A 40 km/hr wind is blowing parallel to a storage facility , as shown in the figure. estimate the normal force on a 0.5-m high, 1-m long window if the front door is left open. assume the windspeed along the outside of thewindow is 1.2 U. Air density is 1.22 kg/m^3. There's a picture that shows a small shed with the wind blowind into an open door with a window on the side od the house. So wind blows straight into the open door and blows parallel to the window. 2. Relevant equations p-p(infinite)=1/2 p(U^2) F=(p-p)A 3. The attempt at a solution ok so i have a solution from my professor but he is know to make mistakes (often) on his homework solutions. His solution: Pressure inside: p-p(infinite)=1/2 p(U^2)=(1/2)1.22((40/3.6)^2)=75.3 N/m^2 P outside window: 1/2 * 1.22 * ((40/3.6)^2) (1-1.2^2)=-33.13 Force on the window is F=change of p *A= (75.3+33.13)*(0.5*1)=54.2N My problem: So my issue is with the pressure outside the window. My solution was to make pressure outside the window 1/2*1.22* ((1.2(40/3.6))^2) Maybe im not understanding the equation but i used 1.2 to multiply by the initial windspeed ((1.2(40/3.6))^2). My prof wrote ((40/3.6)^2) (1-1.2^2). Is he right? i can't figure out why the pressure would be calculated like that.