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Problem 18.54 The 2-kg slender bar and 5-kg block
are released from rest in the position shown. What
minimum coefficient of static friction between the block
and the horizontal surface would be necessary for the
block not to move when the system is released? (See
Example 18.5.)
Solution: This solution is very similar to that of Problem 18.53.
Problem 18.55 As a result of the constant couple M
applied to the 1-kg disk, the angular acceleration of the
0.4-kg slender bar is zero. Determine Mand the coun-
terclockwise angular acceleration of the rolling disk.
1 m
M
0.25 m
40⬚
Solution: There are seven unknowns (M,N,f,O
x,O
y,a,α), six
dynamic equations, and one constraint equation. We use the following
510
Problem 18.56 The slender bar weighs 40 lb and the
crate weighs 80 lb. At the instant shown, the velocity of
the crate is zero and it has an acceleration of 14 ft/s2
toward the left. The horizontal surface is smooth. Deter-
mine the couple Mand the tension in the rope. 6 ft
3 ft 6 ft
M
Solution: There are six unknowns (M,T,N,O
x,O
y,α),five
Problem 18.57 The slender bar weighs 40 lb and the
crate weighs 80 lb. At the instant shown, the veloc-
ity of the crate is zero and it has an acceleration of
14 ft/s2toward the left. The coefficient of kinetic friction
between the horizontal surface and the crate is µk=0.2.
Determine the couple Mand the tension in the rope. 6 ft
3 ft 6 ft
M
Solution: There are seven unknowns (M,T,N,O
x,O
y,α,f),
512
Problem 18.58 Bar AB is rotating with a constant
clockwise angular velocity of 10 rad/s. The 8-kg slender
bar BC slides on the horizontal surface. At the instant
shown, determine the total force (including its weight)
acting on bar BC and the total moment about its center
of mass. 0.4 m
10 rad/s
B
C
y
Problem 18.59 The masses of the slender bars AB
and BC are 10 kg and 12 kg, respectively. The angular
velocities of the bars are zero at the instant shown
and the horizontal force F=150 N. The horizontal
surface is smooth. Determine the angular accelerations
of the bars.
F
0.4 m
0.4 m
0.2 m
AB
C
The FBDs
514
Problem 18.60 Let the total moment of inertia of the
car’s two rear wheels and axle be IR, and let the total
moment of inertia of the two front wheels be IF. The
radius of the tires is R, and the total mass of the car,
including the wheels, is m. If the car’s engine exerts a
torque (couple) Ton the rear wheels and the wheels do
not slip, show that the car’s acceleration is
a=RT
R2m+IR+IF
.
Strategy: Isolate the wheels and draw three free-body
diagrams.
Solution: The free body diagrams are as shown: We shall write
mBg
Problem 18.61 The combined mass of the motorcy-
cle and rider is 160 kg. Each 9-kg wheel has a 330-
mm radius and a moment of inertia I=0.8 kg-m2. The
engine drives the rear wheel by exerting a couple on it.
If the rear wheel exerts a 400-N horizontal force on the
road and you do not neglect the horizontal force exerted
on the road by the front wheel, determine (a) the motor-
cycle’s acceleration and (b) the normal forces exerted on
the road by the rear and front wheels. (The location of
the center of mass of the motorcycle not including its
wheels, is shown.)
1500 mm
649 mm
AB
723 mm
Solution: In the free-body diagrams shown, mw=9 kg and m=
M=−fF(0.33)=Iα.(4)
Bx
516
Problem 18.62 In Problem 18.61, if the front wheel
lifts slightly off the road when the rider accelerates,
determine (a) the motorcycle’s acceleration and (b) the
torque exerted by the engine on the rear wheel.
Problem 18.63 The moment of inertia of the vertical
handle about Ois 0.12 slug-ft2. The object Bweighs
15 lb and rests on a smooth surface. The weight of the
bar AB is negligible (which means that you can treat the
bar as a two-force member). If the person exerts a 0.2-lb
horizontal force on the handle 15 in above O, what is
the resulting angular acceleration of the handle?
A
Solving Equations (1)–(3) with F=0.2 lb, we obtain α=1.06 rad/s2
B
aB
Problem 18.64 The bars are each 1 m in length and
have a mass of 2 kg. They rotate in the horizontal plane.
Bar AB rotates with a constant angular velocity of
4 rad/s in the counterclockwise direction. At the instant
shown, bar BC is rotating in the counterclockwise
direction at 6 rad/s. What is the angular acceleration of
bar BC?
BA
C
aBC
6 rad/s
4 rad/s
518
Problem 18.65 Bars OQ and PQ each weigh 6 lb.
The weight of the collar Pand friction between the
collar and the horizontal bar are negligible. If the system
is released from rest with θ=45◦, what are the angular
accelerations of the two bars? 2 ft
O
Q
P
θ
2 ft
Problem 18.66 In Problem 18.65, what are the angular
accelerations of the two bars if the collar Pweighs 2 lb?
M=(N −P+Qy+Qx)(cos 45◦)=1
Problem 18.67 The 4-kg slender bar is pinned to 2-
kg sliders at Aand B. If friction is negligible and the
system is released from rest in the position shown, what
is the angular acceleration of the bar at that instant?
A
520
Problem 18.68 The mass of the slender bar is mand
the mass of the homogeneous disk is 4m. The system
is released form rest in the position shown. If the disk
rolls and the friction between the bar and the horizon-
tal surface is negligible, show that the disk’s angular
acceleration is α=6g/95Rcounterclockwise.
2R
R
of the disk is
(5) ND−4mg −By=0,
Problem 18.69 Bar AB rotates in the horizontal plane
with a constant angular velocity of 10 rad/s in the coun-
terclockwise direction. The masses of the slender bars
BC and CD are 3 kg and 4.5 kg, respectively. Deter-
mine the xand ycomponents of the forces exerted on
bar BC by the pins at Band Cat the instant shown.
A
B
y
x
C
D
0.2 m 0.2 m
0.2 m
10 rad/s
Solution: First let’s do the kinematics
vC=vB+ωBC ×rC/B
A
D
0.2 m
522
Problem 18.70 The 2-kg bar rotates in the horizontal
plane about the smooth pin. The 6-kg collar Aslides on
the smooth bar. At the instant shown, r=1.2m,ω=
0.4 rad/s, and the collar is sliding outward at 0.5 m/s
relative to the bar. If you neglect the moment of inertia
of the collar (that is, treat the collar as a particle), what
is the bar’s angular acceleration?
Strategy: Draw individual free-body diagrams of the
bar and collar and write Newton’s second law for the
collar in terms of polar coordinates.
r
A
2 m
ω
F=ma:Neθ=md2r
dt2−rω2er+rα +2dr
dt ωeθ.
Equating eθcomponents,
N=mrα +2dr
dt ω=(6)[rα +2(0.5)(0.4)](2).
Solving Equations (1) and (2) with r=1.2 m gives α=−0.255 rad/s2
Problem 18.71 In Problem 18.70, the moment of iner-
tia of the collar about its center of mass is 0.2 kg-m2.
Determine the angular acceleration of the bar, and com-
pare your answer with the answer to Problem 18.70.
Problem 18.72 The axis L0is perpendicular to both
segments of the L-shaped slender bar. The mass of the
bar is 6 kg and the material is homogeneous. Use inte-
gration to determine the moment of inertia of the bar
about L0.
LO
1 m
For the vertical part (Fig. b),
Iv=m
r2dm =1
0
(22+y2)ρA dy
=13
3ρA =26
3kg-m2.
Therefore I0=Ih+Iv=14 kg-m2.
Problem 18.73 Two homogenous slender bars, each of
mass mand length l, are welded together to form the T-
shaped object. Use integration to determine the moment
of inertia of the object about the axis through point O
that is perpendicular to the bars.
l
Ol
524
Problem 18.74 The slender bar lies in the x−yplane.
Its mass is 6 kg and the material is homogeneous. Use
integration to determine its moment of inertia about the
zaxis.
x
y
1 m
2 m
50⬚
x
1 m
Problem 18.75 The slender bar lies in the x−yplane.
Its mass is 6 kg and the material is homogeneous. Use
integration to determine its moment of inertia about the
0
Problem 18.76 The homogeneous thin plate has mass
m=12 kg and dimensions b=1 m and h=2m.
Determine the mass moments of inertia of the plate about
the x,y, and zaxes.
Strategy: The mass moments of inertia of a thin plate
of arbitrary shape are given by Eqs. (18.37)–(18.39) in
terms of the moments of inertia of the cross-sectional
area of the plate. You can obtain the moments of inertia
of the triangular area from Appendix B.
x
y
h
b
526
Problem 18.77 The brass washer is of uniform thick-
ness and mass m.
(a) Determine its moments of inertia about the xand
zaxes.
(b) Let Ri=0, and compare your results with the val-
ues given in Appendix C for a thin circular plate.
Ro
Ri
x
y
Solution:
Ri
Ro
Problem 18.78 The homogenous thin plate is of uni-
form thickness and weighs 20 lb. Determine its moment
of inertia about the yaxis.
y
x
y = 4 – x2 ft
1
–
4
Problem 18.79 Determine the moment of inertia of the
plate in Problem 18.78 about the xaxis.
Solution: The differential mass is dm =W
gA dy dx. The distance
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