The cable in the picture below will break if it\'s put under more than 11,000N o

Question

The cable in the picture below will break if it's put under more than 11,000N of tension. The beam extending from the wall has a mass of 1,000kg and is 5m long. The crate hanging from the beam has mass M = 100kg. The angle of the cable is theta = 30 degree. The value of "L" specifies how far from the end of the beam the crate is placed. What "L" value will cause the cable to experience its maximum tension? Tip: The wall does exert some unknown force on the beam. Should the crate be suspended at a larger or smaller value of "L" to be safe?

Explanation / Answer

Since the system is in equilibrium ,net torque acting on the system is zero

Tnet = 0

Torque due to Tension + Torque due to weight of the beam + Torque due to hanging mass M =0

let the axis of rotation is at the fixed end of the rod

and d be the length of the rod

then

(d*T*sin(theta) - (d/2)*W_beam*sin(90) - (d-L)*W_hanging*sin(90) =0

(5*11000*sin(30)) - ((5/2)*1000*9.8*sin(90)) - ((5-L)*100*9.8*sin(90)) = 0

L = 1.94 m

b) if L is smaller then Torque due to hanging mass will dominate more and it wiil not be safe

So larger L will be safety

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