the point of
projection which is given by the equation
tan a=
€
√1+E+€²
CHAPTER III.
ON UNIFORM ACCELERATING FORCES AND GRAVITY.
ACCORDING to the definition of a uniform accelerating force in
Chapter I., the velocity generated by it in the same time is always
the same, and by the second law of motion, is unaffected by the
previous motion of the body. If we put f= the force, measured
by the velocity it generates in a unit of time, we shall have the
velocity generated in t units=ft. Writing v for this velocity
acquired by the body at the end of the time t from rest, we have
therefore.
v=ft.
9. PROP. To find the relations of the space, time, and force
when a body moves from rest under the action of a uniform
accelerating force.
The velocity of the body is continually increased from 0 up to
ft, if t be the time and f the force. Lets be the whole space
described in the time t and let t be divided into n equal intervals,
The velocities at the end of the times
each
t
n
t 2 t 3t 4t
&c...
(n−1)t
,t,
N ท n n
n
will be respectively
2 3
ft,
4
N
n
N
n
&c. ... f(n − 1)t
fut
n
n
n
får, fut fot
Now if the body moved uniformly during each interval of time with
the velocity it had at the beginning of the interval, from the expres-
sion space = velocity × time, we should have the whole space s
equal to the sum of this series :
DYNAMICS.
135
0+ ft²
2 t2
net
n2
+f
3 t2
no
+ &c. .. +f
(n − 1)ť²
-
n2
=ƒ g{1+2+3+&c. ... + (n−1)}
=
f
n
t²¸n(n−1)
N
X
ft² ft²
2 2n
2
If the body had moved uniformly during each interval of time
with the velocity it had at the end of the interval, we should have
s equal the sum of this series:
12
n2
2f2
+f.
3 t2
no
2
no
2
+
ft² ft²
2
2n®
= ƒ / (1+2+3+ &c. ... +n)
ft² (n+1)n
+ &c. . . . +f(n−1)ť²
2
+fnt?
t2
222
Since the velocity is continually accelerated, the true value of s
will be between these two quantities, however small each interval
may be, or however great n may be; but when n is indefinitely
great, the last terms in each of the above expressions vanish, and
we have therefore
s = { ft².
10. Between the two equations v=ft, and s=ft², we may
eliminate either for t, and thus obtain.
v² = 2fs
s= } v t.
The expression s=vt shows us that the space described from
rest by the action of a uniform accelerating force is one-half of the
space which would have been described in the same time if the
velocity had been constant and equal to its value at the end of the
time.
If we put t=1 in the equation s=ft², we have ƒ= 2s; or f,
136
ELEMFNTARY MECHANICS.
which is the velocity generated by the force in a unit of time, is
measured by twice the space through which the body falls in a
unit of time. It is found that a heavy body in our latitudes falls
through a space of nearly 161 feet in the first second of time;
therefore, if we put g=the accelerating force of gravity, we have
g= velocity of 32.2 feet per second of time, or, with the under-
standing that one second is our unit of time, we write g=32.2
feet.
This value of the force of gravity is only an approximate value
for small heights above the earth's surface.
Sir Isaac Newton's law of universal gravitation is, that every
particle of matter attracts every other particle with a force which
varies directly as the mass of the attracting particle, and inversely
as the square of the distance. It is also shown that a spherical
body equally dense at equal distances from its center attracts a
particle outside its surface as if the matter of the sphere were col-
lected at its center; so that, considering the earth such a sphere,
if g be the force of gravitation at the surface, f the force at any
point exterior to the surface at a distance r from the center, and R
be the earth's radius, we have
or
f:g::
1 1
p2 · R2
R2
f=92
21%2
if μ=gR2
r.2
μ in this expression is called the absolute accelerating force; for it
is the value of ƒ when r=1. When R is taken for unity, r must
be expressed in terms of the earth's radius, and μg=32.2
feet.
When r is very near R, or for small heights above the earth's
surface, we have f=g very nearly, or we may take gravity as a
constant accelerating force in that case. It is found that, on
account of the diurnal rotation of the carth on its axis, its figure
differs sensibly from spherical, being flattened at the poles and
bulging at the equator, or is an oblate spheroid. The centrifugal
DYNAMICS.
137
force (see article 31), being produced by the diurnal rotation, is
greatest at the equator, and is there directly opposed to the force
of gravity. It is also nothing at the poles. The resultant gravi-
tation of a heavy body is affected by the direct action of the cen-
trifugal force, and its indirect action through the change of the
figure of the earth. The ratio for the equator and poles is as
follows:
gravitation at the equator: gravitation at the pole :: 186: 187.
We
e can, for these reasons, only consider gravity as constant for
the same latitude on the earth's surface, and for small altitudes
above it. The direction of gravity at each point on the earth's
surface being perpendicular to the surface taken as that of still
water, does not pass accurately through the center of the oblate
spheroid.
11. PROP. A body being projected with a given velocity u in the
direction in which a uniform accelerating force f acts, to find its
velocity, and the space passed over in a given time.
If v be the velocity, s the space described at the end of the time t,
we shall have, by the second law of motion,
v = velocity of projection + velocity from the action of the force
=u±ft
where the upper sign is to be taken when the force accelerates the
velocity, and the lower when it retards it.
or
In the same way,
+
space described space due to velocity of projection the space.
due to the action of the force
s=ut + 1 ft²
the upper and lower signs to be taken as before.
12. PROP. A body being projected with a given velocity u in the
direction in which a uniformly accelerating force facts, to find its
velocity when it has passed through a given space.
Let v be the velocity when the body has passed through the
space s.
138
ELEMENTARY MECHANICS.
Let h equal the space through which the body must pass to
acquire the velocity u by the action of the force. Then by art. 10,
u² = 2 fh
and for the space h±s we have
v² = 2f(h±s)
=2fh+2fs
=u²±2fs.
The signs to be taken as in the last proposition.
13. PROP. To find the time of flight or of ascent and descent
of a body projected in an opposite direction to the action of the
force.
Let the body be projected from A with a velocity u.
After ascending to some point B, it will return to A
again. From the expression v=u-ft, we have for
t=0, or at time of projection v=u; at B, the highest

И
2u
point, v=0, or t=7=time of ascent; when t=7,
have v=u
-ƒ(24)
we
=-u; or the velocity acquired in
descending during any time is equal to the velocity
lost in the same time in ascending.
B
2
u² — v²
Also, from v²=u²- 2fs, we have s=
—
A
; or s
2f
is the same when v is the same; if A C=s, we have the velocity
at C the same when the body is ascending as when it comes to the
same point again in descending; and the time in passing from C to
B in the ascent is the same as in falling from B to C in the
descent.
Consequently the whole time of flight from leaving A to coming
to it again is t=
2 u
f
If v² > u², s is negative, and must be measured below A.
DYNAMICS.
139
EXAMPLES ON THE DIRECT ACTION OF GRAVITY AS A CONSTANT
ACCELERATING Force.
Ex 1. Find the velocity a stone will acquire in falling during
four seconds by the action of gravity near the earth's surface.
In the expression v=gt, we have here t=4, g=32·2
.. v=4×32·2=128.8
or the stone has acquired a velocity of 128.8 feet per second.
Ex. 2. Find the space the stone in the last question had fallen
through in three seconds.
Using the expression s=1 g t², we have
space required = 1 × 32·2 × 3²
=144·9 feet.
Ex. 3. Find the velocity the same stone had acquired when it
had fallen through a space of 150 feet.
The expression
v²= 2gs gives us
v² = 2 × 32•·2 × 150
=9660
or v=98.3 feet per second nearly, which is rather more than the
velocity which would be acquired in three seconds.
Ex. 4. A heavy body is projected directly downwards with a
velocity of 100 feet per second; what is its velocity at the end of
five seconds?
The formula
v=u+gt gives
v=261 feet per second.
Ex. 5. A heavy body is projected directly upwards with a
velocity of 100 feet per second; find its velocity at the end of five
seconds.
We have now the expression
v=u―g t
140
ELEMENTARY MECHANICS,
and here
v=100-32·2 × 5
-61.
The negative sign shows that the body is coming down again, and
the velocity is 61 feet per second.
Ex. 6. Find how long the body in the last question continued
to ascend.
or
At the highest point the velocity=0
.. 0=100—32·2 × t
100
t= =3.1 seconds.
32.2
Ex. 7. To find the height to which the body in Ex. 5 ascended.
The formula for this case is v²=u² - 2 gs, and at the highest
u2
point v=0..s=
or
2g
space ascended
1002
2×32·2
=155.2 feet.
Ex. 8. A body is projected vertically upwards, and returns to
the same point in ten seconds; show that the velocity of projection
was 161 feet per second.
Ex. 9. Show that when a body falls from rest by the action of
a uniform accelerating force, the spaces described in successive
equal intervals of time are as the series of odd numbers, 1, 3, 5,
7, 9, &c.
Ex. 10. A stone being let fall into a well, it is heard to strike
the water in 7 seconds; required the depth of the well.
Let s be the depth of the well, m the velocity of sound = 1100
feet per second nearly.
DYNAMICS.
141
Then T= time of the stone falling + the time of sound returning.
2 s S
+
9
m
or
m v Q
s+ √s.
= mT
ng
a quadratic equation in ✔s; whence
v
√5= ± √√√ Tm+
m²
m
2g
√2g
which gives s, and the upper sign only is admissible.
Ex. 11. A body is thrown vertically upwards with a velocity u ;
find the time of its being at a given height h.
Let t be the time required,
whence
h=ut - 1 g t²
2 u 2 h
=0
9
g
or
u± √ u² −2 g h
9
The lower sign gives the time as the body ascends, and the upper
sign the time as it comes down again, when at the height h. At
the highest point there will be only one value of t, and
•. √ u²—2gh=0, or h=
as we should find by other modes.
u2
2g
Ex. 12. A body of given elasticity is projected upwards with a
given velocity u to strike a horizontal plane, and in t seconds
returns to the point of projection; required the distance of the
plane from that point.
Let the body be projected from A
with the given velocity u, and strike
the horizontal plane at B with the ve-
locity v.
Let the velocity of rebounding at B
be v', and the modulus of elasticity be e.
B

A
142
ELEMENTARY MECHANICS.
Let A B=s, the distance which is to be found.
Let t, be the time of ascending to B.
رو
te
>>
""
Then
descending from B to A.
and
2
1
t₁ + t₂ = t
2
1
v=u-gt₂
v = ev
v=
= ε(u — gt₁)
s=ut₁ - 1 gt 2
(1)
(2)
2
=v' t₂+ 1 gt2².
Substituting for v' and to their values from (2) and (1), we have a
quadratic equation in t₁ which gives
u +gt±√√ (u+gt)² -
4g
49 (eut + 1 g 1²)
1+ €
t₁ =
2g
Substituting this value of t, in the expression
2
s=ut₁ — — g t₁²,
we have s as required, the lower sign only being admissible.
Ex. 13. Show that a body falling by the action of gravity
acquires a velocity of 1000 feet per second in 31 seconds nearly.
Ex. 14. Show that if a stone be thrown directly upwards with
a velocity of 40 feet per second, it is returning down again after an
interval of 14 seconds.
Ex. 15. Show that the stone of the last example would ascend
to a height of 24.8 feet.
Ex. 16. An imperfectly elastic ball falls upon a hard floor from
a height h, show that it will rebound to a height e² h.
Ex. 17. A stone falling from a steeple passes through the last
of the height in 4th of a second; show that the height of the
g
steeple is 22 (7+4√/3).
n
CHAPTER IV.
ON PROJECTILES.
In the last chapter the motion of a body projected vertically
upwards or downwards, under the action of gravity, was considered,-
the whole motion taking place in the vertical line through the
point of projection. When the direction is any other than vertical,
the path of the body is an arc of the curve called the parabola. By
the second law of motion, gravity produces its full effect indepen-
dent of the motion of projection: and we may consider the latter
as compounded of a horizontal and vertical motion. The latter of
these only can be affected by the action of gravity on the body.
14. PROP. To determine the path of a body projected in a given
direction, with a given velocity, under the action of gravity.
Let v be the velocity of projec-
tion from the point A, in the
direction ATy', and let t be the
time in which the body would
have described the space AT
with the uniform velocity v, if
gravity had not acted.
If TP be the space due to
the action of gravity in the time
t, P will be the actual place of
the body.

A
T
P
S
Now, AT=vt
PT=1g t²
v2
eliminating t,
AT²=
AT-2 PT.
(1)
g
144
ELEMENTARY MECHANICS.
If we draw AVx' a vertical line, and taking A V=PT, com-
plete the parallelogram ATPV, we have A V=x', P V=y', the
oblique co-ordinates of the point P.
Let also h be the height from which the body must fall to
acquire the velocity v, or h=
we have from (1)
v2
2g
12
y¹² = 4 h x'
which, as seen in treatises on conic sections, is the equation to a
parabola whose axis is parallel to Ax, and therefore vertical, Ay'
a tangent at the point A, and h the distance SA of the focus, and
also of the directrix from A. With these data the parabola to
represent the path of the body can be described.
15. PROP. To find the equation to the path of a projectile when
referred to axes of co-ordinates which are horizontal and vertical.
Letv be the velocity of pro-
jection in the direction AT,
which makes with Ax the
angle of elevation of the pro-
jectile TAx=a. Let APB
be the path; and AM=x,
PM=y the co-ordinates of
any point P. Let t be the
time in which the body de- A

M
T
P
B
scribes the arc AP; and let PM produced meet AT in T, we
have
AT-vt
TP=1 g t²
AM-x=vt cos a
PM=y=vt sin a-g t².
Eliminating t, we have from these equations.
y=x tan α-
कर
9
2 v² cos² a
v2
the equation required; or substituting, as in the last Prop., h=2g'
we have
22
y=x tan α-
4h cos² a
DYNAMICS.
145
DEFINITIONS. The horizontal range of a projectile is the
distance AB from the point of projection to the point where it
strikes the horizontal plane in its descent. The time of flight is
the time it takes in describing APB.
16. PROP. To find the time of flight of a projectile on a hori-
zontal plane.
We have generally, as in the last proposition,
y=vt sin a-1 g t²
1119
and if we put y=0, or vt sin a-gt2=0, the result will apply to
the points A and B only; and the values of t are
t=0
at the point A
2 v sin a
t=
g
which is therefore the time of flight required.
at the point B
We should have arrived at this result by the same method as in
article 13, by putting for u the vertical component of the velocity
of projection, v sin a.
17. PROP. To find the range of a projectile on a horizontal
plane.
If we put y=0 in the equation.
we have
or
y=x tan a
x².9
2 v2.cos2 a
9
0=x tan α- x²²
2 v² cos² a
The result, as before, applies to the points A and B, and
x=0
at the point A
2 v² sin a cos a
x=
at the point B
g
v2 sin 2 a
AB=
9
=2h sin 2 a
α
which is the horizontal range required.
This value of AB varies with the angle of elevation a, and is
greatest when sin 2 a=1, or a=45°, v remaining the same.
L
146
ELEMENTARY MECHANICS.
or
Π
sin(+9)=sin(0)
Since
sin 2 (45° +
(45°+)
Ꮎ
+9)=sin 2 (45°-9)
if we put either 45°+2 or 45°.
AB, we have the same re-
sult; or as in the annexed
figure, if A B be the greatest
range of a projectile when
the angle of elevation is 45°,
we shall have A B', a less
horizontal range correspond-
ing to either of the paths
AP'B' or AP" B', when the
elevation of one was as much
A
for a in the equation for
2

p"
above 45° as that of the other was below it.
P
P
B'
B
We see that the horizontal range is the space which would be
described with the uniform horizontal velocity v cos a, in the time
2 v sin a
of flight
9
18. PROP. To find the greatest height which a projectile
attains.
The greatest altitude is evidently the value of y at the middle
point of the path above a horizontal plane, or when the time is
v sin a
g
one-half of the time of flight, or t=
Putting this value of t in the expression
y=vt sina -g t²
2
v² sin² a
we have the greatest altitude=
1
g
v² sin?
9
α
v² sin² a
9
=
h sin² a.
u2
we
Comparing this expression with s=29 see that the
g
DYNAMICS.
147
greatest altitude the projectile attains is that to which it would
rise by the vertical component v sina of the velocity of pro-
jection.
19. PROP. To show that the velocity at each point of the para-
bolic path of a projectile is that which would be acquired in falling
directly from the directrix.
In article 14 it was shown that if h be the height due to the
velocity of projection, it is also the distance of the point of projec-
tion from the directrix of the parabola described. Now if any
point in the parabola were taken for the point of projection, and a
body were projected from it with the same velocity and direction
which it has in the parabola, it would describe the same parabola ;
and therefore what holds for the point of projection holds also for
all other points of the path.
20. PROP. To find the point where a projectile will strike an
inclined plane through the point of projection, and its distance, or the
range on the inclined plane.
Let y=x tan ß be the
equa- 31
tion of the line A C, which is the
intersection of the inclined plane
with the vertical plane in which
the body is projected.
Combining this with the equa-
tion of the path of the projectile A
in article 15, namely,

X2
y=x tan a
4 h cos² a
we have the co-ordinates of the point C
4h
cos a. sin (α-B)
cos B
x = 4 h
cos a.
y = 4 h
sin B. sin (α —ẞ)
2
cos² B
and the distance_ AC=√√√ x²+y²=x sec ß
=4h
cos a. siu (α
cos² B
sin (a—B).
C
H
B
L 2
148
ELEMENTARY MECHANICS.
EXAMPLES IN PROJECTILES.
Ex. 1. Two bodies are projected from the top of a tower with
the same given velocity at different given angles of elevation, and
they strike the horizontal plane at the same point. Required the
height of the tower above the plane.

Let A be the point of pro-
jection, and B the point where
the bodies strike the horizon-
tal plane.
Since the velocity of pro-
jection is given, h, the height
A
from which the bodies must fall to acquire it, is known.
and ẞ be the given angles of elevation.
B
Let a
We have to find -y, the height of the tower, by eliminating x
from the two equations,
which give
22
-y=x tan a
4 h
(1 + tan² a)
x2
-y=xtanß
4 h
(1+tan2 B)
4 h
y=
(tan a + tan ẞ) tan (« +B)
the height required.
Ex. 2. Find the angle of elevation at which a body must
be projected with a given velocity in
order to strike the summit of an ob-
ject whose height and distance are
given.

A
Let A be the point from which
the body is projected to strike the
point B, whose co-ordinates are x=a, y=b.
B
M *
Putting these values in the equation to the path of the projectile,
we find
2h+ √4h(h—b)—a²
tan α =
a
a
DYNAMICS.
149
Ex. 3. Show that the latus rectum of the parabolic path equals
four times the space through which the body must fall to acquire
the horizontal component of the velocity of projection.
Ex. 4. Show that the time of flight on an inclined plane is
given by the expression
4 h sin (a-B)
v cos B
Ex. 5. A body is projected with a velocity due to the height h,
at an elevation a; show that it ranges on a horizontal plane elevated
a height H above the point of projection to a horizontal distance
from that point
= h sin 2
2a{1+√
and that the time of flight is
h
√2 sin a{1
1
H
h sin² a
H
1+
1.
h sin² af
CHAPTER V.
ON CONSTRAINED MOTION.
WHEN a body acted on by any force is constrained to move in
some particular manner, as, for instance, when a body falls down
an inclined plane, or a curve, or swings as a pendulum, we call it
a case of constrained motion.
21. PROP. To determine the relations of the time, space, and
velocity when a body falls by the action of gravity down an inclined
plane.
Let the body fall from the point
B, down the inclined plane AB,
whose inclination to the horizontal
line AC is a.

Let a be its position at any time
t from rest, Ba=s, v=velocity at a.
A
C
g
B
The force f which urges the body down the plane is the resolved
part of gravity in that direction, or
f=g sin a
which being a uniform force, we have only to put the value in the
expressions found in articles 9 and 10, for the required relations
of s, t, and v.
The expressions v=ft, s={ft², v2=2fs become respectively
v=g sina.t, s=g sin a.t2, v²=2g sin a.s, from which the whole
circumstances of the motion can be determined.
22. PROP. To show that the velocity acquired in falling down an
inclined plane is the same as would be acquired in falling down the
perpendicular height directly.
DYNAMICS.
151
When the body arrives at A, the space described is AB;
putting this value for s in the expression v2=2g sina.s, we
have
v²=2g. A B sin a
=2g. BC
α
the expression when the body falls directly through the height
BC of the plane.
23. PROP. To show that the times down any inclined planes
are proportional to the lengths of the planes, when the height is the
same.
Putting AB for S,
s=g sin a. t², we have
BC
and
AB
for sina in the expression
AB
BC
g
ޓރ
AB
BV
or
t=AB
2
9. BC
α AB when B C is constant.
24. PROP. To find the relations of the time, space, and velo-
city when a body is projected directly up or down an inclined
plane.
Putting for f its value g sina, on an inclined plane, in the
expressions of articles 11, 12, and 13, we have
v=u±ft becoming on the inclined plane
s=ut± 3 ft²
v²=u²±2fs
ور
""
"}
>>
>>
""
"
"
دو
v=u±gt sin a
α
s=ut±½g t² sin a
=u²+2g s sin a.
v² = u²
25. PROP. To show that the times of
a body falling down all the chords of a
circle, in a vertical plane, drawn from the
highest or to the lowest point, are the same.
Let AD be the vertical diameter of the
circle, A B a chord down which a body falls
by the action of gravity. Draw B C hori-
zontal. The accelerating force acting on
the body is
BE
A

D
C
E
152
ELEMENTARY MECHANICS.
AC
f=g
AB
Putting s= A B in the expression s=ft², we have
AC
AB = 19 Bt²
AB
2 A B2
or
t2
9 AC
=2AD.
9
AD . . . . since AC: AB:: AB: AD
in a circle; or the time down any chord is the same as that down
the diameter. The same relation evidently holds also for all
chords drawn from the circle to the point D, as, for instance,
DE or B D.
26. PROP. Two bodies hang from the extremities of a cord
passing over a pulley; to determine the motion.
We shall first suppose that the weight of the pulley
may be neglected.
Let P and Q be the weights of the bodies, of which
P
P is the greatest; their masses are
and
respect-
P
g
9
ively. If the weights are equal, the bodies will balance,
and no motion will ensue; but if one is heavier than the
D
other, it will descend and the other rise through an equal
space. The force which produces motion is the difference of the
weights, which, being a moving force, equals the accelerating force
multiplied by the mass moved.
Let f be the accelerating force, we have
or
P-Q=√(P+Q)
P-Q
f=9⋅ P+Q
This being a constant accelerating force, we shall have the
1

DYNAMICS.
153
circumstances of the motion by substituting its value in the ex-
pressions v=ft, s=4ft², v²=2 fs.
When we take into account the inertia of the pulley, we
must add the equivalent mass acting at the cord to the other
masses set in motion. Let I be this inertia or mass at the cord,
we have
P-Q
f=9p+Q+Ig
This is the formula for Atwood's machine, referred to in the
article on the third law of motion. The force being uniform, and
capable of being modified in any manner by changing the relative
and absolute magnitudes of the weights, we are enabled by this
machine to prove in the most satisfactory way, by experiment, the
formulæ for constant accelerating forces.
27. PROP. When a body falls by the action of gravity down any
arc of a smooth curve, the velocity at any point is that due to the
vertical height fallen through.
Let us first suppose that a body falls from
A down a succession of smooth inclined planes.
A B, B C, C D, DE, &c., and loses no part
of its acquired velocity in passing from one
plane to the next. Draw Abcde, a vertical
line, and Bb, Cc, Dd, Ee, &c., horizontal
lines.
E
C
D
A

B
b
d
By article 22, the body will have at B the same velocity as
it would have acquired in falling directly through the vertical
height Ab; and when it has passed without loss of velocity to
the plane B C, it will have at B and all other points the velocity
due to the vertical height fallen through. The same takes place
on each of the other planes, and the velocities at the points
C, D, E, &c. are those due to the vertical heights Ac, Ad, Ae,
&c. respectively. When the number of planes is indefinitely
increased, and the length of each indefinitely diminished, they
form a continuous curve, which, when smooth, acts only perpendi-
cular to the arc at each point, and therefore destroys by its reaction
154
ELEMENTARY MECHANICS.
no part of the acquired velocity as the body passes from one point
to another. Hence the velocity at all points of the curve is that
due to the vertical height fallen through.
If a body be projected up or down a curve, the formulæ of
article 12 hold good; if we put u velocity of projection, v=
velocity after the body has passed through a vertical height h,
we have
v²=u²±2gh.
The upper sign to be taken when the body is projected down, and
the lower one when it is projected up the curve.
28. PROP. When bodies fall by the action of gravity down
any arcs of a circle in a vertical plane, the velocities at the
lowest point are proportional to the lengths of the chords of the
arcs.
Let a body fall from the point B in the figure article 25, down
the arc BD; the velocity at D is that due to the vertical height
fallen through CD
or
vel.2=2 gCD
or
BD2
=2g
AD
2 g
vel. — B D^
AD
α B D.
When the arcs are small, the velocities are nearly proportional
to the lengths of the arcs.
29. PROP. To find the time of a body falling down a cycloidal
arc in a vertical plane with its base horizontal.
C
Р
The cycloid is a curve CAD
described by a point in the cir-
cumference of a circle, whilst P
the circle rolls along the line
CBD, called the base of the
cycloid. The line AB, which
B
D

K
L
p
P
M
N
Q
ZZ
A
DYNAMICS.
155
bisects the base CD perpendicularly, is called the axis of the
cycloid, and is the diameter of the rolling circle when the point
describing the cycloid comes to its central position A, which is
called the vertex of the cycloid.
If a line PpP' be drawn parallel to the base CD, and Ap be
the chord of the circle in its central position; it is shown in
mathematical treatises, that the arcs AP or A P' are equal to
twice the chord Ap, and the tangent at P' is parallel to the
chord Ap.
Let the body fall from the point L to any point P; and take
Q a point near P. Draw the lines LK, PM, QN parallel to
the base of the cycloid, meeting the axis in K, M, N respect-
ively. Describe the semicircle AnmK on AK, cutting P M in
m, Q N in n; and draw the chords Am, An, Km, Kn.
the intersection of Kn and Am.
The velocity at the point P= √2 g. KM, and if the
be indefinitely small, the velocity will be constant in it.
space are AP-arc A Q
the time of describing P Q=
velocity √2gKM
Let o be
arc P Q
Hence
2× chord of circle ApB corresps to P-2 × chord corresps to Q
√2gKM
2 √AB. AM-2√AB.AN
√2g. KM
2 AB
A M
AN
KM
√ KM
AM.AK
g
2 AB
9
{
= 1/2 ABŞ
9
2 AB
9
2 AB
=√²
g
{
{
Am
AK.KM
An
Km Km
m o
Km
}
x angle mKo.
AN.AK
AK.KM
since ultimately no becomes a cir-
cular arc to center A
156
ELEMENTARY MECHANICS.
The same holds for all other points between L and A; and
the whole time of falling from L to A
π
2 AB
9
× sum of all the small angles, as mKn, in the right
angle LKA
2 AB
9
The body acquires in falling to A a velocity which will carry
it through an equal arc on the opposite side of A, which it will
describe in the same time; or if the time of the complete oscil-
lation from L to an equal height on the opposite side of A be t,
we have
t=T
30. It is shown in ma-
thematical treatises, that if
two half cycloids, CO, DO,
equal to AD or A C, be con-
structed with their axes ver-
tical, so that ABO is a straight c
line and B 0=A B, then the
curve COD is the evolute of
the cycloid CAD; or a curve
such that a string of the
length A O being fastened at
2 AB
g

A
B
9
P
D
O and wrapping on the arcs CO or D O, the extremity P of the
string, when drawn tight, will be always in the cycloid CAD,
the length of the arc Oq+q P being always equal to O A.
If a body be suspended at P upon a cord fastened at O,
wrapping and unwrapping on the arcs CO, OD, it will thus
oscillate in a cycloidal arc.
Let l=the length of the cord O A, the time of a complete
oscillation is
t=TA
借
​This is independent of the length of the arc of the cycloid
DYNAMICS.
157
described, and is consequently the same for all arcs. For this
reason the cycloidal pendulum is called isochronous, and from
this property arises the importance of the pendulum in instru-
ments for measuring time; for small arcs near A the cord will
not sensibly wrap upon the arcs of the evolute, and so a pen-
dulum oscillating in small circular arcs has the property of iso-
chronism. The clocks in astronomical observatories have their
pendulums oscillating in very small arcs, which requires the
mechanism of the clockwork to be very accurate, or the clock
would be liable to stop going.
We learn from the expression t=T,
ğ'
that at the same
place on the earth's surface, or when g is constant, tα v
√7;
and at different points of the earth's surface if 7 is invariable,
1
τα ; so that when the times of oscillation of an invariable
√
9
pendulum have been found at different places on the earth's surface,
the force of gravity at those places can be compared.
31. PROP. A body at the extremity of an elastic cord, sup-
posed without weight, describes a circle, with a uniform velocity,
about a fixed point in the cord, as center; to find the tension in
the cord.
Let PQAB be the circle in which the body moves, of which
C is the center.
R
Q
P
32
m
C
B
If the body were suddenly freed from constraint at any point
P, it would, by the first law of motion,
continue moving in the direction it had
at P, and with its uniform velocity, or
it would then move in the tangent PR
with that velocity. The effect of the
tension in the cord is therefore to deflect
the body from the tangent at each point
in the circle, and is called a centripetal
force. The tendency of the body to fly
off by its inertia produces the force
which balances this centripetal force, and is thence called the
centrifugal force.

A
158
ELEMENTARY MECHANICS.
If P Q be an arc described in an indefinitely small time t,
and PR the space which would have been described in the tan-
gent, if the body were free, in the same time; then R and Q
being supposed indefinitely near to P; RQ, by the second law of
motion, becomes the space due to the action of the centripetal
force.
Let PCA be a diameter of the circle, v the uniform velocity
of the body, and draw Qm parallel to R P.
Let f the centripetal accelerating force acting on the body;
we have ultimately
and
RQ=Pm
= 1 ft²
PR=vt.
or
But in the circle,
Q m² = Pm × m A
Q m²
Pm=
m A
PR2
PA'
ultimately
v2 12
1
... ft² = 2. PC
or if r=the radius of circle,
f=22.
v2
γ
If m be the mass of the body at P, the centrifugal moving
force=mf
m v2
=the tension in the cord.
32. DEFINITION. A body suspended by a cord which per-
forms revolutions in a horizontal circle is called the conical
pendulum.
33. PROP. To determine the motion of a body in the conical
pendulum.
DYNAMICS.
159

Let the body be fixed at the ex-
tremity P of the cord AP which is
fastened at A.
Let A C be a vertical line, P C=r
the radius of the circle which the
body describes with the uniform ve-
locity v.
Let the length of the cord
AP, and angle PAC=a.
Since the body is in the same
circumstances at each point of the
P
g
t
گئے
A
Q
circle, the forces acting upon it must balance each other. These
forces are, the weight of the body acting downwards, the centri-
fugal pressure acting horizontally outwards from C, and the
tension in the cord A P.
Resolving horizontally and vertically, we have
t sin a
m v2
0
t cos a
7°
mg=0.
From (2) we have the tension in the cord, t
mg
COS α
the weight of the body
COS a
Eliminating t between (1) and (2), we have
v2
r sin
gr α
=9
Cos a
1
sin² α
COS α
The time of performing one revolution
circumference
velocity
2 Tr
ว
2 T
°T/
COS a
g
AC
(1)
(2)

160
ELEMENTARY MECHANICS.
which depends on the vertical height A C, but not on the length
of the cord.
34. PROP. It is required to find the position of the rails in
the curve of a railway, so that the resultant pressure of a car-
riage passing along the curve with a given velocity may be per-
pendicular to the line drawn from one rail to the other
Let v=
Let APB represent in figure 1 the
curve in a railway, and let the radius
of the curve at P be P C=r.
the velocity of the carriage. If the
reaction of the rails perpendicular to
the line joining them as a b, figure 2,
supplies the place of the tension in
the cord in the last proposition, the
carriage will pass smoothly and safely
along the curve; for we shall then
have the weight of the carriage acting
P
Fig. 1.
A
B
b
p
C
Fig. 2.

个
​a
G
at G, the center of gravity, figure 2, and the centrifugal pressure
acting horizontally outwards giving a resultant pressure perpen-
dicular to the rails which will be destroyed by their reaction.
If a be the angle which the resultant pressure in Gp makes
with a vertical line angle which ab makes with a horizontal
line, we have, from the last proposition,
ΟΙ
v²=gr tan a
2,2
tan α=
gr
which gives the inclination of the line ab to the horizon as re-
quired.
EXAMPLES ON CHAPTER V.
Ex. 1. Two balls fall at the same instant from the common
vertex down two inclined planes which meet the horizontal
plane on which they rest at angles of 30° and 60°; show that
DYNAMICS.
161
the times of falling to the horizontal plane are in the ratio
√3:1.
Ex. 2. To divide the length of a given inclined plane into
three parts, so that the times of descent down them may be
equal.
If the length of the plane be divided into nine equal parts,
the body will descend down one in the first interval of time,
down four in two intervals, and down the whole nine in the
three equal intervals.
Ex. 3. P descending vertically draws Q up an inclined
plane by means of a string passing over a pulley at the vertex ;
find the velocity acquired by P in describing a given space h.
a the
Let P and Q be the weights of the bodies respectively,
inclination of the plane to the horizon, h the space which P
describes ;
the moving force=P-Q sin a
the mass moved, neglecting the pulley,:
P + Q
9
.. the accelerating force=g.
P-Q sin a
and from the general formula v²=2fs we find
P+Q
2gh P-Q sin a
the velocity required=√√2
P+Q
Ex. 4. Twelve pounds weight is so distributed at the ex-
tremities of a cord passing over a pulley, that the more loaded
end descends through seven feet in seven seconds. What is the
weight at each end of the cord?
We find
P=6(1+
P=6(1+17), and Q=6(1-17) pounds.
Ex. 5. If three pendulums suspended in the same vertical
plane have their lengths as the numbers 1, 4, and 9; show that
M
162
ELEMENTARY MECHANICS.
when they commence oscillating together, the first and second
will be together again after four oscillations of the first; the
first and third will be together again after three oscillations of
the first; and the whole will be together again after twelve oscil-
lations of the first.
Ex. 6. The length of the pendulum which beats seconds in
London being 39.138 inches, show that the force of gravity is
represented by 32.19 feet nearly.
Ex. 7. If a body be projected obliquely upon an inclined
plane, show that its path is an arc of a parabola.
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MENTAL PHILOSOPHY.
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CATALOGUE OF WORKS
Lardner's Chemistry for Schools. 170 Illustrations.
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Glossary of Scientific Terms for General Use.
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