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https://what-if.xkcd.com/6/ has been mentioned here before, but I'm questioning whether or not the glass cup with the bottom half as a vacuum would rise at all.

To start with, a vacuum exerts no force. Any perceived "sucking" is actually external pressure pushing into the vacuum. So the only force that could be lifting the glass would be buoyancy of the air around the cup.

Let's ballpark it with a drinking cup that can hold about 500ml. If we consider the cup as an open-top cylinder, an internal radius of 3.8 cm and height of 11 cm gets us a 499ml volume, and the internal surface area is 308 cm^2. Based on a quick Google search, glass is about 2000x more dense than air, so in order for the glass to rise it would need to displace 2000x as much air as there is glass. That suggests that if the glass was sealed (by a weightless forcefield at the top) and was completely empty instead of having some water in it, the total volume of glass would have to be less than 0.5ml, resulting in an average width of 0.016mm. That's thinner than a human hair.

Given that glass cups are significantly thicker than human hair1, is there any truth to the conclusions of that "What If?" Is there some effect that I've misunderstood or underestimated that significantly changes the situation? Or should we conclude, like https://physics.stackexchange.com/a/33642/79374 did with the other vacuum cup, that Randall Munroe either miscalculated or was greatly exaggerating?

1 Citation needed

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For a glass of water, the buoyant force is a small correction to the gravitational force arising because the pressure is different at different altitudes. The timescales in this problem are short enough that we can neglect the center-of-mass motion due to gravity, so we can absolutely neglect the center-of-mass motion due to the buoyant correction.

The net force on the glass will be

\begin{align} \vec F_\text{net} &= \vec F_\text{out} + \vec F_\text{in} \\ &= \int P_\text{air}\ \mathrm d\vec A_\text{outside} + \int P_\text{vacuum}\ \mathrm d\vec A_\text{inside} \end{align}

For a disc with air below and vacuum above, this force works out to be just the air pressure applied to the area of the disc. The drinking glasses in my kitchen have outer diameter $\rm 5\,cm$ at the base, so call the inner area $\rm 15\,cm^2$. (In the limit of a cylinder, the net vertical force on the glass walls is zero.) For a one-atmosphere pressure difference that’s a net force of

$$ P_\text{air} A_\text{base} = 10^5\frac{\rm N}{\rm m^2} \cdot 15\rm\,cm^2 \color{lightgray}{ {}\times\left(\frac{1\rm\,m}{100\rm\,cm}\right)^2} = 150\rm\,N $$

This particular glass from my kitchen has a mass of about $\rm\frac13\,kg$, so with air below and vacuum above it would rocket upwards with an acceleration of about $45g$. (If you’d like, you can re-do the problem without neglecting gravity, and ask whether the correction to $44g$ upward is larger or smaller than our other simplifying assumptions.)

If you consider the glass-plus-liquid as a single system, the no-gravity, no-buoyancy approximation is that the air pressure will symmetrically crush the water and the glass together without changing its center of mass: the water will move down and the glass will move up, so that the total momentum of the system remains at zero.

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  • $\begingroup$ I think I understand where my previous thoughts about buoyancy were wrong. Basically, there's a lot more downward force than I realized on everything even just from the atmosphere. I was right that it's not the vacuum pulling the cup up, but underestimated how much of a sudden lack of force pushing down there would be. As a more intuitive way to think about it, if you put an airtight but otherwise loose seal over the cup and sucked out all the air, it's not hard to imagine being able to lift the cup by pulling on that seal. Maybe there's a buoyancy analogy that would work too. $\endgroup$
    – Rob Watts
    Jun 2, 2022 at 19:21
  • $\begingroup$ As a side note 44g would only be the initial acceleration - once the cup has started moving the pressure on the cup will be reduced. Figuring out the true acceleration over time sounds like a beastly problem though. $\endgroup$
    – Rob Watts
    Jun 2, 2022 at 19:35
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It's not just buoyancy because the slug of liquid can move. Your "force field" at the top of the glass modifies the premise of the XKCD scenario. If you look closely at the illustrations, his slug of liquid moves downward as the glass moves upward. The atmosphere pushes down on the slug of liquid and up on the glass, moving them both. But the center of mass of the system would not move.

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Only if the cup is big enough and resistant to breaking, it will fly up. The volume, with zero mass inside, grows faster than the glass volume (assuming constant glass thickness). So the upward buoyancy force will exceed the total mass at some point.

I only now saw the real problem. If you pull a vacuum in a cylinder with a piston and release the piston, the cylinder will accelerate oppositely. Likewise, the glass with a vacuum on the lower half will jump upward.

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  • $\begingroup$ How big is "big enough"? Is it anywhere near small enough to still be considered a drinking cup? $\endgroup$
    – Rob Watts
    Jun 1, 2022 at 22:20
  • $\begingroup$ @RobWatts It would be a huge cup! I only now see what you actually mean though. Isn't the problem the same as pulling a cylinder vacuum with a piston and then release the piston? $\endgroup$ Jun 2, 2022 at 6:13

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