The door and the seals on an aircraft are subject to a tremendous amount of force during flight...?

Question

At an altitude of 10000m (about 33000 ft.), the air pressure outside the airplane is only 2.7 x 10 to the 4th power N/m squared, while inside is still at normal atmospheric pressure, due to pressurization of the cabin. Calculate the net force due to the air pressures on a door of area 3.0 m squared. If you can show me how you got the answer so I understand it I'd appreciate it.. Thank you

Answer 1

Force = pressure × area

The force on the door from the outside is:

Foutside = pressure × area
= (2.7×10^4 N/m²)(3.0m²)

The force on the door from the inside is:

Finside = pressure × area
= (1.013×10^5 N/m²)(3.0m²)

(Note: 1 atmosphere = 1.013×10^5 N/m².)

The net force is the difference between the inside force and the outside force:

Fnet = Finside−Foutside
= (1.013×10^5 N/m² − 2.7×10^4 N/m²)(3.0m²)

Answer 2

Outside pressure is 27,000 N/m^2
Inside pressure is 100,000 N/m^2
The difference is 73,000 Newtons per square meter.
The door has an area of 3 square meters.
The total force is 219,000 Newtons.

All the measurements in the problem were given in scientific notation with 2-digit accuracy, so we should do the same with our answer.

2.2 x 10^5 N