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Functional anatomy Tissue mechanics: mechanical properties of muscle, tendon, ligament, bone Kinematics: quantification of motion, with no regard for the forces Kinetics: forces, torques Overall course plan

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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

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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?

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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

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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?

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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

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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 / s 2 Human: weight of 154 lbs. – mass of 70 kg – weight of 700 N 6 fig newtons = 1 Newton One small apple ~ 1 Newton

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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.

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Position of C.O.M. in common shapes of uniform density In the middle of a cube In the middle of a circle or sphere In the middle of a square

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C.O.M. in standing person

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A crude method for finding C.O.M. When balanced, the C.O.M. position been found Tips down

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C.O.M. position can move. C.O.M. of whole body: depends on the configuration of the body segments. From Enoka 2.11

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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

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Table 2.1

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Segment mass Table 2.1 in Enoka - weights of segments and the position of the C.O.M. of segments. Example: thigh Thigh weight = 0.127BW - 14.8 Thigh weight and BW are in Newtons If person weighs 700N, then Thigh weight = 74.1 N

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Where is the C.O.M. of the thigh? Proximal: Closer to trunk. Distal: Further from trunk Proximal Distal Thigh C.O.M. is ~ 40% of the distance from the proximal to the distal end of thigh

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C.O.M. can be located outside the body Enoka 2.11

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C.O.M. and the brain

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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

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Newton’s 1 st 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 F=0, then v=0

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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

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Newtons 2 nd 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. F = ma F x = ma x F y = ma y

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a = 5 m/s 2 m = 50 kg F = ma = 250 N a = 5 m/s 2 m = 100 kg F = ma = 500 N 250 N 500 N

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How much force must be exerted to accelerate a 240kg mass to 5.7m/s 2 ? A)1368 kg m/s 2 B)1368 N C)1368 kg D)890 m/s 2 E)none of the above

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Force in angular motion F = ma Tangential a t = ∆v / ∆t F t = ma t = m∆v/∆t Radial a r = v 2 / r F c = mv 2 / r centripetal force F t = m∆v/∆t F c = mv 2 /r

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Force in angular motion: F = ma a r = r 2 a t = r F c = mr 2 F t = mr

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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? A)F t = 140 N, F c = 0 N B)F t = 0 N, F c = 140 rad/s 2 C)F t = 0 N, F c = 140 N D)F t = 140 N, F c = 0 rad/s 2

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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? A) 806675 N B) 806.7 N C)10.995 N D)None of the above

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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

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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.

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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) Ground reaction force mg

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Free body diagram recipe: Simple sketch Outline the defined “system” with ------ Arrow for force of gravity acting on c.o.m. Arrows for GRFs Arrows for other external forces No arrows for internal forces See Ch 2 page 43

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Measurement of ground reaction force Static: a bathroom scale can be used. Dynamic (e.g., jump): a force platform should be used.

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Force platform Can measure forces very quickly (e.g., every 0.0001 s). Can measure the forces in three dimensions – Vertical: F g,y – Horizontal: F g,x – Lateral: F g,z

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Y X Z

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Sign convention for F g,y F g,y Upward is positive Locomotion, jumping, etc.: F g,y > 0 F g,y < 0: not normally possible need suction cup shoes +F g,y

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Sign conventions Positive F g,x : oriented in the direction of motion or toward anterior side. +F g,x

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Analysis of ground reaction forces Can be used to calculate the acceleration of the C.O.M. F= ma F y = ma y F g,y - mg = ma y F g,y = ma y + mg Ground reaction force mg

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Analysis of ground reaction forces Can be used to calculate the acceleration of the C.O.M. F= ma F x = ma x F g,x = ma x +F g,x

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Analysis of ground reaction forces Can be used to calculate the acceleration of the C.O.M. F= ma F y = ma y F x = ma x F g,y - mg = ma y F g,x = ma x F g,y = ma y + mg

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Vertical jump Bodyweight F g,y (N) F g,y = ma y + mg

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RUN 3.9 m/s 2100 1400 700 0 0 0.25 0.50 0.75 Time (s) F g,y (N)

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RUN F g,y Stance

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RUN 3.9 m/s 500 0 -500 F g,x (N) 00.250.500.75 Time (s) Backward Forward

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Time (s) WALK 1.25 m/s 0 700 F g,y (N) 0 0.40.81.2 350 1050

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-210 0 0 0.40.81.2 Time (s) F g,x (N) WALK 1.25 m/s Backward Forward 210

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Walk: 1.25 m/sRun: 3.9 m/s x

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Peak F g,y (kN)

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Suppose a runner experienced the following ground reaction forces at one point in time during the stance phase: F g,x = -286N F g,y = 812N F g,z = 61N Calculate the magnitude and direction for each of the resultant ground reaction force in the sagittal and frontal planes.

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Suppose a runner experienced the following ground reaction forces at one point in time during the stance phase: F g,x =-286N F g,y =812N F g,z =61N Calculate 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 vertical B) 861N at 2.8 rad relative to the horizontal 814 N at 3.07 rad relative to the horizontal C) 861N at 19.4 rad relative to the vertical 814 N at 4.01 rad relative to the vertical D) 861N at 70.6 rad relative to the horizontal 814 N at 85.99 rad relative to the horizontal E) None of the above

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Suppose a runner experienced the following ground reaction forces at one point in time during the stance phase: Fx=-286N Fy=812N Fz=61N Calculate the magnitude and direction of the resultant ground reaction force in the transverse plane. A) 292N at 0.21 rad relative to the forward horizontal B) 292N at 0.21 rad relative to the backward horizontal C) 861N at 0.34 rad relative to the forward horizontal D) 861N at 0.34 rad relative to the forward horizontal

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Dynamic and Static Analyses Dynamic Analysis F= ma Static Analysis F= 0

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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

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Force versus pressure Force: total force measured in Newtons Pressure = force per unit area. – Force is measured in Newtons (N). – Area is measured in m 2 – Pressure: N / m 2 or Pascal (Pa) – P=F/A

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Example of F g versus pressure Running shoe – F g = 2000N. – Sole area = 0.025 m 2. – Average pressure on sole of shoe (F / A) F / A = 2000 N / 0.025 m 2 = 80,000 N/m 2 Cleats (e.g., for soccer) – F g = 2000N. – Cleat total area = 0.005 m 2. – Average pressure on cleats (F / A) F / A = 2000 N / 0.005 m 2 = 400,000 N/m 2

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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

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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

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Rearfoot striker Midfoot striker Point of force application(Center of Pressure) Foot strike Toe off Enoka 2.19

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The center of pressure and point of peak pressure are the same A)True B)False C)It depends

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Possibly confusing terms Force Pressure Center of Pressure Peak pressure Point of force application

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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

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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 F a,x F s = friction force

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Static Friction force (F s ) Friction force depends on: – properties of the surfaces – force acting perpendicular to the surfaces e.g., ground-foot friction: F g,y is the perpendicular force

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Data from Miller and Nelson 1973. Figure 2.18

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F s,max = * F g,y F g,y is vertical ground reaction force is coefficient of friction – dimensionless – depends on the properties of the surfaces – static ( s ) dynamic ( d ) ( s > d ) F a,x F g,y FsFs F a,x = applied force F s = friction force mg

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Static friction F s,max = largest static friction force possible F s,max = s * F g,y Block will not move if F a,x ≤ F s,max s – running shoe - loose gravel : 0.3 – running shoe - grass: 1.5 – bicycle racing tire concrete: 0.8 F a,x F g,y FsFs F a,x = applied force F s = friction force mg

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Dynamic friction Block moves if F a,x > F s,max (max static friction) Moving block experiences dynamic friction F d = d * F g,y d < s F a,x F g,y FsFs F a,x = applied force F s = friction force mg

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Friction and Locomotion Will a person’s shoe slip? Will not slip if vector sum of F g,x & F g,z is less than maximum static friction force. – F g,x & F g,z act parallel to surface

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F g,x F g,z F parallel = (F g,x 2 + F g,z 2 ) 0.5 F parallel

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When will shoe not slip? |F parallel | ≤ F s,max – no slipping | F parallel | ≤ s * F g,y |F parallel | / F g,y ≤ s

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Friction in walking & running Constant velocity walking & running – |F parallel | / F g,y ≤ s – F g,z ~ 0 – |F g,x | / F g,y ≤ s – Walk (1.2 m/s) & run (3.9 m/s): – | F g,x | / F g,y : 0.1 - 0.2 µ s for a running shoe – loose gravel: 0.3 – dry grass: 1.5

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Which will have the greater maximum static friction force? A)A B)B C)The same D)It depends

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Ground reaction force during constant velocity cycling v = 12 m/s (26 mph): F g,z ~ 0 N F g,x = 25 N F g,y = 500 N (half weight of bike & person)

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Bicycle tire friction No slip: Back wheel |F g,x | /F g,y ≤ µ s,max – | F g,x | / F g,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

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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/s 2 – Acceleration force = F g,x under back tire F g,x = ma = 100 kg * 4 m/s 2 = 400 N – | F g,x | / F g,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

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Friction in Object Manipulation

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Friction & FBDs Friction in locomotion allows there to be a horizontal GRF Don’t show GRF AND friction, just GRF

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Problem: Friction force on slope FnFn mg Find maximum friction force in terms of mg, , & µ s.

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Friction force on slope F s,max = F n µ s F n = mg cos F s,max = µ s mg cos F parallel (force pulling downhill parallel to slope) = mg sin FnFn mg

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Friction vs. Gravity force parallel m=70kg µ s = 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 (µ d = 0.4)

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