Is there a buoyancy force or not?

[agtee]

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.

 

[Fightfish]

The "buoyancy force" that you are referring to would be upthrust. Archimedes' Principle states that the upthrust acting on an object submerged at a depth in a fluid is equal to the weight of fluid displaced by the object.

Thus, the upthrust acting on the cylinder would be the weight of fluid displaced by the cylinder (volume of cylinder submerged X density of fluid)

 

[agtee]

but in this case if u look how the cylinder is placed there is no water contact at the bottom of the cylinder as it is attached to the bottom of the tank, so no force(as in terms of pressure) form the bottom side ... so the only pressure from water is only on the circumferential face of the cylinder ... that finally nullifies and gives no upward force .. so how actually buoyancy is acting? Please justify this...

 

[stewartcs]

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.

None.

There is no pressure acting on the bottom of the cylinder so there is no upward force. The only pressure acting on the cylinder would be horizontal to the cylinder which results in no upward force.

CS

 

[agtee]

so what are the limitations for the buoyancy force theory?

 

[Mapes]

None.

There is no pressure acting on the bottom of the cylinder so there is no upward force. The only pressure acting on the cylinder would be horizontal to the cylinder which results in no upward force.

CS

Really??! This is easily falsified experimentally. Just set a lightweight cup in shallow liquid. It will push upward.

When the cylinder is attached to the bottom surface, there is an energy benefit to its upward movement (assuming the average density of the cylinder is less than the liquid density). This is equivalent to an upward force.

You might think of the liquid as a microscopic wedge pushing the cylinder off the bottom surface. If the cylinder isn't perfectly attached to the bottom (by suction, for example), there's no way you can keep this "wedge" from getting underneath.

 

[agtee]

isnt thr ne macroscopic way of explaining this same phenomena as in the terms of forces acting on the cylinder?

 

[russ_watters]

Really??! This is easily falsified experimentally. Just set a lightweight cup in shallow liquid. It will push upward.

Not if it is sealed to the bottom of the container, it won't.

You might think of the liquid as a microscopic wedge pushing the cylinder off the bottom surface. If the cylinder isn't perfectly attached to the bottom (by suction, for example), there's no way you can keep this "wedge" from getting underneath.

That's exactly the situation the OP was after: If you do "perfectly attach" the cylinder to the bottom, then the "wedge" can't get under it.

 

[agtee]

so in that case thrs no upward force?

 

[DaveC426913]

If you do "perfectly attach" the cylinder to the bottom, then the "wedge" can't get under it.

If it is attached to the bottom, then isn't any upward force moot? How would you measure it? Unless, I guess, it is attached so loosely that any amount of bouyancy will overcome the "glue".

 

[agtee]

So now the different case but similar to this one : if there is a hole such that cylinder can just fit into that, so the situation is like lower and the upper part of the cylinder is in the air while middle part is surrounded by water ... now u can measure whether buoyancy is acting or not... what do u think in this case?

 

[vanesch]

The point is that when you PLACE the cylinder in the hole or on the bottom, in order to "squeeze out" the water, you have to apply a certain force onto the cylinder, which will result in a slight elastic deformation of the bottom material, deformation which will correspond to a tension in the material corresponding to about the water pressure at that depth. So the fluid pressure is now replaced by a material tension. If you release the cylinder, this tension in the bottom material will exert an upward force (elasticity of the material) which will be equal or even somewhat larger to the upward force one would experience due to the pressure of the liquid if there weren't that bottom.

However, there will be a slowing-down of the upward movement because at a certain point an upward motion of the cylinder will mean that there needs to be fluid in place, and the surface of influx is the rim of the bottom plate, which is theoretically of 0 thickness. So it might take very long before a small displacement occurs that allows the water to flow underneath the cylinder. Once there is a small film of water, though, the upward motion will not stop.

 

[Studiot]

Edit This is wrong see my post#29

-------------------------------------

Let me start by confirming that there is indeed a bouyancy force given by Archimedes or alternatively as described below.

There are even real world applications of the case described by the OP.
Consider a dam impounding a lake with a cylindrical vertical intake tower standing 'fixed' to the bottom and extending above the surface of the lake.
Both the dam and the tower experience bouyancy forces. Engineers take these into account when designing such structures.

I can even recall an incident where a Contractor's engineer could not come to terms with this and his river diversion pipes ended floating during a flood at great cost to the project.

The alternative to Archimedes is to realize that the surface of water in contact with the dam and tower is not a free surface. You can equate the surface energy of this surface to the bouyancy to derive the 'bouyancy force' in much the same way as you can to derive the 'surface tension force'

 

[stewartcs]

Really??! This is easily falsified experimentally. Just set a lightweight cup in shallow liquid. It will push upward.

When the cylinder is attached to the bottom surface, there is an energy benefit to its upward movement (assuming the average density of the cylinder is less than the liquid density). This is equivalent to an upward force.

You might think of the liquid as a microscopic wedge pushing the cylinder off the bottom surface. If the cylinder isn't perfectly attached to the bottom (by suction, for example), there's no way you can keep this "wedge" from getting underneath.

You're assuming that pressure is on the bottom of the object. I clearly stated that if no pressure was acting on the bottom then there would be no upward force.

If their is no pressure acting on the bottom of the object then no upward force can be generated.

This is easily verified by taking a straw and placing it through a Styrofoam cup and then filling it with water. The straw will not float upward. BTW there are tons of papers on this topic dealing with vertical pipes (risers) in the ocean.

CS

 

[stewartcs]

so in that case thrs no upward force?

No, there is no upward force if there is no pressure on the bottom of the cylinder.

CS

 

[stewartcs]

So now the different case but similar to this one : if there is a hole such that cylinder can just fit into that, so the situation is like lower and the upper part of the cylinder is in the air while middle part is surrounded by water ... now u can measure whether buoyancy is acting or not... what do u think in this case?

Again, if pressure is not acting on the bottom of the object, no buoyant force will be created (presuming this is a perfectly straight cylinder of course with no other pressure ledges).

CS

 

[stewartcs]

Let me start by confirming that there is indeed a bouyancy force given by Archimedes or alternatively as described below.

There are even real world applications of the case described by the OP.
Consider a dam impounding a lake with a cylindrical vertical intake tower standing 'fixed' to the bottom and extending above the surface of the lake.
Both the dam and the tower experience bouyancy forces. Engineers take these into account when designing such structures.

I can even recall an incident where a Contractor's engineer could not come to terms with this and his river diversion pipes ended floating during a flood at great cost to the project.

The alternative to Archimedes is to realize that the surface of water in contact with the dam and tower is not a free surface. You can equate the surface energy of this surface to the bouyancy to derive the 'bouyancy force' in much the same way as you can to derive the 'surface tension force'

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

 

[stewartcs]

If it is attached to the bottom, then isn't any upward force moot? How would you measure it? Unless, I guess, it is attached so loosely that any amount of bouyancy will overcome the "glue".

Actually, no, it's not moot. I deal with very problem almost daily when dealing with the stability of vertical pipes in the ocean. This situation is quite often discussed and even more often misunderstood. People intuitively think that any object in water will weight less. Physics 101 courses don't help by given only the partial picture of what happens (i.e. Archimedes Principle).

Take a long, vertical, and straight steel pipe and cement it in the ocean floor. How much tension would you have to apply to support just the weight of pipe? Intuition would tell you, based on Archimedes Principle, just the wet weight of the steel pipe. However, this is not the case since no pressure is acting on the bottom of the pipe (it is sealed). The tension required would be equal to the air weight of the steel pipe.

CS

 

[Studiot]

Edit This is wrong see my post#29
--------------------------------------
Take a sealed cylinder with flanged flat ends, longer than the depth of water.

Float it on the surface and tow it out horizontally.

Force (yes you will have to ) it to the bottom where you have a preprepared flat steel plate.

With suitable handling equipemnt turn it to the vertical.

Bolt the end flange to to the plate and tighten down.

It will be a cylinder, fixed to the bottom, sticking out of the water as prescibed with no water under it and therefore no water pressure under it.

Now release the bolts.

What happens?

If the plate were the measuring table of a weighbridge, would it measure the same weight as a similar weighbridge on land?

 

[Mapes]

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?

EDIT: This represents just one possible configuration; please see my post #23 below.

 

[elect_eng]

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.

Strickly speaking, you have not fully specified the questions properly. My guess is that you are implying that the cylinder is vertically oriented. In idealized cases, a perfectly vertical cylinder will not have buoyancy force if water can not get under the bottom. Ideal cases can be the example of putting a straw through a cup, or a perfectly rigid container with a perfect seal between the container and cylinder.

However, let's consider non-ideal cases.

If the cylinder is not perfectly vertical then there will be some buoancy force acting on the sides. The force will be proportional to sine of the angle from vertical.

In the vertical case, the surrounding water places a force on the bottom of the ocean/container which is different from what the cylinder is placing on the bottom. Depending on the material of the floor, there can be different responses to stress forces acting on the floor. The cylinder will transmit the atmospheric pressure, plus it's weight per unit area down onto the bottom/floor, while the surrounding water will transmit the atmospheric pressure plus the water depth pressure down on to the bottom. Usually, the pressure on the floor is greater where the cylinder is because the cylinder's density is greater than that of water. There will be stresses and possible consequences of this. If the floor is solid, then there really is no buoyancy force, but if it has some fluid properties, you can argue that the water pressure is partially transmitted through the floor and back onto the bottom of the buoy. In this case, there is some buoyancy force, in a sense. In the case of an ocean, one would need to ask about the material properties of the bottom. Sand saturated with water might behave differently than a solid rock-bed.

It's also interesting to consider the case where the cylinder top is submerged. In this case there will be a downward force exerted on the top surface.

 

[DaveC426913]

Actually, no, it's not moot.CS

Got it. There should be a way of arranging a strain gauge on the bottom of the cylinder so that, if it had a bouyant tendency, it could be measured. All without letting water under the cylinder.

I guess what you've got is a vacuum chamber underneath with a strain gauge inside.

 

[Mapes]

OK, I think I might be able to resolve these opposing views, again considering the energy of the system as a whole. Recall that a force corresponds to a reduction in system energy for a given infinitesimal displacement:

\bold{F_i}=-\left(\frac{dE}{dx_i}\right)\bold{i}

If there is no change in energy, there is no driving force. In my diagram above, there is a lateral contraction of the support as it elongates, so the potential energy of the liquid is decreased as its upper level drops. A buoyancy force is measured.

Studiot's real-life example probably corresponds to my diagram, where a non-rigid attachment connects a hollow object to the basin bottom. In this case, it would be necessary to account for a buoyancy force.

If the support were absent, the container bottom rigid, and the cylinder rigidly attached with no liquid intrusion underneath, it makes no sense to talk about a buoyant force, since it could not be measured even in principle.

If liquid intrusion under the cylinder is allowed, the cylinder will be pushed up due to the lowering of liquid potential energy as its upper level drops.

When a straw is pushed through a cup bottom, there is no energy benefit associated with moving the straw up or down. Thus, there is no buoyancy force. This probably corresponds to the real-life design experiences contributed by stewartcs.

Does this seem reasonable to everybody?

 

[DaveC426913]

If the support were absent, the container bottom rigid, and the cylinder rigidly attached with no liquid intrusion underneath, it makes no sense to talk about a buoyant force, since it could not be measured even in principle.

Consider my modification to add a chamber underneath that liquid cannot penetrate but has room for a strain guage.

Now that I think about it, the chamber should be filled - not with vacuum, but with air at 1 atmo - removing any non-water forces from the equation.


[ EDIT ] No, this cannot work. The pressure differential across the water-air or water-vacuum seal will result in friction forces that cannot be ignored.

If liquid intrusion under the cylinder is allowed, the cylinder will be pushed up due to the lowering of liquid potential energy as its upper level drops.

Perhaps, but the initial force will start at zero, since initially there is no water underneath. If done carefully enough, the duration of this zero force condition could be extended arbitrarily.

 

[DaveC426913]

In idealized cases, a perfectly vertical cylinder will not have buoyancy force if water can not get under the bottom.=

This is the crux of the question being asked. Your response is, no there is no vertical bouyant force.

 

[stewartcs]

This isn't really that complicated. Things don't just arbitrarily have forces exerted on them without being generated from something. That something is the hydrostatic pressure of the fluid acting on the surface of the object thus creating a force. If you resolve all of the forces acting on a submerged object you'll find the resultant force. If the resultant force is upward, it is called a buoyant force. Nothing more to it than that.

Plainly a square box floating in the water fully submerged has a buoyant force (net resultant force) that is equal to the box's weight. If part of the pressure field were removed, say from the bottom side, the net resultant force would change and the box would no longer float. Hence, if there is no pressure acting on the bottom of the box it will not experience a buoyant force and Archimedes Principle will not apply even though it is in the water.

CS

 

[stewartcs]

If there are still non-believers take a look at this old post from Doc Al...It essentially describes this very scenario we are discussing.

https://www.physicsforums.com/showthread.php?t=11531

CS

 

[elect_eng]

This is the crux of the question being asked. Your response is, no there is no vertical bouyant force.

I do believe that is the crux of the question asked, and yes that is my response to it, if that is the case. Still the original question is not explicit about the geometry (whether vertical), nor about the allowed assumptions, whether ideal, or realistic. So, it's important to be as clear as possible when answering.

Perhaps some of the confusion on the issue stems from the differences in non-idealized cases. For example Studiot was bringing in ideas that he is aware of in real applicatons such as dams and other stuctures. Also, the example of the straw is highly idealized because the straw pokes through the cup back into the atmospheric pressure, which is very different than an ocean application with the massive water pressure on the sea-floor.

 

[Studiot]

I have been thinking further about the question an realize I was to hasty.

StewartCS you are right and I was wrong.

There will be a loss of bouyancy over any area from which fluid is excluded.

I apologise to anyone I may have misled.

 

[russ_watters]

Got it. There should be a way of arranging a strain gauge on the bottom of the cylinder so that, if it had a bouyant tendency, it could be measured. All without letting water under the cylinder.

I guess what you've got is a vacuum chamber underneath with a strain gauge inside.

Yes. it all depends on the particulars of the scenario, but forget strain gauges: you could run into a situation where you bolt a structure to the bottom of a tank and if you don't take this issue into account you won't know of the bolts are in tension or compression!

 

[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