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

As it is clear from Fig., `theta_(1)=36.9^(@), theta_(2)=53.1^(@)`.
It `T_(1),T_(2)` are the tensions in the two strings, then for equilibrium along the horizontal,
`T_(1) sin theta_(1)=T_(2) sin theta_(2)`
or ` (T_(1))/(T_(2))=(sin theta_(2))/(sin theta_(1))=(sin 53.1^(@))/(sin36.9^(@))`
`=(0.7404)/(0.5477)=1.3523`
Let d be the distance of centre of gravity C of the bar from the left end.
For rotaional equilibrium about C,
`T_(1) cos theta_(1) xxd=T_(2) cos theta_(2) (2-d)`
`T_(1) cos 36.9^(@)xxd=T_(2) cos 53.1^(@) (2-d)`
`T_(1)xx0.8366 d= T_(2)xx0.6718 (2-d)`
Put `T_(1)=13523 T_(2)` and solve to get
d=0.745 m
`tau=I_(1)alpha_(1)=I_(2)alpha_(2)`
or ` (alpha_(1))/(alpha_(2))=(I_(2))/(I_(1))=((2//5)MR^(2))/(MR^(2))=(2)/(5)`
or ` alpha_(2)=(5)/(2) alpha_(1)`.