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Pressure in a bottle

  1. Oct 14, 2008 #1
    1. The problem statement, all variables and given/known data
    The pressure inside a bottle of champagne is 5.4 atm higher than the air pressure outside. The neck of the bottle has an inner radius of 0.9 cm. What is the frictional force on the cork due to the neck of the bottle?

    2. Relevant equations
    pi x r ^(2) = area of circle
    P = Mass/Area

    3. The attempt at a solution
    I first converted 5.4 atm to SI units (18pa). I did area = pi x 0.09m^(2). I took that and multiplied that by pressure 18pa and found the mass which is 0.477N. I did Fnet = mu x Fnormal.

    18 = mu x 0.477N = 37.7 which is incorrect, anyone have an idea?
  2. jcsd
  3. Oct 15, 2008 #2


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    Hi Kickbladesama! :smile:

    erm … mass is kg, not newtons. :wink:
  4. Oct 15, 2008 #3


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    pressure is NOT "mass divided by area", it is FORCE divided by area. So force is pressure times area.

    Since you converted the pressure from atmospheres to Pascals, You must remember that one Pascal is one Newton per square METER. And 0.9 cm is is 0.009 m, not 0.09 m. Frankly, 0.9 cm seems much too small for a champagn bottle neck but then 9 cm (which is 0.09 m) is much too large.
  5. Oct 15, 2008 #4
    firstly your conversion of atm to pascals is wrong....cos 5.4 atm = 547 155 pascals

    secondly p= force/area
    area = 2.54X10^-4 m2

    thus force = area X pressure = 139 N
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