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`        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. `
7 years ago

28 Points
```										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)
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Nitin Vomani !!!

Regards,
```
7 years ago
amartya
8 Points
```										Sir the answer given by u does not match the answer given in the book the Xcm is right but the Ycm is given as root 3 /4
```
3 years ago
prateek sonker
19 Points
```										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 formula￼In 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
```
10 months ago
Aman
39 Points
```										A(0,0),B(1,0),C(1/2,√3/2)Now as com at x=m1x1+m2x2+m3x3/m1+m2+m3 =1(0)+2(1)+3(1/2)/1+2+3=(2+3/2)6=4+3/12=7/12 com at y=m1y1+m2y2+m3y3/m1+m2+m3              =1(0)+2(0)+3(√3/2)/1+2+3=(3√3/2)/6=3√3/2*6=√3/4COM=√x*x+y*y=√7/12*7/12+√3/4*√3/4           =√49/144+3/16           =√49+27/144            =√76/144            =2√19/12           =   √19/6 m from A
```
one month ago
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