Question

When a boy weighing $20kg$ sits at one end of a $4m$ long see-saw, it gets depressed at this end. How can it be brought to the horizontal position by another boy weighing$40kg$?

• A. By sitting at a distance of $1m$ from the centre on the side opposite to the first boy.
• B. By sitting at a distance of $1m$ from the centre on the same side as the first boy.
• C. By sitting at a distance of $2m$ from the centre on the side opposite to the first boy.
• D. By sitting at a distance of $2m$from the centre on the same side as the first boy.

Answer 1

The correct option is A

By sitting at a distance of $1m$ from the centre on the side opposite to the first boy.

Weights of the two boys : $W_1$ = $20kg$, $W_2$ ​ = $40kg$

Length of see-saw = $4m$
Distance from the centre =$2m$
Moment of Force = Force × perpendicular distance
Then, the principle of moments can be written as follows:
Moment in clockwise direction = Moment in anticlockwise direction
$W_1$​ $\times$ $d_1$​ = $W_2$​ $\times$ $d_2$​

20 $\times$ 2 = 40 $\times$ $d_2$​​

$d_2$​​​ = $\frac{20\times 2}{40}$ = $1m$

Thus, the other boy can sit at a distance of $1m$ from the centre on the side opposite to the first boy, so that the long see - saw can be brought to the horizontal position.