Question

When a boy weighing $20kgf$ sits at 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 $40kgf$

• A. By sitting at distance $1.5m$ from the centre on the side opposite to the boy.
• B. By sitting at distance $1m$ from the centre on the side opposite to the boy.
• C. By sitting at distance $1m$ from the centre on the side of the boy.
• D. By sitting at distance $0.5m$ from the centre on the side opposite to the boy.

Answer 1

The correct option is B By sitting at distance $1m$ from the centre on the side opposite to the boy.

This boy has a perpendicular distance of 2 m from the centre from the point of application of force 20 Kgf. Hence moment of force is $20\times 2=40Kgf$. This must be balanced by horizontal by a man of having perpendicular distance from the centre on the side opposite to the boy is $x$ with a weight of $40Kgf$. Hence moment of force is $40\times x=40$. Hence $x=1m$.