Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dimentionally consistent

  1. Jun 26, 2005 #1


    User Avatar
    Science Advisor
    Gold Member

    Is this expression dimentionally consistant?

    where v is velocity, a is acceleration and x represents a length in meters.
    [tex](m/s)^2=2*m/s^2 * m[/tex]
    [tex]m^2/s^2=2*m/s^2 * m[/tex]
    [tex]m^2/s^2 = 2 * m^2/s^2[/tex]

    I'm not sure if the 2 makes the thing not dimentionally consistant. My guess is since the 2 is unitless, the expression is dimentionally consistant. Any thoughts?
  2. jcsd
  3. Jun 27, 2005 #2
    m^2/s^2 = 2m^2/s^2 is correct both dimensionwise and unit-wise
    Yes, 2 is dimensionless in this equation. The reason why we 1=2 is because this equation is very specific. By doing this we assume acceleration is constant and that the initial velocity is zero. In fact we we multiply both sides by the mass we have

    mv^2 = 2max

    F = ma

    mv^2 = 2fx

    if we assume the force is constant and in the direction of motion we can say
    w = fx

    im sure you see this now

    mv^2/2 = w
    KE = w, so the above equation that we started with is actually a derivation from the work energy theorem. And it basically states that work is the change in kinetic energy. more or less velocity squared is directly proportional to the product of acceleration and displacement. So now that the issue of the 2 has been addressed I just want to clear up something else

    ft/s^2 is dimensionally correct with m/s^2 as the dimension are both L/T^2
    however the equation is not correct by units, in order to make a dimensionally correct equation that has terms with different units into an equation that is correct in both senses, all we need to do is simply multiply by a conversion factor. If you take an advanced dynamics course your EOM's will often have several ugly terms and it is always a good idea to make sure all the terms have the same dimensions. This way if you know the dimensions on a term aren't correct, then you should just accept the fact that the equation in incorrect and there is a mistake in the previous steps, always a good time saver before finalizing any work!
  4. Jun 27, 2005 #3
    2 dimensionally constant and has no dimensions , can be taken as [itex]M^0L^0T^0[/itex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook