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Finding Dimensional Homogeneity

  1. Aug 29, 2011 #1
    1. The problem statement, all variables and given/known data
    Which one of the following equations is dimensionally homogeneous?


    F= force (N)
    m= mass (kg)
    a= acceleration (m/s2)
    V= velocity (m/s)
    R= radius (m)
    t= time (s)

    2. Relevant equations

    1. F=ma
    2. F=m(V2/R)
    3. F(t2-t1)=m(V2-V1)
    4. F=mV
    5. F=m(V2-V1)/(t2-t1)

    3. The attempt at a solution

    Through what I can gather from my textbook and the internet, I started by entering what I know. So:

    From here, I'm not really sure where to go.


    And again, I plug everything in but in my textbook at this point is where they determine if it is or isn't dimensionally homogeneous.

    I would really appreciate any guidance on this, I realize it's a super basic question, but you've got to start somewhere! Thank you.
  2. jcsd
  3. Aug 29, 2011 #2


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    Homogeneous usually means "made up of the same types" or words to that effect.

    If you express each quantity in base units - m , kg , s - you could compare left to right.

    hint: there are two formulas which look very similar, but with one variable different. I would suspect one of those.
  4. Aug 29, 2011 #3


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    Try that one again. It seems you forgot to square something or forgot to divide by something. One of the two. :wink:
  5. Aug 29, 2011 #4


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    Uh, V^2 dimensions are m^2/s^2. You try it again.
  6. Aug 29, 2011 #5
    Okay, so further googling turned up this(http://physics.nist.gov/cuu/Units/units.html) handy chart.

    Am I to understand correctly that 1N= 1kg(m/s2) And then using that I can compare left to right? And if the right side doesn't come out to kg(m/s2) it is NOT dimensionally homogeneous?
  7. Aug 29, 2011 #6


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    Yes. And I don't there is just one that is homogeneous.
  8. Aug 29, 2011 #7
    I think the question may have been mis-typed on my handout. I think it should read "which one of the following is NOT dimensionally homogeneous?"

    When I do the substitutions for F=mV it becomes:

    N=(kg)(m/s) or further:


    Because the (m/s) on the right is not squared as it is on the left, would this be a correct example of an equation that is NOT dimensionally homogeneous?
  9. Aug 30, 2011 #8


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    That is the task for all but the third one, where the left hand side is not simply F, but has time with it.

    I would have the units of F as kgms-2

    Perhaps made clearer as kg m s-2 or kg.m.s-2

    Often these are actually written "dimensionally" using [M] for mass, [T] for time and [L] for length.

    then we would have [M][L][T]-2

    That certainly takes care of countries that use pounds instead of kg, and feet instead of metres.

    Oh and rest assured - only one of the examples is not homogeneous - perhaps you left the word not out of your original post.
  10. Aug 30, 2011 #9


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    certainly would.
  11. Aug 30, 2011 #10
    Thank you all. Your explanations made it click in my head and I think I have it now.
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