Thrust force of an engine

  1. 1. The problem statement, all variables and given/known data

    A plane of mass 1220 kg has an engine failure when flying with an airspeed of 138 km/h at an altitude of 2960 m on a calm day. It then glides at a constant glide angle (which is the direction of flight below the horizontal) 6.80 degrees towards a safe landing at this constant speed of 138 km/h experiencing a drag force of 1330 N that opposes the direction in which the plane is moving.

    (b) Suppose the pilot instead had managed to get the airplane engine started such that he was able to apply full throttle and the airplane climbed along a straight line angled above the horizontal so that it gained altitude at a steady rate of 4.07 m/s. Assuming he was again flying with an airspeed of 138 km/h determine:

    i)The thrust force of the engine (that acts in the direction the plane is moving)

    2. Relevant equations
    g = 9.81 m s-2
    F = ma
    sin theta = Opp/Hyp

    3. The attempt at a solution

    The flight angle above the horizontal the plane is flying is 6.09 degrees
    .
    ...|\
    ...|..\
    .1|....\ 2
    ...|.....\
    ...+----+
    3
    1 =
    Code (Text):
    F_gravity
    2 =
    Code (Text):
    F_thrust
    3 = horizontal

    Using: sin theta = Opp/Hyp
    Hyp = Opp/sin theta
    = (1220*9.81)/sin(6.09)
    = 112811.2053 N

    Actual answer is 2600N
     
    Last edited: Apr 20, 2007
  2. jcsd
  3. HI

    draw a free body diagram first
    as plane is moving with const. velocity
    thrust (in direction of plane)= air resistance + component of gravity in d direction opposite the movement of plane

    thrust= 1330 + M X g X cos(90-6.09)
    = 1330 + 1220 X 9.81 X sin(6.09)
    =1330+1270
    =2600 N

    i guess no problems now Hendrick !:biggrin:
     
  4. AlephZero

    AlephZero 7,298
    Science Advisor
    Homework Helper

    When the plane is gliding at constant velocity there are three forces not two.

    1. Weight vertically down
    2. Drag in the opposite direction to the motion of the plane
    3. Lift from the wings, perpendicular to the motion of the plane.

    Your wrong answer seems to have forgotten about the lift.

    For the second part, the magnitude of the lift force is the same as the first part, because the airspeed is the same.
     
  5. ohh.....
    why they never teach all this at high school?
    but this time that doesn't add any problem to my answer as it will eventually be zero
    neways...Thanks to you ! for correcting me
     
  6. AlephZero

    AlephZero 7,298
    Science Advisor
    Homework Helper

    pcdagr8, when you resolved the forces in the direction the plane is travelling, you got the right answer, because the component of the lift force in that direction is 0.

    But Hendrick didn't do that.

    Anyway, when you draw a free body diagram it's usually best to include ALL the forces :smile:
     
  7. Hi, thanks very much, you really helped me calculation wise.
    Would you be able to show me the vector diagram you used to calculate it? I've tried recreating it but everything seems wrong...

    ---------------------------

    Hi,

    With engine failure (gliding descent):
    .............................^
    ............................/ F_lift (perpendicular to plane)
    .........................../ (angle between horizontal and plane is glide angle at 6.38 degrees)
    .......................~~~~ <---(horizontal)
    F_thrust<-----../plane/..------> F_drag (1330N)
    ..........................|
    ..........................| F_weight
    ..........................V

    Without engine failure (steep climb)
    .......................^
    .........................\ F_lift (perpendicular to plane)
    ..........................\
    .................................
    F_thrust<-----....\plane\..------> F_drag
    .........................~~~~ <---(horizontal) (angle above the horizontal the plane is flying is 6.09 degrees)
    ............................|
    ............................| F_weight
    ............................V

    Would you be able to show me the vector diagrams which produce the calculation which pcdagr8 showed?

    --Thanks
     
  8. The OP first says the flight angle above the horizontal the plane is flying 6.09 degrees.

    I don't understand how he obtained this value :confused:

    I mean, if we draw a triangle, its hypotenuse is the thrust force, its opposite side is the force of gravity. So to find the angle we have to use this

    [tex]sin \theta = \frac{mg}{F_{thrust}}[/tex]

    [tex]=sin^{-1}\frac{11968.2}{4.07}= 54.22[/tex]

    So, why is it that I got 54.22° and not 6.09°? Can anyone show me how to work out the flight angle?
     
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