a uniform bar of mass is supported by support pivoted at top about which bar can swing like simple pendulum. If force F is applied perpendicular to lower end of the bar.How big F must be in order to held the bar in equilibrium.
That will depend upon the angle A that the bar makes with the vertial axis.
The torque due to the weight of the bar, measuered about the pivot, but equal the applied torwue, F*L
F*L = M*g*sin A*(L/2)
(L/2 is the distance of the center of m,ass from the pivot).
Solve for F. L cancels out
F = (M g sinA)/2posted by drwls