Question

The diagram of the leg shows the femur (I) 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 500 N. what is the force of the femur on the kneecap, if the leg is in equilibrium? 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 modelled by a rope and the kneecap acts like a pulley. The tendon above the kneecap makes an angle 0, = 66 degree with respect to the vertical, and the portion of the tendon below the kneecap makes an angle of 0: = 39 degree with respect to the vertical. Enter the .v component, followed by the y component.

Answer 1

Fm= 500 N and Fx and Fy are going to be the force of the femur, in the x component and the y component

first we find the x component:

Fm x cos (66°) - Fx x cos (33°) = 0

so Fx= 242.48 N

next the y component is:

Fm x sin(66°) - Fy x sin(33°)= 0

so the Fy is:

838.67 N

The result of the force is: F= (Fy^2 + Fx^2) ^(1/2) = 873.02N