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 k
Question
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
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2021-07-22T11:36:18+00:00
2021-07-22T11:36:18+00:00 1 Answers
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Answers ( )
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 = ?
Answer:
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.