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Homework Help: Tension force with pulleys and weights

  1. Feb 4, 2009 #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 m = 3.4 kg, and the doctor has decided to hang a 6.0 kg mass from the rope. The boot is held suspended by the ropes and does not touch the bed. ϕ = 10°.


    (a) Determine the amount of tension in the rope by using Newton's laws to analyze the hanging mass.

    (b) The net traction force needs to pull straight out on the leg. What is the proper angle θ for the upper rope?

    (c) What is the net traction force pulling on the leg? (Hint: If the pulleys are frictionless, which we will assume, the tension in the rope is constant from one end to the other.)

    2. Relevant equations


    3. The attempt at a solution

    I was easily able to use W=mg to find the tension in the rope to be 58.8N.
    After that, I am lost.

    I know the Fnet(y) should be zero since the boot is pulling straight out, but I can't get any further than that...

    I also imagine somehow that i will find acceleration along the x-axis, then because the system is in equilibrium (the leg will not move), I will use acceleration and the mass of the boot to find the net traction force.

    I really need some help finding the angle theta. Thanks!

    Attached Files:

  2. jcsd
  3. Feb 4, 2009 #2
    Okay guys, I started out with a free-body diagram showing the forces on the boot. I found the force of weight on the boot to be 33.32N. I then set fnet(y) equal to zero, which showed that 58.8 sin(theta)=33.2+58.8sin(10) and found the angle to be 47.58 degrees, which was correct.

    Now I just can't figure out the third part and have no idea how to go about it. Thanks.
  4. Feb 7, 2010 #3
    The new traction will be the total horizontal force acting on the leg, ie the sum of horizontal components of the forces. We already found the angles for the forces to be balanced. Now the total horizontal force will be,

    Tcos(10) + Tcos(47.558)

    as T = 60N (I take g = 10),

    Total HF = 99.429N
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