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Tension/Traction problem

  1. Oct 1, 2013 #1
    1. The problem statement, all variables and given/known data

    An accident victim with a broken leg is being placed in traction. The patient wears a special boot with a pulley attached to the sole. The foot and boot together have a mass of 4.0 kg, and the doctor has decided to hang a m = 6.9kg mass from the rope. The boot is held suspended by the ropes and does not touch the bed. Find the angle θ.

    Fgw(force of gravity on weight)= 68N
    m1(mass of foot and boot=4kg
    ∅= 17 degrees

    2. Relevant equations
    Fnet=ma
    Fnety=Tpicos∅+Tp2cosθ-Fgw
    3. The attempt at a solution

    I tried to find for the angle between the top rope and horizontal by doing the following;
    Fnety=Tp1cos∅+Tp2cos-Fgw
    Fnety=0
    68/68cos17=Tp2cosθ
    Fnetx=0
    Fnetx=Tp1sin∅+Tp2sinθ=0
    Tp2=Tp1sin∅/sinθ
    Plug tp2 into first equation
    68/68cos17=(68sin17/sinθ)*cosθ
    θ=36.501 degrees
    I'm not sure if this makes sense though and then a later part of the question asks, "what is the traction force?" and it states that the traction force must be pulling outward. Help please! http://puu.sh/4F63M.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 1, 2013 #2

    tiny-tim

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    Hi Minusu! :smile:
    No: you have to decide which body you're adding the forces on.

    In this case, it's the forces on the boot (whose y components have to add to zero) …

    there are only two forces on the boot! :wink:
     
  4. Oct 1, 2013 #3
    Would these two forces be the Fg on the boot and the tension of the ropes?
    So Fnety=-Fgboot+tp1cos∅+tp2cosθ=0
    39.2-68cos17/tp2=cosθ

    And then solve for the x direction so
    Fnetx=Tp1sin17+tp2sinθ=0
    68sin17/sinθ=tp2

    Plug into first

    39.2-68cos17/68sin17=tanθ
    θ=52.4
     
  5. Oct 2, 2013 #4

    tiny-tim

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    Hi Minusu! :smile:

    (just got up :zzz:)
    hmm … i've read the question again, more carefully, and i don't understand how that is compatible with "The boot is held suspended by the ropes and does not touch the bed." :confused:

    if the net traction force (ie the resultant of the two tension forces) is horizontal, then there's nothing to hold the boot up

    i suggest you try ignoring "does not touch the bed", and see if the computer accepts the answer that gives you

    so just solve for the angles needed for the resultant of the two tensions to be horizontal :smile:
     
    Last edited by a moderator: May 6, 2017
  6. Oct 2, 2013 #5

    haruspex

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    Think again about the vertical components of the tensions.
     
  7. Oct 2, 2013 #6

    haruspex

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    I don't see the difficulty. The upper part of the leg is to be taken as a horizontal weightless string. The patient touches the bed, of course.
     
  8. Oct 2, 2013 #7
    Thanks for all the help. I just solved for the angle. The tension in Tp1 and Tp2 is the same, so Fnety=Tp1sin∅+Tp2sinθ-Fg
    Fnety=0
    0=68sin17+68sinθ-39.2
    (39.2+68sin17)/68=sinθ
    .868=sinθ
    θ=sin^-1(.868)
    θ=60.3 degrees.
     
  9. Oct 2, 2013 #8

    haruspex

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    Looks good. Of course, strictly speaking, the leg is a rod with weight distributed along it. There will be a vertical force from the bed at the hip, so not all of the weight of the leg will be taken by the ropes. However, I don't think you have enough information to take that into account.
     
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