International Library of Technology
398
Sciences Pertaining to
Civi. Engineering
167 ILLUSTRATIONS
By
EDITORIAL STAFF
INTERNATIONAL CORRESPONDENCE SCHOOLS
FUNDAMENTAL PRINCIPLES OF MECHANICS
ANALYTIC STATICS
KINEMATICS AND KINETICS
HYDROSTATICS
PNEUMATICS
RUDIMENTS OF ANALYTIC GEOMETRY
Published by
INTERNATIONAL TEXTBOOK COMPANY
SCRANTON, PA.
1927
GJ?
Fundamental Principles of Mechanics: Copyright, 1906, by INTERNATIONAL TEXT-
BOOK COMPANY. Entered at 'Stationers' Hall, London.
Analytic Statics, Parts 1 and 2: Copyright, 1906, by INTERNATIONAL TEXTBOOK
COMPANY. Entered at Stationers' Hall, London.
Kinematics and Kinetics: Copyright, 1906, by INTERNATIONAL TEXTBOOK COMPANY.
Entered at Stationers' Hall, London.
Hydrostatics: Copyright, 1906, by INTERNATIONAL TEXTBOOK COMPANY. Entered at
Stationers' Hall, London.
Pneumatics: Copyright, 190.6, 1901, by INTERNATIONAL TEXTBOOK COMPANY.
Copyright, 1901, 1897, 1895, 1893, by THE COLLIERY ENGINEER COMPANY.
Rudiments of Analytic Geometry: Copyright, 1906, by INTERNATIONAL TEXTBOOK
COMPANY. Entered at Stationers' Hall, London.
All rights reserved
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PREFACE
The volumes of the International Library of Technology are
made up of Instruction Papers, or Sections, comprising the
various courses of instruction for students of the International
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pared by persons thoroughly qualified both technically and by
experience to write with authority, and in many cases they are
regularly employed elsewhere in practical work as experts.
The manuscripts are then carefully edited to make them suit-
able for correspondence instruction. The Instruction Papers
are written clearly and in the simplest language possible, so as
to make them readily understood by all students. Necessary
technical expressions are clearly explained' when introduced.
The great majority of our students wish to prepare them-
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more congenial occupations. Usually they are employed and
able to devote only a few hours a day to study. Therefore
every effort must be made to give them practical and accurate
information in clear and concise form and to make this infor-
mation include all of the essentials but none of the non-
essentials. To make the text clear, illustrations are used
freely. These Illustrations are especially made by our own
Illustrating Department in order to adapt them fully to the
requirements of the text
In the table of contents that immediately follows are given
the titles of the Sections included in this volume, and under
each title are listed the main tppics discussed.
INTERNATIONAL TEXTBOOK COMPANY
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CONTENTS
NOTE. Tills volume is made up of a number of separate Sections, the page
numbeis of which usually begin with 1. To enable the reader to distinguish
between the different Sections, each one is designated by a number preceded by a
Section mark (), which appears at the top of each page, opposite the page number.
In this list of contents, the Section number is given following the title of the Section,
and under each title appears a full synopsis of the subjects treated. This table of
contents will enable the reader to find readily any topic covered.
FUNDAMENTAL PRINCIPLES OF MECHANICS, 27
Pages
Motion 1-10
Introductory Definitions 1-4
Matter; Body, substance, and material; Motion and rest;
Path; Initial and final position; Displacement.
Velocity 5-10
Velocity in uniform motion; Formulas for uniform
motion ; Velocity in variable motion ; Average or mean
velocity; Instantaneous velocity; Direction and magni-
tude of velocity.
Force and Mass 1 1-52
Force ' 11-14
Definition of force; Balanced and unbalanced forces;
Equilibrium; Weight; Measure of force; Magnitude of
a force; Direction of a force; Mechanics; Dynamics;
Kinematics; Statics.
Fundamental Laws of Dynamics 15-35
Law of inertia ; Facts on which the law is founded ;
Galileo's law of the independent effect of forces;
Acceleration; Formulas for uniformly accelerated mo-
tion; Initial and final velocities; Mass; Determination
of the mass of a body; Law of action and reaction;
Stress ; Tension, compression, and 'pressure.
Immediate Consequences of the Preceding Laws 36-52
Important formulas; Space passed over in uniformly
accelerated motion; Modification of the formulas
when the body has an initial velocity; Retardation;
Composition and resolution of forces.
:v
vi CONTENTS
ANALYTIC STATICS, 28, 29
28 Pages
Concurrent Coplanar Forces 1-33
Definitions 1-3
Coplanar and non-coplanar forces; Analytic statics;
Graphic statics ; System of forces ; Equilibrants ; Inter-
nal and external forces.
Fundamental Principles and Formulas 4-10
Moments 11-13
Conditions of Equilibrium 14-23
General statement of the conditions of equilbrium; Equi-
librium of three forces ; Selection of axes.
Stresses in Framed Structures 24-33
Structure ; Machine ; Frames ; Trusses ; Supports ; Reac-
tions ; Struts and ties ; Determination of stresses.
Parallel Forces 34-49
Coplanar Forces 34-45
Two forces ; Theorem of moments ; Definition of a couple ;
Any number of forces.
Non-Coplanar Parallel Forces 46-49
29
Center of Gravity 1-28
Definitions and General Properties 1-6
Important Cases .' 7-28
Symmetrical figures ; Determination of the center of grav-
ity by addition and subtraction; Center of gravity of
polygons; Center of gravity of areas bounded by cir-
cular arcs ; Center of gravity of a plane area bounded
by an irregular curve; Center of gravity of solids.
Coplanar and Non-Concurrent Forces 29-43
Couples 29-35
Effect of a circle; Equivalance and equilibrium of coax-
ial couples.
Equivalence and Equilibrium of Coplanar Non-Concur-
rent Forces ". 36-43
Friction 44-65
Sliding Friction 44-54
Definitions and general principles; Angle and coefficient
of friction.
Resistance to Rolling 1 55-57
The Inclined Plane 58-65
CONTENTS vii
KINEMATICS AND KINETICS, 30 Pages
Composition and Resolution of Velocities 1-7
Uniform Motion in a Circle 8-22
Angular Velocity 8-12
Centripetal and Centrifugal Forces 13-17
Motion of a Train on a Curved Track 18-22
Work and Energy 23-34
Work 23-28
Energy 29-34
HYDROSTATICS, 31
Properties of Liquids : 1-2
Liquid Pressure 3-24
Pascal's Law and Its Applications 3-6
General Theory of Liquid Pressure 7-14
Pressure on an Immersed Surface 15-24
Buoyant Effort of Liquids 25-33
Immersion and Flotation 25-28
Specific Gravity 29-33
PNEUMATICS, 32
Properties of Air and Other Gases 1-16
Vacuum; Barometer; Tension of gases; Mariotte's law;
Manometers and gauges ; Gay-Lussac's law.
The Mixing of Gases. 14-16
Pneumatic Machines 17-41
The Air Pump 17-22
Air Compressors 23-25
The Siphon 26-27
The Locomotive Blast 28
Pumps 29-41
Suction pump; Lifting pump; Force pump; Plunger
pump; Double-acting pumps; Steam pump; Centrif-
ugal pumps; Hydraulic ram; Power necessary to work
a pump.
RUDIMENTS OF ANALYTIC GEOMETRY, 33
Graphs of Equations 1-14
Equations of Lines 15-31
viii CONTENTS
RUDIMENTS OF ANALYTIC GEOMETRY
(Continued') Pages
The Straight Line 16-22
Equation of the straight line, Graph of any equation of
the first degree, Applications, Pressure on the back of
the dam
The Parabola 23-31
Focus; Directrix, Vertex, Axes, Parameter, Applica-
tions , Jet of water issuing from a small orifice , Moment
in a beam.
FUNDAMENTAL PRINCIPLES OF
MECHANICS
MOTION
INTRODUCTORY DEFINITIONS
MATTER-BODY
1. Matter may be defined as either- (a) whatever is
known, or can be supposed capable of being known, through
the sense of touch; or (b) that which occupies space
2. A phenomenon is whatevef happens in nature. The
motions of the heavenly bodies, the rolling of a ball on the
ground, the generation of steam from water, the formation
of rust on the surface of iron, the circulation of the blood,
the beating of the heart, thinking, speaking, walking ali
these are phenomena It will be seen from the definition
and examples just given that the word phenomenon, as
understood and used in science, does not mean "something
extraordinary or wonderful." For science, there is nothing
extraordinary, and a phenomenon is simply a fact anything
that happens.
3. Body, Substance, Material. A body is any
limited portion of matter, as a block of wood, a coin, a
stone, a piece of flesh, an animal. Another meaning of the
word will be given presently.
copvmoHTio BY INTRHNATIONAU TEXTBOOK COMPANY ALL RIHT MMMVBD
27
I I. T 8-3
2 FUNDAMENTAL PRINCIPLES 27
4. All bodies do not possess the same properties, or do
not possess them in the same degree. This fact makes it
necessary to distinguish different kinds of matter, to which
different names are given 1 as, iron, water, air, flesh.
The word substance is often employed in the same sense
as the word matter, but in its more common use, it refers
to matter of a special kind. Thus, water, steel, iron, are
substances, or kinds of matter.
The word body is also used, especially in physics and
chemistry, to denote matter of a particular kind. In this
sense, water, air, hydrogen, gold, are called bodies. In
mechanics, however, the term is usually taken in the sense
given to it in Art. 3.
5. Those substances that are used for works of art are
called materials: steel, iron, brick, stone are examples.
6. Particle. A particle is a body so small that its
dimensions may be disregarded. A particle is accoidmgly
treated as if it were a geometrical point, except that it is
sometimes necessary to consider some of its physical
properties, such as weight. A particle is often called a
material point. All bodies are treated as being com-
posed of an indefinitely large number of particles.
7. It will be noticed in what follows that, in many cases,
properties and principles that have been established and
stated as relating to material points are extended to whole
bodies. Thus, after investigating and explaining the laws
governing the motion of a material point under certain
circumstances, these laws are illustrated by taking as
examples the motions of wheels, cars, engines, etc. The
reason is that in cases of this kind the properties considered
do not depend on the size of the bodies. For example, the
speed of a car moving in a straight track is measured by the
speed of any of its points, and what applies to the speed of
a point applies to the speed of the car.
So, too, the force requj v to move in a :traight track a
train of a certain weigh the same as the force necessary
to move a particle of the same weight; and, therefore, the
27 OF MECHANICS 3
laws obtained m the case of a particle may be applied to the
motion of a tram or any othei body, bo long as the only
thing to be taken into account is the weight of the body.
8. Deformation. When a body is pulled, pressed, or
struck m any direction, its shape is more or less changed,
either permanently or temporarily. A change of this kind
is called deformation, and the body undergoing it is said
to be deformed.
9. A rigid body is a body that is not deformable, that
is, a body whose foim and dimensions remain the same, to
whatever action it may be subjected
There are no absolutely rigid bodies m nature; but it often
simplifies matters to consider bodies at first as being per-
fectly rigid, which is equivalent to neglecting their deform-
ability, the latter being afterwards taken into account.
Besides, in many cases, deformability plays so small a part
that, for practical purposes, it may be neglected entirely.
MOTION AND REST
10. Relative Position. The relative position of a
point with respect to another is determined by the length
and direction of the straight line between the two points
When either the length or the direction, or both the length
and direction, of this line change, the relative position of the
two points is said to change
The relative position of two bodies with respect to each
other is determined by the relative position of the points of
one body with respect to those of the other. When the rela-
tive position of one or more points m one of the bodies with
respect to the points of the other body changes, the relative
position of the two bodies, with respect to each other, is also
said to change.
11. Motion is a change iiT-rtie relative position of two
bodies, and is said to be post* 1 ' M by either body with
respect to the other.
4 FUNDAMENTAL PRINCIPLES 27
12. Rest is the condition of two bodies not in motion
with respect to each other, and is said to belong to either
body with respect to the other. Each body is said to be
fixed with respect to the other body.
13. When the motion or rest of a body is referred to,
without specifying 1 any other body, it is generally understood
that the other body is the earth or its surface. Thus, the
motion of a train, of a steamer, of a horse, as usually spoken
of, means the change of position of the train, steamer, or
horse with respect to the ground that is, to the surface of
the earth, or to objects on that surface.
14. A body may be in motion with respect to another
body, and at rest with respect to a third body. For example,
the smokestack of a locomotive is at rest with respect to the
boiler, since their relative position does not change; yet, both
may be moving with respect to objects on the ground. Like-
wise, a man standing on the deck of a moving vessel is at
rest with respect to the vessel, but in motion with respect to
the water, the shore, etc.
15. Path. The path, or trajectory, of a moving point
is the line over which the point moves, or, in geometrical
language, the line generated by the moving point. The path
of a point may be straight or curved, or a combination of
straight and curved lines.
16. Initial and Final Position. When the motion
of a point during a certain time or over a certain portion of
its path is considered, tha
position occupied by the
point at the beginning of
the time is called the Ini-
tial position of the point,
Flt< ' and the position at the end
of the time is called the final position. Suppose that a
particle starts from A, Fig. 1, and moves so that it describes,
in a certain time, the curve A M B\ this curve is the path of
the particle. With respect to the time considered, A is the
initial and B the final position of the particle,. , , ,
27 OF MECHANICS 5
17. Displacement. The length of the straight line
joining the initial and the final position of a moving point
is called the displacement of the point while the point
passes from the former to the latter position. In Fig 1,
in which the point moves from A to B m the curved path
AM B, its displacement during this motion is the length of
the straight line AB
18. The space passed over, or described, m a certain
time by a moving point is the length of the part of the path
over which the body passes during that time. In Fig 1, the
space described by the particle between A and B is the length
of the curve A MB
19. Direction of Motion In a Curve. By the direc-
tion of a curve at any point is meant the direction of its
tangent at that point So, too, when a point is moving in a
curved path, the direction of motion at any moment is the
direction of the tangent to the path at the point occupied at
that moment by the moving point. In Fig 1, the line PT
just touches the curve AMD at P, it is, therefore, a tangent
to the curve at P, and shows the direction in which the
particle is moving when it reaches the position P
VELOCITY
VELOCITY IN UNIFORM MOTION
20. Uniform Motion. A point is said to move with
uniform motion when it passes over equal distances in
any and every two equal intervals of time The fact, for
instance, that a point moves over 10 feet during the first
second of its motion, and over an equal space during the
fiftieth second, is not enough to define the motion as uni-
form: that the motion may be uniform, the point must
describe the same space (10 feet m this case) in every
second, whatever instant is chosen in which to begin to
count the time; thus, the space described by the point
between the middle of the third second and the middle of
6 FUNDAMENTAL PRINCIPLES 27
the fourth must be the same as the space described in the
twentieth second, or during the second in which the time of
motion changes from 27 15 to 28.15 seconds, or, in short,
during any and every interval of 1 second.
21. Definition of "Velocity. It is a familiar fact that
bodies move more or less rapidly, more or less slowly By
this is meant that they require more or less time to move
from one place to another, or that they move over longer or
shorter distances in a given time. In ordinary language,
the term speed is used to denote the degree of quickness or
rapidity of motion. The measure of the speed, or of the
degree of quickness, of a motion is called velocity. How
this measure is obtained will now be explained.
22. Uniform. Velocity. Let a point M> Fig. 2, move
along the path A B, which may have any form, either curved
or straight; and suppose the
motion to be such that equal
lfIG 2 ""* spaces are passed over in equal
times; that is, suppose the motion to be uniform. Also, let
the number of units of length passed over in a unit of time
(as the number of feet passed over in 1 second) be denoted
by v. It is evident that the magnitude of this number v
depends on how fast the point is moving. Thus, if the point
passes over 10 feet in 1 second, it is moving twice as fast as
if it passed over only 5 feet in 1 second Hence, the num-
ber v may be used to measure the rapidity of the motion, in
which case v is the velocity of the moving point. In general
terms, then, the velocity of a point moving with uniform
motion is the space passed over by the point in a unit of time.
When, as here assumed, the motion is uniform, the veloc-
ity also is said to be uniform, or constant. What is meant
by variable motion and variable velocity will be explained
presently.
Velocity is expressed in units of length per unit of time.
For example, if a body moving uniformly passes over 7 feet
in 1 second, its velocity is 7 feet per second. In 1 minute,
the body passes over a space of 7 X 60 = 420 feet; hence,
27 OF MECHANICS 7
its velocity can be expressed also as 420 feet per minute. A
train that moves uniformly and travels 45 miles an hour has
a velocity of 45 miles per hour.
23. Formulas for Uniform Motion. Let a point or
body moving uniformly pass over a space j during any time t.
Then, it is obvious that the space described in a unit of time
is j t. Hence,
v = S - t (1)
For example, if the point moves over a space of 10 feet in
2 seconds, v = 10 2 = 5 feet per second
If the velocity and the time are known, the preceding
formula gives the space passed over.
s = vt (2)
Similarly, if the velocity and the space are given, the
same formula, or formula 2, gives the time.
t= s - (3)
v
EXAMPLE 1 The wheels of a carnage moving with uniform velocity
make 3,760 revolutions between two stations A and B The diameter
of each wheel is 4 feet, and the time employed by the carnage to
travel from one station to the other is 7 6 hours It is required to find
the velocity of the carnage, m feet per second (By the velocity of
the carriage is meant the velocity of any of its points )
SOLUTION During each revolution, the carriage passes over a
distance equal to the circumference of the wheel, or 4 X 3 1416 ft.
The total distance traveled, or space passed over, m feet, is s = 3,760
X 4 X 3.1416 The time required to travel this distance is 7 6 br.
As the velocity is required in feet per second, this time must be
reduced to seconds, which gives / = 75X60X60 Substituting In
formula 1 the values of s and / just found,
EXAMPLE 2. The distance between two ports is 400 miles What
must be the velocity, in knots, of a vessel that will cover the distance
in 18 hours? (A knot is a velocity of 6,080 feet per hour.)
SOLUTION To find the velocity m feet per hour, we have,
i - 400 mi. = (400 X 5,280) ft , t - 18 hr Therefore, by formula 1,
, = *Lxpo f , per hr . . x M knote . 19 . 298 lnot ,
FUNDAMENTAL PRINCIPLES 27
VELOCITY IK VARIABLE MOTION
24. Variable motion is that motion in which the
moving point, or body, passes over unequal spaces m equal
times, or over equal spaces in unequal times
25. Average or Mean Telocity. It will be remem-
bered that the distinguishing characteristic of uniform
motion is the equality of all distances passed ovei m equal
intervals of time, and that the quotient obtained by dividing
any space 5 by the number / of units in the conesponding
interval of time is a constant quantity (v, formula 1 of
Art 23). Unless all these quotients are equal, the motion,
and therefore the velocity, is not uniform
Suppose, for example, that a tram moves over a distance
of 60 miles in 45 minutes. If it is known beforehand that
the motion is uniform, the velocity may be found by
dividing 60 miles by 46, which gives v = - miles
45 3
per minute. In this case, the space described in any num-
ber t of minutes would be, by formula 2 of Art. 23,
v t = &/ miles.
Suppose, on the contrary, that it is not known whether
the motion has been uniform during the 45 minutes, but
that the distance passed over during the first 5 minutes is
known to be 5.5 miles, and that the distance passed over
between the end of the thirtieth and the end of the fortieth
minute is 13.98 miles Then, the quotients obtained by
dividing the spaces by the numbers measuring the corre-
sponding times are, respectively,
5-5 _ T in 13.98 _ , OQQ 60 _ 4
T " ' ~W ~ lf6W] 46 " 8
Since these quotients are not equal, there is no uniform
motion, and therefore no uniform velocity.
If, however, another train is imagined moving with a uni-
form velocity of $ miles per minute, this tram will pass over
the distance of 60 miles in 45 minutes. For this reason,
\ miles per minute is called mean, or average, velocity
of the actual train considered; it is not a velocity that
27 OF MECHANICS 9
the train really possesses, but rather the equivalent velocity
with which a body moving uniformly would move over
the same space in the same time Likewise, if the motion
of the tram during; the first 5 minutes is considered, its
5 5
mean velocity, foi that interval, is - = 1 10 miles per
5
minute. In general, if a body moves over a space j in the
time /, its mean velocity v m , during that time, is defined
mathematically by the formula
26. Variable Velocity Instantaneous Velocity.
When a point is moving with variable motion, its velocity
at any instant is the velocity that the point would have if,
at that instant, its speed ceased to change, that is, if, after
that instant, the point moved neither faster nor more slowly
than it is moving. Since the motion is variable, the velocity
at any instant is not any velocity with which the body actually
moves over any part of its path, but simply the velocity
with which it would move if its speed became invariable.
The velocity of a moving point or body at any given
moment is called the instantaneous velocity at that
moment. In variable motion, the velocity is said to be
variable, because it changes or varies from instant to
instant.
27. To illustrate the character of variable motion and
velocity, imagine a train A to start from rest and leave a
station at the same time that another train B is passing the
station with a uniform velocity of 60 miles per hour, and
suppose the two trains to run on two parallel tracks. At
first, B will move ahead of A, since A' starts from rest.
vSuppose, however, that A moves faster and faster, so that
to a person in B the motion of B with respect to A will
appear to become slower and slower; there may then be a
time when B will be losing instead of gaining speed, with
respect to A, or when A will appear to be moving past B.
Thus, during the first minute, B will move over 1 mile, and
10 FUNDAMENTAL PRINCIPLES 27
A may move over only 4 mile; but, during: the following
minute, A may move over li miles, and then, it will be
i mile ahead of JS. Since A is at first losing space with
respect to B and then is gaming, there must be an instant
at which A is neither losing nor gaining; at that instant the
two trains are evidently at rest with respect to each other,
and if the speed of A ceased to change, A would continue
to move with the same velocity as , or 60 miles per hour.
The velocity of A at that instant is, therefore, 60 miles per
hour.
It must be understood that, as already stated, the tram A
does not actually move with the velocity of 60 miles per
hour during any interval of time; but this is the velocity with
which A would move if, after the instant considered, its
motion underwent no further change.
28. Direction and Magnitude of Velocity. By the
direction of the velocity of a moving point is meant the
direction in which the point is moving.
29. By the magnitude of a velocity is meant the
numerical value of the velocity, expressed in units of length
per unit of time, as feet per second or miles per hour.
EXAMPLE A float moves down a stream from a point A to o.
point D, a distance of 800 feet, in 2-J minutes The float is observed
at two intermediate points B and C, whose distances from A are,
respectively, 350 and 575 feet, and it Is found that it moves from A
to B in 45 seconds, and from B to Cm 70 seconds. Required, the
mean velocity, m feet per second, of the surface of the stream
(a) between A and D, (6) between A and B t (c) between B and C;
(rf) between C and D
SOLUTION (a) Here s = 800 ft , / = 2| min = 150 sec ; and the
formula of Art 25 gives
v m = r = r^r = 5.33 ft. per sec. Ans.
(6) Here s = 350 ft , t = 45 sec., and, therefore,
v m = 7 = ~r=- = 7 78 ft per sec. Ans.
/45
(c) Here s = 575 ft - 360 ft = 225 ft., t = 70 sec , and, therefore,
225
t ** 70
27 OF MECHANICS 11
(<f) Here s = 800 ft - 575 ft = 225 ft , t = 150 sec - (45 + 70) sec.
= 35 sec , and, therefore,
v m = - = -T- = 6 429 ft per sec. Ans.
EXAMPLES FOB PRACTICE
1 A tram moves with uniform velocity between two stations
375 miles apart, the distance is traveled in 7i hours. What is the
velocity of the train, in feet per second? Ans 73 333 ft per sec
2 A train moves from station A to station , a distance of
90 miles, in 1$ hours, from station B to station C, a distance of
17 miles, m % hour; and from station C to station D, a distance
of 64 miles, in 2 hours Find its mean velocity, m feet per second
(a) between stations A and >, (b) between stations B and C\
(c) between stations B and D. ( (a) 59 012 ft per sec.
Ans { (d) 49 867 ft per sec
[(c) 4752ft. per sec.
FORCE AND MASS
FORCE
DEFINITIONS RELATING TO FORCE MEASURE OF FORCE
30. Definition of Force. It is known from experience
that, when a body is at rest, it can be set in motion by the
action of another body. Thus, if a block of iron is lying on
the ground, it can be set m motion by pulling it with a rope,
by pushing it with the hand, or by placing a strong magnet
near it. Furthermore, the block may be pushed by one per-
son in one direction, while another person pushes it m the
opposite direction; or, the magnet may be placed on one
side of the block, at the same time that the block is pulled
by a string from the opposite side. In cases of this kind,
there may be no motion, on account of the neutralizing
effects of two or more actions on the same body. We say,
however, that there is a tendency to motion; for the
moment one of the opposing actions is removed, the other
causes the body to move
12 FUNDAMENTAL PRINCIPLES 27
31. This action of one body on another, producing, 01
tending to produce, motion in the latter body, is called force,
and the former body is said to exert force on the latter
The characteristic of the action known as force is, then,
that, when exerted on a body originally at rest, and not
influenced by other bodies, it results in the motion of the
first-mentioned body. The nature of the ensuing motion,
and the result of the action in question, when the body
acted on is already in motion or under the action of other
bodies, are complicated phenomena to be either actually deter-
mined by experiment or inferred from experimental data.
32. Balanced and Unbalanced Forces. If two or
more forces act on a body in such a manner that they cause
no motion, owing to the neutralizing effects they tend to pro-
duce, each force is said to be balanced by the combined
action of the others, and is referred to as a balanced force.
When a force acts on a body, and there is no opposing
force preventing the motion of the body, the force is called
an unbalanced force.
These definitions apply to moving bodies as well as to
bodies at rest If two or more forces are balanced when
exerted on a body at rest, they are also balanced when the
body is in motion; that is, they do not affect the motion of
the body.
33. Equilibrium. A body is in equilibrium when it
is under the action of balanced forces. The balanced forces
themselves are also said to be in equilibrium.
It should be noticed that equilibrium and rest are not
equivalent terms A body may move uniformly while acted
on by balanced forces, in which case it is in equilibrium, but
not at rest In this case, however, the motion of the body
is not due to the balanced forces acting on it. This subject
will be better understood after the law of inertia, presently
to be explained, has been studied
34. Weight. Experience teaches that, when a body is
unsupported, it falls to the ground. This fact is ascribed to a
force exerted by the earth on all bodies, which force is known
27 OF MECHANICS 13
by the general name of force of attraction, or simply
attraction, and also force of gravity, or simply gravity.
The atti action of the earth on any particular body is called
the weight of the body The methods of comparing the
weights of bodies are well known.
35. Measure of Force. For engineering purposes,
force is expressed in units of weight, such as pounds, kilo-
grams, etc. The reason for this will be more apparent when
it is considered that every force can be replaced by a weight.
Thus, suppose a body P, Fig. 3, to be suspended from the
extremity A of a per-
fectly symmetrical A Aj B
beam AB resting at
its center on a knife -^
edge F. If we wish / f\ \,Q
to prevent P from
falling, we may pull downwaids on the string at Q, 01
attach at Q a. weight equal to P, or tie a small piece of
iron at Q and place a magnet underneath. In all these cases,
the weight of P balances, and is therefore equivalent to, the
force at Q, by whatever means the latter may be produced.
As the two forces are equivalent, the one may be meas-
ured by the other, and we may say that the force acting at Q
is 20 pounds, kilograms, tons, etc., according as P weighs
20 pounds, kilograms, tons, etc. Suppose, for instance, that
the weight of P is 10 pounds, and that we pull at Q with
sufficient force to keep P from falling, but without moving it.
Then, the force with which we are pulling is 10 pounds. If
P falls, the pull is less than 10 pounds; and if it rises, the
pull is greater than 10 pounds.
36. Magnitude of a Force. By the magnitude of a
force is meant the numerical value of the force, expressed
in units of weight. If, for example, a force is equivalent to
a weight of 12 pounds, its magnitude is 12 pounds.
37. Direction of a Force. The direction of a force
is the direction in which the force moves, or tends to move,
the body on which it acts.
14 FUNDAMENTAL PRINCIPLES 27
DEFINITION AND DIVISIONS OF MECHANICS
38. Mechanics is the science of force and motion. This
science is divided into two general branches, (a) dynamics,
(b) kinematics or phoronomics.
39. Dynamics is the science of force and its effects.
In this science, the forces applied to bodies are given, and
the resulting effects of these forces are determined, or the
conditions of motion are given, and the forces necessary to
produce that motion determined. Rest is considered as a
special case of motion in which the velocity is zero.
40. Kinematics treats of motion alone, without refer-
ence to either force or the physical or mechanical propeities
of bodies. Suppose, for instance, that a body is moving in
a circle with uniform velocity, and that it is required to deter-
mine what force is necessary to preserve this kind of motion,
this is a problem in dynamics, for it relates to force Suppose
that it is required to determine what space the body travels
in a certain time; this is a problem in kinematics, for it can
be solved without having regard to anything but the velocity
and path of the body.
41. Subdivisions of Dynamics. Dynamics is sub-
divided into two branches: (a) kinetics, (b] statics.
42. Kinetics treats of unbalanced forces and their effects.
43. Statics treats of the equivalence and equilibrium of
forces.
NOTE The preceding divisions and definitions are of very recent
origin Formerly, the term dynamics was used in the sense m which
kinetics 13 now used, the latter terra being then unknown Even
today, the old meaning is frequently given to the term dynamics,
the best modern writers, however, use it in the sense just defined
that is, to denote the science of force in general, whether it pro-
duces motion or not
44. Applied mechanics is the application of mechan-
ical principles to the works of human art.
27 OF MECHANICS 16
FUNDAMENTAL LAWS OF DYNAMICS
I-iAW OF INKltTIA
45. Statement of l ho IJHW. In giving the definition
of force (Art. 30), icference was made to the fuel, known
from experience, that a body may act on anothei body,
originally at rest, in such a manner as to pioduce motion
in the latter body. This action was defined as foice That
definition relates only to the effect of force on a body origi-
nally at lest. The law presently to be stated expresses the
general effect of force under any eiieumsUnces; that is,
whether the body on which it is excited is at icst or in
motion. Tins law, known as the lir\v of lm>rltn, or the
first law of motion, may be formulated as follows:
// a body is not acted on by fom\ the body, m> u tvhol<\ /.s
either at test ot inovmg uniformly in a sh (tight faith.
This is not as definite a statement of the law of ineilia as
can be given. But the complete statement requires, in order
to be understood, some familiarity with mechanical piinciples
and conceptions of a high order, with which the beginner
cannot be supposed to be acquainted
By saying that a body as a whole is in motion, it is meant
that all the points of the body are in motion. Each of the
wheels of a locomotive, for instance, has a motion us a
whole, and, if the track is straight, the wheel, as a whole, is
said to be moving in a straight path, whose direction is that
of the track. Any one would understand what was meant
by saying that a billiard ball was rolling in a diiection par-
allel to one of the cushions of the billiard table.
In the simple cases of motion just given, the direction of
the motion of the body as a whole is plainly seen to be flu*
direction in which its center of figure is moving. In the OIKO
of unsymmetrical bodies, the direction of their motion is the
direction in which a certain point in them, called the oonlor
of fifravlty, is moving, For the present, however, the
student may think of the motion of symmetrical bodies
16 FUNDAMENTAL PRINCIPLES 27
only, such as balls, disks, cubes, prisms, etc., all of which
have a center of figure.
46. Facts on Wlilcli the Law is Founded. That a
body does not start to move by itself is a fact so familiar
that it is scarcely necessary to state it That a body, once
set in motion, does not come to rest unless acted on by force
is not so obvious. Bodies often come to rest apparently by
themselves, without any assignable cause, and this gave
rise to the belief, prevalent among the ancients, and still
entertained by some, that rest is the natural condition of
all bodies, and that all moving bodies have a tendency to
come to rest. A little consideration will show that this is
an erroneous view of the nature of motion. If a block is
placed on a stone pavement and struck sidewise, it will
slide for a short distance and soon come to rest; if the
experiment is tried on an asphalt pavement, the block will
move farther and come to rest after a longer time; it will
slide longer and farther on a piece of marble, and longei and
faither still on a sheet of ice. These facts plainly indicate
that the tendency to come to rest is not inherent in the block
itself, but depends on the action of other bodies namely, the
bodies on the surfaces of which the block slides, and we
naturally infer that, were it not for this action (or frlc-
tlonal resistance, as it is called), the block would not
come to rest at all.
Another common obstacle to motion is the resistance of
the medium, by which is meant the action of the substance
through which the body moves. Thus, if a block is placed
on a horizontal suiface, immersed in mercury, and then
struck sidewise, it will move through a very short distance,
If the experiment is repeated m water, the block will describe
a longer space before coming to rest. The space will be
longer in air, and longer still in a space (called a vacuum)
from which the air has been removed by means of an air
pump. Here, as before, the inference is that, did the medium
offer no resistance, or, more properly, were there no medium,
the body would, move forever in a straight path.
27 OF MECHANICS 17
47. In the preceding examples, it will be observed that,
whether, the body is brought to rest by fnctional resistance
or by the resistance of the medium, the diminution of its
velocity is continuous, in other words, the body comes to rest
gradually, and its velocity constantly diminishes while the
resisting force acts
An example of the action of force in increasing the velocity
of a moving body is afforded by the familiar phenomenon of
a falling body Here, the body is constantly acted on by the
attraction of the earth, and its velocity is observed to increase
very lapidly As another example, suppose that a person
pushes a car on a horizontal surface. The car moves at first
with almost no velocity; but, as the person continues to push,
the velocity keeps growing greater and greater.
48. The necessity of applying force to a moving body in
order to change the direction of its motion is also a familiar
fact. If a prismatic block is sliding on a surface in the diiec-
tion of its axis, no change m the direction of its motion will
be observed unless the prism is struck, or pulled, or pushed,
or in some other way interfered with by some other body.
Of course, all foices do not change the direction of motion,
but only those exeited by bodies that are, so to speak, outside
the path of the moving body. Thus, in the example just
given, a pull exerted by a rope in the direction of the axis of
the prism would cause no change in the direction of motion;
but, if the lope were pulled at right angles to the axis of the
prism, the direction of motion would evidently change.
49. Summing up, it may be said that, whenever the motion
of a body has been observed to change \ cither in direction or in
velocity , or in both, it has always been possible to trace the change
to the influence of other bodies ikat ZA, of force. Furthermore ',
bv diminishing' those influences, the changes referred to are corre-
spondingly diminished; and the conclusion has been reached thai,
could these influences be entirely eliminated, the said changes
would not take place at all.
If, then, a body is at rest, and no unbalanced force acts on
it, it will continue at rest; if it is moving 1 in a straight line
ILT398 3
18 FUNDAMENTAL PRINCIPLES 27
with uniform velocity, it will continue to move m the same
straight line with the same velocity If a body moves under
the action of a force for a certain time, its velocity will
constantly change during that time; but, if the force ceases
to act, the body will continue to move uniformly in the
direction and with the velocity it had at the instant the foice
was withdrawn. It is not necessary, however, in order that
the body may continue to move uniformly, to withdraw the
force or forces acting on it. the same result will be obtained
if other forces are introduced equal and opposed to those
already acting. An instance of this is afforded by the
motion of a train- at first, the tractive force of the engine is
greater than the combined resistance of friction and the air,
and the velocity of the train constantly increases, but, as will
be explained elsewhere, the resistance increases with the
velocity, so that there is a moment when the traction of the
engine is balanced by the resistance, and from that moment
on, the train moves uniformly.
50. Conversely, if either the velocity or the direction of
motion of a body changes^ the body must be under the action of
unbalanced forces, for, according to the law of inertia,
rectilinear motion with uniform velocity (in which rest must
be included as a special case) is the only possible motion of
a body not acted on by unbalanced forces. Thus, if a body
moves in a curve, it may be concluded that, whatever its
velocity may be, some unbalanced force or forces must be
constantly acting on the body. The moment these forces
cease to act, the body will continue to move uniformly in
the direction of the tangent to the path at the point occupied
by the body at that moment. For example, if a stone is tied
to a string and swung around, the pull of the string will keep
the stone moving in a circle; but the moment the string
breaks, the stone will fly off on a tangent to this circle. A
train is kept on a curve by the resistance of the rails acting
against the flanges of the wheels; but if this resistance is not
great enough, the wheels get off the rails, and the train, in
leaving the curve, moves straight ahead along the tangent
27 OF MECHANICS 19
51. Inertia of Bodies. The fact that the motions of
all bodies follow the first law of motion is often expiess>ed
by saying that all bodies possess the property of Inez-tin.
In this sense, it is very common in mechanics to refer to
what the motion of a body
is or would be, "by virtue
of the inertia of the body",
that is, according to
the law of inertia. Sup-
pose, for instance, that a -*
particle is moving along the path A B, Fig. 4 It follows
fiom the first law of motion that the particle must be under
the action of unbalanced forces, since the path is not straight.
Suppose, also, that, by some means, it is found that the veloc-
ity of the particle, when at /*, is 6 feet per second. Then,
we say that, if the forces were suddenly withdrawn (or bal-
anced), the particle, "by virtue of its inertia," would move
uniformly along the tangent P T, describing a space of 6 feet
in every second
GALILEO'S LAW OF THE INDEPENDENT EFFECT OF
FORCES
52. Preliminary Explanation. Let A, Fig. 5, be a
body acted on by a force Suppose the force to be such
that, if it acts on the body for a certain time /, and the body
is at rest when the force
A r~~ 7- 1 *" begins to act, the body
/ / moves over the path A A'
I /in that timej so that its
/ / final position is A 1 . Sup-
/ / pose, now, that the body,
/ / instead of being at rest
/, /, when the force is applied,
A PlGlG Al is moving along the path
A A lt in such a manner that, if the force did not act, the body
would move to A l in the time /. It is found from experiment
that if, while the body has this motion, the force referred to
above constantly acts on it during the time *, the direction of
20 FUNDAMENTAL PRINCIPLES 27
the force remaining; unchanged, the final position of the body
after this time is a point A,', such that the line A^AJ is equal
and parallel to A A 1 . In other words, the final position of
the body, with respect to the position the body would have
occupied if the force had not acted, is the same whether the
force starts the body from rest or acts on the body while
the latter is in motion and acted on by other foiceb Had
the body been originally at rest, the effect of the force would
have been to cause the displacement A A', in the direction of
the force, in this case, the final position of the body, had not
the force acted, would have been A. If the force had not
acted and the body had been moving along A A t , its final
position would have been As, the effect of the force is to
cause the displacement A^AJ of the final position, which
displacement is equal and parallel to A A'.
53. Statement of Galileo's Law. From the pre-
ceding facts, and from others of a similar character, is
derived the following general law, which is Galileo's law
of the Independent effects of forces:
The effect of a force on a body is the same whether the body js
at rest or in motion, and, if several forces act simultaneously on
a body, each force produces its effect independently of the othct
forces.
The meaning of this proposition is that each force produces
the same amount of displacement, parallel to its direction, as
if it acted alone and moved the body from a state of rest
Galileo's law may be otherwise stated as follows:
When a force acts on a body, the position of the body at any
time relatively to the Position the body would have if the force
had not acted, is independent of the motion Produced m the body
by any other forces.
54. Experimental Verification. The following
experiment affords an easy verification of Galileo's law In
Fig. 6, a ball e is supported in a cup, the bottom of which is
attached to the lever o in such a manner that a movement of o
will swing the bottom horizontally and allow the ball to drop.
Another ball b rests in a horizontal groove that is provided
27
OF MECHANICS
21
with a slit in the bottom. A swinging arm is actuated by
the spring d m such a manner that, when drawn back, as
shown, and then released, it will strike the lever o and the
ball b at the same time
This gives b an impulse
in a horizontal direction
and swings o so as to
allow c to fall
On trying the experi-
ment, it is found that b
follows a path shown by
the curved dotted line,
and reaches the floor at
the same instant as <?,
which drops vertically
This shows that the
force that gave the first
ball its horizontal move-
ment had no effect on
the vertical force that
compelled both balls to
fall to the floor, the
vertical force producing
the same effect as if the
horizontal force had not
acted.
ACCELERATION
55. Change of
Velocity Caused by
an Unbalanced Uni-
form Force. A uni-
form force is a force
whose magnitude and
dnection do not change. Both from the law of inertia and
fiom the definition of force, it follows that the effect of an
unbalanced uniform force on a body originally at rest is to set
the body m motion and impart to it a continually increasing
22 FUNDAMENTAL PRINCIPLES 27
velocity Galileo's law affords the means to determine
exactly how that increase takes place. Suppose that a uni-
form force acts on a body originally at rest, imparting to it,
in 1 second, a velocity of 10 feet per second This means
that, if the force acts during 1 second and then ceases to act,
and the body is not interfered with by other forces, the body
will, by the law of inertia, continue to move uniformly with a
velocity of 10 feet per second If the force, instead of being
withdrawn at the end of the first second, acts for 1 second
longer, its effect will be independent of the motion already
acquired by the body; that is, during the second second, the
force will impart to the body a velocity of 10 feet per second,
rn addition to the velocity the body had at the end of the first
second, so that, at the end of the second second, the body
will have a velocity of 20 feet per second Likewise, the vel-
ocity at the end of the third second will be 30 feet per second;
and so on.
A similar law of change applies whenever a body is acted
on by an unbalanced uniform force, that is, the velocity of
the body increases by a fixed amount during every unit of
time; or, in other terms, the velocity increases at a constant
rate. If, for example, the change of velocity per second is a
feet, the change m t seconds will be at feet per second. Of
course, it is not necessary to use the second as a unit of time,
nor the foot as a unit of length; any other units may be used.
56. Acceleration. The acceleration of a body
moving in a straight path under the action of an unbalanced
force is the amount by which the velocity of the body in the
direction o the force increases in a unit of time.
57. When a body moves under the action of a uniform
force, its change of velocity is the same for any two equal
intervals of time, as has just been explained. In this case,
the acceleration is said to be uniform, and the motion is
said to be uniformly accelerated.
58. In the example given in Art. 55, the acceleration is
10 feet per second per second The import of this appar-
ently confusing expression is this: J!f at any moment the
I V J
27 OF MECHANICS 23
felocity of the body is v feet per second, this means that, if
at that moment the force ceased to act, the body would con-
tinue to move uniformly with a velocity of v feet per second.
If, however, the uniform force continues to act for 1 second
longer and then ceases to act, the velocity will have increased
to v + 10 feet per second; that is, the body, if left to itself,
will continue to move uniformly with that velocity, descri-
bing v + 10 feet every second, or 10 feet more than before.
Instead of saying that the acceleration is 10 feet per second
per second, it is customary and sufficient to say simply that
the acceleration is 10 feet per second, it being understood
that the velocity is also measured in feet per second. If
the acceleration of a body was said to be 50 miles per
hour, the velocity would be understood to be measured in
miles per hour, and to be, at the end of any one hour,
50 miles per hour greater than at the beginning of that hour.
59. Formulas for Uniformly Accelerated Motion.
Let a body start from rest and move under the action of a
uniform force for t seconds. If the acceleration due to this
force is a, the velocity at the end of 1 second will be a, at
the end of 2 seconds, 2 a; and at the end of / seconds, ta, or
a t. Denoting by v the velocity at the end of / seconds, we
have, then,
v = at (1)
"i (2)
If, at any instant, the body has a velocity z> , the veloc-
ity v t t seconds after that instant, is given by the formula
v z/ + at (3)
From this formula follows
"V 1)a I A\
a - - * (4)
The velocity v t) taken with respect to the interval of time
considered, is called the initial velocity of the body; and v
is called the final velocity. It is immaterial how the
body has acquired the velocity V Q ; in any case, the effect of
the applied force is to produce an increase of velocity equal
24 FUNDAMENTAL PRINCIPLES 27
to at, or v TV feet per second, in t seconds The change
1) -~ i -' 7/
of velocity produced in 1 second is, therefore, - - , as
expressed by formula 4, and this agrees with the general
definition of acceleration In all this, it is assumed that the
force acts in the direction of motion.
If the body starts from rest, v = 0, and formulas 3 and
4 become identical with formulas 1 and 2, respectively.
MASS
60. Force Proportional to Acceleration. Accord-
ing to Galileo's law, if several forces act simultaneously on
a body, each force produces its effect independently of the
others Therefore, if the forces act m the same direction,
they will produce an acceleration equal to the sum of the
accelerations that the forces would produce if each force
acted alone. Furthermore, if the forces are equal, the accel-
eration will be equal to their number multiplied by the
acceleration due to any one of them. It is also evident that
any force may be supposed to be equivalent to any number
of forces acting in the same direction as the given foice, and
whose sum is equal to the given force Thus, a weight of
5 pounds is equivalent to a weight of 3 pounds and one
of 2 pounds, or to a weight of 4 pounds and one of 1 pound,
or to five weights of 1 pound each, etc.
Let a force of 3 pounds act on a body, imparting to it
an acceleration of 5 feet per second. Then, a force of
12 pounds, which is equivalent to four forces each equal to
3 pounds, will produce an acceleration four times as great,
or 4 X 5 = 20 feet per second. Any other force will produce
an acceleration that will be as many times 5 feet per second
as the force contains 3 pounds. For instance, a force of
49
49 pounds will produce an acceleration of -- X 5 = 81.67 feet
3
per second, nearly. In general, let a force F l impart the
acceleration a^ to a certain body. Let F n be another force
imparting the acceleration a n to the same body. If F n is
27 OF MECHANICS 25
twice F lt then will a n be twice ,; and, generally, if F n = nF l
where n is any number, fractional or integral, then a n n ,.
Now,
= = ; s = - = ;
7*i ^*i ! tfl
that is, ~ =
Therefore, /or /// &z#z body, any two forces are to each other
as the accelerations they impart to the body.
From the last equation follows
^ =
a n ai
61. If several forces F lt F a) F,, etc are capable of pro-
ducing, respectively, the accelerations a lt a a , a,, etc. in the
same body, then,
5 = 5 = 5, etc.
fti a, a,
Each of these quotients may be considered to represent
the force necessary to give the body an acceleration of 1 foot
per second (or, in general, 1 unit of length per unit of time).
If the value of this force, or the common value of the pre-
ceding quotients, is denoted by m, then
w = = = 5, etc.;
a a, #o
and hence,
F! = ma^F, = ma,, F a = ma a , etc.
In general, if Fis any force producing in a body the accel-
eration a, we have
m = (1)
a
and F = ma (2)
62. Definition of Mass. Experience teaches that it
requires different forces to produce the same acceleration
m different bodies. Therefore, the value of m is different for
different bodies. When two equal forces produce the same
acceleration in two bodies, the two bodies are said to have
the same mass. If one body requires a greater force than
26 FUNDAMENTAL PRINCIPLES 27
another in order to receive a certain acceleration, the former
body is said to have a greater mass than the latter.
Mass, then, is that property of bodies on which alone the
acceleration they receive when under the action of given
forces depends. The acceleration, of course, depends on
the applied force also, what the definition means is that, so
long as the force remains the same, the acceleration varies
with only that property of bodies called mass
According to this definiton, the mass of a body may be
measured by the force necessary to impart to the body an
acceleration of a unit of length per unit of time For this
reason, the factor m of formulas 1 and 2 of Art. 61 is called
a measure of the mass of the body, or, for shortness, the mass of
the body. It should be remembered that m has different
values for different bodies.
63. Determination of the Mass of a Body Accel-
eration Due to Gravity. The mass of a body has to be
determined experimentally. A known force is applied to
the body for a certain time, and the velocity at the end of the
time is ascertained by some method Then the acceleration
is found by dividing the velocity by the time (formula 2 of
Art. 59), and the mass by dividing the applied force by the
acceleration (formula 1 of Art. 61). The mass is, of
course, expressed m the same units as the force, and its
numerical value further depends on the unit of time and on
the unit of length, these units being involved in acceleration.
In this work, mass will, unless otherwise stated, be referred
to the pound, the foot, and the second as units.
The most convenient force to use for determining the mass
of a body is the force of gravity that is, the weight of the
body. It has been ascertained by actual experiment that a
body falling freely in vacua follows the laws of uniformly
accelerated motion. This might have been anticipated; for,
while the body falls, it is under the action of a uniform force
(gravity), which is measured by the weight of the body.
The velocity of a falling body has been found to increase
at the constant rate of about 32.16 feet per second per second;
27 OF MECHANICS 27
in other words, when acting on a falling body, gravity pro-
duces an acceleration of about 32.16 feet per second. As this
acceleration is due to the weight of the body, we have, deno-
ting this weight by W (from formula 1 of Art. 61),
= W
m 32.16
64. The acceleration produced in a body by the force of
gravity is called the acceleration due to gravity, or the
acceleration of gravity. The magnitude of this accelera-
tion decreases from the pole, where it is about 32.26 feet per
second, toward the equator, where it is about 32.09 feet
per second For the United States, its average value is
32.16 feet per second. This value will be used here, unless
a different value is especially stated.
65. The acceleration due to gravity is usually denoted
by g. If the weight of a body i& W, we have,
W i-t \
m = (1)
and W = mg (2)
The weight vanes proportionately to g t so that the value
of m, as determined from formula 1, is always the same, as
it should be. It is, of course, understood that the weight W
is the weight of the body at the place where the acceleration
of gravity is jr.*
Substituting m formula 2 of Art. 61 the value just found
for m t
r-i K (I ( o \
F (3)
g
which gives the force necessary to produce a given acceler-
ation a in a body having a given weight W.
Both in formula 3 of this article and in formula 2 of
Art. 61, the force F is the force acting in the direction of
*The exact value of g; at any particular place, is determined by
careful observations with a pendulum, to which formulas derived m
advanced mechanics are applied. The following formula gives a
fairly approximate value of ff (feet per second) at any point whose
latitude is L and whose elevation above sea lex el is h (feet):
g = 32.173 - 084 cos L - .000002 h
28 FUNDAMENTAL PRINCIPLES 27
motion. If there is another force acting m an opposite
direction, F must be understood to be the difference
between the two. How forces acting on the same body
or particle are combined will be explained further on.
EXAMPLE 1 A block of any material is placed on a horizontal
smooth surface (that is, a suiface supposed not to offer any factional
resistance, or whose fnctional resistance may be neglected), and Is
pulled by a string for 4 seconds, a registering appaiatus (called a
dynamometer] placed between two portions of the string shows that
the pull is 3 pounds The weight of the block being 20 pounds, it is
required to find (a) the mass of the block, (6) the acceleration of its
motion; (c) its velocity at the end of the fourth second
SOLUTION (a} The mass of the block is found by formula 1.
Taking g equal to 32 16,
JQ
m = sire = 622 Ans -
(6) By formula 2 of Art 61,
F 3
a = ~ = -jjKo = 4 823 ft per sec. Ans.
in ,\j&i
(c) By formula 1 of Art 59,
v = 4 823 X 4 = 19 292 ft per sec Ans
EXAMPLE 2 A uniform force acts on a body originally at rest for
10 seconds, at the end of which the velocity of the body is 40 feet per
second, the force then ceases to act, and a force of 15 pounds is
applied for 5 seconds, at the end of which the velocity of the body is
76 feet per second To find (a} the weight of the body, (/>) the force
that acted during the first 10 seconds
SOLUTION (a) During the second interval (5 seconds), the velocity
has changed from 40 to 75 ft per sec Therefore, the acceleration, by
formula 4 of Art 59, is = = 7 ft. per sec. (= a). The mass of
the body is = -=-, and the weight is
d I
15 X 32.16 _. _, . .
mg = - ~ = 68.914 Ib. Ans.
(d) The acceleration due to the fust force is (formula 2 of
Art. 59), 40 T- 10 = 4 ft per sec By formula 2 of Art. 01, the
force necessarv to produce this acceleration is
m X 4 = x 4 - 8.5714 Ib Ans.
27 OF MECHANICS 29
EXAMPLES FOB PRACTICE
1. Find the force necessary to start a weight of 15 pounds from
rest and give it a velocity of 80 feet per second in 4 seconds
Ans F = 9 3284 Ib.
2 An engine and train of cars having an aggregate weight of
160 tons start from rest and move for 1 minute (60 seconds) over
a level track Assuming the traction of the engine to exceed the
resistances by 2 50 tons, find the velocity of the tram at the end
of the minute Ans v = 32 16 ft per sec
3 A body has an initial velocity of 20 feet per second. A force of
40 pounds is applied, and after 12 seconds the velocity is found to be
85 feet per second Find (a) the acceleration due to the force,
(b) the weight of the body &/() a = 5 4167 ft per sec.
Ans I (d) W = 237.49 Ib
4 A body weighing 980 pounds moves for a certain time under the
action of a uniform force of 50 pounds. At the end of that time,
another force of 50 pounds is applied in the opposite direction After
the application of the second force (the first not being removed) , the
body moves over 300 feet in 15 seconds Find the time during which
the first force acted alone . , v 300-15 , , ,
Ans / == - = - w = 12. 189 sec.
a f
LAW OF ACTION AND REACTION
66. Action and Reaction. Force has been defined as
the action of a body on another, producing, or tending to
produce, a change in the motion of the latter body. When
the weight of a body is spoken of as the action of the earth
on the body, we are concerned only with the effects of this
action on the body in question, and for this reason disregard
whatever action the body may, in its turn, exert on the earth.
In the same manner, we say that a magnet attracts a piece of
iron; this is sufficient for us to know, if the only thing being
dealt with is the condition and motion of the piece of iron.
But it must not be concluded from these common and con-
venient forms of expression that a body can act on another
without itself being acted on by the other body. All action
between bodies is mutual. Take, for instance, a magnet,
hold it in the hand, and bring it near a small piece of iron-
the iron will immediately move toward the magnet and
607 6100
SO FUNDAMENTAL PRINCIPLES 27
adhere to it. Now, place the magnet on the table and
hold the piece of iron in the hand, bringing it near the
magnet: the magnet will move toward the piece of iion
and adhere to it. If both the magnet and the piece of
iron are placed on corks and let float on the .surface of water,
they will both move, each toward the other, and adhere to
each other. Furthermore, in this motion, the acceleiation of
the magnet and that of the iron are found to be in the inverse
ratio of the masses of the two bodies. This shows that the
two bodies are acted on by the same force (Art. (52), 01 that
the action of the magnet on the iron is equal to the action of
the iron on the magnet.
When a body is at rest, it is evident that the rorces rtCting
on it must balance among themselves (law of inertia) Take
a body weighing 4 pounds lying on a flat surface If the
weight of the body were not balanced, the body would move;
the surface, therefore, must exert an upward force equal to
4 pounds, in order to keep the body from falling If we
press against the wall with a force of 10 pounds, the wall
presses on the hand with the same force, but in an opposite
direction. A weight of 20 pounds hanging from a string
will evidently pull the string downwards with a force of
20 pounds; but the string must pull the weight upwards
with the same force, as otherwise the weight would fall.
In general, whenever one body acts on another, the latter
body acts on the former with the same force, but in the
opposite direction.
In considering the mutual action of two bodies, we are
usually concerned with the condition of only one of them; the
force acting on it is called the action, while the force exerted
by it on the other body is called the reaction.
67. Statements of the iJaw of Action and Reaction.
From the facts stated in the preceding article, the following
general law has been derived; it is known as the law of
action and reaction, and was first stated by Newton:
To every action^ there is always an equal and opposite
reaction.
27 OF MECHANICS 31 r ->
68. As the accelerations produced in two bodies by the
same force are inversely proportional to the masses of the
two bodies, the law of action and reaction may be stated in
the following terms
Whenever two bodies aft on each other, they produce > or tend to
produce, in each other accelerations ni opposite directions, -and
these accelerations are inversely propoi tional to the masses of the
two bodies.
This happens whether the two bodies act on each other by
means of a string stretched between the two, or by means of
a rod connecting them, or through some unknown medium,
as in the case of a magnet and a piece of iron. In every case
where a body moves another, the latter moves, or tends to
move, the former
69. Sti-ess. When only the action of one body on
another is considered, this action is called force. If the
mutual action of two bodies, or two parts of a body, is
Jp
lOOlb.. I .20026. .lOOlb. \.100lb.
''
FIG 7
considered, that action is called stress. A stress includes a
pair of equal and opposite forces, an action and a reaction,
a force and a cotmterforce. A stress is measured by the
magnitude of either of the forces of which it consists.
In applied mechanics, the term stress is generally restricted
to the forces acting between two parts of a body Thus,
when a string is pulled at both ends with a force of 10 pounds,
every portion of the string is under a stress of 10 pounds; for,
if the two portions of the string are considered as being
situated on the two sides of a plane perpendicular to its
direction, each portion will be pulled away from the other
with a force of 10 pounds. Therefore, to keep the two por-
tions from separating, each must pull the other with a force
of 10 pounds. Again, consider the bar A B, Fig. 7, pushed
from each end with a force of 100 pounds. Imagine the bar
32 FUNDAMENTAL PRINCIPLES 27
lo be divided by a plane PQ- when the portion AM presses
on B M with a force of 100 pounds toward the right, the por-
tion B M must press on AM with a force of 100 pounds
toward the left.
In these examples, the string is under a stress of 10 pounds,
and the bar under a stress of 100 pounds.
70. Tension, Compression, Pressure. When, as in
the case of the string just considered, the forces tend to pull
each portion of a body from the other, or to lengthen the
body on which they act, the stress is called tension. When,
as in the case of the bar, the forces tend to move each por-
tion toward the other, or to shorten the body on which they
act, the stress is called pressure or compression. The
term pressure, however, is more commonly applied to the
stress between two contiguous bodies, and the term com-
pression to the stress between two contiguous parts of the
same body. In the example of the bar, the stress is ordi-
narily called compression. In the case of a heavy body
resting on a table, the force exerted by the body on the
table and by the table on the body is a pressure.
EXAMPLE 1 Two bodies MI and M,, Fig 8, of masses OTI and m t>
respectively, rest on a smooth horizontal surface, and are connected by
a string S a A force of /''
jf a j * a [ jf t | Sl - pounds is applied to Mi
' ' ' by means of a string 5,
FlG 8 m the direction shown.
To find the acceleration of the resulting motion, and also the tension
in each string.
SOLUTION In all problems of this kind, the strings are supposed
to be rigid and of no mass or weight, if the mass and stretching of
each string were considered, the problem would be more compli-
cated. In almost all cases that occur in practice, the ma<tf of the
string, rope, chain, etc may be neglected without any sensible error;
and the same is true of the extensibility, or stretching capacity, of
the connection For example, the mass of the coupler between two
railroad cars need not be considered in determining the pull between
one car and the other; nor is it necessary to take into account the
amount of stretching m the coupler
Since the two bodies ^/"i and M, are rigidly 1 connected, they must
move with the same acceleration Let this acceleration be a, and let
27 OF MECHANICS 33
the tensions m the strings Si and S a be 7i and T a , respectively Since
the body MI is pulled by the string S\ with a force equal to F, it must
pull the string toward the left with the same force, so that 7~i is
evidently equal to F Also, Af,. acts on M, through the string S, and
imparts to it an acceleration a towaid the right, therefore, the force
exerted by MI on Af a is (from foimula 2 of Art 61) in a a By the
law of action and reaction, Jlf t exerts on Jlf t the hame force, m,a, but
directed toward the left The tension m S, is, therefore, T, = m 9 a
The net force acting on M^ is the difference between F and m a a,
or Fm a a. Consequently, since the resulting acceleration is a, we
have, by formula 2 of Art 01, F w, a = m t a; whence,
a = (1)
7i + Wa
Also, T, = m 3 a = -^-^ (2)
&! + m, v '
The same results may be obtained by assuming at the outset that
the two masses m l and m, are equivalent to one single mass (m l + tti)
acted on by the force F From this, equation (1) follows at once as a
special case of formula 2 of Art 6 1 . Knowing the acceleration of Af t ,
which is a, the unbalanced force T a acting on Hf, is found by for-
mula 2 of Art 61:
~, itt t F
T, = m a a = 7
j + m,
which is the same as before.
EXAMPLE 2 A body Af lt Fig 0, whose weight is W pounds, is
attached to a string, the latter
is passed over a smooth peg P,
and then tied to a second body
7l/ a having a weight of
pounds and free to move along
a smooth horizontal surface
To find the pull m the string
and the resulting acceleration
of the two bodies. Pro. 9
SOLUTION This problem does not diffei from the preceding 1 in
principle, and the method of solving it is identically $he same, but it
has been given in order to caution against a common mistake made
by beginners The object of the peg is simply to change the direc-
tion of the force, it has no effect on the magnitude of the pull, which
is the same in the vertical as In the horizontal portion of the string
It is to be carefully borne in mind, however, that the pull on the
string is not the weight of M^ and that therefore the force pulling M t
is not Wi pounds, as might appear at first sight. The weight W* is
a force acting on the bodies Jtfi and Jtf t , and, like the force F m the
preceding example, has to move these two bodies. If there were
I LT39S-4
34 FUNDAMENTAL PRINCIPLES 27
no body M t , the unbalanced force acting on AF* would he W tt
but, in the present case, this force is diminished by the leaction
of M, transmitted through the string to M^ If the pull in the string
is denoted by T and the common acceleration of the two bodies
is denoted by a, the unbalanced force acting on Jlfi is, as before,
W l - T, and the unbalanced force acting on M, is simply T The
mass of Mi is - (formula 1 of Art. 65), and the mass of Jlf a is .
ff
Substituting in the formulas of the preceding solution, we obtain
a = Wj. w,. "* Wi + w m M
g g
Tff
* or Jjf jy
TXf W = Ef _1_ TV \*J
EXAMPLE 3 A body weighing W pounds lies on the surface of a
horizontal table. Both the table and the body are moving upwards
with an acceleration of a feet per second To find the pressure of
the body on the table.
SOLUTION The pressure of the body on the table is equal to the
pressure of the table on the body, which is the total upward force
acting on the body, call this force P The downward force acting
on the body is W. Therefore, the unbalanced force producing the
W a.
acceleration a is P W, and, as P W => am ^~ t we find, for
the downward pressure exerted by the body,
/.-ZS+Wr-^ + f) (1)
ff g
This shows that the weight of the body is increased by the amount
- . Ifthetableismovingdownwards.ais negative, andjP Ji!.lj!rfv t
If in this case a, ff, then, P = 0, or there is no pressure on the table.
If a is negative (downward) and greater than g t 8 ay a g + a', then,
g
This shows that, if the body remains in contact with the surface of
the table, it will exert an upward pressure equal to -^-A; but this can
only be done by placing the body under the table. In this case, it is
evident that, since the table tends to move faster than the body, the
latter must offer a resistance, or exert an upward pressure on the table.
This problem should be carefully studied, as it is a good illustration
of how the results of mathematical formulas {Ire to be Interpreted.
27
OF MECHANICS
35
EXAMPLES FOB PRACTICE
1. In example 1 of Art 70, the force .F is 500 pounds, and M* and M t
weigh 8 tons and 3 tons, respectively Find (a) the resulting acceler-
ation, in feet per second, (b) the tension on the rope, in pounds (1 ton
= 2,000 pounds) . A ,/()= 731ft per sec
Ans \(t>) T= 13636 Ib.
2 With the weights of M t and Af, as in example 1, find: (a) the
force (tons) necessary to give the two bodies an acceleration of 10 feet
per second, (b) the tension in the rope, m tons
F = 3 4204 tons
T = .9328 ton
Ans
3. In Fig. 10, the three bodies M lt M,, M a , weighing 18 tons,
16 tons, and 20 tons, respectively, are connected by the ropes S L
FIG 10
and S a A force F of 15 tons is, applied to M l as shown. Find,
(a) the resulting acceleration a, m feet per second, (b) the tension T,
in the rope S tt (c) the tension T, in the rope S 3
f a) a = 8 9333 ft per sec.
Ans { b) T a = 5 5556 tons
[ c) T a = 10 000 tons
4 A body that on the surface of the earth weighs 10 pounds, is
weighed m an ascending balloon by means of a spring balance, and
found to weigh 15 pounds With what acceleration is the balloon
ascending? [See equation (1} m example 3 of Art. 70.]
Ans a = 18 08 ft per sec.
5. An engine pulls a tram of
The total force of the locomotive i
train (the engine not included)
resistance of 10 pounds per ton of
during the first minute, find (a)
feet per second, (b) the velocity v
minute, in miles per hour, (c) the
the second car to the first.
seven cars weighing 80 tons each
that is, the total force acting on the
-is 8 tons. Assuming a constant
tram (engine and tender excluded)
the acceleration a of the train, in
of the tram at the end of the first
tension T m the coupler connecting
a = 1 0643 ft per sec
Ans. 4 b) v = 43.541 mi per hr.
T = 6.8571 tons
86 FUNDAMENTAL PRINCIPLES 2?
CMMEDIATE CONSEQUENCES OF THE PRE-
CEDING 1LAW8
IMPORTANT FORMULAS
71. Space Passed Over In Uniformly Accelerated
Motion. Let a body start from rest and move during: t
seconds under the constant action of a force acting in the
direction of motion. Let a be the acceleration of the body,
and v the velocity at the end of the time t. As explained
in Art. 59, v is equal to a t. It can be shown by the use of
advanced mathematics that the space j desciibed by the
body in the time t is the same as if the body had moved din-
ing that time with a constant velocity equal to one-half the
actual final velocity, or |; that is,
2i
.-\* (i)
If, in the second member of this formula, v is replaced
by its equivalent a t, the result is ~ t = a f. Therefoie,
2
j = i/ a (2)
This is one of the most important formulas in mechanics
and should be memorized.
72. If, in formula 1 of Art. 71, t = 1, then, s = $ a;
or, the space described during the first second is numerically
equal to one-half the acceleration.
73. From the general formula v = at, we get t - .
a
This value substituted in formula 1 of Art. 71 gives
t
s = - , whence, v* = 2 as, and
a
v = V2 as
which gives the final velocity at the end of a given space.
74. To find the space and the velocity in terms of the mov-
ing force and the mass or the weight of the moving body, we
27 OF MECHANICS 37
have, by formula 2 of Art. 61, and formula 1 of Art. 65,
Zp
a = = g. These values in formula 2 of Art. 71 give
m W
s - * = -* *r/ a (1)
*~ 2 2 X W*' UJ
The same values in the formula of Art. 73 give
(2)
75. Modification of the Formulas "WTien the Body
Has an Initial Velocity. If, before the force begins to
act, the bodv has an initial velocity v , or, what is the same
thing, if only an interval of time during which the velocity
changes from v a to v is considered, the preceding formulas
must be modified as follows:
As the force has no effect on the motion already acquired
by the body, the latter, having a velocity z/ , will describe,
by virtue of its inertia alone, during the time /, a space equal
to v 1-, to this must be added the space described by the body
in virtue of the action of the force, as given by formula 2 of
Art. 71. Therefore, the total space is given by the formula
J = v,t + \af (1)
or, putting a = v ~ v (formula 4 of Art. 59),
j =.* + *(-.)* = ifa + w.)/ (2)
which is the mean of the initial and the final velocity multi-
plied by the time.
76. The equation a = v ~ v gives / = "L=J!* f Substi-
/ a
tuting this value in formula 2 of Art. 75,
i W
s = + . x -
\ a-
whence, v' v*' => 2 a s, and
v = Vz>o" + 2 a s
This gives the final velocity in terms of the initial velocity,
the space, and the acceleration.
To find s and v in terms of force and mass, it is only
p
necessary to substitute for a.
m
38 FUNDAMENTAL PRINCIPLES 27
77. Falling Bodies. It has been explained (Art. 63)
that, when a body falls freely under the action of gravity, its
acceleration is g feet per second, the value of g being neaily
constant and equal to 32 16. The preceding formulas apply
to falling bodies, for a falling body is simply a body mov-
ing under the action of a force equal to its own weight, with
a uniform acceleration equal to g. It is, therefoie, not
strictly necessary to write new formulas for the motion
of falling bodies. As, however, it is customary to use
the symbol g for the acceleration of a falling body, and h
for the space described, or height fallen through, it is con-
venient to have the general formulas expressed in terms of
these symbols. By putting h for s and g for a in formula 1
of Art. 59, in formulas 2 and 1 of Art. 71, and in the for-
mula given in Art. 73, the following formulas are obtained
for a body falling freely from rest during / seconds:
v=gt (1)
k = iff (2)
h = %vt (3)
v = *hgh (4)
For a body falling for t seconds, after having acquired a
velocity z , formula 3 of Art. 59, formulas 1 and 2 of
Art. 75, and the formula of Art. 76, give
v = v +gt (5)
h = v.t + lgf = (v. + ir/)* (6)
*-*(. + )/ (7)
(8)
78. It should be remembered that, theoretically, these
formulas are exactly correct only for bodies falling m vacua.
When bodies fall through the air, the latter offers a resist-
ance that varies with the surface of the body; hence, all
bodies do not fall in air with exactly the same acceleration^
but in a vacuum they do. This has been verified by the fol-
lowing experiment: A feather and a ball of lead are sup-
ported at the upper (inside) part of a long tube from which
the air has been removed; by a convenient arrangement the
support is quickly turned from the outside, so that both
27 OF MECHANICS 39
bodies will begin to fall at exactly the same time; and it is
found that they reach the bottom of the tube simultaneously;
nor does the ball get ahead of the feather during any portion
of the time of falling. Again, if a paper disk is placed on
top of a dollar, and the latter let fall, the paper will remain
constantly on the dollar and fall in the same time. The
reason is that in this case the paper does not encounter any
more resistance than does the coin.
EXAMPLE 1 What is the velocity at the end of 30 seconds of a
body moving with an acceleration of 8 feet per second?
SOLUTION Substituting the given values in formula 1 of Art 59,
v = 8 X 30 = 240 ft per sec. Ans
EXAMPLE 2 A force of 10 pounds starts a body from rest and
causes it to move over 65 feet If the weight of the body is 40 pounds,
what is the final velocity?
SOLUTION. By formula 3 of Art 65,
Fff 10 X 32 16 . ., ..
a = -^ = 4Q - = 8 04 ft. per sec.
Substituting the known values in the formula of Art 73,
v = V2 as = -N/2 X 8 04 X 65 = 32 33 ft per sec Ans.
EXAMPLE 3. A ball is thrown downwards with a velocity of 5 feet
per second from a tower 200 feet high. What will be its velocity when
it reaches the ground?
SOLUTION Here, the initial velocity v a ~ 6 ft per sec , and the
height h = 200 ft., are given. Substituting these values in formula 8
of Art 77,
v - V6 a + 2 X 32 16 X 200 = 113 53 ft per sec. Ans
EXAMPLE 4 If a ball dropped from a bridge reaches the water in
2 seconds, at what height is the bridge above the surface of the water?
SOLUTION From formula 2 of Art. 77, the height through which
the ball drops is found to be
iX 32.16X2" 6432ft Ans.
EXAMPLES FOR PRACTICE
1. A body starts from a state of rest and moves with an acceleration
of 20 feet per second for 30 seconds. Over what distance has the body
passed? Ans. 9,000 ft.
2. A body weighing 160.8 pounds is started from rest by a force of
40 pounds. Over what space has the body passed if the final velocity is
16 feet per second? Ans. 16 ft.
40 FUNDAMENTAL PRINCIPLES 27
3 A body having an initial velocity of 25 feet per second moves,
with an acceleration of 8 feet pei second, over a distance of 29 feet
Find the final velocity of the body. Ans 33 ft per sec
4 A body falls freely for 2 3 seconds If the initial velocity of the
body is 5 feet per second and its final velocity is 79 feet per second,
through what distance has the body fallen? Ans 96 6 ft
5 A body having an initial velocity of 10 feet per second falls
freely for 12 seconds Through what distance does the body pass?
Ans 2,435 5 ft
RETARDATION
79. Uniformly Retarded Motion. When a body is
moving in a certain direction and a force acts in the opposite
direction, the effect of the force is to diminish the velocity
by equal amounts in equal times, and the motion is said to
be uniformly retai-ded. The amount by which the veloc-
ity is decreased per unit of time is called the retardation
due to the force.
According to Galileo's law, if a force acting 1 on a. body at
rest produces in it an acceleration a, the same force acting
on the same body, when the latter is m motion, will pro-
duce the same acceleration a and the velocity v = a t, in
its own dnection, whatever the previous direction of the
motion may have been Therefoie, if the force acts in a
direction opposite to that in which the body is moving, and if
the latter has an initial velocity v , the velocity after t seconds
will be v, a t. Retardation, then, is nothing but negative
acceleration, and all formulas relating to uniformly acceler-
ated motion apply to uniformly retarded motion, by simply
changing the sign of a. For this reason, the term accelera-
tion, when taken in its most general sense, includes letarda-
tion as a special case, and the latter term is seldom used in
mechanics.
80. Law of Retardation. Let a. body be moving in a
straight line, and let its velocity at a certain moment be TV
If at that moment a force is applied in the opposite direc-
tion, and allowed to act during / seconds, the final velocity v
27 OF MECHANICS 41
and the space s described are given by the following formulas
(See Arts 59 and 75):
v = v at (1)
j = v t t %at* (2)
To find the time in which the body will be brought to rest,
v must be equal to zero; that is, v a t = 0, or / = .
a
Substituting this value in equation (2), and denoting by s 1
the space described during the time /, we have,
/ . &' - J a ! = '-
a a 2 a
If the velocity v a is supposed to have been generated by a
force equal to the retarding force, the time required for the
force to produce this velocity must evidently have been ,
a
which is the same as the time required for the retarding force
to destroy the velocity z/ The space described during the
action of the force, up to the time the velocity v, is attained,
is given by the foimula
j = $af=l;a^=^ = S >
a 2 a
It follows, therefore, that, if a uniform force acts on a
body during a certain time and carries it through a certain dis-
tance, an equal but opposite foice, it applied after the first force,
will destroy the velocity generated by the first force in the same
time and in tlie same space (that is, after the body has described
the same space] required by the first force to generate the velocity
m question.
SOJLDT1ON OF PROBLEMS
81. Fundamental Equations of Motion. The fol-
lowing are the three fundamental equations of motion:
F=ma (1)
v-v a = at 1 ,v
They form two groups, (1) and (2), one consisting of one
equation and the other of two. In group (1), there are three
quantities; and as there is only one equation, two of the
42 FUNDAMENTAL PRINCIPLES 27
quantities must be known before the other can be determined.
If /'is to be determined, m and a must be known, if m is to be
determined, F and a must be known, etc. In group (2) , there
are five quantities, and, as there are only two equations, three
of the quantities must be known in order to determine the other
two; the process of finding these two is simply the process
of solving two equations with two unknown quantities.
If the two groups are taken together, it will be observed
that a is common to both If two of the quantities m group
(1) are given, a becomes known, and to solve group (2),
only two more quantities of that group have to be given
Or, if three of the quantities of group (2) are given, a becomes
known, and only one of the quantities of gioup (1) (either
/'or m) is required to find the other. The process of elimina-
tion is exceedingly simple. When there is no initial velocity,
v = 0.
82. General Solution of Some Important Problems.
As an illustration of the use of the fundamental equations, let
it be required to find F, a, and /, when ?;/, v, v ot and are given.
In group (2), the quantities v, v ot and s are known. The
unknown quantities are a and /. From the first equation of
the group is obtained a = ^-~r^'> and fr m th e second,
a = ^ s ~~ Vt -.
Equating these two values of a, transpo-
sing, and canceling the common factor /,
(v v ) / = 2 (s v t};
whence / = -- -
v + v a
which is the same value that would have been obtained from
formula 2 of Art 75. This value of t substituted in the
equation a = ^-^-^ gives
a _ - Q _ (*LrJ?o) ( v + v ) - ?. "J!s!
~" ~ " ~
Finally, F - ma *
2s
27 OF MECHANICS 43
83. Again, let m, v t v,, and t be given, and let it be
required to find F, s, and a.
The value of a follows at once from the relation v z/
= at, which gives a = ^^. Substituting this value in
the second equation of group (2), formula 2 of Art. 75 is
again found. From group (1) is obtained
F=ma= (-.)
t
The values for F given in this and the preceding article are
very important, and should be memorized.
EXAMPLE 1 A body weighing 250 pounds is moving with a velocitj'
of 25 feet per second, a force acting in the same direction through a
space of 75 feet changes the velocity to 60 feet per second To find
the magnitude of the force
SOLUTION Here, m = SHTB (formula 1 of Art 85), v = 26,
Aft ID
v 50, and s = 75 Substituting these values in the formula given
in Art 82,
-(50 -26")
m(v* - zO 32 16 ^ ; 260 X (2,600 - 626)
2s = 2X75 " " 8216X2X76
= 97.17 Ib. Ans
EXAMPLE 2 A body whose weight is 192 96 pounds moves with a
velocity of 15 feet per second. To find the magnitude of a force neces-
sary to change the velocity to 45 feet per second, that force acting during
15 seconds in the direction of motion
192 96
SOLUTION Here, tn = - pn -' - =6, v 15, v = 45, and / <a 15.
oZ 10
Substituting these values m the formula given in this article,
f = , = - . tt
EXAMPLES FOR PRACTICE
1 A body weighing 100 pounds moves for 6 seconds along a smooth
horizontal surface under the action of a force of 10 pounds Find
the space passed over. Ans. s = 57.888 ft.
2 A body moving with uniformly accelerated motion passes over
100 feet in 12 seconds. What is the acceleration?
Ans. a 1.3880 ft. per sec,
44 FUNDAMENTAL PRINCIPLES 27
3 What force is necessary to give a body weighing 6 tons a velocity
of 10 feet per second in 75 feet?
W-31*
Ans F = ~- = 12438 T = 248 76 Ib
4 A body starts fiom rest and moves with an acceleration of
40 feet per second Find the velocity after the body has passed ovei
75 feet Ant. v = 77 46 ft per sec
5 Find the time in which the body in the preceding example has
described the space of 75 feet Ans t = 1 9365 sec.
6 A body is thrown vertically downwards from a height of 10,000
feet and reaches the ground m 6 seconds Find the velocity with
which the body was projected (Use formula 6 of Ait 77.)
Ans z/o = 1 ,570 2 ft per sec
COMPOSITION AND RESOLUTION OF FORCES
84. Point of ApplI cation, Line of Action, and
Direction of a Force. When a force is applied to a particle
or material point, the particle or point is called the point
of. application of the force.
The line of action of a force is the straight line along
which the force tends to move its point of application The
direction of the force is the direction of this line (see
Art. 37).
85. Graphic Representation of Forces. As a great
many mechanical problems are solved by means of geometry,
c ^ A it is convenient, and almost
FIC, 11 necessaiy, to represent forces
graphically that is, by means of lines. A force is repre-
sented graphically by drawing a line parallel to the line
of action of the force, and of a length proportional to the
magnitude of the force, the direction of the latter being
indicated by an arrowhead marked on the line. In Fig. 11,
O is the point of application of a force, and the force is repre-
sented by O A, whose length is as many units of length as
there are pounds in the force. The arrow indicates that the
force tends to move O from O toward A along the line OA,
If the force is 100 pounds, OA may be made A | A inch, or
mch, or 100 millimeters, etc.; the unit of length used is
27 OF MECHANICS 45
immaterial, provided that the same unit is used for all forces
in the solution of any one problem
It is not necessary, in ordei to represent a foice by a line,
to draw the line through the point of application, or any
other special point. Any line parallel to the line of action of
the force may be used for the purpose
86. Vectors. Any quantity that, like force and velocity,
has magnitude and direction is called a vector qumitlty;
and the line, as OA, Fig 11, by which the quantity is repre-
sented graphically is called a vector. A vector is, then,
simply a line having a length proportional to the magnitude
of a vector quantity, and an airowhead to indicate the direc-
tion of that quantity Such quantities as mass and volume,
which have no direction, cannot be represented by vectors.
87. That extremity of a vector to which the arrowhead
points is called the end of the vector; the other extremity
is the origin. In Fig. 11, O is q
the origin and A the end of the
vector OA.
88. Two vectors having the
same origin or the same end will
here be referred to as being in 11011- Pl 12 B
cyclic order. If the end of one coincides with the origin
of the other, they will be described as being in cyclic oi-der.
In Fig 12, the vectors OA and O B have the common origin O,
PIG 18
and are, therefore, in non-cyclic order; so are the vectors OA
and J3A, Fig. 13, which have the common end A. In
Fig. 14, the end of the vector O A coincides with the origin
of the vector A B, these two vectors are in cyclic order.
46 FUNDAMENTAL PRINCIPLES 27
89. Resultant and Components. When a body is
acted on by several forces, there is usually one single force
that, if acting alone, would produce the same effect as the
several forces combined This single force is called the
resultant of the other forces. With respect to the result-
ant, the combined forces are called components.
90. Composition and Resolution of Forces. The
process of finding the resultant when the components are
known is called the composition of foi-ces. The process
of finding the components when the resultant is known is
called the resolution of forces.
Both the resolution and the composition of forces are of
the utmost importance in dynamics, especially in statics, and
will be explained more fully further on. But here the funda-
mental principles must be stated and explained.
91. Composition of Colllnear Forces. Forces hav-
ing the same line of action are called colllnear forces.
The resultant of several colhnear fotces is equal to thein alge-
braic sum.
For example, if a force of 15 pounds pulls a body upwards,
and a force of 10 pounds pulls the body downwards, along
the same line, the effect will be the same as if the body were
pulled upwards only, with a force of 15 10 = 5 pounds.
If the upward direction is treated as positive, the downward
direction will be negative, and the downward pull of 10
pounds will be represented by ( 10). The resultant is,
therefore, 15 + ( 10) = +5. The same principle applies
to any number of forces. Thus, the resultant of the forces
15 pounds, 37 pounds, 39 pounds, 13 pounds, and 78
pounds is 15 + 37 - 39 + 13 - 78 = 65 - 117 * -52 pounds.
The negative sign indicates that the resultant is a downward
pull of 52 pounds. This principle, which is sufficiently
obvious, has already been applied in this Section.
92. Concurrent and Non-Concurrent Forces. Two
or more forces not having the same line of action, but whose
lines of action meet at a point, are called concurrent
27 OF MECHANICS 47
forces. If the lines of action of several forces do not meet
at one point, the forces are said to be no n- concurrent.
93. Resultant of Two Concurrent Forces: Paral-
lelogram of Forces. Let O t Fig 15, be a particle acted
on by two concurrent forces F l and F, in the directions O A
and OB, respectively. These lines are vectors representing
the two forces. Com-
plete the parallelogram
OA CB, by drawing
AC parallel to OB,
and BC parallel to OA.
Draw the vector C,
forming the diagonal
of the parallelogram.
It can be proved, both Fl<< Jr>
mathematically and by experiment, that this vector represents
the resultant of the forces jf\ and F a ; in other words, that, if
the forces fl and F; are represented in magnitude and direc-
tion by the sides OA and OB of a parallelogram, the result-
ant will be represented in magnitude and direction by the
diagonal C. This principle is called the law of the par-
allelogram of forces, and can be stated in general terms
as follows:
// two concurrent forces are represented in magnitude and
direction by two vectors taken in non-cyclic order, their resultant
will be represented^ in magnitude and direction, by another
vector forming the diagonal of the parallelogram constntcted
on the vectors representing the two forces.
94. Formulas Derived From the Parallelogram of
Forces. When the magnitudes of the forces F,. and F at
Fig. 15, and the angle between their lines of action are
given, the magnitude and direction of the resultant R are
very readily computed by the principles of trigonometry.
In the triangle OA C,
OC = 4O A* + AO~*TOA. A~Cl~wTA
But OC = R, OA = F lt A C * OB = F tt and angle
A = 380 . Substituting these values, and writing
48 FUNDAMENTAL PRINCIPLES U7
cos L for cos (180 Z,), the preceding equation
becomes
R = V/T,' + /,' + 2 /s/; cos Z, (1)
The angle Mi that R makes with F^ is found from the
same triangle by the principle of smes, which gives sm
sin A
A C
= - -; or, replacing sin A by sin L (for A = 180 L),
and writing F a for A C, and R for C,
sin Mi _ F. m
sin Z, R '
z?
whence, sin Mi = -j sin Z, (2)
y?
r
Similarly, sin M, = -~ sin Z, (3)
R
The values of M and yfl/ a may be checked by the i ela-
tion Mi + M a = Z.
EXAMPLE Two concurrent forces having the magnitudes 450 pounds
(= F,.) and 675 pounds (= F,} make an angle of 60 Required
(a) the magnitude jR of their resultant, (fl) the angles MI and M,
made by the line of action of R with the lines of action of FI and f lt
respectively
SOLUTION. (a) Here, /? t = 450 Ib ,/?". = 075 Ib , Z = B0, and
formula 1 gives
A = -\I450 1 + 676 s + 2 X 450 X 675 X cos 60 = 980 75 Ib Ans
(d) Substituting m formulas 2 and 3 the values of F tt F lt A",
and L (angles are given to the nearest 10 seconds),
fl7^
sin Mi = ~Y & X sin 60, whence, M* = 86 SB' 10". Ans
sm M t = 935-75 X sin 60, whence, Jlf a = 23 24' 50". Ans
Checking, M v + Jlf, = 36 SB'- 10" + 23 24' 50" = H0.
95. Rectangular Components. When the two forces
are perpendicular to each other, they are called rectangular-
components. In this case, angle L = 90, sin L = 1,
cos L = 0, and the preceding formulas become
= V/^'+T^ (1)
sin M, = cos M, = 5 = ^^__ (Q)
* V/v + /;
27 OF MECHANICS 49
For this condition, however, the angles MI and M* are
more easilv determined by their tangents or cotangents.
This special case is illustrated in Fig. 16, which is similar to
Fig. 15, except that the
angle A OB is a right ^ ^ f 2 *
angle The parallelo- ^ % V^ fa
gram A CB is a rect-
angle, and the right ^\
triangle O A C gives R '
= OC =
= ^F\* + F t ', as found A
before Also, Fl0 - 16
Similarly, tan M t = = (4)
96. Resolution of a Force Into Two Components.
The next thing to consider is the converse problem of find-
ing the components of a foice when the force is given. This
is called resolving a force into components. We shall
o -Ja B v here confine ourselves
p 7 >- -7 Xz
fa 7 i to the particular prob-
/ lem of resolving a force
/ into two components
whose directions are
given.
Let O C = R, Fig. 17,
be a force acting at O.
17 It is required to find its
components along two given directions. To do this, draw
through O two indefinite lines Xi and OX ti parallel to the
given directions. From C draw a parallel to OX,, meeting
OXt at A, and a parallel to OX lt meeting OX, at B. Then,
OA and OB are the required components Ft. and F a .
To find Ft. and F t by calculation. Since the directions
of Ft. and F, are given, M* and M* are known, and also
L = M, + Af n . Then, from formula 2 of Art. 94,
ILT398 5
FUNDAMENTAL PRINCIPLES 27
_ R sin Mi _ R sin M^ / \
sm L sin (M l -f-
Similarly, f.- .
sin
If one of the components is to make an angle M l with the
>rce R, and the other an angle of 90 yl/i, that is, if the
70 components are to be perpendicular to each other, for-
ula 2 of Art. 95 gives
F. = R sin M, (2)
Similarly, F t = R sin M a = -ff cos M l (3)
The student should again consider Fig. 16, and see how
iese formulas are derived. If the figure is kept in mind
id the geometrical relations of the quantities involved are
imembered, no difficulty in deriving, remembering, and
Dplying the formulas- will be experienced
97. By the component of a force in a certain direction is
sually understood one of two rectangular components. The
>rce is supposed to be resolved into two components at
ght angles to each other, one of which has the direction in
uestion. Such a component is also called the resolute and
le resolved port of the force in the given direction.
EXAMPLES JFOB PRACTICE
1. If, in Fig 17, C represents a force of 40 pounds acting; on a
idy at 0, and the angles that the components Fi and F, make with
ie given force are 46 and 20, respectively, what are the values of
,ese components? . f/^ 15.10 Ib.
Ans 't/?; - 31.21 Ib.
2 A rectangular component F 9 of 60 pounds makes an angle of 40
ith the resultant J?, Find the other component FI and the resultant.
Ana /^ " 60 ' 34 lb -
An8 'lj? . 78.82 lb,
98. The Triangle of Forces, In the construction
lown in Fig. 17, the complete parallelogram OACB has
2en drawn, in order that the principle on which the oonatruc-
on is founded may be understood. But, in practice, it is not
scessary to draw more than one-half of the
27 OF MECHANICS 51
that is, one of the equal triangles OAC, O B C. For, since
OB = AC, the line CA drawn through C parallel to OX, in
order to deteimme the point A, gives at once the length of
the line OB representing the component F a . Also, the
angle M a between F a o
and R is equal to the
angle OCA. Hence
the following construc-
tion (Fig. 18).
When a force R is
given, and it is required
to hnd its components in
two given directions, ^1 RIO is
draw from one extremity, as O, of the vector representing the
given force, a line parallel to one of the given directions, and
from the other extremity, as C, a line parallel to the other
given direction The two sides O A and AC of the triangle
thus formed ate the requited components. The vectors repre-
senting the two components aie in cyclic order with each other,
but in non-cyclic order with the vector representing their
resultant R.
This construction is called the triangle of forces.
Notice that, although A C gives the magnitude and direc-
tion of F t , the force F* must be supposed to be really acting
at O, as shown by the dotted line OB.
The line C B might have been drawn through C, to meet
OX, at B In this case, however, the triangle would have
been as shown by the light dotted lines, but the results
would have been the same as before.
EXAMPLES FOR PRACTICE
1. Two forces, F,. = 1,000 pounds and F, = 8,000 pounds, act
simultaneously on a particle, the lines of action of the two forces
make with each other an angle of 30. Find (a.) their resultant jR;
(t>) the inclination of the resultant to the two components (angles to
nearest 10"). f(a) J? - 3,898.2 Ib.
Ans \ IK f^/t = 22 37' 60"
I (0 > Utf. = 7 22' 10"
52 FUNDAMENTAL PRINCIPLES 27
2, Given a force R = 20 tons, to resolve it into two components /",
and F t making with R angles M* = 30, Jlf a = 45
Aus 1 F * = 14 41 tons
Aus l/> 3 = 10 SHU tons
3 Find- (a) the resultant of two rectangular forces FI = 40 pounds
and F, = 80 pounds, (d) the Inclination of the resultant to the com-
ponents f (a) R = 50 Ib
An s 1 ih\ JM* = JW r >-' 10"
L w l^/, = f)8 7'fiO"
4 A force of 100 tons is inclined to the vertical at an angle of <H)
Resolve it into a horizontal component H and a vertical component /
AriB Iff => Sfi(K)3 tons.
Ans I V = 50 tons.
ANALYTIC STATICS
(PART 1)
CONCURRENT COPLANAR FORCES
DEFINITIONS
1. Coplanar and Non-Coplanar Forces. When the
lines of action of several forces he m the same plane, the
forces are said to be coplanar. If the lines of action are
not in the same plane, the forces are non-coplanar.
2. Analytic Statics and Graphic Statics. So far as
the mathematical treatment of statics is concerned, this
science may be considered as being mainly a branch of
applied geometry. And, as geometrical problems may be
solved either analytically or graphically that is, either
by computation or by construction so, too, statics may be
either analytic or graphic, according to the method of solu-
tion used.
Analytic statics treats of the equilibrium and equivalence
of forces by means of the arithmetical and algebraic relations
existing among the forces, which relations, in the case of
forces represented by vectors, are the same as the relations
existing among the vectors, and depend on the geometric
properties of the figures formed by combinations of such
vectors. In analytic statics, all results are found by
calculation.
Graphic statics treats of the equilibrium and equivalence
of forces by means of geometric figures. In graphic statics,
all results are found by measurement.
COPYNIIIHTBD MY INTBRNATIONAI. TEXTBOOK COMPANY. ALL HlaHTI MMBRVU
28
2 ANALYTIC STATICS 28
3. System of Forces. The aggregate of all the forces
acting on a body or on a group of bodies is called a system
of forces. The simplest sybtem of forces is obviously that
in which there is only one force.
4. Equivalent Systems. Two or more systems of
forces are equivalent when they produce the same effect,
or, what is the same thing, when they have the same
resultant The resultant of any numbei of forces is itself
equivalent to the system formed by the components.
5. Equillbrants. When a system of forces is balanced
by a single force, the latter i& called the equill brant of the
system Conversely, if several
forces balance one single force,
they are termed the equilibrants of
that force.
The equilibrant of several forces
is evidently equal in magnitude,
but opposite in direction, to the re-
sultant of those forces. Hence, it
is also called the antl -resultant.
In Fig. 1, Ft and F a are two
forces acting at O. Their resultant
R is found by the principle of the
parallelogram of forces. The force Q, equal to jR, but acting
in the opposite direction, will evidently balance R, or the sys-
tem of forces F l and F t to which R is equivalent. Therefore,
Q is the equilibrant of F,. and F t . Conversely, Fi and F, are
the equilibrants of Q.
It is also evident that, if the forces F l} F tt and Q are in
equilibrium, any of them may be considered as the equili-
brant of the two others. The same principle applies to any
number of forces.
6. The Kesultaiit Not an Actual Force. It should
be understood that, when several forces act on a body, the
resultant is not an actual force acting on the body, but an
imaginary force whose effect would be the same as the com-
bined effects of the components. Similarly, when a single
28 ANALYTIC STATICS 3
force acts on a body, its components are imaginary forces
that would produce the same effect as the single force.
The resultant may replace the components and the compo-
nents may replace the resultant, but in no case are the com-
ponents and the resultant supposed to act simultaneously.
The equilibrant, on the contrary, is a real force applied in
order to balance a system of other forces; or, if the system
is balanced, any one of the forces is the equilibrant of the
others.
7. Internal and External Forces. With reference to
a body or system of bodies, a force is said to be external
when it is exerted by a body outside of the system. The
forces exerted by parts of a body, or of a system of bodies,
on one another are called internal forces.
8. The terms "internal" and "external" are relative; and
it is evident that a $ p
force may be external _ |
with respect to a body %\ I
and internal with re- A 1, W
72 r
spect to another | |
body, or to a system >, | j Jj,
of which the body in A fb) Q
question forms a part. Fl 2
For instance, the forces F t and F, shown in Fig. 2 (a) are
external to the body A B CD on which they act. If the body
is in equilibrium, any part of it, as AB PQ, to the left of a
section PQ, must push on the other part with a force equal
to F lt while the other part pushes on it with a force F tt = F t .
This mutual push, or stress, between AB PQ and CDQP
consists of two forces, both of which are internal with regard
to the whole body AB CD. But if the condition of only the
part AB PQ is under consideration, the push of CDQP on it
may be regarded as an external force. The portion CDQP
may be removed, as shown in Fig. 2 (), and the portion
AB PQ considered as an independent body acted on by the
external forces Fi and F.
i I
4 ANALYTIC STATICS 28
FUNDAMENTAL PRINCIPLES AND FORMULAS
9. Free Body: Principle of Separate Equilibrium.
When a body is cousideied by itself as a whole disconnected
from other bodies, it is called a free body.
Any pait of a system or body may be treated as a free
body, provided that it is assumed to be under the action of
external foices equal to the internal forces that keep it con-
nected with the rest of the body or system of which it forms
a part. Thus, in Fig. 2 (/>), the portion ABPQ may bfe
treated as a free body, after introducing the force F t) equal
O to the force exerted by the rest of the body A B CD,
that is, by CD QP, on the portion ABPQ.
As another example, take the system formed by a
string OP, Fig 3, carrying a weight W. If it is
desired to study the condition of a part of the string,
as O A, the part A P and the weight W may be sup-
. posed to be removed; but, in order that the condition
of O A may remain unaltered, it is necessary to intro-
duce at A an external force equal to the sum of the
weight Wand the weight of the part AP of the
string.
In general, when a body or a system of bodies is
in equilibrium, every part of it must be in eqtiihb-
Flt * - H rium and may be treated as a separate body by
itself, provided that external forces are introduced to replace
tht; internal forces exerted by the rest of the body or system
on the pait removed. This principle is called the principle
of separate equilibrium; its application, called the free-
body method, is of great value in the solution of static
problems.
10. TruiiHfetiiblllty of the Point of Application.
Concurrent forces have already been defined as those whose
lines of action intersect at a point, They may or may not
be actually applied at their point of intersection. Their
point of application is not essential, provided that the forces
act on a rigid body. This Is expressed by the following
28
ANALYTIC STATICS
5
principle, which is usually known as the principle of the
transmissibdity or transference of force, but will here be called
the principle of the transferability of the point of
application:
When a force acts on a rigid body, the force may be supposed
to be applied at any point on its line of action^ provided that
this point and the body are rigidly connected.
Let a force f, Fig. 4, act on the body M ' N m the direction
Kio 4.
OX. A string may be tied at A, or at^?,, and a pull exerted
from X with the force /''and along the line OX; or it may
be imagined that the string
is tied to a point A, and that
this point is the end of a
rigid rod connected to the
body at A l} A t or to any
other point along OX; no
matter where the string is
tied, and no matter from
what point on OX a pull is
exerted, the effect will be
the same. Again, it may
be imagined that a rigid
strut ,S is attached to the
body and that the force is
applied at A, either by pulling from X or by pushing from O.
It may thus happen that several concurrent forces acting
on a body do not intersect within the body. For example,
in Fig fi, the points of application of the forces F^ F t) and/ 7 ,
actually applied at the points M^ M lt and M, of the body
PQ t may be transferred to the common point of inter-
section O of their lines of action, assuming this point to be
Fio 5
6
ANALYTIC STATICS
28
rigidly connected to the body. Thus, the force F, may be
imagined to be a pull at the end O of a string tied to M,,
and the force F a as a push at the end O of a strut between O
and M 3 .
11. Resultant of Any Number of Concurrent
Forces. The resultant of two concurrent foices can be
easily determined by either the parallelogram or the triangle
of forces, as explained in Fundamental Principles of Mechanics.
The resultant of more than two forces can be found by
successive applications of either of these principles.
Let four concurrent forces F lt F t , F a , F t be represented,
respectively, by the
vectors O A, OB, O C,
OD, Fig. 6. The re-
sultant R? of F l and F,
is found from the par-
allelogram OAEB
constructed on the vec-
tors O A and OB. This
resultant is now com-
bined with another of
the given forces, say
F t , by constructing the
parallelogram OEGC,
which gives the result-
ant R" of R' and F tt
and, therefore, of F lt F tt
and TV Finally, the parallelogram G KD gives the result-
ant R of R" and F tl and, therefore, of F lt F t , F and F 4 .
When analytic methods are used, this process is exceed-
ingly laborious, as it involves the calculation of the magni-
tude and direction of each resultant, A much simpler process
is afforded by the resolution of each force Into two rect-
angular components, as explained in tha next article. If the
problem is solved graphically, the triac&te of forces is used
instead of the parallelogram: A E ia fleawn equal and parallel
to OB or F t ; then, EG equal and pafalllipj to /i; then, GK
Fu 6
I \
28
ANALYTIC STATICS
equal and parallel to F 4 , the vector O If, between O and K,
gives the resultant This method will be further explained
and illustrated in connection with graphic statics. At present,
only analytic methods are under consideration.
12. Method by Rectangular Components. Let fi,
F,, F a , Ft, Fig. 7, be four concurrent forces represented,
r
Of
I
1
f*
1
1
FIG
respectively, by the vectors A,, OA a , OA a , OA 4 . Take any
two lines X f X, Y' Y through O and at right angles to each
other. These lines are used as lines of reference; they are
called coordinate axes, and may be taken in any convenient
position. The axis X' X is usually referred to as the x axis,
or axis of x, and the axis Y' Y t as the y axis, or axis of y.
The angles made by the given forces with the axis of x are
denoted by #,,-#., etc., as shown.
8 ANALYTIC STATICS 28
As explained in Fundamental Principles of Mechanics, the
force /*, can be icsolved into two components in the direc-
tions OX and Y, by drawing from A t a line .-/, C paral-
lel to OX, and a line A,B^ parallel to OY. These lines
determine a parallelogram OB^A^ C, in which OB, and O C,
represent the components of F, in the directions OX and O Y,
respectively. The component OB, is called the x compo-
nent of F,; and O C, the y component. These components
will be denoted by A' and K M respectively. Instead of con-
structing the parallelogram OBiA^C,., it suffices to draw
from A t the line A/?i perpendicular to OX] this determines
the two components of F lt since B^A, = O C, = ]',. The
components of F,, F , F< are similarly found; they will
be denoted by X t and Y 3 , A' a and Y a , etc. It should be
understood that, although K,, for instance, may be repre-
sented by the vector B^A^ that component really ads
through 0, its position being O C. The given forces aic
now reduced to the system formed by the A components ( ) /?,,
OB t , Z? D , #>9,, and the y components O C, OC,, O C,, OC 4 ,
and the resultant of the given forces is the same as thai of
these components. Since the x components act in the same
line, their resultant is equal to their algebraic sum (sou
Fundamental Pnucifilfs of Mechanics}. The same principle
applies to the resultant of the y components. Taking forces
acting toward the right or upwards as positive, and those
acting toward the left or downwards as negative, we luivc,
X, <= OB X t = fl/?., X a = - Off,, A' t = - Oh\
Y, = 0C, Y, = OC, Y, = r)C, K 4 =r - 0(\
Denoting the resultant of the x components ant] that of
the y components by X r and Y ft respectively, we hnve,
X f = A' t + A", + X tt + A". = OB, + OB, - OB, - OJi, [a]
Y,= Y,+ Y.+ K.+ K 4 = OC + (9C + OC - ^f, U)
These resultants are represented in the figure by Oft,
and OC n respectively. The given system of forces has,
therefore, been reduced to the two forces X? O J7 f and K-
=s C r ; and their resultant /?, found by the parallelogram or
by the triangle of forces, is the resultant of the given forces,
In other words, the forces X r and Y r are the rectangular
28 ANALYTIC STATICS 9
components, in the directions OX and O Y, of the required
resultant.
13. The calculation of the magnitude and direction of R
is accomplished as follows. The light tuangle O .?, A* gives:
OB, = OA, cos Jf lt B,A, = OA t sin //
or, since OB^ = X^ OA r = F lt and B l A l = Y lt
X, = F t cos ff lt Y! = F l sm //i
The values of X, and Y a , X e and K n , and X t and Y<. are found
in the same manner It may be well to remembei the pnn-
ciple that the component or resolute of a force hi any direction is
equal to the force multiplied by the cosine of the acute angle
that the line of action of the force makes with that direction.
Thus, the x component of F, is F, cos H a For the other
component, the sine should be used instead of the cosine
Substituting m equations (a) and (b] of the preceding
article the values of A',, K,, etc., the following expressions
are obtained:
X r = F, cos H, + F, cos H a F* cos H 3 F< cos H,.
Y r = Fi sin H, + F* sin H, + F, sin H, - F. sm H*
Having determined X r and Y r , the right triangle OB r A r
gives the magnitude of the resultant R, and the inclination H f
of this resultant to OX-
R = OA r = ^OB r 3 + B r AS = *&' + Y r " (1)
tan H - - - ; (2)
14. As explained in Plane Trigonometry, Part 2, the sym-
bol of summation -V, read sigma, is often used to indicate
the algebraic addition of several quantities denoted by the
one letter affected by subscripts or accents. If, for example,
several quantities are denoted by X^X^X^X^ etc., their
algebraic sum X l + A r , + A' + A' 4 ... is denoted by
the expression - A', read signia x.
With this notation, the algebraic sum of the x components
A', Xv, etc. of any number of forces may be denoted by - X,
and the algebraic sum of the y components by 2 Y. Also,
the sum of the expressions F t cos ff lt F, cos H t , etc. may be
denoted by -i'^cos If, If the components of the resultant
10
ANALYTIC STATICS
are, as before, denoted by A' r and )',., the following equations
apply to any number of forces
X r = 2X = .i'/'cob // (1)
Y r = 2Y = ZF sin // (2)
Having computed X r and K,, the magnitude and dnection
of the resultant R are determined by formulas 1 and 2 of
Art. 13.
EXAMPLE It is required to hml tlie resultant of the three c<>n-
the
l-ll. S
current foices J'\, /'\, J'\ represented in Fitf H, their magnitudes and
relative positions being as shown
SOLUTION Take A" A' inclined nt W to the line of m-lion of /,.
The Inclinations of the other two forces to A'' V aio u-iulily determined
from the given angluH. Thus,
//, = 45", //. = A' ; Os1, - 1HO" - (7fi n H ifi") ** IMI"
//n A'' O A* B HO" - (!()" ^ ^0"
For the JT componontH, with the usual convention iw to
following eqimtioliK uru obtained;
A', = 1(K) )H4ri 11 * 70.711 Ih,
A' =-- - JK) ros UO" -- ~ ir>.(MH) Hi,
A; ~lo() cos L>(] - LlO.IUi lh.
and for they corupononts:
Y, KH)siu4f 11 *. 70.711 lh.
r, m IH) Niu HO" 77,iu n>.
K M * ~ifiHina -fii.:ttiij,
Therefore,
X, i 1 ^ - 70.711 - 4fi.OOO - 14JI.WV - - 115.24 Ib,
K- - A 1 X 70.711 + 77.48 - ftl ,OS 7 ,ST>1 Ib.
28 ANALYTIC STATICS 11
The negative sign of X e indicates that this component of the
resultant R is directed toward the left, as indicated by the vector O B r
Formula 1 of Art 13 now gives,
R = <\/115.24' + 97~3oT n -= 150 86 Ib.
and formula 2 of the same article gives,
tan H r = ?Tj^, Hr = 40 11' 20"
llo U4
In finding the value of tan H r , the negative sign of X r is dis-
regarded, as the only thing that is required is the numerical value of
the angle The signs of X r and Y r show in what quadrant the result-
ant is In the present case, X r (= O B r } is negative and Y f (B,A,}
is positive. This at once shows that R lies in the angle X 1 O Y t and
that H r is, therefore, the angle that R makes with O X'
NOTE In this, as in many other problems In this Course, angle s aie jdyen to the
nearest 10" Thus, an angle of 86 10' 37" is called Bfi 10' 40", an angle of 17 47' 28" is
called 17 47' 20"
MOMENTS
15. Definitions. The moment of a force about u
point, or witli respect to a point, is the product obtained
by multiplying the magnitude of the
force by the perpendicular distance from
the point to the line of action of the
force. In Fig. 9, the moment of F
about the point C is Fp\ about the
point C, it is Fpv
16. The point to which a moment
is referred, or about which a moment is
taken, is called the center of moments,
or origin of moments. The perpen-
dicular p or p l from the origin on the
line of action of the force is called the
lever arm, or simply the arm, of the force with respect
to the origin.
17. A moment is expressed in foot-pounds, inch-tons,
etc., according to the units to which the force and its arm
are referred. If, for instance, the force is 10 pounds and the
arm is 60 feet, the moment is 10 X 60, or 600, foot-pounds;
if the force is 8 tons and the arm is 6 inches, the moment i&
12
ANALYTIC STATICS
8 X 6, or 48, inch-tons. The term foot-pound is used in
kinetics m another sense; but the two meanings are so
different that there is no danger of confusion
18. Sign ami Direction of u Moment. When, look-
ing in the direction of the anowhead of a vector represent-
ing a force, the center of moments lies on the right of the
line of sight, as C and C, Fig. 10, the moment is considered
positive; if the center is at the left, as C v , the moment is
considered negative This may be stated otheiwise thus*
Imagine the level aim p to be levnlvmg about C in the
direction indicated by
../>. the arrowhead. In this
, c , case, / will t evolve from
' left to nght, 01 clock-
wise, and the moment is
considered positive. In
the case of /> on the
contraiy, if this line re-
volvc'd about C, follow-
ing the direction of the
ai low-head, its motion
would be counter-clock-
wise; the moment is
then considered nega-
tive. The direction of the motion just refcned to is called
the direction of th iiioniciil, and is supposed to have
the same sign as the moment itself.
If M l M^ and Jlf, are the moments of /''about C, (',, and C" fl
respectively, then,
M - 4- Fj> - />
//, - + /'/, jffr
/!/ * - />,
19, Itcin'cHentutlon of ti Momont by an Ai-ou, If
the lines CO and CA, Kitf. 10, ure drawn, the area of the
trianffle CO A is
10 Ax CD m |/-> * \M
whence, M * 2(i OA X CD) 2 X area OA C
i'io 10
28 ANALYTIC STATICS 13
The magnitude of the moment M is, therefoie, numeri-
cally equal to twice the aiea of the triangle whose base is
the vector representing; the force F and whose opposite
vertex is the origin of moments.
20. Moment of ti Force In Terms of the Moments
of Its Components. The moment of a force about any point
in //\ plane is equal to the algebraic sum of the moments of
r/<; (omponcnts in any diiections.
Let the foice R, Fig. 11, be resolved into the two com-
ponents Ft and F, in any directions, and let C be the origin
a
* V<*, .
PIG. 11
of moments. The arms of , F l} and F a are, respectively,
p r , Ai and p at as shown. The moments of these three forces
about C will be denoted by M r , M^ and M t , respectively.
Draw CO, and <9 A" perpend ictilai to CO. Denote the angles
made by the lines of action of the three forces with OX "by
If,, #i, .//., as shown, and the resolutes of the forces in the
direction OA'by X,,X lt and JC. Then (Art. 12),
X r = X,. + X 9 ',
that is (Ait. 13),
R cos H r = F! cob HI + F, cos Hj>
whence, multiplying both members of this equation by CO,
ft X CO cos H r = F l X CO cos ff^ + F^X CO cos ff, (a)
1 LT398-
14 ANALYTIC STATICS 2*
But H f = 90 COE, and, therefore, cos /7, = sin CO&
and
CO cos ff, = CO sin (TO^
= C^( triangle C6>>) =A
Also, CO cos // t = C(9 sin C<9 A = A
CO cos // = CO sin COD, = A
Substituting these values in (a),
that is, jWr = ^ + ^/,-
In general, if the algebraic sum of the moments of any
number of concurrent forces is denoted by 2 Af t and the
moment of the resultant by Af rt as above, then,
Mr = XM
CONDITIONS OP EQUILIBRIUM
21. The Absolute Condition. When any number of
concurrent forces are in equilibrium, it is evident that they
can have no rebultant; that is, if their resultant is R, we
must have R 0. This is the general condition, sometimes
called the absolute condition, of equilibrium Expressed
in the form of R 0, however, this condition is of no value
in the solution of problems. It is necessary to express R in
terms of other quantities, or to derive some other relations
by means of which unknown quantities can be detei mined.
22, Condition of lUtsoluton. Since the resultant of
several forces in equilibrium is zero, itb resolutes m any two
directions mUvSt be zero; and, as each resolute is equal to the
algebraic sum of the corresponding resolutes of the given
forces, it follows that the sum of the tesolutes of the forces in
any direction must be zero*
The resolute of a force in a direction perpendicular to
itself is zero. There is, then, one direction (at right angles
to the resultant) ftlongr which the sum of the resolutes of any
number of unbalanced concurrent forces is zero ; but if the
sum of the resolutea is zero for any two directions, the forces
are in equilibrium; for both directions cannot be perpen-
dicular to the line of action of the resultant.
28 ANALYTIC STATICS 15
23. Condition of Moments. Since the resultant is 0,
its moment about any point must be zero; and, as the
moment of the resultant is equal to the sum of the moments
of the components, we must have,
M + M, + M',+ . . . =0,
where, M lt M,, Jlf a , etc. are the moments of the components
about any point in their plane.
It is necessary to specify that the sum of the moments
must be zero, when taken with respect to any point, or every
point in their plane, for, m a system of forces not in equilib-
rium, the sum of their moments about a point on the
resultant is also equal to zero, because in this case the lever
arm of the resultant is zero, and its moment, therefore, is
zero. But, if the forces have a resultant, the sum of their
moments about a point outside of the resultant cannot be
zero. Conversely, if the sum of the moments of the forces
about three points not in the same straight line is zero, the
forces have no resultant; for these three points cannot all lie
on the line of action of the resultant, and the only case m
which the sum of the moments of a system of unbalanced
forces can be zero is that in which the origin of moments is
a point m the line of action of the resultant.
24. Genei-al Statement of the Conditions of Equi-
librium. Summing up: Any balanced system of coplanar
concurrent forces must satisfy the following conditions, each
of which is necessary and sufficient for equilibrium, and
involves the other two:
1. The resultant of the forces must be sero.
2. T/ie sum of the resolutes of the forces in each of any two
directions must be zero.
3. The sum of the moments of the forces about three points in
their plane \ not in the same straight line^ must be zero.
Condition 1 is expressed algebraically by the equation R 0.
Condition 2 may be expressed thus (see Art. 14):
EX = S
Condition 3 may be expressed thus (see Art. 20):
16
ANALYTIC STATICS
1 28
If the lever arms of the forces F n F* t etc are denoted
by Pi t A, etc., the algebraic sum of their moments, or 2 Af,
jt> is equal to F 1 p^ 4- F a /> a + F 3 p* + , etc
Denoting this sum by 2 Fp, formula 2
may be written
*** vpp = (3)
25. Equilibrium of Three
Forces. Problems i elating to the
equilibrium of three forces are usu-
ally most conveniently solved by
means of the triangle of foices. It
is to be observed that the eqiuhbrant
of any number of forces is numer-
ically equal to their resultant, but
acts in the oppobite direction. Let
the forces F lt F yi Fig. 12(tf), act
through O. By di awing through O f ,
us shown at (), the vectors (Y AS and
\-l Aj ) representing the given forces, and completing the
triangle O 1 Aj AJ> the re-
sultant R of /'I and F, is
obtained. In this triangle
of forces, the two compo-
FIG lli
nents are in cyclic order with each other, but in non-cyclic
order with the resultant (see Fundamental Principles of
Mechanics}.
28 ANALYTIC STATICS 17
In Fig. 13 (a), the three forces F^F,, and Q, acting at O,
are supposed to be m equilibrium. The force Q is the equi-
librant of F, and F t ; it is numerically equal to their result-
ant R, but acts in the opposite direction. Consequently,
a triangle O 1 AS A* 1 , Fig. 13 (b), equal to the ti tangle giving
the resultant /?, can be formed with the vectors lepresent-
ing /-" F yt and O\ but the anowheads on Q and 7? must
point in opposite directions, which means that Q must be
taken in cyclic order with F l and F tt .
26. Selection of Axes. When there are more than
three forces in equilibimm, the most convenient method for
determining any of them, when the others aie known, con-
sists m finding expiessions foi the resolutes of all the forces
m two rectangular dnections, and using formulas 1 and 2,
Art 24. A similat method applies to the determination of
the resultant, as aheady explained Theoretically these
directions are entirely aibitrary; but, practically, the direc-
tion of one of the foices, often one whose magnitude is not
known, is almost always a very convenient one to use, as in
this case one of the resolutes of that force is equal to zero,
and the other is equal to the force itself. When this is done,
either direction along the line of action of the force may be
taken as positive and the other as negative; but this does
not necessarily mean that the direction taken as positive is the
direction of the force acting along that line. For example,
forces acting at a point O may be resolved into components
parallel and perpendicular to the line of action, say OK, of
one of the forces; and, for the purpose of this resolution, the
direction OK may be treated as positive and the opposite
direction as negative. This simply means that, if the resolutes
parallel to OK are to be added algebraically, the arithmetical
difference must be taken between those resolutes whose direc-
tion is OK and those whose direction is opposite; if the dif-
ference is positive, the direction of the resultant resolute is
from O toward K\ if negative, from A" toward O.
Similarly, in taking moments, any point can be used as an
origin; but it is convenient to take the origin on the line or
18
ANALYTIC STATICS
28
action of one of the forces, as in this case the moment of
that force is zero, and the force is thereby eliminated from
the equation of moments.
These methods will be better understood by a study of the
following problems:
EXAMPLE 1. A weight of 500 pounds (see Fig. 14) hangs by three
j t
/ I \
W-BOOlb.
FIG. 14
ropes tied to a ring at O, the inclinations of the two slanting ropes O JS t
and O E, to each other and to the horizontal being as shown. It Is
required to find the tensions in the ropes OEi and O E 9 .
SOLUTION BY TRIANGLE OF FORCES. The tensions F,. and F n in
the ropes are represented by the vectors A v and A tt and the weight
by the vector W. These three forces form a balanced system, each
force being the equilibrant of the other two. Considering W an the
28 ANALYTIC STATICS 19
equilibrant of F l and F a , the triangle of forces OBC is obtained by
drawing through B a line parallel to OA lt and through O a line
parallel to O A* In the present case, CO is in the prolongation of
O At The triangle might have been constructed anywhere else; It is
not necessary to draw it m the position here shown Nor is it neces-
sary to draw it accurately to scale, as its only object is to serve as a
guide in the calculation
In this triangle,
K= AiOC = 180 - Ai OA, = 180 - 75;
and, therefore,
sin K = sin (180 - 76) = sm 76
Also, Y being vertical,
Li = 90 - ff m = 00 - 60 = 30
L, = A, O Y = 90 - Hi = 90 - 46 = 46
The triangle O B C now gives
sm L, = -^U sin 30 = 258 82 Ib. Ans.
- ,
sin A sm 7o
F t = -^-sin L 3 = -^osm 45 = 366 03 Ib. Ans.
9 ^ sin 75
SOLUTION BY RESO LUTES Let the vertical line Y and the
horizontal hue OX be taken as axes of coordinates As usual, the
resolutes of Fl will be denoted by A' and YI, and those of F, by X*
and K,, as shown in the figure. They are here drawn for purposes of
illustration, but it is not necessary to draw them in order to apply the
general formulas. The vertical resolute of the weight is W\ the
horizontal resolute is zero.
Placing the sum of the horizontal resolutes equal to zero, we have,
noticing that X l is negative,
- Ft cos Hi + F t cos H t + = 0;
that is, - Fi cos 45 + F a cos 60 - (a)
Similarly, for the vertical resolutes,
Fi sin Hi + F, sm H, - W = 0;
or, Fi sm 45 + F t sin 60 - W = (*)
Prom (a), cos 46
which, substituted in (b), gives
etn ftn
Ft sin 45 + F t cos 45 ^L^_ _
whence, cos w
W _
" sm 46 + cos 45 sm ^ " 8in 46 cos ^ + cos 45 sln
cos 60
W OAO ?f . OAO
sin (45 + 60-) = STW Sm ^ = iSTTS 5 sin 80
as found before.
The value of F t may now be found from (c) .
20
ANALYTIC STATICS
28
The horizontal and vertical resolutes have been used simply to
illustrate the general method, but it will be seen that, by resolving
the forces m these directions, the solution is long and tedious If
however, resolutes perpendicular to the lines of action of the unknown
forces FI and F a are taken, the solution will be very much simplified
First, let the resolutes be taken perpendicular to O J5 y Then, we shall
have, considering the direction from O toward E 3 as positive,
resolute of F a =
resolute of F t = f,. sm 75
resolute of W = -W^smL,= - W sm 30
Therefore (formula 1, Art 24),
/r lS m75- W sin 30 =
whence,
-=a sm 30
sin 75
PIG 1C
i&oo zfc.
In a similar manner, , may
be found by taking the resolutes
perpendicular to FI
This method should be care-
fully studied, as its simplicity
makes it of very great practical
importance.
EXAMPLE 2. A weight W of
1,500 pounds is hung from the
extremity O, Fig 15, of a hori-
zontal bar projecting out of a
vertical wall and held by a
rope ON, Inclined to the vertical
at an angle of 40 Required
the tension T in the rope and
the compression C in the bar
SOLUTION UY TRIANGLE OK
FORCES. It is obvious that W
is the equilibrunt of C and T.
The actual lines of action of W
and T are shown by the vectors
O B and O A. Draw B D per-
pendicular to O //, to meet A' O
produced at D. Then, in the
triangle O B D, the vectors D O and B D represent T and C } respec-
tively. Also, BOD = 40.
C - W tan 40 = 1,500 tan 40 - 1,258.0 Ib. Ans
T - 1 -^so = 1,958.1 Ib Ans.
cos 40 cos 40
SOLUTION BY MOMENTS. If moments are taken about N, the
moment of T will evidently be zero, since //lies on the line of action
28
ANALYTIC STATICS
21
of T, the moment of W will be WxNS=lf>WXEO The
moment of C whose line of action is OE, will be CX EN There-
fore (Art 2O) ,
CXEN = i,mxo,
whence, C = 1,600 J^ = 1,600 tan 40,
as found before
If moments aie taken about E, the moment of C, the resultant,
will be 7cio, that of H' will be the same as befoie, 01 1,500 X E O, and
that of ywill be TX E A negative, because the origin E is on the
A*
FIG 10
left of O A (Art. 18) Equating the sum of the moments of the com-
ponents to the moment of the resultant,
1,500X^0- TxEK~ 0,
v. T -i n EO 1.600 L500
whence, y . 3 ,600 ^. - -^y - ^
as found before
The method of moments is sometimes very convenient. The center
of moments should, if possible, be taken on the line of action of one
of the unknown foices, as that force is thereby eliminated
EXAMPLE 3 To find the equilibrant of the five forces represented
by the full-line vectors in Fig. 10.
SOLUTION Let OA lt the line of action of F,, be taken as the
x direction, and Y, the perpendicular to it at 0, as the y direction of
22 ANALYTIC STATICS 28
resolutes. The angles of F lt F a , F 3> etc , with O X will be denoted
by HI, // a , &3, as in Art 12, and the resolutes of the forces by
X lt X a , y it y,, etc. Here we have H* = 0, H, = 50, ff, = A n O X 1
= 180 - (//, + 90) = 180 - 140 = 40, H< = 60 - H* = 60 - 40
= 20, H, = 30.
Therefoie, JT X - F = 20
X 3 = F, cos 50 = 35 cos 50
X, = - F, cos 40 = - 25 cos 40
X< = - F t cos 20 = - 40 cos 20
X, - F t cos 30 = 45 cos 30
Denoting the x resolute of the equihbrant by X gt we must have
(Art 24),
X f + X,+ X, + X, + Xi + X, =
whence,
X 9 = -(Xi+X, + X, + X< + X*)
= _ 20 - 35 cos 50 + 26 cos 40 + 40 cos 20 - 45 cos 30
- - 24.730 Ib
Similarly, K x -
Y, = - F t sin 50 = - 35 sin 50
Y, = - F 3 sin 40 = - 25 sin 40
K = F* sin 20 40 sin 20
y t = F, sm 30 = 45 sm 30
and y f s -(y 1 + y.+ y.+ y 4 + y 1 )
= 35 sm 50 + 26 sm 40 - 40 sin 20 - 45 sin 30
- 6.700 Ib
The magnitude of the equihbrant Q may be found from the rela-
tion Q = VAV + y,', but it is better to determine first the angle H q
between Q and OX. For this purpose, it is not necessary to take the
sign of X g into account (Art 14)
and, therefore,
Y a 6 700
EXAMPLES FOR PRACTICE
1 Two concurrent forces Fj. = SO tons and F t = 25 tons act at an
angle of 170. Find (a) the magnitude Q of their equilibrant ; (*) the
angle L made by Q with F,. Ano / (a) Q - 6.0132 tons
Ans \(t>)L - 48 64'
2. Resolve a force /? 100 pounds into two equal components Fi
- J" 1 , making an angle of 70. Ans. F l F, 61 039 Ib.
3. Find a general relation between a force F and either of two
equal components F^ making an angle If with each other.
Ans. F" 2-F, costtf
28
ANALYTIC STATICS
23
4 Given the equilibrant Q = 10 tons of the two forces F and F t ,
Fig 17, find the magnitudes of those two forces, the angles being as
shown (The vectors F l and F, in the figure are drawn of arbitrary
lengths, their purpose being to indicate direc-
tions, not magnitudes, the latter being as yet
unknown ) Ans IF,. = 11 154 T.
Ans \F, - 2 9886 T.
6 Find the equilibrant Q of the forces
represented in Fig 18, and the angle H, it
makes with the line of action O X of the 20-ton
force (Find resolutes parallel and perpen-
dicular to O X )
Anc
Ans
W
\Jf f
10 735 tons
5 is/ 50"
6 A weight of 15 tons is supported by two
ropes, one horizontal and the other making an
angle of 45 with the horizon (135 with the
former rope) Find the tensions Ti and T t in
the two ropes, 7i being tension In horizontal
r P e< S
21.213 tons
Q-lOTon*
FIG 17
PIG 18
24 ANALYTIC STATICS
STRESSES IN FRAMED STRUCTURES
DEFINITIONS
NOIE The theoiy of eoncuirent forces finds sunie of its principal
applications in the determination of stiesset, in framed structuies
Although a complete treatment of such a subject would be foreign to
the scope of this Section, a. few simple examples will be given, in order
that the utility of these principles may be seen Before giving those
examples, it will be necessary to define some of the teims employed
27. Structures. A (structure is a statical combina-
tion of parts designed for the transmission of foice. By
saying that a structure is a statical combination, it is meant
that the forces transmitted by the parts are supposed to be
all balanced and produce no motion. Bridges and buildings
are examples
28. MacliluoB. A machine is a combination of paits
designed for the transmission of motion In any special case
in which a machine produces no motion, its vanous parts
being at rest, the machine is to be treated as a stiuctiue.
29. Frames*. The term frame is applied to any combi-
nation of bats, strings, ropes, or other straight paits con-
nected together so that their center lines form a polygon or
a pait of a polygon. Eveiy one of the straight parts so con-
nected is called a member of the frame, and the connection
of two or more members is called a Joint.
Although in some cases the center lines of the various
members connecting at a joint do not meet exactly in a point,
they are considered so to meet, and the forces acting along
the members are treated as concurrent forces meeting at the
center of the joint.
30. Trusses. A trubs is a rigid frame contesting; of
triangles, such as occur in bridges and buildings, and in some
branches of carpentry.
31. Supports. The place or places on which a stuiuture
rests, and to which, therefore, the forces acting on the stiiic 1 -
ture considered as a whole are transmitted, are called
28 ANALYTIC STATICS 25
siipports, because they support, or sustain, the structuie.
Such are the piei s of a bridge and the foundation of an engine.
When a stiucture, as a bridge, rests on diffeient supports,
the point ot application of the resultant pressure on any sup-
port is called <\ point of support.
32. Reactions. At every joint of a frame there are
seveial forces acting, namely:
1 The external forces, or forces applied to the frame
from the outside
2 The mutual reactions, or the forces exerted by the
various members on one another at the joint.
At some joints, there maybe no external foices diiectly
applied, but the membeis may act on one another on account
of external foices acting at other joints.
As to every action there is an equal and opposite reaction,
eveiy suppoit of a stiuctuie exerts on the btiucture a force
equal and opposite to the foice transmitted by the structure to
the support. This foice excited by the support on the struc-
ture is called the reaction at the support, and is to be con-
sidered as one of the external forces acting on the structure.
33. Struts ami Ties. The two kinds of stress ten-
sion and compression that may occur in the members of a
frame are defined in Fundamental Principles of Mechanics. A
tension is often called a pull, and a compression a thrust.
Those members that are designed to resist compression
are called struts; those that are designed to resist tension
are called tie*?.
1>KTKRM I NATION OF STRKS8ES
34. Introductory Explanations. Here those frames
only will be dealt with in which the forces acting on every
member aie applied al its extremities, and m which the con-
ditions are such that the mutual reactions at the joints may
be taken to have lines of action meeting at the common
intersection of the center lines of the membeis.
This being understood, lei A^A, A, A, A,A t Fig. 19, be
three members of a frame, A being the point of intersection
26 ANALYTIC STATICS 28
of the center lines of the members, and A lt A t ,A a , the joints
connecting the three members with other members. It will
be first assumed that there is no external force acting at A.
Let F^P\,F t be the resultants of the forces acting on
A,. A, A, A, and A, A at A^A a , and A tt respectively. By the
principle of separate equilibrium (Art. 9), every member,
considered as a free body, must be in equilibrium; therefore,
the resultant of the forces acting at A on A^A must bal-
ance F^ and that resultant and F must be collinear and act
along A As, so that the resultant at A is a force Fi equal
FIG. 19
in magnitude to F lt but acting in an opposite direction.
Likewise, AA t is kept in equilibrium by the forces F,
and F,, both having the direction of the member, and
F, being the resultant of the forces exerted by the other
members on A, A at A. Similarly for A, A, The vector
representing F, is drawn on one side of the member, in
order to avoid confusion, although F, really acts along
the member A A,. Now, transfer the forces Fi,F, t and F t
to A (Art 10), where they are represented by accented
letters. For convenience, as well as for the sake of uni-
formity, the forces thus transferred are represented by veer
FtftF, 1 , having their common origin at A that is,
28 ANALYTIC STATICS 27
with the arrowheads pointing away iromA. It is evident
that the force F lt being the resultant of the foices acting
directly on A* A at A, is the resultant of F, f and FJ, and,
since F,.' is the equihbrant of F lt it is likewise the equili-
brant of F,' and F a f ; in other words, the forces FS, FJ ', and
Fa' (or their equivalents F^F t , F 3 ) form a system of concur-
rent forces in equilibiium
Substantially the same explanations would apply in case
there weie external forces acting at A. These forces, with
the forces Fi f , F,' t and F a f , would form a system of balanced
concurrent forces.
The problem of determining the forces acting on the
members is thus reduced to the general problem of the equi-
librium of concurrent forces. It is now necessary to ascer-
tain what effect those forces have on the members of the
frame that is, what members are in tension and what
members are in compression.
35. Character of the Stress In a Member. Considui
the condition of A l A It has been explained that this mem-
ber is held in equilibrium by the two forces F l and F[', this
pair of forces constitutes the stress in the member, and the
magnitude of either force is a measure of the stress (see
Fundamental Principles of Mechanics) . It will be observed
that the tendency of the forces 7*1 and FI is to stretch the
member, and that, therefore, the member is m tension. The
stress in A,. A is, therefore, a tension of /i pounds, tons, etc.,
as the case may be. Similarly, the stress in A t A is a com-
pression equal to F, (by which is meant that each of the two
forces constituting the stress has a magnitude equal to F a ),
and the stress m A a A is a tension equal to F*.
36. In determining the character of the stress in a
member, however, it is not necessary to consider the two
forces acting at the extremities of the member. Referring
again to the joint A, where the forces FS, F t f , and F<J (or
their equals F lt F F a ) form a balanced system, it is seen
that FJ and F, f have their arrowheads pointing away from
the ioint; these forces may be said to be pulling at the joint
28
ANALYTIC vSTATIC\S
28
in the direction of the membeis /, / and /,./, icspectively,
and these two members are in tension. The force /*/, on
the contrary, when supposed, as it should be, to be acting
on the joint along ./M, as shown by the dotted vectoi /.y,
pushes against the joint in the direction of the member . / /;
hence, this member is in compulsion.
37. Summing up T/iefonfathuX'il(>ni>aHit'iHbe> Hica^itcs
tension ot compiesvon atconiing "S ? '//<"'/ sttflfiw/ to mt on the
joint through the membo > it pttlh at ot frtewA on the joint.
It has been seen, for instance, that /'V, when supposed to
be acting on the joint thiough the membei /,./ (as shown
by F t ") presses on that joint, and that / . / is in compression.
38. The forces F,', /<"/, Fl (which, for shoitness, are
called the stresses in the membeis) have been represented
by vectors having their common unj>i at the joint. This is
convenient in cornpli-
1 ezited problems in
which the method of
resolutes is used
^ Hut, in almost all
cases, the vectors
may be drawn along
the lines represent-
ing the members, and
then the arrowhead
on any force shows at
once whether the
force is (that is,
measures) a pull or a
thrust.
39, Noot'MHary
Data For Deter-
mining? Htrosaea,
In determining the
stresses in the members meeting at a joint, the external foroas
or the stresses in some of the other members must be known.
Suppose, for instance, that the stress in the member
28 ANALYTIC STATICS 29
Fig. 20, is known to be a compiession equal to F lt repre-
sented by the vectoi A N, and that it is required to find the
stresses F t and F, in the members A, A and A a A Since the
forces F t ,F 3 , and F-, ai e in equilibrium, a vectoi triangle can be
constiucted with the vectoi s lepresenting them Through N
draw A T < I/ parallel to A* A, meeting A, A at Jlf, and mark the
arrowheads on N Jlf and M 'A so that the vectors AN, NAf,
and /If. 1 will be in cyclic order Then, N .If = F 3 ; MA = F n .
As F, pi esses on the joint A, the member A\A is m com-
piession. Similaily, F a) although given in direction by the
vectoi JVM, must be supposed to be acting along A a A, and
it is seen that, when thus applied, it presses on the joint A\
therefoie, 1 3 A also is in compression
40. Opposite Arrowheads 011 a Member. By refer-
ring to Fig. 21, it will be seen that the member connecting
the two joints M and O has two
arrowheads pointing in opposite
directions The student should
not fall into the en or of think-
ing that each arrowhead indi-
cates the direction in which the
part of the member on which it
is marked acts on the other part.
Thus, if the member is cut by
a plane X K, the arrowhead F
does not indicate the direction
in which the part Z acts on
the part MZ\ if that were the
case, the member would be in compression, whereas it is in
tension. The arrowhead F indicates the direction in which
MO acts on O, and, as this action is transmitted from MZ
to Z, the arrowhead F really indicates the direction in which
M Z acts on O Z\ which shows that MZ pulls on OZ, and
that, therefore, OZi& in tension. Likewise, the arrowhead F'
indicates the direction in which the member acts on the
joint M, or the direction in which, a^y part Q the member,
as Z, acts on the part Jtf pelow it. - The forces acting in
ILTJ9S-7 f '
30
ANALYTIC vSTATICS
28
the direction of /'"and JP lire, of course, equal, as each measures
the stress in the member M O\ either may be considered as the
action, and the other as the equal and opposite reaction.
EXAMPI.B. A tiuss consisting of tlnue horizontal membeis PC,
CQ t and B f), Fig liii, and four equal inclined members/*./?, 12 C t CB t
and B Q, carries a weight of ft' pounds hung from the cent IT C. The
truss rests ou two piers /' untl )\ its length, lietwutni thu joints at /*
and Q, IB 2/, and itn height is h. Neglecting thu weight of thts Htrwt:-
ture, it IB required to find the Htreiw in eavh momlier und thts rtruc-tiniiH
at the supports.
SOLUTION BY TKIANOI.R OF FORCRH.- On iiwcmnt of the Hymmetry
of the figure, the angle* PCJ) and Q CH MO eqtwl; tliwy ure both
denoted by //. Also, /; C IH parallel to Ji Q t und D CJt f'/tQ - L,
say. The anglea // and A, however, are nttt ntipptihcd to bo known
and have to be exprowed in terms of / nnd h. The figure
/;t '-~. A <>
(i\
Niu // -
tan // -
m 180
In - aln 2 /^ -
W
28 ANALYTIC STATICS 31
Of the forces that balance at C, it is evident that, owing to the sym-
metry of the tiuss, the foices along CQanA CP are equal and opposite,
and form, therefore, by themselves a balanced system. Hence, the
weight Wmust be balanced by the foices F and f, acting; along the
members CD and C B. The stresses in these two members are
evidently pulls, and it is obvious that /<\ = F,
In the triangle CMN, the vector CM represents the weight W,
i\I ' N is parallel to CB and CN Is in the prolongation of D C
Therefore, MN FI and NC F a . Since F^ => F tt then, also, A",
= A\
Now, #1 = 180 - (90 + H) = 00 - H
N = 180 - (Ki 4- A.) - 180 - 2 A', - 180 - 2 (90 - H) = 2ff
W W
Then, F t -F m - ^sin A', = -sm (90 - H)
W rr
, --- ?- r cos H =
S sm H cos H " 2 sin//
Coming now to the joint B t It is known that the force /*i acts on
It along the member CB, hence, the two forces F a and F^ acting
along DB and QB can be determined Take B M 1 = F v and
draw .fl/'.A" parallel to DB, meeting BQ at A r/ Then, .rt/'JV'
represents the force F* acting along D B t and N 1 B the force F+ acting
along QB The stresses in these members are evidently thrusts
(Art 37) As M> N' is parallel to C Q, the angles B M' N' and B JV' M f
are both equal to B C Q, or H. The triangle B M 1 N 1 being isosceles,
we have F* = FI .
For F, t we have, _
F a = -^.smZ
sin /f
jq/
Putting Fi = - jr> as found above, and sin L = 2 sin H cos .#,
^. = ~ X 2 B
The forces acting at Q are the thrust /?! of the member B Q, the
horizontal force F t along CQ, and the reaction of the pier, which,
as will be shown hereafter, acts vertically upwards Take Q M" to
represent F t (notice that it is not necessary to measure the distance
from Q in the direction of the arrowhead), draw M" N" horizontal
and QN 1 ' vertical, and, starting with F*, whose direction is known,
mark the arrowheads on the sides of the triangle M" Q N" in cyclic
order Then, QN" will represent the reaction /? and N" M" the
force F t in Q C. The latter force acts in the direction Q C, thus
showing the stress in Q C to be a pull. The triangle M" QN" gives
= ZT rr p rr W COS If W t Wl
F.-F.cMjr-FtixHir- 2sln/r - T cotjsr- ^
For the reaction, , ,
JZ - /^ tan // ~ cot /f tan /f = ~
*
32 ANALYTIC STATICS 28
The reaction being equal and opposed to the pressure exerted by
the truss on the pier, or to the part of the load transmitted by the truss
to the pier, it is seen that one half of the load is transmitted to Q, and,
as the truss is symmetrical, the other half is transmitted to the other
pier P This subject will be more fully explained in connection with
the theory of parallel forces By this theoiy, the value of the reaction
can be determined first and the calculation begun at the joint through
which the reaction acts, this is the method used in piactice.
On account of the symmetry of the truss, the stress in DP \^
obviously the same as the stress in B O
SOLUTION BY MOMENTS The solution of this pioblem by the
method of moments is as follows
For joint C, momenta are taken about a point on the line of action
of one of the unknown forces Fj. or F, The equal and opposite forces F t
need not be taken into account If the joint 13 is taken as the origin
of moments, the lever arm of W is fi U = 1 /, and that of F a is
BG = J&Csm L = DC shi L = ~ sin /. = ~* ~
cos H 2 COB H
The moment of W\& negative, and that of F 9 is positive (Art. 18).
Therefore (Art. 24),
- WXBU+F,XBG = 0,
or, writing the values of B U and B G just found,
Wl Ft I sin L
*M
whence,
2 cos//
W cof, H
sin L
This was the value found for F l (which is equal to /",) by using the '
triangle of forces, and may be transformed in the same mannei t
Passing now to joint B, F t is determined by taking moments about D t \
in the line of action of F a . The lever arms of F t is DJ, and the lever f
arm of F Is the perpendicular from t) on Q R produced, but this per- \
pendtcular is equal to B G 1 , because QB and CD are parallel.
Therefore, F L X DJ- F t X B G = 0, i
whence, F* = F* X
because the triangle B CD is isosceles, arid DJ*=BG.
*To find F a , take moments about Q, in the line of F t . The lever
arm of F 9 is Q S = k, that of F^ is Q Z = / sin H. Then,
Fi A - F l I sin H =
' h ~>
W
or, because FI - ^ g (see the solution by the triangle of forces
given above) ,
28
ANALYTIC STATICS
33
Passing now to joint Q, to find F t , take moments about S 1 in the line
of R The lever arm of F t is S/ = S sm H = ~ sin H, that of F,
is (PS = /; Then,
whence,
t I sm //
= 0,
sin H
t . , . \ F t I sin H . ,
but (see above) -j = F at therefore,
F = f = Wj,
To hnd R, take moments about C^m the line of F, The lever arm
of R is Q C = /, that of F t la C T = ~Q Z = / sm JV.
Then, Fl s\n H - I? I = 0, whence,
J? = FI sm /f = /?, sm // = 2~^-X sm // = -^
In practice, it is not necessary to draw the perpendiculars C U, B G,
C T, etc , as their values can be at once written down by means of the
fundamental trigonometric relations among the elements of a right
triangle
EXAMPLES FOR PRACTICE
1. A weight of 2 tons is suspended from a derrick, Fig. 23; the
length of the boom A B is 40 feet; the guy rope CB is fastened at a
point C, 30 feet from A, and the boom is 10 feet out of vertical. Find
FIG 28
the tension FI in the guy rope and the thrust F in the boom.
= .96840 tons
, 2 7542 tons
a
34
ANALYTIC STATICS
28
2 A trapezoidal frame having the dimensions shown in Pig 24
rests on two piers and carries two weights of 5 tons each at the
joints A and B Find the stresses F tt F,, and F 3 in the members A t
A C, and CD, respectively, and the reaction /? ut either pier The
members A B and CD are horizontal, and the inclinations of A C
and B D to CD are equal. [F^ 2 fi tons, thrust
Ans \ F ' = 5 - 5002 tons >
Ans
F a
2 fi tons, pull
5 tons
PARALLEL FORCES
COPLANAR FORCES
TWO FORCES
41. Resultant or Two Parallel Forces Having the
Same Direction. Let two parallel forces /*i and F, Fig. 26,
act on a rigid body A BCD. The lines of action of the two
forces are K^L^ and K*L*, respectively, and the points of
application Ei and , are any two points on those lines (see
Art 10). Draw E^E t . The effect of the forces will not be
changed if any two equal and opposite forces R^ T^ and fi t 7 1 ,,
acting 1 along 1 the line ./?, /?,, are introduced, for these two
forces will evidently balance each other. The two forces F l
and F t may, therefore, be replaced by the four forces f t and
E, 71, F, and /?. T.. The resultant of F, and t T, is -fi 1 , ,S lf
the diagonal of the parallelogram E t A^ 5, 7*,. Similarly,
the resultant of F and E* T, is ^, 5 1 ,. The lines of action
28
ANALYTIC STATICS
35
of ^ ,5*! and E, S, meet at a point E 1 . By the principle of the
transferability of the point of application (Art. 10), the
points of application of E^ S,. and E* S, may be transferred
to ', provided that the point Ef is supposed to be rigidly
connected to the body A CD. In this new position, the
forces are represented by E 1 SJ and E 1 S, f . The force E 1 57
may now be resolved again into its components E' 77, equal
and parallel to j\ 7\, and E 1 'A7, equal and parallel to F^\
PIG 26
and E' S, f into its components E 1 T,', equal and parallel to
E t T, t and & W, equal and parallel to /".. As E' TJ and
E 1 T, f balance each other, only two forces, acting along the
line K L, now remain, whose resultant R is equal to F v -f- F*.
The point of application of this force may be taken any-
where, as at E } on the line K L,
In the similar triangles E f E* and NJJS'SJ we have
EEi E'E
WSJ ~ &
E'E X W 5'
whence
36 ANALYTIC STATICS 28
In the same manner, we get, from the triangles E E' E*
E 1 N,' = E 1 E X N,' S,'
or, because NJ S a f = E 1 T. f = E' 77 = AV SJ t
EE,XE' N.' = E 1 E X A r / S/
This, compared with (), gives
- t X EN! = ^/T 3 X /;' AY,
that is,
", X Fi = ; X /=;, /^ X Vi\ = 1*\ X EE* (b)
, JC> AI j'a f \
whence, - = (c)
i\
From this we obtain, according to the laws of piopoition
(see Geometry),
_
EE, , '
that is,
r = , and F, X E,E* = R X E E, (d}
Likewise,
- 5 = ' and ^ X ^ ^ = /v> X 7 ' : ^
The preceding results may be stated as follows
1. 77w resultant of two paiallcl font's having the same
direction is equal to their turn > its line of attioti ts parallel to
the lines of action of the two foitt's, and its d nation is the same
as the common direct ton of the fwo fanes.
2. The line of action of the tfsidtant is so silimtt'tl that it
divides any line mteicepted between the components into two seg-
ments niveisely proportional to those components [equation (r)].
3. The resultant and the lomponcnts aie so t elated and
situated that, if the points of applitation of thf fhiee are taken
on any straight line (as JK^E^ Fig, 25) tutfi netting then lines
of action^ the produtts of any two of thf three forces by the dis-
tances of their respective points of application fnwi (he point of
application of the third force are equal [equations {) , (</) , (<)].
The last statement applies to distances taken on any line
between the lines of action of the two forces; fur it will be
remembered that E l and , were taken arbitrarily on K* t
28 ANALYTIC STATICS 37
and K a L*. Any other points, as <7, and G a , might have
been taken, and the same reasoning would have led to the
same conclusions
Let /?!/?* = /, REi = l^EE* /. Then, equations (6),
(rf), and (<?) may be written
K /, = /:/. i
F.I = X.
FJ = /?/J
42. Equilibrium of Three Parallel Forces. By
leversing the direction of A*, Fig 25, we have the equili-
brant Q of the forces F t and /'", Arithmetically, the value
of Q is F! -\- Fj. But, if we consider forces acting in one
direction ass positive and those acting in the opposite direc-
tion ab negative, we have
whence, Q + Ft + F a =
This, with any one of equations (/) of Art. 41, gives the
conditions of equilibuum of three parallel forces. The
eqmhbrant must always lie between the other two forces.
43. Theorem of Moments. Let O, Fig. 25, be any
point on the plane of the forces, and draw O P, perpendicular
to their lines of action. Let // M^ Kf r be the moments
of F lt F 9) and A' about O Then,
Mi = FiXOJ\ (a}
M* - F t X ( O P, + A /\) (b)
Afr^RxloPi+PtP)
Adding (a) and ((>) ,
= tf X
or, because /% X A A = JtxPiP (see Ait. 41 ) ,
W + Af.^/tXOPi + JtxPiP^XX (OP, + I\P] = ^/ r
Therefore, the moment of the ) exultant of /wo parallel fotces
about any point in their plane equals the algebiait. sum of the
moments of the components.
The student may take the point O' and verify this prin-
ciple, paying due attention to the signs of the moments,
according: to the convention explained in Art. 18.
38 ANALYTIC STATICS 28
If, instead of the resultant, the equilibrant Q is taken, its
moment M q is equal to M r , and
M, = - Mr = -Mi-M t
whence, M, + M + ^ =
44. Two Pai*allel Forces Acting in Opposite Direc-
tions. When the two component forces act in opposite
directions, as /'land /*" Fig. 26,
the resultant is found as fol-
lows. If a force Q = /*! F a
is applied at a point E in , E^
produced, such that /*",/= Q /
the three forces F lt 1?,, and Q
will be in equilibrium, accord-
ing to the principles stated in
Arts. 42 and 43. Since Q
balances F l and F tt , it must be
FlG ^ equal and opposite to their
resultant R. Therefore, in this case, the resultant is equal to
the difference of the components, and the line E t E l produced is
divided by the three forces F lt F*, and R m the manner indi-
cated by equations (/) of Art. 41. The law of moments applies
in this case also, but due attention must be paid to signs
45. The principle of moments just stated for the case of
two parallel forces and their resultant (or of three balanced
parallel forces) is known as the principle of the lever, or the
law of the lever, as it was first discovered by Archimedes
in the determination of the conditions of equilibrium of a lever.
46. Definition of a Conple. There is an apparent
exception to the foregoing conclusions that must be particu-
larly noticed. If F^ = F,, Fig. 26, the formulas would give
R = F l F, - 0, and, fiom equations (/) of Art. 41,
So, although the resultant is 0, the distance of its point of
application from the points of application of the other two
F I F I
forces cannot be determined, since the fractions ~- and ~~
28 ANALYTIC STATICS 89
do not represent any numbers. This simply means that, in
the case here considered, it is impossible to replace the two
given forces by a single force.
A system of two equal parallel forces acting in opposite
directions but not in the same line, is called a couple.
The theory of couples is veiy important, and will be more
fully treated further on. But here it may be stated, from
what has just been explained, that a couple cannot be either
replaced or balanced by a single force.
47. Resolution of a Force Into Two Parallel Com-
ponents. Equations (/) of Ait. 41 afford a means to
resolve any force into two parallel components passing
through any two given points situated on a line intersecting
the line of action of the given force. For, if ./?, / and /, are
given, we have:
whence, F, = l
* i T *a
Similarly, ft = '
*i T *
If R and one of the distances, as / lt from the line of action
of J? to the line of action of one of the components are given,
and also the magnitude Ft of
this component, the other
component is found from the
relation
Its distance /, from the line
of action of R is determined
as follows:
F.I. = FJ
whence, l* Fi
The distances A and /, may be either perpendicular or
oblique.
EXAMPLE 1. It is required to find the magnitude and line of action
of the resultant R of the parallel forces ^i(* 500 pounds) and
F t (= 260 pounds), the distance / being 4 feet, as shown in Fig. 27.
w +0*
40
ANALYTIC STATICS
828
SOLUTION According to statement 1 of Art 41, the rebitlUnt is
equal to the sura of Fj. and F a , that is, /t" = 500 + 250 = 750 Ib Ans
The distance / t from the hue of action of FI to the line of action of
>?, along the hue /, is given by equations (/) of Art 41 :
l t , fiom which, A = ~-
A
F,l =
From the above, F, = 250, / = -i, and A 1 = 7oO, therefore,
750
500 X 4
"750~
Ans
EXAMPLE 2 In Fig 28, the forces F^= 200 pounds) and /'(= r>0
pounds) act in opposite directions; It is required to find the resultant A'
when the distance I is 6 feet.
_rf_ / -.--<
r^ 'I '
~~~--l-
'/a-
FIG 28
R
Kio '."I
SOLUTION The resultant is equal to the difference between the
forces (Art. 44); that is, R A 1 , - A, = 200 Ib. - 50 Ib = 150 Ib.
Ans.
From equations (/) of Art 4 1 ,
Fii _ 200 X
/.
160
8 ft
COX 6 _ 0f
TBO" " w t<
Ans.
EXAMPLE 3 In Fig 29, resolve the force V\'( 450 pounds) into
two parallel forces FI and F a , when A = 10 feet and /, = 8 feet
SOLUTION. From equations (/) of Art 41,
AV. , ., A^/!
/f,-- 7 ',and^- -/
In this example, A = 450 Ib , / = 10 ft., /, 8 ft., and //, + /
= 10 + 8 - 18 ft. Therefore,
460 X 10
Ans,
260 Ib.
28
ANALYTIC STATECS
41
ANY NUMBER OF FOHCPS
48. Magnitude of the Remiltant. Let /*,, /^ 7s, /%,
Fig 30, be parallel coplanai foices acting through points
A i, A,, etc of a rigid body, their lines of action being,
respectively, K* Z,,, K t L lt etc. By the pimciple of the trans-
ferabihty of the point of application, each force may be sup-
i.
posed to be applied at any point on its line of action; but,
for reasons presently to be explained, it will be assumed that
the points of application are the fixed points /,, .,/,, etc.
In the first place, the i exultant R /.s equal to the algebra it
sum of the components ', and //A I me of action Is parallel to the
lines of action of the components
For, according to Arts. 41 and 44, the icsultant A" of
/'I and F t is parallel to I<\ and /" and equal to /^ + J<\\
the resultant R" of R' and /". is parallel to A" and /;, and
equal to A" + F* /", + / + /'" (the negative sign being
implied); the resultant A? of R" and F t is parallel to R n and
fl, and equal to.-^" + -ft = A + /?", + F, + /;.
42 ANALYTIC STATICS 28
The same reasoning applies to any number of forces; so
that, if their algebraic sum is denoted by 2' F, then R = 2 F.
49. Line of Action of the Resultant. To locate the
line of action K L of the resultant, let N be any point m the
plane of the forces, and NP<. a. perpendicular to the common
direction of the forces, intersecting JK' 1 L l at /*,, A' 3 L a at />
etc.; and KL, the still unknown line of action of the
resultant, at P Let N P* - p NP, = A, etc., and NP = p r
If moments are taken about N, the following equations
obtain (Art 43). '
Moment of JR' = F^ + F.p*
Moment of R" = moment of R' + F,p,
Moment of R - Rp = moment of R" + F t p<
= Ft pi + F t p t + F a A + F t A
In general, if the algebraic sum of the moments of any
number of coplanar parallel forces is denoted by ~ Fp, and
the lever arm of the resultant R by p r , then
Rp, = SFfr
whence, p r = ^ = ^ (1)
This locates the line of action of R with respect to the
point N.
If, instead of the perpendicular distances p t) Ai etc. of
the lines of action of the forces from A'', the oblique dis-
tances NTi. = a,, NT, = a, } etc., along any line NT* (as
when the forces are applied at several points of the same
straight line), are given, the following equations are
obtained:
A = NPi = -AT 71 cos H = a, cos H
p a = NP t = N T, cos H = a, cos H, etc.
A = Af/* = NT cos ff = a r cos .#"
Substituting these values of A, A. etc. in the equation of
moments,
Ra r cos H = /!! cos ff -\- F t a, cos #+ . . . etc.;
whence, * r = *<" + *+ = ( 2 )
28 ANALYTIC STATICS 43
If N is taken on the line of action of one of the forces,
that force is eliminated from formulas 1 and 2, since in this
case both its p and its a are equal to zero Thus, for two
forces, the piecedmg equations become identical with equa-
tions (/) of Art. 41.
It should be carefully borne in mind that, in using the
inclined distances a lt a t , etc., the same rules for signs are to
be observed as for the perpendicular distances A)A> etc -
Thus, F^a,. is to be taken as positive and F a t as negative.
50. Conditions of Equilibrium. When the forces are
in equilibrium, we must have 2 F = 0, and S(Fp) = 0; that
is, the algebraic sum of the forces must be zero, a?id the algebraic
sum of their moments about any point in their plane must be aero
Both of these conditions are necessary for equilibrium
For, in the first place, it is evident that if S F is not zero,
there is a resultant,
and no equilibrium is
possible. In the sec-
ond place, if the
resultant of all the h 8 '
forces acting in one-^1 ^ J*
direction is equal I JL
w
100U>
FIG 81
to the resultant of all
the forces acting in
the opposite direction, the condition -F = obtains, but,
if those two resultants have not the same line of action [in
which case 2(Fp) is not zero] , they form a couple (Art. 46),
and the system cannot be in equilibrium
It should be remembered that the arithmetical meaning of
the expression 2F = is that the arithmetical mm of all the
forces acting in one direction is equal to the arithmetical sum of
all the forces acting in the opposite direction.
EXAMPLE 1 A weight of 100 pounds is hung from the extremity A
of a straight lever A JB } Fig. 31. The lever is suspended by a rope passed
around a smooth peg and carrying a weight W\ and from the extremity
B is hung another weight W*. The dimensions being as shown, it is
required to find the magnitudes of W* and W, that the lever may remain
in equilibrium. (The weight of the lever is neglected.)
44 ANALYTIC STATICS 28
SOLUTION Since JTis the eqmlibrant of H 7 , and 100 lb , the sum
of the moments of the three forces about any point in their plane-must
be zero If moments are taken about 5 1 , which is in the line of action
of W (for W acts upwards at .S 1 ), the foice W will be eliminated, its
moment about 5 being zero The moment of /f, being positive, and
that of the 100-lb weight negative, we have,
6 Wi - 100 X 8 = 0,
Q/V)
whence, W, = = 133 S3 lb Ans
To find W, we have,
100 + Wi - W=Q
W = 100 + /Ft = 100 + 133 33 = 233 33 lb Ans.
Notice that it Is not necessary that the lever should be horizontal
From the mathematical conditions of equilibrium (Art. 5OJ, it follows
that the three weights given above will hold the lever in equlhbnuiu
in any position
EXAMPLE 2. A rigid bar A, Fig 1 32, rests on two supports Si and S lt
OOOlb.
^~~^ son.
G)
O
la
I V* I
*. I ^ Jt f _ - I - //7^ _ J
I ^" " V ~~'^|^ ' ' /{/ *^ \
and carries weights of 600, 300, 250, and 50 pounds placed as shown.
The weight of the bai .being neglected, it is required to find the
reactions J?i and A* a at .5\ and S y , respectively
SOLUTION Here we have a system of balanced parallel forces,
consisting of the given weights, which act downwards, and the two
reactions, which act upwards. Taking moments about >S" U , and
observing the rule of signs, we have,
- 600 X -10 + ^ X 28 - 300 X 23 - 250 X H + 50 X 10 =0;
whence, j?, = --.^ = l,ir>7.1 lb.
To find J! a , we have:
J?i + #, - 800 + 300 + 250 + SO = 1,200,
whence, J? t = ],200 - JK V = 42.9 lb
In order to check these results, the value of Jf may be found by
taking moments about 5^ The moments of the weights at the right
28
ANALYTIC STATICS
45
of Si are positive, the moment of /? and the moment of the weight
at the left of Si are negative. Therefore,
- 600 X 12 + 300 X 5 + ,250 X 20 - ./?, X 28 + 50 X 38 = 0,
whence,
1 200
= -W" = 42 -9 Ib i approximate to tenths.
EXAMPLES FOR PRACTICE
NOTB -lu these examples the weights of bars and ropes are neglected
1 A straight bar is supported at two points S 1 , and 5, 25 feet apart,
and carries five loads between the supports as follows 100 pounds,
placed 8 feet from 5,, 150
pounds, placed 8 feet Vj^
from Sn 200 pounds, |
placed in the middle of
the bar, 300 pounds,
placed 15 feet from J?,,
and 450 pounds, placed FlG M
20 feet from 5\ Find the reactions fa and fa. at Si and S a
Ans. fa = 500 Ib.; fa, = 700 Ib
2. A man carries a weight of 30 pounds hung from a stick resting on
his shoulder, the distance from the weight to the shoulder is 2 5 feet. Find
the pressure P on the shoulder- (a,) when the man holds the stick at a
distance of 1 foot from the shoulder, (b] when he holds the stick at
a distance of 1.5 feet from the shoulder. a / (a) P = 105 Ib
Ans -\(,5) />= 80 Ib
ii and A,, Fig 33, is to
will crush under a pres-
sure exceeding 275
pounds, but it is desired
to have the weight as far
from A t as possible
Find the maximum dis-
tance x at which the
weight can be placed
from A,
Ans x 13 75 ft
4 Five coplanar par-
allel forces act on a body
Aff, Fig 34; magnitudes
and direction of forces,
Find the magnitude and
direction of their resultant fa and the distance a of its line of action
from 0, measured along the hue X.
Ans. R - 350 Ib., acting upwards; a <* 4 875 ft to the right of O.
NOTB First take moments about 0, using the inclined distances instead of the
lever armi then, as a oheok, take moments about P.
I LT 398-8
3 A rigid bar, supported on two piers
carry a weight of 500 pounds. The pier A
FlO 34
and distances along OX, are as shown
46
ANALYTIC STATICS
28
E
5 In example 4, find two forces Ft and F a parallel to the given
forces and equivalent to them, F l to act at a distance of 10 feet to the
right of 0, and F, at a distance of 14 feet to the right of O
A (F t = 798 44 Ib , acting upwards
\F 3 = 448 44 Ib , acting downwards
6 A bar A 1 A 3 , Fig 35, is to carry two equal weights W, sus-
B! ^B 2 pended as shown The
I s bar is supported by two
^Aa ropes AI BI and A*
J I of which AI J3i cannot
r' -j- s' ] be subjected to a tension
greater than 1,000
pounds, and A, B* can-
not be subjected to a ten-
FIG 85 sion greater than 600
pounds. What is the greatest value that W can have, and what are
the tensions F* and F t in the two ropes when the bar carries its
greatest load? f W - 720 Ib
Ans IF* - 840 Ib
[F, = 000 Ib.
NOTE Assume first Fi 1,000, and find Wby taking momenta about As; tlion,
Fa - 2 W Fi As this is greater than 600, start by ausumint ft = 000 pounds, and
proceed to find W and ^"i
NON-COPLANAR PARALLEL FORCES
51. Magnitude of the Resultant of Any Number of
^, A B Parallel Forces.
*"""""" T ^f 1 /** /7* /P 77*
,.-'^- : ~^ -^ a ..._.^ 4 Fl ^- 36, be parallel
N^I--"'" forces acting on a
\ \ ^ 4 body at the points^,,
Ai, etc. The forces
may be either copla-
nar or not. In either
case, they may be
combined in pairs, as
in Art. 48. Hence,
the resultant is equal
to the algebraic sum
of the components.
52. Center of
*io Any System of Par-
allel Forces. According to the principles stated in Art. 41,
R
28 ANALYTIC STATICS 47
the resultant R 1 of F l and F,, Fig. 36, passes through a point A'
P
in the line A t A a , whose distance from A l is A^ A,X-~ f = A t A*
p
X - Since this distance does not depend on the
F l + F 3
inclination of the forces to A^A^ and since the fraction
p
, "--rr remains the same when for F, and F, are substituted
A + F n
any forces proportional to the latter, such as n /\ and n F*
(n being any number, integral or fractional), it follows that,
so long as the points A v and A* remain fixed and the forces
remain parallel and their relative magnitudes and directions
unchanged, the lines of action of the forces may be revolved
about A,, and A, and made parallel to any direction whatever
m space, without changing the position of the point A f ,
through which the resultant R 1 must constantly pass. The
same principle may be proved of the point A" through which
the resultant R" of R' and F a must pass, and of the point A,
traversed by the total resultant R. It follows that, for
every system of parallel forces applied at or, more properly ',
passing 1 through any points whatever^ there zs a fixed point
through which the resultant must pass, whatever the (common}
direction of the forces may be. Furthermore^ that point is the
same for all systems of parallel forces in which the relative
magnitudes of the forces are the same, whatever their absolute
magnitudes may be.
Such a point, as A in Fig. 36, is called the center of the
system of forces considered.
53. Center of Gravity Defined. In the particular
case in which the points of application A^ A,, etc. are the
particles of a tyody, and the forces acting are the weights of
those particlea, the .center of the system of parallel forces
thus formed Is called the center of gravity of the body.
The weight of tihe body, which is the resultant of the
weights of Us $rarttcls> must therefore be treated as a force
whose line of fldifbli 'passes through the center of gravity of
the body. TN |K|^tI^^s for determining the position of the
center of
48 ANALYTIC STATICS 28
54. Coordinates of Center of Parallel Forces.
When all the points of application of the forces lie in one
plane, the center of the forces is easily found as follows:
Let the parallel forces F lt F,, F a act at the points A,, A,, A,,
Fig. 37, and let <9,Yand Y be any two mutually perpen-
dicular lines drawn in the plane containing A 1} A a , A a These
reference lines are called coordinate axes, and the distances
of any point from them are called the coordinate's of that
point Distances perpendicular to O K, or parallel to OX,
are usually denoted by the letter x, sometimes with an accent
or subscript; distances perpendicular to OX, or parallel to
OY, are denoted by the letter y. Distances above OXa.nd
T
JK
/ ,fl| **<2*f J-f&
Pro 87
those to the right of O Y are treated as positive; distances
below X, and those to the left of O Y, as negative.
Let the coordinates of At be O HI K^ A l = x, and O A",
= Jf 1 A l = y lt Similarly, let the coordinates of A, be j. a
and ;/,; those of A a be x a and y a ; and those (still unknown)
of the center A of the forces be x c and y e .
Since the position of A is independent of the common direc-
tion of the forces, the latter may be imagined as acting
in the plane X Kin a direction parallel to OX. If, now,
moments are taken about any point on OX, the lever
arm of F l will evidently be equal to A^ //i = y lt the lever arm
of Ft will be equal to y t , that of A a equal to y,, and that of R
equal to y c . Then, by formula 1 of Art. 49,
\-F t y t -
28 ANALYTIC STATICS 49
If, on the contrary, the forces are imagined as acting in a
direction parallel to O Y t and moments are taken about any
point on O K, the following formula is obtained, by pursuing
the same method as before.
- &*+ F >x> + F 'Xi
ft + K + ft
The coordinates x e and y e locate the center A.
As the foregoing reasoning is evidently applicable to any
number of parallel forces whose points of application all he
in one plane, the following general formulas can be at once
written
(2)
55. Moment About a Line. The product F l y l con-
sidered in the last article is called the moment of the force F t
about the lino OX, or with respect to OX. In general,
when a force is considered as applied to a special point, the
moment of the force about any line perpendicular to the line
of action of (although not necessarily in the same plane with)
the force is the product of the magnitude of the force and the
perpendicular distance of the point of application of the force
from that line.
ANALYTIC STATICS
(PART 2)
CENTER OF GRAVITY
DEFINITIONS AND GENERAL PROPERTIES
1. The center of gravity of a. body has already been
defined as the center of the parallel forces of gravity acting
on the particles of the body, or the center of the weights of
all the particles, each particle being taken as the point of
application of its own weight (see Analytic Statzcs> Part 1).
These forces are considered parallel because they are all
directed toward the earth's center, whose distance from the
surface is so great, compared with the dimensions of ordinary
bodies, that the angle between the lines of action of the
weights of any two particles of a body is practically zero.
(Two terrestrial radii meeting the surface at two points
distant 100 feet from each other make an angle at the center
equal to about 1 second.)
The abbreviation c. g. will here be used to signify center
of gravity. The c. g. of a body is called also center of
mass, center of inertia, and centrold. These terms,
however, will not be used here.
2. Immediate Consequences of the Definition.
Three important consequences follow at once from the
preceding definition, namely:
1. As stated in Analytic Statics, Part 1, the weight of a body
may be treated as a single force acting vertically through the
o/ (fo body, , This is expressed by saying that the
TBXTHOOK COMPANY, ALL MIHT MMBMVKD
529
ANALYTIC STATICS
29
weight of a body may be supposed concentrated at the
c. g of the body
2 The position of the c. g. of a homogeneous body (that is, a
body whose substance is the same throughout, or whose particles
have all the same weight] depends on the form of the body only,
not on the material of which the body is made. For the centei
of a system of parallel forces depends on the relative, not on
the absolute, magnitudes of the forces, and on the relative
positions of their points of application.
3 // the c. g. of a body acted on by its own weight is sup-
ported, the body will remain in equilibrium, whatever its Posi-
tion. For the weight
of a body is equivalent
to a single force acting
through its c. g., and, if
this point is supported
that is, if a force equal
to the weight, but acting
in the opposite direc-
tion, is applied at this
point the two forces
will balance each other.
Conversely, // a body
acted on by its own weight and by other forces is in equilibrium,
the resultant of the other forces must be vertical, act upwards,
and pass through the c. g of the body. Thus, if the triangular
plate ABC, Fig 1, hangs from three strings A O, BO, CO,
and is in equilibrium, its weight W, acting through the center
of gravity G, must be balanced by the resultant of the
reactions of the strings. This resultant must, therefore, be
vertical, act upwards, and pass through G.
3. The Center of Gravity of a Body May be Outulde
the Body. When several forces act on a body, the line of
action of either their resultant or their equilibrant may be
altogether outside the body. For example, in Fig. 2, the
equilibrant Q of the forces F l and F t acting on the lever A^ A t
passes through a point A, outside the lever. This means
FIG
ANALYTIC STATICS
that the only single force that can balance Ft. and F t is the
force Q acting through A 9 ', but, as the point A,, is outside the
lever, that point must be imagined to be rigidly connected to
the lever, in order that Q may act on the latter.
In the same manner, it often happens that the c. g. of a
body is outside the body. Thus, the c. g. of the curved
Q
FIG. 2
FIG 8
rod A /?, Fig. 3, is a point G outside the rod If it is desired
to suspend the rod so that it will be in equilibrium in all
positions, the point G must be connected to the jod by some
means, as by another rod G H. The additional weight of the
rod GH alters the position of the c. g. of the whole system;
but the change may be made small, and its amount easily
calculated, as will be explained presently.
4. Center of Gravity or a Line. Properly speaking,
geometrical lines and surfaces have no c. g , since they have
no weight. By an extension of
the definition, however, every
line and surface is said to have
a c. g., the expression being
taken in the sense now to be
explained.
Consider a body A fi, Fig, 4,
of uny form, and a section, as
PQ, containing the centers of a
number of particles of the body.
The resultant of the weights of
these particles will pass through a fixed point 0, which is the
center of the system of parallel forces constituted by those
weights. Likewise, 0> is the center of the system of parallel
FIG 4
4 ANALYTIC STATICS 29
forces constituted by the weights of the particles in the sec-
tion P'ff. If a sufficient number of sections is taken to
include all the particles of the body, and the centers thus
found are joined, a line A O 'B will be obtained, which will
be intersected by the resultants of the various groups of
particles considered The center of the system formed by
these resultants is the c. g of the body, and is said to be
also the c g of the line ADO'S.
If the body is homogeneous and of uniform cross-section,
all the resultants referred to will be equal, and the number
of them acting on any portion of the line A O' B will be
proportional to the length of that portion. Now, in deter-
mining the center of any system of parallel forces, the
forces may be replaced by any quantities proportional to
them For instance, if the x coordinates of the points of
application of two forces F l and F* are # and jc tt the x
coordinate x t of their center will be given by the formula
(see Analytic Statics, Part 1)
r* i z? Xi -)r -=;Xi
F^XI-\- F, x, _ F 1
If F a and /i are proportional to two given lengths / B and /
F I
that is, if ' = -', we have
A l\
,/.
*i + 7 x,
l lQ it *
'
Therefore, .-+/* = L#* +.**
F> +F. /, + /.
So, in finding the center of a system of parallel forces
whose points of application lie in a line, and whose dis-
tribution along any part of the line is proportional to the
length of that part, that length may be used instead of the
corresponding forces.
5. Center of Gravity of a Plane Area. Take now a
homogeneous prism ACGE, Fig. 5. If all the particles
29
ANALYTIC STATICS
5
FIG. 5
enclosed by a cylindrical surface PQ are considered, the
resultant of their weights will be a vertical force whose line
of action will meet the upper surface of the prism at some
point O. Similarly, other points
may be found where the resultants
of the weights of vertical rows of
pai tides meet the surface ABCD,
thus obtaining 1 a system of parallel
forces whose points of application
may be taken on the plane AB CD,
and whose center is called the c. g.
of the surface A B CD. This point
is evidently the same as the point
where the vertical line through
the c g. of the prism pierces the
plane A B CD. As the forces act-
ing on any such space as P are proportional to the area of
this space, areas may be substituted for forces when dealing 1
with the c. g of a pl?ne surface.
6. General .Definition of the Center of Gravity of
Jjlnes and Surfaces. A line is said to be homogeneous,
or of uniform weight, when parallel forces are distributed
along the line in such a manner that the resultant of the
forces acting on any part of the line is proportional to the
length of that part. The center of the parallel forces thus
acting is the c. g. of the Hue; and their resultant is called
the weight of the line.
The same terms apply to a surface when the parallel forces
acting through it (that is, whose lines of action pierce it) are
such that the resultant of the forces acting through any
part of the surface is proportional to the area of that part.
The center of the parallel forces thus acting is the c. g, of
the surface; and their resultant is called the weight
of the surface.
7. Static Moment. Let Wb& the weight, or any num-
ber (length, area) proportional to the weight, of any figure
line, surface, or solid; and let x be the distance of the center
6 ANALYTIC STATICS 29
of gravity of the figure from any point or line. Then, the
product Wx is called the static moment of the figure with
respect to that point or line
According to the theory of parallel forces, the static
moment of any figure or system of figures about any point or
line is equal to the algebraic sum of the moments of the parts
of which the figure or system is composed If the distances
of the centers of gravity of several areas A^A^ etc from
a given point or line are denoted, respectively, by AH .a,., etc ,
the static moments of these areas about the point or line
are XiA lt x,A a , etc. If the distance of the c g of the system
formed by the aggregate of these areas from the given point
or line is denoted by x c) then,
x e (Ai + A,+ . . . ) = x^Ai + x*A> +
In general, denoting the sum of the areas by 2 A, and
the algebraic sum of their static moments by SxA, the
following equation may be written:
x t 2 A = IxA\
V y. A
whence, x c = =~^
A similar formula applies to lines and volumes. The
formula shows that
1. The algebraic sum of the static moments of the patts
of any figure or system of figures about the c. g. of the figure
or system of figures, or about a line containing that center, is
equal to zero. For, in this case, x e = 0, and, therefore,
SxA (= x t !A) = Conversely,
2. // the algebraic sum of the moments of the parts of a
figure or system of figures about a point is zero, that point is
the c. g. of the figure or system.
29 ANALYTIC STATICS
IMPORTANT CASES
SYMMETRICAL FIGURES
8. Definitions. A figure is symmetrical with respect
to a point O when every straight line passing through that
point meets the figure in pairs of points equidistant from
the point O, the two points of each pair being; on opposite
sides of the point O.
The circle, the ellipse, the sphere, the circular ring, are
each symmetrical with respect to its center. This center is
called the geometric centei', center of figure, or cen-
ter of symmetry of the figure in question. A parallel-
ogram is symmetrical with lespect to the point of intersection
of its diagonals.
9. A figure is symmetrical with respect to a straight
line, called an axis of symmetry, if every straight line
perpendicular to the first-mentioned line meets the figure in
pairs of points equidistant from said line or axis, the two
points of each pair being on opposite sides of the axis
Thus, a rectangle is symmetrical with respect to either of the
lines joining the middle points of two opposite sides. An
ellipse is symmetrical with respect to either of its principal
axes, and n, ciicle with respect to any diameter. The axis of
a right circular cylinder is an axis of symmetry, and so is the
axis of a right circular cone.
10. A solid is symmetrical with respect to a plane,
called a piano of symmetry, when every straight line
perpendicular to the plane meets the solid in pairs of points
equidistant from the plane.
11. Center of Gravity of Symmetrical Figures. It
is evident that, if a figuie is symmetrical with >espect to a point,
a line, or a plane, the c. g. of t/ie figure coincides with the point,
or lies in the line or plane, of symmetry, as the case may be.
Thus, the c. g. of a circle coincides with the center of the
circle; the c. g. of a parallelogram coincides with the inter-
section of the diagonals; the c, g. of an isosceles triangle
8 ANALYTIC STATICS 29
lies in the perpendicular from the vertex to the base (at what
distance will be seen further on), and the c g of a icgular
pyramid lies in a plane through the vertex peipendicular to
the plane of the base
12. Again, if a figure has two axes of symmetry, the c. jg- m
will be at the intersection of those axes, and
7? a solid has three planes of symmetry meeting at a
that point is the c. g. of the solid.
DETERMINATION OF THE CENTER OF GRAVITY BY
ADDITION AND SUBTRACTION
13. Addition Method. If a figure (by which is meant
either a line, a surface, or a solid) or system of figures can
be divided into parts whose centers of gravity are known,
the c. g. of the whole figure or system is easily found by the
principles explained in Analytic Statics, Part 1.
For example, let it be required to find the c. g of a system
consisting of two homogeneous spheres C and C,, Fig. 6,
whose weights are Z-J^,
and whose distance
FlQ fl apart, measured be-
tween centers, is a. This is equivalent to finding 1 the center
of two parallel forces W t and W t acting through d and C,.
Let this center be G; then,
whence, G C, = - ^ a
W, + H7.
14. Subtraction Method. If, on the contrary, the
c. g. of a system is known, and also the c. g. of all its parts
but one, the c. g. of the remaining part is found by the same
general principles just referred to, the only difference being
that subtraction is used instead of addition.
As an illustration, let it be required to find the c. g-. of a
figure obtained by cutting from a rectangle B CDR, Fig. 7,
a rectangular corner EFHI and a circular piece 0, the
29
ANALYTIC STATICS
9
dimensions being as shown Take EDior the axis of x and
EB for the axis of y\ and let x,' and y' be the coordinates
EH' and H 1 G' of the c. g of the rectangle B CDE\ x" and
y, the coordinates EH" and //" O of the c. g. of the circle,
x'" and y'", the coordinates //'" and H'" G 1 " of the c g.
of the rectangle EFH '/; and ^ and j/, the coordinates of G,
the required c. g. According to Art. 5, areas may, when
dealing with surfaces, be substituted for forces in the equa-
tions of moments and of centers of parallel forces. In this
a j{j
sense, the moment of CD with respect to ED is the
same as if the whole area were concentrated at &; that is,
the moment of B CD E with respect to ED is a by'.
The area A of the figure whose center of gravity is required
is a b cii bi TT r*. The c. g. of BCDE is at the inter-
section of the two diagonals, or at the middle point of CE,
and, therefore, xf = EH' = ? andj/' = H' G 1 = .
2 2
Likewise,
x"
EH" = c, y"
and
H" O = b - r,
~ ~~2
The formula of Art. 7 now gives
+ aibt x' 1 '
A-\- ctibi
10
whence,
x e =
ANALYTIC STATICS 29
(A + a, bi + n r')^ - a, b, x"' - x r> x"
a b ! 61 TT r*
Similarly,
whence,
_
y
_ Ay e
ff, b.y" 1
,
ab
,
= a by 1 - a, b,y'" - TT r'y" = 1 \ab* - a, bS -_2_7r r*(b--_r) "I
A 2L ab-atbi-'ni* \
15. The subtraction method, just illustrated, is very
much simplified when the
c. g. of a figure consisting
of two parts and the c. g.
of one of the parts are
known, and it is required
to find the c g. of the other
part. This special case is
illustrated in Fig. 8, where
C, the c. g. of the figure
BCDJS, and the c. g.
of the part C E D, are
known, and it is required
to find G,, the c. g. of the
. .X
PIG 8
part BCE.
Let total area = A\ and area CDE = A,. Then (Art. 13) ,
Ai X G l G, A x G G:
whence,
GG,=
A
EXAMPLE 1 -To find the c g. of the channel section represented
in Fig 9 L
SorurioN.-Owmg to the symmetry of the figure with respect to
the center line OX, its center of gravity G t must be on that line. To
find the distance OG e . *-, of G e from the back of the channel, notice
la " er i8 the dlfference be tween the rectangles
Q D>, whose centers of gravity G and G' are at the
29
ANALYTIC STATICS
11
distances O G = 5 inches, and O G>
- + 5
J 8
i + 5 = ?j
2 8 ID
inches from A D Denoting the area of the section by A, and talcing
moments about A /?,
A x t = (AJtXAD)XOG- (A> JB> X A' D 1 ) X O G>
= 150 X 8 X 16 -36 X 11 X 46
X16
"
I G
I
1 *-
1_ ! *
S \JLZ
i<if. i)
The area of the section is
A = A B X A D - A 1 B> X A 1 W = (M) - ^ X 11
Hence,
.v c t= OG e ** ]
1,870
X 8 X W - x n x45
Km Id
B x_eo_ r _3B_x 11
8
XJJO-36X 11
flT"
in
EXAMPLE 2. To find the c. g. of the angle section represented in
Fig. 10, and having dimensions as shown.
ILT3W-9
12
ANALYTIC STATICS
29
SOLUTION Produce CD to meet HF at / The section is now
divided into two rectangles, BCIH and DEFI The c g of a
rectangle being at the point of intersection of the diagonals, itp
distance from either of two parallel sides is equal to one-half of either
of the other two parallel sides Taking moments about H B, and
denoting the distances of the c g of the section from H B and ffF
by x e and y t , respectively, and the aiea of the section by A,
whence,
Xc = HP =
9 11X13
8" 1 " 16
r 16 '
9 , 11 XJ8
T-CTT - 1.3879"
Moments about H ' F\
whence,
= 2 8879 In Aus.
EXAMPLES FOB PRACTICE
1 Find the center of gravity G 1 of the area obtained by taking the
rectangles ^and D H, Fig 11, from the rectangle A BCD The
FIG 11
dimensions are as shown
.in., nearly
In., nearly
ANALYTIC STATICS
13
2 Find the c g of the area obtained by cutting off the circle A BI C lt
Fig 12, whose radius O l A is 3 inches, from the circle ABC, whose
radius OA is 10 Inches Ans O G = 69231 in = H in., nearly
3 Find the c g of the angle section
represented in Fig 13
= 92500 m. = ff m , nearly
= 2 1750 m = 2Vfc In., nearly
r
L
FIG 13
4 Find the c g of the Z section represented in Fig 14
NOTE In talcing moments about QM, remember tlmt the moments of areas on
opposite sides have opposite sljrns
Ario fOP = .41667 in =
ADS \ PG = l 4167 in _
in , nearly
i( nearly
CENTER OF GRAVITY OF POLYGONS
NOTE In some of the articles that follow, formulas and rules are
given without explaining how they nre obtained This is done when-
ever the processes Involved require the use of advanced mathematics,
or, being elementary, are too long and complicated to be given in
connection with this instruction
16. A Fundamental Principle. // a straight line
divides a plane figute in such a manner that eveiy hue paral-
lel to a hxed direction
meets the perimeter of
the figure at one point
on each side of the first-
mentioned line and is
bisected by said hne>
the c. g. of the figure
lies on that line.
Let OX, Fig. 15,
be a line dividing the
Pro. 10
figure OQXQi into two parts, in such a manner that all lines,
14
ANALYTIC STATICS
29
as QQRR etc., parallel to the fixed line /P"are bisected
at their point of intersection with OX. Then, according to
the principle just stated, the c. g. of the area OQXQ^ lies m
the line OX.
17. Triangle. Let ABC, Fig. 16, be any triangle.
DtawAA' from A to the middle point A' of the opposite
side. Any line, as RS, parallel to B C is bisected at its
intersection J with A A'. Therefore, according to the
proposition of the preceding article, the c g. of the triangle
lies on A A'. For a similar reason, the c. g. must lie
on B B 1 , joining the vertex B and the middle point /}' of
the opposite side A C Therefore, the c g of the triangle
is at the intersection
A of the lines A A 1 and
B B't or of either of
them with the line
CO from C to the
middle of An. It is
shown in geometry
that the distance of
the point G, where
A A 1 , /?/?', and CO
meet, from any of the
vertexes is equal to
two-thirds the length
of the line joining that vertex to the middle point of the
opposite side, or A G = \AA' t BG = Inn 1 , CG - \ CC f .
The lines A A 1 , BB>, CO are called the modlan linen.'
Therefore,
The c. g. of a triangle lies at the intersection of the median
lines, and its distance from any vertex is egual to two-
thirds the length of the median line from that vertex to the
opposite side.
18. The perpendicular distance from the e.g. of a triangle
to any of the sides is equal to one-third the altitude of the tri-
angle, when that side is taken as the base.
29
ANALYTIC STATICS
15
Foi, drawing 1 A ff and < /if perpendicular to B C, Fig. 16,
two similar triangles AA'H and GA'K are formed, which
A A' GA 1
whence, bearing in mind that GA 1 = \AA' or
A A'
GA'
A A 1
19. Sometimes, the distances of the vertexes of a tri-
angle from an axis or line of reference are given, and it is
required to find the distance of the c. g. from the same
line. In Fig. 17, let .1,,
the distances of the
vertexes of the tri-
angle A^AtA* from
the line A' A 7 be y lt
y,, y a . The c. g. of
the triangle is on the
median A* M, and, as
already explained, x ^
GM = \AiM. Let
the distance G P of
the c. g. from X' X be denoted by y e . Draw MN perpen-
dicular and MB parallel to A 7 A'. The figure gives
y e = GP= DP + GD = MN + GD (1)
In the trapezoid A,P t P a A tt the line M N joins the middle
point of A a A* and P t P tt therefore,
In the similar triangles A^M B and GM D,
1:1? YB' that is - -/J- fi !
whence,
GD = ^AijB = ^(AiPi MN} = ijj/, i(y, + j/ a )] (3)
Substituting in equation (1) the values of MN and GD
given by equations (2) and (3), and reducing,
This formula is perfectly general, provided that due atten-
tion is paid to the signs of the coordinates. That is, distances
16
ANALYTIC STATICS
29
measured on one side of the reference line should be consid-
ered positive, and those measured on the opposite side, neg-
ative. Thus, if the vertex A, lies below A'' A", and y t and j/,
are considered positive, y, should be considered negative.
20. Trapezoid. Let B CD , Fig 18, be any trapezoid,
having the bases BE = ,, CD = b,; altitude DPI = /i,
median line Mi M, = m The median line /*/, Af, (that is, a
line through the middle points of the bases) bisects every
line parallel to the bases; therefore, G, the c. gf of the trape-
zoid, must lie on that line (Art 16). It is necessary, there-
fore, only to find the distance, as GP^ of G from either base.
Fin 18
Draw B D; this line divides the trapezoid into two triangles
having the same altitude //, and d l and b, for their bases.
The distance of the c. g. oiBCD from CD is \ h (Art. 18),
and its distance from BE is h h = % h. The distance
of the c. g. of JS D E from B E is h. Denoting the area of
the trapezoid by A, and taking moments about /?/j,
A X GP, = area B CD X I h + area BDE X \ h
whence,
= \ b, h X
6
X i h => (b, + 2 b,)
6'
m
A similar value may be found for /> by simply Inter-
changing fa and b* in this formula.
29 ANALYTIC STATICS 17
Draw M t K* perpendicular to the bases. The similar
triangles M, Mi A" t and GM^P* give
MnM* = MjJCi m = _ h
GM, ~GP\~* GM,
whence, GM, = ^-~- (2)
3 \ 0i + b* /
which gives the distance from the middle point of z to
the c. g., measured along the median line. A similar
expression may be found for GM^ by simply interchanging
&i and b t .
The distance B P l is found by the following formula, in
which N is the angle between CB and CJ, the latter line
being perpendicular to the bases:
+*, + (bl + 2 ^ a) k ^ N ~ ***} (3)
*i + o, J
21. There is another formula that is often useful for the
determination of the c g. of a trapezoid. Let the non-
parallel sides EB and D C, Fig. 19, meet at V. As before,
Mi M, is the line through the
middle points of the bases.
From geometric princi-
ples, it is known that this
line passes through V. Let
d be the c. g. of the triangle
DVE, G, the c g. otCVB,
and G that of the trapezoid
BCDE. Also, let V M,
VM, = a t) VG a. As explained in Art. 17,
= $ a,, and V G t = s . Taking moments about , we
have,
area DVEXG^G = area CVBX G, G\
whence,
t g area C E^g VMf
VMS
since the areas of two similar triangles are proportional to
the squares of their homologous sides. In terms of the
.it U ^
18 ANALYTIC STATICS 29
quantities a lt a,, and a, the preceding proportion may be
written:
a 3 dj. _ af
a "f a, 1"
Solving this equation for a, the following result is finally
obtained:
_ 2 a," - /
a 3 ,' - a,'
22. Trapezoid: Graphic Solution. Produce CD to E'>
Fig. 38, making D E' = d lt draw -fi 7 (7, and produce it to its
intersection d with EB produced. The similar triangles
GM,E' and GMi d give-
fad M.Ef.
GM,'
whence, M 1 d = ~XfaE' (1)
GM,
The value of C^/, is given by formula 2 of Art. 20, and
the value of GM t is given by the same formula, by inter-
changing bt and b,. Therefore,
m J. + 2J.
x a , ,
3 ^ + ^
By construction, fa E' = M t D + D E 1 - i b t + <5 X . Sub-
stituting these values in (1),
fa d = I- 6 / + * X (i*. + ^) = U + ^ a = faj5 + *
t ^a + 0i
therefoie, B d = b,
Hence, the following construction for finding the c. g. of
any trapezoid.
Join the middle points of the bases. Produce the bases in
opposite directions, making the prolonged part of each base equal
to the other base Draw a line joining the extremities of the
prolonged segments The point where thts line intersects the
median line is the required c. g.
23. Any Quadrilateral. Let B CDE, Fig, 20, be any
Quadrilateral. Draw the diagonals BD and EC, and find the
29
ANALYTIC STATICS
19
middle point M of one of them, in this case EC. Take
on D B the distance B N D /, and draw MN. The c. g.
of the quadrilateral is a point G
on MN, obtained by taking on
M 'N a distance AfG = i M ' N.
EXAMPLK 1 To find the c g. of the
channel section represented in Fig 21,
the dimensions being as shown
NOTE The section being symmetrical, the
dimensions of the lower part are the same an
the corresponding: dimensions of the upper
part, and need not be given
SOLUTION The c, g lies on the line
of symmetry O Q drawn through the
middle point of, and perpendicular to,
SJ The only thing to be determined
is the distance O G = x e
FIG 20
In complicated cases like this, it is often convenient to hnd a general
formula first, and then substitute the numerical values. In order to
do this, the dimensions will be represented by the letters written beside
the figures on the diagram Since the channel section to the differ-
ence between the rectangle B CIJ and the trapezoid DE FH> then,
denoting the area of the rectangle by 1?, the area of the trapezoid
by T, that of the section by A, and their centers of gravity by G>, Gt,
and G, respectively,
x t - G, - G G f - \b - G G r (1)
Taking moments about a line YZ drawn parallel to BJ through
the center G r of the rectangle B CIJ, we have, since the moment of
R about G r is 0,
G,_G t
~A
GG,
(2)
20
ANALYTIC STATICS
29
Now,
T = \ (EF+DH) X PQ = i[(A - 2 j - 2/) + (A - 20] (* - **
= (A - $-20 (b -*,)
and (formula 1 of Art 2O) ,
>
X
- 2 j - 20
,
*
iT
b- t, 3A-4s-6t
3 X 2(/;-J-20
^1
S
l\ *'
/.__!p__\/__^
i-v U>- * T
( 1125' I
U*f J
*
&S> ZZlf-^'lZS'i
JS J
T G
Ljff.
A
U
w
y tf ^
FIG 22
Substituting the values of T'tind <7 r G 1 / in (2),
- j - 2 - (* - O (8 A - 4 j - 0]
29
ANALYTIC STATICS
21
Substituting the values of O G, and G G r m (1),
r = J 6 _ <* -M[M*-J- 2
The dimensions, expiessed in sixteenths of an inch, are 6 = 54;
// = 20S, / = (, s = 8, f t = 7 Therefore,
A = b !i - (t> - t,) (// - s - 2 1) = 54 X 208 - (54 - 7) (208 - 8 - 12) ;
and
_ 54 _ ^T)4 - 7) [7(208 - 8 - 12) + $(54 - 7|]
Xf ~ 2 2[fi4 X 208 - (54 - 7) (208 - 8 - 12)]
= 27 -
47 (7 X 188 +
X47\
3 /
27 - 14 187 = 12 HHS sixte. nths
2 X 2,396
= 804 in , or, approximately, x e = f in. Ans.
EXAMPLE 2 To find the c. g of the plane figure represented in
Fig 22, the dimensions being as shown
SOLUTION The figure may be divided into the two rectangle^
B CL M and EFIJ, the trapezoid C 'D K L, and the isosceles tri-
angle FH1 Momenta will be taken about BX and B K The
altitude of the trapezoid is DN ' - CJVtan 60 = (CL - DJf)ta.a 60
= 3 4641 ft The operations are given m the following table, which
needs no explanation.
Area
Lever Arm With Respect to
Moment About
Figure
Square
Feet
BX BY
BX BY
BCLM
ao oooo
i oooo 5.0000
200000 100 OOOO
f
DN CL+aDK
CDhL
\
2 3 CL-\-DK 50000
99.4353 138.5640
= 3 5877
El'IJ
10 1250 <
4 s ' f a 25
-+DJV+CJi J + ED+CX
= 7 7141 L -=5fi350
78 1053 56 PS3I
FIff
i 2656 I
3 5 6350
130853 7iigo
I
= 10 3391
59 1034
210 6l57 303 636l
Having the area and the resultant moments,
,r c = BP
302.6361 _ 10rtl . _ 4 .
,or 5.1205 ft. Aus.
PG
59 1034
210.6167
59.1034
3.5635ft. Ans.
22
ANALYTIC STATICS
29
EXAMPLE FOR PRACTICE
Find the area and the c. g of the Irregular T section represented in
Fig. 23
3 3242 sq in
Ans
\A :
\OP:
\PG--
2 0695 in
7826 in
CENTER OF GRAVITY
OF AREAS BOUNDED
BY CIRCULAR ARCS
24. Circular Sec-
tor. Let O PS V,
Fig 24, be a circular sec-
tor, having the radius
O P = r, and the central
angle P O V = L. As
the sector is symmetrical with respect to the line O S> bisect-
ing the angle L t the c. g. lies on that line, and it is only
necessary to determine the distance OG of the c. g. from
the center of the sector This #
distance is given by the for-
mula p < ^
240 r sin
G
= OG =
In the denominator of this
formula, the angle L is ex-
pressed in degrees and deci-
mals of a degree.
25. For the sectors whose
arcs are, repectively, a semi-
circumference, a quadrant, and a sextant,
formula takes the following special forms:
4 r
Semicircle, y e = - = .4244 r
preceding
(I;
Quadrant, y e = 8 ^ in45 -! = .8002 r (2)
Sextant, y c
^ Bin80 --.6866r (3)
7T
29
ANALYTIC STATICS
23
26. Circular Segment. Let BCD, Fig. 26, be a
circular segment, having c
the radius OB = r, and
the central angle BOD
= L, expressed m de- s \
grees and decimals of a
degree. Here, G, the c. g.
of the segment, lies on
the line of symmetry O C.
The distance y e = <O G is
found by the following
formula:
_ Q G = 240 r sin 3
rcL 180 si
If the length of the choid B D is denoted by c and the area
of the segment B CD by A, the preceding formula can be
reduced to
(2)
y. -
12 A
27. CircularTrapezoid, or Flat Ring. In Fig. 26, let
B CD EH I be a circular trapezoid, or flat ring, having the
radii rt. and r tt and the
central angle L, expressed
in degrees and decimals
of a degree. The c. g.
lies on the line of sym-
metry O C, and its dis-
tance y e from the center O
of the ring is given by the
formula:
= 240 sin Z, r t ' -*,'
Fie,. -jo nL r**-r,*
EXAMPLE To find the c. g. of the area represented In Fig 27
SOLUTION Take the center line O y, bisecting the two circular
arcs, for the axis of y, and the perpendicular OX, through the center
of the arcs, for the axis of A' As usual, x e and y e are the coordinates
of G, the required c g , and A is the area of the figure The diagram
gives A = rectangle B FH T+ segment CD E segment KJ I. The
24
ANALYTIC STATICS
82ft
moment of A with respect to OX\& equal to the algebraic sum of the
moments of these three areas To find the area of the rectangle,
KI must first be found, since TK and IH are given To find the
IS?
JF
S
Pis 27
areas of the segments and their lever arms, the angles Z, and Z, must
be determined The figure gives, expressing angles to the nearest
minute,
K I = 1KQ = 2^lr a a O Q* = 2VJ^ fl ~(r^
= 2Vr 1 " - (r, -~4p = W 967 ft
sin i Z, = , whence Z, = 112 3(
sin
,
-^-, whence Z t =
23'
Having all the required elements, the coordinates .r, nnd y r ure
found by the usual methods, with which the student is now supposed
to be perfectly familiar. In finding x ct however, the operations will
be much shortened by observing that, if a distance K ' H' = / H is
taken, and the line H' F< drawn perpendicular to TK, the figure at
the right of F' H> is symmetrical with respect to O y, and its moment
about Fis consequently zero Therefore, the moment of the whole
area about Xis simply the moment of the rectangle B F 1 H 1 T, and
hence, ' '
A
The results, which should be verified by the student, are-
A = 118 35 sq. ft. Ans.
x t = OP= 18741ft. Ans
y c - PG = 7 7607 ft. Ans.
29
ANALYTIC STATICS
25
EXAMPLES FOR PRACTICE
1. Find the c g of a circular sector whose radius is 10 feet and
whose central angle is 45 y
Ans y c = 6 497 ft j
2 Find the c g of a segment whose
chord is 8 inches and whose radius is B
5 inches Ans. y e = 3 815 m.
3 Find the c g of a circular trapezoid
whose radii aie (i inches and 3 inches, and
whose shorter chord (as E /, Fig, 26) is
3 inches. Ans. y e = 4 466 m.
4. Find the c g of the half circular seg-
ment BCD (Fig 28).
x t = 2.839 m
y e = 13.798 in.
PIQ 28
CENTER OF GRAVITY OP A PLANE AREA BOUNDED BY
AN IRREGULAR CURVE
28. Approximate Analytic Method. To determine
the c. g. of a figure, as /? CD E, Fig. 29,, having- an irregular
contour, proceed as follows:
Draw two lines of reference OX and O Y perpendicular to
each other in any con- Y
venient positions;
preferably, one of the
lines, as OX, should
as nearly as pos-
sible bisect the area.
Divide OX into a
sufficient number of
parts, so that, by
erecting perpendic-
ulars at the points of.
division, the par 01 $he perimeter of the curve intercepted by
two conseGUt$TO;c$rtoates, as HiH^ and /;/ may be treated
Pro. 29
as a straight
a trapezoid.,
oorceflgondlng strip .fifr.ff'././i, as
.#/,//, etc., and the
I i t
26 ANALYTIC STATICS
ordinates ff^H^/if^ etc. Find the distances of the centers
of gravity of the strips, considered as trapezoids, from the
lines OX and. Y. Treat the area of the whole figure BCDE
as equal to the sum of the areas of the trapezoids, and apply
the method of moments, as usual That is, if the sum of the
areas of the trapezoids is denoted by 2 T, and the sum of
their moments about OX and O Y by - Ty_ and 2 Tx t
respectively, the coordinates x t and y e of the c. g. of the
whole figure are given by the formulas
STx
_ _
*'~
Tf
t v ~,
Although this method is only an approximation, it is
sufficiently close for almost all practical purposes. The c. g.
of each trapezoid may usually be taken at the middle point
of its median line
29. Experimental Methods. A very convenient
method of finding the c. g. of an irregular figure (or of any
other figure) consists in drawing the figure to scale on a
piece of cardboard of uniform thickness, then cutting off the
remaining part of the cardboard and balancing the part thus
left on a knife edge in. two positions.
In Fig. 30, let B B' CO be the piece cut out of the card-
board, having the form of the figure
whose c. g. is lequired; let B C and
B 1 C 1 be two positions of the knife
edge, for each of which the piece of
cardboard remains in equilibrium.
Then, their intersection G is evi-
FlG M dently the required c. g.
This method may be used not only for irregular figures,
but also for those figures the determination of whose center
of gravity leads to complicated formulas. The method,
however, is not very accurate, and, as a check, the piece of
cardboard should be balanced in more than two positions
say in six or eight. This will give an idea of the degree of
accuracy attained.
29 ANALYTIC STATICS 27
30. The method just described is also very convenient
for determining the distance of the c g. of a symmetrical
body from a plane perpendicular to a plane or axis of sym-
metry. A locomotive, for instance, is symmetrical with
respect to a. plane midway between and parallel to, the
axis of the cylinders. The distance of the c. g from either
end of the locomotive (that is, from a plane through either
end perpendicular to the axes of the cylinders) is found by
balancing the locomotive on a horizontal rod perpendicular
to the plane of symmetry, passed through ungs attached to
the locomotive and symmetrically located with respect to
that plane. The position of the rod and rings is changed
until the locomotive is found to remain balanced (that is,
with its two ends at the same level) when suspended. The
distance of the rod, when the locomotive is thus balanced,
from either end of the locomotive, is, approximately, the
distance required.
31. Still another method for finding the e.g. of a plane
figure forming: the faces of a thin plate of uniform thickness
consists in suspending the plate by a string and marking on
either face a vertical line that is the prolongation of the
direction of the string; then suspending the plate in a differ-
ent position (that is, tying the string to a different point In
the plate) and marking a line similar to the one marked
before. The intersection of the two lines is the required c. g.
This method can be conveniently used for determining
the c. g. of any body when that center is outside the body
(as an angle section, a bent rod, etc.). Here, when the body
is suspended in one position by a string, a line containing
the c. g. is obtained by making two marks on the body: one
at the point where the string is fastened and another directly
under it, By tyio$ the string to another point of the body,
another lin is( determined, whose intersection with the first
gives the requfrf&jS
28
ANALYTIC STATICS
29
CENTER OF GRAVITY OF SOLIDS
32. Right Cylinder or Prism. The c g. of a homo-
geneous right cylinder or prism evidently coincides with the
middle point of the line joining the centers of the bases.
33. Right Cone or Regular Pyramid. The c. g. of
any right cone or regular pyramid lies
on the perpendicular from the vertex to
the base (line joining vertex with cen-
ter of base) at a distance from the ver-
tex equal to three-quarters of the length
of that perpendicular.
34. Conical Frustum. The c. g.
of a conical frustum, Fig. 31, lies on
the line joining the centers of its bases.
Its distance from the lower base is given
by the formula
PIG 81
v - o G-X r*
y e LA LJ- 7 PV. 7- v
4 (r, + rj
in which r t = radius of lower base;
r, = radius of upper base;
A = altitude of frustum.
29 ANALYTIC STATICS 29
COPLANAR NON-CONCURRENT FORCES
COUPLES
DEFINITIONS-EFFECT OF A COUPLE
35. A statical couple, or simply a couple, has already
been defined as a system of two equal non-collmear parallel
forces having opposite direc- /
tions.
In Fig. 32, the forces F,. and n
F, constitute a couple acting on
the body BCD. Although F,
and F, aie equal in magnitude,
they are denoted by different
letters for convenience in refer-
ring to their lines of action.
36. The lever arm, or
simply the arm, of a couple is FlQ 32
the perpendicular distance (A/ 5 . = p) between the lines of
action of the two forces constituting the couple.
37. The plane of a couple is the plane determined by
the lines of action of the two forces constituting the couple.
38. The axis of a couple is any line perpendicular to
the plane of the couple. Such will be the meaning given to
the term here, although some writers use it in a different
sense. It follows from this definition that all couples whose
planes are either coincident or parallel have the same axis.
39. Coaxial couples are couples having the same axis.
Couples not having the same axis are called non-coaxial.
40. Resultant Moment of the Forces of a Couple.
The resultant moment of the two forces of a couple about any
30 ANALYTIC STATICS 29
point in their plane is constant and equal to the moment of either
fotce about any point on the line of action of the other fotcc.
This is easily shown Paying due attention to bigns, the
resultant moment of the two foices A and 1*\ about any
point 0, Fig 32, is
-F a Xff*0 + F l Xtf l O = -F*(ff t O- //, 0} =
- F. X Hi H* = - F*p = - I\p
For a point <7 between the lines of action of the forces,
the resultant moment is
- F, X ff. 0> - F, X Hi O 1 = -F, (// (V + //, 0} =
-F t p = -Ftp
41. The constant resultant moment of the foices of a
couple about any point m their plane is called the moinout
of the couple, and is numerically equal to the pioduct of
either force by the arm of the couple.
42. Notation. A couple is expressed cither by its
moment (for all coaxial couples having the same moment
are equivalent, as will be shown piesently), or by writing
its two forces in parenthesis with the arm between; thus,
(F^fiiFt). The latter is a more convenient form of expres-
sion for some purposes, and will be often used here.
43. Effect of a Couple. The effect of a couple acting on
a rigid body not acted on by other forces is to turn the body about
an axis passing through its c g.
A full demonstration of this proposition cannot be given
here, as it would be necessary to make use of some kinetic
principles that have not yet
been explained. Moreover, it
is to be observed that, for the
A\ Q. \ IB purposes of statics, it is not
i necessary to know what the
* jf* ~n
V.F, effect of a couple is, for all
the theorems relating to the
FlG 83 equilibrium and equivalence of
couples can be stated and proved without any reference to
what the effect would be if the couples were unbalanced,
The foregoing principle, however, has been stated here, in
29 ANALYTIC STATICS 31
order to caution the student against a common error preva-
lent among beginners. If the couple (F lt p t F a ), Fig 33,
acts on the body A B, its tendency is to turn the body about
an axis passing through its center of gravity G, not about an
axis or point situated between the two forces.
44. Direction tind Sign of a Cotiple. The sign of a
couple will here be treated as positive or negative according
as the moment of either force about a point in the line of
action of the other is positive or negative (see Analytic
Statics, Part 1). In both Fig. 32 and Fig. 33, the moment
of Fi about any point in the line of action of F, t or of F,
about any point in the line of action of F lt is negative, and
the moment of the couple is considered negative.
The motion that the couple (F 1} ^,F a ), Fig 32, tends to
produce is evidently the reverse in direction of the motion
of the hands of a watch with its face placed upwards and its
center at H,. or H t . This motion is said to be counter-clock-
wise, or left-handed Motion similar to that of the hands of
a watch is said to be clockwise, or right-handed. Clockwise
motion is said to have a right-handed direction; counter-
clockwise motion, a left-handed direction.
It will be noticed that the sign of a couple is positive
when the couple tends to produce clockwise motion; other-
wise, however, the couple is negative. This distinction is
of value only when couples are to be combined by algebraic
addition.
The direction and sign of a couple can be very readily
determined by imagining the arm of the couple to be a line
rotating about its center so that each extremity follows the
direction of the force applied at it. Thus, in Fig. 32, if we
conceive Hi ff* to begin to rotate about its center so that H
will follow the direction of F l} and H a the direction of F a , it
will be seen that the rotation will be counter-clockwise.
The couple is, therefore, a left-handed couple, and its sign
is negative.
32 ANALYTIC STATICS 29
EQUIVALENCE AND EQUILIBRIUM OF COAXIAL COUPLES
45. Equilibrant and Equivalent Couples. Since a
couple cannot be replaced by a single force, it follows that
no single force can balance a couple In order to balance a
couple, another couple must be opposed to it. Either couple
is called the equilibi-ant of the other.
Two couples are equivalent when they can each be
balanced by one and the same couple that is, when they
have the same equihbrant.
46. Equilibrium of Two Coaxial Couples. Two
coaxial couples balance each other if they have equal but opposite
moments that is, if their moments are numerically equal but
have opposite directions (or signs}
DEMONSTRATION The demousttation of this principle is given
below Although it is not essential that it be learned or even read, it
affords a very useful and interesting exercise in the composition and
resolution of forces It will be necessary to distinguish three cases, as
follows
Case I. When the two couples ate in the same plane and the fotces
of one are parallel to the forces of the other
Let the couples be (F lt p t F s ) and (F,',p', F a '), Pig. 34 It Is
assumed that, with the usual notation as to signs, F^p = F^f p', or
F,.p + Fj.'p' = On the other hand, the algebraic sum of the four
forces Fi, F*, /V, FJ is equal to
F i zero, or 2F = Therefore, the
ji A a four forces, or the two couples,
a satisfy the two necessary con-
jp * o tf , ditions of equilibrium of parallel
3 ^--^.7 j,. -,Q| forces (Analytic Statics, Part 1).
and so form a balanced system.
\\ f uy It follows from this that a
1 ' couple may be replaced by an-
other couple in the same plane,
IG> M If the two couples have the same
moment and their forces are all parallel, and that the effect of a couple
is not altered by moving the couple In Its plane parallel to itself.
Case 11. When the two couples are in the same plane and the lines
of action of the forces of one intersect the hnes of action of the forces of
the other.
29
ANALYTIC STATICS
Let the couples be (F lt p, F,) and (F^p 1 , F,') , Fig 36. According to
the preceding demonstration, the couple (F^f, F,') may be replaced
by another couple with one of its forces acting through 0,, provided
that the direction of the forces is not changed, and, moreover, the
magnitude of the forces may be changed, provided that the arm is so
changed as to keep the moment constant Thus, if Pi is made
parallel to FJ and F t ', and equal to F lt and if, at the end of the
arm t 0," = p, the force P t is applied, equal in magnitude to />,,
but opposite in direction, then the two couples (F l ',p l ,F, 1 ) and
(/* <9i0,", A) will be equivalent, for, by hypothesis, FJ & is
numerically equal to F t p t and therefore to P t X 0i O," By trans-
ferring the point of application of P, to the intersection / of the lines
A
"*- A.
/ X
I/
A
f i
o a
^ i
yi
FIG 86
of action of P t and F,, and the point of application of F t to the same
point, the two couples (F lt p, F m ] and (FJ.f, F,') are replaced by the
forces P t and F t acting at 1( and P, and F, acting at 7 Since ^
and PI are equal, their resultant must act along the bisector of the
angle />, d F lt which evidently is the same as the bisector of O, O-, O,"
Similarly, the resultant of P, and F, must act along the bisector
of O,fO t , aud, as P a and F t are equal and parallel, respectively,
to PI and FI, the resultants of the two pair of forces are numerically
equal Now, owing to the equality of O^ O a and t O," t the right
triangles (not fully shown) X O a f and t 0,"7 are equal, and the
line 70 t (not shown) is the bisector of both O, 0i 0," and 0,70,".
The two resultants have, therefore, the same line of action, and, as
their algebraic sum is zero, they balance each other.
It follows that a couple may be replaced by any other couple acting
in the same plane, provided that the two have the same moment
34
ANALYTIC STATICS
29
Case III. When the planes of the couples do not coincide, but are
parallel (as they must be, since the couples are supposed to be
coaxial) .
Let PQ and P' Q, Fig 36, be the planes of the two couples, and
let (F^pi.F,) be the couple acting in the plane PQ Since the
moment of the couple in the plane P' Q is F t p lt that couple may be
replaced by another couple (F l ' l pS, F a ') having the same force and arm
as (F lt p lt F a ), and the lines of action of whose forces are parallel to
the lines of action of F l and F 3 This follows from the two cases pre-
viously considered, for the couple (/V.A', FJ) is equivalent to any
other couple acting m the plane P' Q and having the same moment.
Draw 0, OJ and O OJ, meeting at / We may now compound F l
Q
Fit, 30
with FJ, and F t with FJ. Since d O, is equal and parallel to OJ OJ,
the point /is the middle point of O l OJ and O, OS The resultant of
jFi and FJ Is R = F, + F,' = 2 F,, acting through 7 The resultant
of F, and FJ is ' = F t + FJ = 2 F, = 2 F lt acting through /. As
these two resultants act In opposite directions along the same line (for
they are both parallel to the forces of the couples) and have the same
magnitude, they balance each other. Therefore, the two couples to
which the two resultants are equivalent balance each other.
47. It follows from the preceding principle that the only
couple are its moment and its axis.
ver, that the axis is not any
ion. Any line perpendicular
29 ANALYTIC STATICS 35
to the plane of a couple may be taken as its axis; and, con-
versely, if the axis of a couple is given, the couple may be
supposed to act in any plane perpendicular to that axis. The
force and the arm of the couple may be changed atpleasuie,
provided that their product, which is the moment of the
couple, remains unchanged.
48. Resultant and Etinlllbrant of Any Number of
Coaxial Couples. 77/6' resultant of any numhet of coaxial
confilcs is a single couple having the same axis as the component
couples and whose moment is the algebraic sum of the moments of
the component couples. This principle is a consequence of the
one stated in the last article, as can be shown by a process of
mathematical reasoning that it is not necessary to give here.
In general, let M^ M^ M a , etc. be the moments of any
number of coaxial couples, M r the moment of their result-
ant, and M q the moment of their equilibrant Then,
Mr = 2M> and M t = - IM
Also, if several coaxial couples are in equilibrium, the
following equation must obtain:
Mr = SM =
EXAMPLES FOB PRACTICE
1 Find the resultant of the following couples, in which the
sign before each parenthesis indicates the direction of the couple
(10 Ib , 2 ft , 10 Ib ), - (7 Ib , 6 ft , 7 lb.), (25 lb., 12 ft , 25 lb.);
- (8 Ib , 40 ft., 8 lb ) Aus M r = - 42 ft -lb.
2 (fl) Find the lever arm/ of a couple with a foice of 100 pounds
that will balance the following couples: (30 lb , G f t , 30 lb.), (20 lb ,
6ft.,201b.), - (125 lb ,10ft , 1251b ); (70 lb ,4 ft , 70 lb ), - (80 Ib.,
3 ft., 80 lb.). () What is the sign of the balancing couple?
Ana /()/ = 9 3
Ana \ (b) Couple positive
3. One of the forces of a couple acts through a certain point O,
and is equal to 300 pounds, the moment of the couple is 2,574 foot-
pounds. How far from O is the line of action of the other force?
Ans R 58 ft.
4. The moment of a couple is 1,500 foot-pounds. Express it
(a) as a couple having a force of 75 pounds; () as a couple having
an arm of 12 feet. ._ f (a) (75 lb , 20 f t , 75 Jh ^
Ans V v ;, V I25 lb, 12 it., us& It./
36 ANALYTIC STATICS 29
EQUIVALENCE AND EQUILIBRIUM OF COPLANAR
NON-CONCURRENT FORCES
49. To Make the Line of Action of a Force Pass
Through a Given Point. Let F, Fig. 37, be a force act-
ing on a body, and O, a point in the body or rigidly connected
with the body. Let the perpendicular distance OP of O
from the line of action of the force
* be denoted by x. It is obvious that,
if the two forces P and F f , equal
Jijl | to each other and to F, are applied
jfx at 0, they will have no effect on the
I j condition of the body, since they will
2>j- so !o balance each other. The single force
I | Fis, therefore, equivalent to the sys-
I i tern of forces F, F', F', applied as
\-Jf' shown. Now, the two forces F and
1 F' form a couple whose moment is
Fx Therefore, the force F acting
along LK is equivalent to an equal and parallel force F 1
acting through O, together with a couple whose moment is
equal to the moment of F about O. In general,
The line of action of a force may be shifted parallel to itself >
so that it -will pass through any chosen point, provided that a
couple is introduced having a moment equal to the moment of the
force about that point.
50. Resultant of a Couple and a Force In the Plane
of the Couple. Let the force
F, Fig. 38, and a couple (F tt p,
F t ), whose plane contains the
line of action of the force, act
on a rigid body. The couple
may be replaced (Art. 46) by
another couple (F' t x t F')>
whose forces are each numer-
ically equal to F, provided that
the lever arm x is such that F 1 x
29 ANALYTIC STATICS 37
Also, the couple may be so turned and shifted that one of its
forces F 1 will act along the line of action of F, but m an
opposite direction, as shown. The two forces F and F 1
balance each other, so that the system is reduced to the
single force F 1 equal and parallel to F. Therefore,
The resultant of a couple and a single force in the plane of the
couple is a single force equal and parallel to the given force,
acting along a line whose distance from the line of action of the
given force is equal to the moment of the couple divided by the
magnitude of the given force.
The direction in which the distance x, or O P, should be
measured is indicated by the character of the couple. The
moment F' x of the resultant force about any point in the line
of action of the given force must have the same sign as the
moment of the couple. In the figure, the couple is right-
handed, or positive The moment of F 1 about O must, there-
fore, be rig-ht-handed, which indicates that F 1 is on the left of F.
51. Resultant of Any Number of Coplanar Non-
Concurrent Forces. Let F^F^F^F^ Fig. 39, be four
non-concurrent forces having their lines of action in one
plane, and let O be any point in the plane whose perpen-
dicular distances from the lines of action of the forces are,
respectively, A, A, A, A-
As explained m Art. 49,
,
I
*
kjp' S
\ ' X^a 1 ' /"* F l m&y be replaced by a
V I S
">X / force F-!, equal and par-
\ \ allel to F lt acting through
/^i O, combined with a couple
>. whose moment is /^A-
^ The other forces may be
similarly replaced. The
whole system is thus replaced by the four concurrent forces,
/^ F a ',F t f ,F t ', acting through O and equal and parallel,
respectively to F it F,,F tt , /;, together with the four couples
Fip^Fipt, F a p a ,F t p t . The resultant R of the concurrent
forces is found as explained elsewhere; the resultant M r of
the couples is a single couple whose moment is the algebraic
38 ANALYTIC STATICS 29
sum of the moments of the forces about O. Finally, the
resultant of R and M r is found as explained in Ait 50.
If the given forces are resolved into components in two
directions perpendicular to each other, then, with the usual
notation ( see Analytic Statics, Part 1 ) ,
X r =2X=ZFcosff (1)
Y r = IY = 2Fsrnff (2)
R = Vx r +">7 (3)
\r V v"
tan H r = ^ = ^J-- (4)
A r - sL
And also, M r = 2 Fp (5)
52. Conditions of Equilibrium. When the forces
are in equilibrium, both R and M f must be zero. The reso-
lutes of R must, therefore, be zero, and the algebraic condi-
tions of equilibrium are
2X = 2Fcostf = Q (1)
2Y = 2Fsrnff = Q (2)
M r =ZM=!Fp = (3)
These conditions may be stated in words thus:
1 The algebraic sum of the rcsolutes of the fat CM in each
of any two directions at right
angles to each other must be
sero.
i B 2. The algebraic sum of the
\ j / moments of the forces about any
\ \ / and every point in their plane
NO must be scio
I 53. Equilibrium of
I Three Forces. If three
forces F t , F at F , Fig. 40, are
in equilibrium, tmy one of
}fi\ them is the eqtiilibrant of the
other two, and, therefore,
equal and opposite to their
I lesultant. Thus, /'I must be
FlQl 40 equal and opposite to the re-
sultant R of F, and F,. As R passes through the point of
29 ANALYTIC STATICS 39
intersection of the lines of action of F, and F t and ft must
balance V?, the line of action of F, must pass through O. In
general,
// three coplanar foices are in equilibrium, they must be con-
curtent (unless they aie parallel), and if the point of nitersection
of the lines of action of two of the forces is known, the line of
action of the other must pass through that point.
EXAMPLE 1 Four forces, F l = 100 pounds, F a = 200 pounds,
F 3 = 125 pounds, F t = 150 pounds, Fig 41, act on a horizontal
lever O B The inclinations of their lines of action to the horizontal
are as shown, and O A-, - 4 feet, O A a = 10 feet, O A = 15 feet,
O At. = 20 feet Required the magnitude, direction, and line of
action of the resultant A J
SOLUTION Using formulas 1 and 2 of Art 51, we have, since
F, = 100, F tt - 200, F t = 125, F< = ISO, and H, = 45, H* = 50,
77, = 80, 77* = (50,
X r - S F cos H - - F, cos 77 t - F, cos H, + F, cos H 3 + F*. cos ff t
- - 100 cos 4f) - 200 cos 50 + 125 cos 80 + 150 cos 60 = - 102 56
Y r 2 F sin H = 7^ sin ff^. F, sm H? F, sm 7/ a + F t sin ff t
- 100 sin 45 - 200 sin 50 - 125 sin 80 + 150 sin 60 - - 75.694
By formula 3 of Art 51,
A' = VX r + YS = Vl02~5H a + 757694" = 127 47 Ib Ans
By formula 4 of Art 51,
tan H r = % - Jh.rsn; whence H, = 36 25' 40". Ans.
J\f !VM OU
Since p, = O A, sin 45 = 4 sin 45 - 2.8284; />, - O A, sin 50
= 10 sin 50 = 7.6604, p, = OA> sin 80 = 15 sm 80 14.772, and
40 ANALYTIC STATICS 29
p* = OAi sin 60 = 20 sin 60 - 17 321, formula 5 of Art 51 gives
M r = IFp = - ^i A + F, p, + F, p* - F. p.
= - 100 X 2 8284 + 200 X 7 6604 + 125 X 14 772 - 150 X 17 321
= 497 59 ft -lb
The value of Y r shows that the force R acts downwards, its line of
action must, therefore, be on the right of O, since its moment about
this point is positive Hence,
Mr 49769
Mr = Rpr, whence p r = -g 127 47 =
EXAMPLE 2 Two forces, F, = 2,000 pounds and F, = 800 pounds,
Fig 42, act on a horizontal beam 25 feet long resting on two supports
PIQ 42
at its extremities O and B The distances O AI and O A t are, respect-
ively, 8 and 16 feet. The inclinations of FI and F, are as shown.
It is known that the reaction R" at B is vertical Required the magni-
tudes of the two reactions R" and R 1 1 and the inclination //' of R 1 to
the horizontal
SOLUTION Since the moment of R' about O is zero, R" may be
found by taking moments about that point anil using formula 3 of
Art 62, which gives,
Fipi + F* X OA, - R" X OB - 0;
that is, 2,000 X 8 sin 30 + 800 X 16 - A 1 " X 2T> 0;
, _.. 2,000 X 8 sin 30 + 800 X 16 M01K ..
whence, R" OK ~ - = 832 lb. Ans.
it)
By formula 1 of Ai. 52, IF cos H = 0; that is,
R' cos H' - F, cos 30 = 0;
whence,
R 1 cos H 1 = F, cos 30 - 2,000 cos 30 1,732 (1)
Similarly, by formula 2 of Art. 52, 2 F a\n // - 0; that is,
Ri sm If' - F, sin 30 - F, + R" - 0;
whence, R' sin H 1 = ^ sm 80 + F, - R"
2.000 sin 30 + BOO - 832 - 968 (3)
29
ANALYTIC STATICS
Dividing (2) by (1),
R< sin H 1 _ . .,, _ 968
X'cosff' ~ taU ** -l7732 ;
whence, H' = 29 12', nearly Ana.
From (2),
ff
1,984 Ib. Ans.
APPLICATION'S
54. Mutual Reactions. In Fig. 43 (a) are represented
two bars hinged at A,, and A at and to each other at B The
bars are acted on by forces F t and F,. Through the joint B t
the bar A^. B exerts \
on A,B a force R a f ,
which is equal and
opposite to the force
Rt f exerted by A, B on
A* B. This force RJ
R,' is the mutual
reaction between
the two bars at the
joint B. How it is
determined will be
explained presently.
Since the system
formed by the two
bars is m equilibrium
under the action of
the external forces F lt
F t) R lt R tt these forces
form a balanced sys-
tem to which the
general equations of
equilibrium can be
applied. From these Pl i 4S
equations, the reactions Ri. and R, can be found when all
other conditions, such as distances, etc., are known.
The mutual reaction at B is determined by applying the
principle of separate equilibrium (see Analytical Statics,
42
ANALYTIC STATICS
29
Part 1). The part A,B may be removed, and A^B
treated as a free or separate body, provided that a foice is
introduced at B equal to the force exerted by A. B on A* B t
that is, a force equal to RJ. This condition is represented
in Fig '43 (b), where A,B is shown as a free body acted on
by the external forces R i} -ft, and RJ. By applying to the
system constituted by these forces the general equations of
equilibrium, RJ may be determined.
55. Method of Sections. The free-body principle
finds a very useful application in the determination of
stresses in framed structures by the method or sections,
illustrated in Fig. 44. The truss A, B* B* A, rests on piers A,
and A n and is loaded at the joints d, C, and C 91 as shown.
TT,
W
PIG. 44
Let the reactions R* and R, first be determined. Consider-
ing the truss as a whole, the external forces acting on it are
the weights W lt W, and W, and the reactions JR l and R tt
These forces form a balanced system, and, therefore, the
algebraic sum of their moments about any point mubt be
zero (Art. 52). Taking moments about A^ in order to
eliminate R lt
W. XA 1 C*+ WX A, C+ W t XAtCt-R.X A,A t 0,
from which R, can be found. To find ^?i, moments may be
taken about A a ; or, knowing ./?, R l may be found from the
equation J 1 Y = 0, which in this case gives
Ri + R> - W> - W* - W
The stresses in the members may be found by the method
explained in Analytic Statics, Part 1, proceeding joint by joint,
29 ANALYTIC STATICS 43
beginning either atA^ or at/ a , since now R* and R* are known.
Or the method of sections, referred to above, may be used,
as follows:
Let it be required to find the stresses in the member CC a .
Imagine the truss to be cut in two by a plane PQ inter-
secting CC a and the members CB, and B^B*, which meet
at JS 3 .
The part of the truss at the right of PQ may be supposed
to be removed by introducing at the points of separa-
tion E, F, ff, forces 5, T, U, equal to the actions of B* E on
EB^BiF on FC> and C t H on H C, respectively, which are
the measures of the stresses m the three members inter-
sected by the plane The part A^B^EFH may now be
treated as a free body acted on by the external forces R^ W lt
W, S, T, and 7, of which only the last three are unknown.
If moments are taken about B t (the point of intersection
of 6" and T), the forces S and T will be eliminated, and an
equation will be obtained in which the only unknown quantity
will be U.
The same result would have been obtained if the mem-
bers CC at BI G, and B t A n had been cut by the plane PQ.
To find S, moments are taken about C. In every case,
several members should be cut, of which all but one meet
at a joint, and this joint should be, taken as the origin of
moments. If, however, the stresses in some of the mem-
bers cut are known, it is immaterial whether they meet at a
joint or not: the only thing necessary is that, of the mem-
bers whose stresses are unknown, all but one should be
concurrent.
ILT3D8 M
44
ANALYTIC STATICS 29
FRICTION
SLIDING FRICTION
DEFINITIONS AND GENERAL PRINCIPLES
66. Definition of Sliding Friction. Let a block
CDE, Fig. 45, rest on a horizontal surface AB capable of
resisting or balancing the weight W of the block. The
block will then be in equilibrium under the action of the
weight W and the reaction R, the latter being equal and
opposite to W. If, now, a horizontal force F( which, for
convenience, will be
1 supposed to act along
ir ^ a line passing through
the c. g of the block)
is applied, and it is
assumed that no other
force than W, R, and
F acts on the block,
the latter being under
the action of an un-
Fl 45 balanced force, will
move in the direction of F with an acceleration equal to
g (see Fundamental Principles of Mechanics}. So long as
Vv
there is no other force acting on the block, motion will
ensue, however small /'may be.
Experience, however, shows that a small force, whether
horizontal or not, often produces no effect on a body resting
on a surface, and sometimes a very great force is required
before the equilibrium of the body is disturbed. Thus, to
drag a trunk or a box over the floor may require the efforts
of several strong men. We also know that the resistance is
29 ANALYTIC STATICS 46
greater the rougher the surfaces in contact: a rough box Is
not so easily dragged over the sidewalk as a polished stone
block over a smooth wooden floor.
There is, then, a force brought into action whenever there
is a tendency of a body to slide on another. This force,
whose effect is to prevent, or which tends to prevent, motion,
is called sliding- friction, or simply friction. It is
obviously caused by the roughness of the surfaces m con-
tact. No matter how smooth a surface may appear, it
always has small projections or elevations separated by
small depressions. When the surfaces of two bodies are m
contact, the projections of one surface go into the hollows
of the other, the two surfaces thus become more or less
interlocked and cannot slide freely on each other. In order
to cause sliding, a force is necessary, whose magnitude
depends on the roughness of the two surfaces. As this
roughness varies with different bodies, it may be anticipated
(and this is known from experience to be the case) that
friction must be a very variable force, depending both on
the nature and on the conditions of the bodies m contact.
57. Limiting Equilibrium. Given the block CDE,
Fig. 45, resting on the surface A B, the force F may
either move it or leave its equilibrium undisturbed. In
the former case, F must be greater than the friction; in
the latter case, .Fmust be less than the friction Let F m be
the greatest force that can be. applied to the block without
moving it. Then, a force equal and opposite to F m will
represent the maximum friction that can exist between the
block and the surface AB, any force greater than F m will
cause motion, and any force less than F m will be balanced by
the friction. But it is not to be supposed that the friction is
constantly equal to F m : so long as the applied force is less
than F nn the friction is just equal and opposite to the
applied force; the friction grows with the applied force up
to the value F m , beyond which equilibrium ceases to exist
and the body moves under the action of the difference
between the applied force and the maximum friction F m .
46 ANALYTIC STATICS $29
When the block is acted on by a force equal to /v l} it is
said to be in a condition of limiting equilibrium, or on
the point of moving, for the least mciease in the applied
force is sufficient to produce motion In (Ins <MSU, there
exists between the two bodies A />' and C HH tin 1 greatest
possible friction that under the given cucuinstancus can
exist between them. This maximum force of fiietion is
called limiting friction, and will heieaftcr be designated
by POT. Numerically, P m = J ? m .
58. Passive Forces. Friction, like many other resist-
ances, is a passive force that is, a force preventing
motion, but not producing it. The reason foi this is that
friction is brought into action by the application of other
forces to which it is opposed; and, as it can never exceed
those forces, it can never pioduce motion in the direction of
its own line of action. The .same is true of this reactions of
supports. A pier may be capable of exerting n reaction
of 1,000 tons; but, if a stone weighing 1 pound is placed
on it, the pier will exert on the stone an upward pressure
of only 1 pound, or just enough to balance the weight of
the stone.
Forces producing, or tending to produce, motion, are
called active forces.
Passive forces are balancing forces and never acquire
greater magnitudes than the active forces they oppose.
Usually, as in the ca.se of friction and reactions, n passive
force cannot exceed a certain limit; but between Kero and
that limit it can have any value, and so long as the active
force to which the passive force is opposed does not exceed
that limit there will be equilibrium.
59. Shifting of tlio Ijlno of Action of the tteaotlou.
The force of friction is a tangential force, by which is
meant a force acting along the surface of contact of the two
bodies between which it is exerted, In Fig, 46, the force of
friction is not directly opposed to /; but acts along the
surface CE, in the direction EC, How, then, can this force
balance F, not being in line with it?
29
ANALYTIC STATICS
47
The reason is that, on the application of F, the reaction R
is shifted so that it no longer passes through the point where
the vertical through the c g. of the body meets the sup"
porting surface. This is illustrated in Fig. 46. The line of
action of the applied force F meets the vertical line
through G at 0. The resultant of W and F, found in the
usual manner, is F r , whose line of action meets AB at O 1 .
Transferring F r to O 1 (where, for convenience, it is repre-
sented by Fr'), and again resolving it into its components
F 1 F and W = W t it is seen that, in order that there
may be equilibrium, the reaction R must be equal and
PIG. 40
opposite to F r ', and its components Y and P must be,
respectively, equal and opposite to W'(= W} &ndF / (=s F).
The vertical component Y is the resistance of A B to direct
or normal pressure; and at present it will be assumed that
the surfaces in contact are capable of offering this resistance,
whatever the value of Y or W may be. The horizontal
component P is the force of friction acting along the surface
of contact. In the case illustrated in the figure, F is sup-
posed less than F m that is, less than the maximum force
that can be opposed by the friction. Therefore, P, which
can take any value between and F m , will, in this case, be
just equal to F 1 , or F, and balance the latter force.
48
ANALYTIC STATICS
29
It thus appears that the effect of the friction is to shift the
point of application of the reaction from K to O', and to
change the line of action of that reaction from the vertical
direction to the direction O 1 L.
The body CD E may be considered as being acted on by
the two equal and opposite forces R and F r , or by the two
couples - (W,KO f , Y) and (F, OK, P). Let the student
show that the moments of these two couples are numerically
equal.
60. Case In Wnlch F Is Greater Than /*,. We
shall now consider the case in which F is greater than F m ,
or P m In Fig. 47, the block CDE is acted on by its own
Fin 47
weight W and the horizontal force F. As before, F r ' F r is
the resultant of W and F, transferred to the point (V on the
surface of contact; its components are F 1 = F and W W<
As, m this case, /MS greater than F mt the force of friction,
which has its maximum value P, H = F m , is not sufficient to
balance F 1 , and there will be sliding of the block on A B.
The reaction R of the latter surface is the resultant of
Y = W = H^and P m . This reaction is evidently less than
F r ', and its line of action makes with the normal O M to AB
an angle Z less than the angle / made by the line of action
of F r with that normal,
29
ANALYTIC STATICS
49
If, Y (that is, W or W) remaining constant, F is made
less than P m , say equal to O' S, the resultant will be O 1 T,
and the reaction will be O 1 U, whose components are Y = W
and MU equal to the friction />, which in this case, will be
equal to OS.
The conditions of equilibrium may, therefore, be stated
by saying; either that /"must not be greater than P m , or that
/ must not be greater than Z.
ANGLE AND COEFFICIENT OF FRICTION
61. Maximum Resistance. The results of the fore-
going discussion may now be generalized. Let N, Fig. 48, be
the normal force between two bodies whose surface of con-
tact is A B. The force N is the normal component of the
resultant force acting on CD E (or on the other body), when
that resultant, after its point of application has been trans-
"jr
FIG. 48
ferred to the intersection of its line of action with the surface
of contact, is resolved into two components, one parallel, and
one normal, to that surface. Given, besides the normal
component N (usually called the normal pressure), all
other conditions such as the nature of the two bodies,
the extent of their surface of contact, etc, the maximum
50 ANALYTIC STATICS 29
friction P m that can exist between them is determined by
actual experiment Once P m is known, the maximum reac-
tion R, which we shall call the maximum resistance that
can exist between the two surfaces, is
The angle made by the line of action of the reaction with
the normal OS to the surface of contact is given by the,
familiar expression,
Li S fm -fft
If, the normal pressure and all other conditions remaining
constant, a force F greater than R is applied to CDE (in
which case N is the normal component of F) , it is obvious
that there cannot be equilibrium, since the component of F
parallel to A B, which component is equal to M S, is greater
than P m . If, on the contrary, F is less than R t or equal
to R that is, if its line of action falls within the angle Z or
coincides with L equilibrium will obtain.
62. Angle of Friction: Condition of Equilibrium.
The angle Z, Fig 48, is called the angle of friction, and
may be defined as the angle between the line of action of the
maximum resistance and the normal to the surface of contact.
The condition of equilibrium explained in the preceding
articles may be thus stated:
In order that there may not be sliding between two bodies in
contact, it is necessary and sufficient that the resultant of the
applied forces (as F r , Fig. 47} shall not make wtth the normal
to the surface of contact an angle greater than the angle of friction.
63. Coefficient of Friction. Until recently, it was
thought that, for any two given substances, the maximum
friction P m was directly proportional to the normal pres-
sure .Wand independent of all other circumstances. Accord-
ing to this view, if P m ', Pa/'t /V", etc. are the maximum
frictions corresponding to the normal pressures N' % N", N" f ,
pi p tf p tff "
etc., then ^ = = ; = ^. The common value of these
ratios will be denoted by c.
29 ANALYTIC STATICS 61
Having determined the ratio c of the friction to the nor-
mal pressure for any particular case, the friction in any other
D
case could be at once found from the relation = c,
N
whence, P m = c N. Also, since
tan Z = ^2 = c
N
it would follow that the angle of friction was constant for
every two substances sliding on each other.
That there is generally some dependence of the force of
friction on the pressure is a familiar fact. Thus, referring
again to Fig. 45, daily experience shows that, if a pressure Q
is applied to the block, the effort required to drag the block
will increase as the pressure Q increases. The relation
between pressure and friction, however, is not always so
simple as stated above.
The law that friction is proportional to pressure is approxi-
mately true only m some cases, or under certain conditions.
What these conditions are, and how friction varies under
different conditions, are problems to be solved by direct
experiment.
Although the ratio of P m to N is usually variable, it is
customary and convenient to express P m as a fraction of the
normal pressure, and write,
P m = fN (1)
The factor /, or the ratio of the maximum friction to the
normal pressure, is called the coefficient of fi-Iction, or
friction factor. It has been determined experimentally
for various substances under various circumstances; its
values have been tabulated, and the approximate laws of its
variations stated. Here, the theory only of friction will be
considered. In order, however, that a general idea of the
values of / may be obtained, it will be remarked that, in the
majority of cases coming within the practice of the mechanical
engineer, / is usually much less than .5. Thus, for unlubri-
cated metals sliding on one another, the average value of /
is about .18; for dry and smooth wood sliding on the same,
/ averages about .38; and for wood sliding on metal, both
52 ANALYTIC STATICS
dry and smooth, about A. Tn civil-cntfinerring wink, how-
ever, values of /often oceui that e\eeeil ..", ,is in the i ,IHI nf
brick sliding on hiick or on stone, masoniy on hiKk\\otk, civ.
Since tan X ~ / "" 1 , which is the same .is the v.ihut *f /
from foimuhi 1, \vo may write
tan /. ~- f (2)
The angle of friction m.iy, therefore, he defined .is .in
angle whose trigoiuuneti tu tangent is uinuil ti the encHu ieut
of friction, it being understood th.it one of the sides if tins
angle is the normal to the surf.iee of contuet nf llic tw>
bodies whose fiiction is considered.
64. titatlo Frletlon and Kliiftlc l^rlctloiu In
to set a hody in motion over another hody, u fojve is neecs-
sary whose component parallel to tlic snrf.iee of roni;<el i^
greater than P m . Once the body is in motion, it seem** th.it
it should continue in motion, by virtue of its inert ut, if n
force just equal to />, were constantly uiiplied to if; fur this
force would ho sufficient to overcome the frietion. KXJHTI-
ence, however, shows that the force nceessary t Ktiut n
body sliding on another is almost always greater th;ui the
force necessary to keep it sliding, once motion has Itegmi.
In other words, the resistance of friction in greater when
motion is to bo produced, than when it iff to IMJ nminlaiitct).
In the former case, tjie friction IH culled fHetloii of n-nt,
or static fi'totlon; in the latter case, fi'teflon of auttflon,
clyiminle ri-It-tlon, or Idnettc frietion. In cither ease,
the maximum frietion will bo here designated by A,, niul
the coefficient of friction by /, it being undorwtood that. IM
a rale, P m and /have different values for the two
65. In practice, it is necessary to bear in inlmt what the
function of friction is when any particular problem in tp lie
solved. In designing a nmcliine, where niotitm han to 1
produced and maintained, sufficient force should JMJ allowed
to overcome the friction of rest, an othnrwive the machine
could not be started. But, in calculating the c<rriolt*m<y
(A term to be defined elsewhere) of the machine while ia
29
ANALYTIC STATICS
53
motion, the friction of motion should be used. In providing
for the equilibrium of a structure, however, where friction is
a favorable force, the friction of motion should be used; for,
although under ordinary circumstances the structure, being
in a state of rest, will offer a frictional resistance equal to
the maximum friction of rest, the least shock is often suf-
ficient to produce a disturbance of equilibrium; the structure
is, so to speak, started, and then the only resistance prevent-
ing it from continuing to move will be the friction of motion.
This is especially the case m structures subjected to shocks,
such as bridges and engine foundations.
FIG. 49
EXAMPLE A block CD E, Fig 49, whose weight is W, rests on a.
horizontal surface A B The coefficient of friction between the block
and the surface is /. A force F, acting in a. vertical plane containing
the c g of the block, IB applied at an angle H to the horizontal.
Required the magnitude of F, that the block may be on the point
of sliding along A JB.
SOLUTION All problems similar to this may be solved in two
manners, and the result found directly m terms of either / or Z, Of
64 ANALYTIC STATICS 29
course, one result can be transformed algebraically into the other
from the relation tan Z = f
1 Let O be the intersection of the line of action of F with the vertical
through the c g of the block The latter is held in equilibrium by the
forces F, W, and the total reaction J? of A B t whose line of action
O>O must pass through O (Art 53) As explained in Art 62, this,
line of action must make with the normal to A 2?, or with the vertical,
an angle equal to Z Therefore, F r being equal to and colllnear with
R t angle OJT= WOJ = Z The triangle O JT gives
TJ W
W
- WOJ) snZ
W W
= sin (90 + H - Z) sm Z = cos (H - Z) sin Z (1)
2 The line of action of the resultant F r of the forces F and W
meets the surface A B at 0'. Let F and JP be transferred to this
point, as shown. Resolving F into its horizontal and vertical com-
ponents F cos H and F sin H, the resultant normal pressure acting at
N= W-FsinH
Therefore, the resistance of friction is:
P m = Nf = (W-FtinH)f
As this force must balance the horizontal component of F, there
results- ( W - F sin H )f=F cos H\
W f
whence, F = __ , , rr (2)
1 cos /f+ /sin If ^ f
To reduce equation (2) to equation (1), we have
Wf ^ w tan Z
cos ff+fsinH cos /f + tan Z sm ^T
sin Z
*,* cosZ _, sin Z
smZ
cos(H-Z)
EXAMPLES FOR PRACTICE
1. Find the angle of friction, to the nearest minute, corresponding
to each of the following coefficients of friction: (a) /=*.16; (d) f** .25;
(c) f - .50, (d) f = .65.
Ans.
8 32'
2T 14 2'
2T 26 84-
Z - 33 1'
29 ANALYTIC STATICS 55
2. Find the coefficient of friction corresponding to each of the fol-
lowing angles of friction (a) Z => 12 15', (6) Z = 30, (c) Z = 8 35';
(d) Z = 3 17'. f (a) f = .217
A O J(*) f = 577
Ans lW /= .151
[(d) /= 057
3 A block of marble weighing 100 pounds is kept sliding with
uniform velocity on a horizontal pine floor by a force inclined to the
horizontal at an angle H = 30. If the coefficient of kinetic friction
between marble and pine is 45, what must the magnitude of the force
be (a) if H is an angle of depression? (6} if H is an angle of eleva-
tion? & IW F= 70221b.
Ans \(b) F= 41251b.
4. A force of 15 pounds, inclined to the horizontal at an angle of
elevation of 30, is just enough to keep a block of cast iron, weighing
100 pounds, sliding uniformly on a horizontal cast-iron plate Find
the coefficient and the angle of friction . f / = 14
Ans \Z = 7 68'
5. Taking the angle of friction of rest for brick sliding on brick
as 35 30' , with what normal pressure N must a brick weighing
6 pounds be pressed against a vertical brick wall that the brick may
not slide down? (Use only two decimal places for /.)
Ans N = 8 45 Ib.
RESISTANCE TO ROLLING
66. Cause of Resistance to Rolling. Let a homo-
geneous cylinder A D, Fig. 50, of radius r, rest on a hori-
zontal surface X' X. If a horizontal force F is gradually
applied to the cylinder along any line NL perpendicular to
the axis of the cylinder, it will be found that no motion can
be produced before the force F exceeds a certain limit.
Now, the line of action of the weight W of the cylinder
meets the supporting surface at A, directly under the
center O. Did the cylinder touch the surface X' X only at
A, it is obvious that any horizontal force, however small,
would cause motion; for, as the resultant of that force and
the weight could not be vertical, its line of action could not
pass through A, and, therefore, such a resultant could not be
balanced by the reaction of the supporting surface. And,
since experiment shows that it is possible to apply a horizon-
tal force to the cylinder without causing motion, it follows
56
ANALYTIC STATICS
29
that not only the point A, but a part AB of the cylinder
must be in contact with X' X; in which case it is easy to
understand how the
resultant F r of F and
W may be counter-
acted by the icaction
R of the suiface X 1 X.
That contact cannot
take place at the point
it A only is otherwise
evident from the fact
that all substances
are more or less
compressible, so that,
while the weight of
the cylmdei causes a
small depression in
the supporting sur-
face, the reaction of
the latter causes a
small flattening of the cylindrical surface, as shown in the
figure.
67. Coefficient of Boiling: Friction. Suppose that
the cylinder is in a state of limiting equilibrium with respect
to rolling; that is, that any increase of the force F will cause
the cylinder to roll. Let the line of action of the resultant
of Wand -Fmeet the surface of contact at Af t at a horizontal
distance c from the theoretical point of contact A. Resolving
the reaction R into its components F and W, it will be
seen that the cylinder is in equilibrium under the action of
two couples; namely, (F, Afff, -F), and -(W,NH,-W).
Therefore,
Fio 60
But, as the deformation of the surfaces of contact is
very small, we may write, with sufficient approximation,
MH = A N = h; and, therefore, Fh = Wc\ whence
F = W-
h
29 ANALYTIC STATICS 57
It has been ascertained by experiment that the distance c
is independent of the dimensions and weight of the cylinder
and depends only on the materials of the two surfaces in
contact The particular value of c for any two materials
rolling on each other (it is not necessary that one of the
surfaces should be a plane, as X 1 X} is called the coefficient
of rolling friction for those two materials This' coeffi-
cient is not an abstract number, but a length, and its value,
of course, depends on the unit of length used. The follow-
ing are approximate values of the coefficient c.
For elm rolling on oak, c = .032 inch
For iron on iron and steel on steel, c .02 inch.
It appears from the foregoing formula that, for any two
materials, the magnitude of the force F producing limiting
equilibrium depends on its lever arm h. What is necessary
and sufficient in order to produce limiting equilibrium is that
the moment Fh of the applied force should balance the con-
stant moment We The resistance to rolling may, there-
fore, be said to be expressed by a couple We rather than by
a single force.
This constant couple, whose moment is obtained by multi-
plying the normal pressure acting between the two surfaces by the
coefficient of rolling friction , is called a friction couple.
68. To Determine Whether Equilibrium Will Be
Broken by Sliding or by Rolling. Referring again to
Fig. 50, it must be noticed that rolling about M (practically
about A} cannot occur if the cylinder slides before Fh reaches
the limit We (here Wdenotes the sum of all the normal forces
acting between the two surfaces) If / is the coefficient of
sliding friction, ^must not be greater than Wf, or W - must
not be greater than Wf\ therefore, - must not be greater
/i
than /, or c must not be greater than fh.
If c is greater than fh, sliding will begin before rolling can
take place. If c fh, sliding and rolling will begin
simultaneously.
58
ANALYTIC STATICS
29
TUB INCTJNKl)
/ I
/ I
/V
JL
Fifi.6i
69. Definitions. An hiulliuMl piano is, as its
implies, a plane surface inclined to the hon/.on. In Fitf. ftl,
the plane PQRS, making with the horizontal plane /' 7V/.V
(or any other hori-
zontal plane) an angle
' T, is an inclined
plane.
70. Any vertical
plane perpendicular to
an inclined plane is
called a principal plane, and its intersection with the
inclined plane is called a lino of Uot'llvlty. A line of
declivity may alho be defined as a line lyintf in the inclined
plane and perpendicular to the intersection of the ItitU-r plane
with any horizontal plane. In Fig, fil, PQ /", . I />' ( ', A' A 1 V t
being vertical planes perpendicular to /'j^A'A*. aie principal
planes, and the lines Ql\ J1 A } RS are lines of declivity.
71. The angle between an inclined plane tind the horizon-
tal is called the aii^lo of the inclined plane, and is the same us
the angle that any line of declivity imilcea with the horizontal.
Thus, in Fig, 51, the angle of the plane is the common value
of the angles QPT, BAC } .flSf/tnade by the lines of
declivity QP, BA, RS with the horizontal.
72. In practice, an inclined plane is always the surface
of some body, as a plank, an inclined rail, the side of a
hill, etc. If, for any special purposes, it is desirable to take
into account a definite extent of this surface, as /'? RS in
Fig. 61, the line A n (or any other parallel to it and included
between PS and QR) is called the length of the inclined
plane; JB C is the height, and A C the base.
73. Equilibrium of n Body on an Inclined Plane,
Let J t MJ t ', Fig. 62, be a body resting on an inclined
plane AB. The view here represented IB a section made
by a principal plane through the c. g. of the body. All
29
ANALYTIC STATICS
59
forces are supposed to lie in that plane. The angle of the
inclined plane is B A C = H. If only the portion AB were
considered, AB would be the length, BC the height, and
A C the base. But these dimensions are not needed for the
purpose of the present discussion. The line W is a
vertical through the c. g. of the body, W is the weight of
'\
Fib 62
the body, and Fa. force whose line of action meets W
at O. Through O draw YY 1 and XX 1 , the former per-
pendicular, the latter parallel, to A B. Let the line of action
of F make an angle K with Y Y 1 . Denote the equilibrant
of F and W by ?, their resultant by F rt and the angle that
Q and F, make with Y Y' by L. The triangle O T U gives
UT _ sin f/0 r
0tf sin C>r/
Now, UT= F-,OU= W\*\\\UOT = sin ( + #); and
sin 7Y7 sin TOF - sin (180 QOF) sin _
= sin (L + A"). Substituting in the preceding equation,
_^ _ sin ( + #).
whence,
I L T 398-U
sn
+
60 ANALYTIC STATICS 29
The equihbrant Q of /*" and W is the rctiction A 1 oi the
inclined plane, acting through the point / whcic the line ot
action of the resultant 7v meets the plane. When the body
is in a position of limiting 1 equilibiium with respect to sliding,
R makes with the normal J N an angle equal to the angle
of faction Z (Ait. <>2). As JN is parallel to )')'', it
follows that, in this case, L = xf, and the foiou /*' is the
maximum foiec acting; along the line OF that the body can
resist without sliding:. It is often said that this is the foico
that is just enough to start the body moving ovt i r the plane;
but this is not coirect. If the body is aheady moving, this
force, being just enough to balance the friction, will keep the
body moving with constant velocity. If the tootlv is at rest,
the force will simply keep it in a condition of limiting equilib-
rium; the body will not move, but the least increase in the
force will be sufficient to produce motion.
74. To find the value of /"for which the body is in a
condition of limiting equilibrium (or will move with con-
stant velocity, once staited), it is necessary to distinguish
two cases.
CUHO I. The body is on the point of moving up tht plane.
In this case, ATmuat be on the right (in the figure) of Y Y' t
but it may be acute, right, or obtuse. If, however, A' is
acute, it must be greater than //; that is, /''must act on the
right of V V\ for, if /* lay on the left of V l\ the re.sult-
ant F r would act either in the angle VOX' or in the
angle V OX 1 : in the former case, the equilibrant would be
some force directed like Q' t and, in order that the plane
might furnish this equilibrant by its resistance, the reaction JR 1
should have a normal component acting downwards. This
cannot take place, as the plane is not supposed to be capable
of exerting any downward reaction. If the resultant fell in
the angle VOX 1 , the equilibrant would be a force like Q";
this equilibrant might be furnished by the reaction R" of
the plane; but in this case the friction, being: the component
1 of R" parallel to the plane, wouid act upwards, and the body
could not be on the point of moving upwards,
29
ANALYTIC STATICS
61
It being, then, understood that, in the case under con-
sideration, K must be greater than H, the force j^may be
found from the preceding equation by writing Z instead of L\
l (1)
r l ^
p _
This value may be expressed in terms of the coefficient of
friction / as follows:
Sin(Z + ff) _ ^y sin Z cos H + cos Z sm H
sm(Z + K} sm Z cos K + cos Z sin K
PIG fiB
Dividing both terms of the fraction by cos 2", and writing /
instead of ^4 ( = tan Z, Art. 63),
cos Z
/ cos K -f sm K
Leaving: W, ff, and / unchanged, it is seen from for-
mula 1 that, if K is changed, F will have its least value
62 ANALYTIC STATICS 29
when the denominator is the greatest possiblethat is, when
sin (Z+K] = 1; whence,
Z + K = 90, and 90 - K = Z, or FOX = Z
This very important result is expressed by saying that
the best angle of t> action up an inclined plane is the angle of
fnctton.
Case 11. The body is on the point of moving down the
plane (Fig. 53).
In this case, the friction, which is the component parallel
to the plane of the total reaction R, acts upwards. The
line of action of R, and therefore of F ft must lie on the
right (in the figuie) of the normal JN. By a process of
reasoning similar to that employed in the preceding case,
the following formulas are obtained for the present condi-
tions:
F = W !-^L~ f cos *L (4)
sin K f cos K
75. Discussion of Formula 3 of Art. 74 Angle of
epose. So long as H is greater than Z, formula 3 will
give a positive value for F. In this case, F, being the
equilibrant of Q and W> must act outside the angle WOQ,
that is, K must be greater than, Z. The positive value of F
indicates that, if the body is left to itself, it will slide down
the plane, and that, therefore, a force is necessary to keep it
from sliding.
If H= Z, then sin (H - Z) = 0, and, therefore, F=Q.
This means that the body, if left to itself, will rest on the
plane in a state of limiting equilibrium; no force is necessary
to keep the body from sliding; but if the angle of the plane
is increased and no force applied, the body will slide. For
this reason, the angle of friction is often called the angle of
repose, and defined as the greatest angle with the horizon
that the surface of contact of the two bodies to whose friction
it refers can make without the bodies sliding on each other
This relation is made use of in the determination of the
29
ANALYTIC STATICS
63
coefficient of friction. Suppose the plane AB t Fig. 54, to be
the tipper surface of a plank hinged at A, and that it is desired
to find the coefficient of friction between cast iron and marble.
The plank is lined with a plate of cast iron and turned about
the hinge until it is nearly horizontal. A block of marble is
then laid on the plank, and the latter turned upwards until
the block begins to slide. The angle of inclination of the
PIG. 64
plank to the horizontal at which this takes place is the angle
of friction, and its tangent is the coefficient of friction.
When H is less than Z (see Fig 54), the component
of W along the plane is not sufficient to overcome the fric-
tion. Therefore, in order that the body may be on the point
of sliding down, the force F must have a component acting
down the plane; that is, K must be on the left of YY'
In this case,
. / T rr\
(1)
or
sin (Z + JK)
= W -~ c ~^ H ~~- S i5-^.
/ cos K + sm K
(2)
EXAMPLE 1 A wooden box 10 feet long, 6 feet wide, and 4 feet
deep is used for carrying coal up and down an inclined steel-rail track,
the grade of the track being 10 in 100 (which means that the track
rises 10 feet for every 100 feet of length, measured horizontally) . The
weight of coal will be taken as 54 pounds per cubic foot, and the
64
ANALYTIC STATICS
coefficient of kinetic friction between wood and steel as .4. The box
being full, required- (a) the magnitude and inclination of the least
force that will keep the box moving up the plane with constant veloc-
ity; (d) the magnitude of a force parallel to the rails necessary to
produce the same effect; (c) the magnitude of a force parallel to the
rails that will keep the box moving downwards with constant velocity.
(d) If the available force
acting upwards, parallel
to the track, is 4,000
pounds, to what depth
can the box be filled?
SOLUTION (a) For
the weight W we have,
neglecting the weight of
the box, W = 10 X X 4
X 54 = 12,960 Ib The
angle Z of friction, to
the nearest minute, is the
angle whose tangeut is
4, that is, Z = 21 48'.
For the inclination H of
the track we have tan H
= Vnr, whence H = 5
43' Foi the least force
for which the box will
be on the point of mov-
ing upwards, or that will
keep the box, after the
latter has been started, moving upwards with constant velocity, we
must have (Art. 74) K = 90 - Z = 68 12'. These values in formula
1 of Art. 74 give
f = 12,960 sin (5 43' + 21 48') = 6,988 Ib. Ans.
(5) In this case, K = 90, and formula 1 of Art. 74 gives, noticing
that sin (90 + Z] = cos Z,
FIG. 66
(c) As here H is less than Z, formula 1 of Art. 75 is used, making
(see Fig 54) K = Y O X' = 90, and sin (Z+ K] = cos Z.
'-^"^.-^-.^U. An,
(d) Let x be the height of the coal above the bottom of the box;
then, W '= 10 X 6 X .r X 54 Formula 1 of Art 74 gives, noticing
that here F= 4,000 and K = 90,
W
8ln (21
A nrw
4>00a
coa
sin 27 81?
29 ANALYTIC STATICS 65
or, substituting the value of ZTjust given,
10X6X^X54 = 4,000 ^ ^ gp!
whence,
^ = ^X^|^|~ = 2481ft =2ft 5fm, nearly. Ans.
EXAMPLE 2 A piece of rock ABC, Pig 55, lying on the floor of
a mine Is kept from sliding down by a prop M N Weight of A B C
is 3 tons; inclination of CA to horizontal, 60; inclination of NM,
45; coefficient of friction, .75 Required the pressure on the prop.
SOLUTION The pressure on the prop is equal and opposite to
the reaction F of the prop The inclination of NM to the horizontal
being 45, its inclination VO F to the vertical is likewise 45. Here,
H = 60, theiefore,
K => YOF = 60 + 45 = 105, sin K = cos 15, cos K = - sin 15
Formula 4 of Ait 74 gives
sin 60 -3X cos 60 _ 866 - -f X 5
EXAMPLES FOB PRACTICE
1. What force will keep an iron block weighing 4 tons, placed on
an iron plate inclined at 45 to the horizontal, in a condition of limit-
ing equilibrium with respect to upward motion: (a) if the force is
parallel to the plate? (b) if the force makes with the plate an angle
of 30? Take / - 20.
NOTE In this case, it is more convenient to use formulas Involving /, rather
=i 8394T.
= 3.613 T.
2 If, in the preceding example, the force is parallel to the plate
and acts upwards, what must its magnitude be, that the block may
be on the point of moving downwards? Ans F = 2 263 T.
3. Taking the coefficient of kinetic friction between steel and pine
as .16, what is the least force that can keep a steel block having a
weight of 2 tons moving with constant velocity up a pine plank
inclined at 20 to the horizontal? Ans F - 1,944 Ib
4. A block of marble weighing 1,000 pounds is to be kept moving
with constant velocity up an inclined white-pine plank by a force of
600 pounds; what must be the inclination of the plank, assuming that
the best angle of traction is used, and that / = .45?
Ans, H - 12 SSf
KINEMATICS AND KINETICS
COMPOSITION AND RESOLUTION OF
VELOCITIES
1. Graphic Representation of Velocity. Velocity,
like force, is a vector quantity that is, a quantity having
both magnitude and direction and, like all vector quantities,
can be represented by a straight line, called a vector (see
Fundamental Principles of Mechanics] . The vector is drawn
parallel to the direction of the velocity represented; its
length is made, to any convenient scale, equal to the mag-
nitude of that velocity, and
the arrowhead on the vector
is placed so that it will
point in the direction of
the motion under consid-
eration.
In Fig l.let AB be the
path of a moving point,
and v the velocity the point
has when it occupies the
position P on its path.
The direction of the motion at the instant considered is that
of the tangent PTto the curve A B. The velocity v may be
represented by the vector OM drawn parallel to P T through
any convenient point. If v is expressed in feet per second,
and a scale of 5 feet per second to the inch is adopted, the
Similarly for any
o
other scale.
length of the vector O M should be \ .
5
eOPYRIOHTRD Y INTERNATIONAL TIX^BOOK COMPANY, ALL RIHT RBHRVBD
30
2
KINEMATICS AND KINETICS
30
2. Parallelogram ol Velocities. Let a body or par-
ticle O, Fig. 2, be moving in the direction OX, on a flat
surface A B, with a velocity v, relative to that surface, that
is, in such a manner that, if the line OX, is fixed on the sur-
face A B, the particle O will move in that line describing v,
units of length per unit of time. Let P, be the position of
the moving particle after the time /. Then,
OP, = v>t (1)
While the particle has this motion, let the surface A B move
in the direction O A' with uniform velocity z/ l( and let A' B' be
FIG 2
the position of the surface after the time t, O / being the cor-
responding position of 0, and O'X a f , parallel to OX,, the
corresponding position of OX t > Then,
O0 f = v l t (2)
The moving particle will now be at PJ t the distance O'PJ
being equal to P a , or v, t. The figure O O' PJ P. is evidently
a parallelogram.
In the same manner, it can be shown that, after a time /',
in which the moving particle has described the distance
OQ t along the line OX t) while the surface has moved so
30 KINEMATICS AND KINETICS 3
that O" is the position of the starting point O, the final
position Q t ' of the particle will be the end of the diagonal of
the parallelogram O O" Q, 1 Q t) in which
0"Q,' = OQ a = Vm f (3)
O 0" = Q a Q, f = v, f (4)
Dividing (2) by (1) gives
v t OP, a PS
and dividing (4) by (3) gives
2i = "
v, 0QJ
Equating these two values of ,
v,
oa = o o"
OPJ O"Q,'
According to the theory of similar triangles, the last
equation shows that O, Q, f , and /V are in the same straight
line. It follows, therefore, that, at every instant, the mov-
ing particle is on the straight line O P a f or, what is the
same thing, that the particle moves in that line. The spaces
OP,', OQJ described by the particle in times t and /',
respectively, are to each other as O O r is to O O" that is,
as z/,/ is to Vt ^, or as t is to t'. Therefore, the motion of
the particle along OP,' is a uniform motion. If O O" repre-
sents v lt and OQ a represents v, t the diagonal OQJ will
evidently represent the space described by the particle in a
unit of time that is, the velocity v of the particle along its
path O /V . Hence, the following construction:
From any point P, draw two vectors PMi and PJbf,,
representing, to any convenient scale, the velocities z/i and v,,
respectively. Construct a parallelogram PM l MM* on those
two vectors. Then will the diagonal PM represent, to the
scale adopted, the velocity of the moving particle in its path.
3. It will be observed that the velocity v is determined
by the same general method used for finding the resultant
of two concurrent forces. With respect to the velocities v t
and zr., the velocity v, which is the actual velocity of the
KINEMATICS AND KINETICS
30
particle, is called the resultant velocity, and v t and v t are
called the components of v in the directions OX t and OX,,
respectively. Since the particle has the two velocities z^
and v t at the same time, these velocities are said to be
simultaneous .
The principle of the pavallelogfram of velocities, which
is similar to that of the parallelogram of forces, may be
stated as follows:
// two simultaneous velocities of a particle are represented in
magnitude and direction by two vectors drawn from the same
origin, and a parallelogram is constructed on these two vectors,
the resultant velocity is tepresented in magnitude and direction
by that diagonal of the parallelogram that passes through the
common origin of the two vectors, this diagonal being treated a?
a vector having tJie same origin
4. The process of finding the resultant of two or more
simultaneous velocities is called composition of velocities.
As in the case of forces, any velocity may be considered
to be the resultant of two velocities in any given directions.
Thus, the velocity v t
Fig. 3, may be consid-
ered as the resultant
of the simultaneous
velocities v l and v t in
the direction OX, and
OX,, respectively; or
as the resultant of the
simultaneous veloci-
ties vl and v tt ' in the
direction O AV and
OX,', respectively,
etc. To resolve a
PIG 8 . .^ . .
velocity into its com-
ponents in given directions is to find the values of these
components, or to replace the given velocity with these com-
ponents, the process whereby this is accomplished is called
resolution of velocities.
V*.
30
KINEMATICS AND KINETICS
5. Since velocities are combined and resolved in the
same manner as forces, all that has been said relating to the
composition and resolution of forces applies to the composi-
tion and resolution of velocities. Thus, instead of the par-
allelogram of velocities, the triangle of velocities is often
used. In Fig. 2, for example, the resultant v may be deter-
mined by drawing PM^ to represent z/,, and then MM to
represent v,, and drawing PM. The magnitude of PMcan
be ascertained either graphically (by constructing the par-
allelogram or the triangle accurately to scale), or analytically
(by applying the principles
of trigonometry), as in the
case of forces.
EXAMPLE 1 A ship is pro-
pelled by its screw in a north-
east direction at the rate of 15
knots, while the current carries
it due south at the rate of 5
knots Find the resultant
motion of the ship (A knot is
a velocity of 1 nautical mile, or
6,080 feet, per hour )
GRAPHIC SOLUTION Draw
the north-and-south line NS,
Fig 4 From any point A and
to any convenient scale, draw
A B = 15, making an angle of
45 with A N. This will represent the velocity of 15 knots toward the
northeast From B draw B C due south, that is, parallel to N S, and
equal to 5, using the same scale as before Join A C, and mark the
arrowhead so that it will be in non-cyclic order with A and B C.
The length of A C, measured to the scale used for i and v,, will give
the magnitude of the resultant velocity. The direction is determined
by measuring the angle NA C with a protractor. Ans.
FIG 4
ANALYTIC SOLUTION The tuaugle ABC gives
i) = A C = V15" + fi" - 2 X 5 X 15 cos 45 = 12 knots
Also, sin M
. ain 45 5 sin 45
v 12
The angle M, taken to the nearest minute, may be either 17 8' or
180 - 17 8' = 162 52' As B C, opposite M t is the shortest side
of the triangle, the value 17 & must be taken. Then, NA C
6
KINEMATICS AND KINETICS
830
= 45 + 17 ' = (JU H' The ship, therefore, is moving in .1 dm c luui
N (i2 K' E, at the rate of li! knots, nearly Ans.
EXAMPLB 2 A river 4 miles wide h.is a turiunt velmilj of .'I milts
per houi A boat whose paddle wheels can cany it thiough <"> miles
per houi in still water is to cross the rivei fiom a point ,/, Kig. f>, so
as to land at a point />' dneitly
opposite the starting point A, It
is leqmied to hud: (a) the ilutv-
turn in which the hoal must lu:
headed, (ft) the time lequiic-d for
a trip across the i ivei
KOIUTION () Snue I he veloc-
ity of the cm rent and that im-
parted to the boat by the paddle
wheels are both uniform, the boat
must move along the lim; .,-/ //
with a uniform velocity v (to hi
determined). Let /'be th posi-
tion of the boat at any instant,
and I* Q the diieetum in whieh it
is headed, and let i\ he thevelonty
in]]>arted to it in that direotion by
the paddle wheel. In addition to
,,the boat has a velocity v t down the stieam, t-qiuil to the vdoeity of
the current. The velocity v is the resultant of the. velocities f, and v t ,
The triangle PNN lt being right-angled at N, gives
v = PN = V/ 5 7/; r - //, A/ a V, - r,' i V' - .'!
= fi.1110 mi. per lir.
:/, :i 1 ,
Also, sm M -
(&) The time required is
AR 4_
"" ** 5ilOO ""
(5
AIIH.
nearly. Ant.
EXAMPLES JTOB PKACTICK
1. A point has two aimultaneous velocities, nun of 100 fct't pur
second and one of 200 feet per second. The vectors represent ing
these two velocities make an angle of 4fi with cueh other. Kind:
(a] the resultant velocity v ; (b) its inclination M to iho velocity of
200 feet. (Angles are given to the nearest 10
2. A balloon moves upwards with a velocity of fiO feet par necund,
and at the same time the wind carries it la a horizontal direction at the
30 KINEMATICS AND KINETICS 7
rate of 20 feet per second Find (a) the resultant velocity z/, (6) its
inclination M to the vertical . f (a) v = &3 852 ft per sec.
Ans \ (A) M = 21 48' 1C?'
6. Absolute and Kelatlve Velocity. Velocity, like
motion, is always relative, it represents the rate of motion
of a body with respect to another, and the same velocity can
have different values according to the condition of the objects
to which it is referred. Thus, when a locomotive is running,
the velocity of the piston with respect to the cylinder in
which it moves is the space that the piston describes in the
cylinder per unit of time, the velocity of the piston with
respect to the ground is equal to the velocity of the piston
with respect to the cylinder added to or subtracted from the
velocity that, in common with the whole engine, the cylinder
has with reference to the ground added, if piston and engine
are moving in the same direction; otherwise, subtracted.
7. In nearly all practical questions, it is customary to
refer velocities to the surface of the earth. When the
velocity of a body is given without any qualification, it is
usually understood to be the velocity of the body relative
to the ground. This velocity is customarily, although not
properly, called absolute velocity; while the term relative
velocity is restricted to indicate velocity with respect to
objects that are themselves in motion with respect to the
ground. Thus, in the example of the preceding article, the
relative velocity of the piston is its velocity with respect to
the cylinder; while the absolute velocity of the piston is its
velocity with respect to the ground. In the case represented
m Fig. 2, the velocity of the moving particle, relative to the
surface A J3, is v,; the absolute velocity of the particle is v\
the absolute velocity of AB is /. It is obvious that, if a
body moves on another with a certain relative velocity,
while the latter is in motion, the absolute velocity of the
former body is the resultant of its relative velocity and the
absolute velocity of the other body.
KINEMATICS AND KINETICS 30
UNIFORM MOTION IN A CIRCLE
ANGUXAR VELOCITY
8. Angular Displacement. Let a point P, Fig. 6, be
moving uniformly in a circular path of radius r. The center
T of the circle is C, and the velocity of the
moving point is v. If the point moves
from a position P to a position />, in the
time /, the angle H swept over by the
radius in passing from the position CP
to the position CP,. is called the angular
displacement of P with respect to C.
Angular displacement may be measured
PlQi6 in degrees, but is usually measured
in radians. As explained in Geometry, Part 2, the radian
measure of the angle H is arc , or
r
H (in radians) = arc PP -
r
If the length of the arc PP l is denoted by s t
r r
Also, s = rff (2)
9. Angular Velocity. Since the velocity v is uniform,
s = vt\ that is, rH = vt\ whence
tf-H/ (1)
r
The angular displacement is, then, proportional to the
time, and - is evidently the angular displacement per unit of
time, since, when / is made equal to 1 in the preceding
formula, H becomes equal to -. This displacement per unit
30 KINEMATICS AND KINETICS 9
of time is called the angular velocity of the moving 1 point.
The angular velocity may be defined also as the angle
described by the moving point in a unit of time. As already
stated, it is customary to express angular velocity in radians
per unit of time. It is also customary to represent this
quantity by the Greek letter to (o-md-ga). Therefore (see
formula 1),
= ~ (2)
10. Halations Between Angular and Linear Teloc-
ity. The velocity of a moving point in the direction of the
tangent to its path that is, what in previous articles has
been called the velocity of the point is often called linear
or tangential velocity, in order to distinguish it from
angular velocity. When the word velocity is used without
any qualification, linear velocity is meant The velocity in
the direction of the path, especially when the latter is a circle,
is also called circumferential velocity.
From Art. 9, the following' obvious and very important
relation between linear and angular velocity is obtained:
-* (1)
r
v = 1 to (2)
If two points move in two circles of radii r^ and r, with
the same angular velocity to, and linear velocities v and v,,
then, v l = r, to, and v, = r, CD; whence, by division,
= (3)
V, r,
that is, the linear velocities of two points moving in different
circles with the same angular velocity are directly proportional
to the radii of the respective paths.
11. If the linear velocities are the same, and the angular
velocities are to,, and to,, we must have v = r w, = r a to,;
whence,
^i = TJL
to, r*
that is, for the same linear velocity \ the angular velocities of two
moving points are inversely as the corresponding radii of the paths.
I LT 398-13
10 KINEMATICS AND KINETICS 80
12. If, in Fig. 6, the radius CP is imagined to move
with P t all the points in CP will evidently have the same
angular velocity, since they will all describe the same angle
in the same time. If the linear velocity of the point P',
situated at unit's distance from the center, is v r , then,
v f = 1 X a) = of
The angular velocity of P may, therefore, be defined as
being the linear velocity of, or the length of the arc
described in a unit of time by, a point on the radius CP sit-
uated at unit's distance from the center, when the radius is
considered as moving with the point P.
13. Angular Velocity In Terms of Number of Revo-
lutions Per Unit of Time. In practical engineering prob-
lems, it is customary to state the velocity of circular motion
in terms of the number of revolutions per unit of time
(usually per minute); that is, the number of times that the
moving point passes over an entire circumference in a \init
of time. If this number is denoted by n, and the radius of
the circle by r, the space passed over by the point in a unit
of time is 2 n r X n; hence, for the linear velocity of the
point, we have
v = 2nrn
and for its angular velocity (formula 2 of Art. 10),
CD = - = 2;rw
r
It n is the number of revolutions per minute, the veloc-
ities v and oj, referred to the second, are:
Conversely, if v or CD, referred to the second, is given,
the number of revolutions per minute is given by the
formula
n = ^ = 9.5493 ^ = 9.5493 o> (3)
nr r
14. An$rular Velocity of a Rotating Body. When-
ever any body, as a wheel, revolves about a fixed axis, every
30 KINEMATICS AND KINETICS 11
point in the body describes a circle whose radius is equal to
the distance of the point from the axis. Since all points
revolve through the same angle in the same time, they all
have the same angular velocity. This common angular
velocity is called the angular velocity of the rotating body.
EXAMPLB 1 A drum W, Fig. 7, whose radius J? is 1.5 feet is fixed
to a revolving shaft 5". The motion of W is transmitted to another
shaft s by means of a
belt B B passing around
the drum W and a drum
w fixed on the shaft s.
If the velocity of the belt
is 10 feet per second and
the drum w makes 100
revolutions per minute,
it is required to deter-
mine the number N of
revolutions that S or W makes per minute, the radius r of the drum w,
and the angular velocity of the two drums in radians per second. It
is assumed that the circumferential velocity of each drum is the same
as the velocity of the belt
SOLUTION The circumferential velocities of the two drums, in feet
per second, are, respectively, g^ and "on- (Art 13). Since each
of these velocities is equal to the velocity of the belt,
From (1) we get (see also formula 3 of Art. 13), replacing Jf by
its value 1.5,
N - 9.6493 X ~ = 63 602 A ^ s -
1.5
And from (2) (see the same formula) ,
r - 9.5493 X -^ - .95493 ft. Ans.
For the angular velocity Wj (radians per second) of W, formula 1 of
Art. 1O gives
Wt *" 3? * n " 6 6667< AnSi
And for the angular velocity <a, of w (formula 2 of Art. 13).
ta t - 2 - 10.472. Ans.
12
KINEMATICS AND KINETICS
30
EXAJMPLB 2 Two drums D and ZX, Fig 8, of radii r and r 1 feet,
are fixed on a shaft 55 A rope M N L Q is wound around the drums
in opposite directions, so that when MN rises and winds around D t the
FIG 8
portion descends by unwinding from ZX. The rope passes around
a pulley P carrying a weight W. It is required to find the motion of
the weight W when the shaft 5 5 revolves at the rate of revolutions
per minute
NOTE This arrangement is known by the various named of Alfferoutlal
windlass, differential -wheel and axle, and Chinese wheel and axle,
and is used to obtain a slow motion of IV, which, under some conditions, is n
mechanical advantage
SOLUTION. It is assumed that every turn of the rope around the
drums is a circle, and that M N and Q L remain parallel. This is near
enough to the actual conditions occurring in practice
Let the shaft turn in the direction indicated by the curved arrows.
Then N M will rise and Q L will descend For the angular velocity
of the shaft and the drums (formula 2 of Art 13), we have
ta = 10472 n radians per sec. Let v and v 1 be the velocities otNfif
and QL t or of M and Q, respectively, m feet per second. Then (for-
mula 2 of Art. 10),
v sa r ta = 10472 n r
v> s* #-/ at .10472 nr 1
30 KINEMATICS AND KINETICS 13
During any time f, the center O of the pulley, and therefore the
weight W, moves through a distance OO lt which is easily determined.
The length of the rope between M and Q is reduced by an amount
equal to NNi + L L t = 2 OOi But, as during this time a portion v /
of the rope has wound about D, and a portion v' / unwound from -Z7,
2 O O l = v t - v' f,
whence O Oi = i( v')t
Therefore the point O and the load W move uniformly with a
velocity z/a, equal to \(v v 1 } Substituting the values of v and j/
found above,
Vw = \(r -r')ta = t( 10472 n r - .10472 n r')
- 05236 n(r - r 1 ) Ans
EXAMPLES FOR PRACTICE
1. A flywheel 6 feet in diameter makes 76 revolutions per minute
Find 1 (a) its angular velocity ru, in radians per second, (b) the veloc-
ity v of a point on the rim, In feet per second A _ / (a) u> = 7 854
Ana X (d) v - 23.562
2. A point revolves in a circle 8 feet in diameter with an angular
velocity of 4 5 .radians per second Find: (a) its linear velocity v,
(6) the number n of revolutions it makes per minute.
v ~ 18 ft< P er sec>
ft - 42972
3. Two drums A and .Z? on parallel shafts are driven by a belt
traveling at the rate of 1,000 feet per minute; the radius of A is 5 feet,
and B makes 300 revolutions per minute. Find, (a) the number n of
revolutions that A makes per minute; (d) the angular velocities io a
and <y* of the two drums, m radians per second, (c) the radius r of B
\ (a) n = 31.831
Ana J W/ w " 333S
Ans ' d H<w, = 31.416
(c} r - .531 ft.
CENTRIPETAL AND CENTRIFUGAL FORCE
15. Restatement ol the Law of Inertia. A provi-
sional statement of the law of inertia was made in Funda-
mental Principles of Mechanics. Now that the theory of the
center of gravity has been explained, the law can be expressed
in a more complete and definite manner as follows:
// a body is under the action of no force t its center of gravity
is either at rest or moving uniformly in a straight line,
14
KINEMATICS AND KINETICS
30
16. Centripetal Force. Let a body be- moving in such
a manner that its center of gravity G, Fig. 9, travels uni-
formly in a circle A B of radius r. As examples of this kind
of motion may be mentioned a car moving m a curved track,
and the balls of a ball governor. According to the law of
inertia, such motion cannot take place unless the body is
acted on by unbalanced forces, for, were the body under the
action of no force,
the path of its center
of gravity would be a
straight line instead
of a circle It can be
shown by the use of
advanced mathemat-
ics that, in order to
preserve this motion,
the body must be con-
stantly acted on by a
force P whose line
of action is directed
toward the center of
the circle and passes
through the center of gravity of the body. This force,
which is exerted on the revolving body, is called centrip-
etal force.
If the mass of the body is denoted by m, and the linear
velocity of its center of gravity by v, the magnitude of the
centripetal force P is given by the formula
FIG 9
P =
m v
W
Since m = , where W denotes the weight of the body,
and g the acceleration of gravity, we have also,
P-Z& (1)
gr
In applying this formula, v should be expressed in feet
per second, and r in feet, since g is referred to the foot and
second as units.
30 KINEMATICS AND KINETICS 16
If, instead of v t the angular velocity a) is given, we have,
from Art. 10, v = rto, v* = r* a)'. Substituting this value in
formula 1, and reducing,
To express P m terms of the number n of revolutions per
inute, we have (Art
stituting in formula 2,
minute, we have (Art. 13), <a ^ , and, therefore, sub
30
If the revolving body is acted on by any system of forces,
their resultant must pass through the center of gravity of the
body, be directed toward the center of the circle, and have
the magnitude given by any of the preceding formulas.
17. Illustrations of Centripetal Force. A familiar
example of centripetal force is afforded by the circular motion
of a ball tied to a string and swung around, the other end of
the string being held in the hand. The ball constantly tends
to move along the tangent, or "fly off on the tangent," and
would do so, if it were not kept in the circular path by the
pull of the string.
It should be noticed that, while the centripetal force is
always directed toward the center of the circle, it is not
necessarily exerted from that center; in other words, the
body to the action of which the centripetal force is due does
not need to be at the center. In the case of the ball given
above, the hand, which exerts the centripetal force, is placed
at the center of the circle. In the case of a train moving on
a curved track, on the contrary, the centripetal force is] the
resultant of the weight of the car and the pressure of the
rails on the wheels; and this force, although directed toward
the center of the curve, is exerted at the circumference.
18. Centrifugal Force. By the law of action and
reaction, a particle (or body) moving in a circle must exert
on the body to whose action the centripetal force is due a
reaction equal and opposite to that force. This reaction is
16 KINEMATICS AND KINETICS 30
called centrifugal force. Thus, in the case of the ball
considered in the preceding article, the pull exerted on the
hand through the string is the centrifugal force exerted by
the ball on the hand.
Suppose the source of the centripetal force P, Fig. 9, to
be at the center C of the circle. The force P that acts
on the body G may be represented by the vector G M,
while the centrifugal force Q, which is exerted by the
body G, may be represented by the equal and opposite
vector C N.
19* Caution Against Some Common Misconcep-
tions Regarding Centrifugal Force. The following
facts must be clearly understood, as this subject is one
about which students (and even teachers) often fall into
very gross errors:
In studying' the motion of a body in a circular Path, the
centripetal, not the centrifugal, force must be considered as
the only force acting on the body. In Fig. 9, the body G is
under the action of the force P, not under the action of the
force Q.
A body moving in a circular path has no tendency to move
along the radius, its tendency being to move along the tangent.
In Fig. 9, the body G, if left to itself, would move along G T,
not along G C'.
A body moving in a circle is at every instant under the action
of an unbalanced force (the centripetal force}; as, otherwise,
the center of gravity of the body would move in a straight
line.
Although the centripetal and the centrifugal force are
collinear, equal in magnitude and opposite in direction, they
do not act on the same body, and cannot, therefore, balance
each other. It should be kept in mind that, when it is said
that two or more forces balance, the meaning of the state-
ment is that they produce no motion on the body on which they
all act; and, if several forces act on different independent
bodies, it cannot be said that they form either a balanced or
an unbalanced system, as, under such circumstances, the
30
KINEMATICS AND KINETICS
17
expressions balanced system and unbalanced system have
absolutely no meaning.
EXAMPLE A ball J3 t Pig 10, whose weight is W, is suspended by
a string of length / from a point O, and made to revolve uniformly in
a horizontal circle MN, at "
the rate of n revolutions per
minute It is required to
determine (a) the radius r
of the circle M N, (b) the
centripetal force P acting on
the ball, (c) the tension T
m the string
NOTE This contrivance is
called a conical pendulum,
and its operation Is very similar
to tbat of the ball erovomor of an
engine
SOLUTION (a) The two
forces acting on the ball are
the weight Wai the ball and
the tension 7"of the string.
According to the principles
stated in the preceding
articles, the resultant P of
these two forces must be
directed toward the center C
of the circle, and have the magnitude given by formula 3 of Art
that is,
In the triangle BED,
900 jr
P - W tan V,
or. since tan V = tan B O C
(2)
Equating the second members of (1) and (2),
Wjjt'v^ = _Wr_ .
~900> " <>lP-~r*'
whence V/^
Squaring and solving for r,
18 KINEMATICS AND KINETICS 30
(6) Substituting in equation (2) the values ]ust found for V/ 1 r*
and r,
D __
90Q.g- ~ 900
, .
~ * Ans
In the triangle BDE,
T= ED
EXAMPLES FOR PRACTICE
1 A ball weighing 50 pounds revolves uniformly on a smooth
horizontal surface, about an axis to which the ball is connected by a
steel rod (a) If the rod is 10 feet long and the ball makes 00 revolu-
tions per minute, what is the tension in the rod? (&) If the greatest
tension that the rod can stand is 2,500 pounds, what is the greatest
number of revolutions per minute that the ball can make without
breaking the rod? / t a \ 1,381 Ib.
Ans I W 121
2 The height h (= CO, Pig 10) of a conical pendulum Is 2 feet.
How many revolutions per minute can the pendulum make? (Observe
that the number of revolutions is independent of the weight of the
pendulum and of the length of the suspending string.) Ans. 38.3
3 If, in example 2, the weight of the pendulum is 10 pounds and
the length of the string 16 feet, find (a) the centripetal force P\
(5) the tension T in the string. . / (a) 70 4 Ib
Ans.< ) 801b<
MOTION OF A TRAIN ON A CURVED TRACK
20. General Theory. In Fig. 11 are represented the
rails Ri and R, of a curved track, and the flanges F^ and F t of
two directly opposite wheels of a car. The inclination H, or
^i R t N, of the track to the horizontal is shown very much
exaggerated for the sake of clearness. The weight W/that
comes on the two wheels acts through G, at a height a above
30
KINEMATICS AND KINETICS
19
the track equal to the height of the center of gravity of the
car. The difference NR^. in elevation between the outer rail
Rt and the inner rail R t is called the superelevation of the
outer rail When the car is moving uniformly in the curve,
G moves m a horizontal circle whose radius is practically the
same as the radius of the center line of the track. According
to the theory of centripetal force, the resultant P of all the
forces acting on that part of the car whose weight is carried
by the two wheels here considered must pass through G
and be directed toward the center of the circle in which G
moves; that is, P must be horizontal. For convenience
all forces will be resolved into components parallel and
perpendicular to R^R t . The components of P in these
directions are denoted by X and Y t as shown in the tri-
angle G DE.
The forces acting on the car (or on the part here con-
sidered) are W, which is resolved into the component X, and
y., and the pressures of the rails. The components of these
20 KINEMA ^ICS AN ) KINETICS 30
pressures perpendiculs r 'to R l R, e re denoted by Y l and K,;
the component paralle to or alon^ RI R, is denoted by X'.
If, as shown m the figure, X 1 acts toward R a , the pressure is
exerted by the rail R! on the flange ft; if X' acts toward R lt
the pressure is> exerted by R, on the flange F,. In the gen-
eral derivation of the formulas, X' will be treated as acting
toward R a ; if, in any particular case, the value of X 1 is
found to be negative, this means that X 1 acts in the oppo-
site direction, and that it is, therefore, exerted by the lower
rail R,.
Let v = linear velocity of car, in feet per second,
e = superelevation of outer rail, in feet;
b = R^R, = distance between centers of rails, m feet;
r = radius of center line of track, in feet.
According to the principles of statics, we have, since P
is the resultant of the forces acting on the car,
X' + X = X (1)
Y* + y> - Y, = Y (2)
From (1), X' = X-X, (3)
Now,
X = P cos H = cos H (Art. 16) = X
gr gr
gr b
R l R a b
These values in (3) give
X ' = ~(^F-e*-e} (D
b \gr ]
To find Y lt moments are taken about R a . Since the
moments of X' and Y, about R, are each zero, the algebraic
sum of the moments of Y 1 and W must be equal to the
moment of /*; that is,
Y*b- WXKR, = P^GK (4)
The figure gives, MI being horizontal and MS vertical,
KR* = JR t cos H = (MR, - MJ) cos H
= (i6- MG tan H) cos H = i b cos H- a sin H
30 KINEMATICS AND KINETICS 21
Substituting these values in (4), writing -^-^- for P, and
gr
solving for Y 1}
(2)
The value of Y, is found from equation (2) by substi-
tuting the expressions for the values of Y, Y^ and Y a . The
result is
Y, = ** - *- cos H+ iA + a sin tf (3)
b L\ "/ \ gr I J
21. Car Moving Without Exerting Lateral Pres-
sure. The pressure A'' exerted between the rail and the
flange increases with the velocity v, as is evident from
formula 1 of the preceding article. As this pressure is very
injurious both to the rail and to the wheels, the track is so
constructed as to eliminate it, that is, the superelevation e is
so made that, for the greatest velocity of trams moving on the
track, the pressure X' shall be zero. Putting the second
member of formula 1 of the last article equal to zero,
whence, squaring and solving for e,
b^,
(1)
Usually, (---) is a very small fraction. If this fraction is
\ffr/
neglected, the following approximate formula is obtained:
e = b ?*- (2)
gr
22. When A'' is zero, the algebraic sum of the moments
of Yi and Y, about G is zero, since the moments of both
W and P are zero; and, as the lever arms of Ki and Y, are
equal, it follows that Y l == F a . Also, since the resultant P of
Yi, K B> and W is horizontal, the sum of the vertical compo-
nents of yi and X, must be numerically equal to W, and
22 KINEMATICS AND KINETICS 30
the sum of their horizontal components must be equal to 1
If, therefore, a horizontal line A Q is drawn through A, the
vector Q G will represent Xi + Y,, and the vector O si
will represent the horizontal component of Y l + y a , this
component being equal to P It is thus seen that, in this
case, the centripetal force P is obtained by resolving the
weight W into two components one, GQ, peipendicular
to the track, and one, GJS, horizontal: GQ simply bal-
ances the pressure Y l + Y, of the rails, while GE is the
centripetal force.
23. Car on the Point of Upsetting About the
Outer Wheel. If the velocity increases sufficiently, the
pressure X' will become so great that the car will upset by
turning about the rail R-,.. When the car is on the point of
upsetting, the wheel F, is on, the point of being lifted from
the rail J? at and there is, consequently, no pressure between
the wheel and the rail at J? 3 . The relation that exists
between e and v for this condition is obtained by making
Y a = in formula 3 of Art. 20, expressing cos // and
sm H in terms of e and b, and solving for v or e. The pioe-
ess is comparatively complicated, and will not be given
here. There is, however, an approximate solution that is
sufficiently close for practical purposes. In railroad work,
the angle H is always very small, and its cosine can, with-
out any appreciable error, be taken equal to 1. Putting the
second member of formula 3 of Art. 20, equal to zero, and
writing 1 for cos H and 7 for sin H (see Fig, 11),
b
and
.
gr \ gr I b
d(2a
EXAMPLE 1. A curved track is to be constructed for a maximum
train velocity of 45 miles per hour. If the radius of the curve is
2,600 feet, and the distance between centers of rails 6 feet, what must
30 KINEMATICS AND .KINETICS 23
be the superelevation of the outer rail, that there may be no lateral
pressure?
SOLUTION Formula 2 of Art. 21 will be used Here, b 5;
r = 2,500, g 32 16, and v = ' = 66 ft per sec. Therefore,
66 a
e = 5 X Q4> , A _ Knn = .271 ft = 3i in , nearly Ans .
32 16 X 2,500 (_
EXAMPLE 2 With the track constructed as m example 1, what is
the maximum velocity a train can have without upsetting, the height
of the center of gravity of each car above the track being 6 feet?
SOLUTION The formula m Art. 23 gives
.-f *" + **-** persec .
= 134 mi. per hr., nearly. Ans.
EXAMPLES FOR PRACTICE
1. A curved track of 1,600 feet radius is to be built for a maximum
tram velocity of 40 miles per hour, the distance between rails is 5 feet
What must the superelevation be, that there may be no lateral
pressure? Ans 535 ft
2 The radius of a curved track being 1,500 feet, tne superelevation
3| inches, and the distance between, the centers of rails 5 feet, what is
the maximum velocity a tram can Jiav> without exerting any lateral
pressure? Ans. 36.17 mi. per hr.
WORK AND ENERGY
WORK
24, Definition of Work. Let a force F, supposed to
remain constant in magnitude and direction, act on a body
while the point of application of the force undergoes a
certain displacement. This displacement does not neces-
sarily have the direction of the force, nor is the force F
necessarily the only force acting on the body. Let the pro-
jection of the displacement on a line parallel to the direc-
tion of the force be denoted by s. Then, the product Fs is
called the work of the force F for the given displacement.
Thus, in Pig. 12, if the point of application of the force F
moves from A to A f , its displacement is A A'\ the projection
of this displacement on a line parallel to the line of action
24 KINEMATICS AND KINETICS 30
of the force is AS, A' S being perpendicular to the direc-
tion of the force. In this case, then, s = A S, and the
t work of the force, while
B / the point of application
/^X S moves from A to A 1 ,
*\^ fr, is FX AS. The pro-
/ jection A S is called the
MT component of the dis-
f ,--*'* placement A A' parallel
to the line of action of
the force When, as in
Fig. 12, the projection
Pl - 12 of A' on A S moves in
the same direction as the force, the force is said to do work
on the body on which it acts When, as in Fig 13, the pro-
jection of A 1 on A S moves in a direction opposite to that of
the force, the body is said to do work against the force, or the
force is said to do the negative work FX AS on the body.
If the line of action of the s ^
force F, Figs. 12 and 13, coincide / """"--^^
with the path A A' of the point / "~"^'7^
of application, then s = A A'. If / /' \
the point of application moves /
in the direction of the force, the /
work of the force is F X A A 1 } /
otherwise, Fx A A'. / >
If the line of action of F is / /'
perpendicular to A A f , the projec-
tion A S reduces to the point A,
and F X A S = F X = 0; that
is, the work done by the force
is zero. This shows that, when ' Fr0tW
the point of application of a force moves in a direction per-
pendicular to that of the force, the force does no work,
25. Formulas for Work. Denoting the work of the
force F, Fig. 12, by U, we have
U=FXAS (1)
30 KINEMATICS AND KINETICS 25
If Fis resolved into two components Tand N, the former
along A A', and the latter perpendicular to A A', then,
T
T = /''cos M, and F - -
This value in (1) gives
cos y// cos M
This formula shows that the work of a force can be obtained
by multiplying the component of the force in the direction of the
displacement of its point of application, by the length of that
displacement.
26. Work of the Resultant of Several Forces.
When several forces act on a body, the algebraic sum of their
works is equal to the woik of their resultant. This is
an immediate consequence of the principle stated in the
preceding article. Let F he the resultant of several forces,
say F^ J 7 ,, F a . Let the body on which these forces act move
in any direction through a distance x, and let the correspond-
ing work of the forces be U, /i, /, /. Denoting the com-
ponents of the forces in the direction of the displacement by
X, X lt A' a , X 3 , we have, according to the preceding article,
U = Xx, U, = X t x, U* = A',*, U, = A'.*. Therefore,
#++/. = (X t + Xn + X.)x
According 1 to the principles of statics, Ji 4- X, + X a = X\
therefore, 7 X + /, + U, = Xx = U, as stated^ It should
be remembered that the resultant F'\& a force that can replace
the forces F^F*, F,, not a force acting simultaneously with
them.
Since, when seveial forces are in equilibrium, their result-
ant is zero, it follows that, if a body is moving under the
action of balanced forces (in which case the body must be
moving with uniform velocity), the algebraic sum of the
works of the forces for any displacement of the body is equal
to zero.
27. Effort and Resistances. Usually, forces are
applied to bodies in order to produce or to prevent motion.
The forces thus applied are called efforts, and those that
I L T 398-14
26 KINEMATICS AND KINETICS 30
oppose them are called resistances. Thus, when a train is
moved by a locomotive, the pull of the locomotive is the
effort; the friction of the wheels and the resistance of the
air are the resistances. Generally, when the work done on
a body is referred to, the work done by the effort is meant,
although, in some cases, its value is determined by calcula-
ting the work of the resistance. Thus, if a weight W is
raised through a height h, the work of the resistance W
is numerically Wh> which, if the body comes to rest at the
end of the distance h } is also the work of the effort.
28. Unit of "Work. As work is measured by the prod-
uct of a force and a distance, its numerical value depends
on the units of force and length employed. The unit of
work may be defined as the work done by a force equal to
the unit of force acting through a distance equal to the unit
of length. If the unit of force is the pound, and the unit of
length is the foot, work is expressed in foot-pounds; if the
unit of force is the ton, and the unit of length is the inch,
work is expressed in inch-tons; etc.
29. Work Done In Raising a System of Bodies.
It can be shown by the use of advanced mathematics that
the work performed in raising a system of bodies is equal to the
aggregate weight of all the bodies multiplied by the height
through which their center of gravity is raised. Thus, if stones
piled in any manner at the bottom of a tunnel shaft are
raised and formed into a pile on the surface, the work per-
formed is equal to the combined weight of all the stones
multiplied by the vertical distance between the centers of
gravity of the two piles. The stones may be raised one or
more at a time and deposited on the surface in any manner
whatever; the work will always be the same.
EXAMPLE 1. The average pull of a locomotive on the train being
7 5 tons, what is the work done by the locomotive in hauling the
train 1 mile?
SOLUTION. Here F = 7 5 T., J = 1 mi. = 5,280 ft Therefore,
V - 7.6 x 6,280 = 39,600 ft. -tons. Ana.
30 KINEMATICS AND KINETICS 27
EXAMI'LB ->. fiiven an inclined plane CD, Pig 14, 500 feet long
and inclined to the horizontal at an tingle //of .SO", it is reqtmed to
determine the work necessary tn just pull up from C to D a block
whose weight W is SiO pounds The coefficient of friction / between
the block and the plane
is 2fi
SOIUTION The re-
sistances opposing the
motion of the block are
the weight W and the
friction P The work U
of the effort is numeri-
cally equal to the sum
of the works of those re-
sistances Resolving Jf
into two components A r and T, the foimer noinial to the plane and the
lattei along the plane, \ve have (.see Analytic Statics}, T = Wsln H,
N = H r cos //, /' = fN = fW cos //. The work done against I\
in foot-pounds, is /' X CD fiOO / J , and that done against W is
7'X C/) - . r >(H) T. Therefore,
U -= f)(K) T + r>(H) /' = fiOO (T + P) = 5(M) ( W sin // + / W cos //)
= fi(K) W (sin U + /cos //) = fi()0 X SX) (iln + ,2T> X cos M )
= HH.rtM ft -Ib. Ans.
EXAMPI.K U. A cylmdnctil cistern 10 feet in diameter is filled with
water to u distance of fiO feet above the bottom. What work must be
done in pumping all the water into a tank l> r > feet in diameter, whose
bottom is 1(X) foot above the bottom of the cistern? The weight of
water is taken ocjual to (12.5 pounds per cubic foot.
SorimoN. The- volume F"o the water to be raised iu
(* x 4 10 'x>)
cu. ft.
I'or the wuSght of thiw volume wo have
IT V 1(1"
W = l r X (JB.fi , X 50 X 02.5
i
If y is the height of the water In thu tank after pumping;,
r , it X ir>
V*> 4 y
4 X * X J ' X nO
4 ^ * x 4 X W 10" X 50
whence * - x 1B . - ff x 1B - ' - lfi
Tne rtiHtancu of the center of gravity of the water In the clBtern
from the bottom of the cistern is 1'5 ft. The distance of the center
y
>f gravity of the water in the tauk from the bottom of the tank ia ~.
28 KINEMATICS AND KINETICS '30
*or the distance h between the two centers of gravity we have,
therefore,
Then (Art 29),
U=Wh = 1L ^~ X 60 X 62 5 (?5 + ^^) = 21,135,000 ft -Ib
4 N J X 15 I Ans.
NOTE In practical problems, it Is seldom necessary to use moie than five
significant figures
POWER
30. Work, as defined in Art. 24, has no reference to
time. The work performed in raising a given weight through
a given height is the same whether the time occupied is a
second or an hour. However, when the capacities of diffei-
ent agents for doing work are to be compared, time must be
considered; and to indicate the amount of work performed in
a given time, or the rate of doing work, the term power is
used. If an engine does in an hour twice as much work as
another engine, it is said to have twice the power of the
second engine.
31. Horsepower. Power may be expressed in tenns
of any convenient units, as foot-pounds per second, foot-
tons per minute, etc. Thus, if an engine woiking uniformly
raises 40,000 pounds through a distance of 10 feet in 1 min-
ute, the work it performs per minute is 400,000 foot-pounds,
and its power may be stated as 400,000 foot-pounds per min-
ute. The foot-pound per minute is too small a unit for many
purposes, and a larger unit is more generally used by engi-
neers. This unit is called the horsepower, and is equal to
33,000 foot-pounds per minute, or 550 foot-pounds per
second. The abbreviation for horsepower is H. P.
Let F = effort (or resistance), in pounds;
s = distance, in feet, that point of application is
moved;
/ = time occupied, in minutes;
t, = time occupied, in seconds;
ff = horsepower.
30 KINEMATICS AND KINETICS 29
TT PS Fs
8
EXAMPLE 1 What horsepower is required to pull a tram weighing
400 tons at a speed of a mile per minute, if the resistance at this speed
is 12 pounds per ton?
SOLUTION The force F necessary to overcome the resistance is
400 X 12 = 4,800 Ib s = 6,280 ft Substituting In the foimula,
4,800X5,280 _
H= 33,000 X~r = 768H P ' AnS
EXAMPLE 2 A crane hoists a load of 5 tons a height of 22 feet m
20 seconds What horsepower is developed?
SOLUTION The resistance F is 6 X T= 10,000 Ib.; 20 sec. = i mm
Substituting in the formula,
10,000 X_22 H
H ~ 33,000 X * ** H P
10,000 X 22 _
H = ~~- = 2 H p
ENERGY
32. Kinetic Energy. Let a body of mass m and
weight W be moving 1 with a velocity v a . If a force F is
applied to the body in a direction opposite to the direc-
tion of motion, that force will bring the body to rest in a
distance s, such that
F s = a 1 m V* ( 1 )
This formula, which should be committed to memory, is
obtained from the following, derived in Fundamental Prin-
ciples of Mechanics
F^'Jii^-J^l (2)
2s
In the present case, v = 0, and, therefore,
F = ~~, Fs = - i m v?
2s
The negative sign simply indicates that F and v have
opposite directions. When only numerical values are con-
sidered, the sign is disregarded, and the equation written as
in formula 1 .
Notice that Fs is the work done by the body against the
force in the space j. Whatever the force may be, this work
30 KINEMATICS AND KINETICS 30
is always the same, since it is always equal to i m v a ' It
follows, then, that the body can, on account of its velocity
and mass, perform an amount of work numerically equal
to a" m v,".
33. The capacity that an agent has for performing work
is called enei-gy. The energy that a moving body has, on
account of its velocity and mass, is called the kinetic
energy of the body; and, as ]ust explained, is measured by
the product -3 m v a *. Since this quantity is equivalent to
work, it is expressed in units of work, such as foot-pounds
or foot-tons
Denoting the kinetic energy of a moving body by jR" t and
replacing m by (see Fundamental Principles of Mechanics] ,
g
formula 1 of Art. 32 may be written,
Also, Fs = ^p- (2)
EXAMPLE Find the kinetic energy, m foot-tons, of a car weighing
25,000 pounds, and moving at the rate of 30 miles an hour.
SOLUTION Here, W = 25,000 Ib = -^ tons. The velocity of the
tram, in feet per second, is
30X5,280 , .
60 X 60 ^ ^ V '
Therefore (formula 1),
K = \ X |^- X 44* 376 24 ft -tons. Ans.
NOTU The velocity has been reduced to feet per necond, because jf In expressed
in feet per second We might also find v In feet per hour and reduce jf to feet per
hour In all cases, it is necessary to refer v andjr to the same units,
34. Equation of Energy. From formula 2 of Art. 32,
we have, leplacing m by ,
Fs = (z>* zJo") (1)
2.P"
This formula gives the work performed by an unbalanced
force in changing the velocity of a body from z> to v, or its
30 KINEMATICS AND KINETICS 31
kinetic energy from v a " to v". If the force F is the
difference between an effort P and a resistance R, then,
(P R} s = (v* v ")
*f
whence Ps = R s + -^ ( - v, a ) (2)
2.T
This formula is called the equation of energy, and may
be stated thus:
The work done by the effort zs equal to the work done against
the resistance plus the increase of kinetic energy.
EXAMPLE In hoisting coal from a mine, the load to be hoisted,
including cage and car, is 12,000 pounds, the load starts from rest,
and when it is 50 feet from the bottom, it is moving with a velocity of
30 feet per second What is the pull in the hoisting rope?
SOLUTION. The resistance R is the weight, 12,000 Ib. The effort P
is the unknown pull in the rope, v, = 0, v = 30 ft per sec., and
s = 50 ft Substituting in the formula,
PS = tfj + |(- -.-)_ 12,000 x so + 2 x 32i6 (30> " QI)
= 767,910 ft.-lb.
767,910
50
15,358 2 Ib. Ans.
NOTE In the solution of this example It is assumed that, when the load has
reached a height of 60 feet, it is still being accelerated at the average rate of accelera-
tion necessary to fflve it a velocity of 30 feot per second in a space of 60 feet When
the velocity becomes uniform, the pull in the rope is that due to the load only, since
there la then no Increase in the kinetic energy.
35. Potential Energy. As has been explained, a body
may have a capacity for doing 1 work by reason of its velocity.
There are, however, other states or conditions than that of
motion that give a body a capacity for work: (1) Suppose
that a body of weight W is raised through a height h\ in
thus raising the body, the work Wh is dons against the
force W. If, now, the body is permitted to descend through
the same vertical distance h, the weight W becomes the
acting force, and the work Wh is done by it. The body in
its highest position has the work Wh stored in it, and thus
possesses a stock of energy equal also to W h. (2) Suppose
that a spring is extended or compressed; work is done
against the resistance of the spring to change of lengtH.
32 KINEMATICS AND KINETICS 30
If the spring is released, it will return to its original form,
and in so doing can do precisely the amount of woik that
was expended on it. Thus, in the extended or compressed
state, there is an amount of work stored in the spimg, and
the spring possesses energy because of its sti etched 01
compressed condition. Similarly, compressed air possesses
energy merely because it is compressed and can do work in
returning to its original state.
The energy that a body possesses by reason of its position,
state, or condition is called potential energy.
36. It is to be remarked that the potential energy that a
body has, due to its position, is relative to a certain plane
below the body. Thus, if the body weighs W and its
height above the ground is h, its potential energy with
respect to the ground is Wh With lespect to a plane
whose distance below the body is h lt the potential energy
is
37. Conservation of Enerjyy. The law of the con-
servation of eneigy asserts that energy cannot be destroyed.
When energy apparently disappears, it is found that an
equal amount appears somewhere, though perhaps m another
form. Frequently, potential energy changes to kinetic
energy or vice versa. To illustrate, take the case of a
steam hammer or a pile driver. The ram is raised to a
height h above the pile, and, on being released, strikes the
head of the pile and comes to rest. In its highest position,
the ram has the potential energy Wh In falling, it attains
a velocity v that, just at the instant of striking, has the
magnitude ^%ffh, hence, at this instant, the ram has lost
all its potential energy, but h'as a kinetic energy -- = IV h.
**
The loss of one kind is therefore just balanced by the gain
of the other kind. After the blow, the ram comes to rest
and has neither potential nor kinetic energy. Apparently
there is a loss of energy, but really the energy has for the
most part been expended in doing the work of driving the
pile a certain distance into the earth. A small part has
30 KINEMATICS AND KINETICS 33
been expended in heating the ram and head of the pile,
and another part in producing sound.
Heat is a form of energy due to the rapid vibration of the
molecules of bodies. In many cases where energy appar-
ently disappears, it reappears in the foim of heat Thus, the
work clone against frictional resistances appears as heat,
which is usually dissipated into the air as fast as produced
Conversely, there aie examples of heat energy transformed
into other forms. A pound of coal has potential energy,
which, when the coal is burned, is liberated and changed
into heat energy. The heat applied to water produces steam,
and the steam has potential energy. Finally, the steam m
giving up its energy does work in a steam engine, and this
woik is expended in overcoming the friction of shafts, belts,
and machine parts, and ultimately reappears m the form of
heat. However, in all these transformations, the total
amount of energy the sum of the kinetic and potential
energy remains always the same. Such is the law of the
conservation of energy.
EFFICIENCY
38. In every machine, there is an effort or driving force
that produces motion in the machine, and a resistance against
which the machine does work. Take, for example, the rais-
ing of a load by a crane. The effoit is exerted by the work-
man on the crank of the windlass, and the resistance is the
weight of the load lifted. In the case of a machine tool for
cutting metal, the effort is the pull of the belt on the driving
pulley, and the resistance is the resistance of the metal to
the cutting tool. In a given time, say 1 minute, a quantity
of work is done by the effort and another quantity of work is
done against the resistance. Were it not for frictional resist-
ances, these two works would be just equal. In actual cases,
however, only a part of the woik of the effort is usefully
expended in doing work against the resistance. The remain-
der is used in overcoming the friction between the various
sliding 1 surfaces, journals and bearings, etc. The work thus
34 KINEMATICS AND KINETICS 30
expended appears in the form of heat energy, and serves no
useful purpose.
The ratio ful work Qr useful work
total work of effort total energy supplied
the efficiency of the machine.
Representing the efficiency by E, the useful work by /,
and the total work (= energy supplied) by /, we have,
EXAMPLE A water motor receives 150 cubic feet of water per
second, with a velocity of 45 feet per second If the maximum power
that can be obtained from the motor is 459 H P , what is its efficiency?
SOLUTION The weight of the water delivered per second is
150 X 62 5 Ib , the kmet:c energy of which (Art 33) is
150X626X45' .. , , n
2 ft -Ib per sec ( = U)
Since 1 H P = 650 ft -Ib per sec , the useful work per second done
by the motor is (459 X 550) ft -Ib ( = /) . Therefore,
150 X65X 45'
- 8513 ' r ffi 5 per Centl Ans '
HYDROSTATICS
DEFINITIONS PROPERTIES OF LIQUIDS
1. Hydromechanics is that branch of mechanics that
deals with liquids, their properties, and their applications
to engineering. It may be divided into two chief branches:
hydrostatics and hydraulics.
2. Hydrostatics treats of liquids in a state of rest.
3. Hydraulics treats of liquids in motion.
4. In the first of these divisions are included such prob-
lems as the pressure of water on enclosing vessels and sub-
merged surfaces; in the second are included problems relating
to the flow of liquids through orifices, in pipes, and m channels.
5. !Llquid Bodies. A liquid body, or simply a liquid,
is a body whose molecules change their relative positions
easily, being, however, held in such a state of aggregation
that, although the body can freely change its shape, it retains
a definite and invariable volume, provided the pressure and
temperature are not changed. Water and alcohol are exam-
ples of liquid bodies.
6. A perfect liquid is a liquid without internal friction;
that is, one whose particles can move on one another with
absolute freedom. On account of this characteristic property,
a perfect liquid offers no resistance to a change of form.
7. A viscous liquid is a liquid that offers resistance to
rapid change of form on account of internal friction, or vis-
cosity. Tar, molasses, and glycerine are examples of
viscous liquids.
COPYRIaHTBD BY INTIBNATIONAL TEXTBOOK COMPANY ALU KIOHTB
31
2 HYDROSTATICS 31
All liquids are more or less viscous. For the purposes of
hydrostatics, however, water, which is the liquid mainly dealt
with in this work, may be treated as a perfect liquid, its vis-
cosity at ordinary temperatures being too small to be taken
into account.
8. Compressibility. All liquids offer great resistance
to change m volume; that is, they can be compressed but
little. Under the pressure of one atmosphere (about 14 7
pounds per squaie inch), water is compressed about 20067
of its original volume. Foi engineering purposes, it may
be assumed that water is practically incompressible.
9. Density. The density of a homogeneous body is
the mass of the body per unit of volume, and may be obtained
by dividing the total mass of the body by its volume.
If V, W, ;, and D are, respectively, the volume, weight,
mass, and density of a body, then,
or, since m (see Fundamental Principles of Mechanics),
Both from the definition and from this formula, it follows
that the density of any one body varies inversely as the
volume of the body; in other words, when the mass of the
body remains the same, the density is greater the less
the volume of the body, or the space that the body occupies.
This agrees with the ordinary use of the words density and
dense, which are employed to denote the degree of com-
pactness, a body being said to be more or less dense accord-
ing as its molecules are supposed to be more or less close
to one another.
10. Weiglit of "Water. The weight of 1 cubic foot of
water varies with the temperature: at 39.2 F., which is the
temperature of maximum density, it is 62.425 pounds. For
nearly all engineering purposes, 62.5 pounds is used as a
31 HYDROSTATICS 3
convenient and sufficiently approximate value. This value
will be used throughout this work
Since a. column of water 1 square inch in cross-section
and 1 foot high is -j4r cubic foot, its weight is 62 5 -f- 144
= 434 pound. This value is of very frequent application,
and should be memorized.
LIQUID PRESSURE
PASCAL'S LAW AND ITS APPLICATIONS
11. Statement of Pascal's Law. A perfect liquid
transmit* pressure equally in all directions This principle is
Pusc-aPs lu\v, and follows di-
7 A
rectly fiom the definition of a per- "*
feet liquid. i
The difference between a liquid H
and n solid as regards transmis- \
sion of pressure may be illus-
trated as follows: In Fig. 1 are
shown two cylindrical vessels of
the same size. The vessel a is
fitted loosely with a wooden block
of the same size as the vessel.
The vessel b is filled with water,
whose depth is the same as the
length of the wooden block in a. m * l a
Both vessels are fitted with air- FlGl
tight pistons P. For convenience, let the weight of the block
and that of the water be neglected, and suppose that a force
of 100 pounds is applied to each piston. Assume the piston
area to be 10 square inches; then the pressure per square inch
is 100 -T- 10 <= 10 pounds. In the vessel a, this pressure will
be transmitted, undiminished, to the bottom of the vessel;
it is easy to see that there will be no pressure on the sides.
In the vessel b, an entirely different result is obtained. The
pressure on the bottom will be the same as in the other
HYDROSTATICS
31
case, that is, 10 pounds per square inch; but, owing to the
fact that the molecules of water are perfectly free to move,
this pressure of 10 pounds per square inch is transmitted in
every direction with the same intensity; that is to say, the
pressure at all points, such as c, d, e, f,g,h, due to the external
force of 100 pounds, is exactly the same, namely, 10 pounds
per square inch.
12. Verification of Pascal's Law. An experimental
proof of Pascal's law may be effected with the apparatus
shown in Fig. 2 The vessel is filled with water, and is fitted
with pistons of different
diameters. Let a force be
applied to the small piston c ,
as a result, the water in con-
tact with c is subjected to a
certain pressure According
to Pascal's law, this pressure
must be transmitted with
undimimshed intensity in
all directions. For the sake
16 of convenience, let the area
of the piston c be 1 square
inch, and let a force of 6
pounds be applied to it.
The pressure exerted by the
piston on the water is, there-
fore, 5 pounds per square inch. If the area of the piston a
is 40 square inches, and Pascal's law is true, the water must
exert on the face of a a total pressure of 40 X 5 = 200 pounds.
It is found by experiment that a force of 200 pounds must be
applied to the piston a to prevent it from moving outwards.
This shows that the fluid pressure against a is 200 pounds, or
5 pounds per square inch, which is the same as that against c.
Similar results are obtained with the pistons b, d, <?, and /. If
their areas are, respectively, 7, 6, 8, and 4 square inches, the
pressures exerted on them are found to be, respectively, 35,
30, 40, and 20 pounds.
31 HYDROSTATICS 6
Pascal's law may be formally stated as follows: The pres-
sure i>er unit of area exerted anywhere on a mass of liquid is
transmitted undimimshed in all directions, and any surface in
contact with the liquid will be subjected to this pressure in a
direction at right angles to the surface.
13. Application of Pascal's Law. In Fig. 3, let the
area of the piston a be 1 square inch, and that of b, 40 square
inches. According to Pascal's law, 1 pound placed on a
will balance 40 pounds placed on b. If a moves down-
wards 10 inches, then 10 cubic inches of
water will be forced into "the tube b.
This will be distributed in the tube b in
the form of a cylinder whose cubical
contents must be 10 cubic inches, whose
base has an area of 40 square inches,
and whose altitude must therefore be
10 - 40 = } inch. This shows that a
movement of 10 inches of the piston a
causes a movement of i inch of the
piston b. This is an example of the
familiar principle of work that the force, PlQ 3
or effort, multiplied by the distance through which it moves, is
equal to the resistance multiplied by the distance through which
it moves.
14. The Hydraulic Press. The principle just stated
finds an important application in the hydraulic press, which
is shown in Fig. 4. As the lever a is depressed, the piston b
is forced down on the water in the cylinder c. The water is
forced through the bent tube d into a cylinder fitted with
a large piston e> and causes that piston to rise, the plat-
form / is thus lifted, and the bales placed between it and
an upper fixed platform are compressed. Assume the area
of the piston b to be 1 square inch, and that of the piston e
to be 100 square inches. Also, assume the length of the
lever between the hand and the fulcrum to be ten times the
length between the fulcrum and the piston rod b. If the end
of the lever is depressed, say 10 inches, the piston b is
6
HYDROSTATICS
31
depressed one-tenth of 10 inches, or 1 inch, and the piston e
is raised rihr inch, since 1 cubic inch of water is displaced.
If P denotes the force applied and Q the piessure on the
platform /, then P X 10 inches = Q X Toir inch, hence, Q
= 1,000 P, or P = ToW Q. A force of 40 pounds applied
PIG
by hand thus produces a pressure of 40 X 1,000 = 40,000
pounds. But, if the average movement of the hand per
stroke is 10 inches, it will require 1,000 -r 10 = 100 strokes
to raise the platform 1 inch, which again shows that what is
gained in power is lost in speed.
31
HYDROSTATICS
GENERAL THEORY OF LIQUID PRESSURE
15. Downward Pressure. The weight of a mass of
liquid is a force, and will produce a piessure on any surface
in contact with the liquid independently of the pressure pro-
duced by other forces. Thus, in the case of the vessel shown
in Fig. 2, if the vessel is vertical so that the piston d is at
the bottom, d is subjected, in addition to the pressure of
5 pounds per squaie inch transmitted from the piston c, to an
additional pressure due to the weight of the water above it.
Evidently, the piston e will be under an additional pressure
due to the weight of the water, but the pistons a, b, and /,
being above the liquid, will be
subjected only to transmitted
pressuie.
To arrive at the laws govern-
ing the pressure due to the
weight of a liquid, let us con-
sider first the pressure on the
bottom of a vessel containing
the liquid. In Fig. 5 are shown
two vessels of different shape
but with the water level in each
at the same height 24 feet.
Assume the area of the bottom
of a to be 500 square inches, or $M square feet; then the vol-
ume of the water contained in a is -fH X 24 = 83i cubic feet,
and the weight is 83i X 62.5 = 6,208.3 pounds. The pressure
per square inch is 5,208.3 -5- 500 = 10.42 pounds, nearly.
An easier solution, howevei, is effected as follows: The
weight of a column of water 1 square inch in cross-section
and 1 foot long is .484 pound (Art. 10). Now, above
each square inch of the bottom there is a column of water
24 feet high, whose weight is, therefore, 24 X .434 = 10.42
pounds, and this weight is the pressure per square inch on
the bottom of the vessel,
In vessel 6, assume the cross-section of the lower part
to be 600 square inches and that of the upper part to be
ILT39&-1S
PIG 5
8
HYDROSTATICS 31
100 square inches. It is evident that the volume of water
contained is smaller than before, and it might be concluded
that the pressure on the bottom is, consequently, smaller;
but this conclusion would not be true, as will appear from
the following reasoning: Let water be poured into the
empty vessel until the lower part is filled; the pressure on
the bottom is, according to the method of calculation just
given, 10 X .434 = 4.34 pounds per square inch. Imagine,
now, a thin piston or disk fitted tightly into the smaller part
and resting on the surfa'ce of the water below, and then let
water be poured in until the smaller part is full. This
piston acts as the bottom of a vessel 14 feet high and with
a uniform cross-section of 100 square inches. Hence, the
pressure on this piston, due to the weight of the water above,
is 14 X .434 = 6.08 pounds per square inch. According to
Pascal's law, this pressure is transmitted in all directions,
and therefore acts on the bottom of the vessel; hence, the
total pressure on the bottom is 4.34 + 6.08 = 10.42 pounds
per square inch, the same as in the first case. It is evident
that the result will be the same if the piston is left out.
16. From the reasoning used in the preceding paragraph,
it appears that the pressure per unit of area on the bottom of
a vessel depends only on the vertical distance between the
bottom and the liquid surface, and not at all on the shape of
the vessel. This principle is one of great importance.
17. Intensity of Pressure. When a surface is sub-
jected to a uniformly distributed pressure, the pressure per
unit of area is called the intensity of pressure, or unit
pressure. Usually, the unit of force is the pound, and that
of area is either the square inch or the square foot; and so
the intensity of pressure is expressed either in pounds per
square inch or in pounds per square foot.
In general, if P is the total pressure uniformly distributed
over a surface, and A is the area of the surface, the intensity
of pressure p is given by the formula
31 HYDROSTATICS 9
18. Head. The distance from any horizontal layer of a
liquid body to the surface of the liquid is termed the head
on that layer.
Let h = head, in feet, on any horizontal layer;
p = pressure per square inch on the layer, in pounds;
w = weight of a column of liquid 1 foot long and
1 square inch in cross-section.
Then, p = wh (1)
For water, w = .434, and, therefore,
p = .4347* (2)
Also, h => -f~-r; that is,
.4o4
h = 2.304^ (3)
EXAMPLE 1 The depth of water in a stand pipe is 80 feet
(a) What is the pressure per square inch on the bottom ? (b} What
is the pressure per square inch on a horizontal layer 65 feet from the
surface?
SOLUTION. (a) Substituting the value of h in formula 2,
p = 434 X 80 = 34 72 Ib per sq in. Ans
(b) Here // = 65, hence, substituting in formula 2,
p = 434 X 65 = 28 21 Ib. per sq. in. Ans
EXAMPLE 2. What must be the height of water in a stand pipe to
give a pressure of 80 pounds per square inch on the bottom?
SOLUTION. Substituting in formula 3,
h - 2.304 X 80 = 184 32 ft. Ans.
19. Upward and Lateral Pressure. So far, only
downward pressure has been discussed. It is necessary now
to consider upward pressure, as well as lateral (sidewise)
pressure on vertical surfaces. Let the vessel shown in
Fig. 6 (a) be filled with a liquid to the level a. The part
of the liquid in a b rests on the layer at 6, and produces over
that surface an intensity of pressure of w// t pounds per
square inch, where A x is the head, in feet, on the layer at b.
According to Pascal's law, this intensity of pressure is trans-
mitted to all the bounding surfaces below the level b\ hence,
there is a pressure of wh^ pounds per square inch, due to the
liquid in ad, exerted at c,d,e,f t g, and k } at right angles to
10
HYDROSTATICS
31
the surfaces. At any point below the level b there is, how-
ever, additional pressure due to the weight of the liquid
between the point and the level b. At c, which is at the
same level as b, the total upward pressure per unit of area
is w h lt the same as the downward pressure on the layer at b.
Consider now the layer at de at a distance fi a below the
surface a. The pressure on this layer, due to the weight of
the liquid above, is wh t pounds per square inch, and by
Pascal's law this pressure is transmitted to all parts of the
bounding surface below the level de, just as if the layer dc
were a solid piston. The liquid below de can exert no pres-
sure at the points d and e\ hence, at these points th2 pressure
per unit of area is the same as the downward pressure on the
to
layer de, namely, w h, pounds per square inch. The same
reasoning shows that the lateral pressure per unit of area at
the points / and g is w A 3 , where h, is the head on the layer fg.
The following important law, which is a direct consequence
of that of Pascal, may now be stated:
The intensity of pressure at any point of a surface enclosing a
liquid is normal to that surface; it depends only on the depth of
the point below the surface of the liquid^ and is equal to the
intensity of downward pressure on a horizontal layer of the
liquid having' the same head as the point in question.
The formula p = w h is, therefore, general, and expresses
the intensity of pressure at any point whose distance from
the surface of the liquid is A.
31 HYDROSTATICS 11
EXAMPLE In Pig 6, suppose the depth of the various layers below
the level a to be as follows: depth of t>, 10 feet, of de, 17 feet, of fg t
26 feet, and of k, 30 feet. The liquid being water, what are the
pressures per square inch at the points c, d, <?, /, g , kt
SOLUTION Using formula 2 of Art 18 in each case, the pressures
at the various points are found to be as follows
c = 434 X 10 = 4 34 Ib per sq m. Ans.
d = .434 X 17 = 7.38 Ib. per sq. in Ans.
Pressure at, * = <4S4 X 1? = 7 88 lb P er ** in " AnS
Pressure at, f _ 434 x 25 10 85 lb per sq in. Ans.
g = .434 X 25 = 10 85 lb. per sq in. Ans
k = .434 X 30 = 13 02 lb. per sq in Ans.
20. Graphic Determination of Pressure. The
varying intensity of pressure for different points below the
surface of a liquid may be determined graphically as follows:
Let OM, Fig. 6 (), represent, to any convenient scale, the
head h t on the bottom k of the vessel. Draw MN perpen-
dicular to MO, and representing, to any convenient scale,
the intensity of pressure w h* on the bottom of the vessel.
It should be observed that OM represents a distance and
MN a pressure, and that there is no necessary relation
between the two scales. For instance, OMma.j represent h*
to a scale of 2 feet to the inch; and MN } the pressure w 7i 4
to a scale of tsV inch to the pound. Draw ON. Then, any
horizontal line limited by OM and ON will represent the
intensity of pressure at any depth equal to the vertical
distance of that line below O. Thus, CJ3 t whose distance
below is A,, represents w k^ Likewise, D 2? represents w h t .
If the pressure scale is -sV inch to the pound, it will be
found that the line FG measures about H, which represents
11 pounds per square inch. For very accurate work, the scale
should be as large as possible. It should be observed that
the tangent of the angle MONis numerically equal to w,
since
MO hi
21. Pressure Due to Matternal Load, If the surface
of a liquid is subjected to a pressure, this pressure, accord-
ing to Pascal's lav?, is transmitted undiminished to all parts
2 HYDROSTATICS 81
f the enclosing vessel, and must be added to the pressure
ue to the weight of the liquid. Assume the surface a of
'ig. 6 to be subjected to a pressure of 30 pounds per square
ich. In the example of Art. 19, the pressure at c, due to the
ead of water, is 4.34 pounds per square inch. Adding the
D pounds per square inch due to the external force applied
t a, the result is 4.34 + 30 = 34 34 pounds per square inch
.ikewise, the pressure at d and e is 7.38 + 30 = 37 38 pounds
er square inch; the pressure at / and g is 10 85 + 30 = 40.85
ounds per square inch; and the pressure at k is 13.02 + 30
= 43.02 pounds per square inch.
Let G = total load on surface of liquid, in pounds;
A = area of surface loaded, in square inches;
/^
pt = = intensity of pressure on surface, in pounds
A
per square inch;
P = intensity of pressure, in pounds per square inch,
at a point h feet below liquid surface.
Then, p = wh+p* = wh + (1)
When the liquid is water,
p = .434^+A = .434>% + (2)
A
EXAMPLE A vessel filled with salt water weighing 1 03 times as
uch as fresh water has a circular bottom 13 inches in diameter The
p of the vessel is fitted with a piston 3 inches in diameter, on which
laid a weight of 75 pounds. What is the intensity of pressure on the
ttom, if the depth of the vessel is 18 inches?
SOLUTION In this case, w = 1 03 X 434 = .447 lb.; h = 18 in.
1 6 ft , and A = 3 1 X 7864 sq. m By formula 1,
p = .447 X 1 5 + 3 , x ?6 7854 - 11.28 lb per sq. in. Ans,
22. Atmospheric Pressure. When a liquid is exposed
the air, as in the cases represented in Figs. 5 and 6, there
an external pressure, due to the weight of the air, on the
rface of the liquid. This pressure is called atmospheric
essure. It varies according to locality and atmospheri
nditions, but, for nearly all practical purposes, it may
taken as 14.7 pounds per square inch. This pressure is
31
HYDROSTATICS
13
transmitted by the liquid and should be treated as the pres-
sure p n considered in the preceding article. It is customary,
however, in hydrostatic and hydraulic calculations to take the
atmospheric pressure as the zero of reference. When the
pressure of a liquid is referred to, the pressure in excess of
the atmospheric pressure is meant. This is called gauge
pi-essure. When the atmospheric pressure is added to the
gauge pressure, the resultant pressure, which is really the
total pressure on the liquid or surface considered, is called
absolute pressure. Thus, a gauge pressure of 20 pounds
per square inch is equivalent to an absolute pressure of
20 + 14.7 = 34.7 pounds per square inch.
23. Equilibrium of Liquids at Best. Since the
pressure on a horizontal layer due to the weight of a liquid
PIG 7
is dependent only on the height of the liquid, and not on the
shape of the vessel, it follows that, if a vessel has a number
of radiating tubes, as in Fig. 7, the water in each tube will
be at the same level, no matter what may be the shape of
the tubes. For, if the water is higher in one tube than in
the others, the downward pressure at the level , due to the
height of water m this tube, is greater than that due to the
height of the water in the other tubes. This excess of pres-
sure will cause a flow toward the other tubes, which will
continue until there is no further excess that is, until the
14 HYDROSTATICS 31
free surfaces are at the same level. Then the liquid will
come to rest, and will be in equilibrium. The principle
here stated is embodied in the familial saying, Wafer seeks
tts level. This principle explains why city reservoirs are
located on high elevations, and why water, when leaving
the hose nozzles, spouts so high. If there were no resist-
ance by friction and air, the water would rise to a height
equal to the level of the reservoir If a long pipe with a
length equal to the vertical distance between the nozzle and
the level of the water in the reservoir were attached to the
nozzle and held vertically, the water would just reach the end
of the pipe. If the pipe were lowered slightly, the water
would trickle out. Fountains, canal locks, and artesian wells
are examples of the application of this principle.
EXAMPLE 1 The water in a city reservoir is 150 feet above a hydrant
What is the pressure per square inch at the hydrant?
SOLUTION By formula 2, Art 18,
p = 434 X 150 = 65 1 Ib per sq in Ans
EXAMPLE 2 The pressure on a water mam, when the water is not
flowing, is shown by a gauge to be 72 pounds per square inch What
is the elevation of the reservoir above the mam?
SOLUTION. By formula 3, Art 18,
h = 2.304 X 72 = 165.89 ft Ans.
EXAMPUJ8 FOR PRACTICE
1. What is the intensity of pressure on the bottom of a stand pipe
90 feet high? Ans 39 06 Ib, per sq In.
2. A cylindrical vessel 15 inches m diameter is filled with water.
The top of the vessel is fitted with a piston on which is laid a weight
of 300 pounds. The depth of the vessel being 24 inches, determine
the intensity of pressure on the bottom, of the vessel
Ans 2 566 Ib per sq. in.
3. If the intensity of pressure, due to an external load, on a vessel
filled with water is 30 pounds per square inch, what is the intensity of
pressure at a point 12 inches below the surface of the water?
Ans. 30.434 Ib. per sq. in
31
HYDROSTATICS
15
PRESSURE ON AN IMMERSED SURFACE
24. Total Pressure on a Flat Immersed Surface.
Let M N, Fig. 8, be a plate immersed in water. It is pro-
posed to determine the total pressure acting on the upper
surface of this plate. In the first place, the total piessure />,
being the resultant of the pressures acting at all the points
of M N, is perpendicular to MN. The magnitude of P is
determined by means of the following principle, which is
derived by advanced mathematics:
The total pressure on any plane surface immersed in a liquid
is equal to the product of the following quantities ( 1 ) the area
of the surface, (2) the dis-
tance of the center of gravity
of the surface ftom the level
of the liquid, (5) the
weight of the liquid per unit
of -volume.
In applying this princi-
ple, length, area, and vol-
ume should be referred to
the same unit. Thus, if dis-
tances are expressed in
feet, areas should be ex-
pressed in square feet and volumes in cubic feet. The princi-
ple obviously applies to the pressure on the bounding sur-
faces of the vessel or receptacle containing the liquid as well
as to the pressure on the immersed surface.
Let A be the area of the surface M N> Fig 8; G, the center
of gravity of that surface, its distance from the level of the
liquid being h a \ and W, the weight of the liquid per unit of
volume. Then, the foregoing principle may be stated in
symbols thus:
P = A h e W
It should be particularly noticed that the total pressure P
does not act through the center of gravity G. The deter-
mination of the point of application of P will be dealt with
farther on.
Pro. 8
16
HYDROSTATICS
31
EXAMPLB To find the total pressure on the side ML N K oi the
rectangular tank shown in Fig 9, the dimensions being as indicated.
SOLUTION Since ML N K 'is a rectangle, its center of gravity G is
at a distance from the upper side L N equal to 5, or 6 25 ft 2
LI
Also, A = MK X KN = 3 2 X 6.25,
= 3 125 ft.; hence, h r = 3 125 ft
and W = 62 5 Ib
Substituting these values in the formula for P,
P = 3 2 X 6.25 X 3 125 X 62 5 = 3,906 Ib , nearly
Ans.
25. Center ol Pressure. The point where the line
of action of the total or resultant pressure acting on an
immersed surface meets that surface is
called the center of pressure. This
point does not coincide with the center
of gravity of the surface unless the
latter is horizontal In all other cases,
the center of pressure lies below the
center of gravity. Thus, in Fig. 8,
the center of pressure C lies below the
center of gravity G. A general for-
mula for determining the position of
the center of pressure cannot be either
derived or applied without the use of
advanced mathematics. Special for-
mulas applying to some important cases are given in the fol-
lowing articles. The distance of the center of pressure from
the level of the liquid will in all cases be denoted by A as
shown in Fig. 8.
26. Center of Pressure ol a Rectangular Surface.
In Fig. 10 is represented a vessel V containing a liquid.
The bounding surface RSTU is shown inclined to the ver-
tical; but the formulas derived in this and the following two
articles apply both to an inclined and to a vertical surface.
Let K L M N be a rectangular part of the inclined surface,
the sides K L and M N being horizontal The line EFYis
an axis passing through the middle points of K L and M N.
The heads on the points E and F are denoted by hi and Jt t ,
respectively. It should be observed that the formulas given
31
HYDROSTATICS
17
below apply whether the rectangular surface considered forms
part of one of the bounding surfaces of the vessel or whether
it is the surface of any immersed body.
The center of pressure Cis on the line E F, and its depth h c
below the level of the liquid is given by the formula
CD
To find the distance FC, draw FHQ perpendicular to the
JZ
FlQ 10
verticals through F t C t and ., and denote the length EF, or
f* T
NK, by 5. The similar triangles FCH and FEQ give ^
/ y jD
~^r] whence, the following equa-
= -\ that is,
EQ b
tion obtains:
PC
' X
(2)
If the value of h e found by formula 1 is substituted in
formula 2, the result, after several transformations have
been made, is
X* (3)
18
HYDROSTATICS
31
27. If the edge K ' L is flush with the surface of the
water, h* = 0, and formulas 1 and 3 of the preceding article
become, respectively,
A. -iXv- 1 ! = *A, (1)
FC=
'
(2)
28. Center of Pressure of Triangular Surface.
Only the case of an isosceles triangle with its base horizontal
will be treated in this Section. And first, a triangle MFN t
FIG 11
Fig. 11, with its vertex F above its base will be considered
The line EFYis perpendicular to the base MN The alti-
tude FE is denoted by a The rest of the notation is
similar to that in Fig. 10. The center of pressure is on the
line FE, and its depth below the surface of the liquid is
given by the formula
h t = 3/ '' a +-' (1)
The distance PC is determined in the same way as for the
rectangle treated in Art 26, the result being
FC =
fit. A,
(2)
31
HYDROSTATICS
19
29. If the vertex is flush with the surface of the liquid,
//, = 0, and the two formulas of the preceding article become
7 3 til n i / t X
h c - = i /it ( 1 )
4 fli
30. Let, now, the vertex be below the base, as shown
in Fig 12. With the same notation as before,
_ R
.. i- , -.~ T ~" ?s " rm /\
.J'..i _A_____^__ f __/.__\
1 i '
1 i '
! i '/ ,
* ^ /
! /rJ y
S
PIG. 12
3 A. 1
, i
2
i h t
(1)
(2)
31. If the base is on the surface of the liquid, h t 0, and
the two formulas of the preceding article become
h e = %hi (1)
B *xi (2)
32. Center of Pressure of Circular Surface. Let
M ' N, Fig. 13, be a circular surface. The line EFY con-
tains the lowest point E and the highest point F of the circle.
S30
HYDROSTATICS
31
The diameter of the circle is denoted by d. The rest of the
notation is the same as in previous articles.
The center of pressure is on the diameter EF, and the
distances k e and FC are given by the formulas
7B
'. T-T (D
(2)
FC = ' ~ ' X
kt h.
/js'^T^r,^-'^" -^ ,.""'
A 1 ,*: V . " .' x T-.F- -/ -\
r> * .* ,-. i i / /
' - i / /
i \\ !/ / )tf
I H/fc, /
/'l^
PlQ 18
33. If the circle just touches the surface of the liquid,
h t = 0, and formulas 1 and 2 of the preceding article
become
+& (1)
h c =
EXAMPLE. The vertical circular plate a, Fig. 14, 8 inches In diam-
eter, covers an opening of equal size. The center G is 24 inches
below the surface of the -water, and 20 inches below the hinge b,
What horizontal force applied at c is sufficient to move the plate ?
SOLUTION. Area of plate is 50 266 sq. in Head above center of
gravity is 2 ft. Then (Art. 24), the total pressure on the plate is
X 62.5 X 2 = 43.63 Ib.
144
31
HYDROSTATICS
21
Here, A, = 24 + f = 28 in , and A, = 24 - f = 20 m The depth
of the center of pressure below the surface is, by formula 1, Art 32,
h e = 24 167 in , nearly The center of pressure is 167 in below the
center of gravity G, and, therefore, 20 167 m. below the hinge b The
' ;_ _y moment of P, the total pressure, about
r the hinge must be equal to the moment
of .P 1 about the hinge, hence,
FX 40 = P X 20 167 = 43 63
X 20.167 = 879 89 in -Ib ;
whence, F = 879 89 -*- 40 = 22
Ib , nearly Ans.
5
$ 34. Plates 01- Gates With
Liquid Press xi i-e on Both
Sides. The plate A B, Fig. 15
is subjected to water pressure
with the level a' on one side and
a!' on the other. The total pres-
_ _ B ~ -^
FIG 14 FIG. 15
sure on one side is /", acting through the center of pres-
sure C'\ the total pressure on the other side is P' r , acting
through the center of pressure C". The heads on the centers
of pressures O and C" are denoted by h e ' and h e ", respect-
ively; and the difference between the levels a 1 and a" is
denoted by e. The two horizontal forces P' and P" have a
resultant P whose magnitude is P' P". Let h t be the
depth, below a f , of the point of application C of this result-
ant. Taking moments about the point E> we have,
Ph e = /"^'-/>"U," + *);
P' h f P" ( h it 4- e\
p /l < ( + e
whence,
h t =
22
HYDROSTATICS
31
EXAMPLE Let the gate in Fig 15 be rectangular and 6 feet wide;
let the depth BE be 9 feet and the difference *, 2 4 feet. Required
the pressures /*, P", P, and the distance h e
SOLUTION From formula of Art 24, the total pressure on the
right side is
/ J/ = 0X6x625Xo = 15,188 Ib. Ans
The total pressure on the left side is
P" = 6X66x625X-fj- = 8,167.5 Ib Ans.
The resultant force is
P = 15,188 - 8,167 5 = 7,020 5 Ib
From formula 1 of Art 27, h c ' - % X 9 = 6, and h = X 6 6
= 44, also, e = 9 - 6.6 = 2 4. Substituting in formula of Art 34,
. 15,188 X 6 - 8,167 5 X (4 4 + 2 4) ?
7,020.5
35. Assume the plate B, Fig 16, to be entirely below
the water levels a and b,
and, therefore, subjected
to pressure on both sides.
Let the triangle OM ' N be
drawn to represent the
variation in pressure due
to the weight of the water
on the left-hand side (see
Art. 20), and let the tri-
angle 0' M f N' be drawn
to represent the variation
of pressure due to the weight of the water on the right-hand
side Since the angles M O N and M' O 1 N' are equal, the
tangent of each being w (Art. 20), it follows that O 1 N' and
ON are parallel. Take M0 n = M 1 1 , and through O" draw
O"N" parallel to O ' N'\ then, the triangles O 1 M' N 1 and
O" MN" are equal. Now, the intensity of pressure on the
left of the gate, at any level, as x t below the surface, is given
by the intercept K ' E, and that on the right by the intercept
DE = D'E'. The resultant intensity of pressure at that
point is the difference KE DE KD. Since ON and
O" N" are parallel, the horizontal intercepts between them
are equal, which means that the intensity of pressure is the
_ _
N'
M
FIG 16
31 HYDROSTATICS 23
same at all points of the plate B It follows that the center
of pressure for such a submerged plate coincides with its
center of gravity.
The total resultant pressure P on the plate is found as
follows: The intercept SO" represents the intensity of pres-
sure on the left side at the level /?, that is, for the head e,
which is the difference between the levels a and b. Hence,
since the intensity of pressure on B is everywhere equal
to KD, or SO", or we, it follows that, if the area of B
is denoted by A, the total pies&ure is
P = Awe
EXAMPLE In the example of Art 33 (see Fig 14) , suppose there
is water on the right-hand side of the partition If the level of this
water surface is 8 inches lower than that of the level d, find the total
resultant pressure on the plate, and also the force F
SOLUTION The head e = 8 in = f ft The total pressure is
A w e - 50 206 X 434 X ij = 14 54 Ib Ans
This pressure acts through the center of gravity, 20 ill below the
hinge, hence, taking moments about the hinge,
F X 40 = P X 20,
whence, F=\IP=\P= ^~ = 7.27 Ib Ans.
36. Pressure on Surfaces That Are Not Plane. If
the surface sustaining liquid pressure is not plane, but is
(b)
Fio 17
curved or irregular, the total pressure on it in any direction,
neglecting the pressure due to the weight of the liquid, is the
same as the total pressure would be on the projection of the surface
on a plane at light angles to the given direction. To illustrate
this statement, consider the three pistons in Fig. 17. At (a) is
shown a piston with a curved end, at (b), a piston with
an irregular end; and at (f), a piston with a flat end. In
each case, the projection of the surface sustaining pressure,
I L T 398-W
24 HYDROSTATICS 31
on a plane perpendicular to the piston rod, is the circular
cross-section of the cylinder, and if the pressures per unit
of area in the cylinders are the same, the pressure on the
face of the piston is the same in each case. It is assumed
that the pressure per unit of area is the same at all points
of the surface; that is, the change of pressure due to the
varying depth of the liquid is neglected With this restnc-
tion, the law applies to all fluids, both liquid and gaseous.
If the curved or irregular surface forms a part of the wall
of a vessel, or is submerged, so that the pressure on it is
due to the weight of the liquid, the law of Art. 24 is not
true except for a projection on a vertical plane.
37. The whole subject of pressure in any direction on a
curved or irregular surface may be summarized as follows:
1. If the external pressure is so great that the pressure
due merely to the weight of the liquid may be neglected, the
total pressure in any direction is equal .to the product of the
projection of the surface on a plane perpendicular to that
direction, and the pressure per unit of area.
2. When the pressure is due wholly, or in part, to the
weight of the liquid, as m the case of submerged surfaces,
the total pressure in any direction cannot, in general, be
determined except for regularly curved surfaces, such as
spheres, cylinders, cones, etc., and for these the computa-
tions are difficult and must be made by advanced mathematics.
3. In one direction, however the horizontal the total
pressure on a surface is easily found. It is precisely the
same as the horizontal pressure on the projection of the
given surface on a vertical plane.
EXAMPLES FOR PRACTICE
\ What is the pressure on a layer of water 33 feet below tne
surface? Ana 14.32 Ib. per sq. In.
2 What elevation of a reservoir is required to make the pressure
in a. water main 40 pounds per square inch? Ans. 92.16 ft
3. A vertical triangular plate forms the side of a vessel that con-
tains water The base of the tnangle is 4 feet long and lies at the
31 HYDROSTATICS 25
water level; the vertex is 4^ feet below the water level (a) What is
the total pressure on the triangle? (0) What is the distance of the
center of pressure below the water surface? A \ (a) 843 75 Ib
S U*) 2J ft
4. A circular plate 2 feet in diameter is held vertically so that its
upper edge is 5 feet below the liquid surface Find the depth of the
center of pressure Ans ft ^ in
5 In the gate shown in Fig 15, the depth of the head-water is
7 feet, and that of the tail-water is 6 feet (a) What is the resultant
pressure on the gate per foot of length? (f>) How far above the
bottom B is the line of action of the resultant?
An8 /() 760 Ib
Ans \ (6) 3 03 ft
6 The diameter of the plunger of a hydraulic press used m an
engineering establishment is 12 inches. Water is forced into the
cylinder of the press by means of a small pump having a plunger
whose diameter is f inch What pressure is exerted by the large
plunger when the force acting on the small plunger is 125 pounds?
Ans 32,000 Ib.
BUOYANT EFFORT OF LIQUIDS
IMMERSION AND FLOTATION
38. Principle of Archimedes. In a mass of liquid at
rest, suppose a part M N, Fig. 18, of the liquid to become
solid without changing its form or density.
This solid part, having the same density as
before, will be held in equilibrium, or
remain at rest. Let B denote the weight
of the solid part; D, the total downward
pressure; and U, the total upward pressure
on this part. The oidinary static condi-
tions of equilibrium require that the
upward force U should balance the down-
ward forces D and B\ that is, -
U=D + B\ F <o-w
whence, /- D = B (1)
The difference U L) between the upward and the down-
ward pressure is called the buoyant effort of the body MN.
Suppose, now, that the imaginary solid part of the water
is replaced by an actual solid body having precisely the same
26 HYDROSTATICS 31
shape and volume, and let the weight of this body equal W.
The solid will be subjected to the same vertical pressures
D and /as was the part M N of the liquid. Consider, now,
the vertical forces acting on it. The weight \V and the
pressure D are downwards, and tend to cause the body to
sink; the pressure U is upwards, and tends to cause the
body to rise. The resultant downward force is
D+ W- U = W-(U-D) = W-B (2)
The weight B is the weight of a mass of liquid whose
volume is equal to that of the solid, or, what is the same
thing, B is the weight of the liquid displaced by the solid.
The following principle may therefore be stated-
When a solid body is immersed in a liquid, a buoyant effort
equal to the weight of the hqmd displaced acts upwanh and
opposes the action of gravity. The weight of a body, as shown
by a scale, is decreased by an amount equal to the buoyant effot t,
that is, by an amount equal to the -weight of liquid displaicd.
This principle is called the principle of Archimedes, from
the name of its discoverer.
EXAMPLE In Fig 18 is shown a cube immersed in water Ihe
edge is 6 inches, the sides are vertical, and the lateral pressures me
balanced On the upper face, there is a downward pressure I) due
to the head of water of 15 inches, and on the lower face there is an
upward pressure U due to the head of 21 inches. To fancl the buoyant
effort J
SOLUTION. The downward pressure due to the head of 1C in. or
H ft , is ''
6' X 434 X It = 19 53 Ib
The upward pressure on the lower face due to a head of 21 in., or
1-f ft , is
6 X 434 X 1* = 27 34 Ib.
The buoyant effort of the body is, therefore,
27 34 - 19 53 = 7 81 Ib. Ans.
Also, the buoyant effort is equal to the weight of the water dis-
placed by the body The volume of the cube is (j-) or } cu. ft The
weight of the water displaced is
i X 62 5 7 81 Ib. Ans
39. The principle of Archimedes may be experimentally
verified with a beam balance, as shown in Fig. 19. From
one scale pan, suspend a hollow metallic cylinder l> and
31
HYDROSTATICS
27
below that a solid cylinder a of the same size as the hollow
part of the upper cylinder. Put weights m the other scale
pan until they exactly balance the two cylinders If a is
immersed in water, the scale pan containing the weights will
descend, showing that a has lost some of its weight. Now
fill / with water; the volume of water that can be poured
into t is obviously equal to that displaced by a. The scale
pan containing the
weights will rise gradu-
ally until / is filled, when
the scales will balance
again
40. If a body im-
mersed in a liquid has
the same weight as the
liquid it displaces, then
IV = B, the resultant
vertical force IV B is
zero, and the body will
remain at rest at any
depth below the surface.
If the body is heavier
than the liquid it displaces, W is greater than .5, and the
resultant vertical force W B is downwards, hence, the body
will sink to the bottom. If the body is lighter than the
liquid it displaces, B is greater than W, the resultant vertical
force B IV is upwards, and the body will rise to the surface.
41. An interesting experiment in confirmation of the
conclusions just derived may be performed as follows: Drop
an egg into a glass jar filled with fresh water. The mean
density of the egg being a little greater than that of fresh
water, the egg will fall to the bottom. Now dissolve salt
in the water, stirring the mixture; as soon as the salt water
becomes denser than the egg the latter begins to rise. If
fresh water is then ponied in until the egg and water have
the same density, the egg will remain stationary in any
position it may be placed below the surface of the water.
PIG 19
28 HYDROSTATICS ' 31
42. Floating Bodies. A body lighter than a liquid,
bulk for bulk, rises to the surface, when immersed, and
floats. For equilibrium, the buoyant effect B must be just
equal to the weight W of the body. But, since B is the
weight of liquid displaced, it follows that the weight of the
liquid displaced by a floating body is equal to the weight of
the body.
The depth of a floating body m a liquid depends on the
relative weights of equal volumes of the body and the liquid.
If the body is nearly as heavy as the liquid, it will sink until
it displaces nearly its whole volume; if very light, compared
with the liquid, the larger part of the body will be above the
liquid surface. For example, the density of ice being about
nine-tenths that of water, about one-tenth of an iceberg
appears above the surface and nine-tenths is submerged.
The density of pine is about one-half that of water; hence,
about one-half of a floating pine log is submerged, and
one-half is above water.
EXAMPLE 1. Water-tight canvas air bags are used for raising
sunken ships These bags are sunk when collapsed, attached to the
ship by divers and then filled with air from the pumps above, (a) If
the capacity of a bag is 200 cubic feet, what Is the buoyant effort?
(b) How many bags will be required to lift 600 tons?
SOLUTION. (a) The weight of the bag and the enclosed air may
be neglected. The buoyant effort is the weight of the water displaced
by the full bag, that is,
200 X 62.5 - 12,500 Ib. Ans.
(b) To raise 600 tons,
600 X 2,000 __ .
12 500 = ^ are necessarv ' Ans.
EXAMPLE 2 A cast-iron cylinder 14 inches long and 8 inches in
diameter is closed at the ends, and the metal is i inch thick through-
out. Will the cylinder float or sink in water?
SOLUTION. The volume of the entire cylinder is
7854 X 8 1 X 14 = 703.72 cu. in.
The hollow portion has a length of 18 in. and a diameter of 7 in.;
tts volume is, therefore,
.7854 X 7* X 13 - 600.80 cu. !n.
The volume of metal is
703.72 - 600.30 = 203.42 cu. in.
31 HYDROSTATICS 29
Taking the weight of cast iron as 450 Ib per cu. ft., the weight of
the cylinder is
450 X ^|| = 52.97 Ib
If Immersed, the cylinder displaces 703.72 cu. in of water, which
weighs
703 T>
62 5 X j~~ = 25.45 Ib
The buoyant effort being less than the weight, the cylinder will
sink Ans.
SPECIFIC GRAVITY
43. Definition. The specific gravity of a solid or
of a liquid substance is the ratio of the weight of any volume
of that substance to the weight of an equal volume of water.
This ratio varies slightly with temperature, and the values
given by physicists are exactly correct only for a temperature
of about 39 F , at which water has its maximum density.
For practical purposes, however, it is not necessary to take
changes of temperature into account. The abbreviation
Sp. Gr is often used for specific gravity.
44. Specific Gravity of Solids Not Soluble in
Water. The principle of Archimedes affords a very easy
manner of determining 1 the specific gravity of a solid not
soluble in water. The body is first weighed in air; it is then
attached to a scale pan and weighed in water. The difference
between the two weights will be the weight of an equal volume of
water. The ratio of the weight in air to the difference thus found
will be the specific gravity.
Let W be the weight of the body in air and W the weight
in water; then W W is the weight of a volume of water
equal to the volume of the solid, and
Sp. Gr.
** w- W
EXAMPLE. A body weighs in air 364- ounces and in water 30 ounces.
What is its specific gravity?
SOLUTION. Here W 36}, and W 30. Substituting In the
formula,
Sp. *. - ** - f - 6.8. AM.
30 HYDROSTATICS 3J
45. If the body is lighter than water, a piece of iron or
other heavy substance must be attached to it, sufficiently
heavy to sink it. Then the two bodies are weighed together
both in air and in water, both are weighed separately in air,
and the heavier body in water. Subtracting the combined
weight of the bodies in water from then combined weight
in air, the result will be the weight of a volume of water
equal to the volume of the two bodies. The difference
between the weight of the heavy body m air and in water
gives the weight of a volume of water equal to the volume
of the heavy body. Subtracting this last result from the
former, the result will be the weight of a volume of water
equal to the volume of the light body. The weight of the
light body in air divided by the weight of an equal volume
of water, as just determined, is the specific gravity of the
light body.
Let W = weight of both bodies in air;
W = weight of both bodies in water;
w = weight of light body in air;
Wi = weight of heavy body in air;
W, = weight of heavy body in water.
Then, the specific gravity of the light body is given by the
formula
Sp. Gr =
w
(W- W}-(W l - W,)
EXAMPLE. A piece of cork weighs, in air, 4 8 ounces To it is
attached a piece of cast iron weighing 36 ounces in air and 31 ounces
m water The weight of the iron and cork together, in water, is
15 8 ounces, what is the specific gravity (a) of cork? (<J) of cast iron?
SOLUTION. (a) Here w = 4 8, W = 40 8, W = 15 8; W v - 30,
W t = 81 Substituting in the formula,
Spec,fic gravity is (4Q g _ ^*_ (M __ _ ^ ^
(b) To apply formula of Art 44, W = 36, W - 31.
no
Specific gravity of cast iron is Q . ., => 7 2. Ans.
oo SI
46. Specific Gravity of a Liquid. To determine the
specific gravity of a liquid, proceed as follows: Weigh an
empty flask; fill it with water, then weigh it and find the
31 HYDROSTATICS 31
difference between the two results; this will be equal to
the weight of the water Then weigh the flask filled with
the liquid, and subtract the weight of the flask; the result
is the weight of a volume of the liquid equal to the volume
of the water. The weight of the liquid divided by the
weight of the water is the specific gravity of the liquid.
Let w = weight of empty flask,
W = weight of flask when filled with the liquid;
W 1 = weight of flask when filled with water
Then, Sp.Gr. = *-'
W 1 w
EXAMPLE A flask when empty weighs 8 ounces; when filled with
water, 33 ounces, and when filled with alcohol, 28 ounces. What is
the specific gravity of the alcohol?
SOLUTION Here W = 28, w = 8, W = 33 Substituting in the
formula,
no __ Q
Sp Gi. = ^__| = 8 Ans.
47. Nicholson's Hydrometer. Instruments called
hydrometers are in general use for detei-
mining quickly and accurately the specific
gravity of liquids and of some solids. One of
the principal forms is Nicholson's hydrometer,
which is shown in Fig. 20. It consists of a
hollow cylmdei, carrying at its lower end a
basket //, heavy enough to keep the apparatus
upright when placed in water. At the top of
the cylinder is a veitical rod, to which is
attached a shallow pan a The cylinder is made
of such size and weight that the apparatus is
somewhat lighter than water, and a certain
weight W must be placed in the pan to sink Fl
it to a given point c on the rod. The body whose specific
gravity it is desired to find must weigh less than W. It is
placed in the pan a, and enough weight w is added to sink
the point c to the water level. It is evident that the weight
In air of the given body is W w. The body is now
removed from the pan a and placed in the basket d, an
32 HYDROSTATICS 31
additional weight being added to sink the point e to the
water level. Represent the weight now m the pan by W 1 .
The difference W w is the weight of a volume of water
equal to the volume of the body. Hence,
EXAMPLE The weight necessary to sink a hydrometer to the
point c is 16 ounces, the weight necessary when the body is In the
pan a is 7 3 ounces; and when the body is in the basket d, 10 ounces.
What is the specific gravity of the body?
SOLUTION Here W = 16, w = 7 3, W = 10 Substituting in the
formula,
I Q IT Q
Specific gravity = ^ ~ = 3.22 Ans.
1U " i o
48. Volume of an Irregular Solid. The principle of
Archimedes affords a very easy and accurate method of find-
ing the volume of an irregularly shaped body. Let the
weight of the body in air be W, and in water W. The
difference W W is, according to the principle of Archi-
medes, the weight of a volume of water equal to the volume of
the body. If this volume is denoted by F, and the weight of
water per unit of volume is denoted by W , the weight
of the volume V is W, V, and, therefore,
W V = W- W'\
whence, F= W ~ W> (1)
W,
If the volume is expressed in cubic feet, W,is 62.5 pounds,
and, therefore,
V= W ^^~ = -016(^- W) (2)
If the volume is expressed in cubic inches, W* = .03617
pound, and, therefore,
v = "~ = 27<647( w " w ' }
EXAMPLE The weight of a body in air is 96 pounds, and m water,
48 8 pounds What is the volume of the body?
SOLUTION To apply formula 2, we have W = 96 and W * 48.8.
Substituting in the formula,
V .016(98 - 48.8) - .76 cu. ft. Ans.
31 HYDROSTATICS 33
49. If the specific gravity of a body is known, the cubical
contents of the body can be found by dividing its weight by
its specific gravity, and then dividing again by either .08617
or 62 5, according as the volume is desired in cubic inches
or in cubic feet.
EXAMPLE A certain body has a specific gravity of 4 38 and weighs
76 pounds What is the volume of the body m cubic Inches?
76
SOLUTION.- __-_ . 479.7 en. In. Ans.
EXAMPLES FOB PRACTICE
1 If a certain quantity of red lead weighs 5 pounds in air and
4 441 pounds in water, what is its specific gravity? Ans. 8.94
2. A piece of iron weighing 1 pound in air and 861 pound in water
is attached to a piece of wood weighing 1 pound in air. When both
bodies are placed in water they weigh 2 pound. What is the specific
gravity: (a) of the iron? (6) of the wood? Aric /() 7 194
Ans> \() 602
3. An empty flask weighs 13 ounces; when filled with water, it
weighs 22 ounces, and when filled with sulphuric acid, 29.56 ounces.
What is the specific gravity of the acid? Ans. 1 84
4. How many cubic feet of brick having a specific gravity of 1.9
are required to make a total weight of 260 pounds? Ans. 2.39 cu. ft.
PNEUMATICS
PROPERTIES OF AIR AND OTHER
GASES
1. Pneumatics is that branch of science that treats of
the mechanical properties of gases.
2. The distinguishing property of a gas is that, no matter
how small the quantity may be, the gas -will always till the
vessel or vessels that contain it. If
a bladder is partly filled with
air and placed under a glass jar
(called a receiver) from which
the air has been exhausted, the
bladder will immediately
expand, as shown in Fig. 1. The
force that a gas exerts when con-
fined in a limited space is called
tension. In this case, the
word tension means pressure,
and is only used in this sense
when referring to gases. Fm. i
3. As water is the most common type of fluids, so air is
the most common type of gases. It was supposed by the
ancients that .air was "imponderable," by which was meant
that it weighed nothing, and it was not until about the year
1650 that it was proved that air really has weight. A cubic
inch of air, under ordinary conditions, weighs about .31 gram.
The ratio of the weight of air to that of water is about 1 . 774;
that is, air is only T^T as heavy as water.
OOPYMiaHTID BY INTERNATIONAL TKXTMQOK COMPANY. AU. RIOHTft RMKHVtO
532
2 PNEUMATICS 32
It was shown in Hydrostatics that if a body is immersed
in water, and weighs less than the volume of water it dis-
places, the body will rise and project partly out of the water.
The same principle, which is the principle of Archimedes,
applies to gases. If a vessel made of light material is filled
with a gas lighter than air, so that the total weight of the
'. vessel and gas is less than the weight
i of the volume of air they displace, the
\ vessel will nse. It is on this pnnci-
! pie that balloons are made.
! 4. Since air has weight, it is evi-
, dent that the enormous quantity of
air that constitutes the atmosphere
must exert a considerable pressure
on the earth. This is easily proved
by taking a long glass tube, closed at
one end, and filling it with mercury.
If the finger is placed over the open
end, so as to keep the mercury from
running out, and the tube is inverted
and placed in a glass partly filled with
the same liquid, as shown in Fig. 2,
the mercury in the tube will fall, then
rise, and after a few oscillations will
come to rest at a height above the top
of the mercury in the glass equal to
about 30 inches This height will
always be practically the same under
PlQ 2 the same atmospheric conditions
Now, since the atmosphere has weight, it presses on the
upper surface of the mercury in the glass with equal force on
every square unit, except on that part of the surface occupied
by the tube. According to Pascal's law (see Hydrostatics),
this pressure is transmitted in all directions. There being
nothing in the tube, except the mercury, to counterbalance
the upward pressure of the air, the mercury falls in the tube
until it exerts an upward pressure on the upoer surface of
32 PNEUMATICS 3
the mercury in the glass sufficiently great to counterbalance
the downward pressure produced by the atmosphere. In
order that there may be equilibrium, the pie&sure of the air
per unit of area on the upper surface of the mercury in the
glass must equal the pressure (weight) exerted per unit of
area by the mercury inside of the tube. Suppose that the
area of the inside of the tube is 1 square inch, then, since
mercury is 13 6 times as heavy as water, and 1 cubic inch ot
water weighs about 03617 pound, the weight of the mer-
curial column is .03617 X 13.6 X 30 = 14.7574 pounds The
actual height of the mercury is a little less than 30 inches,
and the actual weight of a cubic inch of distilled water is a
little less than .03617 pound. When these considerations
are taken into account, the average weight of the mercmial
column at the level of the sea is 14.696 pounds, or, as it is
usually expressed, 14.7 pounds. Since this weight, exerted
on 1 square inch of the liquid in the glass, just produces
equilibrium, it is plain that the pressure of the outside air
is 14 7 pounds on every square inch of suiface. This pres-
sure is often referred to as one atmosphere. A pressure of
two atmospheres is a pressure of 2 X 14.7 = 29.4 pounds
per square inch, a pressure of three atmospheres is a pres-
sure of 3 X 14.7 = 44.1 pounds per square inch; etc.
5. Vacuum. Referring to Fig. 2, the space between
the upper end of the tube and the upper surface of the mercury
in the tube is called a vacuum, or empty space. If this
space contained a gas of some kind, no matter how small the
quantity might be, the gas would expand and fill the space,
and its tension would, according to the amount present,
cause the column of mercury to fall and become shorter; the
space would then be called a partial vacuum. If the mer-
cury fell 1 inch, so that the column was only 29 inches high,
this would be expressed by saying that there were 29 inches
of vacuum; a fall of 8 inches would be referred to as 22 inches
of vacuum; a fall of 16 inches, as 14 inches of vacuum, etc.
Hence, when the vacuum gauge of a condensing engine
shows 26 inches of vacuum, there i$ enough air in the
PNEUMATICS
32
condenser to produce a pressure of
_ no
X 14.7 =
30
X 14 7
= 1.96 pounds per square inch. In all cases where the mer-
^ __ cunal column is used to measure a vacuum, the
height of the column, in inches, gives the number
of inches of vacuum Thus, if the column were 5
inches high, or the vacuum gauge showed 5 inches,
the vacuum would be 5 inches
If the tube had been filled with water instead
of mercury, the height of the column of water to
balance the pressure of the atmosphere would
have been about 30 X 13.6 = 408 inches = 34 feet
This means that if a tube is filled with water, and
is inverted and placed in a dish of water in a
manner similar to that shown m Fig 2, the result-
ing height of the column of water will be about
34 feet
6. The barometer is an instrument used for
measuring the pressure of the atmosphere There
are two kinds in general use the mercurial
barometer and the aneroid barometer. The latter
was described in Leveling The mercurinl
barometer is shown in Fig 3. The principle is
the same as in the case of the inverted tube shown
in Fig 2 The tube and cup at the bottom are
protected by a brass or an iron casing At the
top of the tube is a graduated scale that can be
read to rsW inch by means of a vernier. Attached
to the casing is an accurate thermometer for deter-
mining the temperature of the outside an at the
time the barometric observation is taken. This
is necessary, since mercury expands when the
temperature rises, and contracts when the temper-
ature falls; for this reason a standard temperature
is assumed, and all barometer readings are reduced to this
temperature. This standard temperature is usually taken as
32 F., at which temperature the height of the mercurial
PlG 3
32 PNEUMATICS 5
column is 30 inches. Another correction is made for the alti-
tude of the place above sea level, and a third correction for
the effects of capillary attraction. It is not necessary here
to go into details regarding these corrections.
7. The pressure of the atmosphere varies with the alti-
tude above sea level, being greater in low than in high
places. At the level of the sea, the height of the mercurial
column is about 30 inches, at 5,000 feet above the sea, it is
24.7 inches; at 10,000 feet above the sea, it is 20 5 inches;
at 15,000 feet above the sea, it is 16.9 inches; at 3 miles,
it is 16 4 inches; and at 6 miles above the sea level, it is 8 9
inches.
The density also varies with the altitude; that is, a cubic
foot of air at an elevation of 5,000 feet above the sea level
does not weigh as much as a cubic foot at sea level. This
is proved conclusively by the fact that at a height of 83" miles
the mercurial column measures but 15 inches, indicating
that half the weight of the entire atmosphere is below that
height It is known that the height of the earth's atmos-
phere is at least 50 miles; hence, the air just before reach-
ing the limit must be in an exceedingly rarefied state. It
is by means of barometers that great heights are measured.
The aneroid barometer has the heights marked on the dial,
so that it can be read directly. With the mercurial barom-
eter, the heights must be calculated from the reading. (See
Leveling.)
8. The atmospheric pressure is everywhere present,
and presses all objects in all directions with equal force. If
a book is laid on the table, the air presses on it in every
direction with an equal average force of about 14.7 pounds
per square inch. It would seem as though it would take
considerable force to raise a book from the table, since, if
the size of the book were 8 inches by 5 inches, the pressure
on it would be 8 X 6 X 14 7 = 588 pounds But there is an
equal pressure beneath the book to counteract the pressure
on the top. It would now seem, as though it would require
a great force to open the book, since there are two pressures
I LT 398-17
6 PNEUMATICS 32
of 588 pounds each, acting in opposite directions, and tending
to crush the book, and so it would but for the fact that there
is a layer of air between each two leaves acting upwards and
downwards with a pressure of 14.7 pounds per square inch.
If two metal plates are made as perfectly smooth and flat as
it is possible to make them, and the edge of one is laid on
the edge of the other, so that one may be slid on the othei,
and the air thus excluded, it will take an immense force,
compared with the weight of the plates, to separate them.
This is because the full pressure of 14 7 pounds per square
inch is then exerted on each plate with no counteracting
equal pressure between the plates
If a piece of flat glass is laid on a flat surface that has been
previously moistened with water, it will require considerable
force to lift it off the surface. This is due to the fact that
the water helps to fill up the pores m the flat surface and
glass, and thus creates a partial vacuum between the glass
and the surface, thereby reducing the counterpressure beneath
the glass.
9. Tension of Gases. In Art. 5, it was said that the
space above the column of mercury in Fig. 2 was a vacuum,
and that if any gas or air were present it would expand, its
tension forcing the column of mercury downwards. If
enough gas is admitted to cause the mercury to stand at
15 inches, the tension of the gas is evidently 14,7 2
= 7 35 pounds per square inch, since the pressure of the out-
side air of 14.7 pounds per square inch only balances 15 inches,
instead of 30 inches, of mercury, that is, it balances only
half as much as it would if there were no gas in the tube;
hence, the pressure (tension) of the gas in the tube is 7.35
pounds. If more gas is admitted until the top of the mercurial
column is just level with the mercury in the cup, the gas m
the tube has then a tension equal to the outside pressure of
the atmosphere. Suppose that the bottom of the tube is
fitted with a piston, and that the total length of the inside of
the tube is 36 inches. If the piston is shoved upwards so
that the space occupied by the gas is 18 inches long, instead
32 PNEUMATICS 7
of 36 inches, the temperature remaining the same as before,
it will be found that the tension of the gas within the tube is
29.4 pounds per square inch. It will be noticed that the
volume occupied by the gas is only half that in the tube
before the piston was moved, while the pressure is twice as
great, since 14.7 X 2 = 29.4 pounds. If the piston is shoved
up so that the space occupied by the gas is only 9 inches,
instead of 18 inches, the temperature still remaining the
same, the pressure will be found to be 58.8 pounds per square
inch. The volume has again been reduced one-half, and the
pressure increased two times, since 29.4 X 2 = 58.8 pounds.
The space now occupied by the gas is 9 inches long, whereas,
before the piston was moved, it was 36 inches long, as the
tube was assumed to be of uniform diameter throughout its
length, the volume is now TUT = i of its original volume, and
CO O
its pressure is -"- = 4 times its original pressure. More-
over, if the temperature of the confined gas remains the
same, the pressure and volume will always vary in a similar
way. The law that states these effects is called Marwtte*s
law, and is as follows:
10. Marietta's Law. The temperature remaining the
same, the volume of a. given quantity of gas varies inversely as
the pressure.
The meaning of this is: If the volume of the gas is dimin-
ished to one-half, one-third, one-fifth, etc. of its former
volume, the tension will be increased two, three, five, etc.
times; or, if the outside pressure is increased two, three, five,
etc. times, the volume of the gas will be diminished to one-
hnlf, one-third, one-fifth, etc. of its original volume, the tem-
perature remaining constant.
Suppose 3 cubic feet of air to be under a pressure of
60 pounds per square inch in a cylinder fitted with a mov-
able piston; then the product of the volume and pressure is
3 X 60 = 180. Let the volume be increased to 6 cubic feet;
then the pressure will be 80 pounds per square inch, and
30 X 6 SB 180, as before, Let the volume be increased to
8 PNEUMATICS 32
24 cubic feet; it is then 24 3 = 8 times the original volume,
and the pressure is one-eighth of the original pressure, or
60 X i = 71 pounds, and 24 X 7i = 180, as in the two pre-
ceding cases. It will now be noticed that, if a gas is allowed
to expand without change of temperature, the pioduct of any
Pressure and the corresponding volume is the same as for any other
pressure and the corresponding volume If the air were com-
pressed, the same result would be obtained.
Let p = pressure corresponding to volume v;
pi = pressure corresponding to volume z^.
Then, p v = p v
Knowing the volume and the pressure for any position of
the piston, and the volume for any other position, the pres-
sure may be calculated; or, if the pressure is known for any
other position, the volume may be calculated.
EXAMPLE 1 If 1 875 cubic feet of air is under a pressure of
72 pounds per square inch, what will be the pressure when the volume
is increased (a) to 2 cubic feet? (b) to 3 cubic feet? (c) to 9 cubic feet?
SOLUTION Solving the last equation for/,, the unknown pressure
gives
. . . p v 72 X 1 875 _, .,
(a) p! = r -- = - = --- = 67 Ib per sq m Ans.
v\ &
... . 72 X 1 875 ._ ,. .
(b) pi = = 45 Ib per sq in Ans.
, . . 72 X 1 875 .,_ ,.
(c) pi = g =15 Ib. per sq. in Ans
EXAMPLE 2 Ten cubic feet of air has a tension of 5.6 pounds per
square inch, what is the volume when the tension is (a) 4 pounds?
(b) 8 pounds? (c) 25 pounds? (d) 100 pounds?
SOLUTION Solving the same equation for v,. gives
(a) , l =^ = 56 * 10 = 14cu ft Ans
pi 4
7 cu ft Ans
,
(c) Vl = 5 6 * 10 = 2 24 cu ft Ans
(<0 i = ^ = 56 cu ft Ans.
11. For the same quantity of gas, the weight per unit
of volume varies inversely as the volume. For example, if
32 PNEUMATICS 9
1 pound of gas occupies n volume of 4 cubic feet, its weight
per cubic foot will be i pound. If it occupies a volume
twice as large, or 8 cubic feet, its weight per cubic foot
will be i pound, or only one-half of what it was before. In
general, it, the temperature of a fixed quantity of gas
remaining the same, the weights per unit of volume when
the gas occupies the volumes v and z>,, respectively, are
denoted by w and w lt then,
w _ v\
Wj. V
and wv z0, v, (1)
Also, since =
v i
and w p l = iv ip (2)
EXAMPLE 1 The weight of 1 cubic foot of air at a temperature of
60 F , and under a pressure of 1 atmosphere (14.7 pounds per
square inch), is 0763 pound, what would be the weight per cubic foot
if the volume were compressed until the tension was 5 atmospheres,
the temperature still being 60 F ?
SOLUTION. Applying formula 2, 1 X w\ = 6 X .0763. Hence,
a/, = .3815 Ib. per cu. ft. Ans.
EXAMPLE 2. If in the last example the air had expanded until the
tension was 5 pounds per square inch, what would have been its
weight per cubic foot?
SOLUTION. Here, p 14 7, p l = 5, and w = 0703 Hence, apply-
ing formula 2, 14.7 X w, = 5 X .0763, whence a/, = ~~y = 02695 Ib.
per cu. ft. Ans.
EXAMPLE 3 If 6.75 cubic feet of air, at a temperature of 60 F.
and a pressure of 1 atmosphere, is compressed to 2 25 cubic feet
(the temperature still remaining 60 P.), what is the weight of a cubic
foot of the compressed air?
SOLUTION Applying formula 1, 6.75 X .0763 = 2 25 X w t . Hence,
w, m - 22 2289 Ib. per cu. ft. An 8 .
12. Manometers and Gauges, There are two ways
of measuring the pressure of a gas: by means -of an instru-
ment called a manometer, and by means of a gauge.
10 PNEUMATICS 32
The manometer generally used is practically the same as a
mercurial barometer, except that the tube is much longer, so
that pressures equal to several atmospheres may be meas-
ured, and is enlarged and bent into a U shape at the lower
end, both the lower and the upper ends are open, the lower
end being connected to the vessel containing the gas whose
pressure it is desired to measure. The gauge is so com-
mon that no description of it will be given here. With both
the manometer and the gauge, the pressure recorded is the
amount by which the piessure being measured exceeds the
atmospheric pressure, and is called gauge pressure. To
find the total pressure, or absolute pressure, the atmos-
pheric must be added to the gauge pressure. In all for-
mulas in which the pressure of a gas or steam occurs, the
absolute pressure must be used, unless the gauge pressure is
distinctly specified as the proper one to use. For conve-
nience, all pressures given in this Section and in the ques--
tions referring to it will be absolute pressures, and the word
"absolute" will be omitted, to avoid repetition.
13. In all that has been said, it has been stated that the
temperature was constant; the reason for this will now be
explained. Suppose 5 cubic feet of air to be confined in a
cylinder placed in a vacuum, so that there will be no pressure
due to the atmosphere, and suppose the cylinder to be fitted
with a piston weighing, say, 100 pounds, and having an
area of 10 square inches. The tension of the air will be
1 r q u a = 10 pounds per square inch. Suppose, now, that the
air, originally at a temperature of 32 F., is heated until its
temperature is 33 F that is, until its temperature is raised
1. It will be found that the piston has risen a certain
amount, and, consequently, the volume has increased, while
the pressure remains the same as befoie, 01 10 pounds per
square inch. If more heat is applied, until the temperature
of the gas is 34 F , it will be found that the piston has again
risen, and the volume again increased, while the pressure
still remains the same. It will be found that for every
increase in temperature there will be a corresponding increase
32 PNEUMATICS 11
of volume. The law that expresses this change is called
Gay-Lussac 's law^ and is as follows:
14. Gay-IJussac's Liaw. // the pressure remains con-
stant, every increase tn temperature of 1 F. produces in a given
quantity of gas an expansion 0/To of its volume at 32 f.
If the pressure remains constant, it will also be found that
every decrease of temperature of 1 F. will cause a decrease
of TU of the volume at 32 F.
Let z/ volume of any quantity of gas at 32 F.;
v = volume of gas at temperature t;
z^ =* volume of gas at temperature /,.
It is assumed that the pressure of the gas remains
unchanged. Then, m passing from the temperature 32 to
the temperature /, the volume of the gas will be increased
J _ QO
algebraically by the amount TUYZ/O (t 32) = jnn v*.
492
Therefore,
or, reducing, v = flo T~- (D
492
Likewise, v, = v, --' (a)
492
Dividing formula 1 by equation (a),
v _ 460 + t_
v\ 460 + t,
whence v - V *M (2)
EXAMPLE If 5 cubic feet of air at a temperature of 45 is heated
under constant pressure up to 177, what is the final volume of the air?
SOLUTION To apply formula 2, we have v^ = 5; , 46; t 177.
Therefore,
460 + 177
v * 5 x 07 cu ' ft AnSi
15. Suppose that a certain volume of gas is confined
in a vessel so that it cannot expand; in other words, sup-
pose that the piston of the cylinder before mentioned is
fastened so that it cannot move. Let a gauge be placed on
ihl
12 PNEUMATICS 32
the cylinder so that the tension of the confined gas can be
registered If the gas is heated, it will be found that, for
every increase of temperature of 1 F , there will be a cor-
responding increase of TUT of the tension That is, the
volume remaining: constant, the tension increases Tn of the
original tension for eveiy degree rise of temperature.
Let p tension of gas at temperature /,
pi = tension of gas at temperature ti.
Then, as in the preceding article,
=
y ^460 + ^
EXAMPLE If a certain quantity of air is heated under constant
volume from 45 to 177, what is the resulting tension, the original
tension being 14 7 pounds per square inch?
SOLUTION Applying the formula,
p = 14 7 X t' 7 - 18 542 lb per sq in Ans
16. According to the modern and now generally accepted
theory of heat, the atoms and molecules of all bodies are in
an incessant state of vibration. The vibratory movement
in liquids is faster than in solids, and in gases faster than
in either of the other two Any increase of heat increases
the vibrations, and a decrease of heat decreases them From
experiments and advanced mathematical investigations, it has
been concluded that at 460 below zero, on the Fahrenheit
scale, all these vibrations cease. This point is called the
absolute zero, and all temperatures reckoned from this
point are called absolute temperatures. The point of
absolute zero has never been reached, nevertheless, it
has a meaning, and is used in many formulas, absolute tem-
peratuies being usually denoted by T. Ordinary temperatures
are denoted by /. When the word temperature alone is used,
it refers to the ordinary way of measuring temperatures, but
when absolute temperature is specified 460 F. must be added
to the ordinary temperature The absolute temperature cor-
responding to 212 F. is 460 -f 212 = 672 F, If the abso-
lute temperature is given, the ordinary temperature may be
found by subtracting 460 from the absolute temperature.
32 PNEUMATICS 13
Thus, if the absolute temperature is 520 P., the ordinary
tempeiature is 520 - 460 = 60.
17. Let p = pressure, in pounds per square inch, of
W pounds of air;
V = volume of air, in cubic feet,
T absolute tempeiature of air.
It is shown in advanced works on the theory of heat that
these quantities are related by the following- equation:
pV = .37 WT
EXAMPLE 1 The pressure on 9 cubic feet of air weighing 1 pound
is 20 pounds per squaie inch, what is the ordinary temperature of the
air?
SOLUTION Here, W= 1. Substituting the other values in the
180
formula gives 20 X 9 = 37 T, hence, T = ~ = 486.6, nearly.
.37
40 5 - 4(50 = 26 6, the ordinary temperature. Ans
EXAMPLE 2 What is the volume of 1 pound of air whose tempera-
ture is 00 F under a pressure of 1 atmosphere?
SOLUTION Here, W - 1, and T = 460 -f 60 - 520, therefore, 14 7
07 \f Knn
X V = .87 X 520, whence V = ~^~ = 18 088 cu. ft Ans.
EXAMPLE 3 If 3 cubic feet of air weighing 35 pound Is under a
pressure of 48 pounds per square inch, what Is the ordinary tem-
perature of the air?
SOLUTION Applying the formula, 48 X 8 - .37 X .35 X T,
JO v ^
whence, T= --%. - 1,112 Then, 1,112 - 460 - 652.
A I X <50
Ans.
EXAMPLK 4. What is the weight of 1 cubic foot of air at a tempera-
ture of 32, and under a pressure of 1 atmosphere?
SOLUTION Here T - 400 + 32 = 492; V = 1; and p = 14 7
(Art. 4). Substituting these values in the formula gives 14 7 X 1
- .37 X 492 X W, whence,
Ans -
If the pressure is taken as 14.696 Ib. per sq. in , the weight of 1 cu. ft.
of air, at 32 and atmospheric pressure, is found to be
- - 08078 lb
14 PNEUMATICS 32
EXAMPLES FOR PRACTICE
1 A vessel contains 25 cubic feet of gas at a pressure of 18 pounds
per square inch, if 125 cubic feet of gas having the same pressure Is
forced into the vessel, what will be the resulting pressure?
Ans 108 Ib. per sq. in.
2 A pound of air has a temperature of 126, and a pressure of 1
atmosphere, what volume does it occupy? Ans. 14 75 cu. ft.
3 A certain quantity of air has a volume of 26 7 cubic feet, a pres-
sure of 19.3 pounds per square inch, and a temperature of 42; what
is its weight? Ans. 2 77 Ib
4 A receiver contains 180 cubic feet of gas at a pressure of 20
pounds per square inch, if a vessel holding 12 cubic feet is filled from
the receiver until its pressure is 20 pounds per square Inch, what will
be the pressure m the receiver? Ans 18$ Ib per sq in.
5 Ten cubic feet of air, having a pressure of 22 pounds per square
inch and a temperature of 75, is heated until the temperature is 300,
the volume remaining the same, what is the new pressure?
Ans 31 25 Ib per sq in.
THE MIXING OF GASES
18. If two liquids that do not act chemically on each
other are mixed together and allowed to stand, it will be
found that after a time the two liquids have separated, and
that the heavier has fallen to the bottom. If two equal ves-
sels containing: gases of different densities are put in com-
munication with each other, it will be found that after a short
time the gases have become mixed in equal proportions.
If one vessel is higher than the other, and the heavier
gas is in the lower vessel, the same result will occur. The
greater the difference of the densities of the two gases,
the faster they will mix. It is assumed that no chemical
action takes place between the two gases. When the two
gases have the same temperature and pressure, the pressure
of the mixture will be the same. This is evident, since the
total volume has not been changed, and, unless the volume
or temperature changes, the pressure cannot change This
property of the mixing of gases is a very valuable one, since,
if gases acted like liquids, carbonic-acid gas (the result
32 PNEUMATICS 15
of combustion), which is li times as heavy as air, would
remain next to the earth, instead of dispersing into the
atmosphere, the result being that no animal life could exist.
19. Mixture of Equal Volumes of Gases
Unequal Pressures. // two gases having equal volumes and
temperatures, but different Pressures, ate mixed in a vessel vu hose
volume equals one of the equal volumes of the gases, the pressure
of the mixture will be equal to the sum of the two pressures,
provided that the temperature remains the same as before.
20. Mixture of Two Gases Having; Unequal Vol-
umes and Pressures. Let v and p be the volume and
pressure, respectively, of one of the gases.
Let v 1 and PI be the volume and pressure, respectively, of
the other gas.
Let f-'and P be the volume and pressure, respectively, of
the mixture. Then, if the temperature remains the same,
yp= vfi + v^
That is, if the temperature is constant, the volume after mix-
ture, multiplied by the resulting pressure, equals the volume of
one gas before mixture multiplied by its pressure, plus the volume
of the other gas multiplied by its pressure*
EXAMPJ.B Two gases at the same temperature, having volumes of
7 cubic feet and 4J cubic feet, and pressures of 27 pounds and
18 pounds per square inch, respectively, are mixed together In o vessel
whose volume is 10 cubic feet. What is the resulting pressure?
SOLUTION. Applying the preceding formula, J V p v + p^ v tl or
/>X 10 - 27 X 7 + 4t X 18. Hence, P - i?-il - 27 Ib. per sq. in.
Aim.
21. Mixture of Two Volumes of Air Having
Unequal Pressures, Volumes, and Temperatures. If
a body of air having a temperature /, a pressure p^ and a vol-
ume v l is mixed with a volume of air having 1 a temperature *
a pressure /> and a volume z>,, to form a volume f having a
pressure /* and a temperature t, then, either the new temper-
ature /, the new volume V, or the new pressure P may be
found, if the other two quantities are known, by the following
16 PNEUMATICS 32
formula, in which T lt T at and Tare the absolute temperatures
corresponding to / / a , and /:
PV= /A^i + A
\Ti T
EXAMPLE Five cubic feet of air having a tension of 30 pounds per
square inch and a temperature of 80 F are compressed, together
with 11 cubic feet of air having a tension of 21 pounds per square inch
and a temperature of 45 F , in a vessel whose cubical contents are
8 cubic feet The new pressure is required to be 45 pounds per square
inch What must be the temperature of the mixture?
SOLUTION Substituting in the formula, 45 X 8 =
Qflft
X T, or 360 = .7352 T Hence, T = ~^ = 489 66, nearly, and
/ = 29 66 Ans
EXAMPLES FOB PRACTICE
1 Two vessels contain air at pressures of 60 and 83 pounds per
square inch The volume of each vessel is 8 47 cubic feet If all the
air in both vessels is removed to another vessel, and the new pressure
is 100 pounds per square inch, what is the volume of the vessel, the
temperature remaining unchanged? Ans. 12 11 cu ft
2 A vessel contains 11 83 cubic feet of air at a pressure of
33 3 pounds per square inch It is desired to increase the pressure to
40 pounds per square inch by supplying air from a second vessel that
contains 19 6 cubic feet of air at a pressure of 60 pounds per square
inch What will be the pressure in the second vessel after the pressure
in the first has been raised to 40 pounds per square inch?
Ans. 55.96 Ib. per sq m.
3 If 4 8 cubic feet of air having a tension of 52 pounds per square
inch and a temperature of 170 is mixed with 13 cubic feet having a
tension of 78 pounds per square inch and a temperature of 265, what
must be the volume of the vessel containing the mixture in older that
the tension of the mixture may be 30 pounds per square inch and the
temperature 80? Ans. 32 31 cu ft.
PNEUMATICS
17
PNEUMATIC MACHINES
THE AIR PUMP
22. The air pump is an instrument for removing air
from an enclosed space. A section of the principal parts is
shown in Fig. 4, and a view of the complete instrument is
given in Fig. 5. The closed vessel R is called the receiver,
FIG 4
and the space it encloses is that from which it is desired to
remove the air. The receiver is usually made of glass, and
the edges are ground so as to be perfectly air-tight. When
made in the form shown, it is called a boll-Jar receiver,
The receiver rests on a horizontal plate, in the center of
which is an opening communicating with the pump cylin-
der C by means of a bent tube /. The pump piston fits the
cylinder accurately, and has a valve V 1 opening upwards.
At the junction of the tube with the cylinder is another
valve V, also opening upwards, When the piston is raised,
1
PNEUMATICS
the valve V closes, and, sinct 1 no .111 can m't into tin- i \liu
dcr from above, the piston leave's a \aamm brliind it. Tin
pressure on top of rbemjj now u-ninvi'd, tin 1 ti'iismu il tli
,ur in llu- rm-iMx IM A'
iMllsi's / * tn MSI-, lh' .111
in tlu- ii'i'i'ivvi thru t'\
pands and uivupu-s thr
sp.u't* U'tt rmpty by
the pistnn, as wi-ll as
tin- spaiv in the- tube /
and tin* IT or i VIM A*.
Tbtj piston is iiuw
pnslu-tl down, tin 1 vahc-
/"i-losfs, tlk 1 \,dvt* / '
opens, and ihr an nt ('
csoapcs. Tilt*
valv /" is snni
suppiirli'tl, as shown in
Kitf. 1, by a nu-tal nxl
passing tlimuuh llu-
piston and itttinu it
somewhat tightly.
When the piston is raised or lowered, this rod moves with it.
A button near the upper end of the rod confines its motion
to within very narrow limits, the piston sliding on the rod
during the greater part of the journey.
S3. JUetfi'ooH und JLlmltH of Kxlvauwtlon, Suppose
that the volume o /? and / together is four times that of ( \
and that there are, ny, 200 grains of nir in 'A* and /, and
50 grains in C, when the piston is at the top of the i-ylimU-i.
At the end of the first stroke, when the piston is a^ain at
the top, 60 grains of air in the cylinder ( ' will have been
removed, and the 200 grains in /? and / will omipy the
spaces jR t /, and C. The ratio between the sum of the
spaces .# and / and the total space jV + / f Tis ^; hence,
j
200 X - 160 grains - the weight of air in / and t after
32
PNEUMATICS
19
the first stroke. After the second stroke, the weight of the
air in R and / will be 200 X X = 200 X (*)' = 200 X iif
= 128 grams. At the end of the third stroke, the weight
will be [200 X ()'] X = 200 X (*)' = 200 X rW = 102 4
grains. At the end of n
strokes, the weight will be
200 X (&)" It is evident
that it Zi impossible by this
method to remove all the air
contained in R and t. It
requires an exceedingly
good air pump to reduce
the tension of the air in R
to -gV inch of mercury.
When the an has become
so rarefied as this, the
valve V will not lift, and,
consequently, no more air
can he exhausted
24. Spreiigel's Air
Pump. In Fig 6, cd is a
glass tube longer than 30
inches, open at both ends,
and connected by means of
India-rubber tubing with a
funnel A filled with mer-
cury and supported by a
stand. Mercury is allowed
to fall into this tube at a
rate regulated by a clamp
at c. The lower end of the
tube c d fits in the flask ./?,
which has a spout at the
side a little higher than the lower end of c d\ the upper part
has a branch at x to which a receiver R can be tightly fixed.
When the clamp a.t c is opened, the first portions of the mer-
cury that run out close the tube and prevent air from enterirg
Fia. 6
20
PNEUMATICS
32
from below. These drops of mercury act like little pistons,
carrying the air in front of them and forcing it out through
the bottom of the tube. The air in R expands to fill the tube
every time that a drop of mercury falls, thus creating a partial
vacuum in R, which becomes more nearly complete as the
process goes on. The escaping mercury falls into the dish H,
from which it can be poured back into the funnel from time
to time. As the exhaustion from R goes on, the meicury
rises in the tubecd until, when the exhaustion is complete, it
forms a continuous column 30 inches high, in other words,
it is a barometer whose vacuum is the receiver R. This
instrument necessarily requires a great deal of time for its
operation, but the results are very complete, a vacuum of
^aiUa inch of mercury being sometimes obtained. By use of
chemicals in addition to the above, a vacuum of TBTMMTO inch
of mercury has been obtained.
25. Magdeburg: Hemispheres. The pressure of the
atmosphere can be made manifest by means of two hollow
hemispheres, such as are shown in Fig. 7.
This contrivance was devised by Otto Von
Guencke, of Magdeburg, and is known as
the Magdeburg hemispheres. One of
the hemispheres is provided with a stop-
cock, by which it can be screwed on to an
air pump. The edges fit accurately and
are well greased, so as to be air-tight.
When the hemispheres contain air, they
can be separated easily; when the air is
pumped out by an air pump, they can
be separated only with great difficulty.
The force required to separate them will
be equal to the area of the largest circle
of the hemisphere (projected area) m
square inches, multiplied by 14.7 pounds.
This force will be the same in whatever
position the hemispheres may be held, .which proves that
the pressure of air on them is the same m all directions.
FIG 7
32
PNEUMATICS
21
26. The Weight Liifter. The pressure of the atmos
phere is shown by means of the apparatus illustrated in
Fig 8. Here, a cylinder fitted with a piston is held in
suspension by a chain. At the top of the cylinder is a
plug a, which can be taken out. This plug is removed and
the piston is pushed up until it touches the
cylinder head. If the plug is then screwed
in, the piston will remain at the top until
a weight has been hung on the rod equal
to the area of the piston multiplied by 14.7
pounds, less the weight of the pibton and
rod. If a force is applied to the rod suffi-
ciently great to push the piston down-
wards, the piston will, on the removal of
the force, raise to the top of the cylinder
any weight that is less than the one men-
tioned. Suppose the weight to be removed,
and the piston to be supported midway
between the top and bottom of the cylinder.
Let the plug be removed, air admitted
above the piston, and the plug screwed
back into its place; if the piston is shoved
upwards, the farther up it goes, the greater
will be the force necessary to push it, on
account of the compression of the air. If
the piston is of large diameter, it will also
require a great force to pull it out of the
cylinder, as a little consideration will show.
For example, let the diameter of the piston
be 20 inches, the length of the cylinder 86
inches, plus the thickness of the piston, and
the weight of the piston and rod 100 pounds. If the piston
is in the middle of the cylinder, there will be 18 inches of
space above it, and 18 inches of space below it. The area
of the piston is 20' X .7864 =* 314.16 square inches, and the
atmospheric pressure on it is 814.16 X 14.7 4,618 pounds,
nearly. In order to shove the piston upwards 9 inches, the
pressure on it must be twice as great, or 9,286 pounds, and
I LT 398-18
Pro 8
22
PNEUMATICS
32
to this must be added 100 pounds, the weight of the piston
and rod, which gives 9,236 + 100 = 9,336 pounds. The
force necessary to cause the piston to move upwards 9 inches
will then be 9,336 4,618 = 4,718 pounds. Now, suppose
the piston to be moved downwards until it is just on the point
of being pulled out of the cylinder The volume above it will
then be twice as great as before, and the pressure one-half as
great, or 4,618 2 = 2,309 pounds. The total upward pies-
sure will be the pressure of the atmosphere less the weight
of the piston and rod, or 4,618 100 = 4,518 pounds, and
the force necessary to pull it downwards to this point will
be 4,518 - 2,309 = 2,209 pounds.
27. Tlie Baroscope. The buoyant effect of air is very
clearly shown by means of an instrument called the baro-
scope, shown in Fig. 9. It con-
sists of a scale beam, from one
extremity of which is suspended
a small weight, and from the
other a hollow copper sphere.
In air, they exactly balance each
other, but when they are placed
under the receiver of an air
pump and the air is exhausted
the sphere sinks, showing that
it is really heavier than the small
weight. Before the air is
exhausted, each body is buoyed
up by the weight of the air it
displaces, and, since the sphere displaces more air, it loses
more weight by reason of this displacement than the small
weight. Suppose that the volume of the sphere exceeds that
of the weight by 10 cubic inches; the weight of this volume
of air is 3.1 grains. If this weight is added to the small
weight, it will overbalance the sphere in air, but will exactly
balance it m a vacuum.
FIG 9
32 PNEUMATICS 23
AIR COMPRESSORS
28. For many purposes, compressed air is preferable to
steam or other gases for use as a motive power; m such
cases, air compressors are used to compiess the air.
These are made in many forms, but the most common one
consists of a cylinder, called the air cylinder, placed in front
of the crosshead of a steam engine, so that the piston of the
air cylinder can be driven by attaching its piston lod to the
crosshead, in a manner similar to a steam pump. A cross-
section of the air cylinder of a compressor of this kind is
shown in Fig 10, in which a is the piston and b is the piston
rod, driven by the crosshead of a steam engine not shown
in the figure. Both ends of the lower half of the cylinder
are fitted with inlet valves d and d', which allow the air to
enter the cylinder, and both ends of the upper half aie fitted
with discharge valves / and f, which allow the air to escape
fiom the cylinder after it has been compressed to the
required pressure.
Suppose the piston a to be moving in the direction of the
arrow; then the inlet valves d in the left-hand end of the
cylinder from which the piston is moving will be forced
inwards by the pressure of the atmosphere, which over-
comes the resistance of the light spring: c t thus allowing
the air to flow in and fill the cylinder. On the other side of
the piston, the air is being compressed, and, consequently,
it acts with the springs ^ to force the inlet valves d' in
the right-hand end of the cylinder to their seats. In the
right-hand end of the cylinder, the discharge valves f are
opened when the pressure of the air in the cylinder is great
enough to overcome the resistance of the light springs f
and the tension of the air in the passages leading to the
discharge pipe h, and the discharge valves / are pressed
against their seats by the springs c and the tension of the air
in the passages. Suppose it is desired to compress the air
to 59 pounds per square inch, and to find at what point of the
stroke the discharge valves will open. Now, a pressure of 59
pounds per square inch equals a pressure of 4 atmospheres,
24
PNEUMATICS
32
very nearly, hence, when the pressure in the cylinder becomes
great enough to force air out through the discharge valves,
the volume must be one-quarter of the volume at atmos-
pheric pressure, or the valves will open when the piston has
traveled three-quarters of its stroke, provided that the air is
compressed at constant temperature.
The air, after being discharged from the cylinder, passes
out through the delivery pipe h, and from there is con-
FIG 10
veyed to its destination. It has been shown that when air or
any other gas is compressed its temperature is increased.
For high pressures, this increase of temperature becomes a
serious consideration, for two reasons: (1) When the air
is discharged at a high temperature, the pressure falls con-
siderably when the air has cooled down to its normal temper-
ature, and this represents a serious loss in the economical
working of the machine. (2) The alternate heating and
32
PNEUMATICS
25
cooling of the compressor cylinder by the hot and cold air
is very destructive to it, and increases the wear to a great
extent. To prevent the air from heating, cooling devices
are lesorted to, the most common one being 1 the so-called
water-Jacket. This is effected in the following manner
The cylinder walls are hollow, as shown in Fifi 10; the cold
water enters this hollow space
in the cylinder wall through the
pipe k k, and flows around the
cylinder, finally passing out
through the discharge pipe /.
The water keeps cold the cyl-
inder walls, which cool the air
as it is compressed.
29. Hei-o's Fountain.
Hero's fountain derives its
name from its inventor, Hero,
who lived at Alexandria about
120 B. C. This fountain, which
is shown in Fig. 11, consists of
a brass dish A and two glass
globes B and C, and depends for
its operation on the elastic prop-
erties of air The dish com-
municates with the lower part
of the globe C by a long tube D,
and another tube E connects
the two globes. A third tube
passes through the dish A to ^
the lower part of the globe B. _
This last tube being taken out,
the globe B is partially filled with water; the tube is then
replaced and water is poured into the dish. The water flows
through the tube D into the lower globe, and expels the air,
which is forced into the upper globe. The air thus com-
pressed acts on the water and makes it jet out through the
shortest tube, as represented in the figure. Were it not for
33
Fie. 11
26
PNEUMATICS
.12
the resistance offered by the atmosphere and by friction,
the issuing water would rise to a height above the water
in the dish equal to the difference of the level of the water m
the two globes.
THE SIPHON
30. The action of the siphon illustrates the effect of
atmospheric pressure. A siphon is simply a bent tube with
unequal branches, open at both ends, and is used to convey a.
liquid from a higher point to a
lower, over an intermediate point
higher than either of the other two.
In Fig 12, a and b are two vessels, b
being lower than a, and a c b is the
bent tube or siphon. Suppose this
tube to be filled with water and
placed in the vessels, as shown,
with the short branch a c in the
vessel a. The watei will flow fiom
the vessel a into , so long as the
level of the water in b is below
the level of the water in a and the
level of the water in a is above
the lower end of the tube ac. Tne
atmospheric pressure on the surfaces of a and b tends to
force the water up the tubes ac and be. When the siphon
is filled with water, each of these pressures is counteracted
in part by the pressure of the water in that branch of the
siphon that is immersed in the water on which the pres-
sure is exerted. The atmospheric pressure opposed to the
weight of the longer column of water will, therefore, be
more resisted than that opposed to the weight of the
shorter column; consequently, the pressure exerted on
the shorter column will be greater than that on the longer
column, and this excess of pressure will produce motion,
The action of the siphon is of great importance, and
should be thoroughly understood. The following considera-
tions and computations will make the subject clear:
PNEUMATICS
27
Let a = area of tube, in square inches;
// = dc = vertical distance, in inches, between the
surface of water in b and highest point
of the center line of tube,
//, = e c = distance, in inches, between the surface of
water in. a and highest point of center
line of tube.
The weight of the water in the short column is .03617 a hi,
and the resultant atmospheric pressuie, tending to force
the water up the short column, is 14 7 X a .03617 a //,.
The weight of the water in the long column is 03617 a/i t
and the icsultant atmospheric pressure, tending to force the
water up the long column, is 14.7 a .03617 a k. The differ-
ence between these two is (14.7 a .03617 ah^} (14.7 a
- 03617 ah} = .03617 a (A - k,) But h - //, = ed = dif-
ference between the levels of the water in the two vessels.
It will be noticed that the short column must not be higher
than 34 feet for water, or the siphon will not work, since the
,r\
PIG 18
pressure of the atmosphere will not support a column of
water that is higher than 84 feet; 28 feet is considered to be
the greatest height for which a siphon will work well.
31. Intermittent Springs.- Sometimes a spring is
observed to flow for a time and then cease; then, after an
28
PNEUMATICS
32
interval, to flow again for a time The generally accepted
explanation of this is that there is an underground reservoir
fed with water through fissures in the earth, as shown in
Fig. 13 The outlet for the water is shaped like a siphon,
as shown. When the water in the reservoir reaches the same
height as the highest point of. outlet, it flows out until the
level of the water in the reservoir falls below the mouth of
the siphon, if the flow of water is greater than the supply to
the reservoir, in which case the flow ceases until the water
in the reservoir again reaches the level of the highest point
of the siphon.
THE LOCOMOTIVE BLAST
32. Fig. 14 shows the front end of a locomotive: B is
the exhaust pipe, the center of which is directly in line with
PIG 14
the center of the smokestack S, T, T are the tubes through
which the hot furnace gases are discharged. The exhaust
32
PNEUMATICS
29
steam has a pressure of about 2 pounds above the atmos-
phere, and rushes through the exhaust pipe E and up the
smokestack 5 1 with a very high velocity, taking the air out
with it, and producing 1 a partial vacuum in the space in
front of the tubes. No air can get in this space except
through the grates of the firebox, consequently, the partial
vacuum created in front of the tubes causes an influx of air
through the grate, and produces the forced draft, or blast.
The faster the engine runs, the greater is the quantity of
air drawn through the grate.
PUMPS
33. The Suction Pump. A section of an ordinary
suction pump is shown in Fig. 15. Suppose the piston
to be at the bottom of
the cylinder and to be
just on the point of
moving upwards in
the direction of the
arrow. As the piston
rises, it leaves a
vacuum behind it.
The air in P then
raises the valve P,
and expands in the
cylinder .#, whereby
its pressure is dimin-
ished below that of
the atmosphere. The
atmospheric pressure
on the surface of the
water in the well
causes the water to rise in the pipe P. When the piston
descends, the air in B escapes through the valves . After
a few strokes, the water fills completely the space under
the piston in cylinder JB, so that, when the piston reaches the
end of its stroke, the water entirely fills the space between the
30 PNEUMATICS 32
bottom of the piston and the bottom of the cylinder and also
the pipe P. The instant that the piston begins its down
stroke, the water in the chamber B tends to fall back into
the well, and its weight forces the valve Fto its seat, thus
preventing any downward flow of the water. The piston
now tends to compress the water in the chamber B, but this
is prevented through the opening of the valves u, u in the
piston. When the piston has reached the end of its down-
ward stroke, the weight of the water above closes the valves
u, u. All the water resting on the top of the piston is then
lifted with the piston on its upward stroke, and discharged
through the spout A, the valve V again opening, and the
water filling the space below the piston as before
It is evident that the distance between the valve V and
the surface of the water m the well must not exceed 34 feet,
the highest column of water that the pressure of the atmos-
phere will sustain, since otherwise the water in the pipe
3 would not reach to the height of the
valve V In practice, this distance
should not exceed 28 feet This is
due to the fact that there is a little
air left between the bottom of the
piston and the bottom of the cylinder,
a little air leaks through the valves,
which are not perfectly air-tight, and
a pressure is needed to raise the valve
against its weight, which, of course,
acts down wards. There are many vari-
eties of the suction pump, differing
principally in the valves and piston,
but the principle is the same in all.
34. The Lifting Pump. A
section of a lifting pump is shown
in Fig 16 These pumps are used
FIG IB when water is to be raised to greater
heights than can be done with the ordinary suction pump.
As will be perceived, it is essentially the same as the suction
32
PNEUMATICS
31
PlO. 17
pump, except that the spout is fitted with a
cock and has a pipe attached to it, leading to
the point of discharge. If it is desired to dis-
charge the water at the spout, the cock may
be opened, otherwise, the cock is closed, and
the water is lifted by the piston up through
the pipe P' to the point of discharge, the
valve c preventing it from falling back into
the pump, and the valve V preventing the
water in the pump from falling back into the
well. It is not necessary that there should
be a second pipe P f , as shown in the figure,
for the pipe P may be continued straight
upwards, as shown in Fig 17. This figure
shows a section of a lifting pump for raising
water from great depths, as from the bottom
of mines to the surface. The pump consists
of a series of pipes connected together, of
which the lower end only is shown in the
figure. That part of the pipe included
between the letters A and B forms the pump
cylinder, in which the piston P works. That
part of the pipe above the highest point of
the piston travel, thiough which the water is
discharged, is called the delivery pipe, and
the part below the lowest point of the piston
travel is called the suction pipe. The
lower end of the suction pipe is expanded,
and has a number of small holes m it, to
keep out solid matter. C is a plate covering
an opening, and may be removed to allow the
suction valve to be repaired. D is a plate
covering a similar opening, through which
the piston and piston valves may be repaired.
The pibton rod, or rather the piston stem, is
made of wrought iron, inserted with wood,
and connected with the piston. The only
limit to the height to which a pump of this
32 PNEUMATICS 32
kind can raise water is the strength of the piston rod Lifting
pumps of this kmd are used to raise water from great depths
to the earth's surface; hence, a very long piston rod is neces-
sary. In the lifting pump shown in Fig. 16, the water is raised
from a point a few feet below the earth's surface to a point
considerably higher. This requires the piston rod to move
through a stuffingbox, as shown at 5 1 , and also necessitates
the rod being round, in order that the water may not leak out.
35. Force Pumps. The force pump differs from the
lifting pump in several important particulars, but chiefly in
the fact that the piston is solid, that is, it has no valves A
section of a suction and force pump is shown in Fig. 18. The
water is drawn up the
suction pipe as be-
H - -T ^ F fore, when the piston
| 5 rises, but when the
piston reverses the
pressure on the water
caused by the descent
of the piston opens
the valve V and
forces the water up
the delivery pipe P'.
When the piston
again begins its up-
ward movement, the
valve V is closed by
Fro I8 the pressure of the
water above it, and the valve V is opened by the pressure of
the atmosphere on the water below it, as in the previous
cases. For an arrangement of this kind, it is not necessary
to have a stuffingbox. The water may be forced to almost
any desired height. The force pump differs again from the
lifting pump in respect to its piston rod, which should not be
longer than is absolutely necessary in order to prevent it
from buckling, while, in the lifting pump, the length of the
piston rod is a matter of indifference.
34
PNEUMATICS
.".2
pressure below the plunger being less than the pressure of
the atmosphere above, the air would rush in instead of being
expelled.
37. Double-Acting- Pumps. In the pumps previously
described, the discharge was intermittent, that is, the pump
could only discharge when the piston was moving in one
direction. In some cases, it is necessary that there should
PIG 20
U a continuous discharge; in all cases, it takes more power
to run the pump with an intermittent discharge, as a little
consideration will show. If the height that the water is to
be raised is considerable, its weight will be very great, and
the entire mass must be put in motion during one stroke
of the piston.
In order to obtain the advantage of a more continuous
discharge, double-acting pumps are used. Fig 20 shows a
part sectional view of such a pump. Two pistons a and 6
32 PNEUMATICS 35
are used, which are operated by one handle c in the manner
shown. The pump has one suction pipe s and one discharge
pipe d The cylinders e and / are separated by a diaphragm ,
so that they cannot communicate with each other above the
pistons In the figure, the handle c is moving to the right,
the piston a upwards, and the piston b downwaids In
moving upwards, the piston a lifts the water above it, causing
it to flow through the delivery valve h into the discharge
pipe d. This upward movement of the piston creates a.
partial vacuum below it m the cylinder <?, and causes the
water to rush up the suction pipe j into the cylinder, as
shown by the arrows In the cylinder /, the downwaid
movement of the piston b raises the piston valve v, and the
weight of the water on the suction valve z keeps it closed.
When the handle c has completed its movement to the right
and begins its return, all the valves on the right-hand side
open except v, and those on the left-hand side close except /;
watei is then discharged into the delivery pipe by the cylin-
der /, and only at the instant of reversal is the flow into the
delivery pipe d stopped.
38. Air Chambers. In order to obtain a continuous
flow of water in the delivery pipe, with as nearly a unifoim
velocity as possible, an air cluimber is usually placed on
the delivery pipe of force pumps as near to the pump cylinder
as the construction of the machine will allow. The air
chambeis aie usually pear-shaped, with the small end con-
nected to the pipe. They are filled with air, which the water
compresses during the discharge. During the suction, the
air thus compressed expands and acts as an accelerating force
on the moving column of water, a force that diminishes with
the expansion of the air, and helps to keep the velocity of
the moving column more nearly uniform. An air chamber
IH sometimes placed on the suction pipe. These air chambers
not only tend to promote a uniform discharge, but they also
equalize the stresses on the pump, and prevent shocks due
to the incompressibility of water. They serve the same pur-
pose in pumps that flywheels do m steam engines. Unless
36
PNEUMATICS
32
the pump moves very slowly, it is absolutely necessary to
have an air chamber on the delivery pipe.
39. Steam Pumps. Steam pumps are force pumps
operated by steam acting on the piston of a steam engine,
directly connected to the pump, and in many cases cast with
the pump. A section of a double-acting steam pump showing 1
the steam and water cylinders, with other details, is illustrated
in Fig. 21. Here G is a steam piston, and R the piston rod,
which is secured at its other end to the plunger P. Fis a
partition cast with the cylinder, which prevents the water in
FIG 21
the left-hand half from communicating with that m the right-
hand half of the cylinder. Suppose the piston to be moving
in the direction of the arrow The volume of the left-hand
half of the pump cylinder will be increased by an amount
equal to the area of the circumference of the plunger multi-
plied by the length of the stroke, and the volume of the right-
hand half of the cylinder will be diminished by a like amount
In consequence of this, a volume of water in the right-hand
half of the cylinder equal to the volume displaced by the
plunger m its forward motion will be forced through the
valves V. V into the air chamber A, through the orifice D,
32 PNEUMATICS 37
and then dischaiged through the delivery pipe H. By reason
of the partial vacuum in the left-hand half of the pump cylin-
der, owing to this movement of the plunger, the water will
be drawn from the reservoir through the suction pipe C into
the chamber K, A", lifting the valves S', S', and filling the
space displaced by the plunger. During the return stroke,
the watei will be drawn through the valves 5, 5 into the
right-hand half of the pump cylinder, and discharged
through the valves V, V in the left-hand half. Each of the
four suction and four discharge valves is kept to its seat,
when not working, by light springs, as shown.
There ate many varieties and makes of steam pumps, the
majority of which are double-acting. In many cases, two
steam pumps are placed side by side, having a common
dehveiy pipe. This arrangement is called a duplex pump.
It is usual so to set the steam pistons of duplex pumps that
when one is completing the stroke the other is in the middle
of its stroke A double-acting duplex pump made to run in
this manner, and having an air chamber of sufficient size,
will deliver water with a nearly uniform velocity
In mine pumps for forcing water to great heights, the
plungers are made solid, and in most cases are extended
through the pump cylinder. In many steam pumps, pistons
are used instead of plungeis, but when very heavy duty is
icquired plungers are preferred.
40. Centrifugal Pumps. Next to the direct-acting
steam pump, the centrifugal pump is the most valuable
instrument for raising water to great heights. As the name
implies, the effects produced by centrifugal force are made
use of. Fig. 22 represents a centrifugal pump with half of
the casing removed. The hub 5* is hollow, and is connected
directly to the suction pipe. The curved arms a, called vanes
or wliiffs, are revolved with a high velocity in the direction
of the arrow, and the air enclosed between them is driven out
through the discharge passage and delivery pipe DD* This
creates a partial vacuum in the casing 1 and suction pipe, and
causes the water to flow in through S, This water is also
I LT 398-19
58
PNEUMATICS
82
nade to revolve with the vanes, and, of course, with the same
velocity. The centrifugal force of the revolving water causes
t to fly outwards toward the end of the vanes, and becomes
jreater the farther away the water gets from the center This
:auses the water to leave the vanes, and finally to leave
he pump by means of the discharge passage and delivery
npe D D. The height to which the water can be forced
depends on the velocity
of the revolving vanes.
In the construction of
a centrifugal pump,
particular care is re-
quired in giving the
correct form to the
vanes, for the effi-
ciency of the machine
depends greatly on this
feature. What is
required is to raise the
water, and the energy
used to drive the pump
hould be devoted as much as possible to this one purpose.
"*he water, when it is raised, should be delivered with as
ittle velocity as possible, for any velocity that the water
hen possesses has been secured at the expense of the energy
sed to drive the pump. The form of the vanes is such
nat the water is delivered at the desired height with the
33 st expenditure of energy.
The number of vanes depends on the size and capacity of
lie pump. It will be noticed that, in the pump shown m
lie figure, the vanes have sharp edges near the hub. The
bject of this is to provide for a free ingress of the water,
nd also to cut any foreign substance that may enter the
ump and prevent it from working properly.
Almost any liquid can be raised with these pumps, but,
rhen they are intended for pumping chemicals, the casing
nd vanes should be made of materials that will not be acted
n by the chemicals.
FIQ 22
32
PNEUMATICS
39
41. Tlie Hydraulic Ram. The construction of a
hydraulic ram is shown in Fig. 23. This machine is used
for raising water from a point below the level of the water
in a spring or reservoir to a point considerably higher, with
no power other than that afforded by the inertia of a moving
column of water. In the figure, a is a pipe called the drive
pipe, connecting the ram with the reservoir; the valve b slides
freely in a guide, and is provided with locknuts to legulate
the distance that it can fall below its seat. When the water
is first turned on by opening the valve , the valve b is
already opened, and the water flows out through c, as shown.
PlO. 28
As the discharge continues, the velocity of the water in the
drive pipe will increase until the upward pressure against
the valve b is sufficient to force the valve to its seat. The
actual closing of the valve takes place very suddenly, and
the momentum of the column of water, which was moving
with an increasing velocity through the drive pipe a, will
very rapidly force some water through the valve d into the
air chamber /. Immediately after this, a rebound takes
place, and for a short interval of time the water flows back
up the drive pipe a and tends to form a vacuum under the
air-chamber valve d\ this opens the snifter valve g and admits
a little air, which accumulates under the valve d and is forced
into the air chamber with the next shock. This air keeps the
40 PNEUMATICS 32
air chamber constantly charged, otherwise, the water, being
under a greater pressure in the air chamber than in the reser-
voir, would soon absorb the air in the chamber and the ram
would cease to work until the chamber was recharged with
air The rebound also takes the pressure off the under side
of the valve b and causes it to drop, and the above-described
operations are repeated. The delivery pipe is shown at c; a
steady flow of water is maintained through it by the pressure
of the air in the chamber /; this air also acts as a cushion
when valve b suddenly closes, and prevents undue shock to
the parts of the ram.
The height to which water can be raised by the hydraulic
ram depends on the weight of the valve b and the velocity
of the water in a.
42. Power Necessary to "Work a Pump.
Principle I. In all pumps, whether lifting, force> steam
single- or double-acting, or centrifugal, the number of foot-
Pounds of power needed to work the pump is equal to the weight
of the water in pounds, multiplied by the vertical distance, in
feet, between the level of the water in the well, or sowcc, and the
point of discharge, plus the work necessary to ovej come the fric-
tion and other resistances.
Principle II. The work done in one stroke of a pump zs
equal to the weight of a volume of water equal to the volume
displaced by the piston during the stroke, multiplied by the total
vertical distance, in feet, through which the water is to be raised,
plus the work necessary to overcome the resistances.
A little consideration will make Principle II evident.
Suppose that the height of the suction is 25 feet; that the
vertical distance between the suction valve and the point
of discharge is 100 feet; that the stroke of the piston is
15 inches, and that its diameter is 10 inches. Let the diam-
eters of the suction pipe and delivery pipe be 4 inches each.
The volume displaced by the pump piston or plunger in one
stroke equals 10 X ; 7 _ 8 5 4 X *- = .68177 cubic foot. The
1,728
weight of an equal volume of water is .68177 X 62.6
32 PNEUMATICS 41
= 42 611 pounds Now, in order to discharge this water,
all the water in the suction and delivery pipes has to be
moved through a certain distance, in feet, equal to 68177
divided by the area of the pipes, in square feet.
4 inches = foot, (i)" X .7854 = ^f- = .0812$ square
y
foot. .68177 -^ .08723- = 7 8126 feet.
The weight of water in the delivery pipe is (^) a X .7854
X 100 X 62.5 = 545 42 pounds.
The weight of water in the suction pipe is (i) X .7854
X 25 X 62 5 = 136.35 pounds.
545 42 + 136 35 = 681 77 pounds, which is the total weight
of water moved in one stroke. The distance that the water
is moved m one stroke is 7.8125 feet; hence, the number of
foot-pounds necessary for one stroke is 681.77 X 7.8125
= 5,326 3 foot-pounds. Had this result been obtained by
Principle II, the process would have been as follows: The
weight of the water displaced by the piston in one stroke
was found to be 42.611 pounds. 42.611 X 125 = 5,326.4
pounds, which is practically the same as the result obtained
by the previous method, and is a great deal shorter. The
slight difference between the two results is due to neglected
decimals.
EXAMPLE. What must be the necessary horsepower of a, double-
acting steam pump if the vertical distance between the point of dis-
charge and the point of suction is 96 feet? The diameter of the
pump cylinder is 8 inches, the stroke is 10 inches, and the number of
strokes per minute is 120. Allow 25 per cent, for friction and other
resistances.
SOLUTION. Since the pump IB double-acting, it raises ft quantity
of water equal to the volume displaced by the plunger at every stroke.
The weight of the volume of water displaced at one stroke la (A)*
X .7854 X itf X 02.5 = 18 18 lb., nearly.
18.18 X 90 X 120 - 209,430 ft.-lb. per minute.
Since 25 per cent. Is to be allowed for friction, the actual number
of foot-pounds per minute is 209,430 + .75 - 279,240. 1 H. P.
= 33,000 ft.-lb. per min,; hence, 8.462 H. P., nearly. Ans.
2 RUDIMENTS OF ANALYTIC GEOMETRY 33
If the acceleration a is supposed to have a fixed value, as
8 feet per second, and different values are assigned to /,
different values will be obtained for s. Here, too, / and .v
are variables /, which is varied at pleasure, is the independent
variable; and s, whose values depend on those of t, is the
dependent variable, or a function of /. The acceleration a,
although represented by a letter, is assumed to be fixed or
invariable, and is therefore a constant.
2. In general, when, for any particular purpose, some of
the quantities represented by letters in an equation are made
to take (or are considered as being such that they can take)
different values, they are called variables. Those to which
values are assigned arbitrarily are called Independent
variables; those whose values depend on the values of
the independent variables are called dependent variables,
or functions of the independent variables. Those quantities
that are supposed to remain fixed are called constants.
A function may also be defined as a quantity whose value
depends on the value or values of one or more other quan-
tities; the very word "depends" indicates that the function
can have different values (that is, can vaiy) according to the
values assigned to other quantities. Thus, the area of a
triangle is a function of the base and altitude; if the base
is assumed to be fixed, it becomes a constant, and the area
is a function of only the altitude. The velocity of a body
moving under the action of an unbalanced force is a function
of the magnitude of the force and the mass of the body; if
the mass is assumed to be fixed, it becomes a constant, and
the velocity is a function of only the force.
3. Graph of an Equation. It was stated in Art. 1
that from the equation A = xi*, a table can be made
giving the values of the function A corresponding to different
values of the independent variable r. Instead of a table, a
diagram may be constructed to represent fte relation between
the values of A and r. Such a diagram; w&ick is the graphic
representation of the equation just givei^, is palled the graph
of that equation, and is constructed as foj
I \
4 RUDIMENTS OF ANALYTIC GEOMETRY $33
value of ns laid off along OX from O to, say, A/, a per-
pendicular is erected at M, intersecting the graph at P.
Then MP will represent the value of A corresponding to
the value OM of r.
4. The perpendicular distances of any point of the graph
from the axes are called the coordinates of that point.
Thus, the coordinates of P are MP (= OM') and M 1 P
(= OM}. The horizontal coordinate OM, or, more gen-
erally, the coordinate representing the independent variable,
is usually called the abscissa, and the other coordinate
OM, the ordinate. It is customary to reckon the abscissa
along the axjs OX, called the axis of x, and the ordinate
on a line parallel to O Y t through the foot of the abscissa.
Thus, the coordinates of P are stated as OM and MP
instead of M'P and MP.
6 RUDIMENTS OF ANALYTIC GEOMETRY 33
connects with the crank OJfby the connecting-! od ff K It
is shown m mechanics that when the crank has described an
angle X from the position OA, which is in line with the axis
of the cylinder, the velocity u of the piston is given approxi-
mately by the formula
/ . v , a sm 2 X\
u = v ( sm X -\ - 1
\ 2/ /
in which v = linear velocity of crankpm K;
I = length of connecting-rod;
a = length of crank.
If the ratio - is represented by <:, the formula may be
If
written,
- = sinAr+sin2AT (1)
v 2
When c is given, the graph of equation (1) can be con-
structed by taking X as the independent variable, and u as
v
L
r
PlQ 3
the function Then, the value of - for any value of X can
v
be found from the graph, and, when v is given, u can be
determined by multiplying by v the value found for --.
v
EXAMPLE 1 To construct the graph of equation (1) when c = $,
for values of X varying from to 180, that Is, for one-half a revolu-
tion of the crank, from the position OA to the position O A', Fig. 3.
SOLUTION Writing, for shortness, y for -, and substituting the
given value of c, equation (1) becomes
y = em X + -rV sm 2 X
By giving to lvalues from to 180, at intervals of 6, the follow-
ing table is obtained:
RUDIMENTS OF ANALYTIC GEOMETRY
EXAMPLES FOR PRACTICE
1 Plat the following equations for values of x varying from
3 to 8 (a) y = X j* - 7 x* + 4 x - 6 (6) v = t x 9 - 1<> x + 4
a) The general form of the guiph is shown in Fig f)
6) The general form of the graph Is shown in Pig
c) The general form of the graph is shown in Fig 7
PIG 5
10 RUDIMENTS OF ANALYTIC GEOMETRY 33
have been proposed to determine the relation between these
two quantities, no exact formula has yet been found. Sup-
pose that a series of experiments on wrought-iron columns
gives the following results:
RATIOS OF LENGTH BREAKING LOAD
TO DIAMETER POUNDS PER SQUARE INCH
6 51,200
7 47,400
8 44,600
9 42,200
10 40,200
11 38,700
12 37,900
13 37,100
14 36,900
These results may be represented graphically, as shown
in Fig 9. Having drawn two coordinate axes OX and O Y,
r\
V,
5
Batio of Length to Diamter
JUT |
FIG.
the values of the ratio of length to diameter, which is the
independent variable, are laid off from O, along OX, to
any convenient scale. Thus, OM* represents the ratio 6;
OJIf represents 7; OM, t 8; etc. The corresponding values
12 RUDIMENTS OF ANALYTIC GEOMETRY 33
values written on the respective projections. Thus, the
point P, is projected at MJ, on which is written 31,500,000,
the population in 1860.
The probable population m any intermediate year can be
readily obtained from the curve. For instance, if it is
desired to find the population in 1878, it will be observed
that 1878 lies between 1870 and 1880, and that the interval
between 1870 and 1878 is .8 of the interval between 1870
and 1880. Therefore, the required population is found by
laying off M t M equal to .8 of M*M t , and drawing the
ordmate MP, which
JT'
P-r represents the approx-
/ imate population in
Pa/ 1878.
P B / 8. In some cases, a
*?' diagram is constructed
m yf |
y rather for the purpose
_3isoopoo_ J:*/ of presenting to the
o -"a p *s e y e > m a striking man-
Pl ^ ner, the variations ol
certain quantities, than
~ ^r f or the purpose of de-
1840 Bo70
Year values, the variations
Pl 10 m the latter being too
irregular. In the examples so far given, the curves are fairly
regular, which shows that the variations in the functions
(ordinates) are not too abrupt, and that the curves may be
depended on to give tolerably approximate values of the
function corresponding to intermediate values of the inde-
pendent variable Example 1 of the following Examples for
Practice is a case in which the curve, on account of its too
great irregularity, could not be depended on to give inter-
mediate values. In such cases, the extremities of the ordi-
nates are joined by straight lines, instead of by a curve.
33 RUDIMENTS OF ANALYTIC GEOMETRY 13
EXAMPLES FOR PRACTICE
1 Draw a. graph representing the average daily water consumption
in the city of Brooklyn between January, 1H71, and December, 1873,
from the following data.
YE.R
MO.TH
January
21,000,000
s.
April . . .
17,500,000
oo
rH
July . . ...
. . 19,600,000
, October .
. . 18,500,000
23,600,000
2
April .
20,500,000
SB'
I-H
July
. 22,500,000
October
. . . 23,000,000
January . .
. . . 29,000,000
CO
April . . .
22,500,000
t- ,
00
July ...
26,500,000
rH
October ...
. 22,500,000
December ...
24,500,000
Ans.
The curve is shown in Pig. 11
o
PIG. 11
2. (a) Taking the hours as abscissas, and the discharges as ordi-
nates, construct a graph for the water discharged by a pipe from the
following observed values for 1 day, (b] determine, by means of the
curve, the probable discharge at 11:30 A. M.; (c] determine the prob-
able discharge at 5:80 F. M.
. The scale of ordlnatoi should be chosen sufficiently large to
show the differences In discharge say 4 Inches to 1 cable loot.
RUDIMENTS OF ANALYTIC GEOMETRY
HOUR
A. M.
DISCHARGE
CUBIC FEBT PER
SKCOND
HOUR
A M
DISCHARGE
CUBIC FEET PER
SECOND
1
2.45
11
2.54
o
2 67
M.
3
2 69
12
233
4
2.62
f M.
5
2.36
1
264
2.30
2
215
7
2.23
3
1 19
8
2.19
4
119
2 2O
5
226
10
2.OO
8
237
Aus.
FIG 12
The general form of the graph is shown in Fig. 12
2 43 cu. ft. per sec.
2 31 cu. ft per sec.
33 RUDIMENTS OF ANALYTIC GEOMETRY 15
EQUATIONS OF LINES
INTRODUCTION
9. Equation of a Line. As already explained, a
graph is constructed from an equation expressing a relation
between two variables, or giving the value of one variable as
a function of the other. In
V
the graph, these variables
are the coordinates of
points on a line, usually
curved. Sometimes, on
the contrary, a line is given,
and it is required to find the
equation of which the line
is the graph. That equa-
tion is called the equation
of the given line, and is a
general expression of the
relation between the two
coordinates of any point of
the line, with reference to two coordinate axes conveniently
chosen.
Take, for instance, a circle of radius r, Fig. 13, and two
rectangular axes OX, OY, passing through its center. Let
P be any point on the circumference, and x and y its coordi-
nates, as shown. The right triangle OMP gives
OM* + MP* = OP 1 -,
that is, x'+y* - r' (1)
This is the equation of the circle referred to two rectangu-
lar axes through the center. In that equation, x and. y
are the coordinates of any point on the circumference. No
16 RUDIMENTS OF ANALYTIC GEOMETRY 33
matter where the point is located, its coordinates satisfy
equation (1). Thus, for the point P lt
x = -OM lt y = -M 1 P 1 ,
and
= OP,' = S
10. The equation of a line is useful in the study of the
geometric properties of the line, and it often serves to
recognize the form of a graph corresponding to a given
equation If, for example, it is found in the solution of a
\f mechanical problem that
the path AB, Fig. 14, of a
moving point is such that
the coordinates x and y of
any of its points, with re-
gard to the axes OX and
v O y, are related by the
equation
x" + y = a'
o = - ' -- x tne quantity a being con-
1 Pl 14 stant, it can be at once
concluded that the path of the point is a circle whose center
is and whose radius is a.
11. Analytic geometry is that branch of mathematics
m which geometric figures are studied by means of their
equations. Surfaces, as well as plane lines, have equations;
but in this Course only a few plane lines will be treated.
THE STRAIGHT LINE
12. Equation of the Straight Line. Let X' X and
y y, Fig. 15, be two axes of coordinates, and A B a straight
line making with X'X an angle H, and intersecting Y' Y
at 7, the distance b (= 01} being known. It should be
understood that the angle H is always measured from the
axis X> X upwards. Thus, if the line were 4iB the
angled would be XJ^B,. Also, b, like the ordinates, is
53 RUDIMENTS OF ANALYTIC GEOMETRY 17
positive upwards and negative downwards. Thus, for the
line A t Bi, the value of b is Of, This being understood,
the equation now to be derived is entirely general, and, with
the symbols interpreted as just explained, applies to all
cases.
The right triangle PMJ gives
PM = JM tan H\
y = (x 4- JO] tan H.
L j. r T
that is,
Now,
therefore, y =[x
- ,
tan H
-- -} tan H = x tan H + b
tan HI
This is the equa-
tion of the line A B.
It is customary to
denote tan H by a,
and write the equa-
tion in the form
y = ax + b
The distances 01
and J are called the
Intercepts of AB
on the axes of y and*,
respectively . 1 1 i s
evident that, at 7, the
abscissa x is 0, and
the ordinate is 01]
and at J the ordi-
nate y is 0, and the
abscissa is OJ.t
13. Graph of Any Equation of the First Degree.
Any equation of the first degree between two variables can
be represented by a straight line; in other words, the graph
of any equation of the first degree between two variables is
a straight line,
Let mx+ ny = P (1)
be any equation of the first degree between the variables
18 RUDIMENTS OF ANALYTIC GEOMETRY 38
x and y, the other quantities m,n,p being constants, either
positive or negative. Solving the equation for y,
n n
Since the tangents of the angles between and 180 con-
tain all possible numbers, both positive and negative, it is
always possible to find an angle whose tangent is . Let
that angle be denoted by H, and denote * by b. Then,
n
= tan H, and equation (2) becomes
n
y = jirtan/f + b (3)
If is positive,
n
H is less than 90; if
negative, H is greater
than 90.
If on the y axis the
distance b is laid off
from O t upwards if b
x is positive, down-
wards if b is negative,
and from the extrem-
ity of b a line is drawn
making with the x
axis an angle equal
to H, that line is the
graph of equation (3) ,
r and, therefore, also
FIG 16 of equation (1). If b
and tan H are positive, the line will have such a position
as A,B^ Fig 16, in which O A = b, and XJ,B, = H. If &
is positive and tan H negative, the graph will have a position
like A t B t > in which OI t = b, and XJ,B t = H. If b is neg-
ative and tan H positive, the graph will be like A, JB,, in
which OI> = b, and XJ,B* = H. If both b and tan H are
negative, the graph will have such a position as A^ B,..
33 RUDIMENTS OF ANALYTIC GEOMETRY 19
14. Because any equation of the first degree between
two variables can be represented by a straight line, formulas
in which the value of a quantity is given in terms of the first
power of another are called straight-line formulas. For
example, the formula
p = 10,000 - 45 -
r
which expresses the intensity p of pressure that a column can
stand, m terms of the ratio - of length to radius, is a straight-
r
line formula. If y is
written instead of p,
and x instead of-, the
r
formula becomes
y = 10,000-45*
= _45*+ 10,000
which is the standard
form of the equation
of a straight line.
15. The simplest
way to draw the
straight line corre-
sponding to an equa-
tion of the first de-
gree is as follows:
Let the equation be
mx -\-ny-\-p = 0.
Select the axes in any convenient position, as X 1 X and V K,
Fig. 17. Making x = in the equation, and solving for y,
the intercept (say <?/,) on the y axis is obtained, and the point
7, where the line intersects that axis is determined. Making
y = 0, and solving for x, the intercept 07, and the point ,/i are
obtained. The line is then drawn through the points 7, and /,.
EXAMPLE 1. To draw the graph of the equation 8# l&y + 20-0.
SOLUTION^ Draw the axes X'X< Y> Y, Fig. 17 Making x In
the equation, we have -16^ + 20 = 0; whence
y - 0/ 1.26
20 RUDIMENTS OF ANALYTIC GEOMETRY 33
Making y = in the equation, we have 8x + 20 = 0, whence
x = OJ= -26
Laying off Of = 1 25, and OJ =25, the line A B drawn through 1
and J is the required graph Ans
EXAMPLE 2 To draw the graph of the equation 10 x + 8y + 40 = J.
SOLUTION Making x 0, and solving for y, Pig. 17,
= 01 = - = -5
Making y = 0, and solving for x,
* - n T 40 _ 4
* ~ l = ~ 10 ~ ~ 4
Laying off OJ* = 4, 07; = 5, and joining/! and J lt the required
graph A! B* is obtained Ans.
EXAMPLE 3 To draw the graph of the equation 4j/ 8x = 0.
SOLUTION It will be noticed that this equation has no term
independent of x and y,
This shows that the line
passes through the origin
of coordinates, or that its
intercepts on the two axes
are zero This follows at
once from Art 13, for
it was there shown that
>-, and, as
n
in this
case,/ = 0, it follows that
b = In this case, the
graph cannot be con-
structed as in the two pre-
ceding examples. Since
the line passes through
the origin O, Pig. 17, it
is only necessary to de-
termine another point.
This is done by assu-
ming any convenient value for x and solving for y Making*- = |, the
equation becomes 4j/ - 4 = 0, whence y = 1 Laying off OM^ i,
and the ordmate MP = 1, the line A, t , drawn through and JP,
is the required graph
FOR PRACTICE
Draw the graphs of the equations (a) 8
(c) 5x+4y = -20. rf I2y - 10 .
Ans
The
are shown, respectively, at
tt A, B, s AtJBt in Pig. 18
33 RUDIMENTS OF ANALYTIC GEOMETRY 21
2 Find the angle that the graph of the equation x + &y 4 =
makes with the x axis. Ans. 141 20> 20"
3 Given the equation x-\- 4y = 20, find (a) the intercept of the
graph on the y axis, (b) the angle that the graph makes with the
x axis (Give seconds in angle to nearest multiple of 10.)
165 57' 60"
Ans
.
{(?)
APPLICATIONS
16. Reactions on a Beam. Let a beam A B t Fig. 19,
resting on the supports A and B> carry a movable load W.
Let the distance of the load from the left support at any
instant be x, and let the length of the beam be denoted
by /. For this position of the load, the reaction Ri at A is
obtained by taking moments about B\ thus,
RJ- W(l-x) = 0;
whence
I
(1)
Since the load W moves, x is a variable, and so is
Equation ( 1 ) gives R l
as a function of x. As
that equation is of the
tirst degree, it can be
represented by a
straight line, Ri being
used instead of the y
used in previous arti-
cles. A convenient
and usual way of
drawing the graph is
as follows: The ori-
gin is taken direct-
ly under A, and the
axis OX is drawn
parallel to A B\ the axis O Y is drawn through O. Equa-
tion (1) shows that the y intercept is W. Therefore, laying
off, to any convenient scale, along O Y the distance Of to
represent W, one point / of the graph is obtained. The
equation also shows that the x intercept (fotind by making
Pio, 19
22 RUDIMENTS OF ANALYTIC GEOMETRY 33
J?, = 0) is /. Therefore, projecting the point B on OX at /,
the point J of the graph is found, and the graph is the
straight line IJ. The ordmate M P represents the left leac-
tion when the load is at W\ the ordmate M' P' represents
the left reaction when the load is at W; etc. If 1 ' O is
drawn parallel to OX, the ordinates M,P, AfJP', etc. will
represent the corresponding values of the right reaction,
since jR, + R* = W= Of. If, for example, the load is
2,000 pounds, and a scale of 1,000 pounds to the inch is
used, Of should be made 2 inches; and, if the ordinate M'P'
is found to measure li inches, the reaction R^ when the
load is at W>, is 1,000 X 1* = 1,500 pounds.
17. Pressure on the Back of a Dam. Let
Fig. 20, be the back or inner face, supposed to be vertical,
o of a dam, the water reach-
ing to the top From
hydrostatics it is known
that the pressure p per
square foot, at any point M
whose depth below the sur-
face is x feet, is given by
the formula
p = wx
-* m which w is the weight
of 1 cubic foot of water.
Fl M As this is an equation of
the first degree, it can be represented by a straight line.
Taking O M,. as the axis of x and as the origin, the latter
point is a point in the graph, since there is no intercept.
Making* = h, the formula gives p = w h t which is the pres-
sure at Mr Laying off, horizontally, M r P t = wh, the line
O P l is the required graph. The intensity of pressure at
any point M is given by the ordinate M P.
If, for instance, the height k is 24 feet, and w is taken
equal to 62.5 pounds, wh = 62 5 X 24 = 1,600 pounds per
square foot. If a scale of 500 pounds per square foot to the
inch is used, M^P^ should be made 1,500 -T- 500 = 3 inches;
33 RUDIMENTS OP ANALYTIC GEOMETRY 23
and, if M ' P measures li inches, the pressure at M is
500 X H = 625 pounds per square foot.
It will be observed that here p is used instead of y, that
the x axis is vertical; and that positive values of x are
counted downwards, and positive values of p toward the
left It is often necessary to make such changes in notation,
so as to adapt the construction to given conditions. The
general principles and methods, however, remain the same.
THK PARABOLA
18. Definitions. A parabola, Fig. 21, is a curve such
that, if any point P on it is taken, the distance P F of that
point from a fixed point F is s r
equal to its distance PN
fiom a fixed line DE. The
fixed point F is called the
focn&j the fixed line DE,
the directrix. The line N, X
passing through the focus
and perpendiculai to the di-
rectrix is called the axis of
the curve, and the point O
where the curve crosses the
axis is called the vertex.
Twice the distance FN<>
from the focus to the di-
rectrix is called the par-
ameter, and is denoted by
2/>; so that 2/ = 2 N.F,
and p = N,F.
The curve is symmetrical with respect to the axis; that is,
to every point P on one side of the axis there corresponds
another point P 1 on the other side, at the same distance from,
and on the same perpendicular to, the axis. Any line, as JPP',
perpendicular to the axis and bounded at its two ends by
the curve is called a double ordluate.
FIG. 21
24 RUDIMENTS OF ANALYTIC GEOMETRY 33
19. Equation of the Parabola Referred to the Axis
and Vertex. Let P, Fig. 21, be any point on the curve,
and O M = x and M ' P y its coordinates, the axes being
the axis OX of the parabola and the perpendicular Y at O
Let the parameter be 2/>. Then, the distance FN from the
focus F to the directrix DE is equal to p, and OF = vfl,
since ON a OF, the point O of the curve being, according
to the definition of a parabola, equally distant from the
directrix DE and the focus F. According to the same
definition, we have
PF= PN
and, therefore,
PF' = PN' (1)
Now,
PF' = PM' + FM' = y 9 + (OM- OF)'
Also,
= N M' = (OM+N t O)' = (x + p)* = A
-\
Substituting in equation (1) these values of PF' and PN',
4 4
whence y' = %px
which is the required equation.
20. Let y^ and y tt be two ordinates corresponding to the
abscissas x* and x, Then, since the equation applies to all
points, y* = 2px lt y' = ZpXi, whence
That is, in any parabola, any two coordinates parallel to the
axis of the curve are to each other as the squares of the corre-
sponding coordinates perpendicular to the axis.
21. It is important that the equation of the parabola
should be so mastered that it can be applied to parabolas in
different positions, and when the notation is different from
the one here used. It should be borne in mind that in the
S33 RUDIMENTS OF ANALYTIC GEOMETRY 25
general equation, the coordinate to be squared is that per-
.pendicular to the axis of the parabola. In Fig. 22 is repre-
sented a parabola with the axis vertical, coinciding with the
y axis. In this case, the equation should be written
Fig. 23 represents a vertical parabola with its axib down-
T x t o
Fio. 22 Pro 23
wards The coordinate axes are denoted by O S and O T,
as shown, and the coordinates of any point P by s and /,
5 being positive downwards, and t positive toward the left
Under such conditions, the equation of the curve should be
written
22. Problem I. To find the parameter and equation of
a parabola, when a double ordinate and the corresponding abscissa
are given.
This is the usual way in which the parabola occurs in
practice. Thus, in road construction, in which the cross-
section of a road is often made parabolic, the width a of the
road, Fig. 24, and the height h of the center above the
ends A and B are given. In order to determine the fall of
the cross-section at different distances from the center, it is
necessary, or at least advisable* to determine the equation
and parameter of the curve A OB.
26 RUDIMENTS OF ANALYTIC GEOMETRY 33
As usual, the parameter will be denoted by 2 p. At the
point M , for which x = -, and y = A, we have, since here
L
the x coordinate is perpendicular to the axis,
whence e lp = ^~
4:h
The general equation of the parabola is, therefore,
*
In practice, points on the curve are determined by assu-
ming values for x, and computing the corresponding values
of y. The equation may, therefore, be more conveniently
written in the form
a' \i aj
Let OM t be divided into any convenient number of equal
Jf,
>F
FIG 25
parts, say n, and give to x successive values correspond-
ing to the points of division; thus, x = ", x = 2 -^-^,
n n
x = 3 -, etc.; or x = , x - 2 , # = 3 , etc. Then,
n n n n
the corresponding values of y beccme
Thus, if OM, is divided into ten equal parts, and OM con-
tains six of those parts,
-Axr-ixH
33 RUDIMENTS OF ANALYTIC GEOMETRY 27
EXAMPLE Given the double ordinate A B = 40 feet, Fig 25, and
the corresponding abscissa OK = 15 feet, it is desired to hnd points
on the curve at intervals of 5 feet on each side of O
SOLUTION Draw O M t parallel to A B, and B MI parallel to KO
Since the points are to be located every 6 ft from O, and O M t
= 20 ft , the latter line should be divided into four equal parts O MI,
Jlf, Mt, etc Here, i0=jx*0 = 20, = ^=4, and h = 1 5.
Therefore, - = ^ =
094, nearly, and the values of y are
SLtM lt Mi PI = 094 X 1
at M., M, /> = 004 X 2
at Af a , M a /> = .094 X 3 1
at Jtf t , Jlf t n = 1 5 ft
09 ft , nearly,
094 X 4 = 38 ft , nearly,
.094 X9 = .85 ft., nearly,
23. Problem II. To construct a parabola when a double
ordinate and the con espotidmg abscissa are given.
In practice, what is usually required is to locate points
of the curve on the % ft
ground, as in the case
of road and street
construction, and in
railro ad curves.
Under such circum-
stances, the ordinates
are calculated as ex-
plained in Art. 22; if
desirable, they may
be platted, and the
curve drawn through
the points thus deter-
mined. A purely
graphic method of
constructing the
curve is explained in
Geometrical Drawing*
Another method,
which is often convenient in the drafting room, is as follows:
Let AB> Fig. 26, be the given double ordinate, and OK
the corresponding abscissa. Bisect ATX at Cand draw CI
parallel to KO and OI perpendicular to K O. Draw /A",
ILT39&-21
Fro 26
28 RUDIMENTS OF ANALYTIC GEOMETRY 33
and IN a perpendicular to it, meeting the axis produced at N .
The line D E, drawn through N perpendicular to OK pro-
duced, is the directrix It is shown here for the purpose
of explanation, but it is not necessary to draw it. Lay off
OF = ON.. The point/' is the focus. From 7V , lay off
along the axis any convenient distances, as N M, N M^ etc ,
and at the points M, M 1} etc. draw indefinite perpendiculars
8Q f > QiQS> to the axis. From F as a center, and with
a radius equal to NM, describe an arc, cutting QQ' at
P and P'\ these are points of the curve. Likewise, the points
P! and P/ are the intersections of ?, Q l / with an arc described
from Fwiih a radius equal to N M* Other points may be
determined in a similar manner, and the curve drawn through
them.
It is usually more convenient to find the points N 9 and F
by calculation, instead of by the geometrical construction
described above. The parameter 2^ is computed as explained
in Art. 22, and then N and OF are laid off each equal
NOTE The correctness of the preceding construction will now be
shown It is not necessary for the student to study the following
demonstration, but he is advised to read it carefully, as it is a good
exercise As explained In Art 22, 2/ = ~ In Fig 26, fl = 2 A K
and A = OK. Therefore,
P
AK*
In the right triangle N, IK, the perpendicular Of, which is equal
to \A K t is a mean proportional between O K and O N a . that is,
(IAK)*= OJV.XOJf,
whence ON, = ^frjj = | [by equation (1)]
Therefore, O N a Is the distance from the vertex to the directrix
(Art 18)
The point P was so determined that FP = N a M = PN. That
point is, therefore, at the same distance from the focus as from the
directrix, and, according to the definition of a parabola, must be a
point of the curve The same reasoning applies to the points P 1 , A, etc.
33 RUDIMENTS OF ANALYTIC GEOMETRY 29
APPLICATIONS
24. Projectiles. In the most general sense of the
term, a projectile is any body thrown into the air The
velocity with which a projectile is thrown is called the initial
velocity, or velocity of projection. Here, only pro-
jectiles thrown horizontally will be considered, and the
resistance of the air will be neglected.
Let a projectile be thrown horizontally in the direc-
tion OX from a point O, Fig 27, with a velocity v. It is
required to determine the path OA of the projectile. The
horizontal line OX and the vertical line O Y will be taken
as axes of coordinates, y being positive downwards. Were
the projectile not acted on by gravity, it would describe, in
any time /, a space OM = vt, which o
will be denoted by x. If it were not
thrown at all, and were allowed to fall
freely from O during the time t, it would
fall through a distance O G = %gf,
denoting, as usual, by g the accelera-
tion due to gravity. This distance O G
will be denoted by y. When the pro- PIG. 27
jectile, after being thrown along OX with the velocity v, is
acted on by giavity, its position P, at the end of the time t,
will be such that M P will be equal and parallel to O G. We
have, therefore,
x = vt
and y = i g t*.
yjt
From the first of these two equations is found /* = -.
v
Substituting this value in the second equation, there results
2 w"
whence x' = y,
#
which is the equation of a parabola with its vertex at O and
axis vertical. This parabola is, therefore, the path followed
2 if
by the projectile. Its parameter is .
30
RUDIMENTS OF ANALYTIC GEOMETRY 33
25. Jet of Water Issuing 1 From & Small Orifice.
One of the most important applications of the theory of pro-
jectiles is to the determination of the velocity of water
flowing from a small orifice. This determination is of much
importance in hydraulics. In Fig. 28 is represented a tank T
from which water flows through a small orifice O. The
water issues horizontally with a velocity v (to be deter-
mined), and each particle, being under the same conditions
as a projectile thrown horizontally with the same velocity,
describes a parabola.
As the jet is narrow,
it may be treated as a
whole as a parabolic
arc OJ. A horizon-
tal string OX is
stretched from O, and
any distance OM^ is
measured. From M*
Pl another string carry-
ing a heavy weight is suspended, and the distance M l P t is
measured. Let OM l = x lt Af l P l = y lm Then,
whence
EXAMPLE. What was the velocity of a jet for which the measured
distances x^. and jy, were, respectively, 6 and 5 feet?
SOLUTION Substituting m the formula the given values of .*-,
and y lt and 32 16 for^-,
x g - = V3 216 X 36 = 10 76 ft. per sec Ans
26. Moment In a Beam. Let a beam A B, Fig. 29, of
length /, carry a movable load W. When the load is at a
distance x from the center of the beam, the left reaction R 1
is found by taking moments about B\ thus,
RJ =
whence R^ =
33 RUDIMENTS OF ANALYTIC GEOMETRY 31
The moment of .#, about I^is called the bending moment
at W. Denoting it by M, we have
The bending moment at the center, which will be denoted
by M^ is obtained by making x = 0, which gives
". - \ L
Equation (1) may p 1
be written
M M
whence
If M - M is de-
noted by y, equation
(1) may be written
/
x
which is the equa-
tion of a parabola
whose axis is perpen-
dicular to A B, and whose parameter is
The curve
is shown at A 1 OB 1 . It may be constructed by points,
assuming values of x and finding the corresponding values
of M and M t M\ or by the method of Art. 23, noticing
that to the double ordinate A' #', or /, corresponds a value
of y (= fCO) equal to M^ since, when x = J/, M 0, and
y = M, M M,. Having drawn the curve, the bending
moment, when the weight is at any point W, is found by
projecting W on A' /?', and measuring the distance QP,
which is equal to the moment at W> to the scale by which
KO represents M,, For -tfie ordinate NP represents y, or
M, M, and, therefore,
M.-M= NP, M = M.-NP = Qfi-NP= QP