special (ie difficult) buoyancy problems

[thebillsfan]

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Shrike posted this many years ago on SDN:
If the density of the fluid above the surface cannot be neglected (i.e., it's a liquid or a very dense gas), then the buoyant force is the sum of the weights of the respective fluids displaced by the portions of the object above and below the surface. In practice, subtract the density of the lighter fluid from the density of the the heavier fluid, use the principle above for solving problems neglecting the upper fluid to get a density of the object, and then add the density of the lighter fluid back in to get the virtual density of the object. (If you're finding how much is below the surface of the heavier fluid, subtract the density of the lighter one from both the heavier fluid and the object, and find the ratio as above.)

This doesn't make sense to me. First off, why would you "subtract the density of the lighter fluid from the density of the the heavier fluid, use the principle above for solving problems neglecting the upper fluid to get a density of the object, and then add the density of the lighter fluid back in to get the virtual density of the object." What's the logic behind that?

 

[thebillsfan]

bumpin to see if anyone can answer this

 

[boaz]

I'm not sure how Shrike got that idea.

I see it this way: Say we have large a box containing equal volumes of two fluids, "A" and "B". A is less dense than B, so A almost exclusively occupies the upper half of the box. Now let's imagine we drop an object, "C", into the box. Assume C is more dense than A but less dense than B. Now imagine for a moment that B is really a solid. If we drop C into the box, C will sink all the way to the "bottom", i.e. the surface of B. Now, B is in fact a fluid, and since it is more dense than C, C will float on B. This case is analogous to the regular case where there is only air (1 atm) above the fluid, with just one important exception: C appears to be more dense than it actually is because of the column of fluid, A, above it. Thus, more of its volume will be submerged in this special case than in a normal case. Since buoyant force is the weight of the volume of fluid displaced, the "true" value will be less than that measured.

Mathematically, we should subtract the weight of the column of B directly above C from the buoyant force. The equation should look like this:

(true buoyant force) = [(density of B)(volume of B displaced)(g)] - [(density of A)(volume of A in column directly above C)(g)]

Again, I don't see how Shrike got the idea that one needs to take into account the volume of A displaced by C. What matters is the volume of the column of A directly above C.

Please let me know what you think!

In case it helps, this is what I wrote on another post re buoyant force:

I also found this confusing for a long time. But I was able to understand it only after stripping down the situation to the bare essentials.

Here's how I see it. Say we have object A and object B. In addition, we have something called Earth, with a huge mass M and which exerts gravitational force on objects A and B.

The problem is that no two objects can occupy the same space at the same time. Since in nature there are no compromises, one object will be above and the other below. Which one? Well, since force depends solely on mass (remember gravity is a constant), we can assume that Earth will prefer the object with the higher mass.

Now, if objects A and B were both solid objects, we would simply say that the object with the greater mass is preferred, no matter what the respective volumes are. However, in our case we have one solid object and one liquid. The solid object is very straightforward: it has a specific mass and so it's trivial to determine the force exerted upon it by Earth. What about object B, the liquid? The key to answering this question is to realize that the liquid's volume is exactly equal to the object's volume. So what determines which is more attracted to the Earth is density, since for a fixed volume, mass is proportional to density. Thus. if the Object A is more dense, it will "sink".

If B is more dense than A, A will "float" because the maximum amount of force it could possibly experience is limited by its finite volume. If such a volume of liquid has more mass than A, A will float.

However, it is not so clear cut. Since water does not have a finite volume, why can't Object A just displace "a little bit" water instead of the whole volume? If we can change the volume such that the masses are equal, Earth won't care which one is bigger or more dense. It turns out that it does just so. Object A in fact displaces exactly enough water such that the mass of water displaced equals the total mass of the object. This is what we call "Buoyant Force". It's a misleading term because it would have you think that a column of water below the object is supporting it; nonsense.

It turns out that the fraction of volume of the object not submerged is equal to the ratio of the density of the object to that of the water.

It boils down to this formula:

D1V1=D2V2

where D1=density of object
V1=total volume of object
D2=density of fluid
V2=volume of fluid displaced

 

[thebillsfan]

(true buoyant force) = [(density of B)(volume of B displaced)(g)] - [(density of A)(volume of A in column directly above C)(g)]

Could you explain a little more where you got the density above the column part? Why wouldn't it be PLUS this column? The buoyant force is supporting the weight of the object AND the weight of the fluid above it, right?

 

[boaz]

It depends on "which" buoyant force we are looking for, which can be either:
1) The "actual" force that is present in this special system; OR
2) The "true" force that would be present in a normal system.

"True" buoyant force is defined as: (density of C)(volume of C that would be submerged in normal case)(g)

 

[boaz]

(true buoyant force) = [(density of B)(volume of B displaced)(g)] - [(density of A)(volume of A in column directly above C)(g)]

Could you explain a little more where you got the density above the column part? Why wouldn't it be PLUS this column? The buoyant force is supporting the weight of the object AND the weight of the fluid above it, right?

I can't illustrate it on SDN but it's kind of analogous to the pictures in the textbooks of the hydrostatic pressure. The column above C has weight, so in effect we can look at the object and the column of fluid above it as one object, weighing down on B.

"Buoyant force" is really an abstract term (see my previous comment above). It does not "support" anything.