Question

Consider the window (of area 0.71 m2) of an airplane. If the pressure inside the plane is atmospheric pressure, while the pressure outside is 17 % of Patm, find the magnitude of the force on the window. Ignore the velocity of the airplane (i.e., ignore Bernoulli’s effect).

Answer 1

The pressure outside of the window is, 17%P,tm 17 100 Put = 0.17(1 atm ) 0.17 atm The difference in the inside and outside pressures is, -1 atm -0.17 atm (1.013x10 Pa 1 atm 0.83 atm|-- 0.84079 x10 Pa The relation between pressure and force is, Therefore, F (AP)A -(0.84079x10 Pa)(0.71 m2) 5.969 x104 N- 6.0x104 N

Answer 2

SOLUTION :

Inside pressure = 1 atmosphere = 101325 Pa

Outside pressure = 0.17 atmosphere = 0.17 * 101325 = 17225.25 Pa 

These pressures act in opposite directions.

So, net pressure on the window = 101325 - 17225.25 = 84099.75 Pa = 84099.75 N/m^2

So, 

Force on the window 

= Net pressure  in N/m^2  * Area in m^2

= 84099.75 * 0.71  N

= 59710.82 N  = 59711 N approx. (ANSWER).