The quadriceps tendon attaches to the tibia at a 30° angle 4 cm from the joint center at the knee. When an 80 N weight is attached to the ankle 28 cm from the knee joint, how much force is required of the quadriceps to maintain the leg in a horizontal position? When the subject mass is 75 kg, what is the magnitude and direction of the reaction force exerted by the femur on the tibia? (Utilize “Foot and leg” in the included chart to obtain weight of the lower leg and location of COM. Neglect the action of other muscles.) 30°0.04 mwt = 80 Ndwt = 0.28 mdp = 0.04 m0.28 m80 N Table 3.1 Anthropometric DataSegmentHandForearmUpper armForearm and handTotal armFootLogThighFoot and legTotal legDefinitionWrist axis/knuckle II middlefingerElbow axis/ulnar styloidGlenohumeral axis elbow axisElbow axis/ulnar styloidGlenohumeral jointularstyloidLateral malleolus/headmetatarsal IIFemoral condyles/modialmallicolusGreater trochanter/femoralcondylesFemoral condyles/medialmalicolusGreater trochanter/medialmalleolusCenter of MassSegment LengthDistal0.494 PSegment Weight/Total Body Weight Proximal0.006 M0.5060.4300.6820.5300.016 M0.028 M0.022 M0.050 M0.0145 M0.0465 M0.100 M0.061 M0.161 M0.4330.4330.6060.4470.570 P0.564 P0.318 P0.470 P0.50 P0.567 P0.567 P0.394 P0.553 PRadius of Gyration/Segment LengthC of O Proximal Distal Density0.2970.587 0.577 M1.1603030.3220.4680.3680.4750.3020.3230.4160.3260.5260.5420.8270.6450.6900.5280.5400.7350.5600.647 M0.645 M0.565 P0.396 P0.690 P0.643 M0.653 M0.572 P0.650 P1.131.071.141.101.091.051.091.06

Given data as per the question weight =80 N Neglect the action of other muscles. considering equlibirium conditionτnet=0wrω−Fmrm=0wrωrω=Fm sin30wrωrω=Fm(12)Fm=2wrωrω=2×80×0.280.04Fm=1120 NNowapplying translational equlibirium conditionsF→m=(−Fmcos30)i+(Fmsin30)j=(−1120cos30)i+(1120sin30)j=(−969.948)i+(560)jandw→=−(80)jNowFnet=F→m+w→+R→=0R→=−(F→m+w→)=−((−969.948)i+(560−80)j)=−(−969.948)i+(480)jR→=(969.948)i−(480)jMagnitude of reaction R→=(969.948)2+(−480)2R→=1082.2195 NDirection of reaction tanθ=RyRxtanθ=480969.948θ=tan−1(480969.948)θ=26.3290 (below the horizontal )