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Outline. Kinetics Linear Forces in human motion Mechanical work, power, & energy Impulse - momentum Angular Torques in human motion Mechanical work, power, & energy Impulse-momentum. Outline. Torques in human motion Definitions

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  1. Outline • Kinetics • Linear • Forces in human motion • Mechanical work, power, & energy • Impulse-momentum • Angular • Torques in human motion • Mechanical work, power, & energy • Impulse-momentum

  2. Outline • Torques in human motion • Definitions • External force ----> muscle force (static analysis) • Review of approach • Mechanical advantage • Musculoskeletal complexity • External force ----> muscle force (dynamic analysis)

  3. Torque (= moment) angular equivalent of force Capability of a force to produce rotation Units: N*m Importance? Muscles cause movement by creating torques about joints.

  4. Torque (T): Capability of a force to produce rotation T = MR * F F= force MR= moment arm (perpendicular distance from the point of rotation to the line of force application) rF = distance to F MR and rFare NOT the same!! F MR rF

  5. Torque (T): Capability of a force to produce rotation T = MR * F F= force MR= moment arm (perpendicular distance from the point of rotation to the line of force application) rF= distance to F F MR rF

  6. What is the Torque (T) due to force F?F=100N; distance to F: rF = 1m, q=30o T = MR * F F 86.6 N 100 Nm 86.6 Nm 50 Nm 50 N MR q rF

  7. Torque is a Vector!Right-Hand Thumb RuleFigure 2.4 r: MR (moment arm) Right-Hand Thumb Rule: 1. align your hand with MR 2. curl your fingers towards F 3. direction of thumb is direction of torque vector

  8. Torque and the Coordinate System Direction of Positive Torque? If using default coordinate system: Use right hand thumb rule Counter-clockwise (CCW) If using flexion/extension terms: Extension is +ve! Be CONSISTENT! y x

  9. What is the Torque (T) due to force F?F=100N; distance to F: rF = 1m, q=30o y F x MR q rF Positive Negative It Depends

  10. Outline • Torques in human motion • Definitions • External force ----> muscle force (static analysis) • Review of approach • Mechanical advantage • Musculoskeletal complexity • External force ----> muscle force (dynamic analysis)

  11. Example A person holds their elbow at 90° with their forearm parallel to the ground. Elbow torque? Step 1: Draw a free body diagram “system” = the forearm + hand Upper arm Forearm Elbow

  12. Factors affecting Elbow Torque: Weight of forearm (Fw) and position of its COM From Table in Enoka(BW = 600 N) Fw (forearm+hand) = 11 N Distance from proximal end to COM is 0.16 m (MR) Fw MR

  13. Elbow torque due to weight of forearm T=MR * F T = 0.16m * 11N T = 1.8 Nm Direction? T = -1.8Nm 11 N 0.16 m

  14. A person holds their forearm so that it is 30° below the horizontal. Elbow torque due to forearm weight? Fw=11N; rF = 0.16m • 1.76 Nm • 1.5 Nm • 0.88 Nm • -1.5 Nm • -0.88 Nm Fw q rF

  15. A person holds their forearm so that it is 30° below the horizontal. Elbow torque due to forearm weight?Fw=11N; rF = 0.16m 11 N MR 30° 0.16 m

  16. Now let’s look at 2 weights

  17. 100 N 11 N A person is holding a 100N weight at a distance of 0.4 m from the elbow. What is the total elbow torque due to external forces? We must consider the effects of 2 forces: forearm (11N) weight being held (100N)

  18. A person is statically holding a 100N weight at a distance of 0.4 m from the elbow. What is the elbow torque due to external forces? • T= (-Tarm) + (-Tbriefcase) • T = Tarm+Tbriefcase • T = (-Tarm) + Tbriefcase • T = Tarm + (- Tbriefcase) • It depends 100 N 11 N 0.16 0.4 m

  19. A person is statically holding a 100N weight at a distance of 0.4 m from the elbow. What is the elbow torque due to external forces? 100 N 11 N 0.16 0.4 m

  20. Torque about shoulder due to external forces when 5 kg briefcase is held with straight arm. Forearm force Upper arm force Briefcase force

  21. Upper arm force Forearm force Briefcase force 20N 0.16 m 15N 15N 0.48 m 49N 0.48 m 0.65 m 31 Nm 20.6 Nm - 41Nm 41 Nm None of the above

  22. 20N 0.16 m 15N 15N 0.48 m 49N 0.48 m 0.65 m

  23. Muscles create torques about joints Upper arm Elbow flexor muscle Biceps force Elbow Forearm T

  24. Static and Dynamic Analyses Statics (acceleration = 0)  F = 0  M = 0 (M is moment or torque) Dynamics (non-zero acceleration)

  25. Static equilibrium F1 F2 R2 Static equilibrium All accelerations are zero Three equations for analysis Fx = 0 Fy = 0  M = 0 R1 Teeter- totter

  26. Static equilibrium F1 F2 R2 R1 Fx = 0: No forces in this direction Fy = 0 F3 - F1 - F2 = 0 Ma = 0 T1 – T2= 0 F1R1 - F2R2 = 0 Convention: Counter Clockwise is positive a F3

  27. Upper arm Question: What muscle force (Fm) is required to support the forearm weight (Fw)? Free-body diagram - static equilibrium Elbow flexor muscle Fm Fw Elbow Forearm

  28. Step 1: Free body diagram. Joint reaction force (Fj): net force generated between adjacent body segments External forces Muscle forces Upper arm Fj Fm Fw Fm Fw Forearm Forearm Elbow Elbow

  29. Segmental Free body diagrams System = forearm+hand Weight Other external forces Muscle force Joint reaction force (Fj): net force generated between adjacent body segments Direction? If you are unsure of the direction a force is acting, draw a POSITIVE vector!! Fw Fj Fm Fw Forearm+hand Elbow

  30. Question: What muscle force (Fm) is required to support the forearm weight (Fw)?Step 1: Free body diagram.Givens: Fw = 11 N, Rw = 0.16 m, Rm = 0.03 mStep 2: Apply appropriate equation Upper arm Fj Fm Fw Fm Fw Forearm Forearm Elbow Elbow

  31. Fj,y Fm Fw Solve for Muscle force, Fm Rm Static equilibrium Melbow = 0 Fj creates no moment at elbow -(Tw) + (Tm) = 0 -(Fw * Rw) + (Fm * Rm) = 0 Fm = (Fw * Rw) / Rm Substitute: Fw= 11 N, Rw = 0.16 m, Rm = 0.03 m Fm = 59 N Rw

  32. If a person holds a 100 N weight in their hand 0.4 m from the elbow, what is the elbow flexor muscle force? Ignore weight of forearm Information Rm = 0.03 m Rext = 0.4 m Step 1: Free body diagram Upper arm Muscle Fext Elbow Forearm

  33. Fext Fm Rm Rext Solve forElbow flexor force Rm = 0.03 m Fext= 100 N Rext= 0.4 m Fj,y Fj,x

  34. When Rm < Rext,muscle force > external force Fm = Fext(Rext / Rm) Last example Fext = 100N Rext > Rm Fm = 1333 N Upper arm Biceps brachialis Fext Rm Elbow Rext

  35. If a person holds a 100 N weight in their hand 0.4 m from the elbow, what is the joint reaction force? (ignore weight of forearm). Free body diagram Apply equations Fj,y Fext Fm Rm Fj,x Rext

  36. If a person holds a 100 N weight in their hand 0.4 m from the elbow, what is the joint reaction force? (ignore weight of forearm). Free body diagram Apply equations: Fy = 0 Fj,y + Fm - Fext = 0 Fm = 1333 N, Fext = 100 N Fj,y = -1233 N Fx = 0 Fj,x = 0 N Fj,y Fext Fm Rm Fj,x Rext

  37. What is the muscle force when a 5 kg briefcase is held with straight arm? Fm Fj Upper arm force Forearm force Briefcase force

  38. 20N 0.16 m 15N T = (49N * 0.65m) + (15N * 0.48m) + (20N * 0.16m) T = (31 Nm) + (7.2 Nm) + (3.2 Nm) T = 41 Nm 15N 0.48 m 49N 0.48 m 0.65 m

  39. Fm Fj 20N 0.16 m Rm = 0.025 m, Fm = ??? 15N 0.48 m 49N 0.65 m -1640 Nm 1640 Nm 1640 N None of the above

  40. Fm Fj 20N 0.16 m Does Fjx = 0? 15N 0.48 m 49N 0.65 m Yes No It depends

  41. At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle torque due to Fg? Step 1: Find moment arm of Fg,x(MRx) & Fg,y(MRy) about ankle. 0.2 m 30° 200N 350N

  42. 0.2 m MRx 30° 200N 350N At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle torque due to Fg? Step1 MRx= 0.2 sin 30° = 0.10 m

  43. MRy 0.2 m MRx 30° 200N 350N At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle torque due to Fg? Step 1 MRx= 0.2 sin 30° = 0.10 m MRy= 0.2 cos 30° = 0.17 m Step 2 T = (Tx) + (Ty) T= (Fg,x*MRx) + (Fg,y * MRy) T = (200 * 0.10) + (350 * 0.17) T = 79.5 Nm

  44. MRy 0.2 m MRx 30° 200N 350N At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle extensor muscle force?MRmusc = 0.05m

  45. Outline: Torque • External force ----> muscle force (statics) • Review of approach • Mechanical advantage • Musculoskeletal complexity • External force ----> muscle force (dynamics)

  46. Mechanical advantage (MA) • Fext = Fm * MA MA = Rm / Rext • MA = 1 Fm = Fext • MA < 1 Fm > Fext • MA > 1 Fm < Fext Upper arm Muscle Fext Rm Rext

  47. MA < 1 • Rmuscle < Rext Fmuscle > Fext Shank Fm Rm Foot Rext Ankle Fext = Fg

  48. MA < 1 Fm Briefcase force

  49. MA > 1 • Rmuscle > Rext • Fmuscle < Fext Fmuscle (splenius capitis) Fext (Fw)

  50. Factive MA > 1 • MA = Ractive / Rext • Ractive > Rext • Factive < Fext Ractive Fext Rext

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