Question

Three point masses `m_(1) = 2Kg, m_(2) = 4kg and m_(3)= 6kg` are kept at the three corners of an equilateral triangle of side 1 m. Find the locationo of their center of mass.

Answer 1

Assume `m_(1)` to be at the origin and X-axis along the line joining `m_(1)` as shown in figure.

From the figure, it is clear that the xy-coordinates of `m_(1)` is `x_(1),y_(1)` = (0,0) that of `m_(2)` is `(x_(2)y_(2)) = (1,0)` and that of `m_(3)` is
`(x_(3)y_(3))` = (1/2,sqrt(3/2))
`X_(CM) = (2 xx 0 + 4 xx 1 + 6 xx 1/2)/(2+4+6) = 7/12m`
`y_(CM) = (2 xx 0 + 4 xx 0+6 xx sqrt(3)//2)/(2+4+6)=7/12 m`
`y_(CM) = (2 xx 0 +4 xx0 +6xxsqrt(3)//2)/(2+4+6) =(3sqrt(3))/2=sqrt(4)/4m`
`therefore` Center of mass is at `(7/12, sqrt(3)/4)`.