A non-uniform bar of weight $W$ and length $L$is suspended by two strings of negligible weight as shown in figure. The angles made by the strings with the vertical are $\lambda {\theta }_1$ and ${\theta }_2$ respectively. The distance $d$ of the centre of gravity of the bar from its left end is

Let $T_1$ and $T_2$ be the tensions in two strings as shown in the figure.

For translational equilibrium along the horizontal direction, we get

$T_1\sin {\theta }_1=T_2\sin {\theta }_2$   …$\left(i\right)$

For rotational equilibrium about $G$

$$\begin{gathered} -T_1\cos {\theta }_{1}d+T_2\cos {\theta }_2(L-d)=0 \\ {T}_1\cos {\theta }_{1}d=T_2\cos {\theta }_2(L-d) \end{gathered}$$   …$\left(ii\right)$

Dividing equation $\left(i\right)$ by $\left(ii\right)$, we get

$$\begin{gathered} \frac{\tan {\theta }_1}{d}=\frac{\tan {\theta }_2}{(L-d)}\text{ or }\frac{L-d}{d}=\frac{\tan {\theta }_2}{\tan {\theta }_1} \\ \left(\frac{L}{d}-1\right)=\frac{\tan {\theta }_2}{\tan {\theta }_1}\Rightarrow \frac{L}{d}=\left(\frac{\tan {\theta }_2}{\tan {\theta }_1}+1\right) \\ \frac{L}{d}=\frac{\tan {\theta }_2+\tan {\theta }_1}{\tan {\theta }_1}\Rightarrow d=L\left(\frac{\tan {\theta }_1}{\tan {\theta }_1+\tan {\theta }_2}\right) \end{gathered}$$