Engineering Mechanics: Statics
Engineering Mechanics: Statics
13th Edition
ISBN: 9780132915540
Author: Russell C. Hibbeler
Publisher: Prentice Hall
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Chapter 5.7, Problem 7FP

The uniform plate has a weight of 500 lb. Determine the tension in each of the supporting cables.

Chapter 5.7, Problem 7FP, The uniform plate has a weight of 500 lb. Determine the tension in each of the supporting cables.

Prob. F5-7

Expert Solution & Answer
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To determine
The tension in each of the supporting cables.

Answer to Problem 7FP

The tension (TA) in cable A is 350lb_, the tension (TB) in cable B is 250lb_, and the tension (TC) in cable C is 100lb_.

Explanation of Solution

Given information:

  • The weight (W) of the plate is 500 lb.
  • The load (P) acting on the plate is 200 lb.

Explanation:

Show the free body diagram of the uniform plate as in Figure (1).

Engineering Mechanics: Statics, Chapter 5.7, Problem 7FP

Use Figure (1).

Apply the equation of equilibrium along the z axis.

Fz=0TA+TB+TCPW=0 (I)

Take the moment about the x axis.

Mx=0TA(x1)+TC(x2)W(x3)P(x4)=0 (II)

Here, the horizontal distance between cables B and A is x1, the horizontal distance between cables B and C is x2, the center of gravity of the plate about x axis is x3, and the horizontal distance between the load and cable B is x4.

Take the moment about the y axis.

My=0TB(y1)TC(y2)W(y3)P(y4)=0 (III)

Here, the vertical distance between cable A and B is y1, the vertical distance between A and C is y2, the center of gravity of the plate about the y axis is y3, the vertical distance of the load acting is y4.

Calculations:

Substitute 200 lb for P and 500 lb for W in Equation (I).

TA+TB+TC200500=0TA+TB+TC=700 (IV)

Substitute 3.0 ft for x1, 3.0 ft for x2, 500 lb for W, 1.5 ft for x3, 200 lb for P, and 3.0 ft for x4 in Equation (II).

TA(3.0)+TC(3.0)500(1.5)200(3)=03TA+3TC=1,350 (V)

Substitute 4.0 ft for y1, 4.0 ft for y2, 500 lb for W, 2.0 ft for y3, 200 lb for P, and 2.0 ft for y4 in Equation (III).

TB(4.0)TC(4.0)+500(2.0)+200(2.0)=04TB4TC=1,4004TB+4TC=1,400 (VI)

Simply Equation (I), multiplying by 3 and it can be written as follows.

3(TA+TB+TC)=3(700)3TA+3TB+3TC=2,100 (VIII)

Conclusion:

Solve Equation (VIII) and (V).

TB=250lb

Thus, the tension in cable B is 250lb_

Substitute 250 lb for TB in Equation (VI).

4(250)+4TC=1,4001,000+4TC=1,400TC=4004TC=100lb

Thus, the tension in cable C is 100lb_

Substitute 250 lb for TB and 100 lb for TC in Equation (IV).

TA+250+100=700TA=700350TA=350lb

Thus, the tension in cable A is 350lb_.

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