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Block floating in between oil and water, what's the FBD?

  1. Mar 25, 2014 #1
    [Mentor note: This thread was moved from General Physics due to it being homework related]

    A friend and I were reviewing problems from a GRE subject which read:
    "A layer of oil with density 800 kg/m^3 floats on top of a volume of water with density 1000 kg/m^3. A block floats at the oil-water interface with 1/4 of its volume in oil and 3/4 of its volume in water, as shown in the figure above (below; i found a similar image). What is the density of the block?"


    I solved it using the formula for the buoyant force, F = ρVg and the weight. So the balance was:
    800gV/4 + 1000g*3V/4 = ρVg
    950 kg/m^3 = ρ
    In this I considered both buoyant forces, from oil and water to be directed upwards.

    However, my friend considered the pressure at the depth of the top of the block to be P0, and at the bottom of the block to be P1. So (being 4h the height of the block, and A its top & bottom area):
    P1 = P0 + ρ_oil*h*g + ρ_water*3h*g
    When I work out the forces on the block (the weight of the oil above, the pressure on the bottom of the block) I'd get a balance:
    P1*A = W + P0*A
    Replacing the values we got the same value of density, 950 kg/m^3.

    What is breaking my brain is the fact that I considered the buoyant force of the oil on the block to be upwards, while my friend considered it to be downwards on the block.

    I want to know what is the real free-body diagram for the block and is the oil pushing down on the block or pulling it up?
  2. jcsd
  3. Mar 25, 2014 #2


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    Oil can't pull up on anything.

    Like a lot of physical problems, a solution can be obtained using more than one approach. You and your friend have encountered such a problem and obtained the same result.

    Remember, the buoyant force on a floating object is produced by the object displacing the fluid in which is floats. This displacement of fluid creates a pressure field which acts over the wetted surface of the object. By Pascal's Law, the pressure acts normal to the wetted surface and the magnitude of the pressure difference is proportional to the depth of the fluid.

    If an object is submerged in a fluid but not resting on the bottom of the container, there will be a hydrostatic pressure acting on the upper surface, just like there is a hydrostatic pressure acting on the bottom surface. The net force acting on the block will be:

    Fnet = Pbot*Abot - Ptop*Atop

    For a wall-sided block, Atop = Abot = A and

    Fnet = Pbot*A - Ptop*A = A*(Pbot - Ptop) = A*ΔP

    We know from Pascal's Law that ΔP = ρgΔH, so

    Fnet = ρgΔH*A = ρgV, where V is the volume of displaced fluid.

    Et voila!, as the French say, we have established Archimedes' Principle.

    If you want to draw a FBD for this problem, you would have one force to represent the weight of the block and two buoyant forces acting in the direction opposite of the weight: one buoyant force is due to the amount of water displaced, and the other is due to the amount of oil displaced. The two buoyant forces added together would equal the weight of the block.
  4. Mar 25, 2014 #3
    Doesn't that equation represent the FBD of the actual vertical forces acting of the block.
    Draw out the block and the vertical forces.

    You would have:
    P1 of the water acts on the bottom surface upwards.
    W of the block due to gravity acts downwards.
    P0 of the oil acts on the top surface of the block downwards.
  5. Mar 25, 2014 #4
    No, he didn't. The pressure exerted by oil on top (times area) is not the buoyant force. The buoyant force comes from a difference in pressure.
    If you want the oil's contribution to the buoyant force you need to take (p2-p0) where p2 is the pressure at the interface. Which you implicitly did when you wrote the buoyant force in your treatment.

    So there is no contradiction between the two methods. His is about overall buoyant force, yours add the components explicitly.
  6. Mar 25, 2014 #5


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    The "buoyant force" is the resultant of the two liquids. The effect of the oil is not 'buoyant'.
    The oil on the top surface of the cylinder is pushing Down. The water on the bottom surface is pushing Up. In the equilibrium condition, the upward force equals the downward force of the oil and the weight of the cylinder. The magnitudes of pressures on the faces, relative to the liquid interface are ρhg. Signs must be taken into account of course, when writing the equation.
  7. Mar 25, 2014 #6
    Thanks to all for your messages.
    So what I've gathered, to sum up the previous comments, is that both liquids (and I'd guess any number of fluids where a solid is inside) generate a net buoyant force which is due to the difference in pressures. This buoyant force can be analyzed in its components which are given by the amount of fluid displaced by each part of the object in the liquids. So I can say a net buoyant force opposes the object's weight and analyze it's (imaginary, non-physical) components. Kind of like when we treat friction as a force on it's own, but it's physically part of the reaction force of the surface.
    On the other hand, and equally valid, I can calculate the net force on the object using the pressure at the bottom and the weight of the oil on top. In fact its mathematical expression is equivalent to the equation using components of the buoyant force, just by rearranging the factors.

    I was wrong thinking that the oil was pulling up on the object, because it isn't physically producing a buoyant effect. It's the whole thing, the oil and the water which produce a net force upwards. In fact, without the oil, the water would be under so much pressure and the object would actually sink a little more.
    Thanks to all for helping me clear that up :)
  8. Mar 25, 2014 #7
    I don't understand what you just said. Something about pressure. Things exhibit buoyant forces because of g.r.a.v.i.t.y., fluid flow, and differences in density. The force is indeed the sum of the weights of (as opposed to "amount of") fluids displaced less the weight of the submerged object. You certainly can use the term "oppose" when describing forces acting in opposition, I suggest you may want to be careful with that since force is a vector quantity and only in the most simplistic cases are forces co-linear (directly in line or directly opposed). Pressure is caused by gravitational force, but it is also caused by other sources of compression. Most fluids are almost incompressible (similar to most solids) and small volume reductions create enormous pressures. Picture a blob of water (or an oil and water blob) in space (free fall, microgravity). Add your block. What will happen?
    Is it correct to say that the blob is under no pressure?
  9. Mar 25, 2014 #8
    I agree what you said about buoyant forces, but in this case it's hydrostatics, so fluid flow is something else. Also, no variation in density is being considered, so that part of the comment doesn't help here.
    I think I can say "oppose" just fine in this case because shear forces do not come up in hydrostatics, so whatever forces appear will generate a net force co-linear with the weight of the object.
    In the part where I said "amount of" I meant it is part of the component, as in it's a factor in the equation.

    In such a case, we can watch the videos from astronauts in the space station such as this one
    Blobs of water tend to stay unified. They're certainly under pressure due to the air around them. In the vacuum you couldn't have that bubble because there wouldn't be anything to keep it under pressure so it just "evaporates".
    From the video you can see that the object inside the bubble doesn't float anywhere, it just stays put.
    Last edited by a moderator: Sep 25, 2014
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