A boy of mass 45 kg and his dog of mass 25 kg stand on a diving board. If the boy stands 2.0 meters from the pivot of the board, and the dog stands 1.2 meters from the pivot, what is the net torque on the board? (g = 9.8 m/s^{2)}

Nm

We use the formula τ = rF_{⊥}. Since the diving board is flat, the forces of the dog and the boy on the board are both perpendicular to it. The perpendicular component of force is equal to F_{g }= mg. Therefore, the torque that the boy applies on the board is equal to τ = (2.0) (45 × 9.8) = 882. Nm, and τ = (1.2) (25 × 9.8) = 294 Nm. Since these two forces are applied I the same direction we add them and get that the magnitude of the net torque on the diving board is 1176 Nm.