Center of Mass: where should additional mass be placed?

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dyampir

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Nov 25, 2006
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I would first like to say hello to all. I am having a really hard time with some of the center of mass problems.

Question


There is a mass of 24 at (-8,0) and mass of 36 at (12,0). Find where a mass of 9 should be placed on the x-axis so that (-3.0) is the center of mass.

The answer is -11/3, but I am not getting that answer.

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This is the work from my first attempt:

Mo = 24(-8) + 36(12) = 240
m = m1 + m2 = 24 + 36 = 60

x = Mo/m ----- x = 240/60 = 4
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This was another thought I had:

Mo = 24(5) + 36(15) .......I got the 5 because -8 is 5 away from -3 and 12 is 15 away from -3. When these are added you get 660.

m = 24(-8) + 36(12) = 240

Mo/m = 660/240 = 11/4 .......this is closer.

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I saw an earlier problem on the forum that deals with basically the same thing and I do not understand that problem either.

Any help on how to get started with this problem will be much appreciated.

Sincerely,
Dyampir
 
1) What an odd problem. Why bother with the 2D coordinates if everything is ON the x-axis?

2) What are you trying to do? I don't see your goal at all. None of your calculations included the 9 mass. Where is it?

3) How sure are you that you have reported the problem correctly? Perhaps the problem statement really says +3?

4) Try this: 24*(-8) + 36*(12) + 9*(x) = 69*(3)

In some circles, it would be called a "weighted average". It's one of the reasons calculus is cool. It can add up infinitely many masses!
 
The problem is from "Technical Mathematics with Calculus (second edition"...chapter 25...section 4....problem #7. The instructions are to find the center of mass of each linear system.

The question: There is a mass of 24 at (-8,0) and a mass of 36 at (12,0). Find where a mass of 9 should be placed on the x-axis so that (3.0) is the center of mass. The answer in the back of the book is -11/3. **you were right about it being (3,0)**

I tried working it out the way you did below and I got -11/3.....I am totally not sure how you knew how to set up the problem. How did you know to multiply the 9 by x on the left......and then add it to the sum of "m" and multiply by 3 on the right? Sorry if this sounds stupid....I am an older female that went back to college and my teacher does NOT explain this stuff to good lol.

thank you so much for your quick response,

sincerely,
dyampir
 
I'm not real sure how to say it otherwise. If you want a center of mass, you must include all the mass. Leaving out the '9'-mass is simply out of the question.
 
Thank you

Lol ok. I went back through and got the next problems right. It really helped to know that it is actually a weighted average.

Thank you so much again.

Dyampir
 
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