Question

A 16.6 kg kid & a 28.6 kg kid are at opposite ends of a 2.21 m long 42.7 kg see-saw, with the fulcrum at the center. You want to add a mass of 25 kg to the see-saw to balance it. How far from the smaller kid should you put it, in meters?

Answer 1

Using Torque balance about the fulcrum at the center:

For see-saw to be balanced, net torque about fulcrum should be zero

fundamentally we know that 25 kg mass should be towards the smaller kid for see-saw to be balanced.

Net torque = 0

W1*(L/2) - W2*(L/2) + W3*(L/2 - x) = 0

W1 = 16.6 kg, W2 = 28.6 kg, and W3 = 25 kg

L = 2.21 m, L/2 = 2.21/2 = 1.105 m

So,

16.6*1.105 - 28.6*1.105 + 25*(1.105 - x) = 0

Solving above equation:

x = [16.6*1.105 - 28.6*1.105 + 25*1.105]/25

x = 0.5746 m from the fulcrum at the center = 1.105 - 0.5746 = 0.5304 m from the smaller kid

So 25 kg mass should be 0.53 m from the smaller kid