Question

A uniform bar 15 meter long is balanced on a pivot placed at it's mid point. a boy boy of mass 55 kilograms sits on one or in the bar at a point 5 meters away from the pivot. What mass can be placed 2meter away from the other end of the bar to keep the bar horizontal

Answer 1

To keep the bar horizontal, the torques on the right and left sides must be equal:

$$F_1d_1=F_2d_2,\\\space\\ F_2=F_1\frac{d_1}{d_2}=(55\cdot9.8)\frac{5}{2}=1347.5\text{ N}.$$

Convert this force of gravity into mass:

$$F_2=m_2g,\\\space\\ m_2=\frac{F_2}{g}=\frac{1347.5}{9.8}=137.5\text{ kg}.$$