125_308 HW- 6_S2018-sol.docx

125_308 HW- 6_S2018-sol.docx - 125:308 Intro to...

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125:308 Intro to Biomechanics Fall 2018 HW: Loads in Knee joint Consider the knee joint handout, pages 24 to end where loads in Quadriceps muscle, femur-tibia joint reaction and patellofemoral joint reaction knee joint at 30 o flexion were computed. Now, consider the case when knee joint angle is 45 o flexion and the external load 150 N is applied vertically down. To find quadriceps force, the FBO is, Take moment at Tibiofemoral joint, moments = 0 ; [(0.05) F-quad] – [(150) (0.4)Cos (45 o )] = 0 F-quad = (150)(0.4) (0.707) / (0.05) = 848.4 N F-patella = F-quad = 848.4 N To find the tibiofemoral joint reaction, FOB is redefined as below.    150N 0.4 m from Knee jt. Patella Tendon Quadriceps muscle, 0.05 m from center
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  • Fall '17
  • quadriceps muscle, joint reaction, Knee jt, reaction knee joint, knee joint handout

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