Check my fluid mechanics work please.

I'm not sure if I should have posted in the advanced section, but this is for the class "fluid mechanics" which is an upper division course. However, the material I'm asking is mainly introductory stuff.

Determine the dimensions of dP/dx.
Where P is a force and x represents distance

So dimensions of P would be ML/T^2
Dimensions of dP/dx would be (ML/T^2)' = M/T^2

Is that right?

Related Advanced Physics Homework Help News on Phys.org
TSny
Homework Helper
Gold Member
Yes, that look's correct.

[Although, I've never seen anyone take a derivative of a dimensional expression as you did! Dimensionally, you can think of a derivative as just a fraction. So, the dimensions of dP/dx is just the dimensions of the numerator (P) divided by the dimensions of the denominator (x). Think of dP as essentially the same as ΔP which is a change in pressure and therefore has the dimensions of pressure. Similarly for dx.]

Last edited:
Hey, thank you! I thought it was a weird problem too. I've never had to do something like this either, so I was a little skeptical about my own method.

I posted in the introductory forum an extension of this question https://www.physicsforums.com/showthread.php?p=4050555#post4050555.
This one really confused me, I've never seen anything like it.

Chestermiller
Mentor
You said that P was force, but in this context, the professor must have meant that P is pressure. It looks like TSny inherently assumed that P is pressure, which is probably correct. The units of pressure can also be considered N/M2, so that dP/dx would have units of N/M3, or force per unit volume. I think that this is what your professor was getting at.

chet

No he stated that P was force.

TSny
Homework Helper
Gold Member
Yes, that look's correct.

[Although, I've never seen anyone take a derivative of a dimensional expression as you did! Dimensionally, you can think of a derivative as just a fraction. So, the dimensions of dP/dx is just the dimensions of the numerator (P) divided by the dimensions of the denominator (x). Think of dP as essentially the same as ΔP which is a change in pressure and therefore has the dimensions of pressure. Similarly for dx.]
Ugh, I mean't to say force instead of pressure. Sorry, Chestermiller and pyroknife.