A wind with speed 50 ~m / s blows parallel to the roof of a house. The area of the roof is 300 ~m ^2 . Assume that the pressure inside the house is atmospheric pressure. Density of air is 1.2 ~kg / m ^3 . The magnitude of the force exerted by the wind on the roof will be

Step-by-step solution

The force magnitude is determined by the pressure difference created by wind flow over the roof using Bernoulli's principle. Applying Bernoulli's equation between a point just above the roof and inside the house, where height differences are negligible: P₁ + 1 2 v₁^2 = P₂ + 1 2 v₂^2 With v₁ = 50 m/s , v₂ = 0 m/s , and = 1.2 kg/m^3 : P₁ + 1 2 (1.2)(50)^2 = P_ atm + 0 P₁ + 1500 Pa = P_ atm The pressure difference is P = P_ atm - P₁ = 1500 Pa . Using F = P A with A = 300 m^2 : F = 1500 300 = 450000 N The magnitude of