Air Pressure and Net Force on a Door in a Tornado

AI Thread Summary
In a tornado, a sudden 15 percent drop in atmospheric pressure can exert a significant net force on a door measuring 195 cm by 91 cm. The normal atmospheric pressure at sea level is approximately 1.0 x 10^5 N/m², leading to a reduced pressure of about 8.5 x 10^3 N/m² during the tornado. The formula P=F/A can be used to calculate the force, but attention to units is crucial, as the result must be in Newtons rather than Pascals. The net force is influenced by the pressure difference on either side of the door, with atmospheric pressure pushing from the outside. Clarifying these calculations and ensuring unit consistency is essential for accurate results.
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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.
 
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Never mind.
I just realized that P=F/A would work.
Whoops xP
 
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?
 
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?
 
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