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Q.
Air flows horizontally with a speed $v =106\, km / hr$. A house has plane roof of area $A=20\, m ^2$. The magnitude of aerodynamic lift of the roof is
Mechanical Properties of Fluids
Solution:
Air flows just above the roof and there is no air flow just below the roof inside the room.
Therefore $v_1=0$ and $v_2=v$.
Applying Bernaulli's theorem at the points inside and outside the roof, we obtain.
$(1 / 2) \rho v_1^2+\rho g h_1+P_1=(1 / 2) \rho v_2{ }^2+\rho g h_2+P_2$.
Since $h_1=h_2=h, v_1=0$ and $v_2=v_1$
$P_1=P_2+1 / 2 \rho v^2 $
$\Rightarrow P_1-P_2=\Delta P=1 / 2 \rho v^2 \text {. }$
Since the area of the roof is $A$, the aerodynamic lift exerted on it $= F =(\Delta P ) A $
$\Rightarrow F =1 / 2 \rho Av ^2$
where $\rho=$ density of air $=1.3\, kg / m ^3$
$ A=20 m ^2, v =29.44\, m / sec .$
$\Rightarrow F =\left\{1 / 2 \times 1.3 \times 20 \times(29.44)^2\right\} N$
$ =1.127 \times 10^4 N .$