When a boy weighing $20 k g f$ 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 $40 k g f$ ?

Open in App

Solution

Let the man sit $x$ away from the center
$∴ 20 \times 4 = 40 x$
$∴ x = 2 m$ away from a center

Suggest Corrections

Similar questions

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?

Q. A person weighing 45 kg sits in one end of a see saw while a boy of 15 kg sits on the other end. If they are separated by 4 m , how far from the boy the centre of mass situated. Neglect the weight of the see saw.

Join BYJU'S Learning Program

When a boy weighing 20kgf 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 a man weighing 40kgf?

Let the man sitx away from the center∴20×4=40x∴x=2m away from a center

When a boy weighing $20 k g f$ 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 $40 k g f$ ?

Let the man sit $x$ away from the center
$∴ 20 \times 4 = 40 x$
$∴ x = 2 m$ away from a center