Question

A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are 36.9∘ and 53.1∘ respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

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Solution

The free body diagram of the bar is shown in the following figure.

Length of the bar, l = 2 m

T1 and T2 are the tensions produced in the left and right strings respectively. At translational equilibrium, we have:

T1sin 36.9∘=T2sin53.1∘T1T2=sin 53.1∘sin 36.9∘=0.8000.600=43⇒T1=43T2

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

T1 cos 36.9∘×d=T2 cos 53.1∘(2−d)⇒T1×0.800 d=T2 0.600(2−d)43×T2×0.800 d=T2[0.600×2−0.600 d]1.067 d+0.6 d=1.2∴ d=1.21.67=0.72 m

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

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