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

Three equal masses of m kg each are fixed at the vertices of an equila

1.	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
Answer» Solution :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 cancelsout . The Y - component survies . `:. F_(R) = 4Gm^(2) hatj-2Gm^(2)hatj=2Gm^(2)hatj`