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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.

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  1. 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, from the pivot).

    Solve for F. L cancels out

    F = (M g sinA)/2

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