Question

An ACL patient is doing leg extensions to strengthen his quadriceps muscles. With no additional weight added, the patient's lower leg (shank+foot) has a mass of 6.25 kg. The center of mass of the lower leg segment is located 32 cm from the axis of rotation of the knee joint. The leg is flexed 25 degree down from horizontal. The quadriceps muscle acts through the patellar tendon, which inserts 5 cm from the axis of rotation of the knee joint and is oriented at an angle of 10 degree with respect to the tibia, as shown. The motion is performed slowly enough that we can assume (quasi) static equilibrium. How much force (F_q) must the quadriceps exert in order to support the weight of the limb? What are the reaction forces at the knee (R_x and R_y)? How do the magnitudes of these force compare with the load imposed by the weight of the limb? Do your values seem reasonable? Why or why not?

Answer 1

Solution: 5 cm 7259 32 cm W = mg (a). Moment at hinge is. Em=0 F, sin 10°(5) -W (32)cos 25° = 0 F, sin 10°(5) =W (32)cos 25° Rearrange the above equation to get Fo. W(32) cos 250 sin 10°(5) mg(32) cos 25° sin 10°(5) (6.25)(9.8) (32)cos 250 sin 10°(5) = 2045.93 N = 2.046kN (b). EF, = 0

R-F, cos 10°sin 65° + Fosin 10° sin 25º = 0 R = Focos 10°sin 65°- Fosin 10°sin 25° = (2045.93 N) cos 10° sin 65°-(2045.93 N) sin 10°sin 25° = 2020.4 -150.14 R = 1870.26N EF, =0 R, - F, cos10°cos 65°+ Fosin 10° cos 25° – mg = 0 R = mg-F [cos 10°cos 65° +sin 10°cos 259] =(6.25)(9.8)-(2045.93 N) [cos10° cos 65°+sin 10°cos 259) = (6.25)(9.8) -2362.64 =-1112.54 R =-1112.54 N The negative sign indicates show the direction of R, is downwards. (C). FR = 2045.93 N W = mg =(6.25) (9.8) = 61.25N Magnitude of Muscle force is greater than the weight of the limb. Fo>>W Hence, the value is not reasonable because the weight of the limb is 61.25 N small but length of the limb is large due to the reason of muscle force that require too high. So, muscle force value is not reasonable.