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Right Hand Rule - Positive or Negative Action

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img41.imageshack.us/i/photolu.jpg/

    2. Relevant equations

    Right Hand Rule

    3. The attempt at a solution

    I have to find the moments at B. I know I have to use the right hand rule to figure out which force has a positive and which force has a negative action. However, I'm not too certain on how to do this. I know how to position my right hand, but I don't know whether to use the "up thumb" or "down thumb" for a specific point.

    Please Help.
     
  2. jcsd
  3. Oct 20, 2009 #2

    tiny-tim

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    Homework Helper

    Hi tornzaer! :smile:

    I find the easiest way to remember the right-hand rule is that if you draw an x,y graph the usual way, then x cross y is along the usual z-axis, drawn upwards. :wink:

    (Alternatively, see http://en.wikipedia.org/wiki/Right_hand_rule)
     
  4. Oct 20, 2009 #3

    Doc Al

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    Staff: Mentor

    I always use the "curled fingers" version of the right hand rule. To find x cross y, I curl the fingers of my right hand from x to y--that automatically makes your thumb point in the direction of the z axis. (At least it does once you get the idea.)

    Here's a site that illustrates how to use that version of the rule for just about every possible combination of vectors (some of the gifs don't work quite so well--the quicktime movies are better): http://physics.syr.edu/courses/video/RightHandRule/index2.html" [Broken]
     
    Last edited by a moderator: May 4, 2017
  5. Oct 20, 2009 #4
    Thanks for the replies. What happens if vectors aren't connected like the ones on the diagram I posted. There are vectors in the same direction and different directions, but they are all separate.

    Also, on question b on the diagram, it asks to find the sum of moments at B. That's what I'm confused about.
     
  6. Oct 20, 2009 #5

    tiny-tim

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    Not sure what you mean by "connected like the ones on the diagram" … they don't look connected to me. :confused:

    If the two original directions are not perpendicular, just move your middle finger until the middle and index finger fit (if necessary, you can bend the middle finger in the middle!).

    The thumb direction always remains perpendicular to the plane of the other two.

    Having said that …

    I never use the right-hand rule, I always use the i j k method instead. :wink:
    For each force the moment (same thing as torque :wink:) is position x force

    so just draw the position vector from B to the point of application of each particular force, and proceed as before. :smile:

    (and also remember that, in a 2D situation, each force will be going either clockwise or anti-clockwise round B, so you can just work out the cross-product direction of one of them, and then all the others will be either in the same direction or in the exact opposite direction, according to whether they're also clockwise or anti-clockwise. :wink:)
     
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