A 30kg child and 20kg child sit on the opposite ends of a seesaw pivoted at its centre. Where should another 20kg child sit in order to balance the seesaw?
To balance the seesaw, the torques on both sides of the pivot point need to be equal. Torque is the product of the force and the distance from the pivot point.
In this case, we have a 30kg child on one end and a 20kg child on the other end. Let's assume the distance from the pivot to the 30kg child is D1, and the distance from the pivot to the 20kg child is D2. The torque exerted by each child is given by Torque1 = force1 * D1 and Torque2 = force2 * D2.
Since the seesaw is balanced, Torque1 = Torque2.
For the 30kg child:
Torque1 = force1 * D1 = 30kg * g * D1, where g is the acceleration due to gravity.
For the 20kg child:
Torque2 = force2 * D2 = 20kg * g * D2.
We need to find the position where we can place another 20kg child so that the torques are balanced. Let's assume this distance is D3.
For the new 20kg child:
Torque3 = force3 * D3 = 20kg * g * D3.
Since the seesaw is balanced, Torque1 = Torque2 = Torque3.
Therefore, we have:
30kg * g * D1 = 20kg * g * D2 = 20kg * g * D3.
We can ignore the gravitational acceleration term 'g' as it cancels out:
30kg * D1 = 20kg * D2 = 20kg * D3.
To balance the seesaw, the distance from the pivot to the new 20kg child should be D3 = (30kg * D1) / 20kg. Similarly, the distance from the pivot to the 20kg child would be D2 = (30kg * D1) / 20kg.
So, the new 20kg child should be placed at a distance from the pivot equal to (30kg * D1) / 20kg on the same side as the 20kg child.