Question

Two children are trying to balance a see-saw. One child has a mass of 40 kg, the other has a mass of 50 kg. If the seesaw is balanced perfectly in the middle and the 40 kg child is sitting 1.6 m from the one end of the see-saw, how far from the center should the 50 kg child be so that the system is perfectly balanced?

Answer 1

To balance the seesaw with a 40 kg child sitting 1.6 m from one end, the 50 kg child needs to sit 1.28 m from the center. This ensures that the torques created by both children are equal and the system is in equilibrium.

Step-by-step explanation:

To balance the seesaw, the torques on both sides must be equal. Torque is calculated by multiplying the force applied by the distance from the pivot point. The formula for torque is:

Torque = Force x Distance

Since the 40 kg child is sitting 1.6 m from the one end of the seesaw, the torque created is:

Torque = (40 kg) x (1.6 m) = 64 Nm

To balance the seesaw, the torque created by the 50 kg child must also be 64 Nm. Let's assume the distance from the center to the 50 kg child is x m. The torque created by the 50 kg child is:

Torque = (50 kg) x (x m)

Setting the two torques equal to each other and solving for x, we get:

64 = (50 kg) x (x m)

x = 1.28 m

So, the 50 kg child should sit 1.28 meters from the center in order to balance the system perfectly.