Find Center of Mass: 3kg at (0,8), 1kg at (12,0), 4kg at (?)

Click For Summary
SUMMARY

The discussion focuses on calculating the coordinates of a 4 kg mass required to achieve a center of mass (CM) at the origin (0,0) for a system comprising a 3 kg mass at (0,8) and a 1 kg mass at (12,0). The participants clarify the use of the center of mass formula, CM_x = (m1*x1 + m2*x2 + m3*x3) / (m1 + m2 + m3) and CM_y = (m1*y1 + m2*y2 + m3*y3) / (m1 + m2 + m3). The correct coordinates for the 4 kg mass are determined to be (-3,-6), ensuring the system's equilibrium.

PREREQUISITES
  • Understanding of center of mass calculations
  • Familiarity with vector components
  • Knowledge of mass and coordinate systems
  • Proficiency in algebraic manipulation of equations
NEXT STEPS
  • Study the derivation of the center of mass formula in physics
  • Learn about vector addition and its application in physics
  • Explore examples of center of mass calculations with multiple masses
  • Investigate the implications of mass distribution on equilibrium in systems
USEFUL FOR

Students in physics, engineers working on mechanics, and anyone interested in understanding mass distribution and center of mass calculations.

Sucks@Physics
Messages
76
Reaction score
0
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?
 
Physics news on Phys.org
Have you tried finding the distance between the points and the origin?
 
This will sound dumb, but I do not remember the distance formula...
 
distance= square root of [(x2-x1)^2 + (y2-y1)^2]
 
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?
 
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.
 
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
 
so the coordinates are oging to (-n,-n) but i still don't see how the formula works
 
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:
  • #10
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 don't think I'm doing this right
 
  • #11
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
that would give me (-8,-12) right?
 
  • #13
Sucks@Physics said:
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

Then use (m1y1+m2y2+m3y3)/(m1+m2+m3) to find the y-coordinate of the CM.
 
  • #14
i don't know m3y3 so is it (m1y1+m2y2)/(m1+m2+m3) = m3y3?
 
  • #15
no, its 0, remember its stated on the problem.. (0,0) is the CM
 
  • #16
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,

<br /> CM{}_x{} = (m{}_1{}x{}_1{} + m{}_2{}x{}_2{} + m{}_3{}x{}_3{}) / (M)<br />

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
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
i just don't understand, i give up
 
Last edited:
  • #19
Sucks@Physics said:
i just don't understand, i give up

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}}<br />

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}}<br />

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
(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
Sucks@Physics said:
(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)

Solve the equations for x3 and y3 before plugging in numbers.
 
  • #22
i don't know why i can't understand this stupid thing
 
  • #23
Sucks@Physics said:
i don't know why i can't understand this stupid thing

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
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
omg i figured it out. thanks simon for not giving up on me! lol
 
  • #26
i set it up liek this...

(3*0)+(1*12)+(4* ?) = 0
(3*8)+(1*0)+(4* ?) = 0

Will that work?
 
  • #27
Sucks@Physics said:
i set it up liek this...

(3*0)+(1*12)+(4* ?) = 0
(3*8)+(1*0)+(4* ?) = 0

Will that work?

Does it match what you know the answer should be?
 
  • #28
yea its (-3,-6)
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
3
Views
2K