Answer to Question #142525 in Mechanics | Relativity for Adam

Question

A non-uniform plank AB of mass 20 kg and length 2 m is pivoted at its midpoint. The plank is in equilibrium in a horizontal position when a particle of mass 5 kg is placed 40 cm from A and another particle of mass 8 kg is placed 30 cm from B. Find the distance of the centre of mass of the plank from A.

Answer 1

if these two particles bring system to equilibrium, center of mass for system of two particles (without the plank) and center of mass of the plank will be on different sides related to midpoint symmetrically

$R_{c}=\frac{m_{particle_a}\times l_{particle_a}+m_{particle_b}\times l_{particle_b}}{m_{particle_a}+m_{particle_b}}$

Calculating it related to midpoint (assume A is on the left from midpoint and B is on the right of midpoing)

$R_{c}=\frac{5\times (-0.6)+8\times 0.7}{5+8}=\frac{-3+5.6}{13}=0.2$

So center of mass of particles is located 20cm on right from midpoint

This means center of mass of plank is 20 cm on left from midpoint so distance from A to center of mass is 80 cm