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Forces and state of equilibrium

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Assuming an aircraft has a force of gravity of 6.6x10^4N [down] acting on it as it travels at a constant velocity of 6.4x10^2 km/h [w]. If the engine is 1.3x10^4N [W] determine:
    (a) The upward lift force on the plane
    (b) The force due to air resistance on the plane


    2. Relevant equations
    Trig


    3. The attempt at a solution
    (a) I drew out a vector diagram which looks very basic with Fg and Fe( force of the engine)
    Because the aircraft is moving at a constant velocity we know that ƩF=0
    I went about it in a very basic was, so i'm not sure of it is correct:
    ƩFx=1.3x10^4N ƩFy= -6.6x10^4N
    ƩF=√(1.3x10^4N)^2 + (-6.6x10^4N)^2
    F=6.7x10^4N
    therefore, the upward life force on the plane is 6.7x10^4N[North of west]
     
  2. jcsd
  3. Oct 3, 2011 #2
    Is the above the pulling force of the engine?
     
  4. Oct 3, 2011 #3
    oh sorry, "If the forward thrust provided by the engine is..."
     
  5. Oct 3, 2011 #4
    OK
    How many forces act on the plane?
     
  6. Oct 3, 2011 #5
    Two forces, Negative force of gravity, and the thrust of the engine.
     
  7. Oct 3, 2011 #6
    If those are the only two forces on the plane then the plane will fall down along the resultant of these two forces.
    THINK again!
     
  8. Oct 3, 2011 #7
    In order for the plane no remain airborne there needs to be a normal force as well to cancel out the force of gravity and air resistance to counter the forward thrust of the engine.
     
  9. Oct 3, 2011 #8
    Correct. Here we better call the normal force by the name 'upward lift'.
    So now you know the answer to(a) and (b).
     
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