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Mass-vectors according to Maxwell

  1. Jun 27, 2003 #1
    From Matter and Motion by James Clerk Maxwell, Article 59,
    "Let us define a mass-vector as the operation of carrying a given mass from the origin to the given point. The direction of the mass-vector is the same as that of the vector of the mass, but its magnitude is the product of the mass into the vector of the mass.
    Thus, if OA is the vector of the mass A, the mass-vector is OA*A"

    I can almost comprehend the idea of a mass-vector. Simply take the "vector of the mass" and multiply it by the mass. But what is the vector of the mass? What would be the point of the mass vector anyway?
    Thanks.
     
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  3. Jun 27, 2003 #2

    Tom Mattson

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    It looks like he means take a mass m (what he calles "A") and take its displacement x (what he calles "OA") from the origin and multiply them to get a new vector mx.

    It would be used in determining the center of mass xCM of a system of particles.

    xCM=(1/M)Σimixi

    where i=index (1,2,3,...) and M=total mass.
     
  4. Jun 28, 2003 #3

    HallsofIvy

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    In modern terms (and not all THAT modern!) multiply the mass and the velocity vector to get the momentum vector.
     
  5. Jun 28, 2003 #4
    And momentum is one of many kinds of mass-vectors? You could use any vector to figure out the center of the mass, such as ma? Thus, according to Tom, the center of mass could be given by the sum of all the forces acting on a system divided by the total mass of the system.
     
  6. Jun 28, 2003 #5

    Tom Mattson

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    No, that approach will not give you the location of the CM. It will give you the acceleration of the CM.
     
  7. Jun 28, 2003 #6
    So to find the CM you have to use the displacement vectors of the masses in the system? If you use the momenta, acceleration, velocity, etc, and apply Tom's formula, you'll get the momentum, acceleration, velocity, etc, of the CM.
     
  8. Jun 28, 2003 #7

    Tom Mattson

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    Yes.

    Yes. To go back to your "acceleration" example, how can I get the sum of miai from the sum of mixi? Answer: by taking the second derivative of xCM with respect to time, like so:

    (d2/dt2)xCM=(1/M)Σimi(d2/dt2)xi
    aCM=Σimiai

    That's the only way I can get the sum of forces from the sum of the mx vectors with the equation for the CM. As you can see, changing from x to a on one side makes it necessary to do it on the other side, so we aren't talking about the location of the CM anymore, but its acceleration.
     
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