A 15-kg child is sitting on a playground teeter-totter, 1.5 m from the pivot. What is the minimum distance, on the other side of the pivot, such that a 220-N force will make the child lift off the ground?

1 m

Explanation:

Given that

Mass of the child, m = 15 kg

Distance of the pivot, d = 1.5 m

Force applied, F = 220 N

To solve this problem, we first need to find the torque around the pivot.

Torque, t = mgd, where

m is the mass of the child

g is the acceleration due to gravity and

d is the distance of the pivot.

Thus, we can say that the torque is

T = 15 * 9.8 * 1.5

T = 220.5 Nm

This torque we have gotten would be used to find the distance, using the inverse of the equation.

T = F * d

d = T / F

d = 220.5 / 220

d = 1 m

Therefore, the minimum distance on the other side of the pivot required is 1 m

the bal will not change its velocity.