**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 $2 \mathrm{~m}$ long. Calculate the distance $d$ of the centre of aravity of the bar from its left end

**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:

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

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

$=\frac{0.800}{0.600}=\frac{4}{3}$

$\Rightarrow T_{1}=\frac{4}{3} T_{2}$

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

$T_{1} \cos 36.9 \times d=T_{2} \cos 53.1(2-d)$

$T_{1} \times 0.800 d=T_{2} 0.600(2-d)$

$\frac{4}{3} \times T_{2} \times 0.800 d=T_{2}[0.600 \times 2-0.600 d]$

$1.067 d+0.6 d=1.2$

$\therefore d=\frac{1.2}{1.67}$

$=0.72 \mathrm{~m}$

Hence, the C.G. (centre of gravity) of the given bar lies $0.72 \mathrm{~m}$ from its left end.