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IIA

LOYELL'S SERIES OF SCHOOL BOOKS.

NOTES AND EXERCISES

MTDBAL PHILOSOPII,

INCLUDING

STATICS, HYDROSTATICS, PNEUMATICS,
DYNAMICS, AND HYDRODYNAMICS,

DESIONBD

FOR THE USE OF NORMAL AND GRAMMAR SCHOOLS, AND
THE HIGHER CLASSES IN COMMON SCHOOLS.

BY JOHN HERBERT SANGSTER, ES(t.,

MATHEMATICAL MASTER AND LECTURER IN CHEMISTRT AND NATURAL

PHILOSOPHY IN THE NORMAL SCHOOL FOR UPPER CANADA ; AND

AUTHOR OF AN ARITHMETIC IN THEORY AND PRACTICE.

indcrrL

PRINTED AND PUBLISHED BY JOHN LOVELL ;
AND SOLD BY R. & A. MILLER.

tloronto :
R. & A. MILLER, 87 YONGE STREET.

1860.

Entered, according to the Act of the Provincial Parliament, in
the year one thousand eight hundred and sixty, by John
LoTBLL, in the Office of the Registrar of the Provlnoe of
Canada. > « a.

.n'ir

PREFACE.

The fpllowing Treatise was originally designed to serve
as a hand-book or companion to the lectures on Natural
Philosophy, delivered to the junior division in the Normal
School. Although numerous text-books on the subject were
already in existence, it was found that they were either too
abstruse and technical for beginners, or too general and
superficial to be of much practical use. The aim of the
present little work is to occupy a position between these
extremes — to present the leading facts of the science in a
form so concise as to be readily remembered, and at the
same time to give that thorough drilling upon the principles
which is absolutely essential to their full comprehension.

As a hand-book to lectures fully illustrated by apparatus,
it was not necessary to introduce many (wood-cuts,^ and
accordingly they have been given only where absolutely
necessary.

The chief peculiarity of the following Treatise is the
introduction to a large extent of problems calculated to
impart that intimate and practical knowledge of the

PREFACE.

facts and principles of the subject, without which the
student^s information on the science of Natural Philosophy
is, comparatively speaking, useless. How frequently do
we meet with a pupil who has read carefully through one
of the common text-books on the subject without acquiring,
to any very great extent, clear and definite ideas of the
science t And what should we say of a work professing to
teach the principles of arithmetic or algebra by mere rules
and explanations, without an appropriate selection of
examples and problems? The exercises are therefore
deemed an important feature in the present little book,
and it is thought that the science may be taught by their
aid more thoroughly and in less time than otherwise.

Toronto, January, 1860.

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ERRATA.

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Page 29, Quea. 44, for 1T60 lbs. read 880 lbs.

Page at, Ques. 68, for lOX3«XlO = 720,read 10x36= lOX "729.

Page 39, Ques. 75, for Ana. 9f , read Am. 181 lbs.

■.""".f ,'

•■ :-:■■. '.{^'f

I''»"

t^; .■■;■'■ .CONTENTS.

CHAPTER I.

Paob

General Sub-divisions of Natural Science, 9

Sub-divisions of Natural Philosophy, 10

Properties of Matter, 10

Table of Tenacity, 13

Attractions, 13

Problems in Attraction of Gravity, 14

CHAPTER n.

Sub-divisions of the Science of General Mechanics,. ... 15

Statics, 15

Parallelogram of Forces, 17

Parallel Forces 18

Centre of Gravity, 19

CHAPTER III.

Mechanical Powers, 20

Virtual Velocities, 20

The Lever, 22

The Compound Lever, 25

The Wheel and Axle, 26

The Diflferential Wheel and Axle, 28

Wheel Work, 30

The Pulley, 33

The Inclined Plane, 38

The Wedge, m..^. 40

''W^.-W-

2 f }

G CONTENTS.

PAOl

The Screw, 41

Tho Differential Screw, 44

The Endless Screw, 45

Friction, 47

Table of Friction, 48

CHAPTER IV.

Unit of Work, 49

Work of different agents, 50

Work on a horizontal plane, 53

Table of traction of tho horse, 54

Work of atmospheric resistance, 57

Work on an inclined plane, 59

Modulus of machines, 64

Table of moduli, 65

The Steam Engine, 65

Work of the Steam Engine, 68

Source of work in the Steam Engine, 71

Pambour's Experimental . Table, 72

CHAPTER V.

Hydrostatics, 76

Liguid pressure, 76

Weight of cubic inch, gallon, and cubic foot of water,.. 77

Pressure against a vertical or inclined surface, 78

Pressure against a vertical or inclined surface at a

given depth 81

Bramah's Hydrostatic Press, 84

Hydrostatic paradox, 87

Hydrostatic Bellows, 87

Specific Gravity, 88

To find the specific gravity of a solid, 89

To find the specific gravity of a liquid, 91 '

■i

CONTENTS. 7

PAOl

To find the specific gravity of a gas, 92

Table of specific gravities, 93

To find the weight of a given mass of any substance,... 93
To find the mass of a given weight of any substance,... 94

CHAPTER VI.

Pneumatics, 95

Composition of Atmospheric Air, 95

Gaseous diffusion, 96

Aqueous vapour, 96

Physical properties of atmospheric air, 97

Weight of air, 97

Density of air, 98

Pressure of air, 99

Mariotte's Law, 100

The Air Pump, 101

Pressure and elasticity of air, ,.102

The Barometer, 103

Use of the barometer as a weather glass, 105

To ascertain the height of mountains, &c., by the bar-
ometer, 105

The Common Pump, 107

The Forcing Pump, 108

The Syphon, 108

CHAPTER VII.

Dynamics, 109

Momentum, ..110

Laws of Motion, /. 114

Reflected Motion, 114

Descent of bodies freely through space, 115

Analysis of the motion of a falling body, 116

Table of formulas for descent of bodies through space,. 117

8 ^ CONTENTS.

ProMoms, 118

DoHCont on inoliiiud plunoH, 124

DcHcont in curvi^H, 125

TubluH of tbrinulafl for doKoont on inclined plnnoH, 12G

ProbloiMH 127

PnyectiloH, 129

Parabolic theory, l.'JO

Modern parabolic theory 131

Velocity of »hot and shell, 133

Circular motion, 1 34

Centrifugal force, 1 34

ProbleniH, 136

Accumulated work, 137

The Pendulum, 139

The centres of suspension and oscillation, 140

Laws of the oscillation of the pendulum, 1 40

Tables of lengths of second's pendulums, 141

Problems, 142

CHAPTER VIII.

Hydrodynamics, 146

Torricelli's theorem, 146

Discharge of water through an orifice, 146

The vena contracta^ 146

Problems, 147

The random of spouting fluids, 149

Velocity of water flowing in a pipe or channel, 149

Upright water-whccls, 160

To find the horse powers of water wliocls, 151

Miscellaneous Problems, 152

Examination Papers, 160

Answers to Examination Papers, 169

Examination Questions, 171

NATURAL PHILOSOPHY.

CHAPTEU I.

SUnDIVISIONS— GENERAL PROPEREIES OP MATTER-
ATTRACTION.

1. Natuml Sciunce, in its widest sense, embraces the
study of all crented oljectB and beings, and t!te laws bj
wliicli tliey are governed.

2. Natural objects are divided into two great classes,
viz : organic and inorganic, the former being distinguished
from the latter by the exhibition of vital power or life.

8. Organic existences are separated into animals and
rrgctahleSf the former distinguished from the latter by the
possession of HenHibility and volition.

4. The difTerent subdivisions of natural science and their
objects are as follows : —

Zoology describes and classifies animals.

Botany teaches the classification, use, habits, structure,
&c., oi plants.

Mineralogy describes and classifies the various mineral
constituents of the earth's crust.

Astronomy investigates the laws, (kc, of celestial phe-
nomena.

Geology has for its object the description, <kc., of the
crust of the earth.

Chemistry teaches us how to unite two or more element-
ary bodies in one compound, or how to decompose comr
pound bodies into their simple elements.

Natural Philosophy or Physics has for its object the
investigation of the general properties of all bodies and
the natural laws by which they are regulated.

PROPERTIES OF MATTER.

[Arts. 5-12.

5. Natural Philosophy is divided into —

I. General Mechanics — including Statics, Hydrostatics,
Dynamics, Hydrodynamics, and Pneumatics.

n. Heat.

HI. Light — including Perspective, Catoptics, Dioptics,
Chromatics, Physical Optics, Polarization, and Actino-
Chemistry.

IV. Electricity — including Statical Electricity, Galvan-
ism, Magnetism, Thermo-Electricity, and Animal Elec-
tricity.

V. Acoustics.

PROPERTIES OP MATTER.

6. Matter exists in three separate forms, — I. JSoUd ;

Vi. Liquid; and HI. Gaseous. ■ .

Note.— The same body may exist in all three forms, as is the case with
ioater, mercury, sulphur, &c. The amount of heat or caloric present deter-
mines the form of the body— if heat be applied, the attraction of cohesion
existing among the particles is gradually overcome, and the body passes
firom a solid to a liquid, and from a liquid to a gas. If heat be abstracted,
the attraction of cohesion gradually draws the particles into closer proximity
ftnd the body passes from a gas to a liauid, and fiHally from a liquid to a
■olid. Hence lieat and cohesion are called antagonistic forces,

7. Matter is distinguished by the possession of certain
distinctive properties.

8. The properties of matter are divided into —

1st. Essential Properties.
2nd. Accessory Properties.

9. The essential properties of matter are those without
which matter could not possibly exist.

10. The essential properties of matter are ExtensioUy
Impenetrabiliti/, Divisibility, Indestnictahilityj Porosity^
Compressibility , Inertia^ and Elasticity,

11. Extension implies that every body must fill a certain
portion of space. ^ ^^

NoTB,— The Dimensions of Extension arc length, breadth, and thickness.

12. Impenetrability implies that no two bodies can oc-
cupy the same portion of spac« at the same time.

AbTS. 13-15.]

PROPERTIES OF MATTER.

NoTB^fixaaiplei of the impenetrability of matter will readily $ugKe8t
tbenaelTw, Amomt the more oommon may be mentioned the imponibfiity
of fllUog a bottle with water until the air is dispUced— the fact that when
the hand is plunged into a veaael Ailed with water, a portion of the liouid
overflows. &o. All instances of the apparent penetrability of matter are
merely examples of displacement. Thus, when a nail is driven into a piece
of wood, it dttplaces the particles of wood, driving them closer together

18. Divisibility is the capability of being c6ntinually
divided and subdivided, and is an essential property only
of masses of matter.

NoTB 1.— The ultimate particles of matter : i. e., those inconceivably
minute molecules which cannot be further subdivided, are termed atoms
( Or, a " not" and temno, " to cut" ; i. e., that which cannot be cut or divided.)

NoTB 2.— The following may be given as examples of the extreme divisi-
bility of matter:—

I. Gold leaf is hammered so thin that 300000 leaves placed one on another
and pressed so as to exclude the air, measure but one inch in thickness. '

II. Wollaston's micrometric wire is so fine that 30000 wires placed side
by side, measure but one inch across— 160 of these wires bound together
do not exceed the diameter of a filament of raw silk, 1 mile of the wire
weighs but a grain, and 7 ounces would reach from Toronto to England.

III. Insects wings are some of them so fine that they do not exceed the
j Q^Qfy^ of an inch in thickness.

IV. The thinnest part of a soap bubble is only the 2500000th part of an
inch in thickness.

V. Blood corpuscles are so small that it requires 50000 corpuscles of human
blood, or 800000 corpuscles of the blood of the musk-deer to cover the head
of a common pin. Yet these corpuscles are compound bodies and may be
resolved, by means of chemistry, into their simple elements.

VI. There are animalcules so minute that millions of them heapeA
together do not equal the bulk of a single grain of sand, and thousands
might swim side by side through the eye of the finest cambric needle. Tet
these creatures possess, in many cases, complicated organs of locomotion
nutrition, &c. '

VII. At Bilin in Bohemia, a hnge mountain consists entirely of shells
so minute that a cubic inch contains 41 billions— a number so vast that
counting as rapidly as possible, day and night without intermission it
would require 780 years to enumerate it. '

VIII. The filament of the spiders web is so fine that 4 miles of it weigh
only about a grain— yet this thread is formed of about 6000 filaments united
together, &c., &c.

14. Indestructahiliti/ implies that it is as impossible for

a finite creature to annihilate as to create matter.

NoTB.— We can change the form of matter at pleasure, but we cannot
destroy it. When fuel, for example, is burned, not a particle is lost, as is
proved by the fact that if we coOect all the products of the combustion*
1. e., the smoke, suot, ashes, &c., and weigh them, we shall find their aggregate
weight exactly equal to that of the wood or coal consumed. We may safely
conclude that there is not a single atom of matter, more or less, attached to
our earth now than at the time of Adam.

15. Porosity implies that the constituent atoms of tmat-
ter do not touch each other, but are separated by small
intervening spaces called pores. : ' '*

l*ttOPERTIKS OF MATTKK.

'Aktb. Ifl-M.

.♦i

NoTB.— The atoms even of the densest bodies are much smaller than the
spaces which separate them. Newton regards them as ii^flniUVy ttnaUer,
as beinc in fact nuere mathematical points, and Sir J. Herschel asks why
the particles of a solid may not be as thinbr distributed through the space
it occupies as the stars that comuose a nebula, and ho compares a ray of
light penetrating glass to a bird tnreadiug the mazes of a forest.

16. Compressibility implies the capability a body pos-
sesses of being forced into a smaller bulk without any di-
minution in the quantity of matter it contains.

NoTB.— Since all matter is porous, it followH, as a necessary consequence,
that all matter must be compressible.

17. Inertia means passiveness or inactivity, or that mat-
ter is incapable of changing its state, either from rest to
motion or from motion to rest.

NoTB.— Bodies moving on or near the surface of the earth soon come to
a state of rest, unless some constant propelling force is applied to them.
This is owing to the action of certain resisting forces, as the resistance of
the atmosphere, friction, and the attraction of gravity.

18. Elasticity is the capability which all bodies possess,
more or less, of recovering their former dimensions after
compression or after having, for a time, been compelled to
assume some other form. " " '

NoTB.— As applied to solids, elasticity is divided into—

1. Elasticity of compression, - r

2. Elasticity of tension, i< .
8. Elasticity of flexure, and

4. Elasticity of torsion.
Some bodies, as putty, seem to possess very little elasticity. In glass all
four kinds appear to exist almost perfect within certain limits— no force
however great or long continued will cause glass to take a set, as it is termed.

19. The accessory properties of matter are those which
merely serve to distinguish one kind of matter from another.

20. The accessory properties of matter are hardness^
softness, flexibility, brittleness, transparency, opacity, mal-
leability, ductility, tenacity, &c.

21. Malleability Qxipve&ses the susceptibility, possessed
by certain kinds of matter, of being hammered out into
thin sheets.

NoTB.— The most malleable metals are gold, silver, iron, copper, and tin

22. Ductility is susceptibility of being drawn out into
fine wire.

NOTB.— 'The most ductile metals are platinum, gold, iron, and copper.

23. Tenacity or toughness implies that a certain force
is necessary to pull the particles of a body asunder.

Arts. 24-27.

ATTRACTION.

Note.— The following table shows the relative tenacity of difTerent
substances. The first column shows the number of pounds weight required
to tear asunder a prism of each substance, having a sectional area of one
square inch, and the second column gives the length of the rod of any
given diameter which, if suspended would be torii asunder by its own
weight :—

TABLE OP TENACITY.

Weight in pounds.

Length in feet.

(Section of rod 1 sq. in.)

(any diameter.)

Metals.

Cast lead, 1824

848

Cast tin, ' 4788
Yellow Brass, 17958

1496

5180

Cast Copper, '- 19072

A008

Cast Iron, 19096

6110

English Malleable Iron, S6872

160S8

Swedish do. 72064

19740

Cast steel. 134266

30465

Woods, , ,

Pine, ' 9640

40600

Elm, * 9720

86800

Oak, 11880

82900

Beech, 12226

88940

Ash, , ! X4180

42080

ATTRACTION.

, ' '^ii* ,.• ."-<-« > ■

24. Attraction is that power in virtue of which particles
and masses of matter are drawn towards each other.

26. Attraction is of several kinds — viz:

I, Attraction of Gravity. '

II. Attraction of Cohesion. . - ; '

III. Attraction of Adhesion. ^

IV. Capillary Attraction.
V. Electrical Attraction.

VI. Magnetic Attraction.
VII. Chemical Attraction. ,. —

28. Attraction of Gravity (Lat. gravitas "weight")
is that force by which masses of matter tend to approach
each other. It is sometimes spoken of as gravitation^ or
when applied to the force by which bodies are drawn
towards the centre of the earth, terrestrial gravity,

27. The intensity of the force of gravity varies directly
as the mass of the bodies, and inversely as the square of
their distance apart.

ATTRACTION.

[ABTB. 28, 29.

, :•

XoTs.— If we suppose two spheres of any kind of matter, lead, for example
to be placed lu presence of each other, and undw such oondiMons that
being themselves flree to move in any dveotlon they are entirely nninftu-
enoed by any other bodies or circumstances they will approach each other
and : —

1st. If their masses are equal their velocities will be equal.

2nd. If one contain twice as much matter as the other, its velocity will

be only half as great as that of the other.
3rd. If one be iniinitely great in comparison with the other its motion will

be inflnitely small in comparison with that of the other ; and
4th. The more nearly they approach each other the more rapid will their

motion become. ,^

28. By 8a3nng the intensity of the force of gravitation
varies inversely as the square of the distance between the
attracting bodies, we merely mean that if the attractive
force exerted between two bodies at any given distance
apart be represented by the unit 1, then, if the distance
apart be doubled, the force of attraction will be reduced to
\ of what it was before ; if the distance between the bodies
be increased to three times what it was, the force of
gravity will be decreased 9 tiraes, or will be only \ of
what it was, &c.

Example 1. — If a body weigh 981 lbs. at the surface of the
earth, what will it weigh 8000 miles from the surface ?

80LX7TIOW.

Here since the distance of the body in the first case is 4000 miles firom
the centre of the earth and in the latter case 1200(Ki. e. 8000+4000) the
distance apart has been trebled. a i > **^ ■< - . , ij =j

Then weight = ^ = ?^ = 109 lbs. Ana.
8* 9

Example 2. — The moon is 240000 miles from the (centre of)
earth, and is attracted to the earth by a certain force. How
much greater would this force become if the moon were at the
surface of the earth ?

SOLVTIOir.

Here ^ !^^, = ^^ = 60, and 602 = 8600 times. An^.
Earths radius 4000 ^

EXERCISES.

3. If a mass of iron weigh 6700 at the surface of the earth how

much would it weigh at the distance of 12000 miles from
. the surface ? Ans. 418| lbs.

4. If a piece of copper weigh 9 lbs. at the distance of 36000

miles from the earth's surface, what would it weigh at the
surface of the earth ? Am, 900 lbs.

29. Attraction of Cohesion is that force by which the
constituent particles of the same body are held together.

AeTS. 30-86.]

STATICS.

NoTB.— The attraction of cohesion acts only at insensible distances ; i. e.,
at distances so minute as to be incapable of measurement. The attracts n
of gravity, on the other hand, acts at sensible distances.

30. Attraction of Adhesion is that force by which the
particles of dissimilar bodies adhere or stick together.

31. Capillar^/ Attraction (Lat, capillar " a hair") is the

force by which fluids rise above their level in confined

situations, such as small tubes, the interstices of porous

substances, <fec.

Note.— It is by capillary attraction that oil and biiming fluid, melteil
tallow, &c., rise up the wick of a lamp or candle.

32. Electrical Attraction is the force developed by fric-
tion on certain substances, as glass, amber, sealing-wax,
&c.

33. Magnetic Attraction is the force by which iron,
nickel, &c., are drawn to the load-stone.

34. Chemical Attraction^ or Chemical AflSnity, is the

force by which two or more dissimilar bodies unite so as

to form a compound essentially dilOferent in its appearance

and properties from either of its constituents.

Thus Potash and Grease unite to form soap— Sulphur and Mercury ui4to
to form Vermillion, &c. ; ur im •j.'titi H -infjft"!'

♦ u.. .CHAPTER II,
STATICS.

35. The Science of general mechanics (Greek michanl^
" a machine") has for its object the investigation of the
action of forces on matter whether they tend to keep it at
rest or to set it in motion.

36. The Science of general mechanics is usually sub-
divided as follows : —

I. Statics, (Greek statos^ " standing,") or the science by
which the conditions of the equilibrium of solids are
determined.
II. Hydrostatics, (Greek hitdor^ *• water," and statos^
*' standing,") or the science by which the conditions
of the equilibrium of liquids are determined.
III. Dynamics (Greek ditnamis " force") or the science by
which the laws that determine the motions ofsolidt
are investigated.

"If' lt lW' W.H W i l'ii l 'i

STATICS.

fART8.87'H.

^iilf

IV. Utdrodtnamios (Greek hMor and ditnamia) or the
Bcionce by which the laws that determine iho motions
of liquids are investigated.
V. Pneumatics (Greek janewTna, " air," and statos, "stand-
ing"J or Pnouma-staticp, tho science by which tlio
conditions of the cquilihirum of clastic fluids^ as
atmospheric air, are investigated. Pneumatics may
bo regarded as a branch of Hydrostatics.

37. A body is said to bo in equilibrium when tho forces
which act upon it mutually counterbalance each other or
arc counterbalanced by some passive force or resistance.

38. Forces that are balanced so as to produce rest are
called statical forces or pressures to distingush them from
woviTJ^, dcjlectingy accelerating or retarding forces.

30. A force has three elements, viz., magnitude^ direc-
tion, and point of application.

40. A force may be represented either by saying it is
equal to a certain number of lbs., oz., &c., or by a line of
definite length. A line has the advantage of completely
defining a force in all its three elements, while a number
can merely represent its magnitude.

41. Whatever number of forces may act upon eno point
of a body, and whatever their direction, they can impart
to tho body only one single motion in one certain direction,

42. When several forces (termed components) act on a
point, tending to produce motion in ditierent directions,
ihey may be incorporated into one force, called the result-
ant, which, acting alone, will have the same mechanical
effect as the several components.

43. When any number offerees act on a point in the
same straight line, tho resultant is equal to their sum, if
they act in the same direction ; but if they act in oppo-
site directions, tho resultant is equal to the difterence be-
tween the sum of those acting in one direction and the
sum of those acting in the other.

44. If two forces acting upon the same point be repre-
sented in magnitude and direction by two lines drawn
thiough that point, then the resultant of such forces will

ABTH. 45 4».|

STATIC?'.

be represented in magnitude and direction by the diagonal
of tha parallelogram, of which these lines are the sides.

46. If any number of forces, A, 13, C, D, &c., act upon
the same point in any direction whatever, and in any plane
whatever, by first finding the resultant of A and B, then of
this resultant and C, then of this resultant and D, and so
on, we shall finally arrive at the determination of a single
force, which will bo mechanically equivalent to, and will
therefore be the resultant of the entire system.

46. If the components act in the same plane, the
resultant is found by means of what is technically termed
the parallelogram of forces, if in different planes by the
parallelojnped of forces.

47. The resultant of two forces, which act on different
points of the same body in parallel lines and in the same
direction, is a single force equal to their sum, acting paral-
lel to them, and in the same direction, at an intermediate
point which divides the line joining the two points of ap-
plication of the components in the inverse ratio of the mag-
nitudes of these components.

48. The resultant of two forces, which act on different
points of the same body in parallel lines but in opposite
directions, is a single force equal to their difference, acting
parallel to them and in the direction of the greater force,
and at a point beyond the greater of the two forces, so
situated, that the point of application of the greater of the
two forces divides the distance between the points of appli-
cation of the smaller force and of the resultant in the
inverse ratio of the magnitudes of the smaller force and of
the resultant. .

49. When any number of parallel forces, A, B, C, D,
♦fee, act on a body, at any point whatever, and in any
planes whatever, by first finding the resultant of A and B,
next of this resultant and C, then of this last resultant and
D, and so on, we shall finally arrive at the determination
of a single force, which will be mechanically equivalent to,
and will therefore be the resultant of the entire system of
parallel forces. - •

STATICS.

[ABTS. 50-6S.

50. When a system of forces consists of two equal,
oppoiite^ and parallel forcesy it is called a Couple,

61. Two equal and parallel forces acting on a body in
contrary directions, have a tendency to make that body
revolve round an axis perpendicular to a plane passing
through the direction of such two parallel and opposite
forces ; and such tendency is proportional to the product
obtained by multiplying the magnitude of the forces by the
distance between their points of application : and, conse-
quently, all couples, in which such products are equal,
and which have their planes parallel, are mechanically
equivalent, provided their tendency is to turn the body
round in the same direction ; but if two such couples have
a tendency to turn the body in contrary directions, then
they have equal and contrary mechanical effects, and
would, if simultaneously applied to the same body, keep it
in equilibrium.

62. If any tWo forces, not parallel in direction, but
which are in the same plane, be applied at any two points
of a body, they admit of a single resultant, which may be
determined by producing the lines, that in magnitude and
direction represent the two forces, until they meet in a
point and then applying the principle of the parallelogram
of forces.

63. If two forces not parallel in direction act in different
planes on two points of a body, they are mechanically
equal to the combined action of a couple and of a single
force, and their effect will be two-fold — 1st, a tendency to
produce revolution ; 2nd, a tendency to produce progres-
sive motion, so that, if not held in equilibrium by some
antagonistic forces, the body will at tHe same time move
forward, and revolve round some determinate axis.

64. The process of incorporating or compounding two
or more forces into one, is called the composition of forces ;
that of separating or resolving a single force into two or
move J h termQd the resolution of forces, ' '

66. As all the molecules of a body may be considered as
gravitating in parallel lines towards the centre of the earth

ABT8. 60-88.]

8TATICS.

move

— these parallel forces may (Art 49) bo compounded into

a single force — which resaltant is equal to the sum o

all the forces aflfocting the particles severally, or, in other

words, to the weight of the mass. The point to which this

resultant is applied, is called the Centre of Gravity^ and the

vertical line m which it acts is termed the Line of Direc*

Hon,

66. Every dense body or solid mass possesses a centre

of gravity.

NoTi.— The centra of mvity is aometimet called the Centr0 cf Inertia
because, if it bo moved, the whole maw !■ moved— it is likewise called the
Centre qf Parallel Foroee, for the raason assigned in Art. 65.

07. The Centre of Gravity may be defined to be that
point in a body, upon which, if the body be supported, it
remains at rest and is balanced in any and every position.

58. If a body, regular or irregular in shape, be freely
suspended by a point, the centre of gravity will inva-
riably lie in the line of suspension. If suspended by seve-
ral points of succession, the lines of suspension will have a
common point of intersection, which point will be the cen-
tre of gravity of the body.

69. The Centre of Gravity is not necessarily in the body
but may be in some adjoining space, as in the case in a
ring, a table, an empty box, kc,

60. The tendency of a body, when free to move in any
direction, is always to rest with the centre of gravity as low
as possible.

61. The Stability of a body resting in any position is
estimated by the magnitude of the force required to disturb
or overturn it, and will therefore depend on the position of
the centre of gravity with reference to the point of sup-
port.

62. A body supported on the centre of gravity is said
to be in a condition of Neutral or Indifferent Equilibrium ;
when the point of support is above the centre of gravity
the body is said to be in a condition of Stable Equilibrium ;
when the point of support is beneath the centre of gravity
the body ia said to be in a condition of Unstable Squili-
hrium.

MECHANICAL POWKllS.

[AUTS. «J-iW.

68. The centre of gravity of two separate bodies may be
found by dividing the line joining their centres in the
inverse ratio of the magnitudes of tno bodies.

CUAPTElt III.
MECHANICAL POWERS.

f ..*>n

64. Tlie object of all Mechanical contrivanceB is

Ist. To gain power at the expense of velocity ; or
2nd. To gain velocity at the sacrifice of force.

66. The relative gain and loss of power and velocity is
regulated by that principle in philosophy known as the
Law of Virtual Velocities, or the Equality of Moments.

66. The Law of Virtual Velocity may be thus enun-
ciated : — '

If in an\f macJdne the power and loelght he in equilibrium
and the wliolc he put in motion, then the power multiplied
hy the units of distance through which it m^ovcs is equal to
the weight multiplied hy the units of distance tJwough
which it mx)ves. ,,. -- .

u.;

Or if F=:power, W=zweight, S=space moved through by P, and
vzuspace through which W moves. ^ , ^ ,-!..«

ThenV : W :: s : S.
W X s

Hence P =

W X 8 P X s

; S = — p — , W = — - — and s

PXS.
' W

ExAMPLK 5. — A weight of 700 lbs. is moved through 90 feet
by a certain power moving through 5100 feet. Required the

power

SOLUTION.

! •<;

Here IK=700, s=«0 and S = 6100.

Hence P = —5— =

Wxs 700 X 90

6100

= 12^ lbs. Ans,

Example 6. — A weight of 500 lbs. is moved by a power of
20 lbs., through how many feet must the power mpve in order to
f^ise tHe weight through 16 feet ?

AuT9. rt7, tw.

MKCIIANICAL I'OWEKS.

s cnun-

80LUTI0N.
Horc »r= 600, P = 20 and » = 10.

Wxa 500X16

HoncR <Sf=: — r^ =

= 400 feet. An«.

P ~ 20

Example 7. — A power of 21 lbs. moving through 76 feci carries
a certain wciglit tlirongli 11 feet. Required tlie welglit?

SOLUTION.

Hero P = 21, 8=75 and j» = 11.

Then 7r=^^ = *^^ = iWi2,ll)8. yJ«*.
s 11 * '

ExAMPiiK 8. — A power of 204 lbs. moving through 30 feet is

made to move a weight of 1000 lbs. Through how many feet

does the weight move ?

SOLUTION.

IIorcr = 20't, ?r=1000andA'=30.

Then s =

VXS 204X30

1000

=8 A «

Ant.
EXERCISES.

9. A power of 7 lbs. is made to move a weight of 1000 lbs.
through 1 1 feet ; through how many feet must the power move ?

Jlns. 1671? feet.

10. A power of 97 lbs. moving through 86 feet raises a certain
weight tlirough 10 feet. Required the weight? ^n». 834^ lbs.

11. A weight of 888 lbs. is raised by a power of 60 lbs. ;
through how many feet must the power move in order to raise
the weight through 1 foot. ^n«. 14|^ feet.

12. A certain power moving through 27 feet is so applied that
it carries a weight of 2600 lbs. through 4 feet. Required the
power ? Ms. 370^^^ lbs.

'67. Any contrivance by which, in accordance with* the
principle of Virtual Velocities, a small force acting through
a large space is converted into a great force acting through
a small space, or vice versd^ is a Machine. Machines are
either simple or complex.

68. In the composition of machinery it is usual to speak
of six mechanical powers — more properly termed Mechan-
ical Elements or Simple Machines, viz : —

The Lever )

The Inclined Plane >• Primary Mechanical Elements,
. The Pulley and Cord)

The Wheel and Axle)

The Wedge >• Secondary Mechanical Mementt.

The Screw )

THK LEVER.

[Abti.6»-7II.

69. In reality however, there are but two simple me-
chanical elements, viz. the Lever and the Inclined Plane,
The Wheel and Axle and the Pulley are merely modifica-
tions of the Icver^ while the Wedge and the Screw are both
formed from the inclined plane.

70. In theoretical mechanics levers are assumed to be
perfectly riff id and imponderable — cords, ropes and chains
are regarded as having neither thickness, stiffness nor
weight, they are assumed to bo mere mathematical lines^
infinitely flexible and infinitely strong. At first no allow-
ance is made for friction, atmospheric resistance, <Src.
After the problem, divested of all these complicating cir-
cumstances has been solved, the result is modified by taking
into consideration the effects of weight, friction, atmosphe-
ric resistance, rigidity of cords, flexibility of bars, &c.

L E V^E R S.

71. The lever is a bar of wood, or iron, moveable about
a fixed point or pivot called the Fulcrum.

72. Levers are either Straight or Bent, Simple or Com-
pound.

73. Of Simple Straight Levers there are three kinds, —
the distinction dependmg upon the relative positions of
the fulcrum, the power, and the weight.

74. In levers of the first class the fulcrum is between

the power and the weight. Fig. 1.

Of this kind of lever, we may
mention as examples, a pair of
rs, pliers or pin
B.the "

scissors, pliers or 1
handle, the beam of a pair of'scales,
Ifoi ■

incers,apump>
Ta pair of scales,
a crowbar when used forpry iug,&c.

■w

76. In levers of the second class the weight is between

the fulcrum and the power.

Nutcrakers, an oar in rowing, a
cowbar when used in lifting, &c., are
examples of levers of the second kind.

Fig. 2.

AnTn.7fl,77.]

THE LEVLH.

isd

76. In levers of tlie third class llie pnwrr is between
the fulcrum and the weight. Fig. 3. P

A pair of common tonKs, nhoop-
Nh(!ani, tun truadio of a foot lathe,
a door whon oiwncd or closed by
placing thu hand near the hincro,
afford examples of levers of the
third cla88.

Note.— In levers of the first class y ■ ■ ' '
the power may bo oithor (greater or
IfiHs than tho weight ; in levers of
the second nlasH. the power is at-
ways leas than tho weight ; and in
lovers of the third class, tho power
is alwayt greater than tho weight.

Hence fevers of tho third class are '

called losing levers, and are used mornly to secure extent of motion. Most
of tho levers in tlie animal economy are lovers of the third kind.

77. That portion of the lever included between the ful-
crum and tho weight is termed the arm of the weight;
that portion between tho fulcrum and tho power is termed
ihQ arm of the power.

The power and the weight in the lever are in equilibrium when
the power is to the weight as the arm of the weight is to the arm
of the power.

Or let P = power ^ W = the weighty A = the arm of the power,
and a = the arm of the weight.

Then P : W :: a : A.

Hence P =

Wxa

W =

PXA

a =

PXW

and A =:

Wxa

Example 13. — The power-arm of a lever is 11 feet long, the
arm of the weight 3 ft. long, the weight is 93 Iba , Required the
power ?

HeroTr=93. A:

SOLUTIOS'.

:11 and a =3.

„ TFXo__ 9?>X3_ „,,>,. J
Then P= — -r— = -— — = 25^\ Ibs.^w*.

Example 14.— The power-arm of a lever ia 17 feet long, thf
arm of the weight is 20 feet long, the power is 110 Ibi. What it
the weight ?

SOLUTION.

Here P = 110 lbs., A= 17 and a = 20.

Then Tr=^^:

110X17
80

=r93ilbi. An$,

THE LEVEU.

[Abts. 78, 79.

ExAMPLK 16. — By means of a lever a power of 4 oz. is made
to balance a weight of 7 lbs. Arour. ; the arm of the weight is 2i
inches long. Required the arm of the power.

SOLUTION.

Here P=:4 oz., Tr= 7 lbs. = 112 oz„ and a == 2\.

Ans.

Then A =2 — i,- = — - — == 70 inches.

r 4

— EXERCISES.

16. The power-arm of a lever is 16 feet long, the arm of the
weight 2 feet long, and the waight is 260 lbs. Required the
power? Ans. 31 i lbs.

The power-arm of a lever is 20 feet long, the arm of the

weight 70 feet ; what power will balance a weight of 5 cwt.

^ Ans. 17J cwt.

The power-arm of a lever is 60 inches long, the arm of the
weight 90 inches long, the power is 76 lbs. Required the
weight? Ans. 50§ lbs.

19. The power-arm of a lever is 17 feet long, the arm of the
weight 19 ft. ; what power will balance a weight of 950 lbs ?

Ans. \QQ\\} \h%.

20. The power-arai of a lever is 12 ft. long, the power is 10 lbs.
and the weight 75 lbs. Required the length of the arm of
the weight. Ans. If feet.

21. By means of a lever a power of I25 lbs. is made to balance
a weight of 93 lbs. ; the arm of the weight being 6J feet,
what is the length of the arm of the power ? Ans. 47^^ ft.

78. When the power and the weight merely balance
each otlier, i, e., when no motion is produced, there is no
difference between the second and third classes of levers
since neither force can be regarded as the mover or the
moved. In order to produce motion, one of these forces
must prevail, and the lever then belongs to the second or
third class, according as the force nearer to or farther from
the fulcrum prevails.

79. If the arms of the lever are curved or bent, their
effective lengths must be ascertained by perpendiculars
drawn from the fulcrum upon the lines of direction of the
power and the weight ; the same rule must; be adopted
when the lever is straight, if the power and weight do not
act parallel with one another.

^BT3. 78, 70.

z. is made
eight is 2 i

irm of the
quired the
IS. 31i lbs.

xm of the
t of 5 cwt.
f. ITJ cwt.

irm of the
iqiiired the
IS. 50§ lbs.

irm of the
of 950 lbs?
061\^\h».

r is 10 lbs.
the arm of
. Ig feet.

to balance
g 6i feet,

. m^ ft.

balance
ere is no
of levers
r or the
se forces
econd or
her from

nt, their
idiculars
)n of the
adopted
t do not

Akts.80,81.] . THE LEVER.

THE COMPOUND LEVER.

80. Two or more simple levers acting upon dne another
constitute what is called a Compound Lever or Corn-

Fig. 4.

▼jt/

"ST

■•

position of Levers. In such a combination the ratio of
the power to the weight, is compounded of the ratios
existing between the several arms of the compound lever.

81. In the compound lever if Wzziweight, V=power, a a' a" the
anns of the ioeight, and A A' A" the arms of the power. Then

?: "W Is a X a' X a" : A X A' X A" .. . ,

^ Wxaxa'xa" PxAxA'xA^

Hence V=:-^^^^ and W = —^^^^^^JT-

Example 22. — In a combination of levers the arms of the power
are 6, 7, and 11 feet, the arms of the weight 2, 3, and 3 J feet, the
weight is 803 lbs. ; what is the power ?

Hero Tr=803lb9.,a-i

SOLUTION.
: 2, a'= 3, a" = 3i. A=z6, ALz^ 7, A" = 11.

mv. D WXaXa"Xa" 803X2X3X3^ „^, ,. ^
^^^^^ ^ = ^X^'X^ ~ = -6^X11- = 3^ ^^'' ^'^'

Example 23. — In a compound lever the power is 17 lbs., the
arms ot the power 9, 7, 6, 5, and 4 ft., and the arms of the weight
2, 3, 1, 1, and i ft. Required the weight.

SOLUTION.

Here P = 17 lbs., A = 9.A'='7, A" = 6, A'" = 5, A"" = 4, a = 2, a' = 8.
a" = 1. a'" = 1, and a"" = i.

._ PXAXA'XA"XA"'XA"" _ 17X9X7X6X5X4 _ 12862«
'Xa">
Alls.

Then W

= 642J0 lbs.

aXa'Xa"Xa'"Xa"" 2X3X1X1X| "~ 2

EXERCISES.

24. In a compound lever the arms of the power are 9 and 17 ft.
the arms of the weight 3 and 4 ft., the power is 19 lbs
What is the weight? jins. 242^ lbs

THE WHEEL AND AXLE.

[ASTS. 82-M.

36. la a compound lever the arms of the power are 6, 8, 10, and
12 ft., the arms of the weight, 7, 6, 3, and 1 ft., the weight
iB 700 lbs. Required the power ? Ans, 12J|.

26. In a compound lever the arms of the weight are 11, 13, and
9 ft., the arms of the power are 4, 7, and 2 ft., the weight is
560 lbs. What is the power? .in». 12870 lbs.

- THE WHEEL AND AXLE.

82. The wheel and axle consists of a wheel with a
cylindrical axle passing through its Fig. 5,
centre, perpendicular to the plane ^
of the wneel. The power is applied
to the circumference of the wheel,
and the weight to the circumfer-
ence of the axle.

83. The wheel and axle is merely
a modification of the lever with un-
equal arms ; the radius of the wheel
corresponding to the arm of the
power and the radius of the axle
to the arm of the weight.

84. The wheel and axle is sometimes called the con-
tinual or perpetwil levevy because the power acts continu-
ally on the weight.

86. The power and weight in the wheel and axle are in
equilihrium when the power is to the weight as the radius
of ike axle is to the radius of the wheel,

86. For the wheel and axle — let F=the powers W = the weight,
r := radius of the axle, B =: radius of the wheel.

Then P : W

Hence P =

Wxr
B

W = -y- ;

HxR

Wxr

r = -T^— ; and E= — p- .

ExAMPLB 27.— In a wheel and axle the radius of the axle is 7
inches, the radius of the wheel is 36 inches, what power will
balance a weight of 643 lbs. ?

,1 ,

[Arts. 82-8<.

6, 8, 10, and
, the weight

e 11, 13, and
he weight is
8. 12870 lbs.

eel with a
5.

the con-
continu-

;fe are in
radius

ie weight,

Wxr
P •

|azle is 1

rer wilj

Abt.86.3

THE WHEEL AND AXLE.

BOLUTIOir.

Here fr= 643 lbs., £ = 30 inches, and r = 7 inches.

Then P =

Wxr^ 643X7 ,„„ ^

Example 28. — In a wheel and axle the radius of the axle is 6
inches, the radius of the wheel is 2*7 inches. What weight will
be balanced by a power of 123 lbs. ?

SOLUTION.

Here P = 123 lbs., ^ = 27 in., and y = 6 in.

Then Jr=?^ = l?2^ = 653i lbs. ^«*. '

Example 29. — By means of a wheel and axle a power of 11 lbs.
is made to balance a weight of 117 lbs., the radius of the axle is
3 inches. Required the radius of the wheel ?

solution. ,1

Here >r= 719 lbs., P = 1 1 lbs., and r = 3 In.

Wxr 719X3 _,„«,. 1. .
~c^ ., — := IW^-v Inches. Ant.

ThenJB= — k^=.

'TT

EXERCISES.

30. In a wheel and axle the radius of the axle is 7 inches, the
radius of the wheel is 70 inches. What power will balance
a weight of 917 lbs.? ^n«. 91^V lbs.

In the whsel and axle the radius of the axle is 5 inches, and
the radius of the wheel 1 7 inches. What power will balance
a weight of 5&'50 lbs. ? Jlns. 1750 lbs.

31.

32.

33.

In a wheel and axle the radius of the axle is 9 inches and
the radius of the wheel is 37 inches. What power will
balance a weight of 925 lbs. ? jins. 225 lbs.

In a wheel and axle the radius of the axle is 11 inches and
the radius of the wheel is 45 inches. What weight will a
power of 17 lbs. balance ? jins. 69^<V lbs.

34. By means of a wheel and axle a power of 37 lbs. balances a
weight of 700 lbs., the radius of the axle being 8 inches,
what is the radius of the wheel? ^ns. 161^^ inches.

35. By means of a wheel and axle a power of 22 lbs. balances a
weight of 870 lbs. If the radius of the wheel be 67 inches
what will be the radius of the axle? Jins, 1^ inches.

THE WHEEL AND AXLE.

[Arts. 87, 88.

1;
id

Fig. 6.

THE DIFFERENTIAL WHEEL AND AXLE.

87. In the differential wheel and axle, the axle consists
of two parts, one thicker than the
other ; the rope by means of which
the weight is raised, winds on the
thicker part, while it rolls off the
thinner. By each revolution of the
wheel the rope rolls once off the
thinner portion and once on the
thickerportion, and is consequently
shortened only by the differences
between the circumferences of the
axles; and the distance through
which the weight isAraised is equal
to half the shortening of the rope.
The effect is therefore the same as if an axle had been
used with a radius equal to half the difference between the
radii of the thicker and thinner parts of the differential
axle.

88. For thk differential tdheel and axle let d = the difference
between the radii of the axles^ R = rOdiusfpf the wheel, P = the
power, and W = the weight.

ThenV : W :: id : R.

Whence P=

R '

W =

PXR

WXjd

and d :

PxR
■ iW

Example 36. — In a dififerential wheel and axle the radius of
the larger axle is 4^ inches, the radius of the smaller axle is 4^
inches, the radius of the wheel is 70 inches. What power will
balance a weight of 1000 lbs. ?

SOLUTION.

Here d = diflference of radii = 4 — i, = At, W = 1000 lbs., JJ = 70 in.

Then P = ^^= lOOOX^^ W = m^ = -A ^bs- ^«*"
ic 70 20

Example 37. — In a differential wheel and axle the radii of the
axles are 2f and 2^ inches, the radius of the wheel is 100 inches.
What power will balance a weight of 7234 lbs. ?

Abt. 59.]

THE WHEEL AND AXLE.

SOLUTIOir.

Here d = \-^ = ^^ in. B = 100, and W = 7234.
Then P = ^ = ?25*^ = im% lbs, ^ns.

Example 38.— In a diflRerential wheel and axle the radii of the
axles are 3i and S^V inches, the radius of the wheel is 86 inches.
What weight will a power of 17 lbs. balance ?

SOLUTION.

Here d = \—{y — ^^ of an inch, 22 = 86 inches, and P = 17 lbs.

Then W-

PXB 17X86 1462

407664 lbs. Ans.

Example 39. — ^In a differential wheel and axle the radius of
the wheel is 32 inches, and a power of 5 lbs. balances a weight
of 729. What is the difference between the radii of the axles ?

soLriiow.
Here W = 729 lbs., P = 5 lbs., and B = 82 inches.

_, , P>^B 6X32 160 „»„,.,, ^
Then d = ,-,■ ■ = . .^ ' = — — = lag of an inch. Ans.
iW i of 729 i|a TT7

EXERCISES.

40. In a differential wheel and axle the radii of the axles are
. 7^ and 7^ inches, and the radius of the wheel is 85 inches,

what power will balance a weight of 6900 lbs. ?

Ms. ff^ lbs.

41. In a differential wheel and axle the radii of the axles are
17 and 16 inches, and the radius of the wheel is 130 inches,
what weight will a power of 17 lbs. balance ?

Ms. 4420 lbs.

42. In a differential wheel and axle, the radii of the axles are
2h and 2} inches, and a power of 23} oz. balances a weight
of 6400 oz. Required the radius of the wheel ?

Jlns. 6^ inches.

43. In a differential wheel and axle, the radii of the axles are
4^ and 5 inches, the radius of the wheel being 120 inches,
what power will balance a weight of 2430 oz. ?

^ns. 8^ff or.
In a differential wheel and axle, the radii of the axles are
If and If feet, the radius of the wheel is 12} feet, what
weight will a power of-i46^1bs. balance. Ms. 146880 lbs.

89. Since the wheel and axle is merely a modification
of the lever, a system of wheels and axles is simply a
modification of the compound lever, and the conditions of

WHEEL WORK.

TAkts. 90-04.

equilibrium are the same, i. e., the ratio of the power to the
weight is compounded of the ratios of the radii of the axles
to the radii of the wheels. In toothed gear^ however,
owing to the difficulty in determining the effective radii of
wheel and axle, the ratio of the power to the weight is
determined by the number of teeth and leaves upon the
wheel and pinion.

90. Axles are made to act on wheels by various methods
— as by the mere friction of their surfaces, by straps or
endless bands, &c. ; but the most common method of
transmitting motion through a train of wheelwork is by
means of teeth or cogs raised upon the circumferences of
the wheels and axles.

91. When cogged wheels and axles are employed, that
part of the axle bearing

the cogs is called sipin- ^^'

ion. The cogs raised

upon the pinion are

called leaves^ those upon

the wheel are termed

teeth.

92. Wheel work may
be used eiiher to con-
centrate or diffuse power.
The power is concen-
trated when the pinions
turn the wheels, as is the
case in the crane^ which
is used to gain power. The power is diffused when the
wheels turn the pinions, as is the case in the fanning mill,
threshing machine, &c., where extent of motion is sought.

98. In a system of toothed wheels and pinions, the conditions of
equilibrium are that, — the power is to the weight as the continued
product of all the leaves is to the continued product of all the teeth.

04. For a train of wheel work let P = the power ^ W = the
weight , t *f' = the teeth of the wheel, and 1 1' T = the leaves of
shepiniott^.

STS, 90-04.

AsT. 94.]

WHEEL WORK.

3r to the
he axles
lowever,
radii of
^eight ia
ipon the

methods
traps or
thod of
rk is by
ences of

ed, that

Then P : W :: 1 X r X \" : t X t' X t".

len the
ig mill,
sought.

itions of
mtinued
he teeth,

V = the
eaves of

Hence P =

wxixi'xi"

and W =

Pxtxt'xt"

txt'xt" ' **'"* " - ixi'xi"

F ExAMPLK 45. — The number of teeth in each of three snccess-
ive wheels is 80 and the number of leaves in each of the pinions
is 5. With this machine what weight will be supported by a
power of 17 lbs.?

soiriiox. ' .

Here P = 17, < = 80, t' = 80, t" = 80, 1 = 8, r = 5 and I" =: 6.

mu xir—PX'Xi^><^'—J^><80X8<»<80_ 8704000 „„„„„,. .
Then Tr=— y^-^^p: ^^^^^ = -nT-^"^^^^ ^^'- ^'^'

Example 46. — In a train of wheel work there are four wheels
and four axles, the iSrst wheel and the fourth axle plain, (t. «. .
without cogs), and having radii respectively of 10 and 2 feet.
The second wheel has 60, the third 90 and the fourth 70 teeth, the
first axle has 7, the second 6 and the third 9 leaves. What
power will hold in equilibrium a weight of 20000 lbs. ?

BOLVTIOir.

Here we have a combination of the simple wheel and axle and a system
of cogged wheels and axles.

?r=20000 lbs. 26=10, r=2, fc=60, <'=90, <"=70, fc=7, Z'=5 and l"=9.

Then cogged wheels and axles acting alone F=

20000X7X8X9

= 161

60X90X70

lbs., and so far as the action of the plain wheel and axles is concerned this
16| lbs., becomes the weight.

Then P =

TTXr 161X2

= -TT=8Hb8.ii>M.

B 10

ExAMPLB 47. — ^In a train of wheel work there are three wheels
and axles, the first wheel and the last axle plain, an . having
a radius of 9 and 3 feet respectively — the cogged wheels have
respectively 80 and 110 teeth, and the pinions 11 and 8 leaves.
What weight will a power of 100 lbs. sustain ?

SOLUTION.

Here P=100 lbs., JB=9, r=3, fc=80, t'=110, 7=11 and V=»,

Pxtxtf 100X80X110

Then for cogged wheel work acting alone 7r=
880000

IXl

11X8

-= 10000 lbs.

T. ii 1. 1 J 1 1 ,^_ PX 22^10000 X9_ 90000 ^
For plain wheel and axle alone 7^2= r= j— = .—— =

80000 lbs. Ans,

WHEEL WOllK.

TAUTS. 'J5 V»s,

EXERCISES.

48. In a system of wheel work there are five wheels and pinions ;
the wheels have respectively 100, 90, 80, 70 and 60 teeth,
and the pinions respectively 9, Y, 11, 9 and T leaves — with
such an appliance, what weight would be sustained by a
power of 11 lbs. ? Jlns. 5333333J lbs.

In a train of four wheels and axles the wheels have respec-
tively 10, 65, 60 and 50 teeth, and the axles respectively 9,
8, 1 and 6 leaves ; with such an instrument, what power
could support a weight of 13000 lbs. ? Jlns. 2|f lbs.

In a train of wheel work there arc three wheels and three
axles, the first wheel and last axle plain, and having radii
respectively 6 and 2 feet. The second and third wheels
have respectively 80 and 50 teeth, and the first and second
pinions respectively 5 and 8 leaves. With such a machine
what weight will be balanced by a power of 11 lbs. ?

Jlns. 33000.

49.

50.

\.f

95. In ordinary wheel work it is usual, in any wheel
and pinion that act on each other, to use numbers of teeth
that are prime to each other so that each tooth of the
pinion may encounter every too«h of the wheel in succes-
sion that thus, if any irregularities exist, they may tend to
diminish one another by constant wear. This odd tooth
in the wheel is termed the hunting cog.

Thus if a pinion contain 10 leaves and the wheel 101 teeth, it is evident
that the wheel must turn round 101 times and the pinion 10 X 101 or 1010
times before the some leaves and the teeth will be again engaged.

96. Wheels are divided into crown, spw and bevelled
gear,

97. The croion wheel has its teeth perpendicular to its
plane ; the spur wheel has its teeth, which are continua-
tions of its radii placed on its rim ; the levelled wj/teeZhas
its teeth obliquely placed, i. e. raised on a surface inclined
at any angle to the plane of the wheel.

98. To communicate motion round parallel axes spur-
gear is employed, bevelled gear is used when the
axes of motion are inclined to one another at any proposed
angle. Where the axes are at right angles to one another
a crown wheel working in a spur pinion or a crown pinion
working in a spur wheel is usually employed.

AKT8.90-10H.1

rilK VULLEY.

to its
;inua-
leZhas
ilined

99. Bevelled wheels are always frusta of cones channel-
led from their apices to their bases.

NoTB.— When bovolled wheels of different diameters are to work together
the sections of the cones of which they are to bo ftiista are found in the
following manner:— -p. „

Let ABbo the dia- * ig- o-

meter of the largo
wheel and B C that
of the smaller. Place
A B and B C so as
to include the pro-
posed angle. Bisect
A B in D and B C in
E. Draw perpen-
diculars D F, E F
meeting in F and
join FA, FB and FO.
Then FAB and FBO
are sections of the
required cones. Also
drawing H G paraU .
lei to A B and G P A
parallel to B C, we
obtain H A B G, and G B C P any required frusta.

THE PULLEY.

100. The Pulley is a circular disc of wood or iron,
grooved on the edge and made to turn on its axis by means
of a cord or rope passing over it.

101. The pulley is merely a modification of the lever
with equal arms, and hence no mechanical advantage is
gained by using it — the theory of its use being just as
perfect if the cord be passed through rings or over perfectly
smooth surfaces. The real advantage of the pulley and
cord as a mechanical power is due to the equal tension
of every part of the cord, i. e., is founded upon the fact
that the same flexible cord, free to run over pulleys or
through smooth rings in every direction must always un-
dergo the same amount of tension in every part of its
length.

102. The pulley is called either fixed or movable ac-
cording as its accis is fixed or moveable. _ .

103. Movable pulleys are used either singly, in which
case they are called runners^ or in combination. Systems
of pulleys are worked either by one cord or by several
cords. Pulleys worked by more than one cord are called
Spanish Bartons.

THE PULLEY.

fARTS. lW-10^.

104. The pulley is often called a 8heaJ\ and the case
in which it turns a block. A block may contain many
sheaves. A combination of ropes, blocks and sheaves is
called a tackle.

106. In the single fixed pulley the power must be equal
to the weight, i. e., a fixed pulley does not concentrate
force at all. And hence the only mechanical advantage
derived from its use is, that it changes the direction of the
power.

Fiff. 9.

106. lib a system of pulley moved by

one cord the conditions of equilibrium are
that the power is to the weight as 1 is to
twice the number of movuhle pulleys.

This is evident from the fact that the weight is
sustained equally by every part of the cord, and,
neglecting the last fold or that to wliich the power is
attached, there are tw6-folds of cord for every mov-
able pulley. Thus in Fig. 9 the weight is sustained
by A and B, each bearing i of it ; and since B passes
over a fixed pulley, the power attached to C must
be equal to the tension exerted on }{ = ^ the
weight.

107. For a system of pulleys moved by one
cord let P = the power, W = the weight and
n = the number of movable pulleys.

Then P : W :: 1 : 2n.

W W

Hence P = 2^, W = P x 2u, n = ^p-

Example 51. — In a system of pulleys worked by a single cord
there are 4 movable pullevs. What power will support a
weight of 804 lbs. ?

SOIUXION.

Here Tr= 804 and » = 4.

Example 52. — In a system of 1 movable pulleys worked by a
single cord, what weight will be supported by a power of Itl bs.?

f SOLUTION.

Here P = 17 and w = 7.

Hence jr=i* X 2 Xnri 17X2X7=^17 X U = 238lbs. Ant

Am. 108.1

THE PULLEY.

tlxAMPLB 63. — 111 a system of movable pulleys worked bj a
single cord a power of 7 lbs. balance a weight of 84 lbs. ; how
many movable pulleys arc there in the combination ?

SOLUTIOK.

Here P=71b«.and ir=84 lbs.

Honcen=~ = -?i=-=0. Ana.
2XP 2X7 H

EXERCISES.

54. In a system of six movable pulleys worked by one cord
the weiglit is 700 lbs. What is the power ? jins. 58^ lbs.

65. In a system of eleven movable pulleys worked by one
cord the weight is 2563 lbs. Required the power ?

Jns. 116i lbs.

56. In a system of 1 movable pulleys worked by one cord,
the power is 37 lbs. Required the weight? Jns. 592 lbs.

57. In a system of eight movable pulleys worked by a single
cord, the power is 13 lbs ; what is the weight ? ./?n». 182 lbs.

58. In a system of movable pul-
leys worked by a single cord, pj_^ ^q
a power of 35 lbs. supports a
weight of 7000 lbs. How
many movable pulleys arc
there in tlie combination ?

jlns. 100.

108. In system of pulleys
such as represented in Fig. 10,
where each movable pulley
hangs by a separate cord, one
extremity of each cord being
attached to a movable pulley
and the other to a hook in a
beam or other fixed support,
each pulley doubles the eflfect,
and the conditions of equi-
librium are that the power is to
the weight as 1 is to 2 raised to
thejyowcr indicated hy the mim-
her of movable pulleys.

Note.— This will become evident by
attentively examiuing the diagram and
following up the several cords. The
figures at the top show the portion of
weight borne by the several' parts of the
beam, those attached to the cords show
the portion of the weight sustained by
each part of the cord.

16 lbs.

THE PULLEY.

rARTS. 100, 110.

109. For a sy stem of pulleys tuch at exemplified in Fig. 10 let
P =: the power, W = the weighl, and n = the number of movable
pulleys,

Then P : W :: 1:2". Hence P = ^t. end W «a P x 2".

Example 50. — In a system of pulleys of'tlio form indicated in
Fig. 10, there arc 5 moveable pulley.'? and a weight of 128 lbs.
What is the power ?

SOLUTION.
Hero n' = 128 lbs. and « = B.

Thou P = ^ = ?# = ~ =4 U.H. Ans.
2" 2" 32

Example 60. — Tn such a systi-m of pulloy.q as is shewn Fig. 10
there are 7 movable pulleys. What weight will a power of 11
lbs. balance ?

SOLUTIOW. ^

Here P=: 11 and ;» = 7.

Ilenco »'= P X 2" = 11 X 2 ' =11 X 128 = 1408 11)8. ^w*.

EXERCISES.

61. In the system of pulleys represented in Fig. 10, where there
are movable pulleys ; what power will sustain a weight
of 8000 lbs.? ./?;?s. 125 l))s,

02. In such a system when there arc 10
movable pulleys, what power will
sustain a weight of 48000 lbs. ?

^ns. 40} lbs.

G3. In such a system when there are 7
movable pulleys, what power will
support a weight of 4564 lbs. ?

Jlns. 355-i lbs.

64. In such a system when there are 3
movable pulleys, what weight will
be sustained by a power of 17 lbs. ?

Jns. 136 lbs.

65. In such a system what weight will a
power of 70 lbs. support when there
are 5 movable pulleys ?

,^ns. 2240 lbs.

66. In such a system what weight^will
a power of 100 lbs. support when
there are 1 1 movable pulleys ?

Jns. 204800 lbs.

110. In a system of pulleys such
as represented in Fig. 11, where the
cord passes over a fixed pulley at-
tached to the beam instead of being

Fig. 11.
2 18

rs. 100. 110.

Ann. Ill, 112.]

THE PULLEY.

t^. 10 let
^ movable

< 2".

catcd in
128 11)3.

Fijf. 10
r of 11

'0 there
weight
25 lbs.

fastened to a hook in the boHm, each movable pulley
triphs the otlect, and tho conditions of equilibrium are that
the power is to the weight <is 1 to 3 raised by the power
imlicated by the number of m<w(it}le pulleys.

This win nppoar plain by a rrforoiu.) to tho accompanying diagram,
whoro tlu! nunib(!rM rcprcHfut titu Haiuc aa in Art. 1U8.

111. Ill a sijHtcin such as is represented in Fig. 11, let P =
■power. W = the weight, and n = the number of movable pullrya.

Then P : W

1 : 3".

Hence P = -- and W = P X S".

ExAMPLR G7. — In tho system of pulleys represented in Pig. 11,
what powcL* will balance a weight of 4500 when there are 4
movable pulleys ?

SOLUTION,

Horo IK=-4600 and >» = 4.

mi D ^^ 4600 __ 4500 ...

Then P=-=^= --=355 lbs.

Ana,

ExAHPLB G8. — In such a system when there are 6 moyable
pulleys, what weight will a power of 10 lbs. support ?

Jlnt. 11 lbs.

SOLUTION.

Horo P = 10, and m:=0.

Then Wz=P X Sn = 10 X 36 SCIO)^ 729 = 7200 lbs. An».

EXERCISES.
69. In the system of pulley represented in figure 11 there are 5
movable pulleys ;what weight may be supported by a power
oflllbs. ? ^ns. 2430 lbs.

10. In such a system there are *I movable pulleys and the

weight is 24057 lbs. Required the power ?
'l. In such a system are all 9

movable pulleys — through

how many feet will the

power descend in order to

raise the weight 10 feet ?

^ns. 196830 feet.

112. If the lines of direc-
tion of the power and weight
make with one [another an
angle greater than 120°, the
power will require to be great-

Fig. 12.

THE INCLIUED PLANE. [Aetb. 113-117.

er than the weight; and as this angle approaches 180°,
the difference between the power and weight will ap-
proach oc . Hence it is impossible for any power -P,
however great, applied at P, to pull the cord ABC
mathematically straight, and that however small the
weight W may be. -

• •*

INCLINED PLANE.

113. The Inclined Plane is regarded in mechanical
science as a perfectly hard, smooth, injlexihle plane, in-
clined obliquely to the weight or resistance.

114. There are two ways of indicating the degree of
inclination of the inclined plane :

Ist. By saying it rises so many feet, inches, &c., in a
certain distance.

2nd. By describing it as rising at some stated angle
with the horizon,

116. In the inclined plane the power may be applied in
any one of three directions :

1st. Parallel to the plane.

2nd. Parallel to the base.

3rd. Inclined at any angle to the base.

116. In the inclined plane the conditions of equilibrium
are as follows : —

Ist. 1/ the power act parallel to the plane: — the power
is to the weight as the height of the plane is to its length.

2nd. If the power act parallel to the base: — the power
is to the weight as the height of the plane is to its base.

NoTB.— The third case does not come within the design of the present
work.

117. For the inclined plane Id P = the power, W = the weight,
L = length of the plane, H = height of the plane and B = base of
theplane.

Aet. 117.]

THE INCLINED PLANE.

ThenF : W :: H : L.
Hence P = — = —

PXL PXL WXH

jihoP : W :: H : B.

/fence P = — g— ; W = — jy— H = ,y and B =r — = — .

Example '72.— On an inclined plane rising 7 feet in 200, what
power acting parallel with the plane will sustain a weight of
4000 lbs.?

SOLUTION. ,

Here W= 4000 lbs., L = 200, and i/= 7
IVxJI 4000X7 28000

ThenP =

200

140 lbs. Am.

Example) 13. — On an inclined plane rising 9 feet in 170— what
weight will support a power of 180 lbs. acting parallel to the
plane?

SOLUTION.

Here P = 180 lbs., X = 170 and i/= 9.

Then W=

PxL 180X170

=3400 lbs. Jft«.

H 9

Example 74.— On an inclined plane a power of II lbs. acting
parallel to the piano supports a weight of 150 lbs.- how much
does the plane rise in ^00 feet?

SOLUTION. . . fr'

Here P = 11 lbs., TV— 150 lbs., L = 200 feet.

Then H=

PX£_11X200
W ~ 160

=14 feet 8 inches. Ans,

Example 75.— The base of an inclined plane is 40 feet and the
height 3 feet, — what power acting parallel to the base will sup-
port a weight of 250 lbs. ?

SOLUTION.

Here Tr= 250 lbs., //= 3, and jB = 40.
TTxfl 250X3

Then P=

="*9iJb8. Anf. 1'^%

Example 76. — On an inclined plane a power of 9 lbs. acting
parallel to the base supports a weight of 700 lbs. — the height of
the plane being 18 feet what is the length of the base ?

' I f

THE INCLINED PLANE. [Akts. 118, 119.

SOLUTION.

Hero P=0 lbs. W- 700 lb8. and /f =18 feot.

Then B =

WxH 700X18

= 1400 feot. Alls,

EXERCICES.

77. On an inclined plane rising 1 foot in 35 feet what power
acting parallel to the plane will support a weight of 17500
lbs. ? Jlns. 500 lbs.

78. On an inclined plane rising 9 feet in 100 feet what power
acting parallel to the plane will sustain a weight of 4237 lbs.?

^ns. 381-i3(fJjylb3.

79. On an inclined plane whose height is 11 feet and base 900
feet what power acting parallel to the base will sustain a
weight of 27900 lbs. ? ^ns. 341 lbs.

80. On an inclined plane rising 7 feet in 91 feet what weight
will be supported by a power of 1300 lbs. acting parallel
with the plane ? Jlns. 16900 lbs.

81. On an inclined plane a power of 2 lbs. acting parallel to the
plane, sustains a weight of 10 lbs.— what is the inclination
of the plane ? »^ns. Plane rises 1 foot in 5 feet.

82. On an inclined plane a power of 7 lbs. acting parallel to the
base sustains a weight of 147 lbs. — if the base of the plane
be ] 7 feet what will its height be ? Jlns. ^ j- feet.

83. On an inclined plane rising 2 feet in 109 feet what weight
will be sustained by a power of 17 lbs. acting parallel to the
plane ? Ms. 926 J lbs.

84. On an inclined plane a power of A} lbs. sustains a weight
of 223-i*f lbs. ; the power acting parallel to the plane what
is the degree of inclination ?

Ms. Plane rises 341 feet in 17189 feet.

85. What weight will be supported by a power of 60 lbs. acting
parallel to the base of an inclined plane whose height is 7
feet and base 15 feet. jins. 1320 lbs.

THE WEDGE.

118. The wedge is merely a movable inclined plane or a
double inclined plane, i. e. two inclined planes joined to-
gether by their bases.

119. The wedge is worked either by pressure or by

jpercussion.

NoTH.— When the wedge is worked hy percussion, the relation between
the power and weight cannot be ascertained since the force of percussion
difTers so completMy from continued forces as to admit of no comparison
with them.

Aaie. 120-122.]

THE SCREW.

feet,
acting
htis 7
Olbs.

120* In the wedge the conditions of equilibrium are
tliat the power is to the weight as half the width of the
hack of the wedge is to its length.

NoTB 1.— IJulike all the other mechanical powers, the practical use of the
wedge depends on Mction, as, were it not prevented by Mction, the wedge
would recoil at every stroke.

NoTB 2.— BAzors, knives, soisson, chisels, awls, pins, needles, ftc, are
exantples of the application of the wedge to practical purposes.

121. For the wedge, let P zz power or pressure, Wz=:the weighty
L = the length of the wedge, and B = the width of the back.

Then P:W::l B:L. Hence P =

WxiB

and Wz=,

PXL

Example 86. The length of a wedge is 24 inr'ic~, and its
thickness at the back 3 inches, what weight would be raised by
a pressure of 750 lbs. ? - ,

SOLVTIOK.

Here P = 760 lbs., £ = 24 inches, and i JB =-: li inches.

Then W -

PxL 780X24

= 750 X 16 = 12000 lbs. Ans.

\B '■ li

Example 87. In a wedge, the length is 17 inches, thickness of
hack 2 inches, and the weight to be raised is 11000 lbs. Re-
quired the pressure to be applied 7

SOLVTIOK.

HerelFssllOOO, L = 17 inches, audi J? = 1 inch. ,• /

mu « Wx^B 11000X1 „^„ , ,. .
Then P = ' = — rz — — 647-^^ lbs. Ans.

, .. EXERCISES.

88. The length of a wedge is 30 inches and the thickness of
its back 1 inch, what weight will be raised by a pressure of
97 lbs. wdPn«. 5820 lbs.

89. The length of a wedge is 19 inches and the thickness of its
back 4 inches, what pressure will be required to raise a
weight of 864 lbs. ? jlns. 90 jf lbs.

90. The length of a wedge is 23 inches and the thickness of its
back 3 inches — ^with this instrument what pressure would
be required to raise a weight of 1771 lbs. ? .Sns. 115J lbs.

THE SCREW.

122. The screw is a modification of the inclined plane

and may be regarded as being formed of an inclined plane

wound round a cylinder.

NoTB.— The lerew bean the same rehition to an ordinary inclined plant
that a circular staircase does to a straight one.

THE SCREW,

[AkTB. 12a-128.

Fig. i3.

128. The threflds of the screw are either triangular or
square. The distance of a thread and a space when the
tljread is square, or the,di8tance between two contiguous
triangular threads, is called the pitch.

124. The screvf is commonly worked by pressure against
the threads of an external screw, called the box or nut.
The power is applied either to turn the screw while the
ljut is fixed, or to turn the nut while the screw is kept
immovable.

126. In practice, the screw is
seldom used asasimpie mechani-
cal power, being nearly always
combined with some one of the
others — usually the lever.

126. The conditions of equili-
brium between the power and the
weight in the. «crew are the same as
for the inclined, plane, where the
Dower acts parallel to the base, i.e.

The power is to the weight as the pitch {i. e. height') is
to the circumference of the base (i. c, length of the plane).

When the screw is worked by means of a lever, the con-
ditions of equilibrium are : —

The power is to the weight as the pitch is to the circum-
ference of the circle described by the power.

\ 127. The eflSciency of the screw as a mechanical power
may be increased by two methods :

Ist. By diminishing the pitch.
• 2nd. By increasing the lengtli of the lever.

128. For the screw, let P = the power, W = the weight, p =
the pitch, and I = hngth of the lever.

Then since the lever forms the radius of the circle described by
the power, and the circumference of a circle is 3*1416 times the
diameter, and the diameter is tunce the radius, P:lV::p:l'X.2x
3-1416.

Wy.p

Hence P =

„, PX?X2X3-1410 , PXiX2X3*1416
W= andp=-

ZX2X3*1416 " P """'' W

NoTB.— The pitch and the length of the lever must be both ezpreesed in
wnita of the same denominations, i. e. both feet, or both inches.

'■*>■

re. 123-128.

Art. 128.]

THE SCREW.

Example 91. What power will sustain a weight of 70000 lbs.
by means of a screw having a pitch of -^(ih of an inch, and th«
lever to which the power is attached 8 ft. 4 in. in length 7

SOLUTION.

Hero 1V= 70000 lbs., p = -j^^ in., and i = 8 ft. 4 in. = 100 in.

Here P—

WX P

7000OX-14-

5000 800000 ,„.. „
=«i;;s::^=sss^=^7 9871bs. Atu.

iX2X31416 100X2X3-M18 628*32 62832

Example 92. — What weight will be sustained by a power of
5 lbs. by means of a screw having a pitch of ^^th of an inch, the
power lever being 50 inches in length ?

SOLUTIOir.

Hero P = 5 lbs.-, p = -jij inch, and 1 = 60 inches.

mt. Mr PX?X2X3-1416 5X50X2X8'1416 1570*8
Then }V= = —

-i^ff

a5708lb8. Am.

Example 93. By means of a screw having a power lefer 6 ft.
10 inches in length, a power of 6 lbs. sustains a weight of '80000
lbs. ; what is the pitch of the screw ?

SOLUTION.

Here P = 6 lbs., W=^ 80000 lbs., and I = 70 inches.
PxiX2X31418 6X70X2X3*1416 2614*944

Then p=:
33

about

1000

W
of an inch. Ans.

80000

=•0326868 inches, or

Example 94. What power will sustain a weight of 96493 lbs.
by u^eans of a screw having a pitch of ]^th of an inch, the power
lever being 25 inches in length?

SOLUTION.

Here W= 96493 lbs., p = -^^^th inch, and I — 25.

ThenP =
108*403 lbs.

WXP

iX2X3*1416
Ans.

96493Xfr
25X2X3*1416

EXERCISES.

'!4

saaiya

157*08

170281764
167*08

95. What power will support a weight of 87000 lbs. by means
of a screw having a pitch of -,%th of an inch, the power
lever being 6 ft. 3 inches long ? ^ins. 31*83 lbs.

96. What weight will be sustained by a power of 200 lbs. acting
on a screw having a pitch of ^%th of an inch — the power
lever being 15 inches long? jin$. 314160 lbs.

97. By means of a screw having a power lever 50 inches in
length, a weight of 9000 lbs. is supported by a power of
12 lbs. Required the pitch of the screw 7

An*. '41888, or rather over f of an inch.

THE DIFFERENTIAL SCREW. [Ahtb. 129, 180.

' ■'

a. Wb^t power will lupport a weight of 11900 lbs. bj means
of a screw having a pitch of ^th of an inch, the power lever
being 10 ft. in length ? ^na. 3-713 lbs.

99. By means of a screw having a power lever 1 ft. 6 inches in
length, a power of 10 lbs. supports a weight of 65400 ;
what is the pitch of the screw ? Ana. -0864 of an inch.

100. What weight will be supported by a power of 60 lbs. act-
ipg on a screw with a pitch of :^th of an inch— the power
lever being 8 ft. 4 inches in length ? Ana. 418880 lbs.

Pig. 14.

THE DIFFERENTIAL SCREW.

129. The differential screw, (invented by Dr. John
Hunter,) like the differential wheel and axle, acts by dimi-
nishing the distance through which the weisfht is moved
in comparison with that traversed by the power.

It consists of two screws of dif-
ferent pitch, wcjrldng one within
the other (Fig. 14), so that at each
revolution of the power lever the
weight is raised through a space
only equal to the d^erence be-
tween the pitch of the exterior
screw and the pitch of the inner
screw. It follows that the mechan-
ical effect of the differential screw
is equal to that of a single screw
having a pitch equal to the differ-
ence of pitch of the two screws.

For iiuiianoe, in Fig. 14, the part B works within the part A. Now, if B
have a pitch of i^h of an inch and A a pitch of iV* then at each revolu-
tion of the handle the weight will be raised through ^—i^d=^ST of an
inch, and th'j whole instrument has the same mechanical effect as a single
screw having a pitch of i itrth of an inch.

130. For the differential acrew, let P =: power, W" = weight,
I ±: length of lever, and d = difference ofpitih df the two acrewa.

Then P : TT:: (i : /X 2X31416. , ' .,

.= . Wxd . .,^ PX/X2X31416

Hence P=^^2X31416^"^^= d •

rs. 129, 130.

AftT. 131.]

THE ENDLESS SCREW.

by means
wrer lever
)-7l3 lbs.

inches in
•f 65400 ;
' an inch.

I lbs. act-
he power
8880 lbs.

h, John
by dimi-
s moved

Now. if B
ch revolu*
^ijs of Ml
B a single

ibeightf
screws.

Example 101.— What power will exert a pressure of 20000 lbs.
by means of a differeatial screw having a power lever 60 inches in
length, the exterior screw a pitch of ^ of an inch, and the inner
screw a pitch of j/^th of an inch ?

* ' SOLUTIOX.

Here W= 20000, 1 - 50 in., and d = A — A = ift/^ - A'o = -<ViJ of an
inch.

_, -, Wxd 2Wmx^a ^W^ 2464.545 246454.54

Then /*=■

60X2X3.1416 314.16

314.16

31416

2X2X3.1416
— 7.81 lbs. Ana.

Example 102.— -What pressure will be exerted by a power of

1000 lbs. acting on x differential screw in which the power lever

is 75 inches long, the pitch of the exterior screw ^Vth of an inch,

and that of the interior screw ^l^th of an inch 7

soLUTioar.

Here P=: 1000 lbs., 1 = 75 inches, and d = f r— -/(T=ii WH— »Vb of
an inch.

1000X75X2X3.1416 471240 400554000

Then

W = ^X^X 2X 3.1416

^Vcr

:=: 120210961! lbs. Ans.

EXERCISES.

103. What power will exert a pressure of 100000 lbs. by means
of a differential screw in which the power lever is 100 inches
long, the pitch of the outer screw ^ of an inch, and that of
the inner screw ^ of an inch ?

^ns. -102 or about iV of a lb.

104. What pressure will be exerted by a power of 20 lbs.
acting on a differential screw in which the power lever is
50 inches long, the pitch of the exterior screw ^f of an inch,
and that of the inner screw ^ of an inch ? Jlns. 3455*76 lbs.

105. What power will give a pressure of 60000 lbs. by means of a
differential screw in which the power lever is 60 inches, the
pitch of the outer screw ^, and that of the inner screw '^ of
an inch ? ^ns. 2'652 lbs.

THE ENDLESS SCREW.

Fig. 15.

131. The Endless Screw, Fig.
15, is an instrument formed by
combining the screw with the
wheel and axle. The teeth of the
wheel are set obliquely so as to
act as tnucli as possible on' the

;,^^i

mri

^^0 f*»**f^«»*

THE ENDLESS SCKEW.

ART8. 132, 13:J

•f,

132. Ill Fig. 15 oacli rovolutioii of tlio Imndio makes
the wheel rovolvc only throupjh tlio simco of ono cog ;
hence if the whole haa 24 cogs, the winch must revolve
24 times in order to make the wheel revolve once.

It follows that ill the nidless or pcrpciiuil screw the con-
ditions of equilibrium are that the power in to the wdght as
the radius of the axle is to the product of the nvmher of
teeth in (he wheel multiplied by the length of the winch ;
i. e., the I'adius of the circle described bi/ the power.

133. For the cadlcsn screw let P-=zpowcr, Wrzzweighl, l-=:ilength
of winch or handle^ t-=znnmher of teeth in the w/jcc/, and I'zrzradius
of axle.

Then P: 1V::r:lxt. Whence P

IVxr

Pxixt

IXt r

Example 106. — la an endless screw the length of the winch
or handle is 25 inches, the wheel has GO cogs, and the axle to
which the weight 19 attached has a radius of 2 inches. What
weight will be sustained by a power of 100 Iba.?

SOLUTIOX, '

= 2 inches, I - 25 inchos, and f ~~ GO.
100X25X60 150000

Hero r=: 100 lbs., r

Then W ~ —

75000 IbK. Ahs.

r 2 2

Example lOY. — In an endless screw the length of the winch is
20 inches, the wheel has 56 tectb and the radius of the axle is
3 inches. What power will support a weight of 14000 lbs, ?

soiiUnox. ,

Hero W ~ 14000 lbs., r~-i inchos, B = 20 uicbcs, and / ~ SO.

Then i>-=

Wxr liOOOX.1 42000

IXt ' 20X36 1120

EXERCISES

---;i7ilbs. Ans.

108. In an endless screw the length of the winch is 18 inches,
tho radius of the axle is 2 inches, the wheel has 48 teeth,
and the power is 120 lbs. Required the weight.

^ns. 51840 lbs.

109. What power will support a weight of a million of lbs. by
means of an endless screw having a winch 25 inches long,
an axle with a radius of 1 inch, and a wheel with 100
teeth ? jins. 400 lbs.

110. What weight will be raised by a power of 40 lbs. by means
of an endless screw in which the winch is 20 inches long,
the raditti of the axle 2 inches, and tho number of teeth in
th« wheel dO ? Mi. 82000 lbs.

Arts. Ml 139.1

FRICTION.

134. Tho tbeorotical results obtained by tbe foregoing
rules are in practice very greatly inodifi«»'' cy fieveral
retarding forces. Thus friction has to be taken into ac-
count in each of the mechanical powers — the weight of the
instrument itself in the lever and in the movable pulley —
the rigidity of cordage in tho pulley and in the wheel and
axle, <feo.

FRICTION.

136. Friction aids the power in supporting the weight,
but opposes the power in moving the weight, and hence
materially affects the conditions of equilibrium in the
mechanical powers.

If P bo thfi power necesnary in tho absencA of all friction and/ the fric-
tion, then the weight will bo hold in equilibrium by any power which is
leoa than P+/, or greater than P—f.

136. Friction is of two kinds : Ist. Sliding Friction.

2nd. Rolling Friction.

137. The fraction which expresses the ratio between
the whole weight and the power necessary to overcome
the friction, is called the coej/icient of friction. The coef-
ficient of sliding friction, in the case of hard bodies, varies
from I to ^. . !

138. On a perfectly level road, power is expended only
for the purpose of overcoming friction, and on the same
road tho ratio between the power and the load is constant,
— varying on common roads, according to their goodness,
from y'y to y^y of the load. On an even railway, however,
it is not more than j\^ to ^l^ of the load, according to the
dampness or dryness of the rail. On a good macadamized
road the coefficient of friction is about ^\, so that a horse
drawinof a load of one ton or 2000 lbs. must draw with a
force of ^V ^^ 2000 lbs. or 66| lbs. ; this is called the force
of traction.

139. Various expedients are in common use for dimin-
ishing the amount of friction, such as crossing the giain,
when wooden surfaces rub on one another, using surfaces
of different materials, as wood on metal^ or one kind of

FRICTION.

fABT. 18».

|i!i"

luotal oil aiiothor kind, and anointing the uurfacc with oil,
tar, or plumbago. Tallow diminishes the friction by one-
half^

The following are the conclusions of Coulomd on the
important subject of sliding friction : —

I. Friction is directly proportional to the pressure.

II. Friction between tlie same two bodies is constant, beina; tminlluonced
by either the extent of surface in contact or tlie velocity of the motion.

III. Friction is greatest between surfaces of the same material.
IT. Friction varies with the nature of the surfaces in contact.

The ft'iotion between surfaces of wood, newly planed— i
The friction between similar metallic surfaces = i

The flriction of a wooden surface on a metallic surface = ^
The ft-iction of iron sliding or iron =^ Ij

The fHction of iron sliding or brass = ^

V. Friotion decreases as the surfaces in contact wear. In wood the
friction is thus reduced from i to \.

VI. Friction is diminished between wooden surfaces by crossing the
fibres. If when the fibres are in the same direction the coonicicnt of fric-
tion is i, it is diminished to k by crossing them.

TII. Friction is greater between rough than between polished surfkces.

Hence arise the use of lubricants in machinery. When the pressure is
small, the most limpid oils are used. At greater nressures, the more viscid
oils are preferred, then tallow, then a mixture of tallow and tar, or tallow
and plumbago, then plumbago alf^ne, and in the heaviest machinery xoop*
ttone has been found to be the most effloacious substance.

NoTK.— At very jrrM^ velocities the friction is perceptibly lessened; when
the pressure is very greatly increased, the ft^iction is not increased in pro>
portion.

BrOLLIVO FBICTI05.

VIII. Friction caused by one body rolling on another is directly proper*
tioual to the pressure, and inversely to the diameter of the rolling body.

That is, if a cylinder rolling along a plane have its pressure doubled, its
friotion will also be doubled ; but if its diameter be doubled, the friction
will be only half of what it was.

The firiction of a wooden cylinder of 32 inches in diameter rolling upon
rollers of wood is ts7 of the pressure.

The friotion of an iron axle turning in a box of brass and well coated with
oilis ^ of the pressure. - ••" " ' ' ' • v- - • - - ,.. >0Uj

Adts. 140, Ul.j

UNIT OP WORK.

OHAPTEH IV.

UNIT OF WORK, WORK OP DIFFERENT AGENTS, HORSl

POWER OF LOCOMOTIVES, STEAM ENGINES,

AND WORK OF STEAM.

UNIT OF WORK.

140. In comparing the work performed by different
iigont9, or by the same agent under diflfcrent circumBtances,
it becomes necessary to make use of some definite and dis-
tinct unit of work. Tlic unit commonly adopted for this
purpose in England and America is the labor requisite to
raise- the weight of one pound through the space of one foot.

Thus in raining 1 lb. through 1 foot. 1 unit of work is prrformed.

If 2 lb. be raiaed 1 ft., or if l lb. be raised 8 ft., 2 units of work are
IMjrformed.

'J 7 lba.be raised through 9 ft., or if 9 lbs. be raised through 7 ft., 6S
units of work are performed, &o.

141. The units of work expended in raising a hody of
n given weight are found hy multiplying the weight of the
hody in lbs. by the vertical space in feet through which it
is raised. \

Example 111. — How many units of work are expended in raia-
inff a weight of 642 lbs. to a height of 70 ft. ?

SOLUTION.

.i/w. Units of work=642 X T0=44940.

fixAMPLB 112. — How many units of work are expended in
raising a weight of 423 lbs. to a height of 267 ft. ? , ?

SOLUTION. ^

.4us. Units of work=423X 267=112941.

Example 113. — How many units of work are expended in
raising 11 tons of coal from a pit whose depth is 140 ft. ?

SOLUTION.

Here, 11 ton8=llX2000c=22000 lbs.
Then 22000X 140 = 3080000 An$.

Example 114.— How many units of work are expended in

raising 7983 gallons of water to the height of 79 ft. ?

SOLUTION.

Here, since a gallon of water weighs 10 lbs., 7983 gal8.=379830 lbs.
Then units of work =3 79830X79 =s 6306570. ilH«.

ExAitPLi 115.— How many units of work are expended ia
raialng'^O )cubie feet of water frdin d wiell irbotae fleptH W»0 feat 7

INIT OF WOKK.

Art. 11!!.

mivtinv.

Hitter A luliic foul of Wilier woIkIis Olii H*'*., •!(» riiliir ffol noiuli 02J <«M)-j:
.1750 Um.
Thou tiMits of woi'U ^37.10 -'im •TlT.'iOO. vi/i.t.

KXKRCISEH.

llrt. How nuirli work would bo required to pinup GOOOO giillonH
ol' wiitor from a niiuo wlio.so Uoptli in HOO ft. 7

Jni. 51(5000000 uiiitH.

llT. How mauy uulta of work would bo cx|)cud4Ml in pumping
8000 cubic foot of wutcr from u luino whoso drpth is G7y
foot? .^Ins. ;i:{9600000 units.

118. JIow much work would bo expended in ralHinpf tho ram of

a pile driving ongino — tho ram woighlng 'J touB, and tho
height to which it ia rai.iod being 29 ft. ?

Jlns. IIGOOO finilH.

119. How much work w"' 'i bo roquirod to raise 17 tons of

coals from a mine whoso depth in 300 foot?

Jlns. 10200000 units.

120. How much work would bo expended in raising GOO cubic

foot of wnter'to tho height of 293 feet ?

.Ins. 10987500 unit,?.

¥■■

142. Tiie iiioi^t important soiuco^ of laboring; I'orco arc
animals, waier^ windy and steam. Tlio laborincf force of
ftnimalft is niodilied by various circumstances, tho most
important of which are tlio duration of tho labor, and the
mode by which it is applied. Tho following tablo shows
the amount of ofFo(;tivo work that can uo performed
under different cirmimiPtaiiccs by the more common living
agents:

TABLE

SHRWLNO TUB WORK DONR PER MINUTE DT VARIOUS AQBNTS.

Duration of labor eight hours per day.

Horse 33000 units

Mule 22000 "

Ass 8250 '

Man, with wheel and axle 2600

drawing horizontally 3200

raising materials with a pulley 1600

throwing earth to the height of 6 ft. . . 660

((

I VI n Of

i'.-
((
U
i(
((

AitT. IIJ.I W(/KK or I.IVINU AOKNTS. 61

Mai), working with liin iirtnH atul h>gH nn in

rowirtg 4000 unitM

" raising water from a well with n pail

and ropo 1 054 "

" raising water from a well with an upright

(haiii pinnp l7.')0 *'

NoTK.- Till) work nNHiKiKMl by Wnlt to tlio liorso per niiitiito WM 33000
iitiitH, hut tlih \h known to Im> nixxit \ too gruat. A liurno of avoraK<*
NtroDKth tii'i'loriiiN iihoiit 2'iOOO tiiiilx of work nor iniiiuto. Tlin niinilMT
KJvoii ill tiio tublr is, liow(<V(<r, Niill uNcd in all caloulatlonn in civil cnd^-

iiccriiiK.

KxAMi*i,E 121. — Ifow many cubic foot of earth, each weighing
100 lbs., will IV niiin tlirow to tho height of 5 feet in a day of H
liourfl?

HOr.UTION.

Sinoo (by tlio Inblo) u iniin tbrowinR earth to tho hoifrht of 5 ft. doos BflO
iniitH of M'urk ))('r iiiinut*'— and from tho oxamplo lio workuHXtKl^lSO
niinnton.

UnltN of work doiio in tho day---500Xiao.

Units of work rnqidrod to throw 1 cubic foot to height of 5 foct :; -100X5.

Thi'ii ■ — 537i <nibic foct. Aug.

100X5 "

Example 122. — How many gallona of water will a man raise

in a day of 8 hoars from a well whoso depth is 10 feet — using a

pail and rope ?

SOLUTION.

Units of work=l05iX«0X8 ; work required to raise 1 gal.--10X70.

mi , i. II 1054X00X8 ^„„, .

Then number of gallons - - — - = 722^ . Ana,

Example 123. — How many gallons of water can a man raise by
means of a chain pump in a day of 8 hours from tho depth of 80
feet ?

SOLUTION.

Units of work jun-forniod by tho man = 1730X00X8.

Ujiits of work required to raise 1 gal. of water — 10X80.

, i. 11 1730X60X8 „.„„ .

Then number of gallons — — ^r— rr— ~ 1038. Ana.

10X80

Example 124. — IIow many tons of earth will a man working
with a wheel and axle raise in a day of 8 hours from a depth of
87 feet?

SOLUTION.

Units of work performed by the man = 2600X60X8.

Units of work required to raise 1 ton to height of 87 ft. r= 2000X 87.

_ . , 2000X60X8 ^ . .

Tonsraisod=^^^j^^- = 7-^^. Ana.

Example 125. — How many gallons of water per hour will an
engine of 7 horse powers raise from a mine whose depth is 1 10
feet?

/•'

WORK 0^ LIVINa AGENTS, &C. CAbt. 140.

BOLUTIOV.

Uidtg of work in one hoiM power==89(|0Q ptr minute.

Units of work in 7 hond powen=sSSOobx7.

TJniti of work performed by ihe'ehidhe per hour=:3S000X7x60.

Units of work i^ulred to ridse 1 gttloh o^ w»tdr to the height of 110 ft.=
10X110. ' I

Hencenumberof 8aUons=:^^^~Sr=i2600. Ant.

ExAMPii 126. — How many horse powers will it require to raise
22 tons of coals per hour from a mine whose depth is 360 feet ?

soLuixoir.

Weight of coals to bo raised = 22 tons = 44000 lbs.

UniM of work required per hour = 44000X.S60.

Units of work in one horse power per hour = 33000X60.

n :^ M i> _44000X360_„ .
Hence.H.P.=g555^— .=8. win,.

BxAMPLi 127. — How many cubic feet of water will an engine
of 15 horse powers pump each hour from a mine whose depth is

^oofeet? ■ ' ' '

BOLUTIOV.

Units of work performed by engine perliour == S3000X60X1S.
Units of jwork requh^ed to raise 1 cubic foot =: 62*6 X 900.

M «- * 1.J * i. _53000X60X16 ■„„ V

Hence, number of cubic feet ==———— — — = 528. An$.

oa o,X800 ■' '

^XAKjPLK 128[. — ^What must be the horse powers of an engine
in brder that working 12 hours per day it may supply 2300 ^mi-
lies with 50 gallons of water each per day — taking the mean
height to which the water is raised as 80 feet, and assuming
that I of the work of the ehgine is lost In transmission ?
•"^^■'roiTJTioir. '"''^"^^''^^'^'"

Weight of water pumped per day = 2SQQX SOX 10.

Unite of work requU^ed di^^

Units of work in one horse power per day = SSOOOX 12X60.

But since i of the work of the engine is lost in transmission,

Useful work of one U,. P. per day=|x33000Xl2X60.

-, _ ^ 2800X 50X10 X80 ■ «. .

Hence. H. P. = --^^5^5-^ = 4.64. Ani.

129. How many cubic feet of earth, each weighing 100 lbs., will

a man raise by means of a pulley from a depth of 30 feet
in a 4ay of 8. hours? Ans. 256 cubic feet.

130. How many cubic feet of water per hour will an engine

of 20 H. P. raise from a* mine Whose depth is 450 roet,
MJWWi^g ^9^ i Pf tfee; vorf^ of tb^ engine if IPAtin
IrapfpiiMjoij? • 4ps.um' oniric ««e^

CAST. 142.

▲mn. 1«, 144.] WORK ON A LEVEL PLANE.

ofllOft.=

e to raise
}Ofeet?

a engine
this

^*Pf

ine

^enffuK

e mean
suming

.will
)0 feet
C ftet.

mgine
feet,
o«rin

131. "Vhrai &Qkt be the H. P. of ain engine in oirdet that it miiy
raise 11 ions of material per hour from a deptii of tOQ ft. ?

Jtu. 1-11 U.F.

132. A forge hammer weighing 890 lbs. makes 50 lifts of 4 feet
eaqh per minute— -What must be the horse powers of the
engine that works the hammer ? Jin*. H. P.=S-39.

133. An engine of 8 horse powers works a forge hammer, cans-
ing it to make 50 lifts per minute, each to the height of
6 feet. What is the weight of the hammer?

Jin»' 9,80 lbs.

134. An engine of 8 horse powers giv^s mptlon to a fpfge ham-
mer, which weight 300 lbs., and makes 30 liifts per minute
of 2 feet each ; and at the same tiI^e r^^ses 2 tons ,of coal
per bpur from the bottom of a mine. Required, tbe. depth
of the mine ? J^m 3990 feet.

Note.— The work of tbe'englne=S3000x 8 units per minnte. From ihls
nibtractthe units of wock reauired ny the hsimner ; thftremaln^.wiU be
the wprk expended per minute in. raising the ooal, : Multiphrbig th|s bi ^
givJBs us the work required per hour for the ooial; and tkis last is the product
of the weight in lbs. by the depth in feet, of which the former is given.

WORK EXPENDED IN MOYINQ A OARRIAQE OR RAIL-
WAY TRAIN ALONG A HORIZONTAL PLANE.

^ 143. In moving a carriage, &c., along a level plane, a
q^rt^ii amount of power is expended in overcoming the
friction of the road. This is rolling friction, and amounts,
as belbre stated (Art. 138), to from j\ to ^j of the entire
load on common roads, and from ^j^ to y}^ of the load
on railway tracks. In the cafle of rfdly^ay trains, friction
is usually tsken as 7 lbs. per ton of 2000 lbs.

144. In running carriages of any descri|>tion, irdik is
employed to overcome the reisistances. These resistances
itfe: —
Ist. Friction — ^which on the same roacl >n4 w|ik the
same load is the same for all yejiocitjes.
Ascent of inclined plan^ — in which, since Umi Ibpd
has to £e lifted vertically through the
height of tho plane, the work is the same,
whatever may be the velocity of the
motion. ,
The Renitdnce of the il<»ton)Ae/'£— which depends
ii^ii the «tteiit of BtitfMe, wA increases
as the square of the velocity.

2(i.

3d.

WORK ON A LEVEL PLANE. rAnis.US.ue.

tfi

146. When a railway train is set in motion, the work
of the locomotive engine at first far exceeds the work of re-
sistances^and the motion is consequently rapidly accelerated.
But as the velocity of the train increases, the atmospheric
resistance also increases, and with such rapidity as very soon
to equalize the work of resistances to the work of the loco-
motive. When this occurs, i. e., when the work applied
by the locomotive is exactly equal to the continued work
of resistances (atmospheric resistance and friction), the
velocity of the train will be uniform. In this case the train
is said to have attained its greatest, or maximum speed.

146. The traction or force with which an animal pulls
depends upon the rate of his motion. A horse, for example,
moving only 2 miles an hour, can draw with a far greater
force than when running at the rate of 6 miles an hour.
The following table shows the relation between the speed
and the traction of a horse : ;- "^'■' '

TABLE, or TRACTION OF A HORSE.

.A,ir..i. spg^a. "■'-'''■• '' . ' j Traction.

A horse moving 2 miles per hour, can draw with a force of 166 lbs.

125 "

104 "

" 83 "

62J "

41§ "

ExAMPLB 135. — What gross load will a horse draw travelling
at the rate of four miles per hour on a road whose friction is i^ of
the whole load ? ,' . . ,

SOLUTION.

Here from the table the traction is 83 lbs., which by the conditious of
the question is ^J^ of the gross load.

Henoe load = 83 X 20 =1660 lbs. il»*. ' / -"- ,; ■^.

EzAUPLB 136. — At what rate will a horse draw a gross load
of 1800 lbs. on a road whose coefficient of friction is -jV?

SOLUTION.

Here, traction =: -^f^ = 100 lbs., whence by the table the rate mu8t be
rather over 3i miles per hour. f ^ . ' ; . {■"

ExAiiPLB 137. — If a horse draw a load of 2500 lbs. upon a
road whose coefficient of friction is -3^, what traction will he
exert and how many units of work will he perform per minute ?

" 5 ' .

'ir. '

«
((
ft

T8. U5,146.

aht.146.] work on a level plane.

he work
nk of re-
elerated.
ospheric
ery soon
ihe I060-
applied
ed work
)n), the
he train
'peed.

ml pulls
xauple,
greater
1 hour,
e speed

TractuMt,

166 lbs.

125 "

104 "

83 "

62j "

41§ «

ivelling
is^of

itious of

IS load

dust be

pon a
ill he
inute ?

SOLUTIOV.

Here, traction == iJ^^a =: 83i lbs., and hence he moveii at a rate of four

miles per hour.

4X6280*
Then distance moved per minute = — ;.. — = 352 feet.

60
Hence units of work = 83i X 362 = 29333i. Aru,

Example 138. — What must be the effective horse powers of a
locomotive engine to carry a train weighing 70 tons upon a
level rail at the steady rate of 40 miles per hour, neglecting at-
mospheric resistance and taking -^his ^^ ^hc coefficient of friction ?

BOLUTIOK.

Here, weight of train = 70 tons =:: 140000 lbs.

Space passed over per minute :

^% miles = ^!^^** = 3520 feet,
140000

Work of flriction to 1 foot = ^^ of 140000 = -^—- = 7«0 units.

Work of friction per minute ■
Units of work in one H. P. =

' 200

3 700X3520 ~ 2464000 units.
33000.

700X3520 2464000

Therefore H. P. of locomotive = — „„„„r- =

= 74*66. Ans.

33000 3:)000

Example 139. — A train weighing 120 tons is carried with a
uniform velocity of 30 milen per hour along a level rail ; assum-
ing the friction to be 11 T'S. \ 1- ton, and neglecting the resistance
of the atmosphere, what ai horse powers of the locomotive ?

' wXION.

30X6280
Space passed over per minute = |§ miles = — — — = 2640 feet.

Work of friction to each foot = 120X11 = 1320 units.
Work of friction per minute — 1320X2640 = 3484800 units.

Hence H. P. = -^~ = 106'6. Ans.
ssooo

Example 140. — At what rate per hour will a train weighing

90 tons be drawn by an engine of 80 horse powers, neglecting

the resistance of the atmosphere and taking 7^17 as the coefficient

of friction ?

SOLUTION.

Work done by the engine per hour = 33000X60X80.

Weight of train in lbs. = 90X2000 = 180000.

Units of work requured to move the train through 1 foot = iy|^ of 180000

= 720.
Work expended in moving the train through 1 mile =720X6280.

. Number of miles per hour ;

38000X60X80

= 41"66, Ans.

720X6280

Example 141. — A train moves on a level rail with the uniform
speed of 35 miles per hour ; assuming the H. P. of the locomotive
to be 50, the friction equal to 9 lbs. per ton, and neglecting
atmospheric resistance, what is the gross weight of the train ?

♦ 52S0 is the number of feet in pne rail?..

I''!'

f I

WORK ON A LEVEL PLANE.

80LUTI0K.-

[AhT. 146.

Work of engine per hour s= 33000X80X50.

Feet moved over per hour = 36X6280.

Work expended per hour in moving 1 ton = 36X5280X9.

. w 1 vi. «A 1 J i. 83000X60X50 ^^ ,„„ .

. • . Weight of train in toni = ^^^ = 59-828. Atis.

ExAMPLB 142. — In what time will an engine of 100 H. P.
move a train of 90 tons weight through a journey of 80 miles
along a level rail, assuming friction to be equal to 10 lbs. per
ton and neglecting atmospheric resistance ?

SOLUTION.

Work expended in moving the train through 1 foot = 90X10 = 900 unitB.
Work expended on whole journey in moving the train ~ 900X5280X80.
Work of engine per minute = 33000X100.

.Number of minutes = -:::r:rrrrmr- = 116^ minutes = 1 hour 56^

minutes. Ans,

33000X100

EXERCISES.

143. What gross load will a horse draw travelling at the rate of
2 miles per hour on a road whose coefficient of friction is i^ ?

Ms. 2988 lbs.

144. What must be the H. P. of a locomotive in order that it
may draw a train whose gross weight is 130 tons, at the
uniform speed of 25 miles per hour, allowing the friction to
be 7 lbs. per ton and neglecting atmospheric resistance ?

Jlns. H. P. 60-66.

145. A train weighs 75 tons and moves with the uniform speed
of 30 miles per hour on a level rail ; taking ^^^j as the coeffi-
cient of friction and neglecting the resistance of the atmos-
phere, what are the horse powers of the engine ?

Jlns. H. P. = 48.

146. In what time will an engine of 160 H. P. moving a train
whose gross weight is 110 tons complete a journey of 150
miles, taking friction to be equal to 7 lbs. per ton, neglect-
ing atmospheric resistance and assuming the rail to be on a
level plane throughout? jins. 1 hour 55| minutes.

147. At what rate per hour will a horse draw a load whose gross
weight is 220C lbs. on a road whose coefficient of friction
is ^V? -Ans. Rather over 3 J miles per hour.

148. From the table given (Art. 145) ascertain at what rate per
Itpur a hprse must travel, when drawing a load, in order to
dp th^e greatest amount of work ? Ms. 3 miles per hour.

149. M what rat^ per hour will a locomotive of 50 H. P. draw
a train whose gross weight is 70 tons, neglecting atmos-
pheric resistance, taking j^^ as the coefficient of friction
and ABBuming the rail to be level? Jnt. 26-78 miles.

\BT. 14(6.

Ants. U>, 148.1 WORlC ON A LEVEL PLANE.

H. P.

30 miles

lbs. per

OOOunitB.
>280X80.

hour 56^

iie rate of
ion is -^ ?
2988 lbs.
jr that it
IS, at the
riction to
tance?
P. 60-66.
trm speed
the coeffi-
ie atmos>

147. When a body moves through the atmosphere or
any other fluid, it encounters a resistance which increases :

1st. In proportion to the surface of the moving body ;

2nd. In proportion to the square of the velocity.

Thus 1st. If a board presenting a surface of 1 sq. foot in moving through
the air meet with a certain resistance, a board haviug a sur*
face of 2 sq. feet will meet with double that resistance ; a
board having a surface of 8 square feet will meet with three
times that resistance, &c.

2nd. If a body moving 2 miles per hour meet with a certain rcsis*
tance, a body of the same size moving 4 miles per hour will

meet with (4)'i or 2', or 4 times that resistaure.
If the velocity be increased 3 times; i. e., to 6 miles per hour,

the resistance will be increased 9 times (i. e., n* times).
If the velocity be increased 7 times, i e., to 14 miles per hour,

the resistance will be increased 7' times, i. «., 49 tunes, &c.

148. In the case of railway trains, the atmospheric re-
sistance is about 33 lbs. when the train is moving at the
rate of 10 miles per hour. It has been found, however, by
recent experiment, that the atmospheric resistance encoun-
tered by a train in motion depends very much upon the
length of the train. • r j

Example 150. — When a train is moving at the rate of 10
miles per hour, it encounters an atmospheric resistance of 33
lbs.; what will be the resistance of the atmosphere when the
train moves at the rate of 60 miles per hour ?

SOLUXIOK.

Here the velocity increases ^ times, i. e., S times. w-r . ,

Hence the resistance increases 6' times —25 times.

. '. Resistance = 33 X 25 = 825 lbs., i. e., 826 units of work are expended
every foot in overcoming the atmospheric resistance.

ExAMPLB 151. — If a train moving 1 miles per hour meet with
an atmospheric resistance equal to 5 lbs. ; what resistance will it
encounter if its speed be increased to 49 miles per hour ?

SOLUTION. ,

Here the velocity increases 7 times, (t. g.. *^). i -

Hence the resistance increases 7' =49 times.

. '. Resistance — 5X49 = 245 lbs. ; L e., 245 units of work are expended
bVery foot in overcoming the atmospheric resistance.

Example 152. If a railway train moving at the rate of 10
miles per hour encounters an atmospheric resistance of 33 lbs.;
what must be the horse powers of the locomotive in order that
the train may move 60 miles per hour, neglecting friction and
asBumixg the rail to be level? .

i i

! t
■ I

WORK ON A LEVEL PLANE.

CAST. 143.

SOIVTIOW.

Here the velocity is increued 6 times, since f ^ = 6.

Then the reftistanee is increased 86 times (Art. 147).

Hence atmospheric resistance =: 33X36 = 1188 lbs. ; 1. e., 1188 units of
work are expended in moving the train through 1 ft.

60X6280

Number of feet train moves through in a minute =:

Units of work required per minute "> _ , , aa v Mon
to overcome atmospheric resistance } ~* **«»'» ^^ «>*«»«.

= 5280.

H.P

«i it 1188X6280 ,^„„ .

of locomotive = - ,„^^^ — = 190*08. Ans.

33000

Example 153. — What must be the H. P. of a locomotive to
more a train at the rate of 40 miles per hour on a level rail,
taking atmospheric pressure as usual, (i. e., 33 lbs. ^hen train
moves 10 miles per hour,) and neglecting friction?

soLxrrioN.

Here velocity increases 4 times, and hence resistance increases 16 times.

Then resistance encountered = 33X16 = 628 = units of work required
per foot.

Feet moved over per hour = 5280X40; hence units of work per hour =:
6280X40X628.

™„ - « « 828X40X6280 „.„, .

Therefore H. P. = 33000^^^ = 56-32. Ans.

Example 154. — What must be the H. P. of a locomotive to
draw a train whose gross weight is 80 tons, along a level rail,
with the uniform velocity of 40 miles per hour, taking atmos-
pheric resistance and friction as usual ?

soLuizoir.

. ^ 40X6280
Feet passed over per minute = — — — = 3620.

Work of flriction per minute = 80X7X3620 = 1971200 units.
Work of atmospheric resi8tanoe=33X16X3520=1858660 units.

Therefore H. P.

Work of friction -f work of atmospheric resistahce

Work of one U. P.
33000 83000

Example 155. — What must be the H. P. of a locomotive to
draw a train, whose gross weight is 125 tons, along a level rail
with the uniform velocity of 42 miles per hour, taking friction as
usual, and assuming that the atmospheric resistance encountered
by the train is equal to 10 lbs. when moving at the rate of 7
miles per hour ?

aoLVTioir.

42XS280

Feet moved over per minute = =lii^i£:? — 3690.

60
Work of friction per minute = 126X7X3696 = 3234000 units.

Work of atmospherio;re8i8tanceperminute=10X36X3696=ri330560 units.

-.„___ Work of ftricMon 4- work of a^ospheric resistance _

inen n.i'. — Work of one ff. 1*. ' ~

8234000+1380660 4864860 ,.„., .

— ^ 1S8.32, Aus.

33000

830OO

iBT. 148.

Abts. 14». ISO.l WORK ON AN INCLINED PLANE.

units of
5280.

lOtive to
3vel rail,
en train

16 times,
t required

er hour =

oaotive to
evel rail,
ig atmos-

resiatatice

notiye to
level rail
riction as
lountered
rate of t

OB60 units,
si stattce _

157.

EXERCISES.

156. If a train •neonnters an atmospheric resistanet of 8 lbs.
when moring at the rate of 6 miles per hour, what re-
sistance will it encounter when its speed is increased to
45 miles per hour ? ytfn«. 648 lbs.

What must be the H. P. of a locomotive to draw a train at
the rate of 30 miles per hour on a level rail, assuming that
the atmospheric resistance is equal to 9 lbs. when the train
moves 6 miles per hour, and neg^'^ct*'* 'riction 7

fn«. H. P = 36.

What must be the H. P. of a locomotive to draw a train
weighing 140 tons along a level rail with the uniform
velocity of 36 miles per hour, taking friction as 7 lbs. per
ton, and the nresistance of the atmosphere 12 lbs. when
the train moves 9 miles per hour? Jlns. H. P. =r 112'512.

A train weighing 200 tons moves along a level rail with a
uniform speed of 30 miles per hour ; what are the H. P.
of the engine—friction and atmospheric resistance being
as usnal? ^ns. H. P. = 135-76.

168.

159.

149. If a body be moved along a surface without fric-

tion or atmospheric resistance, the units of work performed

are found by multiplying the weight of the body in lbs. by

the vertical distance in ^et through which it is raised.

Thus, if a body weighing 12 lbs. be moved 200 feet along an inclined plane
having a rise of 19 feM in 100, the units of work performed will be 12X19X2
= 456, because in moving up the plane 800 feet, the body is raised through
19X2 = 38 feet.

160. When a train is moving along an inclined plane,
and- the inclination is not very great, the pressure on the
plane is very nearly equal to the weight of the body.
Hence we find the work due to friction by Arts. 143-146,
the work due to atmospheric resistance by Art. 148, and
the work due to gravity by Art. 149.

ExAMPLS 160. — A train Weighing 90 tons is drawn up a gradient
having a rise of 3 feet in every 1000 feet^ with the uniform speed
of 40 miles per hour— neglecting friction and atmospheric
resistance, what are the H. P. of .the engine 7

eoLusioir.

Weight of trahi in lbs. =s 00X2000 = 180000.

I'eet travelled per minute = — tz — « 3520.

Vertical distance moved through per minute ss ^^ oi 8620 at lO'fie ft.
tJnits of work due to gravity per minute =s 10'66 X laoooo.

...H.p^J225>^:i. 57-6.-1^. V .

33000 -i A . J

WORK ON AN INCLINED PLANE.

rAm. 150.

I i<

Example 161. — ^A train weighing 140 ions moves up a gradient
having a rise of 3 feet in 1100 feet, with the uniform velocity of
36 miles per hour— neglecting atmospheric resistance and taking
friction as usual, what are the H. P. of the locomotive ?

SOLUTION.

Hereweightof train in lbs. =E 140X2000 =s 280000; audsneod pprminute

86X6280 .,„„, .
= — — — = 8168 feet.
60

The units of work due per minute to friction = 140X7X3168 = 3104640.
Height to which train is raised per minute == j~^ of 3168 = 8*64 ft.
Then units of work due per minute to gravity = 8*04X280000 — - 2410200.
work diio gravity+work due friction 3104640+2419200

.-. H.P.=

6528840

WorkofoneH. P.

33000

83000

= 167*889. Arts.

ExAMPLB 162. — A train weighing 100 tons moves up a gradient
with a uniform velocity of 30 miles per hour, the rise of the plane
being 3 feet in 1000 feet, and taking friction and atmospheric
resistance as usual, ivhat are the H. P. of the locomotive?

SOLUTION.

Here weight of train in lbs. ■■= 100 x 2000 = 200000 ; space passed per
=> 2640 ft., and elevation of train per minute

minute

nfW

ThenH,P.=

. . „ „ 1848000+784080+1684000

of 2640 = 7*92 ft.
Work of friction per minute = 100X7X2640 = 1848000 units.

Work of atmospheric resistance per minute =33X9X2640 =784080 units.

Work of gravity per minute = 7*92X200000 = 1584000 units.

_ Workdu« tofrio. permln.+wnrkdaftoiitinoi. re»Ut. parmln.+workd ueto gr»T. rermlll.
Uoiu of work in one U. If,

4216080 ,„^.^„ .
83000 = IsoUr* ^27*76. ^«,. *

Example 163. — A train weighing 130 tons descends a gradient
having a rise of 1 ft. in 2000 ft. with the uniform velocity of 60
miles per hour — taking atmospheric resistance as usual, and the
coeflBcient of friction 7^0", what are the horse powers of the loco-
motive ?

SOLUTION.

Here weight of train in lbs. = 130X2000 = 260000 ; space passed over per

minute =

60X6280

=6280 ft. ; increase in the velocity = ^§ = 6 ; and verti-
cal fall of train per minute = j^ of 6280 ft. = 18*48 ft.
Then work of friction per minute = j^'X.26(mQX5280 = 1300X5280 ^

6864000 units.
Work of atmospheric resistance per minute=33X 36X6280=6272640 units.

Wbrk of gravity per minute=18*48X 260000=4804800 units.

Then, since the train descends the gradient, ^rravi^j^ acts with the engine.

n -a -a Work of flriction+work of atmos. resist. —work of gravity .

Work of one H. P

H. P. =; 68g*ooo+ogy^g^o-"^Q*so<>

S3000

S8000

km. 150.

radieut
Dcity of
1 taking

r minute

tl0464(«.
*64ft.
2410200.
kl9200

gradient
he plane
ospheric

ussed per

~ rooiT

)80 units.

{r«T. r«rtBlw.

gradient
ty of 60
and the
le loco-

over per
,nd verti-

X5280 ==
340unitft.

e engine.
gravity.

\Jnt.

AEI.150.J WORK ON AX INCLINED PLANE. 01

ExAifPLB 164.— A train weighing 80 tons moves along a
gradient with the uniform speed of 40 mjles per hour— assuming
the inclination of the gradient to be 3 ft. in 1000 ft., and taking
friction and atmospheric resistance as usual, what will be the
H. P. of the locomotive :

1st. If the train move up the gradient, and
2d. If the train move down the gradient ?

BOLUTIOir.

Here weight of train in lbs. := 80X2000=160000 ; space passed over per

minute = — — — = 3520 ft.; velocity is increased f£ = 4 times, and ver.

tical ascent or descent of train j-^^ of 3620 = 10*50 ft.
Worlc of friction = 80X7X3520 — 1971200 units per minute.
Worlc of atmospheric resistance = 33X16X8520 = 1858660 units per min.
Work of gravity = 10'56X 160000 = 1689600 units per minute.

Work of friction+work of atmos.rcsist.iwork of gpravity.

Then H. P. =

Train ascending, H. P. =

Train descending, H. P. =

Work of one H. P.
1071200+18585604-1689600

33000
19712004-1868560— 1689600

6519360

33000
2140160

= 167-253.

= 64-853.

33000 33000

Example 165. — A train weighing 110 tons ascends a gradient
having a rise of | in 100 — taking friction as usual, and
neglecting atmospheric resistance, what is the maximum speed
the train will attain if the li. P. of the locomotive be 120?

80LUTI05.

Hereweightof train in lbs. = 110X2000 = 220000. -

Work of friction in one mile = 110X7X5280 = 4065600 units.

Work of gravity in one mile= Tolf of 5280 =6-6X220000 = 1452000 units.

Total work of resistance in 1 mile = 4065600+1462000 = 6617600 units.

Total work of engine per hour = 33000X60X120 = 237600000 units.

237600000

Number of miles per hour = •

: 43*06 Ans.

6517600

Example 166. — If a horse exert a traction of 120 lbs., what
gross load will he pull up a hill whose rise is 17 feet in 1000 ft.,
assuming the coefficient of friction to be -^^ ?

SOLUTION. ; . .

Workofhorse in moving the load over 1000 ft. = 120xi000 =126000 units.
Work of friction in moving 1 lb. over 1000 ft. = IXiVxiOOO = 100 units.
Work of gravity in moving 1 lb. over 1000 ft. = 1 x 17 — 17 units.
Total work in moving 1 lb. over 1000 ft. = work of friction + work of
gravity = 100+17 = 117 units.
. • . Number of lbs. drawn by horse = i^j)^ — 1025*641. Ans.

Example 167. — What backward pressure is exerted by a horse
in going down a hill which has a rise of 7 feet in 100, with a
load whose gross weight is 2000 lbs,, assuming -gV to be coeffi^
cient of friction?

1, ll

WORK ON AN INCLINED PLANE. [Aii.161.

•oiVTioir.

thtk oilftl«v«lpl»iieth«fHottonwouUlb«|>r of 806« )bf.s=Bri« Ibt.o:
ttiiiUi of work for eaott foot.

Work of RTftvity =: T?iv of 8000 = 140 uniti to eaeh fbot.
Therefore, the baokwMrd preMure ii 144>— S7'14 =3 srM lb*. Am.

EXERCISES.

168. What backward pressure will a horse exert in going down

a hill which has a rise of 9 feet in 100, with a load whos«
gross weight is 1200 lbs., assuming the coefBcient of frio*
tion of the road to be ^? Jm. 68 lbs.

169. What gross load will a horse exerting a traction of 150 lbs.

draw up a hill whose inclination is 3 in 100— assuming
the coefficient of friction to be iV ? ^^n'- 1651*73 lbs.

1 70. What will be the maximum speed attained bj a train

weighing 200 tons, drawn by a locomotive of 160 H. P.
up a gradient having a rise of ^ in 100— taking ftriotion
as usual and neglecting atmospheric resistance 7

Jnt. 29*032 miles per hour.

171. A train weighing 88 tons moves up a gradient having a rise

of i in 100 witn the uniform velooitj of 20 miles per hour
—taking friction and atmospheric resistance as usual,
what are the H. P. of the lo<iomotive ?

w«n«. H. P.= 7M82.

172. A train weighing 95 tons descends a gradient having a fall

of S in 1000 with the uniform speed of 40 miles per hour —
taking friction and atmospheric resistance as usual, what
are the H. P. of the locomotive T Mi. H. P.=r 113*742.

173. A train weighing 1^25 tons moves along a gradient having

a rise of i in 100 with the uniform speed of 25 miles per
hour— taking friction and atmospheric resistance as usual ,
what are the H.P. of the engine,

1st. When the train ascends the gradient?

2d. When the train descends the gradient ?
.i«9. Going up, H.P.=113-t5 ; going down, H.P. =30-416.

151. For finding the H. P., maximum speed, weight of
train, <fec., as in the foregoing examples, by representing
the variable quantities, such as weight, rate of motion, in-
clination of plane, &c , by letters, we may easily deduce
tbrmulas by means of which the work required to soke
such problems will be very materially abbreviated.

ABT. 161.]

W6(tK ON AN INCLINED PLANE.

Thua, ilnce the number of feet moved per minute ig alwayi s

rate per hour in miles x 6280 ^ .... 0280
= rate per hour in milei X -jj-

= rate per hour in miles X 88 ; therefore, whatever maj be the
rate, 88 is a constant multiplier.

Let r "« rato per hour in miles, then 88ra>.r:ite per rain.iu ft.

w 'mm weight of train in tons, then 2000u; >■ weight of
train in lbs.

h mm rise of the plane in every 100 feet.

/— friction per ton.

i^aM given atmospheric resistance at given speed, «.
Then units of work due per minute to friction =/u>X88r.

<♦ '♦ to gravity =s 2000to X Ti^z-

X 88 r =s 20 Aw X 88 r.
Units of work due per min. to atmos. resist. = R i—\ x 88r.
Units of work per min. in given H. P. = H. P. x 33000.
henoeH.P.X33000=s/M>X88r-fil(Yj x 88r±20Awx 88r,

and, factoring this, we get :

H. P. X 33000 =:{fw ■\' R irjf^ 20Aw) 88r.

„/r\« 88r

Therefore H.P.= (> + ^^yj ± 20ftw) -^^^.

Or H. P. = 0+il(y) ± 20 hvf)-^ (I).

From this we obtain by transposition and reduction, and
neglecting atmospheric resistance,
H.P. X3y5
(/'±20A)r ("-^

H. P_21ill

*'-(/±20A)ii, ("^>-

Bince/ is commonly = 7, Ji = 8:i. and » ~ 10, these formulas become
reBpcctively,

H.P.= (7w+-38r«i20Atc)^ (IV.)

H.P.X875 .'^'^ :

(7taO*)r Ki) . .

. H.y.xa76

(7_+20*)ic

(H.)

THE MUDULUH OF MACHINES. ! ARi*. iM, loa

ExAMPLB 174.— A traia weighing 140 tons moves along a
gradient harlng a ride of | in 100 with tho uniform speed of 30
miles per hour ; taking friction and atmospheric rcsintanco us
usual, wiiat are tho H. P. of tlio locomotive ; Ist, when tho truin
moves up the gradient? 2d, when tho train moves down tho
gradient ?

SOLUTION.

Uoro w -is 140, ;• ~ 30, h - |.

U. P. - (7w+'33r2 iidhw)—;^

. . = (Txi-to+'asxao' i mklxiia) ^i^\

— (080-1-207 1 700)-^
1977 X«

577X2

25 20

=- 168'16 or tO'lO. Ans.

ExAMPLB 175.— A train drawn by a locomotive of 80 II. P.
moves along an inclined plane having a rise of j^ in 100 with a
uniform velocity of 45 miles per hour; taking friction as usual
and neglecting atmospheric resistance, what is the weight of the
train ?

SOLUXIOX.

Here H. P. — 80, r - 45, and 7i = ^
Then by formula (V.) to = "~™

80X375

.■iOOOO

30000

soono

.10000

30000

(7:h20xJ)46 (7±3p45

10} X45 *"• -83^4^^ -4flr-°''"n«r"= «*■" *°"» *' ^^^ *"^*" *» 8°*"»

up the gradient, or 181*81 tons if the train is KoiuR down the gradient.

For practice in the application of these formulas, work any of the fore-
going problems.

THE MODULUS OF A MACHINE.

162. Tho Modulus of a raachine is the fraction which
expresses the vahie of the work done compared with the
work applied, the latter being expressed by unity.

Thus if T of tho work applied to a machine be lost in transmitision, the
modulus or useful work of that machine is ^ ; if § be lost in transmission,
the modulus of the machine is f) &c.

153. The amount of work lost depends on fiiotion,
rigidity of cordage, <fec., and in some maciunes is more than
half of the whole work applied. The following table gives
the moduli of machines for raising water :

which
ih the

1, the
aission,

potion,
than
gives

ABia. iH iMi WUKK OF 8TEAM. 66

TADLU or MODULI.
MACIIINR. MODULUB.

Inclinod ohnin pump, |

Upright " i

Ituckot whoul, ^

Aroliimeilian screw, i\

Pumps for draining mines, I

ExAHPLB no.— If 1 H.P. be applied to an upright chain pump,
how many gallons of water will bo raised per hour to the height
of 50 feet?

SOLUTION.

Work applied per hour ~- 33000 X 7 X 60.

Work done =■ 33000 x 7 X 60 X i, since tlio modulus of the upriKht

chain pump in \.

Work ozpondud in raising 1 gallon of water 50 foot = 10 X 60,

XT u • II 33000X7X60X4 ,„„„^ . 'I

.•. Number of gallouH— r^— .;: —13860. Ana.

10 X 60

ExAMPLu 177.— -What must bo the II. P. of an engine to pump
9000 cubic feet of water per hour from a mine whoso depth is
110 feet?

SOLUTION.

Work of raising water per hour = eO00X02ixllO.
EfTectivo work of one H. P. per hour ~ 33000X60X1.
.„_,__ 000X02^ X 110 _ 61876000 _ ^^^. .

■ "•^- - 33000 X 60 X) ~ Imm ~ **'-"'^- ^'"•

WORK OF WATER.

154. When water falls from a height upon the floa*,
boards of a wheel, <fec., the quantity of work it pcrforiifH
is found by multiplying the weisfht of the water by the
height through which it falls. (Sec Chap. VIII.)

STEAM ENGINES AND WORK OF STEAM.

166. A constant power is obtained from the confine-
ment and regulated escape of steam in the various kinds
of steam engines.

166. Steam engines, though diflfering very materially
from one another in detail, are all modifications of two
distioct machines, viz; — " '

WORK OF StSAM.

CABTfl. 167-168.

Ist. The high presiure Bteam eogine,or non-condensing

engine.
2nd. The low pressure steam engine, or condensing
engine,
i 167. The high pressure engine, which is the simpler

form of the two, consists essentially of a strong vessel or
boiler in which tiie steam is generated, a cylinder^ in which
a tightly fitting jM«ton moves backwards and forwards, an
arrangement of vcUvet so a<^usted aa to admit the steam
alternately above and below the piston and also alternately
open and close a way of escap<^ into the air, and lastly
various contrivances by which the oscillations of the piston
may be converted into other kinds of motion suited t<o
the work the engine is to perform.

168. In the low pressure engine, the space into which
the steam drives tke piston is converted, by means of a
eondmsing ehamher^ into a vacuum, so that the motion of
the piston is not resisted by atmospheric pressure, and
steam generated at a low temperature can therefore be used.
168. The varieties of the low pressure engine are chiefly
two,— the ^ngU acting^ and the double acting engine.

160. In the single acting engine the piston is driven
forward by means of steam acting against a vacuum, and
backward by the counterpoising weight of the machinery.
The machine is therefore in action only half the time of the
movement.

161. In the double acting engine the piston is driven
both backward and forward by the steam acting against a
vacuum on the opposite side, and the machine therefore
acts continuously.

162. In the high pressure engine the piston moves both
forwards and backwards against the pressure of the air.

163. The following are the leading ideas that enter into
the coiistruction and operation of the steam engine.

I. When steam is condensed, a vacuum is produced into which the a^jft-
cent bodies have a tendency to rush.

II. When oold water is phMied in contact with steam, it condenses it with
IprMft rapidity, pMMuotng a vaoutim ; and this Vacuum maybe pibduced
without ooolufc the cylinder containing the jteam, if a conimuniefttiou b^
Icept up between this and a vessel containing water,

Alls. 16A-166.]

WORK OF STEABf.

»i

acUft-

III. The Tapour of water exwtlPa (xnuidenlble |iMiM&r#iMii At ooitt]ti»»*
tirejhr low tenipentaM»: tm 6nmi^to,lir MtMT mMUhiMini.

I v. If the preeiure exerted by the piston on • qnantlty,oriteMn'eonflned
in » orlinder be leas than the elastfo IbrM » mi alMAi, the* fteaiA wiU
eraand and ciye motion to ISie piston.

vTlf a yaouiun be nroduoed ma oylihder behind the ptiton, the atmoe*
phericpremue wiU arUre the i^ton backwards.

VI. The same quantity Off lel iTill convert the' sitjtie quantity of water
into steam whatever mij be . le preiMure on iti sUiOMoe.

YII. The higher the preali^ uftdef whidi steam ik Renerated, tbeamaller
its bulk, and the greater its elastic fiaroe.

VIII. The same quintity of #atof dottyerted into steam at any premure
will produce the same mechsittou eflbt; i. A, if fM liMiriire MR>#. the
stelMaa genemtied is Ui^ iii quantity and tfosseosed of obnipaatbehr Itttla
elastic force ; if the pressure be high, the steam generated w (tf smaU quan*
tity, but of high elastic force.

164. High pressure engines are commonly used where
it is desirable to have the engine as simple, chciftp, compact,
and light as possible, as the condensing apparatus renders
the engine more costly and cumbrous. The high pressure
engine is, however, far more liable to burst and get other-'
wise out of repair,

166. The units of work performed per minute by a
steam engine are found by nmUiplyiny together the pret*
sure per square inch on the boiler , the area of the piston in
inches, the length of thx stroke of the piston in feet^ and
the numher of sitroka per minute.

Thus let the pressure exerted on nch square inch of the piston be 80
lbs., and let the piston mue 40 strokes perminute of 8 ft. each, also let
the area of the puiton be 100 sqiiare inches s

Now if a weight of 80 lbs. be placed on each square inch of the surf^e
of the piston, the elastic force of the steam will be Just sufficient to lift the
loaded piston through the length ^f the 8tA)ke in opposition to gravity, then
the work performed oH 1 sq. in. or the piston would be 80X8 for each stroke.
Work performed oh whole piston wotud be 30X8X100 forleach stroke.
Work " " " " " 80X3X100X40 per minute.

166. In the high pressure engine, the pressure of the
atmosphere, about 15 lbs. to the square inch, acts in op-
position to the pressure of the steam; and in the low-pres-
sure or condensmg engine a pressure of about 4 lbs. to the
square inch of the piston is exerted by thn vapour in
the condensing chamber. Besides these, a resistance of 1 lb.
per square inch is commonly allowed for the friction of
the piston. Deducting thdiealbwsnces f)fl>m the total pres-
sure, we obtain the emctive pressure ; and we must further
make an allowance of | of this for the friction of thf
whole engine.

liMlli iiliili..tt

WOKK OF STEAM.

[Aet. 167.

Thus in the high pressure engine : -

Load-^\ load+ + 16 — whole pressure, .

lu the condensing engine :

Load + I load + 1 + 4 ■— tohole pressure.

For cxainplo,— if the whole pressure be B8 lbs. per square inch,
Then for the high pressure engine 68—1—16=42 is the working pres-
sure on the piston, and 42 is f (i. c, load + | load) of the useful pressure*
and hence useful or effectlvo prossure=:42-f-*— 36^ .

Por tl^e low pressure engine 88—1—4 = 63 = working pressure on the
piston, and 63 is i^ of the useful pressure. Therefore useful or efrectlvc
pressure Is 63 -f- f = 46^^

167. For finding the H. P. of astoam engine, let p =
useful pressure in lbs. on each square inch of the piston,
a ma area of piston, / =» length of piston stroke in feet,
and »«*■ number of strokes per minute.

pain «

Then H. 1^.

33000

H. P. X 33000

a In
H. P. X 33000

H. P. X 83000
pal

H. P.X 33000
pan

■ (11.)
• (I")

• (IV.)
.(V.)

Example 1*78. — The piston of an engine bas an area of 260
inches,and makes 110 stroke;^, of 6 feet each, per minute — taking
tbo useful pressure of the steam as 28 T s. per sq. inch, what are
the H. P. of the engine?

soiuTioir.

Here p — 28,a = 260, n = 110, and 1 — 5,

28X250X110X8

Then (formula I.) H. P. »

83000

llOf. Ant.

ExAMPLB 179. — The piston of a high pressure engine bas an
airea of 1200 inches and makes in each minute 30 strokes of 7
feet each — taking the gross pressure of the steam as 48 lbs. per
square inch, what are the H. P. of the engine ?

[Aet. 107.

WVftM Hl/'«"-«»TS-

Aet. 167. 1

WORK OP STEAM.

king pres-
1 prosBuroi

tre on the
r effeotivo

lot p =
> piston,
in feet,

of 260
taking
lat are

has an
98 of 7
bs. per

SOLUTION.

Hero 48 — p+\p'\-U-\-l,or^^p =: 32, and hone© p ~ 32-T-f
Then p — 2»,a — 1200, n — 30, and 1 — 7.

28X1200X30x7

2« lbs.

By formula I., H.r.

33000

— 213-.S1. An9.

Example 180. — The piston of a low pressnre enplnc has a
diameter of 20 ins. and makes 60 strokes of 4 ft. each, per minute
— the pressure of the steam on the boiler is 45 lbs. to the sq.
inch, what are the H. P. of the engine?

SOLUTIOjr.

Hero 43 — p-!-|p+44-l, or fp — 40, and henco p — 40-r-f s= S.*}-
n* -.- 10« x;j-1410 ~ 100X3'1416 = 314'16.
Tlion p -= 35, a --= 314-16, n — 00, and I — 4.
35X314-10X60X4

H. P. =^

33000

= 70*908. Ant.

Example 181. — In a steam engine of 32 horse power, the area of
the piston is 600 inches, the length of the stroke 4 feet, and
the useful pressure of the steam 33 los. to the sq. inch, how
manj strokes does the piston make per minute ?

BOLXTTIOW.

Here, H. P. — 32, a -- 500, 1 = 4, and p = 38

m.. /» .1 T^x H.P.X33000 32X33000

Then (formula IV.) n =

10 Ana,

pal 600X4X33'

Example 182.— In a low pressure steam engine of 190 H. P. the
area of the piston is 1000 inches, the length of stroke 6 feet,
and the number of strokes per minute 110, what is the useful
pressure per square inch on the piston, and also, what is the
gross pressure of the steam ?

SOLUTION.

Here, H. P. = 190, a — 1000, i = 6 and w = 100.

Then (Formula II.) p = i^^g^^o =** ^^'' ~ "'®'"^ pressure.
And pressure on boiler (Art. 166) = 04+ j of 9J+4+1 = 16lf lbs.

Example 183.-^In a high pressure engine the piston has an
area of 800 inches, and makes 40 strokes per minute, of 10 feet
each, what must be the pressure of the steam on the boiler
in order that the engine may pump 120 cubic feet of water
per minute from a mine whose depth is 400 feet — making the
usual allowance for friction and the modulus of the pump 7

* When the diameter of the piston is given, its area i« found by multiply*
ing the square of half the diameter by 8*1416.

Work op stEAHf.

tABT.ier.

I ■■

80LVTI0V.

Here, work done per minute = l<0X6t'8X400 ss 8000000 units. ~

Work applied. «. «.. work of engine = 8000000-f-l = 4BO0O0O units s: H. P.
X88000.

Then by Pormul. II. p ^ H. P. X 88000 ^ 4600000 j. , ibg.=u«,fui

aln 800X10X40 ^»

pressure.

And Art. 166, gross pressure — 1*^+1 of 14^g-fl6+l ^ Sirh ^' ^^*'

ExAMPLi 184.—- The piston of a high pressure engine has an
area of 600 inches, and makes 20 strokes per minute, each 8 ft.
in length, gross pressure of the steam 52 lbs. to the square inch.
How many gallons of water per minute will this engine pump
from a mine whose depth is 600 feet, making the usual allow-
ance for friction and the modulus of the pump ?

BOLVTIOV.

Herea = 600, I = 8,n sr 20, and since 62 =!> + 13> 4* 18 -f 1 ; fP ~ S6,

and p =. 81i.

Work of ei)gine per niittute=:pa2fi = 311X600X8X20 = 8024000.

TTsefUl work per minute s= 30$400pX| =» 20111000.

Work of pumping 1 g^on of water to height of 600 feet = 10X500 ^
6000 units.

. •. No. of gallons pumped per minute = * Vlft»^^ = 408i. Ans.

EXERCISES.

186. The piston of a low pressure steam engine is 40 inches i&
diameter and makes 40 strokes of 6 feet each per minute ;—
the gross pressure of the steam is 37 lbs. per square inch ;
wbAt are tlie &. P. of the engine ? jin*. 2 13'248.

186. The piston of a high-pressure engine is 20 inches in diame-
ter and makes 50 strokes of 4 feet per minute ; taking the
gross pressure of the steam as 40 lbs. per square inch and
making the usual allowance for friction, what are the H. P.
of the engine ? Ant. 39 '984*

187. The piston of an engine has an area of 2400 inches and
makes 16 strokes per minute, each 10 feet in length ; the
useful pressure of the steam on the piston is 20 lbs. per

jr. 'pqjBLMp inch, what are the H. P. of the engine 7

• ,r Jlru. 33;2-72.

188. In a high p.reasure engine of 140 H. P. the piston has an
area of lOOO inches, and makes 20 strokes, of 5 feet each, per
minute ; what is the useful pr^sure of the steam on the
piston and also the g'rose preslrare per square inch?

- Jtni» Useful pressure r: 46*2 lbs. per sq. in.

Gross pressure =: 68'8 ll^s. per iq. in.

I
ii
2
ii
I

Art. 168 J

WOUK OP STEAM.

189. In a low preiisure engine of 100 H. P. the piston has an
area of 200 inches and makes 40 strokes per minute ; the
gross pressure of the steum is 45 lbs. per square inch.
Required the length of the stroke made by the piston?

An^. 11*786 feet.

190. In a high pressure engine of 80 H. P. the piston makes 44
strokes per minute, each 6 feet in length, and the gross pres-
sure of the steam is 66 lbs. per square inch. What is the area
of the piston? Aiu. 286* 714 Bq. in.

191. How many cubic feet of water may be pumped per minute
from a mine whose depth is 600 feet by an engine in which
the piston has an area of 2000 inches, and makes 30 strokes
per minute, each 8 feet in length, the useful pressure of the
steam being 40 lbs. per square inch, and the usual allow-
ance being made for the modulus of the pump?

Jlns. 409*6 cubic feet.

168. In all the modifications of the Bteam engine, the

real source of work is the evaporating power of the boiler;

the amount of work done hj the engine depending not only

upon the rapidity with which the water is evaporated) but

also upon the temperature^ and consequently tne pressure

under which the steam is produced. The foUowmg is a

specimen of an experimental table, given by Pambour,

showing the relation between the pressure, temperature, and

volume of the steam produced by one cubic fiot of water.

By means of this table^ we are enabled to ascertain the

volume of the steam produced by a given quantity of water,

when we know the pressure or temperature under which it

is formed.

NovB 1.— ^e first column gives the inresiore In lbs. to the square inch
Under which the steaiti is produced) the second column shows the correspond*
ing temperature, an indicated by Fahrenheit's thermometer; and the third
OoTumn, the volume of the steam commred with the volume of the water
which produced it. It will be observed that the lower the temperature, or
what amounts to the same thins, the less the pressure under which the
steam is ftarmed. the greater itis volume. Thus under the usual atmospheric

Eressure of 15 lbs. to the square inch (or at the common temperature of
oiling water, 212^ or 213^ Fahr.), a cubic foot of water produces 1609cubib
ISset at steam. If, however, the pressure be deoroasod to l lb. to the square
inch, the steam is formed at the temperature of 103" Fahr, and occupies
20064 enble fSseli ; while if the pressure be increased to 30 lbs. to the square
inch, the temperature required for the production of the steam rises to 251'
Fahr. and the steam only occupies 882 cubic feet.

Non 2.—^ has been shown by nuaierotts experiments that the quantity
of Aiel requisite for the evaporation of a given quantiigr of vafeor is in>
variably the same, no matter what may be the premie under which the
steam u produced. Hence it Is obvious that it irmbst atdHmiaipBr to
employ steam of a bJj|h prewnve.

I .

1 1

I ;

't ■

I f

WORK OF STEAM.

TABLE

[Art, 160

SITOWIZrO TBB VOLUME OF 8TBAM FBODUCBD BT ONE CUBIC BOOT Of
WATBB AT TUB COBRBSPOITDINO FBBBSUBB AND TBMFBRATUBE.

10
16
20
86
30
36
40
46
60

lOS*
161«»
192°
213"
228«
24l»
251«
260»
268»
276°
282«>

3^2

20954

4624

2427

1669

1280

1042

882

765

677

608

652

so £:
MS

60
65
70
75
80
85
90
95
100
106

M » J*

- a;

2880
294»
299"
304"
809®
813"
318«
322"
326«»
330'*
333«'

606
467
434
406
381
359
340
323
307
293
281

169. If we let a s=« area of the piston in square inches.

I = length of stroke made by the piston.

' n=am number of strokes made per minute.

j9 ■« eflfective pressure to each sq. inch
of the piston.

c = cubic feet of water evaporated per
minute.

V a> Yolume of one cubic foot of water
in the form of steam under the
given pressure p.

Then to find a, /, w, p^ c, or v, when the others are given,
we proceed as follows : » :^

When p ig given, v is found by the table.

Art. 160.

Akt. 169. 1

WOBK OF STEAM.

FOOT or
ruBE.

inches.

piston.

minute.

q. incli

ted per

f water
der the

given,

Now the cubic feet of water evaporated per minute = ev,

cubic feet of steam used at each stroke of the piston »- J— -

^ 144*

nn.1

. ' . cubic feet of water used in n strokes >» — ao also

144 *

the steam evaporated or used per minute.

cVy and from this by reduction we obtain
144cv 144CV naJ nal

Hence

nal
"144

144C2;

naJ

na ' al ^ nl

When V is known p may be found by the table.

Example 192. — The piston of a steam engine has an area of
200 square inches and makes a stroke 4 feet in length, the boiler
evaporating -^s of a cubic foot of water per minute, under a
pressure of 40 lbs. to the square inch. What number of strokes
per minute does the piston make ?

SOLUTION.

Here a = 200, ? = 4, c = -^ = '3, and p — ifS', also from table v — 877.
_, lUcv 144X-8X677 ,^.„„ ,„, ^

Then » = -^- = toOX4 = ^ "® ""^ = *«*• -*»'•

Example 193 — The piston of a steam engine has an area of
1000 inches and makes 10 strokes per minute, each 3 feet in
length, the boiler evaporates '4 of a cubic fcot of water per
minute. What is the pressure under which tht) steam is gene-
rated ?

80LUTI0W,

Here a =1000, 2 = 3, n =10, and c = '4.

_, nal 10X1000X3 ,„, a. ». *i. * i,, , -^ x

Then v = rrr- = = 621, whence by the table, p is between

80 and 65, or about 63 lbs.

Example 194. — The piston of a. steam engine has an area of
80 inches, and make 20 strokes per minute ; the boiler evaporates
iV of a cubic foot of water per minute under the pressure of 50
lbs. to the square inch ; required the length of the stroke mad*
by the piston.

80LUTI0W.

Here a=80, n=20, <j=*1 and j9=50 and (table) d=552.

Then?=

144c« 144X*1X652

20X80

=4'968 ft.=4 ft. Hi inchet. Aw.

* We divide by 144 because a, the area of the piston, is given in square
inches, while I, tne length of stroke, is given in feet. To find the oubio
feet of steam we must multiply the len^h of stroke in feet by the area of

the piston in square t^i ; i. e., by -77

144

WORK OF STEAM.

CAST. 170.

\ I

ExAMPLi 196.— 'The boiler of an engine evaporates f of a
cubic foot of water per minute under a pressure of 46 lbs. to the
square inch ; the piston has an area of 260 inches and makes a
stroke 4 feet in length. Required the number of strokes made
by the piston per minute.

SOLUTION.

Here 0=250, l=-4, e='4, p=45, and hence (table) v=608.

=d5'0208, i. e. 35 strokes per minute. An$.

Then n= — r-=

14ACV 144X'4X608

250X4

EXERCISES.

196. The boiler of a steam engine evaporates | of a cubic foot

of water per minute under a pressure of 66 lbs. to the
square inch. If the piston has an area of 144 square
inches and makes strokes 6 feet in length, how many
strokes are made per minute ? ^ru. 69*44.

197. The piston of an engine has an area of 288 inches and

makes 7 strokes per minute. If the boiler evaporates-^
of a cubic fdot of water per minute under the pressure
of 55 lbs. to the square inch, what is the length of the
stroke of the piston ? »^n$. 26-^ feet.

198. The piston of an engine makes 10 strokes of 6 feet each

per minute; the boiler evaporating J a cubic foot of
water per minute under a pressure of 25 lbs. to the square
inch, what is the area of the piston ? Jlns. 1250*4 inches.

199. In a steam engine the piston having an area of 720 inches

makes 20 strokes, of 3 feet each, per minute, what volume
of water converted into steam under a pressure of 20 lbs.
to the square inch, is evaporated per minute by the boiler 7

^ns. ^^ of a cubic foot.

200. The piston of a steam engine has an area of 600 inches

and makes 12 strokes, of 10 feet each, per minute. Now '
if the boiler evaporates 1 cubic foot of water per minute,
what is the volume of the steam produced per minute
and the pressure under which it is generated ?
Jins. Volume = 500 cubic feet.

Pressure = nearly 65 lbs. to the square inch.

170. To find the useful H. P. of an engine when a, n,
?, c, and V are given we proceed as follows :

Find the pressure per square inch of the steam from the Table,
and thence Art, 166 the useful load on each square inch of the
piston ; find also when required any of the other quantities, o, n
or I, and then apply the rules given in Art, 167.

ABT.170.]

WORK OF STEAM.

re inch.

ay n,

Table,
\of the
U, a, n

ExAMPLB 201.— 'What Is the useful load per square inch on the
piston, and what is the eSiBctire horse powers of a high pressure
engine in which the area of the piston is 200 inches, the length
of stroke 6 feet, the eflPective evaporation of the boiler 2 of a
cubic foot per minute and the pressure of tlie steam 70 lbs. to
the square inch ?

SOLUTIOK.

By Art. 166, 70 = f P + 15 + 1 . and hence p = 84rrf = 47*28 = U sefiil load.

144«t; 1<MX -4X406
By Art. 169, n :

al ~ 200X6 -^»"*88-
Hence we have n = 19'488, p = 47'26, a — 200, 1 = 6.

3>a?rt 47.25 X 200 X 6 X 19.488

Then Art. 167, H. P. - ^^rxx =

— 3:J'48. Ant.

33000 33000

Example 202. — What are the eflfective horse powers of a low
pressure engine in which the piston has an area of 288 inches
and makes every minute 16 strokes, the boiler converting \ of a
cubic foot of water per minute into 304 cubic feet of steam?

bOIiUTION.

Since i of a cul)ic foot of water produces 304 cubic feet of steam, 1 cnl)ic
foot of water would produce 608 cubic feet of steam, and hence (Table) tho
gross pressure of the steam is 45 lbs. to the square inch.

Then (Art. 166) 45 =7 jp + 4 + 1 or f P = 40 whence p = 35.

Also (Art. 169) i =

144c« 144X'5X608

:94 ft.

288X16
Then a — 288, i = Oi, n = 16, and p = 35.

X. 1 T **,/.* TT Ti P«^» 35X288X94X16 ^„.,„^ .
I lence Formula I Art. 167. H. P. = ^— = g^^ = 46*42». An$

EXERCISES.

203. What are the effective horse powers of a high pressure

engine in which the piston has an area of 360 inches and
makes 20 strokes per minute, — the boiler evaporating I
of a cubic foot of water per minute under a pressure of
40 lbs. to the square inch ? w5n«. H. P. = 46-528.

204. The piston of a low pressure steam engine has an area of

432 inches and makes strokes 10 feet in length. Now, if
the boiler evaporates -9 of a cubic feet of water per
minute under a pressure of 25 lbs. to the square inch,
what are the useful H. P. of the engine ?

Ana. H. P.= '71-G13,

205. In a high pressure engine the area of the piston is 600

inches, the length of stroke is 6 feet, the effective evapo-
ration of the boiler is i of a cubic foot per minute and
the pressure of the steam in the cylinder 80 lbs. to the
square inch ? Required the H. P. Ans. H. P. = 32-897.

HYDROSTATICS.

[Abii. 171-170.

Ill

CHAPTER V.

HYDROSTATICS.

171. Fluidity consists in tlio transmission of pressuie
in all directions, or, a fluid raay be defined to be a body
whose particles are so free to raove among one another
til. it they yield to any pressure, however small, that may
be applied to them.

172. The term lluid is ooinmonly applied to bodies in
both the liquid and ga«cous slate.

173. Fluids are divided into two classes: —

Ist. Elastic fluids, of which atinosphcric air is the type,
2d. Non-Elastic fluids, of which water is the represen-
tative.

Note.— Water wa.s formerly thought to be absolutely incompressible'
but recent oxpoJ imeuts show that water is diminished in volume t2 olTO
of its bulk for c: .ch atmosph -re of pressure upon it ; or in other words a
pressure of 20i)C atmosphere or 30000 lbs. to the square inch would compress
11 cubic fyet iito 10 cu'jic feet. Alcohol is about twice as compressible
as water.

174. Liquids, by which leini we mean non-ehistic fluids,
differ from gases chiefly in luiving less elasticity and com-
pressibility.

175. Liquids ditt'er from F^olids chiefly in the fact that
their particles are less under the influence of the attraction
of cohesion, and therefore have a freer motion among them-
selves, in consequence of which each atom is drawn sepa-
rately towards the earth by the force of gravity ; hence : —

/. .4 liquid, confined in any vessel, presses equally in all direc-
lions — upwards, downvmrds, and laterally.

11. The surface of a liquid in a state of rest is always level.

111. A liquid rises to the same height in all the tubes connected
with a common reservoir, whatever may be their form or capacity

Note.— The facjt that a liquid exerts a downward pressure is self-evident
and reqires no ilhistration.

The lateral pressure of liquids is shown Viy their spouting from holes
pierced in thi side of the vessel in which they are contained.

The upwar.. pressure is shown by taking a glass cylinder, open at both
ends, and having one end ac(!urately ground. A plate of ground glass i"
held to this end bv nieanH of a piece of string passing through the cylinder
and the closed end of the instrument then emereed in water to a small

AlTi. 17»-178.J

HYDROSTATICS.

at both

glass i«!

cylinder

a small

depth. lT))on lettlriK fo tlio string tho plate Is still hold against the cylin-
dur by the upward pressure of tho water, it wili even sustain any woiKht,
which, together with the plate itself, is not greater than the weight of the
water that would enter the cylinder if the plate were removed.

176. When two liquids of different densities are placed
in the opposite branches of an inverted syphon or bent
tube — their heights in the two legs above the point of con-
tact will be inversely as their densities.

Note.— This may easily be proved by placing mercury and water in a
bent graduated glass tube, when it will be found that tho column of water
will be 13i times as high as th« column of mercury since tho latter is about
\^ times as heavy as the former.

177. The amount of downward pressnre exerted by a
liquid in any vessel is equal to that of a column of the
same liquid, whose base is equal to the an a of the bottom
of the vessel, and whose height is equal to tho depth of
tho liquid, whatever ntay be the form or capacity of tho
vessel.

Note 1.— To illustrate this fact we procure three vcssesl, having bottoms
of the same area, and sides, in tho first pcrpetidicular, in the second conver-
ging towards tho top, and the third diverging towards the top. The bottoms
arc hinged and are iield in their places by a cord passing over a pulley and
terminating in a scale pan in which are placed weights to a certain amount.
Water is then carefully poured into the vessel having the perpendicular
•ides until its downward pressure is just sufficient to force out the bottom
when its depth is accurately measured. Upon using either of the other
vessels it is found that the bottom remains fixed until the water reaches
this depth and is then forced open. This arises from the fact that when
the sides are perpendicular the bottom support tho whole weight of the
water. When the vessel is wider at top than at bottom a portion of the
downward pressure is sustained by the sides, while, when the vessel is
wider at tho bottom than at top, the particles near the bottom are pressed
upon by the whole column of li<iuid above them and their d' \vnward and
lateral pressure is the same as it would be were the column ol' liquid of tho
same dimensions throughout as the base of the vessel.

Note 2,— Care should be taken not to confound weight with pressure, in-
asmuch as tho weight is in proportion to the quantity of liquid but the pres-
sure is in proportionto the extent of base and the perpendicular height of the
liquid. For example, the weight of the water contained in a conical vessel
is found by multiplying the area of the base by one-third of the perpen-
dicular height ; but the pressure, by multiplying tlie area of the base by
whole height. It follows that in a conical vessel the downward pressnre is
equal to three times the weight of the liquid. Hence In a vessel with per-
pendicular sides, the pressure equals the weight ; if the sides diverge up-
wards, the pressure is less than the weight ; and if the sides converge
upwards, the pressure is greater than the weight,

178. A cubic inch of water of the temperature of 60°
Fahr. weighs 0.0?616 lbs. Avoir., a cubic foot at the same
temperature weighs 1000 ounces or 62-5 lbs., and a gal-
lon, 10 lbs.

HTDB08TATICS.

CABT. 178

179. The pressure of a liquid on a vertical or inclined
surface ia equal to the weight of a column of the same
liquid whose base is equal to the area of the surface pressed,
and lieight equal to tue depth of the centre of gravity of
the pressing liquid beneath its level surface.

Or, more simply, the lateral pressure exerted by any liquid on
the side of a vessel is found in lbs. by multiplying the area of the
surface pressed by half the depth of the liquid, and this product by
the weight in lbs. of one cubic foot of that liquid.

Note.— It follows that In a cubical vessel flUod with any liquid the
pr«;ssure on the side in equal to half the weight of the liquid, and hence
the whole prnHsuro exerted by the liquid, downward and laterally, is equal
to three times the weight of the liquid.

APPLICATION OF THE PRINCIPLES CONTAINED
IN ARTS. 176-179.

ExAMPLB 206. — What downward pressure is exerted on the
bottom of an upright cylindrical vessel having a diameter of 20
feet — the water filling it to the depth of 12 feet ?

SOLrTIOK.

Here since the sides are perpendicular, the downward pressure = the
weight.
Area of the bottom — 102 x3"1410 — 100X3-1 ilC = 314-16 foet.
Cubic feet of water = 314-10 X 12 ~ 3769-92.
.'. Weight = 3769-92X 62-5 = 235020 lbs. = pressure. Ans. , '

Example 207. — If olive oil and milk be placed in the two legs
of a bent tube or inverted syphon, when the height of the col-
umn of milk above the point of junction is 20 inches, what
will be the height of the column of oil ?

SOLUTION.

From the table of specific gravities Art. 198, the weight of milk is to that
of olive oil as 1030 : 915.

1010X20

Hence (Art. 176) 915 : 1030 : : 20 : TlC = 22^ inches, Ans.

915

Example 208. — If Mercury and Ether are placed in a bent
tube as in the last example what will be the height of the
column of Mercury when that of the ether is 100 inches high?

SOLITTIOS.

From the table of specific gravities the weight of mercury is to that of

ether as 13596 : 715.
Hence (Art. 176) 13596 : 715

100:

715X100
13596

= Si inches. Ans.

Example 209. — What will be the lateral pressure exerted
against the side of a cistern,— the side being 20 feet long and
the water 12 feet deep ?

▲bt. 179.]

HTDB0STATIC8.

that of

•OLVTIOV.

ArM of the surflMe preM«d=20X 12=240 feet.

Then (Art. 178) lateral pressure = area multiplied by half the depth X
e2'6=M0X6X62-fr=90000lbs. Ant.

ExAUPLi 210. — What is the amount of the pressure exerted
against one side of an upright gate of a canal, tlje gate being
27 feet wide and the water rising on the gate to the height of 8
feet?

80LVTI0V.

Area of the ffatc=27 x 8=216 foet, and half the depth of the water=:4 ft.
Then (Art. 170) pros8uro=:216X'lxa2'S=54000 lbs. Ant.

GxAMPLi 211. — What is the amount of pressure exerted
against a mill-dam whose length is 220 feet, the part submerged
being 9 feet wide, and the water being 7 feet deep ?

BOLUTIOir.

Area ofpartsubmerged=220X 9=1080 feet, and half the depth of water
=3*6 feet.
Then (Art.UO) pres8uro=^1980X 3' 5x62*5=133126 lbs. Ans.

ExAMPLi 212. — If the body of a fish have a surface of 5 square
feet, what will be the aggregate pressure it sustains at the
depth of 100 feet ?

BOLUTIOir.

In this and similar examples the body of the flsh has to sustain a pres-
sure equal to the weight of a column or the water having a base equal in
area to the surface of the fish and a height equal to the depth of the fish
beneath the surftuse of the water.

Then volume of water sustained by the body of the flsh=5X 100=500
cubic feet.

Hence pre8sure=500 X 62*6=31250 lbs. Ana.*

ExAMPLi 213.~-If a man whose body has a surface of 1&
square feet dives in water to the depth of 70 feet, what pressure
does his body sustain ?

80LUTI05.

Column of water sustained by man's body at depth of 70 feet=15X70=
1050 cubic feet.
Hence pres8ure=1050X 62 6=65625 lbs. Ans.

ExAMPLB 214. — To what depth may an empty closed glass
vessel just capable of sustaining a pressure 170 lbs. to the squart
inch be sunk in water before it breaks ?

SOLTTTIOir.

From Art. 17d we find that a cubic inch of water at the common tempe-
rature of 60® Fahr. weighs 0*03616 of a pound Avoirdupois.

Hence the vessel may Tbe sunk as many inches as '03616 lbs. is contained
times ill 170 lbs.

That is depth~l70— 0.03616=4701i inches=301 feet »i inches. Ans.

* In this and following examples involving the same principle, we make
no allowanoe for the Increased pressure at great depths.

tttDttOSTATICS.

[Ant. m.

lil

ill'

V !

ExAMPLi 215.— If an empty corked bottle be sunk to the depth
of 130 feet before the cork is driven in, — what pressure to the
square inch was the cork capable of sustaining before entering
the bottle?

SOLCTIOIT.

Column of wator sustained by each square inch of the cork=130xi2—
1660 cubic inches.

Then weight sustained by each square inch of the cork=1660X 0*03616=^
56-4 lbs. Jtns.

. ^ EXERCISES.

216. What is the amount of pressure exerted against one side

of the upright gate of a canal, — the gate being 24 feet
wide and submerged to the depth of 10 feet ?

^ns. 75000 lbs.

217. What is the amount of pressure exerted against a mill-

dam, — the part submerged being 10 feet wide and 80
feet long and the depth of the water being 8 feet ?

^nj. 200000 lbs.

318. What is the pressure sustained by the sides of a cubical
water tight box placed in water at the depth of 120 feet
beneath the surface, — each edge of the box being 5 feet
long? Jnt. 1125000 lbs.

219. At what depth beneath the surface will a closed glass

vessel, capable of sustaining a pressure of 79 lbs. to the
square inch, break ? ^ns. 182 ft. 0| inch.

220. What pressure is sustained by the body of a man at the

depth of 30 feet, — assuming that his body has a surface
* of 1 J square yards? ^ns. 25312 J lbs.

221. What is the amount of pressure exerted against one side

of the upright gate of a canal, — the gate being 30 feet
wide and submerged to the depth of 5 feet ?

Jlns. 23437J lbs.

222. In a glass tube bent in the form of a syphon a column of
-» turpentine is balanced by means of a column of sea

water,— if the heiVht of the former be 20, 30, or 47
inches what in each case will be the height of the latter ?

Jlns. 16t%, 25| or 39f inches.

223. What is the downward pressure, the pressure on each side

and also the pressure on each end of a rectangular
cistern, — 14 feet long, and 9 feet wide — the water being
10 feet deep ? ^ns. Downward pressures: 78 750 lbs.

Pressure on side=43750 lbs.

Pressure on end=28125 lbs.

124. What amount of pressure is sustained by the body of a
whale the depth of 260 feet, upon the supposition that
his body presents a surface of 200 square yards ?

Am. 29250000 lbs.

iBTH.180.1Sl.l

mroiiostAttcs.

225

In a glass tube bent in the form of a syphon a column of
mercury is balanced in succession by a column of alcohol
and a column of sulphuric acid. If the height of the for-
mer be 10 inches what in each case will be the height of
the latter? »Ans. Alcohol = 111} inches.

Sulphuric acid = 73^ inches.

180. To find the pressure exerted against a vertical of
inclined surface at some given depth beneath the surface
of the water : —

RULE.

^dd the depth of the upper part of the surface to that of the
lower part and divide the sum by 2. The result is the mean height
of the columns of water pressing on that surface.

Then multiply the area of the surface by the mean height of the
water, pressing it, and the result by the weight, in lbs., of one cubic
foot of water.

ExAMPLB 226. — What amount of pressure is sustained by one
square yard of the side of a canal, the upper edge being 10 feet
and the lower edge 12 feet beneath the surface of the water.

, . , , , . aOLUTIOX.

' 10+12

Mean weight of column of water pressing the given surface == — - — — 11

ft., and area of surface = 9 sq. ft.

Then pressure =9X11 = 99 X 62*6 = 6187i lbs. Ana.

ExAMPLB 227. — An upright flood gate is so placed in a canal,
that the water is just level with the top of the gate. — Assuming
the gate to be 30 feet long and 20 feet wide what pressure is
sustained by the lower half of one side ?

SOLUTION.

The upper edge of the half to which the problem refers is 10 feet beneath
the surface, and the lower edge 20 feet, therefore the mean luight of the

column of water pressing against it is — i — = 15 feet.

2
Also area of part of gate given = 30 X 10 = 300 sq. ft.

Hence pressure = 300 X 15 X 62"5 = 281250 lbs. ^n.<5,

181. In problems similar to the last a better rule to
use may be derived from the following consideration.

The pressure on the whole gate is to the pressure on any fraction of it
measured from the top, in the duplicate ratio of 1 to that fraction.
Hence to find the pressure on any part of the gate w»! have the following :

RULE.

First. — If the part of the gate be measured from the top down-
wards.

Find the pressure on the whole gate by Art. 179, and multiply
it by the square of the given fraction.

BTDBOSTATtCS.

CAIT. 181.

!ll^

SicoKD. — If the part of the gate be measured from the bottom
upwards.

Take the given /radian from 1, square the remainder, and sub-
tract it from unity.

Multiply the pressure on the whole gate by the fraction thus ob-
tained and the result will be the pressure on the given fraction.

SxAMPLK 228. The flood-gate of a canal is 16 feet wide and 12
feet deep, and is placed vertically in the canal the water being
on one side onlj and just level with the upper edge of the gate ;
Required thepressure—1'^ On the whole gate.

2"'*. On the upper third of the gate*

S'**. On the lower half of the gate.

4"». On the upper two-fifths of the gate.

6*. On the lower two-eleventhsof the gate.

SOLUTIOir.

Pressure on the>hole gate= 16X12X6X62-5=72000 lbs.
Pressure on upper third = whole pressure X {i)*=72000X^

II.

8000 lbs.

III. Pressure on loMi-er half = whole pressure x f 1— {i)« j=72000Xf=

64000 lbs.

IV. Pressure on upper two-fifth8= whole pressure X (|)* =72000 X ■^^;

= 11620 lbs.
V. Pressure on lower two-elevenths - . whole pressure X f l~(-ft-) ' j ==

2000 X -^jQj- = 23801'6528 lbs.

In III we take the given fraction i from unity this leaves i which we
square and a);ain subtract from unity and thus obtain } for the multiplier.
In y we take the given fraction -^j from unity, this gives us -^^ whioh are

square and again subtract from unity tlius obtaining -^^ for the multi-
plier.

ExAMPLK 229. If a flood gate be placed as in last example
what pressure will be exerted on the upper ^ and what on the
lower I of the gate if it be 10 feet wide and 12 feet deep ?

■ SOLUTIOir.

We Itrst find the pressure on the whole gate by Art. 179.

Then for the upper ^ we multiply the whole pressure by the square of ^.

For the lower 2. we subtract ^ from 1, this gives us g which we square
and thus obtain -^, then we subtract -^i,. from 1 and thus obtain ^.lastly
we multiply the whole pressure by this ^^j

"Whole pressure = 10X12X6X62'5 = 45000 lbs.

Pressure on upper ^ = 45000 X'^ff = 8265^ lbs.

Pressure on lower I = 45000 xM = 28800 lbs.

EXERCISES.

230. The flood-gate of a canal is 30 feet wide and 10 feet deep,
and is placed vertically in the canal, the water being on one
lide only and level with the top, required the presBure — Ist.

ill'

AST. 181.]

HYDROSTATICS.

On the whole gate ; 2Dd. On the upper half of the gate ;
3rd. On the lower half of the gate ; 4th. On the lowest two-
sevenths of the gate.

Am. Pressure on whole gate = 93750 lbs.

" upper half = 23437}

" lower half = 70312}

" lowest two-sevenths = 45918^1 "

231. A hollow globe has a surface of 7 square feet, and is sunk
in water to the depth of 150 feet. Required the total pres-
sure it then sustains. Ans. 65625 lbs.

232. What pressure is exerted against one square yard of an
embankment if the upper edge of the square yard be 11 ft.
and the lower edge 13 feet beneath the surface of ths
water? Ans. 6750.

233. A hollow glass globe is sunk in water to the depth of 400
feet, at which point it breaks. Required the extreme
pressure to the square inch which the vessel was capable of
sustaining. Ans. 173'568 lbs.

234. Required the pressure sustained by the body of a man at a
depth of 100 yards beneath the surface of water — assuming
the man's body to have a surface of 15 square feet?

Ans. 281250 lbs.

235. A flood gate 16 feet long is submerged to the depth of 9

feet in water
of it?

what pressure is exerted against each side

Ans. 40500 lbs.

230. A mill dam is 120 feet long and 11 wide, the water being
exactly level with the top of the dam and the lower edgt of
the dam 7 feet beneath the surface. 1st. What will be thj
pressure exerted against the whole dam. 2nd. What pres-
sure will be exerted agamst the upper part of the drou.
3rd. What pressure will be exerted against the lower iaif
of the dam? Ans. Against whole dam 288750 lbs.

" upper half 72187} lbs.

« lower half 216562} lbs.

237. A flood gate 26 feet wide is submerged perpendicularly to
the depth of 12 feet ; find 1st. The pressure against one side
of the whole part submerged. 2nd. The pressure against the
lower half. 3rd. The pressure against the lowest third.
4th. The pressure against the lowest sixth ?

Alls. 117000 lbs. whole gate.
87750 lbs. lower half.
65000 lbs. lowest third.
35750 lbs. lowest sixth

f^i

Htl>R08TAttCS.

[Ants. l««, 18.1,

162. If wnter bo conflnofl in n vessel siiul n proHMiro to
jiny ftinoiMit bo oxerlinl npoii niiy ono sqimro inch of tba
surface of tlmt w?iter, a pressure lo an e(pnil amount will
bo transmitted to every s-jnani incli of the interior surface
of tbo vessel in wliieli llie water is condniMl.

rii?. If,.

NoTB.— la tlio n^TotiiimiiyiiiK Okh''*'
ciuppoHn tb(» ))istoti /* iiaH tin an-ji of
1 Hqimrn inoh, niut tho piNton >>' an area
of n>0 nquaro iiichcH, thon \f I lb. pros-
nnro h<> appHcMi to /' as wimkIiI or loi)
lb. nuirtt bo appUcnl to p' in ordc^r to
maintain oquilinriuni. It iH this pro-
pcrty of ,'(inal and instant transniix-
Bion of jircHNiiro wiiioli (MiablrH ns to
innko UNO of li.vtlroNtalic prcsHnr<> as
a nio('bani«'al ijowcr, and it is upon tliis
nrlnolplo tliat Uraniaii's Jl.vdrostatic
VroNN IS constnu'tcd.

183. Hram all's II, ydrostatic Tross consists of two stronjr
motaH'o cylinders /I ajid (t, one many times as large as tlur
other, conned ed «. ^^

together by a tube.

The small cylinder jci c3a.

is supplied with L„ .,
a strong forcing
pump *•', and the
larg<>r one with a
tightly fitting pis-
ton ^S^ attached,
a firm jil.rtforin or
strong head /\
Hoth thecvlinders
and the commu-
nicating tube con-
tain water, an<
when downward
]>ressure is applied lo the water in the smaller cylinder,
by means of the attached forcing i)U!np, the piston in the
larger is forced upward by a ])ressure as much greater
tlian the downward pressure in the smaller, as the sectional
area of the larger cylinder is greater than that of the
smaller.

m\fo to V
[ of tlie
Hit Avill
8nrfa<x>

stioiiix
''0 ns tlio

ylindcr,
I in tlic

jnjrcater
ectional

of the

AXTS. IH 185. )

HYDROSTATICS.

For (ixani|)I(N if Ihn Nmallor cylitidRr haro an aroa of half a aquare
inch, and thtf largn ryliniior an ar(>a of ftOO Hquaro innlu'H than thn upward
preNHuro In tho lattur will be lOUti tlniuH a* great as the downward pruMuro
in tho former.

184. liratnali'rt llydroHtatic. Press is used f«)r prcBsinn;
papor, cotton, cloth, gunpowder, and other tilings — also
for tcHting tho strength of ropes, for uprooting trees and
for other pnrpoHCs.

186. lo find tho relation between the force applied and
tho pressure obtained in IJramah's Hydrostatic Press.

RULE.

/, 1/ the power be applied by means of a lever, find the amount oj

downward pressure in the smaller cylinder by the rule in

Jlrt. 11.
If. Divide the sectional area of the lar^e cylinder by that of the

the smaller cylinder and multiply the quotient by the power

applied to the smaller cylinder.

ExAMi'LB 238. — In a Uydroatatic Press the force pump has a
sectional urea of one square inch ; the hirgc cylinder a sectioDal
area of one square foot, tlie force pump is worked by means of
n lever whose arms are to one another aa 21 : 2. If a power of 20
lbs be applied to tho extremity of the lever wliat will be the
upward pressure exerted against the piston in the large cylin-
der?

BOIirTION.

20X21
I'ower a|)i)li(!(l to force i)Uinp ~

210 lbs.

Spctlonal area of sinallor cylinder =- 1 inch and of largw cylinder ~ l^i
inches.
Then UK 1^1111X210- 302410 Ibw Ana.

KxAMi'LB 239. — In a Hydrostatic Press the sectional areas of
the' cylinders are J of an inch and 150 inches, and the power lever
is so divided that its arras are to one another as 7 to 43. What
pressure will be exerted by a power of iOO lbs. applied at the
extremity of the long arm of the lever?

SOLUTION.
,, , , „ ,. , 100X13

Dowuwiml pressure in Huiall cylinder = — = —

Upward i)rcM8urc in largo cylindor — -'- YMii
27«M28fll)s, Ans. >s

nU'j lbs.
-450X614^

Example 240. — Tho area of tho small piston of a Hydrostatic
Press is ^ an inch and that of the larger one 300 inches, the
lever is 30 inches long and tho piston rod is placed 5 inches
from tho fulcrum (so as to form a lever of tho second order)
what power must be applied to the end of the lever in order
to produce au upward pressure in the cylinder of 1000000
lb»?

' I

HYDROSTATICS.

[Abt. 1M.

BOLUTIOK.

300

Downward proMurc in oinKlIor oylinderrrrlOOOOOO lbs — -i- -- 1000000 Ibii.
f- 600 - 10061 Ibu.

•HI

Then power applied - lOflOl lbs. -r -""-IflOOl-: C-STTJ lbs. Ant.

241.

242.

243.

244.

245.

EXERCISES.

In a Hydrostatic Press tlio area of tho Biiinll cylinder is 1
inch, and that of tho largo ono 300 inches, tlio forco
pump is worked by a lever of tho second order 30 inches
long, having the piston rod 2 inchea from tho fulcrum,
if a rn'ssiirc of 60 lbs. bo applied to tho lover what
upwaru pressure will bo produced in the largo cylinder?

.4ns. 225000 Ib.s.

Ir. aTiij'.l;oatatic Press tho forco pump has a sectional area
of half tin inch; the largo cylinder a sectional nrea
.4" ?,00 iiches ; tho force pump is worked by meuna
if ii lov r whose arms are to one another as 1 to 50.
Now !■ i !io60 a forco of 50 lbs. be apj)liod to tho o-x-
tremn; the Ivvcr what will bo the upward proMsuro

exerted against the piston in tho largo cylinder?

Jns. 1000000.

In a Uydrostatic Press tho small cylinder has an area of
one inch, and the largo ono an area of 500 inches, the
pump lever is so divided that its arms are to one another
as 1 to 25 what will be the upward pressure against tho
pioton in tho largo cylinder produced by a forco of
100 lbs. acting at the extremity of tho lever I

Jliis. 1250000.

The area of the small piston of a Hydrostatic Press is J
of an inch and that of tho large one 120 inches — the
arms of tho lover by which the force pump is worked arc
to one another as 40 to 3. Required the upward pres-
sure exerted against the piston of the large cylinder by a
power of 11 lbs. applied at the extrepvty of the lover.

'jns. 3G2GGr, ll'S-

Tho area of tho small piston of a Hydrostati ' 'ress is
IJ inches, and that of the lario one 200 inches — the
arms of tho lever by which the force pump is worked are
to ono another as 20 to li. What power applied at the
extremity of the lever will produce a pressure of Y50000
lbs? ^«s. 421|lbs.

[ I

I- 1

186. Since the pressure of water upon a given base
depends upon the height of tho liquid and not upon its
quantity, it follows that : —

I i

ABT. 1M.

00000 lbs.

(lor is 1
10 force

inclios
fulcrum,
?r what
^Minder?
5000 lbs.

>nal area
mil iirca
Y mciina

1 to 50.

tho ox-

prcHSuro

1000000.

nroa of
hos, the

another
linst tho
brco of

llUfjOOOO.

[rcss is 4
hes — the
•kcd arc
Jrd pros-
Idcr by a
lover.
\6Cy?i lbs.

'rcss is
lies — the
rked are
Id at the

750000
|21| lbs.

in base
)on its

AmTt.l87,lS8.]

HYDROSTATICS.

.1- 1
Any quantity of water how$ver amallj may be made to
balance the pressure of any other quantity however greaty
or to raise any weight however large*

NoTR.— TIiiB is wliat is pommonly called tho Ifydrontatin Paradox. In
ronlitv, howuvcr, thoro Ih iiothiiiK at alS paradoxical in it : Nince, althouich a
nound of water may bn niadu to balance 10 !\)h., or 1000 lbs., or l(M)0O0 Ids,,
it doits it upon procisoly tho same iirinciplo that the powor balances tho
weight ill tho lover and other nicchanioal powern. Thus in order to raise
20 lbs. of water by tho deacf^iidinK force of t \\>., the latter must dcHcwnd 20
inches in ordor to raiio the former 1 inch. Honco what is called the
hydrostatic paradox Ih in strict conformity to tho principle of virtual velo-
cities.

187. This principle is illustrated by an instrument
called tho Hydrostatic Bellows, which consists of a pair
of boards united together by leather as in the common

Fig. 18.

bellows and made water-tight. From
the upper board there rises a long tube,
//, finished with a funnel-shaped termin-
ation, c.

NoTB. — When water is poured into tho tube

an upward pressure is exerted against the

u))per board as much greater than the weight

of the water in the tube as the area of the

board is greater than the sectional area of

the tube.

For oxaniplo, if tho sectional area of tho tube be \
of an inch, and the area of tho board bo 250 inches,
then tho area of tho board will bo 1000 times aH
Kreat as that of tho tube, rnd conscquontly 1 lb. of
water in the tube will exert a pressure of 1000 lbs.
against the upper board of the oellows.

188. To find the ui>ward pressure exerted against the
board of a hydrostatic bellows by the water contained

in the tube.

RULE.
Divide the sectional area of the board by that of the tube^ and
multiply the result by the weight of the water in the tube.

NoTB. — The weight of water in the tube is found by multiply-
ing the sectional area of the tube by the height of the water in inches
and the product J vhich is cubic inches of water ^ by 0.03616 Ibs.^ the
weight of one cubic inch of water .

ExAMPLB 246.— The upper board of a Hydrostatic Bellows hai
an area of 1 foot, the tube has a sectional area of \ an inch
and is filled with water to the height of *l feet. What upward
preasure is eierted againit the top board of the bellows 7

■i— ^,

M ! )

i I

• [

}>■'>

8§

HYDROSTATICS.

[AAXB. 169-199.

144

= 1.61872X288

SOLUTIOir.

Cubic inches of water in the tube =i X 84 = 42.
Weight of water in tube = 0.08616 X 42 =: 1.61872 Ibv

Upward pressure against bellows board =: 1. 61872 X

=437.39 lbs. ii«a,

ExAHPLB 247. — In a Hydrostatic Bellows the board has an
area of 200 inches and the tube a sectional area of | of an inch.
What upward pressure is exerted on the board by 7 lbs. of water
in the tube ?

SOLUTION.
Upward pressure = 7 X*— r* = 7 X 800 = 5000 lbs. An$.

EXERCISES.

248. In a Hydrostatic Bellows the board has an area of 260

inches, the tube has a sectional area of li inches, and
contains 1 1 Iba. of water. What is the amount of upward
pressure exerted against the board of the bellows ?

wfn«. 2200 lbs.

249. The board of a Hydrostatic Bellows has an area of 300

inches, the tube a sectional area of 1 inch and is filled
with water to the height of 10 feet — what pressure will
be exerted against the upper board of the bellows ?
I . Ms. 1301.76 lbs.

250. The tube of a Hydrostatic Bellows has a sectional area of

•72 of an inch and is filled with water to the height of

! ' 50 feet — what weight will be sustained on the bellow's

board if the latter have an area of 3 feet ? jins. 9372.672.

189. A body immersed in any liquid will either float,
sink, or rest in equilibrium, according as it is specifically
lighter, heavier, or the same as the liquid.

190. A floating body displaces a quantity of liquid
equal to its own weight.

191. A body immersed in any liquid loses a portion of
its weight equal to the weight of the liquid displaced, and,
hence, by weighing a body first in air and then in water,
its relative weight or specific gravity may be determined.

192. The specific gravity of a body is its weight as
compared with the weight of an equal bulk or volume of
some other body assumed as a standard.

193. Pure distilled water at the temperature of 60^
Fahr. is taken as the &tandarvl with which to compare all

▲ftfl.lM»10S.]

HTDR08TATICS.

Bolids and liquids, and pure dry atmospheric air al a tempe-
rature of 32<> Fahr., and a barometric pressure of 30 inches
is taken as the standard with which all gases are compared.

194. To find the specific gravity of a solid heavier than
water : —

RULE.

Divide the weight of the body in air by iti lo$i of weight «u
toater, the result will be its Specific Gravity.

Example 251. — A piece of lead weighs 226 grains in air and
only 205 grains in water; required its specific gravity.

SOLUTION.

Loss of weight = 225—205 = 20 grains.

Htnce specifio gravity =: 225-f-20 = 11.250. Ant.

ExAMPLB 252. — A piece of sulphur weighs 97 grains in air
and but 50-5 grains in water; what is its specific gravity?

SOLVTIOK.

Loss of weight in water = 97— 60"6 = 4i6*5 grains.
Then specific gravity = y? -4- 46'6 = 2*008 Ams.

EXEROIOES.

263. A piece of silver weighs 200 grains in air and only 180
grains in water ; required its specific gravity.

^ns. 10.000.

254. A piece of platinum weighs 154J oz. in air and only 147J
OB. in water; required its specifis gravity. jlns. 22.0'71.

255. A piece of glass weighs 193 oz. in air and but 130 oz. in
water ; required its specific gravity. Ans. 3.063.

195. To find the specific gravity of a solid not
i^ufficiently heavy to sink in water.

RULE.

To the body whose specific gravity is sought attach some other
body sufficiently heavy to sink it, and of which the weight in air
and loss of weight in water are known.

Then weigh the united mass in vmter and in air^ from its loss of
weight deduct the loss of weight of the heavier "body in water, arid
divide the absolute weight of the lighter body by the remainder, the
quotient will be the specific gravity of the lighter body.

ExAMPLB 256. — A piece of wood which weighs 55 oz. in air
has attached to it a piece of lead which weighs 45 oz. in air and
41 in water, the united mass weighs 30 oz. in water ; required^
the specific gravity of the piece of wood.

.-M''

UYDROSTATICS.

AST. 19<l.

''ill ^

^'^'U

BOlTTTIOir.

WeL t of united mass In Air=55 + 45 = 100 ox.

waier= 30 "

Loss of woiffht of united maMS in wator=: 70 "
Lossof weight of lead in water = 4 "

Remai uder — • 66=los8 of weight of the wood.

Then 65 -f- 66= '833 = specific gravity of the wood.

Example 257. — A piece of wood which weighs 70 oz. in air
has attached to it a piece of copper which weighs 36 oz. in air
and 31.5 oz. in water, the united mass weighs only 11.7 oz. in
water ; what is the e pecific gravity of the wood ?

Weight of united mass in air
" '• water:

80LU1I0K.
70 + 36:

: 106 oz.

11.7

Loss (^f weight of litedimasa in watmcs 01.) "
Lossc weight of copper " r= 4.5

Loss of weight of wood " == 89.8=los9 of weight of tlie wood.

Then specific gravity of wood=70-f 89.8=: .779. Ans.

EXERCISES.

258. A piece of pine wood which weighs 15 lbs. in air and hag
atta'^hed to it a piece of coppe which weighs 18 lbs. in air
and 16 11,^. in water, the weighi of the united mass in water
is 6 lbs.; required Lhe specific gravity of the pine?

jlns. -600.

269. A piece of cork which weighs 20 oz. in air has attached to
it an iron sinker which weighs 18 oz. in air and 15.73 oz.
in water, the united mass weighs 1 oz. in water ; required
the specific gravity of the cork ? Jlns. -575.

"60. A piece of wood which weighs 33 oz. in air has attached to
it a metal sinker which weighs 21 oz. in air and 18 19 oz.
in water, the united mass weighs 2.5 oz. in water ; what ig
the specific gravity of the wood ? Jns. '677 .

;!:!

196. The specific gravities of liquid maybe determined
in three different ways.

First Mbthod. — A hmall glass flask, which holds precisely 1000
grains of pure distilled water at the temperature q/60® Fahr., is
filled with the liquid in question and accurately weighed, the result
indicates the specific gravity of the liquid.

Second Method. — A piece of substance of known specific gra-
vity is weighed both in and out of the liquid in question. The
difference of weight it multiplied by the Spec. Grav. of the solid.

AAT. 196.

AST. IM.]

HYDROSTATICS.

»f the wood.

oz. in air
oz. in air
1.7 oz. in

of the wood.

ir and Iiag
lbs. in air
3 in water
?
Ins. -GOO.

ttached to
15.73 oz.
required

Ins. '575.

ached to

18 19 oz.

what ig

ns. '677,

terniined

isely 1000
Fahr., is
the result

ecijic grO"
ion. The
the tolidf

and the product divided by the absolute weight of the solid, awl
the result is the Spec, Grav. of the liquid.

That t«» =

w — W

■X*';

where ivz^absolule weight of solid.
w'=z weight in the liquid.
Therefore w — uj'=z loss oficcighl.

s = specific gravity of the liquid,
t' = specific gravity of the s"lid.

.. 19.

Third MBrnoD. — This specific gravity of liquids is
most commonly found in practice by means of an iti-
strument called the Hydrometer, which consists of a
graduated scale rising from a glass or silver bulb,
bener.ih which is a small appendage loaded with shot or
osme other heavy substance. It acts upon the principle
that the greater the density of a liquid the greater will
be its Specific Gravity. The depth to which the instru-
ment sinks in different liquids is shoivn by the gradu-
ated scale, which thus indicates their Specific Gravities.
For liquids specifically lighter than water, the scale is
graduated from the bottom upwards; for those hea-
vier, from the top downwards.

Example 201. — The Thousand-grain Bottle filled
with sulphuric acid weighs 1841 grains.* "What in
the specific gravity of the sulphuric acid ?

SOITTTIOIT.
1811 ~- 1000 = 1.841 Ans.

ExAMPLB 262. — The Thousand-grain Bottle filled with Alcohol
weighs 792 grains, required the specific gravity of Alcohol.

BOIUTIOX.

792 -r 1000 = 792 Ans.
Example 263. — A piece of Zinc (Spec. Grav. 7.190) weighs
27.4 oz. in a certain liquid and 32.7 oz. out of it, required the
Spec. Grav. of the liquid.

soLuirorr.
Here w — 32.7, w' = 27.4. «' = 7.190.

_ w — «j' , 32.7— 27.4^^ ^,„^ 5.3X7.190

Then s = —- — X s' = — :::r;; — x 7.190= — „,_ ■ - 1.165 Ana.

32.7

Example 264.— A piece of Silver (Spec. Grav. 10.500) weighs
47.8 grains in a liquid and 58.2 grains out of it— what is the
Specific Gravity of the liquid ?

* That is not including the weight of the bottle itself.

■>%

^1%."^...

.^JK^J^.

IMAGE EVALUATION
TEST TARGET (MT-3)

1.0 gia 1^

II I.I f '^ i^
11-25 11.4 11.6

Photographic

Sciences

Corporation

33 WEST MAIN STREET

WEBSTER, N.Y. M580

(716) 872-4503

■^

v «.^

1 tj \

ill

d!3

fltDROSTATICS.

SOLVTIOir.

47.8 and «'=^ 10.6.

CASTI.m,19S

Here to = 68.2, w':

Then s = iilli^ X «'=^ifei?::5 x 10.6:-. ''\^^''' =rimAr,..

58.2

EXERCISES.

68.2

265. A piece of copper (Spec. Grav. 8.850) weighs 446.3 grains

in liquid, and 490 grains out of it, required the Specific
Grav. of the liquid. Jln$. .789.

266. The Thousand-grain Bottle filled with Olive oil weighs

915 grains — what is the Specific Gravity of Olive Oil?

^ns. .915.

207. The Thousand-grain Bottle filled with mercury weighs
13596 grains — what is the Specific Gravity of mercury ?

Ans. 13.596.

268. A piece of cast-iron (Spec. Grav. 7.425) weighs 34.61

oz. in a liquid and 40 oz. out of it — what is the Specific
Gravity of the liquid? Am. 1.000 nearly.

269. A piece of gold (Spec. Grav. 19.360) weighs 139.85 grains

in a liquid and 159.7 grains in the air, required the Specific
Gravity of the liquid ? Ans. 2.406.

270. A piece of marble (Spec. Grav. 2.850) weighs 30 lbs. in a

certain liquid, and 35.9 lbs. in the air, required the
Specific Gravity of the liquid ? ^ns. .468.

197. The Specifiic Gravity of gases is found by exhaust-
ing a flask of atmospheric air and filling it with the gas in
question previously well dried. This is accurately weighed
and its weight compared with the weight of the same
volume of dry atmospheric air at the temperature of 60**
Fahr. and under a barometric pressure of 30 inches.

108. The following table gives the Specific Gravities
of the most common substances : —

ASTI.199.200.]

HYDROSTATICS.

TABLE OP SPECIFIC GRAVITIES.

OASES.

Siiu Atmospheric Air,

Hydrogen,

Oxygen,

Nitrogen, •

Ammoniacal Gas, ....
Carbonic Acid Gac, . .
Sulphurous Acid Gas,.
Chlorine,

LIQUIDS.

Distilled Water,

Mercury,

Sulphuric Acid,

Nitric Acid,

Milk

Sea Water,

Wine,

Olive Oil,

Spirits of Turpentine, .

Pure Alcohol,

Ether,

Prussic Acid,

SOLIDS.

Platinum,

Gold, .

Silver,

Lead,

1.000

.069

1.106

.972

.596

1.529

2.234

2.470

1.000

13.596

1.841

1.220

1.030

1.026

.993

.915

.869

.792

.715

.696

22.050
19.360
10.500
11.250

Copper, 8.850

Brass, 8.300

Iron, 7.788

Tin, 7.293

Zinc, 7.190

Diamond, 3.530

Flint Glass, S.3S0

Sulphur, 2.086

Slate 2.840

Brick, 2.000

Common Stone, 2.460

Marble, 2.850

Ivory, 1.825

Phosphorus, 1 . 770

Lignum Vitse, 1.350

Boxwood, 1.320

Potassium, 875

Sodium, 972

Pumice stone, 914

Dry Pine, 657

Dry Poplar 383

Ice, 865

Living Man, 891

Cork, 240

Graphite, 2.500

Bituminous Coal, 1 . 250

Anthracite Coal, 1 . 800

199. A cubic foot of pure distilled water at the tem-
perature of 60° Fahr. weighs exactly 1000 ounces. Hence
if the Specific Gravity of any substance be known the
weight of a cubic foot, &c., may be easily found.

For example.— the Spec. Grav. of Mercury is 13.596 water being 1.000,
and a cubic foot of water weighing lOUO ounces it follows that a cubic foot
of Mercury weighs 13596 ounces.

200. To find the solid contents of a body from its
weight : —

RULE.

Contents in feet ==^ "^i; where w z=z whole weight, andv/:=weight

of a cubic foot as ascertained from its Spec, Grav.

ExAMPLi 271. — How many cubic feet are there in 2240 lbs,
of dry oak (Spec. Grftv, '926.) ?

HYDROSTATICS.

[Abt. SOI.

BOLVTIOir.

Hero

IV 2240 lbs. 35840

to'

—im^cnhictect.

925 oz 025

Example 2Y2. — How many cubic feet are there in a mass of
iron which weighs 17829 lbs. ? ,

SOLUTIOIT.

Specific Gravity of iron — 7*788. Therefore 1 cu. ft. weighs 7788 o».
Then cubic feet in mass = 17829 lbs. -?- 7788 oz. =8G'628 Ans.

201. To find the weight of a body frora its solid con-
tents : —

RULE.
u)=icontents inft.xio'.
Where w and v/ are same as in last rule.

Example 273. — What is the weight of a block of dry oak
10 ft. long, 3 ft. thick, 2 J ft. wide.

Here 10 X 3 X 2i = 75 cubic feet.

Then «j=m)' x 75=925 oz. X 75=69375 oz.=4335 || lbs. Jnt.

Example 274. — What is the weight of a block of marble 8 ft.
long, 2 ft. wide, and IJ ft. thick.

SOLUTION. i'

CJubic feet of marble = 8X2X1^=24.

Spec. Grav. of marble— 2*850. Therefore one cubic foot weighs 2850 ob

Tlicn weight of block — ■ 2850 X 24 =68400 oz. = 4275 lbs. Ans.

EXERCISES.

275. What is the weight of a mass of copper which contains 29

cubic feet ? Jlns. 16040 lbs. 10 oz.

276. How many cubic feet are there in a mass of lead which

weighs seven million pounds? jlns. 9955.55 cub. ft.

277. How many cubic feet of sulphuric acid are there in

78124732 lbs? w«ns. 678976-48 cub. ft.

278. What is the weight of the mercury containe<1 in a rectangu-

lar cistern 6 feet long, 4 feet wide and 1 et deep, the
mercury filling it ? .. 203940 lbs.

279. If a block of zinc be 11 feet long by 3 feet wide and 2 feet

thick, how much does it weigh? ^ns. 29658| lbs.

280. What is the weight of a squared log of dry pine 44 feet

long and 18 inches square. Jlns, 4065 lbs. 3 oz.

[ABT. Ml.

ABTI. 202-20S.]

PNEUMITICS.

masi of

88 ox. '

>lid con-

' dry oak

lbs. jin$.
,rble 8 ft.

i;hs 2850 OS
s.

ntains 29
bs. 10 oz.

id which
cub. ft.

there in
cub. ft.

ectangu-
eep, the
3940 lbs.

nd 2 feet
658} lbs.

44 feet
lbs. 3 oz.

CHAPTER VI.

PNEUMATICS.

202. Pneumatics treats of the mechanical properties
of permanently elastic fluids^ of which atmospheric air
may be taken as the type.

203. The atmosphere (Greek atmoi " gases") or sphere
of gases is the name applied to the gaseous envelope which
surrounds the earth.

204. It is supposed, from certain astronomical con-
siderations that the atmosphere extends to the height of
about 45 miles above the surface of the earth.

NoTB.— The height of the atmosphere is only ^ of the radius of the earth,
BO that upon an artificial globe 12 inches in diameter the atmosphere would
be represented by a covering i^T of an inch in thickness.

206. Atmospheric air is a mechanical mixture chiefly
of two gases, Oxygen and Nitrogen in the proportion of
1 gallon of the former to 4 gallons of the latter. Its
exact composition, omitting the uqueous vapour, is as
follows : —

Composition BY Volume.

Nitrogen, '79*12 per cent.

Oxygen, 20-80

Carbonic acid, '04

Carburetted Hydrogen, -04

Ammonia, Trace.

"S OTIS.— Oxygen is the sustaining principle of animal life and of ordinary
combustion. When an animal is placed in a vessel of pure oxygen its
heart beats with increased energy and rapidity and it very soon dies fh>m
excess of vital action. Many svibstances, also, that are not at ail combus-

In its chemical nature it is distinguished chiefly by its negative properties.
In the atmosphere it serves the imjrortant purpose of diluting the oxygen
and thus fitting it for the function it is designed to perform in the animal
economy.

Carbonic acid is a highly poisonous gas, formed by the union of
oxygen and carbon (charcoal) . It is produced in large quantities during the
process of animal respiration, common combustion, fermentation, volcanic
action and the decay of animal and ve^table substances. Although when
inhaled, it rapidly destroys animal life it constitutes the chief source of food
to the plant. Animals take into the lungs air loaded with oxygen and throw
it off so charged with carbonic acid as to be incapable of again senriiig tov

PNEUMATICS.

[ABtt.206,M7.

tho purposes of respiration. Tho green parts of plants on the contrary,
absorb air, dcconiposo the carbonic acid it contains, retain the carbon and
ffive oir air contaninK no carbonic acid but a largo amount of oxygen. This
IS a most beautiful illustration of tho mutual dependence of tho different
orders of created boinga upon one another. Wero it not for plants the
air would rapidly become so vitiated as to cause the total extinction of
animal life ; were it not for animals, plants would not thrive for want of
the food now supplied in tho form of carbonic acid by the living animal.
As it is, the one order of beings pronarcs the air for the sustenance and
support of tho other, and so admiranly is the matter adjusted that the
composition of the air is, within very narrow limits, invariably the same.

The amount of carbonic acid varies fh)m 3'7 as a minimum to 6*2 as a
maximum in 10000 volumes.

Carburetted Hydrogen is produced during the decay of animal and vege*
table substances. It is one of the chief ingredients of common illuminat-
ing gas, and is poisonous to animals when present in the airs in large quan-
titios.

206. One of the most remarkable characteristics o^
gases, is the property they possess of diffusing themselves
among one another. Thus if a light gas and a heavy one
are once mixed they exhibit no tendency to separate again
and no matter how long they may be allowed to stand at rest,
they arc found upon examination intimately mingled with
each other. Moreover if two vessels be placed one upon
the other, the upper being filled with any light gas (hydro-
gen) and the lower with any heavy gas (carbonic acid)
and if the two gases be allowed to communicate with one
another by a narrow tube, or a porous membrane, a
remarkable interchange rapidly takes place, i, e., in direct
opposition to the attraction of gravity the heavy gas ascends
and the light gas descends until they become perfectly
mixed in both vessels.

NoTE.-The property of gaseotis diffusion has a veiy intimate bearing upon
the composition of the air. If either of the constituents of the air were
to separate from the mass, the extinction of life would soon follow. Besides
were it not for the existence of this property, various vapours would
accumulate in certain lucalitles, as large cities, manufacturing districts,
volcanic regions, &o., in such quantities as to render them totally unin*
habitable.

207. In addition to the gases already mentioned,
atmospheric air always contains more or less water in the
form of invisible vapour. This is derived partly from
combustion, respiration and decay, but chiefly from spon-
taneous evaporation from the surface of the earth. The
amount of invisible vapour thus held in solution depends
upon the temperature of the air being as high as ^^ of
the weight of the air in very hot weather, and as low aa
j\-^ in cold.

.:!

tTI. 106, 107.

AmT0. 806-tlO.]

PNBUtfATICS.

•»

e contrary,
carbon and
xygen. This
lio different

Jlanti the
inction of
for want of
niiK animal,
beuanoe and
ted that the
r the same.

I to 6*2 as a

al and vego*

1 illuminat-

large quan-

ristics o'
lemselves
leavy one
ate again
nd at rest,
^led with
)ne upon
.8 (liydro-
lic acid)
with one
brane, a
in direct
9 ascends
perfectly

aringupon
air were
T. Besides
urs would
: districts.
Ally unin>

ntioned,
in the

y from

spon-

The

lepends

,3V Of

low aa

n
1.

208. The blue color of the sky is due to light that has
suffered prolarization, and which is, therefore reflected
light, like the white light of the clouds. The air appears
to absorb to a certain extent the red rays and yellow rays of
solar li^lit and to reflect the blue rays. In the higher
regions the blue becomes deeper in colour and is mixed
with black. The golden tiuts of sunset depend upon the
largo amount of aqueous vapour held in solution by the air.

209. Air like all other material bodies possesses the

properties of impenetrability, extension, inertia, porosity,

compressibility, elasticity, &c. (See Arts. 11-18.

Note l.— The impenetrability of atmosphorio air is illustrated by
various experiments, among which are the following :

I. If an inverted tumbler be immersed in water the liquid does not rise
in the interior of the tumbler, because the latter is full of air and the water
cannot enter until the air has been displaced.

II. If th« two boards of a bellows be drawn asunder and while in ttaa^
position the nozzle of the bellows be closed, the boards cannot be pressed
together because the bellows is ftiU of air.

III. If an india-rubber bag or a bladder be inflated with air. and prefl>
sure applied, it is found that there is a material something within which
keeps the sides asunder,— that material something is atmospheric air.

Note 2.— The Inertia of atmospheric air is shown :—

I. By the force of wind, which is nothing more than air in motion.

II. By attempting to run on a calm day, carrying an open umbreUa.

III. By the apparent current of wind experienced on a perfectly calm day
by a person standing on the deck of a steamboat, or the platform of a rail-
way car when in rapid motion, which current is caused by the body dis-
placing the air.

IV. By causing a feather and a ball of lead to fall in avacuum, when it
is observed that they fall with the same velocity. In the atmosphere,
however, the ball falls faster than the featRer because it contains a greater
amount of matter with the same extent of surface as the feather, and
meets hence with less resistance fh>m the inertia of the air.

210. Air,. in common with all other forms of matter, is

acted on by the attraction of gravity, and hence possesses

weight.

Note 1.— This is the fundamental fact in the science of pneumatics. To
prove it we take a glass globe capable of containing 100 cubic inches, and after
weighing it accurately, withdraw fW)m it, by means of an air pump, all
the air it contains. When we weigh it again we find that its weight is
about 31 grains less than when filled with air.

1 00 cubic inches of Atmospheric Air weigh 31 grains.

100 " Oxygen " 34 "

100 •• Nitrogen " 80 "

100 " Carbonic Add " 47* "

100 ** Hydrogen " 2 "

Note t.— Although a small quantity of air when examined appears to be
almost imponderable, the aggregate weight of the entire atmoapn?r« is mofi
mouiibeini; equal to :

Ui !

PNEUMATICS.

CASTS. Ill-tlS.

I. Five thouMnd millions of millions of tons, or

II. A glob« of lead 66 miles in diameter, or

III. An ooean of water covering the whole surface of the earth to the
depth of 32 feet, or

IV. A stratum of mercury covering the entire surface of the globe to the
depth of 80 inches.

211. Since the air is ponderable and also compressible,
and since tbe lower stratum has to sustain the pressure of
the superincumbent portion, it necessarily follows that the
air is denser near the surface of the earth than in the
higher regions of the atmosphere.

212. The density of the air decreases in geometrical
progression, while the elevation increases in arithmetical
progression. That is at the height of 2'1 miles, the atmos-
pheric pressure is reduced to one-half, at twice that height
to one- fourth, at three times that height to one-eighth, &c.

NoTB.— The following table exhibits the density, elasticity and pressure
of the aii> at the different elevations given. Halley fixed the height at
which the pressure decreased to o ic-half at 8^ milci, but a more careful
collection, by Biot and Arago, of tho observations made on the Andes and in
balloons, respecting the upward decrease of pressure and temperature, has
led to the adoption of 2*7 miles as the point at which we may say that one-
half of the atmosphere is beneath us.

HTHGHT IS MILES.

DENSITY.

HEIGHT IN INCHES OP
COLUMN OB MERCUEY.

PEE88URE IN Ibs. TO
THE 8Q. INCH.

2.7

7.5

5.4

7.8

8.76

8.1

3.75

1.876

10.8

1.876

.987

18.6

.937

.468

16.2

'ek

.468

.284

18.9

.234

.117

21.6

sis

. .- "^

.058

24.8

' > .058

.029

27.0

To^rr

.029

.014

29.7

fif-r?

.014

•

.007

218. The pressure of the air is a necessary consequence
of its weight, and is equal, at the level of the sea, to about
15 lbs, to the square inch.

1X8.111-118.

uiih to the
riobe to the

pressible,

essure of

that the

1 in the

•metrical
bmetical
e atmos'
it heigfht
hth, (fee.

1 pressure
height at
>re careful
des and in
'atui'e. has
that one>

tr lbs. TO

[KCH.

▲111.214-118. J

PNEUMATICS.

[uence
aboiU

NoTB.— By saying that the pressure of the atmosphere is equal to 15 lbs
to the sq. inch, we mean that it is capable of baUncing a column of meronr/
SO inches in height, and a column of mercury 80 inches in height and having
a sectional area of 1 sq. inch weighs 15 lbs. Or in other words, that a co-
lumn of air having a sectional area of 1 sq. Hch, and extending from the
level of the sea to the top A the atmosphere v/eighs 15 lbs.

214. Air at 60® F. is 810 times as light as water, and
10466 times as light as mercury. It follows that the
pressure of the atmosphere is equal to that of a column of
air of the same density as that at the surface of the
earth 810 times 32 feet or 10466 times 30 inches in
height. That is, if the air were throughout of the same
density that it is at the level of the sea, it would extend to
the height of about 5 miles.

216. The particles of elastic gases, unlike those of
solids or liquids, possess no cohesive attraction, but on the
contrary a powerful repulsion, by means of which they tend
to separate from one another as far as possible.

216. Permanently elastic fluids such as {«tmospheric
air, and certain gases, are chiefly distinguished from non-
elastic fluids, such as water, by the possession of almost
perfect elasticity and compressibility.

Note.— Air and certain gases as Oxygen, Hydrogen, Nitrogen, ^c, are
called permanently elastic to distinguish them fh>m a number of others
as CarDonlo Acid, Nitrous Oxide, &c., which under great pressure and
intense cold pass first into the liquid and finally into tne solia state.

217. If a liquid be placed in a cylinder under the piston
it will remain at the Fame level, no matter to what height
the piston may be raised above it, but if a portion of air or
any other elastic gas bo thus placed in the cylinder and
the piston be air tight, the confined air will expand upon
raising the piston and will always fill the space beneath it,
however great this may become. This expansibility or
tendency to enlarge its volume so as entirely fill the space
in which it is inclosed is termed elasticity.

NoTB.— It is obvious that the elasticity of air is due to the repulsive
power possessed by the particles.

218. The law determining the density and elasticity of
gases under diflerent pressures was investigated by Boyle
in 1060, and afterwards by Mariotte.

100

PNEUMATICS.

LAlT.tl9.

Fig. 20.

' (

NoTB.— To lUuitrat« thif Uw we t»k« a bent glaii
tube Fif^. IuitIdk one limb A muoh loncer than the
other. Trhe longer limb it open »ud the snorter fur-
nished with % nop-coek.

Both ends beinc open a quantity of mrircury is
poured into the tune and of course rises to the same
level in iKtth legs— the surfkco of the morriivy at A a,
sustaininK the weight of a column of air extending to
the top of the atmosphere. We now close the stop*
cook and thus shut on the pressure of the atmosphere
above that point, so that the surfsrce a, cannot bo
affected by tne weight of the atmosphere—/, e., cannot
be influenced by atmospheric pressure. We find,
however, tlmt the mercury in both limbs remains at
the same level, fk'om^which wo infer that the clastic
force of the air conflned above a is equal to the weight
of the whole column on a before the stop cock was
closed.

Hence the elasticity ef tho air is equal to its weight,
which is equal to a column of mercury 80 inches high.

If now wo pour mercury into the tube until ino
air conflned above a is compressed into half its former
volume, i. e., until the mercury rises to b in the shorter
tube, we shall find that the column of mercury bB\H
exactly SO inches in length, or in other words, we have
doubled the pressure on the air con>lned in the shorter
tube and have decreased Its volume to one«half its
former dimensions, and at the same time!doubled its
elastic force since it now reacts against the surface of
the mercury with a force equal to 60 lbs. to the square
inch.

If we increase the height of the mercury in the
longer leg to 60 inches above its height in the shorter
leg, we shall compress the air into one>third its original
volume and at the same time treble its elasticity, and
■o on. Hence the law of Mariotte.

219. Mariotte's law may be thus enun-
ciated.

I. The density and elasticity of a gas vary
directly a$ the pressure to which it is subjected.

II. The volume which a gas occupies under
different pressures varies inversely as the force
of compression.

NoTB.— Becent researches tend to prove that Muriotte's law is true only
within certain limits, and that all gases vary from the law when subjected
to very greatpressures, their density increasing in a greater ratio than their
elasticity. With atmospheric air tne law holds good to a far greater extent
than with any other gas, the correspondence being found to be rigidly exact
when the air in expanded to SOO volumes, and also when it is compressed
into ^ ofits primary volume.

Mariotte's law would require the air to be indefinitely expansible while
we know that there is, beyond all doubt, an upward limit to the atmos-
phere. Dr. Wollaston imagines that when the particles of air are driven
a certain distance apart by their mutual repulsive power, the weight
of the individual particles comes at last to balance this repulsive force
l»pd thxa prevent their further divergence. If this be the case as is

Am. no-ill.]

l»WBTjlfATIC8.

101

probable from various oonsidorations, there li a limit to the rareflM*
tion of a ni, arriving at which tlie gas ceases to expand further and oomes
to have a true uppor surflMo like a liquid. As has been already remarked
this pxaot limit and uppor surftMe of the atmosphere is supposed to be at
an elevation certainly not greater than 46 mites— Blot Axes it at 30 miles.

220. The air pump, as its name implies, is an instrument
used for pumping out or exhausting the air from any closed
vessel.

221. The bell-shaped glass vessel usually attached 'to
the air-pump is called a Receiver^ and when the air is
exhausted as fiir as practicable from this a vacuum is said
to have been produced.

NoTB.— The air pump was invented by Otto Guericke, a celebrated
JhirKoniastcr of Magdnburg, in the year ISOO. At the close of the Imperial
Diet in 1564, ho exhibited hiH first public experiments with it before the
omporor and assemblitd princes aiici nobles of Germany. On this occasion
hoexhaustod the air from two 12-inoh hemispheres fitted together by ground
edges and greatly astonished his noble audience by showing that the oom«
l)incd strength of 12 horses was insufflciont to puu them asunder.

The exhausting syringe of Otto Guericke was so imperfect in its action
that while using it no was compelled to keep it immersed in water to pre>
vent the inward leakage of the air. Since his time, however, the attention
of many eminent men has been directed to the subject, and the form and
construction of the air-pump have been very greatly improved.

222. The exhausting syringe which is the essential part

of an air-pump, consists of a brass cylinder ahcdy supplied

with an air-tight piston efy and an arrangement of valves

hk^ by means of which the air is permitted to pass out

from the receiver q and through the piston ef^ but not in

the contrary direction.

Note.— When the piston ef is raised the valve h closes, and ai the piston
in its ascent produces a partial vacuum be-

neath it, the air contained in the receiver o
opens the valve k by its expansive power and
thus refills the cylinder abed. Now when the

{listen is forced down again, the air contained
n the cylinder tends to rush back into the re-
ceiver but in doing so closes the valve k, and
has therefore no other mode of escape than
through h, thus passing above the piston to bo
lifted out at the next stroke. In this manner
the air continues to bo exhausted until what
remains in the receiver has not sufficient ex-
pansive power to open the valve k, when the
exhaustion is said to be complete.

Pig. 21.

•a

229. The principle upon which the air-pump acts is
the elasticity or expansibility of the air, and since in order
to enable the pump to act, the air contained in the receiver
must possess sufficient elastic force to raise the ralve, it

)lll

109

PNEUMATICS.

UBTt.ni,MO.

followt lliat n perfect vAcuum cnnnot bo Hecured by the

air-pump. Thus, puin|)8 of common cotiHtruction will not

withdraw more than ^^^ of the containod air, but the

improvod form is said to oxhauRt i^^niVr*

NoTR.— ir we suppoie thn oylindcr of thfloxhauNtinfr nrriiifrn tn htvn ilin
iianiP offootive (»p»oUy m thn roorlver, and that the uiHtoii jikmivi at «a<ih
■tmke tho whole Icngih of thn oylindor, it la evidont tliat (n raining tlin
piaton to tliu tup of the oyliiidnr and tlieii dupreMing it again to the l>ottoni.
onn-hair of tho air will have i)aHMfld fhim tho reoolvnr ; tlin muiaining half
ooinpletnlv Ailing it, but having only half an much doniity and ('InHHicity ax
before. Tlio Hooond «troku of thn piHton will nHluoo the quantity, douMity.
and elasticity, to ono-fourth, the third to ono-eighth, and no on an eihlbitou
by tho following table :—

HTROKR. nOKBOUT.

l8t,

and,

3rd,

4th,

Rth,

6th,

7th,

8tb,

9th,

1
I

of
of
of
of

0fT»»
Of Vf
Of if* =»

ofiir =

LBFT IR
VKH8B1.,

BLASTIO rORCR Of THB KBMAtNPKK.

= \
= k

\ of fig =rl»

IR in. of mercury, or 7.35 lbs per sq. in.

7Jin.of

3) in. of

1.876 in. of

0.937 in. of

0.468 in. of

0.234 in. of

0.1l7in. of

0.068 in. of

II
It

or 3.676
or 1.837
or .918
.469
.229
.114
.067
.028

or
or
or
or
or

do.
do.
do.
do.
do.
do.
do.
do.

224. The condensing tyringey which is used for forcinp;
air into a receiver or condensing chamber, differs from an
exhausting syringe only in tiio fact that its valves open
inward towards tlic chamber instead of outward.

226. Tho Air-pump is cinefly employed to illustrate

tho pressure and elasticity of the air.

NoTR 1.— The preamre of the atmosphoro may bo shown by innunierablo
cxperinionts among which are the following :—

I. Whon the air is exhausted fTom the receiver of an air-pump the re«
ceiver is lirnily fastened to the plate and cannot be reraovea until the air
is ro-admitted.

II. The hand placed on the open end of the receiver it pressed inward
with a force sufflcieutly groat to cause pain.

III. Thin square glass*tubes are crushed when the air is exhausted flroui
them.

IV. In the surgical operation ot cupping, the air is removed from a small
cup which is then placed over an opened vein ; tho pressure of the air on
the surrounding parts causes the blood to flow rapidly into the cup.

V. When a cask of boor is tapped, the beer docs not run until a small
hole called the vent-hole bos been made in the upper part of the cask.
Through this the atmospheric air enters and pressing on the surface of the
boer with a force of 16 lbs. to the square inch, forces it through the tap.

VI. The useful small glass instnunents cMed pipettes act upon theprin-
oiple of atmospheric pressure.

Til. A hole is usually made in the lid of a tea-pot so as to bring into play
the presaure of the atmosphere and thua cause tho beTerage to flow more
rapioljr.

lTt.fUillO.

hI by the
I will not V
but the '

to liAVfl tlin
iNOH Rt ua«^h
raiitliiK tliH
llio lM)ttoni,
ittinliiK >ial^
'laHMlcity ax
ity, (liMiNlty.
M eiliiliitoil

INDKK.

poraq. in.
do.
do.
do.

do. ♦
do.
do.
do.
do.

r forcinpf

from All

ves open

llu»trato

iiuniorablo

^ the r«>
ktil th« air

;d inward

sted flroui

ma small
)he air on
ip.

a small
the cask.
,co of the
le tap.

the prin-

into play
low more

ABTi.tM.tr.]

PKEUMATICS.

103

VIII. Flioii walk on fflaas or on thn CftlllnR by prodndniv a facuum under
eaoh foot whloh in thuit prcnnod affaluMt tho Hiirfkoo with a force ■ufllclnnt
to sustain the weight of tlio inm><<t. Thn Om'ko, a Houtli .\mnrlcan litard,
haNanimilar apparatun ntttuOuMl to cacli foot. And wiililn the paat few
yearn a man h«N unccetMltMi In walking acroNn acoiliuic with hid hond down-
ward*, by altornatoly withdrawioR and admitting the air between hiii feet
and tliecelllnR.

IX. Pneumatic chemistry, /. r*., the mode of oolleoting gMci ov«r water
depends upon tlie prim^ipltt of atinosphorlo prossurin

X. If a tumbler or other glasM vohmoI be lllled with water and covered with
apiece of paper, and tln^ linnd bo thru placed tlrnily on tlio uapor anl tin*
whole suddenly and oardfully iuvcrtcd, thn wator does not (low out of the
voNHcl upon roinovluK the hand— bulng hold by the upward pressure of the
ntmoNphoru.

XI. Suction is the effect of atmospheric pressure, as illustrated by Ji'ato •
!nff liquids into the mouth, aUo by thn leather tucker used by boys.

XII. Thn presHurn of the air Ih nIiowii by tho fkMt that it supports or ha*
lancos a column of mercury 30 inches or a column of water 32 feet in lietKht.

XIII. The prnnnure of tho atmosphere retards ebullition or boiling.
Thus if some boiliuR water be partially cooled and thim planed under thn
receiver of an air-pump and tho air oxImuMted, tiin wntnr rei^ommences to
boil, owiuff to the decreasnd nnmsurc. Or if Noine watt^r be boiled in a small
flask and the flask corked wnilo the water Is iMiiiuK, upon allowing the
water to cool partially and then plunging the small flask In a largo vesnel
of cold water, the water in the flask again begins to boil ; thn reason iH, tho
cold water oondenHos the vapour in the upper part of tho flask and thus
produces a partial vacuum.

NuTB 2.— Tho elasticity of tho air uitvy be shown by various oxporim nts
among which arc tho following :—

I. The exhaustion of the receiver of tho air-pump is a proof of the ol tsli-
city of the air.

II. The elasticity of the air is shown by placing a thin square bottle with
its mouth closed, under tho receiver, and exhausting the surrounding air.
tho bottle is broken by tho elastic foroo of tho contained air.

III. When some withered fruit, as apples, flgs, or raisins, with unbroken
skins arc placed under tho recuivor, and tho surrounding air exhausted,
they become plump nrom tho elasticity of the included air.

IV. The elasticity of tho air is shown by tho operation of the air-gun.

v. The elasticity of the air is taking advantage of in applying air as a
stuiling material for cushions, pillows and beds.

226. The barometer (Greek haros " weight" and metreo
"I measure") is an instrument designed to measure the
variations in the amount of atmospheric pressure.

Note.— The barometer was invontod about tho middle of tho 8ev;>n-
toenth century by TorriooUi, a pupil of the celebrated Galileo.

227. The essential parts of a barometer are : —

let. A well formed glass tube 33 or 34 inches long,
closed at one end and having a bore equal throughout, of
two or three lines in diameter. The tube contains pure
mercury only, and is so arranged that the mercury is sup-
ported in the tube by the pressure of the atmosphere; and

^HVmUTiOS.

[▲«n.ttH8l'

I' >

i \

i i

2nd. An Attached graduated scale and various appliances

for protecting the tuDc and ascertaining the exact height

of the column of mercury.

NoTB.— The vacant space between the top of the column of mercury and
the top of the tube is called the Torricellian vacuum, in honor ot the
inventor of the barometer, and in a good instrument is the most perfect
vacuum that can be produced by art.

228. The excellency of a barometer depends principally
upoii the purity of the mercury in the tube and the perfect-
ness of the Torricellian vacuum.

The value of the instrument may be tested :—

1st. Br the brightness of the column of mercury, and the absence of any
■peck, flaw, or dullness on its surface.

2nd. Bs the barometric light ; i. e., flashes of electric light produced in
the dark in the Torricellian vacuum by the friction of the mercury against
ihe glass.

Srd. By the clearness of the ring or clicking sound produced by making
the mercury strike the top of the tube, and which is greatly raodifica
when any particles of air are present above the column.

229. The cause of all the oscillations in the barometer

is to be fouud in ' the unequal and constantly varying dis-

tribalion of heat over the earth's surface. If the air is

much heated at any spot it expands, rises above the mass

of air, and rests upon the colder portions surrounding it.

Tlie ascended air consequently flows off laterally from

above, the pressure of the air is decreased in the warmer

place and the barometer falls. In the colder surrounding

places, however, the barometer rises, because the air that

ascended in the warmer region is diffused over and presses

upon the atmosphere of these cooler parts.

NoTK.— It is found that the fluctuations in the height of the barometer
var^ greatly in extent in different latitudes— being so small in tropical
regions as almost to escape notice, and comparatively so fitful and extreme
in the temperate and firigid zones as to defy all attempts at reducing them
to any system. In our climate the column varies in height fVom a little
over SO inches as a maximum to a little over 27 inches as a minimum.
Within the torrid zone the column of mercunr scarcely ever exhibits any
disturbance greater than what would occur in Canada before a slight thun-
der storm— but such a disturbance is there the sure and rapid precursor of
one of those mighty atmospheric convulsions which sometimes desolate vast
regions and which are frequently as disastrous in their efiTects as the most
violent earthquakes.

230. Besides the irregular fluctuations depending upon
the weather, the barometer is subject to regular semi-diurnal
oscillations depending upon atmospheric tides, caused by
the heat of th« sun— the two maxima of pressure always
occurring at about 9 a.m. and 9 p.m. and the two minima
at about 3 a.m. and 3 p.m.

Abts. 231. 232.]

PNEUMATICS.

105

XoTB.— The semi-diurnal oscillation is greatest at the equator, where it
averages one-tenth of an inch — diminishing to six-hundredtht of an inch
in lat. 30**, beyond which it still decreases, and in our climate becomes
completely masked by the irregular fluctuations peculiar to the temperate
and frigid zones.

231. USE OF THE BAROMETER AS A WEATHER GLASS.

I. The state of the weather to he expected depends not so much
upon the absolute height of the column of mercury as upon the
RAPIDITY AND EXTENT OF ITS MOTION whether rising or falling.

Note.— If the mercury have« convex surface the column is rising; if
the surfiMse is concave the column is falling, when the surface is flat the
column is usually changing ft'om one of these states to the other.

II. A fall in the barometer generally indicates approaching rain, •
high winds, or a thunder storm.

III. j1 rise in the mercury commonly indicates the approach of
fine weather ; sometimes, however, it indicates the approach of a
snow storm.

TV. A rapid rise or fall in the mercury indicates a sudden
change of weather.

V. A steady rise in the column, continued for two or three daysf
is generally followed by a long continuance of fine settled weather.

VI. A steady fall in the column, continued for two or three days,
is commonly followed by a long continuance of rainy weather.

VII. A fluctuating state in the height of the mercury coincides
with unsettled weather.

Note.— The barometer is far more valuably as a means of ascertaining
approaching changes in the state of the wind than in foretelling the ap-
proach of wet or dry weather.

232. To ascertain the height of mountains, &c., by the
barometer. •

HALLEY'S RULE.

I. Find the logarithm corresponding to the number which ex*
presses the height in inches of the column of mercury in the baro-
meter at the level of the sea.

II. Find also the logarithm corresponding to the number which
expressesHn inches the height of the column in the barometer at the
top of the mountain or other given elevation,

III. Subtract the latter of these logarithms from the former i,
multiply the remainder by' the constant number ^ 62170, and the
result will be the elevation in English feet >

Note.— The nimiber 62170 in this rule, and 63046 in the following, were
selected by Halley from certain mathematical reasons into whicn it is
unnecessary to enter.

Example 281. — On the top of a certain mountain the barometer

stands at the height of 21*793 inches, while on the surface of

1
!

1 1 ill

I i ;

I l;i

I-

106

PNEUMATICS.

[ABT. 282.

the oartb. it stands at 29* 780 inches ; required the height of the
mountain? ^ - -^

SOIUTION.

Logarithm of 29*780 = 1*473925 aud logarithm of 21*793 = 1*828317.

Then from 1*473925
Subtract 1*328317

Remainder = '145608X62170 = 9062 feet. Ant.

RULE WITH CORRECTION FOR TEMPERATURE.

I. Obtain^ as before, the difference between the logarithms of
the numbers expressing the heights at which the mercury stands at
the surface of the earth and on the summit of the mountain.

' II. Multiply this difference by the constant number, 63946 — the
result IS the elevation in feet, if the mean temperature of the surface
of the earth and the elevation is 69*68° Fahr.

III. If the mean temperature of the two elevations be not 69*68°
Fahr., add j^t^ of the whole weight found for each degree above
69*68°, or subtract the same quantity if the mean temjterature be
below.

ExAMPLR 282.T-Humboldt found that at the level of the sea,
near the foot of Chimborazo, the mercury stood at the height of
30 inches,'while at the summit of the mountain it was only 14*85
inches. At the same time the temperature at the base of the
mountain was 87'=' Fahr., and at the top 50*40^ Fahr. What is
the height of Chimborazo ? \

SOLUTION.

Log. of 80 = 1*477121, log. of 14.85 = 1*171724 and mean temperature =
87°+50*4'' 137*4" _ „
2 ^^ ~ ^'^^ •

Then 1*477121 — 1*171724 = *306897.

And •306397X63946= 19539 feet.

Since the mean temperature of the two stations is 1^ less tlian 69*68**, we
deduct j^ of the elevation found.

fhs ot 19539 = 40*7 ft. and 19539 — 40'7 == 19498*3 ft. Ans.

LESLIE'S RULE.

rOR HKASmtlNa HBIOBTB BT THK barometer WITHOUT THE USE OF

LOGARITHMS.

I. Note the exact height of the column of mercury at the base
and at the summit of the elevation.

II. Then say, as the sum of the two pressures is to their difference ^
so is the constant number 62000 to the answer in feet.

Example 283. — The barometer in a balloon is observed to stand
at a height of 22 inches, while at the surface of the earth it
stands at 29*8 inches ; what is the elevation of the balloon 7

Abt. 233.]

PNEUMATICS.

107

empcrature =:

Ihan eO'CS", we

THB rSB OF

22+29-8 !
Or, 61-8 : 7*8

29*8

52000

BOLnioir.

- 22 : : 52000 :
52000X7*8

Ans. '
= 7837.8 ft. Ans.

51-8
EXERCISES.

284. At what height would the mercury stand in the barometer

at an elevation of 29'T miles above the earth's surface ?

jlns. 0-0146 inches.

NoTB.— Divide 29*7 by 2*7 (See Art. 212), the quotient is 11, then divide
80 inches by 2", t. e. 2C48, and the result is tlio answer.

285. At what height will the barometer stand in a balloon which

is at an elevation of 16} miles? ^ns. '46875 inches.

286. *It is observed that while the barometer at the base of a

mountain stands at a height of 30 inches, at the top of
the mountain it stands at a height of only 18 inches,

required the height of the mountain.

287. *While the mercury at the base of

a mountain stands at the height
of 29-5 inches, at the summit of
the mountain the barometer indi-
cates a pressure of only 20-4 inches,
what is the height of the moun-
tain ? ^ns. 9482-9 feet.

288. fWhile in a balloon the barometer

indicates a pressure of only 19
inches, at the surface of the earth
the pressure is 29-94 inches — tak-
ing the mean temperature of the
two stations as 72-50">, what is
the elevation of the balloon ?

• ^ns. 12708 feet./

233. The common pump consists
of a barrel SB, a tube AS, which
descends into the writer reservoir, a
piston cdj moving air-tight in the
barrel and two valves, v and .r, which
act in the same manner as in the ex-
jausting syringe of the air pump.

NoTfi 1.— "When the machine begins to act
the piston is raised and produces a vaoiuim
below it in the barrel, and the atmospheric

Jns. 13000 feet.
Fig. 22.

* Use Leslie's rule.

t Use Halley'a rule with correction for temperature; i. e., the second <^f
the rules given.

l08

PNEUMATICS.

[AbTS. 284. 235.

I -v

1 ■

pressure on the water in the reservoir forces it up the tube and throueh
the valve x into the lower part of the barrel. As the piston descends tno
valve .r closes and the water contained in the barrel passes through the
valve V above the piston, to be lifted out at the next stroke. Hence the
common pump is sometimes called a lifting pump.

Note 2.— Since the specific gravity of mercury is 13"596 and the i)ressure
of tlie atmosphere sustains a column of mercury 30 inches in height— it
follows that atmospheric pressure will sustain a column of water SOX 13*590
inches, or 34 ft. in neight. Hence the vertical distance of the valve x above
thesurfaceofthewaterinthereservoirmustbeless Fig. 23.

than 34 feet, or taking the variations in atmospheric
pressure into account, about 82 feet.

234. The forcing pump consists of
a suction pump Aj in which the piston
/* is a solid plug without a valve.
When the piston P descends the
valve V closes and the water is forced
through the valve v' into the chamber
MN". The upper part of this chamber
is filled with compressed air, which,
by the pressure! it exerts against the
surface of the water, ww' drives it
with considerable force through the
pipe or tube HG. "PfF^

NoTK.— Sometimes the forcing pumn is used
without the air chamber, MN. Fig. 23 exhibits the A
arrangement of the valves, &c., in a common ftre
engine with the exception that there is another f
similar forcing pump on the other side of the
air chamber. HO represents the tube leading
to the hose.

235. Tbe Syphon is a bent tube oi glass or other

material having one leg somewhat longer than the other

and is used for transferring liquids from one vessel to

another.

Note.— The machine is set in operation by immers-
ing the shorter leg in the liquid to be decanted, and
sucking the air out of the tube, when the pressure of
the atmosphere forces the liquid into the syphon over
the bend and down through the longer leg instead of
sucking the air out of the syphon, the instrument may
be set m operation by first filling it with the liquid

Fig. 24.

and, while thus full, placing the finger over each end,
and immersing the shorter leg in the liquid.

Note 2.— In order to understand why one limb
must be shorter than the other, it is only necessary to
remember that the pressure of the atmosphere acts as
much at one extremity as at the other. If we raise
the column ofliquid as far as B, by sucking at the ex-
tremity C, and then withdraw the mouth, the water falls back into tho

..».

ABT8. 23A-241.]

DTNAMICS.

109

vessel F. The column will likewise run hook if wc get it no farther than
L, which is the level of the water in the vessel F, because at that point the
upward pressure of the atmosphere prevails over the downward pressure
of the liciuid, but if wo got the column below L, the downward pressure of
the liquid exceeds the upward pressure of the atmosphere and the liquid
will flow.

Thus the motion of the fluid in the syphon is similar to the motion of a
chain haneing over a pulley,— if the two parts of the chain bo equal, the
fluid remains at rest, but if one end be longer than the other, it moves
in the direction of the longer, and fresh linics, so to speak, are added con-
tinuously to the fluid chain by the atmospheric pressure exerted on the
surface of the water.

CHAPTER Vlf. ^

DYNAMICS.

236. When the forces which are the subject of inves-
tigation are balanced, the consideration of them properly
comes under the science of Statics, but when they cease to
be bahinced, and the body acted upon is set in motion
other principles become involved, and the investigation of
these constitutes the more complex science of Dynamics.

237. Statics is a deductive science, ^mcQ all its facts arc
deducible, like those of Arithmetic and Geometry from
abstract truths ; dynamics is an inductive, experimental or
physical science, many of its principles being capable of
proof only by an appeal to the laws of nature.

238. Force may be defined to be the cause of the
change of motion, i. e., force is required : —

1st. To change the state of a body from rest to motion
or from motion to rest.

2nd. To change the velocity of motion.

3rd To change the direction of motion.

239. Forces are either m^tontoneows or continued, and
continued forces are either accelerating, constant or retard-
ing.

240. Motion may be defined to be the opposite of rest
or a continuous changing of place.

241. Motion has two qualities, direction and velocity,
and is of three kinds,

! f

r'l

110

DYNAMICS.

[AST8. 24^251.

Ist. Directf

2nd. Rotatory or Circular ; and

3rd. Vibratory or Oscillatory,

242. An accelerating^ constant or retarding force pro-
duces an accelerated^ uniform or retarded motion.

243^. Velocity is the degree of speed in the motion of
a body and may be either uniform or varied. It is uniform
when all equal spaces great or small, are passed over in
equal times.

244. The principles of the composition and resolution
oi force are equally applicable to motion,

245. Momentum or Motal Force or Quantity of
Motion is the force exerted bv a mass of matter in motion.

246. The momenta of bodies are proportional to their
weights, multiplied by their velocities.

247. When the velocities of two moving bodies are
equal, their momenta are proportional to their masses.

248. When the masses of two moving bodies are
equal, their momenta are proportional to their velocities.

249. When neither the masses nor velocities of two
moving bodies are equal, their momenta are in proportion
to the products of their weights by their velocities.

Note.— When we speak of multiplying a velocity by a weight, we refer
to multiplying^ the number of units of weight by the number of units of
velocity, and it makes no difference what units of each kind are employed
for the product, thus obtained, means nothing by itself, but only by com-
parison with otlier products similarly obtained by the use of the same
units.

For example, when we say that a weight of 11 lbs. moving 6 feet per second,
lias a momentum of 66, all we mean is, that in this case the weight strikes
a body at rest with 66 times the force that a body weighing one lb. and
moving only one foot per second would exert.

250. If a moving body M, having a velocity F, strike
another m at rest, so that the two masses shall coalesce,
and move on together with a velocity v, then Jf X F=
(M-|-7?i)Xv; or whatever momentum may be acquired
by the body m must be lost by M.

251. If a moving body M having a velocity F", strike
another body m moving in the same direction with a
velocity v, so that the two may coalesce, and move on

tI8. 242-261.

AHT8. 252, 258.]

DTNAMICS.

Ill

together with a velocity vel^ — then Myc F+mXvrs
{M^-{'m)Xvelj or \n other words, the two hodies united
nave the same momentum that they separately had before
impact.

252. If a moving body Jlf having a velocity F", strike
another body m moving with a velocity v, in the opposite
direction, so that the two masses shall coalesce and move
on together with a velocity vel — then MXV<^mXv=z
(Jl/4"''*)Xve^ or in other words the body moving with
least force will destroy as much of the momentum of the
other as is equal to its own momentum.

253. If a moving body Jf, having a velocity F, strike
another body m moving obliquely towards it with a velocity
V, so that the two masses shall coalesce and move on together,
then by representing their momenta, just before impact
by lines in the direction of their motion and completing
the parallelogram, the diagonal will represent the quanti-
ty and direction of the momentum of the combined mass.

Example 289. — What is the momentum of a body weighing
78 lbs. and moving with a velocity of 20 feet per second ?

SOLUTION.

Momentum = 78 X 20 := 1560. Ans.

That is, the momentum of such a body is 1560 times as grc9<t as the mo*
mentum of a body weighing only 1 lb<, and moving only 1 ft. per second.

Example 290. — If a body weighing 67 Iba. be moving with
the velocity of 11 feet per second and strike a second body at
rest weighing 33 lbs., so that the two bodies may coalesce, and
move on together, what will be the velocity of the united mass 7

SOLUTION.

Art. 250.— If M be the moving body, V its velocity, m the body at rest
and V the velocity of the united mass ;—

Then (M+m) X « = Mx V and therefore v = ^-V^

M+m
In this example, Jf ;= 67, F = 11 and m = 33.

Thenv :=

itfxr

M-\-m

67X11 737 ».„^. ^ , .

gyqigg = j^ =7*37 feet per second. Ans.

Example 291. — If a body weighing 50 lbs. and moving with
a velocity of 100 ft. per second, come in contact with another
body weighing 40 lbs. and moving in the same direction with
a velocity of 20 feet per second, so that the two bodies coalesce
and move on together, what will be the velocity and momen-
tum of the united mass ?

111^

DYNAMICS.

[Abt. 268.

80LUTI0V.

Art. 2S1.— If ^and m bo the two bodies, and V and v their separate
velooities and tel the velocity of the united mass :—

Then (M+m)X vel = Mx V+mXv. Hence vel = ^^^"*^^

In this example M => tiO, m = ^0, V=* 100 and v = 20.

rri, Tr » _ ^ X F + wt X t>_ BO X 100 + -to X 20 _ BOOO + 80 6800
Then vol ^^- _ ^^— _ - _ —

-■ 64| ft. per see., and momentum = (60 + 40) X 64| = 6800. Ant,

Example 292. — If u body weighing 120 lbs., and moving to
the east with a velocity of 40 feet per second, come into contact
with a second body weighing 90 lbs. and moving to the west,
with a speed of 80 feet per second, so that the two bodies coal-
esce and move onward together, in what direction will they
move, with what velocity, and what will be their momentum ?

SOLUTION.

From Art, 252. if M and m be the bodies, and V and v their respective
velocities, and vel. the velocity of the united mass after impact :—

Then ( Jf + m) x vel. =Mx V r^m Xt» and hence

, af X v^ky.v

vel. = iTT

In tUs example M = 120, w = 90, F = 40 and v = 80.

_. ^ , ilfxr^wtX» (120 X 40) r^ (90 X80)

Thenuei. = - —

Jf+«»

120 + 90

4800^7200
210

2400

= —^ = llf feet per second = the velocity. 11 ^ X (120 + 90) = 11^ X

210 = 2400 = momentum.

And since 90 X 80. the momentum of the body moving to the west is
greater than 120 X 40, the momentum of the body moving lo the east, the
united mass moves to the west.

EXERCISES.

293. What is the momentum of a body weighing 79 lbs. moving

with a velocity of 64 feet per second ? Ans. 5056.

294. Which would strike an object with greatest force, a bullet

weighing one ounce and propelled with a velocity of
2000 feet per second, or a ball weighing 5 lbs. and thrown
with a velocity of 28 feet per second ?

Ans. momentum of bullet = 125.
" of ball = 140.

Therefore the ball would exert
most force of impact.

295. Which has the greatest momentumi a train of cars weighing

1*70 tons and moving at the rate of 40 miles per hour, or
a steamer weighing 790 tons and moving at the rate of 9
miles per hour ? Am. momentum of train = 6800, of

[Abt. 268.

Iieir separate
m

1-800

moying to
ito contact
) the west,
odies coal-
i will they
nentum ?

ir respective

t:—

4800 ^ 7200
210

0) = 11^ X

the west is
le east, the

moving

ns. 5056.

a bullet
locity of
i thrown

t=125.
1 = 140.
d exert

eighing
10 ur, or
ate of 9
800, of

AST. 258.3

DYNAMICS.

113

steamer = 7110, and therefore the latter has most mo-
mentum.

290. If a body weighing GO lbs. and moving at the rate of 86
feet per second, come in contact with another body weigh-
ing 400 lbs., and moving in tlie same direction at the rate
of 12 feet per second, so that the two bodies coalesce and
move on together ; what will be the velocity and mo-
mentum of the united mass?
Arts, velocity = 21^^ feet per second ; momentum =: 9960.

297. If a body weighing 56 lbs. and moving with a velocity of

80 feet per second come in contact with a body at rest,
weighing 70 lbs., so that the two bodies coalesce and
move on together ; what will be the velocity of the united
mass ? jlns. 35-/*u^j feet per second.

298. If a body weighing 77 lbs. and moving from south to north,

with a velocity of 40 feet per second, come in contact
with another body weighing 220 lbs. and moving from
north to south, with a velocity of 14 feet per second, so
that the two bodies coalesce ; in what direction and with
what velocity does the united mass move?
Ans. Their momenta exactly neutralize each other and
the bodies come to a state of rest.

299. If a body weighing 70 lbs., moving to the south with a

velocity of 70 feet per second, come in contact with
another body which weighs 80 lbs. and is moving to the
north with a velocity of 60 feet per second, so that the
two bodies coalesce and move on together ; in what di-
rection will they move and with what velocity and mo-
mentum ? Jins. To the south with velocity of 8 inches
per second. Momentum of united mass = 100.

300. If a body weighing 600 lbs. and moving to the west with a

velocity of 40 per second, come in contact with a second
body weighing 50 lbs. and moving to the east with a
velocity of 20 feet per second, and after the two have
coalesced they come in contact with a third body which
weighs 100 lbs., and is moving in opposite direction with
velocity of 150 feet per second, and the three then coa-
lesce and move on together ; in what direction will their
motion be and what will be the velocity and momentum
of the united mass ? Ans. Direction, west.

Velocity = 10§ feet.

Momentum = 8000.

114

DYNAMICS.

CABTI.U4-SM.

264. When force is communicated by impact to a
body at rest, the body will remain at rest until the force
is dUtrihuted throughout all the atoms of the mass, unless
a fragment be broken off by the force of impact, in which
case this fra^ nent alone moves.

LAWS OF MOTION.

255. The first law of motion. — Every body must persevere
in a state of rest or of uniform motion in a straight line, unless it
be compelled to change that state by force impressed upon it.

256. The second law of motion. — Every change of motion
must be in proportion to the impressed force, and must be in the
direction of that straight line in which the impressed force acts,

267. Tnino law of motion. — ^11 action is attended by a corres-
ponding re-action, which is equal to it in force and opposite in
direction.

These laws are commonly known as Sir I. Newton's laws of motion— in
reality however the flritt ia duo to Kepler, the second to Newton and the
third to Galileo.

258. When a moving elastic body strikes against the
surface of another body, the direction of its motion is
changed, and the motion thus resulting is said to be
reflected. Here : —

Ist. The angle at which the moving body strikes the
surface of the other is called the Angle of Incidence ;

2nd. The angle at which the moving body rebounds is
called the Angle of Reflection ; and

8rd. The Angle of Reflection is always equal to the
Angle of Incidence. i

259. In a vacuum all bodies, whatever may be their
form or density, fall towards the centre of the earth in
vertical lines and with equal rapidity ; but in ordinary
circumstances, i. e., falling through the air, only heavy
bodies fall in vertical lines, and the density and form of a
body materially affect its velocity.

260. The resistance which a body encounters in moving
through the atmosphere or any other fluid, varies : —

1st. Directly as the surface of the moving body.
2nd. Inversely as the square of the velocity of the moving
body (See Art. 147).

kSTI. SS4-M0.

▲bti. sei-t64.]

DTNAHICS.

115

NoTB.— In the oaso of heavy bodies fallinfr through the atr, therosiBtsnce
of the atmosphero producos a considerable dincropancy lM)twoon the actuU
fall of bodies and the distance through which they should theoretically fall.
Thus, it hOM born found by experiment that a ball of lead dropped from
the lantern of St. Paul's Cathedral required 4i seconds to roach the pave-
ment, a distance of !272 foot. Hut in H soconds the ball ought to have lallen
324 feet by theory, the differonco of 02 feet being due to the retarding force
of the atmosphere.

261. A heavy body falling from a lieigLt moves with a
uniformly accelerated motion, since the attraction of gravity
which causes the descent of the body never ceases to act,
and the falling body gains at each moment of its descent a
new impulse, and thus an increase of velocity, so that its
final velocity is the sum of all the infinitely small but equal
increments of velocity thus communicated.

262. Hence the velocity of a falling body at the end of the
second moment of its descent is twice that which it had at the end
of the first second ; at the end of the third second, three times that
which it had at the end of the first ; at the end of the fourth, four
times f ifc. »

203. Hence also a heavy body starting from a state of rest and
falling during any time, acquires a velocity, which would in the
same space of time carry it through twice the space it has passed
over.

264. It has been ascertained by numerous and careful
experiments, that a falling body acquires at the end of the
first second of its descent, a velocity equal to that of 82|
feet per second, and hence during the first second of its
descent a body falls through one half of 32} feet, i. e.,
through 16^3 feet.

Note 1.— The average speed of the falling body is the arithmetical mean
between its initial anof terminal velocities, or in the case of the first second
of its fall, between and 32^, and this is 16^.

Note 2.— In the following exercises wo shall use 32 and 16 in place of
32^ and 16^^, since the fractions materially increase the labour of making
the calculations without illustrating the principles any better than the
whole numbers used alone.

IIG

DYNAMICS.

[A£T8. M5, 860.

200. ANALYSIS OF TU£ MOTION OF A FALLING BODY.

Number
OF Skconds.

Space passed

over each

Second.

Terminal

Velocities.

Total Space.

, 8

100

Note.— Tho nutnbcra in the second, third and fourth columns moan so
many times Itt feet.

From this it is evident that : —

I. The spaces through which the body descends in equally succes-
sive portions of time increase as the odd numbers, 1, 3, 5, 7, 9, £fc.,
and hence the space through which the body falls during any second
of its flight, is found by multiplying 16 feet by the odd number
which corresponds to that second; i. e., one less than twice the
number of the second.

II. The final velocity acquired by a falling body at the end
of successive equal portions of time, varies as the even numbers, 2,
4, 6, 8, ^c, and hence the final velocity acquired by a body at the
end of any second of its fall, is found by multiplying 16 feet by
twice the number of seconds.

III. The whole space passed over by a body falling during
equal successive portions of time, varies as the square of the num-
bers 1, 2, 3, 4, SfC, and hence the whole space passed over during
any given number of seconds, is found by multiplying IQfeet by the
square of the number of seconds.

266. Let t = the time of descent in seconds, v = the terminal velocity,
i, e., the velocity acquired at the end of tho last' second of its fall, a
= whole space passed over, and^ = 32, i. e., the measure of the attraction
of gravity.

Then Art. 2G3, the time is equal to the space di\idcd by half the terminal

velocity, oxt=-

AST8. Bes, 260.
80DY.

KL Space.

1
4

10
26
30
40
04
81
100

kmns mean so

tally succes'

5, 1, 9, 4rc.,

any second

)dd number

twice the

at the end
lumbers, 2,
body at the
6 feet by

ng during
f the num-
ber during
feet by the

il velocity,
its fall, s
attraction

c terminal

ABTI. 267, MS.]

0TKAMIC8.

117

Anln (Art. tflS, III) tho wholo spare paiwpd orcr f« ''nnal to 16, i. f,, half
of the gravity, g, multiplied the square of the time or s -- J^^.

AIho (Art. 2nn, I) the terminal velocity i« equal <o lo, i. «,, ^j multiplied
by twice the time or v = i^r x 2^ =i i/t.

2s
Those three formulas, viz: s =i \ffl*, v =: gt and t ^— arc fundamental

and the remainluK six of the following table are derived from thorn by
transposition and substitution :—

TABLE OF FORMULAS FOR DESCENT OF BODIES FALLIX(i
FREELY THROUGH SPACE.

NO. aiVEN.

//
III

t, 8
Vj g

TO FIND,

JV
V

VII

VIII

g, t
gf »
s, t

FORMULAS.

WHENOi!: DfiUIVKD.

'=l8t'

s=-l
8 =r Itv.

Art. 2G5, m.
From formula V,

From formula VII.

V
V

gt'

V2gs.

Art. 265, /.

From IV and VII by sub-
stituting the value of t

From formula VII.

t=^±

Art. 263.

From formula IV.

From formula I.

267. When a body is thrown vertically upward it
rises with a rcj^ularly retarded motion, losing 32 feet of its
original velocity every second, and it occupies as much
time in rising as it would have required in falling to ac-
quire its initial velocity.

268. If a body be projected upwards or downwards with a given
initial velocity F; and is at the same time acted upon by the force of gravity,
then when the body dccends, in t seconds the initial velocity alone wouf ^
carry it through Vt feet, and gravity alone would carry it throuph ^gf^ feet,
therefore together they carry it through Vt + \gt^ feet, and the terminal
velocity will evidently be F+ tg.

118

DTNAMICS.

[AET.2 .

W-

When the body ascends the initial velocity acting alone would carry it in
t seconds through Vt feet, but in t seconds the force of gravity would
draw it downward through \gt^ feet, and therefore its whoiO ascent will
bo Vi — Iji^, and its terminal velocity will be V—gt. Hence

(X) s = Vt •{■ \gt^ when the body descends.

(XI) a =^ Vt -^ Igt^ when the body ascends.

(XII) « = F + tg when the body descends.

(XIIl) V z=z V— tg when the body ascends.

BxAMPLB 301. — Through how many feet will a body fall dur-
ing the 11th second of its descent?

SOLUTION.

From Art. 265, I. Space = { (11 X 2)— 1 } X 16 = (22—1 ) X 16 = 21 X IC
= 336 feet. Ans.

ExAMPLB 302. — Through hew many feet will a body fall dur-
ing the 17th, the 43rd, and the 61st second of its descent?

SOLUTION.

For the 17th second 17X2= 34—1= 33X16= 528 feet. Ans.
For the 43rd " 43 X 2= 8<' - -1= 85 X 16=1360 feet. Ans.
Forthe61«t "' 61X2=122— 1=121X16=1936 feet. ^m«.

Example 303. — What will be the terminal velocity of a falling
body at the end of the 9th second of its descent ?

SOLUTION.

Formula IV. v = gt ='32 X 9 = 288 feet per second. Ans.

Example 304. — What will be the terminal velocity of a fall-
ing body at the end of the 25th second of its fall, also at the
end of the 33rd second ?

solution.

Formula IV. « = /7^ = 32 X 25 = 800 feet per second at end of 25th second.
v~gt=i 32X33= 1056 feet per second at end of 33rd "

Example 305. — Through how many feet will a body fall
during 5 seconds ?

solution.
Formula I. » = igt^ = i X 32 X 52 = 16 X 25= 400 feet. Ans.

Example 306. — Through how many feet will a body fall in
12 seconds ?

solution.
Formula I. s = hgt^ = 4 X 32 x 12* = 16 X 144 = 2304 feet. Ans,

Example 307. — If a body has fallen until it has acquired a
terminal velocity of 400 feet per second, what is the whole space
through which it has descended ?

SOLUTION.

400

Formula II. « = — = t—t-z =

2X32

160000
64

= 2500 feet. Ant.

! I

[Am. 8 .

ABT. 268.]

DYNAMICS.

. il9

ouldoarryitln
gravity would
olo ascent will
e

)dj' fall dur-

X16=21X1C

ody fall dur-
jscent?

^ of a falling

ty of a fall-
also at the

25th second.
3f33rd "

body fall

ts.

ody fall in

icquired a
hole space

Example 308. — How long must a body fall in order to acquire
a terminal Telocity of 1000 feet "^

Formula VIII. < = — =

SOLTTl^JK.
1000

^ 31i seconds. Ans.

g 32

ExAMPLB 309.— How long miist a body fall in order to acquire
a terminal velocity of 8000 feet per second ?

80I.UTI0W,

Tormula VIII. < = — ^ — rr- = 250 seconds. Aim.
32

Example 310. — What time does a body require to fall through

11200 feet?

SOLUTION.

Formula IX. t =

__ I 2s 12X11200 __ ,

~V7 ^-^/-Si \/7o«

, , „ viMi = 26*45 seconds. An«.

V 32

Example 311. — When a body has descended through 4400
feet, what velocity has it acquired ?

BOLTJTIOIT.

Formula V. « = ^ 2gs = V 2X32X4400* = V 281600 = 630'6 feet per
second.

Example 312.— If an arrow be shot vertically upwards and
reach the ground again after the lapse of 20 seconds, to what
height did it rise ?

SOLTTTION.

From Art. 267 it appears that the anN>w will be as long ascending as de-
scending, and hence the problem is reduced to finding the distance through
which the arrow will fall in half of 20 seconds, i. e., in 10 seconds.

Then formula I. s = \gt^= iX32 X 10* = 16X100 = 1600 feet. Ans.
Example 313. — If a cannon ball be fired vertically with an in-
itial velocity of 1600 feet per second to what height will it rise ?

SOLUTION.

First, the time it ascends is equal to the time it would require if descend-
ing to acquire a terminal velocity of 1600 feet.

By formula VIII. < = — = -r^r = 50 seconds = time of ascent

g 32

Then formula XI. s= Vt-^gt^ = 1600X50 — iX32 X 50* = 80000 — 16
X 2500 = 80000 — 40000 = 40000 feet. Ans.

Example 314. — If a body be shot upward with an initial
Velocity of 1200 feet per second, at what height Will it be at the
end of the 10th second, and also at the end of the 70th second of
its flight ?

SOLUTION.

Formula XI. s=.Vt--\gt^ •=. 1200X 10 — iX32Xlo' = 12000—1600 =
10400 feet = elevation at end of 10th second.

Also 1200X70— ix32X70* =84000 — 16X4900=84000 — '78400= 6600 feet
■s elevation at end of the 70th second.

120

DYNAMICS.

[AST. 268.

ExAMPLB 316. — If a cannon ball be fired vertically with an
initial velocity of 2400 feet per second —

1st. In liow many seconds will it again reach the ground ?
2nd. How far will it rise ?

.3rd. Where will it be at the end of the 40th second ?
* 4th. What will be its terminal velocity ?

» ' 5th. In what other moment of its flight will it have the same

velocity as at the end of the 19th second of its ascent ?

SOLUTION.

Since the initial velocity =r terminal velocity = 2400 ft,

I. Formula VIII. time of ascent = — = — — = 75 seconds, and since

g 32
it IS as long ascending as descending, it again reaches the ground in 160 sec*

II. Formula I. sz=z^gt^ = iX32X75*^ = 16X5625 = 90000 ft. = height
to which it rises.

III. Formula XI. s = Fi{ — i gt^ = 2400X40 — iX 32X 40*=96000— 16X
1600 = 96000 — 25600 = 70400 ft. = elevation at end of 40th second.

IV. Terminal velocity =: initial velocity = 2400 feet per second,

V. Since the whole time of flight = 150 seconds, and, since at all equal
spaces of time from the moment it ceases to ascend and begins to descend,
the velocity is the ikame in rising as in falling, it follows that the moment
in which the body has the same velocity as at the end of the lyth second of
Its ascent is 19 full seconds before it again reaches the grouud, or in 150 ~
19 =: 131st second, i. c., in the end of the 13lst second.

Example 316. — If a body is thrown downwards from an ele-
vation with an initial velocity of TO feet per second, how far
will it descend in 27 seconds ?

SOKJTIOIf.

rormula X. «= Vt+igt^ = 70X27+iX32X27* = 1890+16X729 = 1890
+11664 = 13554 ft. Ans.

Example 317.— If a body is thrown down from an elevation
with an initial velocity of 140 ft. per second, what will be its
velocity at the end of the 30th second?

SOLUTION.

V = V+tg = 140+30X32 = 140+960 = 1100 feet per Second. Ana.

Example 318. — If a body be projected vertically with an ini*
* tial velocity of 400 feet per second, what will be its velocity at
' the end of tlie 12th second?

SOLUTION".

. Formula Xlll.v=V—fgz= 100—12 x 32=400— 384=:16 feet per second. Ans .

Example 319.-- If a cannon ball be fired vertically upward
with an initial velocity of 1800 feet per second : —

1st. In how many seconds will it again reach the ground?
2nd. What will be its terminal velocity ?
3rd. How far will it rise?

4th. Where will it be at the end of the 90th second ?
6th. In what other moment of its flight will it have the same
velocity as at the end of the 27th second of its ascent?

[ABT. 268.

\y with an

le ground ?

nd?

re the same
its ascent ?

ds, and since

id in 160 sec*
ft. r= height

=96000— 16X
cond.

!ond,

3 at all equal
s to descend,
the moment
>th second of
, or in 150 —

om an ele-
d, how far

<729 = 1890

elevation
rill be its

Ans.

ith an ini'>
elocity at

;cond.^«s.
y upward

round ?

Ithe same
;nt?

AbT. 268.]

DYNAMICS.

SOLVTIOir.

121

V ISOO
1. t=z - = ——=661 =:time of ascent or descent, hence whole time

of flight = 56J X 2 = 112i seconds.
II. Terminal velocity = initial velocity =1800 feet per second.
III. Formula 1, S= i flr <« = i X 32 X (56J)2 =16 X 8164.0626=60625 ft.
IV. Formula XI. 8= rt—\g t^ =1800 X 90-i X 32 X 90^ =162000—16
X 8100=162000—129600=32100 ft, = elevation at end of the 90th
second.

V. 112 J— 27= 85i = middle of 86//* second of flight.

ExAUPLB 320. — A stone is dropt into the shaft of a mine and
is heard to strike the bottom in 9 seconds ; allowing sound to
travel at the rate of 1142 ft. per second, and taking gr=: 32^;
required the depth of the shaft.

SOLUTIOX.

Lot a* := time stone takes to fall. Then (9— x) = time sound takes to reach
the top and x^y. 16 iV = depth of shaft = (9— a?) x 1142 feet.

Therefore ^^ = 10278 - 1142x.

198«2 + 13704a? =123336.

148996*2 + 1057948837 + 187799616= 95215392 + 187799616 =
283015008.

386x + 13704 =16823 +

886aJ=3119.

X =: 8.0803 = number of seconds body was falling.

9— a; =9—8.0803 = .9197 = time sound travelled.

And 1142 X .9197 = 1050.2974 feet = depth of shaft.

Example 321. — A body has fallen through m feet when an-
other body begins to fall at a point n feet below it ; required the
distance the latter body will fall before it is passed by the for-
mer?

FIEST SOltrilON. -T- t-

At cud of m ft. < = 1 ^=J-;r' *"^ ^ —Qt—g !2w»-:Vaw^and since

»-= distance to be traversed t=J==; hence S = ^gt^ =

ig X I / — 1 ~ Igx - — = — -. Ans.
"^ \ Nimgf 2mg 4j»

i'lUS*.'! >-'

. ■> I

(1-..I i:;

.( !

!•>.

fi 11(1') :■

'■'■■>[ ■>; '■ :

122 ^

DYNAMICS.

[AST. 268.

iii

III

111

! i

BECOND SOLUTION.

, . ,, . v^ las— lair.and \i(m+n+x)
Let a? = distance. Then (of 2nd body) < = ^y — ^— ^

:s entire time taken by the first body to pass through whole space.

Then J2("'+"+-^)-J'^ = J^ »"*^ multiplying all by V^

V2(m+n+x)—V!J»« = ViJ*.

V'iCw+n+oj) = ^/2x + Vam. and squaring.

2(wi+n+a;) = 2a? + 2j» + i^/imx.
2>» + 2n + 2x = 2x + 2w + 2 V4im;c,
2» = WwJ?.
n = 2 mx.

n^:=Z'imx,

•EXERCISES.

322. Through how many feet will a body fall during the 3tth
second of its descent? ^n«. 1168 ft.

323. Through what space will a body descend in 25 seconds?

^ns. 10000 ft.

324. With what velocity does a body move at the close of the
20th second of its fall ? Ans. 640 ft. per sec.

325. During how many seconds must a body fall in order to ac-
quire a terminal velocity of 1100 ft. per sec? ^ns. 343 sec.

326. Through what space must a falling body pass before it ac-
quires a terminal velocity of 1700 ft. per sec ? jins. 45156i ft.

327. What will be the terminal velocity of a body that has fallen
through 25000 ft. ? Jns. 1264.8 ft.

328. If a body is projected upwards with an initial velocity of
6000 ft. per second, where will it be at the end of the 40th
second? Jlns. At an elevation of 214400 ft.

329. If a body be thrown downward with an initial velocity of
120 feet per second, through how many feet will it fall in 32
seconds? ^n«. 20224 ft.

330. A cannon ball is fired vertically, with an initial velocity of
1936 feet per second : —

* In all cases, when not otherwise directed, use ^ = 82 ft.

[AST. 268.

2(»i+rt+x)

ABT. 268.]

DYNAMICS.

123

taco.

g the 3Yth
\s. 1168 ft.

eccnds?
10000 ft.

ose of the
t. per sec.

der to ac-
342 sec.

fore it ac-
45156i ft.

has fallen
12G4.8 ft.

[elocity of
the 40th
114400 ft.

jloclty of
Ifall in 32
1 20224 ft.

(elocity of

)

1st. How far will it rise?

2nd. Where will it bo at the end of the 6th second?
3rd. In how many seconds will it again reach the ground?
4tli. What will be its terminal velocity?
5th. In what other moment of its flight will it have the same
velocity as at the end of the 13th second of its ascent?

Ans. 1st. 58664 ft.

2nd. At an elevation of 11040 ft.

3rd. 121 seconds.

4th. 1936 ft. per second.

5th. At end of lOSth second of flight.

331. If a body be projected vertically with an initial velocity of
4000 feet per second, taking gravity to 32^ feet : —

1st. How high will the body rise ?
2nd. Where will it be at the end of the 50th second?
3rd. Where will it bo at tlio end of the 100th second?
4th. Where will it bo at the end of the 200th second?
5th. In what time will it again reach the ground ?
Arts. Ist. 248704.66 ft.

2nd. At an elevation of 159790.83 ft.

3rd. " 239166.66 ft.

4th. " 156661.00 ft.

5th. 248.70 seconds.

332. If a cannon ball be lired vertically with an initial velocity
of 1100 feet per second, what will be its velocity at the end
of the 7th second, at the end of the 20th second, and at the
end of the 33rd second ?

Jim. End of 7th sec. vel. = 876 ft.
20th " = 460 ft.
33rd " =44 ft.

333. If a stone be dropped into a well and is seen to strike the
water after the lapse of 5 seconds how deep is the well ?

• Ans. 400 ff.

334. If a stone be thrown downward with an initial velocity of
250 ft. per second, what will be its velocity at the end of
the 3rd, the 9th, the 30th, and the 90th seconds of its des-
cent? w^ns. End of 3rd sec. vel. = 346 ft. per sec.

" 9th " = 538 ft. "
" 30th " =1210 ft.
" 90th «« =3130 ft.

336. A stone is dropt into the shaft of a mine and is heard to
strike the bottom in 12-76 seconds, assuming that sound
travels at the rate of 1100 ft. per second, what is the
depth of the mine? wtfn». 1936 ft.

124 DESCENT ON INCLINED PLANES. [Abtb. 2(»-271.

336. A body has fallen through 400 feet, when another body ;

begins to fall at a point 2500 feet below it ; through what
space will the latter body fall before the former overtakes
it? ^ns. 3906i feet.

337. A body A has fallen during m seconds, when another body

B begins to full, / feet below it ; in what time will A
overtake B'i /».../

Alls.

32 m

DESCENT ON INCLINED PLANES.

269. AVhen a body is descending an inclined plane a
portion of the gravity of the body is expended in pressure
on the plane and the remainder in accelerating the motion
of the descending body.

270. The following are thu laws of the descent of
bodies on inclined planes : —

/. The pressure on the inclined plane .s to the weight of the
body as the base of the plane is to its length.

II. The terminal velocity of the descendinr; body is that
which it would have acquired in falling freely through a distance
equal to the height of the plane.

III, The space passed through by a body falling freely, is to that
gone over an inc ined plane, in equal times, as the length of the
plane is to its height.

IV. If a body which has descended an inclined plane meets at the
foot of it another inclined plane of equal altitude, it will ascend
this plane with the velocity acquired in coming down the former,
it will then descend the second and re-ascend the former plane, and
will thus continue oscill'iting down one plane and up the other.

Note.— The same takes place if the motion be made in a curve instead
of on an indhied plane. In practice, however, the resistance of the atmos-
phere and friction retard the motion very greatly at each oscillation and
very soon bring the body to a state of rest.

271. The final velocity, neglecting friction, on arriving
at the bottom of the plane is dependent solely on the height
of the plane, and will be the same for all planes of equal
h( i<','^t, however various may be their lengths and the
tim' of descent are exactly proportional to the lengths of
the planes.

I H

8. 2d(h271.

ler body
igh what
ivertakes
I06i feet.

her body
e will A

IS. -JL —

32 m

plane a

pressure

motion

jcent of

U of the

is that
distance

to that
of the

ts at the
ascend
former,
ine, and
ther.

instead
e atmos-
tion and

rnving
height
equal
id the
fths of

Arts. 272-275.] DESCENT ON INCLINED PLANES. 125

272. If in a vertical semicircle any number of cords bo
drawn from any points Fig- 25.
whatever and all meeting
in the lowest point of the
semicircle, and a number of
bodies be allowed to stait
along these cords at the
same instant they will all
arrive at the bottom at the
same instant, and at every
instant of their descent
they will all be in the cir-
cumference of a smaller
circle.

Thus in the accompanying figure
if ADP be a semicircle and liP,CP,
DP, HP, FP, any cords, and balls be
allowed to start simultaneously
from A, B, C, D. E and F, they will
all arrive at P at the same instant.
At the end of one-fourth the entire
time they take to fall to P, A will
have arrived at g, and the other
bodies will be in the circumference
flfP; at the end of one-half the time
of descent all will be in the circum-
ference h, &c.

273. Bodies descending curves are subject to the same
law as regards velocity as those on inclined planes, i. e.,
the terminal velocity is due only to the pei^>endicular fall.

274. T\\Q Bmcliystoc1hrone{QvQQk hrachistos,^^ shortest,"
And cJironos, "time,") or curve of quickest descent, is. a
curve somewhat greater than a circular curve, being what
mathematicians denominate a cycloid, or that which is de-
scribed by a point in the circumference of a carriage-wheel
rolling along a plane.

275. Since, Art.. 270, the effect of gravity as an accelerating force on a
body descending an inclined plane is to the effect of gravity on a body
freely falling through the air as the height of the plane is to its length ;
we have accelerating force of gravity or inclined plane: gr:: A:;; and hence

accelerating force of gravity on inclined planes = -r-,

where A=height of plane,

I = length.

g =. effect of gravity=32.

Substituting this value of the effect of gravity in the formulas in Art. 266,

wo get the following formulas for the descent of bodies on inclined planen.

126

DESCENT ON INCLINED PLANES. CABT.27e.

til

FORMULAS FOR DESCENT OP BODIES ON INCLINED

PLANES.

No.

2
3

OIVEW.

g, h, I, t

s, t

S, h, h i

«, V

g, h, I, V
S} ^> h «•

TO
FIND.

FORMULAS.

ght

8 =•

2gh

8 = J<t>

CORRESPONDING
FORMULA IN ART.2G6.

II
III

V =:

2s
T

ght
I

J 2gi

2ghs

_2s

t =

_ J 2ls

— 'J'TT

VI
IV
V

VII

VIII

276. When the body is projected down an inclined

nht^
plane with a given initial velocity V ; 8= Vt + -—r- (10.)

and vz=z Vt-{- ^-(11.) When the body is projected up

an inclined plane with a given initial velocity V ; szuVt

->ti_^ (12) andv = Fit— ^(13.)

Note.— When a body is thrown up an inclined plane, the attraction of
gravity acts as a uniformly retarding force as when a body is projected

CAbt.276.
LINED

'ONDINO
INABT.20d.'

inclined
(10.)
cted up

Sz=Vt

raction of
projected

Abt. 276.] DESCENT ON INCLINED PLANES. 127

vertically Into the air. In the case of the inclined plane the body will

continue to rise with a constantly retarded motion until Vt =^;n~ ^*»®**

it will remain ntationary for an instant and then commence to descend.
It will occupy the same lime in cominx down as in eoing up: its terminal
velocity will bo the same as its initial velocity, and it will have the same
velocity at any given point of the plane both in ascending and descending.

ExAMPLi 338. — Through how many feet will a body fall in
15 seconds on an inclined plane which rises 7 feet in 40 ?

SOLUTIOIf.

Here < = 16, ft, = 7, ? = 40, and flf = 32.

Then a = -^i — = — alTko =030 feet. Antt

ExAMPLB 339.— Through how many feet must a body have
fallen on an inclined plane, having a rise of 3 feet in 32, in ordei*
to acquire a terminal velocity of 1700 feet per second ?

SOLUTION.
Here gr = 32, t> = 1700, ft = 3, ? = 32.

Then » = -—,=

Iv^ 32 X 17002

= 4816663 feet.

Ans,

Igh^ 2 X 32 X 3

Example 340. — ^What will be the velocity at the end of the
20th second, of a body falling down an inclined plane, having
an inclination of 7 feet in 60 ft. ?

SOLUTIOW.

:7,and2 = 60.

__ ght __ 32X7X20 _ ,

Ana.

Herefl' = 32, < = 20.ft:

Then formula 6. t> = •^y—=-^^^^^^^p^ = 74| feet per second.

ExAB(PL<" 341. — On an inclined plane rising 3 ft. in 17, a body
has fallen through one mile, what velocity has it then acquired ?

SOLUTION.

Here a—\ mile = B280 ft., ft = 3, ? = 17 and g — 2,2.

mu e 1-TTT ighs _^ 2X32X3X5280 j

Then formula VI. v =J^ — == J ^-= = V5963294 =s

* V' « V 17

24417 feet per second. Ana.

Example 342. — In what time will a body falling down an
inclined plane, having a rise of 7 feet in 16, acquire a terminal
velocity of 777 feet per second ?

BOLUTlOir. fi.

Here ^r = 32, ft= 7, ? = 16, and v = 111.
Then formula 8.

._?«__ 16X777 _„. , ^

* ~ ft^~"32x7~ ^^* seconds. Ana.

Example 343. — In what time will a body fall through 4780
on an inclined plane, having a rise of 3 feet in 4 ?

Here g — 32, ft
Then formula 9.

SOLUTIOir.
3, i! = 4,ands = 4780.

\2^__
^/ gh~''

2X4X4780

V 32X3

= V398*3 = 19*9 seconds

if:

128 DESCENT ON INCLINED PLANES. TART. 270.

ExAMPLS 344.— If a body bo projected down an inclined
plane, having a rise of 8 feet in 15, with an initial velocity of 80
feet per second, tlirougk what space will it pass in 40 seconds ?

BOLUTIOIT.

Hero i> = 80, flf = 32, A = 8, ? = 16, and < = ^0.

Then fonimlaio, s=n + -2^1 = 10 x 80 -}- ^^5^^^ = 3200+l:W53i

n 2x10

= 18858i ft. Jn#.

Example 345. — If a body be projected up an inclined plane
having a rise 5 feet in 16, with an initial velocity of 2000 ft.
per second :

1st. How far will it rise ?

2nd. When will it again reach the bottom of the plane ?

3rd. What will bo its terminal velocity ?

4th. Where will it be at the end of the 100th second ?

5th. In what other moment of its flight will it have tlic same
velocity as at the end of the 11th second of its ascent?

Hero h

Then formula 8

SOLUTION.
5, 1 = 16, flr = 32 and v = 2000.
?» _ 16X2000
gh

32X5X200»
2X16

„, =: 200 seconds.
6X32

Ist. Formula 12. s=:Vt — ^^ = 200 X 2000 — '■

— 200000 = 200000 ft. Ans.
2nd. Ascent ^ 200 sec. + descent 200 sec. = 400 sec. Ana.
8rd. Terminal velocity := initial velocity = 2000 feet per sec. Ans.
—Vt ^!Mi— • 32X5X1002

-; 400OOO

4th. Formula 12. s

; 100 X 20000 —

=200000

%l " 2X16

— 60000 = 160000 = elevation at end of 100th sec. Ans.

5th. 400 — 11 = 389th second. Am.

EXERCISES.

346. On an inclined plane rising 5 ft. in 19 through what
space will a body descend in half a minute? Ans. 3789^^^ ft.

347. On an inclined plane rising 3 ft. in 13, what velocity will
.iiT: . a descending body acquire in 39 seconds? Ans. 288 ft.

second.

348. What time does a body require to descend through 3800

ft. on a plane rising 19 ft. in 32 ? Ans, 20 seconds.

349. If a body be projected down an inclined plane, having a

fall of 7 in 11 with an initial velocity of 50 feet per

. tf second, what will be its velocity at the end of the 44th

second? Ans. 946 ft. per second.

350. If a body be thrown down an inclined plane having a rise

of 13 feet in 32 with an initial velocity of 100 feet per

second, through how many feet will it descend in 130 sec. ?

Vh.i . .#• Ans. 122850 ft.

TART. 27fl.

ABT8.877,27f).|

PROJECTILES.

129

i inclined
city of 80
seconds ?

200+l.'W53j

icd plane
r 2000 ft.

no?

tlio same

3 nscent?

=•" 400O0O

ins.

-=200000

Ii what

rssfcrft.

ity will
288 ft.

?h 3800
econds.

iving a
eet per
le 44th
second.

a riso

eet per

sec. ?

K850 ft.

H51. If a body be projected up an inclined plane, having a fall
of 6 feet in 8, with an initial velocity of 800 feet per
second :^

1st. How far will it rise ?

2nd. In how many seconds will it again reach bottom of the
plane ?

3rd. What will be its terminal \elocity ?

4th. Where will it be at the end of the 68th second ?

0th. In what other moment of its flight will it have the same
velocity as at the end of the 37th second of its ascent ?

.4ms. 1st. Rise = IGOOO ft. ; 2nd. Time of flight = 80 seconds
3rd. Terminal velocity = 800 feet per second; 4th. Ele-
vation at end of 68th sec. = 8160 ft. ; 0th. At the end
of the 43rd second.

352. A body rolls down an inclined plane, being a riso of 7 ft.
in 20^when it has descended through / feet, another
body commences to descend at a point m feet beneath it.
Through how many feet will the second body descend
before the first body passes it ? 5,^8

PROJECTILES.

277. A projoctilo is a solid body to which a motion
has been communicated near the surface of the earth, by
any force, as muscular exertion, the action of a spring,
the explosive eflfects of gunpowder, &c., which ceases to
act the moment the impulse has been given.

278. A projectile is at once acted upon by two forces : —
1st. The projectile force which tends to make the body

move over equal spaces in equal times ; and
2nd. The force of gravity, which tends to make the body
move towards the centre of the earth over spaces
which are proportional to the squares of the times.
Under the joint influences of these two forces the pro-
jectile describes a curve, which in theory is the parabola,
but which in practice departs very materially from that
figure.

Note l.— The parabola ia that curve which is produced by cutting a
cone parallel to its side.

Note 2.— The parabolic theory is based upon three suppositions, all of
which are more or less inaccurate.

1st. That the force of ifravity is the same in every part of the curve
described by the projectile.

'k *

130

PROJECTILES.

[ABT8. 279, 880.

2n(l. That tho forco of gravity acts In parallel lines. ^

Sril. Tliiit tlio projcc'tilt) inuvos lliroiiKh a noi)-ru,si!«ting modiuni.

1\w (Irst and hi-coiuI of those siippositionH dilFiir soinnonslbly from truth
that thi'y Muiy hi) nsMnnu'd to ht> ahsoliitt'iy ctirroct, hut tho r«»iMtanoo of
till) atrii>>spli'(>r<' so niuturially iill'ft^t.s tho niotiunn of all bodioN, *>Npocially
wluMi llioir velocities aro con'^tidorahU), that it renders tho parabolic thoory
prncticaliy usflesH.

279. Wlieii i\ body is projoctcil horizoiitjiUy forward,

tho liorizonl.il motion dot's not intoifero with the action

of i,'ravity, — tho projoctilo descending witli the same

rapi<lity wiiilo moving forward, tiiat it woukl if acted

upon by (gravity ah)no.

NoTK.— Tilt) aproinpanyhiK fl,i<uro roprcsnntu a tower 141 foot in height.
Now If throe halls a, h, and c, ho nndo to start siinnltaneonsly from p, ono
dropping vortioally, one being projected forward with sutlicient forco to
carry it Ijorizonlally hiilf a mile, and tho tiiird withsutllciont foroo to carry
it horizontally to any other distance, say ono mile, all threo balls will reach
tho gronnd, provided it bo a horizontal piano, at tho same instant. Thus,
each ball will have fallen Ul feet at tho end of tho 1st second, and they will
Bimultaneonsly cross tlio lino dcf. At tho ond of tho 2nd second thoy have
each doscondoJ (it foot, and aro respectively at g, h, and s, &c.

Fig. 26.

^•..#

* 280. According to the parabolic theory :

Ist. The projectile rises to the greatest height, and remains
longest before it again reaches the ground, when
thrown vertically upwards.

2nd. The distance or range over a horizontal plane is
greatest, when the angle of elevation is ko .

3rd. With an initial velocity of 2000 per second, tho pro-
jectile should go about 24 miles.

Note.— The first of these laws is found by experiment to be absolutely
correct, and tho second is not far from tho truth, the greatest range taking
])laco atjku anglo of elevation somewhat less than 45'^.

[AETi. 270, 280.

biy from truth
' n3)jlNtanco of

liOH, <'»MOCiftIlv

traboHc theory

ly forward,
the action
tlie aamo

J if acted

5ot in hclglit.
y from p, orio
■lout fonio to
forco to carry
iIIm will reach
itntit. ThuH.
lind they will
nd tlioy have

AbtI, 881. 282.]

PROJECTILES.

131

remains
J, when

)Iane is

ho pro-

bsolutely
ge taking

The (Ilflforonco betwoon tho third law and the rosnlt of cxpcrlmflnt it pro-
(liKiouM J for no projectile, however Rroat its initial velocity may iiavo been,
liax over been thrown from tlie ^iirfucoof tho earth to a hori'/.ontal diiitance
of 5 niiloH.

281. Whatever may be the initial velocity of projection,

it is speedily reduced by atmospiieric pressure to >i velocity

not exceeding 1280 per second.

Note l.— This arises from tho fact that atmospheric air flown into a
vacuum with a velocity of only 128i» foot per second, mu timt when a ball
moves with a Kfcater velocity tlian this, it leaves a vacnum behind it into
which tho strongly compressed air in front tends powerfully to forco it.

Note 2.— From experiments made witli great care, it has been ascertained
that when tho velocity of a ball or other projectile is 2(MI0 Pftet per second,
tho ball meets with an atmosphorlu roiiistance equal to lUU times its own
weight.

NoTB 3. — Another groat irregularity In tho flring of balls arises from tho
fact that tho ball deviates more or less to tho right or loft, sometimes
crossing tho direct lino several times in a very short course. This deflet^tion
sometimes amounts from J to ^ of the whole rango, or as much as SUO or

400 yards in a railo when there is considorablo wlndago ; 1. c, when tho ball
is too small for tho calibre of tho gun.

282. The motion of projectiles has recently been inves-
tigated with much care, with the view of deducing a new
theory in which tho resistance of the air should be takefi
into account. The following are the most important
results: — : -

WHEN THE BODY IS THROWN VERTICALLY UPWARDS INTO THE AIR.

I. The time of ascent is less thaji the time of descent.

II. The velocity of descent is liss than that of ascent.

III. The terminal velocity is less than the initial velocity.

IV. The velocity of descent is not infinitely accelerated, since
when the velocity becomes very great, the resistance of the atmos-
phere becomes so great as to counterbalance the accelerating force
of gravity, and the velocity of the descending body is thenceforth
uniform.

WHEN THE PROJECTILE IS THROWN AT AN ANGLE OF ELEVATION.

I. The ascending branch of the curve is longer than the descend-
ing branch.

II. The tvne of describing the ascending branch is less than that
of describing the descending branch.

III. The descending velocity is less than the ascending, •

IV. The terminal velocity is less than the initial.

V. The direction of the descending branch is constantly approxi-
mating to a vertical linCf which it never reaches.

132

PROJECTILES.

[Arts. 288-288.

VI. The descending velocity is not infinitely accelerated, hut, as
in case of a body falling vertically, becomes constant after reaching
a certain limit.

VII. The limit of the velocity of descent is different in different
bodies, being greatest when they are dense, and increasing with the
diameter of spherical bodies.

283. The explosive force of gunpowder, fired in a piece

of ordnance, ia equal to 2000 atmospheres, or 30000 lbs. to

the square inch, and it tends to expand itself with a velocity

of 6000 feet per second.

Note.— Gunpowder is an intimate mixture of 6 parts'saltpetre, 1 part
charcoal, and 1 part sulphur. In firing good perfectlj' dry gunpowder, the
ignition takes place in a space of time so short as to appear instantaneous.
1 cubic inch of powder produces 800 cubic inches of cold gas, and, as at the
moment of explosion the gas is red hot, we may safely reckon the expansion
as about 1 into 2000.

284. The greatest initial velocity that can be given to

a cannon ball is little more than 2000 feet per second, and

that only at the moment it leaves the gun.

Note.— The velocity is greatest in the longest pieces ; thus Hutten found
the velocity of a ball of given weight, fired with a given charge of powder,
to be in proportion to the fifth root of the length of the piece.

285. The velocities communicated to balls of equal
weights, from the same piece of ordnance, by unequal
weights of poAvder, are as the square roots of the quantities
of powder.

286. The velocities communicated to balls of different
weights and of the same dimensions, by equal quantities
of powder, are inverselv proportional to the square roots
of the weights of the bans.

287. The depth to which a ball penetrates into an

obstacle is in proportion to the density and diameter of the

ball and the square root of the velocity with • which it

enters.

Note 1.— An 18-pound ball with a velocity of 1200 feet per second pene-
trates 34 inches into dry oak, and a 24-pouna ball with a velocity of 1300 ft.
per second penetrates 13 feet into dry earth.

Note 2.— The length of guns has been much reduced in all possible cases.
Field pieces are now seldom made of greater length than 12 or 14 calibre8(dia-
meter of the ball). The maximum charge of powder has also been dimi-
nished very greatly— now seldom exceeding one-third, and often being as
low as one-twelfth of the weight of the ball.

288. The following rule, obtained from experiment, has
been given, to find the velocity of any shot or shell, when

;AeT8. 283-288.

'ated, but, as
fter reaching

t in different
sing with the

1 in a pieco
'000 lbs. to
1 a velocitv

Itpetre, 1 part
n powder, the
xstantaneous.
and, as at the
bhe expansion

e given to
JCond,an(l

lutton found
of powder,

of equal
unequal
quantities

' different
luantities
are roots

into an
;er of the
which it

!ond pene-
of 1300 ft.

ble cases.

ibres(dia-

)een dimi-

being as

ent, has
1, when

ABT.288.J

PROJECTILES.

133

the weight of the charge of powder and also that of the
shot are known : —

RULE.
Divide three times the weight of powder by the weight of the
shot, multiply the square root of the quotient by 1600, and the pro-
duct will be the velocity per second in feet.

Orif p:=i charge of powder in lbs., w = weight of ball in lbs.,
and V = velocity per second in feet ; then v = 1600 X

Example 353. — What is the velocity of a ball weighing 48
lbs., fired by a charge of 4 lbs. of powder?

Here p = 4 and «<; == 48.

BOLUTIOW.

Then v = 1600 X

V w
X i = 800 feet per second. Ans.

^^ = 1600 X ^ ?^ = 1600 X

= 1600

Example 354. — With what velocity will a charge of 7 lbs. of
powder throw a ball weighing 32 lbs. ?

Hero p = 7 and w = 32.

Then v = 1600 x . ) ^ =
V w

SOLUTION.

I 3X7

; 1600 X J — -
V 82

X "81 = 1296 feet per second. Ans.

= 1600 X

V '65626 = 1600

Example 355. — If 4 lbs. of powder throw a ball 16 lbs. in
weight with a velocity of 1200 ft. per second, vrhat amount of
powder would throw the same ball with a velocity of 600 feet per
second ?

SOLUTION.

Art. 285. vel. : vel. : : V weight of powder : V weight of powder ; or
1 200 : 600 : : V ~4 ' V"^ and hence a? = lib. Ans.

Example 356. — If 3 lbs. of powder throw a ball 6 inches in
diameter and weighing 32 lbs., with a velocity of 850 feet per
second, with what velocity will the same charge throw another
ball of the same dimensions but weighing only 9 lbs. ?

SOLUTION.

Art. 286. VoT VsF: : 850 : a?, or 3 : 5.65 :
And hence x = 1600 feet. Ans.

860 : X.

EXERCISES.

367. With what velocity will a charge of 11 lbs. of powder
throw a cannon ball weighing 24 lbs. ?

Ans. 1875 feet per second.

: f

134

358

CIRCULAR MOTION.

[Abt. 280-202.

With what velocity will a charge of 9 lbs. powder throw a
ball weighing 3G lbs. ? Jns. 1387 feet per second.

359. If 1 lb3. of powder throw a ball with a velocity of 1000

feet per second, what charge will throw the same ball
with a velocity of 1500 feet per second ? Jns. 151 lbs.

360. If a certain charge of powder throw a 10-inch ball weighing

20 lbs. with a velocity 973 feet per second, with what
velocity will the same charge throw a ball. of the same
dimensions weighing only 25 lbs.?

J71S. 863 feet per second.

»>

CIRCULAR MOTION.

289. Centrifugal force (Lat. centrum, *' the center,'
and/i/yto, " I flee") is that force by which a body mov-
ing in a circle tends to fly oft' from the centre.

Note.— Since a body moving in a circle would, if not restrained by other
forces, lly oil' iji :a tangent to that circle, centrifugal force is sometimes
called tangential force,

290. Centripetal force (Lat. centrum, *' the centre," and
"^cto, I seek or rush to ") is that force by which a body
moving in a circle is held or attracted to the centre.

291. Wlicn a body is at once acted upon by both cen-
trifugal and centripetal force, it moves in a curve, and
the form of this curve depends upon the relative intensities
of the two forces: /. e., if the two be equal at all points,
the curve will be a circle, and the velocity of the boily
will be uniform ; but if the centrifugal force, at difttirent
points of the body's orbit, be inversely as the square of the
distance from the centre of gravity, the curve will be an
ellipse, and the velocity of the body will be variable.

292. When a body rotates upon an axis, all its parts
revolve in equal times ; hence the velocity of each particle
increases with its perpendicular distance from the axis,
and so also does its centrifugal force.

Note l.— As long as the centrifugal force is less than the cohesive force
by which the particles are hold together, the body can preserve itself ; but,
as soon as the centrifugal force exceeds the cohesive, the parts of the rotating
mass lly off in directions which are tangents to the circles in which they
were moving.

Note 2. — We have examples of the effects of centrifugal force in the
destructive violence with which rapidly revolving grindstones burst and
fly to pieces— the expulsion of water from a rotating mop, the projection
ol a stone from a sling, the action of the conical pendulum or governor in
regulating the supply of steam in an engine, SiC, olc.

[Am. 289-292.

vder throw a
per second.

city of 1000
le same ball
Ans. 153 lbs.

all weighing
, with what
of the same

t per second.

fie center,"
body mov-

*ained by other

) ia sometimes

entre," and
ich a body
entre.

y both cen-
curvc, and

intensities
b all points,
f the body
at ditfercnt
uarc of the

will be an
iable.

all its parts
ich particle
the axis,

cohesive force
vo itself; but,
)f the rotating
n which they

1 force ill the
nea burst and
he projection
)r governor in

Aets. 293-207.]

CIRCULAR MOTION.

136

293. When the velocity and radius are constant, the
centrifugal force is proportional to the weight.

294. When the radius is constant, the centrifugal force
varies as tlie square of the velocity.

Note. — At the equator the centrifugal force of a particle is uiy of its
gravity or weight and from the equator it diminishfs as wo approach the
poles where it becomes 0. It follows that if the earth were to revolve 17
times faster thau it does, the centrifugal force at tl" <'<iuator would be
equal to gravity, and a body would not fall there a ;l. If the earth
revolved still more rapidly, the water, inhabitant-^ ■<■ ., would be whirled
away into space, and the equatorial regions would ^ uustitute an impassible
zone of sterility.

295. When the velocity is constant, the centrifugal
force is inversely proportional to the radius.

296. When the number of revolutions is constant, the
centrifugal force is directly proportional to the radius.

297. Let c = centrifugal force, v = the velocity per second
in feet, r = radius in feet, g- = 32, w=z weight, and n = the
number of revolutions per second.

wv^ wv^ csr I csr

Then c = — (I), r = — (II), t« = -^ (III), v =^-J; (IV)

gr ^ " eg
Also, since v = rX2X3-1416Xn, tj2 = r« X4X(3-1416)' X n»,

and hence formula I. : c =

wxr" X4X (3-1416)' X^^"
gr

ducing this we get c = wrn" X 1'2345 (V.), ^^= ^^2 v^i.2345

and re-
(VI.),

r ^:z

r^2U5 (^"•)' ^""^ '' = V«;rx 1-2345 <^"^->

Example 361. — ^What is the centrifugal force exerted by a
body weighing 10 lbs. revolving with a velocity of 20 feet per
second in a circle 8 feet in diameter ?

Here «j = 10,« = 20, r =

wi>» 10X20»

Then c -~ -—■-.
gr

SOLUTION.

: 4, and g = 32.
10X400

= 3U lbs. Ans.

32X4 ~ 32X4

fi Example 362. — What centrifugal force is exerted by a body
weighing 15 lbs. revolving in a circle 3 feet in diameter and
making 100 revolutions per minute ?

I »

136

CIRCULAR MOTION.

MOLUTION.

TAbt. S97.

Hero w ~ 13, r = 1'5, « = W = 1*.

Then formula \.: c — wrn^ Xl-2345 ~ ISXl'oX (l!)« X l-23i5 = 77'15628
lbs. Ana.

Example 363. — A body weighing 40 lbs. revolves in a circle
4 feet in diameter ; in order that its centrifugal force may be
1847 lbs., what must be its velocity-and number of revolutions
per second ?

SOLUTION.

Here w = 40 lbs., »• = 2, and c = 1847.

Then formula VIll. : » =

18'li7

— £ I — '.^ — = ViHor9

VwrXl-234i5~ V40X2X 1*2345
= 4'28 = number of revolutions per second, and hence revolutions per
minute = 25t>*8.
Also V = 4iX3-1416X4-32 = 5378 feet per second.

ExAMPLB 364. — The diameter of a grindstone is 4 feet, its
weight half a tbn, and the centrifugal force required to burst it
is 45 tons : with what velocity must it revolve, and how many
revolutions must it make per minute in order to burst ?

Here w = 4, c = 45, and r :

Then formula VIII. :

SOLUTION.
2.

_£ I 45

: 1-2345 ~ V '

EXERCISES.

365. If a ball weighing 4 lbs. be attached to a string 2J feet

long and whirled round in a circle so as to make 120
revolutions per minute, — what must be the strength of
• the string in order to just keep the ball from flying oflf?

Ans. 49-38 lbs.

366. A ball weighing 2 lbs. is attached to a string 3J feet long

and capable of resisting a strain of 200 lbs. ; if the ball
be whirled in a circle with the whole length of the string
as radius, how many revolutions per minute must it make
in order to break the string? Ans. 288^ revolutions.

367. A ball is whirled in a circle, with a velocity of 64 feet per

second, by means of a string 4 feet in length and capa-
ble of resisting a strain of 840 lbs ; what must be the
weight of the ball in order to break the string ?

»/?n». 26i lbs

**~ VtfrX 1-2345 ~ '\/iX2xr2345 " V**'*^'* V
= 6*03 — revolutions per second, and hence 6*03X60 = 361'8 = the revolu-
tiors per minute.
Also velocity =- 4X3*1416X6-03 = 75-775 feet per second.

•fe.

TAXT. S97.

,5 = 77'15626

in a circle
rce may be
revolutions

= Vl8'70i9
olutions per

4 feet, its
[ to burst it

how many
it?

^36-452
the revolu-

ing 2i feet
make 120
trength of
ying off?
49-38 lbs.

feet long

if the ball

the string

ist it make

volutions.

64 feet per
and capa-
st be the
?
». 26^ lbs

Abt8.298.299.] ACCUMULATED WOBK.

137

368. What is the centrifugal force exerted by a body weighing

20 lbs. revolving in a circle 10 feet in diameter nnd mak-
ing 2*8 revolutions per second? Jns. 967-848 lbs.

369. What is the centrifugal force exerted by a body weighing

8 lbs. and revolving in a circle 20 feet in diameter with
a velocity of 100 feet per second ? ^m. 250 lbs.

ACCUMULATED WORK.

298. Work is required to set a body in motion or to
bring a moving body to a state of rest. For example,
when a common engine is first set in action a considerable
portion of the work of the engine goes to give motion to
the fly-wheel and other parts of the machinery ; and before
the engine can come to » state of rest, all of this accumu-
lated work must be destroyed by friction, atmospheric
resistance, &c.

299. To find the work accumulated in a moving

body : — , .

RULE.

/. Find the height in feet from which the body must have fallen
to have acquired the given velocity.

II. Multiply the number thus found by the weight of the body
in pounds.

Or let IT = units of work accumulated^ v = velocity^ w = the
weight in Ibs.^ and g = 32.

Then Art. 266, since s:=zh=z-^ s

tr=Aw=-X

«> =

v*w

ExAifPLB 370.— A ball weighing 10 lbs. is projected on stnooth
ice with a velocity of 100 feet per second : assuming the friction
to be -^ of the weight of the ball, and neglecting atmospheric
resistiince, over what space will it pass before coming to a state
of rest?

SOLUTION.

Here t> =e 100, w — 10, and g = Z2.

Then TJ a= "sir ==

v*to 100* XIO 100000

== 15621 = units of work accu-

se ~ 2X32
mulated in the ball.

Also "^ XlOXl = I = units of work destroyed by friction iu moving

the ball through 1 foot.
Therefore the number of feet s= 1562i -f f = 23431 Ans.

138

ACCUMULATED WORK.

[AbT. 29^.

(

Example 3 '71. —A train weighs 100 tons, and has a velocity
of 40 miles per hour when the steam is turned oS: how far will
it ascend a plane having an inclination of i in 100, taking fric-
tion as 11 lbs. per ton, and neglecting the resistance of the
atmosphere?

SOLUTION.

40X5280
Hero V ^ 40 miles per hour = gyxoo" ^^ ^^^ ^^^^ P®^ second, w = 100
tons = 200UOO Iba., and ^ = 32

Then U = -27 =

«««? (68|)«X200U00 Sm'rX 200000

2X32

= 3411^ X 3125 =

10703555^ = units of work accumulated in the train.

Work of friction = 100 X 1 1 = 1100 units to each foot. -

Work of gravity =: 7uuX2u0000 = 1000 usits to each foot.
Work destroyed by resistances, i. e. friction and gravity, in moving the
train uvur one fout = llou+iOOO = 21U0 units.

10755565^
Therefo"^ > number of feet = — ^;^^ — = 6121*69 fect=nearlyonemile

2100

Example 372. — If a car weighing 3 tons, and moving at the
rate of 10 feet per second on a level rail, pass over 500 feet
before it comes to a state of rest, what is the resistance of fric-
tion per ton ?

SOLUTION,
10«X6000

Work accumulated in car = gxsa
Work of friction = friction X 600.

_ 600000 __ 5375 ^„it3,
~" 04 ""

Therefore friction X 600 = 9376, and hence friction = --— = i82 lbs.

on whole car.
Then friction per ton = 18i-j-3 — 6i lbs. ^JM.

EXERCISES.

<il3. A train weighing 90 tons is moving at the rate of 30 miles
per hour when the steam is shut off: how far will it go
before stopping, on a level plane, assuming the co-effi-
cient of friction to be ji-Q ? ^ns. 6050 feet, or l^g- miles.

374. A train weighing 80 tons has a velocity of 30 miles per

hour when the steam is turned off: how far will it ascend
a plane rising 7 feet in 1000 — taking friction, as usual,
and neglecting atmospheric resistance ?

Jlns. 2880-95 feet.

375. Required the units of work accumulated in a body whose

weight is 29 lbs. and velocity 144 feet per second ? 9396.

[Abt. tai.

a velocity
w far will
iking fric-
icc of the

nd,tc = lOO
5 X 3126 =

moving the

arly one mile

ving at the
er 500 feet
ince of fric-

mits.

= 182 lbs.

lof 30 miles

Ir will it go

Ithe co-effi-

1^8- miles.

miles per
11 it ascend
1, as usual,

)80-95 feet.

lody whose
id? 9396.

AAts. 306-S((S.]

TtlE PENDULUH*

139

376. A ball weighing 15 lbs. is projected on a level plane, with

a velocity of UO feet per second : assuming friction to be
equal to -^a of the weight of the ball, how far will it go
before it comes to a state of rest? ^ns. 1265 625 feet.

377. A train weighing 90 tons has a velocity of 100 feet per

second when the steam is turned off: how far will it go
on a level plane, assuming friction to be equal to 12 lbs.
per ton, and neglecting atmospheric resistance?

^m. 260413 feet.

378. A ball weighing 20 lbs. is thrown along a perfectly smooth

plane of ice with a velocity of 60 feet per second : how
far will it go before stopping if the friction be -^(^ of the
weight? ^ns. 1125 feet.

379. A train weighing 100 tons has a velocity of 25 feet per

second when the steam is turned off: how far will it
descend an incline of 3 in 1000, taking friction to be
equal to 12 lbs. per ton? Jlns. 3255-2 feet.

380. Required the work accumulated in a body which weighs

50 lbs. and which is moving with a velocity of 70 feet
per second? jlns. 3828| units.

381. What work is accumulated in a ram weighing 2000 lbs.

falling with a velocity of 40 feet per second ?

Jlns. 50000 units.

THE PENDULUM.

300. A pendulum consists of a heavy body suspended
by a thread or slender wire, and made to vibrate in a ver-
tical plane.

801. When the body is regarded as a point, and the
thread or wire without weight, the pendulum is called a
Simple Pendulum.

302. A Compound Pendulum or Material Pendulum
consists of a heavy body suspended by a ponderable wire
or thread.

303. The motion of the pendulum from one extremity
to the other of the arc in which it moves, is called an
oscillation or a vibration.

304. The amplitude of the arc of vibration is measured
by the number of degrees, minutes and seconds through
whii-h ihe pendulum oscillates. -'^

305. The duration of a vibration is the space of time
occupied by the pendulum in swinging from one extremity
to the other of the arc of vibration.

140

THE PENDULUM.

[AbtS. 306-310.

306. The length of the pendulum k the distance ^
between the centre of suspension and the centre of oscilla-
tion.

307. The centre of suspension is the point round which
the pendulum moves as centre.

308. The centre of oscillation is that point in a vibrating
body, into which, if all the matter were collected, the
time of vibration would remain unchanged.

Note l.— If a bar of iron or any other substance bo suspended by one
extremity and made to vibrate, it constitutes a compound pendulum.
Now, if tne several particles composing the rod were free to move separately
tiioae nearer the centre of suspension would vibrato more rapidly than
those more remote ; but, since the pendulum is a solid body, all of its
particles must vibrate in the same time, and hence the motion of those
molecules which are nearer the centre of suspension is retarded, while
that of the more remote parts is accelerated. Somewhere in the rod,
however, there must be a point or particle so situated with respect to
the centre of suspension, and the other parts of the rod, that the acco-
leratinR effect of the particles above it is exactly neutralized by the
retarding force of the molecules below it ; and, consequently, this particle
or point vibrates in exactly the same time that it would occupy if liberated
from all connection with the parts above, below and around it, and were
set swinging by an imponderable thread— this point is called the centre
of oscillation.

Note 2.— The centre of oscillation in a vibrating mass coincides with
what is called the centre of-perctusion. The centre of percussion is that
point in a revolving body, which, upon striking against an immovable
obstacle, will cause the whole of the motion accumulated in the revolving
body to be destroyed, so that, at the moment of impact, the body would
have no tendency to move in any direction. In a rod of inappreciable
thickness the centre of percussion is two-tliirds of the length or the rod
from the axis about which it moves.

309. The centres of suspension and oscillation in the
pendulum are interchangeable; i. e., if the pendulum be
inverted and suspended by its centre of oscillation, the
former point of suspension will become the centre of oscil-
lation and the pendulum will vibrate in precisely the same
time.

LAWS OF THE OSCILLATION OF THE PENDULUM.

310. The duration of an oscillation is independent of
its amplitude, provided it does not exceed 4° or 5°.

Note 1.— 'This fact is commonly stated by saying that the vibrations of
the pendulum are isochronous; i. c., equal-timed. Thus, a pendulum of
a given length will oscillate through an arc of 5" in the same time it would
have required to vibrate tlirough an arc of 0*1°, although the amplitude of
the vibration is in the one case 50 times as ^reat as in the other. This arises
from the fact that the pendulum in moving through the larger arc falls
thorough a greater vertical distance, and hence acquires a greater velocity.

). 306-310.

listance
oscilla-

lI which

ibrating
ted, the

led by one
pendulum,
separately
pidly than
, all of ita
)n of those
ded, while
n the rod,
respect to
t the acce-
led by the
(lis particle
If liberated
t, and were
the centre

icides with
aion is that
immovable
e revolving
)ody would
appreciable
or the rod

in in the
ilum be
;ion, the
of oscil-
he same

JM.
Indent of

irations of
kndulum of
lie it would

iplitude of
iThis arises
\r arc falls

' velocity.

ABT8. 811-316.]

THE PENDULUM.

141

Note 2.— Strictly speaking, the oscilUtic .» or the pendulum areisochro'
noMB only when the curve in which they move is a oycloid. When, however,
the common pendulum vibrates in very small arcs, as of 2" or 3*, the oscil-
lations are, for all practical purposes, isochronous.

311. The duration of the vibration is independent of the
weight of the ball and the nature of its substance.

312. Two pendulums of equal lengths perform an equal
number of vibrations in the same period of time.

313. Two pendulums of unequal lengths perform an
unequal number of vibrations in the same period of time —
the longest pendulum performing the smallest number of
oscillations.

314. Pendulums of unequal lengths vibrate in times
which are to one another as the square roots of their lengths.

315. A second^ s pendulum is one that performs exactly
sixty vibrations in a minute, or one vibration in one second.

316. The time occupied by a vibration depends: —
1st. Upon the length of the pendulum ; and
2nd. Upon the intensity of the force of gravity.

Note.— Since the earth is not an exact sphere, being flattened at the poles,
the surface of the earth at the poles is nearer to the centre than at the
equator. Hence the intensity of the force of gravity is less at the equator
than at the poles, and a pendulum that beats seconds at the equator must
be lengthened in order to beat seconds as it is carried towards the poles.
In point of ta,«it, a seconds pendulum at the poles is about one-fifth of an
inch longer than a seconds pendulum at the equator. The following table
shows the length of the seconds' pendulum at difTereat parts of the earth's
surface, and also the magnitude of the force of gravity ; 1. e., the velocity
which the force of gravity will impart to a dense body in falling for one
entire second.

Place.

Jiatitude.

Length of
Seconds* Pen-
dulum.

Telocity acquired

by a body falling

one second.

St. Thomas

0° 24'

7" 55'

40^42'

480 B^r

61" 81'
79» 60' '

39.01 inches

39.09 »

89.10 "

89.12 "

39.13 "
39.21 "

384.986 inches
384.286 "
386.978 « .
386.076 "
886.174 "
386.984 "

Ascension

New York

Paris

London

Spitzbcrgen

142

THE PENDULUM.

CABTi.S17-ttO.

NOTB 2.— Ill Canada tho second's pendulum ii about 30.11 in. in length.

317. The pendulum is applied to three purposes: —
1st. As .1 measure of time;
2nd. As a measure of the force of gravity ; and
3rd. As a standard of measure.

Note.— The pendulum is used as a measure of tiiro by attaching it to
clock-work, which servos the double purpose of rcKistering its oscillation!
and rcNtoriuK to the uondulum tho motion lost in Its vibration by friction
and atmoHnheric resistance. The use of tho pendulum as a standard of
measure will be s(!cn from the following statements, viz:

1st. A pound pressure means that amount of pressure which is exerted
towards the earth, in the latitude of London and at the level of the sea, by
tho quantity of matter called a pound.

2nd. A pound qf matter means a quantity equal to that quantity of pnre
water which, nt tho temperature of G'i deg. Fahrenheit, would occupy
27727 cubic inches.

3rd. A cubic inch is that cube whoso side, taken 30*1393 times, would
measure the cITuctive lenyth of a London seconds pendulum.

4th. A seconds pendulum is that which, by the unassisted and unopposed
effect of its own gravity, would make 8U'M)0 vlbratious in an artificial solar
day, or 8G183'09 in A natural sidereal day.

318. I/t =2 the tine of oscillation, I = the length of the penrlu-
lum, jf = the force of gravity ; i. «., the velocity which the force of
gravity would impart to a dense body falling through one entire
secont, and ^ = 3- 14 16; t. c, the ratio between the diameter of a
circle and its circumference.

Then t=:ir X

t's

(I.) / = -J (II.) and g =

(m.)

l7r*_

g - t'

When t = one second^ formulas (ii) and (iii), respectively become
g
-i (IV.) and g = y^ (V.)

319. To find the time in which a pendulum of given
length will vibrate, or the length of a pendulum that vibrates
in a given time : —

Let I = the length and t = the time. Then since (Art. 314)
the times are as the square roots of the lengths, and in Canada the
seconds pendulum is 39*11 inches in length— we have

* : 1 :: ^/li^JSd-ll ; and henr§

= IJ-

V3911

(VI.), andl = t*XSOU. (vii.)

320. To find the number of vibrations which a pendulum
of given length will lose by decreasing the force of gravity,

11.817-310.

, in lenfth.
08 : —

u:hinR it to
oHcillationi
by friction
itandard of

I is oxortcd
the sea, by

Hty of pnro
uld occupy

mcs, would

unopposed
iificial ttolar

the pendu'
he force of
one entire
meter of a

[III.)

ily become

|of given
vibrates

Irt. 314)
mada the

[ndulum
gravity,

ABT.821.]

THE PENDULUM.

143

i. e., by carrying the pendulum to tbo top of n mouutain
or other elevation.

Let n = the number of vibrationt performed at the earth's
turface in the given time, n' = the number of vibrationt lott in the
tame time, r = the radiut of the earth, = 4000 milet, and h =:
the height of the mourUain in milet or fraction of a mile; then

n' = ~ (viii.), and hence A = — . (ix.)

821. To find the number of vibrations which a pendulum
of given length will gain in a given time by shortening tho
pendulum.

Let I = the given length of the pendulum, and I' = the decreate
in length : also let n = the number of vibrationt performed in the
given time, and n' = the number of vibrationt gained in the tame
time; then

nl' 2n!l

n' = 2"; (X.) and ^' = -;p (xi.)

ExAUPLB 382. — How many vibrations will a pendulum 36
inches long make in one minute 7

SOLUTION.

Formula VI.: <= ^^^;^ = ^^-^^^^ = V9204 = '950 seconds.

Hence tho number of vibrations = 60 -r '059 = 62*56.

Example 383.— Required the length of a pendulum that makes
80 vibrations in a minute ?

SOLUTION. !• . I "

Hero< = f^ = *-

Then formula VII. i = <« X 39-11 = (}) « X 3911 = -^ X 39-11
= 21999 inches.

ExAMPLB 384. — In what time will a pendulum 60 inches long
vibrate ?

SOLUTION.

.^ \ I -

rormula VI. : < = ^^^ = ^^^ = ^I'^S^l = 1-239 sec. Ans.

ExAMPLB 386. — A pendulum which beats seconds is taken to
the top of a mountain one mile high : how many seconds will it
lose in 6 hours ? = v-- '.^s

144

THE PENDULUM.

[Abt. 811.

SOLUTION.
Here n = 6X60X00, A = 1. and r = 4000.

*• # 1 /irtwr\ , ♦»* 6X00X60X1

Taen formula (VIII.) : n' = — = — r-

r 4000

4000

BzAMF.i 388.— If a clock lose 1 minute in 24 hours, how
much must the pendulum be shortened to make it keep true
time?

SOLUTIO.H.
Here w =: 24X60X60, n' = 60. and I =: 89*11.

Then formula XI. : i' = — =: ^^!^^^^'}} - = 00543 or about Xth of
an inch. Ana,

24X00X00

Ejumpli So';. — Through what distance will a heavy body fall
in Canada during one entire second, and what will be its termi-
nal Telocity 7

I SOLUTION

Here < = 1, and ? = 3911.

Then formula V. g =:lir* z= 39-11 X(3'1416)« =: 80.11 X 9-8696506fl =
386*008 inches ~ terminal velocity.

Honce space passed through = -^^^— r = 103.001 inches = 16*0835

feet. Ant,

ExAMPLB 388. — What must be the length of a pendulum in
order to vibrate ten times in a minute 7

SOLUTION.

60
Here f == — = 6 seconds.

Then formula VII. I = <«X39*11 = 10«X89*11 = 100X3911 = 8911 in.
=: 826 feet nearly. Ans.

ExAMPLB 389. — A pendulum which vibrates seconds at the
surface of the earth is taken to the top of a mountain and is
there found to lose 18 seconds in a day of 24 hours : required
the height of the mountain 7

SOLUTION.

Here n' — 24X60X60, n = 18, and r — 4000,

«'»• 18X4000 _, ., ..... . .

Then h=—=z 2^^^^^ = ^ miles = 4400 feet. An,.

ExAMPLi 390. — If a seconds pendulum be shortened !{ inches,
bow many vibrations will it make in one minute 7

111.

...•''u..-,rV "vS .•

[Abt. 811.

ABT.Stl.]

THE PENDULUM.

145

tours, how
keep true

bout -jif th of

y body fall
) its termi-

>-869960S6 =
= 160835

adulam in

= son iu.

ds at the
tin and is
required

flOLUTlON.

Here » = 60, / ~ 30*11, and V = 1-28.

Then formula X.: n'= r-r= =0'058 = thoimmb«rofvibra-

a I ^ X Stf 1 1

tions gained ; hcnco tha number of vibrations made = 00*038. Ant,

ExAMPLi 391.— What will bo the velocity acquired by a heavy
body falling during one entire second in the latitude of Spitz-
bergen ?

SOLUTION.

Here tzsl, and by the tablo Art. 810. 1 = SOZt,
Thenflf = lirt— 39-2lX(3'l'llO)« — 30-21 XOMOCS

EXERCISES.

1 380038 inches. Ant.

i inches,

392. What must be the length of a pendulum in the latitude of

Canada in order that it shall vibrate once in 3 seconds ?

^ns. 351-99 inches.

393. A pendulum that vibrates seconds at the surface of the

earth is carried to the summit of a mountain 3 miles in
height : how many seconds will it lose in 24 hours ?

jint. 64*8.

394. In what time will a pendulum 10 inches in length vibrate ?

Ans. '505 seconds.

395. What velocity will a heavy body falling in the latitude of

New York acquire in one entire second? wfns. 385*903.

396. If a clock lose 10 minutes in 24 hours, how much must

the pendulum be shortened in order that it shall keep
correct time? Jins. -543 or over I of an inch.

397. If a seconds pendulum be shortened 5 inches, how many

vibrations will it make in a minute? ^ns. 63*83.

398. A pendulum which vibrates seconds at the surface of the

earth is carried to the summit of a mountain, where it is
observed to lose 30 seconds in 24 hours : required the
height of the mountain ? Ans. 7333*3 feet,

399. In what time will a pendulum 100 inches long vibrate ','

^ns. 1*69 seconds.

400. Required the length of a pendulum which makes 120 vibra-

tions per minute? ^ns. 9*77 inches.

401. Through how many feet will a body fall in one second, and

what will be its terminal velocity at the end of that por-
tion of time in the latitude of Paris ?

^ns. Terminal velocity = 186*1 in.
' Space passed over =: 16*0875 ft.

146

[Axn. 322-82ft.

HYDRODTNAMICS.

CHAPTER VIII.
HYDRODYNAMICS.

322. Hydrodynamics treats of the motions of liquids and
of tlie forces which they exert upon the bodies when their
action is applied.

323. The particles of a fluid on escaping from an orifice
possess the same veloiiiy as if they had fallen freely in
vacuo from a height eqtial to that of the fluid surface above
the centie of the oriOce.

NoTB.— This is known as Torricelli's theorem. Since all bodies falling
tn vacuo from the same height acquire the »ame velocity, density has no
effect in increasinu: the velocity of a liquid escapinu from an orifice in the
side or in the bottom of a veHScl. Thus water, alcohol, and mercury will all
flow with the same rapidity ; for though the pressure of the mercury is ISj
times greater thatf that of water, it has ISi times as much matter to move

•

324. When a liquid flows from an orifice in a vessel
which is not replenished but the level of which continually
descends, the velocity of the escaping liquid is uniformly
retarded, being as the decreasing series of odd numbers 9,
7, 6, 3, (fee, so that an unreplenished reservoir empties
itself through a given aperture in twice the time the same
quantity of water would have required to flow through the
same aperture had the level been maintained constantly
at the same point.

325. The quantity of fluid discharged from a given
aperture in a given time is found by multiplying the area
of the aperture by the velocity of the escaping liquid.

Note.— Experiments do not agree with this theory as regards the quan-
tity of liquid discharged. The whole subject has been carefully investigated
by Bossut, and he has shown that '

ActtMl discharge : Theoretical discharge : : -62 : 1 or as 6 : 8.

Hence the theoretical discharge must be multiplied by | to obtain the

true quantity.

This discrepancy arises ft*om the fact that the escaping
jet diminishes in diameter just after leaving the vesseL
forming what is known as the vena contracta or contracted
. The minimum diameter of the vein is found at a

Fig. 27.

vein

distance about equal to half the diameter of the aperture
as at c e' Fig. 27. This effect arises from the fact that just
above the orifice the lateral particles of fluid move ai well
as the descending portions.

iiiii

BTI. 822-Stt.

quids and
hen their

nn orifice
freely in
ace above

Ddics falling
181 ty has no
riflce in the
cury will all
sn-ury is 13|
ter to move

•

I a vessel
)ntinually
miformly
imbers 9,
empties
the same
ough the
instantly

a given
the area
uid.

the quan-
ivestigated

:8.
obtain the

ig. 27.

Aaxs.8S6.]

HTDBODTNAMICS.

147

826. Istt Q = the quantity discharged in 1 tecond^ a == area of
aperture, k = height of fluid level above the centre of the or0c9t
g = accelerating force {^gravity, and v ^ velocity.

^ — — Q

Then Jirt. 266 v = V2gA, <i) Q = <hl2gh, (ii) o = •7— , (in)

^2gh
qt -

'*"''*= 2^- (»^> ' '

Note.— Since g — 32, 2g = 64, and V^=: 8. formulas I, II and III become

respectively « 1= sVa Q=Sa^/h,oxida = ^~l^

ExAMPLs 402. — With what velocity does water issue from a
small aperture at the bottom of a vessel filled to the height of
100 feet?

1 SOLUTION.

Fonnula 1 v— sV* = sVlOO = 8X10 - 80 feet per second. Ans.

ExAUPLB 403. — What quantity of water will be discharged ia
one minute from an aperture of half an inch in area — the height
of the water in the vessel being kept constant at 10 feet above
the centre of the orifice ?

SOLUTION.

Hera a=si square inch =s ^^ J ^ of a square foot.

The cubic feet discharged in 1 secoud = SaV^

Cubic feet discharged in 1 minute =60X8axVA = 60 X yItX VlO= 60X
Vs X 3*162 = 6*27 cubic feet — the theoretical quantity, and 6*27x1 = 3*29
cubic feet = true quantity.

Example 404. — What must be the area of an orifice in the side
of a vessel in order that 40 cubic feet of water may issue per
hour — the water in the reservoir being kept constantly at the
level of 20 feet above the centre of the aperture ?

HereQ:

60X60

SOLUTION.
: ^ of a cubic foot, and since this is only f of the

theoretical quantity, Q =: f of ^^ = j^j of a cubic foot. Also h = 2).

Then formula III, a ■■
inch. An9.

_Q_7^

•J5T

V'^ 8V20" = 35=776 =nPj? °^ * '°°*= H *'*"

EzAUPLB 405. — An upright vessel 16 feet deep is filled with
water and just contains 15 cubic feet. Now if a small aperture
i of an inch in area be made in the bottom, in what time will
the vessel empty itself ?

148

HYDRODYNAMICS.

CAbt.827.

' BOLUTIOW. - • V

Iloro h — ld ft., a — i of an iuoh, and Q == 15 cubic foot.
llonco the theoretical quantity =» 16X Jf = 21 cubic foot.

Then velocity at commoncflmont = sV'* = sVw = 32 ft*

Quantity dischargt^d in 1 socend = 32 X ^ i| ^ := ^f^^ = -,i^ of a cubic foot.

Time required to disoharffo 21 cubic foot — 21-f-,'jf = 432 seconds.

lint, Art. 321, ^vhen a vessel onintioa itself, the time required to dischargA

given quantity of water is double tliat requisite for discliarging

quantity when the level is maintained,
llcuc* time — 432X2 = 881 seconds =: 16'-1 minutes. Ans.

EXERCISES.

rging the same

406. With what velocity does water issue from a small aperture

in tho side of a vessel iillud to the height of 25 feet above
the centre of tho orifice ? ^ns, 40 feet per second.

407. With what velocity does water flow from a small aperture

in tho side of a vessel filled with water to the height of
17 fee t above tho centre of tho orifice ?

I ^ns. 32 '984 feet per second.

408. In the last example, if tho water flows into a vacuum, what

is its velocity ? Jins. 56 feet per second.

Note.— Since tho pressure of tho atmosphere is equal to that of a
column of water 32 feet high, the elTective height of the
column of water is 17+32 — 40 feet.

409. How much water is discharged per minute from an aperture

having an area of ^ of an inch— the surface of tho fluid
being kept constant at 36 feet? >dns, 2} cubic feet.

410. What must bo the area of tho aperture in the bottom of a

vessel in order that 90 cubic feet of water may issue per
hour — the level of the water in the vessel being constantly
kept at 20 foet above tho centre of the orifice ?

jlns. *161 or about i^g of an inch.

A vessel contains 20 cubic feet of water, which fills it to
the depth of 30 feet — now if an aperture having an area
of I of an inch bo made in tho bottom of the vessel, in
what time will it empty itself?

»dns. 2 minutes 30) seconds.

411.

327. When waterspouts from several apertures in the
side of a vessel, it is thrown with the greatest random from
the orifice nearest the centre, the jet issuing from the centre
will reach a horizontal distance equal to the entire height
of the liquid, and all jets equally distant from the centre
will be thrown to an equal horizontal distance.

CAbt.827.

)f a cubio foot.

)cond8.

d to disoharRA
;iiig tlio saino

all aperturo
5 foot abovo
per second,
ill aperturo
e height of

per second,
cuum, what
per second.

1 to that of a
leight of the

:in aperture
the fluid
cubic feet.

ottom of a

' issue per
constantly

)f an inch.

mis it to
g an area
vessel, in

seconds.

GS in tlio
om from
le centre
e height
centre

AaiS. 328-330.]

nYDRODTNAMICS.

Fig. 28.

149

NoTW.~Lot VA bo a veiiHol
flilod with water, having its
Hide AH Dor|M)ndicu1ar to tho
horizontal piano HM. On Ali
desoribo tho scinioircle liDA.
lUsoct Ali in C'and in All take
any points 1) and // equally
distant from B, also C aud c'
equally dlHtant from E. Draw
alHo CO, 1)1), HE, &c., porpon-
dicular to A Hand produced to
tho circumforcnco A ItC. Then
if Hmal I oriflcoii t)e pierced i n Ihn
>-. J« of tho voHHol at C',D',E',1)'

. d a, the liquid from E wilL__
upout to twico EE'.-Alt-nM-?k=
tho liquid from U or C" will
BTiout to // — twico VC or C'C and that from Dor D will roach K = twice
DD or D'lJf.

328. When water flows in any bed, as in the channel of
a river or in a , ipe, the velocity becomes constant when
tho lengtli of » d bears a large proportion to i(a sec-
tional area. TI -. ' ■ pipes of more than 100 feet in length
or in rivers whose course is unopposed by natural obstacles,
the velocity of the body of the stream is the same throughout
Wiicn this occurs tho liquid is said to be in train.

329. The velocity of the liquid flowing in a pipe or
channel is not the same in every part of its section^ being
greatest in tho centre of tlie section of tho pipe or in the
middle of the surface of the stream.

NoTB 1.— This arises from tho friction cxertod against the fluid by tho
Interior surface of tho plpo or the banks of the Htroain. !• a str«ain, on
account of the middle part having tho greatest velocity, tho surface Is always
more or less convex.

NoTB 2.-
ways :—

Ist. An open tube bent at right angles is placed in a stream with one of
its logs opposed to tho current and tho other branch vertical— the velocity
of tho stream is measured by tho height to which tho water rises in the
vortical log.

2nd. A float Is thrown into the stream and tho time occupied by It in
passiug over a known distance observed.

Srd. Tho convexity of tho surface may bo measured by a levelling instru*
mcnt, and its velocity thus determined.

330. To fmd the velocity of efflux f and hence the quantity of
water discharged in a given time from a reservoir of given height
through a pipe of given length and diameter: —

-The velocity of a stream may bo determined in throe different

150

HYDRODYNAMICS.

tABT8.8Sl-8S$.

■i

Let d = diameter of pipe, I = lengthf h = height, and v = velocity.
Then, all the dimensions being in feet, v =: 48V ) iTTTi \

Not B.— This is tlio formula of M. Foncelot and la regarded as strictly
accurate.

331. Water is frequently made to drive ma.;hinery by
its weit^ht or rnoinentuin exerted on a vertical watei-wliecl.

332. There are three varieties of vertical water-wheels,
viz : the undershot, the overahot, and the breast wheel.

Fig. 29.

Breast wheel.

Undershot wheel.

Overshot wheel.

NoTK.— The mode in which the water is made to act on these is repre-
sented in Fit?. 29. It will be observed that the undershot wheel is moved by
the moraentuui of the water — the breast wlieel and overshot wheel by its
weight aided by its momentum. An overshot wheel will produce twice
the effect of an undershot wheel, the dimensions, fall, and quantity of
water beiiii< the same. The breast wheel is found to consume twice the
quantity of water required by an overshot wheel to do the same work.

333. In all water-wheels the greatest mechanical effect
is produced when the velocity of the water is 2j- times
that of the wheel.

ITS. 891-3SS.

AST. SS4.3

^TDItODTKAMlCd.

151

-velocity.

^} "

as strictly

inery by
5i-wliecl.

pwheels,
eel.

rcpre-
lovecl by
1 by its
56 twice
ntity of
nee the
ark.

effect
times

334. To find the horse powers of a vertical water-
wheel —

Let b = breadth of stream in feet, d = depth of stream,
V = mean velocity in feet of stream per minute, h = height of fall,
$ =r weight of one cubic foot of water, m ^ modulus of the wheel,
and U = units of work.

«... mbdvsh ,

Then horse powers = ^^^^^ .

ExiuPLK 412.— A water-wheel is worked by a stream 6 feet
wide and 3 feet deep, the velocity of the water is 2i feet per
minute, and the height of the fall 30 feet, required the horse
powers of the wheel, the modulus being -7.

J5r.p.=

SOLUTIOir.
mhdvsh _ <X3X22X30X62'5X7
330UU 33000

= 15'75.iw#.

Example 413.— What is the horse powers of a water-wheel
worked by a stream 2 feet deep, 7 feet wide and having a velocity
of 33 feet per second— the fall being 10 feet and modulus of the
wheel '6 ?

jy.P. =

SOLUTIOBT.
mhdvith •8X7X2X33X6r5X10

33U0U

33U0O

= 4*25 Ans.

ExAUPLB 414.— A water reservoir is 100 feet in height, supplies
water to a city by a pipe 10000 feet in length and 6 inches in
diameter, what is the velocity per second and what quantity
will be discharged in 24 hours ?

Here h = 100, 1

Then Art. 330, v = 48-^ ?

SOLUHOX.

10000, and d = i.

t 10000+54

I 48

-^100 27

Z+54d)

=5 3*38 feet per second =: velocity.

Quantity discharged in 1 second = 3'1416X (J) « X3-36.

Quantity discharged in 24 hours = 3-1416X^J^ X 3-36 X 60 X 60 X 24 —
67001'1904 cubic feet. Ans.

EXERCISES.

415. A water-wheel is worked by a stream 4 feet wide and 3 feet
deep, the velocity of the water is 29 feet per minute, the
fall 20 feet ; required the horse powers of the wheel, its
modulus being '56 ? Ans. 7-38.

152 MISCELLANEOUS PROBLEMS.

416. A water-wheel is worked by a stream 2 feet deep and 4

feet wide, and having a velocity of 50 feet per minute, the
fall is 33 feet and the modulus *84, how many cubic
feet of water per hour will this wheel raise from the depth
of 44 feet? Am. 15120.

417. A water-wheel is worked by a stream 4 feet wide and 3

feet deep, the velocity of the water being 15 feet per
minate, and the fall 27 feet, how many gallons of water
per hour will this wheel raise to a height of 80 feet, the
modulus being -8 7 Ana. 18225 gallons.

418. A water reservoir 80 feet in height supplies water to a city

through a pipe 1 742 fiet in length and 4 inches in diameter,
what is the velocity of the water per second and how
many gallons will it deliver in 10 hours ?

Anf. 115925*04 gallons.

MISCELLANEOUS PROBLEMS.

1. What must be the length of a pendulum in the latitude of

Canada in order to vibrate once in 5 seconds?

2. In a lever the arm of the power is 7 feet long and the arm

of the weight 2 feet 7 inches ; with this instrument what
power will sustain a weight of 743 lbs. ?

3. la a Hydrostatic Press the sectional areas of the cylinders

are to one another as 1427 is to 3, and the force pump is
worked by means of a lever whose arms are to one another
as 27 to 2. Now if a power of 87 lbs. be applied to the
extremity of the lever, what upward pressure will be
exerted against the piston in the larger cylinder ?

4. A cannon ball is fired vertically with an initial velocity of

600 feet per second, it ii required to find —
1st. How far it will rise.

2nd. Where it will be at the end of the 7th second.
3rd. In how many seconds it will again reach the ground.
4th. What will be its terminal velocity.
5th. In what other moment of its flight it will have the same

velocity as at the end of the 5th second of its ascent.

6. A water-wheel is worked by means of a stream 4 feet wide
and 3 J feet deep, the water having a velocity of 27 feet
per minute, and falling from a height of 36 feet, how
many strokes per minute will it give to a forge hammer
weighing 7000 lbs., and the vertical length of the stroke
4 feet ?

6. In a differential wheel and axle the radii of the axles are 3i
and 3f inchesj and a power of 7 pounds sustains a weight
of 1000, what is the radius of the wheel?

> II.*' '^

:t'

i-p and 4
nute, the
ay cubic
;he depth
s. 15120.

ie and 3

feet per

of water

feet, the

gallons.

to a city
liameter,
and how

: gallons.

,titude of

. the arm
ent what

cylinders
pump is
another

>d to the
will be

locity of

(round.

he same
lent.

set wide
27 feet

et, how
ammer
stroke

are 3^
weight

MISCELLANEOUS PROBLEMS.

153

7. How far may an empty vessel capable of sustaining a pres-

sure of 1 59 lbs. to the square inch be sunk in water before
breaking?

8. In a screw the pitch is -j^f of an inch, the power lever 9 feet

2 inches long and the weight is 44000 lbs., what is the
power?

9. How many units of work are ' "tended in raising 70 cubic

feet of water to the L' ' .h^ '3 feet?

10. The piston of a low pressure Sic^m engine has an area of

360 inches and makes 13 strokes of 7 feet each per minute,
the pressure of the steam on the boiler being 40 lbs. to the
square inch, required the horse powers of the engine.

11. Through how ^nany feet will a power of 7 lbs., moving

through 120 feet, carry a weight of 29 lbs. ?

12. A locomotive weighing 75 tons is drawn along an inclined

plane with a uniform velocity of 40 miles per hour, as-
suming the inclination of the plane to be | in 100, and
taking friction and atmospheric pressure as usual, what
is the horse powers of the engine?

1st. If the train is ascending the plane ? ■ ^

2nd. If the train is descending the plane ?

13. If a body weighing 7 lbs. at the surface of the earth be carried

to a distance of 30000 miles from the earth, what will be
its weight?

14. With what velocity per second will water flow from a small

aperture in the side of a vessel, the fluid level being kept
constantly 12 feet above the centre of the orifice?

15. In a Hydrostatic Bellows the tube has a sectional area of 1}

inches, the area of the board is 37 inches, and the tube is
filled with water to the height of 28 feet, what upward
pressure is exerted against the board of the bellows ?

16. In a differential wheel and axle the radii of the axles are If

and 2\ inches, the radius of the wheel is 40 inches, what
power will sustain a weight of 8700 lbs. ?

17. A elock is observed to lose 17 minutes in 24 hours, how

much must 'the pendulum be lengthened in order that it
may keep correct time ?

18. At what height will the mercury stand in a barometer at an

elevation of 30'5 miles ?

19. An upright flood gate of a canal is 17 feet wide and 13 feet

deep, the water being on one side only and level with the

top ; required the pressure.
Ist. On the whole gate.
2nd. On the lowest three-fifths of the gate.

i I

I6i MISCELLANEOUS PBOBLEMS.

3rd. On the middle three<fiftbs of the gate. "' ^

4th. On the upper four-elevenths of the gate.
6th. On the lowest fire-twelfths of the gate.

20. A piece of stone weighs 23 oz. in air and oxily 14'7 oz. in
water ; required its specific gravity.

21 Through how many feet will a body fall in 21 seconds down
an incline of 7 in 16?

22. In a compound lever the arms of the power are 9, 7, 6, and

3 feet, the arms of the weight 3, 2, 1, and i feet, the
' power is 11 lbs. ; required the weight.

23. If mercury and milk are placed together in a bent glass tube

or syphon, and if the column of mercury is 7'9 inches in
length, what will be the length of the column of milk?

24. In a Hydrostatic Press the sectional areas of the cylinders

are to one another as 1111 to 2, the force pump is worked
hj means of a lever whose arms are to one another as 17
to 2 and the power applied is 123 lbs.; what is the up-
ward pressure everted against the piston in the large
cylinder? ^ „

25. In a differential screw the pitch of the exterior screw is f of

an inch, that of the interior screw is -^f of an inch, the
lever is 25 inches long and the power applied is 130 lbs.,
what is the pressure exerted ?

26. A seconds pendulum is observed when carried to the summit

of a mountain to lose 3 seconds in an hour ; what is the
height of the mountain ?

27. Through how many feet will a heavy body fall during the

10th, the 7th, and the 6th seconds of. its descent?

28. In what time will an upright vessel 20 feet high and filled

with water, empty itself through an aperture, in the bot-
tom, three-f fths of an inch in area^ the vessel containing
250 gallons ?

29. A train weighing 80 tons is drawn along a level plane with

a uniform velocity of 20 miles per hour, taking friction
and atmospheric resistance as usual, what is the horse
powers of the locomotive ?

30. What is the weight of the milk contained in a rectangular

vat 11 feet long, 7 feet wide, and 3 feet deep?

31. What would be the height of the mercury in the barometer

at an elevation of 29*7 miles ?

32' What power will support a weight of 666666 by means of an
endless screw having a winch 30 inches long, an axle
with a radius of 2 inches, and a wheel with 120 teeth ? ,

MISCELLANEOUS PROBLEMS.

155

4-7 oz. in

mds down

7, 6, and
I feet, the

iltLBB tube
inches in
)fmilk7

cylinders
is worked
ther as 17
is the up-
the large

ew is f of
inch, the
1 130 lbs.,

le summit
lat is the

iiring the
?

nd filled

the bot-

•ntaining

ane with

friction

le horse

iangular
irometer

ns of an
an axle
eeth?

33. How much work is required to raise 29 tons of coal Arom a

mine 1120 feet deep ?

34. With what volociy does a body move at the close of the

27th second of its descent?

35. What is the entire pressure exerted upon the body of a fish

having a surface of 11 square yards and being at a depth
of 140 feet ?

36. How much water will be discharged in one hour through an

aperture in the side or bottom of a vessel, the water in
the vessel being kept at the constant height of 17 feet
above the centre of the orifice, and the area of the latter
being seven-elevenths of an inch?

37. How many cubic feet of water can a man raise by means of

a chain pump from a depth of 120 feet in a day of 8 hours ?

38. If a stone be thrown down an incline of 11 in 100 with an

initial velocity of 140 feet per second, Vi^hat will be iti
velocity at the 10th second of its descent and through
how many feet will it fall in 21 seconds?

39. At what rate per hour will a train weighing 1 20 tons be

drawn up an incline of i in 100 by an engine of 90 horse
powers, taking friction as usual and neglecting atmos-
pheric resistance ?

40. A water-wheel is driven by a stream 4 feet wide and 3 feet

deep, the fall is 40 feet and the velocity of the stream 38 J
feet per minute — if the modulus of the wheel is "63, what
number of gallons of water will it raise per hour from a
depth of 270 feet?

41. In a system of movable pulleys a power of 2 lbs. sustains a

weight of 64 lbs. ; how many movable pulleys are there ?
1st. If the system be worked by one cord?
2nd. If there are as many cords as moveable pulleys ?

42. At what rate per hour will a horse draw a load whose gross

weight is 1800 lbs. on a road whose coeflBcient of friction

43. In a high pressure engine the piston has an area of 600 inches

and makes 30 strokes of 5 feet each per minute ; what
must be the pressure of the steam on the boiler in order
that the engine may pump 1000 gallons of water per
minute from a mine whose depth is 270 feet — making the
usual allowance for friction and the modulus of the pump ?

44. The oarometer at the summit of a mountain indicates a pres-

sure of 21-73 inches while at the base it indicates a
pressure of 29-44 inches, what is the height of the
mountain, taking the mean temperature of the two stations
as 63-70?

156

MISCELLANEOUS PROBLEMS.

4r ifa stone be thrown vertically upwards and af^ain reaches the
ground after a lapse of 16 seconds, to what height did
it rise ?

46. In a composition of levers the arms of the power are 8, 4, 2,

and 7, tlie arma of the weight are 3, 1, i, and 4 ; what
weight will be sustained by a power of 29 lbs.?

47. A piece of wood which weighs 17 oz. has attached to it a

metal sinlter which weighs 137 oz. in air and 8-G oz. in
■water — the united mass weighs only "6 of an ounce in
water; what is the specific gravity of the wood?

48. What must bo the area of an aperture in the bottom of a

vessel of water 18 feet deep and kept constantly full in
order that 27 cubic feet may be discharged per hour?

49. IIow many tons of coal will be raised per day of ten hours

from a mine whose depth is 400 feet, by a low pressure
engine in which the piston has an area of 1200 inches and
makes 20 strokes of 6 feet each per minute, the pressure
of the steam on the boiler being 45 lbs. to the sq. inch ?

.10. What power vvill support a weight of 70000 by means of a

screw having a pitch of ^\' of an inch and a power lever

9 leet 2 inches in length ?
61. In what time will a pendulum 50 inches long vibrate in the

latitude of Canada ?
52. In a lever whoso power arm is 8i times as long as the arm

of the weight, what power will sustain a weight of

729 lbs.?
63. A train weighing 130 tons is drawn along an incline of i in

100 with a uniform velocity of 25 miles per hour ; taking

fnotion and atmospheric resistance as usual, what is the

hors2 powers of the locomotive : —
1st. If the train is ascending the incline?
2nd. If the train is descending the incline?

54. A seconds pendulum is observed to lose 40 seconds in 24

hours on the summit of a mountain ; required its height.

55. A body is fired vertically with an initial velocity of 2000 feet

per second ; it is required to find —
1st. Where it will be at the end of the 120th second.
2nd. How far it will rise.

3rd. In what space of time it will again reach the ground.
4th. Its terminal velocity.
6th. In what other moment of its flight its velocity will be

the same as at the end of the 49th second.

56. In a wheel and axle the radius of the axle is 3 inches and a

weight of 247 lbs. is sustained by a power of 17 lbs. ;
•what is the radius of the wheel ?

MISCELLANEOUS PRODLEMS.

157

und.
ill be

md a
lbs. ;

6t. With what velocity doea «ratcr flow from a small aperture
in the side or bottom of a vesspl, the fluid level being
kei)t constant at iO feet above the centre of the orifice?

68. In a train of wheel work there are four wheels and four

axles, the flrst wliccl and last axle being plnnc, i. e., without
cogs, and having radii respectively of 12 and 2 feet — the
second wheel has '70, the third 80, the fourth 100 teeth ;
the first axle has 8, the second 1, and the third 1 1 leaves ;
with this machine what weight will bo sustained by a
power of 130 lbs. ?

69. To what depth may a closed empty glass vessel capable of

sustaining a pressure of 200 lbs. to the square inch be
sunk in water before it breaks?

60. Id a differential wheel and axle the radii of the axles are 1^

and 1^ inches ; a power of 2 lbs. sustains a weight of 749,
what is the radius of the wheel ?

61. How many units of work are expended in raising 247 tons

of coal from a depth of 478 ieet ?

62. What are the horse powers of an upright water wheel worked

by a stream 5 feet wide and 2 J feet deep, the velocity of
the water being 110 feet per minute, the fall 6 feet, and
the modulus of the wheel § ?

63. A train weighing 140 tons ascends a gradient having a rise

of I in 100 ; taking friction as usual and neglecting at-
mospheric resistance, what is the maximum speed the train
will attain if the horse powers of the locomotive be 160?

64. A barometer at the summit of a mountain indicates a pres-

sure of 21*4 inches while at the base the pressure is 30*2
inches ; what is the height of the mountain ?

65. On an incline of 7 in 100 what power acting parallel to the

plane will sustain a weight of 947 lbs. ?

66. What centrifugal force is exerted by a ball weighing 40 Ibg.

revolving in a circle 20 feet in diameter and making 140
revolutions per minute ?

67. What is the specific gravity of a piece of metal which weighs

23-47 oz. in air and only 18-12 oz. in water?

68. if a body be thrown vertically upwards and again reaches

the ground in 22 seconds —
Ist. With what vel city was it projected?
2nd. How far did it rise ? , '

69. In a screw the pitch is iV of an inch, the power lever is 40

inches long; what power will sustain a weight of
95000 lbs. ?

70. In what time will an engine of 120 horse powers, moving a

\5Q

MISCELLANEOUS PBOBLEMS.

train whoso gross weight ia 100 tons, complete a journey
of 300 miles, taking friction as usual, neglecting atmos-
pheric resistance, and assuming the rail to ascend
regularly i in 100 ?

71. An engine of 20 horse powers raises 7 tons of coal per minute

from the bottom of a mine 2001 feet deep and at the same
time causes a forgo hammer to make 40 lifts per minute of
3 feet each ; required the weight of the hammer.

72. In u Hydrostatic Press the sectional areas of the cylinders

arc to one another as 1411 to 3, the force pump is worked
b} a lever whoso arms are to one another us 28 to 3, tlio
upward pressure required is 9000 lbs. ; what must bo the
power applied ?

73. In a dilTerential screw the pitch of tho exterior screw is

','a and that of the inner screw tV of an inch, tho power
lever ia G feet 8 inches in length ; what pressure will bo
exerted by a power of 19 lbs. ?

74. A picco of nickel (spec. grav. 7*816) weighs 24 grains in air

and only 16*4 grains in a certain liquid ; required tlie
specific gravity of the liquid.

75. In a differential wheel and axle the radii of the axles are li

and iVVii^ches, the radius of the wheel is 42 inches ; what
weight may be sustained by a power of 23*7 lbs. ?

76. What gross load will a horso draw when travelling at the

rate of 3^ miles per hour on a road whose coefficient of
friction is j^g^?

77. A body has descended through a-\- x feet when a second

body commences to fall at a point 2m feet beneath it ;
what distance will the latter body fall before the former
passes it 7

78. On an incline of ^ in 70 what power acting parallel to the

plane will sustain a weight of 4790 lbs. ?

19. When a body has fallen 7000 feet down an incline of 7 in 20,
what velocity per second has it acquu:ed ?

80. A body weighing 100 lbs. and moving from south to north

with a velocity of 60 feet per second comes into contact
with another body which weighs 430 lbs. and is moving
from north to south with a velocity of 20 feet per second,
and the two bodies coalesce and move on together ;
required the direction and velocity of the motion of the
united mass.

81. An engine of 21 horse powers pumps 40 cubic feet of water

per hour from the bottom of a mine whose depth is 200
feet and at the same time draws coals from the bottom of
the mine ; required the tons of coals drawn up per hour.

MlSdELLlKEOUS PROBLEMS.

169

water
s 200
omof
our.

62. la a fljstom of pulloys worked hy aeveral cords, each attached
by both ends to the pulleys, there are 8 movable pulleys
and as many separate cords ; what weight will be
sustained by a power of 73 lbs. ?

83. A body weighing 20 lbs. and moving at the rate of 47 feet

per second comes in contact with another body weighing
270 lbs. and moving in the same direction with a velocity
of 30 feet per second ; required the velocity and momen-
tum of the united mass.

84. In what time will an engine of 150 horse powers draw a

train whose gross weight is 90 tons through a j- urncy of
220 miles, talcing friction as usual and neglecting atmos-
pheric resistance, one half of the journey to be on a level
plane and the other half up an incline of % in 100 ?

39. In a common wheel and axle a power of 7 lbs. su&tains a
weight of 974 ; the radius of the wheel is 61 inches, what
is the radius of the axle ?

86. At what height would the mercury stand in a barometc

placed at an elevation of 43'2 miles above the level rf
the earth 7

87. If a body be projected down an incline of 7 in 12 with an

initial velocity of 40 feet per second, through how many
feet will it move during the 10th second and over what
space will it have passed in 23 seconds 7

88. In a high pressure engine the piston has an area of 360 inches

and makes 17 strokes of 6 feet each per minute ; taking
the pressure of the steam on the boiler as equal to 56 lbs.
to the square inch,whatare the horse powers of the engine?

89. If a body weighing 1 11 lbs. moving to the east with a velocity

of 90 feet per second come in contact with nnother body
which weighs 70 lbs. and is moving to i\xp sver^t with a
velocity of 40 feet per second, and after tut two have
coalesced they come in contact with a third which weighs
80 lbs. and is moving to the east with a velocity of 20 feet
per second and the three then coalesce and move on
together, what will be the direction, velocity, and momen-
tum of the resulting motion 7

90. What must be the length of n pendulum in the latitude of

Canada in order that it may make 40 vibrations in 1
minute 7

91. What pressure will be exerted upon the body of a man at the

depth of 97 feet beneath the surface of water, the man's
Dody having a surface equal to 14 square feet?

92. A piece of cork which weighs 27*42 grains has attached to

it a sinker which weighs 34*71 grains in air and only

160

EXAMINATION PAPERS.

30*12 grains ia water, the united mass weiK'hs nothing in
water, i. e., ia of the same specific gravity as water ;
required the specific gravity of the cork.

93. What is the weight of a mass of slate which contains 2*7 cubic

feet?

94. How many cubic feet of iron are there in 87 tons?

95. What backward pressure is exerted by a horse in going down

a hill which has a rise of 3 in 40 with a load whose gross
weight is 2100 lbs., assuming friction to be equal to ^j of
the load ?

96. What pressure is exerted against one square yard of an

embankment if the upper edge of the yard be 17 feet and
the lower edge 18 feet beneath the surface of the water ?

97. The length of a wedge is 27 inches and the thickness of the

back 2f inches ; what weight may be raised by a pressure
of 17 lbs.?

98. What is the effective horse powers of a high pressure engine

in which the piston has an area of 420 inches and makes 30
strokes per minute, th^ boiler evaporating t^ of a cubic
foot of water per minute under a pressure of 60 lbs. to the
square inch ?

89. A train drawn by a locomotive of 100 H. P. moves along
an incline of i in 100 with a uniform velocity of 25 miles
per hour ; taking friction as usual and neglecting atmos-
pheric resistance, what is the weight of the train?

1st. If it is ascending the incline ?
2nd. If it is descending the incline ?

EXAMINATION PAPEUS.

' " 1.

1. A railway train weighing 110 tons is drawn along an incline

of i in 100 with a uniform velocity of 42 miles per hour ;
taking friction as usual and atmospheric resistance equal
to 20 lbs. when the train is moving at the rate of 7 mile»
per hour, what are the horse powers of the locomotive ?

1st. If the train is ascending the gradient?

2nd. If the train is descending the gradient?

2. Enunciate the principle of virtual velocities and calculate

through how many feet a weight of 89 7 lbs. will be carried
by a power of 11*7 lbs. moving through 123 feet?

EXAMINATION PAPERS.

161

othing in
s water ;

27 cubic

ing down
ose gross
I to is o*"

rd of an

feet and

J water ?

as of the
pressure

'6 engine
nakes 30
' a cubic
3. to the

»s along

25 miles

; atmos-
?

incline
hour;
equal
mileii

tive?

culatc
arried

3. In a differential wheel and axle the radii of the axles are 3f

and 3^ inches; the radius of the wheel is 42 inches, what
power will sustain a weight of 444-4 lbs. ?

4. Describe the barometer and explain the principles on which

it acts.

5. "What is he weight of a log of boxwood (spec. grar. 1-320)

17 feei long, 1 foot 9 inches wide, and 2 feet 3 inches
thick?

6. The upright gate of a canal is 12 feet wide and 16 feet deep,

the water being on one side only and level with the top ;
• required the pressure :—
1st. On the whole gate ;

2nd. On the lowest five-eighths of the gate ; and
3rd. On the middle seventh of the gate.

7. Give the composition of atmospheric air and state what are

the chief sources of the carbonic acid.

8. The piston of a high pressure engine has an area of 400

inches and makes 12 strokes of 6 feet each per minute, the
pressure of the steam on the boiler is 64 lbs. per square
f. inch ; how many tons of coal per hour will this engine

raise from a mine whose depth is 240 feet?

9. Distinguish between the essential and the accessory properties

of matter and enumerate the former.

10. An upright vessel 17 feet in height is filled with water and
holds just 200 gallons ; in what time will it empty itself
through an aperture in the bottom two fifths of an inch
in area ?

II.

1. A cannon ball is fired vertically with an initial velocity of

800 feet per second; required—
1st. How far it will ascend.

2nd. In what space of time it will again reach the ground.
3rd. Where it will be at the end of the 31st second.
4th. Its terminal velocity.
5th. In what other moment of its Sight it will have the same

velocity as at the close of the 13th second.

2. Enumerate the different kinds of attraction, deilne what is

meant by the attraction of gravity and state by what law
its intensity varies.

3. A piece of stone weighs 73 grains in air and only 35 grains

in water ; required its specific gravity.

162

feXAMIllATION PAPERS^

4. In a Hydrostatic Press the areas of the cylinders are to ond

another as 1347:2, the force pump is worked by means of
a lever whose arms are to one another as 23:2, the power
applied is 120 lbs. ; required the upward pressure exerted
against the piston in the larger cylinder.

5. In a lerer the power arm is T feet long, the arm of the

weight is 5 inches, the power is 11 Ibg. ; required the
weight.

6. Enunciate the principle of the parallelogram of forces and

explain how it is that a force may be more advantageously
represented by a line of given length than by saying it is
equal to a given number of lbs., &c.

7. Name the diflferent kinds of upright water wheels, explain

the difference between them and give the rule for finding
the horse powers of a water wheel.

8. If a closed empty vessel be sunk in water to the depth of

143 feet before it breaks, what was the extreme pressure
to the square inch it was capable of sustaining?

9. Describe what is meant by the vena contracta of escaping

fluids, indicate its position with reference to the orifice of
escape, and give the proportion between the area of the
aperture and the sectional area of the vena contracta.

10. An engine of 50 horse powers draws a train weighing 60 tons
up an incline of i in 100 with a uniform velocity of 20
miles per hour ; taking atmospheric resistance as usual,
what is the friction per ton ?

III.

1. By means of a lever a certain number of lbs. Troy attached

to the arm of the weight balances the same number of
ounces Avoirdupois attached to the arm of the power ;
required the ratio between the arms of the lever, a pound
Troy being to a pounl Avoirdupois as 5700:7000.

2. Enunciate TorricelWs theorem and calculate the velocity

with which a liquid spouts from a small orifice in the side
of a vessel when the level of the fluid is 24 feet above the
centre of the orifice.

3. In a Hydrostatic Bellows the sectional area of the tube is

three-sevenths of an inch and it contains 12 lbs. of water,
the area of the board of the bellows is 3*7 square feet ;
what is the upward pressure exerted against the board of
the bellows ?

EXAMIKATION PAPERS.

161

4. Through hotr many feet will a body fall during the 21st

second of its descent?

5. Define what is meant by specific gravity : Give the rule for

calculating the specilic gravity of a solid not sufficiently
heavy to sink in water and calculate the specific gravity
of cork from the following data : —
A piece of cork which weighs 20 oz. in air has attached to
it an iron sinker which weighs 18 oz. in air and only
15*73 oz. in water; the united mass weighs 1 oz. in
water, req«ired the specific gravity of the cork.

6. What w«ight would be carried through a space of 7 feet by

a power of 5 lbs. moving through 40 feet ?

7. Define what is meant by the centre of gravity of a body and

explain how it may be experimentally determined in a
solid.

8. How many tons of coal per day of ten hours may be raised

from a mine of 660 feet in depth by a low pressure engine
having a piston which has an area of 500 inches, and
makes 20 strokes of 11 feet each per minute, the gross
pressure of the steam on the boiler being 37 lbs. per
square inch?

9. The power arm of a lever is 32 times as long as the arm of

the ^yeight, the power is 97 oz. ; required the weight.

10. A city is supplied with water through a pipe 8 inches in
diameter and 1 mile in length, leading to a reservoir whose
height is 140 feet above the remote end of the pipe ; what
will be the velocity of the water per second and how
much will be discharged in one houi*?

IV.

1. Enunciate the law of decrease in the pressure and density of

the air as we ascend into the higher regions of the
atmosphere.

2. In a Hydrostatic Press the sectional areas of the cylinders

are to one another as 943:2, the force pump is wtrked by
means of a lever whose arms are to one another as 19:3 ;
if the power applied be 87 lbs., what is the upward pres-
sure exerted against the piston in the larger cf linder ?

3. The power arm of a lever is 9 feet long, the arm of the weight

ia 17 feet long and the weight is 6i lbs. ; required the
power.

■OM

164

EXAMINATION PAPERS.

4. Explain when a body is said to be in a condition of itable,

un$table, or indifferent equilibrium.

5. A train weighing 90 tons is drawn along an incline of 2 in

900 with a uniform velocity of 30 miles per hour; taking
friction and atmospheric resistance as usual, what are the
horse powers of the locomotive : —

Ist. If the train is ascending the gradient?

2nd. If the train is descending the gradient?

6. A stone is dropt into a mine and is heard to strike the bot*

torn in lU seconds; required the depth of the mine, if
sound travels at the rate of 1066} feot per second.

7. State the condition of equilibrium in the differential wheel

and axle.

8. What is the weight of the sulphuric acid (specific gravity

1'84I) contained in a rectangular vat 1 feet 4 inches long,
2 feet 5 inches deep, and 3 feet 7 inches wide?

9. At the toptof a mountain a barometer indicates a pressure of

21 inches while at the base the pressure is 29*78 inches —
the temperature at the top is 40'70'' Fahr. and that at the
base is 70'70** Fahr. ; required the height of the mountain.

10. A high pressure steam engine raises 200 cubic feet of water
per minute from a depth of 80 feet , the piston iiasi an area
of 800 inches and makes 10 strokes per minute of 8 feet
each, what is the pressure of the steam on the boiler ?

4.
5.

1. The flood gate of a canal is 10 feet long and 7 feet deep, the

water being on one side and level with the top ; what is

the pressure : —
Ist. On the whole gate ?
2nd. On the lowest two-sevenths of the gate?
3rd. On the middle three-sevenths of the gate?
4th. On the lowest one-ninth of the gate?

2. In a compound lever the arms of the power are 6, 7, and 11

feet, the arms of the weight are 2, 3, and 5 feet ; by means
of this combination what power will sustain a weight of
1000 lbs. ?

3. Enunciate Mariotte's law and ascertain what will be the

density, volume, and elasticity of that amount of atmos-
pheric air, which, under ordinary circumstances, i. e., at
the level of the sea or under a pressure of 15 lbs. to the
square inch, fills a gallon measure, if it be placed in a
piston and subjected to a pressure of 60 lbs. to the sq. inch.

8.
9.

EXAMINATION PAPERS.

165

2 in
taking

4. What power moving through 29 feet will carry a weight of

7 lbs. through 70 feet?

5. An engine of 12 horse powers gives motion to a forge ham-

mer which weighs 400 lbs. and makes 40 lifts of 3 feet
each per minute and at the same time pumps water from
a mine 100 feet deep ; required the number of cubic feet of
water it pumps per hour from the mine.

6. On an inclined plane a power of 341 lbs. acting parallel to

the base sustains a weight of 27900 lbs ; what must be the
length of the base in order that the height may be 11 feet ?

7. Enunciate the three laws of motion commonly known as

Newton's laws, and state to whom they respectively
belong.

8. A piece of sulphur weighs 19 oz. in air and 10 oz. in water ;

required its specific gravity.

9. A ball is thrown up an incline of II in 16 with an initial

velocity of 1100 feet per second ; required —
1st. To what height it will rise,
2nd. Where it will be at the end of the 20th second.
3rd. In what time it will again reach the ground.
4th. Its terminal velocity.
5th. In what other moment of its flight it will have the same

velocity as at the end of the I7th second of its ascent.

10. Required the pressure exerted against a mill-dam 170 feet
long and 16 feet wide, the perpendicular depth of the
water being 12 feet.

\1.

1. When the barometer indicates a pressure 30 inches at the

surface of the earth it is observed to indicate a pressure
of only 13*5 inches in a balloon ; required the approximate
height of the balloon. ,^

2. Giv6 the chief laws connected with the motion of projectiles.
1st. When they are fired vertically, and

2nd. When they are fired at an angle of elevation.

3. Through how many feet will a body fall in 39 seconds ?

4. What are the horse powers of a low pressure engine in which

the piston has an area of 360 inches and makes 11 strokes
of 9 feet each per minute, the gross pressure of the steam
on the boiler being 53 lbs. to the square inch ?

5. What must be the area of the apertare in the side of a yessel

166

EXAMINATION PAPERS.

kept filled with wator to a height of 20 feet above the
centre of the orifice in order that 15 cubic feet of water
may be discharged in one hour ?

6. Describe Bramah's Hydrostatic Press and explain upon what

principle in philosophy its action depends.

7. A piece of wood which weighs 19 oz. has attached to it a

metal sinker which weighs 2*7 oz. in air and 22-'7 oz. in
water— the united mass weighs 11 oz. in water ; required
the specific gravity of the wood.

8. In a compound lever the arms of the power are 7, 8, 9, and

10 feet, the arms of the weight are 2, 3, 4, and 1 feet, the
power is 19 lbs. ; what is the weight?

9. Explain the diflference between the common and the forcing

pump, and also state why the former is sometimes called
the lifting pump.

10.

A train weighing 80 tons is moving at the rate of 30 feet per
second | v/hen the steam is turned off, how far will it
ascend an incline of 3 in 1000, taking friction as usual and
neglecting atmospheric resistance ?

VII.

J. What amount of pressure is exerted against one square yard
of an embankment, the upper edge of the square yard
being 16 ft. 3 in. and the lower edge 19 ft. 9 in. below the
surface of the water ?

2. How much must the pendulum of a clock which gains 1

minute in an hour be shortened in order that it may keep
correct time ?

3. Describe the syphon and give the theory of its action.

4. In a system of eleven movable pulleys worked by a single

cord what weight will a power of 27 lbs. sustain ?

5. In a Hydrostatic Press the large cylinder has a sectional area

of 6i feet, the smaller cylinder a sectional area of 2i
inches, the force pump is worked by means of a lever
whose arms are to one another as 19: IJ ; now if a power
of 100 lbs. be applied to the extremity of the lever what
upward pressure will be exerted against the piston in the
larger cylinder ?

6. Describe the differential screw, and give the conditions of

equilibrium between the power and weight in the common
screw.

EXAMINATION PAPERS.

167

above the
t of water

ipon Tvhat

led to it a
2-7 oz. in
; required

8, 9, and
1 feet, the

le forcing
acs called

JO feet per
far will it
usual and

lare yard
are yard
i)elow the

gains 1
nay keep

a single
?

>nal area
a of 2^
a lever
a power
'^er what
m in the

tions of
common

1. To what depth may an empty glass vessel capable of sus-
taining a pressure of 197 lbs. to the square inch be sunk
in water before it breaks ?

8. In a system of pulleys consisting of eight movable pulleys

worked by eight cords, the upper end of each fastened to
the beam, the power is 7i lbs., what is the weight ?

9. How many gallons of water per hour will an engine of 7

horse powers pump from a mine 67 feet in depth, making
the usual allowance for the modulus of the pump?

10. The piston of a low pressure engi-^e has an area of 400 inches
and makes 20 strokes, each 8 feet in length, per minute, the
boiler evaporates -731 of a cubic foot of water per minute,
what are the useful horse powers of the engine ?

VIII.

1. Explain the difference between the simple and compound

pendulum — also what is meant by the " centre of oscil-
lation" and by the "centre of percussion."

2. What velocity will a heavy body falling freely in the latitude

of London acquire in one entire second, the London
second's pendulum being 39*13 inches long?

3. In a Hydrostatic Bellows the tube is filled with water to the

height of 13 J feet; what uptv^ard pressure is exerted
against the board of the bellows if the area of the latter be
3-,\ feet ?

4. In a differential screw the exterior screw hab a pitch of iV of

an inch, the interior screw a pitch of -5,^ of an inch, the
power lever is 50 inches long; what pressure will be
exerted by a power of 130 lbs. ?

5. A train weighing 100 tons moves up a gradient having an

inclination of f in 100 with a uniform speed of 20 miles
per hour ; taking friction and atmospheric resistance as
usual, what are the horse powers of the locomotive ?

6. When a body has fallen through 2500 feet what velocity has

it acquired ?

7. Explain what is meant by gaseout diffusion and show the

important influence it has in maintaining the composition
of atmospheric air constant at all places.

8. In a common wheel and axle the radius of the axle is 1 1 inches

and the radius of the wheel 47 in. : what power will, with
this machine, sustain a weight of 793 lbs. ?

168

EXAMINATION PAPERS.

9. A flood gate is 22 feet wide and 20 feet deep, the water being
on one side only and level with the top ', required the
pressure—

1st. Against the whole gate.

2nd. Against the lowest three-sevenths.

3rd. Against the upper four-ninths.

4th. Against the middle three-elevenths.

6ih. Against the lowest three-tifths.

10. Give the different rules for finding the specific gravity of
liquids.

IX.

1. In a differential wheel and axle the radii of the axles are 2f

and 2-,3f inches, the radius of the wheel is 90 inches ; what
weight will be sustained by a power of 7 lbs. ?

2. The tube of a Hydrostatic Bellows is filled with water to the

height of. 50 feet; if the board of the bellows has an area
of 6} feet, what upward pressure is exerted against it ?

3. How many vibrations per minute will a pendulum 9 yards

long make ?

4. Give the principal laws of the descent of bodies on inclined

planes.

5. A body has fallen through 3600 feet when another body

begins to fall at a point 4000 feet beneath it ; through
what space will the latter body fall before the first over-
takes it ?

6. The piston of a steam engine has an area of 440 inches and

makes 11 strokes per minute, each 9i^f feet in length,
the boiler evaporates '9 of a cubic foot of water per
minute ; what is the volume of the steam produced per
minute and what is the pressure under which it is gene-
rated ?

1. Give the most important consequences that result from the
fact that each atom of a liquid is separately drawn towards
the centre of the earth by the force of gravity.

8. What gross load will a horse exerting a traction of 74 lbs.

draw on a road whose coefficient of friction is it^f ?

9. What are the conditions of equilibrium between the power

and weight in the inclined plane ?

10. Through how many feet must a body fall in order to acquire
a velocity of 250 feet per second ?

ANSWERS TO EXAMINATION PAPERS. 169

ANSWERS AND

REFERENCES
PAPERS.

TO EXAMINATION

1. H. P. = 29008 or 43-68.

2. Art. 66.

3. -1619 lbs.

4. Arts. 227,229.

5. 5522-34375 lbs.

6. 96000 lbs., 82S00 lbs., and
137l4t lbs.

7. Art. 205.

8. 151-2 tons.

9. Arts. 9, 19, and 10.
10. 125 tons or 750 tons.

II.

1. H.P = 161-28 or 38-08.

6. Art. 44.

2. Arts. 25, 27.

7. Art. 205.

3. 1-921.

8. 6205 lbs. ■ ;

4. 929430 lbs.

9. Arts. 9, 19, and 10.

5. 184} lbs.

10. 8025 lbs. per ton.

III.

1. Power arm 13-^ times

6. 28flbs. .:

great as the arm of the

7. Arts. 57, 58. .

weight.

8. 1400.

2. Arts. 25, 26, and 27.

9. 194 lbs.

3. 14918-4 lbs.

10. Velocity = 6-366 feet per

4. 688 feet.

second.

5. Arts. 192, 195, and -57584.

Quantity = 7962071 cubic

feet per hour.

IV. ':

1. Art. 212.

6. 1600 feet.

2. 259796J lbs.

7. Art. 88.

3. 12^V

8. 730700144 lbs.

4. Art. 62.

9. 9721-2 feet.

5. H.P. = 176-26 or 69-6.

10. 33^ to the square inch.

1. 15312Jlb3.,7500lbs.,6562J
lbs., and3213H|lbs.

2. 64^7f lbs.

3. Art. 219, density 4 times as
great, volume 1 qt. and
elasticity 60 lbs. to the sq.
inch.

4. 16|§lb3.

6. 3340} cubic feet.

7.
8.
9.

257.

10.

900 feet.

Arts. 255, 256,

2111.

27500 feet.

At elevation of 17600 feet.

100 seconds.

1100 feet per second.

At the end of the 33rd sec.

1020000 lbs.

170 ANSWERS TO EXAMINATION PAPERS.

VI.

1. 19724 feet.

2. Art. 282.

3. 24330 feet.

4. 45-36 H. P.

6. ^YA ^^^^ ''^®^'

7.
8.

10.

Arts. 183 and 182, Note.

•618.

3990 Iba.

Arts. U33 and 234.

2163-4 feet.

VII.

1 1040GJ lbs.

Arts. 129, 126.

2. 1-303 incheg.

454 feet.

8 Art. 235.

1920 lbs.

4 594 lbs.

13 791 'J gallons

6. 626933i lbs.

10.

II. P. = 67-87.

VIII.

1. Arts. 301, 302, and 308.

1851? lbs.

2. 386-n inches.

275000 lbs.

3. 3022-68672.

185204A lbs.

4. 14660K5-6 lbs.

64320gV lbs.

6. 133-t>62.

75000 lbs.

6. 400 feet per second.

231000 lbs.

7. Art. 206.

10.

Art. 196. -

1. 8085 lbs.

2. 21479-04 lbs.

3. 20-8

4. Art. 270.

6. llllHe®*'

e. Volume = 339
Pressure = 85

7. Art. 175.

8. 1776 lbs.

9. Art. IIG.
10. 976^6 feet.

'5 cub. feet!

lbs. the sq.

[inch.

EXAMINATION QUESTIONS.

171

rote.

lib

feet;
the sq.
[inch.

QUESTIONS TO BE ANSWERED ORALLY BY THE PUPIL.

Note.— TA« mtmh&rs following the questiong refvr to the articles in
the work where the annwcra nitty be found,

1. What is Natural ScUsncc? (1)

2. Into what classcH an; all iintiirni objncts divided and how arc these
distiiiKuiNhod from each other V (2)

How areaiiiiiialMdisliitKuiiihod from vcgotablos? (3) . ^

What is ZooloK'i' ? (4)

Wl)atis Hot»iiy¥ (4) ;

Wljati8MineraloKy?(4) »

" (4)

8.
4.

6.
6.
7.

Wliat is AHtronomy ?

8. Wliati8Gcoloiry?(4)
0. What is ChcnniHt

try MO
What is tho object of Natural Philosophy? (4)
What am tho Huhdivisious of Natural Philosophy ? (5)
In what separate forms dooH matter exist V (0)
Dollno what is moaiit by tho essential properties of matter. (9)
Enumerate tho ossuiitial properties of matter. (1(»)
What is extension ? ( 1 1 )

10.
11.
12.
1.}.
14.
18.

16. What is impenetrability P Give some illustrations. (12)

17. What is divisibility? (i;{)

18. Docs tho property of divisibility belong to masses or to particles of

matter or to both ? (13)

19. Give some illustrationsoftheoxtrcmc divisibility of matter? (13,Note\
SO. What is Indesfcruotibility ? (14)

21. What is Poro.sity ? {U)

22. What is Compressibility? (10)

23. What is Inertia? (17)

24. If l)0(lies cannot bring themsclvfjs to a state of rest, how is it that all

bodies movin;< upon the earth soou come to rest? (17, Note)
28. What is elasticity ? (18)
26. Name tho dilferent kinds of elasticity as applied to solida. (18, Note)

27. What aro the accessory properties of matter? (19)

28. Enumerate some of the most important of tho accessory properties of

matter. (20)
"" WhatismalleaV)ility ? Which aro thcmost malleable
What is ductility P Name the most ductile metals.

SO.
31.
82.
33.
84.
85.
38

87.
38.
89.
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
60.
61.
62.

Whatistenacit;^'? (23)
is attraction? (24)

•of the metals? (21)
(22)

(25)

(31)

What

Enumerate the dilVercnt kinds of attraction.

What is the attraction of gravity ? (20)

what is the law of variation in the intensity of gravity? (27)

Explain what is meant by saying tho force of gravity varies in/erwly

as the square of the distance. (28)
What is the attraction of cohesion? (29)
What is the attraction of adhesion? (30)
What is capillary attraction ? Give some examples.
What is ehictrical attraction ? (32)
What is magnetic attraction P (33)
What is chemical attraction ? (34)
What is the derivation of the word Statics 7 (36)
what is the object of tho science of Statics? (36)
what is the derivation of the word llydroHatics? (.JO)
What is tlie oVyect of the science of Hydrostatics P (:J«)
What is tho derivation of the word Dynamics? (36)
What is the object of the science of Dynamics? (36)
Wh„t is the derivation of the word Hydrodynamics ? (36)
what is the object of the science of Hydrodynamics? (30)
what is the derivation of the word Pneumatics 1 (36)
What is the object of the science of i'neumatics r (36)

172

EXAMINATION QUESTIONS.

S3.
M.
65.
50.
57.

68.
SO.

60.

fll.
62.

63.

64.
65.

60.
«S.

•49.

70.
71.

73.
74.

75.

.76.
77.

78.
79.
80.

81.
82.
«3.

84.

65.

86.
87.
88.
89.

90.
91.

92.

93.
94.

95.
96.

Whon is a b^y Mid to be in nquillbriura ? (37)

WImt aro Htatical forcoH or pruMSurcii V (:i8)

What aro tli« clMmeiits of a fonio V (39)

Whnt aro tiio dilffri'iit inodcH of roproscntlriR a foro P (40)

Whoii Hcveral forcoH act upon tlio samu iioint of a body, how manj

niotionH oaii thoy givo it V (41 )
DiNtiiiKuiHh hotwo«'n component and remtltant forces. (42)
If N(>v<>ral forcoH act unon a point in tlio humu HtraiKlit lino and in ths

Haiiio diroction, to wliat \» tlivir rcHultant equal V (43)
Wlion Novcral forcott act upon a point in tlio Hanio straight lino but in

opixiHito diroctionH, to what itt their rcHultant c<{ual? (43)
Enunciato the principle of tho paralleloKrani of forccH (4-1)
Wlicn Huvoral forcett act on a point in any direction whatever, state

how the reeiultant may bo found. (45)
What ia tho distinction botwocn the parallelogram of forces and the

parallelopipcd of forcea? (4«J)
Wliat in tiio rosulti

iltant of two parallel forces which act on difTcrent
points of a body but in tho same direction? (47)

what is tho resultant of two parallel forces which act on different
t>ointa of a body and in difi'orent directions P (48)

How do wo find the resultant of any number of parallel forces? (49)

What is a couple I (50)

DistiuKiiish between tho composition of forces and tho resolution of
foro<!S. (5t)

What is the coiltro of gravity of a body ? (57)

Why is tho centre of jfravity callijd also the <'entre of parallel forces? (55)

How may tho centre of gravity of a solid body bo experimentally
determined ? (58)

If a body be free to move in any direction, how will it finally rest with
reference to its centre of gravity? («(»)

How is the stability of a body estimated ? (61)

When is a body said to be in a condition of $table, unstable, or in-
different equilibrium ? (62)

How may the centre of gravity of two scparato bodies bo found ? (68)

What is tho object of all mechanical contiivanoes? (64)

By what law or principle in philosophy is the relative gain or loss of

power and velocity in a macniiie determined? (K5)
Enunciato the principle of virtual velocities? (66)
Wiiat is a machine? (fi?)
How many mechanical elements enter into the composition of

machinery? (68)
Name the primary mechanical elements. (68)
Naino the secondary mechanical elements. (68)
From what mechanical element is the wheel and axle formed P (69)
or what mechanical element are the wedge and screw modifica-
tions? (60)
How are levers, cords, &c., regarded in theoretical mechanics? (70)
What is a lever? (71)
Of how many kinds are levers? (72)
Of simple straight levers how many kinds aro there P (73)
Upon what does the distinction between the three kinds of levers

depend ? (73)
Give examples of lovers of tho first class. (74)
How are the fulcrum, power, and weight placed in levers of the first

class? (75)
How are the fulcrum, power, and weight placed in levers of the second

class? (75)
Oive some examples of levers of the second class. (75)
How are the fulcrum, power, and weight placed in levers of the third

class? (76)
Give some examples of levers of the third class? (76)
In lovers of tho first class, which must bo greatest, the power or the

weight? (76, Note)

EXAMINATION OUESTIONg.

173

OSS of

(70)

' levers

he first
second

le third

or the

07. In InvPTx of the srcond rlawi, which must bo grratrst, the i»owor or the

wi'iKlit? (7«(. Nolo)

08. In loviTK of tho third olasN, wliich miiHt bo the grcatcHt, tlie power or

thn woightV (7tJ. Note)
00. 'What iH tht) arm of the .wriKht ? AVhat is tlin arm of tho power ? (77)

100. Wliiitnre thi< coiniitions of uqiillibriiiia butwuoii the power and the

wciKht In thek'vorV (77)

101. Dt'ducc fortr.uIoM for llniUnfr the power, tlie weight, tlie ami of thn

iM)W(!r or the nrm of the wei^bt whon the other three are given. (77)

102. When thcarniHof the lev<>r are curved or bent, how uiuht their effective

lengths be determined? (70)

103. Wliat iH a compound lever or eompoHitlon of levcrsl' (80)

10k. Deduce rules for tiutliug the power or the weight in a compound
lever. (Hi)

105. Describe the wheel and axle. (h2)

lOrt. Why is thu wheel and axle sometimes eallod a pcrpefual leverP (84)

107. AVhat are the conditions of equilibrium in the wheel and nxb*? (NR)

108. Deduce a set of rules for finding the power, the weight, the radius of

the axleor the ra<liuH of the wheel when the other three are given. (86)

109. Describe the differential wheel and axle? (S7)

110. To what is it, in effect, equivalent? (H7)

111. Dedtice a set of rules for the dillerential wheel and axle.

112. In toothed gear how is tho ratio between the power and the weight

determined? (80)

113. How are axles commonly made to act on wheels? (00)

114. When is wheel work used to concentrate power? Give an example. (92)

115. When is wheel work used to diffuse power? Give an example. (02)

116. What are the conditions of equilibrium in a system of toothed wheels

and pinions ? (OD)

117. What is a pinion 7 what are leaves ? (01)

118. Deduce formulas for finding the power and the weight in a system of

wheels and axles ? (04)

119. Explain what is meant by the hunting cog ? (95)

120. Name the different kinds of wheels? (96)

121. Explain the difference between crown, spur, and bevelled gear? (97)

122. Explain for what purpose crown, spur, or bevelled gear is used? (PS.)
12D. When bevelled wheels of different diameters are to be used together,

show how tho sections of the cones of which they are to be frusta are
found? (00)

184. What is a pulley P (100)

125. Show that from the pulley itself no mechanical advantage is derived *

(101)

126. Wherein consists the real advantage of the pulley and cord as a me-

chanical power? (101)
l->7. When is a pulley said to be fixed ? (102)
128. What is a single movable pulley called ? (103)
120. What are Spanish Bartons? (10;J)

130. Explain the meaning of the worHs sheaf, block, and iachle? (104)

131. What is the only mechanical aflvantage derived from the use of a fixed

pulley? (105)

132. In a system of pulleys worked by a single cord, what are the conditions

of equilibrium ? (106)

133. Deduce a set of rules for a system of pulleys worked by a single cord ?

(107)

134. What are tho conditions of equilibrium in a Spanish Barton when the

separate cords are attached directly to the beam ? (108)

135. What are the conditions of equilibrium when tho separate cords are

attached to the movable pulleys? (109)

136. Deduce in each of these last two cases a set of rules for finding the
ratio between the power and the weight ? (llO and ill)

B';i.f ■t.;'- " ^ imjy, i ^jp i ^f^i.4j i ^^ ^P^

174

EXAMINATION QUESTIONS.

137.

l.'W.
139.
W).
141.
142.
143.
141.
145.

145.

147.

14S.

149.
150.
151.
152.
153.
134.

JOJ.

150.
157.
15S.
IS'J.
lf)0.
161.
Hi2.
103.

164.
165.
166.
1(57.
168.
160.
170.

171.
172.

173.
174.
175.
17.').
177.
178.
17U.

ISO.
IS].
182.

183.

184.

185.

If tho lines of direction of the power and weight malce with one
unotlicr an angle (jroater than 120"^, what is the relation betweon the
power and the weight? (112)

In th(!oretical mechanics now is the Inclined plane regarded ? (1T3)

Wiiat are the modes of indicating the inclination of tlio pl.'uieP (il4)

In the inclined plane how may the power be applied ? (115)

What are the conditions of cqnilibrium in the uiclined plane? (116)

Dcdnco a set of rnl»-s for the inclined plane ? (117)

What is the wedge? (118)

How is the w(Hlge worked P (110)

Wiiat are ih-.' conditions of equilibrium in the wedge when it is worked
l)y proasnre? (120)

In what imi)ortant particular does the wedge differ from all the other
mechanical powers? (120, Note 1)

Give some examples of the application of the wedge to practical pur-
poses? (120, Note 2)

Deduce a set of rules for the wedge P (121)

Describe the screw ? (122)

How is the screw related to an ordinary inclined plane ? (122, Note.)

What is the pitch of the screw ? (123)

How is tlio screw commonly worked ? (124 and 125)

What are the conditions of equilibrium in the screw? (126)

How mav the clficiency of the screw as a mechanical power be increas-
ed ? (127)

Deduce a set of rules for the common screw ? (128)

By whom was the differential screw invented P (129)

Upon what principle does the differential screw act ? (129)

To what is the differential screw, in effect, equivalent ? (129)

Deduce a sot of rules for tlie differential screw P (130)

Describe the endless screw? (131)

What are the conditions of equilibrium in the endless screw P (132)

Deduce a set of rules for the eniUess screw ? (139)

How does friction affect the relation between the power and the
weight in the incchanical elements? (135)

What are the different kinds of friction ? (136)

Wliat is meant by the coefficient of friction ? (137)

What is the (iooilicient of sliding friction ? (138)

AVhat is the coefficient of friction on railways ? (138)

AVhat is the coefficient of friction on good macadamized roads ? (138)

What is meant by the force of traction ? (138)

Enumerate the different expedients in common use for diminishing
friction? (139)

Give Coulomb's conclusions as reirards sliding friction ? (139)

Give Coulomb's ^lonclusions as regards rolling friction? (139)

What is Iho unit of work 1 (140)

How are tli(^ units of work expended in raising a body found ? (141)

What are the most important sources of laboring forces? (112)

How many units of work are there in one horse i)owcr ? (142)

Wliat is m(!ant by the Table in Art. 142 ?

What is tiio true work of tlie horse per minute? (1 12. Note)

In moving a carriage along a horizontal plane, for what purpose is work

expended ?
In the case of railway trains what is the amount of friction ? (143)
In the case of railway trains when does the velocity become uniform?
Upon what does the traction or force with which an animal pulls de-
pend? (Ifti)
At what rate per hour must a horse travel to do most wck ? (146)
Upon what docs the amount of atmospheric resistance ca^ crienced by

a moving body depend P (147)
Explain wliat is meant by tiiis P (147)

ith one

/wn the

(113)
P (iU)

(116)

I worked
he other
ical pur-

, Note.)
iiicreas-

EXAMINATION QUESTIONS.

17$

';*

(132)
md the

(138)
inishing

(141)

; is work

U:\)

iiiform?
nils de-

14(5)
need by

186.

187.

188.

189.

190.
191.
192.

19S.
191.
195.
196.
197.
198.
199.

200.

201.

202.
203.

204.
205.
206.
207.

208.
209.
210.
211.
212.
213.
214.
215.

216.

217.

218.
219.

220.

221.

222.
223.

224.

What is the amount of atmospheric resistance experienced by a train
of medium length moving at the rate of 10 miles per hour P (148)

If a body be moved along a surface without friction or atmospheric
resistance, how may the units of work performed be found P (149)

When a train is moved along an inclined plane, how is the work per-
formed by the locomotive found P (150)

Deduce a set of formulas for finding the horse power, weight, maximum
speed, &c., of trains ? (151 )

What is meant by the modulus of a machine P (162)

Of machines for raising water, which has the greatest modulus P (163)

How is the work performed by water falling from a height found ?
(154)

How is steam converted Into a source of laboring force? (155)

What are the two principle varieties of the steam engine P (156)

What are the essential parts of the high pressure engine P (157)

How does the low pressure differ from the high pressure engine f (168)

What are the varieties of the low pressure engine P ^159)

How do these dillcr from each other P (160, 161)

In the high pressure engine, at what part of the stroke does atmos-
pheric pressure act against the piston ? (162)

Give the leading ideas that enter into the construction of the steam
engine? (163)

In wliat respects is the low pressure engine preferable to the non-con-
densing engine ? (164)

How are the units of work performed by an engine found ? (165)
Knowing the pressure of the steam on the boiler, how do we obtain the

useful pressure on the piston P ( 166)
Give the rules for finding the H. P., &c., of engines ? (167)
What is the real source of work in the steam engine ? (168)
Why is it most advantageous to employ steam of high pressure P (168)
Give formulas for iinding tlie area of the piston, length of stroke, pres-
sure, effective evaporation, &c., in the steam engine P (169)
Define what is meant by ajluicl ? (171)
How is the term fluid commonly applied P (172)
Into what classes are fluids divided ? Name the type of each. (173)
To what extent is water compressible P Alcohol ? (173, Note)
How do liquids chiefly difter from gases ? (174) ;

In what respects do liquids chiefly differ from solids? (175)
Give the most important consequences that flow from this fact? (176)
How would you illustrate tlie upward and lateral pressure of liquids ?

(175, Note)
What relation exists between the respective heights of two liquids of

different densities placed in an inverted syphon? (176)
What is the amount of downward pressure exerted by a liquid confined

in any vessel? (177)
How would you illustrate this fact? (177, Note.)
Show that weight and pressure are not to be confounded with one

another? (177, Note 2.)
What are tho weights respectively of a cubic inch, a cubic foot, and a

gallon of water, at the teinperature of 60° Fahr. ? (178)
To what is the pressure exerted by water on a vertical or inclined

surface equal? (179)
Give a rule for finding tho lateral pressure exerted by water? (179)
How do you find the pressure exerted by water against a vertical or

inclined surface at a ^iveu depth beneath the water ? (180)
How do you find the pressure exerted against any fraction of a vorti-
cal surface when the upper edge is level with tho surface of the
water! a81)

225. Explain what is meant by transmission of pressure by liquids? (182)

226. Describe Braraah's Hydrostatic Press, and lUustratv. by a figure ? (183)

227. Explain the pri ciple upon which Bramah's Press acts F (182, Note)

176

EXAMINATION QUESTIONS.

228. For %hzt purposes is Bramah's Press used ? (184)

S29. How do we And the relation between the power applied and the preS'

sure obtained by Bramah's I'tckh? (185)
2S0. Describe what is meant bjr the Hydrostatic Paradox? (186)

231. Show that it is not in reality a i)ai-adoxf (186, Note)

232. Describe the Hydrostatic Bellows ? (187)

233. Give the rule for linding the upward pressure against the board of a

Hydrostatic Bellows? (188)

234. "When will a body float, sink, or rest in equilibrium in a fluid ? (189)

235. What weight of liquid docs a floating body displace? (190)

236. What portion of its weight is lost by a body immersed in a liquid f

(191)

237. What is the specific gravity of a body? (192)

238. What is the standard of comparison for solids and liquids ? (193)

239. What is the standard of comparison for all gases? (193)

240. How do we find the specific gravity of asolid heavier than water? (1P4)

241. How do we find the specific gravity of a solid not sufficiently heavy to

sink in water? (195)

242. What is the first method of finding the specific gravity of a liquid P

(19«)
2i^3. What is the second method of finding the specific gravity of a liquid ?

(196)

244. How is the specific gravity of a liquid determined by means of the

Hydrometer? (196)

245. Describe the Hydrometer ? (196)

246. What difference is there between liydrometers designed for deter-

mining the specific gravity of liquids specifically lighter than water,
and those for ascertaining the specific gravity of liquids specifically
heavier than water ? (I9t})

247. How is the specific gravity of gases found P (197)

248. How may the weight of a cubic foot of any substance be found when

its specific gravity is known ? (199)

249. How may the solid contents of a body be found from its weight ? (200)

250. How may the weight of a body be found from its solid contents ? (201)

251.
252.
263.
254.
255.
256.
257.
258.
259.
260.

261.

262.
263.

264.

265.
266.

267.
868.
260.
270.
271.
272.

What is Pneumatics P (202)

What is the derivation of the word atmosphere P (203)

What is the atmospliere ? (203)

To what height does the atmosphere extend ? (204)

Give the exact composition of atmospheric air ? (205)

What purpose is served by the oxygen in the air P (205, Note)

What purjiose is served by the nitroeon P (205, Note)

Describe the principal properties of carbonic acid ? (205, Note)

What are the chief sources of carbonic acid ? (205, Note)

What is the maximum and what the minimum amount of carbonic

acid in the air? (205, Note)
Describe the mode by which the air is kept sufficiently pure to sustain

animal life. (205, Note)
Describe the property of gaseous diffusion. (206)
Explain how the property of gaseous ditt'usion affects the composition

or the atmosphere. (206, Note)
Upon whatdoestheauiouutof aqueous vaporpresentin the atmosphere

depend? (207)
What is its maximum amount P What its minimum amount ? (207)
To wliat is the blue color of the sky due ? To what the golden tints of

sunset? (208)
Which of tlie essential properties of matter belong to air P (209)
How would you illustrate the impenetrability of air ? (209, Note)
How would you illustrate the inertia of the air ? (209, Note 2)
Why does air possess weight ? (210)

What may be taken as the fundamental fact of pneumatics ? (210, Note)
What is tne weight of 100 cubic inches of each of the following gases,

viz., oxygen, hydrogen, nitrogen, atmospheric air, carbonic air ?

EXAMINATION QUESTIONS.

177

epres-

rd of a

(189)

liquid?

'P (1P4)
Davy to

liquid?

iquid P

of the

deter-
water,
iiically

when

' (200)
ISJOl)

273. Give some illustrations of the aggregate weight of the atmosphere?

(210, Note 2)

274. How in it that the lower strata of air are denser than the upper P (211)

275. By what law does the density of the atmosphere decrease as we aa^

cendP (212)

•bonic
istafii

sition

phere

[)7)
nts of

)

Jfote)

cases.

276.
277.

278.

279.
280.

m\.

282.

28;j.

284.
285.
286.
287.
288.
289.
290.
291.
292.
293.
294.
295.
296.

297.
298.
299.
300.
301.
802.
303.

304.
SOS.
.%6.
307.
808.

309.

310.
311.
312-

313.
314.
315.

316.

317.
318.
319.

320.

From what does the pressure of the air result ? (213)

What do we mean by sayinR the pressure of the air is equal to 16 lbs.
to the square inch ? (2l3,Note)

If the air were of the same density throughout to what height would
it extend? (214)

How is this known ? (214)

How are perraaiientiy elastic gases chiefly distinguished from non-
elastic gases V (216)

What is meant by permanently elastic gases ? (216, Note)

Illustrate what is meant by tiie elasticity of a gas. (217)

To what is the elasticity of gases due P (217, Note)

Eiuuiciate Mariotte's law ? (219)

Illustrate it by a bent tube as in Art. 218.

To what extent is Mariotte's law true P (219, Note)

What is the air-pump P (220)

By whom and when was it invented ? (221, Note)

Describe the exhausting syringe. (222)

Draw a sketch of the air-pump and describe its mode of action. (222)

Upon what principle does the air-pump act ? (223)

How perfect a vaoium can be secured by the air-pump P (223, Note)

Describe the condensing syringe, (224)

For what purpose is the. air-pump chiefly used? (223)

Give some illustrations of the pressure of the air ¥ (225, Note)

Give some illustrations of the elasticity of the air. (225, Note)

What is the barometer P (226)

By whom and when was it invented P (226, Note)

What are the essential parts of a barometer? (227)

What is meant by the Torricellian vacuum? (227, Note)

How may the excellency of a baronicter be tested? (228)

What is the cause of th» oscillations of the barometer y ?29)

In what regions of the earth are the oscillations of the biu ometer most

fitful and ectensive? (229 Note)
To what regular oscillations is the barometer subject ? (£,;vO";
At what hours are the two maxima of pressure V (2;Wy)
At what hours are the two minima of pressure V (230V
In what region are the serai-diurnal oscillations grciitiibtP (230, Note)
Give some idea of their extent in tropical countries aad cxplaii? why

they arc not observed in our climate. (230, Note>
How may the weather to be expected be foretold u.y ;he oscillatioiis m

the height of the barometric column? (251)
What does a fall in the barometer denote P (231, II.)
What does a rise in the barometer indicate? (231, III.)
What does a sudden change in the height of the mercury in the

barometer denote? (231, IV.)
What does a steady rise in the column denote? (231, V.)
What does a steady fall in the column denote? (231, VI.)
What does a fluctuating state in the height of the column of mercery

denote? (231, VII.)
Give Halley's rule for ascertaining the height of mountains, Ac, bv

the barometer. (232)
Give Halley's rule with correction for temperature. (232)
Give Leslie's rule. (232)
Describe the essential parts of a common pump and illustrate by i

diagram. (233)
Explain why the common pump is sometimoa called a lifting pump.

(233, Note)

178

EXAMINATION QUESTIONS.

32 1 . Explain tho principle upon which the common pump acts. (2.'53, Note 2)
o22. Explain why tlie lower valve must be within 32 feet of tho water in tha

reservoir in order that the pump may act at all times. (234, Note 2)
321. Descrribo the forcing pump. (231')

32 1. Describe tlio essential purts of a Are engine. (234, Note)
325, Describe the syplion. (235)

32(). How is tho syphon set in operation? (235, Note 1)
327. Explain upon what principle the syphon acts. (235, Note 2)

328.

329.
330.
331.
332.
S33.
334.
333,
336,
337.
338,
339.

340,
341.
342.
313,
344,
3i5.

340.

347,

348.

349.
350,
351,
352.
353,
354.

355.
35(5,
357,
358,

359,

3(50,

361.
302,
803,
S64.
305.
366.

Wlien does the consideration of forces come under tho science of

statics V (230)
What kind of forces are considered in dynamics? (236)
Why is statics called a deductive science!' (237)
Why is dynamics called an inductive or experimental science? (237)
What may force be doQned to be? (238)
For wliat purposes is force required? (238)
What are the diflerent kinds of forces as regards duration? (239)
What are the dill'ereut kinds of contirmed forces ? (239^
What may motion be defined to be? (2l0)
What are the qualities of motion ? (241)
What arc the dilforent kinds of motion ? (241)
What kind of a motion is produced by an accelerating, constant, or

retarding force? (2'<'')
What is velocity? (243)
Of how many kinds is velocity ? (243)
Wh«!n is velocity said to be uniform ? (243)
What is momentum or motal force? (245)
To what are the momenta of bodies proportional ? (240)
When tho velocities of two moving bodies are equal, to what are their

momenta proportional ? (247)
When the masses of two moving bodies are equal, to wliat are their

momenta proportional ? (248)
Wlien we speak of multii) lying a velocity by a weight, what do we

mean ? (249, note)
When force is communicated by impact to a body at rest, how long

will the body remain at rest ? (254)
Give the first general law of motion. (255)
Whose law is this? (257, Note)
Give the second law of motion. (256)
Whose law is this? (257, Note)
Give the third law of motion. (257)
Whoso law is this ? (257, Note)

What is reflected motion? (258)

What is the angle of incidence ? (258)

Wliat is the angle of reflection ? (258)

What proportion exists between the angle of incidence and the angle

of reflection? (258)
How would all bodies fall in a vacuum? (259)
Upon what does the resistance encountered by a body moving through

tho atmosphere depend? (20(>)
What is the nature of the motion of a heavy body falling from a

height? (2«n)
WMiat velocity is acquired by a heavy body in falling thrcngh one

second? (2154)
Through how many feet docs a body fall during the first second of its

descent? (2(;5j
Deduce^ a set of formulas for tho descent of bodies freely through

space. (206)
When a body is projected upwards what is tho nature of its

motion? (207)
Give the formulas for the motion of a body projected upwards or

downwards ? (208)

EXAMINATION QUESTIONS.

179

Note 2)
r in the
rote 2)

once of

' (237)
9)

ant, or

3 their
! their
do we
V long

angle

ough
loin a
one
if its
Jugh
' its
s or

367. When ahody is descending an incline how is the gravity expended? (289)
808. What are the laws of descent on inc^litied planes? (270)

369. Upon what is the final velocity of a body falling down an iucline

dependent? (271)

370. What are the laws of descent in curves? (273)

371. What is the braeliystoehrone?

372. What is a cycloid? (274)

373. Deduce a set of formulas for descent on inclines, (27r), 276)

374i.
375.
376.
877.
378.

379.

3S0.
381.
3S2.

383.
384.

385.

386.

387.

388.

389.
390.
391.

392.

393.

894.

395.

396.

Whatisaproji'ctile? (277)

What forces iiillucnco projectiles? (278)

What is the theoretical path of a projectile? (278)

What is a parabola? (278, Note 1)

Upon what erroneous suppositions is the parabolic theory based?

(278, Note 2)
Show that when ahody is projected horizontally forward the horizontal

motion does not interfere with thi; action of gravity. (279. Note)
What are th(! thret; conclusions of the parabolic theory V (2S0)
What is the greatest horizontal range of a projectile? (2S0, Note)
To wliat is tlie velocity of projection speedily reduced, no matter what

it may liave been originally ? (2si) ,

How do you explain tliis? (281, Note 1)
What is th.' atmospheric resistance encountered by a ball or other

projectile having a velocity of 2000 feet per second? (281, Note 2)
When a ball has considerable windage, what is the amount of deflection

in its course? (281, Note 3.)
What are the most important laws regarding the motion of projoctilcB

thrown vertically into the air? (282)
What are the most important laws regarding the motion of projectiles

thrown at an angle of elevatiim? (282)
To what is the explosive force of gunpowder exploded in a cannon

equal ? (283)
With what velocity does exploded gunpowd'ir tend to expand ? (283)
What is the c mposition of gunpowder ? (283, Note)
What is the geatest initial velocity that can be given to a cannon ball ?

(284)
To what is the velocity of a ball of given weight fired with a given

charge of powder projiori ional ? (284, Note)
To what are the velocities of balls of equal weight fired by the same

charge of i»owder proportional ? (2S5)
To what a '. f he velocities of balls of different weights but of the same

dimensiui.s tired by ecjiial quantities of powder proportional ? (28(5)
To what is the depth which a ball penetrates into an obstacle propor-
tional ? (287)
Give the rule for finding the velocity of any shot or shell when its

weight and also that of the charge of powder are known ? (288)

307.
398.
399.
400.

401.
402.
403.
491.

405.
406.

407.

408.
409.

Wliat is centrifugal force ? (2S9)

Wiiyt is sometimes called tang(;ntial force ? (289, Note)

What isoeutiipetal force? (29i))

When does a body move in a circle? (291 )

When does a body move in an ellipse? (2'.)1)

How long can a rotating mass jjreserve itself? (292, Note 1)

Giv(! some examples of the elllv'ts of centrifugal force. (292, Note 2)

If the velocity ami radius are constant, to what is the centrifugal force

proportional? (-I'M)
When the radius is constant how doe»^ the centrifngal force vary? (294)
What is the amount of centrifugal force at the equator? (294, Note)
How rapidly must the earth revolve in order that the centrifugal force

at the equator may equal gravity? (29t, Note)
■W hen the velocity is constant how does the centrifugal force vary? (295)
AVhon the number of revolutions is constant to what is the centrifugal

force proportional? (296)

180

EXAMINATION QUESTIONS.

410. Give a set of formulas for calculating centrifugal force. (297) .

411. Give a rule for liudinK the work accumulated in a moving body. (200)

412. "Wh.it is a pendulum P (300)

413. "What is a simple pendulum ? (301)

414. What is a compound or material pendulum? (302)

415. "What is an oscillation or vibration P (303)

416. "What is the amplitude of the arc of vibration ? (304)

417. "What if t^^ duration of a vibration? (305)

418. "What l:i tht length of a pendulum? (306)

419. "What is the centre of suspension? (307) r
45SO. "What is the centre of oscillation ? (3(i8)

421. What is the centre of percvission ? (308, Note)

422. What is meant by saying the centres of oscillation and suspension are

interchanjjeableP (369)

423. How is the duration of a vibration afTected by its amplitude P (310)

424. What is meant by saying the vibration of the pendulum is isochronous P

(31 i>, Note)

425. vVhal relation exists between the lengths and times of vibrations of

pendulums? (314)

4J16. n.ve the chief laws of the oscillations of the pendulum. (311-316)
427. ^» iiy docs the seconds pendulum vary in length in different latitudes ?

•;n6, Note)
4i>S. "Wiiat is the length of a seconds pendulum in Canada? (316, Note a)
4ii9. '^<.- what purposes is tie pendulum applied? (317)
430, ^ is the pendUlum used as a measure of time? (317, Note)
481. ; ■■■■> V IS the pendulum used as a staudard of measure? (317, Note)

432. il.'v do we find the length of a pendulum to vibrate in a given

time (319)

433. How do we find the number of vibrations lost by a pendulum of given

length when the force of gravity is decreased ? (320)

434. How do we find the number of vibrations gained by a pendulum of

given length when it is shortened? (321)

435. "What isthe science of Hydrodynamics? (322)

436. Enunciate Torricelli's theorem. (323)

437. In what time does a full vessel empty itself through an orifice in the

bottom? (324)

438. How is the quantity of fluid discharged through an orifice of given size

found? (325)
489. What is the vena contracta^ (325, Note)

440. What relation exists between the theoretical discharge and the actual

discharge ? (325, Note)

441. Give the rule for finding the velocity and quantity of fluid discharged

through an aperture of given size. (326)

442. When water spouts from an aperture in the side of a vessel how is the

horizontal distance to which it is thro.vn found? (327)
S43. When a liquid flows through a pipe or channel, which part has the

greatest velocity ? (329)
441. How is the velocity of a strear. ;• iterii. ned ? (329, Note 2)

445. What are the principal varietie^^ of water wheels? ^",32)

446. In water whoels, wlieu is the grc i; *pst mechanical «;i' x ^t produced P (333)

447. Give the rule for finding the.ho:>.>! powers of upright water wheels. (334)

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