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Thread: Dimensional Analysis check

  1. #1
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    Dimensional Analysis check

    Hello all,

    Just want to check what other people get as an answer for this dimensional analysis.

    V=(kLx)/p

    Units are:

    'k' is dimensionless
    'L' is load (N)
    'x' is length (m)
    'p' is (grammes/cm^2)

    Any help and workings shown would be much appreciated.

    Thanks
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  2. #2
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    Quote Originally Posted by OptiEng View Post
    Hello all,

    Just want to check what other people get as an answer for this dimensional analysis.

    V=(kLx)/p

    Units are:

    'k' is dimensionless
    'L' is load (N)
    'x' is length (m)
    'p' is (grammes/cm^2)

    Any help and workings shown would be much appreciated.

    Thanks
    N = M \cdot L \cdot T^{-2}

    x = L

    p = M \cdot L^{-2}

    \frac{M \cdot L \cdot T^{-2} \cdot L}{M \cdot L^{-2}} = L^4 \cdot T^{-2}

    where M is mass, L is length and T is time
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  3. #3
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    Quote Originally Posted by OptiEng View Post
    Hello all,

    Just want to check what other people get as an answer for this dimensional analysis.

    V=(kLx)/p

    Units are:

    'k' is dimensionless
    'L' is load (N)
    'x' is length (m)
    'p' is (grammes/cm^2)

    Any help and workings shown would be much appreciated.

    Thanks
    HI

    What is V ? You need to know what v is so that you can verify whether the equation is dimensionally homogenous .
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  4. #4
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    Quote Originally Posted by mathaddict View Post
    HI

    What is V ? You need to know what v is so that you can verify whether the equation is dimensionally homogenous .
    V is supposed to be Cm^3/cm, and I obtained the same answer as "e^.." so that means the equations can't hold true?
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by OptiEng View Post
    V is supposed to be Cm^3/cm, and I obtained the same answer as "e^.." so that means the equations can't hold true?
    As it stands yes, you'd need to get rid of time
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  6. #6
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    and I also need to lose one of the 'L's too?
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  7. #7
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    is V meant to be cm^3 or cm^3/cm = cm^2?

    Note that acceleration has dimensions L \cdot T^{-2}
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  8. #8
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    Hi,

    V is supposed to be a specific rate of change of volume per cm. It's a strange unit I know. That's why I am questioning it. It is from the Archard's equation of wear used in engineering since the 1950's.

    Thanks for your input
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