7.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. 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.

7.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.7.39. 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.

A free body diagram needs to be drawn.

The length of the bar, l = 2 m
T1 and T2
At translational equilibrium, we have =
(T1 / T2) = ( / = 4/3
T1 = (4/3)T2
For rotational equilibrium, on taking the torque about the centre of gravity, we have
T1 x d = T2 (2-d)
T1 x 0.8d = T2 x 0.6 (2-d)
(4/3)T2 x 0.8
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physics ncert solutions class 11th 2023
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