Physics question!! (please help)

Question

The meterstick shown is 100 cm long. It is free to pivot around its center of gravity (CG), which is at the 50 cm mark. There is a 23.0 N block hanging from the 80 cm mark. Decide where each of the other blocks should be placed, one at a time, to balance out the 23.0 N block. At what mark on the meter stick would you place a 18.0 N block to balance the 23.0 N block? At what mark on the meter stick would you place a 42.0 N block to balance the 23.0 N block?

Answer 1

The torques exerted by the opposing blocks must cancel. Torque is equal to force times the radial distance (the distance between the location of the applied force and the point of rotation).

The torque exerted by the stationary 23N block is (23 N) * (80 cm - 50 cm) = 690 N*cm. Therefore, we must cancel that torque out on the opposite side of the center of gravity (point of rotation) with the other block.

If we use the 18N block, we need to place it at (18 N) * (x cm - 50 cm) = -690 N*cm. Solving for x, we get 11.67 cm.

If we use the 42N block, we need to place it at (42 N) * (x cm - 50 cm) = -690 N*cm. Solving for x, we get 33.57 cm.