Frictional force on a champagne cork with pressure difference

[Kickbladesama]

Homework Statement

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?

Homework Equations

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

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?

 

[tiny-tim]

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?

Hi Kickbladesama!

erm … mass is kg, not Newtons.

 

[HallsofIvy]

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.

 

[arpanroy]

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