What Are the Complexities of Buoyancy and Common Misconceptions?

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    Buoyancy
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Discussion Overview

The discussion revolves around the complexities of buoyancy, addressing common misconceptions and exploring the underlying principles of buoyant force in static fluids. Participants examine the relationship between buoyancy, pressure differentials, and the apparent weight of submerged objects, while also considering the implications of adding different materials to a fluid system.

Discussion Character

  • Exploratory
  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant summarizes their understanding of buoyancy, noting that a submerged object displaces its volume in water and that the buoyant force is equal to the weight of the displaced water.
  • Another participant challenges the consistency of the claims regarding buoyant force and pressure, suggesting that the pressure differential does not change with depth.
  • Some participants discuss the concept of net force, emphasizing that internal forces do not contribute to the net force on an object in isolation.
  • There is a debate about whether adding water to a container produces a buoyant force, with some arguing that it does not, while others assert that buoyancy is present in all fluids.
  • One participant expresses confusion about how buoyant force affects the apparent weight of an object without changing the total system weight.
  • Another participant questions the role of pressure in buoyancy, suggesting that an object can sink until the buoyant force offsets its weight, which is influenced by pressure changes.
  • There is a discussion about the molecular forces in water and how they relate to buoyancy, with one participant proposing that buoyancy might be linked to the attractive and repelling forces at the molecular level.

Areas of Agreement / Disagreement

Participants express differing views on the nature of buoyant force and its implications. Some agree on the basic principles of buoyancy, while others contest specific points, particularly regarding the effects of depth on pressure and buoyancy. The discussion remains unresolved with multiple competing views present.

Contextual Notes

Participants highlight limitations in understanding buoyancy, particularly regarding the definitions of net force and how internal forces interact within a fluid system. There are also unresolved questions about the relationship between buoyancy and molecular forces.

  • #31
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
 
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  • #32
Graeme M said:
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.
 
  • #33
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.
 
  • #34
Graeme M said:
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.
 
  • #35
Graeme M said:
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
 
  • #36
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.
 
  • #37
Graeme M said:
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.
 
  • #38
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.
 
  • #39
Graeme M said:
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.
 
  • #40
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?
 
  • #41
Graeme M said:
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.

Graeme M said:
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.
 
  • #42
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.
 
  • #43
Graeme M said:
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.
 
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  • #44
Graeme M said:
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.
 
  • #45
A.T. said:
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
 
  • #46
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
 
  • #47
Graeme M said:
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.
 
  • #48
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.
 
  • #49
Graeme M said:
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).
 

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