# Wind force

Can someone tell me how to compute the force for a given wind velocity against a solid object per square foot in pounds? For example, 40 mph wind, object 10 square ft...weight per square foot = ? pounds to keep object from being moved by wind.
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You might find <http://timber.ce.wsu.edu/Supplements/WindDesign/PAGE1.asp to be of interest.
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--John
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P = 0.00256V^2 psf, eg 4 psf for V = 40 mph.

With 10 ft^2 facing the wind, the sliding force would be 40 pounds. So if the coefficient of friction with the surface below were 1, it would need to weigh at least 40 pounds to avoid sliding.
If it were (say) 6' tall and symmetrical, the overturning moment would be 40x6/2 = 120 foot-pounds. If it were 4' long in the wind direction and the 6' projection were in the middle, the resisting moment arm would be 4/2 = 2'. If it weighed P pounds, 120 = 2P would make P = 60 pounds min.
Nick
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Thanks, both...I now have the info I needed.
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