How do three unknown forces in a knee fall out of one torque equation, and why is every choice of pivot point equally valid?
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Three-unknown equilibrium setup for the knee
Write the two force-component equations with and the leg weight , then the torque equation about the ligament attachment point, giving three equations in , and .
Vertical offsets and dropping out
Evaluate the three cross products about and record that only the horizontal distances , and survive in the torque equation.
The ligament tension and femur force values
Solve , then from the ratio of the two force equations, then , carrying units to , and .
Equality of torques about any two points
Reproduce the proof: substitute into the torque sum about , pull the constant vector outside, apply , and state .