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Dimensionally correct equations

  1. Sep 6, 2006 #1
    ]Okay, so this is my first physics class ever. I'm a little lost, but I believe I'm on the right track. Am I approaching this problem the right way?


    -------------Dimension------------------------------Dimension-------
    distance (x) (L) Acceleration (a) (L)/(T)^2

    Time (t) (T) Force (F) (M)(L)/(T)^2

    Mass (m) (M) Energy (E) (M)(L)^2/(T)^2

    Speed (v) (L)/(T)


    which of the following equations are dimensionally correct?

    (A) F=ma (d) E=max
    (b) x= at^3/2 (e) v= square root of (Fx/m)
    (c) E= mv/2
     
  2. jcsd
  3. Sep 6, 2006 #2
    You've quoted the correct dimensions for those quantities, which equations do you think are correct?
     
  4. Sep 6, 2006 #3

    force= ma is the only one?
     
  5. Sep 6, 2006 #4
    if, F=ML/T^2 ( L/T^2) can be substituted for (a) thus giving you
    f=ma
     
  6. Sep 6, 2006 #5
    F=ma is correct, but it's not the only one. For the others, multiply the dimensions of the quantities each equation uses together, not forgeting to use the correct powers each has, then compare the answer to the dimensions of the left hand side of the equation.
     
  7. Sep 6, 2006 #6
    I'm so lost. can you help walk me through B?
     
  8. Sep 6, 2006 #7
    OK, on the right we have a*t^3/2. The dimensions of this are [(L)/(T)^2]*[(T)^3/2], which, by cancelling (T)s reduces to (L)/(T)^1/2. On the left we have only (L), so this equation is not dimensionally correct.
     
  9. Sep 6, 2006 #8
    when you work with dimensions, use the fundamental units.

    mass : kilogram: [kg]
    length : meter : [m]
    time : second :

    so speed ; [m]/
    acceleration : [m]/^2
    force in newtons [N] = [kg][m]/^2
    work or energy is in joules [J] = [N][m] = [kg][m]^2 /^2

    When you have a formula, it must be dimensionaly correct, so to verify if it is correct, you use the laws of exponents with the fundamental units. It is a simple algebra.
     
  10. Sep 6, 2006 #9

    Thanks! I think I see what you are doing now. The only part I found confusing is when you said you got (L)/(T)^1/2. You say the (T's) cancel out, but you have T^3 / T^2 so that would give you just T? so wouldn't the answer be LT/2 ?
     
  11. Sep 6, 2006 #10

    berkeman

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    Staff: Mentor

    Just to add a comment for the OP. Carrying units through equations is one of the most fundamental tricks that I learned in college. Luckily I learned it on about the 3rd day of classes, in a first engineering class where we studied problems from all the different kinds of engineering (EE, ME, CE, ChE, etc.). We were working in groups on problems, and we were trying to solve a problem where we were having trouble remembering whether certain quantities belonged in the numerator or denominator. Most of us in our group were baffled at one point, with multiple possible forms of a formula. Then one member spoke up (I still remember this moment clearly, many years later), "Hey guys, this is easy. Just put the units next to each variable in the formula, and when you multiply and divide the variables, do the same to the units. The units in the end have to make sense for the answer."

    We all looked at each other, and then had the correct form of the equation in something less than 10 seconds. To this day, when I have to remember relationships like between velocity, frequency and wavelength, I use units to remember the relationship, and don't bother memorizing it. Like this:

    [tex]\lambda [m] = \frac{v [m/s]}{f [1/s]}[/tex]

    What a great trick. I also use it in very complex calculations, to help me catch my math errors as I go along. Every term on one side of an addition equation has to have the same units, or you cannot add them. And the overall units of the lefthand side of the equation has to equal the units of the righthand side.
     
  12. Sep 6, 2006 #11

    berkeman

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    Staff: Mentor

    BTW, as borisleprof shows, square brackets are usually used to show units. That helps to separate the units from the rest of the terms in the equation.
     
  13. Sep 6, 2006 #12

    You're a freaking genius. I never thought about it that way. It wouldn't make sense to say I run one second per meter. But it sounds right to say I run one meter per second.
     
  14. Sep 6, 2006 #13

    berkeman

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    Staff: Mentor

    That's almost exactly what we all said to that one guy in our workgroup. One of "aha" moments in life!:biggrin:
     
  15. Sep 6, 2006 #14

    Thanks! I think I see what you are doing now. The only part I found confusing is when you said you got (L)/(T)^1/2. You say the (T's) cancel out, but you have T^3 / T^2 so that would give you just T? so wouldn't the answer be LT/2 ?
     
  16. Sep 7, 2006 #15
    Sorry, I thought the equation was a*t^(3/2), not (a*t^3)/2. Yes, if it's t^3, you'd have (L)(T). The 2 has no units.
     
  17. Sep 7, 2006 #16
    thanks


    Thanks, I got the hang of it now. Not too bad.
     
  18. Sep 13, 2009 #17
    Ok, so this is my first physics class and I cannot seem to wrap my head around it. The previous post have helped a lot.
    What about when you have a square root like v(velocity)=sqrt(ax)
     
  19. Sep 13, 2009 #18

    berkeman

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    Welcome to the PF. (BTW, this thead is years old, but whatever...)

    The same always holds, AFAIK, there are no exceptions when it comes to units and math.

    [tex]V [m/s] = \sqrt{a [m/s^2] x [m]} = \sqrt{ax} \sqrt{[m^2/s^2]} = \sqrt{ax} [m/s][/tex]
     
  20. Sep 13, 2009 #19
    Thanks. I was on the right track just have to remember basic algebra.
     
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