Answer: A 350-N, uniform, 1.50-m bar is suspended

Chapter 11, Problem 9E

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QUESTION:

Problem 9E

A 350-N, uniform, 1.50-m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tension of 500.0 N without breaking, and cable B can support up to 400.0 N. You want to place a small weight on this bar. (a) What is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?

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QUESTION:

Problem 9E

A 350-N, uniform, 1.50-m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tension of 500.0 N without breaking, and cable B can support up to 400.0 N. You want to place a small weight on this bar. (a) What is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?

ANSWER:

Solution 9E

Step 1:

a) For equilibrium, the total force will be zero.

So, we can write, FA + FB - Fg - Fx = 0

where , FA - maximum tension of cable A

             FB - maximum tension of cable B

             Fg - Weight of the rod

             Fx - Maximum weight which it can with stand

We are putting negative sign for Fg and Fx because it is acting downward

Therefore, FA + FB -  Fg = Fx

Provided, FA = 500 N

                FB = 400 N

                Fg = 350 N

Therefore, maximum weight the cables can withstand, Fx  = 500 N + 400 N - 350 N = 650 N

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