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)

Question

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)

dls · Answer

@mayankdevnani

dls · Answer

@Yahoo!

anonymous · Answer

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

dls · Answer

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

anonymous · Answer

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

dls · Answer

10-40+90/60 =1?

dls · Answer

i didnt get it what to do

anonymous · Answer

First Find the co-ordinates of m1 m2 and m3

anonymous · Answer

(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...

anonymous · Answer

u got it..)

dls · Answer

thanks :D then?

dls · Answer

@Mashy

anonymous · Answer

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

anonymous · Answer

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