Air flows horizontally with a speed $$v = 106\,km/hr.$$    A house has plane roof of area $$A = 20\,{m^2}.$$   The magnitude of aerodynamic lift of the roof is

Solution :
Air flows just above the roof and there is no air flow just below the roof inside the room. Therefore $${v_1} = 0$$  and $${v_2} = v.$$   Applying Bernaulli’s theorem at the points inside and outside the roof, we obtain.
$$\left( {\frac{1}{2}} \right)\rho v_1^2 + \rho g{h_1} + {P_1} = \left( {\frac{1}{2}} \right)\rho v_2^2 + \rho g{h_2} + {P_2}.$$
Since $${h_1} = {h_2} = h,{v_1} = 0\,{\text{and}}\,{v_2} = {v_1}$$
$${P_1} = {P_2} + \frac{1}{2}\rho {v^2} \Rightarrow {P_1} - {P_2} = \Delta P = \frac{1}{2}\rho {v^2}.$$
Since the area of the roof is $$A,$$ the aerodynamic lift exerted on it $$ = F = \left( {\Delta P} \right)A$$
$$ \Rightarrow F = \frac{1}{2}\rho A{v^2}$$
$$\eqalign{ & {\text{where}}\,\rho = {\text{density}}\,{\text{of}}\,{\text{air}} = 1.3\,kg/{m^3} \cr & A = 20\,{m^2},v = 29.44\,m/\sec . \cr & \Rightarrow F = \left\{ {\frac{1}{2} \times 1.3 \times 20 \times {{\left( {29.44} \right)}^2}} \right\}N \cr & = 1.127 \times {10^4}\,N. \cr} $$