Q. 84.7( 24 Votes )

# 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.

Answer :

Given,

Length of the bar, l = 2 m

Let T_{1} and T_{2} be the tensions produced by the strings.

The free body diagram can be drawn as,

For transitional equilibrium we have,

⇒

∴

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

⇒ T_{1}×0.8×d = T_{2}×0.6×(2-d)

⇒

⇒ 1.067d + 0.6d = 1.2

∴ m = 0.72 m

Hence, Centre of gravity lies at a distance of 0.72 m from the left side of the bar.

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