Question

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=BG=CG=1m`

Answer 1

(a) The angle between GC and the positive x-axis is `30^(@)` ang so is the angle between GB and the negative x-axis. The individual forces in vector notation are

Three equal masses are placea at the three vertices of the `Delta ABC` A mass 2m is placed at the centroid G.
`F_(GA)=(Gm(2m))/(1)hatj`
`F_(GB)=(Gm(2m))/(1)(-haticos 30^(@)-hatjsin 30^(@))`
`F_(GC)=(Gm(2m))/(1)(+hati 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_(QB)+F_(GC)`
`F_(R)=2Gm^(2)hatj+2Gm^(2)(-hati cos 30^(@)-hatj sin 30^(@))`
`+2Gm^(2)(haticos 30^(@)-hatj sin 30^(@))=0`
Alternatively, one expects on the basis of symmetry that the resultant force ougth to be zero.