 Ch 4. Particle Momentum and Impulse Multimedia Engineering Dynamics Impulse & Momentum Consv. Linear Momentum Impact AngularMomentum MassFlow
 Chapter - Particle - 1. General Motion 2. Force & Accel. 3. Energy 4. Momentum - Rigid Body - 5. General Motion 6. Force & Accel. 7. Energy 8. Momentum 9. 3-D Motion 10. Vibrations Appendix Basic Math Units Basic Equations Sections Search eBooks Dynamics Fluids Math Mechanics Statics Thermodynamics Author(s): Kurt Gramoll ©Kurt Gramoll DYNAMICS - CASE STUDY SOLUTION Solution of a) FBD of F-16 Because the takeoff roll is a straight line, this problem can be analyzed using rectangular coordinates. Begin with a free-body diagram. Let the origin of the rectangular coordinate system be located at the point where the plane begins the takeoff roll, with the positive x axis in the direction of the takeoff roll. Assume there is no force or velocity in the z direction. At t = 0, the velocity of the F-16 is 0,       (vx)t=0 = (vy)t=0 = 0 Use the value of the thrust at t = 0 and t = 5 to express the thrust in the x-direction during the first 5 seconds as a function of time, giving      Tt=0 = 22,250i N        Tt=5 = 111,250i N      T(t) = (22,250 + 17,800t)i N         (0 ≤ t ≤ 5 s) The velocity in the y direction is zero before take off and does not effect the velocity in the x direction. Likewise, forces in the y direction are independent of the thrust before take off. The change in momentum in the x direction is equal to the thrust impulse. This relationship can be used to determine the horizontal velocity for first 5 seconds, The total velocity after the first 5 seconds of the takeoff roll is      vt=5 = 22.86i m/s Solution of b) Take-off Output Data From t = 5 seconds to the time of liftoff (tLO), the thrust is constant and can be expressed as an average constant force,      T = Tavg = 111,250i N        (5 ≤ t ≤ tLO s) For an average or constant force, the impulse integral can be simplified to just force multiple the elapsed time.      (t2 - t1) ΣFx-AV = mvx2 - mvx1      (tLO - 5) (Tavg)x = m (vx)t=LO - m (vx)t=5      (tLO - 5) (111,250) = 14,600 (45.75 - 22.86)      tLO = 8.00 s

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