Question

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 ${36.9}^o$ and ${53.1}^o$ respectively. The bar is $2m$ long. Calculate the distance d of the center 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 is ,l=2m

$T_1,T_2$ be the tensions produced in the left and right strings respectively.

At translational equilibrium, we have,

$T_{1}sin{36.9}^{\circ }=T_{2}sin{53.1}^{\circ }$

$⟹\frac{T_1}{T_2}=\frac{4}{3}$

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

$T_1\left(cos{36.9}^{\circ }\right)\times d=T_{2}cos{53.1}^{\circ }(2-d)$

Using both equations,

$d=0.72m$

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