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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.
458308_e6b08580128b43c6b9f7cf261c1ccbf9.png

Solution
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The free body diagram of the bar is shown in the following figure.
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(2d)
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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