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

Open in App

Solution

Weight of boy ( $W_{1}$ ) = 20 kg, $W_{2}$ = 40 kg

Length of see-saw = 4m

Distance from centre = 2m

$W_{1}$ $\times$ $d_{1}$ = $W_{2}$ $\times$ $d_{2}$

20 $\times$ 2 = 40 $\times$ $d_{2}$

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

By sitting at a distance of 1m from two centres on the side opposite to the boy, the long see - saw can be brought to the horizontal position.

Suggest Corrections

Similar questions

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

Q. When a boy weighing

20 k g f

sits at end of a

4 m

long see-saw, it gets depressed at this end. How can it be brought to the horizontal position by a man weighing

40 k g f

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

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

Q. 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?

Join BYJU'S Learning Program