On the other side of the pivot point an adult pushes straight down on the teeter-totter with a force of 190 n. determine the direction the teeter-totter will rotate if the adult applies the force at each of the following distances 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.)

The question seems incomplete. I found a similar version with different details. For the purpose of helping you answering the question, I will use the details from the original questions added with the sub-questions from other question. You can change the details accordingly using the explanation given below:

At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the other side of the pivot point an adult pushes straight down on the teeter-totter with a force of 190 N. Determine the direction the teeter-totter will rotate if the adult applies the force at each of the following distances 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) 3.0 m

(b) 2.5 m

(c) 1.5 m

a) anticlockwise

b) anticlockwise

c) clockwise

Explanation:

To answer this question, we will need the knowledge of moment/torque

Torque, T = r*Fsinθ

r = radial distance from pivot to force, F

F = Force in the direction perpendicular to the distance

θ = angle between force line and r line

Note that the direction of movement will be affected by total torque on the system. Assuming both adult and the child is initially sitting on the same level, we'll have force due to the child weight and adult force going downward and perpendicular to their respective seat on teeter-totter.

Hence,

θ = 90 degrees ->  sinθ = 1

Therefore,

T = r*F(1)

T = r*F

RHS:

Torque of the child,

Tc = 1.8(21*9.8) = 370.44 Nm

LHS:

Torque of adult

Ta = d(190) =190d

with d as the distance of the adult from pivot

a) d = 3.0m

Ta = 190*3 = 570 Nm

Ta > Tc. Therefore the teeter-totter will rotate towards the adult side (LHS - going anti-clockwise)

b) d = 2.5m

Ta = 190*2.5 = 475 Nm

Ta > Tc. Therefore the teeter-totter will rotate towards the adult side (LHS - going anti-clockwise)

c) d = 1.5m

Ta = 190*1.5 = 285

Ta < Tc. Therefore the teeter-totter will rotate towards the child side (RHS - going clockwise)

the force required to move the cart is divided by three or 33.33 n.

it is a negative acceleration