# Homework Help: 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)

21. Nov 13, 2007

### simon1987

Solve the equations for x3 and y3 before plugging in numbers.

22. Nov 13, 2007

### Sucks@Physics

i dont know why i can't understand this stupid thing

23. Nov 13, 2007

### simon1987

Do you not understand the concept? Or how to solve the equations for x3 and y3, respectively? Or what x3 and y3 represent? Or what CMx and CMy represent?

24. Nov 13, 2007

### Sucks@Physics

CMy and CMx are the origin (0,0). x3 and y3 are the coordinates in which i need to find that have the 4kg mass. I guess I don't know how to solve the equation for x3y3. Would it be...

(m1x1+m2x2)/(m1+m2+m3) or (m1x1+m2x2)/(m1+m2) or (m1x1+m2x2+m3)/(m1+m2+m3)

i feel like i've tried everything

25. Nov 13, 2007

### Sucks@Physics

omg i figured it out. thanks simon for not giving up on me! lol

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