How much force must the quadriceps (moment arm of 4.5 cm) exert to maintain the lower leg at 59 degrees from full extension? The combined weight of the lower leg is 60 N, with a center of gravity 30 cm from the axis.
No figure was provided
As figure is not provided let us assume the leg is horizontal when it is fully extended.
Let F_quadricep is the force exerted by quadricep.
As the leg is in equilibrium net force net torque acting on the leg must be zero.
Apply net torque about knee = 0
4.5*F_quadricep - 60*30*sin(90 - 59) = 0
F_quadricep = 60*30*sin(31)/4.5
= 206 N
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