Presentation Transcript

1. Overall course plan Functional anatomy Tissue mechanics: mechanical properties of muscle, tendon, ligament, bone Kinematics: quantification of motion, with no regard for the forces Kinetics: forces, torques
2. Reading Forces & COM: Ch 2: 41-42, 46-54; Ch 3: 115-117 Ground Reaction Forces & Locomotion: Ch 2: 56-59 Pressure, COP, Friction: Ch 2: 60-63 Free-Body Diagrams Ch2:43-44; Ch 3:107-109
3. How do we propel ourselves forward during running? Why does a runner lean on the curve in a track? What keeps an airplane in the air? How does a pitcher throw a curve ball?
4. What is a force? “an agent that produces or tends to produce a change in the state of rest or motion of an object” Kinetics = study of motion resulting from forces Forces are vectors: Magnitude Direction
5. Force is a vector What are the horizontal and vertical components of a force with a magnitude of 72 N acting at 16 degrees above the horizontal?
6. Outline Kinetics: Forces in human motion Newton’s Laws Law of Gravitation Center of Mass (C.O.M.) Laws of motion: First Law: Inertia Second Law: Acceleration Third Law: Action-Reaction Pressure Friction
7. Law of Gravitation All objects with mass attract one another. Earth’s Grav. Force on a human = mg = weight Unit of measurement of force = Newton (N) Newton = kg • m / s2 Human: weight of 154 lbs. mass of 70 kg weight of 700 N 6 fig newtons= 1 Newton One small apple ~ 1 Newton
8. Center of mass (C.O.M.) A point about which all of the mass of an object is evenly distributed. Whole body can be represented by single point mass at C.O.M.
9. Position of C.O.M. in common shapes of uniform density In the middle of a square In the middle of a circle or sphere In the middle of a cube
10. C.O.M. in standing person
11. A crude method for finding C.O.M. Tips down When balanced, the C.O.M. position been found
12. C.O.M. position can move. C.O.M. of whole body: depends on the configuration of the body segments. From Enoka 2.11
13. C.O.M. position depends on the distribution of body weight among and within the body segments Segmental analysis head trunk upper arm fore arm hand thigh shank foot Enoka 2.13
14. Table 2.1
15. Segment mass Table 2.1 in Enoka - weights of segments and the position of the C.O.M. of segments. Example: thigh Thigh weight = 0.127•BW - 14.8 Thigh weight and BW are in Newtons If person weighs 700N, then Thigh weight = 74.1 N
16. Where is the C.O.M. of the thigh? Proximal: Closer to trunk. Distal: Further from trunk C.O.M. is ~ 40% of the distance from the proximal to the distal end of thigh Proximal Thigh Distal
17. C.O.M. can be located outside the body Enoka 2.11
18. C.O.M. and the brain
19. Outline Kinetics: Forces in human motion Newton’s Laws Law of Gravitation Center of Mass (C.O.M.) Laws of motion: First Law: Inertia Second Law: Acceleration Third Law: Action-Reaction Pressure Friction
20. Newton’s 1st Law:Law of Inertia Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by a force impressed upon it Inertia = resistance of an object to motion Directly related to body mass If SF=0, then Dv=0
21. Outline Kinetics: Forces in human motion Newton’s Laws Law of Gravitation Center of Mass (C.O.M.) Laws of motion: First Law: Inertia Second Law: Acceleration Third Law: Action-Reaction Pressure Friction
22. Newtons 2nd law:Law of acceleration A force applied to a body causes an acceleration of a magnitude proportional to the force, in the direction of the force, and inversely proportional to the body’s mass. SF = ma SFx = max SFy = may
23. 250 N a = 5 m/s2 m = 100 kg F = ma = 500 N a = 5 m/s2 m = 50 kg F = ma = 250 N 500 N
24. How much force must be exerted to accelerate a 240kg mass to 5.7m/s2? 1368 kg m/s2 1368 N 1368 kg 890 m/s2 none of the above
25. Force in angular motion F = ma Tangential at = ∆v / ∆t Ft = mat = m∆v/∆t Radial ar = v2 / r Fc = mv2 / r centripetal force Ft = m∆v/∆t Fc = mv2/r
26. at = r Ft = mr Fc = mr2 ar = r2 Force in angular motion: F = ma
27. A 70 kg person is running around a circular path (radius = 2 m) with a constant angular velocity of 1 rad /sec. What are the magnitudes of the average tangential and radial forces are required for this turn? Ft= 140 N, Fc= 0 N Ft= 0 N, Fc= 140 rad/s2 Ft= 0 N, Fc= 140 N Ft= 140 N, Fc= 0 rad/s2
28. Jeff Francis pitches a 250 g baseball with a release velocity of 100MPH (44m/sec). His arm is fully extended at release, and the distance from the ball to his shoulder is 60cm. How much force is needed at the shoulder to keep his arm in place? 806675 N 806.7 N 10.995 N None of the above
29. Outline Kinetics: Forces in human motion Newton’s Laws Law of Gravitation Center of Mass (C.O.M.) Laws of motion: First Law: Inertia Second Law: Acceleration Third Law: Action-Reaction Pressure Friction
30. Law of action-reaction If one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object.
31. mg Ground reaction force mg Example of action-reaction A person exerts a force on the floor equal to their body weight (mg). The floor exerts an equal and opposite force (mg) on the person. “Ground reaction force” (GRF)
32. Free body diagram recipe: Simple sketch Outline the defined “system” with ------ Arrow for force of gravity acting on c.o.m. Arrows forGRFs Arrows for other external forces No arrows for internal forces See Ch 2 page 43
33. Measurement of ground reaction force Static: a bathroom scale can be used. Dynamic (e.g., jump): a force platform should be used.
34. Force platform Can measure forces very quickly (e.g., every 0.0001 s). Can measure the forces in three dimensions Vertical: Fg,y Horizontal: Fg,x Lateral: Fg,z
35. Y X Z
36. Sign convention for Fg,y Fg,y Upward is positive Locomotion, jumping, etc.: Fg,y > 0 Fg,y < 0: not normally possible need suction cup shoes +Fg,y
37. Sign conventions Positive Fg,x: oriented in the direction of motion or toward anterior side. +Fg,x
38. mg Ground reaction force mg Analysis of ground reaction forces Can be used to calculate the acceleration of the C.O.M. SF= ma SFy= may Fg,y - mg = may Fg,y = may + mg
39. Analysis of ground reaction forces Can be used to calculate the acceleration of the C.O.M. SF= ma SFx= max Fg,x = max +Fg,x
40. Analysis of ground reaction forces Can be used to calculate the acceleration of the C.O.M. SF= ma SFy= may SFx= max Fg,y - mg = may Fg,x = max Fg,y = may + mg
41. Vertical jump Fg,y = may + mg Fg,y (N) Bodyweight
42. RUN 3.9 m/s 2100 1400 Fg,y (N) 700 0 0 0.50 0.25 0.75 Time (s)
43. RUN Fg,y Stance
44. RUN 3.9 m/s 500 Forward Fg,x (N) 0 Backward -500 0 0.25 0.50 0.75 Time (s)
45. WALK 1.25 m/s 1050 700 Fg,y (N) 350 0 0 0.4 0.8 1.2 Time (s)
46. 0 0.4 0.8 1.2 Time (s) WALK 1.25 m/s 210 Fg,x (N) Forward 0 Backward -210
47. Walk: 1.25 m/s Run: 3.9 m/s x
48. Peak Fg,y (kN)
49. Suppose a runner experienced the following ground reaction forces at one point in time during the stance phase:Fg,x= -286NFg,y= 812NFg,z= 61NCalculate the magnitude and direction for each of the resultant ground reaction force in the sagittal and frontal planes.
50. Suppose a runner experienced the following ground reaction forces at one point in time during the stance phase:Fg,x=-286NFg,y=812NFg,z=61NCalculate the magnitude and direction for each of these resultant ground reaction force in the sagittal and frontal planes.A) 861N at 0.34 rad relative to the vertical 814 N at 0.07 rad relative to the verticalB) 861N at 2.8 rad relative to the horizontal 814 N at 3.07 rad relative to the horizontalC) 861N at 19.4 rad relative to the vertical 814 N at 4.01 rad relative to the verticalD) 861N at 70.6 rad relative to the horizontal 814 N at 85.99 rad relative to the horizontalE) None of the above
51. Suppose a runner experienced the following ground reaction forces at one point in time during the stance phase:Fx=-286NFy=812NFz=61NCalculate the magnitude and direction of the resultant ground reaction force in the transverse plane.A) 292N at 0.21 rad relative to the forward horizontalB) 292N at 0.21 rad relative to the backward horizontalC) 861N at 0.34 rad relative to the forward horizontalD) 861N at 0.34 rad relative to the forward horizontal
52. Dynamic and Static Analyses Dynamic Analysis SF= ma Static Analysis SF= 0
53. Outline Kinetics: Forces in human motion Newton’s Laws Law of Gravitation Center of Mass (C.O.M.) Laws of motion: First Law: Inertia Second Law: Acceleration Third Law: Action-Reaction Pressure Friction
54. Force versus pressure Force: total force measured in Newtons Pressure = force per unit area. Force is measured in Newtons (N). Area is measured in m2 Pressure: N / m2 or Pascal (Pa) P=F/A
55. Example of Fg versus pressure Running shoe Fg = 2000N. Sole area = 0.025 m2. Average pressure on sole of shoe (F / A) F / A = 2000 N / 0.025 m2 = 80,000 N/m2 Cleats (e.g., for soccer) Fg = 2000N. Cleat total area = 0.005 m2. Average pressure on cleats (F / A) F / A = 2000 N / 0.005 m2 = 400,000 N/m2
56. Distribution of pressure under sole of foot Divide sole into many small squares Find pressure for square (F/A) Use special insoles or pressure sensing mats to do this
57. Distribution of pressure under sole of foot Divide sole into many small squares Find pressure for square (F/A) Use special insoles or pressure sensing mats to do this Center of Pressure: Weighted average of all the downward acting forces indicates the path of the resultant ground reaction force vector
58. Rearfoot striker Point of force application(Center of Pressure) Toe off Foot strike Midfoot striker Enoka 2.19
59. The center of pressure and point of peak pressure are the same True False It depends
60. Possibly confusing terms Force Pressure Center of Pressure Peak pressure Point of force application
61. Outline Kinetics: Forces in human motion Newton’s Laws Law of Gravitation Center of Mass (C.O.M.) Laws of motion: First Law: Inertia Second Law: Acceleration Third Law: Action-Reaction Pressure Friction
62. Friction Resistance to one object sliding, rolling, or flowing over another object or surface. Human movement examples: foot - ground. bicycle tire - ground snowboard on snow friction in joints Friction is a reaction force Fa,x Fs = friction force
63. Static Friction force (Fs) Friction force depends on: properties of the surfaces force acting perpendicular to the surfaces e.g., ground-foot friction: Fg,y is the perpendicular force
64. Figure 2.18 Data from Miller and Nelson 1973.
65. mg Fa,x Fs,max=  * Fg,y Fs Fg,y is vertical ground reaction force  is coefficient of friction dimensionless depends on the properties of the surfaces static (s) dynamic (d) (s > d) Fg,y Fa,x = applied force Fs = friction force
66. mg Fa,x Static friction Fs Fs,max = largest static friction force possible Fs,max = s * Fg,y Block will not move if Fa,x≤Fs,max s running shoe - loose gravel : 0.3 running shoe - grass: 1.5 bicycle racing tire concrete: 0.8 Fg,y Fa,x = applied force Fs = friction force
67. mg Fa,x Dynamic friction Fs Block moves if Fa,x > Fs,max (max static friction) Moving block experiences dynamic friction Fd = d * Fg,y d < s Fg,y Fa,x = applied force Fs = friction force
68. Friction and Locomotion Will a person’s shoe slip? Will not slip if vector sum of Fg,x & Fg,z is less than maximum static friction force. Fg,x & Fg,z act parallel to surface
69. Fg,x Fg,z Fparallel Fparallel = (Fg,x2 + Fg,z2)0.5
70. When will shoe not slip? |Fparallel| ≤ Fs,max no slipping |Fparallel| ≤ s * Fg,y |Fparallel|/ Fg,y ≤ s
71. Friction in walking & running Constant velocity walking & running |Fparallel| / Fg,y ≤ s Fg,z ~ 0 |Fg,x|/ Fg,y ≤ s Walk (1.2 m/s) & run (3.9 m/s): |Fg,x|/ Fg,y: 0.1 - 0.2 µs for a running shoe loose gravel: 0.3 dry grass: 1.5
72. Which will have the greater maximum static friction force? A B The same It depends
73. Ground reaction force during constant velocity cycling v = 12 m/s (26 mph): Fg,z ~ 0 N Fg,x = 25 N Fg,y = 500 N (half weight of bike & person)
74. Bicycle tire friction No slip: Back wheel |Fg,x| /Fg,y ≤ µs,max |Fg,x|/ Fg,y = 25 N / 500 N = 0.05 Static coefficient of friction for bicycle tire: dry concrete: 0.8 wet concrete: 0.5 sand: 0.3
75. Forces parallel to the ground during acceleration in cycling Example: A person on a bicycle (total mass of 100 kg) is accelerating at 4 m/s2 Acceleration force = Fg,x under back tire Fg,x = ma = 100 kg * 4 m/s2 = 400 N |Fg,x|/ Fg,y = 400 N / 500 N = 0.8 Static coefficient of friction for bicycle tire: dry concrete: 0.8 wet concrete: 0.5 sand: 0.3
76. Friction in Object Manipulation
77. Friction & FBDs Friction in locomotion allows there to be a horizontal GRF Don’t show GRF AND friction, just GRF
78. Fn mg q Problem: Friction force on slope Find maximum friction force in terms of mg, q, & µs.
79. Fn mg q Friction force on slope Fs,max = Fn• µs Fn= mg cosq Fs,max = µs • mg cosq Fparallel (force pulling downhill parallel to slope) = mg sin q
80. Friction vs. Gravity force parallel m=70kg µs = 0.5 theta = 30 degrees Solve for static friction force and the component of gravitational force pulling parallel to the slope. Will the block move? Calculate its acceleration (µd = 0.4)