Three equal masses of 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 = 1 m

Explanation

The individual force in vector notation are
`F_(GA) = (G(m)(2m))/(1)hat(j)`
`F_(GB) = (G(m)(2m))/(1) (-hat(l) cos 30^(@) - hat(j) sin 30^(@))`
`F_(GC) = (G(m)(2m))/(1) = (hat(l) = cos 30^(@) - hat(j) sin30^(@))`
From the principle of superposition and the law of vector addition the resultant gravitational force `F_(R)` on (2m) is
`F_(R) = F_(GA) + F_(GB) + F_(GC)`
`F_(R) = 2Gm^(2)hat(j) + 2Gm^(2) (-hat(i) cos 30^(@) - hat(j) sin 30^(@))`
`+2Gm^(2)(hat(i) cos30^(@) - hat(j)sin 30^(@)) = 0`
(b) By symmetry, the X-component of the force cancels out. The Y-component survives.
`therefore F_(R) = 4Gm^(2)hat(j) - 2Gm^(2)hat(j) = 2Gm^(2)hat(j)`