UNIVERSITY OF CALIFORNIA
ANDREW
SMITH
HALLIDIE:
MECHANICS FOR ENGINEERS
MECHANICS FOR
ENGINEERS
A TEXT-BOOK OF INTERMEDIATE
STANDARD
BY
ARTHUR MORLEY
M.Sc., UNIVERSITY SCHOLAR (VICT.)
SENIOR LECTURER IN ENGINEERING IN UNIVERSITY COLLEGE
NOTTINGHAM
WITH 200 DIAGRAMS AND NUMEROUS EXAMPLES
LONGMANS, GREEN, AND CO.
39 PATERNOSTER ROW, LONDON
NEW YORK AND BOMBAY
1905
All rights reserved
PREFACE
ENGINEERING students constitute a fairly large proportion of
those attending the Mechanics classes in technical colleges
and schools, but their needs are not identical with those
of the students of general science. It has recently become
a common practice to provide separate classes in Mathematics,
adapted to the special needs of engineering students, who are
in most institutions sufficiently numerous to justify similar
provision in Mechanics. The aim of this book is to provide
a suitable course in the principles of Mechanics for engineering
students.
With this object in view, the gravitational system of
units has been adopted in the English measures. A serious
injustice is often done to this system in books on Mechanics
by wrongly denning the pound unit of force as a variable
quantity, thereby reducing the system to an irrational one.
With proper premises the gravitational system is just as rational
as that in which the " poundal " is adopted as the unit of
force, whilst it may be pointed out that the use of the latter
system is practically confined to certain text-books and exami-
nation papers, and does not enter into any engineering work.
Teachers of Engineering often find that students who are
learning Mechanics by use of the " poundal " system, fail to
apply the principles to engineering problems stated in the
only units which are used in such cases— the gravitational
<*3
141409
vi Preface
units. The use of the dual system is certainly confusing
to the student, and in addition necessitates much time being
spent on the re-explanation of principles, which might other-
wise be devoted to more technical work.
Graphical methods of solving problems have in some
cases been used, by drawing vectors to scale, and by esti-
mating slopes and areas under curves. It is believed that such
exercises, although often taking more time to work than the
easy arithmetic ones which are specially framed to give exact
numerical answers, compel the student to think of the relations
between the quantities involved, instead of merely performing
operations by fixed rules, and that the principles so illustrated
are more deeply impressed.
The aim has not been to treat a wide range of academic
problems, but rather to select a course through which the
student may work in a reasonable time — say a year — and
the principles have been illustrated, so far as the exclusion
of technical knowledge and terms would allow, by examples
likely to be most useful to the engineer.
In view of the applications of Mechanics to Engineering,
more prominence than usual has been given to such parts
of the subject as energy, work of forces and torques, power,
and graphical statics, while some other parts have received
less attention or have been omitted.
It is usual, in books on Mechanics, to devote a chapter
to the equilibrium of simple machines, the frictional forces
in them being considered negligible : this assumption is so far
from the truth in actual machines as to create a false impres-
sion, and as the subject is very simple when treated experi-
mentally, it is left for consideration in lectures on Applied
Mechanics and in mechanical laboratories.
The calculus has not been used in this book, but the
Preface vii
student is not advised to try to avoid it; if he learns the
elements of Mechanics before the calculus, dynamical illus-
trations of differentiation and integration are most helpful.
It is assumed that the reader is acquainted with algebra to
the progressions, the elements of trigonometry and curve
plotting; in many cases he will doubtless, also, though not
necessarily, have some little previous knowledge of Mechanics.
The ground covered is that required for the Intermediate
(Engineering) Examination of the University of London in
Mechanics, and this includes a portion of the work necessary
for the Mechanics Examination for the Associateship of the
Institution of Civil Engineers and for the Board of Education
Examination in Applied Mechanics.
I wish to thank Professor W. Robinson, M.E., and Pro-
fessor J. Goodman for several valuable suggestions made
with respect to the preparation and publication of this book;
also Mr. G. A. Tomlinson, B.Sc., for much assistance in
correcting proofs and checking examples; in spite of his
careful corrections some errors may remain, and for any
intimation of these I shall be obliged.
ARTHUR MORLEY.
. NOTTINGHAM,
June, 1905'.
CONTENTS
CHAPTER I
KINEMATICS
PAGES
Velocity ; acceleration ; curves of displacement, and velocity ;
falling bodies ; areas under curves ; vectors ; applications to
velocities ; relative velocity ; composition and resolution of
acceleration ; angular displacement, velocity, and acceleration 1-26
CHAPTER II
THE LAWS OF MOTION
First law ; inertia ; weight ; momentum ; second law ; engineers'
units ; c.g.s. system ; triangle and polygon of forces ; impulse ;
third law ; motion of connected bodies ; Atwood's machine 27-47
CHAPTER III
WORK, POWER, AND ENERGY
Work ; units ; graphical method ; power ; moment of a force ;
work of a torque ; energy — potential, kinetic ; principle of work 48-67
CHAPTER IV
MOTION IN A CIRCLE: SIMPLE HARMONIC MOTION
Uniform circular motion ; centripetal and centrifugal force ;
curved track ; conical pendulum ; motion in vertical circle ;
simple harmonic motion ; alternating vectors ; energy in
S.H. motion ; simple pendulum 68-90
x Contents
CHAPTER V
STA TICS—CONCURRENT FORCES— FRICTION
PAGES
Triangle and polygon of forces ; analytical methods ; friction ;
angle of friction ; sliding friction ; action of brakes ;
adhesion ; friction of screw 91-113
CHAPTER VI
STATICS OF RIGID BODIES
Parallel forces ; moments ; moments of resultants ; principle of
moments ; levers ; couples ; reduction of a coplanar system ;
conditions of equilibrium ; smooth bodies ; method of
sections ; equilibrium of three forces 114-139
CHAPTER VII
CENTRE OF INERTIA OR MASS— CENTRE OF GRAVITY
Centre of parallel forces ; centre of mass ; centre of gravity ; two
bodies ; straight rod ; triangular plate ; rectilinear figures ;
lamina with part removed ; cone ; distance of e.g. from
lines and planes ; irregular figures ; circular arc, sector,
segment ; spherical shell ; sector of sphere ; hemisphere . . 140-166
CHAPTER VIII
CENTRE OF GRAVITY— PROPERTIES AND APPLICATIONS
Properties of e.g. ; e.g. of distributed load; body resting on a
plane ; stable, unstable, and neutral equilibrium ; work
done in lifting a body ; theorems of Pappus 167-187
CHAPTER IX
MOMENTS OF INERTIA— ROTATION
Moments of inertia ; particles ; rigid body ; units ; radius of
gyration ; various axes ; moment of inertia of an area ;
circle ; hoop ; cylinder ; kinetic energy of rotation ; changes
in energy and speed ; momentum ; compound pendulum ;
laws of rotation ; torsional oscillation ; rolling bodies . . 188-222
Contents xi
CHAPTER X
ELEMENTS OF. GRAPHICAL STA TICS
PAGES
Bows' notation ; funicular polygon ; conditions of equilibrium,
choice of pole ; parallel forces ; bending moment and
shearing force ; diagrams and scales ; jointed structures ;
stress diagrams ; girders : roofs ; loaded strings and chains . 223-252
APPENDIX . . . . , 253-255
ANSWERS TO EXAMPLES 256-259
EXAMINATION QUESTIONS 260-273
MATHEMATICAL TABLES „..,.. 274-278
INDEX . , . „ . . ...... . .... •. 279-282
MECHANICS FOR ENGINEERS
CHAPTER I
KINEMA TICS
i. KINEMATICS deals with the motion of bodies without
reference to the forces causing motion.
MOTION IN A STRAIGHT LINE.
Velocity.— The velocity of a moving point is the rate of
change of its position.
Uniform Velocity. — When a point passes over equal
spaces in equal times, it is said to have a constant velocity ; the
magnitude is then specified by the number of units of length
traversed in unit time, e.g. if a stone moves 15 feet with a
constant velocity in five seconds, its velocity is 3 feet per
second.
If s = units of space described with constant velocity v in
t units of time, then, since v units are described in each second,
(v x t) units will be described in / seconds, so that —
s = vt
and v = -
Fig. i shows graphically the relation between the space
described and the time taken, for a constant velocity of 3 feet
per second. Note that v=- = - or - or -, a constant
Mechanics for Engineers
velocity of 3 feet per second whatever interval of time is
considered.
12
II
10
9
£ 6
I
1
0
X"
x
x
•^
^
/
-
x
x
J
X
x
x
^
^
x
X"
x
^
1234
Time in> seconds
FIG. i. — Space curve for a uniform velocity of 3 feet per second.
2. Mean Velocity. — The mean or average velocity of a
point in motion is the number of units of length described,
divided by the number of units of time taken.
3. Varying Velocity. — The actual velocity of a moving
point at any instant is the mean velocity during an indefinitely
small interval of time including that instant.
4. The Curve of Spaces or Displacements. — Fig. 2
shows graphically the relation between the space described and
01 T N M
Tifrue in, seconds
FIG. 2.— Space curve for a varying velocity.
the time taken for the case of a body moving with a varying
velocity. At a time ON the displacement is represented by
Kinematics 3
PN, and after an interval NM it has increased by an amount
QR, to QM. Therefore the mean velocity during the interval
OR QR A
NM is represented by -* or ^ or by tan QPR, i.e. by the
tangent of the angle which the chord PQ makes with a hori-
zontal line. If the interval of time NM be reduced indefi-
nitely, the chord PQ becomes the tangent line at P, and the
mean velocity becomes the velocity at the time ON. Hence
the velocity at any instant is represented by the gradient of the
tangent line to the displacement curve at that instant. An upward
slope will represent a velocity in one direction, and a down-
ward slope a velocity in the opposite direction.
5. If the curvature is not great, i.e. if the curve does not bend
sharply, the best way to find the direction of the tangent line
at any point P on a curve such as Fig. 2, is to take two ordi-
nates, QM and ST, at short equal distances from PN, and join
OV
QS; then the slope of QS, viz. -j~, is approximately the same
as that of the tangent at P. This is equivalent to taking the
velocity at P, which corresponds to the middle of the interval
TM, as equal to the mean velocity during the interval of
time TM.
6. Scale of the Diagram. — Measure the slope as the
gradient or ratio of the vertical height, say QV, to the hori-
zontal SV or TM. Let the ratio QV : TM (both being
measured in inches say) be x. Then to determine the velocity
represented, note the velocity corresponding to a slope of
T inch vertical to i inch horizontal, say y feet per second.
Then the slope of QS denotes a velocity of xy feet per
second.
7. Acceleration. — The acceleration of a moving body is
the rate of change of its velocity. When the velocity is in-
creasing the acceleration is reckoned as positive, and when
decreasing as negative. A negative acceleration is also called
a retardation.
8. Uniform Acceleration. — When the velocity of a point
increases by equal amounts in equal times, the acceleration is
said to be uniform or constant : the magnitude is then specified
Mechanics for Engineers
by the number of units of velocity per unit of time ; e.g. if a
point has at a certain instant a velocity of 3 feet per second,
and after an interval of eight seconds its velocity is 19 feet per
second, and the acceleration has been uniform, its magnitude is
increase of velocity 19 -3
tlrneTakelTto-i^rease =~ 8~ = 2 feet per S6COnd m each °f
the eight seconds, i.e. 2 feet per second per second. At the end
of the first, second, and third seconds its velocities would be
(3 + 2)5 (3 + 4), and (3 + 6) feet per second respectively (see
Fig- 3).
20
,0
•
0 1234-5676
Time, in- seconds
FIG. 3. — Uniform acceleration.
9. Mean Acceleration.— The acceleration from 3 feet
per second to 19 feet per second in the last article was sup-
posed uniform, 2 feet per second being added to the velocity
in each second; but if the acceleration is variable, and the
increase of velocity in different seconds is of different amounts,
then the acceleration of 2 feet per second per second during
the eight seconds is merely the mean acceleration during that
increase of velocity
time. The mean acceleration is equal to t^ ^en for increase;
and is in the direction of the change of velocity.
Kinematics
5
The actual acceleration at any instant is the mean
acceleration for an indefinitely small time including that
instant.
10. Fig. 3 shows the curve of velocity at every instant
during the eight seconds, during which a point is uniformly
accelerated from a velocity of 3 feet per second to one of
19 feet per second.
n. Calculations involving Uniform Acceleration. —
If ti = velocity of a point at a particular instant, and/" = uni-
form acceleration, i.e. f units of velocity are added every
second —
then after i second the velocity will be u +/
and „ 2 seconds „ „ // + 2/
}> » 3 '» " " ^ > 3y
„ / „ „ zMvillbefc+/' (i)
e.g. in the case of the body uniformly accelerated 2 feet per
second per second from a velocity of 3 feet per second to a
' velocity of 19 feet per second in eight seconds (as in Art. 8),
the velocity after four seconds is 3 + (2 x 4) = 1 1 feet per
second.
The space described (s) in t seconds may be found as
follows : The initial velocity being u, and the final velocity
being v, and the change being uniform, the mean or average
. . u 4- v
velocity is .
Mean velocity = — = - -f- £- — u -\- \ft
(which is represented by QM in Fig. 3. See also Art. 2).
Hence u + \ft'= j
and s= nt + ±ft2 ... (2)
e.g. in the above numerical case the mean velocity would be —
- = ii feet per second (QM in Fig. 3)
and j = ii X 8 = 88 feet
or s = 3 X 8 + l x 2 x 82 = 24 + 64 = 88 feet
It is sometimes convenient to find the final velocity in
6 Mechanics for Engineers
terms of the initial velocity, the acceleration, and the space
described. We have —
from (i) v = n -\-ft
therefore a* = if + *uft +ft* = ^ + 2/(ut + i//2)
and substituting for (ut + I//2) its value s from (2), we have—
tf = v* + 2/s (3)
The formulse (i), (2), and (3) are useful in the solution of
numerical problems on uniformly accelerated motion.
12. Acceleration of Falling Bodies. — It is found that
bodies falling to the earth (through distances which are small
compared to the radius of the earth), and entirely unresisted,
increase their velocity by about 32*2 feet per second every
second during their fall. The value of this acceleration varies
a little at different parts of the earth's surface, being greater
at places nearer to the centre of the earth, such as high lati-
tudes, and less in equatorial regions. The value of the
" acceleration of gravity " is generally denoted by the letter
g. In foot and second units its value in London is about
32*19, and in centimetre and second units its value is about
981 units.
13. Calculations on Vertical Motion. — A body pro-
jected vertically downwards with an initial velocity u will
in / seconds attain a velocity u + gft and describe a space
tf + fe*8.
In the case of a body projected vertically upward with a
velocity ?/, the velocity after / seconds will be u — gtt and will be
upwards if^/is less than ?/, but downward if gfis greater than
tf. When / is of such a value that gt = u, the downward
acceleration will have just overcome the upward velocity, and
the body will be for an instant at rest : the value of / will then be
-. The space described upward after / seconds will be
it - \/f.
The time taken to rise h feet will be given by the equation —
h = ut - i/?2
This quadratic equation will generally have two roots, the
Kinematics
smaller being the time taken to pass through h feet upward, and
the larger being the time taken until it passes the same point on
its way downward under the influence of gravitation.
The velocity v, after falling through " h " feet from the point
of projection downwards with a velocity #, is given by the
expression tf = ^ll + 2^, and if?/ = o, i.e. if the body be simply
dropped from rest, v* = 2gh, and v = \f 2gh after falling h feet.
14. Properties of the Curve of Velocities. — Fig. 4
shows the velocities at all times in a particular case of a body
FIG. 4.— Varying velocity.
starting from rest and moving with a varying velocity, the
acceleration not being uniform.
(i) Slope of the Curve. — At a time ON the velocity is
PN, and after an interval NM it has increased by an amount
QR to QM; therefore the mean acceleration during the
OR OR
interval NM is represented by or , i.e. by the tangent of
the angle which the chord PQ makes with a horizontal line.
If the interval of time NM be reduced indefinitely, the chord
PQ becomes the tangent line to the curve at P, and the mean
acceleration becomes the acceleration at the time ON. So that
the acceleration at any instant is represented by the gradient of
the tangent line at that instant. The slope will be upward if
the velocity is increasing, downward if it is decreasing ; in the
latter case the gradient is negative. The scale of accelerations
is easily found by the acceleration represented by unit gradient.
If the curve does not bend sharply, the direction of the
8
Mechanics for Engineers
tangent may be found by the method of Art. 5, which is in this
case equivalent to taking the acceleration at P as equal to the
mean acceleration during a small interval of which PN is the
velocity at the middle instant.
(2) The Area under the Curve.— If the velocity is
constant and represented by PN (Fig. 5), then the distance
described in an interval
NM is PN.NM, and there-
fore the area under PQ,
•v viz. the rectangle PQMN,
§ represents the space de-
scribed in the interval
NM.
If the velocity is not
constant, as in Fig. 6, sup-
M pose the interval NM
divided up into a number
of small parts such as
CD. Then AC represents the velocity at the time represented
by OC ; the- velocity is increasing, and therefore in the interval
CD the space described is greater than that represented by the
rectangle AEDC, and less than that represented by the rect-
angle FBDC. The total space described during the interval
N
Tim*,
FIG. 5.
Fi
-^f
— -
^
S
A
E
%x\
/
<0
<^_
0
N CD M
FIG. 6.— Varying velocity.
NM is similarly greater than that represented by a series of
rectangles such as AEDC, and less than that represented by a
series of rectangles such as FBDC. Now, if we consider
the number of rectangles to be increased indefinitely, and
Kinematics
the width of each to be decreased indefinitely, the
PQMN under the curve PQ is the area which lies always
between the sums of the
areas of the two series of
rectangles, however nearly
equal they may be made by
subdividing NM, and the
area PQMN under the curve
therefore represents the space
described in the interval NM.
The area under the curve
is specially simple in the case
of uniform acceleration, for
which the curve of veloci-
ties is a straight line (Fig. 7).
Here the velocity PN being
?/, and NM being / units of time, and the final velocity being
QM =-z>, the area under^PQ is —
V' OF THE
UNIVERSITY
M
FIG. 7.
2
or - x / (as in Art. n)
— u
QR
And if / is the acceleration / = — j— (represented by ~-r-^ or
2|, i.e. by tan QPR),
/.// — v - u
v = n +ft
and the space described - - X / is — X /, which is
2 2
ut -\- J//2 (as in Art. n).
15. Notes on Scales. — If the scale of velocity is i inch
to x feet per second, and the scale of time is i inch to y seconds,
then the area under the curve will represent the distance
described on such a scale that i square inch represents xy
feet.
1 6. In a similar way we may show that the area PQMN
10
Mechanics for Engineers
(Fig. 8) under a curve of accelerations represents the total
increase in velocity in the interval of time NM.
Tinue,
FIG. 8.
M
If the scale of acceleration is i inch to z feet per second
per second, and the scale of time is i inch to y seconds, then
the scale of velocity is i square inch to yz feet per second.
17. Solution of Problems. — Where the motion is of a
simple kind, such as a uniform velocity or uniform acceleration,
direct calculation is usually the easiest and quickest mode of
solution, but where (as is quite usual in practice) the motion is
much more complex and does not admit of simple mathematical
expression as a function of the time taken or distance covered,
a graphical method is recommended. Squared pap'er saves
much time in plotting curves for graphical solutions.
Example i. — A car starting from rest has velocities v feet per
second after t seconds from starting, as given in the following
table :—
/
•v
o
o
4
II'O
£
^
24
44'5
30
49*o
48-9
40
40*6
45
337
26-8
58
24-3
62
24*0
Find the accelerations at all times during the first 60 seconds, and
draw a curve showing the accelerations during this time.
First plot the curve of velocities on squared paper from the
given data, choosing suitable scales. This has been done in
Kinematics
ii
Fig. 9, curve I., the scales being i inch to 10 seconds and
i inch to 20 feet per second.
In the first 10 seconds RQ represents 24-2 feet per second
f 50
£*°
1»
S20
J W
7
Ifi ty 20
-64
30*4
4-0
50
§0
Scale, of Jjvches
FIG. 9.
-2
gain of velocity, and OQ represents 10 seconds; therefore the
acceleration at N 5 seconds from starting is approximately
24*2
--- , or 2*42 feet per second per second. Or thus: unit
gradient i inch vertical in i inch horizontal represents —
20 feet per second
= 2 feet per second Per second
hence =r2lnc =1-21
OQ i inch
hence acceleration at N 1
ic T-OT v o r = 2'42 fe£t Per second per second
lo L 2 L /{ 2
(see Art. 14)
Similarly in the second 10 seconds, SV which is SM — RQ,
represents (39*8 — 24-2), or 15*6 feet per second gain of velocity ;
12
Mechanics for Engineers
therefore the velocity at W 15 seconds from starting is approxi-
mately ^— ,or 1*56 feet per second per second.
10
Continue in this way, finding the acceleration at say 5, 15,
25) 35> 45) and 55 seconds from starting; and if greater ac-
curacy is desired, at 10, 20, 30, 40, 50, and 60 seconds also.
The simplest way is to read off from the curve I. velocities
in tabular form, and by subtraction find the increase, say, in
10 seconds, thus —
/
V
Change in v
for 10 sees.
Acceleration
o
0
\
i3'5
s.X
24-2
2^2
IO
24-2
15
32-8
20
3g'8
12-6
*.
25
45*4
15"
o-92
3°
*3'°
2'1
0'2I
35
47'5
^sT
-0-84
40
40 '6
45
337
50
29-0
- ^
-8-6
-0-86
55
—>'^-
^5^"
-0-51
60
24-1
^
19-3
i '93
15-6
1-56
-13-8
-1-38
-H-6
-,*
From the last line in this table curve II., Fig. 9, has been
plotted, and the acceleration at any instant can be read off
from it.
It will be found that the area under curve II. from the
start to any vertical ordinate is proportional to the correspond-
ing ordinate of curve I. (see Art. 16). The area, when below
the time base-line, must be reckoned as negative.
Example 2. — Find the distance covered from the starting-point
by the car in Example I at all times during the first 60 seconds,
and the average velocity throughout this time.
In the first 10 seconds, the distance covered is found approxi-
mately by multiplying the velocity after 5 seconds by the time, i.e.
13-5 x 10 = 135 feet. This approximation is equivalent to taking
I3'5 feet per second as the mean velocity in the first 10 seconds.
In the next 10 seconds the mean velocity being approximately
32'8 feet per second (corresponding to / = 15 seconds), the distance
covered is 32*8 x 10 — 328 feet, therefore the total distance covered
in the first 20 seconds is 135 + 328 = 463 feet. Proceeding in this
way, taking lo-second intervals throughout the 60 seconds, and
using the tabulated results in Example i, we get the following
results : —
Kinematics
1 '
/ I o
10
20
3^
40
5o
60
Space in j
previous >
0
135
328
454 475
337
251
10 sees. )
Total space
0
135
463
917
1392
1729
1980
from which the curve of displacements, Fig. 10, has been plotted.
2000
1500
IOOO
500
IO 20 30 4-0
Time in seconds
123 +
1,1,1,1
5O
60
Scale of Inches
FIG. 10.
Greater accuracy may be obtained by finding the space
described every 5 instead of every 10 seconds.
The average velocity = space described = 1980 = feet sec
time taken 60
Note that this would be represented on Fig. 9 by a height which is
equal to the total area under curve I. divided by the length of
base to 60 seconds. rc>
EXAMPLES I.
i. A train attains a speed of 50 miles per hour in 4 minutes after starting
from rest. Find the mean acceleration in foot and second units.
14 Mechanics for Engineers
2. A motor car, moving at 30 miles per hour, is subjected to a uniform
retardation of 8 feet per second per second by the action of its brakes.
How long will it take to come to rest, and how far will it travel during this
time?
3. With what velocity must a stream of water be projected vertically
upwards in order to reach a height of 80 feet ?
4. How long will it take for a stone to drop to the bottom of a well
150 feet deep?
5. A stone is projected vertically upward with a velocity of 170 feet per
second. How many feet will it pass over in the third second of its upward
flight ? At what altitude will it be at the end of the fifth second, and also
at the end of the sixth ?
6. A stone is projected vertically upward with a velocity of 140 feet per
second, and two seconds later another is projected on the same path with
an upward velocity of 135 feet per second. When and where will they
meet?
7. A stone is dropped from the top of a tower 100 feet high, and at the
same instant another is projected upward from the ground. If they meet
halfway up the tower, find the velocity of projection of the second stone.
The following Examples are to be worked graphically.
8. A train starting from rest covers the distances s feet in the times /
seconds as follows : —
0
5
II
18
22
27
31
38
46 ( 50
o
10
54
170
260
390
45°
504
550 | 570
1
1
Find the mean velocity during the first 10 seconds, during the first 30
seconds, and during the first 50 seconds. Also find approximately the
actual velocity after 5, 15, 25, 35, and 45 seconds from starting-point, and
plot a curve showing the velocities at all times.
9. Using the curve of velocities from Example 8, find the acceleration
every 5 seconds, and draw the curve of accelerations during the first 40
seconds.
10. A train travelling at 30 miles per hour has steani shut off and
brakes applied ; its speed after / seconds is shown in the following table : —
t
o
4
12
ZO
26
35
42
5°
*' ho™*65 PCr}' 3°'°
26-0
21'5
16-7
I4'O
10-4
77
4-8
1
Kinematics
Find the retardation in foot and second units at 5-second intervals through-
out the whole period, and show the retardation by means of a curve. Read
off from the curve the retardation after 7 seconds and after 32 seconds.
What distance does the train cover in the first 30 seconds after the brakes
are applied ?
ii. A body is lifted vertically from rest, and is known to have the
following accelerations / in feet per second per second after times /
seconds : —
/ ...
r
1
o 0-8
3-0 2-9
1*9
2-85
3-0
2-60
3'9
2 '20
4-8
i'75
6-0
I-36
6-8
1-40
8-0
1-04
8-8
0-97
Find its velocity after each second, and plot a curve showing its velocity at
all times until it has been in motion 8 seconds. How far has it moved in
the 8 seconds, and how long does it take to rise 12 feet ?
VECTORS.
1 8. Many physical quantities can be adequately expressed
by a number denoting so many units, e.g. the weight of a
body, its temperature, arid its value. Such quantities are called
scalar quantities.
Other quantities cannot be fully represented by a number
only, and further information is required, e. g. the velocity of a
ship or the wind has a definite direction as well as numerical
magnitude : quantities of this class are called vector quantities
and are very conveniently represented by vectors.
A Vector is a straight line having definite length and
direction, but not definite position in space.
19. Addition of Vectors. — To find the sum of two vectors
FIG. IT.
db and cd (Fig. n), set out ab of proper length and direction,
and from the end b set out be equal in length and parallel to
i6
Mechanics for Engineers
cd ; join ae. Then ae is the geometric or vector sum of ab and
cd. We may write this —
* ah
= ae
FIG. 12.
or, since be is equal to cd —
ab -f- cd = ae
20. Subtraction of Vectors. — If the vector cd (Fig. 12)
-r is to be subtracted from
the vector ab, we simply
find the sum ae as before,
of a vector ab and second
vector be, which is equal
to cd in magnitude, but is
of opposite sign or direc-
tion ; then —
ae = ab + be = ab — cd
If we had required the difference, cd — ab, the result would
have been ea instead of ae.
21. Applications: Displacements. — A vector has the
two characteristics of a displacement, viz. direction and magni-
tude, and can, therefore, represent it completely. If a body
receives a displacement ab (Fig. n), and then a further dis-
placement completely represented by cd, the total displacement
is evidently represented by ae in magnitude and direction.
22. Relative Displacements. CASE I. Definition. — If
7 a body remains at
rest, and a second
body receives a dis-
placement, the first
body is said to re-
ceive a displacement
of equal amount but
opposite direction re-
la five to the secorid.
CASE II. Where Two Bodies each receive a Displacement. —
If a body A receive a displacement represented by a vector ab
(Fig. 13), and a body B receive a displacement represente'd by
FIG. 13.
Kinematics 1 7
ed, then the displacement of A relative to B is the vector
difference, ab — cd. For if B remained at rest, A would have
a displacement ab relative to it. But on account of B's motion
(cd), A has, relative to B, an additional displacement, dc (Case
I.) ; therefore the total displacement of A relative to B is ab -f- dc
(or, ab —cd) = ab + be = ae (by Art. 20) ; where be is of equal
length and parallel to dc.
23. A Velocity which is displacement per unit time can
evidently be represented fully by a vector ; in direction by the
clinure of the vector, and in magnitude by the number of units
of length in the vector.
24. Triangle and Polygon of Velocities. — A velocity is
said to be the resultant of two others, which are
called components, when it is fully represented
by a vector which is the geometrical sum of two
other vectors representing the two components ;
e.g. if a man walks at a rate of 3 miles per
hour across the deck of a steamer going at 6
miles per hour, the resultant velocity with which
the man is moving over the sea is the vector
sum of 3 and 6 miles per hour taken in the proper
directions. If the steamer were heading due FIG. X4.
north, and the man walking due east, his actual velocity is
shown by ac in Fig. 14 ;
ab — 6 be - 3
ac = ^6a + 32 = A/45 miles per hour
= 671 miles per hour
and the angle 6 which ac makes with ab E. of N. is given by —
tan 0 = % = i 0 = 26° 35'
Resultant velocities may be found by drawing vectors to
scale or by the ordinary rules of trigonometry. If the re-
sultant velocity of more than two components (in the same
plane) is required, two may be compounded, and then a third
with their resultant, and so on, until all the components have
been added. It will be seen (Fig. 15) that the result is repre-
sented by the closing side of an open polygon the sides of
c
1 8 Mechanics for Engineers
which are the component vectors. The order in which the
sides are drawn is immaterial. It is not an essential condition
that all the components should be in the same plane, but if
not, the methods of solid geometry should be employed to
draw the polygon.
Fig. 15 shows the resultant vector af of five co-planor
vectors, ab, be, cd, de, and ef.
If, geometrically, ac = ab -f be
and ad = ac + cd
then ad = ab + be -j- cd
and similarly, adding de and ef- —
af = ab -f be -f cd -f de + ef
In drawing this polygon it is unnecessary to put in the
lines ac, ad, and ae.
25. It is sometimes convenient to resolve a velocity into
two components, i.e. into two other velocities in particular
directions, and such that their vector sum is equal to that
velocity.
Rectangular Components. — The most usual plan is to
resolve velocities into components in two standard directions
at right angles, and in the same plane as the original veloci-
ties : thus, if OX and OY (Fig. 16) are the standard directions,
and a vector ab represents a velocity z/, then the component in
the direction OX is represented by ac, which is equal to ab
cos 6, and represents v cos 0, and that in the direction OY is
represented by cl>3 i.e. by ab sin 0, and is z> sin B.
Kinematics 19
This form of resolution of velocities provides an alternative
method of finding the re- y
sultant of several velocities.
Each velocity may be re-
solved in two standard
directions, OX and OY,
and then all the X com-
ponents added algebraically
and all the Y components
added algebraically. This
reduces the components to
two at right angles, which
may be replaced by a re-
FIG. 16.
sultant R units, such that
the squares of the numerical values of the rectangular com-
ponents is equal to the square of R, e.g. to find the resultant
FIG. 17.
of three velocities V1? Y.2, and V3, making angles a, /3, and y
respectively with some fixed direction OX in their plane
(Fig. 17).
Resolving along OX, the total X component, say X, is —
X = Va cos a + V2 cos /3 + V., cos y
Resolving along OY —
Y = Vt sin « + V., sin /3 + V, sin y
and R2 = X'2 + Y^_
orR = fX*"+ Y2
and it makes with OX an angle 6 such that tan 6 = -.
X
20
Mechanics for Engineers
Fig. 1 7 merely illustrates the process ; no actual drawing
of vectors is required, the method being wholly one of calcu-
lation.
Exercise i.— A steamer is going through the water
at 10 knots per hour, and heading due north. The
current runs north-west at 3 knots per hour. Find
the true velocity of the steamer in magnitude and
direction.
(i) By drawing vectors (Fig. 18).
Set off ab, representing 10 knots per hour, to scale
due north. Then draw be inclined 45° to the direction
ab, and representing 3 knots per hour to the same
scale. Join ac. Then ac, which scales 12*6 knots per
hour when drawn to a large scale, is the true velocity,
^ and the angle cab E of N measures 10°.
FIG. 18. (2) Method by resolving N. and E.
N. component = 10 + 3 cos 45° = 10 + -^ knots per hour,
Or I2'I2
E. „ =3 sin 45° = -4- knots per hour, or 2*12
v 2
Resultant velocity R = >v/(i2'i2)2 + (2'i2)2 = 12 '6 knots per hour
And if 6 is the) 3 / , J$\ 2'I2
angle E.ofN.ran^7"2^\IO+^^^2=75
.'. e = 9° 55'
RELATIVE VELOCITY.
26. The velocity of a point A relative to a point B is the
rate of change of position (or displacement per unit of time)
of A with respect to B.
Let v be the velocity of A, and u that of B.
If A remained stationary, its displacement per unit time
relative to B would be —u (Art. 22). But as A has itself a
velocity v, its total velocity relative to B is v -f ( — ») or v — ?/,
the subtraction to be performed geometrically (Art. 20).
The velocity of B relative to A is of course u — v, equal in
magnitude, but opposite in direction. The subtraction of
velocity v — u may be performed by drawing vectors to scale,
Kinematics
21
by the trigonometrical rules for the solution of triangles, or by
the method of Art. 25.
Example. — Two straight railway lines cross : on the first a
train 10 miles away from the crossing, and due west of it, is ap-
proaching at 50 miles per hour ; on the second a train 20 miles
away, and 15° E. of N., is approaching at 40 miles per hour.
How far from the crossing will each train be when they are nearest
together, and how long after they occupied the above positions?
First set out the two lines at the proper angles, as in the left side
of Fig. 19, and mark the positions A and B of the first and second
FIG. 19.
trains respectively. Now, since the second train B is coming
from 15° E. of N., the first train A has, relative to the second, a
component velocity of 40 miles per hour in a direction E. of N.,
in addition to a component 50, miles per hour due east. The
relative velocity is therefore found by adding the vectors pq 50
miles per hour east, and qr 40 miles per hour, giving the vector pry
which scales 72 miles per hour, and has a direction 57^° E. of N.
Now draw from A a line AD parallel to pr. This gives the posi-
tions of A relative to B (regarded as stationary). The nearest
approach is evidently a distance BD, where BD is perpendicular
to AD. The distance moved by A relative to B is then AD, which
scales 23'2 miles (the trains being then a distance BD, which scales
8*12 miles apart). The time taken to travel relatively 23*2 miles
• • . 23*2
at 72 miles per hour is -• hours = 0*322 hour.
22
Mechanics for Engineers
Hence A will have travelled 50 x 0^322 or 16*1 miles
and B „ „ 40 x 0-322 or 12-9 „
A will then be 6'i miles past the crossing, and
B „ „ 7*1 „ short of the crossing.
FIG. 20.
27. Composition, Resolution, etc., of Accelerations.
— Acceleration being also a vector quantity, the methods of
composition, resolution, etc., of velocities given in Arts. 23
to 26 will also apply to acceleration, which is simply velocity
added per unit of time. It should be noted
that the acceleration of a moving point is not
necessarily in the same direction as its velocity :
this is only the case when a body moves in a
straight line.
If db (Fig. 20) represents the velocity of a
point at a certain instant, and after an interval /
seconds its velocity is represented by ac> then
the change in velocity in / seconds is be, for
ab + be = ac (Art. 19), and be = ac — ab (Art.
20), representing the change in velocity. Then
during the / seconds the mean acceleration is represented by
be 4- /, and is in the direction be.
28. Motion down a Smooth Inclined Plane.— Let a
be the angle of the plane to the horizontal, then the angle
ABC (Fig. 21) to the vertical is (90° - a). Then, since a
body has a downward ver-
tical acceleration g, its
component along BA will
A.
be g cos CBA = g cos
(90° — a) = g sin a, pro-
vided, of course, that there
is nothing to cause a re-
tardation in this direction,
i.e. provided that the plane is perfectly smooth and free from
obstruction. If BC = h feet, AB = h cosec a feet. The
velocity of a body starting from rest at B and sliding down
AB will be at A, \f 2 . g sin 0 X h cosec 6 = V 2g/i, just as if it
had fallen h feet vertically.
FIG. 21.
Kinematics
FIG. 22.
2p. Angular Motion : Angular Displacement. — If
P (Fig. 22) be the posi- p
tion of a point, and Q
a subsequent position
which this point takes
up, then the angle QOP
is the angular displace-
ment of the point about
O. The angular displacement about any other point, such as
O', will generally be a different amount.
30. Angular Velocity. — The angular velocity of a moving
point about some fixed point is the rate of angular displacement
(or rate of change of angular position) about the fixed point ; it
is usually expressed in radians per second, and is commonly
denoted by the letter w. As in the case of linear velocity, it
may be uniform or varying.
A point is said to have a uniform or constant angular
velocity about a point O when it describes equal angles about O
in equal times. The mean angular velocity of a moving point
about a fixed point O is the angle described divided by the
time taken.
If the angular velocity is varying, the actual angular velocity
at any instant is the mean angular velocity during an in-
definitely small interval including that instant.
31. Angular Acceleration is the rate of change of
angular velocity ; it is usually measured in radians per second
per second.
32. The methods of Arts. 4 to n
applicable to angular motion as well as
to linear motion.
33. To find the angular velocity about
O of a point describing a circle of radius
r about O as centre with constant speed.
Let the path PP' (Fig. 23) be de-.
scribed by the moving point in t seconds.
Let v be the velocity (which, although FlG- 23-
constant in magnitude, changes direction). Then angular
c\
velocity about O is w ±= -.
and
to 1 6 are
24
Mechanics for Engineers
But & = - and arc PP' = vt
r
Vt 6 7't V
:. 0 = - and <o = - = -f- / = -
r t r r
This will still be true if O is moving in a straight line with
velocity v as in the case of a rolling wheel, provided that v
is the velocity of P relative to O.
If we consider / as an indefinitely small time, PP' will
be indefinitely short, but the same will remain true, and we
should have <o = - whether the velocity remains constant in
magnitude or varies.
In words, the angular velocity is equal to the linear
velocity divided by the radius, the units of length being the
same in the linear velocity v and the radius /-.
Example. — The cranks of a bicycle are 6\ inches long, and the
bicycle is so geared that one complete rotation of the crank carries
it through a distance equal to the circumference of a wheel 65
inches diameter. When the bicycle is driven at 1 5 miles per hour,
find the absolute velocity of the centre of a pedal — (i) when
vertically above the crank axle ; (2) when vertically below it ;
(3) when above the axle and 30° forward of a vertical line
through it.
The pedal centre describes a circle of 13 inches diameter
relative to the crank axle, i.e. 13* inches, while the bicycle travels
6$ir inches. Hence the velocity of the pedal centre relative to the
crank axle is £ that of the bicycle along the road, or 3 miles per
hour,
15 miles per hour = 22 feet per second
3
= 4'4
(D
Kinematics 25
(1) When vertically above the crank axle, the velocity of pedal
is 22 + 4*4 = 26-4 feet per second.
(2) When vertically below the crank axle, the velocity of pedal
is 22 — 4*4 =17*6 feet per second.
(3) Horizontal velocity X = 22 + 4*4 cos 30° = 22 + 2'2 */$ feet
per second.
Vertical velocity downwards Y = 4*4 x sin 30° = 2 '2 feet per
second.
Resultant velocity being R —
R = 22*S(i + ^\ + f -M2 = 25-8 feet per second
and its direction is at an angle 6 below the horizontal, so that —
2-2 i i
: — — = 0-0852
EXAMPLES II.
1. A point in the connecting rod of a steam engine moves forwards
horizontally at 5 feet per second, and at the same time has a velocity of 3
feet per second in the same vertical plane, but in a direction inclined 1 10°
to that of the horizontal motion. Find the magnitude and direction of its
actual velocity.
2. A stone is projected at an angle of 36° to the horizontal with a
velocity of 500 feet per second. Find its horizontal and vertical velocities.
3. In order to cross at right angles a straight river flowing uniformly at
2 miles per hour, in what direction should a swimmer head if he can
get through still water at 24 miles per hour, and how long will it take him
if the river is 100 yards wide ?
4. A weather vane on a ship's mast points south-west when the ship is
steaming due west at 16 miles per hour. If the velocity of the wind is
20 miles per hour, what is its true direction ?
5. Two ships leave a port at the same time, the first steams north-west
at 15 knots per hour, and the second 30° south of west at 17 knots.
What is the speed of the second relative to the first ? After what time
will they be loo knots apart, and in what direction will the second lie from
the first ?
6. A ship steaming due east at 12 miles per hour crosses the track
of another ship 20 miles away due south and going due north at 16 miles
per hour. After what time will the two ships be a minimum distance apart,
and how far will each have travelled in the interval.
26 Mechanics for Engineers
7. Part of a machine is moving east at 10 feet per second, and after ^
second it is moving south-east at 4 feet per second. What is the amount
and direction of the average acceleration during the ^ second ?
8. How long will it take a body to slide down a smooth plane the
length of which is 20 feet, the upper end being 3*7 feet higher than the
lower one.
9. The minute-hand of a clock is 4 feet long, and the hour-hand is
3 feet long. Find in inches per minute the velocity of the end of the
minute-finger relative to the end of the hour-hand at 3 o'clock and at
12 o'clock.
10. A crank, CB, is I foot long and makes 300 turns clockwise per
minute. When CB is inclined 60° to the line CA, A is moving along AC
at a velocity of 32 feet per second. Find the velocity of the point B rela-
tive to A.
11. If a motor car is travelling at 30 miles per hour, and the wheels
are 30 inches diameter, what is their angular velocity about their axes ? If
the car comes to rest in 100 yards under a uniform retardation, find the
angular retardation of the wheels.
12. A flywheel is making 180 revolutions per minute, and after 20 seconds
it is turning at 140 revolutions per minute. How many revolutions will it
make, and what time will elapse before stopping, if the retardation is
uniform ?
CHAPTER II
THE LAWS OF MOTION
34. NEWTON'S Laws of Motion were first put in their present
form by Sir Isaac Newton, although known before his time.
They form the foundation of the whole subject of dynamics.
35. First Law of Motion. — Every body continues in its
state of rest or uniform motion except in so far as it may be com-
pelled by force to change that state.
We know of no case of a body unacted upon by any force
whatever, so that we have no direct experimental evidence of
this law. In many cases the forces in a particular direction
are small, and in such cases the change in that direction is
small, e.g. a steel ball rolling on a horizontal steel plate. To
such instances the second law is really applicable.
From the first law we may define force as that which tends
to change the motion of bodies either in magnitude or direction.
36. Inertia. — It is a matter of everyday experience that
some bodies take up a given motion more quickly than others
under the same conditions. For example, a small ball of iron
is more easily set in rapid motion by a given push along a
horizontal surface than is a large heavy one. In such a case
the larger ball is said to have more inertia than the small one.
Inertia is, then, the property of resisting the taking up of
motion.
37. Mass is the name given to inertia when expressed as
a measurable quantity. The more matter there is in a body
the greater its mass. The mass of a body depends upon its
volume and its density being proportional to both. We may
define density of a body as being its mass divided by its
volume, or mass per unit volume in suitable units.
28 Mechanics for Engineers
If ;;/ = the mass of a body,
v = its volume,
and p = its density,
then p = "L
V
A common British unit of mass is one pound. This is
often used in commerce, and also in one absolute system
(British) of mechanical units ; but we shall find it more con-
venient to use a unit about 32*2 times as large,? for reasons
to be stated shortly. This unit has no particular name in
general use. It is sometimes called the gravitational unit of
mass, or the " engineer's unit of mass."
In the c.g.s. (centimetre-gramme-second) absolute system,
the unit mass is the gramme, which is about Ib.
453'6
38. The weight of a body is the force with which the
earth attracts it. This is directly proportional to its mass, but
is slightly different at different parts of the earth's surface.
39. Momentum is sometimes called the quantity of
motion of a body. If we consider a body moving with a
certain velocity, it has only half as much motion as two
exactly similar bodies would have when moving at that
velocity, so that the quantity of motion is proportional to the
quantity of matter, i.e. to the mass. Again, if we consider the
body moving with a certain velocity, it has only half the quantity
of motion which it would have if its velocity were doubled, so
that the quantity of motion is proportional also to the velocity.
The quantity of motion of a body is then proportional to
the product (mass) X (velocity), and this quantity is given
the name momentum. The unit of momentum is, then, that
possessed by a body of unit mass moving with unit velocity.
It is evidently a vector quantity, since it is a product of
velocity, which is a vector quantity, and mass, which is a scalar
quantity, and its direction is that of the velocity factor. It
can be resolved and compounded in the same way as can
velocity.
40. Second Law of Motion. — The rate of change of
The Laws of Motion 29
momentum is proportional to the force applied, and takes place in
the direction of the straight line in which the force acts. This
law states a simple relation between momentum and force, and,
as we have seen how momentum is measured, we can proceed
to the measurement of force.
The second law states that if F represents force —
F oc rate of change of (m X v)
where m — mass, v — velocity ;
therefore Fa m X (rate of change v\ if m remains constant
or F oc m X f
where /= acceleration,
and/ oc -
m
where F is the resultant force acting on the mass m ;
hence F = m x f X a constant,
and by a suitable choice of units we may make the constant
unity, viz. by taking as unit force that which gives unit mass
unit acceleration. We may then write —
force = (mass) x (acceleration)
or F = m X /
If we take i Ib. as unit mass, then the force which gives
i Ib. an acceleration of i foot per second per second is called
the poundal. This system of units is sometimes called the
absolute system^ This unit of force is not in general use with
engineers and others concerned in the measurement and calcu-
lation of force and power, the general practice being to take
the weight of i Ib. at a fixed place as the unit of force. We
call this a force of i Ib., meaning a force equal to the weight
of i Ib. As mentioned in Art. 38, the weight of i Ib. of
matter varies slightly at different parts of the earth's surface,
but the variation is not of great amount, and is usually negligible.
1 The gravitational system is also really an absolute system, inasmuch as
all derived units are connected to the fundamental ones by fixed physical
relations. See Appendix.
30 Mechanics for Engineers
41. Gravitational or Engineer's Units.— One pound
of force acting on i Ib. mass of matter (viz. its own weight)
in London1 gives it a vertical acceleration of about 32*2 feet
per second per second, and since acceleration = -, i Ib. of
mass
force will give an acceleration of i foot per second per second
(i.e. 32*2 times less), if it acts on a mass of 32*2 Ibs. Hence,
if we wish to have force defined by the relation —
force = rate of change of momentum,
or force = (mass) x (acceleration)
F = m Xf
we must adopt g Ibs. as our unit of mass, where g is the
acceleration of gravity in feet per second per second in some
fixed place; the number 32^2 is correct enough for most
practical purposes for any latitude. This unit, as previously
stated, is sometimes called the engineers' unit of mass.
Then a body of weight w Ibs. has a mass of W- units
g
and the equation of Art. 40 becomes F = — x /.
Another plan is to merely adopt the relation, force = (mass)
X (acceleration) x constant. The mass is then taken in
pounds, and if the force is to be in pounds weight (and not in
poundals) the constant used is g (32*2). There is a strong
liability to forget to insert the constant g in writing expressions
for quantities involving force, so we shall adopt the former plan
of using 32*2 Ibs. as the unit of mass. The unit of momentum
is, then, that possessed by 32*2 Ibs. moving with a velocity of
i foot per second, and the unit force the weight of i Ib. The
number 32*2 will need slight adjustment for places other than
London, if very great accuracy should be required.
Defining unit force as the weight of i Ib. of matter, we
may define the gravitational unit of mass as that mass which
has unit acceleration under unit force.
42. C.G.S. (centimetre-gramme-second) Units. — In
this absolute system the unit of mass is the gramme ; the
1 The place chosen is sometimes quoted as sea-level at latitude 45°.
The Laws of Motion 31
unit of momentum that in i gramme moving at i centimetre
per second; and the unit of force called the dyne is that
necessary to accelerate i gramme by i centimetre per second
per second. The weight of i gramme is a force of about
981 dynes, since the acceleration of gravity is about 981 centi-
metres per second per second (981 centimetres being equal to
about 32*2 feet).
The weight of one kilogram (1000 grammes) is often used
by Continental engineers as a unit of force.
Example i. — A man pushes a truck weighing 2*5 tons with a
force of 40 Ibs., and the resistance of the track is equivalent to
a constant force of 10 Ibs. How long will it take to attain a
velocity of 10 miles per hour? The constant effective forward
force is 40 — 10 = 30 Ibs., hence the acceleration is —
force 2-5 x 2240
Tr^ss = 3° -* -— 2 — = 0-1725 foot per second per second
10 miles per hour = -8/ or -*/ feet per second
The time to generate this velocity at 0-1725 foot per second per
second is then \4 -f- 0*1725 = 85 seconds, or i minute 25 seconds.
Example 2. — A steam-engine piston, weighing 75 Ibs., is at
rest, and after 0-25 second it has attained a velocity of 10 feet per
second. What is the average accelerating force acting on it
during the 0*25 second ?
Average acceleration = 10 -r- 0-25 = 40 feet per sec.
per sec.
hence average accelerating force is -~ x 40 = 93-2 Ibs.
43. We have seen that by a suitable choice of units the force
acting on a body is numerically equal to its rate of change of
momentum ; the second law further states that the force and
the change of momentum are in the same direction. Mo-
mentum is a vector quantity, and therefore change of momentum
must be estimated as a vector change having magnitude and
direction.
For example, if the momentum of a body is represented by
ab (Fig. 25), and after / seconds it is represented by cd> then
the change of momentum in / seconds is cd — ab — eg (see
32 . Mechanics for Engineers
Art. 20), where ef= cd and gf = ab. Then the average rate
of change of momentum in t seconds is represented by ^ in
magnitude and direction, i.e. the resultant force acting on the
body during the / seconds was in the direction eg. Or Fig. 25
may be taken as a vector diagram of velocities, and eg as
/'(T
representing change of velocity. Then •£_ represents accelera-
tion, and multiplied by the mass of the body it represents the
average force.
Example.— A piece of a machine weighing 20 Ibs. is at a certain
instant moving due east at 10 feet per second, and after 1-25 seconds
it is moving south-east at 5 feet per second. What was the average
force acting on it in the interval ?
The change of momentum per second may be found directly,
or the change of velocity per second may be found, which, when
multiplied by the (constant) mass, will give the force acting.
Using the method of resolution of velocities, the
final component of velocity E. = 5 cos 45° = 4- feet per second
V 2
initial „ „ E. = 10 „ „
hence gain of component \ f $ \ / 5 \
, .4, > = ( -7 10 ) east, or ( 10 f- ) west
velocity / \/v/2 ) \ ^2)
Again, the gain of velocity south is 5 sin 45° = ~~ feet per second
V 2
The Laws of Motion 33
If R = resultant change of velocity—
and R = \/54'3 — 7'37 feet Per second in 1} seconds
Hence acceleration = 7-37 -7- 1*25 = 5-9 feet per second per second,
and average force acting = -^ x 5*9 = 3' 66 Ibs. in a direction
south of west at an angle whose tangent is —j- -*- ( 10 — -4-J
or 0-546, which is an angle of about 28|° south of west (by table
of tangents).
44. Triangle, Polygon, etc., of Forces. — It has been
seen (Art. 27) that acceleration is a vector quantity having
magnitude and direction, and that acceleration can be com-
pounded and resolved by means of vectors. Also (Art. 40)
that force is the product of acceleration and mass, the latter
being a mere magnitude or scalar quantity; hence force is a
vector quantity, and concurrent forces can be compounded by
vector triangles or polygons such as were used in Arts. 19 and
24, and resolved into components as in Arts. 25 and 28.
We are mainly concerned with uniplanar forces, but the
methods of resolution, etc., are equally applicable to forces in
different planes ; the graphical treatment would, however, in-
volve the application of solid geometry.
The particular case of bodies subject to the action of
several forces having a resultant zero constitutes the subject of
Statics-.
The second law of motion is true when the resultant force
is considered or when the components are considered, i.e. the
rate of change of momentum in any particular direction is pro-
portional to the component force in that direction.
45. Impulse. — By the impulse of a constant force in any
interval of time, we mean the product of the force and time.
Thus, if a constant force of F pounds act for / seconds, the
impulse of that force is F X A If this force F has during the
interval / acted without resistance on a mass M, causing its
velocity to be accelerated from i\ to z>2, the change of momentum
D
34
Mechanics for Engineers
during that time will have been from mv^ to mv^, i.e. mv^ — mv^
or m(vz — i>i). And the change of velocity in the interval /
under the constant acceleration f is f X / (Art. IT), therefore
v.2 — Vl. = fft and m(v.2 — vj = m.f.t; but m X/= F, the
accelerating force (by Art. 40), hence m(v^ — z/a) = F/, or, in
words, the change of momentum is equal to the impulse. The
force, impulse, and change of momentum are all to be estimated
in the same direction.
The impulse may be represented graphically as in Fig. 26.
If ON represents t seconds, and PN represents F Ibs. to scale,
M
Time
FIG. 26.
then the area MPNO under the curve MP of constant force
represents F X /*, the impulse, and therefore also the change of
momentum.
Impulse of a Variable Force. — In the case of a
variable force the interval of time is divided into a number of
parts, and the impulse calculated during each as if the force
were constant during each of the smaller intervals, and equal
to some value which it actually has in the interval. The sum
of these impulses is approximately the total impulse during the
whole time. We can make the approximation as near as we
please by taking a sufficiently large number of very small
intervals. The graphical representation will illustrate this
point.
Fig. 27 shows the varying force F at all times during the
interval NM. Suppose the interval NM divided up into a
The Laws of Motion
35
FIG. 27. — Impulse of a variable force.
number of small parts such as CD. Then AC represents the
force at the time OC ; the force is increasing, and therefore in
the interval CD the impulse will be greater than that repre-
sented by the rectangle AEDC, and less than that represented
by the rectangle FBDC.
The total impulse during
the interval NM is simi-
larly greater than that
represented by a series
of rectangles such as
AEDC, and less than
that represented by a
series of rectangles such
as FBDC. Now, if we
consider the number of
rectangles to be indefi-
nitely increased, and the width of each rectangle to be decreased
indefinitely, the area PQMN under the curve PQ is the area
which lies always between the sums of the areas of the two
series of rectangles however far the subdivision may be carried,
and therefore it represents the total impulse in the time NM,
and therefore also the gain of momentum in that time. .
It may be noticed that the above statement agrees exactly
with that made in Art. 16. In Fig. 8 the vertical ordinates
are similar to those in Fig. 27 divided by the mass, and the
gain of velocity represented by the area under PQ in Fig. 8 is
also similar to the gain of momentum divided by the mass.
area POMN
Note that the force represented by - ,^T; — (i.e. by the
length NM
average height of the PQMN) is the mean force or time-
average of the force acting during the interval NM. This
, , - , total impulse
time-average force may be defined as — — .
total time
The area representing the impulse of a negative or opposing
force will lie below the line OM in a diagram such as Fig. 27.
In case of a body such as part of a machine starting from rest
and coming to rest again, the total change of momentum is
zero ; then as much area of the force-time diagram lies below
36 Mechanics for Engineers
the time base line (OM) as above it. The reader should sketch
out such a case, and the velocity-time or momentum-time curve
to be derived from it, by the method of Art. 16, and carefully
consider the meaning of all parts of the diagrams — the slopes,
areas, changes of sign, etc.
The slope of a momentum-time curve represents accelerat-
ing force just as that of a velocity-time represents acceleration
(see Art. 14), the only difference in the case of momentum and
force being that mass is a factor of each.
It is to be noticed that the impulse or change of
momentum in a given interval is a vector quantity having
definite direction. It must be borne in mind that the change
of momentum is in the same direction as the force and
impulse. If the force varies in direction it may be split into
components (Art. 44), and the change of momentum in two
standard directions may be found, and the resultant of these
would give the change of momentum in magnitude and
direction.
46. Impulsive Forces. — Forces which act for a very
short time and yet produce considerable change of momentum
on the bodies on which they act are called impulsive forces.
The forces are large and the time is small. Instances occur in
blows and collisions.
47. The second law of motion has been stated, in Art. 40,
in terms of the rate of change of momentum. It can now be
stated in another form, viz. The change of momentum is equal
to the impulse -of the applied force, and is in the same direction.
Or in symbols, for a mass m —
m(v* - *'i) = F . /
where ^2 and z/j are the final and initial velocities, the sub-
traction being performed geometrically (Art. 20), and F is the
mean force acting during the interval of time /.
Example i. — A body weighing W Ibs. is set in motion by
a uniform net force Px Ibs., and in /x seconds it attains a velocity
V feet per second. It then comes to rest in a further period of
/2 seconds under the action of a uniform retarding force of P2 Ibs.
Find the relation between P,, P2, and V.
The Laws of Motion 37
During the acceleration period the gain of momentum in the
ection of m
is PJ/J, hence-
W
'direction of motion is — .V units, and the impulse in that direction
p*.:2(v
g
During retardation the gain of momentum in the direction
W
of motion is .V units, and the impulse in that direction
is-P2./2; hence—
P /. = ™ V
22 ^'
and finally — . V = P^ = P2/2 = l * fa + /2)
the last relation following algebraically from the two preceding
ones.
Example 2. — If a locomotive exerts a constant draw-bar pull
of 4 tons on a train weighing 200 tons up an incline of i in 120,
and the resistance of the rails, etc., amounts to 10 Ibs. per ton,
how long will it take to attain a velocity of 25 miles per hour from
rest, and how far will it have moved ?
The forces resisting acceleration are —
Ibs.
(a) Gravity T|n of 200 tons (see Art. 28) = - ^^ - 3733
(b) Resistance at 10 Ibs. per ton, 200 x 10 = 2000
Total 5733
The draw-bar pull is 4 x 2240 = 8960 Ibs. ; hence the net
accelerating force is 8960 — 5733 = 3227 Ibs.
Let / be the required time in seconds ; then the impulse is 3227
x / units.
25 miles per hour = fV x 88 feet per second (88 feet per second
= 60 miles per hour)
W
so that the gain of momentum is — . V —
200 X2.2240 x 1 x 88
therefore —
3227 . t = 20° x 224°
from which t = 159 seconds, or 2 minutes 39 seconds
38 Mechanics for Engineers
Since the acceleration has been uniform, the average speed
is half the maximum (Art. 28), and the distance travelled will be in
feet-
\ x y\ x 88 x 159 = 2915 feet
Example 3. — How long would it take the train in Ex. 2 to
go i mile up the incline, starting from rest and coming to rest at
the end without the use of brakes ?
Let tl = time occupied in acceleration,
t.2 = time occupied in retardation.
During the retardation period the retarding force will be as in
Ex. 2, a total of 5733 Ibs. after acceleration ceases. The average
velocity during both periods, and therefore during the whole time,
will be half the maximum velocity attained.
Average velocity = -^— - — feet per second
*1 T f2
5280 r
and maximum velocity = 2 x feet per second
t\ T f-2
200 x 2240 5280
.*. momentum generated = - - x 2 x / units
32 2 tl + t.2
The impulse = 3227/1 = 5733/2
, , 5733,
=
By the second law, change of momentum = impulse.
and substituting for /2 the value found —
200 x 2240 „ 5280 __ 3227, .
32-2 ~ x 2 x F.TT, - 5733 x ^6o(A + V
agreeing with the last result in Ex. i.
hence (/x + /2) = 267 seconds = 4 minutes 27 seconds
Example 4. — A car weighing 12 tons starts from rest, and has
a constant resistance of 500 Ibs. The tractive force, F, on the car
after t seconds is as follows : —
The Laws cf Motion
39
/ ...
o
2
5
8
n
13
16
19
20
F ...
1280
I27O
1 220
I IIO
905
800
720
670
660
I
Find the velocity of the car after 20 seconds from rest, and show
how to find the velocity at any time after starting, and to find the
distance covered up to any time.
Plot the curve of F and /, as in Fig. 28, and read off the force,
PHPO
1200
1000
^ 800
$
• g 600
400
200
0
4
> — -,
--*^
^
X
X
— «
'
—en
24 6 a 10 12 14. 16 18 2(
t. VL secoTids
FIG. 28.
say every 4 seconds, starting from / = 2, and subtract the 500 Ibs.
resistance from each as follows : —
2
6
IO
14
18
F Ibs.
1270
1190
980
760
680
F — 500
770
690
480
260
1 80
The mean accelerating force in the first 4 seconds is approxi-
mately 770 Ibs., and therefore the impulse is 770 x 4, which is also
the gain of momentum,
The mass of the car is " — '— = 835 units
The velocity after 4 seconds =
momentum 770 x 4
mass 835
= 3*69 feet per second
4O Mechanics for Engineers
Similarly, finding the momentum and gain of velocity in each
4 seconds, we have —
/
o
4
8
12
16
20
Gain of momentum \
in 4 seconds . . . \
o
3080
2760
I92O
1040
720
Momentum
0
3080
5840
7760
8800
9520
Velocity, feet per )
second )
o
3-69
7-00
9'3i
10-55
1 1 '41
After 20 seconds the velocity is approximately 11*41 feet per
second. The velocity after any time may be obtained approxi-
mately by plotting a curve of velocities and times from the values
obtained, and reading intermediate values. More points on the
velocity-time curves may be found if greater accuracy be desired.
The space described is represented by the area under the
velocity-time curve, and may be found as in Art. 14.
EXAMPLES III.
1. The moving parts of a forging hammer weigh 2 tons, and are
lifted vertically by steam pressure and then allowed to fall freely. What is
the momentum of the hammer after falling 6 feet ? If the force of the
blow is expended in 0*015 second, what is the average force of the blow ?
2. A mass of 50 Ibs. acquires a velocity of 25 feet per second in
10 seconds, and another of 20 Ibs. acquires a velocity of 32 feet per second
in 6 seconds. Compare the forces acting on the two masses.
3. A constant unresisted force of 7000 dynes acts on a mass of 20 kilo-
grams for 8 seconds. Find the velocity attained in this time.
4. A train weighing 200 tons has a resistance of 15 Ibs. per ton, sup-
posed constant at any speed. What tractive force will be required to
give it a velocity of 30 miles per hour in 1*5 minutes ?
5. A jet of water of circular cross-section and 1-5 inches diameter
impinges on'a flat plate at a velocity of 20 feet per second, and flows off at
right angles to its previous path. How much water reaches the plate per
second ? What change of momentum takes place per second, and what force
does the jet exert on the plate ?
6. A train travelling at 40 miles per hour is brought to rest by a uniform
resisting force in half a mile. Howmuch is the total resisting force in
pounds per ton ?
7. A bullet weighing I oz^ers a block of wood with a velocity of
1800 feet per second, and p^retrates it to a depth- of 8 inches. What is
the average resistance o£/4he wood in pounds to the penetration of the
bullet ?
8. The horizontal thrust on a steam-engine crank-shaft bearing is 10 tons,
The Laws of Motion 41
and the dead weight it supports vertically is 3 tons. Find the magnitude
and direction of the resultant force on the bearing.
9. A bullet weighing I oz. leaves the barrel of a gun 3 feet long with a
velocity of 1500 feet per second. What was the impulse of the force pro-
duced by the discharge? If the bullet took 0-004 second to traverse the
barrel, what was the average force exerted on it ?
10. A car weighing 10 tons starts from rest. During the first 25
seconds the average drawing force on the car is 750 Ibs., and the average
resistance is 40 Ibs. per ton. What is the total impulse of the effective force
at the end of 25 seconds, and what is the speed of the car in miles per hour ?
11. The reciprocating parts of a steam-engine weigh 483 Ibs., and
during one stroke, which occupies 0-3 second, the velocities of these parts
are as follows : —
Time
Velocity 1
O'O
0^025
0-05
O'lOO
0-125
0-I50
0-175
O-200
0-225
0-250
0-275
0*300
in feet|
O'OO
3 '46
6-558-91
10-22
10*90
10-48
9-32
775
6'02
4-14
2'IO
O'OO
per sec. )
Find the force necessary to give the reciprocating parts this motion, and
draw a curve showing its values on a time base throughout the stroke.
Draw a second curve showing the distances described from rest, for every
instant during the stroke. From these two curves a third may be drawn,
showing the accelerating force on the reciprocating parts, on the distance
traversed as a base.
48. Third Law of Motion. — To every action there is an
equal and opposite reaction. By the word " action " here is meant
the exertion of a force. We may state this in another way.
If a body A exerts a certain force on a body B, then B exerts
on A a force of exactly equal magnitude, but in the opposite
direction.
The medium which transmits the equal and opposite forces
is said to be in a state of stress. (It will also be in a state of
strain^ but this term is limited to deformation which matter
undergoes under the influence of stress.)
Suppose A and B (Fig. 29^|-£ connected by some means
(such as a string) suitable to wiflmand tension, and A exerts
a pull T on B. Then B exerts an equal tension T on A.
This will be true whether A moves B or not. Thus A may be
a locomotive, and B a train, or A may be a ship moored to
42 Mechanics for Engineers
a fixed post, B. Whether A moves B or not depends upon
what other forces may be acting on B.
Again, if the connection between A and B can transmit a
B
FIG. 29. — Connection in tension.
thrust (Fig. 30), A may exert a push P on B. Then B exerts
an equal push P' on A. As an example, A may be a gun, and
B a projectile ; the gases between them are in compression.
B
FIG. 30. — Connection in compression.
Or in a case where motion does not take place, A may be
a block of stone resting on the ground B ; then A and B are
in compression at the place of contact.
49. An important consequence of the third law is that the
total momentum of the two bodies is unaltered by any mutual
action between them. For since the force exerted by A on
B is the same as that exerted by B on A, the impulse during
any interval given by A to B is of the same amount as that
given by B to A and in the opposite direction. Hence, if B
gains any momentum A loses exactly the same amount, and
the total change of momentum is zero, and this is true for any
and every direction. This is expressed by the statement that
for any isolated system of bodies momentum is conservative.
Thus when a projectile is fired from a cannon, the impulse or
change of momentum of the shot due to the explosion is of
equal amount to that of the recoiling cannon in the opposite
The Laws of Motion
43
direction. The momentum of the recoil is transmitted to the
earth, and so is that of the shot, the net momentum given to
the earth being also zero.
50. Motion of Two Connected Weights. — Suppose
two weights, W1 Ibs. and W2 Ibs., to be connected by a light
inextensible string passing over a small and perfectly smooth
pulley, as in Fig. 31. If W: is greater
than Wa> with what acceleration (f)
will they move (W1 downwards and W2
upwards), and what will be the tension
(T) of the string ?
(W A
of mass — - ) : the
downward force on it is Wl (its weight),
and the upward force is T, which is the
same throughout the string by the
" third law ; " hence the downward
accelerating force is Wl — T. FIG 3i
Hence (by Art. 40) - -1 ./= Wx - T (i)
<5
Similarly, on W2 the upward accelerating force is T — W2 ;
W
hence — 2./=T-W2 .... (2)
o
adding (i) and (2) —
w.^V-w.-w.
W, - W2
w,
and from (i) —
2\V1W2
" Wj + W2
The acceleration / might have been stated from considering
the two weights and string as one complete system. The
accelerating force on which is W1 — W2, and the mass of
, . . ;. wx + w2
which is -
accelerating force
hence/= -
— W
44
Mechanics for Engineers
As a further example, suppose W.2 instead of being suspended
slides along a perfectly smooth horizontal table as in Fig. 32,
4
W2
t
W
FIG. 32.
the accelerating force is W1} and the mass in motion is
W W
hence the acceleration /=
*
,
, accelerating force on W2 T
and since/ also = - . w - = w .
we have T =
WT+W.
If the motion of W2 were opposed by a horizontal force, F,
the acceleration would be ~-. l "".,, .g.
Wl T W2
We have left out of account the weight of W2 and the
reaction of the table. These are equal and opposite, and
neutralize each other. The reaction of the pulley on the
string is normal to the direction of motion, and has therefore
no accelerating effect.
Atwood's Machine is an apparatus for illustrating the
laws of motion under gravity. It consists essentially of a
light, free pulley and two suspended weights (Fig. 31), which
can be made to differ by known amounts, a scale of lengths,
and clockwork to measure time. Quantitative measurements
of acceleration of known masses under the action of known
accelerating forces can be made. Various corrections are
THE
The Laws of Motion |i UN4VERSIT
necessary, and this method is not the one adopted for m<
the acceleration g.
Example I. — A hammer weighing W Ibs. strikes a nail weigh-
ing w Ibs. with a velocity V feet per second and does not rebound.
The nail is driven into a fixed block of wood which offers a
uniform resistance of P Ibs. to the penetration of the nail. How
far will the nail penetrate the fixed block ?
Let V = initial velocity of nail after blow.
Momentum of hammer before impact = — .V
momentum of hammer and nail after impact = - — . V'
W
.V
Let / = time of penetration.
w
Impulse P/ = — .V (the momentum overcome by P)
_ WV
During the penetration, average velocity = JV (Arts. 11 and 14)
hence distance moved by nail = ^V x /
= 1 W WV
_ lY2- W2
- 2^-p ' w + w
Example 2. — A cannon weighing 30 tons fires a looo-lb. pro-
jectile with a velocity of 1000 feet per second. With what initial
velocity will the cannon recoil ? If the recoil is overcome by a
(time) average force of 60 tons, how far will the cannon travel ?
How long will it take ?
Let V = initial velocity of cannon in feet per second.
Momentum of projectile = x 1000 = momentum of cannon
«*>
1000 30 x 2240
or — - x loco = - — x V
S g
and V - Iooox I00° = 14-87 feet per second
30 x 2240
Let / = time of recoil.
46 Mechanics for Engineers
Impulse of retarding force = 60 x 2240 x t — momentum of shot
1000 x 1000
60 x 2240 x / = -~
and hence / = 0*231 second
14-87 x 0*231
Distance moved = £V x / = - - - - = 1*74 feet
Example 3. — Two weights are connected by a string passing
over a light frictionless pulley. One is 12 Ibs. and the other n Ibs.
They are released from rest, and after 2 seconds 2 Ibs. are removed
from the heavier weight. How soon will they be at rest again,
and how far will they have moved between the instant of release
and that of coming to rest again?
First period.
accelerating force 12—11 g
Acceleration = — - = - - x £" = —
total mass 12+11 - 23
velocity after 2 seconds = 2 x *=— - = 2*8 feet per second
Second period.
Retardation = - x g — —
II + 10 21
2 X £
velocity 23
time to come to rest = - , . = = 2 X f* = 1*825 sec.
retardation g
21
average velocity throughout = \ maximum velocity (Art. 11)
total time = 2 + 1*825 seconds
distance moved = \ x 2*8 x 3*825 = 5*35 feet
EXAMPLES IV.
1. A fireman holds a hose from which a jet of water I inch in diameter
issues at a velocity of 80 feet per second. What thrust will the fireman
have to exert in order to support the jet ?
2. A machine-gun fires 300 bullets per minute, each bullet weighing
1 oz. If the bullets have a horizontal velocity of 1800 feet per second,
find the average force exerted on the gun.
3. A pile-driver weighing W Ibs. falls through // feet and drives a pile
weighing to Ibs. a feet into the ground. Show that the average force of
W2 h
the blow is ^=—. ---- Ibs.
W + w a
4. A weight of 5 cwt. falling freely, drives a pile weighing 600 Ibs..
2 inches into the earth against an average resistance of 25 tons. How far
will the weight have to fall in order to do this?
The Laws of Motion 47
5. A cannon weighing 40 tons projects a shot weighing 1500 Ibs. with
a velocity of 1400 feet per second. With what initial velocity will the
cannon recoil ? What average force will be required to bring it to rest in
3 feet ?
6. A cannon weighing 40 tons has its velocity of recoil destroyed in
2 feet 9 inches by an average force of 70 tons. If the shot weighed 14 cwt.,
find its initial velocity.
7. A lift has an upward acceleration of 3'22 feet per second per second.
What pressure will a man weighing 140 Ibs. exert on the floor of the lift ?
What pressure would he exert if the lift had an acceleration of 3^22 feet per
second per second downward? What upward acceleration would cause
his weight to exert a pressure of 170 Ibs. on the floor ?
8. A pit cage weighs 10 cwt., and on approaching the bottom of the
shaft it is brought to rest, the retardation being at the rate of 4 feet per
second per second. Find the tension in the cable by which the cage is
lowered.
9. Two weights, one of 16 Ibs. and the other of 14 Ibs., hanging
vertically, are connected by a light inextensible string passing over a
perfectly smooth fixed pulley. If they are released from rest, find how far
they will move in 3 seconds. What is the tension of the string ?
10. A weight of 17 grammes and another of 20 grammes are connected
by a fine thread passing over a light frictionless pulley in a vertical plane.
Find what weight must be added to the smaller load 2 seconds after they
are released from rest in order to bring them to rest again in 4 seconds.
How many centimetres will the weights have moved altogether? U~ (» \
11. A weight of 5 Ibs. hangs vertically, and by means of a cord passing
over a pulley it pulls a block of iron weighing 10 Ibs. horizontally along a
table-top against a horizontal resistance of 2 Ibs. Find the acceleration of
the block and tension of the string.
12. What weight hanging vertically, as in the previous question, would
give the lo-lb. block an acceleration of 3 feet per second per second on a
perfectly smooth horizontal table ?
13. A block of wood weighing 50 Ibs. is on a plane inclined 40° to
the horizontal, and its upward motion along the plane is opposed by a
force of 10 Ibs. parallel to the plane. A cord attached to the block, running
parallel to the plane and over a pulley, carries a weight hanging vertically.
What must this weight be if it is to haul the block 10 feet upwards along
the plane in 3 seconds from rest ?
CHAPTER III
WORK, POWER, AND ENERGY
51. Work. — When a force acts upon a body and causes motion,
it is said to do work.
In the case of constant forces, work is measured by the
product of the force and the displacement, one being estimated
by its component in the direction of the other.
One of the commonest examples of a force doing work
is that of a body being lifted against the force of gravity, its
weight. The work is then
measured by the product of
the weight of the body, and
the vertical height through
which it is lifted. If we
draw a diagram (Fig. 33)
setting off the constant force
F by a vertical ordinate, OM,
" N then the work done during
Distance .. .
any displacement represented
by ON is proportional to the
area MPNO, and is represented by that area. If the scale of
force is i inch =/lbs., and the scale of distance is i inch = q
feet, then the scale of work is i square inch = pq foot-lbs.
52. Units of Work. — Work being measured by the
product of force and length, the unit of work is taken as
that done by a unit force acting through unit distance. In
the British gravitational or engineer's system of units, this is
the work done by a force of i Ib. acting through a distance
of i foot. It is called the foot-pound of work. If a weight
0
Work, Power, and Energy
49
W Ibs. be raised vertically through // feet, the work done is
W// foot Ibs.
Occasionally inch-pound units of work are employed,
particularly when the displacements are small.
In the C.G.S. system the unit of work is the erg. This is
the work done by a force of one dyne during a displacement
of i centimetre in its own direction (see Art. 42).
53. Work of a Variable Force. — If the force during
any emplacement varies, we may find the total work done
approximately by splitting the displacement into a number
of parts and finding the work done during each part, as if
the force during the partial displacement were constant and
equal to some value it has during that part, and taking the sum
of all the work so calculated in the partial displacements. We
can make the approximation as near as we please by taking
a sufficiently large number of parts. We may define the work
actually done by the variable force as the limit to which such
a sum tends when the subdivisions of the displacement are
made indefinitely small.
54. Graphical Representation of Work of a Variable
Force. — Fig. 34 is a diagram showing by its vertical ordinates
M
C D
Spa,ce
FIG. 34.
the force acting on a body, and by its horizontal ones the dis-
placements. Thus, when the displacement is represented by
50 Mechanics for Engineers
ON, the force acting on the body is represented by PN.
Suppose the interval ON divided up into a number of small
parts, such as CD. The force acting on the body is represented
by AC when the displacement is that represented by OC.
Since the force is increasing with increase of displacement
the work done during the displacement CD is greater than
that represented by the rectangle AEDC, and less than that
represented by the rectangle FBDC. The total work done
during the displacement will lie between that represented by
the series of smaller rectangles, such as AEDC, and that
represented by the series of larger rectangles, such as FBDC.
The area MPNO under the curve MP will always lie between
these total areas, and if we consider the number of subdivisions
of ON to be carried higher indefinitely, the same remains true
both of the total work done and the area under the curve MP.
Hence the area MPNO under the curve MP represents the
work done by the force during the displacement represented
by ON.
The Indicator Diagram, first introduced by Watt for
use on the steam-engine, is a diagram of the same kind as
Fig. 34. The vertical ordinates are proportional to the total
force exerted by the steam on
the piston, and the horizontal
ones are proportional to the dis-
placement of the piston. The
area of the figure is then pro-
portional to the work done by
the steam on the piston.
In the case of a force vary-
Space, ing uniformly with the displace-
FIG. 35.— Force varying uniformly ment, the CUrVC MP is a Straight
with space. ,. /T-,. ,. , .
line (Fig. 35), and the area
MPNO = P-N X ON, or if the initial force (OM) is Fa
Ibs., and the final one (PN) is F2 Ibs., and the displacement
TT -i- TH"
(ON) is d feet, the work done is - -2 . d foot-lbs.
In stretching an unstrained elastic body, such as a spring,
Work, Power, and Energy 5 i
the force starts from zero (or Fx = o). Then the total work
done is ^F.y/, where F2 is the greatest force exerted, and d is
the amount of stretch.
Average Force. — The whole area MPNO (Figs. 34 and
35) divided by the above ON gives the mean height of the
area; this represents the space- aver age of the force during the
displacement ON. This will not necessarily be the same as
the time-average (Art. 45). We may define the space-average
of a varying force as the work done divided by the displacement.
55. Power.— Power is the rate of doing work, or the
work done per unit of time. .
One foot-pound per second might be chosen as the unit
of power. In practice a unit 550 times larger is used; it is
called the horse-power. It is equal to a rate of 550 foot-lbs.
per second, or 33,000 foot-lbs. per minute. In the C.G.S. system
the unit of power is not usually taken as one erg per second,
but a multiple of this small unit. This larger unit is called
a watt, and it is equal to a rate of io7 ergs per second.
Engineers frequently use a larger unit, the kilowatt, which
is 1000 watts. One horse-power is equal to 746 watts or
0-746 kilowatt.
Example i.— A train ascends a slope of i in 85 at a speed of
20 miles per hour. The total weight of the train is 200 tons, and
resistance of the rails, etc., amounts to 12 Ibs. per ton. Find the
horse-power of the engine.
The total force required to draw the load is —
The number of feet moved through per minute is \ x 88 x 60
= 1760 feet; hence the work done per minute is 1760 x 7670
= 13,500,000 foot-lbs., and since i horse-power = 33,000 foot-lbs.
per minute, the H. P. is 1^H/|{m = 409 horse-power.
Example 2. — A motor-car weighing 15 cwt. just runs freely at
12 miles per hour down a slope of i in 30, the resistance at this
speed just being sufficient to prevent any acceleration. What horse-
power will it have to exert to run up the same slope at the same
speed ?
In running down the slope the propelling force is that of gravity,
which is T^JJ of the weight of the car (Arts. 28 and 44) ; hence the
52 Mechanics for Engineers
resistance of the road is also (at 12 miles per hour) equivalent to
15 X 112
or 56 Ibs.
Up the slope the opposing force to be overcome is 56 Ibs. road
resistance and 56 Ibs. gravity (parallel to the road), and the total
112 Ibs.
The distance travelled per minute at 12 miles per hour is \
mile = 5a£ft or 1056 feet ; hence the work done per minute is
112 x 1056 foot-lbs., and the H.P. is II2 X l°s6 or 3-584 H.P.
33000
Example 3. — The spring of a safety-valve is compressed from
its natural length of 20 inches to a length of 17 inches. It then
exerts a force of 960 Ibs. How much work will have to be done
to compress it another inch, i.e. to a length of 16 inches ?
The force being proportional to the displacement, and being
960 Ibs. for 3 inches, it is fi§ft or 320 Ibs. per inch of compression.
When 1 6 inches long the compression is 4 inches, hence the
force is 4 x 320 or 1280 Ibs. ; hence the work done in compression
is ? - x i, or 1 1 20 inch-lbs. (Art. 54, Fig. 35), or 93-3
foot-lbs.
EXAMPLES V.
1. A locomotive draws a train weighing 150 tons along a level track
at 40 miles per hour, the resistances amounting to 10 Ibs. per ton. What
horse-power is it exerting ? Find also the horse-power necessary to draw
the train at the same speed (a) up an incline of I in 250, (b) down an incline
of i in 250.
2. If a locomotive exerts 700 horse-power when drawing a train of
200 tons up an incline of I in 80 at 30 miles per hour, find the road
resistances in pounds per ton.
3. A motor-car engine can exert usefully on the wheels 8 horse-power.
If the car weighs 16 cwt. , and the road and air resistances be taken at
20 Ibs. per ton, at what speed" can this car ascend a gradient of i in 15 ?
4. A winding engine draws from a coal-mine a cage which with the
coal carried weighs 7 tons ; the cage is drawn up 380 yards in 35 seconds.
Find the average horse-power required. If the highest speed attained is
30 miles per hour, what is the horse-power exerted at that time ?
5. A stream delivers 3000 cubic feet of water per minute to the highest
point of a water-wheel 40 feet diameter. If 65 per cent, of the available
work is usefully employed, what is the horse-power developed by the
wheel ?
6. A bicyclist rides up a gradient of i in 15 at 10 miles per hour. The
Work) Power, and Energy 53
weight of rider and bicycle together is 180 Ibs. If the road and other
resistances are equivalent to Tg0 of this weight, at what fraction of a horse-
power is the cyclist working ?
7. Within certain limits, the force required to stretch a spring is
proportional to the amount of stretch. A spring requires a force of
800 Ibs. to stretch it 5 inches : find the amount of work done in stretching
it 3 inches.
8. A chain 400 feet long and weighing 10 Ibs. per foot, hanging
vertically, is wound up. Draw a diagram of the force required to draw
it up when various amounts have been wound up from o to 400 feet.
From this diagram calculate the work done in winding up (a) the first
100 feet of the chain, (b) the whole chain.
9. A pit cage weighing 1000 Ibs. is suspended by a cable 800 feet long
weighing i'\ Ibs. per foot length. How much work will be done in wind-
ing the cage up to the surface by means of the cable, which is wound on a
drum ?
56. It frequently happens that the different parts of a body
acted upon by several forces move through different distances
in the same time ; an important instance is the case of the
rotating parts of machines generating or transmitting power. It
will be convenient to consider here the work done by forces
which cause rotary motion of a body about a fixed axis.
Moment of a Force. — The moment of a force about a
point is the measure of its turning effect or tendency, about
that point. It is measured by the
product of the force and the per-
pendicular distance from the point
to the line of action of the force.
Thus in Fig. 36, if O is a point, and
AB the line of action of a force F,
both in the plane of the figure, and
OP is the perpendicular from O on
to AB measuring r units of length, FIG g
the moment of F about O is F X r.
The turning tendency of F about O will be in one direction,
or the opposite, according as O lies to the right or left of AB
looking in the direction of the force. If O lies to the right, the
moment is said to be clockwise ; if to the left, contra-clockwise.
In adding moments of forces about O, the clockwise and contra-
clockwise moments must be taken as of opposite sign, and the
54 Mechanics for Engineers
algebraic sum found. Which of the two kinds of moments is
considered positive and which negative is immaterial. If O
lies in the line AB, the moment of F about O is zero.1
The common units for the measurement of moments are
pound-feet. Thus, if a force of i Ib. has its line of action i
foot from a fixed point, its moment about that point is one
pound-foot. In Fig. 36, if the force is F Ibs., and OP represents
r feet, the moment about O is F . r pound-feet.
Moment of a Force about an Axis perpendicular
to its Line of Action. — If we consider a plane perpendicular
to the axis and through the force, it will cut the axis in a point
O ; then the moment of the force about the axis is that of the
force about O, the point of section of the axis by the plane.
The moment of the force about the axis may therefore be
defined as the product of the force and its perpendicular distance
from the axis.
In considering the motion of a body about an axis, it is
necessary to know the moments about that axis of all the
forces acting on the body in planes perpendicular to the axis,
whether all the forces are in the same plane or not. The total
moment is called the torque^ or twisting moment or turning
moment • about the axis. In finding the torque on a body
about a particular axis, the moments must be added algebrai-
cally.
57. Work done by a Constant Torque or Twist-
ing Moment. — Suppose a force F Ibs. (Fig. 37) acts upon a
body which turns about an axis, O, perpendicular to the line
of action of F and distant r feet from it, so that the turning
1 Note that the question whether a moment is clockwise or contra-
clockwise depends upon the aspect of view. Fig. 36 shows a force (F)
having a contra-clockwise moment about O, but this only holds for one
aspect of the figure. If the force F in line AB and the point O be viewed
from the other side of the plane of the figure, the moment would be called
a clockwise one. This will appear clearly if the figure is held up to the
light and viewed from the other side of the page. Similarly, the moment
of a force about an axis will be clockwise or contra-clockwise according
as the force is viewed from one end or the other of the axis. The motion
of the hands of a clock appears contra-clockwise if viewed from the back
through a transparent face.
Work, Power, and Energy
55
moment (M) about O is F . r Ib.-feet Suppose that the
force F acts successively on different parts of the body all
distant r from the axis O about which it rotates, or that the
force acts always on the same
point C, and changes its direc-
tion as C describes its circular
path about the centre O, so as
to always remain tangential to
this circular path ; in either case
the force F is always in the same
direction as the displacement it
is producing, and therefore the
work done is equal to the product
of the force and the displacement
(along the circumference of the
circle CDE). Let the displace-
FIG. 37.
ment about the axis O be through an angle 0 radians correspond-
ing to an arc CD of the circle CDE, so that —
CD
(The angle 0 is 277, if a displacement of one complete cir-
cuit be considered.)
The work done is F X CD = F . rO foot-lbs.
But M = F . r Ib.-feet
therefore the work done = M X 0 foot-lbs.
The work done by each force is, then, the product of the
turning moment and the angular displacement in radians. If
the units of the turning moment are pound-feet, the work will
be in foot-pounds ; if the moment is in pound-inches, the work
will be in inch-pounds, and so on. The same method of calcu-
lating the work done would apply to all the forces acting, and
finally the total work done would be the product of the total
torque or turning moment and the angiilar displacement in
radians.
Again, if w is the angular velocity in radians per second,
the power or work per second is M . w foot-lbs., and the horse-
56 Mechanics for Engineers
power is — ' — , where M is the torque in Ib.-feet ; and if N is
the number of rotations per minute about the axis—
HP _27rN'M
33,000
This method of estimating the work done or the power,
is particularly useful when the turning forces act at different
distances from the axis of rotation.
We may, for purposes of calculation, look upon such a state
of things as replaceable by a certain force at a certain radius,
but the notion of a torque and an angular displacement seems
rather less artificial, and is very useful.
The work done by a variable turning moment during a
given angular displacement may be found by the method of
Arts. 53 and 54. If in Figs. 33, 34, and 35 force be replaced by
turning moment and space by angular displacement, the areas
under the curves still represent the work done.
In twisting an elastic rod from its unstrained position the
twisting moment is proportional to the angle of twist, hence
the average twisting moment is half the maximum twisting
moment ; then, if M = maximum twisting moment, and 6 =
angle of twist in radians —
the work done = ^MO
Example I. — A high-speed steam-turbine shaft has exerted on
it by steam jets a torque of 2100 Ib.-feet. It runs at 750 rotations
per minute. Find the horse-power.
The work done per minute = (torque in Ib.-feet) x (angle turned
through in radians)
= 2100 x 750 x 27r foot-lbs.
2100 X 750 X 27T
horse-power = — — = 300 H.P.
Example 2. — An electro motor generates 5 horse-power, and
runs at 750 revolutions per minute. Find the torque in pound-feet
exerted on the motor spindle.
Horse-power x 33,000 = torque in Ib.-feet x radians per minute
horse-power X 33,000
hence torque m Ib.-feet = — dTa— ey minute-
c x ^,000
= = 35 Ib.-feet
Work, Power, and Energy 57
EXAMPLES VI.
1. The average turning moment on a steam-engine crankshaft is 2000
Ib.-feet, and its speed is 150 revolutions per minute. Find the horse-power
it transmits.
2. A shaft transmitting 50 H.P. runs at 80 revolutions per minute.
Find the average twisting moment in pound-inches exerted on the shaft.
3. A steam turbine develops 250 horse-power at a speed of 200 revolu-
tions per minute. Find the torque exerted upon the shaft by the steam.
4. How much work is required to twist a shaft through 10° if the
stiffness is such that it requires a torque of 40,000 Ib. -inches per radian of
twist ?
5. In winding up a large clock (spring) which has completely "run
down," 8J complete turns of the key are required, and the torque applied
at the finish is 200 Ib. -inches. Assuming the winding effort is always
proportional to the amount of winding that has taken place, how much
work has to be done in winding the clock ? How much is done in the last
two turns ?
6. A water-wheel is turned by a mean tangential force exerted by the
water of half a ton at a radius of lo feet, and makes six turns per minute.
What horse-power is developed ?
58. Energy. — When a body is capable of doing work, it is
said to possess energy. It may possess energy for various
reasons, such as its motion, position, temperature, chemical
composition, etc. ; but we shall only consider two kinds of
mechanical energy.
59. A body is said to have potential energy when it is
capable of doing work by virtue of its position. For example,
when a weight is raised for a given vertical height above datum
level (or zero position), it has work done upon it ; this work is
said to be stored as potential energy. The weight, in returning
to its datum level, is capable of doing work by exerting a force
(equal to its own weight) through a distance equal to the
vertical height through which it was lifted, the amount of
work it is capable of doing being, of course, equal to the amount
of work spent in lifting it. This amount is its potential energy
in its raised position, e.g. suppose a weight W Ibs. is lifted h
feet ; the work is W . h foot-lbs., and the potential energy of the
W Ibs. is then said to be W . h foot-lbs. It is capable of doing
an amount of work W . h foot-lbs. in falling.
58 Mechanics for Engineers
60. Kinetic Energy is the energy which a body has in
virtue of its motion.
We have seen (Art. 40) that the exertion of an unresisted
force on a body gives it momentum equal to the impulse of the
force. The force does work while the body is attaining the
momentum, and the work so done is the measure of the kinetic
energy of the body. By virtue of the momentum it possesses,
the body can, in coming to rest, overcome a resisting force
acting in opposition to its direction of motion, thereby doing
work. The work so done is equal to the kinetic energy of the
body, and therefore also to the work spent in giving the body
its motion.
Suppose, as in Ex. i, Art. 47, a body of weight W Ibs.
is given a velocity V feet per second by the action of a
uniform force Yl Ibs. acting for ^ seconds, and then comes
to rest under a uniform resisting force F2 Ibs. in /2 seconds. We
had, in Art. 47 —
W
Impulse F^ = —V = F24
<b
But, the mean velocity being half the maximum under a
uniform accelerating force, the distance d^ moved in accelerat-
ing, is ^V/j feet, and that 4> moved in coming to rest, is |V/2;
hence the work done in accelerating is —
W w
and work done in coming to rest is —
W w
F2 X JV/a = - V X JV = J-V2
W
hence l-V2 = F^ = F2</2
<j>
These two equalities are exactly the same as those of Ex.
/ W \
i, Art. 47 I viz. —V = F^ = F2/2 ) , with each term multi-
V
plied by — , and problems which were solved from considera-
tions of changes of momentum might often have been (alter-
natively) solved by considerations of change of kinetic energy.
Work, Poiver, and Energy 59
The amount of kinetic energy possessed by a body of
W
weight W Ibs. moving at V feet per second is therefore \— V2
A
foot-lbs.
Again, if the initial velocity had been u feet per second
instead of zero, the change of momentum would have been
W
— (v — ?/), and we should have had —
A
W
Fx/i = — (v — u), v being final velocity
o
ii I ?* \V // "4" v
and the work done = Fj X - - X /t = — (v — u) —^—
= change of kinetic energy
Similarly, in overcoming resistance at the expense of its kinetic
energy, the work done by a body is equal to the change of
kinetic energy whether all or only part of it is lost.
61. Principle of Work. — If a body of weight W Ibs. be
lifted through h feet, it has potential energy \Nh foot-lbs. If
it falls freely, its gain of kinetic energy at any instant is just
equal to the loss of potential energy, so that the sum (potential
energy) -f (kinetic energy) is constant ; e.g. suppose the weight
has fallen freely x feet, its remaining potential energy is
\y(/2 __ x) foot-lbs. It will have acquired a velocity \/ 2gx feet
W
per second (Art. 13), hence its kinetic energy 1— V2, will be
<b
W W
±-X2gx^Nx foot-lbs., hence W(/$-*)-fi-V2 = W/;, which
" & ~ £
is independent of the value of xt and no energy has been lost.
Note that for a particular system of bodies the sum of
potential and kinetic energies is generally not constant. Thus,
although momentum is conservative, mechanical energy is not.
For example^ when a body in motion is brought to rest by a
resisting force of a frictional kind, mechanical energy is lost.
The energy appears in other forms, chiefly that of heat.
Principle of Work.— Further, if certain forces act upon
6o
Mechanics for Engineers
a body, doing work, and other forces, such as frictional ones,
simultaneously resist the motion of the body, the excess of the
work done by the urging forces over that done against the
resistances gives the kinetic energy stored in the body. Or we
may deduct the resisting forces from the urging forces at every
instant, and say that the work done by the effective or net
accelerating forces is equal to the kinetic energy stored. Thus
in Fig. 38, representing the forces and work done graphically as
in Art. 54, if the ordinates of the curve MP represent the
forces urging the body forward, and the ordinates of M'P' re-
present the resistances to the same scale, the area MPNO
represents the work done ; the work lost against resistances is
represented by the area M'P'NO, and the difference between
these two areas, viz. the area MPP'M', represents the kinetic
energy stored during the time that the distance ON has been
traversed. If the body was at rest at position O, MPP'M'
represents the total kinetic energy, and if not, its previous
kinetic energy must be added to obtain the total stored at the
position ON. From a diagram, such as Fig. 38, the velocity
M
FIG. 38.
can be obtained, if the mass of the moving body is known, by
the relation, kinetic energy = J(mass) X (velocity)'2.
Fig. 39 illustrates the case of a body starting from rest and
coming to rest again after a distance O^, such, for example,
as an electric car between two stopping-places. The driving
forces proportional to the ordinates of the curve abec cease
Work, Power, and Energy
61
after a distance oc has been traversed, and (by brakes) the
resisting forces proportional to the ordinates of the curve def
increase. The area abed represents the kinetic energy of the
car after a distance oct and the area efgc represents the work
Distances
FIG. 39.
done by the excess of resisting force over driving force. When
the latter area is equal to the former, the car will have come
to rest.
The kinetic energy which a body possesses in virtue of its
rotation about an axis will be considered in a subsequent
chapter.
Example I.— Find the work done by the charge on a projectile
weighing 800 Ibs., which leaves the mouth of a cannon at a velocity
of 1800 feet per second. What is the kinetic energy of the gun at
the instant it begins to recoil if its weight is 25 tons ?
The work done is equal to the kinetic energy of the projectile —
W
800
K.E. = - x — r x V2 = - x —7— x (i8oo)2 = 40,200,000 foot-lbs.
The momentum of the gun being equal to that of the projectile,
the velocity of the gun is —
1800
~° = 2571 feet per second
and the K.E. = x -
-
x (2571)2 = 577,000 foot-lbs.
It may be noticed that the kinetic energies of the projectile
62
Mechanics for Engineers
and cannon are inversely proportional to their weights. The
i W i W
K.E. is - x -— x V2, or - x - x V x V, which is | x momen-
tum X velocity. The momentum of the gun and that of the pro-
jectile are the same (Art. 52), and therefore their velocities are
inversely proportional to their weights ; and therefore the products
of velocities and half this momentum are inversely proportional
to their respective weights.
Example 2. — A bullet weighing i oz., and moving at a velocity
of 1500 feet per second, overtakes a block of wood moving at
40 feet per second and weighing 5 Ibs. The bullet becomes
embedded in the wood without causing any rotation. Find the
velocity of the wood after the impact, and how much kinetic energy
has been lost.
Let V = velocity of bullet and block after impact.
Momentum of bullet = -^ x -
momentum of block = - x 40 =
hence total momentum before
and after impact
Total momentum after impact =
I = 29375
f X
S g
and therefore V = 7fS~ = 58'! feet per second
Kinetic energy of bullet = -x —x — — x I5oox 1500 = 2183 foot- Ibs.
16 32*2
2
Kinetic energy of block = - x — ^ x 40 x 40
= 124
Total K.E. before impact = 2307 „
Total K.E. after impact = -x^— => x 58-1 x 58-1 = 265 foot-lbs.
Loss of K.E. at impact = 2307 — 265 = 2042 ,,
Example 3. — A car weighs I2'88 tons, and starts from rest ;
the resistance of the rails may be taken as constant and equal to
500 Ibs. After it has moved S feet from rest, the tractive force,
F Ibs., exerted by the motors is as follows : —
S ...
F ...
-
0
1280
20
1270
50
1220
80 no
i no 905
130
800
1 60
720
190
670
200
660
Work, Power, and Energy
Find the velocity of the car after it has gone 200 feet from rest ;
also find the velocity at various intermediate points, and plot a
curve of velocity on a base of space described.
Plot the curve of F and S as in Fig. 40, and read off the force
every 50 feet, say, starting from S = 10, and subtract 500 Ibs.
resistance from each, as follows : —
S ...
10
3°
5°
70
90
1 10
130
ISO
170
190
F ...
1275
1260
1 220
1150
1050
9os
800
740
695
670
F-Soo 775
760
720
650
550
405
300
240
195 170
1200
1000
800
400
200
X
20 40
60 60 100
S z>i feet
FIG. 40.
120
140 160 180 200
The mean accelerating force during the first 20 feet of motion is
approximately equal to that at S = 10, viz. 775 Ibs. ; hence the
work stored as kinetic energy (K.E.), i.e. the gross work done less
that spent against resistance, is —
(1275 x 20) - (500 x 20), or 775 x 20 foot-lbs. = 15,500 foot-lbs.
Then, if V is the velocity after covering S feet, for S = 20 —
W
.. = -x—= 15,500
** «S
and W = I2'88 x 2240 Ibs.
W
therefore — , the mass of the car is -
2240 or 896 units, and —
64
Mechanics for Engineers
i x -V2 = £ x 896 x V2 = 15,500
V = A/34'8 = 5-90 feet per second
Similarly, finding the gain of kinetic energy in each 20 feet, the
square of velocity (V2), and the velocity V, we have from S = 20
to S - 40—
gain of K.E. = 760 x 20 = 15,200 foot-lbs.
/. total K.E. at S = 40 is
15,500 + 15,200 = 30,700 foot-lbs.
and so on, thus —
S
o
!
20 40
60
80
100
1 20
140
160
180
200
Gain of K.E.
in 20 feet,
0
15500
15200 14400
13000
1 1000
8100
6000
4800
3900
3400
foot-lbs.
1
Total K.E.,\!
foot-lbs. /| °
15500
30700
45100
58100
69100
77200
83200
88000
91900
95300
V2 or ^4'
448
o
34-8
68-5
lOO'O
129-4
154-0
172-1
185-5
196*2
204-8 2I2'5
V ft. per sec.
o
5-90
8-28
10-03
11-34
12-40
13-12
13-62
14-01
I4-30
I4-58
These velocities have been plotted on a base of spaces in
Fig. 41-
15
1
b,
I
20 4-0 60 80 IOO 120
S. trt feet
FIG. 41.
180 200
OF THE
Work, Power , and Energy
Example 4, — From the results of Example 3, find in whal
the car travels the distance of 20 feet from S = 80 to S = 100,
and draw a curve showing the space described up to any instant
during the time in which it travels the first 200 feet.
At S = 80, V = ii -34 feet per second
at S = 100, V = 12-40 feet per second
hence the mean velocity for such a short interval may be taken
as approximately —
OF
1 1 -34 + 12*40
, or 1 1 '87 feet per second
Hence the time taken from S = 80 to S = 100 is approximately —
— ~ = 1*685 seconds
Similarly, we may find the time taken to cover each 20 feet, and
so find the total time occupied, by using the results of Ex. 3,
as follows. The curve in Fig. 42 has been plotted from these
numbers.
200
'50
N
100
50
10 15
Time in, seconds
20
25
FIG. 42.
66
Mechanics for Engineers
s
o
20
40
60
80 100
120! 140
i6o| 1 80
200
Mean velocity
1
for last 20 ft.,
o
2'95
7-09
9*15
10-68 11-87
12-76 13-37
1381
I4'i5
14-44
feet per sec.
Time for last
20 feet, se-
o
6780
2-824
2-188
1-872 1-685
1-568
1-496
i "417
i'4i3
I-388
conds
Total time, 1
/ seconds J
0
6780
9-604
11-792
13-66415-349
16-917
18-413
19-850
21-263
22-65I
EXAMPLES VII.
1. Find in foot-pounds the kinetic energy of a projectile weighing
800 Ibs.- moving at 1000 feet per second. If it is brought to rest in 3 feet,
find the space average of the resisting force. I 1- <^X $ o o
2. At what velocity must a body weighing 5 Ibs. be moving in order
to have stored in it 60 foot-lbs. of energy ?
3. What is the kinetic energy in inch-pounds of a bullet weighing I oz.
travelling at 1800 feet per second? If it is fired directly into a suspended
block of wood weighing i'25 lb., how much kinetic energy is lost in the
impact ?
4. A machine-gun fires 300 bullets per minute, each bullet weighing
I oz. and having a muzzle velocity of 1700 feet per second. At what
average horse-power is the gun working? 2- ^~
5. A jet of water issues in a parallel stream at 90 feet per second from
a round nozzle I inch in diameter. What is the horse-power of the jet ?
One cubic foot of water weighs 62*5 Ibs. *7. 0 \.
6. Steam to drive a steam impact turbine issues in a parallel stream
from a jet \ inch diameter at a velocity of 2717 feet per second, and the
density of the steam is such that it occupies 26*5 cubic feet per pound.
Find the horse-power of the jet. [ i°L\~
7. A car weighing 10 tons attains a speed of 15 miles per hour from
rest in 24 seconds, during which it covers 100 yards. If the space-average
of the resistances is 30 Ibs. per ton, find the average horse-power used to
drive the car. £<^. (&
8. How long will it take a car weighing 1 1 tons to accelerate from
10 miles per hour to 15 miles per hour against a resistance of 25 Ibs. per
ton, if the motors exert a uniform tractive force on the wheels and the
horse-power is 25 at the beginning of this period? ^. b
9. A car weighing 12 tons is observed to have the following tractive
forces F Ibs. exerted upon it after it has travelled S feet from rest : —
? ...
0
1440
IO
1390
30
1250
5°
1060
65
910
80
805
94
760
IOO
740
Work, Power, and Energy 67
The constant resistance of the road is equivalent to 600 Ibs. Find the
velocity of the car after it has covered 100 feet. Plot a curve showing
the velocity at all distances for 100 feet from the starting-point. What is
the space-average of the effective or accelerating force on the car ? | o.^ "}
10. From the results of the last question plot a curve showing the
space described at any instant during the time taken to cover the first
100 feet. How long does the car take to cover 100 feet? I ^" ^
11. A machine having all its parts in rigid connection has 70,000 foot-
pounds of kinetic energy when its main spindle is making 49 rotations
per minute. How much extra energy will it store in increasing its speed
to 50 rotations per minute ?
12. A machine stores 10,050 foot-lbs. of kinetic energy when the speed
of its driving-pulley rises from 100 to 101 revolutions per minute. How
much kinetic energy would it have stored in it when its driving-pulley
is making 100 revolutions per minute?
CHAPTER IV
MOTION IN A CIRCLE: SIMPLE HARMONIC
MOTION
62. Uniform Circular Motion. — Suppose a particle de-
scribes about a centre O (Fig. 43), a circle of radius ;• feet
with uniform angular velocity w radians per second. Then
its velocity, ?', at any instant is of magnitude wr (Art. 33), and
its direction is along the tangent to the circle from the point
in the circumference
which it occupies at that
instant. Although its
velocity is always of
magnitude <o/-, its direc-
tion changes. Consider
the change in velocity
between two points, P
and Q, on its path at
an angular distance 0
apart (Fig. 43). Let
the vector cb parallel to the tangent PT represent the linear
velocity 7> at P, and let the vector ab, of equal length to cb and
parallel to QT, the tangent at Q, represent the linear velocity
7' at Q. Then, to find the change of velocity between P and
Q, we must subtract the velocity at P from that at Q; in
vectors —
db — cb = ab + be — ac (Art. 27)
Then the vector ac represents the change of velocity between
the positions P and Q. Now, since abc = PQQ = 0, length
FIG. 43.
Motion in a Circle: Simple Harmonic Motion 69
0 0
#<r = 2a£ . sin -, which represents 2v sin.-, and the time
A
taken between the positions P and Q is - seconds (Art. 33).
o»
Therefore the average change of velocity per second is —
0
60 Sm2
2 v sm - -7- — or uv . /.
2 0> V
2
which is the average acceleration. Now, suppose that Q is
taken indefinitely close to P — that is, that the angle 0 is in-
0
sin - -~|
definitely reduced ; then the ratio — 3 — has a limiting value
unity, and the average change of velocity per second, or
average acceleration during an indefinitely short interval is
v1
wv, or COT or — , since v = wr. This average acceleration
during an indefinitely reduced interval is what we have defined
(Art. 9) as actual acceleration, so that the acceleration at P
ir
is COT or — feet per second per second. And as the angle 6
is diminished indefinitely and Q thereby approaches P, the
vector ab, remaining of the same length, approaches cb (a and c
being always equidistant from b\ and the angle bed increases
and approaches a right angle as 6 approaches zero. Ultimately
the acceleration (wV) is perpendicular to PT, the tangent at
P, i.e. it is towards O.
63. Centripetal and Centrifugal Force. — In the
previous article we have seen that if a small body is describing
a circle of radius ;• feet about a centre O with angular velocity
w radians per second,' it must have an acceleration wV towards
O; hence the force acting upon it must be directed (awards
the centre O and of magnitude equal to its (mass) X o>V or
W
- O>T Ibs., where W is its weight in pounds This force causing
vb
7<D Mechanics for Engineers
the circular motion of the body is sometimes called the centri-
petal force. There is (Art. 51), by the third law of motion,
a reaction of equal magnitude upon the medium which exerts
this centripetal force, and this reaction is called the centrifugal
force. It is directed away from the centre O, and is exerted
W
upon the matter which impresses the equal force — wV upon
the revolving body ; it is not to be reckoned as a force acting
upon the body describing a circular path.
A concrete example will make this clear. If a stone of
weight W Ibs. attached to one end of a string r feet long
describes a horizontal circle with constant angular velocity w
radians per second, and is supported in a vertical direction by
a smooth table, so that the string remains horizontal, the force
W
which the string exerts upon the stone is — o>V towards the
<^
centre of the circle. The stone, on the other hand, exerts on
W
the string an outward pull — trr away from the centre. In
<?}
other cases of circular motion the inward centripetal force
may be supplied by a thrust instead of a tension ; e.g. in the
case of a railway carriage going round a curved line, the centri-
petal thrust is supplied by the rail, and the centrifugal force
is exerted outward on the rail by the train.
64. Motion on a Curved " Banked " Track. — Suppose
a body, P (Fig. 44), is moving with uniform velocity, v, round a
FIG. 44.
smooth circular track of radius OP equal to r feet. At what
angle to the horizontal plane shall the track be inclined or
Motion in a Circle : Simple Harmonic Motion 7 1
"banked" in order that the body shall keep in its circular
path?
There are two forces acting on the body — (i) its own weight,
W ; (2) the reaction R of the track which is perpendicular to
W v2
the smooth track. These two have a horizontal resultant — • —
towards the centre O of the horizontal circle in which the
body moves. If we draw a vector, ab (Fig. 44), vertically, to
represent W, then R is inclined at an angle a to it, where a
is the angle of banking of the track. If a vector, be, be drawn
from b inclined at an angle a to ab, to meet ac, the perpendicular
to ab from a, then be represents R, and ac or (ab + be) represents
W v2
the resultant of W and R, viz. — • -^and—
ac W v* v*
tan a = -T = — • — v W = -
ab g r gr
which gives the angle a required.
65. Railway Curves. — If the lines of a railway curve
be laid at the same level, the centripetal thrust of the rails
on the wheels of trains would act on the flanges of the wheels,
and the centrifugal thrust of the wheel on the track would tend
to push it sideways out of its place. In order to have the action
and reaction normal to the track the outer rail is raised, and the
track thereby inclined to the horizontal. The amount of this
"superelevation" suitable to a given speed is easily calculated.
Let G be the gauge in inches, say, v the velocity in feet
per second, and r the
radius of the curve in
feet. Let AB (Fig. 45)
represent G ; then AC
represents the height
in inches (exaggerated) <^-^ \
which B stands above "^— - '
A, and ABC is the angle
of banking, as in Art. 64. Then AC = AB sin a = AB tan
a nearly, since a is always very small; hence, by Art. 64, AC
represents G tan a, or G inches.
"'' - ^
r
Mechanics for Engineers
FIG. 46.
66, Conical Pendulum.— This name is applied to a
combination consisting of a small weight fastened to one end
of a string, the other end of
cu which is attached to a fixed
point, when the weight keeping
the string taut, describes a
horizontal circle about a centre
vertically under the fixed point.
Fig. 46 represents a conical
pendulum, where a particle, P,
attached by a thread to a fixed
point, O, describes the hori-
zontal circle PQR with con-
stant angular velocity about the centre N vertically under O.
Let T = tension of the string OP in Ibs. ;
W = angular velocity of P about N in radians per second ;
W = weight of particle P in Ibs. ;
r = radius NP of circle PQR in feet ;
/ = length of string OP in feet ;
a = angle which OP makes with ON, viz. PON ;
// = height ON in feet ;
g = acceleration of gravity in feet per second per
second.
At the position shown in Fig. 46 P is acted upon by two
forces— (i) its own weight, W; (2) the tension T of the string
OP. These have a resultant in the line PN (towards N),
the vector diagram being set off as in Art. 64, ab vertical,
representing the weight W, of P, and be the tension T. Then
W
the vector ac = ab -f &r, and represents the resultant force —
0>
X wV along PN ; hence —
W _^ __ o>V
p- (jr
ac
(•%
Also
g
ON or h = NP -f- tan a = r -. = --.,-
ir 0)"
feet
hence the height h of the conical pendulum is dependent only
on the angular velocity about N, being inversely proportional
to the square of that quantity.
Motion in a Circle : Simple Harmonic Motion 73
Since h or / cos a = ~, to2 = f and
A
~ V h
Also the time of one complete revolution of the pendulum is —
angle in a circle 2?r /h
— - — 27T.A / ~
angular velocity cu \/ g
the period of revolution being proportional to the square root
of the height of the pendulum, and the number of revolutions
per minute being therefore inversely proportional to the square
root of the height. This principle is made use of in steam-
engine governors, where a change
in speed, altering the height of a
modified conical pendulum, is
made to regulate the steam
supply.
67. Motion in a Vertical
Circle. — Suppose a particle or
small body to move, say, contra-
clockwise in a vertical circle with
centre O (Fig. 47). It may be
kept in the circular path by a
string attached to O, or by an inward pressure of a circular
track. Taking the latter instance —
Let R = the normal inward pressure of the track ;
W= the weight of the rotating body in pounds;
v = its velocity in feet per second in any position P
such that OP makes an angle 6 to the vertical
OA, A being the lowest point on the circum-
ference j
Z/A = the velocity at A ;
r = the radius of the circle in feet.
W
Then the kinetic energy at A is J— z\2
&
At P the potential energy is W X AN, and the kinetic energy
W
is \— v2, and since there is no work done or lost between A
and P, the total mechanical energy at P is equal to that at A
(Art. 61). Therefore —
FIG. 47.
74 Mechanics for Engineers --
W W
1— .02 +W.AN - \-v*
hence V* + 2^. AN = z/A2 . . . . (i)
Neglecting gravity, the motion in a circle would be uniform, and
W z'2
would cause a reaction — • — from the track (Art. 63). And
in addition the weight has a component W cos 0 in the
direction OP, which increases the inward reaction of the track
by that amount ; hence the total normal pressure —
R=W^2 + Wcos0 . . . . (2)
The value of R at any given point can be found by sub-
stituting for v from equation (i) provided z>A is known. The
least value of R will be at B, the highest point of the circle,
where gravity diminishes it most. If z>A is not sufficient to
make R greater than zero for position B, the particle will
not describe a complete circle. Examining such a case, the
condition, in order that a complete revolution may be made
without change in the sign of R, is —
RB>o
W ?/ 2
U -• - +Wcos i8o°>o
g r
or, since cos 180° = — i —
g r
or £'B2 > gr
and since z>B2 = v\ — 2g- AB = v? — 4gr, substituting for z'B2,
the condition is —
i.e. the velocity at A must be greater than that due to falling
through a height fr, for which the velocity would be »J $gr
(Art. 28). For example, in a centrifugal railway ("looping the
loop ") the necessary velocity on entering the track at the
Motion in a Circle: Simple Harmonic Motion 75
lowest point, making no allowance for frictional resistances,
may be obtained by running down an incline of height greater
than two and a half times the radius of the circular track.
If the centripetal force is capable of changing sign, as in the
case of the pressure of a tubular track, or the force in a light
stiff radius rod supporting the revolving weight, the condition
that the body shall make complete revolutions is that z>B shall
be greater than zero, and since z'B2 = v£ — 4gr, the condition is —
V >
VA. >
i.e. the velocity at A shall be greater than that due to falling
through a height equal to the diameter of the circle. Similarly,
the position at which the body will cease to describe a circular
track (in a forward direction) if z>A is too small for a complete
circuit, when the force can change sign and when it can not,
may be investigated by applying equations (i) and (2), which
will also give the value of R for any position of the body.
The pendulum bob, suspended by a thread, is of course
limited to oscillation of less than a semicircle or to complete
circles.
Example I. — At what speed will a locomotive, going round a
curve of looo-feet radius, exert a horizontal thrust on the outside
rail equal to T^0 of its own weight ?
Let W = the weight of loco,
v = its velocity in feet per second.
W •Jli
Centrifugal thrust = ™ . -'
g 1000
= 17*95 feet per second, equivalent
to 12*22 miles per hour
Example 2. — A uniform disc rotates 250 times per minute
about an axis through its centre and perpendicular to its plane.
It has attached to it two weights, one of 5 Ibs. and the other of 7
Ibs., at an angular distance of 90° apart, the first being I foot
and the second 2 feet from the axis. Find the magnitude and
direction of the resultant centrifugal force on the axis. Find, also,
76 Mechanics for Engineers
where a weight, of 12 Ibs. must be placed on the disc to make the
resultant centrifugal force zero.
The angular velocity is "^—^ ~ radians per second
7 The centrifugal pull F: (Fig. 48) is)
then-i- - ^5-\2_ k= 106 Ibs.
— — I x
and the centrifugal pull F2 is
v 0
32-2- V 3 •
^Fi . hence the resultant R of F! and F2 at right
I' *' angles is —
FIG. 48. R = VIQ62 + 2972 = 315 Ibs.
at an angle tan*1 -— = tan'"1 0*357 = 19-6° to the direction of F2
(Arts. 24 and 44)
To neutralize this, a force of 315 Ibs. will be required in the opposite
direction.
Let x = radius in feet of the 12-lbs. weight placed at 180 — 19*6
or 160-4° contra-clockwise from F2.
3
hence x — 1*23 feet
Example 3.— Find in inches the change in height of a conical
pendulum making 80 revolutions per minute when the speed
increases two per cent.
The increase in speed is r§o x 80 = 1*6 revolutions per minute
to 8r6 revolutions per minute.
The height is ~ (Art. 66), where « is the angular velocity in
radians per second.
At 80 revolutions per minute the angular velocity is —
2ir X 80 Sir ,.
- .-= — radians per second
60 3
hence the height h^ — - 2- =
= 0*4585 foot
Motion in a Circle: Simple Harmonic Motion 77
At 8 1 '6 revolutions per minute the angular velocity is —
2w x 8r6 8-i67T
— -- _ -- radians per second
and the height is h&.§ — ?,, , </ — 0*4411 foot
'
hence the decrease in height is > = foot Qr inch
0*4585 — 0*4411 3
Example 4. — A piece of lead is fastened to the end of a string
2 feet long, the other end of which
is attached to a fixed point. With i\ C.— <— , &>
what velocity must the lead be pro-
jected in order to describe a hori-
zontal circle of 2 feet diameter ?
Let OP, Fig. 49, represent the
string ; then the horizontal line PN
is to be i foot radius.
In the vector triangle abc, ab
represents W, the weight of lead,
be the tension T of the string OP, and ac their resultant ; then —
NP _ tff _ W ^2_1_w= v°
ON ~~ ab~ g ' r ' ~^xi
where v = velocity in feet per second ;
and v - 4*309 feet per second
Exercise 5. — A stone weighing \ Ib. is whirling in a vertical
circle at the extremity of a string 3 feet long. Find the velocity of
the stone and tension of the string — (i) at the highest position, (2)
at lowest, (3) midway between, if the velocity is the least possible
for a complete circle to be described.
If the velocity is the least possible, the string will just be slack
when the stone is at the highest point of the circle.
Let v0 be the velocity at the highest point, where the weight
just supplies the centripetal force ;
(1) Then - x — x — = -
4 A 32*2 3 4
^o2 = 3 x 32'2 = 96*6
and VQ = 9*83 feet per second.
(2) At the lowest point let the velocity be i\ feet per second.
7 8 Mechanics for Engineers
Since there is no loss of mechanical energy, the gain of kinetic
energy is \ x 6 foot-lbs., hence —
ill ill i ,
2-4X*X^=i-4-?* +4'6
and z/i2 = z/o2 + 2 .g . 6
= 96-6 + 386-4 - 483 (or 5x^x3)
vl = V483 = 22 feet per second (nearly)
and the tension is "j ,
i i i 483 > =— + — o^ = i'5lbs., or six times the weight
~~ -f- — « - • . - I 4 I2o'o
of the stone
(3) When the string is horizontal, if v' - velocity in feet per
second —
• Mi11!/ Ill .1
similarly, -.-y/«=-.-.-zV 4- j- 3
' *«'2 = V + <2£ X 3
= 96-6 + 193-2
v' = V289'8 = 17 feet per second
and the tension is \
i i 289/8 > = 075 lb., or three times the weight of the
4" 3^2 x 3 J stone
EXAMPLES VIII.
1. How many circuits per minute must a stone weighing 4 ozs.
make when whirled about in a horizontal circle at the extremity of a string
5 feet long, in order to cause a tension of 2 Ibs. in the string ?
2. At what speed will a locomotive produce a side thrust equal to ^ of
its own weight on the outer rail of a level curved railway line, the radius of
the curve being 750 feet ?
3. What is the least radius of curve round which a truck may run on
level lines at 20 miles per hour without producing a side thrust of more
than Tj^ of its own weight ?
4. How much must the outer rail of a line of 4 feet 8.J inches gauge be
elevated on a curve of 800 fact radius in order that a train may exert a
thrust normal to the track when travelling at 30 miles per hour ?
5. The outer rail of a pair, of 4 feet 8| inches gauge, is elevated 2\ inches,
and a train running at 45 miles per hour has no thrust on the flanges of
either set of wheels. What is the radius of the curve ?
6. At what speed can a train run round a curve of 1000 feet radius
without having any thrust on the wheel flanges when the outer rail is laid
I -5 inches above the inner one, and the gauge is 4 feet 8=] inches ?
7. To what angle should a circular cycle-track of 15 laps to the mile be
Motion in a Circle: Simple Harmonic Motion 79
banked for riding upon at a speed of 30 miles per hour, making no allow-
ance for support from friction ?
8. A string 3 feet long, fixed at one end, has attached to its other
end a stone which describes a horizontal circle, making 40 circuits per
minute. What is the inclination of the string to the vertical ? What is its
tension ?
9. What percentage change of angular speed in a conical pendulum
will correspond to the decrease in height of 3 per cent. ?
10. The revolving ball of a conical pendulum weighs 5 Ibs., and the
height of the pendulum is 8 inches. What is its speed ? If the ball is
acted upon by a vertical downward force of i lb., what is then its speed
when its height is 8 inches? Also what would be its speed in the case of
a vertical upward force of I lb. acting on the ball ?
1 1 . What will be the inclination to the vertical of a string carrying a
weight suspended from the roof of a railway carriage of a train going
round a curve of 1000 feet radius at 40 miles per hour ?
12. A body weighing ;' lb., attached to a string, is moving in a vertical
circle of 6 feet diameter. If its velocity, when passing through the lowest
point, is 40 feet per second, find its velocity and the tension of the string
when it is 2 feet and when it is 5 feet above the lowest point.
68. Simple Harmonic Motion. — This is the simplest
type of reciprocating motion. If a point Q (Fig. 50) describes
a circle AQB with constant angular velocity, and P be the
rectangular projection of Q on a fixed diameter AB of the
circle, then the oscillation to and fro of P along AB is defined
as Simple Harmonic Motion.
Let the length OA of the radius be a feet, called the
amplitude of oscillation.
Let W be the angular velocity of Q in radians per second.
Let 0 be the angle AOQ in radians, denoting any position
ofQ.
Suppose the motion of Q to be, say, contra-clockwise.
A complete vibration or oscillation of P is reckoned in this
country as the path described by P whilst Q describes a
complete circle.
Let T = the period in seconds of one complete vibration ;
then, since this is the same as that for one complete circuit
made by Q —
_ radians in one circle _ 2ir
~ radians described per second ~~ w
8o
Mechanics for Engineers
Let x = distance OP of ]
towards A, the
j \^
' ***,
r*
~L ^ !S
o x ^
^ ^
^ ^
t ^ **
r i s
^
^
s ^
^ ^
^ ^
y
^ 's
° ~
~7 s^
j \
^ ^
£
*•" . 7
•^ ,
^ r
^ ^
•s
> r
p >
^_ \
o
<& L^ ^
t ^
j s 5 1 _
^ — ^ i/
s
1 S > ^
^ r
3s
. .?
X. ^
\ ^ ^
^ Ji
^^
^ ~?
o
s
/
£*
" 3 j
\ ~7 *T
A '
P f
X ^
^ S ^
L "
' ^ .
^ ' . r1
7 s,
£ L. '
^ v
/ j"i
^
^ ^'
s ^
' ^. O V / U ^ !
jM
^^ <" « :
"~ -~~
7^ >
S^
z
j
\
£ i-
, '
r ^
T ^ '
3 V
^ y
^S TE
"^--fc •
OSIH!
of P from O in feet, reckoned positive
= a cos 6 ;
and let v = velocity of P in
feet per second in position 0.
Draw OS perpendicular
to OQ to meet the circum-
ference of the circle AQSB
in S, and draw SM perpen-
dicular to AB to meet it in M.
Then for the position or
phase shown in the figure,
the velocity of Q is wa (Art.
33) in the direction perpen-
dicular to OQ, i.e. parallel to
OS. Resolving this velocity
along the diameter AB, OSM
being a vector triangle, the
component velocity of Q
parallel to AB is -^ X w#,
- &
or ota sin 0, or w . OM. This
is then the velocity of P
towards O, the mid -path.
Since sin 6 =
OM
OS
which gives the velocity of
P in terms of the amplitude
and position.
Or, if OS represents geo-
metrically the velocity of Q,
then OM represents that 01
P to the same scale.
Motion in a Circle: Simple Harmonic Motion 8 1
Acceleration of P. — The acceleration of Q is w2# along
QO towards O (Art. 62). Resolving this acceleration, the
PO
component in direction AB is <o2# X QQ, or o>2<2 . cos 0, or
a/2 . x, towards O ; and it should be noted that at unit distance
from O, when x = i foot, the acceleration of P is w2 feet per
second per second.
The law of acceleration of a body having simple harmonic
motion, then, is, that the acceleration is towards the mid-path
and proportional to its distance from that point. When the
body is at its mid-path, its acceleration is zero ; hence there is
no force acting upon it, and this position is one of equilibrium
if the body has not any store of kinetic energy. Conversely,
if a body has an acceleration proportional to its distance from
a fixed point, O, it will have a simple harmonic motion. If
the acceleration at unit distance from O is p. feet per second
per second (corresponding to o>2 in the case just considered),
by describing a circle with centre O about its path as diameter,
we can easily show that^the body has simple harmonic motion,
and by taking w = vV' P corresponding to w2 in the above
case, we can state its velocity and acceleration at a distance
x from its centre of motion O, and its period of vibration, viz.
velocity v at x feet from O is vV • Va'2 — x\ or J~ji,(a* — x2).
Acceleration at x feet from centre O is p.x, and the time
27T
OI a complete vibration is — ==«
*v.
Alternating Vectors. — We have seen that, the displace-
ment of P being OP, the acceleration is proportional also to
OP, and the velocity to OM ; so that OP and OM are vectors
representing in magnitude and direction the displacement and
velocity of P. Such vectors, having a fixed end, O, and of
length varying according to the position of a rotating vector,
OQ or OS, are called " alternating vectors." It may be noted
that the rate of change of an alternating vector, OP, of ampli-
tude a is represented by another alternating vector, OM, of the
same period, which is the projection of a uniformly rotating
vector of length OS = w . OQ or ua (to a different scale), and
one right angle in advance of the rotating vector OQ, of which
G
82 Mechanics for Engineers
OP is the projection. A little consideration will show that the
rate of change of the alternating vector OM follows the same
law (rate of change of velocity being acceleration), viz. it is
represented by a third alternating vector, ON, of the same
period, which is the projection of a uniformly rotating vector
of length OQ' = to . OS or <o2# (to a different scale), and one
right angle in advance of the rotating vector OS, of which OM
is the projection.
The curves of displacement, velocity, and acceleration of
P on a base of angles are shown to the right hand of Fig. 50.
The base representing angles must also represent time, since
the rotating vectors have uniform angular velocity w. The
Q Q
time t = - seconds, since <o = -. The properties of the curves
of spaces, velocities, and accelerations (Arts. 4, 14, and 16)
are well illustrated by the curves in Fig. 50, which have been
drawn to three scales of space, velocity, and acceleration by
projecting points 90° ahead of Q, S, and Q' on the circle on
the left. The acceleration of P, which is proportional to the
displacement, may properly be considered to be of opposite
sign to the displacement, since the acceleration is to the left
from P to O when the displacement OP is to the right of O.
The curves of displacement and acceleration are called " cosine
curves," the ordinates being proportional to the cosines of angle
POQ, or 0, or w/. Similarly, the curve of velocity is called a
" sine curve." The relations between the three quantities may
be expressed thus —
Displacement (x) : velocity (v) : acceleration
= a cos <o/ : (7w sin w/ : — «<o2 cos <o/
Curved Path. — If the point P follows a curved path
instead of the straight one AB, the curved path having the
same length as the straight one, and if the acceleration of
the point when distant x feet from its mid-path is tangential
to the path and of the same magnitude as that of the point
following the straight path AB when distant x feet from mid-
path, then the velocity is of the same magnitude in each case.
This is evident, for the points attain the same speeds in the
Motion in a Circle: Simple Harmonic Motion 83
same intervals of time, being, under the same acceleration,
always directed in the line of motion in each case. Hence
the periodic times will be the same in each case, viz. -
\V
where p, is the acceleration in feet per second per second
along the curve or the straight line, as the case may be.
69. There are numerous instances in which bodies have
simple harmonic motion or an approximation to it, for in
perfectly elastic bodies the straining force is proportional to
the amount of displacement produced, and most substances
are very nearly perfectly elastic over a limited range.
A common case is that of a body hanging on a relatively
light helical spring and vibrating vertically. The body is
acted upon by an effective accelerating force proportional to
its distance from its equilibrium position, and, since its mass
does not change, it will have an acceleration ( - - ) also
proportional to its displacement from that point (Art. 40), and
therefore it will vibrate with simple harmonic vibration.
Let W = weight of vibrating body in pounds.
e = force in pounds acting upon it at i foot from its
equilibrium position, or per foot of displace-
ment, the total displacement being perhaps
less than i foot. This is sometimes called
the stiffness of the spring.
Then e.x= force in Ibs. x feet from the equilibrium position
and if //, = acceleration in feet per second per second i foot
from the equilibrium position or per foot of displacement
accelerating force _ . W _ eg
U, — — — 6 ~7" — ^fr
mass g W
hence the period of vibration is -^ or 2ir A/ — (Art. 68)
The maximum force, which occurs when the extremities
of the path are reached, is e.a, where a is the amplitude of
84
Mechanics for Engineers
the vibration or distance from equilibrium position to either
extremity of path, in feet.
The crank-pin of a steam engine describes a circle ABC
(Fig. 51), of which the length of crank OC is the radius, with
FIG. 51.
fairly constant angular velocity. The piston P and other
reciprocating parts are attached to the crank-pin by a con-
necting-rod, DC, and usually move to and fro in a straight
line, AP, with a diameter, AB, of the crank-pin circle. If the
connecting-rod is very long compared to the crank-length,
the motion is nearly the same as that of the projection N
of the crank-pin on the diameter AB of the crank-pin circle,
which is simple harmonic. If the connecting-rod is short,
however, its greater obliquity modifies the piston-motion to
a greater extent.
70. Energy stored in Simple Harmonic Motion.—
If e = force in pounds at unit distance, acting on a body of
weight W Ibs. having simple harmonic motion, the force at a
distance x is ex, since it is proportional to the displacement.
Therefore the work done in displacing the body from its equili-
brium through x feet is \ex* (Art. 54 and Fig. 35). This
energy, which is stored in some form other than kinetic energy
when the body is displaced from its equilibrium position,
reaches a maximum \ecfr when the extreme displacement a
(the amplitude) has taken place, and the effective accelerating
force acting on the body is ea. In the mid-position of the
body (x = o), when its velocity is greatest and the force acting
on it is nil, the energy is wholly kinetic, and in other inter-
mediate positions the energy is partly kinetic and partly
otherwise, the total being constant if there are no resistances.
OF
FIG. 52.
Motion in a Circle: Simple Harmonic Motion 8'
Fig. 52 shows a diagram of work stored for various dis-
placements of a body having simple harmonic motion. The
amplitude OA = a, and
therefore the force at A
is ae, which is represented
by AD, and the work done
in moving from O to A is >» ^
represented by the area B 0
AOD (Art. 54 and Fig. 35).
At P, distant x feet from
O, the work done in motion
from O is ^ex2, represented
by the area OHP, and the
kinetic energy at P is
therefore represented by
the area DAPH.
71. Simple Pendulum. — This name refers strictly to a
particle of indefinitely small dimensions and yet having weight,
suspended by a perfectly flexible weightless thread from a fixed
point, about which, as a centre, it swings freely in a circular
arc. In practice, a small piece of
heavy metal, usually called a pendulum
bob, suspended by a moderately long
thin fibre, behaves very nearly indeed
like the ideal pendulum defined above,
the resistances, such as that of the
atmosphere, being small.
Let O, Fig. 53, be the point of
suspension of the particle P of a
simple pendulum.
Let OP, the length of thread, be
/ feet.
Let 0 = angle AOP in radians which OP makes with the
vertical (OA) through O in any position P of the particle.
Draw PT perpendicular to OP, i.e. tangent to the arc of
motion to meet the vertical through O in T.
The tension of the thread has no component along the
direction of motion (PT) at P. The acceleration along PT is
86 Mechanics for Engineers
then g sin 0, since PT is inclined 6 to the horizontal (Art. 28).
If 0 is very small, sin 6 may be taken equal to 0 in radians.
(If 6 does not exceed 5°, the greatest error in this approxi-
mation is less than i part in 800.) Hence the acceleration
arc AP
along PT is gO approximately. And 6 = — -~ — ~p ; therefore
°* X arc AP
acceleration along PT =* -, , and the acceleration is
proportional to the distance AP, along the arc, of P from A,
being i per foot of arc. Hence the time of a complete oscilla-
tion in seconds is —
*** */-,= ^ /-(Art 68)
and the velocity at any point may be found, as in Art. 68, for
any position of the swinging particle.
In an. actual pendulum the pendulum bob has finite dimen-
sions, and the length / will generally be somewhat greater than
that of the fibre by which it is suspended. The ideal simple
pendulum having the same period of swing as an actual pen-
dulum of any form is called its simple equivalent penduhun.
For this ideal pendulum the relation / = STTA/ - holds, and
<b
per
therefore / = --A,, from which its length in feet may be
4?r
calculated for a given time, /, of vibration.
The value of the acceleration of gravity, g, varies at different
parts of the earth's surface, and the pendulum offers a direct
means of measuring the value of this quantity g, viz. by
accurate timing of the period of swing of a pendulum of known
length. The length of an actual pendulum, i.e. of its simple
equivalent pendulum, can be calculated from its dimensions.
Example I. — A weight rests freely on a scale-pan of a spring
balance, which is given a vertical simple harmonic vibration cf
period 0*5 second. What is the greatest amplitude the vibration
may have in order that the weight may not leave the pan ? What
is then the pressure of the weight on the pan in its lowest position ?
Let a = greatest amplitude in feet.
Motion in a Circle : Simple Harmonic Motion 87
The greatest downward force on the body is its own weight,
and therefore its greatest downward acceleration is g, occurring
when the weight is in its highest position and the spring is about
to return. Hence, if the scale-pan and weight do not separate,
the downward acceleration of the pan must not exceed g, and
therefore the acceleration must not exceed - per foot of dis-
placement.
/2ir\2
The acceleration per foot of displacement is ( — j ;
therefore ~ >Sa
i.e. a ^> o'204 feet or 2*448 inches
If the balance has this amplitude of vibration, the pressure
between the pan and weight at the lowest position will be equal to
twice the weight, since there is an acceleration g upwards which
must be caused by an effective force equal to the weight acting
upwards, or a gross pressure of twice the weight from which the
downward gravitational force has to be subtracted.
Example 2. — Part of a machine has a reciprocating motion,
which is simple harmonic in character, making 200 complete oscilla-
tions in a minute ; it weighs 10 Ibs. Find (i) the accelerating force
upon it in pounds and its velocity in feet per second, when it is
3 inches from mid-stroke ; (2) the maximum accelerating force ;
and (3) the maximum velocity if its total stroke is 9 inches, i.e. if
its amplitude of vibration is 4^ inches.
Time of i oscillation = - = 0*3 second
200
therefore the acceleration per foot ) /2 7r\2 4oo7r2
. . > = ( — Is-— feet per second
distance from mid-stroke ] \°'3/ 9
per second
and the accelerating force 0*25 foot from mid-stroke on 10 Ibs. is —
TO AOOir2
- X O*25 X - — = 34'o8 Ibs.
32'2 9
and the maximum accelerating force 4^ inches from mid-stroke is
i '5 times as much as at 3 inches, or 34*08 x 1*5 = 51*12 Ibs.
88
Mechanics for Engineers
The maximum velocity in feet per second occurring at mid-stroke
= amplitude in feet x *J acceleration per foot of displacement
(Art. 68)
= amplitude in feet x
period
= x = = 7'85 feet per second
(Art. 68)
Velocity at 3 inches)
from mid-stroke j ' 4*5
FIG. 54-
= 7-85 x — _— = 5*85 feet per second
Example 3. — The crank of an engine makes 150 revolutions
per minute, and is 1*3 feet long. It is driven by a piston and a very
long connecting rod (Fig. 51), so that the motion of the piston may
be taken as simple harmonic. Find the
piston velocity and the force necessary
to accelerate the piston and recipro-
cating parts, weighing altogether 300
Ibs., (i) when the crank has turned
through 45° from its position (OB) in
line with and nearest to the piston
path ; (2) when the piston has moved
forward 0^65 foot from the end of its
stroke.
Let ABC (Fig. 54) be the circular
path 1-3 feet radius of the crank-pin, CN the perpendicular from
a point C on the diameter AB.
The angular velocity of crank OC is — 7 — — = $* radians per second
(i) The motion of the piston being taken as that of N, the
acceleration of piston when the crank-pin is at C is —
(5?r)2 x 1-3 x cos 45° (wVcos 6t Art. 68)
and the accelerating force is —
^ x (57r)2 x 1-3 x -/- = 21 10 Ibs.
The velocity is —
5 IT x i '3 x sin 45° = H'43 feet per second
Motion in a Circle: Simple Harmonic Motion 89
(2) When BN = 0-65 foot, ON = OB - BN = 1-3 - 0-65 = o 65
,v ON
foot, and CON = cos"1 QC = cos'1 \ = 60°. The accelerating
force is then —
^ x (5*)2 x 1-3 x i = 1493 Ibs.
and the velocity is —
5?r x i '3 x sin 60° = 17*67 feet per second
Example 4. — A light helical spring is found to deflect o'4 inch
when an axial load of 4 Ibs. is hung on it. How many vibrations
per minute will this spring make when carrying a weight of
15 Ibs.?
The force per foot of deflection is 4 -i- — =120 Ibs.
hence the time of vibration is 2ir^ / * 5 "' _ 0*^0,2 second
V 32-2 x 120
and the number of vibrations per minute is — --- = I53'2
Example 5. — Find the length of a clock pendulum which will
make three beats per second. If the clock loses i second per
hour, what change is required in the length of pendulum ?
Let / = length of pendulum in feet.
Time of vibration = £ second
- et = ro inches
The clock loses i second in 3600 seconds, i.e. it makes 3599 x 3
beats instead of 3600 x 3. Since /oc ^2oc —^ where n — number
of beats per hour, therefore —
correct length _ 35992 _
i -09 inches - 36oo2 ~ (I " *«™
= i ~ TsW approximately
therefore shortening required = -ja~* inches = o-ooo6o6 inch
EXAMPLES IX.
I. A point has a simple harmonic motion of amplitude 6 inches and
period 1*5 seconds. Find its velocities and accelerations O'l second, o '2
second, and 0*5 second after it has left one extremity of its path.
go Mechanics for Engineers
2. A weight of 10 Ibs. hangs on a spring, which stretches 0^15 inch
per pound of load. It is set in vibration, and its greatest acceleration
whilst in motion is i6'i feet per second per second. What is the ampli-
tude of vibration ?
3. A point, A, in a machine describes a vertical circle of 3 feet diameter,
making 90 rotations per minute. A portion of the machine weighing 400
Ibs. moves in a horizontal straight line, and is always a fixed distance
horizontally from A, so that it has a stroke of 3 feet. Find the accele-
rating forces on this portion, ( I ) at the end of its stroke ; (2) 9 inches from
the end j and (3) 0-05 second after it has left the end of its stroke.
4. A helical spring deflects \ of an inch per pound of load. How many
vibrations per minute will it make if set in oscillation when carrying a load
of 12 Ibs. ?
5. A weight of 20 Ibs. has a simple harmonic vibration, the period of
which is 2 seconds and the amplitude 1*5 feet. Draw diagrams to stated
scales showing (i) the net force acting on the weight at all points in its
path ; (2) the displacement at all times during the period ; (3) the velocity
at all times during the period ; (4) the force acting at all times during the
period.
6. A light stiff beam deflects 1*145 inches under a load of I ton at
the middle of the span. Find the period of vibration of the beam when so
loaded.
7. A point moves with simple harmonic motion ; when 0*75 foot from
mid-path, its velocity is 1 1 feet per second ; and when 2 feet from the
centre of its path, its velocity is 3 feet per second. Find its period and its
greatest acceleration.
8. How many complete oscillations per minute will be made by a
pendulum 3 feet long ? g — 32 '2.
9. A pendulum makes 3000 beats per hour at the equator, and 3011 per
hour near the pole. Compare the value of g at the two places.
CHAPTER V
STATICS— CONCURRENT FORCES — FRICTION
72. THE particular case of a body under the action of several
forces having a resultant zero, so that the body remains at rest,
is of very common occurrence, and is of sufficient importance
to merit special consideration. The branch of mechanics which
deals with bodies at rest is called Statics.
We shall first consider the statics of a particle, i.e. a body
having weight, yet of indefinitely small dimensions. Many of
the conclusions reached will be applicable to small bodies in
which all the forces acting may be taken without serious error
as acting at the same point, or, in other words, being con-
current forces.
73. Resolution and Composition of Forces in One
Plane. — It will be necessary to recall some of the conclusions
of Art. 44, viz. that any number of concurrent forces can be
replaced by their geometric sum acting at the intersection of
tbe lines of action of the forces, or by components in two
standard directions, which are for convenience almost always
taken at right angles to one another.
Triangle and Polygon of Forces. — If several forces, say four,
as in Fig. 55, act on a particle, and ab, be, cd, de be drawn in
succession to represent the forces of 7, 8, 6, and 10 Ibs. respec-
tively, then ae, their geometric sum (Art. 44), represents a force
which will produce exactly the same effect as the four forces,
i.e. ae represents the resultant of the four forces. If the final
point e of the polygon abcde coincides with the point a, then
the resultant ae is nil, and the four forces are in equilibrium.
This proposition is called the Polygon of Forces, and may be
92
Mechanics for Engineers
stated as follows : If several forces acting on a particle be
represented in magnitude and direction by the sides of a closed
polygon taken in order, they are in equilibrium. By a closed
polygon is meant one the last side of which ends at the point
FIG. 55.
from which the first side started. The intersection of one side
of the polygon with other sides is immaterial.
The polygon of forces may be proved experimentally by
means of a few pieces of string and weights suspended over
almost frictionless pulleys, or by a number of spring balances
and cords.
This proposition enables us to find one force out of several
keeping a body in equilibrium if the remainder are known, viz.
by drawing to scale an open polygon of vectors corresponding
to the known forces, and then a line joining its extremities is
the vector representing in one direction the resultant of the
other forces or in the other direction the remaining force neces-
sary to maintain equilibrium, sometimes called the equilibrant.
For example, if forces Q, R, S, and T (Fig. 56) of given
magnitudes, and one other force keep a particle P in equili-
brium, we can find the remaining one as follows. Set out vectors
ab, be, cd, and de in succession to represent Q, R, S, and T
respectively ; then ae represents their resultant in magnitude
and direction, and ea represents in magnitude and direction the
remaining force which would keep the particle P in equilibrium,
or the equilibrant.
Statics — Concurrent Forces — Friction
93
Similarly, if all the forces keeping a body in equilibrium
except two are known, and the directions of these two are
known, their magnitudes may be found by completing the
FIG. 56.
open vector polygon by two intersecting sides in the given
directions.
In the particular case of three forces keeping a body in
equilibrium, the polygon is a triangle, which is called the
Triangle of Forces. Any triangle having its sides respectively
parallel to three forces which keep a particle in equilibrium
represents by its sides the respective forces, for a three-sided
closed vector polygon (i.e. a triangle) with its sides parallel
and proportional to the forces can always be drawn as directed
for the polygon of forces, and any other triangle with its sides
parallel to those of this vector triangle has its sides also pro-
portional to them, since all triangles with sides respectively
parallel are similar. The corresponding proposition as to any
polygon with sides parallel to the respective forces is not true
for any number of forces but three.
74. Lami's Theorem. — If three forces keep a particle
in equilibrium, each is proportional to the sine of the angle
between the other two.
Let P, Q, and R (Fig. 57) be the three forces in equilibrium
acting at O in the lines OP, OQ, and OR respectively. Draw
any three non-concurrent lines parallel respectively to OP, OQ,
and OR, forming a triangle abc such that ab is parallel to OP, be
to OQ, and ca to OR. Then angle abc - 180 - POQ, angle
94
Mechanics for Engineers
bca = 1 80 — QOR, and angle cab = 180 — ROP, and there-
fore—
sin abc = sin POQ
sin tica = sin QOR
sin cab = sin ROP
In the last article, it was shown that any triangle, such as
FIG.
P Q R
ab be ca
ab be
ca
sin bca sin cab
ab be
sin abc
ca
abc, having sides respectively parallel to OP, OQ, and OR, has
its sides proportional respectively to P, Q, and R, or —
(l)
. (2)
sin QOR sin ROP sin POQ
and multiplying equation (i) by equation (2) —
P Q R
sin QOR "~ sin ROP ~ sin POQ
that is, each of the forces P, Q, and R is proportional to the
sine of the angle between the other two.
This result is sometimes of use in solving problems in
which three forces are in equilibrium.
75. Analytical Methods. — Resultant or equilibrant forces
of a system, being representable by vectors, may be found by
the rules used for resultant velocities, i.e. (i) by drawing
Statics — Concurrent Forces — Friction 95
vectors to scale ; (2) by the rules of trigonometry for the solu-
tions of triangles ; (3) by resolution into components in two
standard directions and subsequent compounding as in Art. 25.
We now proceed to the second and third methods.
To compound two forces P and Q inclined at an angle 6
to each other.
Referring to the vector diagram abc of Fig. 58 (which need
not be drawn, and is used here for the purpose of illustration
and explanation) by the rules of trigonometry for the solution
of triangles —
(ac)* = (ab)* + (&)2 - 2 ab .be cos abc
- (abf + (btf + 2 ab.bc cos 0
hence if ab and be represent P and Q respectively, and R is the
value of their resultant —
R2 = p^ 4. Q^ 4. 2pQ cos o
from which R may be found by extracting the square root, and
its inclination to, say, the direction of Q may be found by
considering the length of the perpendicular ce from c on ad
produced —
Since ec = dc sin 0
and de = dc cos 6
ec dc sin 6 P sin 0
tan cad = — =
ae ad + dc cos <9 ~ Q -f P cos 0
which is the tangent of the angle between the line of action of
the resultant R and that of the force Q.
When the resultant or equilibrant of more than two
concurrent forces is to be found, the method of Art. 25 is
Mechanics for Engineers
sometimes convenient. Suppose, say, three forces F1} F2,
and F3 make angles «, /3, and y respectively with some chosen
fixed direction OX, say that of the line of action of F]} so
that a = .o (Fig. 59).
Y
Fx
Resolve F1? F2, and F3 along OX and along OY perpen-
dicular to OX.
Let Fx be the total of the components along OX,
and let Fy „ „ „ „ „ OY.
Let R be the resultant force, and 9 its inclination to OX ;
then —
Fx = Fj + F2 cos /3 + F, cos y
FY = o + F2 sin /3 + F3 sin y
and compounding Fx and FY, two forces at right angles, R is
proportional to the hypotenuse of a right-angled triangle, the
other sides of which are proportional to Fx and FY ; hence—
R2 = Fx2 + FY2
and R= V"(FX2 + FY2)
The direction of the resultant R is given by the relation —
If the forces of the system are in equilibrium, that is, if
the resultant is nil—
R2 = o
or Fx2 4- FY2 = o
This is only possible if both Fx = o and FY = o.
Statics — Concurrent Forces — Friction
97
The condition of equilibrium, then, is, that the components
in each of two directions at right angles shall be zero. This
corresponds to the former statement, that if the forces are in
equilibrium, the vector polygon of forces shall be closed, as
will be seen by projecting on any two fixed directions at right
angles, the sides of the closed polygon, taking account of the
signs of the projections. The converse statement is true, for
if Fx = o and FY = o, then R = o ; therefore, if the com-
ponents in each of two standard directions are zero, then the
forces form a system in equilibrium, corresponding to the
statement that if the vector polygon is a closed figure,
the forces represented by its sides are in equilibrium.
Example i. — A pole rests vertically with its base on the ground,
and is held in position by five ropes, all in the same horizontal
plane and drawn tight. From the pole the first rope runs due
north, the second 75° west of north, the third 15° south of west, and
the fourth 30° east of south. The tensions of these four are 25 Ibs.,
FIG. 60.
1 5 Ibs., 20 Ibs., and 30 Ibs. respectively. Find the direction of the
fifth rope and its tension.
The directions of the rope have been set out in Fig. 60, which
H
98 Mechanics for Engineers
represents a plan of the arrangement, the pole being at P. The
vector polygon abcde, representing the forces in the order given,
has been set out from a and terminates at e. ae has been
drawn, and measures to scale 18-9 Ibs., and the equilibrant ea is the
pull in the fifth rope, and its direction is 7° north of east from the
pole.
Example 2. — Two forces of 3 Ibs. and 5 Ibs. respectively act on
a particle, and their lines of action are inclined to each other at an
angle of 70°. Find what third force will keep the particle in
equilibrium.
The resultant force R will be of magnitude given by the
relation —
,R2 = 32 + 52+ 2<3>5 cos 70°
= 9 + 25 + (30 x 0-3420) = 34 + 10-26 = 44-26
R = /v/44'42 = 6-65 Ibs.
And R is inclined to the force of 5 Ibs. at an angle the tangent of
which is —
I31bs 3 sin 70° = __UL?'9397_
5 + 3 cos 70° 5 + (3 x 0-3420)
2-8171
- ^ — -r = 0*467
6-026
which is an angle 25°. The equilibrant or
third force required to maintain equilibrium
is, therefore, one of 6*65 Ibs., and its line of
action makes an angle of 180° — 25° or 155°
FIG. 61. with the line of action of the force of 5 Ibs.,
as shown in Fig. 61.
Example 3. — Solve Example I by resolving the forces into
components. Taking an axis PX due east (Fig. 60) and PY due
north, component force along PX —
Fx - - 15 cos 15° - 20 cos 15° + 30 cos 60°
= (— 35 x 0*9659) + (30 x 0-5) = -18-806 Ibs.
Component force along PY—
Fy = 25 + 15 cos 75° - 20 cos 75° - 30 cos 30°
= 25 — (5 x 0*2588) — 30 x o'866o = —2-274 Ibs.
hence R2 = (i8'8i)2 + (2-27)2 = 359*3
= 1 8-96 Ibs.
Statics — Concurrent Forces — Friction 99
R acts outwards from P in a direction south of west, being inclined
to XP at an acute angle, the tangent of which is —
Fv=_2-274 _Q.I2I
Fx 18-806 "
which is the tangent of 6° 54' ; i.e. R acts in a line lying 6° 54'
south of west. The equilibrant is exactly opposite to this, hence
the fifth rope runs outwards from the pole P in a direction 6° 54'
north of east, and has a tension of 18-96 Ibs.
EXAMPLES X.
1. A weight of 20 Ibs. is supported by two strings inclined 30° and 43°
respectively to the horizontal. Find by graphical construction the tension
in each cord.
2. A small ring is situated at the centre of a hexagon, and is supported
by six strings drawn tight, all in the same plane and radiating from the
centre of the ring, and each fastened to a different angular point of the
hexagon. The tensions in four consecutive strings are 2, 7, 9, and 6 Ibs.
respectively. Find the tension in the two remaining strings.
3. Five bars of a steel roof-frame, all in one plane, meet at a point ;
one is a horizontal tie-bar carrying a tension of 40 tons ; the next is also a
tie-bar inclined 60° to the horizontal and sustaining a pull of 30 tons ; the
next (in continuous order) is vertical, and runs upward from the joint, and
carries a thrust of 5 tons ; and the remaining two in the same order radiate
at angles of 135° and 210° to the first bar. Find the stresses in the last
two bars, and state whether they are in tension or compression, i.e. whether
they pull or push at the common joint.
4. A telegraph pole assumed to have no force bending it out of the
vertical has four sets of horizontal wires radiating from it, viz. one due east,
one north-east, one 30° north of west, and one other. The tensions of the
first three sets amount to 400 Ibs., 500 Ibs., and 250 Ibs. respectively. Find,
by resolving the forces north and east, the direction of the fourth set and
the total tension in it.
5. A wheel has five equally spaced radial spokes, all in tension. If the
tensions of three consecutive spokes are 2000 Ibs., 2800 Ibs., and 2400 Ibs.
respectively, find the tensions in the other two.
6. Three ropes, all in the same vertical plane, meet at a point, and there
support a block of stone. They are inclined at angles of 40°, 120°, and
1 60° to a horizontal line in their common plane. The pulls in the first two
ropes are 150 Ibs. and 120 Ibs. respectively. Find the weight of the block
of stone and the tension in the third rope.
76. Friction. — Friction is the name given to that pro-
perty of two bodies in contact, by virtue of which a resistance
ioo Mechanics for Engineers
is offered to any sliding motion between them. The resistance
consists of a force tangential to the surface of each body at
the place of contact, and it acts on each body in such a direction
as to oppose relative motion. As many bodies in equilibrium
are held in their positions partly by frictional forces, it will be
convenient to consider here some of the laws of friction.
77. The laws governing the friction of bodies at rest are
found by experiment to be as follows : —
(1) The force of friction always acts in the direction opposite
to that in which motion would take place ' if it were absent, and
adjusts itself to the amount necessary to maintain equilibrium.
There is, however, a limit to this adjustment and to the
value which the frictional force can reach in any given case.
This maximum value of the force of friction is called the
limiting friction. It follows the second law, viz. —
(2) The limiting friction for a given pair of surfaces depends
upon the nature of the surfaces^ is proportional to the normal
pressure between them^ and independent of the area of the sur-
faces in contact.
For a pair of surfaces of a given kind (i.e. particular sub-
stances in a particular condition), the limiting friction F = //,. R,
where R is the normal pressure between the surfaces, and p, is
a constant called the coefficient of friction for the given surfaces.
This second law, which is deduced from experiment, must be
taken as only holding approximately.
78. Friction during Sliding Motion. — If the limiting
friction between the bodies is too small to prevent motion, and
sliding motion begins, the subsequent value of the frictional
force is somewhat less than that of the statical friction. The
laws of friction of motion, so far as they have been exactly
investigated, are not simple. The friction is affected by other
matter (such as air), which inevitably gets between the two
surfaces. However, for very low velocities of sliding and
moderate normal pressure, the same relations hold approxi-
mately as have been stated for the limiting friction of rest,
viz. —
F = ^R
where F is the frictional force between the two bodies, and R
Statics — Concurrent Forces — Friction
IOI
is the normal pressure between them, and ju, is a constant
coefficient for a given pair of surfaces, and which is less than
that for statical friction between the two bodies. The friction
is also independent of the velocity of rubbing.
79. Angle of Friction. — Suppose a body A (Fig. 62) is
in contact with a body B, and is being pulled, say, to the right,
the pull increasing until the limiting amount of frictional re-
sistance is reached, that is, until, the force of friction reaches a
limiting value F = jooR, where R is the normal pressure between
the two bodies, and j«, is the coefficient of friction. If R and
F, which are at right angles, are compounded, we get the
resultant pressure, S, which B exerts on A. As the friction F
increases with the pull, the inclination 0 of the resultant S of
F and R to the normal of the surface of contact, i.e. to the line
of action of R, will become greater, since its tangent is always
cb F
equal to ^ or -(Art. 75).
Let the extreme inclination to the normal be A when the
friction F has reached its limit, |u,R.
F //,R
tanA = R R =^
This extreme inclination, A, of the resultant force between
102
Mechanics for Engineers
two bodies to the normal of the common surface in contact is
called the angle of friction, and we have seen that it is the angle
the tangent of which is equal to the coefficient of friction —
tan A =
or A = tan
80. Equilibrium of a Body on an Inclined Plane.—
As a simple example of a frictional force, it will be instructive
here to consider Jjie equilibrium of a body resting on an
inclined plane, supported wholly or in part by the friction
between it and the inclined plane.
Let JM, be the coefficient of friction between the body of
weight W and the inclined plane, and let a be the inclination
of the plane to the horizontal plane. We shall in all cases
draw the vector polygon of forces maintaining equilibrium,
not necessarily correctly to scale, and deduce relations between
the forces by the trigonometrical relations between the parts of
the polygon, thus combining the advantages of vector illustra-
tion with algebraic calculation, as in Art. 75. The normal
to the plane is shown dotted in each diagram (Figs. 63-68
inclusive).
i. Body at rest on an inclined plane (Fig. 63).
\N
\
FIG. 63.
If the body remains at rest unaided, there are only two
forces acting on it, viz. its weight, W, and the reaction S of the
plane; these must then be in a straight line, and therefore S
must be vertical, i.e. inclined at an angle « 1o the normal to the
plane. The greatest angle which S can make to the normal
Statics — Concurrent Forces —Friction
103
is A, the angle of friction (Art. 79) ; therefore a cannot exceed
A, the angle of friction, or the body would slide down the
plane. Thus we might also define the angle of friction between
a pair of bodies as the greatest incline on which one body
would remain on the other without sliding.
Proceeding to supported bodies, let an external force, P,
which we will call the effort, act upon the body in stated
directions.
2. Horizontal effort necessary to start the body up the
plane. Fig. 64 shows the forces acting, and a triangle of
forces, abc.
FIG. 64.
When the limit of equilibrium is reached, and the body is
about to slide up the plane, the angle dbc will be equal to A,
the maximum angle which S can make with the normal to the
plane; then —
w = i"=tan(a + A)
or P = W tan (a + A)
which is the horizontal effort necessary to start the body up
the plane.
3. Horizontal effort necessary to start the body sliding
down the plane (Fig. 65).
When the body is about to move down the plane, the angle
cbd will be equal to the angle of friction, A ; then —
P ca
W = ^=tan(A-a)
or P = W tan (A - a)
104
Mechanics for Engineers
If a is greater than A, tkis- can only be negative, i.e. c falls
to the left of a, and the horizontal force P is that necessary
to just support the body on the steep incline on which it cannot
rest unsupported.
4. Effort required parallel to the plane to start the body up
the plane (Fig. 66).
angle rtb - 90° - X
ca = P
When the body is about to slide up the plane, the reaction
S will make its maximum angle A (dbc) to the normal.
T, P _ ac _ sin (A -f- a)
n W " ab ~ sTrT( 9 o° ~- A)
sin (A + «)
or P = W-
COS A
which is the effort parallel to the plane necessary to start the
body moving up the plane.
5. Effort required parallel to the plane to start the body
down the plane (Fig. 67).
Statics — Concurrent Forces — Friction
105
When the body is just about to slide down the plane, cbd= A.
Th P - ca - sin (* ~ a)
W ab sin (90° - A)
cos A
which is the least force parallel to the plane necessary to start
the body moving down the plane. If a is greater than A, this
W
angle acb = 90° — \
ab= W
& = s
ca = P
FIG. 67.
force, P, can only be negative, i.e. c falls between a and d^ and
the force is then that parallel to the plane necessary to just
support the body from sliding down the steep incline.
6. Least force necessary to start the body up the incline.
Draw ab (Fig. 68) to represent W, and a vector, be, of
indefinite length to represent S inclined A to the normal.
Then the vector joining a to the line be is least when it
is perpendicular to be. Then P is least when its line of action
is perpendicular to that of S ; that is, when it is inclined
90° — A to the normal, or A to the plane ; and then —
P ca . /
_=_=s,n(a + X)
Note that when a = o,
P = W sin A
io6
Mechanics for Engineers
which is the least force required to draw a body along the
level.
S
ab- W
6c= S
«i = P
7. Similarly, the least force necessary to start the body
down a plane inclined a to the horizontal is —
P = W sin (A - a)
if A is greater than a. If a is greater than A, P is negative, and
P is the least force which will support the body on the steep
incline. In either case, P is inclined 90° — A to the normal
or A to the plane.
8. Effort required in any assigned direction to start the
body up the plane.
Let 0 be the assigned angle which the effort P makes with
the horizontal (Fig. 69).
angle bac = 90° — 0
acb = 90° — a — A
FIG. 69.
Statics — Concurrent Forces — Friction 107
P _ ca _ s*n (*• H~ a) _ s^n (^ 4- «)
W ' ^ sin ^^ cos 10 — (a + A)}
cos {0 - (A + a)}
which is the effort necessary to start the body up the plane in
the given direction.
9. The effort in any assigned direction necessary to pull
the body down the plane may be similarly found, the resultant
force S between the body and plane acting in this case at an
angle A to the normal, but on the opposite side from that
on which it acts in case 8.
81. Action of Brake = blocks : Adhesion. —A machine
or vehicle is often brought to rest by opposing its motion by
a frictional force at or near the circumference of a wheel or
a drum attached to the wheel. A block is pressed against
the rotating surface, and the frictional force tangential to the
direction of rotation does work in opposing the motion. The
amount of work done at the brake is equal to the diminution
of kinetic energy, and this fact gives a convenient method of
making calculations on the retarding force. The force is not
generally confined to what would usually be called friction, as
frequently considerable abrasion of the surface takes place,
and the blocks wear away. It is usual to make the block of a
material which will wear more rapidly than the wheel or drum
on which it rubs, as it is much more easily renewed. If the
brake is pressed with sufficient force, or the coefficient of
"brake friction" between the block and the wheel is sufficiently
high, the wheel of a vehicle may cease to rotate, and begin to
slide or skid along the track. This limits the useful retarding
force of a brake to that of the sliding friction between the
wheels to which the brake is applied and the track, a quantity
which may be increased by increasing the proportion of weight
on the wheels to which brakes are applied. The coefficient of
sliding friction between the wheels and the track is sometimes
called the adhesion^ or coefficient of adhesion.
82. Work spent in Friction. — If the motion of a body
is opposed by a frictional force, the amount of work done
lo8 Mechanics for Engineers
against friction in foot-pounds is equal to the force in pounds
tangential to the direction of motion, multiplied by the distance
in feet through which the body moves at the point of applica-
tion of the force.
If the frictional force is applied at the circumference of a
cylinder, as in the case of a brake band or that of a shaft or
journal revolving in a bearing, the force is not all in the same
line of action, but is everywhere tangential to the rotating
cylinder, and it is convenient to add the forces together arith-
metically and consider them as one force acting tangentially to
the cylinder in any position, opposing its motion. If the
cylinder makes N rotations per minute, and is R feet radius,
and the tangential frictional force at the circumference of the
cylinder is F Ibs., then the work done in one rotation is 2?rR . F
foot-lbs., and the work done per minute is 2?rRF . N foot-lbs.,
and the power absorbed is - - horse-power (Art. 55).
In the case of a cylindrical journal bearing carrying a
resultant load W Ibs., F = /x,W, where /x, is the coefficient of
friction between the cylinder and its bearing.
83. Friction and Efficiency of a Screw. — The screw
is a simple application of the inclined plane, the thread on
FIG. 70.
either the screw or its socket (or nut) fulfilling the same functions
as a plane of the same slope. For simplicity a square-threaded
screw (Fig. 70) in a vertical position is considered, the diameter
Statics — Concurrent Forces — Friction 109
d inches being reckoned as twice the mean distance of the
thread from the axis.
Let/ = the pitch or axial distance, say in inches, from any
point on the thread to the next corresponding point, so that
when the screw is turned through one complete rotation in its
fixed socket it rises/ inches. Then the tangent of the angle
of slope of the screw thread at its mean distance is — ,, which
corresponds to tan a in Art. 80. Hence, if a tangential hori-
zontal effort P Ibs. be applied to the screw at its mean diameter
in order to raise a weight W Ibs. resting on the top of the
screw —
^ = tan (a + A)
where tan A = \L (Art. 80 (2)) ; or, expanding tan (a + A) —
P tan a -f- tan X _ ird p +
W ~~ i — tan a tan A pp ~ ird — pp
ird
which has the value ±-, or tan a for a frictionless screw.
trd
Again, the work spent per turn of the screw is —
p x ird = W(tan a + X) . ird inch-lbs.
The useful work done is W . p inch-lbs. ; therefore the work
lost in friction is W tan (a + A)?n/ — W/ foot-lbs., an expression
which may be put in various forms by expansion and substitu-
tion. The " efficiency " or proportion of useful work done
to the total expenditure of work is —
W tan a
W tan (a -f X)ird tan (a + A)
which may also be expressed in terms of /, d^ and //,. The
W
quantity -^ is called the mechanical advantage ; it is the ratio
of the load to the effort exerted, and is a function of the
HO
Mechanics for Engineers
dimensions and the friction which usually differs with different
loads.
84. Friction of Machines. — Friction is exerted at all
parts of a machine at which there is relative tangential motion
of the parts. It is found by experiment that its total effects
are such that the relation between the load and the effort,
between the load and the friction, and between the load and
the efficiency generally follow remarkably simple laws between
reasonable limits. The subject is too complex for wholly
theoretical treatment, and is best treated experimentally. It
is an important branch of practical mechanics.
Example i. — A block of wood weighing 12 Ibs. is just pulled
along over a horizontal iron track by a horizontal force of 3^ Ibs.
Find the coefficient of friction between the wood and the iron. How
much force would be required to drag the block horizontally if the
force be inclined upwards at an angle of 30° to the horizontal ?
If /* = the coefficient of friction —
v x 12 = 3! Ibs.
Let P — force required at 30° inclination ;
S = resultant force between the block and the iron track.
CL
12U>s
FIG. 71.
abc (Fig. 71) shows the triangle of forces when the block just
reaches limiting equilibrium. In this triangle, cab — 60°, since P is
inclined 30° to the horizontal ; and —
Statics — Concurrent Forces — Friction 1 1 1
tan abc — ^ - 0*291 or -
hence sin abc — -
-\
i + cot2 aoc}
and cos A = ||
P _ ca _ sin <z3<: _ sin A _ sin A.
rz~ ab sin 0££ sin (A + 60)
J sm A + ^ cos A
= -7_^_2 = Q
7 + 24^3
P = 12 x 0-289 = 3*46 Ibs.
Or thus-
Normal pressure between block \ _ _ p • o
and track / ~
horizontal pull P cos 30°^ ^(12 - P sin 30°)
hence P = 3*46 Ibs.
Example 2. — A train, the weight of which, including locomotive,
is 120 tons, is required to accelerate to 40 miles per hour from rest
in 50 seconds. If the coefficient of adhesion is |, find the necessary
weight on the driving wheels. In what time could the train be
brought to rest from this speed, (i) with continuous brakes (i.e.
on every wheel on the train) ; (2) with brakes on the driving-wheels
only ?
The acceleration is | x 88 x ^ = 1*173 ^eet Per sec- Per sec-
The accelerating force is 1*173 x ~~" = 4'37 tons
The greatest accelerating force obtainable without causing the
driving-wheels to slip is } of the weight on the wheels, therefore
the minimum weight required on the driving-wheels is 7 x 4*37
= 30*6 tons.
(i) The greatest retarding force with continuous brakes is 120 x 1
tons. Hence, if / = number of seconds necessary to bring the train
to rest, the impulse 120 x i x / = — — x ^ x §, the momentum in
ton and second units. Hence —
7 x 88 x 2
t — — = 1 2 '7 5 seconds
3 x 32-2
1 1 2 Mechanics for Engineers
(2) If the brakes are on the driving-wheels only, the retarding
force will be restricted to } of 30^6 tons, i.e. to 4:37 tons, which was
the accelerating force, and consequently the time required to come
to rest will be the same as that required to accelerate, i.e.
50 seconds.
Example 3. — A square-threaded screw 2 inches mean diameter
has two threads per inch of length, the coefficient of friction
between the screw and nut being 0*02. Find the horizontal force
applied at the circumference of the screw necessary to lift a weight
of 3 tons.
The pitch of the screw is \ inch.
If a = angle of the screw, tan a = -5 = 0*0794
27T
and if A. = angle of friction, tan \ = 0*02
Let P = force necessary in tons.
- = tan (a + x) = tan a + tan * = Q-0794 + 0-Q2
3 i - tan o tan \ i — 0*0794 x 0*02
hence P = 0*2987 ton
EXAMPLES XI.
1. A block of iron weighing n Ibs. can be pulled along a horizontal
wooden plank by a horizontal force of 17 Ibs. What is the coefficient of
friction between the iron and the plank? What is the greatest angle to the
horizontal through which the plank can be tilted without the block of iron
sliding oft"? (o . I \~~ <f. ^ » £
2. What is the least force required to drag a block of stone weighing
20 Ibs. along a horizontal path, and what is its direction, the coefficient of
friction between the stone and the path being 0*15 ?
3. What horizontal force is required to start a body weighing 15 Ibs.
moving up a plane inclined 30° to the horizontal, the coefficient of friction
between the body and the plane being 0*25 ? /*/•""•
4. Find the least force in magnitude and direction required to drag a
log up a road inclined 15° to the horizontal if the coefficient of friction
between the log and the road is 0*4. .^
5. With a coefficient friction 0*2, what must be the inclination of a
plane to the horizontal if the work done by the minimum force in dragging
10 Ibs. a vertical distance of 3 feet up the plane is 60 foot Ibs. ?
6. A shaft bearing 6 inches diameter carries a dead load of 3 tons,
and the shaft makes 80 rotations per minute. The coefficient of friction
between the shaft and bearing is 0*012. Find the horse-power absorbed in
friction in the bearing.
Statics — Concurrent Forces — Friction 113
7. If a brake shoe is pressed against the outside of a wheel with a force
of 5 tons, and the coefficient of friction between the wheel and the brake is
o-3, find the horse-power absorbed by the brake if the wheel is travelling
at a uniform speed of 20 miles per hour.
8. A stationary rope passes over part of the circumference of a rotating
pulley, and acts as a brake upon it. The tension of the tight end of the
rope is 120 Ibs., and that of the slack end 25 Ibs., the difference being due
to the frictional force exerted tangentially to the pulley rim. If the pulley
makes 170 rotations per minute, and is 2 feet 6 inches diameter, find the
horse-power absorbed. 3. ^ <yL
9. A block of iron weighing 14 Ibs. is drawn- along a horizontal
wooden table by a weight of 4 Ibs. hanging vertically, and connected to
the block of iron by a string passing over a light pulley. If the coefficient
of friction between the iron and the table is 0*15, find the acceleration of
the block and the tension of the string. J^ 3.^1
10. A locomotive has a total weight of 30 tons on the driving wheels,
and the coefficient of friction between the wheels and rails is 0-15. What
is the greatest pull it can exert on a train ? Assuming the engine to be
sufficiently powerful to exert this pull, how long will it take the train to
attain a speed of 20 miles per hour if the gross weight is 120 tons, and the
resistances amount to 20 Ibs. per ton ? '-. . 31*?
11. A square-threaded screw, I '25 inches mean diameter, has five threads
per inch of length. Find the force in the direction of the axis exerted
by the screw when turned against a resistance, by a handle which exerts a
force equivalent to 500 Ibs. at the circumference of the screw, the co-
efficient of friction being o'o8. %• a. «
CHAPTER VI
STATICS OF RIGID BODIES
85. THE previous chapter dealt with bodies of very small
dimensions, or with others under such conditions that all the
forces acting upon them were concurrent.
In general, however, the forces keeping a rigid body in
equilibrium will not have lines of action all passing through
one point. Before stating the conditions of equilibrium of a
rigid body, it will be necessai y to consider various systems of
non-concurrent forces. We shall assume that two intersecting
forces may be replaced by their geometric sum acting through
the point of intersection of their lines of action ; also that a
force may be considered to act at any point in its line of action.
Its point of application makes no difference to the equilibrium
of the body, although upon it will generally depend the dis-
tribution of internal forces in the body. With the internal
forces or stresses in the body we are not at present concerned.
86. Composition of Parallel Forces. — The following
constructions are somewhat artificial, but we shall immediately
from them find a simpler method of calculating the same
results.
To find the resultant and equilibrant of any two given like
parallel forces, i.e. two acting in the same direction. Let P
and Q (Fig. 72) be the forces of given magnitudes. Draw any
line, AB, to meet the lines of action of P and Q in A and B
respectively. At A and B introduce two equal and opposite
forces, S, acting in the line AB, and applied one at A and the
other at B. Compound S and P at A by adding the vectors
Kd and det which give a vector A^, representing R15 the resultant
Statics of Rigid Bodies 1 1 5
of S and P. Similarly, compound S and Q at B by adding
the vectors Bfandfg, which give a vector sum B^, representing
R.2, the resultant of Q and S. Produce the lines of action of
R: and R.2 to meet in O, and transfer both forces to O. Now
resolve R! and R2 at O into their components again, and we
Vector de represents P.
Vector fg represents Q.
Vectors Kd and B/ represent equal and opposite forces S.
FIG. 72.
have left two equal and opposite forces, S, which have a
resultant nil, and a force P -f- Q acting in the same direction
as P and Q along OC, a line parallel to the lines of action
of P and Q. If a force P + Q acts in the line CO in the
opposite direction to P and Q, it balances their resultant, and
therefore it will balance P and Q, i.e. it is their equilibrant.
Let the line of action of the resultant P + Q cut AB in C.
Since AOC and ked are similar triangles —
CA
OC
ae
Mechanics for Engineers
and since BOC and Eg/" are similar triangles —
CB B/ S
and dividing equation (2) by equation (i) —
CB_ P^
CA~Q
or the point C divides the line AB in the inverse ratio of the
magnitude of the two forces ; and similarly the line of action
OC of the resultant P + Q divides any line meeting the lines
of action of P and Q in the inverse ratio of the forces.
To find the resultant of any two given unlike parallel forces,
i.e. two acting in opposite directions.
Let one of the forces, P, be greater than the other, Q
(Fig. 73). By introducing equal and opposite forces, S, at A
rq
Vector de represents P.
Vector 7^- represents Q.
Vectors Ad and B/" represent equal and opposite forces S.
FIG. 73.
and B, and proceeding exactly as before, we get a force P - Q
acting at O, its line of action cutting AB produced in C.
Since AOC and bed are similar triangles —
Statics of Rigid Bodies i\j
and since BOC and B^/are similar triangles —
CB_g/_^
CO-y^-Q '
Dividing equation (4) by equation (3) —
CB= P
CA~Q
or the line of action of the resultant P - Q divides the line
AB (and any other line cutting the lines of action of P and Q)
externally, in the inverse ratio of the two forces, cutting it
beyond the line of the greater force. If a force of magnitude
P — Q acts in the line CO in the opposite direction to that of
P (i.e. in the same direction as Q), it balances the resultant of
P and Q, and therefore it will balance P and Q ; i.e. it is their
equilibrant.
This process fails if the two unlike forces are equal. The
resultants Ra and R2 are then also parallel, and the point of
intersection O is non-existent. The two equal unlike parallel
forces are not equivalent to, or replaceable by, any single force,
but form what is called a " couple."
More than two parallel forces might be compounded by
successive applications of this method, first to one pair, then to
the resultant and a third force, and so on. We shall, however,
investigate later a simpler method of compounding several
parallel forces.
87. Resolution into Parallel Components. — In the
last article we replaced two
parallel forces, P and Q, acting
at points A and B, by a single
force parallel to P and Q, acting A ^/
at a point C in AB, the posi-
tion of C being such that it
divides AB inversely as the mag-
nitudes of the forces P and Q.
Similarly, a single force may be
J FIG. 74.— Resolution into two like
replaced by two parallel forces parallel components,
acting through any two given points. Let F (Fig. 74) be the
single force, and A and B be the two given points. Join AB
y
1 1 8 Mechanics for Engineers
and let C be the point in which AB cuts the line of action of F.
If, as in Fig. 74, A and B are on opposite sides of F, then F
may be replaced by parallel forces in the same direction as
F, at A and B, the magnitudes of which have a sum F, and
which are in the inverse ratio of their distances from C, viz. a
force F x -j^ at A, and a force F x -r^ at B. The parallel
equilibrants or balancing forces of F acting at A and B are
CR AC1
then forces F X ^ and F X ~^ respectively, acting in the
opposite direction to that of the force F.
If A and B are on the same side of the line of action of the
force F (Fig. 75), then F may be replaced by forces at A and B,
AB
FIG. 75. — Resolution into two unlike parallel components.
the magnitudes of which have a difference F, the larger force
acting through the nearer point A, and in the same direction
as the force F, the smaller force acting through the further
point B, and in the opposite direction to the force F, and the
magnitudes being in the inverse ratio of the distances of the
ffi
forces from C, viz. a force F X X-D at A, in the direction of F,
AJt>
AC
and an opposite force F X at B.
The equilibrants of F at A and B will be F X - in the
osite direction
of F, respectively.
AC
opposite direction to that of F, and F X -H in the direction
Statics of Rigid Bodies
119
As an example of the parallel equilibrants through two
points, A and B, on either side of the line of action of a force,
we may take the vertical up-
ward reactions at the supports
of a beam due to a load con-
centrated at some place on
the beam.
Let W Ibs. (Fig. 76) be
the load at a point C on a
beam of span / feet, C being
x feet from A, the left-hand
support, and therefore I- x feet from the right-hand support, B.
Let RA be the supporting force or reaction at A ;
RB be the supporting force or reaction at B.
FIG. 76.
Then RA = W X =
Ibs.
More complicated examples of the same kind where there
is more than one load will generally be solved by a slightly
different method.
88. Moments. — The moment of a force F Ibs. about a
fixed point, O, was measured (Art. 56) by the product F X d
Ib.-feet, where d was the perpen-
dicular distance in ftet from O
to the line of action of F. Let
ON (Fig. 77) be the perpen-
dicular from O on to the line of
action of a force F.
Set off a vector ab on the
line of action of F to represent
F. Then the product ab . ON,
which is twice the area of the
triangle O^, is proportional to the moment of F about O.
Some convention as to signs of clockwise and contra-clockwise
moments (Art. 56) must be adopted. If the moment of F
about O is contra-clockwise, i.e. if O lies to the left of the line
FIG. 77.
I2O Mechanics for Engineers
of action of F viewed in the direction of the force, it is usual
to reckon the moment and the area Oab representing it as
positive, and if clockwise to reckon them as negative.
89. Moment of a Resultant Force. — This, about any
point in the plane of the resultant and its components, is equal
to the algebraic sum of the
" moments of the components.
Let O (Fig. 78) be any
point in the plane of two
forces, P and Q, the lines
of action of which intersect
at A. Draw Qd parallel to
the force P, cutting the line
of action of Q in c. Let
FIG g the vector Kc represent the
force Q, and set off Kb in
the line of action of P to represent P on the same scale,
P
i.e. such that Kb = Kc X ^.
Complete the parallelogram Kbdc. • Then the vector Kd =
Ac + cd = Kc -f- A£, and represents the resultant R, of P
and Q.
Now, the moment of P about O is represented by twice the
area of triangle KOb (Art. 88), and the moment of Q about
O is represented by twice the area of triangle KOc, and the
moment of R about O is represented by twice the area of
triangle ACM
But the area KQd = area Kcd + area ACV
= area Kbd + area KOc
Kbd and Kcd being each half of the parallelogram Kbdc ;
hence area KOd = area KQb + AO, since AO£ and Kbd are
between the same parallels ; or —
twice area A.Od - twice area KOb + twice area AO.
and these three quantities represent respectively the moments
of R, P, and Q about O. Hence the moment of R about O is
equal to the sum of the moments of P and Q about that point.
If O is to the right of one of the forces instead of to the left
Statics of Rigid Bodies 121
of both, as it is in Fig. 78, there will be a slight modification
in sign ; e.g. if O is to the right of the line of action of Q and to
the left of R and P, the area AO^r and the moment of Q about
O will be negative, but the theorem will remain true for the
algebraic sum of the moments.
Next let the forces P and Q be parallel (Fig. 79). Draw
a line AB from O perpendicular to the lines of action of
P and Q, cutting them in A and B respectively. Then the
resultant R, which is equal to P + Q, cuts AB in C such that
BC _ P
AC ~ Q'
Then P . AC = Q . BC
The sum of moments of P and Q about O is P . OA + Q • OB,
and this is equal to P(OC - AC) -f Q(OC + CB), which is
equal to (P + Q)OC - P . AC + Q . CB = (P + Q)OC, since
P . AC = Q . CB.
And (P + Q)OC is the moment of the resultant R about O.
Hence the moment of the resultant is equal to the sum of
moments of the two component forces. The figure will need
modification if the point O lies between the lines of action of
P and Q, and their moments about O will be of opposite sign,
but the moment of R will remain equal to the algebraic sum
of those of P and Q. The same remark applies to the figure
for two unlike parallel forces.
The force equal and opposite to the resultant, i.e. the
equilibrant, of the two forces (whether parallel or intersecting)
has a moment of equal magnitude and opposite sign to that of
the resultant (Art. 88), and therefore f he equilibrant has a moment
122 Mechanics for Engineers
about any point in the plane of the forces, of equal magnitude and
of opposite sign to the moments of the forces which it balances. In
other words, the algebraic sum of the moments of any two forces
and their equilibrant about any point in their plane is zero.
90. Moment of Forces in Equilibrium. — If several
forces, all in the same plane, act upon a body, the resultant
of any two has about any point O in the plane a moment equal
to that of the two forces (Art. 89). Applying the same theorem
to a third force and the resultant of the first two, the moment
of their resultant (i.e. the resultant of the first three original
forces) is equal to that of the three forces, and so on. By
successive applications of the same theorem, it is obvious that
the moment of the final resultant of all the forces about any
point in their plane is equal to the sum of the moments of all
the separate forces about that point, whether the forces be all
parallel or inclined one to another.
If the body is in equilibrium, the resultant force upon it in
any plane is zero, and therefore the algebraic sum of the moments
of all the separate forces about any point in the plane is zero.
This fact gives a method of finding one or two unknown forces
acting on a body in equilibrium, particularly when their lines
of action are known. When more than one force is unknown,
the clockwise and contra-clockwise moments about any point
in the line of action of one of the unknown forces may most
conveniently be dealt with, for the moment of a force about
any point in its line of action is zero.
The Principle of Moments, i.e. the principle of equation
of the algebraic sum of moments of all forces in a plane acting
on a body in equilibrium to zero, or equation of the clockwise
to the contra-clockwise moments, will be most clearly under-
stood from the three examples at the end of this article.
Levers. — A lever is a bar free to turn about one fixed
point and capable of exerting some force due to the exertion
of an effort on some other part of the bar. The bar may be of
any shape, and the fixed point, which is called the fulcrum,
may be in any position. When an effort applied to the lever
is just sufficient to overcome some given opposing force, the
lever has just passed a condition of equilibrium, and the relation
Statics of Rigid Bodies
123
\4-tons
between the effort, the force exerted by the lever, and the
reaction at the fulcrum may be found by the principle of
moments.
Example i. — A roof- frame is supported by two vertical walls
20 feet apart at points A and B on the same level. The line of the
resultant load of 4 tons on the
frame cuts the line AB 8 feet
from A, at an angle of 75° to the
horizontal, as shown in Fig. 80.
The supporting force at the point
B is a vertical one. Find its
amount.
The supporting force through
the point A is unknown, but its
moment about A is zero. Hence
the clockwise moment of the 4-ton resultant must balance the
contra-clockwise moment of the vertical supporting force RB at B.
Equating the magnitudes of the moments —
- -Sfeet- -
20 feet-
FIG. 80.
4 x 8 sin 75° = 20 x RR (tons-feet)
therefore RB = 32 S^n 75~ = r6 x 0-9659 = 1-545
tons
Example 2. — A light horizontal beam of 12-feet span carries
loads of 7 cwt., 6 cwt., and 9 cwt. at distances of i foot, 5 feet, and
10 feet respectively from the left-hand end. Find the reactions of
the supports of the beam.
If we take moments about the left-hand end A (Fig. 81), the
AC =
AD -
I foot.
5 feet. /
10 feet.
12 feet, >
\7des&
^ 1C
\6curt*
ID
\9cwt.
YE B
AE =
AB =
\
FIG. 81.
vertical loads have a clockwise tendency, and the moment of the
reaction RB at B is contra-clockwise ; hence —
RB x 12 = (7 x i) + (6 x 5) + (9 x 10)
I2RB = 7 + 30 + oo = 127
KB = Mj- = 10-583 cwt.
1 24 Mechanics for Engineers
RA, the supporting force at A, may be found by an equation of
moments about B. Or since —
RB + RA = 7 + 6 + 9 = 22 cwt.
RA = 22 - 10-583 = 11-416 cwt.
Example 3.— An L-shaped lever, of which the long arm is
18 inches long and the short one 10 inches, has its fulcrum at the
right angle. The effort exerted on
the end of the long arm is 20 Ibs.,
inclined 30° to the arm. The short
arm is kept from moving by a cord
attached to its end and perpendicular
to its length. Find the tension of the
chord.
Let T be the tension of the string
in pounds.
Then, taking moments about B
(Fig. 82), since the unknown reaction
of the hinge or fulcrum has no moment
FIG. 82. about that point —
AB sin 30° x 20 = BC x T
18 x \ x 20 = 10 x T
T = 18 Ibs.
EXAMPLES XII.
1. A post 12 feet high stands vertically on the ground. Attached to
the top is a rope, inclined downwards and making an angle of 25° with
the' horizontal. Find what horizontal force, applied to* the post 5 feet
above the ground, will be necessary to keep it upright when the rope
is pulled with a force of 120 Ibs.
2. Four forces of 5, 7, 3, and 4 Ibs. act along the respective directions
AB, BC, DC, and AD of a square, ABCD. Two other forces act,
one in CA, and the other through D. Find their amounts if the six forces
keep a body in equilibrium.
3. A beam of 15-feet span carries loads of 3 tons, \ ton, 5 tons, and
I ton, at distances of 4, 6, 9 and 13 feet respectively from the left-hand end.
Find the pressure on the supports at each end of the beam, which weighs
| ton.
4. A beam 20 feet long rests on two supports 16 feet apart, and over-
hangs the left-hand support 3 feet, and the right-hand support by I foot.
It carries a load of 5 tons at the left-hand end of the beam, and one of
7 tons midway between the supports. The weight of the beam, which may
be looked upon as a load at its centre, is I ton. Find the reactions at the
UNIVERSITY
OF
B
FIG. 83.
Statics of Rigid Bodies
supports, /'.<?. the supporting forces. What upward vertical force at the
right-hand end of the beam would be necessary to tilt the beam 1
5. A straight crowbar, AB, 40 inches long, rests on a fulcrum, C, near to
A, and a force of 80 Ibs. applied at B lifts a weight of 3000 Ibs. at A.
Find the distance AC.
6. A beam 10 feet long rests upon supports at its ends, and carries
a load of 7 cwt. 3 feet from one end. Where must a second load of 19 cwt.
be placed in order that the pressures on the two supports may be equal 1
91. Couples. — In Art. 86 it was stated that two equal
unlike parallel forces are not replace-
able by a single resultant force ; they
cannot then be balanced by a single
force. Such a system is called a couple,
and the perpendicular distance between
the lines of action of the two forces is
called the arm of the couple. Thus,
in Fig. 83, if two equal and opposite
forces F Ibs. act at A and B perpen-
dicular to the line AB, they form a
couple, and the length AB is called the arm of the couple.
92. Moment of a Couple. — This is the tendency to pro-
duce rotation, and is measured by the product of one of the forces
forming the couple and the arm of the couple ; e.g. if the two
equal and opposite forces forming the couple are each forces
of 5 Ibs., and the distance apart of their lines of action is
3 feet, the moment of the couple is 5 x 3, or 15 Ib.-feet ; or
in Fig. 83, the moment of the
couple is F X AB in suitable
units.
The sum of the moments of
the forces of a couple is the
same about any point O in their
plane. Let O (Fig. 84) be any
point. Draw a line OAB per-
pendicular to the lines of action of the forces and meeting
them in A and B. Then the total (contra-clockwise) moment
of the two forces about O is—
F . OB - F . OA = F(OB - OA) = F . AB
FIG. 84.
126
Mechanics for Engineers
This is the value, already stated, of the moment of the
couple, and is independent of the position of O.
A couple is either of clockwise or contra-clockwise ten-
dency, and its moment about any point in its plane is of the
same tendency (viewed from the same aspect) and of the same
magnitude.
93. Equivalent Couples. — Any two couples in a plane
having the same moment are equivalent if they are cf the same
sign or turning tendency, i.e. either both clockwise or both
contra-clockwise ; or, if the
couples are equal in magnitude
and of opposite sign, they
balance or neutralise one
another. The latter form of
the statement is very simply
proved. Let the forces F, F
(Fig. 85) constitute a contra-
clockwise couple, and the forces
F', F' constitute a clock-wise
couple having a moment of the
same magnitude. Let the lines
of action of F, F and those of
F, F intersect in A, B, C, and D, and let AE be the perpen-
dicular from A on BC, and CG the perpendicular from C on
AB. Then, the moments of the two couples being equal —
F x AE = F X GC
F x AB sin ABC = F X CB sin ABC
F X AB = F X CB
F = CB
F AB
Hence CB and AB may, as vectors, fully represent F and F
respectively, acting at B. And since A BCD is a parallelogram,
CD = AB, and the resultant or vector sum of F and F' is in
the line DB, acting through B in the direction DB.
Similarly, the forces F and F' acting at D have an equal
and opposite resultant acting through D in the direction BD.
These two equal and opposite forces in the line of B and D
balance, hence the two couples balance.
FIG. 85.
B
Statics of Rigid Bodies 127
It has been assumed here that the lines of action of F and
F' intersect ; if they do not, equal and opposite forces in the
same straight line may, for the purpose of demonstration, be
introduced and compounded with the forces of one couple
without affecting the moment of that couple or the equilibrium
of any system of which it forms a part.
94. Addition of Couples. — The resultant of several
couples in the same plane and of given moments is a couple
the moment of which is equal to the sum of the moments of
the several couples.
Any couple may be replaced by its equivalent couple
having an arm of length AB (Fig. 93) and forces F15 Flf pro-
vided F! X AB = moment of the
couple. ^ '
Similarly, a second couple may
be replaced by a couple of arm AB
and forces F2, F2, provided F2 X AB
is equal to the moment of this second
couple. In this way clockwise
couples must be replaced by clock-
wise couples of arm AB, and contra-
clockwise couples by contra-clock-
wise couples of arm AB, until finally we have a couple of
moment —
(F, + Fa + Fo -f . . . etc.)AB = F, x AB + F2 X AB + Fsx
AB + . . . etc.
= algebraic sum of moments of
the given couples
the proper sign being given to the various forces.
95. Reduction of a System of Co-planar Forces.—
A system of forces all in the same plane is equivalent to (i) a
single resultant force, or (2) a couple, or (3) a system in equi-
librium, which may be looked upon as a special case of (i),
viz. a single resultant of magnitude zero.
Any two forces of the system which intersect may be
replaced by a single force equal to their geometric sum acting
through the joint of intersection. Continuing the same process
FIG. 86.
128 Mechanics for Engineers
of compounding successive forces with the resultants of others
as far as possible, the system reduces to either a single re-
sultant, including the case of a zero resultant, or to a number
of parallel forces. In the latter case the parallel forces may
be compounded by applying the rules of Art. 86, and reduced
to either a single resultant (including a zero resultant) or to a
couple. Finally, then, the system must reduce to (i) a single
resultant, or (2) a couple, or (3) the system is in equilibrium.
96. Conditions of Equilibrium of a System of
Forces in One Plane. — If such a system of forces is in equi-
librium, the geometric or vector sum of all the forces must be
zero, or, in other words, the force polygon must be a closed
one, for otherwise the resultant would be (Art. 95) a single
force represented by the vector sum of the separate forces.
Also, if the system is in equilibrium (i.e. has a zero re-
sultant), the algebraic sum of all the moments of the forces
about any point in their plane is zero (Art. 90). These are
all the conditions which are necessary, as is evident from
Art. 95, but they may be conveniently stated as three con-
ditions, which are sufficient —
(i) and (2) The sum of the components in each of two
directions must be zero (a single resultant has a zero component
in one direction, viz. that perpendicular to its line of action).
(3) The sum of the moments of all the forces about one
point in the plane is zero.
If conditions (i) and (2) are fulfilled the system cannot
have a single resultant (Art. 75), and if condition (3) is ful-
filled it cannot reduce to a couple (Art. 92), and therefore it
must reduce to a zero resultant (Art. 95), i.e. the system must
be in equilibrium.
These three conditions are obviously necessary, and they
have just been shown to be sufficient, but it should be remem-
bered that the algebraic sum of the moments of all the forces
about every point in the plane is zero. The above three con-
ditions provide for three equations between the magnitudes of
the forces of a system in equilibrium and their relative posi-
tions, and from these equations three unknown quantities may
be found if all other details of the system be known.
Statics of Rigid Bodies 129
97. Solution of Statical Problems. — In finding the
forces acting upon a system of rigid bodies in equilibrium, it
should be remembered that each body is in itself in equi-
librium, and therefore we can obtain three relations (Art. 96)
between the forces acting upon it, viz. we can write three
equations by stating in algebraic form the three conditions of
equilibrium ; that is, we may resolve all the forces in two
directions, preferably at right angles, and equate the com-
ponents in opposite directions, or equate the algebraic sums to
zero, and we may equate the clockwise and contra-clockwise
moments about any point, or equate the algebraic sum of
moments to zero.
The moment about every point in the plane of a system of
co-planar forces in equilibrium is zero, and sometimes it is
more convenient to consider the moments about two . points
and only resolve the forces in one direction, or to take
moments about three points and not resolve the forces. If
more than three equations are formed by taking moments
about other points, they will be found to be not independent
and really a repetition of the relations expressed in the three
equations formed. Some directions of resolution are more
convenient than others, e.g. by resolving perpendicular to some
unknown force, no component of that force enters into the
equation so formed. Again, an unknown force may be elimi-
nated in an equation of moments by taking the moments about
some point in its line of action, about which it will have a zero
moment.
" Smooth " Bodies. — An absolutely smooth body would
be one the reaction of which, on any body pressing against it,
would have no frictional component, i.e. would be normal to
the surface of contact, the angle of friction (Art. 79) being
zero. No actual body would fulfil such a condition, but it
often happens that a body is so smooth that any frictional force
it may exert upon a second body is so small in comparison
with other forces acting upon that body as to be quite negli-
gible, e.g. if a ladder with one end on a rough floor rest against
a horizontal round steel shaft, such as is used to transmit power
in workshops, the reaction of the shaft on the ladder might
K
ISO
Mechanics for Engineers
without serious error be considered perpendicular to the length
of the ladder, i.e. normal to the cylindrical surface of the
shaft.
Example i.— A horizontal rod 3 feet long has a hole in one
end, A, through which a horizontal pin passes forming a hinge.
The other end, B, rests on a smooth roller at the same level. Forces
of 7, 9, and 5 Ibs. act upon the rod, their lines of action, which are
in the same vertical plane, intersecting it at distances of 1 1, 16, and
27 inches respectively from A, and making acute angles of 30°, 75°,
and 45° respectively with AB, the first two sloping downwards
towards A, and the third sloping downwards towards B, as shown
in Fig. 87. Find the magnitude of the supporting forces on the
rod at A and B.
_A ^\lrs\ fe
\ B
\ ~"~-~-.X /
^ ^^ "~~ ~ ,. /
>" ;
(
i
FIG. 87.
Since the end B rests on a smooth roller, the reaction RB at B is
perpendicular to the rod (Art. 97). We can conveniently find this
reaction at B by taking moments about A, to which the unknown
supporting force at A contributes nothing.
The total clockwise moment about A in Ib.-inches is —
= 270*2 Ib.-inches
The total contra-clockwise moment about A is RB x 36. Equating
the moments of opposite sign—
RB x 36 = 270^2 Ib.-inches
Statics of Rigid Bodies 131
The remaining force RA through A may be found by drawing
to scale an open vector polygon with sides representing the forces
7, 9, 5, and 7-5 Ibs. (RB); the closing side then represents RA.
Or we may find RA by resolving all the forces, say, horizontally
and vertically. Let HA be the horizontal component of RA
estimated positively to the right, and VA its vertical component
upwards. Then, by Art. 96, the total horizontal component of all
the forces is zero ; hence —
HA - 7 cos 30° - 9 cos 75° 4- 5 cos 45° = o
HA = 7 x o'866 4- 9 x 0-259 - 5 x 0-707 = 4-85 Ibs.
Also the total vertical component is zero, hence—
VA - 7 sin 30° - 9 sin 75° - 5 sin 45° 4- 7'5 = °
VA = 7 x \ 4- 9 x 0-966 4- 5 x 0-707 - 7-5 = 8*23 Ibs.
Compounding these two rectangular components of RA—
RA = jW&Syr+ (8'23)2} (Art. 75)
RA = V91'2 = 9*54 Ibs.
Example 2. — ABCD is a square, each side being 17-8 inches,
and E is the middle point of AB. Forces of 7, 8, 12, 5, 9, and
6 Ibs. act on a body in the lines and directions AB, EC, BC, BD,
CA, and DE respectively. Find the magnitude, and position
with respect to ABCD, of the single force required to keep the body
in equilibrium.
132 Mechanics for Engineers
Let F be the required force ;
HA be the component of F in the direction AD ;
VA be the component of F in the direction AB ;
p be the perpendicular distance in inches of the force
from A.
Then, resolving in direction AD, the algebraic total component
being zero —
HA + 8 cos OEC +12 + 5 cos 45° - 9 cos 45° \ _
- 6 cos EDA I ~
HA + 8 x -?= + 12 - 4 x i - 6 x -?= = o
Vs V2 Vs
HA + (2 x 0-895) + 12 - 4 x 0-707 = o
HA = — 10*96 Ibs.
Resolving in direction AB—
VA + 7 + 8 cos BEC - 5 cos 45° - 9 cos 45° \ _
+ 6 cos AED I "
VA + 7 + I4 x _L _ I4 x JL = o
VA = -7 - 6*26 + 9-90 = -3-36
then F = V{(io'^6)2 + (3'36)2} = 11-46 Ibs.
and is ipclined to AD at an angle the tangent of which is —
•^4 = 0-3065
- 10-96
i.e. at an angle 180 + 17° or 197°.
Its position remains to be found. We may take moments about
any point, say A. Let p be reckoned positive if F has a contra-
clockwise moment about A.
11-46 x/ + 6 X AD sin ADE - 5 X OA - 12) _
x AB - 8 x AE sin BEC /
106-8 . 89 , 142-4
'
= o
* = FTS = 25'5J !nches
This completes the specification of the force F, which makes
an angle 197° with AD and passes 25-51 inches from A, so as to
have a contra-clockwise moment about A. The position of F is
shown in Fig. 89.
Statics of Rigid Bodies
133
The force might be specified as making 197° with AD and
cutting it at a distance 25-51 -~ sin 197° or -86*5 inches from A ;
z>. 86-5 inches to the left of A.
FIG. 89.
98. Method of Sections. — The principles of the pre-
ceding article may be applied to find the forces acting in the
members of a structure consisting of separate pieces jointed
together. If the structure be divided by an imaginary plane of
section into two parts,
either part may be looked
upon as a body in equi-
librium under certain
forces, some of which are
the forces exerted by
members cut by the plane
of section.
For example, if a
hinged frame such as
ABODE (Fig. 90) is in
equilibrium under given forces at A, B, C, D, and E, and an
imaginary plane of section XX' perpendicular to the plane of
the structure be taken, then the portion P&zyw is in equilibrium
FlG-
134
Mechanics for Engineers
FIG. 91.
under the forces at A and B, and the forces exerted upon it by
the remaining part of the structure, viz. the forces in the bars
BD, BC, and AC. This method of sections is often the
simplest way of finding the forces in the members of a jointed
structure.
Example. — One end of a girder made up of bars jointed
together is shown in Fig. 91. Vertical loads of 3 tons and 5 tons
are carried at B and C respectively, and
the vertical supporting force at H is 12 tons.
The sloping bars are inclined at 60° to the
horizontal. Find the forces in the bars
CD, CE, and FE.
The portion of the girder ACFH cut
off by the vertical plane klm is in equili-
brium under the action of the loads at B
and C, the supporting force at H, and the
forces exerted by the bars CD, CE, and
FE on the joints at C and F. Resolving these forces vertically,
the forces in CD and FE have no vertical component, hence the
downward vertical component force exerted by CE on the left-
hand end of the girder is equal to the excess upward force of the
remaining three, i.e. 12 — 3 — 5 = 4 tons ; hence —
Force in CE x cos 30° = 4 tons
or force in CE = 4 x -'- = 4'6i tons
V3
This, being positive, acts downwards on the left-hand end, i.e. it
acts towards E, or the bar CE pulls at the joint C, hence the bar
CE is in tension to the amount of 4*61 tons. To find the force in
bar FE, take a vertical section plane through C or indefinitely
near to C, and just on the right hand of it. Then, taking moments
about C and reckoning clockwise moments positive —
12 x AC - 3 x BC + x/3 x FE x (force in FE) = o
12 x 2 — 3 x i + x/3 x (force in FE) = o
and force in FE = = = — 12*12 tons
V3
The negative sign indicates that the force in FE acts on F in
the opposite direction to that in which it would have a clockwise
moment about C, i.e. the force pulls at the joint F ; hence the
member is in tension to the extent of 12*12 tons.
OF THE
Statics of Rigid Bodies
C Ofr
Similarly, taking say clockwise moments about E, the
CD is found to be a push of 14*42 tons towards C, i.e. CD has a
compressive force of 14*42 tons in it, as follows : —
12x3-3x2-5x1 + ^3 (force in CD) = o
force in CD = — 14*42
99. Rigid Body kept in Equilibrium by Three
Forces. — If three forces keep a body in equilibrium, they
either all pass through one point (i.e. are concurrent) or are
all parallel. For unless all three forces are parallel two must
intersect, and these are replaceable by a single resultant acting
through their point of intersection. This resultant cannot
balance the third force unless they are equal and opposite and
in the same straight line, in which case the third force passes
through the intersection of the other two, and the three forces
are concurrent.
The fact of either parallelism or concurrence of the three
forces simplifies problems on equilibrium under three forces by
fixing the position of an unknown force, since its line of action
intersects those of the other two forces at their intersection.
The magnitude of the forces can be found by a triangle of
forces, or by the method of resolution into rectangular com-
ponents.
Statical problems can generally be solved in various ways,
some being best solved by one method, and others by different
methods. In the following example four methods of solution
are indicated, three of which depend directly upon the fact that
the three forces are concurrent, which gives a simple method
of determining the direction of the reaction of the rough
ground.
Example I. — A ladder 18 feet long rests with its upper end
against a smooth vertical wall, and its lower end on rough ground
7 feet from the foot of the wall. The weight of the ladder is 40 Ibs.,
which may be looked upon as a vertical force halfway along the
length of the ladder. Find the magnitude and direction of the
forces exerted by the wall and the ground on the ladder.
The weight of 40 Ibs. acts vertically through C (Fig. 92), and
the reaction of the wall Yl is perpendicular to the wall (Art. 97).
These two forces intersect at D. The only remaining force, F2, on
136
Mechanics for Engineers
the ladder is the pressure which the ground exerts on it at B. This
must act through D also (Art. 99), and therefore its line of action
must be BD. Fj may be found by
an equation of the moments about B.
x AE = 40 x
T) = 49 X
= 8-44 Ibs.
And since F2 balances the horizontal
force of 8'44 Ibs. and a vertical force
of 40 Ibs. —
- 40-8 Ibs.
and is inclined to EB at an angle
EBD, the tangent of which is —
AE _ 2 x ^7
which is the tangent of 78*1°.
A second method of solving the problem consists in drawing a
vector triangle, abc (Fig. 92), representing by its vector sides F15
F2, and 40 Ibs. The 4o-lb. force ab being set off to scale, and be
and ca being drawn parallel to F2 and Fl respectively, and the
magnitudes then measured to the same scale. A third method
consists (without drawing to scale) of solving the triangle abc
trigonometrically, thus —
Fj : F2 I 40 = ca \ cb '. ab
=HB:BD:HD
from which F1 and F2 may be easily calculated, viz.
= 8-44 Ibs.
2 x
F2 = 40 x - 40-8 Ibs.
V275
Fourthly, the problem might be solved very simply by resolving
the forces Fj and F2 and 40 Ibs. horizontally and vertically, as in
Statics of Rigid Bodies
137
this particular case the 4o-lb. weight has no component in the
direction of F]? and must exactly equal in magnitude the vertical
component of F2 ; the horizontal component of F2 must also be
just equal to the magnitude of F,.
Example 2. — A light bar, AB, 20 inches long, is hinged at A
so as to be free to move in a vertical plane. The end B is sup-
ported by a cord, BC, so placed that the angle ABC is 145° and
AB is horizontal. A weight of 7 Ibs. is hung on the bar at a
point D in AB 13 inches from A. Find the tension in the cord
and the pressure of the rod on the hinge.
A
D
Bx
\ *^ •>.
x
^
\ *» ^
X
\ ^-^
x s
^^
s
\ ^^
s
\ -^
x
\
E
\
\ x'
\ x
\ x' <
7 Ibs
FIG. 93.
Let T be the tension in the cord, and P be the pressure on the
hinge.
Taking moments about A, through which P passes (Fig. 93)—
T x AF = 7 x AD
T X 20 sin 35° = 7 x 13
H-47T = 91
T = 7 '94 Ibs.
The remaining force on the bar is the reaction of the hinge,
which is equal and opposite to the pressure P of the bar on the
hinge.
The vertical upward component of this is 7 — T sin 35°
= 2-45 Ibs., and the horizontal component is T cos 35° = 6-5 Ibs.
Hence P = V(6'5)2 + (2 45)2 = 6'93 Ibs.
The tangent of the angle DAE is ^ = 0*377, corresponding
to an angle of 20° 40'.
The pressure of the bar on the hinge is then 6 93 Ibs. in a
138 Mechanics for Engineers
direction, AE, inclined downwards to the bar and making an angle
20° 40' with its length.
EXAMPLES XIII.
1. A trap door 3 feet square is held at an inclination of 30° to (and
above) the horizontal plane through its hinges by a cord attached to the
middle of the side opposite the hinges. The other end of the cord, which
is 5 feet long, is attached to a hook vertically above the middle point of the
hinged side of the door. Find the tension in the cord, and the direction
and magnitude of the pressure between the door and its hinges, the weight
of the door being 50 Ibs., which may be taken as acting at the centre of the
door.
2. A ladder 20 feet long rests on rough ground, leaning against a rough
vertical wall, and makes an angle of 60° to the horizontal. The weight of
the ladder is 60 Ibs. , and this may be taken as acting at a point 9 feet from
the lower end. The coefficient of friction between the ladder and ground
is 0*25. If the ladder is just about to slip downwards, find the coefficient
of friction between it and the wall.
3. A ladder, the weight of which may be taken as acting at its centre,
rests against a vertical wall with its lower end on the ground. The
coefficient of friction between the ladder and the ground is J, and that
between the ladder and the wall \. What is the greatest angle to the
vertical at which the ladder will rest ?
4. A rod 3 feet long is hinged by a horizontal pin at one end, and
supported on a horizontal roller at the other. A force of 20 Ibs. inclined
45° to the rod acts upon it at a point 21 inches from the hinged end. Find
the amount of the reactions on the rod at the hinge and at the free end.
5. A triangular roof- frame ABC has a horizontal span AC of 40 feet,
and the angle at the apex B is 120°, AB and BC being of equal length.
The roof is hinged at A, and simply supported on rollers at C. The loads
it bears are as follow : (i) A force of 4000 Ibs. midway along and perpen-
dicular to AB ; (2) a vertical load of 1500 Ibs. at B ; and (3) a vertical
load of 1400 Ibs. midway between B and C. Find the reactions or
supporting forces on the roof at A and C.
6. Draw a 2-inch square ABCD, and find the middle point E of AB.
Forces of 17, 10, 8, 7, and 20 Ibs. act in the directions CB, AB, EC, ED,
and BD respectively. Find the magnitude, direction, and position of the
force required to balance these. Where does it cut the line AD, and what
angle does it make with the direction AD ?
7. A triangular roof-frame ABC has a span AC of 30 feet.- AB is 15
feet, and BC is 24 feet. A force of 2 tons acts normally to AB at its
middle point, and another force of I ton, perpendicular to AB, acts at B.
There is also a vertical load of 5 tons acting downward at B. If the sup-
porting force at A is a vertical one, find its magnitude and the magnitude
and direction of the supporting force at C.
Statics of Rigid Bodies
139
8. A jointed roof- frame, ABCDE, is shown in Fig. 94. AB and BC are
inclined to the horizontal at 30°, EB and DB are inclined at 45° to the
FIG. 94. ' ^
horizontal. The span AC is 40 feet, and B is 10 feet vertically above ED.
Vertical downward loads of 2 tons each are carried at B, at E, and at D.
Find by the method of sections the forces in the members AB, EB,
and ED.
9. A jointed structure, ACD . . . LMB (Fig. 95) is built up of bars all
Fio. 95.
of equal length, and carries loads of 7, 10, and 15 tons at D, F and L re-
spectively. Find by the method of sections the forces on the bars EF, EG,
and DF.
CHAPTER VII
CENTRE OF INERTIA OR MASS — CENTRE OF
GRA VITY
100. Centre of a System of Parallel Forces. — Let
A, B, C, D, E, etc. (Fig. 96), be points at which parallel forces
F1} F2, F3, F4, F5, etc., respectively act. The position of the
resultant force may be found by applying successively the rule
FIG. 96.
of Art. 86. Thus Fx and F2 may be replaced by a force
AX F
FJ + F2, at a point X in AB such that ^ = =? (Art. 86).
This force acting at X, and the force F3 acting at C, may
be replaced by a force Fx + F2 + F8 at a point Y in CX such
XY F
that YC = jTjTjr (Art. 86).
Centre of Inertia or Mass — Centre of Gravity 141
Proceeding in this way to combine the resultant of several
forces with one more force, the whole system may be replaced
by a force equal to the algebraic sum of the several forces
acting at some point G. It may be noticed that the positions
of the points X, Y, Z, and G depend only upon the positions
of the points of application A, B, C, D, and E of the several
forces and the magnitude of the forces, and are independent of
the directions of the forces provided they are parallel. The
point of application G of the resultant is called the centre of the
parallel forces Fx, F2, F3, F4, and F5 acting through A, B, C, D,
and E respectively, whatever direction those parallel forces may
have.
101. Centre of Mass. — If every particle of matter in a
body be acted upon by a force proportional to its mass, and
all the forces be parallel, the centre of such a system of forces
(Art. 100) is called the centre of mass or centre of inertia of
the body. It is quite independent of the direction of the
parallel forces, as we have seen in Art. 100.
Centre of Gravity. — The attraction which the earth
exerts upon every particle of a body is directed towards the
centre of the earth, and in bodies of sizes which are small
compared to that of the earth, these forces may be looked
upon as parallel forces. Hence these gravitational forces have
a centre, and this is called the centre of gravity of the body ; it
is, of course, the same point as the centre of mass.
The resultant of the gravitational forces on all the particles
of a body is called its weight, and in the case of rigid bodies it
acts through the point G, the centre of gravity, whatever the
position of the body. A change of position of the body is
equivalent to a change in direction of the parallel gravitational
forces on its parts, and we have seen (Art. 100) that the centre
of such a system of forces is independent of their direction.
We now proceed to find the centres of gravity in a number of
special cases.
1 02. Centre of gravity of two particles of given weights at
a given distance apart, or of two bodies the centres of gravity
and weights of which are given.
Let A and B (Fig. 97) be the positions of the two particles
142 Mechanics for Engineers
(or centres of gravity of two bodies) of weights w-^ and wa
A G B
u>j ~u>z
FIG. 97.
respectively. The centre of gravity G is (Art. 86) in AB at
such a point that —
or GA —
and GB =
W
. AB
. AB
In the case of two equal weights, AG = GB = JAB.
A convenient method of finding the point G graphically
may be noticed. Set off from A (Fig. 98) a line AC, making
any angle with AB (preferably
at right angles), and proportional
to wz to any scale; from B set
off a line BD parallel to AC
on the opposite side of AB, and
proportional to w^ to the same
scale that AC represents «;2.
Join CD. Then the intersection
of CD with AB determines the
point G. The proof follows simply from the similarity of the
triangles ACG and BDG. *
103. Uniform Straight Thin Rod. — Let AB (Fig. 99)
be the uniform straight rod of length AB : it may be supposed
to be divided into pairs of particles of equal weight situated at
equal distances from the middle
G b o point G of the rod, since there
will be as many such particles
between A and G as between G
and B. The e.g. (centre of gravity) of each pair, such as the
particles at a and £, is midway between them (Art. 102), viz.
at the middle point of the rod, G, hence the e.g. of the whole
rod is at its middle point, G.
>B
FIG. 99.
Centre of Inertia or Mass — Centre of Gravity 143
104. Uniform Triangular Plate or Lamina. — The term
centre of gravity of an area is often used to denote the e.g.
of a thin lamina of uniform material cut in the shape of the
particular area concerned.
We may suppose the lamina ABC (Fig. 100) divided into
an indefinitely large number of strips parallel to the base AC.
The e.g. of each strip, such as PQ,
is at its middle point (Art. 103),
and every e.g. is therefore in the
median BB/ i.e. the line joining B
to the mid-point B' of the base
AC. Hence the e.g. of the whole
triangular lamina is in the median
BB'. Similarly, the e.g. of the A
lamina is in the medians AA' and
CC'. Hence the e.g. of the triangle is at G, the intersection
of the three medians, which are concurrent, meeting at a point
distant from any vertex of the triangle by f of the median
through it. The perpendicular distance of G from any side
of the triangle is \ of the perpendicular distance of the oppo-
site vertex from that side.
Note that the e.g. of the triangular area ABC coincides
with that of three equal particles placed at A, B, and C. For
those at A and C are statically equivalent to two at B', and
the e.g. of two at B' and one at B is at G, which divides BB' in
the ratio 2 : i, or such at B'G = JBB' (Art. 102).
Uniform Parallelogram. — If a lamina be cut in the
shape of a parallelogram, A B
ABCD (Fig. 101), the e.g. of
the triangle ABC is in OB,
and that of the triangle ADC
is in OD, therefore the e.g. of
the whole is in BD. Similarly
it is in AC, and therefore it is D C
, ~ FIG. 101.
at the intersection O.
105. Rectilinear Figures in General. — The e.g. of any
lamina with straight sides may be found by dividing its area up
into triangles, and finding the e.g. and area of each triangle.
144
Mechanics for Engineers
Thus, in Fig. 102, if G13 G.2, and G3 are the centres of gravity
of the triangles ABE, EBD, and
DEC respectively, the e.g. of
the area ABDE is at G4, which
divides the length G^ inversely
as the weights of the triangles
AEB and EDB, and therefore
inversely as their areas. Simi-
larly, the e.g. G of the whole
figure ABCDE divides G.,G4 in-
versely as the areas of the figures
ABDE and BCD. The inverse
division of the lines GjG2 and
of G3G4 may in practice be performed by the graphical method
of Art. 102.
106. Symmetrical Figures. — If a plane figure has an
axis of symmetry, i.e. if a straight line can be drawn dividing it
FIG. 102.
FIG. 103.
into two exactly similar halves, the e.g. of the area of the figure
lies in the axis of symmetry. For the area can be divided into
indefinitely narrow strips, the e.g. of each of which is in the axis
of symmetry (see Fig. 103). If a figure has two or more axes of
FIG. 104.
symmetry, the e.g. must lie in each, hence it is at their intersection,
e.g. the e.g. of a circular area is at its centre. Other examples,
which sufficiently explain themselves, are shown in Fig. 104.
Centre of Inertia or Mass — Centre of Gravity 145
107. Lamina or Solid from which a Part has been
removed. — Fig. 105 represents a lamina from which a piece,
B, has been cut. The centre of gravity of the whole lamina,
including the piece B, is
at G, and the e.g. of the
removed portion B is at g.
The area of the remaining
piece A is a units, and
that of the piece B is b
units. It is required to
find the e.g. of the remain-
ing piece A.
Let G' be the required FIG. 105.
e.g. ; then G is the e.g. of two bodies the centres of gravity of
which are at G' and g, and which are proportional to a and b
respectively. Hence G is in the line G^", and is such that —
GG' : Gg : : b : a (Art. 102)
or GG' = - . Gg
That is, the e.g. G' of the piece A is in the same straight
line ^G as the two centres of gravity of the whole and the part
b
B, at — times their distance, apart beyond the e.g. of the whole
lamina. The point G' divides the line Gg externally in the
ratio ~ . --„ or G'G
a + P
b : a + b.
The same method is ap-
plicable if A is part of a solid
from which a part B has been
removed, provided a repre-
sents the weight of the part A,
and b that of the part B.
Graphical Construc-
tion.— The e.g. of the part A
may be found as follows :
from g draw a line gP (Fig.
1 06) at any angle (preferably
at right angles) to Gg and proportional to a + b. From G
FIG. 106.
146
Mechanics for Engineers
draw GQ parallel to £p and proportional to b. Join PQ, and
produce to meet gG produced in G'. Then G' is the e.g. of
the part A.
1 08. Symmetrical Solids of Uniform Material. — If a
solid is symmetrical about one plane, i.e. if it can be divided
by a plane into two exactly similar halves, the e.g. evidently
lies in the plane, for the solid can be divided into laminae the
FIG. 107.
e.g. of each of which is in the plane of symmetry. Similarly,
if the solid has two planes of symmetry, the e.g. must lie in the
intersection of the two planes, which is an axis of the solid, as
in Fig. 107.
If a solid has three planes of symmetry, the line of inter-
section of any two of them meets the third in the e.g., which is
FIG. 108.
a point common to all three planes, e.g. the sphere, cylinder,
etc. (see Fig. 108).
109. Four Equal Particles not in the Same Plane.—
Let ABCD (Fig. 109) be the positions of the four equal
particles. Join ABCD, forming a triangular pyramid or tetrahe-
dron. The e.g. of the three particles at A, B, and C is at D',
the e.g. of the triangle ABC (Art. 104). Hence the e.g. of the
four particles is at G in DD', and is such that —
D'G : GD = 1:3 (Art. 102)
or D'G = J DD'
Centre of Inertia or Mass — Centre of Gravity 147
Similarly, the e.g. of the four particles is in AA', BB', and CC,
the lines (which are concurrent)
joining A, B, and C to the centres
of gravity of the triangles BCD,
ACD, and ABD respectively. The
distance of the e.g. from any face
of the tetrahedron is J of the per-
pendicular distance of the opposite
vertex from that face.
no. Triangular Pyramid
or Tetrahedron of Uniform FlG- I09-
Material. — Let ABCD (Fig. no) be the triangular pyramid.
Suppose the solid divided into indefinitely thin plates, such
as abc, by planes parallel to the face ABC. Let D' be the
e.g. of the area ABC.
Then DD' will intersect
the plate abc at its e.g.,
viz. at d, and the e.g. of
every plate, and there-
fore of the whole solid,
will be in DD'. Simi-
larly, it will be in AA',
BB', and CC', where A',
B', and C' are the centres
of gravity of the triangles
BCD, CDA, and DAB
respectively. Hence the centre of gravity of the whole solid
coincides with that of four equal particles placed at its vertices
(Art. 109), and it is in DD', and distant \ DD' from D', in CC'
and J CC' from C', and so on. It is, therefore, also distant
from any face, \ of the perpendicular distance of the opposite
vertex from that face.
in. Uniform Pyramid or Cone on a Plane Base. —
If V (Fig. 1 1 1) is the vertex of the cone, and V the e.g. of the
base of the cone, the e.g. of any parallel section or lamina into
which the solid may be divided by plates parallel to the base, will
be in VV. Also if the base be divided into an indefinitely large
number of indefinitely small triangles, the solid is made up of
FIG. no.
148
Mechanics for Engineers
an indefinitely large number of triangular pyramids having the
triangles as bases and a common vertex, V. The e.g. of each
small pyramid is distant from
V f of the distance from its
base to V. Hence the centres
of gravity of all the pyramids
lie in a plane parallel to the
base, and distant from the
vertex, f of the altitude of
the cone.
The e.g. of a right circular
cone is therefore in its axis,
which is the intersection of two
planes of symmetry (Art. 108),
and its distance from the base
is J the height of the cone, or its distance from the vertex is f
of the height of the cone.
Example I. — A solid consists of a right circular cylinder 3 feet
long, and a right cone of altitude 2 feet, the base coinciding with
one end of the cylinder. The cylinder and cone are made of the
same uniform material. Find the e.g. of the solid.
If r '= radius of the cylinder in feet —
the volume of cylinder _ ?rr2 x 3 _ 9
volume of cone nr2 x £ x 2 2
hence the weight of the cylinder is 4*5 times that of the cone.
The e.g. of the cylinder is at A (Fig. 112), the mid-point of its
axis (Art. 108), i.e. 1-5 feet from the plane of the base of the cone.
FIG. in.
AG
FIG. 112.
The e.g. of the cone is at B, \ of the altitude from the base
(Art. in), i.e. 0-5 foot from the common base of the cylinder and
cone. Hence —
AB = AD + DB = 1-5 + 0-5=2 feet
2
And G is therefore in AB, at a distance - — . AB from A
2 + 9
(Art. 102), i.e. AG = ^ of 2 feet = j*T foot, or 4*36 inches.
Centre of Inertia or Mass — Centre of Gravity 149
Example 2.— A quadrilateral consists of two isosceles triangles
on opposite sides of a base 8 inches
long. The larger triangle has two
equal sides each 7 inches long, and
the smaller has its vertex 3 inches
from the 8-inch base. Find the dis-
tance of the e.g. of the quadrilateral
from its 8-inch diagonal.
Let ABCD (Fig. 113) be the
quadrilateral, AC being the 8-inch
diagonal, of which E is the mid-
point ; then —
ED = 3 inches
EB=V72-42= 733 = 5745 inches
The e.g. of the triangle ABC is in EB and \ EB from E ; or,
if G! is the e.g. —
EG: = S~7-^ = 1-915 inches
Similarly, if G2 is the e.g. of the triangle ADC —
EG2 = ^ of 3 inches = I inch
therefore G^ = 1-915 + i = 2-915 inches
This length is divided by G, the e.g. of the quadrilateral, so
that —
G2G _ area of triangle ABC _ BE _ 1-915
G^G ~ areaToTSiangieACD ~ ED i
G2G _ 1-915 _ 1-915
GjGg ~ i + 1-915 ~ ^915
G2G = 1*915 inches
and EG = G2G — G2E = 1-915 — i = 0*915 inch
which is the distance of the e.g. from the 8-inch diagonal.
Example 3. — A pulley weighs 25 Ibs., and it is found that the
e.g. is 0-024 inch from the centre of the pulley. The pulley is
required to have its e.g. at the geometrical centre of the rim, and
to correct the error in its position a hole is drilled in the pulley
with its centre 6 inches from the pulley centre and in the same
diameter as the wrongly placed e.g. How much metal should be
removed by drilling ?
Let x be the weight of metal to be removed, in pounds.
150 Mechanics for Engineers
Then, in Fig. 114, OA being 6 inches and OG 0*024 inch, the
removed weight x Ibs. having its e.g. at A, and the remaining
FIG. 114.
25— .r Ibs. having its e.g. at O, the e.g. G of the two together divides
OA, so that—
O G = __£
GA 25 — 'x
OG x
°r OA = 2~5
hence. = 2*~^ = '5 ' x °-* = o-i Ib.
EXAMPLES XIV.
1. A uniform beam weighing 180 Ibs. is 12 feet long. It carries a
load of 1000 Ibs. uniformly spread over 7 feet of its length, beginning
I foot from one end and extending to a point 4 feet from the other. Find
at what part of the beam a single prop would be sufficient to support it.
2. A lever 4 feet long, weighing 15 Ibs., but of varying cross-section,
is kept in equilibrium on a knife-edge midway between its ends by the
application of a downward force of I '3 Ibs. at its lighter end. How far is
the e.g. of the lever from the knife-edge?
3. The heavy lever of a testing machine weighs 2500 Ibs., and is poised
horizontally on a knife-edge. It sustains a downward pull of 4 tons
3 inches from the knife-edge, and carries a load of I ton on the same side
of the knife-edge and 36 inches from it. How far is the e.g. of the lever
from the knife-edge ?
4. A table in the shape of an equilateral triangle, ABC, of 5 feet sides,
has various articles placed upon its top, and the legs at A, B, and C then
exert pressures of 30, 36, and 40 Ibs. respectively on the floor. Determine
the position of the e.g. Qf the table loaded, and state its horizontal distances-
from the sides AB and BC.
5. Weights of 7, 9, and 12 Ibs. are placed in the vertices A, B, and C
respectively of a triangular plate of metal weighing 10 Ibs., the dimensions
of which are, AB 16 inches, AC 16 inches, and BC II inches. Find the
e.g. of the plate and weights, and state its distances from AB and BC.
6. One-eighth of a board 2 feet square is removed by a straight saw-cut
through the middle points of two adjacent sides. Determine the distance
of the e.g. of the remaining portion from the saw-cut. If the whole board
before part was removed weighed 16 Ibs., what vertical upward force
OF THE
UNIVERSIT
Centre of Inertia or Mass —Centre of Gram^^ji OF
1^/FORHiS
applied at the corner diagonally opposite the saw- cut would be sufficient to
tilt the remaining I of the board out of a horizontal position, if it turned
about the line of the saw-cut as a hinge ?
7. An isosceles triangle, ABC, having AB 10 inches, AC JO inches, and
base BC 4 inches long, has a triangular portion cut off by a line DE,
parallel to the base BC, and 7*5 inches from it, meeting AB and AC in D
and E respectively. Find the e.g. of the trapezium BDEC, and state its
distance from the base BC.
8. The lever of a testing-machine is 15 feet long, and is poised on
a knife-edge 5 feet from one end and 10 feet from the other, and in a
horizontal line, above and below which the beam is symmetrical. The
beam is 16 inches deep at the knife-edge, and tapers uniformly to depths
of 9 inches at each end ; the width of the beam is the same throughout its
length. Find the distance of the e.g. of the beam from the knife-edge.
9. A retaining wall 5 feet high is vertical in front and 9 inches thick
at the top. The back of the wall slopes uniformly, so that the thickness of
the wall at the base is 2 feet 3 inches. Find the e.g. of the cross-section of
the wall, and state its horizontal distance from the vertical face of the
wall.
10. What is the moment of the weight of the wall in Question 9 per
foot length, about the back edge of the base, the weight of the material
being 120 Ibs. per cubic foot? What uniform horizontal pressure per
square foot acting on the vertical face of the wall would be sufficient to
turn it over bodily about the back edge of the base ?
11. The casting for a gas-engine piston maybe taken approximately
as a hollow cylinder of uniform thickness of shell and one flat end of uniform
thickness. Find the e.g. of such a casting if the external diameter is 8
inches, the thickness of shell f inch, that of the end 3 inches, and the
length over all 20 inches. State its distance from the open end.
12. A solid circular cone stands on a base 14 inches diameter, and its
altitude is 20 inches. From the top of this a cone is cut having a base
3' 5 inches diameter, by a plane parallel to the base. Find the distance of
the e.g. of the remaining frustum of the cone from its base.
13. Suppose that in the rough, the metal for making a gun consists
of a frustum of a cone, 10 feet long, 8 inches diameter at one end, and
6 inches at the other, through which there is a cylindrical hole 3 inches
diameter, the axes of the barrel and cone being coincident. How far from
the larger end must this piece of metal be slung on a crane in order to
remain horizontal when lifted ?
14. A pulley weighing 40 Ibs. has its e.g. 0^04 inch from its centre.
This defect is to be rectified by drilling a hole on the heavy side of the
pulley, with its centre 9 inches from the centre of the pulley and in
the radial direction of the centre of gravity. What weight of metal should
be drilled out ?
15. A cast-iron pulley weighs 45 Ibs., and has its e.g. 0^035 inch from
its centre. In order to make the e.g. coincide with the centre of the
152
Mechanics for Engineers
pulley, metal is added to the light side at a distance of 8 inches from the
centre of the pulley and in line with the e.g. What additional weight
is required in this position ? If the weight is added by drilling a hole in
the pulley and then rilling it up to the original surface with lead, how much
iron should be removed, the specific gravity of lead being 1 1 '35, and that of
iron being 7*5 ?
112. Distance from a Fixed Line of the Centre of
Gravity of Two Particles, or Two Bodies, the Centres
of Gravity of which are given.
Let A (Fig. 115) be the position
of a particle of weight w^ and let
B be that of a particle of weight
0/2, or, if the two bodies are of
finite size, let A and B be the
positions of their centres of gravity.
Then the centre of gravity of the
Q
FIG. 115.
M
two weights wv and w% is at G in AB such that —
AG 7£/2 .
-(Art. 102)
or AG =
— . AB
and GB =
— . AB
'i + 0/2
Let the distances of A, B, and G from the line NM be
xly #2, and x respectively, the line NM being in a plane through
the line AB. Then AN = xlt BM = x& and GQ = x.
AG
[OW' BS=AB=z
or GR = — ~ — . BS
w,2
and GQ or x = RQ 4 GR = AN 4
hence x = xl 4-
w.t
/i +
BS
'! 4-
Distance of the e.g. from a Plane. — If x^ and x$ are
the respective distances of A and B from any plane, then NM
Centre of Inertia or Mass — Centre of Gravity 153
may be looked upon as the line joining the feet of perpen-
diculars from A and B upon that plane. Then the distance x
of G from that plane is —
- _ WlXl
(I)
This length x is also called the mean distance of the two
bodies or particles from the plane.
113. Distance of the e.g. of Several Bodies or of
One Complex Body from a Plane.
Let A, B, C, D, and E (Fig. 116) be the positions of 5 par-
ticles weighing wlt w2, w3, «/4, and w5 respectively, or the
FIG. 116.
centres of gravity of five bodies (or parts of one body) of those
weights.
Let the distances of A, B, C, D, and E from some fixed
plane be xlt x.2, x3, x4, and x5 respectively, and let the weights
in those positions be wlt w.2, ws, w^ and w5 respectively. It is
required to find the distance ~x of the e.g. of these five weights
from the plane. We may conveniently consider the plane to
1 54 Mechanics for Engineers
be a horizontal one, but this is not essential ; then xlt x^ x^,
Xto and x5 are the vertical heights of A, B, C, D, and E respec-
tively above the plane. Let a, b, c, d^ and e be the projections
or feet of perpendiculars from A, B, C, D, and E respectively
on the plane, so that Aa, B#, O, T)d, and E<? are equal to x^ x.2,
x3, #4, and xs respectively.
Let G! be the e.g. of u\ and m>, and let gl be its projection
by a vertical line on the plane ; then —
4
Let G2 be the e.g. of (u\ + «;2) and w3, and let ^ be its
projection by a vertical line on the plane ; then G2 divides
GiC so that—
Wm
GiGo == / ; -- \ — i GiC
(w% + w2) 4- w3
and G^ = .
^ + 7£/2 + W3
and substituting the above value of Gi^i —
^1^1 4 ^2*2 4"
wi + ^2 4- a',
Similarly, if G3 is the e.g. of wlt w.2, w3t and w^ and ^ is its
projection on the plane, then —
°3^ = -1 ^ + w! 4- «'8 + «^4 4 ^ and S° °n
and finally —
W&4-
which may be written —
where 5 stands for " the sum of all such terms as." If any
of the points A, B, C, etc., are below the plane, their distances
from the plane must be reckoned as negative.
Centre of Inertia or Mass— Centre of Gravity 155
Plane = moments. — The products u\x^ w2x2, w3xs, etc.,
are sometimes called plane- moments of the weights of the
bodies about the plane considered. The plane-moment of a
body about any given plane is then the weight of the body
multiplied by the distance of its e.g. from that plane.
Then in words the relation (3) may be stated as follows :
" The distance of the e.g. of several bodies (or of a body
divided into parts) from any plane is equal to the algebraic
sum of their several plane-moments about that plane, divided
by the sum of their weights."
And since by (3), ~x X 2(«>) = 2(o>#), we may state that the
plane-moment of a number of weights (or forces) is equal to
the sum of their several plane-moments.
This statement extends to plane-moments the statement
in Art. 90, that the moment of the sum of several forces about
any point is equal to the sum of the moments of the forces
about that point.
It should be remembered that a horizontal plane was chosen
for convenience only, and that the formulae (2} and (3) hold
good for distances from any plane.
114. Distance of the e.g. of an Area or Lamina
from a Line in its Plane.
This is a particular case of the problem of the last article.
Suppose the points A, B, C, D, and E in the last article and
Fig. 1 1 6 all lie in one plane perpendicular to the horizontal
plane, from which their distances are xlt x^ x^ x4, and x5
respectively. Then their projections a, b, c, d, and e on the
horizontal plane all lie in a straight line, which is the inter-
section of the plane containing A, B, C, D, and E with the
horizontal plane, viz. the line OM in Fig. 117.
Thus, if #!, XM XM etc., be the distances of the centres of
gravity of several bodies all in the same plane (or parts of
a lamina) from a fixed line OM in this plane, then the
distance of the e.g. of the bodies (or laminae) from the line
being x —
~~ _ WlX* + w'^ + u>sX* "*" w*x* + • • • ' etc- _ S(otfp)
o>i + ^2 + v>9 4- . . . , etc. " ~iu ^
156 Mechanics for Engineers
This formula may be used to find the position of the e.g. of
a lamina or area by finding its distance from two non-parallel
fixed lines in its plane.
If the lamina is of irregular shape, as in Fig. 118, the dis-
tance of its e.g. from a line OM in its plane may be found
approximately by dividing
it into a number of narrow
strips of equal width by lines
parallel to OM, and taking
the e.g. of each strip as
being midway between the
parallel boundary-lines. The
weight of any strip being
f • "X.
/ \
/ \
\
CL-jr
/ 1
SJ
**3i
^/// ///// //////////////w,^ ^
a
\
^"—— ^
M
FIG.
denoted by w —
w = volume of strip x D
where D = weight of unit volume of the material of the lamina,
or —
w = area of strip x thickness of lamina X D
If the weight of the first, second, third, and fourth strips be
«/!, a>2, wst and w4 respectively, and so on, and their areas be
alt a2, 03, and #4 respectively, the lamina consisting of a material
of uniform thickness /, then w1 = alt.D, u>2 = a.2f.D, and
Centre of Inertia or Mass — Centre of Gravity 157
so on. And if x is the distance of the e.g. of the area from
OM, then by equation (4) —
X —
. . . , etc.
-f-
+ /y /T^ T I pfp
Wjft/*3 -(-..., etc. , .
• (5)
-f- astD + . . . , etc.
or, dividing numerator and denominator by the factor fD —
- ^4^4 + . . . , etc.
tfi + «a + ^3 + «4 + . . . , etc.
= L__2 or — ^ — '
(6)
where A = total area of the lamina, and 2 has the same
meaning as in (3), Art. 113.
Similarly, the distance of the e.g. of the area A from
another straight line may
be found, and then the
position of the e.g. is
completely determined.
Thus in Fig. 119, if
x is the distance of the
e.g. of the lamina from
OM, and y is its distance
from ON, by drawing two
lines, PR and QS, parallel
to OM and ON and dis-
tant x and y from them
respectively, the inter-
section G of the two lines gives the e.g. of the lamina or
area.
Moment of an Area. — The products a-^c^ etc., may be
called moments of the areas a^ etc.
Regular Areas. — If a lamina consists of several parts, the
centres of gravity of which are known, the division into thin
strips adopted as an approximate method for irregular figures
FIG. 119.
158
Mechanics for' Engineers
is unnecessary. The distance x of the e.g. from any line OM is
W or-
_ ^(product of each area and distance of its e.g. from OM)
whole area
or —
S(plane mo. of each area about a plane perpend, to its own)
whole area
The product of an area and the distance of its e.g. from a
line OM may be called the " line moment " of the area about
OM, and we may write—
_ S(line moments of each part of an area)
whole area
For example, in Fig. 120 the area ABECD consists of a
triangle, EEC, and a rectangle, ABCD,
having a common side, BC. Let the
height EF = // ; let AD = / and AB = d.
Then the area ABCD = d X /, and the
area EEC = \ X / X /£, and if Gl is the
e.g. of the triangle EEC, and G2 that of
the rectangle ABCD, the distance x of
the e.g. of the area ABECD from AD is
found thus —
d
FIG. 120.
hd
115. Lamina with Part removed. — Suppose a lamina
(Fig. 121) of area A has a portion of area a, removed. Let
x = distance of e.g. G of A from a line OM in its plane ; let
Xi be the distance of the e.g. of the part a from OM ; and
let x.2 be the distance of the e.g. of the remainder (A — a) from
OM.
Centre of Inertia or Mass — Centre of Gravity 159
x. A =
and x9 =
-^(Art. 114)
-*)
A - a
In this way we can find the distance of the e.g. of the part
A — a from OM, and similarly we can find the distance from
FIG. 121.
any other line in its plane, and so completely determine its
position as in Art. 114. This method is applicable particularly
to regular areas.
1 1 6. Solid with Part removed. — The method used in
the last article to find the e.g. of part of a lamina is applicable
to a solid of which part has
been removed.
If in Fig. 122 A is a
solid of weight W, and a
portion B weighing w is re-
moved, the distance of the
e.g. of the remainder (W — w)
from any plane is x.z where —
W -w
FIG. 122.
by (i) Art. 112 and the method of Art. 115, where x = distance
of e.g. of A from the plane, and xl = distance of e.g. of B from
the plane.
i6o
Mechanics for Engineers
117. Centre of Gravity of a Circular Arc. — Let ABC
(Fig. 123) be the arc, OA being the radius, equal to a units
of length, and the length of arc ABC
being / units. If B is the middle
point of the arc, OB is an axis of
symmetry, and the e.g. of the arc is
in OB. Draw OM parallel to AC.
Let the arc be divided into a
B number of small portions, such as PQ,
each of such small length as to be
sensibly straight. Let the weight of
the arc be w per unit length. The
e.g. of a small portion PQ is at V, its
mid-point. Draw VW parallel to
OM, and join OV. Draw PR and QR
parallel to OM and OB respectively.
Then, if x = distance of e.g. of arc from the line OM, as in
Art. 114 —
_ S(PQ X w x OW) S(PQ X OW)
x = —
S(PQ^OW)
S(PQ) /
Now, since OV, VW, and OW are respectively perpen-
dicular to PQ, RQ, and PR, the triangles PQR and OVW are
similar, and —
PQ= RP
OV ~~ OW
or PQ . OW = OV . RP = a . RP
hence S(PQ . OW) = S(«
and therefore —
= a X AC
AC
-_S(PQ.OW)_* AC
— ~~I — ~~ ~~ 7 * ' /
X a
The e.g. of the arc then lies in OB at a point G such that —
AC chord
OG = OB X -y or radius X -
/ arc
or, if angle AOC = 2a, i.e. if angle AOB = a (radians)—
AC 2AD 2 . a sin a sin a
OG = a X —r- = a X — j— = a X - —7- = a . —
/ / a X A a
Centre of Inertia or Mass — Centre of Gravity 161
FIG. 124.
When the arc is very short, OG is very nearly equal to OB.
118. Centre of Gravity of Circular Sector and
Segment. — Let the sector ABCO (Fig. 124) of a circle
centred at O and of radius a, subtend
an angle 2 a at O. The sector may
be divided into small parts, such as
OPQ, by radial lines from O. Each
such part is virtually triangular when
PQ is so short as to be regarded as a
straight line. The e.g. of the triangle
OPQ is on the median OR, and §0
from O. Similarly, the centres of
gravity of all the constituent triangles,
such as PQO, lie on a concentric arc
abc of radius |-# and subtending an
angle 2 a at O. The e.g. of the sector coincides with
the e.g. of the arc abc, and is therefore in OB and at a
distance §# . - - from O (Art n?) ', f-g- the e.g. of a semi-
circular area of radius " a " is at a distance f # — - or — from
its straight boundary.
The e.g. of the segment cut off by any chord AC (Fig. 124)
may be found by the principles of Art. 115, regarding the
segment as the remainder of the
sector ABCO when the triangle
AOC is removed.
119. Centre of Gravity of
a Zone of a Spherical Shell.
—Let ABC D (Fig. 125) be a zone
of a spherical shell of radius a and
thickness /, and of uniform material
which weighs w per unit volume.
Let the length of axis HF be /.
Divide the zone into a number of
equal smaller zones, such as abed,
by planes perpendicular to the axis OE, so that each has an
axial length //. Then the area of each small zone is the same,
M
FIG. 125.
1 62 Mechanics for Engineers
viz. 27rafi, and the volume of each is then nrah . /, and each
has its e.g. on the axis of symmetry OE, and midway between
the bounding planes, such as ^and be, if h is indefinitely short.
Hence the e.g. of the zone coincides with that of a large
number of small bodies each of weight w . 2irah . t, having their
centres of gravity uniformly spread along the line FH. Hence
the e.g. is at G, the mid-point of the axis FH of the zone, or —
e.g. the distance of the e.g. of a hemispherical shell from the
plane of its rim is half the radius of the shell.
120. Centre of Gravity of a Sector of a Sphere. — Let
O ACB (Fig. 1 2 6) be a spherical sector of radius a. If the sector
. be divided into an indefinitely
great number of equal small
pyramids or cones having a
common vertex O such that their
bases together make up the base
ACB of the sector, the c.g.'s of
the equal pyramids will each be
f # from O, and will therefore be
evenly spread over a portion acb (similar to the surface ACB)
of a spherical surface centred at O and of radius f #. The e.g.
of the sector then coincides with that of a zone, acb^ of a thin
spherical shell of radius f #, and is midway between c and the
plane of the boundary circle ab, i.e. midway between d and c.
Solid Hemisphere. — The hemisphere is a particular case
of a spherical sector, and its e.g. will coincide with that of a
hemispherical shell of f a, where a is the radius of the solid
hemisphere. This is a point on the axis of the solid hemisphere,
and half of f #, or \a from its base.
Example i. — The base of a frustum of a cone is 10 inches
diameter, and the smaller end is 6 inches diameter, the height
being 8 inches. A co-axial cylindrical hole, 4 inches diameter,
is bore&kh rough the frustum. Find the distance of the e.g. of the
ffefciaining solid from the plane of its base.
The solid of which the c g. is required is the remaining portion
Centre of Inertia or Mass — Centre of Gravity 163
of a cone, ABC (Fig. 127), when the upper cone, DBE, and a
cylinder, FGKH, have been removed.
Since the cone diameter decreases 4 inches in a height of
8 inches —
The height BM = 8 + 8xf = 2o inches
and the e.g. of the cone j
AT.^ - : • , >= 5 inches from AC
ABC is £ x 20 inches j
volume of cone ABC = TT . (5)2 . 3J* = «• . &
cubic inches
distance from AC of e.g.)
r i- j T-,~T^TT ( — •> = 4 inches
of cylinder P GKH J
volume of cylinder ) _
FGKH *"
- 22 . 8 = 3277 cubic
inches
volume of cone DBE = TT . 32 . -^ = 3677 cubic
inches A
4- -1/ = ii inches
distance from AC of e.g. \ _
of cone DBE )
then volume of remaining frustum is —
7r(£ga _ 32 - 36) = TT . ^f11 cubic inches
Let h = height of e.g. of this remainder from the base.
Then equating the plane-moments about the base of the three
solids, BDE, FGKH, and the remainder of frustum, to the plane-
moment of the whole cone (Art. 1 13) (and leaving out of both sides
of equation the common factor weight per unit volume) —
TT . &§<>- x 5 = 7r{(32 x 4) + (36 x 11) + (^ x h}}
833'3 = 524 + ^tr^
h = o^o x 309-3 = 3-135 inches
Example 2. — An I-section of a girder is made up of three
rectangles, viz. two flanges having their long sides horizontal, and
one web connecting them having its long side vertical. The top
flange section is 6 inches by i inch, and that of the bottom flange
is 12 inches by 2 inches. The web section is 8 inches deep and
i inch broad. Find the height of the e.g. of the area of cross-section
from the bottom of the lower flange.
Fig. 128 represents the section of the girder.
Let x — height of the e.g. of the whole section.
The height of the e.g. of BCD E is i inch above BE ;
FGHK is 2 + | = 6 inches above BE ;
LMNP is 2 + 8 4- 1 = 10-5 inches above BE.
1 64
Mechanics for Engineers
Equating the sum of the moments of these three areas about
A to the moment of the whole figure about A, we have —
(12 x 2)1 + (8 x 1)6 + (6 x 1)10-5 = I-{(i2X2) + (8xi) + (6xi)}
24 + 48 + 63 = J<24 + 8 + 6)
* = -W = 3'55
;N^
A
/2
FIG. i2E
which is the distance of the c g. from the bottom of the lower
flange.
Example 3. — Find the e.g. of a cast-iron eccentric consisting
of a short cylinder 8 inches in
diameter, having through it a cylin-
drical hole 2-5 inches diameter, the
axis of the hole being parallel to
that of the eccentric and 2 inches
from it. State the distance of the
e.g. of the eccentric from its centre.
This is equivalent to finding the
e.g. of the area of a circular lamina
with a circular hole through it. In
Fig. 129—
AB = 8 inches CD = 2 inches
EF = 2*5 inches
Let the distance of the e.g. from A be x.
If the hole were filled with the same material as the remainder
of the solid, the e.g. of the whole would be at C, its centre.
Centre of Inertia or Mass — Centre of Gravity 165
Equating moments of parts and the whole about A —
AC x (area of circle AB) = (AD x area of circle EF)
+ (x x area of eccentric)
4 x 64 = 6 x 6-25 + ;r(64 - 6-25)
- ^ 256- 37-5 = 3783
5775
hence the distance of the e.g. from C is 4 — 3783 or 0*217 inch.
Example 4. — A hemispherical shell of uni-
form material is 6 inches external radius and
1-5 inches thick. Find its e.g.
Let ABC (Fig. 130) be a solid hemisphere
12 inches diameter, from which a concentric
solid hemisphere abc, 9 inches diameter, has
been cut, leaving a hemispherical shell ACBfaa
1-5 inches thick.
Let x = distance of its e.g. (which is on the
axis of symmetry OC) from O.
Equating moments of volumes about O
(i.e. omitting the factor of weight per unit
volume)—
Volume of solid) , „„ N , ,
ABC 5OC / ~ (v°lume of solid acb x if O<r) + (volume of shell X x)
l7r63 X % X 6 = §TT X (I})3 X J X i + f 7r{63 - (|)3)*
from which x = 2'66 inches
FIG. 130.
The e.g. of the shell is on the axis and 2'66 inches from
centre of the surfaces.
the
EXAMPLES XV.
1. The front wheel of a bicycle is 30 inches diameter and weighs 4 Ibs. ;
the back wheel is 28 inches diameter and weighs 7 Ibs. The remaining
parts of the bicycle weigh 16 Ibs., and their e.g. is 1 8 inches forward of the
back axle and 23 inches above the ground when the steering-wheel is
locked in the plane of the back wheel. Find the e.g. of the whole bicycle ;
state its height above the ground and its distance in front of the back axle
when the machine stands upright on level ground. The wheel centres are
42 inches horizontally apart.
2. A projectile consists of a hollow cylinder 6 inches external and 3
inches internal diameter, and a solid cone on a circular base 6^ inches
diameter, coinciding with one end of the cylinder. The axes of the cone
and cylinder are in line ; the length of the cylinder is 12 inches, and the
1 66
Mechanics for Engineers
height of the cone is 8 inches. Find the distance of the e.g. of the
projectile from its point.
3. A solid of uniform material consists of a cylinder 4 inches diameter
and 10 inches long, with a hemispherical end, the circular face of which
coincides with one end of the cylinder. The other end of the cylinder is
pierced by a cylindrical hole, 2 inches diameter, extending to a depth of
7 inches along the cylinder and co-axial with it. Find the e.g. of the solid.
How far is it from the flat end ?
4. The profile of a crank (Fig. 131) consists of two semicircular ends,
CED and AFB, of 8 inches and 12 inches radii respectively, centred at
points P and O 3 feet apart, and joined by straight
lines AC and BD. The crank is of uniform thick-
ness, perpendicular to the figure, and is pierced
by a hole 10 inches diameter, centred at O. Find
the distance of the c.g of the crank from the axis O.
5. Find the c.g. of a T girder section, the
height over all being 8 inches, and the greatest
width 6 inches, the metal being | inch thick in the
vertical web, and I inch thick in the horizontal
flange.
6. An I-section girder consists of a top flange
6 inches by I inch, a bottom flange IO inches by
I'75 inches, connected by a web 10 inches by I' 15
inches. Find the height of the c.g. of the section
from the lowest edge.
7. A circular lamina 4 inches diameter has two
circular holes cut out of it, one I '5 inches and the
other i inch diameter with their centres I inch and
1*25 inches respectively from the centre of the
lamina, and situated on diameters mutually perpendicular. Find the c.g.
of the remainder of the lamina.
8. A balance weight in the form of a segment of a circle fits inside the
rim of a wheel, the internal diameter of which is 3 feet. If the segment
subtends an angle of 60° at the centre of the wheel, find the distance of its
c.g. from the axis.
9. If two intersecting tangents are drawn from the extremities of a
quadrant of a circle 4 feet diameter, find the distance of the c.g. of the
area enclosed between the tangents and the arc, from either tangent.
10. A balance weight of a crescent shape fits inside the rim of a wheel
of 6 feet internal diameter, and subtends an angle of 60° at its centre. The
inner surface of the weight is curved to twice the radius of the outer surface,
i.e. the centre from which its profile is struck is on the circumference of the
inside of the wheel. The weight being of uniform thickness perpendicular to
the plane of the wheel, find the distance of its c.g. from the axis of the wheel.
N.B. — The profile is equivalent to the sector of a circle plus two
triangles minus a sector of a larger circle.
CHAPTER VIII
CENTRE OF GRAVITY: PROPERTIES AND
APPLICA TIONS
121. Properties of the Centre of Gravity.— Since the
resultant force of gravity always acts through the centre of
gravity, the weight of the various parts of a rigid body may
be looked upon as statically equivalent to a single force equal
to their arithmetic sum acting vertically through the centre of
gravity of the body. Such a single force will produce the same
reactions on the body from its supports ; will have the same
moment about any point (Art. 90) ; may be replaced by the
same statically equivalent forces or components ; and requires the
same equilibrants, as the several forces which are the weights of
the parts. Hence, if a body be supported by being suspended
by a single thread or string, the e.g. of the body is in the same
vertical line as that thread or string. If the same body is
suspended again from a different point in itself, the e.g. is
also in the second vertical line of suspension. If the two lines
can be drawn on or in the body, the e.g., which must lie at
their intersection, can thus be found experimentally. For
example, the e.g. of a lamina may be found by suspending
it from two different points in its perimeter, first from one and
then from the other, so that its plane is in both cases vertical,
and marking upon it two straight lines which are continuations
of the suspension thread in the two positions.
Fig. 132 shows G, the e.g. of a lamina PQRS, lying in both
the lines of suspension PR and QS from P and Q respectively.
The tension of the cord acts vertically upwards on the lamina,
and is equal in magnitude to the vertical downward force of
1 68
Mechanics for Engineers
the weight of the lamina acting through G. The tension can
only balance the weight if it acts through G, for in order that
two forces may keep a body in equilibrium, they must be con-
FIG. 132.
current, equal, and opposite, and therefore in the same straight
line.
A " plumb line," consisting of a heavy weight hanging from
a thin flexible string, serves as a convenient method of obtaining
a vertical line.
122. Centre of Gravity of a Distributed Load. — If
a load is uniformly distributed over the whole span of a beam,
the centre of gravity of the load is at mid-span, and the
reactions of the supports of the beam are the same as would
be produced by the whole load
concentrated at the middle of
the beam. Thus, if in Fig. 133
a beam of 2o-feet span carries a
load of 3 tons per foot of span
(including the weight of the beam)
uniformly spread over its length,
the reactions at the supports A
and B are each the same as would be produced by .a load
of 60 tons acting at C, the middle section of the beam,
viz. 30 tons at each support. Next suppose the load on a
beam is distributed, not evenly, but in some known manner.
Suppose the load per foot of span at various points to be
B
FIG. 133.
Centre of Gravity : Properties and Applications 1 69
shown by the height of a curve ACDEB (Fig. 134). The
load may be supposed to be piled on the beam, so that the
curve ACDEB is its profile, and so that the space occupied is
of constant thickness in a direction perpendicular to the plane
of the figure. Then the e.g. of the load is at the e.g. G of
D
FIG. 134.
the area of a section such as ACDEB in Fig. 134, taken
halfway through the constant thickness. The reactions of
the supports are the same as if the whole load were concen-
trated at the point G. The whole load is equal to the length
of the beam multiplied by the mean load per unit length,
which is represented by the mean ordinate of the curve ACDEB,
i.e. a length equal to the area ACDEB divided by AB.
Example. — As a par-
ticular case of a beam
carrying a distributed load
not evenly spread, take a
beam of 2o-feet span carry-
ing a load the intensity of
which is 5 tons per foot
run at one end, and varying
uniformly to 3 tons per foot
at the other. Fig. 135
represents the distribu-
tion of load. Find the
reactions at A and B.
--»-—
*>
1
t
in- 2O feet
>
•v
I
<Y
FL
B
FIG. 135.
The total load = 20 x mean load per foot = 20 x — - = 80 tons
Let ~x be the distance of the e.g. of area ABCD from BD.
I-(area ACFB + area CDF) = (roxarea ACFB) + (-2^ xarea CDF)
I<3 X 20 + \ . 20 X 2) = (10 X 20 X 3) + ^> X ^- X 2
- = 600 +, 33-3 = 9.1<5feet"
and distance of e.g. from AC = 20 — 9-16 = lo'Sj feet
170 -Mechanics for Engineers
If RA and RB be the reactions at A and B respectively, equating
opposite moments about B of all the forces on the beam —
RA x 20 = 80 x 9'i6
RA — 80 x - - = 36*6 tons
RB = 80 — 36'6 = 43'3 tons
123. Body resting upon a Plane Surface. — As in the
case of a suspended body, the resultant of all the supporting
forces must pass vertically through the e.g. of the body in
order to balance the resultant gravitational forces in that
straight line. The vertical line through the e.g. must then
cut the surface, within the area of the extreme outer polygon
or curved figure which can be formed by joining all the points
of contact with the plane by straight lines. If the vertical
line through the e.g. fall on the perimeter of this polygon
the solid is on the point of overturning, and if it falls outside
that area the solid will topple over unless supported in
some other way. This is sometimes expressed by saying
TG
I
I
FIG. 136.
that a body can only remain at rest on a plane surface if
the vertical line through the e.g. falls within the base. From
what is stated above, the term " base " has a particular mean-
ing, and does not signify only areas of actual contact; e.g.
in Fig. 136 are two solids in equilibrium, with GN, the vertical
line through G, the e.g., falling within the area of contact;
Centre of Gravity : Properties and Applications 1 7 1
but in Fig. 137 a solid is shown in which the vertical through
the e.g. falls outside the area of contact when the solid rests
upright with one end on a horizontal plane. If, however,
it falls within the extreme area ABC, the solid can rest in
equilibrium on a plane.
Plan.
FIG. 137.
Two cases in which equilibrium is impossible are shown in
Fig. 138, the condition stated above being violated. The first
is that of a high cylinder on an inclined plane, and the second
FIG. 138.
that of a waggon-load of produce on the side of a high crowned
road. It will be noticed that a body subjected to tilting will
topple over with less inclination or more, according as its e.g.
is high or low.
Example. — What is the greatest length which a right cylinder
of 8 inches diameter may have in order that it may rest with one
end on a plane inclined 20° to the horizontal ?
Mechanics for Engineers
The limiting height will be reached when the e.g. falls vertically
over the circumference of the base, i.e. when G (Fig. 139) is
FIG. 139.
vertically above A. Then, G being the mid-point of the axis EF,
the half-length of cylinder —
GE = AE cot AGE = AE cot ACD
or GE = AE cot 20° — 4x2 7475 = 10-99 inches
The length of cylinder is therefore 2 x 10*99 = 21*98 inches.
124. Stable, Unstable, and Neutral Equilibrium.—
A body is said to be in stable equilibrium when, if slightly
disturbed from its position, the forces acting upon it tend
to cause it to return to that position.
If, on the other hand, the forces acting upon it after a
slight displacement tend to make it go further from its former
position, the equilibrium is said to be unstable.
If, after a slight displacement, the forces acting upon the
body form, a system in equilibrium, the body tends neither
to return to its former position nor to recede further from it,
and the equilibrium is said to be neutral.
A few cases of equilibrium of various kinds will now be con-
sidered, and the conditions making for stability or otherwise.
125. Solid Hemisphere resting on a Horizontal
Plane. — If a solid hemisphere, ABN (Fig. 140), rests on a
^>
- THE *
Centre of Gravity: Properties and Applicat
horizontal plane, and receives a small tilt, say
angle 0, the e.g., situated at G, f of ON from O and in the
radius ON, takes up the position shown on the right hand of
FIG. 140.
the figure. The forces acting instantaneously on the solid are
then — (i) the weight vertically through G, and (2) the reaction
R in the line MO vertically through M (the new point of
contact between hemisphere and plane) and normal to the
curved surface. These two forces form a " righting couple,"
and evidently tend to rotate the solid into its original posi-
tion. Hence the position shown on the left is one of stable
equilibrium. Note that G lies below O.
• 126. Solid with a Hemispherical End resting on
a Horizontal Plane.' — Suppose a solid consisting of, say,
FIG. 141.
a cylinder with a hemispherical base, the whole being of
homogeneous material, rests on a plane, and the e.g. G (Fig.
141) falls within the cylinder, i.e. beyond the centre O of the
hemispherical end reckoned from N, where the axis cuts the
1/4 Mechanics for Engineers
curved surface. On the left of Fig. 141 the solid is shown in
a vertical position of equilibrium. Now suppose it to receive
a slight angular displacement, as on the right side of the figure.
The weight W, acting vertically downwards through G, along
with the vertical reaction R of the plane, forms a system, the
tendency of which is to move the body so that G moves, not
towards its former position, but away from it. The weight
acting vertically through G and the reaction of the plane acting
vertically through O form an " upsetting couple " instead of a
"righting couple." Hence the position on the left of Fig. 141
is one of unstable equilibrium. Note that in this case G falls
above O. If the upper part of the body were so small that G
is below O, the equilibrium would be stable, as in the case of
the hemisphere above (Art. 125). The lower G is, the greater
is the righting couple (or the greater the stability) for a given
angular disturbance of the body. While in the case of in-
stability, the higher G is, the greater is the upsetting couple or
the greater the instability, and we have seen that such a solid
is stable or unstable according as G falls below or above O.
127. Critical Case of Equilibrium neutral. — If G
coincides with the centre of the hemisphere (Art. 126), the
equilibrium is neither stable nor unstable, but neutral. Suppose
the cylinder is shortened so that G, the e.g. of the whole solid,
falls on O, the centre of
AR /\|R the hemisPhere- Then if
the solid receives a slight
angular displacement, as
in the right side of Fig.
142, the reaction R of
the plane acts vertically
FlG> I42. upwards through O, the
centre of the hemisphere
(being normal to the surface at the point of contact), and
the resultant force of gravity acts vertically downward through
the same point. In this case the two vertical forces balance,
and there is no couple formed, and no tendency to rotate
the body towards or away from its former position. Hence
the equilibrium is neutral.
FlG-
Centre of Gravity : Properties and Applications 175
In each of the above instances the equilibrium as regards
angular displacements is the same whatever the direction of
the displacement. As
further examples of neu-
tral equilibrium, a sphere
or cylinder of uniform
material resting on a
horizontal plane may be
taken. The sphere is
in neutral equilibrium
with regard to angular
displacements in any direction, but the horizontal cylinder
(Fig. 143) is only in neutral equilibrium as regards its rolling
displacements ; in other directions its equilibrium is stable.
Example. — A cone and a hemisphere of the same homogeneous
material have a circular face of i foot radius
in common. Find for what height of the
cone the equilibrium of the compound solid
will be neutral when resting with the hemi-
spherical surface on a horizontal plane.
The equilibrium will be neutral when the
e.g. of the solid is at the centre of the hemi-
sphere, i.e. at the centre O (Fig. 144) of their
common face.
Let h be the height of the cone in feet.
Then its e.g. G1 is \h from O, and its volume
is \h x - x 22 = \irh cubic feet.
The e.g. G2, of the hemisphere is at
volume is §TT cubic feet. Then —
FIG. 144.
foot from O, and its
Gyp _ \h _ weighj^ofhemisphere _ _§7r
G2O § weight of cone
2
and ^h =
h - ,v/3 = 1732 feet
If h is greater than ^3 feet the equilibrium is unstable, and if it
is less than ^3 feet the equilibrium is stable.
Mechanics for Engineers
128. In the case of bodies resting on plane surfaces and
having more than one point of contact, the equilibrium will
be stable if the e.g. falls within the area of the base^ giving
the word the meaning attached to it in Art. 123 for small
angular displacements in any direction. If the e.g. falls on
the perimeter of the base, the equilibrium will be unstable for
displacements which carry the e.g. outside the space vertically
above the " base."
The attraction of the earth tends to pull the e.g. of a body
into the lowest possible position ; hence, speaking generally,
the lower the e.g. of a body the greater is its stability, and
the higher the e.g. the less stable is it.
In the case of a body capable of turning freely about a
horizontal axis, the only position of stable equilibrium will be
that in which the e.g. is vertically below the axis. When it
Unstable
is vertically above, the equilibrium is unstable, and unless the
e.g. is in the axis there are only two positions of equilibrium.
If the e.g. is in the axis, the body can rest in neutral equilibrium
in any position.
Fig. 145 represents a triangular plate mounted on a hori-
zontal axis, C ; it is in unstable, stable, or neutral equilibrium
according as the axis C is below, above, or through G, the e.g.
of the plate.
129. Work done in lifting a Body.— When a body
is lifted, it frequently happens that different parts of it are lifted
through different distances, e.g. when a hanging chain is wound
up, when a rigid body is tilted, or when water is raised from
one vessel to a higher one. The total work done in lifting the
Centre of Gravity: Properties and Applications IJJ
body can be reckoned as follows : Let w}, «>2, w^ w4, etc., be
the weights of the various parts of the body, which is supposed
divided into any number of parts, either large or small, but
such that the whole of one part has exactly the same displace-
ment (this condition will in many cases involve division into
indefinitely small parts). Let the parts wlt w2, ws, etc., be at
heights xlt #2, #3, etc., respectively above some fixed horizontal
plane ; if the parts are not indefinitely small, the distances xlt
x-2> x3, etc., refer to the heights of their centres of gravity.
Then the distance x of the e.g. from the plane is -^VJr
(Art. 113). After the body has been lifted, let x-f, xj, x3, etc.,
be the respective heights above the fixed plane of the parts
weighing wlt w.^ w3t etc. Then the distance x' of the e.g.
above the plane is ,- (Art. 113).
The work done in moving the part weighing w^ is equal to
the weight u\ multiplied by the distance (x± — xj through
which it is lifted ; i.e. the work is w-^(x^ — xj units.
Similarly, the work done in lifting the part weighing 7C2 is
7c.2(x.2 — x.2). Hence the total work done is —
«'i(*i' - *i) + w»(x.! - #2) + W3(x3' - xs) +, etc. »
which is equal to —
(•7^VT/ + W&.1 4- n'axj -f, etc.) - (w^ + WyX.2 + #'3*3 +, etc.)
or 2(wx') — ^(wx)
But ^(wx1) =~x"&(w) and ^(wx) =^(w)
therefore tne work done = x"%(uf) — x$(w)
The first factor, x — xt is the distance through which the
e.g. of the several weights has been raised, and the second
factor, 2(o>), is the total weight of all the parts. Hence the
total work done in lifting a body is equal to the weight of the
body multiplied by the vertical distance through which its e.g.
has been raised.
N
178
Mechanics for Engineers
Example i. — A rectangular tank, 3 feet long, 2 feet wide, and
i '5 feet deep, is filled from a cylindrical tank of 24 square feet
horizontal cross-sectional area. The level of water, before filling
begins, stands 20 feet below
the bottom of the rectangular
tank. How much work is re-
quired to fill the tank, the
weight of i cubic foot of water
being 62-5 Ibs. ?
The water to be lifted is
3 x 2 x 1*5 or 9 cubic feet,
hence the level in the lower
tank will be lowered by ^>4 or
f of a foot, i.e. by a length BC
on Fig. 146. The 9 cubic feet
of water lifted occupies first
the position ABCD, and then
fills the tank EFGH. In the
former position its e.g. is |BC
or j3^ foot below the level AB,
and in the latter position its
e.g. is ^GH or | foot above
the level EH. Hence the
FIG. 146. e.g. is lifted (^ + 20 + f ) feet,
i.e. 2o}f feet, or 20*9375 feet-
The weight of the 9 cubic feet of water lifted is 9 x 62*5
= 562*5 Ibs.
Hence the work done is 562-5 x 20-9375 = 11,770 foot-lbs.
Example 2. — Find the work in foot-pounds necessary to upset
a solid right circular cylinder
3 feet diameter and 7 feet high,
weighing half a ton, which is
resting on one end on a hori-
zontal plane.
Suppose the cylinder (Fig.
147) to turn about a point A on
the circumference of the base.
Then G, the e.g. of the cylinder,
which was formerly 3*5 feet
above the level of the hori-
zontal plane, is raised to a
position G', i.e. to a height A'G'
above the horizontal plane before the cylinder is overthrown.
Centre of Gravity: Properties and Applications 179
The distance the e.g. is lifted is then A'G' - EG—
A'G' =
T
+ EG2) = V(i'52 + 3'S2') = 3*807 feet
The e.g. is lifted 3*807 — 3-5 = 0-307 foot
and the work done is 1120 x 0*307 = 344 foot-lbs.
Example 3. — A chain 600 feet long hangs vertically ; its weight
at the top end is 12 Ibs. per foot, and at the bottom end 9 Ibs. per
foot, the weight per foot varying uniformly
from top to bottom. Find the work necessary
to wind up the chain.
It is first necessary to find the total weight
of the chain and the position of its e.g. The
material of the chain may be considered to be
spread laterally into a sheet of uniform thick-
ness, the length remaining unchanged. The
width of the sheet will then be proportional
to the weight per foot of length ; the total
weight, and the height of the e.g. of the chain,
will not be altered in such a case.
The depth of the e.g. below the highest
point (A) of the chain (Fig. 148) will be the
same as that of a figure made up of a rect-
angle, ACDB, 600 feet long and 9 (feet or other units) broad,
and a right-angled triangle, CED, having sides about the right
angle at C of (CD) 600 feet and (CE) 3 units.
The depth will be —
(600 x 9 x 300) + (\ x 600 x 3 x fljp) ( .
(600 x 9) + (i x 600 x 3)
which is equal to 2857 feet.
The total weight of the chain will be the same as if it were
k- - 9 - *
FIG. 148.
600 feet long and of uniform weight
12
or 10*5 Ibs. per foot,
viz. 600 x 10-5 = 6300 Ibs.
Hence the work done in raising the chain all to the level A is —
6300 x 285*7 = 1,800,000 foot-lbs.
130. Force acting on a Rigid Body rotating uni-
formly about a Fixed Axis.
Let Fig. 149 represent a cross-section of a rigid body of
weight W rotating about a fixed axis, O, perpendicular to the
figure. For simplicity the body will be supposed symmetrical
i8o
Mechanics for Engineers
N
FIG. 149.
about the plane of the figure, which therefore contains G, the
e.g. of the body. In the position shown, let wl be the weight
of a very small portion of the
body (cut parallel to the axis)
situated at a distance r from
O. Let w be the uniform
angular velocity of the body
about the axis O. Then the
force acting upon the small
portion of weight u\ in order
to make it rotate about O is
— 1oj2;-, directed towards O
cS
(Art. 63), and it evidently
acts at the middle of the
length of the portion, i.e. in the plane of the figure. Resolving
this force in any two perpendicular directions, XO and YO,
the components in these two directions are JwV cos 0 and
— wV sin 0 respectively, where 0 is the angle which AO
<b
makes with OX.
These may be written — . w2 . x and — 1w2 . y respectively,
<b O
where x represents r cos 6 and y represents r sin 0, the
projections of r on OX and OY respectively.
Adding the components in the direction XO of the centri-
petal forces acting in the plane of the figure upon all such
portions making up the entire solid, the total component —
(W 2 \ _ W2 W2_
g / g s
and the total component force in the direction YO is
where x and y are the distances of G, the e.g. of the solid
(which is in the plane of the figure), from OY and OX re-
spectively.
Centre of Gravity : Properties and Applications 1 8 1
Hence the resultant force P acting on the solid towards
Ois-
= . W.
where R = x2 4- jv2, the distance of the e.g. from the axis O.
Hence the resultant force acting on the body is of the same
/W \
magnitude as the centripetal force ( — w2R 1 which must act
on a weight W concentrated at a radius R from O in order
that it may rotate uniformly at an angular velocity w. Further,
Tj'
the tangent of the angle which P makes with XO is -^
(Art. 75), which is equal to i or — — where GN is perpen-
dicular to OX. Hence the force P acts in the line GO, and
therefore the resultant force P acting on the rotating body is
in all respects identical with that which would be required to
make an equal weight, W, rotate with the same angular velocity
about O if that weight were concentrated (as a particle) at G,
the e.g. of the body.
It immediately follows, from the third law of motion, that
the centrifugal force exerted by the rotating body on its con-
straints is also of this same magnitude and of opposite direction
in the same straight line.
Example. — Find the force exerted on the axis
uniform rod 5 feet long and weighing 9 Ibs., making
30 revolutions per minute about an axis perpen-
dicular to its length.
The distance from the axis O to G, the c g. of
the rod (Fig. 150), is 2*5 feet, the e.g. being midway
between the ends. The angular velocity of the
rod is - — , — = IT radians per second. The cen-
trifugal pull on O is the same as that of a weight of
9 Ibs. concentrated at 2-5 feet from the axis and describing
about O, TT radians per second, which is —
- x 7T2 x 2-5 = 6-89 Ibs.
1 82 Mechanics for Engineers
131. Theorems of Guldinus or Pappus. — (a) The
area of the surface of revolution swept out by any plane curve
revolving about a given axis in its plane is equal to the length
of the curve multiplied by the length of the path of its e.g.
in describing a circle about the axis. Suppose the curve
ABC (Fig. 151) revolves about the
axis OO', thereby generating a surface
of revolution of which OO' is the axis.
Let S be the length of the curve, and
|3 suppose it to be divided into a large
number of small parts, su s.2, s3, etc.,
each of such short length that if drawn
straight the shape of the curve is not
appreciably altered. Let the distances
of the parts slt s%, s3, etc., from the
axis be xlt x^ x& etc. ; and let G, the
e.g. of the curve which is in the plane of the figure, i.e. the plane
of the curve, be distant x from the axis OO'. The portion sl
generates a surface the length of which is 2irxl and the breadth
sl ; hence the area is 2irxlsl. Similarly, the portion s2 gene-
rates an area 2irx.2 . s>2, and the whole area is the sum —
-\- 2irx3s3 + , etc., or 2ir^(xs)
If the portions slt s2, s3, etc., are of finite length, this result is
only an approximation; but if we understand 2,(xs) to represent
the limiting value of such a sum, when the length of each part
is reduced indefinitely, the result is not a mere approximation.
Now, since 2<(xs) = x X ^(s) = x x S, the whole area of
the surface of revolution is 2irx . S, of which ZTTX is the length
of the path of the e.g. of the curve in describing a circle about
OO', and S is the length of the curve.
(b) The volume of a solid of revolution generated by the
revolution of a plane area about an axis in its plane is equal
to the enclosed revolving area multiplied by the length of the
path of the e.g. of that area in describing a complete circle
about the axis.
Suppose that the area ABC (Fig. 152) revolves about the
axis OO', thereby generating a solid of revolution of which
Centre of Gravity: Properties and Applications 183
OO' is an axis (and which is enclosed by the surface generated
by the perimeter ABC).
Let the area of the plane figure ABC be denoted by A,
and let it be divided into a large
number of indefinitely small parts
#11 #2, #3, etc., situated at distances
#!, #2, xs, etc., from the axis
OO'.
The area a^ in revolving about
OO', generates a solid ring which
has a cross- section a^ and a length
2-irXn and therefore its volume is
27rx1al. Similarly, the volume swept
out by the area a2 is 2irx.2a.2, and
so on. The whole volume swept
out by the area A is the limiting value of the sum of the small
quantities —
or
a1 -f- 2Trx.2a.2 + 2-rrx3a3 -f , etc.,
1 4- a2x2 4- #3*3 + , etc.,) or 2ir2t(ax)
And since ^(ax) = x%(a) = x. A (Art. 114 (6)), the whole
volume is 2irx . A, of which 2irx is the length of the path of
the e.g. of the area in describing a circle about the axis OO',
and A is the area.
Example.— A groove of semicircular section 1^25 inches
radius is cut in a cylinder 8 inches diameter. Find (a) the area of
the curved surface of the groove, A R
and (b) the volume of material
removed.
(a) The distance of the e.g. of
the semicircular arc ABC (Fig.
1 53) from AB is (i'2$ x -^ or —
V 7T / TT
inches. Therefore the distance of
the e.g. of the arc from the axis
O1
FIG. 153.
OO' is ( 4 - --2 ) inches. The
7T /
length of path of this point in making one complete circuit about
1 84
Mechanics for Engineers
OO' is 277(4 - 2-^\ = (877 - 5) inches. The length of arc ABC
\ 7T /
is 1*2577 inches, hence the area of the surface of the semicircular
groove is —
1*2577(877 — 5) square inches = IO772 — 6*2577
= 98*7 — 19*6
= 79' i square inches
(ff) The distance of the e.g. of the area ABC from AB is
4
— x i '25 = 0*530 inch, and therefore the distance of the e.g. from
OO' is 4 — o'53 = 3*47 inches.
The length of path of this point in making one complete circuit
about OO' is 277 x 3*47 = 21*8 inches. The area of the semicircle
is-|-(i'25)277 = 2*453 square inches, hence the volume of the material
removed from the groove is —
21*8 x 2*453 = 53*5 cubic inches
132. Height of the e.g. of a Symmetrical Body,
such as a Carriage, Bicycle, or Locomotive. — It was stated
in Art. 121 that the e.g. of some bodies might conveniently
be found experimentally by suspending the bodies from two
different points in them alternately. This is not always con-
venient, and a method suitable for some other bodies will now
be explained by reference to a particular instance. The e.g.
of a bicycle (which is generally nearly symmetrical about a
FIG. 154.
vertical plane through both wheels) may be determined by first
finding the vertical downward pressure exerted by each wheel
on the level ground, and then by finding the vertical pressures
when one wheel stands at a measured height above the other one.
Suppose that the wheels are the same diameter, and that
the centre of each wheel-axle, A and B (Fig. 154), stands
Centre of Gravity : Properties and Applications 185
at the same height above a level floor, the wheels being locked
in the same vertical plane.
When standing level, let WA = weight exerted by the front
wheel on a weighing machine table ; let WB = weight exerted
by the back wheel on a weighing machine table ; then —
WA + WB = weight of bicycle
Let AB, the horizontal distance apart of the axle centres,
be d inches. If the vertical line through the e.g. G cuts AB in
C, then—
W
Next, let the weight exerted by the front wheel, when A
stands a distance " h " inches (vertically) above B, be Wa ; and
let CG, the distance of the e.g. of the bicycle above AB, be H.
FIG. 155.
Then, since ABE and DGC (Fig. 155) are similar triangles—
GC = BE _
CD AE"
andCD = BC-BD =
WA + WB- WA + WB-
W.-W,
__ __
hence GC or H= ;/ -
In an experiment on a certain bicycle the quantities were
d = 44 inches, h = 6 inches, weight of bicycle = 3 2 '90 Ibs.,
pressure (WA) exerted by the front wheel when the back wheel
1 86 Mechanics for Engineers
was on the same level = 14*50 Ibs., pressure (Wa) exerted
by the front wheel when the back wheel was 6 inches lower
= 13-84 Ibs.
Hence H = _ X x 44
6 32-90
= 6 -5 4 inches
or the height of the e.g. above the ground is 6*54 inches plus
the radius of the wheels. The distance BC of the e.g. horizon-
tally in front of the back axle is ^-^ x 44, or 10-4 inches.
32-90
A similar method may be applied to motor cars or locomotives.
In the latter case, all the wheels on one side rest on a raised
rail on a weighing machine, thus tilting the locomotive sideways.
EXAMPLES XVI.
•M. A beam rests on two supports at the same level and 12 feet apart.
It carries a distributed load which has an intensity of 4 tons per foot-run
at the right-hand support, and decreases uniformly to zero at the left-hand
support. Find the pressures on the supports at the ends.
2. The span of a simply supported horizontal beam is 24 feet, and
along three-quarters of this distance there is a uniformly spread load of
2 tons per foot run, which extends to one end of the beam : the weight of
the beam is 5 tons. Find the vertical supporting forces at the ends.
3. A beam is supported at the two ends 15 feet apart. Reckoning
from the left-hand end, the first 4 feet carry a uniformly spread load of
I ton per foot run ; the first 3 feet starting from the right-hand end carry
a load of 6 tons per foot run evenly distributed, and in the intermediate
portion the intensity of loading varies uniformly from that at the right-
hand end to that at the left-hand end. Find the reaction of the supports.
4. The altitude of a cone of homogeneous material is 18 inches, and
the diameter of its base is 12 inches. What is the greatest inclination on
which it may stand in equilibrium on its base ?
5. A cylinder is to be made to contain 250 cubic inches of material.
What is the greatest height it may have in order to rest with one end on a
plane inclined at 15° to the horizontal, and what is then the diameter of the
base?
6. A solid consists of a hemisphere and a cylinder, each 10 inches
diameter, the centre of the base of the hemisphere being at one end of the
axis of the cylinder. What is the greatest length of cylinder consistent
with stability of equilibrium when the solid is resting with its curved end
on a horizontal plane ?
Centre of Gravity : Properties and Applications 1 87
7. A solid is made up of a hemisphere of iron of 3 inches radius, and
a cylinder of aluminium 6 inches diameter, one end of which coincides
with the plane circular face of the hemisphere. The density of iron being
three times that of aluminium, what must be the length of the cylinder if
the solid is to rest on a horizontal plane with any point of the hemispherical
surface in contact?
8. A uniform chain, 40 feet long and weighing 10 Ibs. per foot, hangs
vertically. How much work is necessary to wind it up ?
9. A chain weighing 12 Ibs. per foot and 70 feet long hangs over a
(frictionless) pulley with one end 20 feet above the other. How much
work is necessary to bring the lower end to within 2 feet of the level of
the higher one ?
10. A chain hanging vertically consists of two parts : the upper portion
is loo feet long and weighs 16 Ibs. per foot, the lower portion is 80 feet
long and weighs 12 Ibs. per foot. Find the work done in winding up
(a) the first 70 feet of the chain, (b) the remainder.
11. A hollow cylindrical boiler shell, 7 feet internal diameter and
25 feet long, is fixed with its axis horizontal. It has to be half filled with
water from a reservoir, the level of which remains constantly 4 feet below
the axis of the boiler. Find how much work is required to lift the water,
its weight being 62*5 Ibs. per cubic foot.
12. A cubical block of stone of 3-feet edge rests with one face on the
ground : the material weighs 150 Ibs. per cubic foot. How much work is
required to tilt the block into a position of unstable equilibrium resting on
one edge?
13. A cone of altitude 2 feet rotates about a diameter of its base at a
uniform speed of 180 revolutions per minute. If the weight of the cone
is 20 Ibs., what centrifugal pull does it exert on the axis about which it
rotates ?
14. A shaft making 150 rotations per minute has attached to it a pulley
weighing 80 Ibs., the e.g. of which is o'l inch from the axis of the shaft.
Find the outward pull which the pulley exerts on the shaft.
15. The arc of a circle of 8 inches radius subtends an angle of 60° at
the centre. Find the area of the surface generated when this arc revolves
about its chord ; find also the volume of the solid generated by the revolu-
tion of the segment about the chord.
16. A groove of V-shaped section, 1*5 inches wide and I inch deep, is
cut in a cylinder 4 inches in diameter. Find the volume of the material
removed.
17. A symmetrical rectangular table, the top of which measures 8 feet
by 3 feet, weighs 150 Ibs., and is supported by castors at the foot of each
leg, each castor resting in contact with a level floor exactly under a corner
of the table top. Two of the legs 3 feet apart are raised 10 inches on to the
plate of a weighing machine, and the pressure exerted by them is 66' 5 Ibs.
Find the height of the e.g. of the table above the floor when the table
stands level.
CHAPTER IX
MOMENTS OF INERTIA — ROTATION
133- Moments of Inertia.
(i) Of a Particle.— -If a particle P (Fig. 156), of weight w
and mass — , is situated at a distance r from an axis OO', then
o
its moment of inertia about that
axis is denned as the quantity
— . r2, or (mass of P) X (distance
S
from OO')2.
0 O1 (2) Of Several Particles -.— -If
FIG. 156.
several particles, P, Q, R, and
S, etc., of weights wlt w2, ws, «/4, etc., be situated at distances
;'u I'M r& and ;-4, etc., respectively from an axis OO' (Fig. 157),
P
s
Q
FIG. 157.
End view of axis OO'.
then the total moment of inertia of the several particles about
that axis is denned as —
Moments of Inertia — Rotation 189
g S S S
2
or 2{(mass of each particle) x (its distance from OO')
(3) Rigid Bodies. — If we regard a rigid body as divisible
into a very large number of parts, each so small as to be
regarded as a particle, then the moment of inertia of the rigid
body about any axis is equal to the moment of inertia of such
a system of particles about that axis. Otherwise, suppose a
body is divided into a large but finite number of parts, and the
mass of each is multiplied by the square of the distance of
some point in it from a line OO' ; the sum of these products
will be an approximation to the moment of inertia of the whole
body. The approximation will be closer the larger the number
of parts into which the body is divided ; as the number of parts
is indefinitely increased, and the mass of each correspondingly
decreased, the sum of the products tends towards a fixed
limiting value, which it does not exceed however far the
subdivision be carried. This limiting sum is the moment of
inertia of the body, which may be written 3(;;/;-2) or 2jf— • r2 \
Units. — The units in which a moment of inertia is stated
depend upon the units of mass and length adopted. No
special names are given to such units. The " engineer's unit"
or gravitational unit is the moment of inertia about an axis of
unit mass (32*2 Ibs.) at a distance of i foot from the axis.
134. Radius of Gyration. — The radius of gyration of a
body about a given axis is that radius at which, if an equal
mass were concentrated, it would have the same moment of
inertia.
Let the moment of inertia ^( ~rrj of a body about some
axis be denoted by I, and let its total weight ^(w) be W, and
/«'\ W
therefore its total mass
Mechanics for Engineers
Let k be its radius of gyration about the same axis. Then,
from the above definition—
135. Moments of Inertia of a Lamina about an
Axis perpendicular to its
Plane.
Let the distances of any
particle, P (Fig. 158), of a
lamina from two perpen-
dicular axes, OY and OX,
in its plane be xl and y\ re-
spectively, and let wt be its
weight, and i\ its distance
from O, so that r? = x? + y^.
Then, if Ix and IY denote the moments of inertia of the
lamina made up of such particles, about OX and OY re-
spectively —
FIG. 158.
and adding —
<1U 2\
—r ), which may be denoted by I<
Then I0 = Ix + I,
(i)
This quantity I0 is by definition the moment of inertia
about an axis OO' perpendicular to the plane of the lamina,
and through O the point of intersection of OX and OY.
Moments of Inertia — Rotation
OF THE
Wr
Hence the sum of the moments of inertia of a lamina
any two mutually perpendicular axes in its plane^ is equal to the
moment of inertia about an axis through the intersection of the
other two axes and perpendicular to the plane of the lamina.
Also, if /£XJ ^Y) an<3 kQ be the radii of gyration about
OX, OY, and OO' respectively, OO' being perpendicular to the
('w'\ W
— ) = — , the mass of the whole
S o
lamina —
W
W
and Ix = >£x2 . —
and IY = /&Y2 . —
W
and therefore, since Ix + IY = £02 . — by (i)
(2)
0'
Or, in words, the sum of the squares of the radii of gyration of
a lamina about two mutually perpendicular axes in its plane,
is equal to the square of its radius of gyration about an axis
through the intersection of the other two axes and perpendicular
to the plane of the lamina.
136. Moments of Inertia of a Lamina about
Parallel Axes in its Plane. — Let P, Fig. 159, be a
constituent particle of weight
7fj of a lamina, distant x1 from
an axis ZZ' in the plane of
the lamina and through G, the
e.g. of the lamina, the distances
being reckoned positive to the
right and negative to the left
of ZZ'. Let OO' be an axis
in the plane of the lamina
parallel to ZZ' and distant d
from it. Then the distance
of P from OO' is d — x^ whether P is to the right or left
of ZZ'.
FIG. 159.
1 92 Mechanics for Engineers
Let I0 be the moment of inertia of the lamina about OO' ;
and let Iz „ „ „ „ ZZ'.
Then—
( ^1 / ?^2 2£/3
I0 = ,++ + ,e
— 2-(o/1^1 -f- «;2#2 4- «'3#j +, etc.)
o
The sum w^ + ^2-^2 + ^3^3 +, etc., is, by Art 114, equal
to
(W-L + 7^2 -f- w3 +, etc.) X (distance of c g. from ZZ')
which is zero, since the second factor is zero. Hence—
(i)
where W is the total weight of the lamina. And dividing each
W
term of this equation by —
(2)
where kQ and kz are the radii of gyration about OO' and ZZ'
respectively.
I37.1 Extension of the Two Previous Articles to
Solid Bodies.— (a) Let ZX and ZY (Fig. 160) represent (by
their traces) two planes perpendicular to the plane of the paper
and to each other, both passing through the e.g. of a solid
body.
Let P be a typical particle of the body, its weight being 7^,
1 This article may be omitted on first reading. The student acquainted
with the integral calculus will readily apply the second theorem to simple
solids.
Moments of Inertia — Rotation
193
and its distances from the planes ZY and ZX being xl and yl
respectively. Then, if r^ is the distance of P from an axis ZZ',
which is the intersection
w
of the planes XZ and
YZ, and passes through
tteag.r^jtf+jfA
Let Iz be the moment
of inertia of the body
about ZZ', and I0 that
about a parallel axis
OO'. Let OO' be distant
d from ZZ', and distant
p and q from planes ZY and ZX respectively. Then/2 4- q"2 = d'1.
Let other constituent particles of the body of weights
a>3, w3t WH etc., be at distances x.2, x3t x^ etc., from the
plane ZY, and distances y.2, y,, j4, etc., from the plane .ZX
respectively, the x distances being reckoned positive to the
right and negative to the left of ZY, and the y distances being
reckoned positive above and negative below ZX. Let ;-2, r3t ;-4,
etc., be the distances of the particles from ZZj. Let u\ + w.2
4- a>3 4-, etc. = 2(a/) or W, the total weight of the body.
By definition —
FIG. 160.
and OP:2 = p
therefore I0 =
02 + (q -
- ^i)2 +
- y 4-
4- W1
etc. —
a/.^ + a/,
4-
3+, etc
) +«',(
a'o^2
,^3 +, etc.)}
4- (w^i2
+, etc.}
'a + , etc.)
> etc-) ~
, etc.)
194 Mechanics for Engineers
= o
since the planes XZ and YZ pass through the e.g. of the body
(Art. 113).
Hence £,=*— </a 4 Iz ..... (0
o>
W
and dividing both sides of (i) by — —
V = ^ 4 k? ...... (2)
where kQ = radius of gyration about OO', and kz = radius of
gyration about ZZ'.
(b) Also—
W-i 9 . Wo a * W« a
Iz = -Vi 4- -Taa + -Jr* +, etc.
<.*> 6 05
h etc.
= -r(^V2 4 w&? 4 u^.? 4, etc.) 4 ^>'iX2 4 w*yf 4
-, etc.)
2 -^yt^^-v j I
z -~w w
which may be written —
(4)
where .*2 and jy2 are the mean squares of the distances of the
body from the planes YZ and XZ respectively. The two
quantities x2 and y* are in many solids easily calculated.
138. Moment of Inertia of an Area. — The moment
of inertia I0 of a lamina about a given axis OO' in its plane
is ^(~^2) (Art. 133), where w is the weight of a constituent
<S
Moments of Inertia — Rotation 195
particle, and r its distance from the axis OO'. This quantity
W
is equal to — . & (Art. 134), where k is the radius of gyration
<b
about this axis OO',.and W is the total weight of the lamina,
so that —
or — ~
In a thin lamina of uniform thickness /, the area a (Fig.
161) occupied by a particle of
weight w is proportional to wt for . ^
w — a . / . D, where D is the weight f <p
per unit volume of the material ;
hence ^(wr2} -
and similarly, W = A . / . D, where
A is the total area of the lamina ; FIG. 161.
hence #:
Thus the thickness and density of a lamina need not be
known in order to find its radius of gyration, and an area may
properly be said to have a radius of gyration about a given
axis.
The quantity 2(ar*) is also spoken of as the moment of
inertia of the area of the lamina about the axis OO' from which
a portion a is distant r.
The double use of this term " moment of inertia " is un-
fortunate. The "moment of inertia of an area" 2(flr2) or
/C'2 . A is not a true moment of inertia in the sense commonly
used in mechanics, viz. that of Art. 133 ; it must be multiplied
by the factor " mass per unit area " to make it a true moment of
inertia. As before mentioned, the area has, however, a radius
of gyration about an axis OO' in its plane defined by the
equation —
196 Mechanics for Engineers
Units. — The units of the geometrical quantity 3(ar*),
called moment of inertia of an area, depend only upon the units
of length employed. If the units of length are inches, a
moment of inertia of an area is written (inches)4.
139, Moment of Inertia of Rectangular Area about
Various Axes. — Let ABCD (Fig. 162) be a rectangle, AB = ^/,
B c • BC = b. The moment of inertia of the
area ABCD about the axis OO' in the side
AD may be found as follows. Suppose AB
divided into a large number n, of equal
parts, and the area ABCD divided into //
equal narrow strips, each of width --. The
FIG 162 °f anv one stl"ip EFGH is practically
at a distance, say, FA from AD, and if
EFGH is the/th strip from AD, FA = / x -.
n
Multiplying the area EFGH, viz. b X -, by the square of
its distance from AD, we have —
(area EFGH) x FA* = b X *- X (^)* =
n \ n J
There are n such strips, and therefore the sum of the
products of the areas multiplied by the squares of their distances
from OO', which may be denoted by S^r2), is —
bd?
6\ n ri*)
v ' n* 6
When n is indefinitely great, ^ = o, and 2= o, and the sum
bdz bd?
^(at2-} becomes -£ X 2 or — • This is the " moment of inertia
of the area " about OO' ; or, the radius of gyration of the area
about OO' being k —
Moments of Inertia — Rotation 197
If ABCD is a lamina of uniform thickness of weight a», its
true moment of inertia about OO' is /£2 = 1 — . d2.
* 0" «J O*
The radius of gyration of the same area ABCD about an
axis PQ (Fig. 163) in the plane of the
figure and parallel to OO' and distant
- from it, dividing the rectangle into
halves, can be found from the formula (2),
Art. 136, viz. —
-Q
I O C iR D O1
FIG. 163.
where k? = radius of gyration about PQ ;
whence /£P2 = (J — J)^/2 = -f^d2
The sum ^(ar1} about PQ is then 2(0) X k\ = bd X — = TV^3
Similarly, if k$ is the radius of gyration of the rectangle
about RS —
and therefore, if kG = radius of gyration about an axis through
G (the e.g.) and perpendicular to the figure —
V = 42 + W or TV(£2 + ^2) (Art. 135 (2))
which is also equal to yjBC2 or JGB2.
Example. — A plane figure consists of a rectangle 8 inches by
4 inches, with a rectangular hole 6 inches Q ^
by 3 inches, cut so that the diagonals of
the two rectangles are in the same straight
lines. Find the geometrical moment of
inertia of this figure, and its radius of
gyration, about one of the short outer
sides.
Let IA be the moment of inertia of the
figure about AD (Fig. 164), and k be its
radius of gyration about AD.
Moment of inertia of abed \
about AD / = T 2varea abcd^ x (Slde **? + (area abc(t)
\
r.
G
a
c
*
FIG. 164.
Moment of inertia of
ABCD about AD
x (JAB)8 (Arts. 139 and 136)
x 4 x &
198 Mechanics for Engineers
Hence IA = moment of inertia of ABCD — moment of inertia of abed
= i.4.83-(^x 63 x3 + 6x3 x42)
= ~34-a ~ (54 + 288) - 340-6 (inches)4
The area of the figure is —
8x4 — 6x3= 14 square inches
therefore & = ~ = 24-33 (inches)2
and k = 4*93 inches
140. Moment of Inertia of a Circular Area about
Various Axes. — (i) About an axis OO' through O, its centre,
and perpendicular to its plane.
Let the radius OS of the circle (Fig. 165) be equal to R.
Suppose the area divided into a large
number ;/, of circular or ring-shaped
-D
strips such as PQ, each of width —
Then the distance of the /th strip from
-n
O is approximately / X — , and its
FIG j6
area is approximately —
"R R R2
27r X radius X width = 2-* x p- • — = mp~f
The moment of inertia of this strip of area about OO' is
then —
- -!>
» n
and adding the sum of all such quantities for all the n strips—
n
R4 «* + 2n* + n~
When ;/ is indefinitely great, - = o and -5 = o, and the
Moments of Inertia - Rotation
199
becomes — , which is the "moment of inertia of
sum
the circular area " about OO'.
And since 2(ar2) about OO' = -— , if we divide each side
of the equation by the area (TrR2) of the circle —
where k0 is the radius of gyration of the circular area about
an axis OO' through its centre and perpendicular to its
plane.
(2) About a diameter.
Again, if £A and kc are the radii of gyration of the same
area about the axes AB and CD
respectively (Fig. 166) —
R2
hence k* = k<? = J . — = —
R2
from which the relations between
the moments of inertia about AB,
DC, and OO' may be found by
multiplying each term by TrR2.
That is, the moment of inertia of
the circular area about a diameter is half that about an axis
through O and perpendicular to its plane.
Example. — Find the radius of gyration of a ring-shaped
area, bounded outside by a circle of radius a, and inside by a
concentric circle of radius £, about a diameter of the outer
circle.
The moment of inertia of the area bounded by the outer circle,
about AB (Fig. 167) is — ; that of the inner circular area about
4
200
Mechanics for Engineers
the same line is " - ; hence that of the ring-shaped area is - (^ - £').
4 4
The area is 7r(a2 - b2} ; hence, if k is the radius of gyration of
the ring-shaped area about AB —
Note that = . = ?_
+
f J ,
so that when # and b
FIG. 167.
are nearly equal, i.e. when « — £
is a small quantity, the radius of
gyration £0, about the axis O, approaches the arithmetic mean
a + b r .
of the inner and outer radii.
141. Moment of Inertia of a Thin Uniform Rod.— The
radius of gyration of a thin rod d units long and of uniform
material, about an axis through one end and perpendicular to
the length of the rod, will evidently be the same as that of a
narrow rectangle d units long, which, by Art. 139, is given by
the relation /£2 = \ d~, where k is the required radius of gyration.
Hence, if the weight of the rod is W Ibs., its moment of inertia
. Wt9 W d*
about one end is — R* or — . — .
g g 3
Similarly, its moment of inertia about an axis through the
W d2
middle point and perpendicular to the length is — • — .
142. Moment of Inertia of a Thin Circular Hoop. —
(i) The radius of the hoop being R, all the matter in it is
at a distance R from the centre of the hoop. Hence the
radius of gyration about an axis through O, the centre of the
hoop, and perpendicular to its plane, is R, and the moment
of inertia about this axis is — . R2, where W is the weight of
the hoop.
Moments of Inertia — Rotation
201
(2) The radius of gyration about diameters OX and OY
(Fig. 1 68) being kx and kv respectively—
(Art. 135(2))
hence
R2
— •
FIG.
and the moment of inertia about
any diameter of the hoop is
W R2
143. Moment of Inertia of
Uniform Solid Cylinder. — (i)
About the axis OO' of the cylinder.
The cylinder may be looked upon as divided into a large
number of circular discs (Fig. 169) by planes perpendicular
to the axis of the cylinder.
The radius of gyration of each
disc about the axis of the cylinder
R2
is given by the relation >£2 = - •— »
where k is radius of gyration of
the disc, and R the outside radius
of the cylinder and discs. If the
weight of any one disc is wt and
that of the whole cylinder is W, the moment of inertia of one
disc is — ,
w R2
FIG. 169.
and that of the whole cylinder is—
R2\ R*
2.-
W R2
o- 2
and the square of the radius of gyration of the cylinder is
R2
(2) About an Axis perpendicular to that of the
Cylinder and through the Centre of One End. — Let OX
(Fig. 170) be the axis about which the moment of inertia
202
Mechanics for Engineers
of the cylinder is required. Let R be the radius of the
cylinder, and / its length.
Let o? = the mean square of the distance of the constituent
particles from the plane YOO'Y' ;
f = the mean square of the distance of the constituent
particles from the plane OXX'O' ;
kQ — the radius of gyration of the cylinder about OO'.
Then ft = P -f ? by Art. 137 (4)
and from the symmetry of the solid, o? = y* •
R2
hence k<? or — •«= 2x2 = if
R2 -
and x2 = — = y
The cylinder being supposed divided into thin parallel rods
all parallel to the axis and / units long, the mean square of the
FIG. 170.
distance of the particles forming the rod from the plane VOX
of one end, is the same as the square of the radius of gyration
of a rod of length / about an axis perpendicular to its length
/2
and through one end, viz - (Art. 141). The axis OX is the
O
intersection of the planes XOO'X' and YOX, the end plane ;
hence, if £x is the radius of gyration about OX —
"D2 71
(Art. 137(2))
Moments of Inertia — Rotation 203
(3) Also, if kG is the radius of gyration about a parallel axis
through G, the e.g. of the cylinder —
The moments of inertia of the cylinder about these various
axes are to be found by multiplying the square of the radius of
gyration about that axis by the mass — , where w is the weight
O
of the cylinder, in accordance with the general relation I
= -J? (Art. 134).
c^
Example. — A solid disc flywheel of cast iron is 10 inches in
diameter and 2 inches thick. If the weight of cast iron is 0*26 Ib.
per cubic inch, find the moment of inertia of the wheel about its
axis in engineers' units.
The volume of the flywheel is TT x 52 x 2 = $OTT cubic inches
the weight is then 0*26 x 5071 — 40*9 Ibs.
and the mass is — — = 1*27 units
32-2
The square of the radius of gyration is i(T5o)2 (feet)2. Therefore
the moment of inertia is—
1*27 x Tffr5ft = o'iio4 unit
EXAMPLES XVII.
1. A girder of I-shaped cross-section has two horizontal flanges 5 inches
broad and I inch thick, connected by a vertical web 9 inches high and I
inch thick. Find the " moment of inertia of the area " of the section about
a horizontal axis through its e.g.
2. Fig. 171 represents the cross-section of a cast-iron girder. AB is 4
inches, BC I inch, EF I inch ; EH is 6 inches, KL is 8 inches, and KN is
i '5 inches. Find the moment of inertia and radius of gyration of the area
of the section about the line NM.
3. Find, from the results of Ex. 2, the moment of inertia and radius of
gyration of the area of section about an axis through the e.g. of the section
and parallel to NM,
204
Mechanics for Engineers
B
H
N
M
FIG. 171.
4. Find the moment of inertia of the area enclosed between two con-
centric circles of 10 inches and 8 inches diameter respectively, about a
diameter of the circles.
5. Find the radius of gyration of the area bounded on the outside by a
circle 12 inches diameter, and on the
inside by a concentric circle of 10
inches diameter, about an axis through
the centre of the figure and perpen-
dicular to its plane.
6. The pendulum of a clock con-
sists of a straight uniform rod, 3 feet
long and weighing 2 Ibs., attached
to which is a disc O'5 foot in diameter
and weighing 4 Ibs. , so that the centre
of the disc is at the end of the rod.
Find the moment of inertia of the
pendulum about an axis perpendicular
to the rod and to the central plane
of the disc, passing through the rod 2'5 feet from the centre of the disc.
7. Find the radius of gyration of a hollow cylinder of outer radius a and
inner radius b about the axis of the cylinder.
8. Find the radius of gyration of a flywheel rim 3 feet in external
diameter and 4 inches thick, about its axis. If the rim is 6 inches broad, and
of cast-iron, what is its moment of inertia about its axis? Cast iron
weighs 0*26 Ib. per cubic inch.
544. Kinetic Energy of Rotation. — If a particle of a
body weighs w-^ Ibs., and is rotating with angular velocity M
about a fixed axis i\ feet from it, its speed is va\ feet
per second (Art. 33), and its kinetic energy is therefore
--- -1 . (otfi)2 foot-lbs. (Art. 60). Similarly, another particle of
the same rigid body situated ra feet from the fixed axis of
rotation, and weighing w2 Ibs., will have kinetic energy equal to
I 1£J
--'-' (w2)2 ; and if the whole body is made up of particles
weighing wlt w2, wst «/4, etc., Ibs., situated at r1} ;-2, r3) r±, etc.,
feet respectively from the axis of rotation, the total kinetic
energy of the body will be—
Jwr( -Vy + ^—r} 4- ^-V32 +, etc. j foot-lbs.
Moments of Inertia — Rotation 205
The quantity (^V + -V + -V 4-, etc.) or s(!?,»)
^ <^T <^T <^T N^ /
has been defined (Art. 133) as the moment of inertia I, of the
body about the axis. Hence the kinetic energy of the body is
W W
JIoo2, or i.-KV, or J-V2 foot-lbs., where K = radius of
o <^
gyration of the body in feet about the axis of rotation, and
V = velocity of the body in feet per second at that radius of
gyration. This is the same as the kinetic energy JMV2 or
W W
-^V2 of a mass M or --, all moving with a linear velocity V.
2A §
The kinetic energy of a body moving at a given linear
velocity is proportional to its mass; that of a body moving
about a fixed axis with given angular velocity is proportional
to its moment of inertia. We look upon the moment of
inertia of a body as its rotational inertia, i.e. the measure of
its inertia with respect to angular motion (see Art. 36).
145. Changes in Energy and Speed. — If a body of
moment of inertia I, is rotating about its axis with an angular
velocity c^, and has a net amount of work E done upon it,
thereby raising its velocity to <o2; then, by the Principle of
Work (Art. 61)—
4I(to22 - oV2) = E
W
or -K22 - W2 = E
or J— (V22 - Va2) = E
&
where K = radius of gyration about the axis of rotation, and
V2 and Vj are the final and initial velocities respectively at
a radius K from the axis.
Hence the change of energy is equal to that of an equal
weight moving with the same final and initial velocities as a
point distant from the axis by the radius of gyration of the
body. If the body rotating with angular velocity o>2 about
the axis is opposed by a tangential force, and does work of
amount E in overcoming this force, its velocity will be reduced
206 Mechanics for Engineers
to wj, the loss of kinetic energy being equal to the amount of
work done (Art. 61).
146. Constant resisting Force. — Suppose a body, such
as a wheel, has a moment of inertia I, and is rotating at an
angular velocity o>2 about an axis, and this rotation is opposed
by a constant tangential force F at a radius r from the axis
of rotation, which passes through the centre of gravity of the
body. Then the resultant centripetal force on the body is
zero (Art. 130). The particles of the body situated at a
distance r from the centre are acted on 'by a resultant or
effective force always in the same straight line with, and in
opposite direction to, their own velocity, and therefore have
a constant retardation in their instantaneous directions of
motion (Art. 40). Hence the particles at a radius r have
their linear velocity, and therefore also their angular velocity,
decreased at a constant rate ; and since, in a rigid body, the
angular velocity of rotation about a fixed axis of every point
is the same, the whole body suffers uniform angular retardation.
Suppose the velocity changes from w.2 to o^ in / seconds,
during which the body turns about the axis through an angle
0 radians. The uniform angular retardation a is ~2 — l.
Also the work done on the wheel is Fr x 0 (Art. 57),
hence —
F . ;-. 8 = il(a>2~ - wj2) = loss of kinetic energy . (i)
The angle turned through during the retardation period is —
e = £1(013*- 0)^-4- F.r
Note that F . ;• is the moment of the resisting force or the
resisting torque.
Again, <o22 — Wl2 = (o>2 + w1)(o>2 — wj
and w.2 — Wj = at
and wj + w2 = twice the average angular velocity
during the retardation
Moments of Inertia — Rotation 207
Hence the relation —
may be written—
or F . r = I . a . . , , . . . . (2)
i.e. the moment of the resisting force about the axis of rotation
is equal to the moment of inertia of the body multiplied by its
angular retardation.
Similarly, if F is a driving instead of a resisting force, the
same relations would hold with regard to the rate of increase
of angular velocity, viz. the moment of the accelerating force
is equal to the moment of inertia of the body multiplied by the
angular acceleration produced. Compare these results with
those of Art. 40 for linear motion.
We next examine rather more generally the relation
between the angular velocity, acceleration, and inertia of a
rigid body.
147. Laws of Rotation of a Rigid Body about an
Axis through its Centre of Gravity. — Let w be the weight
of a constituent particle of the
body situated at P (Fig. 172),
distant r from the axis of rota-
tion O ; let o> be the angular
velocity of the body about O.
Then the velocity v of P is ^r.
Adding the vectors repre-
senting the momenta of all FlG- ^
such particles, we have the total momentum estimated in any
particular direction, such as OX (Fig. 172), viz. —
cos
or $(wr cos 0)
But 3 (wr cos 0) is zero when estimated in any direction if
r cos 0 is measured from a plane through the e.g. Hence the
total linear momentum resolved in any given direction is zero.
Moment of Momentum, or Angular Momentum of
208 Mechanics jFor Engineers
a Rigid Body rotating" about a Fixed Axis. — This is
defined as the sum of the products of the momenta of all the
particles multiplied by their respective distances from the axis,
or the angular momentum is equal to the moment of inertia
(or angular inertia) multiplied by the angular velocity.
Suppose the velocity of P increases from z\ to z'a, the
angular velocity increasing from o^ to w.2, the change of
angular momentum is —
If the change occupies a time / seconds, the mean rate of
change of angular momentum of the whole body is —
where/" is the average acceleration of P during the time /, and
-f or F is the average effective accelerating force on the
particle at P, acting always in its direction of motion, i.e.
acting always tangentially to the circular path of P (see Art. 40).
Also 2 (F . r) is the average total moment of the effective
or net forces acting on the various particles of the body or the
average effective torque on the body.
If these average accelerations and forces be estimated over
indefinitely small intervals of time, the same relations are true,
and ultimately the rate of change of angular momentum is
equal to the moment of the forces producing the change, so
that—
rate of change of Iw = S(Fr) = M
= total algebraic moment of effective
forces, or effective torque
Moments of Inertia — Rotation 209
Also—
rate of change of Iw = I x rate of change of o>
or I . a, where a is the angular acceleration or rate of change of
angular velocity. Hence —
= M = la
a result otherwise obtained for the special case of uniform
acceleration in (2), Art. 146.
Problems can often be solved alternately from equation
(i) or equation (2) (Art. 146), just as in the case of linear
motion the equation of energy (Art. 60) or that of force (Art.
51) can be used (Art. 60).
Example i. — A flywheel weighing 200 Ibs. is carried on a
spindle 2*5 inches diameter. A string is wrapped round the spindle,
to which one end is loosely attached. The other end of the string
carries a weight of 40 Ibs., 4 Ibs. of which is necessary to overcome
the friction (assumed constant) between the spindle and its
bearings. Starting from rest, the weight, pulling the flywheel
round, falls vertically through 3 feet in 7 seconds. Find the
moment of inertia and radius of gyration of the flywheel.
The average velocity of the falling weight is f foot per second,
and since under a uniform force the acceleration is uniform, the
maximum velocity is 2 x 2 or & foot per second.
The net work done by the falling weight, i.e. the whole work
done minus that spent in overcoming friction, is —
(40 - 4)3 foot-lbs. - 1 08 foot-lbs.
The kinetic energy of the falling weight is —
i- 32°2 . (92 = 0-456 fbot-ib.
If I = moment of inertia of the flywheel, and « = its angular
velocity in radians per second. By the principle of work (Art. 61) —
£I«2 4- 0-456 = 108 foot-lbs.
\ rlo.2 = 108 - 0-456 = 107-544 foot-lbs.
The maximum angular velocity o> is equal to the maximum
linear velocity of the string in feet per second divided by the radius
of the spindle in feet, or —
p
2io Mechanics for Engineers
?<v,f x-?- =/2-
12 / 1-25 875
= 8*22 radians per second
therefore \\ x (8'22)2 = 107*544
I07X44 X 2 2K'I
I= "(8 ™y =6^6 =3--
And if k = radius of gyration in feet, since the wheel weighs
200 Ibs. —
20°.^ = 3'i8
32-2
& = 0-518 (foot)2
k — 0716 foot or 8*6 inches
Example 2. — An engine in starting exerts on the crank-shaft
for one minute a constant turning moment of 1000 Ib.-feet, and
there is a uniform moment resisting motion, of 800 Ib.-feet. The
flywheel has a radius of gyration of 5 feet and weighs 2000 Ibs.
Neglecting the inertia of all parts except the flywheel, what speed
will the engine attain during one minute ?
(1) Considering the rate of change of angular momentum —
The effective turning moment is 1000 — 800 = 200 Ib.-feet
The moment of inertia of the flywheel is — - x 52 = 1552 units
Hence if a = angular acceleration in radians per second per second
200 = 15520 (Art. 146 (2))
o = - = 0*1287 radian per second per second
And the angular velocity attained in one minute is —
60 x 0*1287 = 774 radians per second
or *-^ — - = 74 revolutions per minute
27T
(2) Alternatively from considerations of energy.
If u> = angular velocity acquired
- = mean angular velocity
Total angle turned through I . fc x- ^ radians
in one minute
Net work done in one minute = 200 x 3000 foot-lbs.
200 x 30® = il«2
6oooo> — \. 1552. »2
„ — 1^929 = 774 radians per second
J552 as before
Moments of Inertia — Rotation
211
Example 3. — A thin straight rod of uniform material, 4*5 feet
long, is hinged at one end so that it can turn in a vertical plane.
It is placed in a horizontal position, and
then released. Find the velocity of the
free end (i) when it has described an
angle of 30°, (2) when it is vertical.
(i) After describing 30° the centre Q
of gravity G (Fig. 173), which is then 2
at GI, has fallen a vertical distance
ON.
ON = OGX cos 60° =
= 1*125 feet
= 1 X 2 '25
FIG. 173.
If W is the weight of the rod in pounds, the work done by
gravitation is —
W x ri25 foot-lbs.
The moment of inertia of the rod
W (4*5)2 = 27 W
g ' 3 4 ' g
If a>! is the angular velocity of the rod, since the kinetic energy
of the rod must be ri25\V foot-lbs. —
c^2 = I X 7>8f X 32*2 = 1073
o> = 3*28 radians per second
the velocity of A0 in position At is then —
3*28 x 4*5 ~ 1474 feet per second
(2) In describing 90° G falls 2-25 feet, and the kinetic energy is
then 2*25\V foot-lbs.
And if o>.2 is the angular velocity of the rod —
i-^. — ,V= 2'25W
a>.2z = $ X ./f X 32'2 = 21*47
a?., — 4*63 radians per second
and the velocity of A0 in the position A2 is —
4*63 x 4*5 = 20*82 feet per second
212 Mechanics for Engineers
148. Compound Pendulum. — In Art. 71 the motion of
a " simple pendulum " was investigated, and it was stated that
such a pendulum was only approximated to by any actual
pendulum. We now proceed to find the
simple pendulum equivalent (in period)
to an actual pendulum.
Let a body be suspended by means
of a horizontal axis O (Fig. 174) perpen-
dicular to the figure and passing through
the body. Let G be the e.g. of the body
in any position, and let OG make any
angle 6 with the vertical plane (OA)
through O.
FlG- *74- Suppose that the body has been raised
to such a position that G was at B, and then released. Let
the angle AOB be <£, and OG = OB = OA = //.
The body oscillating about the horizontal axis O constitutes
a pendulum.
Let / = length of the simple equivalent pendulum (Art. 71);
I = the moment of inertia of the pendulum about the
axis O ;
kQ = radius of gyration about O ;
kG = radius of gyration about a parallel axis through G.
Let W be the weight of the pendulum, and let M and N
be the points in which horizontal lines through B and G
respectively cut OA.
When G has fallen from B to G, the work done is —
W X MN = W(ON - OM) = W(// cos 0 - h cos <£)
= W/*(cos 0 - cos <t>)
Let the angular velocity of the pendulum in this position
be w, then its kinetic energy is ^-Iw2 (Art. 144), and by the
principle of work (Art. 61), if there are no resistances to
motion the kinetic energy is equal to the work done, or —
-J-Iw2 = W/$(cos 6 - cos 0)
and therefore —
o/2 = — j— (cos B — cos <£) . . . . (i)
Moments of Inertia — Rotation 2 1 3
Similarly, if a particle (Fig. 175) be attached to a point
O' by a flexible thread of length /, and be released from a
position B' such that B'6'A = <£, O'A
being vertical, its velocity v when passing
G' such that G'O'A = 0 is given by—
v"- = 2g. M'N' = 2g/(cos 6 — cos </>)
and its angular velocity w about Of
being ^r—
to2 = ^f(cos 0 - cos <£) ..... (2)
The angular velocity of a particle (or of a simple pendulum)
given by equation (2) is the same as that of G (Fig. 174)
given by equation (i), provided —
g_Vfh^Vfh.g
7 I ~~ W
i.e. provided —
k 2
This length -j- is then the length of the simple pendulum
equivalent to that in Fig. 174, for since the velocity is the same
at any angular position for the simple pendulum of length /
and the actual pendulum, their times of oscillation must be the
same. Also, since —
V = £G2 + & (Art. 137 (2))
kr?
The point C (Fig. 174), distant -j- + h from O, and in the
line OG is called the "centre of oscillation? The expression
k 2 &?
(': + // shows that it is at a distance -j- beyond G from O.
A particle placed at C would oscillate in the same period
about O as does the compound pendulum of Fig. 174.
214 Mechanics for Engineers
Example. — A flywheel having a radius of gyration of 3*25 feet
is balanced upon a knife-edge parallel to the axis of the wheel and
inside the rim at a distance of 3 feet from the axis of the wheel.
If the wheel is slightly displaced in its own plane, find its period of
oscillation about the knife-edge.
The length of the simple equivalent pendulum is —
3 + —3— = 3 + 3'52o8 = 6-5208 feet
Hence the period is 27r\/ - * = 276 seconds
149. The laws of rotation of a body about an axis may be
stated in the same way as Newton's laws of motion as follows :—
Law i. A rigid body constrained to rotate about an axis
continues to rotate about that axis with constant angular
velocity except in so far as it may be compelled to change
that motion by forces having a moment about that axis.
Law 2. The rate of change of angular momentum is pro-
portional to the moment of the applied forces, or torque about
the axis. With a suitable choice of units, the rate of change
of angular momentum is equal to the moment of the Applied
forces, or torque about the axis.
LaW 3. If a body A exerts a twisting moment or torque
about a given axis on a body B, then B exerts an equal and
opposite moment or torque about that axis on the body A.
150. Torsional Simple Harmonic Motion. — If a rigid
•body receives an angular displacement about an axis, and the
moment of the forces acting on it tending to restore equilibrium
is proportional to the angular displacement, then the body
executes a rotary vibration of a simple harmonic kind. Such
a restoring moment is exerted when a body which is suspended
by an elastic wire or rod receives an angular displacement
about the axis of suspension not exceeding a certain limit.
Let M = restoring moment or torque in Ib.-feet per radian
of twist ;
I = moment of inertia of the body about the axis of
suspension in engineer's units ;
//, = angular acceleration of the body in radians per
second per second per radian of twist.
Moments of Inertia — Rotation
21$
Then M = I . /x (Art. 147)
M
Then, following exactly the same method as in Art. 68, if
Q (Fig. 176) rotates uniformly with angular velocity vV in a
circle centred at O and of radius OA, which represents to scale
the greatest angular displacement of
the body, and P is the projection
of Q on OA, then P moves in the
same way as a point distant from O
by a length representing the angular
displacement 0, at any instant to the
same scale that OA represents the
extreme displacement. The whole
argument of Art 68 need not be
repeated here, but the results are — FIG 176.
Angular velocity for an angular displacement 6, represented
by OM, is
Angular acceleration for an angular displacement 0, repre-
sented by PO, is vV'- 0.
27T
T = time of complete vibration = - = seconds
or, sine*
M
T =
Example. — A metal disc is 10 inches diameter and weighs
6 Ibs. It is suspended from its centre by a vertical wire so that
its plane is horizontal, and then twisted. When released, how many
oscillations will it make per minute if the rigidity of the suspension
wire is such that a twisting moment of i Ib.-foot causes an angular
deflection of 10° ?
The twisting moment per radian twist is j
/ TT \ > = 573 Ib.-feet
'-(ito*10) i
The square of radius of gyration is 1(A)2 = °'°862 (foot)2
2i6 Mechanics for Engineers
The moment of inertia is — - x 0*0862 = 0*01615 umt
Hence the time of vibration is 271- . /JL — /,-.. / 0*01615
V M Z*V 573
= 0-337 second
The number of vibrations per minute is \
6o_ = 178
0-337
151. It is evident, from Articles 144 to 150, that the rotation
of a rigid body about an axis bears a close analogy to the
linear motion of a body considered in Chapters I. to IV.
Some comparisons are tabulated below.
Linear. Angular or Rotational.
W
Mass or inertia, — or ;;/. Moment of inertia, I.
0>
Length, /. Angular displacement, 6.
Velocity, v. Angular velocity, w.
Acceleration,/. Angular acceleration, a.
Force, F. Moment of force, or torque, M.
Momentum, — . v or mv. Angular momentum, I . w.
<b
/ fi
Average velocity, - . Average angular velocity, -.
Average acceleration, Vl ^ Average angular acceleration,
Average force, — .— —f — ? or Average moment or torque,
- I(tt" ~ "^
Work of constant force, F . /. Work of constant torque, M . 0.
IJD
Kinetic energy, \ — v* or | mtf. Kinetic energy, Jlw2.
<?>
Period of simple vibration, Period of simple vibration,
2-n-A/ — or 27TA/ --, where 2 +^/ -- where M = torque
e = force per unit displace- per radian displacement.
ment.
Moments of Inertia — Rotation
217
FIG. 177.
The quantities stated as average values have similar mean-
ings when the averages are reckoned over indefinitely small
intervals of time, or, in other words, they have corresponding
limiting values.
152. Kinetic Energy of a Rolling Body.— We shall
limit ourselves to the case of a solid of revolution rolling along
a plane. The e.g. of the
solid will then be in the
axis of revolution about
which the solid will rotate
as it rolls. Let R be
the extreme radius of the V'
body at which rolling
contact with the plane
takes place (Fig. 177);
let the centre O be moving
parallel to the plane with
a velocity V. Then any point P on the outside circumference
of the body is moving with a velocity V relative to O, the
angular velocity of P and of the whole body about O being
^7, or say w radians per second.
Consider the kinetic energy of a particle weighing w Ibs.
at Q, distant OQ or r from the axis of the body. Let OQ
make an angle QOA = 0 with OA, the direction of motion of
O. Then the velocity v of Q is the resultant of a velocity V
parallel to OA, and a velocity <or perpendicular to OQ, and
is such that —
z/2 = (<or)2 + V2 4 2o>;- . V . cos (90 + 0)
Hence the kinetic energy of the particle is —
J^(wV2 + V2 - 2o>rV sin 0)
&
The total kinetic energy of the body is then —
S(— tfa) = 5J— (wV 4- V2 - 2W/-V sin 0)j
21 8 Mechanics for Engineers
Now,S(w. sin 0) = o (Art. 113 (3))
<"W \
~ • r-J - I, the moment of inertia of the solid
'* about the axis O
hence S('— • ^J = |Iw2 + |^
\ 2P" / J ^ <r
= kinetic energy of rotation about O +
kinetic energy of an equal weight
moving with the linear velocity of
the axis.
This may also be written —
where k is the radius of gyration about the axis O. The
W / k^ \
kinetic energy — V2( i ^fTs) ig tnen tne same as tnat of a
*Z
weight W( i + j^2J moving with a velocity V of pure trans-
lation, i.e. without rotation.
In the case of a body rolling down a plane inclined 6 to
the horizontal (Fig. 178),
using the same notation as
in the previous case, the
component force of gravity
through O and parallel to
the direction of motion down
W
the plane is — . sin 6. In
rolling a distance s down
the plane, the work done
FIG. 178. is W sin 0 . s. Hence the
kinetic energy stored after the distance s is —
W / K1 \
i_V2/ i + — ^ ) = W sin 0 . s (Art. 61)
R2
or V2 = 2sg sin ^5T^M
This is the velocity which a body would attain in moving
Moments of Inertia — Rotation 2i<b'"THe ^
&^NIVERS1T\
without rotation a distance s from rest under an accefersttion OF
R2
g sin tf-jST^iT^r Hence the effect of rolling instead of sliding
down the plane is to decrease the linear acceleration and linear
velocity attained by the axis in a given time in the ratio
R2
, kl (see Art. 28).
We may alternatively obtain this result as follows :
Resolving the reaction of the (rough) plane on the body at T
into components N and F, normal to the plane and along
it respectively, the net force acting down the plane on the
body is W sin 6 — F ; and if a = angular acceleration of the
body about O, and/ = linear acceleration down the plane —
But la = FR (Art. 146 (2))
F being the only force which has any moment about O ;
I* I/
hence F = R — -Rj
and the force acting down the plane is W sin 6 — =3-
force acting down the plane /___ . „ I/\ . W
Hence / = — ru , — - = I W sin 9 — ^ » ) T- —
mass of body V R2/ ^
or /as ^ sin 6 x
6 -fgz
R2
Example. — A solid disc rolls down a plane inclined 30° to the
horizontal. How far will it move down the plane in 20 seconds
from rest? What is then the velocity of its centre, and if it weighs
10 Ibs., how much kinetic energy has it ?
The acceleration of the disc will be —
Ti2
32*2 x sin 30° x 2 = 32-2 x i x §
= 1073 feet per second per second
22O Mechanics for Engineers
In 20 seconds it will acquire a velocity of —
20 x 1073 = 214-6 feet per second
Its average velocity throughout this time will be —
214*6
— — = 107-3 feet per second
It will then move —
107-3 x 20 = 2146 feet
corresponding to a vertical fall of 2146 sin 30° or 1073 feet.
The kinetic energy will be equal to the 'work done on it in
falling 1073 feet, />. 1073 x 10 = 10,730 foot-lbs.
EXAMPLES XVIII.
1. What is the moment of inertia in engineer's units of a flywheel
which stores 200,000 foot-lbs. of kinetic energy when rotating 100 times
per minute?
2. A flywheel requires 20,000 foot-lbs. of work to be done upon it
to increase its velocity from 68 to 70 rotations per minute. What is its
moment of inertia in engineer's units ?
3. A flywheel, the weight of which is 2000 Ibs., has a radius of
gyration of 3*22 feet. It is carried on a shaft 3 inches diameter, at the
circumference of which a constant tangential force of 50 Ibs. opposes the
rotation of the wheel. If the wheel is rotating 60 times per minute, how
long will it take to come to rest, and how many rotations will it make in
doing so ?
4. A wheel 6 feet diameter has a moment of inertia of 600 units, and
is turning at a rate of 50 rotations per minute. What opposing force
applied tangentially at the rim of the wheel will bring it to rest in one
minute ?
5. A flywheel weighing 1*5 tons has a radius of gyration of 4 feet.
If it attains a speed of 80 rotations per minute in 40 seconds, find the mean
effective torque exerted upon it in pound-feet ?
6. A weight of 40 Ibs. attached to a cord which is wrapped round the
2-inch spindle of a flywheel descends, and thereby causes the wheel to
rotate. If the weight descends 6 feet in 10 seconds, and the friction of the
bearing is equivalent to a force of 3 Ibs. at the circumference of the spindle,
find the moment of inertia of the flywheel. If it weighs 212 Ibs., what is
its radius of gyration ?
7. If the weight in Question 6, after descending 6 feet, is suddenly
released, how many rotations will the wheel make before coming to rest ?
8. A flywheel weighing 250 Ibs. is mounted on a spindle 2'5 inches
Moments of Inertia — Rotation 221
diameter, and is caused to rotate by a falling weight of 50 Ibs. attached to
a string wrapped round the spindle. After falling 5 feet in 8 seconds, the
weight is detached, and the wheel subsequently makes 100 rotations before
coming to rest. Assuming the tangential frictional resisting force at the
circumference of the axle to be constant throughout the accelerating and
stopping periods, find the radius of gyration of the wheel.
9. A rod is hinged at one end so that it can turn in a vertical plane
about the hinge. The rod is turned into a position of unstable equilibrium
vertically above the hinge and then released. Find the velocity of the
end of the rod (i) when it is horizontal; (2) when passing through its
lowest position, if the rod is 5 feet long and of uniform small section
throughout.
10. A circular cylinder, 3 feet long and 9 inches diameter, is hinged
about an axis which coincides with the diameter of one of the circular ends.
The axis of the cylinder is turned into a horizontal position, and then the
cylinder is released. Find the velocity of the free end of the axis (i) after
it has described an angle of 50°, (2) when the axis is passing through its
vertical position.
11. A flywheel weighs 5 tons, and the internal diameter of its rim is
6 feet. When the inside of the rim is supported upon a knife-edge passing
through the spokes and parallel to its axis, the whole makes, if disturbed,
21 complete oscillations per minute. Find the radius of gyration of the
wheel about its axis, and the moment of inertia about that axis.
12. A cylindrical bar, 18 inches long and 3 inches diameter, is suspended
from an axis through a diameter of one end. If slightly disturbed from
its position of stable equilibrium, how many oscillations per minute will it
make ?
13. A piece of metal is suspended by a vertical wire which passes
through the centre of gravity of the metal. A twist of 8*5° is produced
per pound-foot of twisting moment applied to the wire, and when the
metal is released after giving it a small twist, it makes 150 complete
oscillations a minute. Find the moment of inertia of the piece of metal
in engineer's or gravitational units.
14. A flywheel weighing 3 tons is fastened to one end of a shaft, the
other end of which is fixed, and the torsional rigidity of which is such that
it twists o'4° per ton-foot of twisting moment applied to the flywheel. If
the radius of gyration of the flywheel and shaft combined is 3 feet, find the
number of torsional vibrations per minute which the wheel would make if
slightly twisted and then released.
15. The weight of a waggon is 2 tons, of which the wheels weigh \ ton.
The diameter of the wheels is 2 feet, and the radius of gyration o'g foot.
Find the total kinetic energy of the waggon when travelling at 40 miles
per hour, in foot-tons.
16. A cylinder is placed on a plane inclined 15° to the horizontal, and
is allowed to roll down with its axis horizontal. Find its velocity after
it has traversed 25 feet.
222 Mechanics for Engineers
17. A solid sphere rolls down a plane inclined o to the horizontal.
Find its acceleration. (NOTE.— The square of the radius of gyration of a
sphere of radius R is §R2.)
1 8. A motor car weighs W Ibs., including four wheels, each of which
weigh w Ibs. The radius of each wheel is a feet, and the radius of
gyration about the axis is k feet. Find the total kinetic energy of the
car when moving at v feet per second.
CHAPTER X
ELEMENTS OF GRAPHICAL STATICS
153. IN Chapter VI. we considered and stated the condi-
tions of equilibrium of rigid bodies, limiting ourselves to
those subject to forces in one plane only. In the case of
systems of concurrent forces in equilibrium (Chapter V.), we
solved problems alternatively by analytical methods of resolu-
tion along two rectangular axes, or by means of drawing vector
polygons of forces to scale. We now proceed to apply the
vector methods to a few simple systems of non-concurrent
forces, such as were considered from the analytical point of
view in Chapter VI., and to deduce the vector conditions of
equilibrium.
When statical problems are solved by graphical methods, it is
usually necessary to first draw out a diagram showing correctly
the inclinations of the lines of action of the various known
forces to one another, and, to some scale, their relative posi-
tions. Such a diagram is called a diagram of positions, or
space diagram ; this is not to be confused with the vector
diagram of forces, which gives magnitudes and directions, but
not positions of forces.
154. Bows' Notation. — In this notation the lines of
action of each force in the space diagram are denoted by
two letters placed one on each side of its line of action. Thus
the spaces rather than the lines or intersections have letters
assigned to them, but the limits of a space having a particular
letter to denote it may be different for different forces.
The corresponding force in the vector diagram has the same
two letters at its ends as are given to the spaces separated by
224
Mechanics for Engineers
its line of action in the space diagram. We shall use capital
letters in the space diagram, and the corresponding small letters
to indicate a force in the vector diagram. The notation will
be best understood by reference to an example. It is shown
in Fig. 179, applied to a space diagram and vector polygon for
Space Diagram
5lbs
-.*•* '
6ilbs
FIG. 179.
five concurrent forces in equilibrium (see Chapter V.). The
four forces, AB, BC, CD, DE, of 5 Ibs., 6 Ibs., 5^ Ibs., and
6£ Ibs. respectively, being given, the vectors ab, be, cd, de are
drawn in succession, of lengths representing to scale these
magnitudes and parallel to the lines AB, BC, CD, and DE
respectively, the vector ea, which scales 57 Ibs., represents the
equilibrant of the four forces, and its position in the space
diagram is shown by drawing a line EA parallel to ea from the
common intersection of AB, BC, CD, and DE. (This is ex-
plained in Chapter V., and is given here as an example of the
system of lettering only.) .
155. The Funicular or Link Polygon. — To find
graphically the single resultant or equilibrant of any system of
non-concurrent coplanar forces. Let the four forces AB, BC,
CD, and DE (Fig. 180) be given completely, i.e. their lines of
action (directions and positions) and also their magnitudes.
First draw a vector ab parallel to AB, and representing by its
length the given magnitude of the force AB ; from b draw be
parallel to the line BC, and representing the force BC com-
pletely. Continuing in this way, as in Art. 73, draw the open
Elements of Graphical Statics
225
vector or force polygon abcde; then, as in the case of con-
current forces, Art. 73, the vector ae represents the resultant
(or ea, the equilibrant) in magnitude and direction. The
problem is not yet complete, for the position of the resultant
is unknown. In Chapter VI. its position was determined by
rinding what moment it must have about some fixed point.
The graphical method is as follows (the reader is advised to
FIG. 180.
draw the figure on a sheet of paper as he reads) : Choose any
convenient point o (called a pole) in or about the vector
polygon, and join each vertex a, b, c, d, and e of the polygon
to o ', then in the space diagram, selecting a point P on the
line AB, draw a line PT (which may be called AO) parallel to
ao across the space A. From P across the space B draw a
line BO parallel to bo to meet the line BC in Q. From Q
draw a line CO parallel to co to meet the line CD in R. From
R draw a line DO parallel to do to meet the line DE in S, and,
finally, from S draw a line EO parallel to eo to meet the line
AO (or PT) in T. Then T, the intersection of AO and EO,
is a point in the line of action of EA, the equilibrant, the magni-
tude and inclination of which were found from the vector ca.
Q
226 Mechanics for Engineers
Hence the equilibrant EA or the resultant AE is completely
determined. The closed polygon PQRST, having its vertices
on the lines of action of the forces, is called a funicular or link
polygon. That T must be a point on the line of action of the
resultant is evident from the following considerations. Any
force may be resolved into two components along any two
lines which intersect on its line of action, for it is only neces-
sary for the force to be the geometric sum of the components.
(Art. 75). Let each force, AB, BC, CD, and DE, be resolved
along the two sides of the funicular polygon which meet on
its line of action, viz. AB along TP and QP, BC along PQ
and RQ, and so on. The magnitude of the two components
is given by the corresponding sides of the triangle of forces
in the vector diagram, e.g. AB may be replaced by components
in the lines AO and BO (or TP and QP), represented in magni-
tude by the lengths of the vectors ao and ob respectively, for
in vector addition —
ao + ob = ab (Art. 19)
Similarly, CD is replaced by components in the lines CO and
OD represented by co and od respectively. When this process
is complete, all the forces AB, BC, CD, and DE are replaced
by components, the lines of action of which are the sides TP,
PQ, QR, etc., of the funicular polygon. Of these component
forces, those in the line PQ or BO are represented by the
vectors ob and bo, and therefore have a resultant nil. Similarly,
all the other components balance in pairs, being equal and
opposite in the same straight line, except those in the lines TP
and TS, represented by ao and oe respectively. These two
have a resultant represented by ae (since in vector addition
ao + oe = ae), which acts through the point of intersection T
of their lines of action. Hence finally the resultant of the
whole system acts through T, and is represented in magnitude
and direction by the line ae; the equilibrant is equal and
opposite in the same straight line.
156. Conditions of Equilibrium. — If we include the
equilibrant EA (Fig. 180, Art. 155) with the other four forces,
we have five forces in equilibrium, and (i) the force or vector
Elements of Graphical Statics
227
polygon abcde is closed ; and (2) the funicular polygon
PQRST is a closed figure. Further, if the force polygon is
not closed, the system reduces to a single resultant, which may
be found by the method just described (Art. 155).
It may happen that the force polygon is a closed figure,
and that the funicular polygon is not. Take, for example, a
diagram (Fig. 181) similar to the previous one, and let the
FIG.
forces of the system be AB, BC, CD, DE, and EA, the force
E A not passing through the point T found in Fig. 180, but
through a point V (Fig. 181), in the line TS. If we draw a
line, VW, parallel to oa through V, it will not intersect the line
TP parallel to ao, for TP and VW are then parallel. Re-
placing the original forces by components, the lines of action
of which are in the sides of the funicular polygon, we are left
with two parallel unbalanced components represented by ao
and oa in the lines TP and VW respectively. These form a
couple (Art. 91), and such a system is not in equilibrium nor
reducible to a single resultant. The magnitude of the couple
is equal to the component represented by oa multiplied by the
length represented by the perpendicular distance between the
lines TP and VW. A little consideration will show that it is
also equal to the force EA represented by ea, multiplied by
the distance represented by the perpendicular from T on the
228 Mechanics for Engineers
line VX. Or the resultant of the forces in the lines AB, EC,
CD, and DE is a force represented by ae acting through the
point T; this with the force through V, and represented by
ea, forms a couple.
Hence, for equilibrium it is essential that (i) the polygon
of forces is a closed figure ; (2) that the funicular polygon is a
closed figure.
Compare these with the equivalent statements of the
analytical conditions in Art. 96.
Choice of Pole. — In drawing the funicular polygon, the
pole o (Figs. 1 80 and 181) was chosen in any arbitrary posi-
tion, and the first side of the funicular polygon was drawn
from any point P in the line AB. If the side AO bad been
drawn from any point in AB other than P, the funicular
polygon would have been a similar and similarly situated figure
to PQRST.
The choice of a different pole would give a different
shaped funicular polygon, but the points in the line of action
of the unknown equilibrant obtained from the use of different
poles would all lie in a straight line. This may be best appre-
ciated by trial.
Note that in any polygon the sides are each parallel to a
line radiating from the corresponding pole.
157. Funicular Polygon for Parallel Forces. — To
find the resultant of several parallel forces, we proceed exactly
as in the previous case, but the force polygon has its sides all
in the same straight line ; it is " closed " if, after drawing the
various vectors, the last terminates at the starting-point of the
first. The vector polygon does not enclose a space, but may
be looked upon as a polygon with overlapping sides.
Let the parallel forces (Fig. 182) be AB, BC, CD, and DE
of given magnitudes. Set off the vector ab in the vector
polygon parallel to the line AB, and representing by its length
the magnitude of the force in the line AB. And from b set
off be parallel to the line BC, and representing by its length
the magnitude of the force in the line BC. Then be is evi-
dently in the same straight line as ab, since AB and BC are
parallel. Similarly the vectors cd, de, and the resultant ae of
Elements of Graphical Statics
229
the polygon are all in the same straight line. Choose any
pole o, and join a, b, c, d, and e to o. Then proceed to put
in the funicular polygon in the space diagram as explained in
FIG. 182.
Art. 155. The two extreme sides AO and EO intersect in
T, and the resultant AE, given in magnitude by the vector ae,
acts through this point, and is therefore completely deter-
mined.
158. To find Two Equilibrants in Assigned Lines
of Action to a System of Parallel Forces.
As a simple example, we may take the vertical reactions
a
p
•y
V
--_
b
*X,
F
i A
Sx.
0
~ ~ -
^
f
1 c
r B
—
C ,
. — — -
, D >
E
FCL
,
/>
FIG. 183.
at the ends of a horizontal beam carrying a number of vertical
loads.
Let AB, BC, CD, and DE (Fig. 183) be the lines of action
230
Mechanics for Engineers
of the forces of given magnitudes, being concentrated loads on
a beam, xy, supported by vertical forces, EF and FA, atjy and
x respectively. Choose a pole, 0, as before (Arts. 156 and
157), and draw in the funicular polygon with sides AO, BO,
CO, DO, and EO respectively parallel to ao, bo, eo, do, and eo
in the vector diagram. Let AO meet the line FA (i.e. the
vertical through x) in/, and let q be the point in which EO
meets the line EF (i.e. the vertical through y). Join pq, and
from o draw a parallel line of to meet the line abcde in/. The
magnitude of the upward reaction or supporting force in the
line EF is represented by eft and the other reaction in the line
FA is represented by the vector fa. This may be proved in
the same way as the proposition in Art. 155.
of and fe represent the downward pressure of the beam at
x and y respectively, while fa and ef represent the upward
forces exerted by the supports at these points.
159. In the case of non-parallel forces two equilibrants
can be found — one to have a given line of action, and the
other to pass through a given point, i.e. to fulfil altogether three
conditions (Art. 96).
d
FIG. 184.
Let AB, BC, and CD (Fig. 184) be the lines of action of
given forces represented in magnitude by ab, be, and cd respec-
tively in the vector polygon. Let ED be the line of action of
Elements of Graphical Statics 231
one equilibrant, and p a point in the line of action of the
second. Draw a line, dx> of indefinite length parallel to DE.
Choose a pole, o, and draw in the funicular polygon corre-
sponding to it, but drawing the side AO through the given
point p. Let the last side DO cut ED in q. Then, since the
complete funicular polygon is to be a closed figure, joinjty.
Then the vector oe is found by drawing a line, oe, through o
parallel to pq to meet dx in e. The magnitude of the equili-
brating force in the line DE is represented by the length de,
and the magnitude and direction of the equilibrant EA through
p is given by the length and direction of ea.
1 60. Bending Moment and Shearing Force. — In con-
sidering the equilibrium of a rigid body (Chapter VI.), we have
hitherto generally only considered the body as a whole. The
same conditions of equilibrium must evidently apply to any
part of the body we may consider (see Method of Sections,
Art. 98). For example, if a beam (Fig. 185) carrying loads
Wi, W2, W3, W4, and W5, as shown, be ideally divided into two
! 1
1 1 !
1 :A ;
'////W////////A
W3?
FIG. 185.
k
parts, A and B, by a plane of section at X, perpendicular to
the length of the beam, each part, A and B, may be looked
upon as a rigid body in equilibrium under the action of forces.
The forces acting on the portion A, say, fulfil the conditions of
equilibrium (Art. 96), provided we include in them the forces
which the portion B exerts on the portion A.
Note that the reaction of A on B is equal and opposite to
the action of B on A, so that these internal forces in the beam
make no contribution to the net forces or moment acting on
the beam as a whole.
For convenience of expression, we shall speak of the beam
232 Mechanics for Engineers
as horizontal and the loads and reactions as vertical forces.
Let Rv and RB be the reactions of the supports on the por-
tions A and B respectively.
Considering the equilibrium of the portion A, since the
algebraic sum of the vertical forces on A is zero, B must exert
on A an upward vertical force Wx -f W2 — RA. This force is
called the shearing force at the section X, and may be denoted
by Fx. Then
Fx = W: + W2 - RA, or W, + W2 - RA - Fx = o
If the sum Wx -f \V2 is numerically less than RA, Fx is
negative, i.e. acts downwards on A.
The shearing force at any section of this horizontal beam is
then numerically equal to the algebraic sum of all the vertical
forces acting on either side of the section.
Secondly, since the algebraic sum of all the horizontal forces
on A is zero, the resultant horizontal force exerted by B on
A must be zero, there being no other horizontal force on A.
Again, if xlt d^ and 4 are the horizontal distances of RA, W1}
and W2 respectively from the section X, since W1} W2, and RA
exert on A a clockwise moment in the plane of the figure
about any point in the section X, of magnitude —
RA.^1-W1.^1-W2.4
B must exert on A forces which have a contra- clockwise
moment Mx, say, numerically equal to RA . xl — W^ — W^2,
for the algebraic sum of the moments of all the forces on A is
zero, i.e. —
X= o
or Mx = RA . x, - W,4
This moment cannot be exerted by the force Fx, which has
zero moment in the plane of the figure about any point in the
plane X. Hence, since the horizontal forces exerted by B
on A have a resultant zero, they must form a couple of
contra-clockwise moment, Mx, i.e. any pull exerted by B must
be accompanied by a push of equal magnitude. This couple
MX is called the moment of resistance of the beam at the
section X, and it is numerically equal to the algebraic sum of
Elements of Graphical Statics
233
moments about that section, of all the forces acting to either
side of the section. This algebraic sum of the moments about
the section, of all the forces acting to either side of the section
X, is called the bending moment at the section X.
1 6 1 . Determination of Bending Moments and Shear-
ing Forces from a Funicular Polygon. — Confining our-
selves again to the horizontal beam supported by vertical
forces at each end and carrying vertical loads, it is easy to
show that the vertical height of the funicular polygon at any
distance along the beam is proportional to the bending moment
w,
A '
B
FIG. 186.
at the corresponding section of the beam, and therefore repre-
sents it to scale, e.g. that xl (Pig. 186) represents the bending
moment at the section X.
Let the funicular polygon for any pole o, starting say from
z, be drawn as directed in Arts. 155 and 157, og being drawn
parallel to zp or GO, the closing line of the funicular, so that
R1} the left-hand reaction, is represented by the vector ga and
R2 by fg, while the loads W15 W2, W3, W4, and W5 are repre-
sented by the vectors ab, be, cd, de, and cf respectively. Con-
sider any vertical section, X, of the beam at which the height of
bending-moment diagram is xL Produce xl and the side zw
to meet in y. Also produce the side win of the funicular
234 Mechanics for Engineers
polygon to meet xy in n, and let the next side mq of the
funicular meet xy in /. The sides zwt wm, and mq (or AO, BO,
and CO) are parallel to ao, bo, and co respectively. Draw a
horizontal line, zk, through z to meet xy in k, a horizontal line
through w to meet xy in r, and a horizontal <?H through 0 in
the vector polygon to meet the line abcdef in H. Then in the
two triangles xyz and goo there are three sides in either parallel
respectively to three sides in the other, hence the triangles are
similar, and —
xy __ zy
ag~ ao ^'
Also the triangles zky and oHa are similar, and therefore—
zy zk
ao 0H ^ '
Hence from (i) and (2)—
xy zk a? . zk
- = -£, or xy.M = ag x „*, or Xy = ~-
Therefore, since ag is proportional to Rx, and zk is equal or
proportional to the distance of the line of action of Rj from X,
ag . zk is proportional to the moment of Rj about X, and oH
being an arbitrarily fixed constant, xy is proportional to the
moment of Rj about X.
Similarly —
and therefore yn represents the moment of Wj about X to the
same scale that xy represents the moment of Rj about X.
Similarly, again, nl represents the moment of W2 about X to
the same scale.
Finally, the length xl or (xy — ny — hi) represents the
algebraic sum of the moments of all the forces to the left of
the section X, and therefore represents the bending moment at
the section X (Art. 160).
Elements of Graphical Statics 235
Scales. — If the scale of forces in the vector diagram is —
i inch to/ Ibs.
and the scale of distance in the space diagram is —
i inch to q feet ;
and if <?H is made h inches long, the scale on which xl repre-
sents the bending moment at X is —
i inch to/, q. h. foot-lbs.
A diagram (Fig. 187) showing the shearing force along the
A!B
FIG. 187.
length of the beam may be drawn by using a base line, st, of
the same length as the beam in the space diagram, and in the
horizontal line through g in the force diagram. The shearing
force between the end of the beam s and the line AB is con-
stant and equal to R1} i.e. proportional to ga. The height ga
may be projected from a by a. horizontal line across the space
A. A horizontal line drawn through b gives by its height above
g the shearing force at all sections of the beam in the space B.
Similarly projecting horizontal lines through c, d, e, and f we
get a stepped diagram, the height of which from the base line
st gives, to the same scale as the vector diagram, the shearing
force at every section of the beam.
236 Mechanics for Engineers
EXAMPLES XIX.
1. Draw a square lettered continuously PQRS, each side 2 inches
long. Forces of 9, 7, and 5 Ibs. act in the directions RP, SQ, and QR
respectively. Find by means of a funicular polygon the resultant of these
three forces. State its magnitude in pounds, its perpendicular distance
from P, and its inclination to the direction PQ.
2. Add to the three forces in question I a force of 6 Ibs. in the direction
PQ, and find the resultant as before. Specify it by its magnitude, its
distance from P, and its inclination to PQ.
3. A horizontal beam, 15 feet long, resting on supports at its ends, carries
concentrated vertical loads of 7, 9, 5, and 8 tons at distances of 3, 8, 12,
and 14 feet respectively from the left-hand support. Find graphically the
reactions at the two supports.
4. A horizontal rod AB, 13 feet long, is supported by a horizontal hinge
perpendicular to AB at A, and by a vertical upward force at B. Four
forces of 8, 5, 12, and 17 Ibs. act upon the rod, their lines of action cutting
AB at i, 4, 8, and 12 feet respectively from A, their lines of action making
angles of 70°, 90°, 120°, and 135° respectively with the direction AB, each
estimated in a clockwise direction. Find the pressure exerted on the
hinge, state its magnitude, and its inclination to AB.
5. A simply supported beam rests on supports 17 feet apart, and carries
loads of 7, 4, 2, and 5 tons at distances of 3, 8, 12, and 14 feet respectively
from the left-hand end. Calculate the bending moment at 4, 9, and n feet
from the left-hand end.
6. Draw a diagram to show the bending moments at all parts of the
beam in question 5. State the scales of the diagram, and measure from it
the bending moment at 9, n, 13, and 14 feet from the left-hand support.
7. Calculate the shearing force on a section of the beam in Question 5
at a point 10 feet from the left-hand support ; draw a diagram showing the
shearing force at every transverse section of the beam, and measure from it
the shearing force at 4 and at 13 feet from the left-hand support.
8. A beam of 2O-feet span carries a load of 10 tons evenly spread over
the length of the beam. Find the bending moment and shearing force at
the mid-section and at a section midway between the middle and one end.
162. Equilibrium of Jointed Structures.
Frames. — The name frame is given to a structure consist-
ing of a number of bars fastened together by hinged joints;
the separate bars are called members of the frame. Such
structures are designed to carry loads which are applied mainly
at the joints. We shall only consider frames which have just
a sufficient number of members to prevent deformation or
collapse under the applied loads. Frames having more
Elements of Graphical Statics 237
members than this requirement are treated in books on
Graphical Statics and Theory of Structures. We shall further
limit ourselves mainly to frames all the members of which are
approximately in the same plane and acted upon by forces all
in this same plane and applied at the hinges.
Such a frame is a rigid body, and the forces exerted upon
it when in equilibrium must fulfil the conditions, stated in Art.
96 and in Art. 156. These "external" forces acting on the
frame consist of applied loads and reactions of supports ; they
can be represented in magnitude and direction by the sides of
a closed vector polygon ; also their positions are such that an
indefinite number of closed funicular polygons can be drawn
having their vertices on the lines of action of the external
forces. From these two considerations the complete system
of external forces can be determined from sufficient data, as in
Arts. 155 and 159. The "internal" forces, i.e. the forces
exerted by the members on the joints, may be determined from
the following principle. The pin of each hinged joint is in
equilibrium under the action of several forces which are
practically coplanar and concurrent. These forces are : the
stresses in the members (or the " internal " forces) meeting at
the particular joint, and the "external" forces, i.e. loads and
reactions, if any, which are applied there.
If all the forces, except two internal ones, acting at a given
joint are known, then the two which have their lines of action
in the two bars can he found by completing an open polygon
of forces by lines parallel to those two bars.
If a closed polygon of forces be drawn for each joint in the
structure, the stress in every bar will be determined. In order to
draw such a polygon for any particular joint, all the concurrent
forces acting upon it, except two, must be known, and therefore
a start must be made by drawing a polygon for a joint at which
some external force, previously determined, acts. Remembering
that the forces which any bar exerts on the joints at its two
ends are equal and in opposite directions, the drawing of a
complete polygon for one joint supplies a means of starting
the force polygon for a neighbouring joint for which at least
one side is then known. An example of the determination of
238
Mechanics Jor Engineers
the stresses in the members of a simple frame will make this
more easily understood.
Fig. 1 88 shows the principles of the graphical method of
finding the stresses or internal forces in the members of a
simple frame consisting of five bars, the joints of which have
been denoted at (a) by i, 2, 3, and 4. The frame stands in
the vertical plane, and carries a known vertical load, W, at the
joint 3 ; it rests on supports on the same level at i and 4.
The force W is denoted in Bow's notation by the letters PQ.
The reactions at i and 4, named RP and QR respectively, have
been found by a funicular polygon corresponding to the vector
diagram at ($), as described in Art. 158.
Elements of Graphical Statics 239
Letters S and T have been used for the two remaining
spaces. When the upward vertical force RP at the joint i is
known, the triangle of forces rps at (<r) can be drawn by making
rp proportional to RP as. in (^), and completing the triangle by
sides parallel to PS and SR (i.e. to the bars 12 and 14)
respectively. After this triangle has been drawn, one of the
three forces acting at the joint 2 is known, viz. SP acting in
the bar 12, being equal and opposite to PS in (c). Hence the
triangle of forces spt at (d), for the joint 2 can be drawn. Next
the triangle tpq at (e) for joint 3 can be drawn, tp and pq being
known; the line joining qt will be found parallel to the bar
QT if the previous drawing has been correct ; this is a check
on the accuracy of the results. Finally, the polygon qrst at (/)
for joint 4 may be drawn, for all four sides are known in
magnitude and direction from the previous polygons. The
fact that when drawn to their previously found lengths and
directions they form a closed polygon, constitutes a check to
the correct setting out of the force polygons. The arrow-heads
on the sides of the polygons denote the directions of the forces
on the particular joint to which the polygon refers.
163. Stress Diagrams. — It is to be noticed in Fig. 188
that in the polygons (£), (<:), (d), (<?), and (/), drawn for the
external forces on the frame and the forces at the various joints,
each side, whether representing an external or internal force,
has a line of equal length and the same inclination in some
other polygon.
For example, sr in (c) corresponding to rs in (/), and pt
in (d) with tp in (e). The drawing of entirely separate polygons
for the forces at each joint is unnecessary ; they may all be
included in a single figure, such as (g), which may be regarded
as the previous five polygons superposed, with corresponding
sides coinciding. Such a figure is called a stress diagram for
the given frame under the given system of external loading. It
contains (i) a closed vector polygon for the system of external
forces in the frame, (2) closed vector polygons for the (con-
current) forces at each joint of the structure.
As each vector representing the internal force in a member
of the frame represents two equal and opposite forces,
240 Mechanics for Engineers
arrow-heads on the vectors are useless or misleading, and are
omitted.
Distinction between Tension and Compression
Members of a Frame. — A member which is in tension is
called a " tie," and is subjected by the joints at its ends to a
pull tending to lengthen it. The forces which the member
exerts on the joints at its ends are equal and opposite pulls
tending to bring the joints closer together.
A member which is in compression is called a "strut;"
it has exerted upon it by the joints at its ends two equal and
opposite pushes or thrusts tending to shorten it. The member
exerts on the joints at its ends equal and opposite " outward "
thrusts tending to force the joints apart.
The question whether a particular member is a " tie " or a
" strut " may be decided by finding whether it pulls or thrusts at
a joint at either end. This is easily discovered if the direction of
any of the forces at that joint is known, since the vector polygon
is a closed figure with the last side terminating at the point from
which the first was started. E.g. to find the kind of stress in
the bar 24, or ST (Fig. 188). At joint 4 QR is an upward
force ; hence the forces in the polygon qrst must act in the
-> ->-> ->
directions grt rs, st, and tq ; hence the force ST in bar 24 acts
at joint 4 in the direction s to /, i.e. the bar pulls at joint 4,
or the force in ST is a tension. Similarly, the force in bar 23,
or PT, acts at joint 3 in a direction 1p^ i.e. it pushes at joint
3, or the force in bar 23 is a compressive one.
Another method. — Knowing the direction of the force rp at
joint i (Fig. 1 88), we know that the forces at joint i act in the
directions rp, ps, and sr, or the vertices of the vector polygon
rps lie in the order r — p — s.
The corresponding lines RP, PS, and SR in the space
diagram are in clockwise order round the point i . This order,
clockwise or contra-clockwise (but in this instance clockwise)
is the same for every joint in the frame. If it is clockwise for
joint i, it is also clockwise for joint 2. Then the vertices of
the vector polygon for joint 2 are to be taken in the cyclic
order s— p — /, since the lines SP, PT, and TS lie in clockwise
Elements of Graphical Statics
241
order round the joint 2, e.g. the force in bar 23, or PT, is in
the direction//, i.e. it thrusts at joint 2.
This characteristic order of space letters round the joints is a
very convenient method of picking out the kind of stress in one
member of a complicated frame. Note that it is the character-
istic order of space letters round a joint that is constant — not
the direction of vectors round the various polygons constituting
the stress diagram.
164. Warren Girder. — A second example of a simple
stress diagram is shown in Fig. 189, viz. that of a common type
F/V/V
E\/G\/K\/M
\A
FIG. 189.
of frame called the Warren girder, consisting of a number of
bars jointed together as shown, all members generally being
of the same lengths, some horizontal, and others inclined 60° to
the horizontal.
Two equal loads, AB and BC, have been supposed to act
at the joints i and 2, and the frame is supported by vertical
reactions at 3 and 4, which are found by a funicular polygon.
The remaining forces in the bars are found by completing the
stress diagram abc . . . klm.
Note that the force AB at joint i is downward, i.e. in the
direction db in the vector diagram corresponding to a contra-
clockwise order, A to B, round joint i. This is, then, the
characteristic order (contra-clockwise) for all the joints, e.g. to
find the nature of the stress in KL, the order of letters for
joint 5 is K to L (contra-clockwise), and referring to the vector
242
Mechanics for Engineers
diagram, the direction k to / represents a thrust of the bar KL
on joint 5 ; the bar KL is therefore in compression.
165. Simple Roof=frame. — Fig. 190 shows a simple
roof-frame and its stress diagram when carrying three equal
vertical loads on three joints and supported at the extremities
of the span.
FIG. 190.
The reactions DE and EA at the supports are each obvi-
ously equal to half the total load, i.e. e falls midway between a
and d in the stress diagram. The correct characteristic order
of the letters round the joints (Art. 163) is, with the lettering
here adopted, clockwise.
1 66. Loaded Strings and Chains. — Although not
coming within the general meaning of the word " frame," stress
Elements of Graphical Statics
243
diagrams can be drawn for a structure consisting partly of
perfectly flexible chains or ropes, provided the loads are such
as will cause only tension in flexible members.
Consider a flexible cord or chain, Xi23Y (Fig. 191), sus-
pended from points X and Y, and having vertical loads of
W,
Wi» W2, and W3 suspended from points i, 2, and 3 respectively.
Denoting the spaces according to Bow's notation by the letters
A, B, C, D, and O, as shown above, the tensions in the strings
Xi or AO and i 2 or BO must have a resultant at i equal
to Wj vertically upward, to balance the load at i. If triangles
of forces, abo^ bco, and cdo, be drawn for the points i, 2, and 3
respectively, the sides bo and co appear in two of them, and, as
in Art. 163, the three vector triangles maybe included in a
single vector diagram, as shown at the right-hand by the
figure abcdo.
The lines aot bo, co, and do represent the tensions in the
string crossing the spaces A, B, C, and D respectively. If a
horizontal line, 0H, be drawn from o to meet the line abed in
H, the length of this line represents the horizontal component of
the tensions in the strings, which is evidently constant through-
out the whole. (The tension changes only from one space to
the neighbouring one by the vector addition of the intermediate
vertical load.) The pull on the support X is represented by
244 Mechanics for Engineers
ao, the vertical component of which is aH ; the pull on Y
is represented by od, the vertical component of which is Hd.
A comparison with Art. 157 will show that the various
sections of the string Xi23Y are in the same lines as the sides
of a funicular polygon for the vertical forces Wlf W2, and W3,
corresponding to the pole o. If different lengths of string are
attached to X and Y and carry the same loads, W1} W2, and W3,
in the lines AB, BC, and CD respectively, they will have
different configurations ; the longer the string the steeper will
be its various slopes corresponding to shorter pole distances,
H0, i.e. to smaller horizontal tensions throughout. A short
string will involve a great distance of the pole o from the line
abed, i.e. a great horizontal tension, with smaller inclinations of
the various sections of the string. The reader should sketch for
himself the shape of a string connecting X to Y, with various
values of the horizontal tension Ho, the vertical loads remain-
ing unaltered, in order to appreciate fully how great are the
tensions in a very short string.
A chain with hinged links, carrying vertical loads at the
joints, will occupy the same shape as a string of the same
length carrying the same loads. Such chains are used in sus-
pension bridges.
The shape of the string or chain to carry given loads in
assigned vertical lines of action can readily be found for any
given horizontal tension, H0, by drawing the various sections
parallel to the corresponding lines radiating from o, e,g. AO
or Xi parallel to do (Fig. 191).
Example i.— A string hangs from two points, X and Y, 5 feet
apart, X being 3 feet above Y. Loads of 5, 3, and 4 Ibs. are
attached to the string so that their lines of action are i, 2, and
3 feet respectively from X. If the horizontal tension of the string
is 6 Ibs., draw its shape.
The horizontal distance ZY (Fig. 192) of X from Y is—
V52 - 32 = 4 feet
so that the three loads divide the horizontal span into four equal
parts'.
Let Vx and VY be the vertical components of the tension of the
string at X and Y respectively.
Elements of Graphical Statics
245
The horizontal tension is constant, and equal to 6 Ibs. Taking
moments about Y (Fig. 192) —
Clockwise. Contra-clockwise.
Vx x 4 = (4 x i) + (3 x 2) + (5 x 3) + (6 x 3) Ib.-feet
4VX = 4 + 6+ 15 + 18 = 43
Vx = -^ = 1075 Ibs.
Since the vertical and horizontal components of the tension of
the string at X are known, its direction is known. The direction
of each section of string might similarly be found. Set out the
vector polygon abcd^ and draw the horizontal line Ho to represent
6 Ibs. horizontal tension from H, aH being measured along abed of
such a length as to represent the vertical component 1075 Ibs. of
the string at X. Join o to a, b, c, and d. Starting from X or Y,
draw in the lines across spaces A, B, C, ancl D parallel respectively
to ao, bo, co, and do (as in Art. 157). The funicular polygon so
drawn is the shape of the string.
Example 2. — A chain is attached to two points, X and Y,
X being I foot above Y and 7 feet horizontally from it. Weights
of 20, 27, and 22 Ibs. are to be hung on the chain at horizontal
distances of 2, 4, and 6 feet from X. The chain is to pass through
a point P in the vertical plane of X and Y, 4 feet below, and 3 feet
246
Mechanics for Engineers
horizontally from X. Find the shape of the chain and the tensions
at its ends.
Let Vx and VY be the vertical components of the tension at X
and Y respectively, and let H be the constant horizontal tension
throughout.
FIG. 193.
Taking moments about Y (Fig. 193) —
Clockwise. Contra-clockwise.
Vx x 7 = (H x i) + (20 x 5) + (27 x 3) + (22 x i)
;VX = H + 203 Ibs.-feet (i)
Taking moments about P of the forces on the chain between
X and P—
Clockwise. Contra-clockwise.
Vx x 3 = H x 4 + (20 x i)
3VX = 4H + 20 (2)
and 28VX = 4H + 812 from (i)
hence 25VX = 792
Vx = 3i-681bs.
H = 7VX — 203 = 22176 - 203 = 1876 Ibs.
Draw the open polygon of forces, abed (a straight line), and set
Elements of Graphical Statics 247
off am from a to the same scale, 31*68 Ibs. downwards. From in
set off mo to represent 18*76 Ibs. horizontally to the right of m.
Then the vector ao = am + mo = tension in the string XZ,
which pulls at X in the direction XZ. By drawing XZ parallel
to ao the direction of the first section of the chain is obtained, and
by drawing from Z a line parallel to bo to meet the line of action
BC, the second section is outlined. Similarly, by continuing the
polygon by lines parallel to co and do the complete shape of the
chain between X and Y is obtained.
The tension ao at X scales 37 Ibs., and the tension od at Y
scales 44 Ibs.
167. Distributed Load. — If the number of points at
which the same total load is attached to the string (Fig. 191)
be increased, the funicular polygon corresponding to its shape
will have a larger number of shorter sides, approximating, if
the number of loads be increased indefinitely, to a smooth
curve. This case corresponds to that of a heavy chain or
string hanging between two points with no vertical load but its
own weight. If the dip of the chain from the straight line join-
ing the points of the attachment is small, the load per unit of
horizontal span is nearly uniform provided the weight of chain
per unit length is uniform. In this case an approximation to the
shape of the chain may be found by dividing the span into a
number of sections of equal length and taking the load on each
portion as concentrated at the mid-point of that section. The
funicular polygon for such a system of loads will have one
side more than the number into which the span has been
divided ; the approximation may be made closer by taking
more parts. The true curve has all the sides of all such poly-
gons as tangents, or is the curve inscribed in such a polygon.
The polygons obtained by dividing a span into one, two,
and four equal parts, and the approximate true curve for a
uniform string stretched with a moderate tension, are shown in
Fig. 194.
Note that the dip QP would be less if the tensions OH, OA?
etc., were increased.
1 68. The relations between the dip, weight, and tension
of a stretched string or chain, assuming perfect flexibility, can
248
Mechanics for Engineers
more conveniently be found by ordinary calculation than by
graphical methods.
Assuming that the dip is small and the load per horizontal
Q
FIG. 194.
foot of span is uniform throughout, the equilibrium of a portion
AP (Fig. 195) of horizontal lengths, measured from the lowest
point A, may be considered.
^ l >
rK
FIG. 195.
Let w = weight per unit horizontal length of cord or chain ;
y = vertical height of P above A, viz. PQ (Fig. 195) ;
T = the tension (which is horizontal) at A ;
T = the tension at P acting in a line tangential to the
curve at P.
Elements of Graphical Statics 249
The weight of portion AP is then wx, and the line of action
of the resultant weight is midway between AB and PQ, i.e. at
a distance - from either.
2
Taking moments about the point P —
T x PQ = wx X -
1
wx
This relation shows that the curve of the string is a
parabola.
If d = the total dip AB, and / = the span of the string or
chain, taking moments about N of the forces on the portion
AN—
2
which gives the relation between the dip, the span, and the
horizontal tension.
Returning to the portion AP, if the vector triangle rst be
drawn for the forces acting upon it, the angle 0 which the
tangent to the curve at P makes with the horizontal is given
by the relation —
xw st
-t- = -.= tan0
Also the tension T' at P is T sec 0, or —
r =
and at the ends where x — —•
250
Mechanics for Engineers
And since T = -7, the tension at N or M is
ou
72 -7
which does not greatly exceed ~^j (or T), if -= is small.
Example. — A copper trolley-wire weighs \ Ib. per foot length ;
it is stretched between two poles 50 feet apart, and has a horizontal
tension of 2000 Ibs. Find the dip in the middle of the span.
Let d — the dip in feet.
The weight of the wire in the half-span BC (Fig. 196) is
25 x \ = 12-5 Ibs.
B
2000 Ibs C
•x
\*"^&-
c •""itf.Jfl*
P zooo
FIG. 196.
The distance of the e.g. of the wire BC from B is practically
1 2*5 feet horizontally.
Taking moments about B of the forces on the portion BC —
2000 x d = 12-5 x 12-5
d = 0*07812 foot = 0-938 inch
EXAMPLES XX.
i. A roof principal, shown in Fig. 197, carries loads of 4, 7, and 5 tons
in the positions shown. It is simply supported at the extremities of a span
FIG. 197.
of 40 feet. The total rise of the roof is 14 feet, and the distances FQ and
Elements of Graphical Statics
251
RS are each 5 '4 feet. Draw the stress diagram and find the stress in each
member of the frame.
2. A Warren girder (Fig. 198), made up of bars of equal lengths, carries
a single load of 5 t°ns as shown. Draw the stress diagram and scale off
OF THE ^
UNIVERSITY
FIG. 198.
the forces in each member ; check the results by the method of sections
(Art. 98).
3. Draw the stress diagram for the roof-frame in Fig. 199 under the
given loads. The main rafters are inclined at 30° to the horizontal, and
are each divided by the joints into three equal lengths.
4. A chain connects two points on the same level and 10 feet apart ;
it has suspended from it four loads, each of 50 Ibs., at equal horizontal
intervals along the span. If the tension in the middle section is 90 Ibs.,
draw the shape of the chain, measure the inclination to the horizontal, and
the tension of the end section.
5. Find the shape of a string connecting two points 8 feet horizontally
apart, one being I foot above the other, when it has suspended from it
weights of 5, 7, and 4 Ibs. at horizontal distances of 2, 5, and 6 feet
respectively from the higher end, the horizontal tension of the string being
6 Ibs.
6. A light chain connects two points, X and Y, 12 feet horizontally apart,
X being 2 feet above Y. Loads of 15, 20, and 25 Ibs. are suspended from
the chain at horizontal distances of 3, 5, and 8 feet respectively from X.
The chain passes through a point 7 feet horizontally from X and 4 feet
252 Mechanics for Engineers
below it. Draw the shape of the chain. How far is the point of suspension
of the 15-lb. load from X ?
7. A wire is stretched horizontally, with a tension of 50 Ibs., between
two posts 60 feet apart. If the wire weighs 0-03 Ib. per foot, find the sag
of the wire in inches.
8. A wire weighing 0*01 Ib. per foot is stretched between posts 40 feet
apart. What must be the tension in the wire in order to reduce the sag to
2 inches ?
9. A wire which must not be stretched with a tension exceeding 70 Ibs.
is to be carried on supporting poles, and the sag between two poles is not
to exceed 1*5 inches. If the weight of the wire is 0^025 Ib. per foot, find
the greatest distance the poles may be placed apart.
APPENDIX
UNITS AND THEIR DIMENSIONS
Units. — To express the magnitude of any physical quantity it has
to be stated in terms of a unit of its own kind. Thus by stating
that a stick is 275 feet long, we are using the foot as the unit of
length.
Fundamental and Derived Units. — We have seen that the
different quantities in common use in the science of mechanics
have certain relations to one another. If the units of certain
selected quantities are arbitrarily fixed, it is possible to determine
the units of other quantities by means of their relations to the
selected ones. The units arbitrarily fixed are spoken of as
fundamental units, and those depending upon them as derived
units.
Fundamental Units.— There are two systems of units in
general use in this country. In the C.G.S. system (Art. 42), which
is commonly used in physical science, the units chosen as funda-
mental and arbitrarily fixed are those of length, mass, and time,
viz. the centimetre, gramme, and second.
In the British gravitational system the fundamental units chosen
are those of length, force, and time, viz. the foot, the pound, i.e. the
weight of i Ib. of matter at some standard place, and the mean
solar second.
The latter system of units has every claim to the name
" absolute," for three units are fixed, and the other mechanical
units are derived from them by fixed relations.
The weight of a body of given mass varies at different parts of
the earth's surface in whatever units its mass is measured. The
value of i Ib. force, however, does not vary, since it has been
defined as the weight of a fixed mass at a fixed place.
Dimensions of Derived Units.
(a) Length — Mass — Time Systems. — In any such system other
than, say, the C.G.S. system, let the. unit of length be L centimetres,
the unit of mass M grammes, and the unit of time be T seconds.
254 Mechanics for Engineers
Then the unit of area will be L x L or L2 square centimetres,
i.e. it varies as the square of the magnitude of the unit of length.
Similarly, we may derive the other important mechanical units as
follows : —
Unit volume — L x L x L or L3 cubic centimetres, or unit
volume varies as L3.
Unit velocity is L centimetres in T seconds = ~ centimetres
per second, or LT"1 centimetres per second.
Unit acceleration is = centimetres per second in T seconds
= ~2 centimetres per second, or LT~2 centimetres per
second.
Unit momentum is that of M grammes moving „ centimetres per
ML
second, i.e. *-~- C.G.S. units of momentum, or MLT~]
C.G.S. units.
MLT"1
Unit force is unit change of momentum in T seconds, or — = —
units in one second, or MLT~2 dynes (C.G.S. units of force).
Unit impulse is given by unit force (MLT~2 C.G.S. units) acting
for unit time, T seconds generating a change of momentum
(or impulse) MLT"1 C.G.S. units.
Unit work is that done by unit force (MLT~2 dynes) acting
through L centimetres, i.e. ML2T~2 centimetre-dynes or
ergs.
Unit kinetic energy is that possessed by unit mass, M grammes
moving with unit velocity (LT"1), i.e. m(LT-l)2=lML2T~*
C.G.S. units.
ML2T~2
Unit power is unit work in unit time T seconds, or — — — , or
ML2T~3 ergs per second.
L units of arc
Note that unit angle , : -^~ = i radian, and is inde-
pendent of the units of length, mass, or time.
Unit angular velocity is unit velocity ^ divided by unit radius
L centimetres, or LT-1 -T- L = T-1.
Appendix
255
Unit moment of momentum or angular momentum is unit
momentum MLT—1 at unit perpendicular distance L, or
ML2T-' C.G.S. units.
Unit moment of force is unit force MLT~2 at unit distance L
centimetres, or ML2T~2 C.G.S. units.
Unit rate of change of angular momentum is ML2T-1 C.G.S.
units in unit time T seconds = ML2T~2 C.G.S. units.
Unit moment of inertia is that of unit mass M grammes at unit
distance L centimetres, which is ML2 C.G.S. units.
Thus each derived unit depends on certain powers of the
magnitudes of the fundamental units, or has certain dimensions of
those units.
(b} The dimensions of the same quantities in terms of the three
fundamental units of length, force, and time may be similarly
written as follows : —
Quantity.
[Length.
Force.
vTime.
Velocity.
Acceleration.
Mass.
Momentum.
Impulse.
Work.
Kinetic energy.
Power.
Angular momentum.
Moment of force.
Rate of change of angular
momentum.
Dimensions.
L.
F.
T.
or LT-1.
T or LT-2.
Force
— r- or FL-1T2.
Acceleration
FT.
FT.
FL.
IFL,
FLT-1.
FLT.
FL.
FL.
Symbolical formulae and equations may be checked by testing
if the dimensions of the terms are correct. Each term on either
side of an algebraic equation having a physical meaning must
necessarily be of the same dimensions.
ANSWERS TO EXAMPLES
EXAMPLES I.
(i) 0*305 foot per second per second. (2) 5-5 seconds; 121 feet.
(3)^77 feet per second. (4) 3-053 seconds.
(5) 89-5 feet ; 447-5 feet ; 440-4 feet.
(6) 5-63 seconds after the first projection ; 278 feet.
(7) 567 feet per second, (8) 4-5, 14-6, and 11-4 feet per second.
(10) 0-57 and 0-393 foot per second per second ; 880 feet.
(n) 77-3 feet ; 2*9 seconds.
EXAMPLES II.
(1) 4-88 feet per second ; 35° 23' to the horizontal velocity.
(2) 405 feet per second ; 294 feet per second.
(3) 53° up-stream ; 2 minutes 16-4 seconds. (4) 10° 6 south of west.
(5) J9'54 knots per hour ; 5 hours 7-2 minutes ; 12° 8' west of south.
(6) 48 minutes ; 9*6 miles ; I2"8 miles.
(7) I54'2 feet per second per second ; 2i°'5 south of west.
(8) 2-59 seconds. (9) 5-04; 4716. (ic) 16*83 feet per second,
(u) 35-2 radians per second ; 2*581 radians per second per second.
(12) 135 revolutions and 1*5 minutes from full speed.
EXAMPLES III.
(i) 2735 units; 182,333 lbs- or 81-4 tons. (2) $ or 1-172 to i.
(3) 2 "8 centimetres per second. (4) 9802 Ibs.
(5) I5'33 lt>s- ; 9'53 units per second in direction of jet ; 9*53 Ibs.
(6) 45'3- (7) 4720 Ibs.
(8) 10-43 tons inclined downwards at 16° 40' to horizontal.
(9) 2*91 units ; 727-5 Ibs. (10) 8750 units ; 8-57 miles per hour.
EXAMPLES IV.
(1)67-8 Ibs. (2) 17-48 Ibs. (4) 34-54 feet.
(5) 23'44 feet Per second ; 255,000 Ibs. (6) 1005 feet per second.
(7) 154 Ibs. ; 126 Ibs. ; 6*9 feet per second per second.
(8) ii'243cwt. (9) 9'66feet; 14-93 Ibs.
(10) 4-69 grammes ; 477 centimetres.
(n) 6-44 feet per second per second ; 4 Ibs.
(12) 1-027 Ibs. (13) 48-9 Ibs.
Answers to Examples
EXAMPLES V.
(1) 1 60 horse-power ; 303-36 horse-power
(2) 1575 Ibs. per ton.
(4) 929; 1253.
(6) 0-347 horse-power.
(8) 350,000 foot-lbs. ; 800,000 foot-lbs.
1 6 '64 horse-power.
(3) 22-15 miles per hour.
(5) J47'5 horse-power.
(7) 60 foot-lbs.
(9) 1,360.000 foot-lbs.
EXAMPLES VI.
(i) 57* i horse-power.
(3) 6570 Ib. -feet.
(5) 5340 inch-lbs. ; 2220 inch-lbs.
(2) 39>39° Ib.-feet.
(4) 609 inch-lbs.
(6) 1 2 '8 horse-power.
EXAMPLES VII.
(i) 12,420,000 foot-lbs. ; 4, 140,000 Ibs.
(3) 37,740 inch-lbs. ; 35,940 inch-lbs.
(5) 7*02 horse-power.
(7) 19-6 horse-power.
(9) 10-5 feet per second ; 467 Ibs.
(n) 2886 foot-lbs.
EXAMPLES VIII.
(2) 27-8 feet per second.
(4) 25-5 horse-power.
(6) 7-25 horse-power.
(8) 8-65 seconds.
(10) 15-3 seconds.
(12) 500,000 foot-lbs.
(1)68-5 (2) ii '85 miles per hour.
(4) 4-25 inches. (5) 3052 feet.
(7) 47° to horizontal.
(8) 52°'5 ; 1*64 times the weight of the stone.
(9) !°5 Per cent, increase.
(10) 66 -4 ; 72*7 ; 59-3 revolutions per minute.
(12) 38-33 ; 35-68 feet per second, 7-79 ; 6*28 Ibs.
EXAMPLES IX.
(3) 2672 feet.
(6) 20 miles per hour.
(ii)
(1) o 855, 1-56, 1-81 feet per second ; 8-05, 5-96, 4-4 feet per second per
second.
(2) | inch. (3) 1654, 827, 1474 Ibs.
(4) 153-3. (6) 0-342 second.
(7) 1-103 second ; 67-3 feet per second per second.
(8) 31-23. (9) i to 1-0073.
EXAMPLES X.
(i) 14-65 Ibs. ; 17-9 Ibs. (2) 3 Ibs. ; 13 Ibs.
(3) 9-6 tons tension ; 55*6 tons tension. (4) 4ic>7 south of west ; 720 Ibs.
(5) 2250 Ibs. ; 2890 Ibs. (6) 220 Ibs. ; 58-5 Ibs.
S
258 Mechanics for Engineers
EXAMPLES XI.
(i) 0-154; 8°'8 (2) 2-97 Ibs. ; 8° -5 to horizontal. (3) 14-51 Ibs.
(4) 0'6 times the weight of log ; 36°'8 to horizontal. (5) io0>4
(6) 0*3066 horse-power. (7) 179 horse-power.
(8) 3^84 horse-power.
(9) 3-4 feet per second per second ; 3-57 Ibs.
(10) 4-5 tons; 31-9 seconds, (n) 3820 Ibs.
EXAMPLES XII.
(I) 261 Ibs. (2) 16-97 Ibs. ; 4-12 Ibs.
(3) Left, 5-242 tons ; right, 5-008 tons.
(4) Left, 10 tons ; right, 3 tons ; end, 2^824 tons.
(5) 1*039 inches. (6) 5*737 feet from end.
EXAMPLES XIII.
(1) Tension, 21 '68 Ibs.
(2) 0-1264. (3) 36°.
(4) 15*3 Ibs. at hinge ; 8-25 Ibs. at free end.
(5) 3950 Ibs. at A ; 2954 Ibs. at C.
(6) 1 1 "2 Ibs. cutting AD 2'i inches from A, inclined I9°'3 to DA.
(7) 4-3 tons ; 3-46 tons ; 467° to horizontal.
(8) 8*2 tons compression ; 4-39 tons tension ; 4 tons tension.
(9) 8-78 tons tension ; 25-6 tons compression ; 21 "22 tons tension.
EXAMPLES XIV.
(i) 1*27 feet from middle. (2) 2'o8 inches.
(3) 43 inches. (4) 1*633 feet '> I<225 feet-
(5) 4-18 inches ; 4*08 inches. (6) lo'i inches ; 5'5 Ibs.
(7) 2-98 inches. (8) 27*2 inches.
(9) 975 inches. (10) 1293 Ib.-feet ; 103-5 Ibs. per square foot,
(n) 11-91 inches. (12) 4*82 inches.
(13) 4 feet 5-1 inches. (14) 0*1*7 Ib. (15) 0*197 Ib. ; 0*384 Ib.
EXAMPLES XV.
(i) 19-48 inches ; 16*98 inches. (2) 12*16 inches.
(3) 6*08 inches. (4) 15*4 inches.
(5) 2*52 inches from outside of flange. (6) 4-76 inches.
(7) 0-202 inch from centre. (8) i6'6 inches.
(9) 5-36 inches. (10) 33*99 inches.
Answers to Examples 259
EXAMPLES XVI
(i) 1 6 and 8 tons. (2) 25 and 16 tons.
(3) Left, 16-5 tons ; right, 33-4 tons. (4) 53° 10'.
(5) 16-43 inches; 4'4r inches. (6) 3-53 inches.
(7) 3-67 inches. (8) 8000 foot-lbs.
(9) 1 188 foot-lbs. (10) 140,000 ; 74,400 foot-lbs.
(ii) 75,600 foot-lbs. (12) 2514 foot-lbs.
(13) 110-3 Ibs. (14) 5-11 Ibs.
(15) 37-6 square inches. (16) 7-85 cubic inches.
(17) 4 feet 3-9 inches.
EXAMPLES XVII.
(i) 312 (inches)4. (2) 405 (inches)4; 4-29 inches.
(3) 195 (inches)4 ; 2-98 inches. (4) 290 (inches)4.
(5) 5*523 inches. (6) 0*887 gravitational units.
2
(8) 16*1 inches ; 35*15 gravitational units.
EXAMPLES XVIII.
(i) 3647 gravitational units. (2) 13,215 gravitational units. /
(3) 10 minutes 46 seconds ; 323. (4) 17-48 Ibs.
(5) 35° Ib.-feet. (6) 2*134 gravitational units ; 6*83 inches.
(7) Hi '3- (8) 771 inches.
(9) 22 feet per second ; 31*06 feet per second.
(10) 14*85 feet per second ; 16-94 feet per second.
(11) 3-314 feet; 3819 gravitational units. (12) 537.
(13) 0-0274 units. (14) 125*5.
(15) 117-5 foot-tons. (16) 167 feet per second.
(17) 23 sin o feet per second per second.
EXAMPLES XIX.
(1) 6-47 Ibs. ; 0*016 inch ; IO20<6.
(2) 7-8 Ibs. ; 0-013 inch ; 36°. (3) 17*7 right ; 11*3 left.
(4) 21*6 Ibs. ; 134° measured clockwise.
(5) 30*4. 38-15. 34*85 tons-feet.
(6) 38*15, 34*85, 29-6, 25-95 tons-feet.
(7) I "65 tons; 2*35 tons; 3*65 tons.
(8) 25 tons-feet; nil; 18*75 tons-feet ; 2*5 tons.
EXAMPLES XX.
(4) 48° ; 134*5 lbs- (6) 4'o6 feet. (7) 3*24 inches.
(8) 12 Ibs. (9) 53 feet.
EXAMINATION QUESTIONS
Questions selected from the Mechanics Examinations
Intermediate (Engineering-) Science of London
University.
1. What is implied in the rule : the product of the diameter of
a wheel in feet, and of the revolutions per minute, divided by 28, is
the speed in miles an hour ?
Also in the rule : three times the number of telegraph posts per
minute is the speed in miles an hour ? (1903-)
2. Continuous breaks are now capable of reducing the speed of
a train 3| miles an hour every second, and take 2 seconds to be
applied. Show in a tabular form the length of an emergency stop
at speeds of 3|, 7|, 15, 30, 45, and 60 miles an hour.
Compare the retardation with gravity ; express the resisting
force in pounds per ton ; calculate the coefficient of adhesion of the
break shoe and rail with the wheel ; and sketch the arrangement.
3. Prove that the horse-power required to overcome a resistance
of R Ibs. at a speed of S miles an hour is RS -f- 375. Calculate
the horse-power of a locomotive drawing a train of 2co tons up an
incline of i in 200 at 50 miles an hour, taking the road and air
resistance at this speed at 28 Ibs. a ton. (1903-)
4. If W tons is transported from rest to rest a distance s feet
in / seconds, being accelerated for a distance s1 and time /j by a
force P! tons up to velocity v feet per second, and then brought to
rest by a force P2 tons acting for /2 seconds through s2 feet —
(i.) I - = PI/I =
(ii.)
/••• \ -TI J.> -y
(ill.) -v = 2-1 = 2^ = 2-
*1 /.) ^
>! +P
a
+
Examination Questions 261
A train of 100 tons gross, fitted with continuous breaks, is to
be run on a level line between stations one-third of a mile apart,
at an average speed of 12 miles an hour, including two-thirds of a
minute stop at each station. Prove that the weight on the driving
wheels must exceed 22^ tons, with an adhesion of one-sixth,
neglecting road-resistance and delay in application of the breaks.
(1903.)
5. Give a graphical representation of the relative motion of a
piston and crank, when the connecting rod is long enough for its
obliquity to be neglected ; and prove that at R revolutions a
minute, the piston velocity is ^ tjmes the geometric mean of the
distance from the two ends of the stroke.
Prove that if the piston weighs W Ibs., the force in pounds
which gives its acceleration is —
W ir2R2
— ' — - (distance in feet from mid-point of stroke).
(1903-)
6. Write down the formula for the time of swing of a simple
pendulum, and calculate the percentage of its change due to i per
cent, change in length or gravity, or both.
Prove that the line in Question 4 could be worked principally
by gravity if the road is curved downward between the stations to
a radius of about 11,740 feet, implying a dip of 33 feet between
the stations, a gradient at the stations of I in 13, and a maximum
running velocity of 31 miles an hour. (1903-)
7. Prove that if a hammer weighing W tons falling h feet
drives a pile weighing w tons a feet into the ground, the average
resistance of the ground in tons is —
W2 h
W + w'a
Prove that the energy dissipated at the impact is diminished by
increasing —
W
(1903.)
"W
8. Prove that the total kinetic energy stored up in a train of
railway carriages, weighing W tons gross, when moving at v feet
per second is —
7,2
l— foot-tons
where Wj denotes the weight of the wheels in tons, a their radius,
and k their radius of gyration.
262 Mechanics for Engineers
Prove that W in the equations of Question 4 must be increased
W £2
by — Jp to allow for the rotary inertia of the wheels. (1903.)
9. Determine graphically, by the. funicular polygon, the reaction
of the supports of a horizontal beam, loaded with given weights at
two given points.
Prove that the bending moment at any point of the beam is
represented by the vertical depth of the funicular polygon.
10. A wheel is making 200 revolutions per minute, and after
10 seconds its speed has fallen to 1 50 revolutions per minute. If
the angular retardation be constant, how many more revolutions
will it make before coming to rest ? (1904-)
11. A piston is connected to a flywheel by a crank and con-
necting rod in the usual manner. If the angular velocity of the
flywheel be constant, show that if the connecting rod be
sufficiently long the motion of the piston will be approximately
simple-harmonic ; and find the velocity of the piston in any
position.
If A, B, C, D, E be five equidistant positions of the piston,
A and E being the ends of its stroke, prove that the piston takes
twice as long to move from A to B as it does from B to C.
(1904.)
12. A train whose weight is 250 tons runs at a uniform speed
down an incline of i in 200, the steam being shut off and the
brakes not applied, and on reaching the foot of the incline it runs
800 yards on the level before coming to rest. What was its original
speed in miles per hour ?
[The frictional resistance is supposed to be the same in each
case.] (1904-)
13. A weight A hangs by a string and makes small lateral
oscillations like a pendulum ; another weight, B, is suspended by
a spiral spring, and makes vertical oscillations. Explain why an
addition to B alters the period of its oscillations, whilst an addition
to A does not. Also find exactly how the period of B varies with
the weight. (i9°4-)
14. A steel disc of thickness t and outer radius a is keyed on to
a cylindrical steel shaft of radius b and length /, and the centre of
the disc is at a distance c from one end of the shaft. Find the
distance from this end of the shaft of the mass-centre of the whole.
(1904.)
15. A uniform bar 6 feet long can turn freely in a vertical plane
about a horizontal axis through one end. If it be just started from
Examination Questions 263
the position of unstable equilibrium, find (in feet per second) the
velocity of the free end at the instant of passing through its lowest
position. (1904.)
1 6. Explain why, as a man ascends a ladder, the tendency of
its foot to slip increases.
A man weighing 13 stone stands on the top of a ladder 20 feet
long, its foot being 6 feet from the wall. How much is the
horizontal pressure of the foot on the ground increased by his
presence, the pressure on the wall being assumed to be horizontal ?
(1904.)
17. A horizontal beam 20 feet long, supported at the ends,
carries loads of 3, 2, 5, 4 cwts. at distances of 3, 7, 12, 15 feet
respectively from one end. Find by means of a funicular polygon
(drawn to scale) the pressures on the two ends, and test the
accuracy of your drawing by numerical computation. (1904.)
Questions selected from the Associate Members'
Examinations of the Institution of Civil Engineers.
1. A beam 20 feet long is supported on two supports 3 feet from
each end of the beam ; weights of lolbs. and 20 Ibs. are suspended
from the two ends of the beam. Draw, to scale, the bending-
moment and shearing- force diagrams ; and, in particular, estimate
their values at the central section of the beam.
(I.C.E., February, 1905.)
2. The speed of a motor car is determined by observing the
times of passing a number of marks placed 500 feet apart. The
time of traversing the distance between the first and second posts
was 20 seconds, and between the second and third 19 seconds. If
the acceleration of the car is constant, find its magnitude in feet
per second per second, and also the velocity in miles per hour at
the instant it passes the first post. (I.C.E., February, 1905.)
3. In a bicycle, the length of the cranks is 7 inches, the diameter
of the back wheel is 28 inches, and the gearing is such that the
wheel rotates i\ times as fast as the pedals. If the weight of the
cyclist and machine together is 160 Ibs., estimate the force which
will have to be applied to the pedal to increase the speed uniformly
from 4 to 12 miles an hour in 20 seconds, frictional losses being
neglected. (I.C.E., February, 1905.)
4. A thin circular disc, 12 inches radius, has a projecting axle.
I inch diameter on either side. The ends of this axle rest on two
parallel inclined straight edges inclined at a slope of I in 40, the
264 Mechanics for Engineers
lower part of the disc hanging between the two. The disc rolls
from rest through I foot in 53^ seconds. Neglecting the weight
of the axle and frictional resistances, find the value of^.
(I.C.E., February, 1905.)
5. A gate 6 feet high and 4 feet wide, weighing 100 Ibs., hangs
from a rail by 2 wheels at its upper corners. The left-hand wheel
having seized, skids along the rail with a coefficient of friction of
\. The other wheel is frictionless. Find the horizontal force that
will push the gate steadily along from left to right if applied 2 feet
below the rail. You may solve either analytically or by the force-
and-link polygon. (I.C.E., October, 1904.)
6. A ladder, whose centre of gravity is at the middle of its
length, rests on the ground and against a vertical wall ; the co-
efficients of friction of the ladder against both being \. Find the
ladder's inclination to the ground when just on the point of slipping.
(I.C.E., October, 1904.)
7. The faceplate of a lathe has a rectangular slab of cast iron
bolted to it, and rotates at 480 revolutions per minute. The slab
is 8 inches by 12 inches by 30 inches (the length being radial).
Its outside is flush with the edge of the faceplate, which is 48 inches
diameter. Find the centrifugal force. (Cast iron weighs \ Ib. per
cubic inch.) Where must a circular weight of 300 Ibs. be placed
to balance the slab? (I.C.E., October, 1904.)
8. A boom 30 feet long, weighing 2 tons, is hinged at one end,
and is being lowered by a rope at the other. When just horizontal
the rope snaps. Find the reaction on the hinge.
(I.C.E., October, 1904.)
9. A beam ABCD, whose length, AD, is 50 feet, is supported
at each end, and carries a weight of 2 tons at B, 10 feet from A,
and a weight of 2*5 tons at C, 20 feet from D. Calculate the
shearing force at the centre of the span, and sketch the diagram of
shearing forces. (I. C.E., October, 1904.)
10. Referring to the loaded beam described in the last question,
how much additional load would have to be put on at the point B
in order to reduce the shearing force at the centre of the span to
zero? (I.C.E., October, 1904.)
11. Estimate the super-elevation which ought to be given to the
outer rail when a train moves round a curve of 2000 feet radius at
a speed of 60 miles an hour, the gauge being 4 feet 8j inches.
(I.C.E., February, 1904.)
12. Show that in simple-harmonic motion the acceleration is
proportional to the displacement from the mid-point of the path,
and that the time of a small oscillation of a simple pendulum of
Examination Questions 265
length / is 271-. /-. Deduce the length of the simple pendulum
\r &
which has the same time of oscillation as a uniform rod of length
L suspended at one end. (I.C.E., February, 1904.)
13. To a passenger in a train moving at the rate of 40 miles
an hour, the rain appears to be rushing downwards and towards
him at an angle of 20° with the horizontal. If the rain is actually
falling in a vertical direction, find the velocity of the rain-drops in
feet per second. (I.C.E., February, 1904.)
14. If it take 600 useful horse-power to draw a train of 335 tons
up a gradient of I in 264 at a uniform speed of 40 miles an hour,
estimate the resistance per ton other than that due to ascending
against gravity, and deduce the uniform speed on the level when
developing the above power. (I.C.E., February, 1904.)
15. In a steam-hammer the diameter of the piston is 36 inches,
the total weight of the hammer and piston is 20 tons, and the
effective steam pressure is 40 Ibs. per square inch. Find the
acceleration with which the hammer descends, and its velocity
after descending through a distance of 4 feet. If the hammer then
come in contact with the iron, and compress it through a distance
of i inch, find the mean force of compression.
(I.C.E., February, 1904.)
1 6. A uniform circular plate, i foot in diameter and weighing
4 Ibs., is hung in a horizontal plane by three fine parallel cords
from the ceiling, and when set in small torsional oscillations about
a vertical axis is found to have a period of 3 seconds. A body
whose moment of inertia is required is laid diametrically across it,
and the period is found to be 5 seconds, the weight being 6 Ibs.
Find the moment of inertia of the body about the axis of oscillation.
(I.C.E., February, 1904.)
17. The acceleration of a train running on the level is found
by hanging a short pendulum from the roof of a carriage and
noticing the angle which the pendulum makes with the vertical.
In one experiment the angle of inclination was 5° : estimate the
acceleration of the train in feet per second per second and in miles
per hour per hour. (I.C.E., February, 1904.)
1 8. Two weights, one of 2 Ibs. and the other of i lb., are con-
nected by a massless string which passes over a smooth peg. Find
the tension in the string and the distance moved through by either
weight, from rest, in 2 seconds. (I.C.E., February, 1904.)
19. A solid circular cast-iron disc, 20 inches in diameter and
2 inches thick (weighing 0*25 lb. per cubic inch), is mounted on
ball bearings. A weight of 10 Ibs. is suspended by means of a
266 Mechanics for Engineers
string wound round the axle, which is 3 inches in diameter, and
the weight is released and disconnected after falling 10 feet.
Neglecting friction, find the kinetic energy stored in the wheel,
and the revolutions per minute the wheel is making when the
weight is disconnected, and also the time it would continue to run
against a tangential resistance of \ Ib. applied at the circumference
of the axle. (I.C.E., February, 1904.)
20. A girder 20 feet long carries a distributed load of i ton per
lineal foot over 6 feet of its length, the load commencing at 3 feet
from the left-hand abutment. Sketch the shearing-force and
bending-moment diagrams, and find, independently, the magni-
tude of the maximum bending moment and the section at which
it occurs. (I.C.E., February, 1904.)
21. Explain what is meant by centripetal acceleration, and find
its value when a particle describes a circle of radius r feet with a
velocity of v feet per second. (I.C.E., October, 1903.)
22. In an electric railway the average distance between the
stations is | mile, the running time from start to stop i£ minutes,
and the constant speed between the end of acceleration and
beginning of retardation 25 miles an hour. If the acceleration
and retardation be taken as uniform and numerically equal, find
their values ; and if the weight of the train be 150 tons and the
frictional resistance 1 1 Ibs. per ton, find the tractive force necessary
to start on the level. (I.C.E., October, 1903.)
23. The mass of a flywheel may be assumed concentrated in
the rim. If the diameter is 7 feet and the weight 2\ tons, estimate
its kinetic energy when running at 250 revolutions per minute.
Moreover, if the shaft be 6 inches in diameter and the coefficient
of friction of the shaft in the bearings be 0-09, estimate the number
of revolutions the flywheel will make before coming to rest.
(I.C.E., October, 1903.)
24. A plane inclined at 20° to the horizontal carries a load of
1000 Ibs., and the angle of friction betweeen the load and the plane
is 10°. Obtain "the least force in magnitude and direction which
is necessary to pull the load up the plane.
(I.C.E., October, 1903.)
25. State the second law of motion. A cage weighing 1000 Ibs.
is being lowered down a mine by a cable. Find the tension in the
cable, (i) when the speed is increasing at the rate of 5 feet per
second per second ; (2) when the speed is uniform ; (3) when the
speed is diminishing at the rate of 5 feet per second per second.
The weight of the cable itself may be neglected.
(I.C.E., October, 1903.)
Examination Questions 267
26. Show that when a helical spring vibrates freely under the
action of a weight, its periodic time is the same as that of a simple
pendulum having a length equal to the static extension of the
spring when carrying the weight, the mass of the spring itself
being neglected. (I.C.E., October, 1903.)
27. A flywheel, supported on an axle 2 inches in diameter,
is pulled round by a cord wound round the axle and carrying
a weight. It is found that a weight of 4 Ibs. is just sufficient to
overcome friction. A further weight of 16 Ibs., making 20 Ibs. in
all, is applied, and two seconds after starting from rest it is found
that the weight has descended a distance of 4 feet. Estimate the
moment of inertia of the wheel about the axis of rotation in
gravitational units. (I.C.E., October, 1903.)
28. With an automatic vacuum brake a train weighing 170 tons
and going at 60 miles an hour on a down gradient of i in 100 was
pulled up in a distance of 596 yards. Estimate the total resistance
in pounds per ton ; and if the retardation is uniform, find the time
taken to bring the train to rest. (I.C.E., October, 1903.)
29. A string, ABCD, hangs in a vertical plane, the ends A
and D being fixed. A weight of 10 Ibs. is hung from the point
B, and an unknown weight from the point C. The middle portion
BC is horizontal, and the portions AB and CD are inclined at 30°
and 45° to the horizontal respectively. Determine the unknown
weight and the tensions in the three portions of the string.
(I.C.E , October, 1903.)
30. Two masses, of 10 Ibs. and 20 Ibs. respectively, are attached
to a balanced disc at an angular distance apart of 90° and at radii
2 feet and 3 feet respectively. Find the resultant force on the axis
when the disc is making 200 turns a minute ; and determine the
angular position and magnitude of a mass placed at 2*5 feet radius
which will make the force on the axis zero at all speeds.
(I.C.E., October, 1903.)
31. A crane has a vertical crane-post, AB, 8 feet long, and a
horizontal tie, BC, 6 feet long, AC being the jib. It turns in
bearings at A and B, and the chain supporting the load passes
over pulleys at C and A, and is then led away at 30° to AB. Find
the stresses in the bars and thrusts in the bearings when lifting
i ton at a uniform rate. (I.C.E., February, 1903.)
32. A man ascends a ladder resting on a rough horizontal floor
against a smooth vertical wall. Determine, graphically or other-
wise, the direction of the action between the foot of the ladder and
the floor. (I.C.E., February, 1903.)
33. A train on a horizontal line of rails is accelerating the speed
268 Mechanics for Engineers
uniformly so that a velocity of 60 miles an hour is being acquired
in 176 seconds. A heavy weight is suspended freely from the roof
of a carriage by a string. Calculate, or determine graphically, the
inclination of the string to the vertical.
(I.C.E., February, 1903.)
34. Explain the meaning of the term " centrifugal force." With
what speed must a locomotive be running on level railway lines,
forming a curve of 968 feet radius, if it produce a horizontal thrust
on the outer rail equal to gj of its weight ?
(I.C.E., February, 1903.)
35. Explain how to determine the relative velocity of two
bodies. A is travelling due north at constant speed. When B
is due west of A, and at a distance of 21 miles from it, B starts
travelling north-east with the same constant speed as A. Determine
graphically, or otherwise, the least distance which B attains from A.
(I.C.E., February, 1903.)
36. Two men put a railway-waggon weighing 5 tons into motion
by exerting on it a force of 80 Ibs. The resistance of the waggon
is 10 Ibs. per ton, or altogether 50 Ibs. How far will the waggon
have moved in one minute ? Calculate at what fraction of a horse-
power the men are working at 60 seconds after starting.
(I.C.E., February, 1903.)
37. State and explain fully Newton's third law of motion. A
loo-lb. shot leaves a gun horizontally with a muzzle velocity of
2000 feet per second. The gun and attachments, which recoil,
weigh 4 tons. Find what the resistance must be that the recoil may
be taken up in 4 feet, and compare the energy of recoil with the
energy of translation of the shot. (I.C.E., February, 1903.)
38. An elastic string is used to lift a weight of 20 Ibs. How
much energy must be exerted in raising it 3 feet, supposing the
string to stretch i inch under a tension of I Ib. ? Represent it
graphically. If the work of stretching the string is lost, what is the
efficiency of this method of lifting ? (I.C.E., February, 1903.)
39. Explain how to determine graphically the relative velocity
of two points the magnitudes and directions of whose velocities are
known. Find the true course and velocity of a steamer steering
due north by compass at 12 knots, through a 4-knot current setting
south-west, and determine the alteration of direction by compass in
order that the steamer should make a true northerly course.
(I.C.E., October, 1902.)
40. Find, by graphic construction, the centre of gravity of a
section of an I beam, top flange 4 inches by i inch ; web, between
flanges, 14 inches by \\ inches ; bottom flange 9 inches by 2 inches.
(I.C.E., October, 1902.)
Examination Questions
269
41. A crankshaft, diameter \2\ inches, weighs 12 tons, and it
is pressed against the bearings by a force of 36 tons horizontally.
Find the horse-power lost in friction at 90 revolutions per minute
(coefficient of friction = o'o6.) (I.C.E, October, 1902.)
Questions selected from the Board of Education
Examinations in Applied Mechanics.
{Reprinted by permission of the Controller of His Majesty's
Stationery Office.)
1. A truck, weighing 5 tons without its wheels, rests on 4
wheels, which are circular discs. 40 inches in diameter, each
weighing J ton, and moves down an incline of I in 60. Find the
velocity of the truck in feet per second after moving 100 feet from
rest, if the resistance due to friction is i per cent, of the weight.
What percentage of the original potential energy has been wasted
in friction ? (Stage 3, 1905.)
2. A flywheel is supported on an axle 2^ inches in diameter, and
is rotated by a cord, which is wound round the axle and carries a
weight. It is found by experiment that a weight of 5 Ibs. on the
cord is just sufficient to overcome the friction and maintain steady,
motion. A load of 25 Ibs. is attached to the cord, and 3 seconds
after starting from rest it is found that the weight has descended
5 feet. Find the moment of inertia of this wheel in engineers'
units.
If the wheel is a circular disc 3 feet in diameter, what is its
weight ? (The thickness of the cord may be neglected.)
(Stage 3, 1905.)
3. The angular position D of a rocking shaft at any time / is
measured from a fixed position. Successive positions at intervals
at J second have been determined as follows : —
Time /, se- )
conds )
Position D, |
radians j
O'O
o'io6
0*02
0-208
0-04
Q'337
0-06
0-487
0-08
0-651
O'lO
0-819
0-12
0-978
OT4
rni
0-16
I '201
0-18
I'222
Find the change of angular position during the first interval
from / = o to t = o'02. Calculate the mean angular velocity during
this interval in radians per second, and, on a time base, set this up
as an ordinate at the middle of the interval. Repeat this for the
270
Mechanics *for Engineers
other intervals, tabulating the results and drawing the curve show-
ing approximately angular velocity and time.
In the same way, find a curve showing angular acceleration and
time. Read off the angular acceleration in radians per second per
second, when / = 0*075 second. (Stage 2, 1905.)
4. A motor car moves in a horizontal circle of 300 feet radius.
The track makes sideways an angle of 10° with the horizontal
plane. A plumb-line on the car makes an angle of 12° with what
would be a vertical line on the car if it were at rest on a horizontal
plane. What is the speed of the car ? If the car is just not side-
slipping, what is the coefficient of friction ? (Stage 3, 1904.)
5. A body whose weight is 350 Ibs. is being acted upon by a
variable lifting force F Ibs. when it is at the height x feet from
its position of rest. The mechanism is such that F depends upon
x in the following way ; but the body will stop rising before the
greatest x of the table is reached. Where will it stop ?
X
F
0
530
15
525
25
516
5o
490
70
425
IOO
300
I25
210
150
1 60
180
no
210
90
Where does its velocity cease to increase and begin to diminish ?
(Stage 3, 1904.)
6. Part of a machine weighing I ton is moving northwards at
60 feet per second. At the end of 0*05 second it is found to be
moving to the east at 20 feet per second. What is the average
force (find magnitude and direction) acting upon it during the
interval 0*05 second ? What is meant by " average " in such a
case ? What is meant by force by people who have to make exact
calculations? (Stage 3, 1904.)
7. A flywheel and its shaft weigh 24,000 Ibs. ; its bearings, which
are slack, are 9 inches diameter. If the coefficient of friction is
0*07, how many foot-pounds of work are wasted in overcoming
friction in one revolution ?
If the mean radius (or rather the radius of gyration) is 10 feet,
what is the kinetic energy when the speed is 75 revolutions per
minute ? If it is suddenly disconnected from its engine at this
speed, in how many revolutions will it come to rest ? What is its
average speed in coming to rest? In how many minutes will it
come to rest ? (Stage 2, 1904.)
8. A train, weighing 250 tons, is moving at 40 miles per hour,
and it is stopped in ten seconds. What is the average force during
Examination Questions 271
these ten seconds causing this stoppage? Define what is meant by
force by people who have to make exact calculations.
(Stage 2, 1904.)
9. A tram-car, weighing 15 tons, suddenly had the electric
current cut off. At that instant its velocity was 16 miles per hour.
Reckoning time from that instant, the following velocities, V, and
times, /, were noted : —
V, miles per hour 16 14 12 10
/, seconds o 9*3 21 35
Calculate the average value of the retarding force, and find the
average value of the velocity from / = oto/=35. Also find the
distance travelled between these times. (Advanced, 1903.)
10. A projectile has kinetic energy = 1,670,000 foot-lbs. at a
velocity of 3000 feet per second. Later on its velocity is only
2000 feet per second. How much kinetic energy has it lost?
What is the cause of this loss of energy ? Calculate the kinetic
energy of rotation of the projectile if its weight is 12 Ibs., and its
radius of gyration is 075 inch, and its speed of rotation is 500
revolutions per second. (Advanced, 1903.)
1 1. A weight of 10 Ibs. is hung from a spring, and thereby causes
the spring to elongate to the extent of 0*42 foot. If the weight is
made to oscillate vertically, find the time of a complete vibration
(neglect the mass of the spring itself). (Advanced, 1903.)
12. A flywheel weighs 5 tons and has a radius of gyration of
6 feet. What is its moment of inertia ? It is at the end of a shaft
10 feet long, the other end of which is fixed. It is found that a
torque of 200,000 Ib.-feet is sufficient to turn the wheel through i°.
The wheel is twisted slightly and then released : find the time of a
complete vibration. How many vibrations per minute would it
make ? (Honours, Part I., 1903.)
13. A flywheel of a shearing machine has 150,000 foot-lbs. of
kinetic energy stored in it when its speed is 250 revolutions per
minute. What energy does it part with during a reduction of speed
to 200 revolutions per minute ?
If 82 per cent, of this energy given out is imparted to the shears
during a stroke of 2 inches, what is the average force due to this on
the blade of the shears ? (Advanced, 1902.)
14. A weight of 5 Ibs. is supported by a spring. The stiffness of
the spring is such that putting on or taking off a weight of I Ib.
produces a downward or upward motion of 0^04 foot. What is the
time of a complete oscillation, neglecting the mass of the spring ?
(Advanced, 1902.)
2/2
Mechanics for Engineers
15. A car weighs 10 tons : what is its mass in engineers' units?
It is drawn by the pull P Ibs., varying in the following way, / being
seconds from the time of starting : —
P ...
1020
980
882
720
702
650
7i3-
722
805
t
0
2
5
8 10
13
16
19
22
The retarding force of friction is constant and equal to 410 Ibs.
Plot P — 410 and the time /, and find the time average of this
excess force. What does this represent when it is multiplied by 22
seconds ? What is the speed of the car at the time 22 seconds
from rest? (Advanced, 1902.)
16. A body weighing 1610 Ibs. is lifted vertically by a rope,
there being a damped spring balance to indicate the pulling force
F Ibs. of the rope. When the body had been lifted x feet from its
position of rest, the pulling force was automatically recorded as
follows : —
X
F
0
4010
ii
3915
20
3763
34 1 45
3532 3366
55
3208
66
3100
76
3007
Using squared paper, find the velocity v feet per second for
values of x of 10, 30, 50, 70, and draw a curve showing the probable
values of v for all values of x up to 80. In what time does the
body get from x = 45 to x = 55 ? (Honours, Part I., 1901.)
17. A machine is found to have 300,000 foot-lbs. stored in it as
kinetic energy when its main shaft makes 100 revolutions per
minute. If the speed changes to 98 revolutions per minute, how
much kinetic energy has it lost ? A similar machine (that is, made
to the same drawings, but on a different scale) is made of the same
material, but with all its dimensions 20 per cent, greater. What
will be its store of energy at 70 revolutions per minute ? What
energy will it store in changing from 70 to 71 revolutions per
minute ? (Honours, Part I., 1901.)
1 8. A body of 60 Ibs. has a simple vibration, the total length of
a swing being 3 feet ; there are 200 complete vibrations (or double
swings) per minute. Calculate the forces which act on the body at
the ends of a swing, and show on a diagram to scale what force acts
upon the body in every position. (Advanced, 1901.)
19. An electric tramcar, loaded with 52 passengers, weighs
altogether 10 tons. On a level road it is travelling at a certain
Examination Questions 273
speed. For the purpose of finding the tractive force, the electricity
is suddenly turned off, and an instrument shows that there is a
retardation in speed. How much will this be if the tractive force is
315 Ibs. ? If the tractive force is found on several trials to be, on
the average —
342 Ibs. when the speed is 12 miles per hour
3J5 » » » I0 » »
294 » i> » 8 ?> »»
what is the probable tractive force at 9 miles per hour ?
(Advanced, 1901.)
2/4
LOGARITHMS.
0
1
2
3
4
5
6'
7
8
9
123 4
5
6789
10
0000
0043
0086
0128
0170
0212
0253
0294
0334
0374
4 9 13 17
4 8 12 16
21
20
25 30 34 38
24 28 32 37
11
12
0414
0792
0453
0828
0492
0864
0531
C899
0569
0934
0607
0969
0645
1004
0682
1038
0719
1072
0755
1106
4 8 12 15
4 7 11 15
3 7 11 14
3 7 10 14
19
19
18
17
2327 31 35
22 26 30 33
21 25 28 32
20 24 27 31
13
14
1139
1461
1173
1492
1206
1523
1239
1553
1271
1581
1303
1614
1335
1644
1367
1673
1399
1703
1430
1732
3 7 10 13
3 7 10 12
3 6 9 12
3 6 9 12
16
16
15
15
20 23 26 30
19 22 25 29
18 21 24 28
17 20 23 26
15
1761
1790
1818
1847
1875
1903
1931
1959
1987
2014
3 6 9 11
3 5 8 11
14
14
17 20 23 26
16 19 22 25
16
17
2041
2304
2068
2330
2095
2355
2122
2380
2148
24C5
2175
2430
2201
2455
2227
2480
2253
2504
2279
2529
3 5 8 11
3 5 8 10
3 5 8 10
2 5 7 10
14
13
13
12
16 19 22 24
15 18 21 23
15 18 20 23
15 17 19 22
18
19
2553
2788
2577
2810
2601
2833
2625
2856
2648
2878
2672
2900
2695
2923
2718
2945
2742
2967
2765
2989
2579
2 o 7 9
2479
2468
12
11
11
11
14 16 19 21
14 16 18 21
13 16 18 20
13 15 17 11*
20
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
2468
11
13 15 17 19
21
22
23
24
3222
3424
3617
3802
3243
3444
3636
3820
3263
3461
3655
3838
3284
3483
3674
3c(56
3304
3502
3692
3874
3324
3522
3711
3892
3345
3541
3729
3909
3365
3560
3747
3927
3385
3579
3766
3945
3104
355)8
3784
3962
2468
2468
2467
2457
10
10
9
9
12 14 16 18
12 14 15 17
11 13 15 17
11 12 14 16
25
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
2357
9
10 12 14 15
26
27
28
29
4150
4314
4472
4624
4166
4330
4487
4639
4183
4346
4502
4654
4200
4362
4518
4669
4216
4378
4533
4683
4232
4393
4548
4698
4249
4409
4564
4713
4265
4425
4579
4728
4281
444U
4594
4742
4298
4456
4609
4757
2357
2356
2356
1346
8
8
8
7
1011 13 15
9 11 13 14
9 11 12 14
9 10 12 13
30
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
1346
7
9 10 11 13
31
32
33
34
4914
5051
5185
5315
4928
5065
5193
5328
4942
5079
5211
5340
4955
5092
5224
5353
4969
5105
5237
5366
4983
5119
5250
5378
4997
5132
5263
5391
5011
5145
5276
5103
5024
5159
5289
5116
5038
5172
5302
5428
1346
1345
1345
1345
7
7
6
6
8 10 11 12
8 9 11 12
8 9 10 12
8 9 10 11
35
5441
5453
5165
5478
5490
5502
5514
5527
5539
5551
1245
6
7 9 10 11
36
37
38
39
5563
5682
579b
5911
5575
5694
6809
5922
5587
5705
5821
5933
5599
5717
5832
5944
5611
5729
5843
5955
5623
5740
5855
5966
5635
5752
5866
5977
5647
5763
5877
5988
5658
5775
5888
5999
5670
5786
5899
6010
1245
1235
1235
1234
6
6
6
5
7 8 10 11
7 8 !) 10
7 8 9 iO
7 8 9 10
40
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
1234
5
6 8 9 10
41
42
43
44
6128
6232
6335
6435
6138
6243
6345
6444
6149
6253
6355
6454
6160
6263
6365
6464
6170
6274
6375
6474
6180
6284
6385
6484
6191
6294
6395
6493
6201
6304
6405
6503
6212
6314
6415
6513
6222
6325
6425
6522
1234
1234
1234
1234
5
5
5
5
6789
6789
6789
6789
45
6532
6542
6551
6561
6571
65SO
6590
6599
6609
6618
1234
5
6789
46
47
48
49
6628
6721
6812
6902
6637
6730
6821
6911
6646
6739
6830
6920
6656
6749
6839
6928
6665
6758
6848
6937
6675
6767
6857
6946
6684
6776
6566
6955
6693
6785
6875
6964
6702
6794
6884
6972
6712
6803
6893
6a8l
1234
1234
1234
1234
5
5
4
4
6778
5678
5678
5678
50
6990
6998
7007
7016
7024
7033
7042
7050
7059
7067
1233
4
5678
LOGARITHMS.
275
0
1
2
3
4
5
6
7
8
9
123 4
5
6789
51
52
53
54
7076
7160
7243
7324
7084
7168
7251
7332
7093
7177
7*59
7340
7101
71«5
7267
7348
7110
7193
7275
7356
7118
7202
7284
7364
7126
7210
7292
7372
7135
7218
7300
7380
7143
7226
7308
7388
7152
7235
73i6
7396
1233
1223
122 3
1223
4
4
4
4
5678
5677
5667
5667
55
7404
7412
7419
7427
7435
7443
7451
7459
7466
7474
1223
4
5567
56
57
58
59
7482
7559
7634
7709
7490
7566
7642
7716
7497
7574
7649
7723
7505
7582
7657
7731
7513
7589
7664
7738
7520
7597
7672
7745
7528
7604
7679
7752
7536
7612
7686
7760
7543
7619
7694
7767
7551
7b27
7V01
7774
1223
1123
1123
4
4
4
5567
4 5 b 7
4567
60
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
1123
4
4566
61
62
63
64
7853
7924
7993
8062
7860
7931
8000
8U69
7868
7938
8007
8075
7875
7945
8014
8082
7882
7952
8021
8089
7889
7959
80/8
8096
7896
7966
8035
8102
7903
7973
8041
8109
7910
7980
8048
8116
7917
7987
8Uo5
8122
1123
1123
1123
1123
4
3
3
3
4566
4566
4556
4556
65
8129
8136
8142
8149
8156
8162
8169
8176
8182
818;*
1123
3
4556
66
67
68
69
70
71
72
73
74
8195
8261
8325
8383
8202
8267
8331
8395
8209
8274
8338
8401
8215
8280
8344
8407
8222
8287
8351
8414
8228
8293
8357
8420
8235
8299
8363
8426
8211
8306
837o
8432
8248
8312
8376
8139
8254
8319
8382
8445
1123
1123
1123
1122
3
3
3
3
3
4556
4556
4456
4456
8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
1122
4456
8513
8573
8633
8692
8519
8579
8639
8698
8525
8585
8645
8704
8531
8591
8651
8710
8537
8597
8657
8716
8543
8603
8663
8722
8549
8609
8669
8727
8555
8615
8675
8733
8561
862 L
3681
3739
8567
8627
8686
8745
H 1 2 2
1122
1122
1122
3
3
3
3
4455
4455
4455
4455
75
76
77
78
79
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
1122
3
3
3
3
3455
8808
8865
8921
8976
8814
8871
8927
8982
8820
8876
8932
8i>87
8825
8882
8938
8993
8831
8887
8943
8998
8837
8893
8949
'9004
8842
8899
8954
9009
8848
8904
8960
9015
8854
8910
8965
9020
8859
8915
8971
9025
1122
1122
1122
3455
3445
3445
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
112 2
3
3
3
3445
81
82
83
84
85
9085
9138
9191
9243
9090
9143
9196
92-48
9096
9149
9201
9253
9101
9154
9206
9258
9106
9159
9212
9263
9112
9165
9217
9269
9117
9170
9222
9274
9122
9175
9227
9279
9128
9180
9232
9284
9133
9186
9238
9289
1122
1122
3445
3445
1122
3
3445
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
1122
3
3
2
2
2
3445
86
87
88
89
9345
9395
9445
9494
9350
9400
9450
9499
9355
9405
9455
9504
9360
9410
9460
9509
9365
9415
9465
9513
9370
9420
9469
9518
9375
9425
9474
9523
9380
9430
9479
9528
9385
9435
9484
9533
9390
9440
9489
9538
1122
0 1 U2
0112
0112
3445
3344
3344
3344
3344
3344
3344
3344
3344
90
95i2
9547
9552
9557
9562
9566
9571
9576
9581
9586
0112
2
91
92
93
94
9590
9638
9685
9731
9595
9643
9689
9736
9600
9647
9694
9741
9605
9652
9699
9745
9609
9657
9703
9750
9614
966 L
9708
9754
9619
9666
8713
9759
9624
9671
9717
9763
9628
9675
9722
9768
9633
9680
9727
9773
0112
0112
0112
0112
2
2
2
2
95
9777
9782
97S6
9791
9795
9300
9805
9809
9814
9818
0112
2
3344
3344
3344
3344
3334
96
97
98
99
9823
9868
9912
9956
9827
9872
9917
9961
9832
9877
9921
9965
9836
9881
9926
9969
9841
9886
9930
9974
9845
9890
9934
9978
9850
9894
9939
9983
9854
9899
9943
9987
9859
9903
9948
9991
9863
9908
9952
9996
0112
0112
0112
0112
2
2
2
2
276
ANTILOGARITHMS.
0
1
2
3
4
5
6 '
7
8
9
1234
5
6789
•00
1000
1002
1005
1007
1009
1012
1014
1016
1019
1021
0011
i
1222
•01
1023
1047
1026
1050
1028
1052
1030
1054
1033
1035
1059
1038
1040
1042
1045
0011
i
1222
•03
•04
1074
1096
1074
1099
1076
110^
1079
1104
1081
1107
1084
1109
1086
1112
1089
1114
1091
1117
1094
1119
0011
0111
i
i
1222
2222
•05
1122
1125
1127
1130
1132
1135
1138
1140
1143
1146
0111
i
2222
•06
•07
•08
1148
1175
1202
1230
1151
1178
1205
1233
1153
1180
1208
1156
1183
1211
1159
1186
1213
1161
1189
1216
1164
1191
1219
1167
1194
1222
1169
1197
1225
1172
1199
1227
0111
0111
0111
i
i
i
2222
2222
2223
•10
1259
1262
1263
1268
1271
1274
1276
1279
1282
1235
0111
i
2223
•11
•12
•13
•14
1288
1318
1349
1380
1291
132L
1352
1384
1294
1324
1355
1387
1297
1327
1358
1390
1300
1330
1361
1393
1303
1334
1365
1396
1306
1337
1368
1400
1309
1340
1371
1403
1312
1343
1374
1406
1315
1346
1377
1409
0111
0111
0111
0111
2
2
2
2
2 2 2 3
2223
2233
2233
•15
1413
1416
1419
1422
1426
1423
1432
1435
1439
1442
0111
2
2233
•16
•17
•18
• 1 Q
1445
1479
1514
1549
1449
1483
1517
1552
1452
1486
1521
1455
1489
1524
1560
1459
1493
1528
1462
1496
1531
1567
1466
1500
1535
1469
1503
1538
1472
1507
1542
1476
1510
1545
0111
0111
0111
a
2233
2233
2233
•20
1585
1589
1592
1596
1600
1603
1607
1611
1614
1618
0111
2
2333
•21
•22
•23
•24
1622
1660
1698
1738
1626
1663
170:4
1742
1629
1667
1706
1746
1633
1671
1710
1750
1637
1675
1714
1754
1641
1679
1718
1758
1644
1683
1722
1762
1648
1687
1726
1766
1652
1690
1730
1770
1656
1694
1734
1774
0112
0112
0112
0112
2
2
2
2
2333
2333
2334
2334
•25
1778
1782
1786
1791
1795
1799
1803
1807
1811
1816
0112
2
2334
•26
•27
•28
•29
1820
1862
1905
1950
1824
1866
1910
1954
18>8
1871
1914
1959
1832
1875
1919
1963
1837
1879
1923
1968
1841
1884
1928
1972
1845
1888
1932
1977
1849
1892
1936
1982
1854
1897
1941
1986
1858
1901
1945
1991
0112
0112
0112
0112
2
2
2
2
3334
3334
3344
3344
•30
1995
2000
2004
2009
2014
2018
2023
2028
2032
2037
0112
2
3344
•31
•32
•33
.04
2042
2089
2138
2188
20461
2094
2143
2143
2051
2099
2148
2056
2104
2163
2061
2109
2158
2065
2113
2163
2070
2118
2168
2075
2123
2173
2080
2128
2178
2084
2133
2183
0112
0112
0112
2
2
2
3344
3344
3344
•35
2239
2244
2249
2254
2259
2265
2270
2275
2280
2286
1122
3
3445
•36
•37
•38
•39
2291
2344
2399
2455
2296
2350
2404
2460
2301
2355
2410
2466
2307
2360
2415
2472
2312
2366
2421
2477
2317
2371
2427
2483
2323
2377
2432
2489
2328
2382
2438
2495
2333
2388
2443
2500
2339
2393
2449
2506
1122
1122
1122
1122
3
3
3
3
3445
3445
3445
3455
•40
2512
2518
2523
2529
2535
2541
2547
2553
2559
2564
1122
3
4455
•41
•42
•43
•44
2570
2630
2692
2754
2576
2636
2698
2761
2582
2642
2704
2767
2588
2649
2710
2773
2594
2655
2716
2780
2600
2661
2723
2786
2606
2667
2729
2793
2612
2673
2735
2799
2618
2679
2742
2805
2624
2685
2748
2812
1122
1122
1123
1123
3
3
3
3
4455
4456
4456
4456
•45
2818
2825
2831
2838
2844
2851
2858
2864
2871
2877
1123
3
4556
•46
•47
•48
•49
2884
2951
3020
3090
2891
2958
3027
3097
2897
2965
3034
3105
2904
2972
3041
3112
2911
2979
3048
3119
2917
2985
3055
3126
2924
2992
3062
3133
2931
2999
3069
3141
2938
3006
3076
3148
2914
3013
3083
3155
1123
1123
1123
1123
<}
3
4
4
4556
4556
4566
4566
ANTILOGARITHMS.
27;
0
1
2
3
4
5
6
7
8
9
1 234
5
6789
4567
•50
3162
3170
3177
3184
3192
3199
3206
3214
3221
3228
1123
4
•51
•52
•53
•54
3236
3311
3388
3467
3243
3319
3396
3475
3251
3327
3404
3483
3258
3334
3412
3491
3266
3342
3420
3499
3273
3350
3428
3508
3281
3357
3436
3516
3289
3365
3443
3524
3296
3373
3451
3532
3304
3381
3459
3540
1223
1223
1223
1223
4
4
4
4
5567
5567
5667
6667
•55
3548
3556
3565
3573
3581
3589
3597
3606
3614
3622
1223
4
5677
•56
•57
•58
•59
3631
3715
3802
3890
3639
3724
3811
3899
3648
3733
3819
3908
3656
3741
3828
3917
3664
3750
3837
3926
3673
3758
3846
3936
3681
3767
3855
3945
3690
3776
3864
3954
3698
3784
3873
3963
3707
3793
3882
3972
1233
1233
1234
1234
4
4
4
5
5678
5678
5678
5678
•60
3981
3990
3999
4009
4018
4027
4036
4046
4055
4064
1234
5
6678
•61
•62
•63
•64
4074
4169
4266
4365
4083
4178
4276
4375
4093
4188
4285
4385
4102
4198
4295
4395
4111
4207
4305
4406
4121
4217
4315
4416
4130
4227
4325
4426
4140
4236
4335
4436
4150
4246
4345
4446
4159
4256
4355
4457
1234
1234
1234
1234
5
5
5
5
5
6789
6 7 8 i»
6789
6789
6789
•65
4467
4477
4487
4498
4508
4519
4529
4539
4550
4560
1234
•66
•67
•68
•69
4571
4677
4786
4898
4581
4688
4797
4909
4592
4699
4808
4920
4603
4710
4819
4932
4613
4721
4831
4943
4624
4732
4842
4955
4634
4742
4853
4966
4645
4753
4864
4977
4656
4764
4875
4989
4667
4775
4887
5000
1234
1234
1-234
1235
5
5
6
6
6 7 9 10
7 8 9 10
7 8 9 10
7 8 9 10
•70
5012
5023
5035
5047
5058
5070
5082
5G93
5105
5117
1245
6
7 8 9 11
•71
•72
•73
•74
5129
5248
5370
5495
5140
5260
5383
5508
5152
5272
5395
5521
5164
5284
5408
5534
5176
5297
5420
5546
5188
5309
5433
5559
5200
5321
5445
5572
5212
5333
5458
5585
5224
5346
5470
5598
5236
5358
5483
5610
1245
1245
1345
1345
6
6
6
6
7 8 10 11
7 9 10 11
8 9 10 11
8 9 10 12
•75
f>623
5636
5649
5662
5675
5689
5702
5715
5728
5741
1345
7
8 9 10 12
•76
•77
•78
•79
5754
5888
6026
6166
5768
5902
6039
6180
5781
5916
6053
6194
5794
5929
6067
6209
5808
5943
6081
6223
5821
5957
6095
6237
5834
5970
6109
3252
5848
5984
6124
6266
5861
5998
6138
6281
5875
6012
6152
6295
1345
1345
1346
1346
7
7
7
7
8 9 11 12
8 10 11 12
8 10 11 13
9 10 11 13
•80
6310
6324
6339
6353
6368
6383
6397
6412
6427
6442
1346
7
9 It) 12 13
•81
•82
•83
•84
6457
6607
6761
6918
6471
6622
6776
6934
6486
6637
6792
6950
6501
6653
6808
6966
6516
6668
6823
6982
6531
6683
6839
6998
6546
6699
6855
7015
6561
6714
6871
7031
6577
6730
6887
7047
6592
6745
6902
7063
2356
2356
2356
2356
8
8
8
8
9 11 12 14
9 11 12 14
9 11 13 14
10 11 13 15
•85
7079
7096
7112
7129
7145
7161
7178
7194
7211
7228
2357
8
10 12 13 15
•86
•87
•88
•89
7244
7413
7586
7762
7261
7430
7603
7780
7278
7447
7621
7798
7295
7464
7638
7816
7311
7482
7656
7834
7328
7499
7674
7852
7345
7516
7691
7870
7362
7534
7709
7889
7379
7551
7727
7907
7396
7568
7745
7925
2357
2357
2457
2457
8
9
9
9
10 12 13 15
10 12 14 16
11 12 14 16
11 13 14 16
•90
•91
•92
•93
•94
7943
7962
7980
7998
8017
8035
8054
8072
8091
8110
2467
9
9
10
10
10
11 13 15 17
8128
8318
8511
8710
8147
8337
8531
8730
8166
8356
8551
8750
8185
8375
8570
8770
8204
8395
8590
8790
8222
8414
8610
8810
8241
8433
8630
8831
8260
8453
8650
8851
8279
8472
8670
8872
8299
8492
86HO
8892
2468
2468
2468
2468
11 13 15 17
12 14 15 17
12 14 16 18
12 14 16 18
•95
8913
8933
8954
8974
8995
9016
9036
9057
9078
9099
2468
10
12 15 17 19
•96
•97
•98
•99
9120
9333
9550
9772
9141
9354
9572
9795
9162
9376
9594
9817
9183
9397
9616
9840
9204
9419
9638
9863
9226
9441
9661
9886
9247
9462
9683
9908
9268
9484
9705
9931
9290
9506
9727
9954
9311
9528
9750
9977
2468
2479
2479
2579
11
11
11
11
13 15 17 19
13 15 17 20
13 16 18 20
14 16 18 20
2/8
A
ngle.
Pn
De-
grees.
Radians.
Chord.
Sine.
Tangent.
tangent.
Cosine
0°
0
000
0
0
8
1
1-414
1-5708
90°
1
2
3
4
•0175
•0349
•0524
•0698
•017
•035
•052
•070
•0175
•0349
•0523
•0698
•0175
•0349
•0524
•0699
57-2900
28-6363
19-0811
14-3007
•9998
•9994
•9986
•9976
1-402
1-389
1-377
1-364
1-5533
1-5359
1-5184
1-5010
89
88
87
86
5
•0873
•087
•0872
•0875
11-4301
•9962
1-351
1-4835
85
6
7
8
9
•1047
•1222
•1396
•1571
•105
•122
•140
•157
•1045
•1219
•1392
•1564
•1051
•1228
•1405
•1584
9-5144
8-1443
7-1154
6-3138
•9945
•9925
•9903
•9877
1-338
1-325
1-312
1-299
1-4661
1-4486
1-4312
1-4137
84
83
82
81
10
•1745
•174
•1736
•1763
5-6713
•9848
1-286
T3963
80
11
12
13
14
•1920
•2094
•2269
•2443
•192
•209
•226
•244
•1908
•2079
•2250
•2419
•1944
•2126
•2309
•2493
5-1446
4-7046
4-3315
4-0108
•9816
•9781
•9744
•9703
1-272
1-259
1-245
1-231
1-3788
1 3614
1-3439
1-3265
79
78
77
76
15
•2618
•261
•2588
•2679
3-7321
•9659
1-218
1-3090
75
16
17
18
19
•2793
•2967
•3142
•3316
•278
•296
•313
•330
•2756
•2924
•3090
•3256
•2867
•3057
. '3249
•3443
3-4874
3*2709
3-0777
2-9042
•9613
.'9563
'•9511
•3455
1-204
1-190
1-176
1-161
1-2915
1-2741
1-2566
1-2392
74
73
72
71
20
•3491
•347
•3420
•3640
2-7475
•9397
1-147
T2217
70
21
22
23
24
•3665
•3840
•4014
•4189
•364
•382
•399
•416
•3584
•3746
•3907
•4067
•3839
•4040
•4245
•4452
2-6051
2-4751
2-3559
2-2460
•9336
•9272
•9205
•9135
1-133
1-118
1-104
1-089
1-2043
1-1868
1-1694
1-1519
69
68
67
66
25
•4363
•433
•4226
•4663
2-1445
•9063
1-075
1-1345
65
26
27
28
29
•4538
•4712
•4887
•5061
•450
•467
•484
•501
•4384
•4540
•4695
•4848
•4877
•5095
•5317
•5543
2-0503
1-9626
1 8807
1-8040
•8988
•8910
•8829
•8746
1-060
1-045
1-030
1-015
1-1170
1-0996
1-0821
1-0647
64
63
62
61
30
, '5236
•518
•5000
•5774
1-7321
•8660
1-000
1-0472
60
31
32
33
34
•5411
•5585
•5760
•5934
•534
•551
•568
•585
•5150
•5299
•5446
•5592
•6009
•6249
•6494
•6745
1-6643
1-6003
1-5399
1-4826
•8572
•8480
•8387
•8290
•985
•970
•954
•939
1-0297
1-0123
•9948
•9774
59
58
57
56
35
•6109
•601
•5736
•7002
1-4281
•8192
•923
•9599
55
36
37
38
39
•6283
•6458
•6632
•6807
•618
•635
•651
•668
•5878
•6018
•6157
•6293
•7265
•7536
•7813
•8098
1-3764
1-3270
1-2799
1-2349
•8090
•7986
•7880
•7771
•908
•892
•877
•861
•9425
•9250
•9076
•8901
54
53
52
51
40
•6981
•684
•6428
•8391
1-1918
•7660
•845
•8727
50
41
42
43
44
•7156
•7330
•7505
•7679
•700
•717
•733
•749
•6561
•6691
•6820
•6947
•8693
•9004
•9325
•9657
1-1504
1-1106
1-0724
1-0355
•7547
•7431
•7314
•7193
•829
•813
•797
•781
•8552
•8378
•8203
•8029
49
4$
4?
46
45°
•7854
•765
•7071
1-0000
1-0000
•7071
•765
•7854
45°
Cosine.
Co-
tangent.
Tangent.
Sine.
Chord.
Eadians.
De-
greesu
Angl
INDEX
( The numbers refer to pages)
Acceleration, 3
Adhesion, 107
Alternating vectors, 81
Amplitude, 79
Angular acceleration, 23
momentum, 207
• motion, 23
— velocity, 23
Atwood's machine, 44
Average force (space), 51
force (time), 35
B
Bending moment, 231
diagram, 233
Bicycle, centre of gravity, 184
Bows' notation, 223
Brakes, 107
Centre of gravity, 141-165
of mass, 141
of parallel forces, 140
Centrifugal force, 69, 181
Centripetal force, 69
C.g.s. units, 30
Chains, loaded, 243
Circular arc, 160
motion, 68
sector and segment, 161
Coefficient of adhesion, 107
of friction, 100
Compound pendulum, 212
Conditions of equilibrium, 97, 128,
226
Conical pendulum, 72
Couple, 125
Curve, motion on, 70, 71
Density, 27
Derived units, 253
Displacement curve, 2
, relative, 16
Distributed load, 168, 247
Efficiency of machines, 1 10
of screw, 1 08
Energy, 57
in harmonic motion, 84
, kinetic, 58
Equilibrant, 92
Equilibrium, conditions of, 97, 128,
226
, stability of, 172
280
Mechanics for Engineers
(The numbers refer to pages]
First law of motion, 27
Force, 27, 29
Forces, coplanar, 127
, parallel, 114
, resolution and composition
of, 91
, triangle and polygon of, 33, 91
Frames, 236
Friction, 99
, angle of, 101
, coefficient of, 100
, laws of, 100
of machines, 1 10
of screw, 108
, sliding, 100
, work spent in, 107
Fundamental units, 253
Funicular polygon, 224, 228, 233,
243
Gravitational units, 30, 253
Gravity, acceleration of, 6
Guldinus, 182
H
Harmonic motion, 79
Hemisphere, 162, 172
Horse-power, 51
I
Impulse, 33
Impulsive force, 36
Inclined plane, 102
, smooth, 22
Indicator diagram, 5°
Inertia, 27
Inertia, moment of, 188
, (areas), 194
K
Kinematics, Chapter I.
Kinetic energy, 58
of rotation, 204
of rolling body, 2
Lami's theorem, 93
Laws of motion, Chapter II.
Levers, 122
Lifting, work in, 176
Limiting friction, 100
Load, distributed, 168
Locomotive, centre of gravity, 186
M
Machines, HO
Mass, 27
Mechanical advantages of screw,
109
Method of sections, 133
Moment, 53, 119-122
of an area, 157
of inertia, 188
of areas, 194
of momentum, 207
Momentum, 28
Motion, first law of, 27
of connected weights, 43
, second law of, 28
, simple harmonic, 79
, third law of, 41
Motor-car, centre of gravity, 186
N
Neutral equilibrium, 172, 174
Newton's laws of motion, 27
Index
281
( The numbers refer to pages]
Pappus, 182
Parallel axes, moment of inertia
about, 191, 192
forces, 114
Pendulum, compound, 212
— , conical, 72
, simple, 85
, simple equivalent, 86, 213
Plane-moments, 155
Polygon offerees, 33, 91
, funicular or link, 224
-of velocities, 17
Pound, unit of force, 29, 253
Poundal, 29
Power, 51
Principle of moments, 122
of work, 59
R
Radius of gyration, 189
Railway curve, 71
Reduction of forces, 127
Relative displacement, 16
velocity, 20
Resolution of accelerations, 22
— of forces, 91
of velocity, 18
Rolling body, 217
Roof, 242
Rotation about axis, 179, 204, 207
S
Screw friction, 108
Second law of motion, 28
Sections, method of, 133
Sector of circle, 161
of sphere, 162
Segment of circle, 161
Shearing force, 231
Shearing-force diagram, 235
Simple equivalent pendulum, 86,
213
harmonic motion, 79
, torsional, 214
— pendulum, 85
Smooth body, 129
Space-average force, 51
curve, 2
diagram, 223
Spherical shell, 161
Spring, vibrating, 83
Stable equilibrium, 172
Statics, 91
Stress diagram, 239
, tensile and compressive, 240
String, loaded, 243
polygon, 242
Strut, 240
Theorem of Guldinas or Pappus,
182
Third law of motion, 41
Tie, 240
Time-average force, 35, 36
Torque, 54
Torsional oscillation, 214
Triangle of forces, 33, 91
of velocities, 17
Twisting moment, 54
U
Uniform circular motion, 63
Units, 253
Unstable equilibrium, 172
Vector diagram, 223
Vectors, 15
Velocity, I
282
Mechanics for Engineers
( The numbers refer to pages]
Velocity, angular, 23
, component, 18
curves, 7
— , polygon of, 17
, relative, 20
Vertical circle, motion in, 73
motion, 6
Vibration of spring, 83
W
Warren girder, 133, 241
Weight, 28
Work, 48
in lifting, 176
— of a torque, 54
, principle of, 59
THE END
I'RtNTED BY WILLIAM CLOWES AND SONS, LIMITED, LONDON AND BECCLES.
TEES BOOK IS DUE ON THE LAST DATE
STAMPED BELOW
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OVERDUE.
*.* 75
MAR 13 1942
120ct'51LR
23WIar'56K-0
MAR 9 1956 I
24Nov'57BP
REC'D LD
NOV U 1957
INTERLIBRAR
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UNIV. OF CAL
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