Question

A child of weight 400N and another child of weight 300N play on a seesaw of negligible weight. if the first child sits 2.7m from the pivot of the seesaw , where must the second child sit to make it balance.

Answer 1

Given,

The weight of the 1st child, W=400 N

The weight of the second child, w=300 N

The distance between the point where the 1st child sits and the pivot point, D=2.7 m

The torque is what makes an object rotate. In order to balance the seesaw the torque applied by the weights should be balanced. That is the torque applied by the 1st child should be equal to the torque applied by the second child.

Thus,

`WD=wd`

Where d is the distance from the point to the point where the second child should sit.

Therefore,

`d=(WD)/(w)`

On substituting the known values in the above equation,

`\begin{gathered} d=(400*2.7)/(300) \\ =3.6\text{ m} \end{gathered}`

Thus the second child should sit at a distance of 3.6 m from the pivot.