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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Helga 4 years 2021-07-22T11:36:18+00:00 1 Answers 13 views 0

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    2021-07-22T11:37:43+00:00
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    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.

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