At the local playground, a 20-kg child sits on the right end of a horizontal teeter-totter, 2.0 m from the pivot point. on the left side of the pivot an adult pushes straight down on the teeter-totter with a force of 150 n. in which direction does the teeter-totter rotate if the adult applies the straight down force at a distance of 3.0 m from the pivot? in which direction does the teeter-totter rotate if the adult applies the straight down force at a distance of 2.0 m from the pivot? the adult has 100-kg. how faraway from the pivot needs she to sit on the teeter-totter, without reaching the ground with her feet, for maintaining the plank horizontal?

the whole turn is counterclockwise.   and       x = 0.40 m

Explanation:

This is a problem of rotational equilibrium so we will use the equation

    Σ τ = I α

In this case alf is zero, the torque equation is

     τ = F r

We will take the sense of counterclockwise rotation as positive.

Let's apply these expressions to our exercise.

We have the weight of the child that creates a right turn so it is negative and the force applied to the left of the pivot that creates a left turn the torque is positive

adult     τ₁ = F x

child      τ₂ = W x₂

            τ₁ = 150 3

            τ₁ = 450 N m

            τ₂ = 20 9.8 2.0

            τ₂ = 392 Nm

             τ₁> τ₂

Here we see that the child's side turns in (goes up) and the adult's part goes down, the whole turn is counterclockwise.

Now the adult applies the force to a shorter distance (x = 2.0 m)

     τ₁ = 150 2.0

     τ₁ = 300 N m

     τ₁ < τ₂

The direction of rotation is time, the adult goes up and the child goes down.

The last case the adult sits and asks the distance

    W x = mg x₂

    M g x = m g x₂

    x = x₂ m / M

    x = 2.0 20/100

    x = 0.40 m

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