Flow of pressure in water bottle

Say you would have a plastic bottle partially containing water that is laying on its side. How would applying external pressure through a weight on the side affect the internal pressure. In other words, how well does the pressure transfer from outside to inside the system?

• Do you have any thoughts on what to expect? This sounds a bit like a homework-type question. Generally you're more likely to get the answers you want if you provide all the thought you put into it.
– JMac
Commented Mar 27, 2017 at 9:55
• Yeah sorry you're right. From previously seen formulas I have established that the pressure would definately be higher, but this is based upon the gas laws. If there would there be any possibility to combine the gas and fluid laws the question would be much easier. Commented Mar 27, 2017 at 10:19
• The pressure won't "definitely be higher." If you apply clamping force to the right places on a container of the right shape, you can actually decrease the pressure inside the container by increasing the clamping force. Commented Mar 29, 2017 at 22:32

Based on your comments I'd say you're on the right path on your own.

You already mentioned using the (ideal) gas laws. Beyond that there isn't a whole lot to consider in this problem. (assuming you want to do a cursory analysis of this situation, and not find exact pressure values or anything like that)

It's worth noting that water is approximately incompressible, so it doesn't build pressure quite like the gas does, but as long as the gas is pressurized the water will be at that pressure (and a bit higher the deeper you go, see hydrostatic pressure).

So the basic concept is apply a weight > deform the container > reduced air volume means increased pressure in the bottle due to ideal gas law.

Your rewording of the question is a bit more complicated. The effectiveness of transferring pressure between outside and inside depends quite a bit on the material of the bottle. Depending on the material, it will deform more or less when you apply a force to it.

A weak material, like a balloon very easily deforms, so it's very easy to apply a force to the outside and increase the pressure inside. Something like a steel container doesn't work as well. When you apply a force it does deform a bit, which would increase the pressure; but it barely deforms, so the volume barely changes and pressure barely increases.

• Yes this is basically what I meant. Now though I am looking for a precise way of knowing by exactly how much the internal pressure will be changed. I have previously found the following formula: Qm = C * A * sqrt((2* Delta(Pressure))/diameter of hole) here [link] (physics.stackexchange.com/questions/80277/…). Therefore I would expect to be able to find the internal pressure by finding the Qm, right? Not sure about how well this works. Commented Mar 27, 2017 at 15:45
• @AdrianSie I wouldn't do it that way, that's for sure. Personally it would depend on what I'm doing and what I might know. For example, if you had a good way to determine the volume change, that would help. You could definitely measure the flow through an orifice, but you would need to know about the orifice. There would also be issues with how that would change things compared to having a fully sealed container (as the pressure builds it actually resists more, having an opening to alleviate pressure buildup might make those results irrelevant).
– JMac
Commented Mar 27, 2017 at 15:52
• This is indeed true and what I'm looking for, though would there be a controlled way of finding/formula between external pressure exerted, the amount of stress the bottle can handle and the addition of the internal pressure? The relationship is clear but I might need more precision than a statement :/ Couldn't find too much on internet Commented Mar 28, 2017 at 7:37

It's not so simple. For example, try to open such a bottle and place something light (e.g. coin) on it. Obviously you will have no pressure transfer at all (pressure inside and outside will be the same), but the bottle will not collapse.

• I disagree that the coin wouldn't affect the pressure of the bottle.
– JMac
Commented Mar 27, 2017 at 9:51
• JMac, for an object (coin) so light relative to the "deformability" of the material (plastic), don't you think we can reasonably approximate the bottle as fully rigid and unperturbed by the weight of the coin? Perhaps this may even be better than mere approximation; is deformation continuous with respect to applied force or is there a certain minimum weight that must be applied before a material begins to deform? (I suspect there are various classes of materials, if so let's take the class containing this plastic). Commented Mar 27, 2017 at 17:33
• @electronpusher Deformation is generally continuous on these scales. Many materials have an "elastic" region where there is a linear relationship between stress and deformation. I'm not aware of some lower limit where this fails to be continuous; any weird effects would likely require even far less force than a small coin. A plastic bottle is thin and has very little deformation resistance.
– JMac
Commented Apr 2, 2017 at 15:53

Pressure is $\frac{F}{m^2}$. If we apply a force $W$ on one side, we must apply an equal and opposite force on the other side of the bottle. Therefore, the increase in pressure in the bottle when the weight is applied is $\frac{2W}{\text{Surface Area Of Bottle}}$. If you increase the surface area of the bottle, the increase in pressure will be lower.

• I'm pretty sure it wouldn't increase the pressure in the bottle that simply. A force on the outside of the bottle would cause a deformation to the inside of the bottle, changing the air volume and increasing the pressure. This will be proportional to the force and bottle geometry, but not as directly as you show. The bottle will resist deformation.
– JMac
Commented Mar 27, 2017 at 10:53
• Equal and opposite force? It sounds like you're trying to use Newton's Third Law of Motion. However, that law specifies an action-reaction pair, i.e., the forces that are equal and opposite are for two interacting objects, not both for the same object. The "inside" and "outside" (walls?) of the bottle are not an action-reaction pair, they are the same object. Unless you're talking about an area on one side of the bottle, and an area 180 degrees from it. Either way so, I don't think that analysis is appropriate. Commented Mar 27, 2017 at 17:28
• My analysis is fine. If there is a force acting on one side of the bottle and no net force acting on the bottle(it isn't accelerating), then we can assume there is an equal and opposite force acting on the bottle as well. Since both forces of magnitude $W$ are acting on the bottle, the net amount of force pushing inward on the bottle is $2W$. Commented Mar 27, 2017 at 17:49
• I disagree. I think only one "side" of the bottle needs to be considered (a half-cylinder), and the equal force opposing W is the "tension" of the plastic around the point of depression. I'm not an engineer, but a force of 2W contributing to the pressure seems wrong to me. Perhaps a third party can verify. Commented Apr 2, 2017 at 18:13