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

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?