Question

A wind with speed $40m{{s}^{-1}}$ 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 $({{\rho }_{air}}=1.2kg{{m}^{-3}})$

& 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 \\\ \end{aligned}$$

Answer 1

Force acting on an area is equal to the product of pressure and area, on which the pressure is applied. Total pressure acting on the roof is equal to the difference in pressure outside the roof and the pressure, inside the roof. Both these pressures can be determined using Bernoulli’s principle for fluid dynamics. Direction of force is dependent on the intensity of pressures at two different points.

Formula used:
$1)P+\dfrac{1}{2}\rho {{v}^{2}}=K(const)$
$2)F=({{P}_{atm}}-P)\times A$

Complete step-by-step answer:
Bernoulli’s principle for fluid dynamics states that:
$P+\dfrac{1}{2}\rho {{v}^{2}}=K(const)$
where
$P$ is the static pressure of a fluid on a cross-sectional area
$\rho$ is the density of the fluid
$v$ is the velocity of the fluid
$K(const)$ is any constant
Let this be equation 1.
Coming to our question, we are said that a wind, with speed $40m{{s}^{-1}}$ blows parallel to the roof of a house, whose area is $250{{m}^{2}}$. Assuming that the pressure inside the house is atmospheric pressure, we are required to determine the force exerted by the wind on the roof as well as its direction.
Firstly, let us determine the total pressure acting on the roof. Clearly, pressure is exerted on the roof from the inside as well as outside the roof. If ${{P}_{1}}$ and ${{P}_{2}}$ represent these pressures respectively, then, using equation 1, we can write:
${{P}_{1}}+\dfrac{1}{2}{{\rho }_{air}}{{v}_{1}}^{2}={{P}_{2}}+\dfrac{1}{2}{{\rho }_{air}}{{v}_{2}}^{2}$
where
${{P}_{1}}$ is the pressure inside the roof
${{P}_{2}}$ is the pressure outside the roof
${{\rho }_{air}}$ is the density of air
${{v}_{1}}$ is the velocity of air inside the roof
${{v}_{2}}$ is the velocity of air outside the roof
Let this be equation 2.
Substituting the given values in equation 2, we have

${{P}_{1}}+\dfrac{1}{2}{{\rho }_{air}}{{v}_{1}}^{2}={{P}_{2}}+\dfrac{1}{2}{{\rho }_{air}}{{v}_{2}}^{2}\Rightarrow {{P}_{atm}}+\dfrac{1}{2}{{\rho }_{air}}{{(0)}^{2}}=P+\dfrac{1}{2}{{\rho }_{air}}{{(40)}^{2}}\Rightarrow {{P}_{atm}}-P=\dfrac{1}{2}{{\rho }_{air}}{{(40)}^{2}}=800\times 1.2=960Pa$
where
${{P}_{1}}={{P}_{atm}}$ is the pressure inside the roof, as provided
${{P}_{2}}=P$ is the assumed pressure outside the roof
${{\rho }_{air}}=1.2kg{{m}^{-3}}$ is the density of air, as provided
${{v}_{1}}=0m{{s}^{-1}}$ is the velocity of air inside the roof, because the air is still inside the roof
${{v}_{2}}=40m{{s}^{-1}}$ is the velocity of air outside the roof due to the wind, as provided
Let this be equation 3.
From equation 3, it is clear that the total pressure acting on the roof is given by
${{P}_{atm}}-P=960Pa$
Let this be equation 4.
Now, we know that force acting on an area is equal to the pressure exerted on that area and the surface area, itself. Using this definition, force acting on the roof is given by
$F=({{P}_{atm}}-P)\times A=960Pa\times 250{{m}^{2}}=2.4\times {{10}^{5}}N$
where
$F$ is the force acting on the roof
${{P}_{atm}}-P=960Pa$ is the total pressure exerted on the roof, from equation 4
$A=250{{m}^{2}}$ is the surface area of roof, on which the wind blows, as provided
Let this be equation 5.
Here, we know that pressure inside the roof $(={{P}_{atm}})$ is more than the pressure outside the roof $(=P)$. Therefore, the direction of force will be in the upward direction.
Let this be statement M.
Hence, from equation 5 and statement M, we can conclude that the correct answer is option $A$.

So, the correct answer is “Option A”.

Note: Pressure is a scalar quantity. It does not have direction. Pressure acting on an area is directly proportional to the force acting on that area and inversely proportional to the surface area. Here, force acting on the area is a vector quantity and has both magnitude and direction. Direction of force is dependent on the intensity of pressures at two different points. Force always acts from a region of high pressure to a region of low pressure.