Pressure in Airplane: Understanding P=F/A

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    Airplane Pressure
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Homework Help Overview

The discussion revolves around understanding the concept of pressure in the context of an airplane window, specifically applying the formula P=F/A. The original poster is exploring the implications of pressure differences between the inside and outside of the airplane.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the force required to pull the window inward based on pressure differences. Some participants question the feasibility of a passenger exerting sufficient force to open the window and suggest considering the dimensions involved in the calculations.

Discussion Status

Participants are actively engaging with the problem, providing guidance on calculations and discussing the need for unit conversions. There is an ongoing exploration of the relationship between pressure and force, with some calculations being validated by others.

Contextual Notes

There is a mention of atmospheric pressure values and the need to convert units to Pascals for accurate calculations. The discussion includes assumptions about the absence of latches on the window.

gills
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Homework Statement


I know this is probably a very easy question, but I'm new to this and never worked with it before.

PHY131_chap15_42.jpg







The Attempt at a Solution



so you have 0.25atm pressure going into the window from outside, and 0.75atm from inside pushing out.

Am i assuming that with 0.50atm more pressure coming inward from a person would equal the force of pressure pushing out and therefore any more than that would theoretically pull the window inward?

P = F/A
 
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You're on the right track. Figure out the force required to pull the window open. Is it likely that a passenger can exert such a force?
 
Doc Al said:
You're on the right track. Figure out the force required to pull the window open. Is it likely that a passenger can exert such a force?

Well, if I'm doing this right, it seems pathetically small, but I guess this is assuming there are no latches of some sort. Maybe i need to convert my dimensions?

so P = F/A ---> F = PA

so for F to be able to pull the window in, it must be [tex]\geq[/tex] (0.75 atm/(0.50 * 0.90).

If the outside pressure is 0.25atm, then only 0.50 more is required..(?)

so one must apply a force of (0.50atm) * (0.50 *0.90) = 0.225 N

I thinking the dimensions need to be converted?
 
What is atmospheric pressure?
 
at sea level 1 atm = 14.7 psi = 760mm mercury = 101.3 kPa...a ha, i need to use Pascals...?
 
0.50 atm x 101.3kPa = 50.65 kPa

50.65 kPa * (0.50 * 0.90) = 22.8 kN which is a lot for a human being. Looks good?
 
Looks good.
 
to me, that's a dumb questions.

Thanks again Al
 

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