Question

A 338-N, uniform, 1.62-m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tension of 549.0 N without breaking, and cable B can support up to 417.0 N . You want to place a small weight on this bar.

Here is what I have so far...

Part A

What is the heaviest weight you can put on without breaking either cable?

w =

628

N <--- Correct

Correct

Part B

Where should you put this weight?

d =	.494	m from the cable A <--- Incorrect

Answer 1

The total load distributes equal on the two cables. Half ofthe sum of the maximum load andload of the bar is equal to the tension 417.0 N W834 N W-834 N-338 N-496 N Let d be the distance of the load placed from the cable B. the net torque on the system f the bar is equal to zero if the system is in rotational equilibrium about the axis passing the through the cable B. 1.62 m -(d)WWa+(417 N) (1.62 m)-0 1.62 m (d) (496 N (338N)+(417 N) (1.62 m)-0 d-0.81 m