- #1
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!
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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