Three masses each of mass m are placed at the vertices of an equilateral triangles ABC of side l as shown in figure. The force acting on a mass 2m placed at the centroid O of the triangle is

Options

A. zero
B. `(3Gm^(2))/(l^(2))`
C. `(5Gm^(2))/(l^(2))`
D. `(7Gm^(2))/(l^(2))`

Answer

A

Explanation

(a)
Draw a perpendicular AD to the side BC.
`:. AD= AB sin 60^(@) =(sqrt(3))/(2)l`
Distance AO of the centroid O from A is `(2)/(3)` AD.
`:. AO =(2)/(3)((sqrt(3))/(2)l)=(l)/(sqrt(3))`
By symmetry ` AO=BO=CO=(1)/(sqrt(3))`
Force on mass 2m at O due to mass m at A is
`F_(OA)=(Gm(2m))/((l//sqrt(3))^(2))=(6Gm^(2))/(l^(2))` along OA
Force on mass 2m at O due to mass m at B is
`F_(OB)=(Gm(2m))/((l//sqrt(3))^(2))=(6Gm^(2))/(l^(2))` along OB
Force on mass 2m at O due to mass m at C is
`F_(OC)=(Gm(2m))/((l//sqrt(3))^(2))=(6Gm^(2))/(l^(2))` along OC
Draw a line PQ parallel to BC passing through O. Then
`angleBOP=30^(@)=angleCOQ`
Resolving `vec(F)_(OB) " and " vec(F)_(OC)` into two components.
Components acting along OP and OQ are equal in madnitude and opposite in direction. So, they will cancel out while the component acting along OD will add up.
`:.` The resultant force on the mass 2m at O is
`F_(R)=F_(OA)-(F_(OB) sin30^(@)+F_(OC) sin30^(@))`
`(6Gm^(2))/(l^(2))-((6Gm^(2))/(l^(2))xx(1)/(2)+(6Gm^(2))/(l^(2))xx(1)/(2))=0`