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

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In an evacuated chamber with a 5 cm diameter hole covered by a 0.225 m diameter panel, the force required to remove the panel is influenced by atmospheric pressure acting on the panel and the pressure inside the chamber. The calculations show that the upward force from the air inside the chamber is 45.16 N, while the downward force from atmospheric pressure on the entire panel is 4028 N. The net force equation indicates that the pulling force required to remove the panel is derived from the difference between these forces. There was a confusion regarding the dimensions of the panel and the hole, which was clarified as the panel being 22.5 cm in diameter. The discussion emphasizes the importance of correctly visualizing the forces acting on the panel to arrive at the accurate pulling force.
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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 F_{air} 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 F_{air} 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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