## DLS 2 years ago If the centre of mass of three particles of masses 10,20 and 30g is at the point(1,-2,3)then where should a fourth particle of mass 40g be placed,so that the combined centre of mass of the system is (1,1,1)

1. DLS

@mayankdevnani

2. DLS

@Yahoo!

3. Yahoo!

Position of Cm = m1x1 + m2x2 + m3x3 / (m1+m2+m3)

4. DLS

I didn't understand the co-ordinates,is it 1,0 -2,0 & 3,0?

5. Yahoo!

There are three Co-ordinates....For the 1 St one..CM is @ (1,-2,3)

6. DLS

10-40+90/60 =1?

7. DLS

i didnt get it what to do

8. Yahoo!

First Find the co-ordinates of m1 m2 and m3

9. Yahoo!

(1,-2,3 ) x , y , z x axis of Cm = m1x1 + m2x2 + m3x3 / (m1+m2+m3) y axis of Cm = m1y1 + m2y2 + m3y3 / (m1+m2+m3) etc...

10. Yahoo!

u got it..)

11. DLS

thanks :D then?

12. DLS

@Mashy

13. Mashy

wait .. i know.. on call with gf :P.. 2 mins

14. Mashy

ok.. so let xcm be the centre of mass of the three particle system then \[Mxcm = m1x1+ m2x2 + m3x3 \] now .. when the fourth particle is added.. the centre of mass shifts hence \[M'xcm' = m1x1+ m2x2 + m3x3+ m4x4\] or \[M'xcm' = Mxcm + m4x4\] the only thing you have to find now is x4.. so calculate.. do the same in other 2 axes