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SOLUTIONS
MANUAL
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12–1.
Starting from rest, a particle moving in a straight line has an
acceleration of a = (2t - 6) m>s2, where t is in seconds. What
is the particle’s velocity when t = 6 s, and what is its position
when t = 11 s?
Solution
a = 2t - 6
dv = a dt
L0
v
dv =
L0
t
(2t - 6) dt
v = t 2 - 6t
ds = v dt
L0
s
ds =
s =
L0
t
(t2 - 6t) dt
t3
- 3t2
3
When t = 6 s,
Ans.
v = 0
When t = 11 s,
Ans.
s = 80.7 m
Ans:
s = 80.7 m
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12–94.
A golf ball is struck with a velocity of 80 ft>s as shown.
Determine the speed at which it strikes the ground at B and
the time of flight from A to B.
vA 80 ft/s
B
A
45
10
d
Solution
(vA)x = 80 cos 55° = 44.886
(vA)y = 80 sin 55° = 65.532
+2
1S
s = s0 + v0t
d cos 10° = 0 + 45.886t
1
1+ c 2 s = s0 + v0t + act 2
2
d sin 10° = 0 + 65.532 (t) +
1
( - 32.2) ( t 2 )
2
d = 166 ft
Ans.
t = 3.568 = 3.57 s
(vB)x = (vA)x = 45.886
1 + c2
v = v0 + act
(vB)y = 65.532 - 32.2(3.568)
(vB)y = -49.357
vB = 2(45.886)2 + ( - 49.357)2
Ans.
vB = 67.4 ft>s
Ans:
t = 3.57 s
vB = 67.4 ft>s
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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12–190.
Solve Prob. $12–189 if the particle
has an
angular
#
acceleration u = 5 rad>s2 when u = 4 rad>s at u = p> 2 rad.
y
r
(8 u) ft
r
SOLUTION
u
x
Time Derivatives: Here,
p
r = 8u = 8 a b = 4p ft
2
$
$
r = 8 u = 8(5) = 40 ft>s2
#
#
r = 8 u = 8(4) = 32.0 ft>s
Velocity: Applying Eq. 12–25, we have
#
vr = r = 32.0 ft>s
#
vu = r u = 4p(4) = 50.3 ft>s
Ans.
Ans.
Acceleration: Applying Eq. 12–29, we have
#
$
ar = r - r u2 = 40 - 4p A 42 B = - 161 ft>s2
$
# #
au = r u + 2r u = 4p(5) + 2(32.0)(4) = 319 ft>s2
Ans.
Ans.
Ans:
vr
vu
ar
au
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199
=
=
=
=
32.0 ft>s
50.3 ft>s
- 161 ft>s2
319 ft>s2
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13–55.
Determine the maximum constant speed at which the pilot
can travel around the vertical curve having a radius of
curvature r = 800 m, so that he experiences a maximum
acceleration an = 8g = 78.5 m>s2. If he has a mass of 70 kg,
determine the normal force he exerts on the seat of the
airplane when the plane is traveling at this speed and is at its
lowest point.
r 800 m
Solution
an =
v2
v2
; 78.5 =
r
800
Ans.
v = 251 m>s
+ c ΣFn = man; N - 70(9.81) = 70(78.5)
Ans.
N = 6.18 kN
Ans:
N = 6.18 kN
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*15–24.
The motor pulls on the cable at A with a force F = (e 2t) lb,
where t is in seconds. If the 34-lb crate is originally at rest
on the ground at t = 0, determine the crate’s velocity when
t = 2 s. Neglect the mass of the cable and pulleys. Hint:
First find the time needed to begin lifting the crate.
A
Solution
F = e 2t = 34
t = 1.7632 s for crate to start moving
(+c)
mv1 + Σ
0#
L
Fdt = mv2
2
L1.7632
e 2tdt - 34(2 - 1.7632) =
34
v
32.2 2
2
1 2t
e
- 8.0519 = 1.0559 v2
2
L1.7632
Ans.
v2 = 2.13 m>s
Ans:
v2 = 2.13 m>s
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15–119.
The blade divides the jet of water having a diameter of 3 in.
If one-fourth of the water flows downward while the other
three-fourths flows upwards, and the total flow is
Q = 0.5 ft3>s, determine the horizontal and vertical
components of force exerted on the blade by the jet,
gw = 62.4 lb>ft3.
3 in.
SOLUTION
Equations of Steady Flow: Here, the flow rate Q = 0.5 ft2>s. Then,
Q
0.5
dm
62.4
y =
=
= rw Q =
(0.5) = 0.9689 slug>s.
= 10.19 ft>s. Also,
p
3
2
A
dt
32.2
A B
4
12
Applying Eq. 15–25 we have
©Fx = ©
dm
A youts - yins B ; - Fx = 0 - 0.9689 (10.19)
dt
©Fy = ©
1
dm
3
A youty - yiny B ; Fy = (0.9689)(10.19) + (0.9689)(- 10.19)
dt
4
4
Fx = 9.87 lb
Ans.
Ans.
Fy = 4.93 lb
Ans:
Fx = 9.87 lb
Fy = 4.93 lb
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17–9.
Determine the moment of inertia of the homogeneous
triangular prism with respect to the y axis. Express the result
in terms of the mass m of the prism. Hint: For integration, use
thin plate elements parallel to the x–y plane and having a
thickness dz.
z
–h (x – a)
z = ––
a
h
SOLUTION
dV = bx dz = b(a)(1 -
z
) dz
h
x
b
a
y
x
dIy = dIy + (dm)[( )2 + z2]
2
=
x2
1
dm(x2) + dm( ) + dmz2
12
4
= dm(
x2
+ z2)
3
= [b(a)(1 -
a2
z
z
)dz](r)[ (1 - )2 + z2]
h
3
h
k
Iy = abr
a3 h - z 3
z
(
) + z2(1 - )]dz
h
h
L0 3
= abr[
=
[
3
1
1 1
1
a2
(h4 - h4 + h4 - h4) + ( h4 - h4 )]
2
4
h 3
4
3h3
1
abhr(a2 + h2)
12
m = rV =
1
abhr
2
Thus,
Iy =
m 2
(a + h2)
6
Ans.
Ans:
Iy =
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m 2
( a + h2 )
6
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19–15.
z
A 4-kg disk A is mounted on arm BC, which has a negligible
mass. If a torque of M = 15e0.5t2 N # m, where t is in seconds,
is applied to the arm at C, determine the angular velocity of
BC in 2 s starting from rest. Solve the problem assuming
that (a) the disk is set in a smooth bearing at B so that it
rotates with curvilinear translation, (b) the disk is fixed to
the shaft BC, and (c) the disk is given an initial freely
spinning angular velocity of V D = 5 -80k6 rad>s prior to
application of the torque.
250 mm
A
60 mm
B
M
C
#
(5e0.5 t) N m
SOLUTION
a)
(Hz)1 + ©
L
Mz dt = (Hz)2
2
0+
L0
5e0.5 tdt = 4(vB)(0.25)
5 0.5 t 2
e 2 = vB
0.5
0
vB = 17.18 m>s
Thus,
17.18
= 68.7 rad>s
0.25
vBC =
Ans.
b)
(Hz)1 + ©
2
0 +
L0
L
Mz dt = (Hz)2
1
5e0.5 tdt = 4(vB)(0.25) + c (4)(0.06)2 d vBC
2
Since vB = 0.25 vBC, then
Ans.
vBC = 66.8 rad>s
c)
(Hz)1 + ©
L
Mz dt = (Hz)2
2
1
1
- c (4)(0.06)2 d(80) +
5e0.5 tdt = 4(vB)(0.25) - c (4)(0.06)2 d(80)
2
2
L0
Since vB = 0.25 vBC,
Ans.
vBC = 68.7 rad>s
Ans:
(a) vBC = 68.7 rad>s
(b) vBC = 66.8 rad>s
(c) vBC = 68.7 rad>s
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22–10.
The uniform rod of mass m is supported by a pin at A and a
spring at B. If B is given a small sideward displacement and
released, determine the natural period of vibration.
A
L
Solution
1
Equation of Motion. The mass moment of inertia of the rod about A is IA = mL2.
3
Referring to the FBD. of the rod, Fig. a,
L
1
a + ΣMA = IAa ;
- mg a sin u b - (kx cos u)(L) = a mL2 ba
2
3
B
k
However; x = L sin u. Then
- mgL
1
sin u - kL2 sin u cos u = mL2a
2
3
Using the trigonometry identity sin 2u = 2 sin u cos u,
- mgL
KL2
1
sin u sin 2u = mL2a
2
2
3
$
Here since u is small sin u ≃ u and sin 2u ≃ 2u. Also a = u . Then the above
equation becomes
$
mgL
1
mL2 u + a
+ kL2 bu = 0
3
2
$
3mg + 6kL
u +
u = 0
2mL
Comparing to that of the Standard form, vn =
t =
A
3mg + 6kL
. Then
2mL
2p
2mL
= 2p
vn
A 3mg + 6kL
Ans.
Ans:
t = 2p
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1199
2mL
A 3mg + 6kL