Question

A 2.8 m uniform board weighing 210N lays flat on the ground. A pet chipmunk sits 0.6m from the right. The chipmunk is 520 N. What force is needed to lift the board uniformly at the chipmunk’s end?

Answer 1

To lift the board uniformly at the chipmunk's end, we need to calculate the torque acting on the board.

Torque = force x distance

The force acting on the board is the weight of the board itself, which is 210 N. The distance from the chipmunk to the end of the board is 2.8 m - 0.6 m = 2.2 m.

So, the torque acting on the board is:

Torque = 210 N x 2.2 m = 462 Nm

To lift the board uniformly at the chipmunk's end, we need to apply a force that creates an equal and opposite torque.

So, the force needed to lift the board uniformly at the chipmunk's end is:

Force = Torque / distance

Force = 462 Nm / 0.6 m = 770 N

Therefore, a force of 770 N is needed to lift the board uniformly at the chipmunk's end.