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Origin for formula when an object is submerged in water

  1. Aug 7, 2014 #1
    1. The problem statement, all variables and given/known data

    The formula is w(object)/F(buoy) = [density(object) * V * g] / [density (fluid) * V * g] = [density(object)] / [density(fluid)]

    This is way I thought they did it:

    w(object) = F(buoy)

    w(object) / F(buoy) = 1

    ---

    Vsub/V = density(object) / density(fluid)

    Since Vsub = V

    density(object) / density(fluid) = 1

    Since w(object) / F(buoy) = 1 and density(object) / density(fluid) = 1 they must both equal each other.

    ---

    Finally we know that w(object) = density(object) * V * g

    and F(buoy) = density(object) * V(sub) which is the same as V * g

    ----

    Now that I have explained my reasoning onto the question!

    Does that mean for any object that is sinking that the ratio of the density of the object will always be equal to the density of fluid, because obviously that isn't always the case but according to this formula that is what it says right?
     
  2. jcsd
  3. Aug 7, 2014 #2

    Simon Bridge

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    No - on both counts.

    Think it though from your experience:
    If an object is less dense than the water (i.e. cork) - it floats... where?
    By comparison, an iron nail will sink won't it?
    So how does the density of an iron nail compare with that of water?
    ... therefore, what do you conclude about the density of an object that is sinking?
     
  4. Aug 7, 2014 #3
    Yeah I know from experience that objects with a greater density than a given fluid will sink in said fluid, one's with the same density will float just under the surface of the fluid (but won't sink because Fbuoy and Fg are just equal). But I am just trying to understand how this formula works. According to the formula it contradicts well practical applications doesn't it?
     
  5. Aug 7, 2014 #4

    Simon Bridge

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    ... this is not completely correct. Submarines operate at neutral buoyancy - how deep do they float?

    In what way does the formula contradict your experience?
    afaict it is saying that sinking objects have a density greater than the fluid they are sinking in.
     
  6. Aug 7, 2014 #5
    Well if something has a waay greater density than the fluid in which it is sinking according to this formula Vsub/V = density(object) / density(fluid), as long as the object is fully submerged in water, the ratio of density(object) / density(fluid) can not be greater than one right? (If the object is fully submerged in water, than the volume that has been displaced Vsub has to be equal than the volume of the of the object, V)
     
  7. Aug 7, 2014 #6

    Simon Bridge

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    ... this formula is for a floating object only.
     
  8. Aug 7, 2014 #7
    I had a feeling you were going to say that because I think they used that formula to develop the equation for an object that is completely submerged under water (i.e. sinking) Ok fine then let's start from the beginning. How do you develop this formula: The formula is w(object)/F(buoy) = [density(object) * V * g] / [density (fluid) * V * g] = [density(object)] / [density(fluid)]?
     
  9. Aug 7, 2014 #8
    I was actually wondering that too because the formula that is for "a floating object only" was developed through making Fb = Fg (i.e. floating right) so then why would they used that method to develop another formula for completely submerged objects.
     
  10. Aug 7, 2014 #9

    olivermsun

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    They float at the depth where they have the same density as the surrounding fluid, of course. :wink:
     
  11. Aug 7, 2014 #10

    rude man

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    Here's how you can derive the buoyancy formula:

    1. assume the object is a wholly submerged cube, with sides parallel & perpendicular to the surface. The pressure on the sides is net zero. The pressure on the top is ρg h1. The pressure on the bottom is ρg h2. The buoyancy force is thus A(h2 - h1) (ρg) = weight of displaced fluid.

    A = area of a side
    ρ = density of fluid
    h1 = depth of top surface
    h2 = depth of bottom surface.

    2. If the cube is only partially submerged, the net force is A(ρg h2), again = weight of displaced fluid.

    3. For any arbitrarily shaped object, change the smooth surfaces into small segments of horizontal and vertical sections. (In the limit this perfectly replicates the surfaces.) Now, all vertical sections impart zero buoyancy to the object while all the horizontal sections taken together see a net pressure (from above and below) of ρg(h2 - h1), thus proving Archimede's principle for any general shape.

    P.S. buyancy force does not include the force of gravity on the submerged object. The net force on the object is mg down and Fbuoy. up.
     
  12. Aug 7, 2014 #11
    Couldn't something be fully submerged in water and still not be sinking. In that case Fb = Fg, would my derivation of the formula work then? (Ignore that in the initial question I wrote that it is sinking)
     
  13. Aug 7, 2014 #12

    SteamKing

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    When a submarine is operating submerged, it has what is called 'neutral buoyancy', and like all non-sinking objects, its weight is less than or equal to the weight of the water displaced by the hull. For the particular case of 'neutral buoyancy', the weight of water displaced by the fully submerged hull is equal exactly to the weight of the submarine.

    The submarine can adjust its weight by adding or removing ballast water. In order for the submarine to surface after it has submerged, compressed air is used to remove some ballast water from the hull, making the submarine lighter, and the difference in the buoyancy of the hull and its weight provides a net upward force, causing the submarine to rise to the surface.

    http://en.wikipedia.org/wiki/Submarine

    See the section on 'Technology: submerging and trimming'.
     
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