buoyancy

Understanding Buoyancy: The Limits of Archimedes

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It’s a principle as old as science itself, and remarkably, remains essentially unchanged in the light of revolution after revolution in our understanding of physics. Many of us see it in our first year of physics and never again, as calculus-based treatments of fluid dynamics are usually left for those seeking to specialize in a related field (or are looking to take a crack at a million dollar math problem (see: http://www.claymath.org/millennium-problems/navier-stokes-equation)).

The principle of buoyancy:

“Given an object placed in a fluid in equilibrium, the fluid imparts a force on the object equal and opposite to the weight of the fluid displaced by the object”

In my undergrad treatment of this, we took this principle for granted as somehow arising from pressure of a fluid increasing with depth so that the pressure on the bottom half of an immersed object was larger than on the top half, which amounts to an overall upward force.

But how does this really work?

It wasn’t until I had the privilege of TA-ing a Lagrangian mechanics course that I had the time and motivation to really get to the bottom of this. In order to truly find what that net force was for a general object [itex]\Omega[/itex] (and not some contrived cube or something), I needed to be able to calculate the force on each infinitesimal part of the surface of the object, and to add all those infinitesimal forces up —-I needed to do a surface integral.

Like some of the other ironies of physics, this was a case where vector calculus made things easier to understand. Before we immerse an object into this fluid, we can consider the fluid by itself with spatially varying pressure [itex]P(x,y,z)[/itex] and spatially varying density [itex]\rho(x,y,z)[/itex]. With the fluid in equilibrium, these pressures and densities are assumed to be stable in time. The space that the object shall occupy can be given by the region [itex]\Omega[/itex], and the boundary surface of this region by [itex]\partial\Omega[/itex].

Using the divergence theorem, which relates surface integrals to corresponding volume integrals, one can show that the net force on the volume [itex]\Omega[/itex] of fluid due to the fluid pressure [itex]P[/itex] incident on its surface is expressible as:

[itex]\vec{F}_{net}=-\oint_{\partial\Omega} P(x,y,z)d\vec{a} = -\int_{\Omega}\bigg(\frac{\partial P}{\partial x}\hat{x}+\frac{\partial P}{\partial y}\hat{y}+\frac{\partial P}{\partial z}\hat{z}\bigg) dV[/itex]

where [itex]d\vec{a}[/itex] is an infinitesimal vector with magnitude equal to an infinitesimal surface area and direction pointing directly outward from that surface area.

Next, we need to know how the fluid pressure [itex]P[/itex] changes with height. Indeed, the Barometric equation in fluid mechanics tells us that the rate of pressure increase is proportional to the local fluid density [itex]\rho(x,y,z)[/itex] times the local acceleration due to gravity [itex]\vec{g}(x,y,z)[/itex] .

[itex]\vec{\nabla}P=\rho \vec{g}.[/itex]

When we plug this in, we find for our net force, that:

[itex]\vec{F}_{net}=-\int_{\Omega}\rho\big(g_{x}\hat{x}+g_{y}\hat{y}+g_{z}\hat{z}\big)dV=-\int_{\Omega}\rho \vec{g} dV = -\vec{W}_{fluid}.[/itex].

Finally, no matter what sort of object ends up occupying the volume [itex]\Omega[/itex], this net force will be the same— equal and opposite to the weight of the displaced fluid.

 

What’s really neat about this derivation is that we can understand how far Archimedes’ principle really goes. This is not a principle that applies only to the overly ideal situations you might see in a first year physics class. Here, [itex]\vec{W}_{fluid}[/itex] is the total weight of the fluid displaced as measured by adding up the total force of gravity on every infinitesimal fluid element [itex]\rho dV[/itex]. The fluid density [itex]\rho[/itex] need not be constant, but can be compressible and vary with location. Moreover, we make no assumptions about what form the gravity field [itex]\vec{g}[/itex] takes, whether it is a simple constant vector in space pointing down, or a vector field whose strength decreases with height according to an inverse square law, or even if we want to throw some minuscule cubic corrections from general relativity into the mix. As it is, Archimedes principle is quite possibly the oldest scientific principle that has not been overturned with more advanced treatments over the centuries.

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17 replies
  1. Ken G
    Ken G says:

    Although I agree there is always value in seeing a mathematical derivation, because it informs us about how to do other similar problems, I'd say there is a much simpler and more elegant proof that the buoyancy force is the weight of the fluid displaced.  One simply asserts that the total pressure force on the object does not depend on the object itself, only on the size and shape of its boundary (something used in the above derivation as well).  This means that the object can be made of anything– including whatever is the fluid.  In other words, the object could be the fluid, with an imaginary boundary of arbitrary shape.  Of course any such object must be in force balance, and has a weight, so the buoyancy force must equal that weight.

  2. tech99
    tech99 says:

    I have also realised that if a light object is pulled under the surface and then released, the maximum upward acceleration it can have is -g, because water has to fall under it to create the uplift.

  3. Chestermiller
    Chestermiller says:

    The derivation in this article is not strictly correct mathematically, unless one imagines that the space that is occupied by the solid is replaced by fluid of the same density as the surrounding fluid.  Otherwise, the divergence theorem could not be applied inside the solid, which has a different density.  The state of stress inside an elastic solid would be different from that of a liquid filling the space, and the elastic solid would have to deform (slightly) and develop stress in order to match the pressure distribution at its surface.  In the end, consideration of the stress distribution within the solid would deliver the same result that we obtain if we assume that the space is filled with the original fluid, but it would not be as straightforward as this derivation suggests.

  4. jfizzix
    jfizzix says:

    Yes, in this derivation, one must consider the fluid without an object in it in order for the divergence theorem to work. Doing that, one finds that the force obtained by integrating the pressure over the surface [itex]/partial\Omega[/itex] is equal to integrating the weight of the fluid elements over the volume [itex]\Omega[/itex].

  5. Let'sthink
    Let'sthink says:

    I would like to give an energy-argument to explain Archimedes' Principle in  a simple case of constant gravity acting downwards. There is always a problem about considering pressure at a point as a force per unit area. First I replace that idea by considering pressure as kinetic energy per unit volume. Because there is a constant gravitational force acting or in other words the gravitational potential energy decreases as we go down, we expect the kinetic energy also to increase as we go down, because there s no external force on the fluid. This explains why the pressure down the level increases with depth. For ideal gas this should reflect in the increase of temperature but that does not happen because the ideal gas is infinitely compressible so the density variation counter balances the energy density so that the kinetic energy per molecule remains the same which is really the measure of  temperature.

  6. AgentSmith
    AgentSmith says:

    A submarine in port at the surface displaces its own weight of water. The the ballast tanks are opened, and water floods in. The sub begins to sink until it is completely submerged. It then displaces its own volume of water. While sinking it displaces a mix of its weight and volume (which could be described mathematically.)  These are known facts without the math.   A Spanish marine engineer at Navantia left off a zero somewhere, and the new subs Spain was building would not be able to surface once submerged. A US sub maker had to be consulted to remedy the situation. Presumably there is an opening for a marine engineer at Navantia.  I thought this little story about practical effects of fluid statics would be interesting

  7. David77
    David77 says:

    Water density increases with increasing depth. The water that initially filled the ballast tanks is not as dense as the water deep down around the ocean floor level. Once the sub is deep underneath, you need more than just purging water to overcome those tons of water pressing the sub down. I assume if you analyze the water at a deeper level and figure out how it behaves, use action/reaction mechanism to put it to your use to push the submarine upward.

  8. cjl
    cjl says:

    I have also realised that if a light object is pulled under the surface and then released, the maximum upward acceleration it can have is -g, because water has to fall under it to create the uplift.

    This is in fact not correct – the fluid acceleration does not need to match the upward acceleration of the object, and the object can accelerate faster than g.

  9. Chestermiller
    Chestermiller says:

    Yes, in this derivation, one must consider the fluid without an object in it in order for the divergence theorem to work.
    Doing that, one finds that the force obtained by integrating the pressure over the surface [itex]/partialOmega[/itex] is equal to integrating the weight of the fluid elements over the volume [itex]Omega[/itex].

    Yes. I totally agree. I just wish this had been specifically discussed in the article. Is it too late to edit in a little extra discussion?

  10. jfizzix
    jfizzix says:

    Yes. I totally agree. I just wish this had been specifically discussed in the article. Is it too late to edit in a little extra discussion?

    I can do that.. One sec..

    ..there we go!

  11. Mark Eaton
    Mark Eaton says:

    I believe we are dealing with space or area displacement. The more mass or pressure applied to this surface mass will decrease its buoyancy.
    Inversely related would be a decrease in mass per area of volume, being displaced. changing density. Mass = volume x density. density = mass x volume.
    so the volume will be the mass divided by the density. A sponge displaces less area, than a basketball under water, and has more pressure or gravity applied to it from all angles.

  12. cjl
    cjl says:

    Its the displacement of mass of the fluid not its weight which causes buoyancy.

    Nope, it’s definitely the weight. In half the gravitational field, the buoyant force of an object will be cut in half (all else equal). In zero G, there is no buoyant force at all.

  13. tech99
    tech99 says:

    This is not a correct mechanistic explanation of what is happening.

    Thank you for the comment but can you elaborate a little, as I felt that to accelerate faster than -g it would have no fluid touching the underside to create uplift.

  14. Chestermiller
    Chestermiller says:

    Thank you for the comment but can you elaborate a little, as I felt that to accelerate faster than -g it would have no fluid touching the underside to create uplift.

    What makes you feel that way? For a fluid to cavitate, the pressure at the location in question must drop below the equilibrium vapor pressure of the liquid. Bodies can easily accelerate upward within a liquid with an acceleration than g without causing cavitation in the liquid. In my expert mentor judgment, the arguments you have been putting forth make no sense. Please read the article and the other posts within this thread to improve your background. All the posts in this thread are focusing on the static situation, where the object is not accelerating. So consideration of an acceleration of the object is not on topic. Plus, speculating and guessing in the way that you have done without the solid background is the same as espousing a personal theory, which constitutes misinformation. I strongly urge you to cease such speculation.

    Chet

  15. Chestermiller
    Chestermiller says:

    Water density increases with increasing depth. The water that initially filled the ballast tanks is not as dense as the water deep down around the ocean floor level. Once the sub is deep underneath, you need more than just purging water to overcome those tons of water pressing the sub down. I assume if you analyze the water at a deeper level and figure out how it behaves, use action/reaction mechanism to put it to your use to push the submarine upward.

    This is not quite correct. The buoyant force depends only on the local density at the depth of the sub. Even through there are tons of water pressing down from above, the pressure underneath the sub presses up with nearly equal force. If one takes into account the directionality of the water pressure acting on the sub from all sides, the net force depends only on the weight of the water that is displaced at depth.

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