A15-kg child is sitting on a playground teeter-totter, 1.5 m from the pivot. what is the magnitude of the minimum force, applied 0.30 m on the other side of the pivot, that is needed to make the child lift off the ground?

In order to lift the child off the ground, the torque applied on the other side of the pivot must be equal to the torque produced by the child.

The torque of the child around the pivot is:

where m=15 kg is the mass of the child, g is the gravitational acceleration, d=1.5 m is the distance between the child and the pivot. So, we have


The torque on the other side of the pivot must be equal:

and in this case, the force is applied at a distance of d=0.30 m from the pivot, therefore the intensity of the force must be

From negative to negative