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)

Solution

Physics sir · 4 steps

Faster air on top means lower pressure on top: find that pressure gap and multiply by the wing area.

1. Bernoulli across wing. The top and bottom surfaces are at practically the same height, so ρ g h drops out. Then P + 1/2ρ v² is the same above and below the wing.

   P_(lower) - Pᵤₚₚₑᵣ = 1/2ρ(vᵤₚₚₑᵣ² - v_(lower)²)

2. Pressure difference. vᵤₚₚₑᵣ² - v_(lower)² = 4900 - 4225 = 675 m²s⁻². So Δ P = 1/2 × 1.2 × 675 = 405 Pa.
3. Force on wing. This higher pressure from below acts over the whole wing area, so lift = Δ P × A = 405 × 2.
4. So. The lift on the wing is 810 N.

The common slip: 1620. Dropping the 1/2 in 1/2ρ v² doubles the pressure difference and gives 1620 N instead of 810 N.

The idea to keep. Lift is always pressure difference times area. Get the pressure difference from Bernoulli as 1/2ρ(v₁² - v₂²), and square each speed before subtracting.

Answer from the NTA final answer key. Solution by PrepWiser.