A wind with speed $$40\,m/s$$  blows parallel to the roof of a house. The area of the roof is $$250\,{m^2}.$$  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
$$\left( {{p_{{\text{air}}}} = 1.2\,kg/{m^3}} \right)$$

Solution :
From Bernoulli's theorem $${p_1} + \frac{1}{2}pv_1^2 = {p_2} + \frac{1}{2}pv_2^2$$
where, $${p_1},{p_2}$$  are pressure inside and outside the roof and $${v_1},{v_2}$$  are velocities of wind inside and outside the roof. Neglect the width of the roof. Pressure difference is
$$\eqalign{ & {p_1} - {p_2} = \frac{1}{2}\rho \left( {v_2^2 - v_1^2} \right) = \frac{1}{2} \times 1.2\left( {{{40}^2} - 0} \right) \cr & = 960\,N/{m^2} \cr} $$
Force acting on the roof is given by
$$\eqalign{ & F = \left( {{p_1} - {p_2}} \right)A = 960 \times 250 \cr & = 24 \times {10^4}N = 2.4 \times {10^5}N \cr} $$
As the pressure inside the roof is more than outside to it. So the force will act in the upward direction,
i.e. $$F = 2.4 \times {10^5}N{\text{ - upwards}}{\text{.}}$$