Buoyancy in muddy water 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.

 A: 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).
A: 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).
A: 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.
A: 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. 
A: Buoyancy on a dense and rigid object needs the presence of a gravitational force field. Only then one can talk of 'weight' or 'the apparent loss of weight due to buoyancy' on immersion in a fluid. In any zero-gravity space, one cannot experience any buoyancy.
Fluid dynamics or hydrostatics considerations need ONLY the density of the fluid around the immersed object and the increased fluid pressure proportional to the product of the fluid density and depth (H) of the fluid at the point concerned.
If the fluid density surrounding the object has particles heavier than the fluid, the effective density of the fluid and the hydrostatic pressure difference will manifest as increased buoyancy on the object. Therefore, the tension in the string will be higher when the object will pass through the 'cloud' of the particles-laden fluid.
A: I have to admire @Ben51's answer.
Trying to keep it simple, what happens if you add an additional volume of water with the same density as the original water? The tension on the cord should increase because the level of the water increases, increasing the water pressure at the buoy's level.
Suppose we add something with the same volume as the added water in the first example, but it does not weigh down the buoy. Something that's denser than the water. At equilibrium it will be at the bottom and the level of the water will be higher, increasing the pressure exactly like the first case.
Before it reaches equilibrium will that make any difference? I don't right off see why it ought to. Say you took a lead weight with the same volume and dropped it in the water. It displaces the same volume of water and has the same effect. Say it hangs from a thin cord so at equlibrium it is in the water but not at the bottom. So what? It still displaces the same volume of water. Why would it matter how much mass it has?
As long as the buoy only interacts with water, and has no direct interactions with the weights we throw into the water, then it's only water pressure that matters.
But this is different from @Ben51's answer. How can it be? Two reasons. First, I assume that the falling particles don't hit the buoy and transfer momentum to it that way. Ben51 calculated that force. Second, I assume that the falling particles only affect the buoyancy by increasing the height of the water level.
Can that be right?
What if the sediment was less dense than water. Then it would float on top, and the buoyancy of the buoy would increase by less than the volume of the sediment. The water pressure goes up from the weight of the floating stuff. But if it doesn't float then it's only the volume that raises the water level.
But if we added salt then the density of the water would increase and the lift would be more than the increase in water level. Because the salt and the water would partly occupy the same volume! There we go! When you add something that actually increases the density of the water, then the water is more dense. When you add something that just displaces water, then the water density does not increase. Just the fact that there's something displacing water that's denser than water, doesn't increase the water density. You no more increase the water density by adding silt to it, than you would increase the water density by adding a lead weight.
That's what I think. Am I wrong?
A: 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.
