# What's the hydrostatic pressure in torricelli's experiment?

I've been studying fluid statics, and im getting stuck all the time trying to understand things that seem contradictory for example  the hydrostatic paradox or how a small force can be magnified in Pascal's law.

Anyway, i wanted to ask about the hydrostatic pressure in Torricelli's experiment at container 2 compared to the hydrostatic pressure at container 1

First things first, when a container is open to the atmosphere,then the atmospheric pressure is the same throughout all the container, right? For example at container 1, point 4 has the same atmospheric pressure as points  3,  2 and 1, not the same hydrostatic pressure though.

So what about the atmospheric pressure at container 2 ? Is the atmospheric pressure the same for all the points:  4'  ,  3'  ,  2'  and  1'  ?

If that's true then what is the total pressure (hydrostatic + atmospheric pressure) in each point in the 2nd container?

I'm thinking that maybe the hydrostatic pressure created by the 10,3 meter column in the second container does not spread, does not apply to the whole container. On the other hand, in the first container the hydrostatic pressure created by its 10,3 meter column does indeed spread throughout, points 2 and 1 feel the same pressure.

So what am I thinking wrong?

when a container is open to the atmosphere,then the atmospheric pressure is the same throughout all the container, right?

When a container is open to the atmosphere, then atmospheric pressure is being applied to the surfaces adjacent to the atmosphere. So that pressure is being applied at P4 and P2'.

The fluid will react to that pressure and distribute it throughout the fluid. But I would not say "atmospheric pressure is the same throughout the container". That seems a confusing way to put it.

I'm thinking that maybe the hydrostatic pressure created by the 10,3 meter column in the second container does not spread, does not apply to the whole container.

A (static) fluid cannot resist a pressure in some direction but not others. The pressure for all points at the same height in the fluid must be the same. So the total pressure on P1' and P2' must be identical. If they were not, water would move until it became true.

You can find the pressure at any point by finding a point with a known pressure. P4 and P2' are good ones because they are both at atmospheric pressure. Then from your reference point, you can add or subtract the height of fluid to reach the new point. Lower points have greater pressure, higher points have less pressure.

• Thank you for taking the time to reply🥹 So what whould you say is the pressure at points 1' 2' 3' and 4' Commented Feb 15 at 10:07
• Did you see the final paragraph? P2' is open to the atmosphere, so must have pressure = atm. All the other points you can find the pressure by subtracting $\rho gh$ for the given height difference from the reference point. You're not looking at height below the top of the column (as you would for the first diagram). You're looking at height above P2' (since that's your known pressure). Commented Feb 15 at 10:18
• When i do that dor example for point 4' I find the pressure to be 0, but If the atmospheric pressure is being distributed throughout the container shouldn't it be at least Patm Commented Feb 15 at 10:42
• No, there is no atmosphere at P4'. The diagram in fact is labelled "vacuum" at that point. So the pressure is approximately zero. As I mentioned, trying to say that atmospheric pressure is distributed through the container is not useful and probably confusing. Commented Feb 15 at 17:28
• Is it confusing or is it wrong? And why? When we calculated in simple examples the hydrostatic pressure of an open container, for various depths, we said it's Patm plus whatever the hydrostatic pressure was. For any point we took the same Patm and added ρgh. Why wasn't that confusing? Commented Feb 15 at 17:42