B Pressure of captured air under water

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Summary
How to calculate the air pressure of captured air under water with a weight pushing it downwards
For a construction I am building, I am stumbling on a rather basic physics question regarding pressure.

Let's say I have a cubic bucket of 1m^3 that I place upside down on a water surface. I add a downward force (weight) of say 1000N submerging the bucket under water with the air captured inside.

What is the pressure and volume of the air inside the bucket? My own attempt:

The buoyancy force is also 1000N and is spread over the 1m2 surface, so the pressure is increased with 1000N/1m2 =~ 0.01 atm. Thus the resulting pressure is 1.01 atm, the volume is 1m3 / 1.01 and the waterline inside the bucket would be 1cm above the edge of the bucket.

My confusion is that this would mean that a bucket of the same volume but a smaller width/height would result in higher pressure of the air inside which seems counterintuitative.

Can someone help?
 

A.T.

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My confusion is that this would mean that a bucket of the same volume but a smaller width/height would result in higher pressure of the air inside which seems counterintuitative.
A taller bucket will reach deeper, where the water pressure is higher.
 
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I don't confirm your calculation at all. From a force balance on the bucket, if H is the depth of the top of the bucket and H+L is the depth of the bottom of the bucket, the force balance gives:
$$\rho g A (L+H)-N-\rho g AH=\rho gAh$$where h is the amount that the water rises in the bucked, and the right hand side is the weight of the water in the bucket. From this it follows that $$\rho g A(L-h)=N$$This means that h is 0.9 meters.
 
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Thank you very much. I think I understand your formula.

Although I must say this is even more counterintuative.

I would think that if I attach 100kg to the (idealized 0 mass) bucket placed on the water surface, also 100kg of water would be displaced, so the top of the bucket would sink only 0.1m under water (displacing a volume of 0.1m * 1m2 water). Thus the bottom of the bucket would be 0.9m above the water.

Yet your calculation suggests the waterline would be 0.9m from the top and thus only 0.1m from the bottom of the bucket, compressing the air 10 fold. This also would mean the bucket would sink way deeper. Are we sure we are talking about the same setup?
 
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The forces acting on the bucket are in equilibrium. They are
1. The downward force of the water from above on the outside surface of the top = ##N+\rho g A H##
2. The upward force of the inside air from below on the inside surface of the top = ##\rho gA (H+L)-\rho gAh##
Setting there two forces equal gives $$N=\rho g A(L-h)$$This is just the weight of the displaced water.
My calculation suggests that the water line should be 0.1 m from the top and 0.9 m from the bottom, compressing the air 10 fold.
 
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Thank you for your patience. I am afraid I am still not completely grasping it.

Am I understanding correctly that N equals the downwards force induced by the mass on the bucket (which was given as 1000N)?

So if we set it to zero, h should also be zero because the bucket would float on the surface (H=-1m, H+L=0m, h=0m). This isn't the case in the equation you are giving.

What am I missing?
 
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Thank you for your patience. I am afraid I am still not completely grasping it.

Am I understanding correctly that N equals the downwards force induced by the mass on the bucket (which was given as 1000N)?

So if we set it to zero, h should also be zero because the bucket would float on the surface (H=-1m, H+L=0m, h=0m). This isn't the case in the equation you are giving.

What am I missing?
The equation assumes that N is large enough for the bucket to be fully submerged.
 
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Let's first start by looking at the case where the bucket is not fully submerged. In this case, the added downward force on the top of the bucket N is equal to the gas gauge pressure times the bucket area A. If the depth of submersion of the bottom of the bucket is z (z < L) and the water rises a distance h above the bottom of the bucket, then the gas gauge pressure in the air head space is ##\rho g (z-h) ##. So the force N is given by:
$$N=\rho g A(z-h)$$The air in the head space must satisfy the ideal gas law in terms of absolute pressures. So we must have that $$[p_a+\rho g (z-h)](L-h)=p_aL$$where L is the total length of the bucket and ##p_a## is the atmospheric air pressure.

If we combine these two equations, and solve for the water rise h into the bottom of the bucket, we obtain:
$$\frac{h}{L}=\frac{N}{N+p_aA}$$So as expected, when the applied normal force goes to zero, so also does the height of the water rise into the bottom of the bucket.

If we combine the previous equations, we can also solve for the depth of submersion of the bottom of the bucket as a function of the normal force N: $$\frac{z}{L}=\frac{(N/A)}{\rho g L}+\frac{N}{N+p_aA}$$Note again that, as expected, the submersion depth is zero when N is zero, and the submersion depth increases with increasing N.

OK so far?
 
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Thank you very much for this explanation and my apologies for the confusion, that I realize may have arisen by using the word "submerged" in my question.

OK so far?
Yes!

It's a very interesting result that comparing two buckets, one with A=1m2, L=1m and the other has A=1m2, L=2m, yet both with the same mass, the latter will have a higher rise in the waterline within the bucket, and a corresponding difference in the depth of the bucket.

It's late here and I will need some time to digest the equations, but filling in some variables certainly give sensible values.

Thank you,
Thomas
 
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Are you aware that the force to submerge the bucket N increases with increasing submersion depth until the top of the bucket passes below the surface; from that point on, the force N decreases with further submersion? So the submersion force is maximum when the bucket just becomes fully submerged.

Would you like to see the remainder of the solution, expressed in terms of dimensionless variables?
 

sophiecentaur

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I think I understand your formula.

Although I must say this is even more counterintuative.
Intuition is an unreliable test where Science is concerned. Best to work from something you 'know' and head for a solution.

There is one location in the experiment where you can say something 'certain'. That's the lower surface of the air in the bucket and you can say that it's in equilibrium. The pressures on the surface (up and down) will be the same. It's easiest to assume that the bucket is an exact cylinder so that vertical forces not due to the water surface in the bucket can be ignored. It does work for other shapes, of course.

You can always tell the hydrostatic pressure that's pushing upwards and that's P = ρhg, where h is the distance between the surface of the tank and the surface in the bucket. (Ignore the density of the air; it makes things easier).

The downwards pressure will be due to the weight of water that's been displaced by the air in the invert* plus the weight of the bucket, all distributed over the bottom surface. That's basically a verbal statement of what @Chestermiller wrote as a formula, above.

To find the water level in the bucket, you just equate the two pressures and solve for h. Parts of the bucket sides that are below the surface just contribute to the weight of the bucket. You could use a shorter (same mass) bucket and pump air into it to get the same result.



* That's a statement of Archimedes Principle.
 
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The downwards pressure will be due to the weight of water that's been displaced by the air in the invert* plus the weight of the bucket, all distributed over the bottom surface. That's basically a verbal statement of what @Chestermiller wrote as a formula, above.
In my analysis, I neglected the weight of the bucket.
 

sophiecentaur

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In my analysis, I neglected the weight of the bucket.
I was thinking in terms of the bucket floating at a stable depth in the water. Same thing effectively and 'floating' is an easy thing to feel intuitively. Actually, gas laws aren't necessary and you can just infer that the internal pressure provides the upthrust. Different views. (I always like your use of formulae - at the drop of a hat; an old pro!!!)
 

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