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*per se*, but it is a conceptual difficulty I am having with my physics 1 course, in which we are studying fluid mechanics (moderators please move this post if there is a more appropriate subforum).

## Homework Statement

I was going over the derivation of Torricelli's law using Bernoulli's equation as in my text (Serway & Jewett) and online resources. Just a recap as I understand it: In the diagram below, let our origin be at the hole on the side of the container, so the height at the hole is 0 and the height at the surface is [itex]h[/itex]. Using Bernoulli's equation, with point 1 being at the surface of the water and point 2 being at the hole a distance [itex]h[/itex] down: [itex]P_1+\tfrac{1}{2}\rho v_1^2+\rho gh=P_2+\frac{1}{2}\rho v_2^2\Rightarrow v_2^2-v_1^2=2gh+\frac{2(P_1-P_2)}{\rho}[/itex].http://upload.wikimedia.org/wikipedia/commons/5/5b/TorricelliLaw.svg

Now, since the area at the top is much larger than the area of the hole, we can simply set [itex]v_1=0[/itex]. We also set [itex]P_1=P_{atm}[/itex] the atmospheric pressure.

So far, this makes sense to me. Where I have a hangup is in setting [itex]P_2=P_{atm}[/itex] as well. I think this has something to do with static pressure vs. dynamic pressure, but the definition of pressure as I understand it is a force divided by a cross sectional area. So at the interface of the hole (assuming the hole has area

*A*) I understand that there is a force of magnitude [itex]F=P_{atm}A[/itex] pushing to the left on the water. But is there not also a force from the water pressure pushing to the right on a sliver of water at the interface of the hole? Shouldn't [itex]P_2[/itex] be the difference of these pressures?

## Homework Equations

[itex]P_1+\frac{1}{2}\rho v_1^2+\rho g z_1=P_2+\frac{1}{2}\rho v_2^2+\rho g z_2[/itex]

## The Attempt at a Solution

So, if I concede that [itex]P_2[/itex] should be the same as [itex]P_1[/itex], then it should apply to other problems whose solution is known without Bernoulli's equation. The simplest problem of this type I can think of is just a pipe, open at both ends, containing a column of water of height [itex]h[/itex]. I know from the first part of the semster, that the top of the water will fall at the same speed as the bottom of the water, so [itex]v_1=v_2[/itex]. Then our equation becomes [itex]v_2^2-v_1^2=2gh+\frac{2(P_1-P_2)}{\rho}\Rightarrow 0=2gh+\frac{2(0)}{\rho}\Rightarrow 2gh=0[/itex]. How should I explain this contradiction?