Question

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

A
zero
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B
3Gm2l2
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C
5Gm2l2
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D
7Gm2l2
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Solution

## The correct option is A zeroDraw a perpendicular AD to the side BC.∴AD=ABsin60=√32lDistance AO of the centroid O from A is 23AD.∴AO=23(√32l)=l√3By symmetry, AO=BO=CO=l√3Force on mass 2m at O due to mass m at A isFOA=Gm(2m)(l/√3)2=6Gm2l2 along OAForce on mass 2m at O due to mass m at B isFOB=Gm(2m)(l/√3)2=6Gm2l2 along OBForce on mass 2m at O due to mass m at C isFOC=Gm(2m)(l/√3)2=6Gm2l2 along OCDraw a line PQ parallel to BC passing through O. Then ∠BOP=30=∠COQResolving ¯FOB and ¯FOC into two components.Components acting along OP and OQ are equal in magnitude and opposite in direction. So, they will cancel out while the components acting along OD will add up.∴ The resultant force on the mass 2m at O isFR=FOA−(FOBsin30+FOCsin30)=6Gm2l2−(6Gm2l2×12+6Gm2l2×12)=0

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