Question

Air streams horizontally across an aeroplane wing of area 3 m2 weighing 250 kg.The air speed is 60 m/s and 45 m/s over the top surface and under the bottom surafce respectively. What is the lift
on the wing?(Density of air 1.293g/l)​

Answer 1

Answer:

Explanation:

applying Bernoulli theorem  just above and below the surface of wing  

P1+12ρv12=P2+12ρv22    as  (P1,  P2 are pressure ar top surface and bottom surface)

Hence

P1−P2 =12ρv22 −12ρv12

P2 −P1=ρ2(v12−v22)

now multplying both sides by area

(A)area(P2 −P1)=ρ2(v12−v22)A

upthrust=ρ2(v12−v22)A=1.225×32(3600−2025)

upthrust=2894.06 N