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Homework Help: Easy problem, but cant figure it out

  1. Jun 30, 2008 #1
    I've been trying out this problem for hours and cant figure it out.

    A helicopter is moving horizontally to the right at a constant velocity. The weight of the helicopter is W = 52100 N. The lift force vector L generated by the rotating blade makes an angle of 21.0° with respect to the vertical.

    Find the magnitude of the Lift force. And find the resultant vector R.

    http://img237.imageshack.us/img237/9747/0455vd2.gif [Broken]

    At first i tried doing 52100sin69 for the lift..but that is not the correct answer...STUCK!
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jun 30, 2008 #2

    Doc Al

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    What must the sum of the vertical forces be? What's the vertical component of the lift force? (Call the lift force "L".)
  4. Jun 30, 2008 #3


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    Try resolving the lift into horizontal and vertical components. If the helicopter is moving horizontally, with constant velocity, what does that tell you about the net vertical force? What about the net horizontal force?

    Edit: beaten by a matter of seconds there :biggrin:
  5. Jun 30, 2008 #4
    Vertical component of Lift must be equivalent to the magnitude of the weight.
  6. Jun 30, 2008 #5

    Thats what i tried doing....

    the net vertical force is 0, the horizontal force must then be 52100Ncos(69)..however i inputed this answer and it was marked wrong.
  7. Jun 30, 2008 #6
    By the way I'm so confused now..
  8. Jun 30, 2008 #7

    Doc Al

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    What forces act on the helicopter? (List them.)

    What is the vertical component of each force?

    Add up those vertical components and solve for L.
  9. Jun 30, 2008 #8
    52100Ncos(69) is incorrect....
  10. Jun 30, 2008 #9
    I'm kind of at a loss but I'll try:


    In the x direction: Fx=R-v

  11. Jun 30, 2008 #10
    Velocity is not a force. R is related to the horizontal component of the Lift.

    Your Trig. was also incorrect.
  12. Jun 30, 2008 #11
    I don't understand at all. How should I arrange my equation then?
  13. Jun 30, 2008 #12
    Is the lift force going to be 52100N since the helicopter is not moving up or down?
  14. Jun 30, 2008 #13


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    What is fn?

    v is a velocity, and thus should not be in a force equation. You are missing a term including the lift.

    The easier way to do this, is to forget FyNet and Fx, since everything is in equilibrium. You should resolve the lift (L) into a horizontal and vertical component. Then, since the forces are in equilibrium, you should be able to write two equations:

    Vertical: forces pointing up = forces pointing down
    Horizontal: forces pointing right = forces pointing left.
  15. Jun 30, 2008 #14
    There is no acceleration in either the x or y direction. This means the net force must be zero in both the x and y directions. Therefore, the upwards force must be equivalent to the magnitude(absolute value) of the downforce or weight(weight = M*g). Also, R must be equal to the magnitude of the x component of the lift.

    Use 51200/cos(21) = L to find magnitude of lift

    51200N is y component of Lift.
    Last edited: Jun 30, 2008
  16. Jun 30, 2008 #15

    this problem has been eating away at me for four damn hours now..ive tried all of this and ive read more than need to know about helicopters and i still cant seem to find that missing peice of info.

    forces point up=forced pointing down: L=W

    Horizontal: R=52100cos69

    is that right?
  17. Jun 30, 2008 #16


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    No. There is only a component of L pointing vertically upwards, not the entire force, L. Try drawing a triangle with L as the hypotenuse, and the angle between the veritcal and the hypotenuse is 21 degrees. What is the length of the veritcal side?

    This is wrong since the above equation is wrong.
  18. Jun 30, 2008 #17

    Doc Al

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    No. Try answering each question that I asked exactly as I asked it. Hint: Three forces act on the helicopter, all of which are labeled on your diagram. (The velocity is not one of them--that's why its arrow is drawn in a different color than the force arrows.)
  19. Jun 30, 2008 #18
    Ok that helped alot. Cos(x)=adjacent/hypotneuse so cos 21=52100/L or L=52100/cos21

    is that correct?

    For the second part, how is R related to this triangle?
  20. Jun 30, 2008 #19
    Assuming my above post is correct....

    would the Resultant be: 52100tan21=R
  21. Jun 30, 2008 #20
    Yea, that would be equal to the x-component of L(lift) and -R.
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