Calculating force of the femur on the knee cap.

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SUMMARY

The discussion focuses on calculating the force exerted by the femur on the kneecap using a simplified model of the leg. The quadriceps muscle applies a force of 420 N through a tendon, with angles θ1 = 23° and θ2 = 24° for the tendon above and below the kneecap, respectively. The calculations for the x and y components of the femur's force were attempted, yielding Ffemur = 191.4 N for the x component and Ffemury = -1.67 N for the y component. The solution was confirmed to be correct, with a suggestion to use fewer significant digits for clarity.

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Homework Statement


The diagram of the leg shows the femur (1) and tibia (2). The quadriceps muscle (3) applies a force to the lower leg via a tendon (4) that is embedded with the kneecap (5). If the force applied by the muscle to the tendon is FM = 420 N, what is the force of the femur on the kneecap? A simplified model of the leg is shown next to the diagram. The leg bones are represented by two beams attached by a pin. The tendon is modeled by a rope and the kneecap acts like a pulley. The tendon above the kneecap makes an angle θ1 = 23° with respect to the vertical, and the portion of the tendon below the kneecap makes an angle of θ2 = 24° with respect to the vertical. Enter the x component, followed by the y component.

knee.gif



Homework Equations



fnet=ma



The Attempt at a Solution



I tried:

Fx= max
T1x + T2x + Ffemurx=0
-240sin23 -240sin24 + Ffemur= 0
-93.78 - 97.62 + Ffemur=0
Ffemur= 191.4N

Fy=may
T1y + T2y + Ffemury=0
240cos23 - 240cos24 + Ffemury=0
220.92 - 219.25 + Ffemury=0
Ffemury= -1.67N

I didn't get the right answer..

please help!
 
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Supposed that the tension is the same all along the tendon, and x means horizontal , y means vertical component of force, your solution looks correct. Try to use less significant digits.

ehild
 

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