Position of CM for 3 Cubes Along a Line

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The position of the center of mass (CM) for three cubes with sides l, 2l, and 3l, arranged in a line with the 2l cube in the center, is calculated using the formula CM = (M1*X1 + M2*X2 + M3*X3) / TM. Given that the cubes are uniform, the total mass is 6M, leading to a CM position of 2L. Confusion arose regarding the calculation of total mass and density, but it was clarified that density cancels out in the calculations, and the total length of the arrangement is 6L.

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Question: Three cubes of sides l, 2l, and 3l are placed next to one another (in contact) with their centers along a straight line and the l=2l cube in the center. What is the position, along the line, of the CM of this system? Assume the cubes are made of the same uniform material.

My reasoning: So the equation for CM= M1XI +M2X2 + M3X3/ TM ... where M= mass, and X= distance ...So M(l +2l +3l)/3m = CM
M(6L)/3m = 2L
So, I got the center of Mass is 2L

I don't think my answer is correct, and even if my approach is correct. It is labeled a pretty easy problem, but I get really confused with problems without numbers. Also, I am not sure if you can conclude the total mass is 3M considering that means all the cubes would have to be the same mass ... and I also do not know what the question maker meant when he wrote the l=2l cube in the center ...

Please help!
 
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The coordinates X1, X2, and X3 are the coordinates of the center of mass of each cube. So, all you have to do is place the origin wherever you want and start to calculate. If you place the origin at the beginning of the first cube, then X1 = 0.5 L, and so on..
 
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radou ... thanks for the help but i am still confused... how would you find the total mass of the cubes??
 
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vinny380 said:
radou ... thanks for the help but i am still confused... how would you find the total mass of the cubes??

Well, since the cubes are made of the same uniform material, you may assume the density of the cubes is equal. You know the volume, so, you can calculate the mass.
 
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finding the volume is easy , but how would you go about finding the density? is it simply a known value?
 
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vinny380 said:
finding the volume is easy , but how would you go about finding the density? is it simply a known value?

Yes, call it \rho or something. It will cancel out in the further calculation.
 
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A diagram might help.

center-of-mass.png
 
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is the answer 3.08Lo (thats what i got) ?
 
Last edited:
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According to my calculation, it's 3.83L. But I may be wrong. Nevertheless, it's important you understand the principle. :smile:
 
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  • #10
yeaaa... i just did it again and got 3.83L ... but radou, how does that make sense??
 
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  • #11
vinny380 said:
yeaaa... i just did it again and got 3.83L ... but radou, how does that make sense??

What exactly do you mean?
 
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  • #12
well...if the center of mass is 3.83L ... then wouldn't the center of mass be out of the object given (which is impossible)?
 
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  • #13
vinny380 said:
well...if the center of mass is 3.83L ... then wouldn't the center of mass be out of the object given (which is impossible)?

No it wouldn't, because the total length of the object is L + 2L + 3L = 6L.
 
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  • #14
thanks radou!
 
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