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Resolution of Forces Help Please

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data
    A particles is under the influence of two forces, 500N at 50° from north and 350N at 350° from north. Find the magnitude and direction of the resultant pull on the particle.


    2. Relevant equations

    ?

    3. The attempt at a solution
    [itex]
    x=500cos40-350cos80\\
    x=322.25N\\
    y=500sin40+350cos10\\
    y=666.08N \\

    \sqrt{322.25^2 + 666.08^2}=739.94N \\
    arctan(322.25/666.08)=25.82° from north. \\
    [/itex]
    I am a bit confused on when to use cos or sin. The above is how a classmate pursaded me to change it to but on my first go I had
    [itex]
    x=500cos40-350sin10 \\

    y=500sin40+350cos10 \\
    [/itex]
     
    Last edited: Feb 20, 2013
  2. jcsd
  3. Feb 20, 2013 #2

    rude man

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    Homework Helper
    Gold Member

    Draw the vectors on an x-y plot (+y = North). It should be obvious from that.
     
  4. Feb 20, 2013 #3
    Thanks for your reply. I have already done that and the figure and direction obtained seem correct but I dont know for sure and if my method is correct.

    EDIT: Sorry just realised my mistake. They are both the same but doing it a different way, my calculator was set on radians instead of degrees and that is why I was getting different figure to earlier :redface: sorry.
     
    Last edited: Feb 20, 2013
  5. Feb 20, 2013 #4

    Redbelly98

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    Staff Emeritus
    Science Advisor
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    A very common mistake. Remember to check the setting whenever you have an exam :smile:
     
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