1
$\begingroup$

Lay on the ground in a bright sunny day of spring and look the sky. Feel the air pressure on you. Now lay on your stomach and your back exposed to sky. You won't feel any change in air pressure. You will breathe normally. Even though air above you ( below you actually ) is so less that it cannot exert the same force/pressure on you.

I know that at a horizontal plane, pressure in fluids is same in all directions but why is it that way?

Its like someone is applying a force or exerting a pressure on you from above but you feel the same force/pressure from all directions, even from sideways. (Maybe the same pressure from downwards can be explained by Newton's third law but even from sideways, the same thing?)

I think fluid particles' mobility is the cause of it.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Actually, the weight of the air above you contributes to the pressure that you feel. Because of this, air pressure drops as you go to higher altitude. This change is not so significant on the scale of you turning your head from pointing down to pointing up, so no real change is noticeable.

Why does pressure only depend on altitude and not change when you walk around at the same altitude (in a simplified model)? Since the earth is mostly spherical, the atmosphere is mostly spherical, so at any point on the earth's surface there is the same total weight of air from the atmosphere on top of you.

I think some of your questions are easier to understand when we remember that pressure is force divided by area. To hold up the weight of the atmosphere, the particles at the surface have to push back. The more they push back, the faster they must be moving on average. Now, the direction of their movement can be in three dimensions, so they not only push up by also push horizontally. The average force contributed by these particles is the same in all directions, which is why the pressure is the same in all directions.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Thanks but what would happen if you take a box and take some air in it and then close it. (all of this is done on ground). Now air inside box is not in contact with the atmospheric air. Would the air pressure in box drop? As now, the air inside box is like a different system. It would have its own air pressure variation with depth. $\endgroup$
    – Eclipse239
    Commented Apr 30, 2021 at 7:11
  • $\begingroup$ In that case, the box is still embedded in the surrounding air. So the surrounding air pushes on the box with atmospheric pressure, pushing the walls of the box inward. For the box to not shrink, the air inside the box must be pushing outward with an equal force. (I'm picturing a box with flimsy walls - more like a balloon. A stronger box can probably withstand a bit more pressure on its own and could have a smaller air pressure inside by some amount. But any tiny hole in the box would make the higher-pressure outside air sneak in, so eventually the pressures would equilibrate.) $\endgroup$
    – Quantum Mechanic
    Commented Apr 30, 2021 at 12:38
  • $\begingroup$ The pressure of the air inside the box is same as the pressure of air outside. Because the air inside the box was taken from the air in atmosphere. The air inside the box isn't pressurised by the 'weight' of air above it. Its pressurised due to interactions between the molecules, which are squished onto each other. This squishing was already there when the air was part of atmosphere (and weight of air above it caused this squishing). Why this squishing is so evenly distributed in all directions is what I dont understand. $\endgroup$
    – Rohit Shekhawat
    Commented Feb 24, 2023 at 4:18
  • 1
    $\begingroup$ @RohitShekhawat the "air" inside the box could just as well be another gas completely unrelated to the air in the atmosphere and these ideas would still hold (consider a balloon filled with helium and bobbing at some equilibrium point ~30km above the Earth's surface). Your "squishing" is caused by the random motion of the molecules inside the box, which have no preference which direction they're going, other than a tiny difference in the gravitational pull at the top vs bottom of the box $\endgroup$
    – Quantum Mechanic
    Commented Feb 24, 2023 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.