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Dimension Analysis

  • #1

Homework Statement



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?

Homework Equations



x=α+(2/3)βt^(3/2)

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.
 

Answers and Replies

  • #2
418
28
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.
 
  • #3
993
13
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.
 
  • #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?
 
  • #5
pasmith
Homework Helper
1,740
412
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?
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].
 
  • #6
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
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.
 
  • #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.
 
  • #8
418
28
I mean, I already told you the name of the unit:
meters per root second cubed
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
 

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