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Someone please help me with this cylinder floating upside down in water

  1. Oct 24, 2011 #1
    A hollow cylinder with closed ends is 300 mm in diameter, 450 mm high, has a mass
    of 27 kg and has a small hole in the bottom. With its axial vertical, it is lowered
    slowly into water, and then released. Calculate
    a. The gauge pressure of the air inside it
    b. The height to which the water will rise within it
    c. The depth to which it will sink
    Disregard the effect of the thickness of the cylinder wall but assume that it is uniform
    and that the compression of the air inside is isothermal.

    I already created an equation to find the pressure of the air inside the cylinder.
    P = Patm +γ(h-h')
    where h is the height of the column of the pressurized air in the cylinder and h' is the height of the cylinder that is protruding out of the water.

    Using this I continued to make an equation for the forces on the cylinder. I figured the forces on the cylinder would be an upward buoyant force from the water, a downward weight, and a downward pressure force acting on the surface area of the cylinder. So my equilibrium equation would be...

    Fb = W + PA where P is the air pressure that we said above so,
    Aγ(H-h') = mg + [Patm + γ(h-h')]A once everything is simplified, we get....

    h = H - [mg + APatm]/Aγ

    Apparently this doesn't work because I get a negative value for h, and the height of the air column in the cylinder cannot be negative. My professor provided the calculated answers to these questions but the solutions are provided. someone please give me some insight
     
  2. jcsd
  3. Oct 24, 2011 #2

    NascentOxygen

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    Without looking at this, my first thought is that perhaps it will sink below the surface. Maybe hover at some depth, much like a nautilus. :smile: 27 kg, that's a fair weight. Stability would be a consideration, if it submerges, but may be a separate issue.

    EDIT: just checked. If it is totally filled with air, it will displace 31.8 kg of water, so we can conclude cylinder will float only partially submerged.
     
    Last edited: Oct 24, 2011
  4. Oct 24, 2011 #3

    gneill

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    If you assume for the time being that the cylinder will not sink beneath the waves, then you should be able to calculate the gauge air pressure that would be required to support the weight of the cylinder (since it is thin-walled and has a hole in the bottom, buoyant force is not an issue for the water-filled portion of the cylinder).

    After that, start thinking about PV = nRT and how the volume of the trapped air will have to have changed in order to reach that pressure.
     
  5. Oct 25, 2011 #4
    Well I can assume the compression is isothermal so PiVi = PfVf
    Pi = Patm Vi = HA
    Pf = Patm + γ(h-h') Vf = hA

    Where A is the cross sectional area of the cylinder, h is the height of the column of air, and h' is the height of the column of air above the water surface. Once this is simplified, I get...

    * h^2 + h(10.3 - h') - HPatm/γ = 0 *

    Unfortunately I have two unknowns here so I need another equation. I thought I could use the free body diagram as another equation

    I just need clarification of the free body diagram and its forces.
    I know that there is the force from the weight acting down because the cylinder has weight.
    Now are you saying that the pressure force from the air is not acting down as well? I thought that it would.
    And also, is there a buoyant force in this problem at all?

    I thought the equilibrium expression would be Fb = mg + PA
    is it instead mg = PA
     
  6. Oct 25, 2011 #5

    gneill

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    Yes there is a buoyant force, but because of the geometry of the cylinder it can be resolved as a pressure acting upon the enclosed water/air interface, and transmitted to the underside of the cylinder top via air pressure. In effect, the cylinder is sitting on and supported by a column of air.

    attachment.php?attachmentid=40314&stc=1&d=1319548067.gif

    Note that for calculating the required gauge pressure inside the cylinder that the atmospheric pressure is a baseline (it's a background pressure affecting everything equally) and can be ignored -- it's the "overpressure" Pg that supports the weight of the cylinder. You will have to include it, though, when working with PV = nRT. You should be able to determine the required gauge pressure without requiring any reference to the cylinder's depth of submersion.
     

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  7. Oct 25, 2011 #6
    Ok I think I understand where you are going with the diagram and it is a big help. I just need some clarification in it since you are using different variables than me.

    1. Why is there not an actual buoyant force like there always is? For instance, how come there isn't an Fb = γVsubmerged like usual?

    2. Is Patm + Pd = Patm + γd? and why is it acting upwards? Is it the reaction force from the air pressure from the column of air pushing down?

    3. Is Patm + Pg the absolute pressure inside the column of air?

    I was able to work this out and get the right answers. But I am still a little confused as to how you got all those forces....
     
  8. Oct 25, 2011 #7

    gneill

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    The buoyant force is the result of pressure which operates perpendicular to the submerged surfaces of the object. Pressure on vertical surfaces all cancel out (otherwise the object would move horizontally through the water!), leaving only the horizontal surfaces with unbalanced force. For a solid object the buoyant force is the sum of all the pressures operating vertically, which for a solid cylinder would be the pressure on its bottom face.

    In this case the cylinder is hollow, so its "bottom" is the air/liquid boundary. Only the pressure of the liquid operating there provides the "lifting" force. There are no horizontal surfaces on the cylinder walls, and the bottom of the cylinder with a hole in it contains liquid with the same pressure inside as outside -- so no net forces there.
    Pd is the pressure due to the water at depth d. The absolute pressure at that depth is due to the weight of the water above and the atmosphere above. Pressure in a gas or liquid operates in all directions, but in this case we're looking at the pressure applied to the air/liquid interface.
    Yes.
     
  9. Oct 25, 2011 #8
    You are the greatest. I was able to end up calculating everything and got all the answers. Thanks for much for explaining all this to me. You were much more of a help than my professor
     
  10. Oct 25, 2011 #9

    gneill

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    Glad to be of help! :smile:
     
  11. Oct 25, 2011 #10

    NascentOxygen

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    Typo? Probably too late to edit.
     
  12. Oct 25, 2011 #11

    gneill

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    No, not a typo. Pressure against vertical surfaces create horizontal forces.
     
  13. Oct 26, 2011 #12

    NascentOxygen

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    Of course! There's no mistake at all.

    I was in such awe of your diagram, that it's clear that I wasn't paying close enough attention to the writing.
     
  14. Oct 26, 2011 #13

    gneill

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    :smile: :rofl: :smile:
     
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