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Statics Question with Couple Moments and Right Hand Rule Comprehension

  1. Jul 2, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm having problems with understanding how to use the right hand rule as it applies to forces and determining moments. If I understand correctly, The fingers point in the direction of the force, then curl in the direction of r, but I'm not sure what r would be in this case.


    2. Relevant equations
    M=r X F

    3. The attempt at a solution

    My main struggle is with the force in the upper right corner. If I'm trying to find the moment about point A from this force, I point my fingers in the direction of F, then curl them in the direction of r which I'm considering to be from A to F, and I wind up with my thumb pointing into my computer screen. I'm not sure if this is correct.
  2. jcsd
  3. Jul 2, 2013 #2

    Simon Bridge

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    In that picture there are several forces - each acting at a different place.
    Your first step is to identify the pivot - which you identify as point A.
    The moments will all be "clockwise" or "anticlockwise" - so there is no need for any "hand rule"; right, left, or otherwise. You should be able to tell by just looking at it which way each force should be trying to turn the beam. Just pick one turning direction to be positive.
  4. Jul 2, 2013 #3
    Hi Sara,

    The best way for you to understand how the forces all act relative to point A is to break them down into their rectangular components.

    When I used to solve these types of problems at university I would always draw the following diagram to communicate to the people marking my assignments/exams what I took to be positive i.e. my sign convention.


    Any forces that point in the same direction as either my Y or X arrows I would take as positive forces. Any moments (force x lever arm) that would rotate in the anti-clockwise direction I would also take as positive. When students are first introduced to the concept of torque or bending moments, many tutors use the "right hand rule" to establish a sign convention. Simply raise your right hand in front of you, point your thumb towards your face and rotate your wrist such that your four other fingers rotate in a anti-clockwise motion. Your fingers will rotate in the same direction as the curved arrow in the sketch above. Any forces that try to spin your structure in this direction will be denoted as a positive moment.

    Let's now take the exercise you've been looking at and apply this.

    To better understand how the force of 3kN acting in the "upper right corner" acts on your structure, it's best that you break the force into its rectangular components. In the next diagram I have illustrated this for you.


    I have annotated in red the rectangular components in which the 3kN force at 30 degrees can be represented in the "X-Y" Cartesian plane. Your understanding of basic vectors will allow you to better represent (and calculate both Fx and Fy) these forces as follows:


    With this information I now pose to you the questions:
    1. Is the Force Fy positive or negative relative to your newly formed sign convention?
    2. What is the magnitude of Force Fy?
    3. In what direction will Force Fy cause the structure to spin about point A?
    4. Using our newly formed sign convention, is this spinning in the same direction as our curved arrow/fingers? Hence, is the direction of rotation positive of negative?

    Should you be able to confidently answer these questions for Force Fy, attempt to do the same with Force Fx.
  5. Jul 2, 2013 #4

    Simon Bridge

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    That's very clear and succinct ... however, isn't it more useful to resolve the force into components perpendicular and parallel to the moment arm rather than to the vertical and horizontal - especially, as in this case, when the moment arm makes an angle to these directions?

    Compare, also, your right hand rule above with the right-hand rule more usually applied to cross products - i.e.
  6. Jul 2, 2013 #5


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    It is, but this quickly becomes rather complicated, especially for 3-D problems. The simple way to do this is vectorially, using M = r x F, as stated in the OP.
  7. Jul 3, 2013 #6

    Simon Bridge

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    ... agreed. Even so - you don't get to just add the moments unless they share an axis right?
    Fortunately this case is in 2D ... I was hoping to encourage Engineering01 to refine the discription, or clarify it's concepts, since it had a lot of worthy elements.

    Meantime - waiting for OP feedback :)
    Knows about cross products, therefore does not need hand rule.
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