Air Pressure and Net Force on a Door in a Tornado

[NPM]

Homework Statement

In a tornado, the pressure can be 15 percent below normal atmospheric pressure. Suppose that a tornado occurred outside a door that is 195 cm high and 91 cm wide. What net force would be exerted on the door by a sudden 15 percent drop in normal atmospheric pressure? In what direction would the force be exerted?

Homework Equations

Here lies the problem.
I do no know which formula I should be using.
I know that atmospheric pressure is about 1.0 x 10^5 N/m squared at sea level.
Which I'm inferring is what I would use for the problem.
Could someone post a formula or something?
I think that's all I would need.

 

[NPM]

Never mind.
I just realized that P=F/A would work.
Whoops xP

 

[NPM]

So I tried doing the problem:
P=F/A
Regular pressure = 1.0 x 10^5
15% decrease = 8.5 x 10^3
85000 = F/(195 x 91)
= F/177.45<--- Converted to meters
So I multiplied by the area:
85000 x 177.45=15,083,250 Pa
And that, to me, seems like a completely ridiculous answer.
Would someone care to correct or at least help me?

 

[Redbelly98]

A couple of comments:

1. Remember there is a net force, due 1 full atmosphere of pressure, pushing on the other side of the door.

2. Be careful with the units. It asks for a force, you're answer was in Pa units. Do you see the problem there?