A non-uniform bar of weight `W` is suspended at rest by two strings of negligible weight as shown in Fig. The angles made by the strings with the vertical are `36.9^(@) and 53.1^(@)` respectively. The bar is `2m` long. Calculate the distance `d` of the centre of gravity of the bar from its left end.

Explanation

Let `T_(1) and T_(2)` be the tensions in the strings, then For equilibrium along horizontal
`T_(1) sin 36.9^(@)=T_(2) sin 53.1^(@) " "....(i)`
lengtjh of the bar `l=2m`
for rotational equilibrium,
`[T_(1) cos 39.6^(@)]d=[T_(2)cos53.1^(@)](2-d) " ".....(ii)`
Dividing equation (ii) by equation (i), we get
`d cot 36.9^(@)=(2-d)cot 53.1^(@)`
`1.3318d=0.7508(2-d)`
`2.0826d=1.5016`
`d=0.72m=72 cm`