Question

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?

Answer 1

the whole turn is counterclockwise. and x = 0.40 m

Step-by-step 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