# HELP Center of Mass

1. Nov 12, 2007

### Sucks@Physics

A 3kg mass is positioned at (0,8) and 1 kg mass is positioned at (12,0). What are the coordinates of a 4 kg mass, which will result int he center of mass of the system of three masses being located at the origin (0,0)?

I know that m1x1+m2x2...etc but I don't understand how you can do it with actual coordinates, can someone explain?

2. Nov 12, 2007

### Ginerva123

Have you tried finding the distance between the points and the origin?

3. Nov 12, 2007

### Sucks@Physics

This will sound dumb, but I do not remember the distance formula....

4. Nov 12, 2007

### Ginerva123

distance= square root of [(x2-x1)^2 + (y2-y1)^2]

5. Nov 12, 2007

### Sucks@Physics

so the 3 kg is 8 units away and the 1 kg is 12 units away. would that mean that the 4kg had to be 9 units away? and how do you know which way/what coordinates?

6. Nov 12, 2007

### Ginerva123

You know that the 3kg is due north and the 12kg is to the east of the origin. Therefore, you know that the third mass must be placed south-west of the origin to mantain equilibrium. Bear in mind that this is the y=x line.

7. Nov 12, 2007

### Sucks@Physics

i know to use the m1x1+m2x2+m3x3/(m1+m2+m3) but i still don't understand how you get 2 points from this equation

8. Nov 12, 2007

### Sucks@Physics

so the coordinates are oging to (-n,-n) but i still dont see how the formula works

9. Nov 12, 2007

### fliinghier

you will need to think of the vectors in terms of their components, and realize that the third mass will have a vector with components to cancel out the other two masses' vectors.

edit: also, the post saying it's on the x=y line is misinformed. the answer is not in the form (-n,-n).

Last edited: Nov 12, 2007
10. Nov 12, 2007

### Sucks@Physics

so 3 kg directly up 8 units, 1 kg directly right 12 units and 4 kg in quardrant III. the sum equals 36 so the 4 kg would have to be 9 units away from the origin? I dont think i'm doing this right

11. Nov 12, 2007

### fliinghier

not the sum. the vectors are at right angles. the y component of the 4kg mass equals the 3kg mass's vector, and the x component is the 1kg mass's vector

12. Nov 12, 2007

### Sucks@Physics

that would give me (-8,-12) right?

13. Nov 12, 2007

### simon1987

Then use (m1y1+m2y2+m3y3)/(m1+m2+m3) to find the y-coordinate of the CM.

14. Nov 12, 2007

### Sucks@Physics

i dont know m3y3 so is it (m1y1+m2y2)/(m1+m2+m3) = m3y3?

15. Nov 12, 2007

### aq1q

no, its 0, remember its stated on the problem.. (0,0) is the CM

16. Nov 12, 2007

### simon1987

List everything you know:

m1, x1, y1
m2, x2, y2
CM(x), CM(y)
m3

You need to find x3 and y3.

So set up the equation to find the center of mass,

$$CM{}_x{} = (m{}_1{}x{}_1{} + m{}_2{}x{}_2{} + m{}_3{}x{}_3{}) / (M)$$

Solve it for what you're looking for.

EDIT: To be clearer, a value x1, x2, x3, etc. is the x-coordinate of the position of that mass m1, m2, m3...

17. Nov 12, 2007

### aq1q

maybe this will help you visualize it.. I am not sure if you are familiar of adding vectors using unit vectors--> i(hat) and j(hat). But if you do it like that.. you will have
3(0i+8j) <-- i can multiply by three because mass is a scaler.
1(12i+0j)
-----------
12i + 24j
4(Xi+Yj)
-----------
0i+ 0j <-- center of mass..

Ok, that is more than enough of information.. you should be able to solve it.

18. Nov 13, 2007

### Sucks@Physics

i just don't understand, i give up

Last edited: Nov 13, 2007
19. Nov 13, 2007

### simon1987

It's much easier than you are making it. The equation

$$CM_{x} = \frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3}}{m_{1} + m_{2} + m_{3}}$$

Will give you the x-coordinate for the position of the center of mass: $$CM_{x}$$. You are looking for the position of the 4-kg mass, so solve this equation for x3.

And this second equation

$$CM_{y} = \frac{m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3}}{m_{1} + m_{2} + m_{3}}$$

will give you the y-coordinate for the center of mass ($$CM_{y}$$). Since you're looking for the position of that m3, solve this second equation for y3.

20. Nov 13, 2007

### Sucks@Physics

(3*0)+(1*12)+(4)/(3+1+4)=2
or
(3*8)+(1*12)+(4)/(3+1+4)=5
or
(3*8)+(1*12)/(3+1+4) =4.5/4 = 1.125

this is what i've done and i'm getting nothing close to the answer. the answer is (-3,-6)