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Question
A plane is in level flight at a constant speed and each of its two wings has an area of 25 m2. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m–3).
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Solution 1
The area of the wings of the plane, A = 2 × 25 = 50 m2
Speed of air over the lower wing, V1 = 180 km/h = 50 m/s
Speed of air over the upper wing, V2 = 234 km/h = 65 m/s
Density of air, ρ = 1 kg m–3
Pressure of air over the lower wing = P1
Pressure of air over the upper wing= P2
The upward force on the plane can be obtained using Bernoulli’s equation as:
`P_1 + 1/2 rhoV_1^2 = P_2 + 1/2 rhoV_^2`
`P_1 - P_2= 1/2 rho(V_2^2 - V_1^2)` ...(i)
The upward force (F) on the plane can be calculated as:
`(P_1 - P_2)A`
`= 1/2 rho(V_2^2 - V_1^2)A` Using equation (i)
`= 1/2 xx 1 xxx((65)^2 - (50)^2)xx 50`
= 4400.51 Kg
~ 4400 kg
Hence, the mass of the plane is about 4400 kg.
Solution 2
Here speed of air over lower wing `v_1 = 180 km/h = 180 xx 5/18 = 50 ms^(-1)`
Speed over the upper wing `v_2 = 234 "km/h" = 234 xx 5/18 = 65 ms^(-1)`
:. Pressure difference, `P_1 -P_2 = 1/2 rho (v_2^2 - v_1^2) = 1/2 xx 1(65^2 - 50^2) = 862.5 Pa`
:. Net upward force , `F = (P_1 -P_2)A`
This upward force balances the weight of the plane
:. `mg = F =(P_1 - P_2)A` [A = 25 x 2 = 50 `m^2`]
:. `m = ((P_1 - P_2)A)/g = (862.5xx50)/9.8 = 4400 N`
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