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

AI Thread Summary
The net force on an airplane window at an elevation of 12,500 meters depends on the pressure difference between the inside and outside of the cabin. The outside air pressure at this altitude is significantly lower than standard atmospheric pressure, while the inside is maintained at a higher level for passenger comfort. If the pressures are equal, there is no net force; however, if the inside pressure is greater, a net outward force is exerted on the window. Calculating the exact force requires knowing both the inside and outside pressures. Ultimately, the net force is determined by the pressure differential multiplied by the window's area.
xdarkelf714x
Messages
4
Reaction score
0

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
 
Physics news on Phys.org
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 betwen 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 betwen 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:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top