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

Correct Answer: 810

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: \( F = 810 \, \text{N} \)

Answer 2

Step 1: Apply Bernoulli’s principle.

Pressure difference between lower and upper surfaces: \[ P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2) \] where \( v_1 = 65\,\text{m/s} \) (lower surface), \( v_2 = 70\,\text{m/s} \) (upper surface), and \( \rho = 1.2\,\text{kg/m}^3 \).

Step 2: Substitute values.

\[ P_1 - P_2 = \frac{1}{2} \times 1.2 \times (70^2 - 65^2) \] \[ = 0.6 \times (4900 - 4225) = 0.6 \times 675 = 405\,\text{N/m}^2 \]

Step 3: Calculate lift force.

Lift = Pressure difference × Area \[ F = (P_1 - P_2) \times A = 405 \times 2 = 810\,\text{N} \]

Final Answer:

\[ \boxed{F = 810\,\text{N}} \]