drag e x a m p l e s
DESCRIPTION
DRAG E X A M P L E S. F y =d(mv)/dt =0. A = ? U = 6 m/s. Drag Force = C D ½ U 2 A. Force of Gravity = mg = 120kg x 9.8 ms -2 Drag Force = C D ½ U 2 A = 1.42 ½ 1.23 kg/m 3 (6 m/s) 2 d 2 /4. F D = mg - PowerPoint PPT PresentationTRANSCRIPT
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DRAG
EXAMPLES
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Fy=d(mv)/dt =0A = ?
U = 6 m/s
Drag Force = CD ½ U2A
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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
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FD = CD ½ U2A
Fx = d(mv)/dt
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FD = CD ½ U2A
FD = ma = m(dU/dt) = m(dU/dx)(dx/dt)
CD ½ U2A = m (dU/dx) U
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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
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(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 ?
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(#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)
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Find POWER = F U for this conditionAnd then see if that is enough to win bet.
FT = FR + FD
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air
P = U FT = U(FR + FD)
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8.33 m/s
2.78 m/s
P = U FT = U(FR + FD)
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FD = CD ½ U2A
Total Power = FDU + FRU
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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
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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 ?
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