What Are the Complexities of Buoyancy and Common Misconceptions?

[Graeme M]

I recently posted a question about solids that led to a very productive discussion about solids, weight, gravity and pressure. I learned that many intuitive ideas I had about these concepts were wrong and I now have a better understanding as a result. However, in the course of my reading I came across the buoyant force.

Now of course at a general level I know what buoyancy is, but after reading a number of sources it became apparent that it’s much more complicated than I thought.

I now think I have a fair understanding of this force, but there are a few things I can’t get straight in my head, even after reading a lot.

Here’s my understanding, followed by my question. These points are at a very simplified level – that is, a homogeneous liquid, in this case water, with all other forces removed. I am talking about a static situation so I am disregarding flow, drags, and other complicating factors.

1. A submerged object displaces its volume in water.
2. The weight of the water displaced is equivalent to the upward buoyant force on the object.
3. The buoyant force arises from the difference in pressure between the top of the submerged object and the bottom of the submerged object.
4. The pressure in a column of water increases with depth and hence the buoyant force will increase with depth, all other factors remaining equal.
5. The apparent weight of a submerged object is reduced by the magnitude of the buoyant force. Because weight is a force, the net force applied to the object is the difference between its weight and the buoyant force.
6. If an object’s apparent weight is greater than the buoyant force, the object will sink. And vice versa – if less than the buoyant force, the object will rise.

The total apparent weight of a container of water and an object outside of the container of water is the sum of the two weights. If the object is placed in the container of water, the total weight of the container and water is increased by the weight of the object. However the apparent weight of the object, inside the container, is reduced by the buoyant force.

Assuming that the object’s mass and shape is such that the buoyant force is exactly equal to its weight and the object does not sink, the object’s apparent weight in the container will be zero. However, the total system weight of container, water and object is increased by the weight of the object.

What is happening here? The upward force exactly balances the downward force, yet there is still a nett downward force.

 

[Bystander]

... still a nett downward force.

What is that "net downward force.?" Reread your excellent essay, and think carefully about all you've said.

 

[Chestermiller]

Items 2 and 4 are inconsistent. Item 2 is correct, and item 4 is incorrect.

Regarding your question, if you added water to the container, the total weight of material in the container would increase. If instead you add an object to the container and the object is neutrally buoyant, this would be the same as adding water to the container.

Chet

 

[russ_watters]

What is happening here? The upward force exactly balances the downward force, yet there is still a nett downward force.

We seem to get questions like this a lot. I wonder if schools are doing a poor job of teaching what "net force" is?

1. All forces always come in pairs, if you look hard enough. Typically when people talk of "net force" they are talking about the overall external forces on an object. And when they say the sum is non-zero, that just means they aren't counting the force from inertia and acceleration. So that has nothing to do with the situation here...

2. The internal forces on the floating object don't really need to be considered to determine the the external forces between the container and what is supporting it. You can sum them if you want, to find how the forces are transmitted through the container, but you really don't need to.

3. So if you do want to sum the forces, the upwards force on the object matches the downwards force on the water. Right. But that isn't a "net force" of zero because the two forces are acting on different things -- one acts on the water, the other on the object. Net force is all forces on one "object" summed together. So the water and object in it have other forces on them...

 

[Graeme M]

Chestermiller, I get that in terms of weight for the whole container, adding water or a solid of same weight has the same effect.

However adding water results in no buoyant force whereas adding a solid object does. In the former case there is no other force involved, while in the latter there is but it doesn't matter whether it is nett up, down or neutral. The buoyant force affects apparent weight of the object, but not the whole system.

In other words, the buoyant force can be of any magnitude yet the total system weight is not changed. I can't get my head around that.

 

[Graeme M]

Also, are you saying that my number 4 is wrong because the pressure differential between upper and lower surfaces remains the same regardless of depth and pressure? I did read on some science site somewhere that an object can sink until a point at which the buoyant force is sufficient to offset the weight, so that surely would arise from the change in pressure?

 

[Graeme M]

russ_watters, you say
"So if you do want to sum the forces, the upwards force on the object matches the downwards force on the water. Right. But that isn't a "net force" of zero because the two forces are acting on different things -- one acts on the water, the other on the object. Net force is all forces on one "object" summed together. So the water and object in it have other forces on them..."

But the downwards force is on the object (gravity), and the upwards force is on the object (buoyancy). The upwards force isn't matching a downwards force on the water?

 

[russ_watters]

However adding water results in no buoyant force...

Incorrect. What do you think keeps a parcel of water at the surface from sinking?

Also, are you saying that my number 4 is wrong because the pressure differential between upper and lower surfaces remains the same regardless of depth and pressure?

Right.

I did read on some science site somewhere that an object can sink until a point at which the buoyant force is sufficient to offset the weight, so that surely would arise from the change in pressure?

That would be rare, but it would be because water isn't quite incompressible. So its density increases a little with depth. More common, though, is the opposite: as objects sink, they get compressed and become less buoyant. That's a problem for scuba divers: they have to continuously adjust their buoyancy to avoid sinking to the bottom or shooting to the surface.

But the downwards force is on the object (gravity), and the upwards force is on the object (buoyancy). The upwards force isn't matching a downwards force on the water?

Well sure it is: what force moves the water out of the way for the object to replace it?

 

[Graeme M]

What keeps a parcel of water from sinking?

Hmmm... I wouldn't have thought of it like that! My thinking would say this. Water is a single thing much like a solid. The solid retains its form and shape from the bonds that hold its molecules/atoms in place. I understand that water is similar but that its bonds are not so resilient - that is, water molecules constantly create and break their bonds. So in effect water has the same form as a solid but its shape is flexible. What keeps the water at the top above the water at the bottom is the attractive/repelling forces. In fact I'd have thought that in the absence of currents etc, ie an entirely still column of water, there'd be no specific buoyany force as such, just the same forces that keep a solid together.

So... Are you saying that the buoyant force is simply the same force that keeps molecules/atoms apart? That is, repelling forces at the molecular level? So buoyancy is no more than the effect of two objects with defined form interacting.

That would explain how lowering an object into a container of water reduces the apparent weight of the object while increasing the apparent weight of the container of water. In effect, it would be similar to sliding say a brick across the upper surface of two other bricks - if a a scale were under each of the two lower bricks, the gaining brick will increase in weight proportionally to the rate at which the other brick loses weight.

 

[russ_watters]

No, that has nothing to do with buoyant force. The "parcel of water" thought experiment is simply for this: instead of pouring more water into the container, put the additional water in a plastic bag and put that bag into the container. What happens and why?

 

[Graeme M]

OK... not sure I see the relevance. You have introduced a surface. A plastic bag filled with water is effectively no different to a solid object of equivalent mass. The water in the container cannot know that behind the surface lies more water. The bag filled with water will displace an equivalent volume of water in the container and the buoyant force will equal the weight of that volume of water. Providing your bag has thin walls, the total weight of the bag and water should be very similar to the buoyant force and the bag should float just under the surface. If though your bag has very thick walls, its overall weight may overcome the buoyant force and it will sink (here I am assuming something about the relative densities of plastic and water. And the shape of the bag.).

I don't see that tells me anything about the behaviour of a body of water.

 

[Sanky123]

Good example

 

[russ_watters]

OK... not sure I see the relevance. You have introduced a surface. A plastic bag filled with water is effectively no different to a solid object of equivalent mass. The water in the container cannot know that behind the surface lies more water. The bag filled with water will displace an equivalent volume of water in the container and the buoyant force will equal the weight of that volume of water. Providing your bag has thin walls, the total weight of the bag and water should be very similar to the buoyant force and the bag should float just under the surface. If though your bag has very thick walls, its overall weight may overcome the buoyant force and it will sink (here I am assuming something about the relative densities of plastic and water. And the shape of the bag.).

I don't see that tells me anything about the behaviour of a body of water.

The point of the example is that the parcel of water behaves the same whether the bag is there or not. The bag is just an arbitrary boundary around an arbitrary volume of water, which has no impact on how you conceive the effects of buoyancy.

That idea becomes more useful when the parcel of water becomes warmer or cooler than its surrounding water and its net force changes. Or better yet, air in the atmosphere. Changes in buoyancy/density drive convection, which in the atmosphere drives the weather.

 

[sophiecentaur]

If you are having a problem with the concepts involved then perhaps you could work out a specific example (say a disc of wood of density 0.8) and a 1l beaker of water. (Choose some disc and beaker dimensions.) Put the two, side by side, on the platform of some scales. Now put the disc in the water.and, bearing in mind the total measured mass much be the same, work out the forces on the disc (once it has stopped moving) and the pressures in the water, at the new level. It's all got to work out so just find what you need to add to what to get it to balance. Actually, Archimedes' Principle is a half way house in the argument.
You have either discovered an Earth-shattering fact, hitherto unknown to mankind OR you have got your reasoning wrong, somewhere. Assume the latter. :)

 

[Chestermiller]

Not that Russ needs my help, but I fully support what he is saying. The water surrounding the bag doesn't know whether there is a bag full of water there, a solid object of the same shape there, or water with no bag around it filling the same space. In all three situations, the hydrostatic pressure forces exerted by the surrounding water cause the same upward force on whatever is filling that space.

Chet

 

[Graeme M]

Haha SophieCentaur, if ever I worked out an Earth shattering fact no-one else had, it'd be the greatest miracle in history. I'm flat out figuring out how to get out the door each morning!

Now, to my mind there's a couple of things here, which may be just too hard for me to put into words. Chestermiller I disagree with what you say. Not because I am right and you are wrong, but because what you say makes no sense to me. Of course the problem is I just don't have the breadth of education, but I'm game to take a stab at this.

There is no 'parcel' of water. The buoyant force surely depends on there being a physical surface. If you just have water, there can be no surface - that is, if you examine any section of a still body of water it should look exactly the same as any other.

So the forces involved must be only those forces that inhabit any body. On the other thread, I learned that bonding forces - attractive/repelling - are what keep a solid, solid. A liquid has exactly the same internal forces at the molecular level, except (as I understand it) that the bonds are more short-lived. So in a way, a liquid is a very flexible solid.

What keeps water 'up' is the same thing that keeps the top of a solid 'up', surely? Ignoring currents and temperature and so on, a still cylindrical container of liquid must be broadly the same thing as a solid cylinder. Differences in density of the liquid just give rise to a different profile as the matter responds to gravity.

On the other thread I learned that the atmosphere has a pressure gradient as a response to gravity. So if we had two liquids mixed, one denser than the other, then the more dense will obviously sink to the bottom and the lighter one will be on top. But that is just a response to gravity is it not? It's not the buoyant force at work.

The imaginary parcel is just that. I can see how it's a useful model, but I think it's misleading. I can see how a solid object in a liquid can have a 'buoyant force' operating on it, but I cannot see how an arbitrary chunk of water in a body of water can have anything other than the normal molecular level forces acting on it.

 

[Khashishi]

Regarding your original question, the mass of the full system clearly increases when you add more objects. The mass of the system is simply the sum of all the masses in the system. The apparent weight only matters when measuring the weight submerged in some fluid. If you put your scale outside of the fluid and measure the weight of the whole container, there is no buoyant force so you feel the full weight.

 

[nasu]

It seems that at least part of your original problem relies on the assumption that the zero net force on the body is not compatible with the expected total weight.
In order to see that this is not really a problem, you can take a step back and consider the following configuartion, which have nothing to do with buoyant forces.

1. What weight is measured by the spring scale?
2. What is the net force on the block on the right? (or on the left)

 

[A.T.]

I can see how a solid object in a liquid can have a 'buoyant force' operating on it, but I cannot see how an arbitrary chunk of water in a body of water can have anything other than the normal molecular level forces acting on it.

The total buoyant force on the solid object is just the sum of all molecular level forces exerted on it by the water.

 

[Chestermiller]

Now, to my mind there's a couple of things here, which may be just too hard for me to put into words. Chestermiller I disagree with what you say. Not because I am right and you are wrong, but because what you say makes no sense to me. Of course the problem is I just don't have the breadth of education, but I'm game to take a stab at this.

OK GraemeM, I hear you. Let me first say that I still stand by what Russ and I have been saying and am confident in it, even regarding masses of liquid immersed within a larger body of the same liquid. However, I obviously have not done a good job of explaining the fundamentals of buoyancy to you. So, if it is all right with you, I would like to take a step backwards and, for now, confine attention exclusively to solid bodies of arbitrary shape immersed within a surrounding liquid. We will be focusing on determining the net force exerted by the surrounding liquid on the solid body. We are going to be doing some modeling to quantify the force. (Later we'll come back to fluid parcels).

I assume you are familiar with hydrostatics, and in particular, the equation for the fluid pressure p at a depth z below the air interface: p=p_a+\rho g z
where p_a is the air pressure, ρ is the fluid density, and g is the acceleration of gravity.

Are you also familiar with Pascal's Law indicating that the pressure p acts equally in all directions at a given location in a fluid? This means that if there is an object immersed in a fluid, the pressure acts normal to the surface of the object at all locations on the surface.

I will continue after getting confirmation that you are comfortable with both these concepts.

Chet

 

[Graeme M]

Chet I am not familiar with either of those. However, I do understand that fluid (or gas) pressure acts equally in all directions. And unless I misunderstand your equation entirely, you are saying that fluid pressure increases with depth.

 

[Graeme M]

nasu, I don't know what might be meant by "nett force" in a rigorous sense - I used it in a sort of colloquial sense, that is, what it means to me.

So, the spring scale will measure the total weight of all the suspended mass. Each block is pulled down by gravity but pulled up by the other block. To me then it appears that there is a zero nett force on each block. So I think you are showing me that while each block has zero nett forces, the total system weight remains the same.

 

[Chestermiller]

Chet I am not familiar with either of those. However, I do understand that fluid (or gas) pressure acts equally in all directions. And unless I misunderstand your equation entirely, you are saying that fluid pressure increases with depth.

Is this for a college course or for a HS course? You are telling me that you are not familiar with hydrostatics, but you are trying to understand buoyancy, correct? Please also tell me about your math background (does it include calculus?)

Chet

 

[Graeme M]

Thanks for your interest Chet. High school... I wish. No, I'm 55 years old. I had a pretty basic science education, and was especially bad at math and physics. I just have an interest in things around me and occasionally want to try to nail down certain questions. I see it as a bit of mental exercise - keeping my brain active. I don't have time to go off and do study or even do a lot of reading, so I actually spend a lot of time at nights thinking stuff through rather than sleeping! Buoyancy has kept me awake a lot lately.

This forum has been great. I have learned a lot, even though what I call learning is only pretty basic to you guys.

Now, with buoyancy, I think I understand pretty much the basics. I think I have failed to communicate my question. nasu nailed it with his/her example. In that graphic, the scale is measuring weight because it is directly connected to the blocks.

With buoyancy, the weight of the object is offset by the force of buoyancy. yet the total system weight will always remain the same with a discrete object in the fluid. I guess I am more wondering how the weight of the object is thereby transferred through the fluid. If the force up works against the force down (of the objects weight), how is the force down still transferred in its entirety? There is no direct connection between object and bottom of container as in nasu's example.

That there is a force involved is clear from the following. Place container of liquid on scale and observe weight. Insert finger into liquid. The apparent weight increases. So the buoyant force pushes back against my finger, which is immovable(ish), and thus the container is pushed down. Yet my finger must become lighter by the same amount that the container has increased in weight.

I'm having trouble putting that together in my head.

If though I see a fluid as being a flexible solid, I can 'get' it. Place a solid on a solid on a scale. The repelling molecular forces keep the solids discrete, yet the weight is increased. The weight is transferred through the total system. Buoyancy appears like that to me. My 'bottom' solid is replaced by a liquid (a flexible solid). All that is happening is that the liquid pushes up against the solid object via the exact same forces as the solid on solid example, except that the molecular bonds are not as rigid as in a solid. The weight is transferred through the liquid in much the same way as through a solid.

I think this means that the pressure in my container must increase if an object is added to it. The buoyancy is simply the repelling molecular forces, but the weight is transferred through the liquid and therefore must increase the pressure (as after all, the pressure derives from the weight of the column above and the column above now includes a solid object).

 

[nasu]

nasu, I don't know what might be meant by "nett force" in a rigorous sense - I used it in a sort of colloquial sense, that is, what it means to me.

So, the spring scale will measure the total weight of all the suspended mass. Each block is pulled down by gravity but pulled up by the other block. To me then it appears that there is a zero nett force on each block. So I think you are showing me that while each block has zero nett forces, the total system weight remains the same.

Well, why don't you explain what it means to you? Maybe it will help.

Net force is the same as total force or resultant force. The vector sum of all forces acting on a body or system of bodies.
(No need to put an extra "t". It's net.)

For the block floating in the liquid, even though the net force on the block is zero, the force exerted by the tank with water and floating block on a scale (or on the table) is the sum of the weights. The water acts with a force on the floating block, balancing the weight of the block, like the tension in the cable balances the weight of the blocks in the pulley system.
Imagine that you have the tank with water on a scale. When you add the block, the water level rises so the pressure on the bottom of the tank increases. The scale will "see" this as an increased weight so it will show the sum of the weights (water+ block). Meanwhile the net force on the block itself is zero.

Actually there are even more trivial examples.
If you have a 1 kg block on a scale, the scale will show 1 kg. Now add another 1 kg block on top of the first one. The scale will show 2 kg. But surprise... the net force on the top block is zero. (and so is the net force on the bottom block)
Do you find this surprising? If the answer is yes, try to understand these simpler cases before going into fluids and buoyancy.

 

[Chestermiller]

Hi Graeme M. Now that I know your background, I would like to try to different tack. I think I can do it in a way you can relate to.

First of all, that analogy of a liquid to a solid is not correct. A solid develops internal forces and stresses in proportion to the amount it has deformed. A fluid develops internal forces and stresses, not by the amount it has deformed, but by the rate at which it is deforming. These are totally different mechanisms.

Getting back to your questions. I have an example that I think can help. Suppose you have a tank of water sitting on a scale. The weight of the tank is small compared to the weight of water in it, and can be neglected. You have a solid body immersed in the water within the tank. In Situation 1, the solid body is hanging from a string attached to the ceiling, and there is tension in the string. In Situation 2, there is no string, and the solid body is sinking through the water. To get us started, please tell me which situation you think will result in a higher reading on the scale, Situation 1 or Situation 2. We can continue after you give me your answer to this question.

Please tell me if considering these situations in a little more detail would help solidify your understanding.

Chet

 

[Graeme M]

Chet, let's proceed as you are taking us. I think I understand buoyancy generally, but there are some wrinkles I need to iron out. So I will try to raise these as we go. Is it OK to follow this path on this forum? I don't want to do the wrong thing.

Taking your question above, the solid body is buoyed up in both cases by a force equal to the weight of the volume of water displaced by the object. In both situations, the weight of the body is greater than its buoyancy. In situation 2, the scale will read the total summed weight of water and body. It will read that from the moment the body is completely submerged. In situation 1, the string is taking up that proportion of the body's weight that exceeds its buoyancy. So the scale will read the summed weight less the weight borne by the string. Situation 2 will have the higher reading.

 

[Graeme M]

nasu, I am a little confused by your examples. You say "When you add the block, the water level rises so the pressure on the bottom of the tank increases. The scale will "see" this as an increased weight so it will show the sum of the weights (water+ block)." You imply therefore that the water pressure on the bottom of the tank is equivalent to the weight of the water. Yet pressure is proportional to temperature. If I have a tank of water on a scale and I change the temperature of the water and hence its pressure, will my scale "see" this change in pressure as a change in apparent weight?

 

[Chestermiller]

Chet, let's proceed as you are taking us. I think I understand buoyancy generally, but there are some wrinkles I need to iron out. So I will try to raise these as we go. Is it OK to follow this path on this forum? I don't want to do the wrong thing.

Your analysis below is pretty good. Well done!

Taking your question above, the solid body is buoyed up in both cases by a force equal to the weight of the volume of water displaced by the object. In both situations, the weight of the body is greater than its buoyancy. In situation 2, the scale will read the total summed weight of water and body. It will read that from the moment the body is completely submerged.

Actually, no. If we cut the string in Situation 1, the forces on the solid body will initially be out of balance, and the body will start to accelerate. The water in the tank is not yet aware that this is happening, so the initial reading of the scale after the string is cut will be the same as before the string is cut. As the solid body picks up speed, it begins to experience hydrodynamic drag from the water as a result of the relative motion. This is the same kind of drag force you get when you stick your hand out the car window. The force on the body is upward, and the body exerts a corresponding downward drag force on the water. So the reading on the scale starts to increase. When the body finally reaches terminal velocity, the forces on it are balanced and it is no longer accelerating. The reading on the scale is then equal to the weight of the body plus the weight of the water.

Here's an analysis of the two situations:

Situation 1:
The weight of the body is W_B and the upward force of the water on the body is equal to the weight of the water that has been displaced W_Disp (the latter is the buoyant force on the body). If we do a force balance on the body, we can get the tension T in the string:
T+W_{Disp}-W_B=0
or
T=W_B-W_{Disp}
If we do a force balance on the water, we get:
F_{scale}-W_W-W_{Disp}=0
where W_W is the weight of the water. Solving for the force of the scale, we get
F_{scale}=W_W+W_{Disp}
Thus, the force on the scale is the same as if the region occupied by the solid was filled with water.

Situation 2 (after the body has reached terminal velocity):
If we do a force balance on the body, we get:
F_{Drag}+W_{Disp}-W_B=0
or
F_{Drag}=W_B-W_{Disp}
So the drag force imposed by the water has replaced the tension in the string.
If we do a force balance on the water, we get:
F_{scale}-F_{Drag}-W_{Disp}-W_W=0
or
F_{scale}=F_{Drag}+W_{Disp}+W_W=W_W+W_B

Does any of this make sense to you?

Chet

 

[Graeme M]

Up to a point.

The string tension is a function of the weight, true. But the string is attached to something which is attached to the ground. You are describing only half the system. The string isn't magically taking up and disappearing the weight.

So the summed weight force of the total system including string and thing holding the string must include the weight of the thing holding the string holding the weight of the body as well as the tank of water. With a scale large enough to weigh the complete system we should find no difference in weight regardless of what is happening to the body and the string.

So although total system weight is different, the weight of the solid body is fully borne by the ground in each system. The weight has not gone anywhere.

In the case of the cut string, there is no attachement to anything else, the entire weight is borne by the scale. Case 1 as a complete system is an equivalent situation to case 2.

Therefore the matters of drag and so on seem to me unrelated to weight. The string supports the weight in case 1. yes. But the moment I cut the string all of the weight is transferred to the scale. Else where is that weight going while we wait for it to accelerate to terminal velocity?

The ground will register the same total weight in all cases. In case 2 of the cut string, we can't have the weight suddenly drop while we await the acceleration of the object.

 

[Chestermiller]

Graene M,

I'm sorry to say this, but your comments show a complete lack of knowledge of the most basic fundamentals of Introductory Physics. I don't know of any way to sugarcoat this.

You need to study AP Physics (HS) or Freshman (Introductory) Physics before you can discuss what we have been covering in any meaningful or rational way. I'm hoping you will find it worthwhile to pursue such an endeavor. In my judgement, you should be able to learn it without too much trouble (irrespective of your comment about your terrible skills at math and science).

Until then, I'm afraid I'm going to have to withdraw from our discussion.

Chet

 

[nasu]

nasu, I am a little confused by your examples. You say "When you add the block, the water level rises so the pressure on the bottom of the tank increases. The scale will "see" this as an increased weight so it will show the sum of the weights (water+ block)." You imply therefore that the water pressure on the bottom of the tank is equivalent to the weight of the water. Yet pressure is proportional to temperature. If I have a tank of water on a scale and I change the temperature of the water and hence its pressure, will my scale "see" this change in pressure as a change in apparent weight?

I have the feeling that you have a tendency to complicate the things before you understand the simpler ones.:)

And no, by heating the water the pressure on the bottom does not change. The water expands a little but the density decreases a little too. So the pressure stays the same. You should not operate with generalizations without support. "the pressure depends on temperature" is not a universally true statement.
I suppose you got it from ideal gas law.

 

[Graeme M]

nasu, I am not sure I am over-complicating. The simple fact you've stated is that an increase of pressure on the floor of the container will increase the apparent weight of the container. I can take that at face value, but I immediately think of what conditions could cause an increase in pressure.

To think of it another way, if I have a tank that is wider than it is tall and fill it with water, the total weight will be X, and the pressure Y. But if I make the tank taller than it is wide, will not the pressure on the bottom increase? But equally, the weight will still be X. So how can the change in pressure be read as weight?

Chet, fair enough. I did say I had a very limited background in science. I'd like to have gone a little further before you gave up on me though! :) Thank you for your help so far, anyway.

Are you willing to explain why my question is wrong? If an object is falling in a column of fluid, it must accelerate until its acceleration is offset by drag and it reaches terminal velocity. But its weight is still borne by the fluid the whole time isn't it? How does the fluid know the difference between terminal velocity and something approaching terminal velocity? By your reasoning - it seems to me - a floating object like a small boat, if suddenly allowed to fill with water and sink, must register a reduction in total weight on the scale immediately after it starts to sink.

 

[A.T.]

By your reasoning - it seems to me - a floating object like a small boat, if suddenly allowed to fill with water and sink, must register a reduction in total weight on the scale immediately after it starts to sink.

That is correct. If the total center mass accelerates down, the scale will read a reduced weight.

 

[Chestermiller]

Chet, fair enough. I did say I had a very limited background in science. I'd like to have gone a little further before you gave up on me though! :) Thank you for your help so far, anyway.

Are you willing to explain why my question is wrong? If an object is falling in a column of fluid, it must accelerate until its acceleration is offset by drag and it reaches terminal velocity. But its weight is still borne by the fluid the whole time isn't it?

No. If the object is accelerating, this means that the water is supporting only part of its weight. The air doesn't support the weight of a baseball that is falling through it. Air is just an example of a fluid with a much lower viscosity and density than water.

How does the fluid know the difference between terminal velocity and something approaching terminal velocity?

The drag force is the result of the water deforming (actually the rate at which the water is deforming) in the vicinity of the sinking object. The drag force is proportional to velocity of the object. This force is over and above the buoyant force. If the object is not yet moving, its velocity is zero, and the drag force is zero. At the terminal velocity, the object's acceleration has dropped to zero, and the velocity is such that the drag force is equal to difference between the weight and the buoyant force.

Chet

 

[Graeme M]

Ahhh... I see! And taking nasu's case, I can see now what he's saying when I relate it back to my other thread about weight/pressure.

So when an object is added to a fluid, the displacement of fluid increases the height of the column which increases the bottom pressure which in turn is measured as an increase in weight.

So then.

1. A submerged object displaces its volume in water.
2. The weight of the water displaced is equivalent to the upward buoyant force on the object.
3. The buoyant force arises from the difference in pressure between the top of the submerged object and the bottom of the submerged object.
4. The apparent weight of a submerged object is reduced by the magnitude of the buoyant force. Because weight is a force, the net force applied to the object is the difference between its weight and the buoyant force.
5. If an object’s apparent weight is greater than the buoyant force, the object will sink. And vice versa if less than the buoyant force, the object will rise.
6. The water column effectively increases in height following the addition of the object and hence raises the pressure at the bottom of the column.

And that's it! That answers my original question and the mental model I have makes sense to me now.

Thank you all, this and the other thread I refer to have really opened my eyes and explained so many things I'd often wondered about.

 

[nasu]

nasu, I am not sure I am over-complicating. The simple fact you've stated is that an increase of pressure on the floor of the container will increase the apparent weight of the container. I can take that at face value, but I immediately think of what conditions could cause an increase in pressure.

To think of it another way, if I have a tank that is wider than it is tall and fill it with water, the total weight will be X, and the pressure Y. But if I make the tank taller than it is wide, will not the pressure on the bottom increase? But equally, the weight will still be X. So how can the change in pressure be read as weight?

Yes, you are doing it right now. Was the area of the bottom of the container changing in the original setup? It was not. So how is this story with X and Y even relevant to the problem? If the pressure increases and the area decreases in the same ratio, the force stays the same. I don't understand why are sidetracking all the time instead of thinking a little and trying to understand the basic concepts.

The force acting on the scale is pressure times area. If the area stays constant and the pressure increases the force increases. That's all it is.
When you add the block to the water, the pressure on the bottom increases but the area of the bottom does not change. So the force on the scale increases and the scale measured the increased weight of the water+block.
Same will happen if you just add water to the container. The pressure increases, area stays the same. Again force on the scale increases and you measure the increased weight.

 

[Graeme M]

Gee you guys are so harsh! :)

OK, maybe my discussion style is lacking? But to accuse me of not thinking is a little unfair. It's precisely because I am thinking that I ask these questions.

Look, the basics of buoyancy are simple. I get it. I don't get more advanced aspects of it, however right now I am not interested in the more detailed aspects. I just wanted to know if my basic understanding as described earlier was correct (apparently it is), and if so, I then wanted to see how the weight of the whole system is increased if the upward force completely balances the downward force.

So really I was only interested in the matter of an object completely buoyed up but completely submerged.

You say it is because the column of water is increased in height and hence the bottom pressure increases. OK. I understand that too. But I immediately wonder why that could be so. If the pressure is increased by some other agency, and I assume that is possible, the increased force must then be read as 'weight'. But it wouldn't really be weight. So I am questioning the statement, and my own assumptions.

All I am doing is looking at the logical (to me) permutations of what you are saying to see if it is logically consistent (to me). I am not misunderstanding you - I got it the moment I read it. But I didn't initially agree with it for the reasons stated.

It's the same now when I compare yours and Chet's descriptions. I am happy to accept what has been said and feel I have learned something. But there is now another lurking inconsistency that just nibbles at my mind and will wake me up at 4 AM. None of this matters to you of course, but it's me asking these questions because it matters to me.

So if you are willing to indulge me, here is that inconsistency (to me). And I'll happily admit to these inconsistencies arising because I am as dim as a fencepost. But that shouldn't prevent me having a shot at getting it, should it?

If the object is lowered into the fluid on a string, and completely submerged, it displaces a volume of fluid equal to its own volume. If I understand what you have written correctly, you argue that this increases the height of the column of water and that increases the bottom pressure which then acts as weight force on the scale. OK.

But. To me, it follows that regardless of whether the object is attached to a string, or whether it is just slightly heavier than the buoyant force or a LOT heavier than the buoyant force, the volume of water displaced is the same, and hence the new height of the column is the same, and hence the pressure is the same, and hence the pressure on the scale is the same, for all configurations of object and string or no string.

Once the water has been displaced and the column height increased, it is what it is. The object can do barrel rolls but it won't displace any more volume of water.

Yet Chet has argued for something different. He suggests, and he is backed by AT, that the apparent weight of the tank plus water plus object reduces as the object is accelerated through the water. Unless I misunderstand what has been written, he and AT are saying that the bottom pressure is not related to the system's apparent weight. Because the column of water's height is what it is as soon as the object is completely submerged. Doesn't matter whether the object is accelerating or not.

Clearly, this must be due to a misunderstanding on my part. But until someone show me why, I can't see what that misunderstanding is.

 

[A.T.]

He suggests, and he is backed by AT, that the apparent weight of the tank plus water plus object reduces as the object is accelerated through the water.

If the boat gets a leak and starts to sink, then the water level drops and the boat itself accelerates down. So both centers of mass (boat and water) accelerate down for a short moment. Momentum conservation demands that the scale reads less in that moment.

 

[Graeme M]

Nice! Yes, of course, I didn't think of that. Thank you for the clarification.

But take our example of an object suspended by a string. If it is a solid ball with weight of 10 Newtons and it is lowered completely into the fluid it displaces a volume of water equal to its volume. By nasu's reasoning, this raises the water level and hence the bottom pressure rises, to a value that equates to the additional 10 Newtons of weight. To me this seems to be a fixed value - once the level has risen the pressure has also risen. If the buoyant force in this example is say 6 Newtons, then when we release the string the object will accelerate. You say this will cause a momentary drop in apparent weight on the scale.

Those two thing seem incompatible to me. Once the water is displaced the pressure rises and we then have a new weight on the scale as long as the object is submerged. But you say the weight reading on the scale changes for a short time after the object is released from the string.

What have I missed here?

 

[A.T.]

But take our example of an object suspended by a string. If it is a solid ball with weight of 10 Newtons and it is lowered completely into the fluid it displaces a volume of water equal to its volume. By nasu's reasoning, this raises the water level and hence the bottom pressure rises, to a value that equates to the additional 10 Newtons of weight.

Only if the object has exactly the same density as water, so the string goes slack and doesn't support any of the objects weight anymore.

If the buoyant force in this example is say 6 Newtons,

This is inconsistent with what you wrote above, which implies the buoyant force is 10 Newtons.

 

[Graeme M]

Yes, I haven't described this clearly.

I will try again. By the way, I use Newtons as the measure because I have read this is the correct measure of weight. I would use kilograms in every day situations because that is what a scale is labelled as measuring. I am just trying to describe the force of weight in the proper terms.

Also, I am quite satisfied with the explanation so far for a static situation. I am just bothered by the dynamic situation Chet describes.

Ball has greater density than water.
Ball weighs 10 Newtons out of water.
When submerged it displaces a volume of water of weight 6 Newtons.
Buoyant force is 6 Newtons.
When lowered into water on string, the apparent weight of the ball (ie tension on the string) is 4 Newtons.
Weight of container of water on scale is increased by 6 Newtons.
When released, the ball sinks. It accelerates.

According to nasu, on lowering the ball into the water completely, the water level rises and so too the bottom pressure (I assume he means on the bottom of the container). This pressure is then read by the scale as an increase in weight. This increase can only be equivalent to 6 Newtons (or is he saying it is 10 Newtons?)

When the ball is released, you say the weight on the scale drops for a moment (how long is a moment?). It must then increase until at some point the balls full weight is borne by the container.

So, you argue that the container system weighs X + 6 Newtons with the ball suspended in the water, then something less than that once the ball is released, and then finally X + 10 Newtons when the ball is at rest on the floor of the container.

nasu argues that the container system weighs X + 6 Newtons with the ball suspended in the water because of the change in water level and hence bottom pressure. But in that case, letting the ball drop does not change the water level. So the scale weight must remain as X + 6 Newtons until the ball is at rest on the floor of the container.

 

[Chestermiller]

Nice! Yes, of course, I didn't think of that. Thank you for the clarification.

But take our example of an object suspended by a string. If it is a solid ball with weight of 10 Newtons and it is lowered completely into the fluid it displaces a volume of water equal to its volume. By nasu's reasoning, this raises the water level and hence the bottom pressure rises, to a value that equates to the additional 10 Newtons of weight.

No, it only rises to a value that equates with the weight of the displaced water, which is less than the 10 N weight of the object.

To me this seems to be a fixed value - once the level has risen the pressure has also risen. If the buoyant force in this example is say 6 Newtons, then when we release the string the object will accelerate. You say this will cause a momentary drop in apparent weight on the scale.

No. Just before the string is released, if the object weighs 10 N and the buoyant force is 6 N, the apparent weight on the scale will be the weight of the water in the tank plus 6 N. The instant after the string is cut, the scale reading will not change. But as the weight descends, the scale reading will increase until if reaches the value of the weight of the water in the tank plus the 10 N weight of the object.

 

[A.T.]

But in that case, letting the ball drop does not change the water level. So the scale weight must remain as X + 6

That argument would only work for a static case, not one involving accelerations.

 

[Chestermiller]

That argument would only work for a static case, not one involving accelerations.

To expand on this, the descending object causes the velocity of the water in the tank to no longer be zero. It is pushing water out of its way as it descends, and this water motion propagates (to some extent) to all the water in the tank. This changes the pressure distribution in the tank, including at the bottom of the tank where the pressures are now a little bit higher.

Cher

 

[Chestermiller]

Accelerating the water upward requires a higher force on the bottom than above. Also, when the ball is descending even at a constant rate, the pressure distribution within the tank is affected. So the ball does not have to be accelerating to affect the pressure distribution at the bottom.

Chet

 

[nasu]

Gee you guys are so harsh! :)
You say it is because the column of water is increased in height and hence the bottom pressure increases. OK. I understand that too. But I immediately wonder why that could be so. If the pressure is increased by some other agency, and I assume that is possible, the increased force must then be read as 'weight'. But it wouldn't really be weight. So I am questioning the statement, and my own assumptions.

The scale indicates the force acting on it. It is up the user to interpret that force as weight or some other force. Usually it is used to indicate weight. But if there is a force of a different nature pushing on the scale, it does not make any difference to it (the scale). For example, if you direct a jet of air or water towards the scale, it will show you the force produced by that jet on the scale. It has nothing to do with gravity. You can even push on the scale with your hand and make it show various values of the pushing force. It does not mean that the weight of your hand varies.
In the case of the block submerged, the increased force can be attributed to the increased weight. But tf the pressure is increased due to other forces, the increased reading cannot be attributed to increased weight, of course.
In general, the meaning of the reading on our instrument is up to us. "They" don't know what they are measuring.:)

Regarding the block with string, I think it was already answered by Chestermiller and A.T. in detail.

 

[Graeme M]

Thank you again. As I said, I understand the basics for a static situation and nasu noting the effect of water level and pressure was the missing piece of the puzzle for me. The dynamic case is well beyond me as Chet notes due to me not having the requisite background knowledge. But still intriguing to consider.

 

[A.T.]

The dynamic case is well beyond me

Fluid dynamics is indeed very complex. But there some fundamental laws that still apply to the water+ball system:

- If the the common CoM of water+ball accelerates, then the net force on it is not zero. So the scale doesn't have to show the actual weight of them.

- Unlike in the static case, there is not only the buoyant force, but also the fluid dynamic force (drag) on the ball. Newtons 3rd dictates that this drag also has a reaction on the water, which must be balanced by the bottom pushing the water up (because the water's CoM is not accelerating downwards).