Two children seat themselves on a seesaw. the one on the left has a weight of 400 n while the one on the right weighs 300 n. the fulcrum is at the midpoint of the seesaw. if the child on the left is not at the end but is 1.50 m from the fulcrum and the seesaw is balanced, what is the torque provided by the weight of the child on the right?

600Nm

Explanation:

Hello!

To solve this exercise we must follow the following steps.

1.Draw the free body diagram of the balance (see attached image)

2. For the seesaw to be in balance the torques with respect to point C (see attached image) must be the same but in the opposite direction.

3. Remember that the torque is calculated as the distance by the force perpendicular to the distance.

4. Taking into account the above we propose the following equation

T=Fd

where

T=torque provided by the weight of the child on the righ

F= left child weight =400N

d=distance=1.5m

T=(400N)(1.5m)=600Nm