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  4. In the game of see-saw, what should be the displacement of boy B from right edge to keep the see-saw in equilibrium? (M1=40 kg and M2=60 kg)

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Q. In the game of see-saw, what should be the displacement of boy $B$ from right edge to keep the see-saw in equilibrium? ($M_{1}=40\, kg$ and $M_{2}=60\, kg$)Physics Question Image

System of Particles and Rotational Motion

Solution:

Let $x$ be the distance from centre, then For rotational equilibrium,
image
$(40 \times 10) \times 2 =(60 \times 10) x$
$\Rightarrow x =\frac{8}{6}=\frac{4}{3} m$
So, $60\, kg$ boy has to be displaced
$=2-\frac{4}{3}=\frac{2}{3} m$.