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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.

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#### Solution

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

Length of the bar, l = 2 m

T_{1 }and T_{2} are the tensions produced in the left and right strings respectively.

At translational equilibrium, we have:

`T_1 sin 36.9^@ = T_2 sin 53.1`

`T_1/T_2 = (sin 53.1^@)/(sin 36.9)`

`= (0.800)/0.600 = 4/3`

`=> T_1 = 4/3 T_2`

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

`T_1 cos 36.9 xx d = T_2 cos 53.1 (2-d)`

`T_1 xx 0.800d = T_2 0.600(2-d)`

`4/3xxT_2xx0.800d = T_2[0.600xx2- 0.600d]`

`1.067 d + 0.6d = 1.2`

`∴ d = 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.

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