Force force - a push or pull acting on a body forces are vector quantities –magnitude (size) –direction (line of action) –point of application measured.

Force• force - a push or pull acting on a body• forces are vector quantities

– magnitude (size)– direction (line of action)– point of application

• measured in Newtons (N)– 1 N = (1 kg) (1 m/s2)

• English units (lb)– 1 lb = (1 slug) (1 ft/s2)

BOX

Fpt of application

line of action

Free-Body Diagrams

A free body diagram illustrates all of theexternal forces acting on an object.

FR (GRF)

mg

air resistance

Contact ForcesContact Forces

If whole body is considered to be “the system”Examples of external forces

Weight (gravity)Ground Reaction Force (GRF)FrictionFluid Resistance

Examples of internal forcesJoint Reaction ForceMuscle Force

NEWTON'S LAWS OF MOTIONPhilosophiae Naturalis Principia Mathematica (1686)

Inertia is the resistance of an object to motion - the amount of resistance to linear motion varies directly with the mass of the object.

When an object is in motion its resistance to change in motion is determined by its velocity as well as its mass. Momentum is the mass of an object multiplied by the velocity (p = mv).

Newton’s 1st Law of Motion• Law of Inertia

– a body at rest will remain at rest and a body in motion will remain in motion and move at a constant velocity until a non-zero resultant external force is applied to it.

Newton’s 1st Law of Motion

W

Ry

if Ry = Wthen resultant force = 0if v = 0 and F = 0STATIC EQUILIBRIUM

W

Ry

FpFr

Fr = resistive forceFp = propulsive forceif v = 0 and F = 0DYNAMIC EQUILIBRIUM

V

Newton’s 2nd Law of Motion

• Law of acceleration– when a non-zero resultant external force is

applied to a body, the body will accelerate in the direction of this force. The acceleration is proportional to the force and inversely proportional to the body’s mass

F1

F2

Fa

a F am

a k F

m 1

1

where k1 = constant of proportionality

F k ma where kk

` 2

1

1

m m

F k ma 2

If m is measured in kgand a is measured in m/s2

the SI unit for force is “newton” (N)

1 N = 1 kg x 1 m/s2

where k2=1

F = ma

Alternate Expression of LAW OF ACCELERATION

"The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts."

• F =

• Ft = mv (impulse/momentum relation)

• F = m = mat

mv - mv if

v

t

If an elephant and a feather fall from the same height in the absence of air resistance then the resultant or net force acting on each object is simply their weight. Since W = mg then the acceleration they experience is

F = ma but F = W = mgmg = maora = g

Now consider when air resistance is present. This force would larger on the elephant simply because is a bigger object. But the weight of the elephant is also significantly larger than that of the feather. In fact, relative to weight the air resistance acting on the feather is larger than on the elephant. This affects the resultant force acting on each object such that the resultant force acting on the feather is much closer to 0 N. Thus the feather will have a much lower acceleration.

This example further demonstrates the change in resultant force due to air resistance. Notice that initially air resistance due to the body falling through the air reduces the magnitude of the acceleration but it remains a downward acceleration. Eventually you reach a point where the air resistance equals your body weight. This is known as terminal speed and would be well over 100 mph for a human body.

To allow you to land without hurting yourself you deploy your parachute. This greatly changes the resultant force such that the net force actually points upward such that the acceleration is upward. This successfully decreases your speed to a more manageable level for landing.

Newton’s 3rd Law

• “Law of action and reaction”

• when one body exerts a force on another body, the second exerts an equal but opposite force back on the 1st body

• “for every action there is an equal and opposite reaction”

Rpropulsion

Apropulsion

Rresistance

Aresistance

Note: these forces (action and reaction) are never applied on the same body -- it takes two bodies for a pair of action/reaction forces to exist

LAW OF GRAVITATION

"All bodies attract one another with a force proportional to the product of their masses and inversely proportional to the square of the distance between them."

• FG = G , G = 6.67X10-11 N-m2/kg2 m1*m2

d 2

Mass and Weight

• g depends on the distance from the earth’s center

M

r

M

r

Earth is not a true spherebut rather an ellipsoid: it is fatter at the equator

m = your massM = earth’s massr = radius (distance to (center of earth)G = gravitational constant

Newton’s Law of Gravitation

FgGMm

rFg

rand Fg m

2 2

1

Fg = gravitational force

If m = your mass then Fg = your weight (W)

FgGMm

r

g GM

r

W mg

2

2

Latitude effect = g is smaller at equator larger at poles

Altitude effect = g is smaller at high altitudelarger at low altitude

M

r

Weight

• W = mg– weight is a force– weight = mass x acceleration due to gravity

• Units– N = kg x m/s2

– weight in Newtons = mass in kg x 9.81 m/s2

– a 1 kg mass weighs 9.81 N

The Relevance of Newton’s LawsThe Relevance of Newton’s Laws

These 4 laws allow all motion in the universe to be described and predicted as long as the relative speeds of the objects are small compared to the speed of light.

Ground Reaction Forces

FV

FML

FAP

Force Platform


Impulse - Momentum Relationship


maF amF

t

vvmF if

)( if vvmtF

if mvmvtF

if ppJ

pJ

Impulse-MomentumJ p

“Impulse = change in momentum”Units

smkg

ssmkg

sNtF

/

)/( 2


smkgvm /

Same units!

Impulse/Momentum(Ft = mv)• The impulse/momentum relationship describes the

effects of a force over a period of time.• An impulse (F*t) causes a mass to change its

velocity• A 10 N force is applied to a 2 kg mass for 3

seconds. What is the change in velocity?


Impulse/Momentum

• How long would it take to stop a 7 kg mass that has a velocity of 10 m/s if you are capable of producing a maximum instantaneous force of 35 N?


• It has been determined that a force over 200 N can injure the hand. What is the shortest period of time it will take to stop a 2 kg object traveling at 75 m/s if the hand is to be protected?

Impulse/Momentum


Impulse/Momentum• The braking impulse of a subject running

across a force platform is -10 N-s. The propelling impulse during the same time period is 2 N-s. What is the change in velocity of the subject if she has a mass of 55 kg? Fy

0

0 .8time


• If a 10 kg object is exposed to the above impulse what will be the change in velocity?

5

4

3

2

1

0

2 54310

Time (s)

Force

(N) impulse = area


F

t

Impulse = area under F v. t curve

Jn=negative impulse

Jp=positive impulse

JNET = NET IMPULSEJNET = Jp + Jn


Recall that the net force is

F = GRF - W

and by Newton’s 2nd Law

F = ma

so GRF - W = ma

If GRF > W then a >0

If GRF < W then a < 0

W

GRF

W

GRF

When the jumper initiates the countermovement he/she speeds up in the negative direction

this means that the body experiences a negative acceleration

thus GRF < W

body weightbody weight

GRFGRF

00

W

GRF

As the jumper nears the bottom of the countermovement he/she slows down in the negative direction

this means that the body experiences a positive acceleration

thus GRF > W

lowest point


GRFGRF

00

When the jumper begins the upward portion of the countermovement he/she speeds up in the positive direction

this means that the body experiences a positive acceleration

thus GRF > W

lowest point

takeoff


GRFGRF

00

W

GRF

Total Net Impulse = Total Net Impulse = Negative Net ImpulseNegative Net Impulse + +Positive Net ImpulsePositive Net Impulse

J = mJ = mv = m(vv = m(vff - v - vii))

this can be solved to find the this can be solved to find the takeoff velocity (final velocity takeoff velocity (final velocity of countermovement phase)of countermovement phase)

vvtakeofftakeoff = = J/m since vJ/m since vii = 0 = 0

Knowing vKnowing vtakeofftakeoff allows you to compute jump height allows you to compute jump height

Knowing vKnowing vtakeofftakeoff allows you to allows you to

compute jump heightcompute jump height

vvtakeofftakeoff

vvtoptop = 0 m/s = 0 m/s

vvff22 = v = vii

22 + 2ad + 2ad

wherewhere vvff = v = vtoptop

vvii = v = vtakeofftakeoff

jump heightjump height

SO - jump height is increased bySO - jump height is increased byincreasing the total net increasing the total net impulseimpulse… … not just the net forcenot just the net force

Conservation of Momentum

Momentum = mass x velocity

Total momentumbefore = total momentumafter

for example - if there are 2 objects in the systemm1v1 + m2v2 = m1u1 + m2u2

v = velocity beforeu = velocity aftersubscripts represent object number

Momentum represents the total quantity of motion possessed by a body or system. The momentum of a system cannot be altered without an external force.

Example:

A 100 kg astronaut is moving at a speed of 9 m/s and runs into a stationary astronaut (mass = 150 kg).

Problem:What is the velocity of the astronauts after

the collision?

System = both astronauts

External Forces = none

Therefore the momentum must be conserved.

Total momentum (p):

pbefore =

pafter =

v = ?

A 75 kg rugby player is moving at 2 m/s when he runs into a 100 kg player running at –1.5 m/s. Which direction will the resulting collision travel (-, 0, +)?

If they collide in mid-air (0).

If they collide on the ground the players will be able to exert external forces and the larger player will probably be able to exert larger external forces (+).

pbig fish-before + plittle fish-before = pbig fish-after + plittle fish-after

pbig fish-before + plittle fish-before = pbig fish-after + plittle fish-after

Joint Reaction Force

The net force acting across a jointThe net force acting across a jointe.g. when standing, the thigh exerts a downward forcee.g. when standing, the thigh exerts a downward force

on the shank, conversely, the shank exerts anon the shank, conversely, the shank exerts anupward force on the thigh of equal magnitudeupward force on the thigh of equal magnitude

NOTE: this JRF does not includeNOTE: this JRF does not includethe forces exerted on the jointthe forces exerted on the jointby the muscle crossing theby the muscle crossing thejoint. The force that includesjoint. The force that includesall of the forces crossing the jointall of the forces crossing the jointis known as the is known as the bone-on-bonebone-on-boneforceforce


• There is an interaction between the surface of the block and the table. • The friction force always opposes motion.• The pulling force must be greater than the frictional force to move the block.

Block Pulling Force

Friction Force

Normal Force(weight of the block)


Normal Force = force perpendicular to surface

FN

When surface is horizontal the perpendicular direction is vertical so the normal force is simply the weight of the object

Weight

Normal Force

When surface is inclined the perpendicular direction is NOT vertical so the normal force is only a component of the object’s weight.

• At some point the pulling force will be great enough so that the friction force cannot prevent movement.

Fri

ctio

n F

orce

Applied Force

Static

Dynamic

impending motion

maximal static friction force


• The coefficient of static friction is expressed as:

where s= coefficient of static friction

Fnormal = normal force

Fmax = maximal static friction force

Friction

normals F

Fmax

•The coefficient of friction is a dimensionless number. It is unaffected by the mass of the object or the contact area.

•The greater the magnitude of s the greater the force necessary to move the object.

Friction• As the block moves along the table, there still is a

frictional force that resists motion.

• Sliding and rolling friction are types of dynamic friction.

where d= coefficient of dynamic friction

N = normal force

Ffriction = force resisting motion

Normal

frictiond F

F


Friction

• It has been found experimentally that d< s.

• d depends of the relative speed of the surfaces.

• At speeds from 1 cm/s to several m/s, d is approximately constant.


Why is it easier to pull a desk than push it?

Ph

v

WW

R = W+v

When you push you usually have a downwardcomponent of force -- so the normalforce is increased and therefore thefrictional force is increased.


P

h

v

WW

R = W-v

When you pull you usually have an upwardcomponent of force -- so the normalforce is reduced and therefore the frictional force is reduced.


• The transmission of energy from an object passing through a fluid to the fluid is known as fluid resistance.

• The resistance of an object passing through a fluid increases as the speed of the object increases and as the viscosity of the fluid increases.

Fluid Resistance


• Surface drag is a result of the friction between the surface and the fluid.

• The fluid closest to the object (boundary layer) rubs against the object creating friction.

• Kyle (1989) reported that wearing loose clothing can increase surface drag from 2% to 8%.

Surface and Form Drag


Van Ingen Schenau (1982) reported a 10% reduction in surface drag when a speed skater wears a smooth body suit.

Surface Drag


Form drag occurs when air is driven past an object and is diverted outward creating a low pressure region behind the object.

high pressure

low pressure

Form Drag


The orientation of the object will affect the frontal area and will play an important role in the amount of form drag.

Low form drag

High form drag

Form Drag


frontal area .5m2 (upright)

.42m2 (touring)

.34m2 (racing)

The second cyclist can ride within the low pressure zone of the first cyclist and thus lower the pressure difference and the drag. This is called drafting.


At low velocities laminar flow occurs. The boundary layer remains attached to the surface. During separated flow the boundary layerseparates toward theback of the object and alow pressure regionis formed. During fully turbulent flow the boundary layer becomes turbulent and the size of the pocket is decreased.

laminar

separated

fully turbulent

Flow Type


Factors Affecting Flow Type

• size

• shape

• surface roughness

• viscosity of the fluid

• flow velocity


The particles following the path from D1 to D2 will be more spread out than particles following the path from C1 to C2 because of the greater distance. This creates a low pressure region above the airfoil.

Airfoil

Bernoulli’s Principle, 1738Contact ForcesContact Forces

direction of movement

Lift always acts perpendicular to drag.

Fdrag = 1/2(CdAv2)

Flift = 1/2(ClAv2)

Fair resistance

Fdrag

Flift

Lift


• The lift-to-drag ratio is critical (i.e. the larger the ratio, the more effective the airfoil is in flight).

• L/D ratio is dependent on the angle that the airfoil makes with the incoming air (this is called the ANGLE OF ATTACK).

• Increasing the angle of attack increases the L/D ratio to a point; beyond that point the angle becomes too steep and the airfoil stalls

• typical angles of attack:

airfoil - below 15o

javelin - 10oContact ForcesContact Forces

Angle of Attack Lift Drag Lift

(degrees) (N) (N) Drag

0 0.00 1.17 0.0010 4.33 1.50 2.8920 10.64 4.13 2.5825 12.83 5.79 2.2227 13.80 6.88 2.0128 13.80 7.41 1.8629 11.01 7.94 1.3930 11.21 8.18 1.3735 10.12 8.74 1.1640 8.50 9.55 0.8945 8.90 11.13 0.8050 8.62 12.22 0.7160 6.88 14.98 0.4670 4.77 16.43 0.2980 2.55 16.88 0.1590 0.00 17.73 0.00

Adapted from Aerodynamic Factors Which Influence Discus Flight, Ganslen.

Lift to drag ratios for the discus.


Rotating objects can also create a pressure difference.

low pressurezone

high pressurezone

Direction of air flow

Rotating Objects

Magnus Effect, 1852Contact ForcesContact Forces

low pressurezone

high pressurezone

intended direction of flight

actual directionof flight


The golf club imparts backspin on the golf ball andincreases the length of the drive.


Depth of Dimple Carry Length of Drive(mm) (m) (m)

0.05 107 1340.10 171 1940.15 194 2120.20 204 2180.25 218 2390.30 206 219

From The Mechanics of Sport, E. Bade.

Dimples on a golf ball increase the velocity of the boundary layer and can dramatically influence the length of a drive.


Terminal Speed

An object falling through a fluid reaches its terminal speed when the drag force is equal to its weight. This results in a net force of zero and thus no further acceleration takes place.

weight

drag force


Estimated Terminal Speeds of Selected Spheres

KC DD

2

g

W8

Vg

KT

CD: coefficient of drag: fluid densityD: sphere diameterW: weight of sphereVT: terminal speed

Adapted from SportScience by Peter J. Brancazio.

Weight Diameter K Terminal VBall (N) (cm) (Drag Factor) (m/s)

16-lb shot 71.27 1.86 0.00014 145.28Baseball 1.43 1.14 0.0016 42.47Golf ball 0.45 0.66 0.0018 40.23Basketball 5.84 3.73 0.007 20.12Ping-Pong ball 0.03 0.58 0.04 8.94


Centripetal vs. Centrifugal Force

• Centripetal force (center seeking force) = mass xcentripetal acceleration

• Centrifugal force (center fleeing force) -- reaction to the centripetal force; applied to the other body

rmr

vmmaF t

cc2

2

Consider Newton’s second law of motion:F = ma

Now substitute centripetal acceleration. In centripetal motion the centripetal acceleration is linked to a centripetal force. You can think of this force as being responsible for holding the object in a circular path.

Exampleyou make a right turn in your caryou feel the driver door push on you to the right(toward the center of the curvature of your

curved path)

the door applies a centripetal force to you,you apply a centrifugal force which is

equal and opposite to the centripetal force

door

youFcpFcf

Forces occurring Forces occurring along a curved pathalong a curved path

Pressure

• localized effect of a force being applied to an area of a certain surface

W

Rn

x x x x xx x x x xx x x x x

Bird’s eye viewx = pt of application for a force

Pressure Force

Area

P F

As

if W = 100 lb, A = 500 in2

P lb

inlb in psi 100

50002 02

22. / .

Special ForceSpecial ForceApplicationApplication

40 in2 =0.0258 m2

10 in2 = 0.00645 m2

W = 110 lb *4.45 N/lb = 489.5 N

What is the avg. pressurewhen standing on one foot?

kPa 9.75m 00645.0

N 5.489P

2 kPa 0.19

m 0258.0

N 5.489P

2

In SI Units?

102 54

64 5 0 0062

2

2 2incm

incm m

.. .


Pressure distributions in the foot

While standing, most of the pressure is in the heel and the forefoot

Work• Work is a force applied over a given distance.

• W = F*d

• Units are N-m, 1 N-m = 1 Joule.

• Positive work occurs when the force is applied in the same direction as the motion (concentric contractions).

• Negative work occurs when the force is applied in the opposite direction of the motion (eccentric contractions).


• Only the component of force parallel to the direction of motion is responsible for work being done. If there is no movement there is no work being done.

direction of motion


Internal Work vs External Work

• External work is the result of external forces such as ground reaction forces.

• Internal work is the result of internal forces such as muscle forces.

While running uphill there is internal work done to rotate the segments and external work done to raise the body to a new height.

Energy• Energy is the capacity to do work.

• Performing positive work on an object will increase its energy while performing negative work will decrease its energy.

W = E

• Energy is measured in Joules.


• Kinetic Energy (KE) is the energy that an object possesses because of its velocity.

KE = ½mv2

• Potential Energy (PE) is the energy that an object possesses because of its height. PE = mgh

• Strain Energy (SE) is the energy that an object possesses because of its deformation:

SE = ½kx2


• Kinetic Energy (KE) - energy due to motion

– e.g. A diver (mass = 70 kg) hits the water after a dive from the 10 m tower with a velocity of 14 m/s. How much KE does she possess?

KE mv unit kg m s J 12

2 2 2: /

KE mv kg ms J 1

212

70 14 68602 2( ) ( )

Kinetic Energy


Potential Energy

• Potential Energy (PE) - energy due to gravity– PE = mgh = mass*gravity*height– e.g. A diver (mass = 70 kg) is 10 meters above

the water. How much PE does the diver have?

PE = 70*9.8*10 = 6860 J


Strain Energy

• Strain (SE) - energy due to deformation– SE = ½kx2, k = stiffness, x = deformation– this type of energy arises in compressed springs, squashed

balls ready to rebound, stretched tendons inside the body, and other deformable structures

– e.g. A muscle tendon with a stiffness of 70,000 N/m is stretched by 1 cm. How much strain energy does it have?

SE = ½(70,000)(.01)2 = 3.5 J


Conservation of Energy• Energy can not be created nor destroyed, it can

only change forms (von Helmholtz, 1847).

Energy is often transferredbetween kinetic, potential and strain.

PE = mgh

KE =.5mv2


• If the 4 kg pendulum above is released from a height of .5 meters what is the maximum velocity?PE = 4(9.81).5 = 19.62 J

KE = 19.62 J = .5(4)v2, v = 3.13 m/s

KE = maxPE = 0

KE = 0PE = max

KE = 0PE = max 1 release 3

2

.5m


1.0

1.5

2.0

2.5

3.0 29.4

24.5

19.6

14.7

9.8

0

3.1

4.4

5.4

6.3 19.6

14.7

9.8

4.9

0

Ht (m) PE (J) v (m/s) KE (J)

time Special ForceSpecial ForceApplicationApplication

Conversion of PE to KE and vice-versaStart with no KE b/c v=0

You can use this principle of energy conservation to solve useful problems. For instance – how high would you need to raise the slide in the picture below to successfully prevent the slider from running off the end?

TEinitial = TEfinal TE = PE + KE

PEinitial + KEinitial = PEfinal + KEfinal

m(9.8 m/s2)(4.0 m)+ ½m(8 m/s)2 = m(9.8 m/s2)(h)+ ½m(0 m/s)2

h = 7.6 m

• If you lift a barbell into the air you are performing work on the barbell. You apply a force over a distance.

• By performing work on the barbell you change the amount of PE that the barbell has.

PE = mgh1

PE = mgh2

W = E

Work-Energy Relationship


• If you lift a 300 Newton barbell 1 meter you have a change in energy of 300 J. This is equal to the amount of work done.

PE1 = (300)0 = 0 J

PE2 = (300)1 = 300 J

PE = mgh1

PE = mgh2

W = E


W = TE = PE+ KE+ SE

Work-Energy Relationship

When you ski down a slope you begin with only PE that is converted to KE as go down the hill. Assuming friction is negligible the TE will not change. At the bottom of the hill you have only KE which means you are moving FAST. In order to slow down you have to dissipate this kinetic energy. This requires a change in TE. By the work-energy relationship, work must be performed on the skier to bring him/her to a stop. To perform work you need a force. In this case it is the friction developed between the skier and some unpacked snow at the end of the run.

AAAHHHGGG – I’M BACK IN DRIVER’S ED!

Three people are driving the same type of car.

Driver A is traveling at 10 mph.

Driver B is traveling at 20 mph.

Driver C is traveling at 30 mph.

What can you say about the relationship of the stopping distance between vehicles? (Is it linear or something else?)

The cars have no change in PE so only KE must be changed. Thus the work necessary to bring the cars to rest is equal to the change in KE:

W = KEF * d = ½ m(v)

So the stopping distance (d) is proportional to the square of velocityd v2

Kinetic Energy to Strain Energy• Kinetic energy will be used to

deform the elastic tissues. Strain energy can then be transformed back into kinetic energy during the pushoff phase.

Dawson and Taylor, 1973

• At low speeds, kangaroos use a pentapedal form of locomotion (four feet and a tail).

• At 6-7 km/hr the switch to hopping.

• The rate of O2 consumption increases sharply with speed during pentapedal locomotion.

• When they begin to hop, the rate of O2 consumption decreases with increasing speed until 18 km/hr.

Why?


Kinetic to Strain to Kinetic

• Kangaroos have large Achilles tendons (1.5 cm in diameter and 35 cm in length).

• It is possible that a large amount of strain energy is stored in the tendon during the landing and converted back into kinetic energy as the kangaroo rebounds into the air.


Collisions and Impacts

Elastic

Inelastic

Coefficient of Restitution• The coefficient of restitution (e) determines the

elasticity of an impact.

• If e = 1 the impact is completely elastic. This means that the object contains all of the energy it had before the impact.

• If e = 0 the impact is completely plastic. This means that the object contains none of the energy it had before the impact.


Objects typically deform during an impact

This is referred to as the period of deformation

This is followed by a period of restitutionthus the name coefficient of restitution

Coefficient of Restitution

• e = -

• e = h

h

drop

bounce

Power• Power is the rate of performing work.

P =

P =

P = F *

P = F*v

t

W

t

s*F

t

s


Power

• Units are Joules/second or Watts.

• Greater power must be developed in order to do mechanical work more quickly.

• Power can be positive or negative depending on whether F and v point in same general direction (+ power) or in opposite directions (- power)


Power

• Positive power indicates that energy is being generated and negative power indicates that energy is being absorbed thus:

• Positive power is associated with concentric muscular contractions, while negative power is associated with eccentric muscular contractions.


Percent of Stance

-20

-15

-10

-5

0

5

10

15

0 10 20 30 40 50 60 70 80 90 100

Pow

er (

Wat

ts·k

g-1)

EnergyAbsorption

EnergyGeneration

Knee Power During Running


Power Examplem = 100 kg, g = 9.8 m/s2, h = 2 mW = mgh = 100 (9.8) 2 = 1960 J

now add timeCase 1: raise the barbell slowly -- t = 5 s

W392s 5

J 1960t

WP

Case 2: raise the barbell quickly -- t = 1.5 s

W 1307s 5.1J 1960

tWP


Force force - a push or pull acting on a body forces are vector quantities –magnitude (size) –direction (line of action) –point of application measured.

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