Mechanical Engineering Chapter 14 Problem The Cranes Trolley Moves The Right With Constant Acceleration And The

Problem 14.33 The crane’s trolley at Amoves to the

right with constant acceleration, and the 800-kg load

moves without swinging.

(a) What is the acceleration of the trolley and load?

(b) What is the sum of the tensions in the parallel

cables supporting the load?

5°

A

m,T=mg

cos θ, from which

a=gtan θ=9.81(tan 5◦)=0.858 m/s2

(b) T=800(9.81)

cos 5◦=7878 N

mg

Problem 14.34 The mass of Ais 30 kg and the mass

of Bis 5 kg. The horizontal surface is smooth. The con-

stant force Fcauses the system to accelerate. The angle

θ=20◦is constant. Determine F.

F

A

B

u

Solution: We have four unknowns (F,T,N,a)and four equations

130

Problem 14.35 The mass of Ais 30 kg and the mass

of Bis 5 kg. The coefﬁcient of kinetic friction between

Aand the horizontal surface is µk=0.2. The constant

force Fcauses the system to accelerate. The angle θ=

20◦is constant. Determine F.

F

A

B

u

Solution: We have four unknowns (F,T,N,a)and four equations

Problem 14.36 The 100-lb crate is initially stationary.

The coefﬁcients of friction between the crate and the

inclined surface are µs=0.2 and µk=0.16. Determine

how far the crate moves from its initial position in 2 s

if the horizontal force F=90 lb.

30°

F

Problem 14.37 In Problem 14.36, determine how far

the crate moves from its initial position in 2 s if the

Solution: Use the deﬁnitions of terms given in the solution to

Problem 14.36. For F=30 lb, N=Fsin θ+Wcos θ=101.6 lb, and

Problem 14.38 The crate has a mass of 120 kg, and

the coefﬁcients of friction between it and the sloping

dock are µs=0.6 and µk=0.5.

(a) What tension must the winch exert on the cable to

start the stationary crate sliding up the dock?

(b) If the tension is maintained at the value determined

in part (a), what is the magnitude of the crate’s

velocity when it has moved 2 m up the dock?

30°

Solution: Choose a coordinate system with the xaxis parallel to

the surface. Denote θ=30◦.

move the crate is Fc=T−Wsin θ, from which the tension

(b) After slip begins, the force acting to move the crate is F=

T−Wsin θ−µkN=101.95 N. From Newton’s second law,

F=ma, from which a=F

m=101.95

120 =0.8496 m/s2. The

velocity is v(t) =at =0.8496tm/s, since v(0)=0. The position

is s(t) =a

2t2, since s(0)=0. When the crate has moved

2 m up the slope, t10 =2(2)

a=2.17 s and the velocity is

v=a(2.17)=1.84 m/s.

T

Problem 14.39 The coefﬁcients of friction between the

load Aand the bed of the utility vehicle are µs=0.4

and µk=0.36. If the ﬂoor is level (θ =0), what is the

largest acceleration (in m/s2) of the vehicle for which

the load will not slide on the bed?

A

u

132

Problem 14.40 The coefﬁcients of friction between the

load Aand the bed of the utility vehicle are µs=0.4

and µk=0.36. The angle θ=20◦. Determine the largest

forward and rearward acceleration of the vehicle for

which the load will not slide on the bed.

A

u

Solution: The load is on the verge of slipping. There are two

unknowns (aand N).

Forward: The equations are

Rearward: The equations are

Problem 14.41 The package starts from rest and slides

down the smooth ramp. The hydraulic device Bexerts

a constant 2000-N force and brings the package to rest

in a distance of 100 mm from the point where it makes

contact. What is the mass of the package?

2 m

A

Solution: Set g=9.81 m/s2

a=vdv

ds =gsin 30◦−2000 N

m

0

4.43 m/s

vdv =0.1m

0gsin 30◦−2000 N

mds

0−(4.43 m/s)2

2=gsin 30◦−2000 N

m(0.1 m)

Solving the last equation we ﬁnd m=19.4 kg

30°

N

2000 N

Problem 14.42 The force exerted on the 10-kg mass

by the linear spring is F=−ks, where kis the spring

constant and sis the displacement of the mass relative

to its position when the spring is unstretched. The value

of kis 40 N/m. The mass is in the position s=0 and

is given an initial velocity of 4 m/s toward the right.

Determine the velocity of the mass as a function of s.

Strategy: : Use the chain rule to write the acceleration

as dv

dt=dv

ds

dt=dv

dsv.

k

s

Solution: The equation of motion is −ks =ma

k

0

s2

134

Problem 14.43 The 450-kg boat is moving at 10 m/s

when its engine is shut down. The magnitude of the

hydrodynamic drag force (in newtons) is 40v2, where

vis the magnitude of the velocity in m/s. When the

boat’s velocity has decreased to 1 m/s, what distance

has it moved from its position when the engine was shut

down?

Problem 14.44 A sky diver and his parachute weigh

200 lb. He is falling vertically at 100 ft/s when his

parachute opens. With the parachute open, the magnitude

of the drag force (in pounds) is 0.5v2. (a) What is the

magnitude of the sky diver’s acceleration at the instant

the parachute opens? (b) What is the magnitude of his

velocity when he has descended 20 ft from the point

where his parachute opens?

Problem 14.45 The Panavia Tornado, with a mass of

18,000 kg, lands at a speed of 213 km/h. The decelerat-

ing force (in newtons) exerted on it by its thrust reversers

and aerodynamic drag is 80,000 +2.5v2, where vis the

airplane’s velocity in m/s. What is the length of the air-

plane’s landing roll? (See Example 14.4.)

136

Problem 14.46 A 200-lb bungee jumper jumps from

a bridge 130 ft above a river. The bungee cord has an

unstretched length of 60 ft and has a spring constant

k=14 lb/ft. (a) How far above the river is the jumper

when the cord brings him to a stop? (b) What maximum

force does the cord exert on him?

Problem 14.47 A helicopter weighs 20,500 lb. It takes

off vertically from sea level, and its upward velocity

is given as a function of its altitude hin feet by

v=66 −0.01 hft/s.

(a) How long does it take the helicopter to climb to an

altitude of 4000 ft?

(b) What is the sum of the vertical forces on the heli-

copter when its altitude is 2000 ft?

32.2 ft/s2(−0.46 ft/s2)=−293 lb

Problem 14.48 In a cathode-ray tube, an electron

(mass =9.11 ×10−31 kg) is projected at Owith

velocity v=(2.2×107)i(m/s). While the electron is

between the charged plates, the electric ﬁeld generated

by the plates subjects it to a force F=−eEj, where

the charge of the electron e=1.6×10−19 C (coulombs)

and the electric ﬁeld strength E=15 kN/C. External

forces on the electron are negligible when it is not

between the plates. Where does the electron strike the

screen?

O

x

y

+ + + +

– – – –

Screen

30

mm 100 mm

138

Problem 14.49 In Problem 14.48, determine where

the electron strikes the screen if the electric ﬁeld strength

Solution: Use the solution to Problem 14.48. Assume that the

electron enters the space between the charged plates at t=

9.11 ×10−31 kg j=−j(2.6345 ×1015)

Problem 14.50 An astronaut wants to travel from a

space station to a satellites S that needs repair. She

departs the space station at O. A spring-loaded launch-

ing device gives her maneuvering unit an initial velocity

of 1 m/s (relative to the space station) in the ydirection.

At that instant, the position of the satellite is x=70 m,

y=50 m, z=0, and it is drifting at 2 m/s (relative to

the station) in the xdirection. The astronaut intercepts

the satellite by applying a constant thrust parallel to the

xaxis. The total mass of the astronaut and her maneu-

vering unit is 300 kg. (a) How long does it take the

astronaut to reach the satellite? (b) What is the magni-

tude of the thrust she must apply to make the intercept?

(c) What is the astronaut’s velocity relative to the satel-

lite when she reaches it?

y

x

S

O

Problem 14.51 What is the acceleration of the 8-kg

collar Arelative to the smooth bar?

20°

45°

200 N

A

a=Fc

m=3.63 m/s2.

F

g

8=2.06 m/s2up the bar.

Problem 14.53 The force F=50 lb. What is the mag-

nitude of the acceleration of the 20-lb collar Aalong the

smooth bar at the instant shown?

y

(5, 3, 0) ft

(2, 2, 2) ft

A

32.2 ft/s2a⇒a=22.9 ft/s2

140

c

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Problem 14.54* In Problem 14.53, determine the

magnitude of the acceleration of the 20-lb collar Aalong

the bar at the instant shown if the coefﬁcient of static

friction between the collar and the bar is µk=0.2.

Solution: Use the results from Problem 14.53.

32.2 ft/s2a⇒a=7.89 ft/s2

Problem 14.55 The 6-kg collar starts from rest at posi-

tion A, where the coordinates of its center of mass are

(400, 200, 200) mm, and slides up the smooth bar to

position B, where the coordinates of its center of mass

are (500, 400, 0) mm under the action of a constant force

F=−40i+70j−40k(N). How long does it take to go

from Ato B?

Strategy: There are several ways to work this problem.

One of the most straightforward ways is to note that the

motion is along the straight line from Ato Band that

only the force components parallel to line AB cause

acceleration. Thus, a good plan would be to ﬁnd a unit

vector from Atoward Band to project all of the forces

acting on the collar onto line AB. The resulting constant

force (tangent to the path), will cause the acceleration of

the collar. We then only need to ﬁnd the distance from

Ato Bto be able to analyze the motion of the collar.

x

F

y

z

A

B

Solution: The unit vector from Atoward Bis eAB =et=0.333i

Problem 14.56* In Problem 14.55, how long does the

collar take to go from Ato Bif the coefﬁcient of kinetic

friction between the collar and the bar is µk=0.2?

Strategy: This problem is almost the same as prob-

lem 14.55. The major difference is that now we must

calculate the magnitude of the normal force, N, and then

must add a term µk|N|to the forces tangent to the bar (in

the direction from Btoward A— opposing the motion).

This will give us a new acceleration, which will result

in a longer time for the collar to go from Ato B.

In the direction tangent to the bar, the equation is (F+W)·eAB −

µk|N|=mat.

Problem 14.57 The crate is drawn across the ﬂoor by a

winch that retracts the cable at a constant rate of 0.2 m/s.

The crate’s mass is 120 kg, and the coefﬁcient of kinetic

friction between the crate and the ﬂoor is µk=0.24.

(a) At the instant shown, what is the tension in the cable?

(b) Obtain a “quasi-static” solution for the tension in the

cable by ignoring the crate’s acceleration. Compare this

solution with your result in (a).

4 m

2 m

Solution:

L

2 m

142

Problem 14.58 If y=100 mm, dy

dt =600 mm/s, and

d2y

dt2=−200 mm/s2, what horizontal force is exerted on

the 0.4 kg slider Aby the smooth circular slot?

300 mm

A

y

Problem 14.59 The 1-kg collar Pslides on the vertical

bar and has a pin that slides in the curved slot. The

vertical bar moves with constant velocity v=2 m/s. The

yaxis is vertical. What are the xand ycomponents of

the total force exerted on the collar by the vertical bar

and the slotted bar when x=0.25 m?

y

1 m

P

x

y = 0.2 sin x

π

Solution:

Problem 14.60* The 1360-kg car travels along a

straight road of increasing grade whose vertical proﬁle

is given by the equation shown. The magnitude of the

car’s velocity is a constant 100 km/h. When x=200 m,

what are the xand ycomponents of the total force acting

on the car (including its weight)?

y = 0.0003x2

y

Problem 14.61* The two 100-lb blocks are released

Solution: The relative motion of the blocks is constrained by the

144

Problem 14.62* The two 100-lb blocks are released

from rest. The coefﬁcient of kinetic friction between all

contacting surfaces is µk=0.1. How long does it take

block Ato fall 1 ft?

0.284 s

Problem 14.63 The 3000-lb vehicle has left the ground

after driving over a rise. At the instant shown, it is mov-

ing horizontally at 30 mi/h and the bottoms of its tires

are 24 in above the (approximately) level ground. The

earth-ﬁxed coordinate system is placed with its origin 30

in above the ground, at the height of the vehicle’s center

of mass when the tires ﬁrst contact the ground. (Assume

that the vehicle remains horizontal.) When that occurs,

the vehicle’s center of mass initially continues moving

downward and then rebounds upward due to the ﬂexure

of the suspension system. While the tires are in con-

tact with the ground, the force exerted on them by the

ground is −2400i−18000yj(lb), where yis the vertical

position of the center of mass in feet. When the vehicle

rebounds, what is the vertical component of the velocity

of the center of mass at the instant the wheels leave the

ground? (The wheels leave the ground when the center

of mass is at y=0.)

y

24 in

30 in 24 in

30 in

x

Problem 14.64* A steel sphere in a tank of oil is given

an initial velocity v=2i(m/s) at the origin of the coor-

dinate system shown. The radius of the sphere is 15 mm.

The density of the steel is 8000 kg/m3and the density

of the oil is 980 kg/m3.IfVis the sphere’s volume,

the (upward) buoyancy force on the sphere is equal to

the weight of a volume Vof oil. The magnitude of the

hydrodynamic drag force Don the sphere as it falls

is |D|=1.6|v|N, where |v|is the magnitude of the

sphere’s velocity in m/s. What are the xand ycom-

ponents of the sphere’s velocity at t=0.1 s?

x

y

146

Problem 14.65* In Problem 14.64, what are the xand

ycoordinates of the sphere at t=0.1s?

x=vxo

y=a

b2(ebt −1)−a

bt

x=0.1070 m =107.0mm

Problem 14.66 The boat in Active Example 14.5

weighs 1200 lb with its passengers. Suppose that the

boat is moving at a constant speed of 20 ft/s in a circu-

lar path with radius R=40 ft. Determine the tangential

and normal components of force acting on the boat.

R

Problem 14.67 In preliminary design studies for a sun-

powered car, it is estimated that the mass of the car

and driver will be 100 kg and the torque produced by

the engine will result in a 60-N tangential force on the

car. Suppose that the car starts from rest on the track at

Aand is subjected to a constant 60-N tangential force.

Determine the magnitude of the car’s velocity and the

normal component of force on the car when it reaches B.

B

A

200 m

50 m

Solution: We ﬁrst ﬁnd the tangential acceleration and use that to

Problem 14.68 In a test of a sun-powered car, the

mass of the car and driver is 100 kg. The car starts

from rest on the track at A, moving toward the right.

The tangential force exerted on the car (in newtons) is

given as a function of time by Ft=20 +1.2t. Deter-

mine the magnitude of the car’s velocity and the normal

component of force on the car at t=40 s.

B

A

200 m

50 m

Solution: We ﬁrst ﬁnd the tangential acceleration and use that to

ﬁnd the velocity vand distance sas functions of time.

148

Problem 14.69 An astronaut candidate with a mass of

72 kg is tested in a centrifuge with a radius of 10 m. The

centrifuge rotates in the horizontal plane. It starts from

rest at time t=0 and has a constant angular acceleration

of 0.2 rad/s2. Determine the magnitude of the horizontal

force exerted on him by the centrifuge (a) at t=0; (b) at

t=10 s.

10 m

Solution: The accelerations are

Problem 14.70 The circular disk lies in the horizon-

tal plane. At the instant shown, the disk rotates with

a counterclockwise angular velocity of 4 rad/s and a

counterclockwise angular acceleration of 2 rad/s2. The

0.5-kg slider Ais supported horizontally by the smooth

slot and the string attached at B. Determine the tension

in the string and the magnitude of the horizontal force

exerted on the slider by the slot.

4 rad/s2 rad/s2

0.6 m

B

A

Solution: