A long uniform wooden board (a playground see-saw) has a pivot point at its center. An older child of mass M=32 kg is sitting a distance L

Question

A long uniform wooden board (a playground see-saw) has a pivot point at its center. An older child of mass M=32 kg is sitting a distance L from the pivot. On the other side of the pivot point are two smaller children each of mass M/2. One is sitting a distance L/6 from the pivot. How far from the pivot must the other small child be sitting in order for the system to be balanced?

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Thiên Di 5 years 2021-07-18T07:03:30+00:00 1 Answers 13 views 0

Answers ( )

  1. Thunguyet
    0
    2021-07-18T07:05:23+00:00
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    Answer:\frac{5L}{6}

    Explanation:

    Given

    Wooden board is pivoted at center and

    Older child of mass M=32\ kg is sitting at a distance of L from  center

    if two child of mass \frac{M}{2} is sitting at a distance \frac{L}{6} and x(say) from pivot then net torque about pivot is zero

    i.e.

    \Rightarrow \tau_{net}=MgL-\frac{M}{2}g\frac{L}{6}-\frac{M}{2}gx

    as \tau_{net}=0

    Therefore

    MgL=\frac{M}{2}g\frac{L}{6}+\frac{M}{2}gx

    L-\frac{L}{6}=x

    x=\frac{5L}{6}

    Therefore another child is sitting at a distance of \frac{5L}{6}

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