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# A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in fig. The angles made by the strings with the vertical are 36.9o and 53.1o respectively. The bar is 2 m long. Calculate the distance d of the center of gravity of the bar from its left end.

Solution
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#### The free body diagram of the bar is shown in the following figure.Length of the bar is ,l=2mT1,T2 be the tensions produced in the left and right strings respectively.At translational equilibrium, we have,T1sin36.9∘=T2sin53.1∘⟹T1T2=43For 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=0.72mHence, the centre of gravity of the given bar lies 0.72 m from its left end.

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