Question

A uniform bar is 4m long. It weighs 300N. At one end, 1m from the pivot point, there is a resisting weight of 2000N. What force must be applied at the other end to produce equilibrium?

Answer 1

By equilibrium of torques (around R1):

$$2000·1+300·2-R_2·4=0,\\ R_2=650\text{ N}.$$