Question

A person is doing leg flexion/extension exercises from a sitting position in order to strengthen the person's quadracept muscles. The forces acting in the lower leg are illustrated in the figures below.

W₀ is the weight of the boot, W₁ = 160 N is the weight of the person's lower leg, F_M is the magnitude of the force exerted by the quadricepts muscle on the tibia (shinbone) through the patellar tendon, and F_J is the force due to the tibiofemoral joint. The tibiofemural joint center is located at O. The patellar tendon is attached to the tibia at A, the center of gravity of the lower leg is located at B, and the center of gravity of the weight boot is located at C. The distance between O and A, B, and C for the person are measured to be a = 13 cm, b = 23 cm, and c = 51 cm, respectively. For the position with the lower leg shown, the lorg axis of the tibia makes an angle of β = 45° with the horizontal, and the line of action of the quadriceps muscle makes an angle of θ = 15° with the long axis of the tibia.

1)The training program calls for the force from the quadricepts muscle to be F_M = 1500 N when the lower leg is held fixed. What should be the weight of the boot to meet this training requirement?

F, FM W

Answer 1

Apply net torque about knee = 0

FM*a*sin(theta) - W1*b*sin(90-beta) - Wo*c*sin(90- beta) = 0

1500*13*sin(15) - 160*23*sin(90-45) - Wo*51*sin(90-45) = 0

Wo = (1500*13*sin(15) - 160*23*sin(45))/(51*sin(45) )

= 220 N <<<<<<<<-------------Answer