Understanding the Direction of Pressure in Air and Water Environments

  • Thread starter Thread starter False Prophet
  • Start date Start date
  • Tags Tags
    Direction Pressure
AI Thread Summary
When a person swims out of a submarine at the bottom of the ocean, they experience crushing pressure from all sides, not just from above. This is due to the nature of fluid pressure, which is exerted equally in all directions. The pressure increases with depth, meaning the water below exerts a greater force than the water above. Consequently, while the overall pressure is uniform, the net effect includes a stronger upward force from the water beneath, which supports the weight of the water column above.
False Prophet
Messages
85
Reaction score
0
If you're on a submarine at the bottom of the ocean and you swim out, do you get crushed from all around, or just from the top? I read in a couple places that the pressure is exerted from all directions, but I thought that the pressure is like the "weight" of all the fluid stacked above you. I don't know; please advise.
 
Physics news on Phys.org
There is also the weight of the columns of water above the water beside you. The water on your left and on your right will try to move out from under the columns above them by crushing you.
 
False Prophet said:
If you're on a submarine at the bottom of the ocean and you swim out, do you get crushed from all around, or just from the top?
You'll get crushed from all sides. But since the pressure is a bit higher underneath you, you will be pushed more by the water underneath you. The water pressure, in addition to crushing you, will exert a net upward force on you.
 
False Prophet said:
If you're on a submarine at the bottom of the ocean and you swim out, do you get crushed from all around, or just from the top? I read in a couple places that the pressure is exerted from all directions, but I thought that the pressure is like the "weight" of all the fluid stacked above you. I don't know; please advise.

Yes, this confused me too, but remember that the water below you has to support this "column weight". So as Doc Al said, the pressure below you is even higher than the pressure above.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top