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Homework Help: Newton's 2nd Law confusing!

  1. Sep 25, 2008 #1
    A helicopter is moving horizontally to the right at a constant velocity. The weight of the helicopter is W = 57600 N. The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical.

    (a) What is the magnitude of the lift force in N?

    (b) Determine the magnitude of the air resistance R that opposes the motion.

    I used [tex]\Sigma[/tex]Fx= max--> 0 (because constant velocity means 0 acceleration)= Lsin(21)

    and [tex]\Sigma[/tex]Fy= may-->Lcos21-57600

    now I dont know what to do. What did i do wrong?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 25, 2008 #2
    Hi PhysicsFailure!

    I am a new member and somewhat new to physics compared to most here but I think I can help you out on this one.

    First, we know that the helicopter is not accelerating in either the X OR Y directions (it is moving strictly horizontally). For this reason, the sum of all forces in each direction is 0, not just X.

    Have you tried drawing a simple diagram, with the helicopter represented as a point at the origin, the weight acting downward, and the force of lift acting upward at an angle of 21 degrees to the Y axis?

    I think if you set the Y equation you already have to 0, the force of lift will become apparent. From here, figure what component of that lift on the X axis must be matched by the air resistance for the copter to not have any acceleration (sum of X forces=0)

    Let me know if you get it!
  4. Sep 25, 2008 #3
    So far so good! It looks as though you've done everything right.

    "I used [tex]\Sigma[/tex]Fx= max--> 0 (because constant velocity means 0 acceleration)= Lsin(21)"

    So then what is the force due to air resistance?

    "and [tex]\Sigma[/tex]Fy= may-->Lcos21-57600"

    Part a is asking for L. How would you solve for L in this case?

    Keep in mind that because there is no vertical acceleration, the net force in the y direction will also be zero.

  5. Sep 26, 2008 #4
    haha thanks so much bchandler and mattowander. I actually figured out what I was not doing. I could just solve it! and for some reason i was just not thinking...THANK YOU!!!!!!!!
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