Question

Part II: Solve the following problems 1) A patient rehabilitating a knee injury performs knee extension exercises wearing a 35 N weight boot as show in the figure 1. Calculate the total amount of torque generated at the knee by the weight boot for the four positions shown, given a distance of 0.50 m between the weight boot's CG and the joint center at the knee. Figure 1.

Answer 1

i sine & F o distance hotice angle between Fir Soln Torque = x ² = 1 1 1 TrF sino Given, r=0.50 sa 35N at 0-90° = 2 = 0.50 35 *! Z = 17.5 Nm at 0= 60° - 0.50 x 35 x $/ = 15.15 m) at 0-30 – 2 = 0.50 x 35 x 1 Z=8.75 Non at 0=0 & 2 = 0.50 x 35x0. :=onml