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Equilibrium and Center of Gravity problem

  1. Nov 27, 2006 #1
    I'm having some trouble finding a definitive answer for this problem of my physics lab.

    The large quadricep muscles in the upper leg terminate at the lower end in a tendon attached to the upper end of the tibia. The force on the lower leg when the leg is extended are modeled below (attached jpg), where T is the tension in the tendon, w is the force of gravity acting on the lower leg, and F is the weight of the foot. Find T when the tendon is at an angle of 20 degrees with the tibia, assuming that w = 180N, F = 30.0 N, and the leg is extended at an angle of 45.0 degrees with the vertical. Assume that the center of gravity of the lower leg is at its center, and that the tendon attaches to the lower leg at a point one fifth of the way down the leg.

    I've been looking through some examples to try to come up with a general idea, but this problem seems somewhat different as no lengths or distances are given except that the tendon attaches 1/5 down the leg. So far, I've found forces in x and y components to get

    SUMFx = Tx + wx + Fx = 0
    to get Tx = 0.
    Then in y direction, I'm not sure if i can do this but i found
    Tsin65 - 180 - 30 = 0
    to get T = 231.7.

    Knowing that this is supposed to be a torque-related problem I also used SUMt = tT + tw + tF = 0. Then tT = rTTsinθ. This is where I am stuck. How would I use the torque equilibrium equation if I am not given any distances? Any hints or suggestions would be really helpful!

    Attached Files:

  2. jcsd
  3. Nov 28, 2006 #2


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    Homework Helper

    Welcome to PF p0ke.

    The length of the leg will appear in the calculation of all three torques. So you can cancel it out in "the sum of the torques = to zero". You can then take moments about the knee (pivot point) taking [tex]l[/tex] as the length of the tibia, which then cancels in the equation. The only unknown in the torque equation will then be the tension in the muscle. So you only need to consider this equation. There will also be a reaction force at the knee acting on the tibia, which need not be included in the torque equation if you take the torques about the knee.
    Last edited: Nov 28, 2006
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