## Saturday, December 3, 2016

### Take-off

As I was returning to Colgate from Thanksgiving break, I was sitting on the plane and began to wonder about the physics of flight, specifically how much upward pressure is exerted on the wings at take off. To do this I used the equation:  P= F/A. To find the force in this equation I used the following model: during the plane will accelerate in the positive y-direction (upwards) for the first 500ft. (152.4m) traveled in the y-direction and travel at a constant speed in the y-direction as it passes through the 10,000ft. (3,048m) mark on its way to 28,000ft (8,536.6 m) (the cruising altitude). I will use this model to solve for the acceleration during the first 152.4m using kinematics. I will then use the equation F – mg = ma (where F is the upward force and a is the acceleration just found) to find F. For the area of the plane, the total area of the underside of the plane I used the area fuselage (modeled after a rectangle) plus the area of the wings (provided on the informational page for the plane, a Bombardier CRJ-200). The calculations are as follows:

I estimated the time it took to go from take-off to cruising altitude to be about 8 min. (or 480 sec).

y = yo +v-y(t)+1/2(a-y)(t^2)
(t = time of acceleration)
since yo and v-y = 0,  y = ½(a-y)(t^2)

152.4(2)/(t^2) = a-y

vf = vo + (303.8/(t^2)) t = 303.8/t

(vf is a constant velocity as it passes through the 10,000 ft. mark)

v = y/t

vf = (3048 – 152.4) /  (480 – t)

303.8/t = (3048-152.4) / (480 – t)

t= 45.6s

a-y = .15m/s^2

F – mg = m(a-y)
The maximum take-off weight of the plane is 23,133 Kg

F = (23,133) (9.8 +.15) = 230173.4 N

A = A-fuselage + A-wings

A-fuselage (modeled as a rectangle) = L*W = 26.77*2.69 = 72 m^2
(the length and with of the fuselage are taken from the plane’s informational page)
A-wings (also on the informational page) = 48.35 m^2

A = 72 + 48.35 =120.35 m^2

P = F/A = 230173.4/120.35 = 1912 Pa