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PROBLEM 19.9
A 10-lb block A rests on a 40-lb plate B which is attached to an unstretched
spring of constant k = 60 lb/ft. Plate B is slowly moved 2.4 in. to the left
and released from rest. Assuming that block A does not slip on the plate,
determine (a) the amplitude and frequency of the resulting motion, (b) the
corresponding smallest allowable value of the coefficient of static friction.
SOLUTION
Simple Harmonic Motion:
2
60 lb/ft 6.2161 rad/s
50 lb
32.2 ft/s
n
k
m
ω
== =
6.2161 Hz
22
n
n
f
ω
ππ
==
2
32.2 ft/s
s
g
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.10
A 5-kg fragile glass vase is surrounded by packing material in a cardboard box of negligible weight. The
packing material has negligible damping and a force-deflection relationship as shown. Knowing that the box
is dropped from a height of 1 m and the impact with the ground is perfectly plastic, determine (a) the
amplitude of vibration for the vase, (b) the maximum acceleration the vase experiences in g’s.
SOLUTION
Velocity at end of free fall:
2
2
(2)(9.81 m/s )(1 m) 4.4294 m/s
vgh
v
=
==
Assume that the spring is unstretched during the free fall. Use a simple spring-mass model for the motion of
the vase and the packing material.
5kg
100 N (slope from graph)
10 mm
10 N/m 10000 N/m
m
k
k
=
=
==
Natural frequency: 10000 N/m 44.721 rad/s
5kg
n
k
m
ω
== =
Simple harmonic motion: sin( )
cos( )
mn
nm n
xx t
vx x t
ωφ
ω
ωφ
=+
== +
Let t = 0 at the instant when the box bottom hits the ground.
Then, at t = 0, x = 0 and v = 4.4294 m/s
from which 0
φ
=
and 4.4294 m/s
nm
x
ω
=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.10 (Continued)
(a) Amplitude:
4.4294 m/s 0.099045 m
44.721 rad/s
m
x==
99.0 mm
m
x=
(b) Maximum acceleration:
22
22
(44.721 rad/s) (0.099045 m)
198.087 m/s (20.192)(9.81 m/s )
mnm
ax
ω
==
==
20.2
m
ag=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.11
A 3-lb block is supported as shown by a spring of constant 2 lb/in.k
=
which can act in
tension or compression. The block is in its equilibrium position when it is struck from below
by a hammer which imparts to the block an upward velocity of 90 in./s. Determine (a) the
time required for the block to move 3 in. upward, (b) the corresponding velocity and
acceleration of the block.
SOLUTION
Simple harmonic motion. sin( )
mn
xx t
ω
φ
=+
Natural frequency. , 2 lb/in. 24 lb/ft
n
kk
m
ω
===
()
2
3 lb
32.2 ft/s
24 lb/ft
16.05 rad/s
(0) 0 sin(0 )
0
(0) cos(0 0)
90
(0) 7.5 ft/s
12
n
n
m
mn
xx
xx
x
ω
ω
φ
φ
ω
=
=
== +
=
=+
==
7.5 (16.05) 0.4673 ft
(0.4673)sin(16.05 )(ft/s)
mm
xx
xt
==
= (1)
(a) Time at 3 in. ( 0.25 ft)xx==
()
10.25
0.4673
0.25 0.4673sin(16.05 )
sin
16.05
t
t
−
=
= 0.0352 st=
(b) Velocity and acceleration.
2
cos( )
sin
0.0352
(0.4673)(16.05)cos[(16.05)(0.0352)]
6.34 ft/s
mn n
mn n
xx t
xx t
t
x
x
ω
ω
ωω
=
=−
=
=
=
6.34 ft/s=v
2
2
(0.4673)(16.05) sin[(16.05)(0.0352)]
64.4 ft/s
x=−
=−
2
64.4 ft/s=a
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.12
In Problem 19.11, determine the position, velocity, and acceleration of the block 0.90 s after it has been struck
by the hammer.
SOLUTION
Simple harmonic motion. sin( )
mn
xx t
ω
φ
=
+
Natural frequency.
()
2
3 lb
32.2 ft/s
, 2 lb/in. 24 lb/ft
24 lb/ft
16.05 rad/s
n
n
n
kk
m
ω
ω
ω
===
=
=
(0) 0 sin(0 )
0
90
(0) cos(0 0) (0) 7.5 ft/s
12
7.5 (16.05) 0.4673 ft
(0.4673)sin(16.05 )(ft/s)
m
mn
mm
xx
xx x
xx
xt
φ
φ
ω
== +
=
=+==
==
=
Simple harmonic motion.
2
sin( )
cos( )
sin( )
mn
mn n
mn n
xx t
xx t
xx t
ωφ
ω
ωφ
ω
ωφ
=+
=+
=
−+
At 0.90 s: (0.4673)sin[(16.05)(0.90)] 0.445 ftx
=
= 0.445 ft
=
x
(0.4673)(16.05)cos[(16.05)(0.90)] 2.27 ft/sx
=
=−
2.27 ft/s
=
v
22
(0.4673)(16.05) sin[(16.05)(0.90)] 114.7 ft/sx=− =−
2
114.7 ft/s=a
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.13
The bob of a simple pendulum of length 40 in.l
=
is released from rest when 5.
θ
=+ °
Assuming simple harmonic motion, determine 1.6 s after release (a) the angle ,
θ
(b) the magnitudes of the velocity and acceleration of the bob.
SOLUTION
For simple harmonic motion and 40 in. 3.333 ft:l
=
=
2
32.2 ft/s 3.1082 rad/s
3.333 ft
n
g
l
ω
== =
Angular displacement:
sin( )
mn
t
θ
θωφ
=
+
Initial conditions:
(0) 5 0.08727 rad, and (0) 0:
θθ
=
°= =
(0) 0 cos(0 ) 2
5
(0) 0.08727 rad
180
mn
m
π
θθωφφ
π
θθ
=
=+=
===
5sin (3.1082 rad/s)
180 2
(0.08727 rad) sin (3.1082 rad/s) 2
t
t
ππ
θ
π
⎡⎤
=+
⎢⎥
⎣⎦
⎡
⎤
=+
⎢
⎥
⎣
⎦
(a) At t = 1.6 s.
5sin (3.1082 rad/s)(1.6 s)
180 2
0.022496 rad 1.288
π
π
θ
⎡
⎤
=+
⎢
⎥
⎣
⎦
==°
1.288°
θ
=
(b) Velocity:
cos( )
5(3.1082 rad/s) cos (3.1082 rad/s)(1.6 s)
180 2
0.262074 rad/s
(3.3333 ft)(0.262074 rad/s) 0.874 ft/s
mn n
t
vl
θθω ω φ
π
π
θ
=+
⎡
⎤
=+
⎢
⎥
⎣
⎦
=
== =
0.874 ft/sv=
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.13 (Continued)
Angular acceleration:
22
2
5
sin( ) (3.1082 rad/s) cos (3.1082 rad/s)(1.6 s)
180 2
0.21733 rad/s
mn n
t
π
π
θθω ωφ
⎡⎤
=− + =− +
⎢⎥
⎣⎦
=−
Acceleration:
22
2
222
22 2
() ()
(3.3333 ft)(0.26207 rad/s) 0.22894 ft/s
(3.333 ft)( 0.21733 rad/s ) 0.72443 m/s
nt
n
t
aa a
v
al
l
al
θ
θ
=+
== = =
== − =−
2
0.75974 ft/sa= 2
0.760 ft/sa=
S
O
Da
O
LUTION
ta:
PROB
L
A 150-k
g
the curr
e
and the
s
constant
maximu
m
will occ
u
0.03 s af
t
1
1m=
L
EM 19.14
g
electromag
n
e
nt is turned
o
s
upporting cr
200 kN/m,
d
m
velocity of
t
u
r in the cabl
e
t
er the curren
t
2
50 kg m=
n
et is at rest a
n
o
ff and the st
e
ane have a t
o
d
etermine (a)
t
he resulting
m
e
during the
m
t
is turned off.
100 kg
2
k=
n
d is holding
1
e
el is droppe
d
o
tal stiffness
the frequenc
y
m
otion, (b) th
e
m
otion, (c) t
h
3
2
00 10 N/m×
1
00 kg of scra
p
d
. Knowing t
h
equivalent to
y
, the amplit
u
e
m
inimum t
e
h
e velocity o
f
p
steel when
h
at the cable
a spring of
u
de, and the
e
nsion which
f
the magnet
PROBLEM 19.14 (Continued)
(b) Minimum value of tension occurs when .
m
x
x=−
min 0
12
12
()
m
TTkx
mg m g
mmg
=−
=−
=−
nm n
⎜⎟
⎝⎠
(c) Velocity at 0.03 s.t=
(36.515)(0.03) 1.09545 rad
n
t
ω
==
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 19.15
A 5-kg collar C is released from rest in the position shown and slides without friction on a
vertical rod until it hits a spring of constant k = 720 N/m which it compresses. The velocity of
the collar is reduced to zero and the collar reverses the direction of its motion and returns to its
initial position. The cycle is then repeated. Determine (a) the period of the motion of the
collar, (b) the velocity of the collar 0.4 s after it was released. (Note. This is a periodic motion,
but not simple harmonic motion.)
SOLUTION
With the given properties:
720 N/m 12 rad/s
5 kg
n
k
m
ω
== =
From free fall of the collar
()
02 2 0.5 m 3.132 m/svghg g== ==
The free-fall time is thus:
()
1
20.5 m
2 m 1 0.3193 stggg
== ==
N
ow, to simplify the analysis we measure the displacement from the
p
osition of static displacement of the spring, under the weight of the
collar:
N
ote that the static deflection is:
(
)
(
)
2
5 kg 9.81 m/s
0.068125 m
720 N/m
ST
W
k
δ
== =
Then 5720 0,xx+=
where x is measured positively up from the
p
osition of static deflection. The solution is:
(
)
sin ,
mn
xx t
ω
φ
=+
with velocity
(
)
cos t
mn n
xx
ω
ωφ
=+
N
ow, to determine and ,
m
x
φ
impose the conditions at impact and
count the time from there. Thus:
At impact:
0, 0.068125 m and 3.132 m/s
ST
tx v
δ
=== =− (downward)
or 0.068125 m sin
m
x
φ
=
(
)
3.132 m/s 12 rad/s cos
m
x
φ
−=
