Joint Mechanics What Happens to Muscle Tension?

1. Joint MechanicsWhat Happens to Muscle Tension? • 1. Some attempts to --> joint movement • TORQUE • 2. Some attempts to --> joint, bone stability • The amount that goes to torque and stability depends on • Muscle group • Joint angle • Muscle length and speed of stretch

2. Translating Tension into Torque • Line of pull • Orientation of muscle force vector • Usually along the long axis from insertion to insertion Line of pull

3. Translating Tension into Torque • Angle of pull • Angular relationship between muscle’s line of pull and orientation of bone (a) a

4. Translating Tension into Torque Muscle force can be resolved into 2 vectors 2 roles: 1. Torque --> movement 2. Stabilize joints, bones (Fs) (Fro) Fm Fro Fs

5. Translating Tension into Torque Fm Patella - pulley Fs Fro knee

6. Translating Tension into Torque Knee - quad Fm Fs a Fro sin a = Fro/Fm Fro = Fm • sin a cos a = Fs/Fm Fs = Fm • cos a a = 24°

7. Translating Tension into Torque Fm = 300 N knee Fro Fs a a = 24° Fro = 300N • sin 24° Fro = 122.02 N Fs = 300N • cos 24° Fs = 274.06 N

8. Tension translation to Torque As a (angle of pull) changes; Fro and Fs change also Therefore: Torque production Stability Will change!!

9. Changes in joint angle, a, and translation to torque Fro and therefore torque production will peak somewhere in the ROM Fro or Tm Joint angle

10. Fro and therefore torque production will peak somewhere in the ROM

11. Fro and therefore torque production will peak somewhere in the ROM

12. Fro and therefore Torque production will peak somewhere in the middle of ROM Justin Gatlin: 2004 Olympic Champ - 100m

13. Fro and therefore Torque production will peak somewhere in the middle of ROM Flexing hips, knees increases force & torque Stretches agonist muscles • increased passive tension (rapid) > material properties • increased active tension > increased cross-bridges > increased Ca2+ release - SR > increased stretch reflex *muscle spindles Increased Torque production •a closer to 90° (increased moment arm) --> Fro increased Jump shot Volleyball spike Long jump High jump

14. Changes in joint angle, a, and translation to torque The closer a --> 90°, the greater the Fro When the angle of pull = 900, FRO = FM

15. Changes in joint angle, a, and translation to torque • When a < 900 • Some force ---> stabilization (compression) • Smaller a --> great stabilization Fm Fs a Fro Most joints have a small a throughout their ROM Fs > Fro

16. Changes in joint angle, a, and translation to torque • When a > 900 • Some force ---> dislocation (tension) Happens over a small part of ROM in a few muscles • biceps brachii • gastrocnemius • hamstrings Fro a Fs

17. Elbow - biceps brachii shoul Hand + barbell Fm • When elbow joint q small Fro Fro Fs Fm a q Fs F elbow f = 180° - a a

18. Ankle - Gastrocnemius BW Fro Fm f = 180° - a wheelbarrow Fro a Fm q Fs F ANK a Fs TOES Axis

19. Knee - Hamstrings BW hip Deep Squat Danger Fs --> dislocation torque - long levers shear forces q Fm Fro a knee Fm Fs ankle a Fro F Fs f = 180° - a

20. The anterior deltoid muscle is contracting with a force of 300N. If the angle of pull = 28°, find the rotary and stabilizing portions of the muscle force • Given: Fm = 300N a = 28° • Find: Fro and Fs • Diagram: • Formuls: sin a = Fro/Fm • cos a = Fs/Fm shoulder Fm Fro Fm a Fs a Fro Fs

21. The anterior deltoid muscle is contracting with a force of 300N. If the angle of pull = 28°, find the rotary and stabilizing portions of the muscle force • sin a = Fro/Fm • Fro = Fm • sin a • Fro = 300N • sin 28° • Fro = 300N • 0.469 • Fro = 140.8 N • cos a = Fs/Fm • Fs = 300N • cos 28° • Fs = 300N • 0.883 • Fs = 264.9 N Fro Fm a Fs

22. The hamstrings group is contracting with a force of 400N. If the angle of pull = 105°, find the rotary and stabilizing components. The joint angle is approx. 70° • Given: Fm = 400N a = 105° • Find: Fro and Fs • Diagram: • Formuls: sin f = Fro/Fm • cos f = Fs/Fm hip Fro Fm a knee Fm a Fs Fro f Fs f = 180° - a f = 75°

23. The hamstrings group is contracting with a force of 400N. If the angle of pull = 105°, find the rotary and stabilizing components. The joint angle is approx. 70° • sin f = Fro/Fm • Fro = Fm • sin a • Fro = 400N • sin 75° • Fro = 400N • 0.966 • Fro = 386.4 N • cos f = Fs/Fm • Fs = 400N • cos 75° • Fs = 400N • 0.259 • Fs = 103.5 N (tension or dislocating) Fm knee a Fro f Fs f = 75°

24. T = F • d Tm = Fro • DFA FM FRO Angle of Pull DFA Torque Production at Joints by Muscles (a)

25. The iliopsoas (hip flexors) are contracting with a force of 800 N at an angle of pull of 30°. If the iliospoas insert into the femur 8 cm from the middle of the hip joint, find the torque produced by the hip flexors. • Given: Fm = 800N a = 30° • dFA = 8 cm • Find: Fro and Tm • Diagram: Fs Fro Fm a Fm DFA a Fro Formulas: sin a = Fro/Fm Tm = Fro • DFA

26. The iliopsoas (hip flexors) are contracting with a force of 800 N at an angle of pull of 30°. If the iliospoas insert into the femur 8 cm from the middle of the hip joint, find the torque produced by the hip flexors. • sin a = Fro/Fm • Fro = Fm • sin a • Fro = 800N • sin 30° • Fro = 800N • 0.500 • Fro = 400 N • Tm = Fro • DFA Tm = 400N • 8 cm Tm = 3200 N•cm Tm = 32.0 N-m Fro Fm Fm DFA a Fro