Will the Unsecured Shed Roof Withstand the Storm Winds?

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SUMMARY

The discussion centers on the stability of an unsecured shed roof under wind conditions of 20 m/s. The conclusion is that the roof will indeed fly off due to the pressure dynamics involved. The internal pressure of 1.01 x 10^5 Pa, combined with gravitational forces, must exceed the external pressure, which includes the wind pressure calculated using Bernoulli's principle. The correct interpretation of the equation P + 0.5 * density * v^2 + density * g * y reveals that increased wind velocity results in decreased pressure on the roof.

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Homework Statement


Someone is putting a small shed in their backyard. The roof is not nailed down so gravity alone is holding it down. The wind suddenly flows across the top of the roogf at 20 m/s. The air inside the shed is 1.01 x 10^15 Pa, the normal atmospheric pressure. The roof has an area of 16m^2 and a mass of 250 kg. Density of air is 1.29 kg/m^3. Will the roof fly off?


Homework Equations





The Attempt at a Solution



The answer is that the roof will fly off.

In order for this to be true, pressure inside plus gravity must be greater than pressure outside.

Pushing down I have Pressure_up = 1.01 x 10^5 Pa.

I know that Pressure_down = mg/A + P_wind

I know this must be less than 1.01 x 10^5

However, what is the formula for Pressure_wind? I know the faster the wind, the smaller the pressure. I know formulas such as p + .5*density*v^2 + density*g*y, but it seems like a higher velocity would yield a higher pressure with this formula. It shouldn't be that way. What am I doing wrong?
 
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What is "p + .5*density*v^2 + density*g*y" equal to?

If you use the equation correctly, you will find higher velocity does give lower pressure.
 

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