Answer
The magnitude of the force on the roof is $1.02\times 10^9~N$
Work Step by Step
We can convert the wind speed to units of m/s:
$v_2 = 150~mph\times \frac{1609~m}{1~mi} \times \frac{1~hr}{3600~s} = 67.0~m/s$
Let $P_1$ be the pressure inside the house. We can use Bernoulli's equation to find the pressure difference between the inside and outside of the house:
$P_1 + \frac{1}{2}\rho~v_1^2 = P_2+\frac{1}{2}\rho~v_2^2$
$P_1 + 0 = P_2+\frac{1}{2}\rho~v_2^2$
$P_1 - P_2 = \frac{1}{2}\rho~v_2^2$
$P_1 - P_2 = (\frac{1}{2})(1.225~kg/m^3)(67.0~m/s)^2$
$P_1 - P_2 = 2750~Pa$
We can find the force that this pressure difference exerts on the roof:
$F = \Delta P~A$
$F = (2750~N/m^2)(2000~ft^2)(\frac{1~m}{3.28~ft})^2$
$F = 1.02\times 10^9~N$
The magnitude of the force on the roof is $1.02\times 10^9~N$.
