Three equal masses of m kg each are fixed at the vetrices of an equilateral triangle ABC (a) What is the force acting on a mass 2m placed at the centriod G of the triangle. (b) What is the force if the mass at the vertex A is doubled ? Take AG = BG = 1m

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

1.	Three equal masses of m kg each are fixed at the vetrices of an equilateral triangle ABC (a) What is the force acting on a mass 2m placed at the centriod G of the triangle. (b) What is the force if the mass at the vertex A is doubled ? Take AG = BG = 1m
Answer» Solution :The individual forces in vector notation are `F_(GA)=(G(m)(2M))/1hatj` `F_(GB)=(G(m)(2m))/1 (-hat1 cos 30^@- hatj SIN 30^@)` `F_(GC)=(G(m)(2m))/1 (hat1 cos 30^@ -hatj sin 30^@)` 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^2hatj + 2Gm^2 (-hati cos 30^@ - hatj sin 30^@) + 2Gm^2 (hati cos 30^@ - hatj sin 30^@) =0` (b) By symmetry, the X-component of the force cancels out. The Y-component survives. `therefore F_R=4Gm^2 hatj - 2Gm^2 hatj =2Gm^2 hatj`