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(A) Given:Speed of wind on the upper surface of the wing,$V_{1} = 70 \; m s^{-1}$Speed of wind on the lower surface of the wing,$V_{2} = 63 \; m s^{-1

Vedclass · Accepted Answer

$1.51 	imes 10^{3} \; N$

Vedclass · Answer

(A) Given:Speed of wind on the upper surface of the wing,$V_{1} = 70 \; m s^{-1}$Speed of wind on the lower surface of the wing,$V_{2} = 63 \; m s^{-1}$Area of the wing,$A = 2.5 \; m^{2}$Density of air,$ho = 1.3 \; kg m^{-3}$According to Bernoulli's theorem,the pressure difference between the lower and upper surfaces is given by:$P_{2} - P_{1} = \frac{1}{2} ho (V_{1}^{2} - V_{2}^{2})$The lift on the wing is the force due to this pressure difference:$Lift = (P_{2} - P_{1}) 	imes A$$Lift = \frac{1}{2} ho (V_{1}^{2} - V_{2}^{2}) A$Substituting the values:$Lift = \frac{1}{2} 	imes 1.3 	imes (70^{2} - 63^{2}) 	imes 2.5$$Lift = 0.65 	imes (4900 - 3969) 	imes 2.5$$Lift = 0.65 	imes 931 	imes 2.5$$Lift = 1512.875 \; N$Rounding to two significant figures,the lift is approximately $1.51 	imes 10^{3} \; N$.

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 air to be $1.3 \; kg m^{-3}$.

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