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Fluid mechanics -- fluid in a container moving down an incline

  1. Sep 4, 2014 #1
    I have added a picture of the problem with an attempt for a solution. Please let me know if its right or wrong, and what I have done wrong if its wrong..
     

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  3. Sep 5, 2014 #2

    BvU

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    You can ask yourself what ##\alpha## would be if ##\theta > 45^\circ##...
     
  4. Sep 5, 2014 #3

    HallsofIvy

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    Are we to assume that [itex]\alpha[/itex] is the angle a water level inside the carriage makes with the top of the carriage? The picture looks a bit peculiar. Is the top of the carriage parallel to the bottom?
     
  5. Sep 5, 2014 #4
    Yes, thats why I post the problem here, exactly because I think it seems strange that when [itex]\theta[/itex]=45 then [itex]\alpha[/itex]=90.. I suspect maybe my equation for the x-direction is wrong??

    α is the angle between a pressure surface/line in the water and the ground floor (not the inclined). Yes if there was a top on the carriage it would have been parallell to the incline. But the water inside is certanily not (though I have not confirmed this is true by experiment).

    Thanks!:)
     
    Last edited: Sep 5, 2014
  6. Sep 6, 2014 #5

    BvU

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    As a physicist, I would be much happier with an answer like ##\alpha = \theta##. Without too much thinking: just intuitively (if my beta brain is capable of something like that).

    So now we should go through the relevant equations and your attempt at a solution. More or less as in the template. I suppose ##\vec G## is ##\rho \vec g##. So if ##\vec\nabla p + \rho\vec g = \rho \vec a## I am allowed to re-arrange to ##\vec\nabla p + \rho\left (\vec g -\vec a \right )## and I would be tempted to claim that the water (or whatever is in the carriage) pressure gradient is pointing perpendicular to the incline, therefore has no component along the incline. So isobars are parallel to the incline. The surface is an isobar, so the surface is parallel to the incline....

    Now how about that ? Anyone prove me wrong this late Saturday night ?

    [edit] so your ##{\partial p\over \partial x}## equation misses a term when I look at the line above (you know, the relevant equation you forgot to mention under 2. in the template) ...
     
    Last edited: Sep 6, 2014
  7. Sep 8, 2014 #6
    I have to agree with you ofcourse, but the book says that α>θ.. Either way, it seems that strange things happens when the angles are close to 90 degrees.

    Yes, but if I have that extra term in the equation for the x-direction then I get dp/dx=0, which supports your theory that the pressure lines are parallell with the inclined plane.. So maybe you're right after all?

    Btw the equation for the system should have a minus sign infront the pressure gradient. My bad:( But it doesn't seem to make any different..
     
  8. Sep 9, 2014 #7

    haruspex

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    As a thought experiment, consider θ almost 90°. We have the container and water in free fall, nearly, but with a slight sideways force. Since it is not quite free fall, it is reasonable that the water will settle at the top of the container. The slight sideways force will mean that the surface is tilted from the horizontal. So α is almost 180°.
     
  9. Sep 9, 2014 #8

    BvU

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    @haru: you mean almost 90 ?
     
  10. Sep 9, 2014 #9

    haruspex

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    I was being facetious. The 2θ answer implies α would be near 180, but that would mean the water would be almost upside down while falling more slowly than g, clearly not possible.
     
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