Pressure liquid

  1. what would the pressure of liquid at a depth be in a container which is slanted?
  2. jcsd
  3. i suspect [tex] h.d.g.sinAngle [/tex]. correct me if i am wrong
  4. russ_watters

    Staff: Mentor

  5. in the derivation of the pressure of liquid, the weight is assumed to act perpendicular to the column of liquid. Plz comment
  6. russ_watters

    Staff: Mentor

    Gravity pulls straight down, so...
  7. of course but i find similar case to the inclined plane. can somebody give a reasoning?
  8. russ_watters

    Staff: Mentor

    What do you find similar about an inclined plane? What is YOUR reasoning?
  9. somebody reply
  10. Doc Al

    Staff: Mentor

    The pressure depends on the depth beneath the surface, which is not slanted. As already stated, the shape of the container--whether slanted or vertical--is irrelevant.

    If you want more, give a specific example of what you have in mind with a diagram.
  11. russ_watters

    Staff: Mentor

    Sorry, we don't spoonfeed here. If you want to learn/want help, you need to show some effort at trying to figure it out for yourself. Then when you make a wrong turn, we'll nudge you back in the right direction.
  12. wouldnt this imply that the liquid would accelerate at g in slanted tubes?
  13. Doc Al

    Staff: Mentor

    No. Please describe exactly what you have in mind. Are you talking about hydrostatic pressure? (Which is what I assumed.) Or fluid dynamics?
  14. russ_watters

    Staff: Mentor

    g is acceleration due to gravity. If you block that acceleration, the required force is f=mg. Aka, weight.
  15. DaveC426913

    DaveC426913 16,530
    Gold Member

    Since you're talking about pressure-at-depth then you're talking about a container full of fluid (as opposed to, say, an air-filled container with an amount of water placed in it).

    Water suspended in water is neutrally buoyant, so its not like some arbitrary mass of water is going to start sliding to the bottom, accelerating under gravity.
  16. No. Please describe exactly what you have in mind. Are you talking about hydrostatic pressure? (Which is what I assumed.) Or fluid dynamics?

    I am not familiar with the terms but i guess you are asking whether i am talking about stationary fluids or flowing ones. Eg. in an inclined plane, there is a mass a top, even though its weight acts exactly downward, it would rather move along the plane. And the force with with it moves along the plane is lesser according to its slope. Same for liquids. But as dave said while the liquid is continuous and stationary, the force with with a finite upper part of liquid exerts on the lower part will be the same as for the case of liquids in vertical column. This is not clear to me.

    Eg. lets take a column of liquid standing upright and pour some water into it. And then slant it a bit. Then the depth of the liquid increases even though it is not continuous on the upper part (i hope this is understood). So as the depth increases although not uniformly, the pressure in one side must increase. Help me out with this.
  17. D H

    Staff: Mentor

    Pascal's vases:


    Another set:


    Notice that no matter how weird the shape, the tops of the liquid surfaces in the different containers are at the same level. The pressure difference in some container from bottom to top does not depend on shape. It depends only the height of the liquid.
  18. i havent learned any of hydrodynamics or hydrostatics. So i think this phenomenon is taken as as true. Is there a proof for it?
  19. D H

    Staff: Mentor

    The diagram and image (real) I posted in [post=3643843]post #16[/post] are pretty solid evidence. Hydraulic pumps rely on this principle.
  20. russ_watters

    Staff: Mentor

    This is getting very frustrating for me because a bunch of people are putting a bunch of effort into helping you learn, but it doesn't seem like you are trying at all. For example, you said you don't know terms like "hydrostatic", but that term is defined in the very first sentence of the link i gave you in the first reply!

    "Learning" is not something we can give you: we can tell you where to find the knowledge, but you have to get it into your head.
  21. OP , when I was a kid (well now also I am a kid in class 10th ! :biggrin:) I also used to think the same way as you are currently.

    I used to say :
    Pressure = (Force x sin θ)/Area

    Now I know that I was wrong because - Pressure is "thrust upon area." It is defined this way. Thrust is defined as force acting perpendicularly on a body.

    Hence always ,

    P=hρg x sin 90o or P = hdg x cos 0o

    As sin 90o = cos 0o = 1

    So P=hρg

    Do you get it now ?
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