Question

A wind speed \[40{\text{ m/s}}\] blows parallel to the roof of a house. The area of the roof is \[250{m^2}\], assuming that the pressure inside the house in atmospheric pressure, the force exerted by the wind on the roof, and the direction of the force will be:

Answer 1

Hint:
- In this question, we need to determine the force exerted by the wind on the roof, and the direction of the force. For this, we will use the formula
\[{P_1} + \dfrac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho {v_2}^2 + \rho g{h_2}\].
- Bernoulli's equation states that the sum of the pressure energy, kinetic energy per unit volume, and the potential energy per unit volume is constant.

Complete step by step solution:
Speed of the wind is \[40{\text{ m/s}}\]
The area of the roof is \[250{m^2}\]
Using Bernoulli’s equation, \[P + \dfrac{1}{2}\rho {v^2} + \rho gh = C - - (i) \\ \]
Since the wind is blowing parallel to the roof, hence the change in potential energy becomes negligible \[\rho gh = 0\\ \]
Now, if we assume pressure on the roof to be \[{P_1}\] and velocity \[{V_1} = 40{\text{ m/}}{{\text{s}}^2}\] and the pressure inside the room to be \[{P_0}\], then we can say pressure inside the room and on the roof will be
\[{P_1} + \dfrac{1}{2}\rho {v^2} = {P_0} - - (ii) \\ \]
So we can say the change in pressure will be
\[\Delta P = {P_0} - {P_1} = \dfrac{1}{2}\rho {v^2}\\ \]
Now, as we know, the force exerted on a unit area for changing pressure is given as
 \[F = \Delta P.A - - (iii)\]
Where \[\Delta P = \dfrac{1}{2}\rho {v^2}\] and area of the roof is \[A = 250{m^2}\], hence by substituting these values in equation (iii), we get
\[F = \dfrac{1}{2}\rho {v^2}\left( {250} \right)\\ \]
Where the density of the air is \[{\rho _{Air}} = 1.2{\text{ kg/}}{{\text{m}}^3}\], hence by substituting density and velocity, we get
\[
  F = \dfrac{1}{2}\left( {1.2} \right){\left( {40} \right)^2}\left( {250} \right) \\
   = 1.2 \times 40 \times 40 \times 125 \\
   = 2.4 \times {10^5}N \\
 \]

Therefore the force exerted by the wind on the roof \[ = 2.4 \times {10^5}N\] and since the pressure inside the room is more than the pressure on the roof so the direction of the force will be upwards.

Note:
Students must know that slower-moving fluid creates more pressure than the faster-moving fluid, and the air is also fluid because it can flow and can change its direction.