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Pulley Free Body Diagram

  1. Sep 20, 2015 #1
    1. The problem statement, all variables and given/known data
    A collar with a pulley slides on a frictionless vertical bar GH. A string A B C D is wrapped around, where portion AB of the string is horizontal. A spring with 2.5 lb/in. stiffness is placed between the collar and point H. The spring has 7 in. unstreched length and 5 in. final length. The collar and pulley have a combined weight of 3.1 lb, and the weights of all other components are negligible. Determine the value of P required for equilibrium, and the magnitude of the reaction between the collar bar GH.


    Hint given:
    Before taking any other action, draw a FBD of the collar/pulley. This requires a closed external surface passing though the string at two locations as well as between the contact point between the spring and collar/pulley, and internal surface passing through the interface with the smooth bar. Pay close attention to the direction of the force exerted on the collar/pulley by the spring given the information in the problem statement.

    2. Relevant equations
    ΣFx=0, and ΣFy=0.
    Spring Law: Fs=kδ=k(L-L0)

    3. The attempt at a solution
    Here's my (attempted) FBD: http://imgur.com/YdaoEEW

    I know it's probably incorrect, we never really did an example in class that involved springs. Also I wasn't sure what should have been in the middle B or C? B and C? Only B? Only C? Two body diagrams?

    I know I need to add up all the x and y components then for P figure out the force it takes to make the x components equal 0. Then from there finding the magnitude of HG is just a matter of adding together their components and taking the magnitude.

    I seem to have a really hard time figuring out the components, but when I find them I can pretty easily finish the problem.

    Thank you so much for any help!
  2. jcsd
  3. Sep 20, 2015 #2


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    Having B&C in the middle as you've done is fine. You don't know the radius of the pulley anyway, not that it matters in this case.

    Your diagram is not quite right. Which direction does the spring force act on the collar? Your diagram shows it pulling up on the collar (assuming H is the springs force).
    It's also missing the reaction force from the bar HG.

    You need to balance the y forces to find P, not the x. The reaction force of the bar HG will balance the x forces for whatever P ends up being.
  4. Sep 20, 2015 #3
    Does this look a little better?

    I'm not sure what you meant from the reaction force of bar HG
  5. Sep 20, 2015 #4


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    That's right.
    You can go ahead and balance the Y forces now.
  6. Sep 20, 2015 #5
    Alright so I don't get why the force of the spring would be negative
    FHC=(2.5)lb/in(5-7) which equals -5
  7. Sep 20, 2015 #6
    W = 0 i + -3.1 j
    TCD = TCD(4/5) i+TCD(3/5) j
    TAB = ? i + 0 j
    THC = 0 i + (2.5)(5-7) = -5 j

    These don't seem right :/
  8. Sep 20, 2015 #7


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    That is almost correct.
    Double check the x and y components of TCD.
    What can you say about tension in a string? That is, what relationship do TCD and TAB (and also P) have?
    Fspring would be a better name than THC, T is usually reserved for tensions.

    It's just how Hookes law is defined. In this case it just so happens that for your choice of axes the negative spring force is correct. if the spring was compressed under, rather than above the collar the force would be positive. It's up to you to recognise what sign is appropriate.

    From wiki: https://en.wikipedia.org/wiki/Hooke's_law

    Last edited by a moderator: May 7, 2017
  9. Sep 20, 2015 #8
    I get really confused on this whole triangle thing, could you help me out here?

    They are the only vectors that have x-components?

    Alright this makes sense.
  10. Sep 20, 2015 #9


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    TCD is closer to vertical than horizontal, so should the magnitude of the x or the y component be greater?

    (hint: swap the 3 and the 4 in your expression: TCD = TCD(4/5) i+TCD(3/5) j )

    Nope, the reaction force between the bar GH and the collar also has (only) an x component.
    Tension is equal throughout the string so TCD = TAB =P

  11. Sep 20, 2015 #10
    TCD= TCD (3/5) i + TCD (4/5) j

    So if I find TCD this equals P? How does this play in with the finding the magnitude of the reaction between collar and bar GH?
    Last edited: Sep 20, 2015
  12. Sep 20, 2015 #11


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    In your FBD you have three forces in the x direction, they must sum to zero as the collar isn't accelerating. Once you have TCD you can resolve the two known forces (TCD, X and TAB ) in the x direction to find the one unknown: FGH reaction.
  13. Sep 20, 2015 #12
    So to go about finding TCD... Should I equal all the y-components together then solve for TCD? Then plug that back in to get the components? Then take the magnitude?
  14. Sep 20, 2015 #13


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    Yes. That's the general approach. Good luck.
  15. Sep 20, 2015 #14
    I'm going to assume Tspring is negative because it doesn't make sense any other way.
    ΣFY=-3.1 + TCD (4/5) + 0 - 5
    This results in TCD being 10.125...

    Plugging into (3/5)(10.125)i +(4/5)(10.125)j = (6.075, 8.1)
    The magnitude of this is 10.13lb... which is right

    ΣFX= 0 + 6.075 + x + 0
    x = -6.075... but the x coordinate of TAB is not in the negative direction

    This is so frustrating
    Last edited: Sep 20, 2015
  16. Sep 20, 2015 #15


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    (3/5)(10.125)i ≠ 10.125i

    But you had already found the magnitude of TCD , it's 10.125 lb so that step was unnecessary.
    you will need to find (3/5)(10.125)i though, to find the shaft reaction force.
  17. Sep 20, 2015 #16
    I editted my post, my bad, apparently I cant mutiply
  18. Sep 20, 2015 #17
    I guess what I'm not getting is exactly what the GH reaction is?
  19. Sep 20, 2015 #18


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    TCD, X and T AB are both pulling the collar to the right. But we know the collar doesn't move to the right because we are told it is in equilibrium.
    So what is stopping the collar moving to the right?

    This is a described by Newtons third law 'For every action there is an equal and opposite reaction'.
    For whatever TCD, X and T AB are the bar will applies and equal and opposite force to the collar.

    Just like when you push against a brick wall with your hand, no matter how hard you push, the wall won't move, it just pushes back at you with an equal and opposite force.
  20. Sep 20, 2015 #19
    Didn't I just find TCD, X? 6.075?
  21. Sep 20, 2015 #20


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    That's right.
    So use your FBD and the fact that ∑Fx = 0 to find FGH reaction
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