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Find force to raise load using friction and two bars

  1. Apr 29, 2015 #1
    1. The problem statement, all variables and given/known data

    Calculate the horizontal force P required to raise the 100-kg load. The coefficient of friction between the rope and fixed bars is 0.40.


    2. Relevant equations

    T1/T2 = e^(βμ)

    Where β = the angle of the rope on the log, μ = 0.40, T1 = the bigger tension (P), and T2 = the smaller tension.

    3. The attempt at a solution

    So upon first glance, I assumed that I needed to break this down into a FBD. I broke it into three different diagrams: 1.) the weight, 2.) the lower bar, and 3.) the upper bar. I then tried to solve the tensions from bottom to top.

    FBD of weight
    So this one was pretty simple, the block weighs (100)(9.81) = 981 N, so the tension in the cable has an upward force of 981 N.

    FBD of lower bar

    For this one, my drawing had two forces going forward (due to it moving), but I had the larger force at an angel going towards the upper bar. The smaller force was 981 N.

    To find the angle β, I used (3d)/2 and d/2 and used tan θ = ((3d)/2)/(d/2), which gave me θ = 71.565°. So I multiplied this times π/180 to get 0.39758 rads.

    So to find the larger tension I used the T1/T2 = e^(βμ) equation. T/981 = e^(.4*0.39758) and found T = 1616.771 N

    FBD of upper bar

    For the upper bar, my FBD had 1616.771 N coming from an angle from the lower bar, then going around the upper bar and having P horizontal like it's shown in the problem. For this one P is the bigger force and 1616.771 N is the smaller force.

    To find β, I added 71.565° + 90°, which equals 0.89758π rads.

    So I used the T1/T2 equation again: P/1616.771 = e^(.4*0.89758π), which gave me P = 4994 N, or 4.994 kN.

    This is pretty far off, since the book says the answer is 3.30 kN.
  2. jcsd
  3. Apr 29, 2015 #2


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    Hello there,

    Can you explain the angle you found on the lower bar ? My guess is there's your first glitch.

  4. Apr 29, 2015 #3
    Sorry for the time gap, I've been trying to work away from home using my phone and it wasn't working out too well for me.

    I assumed the angle I was looking for to be that of a triangle that was (d/2) at the base, and (3d/2) height, so I solved with θ = tan^-1([3d/2]/[d/2]) which gave me 71.565°.
  5. Apr 29, 2015 #4
    So I see based on the diagram you provided which makes a lot more sense than what I was seeing in my head/my messed up FBD, 70.1° is way off. and should probably be closer to 45°. I also noticed that I was trying to find the angle using the half the diameter for the base rather than the full diameter.

    If I instead make a triangle out of height = (3d/2) and base = d, I get θ = tan^-1((3d/2)/d), which gives me θ = 56.31 degrees. Then I used 56.31 + 90 degrees for the upper bar, which eventually gives me a final answer of over 4000 N still, so I'm still off.

    I'm thinking my understanding of the geometry to find the angle is way off. Am I doing the steps correctly, aside from the wrong angles?
  6. Apr 29, 2015 #5


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    I don't understand how you are getting tangents into it. The distance between the bars' centres is (twice) the hypotenuse of the triangle I see.
  7. Apr 29, 2015 #6
    So you're using 3d/4 as the hypotenuse and d/2 as the base to find the angle (using cosine) and then subtracting 90 minus that angle to obtain β?
  8. Apr 29, 2015 #7


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    Yes, except that I used arcsin to find the angle the rope crosses the vertical, which will also be β. Comes to the same.
  9. Apr 29, 2015 #8
    Thanks so much haruspex! I ended up getting β = 41.81° and the tension equaling 1313.52, then for the upper bar I got β = 131.81° which gave me P = 3296.7 N for a final answer.

    Also thanks for the visual BvU, I couldn't have pictured the problem without that!
  10. Apr 30, 2015 #9


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    Well done, you night owls ! (I'm at CEST time).

    On a serious note I've been wondering if I could give you something like tips to avoid this kind of glitches in the future. Couldn't find all that much. My consideration to post was twofold: In the first place I was curious because I had never encountered this capstan equation beore, and in the second place I became suspicious at the 71 degrees, when just looking at your picture about half of that would seem more realistic. So I drew a pencil sketch and got around ##\pi/4## and a neater drawing showed it should be a litle less than that.

    Now in this exercise a drawing did good work; in more complicated/abstract cases I can only guess that a little less haste would be good. And a lot of checking.
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