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Explain to me why the net force is found this way?

  1. Jun 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Two crewmen pull a raft through a lock. One crewman pulls with a force of 130N Southwest at an angle of 34 degrees below the horizontal. The second crewman pulls at an angle of 45 degrees above the horizontal Northwest. With what force should the second crewman pull so that the net force of the two crewmen is in the forward direction?


    2. Relevant equations



    3. The attempt at a solution

    I am pretty sure I am missing the obvious here but why would you do:
    Fsin45 = 130sin34 ?
     
  2. jcsd
  3. Jun 10, 2013 #2
    Answer looks correct. Do you have any problem in the method used?
     
  4. Jun 10, 2013 #3
    Yeah, I don't quite understand why you would do it that way.

    I was thinking of Fcos45 = 130cos214 since it's on the x-axis.
     
  5. Jun 10, 2013 #4

    haruspex

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    What is the 'forward direction'? West?
    When you write below/above horizontal, do you mean, respectively, South and North of West?
     
  6. Jun 10, 2013 #5

    CWatters

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    "Cos" gives you the force in the dirction the boat is moving.

    "Sin" gives you the side to side force on the boat. If you want the boat to go straight ahead what must that sum to?

    [STRIKE]The equation..

    Fsin45 = 130sin34

    ..is actually missing two terms (one on each side) that have been cancelled because they are equal.[/STRIKE]

    Edit: Scratch that last bit.
     
    Last edited: Jun 10, 2013
  7. Jun 10, 2013 #6

    CWatters

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    I think a diagram would save time.
     
  8. Jun 10, 2013 #7
  9. Jun 11, 2013 #8

    haruspex

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    If you wanted to calculate the net force in the forward (X) direction you would use cos, but that's not what you're after here. What you need is for the Y components of the two forces to cancel, so the angle of interest in each case is to the Y axis.
     
  10. Jun 11, 2013 #9

    CWatters

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    The diagram clarifies the problem. It confirms it's just a 2D problem not a 3D.

    I have split the vectors into two components which might help. Remember that..

    Sin(∠) = Opposite / Hypotenuse

    so

    Opposite = Hypotenuse * Sin(∠)
     

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    Last edited: Jun 11, 2013
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