Three children want to play on a see saw that is 6 meters long and has a fulcrum in the middle. Two of the children are twins and weigh 40 kg each and sit on the same side at a distance 2m and 3m away from the fulcrum. The other child weighs 80 kg. How far away should he sit from the fulcrum so that the see saw is balanced

Given Information:

mass of child 1 = m₁ = 40 kg

distance from fulcrum of child 1 = d₁ = 2 m

mass of child 2 = m₂ = 40 kg

distance from fulcrum of child 2 = d₂ = 3 m

mass of child 3 = m₃ = 80 kg

Required Information:

distance from fulcrum of child 3 = d₃ = ?

distance from fulcrum of child 3 = 2.5 m

Explanation:

In order to balance the see-saw, the moment of force should be same on both sides of the fulcrum.

Since 2 children are sitting on one side and only 1 on the other side

F₁d₁ + F₂d₂ = F₃d₃

Where Force is given by

F = mg

m₁gd₁ + m₂gd₂ = m₃gd₃

m₁d₁ + m₂d₂ = m₃d₃

Re-arrange the equation for d₃

m₃d₃ = m₁d₁ + m₂d₂

d₃ = (m₁d₁ + m₂d₂)/m₃

d₃ = (40*2 + 40*3)/80

d₃ = 2.5 m

Therefore, the child on the other side should sit 2.5 m from the fulcrum so that the see-saw remains balanced.

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