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 ${\mathrm{\theta }}_1$ and ${\mathrm{\theta }}_2$ respectively. The distance d of the centre of gravity of the bar from its left end is

Let ${\mathrm{T}}_1 \mathrm{and} {\mathrm{T}}_2$ be the tensions in two strings as shown in the figure.
For translational equilibrium along the horizontal direction, we get

${\mathrm{T}}_1 \sin {\mathrm{\theta }}_1 = {\mathrm{T}}_2 \sin {\mathrm{\theta }}_2$-------------(i)
For rotational equilibrium about G
$-{\mathrm{T}}_1 \cos {\mathrm{\theta }}_1\mathrm{d}+{\mathrm{T}}_2 {\mathrm{cos\theta }}_2(\mathrm{L}-\mathrm{d}) = 0$

${\mathrm{T}}_1{\mathrm{cos\theta }}_1\mathrm{d} = {\mathrm{T}}_2 {\mathrm{cos\theta }}_2(\mathrm{L}-\mathrm{d})$
Dividing equation (i) by (ii), we get

$\frac{\tan {\mathrm{\theta }}_1}{\mathrm{d}} = \frac{\tan {\mathrm{\theta }}_2}{(\mathrm{L}-\mathrm{d})} \mathrm{or} \frac{\mathrm{L}-\mathrm{d}}{\mathrm{d}} = \frac{\tan {\mathrm{\theta }}_2}{\tan {\mathrm{\theta }}_1}$

$(\frac{\mathrm{L}}{\mathrm{d}}-1) = \frac{\tan {\mathrm{\theta }}_2}{\tan {\mathrm{\theta }}_1} \Rightarrow \frac{\mathrm{L}}{\mathrm{d}} = (\frac{\tan {\mathrm{\theta }}_2}{\tan {\mathrm{\theta }}_1}+1)$

$\frac{\mathrm{L}}{\mathrm{d}} = \frac{\tan {\mathrm{\theta }}_2+\tan {\mathrm{\theta }}_1}{\tan {\mathrm{\theta }}_1}$

$\mathrm{d} = \mathrm{L}\left(\frac{\tan {\mathrm{\theta }}_1}{\tan {\mathrm{\theta }}_1+\tan {\mathrm{\theta }}_2}\right)$