What is the Aerofoil lift (in newtons) due to Bernoulli's principle on a paper plane of wing area 0.01 m³ if the air passes over the top and bottom surfaces at speeds of 9 m/s and 7 m/s respectively? (Take the density of air as 1.28 kg/m³.) (0.2 N) 


Given:

Area of wing  = A = 0.01 m² 
Speed of the air over the top surface of the wing = v₁ = 9 m s⁻¹ 
Speed of the air over the bottom surface of the wing = v₂ = 7 m s⁻¹
Density of air = ⍴ = 1.28 kg m⁻³


To Find:

Force (Aerofoil lift) = F = ?

Solution:


The Force (Aerofoil lift) F can be calculated by the relation:

F = `\frac {1}{2}` ⍴ (v₂ - v₁) A

by putting values

F = `\frac {1}{2}` ｘ1.28 kg m⁻² ((9 m s⁻¹)² - (7 m s⁻¹)²) 0.01 m³ 


F = `\frac {1}{2}` ï½˜1.28 kg m⁻² (81 m² s⁻² - 49 m² s⁻²) 0.01 m³ 

F = `\frac {1}{2}` ï½˜1.28 kg m⁻² ï½˜32 m² s⁻² ï½˜ 0.01 m³ 

F = 0.2048 kg m s⁻² 

or

F = 0.2 N --------------Ans.



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