Question

A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. 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: T1 sin 36.9° = T2 sin 53.1
$\frac{T_1 }{ T_2}$ = $\frac{sin\, 53.1° }{ 36.9 }$

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

$⇒T_1 = \frac{4 }{ 3} T_2$

For rotational equilibrium, on taking the torque about the centre of gravity, we have : T1 cos 36.9 × d = T2 cos 53.1 (2 - d)

T1 cos 36.9× d=T2 cos 53.1(2-d)

T1 × 0.800d = T2 0.600 (2 - d)

$\frac{4 }{ 3} × T_2$ × 0.800d =T2 [0.600 × 2 - 0.600d]

1.067d + 0.6d = 1.2

$∴d = \frac{1.2 }{1.67}$= 0.72m

Hence, the C.G. (centre of gravity) of the given bar lies 0.72 m from its left end.