Question

A body of mass 100 kg stands on a spring weighing machine inside a lift. The lift starts to
ascend with acceleration of 2.2 m s-2. What is the reading of the machine?​​

Answer 1

Explanation:

Speed of wind on the upper surface of the wing, V1=70m/s

Speed of wind on the lower surface of the wing, V2=63m/s

Area of the wing, A=2.5m2

Densityof air, ρ=1.3kg/m3

According to Bernoullis theorem, we have the relation:

P1+(1/2)ρ(V12)=P2+(1/2)ρ(V22)

P2−P1=(1/2)ρ(V12−V22)

Where,

P1= Pressure on the upper surface of the wing

P2= Pressure on the lower surface of the wing

The pressure difference between the upper and lower surfaces of the wing provides lift to the aeroplane.

Lift on the wing =(P2−P1)A

                           =(1/2)ρ(V12−V22)A

                           =(1/2)1.3[702−632]2.5

                           =1512.87

N=1.51×103 N

Therefore, the lift on the wing of the aeroplane is 1.51×103 N.