A patient rehabilitating a knee injury performs knee extension exercises wearing a 15 N weight boot.

Question

A patient rehabilitating a knee injury performs knee extension exercises wearing a 15 N weight boot. Calculate the amount of torque generated at the knee by the weight boot, given a distance of 0.4 m between the weight boot's CG and the joint center of the knee. 90 degrees, 60 degrees, 30 degrees and 0 degrees.

 

 

Please include how/why "SIN" is used in the finding of the answer. If you are able to attach photos, that would be great! Thank you for your help! 

Answer 1

If θ is the angle between the moment arm r and the applied force F, the magnitude of the torque is the product of the component of the force perpendicular to the moment arm, and the length of the moment arm.  The perpendicular component of the force is F sinθ, so the magnitude of the torque is

 

τ = Fr sinθ

 

F = 15 N

r = 0.4 m

θ = 90°, 60°, 30°, 0°

 

I will do the 60º case.  You can work the others the same way.

 

τ = Fr sinθ = (15 N)(0.4 m)sin60° ≅ 5.196 Nm