Is there a buoyancy force or not?

[russ_watters]

...also, have you guys ever had a pot or dinner plate attach itself to the drain in your sink?

 

[sophiecentaur]

The model of having the rod passing through a frictionless, sealed hole in the bottom of the tank seems to be the easiest one which will eliminate any effect from or on the floor.

 

[Mapes]

To those saying that the buoyant force completely disappears if liquid is excluded from the bottom of the cylinder, please consider again my diagram in post #20, where no liquid contacts the cylinder bottom, but I predict a measurable upward displacement and associated force. Is there an error in this reasoning?

 

[Studiot]

Hello Mapes.

In your post#20

In situation 1
If you welded the box to the floor of the dry container, and then flooded it the strain-force gauge would register the additional weight of the water column above the top of the box. Alternatively you could say it registered the pressure at the top/water interface.

If you were to magically replace the water with the second box in a watertight manner, as in the second diagram the strain gauge would indicate the full weight of the second box.

There would only be bouyancy if the gauge indicated the full weight of the second box minus the weight of the water displaced.

Since we can model water as a frictionless fluid for this purpose there is no other available force to support the full weight of the second box.

This situation actually occurs when a ship grounds on a sandbank. The ship experiences full bouyancy until part of the hull is stuck into the sand. At which point an observable loss of bouyancy occurs. This loss is not total.

 

[DaveC426913]

The model of having the rod passing through a frictionless, sealed hole in the bottom of the tank seems to be the easiest one which will eliminate any effect from or on the floor.

Again:

The pressure differential across the water-air seal will result in friction forces that cannot be ignored.

 

[russ_watters]

To those saying that the buoyant force completely disappears if liquid is excluded from the bottom of the cylinder, please consider again my diagram in post #20, where no liquid contacts the cylinder bottom, but I predict a measurable upward displacement and associated force. Is there an error in this reasoning?

[separate post]I made a diagram with similar reasoning to Studiot. Consider an object with a strain gauge attached, rigidly attached to the container bottom (welded or epoxied, say, so that no liquid can get underneath). I've marked the downward water pressure that acts on the top face of the object; of course, water pressure is exerted on the other faces also.

Now attach an empty cylinder to the object (rigidly by welding or epoxy, so that no water can get underneath). The downward water pressure on the support has now been removed. The support and strain gauge will therefore elongate due to this change in stress.

Does this not represent and measure the buoyant force of the empty cylinder?

No. As you said, "the downward water pressure on the support has removed". That's it. Downward pressure has been removed does not equal upward force, it equals zero force.

 

[russ_watters]

Archimedes principle doesn't apply. If there is no pressure acting on the bottom of this perfectly vertical cylinder then there will be absolutely no buoyant force.

In all of the examples given, pressure has found some way to get under the cylinder and thus provide and upward force. If it does not, there will be no upward force.

CS

My guess, either they didn't figure that the ground underneath the tower was for all intents and purposes liquid. Either that or there was a large pipe underneath the tower that again was buried in the ground under the reservoir and it floated. This would be similar to what happens to a pool if it is left empty or a septic tank if it doesn't have a mound on top of it: it floats.

 

[Loren Booda]

Hi all,
If i attach a cylinder at the bottom of the water tank such that some of its part is out out of water. So in this case what would be the buoyancy force acting on the cylinder?

-agtee.

What would the buoyant force be on such a cylinder if perfectly smooth and undeformable?

Does gravity act in parallel upon the entire curved surface of the cylinder - normal to the end(s) - or do tidal forces affect it?

Can a "pillbox," accustomed to Gaussian surfaces, model this cylinder's forces?

What would the buoyant force be [+,-,0] on such a cylinder if it were immersed in liquid helium rather than water?

 

[Mapes]

No. As you said, "the downward water pressure on the support has removed". That's it. Downward pressure has been removed does not equal upward force, it equals zero force.

Yes, and zero force represents is a decrease in compressive load that will result in the vertical extent of the support increasing and the strain gauge getting longer. This is linear elasticity. You can't reduce the normal load in one axis and not expect an elongation.

EDIT: Perhaps we're getting away the main issue here. My only point is that it's not sufficient to say "The water has nowhere to push up, so no buoyant force can exist." There are configurations (like my diagram in #20) where the water can push sideways (laterally) and cause the vertical extent of a solid support to increase through the Poisson effect. This is equivalent to a buoyant force, since it doesn't occur until the empty cylinder has been attached.

Again, I think the best approach is not to look for forces or locations of force application (since these could be endless), but to consider the total energy of the system with respect to small displacements in the cylinder (if achievable). A decrease in total system energy due to a small upward cylinder displacement is equivalent to a buoyant force.

EDIT 2: Forget about the Poisson effect stuff; it was my poor approach at figuring out this problem. It's much better to consider a free-body diagram of a compliant support under the empty cylinder. If the support elongates when the cylinder is attached (no matter how securely), then the cylinder is exerting a buoyant force. This is shown in post #42 and more precisely in post #44.

 

[russ_watters]

Mapes, boyancy is an actual upwards force, not a lack of downward force. The reading on your second strain gauge is the same whether the water is there or not. It is not a result of the horizontal pressure on the cylinder.

 

[Studiot]

There are configurations (like my diagram in #20) where the water can push sideways (laterally) and cause the vertical extent of a solid support to increase through the Poisson effect. This is equivalent to a buoyant force, since it doesn't occur until the empty cylinder has been attached.

If this were indeed a contributory factor then any bouyancy force would depend upon the structural configuration of the support, rather than the Archimedean principle of the volume of displaced water. You can't have it both ways.

 

[Mapes]

Mapes, boyancy is an actual upwards force, not a lack of downward force.

An actual upwards force by the cylinder exactly offsets the force on the top of support caused to water pressure.

Let me try this tack: if you don't believe that the strain gauge would elongate, then is your problem with my argument that (1) the presence of the empty cylinder changes the stress state on the support from

\left[\begin{array}{ccc}<br /> -p&amp;0&amp;0\\<br /> 0&amp;-p&amp;0\\<br /> 0&amp;0&amp;-p<br /> \end{array}\right]

(assuming the pressure is much larger than the distributed weight of the support or the cylinder) to

\left[\begin{array}{ccc}<br /> -p&amp;0&amp;0\\<br /> 0&amp;-p&amp;0\\<br /> 0&amp;0&amp;0<br /> \end{array}\right]

where the z-axis is vertical? If this is the problem, how would you draw the free-body diagram of the support? Or is your problem with my argument that a change from the first stress state to the second will cause elongation in the z-direction?

If this were indeed a contributory factor then any bouyancy force would depend upon the structural configuration of the support, rather than the Archimedean principle of the volume of displaced water. You can't have it both ways.

I said the elongation of the strain gauge measures the buoyancy force. Let me be clear: The buoyancy force is independent of the support properties. But the strain in the support is not; of course it depends on the support's geometry and material properties. This is true for any strain gauge measurement.

 

[Studiot]

Let me paraphrase your argument, Mapes.
I understand and agree with it, with one proviso: It is not a source of bouyancy, indeed it will act in addition to any bouyancy force.

What I think you are saying is that

The left hand block is subject to triaxial compression. The compressive forces (stresses) on the two horizontal axes are supplied by the pressure of the water on the block's faces. The horizontal stresses will be equal. the vertical stress is supplied by the reaction on base and the weight plus the liquid pressure force on the top surface of the block. This is clearly going to be greater than the pressure force alone, acting on the other two axes so there will be a poisson lengthening in the horizontal direction.

If the water above the block is now replaced by the second box as in the right hand diagram the vertical axis loads and therefore stresses, though still compressive, will be different so there will be a change to the poisson horizontal lengthening of the support block.

All this is agreed. But the reaction force between the base and the support block must equal the weight of the support plus whatever it is supporting. So all your gauge is measuring is the weight of the second box as I said before.

 

[Mapes]

I understand and agree with it, with one proviso: It is not a source of bouyancy, indeed it will act in addition to any bouyancy force.

It is an upward force... It is caused by water displacement... The force is essentially equal to the weight of the water that would have filled the cylinder (assuming a lightweight cylinder)... Does any reference define this phenomenon as anything other than buoyancy? If so, I'd like to look at it; I'm not excluding the possibility that I'm wrong.

But the reaction force between the base and the support block must equal the weight of the support plus whatever it is supporting. So all your gauge is measuring is the weight of the second box as I said before.

For simplicity, I was ignoring the weight of the empty cylinder compared to the weight of the water that it displaced. But yes, to be super precise, the stress state in the support is

\left[\begin{array}{ccc}<br /> -\rho g h(1-\frac{z}{h}) &amp; 0 &amp; 0\\<br /> 0 &amp; -\rho g h(1-\frac{z}{h}) &amp; 0\\<br /> 0 &amp; 0 &amp; -\rho g (h-t_\mathrm{supp})-\frac{W_\mathrm{supp}}{A}(1-\frac{z}{t_\mathrm{supp}})<br /> \end{array}\right]

in the first case (left side of the diagram) and

\left[\begin{array}{ccc}<br /> -\rho g h(1-\frac{z}{h}) &amp; 0 &amp; 0\\<br /> 0 &amp; -\rho g h(1-\frac{z}{h}) &amp; 0\\<br /> 0 &amp; 0 &amp; -\frac{W_\mathrm{cyl}}{A}-\frac{W_\mathrm{supp}}{A}(1-\frac{z}{t_\mathrm{supp}})<br /> \end{array}\right]

in the second case (right side of the diagram), where \rho is the water density, g is gravitational acceleration, h is the depth to the container floor, W_\mathrm{supp} and t_\mathrm{supp} are the weight and height of the support, respectively, A is the cross-sectional area of the cylinder and support, z is the vertical distance from the container floor, and W_\mathrm{cyl} is the weight of the empty cylinder.

I believe it has been assumed in this entire thread that the empty cylinder weighs less than an equal volume of water.

The attachment of the empty cylinder thus corresponds to the addition of a normal tensile stress \rho g(h-t_\mathrm{supp})-W_\mathrm{cyl}/A in the vertical direction. This is buoyancy.

EDIT: Whoops, should have been \rho g(h-t_\mathrm{supp}) instead of \rho g h.

 

[stewartcs]

It is an upward force... It is caused by water displacement... The force is essentially equal to the weight of the water that would have filled the cylinder (assuming a lightweight cylinder)... Does any reference define this phenomenon as anything other than buoyancy? If so, I'd like to look at it; I'm not excluding the possibility that I'm wrong.



For simplicity, I was ignoring the weight of the empty cylinder compared to the weight of the water that it displaced. But yes, to be super precise, the stress state in the support is

\left[\begin{array}{ccc}<br /> -\rho g h(1-\frac{z}{h}) &amp; 0 &amp; 0\\<br /> 0 &amp; -\rho g h(1-\frac{z}{h}) &amp; 0\\<br /> 0 &amp; 0 &amp; -\rho g h-\frac{W_\mathrm{supp}}{A}(1-\frac{z}{t_\mathrm{supp}})<br /> \end{array}\right]

in the first case (left side of the diagram) and

\left[\begin{array}{ccc}<br /> -\rho g h(1-\frac{z}{h}) &amp; 0 &amp; 0\\<br /> 0 &amp; -\rho g h(1-\frac{z}{h}) &amp; 0\\<br /> 0 &amp; 0 &amp; -\frac{W_\mathrm{cyl}}{A}-\frac{W_\mathrm{supp}}{A}(1-\frac{z}{t_\mathrm{supp}})<br /> \end{array}\right]

in the second case (right side of the diagram), where \rho is the water density, g is gravitational acceleration, h is the depth to the container floor, W_\mathrm{supp} and t_\mathrm{supp} are the weight and height of the support, respectively, A is the cross-sectional area of the cylinder and support, z is the vertical distance from the container floor, and W_\mathrm{cyl} is the weight of the empty cylinder.

I believe it has been assumed in this entire thread that the empty cylinder weighs less than an equal volume of water.

The attachment of the empty cylinder thus corresponds to the addition of a normal tensile stress \rho gh-W_\mathrm{cyl}/A in the vertical direction. This is buoyancy.

Mapes,

Are you saying that the hydrostatic pressure is "squeezing" the support thus causing it to push the cylinder up?

CS

 

[Studiot]

It is an upward force

No I clearly showed it is a horizontal strain. No force is involved.

 

[Mapes]

Mapes,

Are you saying that the hydrostatic pressure is "squeezing" the support thus causing it to push the cylinder up?

CS

The hydrostatic load in the x- and y-directions is a constant -\rho g h(1-z/h). However, with the lessening of compressive stress \sigma_z from

-\rho g (h-t_\mathrm{supp})-(W_\mathrm{supp}/A)(1-z/t_\mathrm{supp})

to

-(W_\mathrm{cyl}/A)-(W_\mathrm{supp}/A)(1-z/t_\mathrm{supp})

(shown in post #44), a shrinking of the support dimensions in the x- and y-directions is predicted if the support material has a Poisson's ratio \nu&gt;0 (as essentially all solid materials do). In this sense, the support is squeezed. But I wouldn't say that the cylinder is being pushed up; I would say rather that its buoyancy pulls on the support.

In any case, descriptions like "squeezing" and "pulling" are less precise and more likely to cause confusion than a straightforward look at the equations.

EDIT: Whoops, should have been \rho g (h-t_\mathrm{supp}) instead of \rho g h.

 

[stewartcs]

The hydrostatic load in the x- and y-directions is a constant -\rho g h(1-z/h). However, with the lessening of compressive stress \sigma_z from

-\rho g h-(W_\mathrm{supp}/A)(1-z/t_\mathrm{supp})

to

-(W_\mathrm{cyl}/A)-(W_\mathrm{supp}/A)(1-z/t_\mathrm{supp})

(shown in post #44), a shrinking of the support dimensions in the x- and y-directions is predicted if the support material has a Poisson's ratio \nu&gt;0 (as essentially all solid materials do). In this sense, the support is squeezed. But I wouldn't say that the cylinder is being pushed up; I would say rather that its buoyancy pulls on the support.

In any case, descriptions like "squeezing" and "pulling" are less precise and more likely to cause confusion than a straightforward look at the equations.

OK let's try this: Imagine the support is perfectly rigid and cannot be compressed (idealized case). What upward force is being applied to the cylinder now? Just a normal reaction equal to the air weight of the cylinder. Thus the support is completely supporting the cylinder.

If you try to pick up the cylinder the force required would be equal to the air weight of the cylinder (same as the reaction force prior to lifting it off). At the infinitesimal moment you lift the cylinder off of the support that reaction force is no longer there and the force required to move it up (until the hydrostatic pressure started acting on it) would be the air weight of the cylinder.

Alternately, imagine the support being a perfectly rigid cylinder with a spring scale inside (sealed - frictionless - from the water). Lower the cylinder onto the spring scale and evacuate any water. How much force is registered on the spring scale? The air weight of the cylinder.

Now get rid of any type of support from the bottom of the tank. Place the same cylinder on the bottom of the tank. If no hydrostatic pressure acts on the bottom surface of the cylinder there is no additional external upward force. The only force is the normal reaction of the tank on the bottom of the cylinder.

CS

 

[Mapes]

OK let's try this: Imagine the support is perfectly rigid and cannot be compressed (idealized case). What upward force is being applied to the cylinder now? Just a normal reaction equal to the air weight of the cylinder. Thus the support is completely supporting the cylinder.

If you try to pick up the cylinder the force required would be equal to the air weight of the cylinder (same as the reaction force prior to lifting it off). At the infinitesimal moment you lift the cylinder off of the support that reaction force is no longer there and the force required to move it up (until the hydrostatic pressure started acting on it) would be the air weight of the cylinder.

I can tell you haven't looked at the equations closely, because you're talking about "picking up" the cylinder. You can't "pick up" the cylinder, because it's pulling up with a buoyant force of

\rho g A(h-t_\mathrm{supp})-W_\mathrm{cyl}\approx\rho g Ah

(for t_\mathrm{supp}\ll h and W_\mathrm{cyl}\ll \rho g h A). This is the change in the stress state that the comparisons above tell us. Regardless of strongly felt intuition, is there a problem with these calculations, acquired by free-body diagram?

Part of the problem is that you're thinking of the weight of cylinder in terms of what it would mean at the water's surface. At the surface, an empty cylinder on a support would just produce a net downward force of W_\mathrm{cyl} on the top of the support. But at the bottom of the liquid, you need to compare it with the alternative, a huge hydrostatic pressure \rho g (h-t_\mathrm{supp}) on the top of the support. Replacing it with the weight of the cylinder means you've reduced the pressure on the top of the support. This is equivalent to a buoyant force.

When you say "Thus the support is completely supporting the cylinder.", remember that while it is supporting a downward pressure of W_\mathrm{cyl}/A, before the empty cylinder was attached it supported a downward pressure of \rho g (h-t_\mathrm{supp}). Again, this a reduction in compressive stress, and the magnitude of the reduction is \rho g (h-t_\mathrm{supp})-W_\mathrm{cyl}/A. It is exactly the same as if, by another means, you applied a buoyancy force of \rho g A(h-t_\mathrm{supp})-W_\mathrm{cyl} on the surface of the support.

I'm not quite able to visualize your second thought experiment, the one with the spring scale inside the rigid cylinder. A diagram would help. But please, think about the current thought experiment before switching to a new one, so I can see which specific equation or calculation you disagree with.

 

[stewartcs]

I can tell you haven't looked at the equations closely, because you're talking about "picking up" the cylinder. You can't "pick up" the cylinder, because it's pulling up with a buoyant force of

\rho g A(h-t_\mathrm{supp})-W_\mathrm{cyl}\approx\rho g Ah

(for t_\mathrm{supp}\ll h and W_\mathrm{cyl}\ll \rho g h A). This is the change in the stress state that the comparisons above tell us. Regardless of strongly felt intuition, is there a problem with these calculations, acquired by free-body diagram?

Part of the problem is that you're thinking of the weight of cylinder in terms of what it would mean at the water's surface. At the surface, an empty cylinder on a support would just produce a net downward force of W_\mathrm{cyl} on the top of the support. But at the bottom of the liquid, you need to compare it with the alternative, a huge hydrostatic pressure \rho g (h-t_\mathrm{supp}) on the top of the support. Replacing it with the weight of the cylinder means you've reduced the pressure on the top of the support. This is equivalent to a buoyant force.

When you say "Thus the support is completely supporting the cylinder.", remember that while it is supporting a downward pressure of W_\mathrm{cyl}/A, before the empty cylinder was attached it supported a downward pressure of \rho g (h-t_\mathrm{supp}). Again, this a reduction in compressive stress, and the magnitude of the reduction is \rho g (h-t_\mathrm{supp})-W_\mathrm{cyl}/A. It is exactly the same as if, by another means, you applied a buoyancy force of \rho g A(h-t_\mathrm{supp})-W_\mathrm{cyl} on the surface of the support.

I'm not quite able to visualize your second thought experiment, the one with the spring scale inside the rigid cylinder. A diagram would help. But please, think about the current thought experiment before switching to a new one, so I can see which specific equation or calculation you disagree with.

No, I've not bothered looking closely at the equations because there is no need to.

You're making this way more complicated than it is. How do think Archimedes Principle is derived? It's based on pressure fields (specifically the pressure acting on the surfaces of the object creating external forces that act on the object in the fluid)...not states of stress in elastic members.

Whether the cylinder or supports are elastic or perfectly rigid is irrelevant with respect to buoyant force. A perfectly rigid member that has pressure acting on the bottom surface will still have a buoyant force. If the pressure isn't acting on the bottom it will not (presuming no other pressure ledges again). Same thing for an elastic member.

Remove the elastic members from your example and make them rigid. Do your equations still hold?

Are you still arguing that if there is no hydrostatic pressure acting on the bottom of the cylinder that it will experience a buoyant force?

CS

 

[Studiot]

Haven't you guys been busy while I've been having my dinner?

I owe you an apology, Mapes. I took the weight of your second box to be greater than the water it displaced. If its weight was less it could mean that the vertical stresses acting on the support box are either greater than or less than the horizontal ones.

But consider there is actually no need for your second box at all in the argument.
Once the second box is firmly fixed to the support it effectively becomes part of the support.
So you could consider any horizontal section through the support and ask

As a result of the poisson effect is there a bouyancy force tending to lift the material above the section away (upwards) from the material below?

As an aside. What if the second box overhangs the support?
There will indeed be an upward bouyancy force acting on the second box given by the horizontal area of overhang times the pressure at the overhang.

Similarly suppose your support box was T shaped. There would be a bouyancy force acting on the underside area of the T in contact with the fluid.

Your suggestion is tantamount to saying that a T shaped support would experience a bouyancy force equal to the weight of fluid in its entire volume, not just the exposed flange.

 

[stewartcs]

I'm not quite able to visualize your second thought experiment, the one with the spring scale inside the rigid cylinder. A diagram would help. But please, think about the current thought experiment before switching to a new one, so I can see which specific equation or calculation you disagree with.

Here is a diagram.

Note that the white part is evacuated.

CS

 

[Mapes]

No, I've not bothered looking closely at the equations because there is no need to.

Wow. Well, there's nothing I can do about that.

Whether the cylinder or supports are elastic or perfectly rigid is irrelevant with respect to buoyant force. A perfectly rigid member that has pressure acting on the bottom surface will still have a buoyant force. If the pressure isn't acting on the bottom it will not (presuming no other pressure ledges again).

Agree, agree, respectfully don't agree.

Remove the elastic members from your example and make them rigid. Do your equations still hold?

Yes; the equations do not include the stiffness of any components.

Are you still arguing that if there is no hydrostatic pressure acting on the bottom of the cylinder that it will experience a buoyant force?

Yes, it's what I showed in post #42 and in more detail in post #44 with stress states calculated via a free-body diagram (not shown).

 

[stewartcs]

Wow. Well, there's nothing I can do about that.



Agree, agree, respectfully don't agree.



Yes; the equations do not include the stiffness of any components.



Yes, it's what I showed in post #42 and in more detail in post #44 with stress states calculated via a free-body diagram (not shown).

Sorry didn't mean to sound so rude on that first comment!

I don't see how you can call the reaction at the bottom of the cylinder a buoyant force since it has nothing to do with the hydrostatic pressure acting directly on the other object. Maybe I'm missing your point.

How about this...Using your equations show how the reaction at the bottom (the thing you're calling a buoyant force) is equivalent to the weight of the displaced fluid.

CS

 

[stewartcs]

Yes; the equations do not include the stiffness of any components.

By rigid I mean the shape doesn't change which means there is no Poisson's Effect. Take away that effect and take another look at your position.

CS

 

[stewartcs]

EDIT: Perhaps we're getting away the main issue here. My only point is that it's not sufficient to say "The water has nowhere to push up, so no buoyant force can exist." There are configurations (like my diagram in #20) where the water can push sideways (laterally) and cause the vertical extent of a solid support to increase through the Poisson effect. This is equivalent to a buoyant force, since it doesn't occur until the empty cylinder has been attached.

The problem I have with your argument is that it puts a material dependency on the buoyant force when there is none. What happens if the material properties change? Does the buoyant force change as well?

CS

 

[Mapes]

As a result of the poisson effect is there a bouyancy force tending to lift the material above the section away (upwards) from the material below?

By rigid I mean the shape doesn't change which means there is no Poisson's Effect. Take away that effect and take another look at your position.

That Poisson effect stuff was a wrong approach I took to try to understand the problem. In the end, it has a minimal effect for a sufficiently stiff support. So forget about Poisson contraction or elongation, it's not relevant. I've added a note to the bottom of post #39 to make this clear. Sorry about the detour, I'm been trying to converge toward a solution too. But I stand behind the position that an empty cylinder is buoyant even when its bottom is not exposed to water, based on the comparison of stress states in posts #42 and #44.

Your suggestion is tantamount to saying that a T shaped support would experience a bouyancy force equal to the weight of fluid in its entire volume, not just the exposed flange.

Yes, exactly. The buoyant force depends on the volume of water displaced, not on the shape of the cylinder or its orientation or attachment method. Otherwise you could create an non-conservative force at will simply by attaching it or detaching it, or by rotating it effortlessly. This would constitute a perpetual motion device.

The problem I have with your argument is that it puts a material dependency on the buoyant force when there is none. What happens if the material properties change? Does the buoyant force change as well?

CS

Asked and answered in post #42. The buoyant force that I argue exists is independent of the support stiffness or of any material properties except for cylinder weight. No material properties of the support appear in the buoyant force prediction \rho g A(h-t_\mathrm{supp})-W_\mathrm{cyl}\approx\rho g Ah.

I want to emphasize that the only reason to consider a non-rigid support in the thought experiment is to have some way of measuring a buoyant force. If the support is initially made twice as stiff, the strain gauge will elongate half as much when the empty cylinder is attached, for example. As the support stiffness approaches infinity, the ratio of the strain gauge elongation to the support stiffness remains constant and proportional to the buoyant force.

 

[stewartcs]

Yes, exactly. The buoyant force depends on the volume of water displaced, not on the shape of the cylinder or its orientation or attachment method. Otherwise you could create an non-conservative force at will simply by attaching it or detaching it, or by rotating it effortlessly. This would constitute a perpetual motion device.

The buoyant force doesn't not simply depend on the volume displaced. Take a look at the picture in post #52. The cylinder obviously displaces water, however it will experience no buoyant force.

Asked and answered in post #42. The buoyant force that I argue exists is independent of the support stiffness or of any material properties except for cylinder weight. No material properties of the support appear in the buoyant force prediction \rho g A(h-t_\mathrm{supp})-W_\mathrm{cyl}\approx\rho g Ah.

The object's weight has nothing to do with the buoyant force.

http://hyperphysics.phy-astr.gsu.edu/Hbase/pbuoy.html#bcomp

Although that site gives an incomplete explanation since it only refers to the displaced volume and doesn't explicitly talk about the pressure field.

CS

 

[elect_eng]

To those saying that the buoyant force completely disappears if liquid is excluded from the bottom of the cylinder, please consider again my diagram in post #20, where no liquid contacts the cylinder bottom, but I predict a measurable upward displacement and associated force. Is there an error in this reasoning?

Yes, there is an error in your reasoning. You have stated that an empty cylinder is sitting on the platform. This very statement is an acknowledgment that there is no buoyant force. An empty cylinder would normally float. OK, you glue it then because you think there is buoyant force, but it's not necessary to glue it.

You have correctly noted that the force (or pressure) on the top of the platform has changed, when the cylinder is placed, but buoyant force is all about the force on the cylinder not on container. If there is a gap between the cylinder and the platform, then there is buoyant force on the cylinder, and downward pressure on the top of the platform. The water in between is pushing in both directions- upward on the cylinder and downward on the platform. When the seal is made and no water is in between, both forces are gone. The cylinder has no buoyant force and the platform top no longer feels the water pressure (it feels atmospheric pressure plus the cylinder's weight per unit water-plane area.

Did you read my post #21 that followed yours. I mentioned the stresses on the container. They are indeed important, but they should not be interpreted as buoyant force.

 

[Mapes]

OK, got it. I've been interpreting any stress changes in the system due to the presence of the cylinder as buoyant forces, but this was misguided. And I realize that several of you have worked hard to point this out to me. This has been a great learning experience in visualizing forces and energies. Studiot, stewartcs elect_eng, and russ, thanks for a very stimulating discussion!