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 ($\rho_{air} = 1.2$ kg/$m^3$)

Let $P_1$, $v_1 = 0$ be the pressure and wind speed inside, and $P_2$, $v_2 = 40$ m/s outside (above the roof).
Bernoulli's theorem: $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$
$\Delta P = P_1 - P_2 = \frac{1}{2}\rho v_2^2 = \frac{1}{2}\times 1.2\times 40^2 = 960$ Pa
$F = \Delta P\times A = 960\times 250 = 2.4\times 10^5$ N
Since $P_1 > P_2$ (pressure inside is greater than above the roof), the force acts upwards.
So the force is $2.4\times 10^5$ N, upwards.