Question

Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs. If the child on the left is 3 feet from the pivot, how far must the child on the right be from the pivot fro the seesaw to be balanced.

Answer 1

Given :

Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs.

To Find :

If the child on the left is 3 feet from the pivot, how far must the child on the right be from the pivot from the seesaw to be balanced.

Solution :

We know, for seesaw to balance :

$m_1gd_1=m_2gd_2$

Here, $d_1\ and \ d_2$ is distance from origin from pivot point.

Putting all the values in the equation, we get :

$50\times g\times 3=100\times g\times d_2\\\\d_2=1.5\ feet$

Therefore, distance of right child from pivot to balance the seesaw is 1.5 feet.

Hence, this is the required solution.