Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977305
Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.
Publisher: Mcgraw-hill Education,
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Chapter 13.1, Problem 13.22P

The motor applies a constant downward force F = 1050 lb to the cable used to raise the 4000-lb elevator E shown in the figure. The counterweight has a weight of 2000 lb and the elevator starts from rest. After the elevator travels 20 ft. determine (a) the velocity of the elevator. (b) the velocity of the counterweight.

Chapter 13.1, Problem 13.22P, The motor applies a constant downward force F=1050 lb to the cable used to raise the 4000-lb

Expert Solution
Check Mark
To determine

(a)

The velocity of the elevator.

Answer to Problem 13.22P

vE=4.633ft/s

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 13.1, Problem 13.22P , additional homework tip  1

The constant downward force is equal to F=1050lb.

Weight of elevator E is equal to 4000lb.

Counterweight is equal to 2000lb.

Elevator travels 20ft.

In a dependent motion of particles such as the pulley system shown above,

The total length of the rope is a constant

For example,

2xA+2xB+xC=C

C - Constant

2dxAdt+2dxBdt+dxCdt=0

The kinetic energy of a particle is defined as,

T=12mv2

Principle of work and energy is defined as,

U12=T2T1

Above equation states “If a particle moves from A1 to A2 under an action of force F, the work of the force F is equal to the change in kinetic energy.”

Calculation:

Vector Mechanics For Engineers, Chapter 13.1, Problem 13.22P , additional homework tip  2

Length of both cables is constant, therefore

For cable 1,

yC+2yE=0

Differentiate,

vC+2vE=0

For cable 2,

yE+yW=0

Differentiate,

vE+vW=0

Vector Mechanics For Engineers, Chapter 13.1, Problem 13.22P , additional homework tip  3

For elevator E,

Apply principle of work and energy,

U12=T2T12F(20ft)+T(20ft)(4000lb)(20ft)=12(4000lb32.2ft/s2)vE22(1050lb)(20ft)+T(20ft)(4000lb)(20ft)=12(4000lb32.2ft/s2)vE2

Vector Mechanics For Engineers, Chapter 13.1, Problem 13.22P , additional homework tip  4

For counter weight,

Apply principle of work and energy,

U12=T2T1(2000lb)(20ft)T(20ft)=12(2000lb32.2ft/s2)vW2

But we know that,

vW=vE

Therefore,

(2000lb)(20ft)T(20ft)=12(2000lb32.2ft/s2)(vE)2

Add both equations,

2(1050lb)(20ft)+(2000lb)(20ft)(4000lb)(20ft)=12vE2(4000lb32.2ft/s2+2000lb32.2ft/s2)

Therefore,

vE=4.633ft/s

Conclusion:

The velocity of elevator E is equal to vE=4.633ft/s.

Expert Solution
Check Mark
To determine

(b)

The velocity of the counter weight.

Answer to Problem 13.22P

vW=4.633ft/s

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 13.1, Problem 13.22P , additional homework tip  5

The constant downward force is equal to F=1050lb.

Weight of elevator E is equal to 4000lb.

Counterweight is equal to 2000lb.

Elevator travels 20ft.

In a dependent motion of particles such as the pulley system shown above,

The total length of the rope is a constant

For example,

2xA+2xB+xC=C

C - Constant

2dxAdt+2dxBdt+dxCdt=0

The kinetic energy of a particle is defined as,

T=12mv2

Principle of work and energy is defined as,

U12=T2T1

Above equation states “If a particle moves from A1 to A2 under an action of force F, the work of the force F is equal to the change in kinetic energy.”

Calculation:

According to sub part a,

We have found,

vE+vW=0

And,

vE=4.633ft/s

Therefore, the velocity of counter weight is equal to,

vW=4.633ft/s

Conclusion:

The velocity of counter weight is equal to vW=4.633ft/s

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Chapter 13 Solutions

Vector Mechanics For Engineers

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