movement forces figure reprinted from marey, 1889
TRANSCRIPT
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MovementForces
MovementForces
Figure reprinted from Marey, 1889.
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Objectives Review Newton’s laws of motion, and extend to
angular motion. Expand on the technique of the free body
diagram
Define torque as a rotary force
Describe the forces due to body mass
Explain the forces exerted by the surroundings
To quantify the concept of momentum
To characterize the relation between work and energy
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Newton’s Laws of Motion
Ia. Law of Linear Inertia – An object will remain stationary or move with constant velocity until an unbalanced external force is applied to it
“Constant velocity” – implies straight line (direction)
Inertia – resistance
Inertia quantified by mass (kg)
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Perceiving the Law of Inertia
The Elevator Test: stand in elevator with knees flexed about 20 and press the UP button.
Press the UP button - What happens?
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Perceiving the Law of Inertia
The Elevator Test: stand in elevator with knees flexed about 20 and press the UP button.
Press the UP button - Why do your
legs flex?
Upward force from elevator
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Perceiving the Law of Inertia
The Seat Belt Test: what happens when you press on the brakes as you are driving or if you JAM on the brakes?
Rearward force applied to seat from
wheels through chassis
Trunk accelerates forward relative to thighs and car. The more you JAM on the breaks, the greater the acceleration.
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Law of Inertia – Linear Kinetics
1. Inertia DOES NOT EQUAL Momentum (mv)
i.e. kg is not kg m / s
The downhill skier has inertia (which is constant) but it’s his/her momentum that is important
2. Inertia DOES NOT EQUAL weight (mg)
i.e. kg is not kg*g
Weight is a force vector, inertia is a scalar variable
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Law of Inertia and Inertial Forces
Inertial forces – motion-dependent forces (but really acceleration-dependent forces)
Forces causing acceleration of an object: F = ma
TWO MOST IMPORTANT PROBLEMS:
GROCERY BAG PHENOMENON and WALKING WITH A FULL CUP OF COFFEE
Secondary problem: Back injuries during lifting
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Law of Inertia and Inertial Forces
Inertial Forces important in lifting Effective force: F = ma
F = mg + mavert if avert = 0, F = mg
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Law of Inertia and Inertial Forces
Inertial Forces important in lifting Effective force: F = ma
F = mg + mavert if avert = 0, F = mg
This is a squat from an standing position. When does the descent phase change
to the ascent phase?
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Law of Inertia and Inertial Forces
Inertial Forces important in lifting Effective force: F = ma
F = mg + mavert if avert = 0, F = mg
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Law of Inertia and Inertial Forces
Inertial Forces important in lifting Effective force: F = ma
F = mg + mavert if avert = 0, F = mg
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Ground Reaction Force & Inertia
Inertial Forces important in locomotion (weight ~ 870N)
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Law of Inertia and Inertial ForcesFree Body Diagram of the Foot in Locomotion (no
torques indicated)
∑F = m a
V GRF + A JRF + mg = ma
A JRF = ma - V GRF - mg
Vertical GRF
Ankle JRF
Foot mg
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Law of Inertia and Inertial Forces
Vertical joint reaction forces in walking and running (weight = 780 N)
Note decrease in magnitude as move proximal on the leg.
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Law of Inertia and Inertial Forces
Vertical Joint reaction forces and inertial components in running (weight = 950 N)
Foot inconsequential but trunk is significant
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Inertial Forces in Jumping (like lifting)
One Subject - Low and High Jumps
0
200
400
600
800
1000
1200
1400
1600
1800
0 100 200 300 400 500 600 700
Time (ms)
Fo
rce
(N)
Low Jump
High Jump
BW
This is GRFvertical
1. What type of jump?
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Inertial Forces in Jumping (like lifting)
One Subject - Low and High Jumps
0
200
400
600
800
1000
1200
1400
1600
1800
0 100 200 300 400 500 600 700
Time (ms)
Fo
rce
(N)
Low Jump
High Jump
BW
1. Where is max knee flexion? 2. Where is peak Velocity Vert?
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Newton’s Laws of Motion
Ib. Law of Angular Inertia – An object will remain stationary or rotate with constant angular velocity until an unbalanced external torque is applied to it
Rotational Inertia – resistance
Rotational Inertia = Moment of Inertia = I = mr2
Long, massive objects are hard to rotate – Tight rope walker phenomenon
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Moment of Inertia
Most important application of Moment of Inertia:
Moment of Inertia for individual body segments – the amount of resistance to a change in rotation within each segment.
Affects the rotational motion caused by muscle torques.
Larger people – more mass and longer segments – have larger segment I values (7’ Basketballers)
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Segmental Moment of Inertia
Numerous techniques to calculate Segmental I values
We use Segmental mass as % Body mass from Dempster (1955) and Hanavan model (1964) to predict location of segment center of mass (~43% of total length from proximal end) and segment I values
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Segmental Moment of Inertia
Dempster’s Data: standard set of anthropometric data including segment masses, location of segment center of mass, segment moments of inertia
We use only segment mass as % body mass:
Thigh = 10% Leg = 4.7% Foot = 1.5%
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Segmental Moment of Inertia
Hanavan model – models segments as frustrums of cones (cones with tips cut off)
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Table 2.1 Regression Equations Estimating Body Segment Weights and Locations of the Center of Mass
Proximal end Segment Weight (N) CM location (%) of segment
Head 0.032 Fw + 18.70 66.3 Vertex
Trunk 0.532 Fw – 6.93 52.2 C1
Upper arm 0.022 Fw + 4.76 50.7 Shoulder joint
Forearm 0.013 Fw + 2.41 41.7 Elbow joint
Hand 0.005 Fw + 0.75 51.5 Wrist joint
Thigh 0.127 Fw – 14.82 39.8 Hip joint
Shank 0.044 Fw – 1.75 41.3 Knee joint
Foot 0.009 Fw + 2.48 40.0 Heel
Note. Body segment weights are estimated from total-body weight (Fw), and the segmental center-of-mass (CM)
locations are expressed as a percentage of segment length as measured from the proximal end of the segment.
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Table 2.5 Segment Length, Mass, and Center-of-Mass (CM) Location for Young Adult Women (W) and Men (M)
Segment W M W M W M
Head 20.02 20.33 6.68 6.94 58.94 59.76
Trunk 52.93 53.19 42.57 43.46 41.51 44.86
Upper torso 14.25 17.07 15.45 15.96 20.77 29.99
Middle torso 20.53 21.55 14.65 16.33 45.12 45.02
Lower torso 18.15 14.57 12.47 11.17 49.20 61.15
Upper arm 27.51 28.17 2.55 2.71 57.54 57.72
Forearm 26.43 26.89 1.38 1.62 45.59 45.74
Hand 7.80 8.62 0.56 0.61 74.74 79.00
Thigh 36.85 42.22 14.78 14.16 36.12 40.95
Shank 42.23 43.40 4.81 4.33 44.16 44.59
Foot 22.83 25.81 1.29 1.37 40.14 44.15
Length (cm) Mass (%) CM Location (%)
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Segment W M W M W M
Head 0.0213 0.0296 0.0180 0.0266 0.0167 0.0204
Trunk 0.8484 1.0809 0.9409 1.2302 0.2159 0.3275
Upper torso 0.0489 0.0700 0.1080 0.1740 0.1001 0.1475
Middle torso 0.0479 0.0812 0.0717 0.1286 0.0658 0.1212
Lower torso 0.0411 0.0525 0.0477 0.0654 0.0501 0.0596
Upper arm 0.0081 0.0114 0.0092 0.0128 0.0026 0.0039
Forearm 0.0039 0.0060 0.0040 0.0065 0.0005 0.0022
Hand 0.0004 0.0009 0.0006 0.0013 0.0002 0.0005
Thigh 0.1646 0.1995 0.1692 0.1995 0.0326 0.0409
Shank 0.0397 0.0369 0.0409 0.0387 0.0048 0.0063
Foot 0.0032 0.0040 0.0037 0.0044 0.0008 0.0010
Somersault Cartwheel Twist
Table 2.6 Segmental Moments of Inertia (kg•m2) for Young Adult Women (W) and Men (M) About the Somersault, Cartwheel, and Twist Axes (Frontal, Sagittal, Vertical)
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Law of Inertia and Inertial ForcesFree Body Diagram of the Foot in Locomotion
∑F = m a
V GRF + A JRF + mg = ma
A JRF = ma - V GRF - mg Vertical GRF
Ankle JRF
Foot mg
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Rotational Inertial Torques
T = I ( is angular acceleration of body)
Torque around foot segment in running composed of ankle muscle torque, GRF torque, inertial torque
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Rotational Inertial Torques
Knee angular position and Hip & Knee Torques in Swing phase of Running
Inertial torque (hip on thigh) and applied torque (knee muscles on shank)
+ Torque = flexion- Torque = Extension
Note: hip torque opposite to knee torque.Hip initially flexor to accelerate thigh forward.
What does this do to shank?Note direction of Knee Torque. What
role?What type of muscle activity?
Same in latter half of swing phase.
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Newton’s Laws of Motion
IIIa. Law of Linear Reaction – When one object applies a force on a second object, the second object applies an equal and opposite force onto the first object
“equal and opposite” – equal magnitude and opposite direction
Basis for force platform measurements
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Force Platforms and the Law of Reaction
Platform measures the reaction forces and torques to the forces and torques applied by the person
Forces – reactions to human forces – 3 dimensional
Torques – torques or “free moments” around the center of plate (My, Mx) or around the center of pressure (Mz)
- My, Mx used only to help identify the position of the center of pressure (no biological
information)
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Force Plate in Balance and Falling
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Force Platforms Conventions
Walking direction
Vertical force (Fz and Mz)
Mediolateral force (Fy and My)
Anteroposterior force (Fx and
Mx)
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Force Platform Calibration
Must Calibrate voltage to force and torque:
Sensitivity matrices from manufacturer (theoretical)
and
Applied known forces (experimental)
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Sensitivity Matrices
1. take values from main diagonal of sensitivity matrix (english)
units are ((microvolts/volt) / lb ) or ((microvolts / volt) / ft-lb)
2. multiply sensitivity values by 4000 (amplifier gain)
3. multiply result by 10 (excitation voltage)
4. resultant product in units of microvolt/lb or microvolt/ft-lb
5. convert to SI system and N/Volt or Nm/Volt by:
N/V = ((microvolt / lb) / (1 lb) * (1 lb) / (4.448 N) * (1 volt / 1000000 microvolt))-1
Nm/V = ((microvolt / ft-lb) / (1 ft-lb) * (1 ft-lb) / (0.3048 m * 4.448 N) * (1 volt / 1000000 microvolts))-1
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Sensitivity Matrices
Small Platform (N/v or Nm/v)
Fz=285.8612 Mz=14.68534
Fy=72.77487 My=30.78452
Fx=72.67974 Mx= 30.81251
NEWTONS VOLTAGE ABSOLUTE VOLTAGE N/V(mV) (mV)
0 -5 5 0205.6 -722 722 284.8406.1 -1421 1421 285.8607.0 -2120 2120 286.3807.6 -2818 2818 286.6
1006.5 -3521 3521 285.91200.4 -4190 4190 286.51391.1 -4854 4854 286.61591.8 -5557 5557 286.41783.9 -6236 6236 286.11984.4 -6944 6944 285.82183.9 -7652 7652 285.42381.7 -8335 8335 285.72579.8 -9029 9029 285.72781.9 -9737 9737 285.72980.7 -10439 10439 285.53177.9 -11123 11123 285.75179.7 -18101 18101 286.2
MEAN (N/V) = 285.9
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Empirical Calibration for Voltage to Force
Calibrate vertical force with known weights – straight line is desired result
Must reach force values typically measured
FIG 3. FORCE PLATE CALIBRATION
y = 3.5013x - 4.6142
R2 = 1
0
2000
4000
6000
8000
10000
12000
0 1000 2000 3000 4000
Force (N)
Vo
lt (
mV
)
Large Force Platform Calibration
y = 3.4602x + 0.0341
R2 = 1
0
1000
2000
3000
4000
5000
6000
7000
8000
0 500 1000 1500 2000 2500
Force (N)
Vo
lta
ge
(m
V)
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Center of Pressure
A single point at which the applied GRF will produce the same linear and angular effects on the object
• Force is really applied under the entire object, CoP allows for pin-point application of the force vector• needed for inverse dynamic analysis
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Center of Pressure in Running
CoP from Cavanagh, 1980
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Center of Pressure in Walking
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Accuracy of Center of Pressure
CoP – known distance from plate center to point under the foot
Med
Lat
Ant Post
5 2
6 1 3
7 4
PointAnt/Post Med/Lat Ant/Post Med/Lat Ant/Post Med/Lat
1 -0.050 -0.100 0.107 -0.426 0.157 0.3262 10.550 20.000 11.154 20.448 0.604 0.4483 9.950 0.100 10.328 0.390 0.378 0.2904 9.750 -20.000 10.058 -19.888 0.308 0.1125 -9.450 19.900 -9.171 20.363 0.279 0.4636 -10.250 -0.100 -9.976 0.299 0.274 0.3997 -10.450 -20.000 -10.230 -19.717 0.220 0.283
MEAN: 0.317 0.332SD: 0.144 0.121
Actual Locations Observed Error
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Error in Center of Pressure
Errors of 1 cm cause about 10% error in joint torques
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Calculate Center of Pressure
CoP – known distance from plate center to point under the foot
Digitize the foot and the plate edge
Calculate location of plate center from edge (e.g. large plate is 0.305 m edge to center)
Calculate location of CoP under foot
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Calculate Center of Pressure
CoP calculation – results are the distance between the exact center of plate and the CoP location
My
Fz
Fxdz = .02 m
dx – distance from center of plate
My = Fz(dx) + Fx(dz)
dx = (My – Fx(dz)) / Fz
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Center of Pressure in Locomotion
Data for
walking (solid),
stair ascent (dash),
stair descent (dots)
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Center of Pressure vs. Center of Mass
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Newton’s Laws of Motion
IIIb. Law of Angular Reaction – When one object applies a torque on a second object, the second object applies an equal and opposite torque onto the first object
“equal and opposite” – equal magnitude and opposite direction
Evident in joint or muscle torques
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Law of Angular Reaction
Spring system has equal and opposite torques on levers which would rotate in opposite directions
Equivalent to skeletal joint with muscle torqueWhy does only forearm
rotate then in biceps curl?
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Law of Angular Reaction – Inverse Dynamics & Muscle Torques
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Newton’s Laws of Motion
IIa. Law of Linear Acceleration – a force will accelerate an object in the direction of the force, at a rate inversely proportional to the mass of the object
F = m a
The basis for all biomechanics – forces cause motion
Force – a pushing or pulling effect on an object
Compression vs. tension vs. shear
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Force
Force is a vector
Force measured in Newtons: 1 N = 1 kg m/s2
1 N = 0.225 Lbs. or 1 Lb. = 4.448 N
Force resolution vs. force composition
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Force Resolution
Resolution of force into components:
Hor. and vert. forces in Laboratory reference frame
Stabilizing and rotational in anatomic reference frame
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Force Resolution
Laboratory reference frame for general movement:
High jump
vs
Long jump
4,000 N at 60 or 20 to the horizontal
Force hor = 4,000 N cos Force ver = 4,000 N sin
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Force Resolution
Anatomical reference frame for muscle forces:
= 60°Forearm
Arm Biceps force = 4,000 N
Rotating
StabilizingCalculate stabilizing and rotating components not horizontal and vertical
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Skeletal-Muscle Models & Force Resolution
Skeletal-muscle models used to calculate joint shear, compressive, forces and torque loads for each
muscle force vector.
Position of bony segments and joint centers, lines of muscle force
vectorsGlitsch & Bauman, 1998 Pandy & Shelburne,
1998
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Muscle Forces From Muscle Model
Muscle forces calculated through muscle models then combined for joint loads.
Glitsch & Bauman, 1998
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Muscle Forces to Joint Loads
Resultant joint forces during walking and running from muscle forces and skeletal-muscle model
Glitsch & Bauman, 1998
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Force Composition
Combination of forces into resultant force:
e.g. calculation of total muscle force vectors from component muscles
- calculate the shear force across a joint from each muscle then combine the vectors
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Force Composition in Shoulder and Elbow Muscle Groups
Purpose – predict total muscle force from component vectors
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Force Composition for Shoulder Muscles
Fmc = 2,000 N at 120°
Fms = 2,500 N at 70°
Fres.
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Law of Linear Acceleration & and Linear Impulse-Momentum
Law of Acceleration describes change in momentum of the object, a change in the quantity of motion.
F = m * a
Positive acceleration – increase quantity of motion
Negative acceleration – decrease quantity of motion
Really?
Quantity of motion = Momentum = mass * velocity in kg*m / s
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Law of Linear Acceleration & and Linear Impulse-Momentum
Law of Acceleration restated:
F = m * a
F = m * (vf – vi)/time
F * time = m * (vf – vi) : impulse-momentum equation
F * time = impulse = area under force-time curve = total effect of the accumulated or applied force; measured in Ns = kgm/s2 * s = kgm/s
Impulse Changes Momentum
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Horizontal Impulse in Running
Braking impulse reduces horizontal momentum (i.e. velocity) – impulse & momentum in opposite directions.
Propelling impulse increases horizontal momentum (i.e. velocity) – impulse & momentum in same direction.
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Horizontal Impulse in Running
Runner’s mass = 70 kg Vi = 4.00 m/s
Braking imp. = -18 Ns
Propelling imp = 20 Ns
Velocity at midstance and at toe off?
Calculate for next class-400
-300
-200
-100
0
100
200
300
400
0 0.05 0.1 0.15 0.2
Time (s)
Forc
e (N
)
Imp= -150 N *0.12 s
Imp = 160 N * 0.12 s
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Vertical Impulse in Running
Vertical impulse changes vertical momentum
Initial momentum – down
Final momentum – up
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Vertical Impulse in Jumping
Bodyweight is critical reference
Assess impulse from BW
Calc. velocity at 3 points
mass = 65.7 kg0
200
400
600
800
1000
1200
1400
1600
0 200 400 600 800
Time (ms)
Forc
e (N
)
-59 Ns
59 Ns 196 Ns
V = 0.00 m/s V = -0.91 m/s V = 0.00 m/s V = 3.02 m/s
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Vertical Impulse in Jumping
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Control of Body Momentum
Tan –1 = Vv/Vh
= 25°
= 18°
Vv
Vh
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Comment on Conservation of Momentum
Momentum is conserved in a closed system.
When is the human system closed?
When is a part (e.g. lower extremity) of the human system closed?
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Newton’s Laws of Motion
IIb. Law of Angular Acceleration – a torque will accelerate an object in the direction of the torque, at a rate inversely proportional to the moment of inertia of the object
T = I
Torque – the rotational effect of a force applied at a distance to an axis
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Two Equations for Torque
T = I T = F d
I = mr2
F
d
=
Kinematic – Kinetic Equivalents
I = F d
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Two Calculation Techniques
1) What is the lever arm dist? Biceps attached 3 cm from elbow joint.
= 60°
Forearm
Arm
Biceps force = 4,000 N
0.03 m
Sin 60° = d1/0.03 d1=0.026
T = 4000 N (0.026 m)
= 104 Nm
d1
4,000 N
Use length triangle
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Two Calculation Techniques
2) What is the amount of force perpendicular to lever?
= 60°
Forearm
Arm
Biceps force = 4,000 N
0.03 m
Cos 30° = d1/4000 d1=3464 N
T = 3464 N (0.03 m)
= 104 Nm
d1
4,000 N
Use force triangle
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Law of Angular Acceleration & and Angular Impulse-Momentum
Law of Angular Acceleration restated:
T = I * T = I * (f – i)/time
T * time = I * (f – i) - angular impulse-momentum equation
T * time = angular impulse = area under torque-time curve = total effect of the accumulated or applied torque; measured in Nms = kgm/s2 * m *s = kgm2/s
Angular Impulse Changes Angular Momentum
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Angular Impulse in Movement Analyses
Use area under torque-time curve to assess the total effect of a joint torque
Area sensitive to magnitude and temporal changes
Calculate by one of several methods:
Area = (Point value * Sample rate)
Area = (Point values) * Sample rate
Area =Avg torque in area * total time
0.26 0.23 * 0.17 Nms/kg
0.13 0.14 * 0.33 Nms/kg
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Comment on Conservation of Angular Momentum
Angular momentum is conserved in a closed system.
When is the human system closed?In Diving, vaulting, and figure skating spinning!
The rotating figure skater rotates faster with arms tucked.
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Newton’s Laws of Motion - Summary
Three laws describing linear and angular kinetics
Second law, the law of acceleration, is the basis for most Biomechanics
All six laws apply to all biomechanical situations, but each situation may best be analyzed with a subset of the six laws
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Energy, Work, and Power
An alternative analysis to the dynamic analysis of F=ma for understanding the mechanics of physical systems
Provides insight into motion in terms of a combination of kinematics (displacement) and kinetics (force)
Provides insight into muscle mechanics in terms of contraction types, roles of muscles, sources of movement
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Energy
Energy has many forms – chemical, nuclear, electrical, mechanical, and more
Energy is often transformed from one form to another:Electricity is used to spin CDs
Chemical energy in ATP is used to produce the “power stroke” and slide actin over myosin
Energy is a scalar variable that reflects the “energetic state” of the object
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Energy
Mechanical energy is the capacity to do work and work is the product of force and displacement
Work = Force * Displacement
Mechanical energy is the capacity to move objects
Energy = Zero or positive value (a scalar), Joules = J
1 J is very small – move fingers a few centimeters?
133 J lifts 150 lb (666 N) person up one step (20 cm)
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Forms of Mechanical Energy
Three basic forms of mechanical energy
Potential – position
Kinetic – velocity
Strain - elastic stretch
(or two forms with PE gravitational & strain)
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Potential Energy (or Gravitational Potential
Energy)
Potential Energy = energy of position = energy associated with the weight of an object and its height above the floor
P.E. = mgh in kgm2 / s2 = J
Runner’s body has some P.E.: P.E. = 50 kg (9.81 m/s2) (1 m) = 490 J
Vaulter has more P.E. P.E. = 80 kg (9.81 m/s2) (3 m) = 2,354 J
1 m
3 m
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Potential Energy and Work
How does Potential Energy have the capacity to do work?
Hold a bowling ball 1 m above floor P.E. = 71 kg (9.81 m/s2) (1 m) = 698 J
Drop the ball on your foot.
Did your foot move by the force applied from the bowling ball?
1 m
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Potential Energy and Work
The potential to do work from the Potential Energy is simply held in check by a supporting force onto the object.
The potential to do work inherent within P.E. is a function of the weight of the object and its velocity at impact
(No P.E. in zero gravity)
1 m
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Linear Kinetic Energy
Kinetic Energy = energy of motion = energy associated with the mass and velocity of an object
Linear K.E. = ½ mv2 in kgm2 / s2 = J
Jumper’s body has Linear K.E.: K.E. = ½ (65 kg) (7.4 m/s)2 = 1,780 J
Related to linear momentum = mv
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Kinetic Energy and Work
How does Kinetic Energy have the capacity to do work?
Step in front of the jumper and find out.
The large kinetic energy in her body will cause you to move.
The large kinetic energy in her body will enable her to exert force on you which will cause you to move.
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Rotational Kinetic Energy
Angular position and velocity of body segments during running
Kinetic Energy = energy of motion = energy associated with the moment of inertia & angular velocity of an object
Rotational K.E. = ½ I2 in kgm2 / s2 = J
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Three Components of Energy in Running
Rotational K.E. = ½ I2 Peak values during running: I K.E.
(kgm2) (rad) (J)
Trunk1.09 3.5 6.7 Thigh0.12 8.2 4.0 Leg
0.04 10.2 2.1 Foot0.00 12.0 0.1
Rotational K.E. is not evident on this scale
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Rotational Kinetic Energy and Work
How does Rotational Kinetic Energy have the capacity to do work?
As in linear kinetic energy, the rotational motion can provide the means to apply force on an object.
In most human movement, this “means” is not large and is sometimes completely negligible – it can do only a small amount of work.
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Strain Energy (or Spring Potential Energy)
Energy due to deformation of a spring
Strain Energy = ½ k (x)2 in kgm2 / s2 = J
k = stiffness coefficient – resistance to stretch
x = length of stretch
Spring Force = k (x)
Therefore strain energy related to force and
work
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Total Mechanical Energy
Total work potential in an object – the “energetic state”
Total Energy = P.E. + Linear K.E. + Rotational K.E. = mgh + ½ mv2 + ½ I2
Total
ThighLeg
Trunk
Foot
Segment energies during one cycle of running
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Inability of Gravity to Change Energy
Total Energy = P.E. + K.E. = mgh + ½ mv2 + ½ I2
Constant during flight phases
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Work – Changing Energy
Work represents the change in energy of an object
Work occurs when energy changes
Work occurs when objects are raised or lowered (change in P.E.) or when their velocity changes (change in K.E.)
Work = Total Energy = (mgh + ½ mv2 + ½ I2) in kgm2 / s2 =
J
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Work – Changing Energy
Work = Total Energy = (mgh + ½ mv2 + ½ I2)
= (mghf – mghi) + (½ mv2f –
½ mv2i )
= (61*9.81*1.4 - 61*9.81*1.1) + (0.5*65*3.022 – 0)
= (838 J – 658 J) + (296 J)
= 476 J Energy was increased
Jumper’s mass = 61 kg
CM height = 1.1 m at start & 1.4 m at take off
0 J
0
200
400
600
800
1000
1200
1400
1600
0 200 400 600 800
Time (ms)
Forc
e (N
)
-59 Ns
59 Ns 196 Ns
V = 0.00 m/s V = -0.91 m/s V = 0.00 m/s V = 3.02 m/s
![Page 95: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/95.jpg)
Work – Changing Energy
Work = Total Energy = (mgh + ½ mv2 + ½ I2)
= (mghf – mghi) + (½ mv2f – ½ mv2
i
)
= (65*9.81*1.2 - 65*9.81*1.0) +
(0.5*65*6.982 - 0.5*65*7.42)
= (765J – 638J) + (1583J – 1780J)
= - 70 J Energy was reduced
1 m
1.2 mVf= 6.98 m/s
0 J
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Work – Product of Force & Displacement
Work is performed when a force moves an object
Work = force * displacement in kgm2 / s2 = J
W=Fdcos – calculates the product of the Displacement and the portion of Force in same direction as displ.
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WORK DOES NOT EQUAL TORQUE
Work: force and displacement are parallel to each other
Torque: force and distance are perpendicular to each other
Force * Displacement = Force * Distance
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.
Torque = Force * distance = r x F = rF sin = 0.30 m (40 N) sin 90° = 12 Nm
(cross product – produces a vector) Work = Force * displacement
= d F = dF cos = 0.20 m (40N) cos 0° = 8 Nm = 8 J
(dot product – produces a scalar)
0.30 m
0.20 m
40 N40 N
Work vs. Torque
![Page 99: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/99.jpg)
0.30 m
0.20 m
40 N40 N
distance is a length (static) – a torque is exerted in this position
displacement is a movement (dynamic) – work was performed by lifting
Work vs. Torque
![Page 100: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/100.jpg)
Work – Energy Theorem
Work changes Energy
Work = (mgh + ½ mv2 + ½ I2)
Force * displacement = (mgh + ½ mv2 + ½ I2)
Total system is lifted 0.5 m
2000N*0.5m=(mgh+½ mv2+½I2)
2000 N * 0.5m = 2000 N * h
1000 J = 1000 J added to system
0 J 0 J
![Page 101: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/101.jpg)
Work – Energy Theorem
Work changes Energy
Work = (mgh + ½ mv2 + ½ I2)
Force * displacement = (mgh + ½ mv2 + ½ I2)
Did this force do work?
Did the energy of the box change?
![Page 102: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/102.jpg)
Work By Simultaneous Forces
Double Support Phase in Walking – GRFs under trail limb do positive work (c, toe off force and v+ in “same” direction), under lead limb do negative work (c, heel strike force and v- in “opposite” directions).
Donelan et al. 2002
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Work By Simultaneous Forces
Individual and total work done by each limb.
During double support: Trail leg does positive work. Lead leg does negative work.
Total limb has balance of positive and negative.
Donelan et al. 2002
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Avg lever arm = 0.25 m
Avg Muscle torque = 10 Nm
12 Nm
40 N
While torque is not work, it can do work: Work = Torque *
= angular displacement = 0.78 rad
Work = 10 Nm * 0.80 rad = 8.0 J
(check with linear calculation:
Work=mgh: 40 N(hf) – 40 N(hi)= 8.0 J
hf – hi = 0.20 m)
Work Done By A Torque
![Page 105: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/105.jpg)
Joint torques during stair ascent
Old adults have larger hip torque and this torque performed more work: 0.41 vs. 0.24 J / kg
Young adults have larger knee torque and this torque performed more work: 0.81 vs. 0.56 J / kg
Work Done By Joint Torques
![Page 106: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/106.jpg)
Power – Rate of Work (or Rate of Changing Energy)
Power represents the rate at which work is being done.
Work occurs when energy changes and it occurs at various rates – i.e. fast or slow, high or low
The power used in lifting depends on how fast or slowly the lift occurred.
![Page 107: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/107.jpg)
Power – Rate of Work or Rate of Changing Energy)
P = Work / time = Force * displ. / time = Force * velocity
in kgm / s2 * m / s = kgm2 / s3 = Watts (W)
P = Work / time = Torque * / time = Torque *
in kgm / s2 * m * rad / s = kgm2 / s3 = Watts (W)
![Page 108: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/108.jpg)
0.20 m
40 N
Work = Force * displacement = 0.20 m (40N)
= 8 Nm = 8 J
Lift in 0.5 s: P = Work/time = 16 WLift in 1.0 s: P = Work / time = 8 WLift in 2.0 s: P = Work / time = 4 W
Power During Lifting
![Page 109: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/109.jpg)
Avg lever arm = 0.25 m
Avg Muscle torque = 10 Nm
12 Nm
40 N
Work = Torque *
= 10 Nm * 0.80 rad = 8.0 J
Lift in 0.5 s: P = Work/time = 16 WLift in 1.0 s: P = Work / time = 8 WLift in 2.0 s: P = Work / time = 4 W
Power During Lifting
![Page 110: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/110.jpg)
Elbow joint angular velocity, torque and power
Power = Torque *
Positive power – concentric contraction, positive work, increase energy
Negative power – eccentric contraction, negative work, decrease energy
Joint Power Produced By Joint Torques
![Page 111: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/111.jpg)
Calculate work from power curve:
Work is area under the power curve or a portion of the curve.
Power = Watts = T/s = Nm/s = kgm2/s2 / s = kgm2/s3 * s (for area) = kgm/s2 * m = force * distance = WORK
Joint Power Produced By Joint Torques
![Page 112: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/112.jpg)
Knee power, torque, and angular velocity during stance phase of running.
Knee flexes during brief flexor torque then longer extensor torque – low positive power & work then large negative power & work
Knee extends during long extensor torque then shorter flexor torque – large positive power & work then low negative power & work
Joint Power Produced By Joint Torques
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Knee power, torque, and angular velocity during stance phase of running.
Peak torque at zero velocity – at maximum knee flexion, maximum quadriceps stretch – muscle force maximized early in movement.
Peak power at mid levels of torque and velocity – both torque and velocity contribute to power – muscle work maximized in middle of movements.
Joint Power Produced By Joint Torques
![Page 114: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/114.jpg)
Knee power & torque in stair ascent.
Positive powers dominate by concentric contractions.
Torque and velocity in same direction.
Joint Power Produced By Joint Torques
![Page 115: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/115.jpg)
Knee power & torque in stair descent.
Negative powers dominate by eccentric contractions.
Torque and velocity in opposite directions.
Joint Power Produced By Joint Torques
![Page 116: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/116.jpg)
Positive work equal between groups in ascent.
Work Done By Joint Torques
0.00
0.50
1.00
1.50
2.00
Total Hip Knee Ankle
Wo
rk (
J/kg
)
Old
Young
**
* P < .05
*
-1.50
-1.25
-1.00
-0.75
-0.50
-0.25
0.00
Total Hip Knee Ankle
Wo
rk (
J/k
g)
Old
Young
*
* P < .05*
Negative work not equal between groups in descent.
![Page 117: Movement Forces Figure reprinted from Marey, 1889](https://reader033.vdocument.in/reader033/viewer/2022061501/56649ea05503460f94ba27fc/html5/thumbnails/117.jpg)
Joint torques and powers and muscle activity