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If you have a moment (at the end of a rod)

  1. Mar 18, 2014 #1
    1. The problem statement, all variables and given/known data
    Lets say you have some moment (given) at the end of a rod going clockwise. Do you have to factor this into the equations for forces? My professor has some notes written about it, but I don't know if he's incorrect.

    2. Relevant equations

    3. The attempt at a solution

    So if the moment is 1000 Nm and the distance from a point A on the rod is 10m. Would the force be factored into the equation for the forces in the y-direction (so it would be 100 N)?
    Last edited: Mar 18, 2014
  2. jcsd
  3. Mar 18, 2014 #2
    You might want to draw a free body diagram to clarify exactly what you are talking about but generally the answer is that a moment can be applied anywhere along the rod and it does not affect summing the "forces" in any direction.

    A moment can be considered as a couple: an equal and opposite set of forces F1=-F2 that cancel each other out (∑F=0) but have a distance "d" between them that cause rotation of a body
    ∑M = F1*d = F2*d.
  4. Mar 18, 2014 #3
    Yes, you will have to factor the force applied at the end of the rod if asked to calculate the sum of the forces acting on the rod. For example, if given the moment, distance of the applied force from the pivot, and the weight of the rod, and you were taking positive to be in a downward sense, the sum of the forces, neglecting any air resistance would be: $$∑F = \frac{moment}{distance from pivot} + mg - F_{normal}$$ since the force applied caused the rod to move in a clockwise sense.
    Last edited: Mar 18, 2014
  5. Mar 18, 2014 #4
    A moment cannot be considered a couple unless there is an equal and opposite moment involved. In a moment there is one force involved. This force is not canceled out with an equal and opposite force and would cause the rod to rotate unless there is some type of resistance involved, therefore this is a force applied to the rod in a downward sense that must be included in the sum of the forces.
  6. Mar 18, 2014 #5
    I think you might be assuming the rod is in equilibrium possibly? The OP's example was talking about ∑F in the y-direction and made no mention of a pivot and made no mention about the rod being in equilibrium so I think you have misunderstood his question. Like I said, a free body diagram would help to clarify what he is talking about specifically.

    Also all moments are caused by and only by couples (in fact the words are interchangeable).

    For example, let's say you were to weld another rod that was 90 degrees to the OP's rod but you didn't know how long this new rod was (maybe it was hidden by a curtain). You could only measure the forces caused by the new rod on the OP's rod. If you measured a rotation and no translation then you would conclude that there must be two equal and opposite forces that cancel each other out but are non-concurrent and therefore causing the rotation. The forces could be of any magnitude and any distance apart and located at any distance from the OP's rod but the moment would always be the same that you measured: M = F1*d = F2*d =1000 N*m.
  7. Mar 18, 2014 #6
    If there is a moment on the rod, there is a downward force applied to the rod in the y direction causing a moment in the clockwise sense about the pivot of the rod. Couples and moments are similar but not interchangeable. A moment involves one force, whereas a couple involves two equal and opposite forces; a couple is simply a special case of moment where there is no resultant force.
  8. Mar 18, 2014 #7
    I claim you are wrong and don't really understand what is meant by a moment.

    A moment is a couple and can only be caused by a couple. If there is also a downward force then it is not a pure moment. The OP only talked about applying a 1000 Nm moment not a downward force or any other force for that matter.

    Again, draw yourself a free body diagram and convince yourself I am right.
  9. Mar 18, 2014 #8
    That's the problem. My professor didn't include the moment in his force calculations (which I thought was incorrect). So I brought it up to him after class (he's one of those really defensive professors...) and just kind of agreed with him.
    Last edited: Mar 18, 2014
  10. Mar 18, 2014 #9
    Please take a look at this link, this should help: http://www.differencebetween.com/difference-between-moment-and-vs-couple/

    Moment or more precisely the moment of a force is a measure of the turning effect of a force. Moment of force is measured in Newton meters (Nm) in the SI system, which looks similar to the unit of mechanical work but carries completely different meaning.
    When a force is applied it creates a turning effect about a point other than on the line of action of the force. The amount of this effect or the moment is directly proportional to the magnitude of the force and the perpendicular distance to the force from the point.

    Read more: http://www.differencebetween.com/difference-between-moment-and-vs-couple/#ixzz2wLwuKONN

    When two equal and opposite forces, but with separate lines of action are present in a force system it is called a couple. Both forces create their own moment of force, but the net moment of the couple is independent of the location of the point considered.
    The moment of a couple is given by;

    Read more: http://www.differencebetween.com/difference-between-moment-and-vs-couple/#ixzz2wLx4EVDs
    Last edited: Mar 18, 2014
  11. Mar 18, 2014 #10
    OK, now we are talking about something in equilibrium. In that case we need another couple/moment to balance out the 900N*m couple/moment applied at the end. (Let's ignore the other vertical 600N and 800N forces for now since we are only talking about the effect of a couple/moment.)

    Your professor is right in that you don't include the moment in the force calculations directly. You sum the forces in both direction (∑Fy=0 ∑Fx=0) in order to get a relationship between the forces and then substitute this into summing the moments (∑M=0).

    So for ∑Fx=0 :
    → RAx + RBx =0
    → RAx = -RBx

    and summing moments about C gives:

    -RAx*1.5m - 900 N*m = 0

    and therefore the forces in the resisting couple is:

    RAx = -RBx = (-900N*m)/(-1.5m) = -600 N

    Now it happens in this case that the pinned rod BC can only generate axial forces and therefore there will be a force generated in the y-direction at support C. But this y-direction component is a consequence of the pinned rod and not really because a moment always generates a y-direction force.

    In fact suppose you were to weld the two rods together at B. Then there would be no y-direction forces generated at the supports A and C. You would purely have a couple FA = -FC.
    Last edited: Mar 18, 2014
  12. Mar 18, 2014 #11
    You are confusing a moment as caused by a force (acting at some specified lever arm distance) with a pure moment. In the OP's example the moment was a pure moment of 900N*m. There is no x-direction force and no y-direction force given. Just a 900N*m moment or couple.

    This applied couple generates a reaction at the supports which in turn must be another couple to keep the system in equilibrium. But that's it. In fact, even a moment generated by a force must still ultimately be resolved as a couple somehow in order for the body to remain in equilibrium.

    If the problem mentioned that there was an applied force at some specific location then you would be right in that there would be additional forces generated. But that is not what the problem mentioned.

    Do you see the difference now?
    Last edited: Mar 18, 2014
  13. Mar 18, 2014 #12
    When a moment like that 900 Nm is specified, it implies that there is a distribution of axial stress imposed at the end of the beam. The integral of this axial stress taken over the cross sectional area of the beam is equal to zero (because there are regions of positive stress and negative stress which cancel), so there is no net force acting at the end of the beam (from the moment). So it is not necessary to account for this moment in the force balance. However, if you integrate the stress distribution to get its moment at the end of the beam, it integrates to 900 Nm. So, it must be included in the moment balance. This is just another way of saying what paisiello has been asserting.

  14. Mar 18, 2014 #13
    Perfect, that's the kind of explanation I was looking for. Thanks a lot.
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