A 28.8 kg child sits on a 6.0 m long teeter-totter at a point 1.5 m from the pivot point (at the center of the teeter-totter). On the other side of the pivot point, an adult pushes straight down on the teeter-totter with a force of 180 N. Determine the direction the teeter-totter will rotate if the adult applies the force at a distance of each of the following from the pivot. (Assume the teeter-totter is horizontal when the adult applies the force and that the child's weight applies a clockwise torque.) a.
1. 1.0 m
2. counterclockwise

b.
1. 2.0 m
2. clockwise
3. counterclockwise

c.
1. 3.0m
2. clockwise
3. counterclockwise

case A) tau_net = -243.36 N m, case B)    tau_net = 783.36 N / m,      tau_net = -63.36 N m,  case C)  tau _net = - 963.36 N m,

Explanation:

For this exercise we use Newton's relation for rotation

         Σ τ  = I α

In this exercise the mass of the child is m = 28.8, assuming x = 1.5 m, the force applied by the man is F = 180N

we will assume that the counterclockwise turns are positive.

case a

         tau_net = m g x - F x2

          tau_nett = -28.8 9.8 1.5 + 180 1

         tau_net = -243.36 N m

in this case the man's force is downward and the system rotates clockwise

case b

2 force clockwise, the direction of

 the force is up

          tau_nett = -28.8 9.8 1.5 - 180 2

          tau_net = 783.36 N / m

in case the force is applied upwards

3) counterclockwise

        tau_nett = -28.8 9.8 1.5 + 180 2

         tau_net = -63.36 N m

system rotates clockwise

case c

2 schedule

 tau_nett = -28.8 9.8 1.5 - 180 3

 tau _net = - 963.36 N m

3 counterclockwise

       tau_nett = -28.8 9.8 1.5 + 180 3

       tau_net = 116.64 Nm

the sitam rotated counterclockwise

your answer would be fusion