Question

A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in the figure. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

Answer 1

The free body diagram of the bar is shown in the following figure.

Length of the bar, l = 2 m

T₁ and T₂ are the tensions produced in the left and right strings respectively.

At translational equilibrium, we have:

${\displaystyle T_1{\sin 36.9}^{\circ }=T_2\sin 53.1}$

${\displaystyle \frac{T_1}{T_2}=\frac{{\sin 53.1}^{\circ }}{\sin 36.9}}$

${\displaystyle =\frac{0.800}{0.600}=\frac{4}{3}}$

${\displaystyle \Rightarrow T_1=\frac{4}{3}{T}_2}$

For rotational equilibrium, on taking the torque about the centre of gravity, we have:

${\displaystyle T_1\cos 36.9\times d=T_2\cos 53.1(2-d)}$

${\displaystyle T_1\times 0.800d=T_{2}0.600(2-d)}$

${\displaystyle \frac{4}{3}\times T_2\times 0.800d=T_2[0.600\times 2-0.600d]}$

${\displaystyle 1.067d+0.6d=1.2}$

${\displaystyle \therefore d=\frac{1.2}{1.67}}$

= 0.72 m

Therefore, the center of gravity (C.G.) of the specified bar is located 0.72 meters from its left end.