A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 300$m^2$. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be

$\left({\rho }_{\text{air }}=1.2 \mathrm{kg}/{\mathrm{m}}^3\right)$

 

By Bernoulli's theorem, 

$\underset{\text{inside }}{\underbrace{P_1+\frac{1}{2}\rho v_1^2}}=\underset{\text{outside }}{\underbrace{P_2+\frac{1}{2}\rho v_2^2}}$

Assuming that the roof width is very small Pressure difference,

$P_1-P_2=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)$

$\text{ Here, }\rho =1.2 \mathrm{kg}{ \mathrm{m}}^{-3},v_2=40 \mathrm{m}{ \mathrm{s}}^{-1},v_1=0$

$A=250{ \mathrm{m}}^2$

$P_1-P_2=\frac{1}{2}\times 1.2\left({40}^2-0^2\right)$

$=\frac{1}{2}\times 1.2\times 1600=960 \mathrm{N}{ \mathrm{m}}^{-2}$

$\text{ Force acting on the roof }F=\left(P_1-P_2\right)\times A=960\times 300$

$=2.8\times {10}^5 \mathrm{N}\text{ upwards }$