A wind with speed 40,, m/s blows parallel to | vedclass

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Question

Applying Bernoulli's theorem just above and just below the roof,

$P + \frac{1}{2}\rho {v^2} = {P_0} + 0$

$\left( {{P_0} - P} \right) = \frac

Vedclass · Accepted Answer

$2.4  	imes 10^5 N$, upwards

Vedclass · Answer

Applying Bernoulli's theorem just above and just below the roof,

$P + \frac{1}{2}ho {v^2} = {P_0} + 0$

$\left( {{P_0} - P} ight) = \frac{1}{2}ho {v^2} = \Delta P$

Hence lift of the roof

$F = \Delta P \cdot A = \frac{1}{2}ho A{v^2}$

$ = \frac{1}{2} 	imes 1.2 	imes {\left( {40} ight)^2} 	imes 250 = 2.4 	imes {10^5}N$

as pressure inside the roof is greater than outside the roof. so, force will act upward direction.

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

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