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Pressure + Friction

  1. Nov 17, 2006 #1
    Here's my homework question....

    The pressure inside a bottle of champagne is 4.5 atm higher than the air pressure outside. The neck of the bottle has an inner radius of 1.0 cm (.01 m). What is the frictional force on the cork due to the neck of the bottle that keeps the cork in the bottle?

    1 atm = 101,325 pascles
     
  2. jcsd
  3. Nov 17, 2006 #2
    What have you done on this? do you have a free body diagram for this? do you have a realtionship between force and pressure? Do you have any ideas on why they gave you a radius?
     
  4. Nov 17, 2006 #3
    Force = Pressure x Area.
     
  5. Nov 17, 2006 #4
    Let me see if my conversions are correct first. I assumed the air outside is equal to 1 atm. soo

    Pressure inside the bottle (Pbottle) = (4.5)(101,325) = 455962.5 Pa
    Pressure outside the bottle (Pair) = (1)(101,325) = 101.325 Pa

    If my conversion is wrong, please correct me

    And, since I'm dealing with fluids, I assumed the radius was given to me so that I could plug numbers into Bernouli's equation [(P1-P2)V] or plug numbers into the Continuity equation (A1V1 = A2V2). I first looked at Bernouli's equation since my problem deals with pressure, but I'm not given a velocity. Any hint which equation I should use?

    As for the relationship between force and pressure, I know Pressure = Force/Area. If my conversion above for the Pbottle is correct, I can figure out that Force = Pressure*Area. So the force that the inside of the champagne has on the cork is F = 455962.5*(pi)(.01^2) = 143.24 N. This is how I tried the problem the first time, thinking the friction of the cork had to be equal to or greater than the force being pushed up by my champagne. But that answer seemed to easy to get. More suggestions would be helpful. Thanks!
     
  6. Nov 17, 2006 #5
    Venturing into dealing with velocities shouldn't be necessary. There are basically two forces in your system; the force F1 exterted downwards on the cork due to the outside pressure, and the force F2 exerted upwards due to the internal pressure.

    In order for the cork to stay in place, the frictional force must overcome the net of these two forces, e.g.

    Ff >= F2 - F1.
     
  7. Nov 17, 2006 #6
    But watch the numbers you are using for the pressure, as you state:
    .
    wouldn't that be 5.5 atm then?
     
  8. Nov 17, 2006 #7
    Yeah I saw the same thing...isnt that 5.5atm?
    And yes it's only a force problem.
     
  9. Nov 27, 2006 #8
    it's simple u just hav to draw it u'll find the force of atm outside the bottle and also the friction force directing downwards and lets call the force inside the bottle dueto the pressure "F3" so F1+F2=F3
    where F1 is the atm force
    and F2 is the friction force
    and since F1 and F3 are forces due to pressure then get them from F=P*A
    P:pressure
    A:area

    and about the pressure inside the bottle it's a gauge pressure and outside pressure is given in absolute pressure so as we all know we can't put in the same law different units so u'll have to convert one of them to the other; by the way the atm in gauge is zero pa and in absolute it's 101,325 pa
    and the conversions is as follows
    absolute pressure = gauge pressure + 101,325pa
     
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