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Homework Help: Moment around a point in 3d

  1. Feb 18, 2013 #1
    1. The problem statement, all variables and given/known data

    The threading die is screwed onto the end of the fixed pipe, which is bent through an angle of 20º. Replace the two forces by an equivalent force at O and a couple M. Find M and calculate the magnitude M' of the moment which tends to screw the pipe into the fixed block about its angled axis through O.


    2. Relevant equations

    M=r x f


    3. The attempt at a solution

    I think that F= 30-40 j= -10j, I don´t know if this is right because the book doesn´t give the answer.

    I calculated correctly M=136.46i -679.56k lb*inch, according to the book, but I have some doubts:

    1.- lb inches are not in the SI system, I don´t understand why the force is given in mass unities. I´m used to use Newtons, so I´d need to transform lb to kg, then multiply by 9.8 to get Newtons, then inches to meters and finally I´d get N*m. But Why is this book using mass units to talk about forces?

    2.- The second part says:

    "and calculate the magnitude M' of the moment which tends to screw the pipe into the fixed block about its angled axis through O"

    I did that using vector n=(sin(20), 0, cos(20)) and I got -591.90 lb inch which according to the book was wrong.

    The correct result is 685 lb inch and I deduce from that the fact that they are using this vector n=(sin(20), 0, -cos(20)).

    When you need to project the moment M onto the angled axis you can choose two vectors on the same line but oppossite sense, WHy is correct to use a minus sign on the z component of the vector n, and not the oppossite? How do you know that?

    3.- If I use the general formula for M=Sum r x f, I get the right result. But I tried to do a much more simple calculation and I failed, I´d like to understand why. I have read that the vector M is a free vector, and I think I don´t understand that well, What does free vector means?.

    I thought that I could calculate the moment M around a point, like the origin G of the axis x,y,z and that it should be the same around point O, so I thought M=40*10 - 30*10 k= 100k.

    Obviously that is not the case, that´s false because the distances between G and the forces are different to the distances from O to the forces.

    So ok, I understand that but in this problem:


    I calculated the total moment around point O, calculating the moment created by the 1200 force from point G, and adding the 240 Nm moment vector to that.

    In that case I applied the 240 Nm moment to O without any change, as if O were the point G because I thought that it was a free vector and I could use it in any point and the result would be the same.

    So now it´s a bit confusing to see that I can´t calculate the moment around G and use it around the point O, What´s the difference in both cases?

    As you can see it´s not clear to me what a free vector is, hope your help.

  2. jcsd
  3. Feb 18, 2013 #2
    Observe that the Z-axis is away from the block; you are asked to find the moment that tends to screw the pipe into the block. According to the right hand rule, the moment must be facing into the block. So the angle between these two direction is not 20 degrees, it is 180 - 20 = 160, hence why they have -cos 20.

    Directions are important for moments.

    Equally important is the point about which the moments are computed. Recall that M' = M + R x F, where M is the total moment about some point A, M' is the total moment about some point B, R is vector from B to A, and F is the total force. M' = M only when R is parallel to F.
  4. Feb 18, 2013 #3
    I understand what you say about the angle.

    Well I see that I had the missconception that M was the same around any point, but I still don´t understand if that is true why could I use the moment aroung G in the milling problem to calculate the moment around O?.

    If you have a couple the moment is:

    M= ra x F + rb x -F = (ra - rb) x F = r x F , and it´s the same around any point, I found that in the chapter.

    I tried to transform the 30lb - 40lb into a couple, so the 40lb force is the same as a 30lb force but applied 13,3 inches away from point G.

    Using the above relation:

    r= -10i - 13,3i= -23,3i , inch

    F= 30j lb

    Mcouple=699k, lb inch and it should be the same around any point.

    Obviously that is not correct, but what am I doing wrong?
    Last edited: Feb 18, 2013
  5. Feb 18, 2013 #4
    Yes, the couple does not need a particular point of application; its moment is always the same.

    You can replace the the 30 and 40 lb forces with three forces; two 30 lb and 10 lb. The first two are a couple, so you end up with a force + couple system.
  6. Feb 19, 2013 #5
    Well I don´t know how to substitute the 40lb force for two forces, 30lb and 10 lb, Can you show me where are they applied?:confused:
  7. Feb 19, 2013 #6
    Pretend that instead of the 40 lb force, two forces are applied applied at the same point and in the same direction, one 30 lb and another 10 lb.
  8. Feb 19, 2013 #7

    Thanks, in that way I get the right result.

    Any comment about free vectors? Or about the milling problem were I used M to G and O?
  9. Feb 19, 2013 #8
    As I said in #4, the moment of a pure couple (a system of forces whose resultant is zero) is the same with respect to any point. I am not sure what else I could say on the subject.
  10. Feb 19, 2013 #9
    I see, and that is the case in the milling problem.

    Well thanks for your help, this problem has been useful to learn.
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