Seely F.B. Analytical Mechanics for Engineers
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KINETICS
OF PLANE MOTION
OF RIGID
BODY 341
From
(2),
From
(3),
-805
cos
30+N
=
0,
since
a
y
=Q.
3
1
805 /3
2* 2
X
(5)
(6)
There are four
unknown
quantities
involved
in these
three
equations
and
hence another
equation
is
needed.
Fro
on the kinematics
of the
problem
(Art.
121),
we obtain
the
equation,
By
substituting
the
value
of a from
(7)
in
(6)
and
solving
for
F,
we
obtain the
equation,
Substituting
this value
of F in
(4),
we
obtain,
a
x
=a
=
W.73 ft.
/sec.
2
Hence,
7^=^X10.73
=
134.1
Ib.
Since
the
mass-center
moves
with
uniformly
accelerated rectilinear
motion,
we
may
use
the
equation,
v
Hence,
=
50
+
10.73X10
=
157.3
ft./sec.
386.
The
connecting
rod
of
a
steam
(engine
has a
length
Z=6
ft.
and a
weight
TF
=
220 Ib.
(Fig.
361).
The
length
of the
crank is
r
=
l
ft. and
the
FIG. 361.
engine
runs at a
speed
of
on
=300
r.p.rn.
The
pressure
P
on the
cross-
head,
when
0=30,
is
18,000
Ib. Assume
that the
cross-section of
the
con-
necting
rod is
constant.
Find
the
tangential
crankpin pressure,
Q,
the com-
342
FORCE,
MASS,
AND
ACCELERATION
pression,
S,
in
the
crank
arm,
and the
guide
pressure, N,
which is
assumed
to
act
vertically
(friction
neglected).
Solution.
The
connecting
rod has
plane
motion. Either
equations
(1)
or
equations
(2)
of
Art.
149
may
be
used. Let
equations
(1)
be
chosen and let
the
crosshead end of
the
connecting
rod be selected as the
origin. Further,
let
the
x-
and
y-axes
be chosen horizontal and
vertical
respectively.
The
equations
of
motion then
are,
.......
(1)
.......
(2)
'2T
A
=I
A
a+My(a
A
)
x
-Mx(a
A
) v
,
......
(3)
in
which,
990
(a.0,=0,
Mx=a
A
,
M
=
^~
=6.84
slugs,
7
A
=*JfP-82
slug-ft.
O^i.Z,
Further,
Hence,
Therefore,
sin
30=T
L
2
-.
I sin
(f>
=
r
sin
or
sin
0=4
47'.
sin
=
.0833,
x
=
3 cos
<
=
2.99
ft.,
cos
<f>
=
.9965,
y
=
3
sin
=
0.25
ft.
, a>,
and
a
may
be
found
by
use
of the
semi-graphical
method
discussed
in
Scales
:-
l"
2ft.
l"
30
ft./sec.
1"-400
ft./sec.
2
a
A
936
(scaled)
(4-V
j
r-
FIG.
362.
Prob.
(334).
Thus,
in
Fig.
362,
the
following
two
equations,
(4)
and
(5),
are
solved.
KINETICS
OF
PLANE
MOTION
OF
RIGID
BODY 343
Bl
t
\
V
A)
V
A
=
V
B)
/a
A
=
/O
B
\
and
a
A
are
found,
by scaling
off their
B
A'
\B)
i
\ A)
t
lengths,
to
have
the
following
values:
V
A
=
18
ft./sec.,
^=27.4ft./sec.,
(a^\
=492 ft.
/sec.
2
,
a
A
=936ft./sec.
2
A
\
A"
But,
A
^
A
'
1
Hence,
27 4 492
co=^=4.56rad./sec.
and,
=
-^-
=82 rad./sec.
2
.
6
o
From
(1),
18,000-S
cos
30-Q
cos 60
=
6.84
(936
+0.25
X
82-2.99
X4^56
2
).
From
(2),
N+S
sin
30-Q
sin 60-220=6.84
(0
-2.99X82
-0.25
X4T56
2
)-
From
(3),
-3.42
+4.93
Q
+2.99X220
=
82X82
+6.84X0.25X936.
Solving
for
the three
unknown
forces,
we obtain the
following
values:
Q
=
79501b.,
=9150
lb.,
JV
=
795
Ib.
PROBLEMS
387. A
homogeneous
solid
sphere
rolls
without
slipping
down a
reugh
plane
which
is
inclined
at
an
angle
6
with
the
horizontal.
Show
that
the
acceleration
of
the center
of the
sphere
is
f#
sin
6,
and
that the
ratio
of the
friction
to
the
normal
pressure
must
be
not
less
than
I
tan
6 to
prevent
the
sphere
from
slipping.
344
FORCE,
MASS,
AND
ACCELERATION
388. A solid
sphere
having
a radius
of
8
in. and
a
weight
of
161 Ib.
is
made
to roll
up
a
rough
inclined
plane
(Fig.
363)
by
means of a
flexible
cord,
one
end
of
which is attached
to an axis
through
the center
of
the
sphere.
The
cord
passes
over
a smooth
peg
and has attached to
its other
end
a
suspended
body
B
which
weighs
100 Ib.
Find the
acceleration of
the
body
B
and the
tension in the
cord.
Ans. a
=
1.93 ft.
/sec.
2
FIG.
363.
FIG.
364.
389.
A
homogeneous
cylinder
1
ft. in
diameter
has a flexible cord
wrapped
around
its central
plane.
One end of the
cord is
attached to a fixed
plane
as
shown
in
Fig.
364.
The
cord is taut when the
cylinder
is
allowed to
fall.
Find
(a)
the acceleration
of the
mass-center,
(6)
the
angular
acceleration
of the
cylinder,
and
(c)
the distance traveled
by
the mass-center
in 10
sec.
390. Solve
Prob. 386
by
use of
equations
(2)
of Art.
149.
391.
A
uniform
rod
AB
(Fig.
365)
moves
with
its ends
B
and
A
in
contact
with smooth
planes
which
are
vertical
and horizontal
respectively.
A
variable
horizontal
force, F,
is
applied
at
the end
A. The
weight
of
the rod
is 16.1
Ib.
and its
length
is
8 ft. If the
value of F
for the
position
of the
rod shown
(0
=
60)
is such
that
the
angular
acceleration, a,
of the rod
is
3
rad./sec.
2
,
and the
angular
velocity,
w,
is
2
rad./sec.,
what
are the
values of
F,
NA,
and
NB?
Ans.
2V
4
=
6.171b.;
F=-.031b.;
tf
a-
1.22
ft).
392.
Two solid
cylindrical
discs are
keyed
to an
axle as
shown in
Fig.
366.
A
string
is
wrapped
around
the axle in
its central
plane
and
a
force,
P,
is
exerted
by
the
string
in
a
direction
SECOND
METHOD
OF
ANALYSIS
345
parallel
to the
plane
on which
the
discs roll
and
tangent
to the
under
surface
of
the
axle.
Each
disc
weighs
20 Ib. and
is
2
ft. in
diameter.
The axle
is
6
in.
in
diameter
and
weighs
40
Ib.
The
magnitude
of
the
force
P is
8
Ib.
\\
, 11 the
discs
and axle
roll forward or
backward?
Find
the
acceleration
of
the
central
axis
of the
discs
and axle. Ans.
a
=
1.91
ft.
/sec.
2
393.
If
the
string
in the
pre-
ceding problem
is
wrapped
around
the
axle
in
the
opposite
direction
so
that
the
force
P
is
tangent
to the
top
of
the
axle,
what
is the acceler-
ation
of
the
central
axis of
the
discs and
axle?
394.
A
wheel
has
an
eccentric
weight
attached
near its
rim
so
that
its
mass-center
is
not the
geometric
center
of
the wheel.
If the wheel FIG.
366.
rolls
on
a
horizontal
track,
will
the
wheel
leave
the
track
when
its
velocity
attains some
particular
value?
150. Second
Method
of
Analysis.
Inertia
Forces. In
some
problems
which
deal with
the
plane
motion
of a
rigid
body
under
the
action
of an
unbalanced
force
system,
it is
convenient
to
assume
that
the
reversed
resultant
of the
effective
forces
(inertia
force for
the
body)
acts on
the
body
with
the external
forces,
thereby
forming
a force
system
that
is
in
equilibrium
and thus
reducing
the
kinetics
problem
to
an
equivalent
statics
problem.
In
order
to
use
this
method
of
analysis,
the location
of the
action
line of
the resultant
of
the
effective
forces
must
be
known,
as
well
as the
magnitude
and
sense
of the
resultant
force.
Equations
(1)
of
the
preceding
article
state
that
the
resultant
of
the effective
forces for the
body
(and
hence
also
of
the
external
forces
which
act
on the
body)
is a
force
having
x-
and
^-components
equal
to
M(ao)
x
+Mya
Mxco
2
and
M(ao)
y
Mxa
Mya?,
respect-
ively,
as shown
in
Fig.
367
(a),
and that
the
moment
of
the
resultant
force
with
respect
to the axis
through
the
origin
is
equal
to
Ioa+My(ao)x
Mx(ao)v
.
Equations
(2)
state
that
the resultant
of
the
effective
forces for
the
same
body
is
a
force
having
x-
and
y-
components
equal
to
Max
and
M
a
v ,
respectively,
as
shown in
Fig.
367(6),
and
that
the moment
of
the*
resultant
force with
respect
to
an
axis
through
the mass-center of
the
body
is
Ja.
The resultant
of
the effective
forces
for
any
given
body
as
obtained from each
set
346
FORCE,
MASS,
AND
ACCELERATION
of
equations
is,
of
course,
the same force
1
and its
components
may
intersect
at
any point
on its action line such as
points
A
and A'
in
Fig.
367
(a)
and
(6).
The action line of the
resultant force as
located
by
its moment
arm
q
in
Fig.
367
(a)
or
q
in
Fig.
367(6)
may
be
found
from
the
principle
of moments
by
equating
the
moment of the resultant
of
FIG.
367.
the effective
forces to the sum of the
moments of
the
effective
forces.
The sum
of the moments of
the
effective
forces
about
the
mass-center,
as
already
found,
is
Ja.
And,
the
moment
of
the
resultant force
having
the
components
Ma
x
and
Ma
v
is
Ma-q
(Fig.
3676)
. Thus
the
principle
of
moments
is
expressed
by
the
equation,
Ma-q
=
Ja.
But
7
Therefore,
Ma
1
Thus we
may
write,
M(a
)
x
Or, canceling
the
M's,
we
have the
equations dealing
with
accelerations
as
follows,
which
state
that
the
acceleration of
the
mass-center is
equal
to
the
relative
acceleration
of
the
mass-center with
respect
to
any
axis,
0, plus
the
accelera-
tion
of
O;
a
proposition
which is
discussed in
Art.
128,
SECOND
METHOD OF
ANALYSIS
347
The value
of
qo
(Fig.
367a) may
be found in
a similar
manner.
However,
it is more
convenient,
in
general,
to
resolve
the
single
resultant
force
Ma, having
an
action
line which is located
at the
k
2
a
-
from the
mass-center,
into another force
equal
a
distance
q
and
parallel
to
Ma,
but with
its line of action
passing
through
the
mass-center,
and
a
couple, C, having
a moment
equal
to
Ma-q
(Art.
18)
as shown
in
Fig.
368
(a). But,
as
shown
above, Ma-q
=
Ia
and hence
the
moment
of
the
couple
is
la.
w
(W
FIG.
368.
Therefore,
the resultant of
the
effective forces for
a
rigid
body
having plane
motion
may
be considered to be a force
Ma
having
an action
line
which
passes
through
the mass-center of
the
body
in the
direction of
the
acceleration, a,
of the
mass-center,
and a
couple
having
a
moment
equal
to
la,
the sense of
the
couple being
the
same as
that of
the
angular
acceleration,
,
of
the
body.
The
above facts
together
with D'Alembert's
principle
lead
to a state-
ment
or
proposition
which is of
great
importance
in
the
method
of
reducing
the
kinetics
problem
of
a
rigid body
having plane
motion to an
equivalent problem
in
equilibrium by
introducing
the inertia
force
(or
forces)
for
the
body.
The
proposition may
be stated as
follows:
If a
rigid body
is acted
upon
by
an
unbalanced
force
system
which
gives
to
the
body
a
plane
motion,
the
body may
be
con-
sidered
to
be
in
equilibrium
if
an
additional force and
couple
is
assumed to
act
on the
body
with
the
external
forces;
the addi-
tional
force has
a
magnitude
equal
to
Ma,
its line of
action
348
FORCE,
MASS,
AND
ACCELERATION
passes
through
the mass-center
of
the
body,
and its
direction
is
opposite
to that of
a;
the
couple
has a moment
equal
to
la
and its
direction of
rotation
is
opposite
to that of a.
The
forces of
the
couple
are
not
necessarily
equal
to
Ma
or
parallel
to
a
as
shown
in
Fig.
368
(a),
that
is,
the
couple
may
be
rotated in
the
plane
of motion and the
magnitude
of
the
forces
of
the
couple may
be
changed provided
the moment
of
the
couple
is
always
equal
to
la
(Art. 27).
Thus,
if a force
having
a
magni-
tude
equal
to
Ma,
and
a
couple having
a moment
equal
to
Ma
q
(or
la),
act on the
body
with the external forces
(P
and
W),
as
shown
in
Fig. 368(6),
the
force
system (and
the
body)
would be
in
equilibrium.
ILLUSTRATIVE PROBLEMS
396.
At what
height
should the cushion on a billiard table be
placed
so
that
the billiard
ball in
rebounding
from the
cushion starts
off
without
causing
any
friction on
the
table
top?
Solution. The
billiard ball has
plane
motion and while in
contact
with the
cushion
it is
acted
upon
by
three
forces;
the cushion
pressure
P,
the
weight
W,
and the
table
pressure
N,
as shown
in
Fig.
369.
If
there is no
horizontal friction
force at
A,
then
P
is the resultant
of
the
external
forces
which
act
on
the ball
and it must be
collinear
with the
resultant
of the
effective forces.
The re-
sultant of the effective forces
is a
force
of
magnitude Ma,
the distance of the action
line
from the mass-center
being given
P
by
the
equation,
And since
a
=
ra,
we have
fr
2
2
Hence,
FIG. 369.
396.
Solve
Prob.
386
by introducing
the reversed resultant of the
effective
forces
(inertia
force for
the
rod).
Solution.
The resultant
of
the effective forces is a force of
magnitude
Mo
acting
through
the
mass-center
of the
rod
in
the direction
of
a,
and a
couple
of moment
la,
as
shown in
Fig.
370
(a).
If
G denotes the
mass-center,
then
SECOND
METHOD
OF ANALYSIS
349
a
may
be written
ao
for
convenience.
OG
may
be found from
the
graphic
equation,
_
in which
t-
G
=
\v
A
=
13.7 and
a
=
82.
(See
Prob.
386.)
Hence,
~B B
a
o
fi~2!
4>
3
X
82
4>
986
o
=
62.6
4>
246
4>
986
=
930
ft./sec.
2
,
as scaled
from
the
acceleration
polygon
of
Fig.
370(a).
l"=2f.
l'=400
ft./sec?
Therefore the resultant of
the
effective
forces
is
a
force,
M
and
a
couple,
220
Ma
=X930
=
6360
lb.,
But,
168.1
Ib.-ft.
=
Ma-q
or
168.1=6360g.
.'.
3
=
0.
264
ft.
350
FORCE, MASS,
AND ACCELERATION
Hence
q
=
.264
ft.,
and the
inertia force
for
the rod
is
a
single
force
of
magnitude
6360
Ib.
as
shown
in
Fig. 370(&).
If the inertia force is
assumed to act on the
rod with
the
external
forces,
the rod
may
be
considered to
be in
equilibrium.
The three
unknown forces
may
be found
graphically by
use of the
force and
funicular
polygons
or,
the values
of
/3
and
e
may
be scaled off
and
the
equations
of
equilibrium
used.
The
values
of
ft
and
e
are
found
to be
/3
=
15
35'
and
e
=
0.77
ft.
The
equations
of
equilibrium
for the
non-concurrent forces
in
a
plane
as
shown
in
Fig.
370(6)
are,
2/^
=
0,
18,000-6360
cos
p-S
cos 30-
Q
cos 60=
0,
.
.
(1)
SF
V
=0,
N
-
220
+6360
sin
ft
+S
sin
30
-Q
sin
60
=0,
. .
(2)
SM*=0, 18,000?
sin
<t>-Nl
cos
^,+220X2.99-6360X0.77
=
0,
(3)
in
which
sin
<
and
cos
<f>
have
the same values
as
were
used in Prob.
386.
Thus,
sin
=
.0833,
cos
=
.9965,
sin
/3=sin
15 35'
=
.268,
cos
/3
=
.963,
1
=
6 ft.
Substituting
the
known values
in
the above
equations,
we obtain the
following
equations:
.5Q+.866S-11,875
=
0,
.866Q-.5S-
2282=0,
-5.98N
+4758=0,
the
solution
of which
gives
the
following
values
for the
three unknown
forces:
Q
=
79701b.;
S
=
91001b.;
N
=
7971b.
PROBLEMS
397.
A
solid
homogeneous
disc 3 ft.
in
diameter rolls on
a
straight
hori-
zontal
track
(Fig.
371).
A small
body, B,
is attached to the
disc at a
distance
d
=
0.75 ft.
from
the center
of the
disc.
When
in
the
position
shown in
Fig.
371
the
angular
velocity
of the
disc
is 6 rad.
per
sec.,
and
a force
P is
retarding
the
angular
velocity
at the rate of 4 rad.
per
sec. each second.
The
disc
weighs
60
Ib.
and
B
weighs
8 Ib. Find
the
"f"?
f r
<?
f r *he dfec and for the
body
B.
Also
find the value
of P
and
of the friction
f
orc
e. Ans. P
=
7.91
Ib.;
^
=
1.96
Ib.
FIG. 371.
398.
The
resultant,
R,
of the
forces
acting
on the
connecting
rod
(Fig.
372)
is
320
Ib.
and
its
action line is
located
as
shown.
If the
connecting