The door and the seals on an aircraft are subject to a tremendous amount of force during flight. At an altitude of 10,000 m (about 33,000 ft), the air pressure outside the airplane is only 2.7 x 104 N/m2, while the inside is still at normal atmospheric pressure, due to pressurization of the cabin. Calculate the net force due to the air pressure on a door of area 3.0 m2
To calculate the net force due to air pressure on the door of an aircraft, you can use the formula:
Net Force = Pressure x Area
First, we need to determine the pressure difference between the inside and outside of the aircraft. The pressure difference is equal to the external pressure subtracted from the internal pressure.
Given:
External pressure (P_external) = 2.7 x 10^4 N/m^2
Internal pressure (P_internal) = Normal atmospheric pressure
The normal atmospheric pressure at sea level is approximately 1.013 x 10^5 N/m^2. However, since the inside of the cabin is pressurized to normal atmospheric pressure, the internal pressure is also 1.013 x 10^5 N/m^2.
Therefore, the pressure difference is:
ΔP = P_internal - P_external
= 1.013 x 10^5 N/m^2 - 2.7 x 10^4 N/m^2
= 7.43 x 10^4 N/m^2
Now that we have the pressure difference, we can calculate the net force on the door.
Net Force = Pressure x Area
= ΔP x A
Given:
Area of the door (A) = 3.0 m^2
Pressure difference (ΔP) = 7.43 x 10^4 N/m^2
Net Force = 7.43 x 10^4 N/m^2 x 3.0 m^2
Net Force = 2.23 x 10^5 N
Therefore, the net force due to air pressure on the door of the aircraft is 2.23 x 10^5 N.