Physics Asked by egg destroyer on May 9, 2021
This question is mainly about pressure and how kinetic forces work actually. But I would like to ask it within some certain examples:
Let’s assume I have enough deuterium and tritium at right temperature and pressure to create fusion. You may think the force is applied downwards to these isotopes, and fusion event occurs. Now, if I add another force (as big as first one) at the opposite direction of our primary force, do these isotopes crush and exposure even a higher pressure and force or does the forces annihilate eachother and the fusion event ends?
Actually you can give an answer on our blood vessels too. There is air pressure (let’s assume 1 atm/101325 pascals) and the equilibrium of that as our blood pressure. Does this mean our blood vessels are exposuring ~203000 pascals? or do equal forces/pressures annihilate eachother and means zero?
So, one way to think of pressure is the tendency of an environment to do work on a system when it changes volume. So you stretch something out or compress it, when you do that a certain amount of work is done on that system, if you divide the work by the change in volume you get the pressure. This is why pressure has units of energy per volume, it is not an energy density itself but a rate of energy loss per volume gain.
So, in your first case, the question is whether this new force that you are adding does any new work on the piston if it expands in volume. And the answer is, you stopped just shy of that. So, you want to imagine a cylinder with rigid sides, two movable disks: one at the top, one at the bottom. You have a heavy weight, say, on the top. If I ask you why isn't the bottom plate falling out of the bottom, you have to say something about a constraint force: there is some sort of metal lip of the cylinder which the disc cannot pass through, it is held up by that. So, the piston is overall in a state of force balance, not accelerating or decelerating, because the sides are bolted to the ground (one constraint force) and the bottom cannot fall out beneath a certain height (a different constraint).
Now, you wish to push up on the bottom disk with all the same force as the top, and what you are basically doing right now is providing strain relief. That same force was being provided by the walls of the cylinder, which communicated it to the bolts, which communicated that force to the ground: now you are providing that force directly rather than indirectly. The walls of the piston around that lip that is holding the disc in, breathe a sigh of relief. But no new work is done on the piston if its volume changes, you have stopped before that could happen.
Suppose you keep increasing the force upwards, now what happens? Well, both of the discs start shifting upwards in the piston. If there is a similar lip on the top of the piston holding in the top disc, then when we reach that point the gas inside will compress a little more and will come to a higher pressure. If there is no such lip, then the top disc will pop off the thing and ambient air will rush in and all of the fusion gases will rush out.
I hope you are paying careful attention, because the claim is that this notion of pressure more or less assumes that you get to an equilibrium state where nothing is changing, and then we can ask counterfactual questions, “how would X change if I changed only Y?”. In that equilibrium state, you must have a state of force balance, the external force pressing upwards on the system must be the same as the external force pressing downwards on the system: otherwise, per Newton’s laws, it would be accelerating and therefore not in equilibrium. This causes pressure to be kind of omnidirectional. That downward force from the top could not exist without an upward force from the bottom provided somehow, doesn't matter how. In turn, you don't double the pressure just because you see one force up here and one force down there; those forses are always paired by the state of force balance, so the only question is the work done on the system if it contracts, and if it contracts by a volume $A~delta L$ where $A$ is the cylinder cross-section and $delta L$ is the change of the length measured between the two disks, then you have to divide $delta L$ into a change on the position of the upper disk, say $delta z$, and a change in the position of the lower disk, $delta L-delta z$ adding up to this value. And from there the claim is that since it's the same force $F$ on both the top and the bottom, these two add up to a work $$F~delta z+F~(delta L-delta z)=F~delta L,$$ And from there our definition gives the pressure as $F/A$. You don't double it just because the same force is provided from the bottom, because that’s just not how we define pressure.
Correct answer by CR Drost on May 9, 2021
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