Question

In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are 70 ms–1 and 65 ms–1 respectively. If the wing area is 2 m2 the lift of the wing is _______ N.
(Given density of air = 1.2 kg m-3)

Answer 1

Step 1: Use Bernoulli’s Equation for Lift Force:

$F = \frac{1}{2} \rho (v_1^2 - v_2^2)A$

- Where: - $\rho = 1.2 \, \text{kg/m}^3$ - $v_1 = 70 \, \text{m/s}$ - $v_2 = 65 \, \text{m/s}$ - $A = 2 \, \text{m}^2$

Step 2: Calculate the Lift Force $F$:

$F = \frac{1}{2} \times 1.2 \times (70^2 - 65^2) \times 2$

Step 3: Simplify the Expression:

$F = \frac{1}{2} \times 1.2 \times (4900 - 4225) \times 2$

$F = \frac{1}{2} \times 1.2 \times 675 \times 2$

$F = 810 \, \text{N}$

Conclusion:

So, the correct answer is: $F = 810 \, \text{N}$