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Homework Help: Russell Traction Apparatus

  1. Nov 23, 2016 #1
    1. The problem statement, all variables and given/known data
    The device shown in the figure below (Figure 1) is one version of a Russell traction apparatus. It has two functions: to support the injured leg horizontally and at the same time provide a horizontal traction force on it. This can be done by adjusting the weight W and the angle θ. For this patient, his leg (including his foot) is 81.0 cm long (measured from the hip joint) and has a mass of 16.6 kg . Its center of mass is 41.0 cm from the hip joint. A support strap is attached to the patient's ankle 13.0 cm from the bottom of his foot.

    What weight W(in N) is needed to support the leg horizontally?

    Radiusleg=81.0cm or .810m
    Radiuscm=41.0cm or .410m
    Radiuspulley=13.0cm from foot or 68.0cm from pivot=.680m

    2. Relevant equations
    Στ(torque) = 0
    τ= T⋅R

    3. The attempt at a solution
    Force of the center mass is equal to the mass times the acceleration(downward).
    Fcm = m * a = 16.6kg ⋅ -9.8m/s2
    Fcm = -162.68N
    Fcm * Rcm = -162.68N * .410m = -66.988 N*m = τ

    Now, I sum the torques and solve for Fpulley
    Fpulley * Rpulley - Fcm * Rcm = 0
    Fpulley*.68m = -66.988N*m
    Fpulley = -98.08647N = -98.1N

    I have tried inputing -98.1 and 98.1, both were wrong. Any suggestions? I've looked elsewhere online and this is pretty much the template they provide so maybe I'm getting my numbers wrong somewhere?
  2. jcsd
  3. Nov 23, 2016 #2


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    Hello DU, :welcome:

    For what it's worth: I get the 98.1 N, just like you did (mass is always positive).

    You could try 98.2 (for the case they use g = 9.81 m/s2, but that's nitpicking).
  4. Nov 23, 2016 #3
    No sir, 98.2 did not work.
  5. Nov 23, 2016 #4


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    Is it possible that the pivot point is actually at the foot, the idea being to keep the weight of the leg off of the hip joint?
  6. Dec 1, 2016 #5
    I asked the prof. He checked what the equation they used to solve it. The answer ended up being .5 *(mass of leg)*9.8m/s2. He didnt really understand why. Can anyone else explain? Another part of this problem asked what the maximum weight that could be put on the hanging mass with the answer being 2*(mass of leg)*9.8. Am i misunderstanding the question? I dont feel like theyre using torque at all but this is in the torque section of the book and gives radii.
  7. Dec 1, 2016 #6


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    The data given make this patient's leg very short for its mass, and quite short for a Western adult male generally.
    Maybe the 81cm is supposed to be only from hip to support. That would give a more reasonable total leg length.
    That implies something can be varied to compensate. Changing theta only changes the horizontal tension. Maybe it means varying the position of the support.
  8. Dec 2, 2016 #7
    My guess is that when you talk up supporting it horizontally then it is meaning that it is being held pivot at the hip and pulled outward(at theta) by the two T forces at the foot. For some reason it is asking what the value of one of those T's are because the total mass that needs pulled is the weight of the leg, and then T is half of the weight of the leg. I get that, sure. BUT why doesnt the problem use the radii or torque? Haha.
  9. Dec 2, 2016 #8


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    No, support the injured leg horizontally means supporting it in a horizontal position. That comes from the vertical force.
    It continues:
    That is the net force from the two angled portions of the cable.
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