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A non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in 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.
Solution
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
$T_{1} \sin 36.9^{\circ}=T_{2} \sin 53.1$
$\frac{T_{1}}{T_{2}}=\frac{\sin 53.1^{\circ}}{\sin 36.9}$
$=\frac{0.800}{0.600}=\frac{4}{3}$
$\Rightarrow T_{1}=\frac{4}{3} T_{2}$
For rotational equilibrium, on taking the torque about the centre of gravity, we have
$T_{1} \cos 36.9 \times d=T_{2} \cos 53.1(2-d)$
$T_{1} \times 0.800 d=T_{2} 0.600(2-d)$
$\frac{4}{3} \times T_{2} \times 0.800 d=T_{2}[0.600 \times 2-0.600 d]$
$1.067 d+0.6 d=1.2$
$\therefore d=\frac{1.2}{1.67}$
$=0.72 m$
Hence, the $C.G.$ (centre of gravity) of the given bar lies $0.72 m$ from its left end
