DRAG

EXAMPLES

Fy=d(mv)/dt =0A = ?

U = 6 m/s

Drag Force = CD ½ U2A

mg = 120 kg x 9.8 ms-2

FD = CD ½ U2A

Force of Gravity = mg = 120kg x 9.8 ms-2

Drag Force = CD ½ U2A = 1.42 ½ 1.23 kg/m3 (6 m/s)2 d2/4

FD = mgd = [(8/)(120kg)(9.8 m/s2) (1/1.42)(1/1.23 kg/m3)(1/6m/s)2

d = 6.9 m

U = 6 m/s

FD = CD ½ U2A

Fx = d(mv)/dt

FD = CD ½ U2A

FD = ma = m(dU/dt) = m(dU/dx)(dx/dt)

CD ½ U2A = m (dU/dx) U

CD ½ U2A = m (dU/dx) UdU/U = CD A dx /(2m)

Integrating from x=0

to x=x gives:ln Uf – lnUi = CD Ax /(2m)CD = 2m ln {Uf/Ui} /[ Ax]

CD = 0.299

(A) No wind: max speed = 37km/hrM = 80 kg; FR = 4 N; CD = 1.2; A = 0.25 m2

(B) Head wind = 10 km/hr, maximum ground speed = 30 km/hrIs this possible?

FD = CD ½ U2A

Fx = 0 ?

(#1) No wind: max speed = 37km/hrM = 80 kg; FR = 4 N; CD = 1.2; A = 0.25 m2

(#2) Head wind = 10 km/hr, maximum ground speed = 30 km/hrIs this possible?

FD = CD ½ U2A

Power(#2) < or = Power(#1)

Find POWER = F U for this conditionAnd then see if that is enough to win bet.

FT = FR + FD

air

P = U FT = U(FR + FD)

8.33 m/s

2.78 m/s

P = U FT = U(FR + FD)

FD = CD ½ U2A

Total Power = FDU + FRU

FD = CD ½ U2AA = 23.4 ft2

CD = 0.5FR = 0.015 x 4500 lbf

FD = FR to find U where aerodynamic force = frictional force

Total Power = FDU + FRU

FD = FR

CD ½ U2A = .015xW(0.5)½ 0.00238slug/ft3U2(23.4ft2)

= 0.015 x 4500 lbf

U2 = 67.5/0.0139

U = 69.7 ft/s = 47.5 mph

Where does aerodynamic force = frictional force ?