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Static Equilibrium Problem

  1. Sep 17, 2006 #1
    Here is the problem I am attempting to solve:

    The large quadriceps muscle in the upper leg terminates at its lower end in a tendon attached to the upper end of the tibia. The forces on the lower leg when the leg is extended are modelled as shown where T is the tension in the tendon, C is the force of gravity acting on the lower leg, and F is the force of gravity acting on the foot. Find T when the tendon is at an angle of 25.0° with the tibia, assuming that C = 30.0 N, F = 12.5 N, and the leg is extended at an angle of 40.0° with the vertical. Assume that the centre of gravity of the lower leg is at its centre and that the tendon attaches the lower leg at a point one-fifth of the way down the leg.

    What I am uncertain is my method of solving this question. I'm quite sure that I'm missing something. Here is what I did:
    Fnet = 0
    Angle between quadricep and the dotted line is equal to 40.0°. Therefore, the angle between the quadriceps and T is equal to 65.0°.
    0 = Tsin65.0° - C - F
    T = 46.9 N

    Merci d'avance!
  2. jcsd
  3. Sep 17, 2006 #2


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    HINT: The total moments of force about any point must be zero.
  4. Sep 17, 2006 #3


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    This is not correct, since you could write a scalar equation for the 'x-direction', too, which would be T*cos65 = 0 => T = 0. If you try to sum up the vectors graphically, you'll easily see that there's no state of equilibrium for this set of vectors.
  5. Sep 17, 2006 #4
    If they aren't in the state of equilibrium, does that mean the leg is moving down? Or are there reaction forces that help keep the equilibrium?

    *sighs* Torques...
  6. Sep 17, 2006 #5


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    To be honest, the medical context is killing me. :biggrin:
  7. Sep 17, 2006 #6
    Hmm .. as I'm thinking about this, there would have to be reaction forces to balance all the vectors. But where would they be situated? At the tendon? =S
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