$A$ non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in the figure. 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.

(N/A) The free body diagram of the bar is shown in the figure.
Length of the bar,$l = 2 \; m$.
$T_1$ and $T_2$ are the tensions in the left and right strings respectively.
For translational equilibrium in the horizontal direction:
$T_1 \sin 36.9^{\circ} = T_2 \sin 53.1^{\circ}$
$\frac{T_1}{T_2} = \frac{\sin 53.1^{\circ}}{\sin 36.9^{\circ}} = \frac{0.800}{0.600} = \frac{4}{3}$
$\Rightarrow T_1 = \frac{4}{3} T_2$
For rotational equilibrium,taking the torque about the centre of gravity:
$T_1 \cos 36.9^{\circ} \times d = T_2 \cos 53.1^{\circ} \times (2 - d)$
Substituting $T_1 = \frac{4}{3} T_2$:
$(\frac{4}{3} T_2) \times 0.800 \times d = T_2 \times 0.600 \times (2 - d)$
$\frac{3.2}{3} d = 1.2 - 0.6 d$
$1.067 d + 0.6 d = 1.2$
$1.667 d = 1.2$
$d = \frac{1.2}{1.667} \approx 0.72 \; m$.
Thus,the centre of gravity of the bar lies $0.72 \; m$ from its left end.