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Net External Force

  1. Oct 27, 2013 #1
    1. A crate is pulled to the right with a force of 82 N, to the left with a force of 115N, upward qith a force of 565N and downward with a force of 236N.



    2. what is the magnitude and direction of the net external force



    3. i solved for magnitude(330N) but im not sure how to solve for direction cus i did tan^-1(329/33) and got 84° and i should be getting 96°...?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 27, 2013 #2

    arildno

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    "i solved for magnitude(330N) but im not sure how to solve for direction cus i did tan^-1(329/33) and got 84° and i should be getting 96°...?"
    ----
    On which side of the y-axis does the force vector reside?
    In the first quadrant, or in the second?
     
  4. Oct 27, 2013 #3
    its on the negative y aXIS i think
     
  5. Oct 27, 2013 #4
    oh wait no sorry its in the second quadrant!
     
  6. Oct 28, 2013 #5

    arildno

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    That's right!
    So:
    Since your ordinary inverse tangent function gives values between -90 degrees and +90 degrees, these lie in the first and fourth quadrant.

    Furthermore:
    You chose to suppress the minus sign in the x-value for the tangent value, and therefore your calculator gave you the answer 84 degrees.
    You would then have to add the deviation from 90 degrees in order to get the correct value 96 degrees.

    If you had kept the minus sign, you would have gotten -84 degrees as your answer; adding 180 degrees to this (remember tanget is periodic with 180 degrees as its period!), would also have given the correct answer.
     
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