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Plane in flight - Dynamics Question

  1. Apr 17, 2017 #1
    1. The problem statement, all variables and given/known data
    "Figure 5 depicts a 30 000 kg aircraft climbing at an angle θ = 15˚ when the thrust T = 180 kN. The aircraft’s speed is 300 km/hr and its acceleration is 2m/s2. If the radius of curvature of the path is 20 km (i.e., θ is decreasing), compute the lift and drag forces on the aircraft."

    a = 2m/s^2
    T = 150000N
    Theta = 15 degrees
    Radius of Curvature (K) = 20000m
    v = 300km/hr => v = 83.3 m/s^1

    Screenshot (48).png
    2. Relevant equations
    W=mg, F=ma, a = v^2/K

    3. The attempt at a solution
    I was going to treat this similar to an object moving up an inclined plane. Although, I don't know if this would work.

    I determined W = mg, W= 294,300N.
    I then thought using a = 83.3^2/20000 = 0.35 m/2^1.
    Therefore, F = ma, F = 30,000 * 0.35 = 10,500N.

    Therefore F + F(f) (i.e. Wsin(Theta)) = Drag.

    Upon calculating, I got close to the answer for drag but I believe only by coincidence. As I considered you would have to work out normal and tangential components individually, therefore a= v^2/K would be in the normal direction, not the tangential like I used it for.

    I also am unsure on how to calculate Lift if I cannot treat this an incline plane problem.
     
  2. jcsd
  3. Apr 17, 2017 #2

    kuruman

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    I am not sure I understand what you are doing here. Which F is this and in what direction does it point?
     
  4. Apr 17, 2017 #3

    haruspex

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    That 0.35 is using only the centripetal acceleration. You are given an acceleration of 2ms-2,though it is not clear whether that is the overall acceleration or just the increase in speed. Is this F just the net force up the slope?
    I can't make sense of that. Are you saying F(f) stands for W sin(θ)? Doesn't that add to drag? Where is T in that equation?
     
  5. Apr 18, 2017 #4
    This is where I was confused, as I was using a incorrectly.

    I tried to use a = v^2/K, but as you both noted I wasn't using it corretly but I don't know how to use it.

    So this a would be a(centripetal)? The 2m/s^2 is the overall accleration as far as I'm aware.

    [QUOTE="I can't make sense of that. Are you saying F(f) stands for W sin(θ)? Doesn't that add to drag? Where is T in that equation?[/QUOTE]

    Yeah that's what I was working there. I was going to determine each component then sum them into a net force at the end.
     
  6. Apr 18, 2017 #5
    I ended up redoing some of it. I calculated R = 284,272N, then used a(centripetal) = v^2/K, => a = 0.35 m/s^2.

    I then used F(centripetal) = ma(centripetal), = 10,408N.

    Then I done R - F(centripetal) to calculate lift. I got the answer on the sheet, but is that the correct process?

    I also still can't figure out drag.
     
  7. Apr 18, 2017 #6

    kuruman

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    You are given the magnitude of the acceleration vector a = 2 m/s2. You have calculated the centripetal component to be 0.35 m/s2. Can you figure out from these two numbers the acceleration in the direction of motion? Hint: The direction of motion is perpendicular to the centripetal direction.
     
  8. Apr 18, 2017 #7

    haruspex

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    Yes. I don't see where you defined R, but it looks like it is W cos(θ). Centripetal is the radial component of the resultant, so =R-lift.
     
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