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Landing a 747 with only friction

  1. Feb 8, 2012 #1
    1. The problem statement, all variables and given/known data
    I had this problem floating around in my head and when I worked it out I was a little surprised. A 747 is coming in for a landing and the only force opposing its motion is the force of friction between the ground and its tires how much runway is needed for the plane to come to a complete stop?

    landing speed: 270 kph = 75 m/s
    coefficient of kinetic friction (f) = .7
    mass = 390,000 kg


    3. The attempt at a solution

    I apply the work energy theorem and solve for distance (d)

    .5mv^2=fmgd

    d=v^2/(2fg) ==> (75)^2/(.7*9.8)= 409m

    This is far shorter than the minimum runway needed to land a 747 and I'm ignoring so many other forces. Could someone provide some insight into why this answer is incorrect (or right?) Thanks
     
  2. jcsd
  3. Feb 8, 2012 #2
    Usually, planes are not landed with the majority of breaking force coming from the tires skidding. The wheels typically would roughly match the ground speed, and the a plane would be slowed via breaks on the wheels and aerodynamic forces. There are limits to the amounts of mechanical forces it is advisable to put on the landing gear, and how much force the breaks would be able to exert on the wheels, as well as limits for break heating; the landing gear could catch fire if too much breaking is attempted.
     
    Last edited: Feb 8, 2012
  4. Feb 8, 2012 #3
    Hmm...one possible reason would be that your answer assumes that N = mg immediately upon touching down. It is likely that N is less than mg at the start of the landing due to lift forces.
     
  5. Feb 8, 2012 #4

    ehild

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    The coefficient of kinetic friction you used is valid for sliding friction, when the Boeing were sliding on its belly. But it has got wheels and they roll. :smile: Rolling friction is in the range of 0.01.

    ehild
     
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