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Pascals Principle headache

  1. Sep 29, 2007 #1
    Citations to read:
    http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html#pp (Pascals Principle)
    http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html#hpress (Hydraulic Press)
    http://hyperphysics.phy-astr.gsu.edu/hbase/pflu.html#fp (Static Fluid Pressure)

    I find the Hyperphysics pages much easier to understand, so apologies if you wanted Wikipedia citations.

    My conceptual problems
    Mathematically pascals principle is easy to work with, I dont have much of a problem there, also the hydraulic press principle mathematics is fine by me.

    The main problem is the actual principle itself, I can accept it, but find difficulties, on the "Static Fluid Pressure" citation above one of the diagrams has a small narrow tube exherting a pressure pgh on the large water resivoir below it, this causes an increase in pressure that would be the same with any size of tube above it as pgh is independant of diameter.

    My main problem with it is how the pressure exherted by a narrow stopper, or narrow tube also cause the large basin or bottle below it to have the same Increased pressure, its a purely conceptual problem for me, I'm doing small diagrams of the same thing (narrow tube,with large resivoir below it) and I calculate the downward force of the narrow tube acting on the large resivoir as say, 98N for a 1cm C.S. Area 1m High tube, but the overall force of the fluid below it (lets say, with 10 times the area) would be 980N.

    Could somone please help put it into a nice perspective for me, I know it seems silly, but I refuse to commit anything to long term memory until I'm 100% certain of any questions that I may be asked about it (I tutor lower year students/pupils, and it seems like a pretty valid question, an anacdote to explain it to them would be wonderful, but if I personally cant visualise it, I'm skrewed).

    One of the ways I thought of it, would be like getting a similar shaped steel piston, and exherting a force of 98N down the shaft with my arms, then it would end up exherting a force of 980N on the large side (clearly wouldnt happen) which is where my conceptual confusion comes from, that such a small initial force can make labiathon sized forces in theory.
  2. jcsd
  3. Sep 29, 2007 #2
    Also another conceptual problem is that the extra pressure exherted on the bottom of the fluid would surely increase it's weight massively? This is an easily dispellable argument probably, but as from the previous example that 98N over 1Cm^2 would be == 980N less downward force on the surface its on, giving the effect of an increased weight or what??

    I'm really poor at physics style problems (Chemist here :) )
  4. Sep 29, 2007 #3

    Doc Al

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    I much prefer Hyperphysics whenever possible.

    You have to look at all the forces acting on the fluid. In particular, the top surface of the reservoir pushes down onto the fluid. So, yes, the pressure of the fluid in the small tube gets added to the fluid in the reservoir, but it's not just the weight of that small column that's doing the pushing.

    Ah, but it would exert such a large force on the bottom of the reservoir. Going back to your fluid column example, to support that 98 N column of fluid, the pressure at the top of the reservoir must increase to 98 N/cm^2. And that does mean that the total force on the bottom of the reservoir would increase by 980 N.

    Realize that the additional force of 980 N is exerted on the inside surface of the reservoir bottom, which pushes back with an equal force. But these are all internal forces. If you put the entire reservoir + column onto a scale, the scale would simply read the total weight of fluid + container. But there really is an increased force exerted on the sides of the reservoir--make it out of fragile material that cannot withstand the pressure and that thin column of fluid will cause it to break. Kind of cool!
  5. Sep 29, 2007 #4
    Thanks alot for your lengthy reply, the additional force being internal makes alot of sense now, I just wanted to clarify that at the end of my post, the most helpful part of your reply however was the concept that the larger bottom resivoir is pushing back on my small narrow tube by 98N which sort of made it "Click" in my head.

    But just to clarify, as I love to be 100% sure on these things, that because the larger resivoir is now having to push back with a pressure of "x" the pressure of the entire thing in all directions is now "x", it acts as a sort of reactive force, pressing against my narrow tube in resistance.

    So I guess the best way to explain it is that this new increased pressure of "X" isnt just a downward pressure force, but a pressure force in all directions and is a result of the downward pressure of the column, that is too say a pressure that was once was the source of gravity acting on mass, is now a pressure being created in all directions, to counteract the gravitational pressure created in the first place, and it makes sense, because if the pressure wasnt uniform, there would be a flow of fluid from the high -> low pressure areas in order to equalise the pressure, so the pressure acting upwards on the narrow column MUST be "X", otherwise the downward pressure would prevail, and give rise to a motion downwards?

    And the overall weight of the vessel also is unnaffected for the same reason, since the internal forces pushing outwards are being perfectly balanced out by the vessel itself, it results in no net force, and therefore no net movement or breakage or whatever, so therefore, it has no effect at that point?

    Sorry if I'm musing here, but I just love when science clicks, unless it didnt click, and I sound like some sort of pshycopath looking for his next physics fix.
  6. Sep 30, 2007 #5
    Bump for DocAl's attention or any PF mentor
  7. Sep 30, 2007 #6

    Doc Al

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    I think you've got it. In fluids, and this is the point of Pascal's principle, the pressure acts in all directions.

    Right. All the interactions of vessel with fluid are internal forces within the "vessel+fluid" system, so they cancel out. The weight of the system just depends on its total mass.

    No need to explain yourself in this crowd. :rofl:
  8. Sep 30, 2007 #7
    Ah thanks alot, its very clear now, seems idiotic that I didnt get it in the first place :), also this increased pressure is persistant completely below the narrow column, i'ts not like the column is making everything below it pressure "X" and everything above it, that doesnt make sense, the increased pressure is cumilative and is equal to the depth :)

    Another anaolgy i used to convince myself is by putting a straw in a glass of water, and attempting to make a "Column" of water with the straw, although the water would technically produce a downward pressure of "X" its immediately removed, as the surface of the "Large vessel" (the glass of water) will push upwards with this increased pressure and "Fill" the glass, so the anaology only works if a narrow column is connected to a sealed vessel in which the upward presssure created acts on the vessel rather than air.

    The reason I pulled up this pascals principle stuff was due to the origional design of a barometer, I understood barometers with "U" shaped tubes, where the tube is perfectly cylindrical, so the Atm pressure would create a force F on the mercury, pushing it upwards, but it confused me when the diagrams had an "Open" mercury resivoir below it, and I thought the column would be higher if the resivour had a higher surface area (meaning a net increase of "Force" from the atmosphere).

    The pascals principle therefore states that, although the pressure exherted by the atmosphere is being perfectly counteracted by the mercury pushing back on it (meaning the mercury in the resivor doesnt move) the vaccum column cant push back against mercurys fluid pressure as there isnt an atmosphere within the column to press back?

    EDIT: I think the main confusion was that I was thinking of treating this Resivor/Column approach as an empty vessel containing just air, then pressing a piston down the narrow column, the pressure would increase of course, but this wouldnt be equal to the weight of the piston acting downward, this could be because: dP = nRT/dV, I guess its the uncompressability of liquids that origionally messed me up on this train of thought
    Last edited: Sep 30, 2007
  9. Sep 30, 2007 #8

    Doc Al

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    Exactly. Since the space above the mercury column is vacuum, and since the pressure at any level in a static continuous fluid must be the same (else it would move), the height of the mercury column above the reservoir level is a measure of atmospheric pressure.
  10. Sep 30, 2007 #9
    Mathematical View on it

    Cool, I'm fully convinced :D, but a mathematical analogy is always nice to throw in for, say, somone I might in future tutor, who's having mental problems with the analogy, so how about I used a basic ideal gas layout for the analogy, keep with me on this one though :), I'll lay out the maths here.

    Lets say we have a small tank, the tank has a narrow column protuding out the top of it, the entire thing is air tight, apart from a hole at the top of the afore mentioned narrow column, and the entire apparatus only contains atmospheric air as it hasnt been filled with anything.

    Now to fill in some numbers, lets disreguard the wall thickness of the narrow tube, and say that its internal cross sectional area is 1cm^2, it is 30m high (It's a big column) and the tank at the bottom is also small, lets say its 2cm high by 5cm long and 5cm wide, so 50cm^3 for the sake of argument, this is so the gas compresses and the force balances out for the next part of the analogy, if the tank was massive, then the change in pressure would be small.

    Now we have all the paramaters for the apparatus, one thing is missing, were going to slide a 10 gramme disc that fits perfectly down the narrow tube down it, it will fall due to gravity and will eventually hit an equilibrium point somewhere down the narrow tube where the force exherted on the bottom of the disc cancels out the force exherted by the discs gravity.

    Now for the interesting part, let's start throwing numbers at it, we first convert all the values into more science friendly S.I. Values (meters, pascals and so on) to find:

    [tex] V_{tank} = 5*10^{-5}m^3[/tex]
    [tex] V_{column} = \left[(1*10^{-4})*30\right] = 3*10^{-3}m^3 [/tex]
    [tex] V_{total} = (5*10^{-5})+(3*10^{-3}) = 3.05*10^{-3}m^3 [/tex]​

    From the total volume, we can find the number of moles of gas there is inside the apparatus while it is at atmospheric temperature (I'll use 100kPa to not come out with wierd numbers rather than 1atm (101kPa or whatever it is...)).

    [tex] 3.05*10^{-3} * 1*10^{5} = n * 8.31 * 298 [/tex]
    [tex] n = 0.12316mol[/tex]​
    It doesn't matter what molecules the gas is made of for this particular calculation, now we find the force exherted by the 10g disc from gravity is 0.098N, this force is exherted across 1cm^2 so the pressure below the disc should be 980Pa greater than above the disc in order to suspend it, so:

    [tex] V = \frac{nRT}{P} = \frac{0.12316*8.31*298}{980+1*10^{5}} = 3.02*10m^{-3}[/tex]​

    Volume change = New minus Original so:
    [tex] dV = 3.02*10^{-3}-3.05*10^{-3} = -0.03*10^{-3}m^{3} [/tex]​

    Distance down the pipe before equilibrium: dV = h*A, so 0.03*10^-3/0.0001 = 0.3m down the tube.

    The interesting conclusion is that a 10g disc exherting a force of only 0.098N on the apparatus has effectively risen the pressure by around 980Pa all over it, the 0.098N exhertion of the disc has increased the force produced by the gas by 980 N/M^3 which gives the effect of a multipication of force :).
  11. Oct 5, 2007 #10

    Doc Al

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    I like it! Good example. (My only comment would be pedagogical. I would stick to ratios--Boyle's Law--and not bother with R, T, or calculating n; just use enough detail to make your point.)
  12. Oct 7, 2007 #11
    Ah crap, I always forget how to use the ratios, is it if the initial pressure is say, 100KPa and goes to 100.980Kpa the equation would be:

    [tex] V_{Ratio} = \frac{1}{100.98/100}[/tex]

    [tex]V_{New} = V_{Old} * V_{Ratio}[/tex]
  13. Oct 7, 2007 #12

    Doc Al

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    I'd use Boyle's law:

    [tex]P_1 V_1 = P_2V_2[/tex]

    Which becomes:

    [tex]\frac{P_1}{P_2} = \frac{V_2}{V_1}[/tex]
  14. Oct 7, 2007 #13
    Oh, that old badger :D, Yeah even in this stage of university chemistry, i still never use those ¬_¬, arent they like:

    PV = Constant at constant n, t

    P = 1/V at constant n t
    V = 1/P at constant n t

    Avvagadros Principle
    PV (Proportional) to n, at constant T
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