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Consider the buoyancy force in water with very small but macroscopic particles in it. Such particles (suspension) will very slowly drift downwards and will eventually settle on the bottom. If one did not know that the particles are present there then for calculating the buoyancy force, $F = \rho V g$, one would just use the average density of water with suspended particles in it, which is larger than the density of pure water. Would this be a correct calculation?

Suppose we do an experiment with a cylindrical vessel filled with water and a fully submerged float in it, attached to the bottom with a cord, and then we drop some amount of very fine powder into the water. The powder will form a cloud that will slowly drift downward. What would be the observable effect (if any) on the tension in the cord, once the cloud of particles fully covers the float? Assume that the cloud transverse size is large enough to fully cover the cross-section of the vessel. It is assumed that the dust particles don't stick to the surface of the float.

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Assume spherical particles of radius $r$ and density $\rho_p$. Within the cloud, the volume fraction of particles is $M$ (for muddiness). The radius of the spherical buoy is $R$. The water has density $\rho_w$ and dynamic viscosity $\mu$. Gravitational acceleration is $g$.

Further assume that any particle that makes contact with the buoy stops and then re-enters the water column and resumes sinking (could imagine a horizontal current to sweep the stopped particles off the buoy, or a wiper).

A large concentration of very small particles in otherwise homogeneous fluid would cause convection: heavy fluid overlying light is unstable. Rather than the particles settling through the fluid, this would be the fluid "pulled along" with the particles in sinking plumes. We will ignore this effect. If you like, you can imagine that there is a vertical temperature gradient which keeps things stable even in the presence of the particle cloud.

With that out of the way, there are two things to consider: (1) the increase in buoyancy$^*$; and (2) the force exerted by the impact of the particles on the top of the buoy. The first of these is $$B=\frac{4}{3}\pi R^3 M g (\rho_p-\rho_w).$$

The second--the force transmitted to the buoy by the impacting particles--is their momentum flux: $$F=\pi R^2 M \rho_p v^2,$$ where $v$ is the speed with which the particles sink. Stokes' law is accurate for Reynold's numbers less than one or so, which includes everything one might describe as "particles" (as opposed to "rocks" or "ball bearings"). So the settling speed is given by $$v=\frac{2}{9}\frac{(\rho_p-\rho_w)gr^2}{\mu}.$$

So does the tension increase or decrease? That depends on which dominates: $B$ or $F$. The ratio of their magnitudes is (after a little algebra) $$\frac{B}{F}=\frac{\mu^2 R}{27 \rho_p g r^4 (\rho_p-\rho_w)}.$$

If we plug in typical values for gravity and the properties of water, and choose 100 microns for $r$ and 1.5 g/cm$^3$ for $\rho_p$, we find that the critical buoy radius $R$ is 2 cm. Smaller than that, and the buoy is pushed down by the sinking particles; larger, and the buoyancy wins, increasing the tension.

Note the quartic dependence on particle size: shrinking the particles moderately increases the relative importance of the buoyancy effect dramatically (and vice versa).

$*$ Side note: some previous answers have questioned whether the particles contribute to the buoyancy at all. They do.

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  • $\begingroup$ This solution is correct. The ram pressure strongly decreases with decreasing particles' size, so making them small enough eliminates the ram pressure contribution. An interesting question is how drifting downward particles contribute to pressure distribution in the liquid so the float can feel extra buoyancy force. $\endgroup$ – Maxim Umansky Jan 21 '18 at 22:24
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Introduction

That is an interesting subject. Quotation (my emphasis):

sedimentation holds novel surprises [...] showing that a simple external field like gravity may induce mind-boggling, and theoretically challenging effects

from the paper The unbearable heaviness of colloids: facts, surprises, and puzzles in sedimentation.

I strongly suggest taking a look also at this paper, where they experimentally observe "unexpected effects, such as denser particles floating on top of a lighter fluid" in colloidal mixtures and provide a theoretical approach as well.

Notice, in particular, that "particles have settled" doesn't necessarily mean "they rest at the bottom", but rather that there's a vertical gradient of particle density in the fluid.


First question

Yes, using the average density of the (local) mixture is a good approximation and $F$ will be greater; but only if in the equilibrium state particles are still suspended at the float's height. The approximation is then good: a first correction to it would be to consider the effective volume of the float due to the finite size of the suspended particles, which is of course negligible for a big float.

If, on the other hand (and that actually seems to be the configuration you have in mind), by "macroscopic" you mean big enough to truly completely set at the bottom, then the float could not feel it: for only the total depth will have changed and the buoyancy would be unaffected (as it's determined by the pressure difference over the float).

Experiment

The float will therefore present the same or, if the dust is fine enough, slightly increased buoyancy. But the net result depends on what

once the cloud of particles fully covers the float

exactly means. Particularly, how strongly do these particles adhere to the surface of the floater?

  • If not at all, then the tension in the cord will accordingly stay the same or increase a bit.

  • If they do stick a lot to the float, then the combined (float+particles) density overcomes any extra buoyancy and the tension decreases.

Edit: Now the OP makes clear that the particles do not stick to the float, so the answer to the first question is enough to predict the outcome of the experiment:

  • If the particles completely set at the bottom, leaving only plain water in the bulk of the fluid, then the tension stays the same;

  • If a colloid is formed, then the tension will increase by an additive factor proportional to the particles density and concentration.

A conceptual trap

It might be tempting to think that "when the float is inside of the dust cloud then it is the same as if it were in a liquid with density higher than plain water", but that doesn't happen.

This becomes clear once you remember that the microscopic mechanism behind pressure is simply moment transfer from innumerable collisions with the fluid constituents: and for particles not in equilibrium with the fluid, for particles with are mostly moving downwards, these collisions can't transfer a net upwards moment (the origin of buoyancy), on the contrary, actually, as pointed out in Car Lei's answer.

You can also consider the effect of holding a big piece of lead next to the float: even if the the average density around the float is now how higher, there's obviously no increase in buoyancy. Denser particles mostly falling next to it have exactly the same effect (none, and they push it downwards when falling on top of it).

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  • $\begingroup$ For clarification, I have added a sentence in the statement of the problem "It is assumed that the dust particles don't stick to the surface of the float". What do you mean "If not at all, then the tension in the cord will accordingly stay the same or increase a bit." So will it stay the same or not? And what is "a bit"? Can this be quantified, if all parameters are given? $\endgroup$ – Maxim Umansky Aug 3 '17 at 0:50
  • $\begingroup$ Yes, it can be quantified. I updated my answer, it's hopefully clearer now. $\endgroup$ – stafusa Aug 3 '17 at 1:15
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    $\begingroup$ Yes, when the float is inside of the dust cloud then it is the same as if it were in a liquid with density higher than plain water (since the dust particles are denser than water), so the net effect is increasing the buoyancy force. $\endgroup$ – Maxim Umansky Aug 3 '17 at 1:23
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    $\begingroup$ Dust particles drifting downward impose additional vertical pressure gradient in the water. Just think of force balance for the dust: there is downward force from gravity but acceleration of dust is zero - so there is an opposite force. This force is due to additional vertical pressure gradient in the water, similar to what keeps the water itself from accelerating downwards. This additional pressure $\delta P(z)$ integrated over the float's surface leads to additional vertical buoyancy force. $\endgroup$ – Maxim Umansky Aug 3 '17 at 15:20
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    $\begingroup$ We can neglect the higher water level, if the amount of dust is small in volume. Yes, viscosity is what keeps dust particles from accelerating - but how is this force transmitted to the bottom of the container? It is through additional water pressure! $\endgroup$ – Maxim Umansky Aug 3 '17 at 15:47
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I would think the tension force will increase, as the average liquid density has increased. Note, for example, that the water pressure on the bottom will increase more than it would be if the same volume of water is added. The particles move downwards with a constant speed, so their full weight acts on the water (until they are on the bottom).

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  • $\begingroup$ But the particles will bombard the upper side of the float, so there is a downward force from that, correct? Maybe that is the dominant effect and the tension force will decrease? $\endgroup$ – Maxim Umansky Jul 12 '13 at 5:06
  • $\begingroup$ I don't think there is any 'bombarding' going on. For that to happen you need inertia which, for these slowly settling particles, will be completely absent due to dominant viscous effects. $\endgroup$ – Michiel Jul 12 '13 at 6:03
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My guess is a decrease in the tension.

This is a case where viscosity must not be ignored; the greater-than-buoyancy gravity says the particles shall accelerate downwards, but they ended up at a constant speed. So viscosity, especially in regions around the particles, must not be ignored.

With that in mind let's assume your float has the shape of a slope (the triangle slopes you would find from many textbooks); it still allows particles to fall down, and makes no difference for buoyancy. When particles come into contact, they slide down this slope; the picture is analogous to an object sliding down a slope with water as the lubricant. We know in this case the solid on the other side, which is the slope, is still exerted a force downwards.

What is different from "an object sliding down a slope", is that there is water on the other side than the slope. Drag from that part of water is in the reverse direction of the motion, i.e. upwards in parallel to the slope; it contributes part of the upwards force pushing the particle, and pushing it sideways at the same time. So the slope also gets pushed sideways in the reverse.

EDIT, to elaborate on why the average density explanation will NOT work:

The buoyancy theory from Archimedes relies on the system being static. Only when it is fully settled can you get a well-defined "water pressure", as a function of depth. Inside that cloud of dust there is no well-defined static water pressure; again, because dynamic effects have already affected the outcome in an apparently measurable way, you cannot just say the drag is small. It may be small in numbers as we feel, but not small when compared to the gravity of the particles.

Another important clue is, as you described in your question, that the particles are considered "macroscopic". This means their thermal motion is not enough to overcome the downward pull of gravity, so that they could end up drifting downwards, instead of randomly in all directions. These particles, consequently, cannot "bounce" into the float bottom-up; so no way they could provide any sort of upwards force to the float.

When the particles are indeed microscopic, and forms a uniform solution, they could apparently contribute to buoyancy, as we know they do in something like salt water. You will be able to find salt (as sodium ions and chloride ions) at the bottom surface of the float, bouncing into the float bottom-up, as their thermal motions predict.

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  • $\begingroup$ For a liquid with suspended in it tiny solid particles and a body floating in it, as pointed out in the answer posted by @akhmeteli, it is the effective total density of liquid with suspended particles what matters for the buoyancy force. The downward ram pressure caused by particles bombarding the upper surface of the float exists but it scales as $V^2$ where V is the particles' downward drift velocity. For very small particles V is small so this effect is negligible. $\endgroup$ – Maxim Umansky Aug 2 '17 at 20:30
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The density of the water will increase within the cloud of turbid water. The pressure of the water will increase below the cloud, and should be evenly pressurized in this scenario, give or take the cloud not settling in a uniform disc (which it totally wouldn't). The cloud would slowly diffuse to a greater vertical separation dependent on the specific gravity and surface area of the particles. as the cloud started downward i would anticipate that the tension in the cord would increase, with slight fluctuation caused by harmonics, elasticity, and turbulence. Presuming that the particles are magically repelled by the surface of the float, as the first particles moved past the float, the tension/buoyancy will slowly decrease until all the particles are past, at which point it will be the initial buoyancy less any particles that stay in colloid. The particles will impact the buoy, though the impact will be cushioned by pressure effects and viscous effects dependant on size; impacts will register if they are BB's and will be to small and frequent to measure with the best equipment if they are rockflour particles (which is what i am picturing). Much more importantly, they will accumulate on the buoy's surface, changing everything with their added dead weight.

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