Question

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.

Answer 1

Refer to Fig.

In `DeltaOBN, cos 60^(@) = (ON)/(OB) = (ON)/(0.5)`
`ON = 0.5 cos 60^(@) = 0.25 m`
`sin 60^(@) = (BN)/(OB) = (BN)/(0.5)`
`BN = 0.5 sin 60^(@) = 0.5 (sqrt(3))/(2) = 0.25 sqrt(3)`
`:.` The co-ordinates of `O (0, 0), A (0.5,0)` and `B(0.25, 0.25 sqrt(3))`. The masses of `100g, 150g`, and `200g` are located at the vertices `O,A,B` of the equilateral triangle `OAB`. The co-ordinates of the centre of mass
are `x = (m_(1)x_(1) + m_(2)x_(2) + m_(3)x_(3))/(m_(1)+m_(2) + m_(3))`
`= (100 xx 0 + 150 xx 0.5 + 200 xx0.25)/((100 + 150 + 200))`
`= (75 + 50)/(450) = (125)/(450) = (5)/(18)m`
and `y = (m_(1)y_(1) + m_(2)y_(2) + m_(3)y_(3))/(m_(1) + m_(2) + m_(3))`
`= (100 xx 0 + 150 xx 0+200 xx 0.25 sqrt(3))/(100+150+200)`
`= (50sqrt(3))/(450) = (sqrt(3))/(9) = (1)/(3sqrt(3)) m`
Note that the centre of mass `C` is not the geometric centre of triangle `OAB`. This is because the three vertices of the triangle are not uniformly loaded.