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 $2\, m$ long. Calculate the distance of the centre of gravity (in $cm$ ) of the bar from it left end.

Answer 1

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 $T_{1} \cos \theta+T_{2} \cos \phi=W\,\,\,...(1)$
On balancing the horizontal components,
we get $T_{1} \sin \theta=T_{2} \sin \phi\,\,\,...(2)$
For rotational equilibrium, we balance the torques about the $C G$ of the bar.
Clockwise torque $=$ Anticlockwise torque
$T_{1} \cos \theta \times d=T_{2} \cos \phi \times(2-d)\,\,\,...(3)$
Diving (3) by (2), we get
$d \cot \theta=(2-d) \cot \phi$
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 $16 d=18-9 d$
$d=\frac{18}{25}=0.72\, m =72 \,cm$