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Problem 18.1 A horizontal force F=30 lb is applied
to the 230-lb refrigerator as shown. Friction is negligible.
(a) What is the magnitude of the refrigerator’s accel-
eration?
(b) What normal forces are exerted on the refrigerator
by the floor at Aand B?
60 in
A
14 in
28 in
14 in
F
(a) a=4.2 ft/s2
(b) A=80.7 lb,B=149.3 lb
Since A>0 and B>0 then our assumption is correct.
AB
Problem 18.2 Solve Problem 18.1 if the coefficient of
kinetic friction at Aand Bis µk=0.1.
Solution: Assume sliding without tipping
Problem 18.3 As the 2800-lb airplane begins its take-
off run at t=0, its propeller exerts a horizontal force
T=1000 lb. Neglect horizontal forces exerted on the
wheels by the runway.
(a) What distance has the airplane moved at t=2s?
(b) what normal forces are exerted on the tires at A
and B?
4 ft 3 ft
T
5 ft
W
2 ft
BA
Problem 18.4 The Boeing 747 begins its takeoff run
at time t=0. The normal forces exerted on its tires
at Aand Bare NA=175 kN and NB=2800 kN. If
you assume that these forces are constant and neglect
horizontal forces other than the thrust T, how fast is the
airplanes moving at t=4 s? (See Active Example 18.1.) AB
T
3 m 5 m
470
Problem 18.5 The crane moves to the right with con-
stant acceleration, and the 800-kg load moves without
swinging.
(a) What is the acceleration of the crane and load?
(b) What are the tensions in the cables attached at A
and B?
1.5 m 1.5 m
1 m
5°5°
AB
Solution: From Newton’s second law: Fx=800aN.
Problem 18.6 The total weight of the go-cart and
driver is 240 lb. The location of their combined center of
mass is shown. The rear drive wheels together exert a 24-
lb horizontal force on the track. Neglect the horizontal
forces exerted on the front wheels.
(a) What is the magnitude of the go-cart’s accelera-
tion?
(b) What normal forces are exerted on the tires at A
and B?
15 in
16 in
AB
60 in
6 in 4 in
Solution:
472
Problem 18.7 The total weight of the bicycle and rider
is 160 lb. The location of their combined center of mass
is shown. The dimensions shown are b=21 in.,c =
16 in., and h=38 in. What is the largest acceleration
the bicycle can have without the front wheel leaving the
ground? Neglect the horizontal force exerted on the front
wheel by the road.
Strategy: You want to determine the value of the accel-
eration that causes the normal force exerted on the front
wheel by the road to equal zero.
h
bc
BA
Solution: Given: b=21 in.,c=16 in.,h=38 in.
160 lb
Problem 18.8 The moment of inertia of the disk about
Ois I=20 kg-m2.Att=0, the stationary disk is sub-
jected to a constant 50 N-m torque.
(a) What is the magnitude of the resulting angular
acceleration of the disk?
(b) How fast is the disk rotating (in rpm) at t=4s?
50 N-m
O
Solution:
Problem 18.9 The 10-lb bar is on a smooth horizontal
table. The figure shows the bar viewed from above. Its
moment of inertia about the center of mass is I=0.8
slug-ft2. The bar is stationary when the force F=5lb
is applied in the direction parallel to the yaxis. At that
instant, determine
(a) the acceleration of the center of mass, and
(b) the acceleration of point A.
y
A
F
B
x
2 ft2 ft
Solution:
474
Problem 18.10 The 10-lb bar is on a smooth horizon-
tal table. The figure shows the bar viewed from above.
Its moment of inertia about the center of mass is I=0.8
slug-ft2. The bar is stationary when the force F=5lb
is applied in the direction parallel to the yaxis. At that
instant, determine the acceleration of point B.
y
A
B
2 ft2 ft
Solution:
Problem 18.11 The moment of inertia of the astronaut
and maneuvering unit about the axis through their center
of mass perpendicular to the page is I=40 kg-m2.A
thruster can exert a force T=10 N. For safety, the con-
trol system of his maneuvering unit will not allow his
angular velocity to exceed 15◦per second. If he is ini-
tially not rotating, and at t=0, he activates the thruster
until he is rotating at 15◦per second, through how many
degrees has he rotated at t=10 s? T
300 mm
Problem 18.12 The moment of inertia of the heli-
copter’s rotor is 420 slug-ft2. The rotor starts from rest.
At t=0, the pilot begins advancing the throttle so that
the torque exerted on the rotor by the engine (in ft-lb)
is given as a function of time in seconds by T=200t.
(a) How long does it take the rotor to turn ten revolu-
tions?
(b) What is the rotor’s angular velocity (in rpm) when
it has turned ten revolutions?
Problem 18.13 The moments of inertia of the pulleys
Solution: The tension in each belt changes as it goes around each
Problem 18.14 The moment of inertia of the wind-
tunnel fan is 225 kg-m2. The fan starts from rest. The
torque exerted on it by the engine is given as a func-
tion of the angular velocity of the fan by T=140 −
0.02ω2N-m.
(a) When the fan has turned 620 revolutions, what is its
angular velocity in rpm (revolutions per minute)?
(b) What maximum angular velocity in rpm does the
fan attain?
ω
ωdω
dθ
Problem 18.15 The moment of inertia of the pulley
about its axis is I=0.005 kg-m2. If the 1-kg mass Ais
released from rest, how far does it fall in 0.5 s?
Strategy: Draw individual free-body diagrams of the
pulley and the mass.
100 mm
A
Solution: The two free-body diagrams are shown.
478
Problem 18.16 The radius of the pulley is 125 mm
and the moment of inertia about its axis is I=0.05
kg-m2. If the system is released from rest, how far does
the 20-kg mass fall in 0.5 s? What is the tension in the
rope between the 20-kg mass and the pulley?
20 kg
4 kg
Solution: The free-body diagrams are shown.
Problem 18.17 The moment of inertia of the pulley is
0.4 slug-ft2. The coefficient of kinetic friction between
the 5-lb weight and the horizontal surface is µk=0.2.
Determine the magnitude of the acceleration of the 5-lb
weight in each case.
2 lb
6 in
(a) (b)
5 lb
2 lb
6 in
5 lb
Solution: The free-body diagrams are shown.
32.2 ft/s2a, T2−(2lb)=−2lb
32.2 ft/s2a.
Solving we find
480
Problem 18.18 The 5-kg slender bar is released from
rest in the horizontal position shown. Determine the
bar’s counterclockwise angular acceleration (a) at the
instant it is released, and (b) at the instant when it has
rotated 45◦.
1.2 m
Solution:
Problem 18.19 The 5-kg slender bar is released from
rest in the horizontal position shown. At the instant when
it has rotated 45◦, its angular velocity is 4.16 rad/s. At
that instant, determine the magnitude of the force exerted
on the bar by the pin support. (See Example 18.4.)
1.2 m
Problem 18.20 The 5-kg slender bar is released from
rest in the horizontal position shown. Determine the
magnitude of its angular velocity when it has fallen to
the vertical position.
Strategy: : Draw the free-body diagram of the bar
1.2 m
Problem 18.21 The object consists of the 2-kg slender
bar ABC welded to the 3-kg slender bar BDE. The yaxis
is vertical.
(a) What is the object’s moment of inertia about point
D?
(b) Determine the object’s counterclockwise angular
y
x
A
B
DE
0.2 m
0.2 m 0.4 m 0.2 m
(b)
482
Problem 18.22 The object consists of the 2-kg slender
bar ABC welded to the 3-kg slender bar BDE. The y
axis is vertical. At the instant shown, the object has a
counterclockwise angular velocity of 5 rad/s. Determine
the components of the force exerted on it by the pin
support.
y
x
A
B
C
DE
0.2 m
0.2 m 0.4 m 0.2 m
Solution: The free-body diagram is shown.
0.467 kg-m2=23.1 rad/s2.
From Newton’s Second Law we have
Problem 18.23 The length of the slender bar is l=
4 m and its mass is m=30 kg. It is released from rest
in the position shown.
(a) If x=1 m, what is the bar’s angular acceleration
at the instant it is released?
(b) What value of xresults in the largest angular accel-
eration when the bar is released? What is the angu-
lar acceleration?
x
m
l
(b) To find the critical value for xwe differentiate and set equal to
zero to get
dα
d
Problem 18.24 Model the arm ABC as a single rigid
body. Its mass is 320 kg, and the moment of inertia
about its center of mass is I=360 kg-m2. Point Ais
stationary. If the hydraulic piston exerts a 14-kN force
on the arm at Bwhat is the arm’s angular acceleration?
1.80 m
1.40 m
0.30 m
0.70 m
A
B
C
y
x
Solution: The moment of inertia about the fixed point Ais
1.4m
The rotational equation of motion is now
Problem 18.25 The truck’s bed weighs 8000 lb and
its moment of inertia about Ois 33,000 slug-ft2. At the
instant shown, the coordinates of the center of mass of
the bed are (10, 12) ft and the coordinates of point B
are (15, 11) ft. If the bed has a counterclockwise angular
acceleration of 0.2 rad/s2, what is the magnitude of the
force exerted on the bed at Bby the hydraulic cylin-
der AB ?
B
y
484
Problem 18.26 Arm BC has a mass of 12 kg and the
moment of inertia about its center of mass is 3 kg-m2.
Point Bis stationary and arm BC has a constant coun-
terclockwise angular velocity of 2 rad/s. At the instant
shown, what are the couple and the components of force
exerted on arm BC at B?
x
y
300
mm
40°
700 mm
BA
C
Problem 18.27 Arm BC has a mass of 12 kg and the
moment of inertia about its center of mass is 3 kg-m2.
At the instant shown, arm AB has a constant clockwise
angular velocity of 2 rad/s and arm BC has counter-
clockwise angular velocity of 2 rad/s and a clockwise
angular acceleration of 4 rad/s2. What are the couple
and the components of force exerted on arm BC at B?
x
y
300
mm
40⬚
700 mm
B
A
C
Solution: Because the point Bis accelerating, the equations of
486
Problem 18.28 The space shuttle’s attitude control
engines exert two forces Ff=8 kN and Fr=2 kN. The
force vectors and the center of mass Glie in the x–y
plane of the inertial reference frame. The mass of the
shuttle is 54,000 kg, and its moment of inertia about the
axis through the center of mass that is parallel to the z
axis is 4.5×106kg-m2. Determine the acceleration of
the center of mass and the angular acceleration. (You
can ignore the force on the shuttle due to its weight). 18 m 12m
5°6°
2 m
2 m
G
FfFr
y
x
Problem 18.29 In Problem 18.28, suppose that Ff=
4 kN and you want the shuttle’s angular acceleration
to be zero. Determine the necessary force Frand the
resulting acceleration of the center of mass.
Problem 18.30 Points Band C lie in the x–yplane.
The yaxis is vertical. The center of mass of the 18-
kg arm BC is at the midpoint of the line from Bto
C, and the moment of inertia of the arm about the
axis through the center of mass that is parallel to the
zaxis is 1.5 kg-m2. At the instant shown, the angular
velocity and angular acceleration vectors of arm AB
are ωAB =0.6k(rad/s)and αAB =−0.3k(rad/s2). The
angular velocity and angular acceleration vectors of
arm BC are ωBC =0.4k(rad/s)and αBC =2k(rad/s)2.
Determine the force and couple exerted on arm BC at B.
A15°
760 mm
900 mm
y
x
z
B
50
C
is:
Newton’s second law is
F=Bxi+(By−mg)j=maG:
Bxi+By−(18)(9.81)j=18(−1.059i+0.374j).
Solving, we obtain Bx=−19.1N,By=183.3N.
The equation of angular motion is
MG=IBCαBC:
or (0.45 sin 50◦)Bx−(0.45 cos 50◦)By+MB=(1.5)(2)
Solving for MB, we obtain MB=62.6 N-m.
488
c
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