A boy weighing $20 k g$ is sitting at one end of a see-saw at a distance of $1.2 m$ from the centre. Where should a man weighing $60 k g$ sit on the see-saw, so that it stands balanced?

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Solution

Let, the man weighing $60$ kg will be at a distance of $x$ m from the centre of the see-saw.
Using equilibrium of moment of force,
$60 \times g \times x = 20 \times g \times 1.2$
$x = \frac{20 \times 1.2}{60}$
$x = 0.4 m$ .

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A boy weighing 20 kg is sitting at one end of a see-saw at a distance of 1.2 m from the centre. Where should a man weighing 60 kg sit on the see-saw, so that it stands balanced?

Let, the man weighing 60 kg will be at a distance of x m from the centre of the see-saw.Using equilibrium of moment of force,60×g×x=20×g×1.2x=20×1.260x=0.4m.

A boy weighing $20 k g$ is sitting at one end of a see-saw at a distance of $1.2 m$ from the centre. Where should a man weighing $60 k g$ sit on the see-saw, so that it stands balanced?

Let, the man weighing $60$ kg will be at a distance of $x$ m from the centre of the see-saw.
Using equilibrium of moment of force,
$60 \times g \times x = 20 \times g \times 1.2$
$x = \frac{20 \times 1.2}{60}$
$x = 0.4 m$ .