Two children seat themselves on a seesaw. the one of the left has a weight of 400 n while the one of the right weights 300 n. the fulcrum is at the midpoint of the seesaw. if the child on the left is not at the end but 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?

The equation of torque is:

where r is the distance from the fulcrum and F is the force.  Since the problem states that our seesaw is balanced this means that the angle is zero and so theta = zero and sin(0)=1 therefore we can ignore that part of the equation since it's 1.  Furthermore since the problem states that it is in equilibrium this means that the left side is equal to the right, mathematically :

Let the left be 1 and right be 2.  In order to calculate the amount of torque the right side is providing to the system we need to first find out the distance (r2) that the person on the right is sitting with respect to the fulcrum and so:

And now we will calculate the torque on the right side:

600 mN

+15-35=-20

going in opposite directions