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I've read the answers to the similar question here on StackExchange but didn't find them satisfactory. People's answers were in short that we aren't crushed by the weight of the atmosphere because our internal pressure increases to compensate. People pointed out that you can dive to tremendous depths doubling, tripling and so on the pressure exerted on your skin without ill effect. However, when this question comes up I inevitably think of analogous situations in which the pressure increase is far less drastic but you nonetheless do suffer signfiicantly.

Say your lying down in bed and your annoying little sister comes along and lies on your legs. That's a moderate 1kPa pressure increase at most, far less than the 100kPa if you dive to 10m but you can most definitely feel it nonetheless. What's the explanation, how do you square both situations?

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It is because the pressure is coming from all sides equally, and therefore it does not deform an incompressible body. Compare the human body to a balloon full of water. If you apply a pressure of 400 bar to it, its volume decreases with 1.8%, and its height decreases with 0.6%. This is not noticeable with the naked eye. Now put a cube with 1 kg milk on that same balloon, the pressure is only 0.001 bar, but you see the balloon deform a lot. If the human body does not deform, the cells and blood vessels are not damaged.

In engineering this is called hydro-static pressure. Unless the object has a hollow space, it only causes hydro-static stress, which does not damage an incompressible material. This is because the atoms stay in the same place, and stay next to the same atom as before the pressure. If the pressure comes only from one side, like when you crush a cube from the top, then when the pressure becomes high enough, the material needs to move out of the way of the pressure, and can only go to the sides. When the material has to move to the sides, atoms become separated and the material is damaged.

In engineering for metals, the Von Mises stress criterion is usually used to determine when damage occurs. There is stress in 6 directions, but because the material behaves the same in all directions (isentropic), we can use a single number to determine when it will fail. The formula is:

$$Svm^2=0.5*((S11-S22)^2+(S22-S33)^2+(S33-S11)^2-6(S12^2+S23^2+S31^2))$$

An S with different numbers behind it is a shear-stress. S12 is the shear-stress in one - two direction. They are all zero in this case. An S with two times the same number behind it is a normal stress. S11 is the normal stress in the 1 direction. In the case of hydrostatic pressure, they are all the same S11=S22=S33, therefore the Von Mises stress is zero. This formula shows that a piece of solid steel will not be damaged by pressure, even if you throw it in to the deepest ocean. If the pressure gets high enough, the block of steel will collapse in to a black hole though, Von Mises does not take that in to account.

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  • $\begingroup$ Re, "like when you crush a cube from the top...the material...can only go to the sides." Really? It can't go down? It can't accelerate away from whatever is crushing it from the top? If that is the case, then you aren't applying pressure from only one side. You are applying equal pressure to the top and to the bottom. $\endgroup$ – Solomon Slow Dec 15 '17 at 16:03
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I do not know the answer, but one should consider the fact that the effect of the same force varies significantly depending on how the force is applied. For example, rocks are really robust if they are pushed but break apart more easily if you pull them from different sides.

In more engineering terminology, what I am trying to tell you is that atmospheric pressure, being uniform, should induce almost only normal stress, whereas your sister lying on your legs may induce significant shear stress. If human body is more robust under normal stress than under shear stress (I guess one needs to analyze stress-strain diagram for that), then it makes sense for you to feel pain in the second case.

This is just a guess though, I honestly do not know the answer.

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The explanation that the pressure is equalized by pressures inside is nevertheless correct.

Instead of having people sit on you, consider your own weight while lying flat. It is not insubstantial, but if you can spread it out to half of your body in a waterbed, it's better. Spread it out to your whole body on all surfaces when you float submerged in a pool (pretend you're entirely underwater, neutrally buoyant) and it's even better. But you're not weightless when doing scuba-- the water still presses on you floating, and you can feel this in a drysuit or in waders when you're upright-- there's a definite squeeze on your legs pushing you upwards. There is no getting away from the fact that buoyancy is a pressure, and its force-times-area must equal your weight.

Suppose it's more? Suppose you're down 40 m in a hardhat suit and the pump fails and your helmet valve fails, the line tender isn't paying attention, and the pressure inside your suit falls to becomes that of the surface, while outside the suit, ambient is 4 atm more? The results are horrible and the water squeezes your body up into your helmet and all the blood and soft tissue go up the line to the boat, and not much is left in your suit but bone crushed into the helmet. That used to happen, now and again.

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protected by Qmechanic Dec 15 '17 at 14:54

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