1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dimension Analysis

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data

    In the following expression x is a position and t represents time. What are the physical dimensions of each of the constants α and β? Also, for each what are the corresponding SI units?

    2. Relevant equations

    x=α+(2/3)βt^(3/2)

    3. The attempt at a solution

    I know that since x is a position, it must have the dimension of L and the unit of m. That makes the right have a dimension of L as well. That said [α]=L, but the β is throwing me off because of the t^(3/2). I can't for the life of me figure out how to manipulate that exponent to end of with a dimension of L or L/T or L/T^2.
     
  2. jcsd
  3. Sep 22, 2013 #2
    Basically, the strategy for finding the units of β is to make sure all other units cancel besides one power of [x].

    Since the factor of 2/3 doesn't matter (it is unitless or "dimensionless"), and you already have figured out α, we can write:

    [x]=[βt3/2] = [β] [t]3/2
    where the square brackets mean "the units of" whatever is enclosed. One way to solve for the units of [β] is to use algebra (really some sort of pseudo algebra, but everything works out just fine) like this:

    [x]/[t]3/2 = [β]
    or
    [β]=[x][t]-3/2

    So this means the units of β are units of length times time to the power -3/2, or meters per root second cubed.

    If you plug this in you can verify the result:
    [β t3/2] = [β] [t]3/2 = ([x][t]-3/2) [t]3/2 = [x]
    as desired.
     
  4. Sep 22, 2013 #3
    Try to figure out what must be the dimensions of β so that the dimensions of βt[itex]^{3/2}[/itex] ends up as L.
     
  5. Sep 22, 2013 #4
    I tried this approach and it made sense algebraically. However, I don't know of an SI unit associated with [x][t]^-3/2. Am I missing something?
     
  6. Sep 22, 2013 #5

    pasmith

    User Avatar
    Homework Helper

    Yes; the SI unit is [itex]\mathrm{m}\,\mathrm{s}^{-3/2}[/itex].

    The formula [itex]y = x^a t^b[/itex] makes dimensional sense for any real [itex]a[/itex] and [itex]b[/itex], and if [itex]x[/itex] and [itex]t[/itex] are measured in metres and seconds respectively then [itex]y[/itex] will be measured in units of [itex]\mathrm{m}^a\,\mathrm{s}^b[/itex].
     
  7. Sep 22, 2013 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    It's a tragedy that there isn't an SI unit with dimensions of LT^(-3/2), but it's still OK if a quantity has units which are a mixture of only basic units. There simply aren't enough people to provide names for every combination of basic units.
     
  8. Sep 22, 2013 #7
    Thank you so much. Being a new physics students I made a rookie mistake in assuming the only dimensions were the ones listed on a chart with quantities.
     
  9. Sep 22, 2013 #8
    I mean, I already told you the name of the unit:
    Is it a problem that the units of acceleration are meters per second squared? That is an extremely commonly used one, as in g=9.8 m/s2
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Dimension Analysis
  1. Dimension analysis (Replies: 4)

  2. Dimension analysis (Replies: 5)

  3. Dimension analysis (Replies: 4)

  4. Dimension analysis (Replies: 6)

Loading...