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 2 m 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 = 1 m )

Explanation

(a)
As shown in figure, by taking co-ordinate system such that the centroid of triangle will be origin. `angle"XGC" = angleX.GB=30^@`
Distance of every vertice from the centroid G of a triangle is l = 1 m
The forces on particles A, B and C at the G by the particle on G are respectively
`vecF_(GA)=(Gm(2m))/l^2.hatj`
`vecF_(GB)=(Gm(2m))/l^2(-(1)cos30^(@)hati-(1)sin30^(@) hatj)`
`vecF_(GC)=(Gm(2m))/l^2((1)cos30^(@)hati-(1)sin30^(@) hatj)`
`:.` The resultant force on particle at G,
`vecF=vecF_(GA)+vecF_(GB)+vecF_(GC)`
`= Gm(2m) [(hatj- cos 30^(@) hati + cos 30^(@) hatj- sin 30^(@) hatj-sin30^(@) hatj)]`
`[ :. l = 1 implies l^(2) =1 ]`
`=2 Gm^(2)[hatj-0-2xx1/2hatj][:.sin30^(@) =1/2]`
`= 2 Gm^(2) [ hatj-hatj]`
= 0
`implies` (b) If mass of A is doubled then
Resultant force `F.=vecF_(GA)+vecF_(GB)+vecF_(GC)`
`= 4 Gm^(2) hatj-2Gm^(2) (2 sin 30^(@))hatj`
`F.=4Gm^(2) hatj- 2Gm^(2) hatj-2Gm^(2) hatj" " [:. sin 30^(@) = 1/2]`
`:. F. = 2Gm^(2) hatj`