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Does kgm/s2 give Newton

  1. Feb 11, 2012 #1
    Okay, so if we have the work Formula:
    ##A=Fd## we have units: ##N*m## which gives us joule ##J##,
    but if we have the formula of the force:
    ##F=\frac{mv}{t}## we would have ##\frac{kg * \frac{m}{s}}{s}## which gives us ##kg \frac{m}{s^2}##, so does that give newton?
     
  2. jcsd
  3. Feb 11, 2012 #2
    Yes, because kg = mass and m/s^-2 = acceleration so..... mass x acceleration = Force
     
  4. Feb 11, 2012 #3
    So formula ##F=\frac{mv}{t}## gives us formula ##F=ma## or ##F=ma## gives us ##F=\frac{mv}{t}## ?
     
  5. Feb 11, 2012 #4

    sophiecentaur

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    Looks OK to me. The dimensions are the same on both sides of the equation.
    Why would you doubt this?
    Science is consistent and full of little interesting bits like this. Enjoy.
     
  6. Feb 11, 2012 #5
    That's about right, just the variation is missing. I assume that you know that acceleration is the variation of velocity over time. Dimensional analysis doesn't provide that kind of information.

    Newton's second law is that F ~ dp/dt with p=mv
     
  7. Feb 11, 2012 #6

    sophiecentaur

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    How not??
    Acceleration: ST-2
     
  8. Feb 11, 2012 #7
    Again: it doesn't provide the information about v/t vs. dv/dt.
     
  9. Feb 11, 2012 #8

    sophiecentaur

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    Same units same dimensio. Even same value for uniform acceleration. What probem do you have?
     
  10. Feb 11, 2012 #9
    I have no problem.
     
  11. Feb 11, 2012 #10

    sophiecentaur

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    But you claim that Dimensional Analysis doesn't work in calculus?
     
  12. Feb 11, 2012 #11
    [tex]
    \frac{d v}{d t} = \lim_{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t} = \lim_{\Delta t \rightarrow 0} \frac{v(t + \Delta t) - v(t)}{\Delta t}
    [/tex]

    Now, look at the numerator (angle brackets mean dimension of ...):
    [tex]
    \left[ v(t) \right] = \left[ v(t + \Delta t)\right] = \left[ v \right] \Rightarrow \left[ \Delta v \right] \equiv \left[ v(t + \Delta t) - v(t) \right] = \left[ v \right]
    [/tex]
    where we had used the rule from Dimensional analysis that one can only add and subtract physical quantities with the same dimension, and the result is of the same dimension.

    Next, look at the fraction:
    [tex]
    \left[ \frac{\Delta v}{\Delta t} \right] = \frac{\left[\Delta v \right]}{\left[ \Delta t \right]} = \frac{\left[ v \right]}{\left[ t \right]}
    [/tex]
    where we had used the rule of Dimensional analysis that the dimension of a product or ratio of two physical quantities is the product or ratio of their dimensions.

    Even if you take the limit as [itex]\Delta t \rightarrow 0[/itex], the dimensions of the ratio do not change. So:
    [tex]
    \left[ \frac{d v}{d t} \right] = \left[ \frac{\Delta v}{\Delta t} \right] = \frac{\left[ v \right]}{\left[ t \right]}
    [/tex]

    This is a general rule: The dimensions of a derivative of y w.r.t. x is simply [y]/[x].
     
  13. Feb 11, 2012 #12
    No, not at all. I simply warned the OP that v/t is not identical to dv/dt. For example, if m=1, v=1, t=1, F =/= 1 (except by pure chance) :rolleyes:
     
  14. Feb 11, 2012 #13

    sophiecentaur

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    I think it was you who brought DA into it and you seemed to be 'dissin' it. All you were really doing was questioning the accuracy resulting from some assomptions? I'll buy that.
     
  15. Feb 11, 2012 #14
    No, it wasn't me - and let's hope that the OP gets it. :smile:
     
  16. Feb 11, 2012 #15
    Thank you all for your comments. I got it all
     
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