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Question about dimensional analysis

  • Thread starter Cade
  • Start date
  • #1
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Homework Statement



I have an equation for motion:
x(t) = x(0) + x(0) * k * t[itex]^{1.5}[/itex]
x is in meters and t is in seconds. I have to determine the unit of k.

Homework Equations





The Attempt at a Solution



x(t) = x(0) + x(0) * k * t[itex]^{1.5}[/itex]
[x(t)] = meters
Therefore, [x(0)] = meters and [x(0) * k * t[itex]^{1.5}[/itex]] = meters
meters * [k] * seconds[itex]^{1.5}[/itex] = meters
meters * (1/seconds[itex]^{1.5}[/itex]) * seconds[itex]^{1.5}[/itex] = meters

Does this mean the the unit of k is 1/seconds[itex]^{1.5}[/itex], 1/seconds, or something completely different?
 

Answers and Replies

  • #2
1,506
17
You first option is correct 1/s^1.5
 
  • #3
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Thanks, I didn't know if that was possible. I have another question if you could please check:

A force F is equal to k*x^n, where x is in centimeters.

[k*x^n] = N
[k]*(cm^n) = N
[k] = N/(cm^n)

The unit of k is N/(cm^n), is this correct?
 
Last edited:
  • #4
1,506
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Yes, you seem to have it sorted out!!
 
  • #5
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Great, thanks. I've done far better in more difficult subjects like multivariable calculus, but the lack of a proper instructor for physics has me making stupid mistakes these days. :frown:

Would you happen to know of any online resource where I can practice graphical analysis of equations like these? What I have to do is to take a non-linear data set and convert it to a straight-line equation, determine appropriate units for slope and intercept, and determine values for the constants based on slope and intercepts. I only have two practice problems to work with.
 
  • #6
1,506
17
I have never looked for any online resources for these types of problems.
If you know the power law for the equation (t^1.5 in your first example, n in your second example) Then the graph to plot is x against t^1.5 for the first and F against x^n for the second.
These would give straight lines with gradient k in each case.
If you do not know the power law.... I think that is the case in your second example, you only know it as n then you must take logs :
F = k * x^n
LnF = Lnk + n*Lnx

A graph of LnF against Lnx will be a straight line with gradient n and intercept Lnk from which k can be calculated

Does this make any sense for you, have you met log ~ log graphs
 
  • #7
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I get the hang of those graphs, but I would feel more comfortable with practice. The other problem I have which I can't work is finding the resistivity p of a wire whose resistance R = (4pL)/(pi*d^2)

Plot R on the y-axis and 1/(d^2) on the x-axis to get gradient = 4pL/pi, and then p = gradient/(4L/pi). But I didn't get a correct graph:
https://www.physicsforums.com/showpost.php?p=3618742&postcount=3

Hence, that is why I would feel more comfortable with more to practice with. But I understand what you said about log ~ log graphs, I have this problem (the one I made this thread for, not the previous one) fitted into a linear form properly, all the points rest on a straight line.
 

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