Three equal masses of `m kg` each are fixed at the vertices of an equilateral triangle `ABC`.
a. What is the force acting on a mass `2m` placed at the centroid `G` of the triangle?
b. What is the force if the mass at the vertex `A` is doubled? Take `AG=BG=CG=1m`

Explanation

(i) Refer to Fig. the gravitational force on mass `2 m` at `G` due to mass at `A` is,
`F_(1) = G(m xx 2m)/(1^(2)) = 2 Gm^(2)` along `GA`

Gravitational force on mass `2 m` at `G` due to
mass at `B` is `F_(2) = G(m xx 2m)/(1^(2)) = 2 G m^(2)` along `GB`.
Gravitational force on mass `2 m` at `G` due to
mass at `C` is, `F_(3) = G(m xx 2m)/(1^(2)) = 2 G m^(2)` along `GC`.
Draw `DE` parallel to `BC` passing through point
`G`. Then `angleEGC = 30^(@) = angleDGB`.
Resolving `vec(F_(2))` and `vec(F_(3))` into two rectangular
components, we have
`F_(2) cos 30^(@)` along `GD` and `F_(2) sin 30^(@)` along `GH`,
`F_(3) cos 30^(@)` along `GE` and `F_(3) sin 30^(@)` along `GH`.
Here, `F_(2) cos 30^(@)` and `F_(3) cos 30^(@)` are equal in
magitude and acting in opposite directions, cancel out each other. The resulant force on the mass `2 m` T `G` is
`= F_(1) - (F_(2) sin 30^(@) + F_(3) sin 30^(@))`
`= 2G m^(2) - (2G m^(2) xx(1)/(2) + 2 G m^(2) xx (1)/(2)) = 0`. (ii) When mass at `A` is `2 m`, then gravitational force on mass `2 m` at `G` due to mass `2 m` at `A` is
`F'_(1) = G(2m xx 2m)/(1^(2)) = 4 G m^(2) along GA`
The resultant force on mass `2 m` at `G` due to masses at `A, B` and `C` is
`= F'_(1) - (F_(2) sin 30^(@) + F_(3) sin 30^(@))`
`= 4 Gm^(2) - (2G m^(2) xx(1)/(2) + 2 G m^(2) xx (1)/(2))`
`= 2G m^(2)` along `GA`