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

Explanation

With the x–and y–axes chosen as shown in Fig. 7.9, the coordinates of points O, A and B forming the equilateral triangle are respectively (0,0), (0.5,0), `(0.25,0.25 sqrt(3))`. Let the masses 100 g, 150g and 200g be located at O, A and B be respectively. Then,
`X=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3))/(m_(1)+m_(2)+m_(3))`
`=(100(0)+150(0.5)+200(0.25)gm)/((100+150+200)g)`
`=(75+50)/(450)m=(125)/(450)m=(5)/(18)m`
`Y=(100(0)+150(0)+200(0.25sqrt(3))gm)/(450g)`
`=(50sqrt(3))/(450)m=(sqrt(3))/(9)m=(1)/(3sqrt(3))m`
The centre of mass C is shown in the figure. Note that it is not the geometric centre of the triangle OAB. Why?