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Direction of rotation

  1. Sep 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Assuming positive x direction is counterclockwise, then why would the force 50n and 40n are turned clockwise ? (they have negative value)

    2. Relevant equations


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 14, 2015 #2

    SteamKing

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    It's not that the positive x direction is necessarily CCW, it's that a CCW moment about the z-axis is assumed to be positive.

    A positive (CCW) moment could be the result of a positive force multiplied by a positive distance from the axis of rotation or a negative force multiplied by a negative distance from the axis of rotation.

    These force magnitudes are negative because the force vectors are acting in the negative direction.

    Just like the positive moment direction is assumed to be CCW, positive forces are assumed to act in the positive x or y directions.
     
  4. Sep 15, 2015 #3

    haruspex

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    One quibble: a magnitude is by definition non-negative. Vector components?
     
  5. Sep 15, 2015 #4

    SteamKing

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    That's a fair description.
     
  6. Sep 16, 2015 #5
    ok, i understood that when the force 50n fall on 2m(50x2) ( the resultant force vector is in positive k direction , come out of the page . ) . If I do it in the form of 2 X 50 , then it would be in negative k direction , am i right ? can i do in this way ?
     
    Last edited: Sep 16, 2015
  7. Sep 17, 2015 #6

    SteamKing

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    It's not a force vector you are calculating here: it's a moment vector. Moment vectors result in rotation either clockwise or counter-clockwise w.r.t. a given axis, and one direction, either CW or CCW, is assumed to be positive. In the example problem from the OP, you are told that positive moments are assumed to be CCW w.r.t. the z-axis coming out of the paper, which also implies that positive forces are in the directions of the x and y axes.
     
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