Question

An aircraft cruises at an altitude at which the external air pressure is 0·40 × 105 Pa. The air pressure inside the aircraft cabin is maintained at 1·0 × 105 Pa. The area of an external cabin door is 2·0 m2. What is the outward force on the door due to the pressure difference?

Answer 1

The net pressure on the door is the the difference between external and internal pressure:

$$p = 1\times 10^5Pa - 0.4\times 10^5Pa = 0.6\times 10^5Pa$$

By definition, the pressure is the force per one square meter. Thus, multiplying the pressure $p$ by the area of the door $A = 2m^2$ , we can find the force:

$$F = pA = 0.6\times 10^5Pa \cdot 2m^2 = 1.2\times 10^5N$$

Answer. $1.2\times 10^5N$.