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)1.5 m
C)0.10 m
D)9.8 m
E)2.4 m

Question

Quizplus · Accepted Answer

To lift the child off the ground, the torque on one side of the teeter-totter must be equal and opposite to the torque on the other side. We can set up an equation using the formula for torque:

τ = Fd

where τ is the torque, F is the force, and d is the distance from the pivot. We want to solve for d, so we can rearrange the equation:

d = τ/F

We know that the torque on the child's side is:

τ1 = mgd1

where m is the mass of the child, g is the acceleration due to gravity (approximately 9.8 m/s^2), and d1 is the distance from the pivot to the child. We can plug in the numbers:

τ1 = (15 kg)(9.8 m/s^2)(1.5 m) = 220.5 Nm

Now we can solve for d2, the distance on the other side of the pivot:

d2 = τ1/F = (220.5 Nm)/(220 N) = 1.002 m

So the minimum distance on the other side of the pivot is approximately 1.0 m. Therefore, the answer is A) 1.0 m.

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