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

Mass of the spherical ball at the corner .A. of the equilateral triangle = 1kg
Mass of the spherical ball at the corner .B. of the equilateral triangle = 2kg
Mass of the spherical ball at the comer .C. of the equilateral triangle = 3kg
Side of equilateral triangle = 1m
If .F. is the force of attraction between 1kg, 2kg masses then
`F_1 = G xx (1 xx 2)/((1)^2) rArr F_1 = 2G`
If `.F_2.` is the force of attraction between 1kg, 3kg masses then
` F_2 = G xx (1xx3)/((1)^2) rArr F_2 = 3G`
The angle between the forces `F_1` and `F_2` is `60^@`. If `.F_R.` is the resultant of these two forces, then according to parallelogram law of vectors we can write
`F_R = sqrt(F_1^2 + F_2^2 +2F_1F_2 Cos theta ) rArr F_R = sqrt((2G)^2 + (3G)^2 +2 xx 2G xx 3G xx cos 60^@) rArr F_R = sqrt19 G`