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) The angle between GR and the positive x-axis is `30^(@)`. Similarly the angle between GQ and negative x-axis is `30^(@)`.
Now, `F_(GP)=(G(2m)m)/(1)hat(j)` ………… (i.e., along positive y-axis)
`F_(GQ)=(G(2m)m)/(1)(-hat(i)cos30^(@)-hat(j)sin 30^(@))`
`F_(GR)=(G(2m)m)/(1)(hat(i)cos 30^(@)-hat(j)sin 30^(@))`
By principle of superposition, we get
`vec(F)_(R)=vec(F)_(GP)+vec(F)_(GR)`
`therefore vec(F)_(R)=2Gm^(2)hat(i)+2Gm^(2)(-hat(j)cos 30^(@)-hat(j)sin30^(@))+2Gm^(2)(hat(i)cos 30^(@)-hat(j)sin 30^(@))=0`
(ii) By symmetry all the x-component of the force will cancel out each other.
`therefore vec(F)_(R)=8Gm^(2)hat(j)-2Gm^(2)-sin30^(@)-2Gm^(2)sin 30^(@)`
`=8Gm^(2)hat(j)-4Gm^(2)sin 30^(@)`
`=8Gm^(2)hat(j)-2Gm^(2)hat(j)`
`=6Gm^(2)hat(j)`