Three particles of masses `1kg`, `2kg` and `3kg` are placed at the corners A, B and C respectively of an equilateral triangle ABC of edge `1m`. Find the distance of their centre of mass from A.

Explanation

Assume that 1kg mass is placed at origin as shown in figure.
co-ordinate of A = (0, 0)
co-ordinate of B = (1cos60°,1sin60°) =`(1/2,(sqrt3)/2)`
co-ordinate of C = (1, 0)

Let us assume that position of C.O.M is given by
`vec(r_("com")=x_("com")hati+y_("com")hatj`
Now `x_("com")=(m_(A)x_(A)+m_(B)x_(B)+m_(C)X_(C))/(m_(A)+m_(B)+m_(C))`
`=(1(0)+2(1/2)+3(1))/(1+2+3)=4/6=2/3`
`y_("com")=(1(0)+2((sqrt3)/2)+3(0))/(1+2+3)=(sqrt3)/6`
Position of centre of mass = `(2/3,(sqrt3)/6)`
distance of C.O.M from point A = `sqrt((2/3)^(2)+((sqrt3)/6)^(2))`
`=(sqrt(19))/6`m