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## Homework Statement

An airplane has a 50 m^2 wing that is designed so that air on the top travels 20% faster

than the air on the bottom. The air on the bottom of the wing moves at the plane’s

airspeed and the unloaded airplane has a take-off speed of 90 km/h

A) What is the velocity of the air on top of the wing as the unloaded airplane becomes

airborne?

B) What pressure difference between the top and bottom of the wing as the unloaded

airplane becomes airborne?

C) What is the mass of the unloaded airplane?

D)If on a particular day, the mass of the airplane is increased by 10%, what is the new take-off speed?

## Homework Equations

P1 + .5dv1^2 +dgy1 = P2 + .5dv2^2 + dgy2

## The Attempt at a Solution

A)

90km/h = 25m/s

since the air on the top is 20 percent faster than on the bottom, then vtop = 1.2vbottom = 1.2(25) =30m/s

B)

P_top + .5d(v_top)^2 +dg(y_top) = P_bottom + .5d(v_bottom)^2 + dg(y_bottom)

assuming the difference in height between the top and bottom of the wing is negligible

P_bottom + .5d(v_bottom)^2 = Ptop + .5d(v_top)^2

P_bottom - Ptop = .5d(v_top^2 - v_bottom^2)

if i remember correctly the density of air is 1.29 kg/m^3

deltaP = .5(1.29)[30^2 - 25^2] = 177 pa

C) F = (deltaP)A = (177pa) (50m^2 + 50m^2) = 177(100) = 17700 N

F = mg

17700 = m (9.8)

m =1806 kg

D) this is the part im not too sure about

m = 1806(1.1) = 1987 kg

F = 1987(9.8) = 19473 N

delaP = F/A = 19473/100 = 195

195 = .5(1.29)[(1.2x)^2 - x^2]

195 = .645(1.44x^2-x^2)

302 = x^2(1.44-1)

x^2 = 302/.44 = 686

x = 26 m/s