Question

A non-uniform bar of weight $W$ is suspended at rest by two string of negligible weights as shown in figure. The angle made by the string with the vertical are $36.9^{\circ }$ and $53.1^{\circ }$ respectively. The bar $2m$ long. Calculate the distance of the centre of gravity (in $cm$ ) of the bar from it left end.

Answer 1

Answer: 72

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

For translational equilibrium, on balancing the vertical components of forces,
we get $T_1\cos \theta +T_2\cos \varphi =W...(1)$
On balancing the horizontal components,
we get $T_1\sin \theta =T_2\sin \varphi ...(2)$
For rotational equilibrium, we balance the torques about the $CG$ of the bar.
Clockwise torque $=$ Anticlockwise torque
$T_1\cos \theta \times d=T_2\cos \varphi \times (2-d)...(3)$
Diving (3) by (2), we get
$d\cot \theta =(2-d)\cot \varphi$
or $d\cot 36.9^{\circ }=(2-d)\cot 53.1^{\circ }$
or $d\cot 36.9^{\circ }=(2-d)\tan 36.9^{\circ }$
or $d\times \frac{4}{3}=(2-d)\frac{3}{4}$
or $16d=18-9d$
$d=\frac{18}{25}=0.72m=72cm$