Three equal masses ofkg 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

Three equal masses ofkg each are fixed at the vertices of an equilater

1.	Three equal masses ofkg 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
Answer» Solution :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)`