Two children seat themselves on a seesaw, the one on the left has a weight of 400N whlie the one on the right weighs 300N. The fulcrum is at the midpoint of the seesaw, if the child on the left is not at the end but is 1.5m from the fulcrum, and the seesaw is balanced, what is the torque provided by the weight of the child on the right?

To find the torque provided by the weight of the child on the right, we need to understand the concept of torque. Torque is a measurement of the turning force around an axis or fulcrum. It is calculated by multiplying the force applied perpendicular to the axis of rotation by the distance from the axis.

In this case, since the seesaw is balanced, the torque provided by the child on the left (400N) should be equal to the torque provided by the child on the right (unknown). Let's calculate it.

The torque formula is given by:

Torque = Force × Distance

The distance from the fulcrum to the child on the left is 1.5m. This means that the torque provided by the child on the left is:

Torque_left = Force_left × Distance_left
= 400N × 1.5m
= 600 Nm

Since the seesaw is balanced, the torque provided by the child on the right should be equal to the torque provided by the child on the left:

Torque_right = Torque_left
= 600 Nm

Therefore, the torque provided by the weight of the child on the right is 600 Nm.