Pressure against atmosphere pressure

In summary, the author suggests that we use both pressures on the left and right sides of the volume element when calculating the net force on it. This allows us to more accurately calculate the pressure.
  • #1
athrun200
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Homework Statement


See problem 3


Homework Equations





The Attempt at a Solution


In the solution, the interval for the integral of pressure is from [itex]P_{0}[/itex] to the required P.

However, I wonder that why we don't use net pressure as the intervals.
i.e. From [itex]P_{0}-P_{1}[/itex] to [itex]P-R[/itex]
Where [itex]P_{1}[/itex] is the pressure against atmosphere pressure and [itex]R[/itex] 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 [itex]dp[/itex] which is net pressure.
 

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  • #2


For the integral

[tex]\int dp[/tex]

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)?
 
  • #3


Redbelly98 said:
For the integral

[tex]\int dp[/tex]

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?
 
  • #4


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 [itex]p + dp[/itex]. 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 [itex]p - dp[/itex] for the pressure to the left, then you may do so. Either way, the net force on the volume element is [itex]S \ dp[/itex]
 
  • #5


Redbelly98 said:
If you prefer to use p for the pressure to the right, and [itex]p - dp[/itex] for the pressure to the left, then you may do so. Either way, the net force on the volume element is [itex]S \ dp[/itex]
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?
 
  • #6


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

1. What is atmospheric pressure?

Atmospheric pressure is the force exerted by the weight of the air in the Earth's atmosphere. It is typically measured in units of pressure, such as pounds per square inch (psi) or millibars.

2. How does pressure against atmospheric pressure affect weather?

The difference in pressure between areas of high and low atmospheric pressure is what causes wind to blow and ultimately influences weather patterns. High pressure areas typically bring clear skies and fair weather, while low pressure areas often bring storms and precipitation.

3. How does pressure against atmospheric pressure affect the human body?

Our bodies are adapted to function at normal atmospheric pressure, and changes in pressure can affect our health. For example, ascending to high altitudes can cause a decrease in atmospheric pressure, which can lead to altitude sickness. On the other hand, diving to great depths can cause an increase in pressure, which can lead to decompression sickness.

4. What are some devices used to measure pressure against atmospheric pressure?

Some common devices used to measure pressure against atmospheric pressure include barometers, manometers, and aneroid gauges. These devices use different mechanisms to measure and display pressure, and they are often calibrated in different units.

5. How is pressure against atmospheric pressure related to the gas laws?

According to the ideal gas law, the pressure of a gas is directly proportional to its temperature and the number of gas molecules present, and inversely proportional to the volume of the container. This means that as atmospheric pressure changes, the behavior of gases will also change accordingly.

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