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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.

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Solution

## From Figure, the gravitational force on mass 2m at G due to mass at A isF1=Gm×2m12=2Gm2 along GAGravitational force on mass 2m at G due to mass at B isF2=Gm×2m12=2Gm2 along BGGravitational force on mass 2m at G due to mass at C is F3=Gm×2m12=2Gm2 along GCDraw DE parallel to BC passing through point G. Then ∠EGC=30o=∠DGB.Resolving −→F2 and −→F3 into two rectangular components, we haveF2cos30o along GD and F2sin30o along GHF3cos30o along GE and F2sin30o along GHHere, F2cos30o and F3sin30o are equal in magnitude and acting in opposite directions, and cancel out each other. The resultant force on mass 2m at G isF1−(F2sin30o+F3sin30o)=2Gm2−(2Gm2×12+2Gm2×12)=0b. When mass at A is 2m, then gravitational force on mass 2m at G due to mass 2m at A isF′1=G2m×2m12=4Gm2 along GAThe resultant force on mass 2m at G due to masses at A, B and C isF1=(F2sin30o+F3sin30o)=4Gm2−(2Gm2×12+2Gm2×12)=2Gm2 along GA.

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