Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100g, 150g, and 200g respectively. Each side of the equilateral triangle is 0.5m long.

Explanation

Suppose mass of particle located at O, `m_(1)` = `100 g` and its coordinates `vec(r_(1))=(0,0)`.
The mass of particle located at `A,m_(2)=150g` and its coordinates `vec(r_(2))=(0.5,0)`.
The mass of particle located at `B,m_(3)=200g` and its coordinates `vec(r_(3))=10.5cos60^(@),0.5sin60^(@)`.
`=(0.5xx(1)/(2),0.5xx(sqrt(3))/(2))`
`=(0.25,0.25sqrt(3))`
Coordinates of centre of mass
`r_(cm)^(rarr)=(m_(1)vecr_(1)+m_(2)vecr_(2)+m_(3)vecr_(3))/(m_(1)+m_(2)+m_(3))`
`=(100(0,0)+150(0.5,0)+200(0.25,0.25sqrt(3)))/(100+150+200)`
`=((0,0)+(75,0)+(50,50sqrt(3)))/(450)`
`=(("125,"50sqrt(3)))/(450)`
`=((5)/(18),(1)/(3sqrt(3)))m`
`=(0.28,0.19)m`