forces chapter 6 pages: 116-147. force a force is a push or pull upon an object resulting from the...

ForcesForces

Chapter 6Chapter 6

Pages: 116-147Pages: 116-147

ForceForce

A force is a A force is a pushpush or or pullpull upon an object resulting upon an object resulting

from the object's interaction from the object's interaction with another object. with another object.

Contact ForcesContact ForcesLong-Range Forces Long-Range Forces

Contact ForcesContact ForcesContact forcesContact forces are types are types of forces in which the two of forces in which the two interacting objects are interacting objects are physically in contact with physically in contact with each other. each other.

FrictionFriction

TensionTension

Air

Air

Resi

stance

Resi

stance

Long-Range ForcesLong-Range ForcesLong-Range Forces are types of Long-Range Forces are types of

forces in which the two interacting forces in which the two interacting objects are not in physical contact objects are not in physical contact

with each other, but are able to with each other, but are able to exert a push or pull despite the exert a push or pull despite the

physical separation. physical separation.

Grav

ity

Magne

tism

Electri

cal

ForceForce

ForceForceFF for use in equations. for use in equations.

NewtonNewton is the unit for Force. is the unit for Force.

NN abbreviation for Newton. abbreviation for Newton.

Net ForceNet Force can can accelerateaccelerate..

N = kg m/sN = kg m/s22

ForceForce is a is a VectorVector QuantityQuantity

MagnitudeMagnitudeDirectionDirection

4000lb

Read Pages 118-119Read Pages 118-119Answer Question 1 in Answer Question 1 in

NotebookNotebook

HomeworkHomework

Page: 124Page: 124

Questions: 7-11Questions: 7-11

Free Body DiagramsFree Body Diagrams

The purpose of a free-body force diagram is to assist you in trying to determine the net force acting on a body.

Net ForceNet Force

The purpose of a free-body force diagram is to assist you in trying to determine the net force acting on a body.

Free Body DiagramsFree Body Diagrams

The net force is the vector sum of all the individual forces acting on a system.

Fnet = F1 ± F2 ± F3 ± F4 …

Constructing Constructing “free-body force diagram” “free-body force diagram”

1.1. Identify the object(s) you Identify the object(s) you will draw a diagram for.  will draw a diagram for. 

Constructing Constructing “free-body force diagram” “free-body force diagram”

2.2. Identify all the forces acting Identify all the forces acting directlydirectly on the object and on the object and the object exerting them.the object exerting them.

GravityGravityTable Table

Constructing Constructing “free-body force diagram” “free-body force diagram”

3.3.Draw a dot to represent Draw a dot to represent the object of interest. the object of interest.

Constructing Constructing “free-body force diagram” “free-body force diagram”

4. 4. Draw a vector to Draw a vector to represent each force.represent each force.

Gravity

Table

Constructing Constructing “free-body force diagram” “free-body force diagram”

5. 5. If the object is If the object is stationary or is moving stationary or is moving at a constant velocity, at a constant velocity, the vectors should the vectors should graphically add up to graphically add up to zero. zero. 

Constructing Constructing “free-body force diagram” “free-body force diagram”

5. 5.   If the object is   If the object is accelerating, the sum of accelerating, the sum of the vectors should the vectors should produce a vector in the produce a vector in the same direction as the same direction as the acceleration. acceleration.

Constructing Constructing “free-body force diagram” “free-body force diagram”

Gravity

Floor

Standing on Standing on FloorFloor

FFfloorfloor=F=Fgravitygravity

Constructing Constructing “free-body force diagram” “free-body force diagram”

Gravity

Muscle

JumpingJumping FFmusclemuscle>F>Fgravitygravity

Constructing Constructing “free-body force diagram” “free-body force diagram”

Gravity

In the AirIn the Air FFgravitygravity

Types of MotionTypes of Motion

No MotionNo Motion

Gravity

RoadBrakesMotor

FFRoadRoad=F=FGravityGravity

FFMotorMotor=F=FBrakesBrakesNo Net No Net ForceForce

Types of MotionTypes of Motion

Constant VelocityConstant Velocity

Gravity

RoadFrictionMotor

FFRoadRoad=F=FGravityGravity

FFMotorMotor=F=FFrictionFrictionNo Net No Net ForceForce

Types of MotionTypes of Motion

Speeding UpSpeeding Up

Gravity

RoadFrictionMotor

FFRoadRoad=F=FGravityGravity

FFMotorMotor>F>FFrictionFriction Net ForceNet Force

Types of MotionTypes of Motion

Slowing DownSlowing Down

Gravity

RoadFrictionMotor

FFRoadRoad=F=FGravityGravity

FFMotorMotor<F<FFrictionFriction Net ForceNet Force

Finding Net ForceFinding Net Force

FG=4000N

FR=4000NFF=400NFM=400N

VerticalVerticalFFnetnet = F = FRR - F - FGG

FFnetnet=4000N–=4000N–4000N4000N

FFnetnet= 0N= 0N

HorizontalHorizontalFFnetnet = F = FFF - F - FMM

FFnetnet=400N–400N=400N–400N

FFnetnet= 0N= 0N

Finding Net ForceFinding Net Force

FG=4000N

FR=4000NFF=400NFM=400N

Not Moving Not Moving OrOr

Constant Velocity Constant Velocity

Finding Net ForceFinding Net Force

FG=4000N

FR=4000NFF=100NFM=400N

VerticalVerticalFFnetnet = F = FRR - F - FGG

FFnetnet=4000N–=4000N–4000N4000N

FFnetnet= 0N= 0N

HorizontalHorizontalFFnetnet = F = FFF - F - FMM

FFnetnet=100N–400N=100N–400N

FFnetnet= -300N= -300N

Finding Net ForceFinding Net Force

FG=4000N

FR=4000NFF=100NFM=400N

AcceleratingAcceleratingto the Left. to the Left.

Newton’s Second Law of Newton’s Second Law of MotionMotion

Law Law

ofof

AccelerationAcceleration

Newton’s Second Law of Newton’s Second Law of MotionMotion

The acceleration of an object as produced by a net force is directly proportional to the

magnitude of the net force, in the same direction as the net

force, and inversely proportional to the mass of the

object.

Newton’s Second Law of Newton’s Second Law of MotionMotion

FFff = = 40N40N

FFPP = = 41N41N

FFNetNet = = 1N1N

AccelerateAcceleratess

Newton’s Second Law of Newton’s Second Law of MotionMotion

FFNetNet = ma = maFFmm

aa

Newton’s First Law of MotionNewton’s First Law of Motion

Law Law

ofof

InertiaInertia

InertiaInertia

Inertia is the resistance an object has to a change in its

state of motion.

MassMass

Newton’s First Law of MotionNewton’s First Law of Motion

Seat

Seat

Belt!!!!

Belt!!!!

Newton’s First Law of Newton’s First Law of MotionMotion

An object at rest tends to An object at rest tends to stay at rest and an object in stay at rest and an object in motion tends to stay in motion tends to stay in motion with the same motion with the same speed and in the same speed and in the same direction direction unless acted upon unless acted upon by an unbalanced forceby an unbalanced force..

Mass vs. WeightMass vs. Weight

MassMass is the amount of stuff is the amount of stuff you are made up of. (kg or you are made up of. (kg or slugs) Does not change!!!!slugs) Does not change!!!!

Mass vs. WeightMass vs. WeightWeightWeight depends on how much depends on how much gravity is acting on you at the gravity is acting on you at the moment; you'd weigh less on moment; you'd weigh less on

the moon than on Earth. the moon than on Earth.

(newtons or pounds)(newtons or pounds)

Mass vs. WeightMass vs. Weight

WeightWeight

Mass vs. WeightMass vs. Weight

MasMasss

WeightWeight

To calculate weight use the To calculate weight use the acceleration due to gravity acceleration due to gravity (9.8m/s(9.8m/s22). This will be called g.). This will be called g.

F=ma Fg=mg

Weight ProblemsWeight Problems

Mr. Clune has a mass of Mr. Clune has a mass of 110kg. How much does he 110kg. How much does he weight?weight?Given: m=110kgGiven: m=110kg

g=9.8m/sg=9.8m/s22

Find: FFind: Fgg=?=?

Equation: FEquation: Fgg=mg=mg

=(110kg)(9.8m/s=(110kg)(9.8m/s22))

FFgg=1078N=1078N

F=ma ProblemsF=ma ProblemsA boy pulls a sled that has a mass of A boy pulls a sled that has a mass of

5kg across the snow. The sled 5kg across the snow. The sled accelerates at a rate of 0.5m/saccelerates at a rate of 0.5m/s22. What . What

is the net force of on the sled?is the net force of on the sled?

a=0.5m/sa=0.5m/s22

FFnetnet

Equation: FEquation: Fnetnet=mg=mg

=(5kg)(0.5m/s=(5kg)(0.5m/s22))

Given: m=5kgGiven: m=5kg

a=0.5m/a=0.5m/ss22

Find: FFind: Fnetnet=?=?

FFnetnet=2.5N=2.5N

F=ma ProblemsF=ma Problems

A rock with a mass of 10kg fell A rock with a mass of 10kg fell off a cliff. At a specific time off a cliff. At a specific time

during its’ fall it had an during its’ fall it had an acceleration of 3m/sacceleration of 3m/s22, due to , due to

air resistance. What is the force air resistance. What is the force of air on this rock at this time?of air on this rock at this time?

++

FF

gg

FFairair

aaFFnetnet

FFnetnet== FFair air + + FFg g FFnetnet= =

ma ma FFgg= = mg mg

FFairair== FFnet net - F- Fg g

FFairair== mama - - mgmgFFairair== m(am(a – – g)g) FFairair==

10kg{(-3m/s10kg{(-3m/s22)–(-9.8m/s)–(-9.8m/s22)})}FFairair==

10kg{(-3m/s10kg{(-3m/s22)+(9.8m/s)+(9.8m/s22)})}FFairair== 10kg(6.8m/s10kg(6.8m/s22))

FFairair== 68N68N

HomeworkHomework

Page: 147Questions: 22, 27,29

Due: 10/25/06

FactorsFactors that determine that determine Friction Friction

Wei

ght

Wei

ght Moving

MovingSta

tionary

Sta

tionarySurface

Surface

Friction ForcesFriction Forces

Fg

FT

FPFf

FN

FFNN – Normal Force: This force – Normal Force: This force which will affect frictional which will affect frictional resistance is the component of resistance is the component of applied force which acts applied force which acts perpendicular or "normal" to perpendicular or "normal" to the surfaces which are in the surfaces which are in contact and is typically contact and is typically referred to as the normal force.referred to as the normal force.

Friction ForcesFriction Forces

FFTT – Surface Force: This – Surface Force: This force opposite the normal force opposite the normal force which is equal to this force which is equal to this force. force.

Friction ForcesFriction Forces

FFPP – Push or Pull Force: This – Push or Pull Force: This force is pushing or pulling force is pushing or pulling the object. the object.

Friction ForcesFriction Forces

FFff – Friction Force: – Friction Force: Frictional Frictional resistance to the relative resistance to the relative motion of two solid objects. motion of two solid objects.

Friction ForcesFriction Forces

FFfsfs – – StaticStatic Friction Force: Friction Force: Static frictional forces are Static frictional forces are non-moving forcesnon-moving forces between between two surfaces. It will increase two surfaces. It will increase to prevent any relative motion to prevent any relative motion up until some limit where up until some limit where motion occurs. motion occurs.

Friction ForcesFriction Forces

FFfkfk – – KineticKinetic Friction Force: Friction Force: The force between two The force between two surfaces that are surfaces that are movingmoving with respect to one another, with respect to one another, the frictional resistance is the frictional resistance is almost constant over a wide almost constant over a wide range of low speeds. range of low speeds.

Friction ForcesFriction Forces

μμ – – Coefficient of Coefficient of Friction: Friction: The The ratioratio of the of the force of friction (Fforce of friction (Fff)) between two bodies and between two bodies and the the force pressing them force pressing them together (Ftogether (FNN))..

Friction ForcesFriction Forces

Coefficient Coefficient of Friction of Friction

Ff

FN

μs= Ffs

FN

Coefficient Coefficient of Friction of Friction

Ff

FN

μk= Ffk

FN

Friction ProblemFriction Problem

A refrigerator of total weight 400N A refrigerator of total weight 400N is pushed at a constant speed is pushed at a constant speed across a room by pushing across a room by pushing horizontally on one side with a horizontally on one side with a force of 160N. What is the force of 160N. What is the coefficient of kinetic friction?coefficient of kinetic friction?

FfkFN

Ffk = 160NFN = 400Nμk = ?

μk=

Ffk

FN

μk=

140N400N

μk= 0.35

If the coefficient of static friction between If the coefficient of static friction between the floor and the refrigerator was 0.6, the floor and the refrigerator was 0.6, how much force would be needed to how much force would be needed to start the refrigerator moving?start the refrigerator moving?

Ffs = ?

FN = 400Nμs = 0.6 μs=

Ffs

FN

= μk Ffs FN

= (0.6) Ffs(400N)

= μk Ffs FN

= 240N Ffs

HomeworkHomeworkPage: 133

Questions: 14,15 Page: 145

Questions: 33-35Due: 11/2/06

Newton’s Third LawNewton’s Third Law

"For every action, "For every action, there is an equal there is an equal

and opposite and opposite reaction." reaction."

While driving, Anna Litical While driving, Anna Litical observed a bug striking the observed a bug striking the

windshield of her car. Obviously, windshield of her car. Obviously, a case of Newton's third law of a case of Newton's third law of

motion. The bug hit the motion. The bug hit the windshield and the windshield hit windshield and the windshield hit the bug. Which of the two forces the bug. Which of the two forces is greater: the force on the bug is greater: the force on the bug or the force on the windshield? or the force on the windshield?

Rockets are unable to accelerate in Rockets are unable to accelerate in space because ...space because ... There is no air in space for the There is no air in space for the rockets to push off of. rockets to push off of.

There is no gravity is in space. There is no gravity is in space.

There is no air resistance in There is no air resistance in space. space.

... nonsense! Rockets do ... nonsense! Rockets do accelerate in space. accelerate in space.

A gun recoils when it is fired. The recoil is the result of action-reaction force pairs. As the gases from the gunpowder explosion expand, the gun pushes the bullet forwards and the bullet pushes the gun backwards. The acceleration of the recoiling gun is ...

a.greater than the acceleration of the bullet. b.smaller than the acceleration of the bullet. c.the same size as the acceleration of the bullet.

In the top picture, a physics student is pulling upon a rope

which is attached to a wall. In the bottom picture, the physics

student is pulling upon a rope which is held by the Strongman.

In each case, the force scale reads 500 Newtons. The physics

student is pulling…

with more force when the rope is attached to the wall.

with more force when the rope is attached to the Strongman. the same force in each case.