Mechanical Engineering Chapter 14 Problem Active Example Suppose That The Coefcient Kinetic Friction Between The Crate

Problem 14.1 In Active Example 14.1, suppose that

the coefﬁcient of kinetic friction between the crate and

the inclined surface is µk=0.12. Determine the dis-

tance the crate has moved down the inclined surface at

t=1s.

20⬚

Solution: There are three unknowns in this problem: N,f, and

a. We will ﬁrst assume that the crate does not slip. The governing

equations are

110

Problem 14.2 The mass of the Sikorsky UH-60A heli-

copter is 9300 kg. It takes off vertically with its rotor

exerting a constant upward thrust of 112 kN.

(a) How fast is the helicopter rising 3 s after it takes

off?

(b) How high has it risen 3 s after it takes off?

Strategy: Be sure to draw the free-body diagram of

the helicopter.

Solution: The equation of motion is

Problem 14.3 The mass of the Sikorsky UH-60A heli-

copter is 9,300 kg. It takes off vertically at t=0. The

pilot advances the throttle so that the upward thrust of its

engine (in kN) is given as a function of time in seconds

by T=100 +2t2.

(a) How fast is the helicopter rising 3 s after it takes

off?

(b) How high has it risen 3 s after it takes off?

Solution: The equation of motion is

t

Problem 14.4 The horizontal surface is smooth. The

30-lb box is at rest when the constant force Fis applied.

Two seconds later, the box is moving to the right at

20 ft/s. Determine F.

F

20⬚

Solution: We use one governing equation and one kinematic

relation

Problem 14.5 The coefﬁcient of kinetic friction between

the 30-lb box and the horizontal surface is µk=0.1. The

box is at rest when the constant force Fis applied. Two

seconds later, the box is moving to the right at 20 ft/s.

Determine F.

F

20⬚

Solution: We use two governing equations, one slip equation, and

Fx:Fcos 20◦−f=30 lb

32.2 ft/s2a,

Fy:N−Fsin 20◦−30 lb =0,

f=(0.1)N,

v=(20 ft/s)=a(2s).

Solving, we ﬁnd

a=10 ft/s2,N =34.7lb,f =3.47 lb,F=13.6lb.

F = 10 N 78,48 N

N

Fr

Problem 14.6 The inclined surface is smooth. The

velocity of the 114-kg box is zero when it is subjected

to a constant horizontal force F=20 N. What is the

velocity of the box two seconds later?

F

20⬚

112

Problem 14.7 The coefﬁcient of kinetic friction between

the 14-kg box and the inclined surface is µk=0.1. The

velocity of the box is zero when it is subjected to a con-

stant horizontal force F=20 N. What is the velocity of

the box two seconds later?

F

20⬚

Solution: From the governing equations and the slip equation, we

can ﬁnd the acceleration of the box (assumed to be down the incline).

Problem 14.8 The 170-lb skier is schussing on a 25◦

slope. At the instant shown, he is moving at 40 ft/s. The

kinetic coefﬁcient of friction between his skis and the

snow is µk=0.08. If he makes no attempt to check his

speed, how long does it take for it to increase to 60 ft/s?

Solution: The governing equations and the slip equation are used

to ﬁnd the acceleration

F ր:N−(170 lb)cos 25◦=0,

32.2 ft/s2a.

a=11.3 ft/s2,N =154 lb,f =12.3lb.

114

Problem 14.9 The 170-lb skier is schussing on a 25◦

slope. At the instant shown, he is moving at 40 ft/s. The

kinetic coefﬁcient of friction between his skis and the

snow is µk=0.08. Aerodynamic drag exerts a resisting

force on him of magnitude 0.015v2, where vis the mag-

nitude of his velocity. If he makes no attempt to check

his speed, how long does it take for it to increase to

60 ft/s?

32.2 ft/s2a.

60 ft/s

dv

dt=t

Problem 14.10 The total external force on the 10-kg

object is constant and equal to F=90i−60j+20k(N).

At time t=0, its velocity is v=−14i+26j+32k(m/s).

What is its velocity at t=4 s? (See Active Example 14.2.)

y

兺 F

Solution:

Problem 14.11 The total external force on the 10-kg

object shown in Problem 14.10 is given as a function of

time by F=(−20t+90)i−60j+(10t+40)k(N).

At time t=0, its position is r=40i+30j−360k(m)

and its velocity is v=−14i+26j+32k(m/s). What is

its position at t=4s?

Problem 14.12 The position of the 10-kg object shown

in Problem 14.10 is given as a function of time by r=

(20t3−300)i+60t2j+(6t4−40t2)k(m). What is the

total external force on the object at t=2s?

116

Problem 14.13 The total force exerted on the 80,000-

lb launch vehicle by the thrust of its engine, its weight,

and aerodynamic forces during the interval of time from

t=2stot=4 s is given as a function of time by F=

(2000 −400t2)i+(5200 +440t)j+(800 +60t2)k(lb).

At t=2 s, its velocity is v=12i+220j−30k(ft/s).

What is its velocity at t=4s?

m=(2000 −400t2)lb

32.2 ft/s2=(0.805 −0.161t2)ft/s2,

3[0.0242][43−23]−30ft/s =−28.9 ft/s,

Problem 14.14 At the instant shown, the horizontal

component of acceleration of the 26,000-lb airplane due

to the sum of the external forces acting on it is 14 ft/s2.If

the pilot suddenly increases the magnitude of the thrust

Tby 4000 lb, what is the horizontal component of the

plane’s acceleration immediately afterward?

15°

x

T

y

Solution: Before

Problem 14.15 At the instant shown, the rocket is trav-

eling straight up at 100 m/s. Its mass is 90,000 kg and

the thrust of its engine is 2400 kN. Aerodynamic drag

exerts a resisting force (in newtons) of magnitude 0.8v2,

where vis the magnitude of the velocity. How long does

it take for the rocket’s velocity to increase to 200 m/s?

200 m/s

dv

dt=t

118

Problem 14.16 A 2-kg cart containing 8 kg of water

is initially stationary (Fig. a). The center of mass of the

“object” consisting of the cart and water is at x=0. The

cart is subjected to the time-dependent force shown in

Fig. b, where F0=5 N and t0=2 s. Assume that no

water spills out of the cart and that the horizontal forces

exerted on the wheels by the ﬂoor are negligible.

(a) Do you know the acceleration of the cart during

the period 0 <t<t

0?

(b) Do you know the acceleration of the center of mass

of the “object” consisting of the cart and water

during the period 0 <t<t

0?

(c) What is the x-coordinate of the center of mass of

the “object” when t>2t0?

t

F

F0

y

x

(a)

Problem 14.17 The combined weight of the motorcy-

cle and rider is 360 lb. The coefﬁcient of kinetic friction

between the tires and the road is µk=0.8. The rider

starts from rest, spinning the rear wheel. Neglect the

horizontal force exerted on the front wheel by the road.

In two seconds, the motorcycle moves 35 ft. What was

the normal force between the rear wheel and the road?

32.2 ft/s2(17.5 ft/s2)=195.6 lb

Problem 14.18 The mass of the bucket Bis 180 kg.

From t=0tot=2 s, the xand ycoordinates of the

center of mass of the bucket are

x=−0.2t3+0.05t2+10 m,

y=0.1t2+0.4t+6m.

Determine the xand ycomponents of the force exerted

on the bucket by its supports at t=1s.

x

B

y

Solution:

Problem 14.19 During a test ﬂight in which a 9000-kg

helicopter starts from rest at t=0, the acceleration of its

center of mass from t=0tot=10 s is a=(0.6t)i+

(1.8−0.36t)jm/s2. What is the magnitude of the total

external force on the helicopter (including its weight) at

t=6s?

T

L

Path

y

Solution: From Newton’s second law: F=ma. The sum of

the external forces is F=F−W=9000[(0.6t)i+(1.8−0.36t)

120

Problem 14.20 The engineers conducting the test

described in Problem 14.19 want to express the total

force on the helicopter at t=6 s in terms of three forces:

the weight W, a component Ttangent to the path, and

a component Lnormal to the path. What are the values

of W,T, and L?

Solution: Integrate the acceleration: v=(0.3t2)i+(1.8t−0.18

t2)j, since the helicopter starts from rest. The instantaneous ﬂight

0.3(6)2=21.8◦. A unit vector

tangent to this path is et=icos β+jsin β. A unit vector normal to

this path en=−isin β+jcos β. The weight acts downward:

Problem 14.21 At the instant shown, the 11,000-kg

airplane’s velocity is v=270 im/s. The forces acting

on the plane are its weight, the thrust T=110 kN, the

lift L=260 kN, and the drag D=34 kN. (The x-axis is

parallel to the airplane’s path.) Determine the magnitude

of the airplane’s acceleration.

y

x

LT

Horizontal

Path

15°

D

mg

Problem 14.22 At the instant shown, the 11,000-kg

airplane’s velocity is v=300i(m/s). The rate of change

of the magnitude of the velocity is dv/dt =5 m/s2. The

radius of curvature of the airplane’s path is 4500 m, and

the yaxis points toward the concave side of the path.

The thrust is T=120,000 N. Determine the lift Land

drag D.

Solution:

g=9.81 m/s2

T=120000 N

Fx:Tcos 15◦−D−mg sin 15◦=max

Fy:L+Tsin 15◦−mg cos 15◦=may

ay=V2/ρ

Solving, D=33.0kN,L=293 kN

L

T

y

x

15°

mg

D

T

15°

Problem 14.23 The coordinates in meters of the 360-

kg sport plane’s center of mass relative to an earth-

ﬁxed reference frame during an interval of time are x=

20t−1.63t2,y=35t−0.15t3, and z=−20t+1.38t2,

where tis the time in seconds. The y– axis points upward.

The forces exerted on the plane are its weight, the thrust

vector Texerted by its engine, the lift force vector

L, and the drag force vector D.Att=4 s, determine

T+L+D.

y

x

z

122

Problem 14.24 The force in newtons exerted on the

360-kg sport plane in Problem 14.23 by its engine, the

lift force, and the drag force during an interval of time

is T+L+D=(−1000 +280t)i+(4000 −430t)j+

(720 +200t)k, where tis the time in seconds. If the

coordinates of the plane’s center of mass are (0, 0, 0)

and its velocity is 20i+35j−20k(m/s) at t=0, what

are the coordinates of the center of mass at t=4s?

Problem 14.25 The robot manipulator is programmed

so that x=40 +24t2mm, y=4t3mm, and z=0 dur-

ing the interval of time from t=0tot=4 s. The y

axis points upward. What are the xand ycomponents

of the total force exerted by the jaws of the manipulator

on the 2-kg widget Aat t=3s? A

x

y

x

Problem 14.26 The robot manipulator described in

Problem 14.25 is reprogrammed so that it is stationary

at t=0 and the components of its acceleration are ax=

400 −0.8vxmm/s2,ay=200 −0.4vymm/s2from t=

0tot=2 s, where vxand vyare the components of

robot’s velocity in mm/s. The yaxis points upward.

What are the xand ycomponents of the total force

exerted by the jaws of the manipulator on the 2-kg

widget Aat t=1s?

124

Problem 14.27 In the sport of curling, the object is to

slide a “stone” weighting 44 lb into the center of a target

located 31 yards from the point of release. In terms of

the coordinate system shown, the point of release is at

x=0,y =0. Suppose that a shot comes to rest at x=

31.0 yards, y=1 yard. Assume that the coefﬁcient of

kinetic friction is constant and equal to µk=0.01. What

were the xand ycomponents of the stone’s velocity at

release?

y

31 yd

Curling

stone

Solution: The stone travels at an angle relative to the xaxis.

Problem 14.28 The two masses are released from rest.

How fast are they moving at t=0.5 s? (See Example 14.3.)

5 kg

2 kg

Solution: The free-body diagrams are shown. The governing

equations are

Problem 14.29 The two weights are released from rest.

The horizontal surface is smooth. (a) What is the ten-

sion in the cable after the weights are released? (b) How

fast are the weights moving one second after they are

released?

5 lb

Problem 14.30 The two weights are released from rest.

The coefﬁcient of kinetic friction between the horizon-

tal surface and the 5-lb weight is µk=0.18. (a) What is

the tension in the cable after the weights are released?

(b) How fast are the weights moving one second after

they are released?

5 lb

Problem 14.31 The mass of each box is 14 kg. One

second after they are released from rest, they have moved

0.3 m from their initial positions. What is the coefﬁcient

of kinetic friction between the boxes and the surface?

30⬚

Solution: We will ﬁrst use the kinematic information to ﬁnd the

From the free-body diagrams we have four equations of motion:

128

Problem 14.32 The masses mA=15 kg and mB=30 kg,

and the coefﬁcients of friction between all of the surfaces

are µs=0.4 and µk=0.35. The blocks are station-

ary when the constant force Fis applied. Determine

the resulting acceleration of block Bif (a) F=200 N;

(b) F=400 N.

F

B

A

Solution: Assume that no motion occurs anywhere. Then