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Question

In figure, the bar is uniform and weighing 500 N. How large must W in N be if T1 and T2 are to be equal?


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Solution

Taking torque about the attachment point for W , we get
T1(0.4 L)+T2(0.3 L)+500(0.2 L)=0
where T1=T2=T
T(0.1)=100T=1000 N
By balancing forces
Fy=02TW500=0
W=1500 N

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