The seesaw in the figure below is 4.5 m long. Its mass of 20 kg is uniformly distributed. The child on the left end has a mass of 14 kg and is a distance of 1.4 m from the pivot point while a second child of mass 39 kg stands a distance L from the pivot point, keeping the seesaw at rest. Find L the distance of the child on the right.

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  1. so, the mass of the seesaw can be ignored, since it is uniformly distributed. Now for the loads and their moments. They must balance on both sides, so

    14*1.4 = 39*L

    Simple, no?

    Actually, since I have no diagram, I can't say that the seesaw mass may be ignored. If the pivot point is not at the center, but rather at some arbitrary distance d from the left), then we have to allow for the different mass of the seesaw needing to be balanced. In that case,

    20(d/4.5)+14*1.4 = 20(1 - d/4.5) + 39L

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  2. Oops. I just gave the mass of the seesaw on each side. That must be multiplied by the distance of its center of mass:

    20(d/4.5)(d/2)+14*1.4 = 20(1 - d/4.5)((4.5-d)/2) + 39L

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  3. Take moment about the left end
    total mass = 20 + 14 + 39 = 73 kg

    That is force up from pivot
    (I am going to do force in kilograms because g will cancel everywhere.)
    pivot distance from left = 1.4
    Distance left to 39 kg = x
    (our L is (x -1.4) )
    moments about left end
    14 * 0
    - 73 * 1.4 up from piviot
    + 20 * 2.25 mass of plank halfway along
    + 39 * x
    39 x + 45 = 102
    x = 1.47
    so L = .07 from the pivot
    In other words the little kid just about balanced the plank.

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