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MECHANICS
AND
HYDEOSTATICS
FOR BEGINNERS
BY
S. L. LONEY. M.A.
LATE PROFESSOR OF MATHEMATICS AT THE ROYAL HOLLOWAY
COLLEGE (UNIVERSITY OF LONDON),
SOMETIME FELLOW OF SIDNEY SUSSEX COLLEGE, CAMBRIDGE.
CAMBRIDGE :
AT THE UNIVERSITY PRESS
1922
Published Jan. 1893, Reprinted Oct. 1893.
Second Edition Jan. 1894.
Third Edition Nov. 1894.
Reprinted Oct. 1896. Jan. 1898, 1900. 1902, 1903, 1905, 1907,
1909, 1911, 1913, 1915, 1917, 1919, 1920, 1922
PRINTED N GREAT BRITAIN.
PKEFACE.
THIS little book is of a strictly elementary character,
and is intended for the use of students whose
knowledge of Geometry and Algebra is not presumed
to extend beyond the first two Books of Euclid and the
solution of simple Quadratic Equations.
A student who is not acquainted with the first few
propositions of Euclid's Sixth Book and the elements of
Trigonometry, is recommended to commence with the
Appendix at the end of the book. In this Appendix
will be found the very few propositions in Elementary
Trigonometry that are used in the text.
A few articles, with an asterisk prefixed, may be
omitted on ^ first reading, and the Test Examination
Papers may be taken at the end of the chapters to
which they refer.
For any corrections, or suggestions for improvement,
I shall be gratefuL
S. L. LONEY.
Royal Hollow at College,
Egham, Surrey.
December, 1892.
In the Second Edition, in deference to those friends
who have criticised the book, I have measured Hydro-
statical Pressure in lbs. weight instead of in poundals.
In Chapters XIX. to XXIII. will therefore now be
found w in the place of gp.
November, 1893.
CONTENTS.
STATICS.
chap. page
I. Introduction ......... 1
II. Composition and Resolution of Forces .... 6
HI. Composition and Resolution of Forces (continued) 20
IV. Parallel Forces 27
V. Moments 85
VI. Couples , . . . . 45
VII. Equilibrium of a rigid body acted on by three
FORCES IN ONE PLANE ...... 50
Ylil. Centre of Gravity . 55
Centre of gravity of a Triangle 58
General formulae for the determination of the centre
of gravity 60
Properties of the centre of gravity .... 71
Stable and unstable equilibrium , .... 74
IX. Machines 78
The Lever 78
Pulleys 82
The Inclined Plane 92
The Wheel and Axle 96
The Common Balance . . .99
The Steelyards 108
viii CONTENTS.
CHAP. * AQJS
X. Pbiotioh 107
XI. Wok* 114
Principle of Work 116
The Screw 117
DYNAMICS.
YTT, Velocity and Acceleration, Rectilinear Motion . . 123
Motion in vertical lines under gravity . . .129
XIII. The Laws of Motion 135
The relation P=mf 137
XIV. The Laws of Motion. Application to Simple problems 145
Particles connected by strings 145
Motion on inclined planes ...... 148
Atwood's Machine ....... 155
XV. Impulse, Work and Energy ...... 159
Motion of a shot and gun 161
Kinetic and Potential Energy 164
XVI. Composition of Velocities and Accelerations . . 168
Projectiles 177
Uniform motion in a circle 180
HYDROSTATICS.
XVII. Fluid Pressure 183
Bramah's Press and the Safety Valve . . .188
Xviu. Density and Specific Gravity 192
XIX. Pressure at different points of a homogeneous fluid
AT REST 198
Whole Pressure 204
Centre of Pressure ....... 208
CONTENTS. ix
CHAP. PAGE
XX. Resultant Vertical Pressure . 211
Floating Bodies 214
XXI. Methods of determining Specific Gravity . . 228
Specific Gravity Bottle 228
Hydrostatic Balance 230
Hydrometers 234
XXII. Pressure op Gases 240
Atmospheric pressure and Barometers . . . 240
Boyle's Law 245
Gay-Lussao's Law and Absolute Temperature . 252
XXTTI. Machines and Instruments 256
Diving Bell 256
Water-Pumps 260
Air-Pumps 266
The Siphon 272
Test Examination Papers 275
Appendix I. Trigonometry 281
Appendix II. Formula concerning areas, volumes, etc. 291
Answers.
MECHANICS AND HYDKOSTATICS
FOR BEGINNERa
CHAPTER L
INTRODUCTION.
1. Thb present book is divided into three portions.
The first portion will treat of the action of forces on
bodies, the forces being so arranged that the bodies are at
rest. This is the subject of Statics.
The second portion will deal with the action of forces on
bodies in motion. This is the subject of Dynamics.
The third portion will deal with the properties of
liquids and gases and of the effect of forces on them when
they are at rest. This is the subject of Hydrostatics.
The title Dynamics is often used to include all three of
these subdivisions.
2. A Body is a portion of matter limited in every
direction.
3. Force is anything which changes, or tends to
change, the state of rest, or uniform motion, of a body.
4. Rest. A body is said to be at rest when it does
not change its position with respect to surrounding objects.
5. A Particle is a portion of matter which is in-
definitely small in size.
6. A Rigid Body is a body whose parts always
preserve an invariable position with respect to one another.
In nature no body is perfectly rigid. Every body yields,
perhaps only very slightly, if force be applied to it. If a
rod, made of wood, have one end firmly fixed and the other
end be pulled, the wood stretches slightly ; if the rod be
made of iron the deformation is very much less.
L. M. H. 1
2 STATICS.
To simplify our enquiry we shall assume, in the first
two divisions of our subject, that all the bodies with which
we have to deal are perfectly rigid.
7. Equal Forces. Two forces are said to be equal
when, if they act on a particle in opposite directions, the
particle remains at rest.
8. Mass. The mass of a body is the quantity of
matter in the body. The unit of mass used in England is
a pound and is denned to be the mass of a certain piece of
platinum kept in the Exchequer Office.
Hence the mass of a body is two, three, four... lbs.,
when it contains two, three, four... times as much matter
as the standard lump of platinum.
9. Weight. The idea of weight is one with which
everyone is familiar. We all know that a certain amount
of exertion is required to prevent any body from falling to
the ground. The earth attracts every body to itself with
a force which, as we shall see in Dynamics, is proportional
to the mass of the body.
The force with which the earth attracts any body to
itself is called the weight of the body.
10. Measurement of Force. We shall choose, as our
unit of force in Statics, the weight of one pound. The unit
of force is therefore equal to the force which would just
support a mass of one pound when hanging freely.
We shall find in Dynamics that the weight of one
pound is not quite the same at different points of the
earth's surface.
In Statics, however, we shall not have to compare forces
at different points of the earth's surface, so that this variation
in the weight of a pound is of no practical importance ; we
shall therefore neglect this variation and assume the weight
of a pound to be constant.
11. In practice the expression "weight of one pound"
is, in Statics, often shortened into "one pound." The
student will therefore understand that "a force of 10 lbs."
means " a force equal to the weight of 10 lbs."
INTRODUCTION. 3
12. Forces represented by straight lines. A force will
be completely known when we know (i) its magnitude,
(ii) its direction, and (iii) its point of application, i.e. the
point of the body at which the force acts.
Hence we can conveniently represent a force by a
straight line drawn through its point of application; for
a straight line has both magnitude and direction.
Thus suppose a straight line OA represents a force,
equal to 10 lbs. weight, acting at a point 0. A force of
5 lbs. weight acting in the same direction would be repre-
sented by OB, where B bisects the distance OA, whilst a
force, equal to 20 lbs. weight, would be represented by 00,
where OA is produced till AC equals OA.
An arrowhead is often used to denote the direction in
which a force acts,
13. Subdivisions of Force. There are three different
forms under which a force may appear when applied to a
mass, viz. as (i) an attraction, (ii) a tension, and (iii) a
reaction.
14. Attraction. An attraction is a force exerted by
one body on another without the intervention of any
visible instrument and without the bodies being necessarily
in contact. The only example we shall have in this book
is the attraction which the earth has for every body ; this
attraction is (Art. 9) called its weight.
15. Tension. If we tie one end of a string to any
point of a body and pull at the other end of the string, we
exert a force on the body ; such a force, exerted by means
of a string or rod, is called a tension.
If the string be light [i.e. one whose weight is so small
that it may be neglectedl the force exerted by the string is
the same throughout its length.
1-2
4 STATICS.
For example, if a weight W be supported by means of
a light string passing over the edge of a table it is found
a
w
that the same force must be applied to the string whatever
be the point, A, B, or C, of the string at which the force is
applied.
Now the force at A required to support the weight
is the same in each case ; hence it is clear that the effect
at A is the same whatever be the point of the string to
which the tension is applied and that the tension of the
string is therefore the same throughout its length.
Again, if the weight W be supported by a light string
passing round a smooth peg ^4, it is found that the same
force must be exerted at the other end of the string what-
ever be the direction (AB, AC, or AD) in which the string
is pulled and that this force is equal to the weight W.
Hence the tension of a light string passing round a
smooth peg is the same throughout its length.
If two or more strings be knotted together the tensions
are not necessarily the same in each string.
The student must carefully notice that the tension of a string is
not proportional to its length. It is a common error to suppose that
the longer a string the greater* is its tension; it is true that we can
often apply our force more advantageously if we use a longer piece of
string, and hence a beginner often assumes that, other things being
equal, the longer airing has the greater tension.
INTRODUCTION. 5
16. Reaction. If one body lean, or be pressed, against
another body, each body experiences a force at the point of
contact ; such a force is called a reaction.
The force, or action, that one body exerts on a second
body is equal and opposite to the force, or reaction, that
the second body exerts on the first.
This statement will be found to be included in Newton's
Third Law of Motion.
Example. If a ladder lean against a wall the force
exerted by the end of the ladder upon the wall is equal and
opposite to that exerted by the wall upon the end of the
ladder.
17. Equilibrium. When two or more forces act
upon a body and are so arranged that the body remains at
rest, the forces are said to be in equilibrium.
18. Smooth bodies. If we place a piece of smooth
polished wood, having a plane face, upon a table whose top
is made as smooth as possible we shall find that, if we
attempt to move the block along the surface of the table,
some resistance is experienced. There is always some
force, however small, between the wood and tho surface
of the table.
If the bodies were perfectly smooth there would be
no force, parallel to the surface of the table, between the
block and the table; the only force between them would
be perpendicular to the table.
Def. When two bodies, which are in contact, are
perfectly smooth the force, or reaction, between them is
perpendicular to their common surface at the point of
contact.
CHAPTER IL
COMPOSITION AND RESOLUTION OF FORCES.
19. Suppose a flat piece of wood is resting on a
smooth table and that it is pulled by means of three
strings attached to three of its corners, the forces exerted
by the strings being horizontal ; if the tensions of the
strings be so adjusted that the wood remains at rest it
follows that the three forces are in equilibrium.
Hence two of the forces must together exert a force
equal and opposite to the third. This force, equal and
opposite to the third, is called the resultant of the first
two.
Resultant. Def. If two or more forces P, Q t S...
act upon a rigid body and if a single force, R, can be
found whose effect upon the body is tte same as that of the
forces P, Q t S... this single force R is called the resultant of
the other forces and the forces P, Q, S... are called tJie com-
ponents of R.
20. Resultant of forces acting in the same straight line.
If two forces act on a body in the same direction their
resultant is clearly equal to their sum ; thus two forces
acting in the same direction, equal to 5 and 7 lbs. weight
respectively, are equivalent to a force of 12 lbs. weight
acting in the same direction as the two forces.
If two forces act on a body in opposite directions their
resultant is equal to their difference and acts in the direction
of the greater; thus two forces acting in opposite directions
and equal to 9 and 4 lbs. weight respectively are equivalent
to a force of 5 lbs. weight acting in the direction of the first
of the two forces.
21. When two forces act at a point of a rigid body
in different directions their resultant may be obtained by
means of the following
COMPOSITION AND RESOLUTION OF FORCES. 7
Theorem. Parallelogram of Forces. If two
forces, acting at a point, be represented in magnitude and
direction by the two sides of a parallelogram drawn from one
of its angular points, their resultant is represented both in
magnitude and direction by the diagonal of the parallelogram
passing through that angular point.
In the following article we shall give an experimental
proof.
In Chapter XVI. will be found a proof founded on
Newton's Laws of Motion.
i
22. Experimental proof.
Let L, M, and N be three small smooth pegs over which
pass light strings supporting masses P lbs., Q lbs., and R lbs.
respectively. Let one end of each of these strings be tied
together at a point ; then [unless two of the weights are
together less than the third] the system will take up some
such position as that in the figure. The tension of the
string OL is unaltered by passing round the smooth peg L
and is therefore equal to the weight of P lbs. ; so the
tensions in the strings M and ON are respectively equal to
the weights of Q and R lbs.
Hence the point is in equilibrium under the action of
forces which are equal respectively to the weights of P, Q,
and R lbs.
8 STATICS.
Along OL % OM y and ON measure off* distances 0A f OB,
and OC proportional to P, Q, and E respectively and com-
plete the parallelogram OADB.
Then it will be found that OD is exactly equal in mag-
nitude, and opposite in direction, to OC.
But the effect of the forces OA and OB is equal and
opposite to that of OC. Hence the effect of the force OB
is exactly the same as that of the forces OA and OB.
This will be found to be true whatever be the relative
magnitudes of P, Q, and R, provided only that one of them
is not greater than the sum of the other two.
Hence we conclude that the theorem enunciated is
always true.
23. The pegs of the above experiment may be advantageously
replaced by light smooth pulleys [Art. 100] or we may use three
Salter's Spring Balances furnished with hooks at their ends as in
the annexed figure.
Each of these Balances shows, by a pointer which travels up and
down a graduated face, what force is applied to the hook at its end.
The three hooks are fastened together and forces are applied to the
rings at the other ends of the instruments and they are allowed to
take up their position of equilibrium. The forces, which the pointers
denote, replace the tensions of the strings in the preceding experiment
and the rest of the construction follows as before.
24. To find the direction and magnitude of the re-
sultant of two forces, we have to find the direction and
magnitude of the diagonal of a parallelogram of which the
two sides represent the forces.
COMPOSITION AND RESOLUTION OF FORCES. 9
Ex. 1. Find the resultant of forces respectively equal to 12 and 5
lbs. weight and acting at right angle*.
Let OA and OB represent the forces so that OA is 12 units of
length and OB is 6 units of length ; complete the rectangle OACB.
Then OC*=OA 2 + JC*=12*+5* = 169. .'. 0(7=13.
AC
Also t*nC0A = ^ = &.
Hence the resultant is a force equal to 13 lbs. weight making with
the first force an angle whose tangent is ^, i.e. about 22 37'.
Ex. 2. Find the resultant of forces equal to the weights of 5 and
3 lbs. respectively acting at an angle of 60.
Let OA and OB represent the forces, so that OA is 6 units and OB
3 units of length; alBO let the angle AOB be 60.
Complete the parallelogram OACB and draw CD perpendicular to
OA. Then OC represents th6 required resultant.
Now ^D=JCcosCMD = 3cos60 o =t; .:OD = ^.
Also DC=AC sin 60= S ^
2
and
/. OC= N /OD 2 + DC a = N / t P + -V = >/'49=7,
Hence the resultant is a force equal to 7 lbs. weight in a direction
making with OD an angle whose tangent is - , i.e. about 21 47'.
25. The resultant, it, of two forces P and Q acting at
an angle a may be easily obtained by Trigonometry.
10 STATICS,
For let OA and OB represent the forces P and Q acting
B, C * yfi
O P AD O'pOA
at an angle a. Complete the parallelogram OACB and draw
CD perpendicular to OA, produced if necessary.
Let R denote the magnitude of the resultant.
Then OD = OA + AD = OA + AC cos DAC
= P+ Q cob BOD = P + Q cos a.
[If D fall between O and A, as in the second figure, we have
OD = OA - DA = OA - AC cob DAC-P-Q cos (180- a) = P+Q cos a.]
Also DC = AC sin DAC = Q sin a.
.\ 7? 8 = 0C 2 = 0D*+CD* = (P+Q cos a) 2 + (# sin a) 2
= J P 8 +# 8 + 27^008 0.
.\ R = VP2+Q2+2FQcosa (i).
Also t&n COD = 7rF - = J ^-,. (n .
OD P + Q cos a v
These two equations give the required magnitude and
direction of the resultant.
Cor. 1. If the forces be at right angles, we have a =90,
so that R = V/ >2 +^ s , and tan CO A = Q
Cor. 2. (1) Wheno = 0, R = P + Q,
(2) When a = 30, R = JI+Q 2 +J3PQ,
(3) When a = 45, R = JP* + Q>+J2PQ,
(4) When o = 60, R = JP* + Q* + PQ,
(5) When a = 1 20, R = JP'+Q^PQ,
(6) When a = 135, R = JP* + Q*~J2PQ,
(7) When a = 150, R = JP^Q'-^PQ,
(8) When a - 1 80, R = P-Q.
COMPOSITION AND RESOLUTION OF FORCES, 11
EXAMPLES. L
1. In the following six examples P and Q denote two compo-
nent forces acting at an angle a and R denotes their resultant.
[The remits should also be verified by a graph and accurate
measurement.]
(i). IfP = 24; Q= 7; a= 90; find P.
(ii). IfP=13; 11 = 14; o= 90; find Q.
(iii). IfP= 7; Q= 8; a= 60; find R.
(iv). IfP= 5; Q- 9; a=120; find R.
(v). IfP= 3; Q= 5; P= 7; find a.
(vi). IfP= 5; R= 7; a= 60 ; find Q.
2. Find the greatest and least resultants of two forces whose
magnitudes are 12 and 8 lbs. weight respectively.
3. Forces equal respectively to 3, 4, 5, and 6 lbs. weight act on a
particle in directions respectively north, south, east, and west ; find
the direction and magnitude of their resultant.
4. Forces of 84 and 187 lbs. weight act at right angles ; find their
resultant.
5. Two forces whose magnitudes are P and PJ2 lbs. weight act
on a particle in directions inclined at an angle of 135 to each other ;
find the magnitude and direction of the resultant.
6. Two forces acting at an angle of 60 have a resultant equal to
2^3 lbs. weight ; if one of the forces be 2 lbs. weight, find the other
force.
7. Two equal forces act on a particle; find the angle between
them when the square of their resultant is equal to three times their
product.
8. Find the magnitude of two forces such that, if they act at
right angles, their resultant is ^10 lbs. weight, whilst when they act at
an angle of 60 their resultant is ^13 lbs. weight.
9. Find the angle between two equal forces P when their resultant
is equal to P.
10. Two given forces act on a particle ; find in what direction a
third force of given magnitude must act so that the resultant of the
three may be as great as possible.
11. If one of two forces be double the other and the resultant be
equal to the greater force, find the cosine of the angle between the
forces.
12. Two forces equal to 2P and P respectively act on a particle ;
if the first be doubled and the second be increased by 12 lbs. weight
the direction of the resultant is unaltered ; find the value of P.
13. The resultant of forces P and Q is R ; if Q be doubled, R is
doubled ; if Q be reversed, R is again doubled ; prove that
P : Q : R :: ^2 : ^8 : J2.
12
STATICS.
26. Two forces, given in magnitude and direction, have
only one resultant ; for only one parallelogram can be con-
structed having two lines OA and OB (Fig. Art. 25) as
adjacent sides.
A force may be resolved into two components in
an infinite number of ways; for an infinite number of
parallelograms can be constructed having OC as a diagonal
and each of these parallelograms would give a pair of such
components.
27. The most important case of the resolution of forces
occurs when we resolve a force into two components at
right angles to one another.
Suppose we wish to resolve a force F y represented by
OC, into two components, one of which is in the direction
OA and the other is perpendicular to OA.
Draw CM perpendicular to OA and complete the paral-
lelogram OMGN. The forces represented by OM ami ON
have as their resultant the force OC, so that OM and ON
are the required components.
Let the angle AOC be a.
Then 0M= OC cosa = Fcosa,
and 0N= MC = OC sin a = Fain a.
[If the point M lie in OA produced backwards, as in the second
figure, the component of F in the direction OA
= -0M= - OC oos COM =OC cos a = F cob a.
Also the component perpendicular to OA
= ON = MC=OCainCOM=Fsma.]
Hence, in each case, the required components are
F oos a and F sin a.
COMPOSITION AND RESOLUTION OF FORCES. 13
Thus a force equal to 10 lbs. weight acting at an angle of 60 with
the horizontal is equivalent to 10 oos 60 ( = 10 x = 5 lbs. weight) in a
/3
horizontal direction, and 10 sin 60 (=10 x^r-= 5 x 1-732=8-66 lbs.
weight) in a vertical direction.
28. Def. The Resolved Part of a given force in a
given direction is the component in the given direction
which, with a component in a direction perpendicular to the
given direction, is equivalent to the given force.
Thus in the previous article the resolved part of the
force F in the direction OA is F cos a. Hence
The Resolved Part of a given force in a given direction is
obtained by multiplying the given force by the cosine of the
angle between the given force and the given direction.
29. A force cannot produce any effect in a direction
perpendicular to its own line of action. For (Fig. Art. 27)
there is no reason why the force ON should have any
tendency to make a particle at move in the direction OA
rather than to make it move in the direction AO produced;
hence the force ON cannot have any tendency to make the
particle move in either the direction OA or AO produced.
For example, if a railway carriage be standing at rest
on a railway line it cannot be made to move along the rails
by any force which is acting horizontally and in a direction
perpendicular to the rails.
EXAMPLES. IL
1. A force equal to 10 lbs. weight is inclined at an angle of 30 to
the horizontal; find its resolved parts in a horizontal and vertical
direction respectively.
2. Find the resolved part of a force P in a direction making an
angle of 45 with its direction.
3. A truck is at rest on a raDway line and is pulled by a hori-
zontal force equal to the weight of 100 lbs. in a direction making an
angle of 60 with the direction of the rails ; what is the force tending
to urge the truck forwards?
4. A body, of weight 20 lbs., is placed on an inclined plane
whose height is 4 feet and whose length is 5 feet ; find the resolved
parts of its weight along and perpendicular to the plane.
30. Triangle of Force*. If thru* forces^ acting at
14 STATICS.
a point, be represented in magnitude and direction by the
sides of a triangle, taken in order, that is, taken the same
way round, they will be in equilibrium.
Let the forces t, Q, and R acting at the point be
represented in magnitude and direction by the sides AB,
,_-C
BC, and GA of the triangle ABC', they shall be in equi-
librium.
Complete the parallelogram ABCD.
The forces represented by BC and AD are the same,
since BC and AD are equal and parallel.
Now the resultant of the forces AB and AD is, by the
parallelogram of forces, represented by AC.
Hence the resultant of AB, BC and CA is equal to the
resultant of forces AC and CA, and is therefore zero.
Hence the three forces P, Q f and R are in equilibrium.
Cor. Since forces represented by AB, BC, and CA are in equi-
librium, and since, when three forces are in equilibrium, each is
equal and opposite to the resultant of the other two, it follows that
the resultant of AB and BC is equal and opposite to CA, i.e. their
resultant is represented by AC.
Hence the resultant of two forces, acting at a point and represented
by the sides AB and BC of a triangle, is represented by the third
side AC.
31. In the Triangle of Forces the student must carefully note
that the forces must be parallel to the sides of a triangle taken in
order.
For example, if the first force act in the direction AB, the second
must act in the direction BC, and the third in the direction CA ; if
the second force were in the direction GB, instead of BC, the forces
would not be in equilibrium.
The three forces must also act at a point; if the lines of action of
the forces were BC, CA, and AB they would not be in equilibrium;
for the foroes AB and BC would have a resultant, acting at B, equal
and parallel to AC. The system of forces would then reduce to two
equal and parallel forces acting in opposite directions, and, as we
shall see in a later chapter, such a pair of forces could not be in
equilibrium.
COMPOSITION AND RESOLUTION OF FORCES. 15
32. The converse of the Triangle of Forces is also
true, viz. that If three forces acting at a point be in equi-
librium they can be represented in magnitude and direction
by the sides of any triangle which is drawn so as to have its
sides respectively parallel to ike directions of the forces.
Let the three forces P t Q, and it?, acting at a point 0,
be in equilibrium. Measure off lengths OX and M along
the directions of P and Q to represent these forces respec-
tively.
Complete the parallelogram OLNM and join ON.
Since the three forces P, Q and R are in equilibrium,
each must be equal and opposite to the resultant of the
other two. Hence R must be equal and opposite to the
resultant of P and Q, and must therefore be represented
by NO. Also LN is equal and parallel to OM.
Hence the three forces P, Q and R are parallel and
proportional to the sides OL, LN and NO of the triangle
OLN.
Any other triangle, whose sides are parallel to those of
the triangle OLN, will have its sides proportional to those
of OLN and therefore proportional to the forces.
Again any triangle, whose sides are respectively per-
pendicular to those of the triangle OLN, will have its sides
proportional to the sides of OLN and therefore proportional
to the forces.
33. LamPs Theorem. If three forces acting on a
particle keep it in equilibrium, each is proportional to the
sine of the angle between the other two.
Taking Fig., Art. 32, let the forces P t Q and R be in
equilibrium. As before, measure off lengths OL and OM to
16 STATICS.
represent the forces P and Q, and complete the parallelo-
gram OLNM. Then NO represents R.
Since the sides of the triangle OLN are proportional to
the sines of the opposite angles, we have ^T-
PL LN NO
sin LNO ~ sin LON ~ sin OLN'
But
sin LNO = sin NOM= sin (180 - QOR) = sin QOR,
sin L0N= sin (180 - LOR) = sin ROP,
and sin 0LN= sin (180 - POQ) = sin POQ.
Also LN=0M.
M 0J/ iTO
Hence
i.e.
sin #0i2 sin ROP sin P0 '
P Q R
sin #0i2 sin ROP sin P<9# *
Ex. The resultant of two forces acting at an angle of 150 is perpen-
dicular to the smaller of these forces. The greater component being
equal to 30 lbs. weight, find the other component and the resultant.
Taking the figure of Art. 32, we have
P=30 and POQ =150.
Also MON is 90, so that, if R be equal and opposite to the
required resultant, then QOR=9Q.
Hence Lami's theorem gives
30 _ Q R
sin 90 ~ sin 120 ~ sin 150 '
,-, an- Q - R
2
.-.Q = 15V31bs. wt.,
and 12 = 15 lbs. wt.
34. Polygon of Forces. If any number of forces,
acting on a particle, be represented, in magnitude and
direction, by the sides of a polygon, taken in order, the
forces shall be in equilibrium.
Let the sides AB, BO, CD, BE, EF and FA of the
polygon ABODE F represent the forces acting on a particle
0. Join AC, AD and AE.
COMPOSITION AND RESOL UTION OF FORGES. 17
By the corollary to Art 30, the resultant of forces AB
and BC is represented by AG,
Similarly the resultant of forces AG and GD is repre-
sented by AD; the resultant of forces AD and DE by AE\
and the resultant of forces A E and EF by AF.
Hence the resultant of all the forces is equal to the
resultant of AF and FA, i.e. the resultant vanishes.
Hence the forces are in equilibrium.
A similar method of proof will apply whatever be the
number of forces. It is also clear from the proof that the
sides of the polygon need not be in the same plane.
#86. The converse of the Polygon of Forces is not true; for the
ratios of the sides of a polygon are not known when the directions of
the sides are known. For example, in the above figure, we might
take any point A' on AB and draw A'F' parallel to AF to meet EF in
F 1 ; the new polygon A'BCDEF' has its sides respectively parallel to
those of the polygon ABC DBF but the corresponding sides are clearly
not proportional. > <~
#36. The resultant of two forces, acting at a point
in directions OA and OB and represented in magnitude by
A . OA and /x . OB, is represented by (\ + /a) . OC, where C is
a point in AB such that X . CA = p . CB.
For let G divide the line AB, such that
\.CA=p.CB.
Complete the parallelograms OGAD and OCBE.
By the parallelogram of forces the force X. OA is
equivalent to forces represented by X.OG and X . OD.
Also the force /* . OB is equivalent to forces represented
by fji.OCa.ndfi.OE.
L. m h. 2
18 STATICS.
Hence the forces X . OA and ft . OB are together equiva-
lent to a force (X + fi) OC together with forces X . OD and
p.OE.
E<
But, (since \.OD = \.CA=fi.CB = fi. OB) these two
latter forces are equal and opposite and therefore are in
equilibrium.
Hence the resultant is (X + fx) . OC.
Cor. The resultant of forces represented by OA and
OB is 20(7, where C is the middle point of A B.
This is also clear from the fact that OC is half the
diagonal OB of the parallelogram of which OA and OB are
adjacent sides.
EXAMPLES, m.
1. Three forces acting at a point are in equilibrium ; if they
make angles of 120 with one another, shew that they are equal.
If the angles are 60, 150, and 150, in what proportions are the
forces?
2. Three forces acting on a particle are in equilibrium ; the angle
between the first and second is 90 and that between the second and
third is 120 ; find the ratios of the forces.
3. Forces equal to 7P, 5P, and 8P acting on a particle are in
equilibrium ; find the angle between the latter pair of forces.
4. Two forces act at an angle of 120. The greater is represented
by 80, and the resultant is at right angles to the less. Find the
latter.
5. Two foroes acting on a particle are at right angles, and are
balanced by a third force making an angle of 160 with one of them.
The greater of the two forces being 3 lbs. weight, what must be the
values of the others ?
6. The magnitudes of two forces are as 3 : 5, and the direction of
the resultant is at right angles to that of the smaller force ; compare
the magnitudes of the greater force and the resultant.
COMPOSITION AND RESOLUTION OF FORCES. 19
7. The sum of two forces is 18, and the resultant, whose direc-
tion is perpendicular to the lesser of the two forces, is 12; find the
magnitudes of the forces.
8. If two forces P and Q act at suoh an angle that R=P, shew
that, if P be doubled, the new resultant is at right angles to Q.
9. The resultant of two forces P and Q is equal to Q^/3 and
makes an angle of 30 with the direction of P; prove that P is either
equal to, or double of, Q.
10. Construct geometrically the directions of two forces 2P and
3P which make equilibrium with a force of 4P whose direction is
given.
11. The sides AB and AC of a triangle A EC are bisected in D and
E ; shew that the resultant of forces represented by BE and DC is
represented in magnitude and direction by BC.
12. P is a particle acted on by forces represented by X . AP and
X . PB where A and B are two fixed points ; shew that their resultant
is constant in magnitude and direction wherever the point P may be.
13. ABCD is a parallelogram ; a particle P is attracted towards A
and G by forces which are proportional to PA and PC respectively and
repelled from B and D by forces proportional to PB and PD ; shew
that P is in equilibrium wherever it is situated.
The following are to be solved by geometric construction and
measurement. In each case P and Q are two forces inclined at an
angle a, and R is their resultant making an angle 6 with P.
14. P=50 kilog., Q = 60 kilog. and P = 70 kilog.; find a and 6.
15. P=30, P. = 40 and a = 130 ; find Q and d.
16. P=60, o=75 and 0=40 ; find Q and R.
17. P=60, P = 40 and 0=50 ; find Q and a.
18. P = 80, a = 55 and R = 100 ; find Q and 0.
19. P=25 lbs. wt.. Q = 20 lbs. wt. and = 35; find R and a.
_2
CHAPTER m.
COMPOSITION AND RESOLUTION OF FORCES (continued).
37. To find the resultant of any number of forces in one
plane acting upon a particle.
Let the forces P t Q, R... act upon a particle at 0,
Through draw a fixed line OX and a line 7 at right
angles to OX.
Let the forces P, Q, R,... make angles a, /?, y... with
OX.
The components of the force P in the directions OX
and OY are, by Art. 27, P cos a and Psina respectively;
similarly, the components of Q are Q cos ft and Q sin ft ;
similarly for the other forces.
Hence the forces are equivalent to a component,
Pcosa+ <2cos/J + jRcosy... along OX,
and a component,
Psina+@sin/J + JJsiny... along OY.
Let these components be X and Y respectively, and let
F be their resultant inclined at an angle to OX.
Since F is equivalent to F co&6 along 0X t and ^sin 6
along OF, we have,
Fcos6=X (1%
Fam6=Y (2).
COMPOSITION AND RESOLUTION OF FORCES. 21
Hence, by squaring and adding,
F* = X*+Y\
Y
Also, by division, tan = -=
These two equations give F and 6, i.e. the magnitude
and direction of the required resultant.
38* Graphical Construction. The resultant of
a system of forces acting at a point may also be obtained
by means of the Polygon of Forces. For (Fig. Art. 34),
forces acting at a point and represented in magnitude
and direction by the sides of the polygon ABCDEF are in
equilibrium. Hence the resultant of forces represented by
AB, BC, CD, DE and EF must be equal and opposite to
the remaining force FA, i.e. the resultant must be repre-
sented by AF.
It follows that the resultant of forces P, Q, R, S
and T acting on a particle may be obtained thus ; take a
point A and draw AB parallel and proportional to P, and
in succession BC, CD, DE and EF parallel and proportional
respectively to Q, R, S, and T ; the required resultant will
be represented in magnitude and direction by the line AF.
The same construction would clearly apply for any
number of forces.
39. Ex. 1. A particle is acted upon by three forces, in one plane,
equal to 2, 2^/2, and 1 lbs. weight respectively; the first is horizontal,
the second acts at 45 to the horizon, and the third is vertical; find
their resultant.
Here Z=2 + 2, v /2co845 o +0=2 + 2 x /2 . 4o =4
7=0 + 2^2 Bin 45+l=2 N /2. -4+1=3.
Hence Foot $=4, and JF sin 0=3;
.*. F=.J+&=5, and tan = .
The resultant is therefore a force equal to 5 lbs. weight acting at
an angle with the horizontal whose tangent is f, i.e. 36 52*.
Ex. 2. A particle is acted upon by forces represented by P,
2P, 3JSP, and 4P; the angles between the first and second, the
second and third, and the third and fourth are 60, 90 and 150
respectively. Shew that the resultant is a force P in a direction
inclined at an angle of 120 to that of the tint force.
22 STATICS,
In this example it will be a simplification if we take the fixed
Y
line OX to coincide with the direction of the first force P ; let XOX
and YOY' be the two fixed lines at right angles.
The second, third, and fourth forces are respectively in the first,
second, and fourth quadrants, and we have clearly
POZ=60, COX'=30, and DOZ=60.
The first force has no component along OY.
The second force is equivalent to components 2P cos 60 and
2P sin 60 along OX and OY respectively.
The third force is equivalent to forces
3,y3Pcos30 o and S^PsinSO
along OX and OY respectively, i.e. to forces -S^/SPcosSO and
BJ3P sin 30 along OX and OY.
So the fourth force is equivalent to 4P cos 60 and 4P sin 60
along OX and Or, i.e. to 4P cos 60 and - 4P sin 60 along OX and OY.
Hence X= P + 2P cos 60 - 3^/3P cos 30 + 4P cos 60
T +2jP - 2'
and
=P+P
7= + 2P sin 60 + SJBP sin 30 - 4P sin 60
P N /3 + 3 4 3 P-4P.^=f P.
and
Hence, if F be the resultant at an angle 9 with OX, we have
F=Jx*+Y*=P,
Y
tan = ^.= -^3 = tan 120,
so that the resultant ia k force P at an angle of 120 with the first
force.
COMPOSITION AND RESOLUTION OF FORCES. 23
EXAMPLES. IV.
[Questions 2, 3, 4, 5 and 8 are suitable for graphic solutions.]
1. Forces of 1, 2, and ^3 lbs. weight act at a point A in
directions AP, AQ, and AR, the angle PAQ being 60 and PAR a
right angle ; find their resultant.
2. A particle is acted on by forces of 5 and 3 lbs. weight which
are at right angles and by a force of 4 lbs. weight bisecting the angle
between them ; find the magnitude of the force that will keep it at
rest.
3. Three equal forces, P, diverge from a point, the middle one
being inclined at an angle of 60 to each of the others. Find the
resultant of the three.
4. Three forces 5P, 10P, and 13P act in one plane on a particle,
the angle between any two of their directions being 120. Find the
magnitude of their resultant.
5. Forces 2P, 3P, and 4P act at a point in directions parallel to
the sides of an equilateral triangle taken in order ; find the magnitude
and line of action of the resultant.
6. Two forces equal respectively to 9 and 12 lbs. weight act at an
angle of 135 on a particle ; a third force, equal to 10 lbs. weight,
aots on the particle, its direction being between the first two and at
30 to the first force; find the magnitude of the resultant of these
forces.
7. ABCD is a square ; forces of 1 lb. wt., 6 lbs. wt., and 9 lbs. wt.
aot in the directions AB,AC, and AD respectively; find the magnitude
of their resultant correct to two places of decimals.
8. Five forces, acting at a point, are in equilibrium; four of
them, whose respective magnitudes are 4, 4, 1, and 3 lbs. weight make,
in succession, angles of 60 with one another. Find the magnitude
of the fifth force.
40. To find ths conditions of equilibrium of any number
of forces acting upon, a particle.
Let the forces act upon a particle as in Art. 37.
If the forces balance one another the resultant must
vanish, i.e. F must be zero.
Hence X*+Y* = 0.
Now the sum of the squares of two real quantities
cannot be zero unless each quantity is separately zero ;
.-. X = 0, and F=0.
Hence, if the forces acting on a particle be in equi-
librium then the algebraic sum of their resolved parts in
two directions at right angles are separately zero.
Conversely, if the sum of their resolved parts in two
24
STATICS.
directions at right angles separately vanish, the forces are
in equilibrium.
For, in this case, both X and Y are zero, and therefore
F is zero also.
Hence, since the resultant of the forces vanishes, the
forces are in equilibrium.
41. When there are only three forces acting on a
particle the conditions of equilibrium are often most easily
found by applying Lami's Theorem (Art. 33).
42. Ex. 1. A body of 65 lbs. weight is suspended by two strings
of lengths 5 and 12 feet attached to two points in the same horizontal
line whose distance apart is 13 feet ; find the tensions of the strings.
Let AG and BG be the two strings, so that
AC =5 ft., J3C=12 ft., and AB=U ft.
A
D
je^
T i\
65
E
Since 13 a = 12*+ 5*, the angle AGB is a right angle.
Let the direction GE of the weight be produced to meet AB in D ;
also let the angle CBA be 0, so that
IACD = 90- lBCD= lCBD = 0.
Let T x and T t be the tensions of the strings. By Lami's theorem
we have
65
Bin EGB sin EC A
sin AC B
65
= sin90 ;
But
COS B:
sin BCD sin
r i= =65 cos 0, and T a =65 sin 0.
BG 12 . . AC 5
: ^=i3' and8m '=zzri3*
.*. r 1 = 60, and r,=251bs. wt.
Otherwise thus; The triangle AGB has its sides respectively per-
pendicular to the directions of the forces T x , T a , and 65 j
BG
CA
65
AB
*i*
a-".
and
*.-2S=-
COMPOSITION AND RESOLUTION OF FORCES. 25
Or again, we may apply the result of Art. 40. Equating to zero
the sum of the resolved parts in the horizontal and vertical directions,
we have
T, cos OB A - Tj cos GAB = 0,
and T a sin CBA + T x sin CAB - 65 = 0.
,, CB 12 , . __ A CA 5
But cob CBA = jg = 7a and sin CBA=-= = ^,
Also cos CAB = , and sin CAB = j=
The above equations therefore become
and STj+m^esxia.
Solving, we have T, = 60 and r 3 =26.
EXAMPLES. V.
1. Two men carry a weight IF between them by means of two
ropes fixed to the weight ; one rope is inclined at 45 to the vertical
and the other at 30 ; find the tension of each rope.
2. A body, of mass 2 lbs., is fastened to a fixed point by means
of a string of length 25 inches ; it is acted on by a horizontal force F
and rests at a distance of 20 inches from the vertical line through
the fixed point ; find the value of F and the tension of the string.
3. A body, of mass 130 lbs., is suspended from a horizontal beam
by strings, whose lengths are respectively 1 ft. 4 ins. and 5 ft. 3 ins.,
the strings being fastened to the beam at two points 5 ft. 5 ins. apart.
What are the tensions of the strings?
4. A body, of mass 70 lbs., is suspended by strings, whose lengths
are 6 and 8 feet respectively, from two points in a horizontal line
whose distance apart is 10 feet ; find the tensions of the strings.
5. A mass of 60 lbs. is suspended by two strings of lengths
9 and 12 feet respectively, the other ends of the strings being attached
to two points in a horizontal line at a distance of 15 feet apart ; find
the tensions of the strings.
6. A string suspended from a ceiling supports three bodies, each
of mass 4 lbs., one at its lowest point and each of the others at
equal distances from its extremities; find the tensions of the parts
into which the string is divided.
7. Two equal masses, of weight W, are attached to the extremities
of a thin string which passes over 3 tacks in a wall arranged in the
form of an isosceles triangle, with the base horizontal and with a
vertical angle of 120; find the pressure on each tack.
26 STATICS. Exs. V.
8. A stream is 96 feet wide and a boat is dragged down the middle
of the Btream by two men on opposite banks, each of whom polls
with a force equal to 100 lbs. wt. ; if the ropes be attached to the same
point of the boat and each be of length 60 feet, find the resultant
pressure on the boat.
9. Two masses, each equal to 112 lbs., are joined by a string
which passes over two small smooth pegs, A and B, in the same
horizontal plane ; if a mass of 6 lbs. be attached to the string halfway
between A and B, find in inches the depth to which it will descend
below the level of AB, supposing AB to be 10 feet.
What would happen if the small mass were attaohed to any other
point of the string?
10. A heavy chain has weights of 10 and 16 lbs. attached to its
ends and hangs in equilibrium over a smooth pulley ; if the greatest
tension of the chain be 20 lbs. wt., find the weight of the chain.
11. A heavy chain, of length 8 ft. 9 ins. and weighing 15 lbs.,
has a weight of 7 lbs. attached to one end and is in equilibrium
hanging over a smooth peg. What length of the chain is on each
side?
12. A body is free to slide on a smooth vertical circular wire and
is connected by a string, equal in length to the radius of the circle,
to the highest point of the circle ; find the tension of the string and
the pressure on the circle.
13. Explain how the force of the current may be used to urge
a ferry-boat across the river, assuming that the centre of the boat
is attaohed by a long rope to a fixed point in the middle of the
stream.
14. Explain how a vessel is enabled to sail in a direction nearly
opposite to that of the wind.
[Let AB be the direction of the keel and therefore that of the
ship's motion, and OA the apparent direction of the wind, the angle
OAB being acute and equal to a. Let iC bo the direction of the
sail, AG being between OA and AB and the angle BAC being 6.
Let P be the force exerted by the wind in a direction perpen-
dicular to the sail. Resolve it into two components, P cos 6
perpendicular to AB and P sin B along AB. The former component
produces leeway (i.e. motion sideways). The latter is never zero
unless d or P vanishes. Also P never entirely vanishes unless the
direction of the wind coincides with that of the sail]
CHAPTER IV.
PARALLEL FORCES.
43. Introduction, or removal, of equal and opposite
forces. We shall assume that if at any point of a rigid
body we apply two equal and opposite forces, they will
have no effect on the equilibrium of the body; similarly,
that if at any point of a body two equal and opposite
forces are acting they may be removed.
44. Principle of the Transmissibility of Force. If a
force act at any jjoint of a rigid body, it may be considered
to act at any other point m its line of action provided that
this latter point be rigidly connected with ike body.
Let a force F act at a point A of a body in a direction
AX. Take any point B in AX and at B introduce two
equal and opposite forces, each equal to F, acting in the
directions BA and BX -, these will have no effect on the
equilibrium of the body.
The forces F acting at A in the direction AB, and F at
B in the direction BA, are equal and opposite; we shall
assume that they neutralise one another and hence that
they may be removed.
We have thus left the force F at B acting in the
direction BX, and its effect is the same as that of the
original force F at A.
28 STATICS.
The internal forces in the above body would be different
according as the force F is supposed applied at A or B ;
of the internal forces, however, we do not treat in the
present book.
45. In Chapters II. and in. we have shewn how to
find the resultant of forces which meet in a point. In
the present chapter we shall consider the composition of
parallel forces.
In the ordinary statical problems of every-day life
parallel forces are of constant occurrence.
46. Def. Two parallel forces are said to be like when
they act in the same direction ; when they act in opposite
parallel directions they are said to be unlike.
47. To find the resultant, of two parallel forces acting
upon a rigid body.
Case I. Let the forces be like.
Let P and Q be the forces acting at points A and B of
the body, and let them be represented by the lines AL and
BM.
Join AB and at A and B apply two equal and opposite
forces each equal to S and acting in the directions BA and
AB respectively. Let these forces be represented by AD
and BE. These two forces balance one another and have
no effect upon the equilibrium of the body.
Complete the parallelograms ALFD and BMGE ; let
the diagonals FA and GB be produced to meet in 0. Draw
OC parallel to AL or BM to meet AB in C.
The forces P and S at A have a resultant P lt repre-
sented by AF. Let its point of application be removed to
0.
So the forces Q and at B have a resultant <2 X repre-
sented by BG. Let its point of application be transferred
to 0.
The force P l at may be resolved into two forces,
S parallel to A J), and P in the direction OC.
So the force Q x at may be resolved into two forces, S
parallel to BE, and Q in the direction OC.
Also these two forces S acting at are in equilibrium.
PARALLEL FORCES.
29
Hence the original forces P and Q are equivalent to
a force (P + Q) acting along OC, i.e. acting at C parallel to
the original directions of P and Q.
J*.
*7*
To determine the position of the point C. The triangle
OCA is, by construction, equiangular with the triangle ALF;
OC AL P
- CA x zF = S ( Eua **" 4 ' r -^PP 01 " 1 ** L Art - 2 ) ( X )*
So, since the triangles OCB and 2?J/# are equiangular,
we have
OC BM Q
Kfh
i <
CB MG S
Hence, from (1) and (2), by division,
CA Q
CB P'
i.e. C divides the line AB internally in the inverse ratio of
the forces.
Case II. Let the forces be unlike.
Let P and Q be the forces (P being the greater) acting
at points A and B of the body, and let them be represented
by the lines AL and BM.
Join AB, and at A and B apply two equal and opposite
forces, each equal to S t and acting in the directions BA
30 STATICS.
and A B respectively. Let these forces be represented by
AD and BE respectively; they balance one another and
have no effect on the equilibrium of the body.
Complete the parallelograms ALFD and BMGE, and
produce the diagonals AF and GB to meet in 0.
[These diagonals will always meet unless they be parallel, in
which ease the forces P and Q will be equal.]
Draw OC parallel to AL or BM to meet AB 'yd. C.
The forces P and S acting at A have a resultant P x
represented by AF. Let its point of application be trans-
ferred to 0.
So the forces Q and S acting at B have a resultant Q x
represented by BG. Let its point of application be trans-
ferred to 0.
The force P x at may be resolved into two forces,
parallel to AD, and P in the direction CO produced.
So the forces Q x at may be resolved into two forces,
S parallel to BE, and Q in the direction OC.
Also these two forces S acting at are in equilibrium.
Hence the original forces P and Q are equivalent to
a force PQ acting in the direction CO produced, i.e. acting
at C in a direction parallel to that of P.
To determine the position of the point C. The triangle
OCA is, by construction, equiangular with the triangle FDA;
OC FD AL P v ^ . ,. A ^ on m
' CA = DA = AD = S [ Euc - TI - 4 or A PP endlx > Art - 2 3 ( x )-
PARALLEL FORCES. 31
So, since the triangles OGB and BMG are equiangular,
we have
OC_BM_Q m
CB~ MG S Kh
CA
Hence, from (1) and (2), by division, ^ = pi * C
divides the line AB externally in the inverse ratio of the
forces.
To sum up ; If two parallel forces, P and Q, act at
points A and B of a rigid body,
(i) their resultant is a force whose line of action is
parallel to the lines of action of the component forces ;
also, when the component forces are like, its direction is
the same as that of the two forces, and, when the forces
are unlike, its direction is the same as that of the greater
component.
(ii) the point of application is a point C in i^ such
that
P.AC = Q.BC.
(iii) the magnitude of the resultant is the sum of the
two component forces when the forces are like, and the
difference of the two component forces when they are
unlike.
48. Case of failure of the preceding construction.
In the second figure of the last article, if the forces
P and Q be equal, the triangles FDA and GEB are equal
in all respects, and hence the angles DAF and EBG will be
equal.
In this case the lines AF and GB will be parallel and
will not meet in any such point as ; hence the construction
fails.
Hence there is no single force which is equivalent to two
equal unlike parallel forces.
We shall return to the consideration of this case in
Chapter vi.
49. If we have a number of like parallel forces acting
on a rigid body we can find their resultant by successive
applications of Art. 47. We must find the resultant of the
32
STATICS.
first and second, and then the resultant of this resultant
and the third, and so on.
The magnitude of the final resultant is the sum of the
forces.
If the parallel forces be not all like, the magnitude of
the resultant will be found to be the algebraic sum of the
forces each with its proper sign prefixed.
60. Ex. A horizontal rod, 6 feet long, whose weight may be neglected,
rests on two supports at its extremities; a body, of weight 6 cwt., is
suspended from the rod at a distance of 2J feet from one end; find the
reaction at each point of support. If one support could only bear a
pressure equal to the weight of 1 cwt., what is the greatest distance from
the other support at which the body could be suspended t
Let A B be the rod and R and S the pressures at the points of sap-
port. Let C be the point at which the body is suspended bo that
A
A
s
A
Qcwt
AG =3 and CB=2\ feet. For equilibrium the resultant of R and S
must balance 6 cwt. Hence, by Art. 47,
JS + Sf=6.... (1),
"* 8~ AC-B-7 {2) '
r 7
Solving (1) and (2), we have -R^S' and 8= 2' Hence tne P re8sures
are 2 and 3 cwt. respectively.
If the reaction at A can only be equal to 1 cwt., 8 must be 5 owt.
Hence, if AC be x, we have
lBCQ-x
6~ AC~ x '
.*. x=5 feet.
Hence BO is 1 foot.
EXAMPLES. VL
In the four following examples A and B denote the points of appli-
cation of parallel forces P and Q, and C is the point in which their
resultant R meets AB.
PARALLEL FORCES, 33
1. Find the magnitude and position of the resultant (the forces
being like) when
(i) P=4; Q = 7; AB-U inches;
(ii) P=ll; Q = 19; .45=2$ feet;
(iii) P=5; Q = 5; AB = S feet.
2. Find the magnitude and position of the resultant (the forces
being unlike) when
(i) P=17; Q = 25; AB=S inches;
(ii) P=23; Q = 15; J5=40 inches;
(iii) P=26; Q = 9; AB = S feet
3. The forces being like,
(i) ifP=8; P = 17; AC =4$ inches; find Q and 4P ;
(ii) if Q = 11 ; 4C= 7 inches ; .4P= 8 inches ; find P and R ;
(iii) if P=6 ; AC =9 inches ; CP=8 inches; find Q and P.
4. The forces being unlike,
(i) if P=8; P = 17; AC=\ inches; find Q and AB;
(ii) if Q=ll; AC= -7 inches; JP = 8f inches; find P and P;
(iii) if P=6 ; 4C= - 9 inches ; AB=12 inches ; find Q and P.
5. Find two like parallel foroes acting at a distance of 2 feet
apart, which are equivalent to a given force of 20 lbs. wt., the line
of action of one being at a distance of 6 inches from the given force.
6. Find two unlike parallel forces acting at a distance of 18
inches apart which are equivalent to a force of 30 lbs. wt., the greater
of the two forces being at a distance of 8 inches from the given force.
7. Two men carry a heavy cask of weight 1J cwt., which hangs
from a light pole, of length 6 feet, each end of which rests on a
shoulder of one of the men. The point from which the cask is hung
is one foot nearer to one man than to the other. What is the pressure
on each shoulder?
8. Two men, one stronger than the other, have to remove a
block of stone weighing 270 lbs. by means of a light plank whose
length is 6 feet ; the stronger man is able to carry 180 lbs. ; how must
the block be placed so as to allow him that share of the weight?
9. A uniform rod, 12 feet long and weighing 17 lbs., can turn
freely about a point in it and the rod is in equilibrium when a weight
of 7 lbs. is hung at one end ; how far from the end is the point about
which it can turn?
N.B. The weight of a uniform rod may be taken to act at it$
middle point.
10. A straight uniform rod is 3 feet long ; when a load of 5 lbs.
is placed at one end it balances about a point 3 inches from that end ;
find the weight of the rod.
L. M. H. 3
34 STATICS. Exs. VI.
11. A uniform bar, of weight 8 lbs. and length 4 feet, passes over
a prop and is supported in a horizontal position by a force equal to
1 lb. wt. acting vertically upwards at the other end ; find the distance
of the prop from the oentre of the beam.
12. A heavy uniform rod, 4 feet long, rests horizontally on two
pegs whioh are one foot apart ; a weight of 10 lbs. suspended from
one end, or a weight of 4 lbs. suspended from the other end, will just
tilt the rod up ; find the weight of the rod and the distances of the
pegs from the centre of the rod.
13. A uniform iron rod, 2$ feet long and of weight 8 lbs., is
placed on two rails fixed at two points, A and B, in a vertical wall.
A B is horizontal and 5 inches long ; find the distances at which the
ends of the rod extend beyond the rails if the difference of the pres-
sures on the rails be 6 lbs. wt.
14. A uniform beam, 4 feet long, is supported in a horizontal
position by two props, which are 3 feet apart, bo that the beam pro-
jects one foot beyond one of the props ; shew that the pressure on
one prop is double that on the other.
15. One end of a heavy uniform rod, of weight W, rests on a
smooth horizontal plane, and a string tied to the other end of the
rod is fastened to a fixed point above the plane; find the tension
of the string.
16. A man carries a weight of 50 lbs. at the end of a stick, 3 feet
long, resting on his shoulder. He regulates the stick so that the
length between his shoulder and his hands is (1) 12, (2) 18 and (3) 24
inches ; how great are the forces exerted by his hand and the pressures
on his shoulder in eaoh case ?
CHAPTER V.
MOMENTS.
51. Def. The moment of a force abotti a given point
is the product of the force and the perpendicular drawn
from the given point upon the line of action of the force.
Thus the moment of a force F about a given point is
F x ON, where ON is the perpendicular drawn from upon
the line of action of F.
It will be noted that the moment of a force F about
a given point never vanishes unless either the force
vanishes or the force passes through the point about which
the moment is taken.
52. Geometrical representation of a moment.
Suppose the force F to be represented in magnitude,
direction, and line of action by the line AB. Let be any
given point and O^the perpendicular from upon AB or
AB produced.
Join OA and OB.
By definition the moment of F about is F x ON, i.e.
AB x ON. But AB x ON is equal to twice the area of the
triangle OAB [for it is equal to the area of a rectangle
whose base is AB and whose height is ON]. Hence the
S 2
36
STATICS.
moment of the force F about the point is represented by
twice the area of the triangle OAB t i.e. by twice the a/rea of
yO
/
F N
B
the triangle whose base is the line representing the force and
whose vertex is the point about which the moment is taken.
53. Physical meaning of the moment of a force about a
Suppose the body in the figure of Art. 51 to be a plane
lamina [i.e. a body of very small thickness, such as a piece
of sheet-tin or a thin piece of board] resting on a smooth
table and suppose the point of the body to be fixed.
The effect of a force F acting on the body would be to
cause it to turn about the point as a centre, and this
effect would not be zero unless (1) the force F were zero, or
(2) the force ^passed through 0, in which case the distance
ON would vanish. Hence the product F x ON would seem
to be a fitting measure of the tendency of F to turn the
body about 0. This may be experimentally verified as
follows ;
Let the lamina be at rest under the action of two forces
MOMENTS.
37
F and F , whose lines of action lie in the plane of the
lamina. Let ON and 0N X be the perpendiculars drawn
from the fixed point upon the lines of action of F
and F x .
If we measure the lengths ON and 0N X and also the
forces ^and F xi it will be found that the product F . ON
is always equal to the product F x . 0N X .
Hence the two forces, F and F ly will have equal but
opposite tendencies to turn the body about if their
moments about have the same magnitude.
54. Positive and negative moments. In Art. 53 the
force F would, if it were the only force acting on the
lamina, make it turn in a direction opposite to that in
which the hands of a watch move, when the watch is laid
on the table with its face upwards.
The force F x would, if it were the only force acting on
the lamina, make it turn in the same direction as that in
which the hands of the watch move.
The moment of F about is said to be positive, and the
moment of F x about is said to be negative.
55. Algebraic sum of moments. The algebraic sum of
the moments of a set of forces about a given point is the
sum of the moments of the forces, each moment having its
proper sign prefixed to it.
Bx. ABCD is a square; along the sides AB, CB, DC, and DA forces
act equal respectively to 6, 5, 8, and 12 lbs. tot. Find the algebraic
sum of their moments about the centre, 0, of the square, if the side of
the square be 4 feet.
38
STATICS.
The foroes along DA and AB tend to turn the square about in
the positive direction whilst the forces along the sides DC and CB tend
to turn it in the negative direction.
The perpendicular distance of O from each force is 2 feet.
Hence the moments of the forces are respectively
+ 6x2, -6x2, -8x2, and +12x2.
Their algebraic sum is therefore 2[6-5-8+12] or 10 units of mo-
ment, t. e. 10 times the moment of a force equal to 1 lb. wt. acting at
the distance of 1 foot from O.
56. Theorem. The algebraic sum of the moments of
any two forces about any point in their plane is equal to the
moment of their resultant about the same point
Case I. Let the forces meet in a point.
C O C O
\-
Let P and Q acting at the point A be the two forces and
the point about which the moments are taken. Draw OC
parallel to x the direction of P to meet the line of action of
Q in the point C.
Let AC represent Q in magnitude and on the same
scale let AB represent P \ complete the parallelogram
ABDC, and join OA and OB, Then, by the Parallelogram
of Forces, AD represents the resultant, R, of P and Q.
(a) If be without the angle DAC, as in the first
figure, we have to shew that
2AOAB + 2AOAC = 2AOA&.
Since AB and OB are parallel, we have
A OAB - A DAB = A A CD. [Euc. i. 37]
9 \ 2AOAB+2AOAC=2&ACD + 2AOAC = 2AOAD.
(ft) If be within the angle CAD, as in the second
figure, we have to shew that
2AA0B-2AA0C = 2AA0D.
MOMENTS. 39
As in (a), we have
AAOB = ADAB = AAGD.
/. 2aAOB-2aAOC=2aACD-2aOAC = 2aOAD.
Case II. Let the forces be parallel.
O
Let P and Q be two parallel forces and R (=P+ Q)
their resultant.
From any point in their plane draw OACB perpen-
dicular to the forces to meet them in A, C, and B respec-
tively.
By Art. 47 we have P.AC=Q.CB (1)
.'. the sum of the moments of P and Q about
= Q.OB + P.OA
= Q(OC+CB) + P(OC-AC)
= (P+Q)OC + Q.CB-P.AC
*= (P+ Q) . OC, by equation (1),
= moment of the resultant about 0.
If the point about which the moments are taken be
between the forces, as 0,, the moments of P and Q have
opposite signs.
In this case we have
Algebraic sum of moments of P and Q about 0,
= P.0 1 A-Q.0 1 B
= P(0 1 C + CA)-Q(CB-0 1 C)
^(P + Q).0 1 C^P.CA-Q.CB
= (P+Q). x C f by equation (1).
The case when the point has any other position, as also
40 STATICS.
the case when the forces have opposite parallel directions,
are left for the student to prove for himself.
67. Case I. of the preceding proposition may be otherwise proved
in the following manner :
Let the two forces, P and Q, be represented by AB and AC re-
spectively and let AD represent the resultant R so that ABDG is a
parallelogram.
Let be any point in the plane of the forces. Join OA and draw
BL and CM, parallel to OA, to meet AB in L and M respectively.
Since the sides of the triangle A CM are respectively parallel to the
sides of the triangle DBL, and since AC is equal to BD,
:. AM=LD,
.'. a OAM = a OLD. [Euc. i. 38]
First, let fall without the angle CAD, as in the first figure.
Then 2aOAB+2aOAG
= 2 a OAL + 2 a OAM [Euc. l 37]
= 2a04L + 2aOLD
= 2aOAD.
Hence the sum of the moments of P and Q is equal to that of R.
Secondly, let fall within the angle CAD, as in the second figure.
The algebraic sum of the moments of P and Q about
=2aOAB-2aOAC
as 2 A OAL - 2 A OAM [Euc. I. 37]
s=2aO^L-2aOLD
=2aOAD
= moment of R about O.
MOMENTS. 41
58. If the point about which the moments are taken
lie on the resultant the moment of the resultant about the
point vanishes. In this case the algebraic sum of the
moments of the component forces about the given point
vanishes, i.e. The moments of two forces about any point on
the line of action of their resultant are equal and of opposite
sign.
The student will easily be able to prove this theorem
independently from a figure ; for, in Art. 56, the point
will be found to coincide with the point D and we have
only to shew that the triangles ACO and ABO are now
equal, and this is obviously true. [Euc, I. 34.]
59. Generalised theorem of moments. If any
number of forces in one plane acting on a rigid body have a
resultant, the algebraic sum of their moments about any point
in their plane is equal to the moment of their resultant.
For let the forces be P, Q, R, S,... and let be the
point about which the moments are taken.
Let P x be the resultant of P and Q,
P, be the resultant of P x and 72,
P 8 be the resultant of P 2 an d >
and so on till the final resultant is obtained.
Then the moment of i\ about = sum of the moments
of P and Q (Art. 56) ;
Also the moment of P s about = sum of the moments
of Pi and R
= sum of the moments of P, Q } and R.
So the moment of P 3 about
= sum of the moments of P 2 and S
= sum of the moments of P, Q, R, and S t
and so on until all the forces have been taken.
Hence the moment of the final resultant
= algebraic sum of the moments of the component forces.
Cor. It follows, similarly as in Art. 58, that the alge-
braic sum of the moments of any number of forces about a
point on the line of action of their resultant is zero; so,
conversely, if the algebraic sum of the moments of any
42
STATICS.
number of forces about any point in their plane vanishes,
then, ett/ier their resultant is zero (in which case the forces
are in equilibrium), or the resultant passes through the
point about which the moments are taken.
60. Ex. A rod, 5 feet long, supported by two vertical string*
attached to its ends has weights of 4, 6, 8 and 10 lbs. hung from the
rod at distances of 1, 2, 3 and 4 feet from one end. If the weight of
the rod be 2 lbs., what are the tensions of the strings t
Let AF be the rod, B, C, D and E the points at which the weights
R-
C G D
l. LI l. I
8 +10
2
are hung ; let G be the middle point ; we shall assume that the weight
of the rod acts here.
Let B and S be the tensions of the strings. Since the resultant of
the forces is zero, its moment about A must be zero.
Hence, by Art. 59, the algebraic sum of the moments about A
must vanish.
Therefore 4x 1 + 6x2 + 2 x 2 + 8x 8 + 10x4- 5x5=0,
.-. 53=4 + 12 + 5 + 24 + 40 = 85,
.-. S=17.
Similarly, taking moments about F, we have
5U = 10 x 1 + 8 x 2 + 2x2^ + 6x8 + 4x4 = 65,
.\ 2? =13.
The reaction B may be otherwise obtained. For the resultant of the
weights is a weight equal to 30 lbs. and that of B and S is a force
equal to B + S. But these resultants balance one another.
.-. B + S=30;
,\ B = 30-3=30- 17 = 13.
EXAMPLES. VH
1. The side of a square ABGD is 4 feet ; along the line* CB, BA,
DA and DB. respectively act forces equal to 4, 3, 2 and 5 lbs. weight ;
find to the nearest decimal of a foot-pound the algebraic sum of the
moments of the forces about G.
2. A pole of 20 feet length is placed with its end on a horizontal
plane and is pulled by a string, attached to its upper end and inclined
at 30 to the horizon, whose tension is equal to SO lbs. wt. ; find the
MOMENTS. 43
horizontal foroe which applied at a point 4 feet above the ground will
keep the pole in a vertical position.
3. A uniform iron rod is of length 6 feet and mass 9 lbs. , and
from its extremities are suspended masses of 6 and 12 lbs. respec-
tively; from what point must the rod be suspended so that it may
remain in a horizontal position ?
4. A uniform beam is of length 12 feet and weight 50 lbs., and
from its ends are suspended bodies of weights 20 and 30 lbs. respec-
tively; at what point must the beam be supported so that it may
remain in equilibrium ?
5. Masses of 1 lb., 2 lbs., 3 lbs., and 4 lbs. are suspended from a
uniform rod, of length 5 ft., at distances of 1 ft., 2 ft., 3 ft., and 4 ft.
respectively from one end. If the mass of the rod be 4 lbs., find the
position of the point about which it will balance.
6. A uniform rod, 4 ft. in length and weighing 2 lbs., turns freely
about a point distant one foot from one end and from that end a
weight of 10 lbs. is suspended. What weight must be placed at the
other end to produce equilibrium ?
7. A heavy uniform beam, 10 feet long, whose mass is 10 lbs., is
supported at a point 4 feet from one end ; at this end a mass of 6 lbs.
is placed ; find the mass which, placed at the other end, would give
equilibrium.
8. The horizontal roadway of a bridge is 30 feet long, weighs
6 tons, and rests on similar supports at its ends. What is the pressure
borne by each support when a carriage, of weight 2 tons, is (1) half-
way across, (2) two-thirds of the way across ?
9. A light rod, AB, 20 inches long, rests on two pegs whose
distance apart is 10 inches. How must it be placed so that the
pressures on the pegs may be equal when weights of 2W and 3W
respectively are suspended from A and B ?
10. A light rod, of length 3 feet, has equal weights attached to it,
one at 9 inches from one end and the other at 15 inches from the other
end ; if it be supported by two vertical strings attached to its ends and
if the strings cannot support a tension greater than the weight of
1 cwt. , what is the greatest magnitude of the equal weights ?
11. A heavy uniform beam, whose mass is 40 lbs., is suspended
in a horizontal position by two vertical strings each of which can
sustain a tension of 35 lbs. weight. How far from the centre of the
beam must a body, of mass 20 lbs., be placed so that one of the strings
may just break ?
12. A rod, 16 inches long, rests on two pegs, 9 inches apart, with
its centre midway between them. The greatest masses that can be
suspended in succession from the two ends without disturbing the
equilibrium are 4 lbs. and 5 lbs. respectively. Find the weight of the
rod and the position of the point at vrh\oh it* weight acts.
44 STATICS. Exs. VII.
13. A straight rod, 2 feet long, is movable about a hinge at one
end and is kept in a horizontal position by a thin vertical string
attaohed to the rod at a distance of 8 inches from the hinge and
fastened to a fixed point above the rod ; if the string can just support
a mass of 9 ozs. without breaking, find the greatest mass that can
be suspended from the other end of the rod, neglecting the weight of
the rod.
14. A tricycle, weighing 5 stone 4 lbs., has a small wheel sym-
metrically placed 3 feet behind two large wheels which are 3 feet apart ;
if the centre of gravity of the machine be at a horizontal distance of
9 inches behind the front wheels and that of the rider, whose weight
is 9 stone, be 3 inches behind the frout wheels, find the pressures on
the ground of the different wheels.
15. A front-steering tricycle, of weight 6 stone, has a small wheel
symmetrically placed 3 ft. 6 ins. in front of the line joining the two
large wheels which are 3 feet apart ; if the centre of gravity of the
machine be distant horizontally 1 foot in front of the hind wheels and
that of the rider, whose weight is 11 stone, be 6 inches in front
of the hind wheels, find how the weight is distributed on the different
wheels.
16. A dog-cart, loaded with 4 owt., exerts a pressure on the horse's
back equal to 10 lbs. wt. ; find the position of the centre of gravity of
the load if the distance between the pad and the axle be 6 feet.
17. The wire passing round a telegraph pole is horizontal and
the two portions attaohed to the pole are inclined at an angle of 60
to one another. The pole is supported by a wire attaohed to the
middle point of the pole and inclined at 60 to the horizon ; shew that
the tension of this wire is 4 v '3 times that of the telegraph wire.
18. A cyclist, whose weight is 160 lbs., puts all his weight upon
one pedal of his bicycle when the crank is horizontal and the bicycle
is prevented from moving forwards. If the length of the crank is
6 inches and the radius of the chain wheel is 4 inches, shew that the
tension of the chain is 225 lbs. wt.
## CHAPTER VL
COUPLES.
61. Def. Two equal unlike parallel forces, whose
lines of action are not the same, form a couple.
The Arm of a couple is the perpendicular distance
between the lines of action of the two forces which form
the couple, i.e. is the perpendicular drawn from any point
lying on the line of action of one of the forces upon the
line of action of the other. Thus the arm of the couple
(P, P) is the length AB.
The Moment of a couple is the product of one of the
forces forming the couple and the arm of the couple.
In the figure the moment of the couple is P x AB.
62. Theorem. The algebraic sum of the moments of
the two forces forming a couple about any point in their
plane is constant, and equal to the moment of the couple.
Let the couple consist of two forces, each equal to P,
and let be any point in their plane.
Draw OAB perpendicular to the lines of action of the
forces to meet them in A and B respectively.
46
STATICS.
The algebraic sum of the moments of the forces about
= P.O-P.OA = P(OB-OA) = P.AB
= the moment of the couple, and is therefore the same
whatever be the point about which the moments are
taken.
63. Theorem. Two couples, acting in one plane upon
a rigid body, whose moments cure equal and opposite^ balance
one another.
Let one couple consist of two forces (P t P), acting at
the ends of an arm p, and let the other couple consist of
two forces (Q, Q), acting at the ends of an arm q.
Case I. Let one of the forces P meet one of the forces
Q in a point 0, and let the other two forces meet in 0'.
From 0' draw perpendiculars, O'M and 0'i\T, upon the
forces which do not pass through 0\ so that the lengths of
these perpendiculars are p and q respectively.
Since the moments of the couples are equal in magni-
tude, we have
P.p = Q.q, i.e., P . 0'M= Q . 0N.
tf\
COUPLES.
47
Hence, (Art. 58), 0' is on the line of action of the
resultant of P and Q acting at 0, so that 00' is the
direction of this resultant.
Similarly, the resultant of P and Q at 0' is in the
direction OO.
Also these resultants are equal in magnitude; for the
forces at are respectively equal to, and act at the same
angle as, the forces at 0'.
Hence these two resultants destroy one another, and
therefore the four forces composing the two couples are in
equilibrium.
Case II. Let the forces composing the couples be all
parallel, and let any straight line perpendicular to their
directions meet them in the points A t fi, C and D, as in
the figure, so that we have
P.AB = Q.CB (i).
Let L be the point of application of the resultant of Q
at C and P at , so that
P.BL = Q.CL (ii).
By subtracting (ii) from (i), we have
P.AL = Q.LD,
so that L is the point of application of the resultant of P
at A, and Q at D.
O
pr
But the magnitude of each of these resultants is
(P + Q\ and they have opposite directions ; hence they are
in equilibrium.
Therefore the four forces composing the two couples
balance.
64. Since two couples in the same plane, of equal but
48 STATICS.
opposite moments, balance, it follows, by reversing the
directions of the forces composing one of the couples, that
Any two couples of equal moment in tfa same plane are
equivalent.
It follows also that two like couples of equal moment
are equivalent to a couple of double the moment.
65. Theorem. Any number of couples in the same
plane acting on a rigid body a/re equivalent to a single
couple, whose moment is equal to the algebraic sum, of the
moments of the couples.
For let the couples consist of forces (P, P) whose arm
* s P) (Qi Q) w ho se arm is q, (R, R) whose arm is r, etc.
Replace the couple (Q, Q) by a couple whose components
have the same lines of action as the forces (P f P). The
magnitude of each of the forces of this latter couple will
be X, where X.p = Q.q, (Art. 64)
so that X=Q-.
P
So let the couple (R, R) be replaced by a couple
(%'*;)
whose forces act in the same lines as the
P ' P>
forces (P, P).
Similarly for the other couples.
Hence all the couples are equivalent to a couple, each of
a r
whose forces i&P + Q- + R+... acting at an arm p.
P P
The moment of this couple is
( p+ Q q r B r-)-p>
i.e., P.p + Q.q + R .r+ ....
Hence the original couples are equivalent to a single
couple, whose moment is equal to the sum of their moments.
If all the component couples have not the same sign we
must give to each moment its proper sign, and the same
proof will apply.
Ex. ABGD is a square; along AB and CD act forces of 3 lbs.
wt. , and along AD and CB forces of 4 lbs. wt., whilst at A and G are
applied forces, parallel respectively to BD and DB, each equal to
COUPLES. 49
5^2 lbs. wt. Find the moment of the couple to which these are equiva-
lent, if the aide of the square be 2 feet.
By Art. 54 the moment of the first couple is positive and those of
the other two are negative.
The distance AC= ^2* + 2* =2^/2.
Hence the required moment, by the last article,
3x2-4x2-5 N /2x^(7
= 6- 8- 20= -22.
Hence the equivalent couLe is one whose moment is negative and
equal to 22 ft. lbs. wt,
EXAMPLES. VTTT,
1. ABCD is a square whose side is 2 feet; along AB, BC, CD and
DA act forces equal to 1, 2, 8, and 5 lbs. wt., and along AG and DB
forces equal to 5 *J2 and 2 ^2 lbs. wt. ; shew that they are equivalent
to a couple whose moment is equal to 16 foot-pounds weight.
2. Along the sides AB and CD of a square ABCD act forces each
equal to 2 lbs. weight, whilst along the sides AD and CB act forces
each equal to 5 lbs. weight ; if the side of the square be 3 feet, find
the moment of the couple that will give equilibrium.
3. ABCDEF is a regular hexagon ; along the sides AB, CB, DE
and FE act forces respectively equal to 5, 11, 5, and 11 lbs. weight,
and along CD and FA act forces, each equal to x lbs. weight. Find
x, if the forces be in equilibrium.
4. A horizontal bar AB, without weight, is acted upon by a
vertical downward force of 1 lb. weight at A, a vertical upward force
of 1 lb. weight at B, and a downward force of 5 lbs. weight at a
given point C inclined to the bar at an angle of 30. Find at what
point of the bar a force must be applied to balance these, and find
also its magnitude and direction.
L. M. H.
CHAPTER VII
EQUILIBRIUM OF A RIGID BODY ACTED ON BY THREE
FORCES IN A PLANE.
66. In the present chapter we shall discuss some
simple cases of the equilibrium of a rigid body acted upon
by three forces lying in a plane.
By the help of the theorem of the next article we shall
find that the conditions of equilibrium reduce to those of a
single particle.
67. Theorem. If three forces, acting in one plane
upon a rigid body, keep it in equilibrium, they must either
meet in a point or be parallel.
If the forces be not all parallel, at least two of them
p \
must meet ; let these two be P and Q, and let their direc-
tions meet in 0.
The third force R shall then pass through the point 0.
Since the algebraic sum of the moments of any number
of forces about a point in their plane is equal to the moment
of their resultant,
therefore the sum of the moments of P, Q, and R about
is equal to the moment of their resultant.
But this resultant vanishes since the forces are in equi-
librium.
Hence the sum of the moments of P t Q, and R about
is zero.
THREE FORCES ACTING ON A BODY. 51
But, since P and Q both pass through 0, their momenta
about vanish.
Hence the moment of R about vanishes.
Hence by Art. 58, since R is not zero, its line of action
must pass through 0.
Hence the forces meet in a point.
Otherwise. The resultant of P and Q must be some force passing
through 0.
But, since the forces P, Q, and R are in equilibrium, this resultant
must balance R.
But two forces cannot balance unless they have the same line of
action.
Hence the line of action of R must pass through 0.
68. By the preceding theorem we see that the con-
ditions of equilibrium of three forces, acting in one plane,
are easily obtained. For the three forces must meet in a
point ; and by using Lami's Theorem, (Art. 33), or by
resolving the forces in two directions at right angles, (Art.
37), we can obtain the required conditions.
Ex. 1. A heavy uniform rod AB is hinged at A to a fixed point,
and rests in a position inclined at 60 to the horizontal^ being acted
upon by a horizontal force F applied at the lower end B : find the
action at the hinge and the magnitude of F.
Let the vertical through C, the middle point of the rod, meet the
horizontal line through B in the point D and let the weight of the
rod be W.
There are only three forces acting on the rod, viz., the force F, the
weight W, and the unknown reaotion, P, of the binge.
W
These three forces must therefore meet in a point.
43
52
STATICS.
Now F and W meet *t D; hence the direction of the aotion at
the hinge muat be the line DA.
Draw AE perpendicular to EB.
hetAC=CB = a.
Henoe
BE = AB oos 60 s 2a x \ = a.
4.E = jAW-BE^aJd,
s
and ^D= ,JAE*+DE*= ^J 3a 2 +^ = ^n/13.
Since the triangle ADE has its sides respectively parallel to the
forces F t P, and W, we have
DA~ ED~~ AE'
P _F _ W
" ^13" 1 -2^/3'
Bx. 2. A uniform rod, AB, is inclined at an angle of 60 to the
vertical with one end A resting against a smooth vertical wall, being
supported by a string attached to a point G of the rod, distant 1 foot
from B, and also to a ring in the wall vertically above A ; if the length
of the rod be 4 feet, find the position of the ring and the inclination and
tension of the string.
Let the perpendicular to the wall through A and the vertical line
through the middle point, G, of the rod meet in O.
The third force, the tension T of the string, must therefore pass
through 0. Hence CO produced must pass through D, the position of
the ring,
Let the angle CDA be B, and draw CEF horizontal to meet 00 in
E and the wall in jP.
Then
^t, C^ CG sin CGE
tentf = tanC70E = = -^
1 . sin 6
8 . oo t0
1^
J*
THREE FORCES ACTING ON A BODY, 53
.-. 0=30.
.'. A CD = 60 -0 = 30.
Hence AD=AC=S feet, giving the position of the ring.
If B be the reaction of the wall, and W be the weight of the
oeam, we have, since the forces are proportional to the sides of the
triangle AOD t
2^_ J R _ W
OD~AO~DA'
. T - W 9R x _W_-w
"DA cos 30 V 3 '
Ad 1
and R=W=W tan 30 = JT.-^.
EXAMPLES. IX.
1. A uniform rod, AB % of weight W, is movable in a vertical
plane about a hinge at A, and is sustained in equilibrium by a weight
P attached to a string BCP passing over a smooth peg C, AC being
vertical; if AC be equal to AB, shew that PssJFcos ACB, and that
the action at the hinge is W sin ACB.
2. A uniform rod can turn freely about one of its ends, and is
pulled aside from the vertical by a horizontal force acting at the
other end of the rod and equal to half its weight ; at what inclination
to the vertical will the rod rest ?
3. A rod AB, hinged at A, is supported in a horizontal position
by a string BC, making an angle of 45 with the rod, and the rod has
a mass of 10 lbs. suspended from B. Neglecting the weight of the
rod, find the tension of the string and the action at the hinge.
4. A uniform heavy rod AB has the end A in contact with a
smooth vertical wall, and one end of a string is fastened to the rod
at a point C, such that AC=AB, and the other end of the string is
fastened to the wall ; find the length of the string, if the rod rest in
a position inclined at an angle to the vertical.
5. ACB is a uniform rod, of weight W; it is supported (B being
uppermost) with its end A against a smooth vertical wall AD by means
of a string CD, DB being horizontal and CD inclined to the wall at
an angle of 30. Find the tension of the string and the pressure on
the wall, and prove that ACx^AB.
6. A uniform rod, AB, resting with one end A against a smooth
vertical wall is supported by a string BC which is tied to a point C
vertically above A and to the other end B of the rod. Draw a diagram
shewing the lines of action of the forces which keep the rod in equi-
librium, and shew that the tension of the string is greater than the
weight of the rod.
7. A ladder, 14 feet long and weighing 50 lbs., rests with one
54 STATICS. Exs. IX.
end against the foot of a vertical wall and from a point 4 feet from
the upper end a cord which is horizontal runs to a point 6 feet above
the foot of the wall. Find the tension of the cord and the reaction at
the lower end of the ladder.
8. A smooth hemispherical bowl, of diameter a, is placed so that
its edge touches a smooth vertical wall ; a heavy rod is in equilibrium,
inclined at 60 to the horizon, with one end resting on the inner
surface of the bowl, and the other end resting against the wall ; shew
that the length of the rod must be a + -j*= .
9. A sphere, of given weight W, rests between two smooth
planes, one vertical and the other inclined at a given angle a to the
vertical ; find the reactions of the planes.
10. A solid sphere rests upon two parallel bars which are in the
same horizontal plane, the distance between the bars being equal to
the radius of the sphere ; find the reaction of each bar.
11. A smooth sphere is supported in contact with a smooth
vertical wall by a string fastened to a point on its surface, the other
end being attached to a point in the wall ; if the length of the string
be equal to the radius of the sphere, find the inclination of the string
to the vertical, the tension of the string, and the reaction of the wall.
12. A picture of given weight, hanging vertically against a smooth
wall, is supported by a string passing over a smooth peg driven into
the wall ; the ends of the string are fastened to two points in the
upper rim of the frame which are equidistant from the centre of the
rim, and the angle at the peg is 60; compare the tension in this case
with what it will be when the string is shortened to two-thirds of its
length.
13. A picture, of 40 lbs. wt., is hung, with its upper and lower
edges horizontal, by a cord fastened to the two upper corners and
passing over a nail, so that the parts of the cord at the two sides of
the nail are inclined to one another at an angle of 60. Find the
tension of the cord in lbs. weight.
14. A picture hangs symmetrically by means of a string passing
over a nail and attached to two rings in the picture ; what is the
tension of the string when the picture weighs 10 lbs., if the string be
4 feet long and the nail distant 1 ft. 6 inches from the horizontal line
joining the rings ?
CHAPTER VIIL
CENTRE OF GRAVITY.
69. Every particle of matter is attracted to the
centre of the Earth, and the force with which the Earth
attracts any particle to itself is, as we shall see in
Dynamics, proportional to the mass of the particle.
Any body may be considered as an agglomeration of
particles.
If the body be small, compared with the Earth, the
lines joining its component particles to the centre of the
Earth will be very approximately parallel, and, within the
limits of this book, we shall consider them to be absolutely
parallel
On every particle, therefore, of a rigid body there is
acting a force vertically downwards which we call its
weight.
These forces may by the process of compounding
parallel forces, (Art. 49), be compounded into a single
force, equal to the sum of the weights of the particles,
acting at some definite point of the body. Such a point
is called the centre of gravity of the body.
Centre of gravity. Def. The centre of gravity of a
body, or system of particles rigidly connected together, is
that point through which tlie line of action of the v)eight of
the body alv)ay$ passes, in wluttever position the body ts
placed.
70. Every body, or system of particles rigidly connected
toget/ter, has a centre of gravity.
Let A, B, G, D... be a system of particles whose
weights are w^ />,, w s ... .
Join AB, and divide it at G. so that
AG X : G X B :: to, : tOj.
Then parallel forces w x and w? acting at A and B, are,
by Art. 47, equivalent to a force (to l + 1/> 2 ) acting at 6r ? .
56 STATICS.
Join G X C, and divide it at G % so that
G x Gi : G % :: w 3 : t^ + tr
Tlien parallel forces, (tt^ + w^) at G x and w s at C, are
equivalent to a force (w l + w a + w s ) at G % .
Hence the forces w u w 2 and w % may be supposed to be
applied at G 9 without altering their effect.
Similarly, dividing G*D in G a so that
G 2 G 3 : G 3 D :: w 4 : w x + w 2 + tv 3 ,
we see that the resultant of the four weights at A, B, C,
and D is equivalent to a vertical force, to x + 10 3 + w> 8 + u> 4 ,
acting at (? 8 .
Proceeding in this way, we see that the weights of any
number of particles composing any body may be supposed
to be applied at some point of the body without altering
their effect.
##7l. Since the construction for the position of the resultant of
parallel forces depends only on the point of application and magni-
tude, and not on the direction of the forces, the point we finally arrive
at is the same if the body be turned through any angle ; for the
weights of the portions of the body are still parallel, although they
have not the same direction, relative to the body, in the two posi-
tions.
We can hence shew that a body can only have one centre of
gravity. For, if possible, let it have two centres of gravity Q and G v
Let the body be turned, if necessary, until GQ X be horizontal. We
shall then have the resultant of a system of vertical, forces acting
both through G and through G v But the resultant force, being itself
necessarily vertical, cannot act in the horizontal line GQ X .
Hence there can be only one centre of gravity.
72. Centre of gravity of a uniform rod. Let
AB be a uniform rod, and G its middle point.
p Q
A- 1 , 1 B
G
CENTRE OF GRA VITY, 57
Take any point P of the rod between G and A, and a
point Q in GB, such that
GQ = GP.
The centre of gravity of equal particles at P and Q
is clearly G ; also, for every particle between G and A,
there is an equal particle at an equal distance from G,
lying between G and B.
The centre of gravity of each of these pairs of particles
is at G ; therefore the centre of gravity of the whole rod
is at G.
73. Centre of gravity of a uniform parallelo-
gram. Let ABCD be a paral- E
lelogram, and let E and F be p /*" & h J
the middle points of AD and
BC.
Divide the parallelogram in- B
to a very large number of strips,
by means of lines parallel to AD y of which PR and QS
are any consecutive pair. Then PQSR may be considered
to be a uniform straight line, whose centre of gravity is
at its middle point G x .
So the centre of gravity of all the other strips lies on
EF t and hence the centre of gravity of the whole figure
lies on EF.
So, by dividing the parallelogram by lines parallel to
AB, we see that the centre of gravity lies on the line
joining the middle points of the sides AB and CD.
Hence the centre of gravity is at G the point of inter-
section of these two lines.
G is clearly also the point of intersection of the diagonals
of the parallelogram.
74. It is clear from the method of the two previous
articles that, if in a uniform body we can find a point G
such that the body can be divided into pairs of particles
balancing about it, then G must be the centre of gravity
of the body.
The centre of gravity of a uniform circle, or uniform
sphere, is therefore its centre.
58 STATICS.
It is also clear that if we can divide a lamina into
strips, the centres of gravity of which all lie on a straight
line, then the centre of gravity of the lamina must lie on
that line.
Similarly, if a body can be divided into portions, the
centres of gravity of which lie in a plane, the centre of
gravity of the whole must lie in that plane.
75. Centre of gravity of a uniform triangular
lamina. Let ABC be the tri-
angular lamina and let D and E
be the middle points of the sides
BC and CA. Join AD and BE,
and let them meet in G. Then G
shall be the centre of gravity of
the triangle.
Let B X C X be any line parallel to the base BC meeting
AD in D x .
As in the case of the parallelogram, the triangle may
be considered to be made up of a very large number of
strips, such as B X C X , all parallel to the base BC.
Since B X C X and BC are parallel, the triangles AB } D X
and ABD are equiangular; so also the triangles AD X C X
and ADC are equiangular.
B X D X AD X D X C X /T31 . A T . . #%
Hence in* ib = ~w ( VL PP ' ' 3)
But BD = DC; therefore B X D X = D X C X . Hence the
centre of gravity of the strip B X C Z lies on AD.
So the centres of gravity of all the other strips lie on
AD, and hence the centre of gravity of the triangle lies
on AD.
Join BE, and let it meet A D in G.
By dividing the triangle into strips parallel to iC we
see, similarly, that the centre of gravity lies on BE.
Hence the required centre of gravity must be at G.
Since D is the middle point of BC and E is the middle
point of CA, therefore DE is parallel to AB.
Hence the triangles GDE and GAB are equiangular,
GD DECE _i
" GA ~ AB ~CA~^
CENTRE OF GRA VITT. 59
so that 2GD = GA, and 3GD = GA + GD = AD.
:. GD = \AD.
Hence the centre of gravity of a triangle is on the
line joining the middle point of any side to the opposite
vertex at one-third the distance of the vertex from that
side.
#76. The centre of gravity of any triangular lamina is the same
at that of three equal particles placed at the vertices of the triangle.
Taking the figure of Art. 75, the centre of gravity of two equal
particles, each equal to w, at B and C, is at D the middle point of B C ;
also the centre of gravity of 2w at D and w at A divides the line DA
in the ratio of 1 : 2 [Art 47.]. But G, the centre of gravity of the lamina,
divides DA in the ratio of 1 : 2.
Hence the centre of gravity of the three particles is the same as
that of the lamina.
#77. The position of the centre of gravity of some other bodies
may be stated here.
The centre of gravity of a pyramid on any base is on the line
joining the vertex to the centre of gravity of the base and divides this
line in the ratio of 3 : 1.
The centre of gravity of a solid cone is on its axis at a distance
from the base equal to of its altitude; if the cone be hollow, the
distance is of the altitude.
The centre of gravity of a solid hemisphere of radius r is on that
radius which is perpendicular to its plane face at a distance from
o
the centre. If the hemisphere be hollow, this distance is = .
EXAMPLES. X.
1. An isosceles triangle has its equal sides of length 5 feet and
its base of length 6 feet ; find the distance of the centre of gravity
from each of its angular points.
2. The sides of a triangular lamina are 6, 8, and 10 feet in
length; find the distance of the centre of gravity from each of its
angular points.
3. D is the middle point of the base BC of a triangle ABC ; shew
that the distance between the centres of gravity of the triangles ABD
and ACD is \BC.
4. A heavy triangular plate ABC lies on the ground ; if a vertical
force applied at the point A be jnst great enough to begin to lift that
vertex from the ground, shew that the same force will sufiice, if applied
at B or C.
60 STATICS. Exs. X.
5. The base of a triangle is fixed, and its vertex moves on a given
straight line ; shew that the centre of gravity also moves on a straight
line.
6. A uniform equilateral triangular plate is suspended by a
string attached to a point in one of its sides, which divides the side
in the ratio 2:1; find the inclination of this side to the vertical.
78. General formulae for the determination
of the centre of gravity.
In the following articles will be obtained formulae
giving the position of the centre of gravity of any system
of particles, whose position and weights are known.
Theorem. If a system of particles whose weights are
w Xy w if ... to n be on a straight line, and if their distance*
measured from a fixed powU in the line be
Oq, asa, ... x nf
the distance, x, of their centre of gravity from the fixed point
is given by
_ _ Wfa + WtfC], + .,. + w n x n
to x + to i + ... + w n
Let A, B, C, D... be the particles and let the centre of
gravity of w x and i# 2 at A and B be G x ; let the centre of
O A B C D
r 3 ^
J ^3 W *
gravity of {w x + w> 2 ) at G x and w s at C be G it and so for the
other particles of the system.
By Art. 70, we have w x . AG X = w, . G X B.
:. w x (OG x -OA) = w a (OB-OG 1 ).
Hence (w x + w t ) . OG x = w x . OA + to, . OB,
U , 06, = ^^^' (1).
Similarly, since G u is the centre of gravity of (w x + w?)
at G x and w 3 at (7, we have
OG A w i + w *)' 0G i + w *> 0C
* (w x + w z ) + w 3
CENTRE OF 6RA VITT, 61
W>! + W z + W z '
go 0(9 = (V^ + ^'^i + ' Qi>
* (lOj + > 2 + to 3 ) + w 4
_ tg^ + tc^ + tfl^ 4- w& 4
Proceeding in this manner we easily have
_ WjXi + M?aa; a + . .. + to n x n
t0 1 + tt> 2 + ... +w n *
whatever be the number of the particles in the system.
Otherwise, The above formula may be obtained by the use of
Article 69. For the weights of the particles form a system of parallel
forces whose resultant is equal to their sum, viz. w x + w^+. r +w^.
Also the sum of the moments of these forces about any point in their
plane is the same as the moment of their resultant. But the sum of
the moments of the forces about the fixed point O is
w lXl + w&+.+w n x n .
Also, if x be the distance of the centre of gravity from 0, the moment
of the resultant is
{tr 1 + tt> 3 +...+tc n )x5.
Hence x(w 1 -\-w i +...+w. n )=w 1 x 1 +w 3 x i +...+w n x n ;
MA + wT a +. -.+> *,
t.C., Jj .
10. Ex. 1. A rod AB, 2 feet in length, and of weight 5 lbs., is
trisected in the points G and D, and at the points A, G, D and B are
placed particles of 1, 2, 3 and 4 lbs. weight respectively ; find what
point of the rod must be supported so that the rod may rest in any
position, i.e., find the centre of gravity of the system.
Let G be the middle point of the rod and let the fixed point of
the previous article be taken to coincide with the end A of the rod.
The quantities x^, x 2 , x 3 , x 4 and x B are in this case 0, 8, 12, 16, and
24 inches respectively.
Hence, if X be the point required, we have
1.0 + 2.8 + 5.12 + 3.16 + 4.24
1+2+5+3+4
= -=-=-= 14$ inches.
15
Hx. a. If, in the previous question, the body at B be removed and
another body be substituted, find the weight of this unknown body so that
the new centre of gravity may be at the middle point of the rod.
Let X Iba. be the required weight.
62 STATICS.
Since the distance of the new centre of gravity from A is to be
12 inches, we have
1.0 + 2.8 + 5.12 + 3.16 + X.24 _ 124 + 24X
1+2+6+3+X ~ 11 + X '
A 132 + 12X = 124 + 24X.
.-. X=|lb.
Ex. a. To the end of a rod, whose length is 2 feet and whose weight
is 3 lbs., is attached a sphere, of radius 2 inches and weight 10 lbs.;
find the position of the centre of gravity of the compound body.
Let OA be the rod, G 1 its middle point, % the centre of the sphere,
and G the required point.
Bnt OGfj= 12 inches and OG 2 = 26 inches.
EXAMPLES. XL
1. A straight rod, 1 foot in length and of mass 1 ounoe, has an
ounce of lead fastened to it at one end, and another ounce fastened to
it at a distance from the other end equal to one-third of its length ,
find the centre of gravity of the system.
2. A uniform bar, 3 feet in length and of mass 6 ounces, has
3 rings, each of mass 3 ounces, at distances 3, 15 and 21 inches from
one end. About what point of the bar will the system balance ?
3. A uniform rod AB is 4 feet long and weighs 3 lbs. One lb. is
attached at A, 2 lbs. at a point distant 1 foot from A, 3 lbs. at 2 feet
from A, 4 lbs. at 3 feet from A, and 5 lbs. at B. Find the distance
from A of the centre of gravity of the system.
4. A telescope consists of 3 tubes, each 10 inches in length,
one within the other, and of weights 8, 7, and 6 ounces. Find the
position of the centre of gravity when the tubes are drawn out at full
length.
5. Twelve heavy particles at equal intervals of one inoh along a
straight rod weigh 1, 2, 3,... 12 grains respectively; find their centre
of gravity, neglecting the weight of the rod.
6. The four silver coins from one shilling downwards are placed
in a straight line with equal distances of 6 inches between their cen-
tres. Find their centre of gravity.
7. A rod, of uniform thickness, has one-half of its length com-
posed of one metal and the other half composed of a different metal,
and the rod balances about a point distant one-third of its whole
length from one end ; oompare the weight of equal quantities of the
metal.
CENTRE OF GRAVITY.
63
8. A cylindrical vessel one foot in diameter and one foot high is
made of thin sheet metal of uniform thickness. If it be half tilled
with water where will be the common centre of gravity of the vessel
and water, assuming the weight of the vessel to be th of the con-
tained water?
9. A rod, 12 feet long, has a mass of 1 lb. suspended from one
end, and, when 15 lbs. is suspended from the other end, it balances
about a point distant 3 ft. from that end; if 8 lbs. be suspended there,
it balances about a point 4 ft. from that end. Find the weight of the
rod and the position of its centre of gravity.
80. Theorem. If a system of particles, whose
weights are w 1 , w z , ... w n , He in a plane, and if OX and OY
be two fixed straight lines in the plane at right angles, and if
the distances of the particles from OX be y x , y z , ... y n , and
the distance of their centre of gravity be y, then
y =
v>yyi + y>&* + ~> + Wny*.
W x + W 2 + ... +W n
Similarly, if the distances of the particles from OY be
x 1 , x it ... x n and that of their centre of gravity be x, then
_ w& + W.& + ... + w n x n
w x + w 2 + ... + w H
Let A, B, C, ... be the particles, and AL, BM, CN... the
perpendiculars on OX.
Let G x be the centre of gravity of w x and v> 2 , G 2 the
centre of gravity of (w l + w a ) at G x and w 3 at C, and so on.
Also let G be the final point thus arrived at, ie. the
centre of gravity of all the particles.
64
STATICS.
Since the resultant weight {w x + to, + . . . + tv n ) acting at
G is equivalent to the component forces w li to a ,... the
resultant would, if the line OX be supposed to be a fixed
axis, have the same moment about this fixed axis that the
component weights have.
But the moment of the resultant is
and the sum of the moments of the weights is
Hence
w 1 y 1 + w 3 2/ a + ... + 10,^
w 1 + w 1 + ... + to,
In a similar manner we should have
_ W& + W& + .
x
U\ + W % + ... + VJ n
The theorem of this article may be put somewhat
differently as follows ;
The distance of tlie centre of gravity from any line in
the plans of the particles is equal to a fraction, whose
numerator is the sum of the products of each weight into its
distance from the given line, and whose denominator is the
sum of the weights.
81. Ex. 1. A square lamina, whose weight is 10 lbs., has attached
to its angular points particle* whose weights, taken in order, are 3, 6, 6
and 1 lbs. respectively. Find the. position of the centre of gravity of
the system, if the side of the lamina be 25 inches.
Let the particles be placed at the angular points 0, A, B and C.
Let the two fixed lines from which the distances are measured be OA
and 0(7.
CENTRE OF GRAVITY. 65
The weight of the lamina acts at its centre D. Let G be the
required centre of gravity and draw DL and GM perpendicular
to OX.
The distances of the points 0, A, B, G and D from OX are clearly
0, 0, 25, 25, and 12 inches respectively.
__. _ 3. + 6. + 5. 25 + 1. 25 + 10. 12 275 * .
H MG= y = 3 + 6 + 5 + 1 + 10 - = -25= llm *'
So the distances of the particles from OY are 0, 25, 25, and
12 inches respectively.
rt __ _ 3.0 + 6.25 + 5.25 + 1.0 + 10.121 400 1C .
OM=x= 5 ~ = = 77T = 77^=16 ms.
3 + 6 + 5 + 1 + 10 25
Hence the required point may be obtained by measuring 16 inches
from along OA and then erecting a perpendicular of length 11 inches.
Ex. 2. OAB is an isosceles weightless triangle, whose base OA is
6 inches and whose sides are each 5 inches ; at the points 0, A and
are placed particles of weights 1, 2, and 3 lbs. ; find their centre of
gravity.
Let the fixed line OX coincide with OA and let OF be a perpen-
dicular to OA through the point O.
If BL be drawn perpendicular to OA, then (XL = 3 ins., and
LB= v /5 2 -3 2 =4ins.
Hence, if G be the required centre of gravity and GM be drawn
perpendicular to OX, we have
1.0 + 2.6 + 3.3 21 Q1 . ,
0M IT2T3 = T= 3 * mches '
- 1.0 + 2.0 + 3.4 12 . .
and MG=0 j^^ = ^=2 inches.
Hence the required point is obtained by measuring a distance
3J inches from along OA and then erecting a perpendicular of
length 2 inches.
82. Centre of Parallel forces.
The methods and formulae of Arts. 78 and 80 will
apply not only to weights, but also to any system of parallel
forces and will determine the position of the resultant of
any such system. The magnitude of the resultant is the
sum of the forces. Each force must, of course, be taken
with its proper sign prefixed.
There is one case in which we obtain no satisfactory
result; if the algebraic sum of the forces be zero, the
resultant force is zero, and the formulae of Art. 80 give
x = oo , and y = oo .
l. m. n. 5
66 STATICS. Exs.
In this case the system of parallel forces is, as in
Art. 61, equivalent to a couple.
EXAMPLES. XII.
1. Particles of 1, 2, 3, and 4 lbs. weight are placed at the angular
points of a square ; find the distance of their c.o. from the centre of
the square.
2. At two opposite corners A and C of a square ABCD weights
of 2 lbs. each are placed, and at B and D are placed 1 and 7 lbs.
respectively ; find their centre of gravity.
3. Particles of 5, 6, 9 and 7 lbs. respectively are placed at the
corners A, B, G and D of a horizontal square, the length of whose side
is 27 inches ; find where a single force must be applied to preserve
equilibrium.
4. Five masses of 1, 2, 3, 4, and 5 ounces respectively are placed
on a square table. The distances from one edge of the table are 2, 4,
6, 8, and 10 inches and from the adjacent edge 3, 5, 7, 9, and 11 inches
respectively. Find the distance of the centre of gravity from the two
edges.
5. Weights proportional to 1, 2, and 3 are placed at the corners
of an equilateral triangle, whose side is of length a ; find the distance
of their centre of gravity from the first weight.
Find the distance also if the weights be proportional to 11, 13,
and 6.
6. ABC is an equilateral triangle of side 2 feet. At A y B, and C
are placed weights proportional to 5, 1, and 3, and at the middle
points of the sides BC, GA, and AB weights proportional to 2, 4,
and 6 ; shew that their centre of gravity is distant 16 inches from B.
7. Equal masses, each 1 oz., are placed at the angular points of
a heavy triangular lamina, and also at the middle points of its sides ;
find the position of the centre of gravity of the masses.
8. ABC is a triangle right angled at A, AB being 12 and AG
15 inches; weights proportional to 2, 3, and 4 respectively are placed
at A, C, and B ; find the distances of their centre of gravity from B
and C.
9. Particles, of mass 4, 1, and 1 lbs., are placed at the angular
points of a triangle ; shew that the centre of gravity of the particles
bisects the distance between the centre of gravity and one of the
vertices of the triangle.
10. To the vertices A, B, and G of a uniform triangular plate,
whose mass is 3 lbs. and whose centre of gravity is G, particles of
masses 2 lbs., 2 lbs., and 11 lbs., are attached ; shew that the centre
of gravity of the system is the middle point of GG.
11. Find the centre of parallel forces equal respectively to P, 2P,
XII.
CENTRE OF GRAVITY.
67
3P, 4P, 5P, and 6P, the points of application of the forces being at
distances 1, 2, 3, 4, 5, and 6 inches respectively from a given point A
measured along a given line AB.
12. At the angular points of a square, taken in order, there act
parallel forces in the ratio 1:3:5:7; find the distance from the
centre of the square of the point at which their resultant acts.
13. A, B, C, and D are the angles of a parallelogram taken in
order ; like parallel forces proportional to 6, 10, 14, and 10 respectively
act at A, B, C and D ; shew that the centre and resultant of these
parallel forces remain the same, if, instead of these forces, parallel
forces, proportional to 8, 12, 16 and 4, act at the points of bisection
of the sides AB, BG, CD, and DA respectively.
83. Given the centre of. gravity of the two portions of a
body t to find the centre of gravity of the whole body.
Let the given centres of gravity be G x and G 2i and let
the weights of the two portions be W x and W 2 ; the re-
quired point G, by Art. 70, divides G X G^ so that
G X G : GG % :: W % : W x .
The point G may also be obtained by the use of
Art. 78.
Ex. On the same bate AB, and on opposite sides of it, isosceles
triangles CAB and DAB are described whose altitudes are 12 inches
and 6 inches respectively. Find the distance from AB of the centre
of gravity of the quadrilateral GADB.
Let CLD be the perpendicular to AB, meeting it in L, and let (ij
and G 3 be the centres of gravity of the two triangles CAB and DAB
respectively. Hence
<7G,=$. CX=8, and CG 2 =C7L + I,G 2 =12 + 2 = 14.
The weights of the triangles are proportional to their areas, i.e., to
^AB . 12 and \AB . 6.
If G be the centre of gravity of the whole figure, we have
62
68 STATICS.
cr _ A CAB x CQ t + aDAB x CG S
ACAB+ADAB
Henoe LO=CL-CG=2 inches.
84. Given the centre of gravity of the whole of a body
and of a portion of the body, to find the centre of gravity of
the remamder.
Let G be the centre of gravity of a body ABCD, and G x
that of the portion ADC.
Let W be the weight of the whole body and W^ that of
the portion ACD t so that W s (= W- W x ) is the weight of
the portion ABC.
Let G t be the centre of gravity of the portion ABC.
Since the two portions of the body make up the whole,
therefore W x at G x and W 9 at G t must have their centre of
gravity at G.
Hence G must lie on Q-fi % and be such that
W 1 .GG 1 = W 9 .GG % .
Hence, given G and G lt we obtain G % by producing G X G
to G %% so that
GG 2 =^.GG X
Wl GQ X .
W-W x
The required point may be also obtained by means of
Art. 78.
Ex. 1. From a circular disc, of radius r, t* cut out a circle, whose
diameter is a radius of the disc; find the centre of gravity of the
remainder.
CENTRE OF GRAVITY. 69
Since the areas of circles are to one another as the squares of their
radii [App. II.], we have
area of the portion cut out : area of the whole circle
'"
:: 1 : 4.
Hence the portion cut off is one-quarter, and the portion remain-
ing is three-quarters, of the whole, so that W X =^W..
Now the portions W x and W 2 make up the whole disc, and therefore
balance about 0.
Hence W 2 . OG 2 = W x . OG x =$W a x \r.
.'. OQ t =\r.
# Ex. 2. From a triangular lamina ABC is cut off, by a line parallel
to its base BO, one-quarter of its area ; find tlie centre of gravity of the
remainder.
Let AB X G X be the portion out off. so that
aAB x C x : A ABC :: 1 : 4.
7 \
/Qi
J2i L
o
By Euc. yi. 19, since the triangles AB X C X and ABC are similar, we
have
AAB X C X : a ABC :: AB X * : AB\
:. AB* : AB* : : 1 : 4,
and hence AB X ^AB.
The line B X C X therefore biseots AB, AC, and AD.
Let G and G x be the centres of gravity of the triangles ABC and
AB X C X respectively ; also let W x and W % be the respective weights of
the portion out off and the portion remaining, so that W 2 =ZW X .
70 STATICS. Exs.
Since JP 5 at 2 and W x at G x balanoe about Q, we have, by Art. 78,
DQ _ W 1 . DG 1+ W 2 . DG 2 _ DG l + BDG a
U "~ W x +W 2 4 - (1} *
But DG-\DA=^DD lt
and DG l =DD l + \D 1 A = DD 1 + \DD r =$DD l .
Hence (i) is 4 x |2)2) 1 =D2) 1 +32)(? 2 .
EXAMPLES. XIH
[Ext. 1, 2 and 4 9 are suitable for verification by experiment. ]
1. A uniform rod, 1 foot in length, is broken into two parts, of
lengths 5 and 7 inches, which are placed so as to form the letter T> the
longer portion being vertical ; find the centre of gravity of the system.
2. Two rectangular pieces of the same cardboard, of lengths 6 and
8 inches and breadths 2 and 2 inches respectively, are placed touching,
but not overlapping, one another on a table so as to form a T-shaped
figure. Find the position of its centre of gravity.
3. A heavy beam consists of two portions, whose lengths are as
8 : 5, and whose weights are as 3:1; find the position of its centre
of gravity.
4. Two sides of a rectangle are double of the other two, and on
one of the longer sides an equilateral triangle is described ; find
the centre of gravity of the lamina made up of the rectangle and
the triangle.
5. A piece of cardboard is in the shape of a square ABGD with
an isosceles triangle described on the side BG ; if the side of the
square be 12 inches and the height of the triangle be 6 inches, find the
distance of the centre of gravity of the cardboard from the line AD.
6. From a parallelogram is out one of the four portions into
which it is divided by its diagonals ; find the centre of gravity of the
remainder.
7. A parallelogram is divided into four parts, by joining the
middle points of opposite sides, and one part is out away; find the
centre of gravity of the remainder.
8. From a square a triangular portion is cut off, by cutting the
square along a line joining the middle points of two adjacent sides;
find the centre of gravity of the remainder.
9. From a triangle is out off $th of its area by a straight line
parallel to its base. Find the position of the centre of gravity of the
remainder.
10. A piece of thin uniform wire is bent into the form of a four-
sided figure, ABGD, of which the sides AB and CD are parallel, and
BG and DA are equally inclined to AB. If AB be 18 inches, CD
12 inohes, and BG and DA each 5 inches, find the distance from AB
of the centre of gravity of the wire.
XIII. CENTRE OF GRAVITY. 71
11. A uniform plate of metal, 10 inches square, has a hole of area
3 square inches cut out of it, the centre of the hole being 2 inches from
the centre of the plate ; find the position of the centre of gravity of
the remainder of the plate.
12. Where must a circular hole, of 1 foot radius, be punched out
of a circular disc, of 3 feet radius, so that the centre of gravity of
the remainder may be 2 inches from the centre of the disc ?
13. Two uniform spheres, composed of the same materials, and
whose diameters are 6 and 12 inches respectively, are firmly united ;
find the position of their oentre of gravity. [The volumes of two
spheres are in the ratio of the cubes of their radii.]
14. A solid right circular cone of homogeneous iron, of height
64 inches and mass 8192 lbs., is cut by a plane perpendicular to its
axis so that the mass of the small cone removed is 686 lbs. Find the
height of the centre of gravity of the truncated portion above the base
of the cone.
PROPERTIES OF THE CENTRE OF GRAVITY.
85. If a rigid body be in equilibrium, one point only
of the body being fixed, the centre of gravity of the body will
be in the vertical line passing through the fixed point of t?ie
body.
Let be the fixed point of the body, and G its centre of
gravity.
The forces acting on the body are the reaction at
the fixed point of support of the body, and the weights of
the component parts of the body.
The weights of these component parts are equivalent to
a single vertical force through the centre of gravity of the
body.
Also, when two forces keep a body in equilibrium, they
72 STATICS.
must be equal and opposite and have the same line of action.
But the lines of action cannot be the same unless the vertical
line through G passes through the point 0.
Two cases arise, the first, in which the centre of gravity
G is below the point of suspension 0, and the second, in
which G is above 0.
In the first case, the body, if slightly displaced from its
position of equilibrium, will tend to return to this position;
in the second case, the body will not tend to return to its
position of equilibrium.
86. To find, by experiment, the centre of gravity of
a body of any shape.
Fii one point of the body and let the body assume
its position of equilibrium. Take a point A of the body
vertically below 0; then, by the last article, the centre of
gravity is somewhere in the line OA.
Secondly, release the point of the body, and fix a
second point 0' (not in the straight line OA) ; let the body
take up its new position of equilibrium. Take a point A'
in the body, vertically below 0', so that the centre of gravity
is somewhere in the line O'A'.
The required centre of gravity will therefore be the
point of intersection of the lines OA and O'A'.
The student should apply this method in the case of
a body such as an irregularly shaped piece of paper ; the
points and 0' can be easily supported by means of a pin
put through the paper.
87. If a body be placed with its base in contact with
a horizontal plane, it will stand, or fall, according as the
vertical line drawn through the centre of gravity of the body
meets the plane within, or without, the base.
The forces acting on the body are its weight, which acts
at its centre of gravity G, and the reactions of the plane,
acting at different points of the base of the body. These
reactions are all vertical, and hence they may be com-
pounded into a single vertical force acting at some point
of the base (Art. 49).
Since the resultant of two like parallel forces acts
always at a point between the forces, it follows that the
CENTRE OF GRAVITY.
73
resultant of all the pressures on the base of the body
cannot act through a point outside the base.
Hence, if the vertical line through the centre of gravity
of the body meet the plane at a point outside the base,
it cannot be balanced by the resultant pressure, and the
body cannot therefore be in equilibrium, but must fall over.
If the base of the body be a figure having a re-entrant
angle, as in the above figure, we must extend the meaning
of the word " base " in the enunciation to mean the area
included in the figure obtained by drawing a piece of thread
tightly round the geometrical base. In the above figure
the " base " therefore means the area ABDEFA.
For example, the point C, at which the resultant pres-
sure acts, may lie within the area AHB, but it cannot
lie without the dotted line AB.
If the point C were on the line AB, between A and B,
the body would be on the point of falling over.
88. Ex. A cylinder, of height h, and the radius of whose base is
r, is placed on an inclined plane and prevented from sliding; if the
inclination of the plane be gradually increased, find when tlie cylinder
will topple.
74
STATICS.
Let the figure represent the section of the cylinder when it is on
the point of toppling over ; the vertical
line through the centre of gravity G of
the body must therefore just pass through
the end A of the base. Hence GAD must
be equal to the angle of inclination, a,
of the plane.
GD 2r
Hence tan a = tan GAD = 7 = -=- ,
DA n
giving the required inclination of the
plane.
Stable, unstable, and neutral equilibrium.
89. We have pointed out in Art. 85 that the body
in the first figure of that article would, if slightly dis-
placed, tend to return to its position of equilibrium, and
that the body in the second figure would not tend to return
to its original position of equilibrium, but would recede
still further from that position.
These two bodies are said to be in stable and unstable
equilibrium respectively.
Again, a cone, resting with its flat circular base in
contact with a horizontal plane, would, if slightly displaced,
return to its position of equilibrium; if resting with its
vertex in contact with the plane it would, if slightly dis-
placed, recede still further from its position of equilibrium ;
whilst, if placed with its slant side in contact with the
plane, it will remain in equilibrium in any position. The
equilibrium in the latter case is said to be neutral.
90. Consider, again, the case of a heavy sphere, rest-
ing on a horizontal plane, whose centre of gravity is not at
its centre.
Let the first figure represent the position of equilibrium,
the centre of gravity being either below the centre 0, as G l9
CENTRE OF GRAVITY. 75
or above, as G t . Let the second figure represent the sphere
turned through a small angle, so that B is now the point of
contact with the plane.
The pressure of the plane still acts through the centre
of the sphere.
If the weight of the body act through G ly it is clear that
the body will return towards its original position of equi-
librium, and therefore the body was originally in stable
equilibrium.
If the weight act through G 9J the body will move still
further from its original position of equilibrium, and there-
fore it was originally in unstable equilibrium.
If however the centre of gravity of the body had been
at 0, then, in the case of the second figure, the weight
would still be balanced by the pressure of the plane ; the
body would thus remain in the new position, and the
equilibrium would be called neutral.
91. Def. A body is said to be in stable equi-
librium when, if it be slightly displaced from its position
of equilibrium, the forces acting on the body tend to
make it return towards its position of equilibrium ; it is
in unstable equilibrium when, if it be slightly displaced,
the forces tend to move it still further from its position of
equilibrium; it is in neutral equilibrium, if the forces
acting on it in its displaced position be in equilibrium.
#92. Ex. A homogeneous body, consisting of a cylinder and a
hemisphere joined at their bases, is placed with the hemispherical end
on a horizontal table ; is the equilibrium stable or unstable ?
Let Gj and G 3 be the centres of gravity of the hemisphere and
76 STATICS. Exs.
cylinder, let A be the point of the body which is initially in
contaot with the table, and let be the centre of the base of the
hemisphere.
If h be the height of the cylinder, and r be the radios of the base,
we have
OQ x =%r (Art. 77), and OG,=^.
Also the weights of the hemisphere and cylinder are proportional
to %irr* and T.7*h. [App. II.]
The reaction of the plane, in the displaced position of the body,
always passes through the centre 0.
The equilibrium is stable or unstable according as O, the centre
of gravity of the compound body, is below or above 0,
i.e., according as
OG-y x wt. of hemisphere is > 00 t x wt. of cylinder,
i.e., according as
SrxfirrMs^x*-^
i.e., according as
$*;
i.e., according as
r is =V.
EXAMPLES. XIV.
1. A carpenter's rule, 2 feet in length, is bent into two parts at
right angles to one another, the length of the shorter portion being
8 inches. If the shorter be placed on a smooth horizontal table, what
is the length of the least portion on the table that there may be equi-
librium ?
2. A cylinder, whose base is a circle of one foot diameter and
whose height is 3 feet, rests on a horizontal plane with its axis
vertical. Find how high one edge of the base can be raised before
the cylinder overturns.
3. A hollow vertical cylinder, of radius 2a and height 3a, rests
on a horizontal table, and a rod is placed within it with its lower end
resting on the circumference of the base ; if the weight of the rod be
equal to that of the cylinder, how long must the rod be so that it
may just cause the cylinder to topple over ?
4. A square table stands on four legs placed respectively at the
middle points of its sides ; find the greatest weight that can be put at
one of the corners without upsetting the table.
5. A square four-legged table has lost one leg ; where on the table
should a weight, equal to the weight of the table, be placed, so that
the pressures on the three remaining legs of the table may be equal ?
XIV. CENTRE OF GRAVITY. 77
6. A square table, of weight 20 lbs., has legs at the middle points
of its sides, and three eqnal weights, each equal to the weight of the
table, are placed at three of the angular points. What is the greatest
weight that ean be placed at the fourth corner so that equilibrium
may be preserved ?
7. The side CD of a uniform square plate ABGD, whose weight
is W, is bisected at E and the triangle AEB is out off. The plate
ABCEA is placed in a vertical position with the side GE on a hori-
zontal plane. What is the greatest weight that can be placed at A
without upsetting the plate?
8. ABC is a flat board, A being a right angle and AC in contact
with a flat table; D is the middle point of AC and the triangle ABD
is out away; shew that the triangle is just on the point of falling
over.
9. ABC is an isosceles triangle, of weight TP, of which the angle
A is 120, and the side AB rests on a smooth horizontal table, the
W
plane of the triangle being vertical ; if a weight -~- be hung on at C,
9
shew that the triangle will just be on the point of toppling over.
10. A number of bricks, each 9 inches long, 4 inches wide, and
3 inches thick, are placed one on another so that, whilst their narrowest
surfaces, or thicknesses, are in the same vertical plane, each brick
overlaps the one underneath it by half an inch; the lowest brick
being placed on a table, how many bricks can be so placed without
their falling over?
11 A solid uniform hemisphere rests upon a horizontal surface
with its flat surface horizontal and uppermost. Show that it is in
stable equilibrium.
CHAPTER IX.
MACHINES.
93. In the following chapter we shall explain and
discuss the equilibrium of some of the simpler machines,
viz., (1) The Lever, (2) The Pulley and Systems of Pulleys,
(3) The Inclined Plane, U) The Wheel and Axle, (5) The
Common Balance, and (6) The Steelyards.
We shall suppose the different portions of these
machines to be smooth, and that the forces acting on
them always balance, so that the machines are at rest.
94. When two external forces applied to a machine
balance, one is called the Power and the other is called the
Weight.
A machine is always used in practice to overcome some
resistance; the force we exert on the machine is the power;
the resistance to be overcome, in whatever form it may
appear, is called the Weight.
95. Mechanical Advantage. If in any machine
a power P balance a weight W, the ratio W : P is called
the mechanical advantage of the machine, so that
Weight = Power x Mechanical Advantage.
Almost all machines are constructed so that the me-
chanical advantage is a ratio greater than unity.
If in any machine the mechanical advantage be less
than unity, it may, with more accuracy, be called me-
chanical disadvantage.
I. The Lever.
96. The Lever consists essentially of a rigid bar,
straight or bent, which has one point fixed about which
MACHINES. THE LEVER.
79
the rest of the lever can turn. This fixed point is called
the Fulcrum, and the perpendicular, distances between the
fulcrum and the lines of action of the power and the weight
are called the arms of the lever.
When the lever is straight, and the power and weight
act perpendicular to the lever, it is usual to distinguish
three classes or orders.
Class I. Here the power P
and the weight W act on op-
posite sides of the fulcrum C.
Class II. Here the power P
and the weight W act on the
same side of the fulcrum G, but
the former acts at a greater dis-
tance than the latter from the
fulcrum.
Class III. Here the power
P and the weight W act on the
same side of the fulcrum (7, but
the former acts at a less dis-
tance than the latter from the
fulcrum.
mm eummMitt'iMMmi G
W
97. Conditions of equilibrium of a straight lever.
In each case we have three parallel forces acting on
the body, so that the reaction, R, at the fulcrum must
be equal and opposite to the resultant of P and W.
In the first class P and W are like parallel forces, so
that their resultant is P+ W. Hence
R = P+W.
In the second class P and W are unlike parallel forces,
so that P+R~ TT, i.e. R= W - P.
So in the third class R + W= P, i.e. R = P-W.
In the first and third cases we see that R and P act in
80 STATICS.
opposite directions; in the second class they act in the
same direction.
In all three classes, since the resultant of P and W
passes through (7, we have, as in Art. 47,
P. AC=W.BC,
i.e. P x the arm of P= W x the arm of W.
Since -~ = T~ir > we observe that generally in
P arm of W to J
Class I., and always in Class II., there is mechanical
advantage, but that in Class III. there is mechanical
disadvantage.
The practical use of levers of the latter class is to
apply a force at some point at which it is not easy to apply
the force directly.
98. Examples of the different classes of levers are ;
Class I. A Poker (when used to stir the fire, the bar
of the grate being the fulcrum) ; A Claw-hammer (when
used to extract nails) ; A Crowbar (when used with a point
in it resting on a fixed support) ; A Pair of Scales ; The
Brake of a Pump.
Double levers of this class are ; A Pair of Scissors, A
Pair of Pincers.
Class II. A Wheelbarrow; A Cork Squeezer; A
Crowbar (with one end in contact with the ground)) An
Oar (assuming the end of the oar in contact with the water
to be at rest).
A Pair of Nutcrackers is a double lever of this class.
Class III. The Treadle of a Lathe; The Human
Forearm (when the latter is used to support a weight placed
on the palm of the hand. The Fulcrum is the elbow t and
the tension exerted by the muscles is the power).
A Pair of Sugar-tongs is a double lever of this class.
99. In Art. 97 we have neglected the weight of
the bar itself. If the weight be taken into consideration,
or if the lever be bent, we must obtain the conditions of
equilibrium by equating to zero the algebraic sum of the
moments of the forces about the fulcrum.
MACHINES. THE LEVER. 81
Ex. If two weights balance, about a fixed fulcrum, at the extremi-
ties of a straight lever, in any position inclined to the vertical, they
will balance in any other position.
Let AB be the lever, of weight W, and let its centre of gravity be
O. Let the lever balance about a
fulcrum in any position inclined
at an angle 6 to the horizontal, the
weights at A and B being P and W
respectively.
Through draw a horizontal line
LONM to meet the lines of action of
P, W, and W in L, N, and M re- A< r^ ^W ^W
spectively.
Since the forces balance about 0,
we have
P.OL=W.OM+W'.ON.
Vp
;. P. OAcosd = W. OB coad+W .OGgob$.
,\ P.OA = W.OB + W'.OG.
This condition of equilibrium is independent of the inclination $
of the lever to the horizontal; hence in any other position of the
lever the condition would be the same.
Hence, if the lever be in equilibrium in one position, it will be in
equilibrium in all positions.
EXAMPLES. XV.
1. In a weightless lever, if one of the forces be equal to 10 lbs. wt.
and the pressure on the fulcrum be equal to 16 lbs. wt. , and the length
of the shorter arm be 3 feet, find the length of the longer arm.
2. Where must the fulcrum be so that a weight of 6 lbs. may
balance a weight of 8 lbs. on a straight weightless lever, 7 feet long ?
If each weight be increased by 1 lb., in what direction will the
lever turn?
3. If two forces, applied to a weightless lever, balance, and if the
pressure on the fulcrum be ten times the difference of the forces, find
the ratio of the arms.
4. A lever, 1 yard long, has weights of 6 and 20 lbs. fastened to
its ends, and balances about a point distant 9 inches from one end ;
find its weight.
5. A straight lever, AB, 12 feet long, balances about a point,
I foot from A, when a weight of 13 lbs. is suspended from A. It will
balance about a point, which is 1 foot from B, when a weight of
II lbs. is suspended from B. Shew that the centre of gravity of the
lever is 5 inches from the middle point of the lever.
6. A uniform lever is 18 inches long and is of weight 18 ounces;
find the position of the fulcrum when a weight of 27 ounces at one
end of the lever balances one of 9 ounces at the other.
L. M. H. 6
82 STATICS. Exs. XV.
If the lesser weight be doubled, by how much must the position of
the fulcrum be shifted so as to preserve equilibrium ?
7. The short arm of one lever is hinged to the long arm of a
second lever, and the short arm of the latter is attached to a press;
the long arms being each 3 feet in length, and the short arms 6 inches,
find what pressure will be produced on the press by a force, equal to
10 stone weight, applied to the long end of the first lever.
8. The arms of a bent lever are at right angles to one another,
and they are in the ratio of 5 to 1. The longer arm is inclined
to the horizon at an angle of 45, and carries at its end a weight of
10 lbs. ; the end of the shorter arm presses against a horizontal plane ;
find the pressure on the plane.
9. Shew that the propelling force on an eight-oared boat is
224 lbs. weight, supposing each man to pull his oar with a force of
56 lbs. weight, and that the length of the oar from the middle of the
blade to the handle is three times that from the handle to the row-
lock.
10. In a pair of nutcrackers, 6 inches long, if the nut be placed
at a distance of | inch from the hinge, a pressure of 3 lbs. applied to
the ends of the arms will crack the nut. What weight placed on the
top of the nut will crack it?
11. A man raises a 3-foot cube of stone, weighing 2 tons, by
means of a crowbar, 4 feet long, after having thrust one end of the
bar under the stone to a distance of 6 inches ; what force must be
applied at the other end of the bar to raise the stone?
II. Pulleys.
100. A pulley is composed of a wheel of wood, or
metal, grooved along its circumference to receive a string
or rope ; it can turn freely about an axle passing through
its centre perpendicular to its plane, the ends of this axle
being supported by a frame of wood called the block.
A pulley is said to be movable or fixed according as its
block is movable or fixed.
The weight of the pulley is often so small, compared
with the weights which it supports, that it may be neg-
lected ; such a pulley is called a weightless pulley.
We shall always neglect the weight of the string or
rope which passes round the pulley.
We shall also always consider the pulley to be perfectly
smooth, so that the tension of a string which passes round
a pulley is constant throughout its length.
MACHINES. TEE PULLEY.
83
101. Single Pulley. The use of a single pulley is to
apply a power in a different direction from that in which it
is convenient to us to apply the power.
Thus, in the first figure, a man standing on the ground
and pulling vertically at one end of the rope might support
a weight W hanging at the other end ; in the second figure
the same man pulling sideways might support the weight.
In each case the tension of the string passing round
the pulley is unaltered ; the power P is therefore equal to
the weight W.
In the first figure the action on the fixed support to
which the block is attached must balance the other forces
on the pulley-block, and must therefore be equal to
W+P + w,
i.e., 2W+w, where w is the weight of the pulley -block.
ffl
W
VP
In the second figure, neglecting the weight of the
pulley, the power P, and the weight W t being equal,
must be equally inclined to the line OA.
Hence, if T be the tension of the supporting string OB
and 20 the angle between the directions of P and W t we
have
T=PcoaO+ WcoaO = 2WcoaO.
102. We shall discuss three systems of pulleys and
shall follow the usual order ; there is no particular reason
for this order, but it is convenient to retain it for purposes
of reference.
62
84
STATICS.
First y stem of Pulleys. Each string attached to
the supporting beam. To find the relation between the power
and the weight.
In this system of pulleys the weight is attached to the
lowest pulley, and the string passing round it has one
end attached to the fixed beam, and the other end attached
to the next highest pulley; the string passing round the
latter pulley has one end attached to the fixed beam, and
the other to the next pulley, and so on; the power is
applied to the free end of the last string.
Often there is an additional fixed pulley over which the
free end of the last string passes ; the power may then be
applied as a downward force.
Let A lt A it ... be the pulleys, beginning from the
lowest, and let the tensions of the
strings passing round them be T u
T ti ... . Let Wbe the weight and
P the power.
[N.B. The string passing round any
pulley, A say, pulls A % vertically up-
wards, and pulls A 9 downward*.]
I. Let the weights of the pulleys
be neglected.
From the equilibrium of the
pulleys A lt A 9 , ... , taken in order,
we have
2Ti= W; :.T X = \W.
2T t = T i; .'.T^T^W.
1
2T t =T,; :.T S = $T, = -W.
2T< = T t ; .'.T 4 = iT t = ^W.
But, with our figure, T t = P.
' 2*
MACHINES. THE PULLEY. 85
Similarly, if there were n pulleys, we should have
Hence, in this system of pulleys, the mechanical ad-
vantage
P '
II. Let the weights of the pulleys in succession, be-
ginning from the lowest, be w x , w 2 , ....
In this case we have an additional downward force
on each pulley.
As before, we have
2T X = W + w lt
2T s = T 2 + w 3 ,
2 ~2 + 2 '
T -IT + W *- W + Wl + W * + Wi
and P-T -ir*- 4 -^- 1 *- 8 -^ 8 -^ 4
Similarly, if there were n pulleys, we should have
AT
It follows that the mechanical advantage, -p , depends
on the weight of the pulleys.
In this system of pulleys we observe that the greater
the weight of the pulleys, the greater must P be to support
a given weight W ' \ the weights of the pulleys oppose the
power, and the pulleys should therefore be made as light as
is consistent with the required strength.
86 STATICS. Exs.
Stress on the beam from which the pulleys are hung.
Let R be the stress on the beam. Since B, together
with the power P, supports the system of pulleys, together
with the weight W t we have
R + P = W+w x + w t + ... +w n .
From this equation and (1) we easily obtain R.
Ex. If there be 4 movable pulleys, whose weights, commencing
with the lowest, are 4, 6, 6, and 7 lbs. , what power will support a body
of weight 1 cwt., and what is the stress on the beam t
Using the notation of the previous article, we have
2^=112 + 4; /. ^=58.
2T 2 = T 1 + 5 = 63; .-. r a =31.
2r 3 =T 3 + 6=37i; A r,= 18|.
2P =T 8 + 7 = 25f ; /. P =12 lbs. wt.
Also +P=112+4 + 5 + 6+7=134.
.'. B=121ilbs. wt.
EXAMPLES. XVL
1. In the following cases, the movable pulleys are weightless,
their number is n, the weight is W, and the power is P ;
(1) If n=4 and P=20 lbs. wt., find W;
(2) If=4and JF=lowt., findP;
(3) If W=56 lbs. wt. and P=7 lbs. wt. , find n.
2. In the following cases, the movable pulleys are of equal weight
w, and are n in number, P is the power, and W is the weight;
(1) If n=4, w=l lb. wt., and JF=97 lbs. wt., find P;
(2) If n=3, w=l$ lbs. wt., and P= 7 lbs. wt., find W\
(3) If n=5, JF=776 lbs. wt., and P=31 lbs. wt., find w ;
(4) If 1^=107 lbs. wt., P=2 lbs. wt., and tc=| lbs. wt., find n.
3. In the first system of pulleys, if there be 4 pulleys, each of
weight 2 lbs., what weight can be raised by a power equal to the
weight of 20 lbs. ?
4. If there be 3 movable pulleys, whose weights, commencing
with the lowest, are 9, 2, and 1 lbs. respectively, what power will sup-
port a weight of 69 lbs. ?
5. If there be 4 movable pulleys, whose weights, commencing
with the lowest, are 4, 3, 2, and 1 lbs. respectively, what power will
support a weight of 54 lbs. ?
XVI.
MACHINES. THE PULLEY.
87
6. If there be 3 movable pulleys and their weights beginning from
the lowest be 4, 2, and 1 lbs. respectively, what power will be required
to support a weight of 28 lbs. ?
7. A system consists of 4 pulleys, arranged so that each
hangs by a separate string, one end being fastened to the upper block,
and all the free ends being vertical. If the weights of the pulleys,
beginning at the lowest, be w, 2w, Sw, and 4m?, find the power
necessary to support a weight low, and the magnitude of the single
force necessary to support the beam to which the other ends of the
string are attached.
8. A man, of 12 stone weight, is suspended from the lowest of a
system of 4 weightless pulleys, in which each hangs by a separate
string, and supports himself by pulling at the end of the string which
passes over a fixed pulley. Find the amount of his pull on this
string.
9. A man, whose weight is 156 lbs., is suspended from the
lowest of a system of 4 pulleys, each being of weight 10 lbs., and
supports himself by pulling at the end of the string which passes over
the fixed pulley. Find the force which he exerts on the string, sup-
posing all the strings to be vertical.
103. Second system of pulleys. The same spring
passing round all the pulleys. To find the relation be-
tween the power and the weight.
p kP-
88 STATICS. Exs.
In this system there are two blocks, each containing
pulleys, the upper block being fixed and the lower block
movable. The same string passes round all the pulleys
as in the figures.
If the number of pulleys in the upper block be the
same as in the lower block (Fig. 1), one end of the string
must be fastened to the upper block ; if the number in
the upper block be greater by one than the number in
the lower block (Fig. 2), the end of the string must be
attached to the lower block.
In the first case, the number of portions of string con-
necting the blocks is even ; in the second case, the number
is odd.
In either case, let n be the number of portions of string
at the lower block. Since we have only one string passing
over smooth pulleys, the tension of each of these portions
is P, so that the total upward force at the lower block
isw.P.
Let W be the weight supported, and w the weight of
the lower block.
Hence W + w = nP, giving the relation required.
In practice the pulleys of each block are often placed
parallel to one another, so that the strings are not mathe-
matically parallel; they are, however, very approximately
parallel, so that the above relation is still very approxi-
mately true.
EXAMPLES. XVIL
1. If a weight of 6 lbs. support a weight of 24 lbs., find the
weight of the lower block, when there are 3 pulleys in eaeh block.
2. If weights of 5 and 6 lbs. respectively at the free ends of the
string support weights of 18 and 22 lbs. at the lower block, find the
number of the strings and the weight of the lower block.
3. If weights of 4 lbs. and 5 lbs. support weights of 5 lbs. and
18 lbs. respectively, what is the weight of the lower block, and how
many pulleys are there in it?
4. A power of 6 lbs. just supports a weight of 28 lbs., and a power
of 8 lbs. just supports a weight of 42 lbs. ; find the number of strings
and the weight of the lower block.
5. In the second system of pulleys, if a basket be suspended from
the lower block and a man in the basket support himseli and the
XVII.
MACHINES. THE PULLEY.
89
basket, by palling at the free end of the string, find the tension he
exerts, neglecting the inclination of the string to the vertical, and
assuming the weight of the man and basket to be W.
6. A man, whose weight is 12 stone, raises 3 owt. by means of a
system of pulleys in which the same string passes round all the
pulleys, there being 4 in each block, and the string being attached to
the upper block; neglecting the weights of the pulleys, find what
will be his pressure on the ground if he pull vertically downwards.
104. Third system of pulleys. All the strings
attached to the weight. To find the relation between Uie
power and the weight.
In this system the string passing round any pulley
is attached at one end to a bar, from
which the weight is suspended, and at c
the other end to the next lower pulley;
the string round the lowest pulley is
attached at one end to the bar, whilst
at the other end of this string the
power is applied.
In this system the upper pulley is a
fixed pulley.
Let A lt A 2 , A z ... be the movable
pulleys, beginning from the lowest, and
let the tensions of the strings passing
round these pulleys respectively be
T lf T T 3 ....
If the power be P, we have clearly
T^P.
I. Let the weights of the pulleys be
For the equilibrium of the pulleys, taken in order and
commencing from the lowest, we have
T 2 = 2T l = 2P i
T 3 =2T t =2>P,
and T< = 2T 8 = 2*P.
But, since the bar, from which W is suspended, is in
equilibrium, we have
W=T l + T t T-T 3 + T 4
90 STATICS.
= P+ 2P+2 a jP + 2 8 i>
= ^|^ = ^v2 4 -l) 0).
If there were n pulleys, of which (n - 1) would be
movable, we should have, similarly,
W=T l +T 2 + T a + ... +T H
= P+2P+2*P+... + 2 n ~ 1 P
-*[fca-
by summing the geometrical progression,
= P(2 n -l) (2).
Hence the mechanical advantage is 2 n 1.
II. Let the weights of the movable pulleys, taken in
order and commencing ivith the lowest, be w lt w 2f ....
Considering the equilibrium of the pulleys in order,
we have
T i = 2T 1 + w 1 = 2P + w 1 ,
T 9 = 2T i + w % = 2 i P+2w 1 + w a1
T 4 =2T s + to 9 =2 9 P + 2 a w l + 2w 2 + w t .
But, from the equilibrium of the bar,
W=T A +T t + T t +T 1
= (2 8 + 2 a + 2 + 1) P + (2 s + 2 + 1) w x + (2 + 1) w t + w t
2 4 -l _ 2 8 -l 2 3 -l
== 23T P+ 23T M?1 + 13r M,a + M,
= (2<-l)P + (2-l)u, 1 + (2*-l)u> t + u> s (3).
If there were n pulleys, of which (n - 1) would be
movable, we should have, similarly,
JT=(2-l)P + (2- 1 -l)> 1 + (2- 8 -l)w,+ ...
+ (2-l) *-. + (* -l)*-i (4).
Stress on the supporting beam. This stress balances the
power, the weight, and the weight of the pulleys, and there-
fore equals
P+ W+w 1 + w i + ... +w n ,
and hence is easily found.
MACHINES. THE PULLEY. 91
Ex. If there be 4 pulleys, whose weights, commencing with the
lowest, are 4, 6, 6, and 7 lbs., what power will support a body of weight
1 ewt. t
Using the notation of the previous article, we have
T t =2P+4,
T 8 =2T, + 5 = 4P+13,
T 4 =2r s + 6 = 8P + 32.
Also 112 = r 4 +T 3 + T a +P=15P + 49.
.-. P=?|=4$l08. wt.
# 105. In this system we observe that, the greater the weight of
each pulley, the less is P required to be in order that it may support
a given weight W. Hence the weights of the pulleys assist the power.
If the weights of the pulleys be properly chosen, the system will
remain in equilibrium without the application of any power whatever.
For example, suppose we have 3 movable pulleys, each of weight
to, the relation (3) of the last article will become
W=15P + llw.
Hence, if llw = W, we have P zero, .., if the weight to be sup-
ported be eleven times the weight of each of the three movable pulleys,
no power need be applied at the free end of the string to preserve
equilibrium.
#106. In the third system of pulleys, the bar supporting the
weight W will not remain horizontal, unless the point at which the
weight is attached be properly chosen.
In any particular case the proper point of attachment can be
easily found.
Taking the figure of Art. 104 let there be three movable pulleys,
whose weights are negligible. Let the distances between the points
D, E, F and at which the strings are attached, be successively a,
and let the point at which the weight is attached be X.
The resultant of T lt T tt T 8 and T 4 must pass through X.
Henoe by Art. 78,
x _ T 4 x 0+ T 8 x a + T a x 2a+ T x x 3a
_ 4P . a + 2P . 2a + P . 3a _ 11a
~ 8P+4P + 2P + P ~ 15
.% DX=ftDE, giving the position of X.
EXAMPLES. XVm,
1. In the following cases, the pulleys are weightless and n in
number, P is the power, and W the weight ;
(1) If n=4 and P=2 lbs. wt., find W;
92 STATICS. Exs. XVHI.
(2) If n=5 and fT=124 lbs. wt., find P;
(3) If JP=105 lbs. and P=7 lbs. wt., find n.
2. In the following oases, the pulleys are equal and each of weight
w, P is the power, and W is the weight ;
(1) If n=4, u>= 1 lb. wt., and P= 10 lbs. wt., find W\
(2) If n=3, ic= i lb. wt., and JP=114 lbs. wt., find P;
(3) If n= 5, P- 3 lbs. wt., and W- 106 lbs. wt., find w ;
(4) If P=4 lbs. wt., JP=137 lbs. wt., and v>=$ lb. wt., find n.
3. If there be 5 pulleys, each of weight 1 lb., what power is re-
quired to support 3 owt. ?
If the pulleys be of equal size, find to what point of the bar the
weight must be attached, so that the beam may be always hori-
zontal.
4. If the strings passing round a system of 4 weightless pulleys
be fastened to a rod without weight at distances successively an inch
apart, find to what point of the rod the weight must be attached, so
that the rod may be always horizontal.
5. Find the mechanical advantage, when the pulleys are 4 in
number, and each is of weight - G \th that of the weight.
6. In a system of 8 weightless pulleys, in which each string is
attached to a bar which carries the weight, if the diameter of each
pulley be 2 inches, find to what point of the bar the weight should be
attached so that the bar may be always horizontal.
7. In the third system of 3 pulleys, if the weights of the pulleys
be all equal, find the relation of the power to the weight when equi-
librium is established. If each pulley weigh 2 ounces, what weight
would be supported by the pulleys only?
If the weight supported be 25 lbs. wt., and the power be 3 lbs. wt.,
find what must be the weight of each pulley.
8. In the third system of weightless pulleys, the weight is sup-
ported by a power of 70 lbs. The hook by which one of the strings is
attached to the weight breaks, and the string is then attached to the
pulley which it passed over, and a power of 150 lbs. is now required.
Find the number of pulleys and the weight supported.
III. The Inclined Plane.
107. The Inclined Plane, considered as a mechanical
power, is a rigid plane inclined at an angle to the horizon.
It is used to facilitate the raising of heavy bodies.
In the present chapter we shall only consider the case
of a body resting on the plane, and acted upon by forces
in a plane perpendicular to the intersection of the inclined
MACHINES. THE INCLINED PLANE. 93
plane and the horizontal, i.e., in a vertical plane through
the line of greatest slope.
The reader can picture to himself the line of greatest slope on an
inclined plane in the following manner:
take a rectangular sheet of cardboard
ABCD, and place it at an angle to the
horizontal, so that the line AB is in con-
tact with a horizontal table: take any
point P on the cardboard and draw PM
perpendicular to the line AB; PM is the
line of greatest slope passing through the
point P.
From G draw CE perpendicular to the horizontal plane through
AB, and join BE. The lines BG, BE, and CE are called respec-
tively the length, base, and height of the inclined plane; also the
angle CBE is the inclination of the plane to the horizon.
108. The inclined plane is supposed to be smooth, so
that the only reaction between it and any body resting
on it is perpendicular to the inclined plane.
Since the plane is rigid, it is capable of exerting any
reaction, however great, that may be necessary to give
equilibrium.
109. A body, of given weight, rests on a smooth inclined
plane; to determine the relations between the power, the
weight, and the reaction of the plane.
Let W be the weight of the body, P the power, and
B the reaction of the plane ; also let a be the inclination
of the plane to the horizon.
Case I. Let the power act up the plane along the line of
greatest slope.
Let AC be the inclined plane, AB the horizontal line
through A, DE a vertical line, and let
the perpendicular to the plane through
D meet AB in F, and the vertical line
through C in the point G.
Hence i FCC = L FDE = a.
Also the right angles GDC and ABC
are equal.
Hence the triangles GDC and ABC are equiangular,
so that DC :DG : GC :: BC : AB : AC
(Euc. VI., 4 or App., Art. 2),
94
STATICS.
Now the forces P, R, and W are parallel to the sides
DC, DG, and GC of the triangle DGC and are therefore
proportional to them,
:. P:R: W::DC:DG: GC
::BC :AB:AC
: : Height of Plane : Base of Plane : Length of Plane.
Otherwise thus: Resolve W along and perpendicular to the
plane ; its components are
W cos ADE, i.e., W sin a, along DA,
and W sin ADE, i.e., W cos a, along DF.
Hence P=JPsina, and R=W cos a.
Case II. Let the power act horizontally.
[In this case we must imagine a small hole
in the plane at D through which a string is
passed and attached to the body, or else that
the body is pushed toward the plane by a hori-
zontal force.]
Let the direction of P meet the verti-
cal line through C in the point H.
As in Case I. the triangles GDH and
ACB are equiangular,
so that HD.DG: GH :: BC : CA : AB
(Euc. VI., 4 or App., Art. 2).
But the forces P, R, and W are parallel to the sides
DH, GD and EG of the triangle HDG and are therefore
proportional to them.
/. PiRiWv.HDiQDiHQ
:: BC :CA: AB.
:: Height of Plane : Length of Plane : Base of Plane.
Otherwise thus : The components of W along and perpendicular
to the plane are W sin a and W cos a; the components of P, similarly,
are P cos a and P sin a.
.*. Pcos a=FTsino,
and
"8in 2 a
K = P sin a+ W cos a= W |~
Lcos a
/. P= W tan a, and
]=
W
sin 2 a + cos 2 a
cos a
= IFseoa.
= Wseca.
MACHINES. THE INCLINED PLANE. 95
#Case III. Let the power act at an angle 6 with the
inclined plane.
By Lami's Theorem we have
P R
W
%.e.
sin (R, W)
P
' sin (180- a) =
P
i.e. t -
sin a
sin
sin ( W, P)
R
sin (P, R)
W
sin (90 +0+ a) sin (90 -0)'
R W
cos (6 + a) "" cos $ '
cos 6 cos
The results of Cases II. and III. might in a similar
manner have been obtained by a direct application of
Land's Theorem.
It will be noted that Case III. includes both Cases I.
and II. ; if we make zero, we obtain Case L ; if we put
6 equal to ( - a), we have Case II.
EXAMPLES. TTY
1. What force, acting horizontally, could keep a mass of 16 lbs.
at rest on a smooth inclined plane, whose height is 3 feet and length
of base 4 feet, and what is the pressure on the plane ?
2. A body rests on an inclined plane, being supported by a force
acting up the plane equal to half its weight. Find the inclination of
the plane to the horizon and the reaction of the plane.
3. A rope, whose inclination to the vertical is 30, is just strong
enough to support a weight of 180 lbs. on a smooth plane, whose
inclination to the horizon is 30. Find approximately the greatest
tensiou that the rope could exert.
4. A body rests on a plane, inclined at an angle of 60 to the
horizon, and is supported by a force inclined at an angle of 30 to the
96
STATICS.
Exs. XIX.
horizon ; shew that the force and the reaction of the plane are each
equal to the weight of the body.
5. A body rests on a plane, inclined to the horizon at an angle
of 30, being supported by a power inclined at 80 to the plane ; find
the ratio of the weight of the body to the power.
6. A body, of weight 2P, is kept in equilibrium on an inclined
plane by a horizontal force P, together with a force P acting parallel
to the plane ; find the ratio of the base of the plane to the height and
also the pressure on the plane.
7. A body, of 5 lbs. wt., is placed on a smooth plane inclined at
30 to the horizon, and is acted on by two forces, one equal to the
weight of 2 lbs. and acting parallel to the plane and upwards, and the
other equal to P and acting at an angle of 30 with the plane. Find
P and the pressure on the plane.
8. Find the force which acting up an inclined plane will keep a
body, of 10 lbs. weight, in equilibrium, it being given that the force,
the pressure on the plane, and the weight of the body are in
arithmetical progression.
9. A number of loaded trucks, each containing 1 ton, on one
part of a tramway inclined at an angle a to the horizon supports
an equal number of empty trucks on another part whose inclination is
/3. Find the weight of a truck.
IV. The Wheel and Axle.
110. This machine consists of a strong circular
cylinder, or axle, terminating in two pivots, A and B,
which can turn freely on fixed supports. To the cylinder
is rigidly attached a wheel, CD, the plane of the wheel
being perpendicular to the axle.
Round the axle is coiled a rope, one end of which is
MACHINES. THE WHEEL AND AXLE. 97
firmly attached to the axle, and the other end of which is
attached to the weight.
Round the circumference of the wheel, in a direction
opposite to that of the first rope, is coiled a second rope,
having one end firmly attached to the wheel, and having
the power applied at its other end. The circumference of
the wheel is grooved to prevent the rope from slipping off.
111. To find the relation between the power and tlie
weight.
A body, which can turn freely about a fixed axis, is in
equilibrium if the algebraic sum of the moments of the
forces about the axis vanishes. In this case, the only forces
acting on the machine are the power P and the weight TF,
which tend to turn the machine in opposite directions.
Hence, if a be the radius of the axle, and b be the radius of
the wheel, the condition of equilibrium is
P.b=W.a.
W
Hence the mechanical advantage = -ft-
_ b radius of the wheel
a radius of the axle
119. Theoretically, by making the quantity- very large, we can
make the mechanical advantage as great as we please ; practically
however there are limits. Since the pressure of the fixed supports on
the axle must balance P and W, it follows that the thickness of the
axle, i.e., 2a, must not be reduced unduly, for then the axle would
break. Neither can the radius of the wheel in practice become very
large, for then the machine would be unwieldy. Hence the possible
values of the mechanical advantage are bounded, in one direction by
the strength of our materials, and in the other direction by the
necessity of keeping the size of the machine within reasonable limits.
11 a. In Art. Ill we have neglected the thicknesses of the ropes.
If, however, they are too great to be neglected, compared with the
radii of the wheel and axle, we may take them into consideration by
supposing the tensions of the ropes to act along their middle threads.
Suppose the radii of the ropes which pass round the axle and
wheel to be a: and y respectively ; the distances from the line joining
the pivots at which the tensions now act are (a + ) and (b+y) re-
spectively. Hence the condition of equilibrium is
LK.H. 7
STATICS.
Exb.
bo that
F(b+y)=W(a + x),
P _ gum of the radii of the axle and its rope
W ~~ Bum of the radii of the wheel and its rope *
114. Other forms of the Wheel and Axle are the
Windlass and Capstan. In these machines the power
instead of being applied, as in Art. 110, by means of a rope
passing round a cylinder, is applied at the ends of a spoke,
or spokes, which are inserted in a plane perpendicular to
the axle.
The Windlass is often used for raising water from a well
or for lifting building mate-
rials from the ground to a
scaffold. The Capstan is em-
ployed on ships for lifting
anchors.
In the Windlass the axle is
horizontal, and in the Capstan
it is vertical
In the latter case the
" weight " consists of the ten-
sion T of the rope round the
axle, and the power consists of
the forces applied at the ends
of bars inserted into sockets at the point A of the axle.
The condition of equilibrium may be obtained as in Art.
111.
EXAMPLES. XX.
1. If the radii of the wheel and axle be respectively 2 feet and
3 inches, find what power must be applied to raise a weight of
56 lbs.
2. If the radii of the wheel and axle be respectively 30 inches
and 5 inches, find what weight would be supported by a force equal
to the weight of 20 lbs., and find also the pressures on the supports on
which the axle rests.
If the thickness of the ropes be each 1 inch, find what weight would
now be supported.
3. If by means of a wheel and axle a power equal to 3 lbs. weight
balance a weight of 30 lbs., and if the radius of the axle be 2 inches,
what is the radius of the wheel ?
XX. MACHINES. THE COMMON BALANCE. 99
4. The axle of a capstan is 16 inches in diameter and there are
8 bars. At what distance from the axis most 8 men push, 1 at each
bar and each exerting a force equal to the weight of 26 lbs., in order
that they must jnst produce a strain sufficient to raise the weight of
1 ton?
5. Four sailors raise an anchor by meanB of a capstan, the radius
of which is 4 ins. and the length of the spokes 6 feet from the capstan ;
if each man exert a force equal to the weight of 112 lbs., find the
weight of the anchor.
6. Four wheels and axles, in each of which the radii are in the
ratio of 5 : 1, are arranged so that the circumference of each axle is
applied to the circumference of the next wheel ; what power is required
to support a weight of 1875 lbs. ?
V. The Common Balance.
115. The Common Balance consists of a rigid beam
AB, carrying a scale-pan suspended from each end, which
can turn freely about a fulcrum outside the beam. The
fulcrum and the beam are rigidly connected and, if the
balance be well constructed, at the point is a hard steel
wedge, whose edge is turned downward and rests on a small
plate of agate.
F
B
H
The body to be weighed is placed in one scale-pan and
in the other are placed weights, whose magnitudes are
known ; these weights are adjusted until the beam of the
balance rests in a horizontal position. If Off be perpen-
dicular to the beam, and the arms HA and HB be of equal
length, and if the centre of gravity G of the beam lie in
the line OH, and the scale-pans be of equal weight, then
the weight of the body is the same as the sum of the
weights placed in the other scale-pan.
72
100 STATICS.
If the weight of the body be not equal to the sum of
the weights placed in the other scale-pan, the balance
will not rest with its beam horizontal, but will rest with
the beam inclined to the horizon.
In the best balances the beam is usually provided with
a long pointer attached to the beam at H. The end of
this pointer travels along a graduated scale and, when
the beam is horizontal, the pointer is vertical and points
to the zero graduation on the scale.
116. Requisites of a good balance.
(1) The balance must be true.
This will be the case if the arms of the balance be
equal, if the weights of the scale-pans be equal, and if
the centre of gravity of the beam be on the line through
the fulcrum perpendicular to the beam ; for the beam will
now be horizontal when equal weights are placed in the
scale-pans. To test whether the balance is true, first see
if the beam is horizontal when the scale-pans are empty ;
then make the beam horizontal by putting sufficient
weights in one scale-pan to balance the weight of a body
placed in the other; now interchange the body and the
weights; if they still balance one another, the balance
must be true ; if the beam assumes any position inclined to
the vertical, the balance is not true.
(2) The balance should be sensitive, i.e., the beam
must, for any difference, however small, between the weights
in the scale-pans, be inclined at an appreciable angle to
the horizon.
(3) The balance should be stable and should quickly
take up its position of equilibrium.
It is found that the balance is most sensitive when the
distances of the points and G from the beam AB are
very small and that it is most stable when these distances
are great.
Hence we see that in any balance great sensitiveness
and quick weighing are to a certain extent incompatible.
In practice this is not very important; for in balances
where great sensitiveness is required (such as balances used
in a laboratory) we can afford to sacrifice quickness of
MACHINES. THE COMMON BALANCE. 101
weighing ; the opposite is the case when the balance is used
for ordinary commercial purposes.
To insure as much as possible both the qualities of
sensitiveness and quick weighing, the balance should be
made with fairly light long arms, and at the same time
the distance of the fulcrum from the beam should be
considerable.
117. Double weighing. By this method the weight
of a body may be accurately determined even if the balance
be not accurate.
Place the body to be weighed in one scale-pan and in
the other pan put sand, or other suitable material, sufficient
to balance the body. Next remove the body, and in its
place put known weights sufficient to again balance the
sand. The weight of the body is now clearly equal to the
sum of the weights.
This method is used even in the case of extremely good
machines when very great accuracy is desired.
118. Ex. 1. The arms of a balance are equal in length but the
beam is unjustly loaded ; if a body be placed in each scale-pan in suc-
cession and weighed, shew that its true weight is the arithmetic mean
between its apparent weights.
For let the length of the arms be a, and let the horizontal distance
of the centre of gravity of the beam from the fulcrum be x.
Let a body, whose true weight is W, appear to weigh JFj and W t
successively.
If W be the weight of the beam, we have
W .a=W.x+W 1 .a,
and W i .a = W.x+W .a. (Art. 59. Oor.)
Hence, by subtraction,
(W-WJa=(W 1 -W)a.
= arithmetic mean between the
apparent weights.
Ex. 2. The arms of a balance are of unequal length, but the beam
remains in a horizontal position when the scale-pans are not loaded ;
shew that, if a body be placed successively in each scale-pan, its true
weight is the geometrical mean between its apparent weights.
Since the beam remains horizontal when there are no weights in
the scale- pans, it follows that the centre of gravity of the beam and
scale- pans must be vertically under the fulcrum.
102 STATICS. Ess.
Let a and b be the lengths of the arms of the beam and let a body,
whose true weight is W, appear to weigh W x and JP 3 successively.
Henoe W ,a = W l .b (1),
and W % . a-W . b (2).
Henoe, by multiplication, we have
W^.ab^W-JV^.ab.
i.e., the true weight is the geometrical mean between the apparent
weights.
Ex. 3. If in the previous question the arms be of lengths 11 and
12 inches and if a grocer appear to weigh out 132 lbs. of tea, using
alternatively each of the scale-pans, prove that he wiU defraud himself
by half a lb.
The nominal quantity weighed is 66 lbs. from each scale-pan.
But, by equations (1) and (2) of the previous example the quanti-
ties really weighed are | . 66 and |f . 66 lbs. i.e. 60$ and 72 lbs., so
that altogether he weighs out 132$ lbs. instead of 132 lbs.
EXAMPLES. TTEX
1. The only fault in a balance being the unequalness in weight
of the scale-pans, what is the real weight of a body which balances
10 lbs. when placed in one scale-pan, and 12 lbs. when placed in the
other?
2. The arms of a balance are 8f and 9 ins. respectively, the goods
to be weighed being suspended from the longer arm ; find the real
weight of goods whose apparent weight is 27 lbs.
3. One scale of a common balance is loaded so that the apparent
weight of a body, whose true weight is 18 ounces, is 20 ounces ; find
the weight with which the scale is loaded.
4. A substance, weighed from the two arms successively of a
balance, has apparent weights 9 and 4 lbs. Find the ratio of the
lengths of the arms and the true weight of the body.
5. A body, when placed in one scale-pan, appears to weigh 24 lbs.
and, when placed in the other, 25 lbs. Find its true weight to three
places of decimals, assuming the arms of the scale-pans to be of
unequal length.
6. A piece of lead in one pan A of a balance is counterpoised by
100 grains in the pan B ; when the same piece of lead is put into the
pan B it requires 104 grains in A to balance it ; what is the ratio of
the length of the arms of the balance ?
7. A body, placed in a scale-pan, is balanced by 10 lbs. placed in
the other pan ; when the position of the body and the weights are
interchanged, 11 lbs. are required to balanoe the body. If the length
XXI. MACHINES. THE STEELYARDS. 103
of the shorter arm be 12 ins., find the length of the longer arm and
the weight of the body.
8. The arms of a false balance, whose weight is neglected, are in
the ratio of 10 : 9. If goods be alternately weighed from each arm,
shew that the seller loses $ th per cent.
9. If the arms of a false balance be 8 and 9 ins. long respectively,
find the prices really paid by a person for tea at two shillings per lb.,
if the tea be weighed oat from the end of (1) the longer, (2) the shorter
arm.
10. A dealer has correct weights, bat one arm of his balance is
jJ&th part shorter than the other. If he sell two quantities of a
certain drug, each apparently weighing 9 lbs., at 40*. per lb., weigh-
ing one in one scale and the other in the other, what will he gain or
lose ?
VI. The Steelyards.
119. The Common, or Roman, Steelyard is a machine
for weighing bodies and consists of a rod, AB, movable
about a fixed fulcrum at a point C.
At the point A is attached a scale-pan which contains
the body to be weighed, and on the arm CB slides a movable
GC O
*lXXj
T,
w
weight P. The point at which P must be placed, in order
that the beam may rest in a horizontal position, determines
the weight of the body in the scale-pan. The arm CB has
numbers engraved on it at different points of its length, so
that the graduation at which the weight P rests gives the
weight of the body.
120. To graduate the Steelyard. Let W be the weight
of the steelyard and the scale-pan, and let G be the point
of the beam through which W acts. The beam is usually
constructed so that O lies in the shorter arm AC.
When there is no weight in the scale-pan, let O be the
point in CB at which the movable weight P must be placed
to balance W.
104 STATICS.
Taking moments about (7, we have
W .OC = P.G0 (i).
This condition determines the position of the point
which is the zero of graduation.
When the weight in the scale-pan is W, let X be the
point at which P must be placed. Taking moments, we
have
W.CA + W . GC = P. CX (ii).
By subtracting equation (i) from equation (ii), we have
W . CA = P . OX.
:. ox^.ca (iii).
First, let W-P ; then, by (iii), we have
0X=CA.
Hence, if from we measure off a distance 0X X (= CA),
and if we mark the point X x with the figure 1, then,
when the movable weight rests here, the body in the
scale-pan is P lbs.
Secondly, let W=2P; then, from (iii), 0X=WA.
Hence from mark off a distance 2 CM, and at the
extremity put the figure 2. Thirdly, let W=3P; then,
from (iii), OX= 3CA, and we therefore mark off a distance
from equal to 3 CM, and mark the extremity with the
figure 3.
Hence, to graduate the steelyard, we must mark off from
successive distances CA, 2 CM, 3CM,... and at their ex-
tremities put the figures 1, 2, 3, 4,.... The intermediate
spaces can be subdivided to shew fractions of P lbs.
If the movable weight be 1 lb., the graduations will
shew pounds.
121. The Danish steelyard consists of a bar AB, ter-
minating in a heavy knob, or ball, B. At A is attached a
scale-pan in which is placed the body to be weighed.
The weight of the body is determined by observing
about what point of the bar the machine balances.
[This is usually done by having a loop of string, which can slide
along the bar, and finding where the loop must be to give equi-
librium.]
MACHINES. THE STEELYARDS. 105
122, To graduate the Danish steelyard. Let P be the
weight of the bar and scale-pan, and let G be their common
a , , **' , ' Q r \
V P
WW
centre of gravity. When a body of weight W is placed in
the scale-pan, let X be the position of the fulcrum.
By taking moments about X, we have
AX. W=XG .P = (AG-AX).P.
;. AX(P+W) = P.AG.
: ' AX =p-^w AG
First, let W=P; then AX=\AG.
Hence bisect AG and at the middle point, X lt engrave
the figure 1 ; when the steelyard balances about this point
the weight of the body in the scale-pan is P.
Secondly, let W=2P; then AX = ^AG.
Take a point at a distance from A equal to \AG and
mark it 2.
Next, let W in succession be equal to 3P, 4JP, . . . ; from
(i), the corresponding values of AX are \A G, \AG, Take
points of the bar at these distances from A and mark them
3, 4,...
Finally, let W = $P; then, from (i), AX = %AG ;
and let W=$P; then, from (i), AX= \AG.
Take points whose distances from A are \AG> \AG y
^AG,..., and mark them \, J, {,....
It will be noticed that the point G can be easily de-
termined; for it is the position of the fulcrum when the
steelyard balances without any weight in the scale-pan.
Ex. A Danish steelyard weighs 6 lbs., and the distance of its
centre of gravity from the scale-pan is 3 feet ; find the distances of the
successive points of graduation from the fulcrum.
Taking the notation of the preceding article, we have P = 6, and
AG = Bteet.
106 STATICS, Bxs. XXIL
Hence, when W=l,AX l = ^--2^ feet,
when W=2, AX 2 =^ = 2\ feet,
when W=3, AX 3 = ty = 2 feet,
18
when TF=, ^^^=2^ feet,
and so on.
These give the required graduations.
EXAMPLES. XXTT.
1. A common steelyard weighs 10 lbs. ; the weight is suspended
from a point 4 inches from the fulcrum, and the centre of gravity of
the steelyard is 3 inches on the other side of the fulcrum ; the movable
weight is 12 lbs. ; where should the graduation corresponding to 1 cwt.
be situated?
2. A heavy tapering rod, 14 inches long and of weight 3 lbs.,
has its centre of gravity If inches from the thick end and is used as
a steelyard with a movable weight of 2 lbs. ; where must the fulcrum
be placed, so that it may weigh up to 12 lbs., and what are the inter-
vals between the graduations that denote pounds ?
3. In a steelyard, in which the distance of the fulcrum from the
point of suspension of the weight is one inch and the movable weight
is 6 ozs., to weigh 15 lbs. the weight must be placed 8 inches from the
fulcrum ; where must it be placed to weigh 24 lbs. ?
4. A steelyard, AB, 4 feet long, has its centre of gravity 11 inches,
and its fulcrum 8 inches, from A. If the weight of the machine be
4 lbs. and the movable weight be 3 lbs., find how many inohes from B
is the graduation marking 15 lbs.
5. A uniform rod, 2 feet long and of weight 3 lbs., is used as
a steelyard, whose fulcrum is 2 inches from one end, the sliding
weight being 1 lb. Find the greatest and the least weights that can
be measured.
Where should the sliding weight be to shew 20 lbs. ?
6. In a Danish steelyard the distance between the zero gradu-
ation and the end of the instrument is divided into 20 equal parts and
the greatest weight that can be weighed is 8 lbs. 9 ozs. ; find the weight
of the instrument.
7. Find the length of the graduated arm of a Danish steelyard,
whose weight is 1 lb., and in which the distance between the gradua-
tions denoting 4 and 5 lbs. is one inch.
8. In a Danish steelyard the fulcrum rests halfway between the
first and second graduation ; shew that the weight in the scale-pan is
|ths of the weight of the bar.
CHAPTER X.
I
FRICTION.
123. In Art. 18 we defined smooth bodies to be bodies
such that, if they be in contact, the only action between
them is perpendicular to both surfaces at the point of con-
tact. With smooth bodies, therefore, there is no force
tending to prevent one body sliding over the other. If
a perfectly smooth body be placed on a perfectly smooth
inclined plane, there is no action between the plane and
the body to prevent the latter from sliding down the plane,
and hence the body will not remain at rest on the plane
unless some external force be applied to it.
Practically, however, there are no bodies which are
perfectly smooth ; there is always some force between two
bodies in contact to prevent one sliding upon the other.
Such a force is called the force of friction.
Friction. Def. If two bodies be in contact with one
ctnother, the property of the two bodies, by virtue of which
a force is exerted between the two bodies at their point of
contact to prevent one body sliding on the other, is called
friction; also the force exerted is called the force of friction.
124. Friction is a self-adjusting force ; no more friction
is called into play than is sufficient to prevent motion.
Let a heavy slab of iron with a plane base be placed on
a horizontal table. If we attach a piece of string to some
point of the body, and pull in a horizontal direction passing
through the centre of gravity of the slab, a resistance is felt
which prevents our moving the body; this resistance is
exactly equal to the force which we exert on the body.
If we now stop pulling, the force of friction also ceases
to act ; for, if the force of friction did not cease to act, the
body would move.
The amount of friction which can be exerted between
108 STATICS,
two bodies is not, however, unlimited. If we continually
increase the force which we exert on the slab, we find that
finally the friction is not sufficient to overcome this force,
and the body moves,
125. Friction plays an important part in the me-
chanical problems of ordinary life. If there were no friction
between our boots and the ground, we should not be able
to walk ; if there were no friction between a ladder and
the ground, the ladder would not rest, unless held, in any
position inclined to the vertical.
126. The laws of statical friction are as follows:
Law I. When two bodies are in contact, the direction
of the friction on one of them at its point of contact is oppo-
site to the direction in which this point of contact would com-
mence to move.
Law II. The magnitude of the friction is, when there
is equilibrium, just sufficient to prevent the body from moving.
Suppose, in Art. 109, Case L, the plane to be rough,
and that the body, instead of being supported by a
power, rested freely on the plane. In this case the power
P is replaced by the friction, which is therefore equal to
TTsina.
Ex. l. In what direction does the force of friction act in the
case of the feet of a man who is walking ?
Ex. 2. A body, of weight 30 lbs., rests on a rough horizontal
plane and is acted upon by a force, equal to 10 lbs. wt., making an
angle of 30 with the horizontal ; shew that the foroe of friction is
equal to about 8*66 lbs. wt.
Ex. 3. A body, resting on a rough horizontal plane, is acted on
by two horizontal forces, equal respectively to 7 and 8 lbs. wt., and
acting at an angle of 60; shew that the foroe of friction is equal in
magnitude to 13 lbs. wt.
Ex. 4. A body, of weight 40 lbs., rests on a rough plane inclined
at 30 to the horizon, and is supported by (1) a force of 14 lbs. wt.
acting up the plane, (2) a force of 25 lbs. acting up the plane, (3) a
horizontal force equal to 20 lbs. wt., (4) a force equal to 30 lbs. wt.
making an angle of 30 with the plane.
Find the force of friction in each case.
Ant. (1) 6 lbs. wt. up the plane; (2) 5 lbs. wt. down the plane;
(3) 2-68 lbs. wt. up the plane ; (4) 5 '98 lbs. wt. down the plane.
FRICTION. 109
127. The above laws hold good, in general; but the
amount of friction that can be exerted is limited, and equi-
librium is sometimes on the point of being destroyed
Limiting Friction. Def. When one body is just on
the point of sliding upon (mother body, the equilibrium is
said to be limiting, and the friction then exerted is called
limiting friction.
128. The direction of the limiting friction is given by
Law I. (Art. 126).
The magnitude of the limiting friction is given by the
three following laws.
Law III. The limiting friction always bears a constant
ratio to the normal reaction, and this ratio depends only on
the substances of which the bodies are composed.
Law IV. The limiting friction is independent of the
extent and shape of the surfaces in contact^ so long as the
normal reaction is unaltered.
Law V. When motion ensues, by one body sliding over
the other, the direction of friction is opposite to the direction
of motion; the magnitude of the friction is independent of
the velocity, but the ratio of the friction to the normal reaction
is slightly less than when the body is at rest and just on the
point of motion.
The above laws are experimental, and cannot be ac-
cepted as rigorously accurate ; they represent, however, to
a fair degree of accuracy the actual circumstances.
129. Coefficient of Friction. The constant ratio
of the limiting friction to the normal pressure is called the
coefficient of friction, and is generally denoted by /x ; hence,
if F be the friction, and R the normal pressure, between
two bodies when equilibrium is on the point of being
destroyed, we have - = /*, and hence F=fiR.
The values of p are widely different for different pairs
of substances in contact ; no pairs of substances are, how-
ever, known for which the coefficient of friction is as great
as unity.
110
STATICS.
130. Angle of Friction. When the equilibrium is
limiting, if the friction and the normal reaction be com-
pounded into one single force, the angle which this force
makes with the normal is called the angle of friction, and
the single force is called the resultant reaction.
Let A be the point of contact of the two bodies, and
let AB and AC be the directions of the normal force R
and the friction /xR.
Let AD be the direction of the resultant reaction S t so
that the angle of friction is BAD. Let this angle be A.
Since R and //J? are the components of S, we have
SgobX = R,
and S sin X = llR.
Hence, by squaring and adding, we B \ ^8
have fv^~. / ^^O
and, by division, tan X = /*.
Hence we see that the coefficient of friction is equal to
the tangent of the angle of friction.
131. If a body be placed upon a rough inclined plane,
and be on the point of sliding down the plane under the
action of its weight and the reactions of the plane only, the
angle of inclination of the plane to the horizon is equal to
the angle of friction.
Let 6 be the inclination of the plane to the horizon, W
the weight of the body, and R the normal reaction.
Since the body is on the point of motion down the
plane, the friction acts up the plane and is equal to /xR.
FRICTION. Ill
Resolving perpendicular and parallel to the plane, we
have Wcos6 = B t
and W sin = pB.
Hence, by division,
tan 6 = /jl = tan (angle of friction),
.'. $ = the angle of friction.
This may be shewn otherwise thus :
Since the body is in equilibrium under the action of its weight and
the resultant reaction, the latter must be vertical; but, since the
equilibrium is limiting, the resultant reaction makes with the normal
the angle of friction.
Hence the angle between the normal and the vertical is the angle
of friction, i.e. t the inclination EAD of the plane to the horizon,
which is equal to the angle EDF, is the angle of friction.
N.B. The student must carefully notice that, when the
body rests on the inclined plane supported by cm external force,
it must not be assumed that the coefficient of friction is equal
to the tangent of inclination of the plane to the horizon.
132. To determine the coefficient of friction experiment-
ally,
By means of the theorem of the previous article the
coefficient of friction between two bodies may be experi-
mentally obtained.
For let an inclined plane be made of one of the sub-
stances and let its face be made as smooth as is possible ;
on this face let there be placed a slab, having a plane face,
composed of the other substance.
If the angle of inclination of the plane be gradually
increased, until the slab just slides, the tangent of the
angle of inclination is the coefficient of friction.
To obtain the result as accurately as possible, the ex-
periment should be performed a large number of times
with the same substances, and the mean of all the results
taken.
133. Equilibrium on a rough inclined plane.
Ex. 1. A body, of mass 20 lbs. t is placed on a rough horizontal
plane and is acted on by a force F in a direction making an angle of
60 with the plane ; if the coefficient of friction be and the body be on
the point of motion, find the value ofF and the reaction of the plane.
112 STATICS.
Let R be the reaction of the plane, so that the friction is \R.
Since the body is in equilibrium the vertical components of the
forces acting on it must balance and so also must the horizontal com-
ponents (Art. 40).
A P. + P sin 60 =20,
and Fco8 60=iJ.
Hence JR + P^? = 20 (1),
m
and ?=ii2 ... (2).
The equation (2) gives F=R, and then (1) gives R ( 1 + ^- J = 20
40
A R=^ -=40 (2 -JS) = A0 (2 -1-73205) = 10-718.
Z + ^Jo
Hence the force and the reaction are each equal to 10-718 lbs. wt.
nearly.
Ex. 2. A body, of mass 30 lbs., is placed on a rough inclined plane
whose inclination to the horizon is 60 and is kept in equilibrium by a
force acting upward along the surface of the plane ; if the coefficient of
friction be j^find the magnitude of this force when the body is on the
v 3
point of sliding (1) up, (2) down, the plane.
Take the figure of Art. 131, and let P be the required force.
(1) When the body is on the point of moving up the plane the
friction acts down the plane and is equal to -j^ R, where R is the
reaction of the plane.
Hence, resolving along and perpendicular to the plane, we have
P ^ P.=80 sin 60= 15^/3 (i)
and P=30 cos 60=15 (ii)
.-. P=^ + 15^3 = 20^3 = 34-64 lbs. wt. nearly.
(2) When the body is on the point of sliding down the plane, the
friction acts up the plane.
In this case we have P + -rr fi=30 sin 60 =15^3,
is/ 6
andi*=30cos60=15.
A P= 15 J3 -5JS = 10^3 = 17-32 lbs. wt. nearly.
If the force P have any value between the two values then found
the body will be in equilibrium.
Exs. TTTTT. FRICTION. 113
EXAMPLES. XXm.
1. A body, of weight 40 lbs., rests on a rough horizontal plane
whose coefficient of friction is -25 ; find the least force which acting
horizontally would move the body.
Determine the direction and magnitude of the resultant pressure
of the plane in each case.
2. A heavy block with a plane base is resting on a rough hori-
zontal plane. It is acted on by a force at an inclination of 45 to the
plane, and the force is gradually inoreased till the block is just going
to slide. If the coefficient of friction be '5, compare the force with
the weight of the block.
3. A mass of 30 lbs. is resting on a rough horizontal plane and
can be just moved by a force of 10 lbs. wt. acting horizontally ; find
the coefficient of friction and the direction and magnitude of the
resultant reaction of the plane.
4. The height of a rough plane is 5 feet and its length is 13 feet ;
shew that, if the coefficient of friction be \, the least force, acting
parallel to the plane, that will support 1 cwt. placed on the plane is
8^ lbs. wt. ; shew also that the force that would be on the point of
moving the body up the plane is 77^ lbs. wt.
5. The base of an inclined plane is 4 feet in length and the height
is 8 feet; a force of 8 lbs., acting parallel to the plane, will just prevent
a weight of 20 lbs. from sliding down j find the coefficient of friction.
6. A body, of weight 4 lbs., rests in limiting equilibrium on a
rough plane whose slope is 30 ; the plane being raised to a slope of
60, find the force along the plane required to support the body.
7. A weight of 30 lbs. just rests on a rough inclined plane, the
height of the plane being f ths of its length. Shew that it will require
a force of 36 lbs. wt. acting parallel to the plane just to be on the
point of moving the weight up the plane.
8. A weight of 60 lbs. is on the point of motion down a rough
inclined plane when supported by a force of 24 lbs. wt. acting parallel
to the plane, and is on the point of motion up the plane when under
the influence of a force of 36 lbs. wt. parallel to the plane; find the
coefficient of friction.
L. M. H.
CHAPTER XI
WORK.
134. Work. Def. A force is said to do work when
its point of application moves in the direction of the force.
The force exerted by a horse, in dragging a waggon, does work.
The force exerted by a man, in raising a weight, does work.
The pressure of the steam, in moving the piston of an engine,
does work.
The measure of the work done by a force is the product
of the force and the distance through which it moves its
point of application in the direction of the force.
Suppose that a force acting at a point A of a body
A D ^ B
moves the point A to Z>, then the work done by P is
measured by the product of P and AD.
If the point D be on the side of A toward which the
force acts, this work is positive ; if D lie on the opposite
side, the work is negative.
Next, suppose that the point of application of the force
is moved to a point C, which does not lie on the line AB.
Draw CD perpendicular to AB, or AB produced. Then
AD is the distance through which the point of application
is moved in the direction of the force. Hence in the first
figure the work done is P x AD ; in the second figure the
C C
/r: .
B O A P B
work done is P x AD. When the work done by the
force is negative, this is sometimes expressed by saying
that the force has work done against it.
WORK. 115
135. In the case when A C is at right angles to AB,
the points A and D coincide, and the work done by the
force P vanishes.
As an example, if a body be moved about on a hori-
zontal table the work done by its weight is zero. So, again,
if a body be moved on an inclined plane, no work is done
by the normal reaction of the plane.
136. The unit of work, used in Statics, is called a
Foot-Pound, and is the work done by a force, equal to the
weight of a pound, when it moves its point of application
through one foot in its own direction. A better, though
more clumsy, term than "Foot-Pound" would be Foot-
Pound weight.
Thus, the work done by the weight of a body of 10
pounds, whilst the body falls through a distance of 4 feet,
is 10 x 4 foot-pounds.
The work done by the weight of the body, if it were
raised through a vertical distance of 4 feet, would be
10 x 4 foot-pounds.
137. It will be noticed that the definition of work,
given in Art. 134, necessarily implies motion. A man
may use great exertion in attempting to move a body, and
yet do no work on the body.
For example, suppose a man pulls at the shafts of a
heavily-loaded van, which he cannot move. He may pull
to the utmost of his power, but, since the force which he
exerts does not move its point of application, he does no
work (in the technical sense of the word).
138. To find the work done in dragging a body up a
smooth inclined plane.
Taking the figure of Art. 109, Case I., the work done
by the force P in dragging the body from A to C is
PxAC.
But P=TTsina.
Therefore the work done is Wain ax AC,
i.e. t Wx AC am a, .., WxBG.
Hence the work done is the same as that which would
82
116 STATICS.
be done in lifting the weight of the body through the same
height without the intervention of the inclined plane.
139. The previous article is one of the simplest ex-
amples of what we shall find to be a universal principle,
viz., Whatever be the machine we use, provided that there
be no friction and that the weight of the machine be neglected,
the work done by the power is always equivalent to the work
done against the weight.
Assuming that the machine we are using gives me-
chanical advantage, so that the power is less than the
weight, the distance moved through by the power is there-
fore greater than the distance moved through by the weight
in the same proportion. This is sometimes expressed in
popular language in the form ; What is gained in power is
lost in speed.
140. In any machine if there be any roughness (as in
practice there always is) the work done by the power is
more than the work done against the weight. The
principle may be expressed thus,
In any machine, the work done by the power is equal to
the work done against the weight, togetJier with the work done
against the frictional resistances of the machine, and the
work done against the weights of the component parts of the
machine.
The ratio of the work done on the weight to the work
done by the power is, for any machine, called the modu-
lus or efficiency of the machine. We can never get rid
entirely of frictional resistances, or make our machine
without weight, so that some work must always be lost
through these two causes. Hence the modulus of the
machine can never be so great as unity. The more nearly
the modulus approaches to unity, the better is the machine.
#141. The student can verify the truth of the principle, enun-
ciated in Art. 139, for all machines; we shall consider only a few
First tyshnu of pulleys (Art. 102).
Neglecting ibe weights of the pulleys we have, if there be four
pulleys,
WORK. 117
If the weight W be raised through a distance x, the pulley A t
would, if the distance A x A t remained unchanged, rise a distance x ;
but, at the same time, the length of the string joining A x to the beam
is shortened by a, and a portion x of the string therefore slips round
A x ; hence, altogether, the pulley A t rises through a distance 2x.
Similarly, the pulley A 3 rises a distance 4x, and the pulley A 4 a
distance 8x.
Since A 4 rises a distance 8x, the strings joining it to the beam and
to the point at which P is applied both shorten by 8x.
Hence, since the slack string runs round the pulley A 4 , the point
of application of P rises through l&r.
Hence
work done by the power P. 16s 2* * _ W.x
work done against the weight W.x ~ W.x ~W.x~~
Hence the principle is verified.
Third system of pulleys (Art. 104).
Suppose the weight W to ascend through a space *. The string
joining B to the bar shortens by , and hence the pulley A ? descends
a distance x. Since the pulley A s descends x and the bar rises x, the
string joining A* to the bar shortens by 2x, and this portion slides
over A. ; hence the pulley A s descends a distance equal to 2x together
with the distance through which A 3 descends, i.e., A descends a
distance 2x + x, or Sx. Hence the string A^F shortens by 4a:, which
slips over the pulley A t , so that the pulley A-, descends a distance 4x
together with the distance through which A % descends, i.e., 4x + 3x, or
7x. Hence the string A X G shortens by 8x, and A x itself descends 7x,
so that the point of application of P descends 15x.
Neglecting the weight of the pulleys, the work done by P therefore
*16* . P=x (2* - 1) P=x . W by equation (1), Art. 104,
acwork done on the weight W.
Smooth inclined plane (Art. 109, Case III.).
Let the body move a distance x along the plane; the distance
through which the point of application of P moves, measured along
its direction of application, is clearly x cos 6 ; also the vertical distance
through which the weight moves is x sin a.
Hence the work done by the power is P . x cos 6, and that done
against the weight is W.x Bin a.. These are equal by the relation
proved in Art. 109.
142. Assuming the Principle of work enunciated in
Art. 139 we shall use it to find the conditions of equili-
brium of a smooth screw.
A Screw consists of a cylinder of metal round the out-
side of which runs a protuberant thread of metal.
IIS STATICS.
Let A BCD be a solid cylinder, and let EFGH be a
DfG Hi iQ
1
rectangle, whose base EF is equal to the circumference
of the solid cylinder. On EH and FO take points
L,JSr, Q... and K t M , P.
such that -tfZ, #, . . ,FK, KM, MP
are all equal, and join EK y LM, NP,....
Wrap the rectangle round the cylinder, so that the
point E coincides with A and EH with the line AD. On
being wrapped round the cylinder the point F will coincide
with E at A,
The lines EK, LM, NP ... will now become a con-
tinuous spiral line on the surface of the cylinder and, if
we imagine the metal along this spiral line to become
protuberant, we shall have the thread of a screw.
It is evident, by the method of construction, that the
thread is an inclined plane running round the cylinder
and that its inclination to the horizon is the same every-
where. This inclination is often called the angle of the
screw, and the distance between two consecutive threads,
measured parallel to the axis, is called the pitch of the
screw.
The section of the thread of the screw has, in practice,
various shapes. The only kind that we shall consider has
the section rectangular.
143. The screw usually works in a fixed support, along
the inside of which is cut out a hollow of the same shape
as the thread of the screw, and along which the thread
slides. The only movement admissible to the screw is to
WORK.
119
revolve about its axis, and at the same time to move in
a direction parallel to its length.
If the screw were placed in an upright position, and
a weight placed on its top, the screw would revolve and
descend since there is supposed to be no friction between
the screw and its support. Hence, if the screw is to
remain in equilibrium, some power must be applied to it ;
this power is usually applied at one end of a horizontal
arm, the other end of which is rigidly attached to the
screw.
144. In a smooth screw, to find the relation between the
power and the weight.
Let b be the distance, AJB, from the axis of the screw,
of the point at which the power P is applied.
YW
Let the arm at the end of which P acts make a
complete revolution. The distance through which the
point of application of P moves
= circumference of a circle of radius b
= 2tt6.
Hence the work done by P is P x 2vb.
In the same time the screw rises by a distance equal to
120 STATICS. Exs. XXIV.
that between consecutive threads, i.e. the pitch of screw, so
that the work done against the weight is
W x the pitch of the screw.
Hence, by the Principle of work,
P x 2irb = W x pitch of the screw.
W
The mechanical advantage = -=j
circumference of a circle whose radius is the power-arm
distance between consecutive threads.
Theoretically, the mechanical advantage in the case
of the screw can be made as large as we please, by
decreasing sufficiently the distance between the threads
of the screw. In practice, however, this is impossible;
for, if we diminish the distance between the threads to
too small a quantity, the threads themselves would not
be sufficiently strong to bear the strain put upon them.
EXAMPLES. XXIV.
1. Find what mass can be lifted by a smooth vertical screw of
1\ ins. pitch, if the power be a force of 25 lbs. wt. acting at the end of
an arm, 3 \ feet long.
2. What must be the length of the power-arm of a screw, having
6 threads to the inch, so that the mechanical efficiency may be 216 ?
3. What force applied to the end of an arm, 18 ins. long, will
produce a pressure of 1100 lbs. wt. upon the head of a screw, when
seven turns cause the screw to advance through frds of an inch ?
4. A screw, whose pitch is \ inch, is turned by means of a lever,
4 feet long ; find the power which will raise 15 cwt.
5. The arm of a screw-jack is 1 yard long, and the screw has
2 threads to the inch. What force must be applied to the arm to
raise 1 ton ?
6. What is the thrust caused by a screw, having 4 threads to the
inch, when a force of 50 lbs. wt. is applied to the end of an arm, 2 feet
long?
145. Theorem. To shew that the work done m
raising a number of particles from one position to another is
Wh, where W is the total weight of the particles, and h is the
distance through which the centre of gravity of the particles
has been raised.
WORK. 121
Let ioj, w a , w^...w n be the weights of the particles ; in
the initial position let Xj f a;,, tCg,...a5 n be their heights above
a horizontal plane, and x that of their centre of gravity, so
that, as in Art. 80, we have
_ W 1 X l + W^C s +... + w n x n
a? = (1).
Wx + Wt-f ... + w n
In the final position let a?/, x 9 \ ...#' be the heights of
the different particles, and x' the height of the new centre
of gravity, so that
_, WyXf + w&J + . . . w n x n '
X = IZ).
w 1 + w % + ...w n v
But, since W! + to 2 + ... = W, equations (1) and (2) give
wfa + w&i + ... = W . x,
and w^ + wfa + ... = W . x'.
By subtraction we have
w i {vi -Vi) + W9 ( x 2 - a?) + = W (#' - *)
But the left-hand member of this equation gives the
total work done in raising the different particles of the
system from their initial position to their final position ;
also the right-hand side
W x height through which the centre of gravity has been
raised
Hence the proposition is proved.
146. Power. Def. The power of an agent is (he
amount of work tlwl would be done by the agent if working
uniformly for the unit of time.
The unit of power used by engineers is called a Horse -
Power. An agent is said to be working with one horse-
power when it performs 33,000 foot-pounds in a minute, i.e.,
when it would raise 33,000 lbs. through a foot in a minute,
or when it would raise 330 lbs. through 100 feet in a
minute, or 33 lbs. through 1000 feet in a minute.
This estimate of the power of a horse was made by
"Watt, but is rather above the capacity of ordinary horses.
The word Horse-power is usually abbreviated into H. p.
122 STATICS. Ex*. XXV.
Bx. A well, of which the section i$ a square whose side is 4
feet, and whose depth is 800 feet , is full of water ; find the work done,
in foot-pounds, in pumping the water to the level of the top of the well.
Find also the H. P. of the engine which would just accomplish this
work in one hour.
[N.B. A oubio foot of water weighs 1000 ounces.]
Initially the height of the centre of gravity of the water above the
bottom of the well was 150 feet and finally it is 300 feet, so that the
height through which the centre of gravity has been raised is 150 feet.
The volume of the water =4 x 4 x 300 cubic feet.
Therefore its weight = 4 x 4 x 300 x m lbs. = 300,000 lbs.
Hence the work done = 300,000 x 150 ft. -lbs. =45,000,000 ft. -lbs.
Let x be the required h. p. Then the work done by the engine in
one hour = x x 60 x 33,000.
Hence we have x x 60 x 33,000 = 45,000,000 ;
EXAMPLES. XXV.
1. How many cubic feet of water will an engine of 100 h. p. raise
in one hour from a depth of 150 feet ?
2. A tank, 24 feet long, 12 feet broad, and 16 feet deep, is filled
by water from a well the surface of which is always 80 feet, below the
top of the tank ; find the work done in filling the tank, and the h. p.
of an engine, whose modulus is *5, that will fill the tank in 4 hours.
3. A chain, whose mass is 8 lbs. per foot, is wound up from a
shaft by the expenditure of four millions units of work; find the
length of the chain.
4. In how many hours would an engine of 18 h.p. empty a vertical
shaft full of water if the diameter of the shaft be 9 feet, and the depth
420 feet?
5. Find the h. p. of an engine that would empty a cylindrical
shaft full of water in 82 hours, if the diameter of the shaft be 8 feet
and its depth 600 feet.
6. A tower is to be built of brickwork, the base being a rectangle
whose external measurements are 22 ft. by 9 ft., the height of the
tower 66 feet, and the walls two feet thick ; find the number of hours
in which an engine of 3 h. p. would raise the bricks from the ground,
the weight of a oubio foot of brickwork being 112 lbs.
7. At the bottom of a coal mine, 275 feet deep, there is an iron
cage containing coal weighing 14 cwt., the cage itself weighing 4 cwt.
109 lbs., and the wire rope that raises it 6 lbs. per yard. Find the
work done when the load has been lifted to the surface, and the h. p.
of the engine that can do this work in 40 seconds.
v 8. In a wheel and axle, if the radius of the wheel be six times
that of the axle, and if by means of a power equal to 5 lbs. wt. a
body be lifted through 50 feet, find the amount of work expended.
DYNAMICS.
CHAPTER XII.
VELOCITY AND ACCELERATION. RECTILINEAR MOTION.
147. A point is said to be in motion when it changes
its position.
The path of a moving point is the line, straight or
curved, which would pass through all the successive positions
of the point.
148. If at any instant the position of a moving point
be P and at any subsequent instant it be Q, then PQ is
its displacement, or change of position, in the intervening
time.
To know the displacement of a moving point we must
know both the length and the direction of the line joining the
two positions of the point. Hence the displacement of a
point involves both magnitude and direction.
149. Velocity. Def. The velocity of a moving
point is the rate of its displacement, t.e., the rate at which it
changes its position.
A velocity therefore possesses both magnitude and
direction. We have not fully specified the velocity of a
moving point unless we have stated both its rate and its
direction of motion.
In Chapters XII. XV. we shall consider only the
cases in which the direction of the velocity of a moving
point is constant.
150. A point is said to be moving with uniform velo-
city when it is moving in a constant direction and passes
over equal lengths in equal times, however small these times
may be.
124 DYNAMICS.
Suppose a train described 30 miles in each of three consecutive
hours. We are not justified in saying that its velocity is uniform
unless we know that it describes half a mile in each minute, 44 feet in
each second, one-millionth of 30 miles in each one-millionth of an
hour, and so on.
When uniform, the velocity of a moving point ia
measured by its displacement in each unit of time.
When variable, the velocity is measured at any instant
by the displacement which the point would have if it
moved during that unit of time with the velocity which it
had at the instant under consideration.
By saying that a train is moving with a velocity of 40 miles per
hour, we do not mean that it has gone 40 miles in the last hour, or
that it will go 40 miles in the next hour, but that, if its velocity
remained constant for one hour, then it would describe 40 miles in
that hour.
151. The unit of velocity is the velocity of a moving
point which has a displacement of a unit of length in each
unit of time. In England the units of length and time
usually employed are a foot and a second, so that the unit
of velocity is the velocity of one foot per second.
In scientific measurements the unit of length usually employed is
a centimetre, so that the corresponding unit of velocity is one centi-
metre per second.
The centimetre is one-hundredth part of a unit which is called
a metre and is equal to 39*37 inches approximately. A decimetre is
^th and a rnillimetre r^th of a metre.
Since the unit of velocity depends on the units of length
and of time, it follows that, if either or both of these be
altered, the unit of velocity will also, in general, be altered.
152. If a point be moving with velocity w, then in
each unit of time the point moves over u units of length.
Hence in t units of time the point passes over u . t units
of length.
Therefore the distance s passed over by a point which
moves for time t with velocity u is given by = u . t.
153. Acceleration. Def. The acceleration of a
moving point is the rate of change of its velocity.
The acceleration is uniform when equal changes of
velocity take place in equal intervals of time, however
small these times may be.
ACCELERATION. 125
When uniform, the acceleration is measured by the
change in the velocity per unit of time ; when variable, it is
measured at any instant by what would be the change of
the velocity in a unit of time, if during that unit of time
the acceleration remained the same as at the instant under
consideration.
154. The magnitude of the unit of acceleration is the
acceleration of a point which moves so that its velocity is
changed by the unit of velocity in each unit of time.
Hence a point is moving with n units of acceleration
when its velocity is changed by n units of velocity in each
unit of time.
155. In the simple case of motion in a straight line
the only change that the velocity can have is either an
increase or a diminution. In the former case the accelera-
tion is positive ; in the latter case it is negative and is often
called a retardation.
For example suppose a train to be always running due south and
that in ten minutes its velocity is uniformly diminished from 30 miles
per hour, i.e. 44 feet per second, to 15 miles per hour, i.e. 22 feet per
second. In 600 seconds the loss of velocity is 22 feet per second.
Hence in 1 second the loss of velocity is ffi feet per second. This is
expressed by saying that its acceleration is - ^& foot-second units.
If in this time the velocity had increased from 15 miles per hour
to 30 miles per hour, the acceleration would have been positive and
equal to ^ foot-second units.
156. Theorem. A point moves in a straight line,
starting with velocity u, and moving with constant accelera-
tion/ m its direction of motion ; if v be its velocity at the
end of time t, and s be its distance at that instant from Us
starting point, then
(1) V~U + ft,
(2) 8 = ut + ift 3 ,
and (3) V a = U 2 + 2fs.
(1) Since f denotes the acceleration, i.e., the change in
the velocity per unit of time Jt /l denotes the change in the
velocity in t units of time.
But, since the particle possessed u units of velocity
initially, at the end of time t it must possess u+fl units
of velocity, i.e. v=u+Jl
V.t
126 DYNAMICS.
* (2) Let V be the velocity at the middle of the interval
so that, by(l), F = *+/..
Now the velocity changes uniformly throughout the
interval 6. Hence the velocity at any instant, preceding
the middle of the interval by time T, is as much less than
F, as the velocity at the same time T after the middle
of the interval is greater than V.
Hence, since the time t could be divided into pairs of
such equal moments, the space described is the same as if
the point moved for time t with velocity V.
(3) The third relation can be easily deduced from the
first two by eliminating t between them.
For, from (1), J = (u +fb)*
= u M + 2f(ut + $j?).
Hence, by (2), w* = u' + 2/*.
167. As an illustration of the method of proof nsed in the
preceding article oonsider the case of a train which moves on a
straight line of rails ; let its velocity at 12 noon be 30 feet per second
and at 1 p.m. let it be 60 feet per second, and let it have uniformly
increased its velocity during the hour, i.e. let it have moved with
uniform acceleration. At 12.30, the middle of the interval, the
velocity was 45 feet per second ; at 12.20 and 12.40 the velocity was
40 and 50 feet per second respectively. Clearly 40 is as much less
than 45 as 50 is greater than 45. Hence in a short space of time
following 12.20 and another equal short space following 12.40, the
distance described would be just twice that described in an equal
short space of time following 12.30. By reasoning in this manner we
see that the total distance described in the hour is the same as what
would have been described had the velocity always been what it was
at 12.30.
158. When the moving point starts from rest we have
w = 0, and the formulae of Art. 156 take the simpler forms
and t>*=2/.
169. Bx. 1. A point $tarU with a velocity of 4 feet per $eeond
and with an acceleration of I foot-*r<mA unit. What i* its velocity at
* Vox an alternative proof see Page 293.
ACCELERATION. 127
the end of the first minute and how far ha* it then gone t What it iU
velocity when it has described 200 feet ?
At the end of 60 seconds its velocity, by Art. 156 (1),
=s 4 + 1 . 60 = 64 feet per second.
The distance described, by (2),
=4 . 60+ * . 1 . 60= 240 + 1800* 2040 feet.
Its velocity when it has described 200 feet, by (3),
= a/4 2 +2. 1.200= JilQ=s 20 . 4 feet per second nearly.
Ex. a. A train, which i$ moving at the rate of 60 miles per
hour, is brought to rest in 3 minutes with a uniform retardation ; find
this retardation, and also ' the distance that the train travels before
coming to rest.
n v 60x1760x3 __. . ,
60 miles per hours ^ =88 feet per second.
oU x bO
Let / be the acceleration with which the train moves.
Since in 180 seconds a velocity of 88 feet per second is destroyed,
we have (by formula (1), Art. 156)
0=88+/ (180).
22
A /s - t= ft.-sec. units.
45
[N.B. /has a negative value because it is a retardation.]
Let x be the distance described. By formula (3), we have
0=88+2/-??y*.
(-
A =88*x ^=7920 feet.
44
160. Space described in any particular second.
[The student will notice carefully that the formula (2) of Art. 156
gives, not the space traversed in the t* second, but that traversed in
t seconds.]
The space described in the <* second
= space described in t seconds space described in (t 1)
seconds
-[*+i/n-[(-i)+i/(-iy]
,2<-l
+/-J--
Hence the spaces described in the first, second, third,
....nth seconds of the motion are
i * m * 2w - 1 .
tt + i/ tt + / ... M + 3 /
128 DYNAMICS. Exs.
Bx. A -point is moving with uniform acceleration ; in the eleventh
and fifteenth seconds from the commencement it moves through 24 and
32 feet respectively ; find its initial velocity, and the acceleration with
which it moves.
Let u be the initial velocity, and / the acceleration.
Then 24= distance described in the eleventh second
=[u.ll + i/.ll 2 )-[tt.l0 + i/\10 2 ].
.-. 24=m+V/- (1).
So 32 = [u.l5+i/.15 a ]-[tt.l4 + i/.14 3 ].
A 32=m + V/- ~ ~ ( 2 )-
Solving (1) and (2), we have u = 3, and/= 2.
Hence the point started with a velocity of 3 feet per second, and
moved with an acceleration of 2 ft. -sec units.
EXAMPLES. XXVL
1. The quantities u, /, v, s, and t having the meanings assigned
to them in Art. 156,
(1) Given u= 2, /= 3, t= 5, find v and r\
(2) Given u= 7, /=-l, t= 7, find v and * ;
(3) Given m= 8, v= 3, *= 9, find/ and f t
(4) Given v- - 6, s * - 9, /= - f , find u and t.
The units of length and time are a foot and a second.
2. A body, starting from rest, moves with an acceleration equal to
2 ft. -sec. units ; find the velocity at the end of 20 seconds, and the
distance described in that time.
3. In what time would a body acquire a velocity of 30 miles per
hour, if it started with a velocity of 4 feet per second and moved with
the ft. -sec. unit of acceleration?
4. With what uniform acceleration does a body, starting from
rest, describe 1000 feet in 10 seconds ?
5. A body, starting from rest, moves with an acceleration of 3
centimetre-second units; in what time will it acquire a velocity of
30 centimeties per second, and what distance does it traverse in that
time?
6. A point starts with a velocity of 100 cms. per second and
moves with - 2 centimetre-second units of acceleration. When will
its velocity be zero, and how far will it have gone ?
7. A body, starting from rest and moving with uniform accelera-
tion, describes 171 feet in the tenth second ; find its acceleration.
8. A particle is moving with uniform acceleration ; in the eighth
and thirteenth second after starling it moves through 8^ and 7 feet
respectively ; find its initial velocity and its acceleration.
K'\
XXVI. ACCELERATION. 129
9. In two successive seconds a particle moves through 20$ and
23 feet respectively; assuming that it was moving with uniform
acceleration, find its velocity at the commencement of the first of these
two seconds and its acceleration. Find also how far it had moved
from rest before the commencement of the first second.
10. A point, moving with uniform acceleration, describes in the
last second of its motion ^ths of the whole distance. If it started
from rest, how long was it in motion and through what distance did
it move, if it described 6 inches in the first second ?
11. A point, moving with uniform acceleration, describes 25 feet
in the half second whioh elapses after the first second of its motion,
and 198 feet in the eleventh second of its motion ; find the acceleration
of the point and its initial velocity.
12. A body moves for 3 seconds with a constant acceleration
during which time it describes 81 feet; the acceleration then ceases
and during the next 8 seconds it describes 72 feet; find its initial
velocity and its acceleration.
13. The speed of a train is reduced from 40 miles an hour to 10
miles per hour whilst it travels a distance of 150 yards; if the re-
tardation be uniform, find how much further it will travel before
coming to rest.
14. A point starts from rest and moves with a uniform accelera-
tion of 18 ft. -sec. units ; find the time taken by it to traverse the first,
second, and third feet respectively.
15. A particle starts from a point O with a uniform velocity of
4 feet per second, and after 2 seconds another particle leaves O in the
same direction with a velocity of 5 feet per second and with an
acceleration equal to 3 ft. -sec. units. Find when and where it will
overtake the first particle.
16. A point moves over 7 feet in the first second during which it
is observed, and over 11 and 17 feet in the third and 3ixth seconds
respectively ; is this consistent with the supposition that it is subject
to a uniform acceleration ?
Motion under Gravity.
161. Acceleration of falling bodies. When a
heavy body of any kind falls toward the earth, it is a
matter of everyday experience that it goes quicker and
quicker as it falls, or, in other words, that it moves with
an acceleration. That it moves with a constant acceleration
may be shewn by various experiments, one of which will be
explained in Art. 192.
From the results of these experiments we learn that, if
a body be let fall towards the earth in vacuo^ it will move
with an acceleration which is always the same at the same
L. m. h. 9
130 DYNAMICS.
place on the earth, but which varies slightly for different
places.
The value of this acceleration, which is called the
"acceleration due to gravity," is always denoted by the
letter "g."
When foot-second units are used, the value of g varies
from about 32*091 at the equator to about 32*252 at the
poles. In the latitude of London its value is about 32*19 ;
in other words in the latitude of London the velocity of a
body falling in vacua is increased in each second by 32*19
feet per second.
When centimetre-second units are used, the extreme
limits are about 978 and 983 respectively, and in the
latitude of London the value is about 981.
[In all numerical examples, unless it is otherwise stated,
the motion may be supposed to be in vacuo, and the value
of g taken to be 32 when foot-second units, and 981 when
centimetre-second units, are used.]
162. Vertical motion under gravity. Suppose a
body is projected vertically from a point on the earth's sur-
face so that it starts with velocity u. The acceleration of
the body is opposite to the initial direction of motion, and is
therefore denoted by g. Hence the velocity of the body
continually gets less and less until it vanishes; the body is
then for an instant at rest, but immediately begins to acquire
a velocity in a downward direction, and retraces its steps.
163. Time to a given height. The height h at which
a body has arrived in time t is given by substituting g
for/ in equation (2) of Art. 156, and is therefore given by
h = ut-yp.
This is a quadratic equation with both roots positive ; the
lesser root gives the time at which the body is at the given
height on the way up, and the greater the time at which it
is at the same height on the way down.
Thus the time that elapses before a body, which starts with a
velocity of 64 feet per second, is at a height of 28 feet is given by
28=64t - 16t a , whence =$ or .
Hence the particle is at the given height in half a second from the
commencement of its motion, and again in 3 seconds afterwards.
ACCELERATION. 131
164. Velocity at a given height.
The velocity v at a given height h is, by equation (3) of
Art. 156, given by
v*=u i -2gh.
Hence the velocity at a given height is independent of the
time, and is therefore the same at the same point whether
the body be going upwards or downwards.
165. Greatest height attained.
At the highest point the velocity is just zero ; hence, if
x be the greatest height attained, we have
= u*-2gx.
Hence the greatest height attained = ^- .
Also the time T to the greatest height is given by
Q = u-gT.
g'
166. Velocity due to a given vertical fall from rest.
If a body be dropped from rest, its velocity after falling
through a height h is obtained by substituting 0, g, and h for
u ># /and 8 in equation (3) of Art. 156;
.'. v = J%jh.
187. Ex. 1. A particle is projected vertically into the air with
a velocity of 80 feet per second ; find (i) what times elapse before it is
at a height of 64 feet, (ii) its velocity when at a height of 40 feet, and
(hi) the greatest height it attains.
(i) The required time t is given by
64=804 -yt*,
U. by t a -6<+4=0.
.'. t = i or 1 seconds.
(ii) The required velocity = J8QP - 2 . 32 . 40 = ^6400 - 2560
=^3840= nearly 62 feet per second.
(hi) The greatest height h is given by
0=80*- 2. y. ft.
fc 80 2
^100 feet.
93
132 DYNAMICS. Exs.
Ex. 2. A cage in a mine-$haft descends with 2 ft.-stc. unit* of
acceleration. After it has been in motion for 10 seconds a particle is
dropped on it from the top of the shaft. What time elapses before the
particle hits the cage t
Let T be the time that elapses after the second particle starts.
The distance it has fallen through is therefore fc^T*. The cage has
'* been in motion for (T+10) seconds, and therefore the distance it has
fallen through is
$.2(T+10) a , i.e. (T+10) 8 .
Hence we have (T + 10) a = ^T 3 a 16T*.
A T+10=4T.
A T=S seoonds.
Ex. 8. A stone is thrown vertically with the velocity which would
just carry it to a height of 100 feet. Two seconds later another stone is
projected vertically from the same place with the same velocity ; when
and where will they meet t
Let u be the initial velocity of projection. Since the greatest
height is 100 feet, we have
0=t* 9 - 20.100.
A u=>J2g. 100=80.
Let T be the time after the first stone starts before the two stones
meet.
Then the distance traversed by the first stone in time T
= distance traversed by the second stone in time (T - 2),
. 802 T -&T*=80(T-2)-i fl r(2 , -2)
= 80T - 160 - J? (T - 4T + 4).
A 160=&(4T-4) = 16(4T-4).
A T =3 seconds.
Also the height at which they meet = SOT- yi*
=280- 196=84 feet
The first stone will be coming down and the second stone going
upwards.
EXAMPLES. Tx vil
1. A body is projected from the earth vertically with a velocity of
40 feet per second ; find (1) how high it will go before coming to rest,
(2) what times will elapse before it is at a height of 9 feet.
2. A particle is projeoted vertically upwards with a velooity of
40 feet per second. Find (i) when its velocity will be 25 feet per
second, and (ii) when it will be 25 feet above the point of pro-
jection.
3. A stone is thrown vertically upwards with a velocity of 60 feet
per second. After what times will its velooity be 20 feet per second,
and at what height will it then be ?
XXVII. ACCELERATION. 133
4. Find (1) the distance fallen from rest by a body in 10 seconds,
(2) the time of falling 10 feet, (3) the initial vertical velocity when the
body describes 1000 feet downwards in 10 seconds.
& A stone is thrown vertically into a mine-shaft with a velocity
of 96 feet per second, and reaches the bottom in 3 seconds ; find the
depth of the shaft.
' 6. A body is projected from the bottom of a mine, whose depth
is 88 g feet, with a velocity of 24 g feet per second ; find the time in
which the body, after rising to its greatest height, will return to the
surface of the earth again.
7. The greatest height attained by a particle projected vertically
npwards is 225 feet ; find how soon after projection the particle will
be at a height of 176 feet. .
8. A body moving in a vertical direction passes a point at a
height of 54-5 centimetres with a velocity of 436 centimetres per
second; with what initial velocity was it thrown up, and for how
much longer will it rise?
9. A particle passes a given point moving downwards with a
velocity of fifty metres per second ; how long before this was it moving
upwards at the same rate ?
10. A body is projected vertically upwards with a velocity of
6540 centimetres per second ; how high does it rise, and for how long
is it moving upwards ?
11. Given that a body falling freely passes through 176*99 feet in
the sixth second, find the value of g. *
12. A falling particle in the last second of its fall passes through
224 feel Find the height from which it fell, and the time of its
falling.
13. A body falls freely from the top of a tower, and during the
last second of its flight falls |f ths of the whole distance. Find the
height of the tower.
14. Assuming the acceleration of a falling body at the surface of
the moon to be one-sixth of its value on the earth's surface, find the
height to which a particle will rise if it be projected vertically upward
from the surface of the moon with a velocity of 40 feet per second.
15. A stone A is thrown vertically upwards with a velocity of
96 feet per second ; find how high it will rise. After 4 seconds from
the projection of A , another stone B is let fall from the same point.
Shew that A will overtake B after 4 seconds more.
16. A body is projected upwards with a certain velocity, and it is
found that when in its ascent it is 960 feet from the ground it takes
4 seconds to return to the same point again ; find the velocity of pro-
jection and the whole height ascended.
17. A body projected vertically downwards desoribed 720 feet in
t seconds, and 2240 feet in 2t seconds ; find t, and the velocity of pro-
jection.
134 DYNAMICS. Exs. XXVII.
18, A stone is dropped into a well, and the sound of the splash is
heard in 7& seconds ; if the velocity of sound be 1120 feet per second,
find the depth of the well.
19, A stone is dropped into a well and reaches the bottom with a
velocity of 96 feet per second, and the sound of the splash on the water
reaches the top of the well in 3^ seconds from the time the stone
starts ; find the velocity of sound.
20, From a balloon, ascending with a velocity of 32 ft. per second,
a stone is let fall and reaches the ground in 17 seconds ; how high
was the balloon when the stone was dropped?
21, A balloon has been ascending vertically at a uniform rate
for 4-5 seconds and a stone let fall from it reaches the ground in
7 seconds. Find the velocity of the balloon and the height from
which the stone fell.
22, If a body be let fall from a height of 64 feet at the same
instant that another is sent vertically from the foot of the height
with a velocity of 64 feet per second, what time elapses before they
meet?
If the first body starts 1 Bee. later than the other, what time will
elapse?
CHAPTER XIII.
THE LAWS OF MOTION.
168. In the first chapter we stated that the mass of
a body was the quantity of matter in the body.
Matter is " that which can be perceived by the senses "
or " that which can be acted upon by, or can exert, force."
No definition can however be given that would convey
an idea of what matter is to anyone who did not already
possess that idea. It, like time and space, is a primary
conception.
If we have a small portion of any substance, say iron,
resting on a smooth table, we may by a push be able to
move it fairly easily ; if we take a larger quantity of the
same iron, the same effort on our part will be able to move
it less easily. Again, if we take two portions of platinum
and wood of exactly the same size and shape, the effect
produced on these two substances by equal efforts on our
part will be quite different. Thus common experience
shews us that the same effort applied to different bodies,
under seemingly the same conditions, does not always pro-
duce the same result. This is because the masses of the
bodies are different.
169. The British unit of mass is called the Imperial
Pound, and consists of a lump of platinum deposited in the
Exchequer Office, of which there are in addition several
accurate copies kept in other places of safety.
The French, or scientino, unit of mass is called a gramme, and is
the one-thousandth part of a certain quantity of platinum deposited in
the Archives. The gramme was meant to be denned as the mass of a
cubic centimetre of pure water at a temperature of 4 0.
It is a much smaller unit than a Pound.
One Gramme= about 15*432 grains.
One Pound m about 453-6 grammes.
136 DYNAMICS.
The system of units in which a centimetre, gramme, and second,
are respectively the units of length, masB, and time, is generally called
the o.o.s. system of units.
170. The Momentum, or Quantity of Motion, of a
body is equal to the product of the mass and the velocity
of the body. Thus mv is the momentum of a body whose
mass is m and which moves with velocity v.
171. We shall now enunciate what are commonly
called Newton's Laws of Motion.
They are ;
Law I. Every body continues in its state of rest, or of
uniform motion in a straight line, except in so far as it be
compelled by external impressed force to change that state.
Law II. The rate of change of momentum is propor-
tional to the impressed force, and takes place in the direction
of the straight line in which the force acts.
Law III. To every action there is an equal and opposite
reaction.
No formal proof, experimental or otherwise, can be given
of these three laws. On them however is based the whole
system of Dynamics, and on Dynamics the whole theory of
Astronomy. Now the results obtained, and the predictions
made, from the theory of Astronomy agree so well with the
actual observed facts of Astronomy that it is inconceivable
that the fundamental laws on which the subject is based
should be erroneous. For example, the Nautical Almanac
is published four years beforehand, and the predictions in
it are always correct. Hence we believe in the truth of the
above three laws of motion because the conclusions drawn
from them agree with our experience.
172. Law I. We never see this law practically ex-
emplified in nature because it is impossible ever to get
rid of all forces during the motion of the body. It may
be seen approximately in operation in the case of a piece
of dry, hard ice projected along the surface of dry, well
swept, ice. The only forces acting on the fragment of ice,
in the direction of its motion, are the friction between the
two portions of ice and the resistance of the air. The
THE LAWS OF MOTION. 137
smoother the surface of the ice the further the small
portion will go, and the less the resistance of the air the
further it will go. The above law asserts that if the ice
were perfectly smooth and If there were no resistance of
the air and no other forces acting on the body, then it
would go on for ever in a straight line with uniform
velocity.
The law states a principle sometimes called the Prin-
ciple of Inertia, viz. that a body has no innate ten-
dency to change its state of rest or of uniform motion in
a straight line. If a portion of metal attached to a piece
of string be swung round on a smooth horizontal table,
then, if the string break, the metal, having no longer any
force acting on it, proceeds to move in a straight line, viz.
the tangent to the circle at the point at which its circular
motion ceased.
If a man step out of a rapidly moving train he is
generally thrown to the ground ; his feet on touching the
ground are brought to rest; but, as no force acts on the
upper part of his body, it continues its motion as before,
and the man falls to the ground.
If a man be riding on a horse which is galloping at a
fairly rapid pace and the horse suddenly stops, the rider is
in danger of being thrown over the horse's head.
If a man be seated upon the back seat of a dog-cart,
and the latter suddenly start, the man is very likely to be
left behind.
173. Law II. From this law we derive our method
of measuring force.
Let m be the mass of a body, and f the acceleration
produced in it by the action of a force whose measure
is P.
Then, by the second law of motion,
Pec rate of change of momentum,
oo rate of change of mv,
oo m x rate of change of v (m being unaltered),
cc m.f.
\ P= k . m/j where X is some constant.
138 DYNAMICS.
Now let the unit . of force be so chosen that it may
produce in unit mass the unit of acceleration.
Hence, when m= 1 andy= 1, we have P 1,
and therefore X= 1.
The unit of force being thus chosen, we have
P = m.f.
Therefore, when proper units are chosen, the measure
of the force is equal to the measure of the rate of change
of the momentum.
174. From the preceding article it follows that the
magnitude of the .unit of force used in Dynamics depends
on the units of mass, and acceleration, that we use. The
unit of acceleration, again, depends, by Arts. 151 and 154,
on the units of length and time. Hence the unit of force
depends on our units of mass, length, and time. When
these latter units are given the unit of force is a deter-
minate quantity.
When a pound, a foot, and a second are respectively the
units of mass, length, and time, the corresponding unit of
force is called a Poundal.
Hence the equation P = mf is a true relation,
m being the number of pounds in the body, P the
number of poundals in the force acting on it, and
f the number of units of acceleration produced in
the mass m by the action of the force P on it.
This relation is sometimes expressed in the form
. ,. Moving Force
Acceleration = -^r= - j- .
Mass moved
N.B. All through this book the unit of force used will
be a poundal, unless it is otherwise stated. Thus, when we
say that the tension of a string is T t we mean T poundals.
# 176. When a gramme, a centimetre, and a second are respec-
tively the units of mass, length, and time the corresponding unit of
force is called a Dyne.
Hence when the equation P=mf is used in this system the force
must be expressed in dynes, the mass in grammes, and the acceleration
in centimetre-second units.
THE LAWS OF MOTION. 139
176. Connection between the unit of force
and the weight of the unit of mass. As explained in
Art. 161, we know that, when a body drops freely in vacuo,
it moves with an acceleration which we denote by "g";
also the force which causes this acceleration is what we call
its weight.
Now the unit of force acting on the unit of mass pro-
duces in it the unit of acceleration.
Therefore g units of force acting on the unit of mass
produce in it g units of acceleration (by the second law).
But the weight of the unit of mass is that whicji pro-
duces in it g units of acceleration.
Hence the weight of the unit of mass = g units of force.
177. FootrPound-Second System of units. In this sys-
tem g is equal to 32*2 approximately.
Therefore the weight of one pound is equal to g units of
force, i.e. to g . poundals, where g=32'2 approximately.
Hence a poundal is approximately equal to ^r-jr times
the weight of a pound, i.e. to about the weight of half an
ounce.
Since g has different values at different points of the
earth's surface, and since a poundal is a force which is the
same everywhere, it follows that the weight of a pound
is not constant, but has different values at dif-
ferent points of the earth's surface.
#178. Gentimetre-Chramme-Seeond System of units. In this sys-
tem g is equal to 981 approximately.
Therefore the weight of one gramme is eqnal to g units of force, i.e.
to g . dynes, where gss 981 approximately.
Hence a dyne is equal to the weight of about 3^ of a gramme.
The dyne is a muoh smaller unit than a poundal. The ap-
proximate relation between them may be easily found as follows :
One Poundal _ 32^2 wt - 0fftp0PDd
One Dyne 1 _. .
r^ wt. of a gramme
981 one pound 981 Aet% a mmM
= o n -a x = sir* x 453 *6 (by Art. 169).
32*2 one gramme 82*2 x J '
Henoe One Poundal = about 18800 dynes.
HO DYNAMICS, Exs,
Ex. 1. A mass of 20 pounds is acted on by a constant force which
in 5 seconds produces a velocity of 15 feet per second. Find the force,
if the mass was initially at rest.
From the equation v=u+ft, we have /=^=8.
Also, if P be the force expressed in poundals, we have
P = 20 x 3 = 60 poundals.
Hence P is equal to the weight of about |$, i.e. 1 J, pounds.
Ex. 2. A mass of 10 pounds is placed on a smooth horizontal
plane, and is acted on by a force equal to the weight of 3 pounds ;
find the distance described by it in 10 seconds.
Here moving force = weight of 3 lbs. = 3^ poundals j
and mass moved = 10 pounds.
Hence, using ft. -sec. units, the acceleration =t?
80 that the distance required = . j| . 10'= 480 feet.
Ex. 3. Find the magnitude of the force which, acting on a kilo-
gramme for 5 seconds, produces in it a velocity of one metre per
second.
Here the velocity acquired = 100 oms. per sec.
Hence the acceleration = 20 o. a. e. units.
1000 x 20
Hence the force = 1000 x 20 dynes = weight of about ^ or
20*4 grammes.
EXAMPLES. XXVIII.
1. Find the acceleration produced when
(1) A force of 5 poundals acts on a mass of 10 pounds.
(2) A force equal to the weight of 5 pounds acts on a mass of
10 pounds.
(3) A force of 50 pounds weight acts on a mass of 10 tons.
2. Find the force expressed (1) in poundals, (2) in terms of the
weight of a pound, that will produce in a mass of 20 pounds an
acceleration of 10 foot- second units. .
3. Find the force whioh, acting horizontally for 5 seconds on a
mass of 160 pounds placed on a smooth table, will generate in it a
velocity of 15 feet per second.
4. Find the magnitude of the force whioh, acting on a mass of
10 cwt. for 10 seconds, will generate in it a velocity of 3 miles per
hour.
5. A force, equal to the weight of 2 lbs., acts on a mass of 40 lbs.
for half a minute ; find the velooity acquired, and the spaoe moved
through, in this time.
XXVIII. TH&LA tf& OF MOTION. 141
6. A body, acted upon by a uniform force, in ten seconds describes
a distance of 25 feet ; compare the force with the weight of the body,
and find the velocity acquired.
7. In what time will a force, which is equal to the weight of a
pound, move a mass of 18 lbs. through 50 feet along a smooth
horizontal plane, and what will be the velocity acquired by the
mass?
8. A body, of mass 200 tons, is acted on by a force equal to w
112000 poundals ; how long will it take to acquire a velocity of 30 miles C^
per hour?
9. In what time will a force, equal to the weight of 10 lbs., acting
on a mass of 1 ton move it thungh 14 feet? v
10. A mass of 224 lbs. ft^laced on a smooth horizontal plane,
and a uniform pressure acting on it parallel to the table for 5 seconds
causes it to describe 50 feet in that time ; shew that the pressure is
equal to about 28 lbs. weight.
11. A heavy truck, of mass 16 tons, is standing at rest on a smooth
line of rails. A horse now pulls at it steadily in the direction of the
line of rails with a force equal to the weight of 1 owt. How far
will it move in 1 minute ?
12. A 80-ton mass is moving on smooth horizontal rails at the
rate of 20 miles per hour ; what force would stop it in (1) half a
minute, and (2) in half a mile.
13. A force equal to the weight of 10 grammes acts on a mass
of 27 grammes for 1 second ; find the velocity of the mass and the
distance it has travelled over. At the end of the first second the
force ceases to act ; how far will the body travel in the next minute ?
14. A pressure equal to the weight of a kilogramme acts on a
body continuously for 10 seconds, and causes it to describe 10 metres
in that time ; find the mass of the body.
15. A mass which starts from rest is acted upon by a force
which in a-foth of a second communicates to it a velocity of 3 miles
per hour ; find the ratio of the force to the weight of the mass.
16. A horizontal pressure equal to the weight of 9 lbs. acts on a
mass along a smooth horizontal plane ; after moving through a space
of 25 feet the mass has acquired a velocity of 10 feet per second ; find
its magnitude.
17. A body is placed on a smooth table and a pressure equal to
the weight of 6 lbs. acts continuously on it ; at tbe end of 3 seconds
the body is moving at the rate of 48 feet per second ; find its mass.
18. A body, of mass 3 lbs., is falling under gravity at the rate of
100 feet per second. What is the uniform force that will stop it (1) in
2 seconds, (2) in 2 feet ?
19. Of two forces, one acts on a mass of 5 lbs. and in one-eleventh
of a second produces in it a velocity of 5 feet per second, and the other
acting on a mass of 625 lbs. in 1 minute produces in it a velocity of
18 miles per hour; compare the two forces.
142 DYNAMICS. Exs. XXVIII
20. -A- mass of 10 lbs. falls 10 feet from rest, and is then brought
to rest by penetrating 1 foot into some sand ; find the average pressure
of the sand on it.
21. A ballet moving at the rate of 200 feet per second is fired into
a trunk of wood into which it penetrates 9 inches ; if a bullet moving
with the same velocity were fired into a similar piece of wood 5 inches
thick, with what velocity would it emerge, supposing the resistance to
be uniform?
179. A poundal and a dyne are called Absolute Units
because their values are not dependent on the value of g,
which varies at different places on the earth's surface. The
weight of a pound and of a gramme do depend on this
value. Hence they are called Gravitation Units.
180. The weight of a body is proportional to its mass
and is independent of the kind of matter of which it is com-
posed. The following is an experimental fact : If we have
an air-tight receiver, and if we allow to drop at the same
instant, from the same height, portions of matter of any
kind whatever, such as a piece of metal, a feather, a piece
of paper etc., all these substances will be found to have
always fallen through the same distance, and to hit the base
of the receiver at the same time, whatever be the sub-
stances, or the height from which they are allowed to fall.
Since these bodies always fall through the same height in
the same time, therefore their velocities [rates of change of
space,] and their accelerations [rates of change of velocity,]
must be always the same.
The student can approximately perform the above experiment
without creating a vacuum. Take a penny and a light substance,
say a small piece of paper; place the paper on the penny, held
horizontally, and allow both to drop. They will be found to keep
together in their fall, although, if they be dropped separately, the
penny will reach the ground much quicker than the paper. The
penny dears the air out of the way of the paper and so the same
result is produced as would be the ease if there were no air.
Let W x and W % poundals be the weights of any two of
these bodies, m and m* their masses. Then since their
accelerations are the same and equal to </, we have
W 1 = m l g )
and IT, = rr^g ;
/. W x ; r, :: m, : m*
or the weight of a body is proportional to its mass.
THE LAWS OF MOTION. 143
Hence bodies whose weights are equal have equal
masses; so also the ratio of the masses of two bodies is
known when the ratio of their weights is known.
N.B. The equation W=mg is a numerical one, and
means that the number of units of force in the weight of
a body is equal to the product of the number of> units of
mass in the mass of the body, and the number of units of
acceleration produced in the body by its weight.
181. Distinction between mass and weight. The student mast
carefully notice the difference between the mass and the weight of a
body. He has probably been so accustomed to estimate the masses of
bodies by means of their weights that he has not clearly distinguished
between the two. If it were possible to have a cannon-ball at the
centre of the earth it would have no weight there ; for the attraction
of the earth on a particle at its centre is zero. If, however, it were in
motion, the same force would be required to stop it as would be
necessary under similar conditions at the surface of the earth. Hence
we see that it might be possible for a body to have no weight ; its mass
however remains unaltered.
The confusion is probably to a great extent caused by the fact that
the word "pound" is used in two senses which are scientifically
different ; it is used to denote both what we more properly call " the
mass of one pound" and "the weight of one pound." It cannot be
too strongly impressed on the student that, strictly speaking, a pound
is a mass and a mass only ; when we wish to speak of the force with
which the earth attracts this mass we ought to speak of the " weight
of a pound." This latter phrase is often shortened into " a pound,"
but care must be taken to see in which sense this word is used.
It may also be noted here that the expression "a ball of lead
weighing 20 lbs." is, strictly speaking, an abbreviation for " a ball of
lead whose weight is equal to the weight of 20 lbs." The mass of the
lead is 20 lbs. ; its weight is 20g poundals.
182. Weighing by Scales and a Spring Balance. We have pointed
out (Art. 161) that the acceleration due to gravity, i.e. the value of g t
varies slightly as we proceed from point to point of the earth's surface.
When we weigh out a substance (say tea) by means of a pair of scales,
we adjust the tea until the weight of the tea is the same as the weight
of sundry pieces of metal whose masses are known, and then, by
Art. 180, we know that the mass of the tea is the same as the mass of
the metal. Hence a pair of scales really measures masses and not
weights, and so the apparent weight of the tea is the same every-
where.
When we use a spring balance, we compare the weight of the tea
with the force necessary to keep the spring stretched through a certain
distance. If then we move our tea and spring balance to another
place, say from London to Paris, the weight of the tea will be different,
whilst the force necessary to keep the spring stretohed through the
144 DYNAMICS.
same distance as before will be the same. Hence the weight of the
tea will pull the spring through a distance different from the former
distance, and hence its apparent weight as shewn by the instrument
will be different.
If we have two places, A and B, at the first of which the numerical
value of g is greater than at the second, then a given mass of tea will
[as tested by the spring balance,] appear to weigh more at A than it
does at B.
Ex. 1. At the equator the value of g is 32*09 and in London the
value is 32*2; a merchant buys tea at the equator, at a shilling
per pound, and sells in London ; at what price per pound (apparent)
must he sell so that he may neither gain nor lose, if he use the same
spring balance for both transactions ?
A quantity of tea which weighs 1 lb. at the equator will appear
32*2 32*2
to weigh lbs. in London. Hence he should sell 5^- lbs. for
3209
one shilling, or at the rate of ^7^ shillings per pound.
Ex. a. At a place A, g = Z2-2i, and at a place B, g-32- 12. A
merchant buys goods at 10 per cwt. at A and sells at B, using the
same spring balance. If he is to neither gain nor lose, shew that his
selling price must be 10. 0*. 9(2. per cwt. nearly.
183. Law III. To every action there is an equal and
opposite reaction.
Every exertion of force consists of a mutual action
between two bodies. This mutual action is called the stress
between the two bodies, so that the Action and Reaction
of Newton together form the Stress.
UlTistratlons. 1, If a book rest on a table, the book presses the
table with a force equal and opposite to that which the table exerts on
the book.
2. If a man raise a weight by means of a string tied to it,
the string exerts on the man's hand exactly the same force that it
exerts on the weight, but in the opposite direction.
3. The attraction of the earth on a body is its weight, and
the body attracts the earth with a force equal and opposite to
its own weight.
4. When a horse drags a canal-boat by means of a rope, the
rope drags the horse back with a force equal to that with which
it drags the boat forward. [The weight of the rope is neglected.]
[The horse begins to move because the force he exerts is greater
than the force that the rope exerts on him ; the boat begins to move
because the force exerted by the rope on it is greater than the
resistance that the water offers to its motion. In other words, at the
beginning of the motion the force exerted by the horse > the tension
of the rope > the resistance of the water.]
CHAPTER XIV.
y&
WS OF MOTION (CONTINUED). APPLICATION TO
SIMPLE PROBLEMS.
184. Motion of two particles connected by
a string.
Two particles, of masses m^ and m,, are connected by a
light mextensible string which passes over a
small smooth pulley. If my be-> m?, find the r\
resulting motion of the system^ and the tension
of the string.
Let the tension of the string be T
poundals; the pulley being smooth, this will
be the same throughout the string.
Since the string is inextensible, the velo-
city of wij upwards must, throughout the
motion, be the same as that of my downwards.
Hence their accelerations [rates of change
of velocity] must be the same in magnitude.
Let the magnitude of the common accelera-
tion be /.
Now the force on my downwards is m^g T poundals.
Hence myg-T=mJ (1).
So the force on m, upwards is T m^g poundals ;
/. T-m t g = m x f (2).
T.
aT
i \7fl.
Adding (1) and (2), we gave f= * ^
common acceleration.
Also, from (2), T^m* (f+g)
Im^m^
9, giving
the
g poundals.
my + mi * r
Since the acceleration is known and constant, the
equations of Art 156 give the space moved through and
the velocity acquired in any given time.
I* M. H. 10
146 DYNAMICS.
185. Two particles, of masses m^ and m^, are connected
by a light inextensible string; m^ m 2 j
is placed on a smooth horizontal
table and the string passes over
the edge of the table, mj hanging
freely ; find the resulting motion.
Let the tension of the string be T poundals.
The velocity and acceleration of m a along the table must
be equal to the velocity and acceleration of m^ in a vertical
direction.
Let f be the common acceleration of the masses
The force on m x downward is
rrhg-T;
:. m.g-T^mJ (1)
The only horizontal force acting on m 2 is the tension T;
[for the weight of m^ is balanced by the reaction of the
table].
.'. T^mJ (2)
Adding (1) and (2), we have
m } g = (m 1 + m 2 )/
" J m^ + rn^ 9 '
giving the required acceleration.
Hence, from (2)
body whose mass is
ical
Hence, from (2), T g poundals = weight of a
m 1 m a
mj + m,'
180. Ex. Two particles, of masses 11 and 13 lbs., are connected
by a light string passing over a small smooth pulley. Find (1) the
velocity at the end of 4 seconds, and (2) the space described in 4 seconds.
If at the end of 4 seconds the string be cut, find the distance described
by each in the next 2 seconds.
If T poundals be the tension of the string and / the common
acceleration, we have lSq-T=lSf
and T- llg =11/.
Hence, by addition /=tL
Exs. XXIX. THE LA WS OF MOTION, 147
4x 32
The common velocity at the end of 4 seconds = 4/= = 10| feet
per second.
The space described in this time ./.4 s
-8xg=H|feet
If the string be now ont the larger mass starts downward with
velocity 10| and acceleration g\ the smaller starts upward with
velocity 10$ and with an acceleration -g.
The spaoe described by the larger mass in the next 2 seconds
= lO$x2 + i.0.2 2 =21i + 64=85*feet.
The space described by the smaller mass
= lOf x 2 - g . 2 2 =21 - 64= - 42$ feet.
In these two seconds the upward velocity of the smaller mass has
been destroyed and it has fallen to a point which is 42| feet below the
point at which it was when the string was cut.
EXAMPLES. y*TT,
1. A mass of 9 lbs., descending vertically, drags up a mass of
6 lbs. by means of a string passing over a smooth pulley ; rind the
acceleration of the system and the tension of the string.
2. Two particles, of masses 7 and 9. lbs., are connected by a light
string passing over a smooth pulley. Find (1) their common accelera-
tion, (2) the tension of the string, (3) the velocity at the end of
5 seconds, (4) the distance described in 5 seconds.
3. Masses of 14 and 18 ounces are connected by a thread passing
over a light pulley ; how far do they go in the first 3 seconds of the
motion, and what is the tension of the string?
4. To the two ends of a light string passing over a small smooth
pulley are attached masses of 977 grammes and x grammes ; find x so
that the former mass may rise through 200 centimetres in 10 seconds.
5. Two masses of 50 and 70 grammes are fastened to the ends of
a cord passing over a frictionless pulley supported by a hook. When
they are free to move, shew that the pull on the hook is equal to
116 grammes' weight.
6. Two equal masses, of 3 lbs. each, are connected by a light
string hanging over a smooth peg ; if a third mass of 3 lbs. be laid on
one of them, by how much is the pressure on the peg increased ?
7. Two masses, each equal to m, are connected by a string passing
over a smooth pulley ; what mass must be taken from one and added
to the other, so that the system may describe 200 feet in 5 seconds ?
102
148 DYNAMICS. Exs. XXIX.
8. A mass of 8 lbs., descending vertically, draws up a mass of
2 lbs. by means of a light string passing over a pulley ; at the end of
5 seconds the string breaks ; find how much higher the 2 lb. mass
will go.
9. A body, of mass 9 lbs., is plaoed on a smooth table at a
distance of 8 feet from its edge, and is connected, by a string passing
over the edge, with a body of mass 1 lb. ; find
(1) the common acceleration,
(2) the time that elapses before the body reaches the edge of
the table,
and (3) its velocity on leaving the table.
10. A mass of 19 ounces is placed on a smooth table and
connected by a light string passing over the edge of the table with a
mass of 5 ounces which hangs vertically ; find the acceleration of the
masses and the tension of the string.
11. A mass of 70 lbs. is placed on a smooth table at a distance of
8 feet from its edge and connected by a light string passing over the
edge with a mass of 10 lbs. hanging freely; what time will elapse
before the first mass will leave the table ?
12. A mass of 100 grammes is attached by a string passing over
a smooth pulley to a larger mass ; find the magnitude of the latter,
so that if after the motion has continued 3 seconds the string be cut,
the former will ascend 54*5 centimetres before descending.
13. Two scale-pans, of mass 3 lbs. each, are connected by a string
passing over a smooth pulley ; shew how to divide a mass of 12 lbs.
between the two scale-pans bo that the heavier may descend a distance
of 50 feet in the first 5 seconds.
14. Two strings pass over a smooth pulley ; on one side they are
attached to masses of 3 and 4 lbs. respectively, and on the other to
one of 5 lbs. ; find the tensions of the strings and the acceleration of
the system.
15. A string hung over a pulley has at one end a weight of 10 lbs.
and at the other end weights of 8 and 4 lbs. respectively ; after being
in motion for 5 seconds the 4 lb. weight is taken off ; find how much
further the weights go before they first come to rest.
187. Motion down a smooth inclined plane.
Let a be the inclination of the plane to the horizon. If
a particle be sliding down the plane the only forces acting
on it are its weight mg vertically downwards and the
normal reaction of the plana
The weight mg may be resolved (as in Art. 27) into
mg sin a down the plane and mg cos a perpendicular to the
plane. The latter force is balanced by the normal reaction,
THE LAWS OF MOTION. 149
and the former produces the acceleration / down the
plane.
Hence mg sin a = mf
The acceleration of the particle down the plane is
therefore g sin a.
The velocity of the particle after it has described a
length I of the plane = J2g sin a . I = J2gh, where h is the
height of the plane, and is therefore the same as that of a
particle which has dropped vertically through a distance
equal to the height of the plane.
188. Two masses, m^ and m^, are connected by a string;
wij is placed on a smooth plane inclined at an angle a to the
horizon, and the string, after passing
over a small smooth pulley at the top
of the plane, supports m^, which hangs
vertically ; if m^ descend, Jmd the re-
sulting motion.
Let the tension of the string be
T poundals. The velocity and accele- ^ m
ration of m, up the plane are clearly equal to the velocity
and acceleration of m^ vertically.
Let f be this common acceleration. For the motion of
m x , we have
m^-T=mJ (1).
The weight of m a is rn^g vertically downwards.
The resolved part of rn^g perpendicular to the inclined
plane is balanced by the reaction R of the plane, since m t
has no acceleration perpendicular to the plane.
The resolved part of the weight down the inclined plane
is m^g sin a, and hence the total force up the plane is
Tm^g sin a.
Hence T-m^g sin a^m^f (2).
Adding (1) and (2), we easily have
-_m 1 -m, sin a
rn^-t m^
Also, substituting in (1),
150 DYNAMICS.
T = m x (g-f) = rr h g]\ -
rn^m.2 (1 + sin a)
mj + fw,
g poundals,
]
giving the tension of the string.
189. Motion on a rough plane. A particle slides
down a rough inclined plane inclined to the horizon at an
angle a; if fx be the coefficient o//riction } to determine the
motion.
Let m be the mass of the particle, so that its weight is
mg poundals ; let R be the reaction, and jjlR the friction.
The total force perpendicular to the plane is
(R - mg cos a) poundals.
The total force down the plane is (mg sin a fxR)
poundals.
Now perpendicular to the plane there cannot be any
motion, and hence there is no change of motion.
Hence the acceleration, and therefore the total force, in
that direction is zero.
.'. R mg cos a = (1).
Also the acceleration down the plane
moving force ma sin a - uR . . x , *
= 5- = = g (sin a - u cos a), by (1).
mass moved m x n '
Hence the velocity of the particle after it has moved
THE LAWS OF MOTION. 151
from rest over a length I of the plane is, by Art. 158, equal
to J%gl (sin a ft cos a).
Similarly, if the particle were projected up the plane,
we have to change the sign of //,, and its acceleration in
a direction opposite to that of its motion is
g (sin a + /* cos a).
190. A train, of mass 50 tons, is ascending an incline of
1 in 100; the engine exerts a constant tractive force equal to
the weight of 1 ton, and the resistance due to friction etc. may
be taken at 8 lbs. weight per ton ; find the acceleration with
which the train ascends the incline.
The train is retarded by the resolved part of its weight
down the incline, and by the resistance of friction.
The latter is equal to 8 x 50 or 400 lbs. wt.
The incline is at an angle a to the horizon, where
sinc^y^.
The resolved part of the weight down the incline there-
fore
= ^8^(1 = 50x2240x3^108. wt.
= 1120 lbs. wt.
Hence the total force to retard the train = 1520 lbs. wt.
But the engine pulls with a force equal to 2240 lbs.
weight.
Therefore the total force to increase the speed equals
(2240-1520) or 720 lbs. weight, i.e. 720g poundals.
Also the mass moved is 50 x 2240 lbs.
Hence the acceleration = ^ ^Trm
50x2240
ft. -sec. units.
1400
Since the acceleration is known, we can, by Art. 156,
find the velocity acquired, and the space described, in a
given time, etc.
EXAMPLES. TTT
1, A body is projeoted with a velocity of 80 feet per second up
a smooth inclined plane, whose inclination is 30; find the space
described, and the tune that elapses, before it oomes to rest.
152 DYNAMICS. Em.
2. A heavy particle slides from rest down a smooth inclined
plane which is 15 feet long and 12 feet high. What is its velocity
when it reaches the ground, and how long does it take ?
3. A particle sliding down a smooth plane, 16 feet long, acquires
a velocity of 16^/2 feet per second ; find the inclination of the plane.
4. What is the ratio of the height to the length of a smooth
inclined plane, so that a body may be four times as long in sliding
down the plane as in falling freely down the height of the plane start-
ing from rest ?
5. A heavy body slides from rest down a smooth plane inclined
at 30 to the horizon. How many seconds will it be in sliding
240 feet down the plane and what will be its velocity when it has
described this distance?
6. A partiole slides without friction down an inclined plane, and
in the 5th second after starting passes over a distance of 2207-25 centi-
metres ; find the inclination of the plane to the horizon.
7. A particle, of mass 5 lbs., is placed on a smooth plane inclined
at 30 to the horizon and connected by a string passing over the top
of the plane with a particle of mass 3 lbs. which hangs vertically ;
find (1) the common acceleration, (2) the tension of the string, (3) the
velocity at the end of 3 seconds, (4) the space described in 3 seconds.
8. A body, of mass 12 lbs., is placed on an inclined plane, whose
height is half its length, and is connected by a light string passing
over a pulley at the top of the plane with a mass of 8 lbs. which
hangs freely ; find the distance described by the masses in 5 seconds.
9. A mass of 6 ounces slides down a smooth inclined plane
whose height is half its length and draws another mass from rest over
a distance of 3 feet in 5 seconds along a horizontal table which is
level with the top of the plane over which the string passes ; find the
mass on the table.
10. If a train of 200 tons, moving at the rate of 30 miles per
hour, can be stopped in 60 yards, compare the friction with the weight
of a ton.
11. A train is running on horizontal rails at the rate of 30 miles
per hour, the resistance due to friction, eto. being 10 lbs. wt. per ton ;
if the steam be shut off, find (1) the time that elapses before the train
comes to rest, (2) the distance described in this time.
12. In the previous question if the train be ascending an incline
of 1 in 112, find the corresponding time and distance.
13. A train of mass 200 tons is running at the rate of 40 miles
per hour down an incline of 1 in 120 ; find the resistance necessary to
stop it in half a mile.
14. A train runs from rest for 1 mile down a plane whose descent
is 1 foot vertically for each 100 feet of its length ; if the resistances
be equal to 8 lbs. per ton, how far will the train be carried along the
horizontal level at the foot of the incline?
XXX. THE LAWS OF MOTION. 153
15. A train of mass 140 tons, travelling at the rate of 15 miles
per hoar, comes to the top of an incline of 1 in 128, the length of the
incline being half a mile, and steam is then shnt off; taking the
resistance due to friction, etc. as 10 lbs. wt. per ton, find the distance
it describes on a horizontal line at the foot of the incline before
coming to rest.
16. A mass of 5 lbs. on a rough horizontal table is connected by a
string with a mass of 8 lbs. which hangs over the edge of the table ; if
the coefficient of friction be J, find the resultant acceleration.
Find also the coefficient of friction if the acceleration be half that
of a freely falling body.
17. A mass of 20 lbs. is moved along a rough horizontal table by
means of a string which is attached to a mass of 4 lbs. hanging over
the edge of the table ; if the masses take twice the time to acquire the
same velocity from rest that they do when the table is smooth, find the
coefficient of friction.
18. A body, of mass 10 lbs., is placed on a rough plane, whose
coefficient of friction is -7= and whose inclination to the horizon is
30 ; if the length of the plane be 4 feet and the body be acted on by a
force, parallel to the plane, equal to 15 lbs. weight, find the time that
elapses before it reaches the top of the plane and its velocity there.
19. If in the previous question the body be connected with a
mass of 15 lbs., hanging freely, by means of a string passing over the
top of the plane, find the time and velocity.
20. A particle slides down a rough inclined plane, whose inclina-
tion to the horizon is 45 and whose coefficient of friction is j ; shew
that the time of descending any space is twice what it would be if
the plane were perfectly smooth.
191. A body, of mass m lbs., is placed on a horizontal
plane which is in motion with a vertical wpward accelera-
tion/; find the presswre between the body and the plane.
Let R be the pressure between
the body and the plane. ir
Since the acceleration is ver-
tically upwards, the total force
acting on the body must be ver- .
tically upwards.
The only force, besides R,
acting on the body is its weight wm?
mg acting vertically downwards.
Hence the total force is R mg vertically upwards, and
154 DYNAMICS.
this produces an acceleration/; hence
R rng = mf, giving R.
In a similar manner it may be shewn that, if the body
be moving with a downward acceleration f, the pressure R y
is given by
mg R x = mf.
We note that the pressure is greater or less than the
weight of the body, according as the acceleration of the
body is upwards or downwarda
Ex. 1. The body is of mass 20 lbs. and is moving with (1) an
upward acceleration of 12 ft.-se.c. units, (2) a downward acceleration of
the same magnitude ; find the pressures.
In the first case we have
.8-20.0=20.12.
:. 12=20 (32 + 12) poundals= wt. of 27 lbs.
In the second case we have
20.0 -1^ = 20. 12.
.-. J2=20 (32 - 12) ponndals= wt. of 12 lbs.
Ex. a. Two scale-pans, each of mass 3 ounces, are connected by
a light string passing over a smooth pulley. If masses of 4 and 6
ounces respectively be placed in the pans, find the pressures on the pans
during the. subsequent motion.
On one side the total mass will be 9 ounces and on the other side
9 7 a
7 ounces. Hence, by Art. 184, the acceleration /= ^ = #=| .
9 + 7 o
Let P poundals be the pressure on the 4 oz. mass. The total
4
force on this 4oz. mass therefore=P-^0 poundals upwards,
so that p-*_ g= *;f ;
4 4 9
/. p =iQ( a +f)=lQ'Q9= weight of 4$ ounces.
If F poundals be the pressure on the other mass, the total foroe
on it is jg g - P f downwards,
le'-^ie'-
6 7
,\ P = jg . g g = 5 ounces weigh*.
THE LAWS OF MOTION.
155
192. Atwood's Machine. This machine in its sim-
plest form consists of a vertical pillar
AE, of about 8 feet in height, firmly
clamped to the ground, and carrying
at its top a light pulley which will
move very freely. This pillar is gradu-
ated and carries two platforms, D and
F, and a ring E, all of which can be
affixed by screws at any height de-
sired. The platform D can also be
instantaneously dropped. Over the
pulley passes a fine cord supporting at
its ends two long thin equal weights,
one of which, P y can freely pass through
the ring F. Another small weight Q
is provided, which can be laid upon
the weight jP, but which cannot pass
through the ring E.
The weight Q is laid upon P and
the platform D is dropped and motion
ensues; the weight Q is left behind
as the weight P passes through the ring; the weight P
then traverses the distance EF with constant velocity, and
the time T which it takes to describe this distance is care-
fully measured.
By Art. 184 the acceleration of the system as the weight
falls from D to E is
(Q + P)-P . Q
(Q + P) + P g ' % '*' Q + 2P 9 '
Denote this by/, and let DE=h.
Then the velocity v on arriving at E is given by
After passing E % the distance EF is described with
constant velocity v.
Hence, if EF= h^ we have
*-&
v =
Q + 2P
ghTK
156 DYNAMICS.
Since all the quantities involved are known, this relation
gives us the value of g.
By giving different values to P, Q, h and Aj, we can
in this manner verify all the fundamental laws of motion.
In practice, the value of g cannot by this method be
found to any great degree of accuracy ; the chief causes of
discrepancy being the mass of the pulley, which cannot be
neglected, the friction of the pivot on which the wheel
turns, and the resistance of the air.
The friction of the pivot may be minimised if its ends
do not rest on fixed supports, but on the circumferences of
four light wheels, called friction wheels, two on each side,
which turn very freely.
There are other pieces of apparatus for securing the
accuracy of the experiment as far as possible, e.g. for
instantaneously withdrawing the platform D at the re-
quired moment, and a clock for beating seconds accu-
rately.
193. By using Atwood's machine to shew that the acceleration of a
given mass is proportional to the force acting on it.
We shall assume that the statement is true and see whether the
results we deduce therefrom are verified by experiment.
To explain the method of procedure we shall take a numerical
example.
Let P be 49 ozs. and Q 1 oz. so that the mass moved is 100 ozs.
and the moving force is the weight of 1 oz.
The acceleration of the system therefore =ihi9 (Art 184 )-
Let the distance DE be one foot so that the velooity when Q is
taken off = a/ 2 . ~ . 1= A ** P 61 Beo - a i for simplicity, we take g
equal to 32.
Let the platform F be carefully placed at such a point that the
mass will move from E to F in some definite time, say 2 sees.
Then EF=& . 2=| feet.
Now alter the conditions. Make P equal to 48 and Q equal to 4 ozs.
The mass moved is still 100 ozs. and the moving force is now the
weight of 4 ozs.
The acceleration is now ~ and the velocity at E
-V
4(7
2 . ~ . l=f ft. per second.
In 2 seconds the mass would now describe V ieet > B0 '^at, if our
Exs-XXXI. THE LAWS OF MOTION. 157
hypothesis be eorrect, the platform F must be twice as far from E as
before. This is found on trial to be eorrect.
Similarly if we make P=45 ozs. and Q=9 ozs., so that the mass
moved is still 100 ozs., the theory would give us that EF should be
^ feet, and this would be found to be correct.
The experiment should now be tried over again ab initio and P and
Q be given different values from the above ; alterations should then be
made in their values so that 2P+ Q is constant.
By the tame method to shew that the force varies as the mass when
the acceleration is constant.
As before let P=49 ozs. and Q-l oz. so that, as in the last
experiment, we have EF= $ feet.
Secondly, let P=99 ozs. and Q 2 ozs., so that the moving force
is doubled and the mass moved is doubled. Hence, if our enunciation
be correct, the acceleration should be the same, since
second moving force _ first moving force ~
second mass moved ,. first mass moved
The distance EF moved through in 2 seconds should therefore be
the same as before, and this, on trial, is found to be the case.
Similarly if we make P=148 ozs. and <? = 3 ozs. the same result
would be found to follow.
EXAMPLES. XXXL
1. If I jump off a table with a twenty-pound weight in my hand,
what is the pressure of the weight on my hand ?
2. A mass of 20 lbs. rests on a horizontal plane which is made to
ascend (1) with a constant velocity of 1 foot per second, (2) with a
constant acceleration of 1 foot per second per second ; find in eaoh
case the pressure on the plane.
3. A man, whose mass is 8 stone, stands on a lift which moves
with a uniform acceleration of 12 ft. -sec. units; find the pressure
on the floor when the lift is (1) ascending, (2) descending.
4. A bucket containing 1 owt. of coal is drawn up the shaft of a
coal-pit, and the pressure of the coal on the bottom of the bucket
is equal to the weight of 126 lbs. Find the acceleration of the
bucket.
5. A balloon ascends with a uniformly accelerated velocity, so
that a mass of 1 cwt. produces on the floor of the balloon the same
pressure which 116 lbs. would produce on the earth's surface; find the
height which the balloon will have attained in one minute from the
time of starting.
6. Two scale-pans, each of mass 2 ounces, are suspended by a
weightless string passing over a smooth pulley ; a mass of 10 ounces
is placed in the one, and 4 ounces in the other. Find the tension of
the string and the pressures on the scale-pans.
158 DYNAMICS. Exs. XXXI.
7. A string, passing over a smooth pulley, supports two scale-
pans at its ends, the mass of each scale-pan being 1 ounce. If masses
of 2 and 4 ounces respectively be placed in the scale-pans, find the
acceleration of the system, the tension of the string, and the pressures
between the masses and the scale-pans.
8. The two masses in an Atwood's machine are each 240 grammes,
and an additional mass of 10 grammes being placed on one of them
it is observed to descend through 10 metres in 10 seconds; hence
shew that #=980.
9. Explain how to use Atwood's machine to shew that a body
acted on by a constant force moves with constant acceleration.
CHAPTER XV.
IMPULSE, WORK, AND ENERGY.
194. Impulse. Def. The impulse of a force in a
given time is equal to the product of the force (%f constant,
s and the mean value of the force if variable) and the time
divring which it acts.
The impulse of a force P acting for a time t is therefore
P.t.
The impulse of a force is also equal to the momentum
generated by the force in the given time. For suppose a
particle, of mass m, moving initially with velocity u is
acted on by a constant force P for time t. If / be the
resulting acceleration, we have P = mf
But, if v be the velocity of the particle at the end of
time t, we have v = u +ft.
Hence the impulse =Pt = mfl = mv mu
the momentum generated in the given time.
The same result is also true if the force be variable.
Hence it follows that the second law of motion might
have been enunciated in the following form ;
The change of momentum of a particle in a
given time is equal to the impulse of the force
which produces it and is in the same direction.
195. Impulsive Forces. Suppose we have a force
P acting for a time t on a body whose mass is m, and let
the velocities of the mass at the beginning and end of this
time be u and v. Then by the last article
Pt = m (v - u).
Let now the force become bigger and bigger, and the time
t smaUer and smaller. Then ultimately P will be almost
infinitely big and t almost infinitely small, and yet their
product may be finite. For example P may be equal to
160 DYNAMICS.
10 7 poundals, t equal to j~ seconds, and m equal to one
pound, in which case the change of velocity produced is
the unit of velocity.
To find the whole effect of a finite force acting for a
finite time we have to find two things, (1) the change in
the velocity of the particle produced by the force during
the time it acts, and (2) the change in the position of
the particle during this time. Now in the case of an
infinitely large force acting for an infinitely short time,
the body moves only a very short distance whilst the force
is acting, so that this change of position of the particle
may be neglected. Hence the total effect of such a force
is known when we know the change of momentum which
it produces.
Such a force is called an impulsive force. Hence
Def. An impulsive force is a very great force acting
for a very short time, so that the change in the position oj
the particle during the time the force acts on it may be
neglected. Its whole effect is measured by its impulse, or
the change of momentum produced.
In actual practice we never have any experience of an
infinitely great force acting for an infinitely short time.
Approximate examples are, however, the blow of a hammer,
and the collision of two billiard balls.
The above will be true even if the force be not uniform.
In the ordinary case of the collision of two billiard balls
the force generally varies very considerably.
Ex. 1. A body, whose mass is 9 lbs., is acted on by a force which
changes its velocity from 20 miles per hour to 30 miles per hour. Find
the impulse of the force.
Ans. 132 units of impulse.
Ex. 2. A mass of 2 lbs. at rest is struck and starts off with a
velocity of 10 feet per second ; assuming the time during which the
blow lasts to be tW> nn d the average value of the force acting on the
Am. 2000 poundals.
Ex. 3. A glass marble, whose mass is 1 ounce, falls from a height
of 25 feet, and rebounds to a height of 16 feet ; find the impulse, and
the average force between the marble and the floor if the time during
which they are in contact be ^".
Ans. 4Jj units of impulse ; 47 poundals.
IMPULSE, WORK, AND ENERGY. 161
196. Impact of two bodies. When two masses
A and B impinge, then, by the third law of motion, the
action of A on B is, at each instant during which they are
in contact, equal and opposite to that of B on A.
Hence the impulse of the action of A on B is equal and
opposite to the impulse of the action of B on A.
It follows that the change in the momentum of B is
equal and opposite to the change in the momentum of A,
and therefore the sum of these changes, measured in the
same direction, is zero.
Hence the sum of the momenta of the two masses,
measured in the same direction, is unaltered by their
impact.
Ex. 1. A body, of mass 3 lbs. , moving with velocity 13 feet per
second overtakes a body, of mass 2 lbs., moving with velocity 3 feet per
second in the same straight line, and they coalesce and form one body ;
find the velocity of this single body.
Let V be the required velocity. Then, since the sum of the momenta
of the two bodies is unaltered by the impact, we have
<3 + 2)F=3xl3 + 2x3 = 45 units of momentum.
,% F=9 ft. per sec.
fix. a. If in the last example the second body be moving in the
direction opposite to that of the first, find the resulting velocity.
In this case the momentum of the first body is represented by 3 x 13
and that of the second by -2x3. Hence, if V x be the required
velocity, we have
(3 + 2) V X =Z x 13 - 2 x 3=33 units of momentum.
V x =*i = ty ft. per seo.
197. Motion of a shot and gun. When a gun is
fired, the powder is almost instantaneously converted into
a gas at a very high pressure, which by its expansion
forces the shot out. The action of the gas is similar to
that of a compressed spring trying to recover its natural
position. The force exerted on the shot forwards is, at any
instant before the shot leaves the gun, equal and opposite
to that exerted on the gun backwards, and therefore the
impulse of this force on the shot is equal and opposite to
the impulse of the force on the gun. Hence the momen-
tum generated in the shot is equal and opposite to that
generated in the gun, if the latter be free to move.
L. M. EL 11
162 DYNAMICS. Exs. XXXII.
Bx. A shot, whose matt it 400 Ibt., it projected from a gun, of ma*t
60 tont, with a velocity of 900 feet per tecond ; find the retulting velo-
city of the gun.
Since the momentum of the gun is equal and opposite to that of
the shot we have, if v be the velocity communicated to the gun,
50 x 2240 xt> = 400x900.
/. v=8^f ft. per 8eo.
EXAMPLES. XXXH
1. A body of mass 7 lbs., moving with a velocity of 10 feet per
second, overtakes a body, of mass 20 lbs., moving with a velocity of
2 feet per second in the same direotion as the first ; if after the impact
they move forward with a oommon velocity, find its magnitude.
2. A body, of mass 8 lbs., moving with a velocity of 6 feet per
second overtakes a body, of mass 24 lbs., moving with a velocity of
2 feet per second in the same direction as the first ; if after the impact
they coalesce into one body, shew that the velocity of the compound
body is 3 feet per second.
If they were moving in opposite directions, shew that after impact
the compound body is at rest.
3. A body, of mass 10 lbs., moving with velocity 4 feet per second
meets a body of mass 12 lbs. moving in the opposite direction with a
velocity of 7 feet per second ; if they coalesce into one body, shew that
it will have a velocity of 2 feet per second in the direction in which the
larger body was originally moving.
4. A shot, of mass 1 ounce, is projected with a velocity of 1000 feet
per second from a gun of mass 10 lbs. ; find the velocity with which
the latter begins to recoil.
5. A shot of 800 lbs. is projected from a 40-ton gun with a velocity
of 2000 feet per second ; find the velocity with which the gun would
commence to recoil, if free to move in the line of projection.
6. A shot, of mass 700 lbs., is fired with a velocity of 1700 feet per
second from a gun of mass 38 tons ; if the recoil be resisted by a
constant pressure equal to the weight of 17 tons, through how many
feet will the gun recoil ?
7. A gun, of mass 1 ton, fires a shot of mass 28 lbs. and recoils
up a smooth inclined plane, rising to a height of 5 feet ; find the initial
velocity of the projectile.
8. A hammer, of mass 4 owt., falls through 4 feet and comes to
rest after striking a mass of iron, the duration of the blow being ^th
of a second ; find the pressure, supposing it to be uniform, which is
exerted by the hammer on the iron.
9. Masses m and 2m are connected by a string passing over a
smooth pulley; at the end of 8 seconds a mass m is picked up by
the ascending body; find the resulting motion.
IMPULSE, WORK, AND ENERGY. 163
198. Work. We have pointed out in Art. 136, that
the unit of work used by engineers is a Foot-Pound, 'which
is the work done in raising the weight of one pound through
one foot.
The British absolute unit of work is the work done
by a poundal in moving its point of application through
one foot.
This unit of work is called a Foot-Poundal.
With this unit of work the work done by a force of P
poundals in moving its point of application through 8 feet
is P. 8 foot-poundals.
Since the weight of a pound is equal to ^-poundals, it
follows that a Foot-Pound is equal to g Foot-Poundals.
The c.g.b. unit of work is that done by a dyne in moving its point
of application through a centimetre, and is called an Erg.
One Foot-Poundal =421390 Ergs nearly.
199. Bx 1. What is the H.P. of an engine which keeps a train, of
mass 150 tons, moving at a uniform rate of 60 miles per hour, the resist-
ances to the motion due to friction, the resistance of the air, etc. being
taken at 10 lbs. weight per ton.
The force to stop the train is equal to the weight of 150 x 10, i.e.
1500, lbs. weight.
Now 60 miles per hour is equal to 88 feet per second.
Hence a force, equal to 1500 lbs. wt., has its point of application
moved through 88 feet in a second, and hence the work done is
1500x88 foot-pounds per second. -
If x be the h.p. of the engine, the work it does per minute is
x x 33000 foot-lbs., and hence the work per second is * x 550 foot-lbs.
/. a? x 650= 1500x88.
.-. *=240.
Ex. 2. Find the least H.P. of an engine which is able in 4 minutes
to generate in a train, of mass 100 tons, a velocity of 30 miles per hour
on a level line, the resistances due to friction, etc. being equal to 8 lbs.
weight per ton, and the putt of the engine being assumed constant.
Since in 240 seconds a velocity of 44 feet per second is generated
44 11
the acceleration of the train must be ^ or ^ foot-second units.
Let the force exerted by the engine be P poundals.
The resistance due to friction is equal to 800 pounds' weight;
hence the total force on the train is P - SOOg poundals.
11
Hence P - 8Q0g = 100 x 2940 x
112
164 DYNAMICS. Exs. XXXTH
P=800 (9+^) poundals=800(l + g^^ lbs. weight
=800 xH? lbs. weight.
4o
When the train is moving at the rate of SO miles per hour, the wort
125
done per second must be 800 x -j^- x 44 foot-lbs.
Hence, if * be the h.p. of the engine, we have
ax 650=800 x^x 44.
.-. *:166.
EXAMPLES. "x"x*y m,
1. A train, of mass 50 tons, is kept moving at the uniform rate of
30 miles per hour on the level, the resistance of air, friction, eto.,
being 40 lbs. weight per ton. Find the h.p. of the engine.
2. What is the horse-power of an engine which keeps a train
going at the rate of 40 miles per hour against a resistance equal
to 2000 lbs. weight ?
3. A train, of mass 100 tons, travels at 40 miles per hour up an
incline of 1 in 200. Find the h.p. of the engine that will draw the
train, neglecting all resistances except that of gravity.
4. A train of mass 200 tons, including the engine, is drawn
up an incline of 3 in 500 at the rate of 40 miles per hour by an engine
of 600 h.p.; find the resistance per ton due co friction, etc.
5. Find the h.p. of an engine which can travel at the rate of
25 miles per hour up an incline of 1 in 100, the mass of the engine
and load being 10 tons, and the resistances due to friction, eto. being
10 lbs. weight per ton.
6. Determine the rate in h.p. at which an engine must be able to
work in order to generate a velocity of 20 miles per hour on the level
in a train of mass 60 tons in 3 minutes after starting, the resistances
to the motion being taken at 10 lbs. per ton and the force exerted by
the train being assumed to be constant.
7. Find the work done by gravity on a stone having a mass of
lb. during the tenth second of its fall from rest.
*200. Energy. Def. The Energy of a body is its
capacity for doing work and is of two kinds, Kinetic and
Potential.
The Kinetic Energy of a body is the energy which it
possesses by virtue of its motion, and is measured by the
amount of work that the body can perform against the im-
pressed forces before its velocity is destroyed.
IMPULSE, WORK, AND ENERGY. 165
A falling body, a swinging pendulum, and a cannon-
ball in motion all possess kinetic energy.
Consider the case of a particle, of mass m, moving with
velocity w, and let us find the work done by it before it
comes to rest.
Suppose it brought to rest by a constant force P re-
sisting its motion, which produces in it an acceleration f
given by P = mf.
Let x be the space described by the particle before it
comes to rest, so that = u % + 2 (-/) . x.
:.fx = lu\
Hence the kinetic energy of the particle
= work done by it before it comes to rest
* Px = mfx = \mu % .
Hence the kinetic energy of a particle is equal to the product
of its mass and one half the square of its velocity.
*201. Theorem. To shev) that the change of kinetic
energy per unit of space is equal to the acting force.
If a force P, acting on a particle of mass m, change its
velocity from u to v in time t whilst the particle moves
through a space s, we have v* u % 2fs, where f is the
acceleration produced.
.-. kant^m* =m/=P ( i).
This equation proves the proposition when the force is
constant.
Cor. It follows from equation (1) that the change in
the kinetic energy of a particle is equal to the work done
on it.
202 . The Potential Energy of a body is the work it
can do by means of its position in passing from its present
configuration to some standard configuration (usually called
its zero position).
A bent spring has potential energy, viz. the work it
can do in recovering its natural shape. A body raised to
a height above the ground has potential energy, viz. the
work its weight can do as it falls to the earth's surface.
166 DYNAMICS.
Compressed air has potential energy, viz. the work it can
do in expanding to the volume it would occupy in the
atmosphere.
The following example is important :
*203. A particle of mass m falls from rest at a height h
above the ground ; to shew that the sum of its potential and
kinetic energies is constant throughout the motion.
Let H be the point from which the particle starts, and
the point where it reaches the ground.
Let v be its velocity when it has fallen through a
distance HP (= x\ so that i^ = 2gx.
Its kinetic energy at P = \mv* = rngx.
Also its potential energy at P
= the work its weight can do as it falls from P to
= mg . OP mg (h x).
Hence the sum of its kinetic and potential energies at P
=*mgh.
But its potential energy when at H is mgh, and its kinetic
energy there is zero.
Hence the sum of the potential and kinetic energies is
the same at P as at H ; and, since P is any point, it follows
that the sum of these two quantities is the same throughout
the motion.
#204. The statement proved in the previous article,
viz., that the sum of the kinetic and potential energies is
constant throughout the motion, is found to be true for all
cases of motion where there is no friction or resistance of
the air, nor any impacts. This is an elementary illustration
of the Principle known as that of the Conservation of Energy,
which states that Energy is indestructible. It maybe changed
into different forms but can never be destroyed. When a
body slides along a rough plane some of its mechanical
energy becomes transformed and reappears in the form of
heat partly in the moving body and partly in the plane.
Ex. A bullet, of mass 4 ozs., is fired into a target vrith a velocity
of 1200 feet per second. The mass of the target is 20 lbs. and it is free
to move : find the loss of kinetic energy in foot-pounds.
Exs. XXXIV. IMPULSE, WORK, AND ENERGY. 167
Let V be the resulting common velocity of the shot and target.
Since no momentum is lost (Art. 196) we have
( 20 4) F 4 xim
' r ~ 27 *
The original kinetic energy = ^ jg 1200 a = 180000 foot-poundals.
The final kinetio energy = ^ ( 20 + Ta ) v%
20000 ,. , ,
= q foot-ponndals.
m i i -cmn 20 1600000, . ..
The energy lost a 180000 5 = 5 foot-poundali
=^? ft ..ib..
It will be noted that, in this ease, although no momentum is lost
80
by the impact, yet ^r- ths. of the energy is destroyed.
EXAMPLES. XXXIV.
1. A body, of mass 10 lbs., is thrown up vertically with a velocity
of 32 feet per second ; what is its kinetic energy (1) at the moment of
propulsion, (2) after half a second, (3) after one second?
2. Find the kinetic energy measured in foot-pounds of a cannon-
ball of mass 25 pounds discharged with a velocity of 200 feet per
second.
3. Find the kinetic energy in ergs of a cannon-ball of 10000
grammes discharged with a velocity of 5000 centimetres per second.
4. What is the horse power of an engine that can project 10000
lbs. of water per minute with a velocity of 80 feet per second, twenty
per cent, of the whole work being wasted by friction ?
5. A cannon-ball, of mass 5000 grammes, is discharged with a
velocity of 500 metres per second. Find its kinetic energy in ergs, and,
if the cannon be free to move, and have a mass of 100 kilogrammes,
find the energy of the recoil.
6. A bullet, of mass 2 ounces, is fired into a target with a velocity
of 1280 feet per second. The mass of the target is 10 lbs. and it is
free to move ; find the loss of kinetic energy by the impact in foot-
pounds.
7. Equal forces act for the same time upon unequal masses M
and m ; what is the relation between (1) the momenta generated by
the ioroes and (2) the amounts of work done by them.
CHAPTER XVI.
COMPOSITION OF VELOCITIES AND ACCELERATIONS.
PROJECTILES.
205. Since the velocity of a point is known when its
direction and magnitude are both known, we can con-
veniently represent the velocity of a moving point by a
straight line AB ; thus, when we say that the velocities of
two moving points are represented in magnitude and
direction by the straight lines AB and CD, we mean that
they move in directions parallel to the lines drawn from
A to B y and C to D respectively, and with velocities which
are proportional to the lengths AB and CD.
206. A body may have simultaneously velocities in
two, or more, different directions. One of the simplest
examples of this is when a person walks on the deck of a
moving ship from one point of the deck to another. He
has a motion with the ship, and one along the deck of
the ship, and his motion in space is clearly different from
what it would have been had either the ship remained at
rest, or had the man stayed at his original position on the
deck.
Again, consider the case of a ship steaming with its bow
pointing in a constant direction, say due north, whilst a
current carries it in a different direction, say south-east,
and suppose a sailor is climbing a vertical mast of the
ship. The actual change of position and the velocity of
the sailor clearly depend on three quantities, viz., the rate
and direction of the ship's sailing, the rate and direction of
the current, and the rate at which he climbs the mast.
His actual velocity is said to be "compounded" of these
three velocities.
In the following article we shew how to find the
velocity which is equivalent to two velocities given in
magnitude and direction.
VELOCITY. 169
207. Theorem. Parallelogram of Velocities.
If a moving point possess simultaneously velocities which
are represented in magnitude and direction by the two sides
of a parallelogram drawn from a point, they are equi-
valent to a velocity which is represented in magnitude and
direction by the diagonal of the parallelogram passing
through the point.
Let the two simultaneous velo-
cities be represented by the lines
AB and AC, and let their magni-
tudes be u and v.
Complete the parallelogram
BACD.
Then we may imagine the motion of the point to be
along the line AB with the velocity u, whilst the line AB
moves parallel to the foot of the page so that its end A
describes the line AC with velocity v. In the unit of time
the moving point will have moved through a distance AB
along the line AB, and the line AB will have in the same
time moved into the position CD, so that at the end of the
unit of time the moving point will be at D.
Now, since the two coexistent velocities are constant
in magnitude and direction, the velocity of the point from
A to D must also be constant in magnitude and direction ;
hence AD is the path described by the moving point in
the unit of time.
Hence AD represents in magnitude and direction the
velocity which is equivalent to the velocities represented
by AB and AC.
To facilitate his understanding of the previous article, the student
may look on AC as the direction of motion of a steamer, whilst AB is
a chalked line, drawn along the deok of the ship, along which a man
is walking at a uniform rate.
208. Def. The velocity which is equivalent to two or
more velocities is called their resultant and these velocities
are called the components of this resultant.
Since the resultant of two velocities are found in the
same way as the resultant of two forces, it can be shewn
similarly as in Art. 25 that the resultant of two velocities u
170 DYNAMICS.
and v acting at an angle a is
Ju 2 + ^ + 2uv cos a.
209. A velocity can be resolved into two component
velocities in an infinite number of ways. For an infinite
number of parallelograms can be described having a given
line AD as diagonal ; and, if ABDG be any one of these,
the velocity AD is equivalent to the two component veloci-
ties AB and AC.
The most important case is when a velocity is to be
resolved into two velocities in two directions at right angles,
one of these directions being given. When we speak of
the component of a velocity in a given direction it is under-
stood that the other direction in which the given velocity
is to be resolved is perpendicular to this given direction.
Thus, suppose we wish to resolve a velocity u, repre-
sented by AD, into two components q q
at right angles to one another, one
of these components being along a
line AB making an angle with
AD.
Draw DB perpendicular to AB,
and complete the rectangle ABDC. A B
Then the velocity AD is equivalent to the two com-
ponent velocities AB and AC.
Also AB = AD gq&$ = u cos 6,
and AC = BD = AD BmO = uan$.
We thus have the following important
Theorem. A velocity u is equivalent to a velocity
u cos 6 along a line making an angle 6 with its own direction
together with a velocity u sin $ perpendicular to the direction
of the fir st component.
The case in which the angle 6 is greater than a right
angle may be considered as in Art. 27.
Ex. 1. A man is walking in a north-easterly direction with a
velocity of 4 miles per hour ; find the components of his velocity in
directions dne north and due east respectively.
Ana. Each is "ij'i miles per hoar.
VELOCITY. 171
Bx. 3. A point is moving in a straight line with a velocity of
10 feet per second ; find the component of its velocity in a direction
inclined at an angle of 30 to its direction of motion.
Ana. 5^/3 feet per second.
Bx. 8. A body is sliding down an inclined plane whose inclina-
tion to the horizontal is 60 ; find the components of its velocity in
the horizontal and vertical directions.
u kV3
Ana. j: and u ~ , where u is the velocity of the body.
A A
210. Triangle of Velocities. If a moving point
possess simultaneously velocities represented by the two sides
AB and BC of a triangle taken in order, they are equivalent
to a velocity represented by AC.
For, completing the parallelogram ABCD, the lines AB
and BC represent the same velocities as AB and AD and
hence have as their resultant the velocity represented by AC.
Cor. If there be simultaneously impressed on a point three
velocities represented by the sides of a triangle taken in order, the
point will be at rest.
211. The resultant of two more velocities is generally
most conveniently found by resolving along two directions
at right angles. The method is the same as that for forces
in Art. 37.
Ex. 1. A vessel steams with its bow pointed due north with a velo-
city of 15 miles an hour, and is carried by a current which flows in a
south-easterly direction at the rate of 3^/2 miles per hour. At the end
of an hour find its distance and bearing from the point from which it
started.
The ship has two velocities, one being 15 miles per hoar north-
wards, and the other 3 y'2 miles per hoar south-east.
Now the latter velocity is equivalent to
3 J2 cos 45, that is, 3 miles per hoar eastward,
and 8 J2 sin 45, that is, 3 miles per hour southward.
Hence the total velocity of the ship is 12 miles per hoar northwards
and 3 miles per hour eastward.
Hence its resultant velocity is ^/l2*+3 a or ^153 miles per hour
in a direction inclined at an angle, whose tangent is \, to the north,
i.e., 12-37 miles per hour at 14 2' east of north.
Ex. a. A point possesses simultaneously velocities whose measures
are 4, 3, 2 and 1 ; the angle between the first and second is 30, between
the second and third 90, and between the third and fourth 120 ; find
thrtr resultant.
172 DYNAMICS. Exs.
Take OX along 4h8 direction of the first velocity and OY perpen-
dicular to it.
The angles which the velocities make with OX are respectively
0, 30, 120, and 240.
Hence, if V be the resultant velocity inclined at an angle $ to OX,
we have
V cos = 4 + 3 cos 30 + 2 cos 120 + 1 . cos 240,
and V sin 0= 3 sin 30 + 2 sin 120 + 1 . sin 240.
We therefore have
r.^ + *.f + ,(-J) + i(-|)-|^,
and r 8 i=S.^2.f-l.f = 3 ^?.
Hence, by squaring and adding,
F a =16 + V3,
3+ /3
and, by division, tan d = g v . =2^/3- 8.
Hence the resultant is a velocity ^16 + 9^3 inclined at an angle
whose tangent is (2^3 - 3), i.e., 5*62 at 24 54', to the direction of the
first velocity.
EXAMPLES. XXXV.
1. The velocity of a ship is 8^ miles per hour, and a ball is
bowled aoross the ship perpendicular to the direction of the ship
with a velocity of 3 yards per second ; describe the path of the ball
in space and shew that it passes over 45 feet in 3 seconds.
2. A boat is rowed with a velocity of 6 miles per hour straight
across a river which flows at the rate of 2 miles per hour. If its
breadth be 300 feet, find how far down the river the boat will reach the
opposite bank below the point at which it was originally directed.
3. A man wishes to cross a river to an exactly opposite point on
the other bank ; if he can pull his boat with twice the velocity of the
current, find at what inclination to the current he must keep the
boat pointed.
4. A boat is rowed on a river so that its speed in still water
would be 6 miles per hour. If the river flow at the rate of 4 miles
per hour, draw a figure to shew the direction in which the head of the
boat must point so that the motion of the boat may be at right
angles to the current.
5. A stream runs with a velocity of 1 miles per hour ; find in
what direction a swimmer, whose velocity is 2 miles per hour, should
start in order to cross the stream perpendicularly.
What direction should be taken in order to cross in the shortest
timet
XXXV. VELOCITY. 173
6. A ship is steaming in a direction due north across a current
running due west. At the end of one hour it is found that the ship
has made 8 s /3 miles in a direction 30 west of north. Find the
velocity of the current, and the rate at which the ship is steaming.
7. A ship is sailing north at the rate of 4 feet per second ; the
current is taking it east at the rate of 3 feet per second, and a sailor
is climbing a vertical pole at the rate of 2 feet per second ; find the
velocity and direction of the sailor in space.
8. A point which possesses velocities represented by 7, 8, and 13
is at rest; find the angle between the directions of the two smaller
velocities.
9. A point possesses velocities represented by 3, 19, and 9
inclined at angles of 120 to one another; find their resultant.
10. A point possesses simultaneously velocities represented by u,
2u, S,JSu, and 4u ; the angles between the first and second, the second
and third, and the third and fourth, are respectively 60, 90, and 150;
shew that the resultant is u in a direction inclined at an angle of 120
to that of the first velocity.
212. Change of Velocity. Suppose a point at any
instant to be moving with a
velocity represented by OA,
and that at some subsequent
time its velocity is represented
by OB.
Join AB, and complete the
parallelogram OABC.
Then velocities represented
by OA and OC are equivalent to the velocity OB. Hence
the velocity 00 is the velocity which must be compounded
with OA to produce the velocity OB. The velocity OC is
therefore the change of velocity in the given time.
Thus the change of velocity is not, in general, the
difference in magnitude between the magnitudes of the
two velocities, but is that velocity which compounded with
the original velocity gives the final velocity.
The change of velocity is not constant unless the change
is constant both in magnitude and direction,
EXAMPLES. XXXVL
1. A point is moving with a velocity of 10 feet per second, and
at a subsequent instant it is moving at the same rate in a direction
inclined at 30 to the former direction ; find the change of velocity.
174
DYNAMICS.
Ejb. XXXIV.
On drawing the figure, as in the last article, we have OA = OB = 10,
and the angle AOB = SQ.
Since OA-OB, we have/ 0-4.8= 75, and therefore L A OG =105.
The change in the velocity, i.e., 0(7,
=AB= Jl0*+l(P-2. 10. 10cos80=5 ^8-4^3=5 (V6-V),
and is in a direotion inclined at 105 to the original direction of
motion.
2. A point is moving with a velocity of 5 feet per second, and at
a subsequent instant it is moving at the same rate in a direction
inclined at 60 to its former direction ; find the change of velocity.
3. A point is moving eastward with a velocity of 20 feet per
second, and one hour afterwards it is moving north-east with the same
speed ; find the change of velocity.
4. A point is describing with uniform speed a circle, of radius
7 yards, in 11 seconds, starting from the end of a fixed diameter ; find
the change in its velocity after it has described one-sixth of the cir-
cumference.
213. Theorem. Parallelogram of Accelera-
tions. If a moving point have simultaneously two accelera-
tions represented in magnitude and direction by two sides of
a parallelogram drawn from a point, they are equivalent to
an acceleration represented by the diagonal of the parallelo-
gram passing through that angular point.
Let the accelerations be represented by the sides A B and
AG of the parallelogram ABDG, i.e. let AB and AG repre-
sent the velocities added to the velocity of the point in a
unit of time. On the same scale let EF represent the
velocity which the particle has at any instant. Draw the
parallelogram EKFL having its sides parallel to AB and
ACCELERATION. 175
AC; produce EK to M, and EL to N, so that Z2f and
LN are equal to -45 and AC respectively. Complete the
parallelograms as in the above figure.
Then the velocity EF is equivalent to velocities EK and
EL. But in the unit of time the velocities KM and LN are
the changes of velocity.
Therefore at the end of a unit of time the component
velocities are equivalent to EM and EN, which are equi-
valent to EO, and this latter velocity is equivalent to velo-
cities EF and FO. (Art. 209.)
Hence in the unit of time FO is the change of velocity
of the moving point, i.e. FO is the resultant acceleration of
the point.
But FO is equal and parallel to AD.
Hence AD represents the acceleration which is equi-
valent to the accelerations AB and AC, i.e. AD is the
resultant of the accelerations AB and AC.
Hence accelerations are resolved and compounded in
the same way as velocities.
214. Parallelogram of Forces. We have shewn
in the last article that if a particle of mass m have accele-
rations f x and/, represented in magnitude and direction by
lines AB and AC, then its resultant acceleration f % is repre-
sented in magnitude and direction by AD, the diagonal of
the parallelogram of which AB and AC are adjacent sides.
Since the particle has an acceleration f x in the direction
AB there must be a force F x (= m/j) in that direction, and
similarly a force P t (=#/,) in the direction AC. Let
AB X and AC X represent these forces in magnitude and
176 DYNAMICS.
direction. Complete the parallelogram AB X D X C X . Then
since the forces in the directions AB X and AC X are propor-
tional to the accelerations in those directions,
.*. AB X :AB:: B X D X : B>.
Hence, by simple geometry, we have A, D and D x in a
straight line, and
AB X :AB:: AC, : AC
:'.B X D X :BD.
Hence AD X represents the force which produces the
acceleration represented by AD, and hence is the force
which is equivalent to the forces represented by AB X and
AC X .
Hence we infer the truth of the Parallelogram of Forces.
215. Physical Independence of Forces. The
latter part of the Second Law of Motion states that the
change of motion produced by a force is in the direction
in which the force acts.
Suppose we have a particle in motion in the direction
AB and a force acting on it in the direction AC; then
the law states that the velocity in the direction AB is
unchanged, and that the only change of velocity is in the
direction AC; so that to find the real velocity of the
particle at the end of a unit of time, we must compound
its velocity in the direction AB with the velocity generated
in that unit of time by the force in the direction AC.
The same reasoning would hold if we had a second force
acting on the particle in some other direction, and so for
any system of forces. Hence if a set of forces act on a
particle at rest, or in motion, their combined effect is
found by considering the effect of each force on the particle
just as if the other forces did not exist, and as if the particle
were at rest, and then compounding these effects. This
principle is often referred to as that of the Physical Inde-
pendence of Forces.
As an illustration of this principle consider the motion of a ball
allowed to fall from the hand of a passenger in a train which is
travelling rapidly. It will be found to hit the floor of the carriage at
exactly the same spot as it would have done if the carriage had been
at rest. This shews that the ball must have continued to move
PROJECTILES.
177
forward with the same velocity that the train had, or, in other words,
the weight of the body only altered the motion in the vertical direction,
and had no influence on the horizontal velocity of the particle.
Again, a circus-rider, who wishes to jump through a hoop, springs
in a vertical direction from the horse's back ; his horizontal velocity
is the same as that of the horse and remains unaltered ; he therefore
alights on the horse's back at the spot from which he star led.
Projectile!.
216. By the use of the principle of the last article we
can determine the motion of a particle which is thrown into
the air, not necessarily in a vertical line, but in any
direction whatever.
Let the particle be projected from a point P with velocity
u in a direction making an angle a with the horizon ; also
let PAP' be the path of the particle, A being the highest
point, and P the point where the path again meets the hori-
zontal plane through P. The distance PP is called the
range on the horizontal plane through P,
Now the weight of the body only has effect on the
motion of the body in the vertical direction; it therefore
has no effect on the velocity of the body in the horizontal
direction, and this horizontal velocity therefore remains un-
altered (since the resistance of the air is disregarded).
The horizontal and vertical components of the initial
velocity of the particle are u cos a and u sin a respectively.
The horizontal velocity is, therefore, throughout the
motion equal to u cos a.
In the vertical direction the initial velocity is u sin a
and the acceleration is g, [for the acceleration due to
L. VL U.
12
178 DYNAMICS.
gravity is g vertically downwards, and we are measuring
our positive direction upwards]. Hence the vertical motion
is the same as that of a particle projected vertically up-
wards with velocity u sin a, and moving with acceleration
-9-
The resultant motion of the particle is the same as that
of a particle projected with a vertical velocity u sin a inside
a vertical tube of small bore, whilst the tube moves in a
horizontal direction with velocity u cos a.
The time of flight, t, i.e. the time the projectile is in
the air, is therefore twice the time in which a vertical
velocity u sin a is destroyed by g, i.e. t = 2 . Also the
__ . . _ nrtl ft t* s sinacosa
horizontal range PF = u cos a x t = 2 .
9
217. Bz. 1. A cannon ball is 'projected horizontally from the
top of a tower \ 49 feet high, with a velocity of 200 feet per second. Find
(1) the time of flight,
(2) the distance from the foot of the tower of the point at which it
hits the ground, and
(8) its velocity when it hits the ground.
(1) The initial vertical velocity of the ball is zero, and hence t,
the time of flight, is the time in which a body, falling freely under
gravity, would describe 49 feet.
Hence, by Art. 156, 49=& . i*=16*.
.. t=.\ second.
(2) During this time the horizontal velocity remains constant,
and therefore the required distance from the foot of the tower
= 200x=350 feet.
(3) The vertical velocity at the end of { second = J x 32=66 feet
per second, and the horizontal velocity is 200 feet per second.
Hence the required velocity = V 2 a + 66 *= 8>/674= 207-7 feet
nearly.
Ex. 8. From the top of a cliff, 80 feet high, a stone is thrown so
that it starts with a velocity of 128 feet per second, at an angle of 30
with the horizon ; find where it hits the ground at the bottom of the
cliff.
The initial vertical velocity is 128 sin 30, or 64, feet per second,
and the initial horizontal velocity is 128 cos 30, or 64^3, feet per
second.
Let T be the time that elapses before the stone hits the ground.
Exs. XXXVII. PROJECTILES, 179
Then T is the time in which a stone, projected with vertical
velocity 64 and moving with acceleration - g, describes a distance
- 80 feet.
,\ -8O=64T-i0T*.
Hence T=5 seconds.
During this time the horizontal velocity remains unaltered, and
hence the distance of the point, where the stone hits the ground, from
the foot of the cliff = 320^/3= about 554 feet.
Ex. 3. A bullet is projected, with a velocity of 640 feet per
second, at an angle of 30 with the horizontal; find (1) the greatest
height attained, and (2) the range on a horizontal plane and the time
of flight.
The initial horizontal velocity
= 640 cos 30= 640 x ^r- = 320^3 feet per second.
m
The initial vertical velocity =640 sin 30 =320 feet per second.
(1) If h be the greatest height attained, then h is the distance
through which a particle, starting with velocity 320 and moving with
acceleration - g, goes before it comes to rest.
;. 0=3203-20*;
' =S= 1600fo6 '-
(2) If t be the lime of flight, the vertical distance described in
time t is zero.
.. 0=320* -J0*>;
640
'. t = 20 seconds.
9
The horizontal range = the distance described in 20 seconds by a
particle moving with a constant velocity of 320^/3 ft. per sec.
20x820^/3 = 11085 feet approximately.
EXAMPLES. XXXVH
1. A particle is projected at an angle a to the horizon with a
velocity of u feet per second ; find the greatest height attained, the
time of flight, and the range on a horizontal plane, when
(1) u=64, o=30;
(2) m=80, a=60.
2. A projectile is fired horizontally from a height of 9 feet from
the ground, and reaohes the ground at a horizontal distance of 1000
feet. Find its initial velocity.
3. A stone is dropped from a height of 9 feet above the floor of a
railway carriage which is travelling at the rate of 30 miles per hour.
Find the velocity and direction of the particle in space at the instant
when it meets the floor of the carriage.
122
180 DYNAMICS. Em. XXXVII.
4. A Bhip is moving with a velocity of 16 feet per second, and
a body is allowed to fall from the top of its mast, which is 144 feet
high; find the velocity and direction of motion of the body, (1) at
the end of two seconds, (2) when it hits the deck.
5. A particle is projected horizontally from the top of a tower at
the rate of 10 miles per hour and falls under the action of gravity.
Assuming that no other forces are acting draw a figure to represent
its position at the end of 1, 1, 2, and 3 seconds.
6. A balloon is carried along at a height of 100 feet from the
ground at the rate of 40 miles per hour and a stone is dropped from
it ; find the time that elapses before it reaches the ground and the
distance from the point where it reaches the ground to the point
vertically below the point where it left the balloon.
7. A stone is thrown horizontally, with velocity j2gh, from the
top of a tower of height h. Find where it will strike the level ground
through the foot of the tower. What will be its striking velocity?
S 8. A shot is fired from a gun on the top of a cliff, 400 feet high,
with a velocity of 768 feet per second, at an elevation of 30. Find
the horizontal distance from the vertical line through the gun of the
point where the shot strikes the water.
9. From the top of a vertical tower, whose height is ^-g feet, a
particle is projected, the vertical and horizontal components of its
initial velocity being 6g and 8g respectively ; find the time of flight,
and the distance from the foot of the tower of the point at which it
strikes the ground.
Uniform motion in a circle.
218. We have learnt from the First law of Motion
that every particle, once in motion, will, unless it be
prevented from so doing, continue to move in a straight
line with unchanged velocity. Hence a particle will not
describe a circle, or any cwrved path, unless it be compelled
to do so. When a particle is describing a circle, in such a
manner that the magnitude of its velocity is constant, the
direction of its velocity is continually changing. There is
therefore a change in its velocity (Art. 212) and so it moves
with an acceleration.
219. If a particle describes a circle of radius r so that
the magnitude of its velocity is v, it can be proved (Ele-
ments of Dynamics, Art 135) that its acceleration / is
always equal to and that it is always in a direction
UNIFORM MOTION IN A CIRCLE. 181
towards the centre of the circle. We shall assume this
result.
Hence / ~
Now, by the Second law of Motion, wherever there is
acceleration, there must be force to produce it. Also, by
Art. 173, we know that the force P required to produce in
a mass m an acceleration f is given by P mf.
Hence, if a particle describe with velocity v a circle of
radius r, it must be acted on by a force P directed toward
the centre of the circle, such that
P = mf= m .
9 r
220. The force spoken of in the preceding article
exhibits itself in various forms.
As a simple example consider the case of a particle tied
to one end of a string, the other end of which is attached
to a point of a smooth horizontal table. Let the string be
stretched out and laid flat on the table and let the particle
be struck so as to start moving on the table in a direction
at right angles to the string.
It will describe a circle about the fixed end of the string
as centre. In this case the tension of the string supplies
the force requisite to make the particle move in a circle.
Ex. A particle, of mass 3 lbs., moves on a smooth table with a
velocity of 4 feet per second, being attached to a fixed point on the
table by a string of length 5 feet ; find the tension of the string.
Here t>=4, and r5.
Hence, by the last article, the acceleration of the particle is
towards the fixed point and equals , i.e. ft.-seo. units.
r 5
Henoe the tension of the string
v* . 16 48
= wi - = 3xy = j poundals.
* **" f 5x32 ' *"' 10 ' f * ^ nnd '
If the string were so weak that it could not exert this tension it
would break ; the particle would then proceed to describe a straight
line on the table.
182 DYNAMICS. Exs. XXXVIII.
221. In the case of a locomotive engine moving on
horizontal rails round a curve the required force is provided
by the pressure between the rails and the flanges of the
wheels. If the rails were horizontal and at the same level
and if there were no flanges to the wheels, the engine
would not describe a curved portion of the line but would
leave the rails.
Ex. A locomotive engine, of mass 10 tons, moves on a curve, whose
radius is 600 feet, with a velocity of 15 miles per hour; what force
must be exerted by the rails t
15 miles per hour = 22 feet per second.
_ t> 3 22 2
Henoe 7-656-
Also the mass of the engine = 2240 x 10 lbs.
v 3
Hence the force required = m
22 3
= 2240 x 10 x qqq poundals
= 32 X 600 tOn8Weight
121 . ._ .
= 486 ' ie ' nearly *' t0n Wt *
EXAMPLES. XXXVIII.
1. A body, of mass 20 lbs. describes a circle of radius 10 ft. with a
velocity of 15 ft. per second. Find the force required to make it do so.
2. What must be the force that acts toward the centre of a circle,
whose radius is 5 ft., to make a body of mass 10 lbs. describe the
circle with a velocity of 20 ft. per second ?
3. With what velocity must a mass of 10 grammes revolve hori-
zontally at the end of a string, half a metre long, to cause the same
tension in the string as would be caused by a mass of one gramme
hanging vertically at the end of a similar string?
4. A string, 5 ft. long, can just support a weight of 16 lbs. and
has one end tied to a point on a smooth horizontal table. At the
other end is tied a mass of 10 lbs. What is the greatest velooity with
which the mass can be projected on the table so that the string may
not break ?
5. An engine, of mass 9 tons, passes round a curve, half a mile
in radius, with a velocity of 35 miles per hour. What pressure
tending towards the centre of the curve must be exerted by the rails ?
6. If in the previous question the mass of the engine be 12 tons,
its velocity 60 miles per hour, and the radius of the curve be 400 yds.,
what is the required force ?
HYDKOSTATICS.
CHAPTER XVII.
FLUID PRESSURE.
222. In Statics we have considered the equilibrium
of rigid bodies only, and we have denned a rigid body as
one the particles of which always retain the same position
with respect to one another. A rigid body possesses
therefore a definite size and shape. We have pointed out
that there are no such bodies in Nature, but that there are
good approximations thereto.
In Hydrostatics we consider the equilibrium of such
bodies as water, oils, and gases. The common distinguish-
ing property of such bodies is the ease with which their
portions can be separated from one another.
If a very thin lamina be pushed edgeways through water
the resistance to its motion is very small, so that the force
of the nature of friction, i.e. along the surface of the
lamina, must be very small. There are no fluids in which
this force quite vanishes, but throughout this book we
shall assume that no such force exists in the fluids we have
to deal with. Such a hypothetical fluid is called a perfect
fluid, the definition of which may be formally given as
follows ;
223. Perfect Fluid. Def. A perfect fluid is a
substance such that its shape can be altered by any tan-
gential force however small, if applied long enough, of
which portions can be easily separated from the rest of the
mass, and between different portions of which there is no
tangential, i.e. rubbing, force of the nature of friction. The
difference between a perfect fluid and a body like water is
chiefly seen in the case of the motion of the water.
For example, if we set water revolving in a cup, the
184 HYDROSTATICS.
frictional resistances between the water and the cup and
between different portions of the water soon reduces it to
rest. When water is at rest it practically is equivalent to
a perfect fluid.
224. Fluids are again subdivided into two classes, viz.
Liquids and Gases.
Liquids are substances such as water and oils. They
are almost entirely incompressible. An incompressible
body is one whose total volume, i.e. the space it occupies,
cannot be increased or diminished by the application of
any force, however great, although any force, however
small, would change its shape. All liquids are really
compressible under very great pressure but only to a very
slight degree. For example, a pressure equal to about
200 times that of the atmosphere will only reduce the
volume of a quantity of water by a one-hundredth part.
This compressibility we shall neglect and therefore define
liquids, as those fluids which are incompressible.
Gases, on the other hand, are fluids which can easily be
made to change their total volume, i.e. which are, more or
less easily, compressible.
If a child's air-ball be placed under the receiver of an
air-pump from which the air has been excluded it will
increase very much in size. If the skin of the air-ball be
broken the air will expand and fill the receiver whatever
be the size of the latter.
225. The definitions of a liquid and gas may be
formally stated as follows;
A perfect liquid is a fluid which is absolutely incom-
pressible.
A gas is a fluid such thai a flnite quantity of it will, if
the pressure to which it is subjected be sufficiently diminished
expand so as to fill any space however great.
226. The differences between a rigid body, a liquid,
and a gas may be thus expressed ;
A perfectly rigid body has a definite size and a definite
shape.
A perfect liquid has a definite size but no definite shape.
A perfect gas has no definite size and no definite shape.
FLUID PRESSURE. 185
227. Viscous fluids. No fluids are perfect. Many
fluids, such as treacle, honey, and tar, offer a considerable
resistance to forces which tend to alter their shape. Such
fluids, in which the tangential or rubbing action between
layers in contact cannot be neglected, are called viscous
fluids.
228. Pressure at a point. Suppose a hole to be
made in the side of a vessel containing fluid, and that this
hole is covered by a plate which exactly fits the hole. The
plate will not remain at rest unless it be kept at rest by
the application of some force; in other words the fluid
exerts a force on the plate.
Also the fluid can, by definition, only exert a force
perpendicular to each element of area of this plate.
If the force exerted by the fluid on each equal element
of area of the plate be the same, the pressure at any point
of the plate is the force which the fluid exerts on the unit
of area surrounding that point.
If, however, the force exerted by the fluid on each
equal element of the area of the plate be
not the same, as in the case of the plate
CD, the pressure at any point P of this
plate is that force which the fluid would > *^^
exert on a unit of area at P if on this A b
unit of area the pressure were uniform
and the same as it is on an indefinitely small area at P.
The pressure at any point within the fluid, such as Q,
is thus obtained. Suppose an indefinitely small rigid plate
placed at Q so as to contain Q and let its area be a square
feet Imagine all the fluid on one side of this plate
removed and that, to keep the plate at rest, a force of X
lbs. wt. must be applied to it. The pressure at the point
Q is then lbs. wt. per square foot.
229. The theoretical unit of pressure is, in the foot-pound system
of units, one poundal per square foot. In the o.o.s. system the
corresponding unit is one dyne per square centimetre.
In practice the pressure at any point of a fluid is not usually
expressed in poundaU per square foot but in lbs. wt. per square inch.
186 HYDROSTATICS.
The former measure is however the best for theoretical calculation
and may be easily converted into the latter.
230. Transmission of fluid pressure. If any
presswre be applied to the surface of a fluid it is transmitted
equally to all parts of the fluid.
This proposition may be proved experimentally as
follows ;
Let fluid be contained in a vessel of any shape and
in the vessel let there be
holes A,B,C,D...oi various
sizes, which are stopped by
tight-fitting pistons to which
forces can be applied.
Let the areas of these """^fp
pistons be a, ft, c, c?,... square
feet, and let the pistons
be kept in equilibrium by J
forces applied to them.
If an additional force
p . a be applied to A [i.e. an additional pressure of p
lbs. wt. per unit of area of A] it is found that an addi-
tional force of p . b lbs. wt. must be applied to B, one of
p . c lbs. wt. to C, one of p . d lbs. wt. to D, and so on,
whatever be the number of pistons. Hence an additional
pressure of p, per unit of area, applied to A causes an
additional pressure of p, per unit of area, on B, and of the
same additional pressure per unit of area on each of the
other pistons G, D y
Hence the proposition is proved.
231. The pressure at any point of a fluid at rest is the
same in all directions.
This may be proved experimentally by a modification of
the experiment of the last article.
For suppose any one of the pistons, 2), to be so
arranged that it may be turned into any other position, i.e.
so that its plane may be made parallel to the planes of
either A, B, or C or be made to take any other position
whatever. It is found that the application of an additional
pressure at A t of p per unit of area, produces the same
FLUID PRESSURE.
187
additional pressure, of p per unit of area, at D whatever
be the position that D is made to occupy.
#282. The foregoing proposition may be deduced from the funda-
mental fact that the pressure of a fluid is always perpendicular to any
surface with which it is in contact.
Consider any portion of fluid in the shape of a triangular prism
having its base AGC'A' horizontal, and its faces ABB 1 A' and its two
triangular ends ABC and A'B'C all vertical.
Let the length AA\ the breadth AG, and the height AB be all
very small and let P, Q t and B be respectively the middle points of
AA', B&, and CC.
Px X x QR
P
iw % xyz
Let the lengths of AA' t AB, and AC be x, y> and z respectively.
Since the edges x and z are very small the pressure on the face
AA'CC may be considered to be uniform, so that, if p be the pressure
on it per unit of area, the force exerted by the fluid on it is p x xz
and acts at the middle point of PR.
So if p' and p" be the pressures per unit of area on AA'BB 1 and
BCC'B' respectively, the forces on these areas are p' x xy and p" x x . QR
acting at the middle points of PQ and QR respectively.
If w poundals be the weight of the fluid per unit of volume, then,
since the volume of the prism is A A' x area ABC, i.e. xx\yz, the
weight of the fluid prism is v> x \xyz and acts vertically through the
centre of gravity of the triangle PQR.
This weight and the three forces exerted on the faces must be
a system of forces in equilibrium; for otherwise the prism would
move.
Hence, resolving the forces horizontally, we have
p' . xy-p" x x . QR x cos (90 - R) =p" x * . QR x sin R [App. 1., Art. 8.]
=p"xx.PQ=p".xy,
so that p'=p" _ (1).
Again, resolving vertically, we have
p x xz - tc x xyz =p" x * . QR x sin (90 - R)
=p" xx.QRxcoaR =p" xx.PR =p" x xz,
.-. p-p"=tr.iy (2).
188
HYDROSTATICS.
Now let the sides of the prism be taken indefinitely small (in
which case p, p', and p" are the pressures at the point P in directions
perpendicular respectively to PR> PQ, and QR). The quantity to . y
now becomes indefinitely small and therefore negligible.
The equation (2) then becomes
so that pssp'ssp".
Now the direction of BC is any that we may choose, so that it
follows that the pressure of the fluid at P is the same in all directions.
233. Bramah's or the Hydrostatic Press.
Bramah's press affords a simple example of the transmission
of fluid pressures.
In its simplest shape it consist* of two cylinders ABCD
and EFGH both containing water, the two cylinders being
connected by a tube CG. The section of one cylinder is
very much greater than that of the other.
In each cylinder is a closely fitting water-tight piston,
the areas of the sections of which are X and x.
To the smaller piston a pressure equal to P lbs. wt. per
unit of its area is applied, so that the total force applied to
it is P . x lbs. wt.
By Art. 230 a pressure of P lbs. wt. per unit of area
will be transmitted throughout the fluid, so that the force
exerted by the fluid on the piston in the larger cylinder
will be P. X lbs. wt.
This latter force would support on the upper surface of
the piston a body whose weight is P . X lbs.
Hence
weight of the body supported _ P . X _ X
force applied ~ P.x ~~ x '
FLUID PRESSURE. 189
so that the force applied becomes multiplied in the ratio of
X to x, i.e. in the ratio of the areas of the two cylinders.
In the above investigation the weights of the two
pistons have been neglected and also the difference between
the levels of the fluid in the two cylinders.
The pressure is usually applied to the smaller piston by
means of a lever KLM which can turn freely about its end
K which is fixed. At M the power is applied and the
point L is attached to the smaller piston by a rigid rod.
Theoretically we could by making the small piston
small enough and the large piston big enough multiply
to any extent the force applied. Practically this multipli-
cation is limited by the fact that the sides of the vessel
would have to be immensely strong to support the pressures
that would be put upon them.
234. Ex. If the area of the small piston in a Bramah's Press be
& sq. inch and that of the large piston be 2 square feet, what weight would
be supported by the application of 20 lbs. wt. to the smaller piston f
The pressure at each point of the fluid in contact with the small
piaion is 20-=- , i.e. 60 lbs. wt. per square inch.
This pressure is (by Art. 230) applied to each square inch of the
larger piston whose area is 288 sq. ins.
Hence the total force exerted on the large piston is 288 x 60, i.e.
17280 lbs. wt., i.e. 7f ton's wt.
A weight of 7f tons would therefore be supported by the larger
piston.
235. Bramah's Press forms a good example of the
Principle of Work as enunciated in Art. 139.
For, since the decrease in the volume of the water in
the small cylinder is equal to the increase of the water in
the large cylinder, it follows that
X.Y=x.y,
where Y and y are the respective distances through which
the large and small pistons move.
Hence ^ = ^.
x Y
. force exerted by large piston _ X _ y
force exerted by small piston x ~~ Y*
.*. force exerted by large piston x Y
* force exerted by small piston x y,
190 HYDROSTATICS.
i.e. the work done by large piston is the same as that done
on the small piston. Hence the Principle of Work holds.
236. Safety Valve. The safety valve affords
another example of the pressure exerted by fluids. In the
case of a boiler with steam inside it the pressure of the
steam might often become too great for the strength of the
boiler and there would be danger of its bursting. The use
of the safety valve is to allow the steam to escape when
the pressure is greater than what is considered to be safe.
In one of its forms it consists of a circular hole D in
the side of the boiler into which there
fits a plug. This plug is attached at
B to an arm ABC, one end of which,
A, is jointed to a fixed part of the
machine. The arm ABC can turn
about A and at the other end C can
be attached whatever weights are desired.
It is clear that the pressure of the steam and the
weight at C tend to turn the arm in different directions.
When the moment of the pressure of the steam about A is
greater than the moment of the weight at C, the plug D will
rise and allow steam to escape, thus reducing the pressure.
In other valves there is no lever ABC and the plug is
replaced by a circular valve, which is weighted and which
can turn about a point in its circumference.
Ex. The arms of the lever of a safety valve are 1 inch and
18 inches and at the end of the longer arm is hung a weight of 20 lbs.
If the area of the valve be 1 square inches, what is the maximum
pressure of the steam which is allowed f
If p lbs. wt. per square inch be the required pressure, the total
force exerted on the valve by the steam =p x $ lbs. wt.
When the valve is just going to rise the two forces -J- and 20 lbs.
wt. balance at the ends of arms 1 and 18 inches.
Hence ^x 1=20x18.
/. j=2401bs. wt.
Exs. XXXIX. FLUID PRESSURE. 191
EXAMPLE& XXXIX.
1. In a Bramah's Press the diameters of the large and small
piston are respectively 2 decimetres and 2 centimetres ; a kilogram
is placed on the top of the small piston ; fiad the mass which it will
support on the large piston.
2. In a Bramah's Press the area of the larger piston is 100 square
inches and that of the smaller one is square inch ; find the force
that must be applied to the latter so that the former may lift 1 ton.
3. A water cistern, which is full of water and closed, can just
bear a pressure of 1500 lbs. wt. per square foot without bursting.
If a pipe whose section is square inch communicate with it
and be filled with water, find the greatest weight that can safely be
placed on a piston fitting this pipe.
4. In a Bramah's Press if a pressure of 1 ton wt. be produced
by a power of 5 lbs. wt. and the diameters of the pistons be in the
ratio of 8 to 1, find the ratio of the lengths of the arms of the lever
employed to work the piston.
5. In a hydraulic press the radii of the cylinders are 3 inches
and 6 feet respectively. The power is applied at the end of a lever
whose length is 2 feet, the piston being attached at a distance of
2 inches from the fulcrum. If a body weighing 10 tons be placed upon
the large piston, find the power that must be applied to the lever.
If the materials of the press will only bear a pressure of 150 lbs.
wt. to the square inch, find what is the greatest weight that can be
lifted.
6. A vessel full of water is fitted with a tight cork. How is it
that a slight blow on the cork may be sufficient to break the vessel ?
7. The arms of the lever of a safety valve are of lengths 2 inches
and 2 feet, and at the end of the longer arm is suspended a weight of
12 lbs. If the area of the valve be 1 square inch, what is the pressure
of the steam in the boiler when the valve is raised ?
8. Find the pressure of steam in a boiler when it is just sufficient
to raise a circular safety-valve which has a diameter of \ inch and
is loaded so as to weigh \ lb.
9. The weight of the safety-valve of a steam boiler is 16 lbs. and
its section is \ of a square inch. Find the pressure of the steam in
the boiler that will just lift the safety-valve.
CHAPTER XVIII.
DENSITY AND SPECIFIC GRAVITY.
237. Density. Def. The density of a homogeneous
body is the mass of a unit volume of the body.
The mass of a cubic foot of water is found to be about
1000 ozs. i.e. 62 J lbs. Hence the density of water is 62 J lbs.
per cubic foot.
A gramme is the mass of the water at 4 C. which would
fill a cubic centimetre. Hence the density of water at 4 0.
is one gramme per cubic centimetre.
The reason why a certain temperature is taken when we define
a gramme is that the volume of a given mass of water alters with the
temperature of the water. If we take a given mass (say 1 lb.) of
water and cool it gradually from the boiling point 100 C. [i.e. 212 P.],
it is found to occupy less and less space until the temperature is
reduced to 4 C. [about 39*2 P.], If the temperature be continually
lowered still further the volume occupied by the pound of water is
now found to increase until the water arrives at its freezing point.
Hence the pound of water occupies less space at 4C. than at any
other temperature, i.e. there is more water in a given volume at 4 0.
than at any other temperature, i.e. the density is greatest at 4 0.
The mass of a cubic foot of mercury is found to be
13*596 times that of a cubic foot of water. Hence the
density of mercury is nearly 13*596 x 62J lbs. per cubic
foot.
If we use centimetre-gramme units the density of
mercury is 13*596 grammes per cubic centimetre.
238. If W be the weight of a given substance in poundals,
p its density in lbs. per cubic foot, V its volume in cubic feet,
and g the acceleration due to gravity measured in foot-second
units, then W=gpV.
For, if M be the mass of the substance, we have by
Art. 180,
W=Mg.
DENSITY AND SPECIFIC GRAVITY. 193
Also M= mass of V cubic feet of the substance
= Fx mass of one cubic foot
= V. P .
.: w= 9p v.
A similar relation is true if W be expressed in dynes, p
in grammes per cubic centimetre, V in cubic centimetres,
and g in centimetre-second units.
239. Specific Gravity. Def. Tiie specific gravity
of a given substance is the number which expresses the
ratio which the weight of any volume of the substance bears
to the weight of an equal volume of the standard substance.
[N.B. The term specific* gravity is often shortened to
sp. gr.]
For convenience the standard substance usually taken is
pure water at a temperature of 4* C.
Since the weight of a cubic foot of mercury is found to
be 13*596 times that of a cubic foot of water, the specific
gravity of mercury is the number 13*596.
When we say that the specific gravity of gold is 19*25,
water is the standard substance, and hence we mean that a
cubic foot of gold would weigh 19 25 times as much as a
cubic foot of water, i.e. about 19*25 x 62 J lbs., i.e. about
1203J lbs. wt
240. Specific gravity of gates. Since gases are very light com-
pared with water, their sp.gr. is often referred to air at a temperature
of 0C. and with the mercury-barometer [Art. 289] standing at a
height of 30 inches. The mass of a cubic foot of air under these
conditions is about 1*25 ozs.
241. The following table gives the approximate
specific gravities of some important substances.
Solids.
Platinum 21*5 Glass (Crown) 2*5 to 2*7
Gold 19*25 (Flint) 3*0 to 3*5
Lead 11-3 Ivory 1*9
Silver 10*5 Oak *7 to 10
Copper 8*9 Cedar *6
Brass 8*4 Poplar 4
Iron 7-8 Cork -24
L M. H. 13
194 HYDROSTATICS.
Liquids at C.
Mercury 13596 Milk 103
Sulphuric Acid 1*85 Alcohol "8
Glycerine 1-27 Ether -73
242. If W be the weight of a volume V of a body whose
specific gravity is , and w be the weight of a unit volume of
the standard substance, then W= V .s .w.
_ weight of a unit volume of the body
weight of a unit volume of the standard substance *
.'. wt. of unit volume of the body = s . w.
.'. wt. of V units of volume of the body = V.s.w.
.'.W = V.s.w.
Ex. If a cubic foot of water weigh 62J lbs., what it the weight of
4 cub. yards of copper, the sp. gr. of copper being 8*8 ?
Here u*=62 lbs. wt., F=108 oub. ft., and =8*8.
^ W= 108 x 8-8 x 62$ m 59400 lbs. wt.
= 26f tons wt.
EXAMPLES. XL.
1. What is the weight of a cubic foot of iron (sp. gr. - 9) ?
2. The sp. gr. of brass is 8 ; what is its density in ounces per
cubic inch, given that the density of water is 1000 ozs. per cubio
foot?
3. A gallon of water weighs 10 lbs. and the sp. gr. of mercury is
13-598. What is the weight of a gallon of meroury f
4. Find the weight of a litre (a oub. decimetre or 1000 cub. cms.)
of meroury at the standard temperature when its sp. gr. is 13*6.
5. If 13 oub. ins. of gold weigh as much as 96 cub. ins. of quartz
and the sp. gr. of gold be 19*25, find that of quartz.
6. The sp. gr. of gold being 19*25, how many oubic feet of- gold
will weigh a ton ?
7. The sp. gr. of cast copper is 8*88 and that of copper wire is
8*79. What change of volume does a kilogramme of cast copper
undergo in being drawn into wire?
8. If a foot length of iron pipe weigh 64*4 lbs. when the diameter
of the bore is 4 ins. and the thickness of the metal is 1\ ins., what is
the sp. gr. of the iron ?
9. A rod 18 ins. long and of uniform cross-section weighs 3 ozs.
and its sp. gr. is 8*8. What fraction of a square inch is its area ?
DENSITY AND SPECIFIC GRAVITY. 195
243. Specific gravities and densities of mix-
tures. To find the specific gravity of a mixture of given
volumes of different fluids whose specific gravities are given.
Let V lt F a , V t ... be the volumes of the different fluids
and 8 l9 * 2 , 8 t ... their specific gravities, so that the weights of
the different fluids are V&w, V&to, JV,w, . . . where to is the
weight of a unit volume of the standard substance.
(1) When the fluids are mixed let there be no dimi-
nution of volume, so that the final volume is V 1 +V t +V t +...
Let be the new specific gravity, so that the sum of the
weights of the fluid is
(V 1 + V 2 + r 9 +...)s.io.
Since the sum of the weights must be unaltered, we
have
\V 1 + V* + F,+ ...]s. u>=V 1 8 1 w + V38 9 w+VgS a w+ ...
F 1 + F 3 + F 8 +... *
(2) When there is a loss of volume on mixing the
fluids together, as sometimes happens, let the final volume
be n times the sum of the original volumes, where n is a
proper fraction.
In this case we have
n [ V x + F 2 + F, + . . .] iw = Vfaw + V&w + ...,
so that * = * - ;& ? r .
n(r, + p;+...)
Similar formulae will hold if the densities instead of the
specific gravities be given. For the original specific gravities
8 ly s a ,... we must substitute the original densities p lt p a ,...
and for the final specific gravity i we must substitute the
final density p.
Ex. Volumes proportional to the numbers 1, 2 and 3 of three liquids
whose sp. grs. are proportional to 1% 1*4 and 1-6 are mixed together;
find the sp. gr. of the mixture.
Let the volumes of the liquids be *, 2x, and 3x. Their weights
are therefore
wx x 1*2, 2wx x 1*4, and 3wx x 1*6.
Also, if * be the sp. gr. of the mixture, the total weight is
ws(x + ix + 'ix).
13 %
196 HYDROSTATICS,
Equating these two, we have
6we .x=wxx8'8,
/. i=*x 8-8 = 1-46.
244. To find the specific gravity of a mixture of given
weights of different fluids whose specific gravities are given.
Let W lt W s ,... be the weights of the given quantities of
fluid, *!,*... their specific gravities, and w the weight of a
unit volume of the standard substance.
By Art. 242, the volumes of the different fluids are
Wi Is
SjW* s^w'"'
If no loss of volume takes place when the mixture is
made, the new volume is - + ? + ...
SjW 8,10
Hence, if * be the new specific gravity, the sum of the
weights of the fluids is, by Art. 242,
( + + ... J 8W, 1.6. ( + - + ... ) *.
\8jW S t W J \*i S, /
Hence, since the sum of the weights necessarily remains
the same, we have
so that m
! + ?!
8. 8,
If there be a contraction of volume so that the final
volume is n times the sum of the original volumes, then, as
in the last article, we have
.r* + %....T
L *i h J
A similar formula gives the final density in terms of the
weights and the original densities.
Bx. 10 lbs. tot. of a liquid, of sp. gr. 1*25, is mixed with 6 lbs. wt.
of a liquid of sp. gr. 1*15. Wluit is the sp. gr. of tlie mixture t
DENSITY AND SPECIFIC GRAVITY. 197
If to be the weight of a cubic foot of water the respective volume*
of the two fluids are, by Art. 242,
j and r-^-z cub. ft.
1-25 xte lloxt*
Hence, if i be the required sp. gr., we have
( i + ^-\ i .w=total weight = 10 + 6.
\l-2oxw 1*15 x w)
EXAMPLES. XLL
1. The sp. gr. of a liquid being -8, in what proportion by volume
must water be mixed with it to give a liquid of sp. gr. -85 ?
2. What is the volume of a mass of wood of sp. gr. *5 so that
when attached to 500 ozs. of iron of sp. gr. 7, the mean sp. gr. of the
whole may be unity ?
3. What weight of water must be added to 27 ozs. of a salt solution
whose sp. gr. is 1*08 so that the sp. gr. of the mixture may be 1*05 ?
4. Three equal vessels, A, B, and C, are half full of liquids of
densities ft, ft, and ft respectively. If now B be filled from A and
then G from B, find the density of the liquid now contained in C, the
liquids being supposed to mix completely.
5. When equal volumes of two substances are mixed the sp. gr.
of the mixture is 4; when equal weights of the same substances
are mixed the sp. gr. of the mixture is 3. Find the sp. gr. of the
substances.
6. When equal volumes of alcohol (sp. gr. = *8) and distilled water
are mixed together the volume of the mixture (after it has returned to
its original temperature) is found to fall short of the sum of the
volumes of its constituents by 4 per cent. Find the sp. gr. of the
mixture.
7. A mixture is made of 7 cub. cms. of sulphuric acid (sp. gr.
= 1 -843) and 3 cub. cms. of distilled water and its sp. gr. when cold
is found to be 1*615. What contraction has taken place ?
CHAPTER XIX.
PRESSURES AT DIFFERENT POINTS OF A HOMOGENEOUS
FLUID WHICH IS AT REST.
245. A fluid is said to be homogeneous when, if any
equal volumes, however small, be taken from different
portions of the fluid, the masses of all these equal volumes
are equal.
246. The pressure of a heavy homogeneous fluid at all
points in the sayne horizontal plane is the same.
Consider two points of a fluid, P and Q, which are in
the same horizontal plane.
Join PQ and consider a small por-
tion of the fluid whose shape is a very
thin cylinder having PQ as its axis.
The only forces acting on this
cylinder in the direction of the axis
PQ are the two pressures on the plane ends of the cylinder.
[For all the other forces acting on this cylinder are perpendicular
to PQ and therefore have no effect in the direction PQ.]
Hence, for equilibrium, these pressures must be equal
and opposite.
Let the plane ends of this cylinder be taken very small
so that the pressures on them per unit of area may be taken
to be constant and equal respectively to the pressures at P
and Q.
Hence pressure at P x area of the plane end at P
= pressure at Q x area of the plane end at Q.
.'. pressure at P = pressure at Q.
><I3P^pQE
PRESSURE AT A GIVEN DEPTH. 199
247. To find the pressure at any given depth of a heavy
homogeneous liquid, the pressure of the atmosphere being
Take any point P in the liquid and draw a vertical line
PA to meet the surface of the
liquid in the point A.
Consider a very thin cylinder
of liquid whose axis is PA. This
cylinder is in equilibrium under
the forces which act upon it.
The only vertical forces acting
on it are its weight and the force
exerted by the rest of the fluid upon the plane face at P.
If a be the area of the plane face and x the depth AP,
the weight of this small cylinder of liquid is w x a x x,
where w is a weight of a unit volume of the liquid.
Also the vertical force exerted on the plane end at P is
p x a, where p is the pressure at P per unit of area.
Hence p . a = w . a . x.
.*. p w.x.
Cor. Since the pressure at any point of a liquid depends
only on the depth of the point, the necessary strength of the
embankment of a reservoir depends only on the depth of
the water and not at all on the area of the surface of the
water.
248. In the above expression for, the pressure care
must be taken as to the units in which the quantities are
measured. If British units be used, x is the depth in feet,
w is the weight of a cubic foot of the liquid, and p is the
pressure expressed in lbs. wt. per square foot.
If o.G.s. units be used, x is the depth in centimetres,
w is the weight of a cubic centimetre of the liquid, and p
is the pressure expressed in grammes weight per square
centimetre. If the liquid be water, it should be noted that
w equals the weight of one gramme. [Art. 237.]
249. The theorem of Art. 247 may be verified experimentally.
PQ is a hollow cylinder one end of which Q is closed by a thin light
fiat disc which fits tightly against this end.
The cylinder and disc are then pushed into the water, the
200
HYDROSTATICS.
p
oylinder remaining always in a vertical position. The disc does not
fall, being supported by the pressure of the
water.
Into the upper end of the oylinder water
is now poured very slowly and carefully.
The disc is not found to fall until the water
inside the cylinder stands at the same height
as it does outside.
If h be the depth of the point Q, and A be
the area of the cylinder, the pressure on Q of
the external fluid must balance the weight of
the internal fluid, and this weight is A.hxw, i.e. Axwh. Hence
the external pressure at Q per unit of area must be wh.
250. In Art. 247 we have neglected the pressure of
the atmosphere, i.e. we have assumed the pressure at A to
be zero.
If this pressure be taken into consideration and denoted
by II, the equation of that article should be
p. a w. a.sc + n .a,
i.e. p = wx + H.
The pressure of the atmosphere is roughly equal to about
15 lbs. wt. per sq. inch. [This pressure is often called
" 15 lbs. per square inch." See Arts. 11 and 181.]
Instead of giving the atmospheric pressure in lbs. wt.
per sq. inch it is often expressed by saying that it is the
same as that of a column of water, or mercury, of a given
height.
This, as we shall see in Chapter 22, is the same as
telling us the height of the barometer made of that liquid.
For example, if we are told that the height of the water-
barometer is 34 feet we know that the pressure of the
atmosphere per squcure foot = weight of a column of water
whose base is a square foot and whose height is 34 feet
= wt. of 34 cubic feet of water
= 34 x 62 J lbs. wt.
Hence the pressure of the atmosphere per square inch
34 x 62^ ,v *
I4109 lbs. wt.
PRESSURE AT A GIVEN DEPTH.
201
Ex. Find the presture in water at a depth of 222 feet, the height
of the tcater-barometer being 34 feet.
If w be the weight of a cubic foot of water, we have
n=tc.841bs. wt.
A j>=n + t0fc=w. 84 + 1*. 222 = 256 to per sq.ft.
= 256 x 62 Jj lbs. wt. per sq. ft.
16000., . ,
. lbs. wt. per. sq. inch
= 111} lbs. wt. per sq. inch.
251. The surface of a heavy liquid at rest is horizontal.
Q
Take any two points, P and Q, of the liquid which are
in the same horizontal plane. Draw vertical lines PA and
QB to meet the surface of the liquid in A and B.
Then, by Art. 246, the pressure at P
= the pressure at Q.
Hence, by Art. 250, II + w . PA = II + to. QB.
:. PA = QB.
Hence, since PQ is horizontal, the line AB must be
horizontal also.
Since P and Q are any two points in the same horizontal
line, it follows that any line AB drawn in the surface of the
liquid must be horizontal also.
Hence the surface is a horizontal plane.
252. In the preceding proofs we have assumed that
the weights of different portions of the fluids act vertically
downwards in parallel directions. This assumption, as was
pointed out in Art. 69, is only true when the body spoken
of is small compared with the size of the earth.
202
HYDROSTATICS.
If the body be comparable with the earth in size we
cannot neglect the fact that, strictly speaking, the weights
of the different portions of the body do not act in parallel
lines but along lines directed toward the centre of the
earth.
The theorem of the preceding article would not therefore
apply to the surface of the sea, even if the latter were
entirely at rest.
253. The proposition of Art. 246 can be proved for
the case when it is impossible to connect the two points by
a horizontal line which lies wholly within the fluid.
For the two points P and Q can be connected by vertical
and horizontal lines such as PA, AB, and BQ in the figure.
We then have
pressure at A = pressure at B.
But pressure at A pressure at P + w.AP,
and pressure at B = pressure at Q + w . BQ.
But, since P and Q are in the same horizontal plane,
AP=BQ.
Hence the pressure at P = pressure at Q.
Similarly for a vessel of the above shape the proposition
is true for any two points at the same level. Hence
the surface of the fluid will always stand at the same level
provided the fluid be at rest.
For example, the level of the tea inside a teapot and in
the spout of the teapot is always at the same height.
The statement that the surface of a liquid at rest is a
horizontal plane is sometimes expressed in the form "water
finds its own level."
PRESSURE AT A GIVEN DEPTH. 203
It is this property of a liquid which enables water to be
supplied to a town. A reservoir is constructed on some
elevation which is higher than any part of the district to be
supplied. Main pipes starting from the reservoir are laid
along the chief roads and smaller pipes branch off from
these mains to the houses to be supplied. If the whole of
the water in the reservoir and pipes be at rest the surface
of the water would, if it were possible, be at the same level
in the pipes as it is in the reservoir. The mains and side-
pipes may rise and fall, in whatever manner is convenient,
provided that no portion of such main or pipe is higher than
the surface of the water in the reservoir.
EXAMPLES. XLTL
1. If a cubic foot of water weigh 1000 ozs. what is the pressure
per sq. inch, at the depth of a mile below the surface of the water f
2. Find the depth in water at which the pressure is 100 lbs. wt.
per sq. inch, assuming the atmospherio pressure to be 15 lbs. wt. per
sq. inch.
3. The sp. gr. of a certain fluid is 1*56 and the pressure at a
point in the fluid is 12090 ozs. ; find the depth of the point, a foot
being the unit of length.
4. The pressure in the water-pipe at the basement of a building
is 34 lbs. wt. to the sq. inch and at the third floor it is 18 lbs. wt. to
the sq. inch. Find the height of this floor above the basement.
5. If the atmospheric pressure be 14 lbs. wt. per sq. in. and the
sp. gr. of air be -00125, find the height of a column of air of the same
uniform density which will produce the same pressure as the actual
atmosphere produces.
6. If the force exerted by the atmosphere on a plane area be
equal to that of a column of water 34 feet high, find the force
exerted by the atmosphere on a window-pane 16 inches high and one
foot wide.
7. The pressure at the bottom of a well is four times that at a
depth of 2 feet ; what is the depth of the well if the pressure of the
atmosphere be equivalent to that of 30 feet of water ?
8. If the height of the water-barometer be 34 ft., find the depth
of a point below the surface of water such that the pressure at it may
be twice the pressure at a point 10 ft. below the surface.
9. If the pressure at a point 5 feet below the surface of a lake be
one-half of the pressure at a point 44 feet below the surface, what
must be the atmospherio pressure in lbs. wt. per sq. inch ?
204 HYDROSTATICS. Exs. XLII.
10. A vessel whose bottom is horizontal contains mercury whose
depth is 30 inches and water floats on the mercury to a depth of 24
inches; find the pressure at a point on the bottom of the vessel in
lbs. wt. per. sq. inch, the sp. gr. of mercury being 13 '6.
11. A vessel is partly filled with water and then oil is poured in
till it forms a layer 6 inches deep ; find the pressure per sq. inch due
to the weight of the liquids at a point 8-6 inches below the upper
surface of the oil, assuming the sp. gr. of the oil to be *92 and the
weight of a oubio inch of water to be 252 grains.
12. A vessel contains water and mercury, the depth of the water
being two feet. It being given that the sp. gr. of mercury is 13*568
and that the mass of a cubic foot of water is 1000 ozs. , find, in lbs.
wt. per sq. in., the pressure at a depth of two inches below the com-
mon surface of the water and mercury.
13. The lower ends of two vertical tubes, whose cross sections
are 1 and *1 sq. inches respectively, are connected by a tube. The
tube contains mercury of sp. gr. 13*596. How much water must be
poured into the larger tube to raise the level in the smaller tube by
one inch ?
254. Whole Pressure. Def. If for every small
element of area of a material surface immersed in fluid the
pressure perpendicular to this small element be found, the
sum of all such pressures is called the whole pressure, or
thrust, of the fluid upon the given surface.
In the following article it will be shewn how this
thrust may be calculated.
Theorem. If a plane surface be immersed in a liquid,
the whole pressure on it is equal to wS . z\ where S is the
area of the plane surface and z is the depth of its centre of
gravity below the surface of the liquid, the pressure of the
air being neglected.
For consider any plane area, horizontal or inclined to
the horizon, which is immersed in A
Consider any small element |!|g=|gz^ ==\g||
Oj of the plane surface situated r^irr - - ^s ^zz ~
at P and draw PA vertical to ~5aa^.rz: :
meet the surface of the liquid in -rzrr zrz^SzJrzi ' .
A, and let PA be z, .
The pressure on this small area is therefore wa x z .
(Art. 247).
Similarly if a,, ag,...be any other elements of the plane
WHOLE PRESSURE OR THRUST. 205
surface whose depths are a , *, ... the pressures on them are
w*# %y wa# ti
Hence the whole pressure
= w[a 1 z l + a& t + ].
But, if % be the depth of the centre of gravity of the
given plane area, we have, as in Art. 80,
a 1 1 + a^ a + ...
% B
^ + 0,+ ... '
.\ a 1 z 1 +a 2 s,+ =*(oi + a*+ ...)=. S.
Hence the whole pressure = wz.S = area of the surface
multiplied by the pressure at its centre of gravity, i.e. the
whole pressure is equal to the weight of a column of liquid
whose base is equal to the area of the given plane surface,
and whose height is equal to the depth below the surface of
the liquid of the centre of gravity of the given plane
surface.
The preceding proposition is true when the area is not
plane and the proof of the preceding article holds, but in
this case the whole pressure is not the resultant thrust on
the area.
365. Ex. 1. A square plate, whose edge is 8 inches, is immersed
in sea-water, its upper edge being horizontal and at a depth of 12
inches below the surface of the water. Find the whole pressure of the
water on the surface of the plate when it is inclined at 45 to the
horizon, the mass of a cubic foot of sea-water being 64 lbs.
The depth of the centre of gravity of the plate
= 12 + 4 cos 45 =(12 + 2^2) inches =^^ ft.
o
Also the area of the plate = f 5 J sq. ft.
Hence the whole pressure, or thrust,
= | x ^? x6 41bB.w.
= 35149 lbs. wt. nearly.
Ex. 2. A vessel in the form of a cube, each of whose edges is
2 feet y is half filled with mercury and half with water. If the sp. gr.
of the mercury be 13*6, find the whole pressure or thrust on one of its
vertical faces.
The formula of Art. 254 does not exaotly suit this question, but by
means of an artifice it may be made applicable.
For the thrust on the vertical face may be considered due to the
206 HYDROSTATICS. Exs.
thnist of two liquids, one, via., water, of sp. gr. 1, filling the whole
vessel, and the other of density 12-6 [i.e. 13-6 - 1] filling the lower
half of the vessel. The required thrust is the sum of the thrusts due
to the two liquids.
The thrust due to the first
=wt. of 2 s x 1 x 1 cubic ft. of water
= wt. of 4 cubic ft. of water.
The thrust due to the second
=wt. of [2 x 1] x x 12-6 cubic ft. of water
= wt. of 12*6 cubio ft. of water.
The sum of these
=wt. of 16-6 cubic ft. of water =16600 ozs. = 1037 lbs. wt.
Ex. 3. ' A hollow cone stands with its base on a horizontal table.
The area of the base is 100 sq. inches and the height of the cone is
8-64 inches and it is filled with water. Find the thrust on the base
of the cone and its ratio to the weight of the water in the cone.
The thrust = wt. of 100 x 8" 64 cubic inches of water
864
= 1728 X 100 0ZS * Wt * ~ 50 ZB# Wt '
= 31-25 lbs. wt.
Since the volume of the cone is one-third the product of the
height and the area of the base, the weight of the contained water
= wi of \ x 100 x 8*64 cubic inchea
=4 x 31-25 lbs. wt.
Hence, the thrust on the base of the cone
a= three times the weight of the contained water.
This result, which at first sight seems impossible, is explained by
the fact that the upward thrust of the base has to balance both the
weight of the liquid and also the vertical component of the thrust
of the curved surface of the cone upon the contained fluid, and this
component aots downward and could be proved to be equal to twice
the weight of the contained fluid.
EXAMPLES. XLIIL
1. A cube, each of whose edges is 2 ft. long, stands on one of its
faces on the bottom of a vessel containing water 4 ft. deep. Find
the whole pressure, or thrust, of the water on one of its upright faces.
2. Water is supplied from a reservoir which is 400 ft. above the
level of the sea, A tap in one of the houses supplied is at a height of
150 ft. above the sea-level and has an area of 1 sq. ins. Find the
whole pressure, or thrust, on the tap.
3. A oube of one foot edge is suspended in water with its upper
face horizontal and at a depth of 2 ft. below the surface. Find the
whole pressure, or thrust, on each face of the cube.
XLIII. WHOLE PRESSURE OR THRUST. 207
4. A hole, six ins. sq., is made in a ship's bottom 20 ft. below the
water-line. What force must be exerted to keep the water out by
holding a piece of wood against the hole, assuming that a cubic ft.
of sea-water contains 64 lbs. f
5. The pressure of the atmosphere being 14 lbs. per sq. in.,
find in owts. the thrust on a horizontal area of 7 sq. ft. placed in
water at a depth of 32 ft.
6. A cubical vessel, whose side is one foot, is filled with water.
Find the thrust on its surface.
7. The area of one face of a deep-sea thermometer is 54 sq. ins.
Prove that when it is sunk to a depth of 6000 yards, the thrust on it
is about 192 tons wt., assuming a cubic fathom of sea-water to weigh
6-15 tons.
8. Find the resultant thrust on either side of a vertical wall,
whose breadth is 8 ft. and depth 12 ft., which is built in the water
with its upper edge in the surface, the height of the water-barometer
being 33 ft.
9. A vessel, whose base is 6 ins. sq. and whose height is 6 ins.,
has a neck of section 4 sq. ins. and of height 3 ins. ; if it be filled with
water, find the thrust on the base of the vessel.
10. The dam of a reservoir is 200 yards long and its face towards
the water is rectangular and inclined at 30 to the horizon. Find the
thrust acting on the dam when the water is 30 ft. deep.
Has the size of the surface of the water in the reservoir any effect
on this thrust ?
11. A vessel shaped like a portion of a cone is filled with water.
It is one inch in diameter at the top and eight inches in diameter at
the bottom and is 12 ins. high. Find the pressure in lbs. wt. per
sq. in. at the centre of the base and also the thrust on the base.
12. A square is placed in liquid with one side in the surface.
Shew how to draw a horizontal line in the square dividing it into two
portions, the thrusts on which are the same.
13. A vessel , in the shape of a cube whose side is one decimetre,
is filled to one-third of its height with mercury whose sp. gr. is 13*6
while the rest is filled with water. Find the thrust against one of its
sides in kilogrammes wt.
14. A rectangular vessel, one face of which is of height two feet
and width one foot, is half filled with mercury and half with water.
Find the thrust on this face, given that the sp. gr. of mercury is 13 *5.
15. A cylindrical vessel, whose height is 5 ft. and diameter 1 ft.,
is filled with water and held so that the line joining its centre to a
point on the rim of one of its plane faces is vertical ; find the thrust
on each of its plane faces.
208 HYDROSTATICS. Exs. XLIII.
16. If in the last example the vessel is held with (1) its axis
horizontal, (2) its axis vertical, find the corresponding thrusts.
17. Two equal small areas are marked on the side of & reservoir
at different depths below the surface of the water. If the thrust on A
be four times that of B and if water be drawn off so that the surface
of the water in the reservoir falls one foot, the thrust on A is now
nine times that on B. What were the original depths of A and B
below the surface of the water?
18. A cubical box, whose edge measures 1 ft., has a pipe com-
municating with it whioh rises to a vertical height of 20 ft. above the
lid. It is filled with water to the top of the pipe. Find the upward
thrust on the lid and the downward thrust on the base and show that
their difference is equal to the weight of the water in the box.
How do you explain the fact that the thrust on the base is greater
than the weight of the liquid it contains?
19. An artificial lake, mile long and 100 yards broad, with a
gradually shelving bottom whose depth varies from nothing at one
end to 88 ft. at the other, is dammed at the deep end by a masonry
wall across its entire breadth. If the weight of the water be | ton
weight per oubio yard, prove that the thrust on the embankment is
32266} tons weight, and that the total weight of the water in the lake
is 484000 tons.
256. Centre of pressure of a plane area.
If a plane area be immersed in liquid the pressure at
any point of it is perpendicular to the plane area and is
proportional to the depth of the point.
The pressures at all the points of this area therefore
constitute a system of parallel forces whose magnitudes are
known.
By Art. 49 it follows that all these parallel forces can
be compounded into one single force acting at some definite
point of the plane area.
This single force is called the resultant fluid thrust
and is in the case of a plane area the same as the whole
pressure, and the point of the area at which it acts is
called the centre of pressure of the given area.
The determination of the centre of pressure in any given
case is a question of some difficulty.
We shall not discuss it here but shall state the position
of the centre of pressure in one or two simple cases.
CENTRE OF PRESSURE.
209
(1) A rectangle ABCD is immersed with one side AB
in the surface. If L and M be the middle points of AB
and CD, the centre of pressure is at F, where LF = \L M.
(2) A triangle ABC is immersed with an angular
point A in the surface and the base BC horizontal. If D
be the middle point of BC, the centre of pressure F lies
on AD, such that AF =\AD.
(3) A triangle ABC is immersed with its base BC in
the surface. If D be the middle point of the base, the
centre of pressure F bisects DA.
ttz-jk!
Ex. A rectangular hole ABCD, whose lower side CD is horizontal,
is made in the bide of a reservoir, and is closed by a door whose plane is
vertical, and the door can turn freely about a hinge coinciding with AB.
What force must be applied to the middle point of CD to keep the door
shut if AB be one foot and AD 12 feet long, and 'if the water rise to the
level of AB}
L. M. EL
U
210 HYDROSTATICS.
If P be the required force then its moment about AB and the
moment of the pressure of the water about AB must be equal.
The pressure of the water, by Art. 254,
1 x 12 x 6 x J^A lbs. wt. =4500 lbs. wt.
Also, by (1), it acts at a point whose distance from AB
= |^D = 8feet.
Hence, by taking moments about AB,
Px 12 = 4500x8
,\ P = 3000 lbs. wt
CHAPTER XX.
RESULTANT VERTICAL THRUST.
257. If a portion of a curved surface be immersed in
a heavy fluid, as in the figure of the next article, the deter-
mination of the total effect of the pressure of the fluid on
it, i.e. of the resultant fluid thrust, is a matter of some
difficulty. For the pressures at different points of the sur-
face act in different directions and in different planes.
In the present book we shall be only concerned with
the total vertical force exerted by the fluid on the curved
surface. This force is called the Resultant Vertical
Thrust. It is equal to the sum of the vertical com-
ponents of the pressures which act at the different points
of the given surface. For these vertical components
compound, since they are parallel forces, into one single
vertical force.
In the next article it will be shewn how this resultant
vertical thrust may be found.
258. Resultant vertical thrust on a surface immersed
in a heavy fluid.
Consider a portion of surface PEQS immersed in the
fluid. Through each point n
of the bounding edge of
this surface conceive a
vertical line to be drawn,
and let the points in which
these vertical lines meet
the surface of the fluid
form the curve ACBD.
Consider the equili-
brium of the portion of
the fluid enclosed by these
vertical lines, by the surface PEQS, and by the plane surface
ACBD. Since, as in Art. 247, the vertical thrust of each
H 2
^Ef^E2^3^E^rr^^E^=^^:
212 HYDROSTATICS.
small element of surface of PSQR balances the weight of
the corresponding thin superincumbent cylinder of fluid,
therefore the resultant of all these elementary vertical
thrusts (i.e. the resultant vertical thrust) must be equal
and opposite to, and in the same line of action as, the
resultant of the weights of these thin cylinders.
But this latter resultant is the weight of the liquid
PRQSDACB and acts at its centre of gravity.
Also the thrust of the surface upon the fluid is equal
and opposite to that of the fluid upon the surface.
Hence, " The resultant vertical thrust on any surface
immersed in any heavy fluid is equal to the weight of the
superincumbent fluid and acts through the centre of gravity
of this superincumbent fluid."
259. If the fluid, instead of pressing the surface
downwards, press it upward as in the adjoining figure, the
same construction should be made as in the last article.
The pressure at any point of the surface PRQS depends
only on the depth of the point below
the surface of the fluid. c a\ /
Hence, in our case, the pressure ^j "bj ||f|f
is, at any point, equal in magnitude jLfMBfcssv
but opposite in direction to what iflSferH^?^
the pressure will be if the fluid f^^^~z~^^\
inside the vessel be removed and t^^^rF-^^^J
instead fluid be placed outside the K^EzE^i^y
vessel so that AB is its surface. >^T "^r
In the latter case the resultant
vertical thrust will be the weight of the fluid PQAB.
Hence, in our case, The resultant vertical thrust on the
given portion of surface is equal to the weight of the fluid
that could lie upon it up to the level of the surface of the fluid
and acts vertically upwards through the centre of gravity
of this fluid.
260. The resultant vertical thrust on a body immersed,
wholly or partly , in a fluid is equal to the weight of the fluid
displaced.
RESULTANT VERTICAL THRUST. 213
Consider the body PTQU wholly immersed in a fluid.
Let a vertical line be con- p
ceived to travel round the surface J^-jst ^ ^ '^^ be^l
of the body, touching it in the gl^^iill||i^gi||l|gl
curve PRQS and meeting the ==E^=^y^3E^=lly^~
surface of the fluid in the curve E5^^t-rfe^^_-f^^E
The resultant vertical thrust zr?Eijp^^^^^^-??r~
on the surface PTQRP is equal z:z~::i-jzislzzzz7"JL
to the weight of the fluid that
would occupy the space PTQBA and acts vertically
upwards.
The resultant vertical thrust on the surface PUQRP
is equal to the weight of the fluid that would occupy the
space PUQBA and acts vertically downwards.
The resultant vertical thrust on the whole body is
equal to the resultant of these two thrusts, and is there-
fore equal to the weight of the fluid that would occupy the
space PTQU and acts upwards through the centre of
gravity of the space PTQU.
Hence, The resultant vertical thrust on a body totally
immersed is equal to the weight of the displaced fluid and
acts vertically through the centre of gravity of the displaced
fluid.
This centre of gravity of the displaced fluid is often
called the centre of buoyancy of the body.
The important result just enunciated is known as the
Principle of Archimedes.
261. The same theorem holds if the body be partially
immersed, as may be easily seen.
If the shape of the body be somewhat irregular, as in
the figure on the next page, the total vertical thrust is
equal to the weight of the fluid APQB acting upwards, less
the weights of the fluid SDBQ and RGAP acting down-
wards, plus the weights of the fluid DSM and GLR acting
upwards, i.e., is equivalent to the weight of the fluid that
could be contained in LRPQSM acting upwards through its
centre of gravity.
214
HYDROSTATICS,
Floating Bodies.
262. Conditions of equilibrium of a body freely
floating in a liquid.
Consider the equilibrium of a body floating wholly or
partly immersed in a liquid.
There are two, and only two, vertical forces acting on
the body, (1) its weight acting through the centre of
gravity G of the body, and (2) the resultant vertical thrust
on the body which is equal to the weight of the displaced
fluid and acts through the centre of buoyancy, i.e. the centre
of gravity G' of the displaced liquid.
For equilibrium these two forces must be equal and act
in opposite directions in the same vertical line.
FLOATING BODIES. 215
Hence the required conditions are :
(1) The weight of the displaced fluid must be equal to
the weight of the body, and
(2) The centres of gravity of the body and the
displaced fluid must be in the same vertical line.
263. Ex. 1. A cylinder of wood, of height 6 feet and weight
50 lbs., floats in water. If its sp. gr. be %, find how much it will be
depressed if a weight of 10 lbs. be placed on its upper surface.
Let A be the area of the section of the cylinder. Then
50=^.6.f.u>=^.6.|.62$,
so that A = &Bq. ft.
Let * be the distance through whioh the wood is depressed when
10 lbs. are placed on it. The weight of the water which would occupy
a cylinder, of section A and height x, must therefore be 10 lbs.
/. 10= A .x.w^-is .x.62$.
.'.=:* f*.
Ex. 2. A man, whose weight is equal to 160 lbs. and whose sp. gr.
is 11, can just float in water with his head above the surface by the aid
of a piece of cork which is wholly immersed. Having given that the
volume of his head is one-sixteenth of his whole volume and that the
sp. gr. of cork is -24, find the volume of the cork.
Taking the wt. of a cubic ft. of water to be 62 lbs. we have, if V
be the volume of the man,
160=FxBx62,
so that F=VP CQD - ft -
Again, since the weight of the man and the cork must be equal to
the weight of the fluid displaced, we have, if V be the volume of the
cork in cubic feet,
160+rx -24x62=(H^+ 0-1.62*.
Arx.76x62^160-i|. F .62i=160-^.^.^.
96 F'-160 1500 - 260
_, 2 260 104 . _.
F -95 X lT = 209 CUb - ft -
Ex. 8. A loaded piece of wood and an elastic bladder containing
air just float at the surface of the sea; what will happen if they be
both plunged to a great depth in the sea and then released ?
The resultant upward thrust of a homogeneous liquid on a body
is always the same, whatever be its depth below the surface of the
liquid, provided that the volume of the body remains unaltered.
216 HYDROSTATICS. Exs.
In the case of the wood, which we assume to be incompressible,
the resultant thrust on it at a great depth is the same as at the
surface and therefore the body just floats.
In the case of the elastic bladder the pressure of the sea at a
great depth compresses the bladder and it therefore displaces much
less liquid than at the surface of the sea. The resultant vertical
thrust therefore is much diminished; and, as the bladder only just
floated at the surface, it will now sink,
EXAMPLES. XLIV.
1. A man, of weight 160 lbs., floats in water with 4 cubio inches
of his body above the surface. What is his volume in cubic feet ?
2. A glass tumbler weighs 8ozs.; its external radius is 1^ ins.
and its height is 4$ ins. ; if it be allowed to float in water with its axis
vertical, find what additional weight must be placed on it to sink it.
3. What volume of iron must be attached to a wooden beam, of
length 10 ft., breadth 2 ft., and depth 5 ins., in order to sink it?
[sp. gr. of iron =7-2; sp. gr. of wood = -7.]
4. A certain body just floats in water. On placing it in sul-
phuric acid, of sp. gr. 1-85, it requires the addition of a weight of 42-5
grammes to immerse it. Find its volume.
5. A oubic foot of air weighs 1*2 ozs. A balloon so thin that the
volume of its substance may be neglected contains 1*5 cubio ft. of
coal-gas, and the envelope together with the car and appendages
weighs 1 oz. The balloon just floats in the middle of a room without
ascending or descending ; find the sp. gr. of the gas compared with
(1) air, and (2) water.
6. The mass of a litre (t.., a cubic decimetre) of air is 1-2
grammes and that of a litre of hydrogen is "089 grammes. The
material of a balloon weighs 50 kilogrammes; what must be its
volume so that it may just rise when filled with hydrogen f
7. What must be the internal volume of a balloon if the whole
mass to be raised is 500 lbs. (occupying 5 oubic ft.), the mass of a
cubio ft. of air being *081 lb. and that of a cubic ft. of the gas with
which the balloon is filled being -0054 lb.?
8. A body floats with one- tenth of its volume above the surface
of pure water. What fraction of its volume would project above the
surface if it floated in a liquid of sp. gr. 1-25?
9. A piece of iron weighing 275 grammes floats in mercury of
sp. gr. 15*59 with |ths of its volume immersed. Find the volume
and sp. gr. of the iron.
10. If an iceberg be in the form of a cube and float with a height
of 30 ft. above the surface of the water, what depth will it have below
the surface of the water, given thai the densities of ice and sea-water
areas -918 to 1-025?
XLIV. FLOATING BODIES. 217
11. A ship, of mass 1000 tons, goes from fresh water to salt
water. If the area of the section of the ship at the water line be
15000 sq. ft. and her sides vertical where they cut the water, find
how much the ship will rise, taking the sp. gr. of sea water to be
1*026..
12. A cubical block of wood of sp. gr. -8, whose edge is one foot,
floats with two faces horizontal down a fresh water river and out to
sea where a fall of snow occurs causing the block to sink to the same
depth as in the river. If the sp. gr. of sea water be 1 *025, shew that
the weight of the snow on the block is 20 ozs.
13. A wine- bottle, which below the neck is perfectly cylindrical
and has % flat bottom, is placed in pure water and is found to float
upright with 4$ inches immersed. The bottle is now removed from
the water, is dried, and immersed in oil of sp. gr. '915. How much
of it will be immersed ?
14. A piece of pomegranate wood, whose sp. gr. is 135, is
fastened to a block of lignum vitse, whose sp. gr. is '65, and the
combination will then just float in water ; shew that the volumes of
the portions of wood are equal.
15. A piece of cork, whose weight is 19 ozs., is attached to a bar
of silver weighing 63 ozs. and the two together just float in water ; if
the sp. gr. of silver be 10*5, find the sp. gr. of cork.
16. A piece of box-wood whose sp. gr. is 1*32 is fastened to a
piece of walnut-wood of sp. gr. -68 and the two together just float in
water ; compare the volumes of the two woods.
17. A rod of uniform section is formed partly of platinum
(sp. gr. = 21) and partly of iron (sp. gr. =7*5). The platinum portion
being 2 ins. long, what will be the length of the iron portion when
the whole floats in mercury (sp. gr. = 13*5) with one inch above the
surface?
18. A piece of gold, of sp. gr. 19*25, weighs 96*25 grammes, and
when immersed in water displaces 6 grammes. Examine whether
the gold be hollow and, if it be, find the size of the cavity.
19. A man, whose weight is ten stone and whose sp. gr. is 11,
just floats in water by holding under the water a quantity of cork.
If the sp. gr. of the oork be '24, find its volume.
20. A cube of wood floating in water supports a weight of
480 ozs. On the weight being removed it rises one inch. Find the
size of the cube.
21. A block of wood floats in liquid with |ths of its volume
immersed. In another liquid it floats with |rds of its volume
immersed. If the liquids be mixed together in equal quantities by
weight, what fraction of the volume of the wood would now be
immersed ?
218 HYDROSTATICS. Exs. XLIV.
22. A piece of iron, the mass of which is 26 lbs., is placed on the
top of a cubical block of wood, floating in water, and sinks it so that
the upper surface of the wood is level with the surface of the water.
The iron is then removed. Find the mass of the iron that must be
attached to the bottom of the wood so that the top may be as -before
in the surface of the water.
[Sp. gr. of iron =7*5.]
23. A cubical box of one foot external dimensions is made of
material of thickness one inch and floats in water immersed to a
depth of 3 inches. How many cubic ins. of water must be poured
in so that the water outside and inside may stand at the same level?
How deep in the water will the box then be?
24. A thin uniform rod, of weight W, is loaded at one end with
a weight P of insignificant volume. If the rod float in an inclined
position with -th of its length out of the water, prove that
[n-l)P=W.
25. An ordinary bottle containing air and water floats in water
neck downwards. Shew that if it be immersed in water to a sufficient
depth and left to itself it will sink to the bottom. What condition
determines the point at whioh it would neither rise nor sink ?
264. A body floats with part of its volume immersed
in one fluid and with the rest in another fluid ; to determine
the conditions of equilibrium.
The weight of the body must clearly be equal to the
resultant vertical thrust of the two fluids, i.e. to the sum
of the weights of the displaced portions of the two fluids.
Also the centres of gravity of the body and of the displaced
fluid must be in the same vertical line.
This includes the case of a body floating partly im-
mersed in liquid and partly in air.
266. Ex. 1. A vessel contains water and mercury. A cube of iron,
5 cms. along each edge, is in equilibrium in the fluids with its faces
vertical and horizontal. Find how much of it is in each liquid, the
specific gravities of iron and mercury being 7*7 and 13*6.
Let x cms. be the height of the part in the mercury and therefore
(5 - x) cms. that of the part in the water.
Since the weight of the iron is equal to the sum of the weights
of the displaced mercury and water, therefore
5x7-7 = *xl3-6+(5-*)xl.
/. *=2 T 8 ^oms.
Bx. 2. A piece of wood floats in a beaker of water with -fifths
of its volume inane red. When the beaker is put under the receiver
of an air-pump and the air withdrawn, how is the immersion of the
wood ujf'txud if the sp. gr. of air be "0013 ?
FLOATING BODIES. 219
Let V be the volume of the wood and xV the volume immersed
when the air is withdrawn.
9V V
The wt. of y of water together with that of =^ of air must equal
the wt. of xV of water. For each is equal to the wt. of the wood.
QV V
...T.i + f xooi3=*r.i.
.*. *= -90013,
so that the volume immersed in water is increased from *9F to
90013 P.
EXAMPLES. XLV.
1. A circular cylinder floats in water with its axis vertical, half
its axis being immersed ; find the sp. gr. of the cylinder if the sp. gr.
of the air be '0013.
2. An inch cube of a substance, of sp. gr. 1*2, is immersed in a
vessel containing two fluids which do not mix. The sp. grs. of the
fluids are 1*0 and 1*5. Find how muoh of the solid will be immersed
in the lower fluid.
3. A cub. ft. of water weighs 1000 ozs. and a cub. ft. of oil
840 ozs. The oil is poured on the top of the water without mixing
and a sphere whose volume is 36 cub. ins. and whose mass is 19
ozs. is placed in the mixture. How many cub. ins. of its volume
will be below the surface of the water, the layer of oil being suffici-
ently deep for complete immersion of the sphere?
4. A uniform cylinder floats in mercury with 5*1432 ins. of the
axis immersed. Water is then poured on the mercury to a depth of
one inch and it is found that 5*0697 ins. of the axis is below the
surface of the mercury. Find the sp. gr. of the mercury.
5. A cylinder of wood floats in water with its axis vertical and
having three-fourths of its length immersed. Oil, whose sp. gr. is
half that of water, is then poured into the vessel to a sufficient depth
to cover the cylinder. How muoh of the cylinder will now be
immersed in the water?
6. If a body be floating partially immersed in a fluid and the air
in contact with it be suddenly removed, will the body rise or sink ?
7. If a body be floating partially immersed in fluid under the
exhausted receiver of an air-pump and the air be suddenly admitted,
will the body rise or sink ?
8. A piece of wood floats partly immersed in water and oil is
poured on the water until the wood is completely covered. What
change, if any, will this make in the volume of the portion of the
wood below the water ?
220 HYDROSTATICS. Exs. XLV.
9. A body floats in water contained in a vessel placed nnder an
exhausted receiver with half its volume immersed. Air is then
forced into the receiver until its density is 80 times that of air at
atmospheric pressure. Prove that the volume immersed in water
will then be ths of the whole volume, assuming the sp. gr. of air at
atmospheric pressure to be '00125.
10. A cube floats in distilled water with % ths of its volume .
immersed. It is now placed inside a condenser where the pressure is
that of ten atmospheres ; find the alteration in the depth of immersion,
the sp. gr. of the air at atmospheric pressure being *0013.
266. A body rests totally immersed in a given fluid,
being supported by a string ; to find the tension of the string.
The vertical upward forces acting on the body are the
tension of the string and the resultant vertical thrust of
the fluid which, by Art. 260, is equal to the weight of the
displaced fluid. The vertical downward force is the weight
of the body.
Hence, for equilibrium, we have
Tension of the string + wt. of displaced fluid = wt. of the
body, so that
Tension of the string = wt. of the body wt. of the
displaced fluid.
267. The tension of the string in the previous article
is the apparent weight of the body in the given fluid, so that
the apparent weight of the body in the given fluid is less
than its real weight by the weight of the fluid which it
displaces.
If a body of weight W and sp. gr. s be immersed in water the
W W
weight of the water displaced is , so that is the apparent loss
of weight. If it be immersed in a liquid of sp. gr. *' the apparent
loss of weight is W. -
This fact is of some importance when we are " weighing n
a given body by means of a balance or otherwise. To
obtain a perfectly accurate result the weighing should be
performed in vacuo. Otherwise there will be a slight
discrepancy arising from the fact that the quantities of air
displaced by the body and by the " weights " that we use
are different. Since however the weights of the displaced
FLOATING BODIES. 221
air are in general very small compared with that of the
body this discrepancy is not very great.
If great accuracy be desired the volumes of the body
weighed and the " weights " must be determined and the
weights of the displaced air allowed for at the rate of \\ ozs.
per cubic foot.
268. Ex. An accurate balance is completely immersed in a vessel
of water. In one scale-pan some glass (sp. gr. = 2*5) is being weighed
and is balanced by a one-pound weight, whose sp. gr. is 8, which is
placed in the other scale-pan. Find the real weight of the glass.
Let the real weight of the glass be W lbs. The weight of the
water which the glass displaces therefore = ^ W%W.
The tension of the string holding the scale-pan in which is the
glass therefore
= W-\W=\W.
Again, the weight of the water displaced by the lb. wt. =lb. wt.,
so that the tension of the string supporting the scale-pan in which
is the "weight"
= llb. wt.-ilb. wt.=flb. wt.
Since the beam of the balance is horizontal, the tensions of these
two strings must be the same.
so that TF=ff = HI lbs. wt.
This is the real weight of the glass.
EXAMPLES. XLVI.
1. A body, whose wt. is 18 lbs. and whose sp. gr. is 3, is
suspended by a string. What is the tension of the string when the
body is suspended (1) in water, (2) in a liquid whose sp. gr. is 2?
2. Water floats upon mercury whose sp. gr. is 13, and a mass of
platinum whose sp. gr. is 21 is held suspended by a string so that
f ths of its volume is immersed in the mercury and the remainder of
its volume in the water. Prove that the tension of the string is half
the weight of the platinum.
3. A vessel contains mercury and water resting on the surface of
the mercury. A mass of solid gold, wholly immersed in the fluids, is
held suspended in the vessel by a fine string, the volumes immersed
in the mercury and water being as 17 : 7. Prove that the tension of
the string will be half the weight of the gold, the sp. grs. of gold and
mercury being 19 and 13 respectively.
222 HYDROSTATICS. Exs. XLVL
4. A piece of copper and a piece of silver are fastened to the ends
of a string passing over a pulley and hang in equilibrium when
entirely immersed in a liquid whose sp. gr. is 1-15. Find the ratio of
the volumes of the metals, the sp. grs. of silver and copper being 10-5
and 8-9.
5. A piece of silver and a piece of gold are suspended from the
two ends of a balance beam which is in equilibrium when the silver
is immersed in alcohol (sp. gr. = -85) and the gold in nitric acid
(sp. gr. = l*5). The sp. grs. of silver and gold being 10*5 and 19*8
respectively, find the ratio of their masses.
6. If the sp. gr. of iron be 7*6, what will be the apparent weight
of 1 owt. of iron when weighed in water, and how many lbs. of wood
of sp. gr. *6 will be required to be attached to it so that the joint body
may just float ?
7. A gold coin weighing half an ounce rests at the bottom of a
glass of water ; find the thrust on the bottom of the glass if the
sp. gr. of the coin be 18*25.
8. A solid, of weight 1 oz. , rests on the bottom of a vessel of
water ; if the thrust of the body on the bottom be }| oz. find its
sp. gr.
9. A compound of gold and alloy weighs 67 grains in air and 63
in water; the sp. gr. of the gold is 19 and that of the alloy is 10;
find the weight of gold in the compound.
10. The weights in air and water of a mixture composed of copper
and lead are as 10:9; if the sp. grs. of copper and lead be 9 and
11*5, compare the weights of the metals forming the mixture.
11. A piece of lead and a piece of wood balance one another
when weighed in air; which will really weigh the most and why?
12. The mass of a body A is twice that of a body B, but their
apparent weights in water are the same. Given that the sp. gr. of A
is $, find that of B.
13. A block of wood, of volume 26 cub. ins., floats in water with
two-thirds of its volume immersed; find the volume of a piece of
metal, whose sp. gr. is 8 times that of the wood, which, when sus-
pended from the lower part of the wood, would cause it to be just
totally immersed. "When this is the case find the upward force which
would hold the combined body just half immersed.
14. A cylindrical bucket, 10 ins. in diameter and one foot high,
is half filled with water. A half hundred- weight of iron is suspended
by a thin wire and held so that it is completely immersed in the water
without touching the bottom of the bucket. Subsequently the wire
is removed and the iron is allowed to rest on the bottom of the
bucket. By how much will the thrust on the bottom be increased in
each case by the presence of the iron ?
[The mass of a cubic foot of iron is 440 lbs.]
FLOATING BODIES. 223
269. If a body be totally immersed in a fluid whose
specific gravity is greater than that of the body, the
resultant vertical thrust on the body is greater than its
weight, and the body will ascend unless prevented from
doing so.
Ex. 1. A piece of wood, of weight 12 lbs. and sp. gr. J, is tied by
a string to the bottom of a vessel of water so as to be totally immersed.
What is the tension of the string f
Since
wt. of water displaced by the wood _ sp. gr. of water
wt. of the wood ~ sp. gr. of the wood
.*. wt. of the displaced water = $ x 12 lbs. wt. = 16 lbs. wt
For equilibrium we must have
Tension of the string + wt. of the wood
=wt. of the displaced water.
/. tension of the string = 16 - 12=4 lbs. wt.
Dx. 2. The mass of a balloon and the gas which it contains is
3500 lbs. If the balloon displace 48000 cub. ft. of air and the mass of
a cub. ft. of air be 1*25 ozs., find the acceleration with which the
balloon commences to ascend.
The weight of the air displaced by the balloon = 48000 x 1-25 oz. wt.
=3750 lbs. wt.
Hence the upward force on the balloon
= wt. of displaced air - wt. of balloon
=250 lbs. wt.=250 g poundals.
:. initial acceleration of balloon ~ - T = - ^ = ^ .
mass moved 3500 14
EXAMPLES. XLVTE.
1. A piece of cork, weighing one ounce, is attached by a string to
the bottom of a vessel filled with water so that the cork is wholly
immersed. If the sp. gr. of the oork be '25, find the tension of the
string.
2. A piece of oork weighs 12 ozs. and its cp. gr. is '24. What is
the least weight that will immerse it in water?
3. A block of wood, whose sp. gr. is *8 and weight 6 lbs., is
attached by a string, which cannot bear a strain of more than 2 lbs.
wt., to the bottom of a barrel partly filled with water in which the
block is wholly immersed. Fluid whose sp. gr. is 1*2 is now poured
into the barrel, so as to mix with the water, until the barrel is full.
Prove that the string will break if the barrel were less than two-thirds
full of water.
224 HYDROSTATICS. Exs. XLVH.
4. A block of wood, whose sp. gr. ie 1 '2 and whose weight is 6 lbs.,
is supported by a string, which cannot bear a tension of more than
1-5 lbs. wt., in a barrel partly filled with water in which the block is
wholly immersed. Fluid whose sp. gr. is *7 is now poured into the
barrel, so as to mix with the water, until it is filled. Shew that the
string would break if the barrel were less than two-thirds full of
water.
5. A cylinder of wood, whose weight is 15 lbs. and length 3 ft.,
floats in water with its axis vertical and half immersed in water.
What force will be required to depress it six inches more ?
6. A litre of air contains 1*29 grms. and a litre of coal-gas -62
grin. A balloon contains 4,000,000 litres of coal-gas and the mass of
the envelope and its appendages is 1,500,000 grms. What additional
weight will it be able to sustain in the air ?
7. A balloon containing 10 cub. ft. of hydrogen is prevented
from rising by a string attached to it. Find the tension of the
Btring, a cub. ft. of air being assumed to weigh 1-25 ozs. and the
sp. gr. of air being 14-6 times that of hydrogen.
8. The volume of a balloon and its appendages is 64,000 cub. ft.
and its mass together with that of the gas it contains is 2 tons j with
what acceleration will it commence to ascend if the mass of a cub. ft.
of air be 1*24 ozs.?
270. Conditions of equilibrium of a body partly im-
mersed in a fluid and. supported by a string attached to some
point of it.
Ti
Let P be the point of the body at which the string is
attached and let T poundals be its tension.
Let V be the volume of the body, to its wt. per unit of
volume, and G its centre of gravity.
Let V be the volume of the displaced fluid, to' its wt.
per unit of volume, and G' its centre of gravity.
Let the vertical lines through P, G, and G' meet the
surface of the fluid in the points A, B, and C.
FLOATING BODIES. 225
The vertical forces acting on the body are
(1) the tension T acting upwards through A>
(2) the weight Vw acting downwards through B,
(Art. 238)
and (3) the resultant vertical thrust FV acting up-
wards through G. (Art. 260.)
Since these three forces are in equilibrium the points A,
B, and G must be in the same straight line, and also, by
Art 47, we must have
T+ V'w'=Vw (i),
and V'u/xAC=VwxAB (ii).
Ex. 1. A uniform rod, of length 2a, floats partly immersed in a
liquid, being supported by a string fastened to one of its ends. If the
density of the liquid be times that of the rod, prove that the rod will
rest with half its length out of the liquid.
Find also the tension of the string.
Let LM be the rod, N the point where it meets the liquid, G' the
middle point of MN, and G the middle
point of the rod.
Let w be the weight of a unit volume
of the rod and \w that of the liquid.
Let the length of the immersed por- 4 Jj?^
tion of the rod be x and k the sectional
area of the rod.
The weight of the rod is fc.2a.to and ":?i":r?ir:r:?. : . : j:
that of the displaced liquid is k . x . \w.
If T be the tension of the string, the conditions of equilibrium are
T+h.x.$w=2a.k.w (1),
and k.x.$wxAC=2a.h.wxAB _ (2).
The second equation gives
2x _ AB _ LG _ a
9a~AC~LG'~2a-x t
.'. a5 3 -4ax + 3a=0.
Hence x=a, the larger solution 3a of this equation being clearly
inadmissible.
Hence half the rod is immersed.
Also, substituting this value in (1), we have
T=f ft.a.u?=wt. of the rod.
Ex. 2. A uniform rod, six feet long, can move about a fulcrum
which is above the surface of some water. In the position of equili-
brium four feet of the rod are immersed ; prove that its ep. gr. is .
Ex. 3. A uniform rod is suspended by two vertioal string?
L. M. H. 15
226
HYDROSTATICS.
attached to its extremities and half of it is immersed in water ; if ita
sp. gr. be 2 '5, prove that the tensions of the strings will be as 9 : 7.
Ex. 4. A uniform rod capable of turning about one of its ends,
which is out of the water, rests inclined to the vertical with one-
third of its length in the water ; prove that its sp. gr. is .
Stability of equilibrium.
271. When a body is floating in liquid we have shewn
that its centre of gravity G and the centre of buoyancy H
must be in the same vertical line. [Art. 262.1
Now let the body be slightly turned round, so that the
line HG becomes inclined to the vertical. The thrust of
the liquid in the new position may tend to bring the body
back into its original position, in which case the equilibrium
was stable, or it may tend to send the body still further
from its original position, in which case the equilibrium
was unstable.
atjjL_aLjiis E - ^4ii^9
Fig. 1.
'm
1=A Hjf
h'
JM1
Tyi^'jhJ-^z
ivrIV==
Fig. 2. Fig. S.
The different cases are shewn in the annexed figures.
Fig. 1 shews the body in its original position of equilibrium ;
in Figs. 2 and 3 it is shewn twisted through a small angle.
In each case H is the new centre of buoyancy and WM is
drawn vertical to meet HG in M.
In Fig. 2, where the point M is above G, the tendency
of the forces is to turn the body in a direction opposite to
STABILITY OF EQUILIBRIUM. 227
that in which the hands of a watch rotate. The body will
therefore return toward its original position and the equi-
librium was stable.
In Fig. 3, where the point if is below G t the tendency of
the forces is opposite to that of Fig. 2. The body will
therefore go further away from its original position and the
equilibrium was unstable.
[We have assumed that, in the above figures, the
vertical line through H' meets HG ; this is generally the
case for symmetrical bodies.]
It follows that the stability of the equilibrium of the
above body depends on the position of M with respect to G.
On account of its importance the point M has a name and
is called the Metacentre. It may be formally defined as
follows :
272. Metacentre. Def. If a body float freely, and
be slightly displaced so that it displaces the same quantity of
liquid as before, the point in which the vertical line through
the new centre of buoyancy meets the line joining the centre
of gravity of the body to the original centre of buoyancy is
called the Metacentre.
The body is in stable or unstable equilibrium according
as the Metacentre is above or below the centre of gravity
of the body.
It follows therefore that, to insure the stability of a
floating body, its centre of gravity must be kept as low as
possible. Hence we see why a ship often carries ballast,
and why it is necessary to load a hydrometer (Art. 280) at
its lower end.
In any given case the determination of the position
of the Metacentre is a matter of considerable difficulty.
This position depends chiefly on the shape of the vesseL
If the portion of the solid which is in contact with the
liquid be spherical, it is clear that the pressure at each
point of this spherical surface is perpendicular to the
surface and so passes through the centre of the spherical
surface. Hence the resultant thrust on the surface passes
through the centre, which is therefore the Metacentre in
this case.
152
CHAPTER XXI.
METHODS OF FINDING THE SPECIFIC GRAVITY
OF BODIES.
273. In the present chapter we shall discuss some
ways of obtaining the specific gravity of substances.
To find the specific gravity of any substance with
respect to water we have to compare its weight with that
of an equal volume of water.
The principal methods are by the use of (1) The Specific
Gravity Bottle, (2) The Hydrostatic Balance, and (3) Hydro-
meters. We shall consider these three in order.
274. Specific Gravity Bottle. This is a bottle
capable of holding a known quantity of liquid and closed
by a tightly-fitting glass stopper, which is pierced by a small
hole to allow the superfluous liquid to spirt out when the
stopper is pushed home.
(1) To find the specific gravity of a given liquid.
Let the weight of the bottle when exhausted of air be w.
When filled with water and the stopper put in let the
weight be w'.
When filled with the given liquid let its weight be w".
Then
w' w weight of the water that would fill the bottle, and
w" -w= weight of the fluid that would fill the bottle.
Since w" w and w' w are the weights of equal quan-
tities of the given liquid and water, therefore, by Art. 239,
the sp. gr. of the liquid is
w"-w
w' -w'
(2) To find the specific gravity of a given solid which is
insoluble in water.
SPECIFIC GRAVITY BOTTLE. 229
Let the solid be broken into pieces small enough to go
into the bottle and let the total weight of the pieces be W.
Put the solid into the bottle, fill it with water and put
in the stopper, and weigh. Let the resulting weight be w".
Let the weight of the bottle when filled with water only
be w.
Then
TF+ w' = sum of the weights of the solid and of the bottle
when filled with water.
Also
to" = total weight of the solid and of the bottle when filled
with water - weight of the water displaced by the
solid.
Hence, by subtraction,
W+ to' - w" = weight of the water displaced by the solid.
Therefore W and W+w'-w" are the weights of equal
volumes of the solid and water, so that the required sp. gr.
W
~ W+w'-tv"'
In performing the operations described some precautions mnst
be taken and corrections made. The water should be at some definite
temperature; a convenient temperature is 20 C.
If it were convenient the weighings should take place in vacuo.
For, as explained in Art. 267, the air displaced by the weights and the
bodies weighed has some effect on the result of a delicate experiment.
In practice the weighings take place in air and corrections are applied
to the results obtained.
275. If the body be, like sugar, soluble in water it
must be placed in a liquid in which it is insoluble. In the
case of sugar alcohol is a suitable liquid.
Again, potassium decomposes water ; it therefore should
be weighed in naphtha.
Using the notation of the last article, we have in these
cases
sp. gr. of the solid _ W
sp. gr. of the liquid ~" W + w' uP'
278. Ex. A sp. gr. bottle when filled with water weighs 1000
grain*. If 350 grain* of a powdered substance be introduced into the
bottle it weighs 1250 grains. Find the sp. gr. of the powder.
230 HYDROSTATICS.
Here 1250= 1000 + wt. of substance - wt. of the water it displaces.
.*. wt. of water displaced = wt. of substance - 250 = 100 grains.
. , wt. of substance 350 n _
A required sp. gr. = s-p -. , . . , = - = 3-5.
wt. of displaced fluid 100
EXAMPLES. XLVIH.
1. A given sp. gr. bottle weighs 7*95 grains ; when full of water
it weighs 187*63 grains and when full of another liquid 142*71 grains.
Find the sp. gr. of the latter liquid.
2. When a sp. gr. bottle is filled with water it is counterpoised
by 983 grains in addition to the counterpoise of the empty bottle and
by 773 grains when filled with aloohol; what is the sp. gr. of the
latter?
3. A sp. gr. bottle, full of water, weighs 44 grms. and when some
pieces of iron, weighing 10 grms. in air, are introduced into the bottle
and the bottle is again filled up with water the combined weight is
62*7 grms. Find the sp. gr. of iron.
4. A sp. gr. bottle completely full of water weighs 38-4 grms. , and
when 22-3 grms. of a certain solid have been introduced it weighs
49 '8 grms. Find the sp. gr. of the solid.
5. A sp. gr. bottle weighs 212 grains when it is filled with water;
60 grains of metal are put into it ; the overflow of water is removed
and the bottle now weighs 254 grains. Find the sp. gr. of the
metal.
277. The Hydrostatic Balance. This is an
ordinary balance except that it has one of its pans sus-
pended by shorter suspending arms than the other, and
that it has a hook attached to this pan to which any
substance can be attached.
(1) To find the specific gravity of a body which would
sink in water.
Let W be the weight of the body when weighed in the
ordinary manner. Suspend the body by a light thread or
wire attached to the hook of the shorter arm of the scale-
pan, and let the body be totally immersed in a vessel filled
with water.
Put weights into the other scale-pan until the beam of
the balance is again horizontal and let w be the sum of these
weights.
HYDROSTATIC BALANCE. 231
Then
w = apparent weight of the solid in water
= real weight of the body - the weight of the water it
displaces
c W - wt. of the displaced water.
.*. W-w vft. of the displaced water.
Also W wt. of the solid.
. w , A
"F^ = requl p * gr *
If the liquid be not water, but some other liquid, then
W sp. gr. of the body
W w ~ sp. gr. of the liquid '
i.e. the ratio of the sp. grs. of the body and liquid is the
ratio of the real weight of the body to the difference
between the real weight of the body and its apparent
weight when placed in the given liquid.
(2) To find the specific gravity of a body which would
float in water.
In this case the body must be attached to another body,
called a sinker, of such a kind that the two together
would sink in water.
Let W be the weight of the body alone,
W the weight of the sinker alone,
to the weight of the sinker and body together when
placed in water,
and to' the weight of the sinker alone in water.
Then
w = real wt. of the sinker and body wt. of the water
displaced by the sinker and body. (Art. 267).
= W + W' wt. of water displaced by the sinker and body.
.". W + W w = wt. of water displaced by the sinker and
body. (1)
So W to' = wt. of water displaced by the sinker
alone. (2)
Hence, by subtraction,
W- w + w' = wt. of water displaced by the body alone.
Also W real wt. of the body.
W
.*. r= ; = sp. gr. of the body.
W-w+w r a '
232 HYDROSTATICS.
It will be noted that the result does not contain W, which is the
weight of the sinker, so that in practice this weight is not required.
(3) To find the specific gravity of a given liquid.
Take a body which is insoluble in the given liquid and
in water and let its actual weight be W.
When suspended from the short arm of the hydrostatic
balance and placed in water let its apparent weight be w.
When the given liquid is substituted for water let the
apparent weight be w'.
Then w = wt. of the body wt. of the water it displaces.
Hence wt. of the water displaced = W - w.
So wt. of the liquid displaced = W to'.
Hence W w' and Ww are the weights of equal
volumes of the liquid and water.
* -^7^ = squired sp.gr.
278. Ex. 1. A pxece of copper weighs 9000 grm*. in air and
7987*5 grins, when weighed in water. Find its specific gravity.
Here
7987*5= 9000 -wt. of water displaced by the copper.
.*. wt. of displaced water =1012-5.
/. required sp. gr. = jqj^ = 8-8.
Ex. S. A piece of cork weighs 30 grms. in air; when a piece of
lead is attached the combined weight in water is 6 grms. ; if the weight
of the lead in water be 96 grms. , what is the sp. gr. of the cork t
If w be the wt. of lead in air, the wt. of water displaced by the lead
and cork = to + 30 - combined wt. in water = w + 30 - 6 = w + 24.
So wt. of water displaced by the lead = w 96.
Hence the wt. of water displaced by the cork
= (w + 24) -(to -96) = 120.
A sp. gr. of the cork = ^ = j .
Ex. 3. If a ball of platinum weigh 20*86 ozs. in air, 19*86 in
water, and 19*36 in sulphuric acid, find the sp. gr. of the platinum and
the sulphuric acid.
Wt. of the water displaced by the platinum
=20*86-19*86=1 oz.
Wt. of the sulphurio acid displaced by the platinum
^20-86- 19*36=1-5 ozs.
HYDROSTATIC BALANCE. 233
Hence the sp. gr. of the platinum = = ' = 20*86,
and the sp. gr. of sulphuric acid = -y " = 1*5.
EXAMPLES. XLIX.
1. If a body weigh 732 grins, in air and 252 grms. in water, what
is its sp. gr.?
2. A piece of flint-glass weighs 2-4 ozs. in air and 1*6 ozs. in
water ; find its sp. gr.
3. A piece of cupric sulphate weighs 3 ozs. in air and 1 -86 ozs. in
oil of turpentine. What is the sp. gr. of the cupric sulphate, that of
oil of turpentine being *88 ?
4. It is required to find the sp. gr. of potassium which de-
composes water. A piece of potassium weighing 432*5 grms. in air is
weighed in naphtha, the sp. gr. of which is *847, and is found to weigh
9 grms. What is the sp. gr. of potassium ?
5. A piece of lead weighs 30 grains in water. A piece of wood
weighs 120 grains in air and when fastened to the wood the two
together weigh 20 grains in water. Find the sp. gr. of the wood.
6. A solid, which would float in water, weighs 4 lbs., and when
the solid is attached to a heavy piece of metal the whole weighs 6 lbs.
in water; the weight of the metal in water being 8 lbs., find the
sp. gr. of the solid.
7. A body of weight 300 grms. and of sp. gr. 5 has 200 grms. of
another body attached to it and the joint weight in water is 200 grms.
Find the sp. gr. of the attached substance.
8. A piece of glass weighs 47 grms. in air, 22 grms. in water, and
25-8 grms. in alcohol. Find the sp. gr. of the alcohol.
9. A bullet of lead, whose sp. gr. is 11-4, weighs 1*09 ozs. in air
and 1 oz. in olive oil. Find the sp. gr. of the olive oil.
10. A ball of glass weighs 665*8 grains in air, 465*8 grains in
water and 297*6 grains in sulphuric acid. What is the sp. gr. of the
latter?
11. A pieoe of marble, of sp. gr. 2-84, weighs 92 grms. in water
and 98*5 grms. in oil of turpentine. Find the sp. gr. of the oil and
the volume of the marble
12. A body is weighed in two liquids, the first of sp. gr. *8 and
the other of sp. gr. 1*2. In the two oases its apparent weights are 18
and 12 grms. respectively. Find its true weight and sp. gr.
234 HYDROSTATICS. Ezs. XLIX.
13. The apparent weight of a sinker when weighed in water is 5
times the weight in vacuo of a portion of a certain substance; the
apparent weight of the sinker and substance together is 4 times the
same weight ; find the sp. gr. of the substance.
14. A given body weighs 4 times as much in air as in water and
one-third as much again in water as in another liquid. Find the sp.
gr. of the latter liquid.
279. Hydrometers. A hydrometer is an instrument
which, by being floated in any liquid, determines the
sp. gr. of the liquid. There are various forms of the
hydrometer; we shall consider two, viz. the Common
Hydrometer and Nicholson's Hydrometer.
280. Common Hydrometer. This consists of a
straight glass stem ending in a bulb, or
bulbs, the lower of which is loaded with
mercury to make the instrument float
with its stem vertical.
A
HO
ill
To find the specific gravity of a
given liquid.
When immersed in the given liquid
let the instrument float with the point
D of the stem at the surface of the
liquid.
When immersed in water let it float with the point G
of the stem in the surface of the water.
Let V be the total volume of the instrument and a the
area of the section of the stem, this section being constant
throughout the stem.
When immersed in the first liquid the portion of the
stem, whose length is AD and whose volume is a. AD, is
out of the liquid. The volume immersed is therefore
V-a.AD.
Similarly, when placed in water, the volume immersed is
V-a.AC.
In each case the weights of the displaced fluids are equal
to the weight of the instrument, so that the weights of the
fluids are the same.
Hence, if 8 be the sp. gr. of the liquid, we have
COMMON HYDROMETER. 235
8(V-a.AD) = V-a.AC.
V-a.AC
8 =
AD
In practice the instrument maker sends out the common
hydrometer graduated by marking along its stem at different
points the sp. grs. of the liquids in which the given instru-
ment would sink to these points.
A hydrometer to shew the sp. grs. of liquids of all
densities would have to be inconveniently long. Hydro-
meters are therefore usually made in sets to be used for
liquids specifically lighter than water, for medium liquids,
and for very heavy liquids respectively.
S81. Ex. 1. The whole volume of a common hydrometer is
6 cubic inches and the stem, which is a square, is $ inch in breadth; it
floats in one liquid with 2 inches of its stem above the surface and in
another with 4 inches above the surface. Compare the specific gravities
of the liquids.
In the first liquid the volume immersed
==6 - 2 -8' = l2- OUb - m8 '
In the second liquid the volume immersed
= 6 -8i = 32- Cllb * mS -
Hence, if x and * 2 be the required specific gravities, we have
191 _ 190
"32"'* 1 " 32 '**
*i
Ex. 2. The stem of a common hydrometer is cylindrical and the
highest graduation corresponds to a specific gravity of 1 whilst the
lowest corresponds to 1*2. What specific gravity will correspond to a
point exactly midway between these divisions t
Let V be the volume of that portion of the hydrometer which is
below the highest graduation, W its weight, A the area of the section
of its stem, x the distance between its extreme graduations and w
the wt. of a unit volume of water, so that
V.l.w=W (1)
(V-A.x)tfw=W (2).
.*. r= , and4.a;= .
6 HYDROSTATICS.
Hence, if s be the required sp. gr., we have
[V-bA.x]$.w=W.
= H = 1* = 1.6.
(V-%Ax)w~ W-&W
This, it will be noted, is not half way between 1 and 1-2. On this
hydrometer the distances between marks denoting equal increments
of sp. gr. are not equal.
282. Nicholson's Hydrometer. This hydrometer
consists of a hollow metal vessel A - ^ B
which supports by a thin stem a p |
small pan B on which weights can |S|lpH? I||pggppl
be placed. At its lower end is a =1I~^- ^pE=
small heavy cup or basket C, which zzz A l-^fE.
should be heavy enough to ensure ~~ -
stable equilibrium when the instru- " y^ V ~
ment is floated in a liquid. ~\/- -
The instrument may be used to I T <h^~
compare the sp. grs. of two liquids ir~~ *. G /-V."7r.
and also to find the sp. gr. of a solid.
On the stem is a well-defined mark D, and the method
consists of loading the instrument till it sinks so that this
mark is in the surface of the two liquids to be compared.
(1) To find the sp. gr. of a given liquid.
Let W be the weight of the instrument. Let w be the
weight that must be placed on the pan B, so that the point D
of the instrument may float in the level of the given liquid.
Let W\ be the weight required when water is substituted
for the given liquid.
In the first case it follows, by Art. 262, that W + to is
the weight of the liquid displaced by the instrument.
So W+Wiis the weight of the water displaced by the
instrument.
Hence W+w and W+Wx are the weights of equal
volumes of the given liquid and water.
The required sp. gr. therefore = = .
(2) To find the sp. gr. of a solid body.
Let w 1 be the weight which when placed in the pan B
will sink the instrument in water to the point D.
NICHOLSON'S HYDROMETER. 237
Place the solid upon the pan and let the weight now
required to sink the instrument to D be w t .
The weight of the solid therefore = i0j ?,.
Now place the solid in the cup C underneath the water
and let w z be the weight that must be placed in B to sink
the instrument to D.
The wt. of the solid together with w 2 have therefore the
same effect as the wt. of the solid in water together
with 10,.
.'. wt. of the solid + w*
= wt. of the solid in water + to 3 .
.". t0, w % = wt. of the solid - wt. of the solid in water.
= wt. of the water displaced by the solid.
(Art. 267).
Also w x - w % = wt. of the solid.
. . = the required sp. gr.
It will be noted that a Nicholson's Hydrometer always
displaces a constant volume of liquid, whilst the Common
Hydrometer always displaces a constant weight of liquid.
283. Ex. A Nicholson's Hydrometer when loaded with 200
grains in the upper pan sinks to the marked point in water ; a stone
is placed in the upper pan and the weight required to sink it to the
same point is 80 grains ; the stone is then placed in the lower pan and
the weight required is 128 grains; find the sp. gr. of the stone.
The weight of the hydrometer being W grains, the weight of the
fluid displaced is equal to (i) JP+200, (ii) JF+80 + wt. of stone,
and (iii) JP+128 + wt. of stone in water.
, A W+ 200 = W + 80 + wt. of stone
= W+ 128 + wt. of stone in water.
A 120=wt. of stone _ (1)
72 = wt. of stone in water
= 120- wt. of water displaced by stone (2).
. , wt.of stone 120
A required sp. gr.= = -
wt. of water displaced by stone 120 - 72
48 2 "
238 HYDROSTATICS.
EXAMPLES. L.
1. A common hydrometer weighs 2 ozs. and is graduated for
sp. grs. varying from 1 to 1*2. What should be the volumes in oubie
ins. of the portions of the instrument below the graduations 1, 1*1,
and 1*2 respectively?
2. When a common hydrometer floats in water ^ths of its
volume is immersed, and when it floats in milk ,*& f i* 8 volume
is immersed; what is the sp. gr. of milk?
3. A common hydrometer has a small portion of its bulb rubbed
off from frequent use. In consequence when placed in the water it
appears to indicate that the sp. gr. of water is 1-002; find what
fraction of its weight has been lost, if * be the specific gravity of the
substance of which the bulb is made.
4. A Nicholson's Hydrometer weighs 8 ozs. The addition of
2 ozs. to the upper pan causes it to sink in one liquid to the marked
point, while 5 ozs. are required to produce the same result in another
liquid. Compare the sp. grs. of the liquids.
5. A Nicholson's Hydrometer, of weight 4J ozs., requires weights
of 2 and 2| ozs. respectively to sink it to the fixed mark in two different
fluids. Compare the sp. grs. of the two fluids.
6. A Nicholson's Hydrometer is of weight 3| ozs., and a weight of
If ozs. is necessary to sink it to the fixed mark in water. What
weight will be required to sink it to the fixed mark in a liquid of
density 2-2?
7. A certain solid is placed in the upper cup of a Nicholson's
Hydrometer, and it is then found that 12 grains are required to sink
the instrument to the fixed mark; when the solid is placed in the
lower cup 16 grains are required, and when the solid is taken away
altogether 22 grains are required ; find the sp. gr. of the solid.
8. The standard weight of a Nicholson's Hydrometer is 1250
grains. A small substance is placed in the upper pan and it is found
that 530 grains are needed to sink the instrument to the standard
point ; when the substance is placed in the lower pan 620 grains are
required. What is the sp. gr. of the substance ?
9. In a Nicholson's Hydrometer if the sp. gr. of the weights be
8, what weight must be placed in the lower pan to produce the same
effect as 2 ozs. in the upper pan?
284. If two liquids do not mix there is another
method, by using a bent tube, of comparing their sp. grs.
ABC is a bent tube having a uniform section and
straight legs.
The two liquids are poured into the two legs and rest
U TUBES.
239
with their common surfaces at D, and with the surfaces in
contact with the air at P and Q.
Let D be in the leg BC and E the point of
the leg AB which is at the same level as D.
Let ! and * a be the sp. grs. of the fluids.
If w be the weight of a unit volume of
the standard substance, the pressures at E and
D are respectively
. 8 1 .w.EP + IL and a t .w. DQ + 11,
where II is the pressure of the air.
For equilibrium these two pressures must
be the same.
.*. 8 1 . w . EP + n = s a . w . DQ + n.
. 8 J= DQ
"* 3 EP 1
i.e. the sp. grs. of the two fluids are inversely as the heights
of their respective columns above the common surface.
EXAMPLES. LL
1. The lower portion of a U tube contains mercury. How many
inches of water must be poured into one leg of the tube to raise the
mercury one inch in the other, assuming the sp. gr. of mercury to be
13-6?
2. Water is poured into a U tube, the legs of which are 8 inohes
long, till they are half full. As much oil as possible is then poured
into one of the legs. What length of the tube does it occupy, the sp.
gr. of oil being | ?
3. A uniform bent tube oonsists of two vertical branches and of
a horizontal portion uniting the lower ends of the vertical portions.
Enough water is poured in to occupy 6 inches of the tube and then
enough oil to occupy 5 inches is poured in at the other end. If the
sp. gr. of the oil be , and the length of the horizontal part be 2
inches, find where the common surface of the oil and water is
situated.
CHAPTER XXII.
PRESSURE OF GASES.
285. We have pointed out in Art. 224 that the
essential difference between gases and liquids is that the
latter are practically incompressible whilst the former are
very easily compressible.
The pressure of a gas is measured in the same way as
the pressure of a liquid. In the case of a liquid the pressure
is due to its weight and to any external pressure that may
be applied to it. In the case of a gas, however, the external
pressure to which the gas is subjected is, in general, the
chief cause which determines the amount of its pressure.
286. Air has weight. This may be shewn ex-
perimentally as follows :
Take a hollow glass globe, closed by a stopcock, and by
means of an air-pump (Art. 314) or otherwise (e.g. by
boiling a small quantity of water in the globe) exhaust it of
air and weigh the globe very carefully.
Now open the stopcock and allow air at atmospheric
pressure to enter the globe and again weigh the globe very
carefully.
The globe appears to weigh more in the second case
than it does in the first case. This increase in the weight
is due to the weight of the air contained in the globe.
The sp. gr. of air referred to water is found to be
001293, i.e. the weight of a cubic foot of air is about
1*293 ounces.
287. Pressure of the atmosphere.
Take a glass tube, 3 or 4 feet long, closed at one end B
and open at the other end A. Fill it carefully with mer-
cury. Invert it and put the open end A into a vessel
DEFG of mercury, whose upper surface is exposed to the
atmosphere. Let the tube be vertical.
PRESSURE OF THE AIR. 241
The mercury inside the tube will be found to descend
till the surface of the mercury inside the
tube is at a point C whose height above J G
the level H of the mercury in the vessel
is about 29 or 30 inches.
For clearness suppose the height to
be 30 inches. The pressure on each
square inch just inside the tube at H
is therefore equal to the weight of the
superincumbent 30 inches of mercury.
But the pressure at H just inside 'f
the tube is equal to the pressure at the surface of the
mercury in the vessel, which again is equal to the pressure
of the atmosphere in contact with it.
Hence in our case the pressure of the atmosphere is
the same as that produced by a column of mercury
30 inches high.
This experiment is often referred to as Torricelli's
experiment.
If the height of the column of mercury inside the tube
be carefully noted it will be found to be continually
changing. Hence it follows that the pressure of the at-
mosphere is continually changing. It is, in general, less
when the atmosphere has in it a large quantity of vapour.
288. The pressure of the atmosphere may, when the
height of the column HO is known, be easily expressed in
lbs. wt. per sq. foot or sq. inch.
For the density of pure mercury is 13*596 times that of
water, i.6. it is 13596 ounces per cubic foot.
When the height of the column HO is 30 inches the
pressure of the atmosphere per sq. inch
= wt. of 30 cubic inches of mercury,
i=30* 13596 lbn wt
= 30x 1728xl6 lb8 - wt
= 14-75. .. lbs. wt.
Similarly in centimetre-gramme units, if the height of
the column be 76 cms. the pressure of the atmosphere per
bq. cm. = wt. of 76 cub. cms. of mercury
l> m. h. 16
242
HYDROSTATICS.
= wt. of 76 x 13-596 cub. cms. of water
= 76 x 13*596 grammes wt.
= 1033*296 grammes wt.
= 1013663-376 dynes.
289. Barometer. The barometer is an instrument
for measuring the pressure of the air. In its simplest form
it consists of a tube and reservoir similar to that used in
the experiment of Art. 287 and contains liquid supported
by atmospheric air. The pressure is measured by the height
of the liquid inside the tube above the level of the liquid in
the reservoir.
The liquid generally employed is mercury, on account of
its great density. Glycerine is sometimes used instead.
The ordinary height of the mercury barometer is between
29 and 30 inches.
If water were used the height would be about 33 to
34 feet.
290. Siphon Barometer. The usual form
barometer in practice is a bent tube ABC, the
diameter of the long part AB being considerably
smaller than that of the short part BC. It is
placed so that the two portions of the tube
are vertical.
The end of the short limb is exposed to the
atmosphere and the end A of the long limb
is closed. The long limb is usually about
3 feet long and inside the tube is a quantity
of mercury. Above the mercury in the long
tube there is a vacuum.
When the surfaces of the mercury in the
are at P and C respectively, the pressure of
measured by the weight of a column of mercury whose
height is equal to the vertical distance between P and C,
i.e. to the vertical distance PD where D is a point on the
long limb at the same level as C.
For, since there is a vacuum above P, the pressure at D
is equal to the weight of a column of mercury of height DP.
THE BAROMETER. 243
Again, since C and D are at the same level the pressures
at these two points are the same ; also the pressure of the
mercury at C is equal to the pressure of the atmosphere.
Hence the pressure of the atmosphere is equal to the
weight of the column DP.
The tube DP is marked at regular intervals with num-
bers shewing the height of the barometer corresponding to
each graduation.
291. Graduation of a barometer. In graduating a
barometer there is one important point to be taken into
consideration, and that is that if the level of the mercury in
BA rises the level of that in BO must fall. The required
height of the barometric column is always the difference
between these two levels.
Suppose the section of the part BA to be uniform and
equal to x^th of a square inch and that the section of the
larger tube near C is uniform and equal to one square inch.
Also suppose that the level of the mercury in the smaller
tube appears to rise one inch. Since the increase of the
volume of mercury in one tube corresponds to a decrease
in the other, it follows that the level of the mercury in the
shorter tube has fallen y^th of an inch.
Hence the height between the two levels has increased
Dv (1 + tu) or Tff*'* 18 f an inch. Therefore an apparent
increase of one inch in the height of the mercury does, in
our case, correspond to a real increase of yjths of an inch.
So an apparent increase of yy ^ J1C ^ 1 corresponds to a real
increase of one inch.
To avoid the trouble of having to make this correction,
the tube BA is divided into intervals of \% inch, and the
markings are made as if these intervals are really inches.
More generally. Let the smaller tube be of sectional
area A and the bigger of sectional area A\ and suppose
both A and A' to be constant.
A rise of a; in the level of the mercury in the long tube
would cause a fall of -r, x in the short tube.
A
Hence an apparent increase of as in the height of the
162
244 HYDROSTATICS.
barometric column would correspond to a real increase of
A . .A + A'
& + "J", , * Of 77 *
So an apparent increase of . j, x would correspond to
a real increase of x.
Hence, to ensure correctness, the distances between the
successive graduations in the long tube are shorter than
they are marked in the ratio A' : A + A'.
Since the sp. gr. of mercury, like other liquids, alters
with its temperature, the latter must be observed in making
an accurate reading of the barometer. The length of the
corresponding column at the standard temperature C. can
then be calculated.
EXAMPLES. UL
1. At the bottom of a mine a mercurial barometer stands at
77-4 cms. ; what would be the height of an oil barometer at the same
place, the sp. grs. of mercury and oil being 13*596 and *9 ?
2. If the height of the water 'barometer be 1033 cms., what will
be the pressure on a circular disc whose radius is 7 cms. when it is
sunk to a depth of 50 metres in water ?
3. When the barometer is standing at 30 ins. find the total
pressure of the air on the surface of a man, assuming that the area
of this surface is 18 sq. ft. and that the sp. gr. of mercury is 13*596.
4. Glycerine rises in a barometer tube to a height of 26 ft. when
the mercury barometer stands at 30 ins. The sp. gr. of mercury
being 13*6, find that of glycerine.
If an iron bullet be allowed to float on the mercury in a barometer,
how would the height of the mercury be affected ?
5. The diameter of the tube of a mercurial barometer is 1 cm.
and that of the cistern is 4*5 cms. If the surface of the mercury in
the tube rise through 2*5 cms., find the real alteration in the height
of the barometer.
6. The diameter of the tube of a mercurial barometer is in.
and that of the cistern is 1 ins. When the surface of the mercury
rises 1 in., find the real alteration in the height of the barometer.
292. Connection between the pressure and density of a
gas.
It is easy to shew that the density of a gas alters when
its pressure alters.
BOYLE'S LAW. 245
Take an ordinary glass tnmbler and immerse it mouth
downwards in water, taking care always to keep it vertical.
As the tumbler is pushed down into the water the latter
rises inside the tumbler, shewing that the volume of the
air has been reduced.
Also the pressure of the contained air, being equal to
the pressure of the water with which it is in contact, is
greater than the pressure at the surface of the water. Also
the pressure at the surface of the water is equal to atmo-
spheric pressure, which was the original pressure of the
contained air. Hence we see that whilst the contained "air
is compressed its pressure is increased.
Consider again the case of a boy's pop-gun. To expel
the bullet the boy sharply pushes in the piston of the gun,
thereby reducing the volume of the air considerably ; since
the bullet is expelled with some velocity the pressure of the
air behind it must be increased when the volume of the air
is reduced.
As another example take a bladder with very little air
in it but tied so that this air cannot escape. Place the
bladder under the receiver of an air-pump and exhaust the
air. As the air gets drawn out its pressure on the bladder
becomes less ; the air inside the bladder is therefore subject
to less pressure, and in consequence expands and causes the
bladder to swell out.
293. The relation between the pressure and the volume
of a gas is given by an experimental law known as Boyle's
Law, which says that
The pressure of a given quantity of gas, whose temperature
remains unaltered, varies inversely as its volume.
Hence, if the volume of a given quantity of gas be doubled, its
pressure is halved ; if the volume be trebled, its pressure is one-third
of what it was originally; if its volume be halved, its pressure is
doubled.
In general, if its volume be increased n times, the corresponding
pressure becomes divided by n. This is the meaning of the expression
" varies inversely as."
This Law is generally known on the Continent under
the name of Marriotte'a Law.
i6
HYDROSTATICS.
294. In the case of air the law may be verified experi-
mentally as follows :
ABC is a bent tube of uniform bore of
which the arms BA and BG are straight. The
arm BC is much longer than BA.
At A let there be a small plug or cap
which can be screwed in so as to render the
tube BA air-tight.
First let this cap be unscrewed. Pour in
mercury at G until the surface is at the same
level D and E in the two tubes.
Screw in the cap at A tightly so that a
quantity of air is enclosed at atmospheric
pressure. b
Pour in more mercury at C until the level of the
mercury in the longer arm stands at G. The level of the
mercury in the shorter arm will be found to have risen to
some such point as F, which however is below G. It follows
that the air in the shorter arm has been diminished in
volume.
Let h be the height of the mercury barometer at the
time, and let H be the point on the longer tube at the same
level as F. Then the pressure of the enclosed air
= pressure at F
= pressure at H
= wt. of column HG + pressure at G
wt. of column HG + wt. of column h
= wt. of a column (HG + h).
final pressure _ wt. of a column (HG + h) _ HG + h
original pressure wt. of a column h h
original volume of the air _ DA
final volume of the air FA
It iafownd, when careful measurements are made, that
HG+h DA
h ~ FA'
final pressure _ original volume
original pressure final volume '
BOYLE 'S LAW.
247
i.e. final pressure : original pressure
1 1
final volume ' original volume *
This proves the law for a diminution in the volume of
the air.
295. Boyle's Law may also be verified by the follow-
ing method, which is a modifi-
cation of that of Art. 294, and
is applicable to both increases
and decreases of the volume of
the air.
AB and CD are two glass
tubes which are connected by
flexible rubber tubing and are
attached to a vertical stand.
AB is closed at the top but CD
is open. A vertical scale is fixed
to the stand, and CD can move
in a vertical direction parallel
to this scale. The rubber tubing
and the lower part of the glass
tubes are filled with mercury.
The upper part of the tube
AB is tilled with air, and its
pressure at any time is measured
by h + ED, where E is at the
same level as B and h is the
height of the mercury barometer. Raise or lower the
movable tube CD. Then in all cases it will be found that
AB
1
h + ED'
i.e. that volume oc
pressure
296. We have spoken as if Boyle's Law were perfectly accurate ;
until comparatively recent times it was supposed to be so. More
accurate experiments by Despretz and Kegnault have shewn that it
248 HYDROSTATICS.
is not strictly accurate for all gases. It is however extremely near
the truth for gases which are very hard to liquefy, such as air, oxygen,
hydrogen, and nitrogen. Most gases are rather more compressible
than Boyle's Law would imply.
A gas which accurately obeyed Boyle's Law would be called a
Perfect Gas. The above-mentioned gases are nearly perfect gases.
297. Let p be the original pressure, v' the original
volume, and p the original density of a given mass of gas.
When the volume of this gas has been altered, the
temperature remaining constant, let p be the new pressure,
v the new volume, and p the new density of the gas.
Boyle's Law states that
p _ v'
p v
i.e. p.v=p' .v' (1).
Now p . v and p . v' are each equal to the given mass of
the gas which cannot be altered.
.*. p.v = p' .v (2).
From (1) and (2), by division,
p-p"
Hence - is always the same for a given gas. Let its
value be denoted by k, so that p = kp.
Ex. Assuming the sp. gr. of air to be -0013 when the height of the
mercury barometer is 30 inches, the sp. gr. of mercury to be 13*596,
and the value of g to be 32-2, prove that the value of k, for foot-second
units, is 841906 nearly.
30
For p = zr^z x 13 *596 x g x 62 & poundals per square foot,
12
and / >=-0013x62$lbs.
= 10944780 =841906neftrly _
Id
298. Ex. 1. The sp. gr. of mercury is 13-6 and the barometer
stands at 30 ins. A bubble of gas, the volume of which is 1 cub. in.
when it is at the bottom of a lake 170 ft. deep, rises to the surface.
What will be its volume when it reaches the surface f
BOYLE'S LAW. 249
If w be the weight of a cab. ft. of the water, the pressure per
sq. ft. at the bottom of the lake
= 170 w + 13-6 x2\w
=204*.
Also the pressure at the top of the lake = 13-6 x 2Jw
=34 tc.
Hence, if x be the required volume, we have
sx34u7 = lx2O4i0.
.-. *=6 cub. ins.
Ex. 2. At what depth in water would a bubble of air sink, given
that the weights of a cub. ft. of water and air are respectively 1000 and
1J ozs. t and that the height of the water barometer is 34 ft. ?
Let x be the depth at which the bubble would just float. Then,
by Boyle's Law,
x + 34 _ density of air at depth x
34 ' atmospheric density
density of water
atmospheric density
-5=300.
, since the bubble just floats,
.-. a; =27166 ft. = slightly greater than 6 miles.
Below this depth the bubble would sink ; above this depth it would
rise.
Ex. 3. 10 cub. cms. of air at atmospheric pressure are measured
off. When introduced into the vacuum of a barometer they depress the
mercury, which originally stood at 76 cms., and occupy a volume of
15 cub. cms. What is the final height of the barometer ?
Let II denote the atmospheric pressure. By Boyle's Law we have
final pressure of the air _ original volume _ 10 _ 2
II " final volume 15 "" 8 *
.*. final pressure of the air = | II.
The pressure above the column of mercury is now of atmospheric
pressure, so that the length of the column of mercury is only } of its
original length and is therefore 25$ cm.
Ex. 4. When the reading of the true barometer is 30 ins. the
reading of a barometer, the tube of which contains a small quantity
of air whose length is then 3 & ins., is 28 ins. If the reading of the
true barometer fall to 29 ins. prove that the reading of the faulty
barometer will be 27$ ins.
At an atmospheric pressure of 30 ins. of mercury, let x ins. be
250 HYDROSTATICS. Exs.
the length of the column of air. When the length is 3 \ ins. its
pressure per square inch
= x atmospheric pressure = . w . 30,
where w is the weight of a cubio inch of mercury.
Hence, for the equilibrium of the faulty barometer, we have
5- . tc . 30 + %o . 28 = atmospheric pressure
= u>.30.
A = *.
When the real barometer pressure is 29 ins., let the height of the
faulty barometer be y ins., so that the pressure of the air per sq. inch
xx 30
=5 10
31* -y
29 =^ + 31^-y X8 = + 9nV
A t/ = 27*,
the other solution of this equation, viz. 33, being clearly inadmissible.
EXAMPLES. LIII.
1. What is the sp. gr. of the air at standard pressure (760 mms.
of mercury) when the sp. gr. of air at a pressure of 700 mms. of
mercury, referred to water at 4 0. as standard, is found to be
00119 ?
2. When the height of the mercurial barometer changes from
29*4=5 ins. to 3023 ins., what is the change in the weight of 1000 cub.
ins. of air, assuming that 100 cub. ins. of air weigh 31 grains at the
former pressure ?
3. Assuming that 100 cub. ins. of air weigh 35 grains when the
barometer stands at 30 ins., find the weight of the air that gets out
of a room when the barometer falls from 30 ins. to 29 ins., the
dimensions of the room being 30 ft. by 20 ft. by 15 ft.
4. When the water barometer is standing at 33 ft. a bubble at
a depth of 10 ft. from the surface of water has a volume of 3 cub. ins.
At what depth will its volume be 2 cub. ins. ?
5. An air-bubble at the bottom of a pond, 10 ft. deep, has a
volume of '00006 of a cub. in. Find what its volume becomes when
it just reaches the surface, the barometer standing at 30 ins. and
mercury being 13*6 times as heavy as water.
6. Assuming the height of the water barometer to be 34 ft., find
to what depth an inverted tumbler must be submerged so that the
volume of the air inside may be reduced to one-third of its original
volume.
mi. PRESSURE OF GASES. 251
7. A cylindrical test tube is held in a vertical position and im-
mersed mouth downward in water. When the middle of the tube is
at a depth of 32-75 ft. it is found that the water has risen halfway
up the tube. Find the height of the water barometer.
8. A uniform tube closed at the top and open at the bottom is
plunged into mercury, bo that 25 cms. of its length is occupied by
gas at an atmospheric pressure of 76 cms. of mercury ; the tube is
now raised till the gas occupies 50 oms. ; by how much has it been
raised?
9. What are the uses of the small hole which is made in the lid
of a teapot and of the vent-peg of a beer barrel ?
10. A hollow closed cylinder, of length 2 ft., is full of air at the
atmospheric pressure of 15 lbs. per square inch when a piston is
12 ins. from the base of the cylinder ; more air is forced in through
a hole in the base of the cylinder till there is altogether three times
as much air in the cylinder as at first ; if the piston be now allowed to
rise 4 ins., what is the pressure of the air on each side of the piston ?
Through how many inches must the piston move from its original
position to be again in equilibrium ?
11. A balloon half filled with coal-gas just floats in the air when
the mercury barometer stands at 30 ins. What will happen if the
barometer sinks to 28 ins. ?
What would happen if the balloon had been quite full of gas at
the higher pressure ?
12. Why does a small quantity of air introduced into the upper
part of a barometer tube depress the mercury considerably whilst a
small portion of iron floating on the mercury does not depress it ?
13. A barometer stands at 30 ins. The vacuum above the
mercury is perfect, and the area of the cross- section of the tube is a
quarter of a sq. in. If a quarter of a cub. in. of the external air be
allowed to get into the barometer, and the mercury then fall 4 ins.,
what was the volume of the original vacuum ?
14. A bubble of air having a volume of 1 cub. in. at a pressure
of 30 ins. of mercury escapes up a barometer tube, whose cross- section
is 1 sq. in. and whose vacuum is 1 in. long. How much will the
mercury descend?
15. The top of a uniform barometer tube is 33 ins. above the
mercury in the tank, but on account of air in the tube the barometer
registers 28*6 ins. when the atmospheric pressure is equivalent to that
of 29 in. of mercury. What will be the true height of the barometer
when the height registered is 29*48 ins. ?
16. The top of a uniform barometer tube is 36 ins. above the
surface of the mercury in the tank. In consequence of the pressure
of air above the mercury the barometer reads 27 in. when it should
read 28-5 ins. What will be the true height when the reading of the
barometer is 30 ins. ?
252 HYDROSTATICS. Exs. LIII.
17. The readings of a true barometer and of a barometer which
contains a small quantity of air in the upper portion of the tube are
respectively 30 and 28 ins. When both barometers are placed under
the receiver of an air-pump from which the air is partially exhausted,
the readings are observed to be 15 and 14*6 ins. respectively.
Prove that the length of the tube of the faulty barometer measured
from the surface of the mercury in the basin is 31-35 ins.
18. The two limbs of a Marriotte's tube are graduated in inches.
The mercury in the shorter tube stands at the graduation 4, and
5 ins. of air are enclosed above it. The mercury in the other limb
stands at the graduation 38, and the barometer at the time indicates
a pressure of 29'5 ins. Find to what pressure the 5 ins. of air are
subjected and also the length of the tube they would occupy under
barometric pressure alone.
19. A gas-holder consists of a cylindrical vessel inverted over
water. Its diameter is 2 ft. and its weight 60 lbs. Find what part
of the weight of the cylinder must be counterpoised to make it supply
gas at a pressure equivalent to that of 1 in. of water.
20. A pint bottle containing atmospheric air just floats in water
when it is weighted with 5 ozs. The weight is then removed and the
bottle immersed neck downwards and gently pressed down.
Shew that it will just float freely when the level of the water inside
the bottle is 11 ft. below the surface, and will sink if lowered further,
and rise if raised higher. The water barometer stands at 33 ft. and
a pint of water weighs 20 ozs.
21. A closed air-tight cylinder, of height 2a, is half full of water
and half full of air at atmospheric pressure, which is equal to that
of a column, of height h, of the water. Water is introduced with-
out letting the air escape so as to fill an additional height k of
the cylinder, and the pressure of the base is thereby doubled. Prove
that
k=a+h- Jah+h*.
22. In a vertical cylinder, the horizontal section of which is a
square of side 1 ft., is fitted a weightless piston. Initially the air
below the piston occupies a space 7 ft. in length and is at the same
pressure as the external air. 6 cub. ft. of water are taken and two-
thirds of a cub. ft. of iron. If the iron be placed on the piston
it sinks 1 ft. If the water also be then poured on it, it sinks
through $ ft. Find the sp. gr. of iron and the height of the water
barometer.
299. Relations between the pressure, temperature, and
density of a gas.
It can be shewn experimentally that a given mass of
gas, for each increase of 1 C. in its temperature, has its
volume increased (provided its pressure remain constant) by
PRESSURE OF GASES. 253
an amount which is equal to *003665 times its volume at
0C.
Thus, if P be the volume of the given mass of gas at
temperature C. and a stand for '003665, the increase in
volume for each degree Centigrade of temperature is aV .
Hence the increase for t C. is aV . t, so that if V be the
volume of this air at temperature t C, then
r=V +aV t=V (l + at).
If p and p be the respective densities at the temperatures
t C. and 0* C, then, since
pr=PoV ,
y
we have - = w = 1 + a*.
P V*
* Po = P (1 + <**)
The above law is sometimes known as Gay-Lussac's
and sometimes as Charles'.
300. A relation similar to that of the previous article
holds for all gases. For those approximating to perfect
gases a is very nearly the same quantity.
If the temperature be measured by the Fahrenheit
thermometer and not the Centigrade, the value of a is
t x vts near ly [for 180 degrees on the Fahrenheit scale
equal 100 degrees on the Centigrade scale, i.e. 1 F. = | C]
301. Ex. 1. If the volume of a certain quantity of air at a
temperature of 10 C. be 300 cub. cms., what will be its volume (at the
same pressure) when its temperature is 20 C. t
If V be its volume at C, then
Hence the volume at 20 C. = F+ 20 . . V
aim
Ex. 2. The volume of a certain quantity of gas at 15 G. is 400 cub.
cms. ; if the pressure be unaltered, at what temperature will its volume
be 500 cub. cmt. t
254
HYDROSTATICS.
Let t be the required temperature. Then
500
400
volume at temp. t 0.
volume at temp. 15 0.
1 +
1 +
t
273
15
273
273 + t
'' 288
t = 87G.
302. Suppose the gas at a temperature 0* C. to be
confined in a cylinder, and to support a piston of
such. a weight that the pressure of the gas is p,
and let the density of the gas be o , so that
p~** a).
Let heat be applied to the cylinder till the
temperature of the gas is raised to t C, and let
the density then be p.
By Gay-Lussac's law we have then
p = p (1+a*) (2).
From (1) and (2), we have
p = kp(l+at),
giving the relation between the pressure, density, and
temperature of the gas.
303. Absolute temperature. If a gas were con-
tinually cooled till its temperature was far below C. and
if it did not liquefy and continued to obey Gay-Lussac's
and Boyle's Laws, the pressure would be zero for a tempe-
rature t, such that
1 + at = 0,
i.e. when
t
- = -273.
This temperature 273" is called the absolute zero of
the Air Thermometer and the temperature of the gas
measured from this zero is called the absolute temperature.
The absolute temperature is generally denoted by T t so that
T = -+6,
a
FMESSUEE OF GASES. 255
Hence
p = kp (1 + at) = kpa ( - + t\ = fyxiT 7 .
Therefore, if F be the volume of a certain quantity of
gas, we have
p. V
^~- = ka.[V.p] = kax mass of the gas = a constant.
Hence the product of the pressure and volume of any
given mass of gas is proportional to its absolute temperature.
Ex. The radius of a sphere containing air is doubled and the
temperature raised from C. to 91 G. Prove that the pressure of the
air is reduced to one-sixth of its original value, the coefficient of expan-
sion per 1 C. being j-fy .
Let p be the original and p' the final pressure, p the original and
p' the final density.
Since the radius of the sphere is doubled, the final volume is 8
times the original volume.
.-. p'=\p.
j>'_ y(l + q.91) _lr 91-1
* j>~ kp 8L 273J
1 364 _1
= 8*273~6*
EXAMPLES. LIV.
In the following examples take a as ^=^ .
1. If a quantity of gas under a pressure of 57 ins. of mercury and
at a temperature of 69 0. occupy a volume of 9 cub. ins., what volume
will it occupy under a pressure of 51 ins. of mercury and at a tempe-
rature of 16 0. ?
2. A mass of air at a temperature of 39 0. and a pressure of
32 ins. of mercury occupies a volume of 15 cub. ins. What volume
will it occupy at a temperature of 78 0. under a pressure of 54 ins. of
mercury?
3. At the sea-level the barometer stands at 750 mm. and the
temperature is 7 C, while on the top of a mountain it stands at
400 mm. and the temperature is 13 0. ; compare the weights of a
cub. metre of air at the two places.
4. A cylinder contains two gases which are separated from each
other by a movable piston. The gases are both at 0C. and the
volume of one gas is double that of the other. If the temperature of
the first be raised t, prove that the piston will move through a space
2lat
^ , where I is the length of the oylinder, and a is the coefficient
y + oat
of expansion per 1 0.
CHAPTER XXIII.
MACHINES AND INSTRUMENTS ILLUSTRATING THE
PROPERTIES OF FLUIDS.
304. Diving Bell. This machine consists of a
heavy hollow bell-shaped vessel constructed of metal and
open at its lower end. It is heavy enough to sink under
its own weight, carrying down with it the air which it
contains. Its use is to enable divers to go to the bottom
of deep water and to perform there what operations they
wish. The bell is lowered into the water by means of a
chain attached to its upper end.
As the bell sinks into the water the pressure of the
contained air, which is always equal to that of the water
with which it is in contact, gradually increases. The
volume of the contained air, by Boyle's Law, therefore
gradually diminishes and the water will rise within the bell.
To overcome this compression of the air, a tube com-
municates from the upper surface of the bell to the surface
of the water and by this tube pure air is forced down into
the bell, so that the surface of the water inside it is always
kept at any desired level. A second tube leads from the
bell to the surface of the water so that the vitiated air
may be removed.
The tension of the chain which supports the bell is
equal to the weight of the bell less the weight of the
quantity of water that it displaces. If no additional air
be pumped in as the bell descends, the air becomes more
and more compressed and therefore the amount of water
displaced continually diminishes. Hence, in this case, the
tension of the chain becomes greater and greater.
305. A diving bell is lowered into water of given
density. If no air be supplied from above find (1) the
compression of the air at a given depth a, (2) tiie tension of
B
----- &-.::
DIVING BELL. 257
the chain at this depth, and (3) the amount of air at atmo-
spheric pressure that must be forced in so tliat at this depth
the water may not rise within the bell.
(1) Let b be the height of the bell. At a depth a let
x be the length of the bell occupied by
the air and let h be the height of the
water-barometer in atmospheric air. a
Let II be the pressure of the atmo- SH=
sphere, II' the pressure of the air inside ^rrfr^fe -
the bell, and w the weight of a unit rjr:r j"""p"]f ='
volume of water,
so that II = wh,
and n . b = II' . x, by Boyle's Law. z
Hence n' = w .
x
Now the pressure of the air and the water at their com-
mon surface inside the bell must be equal for equilibrium.
.*. II' = pressure at G = ?( + a) + II
= w (x + a + h).
Equating these two values of IT, we have
hb . TX
w =w(x + a + h)
so that a? + (a + h) x - hb - 0.
This is a quadratic equation having one positive and
one negative root. The positive root is the one we require.
The compression is then b - x.
(2) If A be the area of the section of the bell, the
amount of water displaced is A . x and its weight is there-
fore wAx. Hence, if FT be the weight of the bell, the
tension of the chain
= W-wAx.
(3) Let V be the volume of the diving bell and V the
volume of atmospheric air that must be forced in to keep
the water level at D.
In this case the pressure of the air within the bell
= pressure of the water at D
= w (b + a) + II = w (a + b + h).
L. m. h. 17
25S HYDROSTATICS. Exs.
Hence a volume (V+ V) at atmospheric pressure II
(i.e. wh) must occupy a volume V at pressure to (a + b + h).
Therefore, by Boyle's Law,
(V+r)h=V(a + b+h).
.'. V = V j , giving the required volume.
306. Ex. 1. A cylindrical diving bell weighs 2 ton* and has an
internal capacity of 200 cubic feet while the volume of the material com-
posing it is 20 cubic feet. The bell is made to sink by attached weights.
At what depth may the weights be removed and the bell just not ascend,
the height of the water barometer being S3 feet?
Let x be the required depth, so that the pressure of the air con-
tained =w (x + 33), where w is the weight of a cubio foot of water.
The volume of the air then x w (as + 33)
= 200 x w . 33, by Boyle's Law.
Hence the volume of the water displaced (in oubio feet)
~ M + * + 33 '
Also the weight of this water must be 2 tons.
* "- (*)'
giving *= 94^ ft.
N.B. In this example the difference between the pressure of the
water in contact with the air inside the bell and the pressure of the
water at the bottom of the bell has been neglected.
Ex. 2. How is the tension of the rope of a diving bell affected by
opening a bottle of soda water inside the bell t
Assuming that the pressure of the air inside the soda-water bottle
is greater than that of the air in the bell, the air inside the bottle will
expand after it is released. Hence the level of the water inside the
bell will be lowered. The quantity of water displaced will therefore
become greater and the tension of the rope will be diminished.
EXAMPLES. LV.
1. A cylindrical diving bell, whose height is 6 feet, is let down
till its top is at a depth of 80 feet ; find the pressure of the contained
air, the height of the water barometer being 33$ feet.
2. How far must a diving bell descend so that the height of a
barometer within it may change from 80 to 31 inches, assuming the
sp. gr. of mercury to be 13& and the bell to be kept full of air ?
LV. DIVING BELL. 259
3. If the mercury in the barometer within a diving bell were to
rise 12 inches, at what depth below the surface would the diving bell
be ? (sp. gr. of mercury = 13 "6).
4. A diving bell with a capacity of 200 cubic feet rests on the
bottom in water of 150 feet depth. If the height of the mercury
barometer be 29*5 inches, and the sp. gr. of mercury be 13*6, find
how many cubio feet of air at atmospheric pressure are required to nil
the bell.
5. A cylindrical diving bell, whose height is 9 feet, is lowered
till the level of the water in the bell is 17 feet below the surface. The
height of the water barometer being 34 feet, find the depth of the
bottom of the bell. If the area of the section of the bell be 25 square
feet, find how much air at atmospheric pressure must be pumped
into the bell to drive out all the water.
6. A diving bell having a capacity of 125 oubio feet is sunk in
salt water to a depth of 100 feet. If the sp. gr. of salt water be 1*02
and the height of the water barometer be 34 feet, find the total
quantity of air at atmospheric pressure that is required to fill the
bell.
7. The bottom of a cylindrical diving bell is at rest at 17 feet
below the surface of water, and the water is completely excluded by
air pumped in from above. Compare the quantity of air with that
which it would contain at the atmospheric pressure, the water
barometer standing at 34 feet.
8. A cylindrical diving bell, 10 feet high, is sunk to a certain
depth and the water is observed to rise 2 feet in the bell. As much
air is then pumped in as would fill 4 Vo tns ' 'he bell at atmospheric
pressure and the surface of the water in the bell sinks one foot.
Find the depth of the top of the bell and the height of the water
barometer.
9. The height of the water barometer being 33 feet 9 inches and
the sp. gr. of mercury 13*5, find at what height a common barometer
will stand in a cylindrical diving bell which is lowered till the water
fills one-tenth of the bell How far will the surface of the water in
the bell be below the external surface of the water ?
10. A diving bell is lowered into water at a uniform rate and air
is supplied to it by a force pump so as just to keep the bell full
without allowing any air to escape. How must the quantity, i.e.
mass, of air supplied per second vary as the bell descends ?
11. A cylindrical diving bell of height T is sunk into water till its
4
lower end is at a depth nh below the surface ; if the water fill |th of
the bell, prove that the bell contains air whose volume at atmospheric
pressure would be g ( n + ^ J V, where V is the volume of the bell
and h is the height of the water barometer.
172
260
HYDROSTATICS.
Exs. LV.
This
12. A cylindrical diving bell is lowered in water and it is
observed that the depth of the top when the water fills \ of the inside
is 3J times the depth when the water fills of the inside ; prove that
the height of the cylinder is \ of the height of the water barometer.
13. A cylindrical diving bell, of height 10 feet and internal
radius 3 feet, is immersed in water so that the depth of the top is
100 feet. Prove that, if the temperature of the air in the bell be now
lowered from 20 0. to 15 C. and if 30 feet be the height of the water
barometer at that time, the tension of the chain is increased by about
67 lbs.
14. A small hole is made in the top of a diving bell; will the
water flow in or will the air flow out?
15. A diving bell is stationary at a certain depth under water
when a body falls into the water from a shelf inside the bell and
remains under the bell ; prove that the water will rise inside the bell
but that the bell will contain less water than before.
16. If the density of the air in a closed vessel be double that of
atmospheric air and the vessel be lowered into a lake, explain what
will happen if a hole be made in the bottom of the vessel, when its
depth is (1) less than, (2) equal to, (3) greater than 34 feet, which is
then the height of the water barometer.
307. The Common or Suction Pump.
pump consists of two cylinders, AB
and BC, the upper cylinder being of
larger sectional area than the lower,
and the lower cylinder being long
and terminating beneath the surface
of the water which is to be raised.
Inside the upper cylinder works
a vertical rod terminating in a pis-
ton DE, fitted with a valve F which
only opens upwards.
, This piston can move vertically
from B to L where the spout of the
pump is. At B the junction of the
two cylinders there is a valve N
which also only opens upwards. 55S^^=UY^r^Hf|
The rod is worked by a lever
GHK, straight or bent, H being the
fulcrum and K the end to which the force is applied.
Action of the Pump. Suppose the piston to be at the
lower extremity of the upper cylinder and that the water
has not risen inside the lower cylinder.
COMMON PUMP. 261
By a vertical force applied at K the piston DE is raised
the valve ^therefore remaining closed. The air between
the piston and the valve N becomes rarefied and its
pressure therefore less than that of the air in BC.
The valve N therefore rises and air goes from BC into
the upper cylinder. The air in BC in turn becomes rare-
fied, its pressure becomes less than atmospheric pressure,
and water from the reservoir rises into the cylinder CB.
When the piston reaches L its motion is reversed. The
air between it and N becomes compressed and shuts down
the valve N. When this air has been compressed, so that
its pressure is greater than that of the atmosphere, it
pushes the valve F upwards and escapes. This continues
till the piston is at B when the first complete stroke is
finished.
Other complete strokes follow, the water rising higher
and higher in the cylinder CB until its level comes above
B, provided that the height CB be less than the height of
the water barometer. This is the one absolutely essential
condition for the working of the pump.
[In practice, on account of unavoidable leakage at the valves, the
height CB must be a few feet less than the height of the water
barometer.]
At the next stroke of the piston some water is raised
above it and flows out through the spout LM. At the
same time the water below the piston will follow it up to
Z, provided the height CL be less than that of the water
barometer.
[If this latter oondition be not satisfied the water will rise only to
some point P between B and L and in the succeeding strokes only
the amount of water occupying the distance BP will be raised.]
308. The two cylinders spoken of in the previous
article may be replaced by one cylinder provided that a
valve, opening upwards, be placed somewhat below the
lowest point of the range of the piston.
The lower cylinder need not be straight but may be of
any shape whatever, provided that the height of its upper
end B above the level of the water be less than the height
of the water barometer.
262 HYDROSTATICS.
The height of the water barometer being usually about
33 feet, the lowest point of the range of the piston must be
at a somewhat less height than this above the reservoir so
that the pump may work.
# 809. Tension of the Piston rod.
Let a be the area of the piston, h the height of the water baro-
meter, and w the weight of a unit volume of water.
The tension of the piston rod must overcome the difference of the
pressures on the upper and lower surfaces of the piston.
First, let the water not have risen to the point B but let its level
beQ.
The pressure of the air above Q
= pressure of the water at Q
=pressure at C - w . CQ = to (h - CQ).
The pressure on the lower surface of the piston therefore equals
axw(h- CQ) and that on the upper is equal to axwh. Hence, if
T be the required tension, we have
T + a.w.(h-CQ) = a .w.h.
.'. T=axw.CQ.
Secondly, let the water have risen to a point P which is above the
valve N.
The pressure at a point on the upper surface of the piston
=w . DP+wh-w (h+DP).
The pressure at a point on the lower surface
=toft - w . CD-w (h - CD).
Hence we have
T+a.w{h-CD)=a.w(h+DP)
.'. T=a.w.CP.
Hence, in both cases, the tension of the rod is equal to the weight
of a column of water whose area is equal to that of the piston and
whose height is equal to the distance between the levels of the water
within and without the pump.
810. Ex. If the barrel of a common pump be 18 inches long and
its lower end 21 feet above the surface of the water and if the section of
the pipe be f^ths of that of the barrel, find the height of the water in
the pipe at the end of the first stroke, assuming the height of the water
barometer to be 32 feet.
Let A and J X A be the areas of the sections of the barrel and pipe
respectively, and let * feet be the required height. The original
/ 8 \ 9 A
volume of the air in the pump = (^.4x21) cub. ft. = cub. ft. At
the end of the first up stroke the volume
FORCING PUMP. 263
Its pressure then, by Boyle's Law,
9A
2 21
=n , * , =n -=-
( 6 -n)
where II is the external atmospherio pressure. Hence a column x of
water is supported, the pressure at the bottom being n and that at
21
the top being n .
i8 X
TT 21
But n=tt>32.
A (32 -a?) (28 -x) = 21x32.
A a: 8 -60* + 224=0.
.*. x = 4 feet nearly.
311. Lifting Pump. This is a modification of the
common pump. The top of the pump-barrel is in this case
closed and the piston rod works through a tight collar
which will allow neither air nor water to pass.
The spout is made of smaller section than in the
common pump ; instead of turning downwards it turns up
and conducts the water through a vertical pipe to the
height required.
The spout is furnished at L with a valve which opens
outwards.
As the piston rises this valve opens and the water
enters the spout. When the piston descends this valve
closes and opens again at the next upward stroke.
By this process the water can be lifted to a great
height provided the pump be strong enough.
312. Forcing Pump. In this pump the piston DE
is solid and has no valve. The lower barrel BC has a valve
at B opening upward as in the common pump.
There is a second valve F at the bottom of the upper
barrel opening outward and leading to a vertical pipe GH.
In its descending stroke the piston drives the air
through F, and in its ascending stroke the valve F is
closed, N" is opened, and the water rises in CB as in the
common pump.
When the level of the water is above B the piston in
264
HYDROSTATICS.
its descending stroke drives the water through F up into
the tube GH. In the ascending stroke of the piston the
valve F closes and prevents the water in GH from return-
ing.
In this manner after a succession of strokes the water is
raised to a height which depends only on the pressure on
the piston and the strength of the pump.
The flow in the forcing pump as just described will be
intermittent, the water only flowing during the downward
stroke of the piston.
To obtain a continuous stream the pipe from F leads
into another chamber partially filled with air. From this
chamber a tube LM t whose end is well below the air in the
chamber, leads up to the height required.
When the piston DE is on its downward stroke the air
in this chamber is being compressed at the same time that
water is being forced up the tube LM.
When the piston is on its upward stroke and the valve
F therefore closed, this air being no longer subjected to the
pressure caused by the piston endeavours to recover its
original volume. In so doing it keeps up a continuous
pressure on the water in the air chamber and forces this
water up the tube, thus keeping up a continuous flow.
313. Fire-engine. The "manual" fire-engine is
essentially a forcing pump with an air chamber.
There are however two barrels AB and A'B' each con-
FIRE-ENGINE.
265
necting with the air vessel, and two pistons, D and Z) 7 , one
of which goes down whilst the other
goes up.
The ends, T and T t of the piston
rods are attached to the ends of a bar
TMT, which can turn about a fixed
fulcrum at M.
A practically constant stream is thus
obtained; for the air chamber main-
tains the flow at the instants when the
pistons reverse their motion.
EXAMPLES. LVL
1. The height of the barometer column varies from 28 to 31
inches. What is the corresponding variation in the height to which
water can be raised by the common pomp, assuming the sp. gr. of
mercury to be 13*6 ?
2. If the water barometer stand at 33 ft. 8 ins. and if a common
pump is to be used to raise petroleum from an oil-well, find the
greatest height at which the lower valve of the pump can be placed
above the surface of the oil in the well. The sp. gr. of petroleum
is -8.
3. A tank on the sea-shore is filled by the tide whose sp. gr. is
1*025. It is desired to empty it at low tide by means of a common
pump whose lower valve is on the same level as the top of the tank.
Find the greatest depth which the tank can have so that this may be
possible when the water barometer stands at 34 ft. 2 ins.
4. One foot of the barrel of a pump contains 1 gallon (10 lbs.).
At each stroke the piston works through 4 inches. The spout is
24 feet above the surface of the water in the well ; how many foot-
pounds of work are done per stroke?
5. If the fixed valve of a pump be 29 feet above the surface of
the water, and the piston, the entire length of whose stroke is 6 inches,
be when at the lowest point of its stroke '4 inches from the fixed
valve, find whether the water will reach the pump barrel, the height
of the water barometer being 32 feet.
6. If the length of the lower pipe of a common pump above the
surface of the water be 16 feet and the area of the barrel of the pump
16 times that of the pipe, find the length of the stroke so that the
water may just rise into the barrel at the end of the first stroke, the
266
HYDROSTATICS.
Exs. LVI.
water barometer standing at 32 feet. If the length of the stroke of
the piston be one foot, find the height to which the water will rise at
the end of the first stroke.
7. A lift pump is employed to raise water through a vertical
height of 200 feet. If the area of the piston be 100 square inches,
what is the greatest force, in addition to its own weight, that will be
required to lift the piston ?
8. The area of the piston in a force pump is 10 square inches
and the water is raised to a height of 60 feet above the piston. Find
the force required to work the piston.
9. A forcing pump, the diameter of whose piston is 6 inches, is
employed to raise water from a well to a tank. If the bottom of the
piston be 20 feet above the surface of the water in the well and
100 feet below that of the water in the tank, find the least force to (1)
raise, (2) depress the piston, the friction and weights of the valves
being neglected, and the height of the water barometer being 32 feet.
314. Air pumps form another class of machines.
Their use is to pump the air out of a vessel in which a
vacuum is desired.
Smeaton's Air-Pump, This Pump consists of a
cylinder GB having valves open-
ing upwards at C and B, within
which there works a piston D
having a valve which also opens
upwards.
The valves must be very care-
fully constructed to be as air-
tight as possible.
The lower end B is connected
by a pipe with the vessel, or re-
ceiver, A, which is to be emptied
of air.
Suppose the working to commence with the piston at
B. The piston is raised and a partial vacuum thus formed
between it and B; the pressure of the air below B opens
the valve at B and air from the receiver follows the piston.
At the same time the air above D becomes condensed,
opens the valve at C, and passes out into the atmosphere.
When the piston is at C its motion is reversed; the
air between it and B becomes compressed, shuts the valve
AIR-PUMPS. 267
B, and opens the valve at D. The air that was between
the piston and B therefore passes through the piston valve
and occupies the space above the piston.
Thus in one complete stroke a quantity of air has been
removed from below B.
In each succeeding stroke the same volume of air (but
at a diminishing pressure) is removed, and the process can
be continued until the pressure of the air left in the
receiver is insufficient to raise the valves.
The advantage of the valve at C is that during the
downward stroke of the piston the pressure of the air
above it becomes much less than atmospheric pressure, and
hence the piston-valve is more easily raised than would
otherwise be the case.
Also the work which the piston has to do during its
upward stroke is considerably lessened.
315. Bate of Exhaustion of the Air. Let V be the
volume of the receiver (including the passage leading from
the receiver to the lower valve of the cylinder), and V be
the volume of the cylinder between its higher and lower
valves.
Let p be the original density of the air in the receiver
and ft the density after the first half stroke. The air
which originally occupied a volume V of density p now
occupies a volume ( F + V) and is of density p l .
* V. p = (F+ T) ft, by Boyle's Law,
7 n\
* K=T^f'P 0)-
When the piston has descended to B again a volume V
has escaped, so that we now have a volume F in the
receiver of density ft.
The process is now repeated. Hence, if p s be the
density in the receiver after the second complete stroke,
then
P*- V+r
^(rrr)*
268 HYDROSTATICS.
So the density after the third complete stroke
and the density after the nth stroke = ( ^ - ) p.
This density is never zero, so that, even theoretically, a
complete vacuum can never be obtained.
Ex. If the receiver be 6 timet as large as the barrel, find how
many strokes must be made till the density of the air is less than fc of
the original density.
Here _?L aE __ =: ?
r+r 6+1 7
6 /6\ a 36 /6\ 3 216
" * = 7 p; ^{l) P = 49 p} *=[$) '=*&"
/6\* 1296 /6\ 8 7776
*-\j) pss mi pip ' ,s \y p =im7 p -
' P*>hP, and p 6 <y.
Therefore 5 strokes must be made.
316. The double-barrelled or Hawksbee's Air-
pump. This machine consists of two cylinders, each
y
similar to the single cylinder in Smeaton's Pump and each
furnished with a piston. These two pistons are both turned
by a toothed wheel U, the teeth of which catch in suitable
teeth provided in the pistons.
AIR-PUMPS. 269
This wheel is turned by a handle Ft*.
As one piston goes up the other goes down. In the
figure the left-hand piston is descending and the right-hand
piston is ascending.
One advantage of this form of machine is that the
resistance of the air which retards one piston has the effect
of assisting the descent of the other.
The rate of exhaustion in Hawksbee's Pump can be
calculated in a similar manner to that for Smeaton's
Pump. In this case V is the volume of each cylinder
and n is the number of half strokes made by each piston,
i.e. the number of times either piston traverses its cylinder,
motions both in an upward and downward direction being
counted.
317. Mercury Gauge.
The pressure of the air in the receiver is shown at any
instant by an instrument called the mercury gauge.
This has two common forms.
In one form it is a small siphon barometer, consisting
of a small bent tube with almost equal arms. One arm
has a vacuum above the mercury and the other arm is open
and connected with the air in the receiver. As the pres-
sure in the receiver diminishes the height of the mercury
in the vacuum tube diminishes also, and the pressure of the
air in the receiver is measured by the difference of the
levels in the two arms of the gauge.
In another form it consists of a straight barometer tube,
the upper end of which communicates with the receiver,
and the lower end of which is immersed in a vessel of
mercury open to the atmosphere. As the pressure of the
air in the receiver diminishes the mercury is forced up this
tube, and the height of the mercury in the tube measures
the excess of the atmosphere pressure over the pressure of
the air in the receiver.
318. The Air-condenser or Condensing Air-
pump. The object of this instrument is exactly opposite
to that of the Air-pump, viz. to increase the pressure of
the air in a vessel instead of diminishing it.
270
HYDROSTATICS.
The condenser consists of a vessel A, to which is
attached a cylinder CB, in which
works a piston D. In the piston
D and at B (between D and the
vessel A) are valves, both of which
open downwards.
When the piston is pressed
down, the air between D and B
becomes condensed, opens the valve
B, and is forced into the vessel A.
When the piston gets to B its
action is reversed, the atmosphere
outside presses the valve D open,
and the pressure inside A, being
now greater than that of the air
between B and the piston, shuts
the valve B.
When the piston gets to the
highest point of its range the motion is again reversed and
more air is forced into A.
The vessel A is provided with a stop-cock E % which can
be used to close A when it is desired.
319. Density of the Air in the Condenser. Let V be
the volume of the vessel A, including that portion of the
cylinder below the valve j9, and V that of the cylinder
between the valve B and the highest point of the range of
the piston. In each stroke of the piston a volume V of
air at atmospheric pressure is forced into the condenser.
Hence at the end of n strokes there is in the condenser
a quantity of air which would occupy a volume V+ nP'at
atmospheric pressura
If p be the original density of the air and p H the density
after n strokes, we have
V + nV
P = y~ P>
Ex. A condenser and a Smeatorit Air-pump have equal barrels and
the same receiver, the volume of either barrel being one-tenth of that of
the receiver; if the condenser be worked for 8 strokes and then the
AIB-PUMPS. 271
pump for 6 strokes, prove that the density of the air in the receiver will
be approximately unaltered.
If P be the original density, the density at the end of 8 strokes of
the condenser = x " p = p.
Also the density at the end of 6 strokes of the pomp
10
18 / V \ 18
*!!
1800000
177156r 1016 >-
Henoe the final density is very nearly equal to the original
density.
EXAMPLES. LVII.
1. Find the ratio of the receiver of Smeaton's Air-pomp to that
of the barrel, if at the end of the fourth stroke the density of the air
in the receiver is to its original density as 81 : 256.
2. The cylinder of a single-barrelled air-pump has a sectional
area of 1 square inch, and the length of the stroke is 4 inches. The
pump is attached to a receiver whose capacity is 36 cubic inches.
After eight complete strokes compare the pressure of the air in the
cylinder with its original pressure.
3. In one air-pump the volume of the barrel is ^th of that of the
receiver and in another it is th of the receiver. Shew that after
three ascents of the piston the densities of the air in the two receivers
are as 1728: 1331.
4. If each of the barrels of a double-barrelled air-pump has a
volume equal to one-tenth of that of the receiver, what diminution of
pressure will be produced in the receiver after four complete strokes
of the handle of the pump?
5. A bladder is one-eighth filled with atmospheric air and placed
under the receiver of an air-pump ; if the capacity of the receiver be
twice that of the barrel, prove that it will be fully distended before the
completion of the sixth stroke.
6. In the process of exhausting a certain receiver after ten strokes
of the pump the mercury in a siphon gauge connected with the receiver
stands at 20 inches, the barometer standing at 30 inches. At what
height will the mercury in the gauge stand after 20 more strokes ?
7. If the piston of an air-pump have a range of 6 inches and at
its highest and lowest positions be one-fourth of an inch from the top
and bottom of the barrel respectively, prove that the pressure of the
air in the receiver cannot be reduced below ^th of atmospheric
pressure.
[Tho portion of the barrel which is untraversed by the piston is
called tne " olearance."]
272
HYDROSTATICS.
8. If the capacity of the barrel of a condensing air-pump be
80 cubic cms. and the capacity of the receiver 1000 cubic cms., how
many strokes will be required to raise the pressure of the air in the
receiver from one to four atmospheres?
9. The volume of the receiver of a condenser being 8 times that
of the barrel, after how many strokes will the density of the air in the
receiver be twice that of the external air ?
10. If the volume of the receiver be 5 times that of the barrel,
how many strokes must be made to increase the pressure in the
receiver to 5 times the original pressure ?
11. In a condenser the area of the piston is 5 square inches and
the volume of the receiver is ten times as great as the volume of the
range of the piston. If the greatest intensity of the force that can be
used to make the piston move be 165 lbs. wt., find the greatest number
of complete strokes that can be made, the pressure of the atmosphere
being taken to be 15 lbs. wt. per sq. in.
12. If of the volume B of the cylinder of a condenser only C is
traversed by the piston, prove that the pressure in the receiver cannot
D
be made to exceed = ~ atmospheres.
Siphon. The siphon is an instrument used for
vessels containing
320.
emptying
liquid. It consists of a bent
tube ABC, one arm AB being
longer than the other BC.
The siphon is filled with the
liquid and, the ends A and G
being stopped, is inverted, the
end C of the shorter arm
being placed under the level
of the liquid in the vessel.
The instrument must be
held so that the end A is
below the level of the liquid
in the vessel.
If the ends A and C be
now opened the liquid will begin to flow at A } and will
continue to do so as long as the end A is below the surface
of the liquid.
To explain the action of the instrument. Let B be the
highest point of the siphon. Draw a line BMN vertically
downwards to meet the level of the surface of the liquid in
M and a horizontal line through A in N.
THE SIPHON. 273
Let Q be the point in which the horizontal plane through
P meets the limb BA.
Consider the forces acting on the liquid in the siphon
jnst before any motion takes place.
The pressure at Q = pressure at P
pressure of the atmosphere.
Also pressure of the liquid at A
m pressure at Q + wt. of column NM.
Hence the pressure of the liquid at A is greater than
atmospheric pressure, and therefore the fluid at A will flow
out and the liquid in the limb BA will follow.
A partial vacuum would tend to be formed at B and,
provided the height MB be less than h, the height of the
barometer formed by the liquid, liquid would be forced from
the vessel up the tube CB and a continuous flow would
take place.
321. The two conditions which must hold so that the
siphon can act are :
(1) The end A (or the level of the liquid into which A
dips) must be below the level of the liquid in the vessel
which is to be emptied. Otherwise the pressure of the
liquid at A would be less, instead of greater, than atmo-
spheric pressure, and the fluid would not flow out at A.
(2) The height of the top of the siphon above the
liquid at P must be less than the height of the correspond-
ing liquid barometer. For otherwise the pressure of the
atmosphere could not support a column so high as PB.
In the case of water the greatest height of B above P
is about 34 ft., for mercury it is about 30 ins.
822. Ex. Water is flowing out of a vessel through a siphon.
What would take place if the pressure of the atmosphere were removed
and afterwards restored (1) when the lower end is immersed in water,
(2) when it is not f
In the first case the water in the two arms of the siphon would fall
back into the two vessels and a vacuum would be left in the siphon.
On the restoration of atmospherio pressure the siphon would resume
its action.
In the second case the two arms would empty themselves as
before ; on the restoration of the air the latter would now enter the
open end of the siphon and fill it ; consequently no action would now
take place.
L. M. H. 18
274 HYDROSTATICS.
EXAMPLES. LVm.
1. Over what height can water be carried by a siphon when the
mercurial barometer stands at 30 inches (sp. gr. mercury = 136) ?
2. What is the greatest height over which a fluid (of sp. gr. 1*5)
can be carried by a siphon when the mercury stands at 30 inches, the
sp. gr. of mercury being 13*6?
3. An experimenter wishes to use a siphon to remove mercury
from a vessel 3 feet deep. Why will he not be able to remove all
of it by this means?
4. A cylindrical vessel, whose height is that of the water baro-
meter, is three-quarters full of water and is fitted with an air-tight lid.
If a siphon, whose highest point is in the surface of the lid and the
end of whose longer arm is on a level with the bottom of the vessel,
be inserted through an air-tight hole in the lid, prove that one-third
of the water may be removed by the action of the siphon.
5. What would happen if a small hole were made in (1) the
shorter limb, (2) the longer limb of a siphon in action?
TEST EXAMINATION PAPERS.
A. (Chaps. L VIL)
1. State and prove the converse of the Triangle of Forces.
Apply the Polygon of forces to shew that forces of 4 lbs. wt. acting
E., 2 lbs. wt. acting S */2 lbs. wt. acting S.W. and 3 v /2 lbs. wt. acting
N.W. are in equilibrium.
2. Find the magnitude and direction of the resultant of two
forces P and Q whose directions meet at an angle a.
What is the magnitude of the resultant of two forces, equal
respectively to 7 and 8 lbs. wt., which act on a particle at an angle of
60?
3. Find the resultant of two unequal unlike parallel forces.
A rod, 10 ft. long, whose weight may be neglected, has a mass
attached to each end and balances about a point, the pressure on
which is 12 lbs. wt. The mass at one end is 7 lbs. What is the other
mass and where is the point ?
4. Two given forces meet in a point. Prove that the algebraic
sum of their moments about any point in their plane is equal to the
moment of their resultant about the same point.
Verify this, numerically, in the case in which the forces are repre-
sented by two of the sides of a square and the point bisects one of the
other sides.
5. Define a Couple, and prove that two couples are equivalent if
their moments are algebraically the same.
Prove that any number of couples in one plane are equivalent to a
single couple whose moment is equal to the algebraic sum of the
moments of the given oouples.
6. If three forces in one plane keep a body in equilibrium, prove
that they must meet in a point or be parallel.
A rod AB is horizontal and is supported by two strings, tied to it
at A and B, which are inclined at 60 and 30 respectively to the ver-
tical. Prove that the weight of the rod outs through a point C in AB,
such that AC ^CB. Find also the tensions of the strings in terms of
the weight of the rod.
182
276 EXAMINATION PAPERS.
B. (Chaps. VUl- XL)
1. Define the Centre of Gravity of a body, and find its position in
the case of a uniform triangular lamina.
From a thin uniform rectangle, whose sides are 6 and 8 inches
respectively, a square is removed, at one corner, of side 4 inches.
Find the distance of the centre of gravity of the remainder from the
two sides of the rectangle that are uncut.
2. If a body be suspended from a point about which it can turn
freely, prove that the centre of gravity will be vertically below the
point of support.
A piece of wire, 3 ft. long, is bent into the form of 3 sides of a
square and is hung up by one end. If the side attached to the point
of support be inclined at an angle a to the horizon, prove that
4
tana^.
3. Find the mechanical advantage in that system of pulleys in
which a separate string passes under eaoh pulley and has one end
attached to the beam from which the system is suspended; the
strings are all supposed parallel and the weights of the pulleys are
neglected.
If there be 3 movable pulleys in the above system and each pulley
weigh 8 oz., what power is required to support a weight of 16 lbs.?
4. Find what horizontal force will support a body, of weight W,
on a smooth plane which is inclined to the horizon at an angle a.
A weight of 7 lbs. lies on a smooth plane inclined to the horizon
at an angle of 60. A string, attached to this weight, passes over a
pulley at the top of the plane. What is the greatest number of
weights of 1 ounce each that can be attached to the free end of the
string without making the body move up the plane ?
5. Describe the Common Balance, and state what are the requisites
of a good balance.
The arms of a balance are unequal in length but its beam is
horizontal when the scale-pans are empty ; find the real weight of a
body which, placed successively in the two scale-pans, appears to
weigh 8 and 9 lbs.
6. Enunciate the Principle of Work, and prove that the work
done in raising any number of material particles is the same as that
done in raising a particle, equal in weight to their sum, through a
distance equal to the vertical distance between the initial and final
positions of their centre of gravity.
By the principle of the equivalence of work find the force required
to move a truck, of weight 5 tons, up a smooth incline of 1 in 50.
EXAMINATION PAPERS. 277
0. (Chaps. XII. and Xm.)
1. Define Velocity and Acceleration, and prove the formula
t7=u+/t, and =tt + i/t a , explaining the meaning of the symbols
involved.
A certain particle, starting with a velocity of 2 feet per second
and moving in a straight line, moves through 35 feet in the sixth
second of its motion ; determine its acceleration, assuming it to be
uniform.
2. Explain carefully what is meant by the expression " = 32."
A man ascends the Eiffel Tower to a certain height and drops
a stone. He then ascends another 100 feet and drops another stone.
The latter takes half a second longer than the former to reach the
ground. Neglecting the resistance of the air, find the elevation
of the man when he dropped the first stone and the time it took
to drop.
3. Define the terms Mass, Gramme, and Momentum. State
the three laws of Motion and give some illustration of the First
Law.
What is meant by the Frinoiple of Inertia ?
On what grounds do we accept the truth of the Laws of Motion ?
4. Prove the relation Pmf, stating carefully the meanings of the
symbols, and the units in terms of which they are measured.
Define a Poundal and a Dyne, and obtain the relations between
them and the weights of their corresponding units of mass.
How long would it take a poundal to stop a train whose mass
is 12 tons and which has a velocity of 20 miles per hour ?
5. Distinguish between Mass and Weight. Give the experiment
and the reasoning by which we shew that the weights of two bodies
are, at the same place, proportional to their masses.
How is it that the weight of a quantity of tea appears to be
the same at all points of the earth's surface when a pair of scales
is used, but that this is not the case when a spring balance is used
instead?
6. Give examples of the Third Law of Motion, explaining care-
fully the application of the Law.
278 EXAMINATION PAPEItS.
D. (Chaps, XIV. XTL)
1, Two masses, of 2 and 5 lbs. respectively, are connected by a
light string hanging over a small pulley. Without using formula
find the acceleration of the system.
If the smaller mass be placed on a smooth table and the string be
laid on the table at right angles to the edge with the larger mass
hanging freely, find the acceleration.
2. Describe Atwood's Machine. By its use shew how to prove
that the acceleration of a given body is proportional to the force
acting on it.
3. . Define Impulse and Impulsive Force. If a shot be fired from
a gun prove that the initial momentum of the shot is equal and oppo-
site to that of the gun.
A bullet, of mass m, moving with horizontal velocity v, strikes, at
the centre of one of its plane faces, a cubical block of wood, of mass
M t which is placed on a smooth table and remains imbedded in it.
Find the velocity with which the block commences to move.
4. Define Kinetio and Potential Energy, and give illustrations.
What is meant by the Principle of the Conservation of Energy ?
Prove it for the case of a partiole falling freely under gravity.
5. Enunciate the proposition known as the Parallelogram of
Accelerations and deduce from it the Parallelogram of Forces.
6. Explain the principle of the Physical Independence of Forces,
and give illustrations.
Apply this principle to find the velocity of projection of a ball
which is thrown into the air and reaches the ground again in 3 seconds
at a distance of 108 feet from the point of projection.
7. Explain from simple dynamical principles why a tricycle is
very liable to be upset when it is ridden quickly round a corner of a
street.
A mass of 4 lbs. revolves on a smooth table, being tied to the end
of a string the other end of which is attached to the table. If the
length of the string be 30 inches and the velocity of the mass 10 feet
per second, what is the tension of the string in poundals ?
EXAMINATION PAPERS. 279
E. (Chaps. XVIL XX.)
1. Define Fluid, Liquid, Gas, and Pressure at a point.
How is it proved experimentally that pressure is transmitted
equally to all parts of a fluid, and that the pressure at any point of a
fluid at rest is the same in all directions ?
2. Define Density and Specific Gravity, and shew how they are
measured. How is the specific gravity of a mixture, consisting of
known weights of fluids of known specific gravities, obtained.
The sp. grs. of two liquids are respectively 1*3 and -8. Three lbs.
wt. of the former are added to one lb. wt. of the latter. Find the sp.
gr. of the resulting mixture.
3. Prove that the surface of a heavy liquid, whioh is at rest, is
always a horizontal plane, whatever be the shape of the containing
vesseL
4. Prove that the whole pressure on any material surface exposed
to liquid pressure is equal to the weight of a cylinder of liquid whose
base is equal to the area of the given surface, and whose height is
equal to the depth of the centre of gravity of the surface below the
surface of the liquid.
Explain how it is that the total force exerted upon the side of any
cubical cistern containing water is not proportional to the depth of the
water in the cistern.
5. Prove that the thrust exerted by a fluid on any body im-
mersed in it is equal to the weight of the fluid displaced by the body,
and acts through the centre of gravity of this displaced ilmd.
A piece of metal, of weight 10 lbs., floats in mercury of density
13*5 with f ths. of its volume immersed. Find the volume and density
of the metal.
6. Define Stability of Equilibrium and Metacentre. Explain by
figures the relation between the positions of the Metacentre and Centre
of Gravity of a body and its Stability.
Why does an ordinary plank of wood always float in water with its
length horizontal and not with its length vertical?
280 EXAMINATION PAPERS.
F. (Chaps. XXI- XXIII.)
1. Explain how we obtain the sp. gr. of a liquid or solid by means
of the Hydrostatic Balance.
If we wished to accurately determine the weight of some body,
whose sp. gr. is very small, by using Salter's Spring Balance, what
corrections should we have to apply ?
2. Shew how to find the sp. gr. of a given liquid by the use of
Nicholson's Hydrometer.
A solid is placed in the upper cup of a Nicholson's Hydrometer and
it is then found that 5 ozs. are required to sink the instrument to the
fixed point; when the solid is placed in the lower cup 7 ozs. are
wanted, and when the solid is taken away altogether 10 ozs. are re-
quired. What is the sp. gr. of the substance f
3. Explain Boyle's Law whioh connects the pressure and volume
of a gas whose temperature remains constant, and shew how it can be
verified in the case of the expansion of a gas.
How is it that a firmly-corked bottle full of air and immersed to a
great depth in the sea will have its cork driven in ?
Explain why an elastic bladder full of air would, if sunk deep
enough, then sink still further if left to itself.
4. If a diving-bell be sunk into water and no additional air be
supplied to it, prove that the tension of the supporting chain increases
with the depth.
The height of the water barometer is 84 feet and the depth below
the surface of the water of the lowest point of a diving-bell is 68 feet.
If it be now full of air, how much of this air will escape as the bell is
drawn up to the surface ?
5. Describe the single-barrelled Air-pump. What circumstances
limit the degree of exhaustion attainable with such a pump ?
6. Describe, and explain the action of, the Siphon.
What are the conditions that must hold so that it may act ?
What would be the effect of piercing a small hole at the highest
point of the siphon ?
Why cannot a siphon be used to empty the water from the hold of
a vessel which is at rest in a harbour f
APPENDIX L
Similar Triangles.
1. Two triangles are said to be equiangular when
the angles of one are respectively equal to the angles of
the other.
Thus if the three triangles ABC, A-^B-fi-n and A*B 2 C t
have (1) the angles A, A x and A 2 all equal, (2) the angles
B, B lt and J5 a all equal, and therefore (3) the angles C,
Cj, and (7, all equal, the triangles are equiangular.
2. The fundamental property of equiangular triangles
which has been used in several articles of the previous book
is " If two triangles are equiangular, the sides opposite the
equal angles in each are proportional."
For example in the above triangles
AB__BC^_ CA
A x Br B X C X " C X A X '
and AJB % ~ B % - C*AJ
The proof of this is in Euc. VI. 5.
APPENDIX.
3. As a particular case of the foregoing doctrine, con-
sider a triangle ABC in which is
drawn a line DE parallel to the base.
Since DE and BC are parallel,
we have
iADE = i ABC.
So lAED= lACB.
The triangles ADE and ABC are
therefore equiangular, so that
ADAE DE
AB~ AC~ BC
This is Euc. VI. 2.
Trigonometry.
4. In Geometry angles are measured in terms of a right
angle. This however is an inconvenient unit of measure-
ment on account of its size.
We therefore subdivide a right angle into 90 equal
parts called degrees, each degree into 60 equal parts
called minutes, and each minute into 60 equal parts called
seconds.
The symbols used for a degree, a minute, and a second
are 1, 1', and 1".
Thus 10 11 ' 12" means an angle which is equal to
10 degrees, 11 minutes and 12 seconds.
5. Trigonometrical Ratios. Def. Let a revolving
line OP start from the fixed line OA and trace out the
M
H*J. Fig. 2.
angle AOP. In the revolving line take any point P and
draw PM perpendicular to the initial line OA t produced if
necessary, and let it meet it in the point M.
TRIGONOMETRY. 283
In the triangle MOP, OP is the hypothenuse, PM is
the perpendicular, and OM the base.
Then
jyp Le. tt is called the Sine of the angle AOP.
OM. Base _ .
i.e. ^ Cosine
OP * Hyp.
MP. Perp. m
OJ/^Base" Tangent
OP . Hyp. _
I? L *Per| Cosecant.
OP . Hyp. a
m^B^e ^eca D t....
OM . Base
i.e. ^ (Cotangent.
MP ' ' Perp.
These six quantities are called the trigonometrical ratios
of the angle AOP; the three latter are not so important as
the first three and have not been used in this book ; we
shall not refer to them any more.
6. If AOP be called 0, it is clear that
i 14 .* ^P 9 OM 9 OM* + MP* t/ _
sin' 6 + coi' ^-Qp+Qp^ Qpi = 1 (Euc. 1. 47),
... sin0 MP OM MP . L
and that ^ = 7 ^ fi + -^ = 7rr>= tan 6.
cos 6 OP OP OM
These two relations are very important. It follows
that tan 8 is known when sin 6 and cos $ are known.
Values of the trigonometrical ratios in some
useful cases.
7. Angle of 45\
Let the angle A OP traced out
be 45.
Then, since the three angles of
a triangle are together equal to two
right angles,
l OPM = 180 - l POM- l PMO
- 18(T - 45' - 90* - 45* - i POM.
284
APPENDIX.
and
OM=MP = a (say),
0P= JOM*TMF = J2.a.
MP a I
0P~ J2.a~j2
;. sin 45 =
and
cos 45"
tan 45*
OM
OP
1.
a
^27a
1
7*
Angle of 30*.
Let the angle AOP traced
out be 30.
Produce PM to F making
MF equal to PM.
The two triangles OMP and
OMP' have their sides OM and
MP equal to OM and MP and
the contained angles equal.
.\0P'=0P,and i 0FP = l OPF
= 60, so that the triangle FOP
is equilateral.
Hence OF = PF* = 4iW a
where if equals a.
40P a -4a 9 ,
and
.\ OP
2a
= -^,and J/P =
= J0P =
a
;. sin 30
JfP 1
" 0P~2'
cos 30 =
OM 2a
-0P =a +J3-
x/3
" 2
tan 30 =
sin 30*
"cos 30* ~
1
73'
Angle of 60.
Let the angle AOP traced out be 60.
Take a point N on 0A t so that
2fir=aW=a(say).
The two triangles OMP and iO/P have now the sides
TRIGONOMETRY.
285
OM and MP equal to NM and
MP respectively, and the in-
cluded angles equal, so that
the triangles are eqnal
.\ PN= OP, and
iPNM=*lPOM=W.
The triangle OPN is there-
fore equilateral, and hence
OP=ON=20M=2a.
.*. J/P= JOP % -OM i = Jitf^tf= J3.a.
MP J3a_J3
OP 2a 2 '
OM a 1
Hence
sin 60"
cos 60 = -=r^ = -s- =
and
tan 60
a
OP ~Ya
sin 60
2
JS.
M
cos60 c
4wgr^ o/0\
Let the revolving line OP have turned through a very
small angle, so that the angle
MOP is very small.
The magnitude of MP is
then very small and initially,
before OP had turned through an angle big enough to be
perceived, the quantity MP was smaller than any quantity
we could assign, i.e. was what we denote by 0.
Also, in this case, the two points M and P very nearly
coincide, and the smaller the angle AOP the more nearly
do they coincide.
Hence, when the angle AOP is actually zero, the two
lengths OM and OP are equal and MP is zero.
AO MP A
Hence sin = ^ = ^ = 0,
cos =
OM OP
OP
and tan = $ = 0.
OP
= 1,
286
APPENDIX.
OM
Angle of 90\
Let the angle A OP be very nearly, but
not quite, a right angle.
When OP has actually described a
right angle the point M coincides with 0,
so that then OM is zero and OP and MP
are equal.
. A . MP OP ,
Hence sin 90 =op = op == '
cos 90' = ^=^ = 0,
OP OP '
and tan 90 = = what we call infinity
8. To shew that gin (90 - 6) = cos 0,
awe? cos (90 - #) = sin 0.
Let the angle A0P be 0.
By Euc. I. 32, z OPJf
= 180 - l POM- l 0MP
= 18O o -0-9O' = 9O o -0.
[When the angle 0PM is
considered, the line MO is the
" perpendicular " and MP is the
"base."]
Hence sin (90* - 0) = ^ =
PM
and cos (90 s - 0) = ^ =
Two angles, such as 6 and 90 6, whose sum is a
right angle are said to be complementary,
9. In the figures of Art. 5 lines measured horizontally
from towards the right are said to be positive, whilst
those measured from towards the left are negative.
Thus, in Fig. 1, OM is positive, whilst in Fig. 2 it is
negative.
Hence, for an acute angle AOP, as in Fig. 1, the cosine
is positive ; for an obtuse angle, as in Fig. 2, it is negative.
Similarly, for lines in a perpendicular direction, those
cos MOP = cos f
sin MOP = sin 0.
TRIGONOMETRY.
287
drawn towards the top of the page are positive and those
towards the bottom are negative.
10. To shew that sin (180 - 0) = sin 0,
and cos (180 - 0) = - cos 0.
We take the case only when 6 is < 90.
M A
Let the angle A OP be 6. Produce AO to A' and make
A'OF equal to 0, so that
l AOF = 180 - l A' OF = 180 - B.
On OF take F such that OF equals OP and draw
FM' perpeudicular to A0A'.
The triangles MOP and M'OF are equal in all respects,
but 0M' is negative whilst 0M is positive.
/. M'F = + MP,
And 0M' =-0M.
Hence sin (180 -0)
M'F
op =Bmd >
and
So
Exs.
Oi*
cos(18O'-0) = ^ = -^ = -cos0.
tan (180 -$) = - tan 0.
sin 150 = sin (180 - 30) = sin 30 = ,
cos 120 = cos (180 - 60) = - cos 60 = - J,
1
sin 135 = sin (180 - 45) = sin 45 =
J*'
and
cos 135 = cos (180 - 45) = - cos 45 = - -L ,
V 2
cos 150 = cos (180 -30) = -cos 30 = - ^?,
288
APPENDIX.
11. The student is advised to make himself familiar
with the following table.
30 45 60 90* 120
cosine
tangent
135 c
150 180
1
2
1
*/2
n/3
2
1
n/3
2
1
V2
1
2
1
2
1
s/2
1
2
1
2
1
">/2
^3
2
-1
1
n/3
1
^3
00
-n/3
-1
1
n/3
If the portion of the table included between the thick
lines be accurately committed to memory the rest of it may
be easily reproduced.
For, by Art. 8,
(1) the sine of 60 and 90 are respectively the cosines
of 30 and 0.
(2) the cosines of 60 and 90 are respectively the
sines of 30 and 0.
Also, by Art. 10, the sines and cosines of angles
between 90 and 180 are reduced to those of angles
between and 90.
Finally, the third line can be obtained from the other
two since the tangent is always the sine divided by the
cosine.
12. The sines and cosines of all angles between 0* and
45 are tabulated and appear in books of mathematical
tables.
Hence, when the angle is known, its sine can be found
from the tables ; so, when the sine of an angle is known,
a value for the angle itself, less than a right angle, can be
found. Similarly for the cosine or tangent.
13. If ABC be any triangle, the sides BC, CA, and AB
which are opposite to the angles A, B, and C of the triangle
are respectively denoted by a, 6, and c.
TRIGONOMETRY.
289
In any triangle ABC, to prove that
b 2 = c 2 + a 2 - 2 ca cos B.
B "DC
Fig. 1.
Draw AD perpendicular to BC.
Case I. Let B be acute.
By Euc. II. 13, we have
AC=CB* + BA'-2CB. BD,
i.e. b t = a i +c a -2a.BD.
ED
But = cos B, so that BD = c cos B.
c
.*. 6 3 =c J + a a -2accos J ff.
Case II. Let C be obtuse, as in Fig. 2.
By Euc. II. 12, we have
AC>= C& + BA> + 2CB . BD %
id. b* = a* + c i +2a.BD.
BD
But = cos ABD = cos (180 - B) = - cos J5, (Art. 10).
,\ b* = c* + a i -2accos B.
Whether B be acute or obtuse we therefore have the
same relation.
Similarly we could prove that
c 2 = a a + 6 3 -2a6cos C,
and a* = 6* + c* - 26c cos 4.
L. M. H. 19
290
APPENDIX.
14. In any triangle to prove that the sines of the angle*
are proportional to the opposite sides, i.e. that
in A sin B gin C
a " b == ""c--'
AD
In Fig 1. we have -j = sin C.
AD
In Fig. 2, we have ^- = sin ACD =* sin (180 - C)
= sin C (Art. 10).
A A
a DO
Fig.l.
In either case, AD = b sin C.
Also = sin B, so that AD = c sin B.
o
.'. c sin B = b sin C.
# sin J? _ sin C
b o
Similarly it can be shewn that each of these is equal to
sin A
APPENDIX II.
TABLE OF LENGTHS, AREAS, VOLUMES ETC.
1. The area of a Triangle = basexperp. on it from opposite
vertex = half the product of any two sides and the sine of the inolnded
angle.
2. The length of the circumference of a Circle of radius rstSvr,
where
*:* 314159265...
[An approximation to the value of tc is ^-, and this is the value that
has been used throughout this book.]
3. The area of a Circle of radius tti*.
4. The area of the surface of a Cylinder = Product of its height
and the perimeter of its base.
5. Volume of a Cylinder = Product of its height and the area of its
6. The area of the surface of a Sphere of radius r=4r 8 .
7. The Volume of a Sphere =7i-r 3 .
8. The Area of the surface of a Cone = One half the product
of the perimeter of its base and its slant side = ml.
9. The Volume of a Cone = One third the product of the height
and the area of the base =\m*h.
Values of " g ".
Plact, Ft.-Sec. units. Cm.-Sec. units.
The Equator 32-091 97810
London 32-19 981-17
North Pole 82-252 983-11
1 centimetre =-39370 inches = -032809 feet.
1 foot =30 -4797 centimetres.
1 litre = 1 cubic decimetre =1000 cubic centimetres.
1 gramme = 15-432 grains = -0022046 lb.
1 lb. =453-59 grammes.
1 poundal= 13825 dynes.
The mass of a cubic foot of water is very nearly 1000 ozs. i.e. 62
lbs.
The mass of a oubio centimetre of water at 4C. is one gramme.
The ep. gr. of mercury is 13-596.
192
APPENDIX III.
Alternative proof of equation (2) of Art. 156.
Let the time t be divided into n equal intervals, each equal to x,
so that t = nx.
The velocities of the point at the beginnings of these successive
intervals are
u, u+/x, u+2/x, u+(n-l)/x.
Hence the space b^ which would be moved through by the point, if
it moved during each of these intervals * with the velocity which
it has at the beginning of each, is
I= u . x + (u+/x) .x + {u+2fx) .x + + [u+f(n-l)x].x
= n.ux+/*2 [1 + 2 + 3 + + (n-l)]
m nux +/x 2 ^ - , on summing the arithmetic progression,
=rut + /* ( 1 - - J , since x -
Also the velocities at the ends of these successive intervals are
u+/x, u + 2/x, u + Sfx, u + nfx.
Hence the space * a which would be moved through by the point, if
it moved during each of these intervals x with the velocity which it
has at the end of each, is
,= (u+fx) . x+ {u + 2fx) . x+ (u + 3/x) .x + ~.~ + {u+nfx).x
= n.ux+/x 2 [1 + 2 + 3 + +n]
, , ,n(n + l)
^nux + fx*-^ -'
s=ut + /t a ( 1+- j , as before.
Now the true space * is intermediate between i and * 2 ; also the
larger we make n and therefore the smaller the intervals x become,
the more nearly do the two hypotheses approach to coincidence.
If we make n infinitely great, and therefore - infinitely small, the
values of s 1 and 2 both become ut + /i 2 .
Hence *=ut + bft*.
ANSWERS TO THE EXAMPLES.
I. (Page 11.)
1. (i) 25; (ii) 3^/3; (iii) 13; (iv) ^61 ; (v) 60; (vi) 8.
2. 20 lbs. wt.; 4 lbs. wt.
3. \/2 lbs. wt. in a direction south-west. 4. 205 lbs. wt.
5. P lbs. wt. at right angles to the first component.
6. 2 lbs. wt. 7. 60. 8. 3 lbs. wt. ; 1 lb. wt. 9. 120.
10. In the direction of the resultant of the two given forces.
11. -i. 12. 12 lbs. wt.
H. (Page 13.)
1. 5^3 and 5 lbs. wt. 2. *V2. 3. 50 lbs. wt.
4. 16 and 12 lbs. wt.
III. (Pages 18, 19.)
1. 1:1:^3. 2. ^3:1:2.
5. 2^/3 and ^3 lbs. wt.
14. 101 ; 57. 15. 52 ; 95.
17. 46; 138. 18. 29-6; 14.
19. ^=34-4 lbs. wt. ; ai =81 ; B a
IV. (Page 23.)
1. 4 lbs. wt. in the direction AQ.
2. 9.76 lbs. wt. nearly.
3. 2P in the direction of the middle force.
4. 7P. 5. \/3P at 30 with the third force.
6. s/S2o + 90^/3 - 48^/2 - 60^/6 = 16-3 lbs. wt.
7. 14-24 lbs. wt. 8. 5 lbs. wt. opposite the second force,
V. (Pages 25, 26.)
1. y (n/6 - */2) ; W U/3 - 1). 2. 2| and 3 lbs. wt.
6.
3.
5:4.
16. e
120.
7.
72 ; 101.
4. 40.
5 and 13.
2~
;6-5 lbs.
Wt.; a 2 =169.
ANSWERS
8.
120 lbs. <
10.
4. 56 and 42 lbs. wt.
6. 4, 8, and 12 lbs. wt.
wt.
14 lbs. wt.
11
3. 126 and 32 lbs. wt.
5. 48 and 36 lbs. wt.
7. W.
9. 1-34 inches.
11. 6 ft. 5 ins. ; 2 ft. 4 ins.
12. They are each equal to the weight of the body.
VI. (Pages 3234.)
1. (i) B=ll, AG =7 ins.; (ii) 22 = 30, AG =1 ft. 7 ins.;
(iii) 22 = 10, AG =1 ft. 6 ins.
2. (i) 12 = 8, AG = 25 ins.; (ii) 22 = 8, AG= -75 ins.;
(iii) 22=17, AG= - 19^ ins.
3. (i) Q = 9, AB = Siine.; (ii) P=2|, 22 = 13|;
(iii)Q = 6|,22 = 12|.
4. (i) Q=25, AB = S& ins.; (ii) P=24|, 22 = 13|;
(iii)Q = 2f,22 = 3f.
5. 15 and 5 lbs. wt. 6. 43 and 13 lbs. wt.
7. 98 and 70 lbs. wt.
8. The block must be 2 ft. from the stronger man.
9. 4 ft. 3 ins. 10. lib. wt. 11. 1 foot.
12. 20 lbs. ; 4 ins. ; 8 ins. 13. 14$ ins. ; 10f ins.
15. hW.
16. (i) 100 and 150 lbs. wt.; (ii) 50 and 100 lbs. wt.; (iii) 25 and
75 lbs. wt.
VII. (Pages 4244.)
1. 10-1. 2. 75^3 = 129-9^3. wi
3. 3 ft. 8 ins. from the 6 lb. wt.
4. At a point distant 6 '6 feet from the 20 lbs.
5. 2f ft. from the end. 6. 2| lbs. 7. 2lbs.
8. (i) 4 tons wt. each ; (ii) 4^ tons wt. , 3 tons wt.
9. B is 3 inches from the nearest peg. 10. f cwt.
11. One-quarter of the length of the beam.
12. The weight is 3lbs. and the point is 8 ins. from the 5 lb. wt.
13. 3 ozs. 14. 85, 85, and 29 lbs. wt.
15. 96, 96, and 46 lbs. wt. 16. 1H ins. from the axle.
ANSWERS. iU
VIII. (Page 49.)
2. 9ft.-lbs. 3. 6.
4. A force equal, parallel, and opposite, to the force at C, and
acting at a point C" in AC, such that CC is %AB.
IX. (Pages 53, 54.)
2. 45. 3. lOx/2 and 10 lbs. wt.
4. The length of the string is AG. 5.1 WV 3 ; a Tfy3.
7. 46f and V \/421 ( = 68-4) lbs. wt.
9. W cosec a and JP cot a.
10. i^V3 where TP" is the weight of the sphere.
11. 30; WJ3; \WJ$. 12. ^7:2^3.
13. W 3 ( = 23-094) lbs. wt. 14. 6 lbs. wt.
X. (Pages 59, 60.)
1. 2|ft. ; K/97 (=3-283...) ft.
2. 3^ ft. ; |V73( = 5-696)ft.; f^/ll ( = 4-807) ft. 6. 60.
XI. (Pages 62, 6a)
1. 4| inches from the end. 2. 15 inches from the end.
3. 2 feet. 4. fr ^^ fr m the middle.
5. 7 ins. from the first particle.
6. 5^$ ins. from the centre of the shilling. 7. 5:1,
8. 3*5 ins. from the base.
10. 12 lbs. ; the middle point of the rod.
XH. (Pages 66, 67.)
1. One- fifth of the side of the square.
2. It is distant -r from AB and 7 from AD, where a is the side
4 4
of the square.
3. At a point whose distances from AB and AB are respectively
16 and 15 inches.
4. 11 and 8} inches. 5. ^19 ; J283.
iv ANSWERS.
7. At the centre of gravity of the lamina.
8. 8 \ and 11$ inches. H. 4 inches from A.
12. One-quarter of the side of the square.
XIIL (Pages 70, 71.)
1. 2^ T inches from the joint.
2. 5 inches from the lower end of the figure.
3. It divides the beam in the ratio of 5 : 11.
4. At the centre of the base of the triangle.
5. 7$ inches.
6. Its distance from the centre of the parallelogram is one-ninth
of a side.
7. The distance from the centre is one-twelfth of a diagonal.
8. The distance from the centre is ^th of the diagonal.
9. It divides the line joining the middle points of the opposite
parallel sides in the ratio of 5 : 7.
10. 1& inch. 11. ^ inch from the centre.
12. The centre of the hole must be 16 inches from the centre of
the disc.
13. It is one inch from the centre of the larger sphere.
14. 13-532 inches.
XIV. (Pages 76, 77.)
1. 6 inches. 2. rV \A0 feet = 3*8 inches nearly.
3. 15a. 4. The weight of the table.
5. On the line joining the centre to the leg which is opposite to
the missing leg and at a distance from the centre equal to one-third
of the diagonal of the square.
6. 120 lbs. 7. y.
10. 18 if the bricks overlap in the direction of their lengths, and 8
if in the direction of their breadths.
XV. (Pages 81, 82.)
1. 5 feet. 2. 4 feet from the first weight; toward the first
weight. 3. 11 : 9. 4. 2 lbs.
6. 6 ins. from the 27 ounces ; If inch. 7. 360 stone wt.
8. 50 lbs. wt. 10. 20 lbs. wt.
11. A force equal to the weight of 2 cwt.
ANSWERS, v
XVI. (Pages 86, 87.)
1. (i) 320; (ii) 7; (iii) 3. 2. (i) 7; (ii) 45*; (iii) 7; (iv) 6.
3. 290 lbs. 4. lOflbs. 5. 5 lbs. 6. 5 lbs.
7. 4w ; 21w. 8. 9H lbs. wt. Q. 18 lbs. wt.
XVII. (Pages 88, 89.)
1. 6 lbs. 2. 4 strings ; 2 lbs.
3. 47 lbs.; 6 pulleys. 4. 7 strings ; 14 lbs.
W
5. = , where n is the number of strings. 6. 9 stone wt.
n + 1
XVHE. (Pages 91, 92.)
1. (i) 30 lbs.; (ii) 4 lbs.; (iii) 4.
2. (i) 161 lbs.; (ii) 16 lbs. wt.; (iii) lb. ; (iv) 5.
3. 10 lbs. wt.; the point required divides the distance between
the tirst two strings in the ratio of 23 : 5.
4. | inch from the end. 5. 18 T V
6. inch from the end. 7. W=7P + 4w ; 8 ozs. ; 1 lb. wt
8. 4; 1050 lbs.
XIX. (Pages 95, 96.)
1. 12 lbs. wt.; 20 lbs. wt. 2. 30; ^~ .
3. 103-92 lbs. wt. 5. v/3 : 1. 6.3:4; 2P.
7. -^rlbs.wt.; ^3 lbs. wt. 8. 6 lbs. wt.
tons.
Bin /3- sin a
XX. (Pages 98, 99.)
1. 7 lbs. wt. 2. 120 lbs. wt.; 140 lbs. wt.; 110}f lbs. wt.
3. 20 inches. 4. 7 feet. 5. 3| tons.
6. 3 lbs. wt.
XXI. (Pages 102, 103.)
1. 11 lbs. 2. 264 lbs. 3. 2 ozs.
4. 2 : 3 ; 6 lbs. 5. 24-494 lbs. 6. 5 : ^26.
7. WHO inches ; Jl 10 lbs. 9. 2s. Bd. ; Is. 9|d.
10. He will lose one shilling.
195
vi ANSWERS.
1.
2.
3.
4.
5.
6.
XXII. (Page 106.)
34$ inches from the fulcrum.
2 inches from the end ; 1 inch.
32 inches from the fulcrum.
4 inches.
26 lbs. ; 14 lbs. ; 10 ins. from the fulcrum,
3 ozs. 7. 30 inches.
XXHI. (Page 113.)
1, 10 lbs. wt. ; 10^/17 lbs. wt. at an angle, whose tangent is 4,
with the horizontal.
Z ' W~ 3 ~ 4 ' 14 *
3. 10 ^10 lbs. wt. at an angle, whose tangent is 3, with the horizon.
5.
i. 6. W3 1bs.wt. 8. fj.
1.
4.
XXIV. (Page 120.)
4400 lbs. 2. 5^ inches. 3. ff lbs. wt.
Iffc lbs. wt. 5. 4f lbs. wt. 6. 13if tons wt.
1.
3.
6.
8.
XXV. (Page 122.)
21120. 2. 23,040,000 ft.-lbs. ; 5& h.p.
1000 feet. 4. 9H hours. 5. &&.
4-4352. 7. 660,000 ft.-lbs.; 30 h.p.
1500 ft.-lbs.
XXVI. (Pages 128, 129.)
1, (1) 17 ft. per sec. ; 47 feet. (2) ; 24 feet.
(3) - ff ; 1-j 7 !- sees. (4) 3 ft. per sec. ; 6 sees.
2. 40 ft. per sec. ; 400 ft. 3. 40 sees. 4. 20 ft. -sec. units.
5. 10 sees, ; 150 cms. 6. I n 50 sees. ; 25 metres.
7. 18 ft.-sec. units. 8. 10 ft. per sec. ;-{ ft.-sec. unit.
9. 19 ft. per sec. ; 3 ft.-sec. units; 60 ft. 10. 5 sees. ; 12 ft.
11. 16 ft.-sec. units; 30 ft. per see.
12. 30 ft. per sec. ; - 2 ft.-sec. units.
13. 30 ft. 14. I , ^f^ , and ^ 3 ~^ 2 sees, respectively.
DO O
15. In 2 sees, at 16 ft. from 0. 16. Yes.
ANSWERS. vii
XXVn. (Pages 132134)
1. 25 ft. ; \ sec. and 2\ sees.
2. (i) In sees. ; (ii) in l\ sees.
3. In 1\ and 1\ sees. ; 50 ft.
4. (1) 1600ft.; (2) \JIQ sec.; (3) 60 ft. per sec. upwards.
5. 432 ft. 6. 44 sees. 7. 2 sees, or 5^ sees.
8. 545 ems. per sec. ; sec. 9. 10*2 sees.
10. 218 metres; 6f sees. 11. 3218. 12. 900 ft. ; 7 sees.
13. 100 ft. 14. 150 ft. 15. 144 ft.
16. 256 ft. per sec. ; 1024 ft. 17. t=5; 64 ft. per sec.
18. 784 ft. 19. 1120 ft. per sec. 20. 4080 ft.
21. 68 5 V ft. per sec. ; 306f ft. 22. 1 sec. ; 1 sees.
XXVIII. (Pages 140142.)
1. (1) I, (2)|, (^^ft.-sec. units.
2. (1) 200poundals; (2) 6ilbs. wt. 3. 15 lbs. wt.
4. 15| lbs. weight. 5. 48 ft.-sec. units ; 720 feet.
6. 1 : 64 ; 5 ft. per sec. 7. 7 sees. ; 13 ft. per sec.
8. 2 min. 56 sees. 9. 14 sees. H. 180 feet
12. H tons wt - ; H tons w *-
13. 363^ cms. per sec.; 181f cms.; 21800 cms.
14. 49-05 kilogrammes. 15. 55: 2. 16. 144 lbs.
17. 12 lbs. 18. 7H lbs. wt.; 237$ lbs. weight.
19. They are equal. 20. 110 lbs. wt. 21. 133 ft. per sec
XXIX. (Pages 147, 148.)
1. g; 7lb8.wt.
2. (1) 4 ft. -sec. units; (2) 7J lbs. wt. ; (3) 20 ft. per sec.; (4) 50 ft.
3. 18 ft.; 15f oz. wt. 4. a: = 985. 6. By 2 lbs. wt.
7. f . 8. 16 ft.
9. (1) ^ ; (2) ^5 sees.; (3) ^V 5 ft - P^ sec -
iii ANSWERS.
10. gf; 3fozs. wt. 11. 2 sees. 12. 125 grammes.
13. In the ratio 19 : 13. 14. 2 and 3 lbs. wt.; f .
o
15. 29 ft. 9 ins. nearly.
XXX. (Pages 151153.)
1. 200 ft.; 5 sees. 2. 16^/3 ft. per sec. ; f^ 3 secs - 3. 30.
4. 1 : 4. 5. ^30 sees. ; 16^/30 ft. per sec. 6. 30.
7. (1) 2 ft.-sec. units ; (2) 2^f lbs. wt. ; (3) 6 ft. per sec. ; (4) 9 ft.
8. 40 ft. 9. 24 lbs. 10 ozs. 10. 605 : 18.
11. (i) 5 min. 8 sees.; (ii) 6776 feet.
12. 1 min. 42$ sees. ; 2258| feet. 13. 5ff tons wt.
14. 1 mile 1408 yds. 15. 1224fyds. 16. &9\ 3.
17. 'I. 18. ^ sec. ; 8^/2 ft. per sea.
19. W 5 secs - ; W 5 ft- P er se
XXXI. (Pages 157, 158.)
1. Nothing. 2. (1)20 lbs. wt.; (2) 20flbs. wt.
3. (1) 154 lbs. wt.; (2) 70 lbs. wt. 4. |. 5. 2057$ feet.
6. lb. wt. ; & lb. wt. ; lb. wt.
7, 3$ ozs. wt. ; j ; 1\ ozs. wt. j 3 ozs. wt.
XXXH. (Page 162.)
1. 4 7 ft. per sec. 4. 6J ft. per seo. 5. 17* ft- per sec.
6. 6*8. ..ft. 7. 1431 ft. per sec. nearly. 8. wt. ofl04cwt.
9. The masses move with a velocity of 24 ft. per seo.
XXXni. (Page 164.)
1. 160. 2. 213. 3. 119*46. 4. 14-685 lbs. wt.
5. 21f. 6. 68&V 7. 152 ft -lbs.
ANSWERS. **
XXXIV. (Page 167.)
1. (1) 5120, (2) 1280, (3) 0, units of energy. 2. 15625.
3. 125 xlO 9 . 4. 87ff 5. 625 xlO 10 ; 3125 x10 s .
6. 3160|^ ft. -lbs.
7. (1) They are equal ; (2) They are in the ratio m : M.
XXXV. (Pages 172, 173.)
2. 100 ft. 3. 120.
5. At an angle whose cosine is - $, i.e. 126 52', with the current;
perpendicular to the current so that his resultant direction makes an
angle whose tangent is $, i.e. 59 2 / , with the current.
6. 4^3 miles per hour ; 12 miles per hour.
7. sJ29 at an angle of elevation, whose tangent is #, above a
horizontal line which is inclined at an angle, whose tangent is $,
north of east.
8. 60.
9. 14 at an angle whose cosine is with the greatest velocity.
XXXVI. (Pages 173, 174.)
2. 5 ft. per sec. at 120 with its original direction.
3. 20^/2 -as/2 ft. per sec. towards N.N.W.
4. 12 ft. per sec. at 120 with the original direction.
XXXVII. (Pages 179, 180.)
1. (1) 16 ft. ; 2 sees. ; 1109 ft. (2) 75 ft. ; 4-33... sees. ; 173-2 ft.
2. 1333$ ft. per sec.
3. 50- 1 at an angle, whose tangent is & , to the horizon.
4. (1) 16 Jvj (=65-97) ft. per sec. at an angle, whose tangent
is 4, to the horizon.
(2) 16^/37 ( = 97-32) ft. per sec. at an angle whose tangent is
6, to the horizon.
6. 2^ sees. ; 1461 ft. 7. 2h; 2*J~gh. 8. 5543 yards nearly.
9. 13 sees. ; 3328 ft.
ANSWERS.
XXXVIII. (Page 182.)
1. 14^ lbs. wt. 2. 25 lbs. wt.
3. \/4905, i.e. about 70, cms. per sec 4. 16 ft. per seo.
5. 628f lbs. wt. 6. 2f tons wt.
XXXIX. (Page 191.)
1. 156-25 kilos. 2. 5-6 lbs. wt. 3. 2ff lbs. wt.
4. 7 : 1. 5. 3H lbs. wt. ; *^tt = 1091^ tons' wt.
7. 144 lbs. wt. per sq. inch.
128
8. = 40 T 8 r lbs. wt. per sq. inch.
9. 80 lbs. wt. per sq. inch.
XL. (Page 194.)
1. 562^ lbs. wt. 2. 4-629... 3. 135*98 lbs. wt.
4. 13600 grammes wt. 5. 2-6. 6. l&H cub - ft -
7. Its volume is increased by 1-153... cub. cms.
XLI. (Page 197.)
1.
In ratio 1 : 3.
2. fcub.fi
3. 15 ozs.
4.
i(ft + ft + 2ft).
5. 6 and 2.
6. '9375.
7.
tWV CUD - cms -
XLII. (Pages 203, 204.)
1. 2291f lbs. wt. 2. 195-84 ft. 3. 7| ft. 4. 36-864 ft.
5. 4 miles 1561-6 yds. 6. 2833J lbs. wt. 7. 98 ft.
8. 54 ft. 9. 14H|. 10. 15f. 11. 2021-04 grains wt.
12. lfH. 13. 14-9556 cub. ins.
XLIII. (Pages 206208.)
1. 750 lbs. wt. 2. 162f| lbs. wt.
3. 156 lbs. wt. on the upper face ; 218f lbs. wt. on the lower
face; 187J lbs. wt. on each vertical face.
ANSWERS. xi
4. 320 lbs. wt. 5. 255| cwt. 6. 187 lbs. wt.
8. 104^1 tons wt. 9. Hf f lbs. wt. 10. 15066|| tons wt.
11. *ft lbs. wt. per sq. in.; W= 21 H lbs - **
12. It divides the vertical sides in the ratio 1 : ^2 - 1.
13. 1*2 kilogr. wt. 14. 515f lbs.
15. . X ^ 26 lb B .wt.; TX 6 -?^lbs.wt.
41o 416
16. i^7r = 24^ lbs. wt.; A|i tt = 245H lbs. wt.
17. 6|- ft. ; 1| ft. 18. 1250 and 1312 lbs. wt. respectively.
XLIV. (Pages 216218.)
1. 2tS#& cub. ft. 2. 10vr 9 A ozs., taking 7r=^ 2 .
3. U cub. ft. 4. 50 cub. cms. 5. 4 ; -00053.
6. l&rhrr cub. metres. 7. 6608-4 cub. ft. 8. A-
9. 31HHcub. cms.; 866i. 10. 257 x Vr ft.
11. -726... inch. 13. 4|f ins. 15. '25.
16. They are equal. 17. 4$ ins.
18. There is a cavity of volume 1 cub. cm.
19. 463^ cub. ins. 20. The edge of the cube is 28*8 ins.
21. H- 22. 30 lbs. 23. 900 cub. ins.; 10 ins.
XLV. (Pages 219, 220.)
1. -50065. 2. I cub. in. 3. 18-9 cub. inch.
4. 13-6054... 5. One half. 6. It will sink.
7. It will rise. 8. It will be lessened.
10. The new depth of immersion is to the original depth as
3935 : 3948.
XLVL (Pages 221, 222.)
1. (1) 12 lbs. wt. ; (2) 6 lbs. wt. 4. 155 : 187.
5. 37380 : 37249. 6. 97 A lbs. wt. ; 145^ lbs. wt.
7. t*& oz. 8. 18-5, 9. 57 grains. 10. 27 : 23.
11. The piece of wood. 12. 5.
13. 2 cub. ins. ; && lbs. wt. 14. 7U lbs. wt.; 56 lbs. wt.
xii ANSWERS,
XL VII. (Pages 223, 224.)
1. 3 oz. wt. 2. 2 lbs. 6 ozs. 5. 5 lbs. wt.
6. 1580000 grammes. 7. HH oz - wt - 8. A 0-
XL VIII. (Page 230.)
1. -75. 2. '7864 nearly. 3. 7A- 4. 2-0458...
5. 6.
XLIX. (Pages 233, 234.)
1. 1-525. 2. 3. 3. 2 A- 4. 865. 5. H-
6. f 7. f 8. 848. 9. '9413 nearly.
10. 1-841. 11. -87 ; 50 cub. cms. 12. 30 grms. wt. ; 2.
13. '5. 14. lA-
L. (Page 238.)
1. 3-456, 3-14i8 and 2-88. 2. 1*03. 8. ggpJTT'
4. 10 : 13. 5. 18 : 19. 6. 8A oz. 7. 2-5.
8, 8. 9. 2|oz.
LI. (Page 239.)
1. 27-2. 2. 6 inches.
3. At the bottom of the vertical tube containing the oiL
LH. (Page 244.)
1. 1169-256 cms. 2. 929082 grms. wt., taking f-*V*
3. 17AV* tons wt.
4. 1tV> tne height would be lessened by a distance x, such that
the weight of the mercury in a length x of the tube would equal the
weight of the bullet. This assumes that the bullet fits the tube. If
it floats in the mercury there would be no alteration in the height.
5. 2-623... cms. 6. 1A inch.
LHI. (Pages 250 252.)
1. -001292. 2. An increase of S r % grains wt
3. 25-92 lbs. 4. 31-5 feet. 5. -00007764... cub. in.
6. Till the level of the water inside is 68 feet below the surface
of the water.
7. 32-75 ft. 8. 63 cms.
10. The pressures on the two faces are 56 and 22 lbs. wt. per
square inch ; 8 inches.
11. (1) It would float; (2) it would sink. 13. \ oub. inoh.
ANSWERS. xiii
14. 5 inches. 15. 29*98 inches. 16. 32 inches.
18. The pressure is that due to 63 inches of mercury ; 10$ ins.
19. 34-4 lbs. wt. nearly. 22. 7*5 ; 30 ft.
LIV. (Page 255.)
1. 8J cub. inches. 2. 10 cub. inches. 3. 429 : 224.
LV. (Pages 258260.)
1. 8 4 5... atmospheres, nearly. 2. li ft. 3. 14 ft.
4. 1097JWV- 5. 20 ft.; 132^^. ft. 6. 500 cub. ft.
7. The quantities are as 3 : 2.
8. The depth of the top of the bell is 3 inches ; the height of the
water-barometer is 33 ft.
9. 33 J ins. ; 3 ft. 9 ins. 10. It remains constant.
14. The air will flow out.
16, (1) Some air will flow out ; (2) there will be equilibrium ;
(3) some water will flow in.
LVI. (Pages 265, 266.)
1. The height varies from 31-73 to 35-13 feet.
2. 42 ft. 1 in. 3. 33 ft. 4 ins. 4. 80. 5. It will.
6. 2 feet ; 32 - 16 ^2 = 9*37 feet nearly. 7. 8680$ lbs. wt.
8. 260^ lbs. wt.
9. x lbs. wt. ; - lbs. wt.
o o
LVII. (Pages 271, 272.)
1. 8:1. 2. They are as 9 8 : 10 8 .
4. The final pressure is to the original pressure as 10 8 : ll 8 , i.e.
nearly as 10 : 21.
6. 8| ins. 8. Between 37 and 38. 9. 8. 10. 20.
11. 22.
LVIH. (Page 274.)
1. 34 feet 2. 22 ft. 8 ins.
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