What is the net force on an airplane window at an elevation of 12500 m?

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Homework Help Overview

The problem involves calculating the net force on an airplane window at an elevation of 12500 m, given the dimensions of the window and some pressure values. The subject area includes fluid dynamics and pressure differentials in aviation contexts.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply pressure equations to find the net force, considering both the top and bottom pressures on the window. Some participants question the completeness of the information provided, particularly regarding outside air pressure at the specified elevation and the internal cabin pressure.

Discussion Status

The discussion is ongoing, with participants exploring different assumptions about the pressures involved. Some guidance has been offered regarding the need to consider the pressure difference between the inside and outside of the airplane, as well as the implications of assuming standard atmospheric conditions.

Contextual Notes

There is a noted lack of information about the outside air pressure at 12500 m and the internal cabin pressure, which are crucial for determining the net force on the window. Participants are also discussing the implications of assuming standard conditions versus an unpressurized aircraft.

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Homework Statement


An airplane is flying at an elevation of 12500 m. The window on an airplane is a square of length 0.115 m on each side. What is the net force on the window?


Homework Equations


P = Po + pgh
p = 1.26 kg/m^3
Pair = 101.3 kPa


The Attempt at a Solution


I thought about using the elevation in the first equation, but I remembered that it applied to the displaced air so I used the height of the window. I had the force from the air pressure on the top of the window and the force from the air pressure on the bottom of the window.

P = 101300 Pa + (1.26 kg/m^3)(9.8 m/s^2)(-0.115 m)
F = P * A
F = 101299 Pa * (0.115 m)^2
F = 1340 N downwards
 
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You seem to have missing information. You need to know the pressure of the outside air at 12500m, as well as the pressure of the air inside the plane (which is pressurized to a certain level for the comfort and survival of the passengers). If the pressure inside and outside the cabin is the same, there is no net force on the window. it's the difference in pressure that causes the net force. Also, you don't have to compute the small difference in pressure between the top and bottom of the window. Also, check your direction of the net force.
 
Unfortunately, that was the only information I got for the quiz.
PhanthomJay said:
You seem to have missing information. You need to know the pressure of the outside air at 12500m, as well as the pressure of the air inside the plane (which is pressurized to a certain level for the comfort and survival of the passengers). If the pressure inside and outside the cabin is the same, there is no net force on the window. it's the difference in pressure that causes the net force. Also, you don't have to compute the small difference in pressure between the top and bottom of the window. Also, check your direction of the net force.
 
Well then i don't know what you are supposed to assume, perhaps the inside of the plane at standard sea level atmospheric pressure? I'm not much into the metric system, but in USA units, the outside air pressure at 12500m (40000 feet or so) might be about only 2psi outside the plane, but maintained at 10 psi inside the plane, resulting in a net outward pressure of 8psi, and a net outward force (directed perpendicular to the window pointed away from the plane) of 8psi times the area of the window.
Or maybe you're supposed to assume the aircraft is unpressurized, in which case I already hinted at the answer.
 
Last edited:

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