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Archimedes Principle proofs

  1. Aug 24, 2004 #1
    Ok, I've seen many proofs of this, all being the same, but the closest I could find online was here: http://freespace.virgin.net/mark.davidson3/IMS2121/buoyancy/Buoyancy.html

    Basically the idea is you mess around with the formulas for pressure and hey bingo. However, I have one question - the general idea seems that the up force is a reaction force due to the force downwards of the water above the cube + the weight of the cube, with would indeed be h2pgA IF the cube was of water! Otherwise, the density is not the same and therefore you would have to bring more variables into the proof. Do you get me? Can anyone help and elaborate on the proof?
    Last edited by a moderator: Apr 21, 2017
  2. jcsd
  3. Aug 24, 2004 #2
    The weight of the cube is not involved in the boyant force, it is only involved in whether the cube sinks or floats. The boyant force is due to the pressure variations at different depths in the water. Since the water pressure at depth h is [tex] \rho g h[/tex], where h is the depth from the surface and g is 9.8m/s^2 and rho is the symbol for the density of the liquid, in this case its water. Since pressure is F/A, the pressure times the area (on to which the pressure is being applied) equals the force being applied onto that area. In the web site, you have a cube with areas h2^2 on the top and bottom. The bottom is at depth (h1 + h2) while the top is at depth h1. When you see the derivation done in the web site, you notice that the boyant force is actually equal to the weight of the water displaced, since [tex] F_b = \rho gh_2^3[/tex], the h2^3 is the volume of the cube, and therefore the volume of the water displaced. The volume times the density would give you the mass of the water, and mass time g is mg, which = the weight of the water. For the cube to float, then the weight of THE CUBE must be less than the weight of the water displaced (of less than the boyant force, which is the same thing). And for the cube to sink, the weight of the cube must be more than the weight of the water displaced. Also, just to simplify things...since the weight is determined by the mass times g, you can cancel the g from both and therefore tell if something will float/sink by the mass of water displaced and the mass of the cube.
    Last edited: Aug 24, 2004
  4. Aug 24, 2004 #3

    Doc Al

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    proof of Archimedes principle

    The buoyant force on that cube is entirely due to the difference in pressure on the top and bottom sides of that cube. It it not a reaction force to the weight of the cube: The weight of the cube is irrelevant to calculating the buoyant force.

    However, the weight of the cube is relevant to a simple proof of Archimedes principle. Consider an imaginary cube (or any shape) inside a fluid that is at equilibrium. The net force on that imaginary cube must be zero. The only forces on it are the buoyant force and gravity. Thus, the buoyant force equals the weight of the cube: which is the weight of the displaced fluid. Make sense?
  5. Aug 25, 2004 #4
    But why does the water exert a force UPWARDS on the cube? It makes sense to say that Fdownwards = h1gpA, but why should the water below the cube exert a force of h2gpA upwards?
    Thanks. :smile:
  6. Aug 25, 2004 #5


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    Because water pressure increases with depth, thus higher pressure below than above.
  7. Aug 25, 2004 #6


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    Perhaps it would be clearer this way:

    The water is being displaced by the cube. The displaced water cannot go down because the container in which it rests has a bottom. It cannot go sideways because the container has sides. The water level in the container must go up.

    So the cube and the water are both applying upward force on one another. Like two children on a sea-saw, whichever one is heavier will sink down, and the other will be forced up (barring the input of some outside force, that is). If it is the cube that sinks, then the weight of the water it displaces must be subtracted from the weight of the cube, since gravity is still trying to pull that water back down to where it would be if the cube were absent.
  8. Aug 26, 2004 #7


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    This is from the website in Cheman's link:
    I don't want to appear smug, but it should be: "the upthrust on the body is equal to the weight of the displaced fluid."
    The above suggest you should always work with the weight of water. Denser materials give a higher buoyant force, that's why lead floats in mercury, but sinks in water.
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