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I know that the pressure increases linearly between A and B, once the mug is at B there will be a jump in pressure, and then the pressure will be constant between B and C. I'm just not sure if this sudden jump in pressure is increasing or decreasing. I think the pressure would suddenly increase, but I'm not sure (I inserted what I think it looks like below). Could someone please help me reason through it?

What I think it looks like

John
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    Of course the pressure increases (since the gas compresses), but what makes you think there will be such a discontinuous change in pressure? – joseph h Dec 05 '22 at 04:21
  • @josephh I was figuring that would happen because once the mug is entirely submerged isn't the water sucked in? – John Dec 05 '22 at 04:26
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    It looks to me that in this experiment, you are lowering the mug into the water at a steady rate. Is this what's happening? You should actually explain what is happening here. A diagram with no explanation is not usually helpful. – joseph h Dec 05 '22 at 04:29
  • @josephh I'm lowering an upside down mug with air in it in water along the dashed line in the diagram. – John Dec 05 '22 at 04:30
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    Then there is no reason to think that there would be such a discontinuity. – joseph h Dec 05 '22 at 04:33
  • @josephh so it would just be a linear increase in pressure? – John Dec 05 '22 at 04:35
  • In general, pressure varies linearly with depth, according to $p=\rho gh$. If the cup is moved smoothly from one depth to another, there is no reason to think otherwise. – joseph h Dec 05 '22 at 04:40
  • @josephh doesn't the fact that the mug is filled with air not water change anything? – John Dec 05 '22 at 04:44
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    Is the mug being lowered through these positions, or is the mug being held in place and you just want to know the fluid pressure at A, B, C & D regardless of if it's the air in the mug, or just water in the cup? – JMac Dec 05 '22 at 14:11

2 Answers2

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Your diagram is essentially correct for a stationary mug positioned as shown in the figure (which to me is clearly your intention despite some apparent confusion on that point).

If you consider a line to the side of the mug, one that is only in water, then the pressure simply increases linearly from top to bottom by $\Delta P = \rho g \Delta h$. That determines the pressure in the water at A, C, and D.

The pressure in the air at C is equal to the pressure in the water at C. Otherwise the water and the air would be accelerating.

Because air is not very dense ($\rho \approx 0$) the pressure in the air at B is nearly the same as the pressure in the air at C ($\Delta P=\rho g \Delta h \approx 0$). This difference in pressure between the air inside pushing up on the mug and the water outside pushing down is what causes the mug to be buoyant.

So your graph is essentially correct.

Dale
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  • Then what causes the almost instantaneous rise in pressure at B (on the graph)? – joseph h Dec 05 '22 at 05:01
  • @josephh I have updated the answer to be even more explicit about what causes it – Dale Dec 05 '22 at 05:39
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    Isn't the OP already assuming that the pressure of the air in the cup is constant at all points? And if the cup is pushed into the water, he is graphing how the pressure of this air changes with depth? I'd expect a continuous linear increase in pressure. – joseph h Dec 05 '22 at 05:51
  • @josephh I don’t know what to say. Your expectation is incorrect. The answer already explains why. A continuous linear function would violate $\Delta P= \rho g \Delta h$ in the air. What step in the answer is unclear to you? – Dale Dec 05 '22 at 12:58
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    @Dale I think this is an ambiguity in OPs question. Myself and others are reading it as "the pressure in the mug is measured as the mug is lowered through the water, what does the pressure look like as a function of the mugs position while being lowered", whereas you are considering the fluid pressure at each point when the mug is stationary. IMO OPs wording could mean either. – JMac Dec 05 '22 at 13:01
  • @JMac you should clarify with the OP. To me it is pretty clearly not asking that. Particularly given the fact that the OP's pressure diagram is correct if you do not read it the way you mention. – Dale Dec 05 '22 at 13:56
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    @Dale OPs graph suggests stationary, but the wording of the problem, and their comments, confuse the matter quite a bit. OP says in a comment " I'm lowering an upside down mug with air in it in water along the dashed line in the diagram.", whereas you are describing a mug being held at one point on the line, which would not be "lowering along the line". Anyways, the point of my comment was just to recognize that you and Joseph were talking about different graphs. – JMac Dec 05 '22 at 14:09
  • @JMac then rather than discuss it between me and you perhaps we should let josephh and John resolve it. I think it is clear that josephh is misunderstanding John's question, but that is really something for them to clarify in comments to the question, not for you and me to discuss in comments to my answer – Dale Dec 05 '22 at 14:19
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    I will say. If a question is ambiguous and can be read one of two ways (one way makes a figure in the question right and the other makes a figure in the question wrong) then usually the intended way is the one that makes the figure right. – Dale Dec 05 '22 at 14:28
  • Okay, in the future I'll try to remember not to help clear things up if I can make sense of the misunderstanding between two people in the comments on your answers. Sorry I tried to bridge the gap between you two talking about different things. – JMac Dec 05 '22 at 14:41
  • @JMac I must say that you helped clear up the confusion for me, I didn't even consider that the mug would be stationary because OP's wording implied otherwise. But the graph matches the stationary case. So thanks for that, and I think this answer could be improved by making it clear it considers the stationary case. – jpa Dec 05 '22 at 17:31
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    @jpa I have clarified in the first paragraph – Dale Dec 05 '22 at 17:38
  • @Dale As I was saying, the answer you have is correct if you look at points along the cup, but I'm still not convinced this was the OPs intent, also as stated earlier. Of course the OP is ambiguous and even more so in the comments he made above. Also, the OP for some reason has not yet clarified the intent of the question, despite numerous attempts by myself and Jmac. Thanks. – joseph h Dec 05 '22 at 21:41
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    @josephh you need to take that up with them. I simply don’t see the ambiguity in the question that you do, particularly given that the graph is correct with the question interpreted in the way I think is clearly intended – Dale Dec 05 '22 at 21:57
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    Yea, I also see the graph immediately as measuring pressure at various points of the setup, while the mug is already lowered and is stationary. – justhalf Dec 06 '22 at 04:59
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The thing that cup is filled with an air will not change anything. Pressure will increase gradually and continuously due to bigger depths and air in the cup will get constantly compressed more and more, due to fact that water pressure will equate to ideal gas pressure in the cup:

$$ \rho g h = \frac {nRT}{V} \tag 1$$

From (1) you can see that as depth increases, volume of gas decreases (it gets compressed by water pressure), assuming isothermal process.

If something would force process to be isochoric (of a constant volume), then along with depth- temperature of gas inside a cup must increase to compensate for water pressure changes. In any case, I do not believe that process will be isobaric between points BC (as drawn in a chart), because for that to happen, while gas volume decreases something must cool the gas to lower its temperature constantly- which is highly unlikely.

Agnius Vasiliauskas
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