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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