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Statics and dynamics for engineering stu
3 1924 004 689 166
STATICS AND DYNAMICS
FOR
ENGINEERING STUDENTS.
IRVING P. CHURCH, C.E.,
ASST. PROFESSOR OF CIVIL ENGINEERING, CORNELL UNIVERSITY.
SECOND EDITION.
NEW YORK:
JOHN WILEY & SONS.
1886.
UNIVERSITY
\LIBRARY
Copyright, 1886,
By IRVING P. CHURCH.
. Press of J. J. Little & Co.,
Nos. 10 to 20 Astor Place, New York.
PREFACE.
For the Engineering student, pursuing the study of Applied
Mechanics as part of his professional training and not as ad-
ditional mathematical culture, not only is a thoroughly system-
atic, clear, and logical treatment of the subject quite essential,
but one which presents the quantities and conceptions involved
in as practical and concrete a form as possible, with all the aids
of the printer's and engraver's arts ; and especially one which,
besides showing the derivation of formulae from principles, in-
culcates, illustrates, and lays stress on correct numerical substi-
tution, and the proper use of units ; for without this no reliable
numerical results can be reached, and the principal object of
those formulae is frustrated.
With these requirements in view, an experience of ten years
in teaching the Mechanics of Engineering at this institution has
led the writer to prepare the present work, embodying the fol-
lowing special features :
1. The diagrams are very numerous (about one to every
page ; an appeal to the eye is often worth a page of description).
2. The diagrams are very full and explicit, thus saving time
and mental effort to the student. In problems in Dynamics
three kinds of arrows are used, to distinguish forces, velocities,
and accelerations, respectively.
3. Illustrations and examples of a practical nature, both
algebraic and numerical, are of frequent occurrence.
4. Formulae are divided into two classes ; those (homo-
geneous) admitting the use of any system of units whatever for
measurements of force, space, mass, and time, in numerical
substitution ; and those which are true only for specified units.
Attention is repeatedly called to the matter of correct numerical
substitution, especially in Dynamics, where time and mass, as
well as force and space, are among the quantities considered.
5. The general theorem of Work and Energy in machines is
developed gradually by definite and limited steps, in preference
to giving a single demonstration which, from its generality,
might be too vague and abstruse to be readily grasped by the
student.
6. As to the vexed question of " Centrifugal force," a con-
sistent application of Newton's Laws is made, necessitating the
result that the centrifugal force, properly so called, is not an
action (force) upon the body constrained to move in the curve,
but upon the constraining body.
7. The definition of force is made to include and illustrate
Newton's law of action and reaction, the misunderstanding of
which leads to such lengthy and unprofitable discussions in
technical journals every few years.
Cobnell University, Ithaca, N. T.,
February, 1886.
TABLE OF CONTENTS.
PRELIMINARY CHAPTER.
PAGE
§§ 1-15(1. Definitions. Kinds of Quantity. Homogeneous Equa-
tions. Parallelogram of Forces 1
PAET I. STATICS.
Chapter I. Statics op a Material Point.
§§ 16-19. Composition and Equilibrium of Concurrent Forces 8
Chapter II. Parallel Forces and the Centre op Gravity.
§§ 20-22. Parallel Forces 13
§§ 23-27*. Centre of Gravity. Problems. Centrobaric Method. . . 18
Chapter III. Statics op a Rigid Body.
§§28-34. Couples 27
§§ 35^-39. Composition and Equilibrium of Non-concurrent Forces. 31
Chapter IV. Statics of Flexible Cords.
§§ 40-48. Postulates. Suspended Weights. Parabolic Cord. Cat-
enary 42
PAET II. DYNAMICS.
Chapter I. Rectilinear Motion op a Material Point.
§§ 49-55. Uniform Motion. Falling Bodies. Newton's Laws.
Mass 49
§§ 56-60. Uniformly Accelerated Motion. Graphic Representa-
tions. Variably Accelerated Motions. Impact... 54
Chapter II. Virtual Velocities.
§§ 61-69. Definitions and Propositions. Connecting-rod. Prob-
lems 67
VI CONTENTS.
Chapter III. Curvilinear Motiox op a Material Point.
PAGE
§§ 70-74. Composition of Motions, of Velocities, etc. General
Equations 72
§§ 75-84, Normal Acceleration. Centripetal and Centrifugal
Forces. Simple Pendulums. Projectiles. Rela-
tive Motion ; 77
Chapter IV. Moment of Inertia.
§§ 85-94. Plane Figures. Rigid Bodies. Reduction Formulae.
The Rectangle, Triangle, etc. Compound Plane
Figures. Polar Moment of Inertia 91
§§ 95-104. Rod. Thin Plates. Geometric Solids 98
§§ 105-107. Numerical Substitution. Ellipsoid of Inertia 102
Chapter V. Dynamics of a Rigid Body.
§§ 108-115. Translation. Rotation about a Fixed Axis. Centre of
Percussion 105
§§ 116-121. Torsion Balance. Compound Pendulum. The Fly-
wheel 116
§§122-123. Uniform Rotation. " Centrifugal Action." Free Axes. 125
§§ 124-126. Rolling Motions. Parallel Rod of Locomotive 130
Chapter VI. Work, Energy, and Power.
§§ 127-134. Work. Power. Horse-power. Kinetic Energy 133
§§135-138. Steam-hammer. Pile-driving. Inelastic Impact 138
§§ 139-141. Rotary Motion. Equivalent Systems of Forces. Any
Motion of a Rigid Body 143
§§ 142-146. Work and Kinetic Energy in a Moving Machine of
Rigid Parts 147
§§ 147-155. Power of Motors. Potential Energy. Heat, etc. Dy-
namometers. Atwood's Machine. Boat-rowing.
Examples 153
Chapter VII. Friction.
§§156-164. Sliding Friction. Its Laws. Bent Lever 164
§§ 165-171. Axle - friction. Friction Wheels. Pivots. Belting.
Transmission of Power by Belting , 175
§§ 172-177. Rolling Friction. Brakes. Engine-friction. Anoma-
lies in Friction. Rigidity of Cordage. Examples. 186
MECHANICS OF ENGINEERING.
PRELIMINARY CHAPTER.
1. Mechanics treats of the mutual actions and relative mo-
tions of material bodies, solid, liquid, and gaseous ; and by
Mechanics of Engineering is meant a presentment of those
principles of pure mechanics, and their applications, which are
of special service in engineering problems.
2. Kinds of Quantity. — Mechanics involves the following
fundamental kinds of quantity : Space, of one, two, or three
dimensions, i.e., length, surface, or volume, respectively ; time,
which needs no definition here; force and mass, as defined be-
low ; and abstract numbers, whose values are independent of
arbitrary units, as, for example, a ratio.
3. Force. — A force is one of a pair of equal, opposite, and
simultaneous actions between two bodies, by which the state
of their motions is altered or a change of form in the bodies
themselves is effected. Pressure, attraction, repulsion, and
traction are instances in point. Muscular sensation conveys
the idea of force, while a spring-balance gives an absolute
measure of it, a beam-balance only a relative measure. In
accordance with Newton's third law of motion, that action and
reaction are equal, opposite, and simultaneous, forces always
occur in pairs; thus, if a pressure of 40 lbs. exists between
bodies A and B, if A is considered by itself (i.e., "free"),
apart from all other bodies whose actions upon it are called
forces, among these forces will be one of 40 lbs. directed from
B toward A. Similarly, if B is under consideration, a force
2 MECHANICS OF ENGINEERING.
of 4:0 lbs. directed from A toward B takes its place among the
forces acting on B. This is the interpretation of Newton's
third law.
In conceiving of a force as applied at a certain point of a
body it is useful to imagine one end of an imponderable spiral
spring in a state of compression (or tension) as attached at the
given point, the axis of the spring having the given direction
of the force.
4. Mass is the quantity of matter in a body. The masses of
several bodies being proportional to their weights at the same
locality on the earth's surface, in physics the weight is taken
as the mass, but in practical engineering another mode is used
for measuring it (as explained in a subsequent chapter), viz.:
the mass of a body is equal to its weight divided by the ac-
celeration of gravity in tlie locality where the weight is taken,
or, symbolically, _3£ = G -s- g. This quotient is a constant
quantity, as it should be, siuce the mass of a body is invariable
wherever the body be carried.
5. Derived Quantities. — All kinds of quantity besides the
fundamental ones just mentioned are compounds of the latter,
formed by multiplication or division, such as velocity, accele-
ration, momentum, work, energy, moment, power, and force-
distribution. Some of these are merely names given for
convenience to certain combinations of factors which come
together not in dealing with first principles, but as a result of
common algebraic transformations.
6. Homogeneous Equations are those of such a form that they
are true for any arbitrary system of units, and in which all
terms combined by algebraic addition are of the same kind.
Thus, the equation 8 — -$■ (in which g = the acceleration of
gravity and t the time of vertical fall of a body in vacuo,
from rest) will give the distance fallen through, s, whatever
units be adopted for measuring time and distance. But if for
PRELIMINARY CHAPTER. 3
g we write the numerical value 32.2, which it assumes when
time is measured in seconds and distance in feet, the equation
s = 16. If is true for those units alone, and the equation is not
of homogeneous form. Algebraic combination of homogeneous
equations should always produce homogeneous equations ; if
not, some error has been made in the algebraic work. If any
equation derived or proposed for practical use is not homogene-
ous, an explicit statement should be made in the context as to
the proper units to be employed.
7. Heaviness. — By heaviness of a substance is meant the
weight of a cubic unit of the substance. E.g. the heaviness of
fresh water is 62.5, in case the unit of force is the pound,
and the foot the unit of space; i.e., a cubic foot of fresh
water weighs 62.5 lbs. In case the substance is not uniform
in composition, the heaviness varies from point to point. If
the weight of a homogeneous body be denoted by G, its volume
by V, and the heaviness of its substance by y, then G = Vy.
Weight in Pounds of a Cubic Foot (i.e., the heaviness) op various
Materials.
Anthracite, solid 100
" broken 57
Brick, common hard 125
" soft 100
Brick-work, common 112
Concrete 125
Earth, loose 72
as mud 102
Granite 164 to 172
Ice 58
Iron, cast 450
" wrought 480
Masonry, dry rubble 138
" dressed granite or
limestone 165
Mortar 100
Petroleum 55
Snow 7
" wet 15 to 50
Steel 490
Timber 25 to 60
Water, fresh 62.5
" sea 64.0
8. Specific Gravity is the ratio of the heaviness of a material
to that of water, and is therefore an abstract number.
9. A Material Point is a solid body, or small particle, whose
dimensions are practically nothing, compared with its range of
motion.
4 MECHANICS OF ENGINEERING.
10. A Rigid Body is a solid, whose distortion or change of
form under any system of forces to be brought upon it in
practice is, for certain purposes, insensible.
11. Equilibrium. — "When a system of forces applied to a
body produces the same effect as if no force acted, so far as
the state of motion of the body is concerned, they are said to
be balanced, or to be in equilibrium.
12. Division of the Subject. — Statics will treat of bodies at
rest, i.e., of balanced forces or equilibrium; dynamics, of
bodies in motion ; strength of matenals will treat of the effect
of forces in distorting bodies ; hydraulics, of the mechanics
of liquids ; pneumatics, of the mechanics of gases.
13. Parallelogram of Forces. — Duchayla's Proof. To fully
determine a force we must have given its amount, its direc-
tion, and its point of application in the body. It is generally
denoted in diagrams by an arrow. It is a matter of experience
that besides the point of application already spoken of any
other may be chosen in the line of action of the force. This
is called the transmissibility of force ; i.e., so far as the state of
motion of the body is concerned, a force may be applied any-
where in its line of action.
The Resultant of two forces (called its components) applied
at a. point of a body is a single force applied at the same point,
which will replace them. To prove that this resultant is given
in amount and position by the diagonal of the parallelogram
formed on the two given forces (conceived as. laid off to some
scale, so many pounds to the inch, say), Duchayla's method
requires four postulates, viz. : (1) the resultant of two forces
must lie in the same plane with them ; (2) the resultant of two
equal forces must bisect the angle between them ; (3) if one of
the two forces be increased, the angle between the other force
and the resultant will be greater than before ; and (4) the trans
missibility of force, already mentioned. Granting these, we
proceed as follows (Fig. 1) : Given the two forces P and Q =
PRELIMINARY CHAPTER. 5
P' + P" (P' and P" being each equal to P, so that Q = 2P),
applied at 0. Transmit P" to A. Draw the parallelograms
OP and AD; OP will also be a parallelogram. By postulate
(2), since OB is a rhombus, P and P' at may be replaced by
a single force R' acting through B. Transmit R' to B and
replace it by P and P'. Transmit P from B to J., P' from
B to P. Similarly P and P", at ^4, may be replaced by a
single force R" passing through P; transmit it there and re-
solve it into P and P". P' is already at P. Hence P and
P' + P", acting at P, are equivalent to P and P' + P" act-
ing at 0, in their respective directions. Therefore the result-
ant of P and P' -j- P" must lie in the line OP, the diagonal
of the parallelogram formed on P and Q = 2P at 0. Similarly
C/ F? /B
Fig. 1.
this may be proved (that the diagonal gives the direction of
the resultant) for any two forces P and raP; and for any two
forces nP and mP, m and n being any two whole numbers,
i.e., for any two commensurable forces. "When the forces are
incommensurable (Fig. 2), P and Q being the given forces,
we may use a reduetio ad absurdum, thus : Form the parallelo-
gram OP on P and Q applied at 0. Suppose for an instant
that R the resultant of P and Q does not follow the diagonal
OP, but some other direction, as OP'. Note the intersection
H, and draw HG parallel to PB. Divide P into equal parts,
each less than HP ; then in laying off parts equal to these from
along OB, a point of division will come at some point F
between C and B. Complete the parallelogram OFEO. The
force Q" = OF is commensurable with P, and hence their
6 MECHANICS OP ENGINEERING.
resultant acts along OE. Now Q is greater than Q", while P
makes a less angle with P than OE, which is contrary to pos-
tulate (3); therefore R cannot lie outside of the line OD.
Q.E.D.
It still remains to prove that the resultant is represented in
a/mount, as well as position, by the diagonal. OD (Fig. 3) is
\/r' the direction of P the resultant of P and
/F\ Q; required its amount. If P' be a force
/rv 7\ 7 equal and opposite to P it will balance P
\ /£ \q/ an ^ Q 5 i- e > tne resultant of P' and P
p (j'a must lie in the line QO prolonged (besides
FlG ' 3 ' being equal to Q). We can therefore de-
termine P' by drawing PA parallel to DO to intersect QO
prolonged in A ; and then complete the parallelogram on
PA and PO. Since OFAP is a parallelogram P' must =PA,
and since OAPD is a parallelogram BA=OD; therefore
K'= OD and also P= OD. Q. E. D.
Corollary. — The resultant of three forces applied at the same
point is the diagonal of the parallelopiped formed on the three
forces.
14. Concurrent forces are those whose lines of action intersect
in a common point, while non-concurrent forces are those which
do not so intersect ; results obtained for a system of concurrent
forces are really derivable, as particular cases, from those per-
taining to a system of non-concurrent forces.
15. Resultant. — A single force, the action of which, as re-
gards the state of motion of the body acted on, is equivalent to
that of a number of forces forming a system, is said to be the
Resultant of that system, and may replace the system ; and con-
versely a force which is equal and opposite to the resultant of
a system will balance that system, or, in other words, when it
is combined with that system there will result a new system in
equilibrium.
In general, as will be seen, a given system of forces can al-
PRELIMINARY CHAPTER. 7
ways De replaced by two single forces, but these two can be
combined into a single resultant only in particular cases.
15a. Equivalent Systems are those which may be replaced by
the same set of two single forces — or, in other words, those
which have the same effect, as to state of motion, upon the
given body.
15b. Formulae. — If in Pig. 3 the forces P and Q and the angle a =
PO Q are given, we have, for the resultant,
B—OB = ^ P* + Q' + 2 PQ cos a.
(If a is > 90° its cosine is negative.) In general, given any three parts
of either plane triangle OBQ, or OB B, the other three may be obtained
by ordinary trigonometry. Evidently if a = 0, R = P + Q; if a =
180°, B = P- Q; and if a = 90°, B- V P a + Q\
15c. Varieties Of Forces. — Great care should be used in deciding
what may properly be called forces. The latter may be divided into ac-
tions by contact, and actions at a distance. If pressure exists between two
bodies and they are perfectly smooth at the surface of contact, the pressure
(or thrust, or compressive action), of one against the other constitutes a force,
whose direction is normal to the tangent plane at any point of contact (a
matter of experience) ; while if those surfaces are not smooth there may also
exist mutual tangential actions or friction. (If the bodies really form a
continuous substance at the surface considered, these tangential actions are
called shearing forces.) Again, when a rod or wire is subjected to tension,
any portion of it is said to exert a pull or tensile force upon the remainder ;
the ability to do this depends on the property of cohesion. The foregoing
are examples of actions by contact.
Actions at a distance are exemplified in the mysterious attractions, or re-
pulsions, observable in the phenomena of gravitation, electricity, and mag-
netism, where the bodies concerned are not necessarily in contact. By the
term weight we shall always mean the force of the earth's attraction on the
body in question, and not the amount of matter in it.
[Note. — In some common phrases, such as "The tremendous force " of a heavy hody in
rapid motion, the word force is not used in a technical sense, but signifies energy (as ex-
plained in Chap. VI.). The mere fact that a hody is in motion, whatever its mass and
velocity, does not imply that it is under the action of any force, necessarily. For instance,
at any point in the path of a cannon ball through the air, the only forces acting on it are
the resistance of the air and the attraction of the earth, the latter having a vertical and
downward direction.]
PART I.-STATICS.
CHAPTER I.
STATICS OF A MATERIAL POINT.
16. Composition of Concurrent Forces. — A system of forces
acting on a material point is necessarily composed of concurrent
forces.
Case I. — All the forces in One Plane. Let be the
material point, the common point of application of all the
forces ; P t , JP a , etc., the given forces, making
angles «„ oc v etc., with the axis X. By the
parallelogram of forces P l may be resolved
into and replaced by its components, .P,cos a
acting along X, and P r sin a along Y.
fig. 4. Similarly all the remaining forces may be re-
placed by their X and Y components. "We have now a new
system, the equivalent of that first given, consisting of a set of
X forces, having the same line of application (axis X), and a
set of Y forces, all acting in the line Y. The resultant of the
X forces being their algebraic sum (denoted by 2X) (since
they have the same line of application) we have
2X — P 1 cos a l -J- P t cos ot t -j- etc. = 2(P cos a),
and similarly
2Y= P, sin a s -\- P 3 sin a t -\- etc. = 2(P sin a).
These two forces, 2X and 27, may be combined by the
parallelogram of forces, giving R — {/(2XY -\- (2Y)' as the
single resultant of the whole system, and its direction is deter-
mined by the angle or; thus, tan a = -^y-; see Fig. 5. For
equilibrium to .exist, P must = 0, which requires, separately,
STATICS OF A MATERIAL POINT.
9
2X = 0, and 27- (for the two squares (2X)' and
(2 7") a can neither of them be negative quantities).
Case II. — The forces having any directions in space,
but all applied at 0, the material point. Let P J} P 2 ,
etc., be the given forces, j 3 , making the angles or,, /?„ and y„
respectively, with three arbitrary axes, X, Y, and Z (Fig. 6),
at right angles to each other and intersecting at 0, the origin.
Similarly let a v fl v y t1 be the angles made by _P a with these
axes, and so on for all the forces. By the parallelopiped of
forces, Pi may be replaced by its components.
Xi = Pi cos a u Yi — Pi cos /?„ and Z x = P, cos y x ; and
Fig. 6.
Fig. 7.
similarly for all the forces, so that the entire system is now
replaced by the three forces,
2X = Pi cos «, + P 2 cos a t -f etc ;
2Y=Pi cos A + P, cos J3, + etc;
2Z — P x cos Xj + P a cos x, + etc ;
and finally by the single resultant
B = V(2Xy + (27)' + (2Z) T ,
Therefore, for equilibrium we must have separately,
2J= 0, 27= 0, and 2Z = 0.
P's position may be determined by its direction cosines, viz.,
2X a
cos a = -Q- ; cos />
27
2Z
— ;cmy = —
17. Conditions of Equilibrium. — Evidently, in dealing with
a system of concurrent forces, it would be a simple matter to
10 MECHANICS OF ENGINEERING.
replace any two of the forces by their resultant (diagonal
formed on them), then to combine this resultant with a third
force, and so on until all the forces had been combined, the
last resultant being the resultant of the whole system. The
foregoing treatment, however, is useful in showing that for
equilibrium of concurrent forces in a plane only two conditions
are necessary, viz., 2X = and 2Y ' = ; while in space
there are three, 2X = 0, 2J"= 0, and 2Z = 0. In Case I.,
then, we have conditions enough for determining two unknown
quantities ; in Case II., three.
18. Problems involving equilibrium of concurrent forces.
(A rigid body in equilibrium under no more than three forces
may be treated as a material point, since the (two or) three,
forces are necessarily concurrent.)
Peoblem 1. — A body weighing G lbs. rests on a horizontal
table : required the pressure between it and the table. Fig. 8.
Consider the body free, i.e., conceive all other bodies removed
, (the table in this instance), being replaced by the
1 forces which they exert on the first body. Taking
G the axis ^vertical and positive upward, and not
0, —
•-+X assuming in advance either the amount or direc-
Tjq tion of JV, the pressure of the table against the
body, but knowing that G, the action of the earth
on the body, is vertical and downward, we have
here a system of concurrent forces in equilibrium, in which
the X and Y components of G are known (being and —
G respectively), while those, iT x and iT T , of N are unknown.
Putting 2X — 0, we have _ZT X + = ; i.e., iVhas no hori-
zontal component, .•. _ZV is vertical. Putting 2Y=0, we
have 2F T — G = 0, .-. JV T = -\- G ; or the vertical component
of N, i.e., N itself, is positive (upward in this case), and is
numerically equal to G.
Peoblem 2. — Fig. 9. A body of weight G (lbs.) is moving
in a straight line over a rough horizontal table with a uniform
velocity o (feet per second) to the right. The tension in an
oblique cord by which it is pulled is given, and = P (lbs.),
STATIC8 OF A MATERIAL POINT. 11
which remains constant, the cord making a given angle of
elevation, a, with the path of the body. Required the vertical
pressure JV (lbs.) of the table, and also its + y
horizontal action F (friction) (lbs.) against
the body.
Referring by anticipation to Newton's first
law of motion, viz., a material point acted
on by no force or by balanced forces is either Fl °- 9 -
at rest or moving uniformly in a straight line, we see that this
problem is a case of balanced forces, i.e., of equilibrium. Since
there are only two unknown quantities, N and F, we may
determine them by the two equations of Case I., taking the
axes X and Y as before. Here let us leave the direction of
iVas well as its amount to be determined by the analysis. As
F must evidently point toward the left, treat it as negative in
summing the X components ; the analysis, therefore, can be
expected to give only its numerical value.
2X = gives P cos a — F= 0. .-. F ' = P cos a.
^JT^Ogivesir+Psina:- G = 0. .-. 1ST = G — P sin a.
.'. iV is upward or downward according as G is > or < P
sin a. For JY to be a downward pressure upon the body would
require the surface of the table to be above it. The ratio of the
friction F to the pressure JV which produces it can now be
obtained, and is called the coefficient of friction. It may vary
slightly with the velocity.
This problem may be looked upon as arising from an experi-
ment made to determine the coefficient of friction between the
given surfaces at the given uniform velocity.
19. The Free-Body Method. — The foregoing rather labored so-
lutions of very simple problems have been made such to illus-
trate what may be called the free-body method of treating any
problem involving a body acted on by a system of forces. It
consists in conceiving the body isolated from all others which
act on it in any way, those actions being introduced as so many
forces, known or unknown, in amount and position. The sys-
tem of forces thus formed may be made to yield certain equa-
12 MECHANICS OF ENGINEERING.
tions, whose character and number depend on circumstances,
such as the behavior of the body, whether the forces are con-
fined to a plane or not, etc., and which are therefore theoreti-
cally available for determining an equal number of unknown
quantities, whether these be forces, masses, spaces, times, or
abstract numbers. Of course in some instances the unknown
quantities may enter these equations with such high powers
that the elimination may be impossible ; but this is a matter
of algebra, not of mechanics.
PARALLEL FORCES AND THE CENTRE OF GRAVITY. 13
CHAPTER II.
PARALLEL FORCES AND THE CENTRE OF GRAVITY.
20. Preliminary Remarks. — Although by its title this section
should be restricted to a treatment of the equilibrium of forces^
certain propositions involving the composition and resolution
of forces, without reference to the behavior of the body under
their action, will be found necessary as preliminary to the prin-
cipal object in view.
As a rigid body possesses extension in three dimensions, to
deal with a system of forces acting on it we require three co-
ordinate axes : in other words, the system consists of " forces
in space," and in general the forces are non-concurrent. In
most problems in statics, however, the forces acting are in one
plane: we accordingly begin by considering non-concurrent
forces in a plane, of which the simplest case is that of two
parallel forces. For the present the body on which the forces
act will not be shown in the figure, but must be understood to
be there (since we have no conception of forces independently
of material bodies). The device will frequently be adopted of
introducing into the given system two opposite and equal forces
acting in the same line: evidently this will not alter the effect
of the given system, as regards the rest or motion of the body.
pK — TP
21. Resultant of two Parallel
Forces. »
* jf..-4o
Case I. — The two forces have
{-Or-
O S A\ D /B S
the same direction. Fig. 10.
Let P and Q be the given forces,
and AB a line perpendicular to
them (P and Q are supposed to have
been transferred to the intersections Fl °- 10 -
A and B). Put in at A and B two equal and opposite
forces S and S, combining them with P and Q to form P'
C\
Q'
14 MECHANICS OF ENGINEERING.
and Q'. Transfer P' and Q' to their intersection at C, and there
resolve them again into 8 and P, /Sand Q. 8 and 8 annul each
other at C; therefore P and Q, acting along a common line CD,
replace the P and Q first given ; i.e., the resultant of the origi-
nal two forces is a force P =P -f- Q, acting parallel to them
through the point D, whose position must now be determined.
The triangle CAD is similar to the triangle shaded by lines,
.: P : 8 :: CD : x; and CDB being similar to the triangle
shaded by dots, .•. 8 : Q :: a — x : CD. Combining these, we
P a — x j Qa Qa
have -Tj = and .-. x — -^— _ = -5_ N ow W ntethis
i?o: = Qa, and add i?c, i.e., Pc -(- (?c, to each member, c being
the distance of (Fig. 10), any point in AB produced, from
A. This will give P(x-\-c) — Pc -\-Q{a -\- c), in which c,
a-{-c, and x-{-c are respectively the lengths of perpendiculars
let fall from upon P, Q, and their resultant P. Any one of
these products, such as Pc. is for convenience (since products of
this form occur so frequently in Mechanics as a result of alge-
braic transformation) called the Moment of the force about the
arbitrary point 0. Hence the resultant of two parallel forces of
the same direction is equal to their sum, acts in their plane, in
a line parallel to them, and at such a distance from any arbi-
trary point in their plane as may be determined by writing
its moment about equal to the sum of the moments of the
two forces about 0. is called a centre of moments, and each
of the perpendiculars a lever-arm.
Case II. — Two parallel forces P and Q of opposite direc-
tions. Fig. 11. By a process similar to the foregoing, we
obtain P = P — Q and \P — Qp
= Qa, i.e., Px = Qa. Subtract
each member of the last equation
from Pc (i.e., Pc— Qc), in which c
\ is the distance, from A, of any arbi-
>vp s trarv points in AB produced. This
c a 7"^Kl £ ives R{° — x) = Pc— Q(a + c).
— a,! — 3 But (c — x), c, and (a-\-c) are re-
Fio. n. spectively the perpendiculars, from
PARALLEL FORCES AND THE CENTRE OF GRAVITY. 15
0, upon It, P, and Q. That is, R{c — x) is the moment of P
about ; Pc, that of P about ; and Q(a -\- c), that of Q
about „ in which b, 5„
and b, are the oblique distances of the three lines of action
from any point in their plane, and lie on the same straight
line ; P is the resultant of the parallel forces P l and P,.
22. Resultant of any System of Parallel Forces in Space. —
LetP„ P v P w etc., be the forces of the system, and x„ y„
s u x ti Vv z v etc -> tne co-ordinates of their points of application
as referred to an arbitrary set of three co-ordinate axes X, Y,
and Z, perpendicular to each other. Each force is here re-
16 MECHANICS OP ENGINEERING.
stricted to a definite point of application in its line of action
(with reference to establishing more directly the fundamental
equations for the co-ordinates of the centre of gravity of a
body). The resultant P' of any two of the forces, as
P l and P„ is = P t -j- P„ and may be applied at C, the in-
tersection of its own line of action with a line BD joining
the points of application of P t and P„ its components.
Produce the latter line to A, where it pierces the plane JT Y,
and let J„ V, and 5„, respectively, be the distances of B, C,
D, from A ; then from the corollary of the last article we have
P'V = PA + PA;
but from similar triangles
V : \ : \ : : z' : a, : z„ .: P'z' = P A + P A .
Now combine P', applied at C, with P„ applied at E, calling
their resultant P" and its vertical co-ordinate z", and we obtain
P"z" = P'z' + P 3 z s , i.e., P"z" = P l3l + P A + P s z 3 ,
also
Proceeding thus until all the forces have been considered, we
shall have finally, for the resultant of the whole system,
5 = P I + P,+ P i + eto.;
and for the vertical co-ordinate of its point of application,
which we may write z,
Bz = P A + P A + P 3 z, + etc ;
and similarly for the other co-ordinates.
- 2(Px) , - 2(Py) .
« = -^p- ; and y = -A^.
In these equations, in the general case, such products as P&,
etc., cannot strictly be called moments. The point whose co-
-w.
0! +X
PARALLEL FORCES AND THE CENTRE OF GRAVITY. 17
ordinates are the x, y, and s, just obtained, is called the Centre
of Parallel Forces, and its position is independent of the {com-
mon) direction of the forces concerned.
Example. — If the parallel forces are contained in one plane,
and the axis Y be assumed parallel to the direction of the
forces, then each product like P 1 x 1 will be a moment, as de-
fined in § 21 ; and it will be noticed in the accompanying nu-
merical example, Fig. 14, that a detailed substitution in the
i
Ex — P.x, + Pjo t + etc., . . . (1)
having regard to the proper sign of each
force and of each abscissa, gives the same " fiq. 14
result as if each product Px were first obtained numerically,
and a sign affixed to the product considered as a moment
about the point 0. Let P, = — 1 lb.; P 2 = + 2 lbs.; P 3 =
+ 3 lbs.; P, = — 6 lbs.; x, = + 1 ft.; x, = + 3 ft.; x, = — 2
ft; and x t = — 1 ft. Required the amount and position of the
resultant E. In amount E = 2P = — 1 + 2 + 3 — 6 = — 2
lbs.; i.e., it is a downward force of 2 lbs. As to its position,
Ex= 2{Px) gives ( - 2)» = (- 1) X (+ 1) + 2 X 3 +
3 X (- 2) + (-6) X (- 1) = - 1 + 6 - 6 + 6. Now from
the figure, by inspection, it is evident that the moment of P,
about is negative (with the hands of a watch), and is numer-
ically = 1, i.e., its moment = — 1 ; similarly, by inspection,
that of P^ is seen to be positive, that of P 3 negative, that of
P t positive; which agree with the results just found, that
(— 2)» = — 1 + 6— 6+6 = + 5ft. lbs. (Since a moment
is a product of a force (lbs.) by a length (ft.), it may be called
so many foot-pounds.) Next, solving for x, we obtain
x = (+ 5) -=- (— 2) = — 2.5 ft.; i.e., the resultant of the given
forces is a downward force of 2 lbs. acting in a vertical line
2.5 ft. to the left of the origin. Hence, if the body in question
be a horizontal rod whose weight has been already included in
the statement of forces, a support placed 2.5 ft. to the left of
and capable of resisting at least 2 lbs. downward pressnre
will preserve equilibrium ; and the pressure which it exerts
18 MECHANICS OF ENGINEERING.
against the rod must be an upward force, P„ of 2 lbs., i.e., the
equal and opposite of the resultant of P„ P„ P„ P t -
Fig. 15 shows the rod as a free body in equilibrium under
the five forces. P 6 = + 2 lbs. = the reaction of the support.
Of course _P 5 is one of a pair of equal
and opposite forces ; the other one
is the pressure of the rod against the
J!
P s \ — as — lo support, and would take its place among
fk. is. the forces acting on the support.
23. Centre of Gravity. — Among the forces acting on any
rigid body at the surface of the earth is the so-called attraction
of the latter (i.e., gravitation), as shown by a spring-balance,
which indicates the weight of the body hung upon it. The
weights of the different particles of any rigid body constitute a
system of parallel forces (practically so, though actually slightly
convergent). The point of application of the resultant of these
forces is called the centre of gravity of the body, and may also
be considered the centre of mass, the body being of very small
dimensions compared with the earth's radius.
If x, y, and z denote the co-ordinates of the centre of gravity
of a body referred to three co-ordinate axes, the equations
derived for them in § 22 are directly applicable, with slight
changes in notation.
Denote the weight of any particle of the body by dG, its
volume by dV, by y its heaviness (rate of weight, see § T) and
its co-ordinates by x, y, and z ; then, using the integral sign as
indicating a summation of like terms for all the particles of the
body, we have, for heterogeneous bodies,
fyxdV _ - fyydV _ - _fyzdV.
X ~ fydV y - fydV Z -fydV> ' ' W
while, if the body is homogeneous, y is the same for all its ele-
ments, and being therefore placed outside the sign of summa-
tion, is cancelled out, leaving for homogeneous bodies ( V de-
noting the total volume)
- fxdV - fydV - fzdY
* --^r", v = —v-> and s --V-- ■ • ( 2 )
PARALLEL FORCES AND THE CENTRE OP GRAVITY. 19
Corollary. — It is also evident that if a homogeneous body is
for convenience considered as made up of several finite parts,
whose volumes are V„ V„ etc., and whose gravity co-ordinates
are as,, y„ s 1 ; x„ y„ z, ; etc., we may write
- _ v*+ r£+...
x - 7; + r, + . . . • • . • • w
If the body is heterogeneous, put 6*, (weights), etc., instead
of V„ etc., in equation (3).
If the body is an infinitely thin homogeneous shell of uni-
form thickness = h, then dV= hdF(dF denoting an element,
and F the whole area of one surface) and equations (2) become,
after cancellation,
v-Ml. T s - fydF - i-Ml u)
Similarly, for a homogeneous wire of constant small cross-
section (i.e.. a geometrical line, having weight), its length
being s, and an element of length ds, we obtain
m= Jj^.-^.-^ ^ (5)
It is often convenient to find the centre of gravity of a thin
plate by experiment, balancing it on a needle-point; other
shapes are not so easily dealt with.
24. Symmetry. — Considerations of symmetry of form often
determine the centre of gravity of homogeneous solids without
analysis, or limit it to a certain line or plane. Thus the centre
of gravity of a sphere, or any regular polyedron, is at its centre
of figure ; of a right cylinder, in the middle of its axis ; of a
thin plate of the form of a circle or regular polygon, in the
centre of figure ; of a straight wire of uniform cross-section, in
the middle of its length.
Again, if a homogeneous body is symmetrical about a plane,
the centre of gravity must lie in that plane, called a plane of
20 MECHANICS OP ENGINEERING.
gravity ; if about a line, in that line called a line of gravity ;
if about a point, in that point.
25. By considering certain modes of subdivision of a homo-
geneous body, lines or planes of gravity are often made appar-
ent. E.g., a line joining the middle of the bases of a trape-
zoidal plate is a line of gravity, since it bisects all the strips
of uniform width determined by drawing parallels to the
bases ; similarly, a line joining the apex of a triangular plate to
the middle of the opposite side is a line of gravity. Other
cases can easily be suggested by the student.
26. Problems. — (1) Required the position of the centre of
A gravity of a, fine homogeneous wire of the
-j-3 form of a circular arc,AB, Fig. 16. Take
\ the origin at the centre of the circle, and
! the axis 21 bisecting the wire. Let the
u \ A dx j length of the wire, s, = 2s, ; ds = ele-
^v - ?// 1 mentofarc. "We need determine only the
§%/ x, since evidently y = 0. Equations (5),
, fsads
fig. 16. g 23, are applicable here, i.e., x — .
s
From similar triangles we have
ds : dy :: r : x; .: ds
rdy
x
,y = + <» 2ra
>
— r n y ■** " zra
:.x = — / dy — q — , i.e., = chord X radius — length of
wire. For a semicircular wire, this reduces to x = 2r -5- it.
Peoblem 2. Centre of gravity of trapezoidal {and trian-
gular) thin plates, homogeneous, etc. — Prolong the non-parallel
sides of the trapezoid to intersect at 0, which take as an origin,
making the axis X perpendicular to the bases and b t . "We
may here use equations (4), § 23, and may take a vertical strip
for our element of area, dF, in determining x ; for each point
of such a strip has the same x. Now dF = (y -\- y')dx, and
PARALLEL FORCES AND THE CENTRE OF GRAVITY. 21
from similar triangles y -f- y' = j x. Hence F = w {bh — bji^)
can be written -^ 7 (h*
2A
h'), and x = -—p — becomes
b ph
hJiH
x'dx
1 I
+ - 7 W- K) = -
J ' 2 A
2 U
3 A a
~ LA
for the trapezoid.
— 2
For a triangle A, = 0, and we bave x = h; that is, the
o
centre of gravity of a triangle is one third the altitude from the
base. The centre of gravity is finally determined by knowing
Fig. 17.
Fio. 18.
that a line joining the middles of I and b l is a line of gravity ;
or joining and the middle of b in the case of a triangle.
Problem 3. Sector of a circle. Thin plate, etc. — Let the
notation, axes, etc., be as in Fig. 18. Angle of sector = 2a;
x = ? Using polar co-ordinates, the element of area dF (a
small rectangle) = pd J p' dp \dq>,
is that portion of the summation / / cos q>p , dpd(p which
belongs to a single elementary sector (triangle), since all its
elements (rectangles), from centre to circumference, have the
same
.)
That is,
- 1 /•* P+ a r* r+" 2 r sin a
a — 4 r sin £ fl
or, putting p = 2a = total angle of sector, a; = ^ g — •
— 4t*
For a semicircular plate this' reduces to x = ^— .
[iTofe. — In numerical substitution the arcs « and yS used
above (unless sin or cos is prefixed) are understood to be ex-
pressed in circular measure (^-measure) ; e.g., for a quad-
rant, /? = | = 1.5707 + ; for 30°,/? = ^; or, in general, if /J
. , 180° , n~\
In degrees = , then p in ^-measure — — .
^\\ Problem 4. Sector of a flat ring ; thin
F A* P^te) e * c - — Treatment similar to that of
^
the product of this arm, by one of the <^j **^^ ^4^^
forces. The axis of a couple is an ^^^^^^^
imaginary line drawn perpendicular to j^g. 2 4.
its plane on that side from which the rotation appears positive
(against the hands of a watch). (An ideal rotation is meant,
suggested by the position of the arrows ; any actual rotation
of the rigid body is a subject for future consideration.) In
dealing with two or more couples the lengths of their axes are
made proportional to their moments; in fact, by selecting a
proper scale, numerically equal to these moments. E.g., in Fig.
24, the moments of the two couples there shown are Pa and
Qb; their axes p and q so laid off that Pa : Qb :: p : q, and
that the ideal rotation may appear positive, viewed from the
outer end of the axis.
29. No single force can balance a couple. — For suppose the
couple P, P, could be balanced by a force JZ', then this, acting
n> at some point C, ought to hold the couple
D | jC^B £_ in equilibrium. Draw CO through 0, the
»T /p $+ centre of symmetry of the couple, and
Fio. 25. make OP = OC. At D put in two op-
posite and equal forces, 8 and T, equal and parallel to R'.
The supposed equilibrium is undisturbed. But if R', P, and
28 MECHANICS OF ENGINEERING.
P are in equilibrium, so ought (by symmetry about 0) S, P,
and P to be in equilibrium, and they may be removed without
disturbing equilibrium. But we have left .Z'and P', which are
evidently not in equilibrium ; .\ the proposition is proved by
this reductio ad ahsurdum. Conversely a couple has no single
resultant.
30. A couple may he transferred anywhere in its own plane.
— First, it may be turned through any angle a, about any
pi point of its arm, or of its arm produced.
G " ; ^'"| jjf -|- Let (P, P')be a couple, G any point of its
\-'Nci jP' arm (produced), and a any angle. Make
^J-*^.... | GC = GA, CD = AB, and put in at C,
% \ 2 \. | | P, and P 3 equal to P (or P'), opposite to
d\__P« --"Kjeaeh other and perpendicular to GO; and
8 > r* P 3 and P 4 similarly at D. Now apply and
Fig. 26. combine P and P, at 0, P' and P 4 at 0'\
then evidently E and R' neutralize each other, leaving P„ and
P, equivalent to the original couple (P, P'). The arm
CD = AB. Secondly, if G be at infinity, and a = 0, the
same proof applies, i.e., a couple may be moved parallel to
itself in its own plane. Therefore, by a combination of the
two transferrals, the proposition is established for any trans-
ferral in the plane.
31. A couple may he replaced by another of equal moment
in a parallel plane.— Let (P, P') be a couple. Let CD, in a
parallel plane, be parallel to AB. At D put in a pair of equal
and opposite forces, # 3 and S., parallel to P and each = "==P.
ED
Similarly at C, S t and S, parallel to P and each = —P,
EG
But, from similar triangles,
AE_BE
ED ~ EO ; •'• "' = S ' = S > — S *-
STATICS OF A BIGID BODY. 29
[Note.— The above values are so chosen that the intersection point E
may be the point of application of (P + £,), the resultant of P and B,;
and also of (P+ S 3 ), the resultant of Pand S,, as follows from § 21; thus
(Pig. 28), Ji, the resultant of the two parallel forces Pand B 3 , is = P + B,,
and its moment about any centre of moments, as E, its own point of ap-
plication, should equal the (algebraic) sum of the moments of its com-
ponents about E; i.e., Ji X zero = P . AE — S, , HE; . : S a = = . P.]
1)E J
(P-S 3 ) 4 P
B
t»
S,i Sji./-- - I
2 s ^(pvs 2 > -i [ -1-
Fig. 27. Fig. 28.
Replacing P' and S, by (P' + $,), and P and £ a by
(P -J- #,), the latter resultants cancel each other at P, leaving
the couple (S v S t ) with an arm CD, equivalent to the original
couple P, P' with an arm AB. But, since S, = = . P =
__ EC
AB
~~ . P, we have S 1 X CP = PxAB ; that is, their moments
L/-D
are equal.
32. Transferral and Transformation of Conples. — In view of
the foregoing, we may state, in general, that a couple acting on
a rigid body may be transferred to any position in any parallel
plane, and may have the values of its forces and arm changed
in any way so long as its moment is kept unchanged, and still
have the same effect on the rigid body (as to rest or motion,
not in distorting it).
Corollaries. — A couple may be replaced by another in any
position so long as their axes are equal and parallel and simi-
larly situated with respect to their planes.
A couple can be balanced only by another couple whose axis
is equal and parallel to that of the first, and dissimilarly situ-
ated. For example, Fig. 29, Pa being = Qb, the rigid body
AB (here supposed without weight) is in equilibrium in each
30
MECHANICS OF ENGINEERING.
ease shown. By " reduction of a couple to a certain arm a"
is meant that for the original couple whose arm is a', with
forces each = P', a new couple is substituted whose arm shall
he = a, and the value of whose forces P and P must be com-
puted from the condition
Pa = P'a', i.e., P = P'a! -=- a.
Fig.
Fiq. 30.
33. Composition of Couples. — Let (P, P') and (Q, Q') be two
couples in different planes reduced to the same arm AB = a,
which is a portion of the line of intersection of their planes.
That is, whatever the original values of the individual forces
and arms of the two couples were, they have been transferred
and replaced in accordance with § 32, so that P.AB, the
moment of the first couple, and the direction of its ax\s,p,
have remained unchanged ; similarly for the other couple.
Combining P with Q and P' with Q', we have a resultant
couple (P, P') whose arm is also AB. The axes p and q of
the component couples are proportional to P . AB and Q . AB,
i.e., to P and Q, and contain the same angle as P and Q.
Therefore the parallelogram p . . . q is similar to the parallelo-
gram P . . . Q; whence p : q : r : : P : Q : R , or p : q : r : :
Pa : Qa : Pa. Also r is evidently perpendicular to the plane
of the resultant couple (P, P'), whose moment is Pa. Hence
r, the diagonal of the parallelogram on p and q, is the axis of
the resultant couple. To combine two couples, therefore, we
have only to combine their axes, as if they were forces, by a
parallelogram, the diagonal being the axis of the resultant
couple ; the plane of this couple will be perpendicular to the
STATICS OP A RIGID BODY.
31
axis just found, and its moment bears the same relation to the
moments of the component couples as the diagonal axis to the
two component axes. Thus, if two couples, of moments Pa
and Qb, lie in planes perpendicular to each other, their result-
ant couple has a moment Be — i/(PaY -\- (Qbf-
If three couples in different planes are to be combined, the
axis of their resultant couple is the diagonal of the parallelo-
piped formed on the axes, laid off to the same scale and point-
ing in the proper directions, the proper direction of an axis
being away from the plane of its couple, on the side from
which the couple appears of positive direction.
34. If several couples lie in the same plane their axes are
parallel and the axis of the resultant couple is their algebraic
sum ; and a similar relation holds for the moments : thus, in
Fig. 24, the resultant of the two couples has a moment = Qb
— Pa, which shows us that a convenient way of combining
couples, when all in one plane, is to call the moments positive
or negative, according as the ideal rotations are against, or with,
the hands of a watch, as seen from the same side of the plane ;
the sign of the algebraic sum will then show the ideal rotation
of the resultant couple.
35. Composition of Non-concurrent Forces in a Plane. — Let
P lS P a , etc., be the forces of the system ; x x , y it x„ y„ etc., the
xr
It!
+ Yi'
co-ordinates of their points of application ; and «„ a„ . . . etc.,
their angles with the axis X. Eeplace P, by its components
X, and Y x , parallel to the arbitrary axes of reference. At the
orio-in put in two forces, opposite to each other and equal and
parallel toX, ; similarly for T,. (Of course X, = P, cos a and
y = P t sin a.) We now have P, replaced by two forces X,
32 MECHANICS OF ENGINEERING.
and i; at the origin, and two couples, in the same plane, -whose
moments are respectively - X,y, and + !>„ and are there-
fore (§ 34) equivalent to a single couple, in the same plane with
a moment = (7&—XJ/.).
Treating all the remaining forces in the same way, the whole
system of forces is replaced by
the force 2(X) =X X + X, + ... at the origin, along the axis X;
the force 2(7) - 7 V + 7, + ... at the origin, along the axis T;
and the couple whose mom. G = 2 ( Yx — Xy), which may be
called the couple (see Fig. 32), and may be placed anywhere
in the plane. Now 2(X) and 2( 7) may be combined into a
force E ; i.e.,
. 2X
E = Y(2Xy -+- 2 7f and its direction-cosine is cos a = — rr-
Since, then, the whole system reduces to C and E, we must
have for equilibrium E — 0, and G = ; i.e., for equilibrium
2X= 0, 27= 0, and 2(7x-Xy) = 0. . eq. (1)
If R alone = 0, the system reduces to a couple whose mo-
ment is G = 2( 7x—Xy) ; and if G alone = the system re-
duces to a single force E, applied at the origin. If, in general,
neither E nor G = 0, the system is still equivalent to a single
force, but not applied at the origin (as could hardly be ex-
pected, since the origin is arbitrary) ; as follows (see Fig. 33) :
Replace the couple C by one of equal moment, G, with each
force = E. Its arm will therefore be -^. Move this couple
in the plane so that one of its forces E may cancel the E al-
ready at the origin, thus leaving a single resultant E for the
wljple system, applied in a line at a perpendicular distance,
c = -55 , from the origin, and making an angle a whose cosine =
2X . , , . v
~s~, with the axis X.
36. More convenient form for the equations of equilibrium
of non-concurrent forces in a plane. — In (I.), Fig. 34, O being
STATICS OF A BIGUD BODY. 33
any point and a its perpendicular distance from a force P ;
put in at two equal and opposite forces P and P' = and ||
to P, and we have P replaced by an equal single force P' at
O, and a couple whose moment is + Pa. (II.) shows a simi-
lar construction, dealing with the Jf and ^components of P,
so that in (II.) P is replaced by single forces X' and Y' at
•/P ! v. -am
Fio. 33.
(and they are equivalent to a resultant P', at 0, as in (I.), and
two couples whose moments are -|- Yx and — Xy.
Hence, being the same point in both cases, the couple Pa
is equivalent to the two last mentioned, and, their axes being
parallel, we must have Pa = Yx — Xy. Equations (1),
§ 35, for equilibrium, may now be written
2X —0, 2 Y = 0, and 2{Pa) = 0. . . (2)
In problems involving the equilibrium of non-concurrent
forces in a plane, we have three independent conditions, or
equations, and can determine at most three unknown quantities.
For practical solution, then, the rigid body having been made
free (by conceiving the actions of all other bodies as repre-
sented by forces), and being in equilibrium (which it must be
if at rest), we apply equations (2) literally ; i.e., assuming an
origin and two axes, equate the sum of the JT components of
all the forces to zero ; similarly for the Y components ; and
then for the "moment-equation," having dropped a perpen-
dicular from the origin upon each force, write the algebraic
sum of the products {moments) obtained by multiplying each
force by its perpendicular, or "lever-arm," equal to zero, call-
ing each product + or — according as the ideal rotation ap-
pears'against, or with, the hands of a watch, as seen from the
same side of the plane. (The converse convention would do as
well.)
34 MECHANICS OP ENGINEERING.
Sometimes it is convenient to use three moment equations,
taking a new origin each time, and then the 2X= and 2 Y
= are superfluous, as they would not be independent equa-
tions.
37. Problems involving Non-concurrent Forces in a Plane. — ■
Remarks. The weight of a rigid body is a vertical force
through its centre of gravity, downwards.
If the surface of contact of two bodies is smooth the action
(pressure, or force) of one on the other is perpendicular to the
surface at the point of contact. If a cord must be imagined
cut, to make a body free, its tension must be inserted in the
line of the cord, and in such a direction as to keep taut the
small portion still fastened to the body. In case the' pin of
a hinge must be removed, to make the body free, its pressure
against the ring being unknown in direction and amount, it is
most convenient to represent it by its unknown components X
and Y, in known directions. In the following problems there
is supposed to be no friction. If the line of action of an un-
known force is known, but not its direction (forward or back
ward), assume a direction for it and adhere to it in all the three
equations, and if the assumption is correct the value of the
force, after elimination, will be positive ; if incorrect, negative.
Problem 1. — Fig. 35. Given an oblique rigid rod, with two
loads G t (its own weight) and G t ; required the reaction of the
smooth vertical wall at A, and the direction and amount of the
SAin^-pressure at 0. The reaction at A
■
i ' a 9Y ...
//— ' must be horizontal ; call it X. The pres-
jj-j sure at 0, being unknown in direction, will
have both its X and Y components un-
known. The three unknowns, then, are
X , X, and Y„ while G„ G„ a,, a„ and
h are known. The figure shows the rod
as a free body, all the forces acting on it
have been put in, and, since the rod is at rest, constitute a sys-
tem of non-concurrent forces in a plane, ready for the condi-
tions of equilibrium. Taking origin and axes as in the figure,
STATICS OF A RIGID BODY.
35
2X = gives +X - X' = ; 2Y =
— G, = 0; while 2 (Pa) = 0, about
G k a t — G^a t = 0. (The moments of
each, = zero.) By elimination
are,
G,;X e = X-,--[G l a l +G>a 1 ]
gives + Y a - G x
0, gives -\- Xh —
X„ and r o about 6>
we obtain Y = 3P, + P v Hence
A must point to the left, i.e., is a thrust or compression, and is
^F.-SP-PJ.
Similarly, taking moments about 0„ the intersection of A
and B, we have an equation in which the only unknown is C,
viz., (Oh) - Yfr + P x a = 0. .-. (Oh) = £a[3 F, - 2P,],
STATICS OF A RIGID BODY.
37
a positive moment, since 3 V x is >2P, ; .-. Cmust point to the
right, i.e., is a tension, and = ^3 V, — 2P 1
Finally, to obtain B, put 2(vert. comps.) = 0; i.e. (Pcos [-JP, -4- \PX
B = [P + P %
F] + cos * = -^JF[P, + p - rj.
Problem 4. — Given the weight (7, of rod, the weight 0„
and all the geometrical elements (the student will assume a
G, AG 2
Fig. 40. Fig. 41.
convenient notation); required the tension in the cord, and the
amount and direction of pressure on hinge-pin.
Problem 5. — Roof-truss ; pin-connection ; all loads at joints ;
wind-pressures W and W, normal to OA ; required the three
reactions or supporting forces (of the two horizontal surfaces
and one vertical surface), and the
stress in each piece. All geomet-
rical elements are given ; also P,
P, P, W.
38. Composition of Non-concur-
rent Forces in Space. — Let P„ P„
etc., be the given forces, and a*„ y„
s„ x„ y v z„ etc., their points of ap-
plication referred to an arbitrary
origin and axes; or,, /?„ y x , etc.,
the angles made by their lines of application with X, T, and Z.
Fig. 48.
38
MECHANICS OE ENGINEERING.
Considering the first force JP„ replace it by its three com-
ponents parallel to the axes, X l = I 3 1 cos a^; Y 1 = JP l cos yff,;
and Z^ = P 1 cos y l (P x itself is not shown in the figure). At
0, and also at A, put a pair of equal and opposite forces,
each equal and parallel to Z,; -Z, is now replaced by a single
force Z x acting upward at the origin, and two couples, one
in a plane parallel to YZ and having a moment = — -Z,y, (as
we see it looking toward from a remote point on the axis
-\- X), the other in a plane parallel to XZ and having a mo-
ment = -|- Z t x t (seen from a remote point on the axis -f- Y).
Similarly at and O put in pairs of forces equal and parallel
to X„ and we have X„ at B, replaced by the single force X x
at the origin, and the couples, one in a plane parallel to XY t
and having a moment -j- X t y„ seen from a remote point on
the axis -\- Z, the other in a plane parallel to XZ, and of a
moment =—X 1 s 1 , seen from a remote point on the axis -\-Y\
and finally, by a similar device, Y 1 at B is replaced by a force
Y 1 at the origin and two couples, parallel to the planes XY
and YZ, and having moments — Y t x 1 and + Zjz,, respective-
ly. (In Fig. 42 the single forces at the origin are broken
lines, while the two forces constituting any one of the six
couples may be recognized as being
equal and parallel, of opposite di-
rections, and both continuous, or
both dotted.) We have, therefore,
replaced the force _P a by three
forces X v , Y t , Z„ at 0, and six
couples (shown more clearly in
Fig. 43; the couples have been
transferred to symmetrical posi-
tions). Combining each two couples
whose axes are parallel to X, Y,
Fio. 43.
or Z, they can be reduced to three, viz.,
one with an X axis and a moment = T",2, — Z l y l ;
one with a Zaxis and a moment == Z x x x — X x z^\
one with a Z axis and a moment = X,y 1 — Y x x v
STATICS OF A BIGID BODY. 39
Dealing with each of the other forces P„ J°„ etc., in the same
manner, the whole system may finally be replaced by three
forces 2X, 2 7, and 2Z, at the origin and three couples
whose moments are, respectively,
L = 2( Yz — Zy) with its axis parallel to X ;
3f = 2{Zx — Xz) with its axis parallel to Y;
N = 2(Xy — Yx) with its axis parallel to Z.
The " axes" of these couples, being parallel to the respective
co-ordinate axes X, Y, and Z, and proportional to the mo-
ments L, M, and N, respectively, the axis of their resultant
C, whose moment is G, must be the diagonal of a parallelo-
pipedon constructed on the three component axes (propor-
tional to) L, M, and N. Therefore, G = VIF+jF+lF,
while the resultant of 2X, 2 Y, and 2Z is
b = V{zxy + (2 Yy + (2zy
acting at the origin. If a, (3, and y are the direction-angles
2j ^ Y ~SZ
of R, we have cos a = -^-, cos fi = ~^-, and cos y = -^ ;
while if A, M, an( i v are those of the axis of the couple G, we
L M 3 N
have cos A. = 77, cos p. = -q, and cos v = -g.
For equilibrium we have both G = and R = ; i.e.,
separately, six conditions, viz., •
2X=0,2Y = 0, 2Z=0 ; and 2=0, 3/=0, i7=0 . (1)
Now, noting that 2X =0,27=0, and 2{Xy — Yx)=0
are the conditions for equilibrium of the system of non-concur-
rent forces which would be formed by projecting each force of
our actual system upon the plane XY, and similar relations
for the planes YZ and XZ, we may restate equations (1) in
another form, more serviceable in practical problems, viz. :
Hote. Xf a system of non-concurrent forces in space is in
equilibrium, the plane systems formed by projecting the given
system upon each of three arbitrary co-ordinate planes will also
be m equilibrium. But we can obtain only six independent
40 MECHANICS OF ENGINEERING.
equations in any case, available for six unknowns. If H alone
= 0, we have the system equivalent to a couple O, whose
moment = G ; if G alone = 0, the system has a single re-
sultant It applied at the origin. In general, neither It nor G
being = 0, we cannot further combine H and C(as was done
with non-concurrent forces in a plane) to produce a single re-
sultant unless It and Care in the same plane; i.e., when the
angle between R and the axis of G is = 90°. Call that angle
6. If, then, cos 6 = cos a cos A. -j- cos ft cos jn -f- cos y cos v
is = = cos 90°, we may combine It and C to produce a
single resultant for the whole system ; acting in a plane con-
taining It and parallel to the plane of C in a direction parallel
to It, at a perpendicular distance c = -5 from the origin and
= It in intensity. The condition that a system of forces in
space have a single resultant is, therefore, substituting the
previously derived values of the cosines, (2J£) . L -J- (2 Y) . M
+ (2Z) . N = 0.
This includes the cases when It is zero and when the system
reduces to a couple.
To return to the general case, It and C not being in the
same plane, the composition of forces in space cannot be
further simplified. Still we can give any value we please to
P, one of the forces of the couple C, calculate the correspond-
G
ing arm a = -p, then transfer C until one of the _P's has the
same point of application as It, and combine them by the
parallelogram of forces. "We thus have the whole system
equivalent to two forces, viz., the second P, and the resultant
of It and the first P- These two forces are not in the same
plane, and therefore cannot be replaced by a single resultant.
39. Problem. (Non-con current forces in space.) — Given all
geometrical elements (including or, /3, y, angles of P), also the
weight of Q, and weight of apparatus G ; A being a hinge whose
pin is in the axis Y, O a ball-and-socket joint : required the
amount of P (lbs.) to preserve equilibrium, also the pressures
STATICS OP A RIGID BODY.
41
{amount ana direction) at A and ; no friction. Eeplace P
by its X, Y, and Z components. The pressure at A will have
Fig. 44.
Z and X components ; that at 0, X, Y, and Z components.
The body is now free, and there are six unknowns.
2X, 2 Y, and 2Z give, respectively,
P cos a + X, + X = ;
P cos /? + 7", = ; and Z, + Z — Q — Q - P cps y = 0.
As for moment-equations (see note in last paragraph), project-
ing the system upon YZ and putting 2{Pa) about O = 0,
we have
— Z l l+Qd+ Ge + (P cos y)h-Jr(Peo$0)c = 0;
projecting it upon XZ, and putting 2{Pa) about = 0, we
have Qr — (P cos «)c — (P cos ^)
H
y
the differential equation of the curve;
yy I xdx = 4t • % ! or x* = y, the equation of a
H*
Note.— The same result, ^ = ~- , may be obtained by considering that
parabola whose vertex is at and axis vertical
dp
dx ~ Ho
we have here (Prin. II.) a, free rigid body acted on
by three forces, T, Bo, and It = qx, acting verti-
"■ cally through the middle of the abscissa x; the
resultant of Eo and B must be equal and oppo-
V 1 v R dv
I ! * site to T, Fig. 52. .-. tans ds r>
dx = vft^- •"• x = C J« vFT^ = lo lo ^< s + Vs *+ ^
and x = e . log, [(« + #+7)^-c], ... (2)
a relation between the horizontal abscissa and length of curve.
Again, in eq. (1) put dx* = ds 1 — dy', and solve for dy.
This gives dy = -^L,- = \ . fq^- Therefore
y = ijfV + s*y *<%( - f) -=- 2y (4)
Example. — A 40-foot chain weighs 240 lbs., and is so hung
from two points at the same level that the deflection is 10
feet. Here, for s — 20 ft., y = 10; hence eq. (4) gives the
parameter, c = (400 — 100) -=- 20 = 15 feet, q = 240 -=- 40
= 6 lbs. per foot. .-. the tension at the middle is JI = qc
— 6 X 15 = 90 lbs.; while the greatest tension is at either
support and = \'W + 12~0 2 = 150 lbs.
Knowing c = 15 feet, and putting s = 20 feet = half
length of chain, we may compute the corresponding value of
x from eq. (2) ; this will be the half-span [log, m — 2.30258
X (common log m)]. To derive s in terms of x, transform
eq. (2) in the sense in which n = log, m may be transformed
into e n = m, clear of radicals, and solve for s, which gives
= &[f -
(4)
Again, eliminate s from (2) by substitution from (3), trans-
form as above, clear of radicals, and solve for y -4- c, whence
r- X _ X ~
y -J- o = ic\j d + e 5 J, (5)
48 MECHANICS OF ENGINEERING.
which is the equation of a catenary with axes as in Fig. 54.
If the horizontal axis be taken a distance = c below the ver-
tex, the new ordinate y' — y -\- o, while x remains the same ;
the last equation is simplified.
If the span and length of chain are given, or if the span
and deflection are given, c can be determined from (4) or (5)
only by successive assumptions and approximations.
PART II -DYNAMICS.
CHAPTER I.
RECTILINEAR MOTION OF A MATERIAL POINT.
49. Uniform Motion implies that the moving point passes
over equal distances in equal times ; variable motion, that un-
equal distances are passed over in equal times. In uniform
motion the distance passed over in a unit of time, as one sec-
ond, is called the velocity (= v), which may also be obtained
by dividing the length of any portion {= s) of the path by
the time (= t) taken to describe that portion, however small or
great ; in variable motion, however, the velocity varies from
point to point, its value at any point being expressed as the
quotient of ds (an infinitely small distance containing the
given point) by dt (the infinitely small portion of time in
which ds is described).
49a. By acceleration is meant the rate at which the velocity
of a variable motion is changing at any point, and may be a
uniform acceleration, in which case it equals the total change
of velocity between any two points, however far apart, divided
by the time of passage ; or a variable acceleration, having a
different value at every point, this value then being obtained
by dividing the velocity-increment, dv, or gain of velocity
in passing from the given point to one infinitely near to it, by
dt, the time occupied in acquiring the gain. (Acceleration
must be understood in an algebraic sense, a negative accelera-
tion implying a decreasing velocity, or else that the velocity in
a negative direction is increasing.) The foregoing applies to
motion in a path or line of any form whatever, the distances
mentioned being portions of the path, and therefore measured
along the path.
4
50 MECHANICS OP ENGINEERING.
50. Rectilinear Motion, or motion in a straight line. — The
general relations of the quantities involved may be thus stated
(see Fig. 55) : Let v = velocity #f the body at any instant ;
-S o : . s _ v .. dsds- +S t hen f» = gain of velocity
• \~^. * T~~ hi an instant of time dt. Let
• dt'^du i t = time elapsed since the
FlG ' 55 ' body left a given fixed point,
which will be taken as an origin, 0. Let s = distance (-(- or
— ) of the body, at any instant, from the origin ; then ds =
distance traversed in a time dt. Let^> = acceleration = rate
at which v is increasing at any instant. All these may be
variable ; and t is taken as the independent variable, i.e., time
is conceived to elapse by equal small increments, each = dt;
hence two consecutive <#s's will not in general be equal, their
difference being called d's. Evidently d't is = zero, i.e., dt is
constant.
Since -=- = number of instants in one second, the velocity at
any instant (i.e., the distance which would be described at that
rate in one second) is v = as . -j- ; .". v = -57 (1.)
1 / (ds\ d?s\
Similarly, p = dv . -^, and ^since dv = dy^J = -%)'
dv d*s /TT .
■■* Z =-di=d? ( IL >
Eliminating dt, we have also vdv =jpds (HI.)
These are the fundamental differential formulae of rectilinear
motion (for curvilinear motion we have these and some in ad-
dition) as far as kinematics, i.e., as far as space and time, is
concerned. The consideration of the mass of the material
point and the forces acting upon it will give still another rela-
tion (see § 55).
51. Rectilinear Motion due to Gravity. — If a material point
fall freely in vacuo, no initial direction other than vertical
having been given to its motion, many experiments have
RECTILINEAR MOTION OF A MATERIAL POINT. 51
shown that this is a uniformly accelerated rectilinear motion
in a vertical line having an acceleration (called the accelera-
tion of gravity) equal to 32.2 feet per square second, or 9.81
metres per square second ; i.e., the velocity increases at this
constant rate in a downward direction, or decreases in an up-
ward direction.
[Note. — By " square second " it is meant to lay stress on the fact that an
acceleration (beiDg = d 2 s -t- df) is in quality equal to one dimension of
length divided by two dimensions of time. E.g., if instead of using the
foot and second as units of space and time we use the foot and the minute,
g will = 32.2 X 3600; whereas a velocity of say six feet per second would
= 6 X 60 feet per minute. The value of g = 32.2 implies the units foot
and second, and is sufficiently exact for practical purposes.]
52. Free Fall in Vacuo.— Fig. 56. Let the body start at
with an initial downward velocity = c. The accelera- _s
tion is constant and = -(- g. Reckoning both time and j
distance (+ downwards) from 0, required the values of *° I
the variables s and v after any time t. From eq. (II.), j I c
§ 50, we have + g = dv -f- dt ; .-.dv = gdt, in which the v h
two variables are separated. I ^ m
Hence J o dv = gj dt; i.e., \jo = g\j; or v - c = ^
gt - ; and finally, v = c + gt (1) Fl °- 66 -
(Notice the correspondence of the limits in the foregoing
operation ; when t = 0, v = -f- c.)
From eq. (I.), § 50, v = ds -f- dt ; /. substituting from (1),
ds — (c -f- gt)dt, in which the two variables s and t are sepa-
rated.
/ s pt pt r*
ds = cj dt + gj a tdt\\.*.,\j =
t rtf
t+ff
M)
Lo2'
or s=ct + kgf (2)
Again, eq. (HI.), § 50, vdv = go's, in which the variables v
and s are already separated.
.-. £ "vdv =g£*ds ; or [V = g\J ; i- e -> £0" - O = gs,
or
"'-"■ (3)
2?
52 MECHANICS OF ENGINEERING.
If the initial velocity = zero, i.e., if the body falls from rest,
eq. (3) gives s—^-andv— ^2gh. [From the frequent re-
v'
currence of these forms, especially in hydraulics, x-is called the
"height due to the velocity «," i.e., the vertical height through
which the body must fall from rest to acquire the velocity v ;
while, conversely, Y^gh j s called the velocity due to the height
or head A.J
By eliminating g between (1) and (3), we may derive another
formula between three variables, s, v, and t, viz.,
» = Xp + v)6 (4)
53. Upward Throw. — If the initial velocity were in an up-
ward direction in Fig. 56 we might call it — c, and introduce it
with a negative sign in equations (1) to (4), just derived ; but
for variety let us call the upward direction +, in which case
an upward initial velocity would = -f- c, while the acceleration
= — g, constant, as before. (The motion is supposed confined
within such a small range that g does not sensibly vary.) Fig.
lie
| 57. From p = dv -f- dt we have dv = — gdt and
mo t° /*" , /*'
" ._ J. dv = ~ ffJo dt;.:v-c = - gt;orv = c - gt. (l)a
s * From v = ds ,-f- dt, ds = cdt — gtdt,
/>» pt nt
ac i-e., J ds = cJ o dt - gj tdt ; or s = ct - \gt\ (2)a
T ' (*» n»
— S vdv = pds gives / vdv = — g I ds, whence
Fio. 57. « /c « /0
& — V 1
§ (v' —
c ~ g, v is negative, showing a downward motion ; for t >
2c -7- g, s is negative, i.e., the body is below the starting-point
while the rate of change of v is constant and = — g at all
points.
54. Newton's Laws. — As showing the relations existing in
general between the motion of a material point and the actions
(forces) of other bodies upon it, experience furnishes the fol-
lowing three laws or statements as a basis for dynamics :
(1) A material point under no forces, or under balanced
forces, remains in a state of rest or of uniform motion in a
right line. (This property is often called Inertia.)
(2) If the forces acting on a material point are unbalanced,
an acceleration of motion is produced, proportional to the re-
sultant force and in its direction.
(3) Every action (force) of one body on another is always
accompanied by an equal, opposite, and simultaneous reaction.
(This was interpreted in § 3.)
As all bodies are made up of material points, the results ob-
tained in Dynamics of a Material Point serve as a basis for the
Dynamics of a Kigid Body, of Liquids, and of Gases.
55. Mass. — If a body is to continue moving in a right line,
the resultant force P at all instants must be directed along that
line (otherwise it would have a component deflecting the body
from its straight course).
In accordance with Newton's second law, denoting by^j the
acceleration produced by the resultant force {G being the
body's weight), we must have the proportion P : Q : : jp : g ;
i.e.
P = -.p , ovP = Mj>. . . (IV.)
9
Eq. IV. and (I.), (II.), (III.) of § 50 are the fundamental
equations of Dynamics. Since the quotient O -e- g is invaria-
54 MECHANICS OF ENGINEERING.
ble, wherever the body be moved on the earth's surface (G and
g changing in the same ratio), it will be used as the measure
of the mass J/^or quantity of matter in the body. In this way
it will frequently happen that the quantities G and g will ap-
pear in problems where the weight of the body, i.e., the force
of the earth's attraction upon it, and the acceleration of gravity
have no direct connection with the circumstances. No name
will be given to the unit of mass, it being always understood
that the fraction G-i-g will be put for M before any numeri-
cal substitution is made. From (IY.) we have, in words,
( accelerating force = mass X acceleration;
\ also, acceleration — accelerating force -=- mass.
56. Uniformly Accelerated Motion. — If the resultant force is
constant as time elapses, the acceleration must be constant (from
eq. (IV.), since of course M is constant) and =I > -~lf. The
motion therefore will be uniformly accelerated, and we have
only to substitute -j- ^> (constant) for g in eqs. (1) to (4) of
§ 52 for the equations of this motion, the initial velocity being
= c (in the line of the force).
v = c+pt . . . (1); s=ct + ipf; ... (2)
(v 2 — c 2 )
s = — ^— ; . . . (3), and » = |(c + v )t . . . (4)
If the force is in a negative direction, the acceleration will
be negative, and may be called a retardation; the initial veloc-
ity should be made negative if its direction requires it.
57. Examples of TInif. Ace. Motion.— Example 1. Fig. 58.
A small block whose weight is \ lb. has already described a
-S c P M ,„ distance Ao — 48 inches over a
A SMOOTH > ^^^i^ m | > *"^ "T" o i |
£2 J ~ 6 misr \ smooth portion of a horizontal
fio. 58. table in two seconds; at it en-
counters a rough portion, and a consequent constant friction of
2 oz. Eequired the distance described beyond 0, and the time
occupied in coming to rest. Since we shall use 32.2 for g,
times must be in seconds, and distances in feet ; as to the unit
RECTILINEAR MOTION OF A MATERIAL POINT. 55
of force, as that is still arbitrary, say ounces. Since AO was
smooth, it must have been described with a uniform motion
(the resistance of the air being neglected); hence with a veloc-
ity = 4 ft. -f- 2 sec. = 2 ft. per sec. The initial velocity for
the retarded motion, then, is c — -\- 2 at 0. At any point be-
yond the acceleration = force -=- mass = (— 2 oz.) -=- (8 oz.
-i- 32.2) = — 8.05 ft. per square second, i.e., p — — 8.05 =
constant ; hence the motion is uniformly accelerated (retarded
here), and we may use the formulae of § 56 with c = -f- 2, p =
— 8.05. At the end of the motion v must be zero, and the
corresponding values of s and t may be found by putting v =
in equations (3) and (1), and solving for s and t respectively :
thus from (3),* =i(-4)-f- (— 8 - 05 )> i- e ->* = 0-248 +, which
must be feet ; while from (1), t=(—2) + (— 8.05) = 0.248 +,
which must be seconds.
Example 2. (Algebraic.) — Fig. 59. The two masses M 1 =
G 1 -v- g and M = G -e- g are connected by a flexible, inexten-
sible cord. Table smooth. Required the acceleration common
to the two rectilinear motions, and the tension in the string S,
-&
^_i
Fiq. 59. Fig. 60.
there being no friction under G„ none at the pulley, and no
mass in the latter or in the cord. At any instant of the mo-
tion consider #, free (Fig. 60), N being the pressure of the
table against (l-±).
(1)
RECTILINEAR MOTION OF A MATERIAL POINT. 61
Evidently v decreases, as it should. Now inquire how small
a value c may have that the body shall never return; i.e.,
that v shall not = until s = oo. Put v = and s = oo in
(1) and solve for c ; and we have
N,"
/
J.A
I/////'/
\ 1
\ si
ts-
%
in
1
[<-A+-G->
FhT "!r
c = 4/2^ = V2 X 32.2 X 21000000,
= about 36800 ft. per sec. or nearly 7 miles per sec. Con-
versely, if a body be allowed to fall, from rest, toward the
earth, the velocity with which it would strike the surface
would be less than seven miles per second through whatever
distance it may have fallen.
If a body were allowed to fall through a straight opening in
the earth passing through the centre, the motion would be har-
monic, since the attraction and consequent acceleration now
vary directly with the distance from the centre. See Prob. 2.
This supposes the earth homogeneous.
Problem 4. — Steam working expansively and raising a weight.
Fig. 67. — A piston works without
friction in a vertical cylinder. Let
S = total steam-pressure on the
under side of the piston ; the weight
G, of the mass G -r- g (which in-
cludes the piston itself) and an
atmospheric pressure = A, con-
stitute a constant back-pressure.
Through the portion OB = s % , of rio. 67.
the stroke, S is constant = S„ while beyond B, boiler com-
munication being " cut off," S diminishes with Boyle's law, i.e.,
in this case, for any point above B, we have, neglecting the
" clearance", F being the cross-section of the cylinder,
S:S x \'.Fa x '.F*\ or £=$»,-*-*.
Pull length of stroke = ON = s n . Given, then, the forces
S and A, the distances s, and s n , and the velocities at O and
fit N both = (i.e., the mass M = O -r- g is to start from rest at
O and to come to rest at N), required the proper weight G to
62 MECHANICS OF ENGINEERING-.
fulfil these conditions, 8 varying as already stated. The accel-
eration at any point will be
p = [S - A- G] -i- M. . . . . (1)
Hence (eq. III.) Mvdv = [8 — A — G]ds, and .\ for the
whole stroke
Mf o vdv=f o 18 — A- G\ds; i.e.,
= S,£ds + 8, S ,£j - *r* - *£*>
or 8,s, [l + log e S fJ = As n + Gs n . ... (2)
Since 8 = 8, = constant, from (? to .5, and variable, =
S,s, -7- s, from B to iV, we have had to write the summation
X
■N
Sds in two parts.
From (2), G becomes known, and .\ M also (= G -f- g).
Required, further, the time occupied in this upward stroke.
From to B (the point of cut-off) the motion is uniformly
accelerated, since p is constant (S being = 8, is eq. (1) ),
with the initial velocity zero; hence, from eq. (3), § 56,
the velocity at B = v, — V2 [8, — A — G]s, -f- Mis known ;
.•. the time t, =± 2s, -f- v, becomes known (eq. (4), § 56) of de-
scribing OB. At any point beyond B the velocity v may be ob-
tained thus : From (III.) vdv = pds, and eq. (1) we have,
summing between B and any point above,
Mf\dv = S,s,f s ° ~-{A + G)/^ ; i.e.,
Qlv 1 —v*\ «
f L — 2 — = %'* l0 £. ,* ~ i A + G) (» - *,)•
This gives the relation between the two variables v and s
anywhere between B and iV"; if we solve for v and insert its
value in dt — ds -f- «, we shall have dt = a function of s and
">•> M l C ~ <0 = +f t Pdt. . (3)
RECTILINEAR MOTION OE A MATERIAL POINT. 65
The two integrals are identical, numerically, term by term,
since the pressure which at any instant accelerates M, is nu-
merically equal to that which retards M 1 ; hence, though we do
not know how P varies with the time, we can eliminate the
definite integral between. (2) and (3) and solve for C. If
the impact is inelastic (i.e., no power of restitution in either
body, either on account of their total inelasticity or damaging
effect of the pressure at the surfaces of contact), they continue
to move with this common velocity, which is therefore their
final velocity. Solving, we have
6 ~ M,+M, W
Next, supposing that the impact is partially elastic, that the
bodies are of the same material, and that the summation
I Pdt for the second period of the impact bears a ratio, e,
t/t' i
to that / Pdt, already used, a ratio peculiar to the material,
if the impact is not too severe, we have, summing equations
(1) for the second period (letting F, and V, = the velocities
after impact),
M t fjdv, = - fJ'Pdt, i.e., Mi V,-0) = - ef'pdt ; (5)
M, f/dv, = + fJ'Pdt, i.e., Ml V-C) = -\- ef'Pdt. (6)
e is called the coefficient of restitution.
Having determined the value of / Pdt from (2) and (3) in
terms of the masses and initial velocities, substitute it and that
of C, from (4), in (5), and we have (for the final velocities)
V, = \M fil + J/a- eM^-c,)} - [Jf, + Jf J; . (7)
and similarly
V^lM^ + M^+eM^-c^ + lM. + M^ . (8)
For e = 0, i.e., for inelastic impact, V,= V,= Cm eq. (4) ; for
66 MECHANICS OP ENGINEERING.
e = 1, or elastic impact, (7) and (8) become somewhat simpli-
fied.
To determine e experimentally, let a ball (M t ) of the sub-
stance fall upon a very large slab (Jf s ) of the same substance,
noting both the height of fall h x , and the height of rebound H x .
Considering M, as = co, with
Ci= V 2gh x , V x = - V 2g£T x , and c„ = o,
eq. (7) gives
- Vlp?;, =- e ^2g~h x ;.:e= VB i -i-h 1 .
Let the student prove the following from equations (2), (3),
(5), and (6) :
(a) For any direct central impact whatever,
M fi , + M,e t = M l V,-\- M, Y v
[The product of a mass by its velocity being sometimes
called its momentum, this result may be stated thus :
In any direct central impact the sum of the momenta before
impact is equal to that after impact (or at any instant during
impact). This principle is called the Conservation of Momen-
tum. The present is only a particular case of a more general
proposition.
It may be proved C, eq. (4), is the velocity of the centre of
gravity of the two masses before impact ; the conservation of
momentum, then, asserts that this velocity is unchanged by the
impact, i.e., by the mutual actions of the two bodies.]
(5) The loss of velocity of M x , and the gain of velocity of
J/j, are twice as great when the impact is elastic as when in-
elastic.
(c) If e = 1, and M x = M„ then V x = + c„ and V % = c x .
Example. — Let M x and M* be perfectly elastic, having weights = 4 and
5 lbs. respectively, and let Ci = 10 ft. per sec. and e a = — 6 ft. per sec.
(i. e., before impact Mi is moving im a direction contrary to that of M x ).
By substituting in eqs. (7) and (8), with e = 1, Mi =4 + y, and M t — 5-i-g,
we have
7i = I[~4 x 10 + 5 x (- 6) - 5 (lO - (- 6))1= - 7.7 ft. per sec.
Vi =z I[~4 x 10 + 5 x (- 6) + 4 (lO - (- 6))] = + 8.2 ft. per sec.
nfl flip vftlnnittfis aftftr lTnTiftnt. Nntip.fi thfiir directions, as indicated hv their
"virtual velocities." 67
CHAPTER II.
" VIRTUAL VELOCITIES."
61. Definitions. — If a material point is moving in a direction
not coincident with that of the resultant force acting (as in
curvilinear motion in the next chapter), and any element of its
path, ds, projected upon this force ; the length of this projec-
tion, du, Fig. 69, is called the "Yiktual "Velocity" of the
force, since du -f- dt may be considered the veloc-
ity of the force at this instant, just as ds -e- dt is^ M .
that of the point. The product of the force by o^J\
its du will be called its virtual moment, reckoned % Pj cos arj-4-Pj cos « 2 . Hence P(ds cos a)=
^'s' :Si v P t (ds cos <*,) + P\(ds cos «„). But ds cos a
~"~X = the projection of ds upon P, i.e., = du ;
Fig. 70. fa cos a^ = ^ Mi) e tc. ; .\ Pdu = P x du^ -\-
P t du t . If in Fig. 70 a, were > 90°, evidently we would
have Pdu — — P^du, + P^du,, i.e., P t du x would then be
negative, and OP s would fall behind 0; hence the definition
of -L. and. — in § 61. For any number of components the
proof would be similar, and is equally applicable whether they
are in one plane or not.
63. Prop. II. — The sum of the virtual moments equals zero,
for concurrent forces in equilibrium.
68 MECHANICS OP ENGINEERING.
(If the forces are balanced, the material point is moving in
a straight line if moving at all.) The resultant force is zero.
Hence, from § 62, P t du, -\- P i du % -\- etc. = 0, having proper
regard to sign, i.e., 2{Pdu) = 0.
64. Prop. III. — The sum of the virtual moments egualszero
for any small displacement or motion of a rigid body in equi-
librium under non-concurrent forces in a plane; all points of
the body moving parallel to this plane. (Although the kinds
of motion of a given rigid body which are consistent with
balanced non-concurrent forces have not yet been investigated,
we may imagine any slight motion for the sake of the alge-
braic relations between the different du\ and forces.)
. First, let the motion be a translation, all
,. — ^/ points of the body describing equal parallel
xj />~ X lengths = ds. Take .Z" parallel to ds ; let a„
<\d?..f s \jr etc., be the angles of the forces with X.
> ' Sg* Then(§35)2(Pcosa) = 0;.\ /';.:.:.,- ) " = ; but ds cos a, = du x ; ds cos a t = du t ;
S(3V etc. ; .-. 2{Pdu) = 0. Q. E. D.
4- Secondly, let the motion be a rotation
fig. 71. through a small angle dd in the plane of the
forces about any point (9 in that plane, Fig. 72. With as a pole
let p, be the radius-vector of the point of application of _P„ and
«, its lever-arm from O; similarly for the
other forces. In the rotation each point of
application describes a small arc, ds v ds„
etc., proportional to p„ p„ etc., since ds t
= p^dd, ds, = p^dd, etc. From § 36, , •:'••;
P& + etc. = ; but from similar triangles < " : -
dst : du t :: p, : a, ; .-. a, = p$u x -=- ds l
= du, -=- dd ; similarly a 2 = du, ~ dd, etc. F,G - 72 -
Hence we must have [P^du, -j- P^du, +...]-- dd — 0, i.e.,
2{Pdu) = 0. Q. E. D.
Now since any slight displacement or motion of a body may
be conceived to be accomplished by a small translation fol-
lowed by a rotation through a small angle, and since the fore-
"virtual velocities." 69
going deals only with projections of paths, the proposition is
established and is called the Principle of Virtual Velocities.
[A similar proof may be used for any slight motion what-
ever in space when a system of non-concurrent forces is bal-
anced.] Evidently if the path (ds) of a point of application is
perpendicular to the force, the virtual velocity (du), and con-
sequently the virtual moment (Pdu) of the force are zero.
Hence we may frequently make the displacement of such a
character in a problem that one or more of the forces may be
excluded from the summation of virtual moments.
65. Conneoting-Rod by Virtual Velocities. — Let the effective
steam-pressure P be the means, through the connecting-rod
and crank (i.e., two links), of raising the weight G very slowly;
neglect friction and the weight of the links themselves. Con-
sider AB as free (see (b) in Fig. 73), ,Z?Calso, at (c); let the
c
G
-.C
(c.) ■■
"small displacements" of both be simultaneous small portions
of their ordinary motion in the apparatus. A has moved to A^
through dx ; B to B„ through ds, a small arc ; has not
moved. The forces acting on AB are P (steam-pressure), JV
(vertical reaction of guide), and N' and T (the tangential and
normal components of the crank-pin pressure). Those on BG
are N' and T (reversed), the weight G, and the oblique pressure
of bearing P'. The motion being slow (or rather the accelera-
tion being small), each of these two systems will be considered as
balanced. Now put 2(Pdu) — for AB, and we have
P& + JX0 + J'X0-T& = 0. . . (1)
For the simultaneous and corresponding motion of BG,
2(Pdu) = gives
70
MECHANICS OF ENGINEEBING.
W X0-\- Tds- Gdh + P' X0 = 0,
(2)
dh being the vertical projection of G's motion.
From (1) and (2) we have, easily, Pdx — Gdh = 0, . (3)
..-/••?i which is the same as we might have
j. N y^a» obtained by putting 2(Pdu) = Ofor
■ Ig^p' the two links together, regarded col-
* \ lectively as a free body, and describ-
Fl °- 74 - ing a small portion of the motion
they really have in the mechanism, viz.,
"A,
• Pdx + JVxO-Gdh + P'xO = 0.
We may therefore announce the —
(4)
66. Generality of the Principle of Virtual Velocities. — If any
mechanism of flexible inextensible cords, or of rigid bodies
jointed together, or both, at rest, or in motion with very small
accelerations, be considered free collectively (or any portion of
it), and all the external, forces put in; then (disregarding
mutual frictions) for a small portion of its prescribed motion,
2(Pdu) must = 0, in which the du, or virtual velocity, of
each force, P, is the projection of the path of the point of
application upon the force (the product, Pdu, being + or —
according to § 61).
67. Exa?nple. — In the problem of § 65, having given the
weight G, required the proper steam-pressure (effective) P to
hold G in equilibrium, or to raise it uniformly, if already in
motion, for a given position of the links. That is. Fig. 75,
given a, r, c, a,
and /?,
re-
a
quired the ratio dh: dx; for,
from equation (3), § 65, P
= G(dh : dx). The projec-
tions of dx and ds upon AB
will be equal, since AB =
-4,-Z?,, and makes an (infinitely) small angle with A x B a i.e.,
dx cos a = ds cos (/? — a). Also, dh = (c : r)ds sin /J.
dx A,
Fiq. 75.
"VIRTUAL VELOCITIES." 71
Eliminating ds, we have,
dh _ c sin /? cos a c sin /J cos a
dx ~ r cos (ft — a) ' ' " ~ ^ cos (/3 — a)'
68. When the acceleration of the parts of the mechanism is
not practically zero, 2(_Pdu) will not = 0, but a function of
the masses and velocities to be explained in the chapter on
Work, Energy, and Power. If friction occurs at moving joints,
enough " free bodies" should be considered that no free body
extend beyond such a joint ; it will be found that this friction
cannot be eliminated in the way in which T and W were, in
§ 65.
69. Additional Problems; to be solved by " virtual velocities."
Problem 1.- — Find relations between the forces acting on a
straight lever in equilibrium ; also, on a bent lever.
Problem 2. — When an ordinary copying-press is in equilib-
rium, find the relation between the force applied horizontally
and tangentially at the circumference of the wheel, and the
vertical resistance under the screw-shaft.
72
MECHANICS OF ENGINEERING.
CHAPTER III.
CURVILINEAR MOTION OF A MATERIAL POINT.
Fig. 76.
[Motion in a plane, only, will be considered in this chapter.]
70. Parallelogram of Motions. — It is convenient to regard
the curvilinear motion of a point in a plane as compounded, or
made up, of two independent rectilinear motions parallel
respectively to two co-ordinate axes X. and Y, as may be ex-
plained thus : Fig. 76. Consider the
drawing-board CD as fixed, and let the
head of a T-square move from A
toward B along the edge according to
any law whatever, while a pencil moves
from M toward Q along the blade. The
result is a curved line on the board, whose
form depends on the character of the
two JTand Y component motions, as they may be called. If
in a time £, the T-square head has moved an ^"distance = MN,
and the pencil simultaneously a Y distance = MP, by com-
pleting the parallelogram on these lines, we obtain H, the
position of the point on the board at the end of the time t t .
Similarly, at the end of the time £/ we find the point at B'.
71. Parallelogram of Velocities. — Let the X and Amotions
be uniform, required the resulting motion. Fig. 77. Let c„
and c y be the constant uniform Xand ^velocities. Then in
any time, t, we have x — cj and y =
c y t ; whence we have, eliminating t,
® -s- y = c a -~ c v = constant, i.e., x is
proportional to y, i.e., the path is a
straight line. Laying off OA = c m
and AB = c v , Bis a, point of the path,
and OB is the distance described by the point in the first
CURVILINEAR MOTION OE A MATERIAL POINT. 73
second. Jsince by similar triangles OR : x : : OB : c x , we
Lave also OR = OB . t ; hence the resultant motion is uniform,
and its velocity, ~OB = c, is the diagonal of the parallelogram
formed on the two component velocities.
Corollary. — If the resultant motion is curved, the direction
and velocity of the motion at any point will be given by the
diagonal formed on the component velocities at that instant.
The direction of motion is, of course, a tangent to the curve.
72. Uniformly Accelerated X and Y Motions.— The initial
velocities of both being zero. Required the resultant motion.
Fig. 78. From § 56, eq. (2) (both e x andc„ /
being — 0), we have x = \pj? and y = //"" ■^Jr
ip v f, whence x ~ y = p x -+p y = constant, 4fv^7 /
and the resultant motion is in a straight jL^fy- r ' x
line. Conceive lines laid off from on X 0< x _>
and Y to represent ^ and^, to scale, and Fia 78-
form a parallelogram on them. From similar triangles {OB
being the distance described in the resultant motion in any
time t),OR : x : : 'OB : p x ; .-. OR~- \OBl'. Hence, from the
form of this last equation, the resnltant motion is uniformly
accelerated, and its acceleration is OB = p, (on the same scale
as^. andj? B ).
This might be called the parallelogram of accelerations, but
is really a parallelogram of forces, if we consider that a free
material point, initially at rest at 0, and acted on simulta-
neously by constant forces P x and P y (so that p x = _P X -^ M
and^, = P y -~ M), must begin a uniformly accelerated recti-
linear motion in the direction of the resultant force, having no
initial velocity in any direction.
73. In general, considering the point hitherto spoken of as a
free material point, under the action of one or more forces, in
view of the foregoing, and of Newton's second law, given the
initial velocity in amount and direction, the starting-point,
the initial amounts and directions of the acting forces and the
74 MECHANICS OF ENGINEERING.
laws of their variation if they are not constant, we can resolve
them into a single X and a single T force at any instant,
determine the X and Amotions independently, and afterwards
the resultant motion. The resultant force is never in the direc-
.tion of the tangent to the path (except at a point of inflection).
The relations which its amount and direction at any instant
bear to the velocity, the rate of change of that velocity, and
the radius of curvature of the path will appear in the next
paragraph.
74. General Equations for the curvilinear motion of a ma-
terial point in a plane. — The motion will be considered result-
J\\ ing from the composition of
xi- ^ v\r independent X and JT motions,
I 1*> X and Y being perpendicular to
L--" each other. Fig. 79. In two
M consecutive equal times (each
i\ = dt), let dx and dx' = small
spaces due to the X motion;
A^^ d 'P ' , "'1 and dy and CK = dy', due to
I the 1 motion. Ihen ds and
J «, = the acceleration of v, i.e., the tangential acceleration.
CURVILINEAR MOTION OP A MATERIAL POINT. 75
then jp t = d*s -f- dt , and, since d's — the sum of the projec-
tions of FF&ud GF on BC, i.e., d*s = d'x cos a -f d*y sin a,
we have
jj» = -^ cos a +^5 sin a; i.e., _p t = ^ cos a -fj?„ sin a. (2)
By Normal Acceleration we mean the rate of change of the
velocity in the direction of the normal. In describing the ele-
ment AB = ds, no progress lias been made in the direction of
the normal BLTi.e., there is no velocity in the direction of the
normal; but in describing B G (on account of the new direc-
tion of path) the point has progressed a distance GL (call it
d'n) in the direction of the old normal BH (though none in
that of the new normal CI). Hence, just as the tang. ace.
ds' — ds d's , . . GL — zero d'n
= sr- = dt» so the nomal acceL = sr^ = w
It now remains to express this normal acceleration (=j> n ) in
terms of the X and Y accelerations. From the figure, GL
= CM- ML, i.e.,
d'n = d?y cos a — d?x sin a {since EF = d'x\ ;
d'n d'y d 2 x .
'■If =-df C0Sa -df Sma -
Hence Pn=Py cos a — ^.sin a (3)
The norm. ace. may also be expressed in terms of the tang,
velocity v, and the radius of curvature r, as follows :
ds' = rda, or da — ds' -~- r ; also d'n = ds'da, — ds'' ~ r,
. d'n (ds'Vl V
If now, Fig. 80, we resolve the forces X = Mp x and Y
Y = Mp v , which at this instant account for the
X and Y accelerations (M = mass of the
,/' i**** material point), into components along the
K' \ tangent and normal to the curved path, we
..•-' shall have, as their equivalent, a tangential
force
?K#
•M\
Fis. SO.
T = J^a, cos « + Mf y sin «,
76 MECHANICS OP ENGINEERING.
and a normal force
JST = Mp y cos a — Mp x sin a.
But [see equations (2), (3), and (4)] we may also write
T = Mp t = M Tt ; and W=Mp n = M-. . (5)
Hence, if a free material point is moving in a curved path,
the sum of the tangential components of the acting forces must
equal (the mass) X tang. acceL; that of the normal components,
= (the mass) X normal accel. = (mass) X (square of veloc. in
path) -=- (rad. curv.).
It is evident, therefore, that the resultant force (= diagonal
on T and N or on JTand Y, Fig. 80) does not act along the tan-
gent at any point, but toward the concave side of the path ; un-
less r = 00.
Radius of curvature. — From the line above eq. (4) we
have d'n = ds'-^r; hence (line above eq. (3)), ds* ~ r =
d'y cos a — d'x sin a ; but cos a=dx-i-ds, and sin a=dy-i-ds,
ds* dx dy ds' . Tdxd'y — dyd'xl
i.e., -y = dx*d\_ ||] = dx'd (tan a),
•' r_ \dtj ■ i\dtl dt~J>
, r „<^ tan «1
r =*-*-l>-3r-J ( 6 )
which is equally true if, for v x and tan a, we put v v and
tan (90° — a), respectively.
Change in the velocity square. — Since the tangential accelera-
tion -£ =p t , we have ds-j- —p t ds; i.e.,
^dv=p t ds, or vdv—p t ds and .-. — ~^- = Jp t ds. (7)
having integrated between any initial point of the curve where
v = c, and any other point where v = v. This is nothing
more than equation (III.), of § 50.
CURVILINEAR MOTION OF A MATERIAL POINT.
77
-Fig.
Let
81. Let
OB' be a
75. Normal Acceleration. Second Method. -
C be the centre of curvature and OB = 2r.
portion of the oscillatory parabola (vertex at j^-..
; any osculatory curve will serve). When \"
ds is described, the distance passed over in
the direction of the normal is AB ; for 2ds,
it would be A'B' = iAB (i.e., as the
square of OB'; property of a parabola),
and so on. Hence the motion along the normal is uniformly
accelerated with initial velocity = 0, since the distance AB,
varies as the square of the time (considering the motion along
the curve of uniform velocity, so that the distance OB is di-
rectly as the time). If p n denote the accel. of this uniformly
accelerated motion, its initial velocity being = 0, we have (eq.
2, § 56) AB = %p n df, i.e., p n = 2AB -^ ^._Bnt from the
similar triangles ODB and OAB we have, AB : ds : : ds : 2r,
hence 2 A B = ds' ~r, .:_p n = ds' -f- rdt' = v' -f- r.
76. Uniform Circular Motion. Centripetal Force.— The ve-
locity being constant,^ must be = 0, and .\ T(ov 2Tii there
are several forces) must = 0. The resultant of all the forces,
therefore, must be a normal force = (Mc' -=- r) = a con-
stant (eq. 5, § T4). This is called the " deviating force,"
or "centripetal force ;" without it the body would continue
in a straight line. Since forces always occur in pairs (§3),
a "centrifugal force," equal and opposite to the "centri-
petal" (one being the reaction of the other), will be found
among the forces acting on the body to whose constraint the
deviation of the first body from its natural straight course is
due. Tor example, the attraction of the earth on the moon
acts as a centripetal or deviating force on the latter, while the
equal and opposite force acting on the earth may be called
the centrifugal. If a small block moving on a
smooth horizontal table is gradually turned from
its straight course AB by a fixed circular guide,
tangent to AB at B, the pressure of the guide
against the block is the centripetal force Mc'-i- r
directed toward the centre of curvature, while
78
MECHANICS OF ENGINEERING.
Fig. 83.
the centrifugal force Mc 1 -— r is the pressure of the block
against the guide, directed away from that centre. The cen-
trifugal force, then, is never found among the forces acting on
the body whose circular motion we are dealing with.
The Conical Pendulum, or governor-ball. — Fig. 82. If a
material point of mass = M= G -4- g, suspended on a cord of
length = I, is to maintain a uniform cir-
cular motion in a horizontal plane, with a
given radius r, under tbe action of gravity
and the cord, required the velocity c to be
given it. At JS we have the body free.
The only forces acting are G and the cord-
tension P. The sum of their normal com-
ponents, i.e., 2JV, must = M& -=- r, i.e., P sin a = Mc' -f- r ;
but, since 2 (vert, comps.) = 0, P cos a = G. Hence
G tan a = Gc 1 ^- gr ; .•. c = Vgr tan a. Let u = number of
revolutions per unit of time, then u = c -f- 27tr = Vg -f- 2?r V%;
i.e., is inversely proportional to the (vertical projection )* of
the cord-length. The time of one revolution is = 1 ~ u.
Elevation of the outer rail on railroad curves (considera-
tions of traction disregarded). — Consider a single car as a
material point, and free, having a given
velocity = c. P is the rail-pressure
against the wheels. So long as the car £_-
follows the track the resultant R of P
and G must point toward the centre of w .-~
curvature and have a value = Mc* —- r.
But i?= G tan a, whence tan a = c 2 -=- gr.
If therefore the ties are placed at this
angle a with the horizontal, the pressure
will come upon the tread and not on the flanges of the wheels ;
in other words, the car will not leave the track. (This is really
the same problem as the preceding )
Apparent weight of a lody at the equator. — This is less than
the true weight or attraction of the earth, on account of the
uniform circular motion of the body with the earth in its
diurnal rotation. If the body hangs from a spring-balance,
Fio. 84.
CURVILINEAR MOTION OP A MATERIAL POINT. 79
whose indication is G lbs. (apparent weight), while the true
attraction is G' lbs., we have G' — G = Mc* ~ r. For M
we may use G -f- g (apparent values); for r about 20,000,000
ft.; for c, 25,000 miles in 2-± hrs„ reduced to feet per second.
It results from this that G is < G' by -^G' nearly, and
(since IT = 2S9) hence if the earth revolved on its axis seven-
teen times as fast as at present, G would = 0, i.e., bodies
would apparently have no weight, the earth's attraction on
them being just equal to the necessary centripetal or deviating
force necessary to keep the body in its orbit.
Centripetal force at any latitude. — If the earth were a ho-
mogeneous liquid, and at rest, its form would be spherical ; but
when revolving uniformly about the polar diameter, its form
of relative equilibrium (i.e., no motion of the particles relatively
to each other) is nearly ellipsoidal, the polar diameter being an
axis of symmetry.
Lines of attraction on bodies at its surface do not intersect
in a common point, and the centripetal force requisite to keep
a suspended body in its orbit (a small circle of the ellipsoid),
at any latitude /? is the resultant of the attraction or true
weight G' directed (nearly) toward the centre, and of G the
tension of the string. Fig. 85. G is the apparent weight, in-
dicated by a spring-balance and MA is its ^^___ ,.--/ G
line of action (plumb-line) normal to the ^ ' M
ocean surface. Evidently the apparent
weight, and consequently g, are less than
the true values, since N must be perpen- \^ Ss - f __I
dicnlar to the polar axis, while the true
values themselves, varying inversely as fnTIT
the square of MO, decrease toward the equator, hence the ap-
parent values decrease still more rapidly as the latitude dimin-
ishes. The following equation gives the apparent g for any
latitude /?, very nearly (units, foot and second):
g = 32.1808 — 0.0821 cos 2/3.
(The value 32.2 is accurate enough for practical purposes.)
Since the earth's axis is really not at rest, but moving about
80
MECHANICS OF ENGINEERING.
the sun, and also about the centre of gravity of the moon and
earth, the form of the ocean surface is periodically varied, i.e.,
the phenomena of the tides are produced.
77. Cycloidal Pendulum. — This consists of a material point
at the extremity of an imponderable, flexible, and inextensible
cord of length = I, confined to the arc of a cycloid in a ver-
tical plane by the cycloidal evolutes shown in Fig. 86. Let
the oscillation . begin (from rest) at A, a height = h above
the vertex. On reaching any lower point, as B (height = z
above 0), the point has acquired some velocity v, which is at
this instant increasing at some rate = p t . Now consider the
point free, Fig. 87; the forces acting are P the cord-tension,
normal to path, and G the weight, at an angle + P cos 90° = (G -j- g)p t ; .-.p t = g cos (p
Hence (eq. (7), § 74), vdv = p t ds gives
vdv — g cos cpds ; but ds cos
. t eneral E is different iti amount and direc-
&{>■/ -i. . v tion for each ds of the path, but du is the
nL.____\s!r ""-■■... distance through which E acts, in its own
fig. 89. direction, while the body describes any ds ;
Edu is called the work done by E when ds is described by the
body. The above equation is read : The difference between the
initial and final kinetic energy of a body = the work done by
the resultant force in that portion of the path.
(These phrases will be further spoken of in Chap. YI.)
Application of equation (a) to a planet in its orbit about
the sun. — Fig. 90. Here the only force at any instant is the at-
traction of the sun E = C -=- u* (see Prob. 3, § 59),
where C is a constant and u the variable radius
vector. As u diminishes, v increases, therefore
dv and du have contrary signs; hence equation ,
(a) gives (c being the velocity at some initial
point 0)
\Mv; -Imc^-0 /*** = o\±- ! n ; (b)
& 2 i/u, u ]_u l u a S
I 2 Of 1 T n
'"' Vl ~y G '~^"M[_u~ l ~ —|> whicl1 is depend-
ent of the direction of the initial velocity c.
Note.— If u„ were = infinity, the last member of equation (b) would re-
duce to O ■*- «„ and is numerically the quantity called potential in the
theory of electricity.
CTJRVILINEAK MOTION OF A MATERIAL POINT. 83
Application of eq. (a) to a projectile in vacuo. — G, the
body's weight, is the only force acting, and O ■
therefore = P, while M — G -j- g. There-
fore equation (a) gives
— GJ dy = Gy»
dy-du
9
2
G=R
.'. v 1 — Vc* -\- %gy\, which is independent of Fl0i 91 ,
the angle, a, of projection.
Application of equation (a) to a body sliding, without fric-
tion, on a fixed curved guide in a vertical plane ; initial velo-
city = c at 0. — Since there is some pressure at each point be-
tween the body and the guide, to consider the body free in
space, we must consider the guide removed and that the body
describes the given curve as a re-
sult of the action of the two forces,
its weight G, and the pressure P,
of the guide against the body. G
is constant, while P varies from
point to point, though always (since
there is no friction) normal to curve.
At any point, P being the resultant
of G and P, project ds upon P, thus obtaining du ; on G,
thus obtaining dy ; on P, thus obtaining zero. But by the
principle of virtual velocities (see § 62) we have Pdu = Gdy
-j- P X zero = Gdy, which substituted in eq. (a) gives
- zr(« ' - J) = / So
and therefore depends only on the vertical distance fallen
through and the initial velocity, i.e., is independent of the
form of the guide.
As to the value of P, the mutual pressure between the guide
and body at any point, since ~2Df must equal Mv* ^-r,r being
the variable radius of curvature, we have, as in § 77,
P — G sin q> = Mv* ~r; .: P = G[sin and r are different from point to point of
84 MECHANICS OP ENGINEERING.
the path, P is not constant. (The student will explain how P
may be negative on parts of the curve, and the meaning of
this circumstance.)
80. Projectiles in Vacuo. — A ball is projected into the air
(whose resistance is neglected, hence the
phrase, in vacuo) at an angle = a with the
"J
Cx
G^i horizontal; required its path; assuming it
j confined to a vertical plane. Resolve the
motion into independent horizontal (X)
fig. 93. a nd vertical ( Y) motions, G, the weight,
the only force acting, being correspondingly replaced by its
horizontal component = zero, and its vertical component
— — G. Similarly the initial velocity along X= c x = c cos a ,
along Y, = c y — csin «„. The 2T acceleration —j) x = -^- M
= 0, i.e., the X motion is uniform, the velocity v„ remains
= c x = c cos #„ at all points, hence, reckoning the time from 0,
at the end of any time t we have
x = c(cos a )t (1)
In the Y motion,^ = (— G) ■— M= — g, i.e., it is uniformly
retarded, the initial velocity being o y = c sin a ; hence, after
any time t, the T^ velocity will be (see § 56) v y = c sin a — gt,
while the distance
y = c(sin a a )t — \gf (2)
Between (1) and (2) we may eliminate t, and obtain as the
equation of the trajectory or path
ax*
y = x tan a — — -^ — - — .
u * 2o 2 cos 3 a
For brevity put c* = 2gh, h being the ideal height due to the
velocity c, i.e., o 2 -j- 2g (see § 53 ; if the ball were directed ver-
tically upward, a height h = o* -f- 2g would be actually at-
tained, or being = 90°), and we have
x*
y = xta.na e — ~j 5 (3)
in cos a v '
This is easily shown to be the equation of a parabola, with its
axis vertical.
CURVILINEAR MOTION OE A MATERIAL POINT. 85
The horizontal range.— Fig. 94. Putting y = in equa-
tion (3), we obtain
as tan a, — Jr -^~, — 1 = 0, y| c
L 4/i cos a-„J ' ; £....
which is satisfied both by x = (i.e., at the \K°-°
origin), and by x = 4A cos « sin a . Hence ^—Illv-
the horizontal range for a given o and a a is fig. 94.
x r = 4A cos a sin «„ = 2A sin 2a .
For «-„ = 45° this is a maximum (0 remaining the same),
being then = 2A. Also, since sin 2« = sin (180° — 2a ) =
sin 2(90° — or,), therefore any two complementary angles of
projection give the same horizontal range.
Greatest height of ascent ; that is, the value of y maximum,
= Vm-— Fig. 94. Differentiate (3), obtaining
<& — ° — 2A cos 2 «„'
which, put = 0, gives x = 2A sin a cos ar , and this value of
x in (3) gives y m = h sin a «„.
(Let the student obtain this more simply by considering the
Y motion separately.)
To strike a given point ; c being given and a a required. —
Let x' and y' be the co-ordinates of the given point, and a/
the unknown angle of projection. Substitute these in equa-
tion (3), h being known = c" -=- 2g, and we have
y' = x' tan a' — ~n 5 — 7. Put cos' a' = - — ■
" " 4A cos" a J ' 1 -j- tan a « "
and solve for tan a J, whence
tan a\ = [2A ± 4/4A 2 - a;' 2 - 4%'] ^a/.. . (4)
Evidently, if the quantity under the radical in (4) is negative,
tan «/ is imaginary, i.e., the given point is out of range with
the given velocity of projection c= V2gh; if positive, tan a/
has two values, i.e., two trajectories may be passed through
the point ; while if it is zero, tan ce ' has but one value.
The envelope, for all possible trajectories having the same
86
MECHANICS OF ENGINEERING.
initial velocity c (and hence the same h) ; i.e., the curve tan-
gent to them all, has but one point of contact with any one of
them ; hence each point of the envelope, Fig. 95, must have
co-ordinates satisfying the con-
dition, 4k' — x" — 4hy' = ; i.e.
(see equation (4) ), that there is
but one trajectory belonging to it.
Hence, dropping primes, the
equation of the envelope is 4A 2 —
x* — 4hy = 0. Now take 0" as a
new origin, a new horizontal axis X." , and reckon y" positive
.S j
0'
f
"^N
+
11 !
/ *1
\ 1
/ /
\ \.
Fig. 95.
downwards; i.e., substitute
and y = h — y''
The
equation now becomes x' n = 4hy" ; evidently the equation of
a parabola whose axis is vertical, whose vertex is at 0", and
whose parameter = 4A = double the maximum horizontal
range. is therefore its focus.
The range on an inclined plane. — -Fig. 96. Let OG be
the trace of the inclined plane ; its equation y
is y = x tan /?, which, combined with the
equation of the trajectory (eq. 3), will give
the co-ordinates of their intersection C.
That is, substitute y = x tan /3 in (3) and
solve for x, which will be the abscissa x x , of G.
sin /3
OS
x.
4A cos 2 a
.-. x, = 4A cos a,
= tan a — tan /? =
sin «„
cos oc„
#1 -r
Fig. 96.
This gives
sin («„ — /?)
cos /3 cos a cos /3 '
sin («„ — P) — cos /?, and the range OG,
which = x t -=- cos /?, is = (4A -e- cos 2 /?) cos a sin (a — yS). (5)
7%e maximum range on a given inclined plane, /?, c (and
.'. A), remaining constant, while at Q varies. — That is, required
the value of a which renders OG a maximum. Differentiat-
ing (5) with respect to a , putting this derivative = 0, we have
[4A -=- cos 2 /?] [cos « cos («„ — /3) — sin a sin (a a — /S)] = ;
whence cos [>„ -f- (<*„ — $] = ; i.e., 2« — /3 = 90° ; or,
«„ = 45° -f- |/J, for a maximum range. By substitution this
maximum becomes known.
TAe velocity at any point of the path is v = Vv x * + 1)^ =
CURVILINEAR MOTION OP A MATERIAL POINT. 87
Vfdt = ff + 9*. (5)
The sfope, tan a, at any point = « s -f- 1>,. = (4£ ! — 9) -+■ 12t,
. d tan a 4f + 9
and •■■ ST = '-13T ( 6 )
The radius of curvature at any point (§ 74, eq. (6) ), sub-
stituting v x = 12t, also from (4) and (6), is
r = v
[V^] = > + V. ■ ■ TO
and the normal acceleration = v' -i-r (eq. (4), § 74), becomes
from (4) and (7) p n — 12 (ft. per square second), a constant.
Hence the centripetal or deviating force at any point, i.e., the
CURVILINEAR MOTION OF A MATERIAL POINT. 89
^N of the forces X and Y, is the same at all points, and =
Mv* ~r — 12M.
From equation (3) it is evident that the curve is symmetrical
ahout the axis X. Negative values of t and s would apply to
points on the dotted portion in Fig. 97, since the body may be
considered as having started at any point whatever, so long as
all the variables have their proper values for that point.
(Let the student determine how the conditions of this motion
could be approximated to experimentally.)
83. Relative and Absolute Velocities. — Fig. 98. Let M be a
material point having a uniform motion of velocity v % along a
straight groove cut in the deck of a steamer, which itself has
a uniform motion of translation, of velocity „ the angle of aberration SMT, Fig. 98, will not exceed
20 seconds of arc ; while it is zero when w and -y, are parallel.
Returning to the wind and sail-boat, it will be seen from
Fig. 98 that when v s = or even > w, it is still possible for v,
to be of such an amount and direction as to give, on a sail
properly placed, a small wind-pressure, having a small fore-and
aft component, which in the case of an ice-boat may exceed
the small fore-and-aft resistance of such a craft, and thus v l will
be still further increased ; i.e., an ice-boat may sometimes travel
faster than the wind which drives it. This has often been
proved experimentally on the Hudson River.
MOMENT OF INERTIA. 91
CHAPTER IV.
MOMENT OF INERTIA.
[Note. — For the propriety of this term and its use in Mechanics, see
§114; for the present we are only concerned with its geometrical nature.]
85. Plane Figures. — Jnst as in dealing with the centre of
gravity of a plane figure (§ 23), we had occasion to sum the
series fzdF, z being the distance of any element of area, dF,
from an axis ; so in subsequent chapters it will be necessary to
know the value of the series fz*dF for plane figures of various
shapes referred to various axes. This summation J'z'dF of
the products arising from multiplying each elementary area of
the figure by the square of its distance from an axis is called
the moment of inertia of the plane figure with respect to the
axis in question • its symbol will be I. If the axis is perpen-
dicular to the plane of the figure, it may be named the polar
mom. of inertia (§ 94) ; if the axis lies in the plane, the rec-
tangular mom. of inertia (§§ 90-93). Since the / of a plane
figure evidently consists of four dimensions of length, it may
always be resolved into two factors, thus /= Fk\ in which
F— total area of the figure, while h = VI '-f- F, is called the
radius of gyration, because if all the elements of area were
situated at the same radial distance, h, from the axis, the
moment of inertia would still be the same, viz.,
I = JVdF = hfdF = Fh\
86. Rigid Bodies. — Similarly, in dealing with the rotary
motion of a rigid body, we shall need the sum of the series
fp'dM, meaning the summation of the products arising from
multiplying the mass dM of each elementary volume dVof a
92 MECHANICS OF ENGINEERING.
rigid body by the square of its distance from a specified axis.
This will be called the moment of inertia of the hody with
respect to the particular axis mentioned (ofteii indicated by a
subscript), and will be denoted by /. As before, it can often
be conveniently written MJe 1 , in which M is the whole mass,
and k its "radius of gyration" for the axis used, k being
= Vl-i- M. If the body is homogeneous, the heaviness, y, of
all its particles will be the same, and we may write
l=f P >dM= ( r -=- g)fp'dV= ( r - g) Vk\
87. If the body is a homogeneous plate of an infinitely small
thickness = r, and of area = F, we have I = {y -=- g)fp'd V
= iy -T- g)rfp'dF; i.e., = {y -=- g) X thickness X mom. iner-
tia of the plane figure.
88. Two Parallel Axes. Reduction Formula.— Fia\ 99. Let
Z and Z be two parallel axes. Then I z
=fp 2 dM, and I z ,=fp"dM. But d being
the distances between the axes, so that a*
+ b 2 = d\ we have p"= (x - a)'+(y—by
=(x' + f) + d 3 - lax - 2by, and .-.
I z , =fp>dM-\-d*fdM- 2afxdM
-2bfydM. . (1)
But fp'dM = I z ,fdM= M, and from the
theory of tlie centre of gravity (see § 23, eq. (1), knowing that
dM = yd F-f- g, and .-. that [fyd V] -r- g= M ) we h&vefxdM
= Mx and fydM = My ; hence (1) becomes
l z , = I Z + Md' - 2ax~ - 2b~y), .... (2)
in which a and b are the x and y of the axis Z; x and y refer
to the centre of gravity of the body. If Z is a gravity-axis
(call it g), both x and y = 0, and (2) becomes
I z ,=I g + Md* or fe 2 =V+^- • • (3)
It is therefore evident that the mom. of inertia about a grav-
ity-axis is smaller than about any other parallel axis.
Eq. (3) includes the particular case of a plane figure, by
MOMENT OF INERTIA.
93
writing area instead of mass, i.e., when Z (now g) is a gravity-
axis,
I*=I g + Fd? (4 )
89. Other Reduction Formulae ; for Plane Figures.— (The axes
here mentioned lie in the plane of the figure.) For two sets
of rectangular axes, Laving the same origin, the following holds
good. Fig. 100. Since
Ix=ftfdF, and I r =fx*dF,
we have I x + I Y =f(a? _j_ y ^dF.
Similarly, I v + J r = yjy _|_ u y,ft
But since the x and y of any dF\\?iv& the same hypothennse as
the u and v, we have v* + it? = ai 2 + y'; . -. 4. -|_ / F - J p _j_ / r .
Y
Fig. 100.
Fio. 100a.
Z^ .JO = ibh\
1
dF= bdz .: I g =/z 1 dF= b /* s
Z'dz:
&bh\
Thirdly, about any other axis in its plane. Use the results
already obtained in connection with the reduction-formulae of
§§88, 89.
90a. The Triangle. — First, about an axis through the vertex
and parallel to the base ; i.e., I v , & v < ^_
in Fig. 103. Here the length
of the strip is variable ; call it y. ^
From similar triangles
y = (b h- h)z;
I v =.J?dF= fz'ydz = (b -^ fyj^z'dz
Fig. 104.
\bh\
Secondly, about g, a gravity-axis parallel to the base. Fig.
104. From § 88, eq. (4), we have, since F— %bh and
d = %h, I„ = I r - Fd> = $bh> - &h . ih' = &bh>.
MOMENT OF INEETIA.
95
Thirdly, Fig. 104, about the base ; I B = ? From § 88, eq.
(4), I B = I„-\- Fd\ with d=ih; hence
/* = ?\bh 3 + #A . %h* = ^bh\
91. The Circle.— About any diameter, as g, Fig. 105. Polar
co-ordinates, /„ = fz'dF. Here we take dF — area of an ele-
mentary rectangle = pdcp . dp, while z = p sin cp.
-ffe r
-i-V
Fio. 105.
Fiq. 106.
7 ff = / / (p sin cpfpdcpdp = I sin" h° - &A']. • • • (§ 90 )- • ■ C 1 )
./ifo«£ " fio. 109.
= x 1 -f- y% for each dF; hence
I v =f(x> + y >)dF=fx*dF+fy*dF = I T +I X .
7
98 MECHANICS OF ENGINEERING.
i.e., the polar moment of inertia about any given point m
the plane equals the sum of the rectangular moments of iner-
tia about any two axes of the plane figure, which intersect at
right angles in the given point. "We have therefore for the
circle about its centre
I v = l?tr l + \nr* = \nr" ;
For a ring of radii r t and r„
i, = t« - O ;
For the rectangle about its centre,
I p = &bh° + &hV = &bh(b> + A 2 ) ;
For the square, this reduces to
j p — s° •
(See §§90 and 91.)
95. Slender, Prismatic, Homogeneous Rod. — "Returning to the
moment of inertia of rigid bodies, or solids, we begin with that
of a material line, as it might be called, about
tfyjis ''- an axis through its extremity making some an-
: sj/s* ,/'* gle ol with the rod. Let I = length of the rod,
/y x \X' F\\& cross-section (very small, the result being
t ..--'' strictly true only when F = 0). Subdivide
fig. no. the rod into an infinite number of small prisms,
each having i^as a base, and an altitude = ds. Let y = the
heaviness of the material ; then the mass of an elementary
prism, or dM, = (y -r- g)Fds, while its distance from the axis
Z is p = s sin a. Hence the moment of inertia of the rod
with respect to Z as an axis is
I z =fp>dM = {y -T- g)F sin' a J s'ds = $(y -f- g)FV sin' a.
But yFl -=- g = mass of rod and I sin a = a, the distance of
the further extremity from the axis ; hence I z = \Ma? and
the radius of gyration, or Jc, is found bj writing \Mo?= Mk' ;
.: ¥ = %a% or Tc = V%a (see § 86). If a = 90°, a — I.
96. Thin Plates. Axis in the Plate.— Let the plates be homo-
geneous and of small constant thickness = r. If the surface of
MOMENT OF INEETIA. 99
the plate be = F, and its heaviness y, then its mass = yFr -~ g.
From § 87 we have for the plate, about any axis,
I — (v -*- 9) r X mom. of inertia of the plane figure formed by
the shape of the plate (1)
Rectangular plate. Gravity-axis parallel to base. — Dimen-
sions b and h. From eq. (1) and § 90 we have
i a ={y - g)r ■ tW' 8 = (r^r + g) r \w= r \Mh% .-. v = ^h\
Similarly, if the base is the axis, I B = %Mh% .-. ¥ = $¥.
Triangular plate. Axis through vertex parallel to base. —
From eq. (1) and § 90a, dimensions being b and h,
Iv = (y^r g)r\W = {j\bhr -=- g)\¥ = \M¥; .: ¥ =± \h\
Circular plate, with any diameter as axis. — From eq. (1)
and § 91 we have
Ig = (y "=- 9) r \ nrK = (yxr'r ~ g)^ = \Mr*\ :. ¥ = %r\
97. Plates or Right Prisms of any Thickness (or Altitude).
Axis Perpendicular to Surface (or Base). — As before, the solid is
homogeneous, i.e., of constant heaviness y^ z
let the altitude = h. Consider an elementary
prism, Fig. Ill, whose length is parallel to the
axis of reference Z. Its altitude = h = that
of the whole solid ; its base = dF = an element
of F \\\q area of the base of solid; and each
point of it lias the same p. Hence we may fio. hi.
take its mass, = yhdF-t- g, as the dM in summing the series
l z =f f MM;
.:I z =(yh+g)f P 'dF
= (yh -=- g) X polar mom. of inertia of base. . . (2)
By the use of eq. (2) and the results in § 94 we obtain the
following:
Circular plate, or right circular cylinder, about the geo-
metrical axis, r = radius, h = altitude.
I g = {yh -r- g)knr l = (yhnr* ~- g)\r" = \Mr>; .: ¥ = *r\
Right parallelopiped or rectangular plate. — Fig. 112,
I g = (yh ~ g)^HK + V) = *W', .: ¥ = -&?,
100
MECHANICS OP ENGINEERING.
For a hollow cylinder, about its geometric axis,
Fig 112.
Fig. 113.
98. Circular Wire. — Fig. 113 (perspective). Let Z be a
gravity-axis perpendicular to the plane of the wire ; X and Y
lie in this plane, intersecting at right angles in the centre 0.
The wire is homogeneous and of constant (small) cross-section.
Since, referred to Z, each dM has the same p = r, we have
I z =fr'dM= Mr\ Now I x must equal I T , and (§ 94) their
sum = I z ,
.; I x , or I T , = \Mr*, and k x , or ky = %r*.
99. Homogeneous Solid Cylinder, about a diameter of its base.
— Fig. 114. I x = ? Divide the cylinder into an infinite num-
ber of laminae, or thin plates, parallel to the
base. Each is some distance z from X, of
thickness dz, and of radius r (constant). In
each draw a gravity-axis (of its own) parallel to
fig. 114. X. We may now obtain the I x of the whole
cylinder by adding the I x s of all the laminae. The I„ of
one lamina (§ 96, circular plate) = its mass X ^ ',
I x (eq. (3), § 88) = its I g -f (its mass) X z\ Hence for the
whole cylinder
/x = f [(rdznr* -=- g)(& + s')]
g ^ any
hence its
i.e..
I x = (nr'hy + g){^ + ^) = M& + &>)■
100. Let the student prove (1) that if Fig. 114 represent
any right prism, and lc F denote the radius of gyration of any
one lamina, referred to its gravitj'-axis parallel to X, then the
I x of whole prism = M(k F 2 -f £A 3 ) ; and (2) that the moment
MOMENT OF INERTIA.
101
of inertia of the cylinder about a gravity-axis parallel to the
base is = M(%r° + -^h').
101. Homogeneous Right Cone. — Fig. 115. First, about an
axis V, through the vertex and parallel to the base. As before,
divide into laminae parallel to the base. Each is a
circular thin plate, but its radius, x, is not = r, but, *\
from proportion, is x = (r -=- h)z.
The I oi any lamina referred to its own gravity-
axis parallel to I 7 " is (§96) = (its mass) X ■& and (G, ~ gW^ + r,'),
respectively (§ 97), while if the arms are
not very thick compared with their length, we have for them
(§§ 95 and 88)
4 ( an axis perpen-
dicular to paper through 0) and we have — Ra = —fdMpz
= —pfdMz = — pMz (§88), i.e., a = ~z~. A similar result
may be proved as regards y. Hence, if a rigid "body has a
motion of translation, the resultant force must act in a line
through the centre of gravity (here more properly called the
centre of mass), and parallel to the direction of motion. Or,
practically, in dealing with a rigid body having a motion of
translation, we may consider it concentrated at its centre of
mass. If the velocity of translation is uniform, R — Jf X
= 0, i.e., the forces are balanced.
DYNAMICS OF A RIGID BODY. 107
110. Rotation about a Fixed Axis. — First, as to the elements
of space and time involved. Fig. 123. Let be the axis of
rotation (perpendicular to paper), OY a fixed -> ^ (J
line of reference, and OA a convenient line of r^ ~/v\
the rotating body, passing through the axis and / X. a )
perpendicular to it, accompanying the body in ) o — — T~
its angular motion, which is the same as that of V^_^ • —
OA. Just as in linear motion we dealt with FlG - m
linear space (s), linear velocity (v), and linear acceleration (p),
so here we distinguish at any instant ;
a, the angular space between OY and OA ;
co = j , the angular velocity, or rate at which a is changing ;
and
6 = -j- = --^, the angular acceleration, or rate at which co
is changing.
These are all reckoned in 7r-measure and may be -f- or — ,
according to their direction against or with the hands of a
watch.
(Let the student interpret the following cases: (1) at a cer-
tain instant oo is +, and 6 — ; (2) co is — , and 6 +; (3) a is
— , co and 6 both +; (4) a-\-,a> and 6 both — .) For rotary
motion we have therefore, in general,
dot ._„, a dco d
co =
dCO a a ,-c-r-r-r -
dt> ^ Y x -> dt ~ dt
and •"• ™dco — Oda ; (VIII.)
corresponding to eqs. (I.), (IL), and (III.) in § 50, for rectilinear
motion.
Hence, for uniform rotary motion, co being constant and
6 = 0, we have a = cot, t being reckoned from the instant
when a = 0.
For uniformly accelerated rotary motion 6 is constant, and
108 MECHANICS OF ENGINEERING.
if gj„ denote the initial angular velocity (when a and t = 0),
we may derive, precisely as in §56,
w = go, + et ; . . (i) « = gv! + i^ 3 ; • • (2)
« = C °°~ d a>i ' ] . . (3) and « = *(«. + «)* . . (4)
If iu any problem in rotary motion Q, go, and a have been
determined for any instant, the corresponding linear values for
any point of the body whose radial distance from the axis is p,
will be s = ap (= distance described by the point measured
along its circular path from its initial position), v = cop = its
velocity, and^? t = 6p its tangential acceleration, at the instant
in question.
Examples. — (1) "What value of go, the angular velocity, is
implied in the statement that a pulley is revolving at the rate
of 100 revolutions per minute ?
100 revolutions per minute is at the rate of 2?r X 100
= 628.32 (^-measure units) of angular space per minute
= 10.472 per second ; .-. go = 628.32 per minute or 10.472
per second.
(2) A grindstone whose initial speed of rotation is 90 revo-
lutions per minute is brought to rest in 30 seconds, the an-
gular retardation (or negative angular acceleration) being con-
stant ; required the angular acceleration, 8, and the angular
space a described. Use the second as unit of time.
go, = 2tt| £ = 9.4248 per second ; .-. from eq. (1)
GO — GO
= ~ = ~ 9A2i "*" 30 = — °- 3141 (""-measure units)
per " square second." The angular space, from eq. (2) is
a = a>.t + ±df = 30 X 9.42 - £(0.314)900 = 141.3
(^-measure units), i.e., the stone has made 22.4 revolutions in
coming to rest and a point 2 ft. from the axis has described a
distance s = ap - 141.3 X 2 = 282.6 ft. in its circular path.
111. Rotation. Preliminary Problem. Axis Fixed.— For
clearness in subsequent matter we now consider the following
DYNAMICS OF A RIGID BODY.
109
FlQ. 124.
simple case. Fig. 124 shows a rigid body, consisting of a
drum, an axle, a projecting arm, all
of which are imponderable, and a
single material point, whose weight
is G and mass M. An imponderable
flexible cord, in which the tension is
kept constant and = P, unwinds
from the drum. The axle coincides
with the vertical axis Z, while the cord
is always parallel to Y. Initially (i.e., when t = 0) M lies at
rest in the plane ZY. Required its position at the end of any
time t (i.e., at any instant) and also the reactions of the bearings
at and O x , supposing no vertical pressure to exist at O x , and
that P and M are at the same level. No friction. At any in-
stant the eight unknowns, a, oa, 6, X„ Y„, Z„ X„ and Y„ may
be found from the six equations formed by putting 2X, etc.,
of the system of forces in Fig. 124, equal, respectively, to the
2X, etc., of the imaginary equivalent system in Fig. 125, and
two others to be mentioned subsequently. Since, at this in-
stant, the velocity of M must be v = cop and its tangential ac-
celeration p t = dp, its circular motion
could be produced, considering it free (eq.
(5), §74), by a tangential force T — mass
X Pt = Mdp, and a normal centripetal
force N=Mv* -=- p=M(wpf -=- p=w>Mp.
Hence the system in Fig. 125 is equivalent
to that of Fig. 124, and from putting the 2 (mom.) z of one
= that of the other, we derive
Pa=Tp; i.e., Pa = 6Mft, .... (1)
whence 6 becomes known, and is evidently constant, since P,
a, M, and p are such. .\ the angular motion is uniformly ac-
celerated, and from eqs. (1) and (2), § 110, go and a become
known -,
i.e., oo=6t, . . . (2) and a = $dt' (3)
Putting (2Z of 124) = (2Z of 125), gives
Z.-O = 0) i.e., Z = G (4)
Fig. 125.
110 MECHANICS OF ENGINEERING.
Proceeding similarly with the .SXof each system,
X + X, = Tcos a — Xsin a — dMp cos a — ofMpsm a, (5)
and with the 2 Y of each,
P-\-Y e -\-Y l = — Tsm a — Xcos a = — dMp sin a
— oo'Mp cos a ; (6)
while with the 2 (mom.) x we have, conceiving all the forces- in
each system projected on the plane ZY(see §38), and noting
that y = p cos a and x = p sin a,
-f Op cos a + Yl + Pb = — (0Jf/> sin a)h—(ot?Mp cosa)b,(7)
and with the 2 (mom.) r ,
— Gp sin a — X,Z = — (dMp cos a)J + (<^Mp sin a)J. . (8)
From (7) we may find T",; from (8), X,; then X and Z,
from (5) and (6). It will be noted that as the motion proceeds
8 remains constant ; oo increases with the time, a with the
square of the time ; Z, is constant, = G ; while X„, Y„, X„
and Y 1 have variable values dependent on p cos a and p sin ar,
i.e., on the co-ordinates y and a? of the moving material point.
112. Particular Supposition in the Preceding Problem with
Numerical Substitution. — Suppose we have given (using the
foot-pound-second system of units in which g = 32.2) G = 64.4
lbs., whence
M= (G + g) = 2; P = 4 lbs., I = 4ft., b = 2 ft, a = 2ft.,
and p = 4 ft.; and that J/ is just passing through the plane
ZX, i.e., that a — \n. "We obtain, first, the angular accelera-
tion, eq. (1),
d = Pa~ Mp" = 8 -f- 32 = 0.25 = \.
From eqs. (2) and (3) we have at the instant mentioned (not-
ing that when a was = 0, t was = 0)
go' = 2«0 = \n = 0.7854 -+-,
while (2) gives, for the time of describing the quadrant,
t = go — d = 3.544. . . . seconds.
Since at this instant cos a = and sin a = 1, we have, from
(7),
+ 0+ r,x4 + 4x2^ - JX2X4X2; .-. T 1= -31bs.
DYNAMICS OF A RIGID BODY.
Ill
The minus sign shows it should point in a direction contrary
to that in which it is drawn in Fig. 124. Eq. (8) gives
- 64.4 x i-X 1 X 4=-0+i7T X 2x4x2 ;.-.X l = - 67.54 lbs.
And similarly, knowing Yi and X x , we have from (5) and (6),
X„ = + 61.26 lbs., and F = - 3.00 lbs.
The resultant of X, and Y xl also that of X„ Y„ and Z„ can
now be found by the parallelogram (and pafallelopipedon) of
forces, both in amount and position, noting carefully the direc-
tions of the components. These resultants are the actions of
the supports upon the ends of the axle ; their equals and
opposites would be the actions or pressures of the axle against
the supports, at the instant considered (when M is passing
through the plane ZX; i.e., with a = \tz). (At the same in-
stant, suppose the string to break; what would be the effect on
the eight quantities mentioned?)
113. Centre of Percussion of a Rod suspended from one End. —
Fig. 126. The rod is initially at rest (see (I.) in figure), is straight,
homogeneous, and of constant
(small) cross-section. Neglect its
weight. A horizontal force or
pressure, P, due to a blow (and
varying in amount during the
blow), now acts upon it from the
left, perpendicularly to the axis,
Z, of suspension. An accelerated
rotary motion begins about the fixed axis Z.
free, at a certain instant, with the reactions X and Y put in
at O . (III.) shows an imaginary system which would produce
the same effect at this instant, and consisting of a (IT = dMdp,
and a dM= 6I Z ,
where I z is the moment of inertia of the rod about Z, and from
§95 = \Ml\ Now p-il; hence, finally,
*■.=*[>-!• r]-
Jf now Y t is to = 0, i.e., if there is to be no shock between
the rod and axis, we need only apply P at a point whose dis-
tance a = f I from the axis ; for then Y = 0. This point is
called the centre of percussion for the given rod and axis. It
and the point of suspension are interchangeable (see § 118).
(Lay a pencil on a table; tap it at a point distant one third of
the length from one end ; it will begin to rotate about a vertical
axis through the farther end. Tap it at one end ; it will begin
to rotate about a vertical axis through the point first mentioned.
Such an axis of rotation is called an axis of instantaneous rota-
tion, and is different for each point of impact — just as the
point of contact of a wheel and rail is the one point of the
wheel which is momentarily at rest, and about which, therefore,
all the others are turning for the instant. Tap the pencil at
its centre of gravity, and a motion of translation begins; see
§109.)
114. Rotation. Axis Fixed. General Formulae. — Consider-
Fio. 127. Fig. 128.
ing now a rigid body of any shape whatever, let Fig. 127 indi-
cate the system of forces acting at any given instant, Z being
DYNAMICS OF A RIGID BODY. 113
the fixed axis of rotation, go and 6 the angular velocity and
angular acceleration, at the given instant. X and Y are two
axes, at right angles to each other and to Z, fixed in space. At
this instant each dM oi the body has a definite x, y, and cp
(see Fig. 128), which will change, and also a p, and z, which will
not change, as the motion progresses, and is pursuing a circu-
lar path with a velocity = cap and a tangential acceleration
= dp. Hence, if to each dM of the body (see Fig. 128) we
imagine a tangential force dT =■ dMdp and a normal force
.— dM(aop)' -fp= ao'dMp to be applied (eq. (5), §74), and
these alone, we have a system comprising an infinite number of
forces, all parallel to XY, and equivalent to the actual system
in Fig. 127. Let 2X, etc., represent the sums (six) for Fig.
127, whatever they may be in any particular case, while for
128 we shall write the corresponding sums in detail. Noting
that
fdW cos cp = cofdMp cos cp = wfdMy = = da = 7.68 feet per square second for the uniform-
ly accelerated translation.
Nothing has yet been said of the velocities and initial condi-
tions of the motions ; for what we have derived so far applies
to any point of time. Suppose, then, that the angular velocity
co = zero when the time, t = ; and correspondingly the ve-
locity, v — ooa, of translation- of _P,, be also = when t = 0.
At the end of any time t, co — 6t (§§ 56 and 110) and v =pt
= dot ; also the angular space, a = $0f, described by the par-
allel opiped during the time t, and the linear space s = \pt
= \datf, through which the weight J 3 l has sunk vertically.
For example, during the first second the parallelopiped has ro-
tated through an angle a = \dt = iX 15.36x1 = 7.68 units,
?r-measure, i.e., (7.68 -e- 27r) = 1.22 revolutions, while P 1 has
sunk through * = \6a£ = 3.84 ft., vertically.
The tension in the cord, from (2), is
P = 48(1 - 15.36 X i H- g) = 48(1 - 0.24) = 36.48 lbs.
The pressures at the bearings will be as follows, at any in-
stant : from (4) and (8), X t and X, must individually be zero ;
from (6) Z,= , and a, in angular motion
correspond precisely to those (p, v, and s) of rectilinear motion,
it is evident that the present is a case of harmonic motion,
already discussed in Problem 2 of § 59. Applying the results
there obtained, since B of eq. (1) corresponds to the a of that
problem, we find that the oscillations are isochronal, i.e., their
118 MECHANICS OF ENGINEERING.
durations are the same whatever the amplitude (provided the
elasticity of the rod is not impaired), and that the duration of
one oscillation (from one extreme position to the other) is
t' = n -f- VJ?, or finally,
t'=n VaJ z -T- QA- (2)
117. The Compound Pendulum is any rigid body allowed to
oscillate without friction under the action of gravity when
mounted on a horizontal axis. Fig. 131 shows the
body free, in any position during the progress of
the oscillation. Cis the centre of gravity; let OG
= s. From (XIV.), § 114, we have 2 (mom. about
fixed axis)
= angul. ace. X mom. of inertia.
.-. — Gs sin a = 9I„
and 6 = — Gs sin a -^ /„ = — Mgs sin a -±- Mk\,
i.e., 6 = — gs sin a -;- k* (1)
Hence 9 is variable, proportional to sin a. Let us see what
the length I = OK, of a simple circular pendulum, must be, to
have at this instant (i.e., for this value of «) the same angular
acceleration as the rigid body. The linear (tangential) accelera-
tions of K, the extremity of the required simple pendulum
would be (§ Y7)^>( = g sin a, and hence its angular acceleration
would = g sin a-^-l. Writing this equal to d in eq. (1), we
obtain
l = K + s (2)
But this is independent of a ; therefore the length of the sim-
ple pendulum having an angular acceleration equal to that of
the oscillating body is the same in all positions of the latter,
and if the two begin to oscillate simultaneously from a position
of rest at any given angle a 1 with the vertical, they will keep
abreast of each other during the whole motion, and hence have
DYNAMICS OF A KIGID BODY.
119
the same duration of oscillation ; which is .•. , for small ampli-
tudes (§ 78),
t'=n VT^Tg = n Vk?Tjp, .... (3)
iST is called the centre of oscillation corresponding to the given
centre of suspension 0, and is identical with the centre of per-
cussion (§113).
Example.— Required the time of oscillation of a cast-iron
cylinder, whose diameter is 2 in. and length 10 in., if the axis
of suspension is taken 4 in. above its centre. If we use 32.2
for g, all linear dimensions should be in feet and times in
seconds. From § 100, we have
From eq. (3), § 88,
m-
i
ITT
+ tV-^) :
^T*T'
Xk
I*/-
Jf[ T h--W + i] = -«'X 0.170;
.-. k; — 0.170 sq. ft.; .-. t'= it V0.170 -=- 32.2 X i = 0.395 sec.
118. The Centres of Oscillation and Suspension are Inter-
changeable. — (Strictly speaking, these centres are points in the
line through the centre of gravity perpendicular to the axis of
suspension.) Eefer the centre of oscillation K to the centre
of gravity, thus (Fig. 132, at (I.)) :
= l-s =
Ms
Mh a *-\-Ms* A '' ...
Ms S = T- (1)
Now invert the body and suspend it at K;
required CK X , or «„, to find the centre of F
oscillation corresponding to K as centre of j"
suspension. By analogy from (1) we have «,;
s 2 = he -*- s l ; but from (1), h c ' -=-«, = * .:
s 2 = s ; in other words, K r is identical with (i.) (n.)
O. Hence the proposition is proved. Fia m
Advantage may be taken of this to determine the length Z
of the theoretical simple pendulum vibrating seconds, and thus
finally the acceleration of gravity from formula (3), § 117, viz.,
120 MECHANICS OF ENGINEERING.
when f = 1.0 and I (now = L) has been determined experi-
mentally, we have
g (in ft. per sq. second) = L (in ft.) X n*. . . (2)
This most accurate method of determining g at any locality
requires the use of a bar of metal, furnished with a sliding
weight for shifting the centre of gravity, and with two project-
ing blocks provided with knife-edges. These blocks can also
be shifted and clamped. By suspending the bar by one knife-
edge on a proper support, the duration of an oscillation is com-
puted by counting the total number in as long a period of
time as possible; it is then reversed and suspended on the
other with like observations. By shifting the blocks between
successive experiments, the duration of the oscillation in one
position is made the same as in the other, i.e., the distance be-
tween the knife-edges is the length, I, of the simple pendulum
vibrating in the computed time (if the knife-edges are not equi-
distant from the centre of gravity), and is carefully measured.
The I and t' of eq. (3), § 117, being thus known, g may be com-
puted. Professor Bartlett gives as the length of the simple
pendulum vibrating seconds at any latitude /3
L (in feet) = 3.26058 — 0.008318 cos 2/3.
119. Isochronal Axes of Suspension. — In any compound
pendulum, for any axis of suspension, there are always three
others, parallel to it in the same gravity plane, for which the
oscillations are made in the same time as for the first. For
any assigned time of oscillation t', eq. (3), § 117, compute the
corresponding distance CO = s of O from C;
i.e., from t = jtSt— = — - — i~ ,
Mgs Mgs '
we have *= (gt»-i-%n*)± ^V^-M:* 4 ) — jfe/. . . (1)
Hence for a given f, there are two positions for the axis O
parallel to any axis through C, in any gravity-plane, on both
sides; i.e., four parallel axes of suspension, in any gravity-
plane, giving equal times of vibration ; for two of these axes
DYNAMICS OF A RIGID BODY.
121
we must reverse the body. E.g., if a slender, homogeneous,
prismatic rod be marked off into thirds, the (small) vibrations
will be of the same duration, if the centre of suspension is
taken at either extremity, or at either point of division.
Example. — Required the positions of the axes of suspension,
parallel to the base, of a right cone of brass, whose altitude is
six inches, radius of base, 1.20 inches, and weight per cubic inch
is 0.304 lbs., so that the time of oscillation may be a half-
second. (N.B. For variety, use the inch-pound-second system
of units, first consulting § 51.)
120. The Fly-Wheel in Fig. 133 at any instant experiences
a pressure P' against its crank-pin from the connecting-rod
and a resisting pressure P" from the teeth of a spur-wheel with
Fig 133.
which it gears
Its weight G acts through C (nearly), and
there are pressures at the bearings, but these latter and G have
no moments about the axis C (perpendicular to paper). The
fio-nre shows it free, P" being assumed constant (in practice
this depends on the resistances met by the machines which D
drives, and the fluctuation of velocity of their moving parts).
P', and therefore T its tangential component, are variable,
depending on the effective steam-pressure on the piston at any
instant, on the obliquity of the connecting-rod, and in high-
speed engines on the masses and motions of the piston and con-
necting-rod. Let r = radius of crank-pin circle, and a the
perpendicular from on P". From eq. (XIV.), § 114, we
have
Tr - P"a = 6I CX .-. 6 = (Tr- P"a) H- J a ■ (1
122 MECHANICS OF ENGINEERING.
as the angular acceleration at any instant ; substituting which in
the general equation (VIII.), § 110, we obtain
Iccodoo = Trda — P"ada (2)
From (1) it is evident that if at any position of the crank-pin
the variable Tr is equal to the constant P"a, 6 is zero, and
consequently the angular velocity co is either a maximum or a
minimum. Suppose this is known to be the case both at to
and n; i.e., suppose T, which was zero at the dead-point A,
has been gradually increasing, till at n, Tr = P"a ; and there-
after increases still further, then begins to diminish, until nt to
Tr again = P"a, and continues to diminish toward the dead-
point B. The angular velocity co, whatever it may have been
on passing the dead-point A, diminishes, since 6 is negative,
from A to n, where it is co n , a minimum ; increases from n to
m, where it reaches a maximum value, co m . n and to being-
known points, and supposing co n known, let us inquire what
oo m will be. From eq. (2) we have
IcJ G3da>=j n Trda — P" J ada. . . (3)
But rda — ds' = an element of the path of the crank-pin, and
also the " virtual velocity" of the force T, and ada = ds", an
element of the path of a point in the pitch-circle of the fly-
wheel, the small space through which P" is overcome in dt.
Hence (3) becomes
Ic\{^ - O =J n Tds - P" X linear arc EF. (4)
To determine / Tds we might, by a knowledge of the vary-
ing steam-pressure, the varying obliquity of the connecting-rod,
etc., determine T for a number of points equally spaced along
the curve nm, and obtain an approximate value of this sum by
Simpson's Rule ; but a simpler method is possible by noting
(see eq. (1), § 65) that each term Tds of this sum = the corre-
sponding term Pdx in the series / Pdx, in which P = the
DYNAMICS OP A RIGID BODY. 123
effective steam-pressure on the piston in the cylinder at any in-
stant, dx the small distance described by the piston while the
crank-pin describes any ds, and n' and ml the positions of the
piston (or of cross-head, as in Fig. 133) when the crank-pin is
at n and m respectively. (4) may now be written
Pdx - P" X linear arc EF, (5)
from which a> m may be found as proposed. More generally, it
is available, alone (or with other equations), to determine any
one (or more, according to the number of equations) unknown
quantity. This problem, in rotary motion, is analogous to that
in § 59 (Prob. 4) for rectilinear motion. Friction and the in-
ertia of piston and connecting-rod have been neglected. As
to the time of describing the arc nm, from equations similar to
(5), we may determine values of oo for points along nm, divid-
ing it into an even number of equal parts, calling them t»„ oo„
etc., and then employ Simpson's Rule for an approximate value
I - ™ 1 p m da
of the sum \ t= I — (from eq. (VI.), § 110) ; e.g., with
I . f| t/ ti CO
four parts, we would have
c
1
t = ^ (angle w Cm, tineas.)
Lg?„ oo ' co ' a>. w m -\
.(6)
121. Numerical Example. Fly-Wheel.— (See Fig. 133 and
the equations of § 120.) Suppose the engine is non-condensing
and non-expansive (i.e., that P is constant), and that
P= 5500 lbs., r = 6in. =fft., a = 2ft.,
and also that the wheel is to make 120 revolutions per minute,
i.e., that its mean angular velocity is to be
oo' = Ytj 1 X 2-7T, i.e., go' = 4?r.
First, required the amount of the resistance P" (constant)
that there shall be no permanent change of speed, i.e., that the
ano-nlar velocity shall have the same value at the end of a com-
plete revolution as at the beginning. Since an equation of the
form of eq. (5) holds good for any range of the motion, let
124 MECHANICS OP ENGINEERING.
that range be a complete revolution, and we shall have zero as
the left-hand member ; /Pdx - P X 2 f t. = 5500 lbs. X 2 ft.,
or 11,000 foot-pounds (as it may be called); while P" is un-
known, and instead of lin. arc EF we have a whole circumfer-
ence of 2 ft. radius, i.e., 4^ ft.;
.-. = 11,000 - P" X 4 X 3.1416; whence P" = 875 lbs.
Secondly, required the proper mass to be given to the fly-
wheel of 2 ft. radius that in the forward stroke (i.e., while the
crank-pin is describing its upper semicircle) the max. angular
velocity w m shall exceed the minimum oo n by only-j^a/, assum-
ing (which is nearly true) that i(oo m -)- oo n ) = oo . There be-
ing now three unknowns, we require three equations, which
are, including eq. (5) of § 120, viz.:
Mk c *i(Go m + oo n ){oo m - oa n )
/>m'
= / Pdx - P" X linear arc EF; (5)
i(^m+ ».)= «/= 4* ; (7) and eo m -co n = T \m' = f jr. (8)
Tlie points n and m are found most easily and with sufficient
accuracy by a graphic process. Laying off the dimensions to
scale, by trial such positions of the crank- pin are found that
T, the tangential component of the thrust P' produced in the
connecting-rod by the steam-pressure P (which may be resolved
into two components, along the connecting-rod and a normal
to itself) is —(a -j- r)P", i.e., is = 3500 lbs. These points will
be n and m (and two others on the lower semicircle). The
positions of the piston nf and m', corresponding to n and m of
the crank-pin, are also found gi'aphically in an obvious manner.
"We thus determine the angle nCm to be 100°, so that linear
arc EF— Iflftn X 2 ft. = i$-n, while
/ Pdx = 5500 lbs. X / dx = 5500 Xn'm'= 5500 X 0.77 ft.,
On' e/n'
n'm' being scaled from the draft.
Now substitute from (7) and (8) in (5), and we have, with
h c = 2 ft. (which assumes that the mass of the fly-wheel is con-
centrated in the rim),
DYNAMICS OF A KIGID BODY.
125
(# -7- g) X 4 X 4tt x |» = 5500 X 0.77 - 875 X *+*,
which being solved for G (with ^ = 32.2 ; since we Lave used
the foot and second), gives G = 600.7 lbs.
The points of max. and min. angular velocity on the back-
stroke may be found similarly, and their values for the fly-
wheel as now determined ; they will differ but slightly from
the oa m and oo n of the forward stroke. Professor Cotterill says
that the rim of a fly-wheel should never have a max. velocity
> 80 ft. per sec; and that if made in segments, not more than
40 to 50 feet per second. In the present example we have for
the forward stroke, from eqs. (7) and (8), w m = 13.2 (^-measure
units) per second ; i.e., the corresponding velocity of the wheel-
rim is v m — co m a — 26.4 feet per second.
122. Angular Velocity Constant. Fixed Axis.— If go is con-
stant, the angular acceleration, 6, must be = zero at all times,
which requires 2 (mom.) about the axis of
rotation to be = (eq. (XIV.), § 114). An
instance of this occurs when the only forces
acting are the reactions at the bearings on
the axis, and the body's weight, parallel to
or intersecting the axis ; the values of these
reactions are now to be determined for dif- /'
ferent forms of bodies, in various positions fiq. 134.
relatively to the axis. (The opposites and equals of these reac-
tions, i.e., the forces with which the axis acts upon the bearings,
are sometimes stated to be due to the " centrifugal forces" or
"centrifugal action," of the revolving body.)
Take the axis of rotation for Z, then, with = 0, the equa-
tions of § 114 reduce to
2J= — cd'Mx; .
27= --afMy; .
2Z= ; . . .
2 moms. x = — oofdMyz ;
2 moms. j- = -j- oofdMxz ;
2 moms.# = 0. . . .
(IXa.)
(Xa.)
(XIa.)
(XITa.)
(Xllla.)
(XlVa.)
126 MECHANICS OP ENGINEERING.
For greater convenience, let us suppose the axes X and Y
(since their position is arbitrary so long as they are perpen-
dicular to each other and to Z) to revolve with the tody in its
uniform rotation.
122a. If a homogeneous body have a plane of symmetry
and rotate uniformly about any axis Z perpendicular to that
plane {intersecting it at 0), then the acting forces are equiva-
___ lent to a single force, — oa'Mp, applied
" p at and acting in a gravity-line, but
p 0^ J-—-"' j^> directed away from, the centre of
/^^3j^'22§i_ gravity. It is evident that such a
y - Nj^--^' * force P = oa'Mp, applied as stated
FIG - m (see Fig. 135), will satisfy all six con-
ditions expressed in the foregoing equations, taking X through
the centre of gravity, so that x = p. For, from (IXa.), P must
= Go*Mp, while in each of the other summations the left-
hand member will be zero, since P lies in the axis of X; and
as their right-hand members will also be zero for the present
body (y = ; and each of the sums fdMyz smdfdMxz is zero,
since for each term dMy{ -\- z) there is another dMy( — z)
to cancel it ; and similarly, f 'or fdMxz), they also are satisfied ;
Q. E. D. Hence a single point of support at will suffice to
maintain the uniform motion of the body, and the pressure
against it will be equal and opposite to P.
First Example. — Fig. 136. Supposing (for greater safety)
that the uniform rotation of 210 revolutions
per minute of each segment of a fly-wheel is ^--""
maintained solely by the tension in the cor- JL ^'
responding arm, P ; required the value of P * f
if the segment and arm together weigh -fa of
a ton, and the distance of their centre of FlG - 1S6 -
gravity from the axis is p = 20 in., i.e., = -| ft. With the foot-
ton-second system of units, with g = 32.2, we have
P = oo'Mp = [^>- X ZttJ X [^V -H 32.2] X 1 = 0.83 tons,
or 1660 lbs.
DYNAMICS OP A ltlGID BODY.
127
Second Example. — Fig. 137. Suppose the uniform rotation
of the same fly-wheel depends solely on the tension in the rim,
required its amount. The figure shows the half-
rim free, with the two equal tensions, P', put in at
the surfaces exposed. Here it is assumed that the
arms exert no tension on the rim. From § 122a we
have 2P' = a?Mp, where M is the mass of the half- P '
rim, and p its gravity co-ordinate, which may be ob- fig. 137.
tained approximately by § 26, Problem 1, considering the rim
as a circular wire, viz., p = 2r -=- n.
Let M= (180 lbs.) -=- g, with r = 2 ft. We have then
P' = -J(22) 2 (180 -f- 32.2)(4 -=- it) = 1718.0 lbs.
(In realitj' neither the arms nor the rim sustain the tensions
just computed ; in treating the arms we have supposed no duty
done by the rim, and vice versa. The actual stresses are less,
and depend on the yielding of the parts. Then, too, we have
supposed the wheel to take no part in the transmission of mo-
tion by belting or gearing, which would cause a bending of the
arms, and have neglected its weight.)
122b. If a homogeneous body have a line of symmetry and
rotate uniformly about an axis parallel to it (0 being the foot
of the perpendicular from the centre of gravity on the axis),
t/ien the acting forces are equivalent to a single force P
= oo'Mp, applied at and acting in a gravity-line away
from the centre of gravity.
Taking the axis X through the
g,M centre of gravity, Z being the
"~ v; "l axis of rotation, Fig. 138, while
w ] Z' is the line of symmetry, pass
an auxiliary plane Z' Y' parallel
to ZT. Then the sum fdMxz
may be written fdM(p + x')z
which = ~pfdMz + fdMx'z 1
But fdMz = Mz = 0, since z
Fig. 138.
= 0, and every term dM(-\- x')z is cancelled by a numerically
128
MECHANICS OF ENGINEERING.
equal term dM{— x')z of opposite sign. Hence fdMxz = 0.
Also fdMyz = 0, since eacli positive product is annulled by an
equal negative one (from symmetry about Z'). Since, also,
y = 0, all six conditions in § 122 are satisfied. Q. E. D.
If the homogeneous body is any solid of revolution whose
geometrical axis is parallel to the axis of rotation, the forego-
ing is directly applicable.
122c. If a homogeneous body revolve uniformly about any
axis lying in a plane of symmetry, the acting forces are equiv-
alent to a single force P = oo'Mp, acting parallel to the grav-
ity-line which is perpendicular to the axis (Z), and away
from the centre of gravity, its distance from any origin in
the axis Z being = [fdMxz'] -e- Mp {the plane ZX being a
gravity-plane). — Fig. 139. From the position of the body we
have p = x, and y = ; hence if a
value co'Mp be given to P and it be
made to act through Z and parallel to
X, and away from the centre of gravity,
all the conditions of § 122 are satisfied
except (Xlla.) and (XHIa.). But
symmetry about the plane XZ makes
fdMyz = 0, and satisfies (X1I«.), and
Mp from along Z
Fig. 139.
by placing P at a distance a = fdMxz
we satisfy (XHIa.). Q. E. D.
Example. — A slender, homogeneous, prismatic rod, of length
= I, is to have a uniform motion, about a ver- q
tical axis passing through one extremity,
maintained by a cord-connection with a fixed
point in this axis. Fig. 140. Given co, cp, I,
(p = \l cos ~ a')-].
(If the body slid without friction,^ would = gsin /?.) Hence
for a cylinder (§ 97), k z * being = ^a', we havep = \g sin /?;
and for a sphere (§ 103) p — %-g sin /J.
(If the plane is so steep or so smooth that both rolling and
slipping occur, then da no longer — p, but the ratio of P to N
is known from experiments on sliding friction ; hence there are
still four equations.)
The motion of translation being thus found to be uniformly
accelerated, we may use the equations of § 56 for finding dis-
tance, time, etc.
Query. — How may we distinguish two spheres by allowing
them to roll down the same inclined plane, if one of them is
silver and solid, while the other is of gold, but silvered and
hollow, so as to be the same as the first in diameter, weight,
and appearance?
125. Parallel-Rod of a Locomotive. — "When the locomotive
moves uniformly, each dM of the rod between the two (or
three) driving-wheels rotates with | \ ;
uniform velocity about a centre of its J B I l \ P f ?r /
own on the line BD, Fig. 143, and \J^y (i.) \J_^X
with a velocity v and radius r common ^__.
to all, and likewise has a horizontal \^j)\ij y UKjJ
wniformmotion of translation. Hence en.)'
if we inquire what are the reactions P _ _ FlG - 14B -
of its supports, as induced solely by its weight and motion,
when in its lowest position (independently of any thrust along
132 MECHANICS OE ENGINEERING.
the rod), we put 2Y of (I.) = 27of (II.) (II. shows the
imaginary equivalent system), and obtain
2P — G -fdN =fdMo* -f- r = (v* -=- r)/rfJf = Mv' -~ r.
Example. — Let the velocity of translation — 50 miles per
hour, the radius of the-pins be 18 in. = f ft., and = half that
of the driving-wheels, while the weight of the rod is 200 lbs.
With g = 32.2, we must use the foot and second, and obtain
v = £[50 X 5280 -f- 3600] ft. per second = 36.6;
while M = 200 -=- 32.2 = 200 X .0310 = 6.20 ;
and finally P = £[200 + 6.2(36.6)'-=- f] = 2868.3 lbs.,
or nearly 1£ tons, about thirty times that due to the weight
alone.
126. So far in this chapter the motion has been prescribed,
and the necessary conditions determined, to be fulfilled by the
acting forces at any instant. Problems of a converse nature,
i.e., where the initial state of the body and the acting forces
are given while the resulting motion is required, are of much
greater complexity, but of rare occurrence in practice. The
reader is referred to Rankine's Applied Mechanics. A treat-
ment of the Gyroscope will be found in the American Journal
of Science for 1857, and in the article of that name in Johnson's
Cyclopaedia.
WOKK, ENERGY, AND POWER. 133
CHAPTER YT.
WORK, ENERGY, AND POWER.
127. Remark. — These quantities as defined and developed
in this chapter, though compounded of the fundamental ideas
of matter, force, space, and time, enter into theorems of such
wide application and practical use as to more than justify their
consideration as separate kinds of quantity.
128. Work in a Uniform Translation. Definition of Work. —
Let Fig. 144 represent a rigid body having a motion of trans-
lation parallel to X, acted on by a
system of forces P„ P 2 , P 3 , and P„
which remain constant.
Let s be any distance described by
the body during its motion ; then 2X ~al
must be zero (§ 109), i.e., noting that
jR, and P 4 have negative X com-
ponents (the supplements of their
angles with .STare used),
P, cos a, + P, cos a-, — P, cos a, — P, cos a k = ;
or, multiplying by s and transposing, we have (noting that
s cos a 1 — s l the projection of s on P„ that s cos a 2 = s a , the
projection of s on P 2 , and so on),
p,*, + Pa = J?a + J?a («)
The projections s„ s„ etc., may be called the distances de-
scribed in their respective directions by the forces P„ P„ etc.;
P and P, having moved forward, since 5, and &, fall in front
of the initial position of their points of application ; P 3 and P,
backward, since s 3 and s, fall 5c 2 "o,
gins, by means of which they are 14- -+l
brought to a common velocity at the " M, M 2
instant of greatest tension R', and fig. mi.
elongation s' of the chain) are c 1 = c„ and c 2 — 0.
During the stretching, i.e., the first period of the impact, the
kinetic energy lost by the masses has been expended in stretch-
ing the chain, i.e., in doing the work £i?V ; hence we may
write (the elasticity of the chain not being impaired) (see eq. (1) )
in which the different symbols have the same meaning as in
Example 1, in which the rod corresponds to the chain of this
example.
(Let the student explain why the stipulation is not made here
that one end of the chain shall remain fixed.)
In numerical substitution, 32.2 for g requires the use of the
units foot and second for space and time, while the unit of
force may be anything convenient.
139. Work and Energy in Eotary Motion. Axis Fixed.—
The rio-id body being considered free, let an axis through
perpendicular to the paper be the axis of rotation, and resolve
all forces not intersecting the axis into components parallel
144
MECHANICS OF ENGINEERING.
Fiq. 152.
and perpendicular to the axis, and the latter again into com-
ponents tangent and normal to the circular path of the point
of application. These tangential com-
ponents are evidently the only ones
of the three sets mentioned which
have moments about the axis, those
having moments of the same sign as
oo (the angular velocity at any instant)
being called working forces, T„ T„
etc. ; those of opposite sign, resist-
ances, T/, T 2 ', etc.; for when in time
dt the point of application B„ of T x , describes the small arc
ds i = a x da, whose projection on T x is = ds lt this projection
falls ahead (i.e., in direction of force) of the position of the
point at the beginning of dt, while the reverse is true for T/.
From eq. (XIV.), § 114, we have for 8 (angul. accel.)
Q = (?>, + r,g, + ••■•) - (r, y + r, y + ••• •) (1)
which substituted in oodao = dda (from § 110) gives (remem-
bering that a.dat = ds x , etc.), after integration and transposition,
f U Tfa +f U Trfs, + etc.
=£ T/ds/ +£ T:dsJ + etc. + [*«,„•/ - io,,'/], (2)
where and n refer to any two (initial and final) positions of
the rotating body. Eq. (4), § 120, is an example of this.
Now ico n 'I= ia> n ydMp q =f\dM{w n p)\ which, since w n p
is the actual velocity of any dM&t this (final) instant, is nothing
more than the sum of the amounts of kinetic energy possessed
at this instant by all the particles of the body ; a similar state-
ment may be made for £<*>„"-£
Eq. (2) therefore may be put into words as follows :
Between any two positions of a rigid body rotating about a
fixed axis, the worh done by the working forces is partly used
in overcoming the resistances, and the remainder in changing
the kinetic energy of the individual particles. If in any case
"WORK, ENERGY, AND POWER. 145
this remainder is negative, the final kinetic energy is less than
the initial, i.e., the work done by the working forces is less than
that necessary to overcome the resistances through their respec-
tive spaces, and the deficiency is made up by the restoring of
some of the initial kinetic energy of the rotating body. A
moving fly-wheel, then, is a reservoir of kinetic energy.
Eq. (2) has already been illustrated numerically in §121,
where the additional relation was utilized (for a connecting-rod
and piston of small mass), that the work done in the steam-
cylinder is the same as that done directly at the crank-pin by
the working-force there.
140. "Work of Equivalent Systems the Same. — If two plane
systems of forces acting on a rigid body are equivalent (§ 1 5a),
the aggregate work done by either of them during a given slight
displacement or motion of the body parallel to their plane is
the same. By aggregate work is meant what has already been
defined as the sum of the " virtual moments" (§§ 61 to 64), in
any small displacement of the body, viz., the algebraic sum of
the products, -2 (Pdu), obtained by multiplying each force by
the projection (du) of the displacement of (or small space
described by) its point of application upon the force. (We
here class resistances as negative working forces.)
Call the systems A and B; then, if all the forces of B were
reversed in direction and applied to the body along with those
of A, the compound system would be a balanced system, and
hence we would have (§ 64), for a small motion parallel to the
plane of the forces,
2(Pdu) = 0, i.e., 2(Pdu) for A - 2(Pdu) for B = 0,
or _|_ 2(Pdu) for A = + 2(Pdu) for B.
But + 2 (Pdu) for A is the aggregate work done by the forces
of A during the given motion, and + 2(Pdu) for B is a
similar quantity for the forces of B (not reversed) during the
same small motion if B acted alone. Hence the theorem is
proved, and could easily be extended to space of three dimen-
sions.
j0
146
MECHANICS OF ENGINEERING.
141. Relation of "Work and Kinetic Energy for any Extended
Motion of a Rigid Body Parallel to a Plane.— (If at any instant
any of the forces acting are not
parallel to the plane mentioned,
their components lying in or
parallel to that plane, will be used
instead, since the other compo-
nents obviously would be neither
working forces nor resistances.)
Fig. 153 shows an initial position,
o, of the body ; a final, n ; and any intermediate, as q. The
forces of the system acting may vary in any manner during
the motion.
In this motion each dM describes a curve of its own with
varying velocity v, tangential acceleration p t , and radius of
curvature r ; hence in any position q, an imaginary system B
(see Fig. 154), equivalent to the actual system A (at q in Fig.
153), would be formed by applying to each dM a
tangential force dT = dMp t , and a normal force
dN = dMv* -J- r. By an infinite number of con-
secutive small displacements, the body passes from
o to n. In the small displacement of which q is the
initial position, each dM describes a space ds, and
dT does the work dTds = dMvdv, while dN does
dN X = 0. Hence the total work done by B in the small
displacement at q would be
dT
\
■c
N
Fiq. 154.
the work-
= dM'v'dv' + dM"v"dv" + etc.,
(1)
including all the dM 's of the body and their respective veloci-
ties at this instant.
But the work at q in Fig. 153 by the actual forces (i.e., of
system A) during the same small displacement must (by § 140)
be equal to that done by JB, hence
Prfu, + P.du, + etc. = dM'v'dv' + dM"v"dv" + etc. (g)
Now conceive an equation like (q) written out for each of
WORK, ENERGY, AND POWER. 147
the small consecutive displacements between positions o and
n and corresponding terms to be added ; this will give
P/lu^+J P.,<£w, -f- etc.
= dM'f n v'dv' + dM"f n v"dv" + etc.
= \dM' (» n " - v ") + idM"(v n "> - v,'") + etc.
The second member may be rewritten so as to give, finally,
/ P l dru l + / P^u,-\-eta.=2(idMv n ')-2(idI£v. t ),(XV.)
or, in words, the work done by the acting forces (treating a re-
sistance as a negative working force) between any two posi-
tions is equal to the gain (or loss) in the aggregate kinetic
energy of the particles of the body between the two positions.
To avoid confusion, 2 has been used instead of the sign/" in
one member of (XV.), in which v n is the final velocity of any
dJf (not the same for all necessarily) and v the initial.
(The same method of proof can be extended to three dimen-
sions.)
Since kinetic energy is always essentially positive, if an ex-
pression for it comes out negative as the solution of a problem,
some impossible conditions have been imposed.
142. Work and Kinetic Energy in a Moving Machine. —
Defining a mechanism or machine as a series of rigid bodies
jointed or connected together, so that working-forces applied
to one or more may be the means of overcoming resistances
occurring anywhere in the system, and also of changing the
amount of kinetic energy of the moving masses, let us for
simplicity consider a machine the motions of whose parts are
all parallel to a plane, and let all the forces acting on any one
piece, considered free, at any instant be parallel to the same
plane.
Now consider each piece of the machine, or of any series of
its pieces, as a free body, and write out eq. (XV.) for it be-
tween any two positions (whatever initial and final positions are
148
MECHANICS OF ENGINEEKING.
selected for the first piece, those of the others must be corre-
sponding initial and corresponding final positions), and it will
be found, on adding np corresponding members of these equa-
tions, that the terms involving those components of the mutual
pressures (between the pieces considered) which are normal
to the rubbing surfaces at any instant will cancel out, while
their components tangential to the rubbing surfaces {i.e., fric-
tion, since if the surfaces are perfectly smooth there can be
no tangential action) will appear in the algebraic addition as
resistances multiplied by the distances rubbed through, meas-
ured on the rubbing surfaces. For example, Fig. 155, where
one rotating piece both presses and rubs on another. Let the
normal pressure between them at A be B 2 = JP t ; it is a work-
ing force for the body of mass M", but a resistance for Jf,
hence the separate symbols for the numerically equal forces
(action and reaction).
Similarly, the friction at A is i? 3 = JP t ; a resistance for M' ,
a working-force for M" . (In some cases, of course, friction
may be a resistance for both bodies.) For a small motion, A
describes the small arc AA' about 0' in dealing with M', but
for M" it describes the arc A A" about 0", A' A" being
parallel to the surface of contact AD, while AB is perpen-
Fio. 156.
Fig. 157.
Fig. 155.
dicular to A' A". In Figs. 156 and 157 we see M and M"
free, and their corresponding small rotations indicated. During
these motions the kinetic energy (K. E.) of each mass has
changed by amounts d(K. E.)^ and d(K. E.)^,, respectively, and
hence eq. (XV.) gives, for each free body in turn,
Pfla' - B,AB - B 3 A 7 B = d(K. E.)jr . (1)
- RW+ P.AB + P^A^B = d(K. E.V. . . (2)
WORK, ENERGY, AND POWER. 149
Now add (1) and (2), member to member, remembering that
P 3 = P, and JP t =z _R a = F t = friction, and we have
P x oa!~- F % A'A" - RJ>b" = d(K. E.)jp + d{K. E. V, (3)
in which the mutual actions of M' and M" do not appear,
except the friction, the work done in overcoming which, when
the two bodies are thus considered collectively, is the product
of the friction by the distance A' A" of actual rubbing meas-
ured on the rubbing surface. For any number of pieces, then,
considered free collectively, the assertion made at the beginning
of this article is true, since any finite motion consists of an
infinite number of small motions to each one of which an equa-
tion like (3) is applicable.
Summing the corresponding terms of all such equations, we
have
C P x du, + /'"p.rfw.+ etc. = 2(K.E.) K - 2(K. E.)..(XVI.)
This is of the same form as (XV.), but instead of applying to a
single rigid body, deals with any assemblage of rigid parts
forming a machine, or any part of a machine (a similar proof
will apply to three dimensions of space); but it must be remem-
bered that it excludes all the mutual actions of the pieces con-
sidered except friction, which is to be introduced in the manner
just illustrated. A flexible inextensible cord may be considered
as made up of a great number of short rigid bodies jointed
without friction, and hence may form part of a machine with-
out vitiating the truth of (XVI.).
2(K. E.) n signifies the sum obtained by adding the amounts
of kinetic energy (%dMv n * for each elementary mass) possessed
by all the particles of all the rigid bodies at their final posi-
tions ; 2(K. E.)„, a similar sum at their initial positions. For
example, the K. E. of a rigid body having a motion of transla-
tion of velocity v, = IvfdM = \ Mv 2 ; that of a rigid body
having an angular velocity a> about a fixed axis Z, = WIz
(§ 139) ; while, if it has an angular velocity oa about a gravity-
150 MECHANICS OF ENGINEERING.
axis Z, which has a velocity v z of translation at right angles to
itself, the (K. E.) at this instant may be proved to be
iMvz' + WIz,
i.e.-, is the sum of the amounts due to the two motions sepa-
rately.
143. K. E. of Combined Rotation and Translation. — The last
statement may be thus proved. Fig. 157.
At a given instant the velocity of any dM is
~^ < —~ ^.^ v, the diagonal formed on the velocity v z of
T~tfM\ **' translation, and the rotary velocity cop rela-
— ~/ \ tively to the moving gravity-axis Z (per-
pendicular to paper) (see § 71),
Z Fig. 158. i.e., «' = ^Z + (<»/>)' ~ K W P) V Z COS cp ;
hence we have K. E., at this instant,
= f\dMv i =\oifd,ll + WfdMft - wv z fdMp cos = y, and fdMy = Jfy = 0, since Z is a gravity-
axis,
.-. K E. = £Jf — FJ' — F'nr" - 6™ no, sin ft — Fjta, . . (1)
provided the respective distances are actually described by
these forces, i.e., if the masses have sufficient initial kinetic
energy to carry the crank-pin beyond the point of minimum
velocity, with the aid of the working force P a whose effect is
small up to that instant.
As for the total initial kinetic energy, i.e., 2(K. E.)„, let us
express it in terms of the velocity of crank-pin at o, viz., F„.
The (K.E.). of M is nothing ; that of M", which at this in-
stant is rotating about its right extremity (fixed for the instant)
with angular velocity co" = F. -*- I", is WV; that of JT
= W' */„'", in which oo'" = V + r; that of M» (translation)
= Im^v '-, in which v» = {a+r) V.. 2(JL E.)„ is expressed
152
MECHANICS OP ENGINEERING.
in a corresponding manner with V n (final velocity of crank-pin)
instead of V„. Hence the right-hand member of (XVI.) will
give (putting the radius of gyration of M" about 0" = h",
and that of M" about C — h)
(2)
By writing (1)=(2), we have an equation of condition, capa-
ble of solution for any one unknown quantity, to be satisfied
for the extent of motion considered. It is understood that the
chain is always taut, and that its weight and mass are neg-
lected.
145. Numerical Case of the Foregoing. — (Foot-pound-second
system of units for space, force, and time; this requires g
- 32.2.)
Suppose the following data :
Feet.
Lbs.
Lbs.
Mass Units.
V = 2.0
I" = 4.0
a = 1.5
r = 1.0
k = 1.8
k" = 2.3
r" = 0.1
Pi =
F, =
V" (av'ge) =
F t =
6000
200
400
300
O' = 60
G" = 50
= 13.4
M" = 100.0
Also let To = 4
ft. per sec. ; /3=30°
Denote (1) by IT and the large bracket in (2) by M (this by
some is called the total mass " reduced" to the crank-pin).
Putting (1) = (2) we have, solving for the unknown V m
/2W
-HI?.
For above values,
W = 12,000 - 400
= 2467 foot-pounds ;
while M = 0.5 + 40.3 + 225.0
whence
M
125.7 — 7590.0 - 1417.3
265.8 mass-units ;
(3)
V n = 4/18.56 + 16 = 4/34.56 = 5.88 ft. per second.
WORK, ENERGY, AND POWER. 153
As to whether the crank-pin actually reaches the dead-point
n, requires separate investigations to see whether V becomes
zero or negative between o and n (a negative value is inad-
missible, since a reversal of direction implies a different
value for W), i.e., whether the proposed extent of motion is
realized ; and these are made by assigning some other inter-
mediate position m, as a final one, and computing V m , remem-
bering that when m is not a dead-point the (K. E.) m of M' is not
zero, and must be expressed in terms of V m , and that the
(K. E.) m of the connecting-rod Jf'must be obtained from § 143.
146. Regulation of Machines. — As already illustrated in
several examples (§ 121), a fly-wheel of sufficient weight and
radius may prevent too great fluctuation of speed in a single
stroke of an engine ; but to prevent a permanent change, which
must occur if the work of the working force or forces (such as
the steam-pressure on a piston, or water-impulse in a turbine)
exceeds for several successive strokes or revolutions the work
required to overcome resistances (such as friction, gravity, re-
sistance at the teeth of saws, etc., etc.) through their respective
spaces, automatic governors are employed to diminish the
working force, or the distance through which it acts per stroke,
until the normal speed is restored ; or vice versa, if the speed
slackens, as when new resistances are temporarily brought into
play. Hence when several successive periods, strokes (or other
cycle), are considered, the kinetic energy of the moving parts
will disappear from eq. (XVI.), leaving it in this form :
work of working-fwces = work done upon resistances.
147. Power of Motors. — In a mill where the same number of
machines are run continuously at a constant speed proper for
their work, turning out per hour the same number of barrels
of flour, feet of lumber, or other commodity, the motor (e.g.,
a steam-engine, or turbine) works at a constant rate, i.e., de-
velops a definite horse-power (H.P.), which is thus found in
the case of steam-engines (double-acting) :
154 MECHANICS OP ENGINEERING.
H.P. = total mean effective \ ! distance in feet j
steam-pressure on I X ] travelled by pis- V -r- 550,
piston in lbs. ) ( ton per second. )
i.e., the work (in ft.-lbs) done per second by the working force
divided by 550 (see § 132). The total effective pressure at any
instant is the excess of the forward over the back-pressure,
and by its mean value (since steam is usually used expansively)
is meant such a value P' as, multiplied by the length of stroke
Z, shall give
P'l=fPdx,
where P is the variable effective pressure and dx an element
of its path. If u is the number of strokes per second, we may
also write {foot-pound- second system)
H.P. = P'lu -=- 550 = \_fpdx~ju + 550. (XYII.)
Yery often the number of revolutions per minute, m, of the
crank is given, and then
H.P. = P' (lbs.) X 2Z (feet) X m -=- 33,000.
If F '= area of piston we may also write P' = Fp', where p'
is the mean effective steam-pressure per unit of area. Evi-
dently, to obtain P' in lbs., we multiply i^in sq. in. by^' in
lbs. per sq. in., or F in sq. ft. by^>' in lbs. per sq. foot ; the
former is customary, p' in practice is obtained by measure-
ments and computations from " indicator-cards" (see § 135, in
which (P 2 — P,) corresponds to P of this section) ; or P'l, i.e.,
/ Pdx, may be computed theoretically as in § 59, Problem 4.
The power as thus found is expended in overcoming the
friction of all moving parts (which is sometimes a large item),
and the resistances peculiar to the kind of work done by the ma-
chines. The work periodically stored in the increased kinetic
energy of the moving masses is restored as they periodically
resume their minimum velocities.
"WORK, ENERGY, AND POWER. 155
148. Potential Energy. — There are other ways in which work
or energy is stored and then restored, as follows :
First. In raising a weight G through a height h, an amount
of work = Gh is done upon G, as a resistance, and if at any
subsequent time the weight is allowed to descend through the
same vertical distance h (the form of path is of no account), G,
now a worlcing force, does the work Gh, and thus in aiding the
motor repays, or restores, the Gh expended by the motor in
raising it. If h is the vertical height through which the centre
of gravity rises and sinks periodically in the motion of the
machine, the force G may be left out of account in reckoning
the expenditure of the motor's work, and the body when at its
highest point is said to possess an amount Gh of potential
energy, i.e., energy of position, since it is capable of doing the
work Gh in sinking through its vertical range of motion.
Second. So far, all bodies considered have been by express
stipulation rigid, i.e., incapable of changing shape. To see
the effect of a lack of rigidity as affecting the principle of
work and energy in machines, -<-~-^Z^r^r~ ^... ) .
take the simple case in Fig. 160. p f^f^mii^P
A helical spring at a given in- */~^ww&^*J d *
stant is acted on at each end by t a M b hd-J3 *"
a force P in an axial direction / J
(they are equal, supposing the Fio. wo.
mass of the spring small). As the machine operates of which
it is a member, it moves to a new consecutive position B,
suffering a further elongation d\ in its length (if P is increas-
ing. P on the right, a working force, does the work Pdx';
how is this expended ? P on the left has the work Pdx done
upon it, and the mass is too small to absorb kinetic energy or
to bring its weight into consideration. The remainder, Pdx'
— Pdx = PdX, is expended in stretching the spring an addi-
tional amount d\, and is capable of restoration if the spring
retains its elasticity. Hence the work done in changing the
form of bodies if they are elastic is said to be stored in the
form of potential energy. That is, in the operation of ma-
chines, the name potential energy is also given to the energy
156 MECHANICS OF ENGINEERING.
stored and restored periodically in the changing and regaining
of form of elastic bodies.
149. Other Forms of Energy. — Numerous experiments with
various kinds of apparatus have proved that for every 772
(about) f t.-lbs. of work spent in overcoming friction, one British
unit of heat is produced (viz., the quantity of heat necessary to
raise the temperature of one pound of water from 32° to 33°
Fahrenheit) ; while from converse experiments, in which the
amount of heat used in operating a steam-engine was all carefully
estimated, the disappearance of a certain portion of it could only
be accounted for by assuming that it had been converted into
work at the same rate of (about) 772 ft.-lbs. of work to each
unit of heat (or 425 kilogrammetres to each French unit of
heat). This number 772, or 425, according to the system of
units employed, is called the Mechanical Equivalent of Heat,
first discovered by Joule and confirmed by Hirn.
Heat then is energy, and is supposed to be of the kinetic
form due to the rapid motion or vibration of the molecules of
a substance. A similar agitation among the molecules of the
(hypothetical) ether diffused through space is supposed to pro-
duce the phenomena of light, electricity, and magnetism.
Chemical action being also considered a method of transform-
ing energy (its possible future occurrence as in the case of coal
and oxygen being called potential energy), the well-known
doctrine of the Conservation of Energy, in accordance with
which energy is indestructible, and the doing of work is simply
the conversion of one or more kinds of energy into equivalent
amounts of others, is now one of the accepted hypotheses of
physics.
"Work consumed in friction, though practically lost, still re-
mains in the universe as heat, electricity, or some other subtile
form of energy.
150. Power Required for Individual Machines. Dynamome-
ters of Transmission. — If a machine is driven by an endless
belt from the main-shaft, J., Fig. 161, being the driving-pulley
WOKE, ENERGY, AND POWEE.
157
Fig. 161.
on the machine, the working force which drives the machine
in other words the " grip" with which the
belt takes hold of the pulley tangentially p~
= P — P', P and P' being the tensions
in the "driving" and "following" sides of
the belt respectively. The belt is supposed
not to slip on the pulley. If v is the ve-
locity of the pulley- circumference, the
work expended on the machine per second, i.e., the power, is
Z = (P-P> (1)
To measure the force (P - P'), an apparatus called a Dy-
namometer of Transmission may be placed between the main
shaft and the machine, and the belt made to pass through it in
such a way as to measure the tensions P and P, or princi-
pally their difference, without meeting any resistance in so do-
ing ; that is, the power is transmitted, not absorbed, by the
apparatus. One invention for this purpose (mentioned in the
Jou rnal of the F ranklin Institute some years ago) is shown
(in principle) in Fig. 162. A ver-
tical plate carrying four pulleys and
a scale-pan is first balanced on the
pivot C. The belt being then ad-
justed, as shown, and the power
turned on, a sufficient weight G is
placed in the scale-pan to balance
Fig. 163. the plate again, for whose equilib-
rium we must have Gb = Pa — P'a, since the P and P' on
the right are purposely given no leverage about 0. The ve-
locity of belt, v, is obtained by a simple counting device.
Hence (P — P') and v become known, and .\ L from (1).
Many other forms of transmission-dynamometers are in use,
some applicable whether the machine is driven by belting or
gearing from the main shaft. Emerson's Hydrodynamics de-
scribes his own invention on p. 283, and gives results of meas-
urements with it ; e.g., at Lowell, Mass., the power required
to drive 112 looms, weaving 36-inch sheetings, No. 20 yarn,
158 MECHANICS OF ENGINEERING.
60 threads to the inch, speed 130 picks to the minute, was
found to be 16 H.P., i.e., \ H.P. to each loom (p. 335).
151. Dynamometers of Absorption. — These are so named
since they furnish in themselves the resistance (friction or a
weight) in the overcoming (or raising) of which the power is
expended or absorbed. Of these the Prony Friction Brake
is the most common, and is used for measuring the power
developed by a given motor (e.g., a steam-engine or turbine)
not absorbed in the friction of the motor itself. Fig. 163
Fig. 163.
shows one fitted to a vertical pulley driven by the motor. By
tightening the bolt JB, the velocity v of pulley-rim may be
made constant at any desired value (within certain limits) by
the consequent friction, v is measured by a counting appara-
tus, while the friction (or tangential components of action be-
tween pulley and brake), = F, becomes known by noting the
weight O which must be placed in the scale pan to balance the
arm between the checks ; then
Fa =Gh, (1)
for the equilibrium of the brake (supposing the weight of
brake and scale-pan previously balanced on C) and the work
done per unit of time, or power, is
Z = Fv (2)
A " dash-pot " is frequently connected with the arm to prevent
sudden oscillations. In case the pulley is horizontal, a bell-
crank lever is added between the arm and the scale-pan, and
then eq. (1) will contain two additional lever-arms.
WORK, ENERGY, AND POWER.
159
■<-
>s^
I - |
\ 1
p,
p 2
Pa
P4
Pu
T j
p.
152. The Indicator, used with steam and other fluid engines,
is a special kind of dynamometer in which the automatic mo-
tion of a pencil describes a curve
on paper whose ordinates are
proportional to the fluid pres-
sures exerted in the cylinder at
successive points of the stroke.
Thus, Fig. 164, the back-pres-
sure being constant and = P b ,
the ordinates P , 7*,, etc., represent the effective pressures at
equally spaced points of division. The mean effective pressure
7" (see § 147) is, for this figure, by Simpson's Rule (six equal
spaces),
P' = A[P. + 4(P, + P, + P t ) + 2(P a + P 4 ) + PJ.
IP,,
Fig. 164.
This gives a near approximation.
§147.
The power is now found by
153. The theory of Atwood's Machine is most directly ex-
pressed by the principle of work and energy ; i.e., by eq.
(XVI.), §142. Fig. 165. The parts
considered free, collectively, are the
rigid bodies P, Q, G, and four friction-
wheels like G t ; and the flexible cord,
which does not slip on the upper pul-
ley. There is no slipping at 7>, hence
no sliding friction there. The actions
of external bodies on these eight consist
of the working force P, the resistances
Q and the four F's (at bearings of f ric-
Fio 165. * tion-wheel axles); all others {G, iG>,
and the four i?'s) are neutral. Since there is no rubbing be-
tween any two of the eight bodies, no mutual actions whatever
will enter the equation. Let P > Q, and /and 7, be the mo-
ments of inertia of G and G„ respectively, about their respec-
tive axes of figure. Let the apparatus start from rest, then
when P has descended through any vertical distance «, and ac-
160 MECHANICS OF ENGINEERING.
quired the velocity v, Q has been drawn up an equal distance
and acquired the same velocity, while the pulley G has ac-
quired an angular velocity go = v -=- a, each friction-pulley an
angular velocity oo l = (r : a)v -=- a v As to the forces, P has
done the work Ps, Q has had the work Qs done upon it, while
each ,77 has been overcome through the space (r 1 : a,)(r : a)s ;
all the other forces are neutral. Hence, from eq. (XVI.), § 142
(see also § 139), we have
p s -Qs- 4F^ . -s
_VP , (TK i li? T-i L ll Hat
~Lg + g J2 + 2 a" + 2 " a 2 " a, 2 * 7 ' °"
Evidently v = Vs X constant, i.e., the motion of P and Q is
uniformly accelerated. If, after the observed space s has been
described, P is suddenly diminished to such a value P' that
the motion continues with a constant velocity = v, we shall
have, for any further space s',
P's'-Qs'-iF^.-s' = 0,
from which .77 can be obtained (nearly) ; while if if be the ob-
served time of describing s', v = s' -=- t' becomes known.
Also we may write 7"= (G -=- ^)F and 7j = (6^ -f- g)k?, and
thus finally compute the acceleration of gravity, g, from our
first equation above.
154. Boat-Rowing. — Fig. 166. During the stroke proper,
let P = mean pressure on one oar-handle ; hence the pressures
on the foot-rest are 2P, resistances. Let J!f=massof boat
and load, v„ and v n its velocities at beginning and end of stroke.
P 1 — pressures between oar-blade and water. R = mean re-
sistance of water to the boat's passage at this (mean) speed.
These are the only (horizontal) forces to be considered as act-
ing on the boat and two oars, considered free collectively.
During the stroke the boat describes the space s 3 = CD, the
oar-handle the space « 3 = AB, while the oar-blade slips back-
WORK, ENERGY, AND POWER.
161
ward through the small space (the " slip") = s, (average).
Hence by eq. (XVI.), § 142,
2P Sl - 2Ps 3 -U s, - 2P A = £ M{v: - v,*) ;
i.e., 2P(e t -8 t )=2PxAE=z2Pa =Ps 3 +2P A + fKK'-w.");
or, in words, the product of the oar-haudle pressures into the
distance described by them measured on the boat, i.e., the work
done by these pressures relatively to the boat, is entirely ac-
counted for in the work of slip and of liquid-resistance, and in-
Fio. 166.
creasing the kinetic energy of the mass. (The useless work
due to slip is inevitable in all paddle or screw propulsion, as
well as a certain amount lost in machine-friction, not considered
in the present problem.) During the " recover" the velocity
decreases again to v„.
155. Examples. — 1. "What work is done on a level track, in
bringing up the velocity of a train weighing 200 tons, from
zero to 30 miles per hour, if the total frictional resistance (at
any velocity, say) is 10 lbs. per ton, and if the change of speed
is accomplished in a length of 3000 feet ?
{Foot-ton-second system.) 30 miles per hour = 44 ft. per
sec. The mass
= 200 -=- 32.2 = 6.2 ;
.-. the change in kinetic energy,
(=iMv--iM X 2 ),
= £( 6 - 2 ) X 44 3 = 6001.6 ft.-tons.
11
162 MECHANICS OP ENGINEERING.
The work done in overcoming friction = Fs, i.e.,
= 200 X 10 X 3000 = 6,000,000 f t.-lbs. = 3000 ft.-tons ;
.-. total work = 6001.6 + 3000 = 9001.6 ft.-tons.
(If the track were an up-grade, 1 in 100 say, the item of
200 X 30 = 6000 ft.-tons would be added.)
Example 2. — Bequired the rate of work, or power, in Ex-
ample 1. The power is variable, depending on the velocity of
the train at any instant. Assume the motion to be uniformly
accelerated, then the working force is constant ; call it P.
The acceleration (§ 56) will be ^=-y a -=-2s=1936-j-6000= 0.322
ft. per sq. sec; and since P — F '= Mp, we have
P = 1 ton -t- (200 -T- 32.2) X 0.322 = 3 tons,
which is 6000 -f- 200 = 30 lbs. per ton of train, of which 20 is
due to its inertia, since when the speed becomes uniform the
work of the engine is expended on friction alone.
Hence when the velocity is 44 ft. per sec, the engine is
working at the rate of Pv = 264,000 f t.-lbs. per sec, i.e., at the
rate of 480 H. P.;
At J of 3000 ft. from the start, at the rate of 240 H. P., half
as much ;
At a uniform speed of 30 miles an hour the power would be
simply 1 X 44 = 44 ft. -tons per sec. = 160 H. P.
Example 3. — The resistance offered by still water to the
passage of a certain steamer at 10 knots an hour is 15,000 lbs.
What power must be developed by its engines, at this uniform
speed, considering no loss in "slip" nor in friction of ma-
chinery ?
Example 4. — Same as 3, except that the speed is to be 15
knots (i.e., nautical miles ; each = 6086 feet) an hour, assum-
ing that the resistances are as the square of the speed (approxi-
mately true).
Example 5. — Same as 3, except that 12$ of the power is ab-
sorbed in the " slip" (i.e., in pushing aside and backwards the
water acted on by the screw or paddle), and 8% in friction of
machinery.
Example 6.— In Example 3, if the crank-shaft makes 60
"WORK, ENERGY, AND POWER. 163
revolutions per minute, tlie crank-pin describing a circle of 18
inches radius, required the average value of the tangential
component of the thrust (or pull) of the connecting-rod against
the crank-pin.
Example 7. — A solid sphere of cast-iron is rolling up an in-
cline of 30°, and at a certain instant its centre Las a velocity of
36 inches per second. Neglecting friction of all kinds, how
much further will the ball mount the incline (see § 143) ?
Example 8. — In Fig. 163, with b = 4 f t. and a = 16 inches,
it is found in one experiment that the friction which keeps the
speed of the pulley at 120 revolutions per minute is balanced
by a weight O = 160 lbs. Required the power thus measured.
Although in Examples 1 to 6 the steam cylinder is itself in
motion, the work per stroke is still = mean effective steam-
pressure on piston X length of stroke, for this is the final form
to which the separate amounts of work done by, or upon, the
two cylinder heads and the two sides of the piston will re-
duce, when added algebraically. See § 154.
164 MECHANICS OE ENGINEEBING.
CHAPTEE TIL
FRICTION.
156. Sliding Friction. — "When the surfaces of contact of two
bodies are perfectly smooth, the direction of the pressure or pair
of forces between them is normal to these surfaces, i.e., to their
, tangent-plane ; but when they are rough, and
\"? moving one on the other, the forces or ac-
pY iH tions between them incline away from the
-fV — — i^ p normal, each on the side opposite to the di-
, ■■ J . : c^-) . r"' ', rection of the (relative) motion of the body
' \g '"-'^ on which it acts. Thus, Fig. 167, a block
Fio. 167. whose weight is G, is drawn on a rough
horizontal table by a horizontal cord, the tension in which is
P. On account of the roughness of one or both bodies the ac-
tion of the table upon the block is a force _P„ inclined to the
normal (which is vertical in this case) at an angle = is
called the angle of friction, while the tangential component of
jP, is called the friction = F. The normal component JV,
which in this case is equal and opposite to G the weight of the
body, is called the normal pressure.
Obviously F = iVtan bjf, we have
F=fir. (i)
yis called the coefficient of friction, and may also be defined
as the ratio of the friction F to the normal pressure N which
produces it.
FRICTIOK. 165
In Fig. 167, if the motion is accelerated (ace. = p), we have
(eq. (IV.), § 55) P - F = Mp ; if uniform, P-F=0; from
which equations (see also (1))/ may be computed. In the
latter case/ may be found to be different with different veloci-
ties (the surfaces retaining the same character of course), and
then a uniformly accelerated motion is impossible unless P
were to vary as F.
As for the lower block or table, forces the equals and op-
posites of W &x\AF{oy a single force equal and opposite to P,)
are comprised in the system of forces acting upon it.
As to whether F is a working force or a resistance, when
either of the two bodies is considered free, depends on the cir-
cumstances of its motion. For example, in friction-gearing
the tangential action between the two pulleys is a resistance
for one, a working force for the other.
If tjie force P, Fig. 167, is just sufficient to start the body,
or is just on the point of starting it (this will be called impending
motion), Fis, called the friction of rest. If the body is at rest
and P is not sufficient to start it, the tangential component will
then be < the friction of rest, viz., just = P. AsP increases,
this component continually equals it in value, and P 1 acquires
a direction more and more inclined from the normal, until the
instant of impending motion, when the tangential component
=f]F= the friction of rest. When motion is once in prog-
ress, the friction, called then the friction of motion, = fN,
in which/ is not necessarily the same as in the friction of rest.
157. Laws of Sliding Friction. — Experiment has demon-
strated the following relations approximately, for two given
rubbing surfaces :
(1) The coefficient,/ is independent of the normal pressure
N.
(2) The coefficient,/ for friction of motion, is the same at
all velocities.
(3) The coefficient, / for friction of rest (i.e., impending
motion) is usually greater than that for friction of motion
(probably on account of adhesion).
166
MECHANICS OF ENGINEERING.
(4) The coefficient,/", is independent of the extent of rub-
bing surface.
(5) The interposition of an unguent (such as oil, lard, tallow,
etc.) diminishes the friction very considerably.
158. Experiments on Sliding Friction. — These may be made
with simple apparatus. If a block of weight = G, Fig. 168,
be placed on an inclined plane of uniformly rough surface,
and the latter be gradually more and more inclined from the
horizontal until the block begins to move, the value of /3 at
Fig. 168.
Fig. 169.
this instant = 2s
f--G\-—G—--gf (3)
If G, Fig. 59, is made just sufficient to start the block, or
sledge, G„ we have for the friction of rest
/=! «
160. Kesults of Experiments on Sliding Friction.— Professor
Thurston in his article on Friction (which the student will do
well to read) in Johnson's Cyclopaedia gives the following
epitome of results from General Morin's experiments (made
for the French Government in 1833) :
168
MECHANICS OF ENGINEERING.
TABLE FOR FRICTION OF MOTION.
No.
Surfaces.
Unguent.
Angle .
/ = tan <£>.
1
Wood on wood.
None.
14° to26£°
0.25 to 0.50
2
Wood on wood.
Soap.
2° to Hi"
0.04 to 0.20
3
Metal on wood.
None.
26£° to 31£°
0.50 to 0.60
4
Metal on wood.
Water.
15° to 20°
0.25 to 0.35
5
Metal on wood.
Soap.
111°
0.20
6
Leather on metal.
None.
29*°
0.56
7
Leather on metal.
Greased.
13°
0.23
8
Leather on metal.
Water.
20°
0.36
9
Leather on metal.
Oil.
8F
0.15
10
Smoothest and best
lubricated surfaces.
1|° to 2°
03 to 036
For friction of rest, about 40$ may be added to the coeffi-
cients in the above table.
In dealing with the stone blocks of an arch-ring, 30° with the normal to the joint
(see § 161) slipping may take place (the adhesion of cement
being neglected).
General Morin states that for a sledge on dry ground/" =
about 0.66.
Weisbach gives for metal on metal, dry (R. E. brakes for
example),/ = from 0.15 to 0.24. Trautwine's Pocket-Book
gives values of/ for numerous cases of friction.
161. Cone of Friction.— Fig. 170. Let A and B be two
rough blocks, of which B is immovable, and P the resultant
-.j*/ c of all the forces acting on A, except the pres-
sure from B. B can furnish any required
normal pressure N to balance P cos /?, but
the limit of its tangential resistance is fN.
So long then as /? is < q> the angle of fric-
tion, or in other words, so long as the line of
fio. 170. action of P is within the " cone of friction"
generated by revolving OC about ON, the block A will not
FEIOTION.
169
slip on B, and the tangential resistance of B is simply P sin
§ ; but if /? is > cp, this tangential resistance being only/iV
and < P sin /J, A will begin to slip, with an acceleration.
162. Problems in Sliding Friction.— In the following prob-
lems/is supposed known at points where rubbing occurs, or
is impending. As to the pressure N to which the friction is
due, it is generally to be considered unknown until determined
by the conditions of the problem. Sometimes it may be an
advantage to deal with the single unknown force P (resultant
of JFand/JV) acting in a line making the known angle cp with
the normal (on the side away from the motion).
Problem 1. — Kequired the value of the weight P, Fig. 171,
the slightest addition to which will cause motion of the hori-
zontal rod OB, resting on rough planes at 45°. The weight
G of the rod may be applied at the
middle. Consider the rod free ; at
each point of contact there is an un-
known JV and a friction due to it
yiV; the tension in the cord will be
= P, since there is no acceleration
and no friction at pulley. Notice
the direction of the frictions, both opposing the impending
motion. [The student should not rush to the conclusion that
iVand JV X are equal, and are the same as would be produced by
the components of G if the latter were transferred to A and
resolved along AO and AB ; but should await the legitimate
results deduced by algebra, from the equations of condition
for the equilibrium of a system of forces in a plane. Few
problems in Mechanics are so simple as to admit of an imme-
diate mental solution on inspection ; and guess-work should be
carefully avoided.]
Taking an origin and two axes as in figure, we have (eqs.
(2), § 36), denoting the sine of 45° by m,
2X.... f^ + mG-JV -P = 0;. . (1)
2Y.... N^ + ftf- mG = 0;. . (2)
~2{Pa) fNa + Na- Gb = 0. . . (3)
Fig. 171.
170
MECHANICS OP ENGINEERING.
The three unknowns P, JV, and JV, can now be found.
Divide (3) by a, remembering that b : a = m, and solve for
JV; substitute it in (2) and JV, also becomes known ; while P
is then found from (1) and is
P =
2mfG fV2
G.
Fia. 173.
Peoblem 2. — Fig. 172. A rod, centre of gravity at middle,
leans against a rongh wall, and rests on an equally rough floor;
\m, how small may the angle a become before it
P 2 p slips ? Let a = the half-length. The figure
90 shows the rod free, and following the sugges-
P tion of § 162, a single unknown force P,
P making a known angle
.
But BE, which = 2a sin a, = BE— AB ;
.•. 2a sin a = a cos a [cot cp — tan cp~\. . . (1)
Dividing by cos a, and noting that tan
" to transmit 2 H.P.;
"~" Z/, p <;'- -~jfc ) if a = 45°', f for impending (relative)
a U /n ^ motion, i.e., for impending slipping =
fig. 175. 0'40, and the velocity of the pulley-rim
is 9 ft. per second ?
The limit-value of the tangential " grip"
T - 2fN= 2 X 0.40 X P sin 45°,
2 H. P. = 2 X 550 = 1100 ft.-lbs. per second.
Putting T X 9 f t. = 1100, we have
2 X 0.40 X4XPX9 = 1100 ; .-. P = 215 lbs.
Problem 6. — A block of weight G lies on a rough plane,
inclined an angle /3 from the horizontal ; find the pull P, mak-
ing an angle a with the first plane, which will maintain a uni-
form motion up the plane.
FRICTION. 173
Problem 7.— Same as 6, except that the pull P is to permit
a uniform motion down the plane.
Peoblem 8. — The thrust of a screw-propeller is 15 tons.
The ring against which it is exerted has a mean radius of 8
inches, the shaft makes one revolution per second, and/= 0.06.
Kequired the PI. P. lost in friction from this cause.
163. The Bent-Lever with Friction. Worn Bearing. — Fi°\
176. Neglect the weight of the lever, and suppose the plumb-
er-block so worn that there is Q r HEp,
contact along one element only of / \ \ / / j f~? *
the shaft. Given the amount and q/ )/\/L I I
line of action of the resistance B, / K \ \ / / '-•»
and the line of action of P, re- /\^V^t^X-J
quired the amount of the latter for / ^\*7h .-^\
impending slipping in the direction * R
of the dotted arrow. As P grad-
ually increases, the shaft of the
lever (or gear-wheel) rolls on its FlG . 17
bearing until the line of contact has reached some position A,
when rolling ceases and slipping begins. To find A, and the
value of P, note that the total action of the bearing upon the
lever is some force P„ applied at A and making a known
angle
than for a flat-ended pivot ; for, on account
of the wedge-like action of the bodies, the
pressure causing friction is greater. The sum of
the moments of these resultant frictions about
A is the same as if only two elements of the
cone received pressure (each = N = %R -~ sin a). Hence the
FRICTION.
181
moment of friction of the pivot, i.e., the moment of the force
necessary to maintain uniform rotation, is
2
3'' '
P*=f*N^=f*
sin a 3 "
and work lost per revolution = ^nf r
3 •'sin a ,-
By making r, small enough, these values may be made less
than those for a flat-ended pivot of the same diameter = 2r.
In Schiele's " anti-friction" pivots the outline is designed
according to the following theory for securing uniform vertical
wear. Let^» = the pressure per
horizontal unit of area (i.e.,
= -ff -j- horizontal projection of /?*-,..
the actual rubbing surface) ;
this is assumed constant. Let
the unit of area be small, for N
algebraic simplicity. The fric-
tion on the rubbing surface, whose horizontal projection = unity,
is = /iV" =/(p -=- sin a) (see Fig. 182; the horizontal com-
ponent of p is annulled by a corresponding one opposite). The
work per revolution in producing wear on this area =f]¥27ty.
But the vertical depth of wear per revolution is to be the same
at all parts of the surface; and this implies that the same
volume of material is worn away under each horizontal unit of
area. Hence fNlny, i.e.,f- 2^y, is to be constant for all
SHI Ol
values of y ; and since fp and 2ar are constant, we must have,
as the law of the curve,
V
sin a
-, i.e., the tangent SO = the same at all points.
This curve is called the " tractrixP Schiele's pivots give a
very uniform wear at high speeds. The smoothness of wear
prevents leakage in the case of cocks and faucets.
169. Normal Pressure of Belting. — "When a perfectly flexible
cord, or belt, is stretched over a smooth cylinder, both at rest,
182
MECHANICS OF ENGINEERING.
the action between tliem is normal at every point. As to its
j&. » ♦ \ | s amount, p, per linear unit of arc, the f ol-
C\ V^= ? ** lowing will determine. Consider a semi-
circle of the cord free, neglecting its weight.
Fig. 183. The force holding it in equilib-
rium are the tensions at the two ends (these
are equal, manifestly, the cylinder being
_ smooth ; for they are the only two forces
/Tjp *" having moments about C, and each has the
fig. is3. same lever-arm), and the normal pressures,
which are infinite in number, but have an intensity, p, per
linear unit, which must be constant along the curve since S is
the same at all points. The normal pressure on a single ele-
ment, ds, of the cord is = pds, and its X component =
pds cos 6 = prdO cos 0. Hence 2^= gives
cos Bdd — 2S = 0, i.e., rp\ sin 6 = 28;
*•* L-iir
.:rp\l -(-1)] = 2S or p = §. . . . . (1)
170. Belt on Sough Cylinder. Impending Slipping. — If fric-
tion is possible between the two bodies, the tension may vary
along the arc of contact, so that p also varies, and consequently
Fig. 184.
pds
0\
Fig. 185.
the friction on an element ds being =fpds =f(S-t-r)ds, also
varies. If slipping is impending, the law of variation of the
tension S may be found, as follows : Fig. 184, in which the
FKICTION. 183
impending slipping is toward the left, shows the cord free.
For any element, ds, of the cord, we have, putting 2 (moms,
about O) = (Fig. 185),
(S+ dS)r = Sr + dFr ; i.e., dF= dS,
or (see above) dS =f{8 ~ r)ds.
But ds = rdd ; hence, after transforming,
„,_, dS
f dd = ~S W
In (1) the two variables d and S are separated ; (1) is there-
fore ready for integration.
fa = log e S n - log, 8 a = log e [|]. (2)
Or, by inversion, S t e' a — S n , (3)
e, denoting the Naperian base, = 2.71828 4-; a of course is in
jr-measure.
Since 8 n evidently increases very rapidly as a becomes
larger, 5„ remaining the same, we have the explanation of the
well-known fact that a comparatively small tension, S„ exerted
by a man, is able to prevent the slipping of a rope around a
pile-head, when the further end is under the great tension S n
due to the stopping of a moving steamer. For example, with
f = -J, we have (Weisbach)
f or a — £ turn, or a = %n, S n = 1.695, ;
= -J turn, or a = tt, 5 n = 2.855, ;
= 1 turn, or a = 2ff, 5 n — 8.125, ;
= 2 turns, or a = 4tt, S n = 65.945, ;
= 4 turns, or a = 8rr, 5„ = 4348.565,.
If slipping actually occurs, we must use a value of /for fric-
tion of motion.
Example. — A leather belt drives an iron pulley, covering
one half the circumference. What is the limiting value of the
184
MECHANICS OP ENGINEERING.
ratio of S n (tension on driving-side) to S (tension on follow-
ing side) if the belt is not to slip, taking the low value of
f = 0.25 for leather on iron ?
"We have given fa = 0.25 X n — .7854, which by eq. (2) is
the JSTaperian log. of (S n : S t ) when slipping occurs. Hence the
common log. of (S n : 8 a ) = 0.7854 X 0.43429 = 0.34109 ; i.e.,
if
(S n : S.) = 2.193, say 2.2,
the belt will slip (for/ = 0.25).
(0.43429 is the modulus of the common system of loga-
rithms, and = 1 : 2.30258. See example in § 48.)
At very high speeds the relation p = S -f- r (in § 169) is not
strictly true, since the tensions at the two ends of an element
ds are partly employed in furnishing the necessary deviating
force to keep the element of the cord in its circular path, the
remainder producing normal pressure.
171. Transmission of Power by Belting or "Wire Hope. — In the
simple design in Fig. 186; it is required to find the motive
weight G, necessary to overcome the given resistance It at a
Fig. 186.
uniform velocity = *>,j also the proper stationary tension
weight 6r to prevent slipping of the belt on its pulleys, and
the amount of power, Z, transmitted.
In other words,
Given ;
j R, a, r, a„ r,; ac = it for both pulleys, )
( i>,; andy for both pulleys ; f
Kequired • \ ^ ' ^' to f nrnisn ^ '■> &« * or no S ^P 5 v tne velocity
' I of O ; v' tlrat of belt ; and the tensions in belt.
FRICTION.
185
Neglecting axle-friction and the rigidity of the belting, the
power transmitted is that required to overcome R through a
distance = v t every second, i.e.,
Z = Rv, (1)
Since (if the belts do not slip)
a : r : : v' : v, and a l : r i : : v' : v t ,
we have
v — —v..
r.
and v = v..
a r, '
(2)
Neglecting the mass of the belt, and assuming that each pul-
ley revolves on a gravity-axis, we obtain the following, by con-
sidering the free bodies in Fig. 187 :
CA toe)
(B free)
Fig. 187.
(B aad f r_uck fr.e.S)
2 (moms.) = in A free gives Rr x = (S n — S )a t ;
2 (moms.) = in B free gives Gr = (S n — S )a ;
(3)
(4)
a= a -. r ^R.
r a.
whence we readily find
Evidently R and G are inversely proportional to their velo-
cities v t and v ; see (2). This ought to be true, since in Fig.
186 G is the only working-force, R the only resistance, and
the motions are uniform ; hence (from eq. (XVI.), § 142)
Gv - Rv, = 0.
2X = 0, for B and truck free, gives
G = S n + S , (5)
while, for impending slip,
8, = S a ef" (6)
186 MECHANICS OP ENGINEERING.
By elimination between (4), (5), and (6), we obtain
r _ ff 6 Jl±l - z ■ ** + 1 m
^ — a " «> - 1 _ x? ' ~ef" — V ' ' ' Kl)
L e?*
and S * — v'' e>— 1 ( 8 )
Hence 6 s , and S n vary directly as the power transmitted and
inversely as the velocity of the belt. For safety G should be
made > the above value in (7); corresponding values of the
two tensions may then be found from (5), and from the rela-
tion (see § 150)
{S n -S )v' = Z (6a)
These new values of the tensions will be found to satisfy the
condition of no slip, viz.,
(&:#.)<«*(§ 170).
For leather on iron, ef" = 2.2 (see example in § 170), as a
low value. The belt should be made strong enough to with-
stand S n safely.
As the belt is more tightly stretched, and hence elongated,
on the driving than on the following side, it " creeps" back-
ward on the driving and forward on the driven pulley, so that
the former moves slightly faster than the latter. The loss of
work due to this cause does not exceed 2 per cent with ordinary
belting (Cotterill).
In the foregoing it is evident that the sum of the tensions in
the two sides = 0„ i.e., is the same, whether the power is
being transmitted or not ; and this is found to be true, both in
theory and by experiment, when a tension-weight is not used,
viz., when an initial tension S is produced in the whole belt
before transmitting the power, then after turning on the latter
the sum of the two tensions (driving and following) always
= 2S, since one side elongates as much as the other contracts ;
it being understood that the pulley-axles preserve a constant
distance apart.
172. Rolling Friction. — The few experiments which have
been made to determine the resistance offered by a level road-
FEICTION. 187
way to the uniform motion of a roller or wheel rolling upon it
corroborate approximately the following theory. The word
friction is hardly appropriate in this connection (except when
the roadway is perfectly elastic, as will be seen), but is sanctioned
by usage.
First, let the roadway or track be compressible, but inelastic,
G the weight of the roller and its load, and P the horizontal
force necessary to preserve a uniform motion
(both of translation and rotation). The track
(or roller itself) being compressed just in
front, and not reacting symmetrically from
behind, its resultant pressure against the
roller is not at vertically under the centre,
but some small distance, OD = b, in front. (The successive
crushing of small projecting particles has the same effect.)
Since for this case of motion the forces have the same relations
as if balanced (see § 124), we may put 2 moms, about D = 0,
. .:Pr=Gb; or, P = h ~G (1)
Coulomb found for
Boilers of lignum-vitse on an oak track, b = 0.0189 inches;
Eollers of elm on an oak track, b = 0.0320 inches.
"Weisbach's experiments give, for cast-iron wheels 20 inches in
diameter on cast-iron rails,
b = 0.0183 inches ;
and Eittinger, for the same, b = 0.0193 inches.
Pambour gives, for iron railroad wheels 39.4 inches in diameter,
b = 0.0196 to
0.0216 inches.
According to the foregoing theory, P, the " rolling friction"
(see eq. (1)), is directly proportional to G, and inversely to the
radius, if b is constant. The experiments of General Morin and
others confirm this, while those of Dupuit, Poiree. and Sauvage
indicate it to be proportional directly to G, and inversely to the
square root of the radius.
188 MECHANICS OP ENGINEERING.
Although b is a distance to be expressed in linear units, and
not an abstract number like the f and f for sliding and axle-
friction, it is sometimes called a " coefficient of rolling fric-
tion." In eq. (1), b and r should be expressed in the same
unit.
Of course if P is applied at the top of the roller its lever-
arm about D is 2r instead of r, with a corresponding change
in eq. (1).
With ordinary railroad cars the resistance due to axle and
rolling frictions combined is about 8 lbs. per ton of weight on
a level track. For wagons on macadamized roads b = \ inch,
but on soft ground from 2 to 3 inches.
Secondly, when the roadway is perfectly elastic. This is
chiefly of theoretic interest, since at first sight no force would
be considered necessary to maintain a uniform rolling motion.
But, as the material of the roadway is compressed under the
roller its surface is first elongated and then recovers its former
state ; hence some rubbing and consequent sliding friction must
occur. Fig. 189 gives an exaggerated view of the circum-
stances, P being the horizontal force applied at the centre
necessary to maintain a uniform motion. The roadway (rub-
ber for instance) is heaped up both in front and behind the
roller, being the point of greatest pressure and elongation
of the surface. The forces acting are G, P, the normal
pressures, and the frictions due to them, and must form a
balanced system. Hence, since G and P, and also the normal
pressures, pass through G, the resultant of the frictions must
also pass through G; therefore the frictions, or tangential
actions, on the roller must be some forward and some backward
FRICTION.
189
(and not all in one direction, as seems to be asserted on p. 260
of Cotterill's Applied Mechanics, where Professor Eeynolds'
explanation is cited). The resultant action of the roadway
upon the roller acts, then, through some point D, a distance
OD = b ahead of 6>, and in the direction DC, and we have as
before, with D as a centre of moments,
JPr=Gb, or P= b -G.
r
If rolling friction is encountered above as
well as below the rollers, Fig. 190, the
student may easily prove, by considering
three separate free bodies, that for uniform
motion
* = '-£*« ■
where b and 5, are the respective " coefficients of rolling fric-
tion" for the upper and lower contacts.
~~^Examjple 1. — If it is found that a train of cars will move
uniformly down an incline of 1 in 200, gravity being the only
working force, and friction (both rolling and axle) the only
resistance, required the coefficient, f, of axle-friction, the
diameter of all the wheels being 2r = 30 inches, that of the
journals 2a = 3 inches, taking b = 0.02 inch for the rolling
friction. Let us use equation (XYI.) (§ 142), noting that while
the train moves a distance s measured on the incline, its weight
G does the work G ^-r s, the rolling friction - G (at the axles)
has been overcome through the distance s, and the axle-friction
(total) through the (relative) distance - s in the journal boxes ;
whence, the change in kinetic energy being zero,
~ Gs- l Gs- a -fGs = 0.
200 r r J
Gs cancels out, the ratios b : r and a : r are = -^-^ and -^
respectively (being ratios or abstract numbers they have the
190 MECHANICS OF ENGINEERING.
sawie numerical values, whether the inch or foot is used), and
solving, we have
/ = 0.05 - 0.0133 = 0.036.
Example 2. — How many pounds of tractive effort per ton
of load would the train in Example 1 require for uniform mo-
tion on a level track ?
173. Railroad Brakes. — During the uniform motion of a
railroad car the tangential action between the track and each
wheel is small. Thus, in Example 1, just cited, if ten cars of
eight wheels each make up the train, each car weighing 20 tons,
the backward tangential action of the rails upon each wheel is
only 25 lbs. "When the brakes are applied to stop the train
this action is much increased, and is the only agency by which
the rails can retard the train, directly or indirectly : directly,
when the pressure of the brakes is so great as to prevent the
wheels from turning, thereby causing them to "skid" (i.e.,
slide) on the rails ; indirectly, when the brake-pressure is of
such a value as still to permit perfect rolling of the wheel, in
which case the rubbing (and heating) occurs between the brake
and wheel, and the tangential action of the rail has a value
equal to or less than the friction of rest. In the first case,
then (skidding), the retarding influence of the rails is the fric-
tion of motion between rail and wheel ; in the second, a force
which may be made as great as the friction of rest between rail
and wheel. Hence, aside from the fact that skidding produces
objectionable flat places on the wheel-tread, the brakes are
more effective if so applied that skidding is impending, but
not actually prodnced ; for the friction of rest is usually greater
than that of actual slipping (§ 160). This has been proved
experimentally in England. The retarding effect of axle and
rolling friction has been neglected in the above theory.
Example 1. — A twenty-ton car with an initial velocity of 80
feet per second (nearly a mile a minute) is to be stopped on a
level within 1000 feet ; required the necessary friction on each
of the eight wheels.
Supposing the wheels not to skid, the friction will occur
FRICTION. 191
between the brakes and wheels, and is overcome through the
(relative) distance 1000 feet. Eq. (XYL), § 142, gives (foot-
lb.-second system)
-8^X1000=0- I ^f(80)',
from which F ( = friction at circumference of each wheel)
= 496 lbs.
Example 2. — Suppose skidding to be impending in the fore-
going, and the coefficient of friction of rest (i.e., impending
slipping) between rail and wheel to he f =0.20. In what
distance will the car be stopped ?
Example 3. — Suppose the car in Example 1 to be on an up-
grade of 60 feet to the mile. (In applying eq. (XVI.) here,
the weight 20 tons will enter as a resistance.)
Example 4. — In Example 3, consider all four resistances,
viz., gravity, rolling friction, and brake and axle frictions, the
distance being 1000 ft., and F the unknown quantity.
174. Estimation of Engine and Machinery Friction. — Accord-
ing to Professor Cotterill, a convenient way of estimating the
work lost in friction in a steam-engine and machinery driven
by it is the following :
Let fi m = mean effective steam-pressure per unit of area of
piston, and conceive this composed of three por-
tions, viz.,
2> a = the necessary pressure to drive the engine alone un-
loaded, at the proper speed ;
n' m = pressure necessary to overcome the resistance caused
by the useful work of the machines ;
ep' m = pressure necessary to overcome the friction of the
machinery, and that of the engine over and above
its friction when unloaded. This is about 15$ of
p' m (i.e., e = 0.15), except in large engines, and
then rather less.
That is, by formula, F being the piston-area and I the length
of stroke, the work per stroke is thus distributed :
FpJ = F[(l + ey m +_p -]l,
192 MECHANICS OF ENGINEERING.
p is " from 1 to 1J, or in marine engines 2 lbs. or more per
square inch."
175. Anomalies in Friction. — Experiment has shown consid-
erable deviation under certain circumstances from the laws of
friction, as stated in § 157 for sliding friction. At pressures
below J lb. per sq. inch the coefficient f increases when the
pressure decreases, while above 500 lbs. (Rennie, with iron and
steel) it increases with the pressure. With high velocities, how-
ever, above 10 ft. per second, f is mucli smaller as the velocity
increases (Bochet, 1858).
As for axle-friction, experiments instituted by the Society of
Mechanical Engineers in England (see the London Engineer
for March 1 and 21, 1884) gave values for f less than -j-J-g-
when a " bath" of the lubricant was employed. These values
diminished with increase of pressure, and increased with the
velocity (see below, Hirn's statement).
Professor Cotterill says, " It cannot be doubted that for
values of jpv (see § 166) > 5000 the coefficient of friction of
well-lubricated bearings of good construction diminishes with
the pressure, and may be much less than the value at low speeds
as determined by Morin" (p. 259 of his Applied Mechanics).
Professor Thurston's experiments confirmed those of Hirn as
to the following relation : "The friction of lubricated surfaces
is nearly proportional to the square root of the area and pres-
sure." Hirn also maintained that, "in ordinary machinery,
friction varies as the square root of the velocity."
176. Rigidity of Ropes. — If a rope or wire cable passes over
a pulley or sheave, a force P is required on one side greater
than the resistance Q on the other for uniform motion, aside
from axle-friction. Since in a given time both P and Q
describe the same space s, if P is > Q, then Psh > Qs, i.e.,
the work done by P is > than that expended upon Q. This
is because some of the work Ps has been expended in bending
the stiff rope or cable, and in overcoming friction between the
strands, both where the rope passes upon and where it leaves
FRICTION.
193
191, the material being
the pulley. "With hemp ropes, Fig
nearly inelastic, the energy spent in bending it on at D is
nearly all lost, and energy must also be spent in straightening
Fig. 191.
it at E; but with a wire rope or cable some of this energy is
restored by the elasticity of the material. The energy spent
in friction or rubbing of strands, however, is lost in both cases.
The figure shows geometrically why P must be > Q for a
uniform motion, for the lever-arm, a, of P is evidently < h
that of Q. If axle-friction is also considered, we must have
Pa=Qb+f{P+Q)r,
r being the radius of the journal.
Experiments with cordage have been made by Prony, Cou-
lomb, Eytelwein, and "Weisbach, with considerable variation in
the results and formulae proposed. (See Coxe's translation of
vol. i., Weisbach's Mechanics.)
With pulleys of large diameter the effect of rigidity is very
slight. For instance, "Weisbach gives an example of a pulley
five feet in diameter, with which, Q being = 1200 lbs., P
= 1219. A wire rope f in. in diameter was used. Of this
difference, 19 lbs., only 5 lbs. was due to rigidity, the remainder,
U lbs., being caused by axle-friction. When a hemp-rope .1.6
inches in diameter was substituted for the wire one, P—