Where Should the Third Particle Be Placed for Desired Center of Mass?

In summary, to place a 3.74 kg particle at the center of mass of a three-particle system, you need to solve for x and y coordinates using the equations -.565 = (-5.4912+1.0982+3.74x), and -.601 = (-3.12(.652)+2.89(-.636)+3.74y).
  • #1
AnkhUNC
91
0
[SOLVED] Center of Mass problem

Homework Statement


A 3.12 kg particle has the xy coordinates (-1.76 m, 0.652 m), and a 2.89 kg particle has the xy coordinates (0.380 m, -0.636 m). Both lie on a horizontal plane. At what (a)x and (b)y coordinates must you place a 3.74 kg particle such that the center of mass of the three-particle system has the coordinates (-0.565 m, -0.601 m)?


Homework Equations





The Attempt at a Solution


I know this problem is extremely easy but I'm just having trouble getting it. I solve for x(com) and y(com) =

-.565 = (3.12(-1.76)+2.89(.380)+3.74(x))/3.12+2.89+3.74
-.601 = (3.12(.652)+2.89(-.636)+3.74(y))/3.12+2.89+3.74

This is correct right? So how do I pull out to solve for x and y?
 
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  • #2
Try simplifying your expressions and cross-multiplying?
 
  • #3
I've tried that and I'm just doing something wrong I think
 
  • #4
There is nothing wrong with your approach. Maybe if you could post the steps to get the values of x and y someone could figure out where you are going wrong.
 
  • #5
M total = 9.75

So -.565 = (-5.4912+1.0982+3.74(x)) / 9.75 -> (-.653(x))/9.75 -> -.066974359(x) = -.565

So x = 8.436064315 but I did it wrong apparently.
 
  • #6
In the numerator you have (-5.4912 + 1.0982 + 3.74x).
You can't add the numbers together to get (-.653x).

Instead you have (-4.393 + 3.74x).
Only the last number is multiplied by x.
 
  • #7
Can I divide the 3.74x by 9 though?
 
  • #8
Why would you want to? As mace2 pointed out the equation you want to solve is (3.74*x-4.393)/9.75=-.565.
 
  • #9
So what should my next step be? Sorry my algebra is horrible but I have no problem in calculus. Go figure.
 
  • #10
I would multiply both sides of the equation by 9.75. Shouldn't you? Clear the fraction first. BTW problems in algebra should be causing you HUGE problems in calculus. I've never seen the reverse. Go figure.
 
  • #11
Yeah I was just about to ask that. Thanks for the help got it figured out :P
 

FAQ: Where Should the Third Particle Be Placed for Desired Center of Mass?

What is the Center of Mass problem?

The Center of Mass problem is a physics concept that involves finding the point at which an object's mass is evenly distributed in all directions. This point is also known as the center of gravity.

How is the Center of Mass calculated?

The Center of Mass is calculated by taking the weighted average of the positions of all the individual particles that make up an object. This means multiplying each particle's mass by its distance from a chosen reference point, and then dividing the sum of these values by the total mass of the object.

Why is the Center of Mass important?

The Center of Mass is important because it helps us understand how an object will behave when subjected to external forces. It is also useful for analyzing the stability and balance of objects and systems.

Can the Center of Mass be outside of an object?

No, the Center of Mass cannot be outside of an object. It is always located within the physical boundaries of the object, and can be at the very edge or even outside of it.

How does the Center of Mass change with different objects?

The location of the Center of Mass can vary depending on the shape, size, and distribution of mass within an object. For example, a long, thin object will have its Center of Mass closer to one end, while a symmetrical object will have its Center of Mass in the middle.

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