Question

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.

Answer 1

As is clear from Fig. `theta_(1) = 36.9^(@), theta_(2) = 53.1^(@)`
If `T_(1),T_(2)` are the tensions in the strings, then for equilibrium along the horizontal, `T_(1)sin theta_(1) sin theta_(2)`
or `(T_(1))/(T_(2)) = (sin theta_(2))/(sin theta_(1)) = (sin 53.1^(@))/(sin 36.9^(@)) = (0.7407)/(0.5477) = 1.3523`
Let `d` be the distance of centre of gravity `C` of the bar from the left end.
For rotational equilibrium about `C`,
`T_(1) cos theta_(1) xx d = T_(2) cos theta_(2)(2 - d)`
`T_(1)cos 36.9^(@) xx d = T_(2)cos 53.1^(@) (2 - d)`
`T_(1)xx 0.8366 d= T_(2) xx 0.6718 (2-d)`
Put `T_(1) = 1.3523 T_(2)` and solve to get `d = 0.645 m`