Download Aging and Stiffness

Movement
Forces
Figure reprinted from Marey, 1889.
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
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)
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?
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
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?
Trunk accelerates
Rearward force
applied to seat from
wheels through
chassis
forward relative to
thighs and car. The
more you JAM on the
breaks, the greater the
acceleration.
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
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
Law of Inertia and Inertial Forces
Inertial Forces important in lifting
Effective force: F = ma
F = mg + mavert if avert = 0, F = mg
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?
Law of Inertia and Inertial Forces
Inertial Forces important in lifting
Effective force: F = ma
F = mg + mavert if avert = 0, F = mg
Law of Inertia and Inertial Forces
Inertial Forces important in lifting
Effective force: F = ma
F = mg + mavert if avert = 0, F = mg
Ground Reaction Force & Inertia
Inertial Forces important in locomotion (weight ~ 870N)
Law of Inertia and Inertial Forces
Free Body Diagram of the Foot in Locomotion (no
torques indicated)
Ankle JRF
∑F = m a
V GRF + A JRF + mg = ma
Foot mg
A JRF = ma - V GRF - mg
Vertical GRF
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.
Law of Inertia and Inertial Forces
Vertical Joint reaction
forces and inertial
components in running
(weight = 950 N)
Foot inconsequential but
trunk is significant
Inertial Forces in Jumping
(like lifting)
This is GRFvertical
One Subject - Low and High Jumps
1. What type of jump?
1800
Force (N)
1600
1400
Low Jump
1200
High Jump
1000
800
600
BW
400
200
0
0
100
200
300
400
Time (ms)
500
600
700
Inertial Forces in Jumping
1.
(like lifting)
Where is max knee flexion?
One Subject - Low and High Jumps 2. Where is peak Velocity Vert?
1800
Force (N)
1600
1400
Low Jump
1200
High Jump
1000
800
600
BW
400
200
0
0
100
200
300
400
Time (ms)
500
600
700
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
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)
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
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%
Segmental Moment of Inertia
Hanavan model –
models
segments as
frustrums of
cones (cones
with tips cut
off)
Table 2.1 Regression Equations Estimating Body Segment
Weights and Locations of the Center of Mass
CM location (%)
Proximal end
of segment
Segment
Weight (N)
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.
Table 2.5 Segment Length, Mass, and Center-of-Mass (CM)
Location for Young Adult Women (W) and Men (M)
Length (cm)
Mass (%)
Segment
W
M
W
Head
20.02
20.33
6.68
Trunk
52.93
53.19
Upper torso
14.25
Middle torso
M
CM Location (%)
W
M
6.94
58.94
59.76
42.57
43.46
41.51
44.86
17.07
15.45
15.96
20.77
29.99
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
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)
Segment
Somersault
W
M
Cartwheel
W
M
Twist
M
W
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
Law of Inertia and Inertial Forces
Free Body Diagram of the Foot in Locomotion
∑F = m a
Ankle JRF
V GRF + A JRF + mg = ma
Foot mg
A JRF = ma - V GRF - mg
Vertical GRF
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
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.
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
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)
Force Plate in Balance and Falling
Force Platforms Conventions
Vertical force (Fz and Mz)
Anteroposterior
force (Fx and
Mx)
Mediolateral force (Fy and My)
Walking direction
Force Platform Calibration
Must Calibrate voltage to force and torque:
Sensitivity matrices from manufacturer (theoretical)
and
Applied known forces (experimental)
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
Sensitivity Matrices
NEWTONS
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
0
205.6
406.1
607.0
807.6
1006.5
1200.4
1391.1
1591.8
1783.9
1984.4
2183.9
2381.7
2579.8
2781.9
2980.7
3177.9
5179.7
VOLTAGE
(mV)
-5
-722
-1421
-2120
-2818
-3521
-4190
-4854
-5557
-6236
-6944
-7652
-8335
-9029
-9737
-10439
-11123
-18101
MEAN (N/V) =
ABSOLUTE VOLTAGE
(mV)
5
722
1421
2120
2818
3521
4190
4854
5557
6236
6944
7652
8335
9029
9737
10439
11123
18101
N/V
0
284.8
285.8
286.3
286.6
285.9
286.5
286.6
286.4
286.1
285.8
285.4
285.7
285.7
285.7
285.5
285.7
286.2
285.9
Empirical Calibration for Voltage to Force
FIG 3. FORCE PLATE CALIBRATION
Calibrate vertical force with known
weights – straight line is desired
result
12000
8000
6000
Must reach force values typically
measured
y = 3.5013x - 4.6142
2
R =1
4000
2000
Large Force Platform Calibration
0
0
8000
1000 2000 3000 4000
Force (N)
y = 3.4602x + 0.0341
R2 = 1
7000
Voltage (mV)
Volt (mV)
10000
6000
5000
4000
3000
2000
1000
0
0
500
1000
1500
Force (N)
2000
2500
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
Center of Pressure in Running
CoP from Cavanagh,
1980
Center of Pressure in Walking
Accuracy of Center of Pressure
CoP – known distance from plate
center to point under the foot
Point
Med
5
6
7
2
1
3
4
Lat
Ant
Post
1
2
3
4
5
6
7
Actual Locations
Observed
Error
Ant/Post Med/Lat Ant/Post Med/Lat Ant/Post Med/Lat
-0.050
10.550
9.950
9.750
-9.450
-10.250
-10.450
-0.100
20.000
0.100
-20.000
19.900
-0.100
-20.000
0.107
11.154
10.328
10.058
-9.171
-9.976
-10.230
-0.426
20.448
0.390
-19.888
20.363
0.299
-19.717
MEAN:
SD:
0.157
0.604
0.378
0.308
0.279
0.274
0.220
0.326
0.448
0.290
0.112
0.463
0.399
0.283
0.317
0.144
0.332
0.121
Error in Center of Pressure
Errors of 1 cm cause
about 10% error in
joint torques
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
Calculate Center of Pressure
CoP calculation – results are the distance between
the exact center of plate and the CoP location
Fz
Fx
dz = .02 m
My = Fz(dx) + Fx(dz)
dx = (My – Fx(dz)) / Fz
My
dx – distance from
center of plate
Center of Pressure in Locomotion
Data for
walking (solid),
stair ascent (dash),
stair descent (dots)
Center of Pressure vs. Center of Mass
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
Law of Angular Reaction
Spring system has equal
and opposite torques
on levers which would
rotate in opposite
directions
Why does only forearm
rotate then in biceps
curl?
Equivalent to skeletal joint
with muscle torque
Law of Angular Reaction – Inverse
Dynamics & Muscle Torques
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=ma
The basis for all biomechanics – forces cause motion
Force – a pushing or pulling effect on an object
Compression vs. tension vs. shear
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
Force Resolution
Resolution of force into components:
Hor. and vert. forces in Laboratory reference frame
Stabilizing and rotational in anatomic reference frame
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 

Force Resolution
Anatomical reference frame for muscle forces:
Arm
Biceps force = 4,000 N
Rotating
Stabilizing
 = 60°
Forearm
Calculate stabilizing
and rotating
components not
horizontal and vertical
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 vectors
Glitsch & Bauman,
1998
Pandy & Shelburne,
1998
Muscle Forces From Muscle Model
Muscle forces
calculated
through muscle
models then
combined for
joint loads.
Glitsch & Bauman, 1998
Muscle Forces to Joint Loads
Resultant joint forces
during walking and
running from
muscle forces and
skeletal-muscle
model
Glitsch & Bauman, 1998
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
Force Composition in Shoulder and
Elbow Muscle Groups
Purpose – predict total muscle force
from component vectors
Force Composition for Shoulder Muscles
Fmc = 2,000 N at 120°
Fms = 2,500 N at 70°


Fres.
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
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
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.
Horizontal Impulse in Running
Runner’s mass = 70 kg
Vi = 4.00 m/s
Braking imp. = -18 Ns
Propelling imp = 20 Ns
400
300
Force (N)
200
Imp= -150 N *0.12 s
100
0
-100
-200
0
0.05
0.1
0.15
0.2
Imp = 160 N * 0.12 s
Velocity at midstance
and at toe off?
-300
-400
Time (s)
Calculate for next class
Vertical Impulse in Running
Vertical impulse changes vertical
momentum
Initial momentum – down
Final momentum – up
Vertical Impulse in Jumping
V = 0.00 m/s
V = -0.91 m/s
V = 0.00 m/s
V = 3.02 m/s
1600
1400
Force (N)
1200
Bodyweight is
critical reference
Assess impulse
from BW
1000
59 Ns
800
600
196 Ns
-59 Ns
400
200
0
0
200
400
Time (ms)
600
800
Calc. velocity at
3 points
mass = 65.7 kg
Vertical Impulse in Jumping
Control of Body Momentum
Tan –1 = Vv/Vh
 = 25°
 = 18°

Vh
Vv
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?
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
Two Equations for Torque
T=I
T=Fd

F
=
I = mr2
d
Kinematic – Kinetic Equivalents
I=Fd
Two Calculation Techniques
Arm
Biceps force =
4,000 N
1) What is the lever arm dist?
Biceps attached 3 cm
from elbow joint.
4,000 N
Sin 60° = d1/0.03
d1=0.026
 = 60°
d1
T = 4000 N (0.026 m)
Forearm
= 104 Nm
0.03 m
Use length triangle
Two Calculation Techniques
Arm
Biceps force =
4,000 N
2) What is the amount of
force perpendicular to
lever?
4,000 N
Cos 30° = d1/4000
d1=3464 N
 = 60°
T = 3464 N (0.03 m)
Forearm
= 104 Nm
d1
0.03 m
Use force triangle
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
Angular Impulse in Movement Analyses
0.26
0.23
* 0.17 Nms/kg
0.13
0.14
* 0.33 Nms/kg
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
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.
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
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
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
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)
Forms of Mechanical Energy
Three basic forms of
mechanical energy
Potential – position
Kinetic – velocity
Strain - elastic stretch
(or two forms with PE 
gravitational & strain)
Potential Energy (or Gravitational Potential
Energy)
1m
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
3m
Vaulter has more P.E.
P.E. = 80 kg (9.81 m/s2) (3 m) = 2,354 J
Potential Energy and Work
How does Potential Energy have the
capacity to do work?
1m
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?
Potential Energy and Work
1m
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)
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
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.
Rotational Kinetic Energy
Kinetic Energy = energy of
motion = energy associated
with the moment of inertia &
angular velocity of an object
Rotational K.E. = ½ I2
in kgm2 / s2 = J
Angular position and velocity of
body segments during running
Three Components of Energy in Running
Rotational K.E. = ½ I2
Peak values during running:
I
 K.E.
(kgm2) (rad)
Trunk
Thigh
Leg
Foot
1.09
0.12
0.04
0.00
3.5
8.2
10.2
12.0
(J)
6.7
4.0
2.1
0.1
Rotational K.E. is not evident
on this scale
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.
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
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 I2
Total
Trunk
Thigh
Leg
Foot
Segment
energies
during
one cycle
of
running
Inability of Gravity to Change Energy
Total Energy
= P.E. +
K.E.
= mgh +
½ mv2 +
½ I2
Constant
during
flight
phases
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 I2)
in kgm2 / s2 = J
Work – Changing Energy
V = 0.00 m/s
V = -0.91 m/s
V = 0.00 m/s
V = 3.02 m/s
1600
1400
Force (N)
1200
0J
Work = Total Energy
=  (mgh + ½ mv2 + ½ I2)
1000
59 Ns
800
600
196 Ns
= (mghf – mghi) + (½ mv2f –
½ mv2i )
-59 Ns
400
200
0
0
200
400
600
800
Time (ms)
Jumper’s mass = 61 kg
CM height = 1.1 m at start &
1.4 m at take off
= (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
Work – Changing Energy
1m
Work = Total Energy =  (mgh + ½ mv2 +
½ I2)
0J
= (mghf – mghi) + (½ mv2f – ½ mv2i )
= (65*9.81*1.2 - 65*9.81*1.0) +
(0.5*65*6.982 - 0.5*65*7.42)
= (765J – 638J) + (1583J – 1780J)
1.2 m
Vf= 6.98 m/s
= - 70 J
Energy was reduced
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.
Force * Displacement = Force * Distance
WORK DOES NOT EQUAL TORQUE
Work: force and displacement are parallel to each other
Torque: force and distance are perpendicular to each
other
Work vs. Torque
0.30 m
40 N
Torque = Force * distance
= r x F = rF sin 
= 0.30 m (40 N) sin 90°
0.20 m
= 12 Nm
(cross product – produces a vector)
Work = Force * displacement
40 N
= d F = dF cos 
= 0.20 m (40N) cos 0°
= 8 Nm = 8 J
(dot product – produces a scalar)
.
Work vs. Torque
distance is a length (static) – a
torque is exerted in this position
0.30 m
0.20 m
40 N 40 N
displacement is a movement
(dynamic) – work was performed
by lifting
Work – Energy Theorem
Work changes Energy
Work =  (mgh + ½ mv2 + ½ I2)
Force * displacement =  (mgh + ½ mv2 + ½ I2)
Total system is lifted 0.5 m
2000N*0.5m=(mgh+½
0J
0J
mv2+½I2)
2000 N * 0.5m = 2000 N * h
1000 J = 1000 J added to system
Work – Energy Theorem
Work changes Energy
Work =  (mgh + ½ mv2 + ½ I2)
Force * displacement =  (mgh + ½ mv2 + ½ I2)
Did this force do work?
Did the energy of the
box change?
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
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.
Donelan et al. 2002
Total limb has balance of
positive and negative.
Work Done By A Torque
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
12 Nm
40 N
Avg lever arm = 0.25 m
Avg Muscle torque = 10 Nm
(check with linear calculation:
Work=mgh: 40 N(hf) – 40 N(hi)= 8.0 J
hf – hi = 0.20 m)
Work Done By Joint Torques
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
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.
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)
Power During Lifting
Work = Force * displacement
= 0.20 m (40N)
= 8 Nm = 8 J
0.20 m
40 N
Lift in 0.5 s: P = Work/time = 16 W
Lift in 1.0 s: P = Work / time = 8 W
Lift in 2.0 s: P = Work / time = 4 W
Power During Lifting
Work = Torque * 
= 10 Nm * 0.80 rad = 8.0 J

12 Nm
40 N
Avg lever arm = 0.25 m
Avg Muscle torque = 10 Nm
Lift in 0.5 s: P = Work/time = 16 W
Lift in 1.0 s: P = Work / time = 8 W
Lift in 2.0 s: P = Work / time = 4 W
Joint Power Produced By Joint Torques
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
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
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
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
Knee power &
torque in stair
ascent.
Positive powers
dominate by
concentric
contractions.
Torque and
velocity in same
direction.
Joint Power Produced By Joint Torques
Knee power &
torque in stair
descent.
Negative powers
dominate by
eccentric
contractions.
Torque and
velocity in
opposite
directions.
Work Done By Joint Torques
2.00
* P < .05
Old
Young
1.00
*
Positive work equal between
groups in ascent.
*
*
0.50
Total
Hip
Knee
Ankle
0.00
0.00
Total
Hip
Knee
-0.25
Ankle
Work (J/kg)
Work (J/kg)
1.50
Negative work not equal
between groups in descent.
-0.50
-0.75
*
-1.00
-1.25
-1.50
*
* P < .05
Old
Young
Joint torques and
powers and muscle
activity