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Moment of force from an axis - 3D problem

  1. May 27, 2015 #1
    1. The problem statement, all variables and given/known data
    An OAB pipe is embedded in O. A cable BC = 1000N and F = ( 0,-f , 0) act on B.
    Question:
    Determine the magnitude of the force F as the resultant moment about the axis OA is canceled.

    Picture in attachment:

    2. Relevant equations


    3. The attempt at a solution
    O=(0 0 0)
    A=(1.5 0 0)
    B=(1.5 -sin(40)*0.75 cos(40)*0.75)
    C=(0 0.6 1.2)
    VTBC= 1000*unit(C-B)
    VF=(0 -f 0)
    VO=(0 Oy Oz) ==translation forces at o
    vMo=(0 momenty momentz) == rotational forces at o
    translation equilibrium = VO + VTBC + VF = (0 0 0)
    VFxrOB*uAO => crossP(b-o,vf)*unitV(a-o)= (0 0 0)
    crossp function = cross product of two vectors.
    unitV function = unit vector
    F = ??

    Calculating moment from an axis is confusing me a lot.
    I thank you in advance for helping me.
     

    Attached Files:

    Last edited: May 28, 2015
  2. jcsd
  3. May 28, 2015 #2

    haruspex

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    OB? It says:
    Not sure why you have unitV(a-b) in there. The cross product produces a vector, and you need a vector result. What is the multiplication by unitV(a-b) doing?

    Also, the cancellation should be between the torque of F about OA and the torque of the cable tension about OA. You don't seem to have that in your equation at all.
     
  4. May 28, 2015 #3
    sorry, its unitv(a-o) and not a-b

    I could simply do crossP(b-a,vf)*crossP(b-a,vtbc)?
    that gives me a weird answer if I dont include unitv(a-o), that part is to use the axis ob
     
  5. May 28, 2015 #4

    haruspex

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    OK, that makes more sense. You are taking the dot product with that to extract the torque component in the AO direction.
    You don't cancel two vectors by multiplying them together.
    crossP(b-a,vf) is the torque of F about OB. Do as you suggested, taking the dot product with unitv(a-o).
    Now do all the same with the cable tension as you did with F.
     
  6. May 28, 2015 #5
    Yes, this is what I have now: dotP(crossP(b-o,vf)+crossP(b-a,vtbc),unitV(a-o))= 0
    f= -585N
     
  7. May 28, 2015 #6

    haruspex

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    Is there a typo in there?
     
  8. May 28, 2015 #7
    yes :(
    I am sleepy.
    dotP(crossP(b-a,vf)+crossP(b-a,vtbc),unitV(a-o))= 0
    I hope no more typo now
     
  9. May 28, 2015 #8

    haruspex

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    Yes, that looks right.
     
  10. May 28, 2015 #9
    Question 2) With the strength found in a) , determine the components along y and z according to the time which is O.

    solve(crossp(b-a,vf)+crossp(b-a,vtbc)+crossp(o-a,vmo))= (0 0 0)
    momx = 0N
    momy= -246.9N
    momz = 294.26N

    Is that correct?
     
  11. May 28, 2015 #10

    haruspex

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    I have no idea what that means. Is this a translation?
    What's this term? The cross product of a distance vector and a moment? What does that produce?
     
  12. May 28, 2015 #11
    Basically I have to find the moments at point O from point A(I decided that point)
     
  13. May 28, 2015 #12

    haruspex

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    OK, but moments of what about O? F? F+cable tension?
    (And I don't understand what you mean by "from point A")
     
  14. May 28, 2015 #13
    I am not really sure, can I just do it from point o?
    and do this
    solve(crossp(b-o,vf)+crossp(b-o,vtbc)+vmo= (0 0 0))
    That make more sense I think now.
     
  15. May 28, 2015 #14

    haruspex

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    That makes more sense, yes. It assumes you want the total moment, not just that from F, but which is wanted is not clear to me from the question.
    One error though - you are asked for the moment, not for what moment you would have to add to get a net zero moment.
     
  16. May 28, 2015 #15
    solve(crossp(b-o,vf)+crossp(b-o,vtbc)= (0 0 0))
    = 0 -921.91 -773.575
     
  17. May 28, 2015 #16

    haruspex

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    ?
     
  18. May 28, 2015 #17
    sorry, I did not put that in my calculator actually.
    (crossp(b-o,vf)+crossp(b-o,vtbc) right here.
     
  19. May 29, 2015 #18
    Last question:
    Determine the three components of the reaction force in O.

    I would simply do this ? (vtbc+vf)
    and this gives me (-768 -268 320)

    Also thank you very much for helping it really do help me a lot.
     
  20. May 29, 2015 #19

    haruspex

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    Yes, except that this time you are looking for a force that gives a net zero. It's the reaction force you are asked for, i.e. the force exerted on the pipe by its wall support.
     
  21. May 29, 2015 #20
    do I have to include vo in my equation?
    (vtbc+vf+vo)=(0 0 0) ?
     
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