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 T1 and T2 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,



T1×0.8×d = T2×0.6×(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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