Center of mass

[jovial sand]

Hello, I need some help with this question.

Translation:
"Determine the center of mass for 3 identical point masses m = 1 g with location vectors
r1 = (0, 1), r2 = (8, 1) and r3 = (2, 4). <- I got the center of mass to (3.3, 2) correct?

And now for the second part:
ii. We now add another 6 g where the point masses above are already located.
How should we distribute this mass so that the center of mass is moved to the point (2, 2)

I don't know how to start on ii.

Thanks in advance!

[cloud current]

Well, suppose that the initial center of mass is R and the final center of mass is R'.
Also, let the total mass of the initial system be M and the additional mass be m (and its position be r). Then:
R' = (MR + mr)/(M + m)
So, we express r from here.
MR + mr = (M + m)R'
mr = (M + m)R' - MR = mR' + M(R' - R)
r = R' + (M/m)(R' - R)
All that's left is to substitute m, M, R and R'.

[cloud current]

Oh, and (a) is correct, but don't round. R = (10/3, 2).

[jovial sand]

So
R = 1g
R´ = 9g

M = R
m = R´
r = (2, 2)?

[cloud current]

What? That doesn't make sense. You can't equate vectors and scalars.
r = R' + (M/m)(R' - R) = {2, 2} + (3 g/(6 g))({2, 2} - {10/3, 2}) = {2, 2} + (1/2){-4/3, 0} = {4/3, 2}

[jovial sand]

Ye it doesn't lol, idk why I did that but I solved it!

[cloud current]

Good to hear!