1

Linear KineticsmdashIntroduction

This week wersquoll discuss Newtonrsquos 3 laws of motion and the universal law of gravitation Related to each of the three laws of motion we will focus on the following concepts respectively and relate them to human movement (1) conservation of momentum (2) impulse-momentum relationship and (3) reaction forces

Introductory Video

Newtonrsquos 1st Law

Law of Inertia a body will remain in motion unless it is acted on by a net external force that does not equal zero Recall that inertia is quantified via mass and is independent of position (unlike weight)

Newtonrsquos 1st LawIn other words objects will remain in motion if ΣF = 0 (static equilibrium)

Recall that this does not require velocity to be zero Two relatively horizontal examples (BYU and Bountiful) and a relatively vertical example (a dumbbell curl)

Momentum (L = m V)

Before we go further today remember that momentum (L) is a simultaneous consideration of mass and velocity

Why and when does momentum matter For what activities is momentum most important

Quick Practice If a 120-kg middle linebacker is moving at 6 ms how much momentum does he haveAnswer 720 kg ms

Conservation of MomentumThe Law of Inertia provides a basis for the idea of conservation of momentum ie if net external force acting on a system equals zero (ΣF=0) momentum is conserved

We can use this idea to estimate some implications of certain impactscollisions

Some Common Characteristics of Collisions

occur over a relatively short duration

involve relatively high-magnitude contact forces

result in rapid changes in momentum for one or all colliding objects

often involve deformation and restitution

2

More common collisions

3D motion analysisLanding from a jumphellip

Knowing the first law and a little bit about the nature or quality of the collision can help us better understand some implications of the collision including post-impact velocity Why does post-impact velocity matter

And an unusual collisionhellip

Nature of CollisionsElastic collision ndash kinetic energy is mostly conserved for the colliding object

ndashbilliard ball collision

ndashbouncing superballs

Inelastic collision ndash kinetic energy is mostly lost for the colliding object

ndashclay hitting floor

ndashhead-on car collision

Coefficient of Restitution (e)Coefficient of restitution (e) a value that is used to describe the quality of a collision and depends on the nature of the colliding objects

e = 1 perfectly elastic collision

e = 0 perfectly inelastic collision

e is a ratio of post- and pre-collision velocities

In other words

ball velocities before impact

ball velocities after impact

u1 u2

v1 v2

1 2

or

This derivation works if motion is vertical and one of the colliding objects is stationary pre- and post-collision

hb = bounce height

hd = drop height

Impactscollisions Coefficients of restitution (e) of various balls dropped from 72 inches onto a hardwood floor

type of ball height bounced e

Superball 568 in 089

basketball 418 in 076

volleyball 398 in 074

lacrosse 275 in 062

softball 73 in 032

cricket 70 in 031

3

Coefficients of restitution (e) of a volleyball dropped from 72 inches onto various surfaces

type of surface height bounced e

astroturf 413 in 076

wood 407 in 075

concrete 395 in 074

tumbling mat 327 in 067

gravel 135 in 043

grass 130 in 042

Coefficients of restitution (e) of different balls dropped from 72 inches at various temperatures

type of ball height bounced e

baseballcold 18 in 050

normal 20 in 053

heated 22 in 055

golf ballcold 32 in 067

normal 46 in 080

heated 51 in 084

Applicationsbull Running shoes ndash advertising claims

bull Basketball ndash proper ball inflation

bull Golf ndash ldquohotrdquo golf balls 083 is the legal max

Momentum of player + ball before the catch is equal to momentum of player + ball after the catch

Practice problem 1

vBALL = 20 ms

vPLAYER = 0 msvPLAYER+BALL =

LBALL = (05 kg)(20 ms) = 10 kgms

LPLAYER = (90 kg)(0 ms) = 0 kgms

LTOTAL = MBALL+MPLAYER = 10 kgms

LBALL+PLAYER = mBALL+PLAYER vBALL+PLAYER

10 kgms = (905 kg)(vBALL+PLAYER)

vBALL+PLAYER = 011 ms

use L = mvand

Practice problem 2

A golf ball (mass = 46 g) is struck by a golf club (mass = 210 g) The club headrsquos velocity immediately prior to impact is 50 ms If the coefficient of restitution is 080 how fast are the ball and club moving after impact See Chapter 3 of the McGinnis text for the solution

Use the following equations

4

Newtonrsquos 2nd LawWersquove now discussed what results when ΣF = 0 however what results when ΣF ne 0

The Law of Acceleration A force that is applied to a body will cause an acceleration of that body of a magnitude that is proportional to the force in the direction of the force and inversely proportional to the bodyrsquos mass

Newtonrsquos 2nd Law

One Newton (N) will accelerate 1 kg at a rate of 1 ms2 [1 N = 1 (kg)(1ms2)]

Some basic practice A 10 kg mass is struck with a force of 10 N for 1 sec If the mass was initially at rest what is itrsquos velocity at the end of 1 sec

10 kg10 N

Newtonrsquos Laws ndash 2nd Law

Because a = Fm a = (10N10kg) = 1 ms2

bull Also a = (v2 - v1) then 1 ms2 = (v2 - 0 ms)t 1s

bull The velocity after 1 second (v2) is 1 ms

10 kg10 N

Newtonrsquos Laws ndash 2nd Law

Some more basic practice

Now suppose that the same force (10 N) is applied for 5 sec and the mass is 20kg How fast will the mass be moving after 5 sec Use F = ma to get the answer which is 25 ms

20 kg10 N

Newtonrsquos Laws ndash 2nd Law

Last bit of basic practicehellip

Now suppose the same mass has a vertical velocity of 10ms and is hit from the side with a the same 10-N force for the same duration (5 s) If we ignore gravity for now what will the horizontal velocity be after 5 s 25 ms vertical velocity 10 ms resultant velocity 103 ms

Vertical Velocity = 10 ms

20 kg10 N

But what about a force that is applied throughout a duration

ΣF and a in ΣF = ma are instantaneous measures however we often need to consider the effect of a net force throughout a duration of time To do this impulse (ΣFΔt) is necessary

5

Momentum amp ImpulseImpulse (FmiddotΔt) indicates the effect of a force over a duration of force application and equals the product of the (1) average force and (2) duration of the force application impulse is the time integral of the force time plot

An impulse causes (1) change in momentum and if mass is constant (2) change in velocityImpulse data set 1

Impulse data set 2

Functional Asymmetry

∆Momentum Impulse amp Force

The same impulse and resulting change in momentum can occur in very different waysndash large forces over small or large times

ndash small forces over small or large times

Newtonrsquos Laws ndash 3rd LawLaw of Reaction when a body exerts a force on another body the second body exerts a reaction force on the first body that is equal in magnitude and opposite in direction

The ground reaction force (GRF) reflects total body center of mass acceleration not acceleration of any single body segment

Walking amp Running GRFThe resultant GRF is often resolved into three perpendicular components that have functional significance during walking and running (1) Rz or vertical GRF (support) (2) Rx or anterior-posterior GRF (braking amp propulsion) and (3) Ry or medial-lateral GRF (balance amp stability)

Rz

1

Rx

Ry

Anterior Knee Pain and GRF Anterior Knee Pain TENS Therapy and GRF

6

Running GRFsCenter of Pressure and Strike Index

VGRF COP trajectory and strike index for rear-foot and mid-foot strikers during running

We recently used COP trajectory to better understand functional ankle instability subjects with chronic ankle instability exhibit laterally-deviated COP

Law of Gravitation

All bodies are attracted to one another with a force that is proportional to the product of the masses and inversely proportional to the square of the distance between them

Fg = G (m1m2)

d2

Fg minus the attraction force (N)G minus the gravitational constant (66710-11 Nm2kg2)m1 amp m2 minus masses of the two bodies (kg)d minus the distance between them (m)

Newtonrsquos Laws ndash Gravity

Law of Gravitation

How attracted are you to your classmates

G = 66710-11 Nm2kg2

m1 = 65 kgm2 = 85 kgd = 15 m

Fg = G (m1m2) = 66710-11 (65 85) =

d2 (15)2

Newtonrsquos Laws ndash Gravity

16 times10-7 N

Not very

Law of Gravitation

How attracted are you to the Earth (m1)

G = 66710-11 Nm2kg2

m1 = 5981024 kgm2 = 75 kgd = 637106 m

Fg = G (m1m2) = 66710-11 (5981024 75)

d2 (637106)2

Newtonrsquos Laws ndash Gravity

Fg = 737 N (166 lb) This is what we call your weight

7

bull If m1 = mass of the earth and m2 = mass of another object m2 g = G(m1m2 d2) then g is the acceleration due to the earthrsquos gravity

bull m2 appears on both sides of equation so we can divide by m2 and cancel it out g = G(m1 d2)

bull Acceleration of an object due to gravity is essentially a constant (independent of mass)

bull Given the earthrsquos radius of 3900 miles small changes in altitude have little effect on ldquogrdquo which is equal to approximately minus981 ms2

Newtonrsquos Laws ndash Gravity Chapter 3 Summarybull Linear kinetics is the study of forces related to linear

motion and can be explained by Newtonrsquos Laws

bull Law 1 helps explain why momentum is conserved when ΣF = 0 Collisions can be better understood by understanding this law and the coefficient of restitution

bull Law 2 helps explain the relationship between impulse and momentum Momentum is not conserved when the sum of force does not equal zero

bull Law 3 underlies almost all human motion including human locomotion Think ground reaction force

bull Law of gravitation is the basis for gravity

Some more practicehellipA sprinter experiences a peak instantaneous resultant GRF of 2800 N that is oriented 60deg above the ground The GRF acts through the heel not long after heel strike

1 What are the horizontal and vertical components of force acting on the heel 1400 amp 2425 N

2 If peak vertical GRFs during running are generally about 43 times body weight what is the runnerrsquos body mass 575 kg

3 What are the corresponding instantaneous horizontal and vertical accelerations of the whole-body center of mass ~24 mss amp ~32 mss

Some more practicehellipTwo impulses are applied to a high jumper GRF and gravity As a result the high jumper leaves the ground with a vertical velocity of 63 ms If the average GRF is 5000 N and application duration is 01 s

1 What must the jumperrsquos mass equal 6867 kg

2 Assuming equal release and landing heights what will the jumpers time of flight be 128 s

3 What will the jumperrsquos maximal height be 202 m

2

More common collisions

3D motion analysisLanding from a jumphellip

Knowing the first law and a little bit about the nature or quality of the collision can help us better understand some implications of the collision including post-impact velocity Why does post-impact velocity matter

And an unusual collisionhellip

Nature of CollisionsElastic collision ndash kinetic energy is mostly conserved for the colliding object

ndashbilliard ball collision

ndashbouncing superballs

Inelastic collision ndash kinetic energy is mostly lost for the colliding object

ndashclay hitting floor

ndashhead-on car collision

Coefficient of Restitution (e)Coefficient of restitution (e) a value that is used to describe the quality of a collision and depends on the nature of the colliding objects

e = 1 perfectly elastic collision

e = 0 perfectly inelastic collision

e is a ratio of post- and pre-collision velocities

In other words

ball velocities before impact

ball velocities after impact

u1 u2

v1 v2

1 2

or

This derivation works if motion is vertical and one of the colliding objects is stationary pre- and post-collision

hb = bounce height

hd = drop height

Impactscollisions Coefficients of restitution (e) of various balls dropped from 72 inches onto a hardwood floor

type of ball height bounced e

Superball 568 in 089

basketball 418 in 076

volleyball 398 in 074

lacrosse 275 in 062

softball 73 in 032

cricket 70 in 031

3

Coefficients of restitution (e) of a volleyball dropped from 72 inches onto various surfaces

type of surface height bounced e

astroturf 413 in 076

wood 407 in 075

concrete 395 in 074

tumbling mat 327 in 067

gravel 135 in 043

grass 130 in 042

Coefficients of restitution (e) of different balls dropped from 72 inches at various temperatures

type of ball height bounced e

baseballcold 18 in 050

normal 20 in 053

heated 22 in 055

golf ballcold 32 in 067

normal 46 in 080

heated 51 in 084

Applicationsbull Running shoes ndash advertising claims

bull Basketball ndash proper ball inflation

bull Golf ndash ldquohotrdquo golf balls 083 is the legal max

Momentum of player + ball before the catch is equal to momentum of player + ball after the catch

Practice problem 1

vBALL = 20 ms

vPLAYER = 0 msvPLAYER+BALL =

LBALL = (05 kg)(20 ms) = 10 kgms

LPLAYER = (90 kg)(0 ms) = 0 kgms

LTOTAL = MBALL+MPLAYER = 10 kgms

LBALL+PLAYER = mBALL+PLAYER vBALL+PLAYER

10 kgms = (905 kg)(vBALL+PLAYER)

vBALL+PLAYER = 011 ms

use L = mvand

Practice problem 2

A golf ball (mass = 46 g) is struck by a golf club (mass = 210 g) The club headrsquos velocity immediately prior to impact is 50 ms If the coefficient of restitution is 080 how fast are the ball and club moving after impact See Chapter 3 of the McGinnis text for the solution

Use the following equations

4

Newtonrsquos 2nd LawWersquove now discussed what results when ΣF = 0 however what results when ΣF ne 0

The Law of Acceleration A force that is applied to a body will cause an acceleration of that body of a magnitude that is proportional to the force in the direction of the force and inversely proportional to the bodyrsquos mass

Newtonrsquos 2nd Law

One Newton (N) will accelerate 1 kg at a rate of 1 ms2 [1 N = 1 (kg)(1ms2)]

Some basic practice A 10 kg mass is struck with a force of 10 N for 1 sec If the mass was initially at rest what is itrsquos velocity at the end of 1 sec

10 kg10 N

Newtonrsquos Laws ndash 2nd Law

Because a = Fm a = (10N10kg) = 1 ms2

bull Also a = (v2 - v1) then 1 ms2 = (v2 - 0 ms)t 1s

bull The velocity after 1 second (v2) is 1 ms

10 kg10 N

Newtonrsquos Laws ndash 2nd Law

Some more basic practice

Now suppose that the same force (10 N) is applied for 5 sec and the mass is 20kg How fast will the mass be moving after 5 sec Use F = ma to get the answer which is 25 ms

20 kg10 N

Newtonrsquos Laws ndash 2nd Law

Last bit of basic practicehellip

Now suppose the same mass has a vertical velocity of 10ms and is hit from the side with a the same 10-N force for the same duration (5 s) If we ignore gravity for now what will the horizontal velocity be after 5 s 25 ms vertical velocity 10 ms resultant velocity 103 ms

Vertical Velocity = 10 ms

20 kg10 N

But what about a force that is applied throughout a duration

ΣF and a in ΣF = ma are instantaneous measures however we often need to consider the effect of a net force throughout a duration of time To do this impulse (ΣFΔt) is necessary

5

Momentum amp ImpulseImpulse (FmiddotΔt) indicates the effect of a force over a duration of force application and equals the product of the (1) average force and (2) duration of the force application impulse is the time integral of the force time plot

An impulse causes (1) change in momentum and if mass is constant (2) change in velocityImpulse data set 1

Impulse data set 2

Functional Asymmetry

∆Momentum Impulse amp Force

The same impulse and resulting change in momentum can occur in very different waysndash large forces over small or large times

ndash small forces over small or large times

Newtonrsquos Laws ndash 3rd LawLaw of Reaction when a body exerts a force on another body the second body exerts a reaction force on the first body that is equal in magnitude and opposite in direction

The ground reaction force (GRF) reflects total body center of mass acceleration not acceleration of any single body segment

Walking amp Running GRFThe resultant GRF is often resolved into three perpendicular components that have functional significance during walking and running (1) Rz or vertical GRF (support) (2) Rx or anterior-posterior GRF (braking amp propulsion) and (3) Ry or medial-lateral GRF (balance amp stability)

Rz

1

Rx

Ry

Anterior Knee Pain and GRF Anterior Knee Pain TENS Therapy and GRF

6

Running GRFsCenter of Pressure and Strike Index

VGRF COP trajectory and strike index for rear-foot and mid-foot strikers during running

We recently used COP trajectory to better understand functional ankle instability subjects with chronic ankle instability exhibit laterally-deviated COP

Law of Gravitation

All bodies are attracted to one another with a force that is proportional to the product of the masses and inversely proportional to the square of the distance between them

Fg = G (m1m2)

d2

Fg minus the attraction force (N)G minus the gravitational constant (66710-11 Nm2kg2)m1 amp m2 minus masses of the two bodies (kg)d minus the distance between them (m)

Newtonrsquos Laws ndash Gravity

Law of Gravitation

How attracted are you to your classmates

G = 66710-11 Nm2kg2

m1 = 65 kgm2 = 85 kgd = 15 m

Fg = G (m1m2) = 66710-11 (65 85) =

d2 (15)2

Newtonrsquos Laws ndash Gravity

16 times10-7 N

Not very

Law of Gravitation

How attracted are you to the Earth (m1)

G = 66710-11 Nm2kg2

m1 = 5981024 kgm2 = 75 kgd = 637106 m

Fg = G (m1m2) = 66710-11 (5981024 75)

d2 (637106)2

Newtonrsquos Laws ndash Gravity

Fg = 737 N (166 lb) This is what we call your weight

7

bull If m1 = mass of the earth and m2 = mass of another object m2 g = G(m1m2 d2) then g is the acceleration due to the earthrsquos gravity

bull m2 appears on both sides of equation so we can divide by m2 and cancel it out g = G(m1 d2)

bull Acceleration of an object due to gravity is essentially a constant (independent of mass)

bull Given the earthrsquos radius of 3900 miles small changes in altitude have little effect on ldquogrdquo which is equal to approximately minus981 ms2

Newtonrsquos Laws ndash Gravity Chapter 3 Summarybull Linear kinetics is the study of forces related to linear

motion and can be explained by Newtonrsquos Laws

bull Law 1 helps explain why momentum is conserved when ΣF = 0 Collisions can be better understood by understanding this law and the coefficient of restitution

bull Law 2 helps explain the relationship between impulse and momentum Momentum is not conserved when the sum of force does not equal zero

bull Law 3 underlies almost all human motion including human locomotion Think ground reaction force

bull Law of gravitation is the basis for gravity

Some more practicehellipA sprinter experiences a peak instantaneous resultant GRF of 2800 N that is oriented 60deg above the ground The GRF acts through the heel not long after heel strike

1 What are the horizontal and vertical components of force acting on the heel 1400 amp 2425 N

2 If peak vertical GRFs during running are generally about 43 times body weight what is the runnerrsquos body mass 575 kg

3 What are the corresponding instantaneous horizontal and vertical accelerations of the whole-body center of mass ~24 mss amp ~32 mss

Some more practicehellipTwo impulses are applied to a high jumper GRF and gravity As a result the high jumper leaves the ground with a vertical velocity of 63 ms If the average GRF is 5000 N and application duration is 01 s

1 What must the jumperrsquos mass equal 6867 kg

2 Assuming equal release and landing heights what will the jumpers time of flight be 128 s

3 What will the jumperrsquos maximal height be 202 m

3

Coefficients of restitution (e) of a volleyball dropped from 72 inches onto various surfaces

type of surface height bounced e

astroturf 413 in 076

wood 407 in 075

concrete 395 in 074

tumbling mat 327 in 067

gravel 135 in 043

grass 130 in 042

Coefficients of restitution (e) of different balls dropped from 72 inches at various temperatures

type of ball height bounced e

baseballcold 18 in 050

normal 20 in 053

heated 22 in 055

golf ballcold 32 in 067

normal 46 in 080

heated 51 in 084

Applicationsbull Running shoes ndash advertising claims

bull Basketball ndash proper ball inflation

bull Golf ndash ldquohotrdquo golf balls 083 is the legal max

Momentum of player + ball before the catch is equal to momentum of player + ball after the catch

Practice problem 1

vBALL = 20 ms

vPLAYER = 0 msvPLAYER+BALL =

LBALL = (05 kg)(20 ms) = 10 kgms

LPLAYER = (90 kg)(0 ms) = 0 kgms

LTOTAL = MBALL+MPLAYER = 10 kgms

LBALL+PLAYER = mBALL+PLAYER vBALL+PLAYER

10 kgms = (905 kg)(vBALL+PLAYER)

vBALL+PLAYER = 011 ms

use L = mvand

Practice problem 2

A golf ball (mass = 46 g) is struck by a golf club (mass = 210 g) The club headrsquos velocity immediately prior to impact is 50 ms If the coefficient of restitution is 080 how fast are the ball and club moving after impact See Chapter 3 of the McGinnis text for the solution

Use the following equations

4

Newtonrsquos 2nd LawWersquove now discussed what results when ΣF = 0 however what results when ΣF ne 0

The Law of Acceleration A force that is applied to a body will cause an acceleration of that body of a magnitude that is proportional to the force in the direction of the force and inversely proportional to the bodyrsquos mass

Newtonrsquos 2nd Law

One Newton (N) will accelerate 1 kg at a rate of 1 ms2 [1 N = 1 (kg)(1ms2)]

Some basic practice A 10 kg mass is struck with a force of 10 N for 1 sec If the mass was initially at rest what is itrsquos velocity at the end of 1 sec

10 kg10 N

Newtonrsquos Laws ndash 2nd Law

Because a = Fm a = (10N10kg) = 1 ms2

bull Also a = (v2 - v1) then 1 ms2 = (v2 - 0 ms)t 1s

bull The velocity after 1 second (v2) is 1 ms

10 kg10 N

Newtonrsquos Laws ndash 2nd Law

Some more basic practice

Now suppose that the same force (10 N) is applied for 5 sec and the mass is 20kg How fast will the mass be moving after 5 sec Use F = ma to get the answer which is 25 ms

20 kg10 N

Newtonrsquos Laws ndash 2nd Law

Last bit of basic practicehellip

Now suppose the same mass has a vertical velocity of 10ms and is hit from the side with a the same 10-N force for the same duration (5 s) If we ignore gravity for now what will the horizontal velocity be after 5 s 25 ms vertical velocity 10 ms resultant velocity 103 ms

Vertical Velocity = 10 ms

20 kg10 N

But what about a force that is applied throughout a duration

ΣF and a in ΣF = ma are instantaneous measures however we often need to consider the effect of a net force throughout a duration of time To do this impulse (ΣFΔt) is necessary

5

Momentum amp ImpulseImpulse (FmiddotΔt) indicates the effect of a force over a duration of force application and equals the product of the (1) average force and (2) duration of the force application impulse is the time integral of the force time plot

An impulse causes (1) change in momentum and if mass is constant (2) change in velocityImpulse data set 1

Impulse data set 2

Functional Asymmetry

∆Momentum Impulse amp Force

The same impulse and resulting change in momentum can occur in very different waysndash large forces over small or large times

ndash small forces over small or large times

Newtonrsquos Laws ndash 3rd LawLaw of Reaction when a body exerts a force on another body the second body exerts a reaction force on the first body that is equal in magnitude and opposite in direction

The ground reaction force (GRF) reflects total body center of mass acceleration not acceleration of any single body segment

Walking amp Running GRFThe resultant GRF is often resolved into three perpendicular components that have functional significance during walking and running (1) Rz or vertical GRF (support) (2) Rx or anterior-posterior GRF (braking amp propulsion) and (3) Ry or medial-lateral GRF (balance amp stability)

Rz

1

Rx

Ry

Anterior Knee Pain and GRF Anterior Knee Pain TENS Therapy and GRF

6

Running GRFsCenter of Pressure and Strike Index

VGRF COP trajectory and strike index for rear-foot and mid-foot strikers during running

We recently used COP trajectory to better understand functional ankle instability subjects with chronic ankle instability exhibit laterally-deviated COP

Law of Gravitation

All bodies are attracted to one another with a force that is proportional to the product of the masses and inversely proportional to the square of the distance between them

Fg = G (m1m2)

d2

Fg minus the attraction force (N)G minus the gravitational constant (66710-11 Nm2kg2)m1 amp m2 minus masses of the two bodies (kg)d minus the distance between them (m)

Newtonrsquos Laws ndash Gravity

Law of Gravitation

How attracted are you to your classmates

G = 66710-11 Nm2kg2

m1 = 65 kgm2 = 85 kgd = 15 m

Fg = G (m1m2) = 66710-11 (65 85) =

d2 (15)2

Newtonrsquos Laws ndash Gravity

16 times10-7 N

Not very

Law of Gravitation

How attracted are you to the Earth (m1)

G = 66710-11 Nm2kg2

m1 = 5981024 kgm2 = 75 kgd = 637106 m

Fg = G (m1m2) = 66710-11 (5981024 75)

d2 (637106)2

Newtonrsquos Laws ndash Gravity

Fg = 737 N (166 lb) This is what we call your weight

7

bull If m1 = mass of the earth and m2 = mass of another object m2 g = G(m1m2 d2) then g is the acceleration due to the earthrsquos gravity

bull m2 appears on both sides of equation so we can divide by m2 and cancel it out g = G(m1 d2)

bull Acceleration of an object due to gravity is essentially a constant (independent of mass)

bull Given the earthrsquos radius of 3900 miles small changes in altitude have little effect on ldquogrdquo which is equal to approximately minus981 ms2

Newtonrsquos Laws ndash Gravity Chapter 3 Summarybull Linear kinetics is the study of forces related to linear

motion and can be explained by Newtonrsquos Laws

bull Law 1 helps explain why momentum is conserved when ΣF = 0 Collisions can be better understood by understanding this law and the coefficient of restitution

bull Law 2 helps explain the relationship between impulse and momentum Momentum is not conserved when the sum of force does not equal zero

bull Law 3 underlies almost all human motion including human locomotion Think ground reaction force

bull Law of gravitation is the basis for gravity

Some more practicehellipA sprinter experiences a peak instantaneous resultant GRF of 2800 N that is oriented 60deg above the ground The GRF acts through the heel not long after heel strike

1 What are the horizontal and vertical components of force acting on the heel 1400 amp 2425 N

2 If peak vertical GRFs during running are generally about 43 times body weight what is the runnerrsquos body mass 575 kg

3 What are the corresponding instantaneous horizontal and vertical accelerations of the whole-body center of mass ~24 mss amp ~32 mss

Some more practicehellipTwo impulses are applied to a high jumper GRF and gravity As a result the high jumper leaves the ground with a vertical velocity of 63 ms If the average GRF is 5000 N and application duration is 01 s

1 What must the jumperrsquos mass equal 6867 kg

2 Assuming equal release and landing heights what will the jumpers time of flight be 128 s

3 What will the jumperrsquos maximal height be 202 m

4

Newtonrsquos 2nd LawWersquove now discussed what results when ΣF = 0 however what results when ΣF ne 0

The Law of Acceleration A force that is applied to a body will cause an acceleration of that body of a magnitude that is proportional to the force in the direction of the force and inversely proportional to the bodyrsquos mass

Newtonrsquos 2nd Law

One Newton (N) will accelerate 1 kg at a rate of 1 ms2 [1 N = 1 (kg)(1ms2)]

Some basic practice A 10 kg mass is struck with a force of 10 N for 1 sec If the mass was initially at rest what is itrsquos velocity at the end of 1 sec

10 kg10 N

Newtonrsquos Laws ndash 2nd Law

Because a = Fm a = (10N10kg) = 1 ms2

bull Also a = (v2 - v1) then 1 ms2 = (v2 - 0 ms)t 1s

bull The velocity after 1 second (v2) is 1 ms

10 kg10 N

Newtonrsquos Laws ndash 2nd Law

Some more basic practice

Now suppose that the same force (10 N) is applied for 5 sec and the mass is 20kg How fast will the mass be moving after 5 sec Use F = ma to get the answer which is 25 ms

20 kg10 N

Newtonrsquos Laws ndash 2nd Law

Last bit of basic practicehellip

Now suppose the same mass has a vertical velocity of 10ms and is hit from the side with a the same 10-N force for the same duration (5 s) If we ignore gravity for now what will the horizontal velocity be after 5 s 25 ms vertical velocity 10 ms resultant velocity 103 ms

Vertical Velocity = 10 ms

20 kg10 N

But what about a force that is applied throughout a duration

ΣF and a in ΣF = ma are instantaneous measures however we often need to consider the effect of a net force throughout a duration of time To do this impulse (ΣFΔt) is necessary

5

Momentum amp ImpulseImpulse (FmiddotΔt) indicates the effect of a force over a duration of force application and equals the product of the (1) average force and (2) duration of the force application impulse is the time integral of the force time plot

An impulse causes (1) change in momentum and if mass is constant (2) change in velocityImpulse data set 1

Impulse data set 2

Functional Asymmetry

∆Momentum Impulse amp Force

The same impulse and resulting change in momentum can occur in very different waysndash large forces over small or large times

ndash small forces over small or large times

Newtonrsquos Laws ndash 3rd LawLaw of Reaction when a body exerts a force on another body the second body exerts a reaction force on the first body that is equal in magnitude and opposite in direction

The ground reaction force (GRF) reflects total body center of mass acceleration not acceleration of any single body segment

Walking amp Running GRFThe resultant GRF is often resolved into three perpendicular components that have functional significance during walking and running (1) Rz or vertical GRF (support) (2) Rx or anterior-posterior GRF (braking amp propulsion) and (3) Ry or medial-lateral GRF (balance amp stability)

Rz

1

Rx

Ry

Anterior Knee Pain and GRF Anterior Knee Pain TENS Therapy and GRF

6

Running GRFsCenter of Pressure and Strike Index

VGRF COP trajectory and strike index for rear-foot and mid-foot strikers during running

We recently used COP trajectory to better understand functional ankle instability subjects with chronic ankle instability exhibit laterally-deviated COP

Law of Gravitation

All bodies are attracted to one another with a force that is proportional to the product of the masses and inversely proportional to the square of the distance between them

Fg = G (m1m2)

d2

Fg minus the attraction force (N)G minus the gravitational constant (66710-11 Nm2kg2)m1 amp m2 minus masses of the two bodies (kg)d minus the distance between them (m)

Newtonrsquos Laws ndash Gravity

Law of Gravitation

How attracted are you to your classmates

G = 66710-11 Nm2kg2

m1 = 65 kgm2 = 85 kgd = 15 m

Fg = G (m1m2) = 66710-11 (65 85) =

d2 (15)2

Newtonrsquos Laws ndash Gravity

16 times10-7 N

Not very

Law of Gravitation

How attracted are you to the Earth (m1)

G = 66710-11 Nm2kg2

m1 = 5981024 kgm2 = 75 kgd = 637106 m

Fg = G (m1m2) = 66710-11 (5981024 75)

d2 (637106)2

Newtonrsquos Laws ndash Gravity

Fg = 737 N (166 lb) This is what we call your weight

7

bull If m1 = mass of the earth and m2 = mass of another object m2 g = G(m1m2 d2) then g is the acceleration due to the earthrsquos gravity

bull m2 appears on both sides of equation so we can divide by m2 and cancel it out g = G(m1 d2)

bull Acceleration of an object due to gravity is essentially a constant (independent of mass)

bull Given the earthrsquos radius of 3900 miles small changes in altitude have little effect on ldquogrdquo which is equal to approximately minus981 ms2

Newtonrsquos Laws ndash Gravity Chapter 3 Summarybull Linear kinetics is the study of forces related to linear

motion and can be explained by Newtonrsquos Laws

bull Law 1 helps explain why momentum is conserved when ΣF = 0 Collisions can be better understood by understanding this law and the coefficient of restitution

bull Law 2 helps explain the relationship between impulse and momentum Momentum is not conserved when the sum of force does not equal zero

bull Law 3 underlies almost all human motion including human locomotion Think ground reaction force

bull Law of gravitation is the basis for gravity

Some more practicehellipA sprinter experiences a peak instantaneous resultant GRF of 2800 N that is oriented 60deg above the ground The GRF acts through the heel not long after heel strike

1 What are the horizontal and vertical components of force acting on the heel 1400 amp 2425 N

2 If peak vertical GRFs during running are generally about 43 times body weight what is the runnerrsquos body mass 575 kg

3 What are the corresponding instantaneous horizontal and vertical accelerations of the whole-body center of mass ~24 mss amp ~32 mss

Some more practicehellipTwo impulses are applied to a high jumper GRF and gravity As a result the high jumper leaves the ground with a vertical velocity of 63 ms If the average GRF is 5000 N and application duration is 01 s

1 What must the jumperrsquos mass equal 6867 kg

2 Assuming equal release and landing heights what will the jumpers time of flight be 128 s

3 What will the jumperrsquos maximal height be 202 m

5

Momentum amp ImpulseImpulse (FmiddotΔt) indicates the effect of a force over a duration of force application and equals the product of the (1) average force and (2) duration of the force application impulse is the time integral of the force time plot

An impulse causes (1) change in momentum and if mass is constant (2) change in velocityImpulse data set 1

Impulse data set 2

Functional Asymmetry

∆Momentum Impulse amp Force

The same impulse and resulting change in momentum can occur in very different waysndash large forces over small or large times

ndash small forces over small or large times

Newtonrsquos Laws ndash 3rd LawLaw of Reaction when a body exerts a force on another body the second body exerts a reaction force on the first body that is equal in magnitude and opposite in direction

The ground reaction force (GRF) reflects total body center of mass acceleration not acceleration of any single body segment

Walking amp Running GRFThe resultant GRF is often resolved into three perpendicular components that have functional significance during walking and running (1) Rz or vertical GRF (support) (2) Rx or anterior-posterior GRF (braking amp propulsion) and (3) Ry or medial-lateral GRF (balance amp stability)

Rz

1

Rx

Ry

Anterior Knee Pain and GRF Anterior Knee Pain TENS Therapy and GRF

6

Running GRFsCenter of Pressure and Strike Index

VGRF COP trajectory and strike index for rear-foot and mid-foot strikers during running

We recently used COP trajectory to better understand functional ankle instability subjects with chronic ankle instability exhibit laterally-deviated COP

Law of Gravitation

All bodies are attracted to one another with a force that is proportional to the product of the masses and inversely proportional to the square of the distance between them

Fg = G (m1m2)

d2

Fg minus the attraction force (N)G minus the gravitational constant (66710-11 Nm2kg2)m1 amp m2 minus masses of the two bodies (kg)d minus the distance between them (m)

Newtonrsquos Laws ndash Gravity

Law of Gravitation

How attracted are you to your classmates

G = 66710-11 Nm2kg2

m1 = 65 kgm2 = 85 kgd = 15 m

Fg = G (m1m2) = 66710-11 (65 85) =

d2 (15)2

Newtonrsquos Laws ndash Gravity

16 times10-7 N

Not very

Law of Gravitation

How attracted are you to the Earth (m1)

G = 66710-11 Nm2kg2

m1 = 5981024 kgm2 = 75 kgd = 637106 m

Fg = G (m1m2) = 66710-11 (5981024 75)

d2 (637106)2

Newtonrsquos Laws ndash Gravity

Fg = 737 N (166 lb) This is what we call your weight

7

bull If m1 = mass of the earth and m2 = mass of another object m2 g = G(m1m2 d2) then g is the acceleration due to the earthrsquos gravity

bull m2 appears on both sides of equation so we can divide by m2 and cancel it out g = G(m1 d2)

bull Acceleration of an object due to gravity is essentially a constant (independent of mass)

bull Given the earthrsquos radius of 3900 miles small changes in altitude have little effect on ldquogrdquo which is equal to approximately minus981 ms2

Newtonrsquos Laws ndash Gravity Chapter 3 Summarybull Linear kinetics is the study of forces related to linear

motion and can be explained by Newtonrsquos Laws

bull Law 1 helps explain why momentum is conserved when ΣF = 0 Collisions can be better understood by understanding this law and the coefficient of restitution

bull Law 2 helps explain the relationship between impulse and momentum Momentum is not conserved when the sum of force does not equal zero

bull Law 3 underlies almost all human motion including human locomotion Think ground reaction force

bull Law of gravitation is the basis for gravity

Some more practicehellipA sprinter experiences a peak instantaneous resultant GRF of 2800 N that is oriented 60deg above the ground The GRF acts through the heel not long after heel strike

1 What are the horizontal and vertical components of force acting on the heel 1400 amp 2425 N

2 If peak vertical GRFs during running are generally about 43 times body weight what is the runnerrsquos body mass 575 kg

3 What are the corresponding instantaneous horizontal and vertical accelerations of the whole-body center of mass ~24 mss amp ~32 mss

Some more practicehellipTwo impulses are applied to a high jumper GRF and gravity As a result the high jumper leaves the ground with a vertical velocity of 63 ms If the average GRF is 5000 N and application duration is 01 s

1 What must the jumperrsquos mass equal 6867 kg

2 Assuming equal release and landing heights what will the jumpers time of flight be 128 s

3 What will the jumperrsquos maximal height be 202 m

6

Running GRFsCenter of Pressure and Strike Index

VGRF COP trajectory and strike index for rear-foot and mid-foot strikers during running

We recently used COP trajectory to better understand functional ankle instability subjects with chronic ankle instability exhibit laterally-deviated COP

Law of Gravitation

All bodies are attracted to one another with a force that is proportional to the product of the masses and inversely proportional to the square of the distance between them

Fg = G (m1m2)

d2

Fg minus the attraction force (N)G minus the gravitational constant (66710-11 Nm2kg2)m1 amp m2 minus masses of the two bodies (kg)d minus the distance between them (m)

Newtonrsquos Laws ndash Gravity

Law of Gravitation

How attracted are you to your classmates

G = 66710-11 Nm2kg2

m1 = 65 kgm2 = 85 kgd = 15 m

Fg = G (m1m2) = 66710-11 (65 85) =

d2 (15)2

Newtonrsquos Laws ndash Gravity

16 times10-7 N

Not very

Law of Gravitation

How attracted are you to the Earth (m1)

G = 66710-11 Nm2kg2

m1 = 5981024 kgm2 = 75 kgd = 637106 m

Fg = G (m1m2) = 66710-11 (5981024 75)

d2 (637106)2

Newtonrsquos Laws ndash Gravity

Fg = 737 N (166 lb) This is what we call your weight

7

bull If m1 = mass of the earth and m2 = mass of another object m2 g = G(m1m2 d2) then g is the acceleration due to the earthrsquos gravity

bull m2 appears on both sides of equation so we can divide by m2 and cancel it out g = G(m1 d2)

bull Acceleration of an object due to gravity is essentially a constant (independent of mass)

bull Given the earthrsquos radius of 3900 miles small changes in altitude have little effect on ldquogrdquo which is equal to approximately minus981 ms2

Newtonrsquos Laws ndash Gravity Chapter 3 Summarybull Linear kinetics is the study of forces related to linear

motion and can be explained by Newtonrsquos Laws

bull Law 1 helps explain why momentum is conserved when ΣF = 0 Collisions can be better understood by understanding this law and the coefficient of restitution

bull Law 2 helps explain the relationship between impulse and momentum Momentum is not conserved when the sum of force does not equal zero

bull Law 3 underlies almost all human motion including human locomotion Think ground reaction force

bull Law of gravitation is the basis for gravity

Some more practicehellipA sprinter experiences a peak instantaneous resultant GRF of 2800 N that is oriented 60deg above the ground The GRF acts through the heel not long after heel strike

1 What are the horizontal and vertical components of force acting on the heel 1400 amp 2425 N

2 If peak vertical GRFs during running are generally about 43 times body weight what is the runnerrsquos body mass 575 kg

3 What are the corresponding instantaneous horizontal and vertical accelerations of the whole-body center of mass ~24 mss amp ~32 mss

Some more practicehellipTwo impulses are applied to a high jumper GRF and gravity As a result the high jumper leaves the ground with a vertical velocity of 63 ms If the average GRF is 5000 N and application duration is 01 s

1 What must the jumperrsquos mass equal 6867 kg

2 Assuming equal release and landing heights what will the jumpers time of flight be 128 s

3 What will the jumperrsquos maximal height be 202 m

7

bull If m1 = mass of the earth and m2 = mass of another object m2 g = G(m1m2 d2) then g is the acceleration due to the earthrsquos gravity

bull m2 appears on both sides of equation so we can divide by m2 and cancel it out g = G(m1 d2)

bull Acceleration of an object due to gravity is essentially a constant (independent of mass)

bull Given the earthrsquos radius of 3900 miles small changes in altitude have little effect on ldquogrdquo which is equal to approximately minus981 ms2

Newtonrsquos Laws ndash Gravity Chapter 3 Summarybull Linear kinetics is the study of forces related to linear

motion and can be explained by Newtonrsquos Laws

bull Law 1 helps explain why momentum is conserved when ΣF = 0 Collisions can be better understood by understanding this law and the coefficient of restitution

bull Law 2 helps explain the relationship between impulse and momentum Momentum is not conserved when the sum of force does not equal zero

bull Law 3 underlies almost all human motion including human locomotion Think ground reaction force

bull Law of gravitation is the basis for gravity

Some more practicehellipA sprinter experiences a peak instantaneous resultant GRF of 2800 N that is oriented 60deg above the ground The GRF acts through the heel not long after heel strike

1 What are the horizontal and vertical components of force acting on the heel 1400 amp 2425 N

2 If peak vertical GRFs during running are generally about 43 times body weight what is the runnerrsquos body mass 575 kg

3 What are the corresponding instantaneous horizontal and vertical accelerations of the whole-body center of mass ~24 mss amp ~32 mss

Some more practicehellipTwo impulses are applied to a high jumper GRF and gravity As a result the high jumper leaves the ground with a vertical velocity of 63 ms If the average GRF is 5000 N and application duration is 01 s

1 What must the jumperrsquos mass equal 6867 kg

2 Assuming equal release and landing heights what will the jumpers time of flight be 128 s

3 What will the jumperrsquos maximal height be 202 m