Question

A wind with speed $40\text{m/s}$ blows parallel to the roof of a house. The area of the roof is $250{\text{m}}^2$. Assuming that the pressure inside the house is equal to atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be $({\rho }_{\text{air}}=1.2{\text{kg/m}}^3)$

• A. $2.4\times {10}^5\text{N, downwards}$
• B. $4.8\times {10}^5\text{N, downwards}$
• C. $4.8\times {10}^5\text{N, upwards}$
• D. $2.4\times {10}^5\text{N, upwards}$

Answer 1

The correct option is D $2.4\times {10}^5\text{N, upwards}$

Applying Bernoulli's equation just above and below the roof,
$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2...\left(i\right)$
$h=\text{constant}$, since pressure $P_1\&P_2$ have been taken just above and below a point on the roof (elevation is constant)
$P_1=P_{\text{atm}}=P_0$ is the pressure inside the house or just below the roof.


From equation $\left(i\right)$, we get
$\Delta P=P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)...\left(ii\right)$
Net force on the roof is due to the pressure differential.
$\Rightarrow F=\Delta P\times A...\left(iii\right)$

From Eq.$\left(ii\right)\&\left(iii\right)$,
$\frac{1}{2}\rho (v_2^2-v_1^2)=\frac{F}{A}$
$\frac{1}{2}\times 1.2\times ({40}^2-0^2)=\frac{F}{250}$
($\because P_1>P_2$, hence net pressure differential is along upwards direction)
$\Rightarrow F=2.4\times {10}^5\text{N, upwards}$