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 = BH = 1m

Explanation

The individual forces in vector notation are
`F_(GA) = (G(m)(2m))/1 hatj`
`F_(GB)=(G(m)(2m))/1(-haticos 30^@-hatjsin30^@)`
`F_(GC)=(G(m)(2m))/1(haticos 30^@-hatjsin30^@)`
From the principle of superposition and law of vector addition the resultant gravitational force `F_R ` on (2m) is
`F_(R) = F_(GA) + F_(GC)`
`F_(R)= 2Gm^(2) hatj+2Gm^(2)(-haticos30^@-hatjsin30^@)`
`+ 2GM^(2)(haticos 30^(@)-hatjsin30^@) = 0 `
b) By symmetry , the X - component of the force cancels out . The Y - component survies .
`:. F_(R) = 4Gm^(2) hatj-2Gm^(2)hatj=2Gm^(2)hatj`