Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC. (a) What is the force acting ona 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 )

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 ona 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 )
Answer» Solution :(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`