A non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in fig. The angles made by the strings with the vertical are $36 . 9^{o}$ and $53 . 1^{o}$ respectively. The bar is $2 m$ long. Calculate the distance d of the center of gravity of the bar from its left end.

Question

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Answer 1

The free body diagram of the bar is shown in the following figure.
Length of the bar is ,l=2m
$T_{1}, T_{2}$ be the tensions produced in the left and right strings respectively.
At translational equilibrium, we have,
$T_{1} s i n 36 . 9^{\circ} = T_{2} s i n 53 . 1^{\circ}$
$⟹ \frac{T _{1}}{T _{2}} = \frac{4}{3}$
For rotational equilibrium, on taking the torque about the centre of gravity, we have:

$T_{1} (c o s 36 . 9^{\circ}) \times d = T_{2} c o s 53 . 1^{\circ} (2 - d)$
Using both equations,
$d = 0.72 m$
Hence, the centre of gravity of the given bar lies 0.72 m from its left end.

Answer 2

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

Length of the bar is ,l=2m

T_{1}, T_{2}

be the tensions produced in the left and right strings respectively.

At translational equilibrium, we have,

T_{1} s i n 36 . 9^{\circ} = T_{2} s i n 53 . 1^{\circ}

⟹ \frac{T _{1}}{T _{2}} = \frac{4}{3}

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

T_{1} (c o s 36 . 9^{\circ}) \times d = T_{2} c o s 53 . 1^{\circ} (2 - d)

Using both equations,

d = 0.72 m

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