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 ${36.9}^{\circ }$ and ${53.1}^{\circ }$ 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
$T_1$ and $T_2$ are the tensions produced in the left and right strings respectively. At translational equilibrium, we have:
$T_{1}sin{36.9}^{\circ }=T_{2}sin{53.1}^{\circ }\frac{T_1}{T_2}=\frac{sin{53.1}^{\circ }}{sin{36.9}^{\circ }}=\frac{0.800}{0.600}=\frac{4}{3}\Rightarrow T_1=\frac{4}{3}{T}_2$
For rotational equilibrium, on taking the torque about the centre of gravity, we have:
$T_{1}cos{36.9}^{\circ }\times d=T_{2}cos{53.1}^{\circ }(2-d)\Rightarrow T_1\times 0.800d=T_{2}0.600(2-d)\frac{4}{3}\times T_2\times 0.800d=T_2[0.600\times 2-0.600d]1.067d+0.6d=1.2\therefore d=\frac{1.2}{1.67}=0.72m$
Hence, the C.G. (centre of gravity) of the given bar lies 0.72 m from its left end.