Question

A meter stick on a horizontal frictionless table top is pivoted at the 80-cm mark. A force is applied perpendicularly to the end of the stick at 0 cm, as shown. A second force (not shown) is applied perpendicularly at the 100-cm end of the stick. The forces are horizontal. If the stick does not move, the force exerted by the pivot on the stick:

Answer 1

The stick is in equilibrium:
the net force and net torque on the stick will be zero.
i.e,
Force exerted by the pivot on the stick=F1+F2...........(1)
and F1*.80=F2*.20=>F2=4F1.
hence from (1)
Force exerted by the pivot on the stick=F1+F2=F1+4F1=5F1 answer.