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.

Answer 1

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:

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

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