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Finding the direction of the resultant

  1. Oct 4, 2011 #1
    A force Farrowbold1 of magnitude 5.30 units acts on an object at the origin in a direction θ = 51.0° above the positive x-axis. (See the figure below.) A second force Farrowbold2 of magnitude 5.00 units acts on the object in the direction of the positive y-axis. Find graphically the magnitude and direction of the resultant force Farrowbold1 + Farrowbold2.

    the magnitude is 9.71
    and the direction is 69.9 ° counterclockwise from the +x axis

    where i am stuck is the direction part. I dont know what angle they are talking about to find it.
    Could you please help explain how to find the direction of the resultant force counterclockwise from the +x axis?
     
  2. jcsd
  3. Oct 4, 2011 #2

    gneill

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    Staff: Mentor

    The angle is the angle that the resultant vector makes with the positive x-axis. If you sketch the vector on a set of coordinate axes (using its x and y components) then you should be able to identify a trig function or two that will give you the required angle.

    Note: You have to be a bit careful when the vector lies in quadrants where one or both components may have negative values.
     
  4. Oct 4, 2011 #3
    Could you tell me what you did. I closed this problem and a new practice problem appeared but the same concept. I still cannot get the correct direction. I thought it was suppose to be tan^-1(opposite length/adjacent length) to find the angle but it keeps on telling me my answer is off...
    i used the resultant triangle and found tan^-1(y/x) but it is incorrect.


    Edit: they keep saying that my answer is within 10% of the correct answer. It could be my rounding off, but if you tell me the correct way of finding it would help
     
  5. Oct 4, 2011 #4
    the magnitude is off a bit because i think of my rounding off.
    but please look at my picture and where do i go from there to find the angle of the resultant counterclockwise from the +x axis


    oops in my picture i put R^1/2, instead it should be ^2
     

    Attached Files:

  6. Oct 4, 2011 #5
    nevermind, i found my huge mistake.
     
  7. Oct 4, 2011 #6

    gneill

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    Staff: Mentor

    For any vector given in x and y components you should first determine which quadrant of the Cartesian plane the vector lies in. This is done by looking at the signs of the components. A quick sketch on a set of coordinate axes will make it clear.

    Then you want to find the angle. How you do this will depend upon which quadrant the vector lies in. Sometimes it will be easier to find the angle with respect to some other axis first, then add or subtract appropriate multiples of 90 degrees to make up the correct offset. It should be clear if you make the sketch.

    In the present case, both components of the resultant are positive, so the vector lies in the 1st quadrant. The positive x-axis lies below it, so the angle is simple atan(y/x), where y and x are the vector components. (note: atan(y/x) is the same as tan-1(y/x) ). So you were working correctly.

    If the angle you're getting looks incorrect, check to make sure that your calculator is returning angles in degrees rather than radians. It's also best to carry a few extra decimal places for intermediate results. Round final results to the appropriate number of significant figures.
     
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