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  4. Two small kids weighing 10 kg and 15 kg, respectively, are trying to balance a see saw of total length 5.0 m, with the fulcrum at the centre. If kid of mass 10 kg is sitting at an end, at what distance from fulcrum on other side (in m) should the other kid sit?

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Q. Two small kids weighing $10 \,kg$ and $15\, kg$, respectively, are trying to balance a see saw of total length $5.0 \,m$, with the fulcrum at the centre. If kid of mass $10 \,kg$ is sitting at an end, at what distance from fulcrum on other side (in m) should the other kid sit?Physics Question Image

System of Particles and Rotational Motion

Solution:

If the kid of mass $15 \,kg$ sits at the end, he will produce maximum torque about fulcrum. The see-saw will not be balance for any position of kid of mass $10 \, kg$. It is clear that the $10 \, kg$ kid should sit at the end and the $15 \,kg$ kid should sit closer to the centre. Suppose his distance from the centre is $x$.
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As the kids are in equilibrium, the normal force between a kid and the see-saw equals the weight of that kid. Considering the rotational equilibrium of the see-saw, the torque of the forces acting on it should add to zero. The forces are as follows:
(a) $(15\, kg ) g=150\, N$ downwards by the $15 \,kg$ kid,
(b) $(10\, kg ) g=100 \,N$ downwards by the $10\, kg$ kid,
(c) weight of the see-saw, and
(d) the normal force by the fulcrum.
Taking torques about the fulcrum, $150 \times x=100 \times 2.5$
or $x=1.7 \,m$