Question

For problems 5-8, the joint is moving through its range of motion. As it does this, the angle between the muscle and the long axis of the bone, theta, changes. Assume that the muscle force is a constant 500 N and the distance from axis of rotation to the muscle insertion, d, is 0.02 m. Find the torque for each angle theta. Refer to the figure below. (HINT: Calculate the torque using T = F * sin(angle) * d.) 5. At an angle of 30°. Show your work. (5 points) 6. At an angle of 60°. Show your work. (5 points) _7. At an angle of 90°. Show your work. (5 points) 8. At which angle is the maximum torque produced? Why? (5 points)

Answer 1

Force; F= 500N. de 0.02m Torque, t = F sino.d T = F.disina - 5. 6 0 = 30° = sino = sin 30º = { is T = F.d sin 300 = 500 x 0.02% TE 5 N-m ] 0= 60° → sin Oz sin 60°= VE = T= F. di sin 60° = 500 0.02 x 12 .=553= 5x1.732= 8.66 N-m I=8.66 N-m. 0=90° → sin Ozsin o = 1 - Ta F-disin 90° = 500X0.02 x 1 =10 Z=10 Nom ] 7. Torque will be maximum when angle o=90° and sinoz singo=1 because the maximum value of sino is 1.