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)

Question

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) (A) (4/3) m (B) 1 m (C) (2/3) m (D) Zero

Tardigrade Admin · Accepted Answer

Let x be the distance from centre, then For rotational equilibrium, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/25ae7191724ed9e7113ce6f2b93d339d-.png" /> (40 × 10) × 2 =(60 × 10) x ⇒ x =(8/6)=(4/3) m So, 60 kg boy has to be displaced =2-(4/3)=(2/3) m.

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? ( $M_{1} = 40 k g$ and $M_{2} = 60 k g$ )

$\frac{4}{3} m$

$1 m$

$\frac{2}{3} m$

Zero

Let $x$ be the distance from centre, then For rotational equilibrium,

$(40 \times 10) \times 2 = (60 \times 10) x$
$\Rightarrow x = \frac{8}{6} = \frac{4}{3} m$
So, $60 k g$ boy has to be displaced
$= 2 - \frac{4}{3} = \frac{2}{3} m$ .

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? (M1​=40kg and M2​=60kg)

34​m

1m

32​m