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Specify force magnitude and angle

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  1. Nov 2, 2015 #1
    1. The problem statement, all variables and given/known data
    See picture.

    2. Relevant equations


    3. The attempt at a solution
    I tried drawing a parallelogram, but it didn't help, so I'm stuck. I'd need an idea in the right direction, if possible.
     

    Attached Files:

  2. jcsd
  3. Nov 2, 2015 #2
     
  4. Nov 2, 2015 #3

    gneill

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    I'm not seeing how you've accomplished your decomposition of F1. The magnitude of the vector comprised by those components would be about 367 N, which is larger than the given magnitude of F1. So you might want to re-think your method (whatever that was...).

    The diagram is very helpful because it gives you the angles between F1 and each of the coordinate axes. This should make finding the projections of F1 on each of the axes easy to find.
     
  5. Nov 2, 2015 #4

    SteamKing

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    I don't know what sort of parallelogram you drew, but this is a 3-D kinda problem.
     
  6. Nov 2, 2015 #5
    yes, I was going to draw a parallelepiped to see if it could help me.
     
  7. Nov 2, 2015 #6
    F1z = cos 60 * 300N = 150N
    F1x = cos 45 * 300N = 150*sqrt2 N
    F1y = cos 60 * 300N = 150 N

    Would this be correct?
    I'm confused about the next step..
     
  8. Nov 2, 2015 #7

    gneill

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    For the Z component the angle was not 60 degrees. What does the diagram show it to be? Do you expect the Z-compoennt to be less than or greater than zero?

    Check your value for cosine 45 degrees. A cosine should not be > 1.*

    * EDIT: Oops. I didn't process the fact that the 300 N was divided by two giving 150 N, so the cosine being ##\sqrt{2} / 2## is satisfied! Sorry about that!
    You'll want to write a vector equation that expresses your desired result. Fix your F1 vector first though.
     
    Last edited: Nov 2, 2015
  9. Nov 2, 2015 #8
    F1z = cos 120 * 300N = -0.5 * 300N = -150N (but how does that make sense?)

    cos 45 = (√2)/2
    so F1x = 300 N * cos 45 = 150 N * √2 = 150√2 N

    F1y = cos 60 * 300N = 150 N

    Is this right now?
     
  10. Nov 2, 2015 #9

    gneill

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    Yes, that looks right. The z-component is negative because the vector F1 is directed below the x-y plane.
     
  11. Nov 2, 2015 #10
    so the idea with F2 would be to cancel out the F1z and F1x and amplify F1y by 650N?
    So I need a
    F2z = 150 N
    F2x = -150√2 N
    F2z = 650 N

    Am I going in the right direction?
     
  12. Nov 2, 2015 #11

    gneill

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    Yes, but sort out your component subscripts. You've listed F2z twice. It would help if you put them in standard order, too: x, y, z. Otherwise, I think you're there.
     
  13. Nov 2, 2015 #12
    F2x = -150√2 N
    F2y = 650 N
    F2z = 150 N

    How do I calculate F2?
     
  14. Nov 2, 2015 #13

    SteamKing

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    You don't know how to calculate the magnitude from the three force components?
     
  15. Nov 2, 2015 #14
    F2 = √((-150√2)^2 + 150^2 + 650^2) = 700 N

    To calculate the angles of F2 I used:

    cos x = (F2x/F2) = (-150√2 N / 700 N) ---> x = 72.36°
    cos y = (F2y/F2) = (650N / 700 N) ---> y = 21.79°
    cos z = (F2z/F2) = (150N / 700 N) ---> z = 77.63°

    Correct?
     
  16. Nov 2, 2015 #15
    Sorry, messed up x:
    cos x = (F2x/F2) = (-150√2 N / 700 N) ---> x = 107.64
     
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