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Mechanics problem involving trig and forces.

  1. Aug 1, 2010 #1
    A fully loaded Cessna-182 airplane of mass 1260 kg has an engine failure when flying with an airspeed of 126 km/h at an altitude of 2870 m on a calm day. It then glides at a constant glide angle (which is the direction of flight below the horizontal) towards a safe landing at this constant speed of 126 km/h experiencing a drag force of 1280 N that opposes the direction in which the plane is moving.
    Please use: g = 9.81 m s-2


    [PLAIN]http://img805.imageshack.us/img805/5625 [Broken]

    I would like to know what equations and values i should be using to find the glide angle of the plane.

    I tried using the forces given, weight and drag, with trig but as I know that the drag force is not a component force of weight I didn't have high hopes. So of course i got the wrong answer.

    Will upload the working once I draw the diagram I drew in paint.

    I know the answer is 5.94 degrees.

    Thanking you in advance. Been banging my head around this problem I just don't see what to use to work out the glide angle.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 2, 2010 #2

    Lok

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    If something is falling, what equations do we use?

    If you have a Force (be it Drag) exerted over a distance what can you get out of it?
     
  4. Aug 2, 2010 #3
    hey lok i presume we are talking about the kinematics equations. I will have a look back and report to you in a min but i dont see how i can get the angle using those equations.

    actually its late in NZ so imma c this tomorrow but i think i got it, just need to manipulate the equations properly.
     
    Last edited: Aug 2, 2010
  5. Aug 2, 2010 #4

    Lok

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    Well you know the speed it moves forward (or at the unknown angle) and all you have to find is the speed it is falling considering it is dropping at a constant speed attracted by gravity and slowed down to a constant speed by friction.

    And then there is a bit of Trig...
     
    Last edited: Aug 2, 2010
  6. Aug 2, 2010 #5
    Hey lok sorry for the late reply i did work it out last night. just been busy and havent been able to put it up i will in an hour or 2 though thanks for the help.
     
  7. Aug 3, 2010 #6

    Lok

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    Np. Anytime.
     
  8. Aug 3, 2010 #7
    I used the drag force divided by the weight force, and then took the inverse sin of the answer.

    Θ = sin-1 ((drag force)/(mg))


    Unfortunately i cant explain y i used this as that will require me to do a diagram which i find time consuming, and as i'm short on time i cant help it SORRY !
     
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