Calculating Pressure Inside Airtight Container

[jinman]

Homework Statement

A Partially evacuated airtight container has a tight-fitting lid of surface area 77m^2 and negligible mass. If the force required to remove the lid is 480N and the atmospheric pressure is 1x10^5 Pa, what is the air pressure inside the container?

Homework Equations

p=F/A

The Attempt at a Solution

p=F/A
p=480N/77m^2

p=6.23Pa

This did not make sense to me, so i thought maybe it was the difference in pressures:

Po-Pd=Pf
1x10^5-6.23Pa=99993.78Pa??

I don't think that is right either. Where I am going wrong?

 

[oedipa maas]

It always helps to start off by balancing forces. There is a downward force on the lid due to the motion of the particles in the atmosphere (F_A). There is an upward force on the lid due to the motions of the particles in the container (F_C). And, finally there is a normal force exerted by the walls of the container on the lid - also in the upward direction (F_N). Maybe you can imagine this more easily by covering the container with a flat metal sheet instead of with a screw-top lid.

In the static case where you aren't doing anything to the container the forces are balanced:
F_C + F_N - F_A = 0

In order to remove the lid you need to apply an additional upward force (F_you)

If F_you is too small, you will only decrease the size of the normal force that the container walls have to exert to support the atmosphere pressing down on the lid: F_N (F_N goes to F'_N):

F_C + F_you + F'_N - F_A = 0

What happens when F_you is big enough when F'_N goes to 0?

 

[jinman]

Would the lid be removed at that point because F_you cancels out F_N?

 

[jinman]

I'm lost on this. Any other suggestions?

 

[Redbelly98]

I think jinman's method is correct, but the input numbers as given are wrong somewhere.

77m^2 is an awfuly large size for a container lid. Reread the question in the book (or wherever you got it from) to see if that figure should really be something else.