Question

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

• A. $2.4\times {10}^{5}N$, upward
• B. $2.4\times {10}^{5}N$, downward
• C. $4.8\times {10}^{5}N$, downward
• D. $4.8\times {10}^{5}N$, upward

Answer 1

Answer: (A)

$2.4\times {10}^{5}N$, upward

Applying Bernoulli's theorem to two points just inside the house and outside the house,

$P+\frac{1}{2}\rho v^2=constant$

Inside the pressure is $P_{atm}$

The pressure outside is $P$

v is zero inside.

$\therefore P-P_{atm}=\frac{1}{2}\rho v^2=\frac{1}{2}\times 1.2\times {40}^2=960$

Force is : $F=(P-P_{atm})\times Area=960\times 250=2.4\times {10}^{5}N$ in upward direction as pressure goes from higher to lower pressure area.