Calculating Outward Force of a Tornado on a Wall

AI Thread Summary
To calculate the outward force on a wall during a tornado, the pressure difference between the inside and outside of the house must be considered. The external pressure of the tornado is given as 0.2 atm, which converts to 20260 Pa, while the atmospheric pressure is 101300 Pa. The area of the wall is 176 m², leading to an initial calculation of 3.56576 x 10^6 N for the force. However, this calculation only accounts for the external pressure; the internal pressure must also be factored in to determine the net outward force. The problem implies that the internal pressure is atmospheric, necessitating a correction for accurate force calculation.
kiwikahuna
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Homework Statement


Air within the funnel of a large tornado may have a pressure of only 0.2 atm. Atmospheric pressure is 101300 Pa. What is the approximate outward force F on a 11m X 16 m wall if a tornado suddenly envelopes the house? Answer in units of N.


Homework Equations


P = F/A


The Attempt at a Solution



I converted 0.2 atm to 20260 Pa. The area is 11m X 16 m = 176 m^2.

F = (20260 Pa) (176 m^2)
F = 3.56576 X 10^6 N

The answer is wrong. What did I overlook?
 
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is there 0 pressure inside the house?
 
The problem doesn't say.
 
kiwikahuna said:
I converted 0.2 atm to 20260 Pa. The area is 11m X 16 m = 176 m^2.

F = (20260 Pa) (176 m^2)
F = 3.56576 X 10^6 N

The answer is wrong. What did I overlook?
For one thing, you only calculated the inward force exerted by the air outside the house. As ice109 says, you forgot that it's the air inside the house that pushes the walls outward. (Find the net force.)
 
kiwikahuna said:
The problem doesn't say.

you have to infer it. if this house is inside the atmosphere is there zero pressure inside the house?
 
Doc Al said:
For one thing, you only calculated the inward force exerted by the air outside the house. As ice109 says, you forgot that it's the air inside the house that pushes the walls outward. (Find the net force.)

cheater...
 
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