movement forces figure reprinted from marey, 1889

MovementForces

MovementForces

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?


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


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.

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


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?

Ground Reaction Force & Inertia

Inertial Forces important in locomotion (weight ~ 870N)

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


Vertical joint reaction forces in walking and running (weight = 780 N)

Note decrease in magnitude as move proximal on the leg.


Vertical Joint reaction forces and inertial components in running (weight = 950 N)

Foot inconsequential but trunk is significant

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?

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?


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


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%


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

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.

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 (%)

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)

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

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.


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

Walking direction

Vertical force (Fz and Mz)

Mediolateral force (Fy and My)

Anteroposterior force (Fx and

Mx)

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

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

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)

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

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

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

My

Fz

Fxdz = .02 m

dx – distance from center of plate

My = Fz(dx) + Fx(dz)

dx = (My – Fx(dz)) / Fz

Center of Pressure in Locomotion

Data for

walking (solid),

stair ascent (dash),

stair descent (dots)

Center of Pressure vs. Center of Mass


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

Equivalent to skeletal joint with muscle torqueWhy does only forearm

rotate then in biceps curl?

Law of Angular Reaction – Inverse Dynamics & Muscle Torques


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

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:

= 60°Forearm

Arm Biceps force = 4,000 N

Rotating

StabilizingCalculate 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

vectorsGlitsch & 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

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

Vertical Impulse in Running

Vertical impulse changes vertical momentum

Initial momentum – down

Final momentum – up

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

Vertical Impulse in Jumping

Control of Body Momentum

Tan –1 = Vv/Vh

= 25°

= 18°

Vv

Vh

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?


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 = F d

I = mr2

F

d

=

Kinematic – Kinetic Equivalents

I = F d

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

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

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

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

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)

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

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

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

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

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

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

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 I2

Total

ThighLeg

Trunk

Foot

Segment energies during one cycle of running

Inability of Gravity to Change Energy

Total Energy = P.E. + K.E. = mgh + ½ mv2 + ½ I2

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 I2) in kgm2 / s2 =

J


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


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

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.

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

.

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

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

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

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?

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.

Total limb has balance of positive and negative.

Donelan et al. 2002

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

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

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)

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

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

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


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


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.


Knee power & torque in stair ascent.

Positive powers dominate by concentric contractions.

Torque and velocity in same direction.


Knee power & torque in stair descent.

Negative powers dominate by eccentric contractions.

Torque and velocity in opposite directions.


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.

Joint torques and powers and muscle activity

movement forces figure reprinted from marey, 1889

Documents

law of linear inertia

liftinglaw of inertia

nlaw of inertia

effective force

significantinertial

upward force

force vector

force datasubject