Mechanical Engineering Chapter 18 Problem Horizontal Force Applied The Refrigerator Shown Friction Negligible What The Magnitude

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/s2a, T2−(2lb)=−2lb

32.2 ft/s2a.

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

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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