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Resolving and reducing forces.

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data
    I am trying to find the resultant forces in the attached picture.


    2. Relevant equations



    3. The attempt at a solution

    The resultant of a couple of forces = 0
    F in x direction = 5cos(45)
    F in y direction = -5sin(45) - 2

    Therefore:
    R = ((5√2)/2)i + ((-(5√2)/2)-2)j
    R = √(3.53^2+(-5.53)^2) = 6.56kN

    is this correct does the distance of the 2kN force come in to play somewhere. If so where?
     

    Attached Files:

  2. jcsd
  3. Apr 19, 2012 #2

    tiny-tim

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    hi mm391! :smile:

    looks ok so far …

    now you need to find the line along which it acts (of all the possible parallel lines) :wink:
     
  4. Apr 20, 2012 #3
    Thanks Tiny Tim but I am still al ittle confused.

    What do you mean "find the line along which it acts" ? What line. All I can see it that the 2kN force acts 6 meters from point P but I am not sure how to incorporate that into the question.

    Thanks
     
  5. Apr 20, 2012 #4

    tiny-tim

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    a force that is (say) 10 N north-east can be applied anywhere along a rigid body

    if applied on the left side, it will turn it clockwise, if applied on the right side, it will turn it anti-clockwise

    the point of application (or the line of application, through that point) matters, and when specifying the resultant of forces, you must also specify the line of application :wink:
     
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