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Expressing force in cartesian vector form

  1. Sep 20, 2016 #1
    1. The problem statement, all variables and given/known data

    Hibbeler14.ch2.p65.jpg

    Express F1 and F2 in cartesian vector form
    2. Relevant equations


    3. The attempt at a solution
    I feel fairly confident in my work for F2 but F1 not so much. Especially the j component of F1. I assume my F(1x) component runs from the z axis to the tip of F1 and my F(1k) component runs from the tip of F1 to the line below (which forms the light blue triangle). F(1y) runs along the y axis and forms the base of the dark blue triangle (I think). Any help would be much appreciated!

    F(1x)= -300cos60 = -150lb
    F(1y) = 300cos60sin45 = 106lb
    F(1z) = 300sin60 = 260lb
    F1 = -150i + 106J + 260k

    F2 = 50cos60i + 500cos45j + 500cos120 = 250i + 354j - 250k
     
  2. jcsd
  3. Sep 20, 2016 #2

    TSny

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    I don't think this is correct.

    All of your other components for F1 and all of your components for F2 look correct to me.
     
  4. Sep 20, 2016 #3
    Are you saying my F(1x) does not look correct?
     
  5. Sep 20, 2016 #4

    SammyS

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    It looks to me that the base of the F1 parallelogram (rectangle) in the figure lies in the x-y plane and makes an angle of 45° with the y-axis.

    The F1 makes an angle of 60° with the base of the parallelogram (rectangle).
     
  6. Sep 20, 2016 #5
    I agree. What part of my work do you not think is correct?
     
  7. Sep 20, 2016 #6

    SammyS

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    The magnitude of your vector is not 300 N.
     
  8. Sep 20, 2016 #7

    TSny

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    Yes, I don't agree with your result for F1x. I think it should be similar to your calculation for F1y.
    For your calculation of F1y, why did you use the sine of 45 rather than the cosine of 45?
     
  9. Sep 20, 2016 #8
    I figured the cosine would give the base of the parallelogram. I thought I needed the base of the dark blue triangle along the y axis to give the y component. I though 300 cos30 was the hypotenuse of the dark blue triangle.
     
  10. Sep 20, 2016 #9

    TSny

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    Yes. But isn't the base of the dark blue triangle (along the y axis) the "side adjacent" to the angle marked 45 degrees in the triangle? So, I would think you would use the cosine rather than the sine of 45 degrees. (Of course they happen to be equal for 45 degrees, but if it had been 40 degrees you would need to choose the correct trig function.)
    Yes, that's right. Isn't the x component of F1 equal to the other leg of the dark blue triangle?

    Maybe my old eyes aren't seeing the diagram correctly.
     
  11. Sep 20, 2016 #10

    SammyS

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    I agree, (except for the sign).
     
  12. Sep 20, 2016 #11

    TSny

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    Yes. The sign needs to be considered!
     
  13. Sep 21, 2016 #12
    So, F(1x) = -300cos60sin45
    F(1y) = 300cos60cos45
    F(1z) = 300sin60
     
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