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Magnitude and direction

  1. Jan 26, 2004 #1
    My problem is to determine the magnitude and direction of the resultant force FR=F1+F2 and it's direction, measured counterclockwise from the positive x direction.

    f1=250 lb @ 60 degrees from x
    f2= 375 lb @ -45 degrees from x

    Ok I got the magnitude by doing this:
    (360-2(255))/2=-75 degrees

    fr=sqroot of (250^2+375^2-2(250)(375)cos(75)
    fr=393.188~ 393

    then I got the angle by this:
    375/sin x = 393.188/sin 75
    how do i get the resultant angle, what do I need to add to the 67 degrees?
  2. jcsd
  3. Jan 26, 2004 #2
    breakdown x and y components of each by the following

    X= r cos(angle)
    Y= r sin(angle)

    Add the two x and y compents to get the x and y of the resultant. Use the pathagorean theroum to get the resultant magnitude and use tan^(-1) (y/x) to get resultant directions.

  4. Jan 26, 2004 #3
    Preview Rough fig in attachment so

    [tex] tan{\alpha} = \frac{ysin{\theta}}{x+ycos{\theta}}[/tex]

    [tex] tan{\alpha} = \frac{ysin{\theta}}{x+ycos{\theta}}[/tex]

    Attached Files:

    Last edited: Jan 26, 2004
  5. Jan 26, 2004 #4
    Thank You

    Thank you both for your help.
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