High School Calculating Leverage & Force in Lower Leg Squat

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To calculate leverage on the lower leg during a squat, one must consider the effective lever arm of the quadriceps, which varies with knee flexion angles and is approximately 5 cm. The external torque can be determined using ground reaction forces or knee position, although some prefer to focus on force distribution through trigonometry. There are concerns about the realism of the movement model being applied, as it may not accurately reflect practical scenarios. The discussion highlights the complexity of biomechanics in squatting and the challenges in understanding the forces involved. Overall, calculating these dynamics requires a careful analysis of various factors influencing the squat.
tomlib
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How can I calculate the leverage on the lower leg during a squat and the force that the thigh muscle must exert in order for the body to perform the squat. Can you help me?
 
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tomlib said:
How can I calculate the leverage on the lower leg during a squat and the force that the thigh muscle must exert in order for the body to perform the squat. Can you help me?
You have to look up the effective lever arm of the quadriceps at different knee flexion angles. It varies based on patella anatomy, but is roughly around 5 cm. The external torque can be computed from the ground reaction force, by inverse dynamics (if you care about the shank and foot mass), or just from the knee position.
 
I would not count the torque, I would like to stick to the classic distribution of forces by vectors using trigonometry and crank calculation. Unfortunately, I don't have such an opportunity to understand everything. This movement seems unrealistic to me.

I saw how much work it can be to raise a sail for windsurfing.
When applying this model, the rope is actually in line with the boom, such a force is impossible.
 
tomlib said:
This movement seems unrealistic to me.
A simple squat?
 
I built a device designed to brake angular velocity which seems to work based on below, i used a flexible shaft that could bow up and down so i could visually see what was happening for the prototypes. If you spin two wheels in opposite directions each with a magnitude of angular momentum L on a rigid shaft (equal magnitude opposite directions), then rotate the shaft at 90 degrees to the momentum vectors at constant angular velocity omega, then the resulting torques oppose each other...

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