Pressure against atmosphere pressure

[athrun200]

Homework Statement

See problem 3

Homework Equations

The Attempt at a Solution

In the solution, the interval for the integral of pressure is from $P_{0}$ to the required P.

However, I wonder that why we don't use net pressure as the intervals.
i.e. From $P_{0}-P_{1}$ to $P-R$
Where $P_{1}$ is the pressure against atmosphere pressure and $R$ is the reaction force act on liquid by the bottom of the tube.

It seems the interval made by me make more sense than the solution since the interval match the meaning of $dp$ which is net pressure.

 

[Redbelly98]

For the integral

$$\int dp$$

Since the integral contains the term dp, then the variable of integration is p, the pressure. Hence the limits of integration should be values of the pressure.

Also, I'm not sure what you mean by "net" pressure. Do you mean absolute pressure minus atmospheric (what is usually called gauge pressure)?

 

[athrun200]

For the integral

$$\int dp$$

Since the integral contains the term dp, then the variable of integration is p, the pressure. Hence the limits of integration should be values of the pressure.

Also, I'm not sure what you mean by "net" pressure. Do you mean absolute pressure minus atmospheric (what is usually called gauge pressure)?

Since for any mass element of the liquid, there are two pressure acting on them. One it from the left to right, another one is from the right to left.

Take the mass element of the liquid closest to the center as an example.
Atmosphere pressure act from the left to it.
Another pressure act from the rest of the liquid to it in order to provide the force needed for circular motion.(Is it what we call gauge pressure? I can't remember the terms)

So each mass element has 2 pressure acting on it. Why we only choose the one on the left as the interval? Can I still obtain the answer if the pick another one as the interval? If no, why not?

 

[Redbelly98]

Okay, I think I better understand what you are asking.

We do use both pressures, on the left and right sides of the volume element. Letting p be the pressure to the left of the element, then the pressure to the right is $p + dp$. Both of these pressures are used to come up with the net force acting on the volume element.

If you prefer to use p for the pressure to the right, and $p - dp$ for the pressure to the left, then you may do so. Either way, the net force on the volume element is $S \ dp$

 

[Redbelly98]

If you prefer to use p for the pressure to the right, and $p - dp$ for the pressure to the left, then you may do so. Either way, the net force on the volume element is $S \ dp$

Well, I realize I should add something more to this. The pressure p is a function of position x, i.e. it is p(x). So if you use p as the pressure at the right edge of the volume element, then x would be the position at the right edge of the volume element, not the left edge as shown in the solution you posted. The left edge would be located at x-dx in that case.

Hope that addresses your question -- though I am not quite certain that it does?

 

[athrun200]

I think I understand a bit more now, thank you very much!