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.

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8 months ago

A free body diagram needs to be drawn.

 

The length of the bar, l = 2 m

T1 and T2

At translational equilibrium, we have Lq? =LR?  = T1

(T1 / T2) = ( T2 / T1sin? 36.9° = 4/3

T1 = (4/3)T2

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

T1 T2sin? 53.1° x d = T2 sin? 53.1°  (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

physics ncert solutions class 11th 2023

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