Question

The pressure inside a commercial airliner is maintained at 1ATM (1.01x10⁵ N/m²). What is the outward force exerted on a 1mx2m cabin door if the outside pressure (at 10km high) is .3 ATM?

Answer 1

There are two pressures at work here, and you can think of them both as creating their own forces. You're after the combination of forces, or the "resultant" force.

Inside . Outside
-------->|<--
-------->|<--
-------->|<--
-------->|<--
1 atm . 0.3 atm

First, let's convert the pressures to metric units. The metric unit for pressure is a Pascal (Pa), equal to N/m^2. You're given 1.00 atm = 105 Pa, but that's actually not correct. 1 atm = 101,325 Pa, or 101.325 kPa.

So, 0.3 atm * (101 kPa/1.00 atm) = 30.3 kPa.

Now, let's find the force on the inside and outside of the door. Your equation, P = F/A is right. Let's solve it for F.

P = F/A
F = P*A
Fin = Pin * A
Fin = 101 kPa * A
Fin = 101 kN/m^2 * (1 m * 2 m)
Fin = 202 kN

Fout = Pout * A
Fout = 30.3 kPa * A
Fout = 30.3 kN/m^2 * (1 m * 2 m)
Fout = 60.6 kN

So, we have

............|
202 kN..|
---------->|<--- 60.6 kN
............|
............|

The two forces act in opposite directions, so we subtract them. So, the total, or resultant force acting on the door is

F = 202 kN - 60.6 kN
F = 141.4 kN

Now, a kN is 1000 N, so

F = 141400 N