How Is the Force Distributed in an Evacuated Chamber with a Panel Cover?

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Homework Help Overview

The discussion revolves around a physics problem involving an evacuated chamber with a panel cover. The original poster is attempting to calculate the force required to remove the panel, given the pressure inside the chamber and the dimensions of the hole and panel.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster describes their method of calculating forces acting on the panel, including the atmospheric pressure and the pressure from the chamber. They express uncertainty about their calculations and seek clarification on the forces involved.

Discussion Status

Some participants affirm the original poster's approach, while others raise questions about the dimensions provided, indicating potential confusion regarding the setup. There is an ongoing exploration of the assumptions related to the area of force application.

Contextual Notes

Participants note discrepancies in the dimensions of the panel and hole, which may affect the calculations. The original poster also mentions a requirement for significant digits in their final answer.

Arctangent
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Fluids -- evacuated chambers

I've worked on this problem in a few different ways. Maybe I'm having trouble with the direction of my forces:

We have a
an evacuated chamber
the pressure inside of the chamber is 23.0 kPa
a 5cm diameter hole covered by a
0.225m diameter circular panel (mass is negligable)

And the question asks for the force required to pull off the panel.

My first thinking was that because the "vacuum" is being caused by the pressure of the atmosphere,
I'm calling it Fair,
is the force pointing down on the panel
The force of the air inside the chamber is pushing up through the 5cm diameter hole on a small section of panel
Fc,
and the pulling force required to take off the panel is upwards,
Fp.


So,
Fnet = Fp + Fc - Fair = 0
Fp = Fair - Fc

Fc = PA = 23000*pi*r^2 = 45.16 N
where r is the radius of the hole in the chamber

Fair = 101300*pi*R^2 = 4028 N
where R is the radius of the panel

And then I'd solve for Fp. But my answer -- 3982.61 N -- was incorrect.

I'm obviously missing something critical. Would you have any ideas of what I'm doing wrong, and maybe point me in the right direction?

Thanks!
 
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Your approach is correct. I get 4026 N for [itex]F_{air}[/itex] which gives 3981 N.

AM
 
"a 5cm diameter hole covered by a
0.225m diameter circular panel (mass is negligable"

I am trying to visualize this. A 5 cm hole covered by a 2.25 cm panel? What am I missing?
 
Gamma said:
"a 5cm diameter hole covered by a
0.225m diameter circular panel (mass is negligable"

I am trying to visualize this. A 5 cm hole covered by a 2.25 cm panel? What am I missing?

.225m = 22.5 cm

.05m = 5.0 cm

just mixed up the conversion a little, that's all.
 
Andrew Mason said:
Your approach is correct. I get 4026 N for [itex]F_{air}[/itex] which gives 3981 N.

AM

Are you sure? I think I have something wrong here. My final answer is supposed to be in 3 significant digits.

Would the air from the chamber only be pushing up to that 5 cm diameter area or would it be pushing up on the whole panel?
 

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