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Magnitude and direction of Vectors using head to tail rule

  1. Jan 2, 2013 #1
  2. jcsd
  3. Jan 2, 2013 #2
    I think you've got the directions wrong. The x-component of forces are in opposite directions, so you need to account for that.

    You've calculated the y-component of the resultant correctly, but even there you haven't specified the downward direction which could cause you to lose marks sometimes.

    Since, your calculation for x-components is wrong, the magnitude and direction of the resultant is also wrong...:smile:
     
  4. Jan 2, 2013 #3

    jedishrfu

    Staff: Mentor

    the first two look okay the last page #3 you used vector length value so your tan angle is wrong.

    A quick way to check your solution is to graph it and take measurements for the angle and length to see that your results agree.
     
  5. Jan 2, 2013 #4
    Shouldn't Fx=-6000cos(60°)+2000cos(45°)?
     
  6. Jan 2, 2013 #5

    jedishrfu

    Staff: Mentor

    Yes, I think you're right. I didn't notice that.
     
  7. Jan 2, 2013 #6
    so after doing Fx=-6000cos(60°)+2000cos(45°)
    resultant force is 6797.915 correct?
     
  8. Jan 2, 2013 #7
    Yes. And what will be the angle made by that resultant with the positive x-axis??
     
  9. Jan 2, 2013 #8
    [itex]tan\theta = f_y/F_x[/itex]
    76.51(i didn't use minus sign in f_y)
    180-76.51= 103.48
    right?
     
  10. Jan 2, 2013 #9
    Correct! :)
     
  11. Jan 2, 2013 #10
    Thanks for the help
     
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