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Analyzing Drag Force and Maximum Speed in Aerodynamic Conditions

This analysis investigates the drag force acting on an object in motion and how it affects the maximum speed achievable under various conditions. Using the drag equation, we calculate the forces of gravity and drag, considering parameters such as mass, cross-sectional area, and drag coefficient. We explore scenarios with different wind conditions, determining the feasibility of achieving specified ground speeds. Additionally, this study delves into the necessary power requirements to overcome these forces, providing insights into aerodynamic performance in real-world applications.

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Analyzing Drag Force and Maximum Speed in Aerodynamic Conditions

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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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