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Simple Free Body Diagram question

  1. Oct 10, 2004 #1
    I have a question in my text book that I can't answer for the life of me.

    "A 6500-kg helicopter accelerates upward at .6 m/s^2 while lifting a 1200-kg car. (a) what is the lift force exerted by the air on the rotors?"

    Now, my diagram has a vector pointing up with a force of (6500*9.8)N. There is also a vector pointing down at (1200*9.8)N. Wouldn't the net force then be equal to the force of the pull minus the force of the weight? Fn-(6500*9.8)-(1200*9.8) = net force

    Or am I completely wrong? At the moment, having read this section multiple times, I'm still having trouble grasping this whole natural force and the summation of the forces = ma. My teacher doesn't really help much as she isn't the greatest teacher either...
  2. jcsd
  3. Oct 10, 2004 #2

    Doc Al

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    Staff: Mentor

    weight acts down

    Why would the weight of the helicopter point up?
    Yes, if by Fn you mean the force of the air on the rotors. (It looks like you changed your mind about the direction of the helicopter's weight!)

    Your force diagram should show the weight of both acting down, and the force of the air acting up. The net force acts up, since we know the acceleration is up.
  4. Oct 10, 2004 #3
    Oooohhhhhhhh......I see what I did wrong! For some reason, I was thinking the weight of the helicopter was...um...tension up or something. I don't know, but thank you very much!
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