In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are $70 \;m s ^{-1}$ and $63\; m s ^{-1}$ respectively. What is the lift on the wing if its area is $2.5 \;m ^{2}$ ? Take the density of atr to be $1.3\; kg m ^{-3} .$
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Speed of wind on the upper surface of the wing, $V_{1}=70 m / s$ Speed of wind on the lower surface of the wing, $V_{2}=63 m / s$ Area of the wing, $A=2.5 m ^{2}$ Density of air, $\rho=1.3 kg m ^{-3}$
According to Bernoulli's theorem, we have the relation:
$P_{1}+\frac{1}{2} \rho V_{1}^{2}=P_{2}+\frac{1}{2} \rho V_{2}^{2}$
$P_{2}-P_{1}=\frac{1}{2} \rho\left(V_{1}^{2}-V_{2}^{2}\right)$
Where,
$P_{1}=$ Pressure on the upper surface of the wing
$P_{2}=$ Pressure on the lower surface of the wing
The pressure difference between the upper and lower surfaces of the wing provides lift to the aeroplane.
Lift on the wing $=\left(P_{2}-P_{1}\right) A$
$=\frac{1}{2} \rho\left(V_{1}^{2}-V_{2}^{2}\right) A$
$=\frac{1}{2} 1.3\left((70)^{2}-(63)^{2}\right) \times 2.5$
$=1512.87$
$=1.51 \times 10^{3} N$
Therefore, the lift on the wing of the aeroplane is $1.51 \times 10^{3} \;N$.
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