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Engineering, Statics -- A pulley is free to move on the cable

  1. Mar 4, 2015 #1
    1. The problem statement, all variables and given/known data
    1. Pulley B is free to move on the cable ABC which is fastened at A and being pulled by force F through the pulley C. Neglecting friction:
    a. Determine location of point B as a function of d.
    b. Express the force F required to maintain equilibrium in terms of P, d, and h.

    3. The attempt at a solution
    I have no idea where to even start. I thought about creating an FBD of the weight hanging on the cable for part b, but I can't figure out how to tie that to h and d.
     

    Attached Files:

  2. jcsd
  3. Mar 4, 2015 #2

    billy_joule

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    A FBD of the weight is a good start. h and d define the angle of the cable which gives the angle of the force for the FBD of the weight.

    Attempt a FBD and we can go from there.
     
  4. Mar 4, 2015 #3
    OH, okay, I'll try this. There are four problems after this one that I did no problem.

    Not sure why I didn't see that.. I'll work on it a bit. Thank you.
     
  5. Mar 5, 2015 #4
    Okay, so I tried drawing a FBD of point B, but I'm still not sure how to equate the measurements.

    My friend recently mentioned to me that the pulley may be hanging directly in the middle of the cable. If that's the case I can use the tangent value of alpha to put d/2 and h in it, but I'm still not sure how to write point B as a location.

    In the picture i just uploaded is theta supposed to be equal to beta? If they're related, then I don't see how theta affects the amount of force needed to keep the system in equilibrium.
     

    Attached Files:

  6. Mar 6, 2015 #5

    billy_joule

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    At equilibrium B will be at d/2.
    Theta is not equal to alpha, draw some right angle triangles to find their relationship. What do the angles of triangles add to?
    Separate F and F dash into their horizontal and vertical components to find their magnitude in terms of P and theta **.(**EDIT: beta or alpha, not theta)
     
    Last edited: Mar 6, 2015
  7. Mar 6, 2015 #6
    But is theta equal to beta? I just realized that I didn't include my updated drawing.
    I named the angle, that is adjacent to theta, beta.
     

    Attached Files:

  8. Mar 6, 2015 #7

    billy_joule

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    No, theta is a fixed and not relevant to your calculations, I'm not sure why it's included on the diagram.
    Beta is a function of h and d.
     
    Last edited: Mar 6, 2015
  9. Mar 6, 2015 #8
    Sorry.. I used the word adjacent incorrectly. Here is what I drew so far.

    I still don't understand how beta can be related to theta. I'm attempting to imagine the pulley setup, and it seems like no matter what angle theta is, the forces F and F' will be equal and unaffected.
     

    Attached Files:

  10. Mar 6, 2015 #9

    billy_joule

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    You are correct - In post #5 I should have said alpha or beta, not theta. I've edited in a correction.
     
  11. Mar 6, 2015 #10
    Okay, so this is what I got so far:

    F_x = F * sin(α)
    F_y = F * cos(α)

    F'_x = F * sin(α)
    F'_y = F * cos(α)

    P_x = 0
    P_y = P

    ΣF_x = Fsinα - Fsinα = 0
    0 = 0

    ΣF_y = Fcosα + cosα - P = 0
    2Fcosα = P
    F = P/(2cosα)

    I know I'm somehow supposed to substitute cosα for known values but all I can think of to substitute it with is h/l. Lower case L being the unknown.
    I'm still also not sure how to determine the location of point B as a function of d.
     
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