InterviewSolution
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InterviewSolution
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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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Answer» (a) Refer to Figure , `/_CBC = 30^(@) = /_CDX` `AD = (2)/(3) AF = (2)/(3) xx a sin 60^(@)` `= (2)/(3) xx a (sqrt(3))/(2) = (a)/(sqrt(3)) = BD = CD` Force acting on mass at `D` due to mass at `A` is `vec(F_(DA)) = (Gm xx 2m)/((a//sqrt(3))^(2)) hatj = (6Gm^(2))/(a^(2))hatj` Force on mass at `D` due to mass at `B` is `vec(F_(DB)) = (Gm xx 2m)/((a//sqrt(3))^(2)) (-hati cos 30^(@) - hatj sin 30^(@))` Force on mass at `D` due to mass at `C` is `vec(F_(DC)) = (Gm xx 2m)/((a//sqrt(3))^(2)) (-hati cos 30^(@) - hatj sin 30^(@))` Resulting force on a mass at `D` is `vec(F_(0)) = vec(F_(DA)) = vec(F_(DB)) = vec(F_(DC))` `= (6 Gm^(2))/(a^(2)) hatj + (6m^(2))/(a^(2)) (-hati cos 30^ (@) - hatj sin 30^(@))` `+ (6m^(2))/(a^(2)) (-hati cos 30^(@) - hatj sin 30^(@)) = 0` (b) If mass at `A` is doubled, then finding `vec(F_(DA)), vec(F_(DB))` and `vec(F_(DC))` in terms of rectagular components along `X`-axis and `Y`-axis, we note that the componet force along `X`-axis cancel out but the component force along `Y`-axis will survive. Then `vec(F_(R)) = (12 Gm^(2))/(a^(2)) hatj - (6Gm^(2))/(a^(2)) hatj = (6Gm^(2))/(a^(2)) hatj` |
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