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DRAG E X A M P L E S

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

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DRAG E X A M P L E S

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  1. DRAG E X A M P L E S

  2.  Fy=d(mv)/dt =0 A = ? U = 6 m/s Drag Force = CD ½  U2A

  3. 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 = mg d = [(8/)(120kg)(9.8 m/s2) (1/1.42)(1/1.23 kg/m3)(1/6m/s)2 d = 6.9 m FD = CD ½  U2A mg = 120 kg x 9.8 ms-2 U = 6 m/s

  4. FD = CD ½  U2A  Fx = d(mv)/dt

  5. FD = CD ½  U2A FD = ma = m(dU/dt) = m(dU/dx)(dx/dt) CD ½  U2A = m (dU/dx) U

  6. CD ½  U2A = m (dU/dx) U dU/U = CD  A dx /(2m) Integrating from x=0 to x=x gives: ln Uf – lnUi = CD  Ax /(2m) CD = 2m ln {Uf/Ui} /[ Ax] CD = 0.299

  7. (A) No wind: max speed = 37km/hr M = 80 kg; FR = 4 N; CD = 1.2; A = 0.25 m2 (B) Head wind = 10 km/hr, maximum ground speed = 30 km/hr Is this possible? • FD = CD ½  U2A • Fx = 0 ?

  8. (#1) No wind: max speed = 37km/hr M = 80 kg; FR = 4 N; CD = 1.2; A = 0.25 m2 (#2) Head wind = 10 km/hr, maximum ground speed = 30 km/hr Is this possible? FD = CD ½  U2A Power(#2) < or = Power(#1)

  9. FT = FR + FD Find POWER = F U for this condition And then see if that is enough to win bet.

  10. P = U FT = U(FR + FD) air

  11. 8.33 m/s 2.78 m/s P = U FT = U(FR + FD)

  12. FD = CD ½  U2A Total Power = FDU + FRU

  13. FD = CD ½  U2A A = 23.4 ft2 CD = 0.5 FR = 0.015 x 4500 lbf FD = FR to find U where aerodynamic force = frictional force Total Power = FDU + FRU

  14. Where does aerodynamic force = frictional force ? 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

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