Question

Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3 m, at what distance from the pivot point is the small child sitting in order to maintain the balance?

Answer 1

Answer:

Explanation:

Given

mass of first child $m_1=20\ kg$

mass of second child $m_2=30\ kg$

Distance between two children is $d=3\ m$

Suppose light weight child is placed at a distance of x m from Pivot point

therefore

Torque due to heavy child $T_1=m_2g\times (3-x)$

Torque due to small child $T_2=m_1g\times x$

Net Torque about Pivot must be zero

Therefore $T_1=T_2$

$30\times g\times (3-x)=20\times g\times x$

$9-3x=2x$

$9=5x$

$x=\frac{9}{5}$

$x=1.8\ m$