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Identifying forces in a friction problem

  1. Sep 26, 2016 #1
    1. The problem statement, all variables and given/known data
    Determine the smallest force P that will cause the spool to have impending motion.

    1.jpg

    2. Relevant equations

    Equations of equilibrium, and static frictional equations.

    ##ΣFx = 0;##
    ##ΣFy = 0;##
    ##ΣMb = 0;##
    ##Fsa = μNa##
    ##Fsb = μNb##

    3. The attempt at a solution

    My first problem here is properly identifying the forces at work, specifically, the forces at point B. I've made a drawing that shows how I think the forces apply at these points:

    2.jpg

    ##G = mg = 150 * 9.81 = 1471.5N##

    This makes my equations of equilibrium as follows:

    ##ΣFx = 0; P - Fsa - Fsb cos 51.3 - Nb sin 51.3 = 0##
    ##ΣFy = 0; Na + Nb cos 51.3 - Fsb sin 51.3 - 1471.5N = 0##
    ##ΣMb = 0; P(0.45) + Fsa(0.15) + Na(0.312) - 1471.5(0.312) = 0##
    ##Fsa = 0.4 * Na##
    ##Fsb = 0.2 * Nb##

    I selected the moment equation to be set around point B, as this seems to give me the equation with the least amount of unknowns, but I've also tried to solve the problem where I look at the moment around point A instead.

    So as for the first problem I'm facing: Have I correctly identified the forces at play?
    If I have, do my equations of equilibrium look correct? I've failed to be able to work with the equations to find any of the unknows I need to continue.

    Thanks for any help.
     
  2. jcsd
  3. Sep 26, 2016 #2

    andrevdh

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    I would have thought that the point where P is large enough the forces at A would disappear, reducing the problem to balancing the torques around B?
     
  4. Sep 26, 2016 #3
    Hmm, I did consider that at first, but it felt weird considering I can then "skip" a lot of the information that the assignment provides. For instance, would the assignment then need to provide me with μb, when it can be completely ignored in this case?

    If I understand you correctly, the task can then be solved solely by:

    ##ΣMb = 0; P(0.45) - G(0.312)##
    ##P = \frac {1471.5 * 0.312} {0.45} = 1020.2N##

    So from this, at 1020.2N the spool will have impending motion, and if you add more force it will have motion. Was that really all I had to do here? =/
     
    Last edited: Sep 26, 2016
  5. Sep 26, 2016 #4

    CWatters

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    μ
    I'm not so sure.

    If μa and μb were zero the forces at A and B would act through the centre of the spool. There would be nothing to stop the spool rotating about it's centre (without climbing).

    Ok so they aren't zero but does that mean we rule out the possibility that rotation about the centre of the spool is about to occur? Does that also count as "impending motion" ?
     
  6. Sep 26, 2016 #5
    The problem doesn't specify that the spool has to be about to travel up the step in order for impending motion to occur, so I believe it spinning around its own axis would also apply.
     
  7. Sep 27, 2016 #6

    andrevdh

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    Aaah, ok, so you think they want you to consider it not succeeding in climbing up but staying down there and only spinning in the corner down there and only when the midpoint of the structure is in line with the string it can be pulled up and over the corner.
     
  8. Sep 27, 2016 #7
    That's quite possible. I'd guess that it's one of the things that the assignment wants me to check, whether the spool will climb, or stay in place when pulling it with force P.

    But I'm not sure how to handle the different situations. Are there different calculation methods when impending motion is about to cause the spool to spin, rather than climb?

    I tried to visualize how the forces work: If the spool spins in place, will Fsa and Fsb work in the same direction as my drawing, or would they actually work in the opposite direction? I'm having some problems visualizing how that part works.
     
  9. Sep 27, 2016 #8

    andrevdh

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    No, friction opposes the motion, that is P wants to rotate the spool clockwise, so the forces would be in the opposite direction.
     
    Last edited: Sep 27, 2016
  10. Sep 27, 2016 #9

    haruspex

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    Yes. Each of the two motions, if just about to occur, allows you to apply an extra equation. But the extra equations are different for the two cases.
    What equation can you write for the case where something is about to slide?
    What equation can you write for the case where something is about to rotate?
     
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