Center of Gravity (COG)

1. 10 N 20 N CG 2m 1m The weights are balanced, creating equal torques on either side of the fulcrum. Center of Gravity (COG) • the point around which a body’s weight is equally distributed in all directions

2. Center of Gravity (COG) • COG ~ COM • Geometric center? • Fixed? • Located outside of the body? • Moves in the direction of added mass

3. Center of Gravity (COG) • COG ~ COM • Geometric center? • Fixed? • Located outside of the body? • Moves in the direction of added mass

4. Applications • Muscle Function – Rotational Effects • Segmental Alignments • Whole Body Stability

5. Basic Kinetic Concepts

6. Course Content • Introduction to the Course • Biomechanical Concepts Related to Human Movement • Anatomical Concepts Related to Human Movement • Applications in Human Movement

7. Biomechanical Concepts • Basic Kinematic Concepts • Vector Algebra • Basic Kinetic Concepts

8. What is kinetics? • The study of forces tending to cause, causing, or resulting from motion. • Can identify and alter forces to change motion for desirable results.

9. Basic Kinetic Concepts • Force • Torque • Free Body Diagrams

10. Basic Kinetic Concepts • Force • Torque • Free Body Diagrams

11. Force • A push or pull • Vector quantity with 4 characteristics • Magnitude • Direction • Point of application • Line of force • The interaction of an object with its surroundings

12. For a force to cause acceleration: • Must overcome opposing forces (net force) • Must overcome inertia (mass) • Even if acceleration does not occur, deformation of the object will occur.

13. Common Forces • Weight • Reaction forces • GRF • Friction • JRF • Muscle force • Elastic force • Intraabdominal pressure • Inertial force • Fluid force

14. Basic Kinetic Concepts • Force • Torque • Free Body Diagrams

15. Torque • AKA moment of force, or moment • The ability of a force to create rotation • moment arm - perpendicular distance from the line of force to the axis of rotation • Must have a moment arm (be an eccentric force)

16. For a torque to cause angularacceleration: • Must overcome opposing torques (net torque) • Must overcome moment of inertia (mass, length of the rotating body)

17. Which way will the door turn? FC = 200 N FB = 100 N dC = .05 m dB = 1 m TC = FC*dC TB = FB*dB TC = 200 N * .05 m TB = 100 N * 1 m TC = 10 Nm TB = 100 Nm TN = 100 Nm + (-10 Nm) TN = 90 Nm

18. What must be done for the class to win? FC = 200 N FB = 100 N dC = .05 m dB = 1 m  TC or decrease  TB sufficiently. How? TC  FC   dC

19. Increase the magnitude of FC. FC = 300 N TC = FC*dC TB = FB*dB TC = 300 N * .05 m TB = 100 N * 1 m TC = 15 Nm TB = 100 Nm TN = 100 Nm + (-15 Nm) TN = 85 Nm

20. It would take 2000 N of force to get 100 Nm of torque. TC = FC*dC TB = FB*dB TC = 2000 N * .05 m TB = 100 N * 1 m TC = 100 Nm TB = 100 Nm TN = 100 Nm + (-100 Nm) TN = 0 Nm

21. Increase the magnitude of dC by changing the point of force application. FC = 200 N dC = .15 m TC = FC*dC TB = FB*dB TC = 200 N * .15 m TB = 100 N * 1 m TC = 30 Nm TB = 100 Nm TN = 100 Nm + (-30 Nm) TN = 70 Nm FB = 100 N

22. It would take a moment arm of .5 m to get 100 Nm of torque. We can not apply the force far enough away to get a moment arm this large. We run out of door! TC = FC*dC TB = FB*dB TC = 200 N * .5 m TB = 100 N * 1 m TC = 100 Nm TB = 100 Nm TN = 100 Nm + (-100 Nm) TN = 0 Nm

23. Increase the magnitude of dC by changing the direction of force application. TC = FC*dC TB = FB*dB TC = 200 N * .20 m TB = 100 N * 1 m TC = 40 Nm TB = 100 Nm TN = 100 Nm + (-40 Nm) TN = 60 Nm

24. We would not be able to change the angle enough to get 100 Nm of torque.

25. It will take a combination of all three, or else TB would have to be reduced in a similar (but opposite) manner.

26. How would the door rotate if a force were applied to the door in this manner?

27. Summary • Net torque determines rotation • Can change net torque by changing one or more individual torques • Can change individual torques by changing • Magnitude of force • Direction of force (moment arm) • Point of application of force (moment arm) • Often takes a combination of all of these

28. Applications of Torque in Human Movement • Muscle Function • Movement Analysis

29. What kind of torque does the biceps brachii create? What kind of torque does gravity create?

30. Which way will the arm rotate?

31. What conditions have to be true for the arm to flex? For the arm to extend? What role does the triceps brachii play?

32. What is the torque output of this muscle in the frontal plane?

33. What is the torque output of this muscle in the frontal plane?

34. Can this muscle cause frontal plane rotation of the scapula? What are the angular movements of the scapula in the frontal plane called?

35. Yes, if it has a moment arm for that axis of rotation! Will this muscle cause downward rotation?

36. Yes, if its torque is larger than the opposing torques.

37. What happens to the length of the moment arm of the muscle throughout the ROM? What about muscle torque? Does muscle force stay constant through the ROM?

38. What happens to the length of the moment arm of the muscle throughout the ROM? What about muscle torque? Does muscle force stay constant through the ROM?

39. Applications of Torque in Human Movement • Muscle Function • Movement Analysis

40. What happens to the resistive force throughout the ROM? What about the resistive torque?

41. What happens to the resistive force and resistive torque as she goes through the ROM? d W d W

42. What would happen to resistive torque if she put her hands behind her head? d W d W

45. Basic Kinetic Concepts • Force • Torque • Free Body Diagrams

49. Summary • Net force determines magnitude and direction of acceleration for linear motion. • Net torque determines magnitude and direction of acceleration for angular motion. • FBDs are a necessary first step in analyzing human motion.

50. For the next lecture unit: • Lecture Topic #3 • Subtopic A – The Skeletal System