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Graphical method to find magnitude and direction

  1. Feb 4, 2010 #1
    1. The problem statement, all variables and given/known data

    The force vector FA has a magnitude of 90.0 newtons and points
    due east. The force vector FB has a magnitude of 135 newtons and
    points 75° north of east. Use the graphical method and find the
    magnitude and direction of (a) FA - FB (give the direction with respect
    to due east) and (b) FB + FA (give the direction with respect
    to due west).

    3. The attempt at a solution
    So confused here. I have tried this all kinds of ways and cannot get the answer the book came up with.
    I have tried:
    a. R = sqrt((90)^2 + (135)^2) = 162 N
    For direction I did tan-1 (135/90) = 56.3 degrees South of east.
    B. Same answer just N. of east.

    The book sais the answer are 142 N for both and 67 degrees.
    Help!!!!! I need step by step because I just don't get it. This is not for a homework assignment I am just doing the odds out of the back of the book for practice for our test.
     
  2. jcsd
  3. Feb 4, 2010 #2
    It is talking about a graphical method. What about drawing the vectors, adding/substracting them and measure the result. You add two vectors by starting the second at the end of the first, and you substract two vectors by ending the second at the end of the first.

    For a coordinate geometry solution you should express x and y coordinates of the vectors, and add/substract them.

    You need definition of sine and cosine to convert from (length,angle) to (x,y). You need pythagoras and definition of tangent to convert back.
     
  4. Feb 4, 2010 #3

    tiny-tim

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    Welcome to PF!

    Hi xupe33jrm! Welcome to PF! :smile:

    (have a square-root: √ and try using the X2 tag just above the Reply box :wink:)
    No, that would be if the 90 and 135 were perpendicular.

    Draw a vector triangle …

    two of the sides will be 90 and 135 …

    now use the cosine and/or sine formula.

    What do you get? :smile:
     
  5. Feb 4, 2010 #4
    Could someone maybe draw me a diagram on how to set it up, because I am just not understanding. This problem is driving me nuts. If I knew how to draw it maybe I can figure out the math! thanks!
     
  6. Feb 4, 2010 #5

    tiny-tim

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    Hi xupe33jrm!! :smile:

    FA: 135 "horizontal", arrow pointing right.

    FB: 90 slanting up and to the right, almost "vertical", arrow slanting up and to the right.

    FA - FB: Start both FA and FB at the same place, then FA - FB is the unknown length joining the ends, arrow pointing down (so that you follow the arrows: FB + (FA - FB) = FA :wink:).

    FA + FB: Start FB from the end of FA (not the beginning), then FA + FB is the unknown length joining the ends, arrow pointing up (so that you follow the arrows: FA + FB = (FA + FB))
     
  7. Feb 4, 2010 #6
    Why would 90 not be on the east horizontal since it points due east? Also when trying to find the answer what equation do I use. I am really confused and I know it is probably a simple problem.
    Is the drawing I included look right and if so am I trying to find the green side?
     

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  8. Feb 4, 2010 #7

    tiny-tim

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    oh sorry, i got the 90 and the 135 the wrong way round. :redface:
    Yes, but you should always put the arrows in, or you may end up drawing the wrong triangle.

    Now use the cosine formula.
     
  9. Feb 4, 2010 #8
    I got the answer, thanks so much guys. I was getting very angry with this problem, but know it seems so easy!!!
     
  10. Sep 14, 2011 #9
    I still don't get this problem can someone explain please!!!
     
  11. Sep 15, 2011 #10

    tiny-tim

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    welcome to pf!

    hi camrik55! welcome to pf! :wink:

    show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  12. Sep 15, 2011 #11
    I finally found it out. Thank you though. I am sure I will have more questions very soon!
     
  13. Sep 19, 2011 #12
    I've been trying to figure this out for hours and I know it is easy but it keeps tricking me up. I know you need to use the equation W=(F cos theta) but my answer doesn't seem to come out. Here is my problem

    A force does +170 J of work when it acts on a moving object and its direction is in the same direction as the object’s displacement. How much work does this force do when the angle between it and the object’s displacement is 56°?
     
  14. Sep 20, 2011 #13

    tiny-tim

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    hi camrik55! :smile:

    (have a theta: θ :wink:)
    the force, F, and the https://www.physicsforums.com/library.php?do=view_item&itemid=378", d, are the same in both cases …

    so write out the equation for F d and 170, and then see how it changes if you change the angle to 56° (from 0°) :smile:
     
    Last edited by a moderator: Apr 26, 2017
  15. Jun 17, 2012 #14
    i dont get this problem.
    what is the formula for this problem?
     
  16. Jun 17, 2012 #15

    cepheid

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    The relevant equation is W = Fdcos(theta) as someone already mentioned.
     
  17. Jun 17, 2012 #16
    based to my friend the graph is like this..?
     

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  18. Jun 17, 2012 #17
    My mistake, yes the graph is correct.
     
  19. Jun 17, 2012 #18

    tiny-tim

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    for FA + FB, yes :smile:
     
  20. Jun 17, 2012 #19
    what graph is correct?

    is this problem use pythagorean theorem or component method?
     
  21. Jun 17, 2012 #20

    tiny-tim

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    you can always use either the cosine formula or the component method

    (pythagoras only works for right-angled triangles … for general triangles you need the cosine formula)
     
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