Finding Tension in a Quadriceps Tendon: A Static Equilibrium Problem

AI Thread Summary
The discussion revolves around calculating the tension in the quadriceps tendon when the leg is extended, considering the forces acting on the lower leg and foot. The initial calculation suggests a tension of 46.9 N, but it is deemed incorrect due to the lack of equilibrium in the force vectors. Participants highlight the importance of considering both the vertical and horizontal components of the forces to achieve a proper equilibrium state. There is uncertainty about the presence and location of reaction forces that would maintain equilibrium. The conversation emphasizes the complexity of the problem, particularly in a medical context.
bjgawp
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Here is the problem I am attempting to solve:
http://img227.imageshack.us/img227/2747/problempr4.jpg

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 modeled 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!
 
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HINT: The total moments of force about any point must be zero.
 
bjgawp said:
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

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.
 
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...
 
To be honest, the medical context is killing me. :biggrin:
 
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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