Compare and explain in complete sentences and formulas

how the third Newton’s law is applied to find

the resultant force.

Forces:

Maintaining Equilibrium or Changing Motion

FINISH 50 % OF THE PROJECT

ForceForce: a push or pull acting on a body that

causes or tends to cause a change in the linear motion of the body Characteristics of a force

magnitude direction point of application. line of action

Net Force: resultant force (overall effect of multiple forces acting on a body) Example: push from side and front = at angle

ForceFree body diagram - sketch that shows a

defined system in isolation with all the force vectors acting on the system.

Classifying ForcesInternal Force: acts within the object or

system whose motion is being investigatedaction / reaction forces both act on different

parts of the system tensile-internal pulling forces when the structure is

under tension compressive- internal pushing (squeezing) forces act

on the ends of an internal structure

do not accelerate the body Orientate segments, maintain structural integrity

Internal ForcesExamples

Contraction of musclesDo not accelerate the body

Classifying ForcesExternal Force: acts on object as a result

of interaction with the environment surrounding itnon-contact - occur even if objects are not

touching each other gravity, magnetic

contact - occur between objects in contact fluid (air & water resistance) reaction forces with another body (ground,

implement) vertical (normal) reaction force

acts perpendicular to bodies in contact shear reaction force

acts parallel to surfaces in contact (friction)

F = maForce may also be defined as the product of a

body's mass and the acceleration of that body resulting from the application of the force.

Units of force are units of mass multiplied by units of acceleration.

Units of ForceMetric system (systeme internationale -SI)

Newton (N) the amount of force necessary to accelerate a

mass of 1 kg at 1 m/s2

English systempound (lb)

the amount of force necessary to accelerate a mass of 1 slug at 1 ft/s2

equal to 4.45 N

Weight (external force)Weight - the amount of gravitational force

exerted on a body. wt=mag.

Since weight is a force, units of weight are units of force - either N or lb.

As the mass of a body increases, its weight increases proportionally.

WeightThe factor of proportionality is the

acceleration of gravity, which is -9.81m/s2 or - 32 ft/s2.

The negative sign indicates that the acceleration of gravity is directed downward or toward the center of the earth.

WeightOn the moon or another planet with a

different gravitational acceleration, a body's weight would be different, although its mass would remain the same.Space station example

WeightBecause weight is a force, it is also

characterized by magnitude, direction, and point of application.

The direction in which weight acts is always toward the center of the earth.

Center of Weight (Gravity) The point at which weight is assumed to act

on a body is the body's center of gravity.

Friction (external force)Component of a contact force that acts

parallel to the surface in contactacts opposite to motion or motion tendencyreflects interaction between molecules in

contactreflects force “squeezing” surfaces togetheracts at the area of contact between two

surfaces

Static friction: surfaces not moving relative to each otherMaximum static friction: maximum amount of

friction that can be generated between two static surfaces

Dynamic friction: surfaces move relative to each otherconstant magnitude friction during motion

always less than maximum static friction

Friction

F = NFriction depends on

Nature of materials in contact ()

Force squeezing bodies together (N)Known as the normal contact (reaction) force

Friction is FUN

Nature of materials in contactCoefficient of friction ()

value serves as an index of the interaction between two surfaces in contact.

N

F

Coefficient of Friction ()

Factors Affecting FrictionThe greater the coefficient of friction, the

greater the friction

The greater the normal contact force, the greater the friction

Normal reaction force (NRF)AKA - normal contact forceforce acting perpendicular to two surfaces in

contact.magnitude intentionally altered to increase or

decrease the amount of friction present in a particular situation football coach on sled push (pull) upward to slide object pivot turn on ball of foot

Reaction Force

Examples of manipulating and Nc

shoe design

Examples of manipulating and Nc

•shoe design•grips (gloves, tape, sprays, chalk)•skiing: decrease for speed, increase for safety•curling•your examples???

Friction and Surface AreaFriction force is proportional to the normal

contact forceFriction is not affected by the size of the

surface area in contactnormal contact force distributed over the area

in contactFriction is affected by the nature of the

materials in contact

Friction and Surface AreaWith dry friction, the amount of surface area

in contact does not impact the amount of friction.

Same force acting over more area

Same force acting over smaller area

Coefficient of FrictionHot, soft, rough surfaces have higher

coefficients of frictionTires, concrete, etc

Cold, hard, smooth surfaces have lower coefficients of frictionIce, marble, etc

FrictionCalculate the force of friction when you slide

on ice.

Given = Normal force of 1000NC of F = 0.02

SolutionCalculate the force of friction when you slide

on ice.Given = Normal force of 1000NC of F = 0.02F = µ * NF = (0.02) * (1000N)F = 20 N

ForceForce: a push or pull acting on a body that

causes or tends to cause a change in the linear motion of the body (an acceleration of the body) Characteristics of a force

magnitude direction point of application line of action sense (push or pull along the line of action)

Vector represented with an arrow

RecallConcept of Net External Force

Must add all forces acting on an object together

Free body diagramFree body diagram - sketch that shows a

defined system in isolation with all the force vectors acting on the systemdefined system: the body of interestvector: arrow to represent a force

length: size of the force tip: indicates direction location: point of application

Free body diagramMe, at rest in front of class

sagittal plane view

What are the names of the forces? How big are the forces?What direction are the forces?Where are the forces applied?

Note: link contains practice problems

Free body diagram ==>static analysis

Me (mass= 75 kg): at rest (a = 0) in front of class (assume gravity at -10m/s/s)sagittal plane view

Wt & vGRF C of M, at feet750N, ?????- & +

F = m aF = 0Wt + vGRF = 0vGRF = - WtvGRF = - (-750)vGRF = 750 N

Note: link contains practice problems

Addition of Forces(calculating the net [resultant] force)

Net force = vector sum of all external forces acting on the object (body)account for magnitude and directione.g., Add force of 100N and 200N

Addition of Forces(calculating the net [resultant] force)Net force = vector sum of all external forces

acting on the object (body)account for magnitude and directione.g., Add force of 100N and 200N

act in same direction??? Act in opposite direction??? Act orthogonal to each other??? Act at angles to each other???

Colinear forcesForces have the same line of actionMay act in same or different directions

ie tug of war teammates: 100N, 200N, 400N show force on rope graphically

Colinear forcesForces have the same line of actionMay act in same or different directions

ie tug of war teammates: 100N, 200N, 400Ntug of war opponents: 200N, 200N, 200N

show force on rope graphically

Colinear forcesForces have the same line of actionMay act in same or different directions

ie tug of war teammates: 100N, 200N, 400Ntug of war opponents: 200N, 200N, 200Ncalculate resultant of the two teams

show force on rope graphically calculate algebraically

Free body diagram ==>static analysisWeightlifter (mass 80 kg)100 kg bar overhead at rest (a = 0)sagittal plane view

What are the forces? Where are the forces applied?How big are the forces?What direction are the forces?

Free body diagramsa) Weightlifterb) bar

Fig 1.19,

p 42

Free body diagram ==>static analysisWeightlifter (80 kg)100 kg bar overhead at rest (a = 0)sagittal plane view

What are the forces? Where are the forces applied?How big are the forces?What direction are the forces?

•Wt & vGRF •CofM, at feet•800N, ?????•- & +

Free body diagram ==>static analysisWeightlifter (80 kg)100 kg bar overhead at rest (a = 0)sagittal plane view

Wt, bar & vGRF CofM, on hands, at feet800N, 1000N, ?????- , -, +

F = m aF = 0Wt + bar + vGRF = 0vGRF = - Wt - barvGRF = - (-800) - (-1000)vGRF = + 1800 N

Concurrent ForcesForces do not act along same line, but do

act through the same pointie gymnast jumps up to grab bar. Coach

stops swinging by applying force to front and back of torso.20 N posterior directed push on front of torso30 N anterior directed push on back of torso550N force from bar on gymnast’s handsgymnast mass 50 kg

Concurrent Forcesgymnast hanging from grab

bar. Coach applies force to front and back of torso to stop swing.20 N posterior directed push on

front of torso30 N anterior directed push on

back of torso550N force from bar on

gymnast’s handsgymnast mass 50 kg

Page 29 in book

Concurrent Forcesgymnast hanging from grab bar. Coach

applies force to front and back of torso to stop swing.20 N posterior directed push on front of

torso30 N anterior directed push on back of torso550N force from bar on gymnast’s handsgymnast mass 50 kg

What is the resultant force?Tip to tail method (fig 1.8 & 1.9 in text)separate algebraic summation of horizontal

and negative forces Pythagorean theorem to solve resultant

magnitude Inverse tangent to solve direction (angle)

Horizontal forces:

20N – 30N = -10N

Vertical forces:

-500N + 550N = 50N

Resultant force:

a2 + b2 = c2

(-10N)2 + (50N)2 = c2

(100N2) + (2500N2) = c2

2600N2 = c2

C = 51N

Quantifying KineticsVector composition - process of determining

a single vector from 2 or more vectors through vector addition.

Resultant - single vector that results from vector composition.

Quantifying KineticsVector resolution - breaking down a resultant

vector into its horizontal and vertical components.Graphic method.Trigonometric method.

Graphic Method

Trigonometry: SOH, CAH, TOA

Key Equations: sin = opp/hypcos = adj/hyptan = opp/adj

Pythagoras theorema2 + b2 = c2

hyp (c)

adj

opp

Check this site.

Vector CompositionSample: = 30 degreesadj = 100 NFind opp and hyp

hyp (c)

adj

opp

Vector Compositionusecos = adj/hyptan = opp/adj hyp (c)

adj

opp

Vector Compositioncos = adj/hypcos 30 = 100/hyphyp = 100/.866hyp = 115.47 N

tan = opp/adjtan 30 = opp/100opp = .5774 x 100opp = 57.74 N

hyp (c)

adj

opp

Vector ResolutionSample = 35 degreeshyp = 120 NFind opp and adj

hyp (c)

adj

opp

Trigonometric CalculationsUsesin = opp/hypcos = adj/hyp

hyp (c)

adj

opp

Vector Resolutionsin = opp/hypsin 35 = opp/120opp = 120 X .5736opp = 68.83 N

cos = adj/hypcos 35 = adj/120adj = 120 X .8192adj = 98.30 N

hyp (c)

adj

opp

Note orientation of stance leg.

Note orientation of stance leg.

Note orientation of stance leg.

Resolution of ForcesForces are not colinear and not Hor & Vert?

ie figure 1.1: forces on shot 100 N from shot-putters hand mass of shot = 4 kg

What is the resultant force on shot? Draw Components of 100N force

solve graphically: tedious & imprecise trigonometric technique

100N wt

Resolution of ForcesForces are not colinear and not concurrent

ie figure 1.1: forces on shot 100 N from shot-putters hand mass of shot = 4 kg W = mg = (4kg)(-10m/s/s)= -40 N

What is the resultant force on shot? Draw Components of 100N force

solve graphically: tedious & imprecise trigonometric technique

100Nwt-40N

60o

SolutionGivenHyp = 100NAngle = 60 degreesSin angle = opp/hypSin 60 = opp/hypSin 60*hyp = opp(0.866)*(100N)=

86.6N

GivenHyp = 100NAngle = 60 degCos angle = adj/hypCos 60 = adj/hypCos 60*hyp = adj(0.500)*(100N) = 50N

Free body diagram ==>static analysisChild on swing (20 kg)Mother’s force

40 N horizontal10 N upward

at rest (a = 0)force of swing on child?

Free body diagram ==>static analysisChild on swing (mass = 20 kg)Mother’s force

40 N horizontal10 N upward

at rest (a = 0)force of swing on child?

Fx = m ax Fy = m ay

SolutionFx = Rx + 40N = 0

Rx = -40N

Fy = Ry + 10N + (-200N) = 0

Ry = 190N