A uniform L shaped rod of mass 3 m is hinged at point O. length OB is two times the length OA. It is in equilibrium.
Find
(a). Relation between `alpha` and `beta`
(b). Net hinge force.

Explanation

(a). Length OB is two times the length OA. Therefore, mass of OB is 2m and that of OA is m ad their weighht will act at their centres (as the rod is uniform)
If total length is 3l then
`OA=l` `impliesOC_(1)=(OA)/(2)=(l)/(2)`
and `r_(1)=OC_(1)sinalpha=(l)/(2)sinalpha`
`OB=2l` `impliesOC_(2)=(OB)/(2)=l`
and `r_(2)=OC_(2)sinbeta=lsinbeta`
net torque about `O=0`
`implies` anticlockwise torque of `mg=` clockwise torque of 2 mg
`implies(mg)r_(1)=(2mg)r_(2)`
`impliesmg((l)/(2)sinalpha)=(2mg)lsinbeta`
or `(sinalpha)/(sinbeta)=4` is the required relation between `alpha` and `beta`
(b). net force on the rod is also zero. Therefore, hinge force is 3 mg `(=mg+2mg)` in upward direction.