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Force on a helicopter

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data
    A 13000- kg helicopter is lifting a 7100- kg truck with an upward acceleration of 2.1 m/s2. Calculate the force the air exerts on the helicopter blades.

    Do I use the apparent weight equation for this?

    Wapp = m(g + ay)
     
  2. jcsd
  3. Sep 16, 2007 #2
    No it's newton's second law.

    Draw the force diagram, you can assume that the truck and helicopter are 1 object with the mass of truck+helicopter, since you dont need to know what force the helicopter exerts on the truck.

    [tex]\Sigma\vec{F}=ma[/tex]

    you know a, the m, and there is 1 force you know that is acting against the helicopter. and then there's the uknown force that the helicopter+truck exert upward in order to attain the acceleration of 2.1 m/s^2 which exerts some force on the air,
    and by newton's third law exerts some force on the helicopter.
     
  4. Sep 16, 2007 #3
    so...
    F = (20100kg)(2.1m/s^2)
     
  5. Sep 16, 2007 #4
    [tex]\Sigma\vec{F}=20100kg * 2.1 \frac{m}{s^2}[/tex]
    yes but what's going to be on the left hand side?
     
  6. Sep 16, 2007 #5
    the force the air exerts and is there tension force since the plane is carrying the truck by cables?
     
  7. Sep 16, 2007 #6
    well yea the force the air exerts but you might want to think of it as the force the helicopter+truck exert upwards against gravity.

    and the cables might be necessary if you were determining the force exerted onto the truck by the helicopter.

    But there's one force that's pulling the helicopter down. And the force that the helicopter exerts up overcomes this force and gives it an acceleration of 2.1m/s^2
     
  8. Sep 16, 2007 #7
    so we have the force of gravity pulling it down...
    H - W = 21000 kg x 2.1 m/s^2
    H being the force the helicopter+truck exerts?
     
  9. Sep 16, 2007 #8
    yes what you want is that H.
    but that H is going up, the force the blades exert on the air. You want the force that is opposite of that. Using newton's 3rd law, the action or reaction of the pair.
     
  10. Sep 16, 2007 #9

    learningphysics

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    You want the force the air exerts on the helicopter+truck... that's an upwards force. We need the forces on the system. Not the forces by the system.
     
  11. Sep 16, 2007 #10
    so it would be - H
    21000 kg x 2.1 m/s^2 + W = -H
    -21000 kg x - 2.1 m/s^2 - W = H
     
  12. Sep 16, 2007 #11

    learningphysics

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    No. The way you had it the first time was right... H is the upward force the air exerts on the helicopter/truck system.
     
  13. Sep 16, 2007 #12
    ok ok I get it now o.o >< forgot how helicopters worked >.>

    Yes you had it right just add W to both sides of that equation and you'll get H as a positive #
     
  14. Sep 16, 2007 #13
    so if it was by the system it would be - H?
    and now I have...
    21000 kg x 2.1 m/s^2 + 206010 N = 476721 N
    thats big! Is that right or did I miss something?
     
  15. Sep 17, 2007 #14

    learningphysics

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    Well, the answer will be big... but you made a mistake with the mass... it should be 20100 not 21000
     
  16. Sep 17, 2007 #15
    hm..it's not 21000kg it's 20100 hehe.

    It seems right, it's a huge weight O.O like ~27 tons ~ 54k lb's being lifted up, requires huge force, and to accelerate ...
     
  17. Sep 17, 2007 #16
    Okay my mistake...
     
  18. Sep 17, 2007 #17
    Now I have to find the tension in the cable holding the truck so would I just do...
    -T = m x a + W - H
     
  19. Sep 17, 2007 #18

    learningphysics

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    Which body are you taking a freebody diagram of? Deciding that is always the first step.
     
  20. Sep 17, 2007 #19
    the helicopter
     
  21. Sep 17, 2007 #20

    learningphysics

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    Oh... ok, then your equation looks right.
     
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