Precalculus with Limits: A Graphing Approach
Precalculus with Limits: A Graphing Approach
7th Edition
ISBN: 9781305071711
Author: Ron Larson
Publisher: Brooks/Cole
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Chapter 10, Problem 42RE
To determine

To find: The tension in each of the supporting cableshandling a load of 200 pounds.

Expert Solution & Answer
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Answer to Problem 42RE

The tension ineach of the supporting cables are 433.17N , 312.5N and 312.5N respectively.

Explanation of Solution

Given information:

A load of 200 pounds is supported by three cables, as shown in the figure(1).

  Precalculus with Limits: A Graphing Approach, Chapter 10, Problem 42RE

Figure(1)

Calculation:

Three upward forces OA , OB and OC are acting on the load. The forces are in the form vector.

The vector OA is,

  OA→=〈0,10,10〉

The vector OB is,

  OB→=〈−4,−6,10〉

The vector OC is,

  OC→=〈4,−6,10〉

One downward force is also acting on the load that is the weight 200 pounds.Let w be the force and its vector form is:

  w=〈0,0,−200〉

The unit vector in direction of OA is OA→‖OA→‖ . Let a be the force in direction of OA can be calculated as:

  a=‖a‖OA→‖OA→‖=‖a‖〈0,10,10〉0+100+100=‖a‖〈0,10,10〉102=‖a‖〈0,0.707,0.707〉

The unit vector in direction of OB is OB→‖OB→‖ . Let b be the force in direction of OB can be calculated as:

  b=‖b‖OB→‖OB→‖=‖b‖〈−4,−6,10〉16+36+100=‖b‖〈−4,−6,10〉12.33=‖b‖〈−0.32,−0.49,0.81〉

The unit vector in direction of OC is OC→‖OC→‖ . Let c be the force in direction of OC can be calculated as:

  c=‖c‖OC→‖OC→‖=‖c‖〈4,−6,10〉16+36+100=‖c‖〈4,−6,10〉12.33=‖c‖〈0.32,−0.49,0.81〉

At equilibrium, the resultant acting on load is zero.Therefore,

  a+b+c−w=0

Now, the equations formedare

  0.707‖a‖−0.81‖b‖=00.707‖a‖−0.81‖b‖−0.81‖c‖=00.707‖a‖−0.81‖b‖−0.81‖c‖=−200

Solve the above equations.

  ‖b‖=‖c‖=312.5 N‖a‖=433.17 N

Therefore,the tension ineach of the supporting cables are 433.17N , 312.5N and 312.5N respectively.

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Chapter 10, Problem 41RE
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Chapter 10, Problem 43RE

Chapter 10 Solutions

Precalculus with Limits: A Graphing Approach

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