Three particles of masses 1.0kg,2.0kg and 3.0kg are placed at the corners A,B and C respectively of an equilateral triangle ABC of edge 1 m.Locate the centre of mass of the system.

Dear Nitin Vomani,

Ans:- Let the trangle ABC  is placed in such a way in the co ord plane so that the mid point of the side BC is the origin

point A=1kg

point B=2kg

point C=3kg

hence the X co ord is=[(2×(-1/2)+(3×1/2)+(1×0)]/(2+3+1)=1/12

The Y co ord is=[(2×0)+(3×0)+(1×√3/2)]/(1+2+3)=√3/12

Hence the COM is At=( 1/12  , √3/12)

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Soumyajit Das IIT Kharagpur

To find the center of mass:Fix one vertex of triangle as origin and give coordinates to other 2 vertices.apply the center of mass formulaIn the figure shown above A(0, 0) , B (1,0) and C(1/2, sqrt(3)/2)Xcm = M1X1+M2X2+M3X3/(M1+M2+M3) = 1(0)+2(1)+3(1/2)/(1+2+3) = 7/12Ycm = M1Y1+M2Y2+M3Y3/(M1+M2+M3) = 1(0)+2(0)+3(sqrt(3)/2)/(1+2+3) = sqrt(3)/4