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Question: A wind speed \[40{\text{ m/s}}\] blows parallel to the roof of a house. The area of the roof is \[25...

A wind speed 40 m/s40{\text{ m/s}} blows parallel to the roof of a house. The area of the roof is 250m2250{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:

Explanation

Solution

- 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
P1+12ρv12+ρgh1=P2+12ρv22+ρgh2{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 m/s40{\text{ m/s}}
The area of the roof is 250m2250{m^2}
Using Bernoulli’s equation, P+12ρv2+ρgh=C(i) 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 ρgh=0 \rho gh = 0\\\
Now, if we assume pressure on the roof to be P1{P_1} and velocity V1=40 m/s2{V_1} = 40{\text{ m/}}{{\text{s}}^2} and the pressure inside the room to be P0{P_0}, then we can say pressure inside the room and on the roof will be
P1+12ρv2=P0(ii) {P_1} + \dfrac{1}{2}\rho {v^2} = {P_0} - - (ii) \\\
So we can say the change in pressure will be
ΔP=P0P1=12ρv2 \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=ΔP.A(iii)F = \Delta P.A - - (iii)
Where ΔP=12ρv2\Delta P = \dfrac{1}{2}\rho {v^2} and area of the roof is A=250m2A = 250{m^2}, hence by substituting these values in equation (iii), we get
F=12ρv2(250) F = \dfrac{1}{2}\rho {v^2}\left( {250} \right)\\\
Where the density of the air is ρAir=1.2 kg/m3{\rho _{Air}} = 1.2{\text{ kg/}}{{\text{m}}^3}, hence by substituting density and velocity, we get

F=12(1.2)(40)2(250) =1.2×40×40×125 =2.4×105N  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 =2.4×105N = 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.