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Need Help with a Basic Fluids Problem - Buoyancy and Archimedes' principle

  1. Apr 6, 2009 #1
    Good afternoon, I am having some trouble solving this basic fluids problem, and I am hoping you can look over my work, and offer some assistance.

    1. The problem statement, all variables and given/known data

    A submarine has a total mass of 2.1 multiplied by 10^6 kg, including crew and equipment. The vessel consists of two parts, the pressure hull, which has a volume of 2 multiplied by 10^3 m^3, and the ballast tanks, which have a volume of 6 multiplied by 10^2 m^3. When the sub cruises on the surface, the ballast tanks are filled with air. When cruising below the surface, seawater is admitted into the tanks. (Neglect the mass of air in the tanks and use 1.025 as the specific gravity of seawater.)

    What fraction of the submarine's volume is above the water surface when the tanks are filled with air? (%)

    2. Relevant equations

    0da27be206af8796ac55da6eb87dd2e8.png

    3. The attempt at a solution

    First I set the right hand side of the equation to 1, since the apparent weight will be zero because the submarine is floating on the surface and their will be no experienced force.

    Therefore I have

    p(sub) = p(sea water)

    m(sub) / Volume (sub) = density of water

    m(sub) / density of water = Volume of Sub Submerged in Water.

    or

    Volume Sub Submerged in water (m^3) = 2.1*10^9 g / (1.025*(1000*1000*1000) g/m^3)

    But I get a Volume of 2.04 m^3,

    or ~1% of the sub is submerged and ~99% of the sub is above water.

    This answer is incorrect, and is intuitively wrong as I don't think a 200,000 kg 2m^3 block would float.

    Thanks in advance for your help
     
  2. jcsd
  3. Apr 6, 2009 #2

    LowlyPion

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    Homework Helper

    Welcome to PF.

    I would think about it slightly differently.

    Your submarine has a total volume and the weight of the water it displaces is your buoyant force available.

    Subtracting your dead weight then yields the amount of water that the submarine would need to submerge.

    When completely filled with air then dead weight/total water weight displaced is your % of volume below the waterline then isn't it?
     
  4. Apr 6, 2009 #3
    Thanks for the reply, unfortunately I am having a bit of trouble following you.

    Here is how I am interpreting your strategy: I should find the total buoyant force when the sub is filled with enough water such that the sub's gravitational force is canceled out by the buoyant force.

    I use the weight of water to find the dead weight (what do you mean by dead weight)?

    Then I divide the dead-weight by the weight of the total displaced water?

    Thanks again!
     
  5. Apr 6, 2009 #4

    LowlyPion

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    Homework Helper

    Not exactly ... I think.

    Isn't the total volume times the density of sea water going to be equal to the amount of force available ... regardless of what weight may be inside?
    If it's 0 weight inside then isn't the whole thing going to be floating? If the weight inside is say 2/3 then won't 2/3's be submerged and 1/3 out?

    Given that thought, then what is your total volume? 2000 of the Pressurized Hull and 600 in the ballast tanks? = 2600 m3 ?

    Isn't that volume times the sea water weight going to be it's total possible buoyant force? And you are only using just the carry weight when air is in the ballast?
     
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