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Buoyancy Force Question

  1. Oct 20, 2008 #1
    This is my first post here so please be gentle! I am looking to investigate the buoyancy force of a cuboid. The dimensions of the cuboid are 30cm x 20cm x 15 cm. When the cuboid is placed in the water (the face 30cm x 20cm faces the water), 4cm of the cuboid will be submerged. If I push the cuboid so that the top face is 15mm below the surface of the water, I will have pushed the cuboid (15 - 4) + 1.5 = 12.5 cm. Therefore, using some notes that I have I have worked out the buoyancy force to be:

    F = p*A*y*g

    where p = pressure of the water (998.2 kg/m^3, pressure of tap water at 20oC), A is the cross sectional area to face the water (0.06 m), y is the overall displacement (0.125 m) and g = 9.8m/s, acceleration due to gravity. If I work this out, I get: 73.37 Newtons.

    First things first, is this correct? Please note that this is NOT homework, just a problem that I am looking into. I have looked at some notes where I was placing a bottle filled with sand in water, and then pushing it to a new depth. We used F = p*A*y*g to work out the buoyancy force in that case, however I dont believe that we considered the case if the bottle was pushed so far into the water that it was fully submerged (as in this case). Therefore, I think that the equation is correct so far as I push the cuboid down to a depth of 11cm so that the top face of the cuboid will be level with the top of the water, however I am unsure if I can carry the above reasoning through to when the object is fully submerged in water. I then want to consider pushing the cuboid so that the top face is 15cm and then 30cm deep, if I know that I can apply the above principle at 15mm then I know that I can extend it to these two cases.

    What I am really confused with (the reason i dont think that this works past 11cm) is that Archimedes said the Buoyancy Force is equal to the mass of the water displaced. However, when fully submerged, surely there wont be any more water displaced if we push the cuboid deeper and deeper?!?

    Any help on this would be greatly appreciated!

    Cheers in advance,
  2. jcsd
  3. Oct 21, 2008 #2
    Correct, once an object is fully submerged, the volume of water displaced is constant...hence the bouyant force, if it exists, remains constant...
  4. Oct 21, 2008 #3
    It looks like you are using the correct formula. The p in your formula actually is the greek letter, 'rho' which stands for density.

    Keep in mind that if the cube has a mass associated with it, then the buoyant force isn't the only force which needs to be considered.

    You're also right about Archimedes principle - after the object is completely submerged, the buoyant force will be the same at any depth. (I suppose some approximations are being made here - density of water constant regardless of depth, etc..) This can all be derived by first showing that the pressure in water varies with depth something like:

    P(depth) = Pair + density(water)*Area*depth*g.

    Next, use F=PA to sum up the forces acting above and below the cube. At this point, you basically have Archimedes law.
  5. Oct 21, 2008 #4

    Thank you Naty1 and spaceboy033 for the replies. I denoted p as 'rho' in my equation. When you say that if the cube has mass we must also consider another force, do you mean m*g? If so, the weight of the cube is 3kg. Using wikipedia, I noticed that the formula:

    m*g - p*A*y*g

    can be used to determine the net buoyancy force. However, if I take away 3 from 64.57 N (amended buoyancy force, after I changed the depth to 0.11m) I will get a number that is negative. So, should I just consider 64.57 - 3 = 61.57 Newtons here to be the net buoyancy force?
  6. Oct 21, 2008 #5
    I would say the buoyant force is simply the force due to the displacement of water, Fb= +pAyg. Since you mention that the cube floats, this buoyant force must be opposing gravity, Fg= -mg.

    To double check your calculations, you might consider the equilibrium situation. When the cube is floating stationary at 4cm depth or whatever it is, the -mg force should be balancing out the buoyant force exactly. At other depths, you should expect your cube to accelerate, and perhaps bob up and down.
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