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Length of the bar is ,l=2m

T1,T2 be the tensions produced in the left and right strings respectively.

At translational equilibrium, we have,

T1sin36.9∘=T2sin53.1∘

⟹T1T2=43

For rotational equilibrium, on taking the torque about the centre of gravity, we have:

T1(cos36.9∘)×d=T2cos53.1∘(2−d)

Using both equations,

d=0.72m

Hence, the centre of gravity of the given bar lies 0.72 m from its left end.

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