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Fluid Mechanics, Depth, Pressure

  1. Apr 6, 2012 #1
    I'm having a little problem with my book as I was reading about fluid mechanics. The book seems to have skipped a bit of some crucial part (at least for me) during the derivation for fluid pressure at certain depths (where the weight of the fluid is not neglected).

    Here, I'll try to reconstruct the derivation and try to point out the parts where it confused me.

    First suppose we have a fluid with definite volume where its density is the same throughout, hence it's uniform. Now, if we take an element fluid with thickness dy then and its top and bottom surfaces are the same, say A. Its volume is dV=A*dy, it's mass dm=(rho)*dV=(rho)*A*dy, and its weight w is dmg=(rho)*g*A*dy.

    Where rho is the density of the fluid.

    When the book gave an analysis of the forces on the y-component of that certain element fluid, the upward force is given by F(upward) = pA. I understand that part since there is pressure (p) pressing the fluid at its bottom area. Now, when the book gave the downward forces, it's given by F(downward) = (p+dp)*A and this confused me, where did the additional dp come from? There is the p that presses the upper area of the fluid but what about dp? Also the other downward force is the weight, but the fluid is in equilibrium, so:

    (Sum)Fy = pA-(p+dp)*A-W=0 *The fact that there is force p*A upward and downward plus the weight is also non intuitive for me.*

    So yeah, the biggest question for me is dp in the derivation, what is that?
     
  2. jcsd
  3. Apr 6, 2012 #2

    tiny-tim

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    hi mathsciguy! :smile:

    (have a rho: ρ :wink:)
    it's more intuitive if we write p(y) instead of just p, where y is height

    then the force upward at height y is Ap(y),

    and the force downward at height y+dy is Ap(y+dy), = A{p(y) + dp} (where dp = (dp/dy) dy)

    so the total force downward (excluding gravity) is A{p(y) + dp} - Ap(y) = Adp​

    (and dp will be negative because y is height, not depth :wink:)
     
  4. Apr 6, 2012 #3
    Thanks, I thought no one would reply since my post was kinda messy.

    So the pressure really is different at differing altitudes, but I thought that was caused by considering the force done by the gravity I.e. the weight. Since there is weight then it gives a downward pressure at the top surface (of course I know that if I think of it like that then the element fluid will not be in equilibrium, but I think that's another question).

    I've thought of that, since from what I know, if we consider the case where there is negligible force done by the gravity then the pressure is the same throughout the volume of the fluid, it's the same even if we consider the pressure done by fluid onto a part of itself.

    Edit: I think, my confusion is caused by not being sure of why pressure vary in altitude. Would it vary in altitude if we neglect the weight of the fluid?
     
    Last edited: Apr 6, 2012
  5. Apr 6, 2012 #4

    tiny-tim

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    yes, that's correct …

    pressure = force per area, and the only force downward is the weight of the fluid above :smile:
    not following you … the slab of height dy is in equilibrium :confused:
    no, eg air is a fluid, and its density is negligible (for most bodies, not for balloons etc), which is why we usually regard atmospheric pressure as the same throughout the lab, regardless of height :wink:
     
    Last edited: Apr 6, 2012
  6. Apr 6, 2012 #5
    Sorry, what actually goes through my mind is this:
    Not considering dy, Fnet = Ap(y)-w-Ap(y) = ma *Well, because I thought at both top and bottom A there is that same pressure p, we just got an additional pressure done by w.

    I kind of get it now though, if we solve for the net force on parallel to gravity we could see that Adp=-W, and this makes sense to me now.

    Many thanks sir tiny-tim.
     
    Last edited: Apr 6, 2012
  7. Apr 6, 2012 #6
    Sorry, what actually goes through my mind is this:
    Not considering dy, Fnet = Ap(y)-w-Ap(y) = ma *Well, because I thought at both top and bottom A there is that same pressure p, we just got an additional pressure done by w.

    I kind of get it now though, if we solve for the net force on the y-axis we could see that Adp=-W, and this makes sense to me now. It's kind of not very intuitive to see that Adp is an upward force. (I might be wrong all together on this, I hope not)

    Many thanks sir tiny-tim.
     
  8. Apr 6, 2012 #7

    tiny-tim

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    it is! … there's more pressure underneath than on top, and that's the bouyant force that helps that slab of fluid to float! :biggrin:
     
  9. Apr 6, 2012 #8
    Thanks, I haven't read about buoyant force in depth yet, it's a good thing I've got some idea.
     
  10. Apr 6, 2012 #9

    tiny-tim

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    well, that's where to find it! :biggrin:
     
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