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Full text of "Mechanics and hydrostatics for beginners"






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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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