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I am looking at the following textbook problem (please note that I am not asking for a solution to the problem, only using it as a motivating example):

A host pours the remnants of several bottles of wine into a jug after a party. The host then inserts a cork with a $2.00$-cm diameter into the bottle, placing it in direct contact with the wine. The host is amazed when the host pounds the cork into place and the bottom of the jug (with a $14.0$-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a $120$-N force.

(source)

The idea here is that the glass of the bottle will break if it is subjected to a certain force, and while the glass may be able to withstand the $120$-N force at the top, by Pascal's principle, the force gets multiplied at the bottom by the ratio of areas.

My example

So, let's say I have an enormous, cube-shaped aquarium that measures $10$ m on each side and is completely sealed. I drill a hole of area $1 \text{ mm}^2$ into the top and push an appropriately sized pin into the hole with $50$ N of force.

Then the force exerted against the bottom is

$$F_2 = \frac{A_2}{A_1} F_1 = \frac{100 \text{ m}^2}{(.001\text{ m})^2} 50 \text{ N} = 5 \times 10^{10}\text{ N}$$

which suggests that the aquarium would break, just like the bottle. My intuition says that this is not the case. But why wouldn't it be?

My first thought was that it isn't the change in force, but the increased pressure that causes the bottle to break in the problem above (despite how it's worded). But in the aquarium, the change in pressure is $F/A = 5\times 10^7$ Pa, which is still significant when compared to the pressure at the bottom of the aquarium $\rho g h = 1000 \cdot 9.81 \cdot 10 = 98100$ Pa.

My second thought was that maybe it has to do with how the shape of the bottle directs all of the force downward, whereas in the aquarium it might get transferred in every direction from the pinhole. But we could modify the aquarium to have a tapered cone shape, just like the bottle, and the same equations would apply. (Except that $h$ would be higher, so the baseline pressure would be higher.)

Questions

Is it correct to say that a suitably small pin pushed with $50$ N of force can exert an arbitrary extra force on the bottom of the bottle?

Likewise, is it correct to say that the same setup could cause an arbitrary increase in the pressure of the fluid it holds?

If so, why can't I use this technique to destroy all sorts of airtight containers?

Is it the pressure or the force that ultimately causes the container to break?

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  • $\begingroup$ It sounds like you re-invented Pascal's Barrel $\endgroup$
    – Ruben Verresen
    Commented Jun 11, 2020 at 6:20
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    $\begingroup$ Wow. So it really would break ...? Then how come we can keep mantis shrimps, which can punch with 1500 N of force, in an aquarium? And is it the pressure or the force that causes the break? $\endgroup$
    – Max
    Commented Jun 11, 2020 at 6:38
  • $\begingroup$ To increase the pressure in the aquarium, you really need to compress the volume of liquid within the aquarium. The pin won't be enough to do this enough, and its force won't be 50 N. $\endgroup$
    – Chet Miller
    Commented Jun 11, 2020 at 12:24
  • $\begingroup$ This confuses me more. I thought water was nearly incompressible. And I can apply 50 N of force to a pin, e.g. by setting a 5 kg brick on it. $\endgroup$
    – Max
    Commented Jun 11, 2020 at 12:30
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    $\begingroup$ More to the point: the sides of the aquarium are slightly flexible. $\endgroup$
    – S. McGrew
    Commented Jun 11, 2020 at 14:37

1 Answer 1

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Suppose that the pin is 1 mm in diameter and 5 cm long. So its insertion volume is $5\times 10^{-5}\ m^3$. A 10 gallon tank contains 0.0379 m^3, so the fractional volume change is 0.00132. The bulk modulus of water is 2.1 GPa. So the pressure increase would be 1320 kPa. This would be assuming that the tank walls are rigid, and don't flex. But, of course, they would.

Still, I stand corrected.

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  • $\begingroup$ In my original post, I actually made an error in converting from $\text{mm}^2$. I've fixed my post now, but I think you have carried my error over. Also, in my argument, the volume of the pin doesn't matter, because the change in pressure occurs at the instant the pin is pushed. The compressibility of the water dictates only how much resistance it gives to my pushing (although this quantity may increase when the pin is further in). Nonetheless, I think the essence of your argument, that the walls are flexible, correctly addresses my question, but I'd like to wait for corroboration. $\endgroup$
    – Max
    Commented Jun 12, 2020 at 0:27
  • $\begingroup$ For the case of a rigid container, I stand by my Answer. $\endgroup$
    – Chet Miller
    Commented Jun 12, 2020 at 1:49

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