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 12.1, Problem 12.29P
To determine

(a)

The acceleration of the panel and the tension in the cord after the system released from rest.

Expert Solution
Check Mark

Answer to Problem 12.29P

Acceleration of the panel a=8.944ft/s2

Tension in the cord T=18.056lb

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.29P , additional homework tip  1

Weight of panel =40lb

Weight of the counterweight =25lb

First, we draw the free body diagram and kinetic diagram of the panel:

F = force exerted by counterweight

Weight W=mg

m=Wg

Now, force F=ma

F=Wg×a

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.29P , additional homework tip  2

Fx=ma

TF=ma

TF=40ga_______(1)

Now, computing the forces for counterweight A,

For this we draw the free body and kinetic diagram of the counterweight A;

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.29P , additional homework tip  3

Hence, from the above diagram it is clear that the acceleration component of counterweight A has two components.

Now, aA=aP+aA/PaA=a+a

Fx=max

F=25ga______(2)

Fg=mag

WT=25ga25T=25ga________(3)

Now, adding equation (1), (2) and (3);

TF+F+25T=40+25+25ga25=90gaa=2590ga=2590(32.2)a=8.944ft/s2

Now, Tension in the cord from equation (3),

25T=25ga

25T=2532.2×8.94425T=6.944T=18.056lb

To determine

(b)

The acceleration of the panel and the tension in the cord after the system released from rest.

Expert Solution
Check Mark

Answer to Problem 12.29P

Acceleration of the panel a=12.38ft/s2

Tension in the cord T=15.38lb

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.29P , additional homework tip  4

Weight of panel =40lb

Weight of the counterweight =25lb

First, we draw the free body diagram and kinetic diagram of the panel:

F = force exerted by counterweight.

Weight W=mg

m=Wg

Now, force F=ma

F=Wg×a

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.29P , additional homework tip  5

Fy=ma

T=ma

T=40ga_______(1)

Now, computing the forces for counterweight A,

For this we draw the free body and kinetic diagram of the counterweight A;

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.29P , additional homework tip  6

Fy=ma

25T=25ga______(2)

Now, adding equation (1) and (2);

T+25T=40+25ga25=65gaa=2565ga=2565(32.2)a=12.38ft/s2

Now, Tension in the cord from equation (1),

T=40ga

T=4032.2×12.38T=15.38lb

To determine

(c)

The acceleration of the panel and the tension in the cord after the system released from rest.

Expert Solution
Check Mark

Answer to Problem 12.29P

Acceleration of the panel a=12.38ft/s2

Tension in the cord T=15.38lb

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.29P , additional homework tip  7

Weight of panel =40lb

Weight of the counterweight =25lb

First, we draw the free body diagram and kinetic diagram of the panel:

F = force exerted by counterweight

Weight W=mg

m=Wg

Now, force F=ma

F=Wg×a

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.29P , additional homework tip  8

Since, panel is accelerated to the left, there is no force exerted by the panel on counterweight hence;

Fx=ma

T=ma

T=40ga_______(1)

Now, computing the forces for counterweight A,

For this, we draw the free body and kinetic diagram of the counterweight A;

Vector Mechanics For Engineers, Chapter 12.1, Problem 12.29P , additional homework tip  9

Fy=ma

25T=25ga______(2)

Now, adding equation (1) and (2);

T+25T=40+25ga25=65gaa=2565ga=2565(32.2)a=12.38ft/s2

Now, Tension in the cord from equation (1),

T=40ga

T=4032.2×12.38T=15.38lb

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Chapter 12 Solutions

Vector Mechanics For Engineers

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