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?
A) 1.0 m
B) 0.10 m
C) 1.5 m
D) 2.4 m
E) 9.8 m

Question

Quizplus · Accepted Answer

To lift the child off the ground, the torque produced by the 220-N force must equal or exceed the torque produced by the child's weight. The child's weight is $15\, \text{kg} \times 9.8\, \text{m/s}^2 = 147\, \text{N}$. The torque due to the child's weight is $147\, \text{N} \times 1.5\, \text{m} = 220.5\, \text{Nm}$. To find the minimum distance ($d$) on the other side of the pivot for the 220-N force, we set the torques equal: $220\, \text{N} \times d = 220.5\, \text{Nm}$. Solving for $d$ gives $d = 220.5\, \text{Nm} / 220\, \text{N} = 1.0023\, \text{m}$, which rounds to 1.0 m, the closest answer provided.

A 15-Kg Child Is Sitting on a Playground Teeter-Totter, 1