Newton’s Laws MONDAY, September 14, Re-introduction to Newton’s 3 Laws

Newton’s LawsNewton’s LawsNewton’s Newton’s

LawsLaws

MONDAY, September 14, MONDAY, September 14,

Re-introduction toRe-introduction to

Newton’s 3 LawsNewton’s 3 Laws

What is Force?What is Force? A push or pull on an A push or pull on an object.object.

Unbalanced forces Unbalanced forces cause an object to cause an object to accelerate…accelerate… To speed upTo speed up To slow downTo slow down To change directionTo change direction

Force is a vector!Force is a vector!

Types of ForcesTypes of Forces Contact forces: involve contact Contact forces: involve contact

between bodies.between bodies. Normal, FrictionNormal, Friction

Field forces: act without necessity of Field forces: act without necessity of contact.contact. Gravity, Electromagnetic, Strong, WeakGravity, Electromagnetic, Strong, Weak Dark side, light side, etc.Dark side, light side, etc.

Question: Is there Question: Is there reallyreally any any such thing as a contact force?such thing as a contact force?

A.A. yea. yea.

B.B. No.No.

C.C. I think this is a trick so I’m not I think this is a trick so I’m not answering.answering.

If the If the net forcenet force on a body is on a body is zero then it must….zero then it must….

A.A. Be At restBe At rest

B.B. Have a Constant VelocityHave a Constant Velocity

C.C. Not be touching the groundNot be touching the ground

D.D. Not be a jediNot be a jedi

Forces and EquilibriumForces and Equilibrium

If the If the net forcenet force on a body is zero, it is in on a body is zero, it is in equilibriumequilibrium..

An object in equilibrium may be moving An object in equilibrium may be moving relative to us (dynamic equilibrium).relative to us (dynamic equilibrium).

An object in equilibrium may appear to An object in equilibrium may appear to be at rest (static equilibrium).be at rest (static equilibrium).

Galileo’s Thought Galileo’s Thought ExperimentExperiment

This thought experiment lead to Newton’s First Law.

Galileo’s Thought Galileo’s Thought ExperimentExperiment

© The Physics Classroom, Tom Henderson 1996-2007

Newton’s First LawNewton’s First Law

The Law of Inertia.The Law of Inertia. A body at rest stays at rest unless A body at rest stays at rest unless

acted upon by an external forceacted upon by an external force A body in motion stays in motion in a A body in motion stays in motion in a

straight line unless acted upon by an straight line unless acted upon by an external force.external force.

Or.. Or.. Mass wants to do its thing.Mass wants to do its thing. This law is commonly applied to the This law is commonly applied to the

horizontal component of velocity, which is horizontal component of velocity, which is assumed not to change during the flight of assumed not to change during the flight of a projectile.a projectile.

Note on InertiaNote on Inertia Inertia is a property of matter.Inertia is a property of matter. Forget any definition of mass that you Forget any definition of mass that you

learned in Chemistry.learned in Chemistry. Mass is a measure of an objects inertia.Mass is a measure of an objects inertia. It is the resistance to changes in motionIt is the resistance to changes in motion

An objects mass cannot change An objects mass cannot change unless you actually change the unless you actually change the

amount of matter presentamount of matter present

A.A. True- I believe in conservation of massTrue- I believe in conservation of mass

B.B. False- I am a non-believer and think it’s False- I am a non-believer and think it’s more complicated than that.more complicated than that.

Newton’s Second LawNewton’s Second Law

A body accelerates when acted upon A body accelerates when acted upon by a net external force by a net external force

The acceleration is proportional to The acceleration is proportional to the the netnet (or resultant) force (or resultant) force andand is in is in the direction which the net force acts.the direction which the net force acts.

This law is commonly applied to the This law is commonly applied to the vertical component of velocity.vertical component of velocity.

FF = m = maa

Newton’s Third LawNewton’s Third Law

For every action there exists an For every action there exists an equal and opposite reaction.equal and opposite reaction.

If A exerts a force If A exerts a force FF on B, then B on B, then B exerts a force of -exerts a force of -FF on A. on A.

Or Or Action Force = -Reaction ForceAction Force = -Reaction Force

Commonly Confused TermsCommonly Confused Terms

Inertia:Inertia: or the resistance of an object to or the resistance of an object to being acceleratedbeing accelerated

Mass:Mass: the same thing as inertia (to a the same thing as inertia (to a physicist).physicist).

Weight:Weight: gravitational attraction gravitational attraction

We We normalnormally use the term “weightless” to ly use the term “weightless” to describe the feeling that astronauts get describe the feeling that astronauts get

when they are in orbit or in simulators like when they are in orbit or in simulators like the Vomit Comet. What does “weightless” the Vomit Comet. What does “weightless”

really mean when applied to these really mean when applied to these situations?situations?

A.A. You have no weightYou have no weightB.B. You have weight but it is not opposed by You have weight but it is not opposed by

another force.another force.C.C. You have no massYou have no massD.D. You have been eating dehydrated space You have been eating dehydrated space

food so you don’t weigh as much.food so you don’t weigh as much.

Only Two forces, FOnly Two forces, F11 and F and F22 act upon a act upon a body of mass 3.0 kg. The body moves body of mass 3.0 kg. The body moves at constant speed. What do you know at constant speed. What do you know

must be true?must be true?

A.A. The net force must point in the The net force must point in the direction of motiondirection of motion

B.B. The object must be weightless.The object must be weightless.

C.C. The vectors FThe vectors F11 and F and F22 must be 180 must be 180° ° opposite each other.opposite each other.

D.D. The sum of forces FThe sum of forces F11 and F and F22 must must be zerobe zero

Two forces, Two forces, FF11 = (4i – 6j + k) N and F = (4i – 6j + k) N and F22 = (i – 2j - 8k) N, act = (i – 2j - 8k) N, act upon a body of mass 3.0 kg. No other forces upon a body of mass 3.0 kg. No other forces act upon the body at this time. What do you act upon the body at this time. What do you

know must be true?know must be true?

A.A. The body’s motion will not change.The body’s motion will not change.

B.B. The body’s motion will change.The body’s motion will change.

C.C. Today is Saturday.Today is Saturday.

D.D. There must be another force.There must be another force.

A tug-of-war team ties a rope to a tree and pulls A tug-of-war team ties a rope to a tree and pulls horizontally to create a tension of 30,000 N in the horizontally to create a tension of 30,000 N in the rope. Suppose the team pulls equally hard when, rope. Suppose the team pulls equally hard when,

instead of a tree, the other end of the rope is being instead of a tree, the other end of the rope is being pulled by another tug-of-war team such that no pulled by another tug-of-war team such that no

movement occurs. What is the tension in the rope movement occurs. What is the tension in the rope in the second case?in the second case?

A.A. Less than 30,000 NLess than 30,000 N

B.B. 30,000 N30,000 N

C.C. More than 30,000 NMore than 30,000 N

D.D. The rope will breakThe rope will break

Working a 2Working a 2ndnd Law Problem Law Problem

Working Newton’s 2Working Newton’s 2ndnd Law Problems Law Problems is best accomplished in a systematic is best accomplished in a systematic fashion.fashion.

The more complicated the problem, The more complicated the problem, the more important it is to have a the more important it is to have a general procedure to follow in general procedure to follow in working it.working it.

22ndnd Law Procedure Law Procedure

1.1. Identify the body to be analyzed.Identify the body to be analyzed.2.2. Select a reference frame, stationary or Select a reference frame, stationary or

moving, but moving, but not acceleratingnot accelerating (2 (2ndnd law law holds for only inertial reference frames).holds for only inertial reference frames).

3.3. Draw a force or free body diagram.Draw a force or free body diagram.4.4. Set up Set up ΣΣFF = m = maa equations for each equations for each

dimension.dimension.5.5. Use kinematics or calculus where Use kinematics or calculus where

necessary to obtain acceleration.necessary to obtain acceleration.6.6. Substitute known quantities.Substitute known quantities.7.7. Calculate the unknown quantities.Calculate the unknown quantities.

A 5.00-g bullet leaves the muzzle of a rifle with a A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. The bullet is accelerated by speed of 320 m/s. The bullet is accelerated by

expanding gases while it travels down the 0.820 m expanding gases while it travels down the 0.820 m long barrel. Assuming constant acceleration and long barrel. Assuming constant acceleration and

negligible friction, what is the force on the bullet?negligible friction, what is the force on the bullet?

A.A. 160 N160 N

B.B. 320 N320 N

C.C. 640 N640 N

D.D. 1280 N1280 N

A 3.00 kg mass undergoes an acceleration given A 3.00 kg mass undergoes an acceleration given byby

aa = (2.50 = (2.50ii + 4.10 + 4.10jj) m/s) m/s22.. Find the resultant force Find the resultant force FF and its magnitude. and its magnitude.

A.A. 4.8 N 594.8 N 59° from horizontal° from horizontal

B.B. 14.4 N 3114.4 N 31° from horizontal° from horizontal

C.C. 4.8 N 314.8 N 31° from horizontal° from horizontal

D.D. 14.4 N 5914.4 N 59° from horizontal° from horizontal

Wednesday, September 17, Wednesday, September 17, 20092009

Types of Forces Commonly Types of Forces Commonly Found in Newton’s Law Found in Newton’s Law Problems and PracticeProblems and Practice

Normal forceNormal force

The force that keeps one object from The force that keeps one object from “invading” another object.“invading” another object. Ex: weight is the force of attraction of Ex: weight is the force of attraction of

our body for the center of the planet.our body for the center of the planet. We don’t fall to the center of the planet.We don’t fall to the center of the planet. The normal force keeps us up.The normal force keeps us up.

It It ALWAYSALWAYS acts acts perpendicularperpendicular to to contact!!! NOT always vertical!!!!contact!!! NOT always vertical!!!!

Normal ForceNormal Forceon Flat surfaceon Flat surface

Normal Forceon Flat surface

Fn = -(mg) for objects resting on horizontal surfaces (no vertical acceleration).

mg

N

Normal Force on Ramp

Normal Force on Ramp

mg

The normal force is perpendicular to angled

ramps as well. It’s usually equal to the

component of weight perpendicular to the

surface.

N

mgcos

mgsin

N = -(mgcos)

What will acceleration be in this situation?

A. mgsinB. mgcosC. gsinD. gcosE. g

N

mg

mgcos

mgsin

Normal Force on RampNormal Force on Ramp

mg

How could you keep the block from accelerating?

N

mgcos

mgsin

N = -(mgcos)

F

TensionTension

A pulling force.A pulling force. Generally exists in a rope, string, Generally exists in a rope, string,

or cable.or cable. Arises at the molecular level, Arises at the molecular level,

when a rope, string, or cable when a rope, string, or cable resists being pulled apart.resists being pulled apart.

Tension (static 2D)Tension (static 2D)

The horizontal and vertical components The horizontal and vertical components of the tension are equal to zero if the of the tension are equal to zero if the system is not accelerating.system is not accelerating.

15 kg

30o 45o

32 1

Tension (static 2D)Tension (static 2D)

The horizontal and vertical components The horizontal and vertical components of the tension are equal to zero if the of the tension are equal to zero if the system is not accelerating.system is not accelerating.

15 kg

30o 45o

T1T2

T3

Fx = 0Fy = 0-T3

mg

32 1

A Mass m hangs by a cable from the A Mass m hangs by a cable from the bottom of an elevator. What is the bottom of an elevator. What is the

tension in the cable when the elevator tension in the cable when the elevator is stopped at a floor?is stopped at a floor?

A.A. ggB.B. mgmgC.C. Greater in Greater in

magnitude than magnitude than mgmg

D.D. Less magnitude Less magnitude than mgthan mg

E.E. 00m

Tension (elevator)Tension (elevator)

The sum of The sum of the forces the forces

must be zero must be zero or the mass or the mass

would would accelerate!accelerate!M

T

Mg


tension in the cable when the elevator tension in the cable when the elevator accelerates upward?accelerates upward?




E.E. 00m


T must be T must be greater than greater than mg in order mg in order to result in to result in an upward an upward

acceleration!acceleration!M

T

Mg


tension in the cable when the elevator tension in the cable when the elevator is moving at constant speed between is moving at constant speed between

floors?floors?A.A. ggB.B. mgmgC.C. Greater in Greater in



E.E. 00m


tension in the cable when the elevator tension in the cable when the elevator is slowing down at the top floor?is slowing down at the top floor?




E.E. 00m


Net force Net force must point must point

down in order down in order to accelerate to accelerate

downdownM

T

Mg

A Mass m hangs by a cable from the A Mass m hangs by a cable from the bottom of an elevator. What will the bottom of an elevator. What will the

tension in the cable be if it comes tension in the cable be if it comes attached from the elevator?attached from the elevator?




E.E. 00m


No opposing No opposing force! force!

(weightless???)(weightless???)

M Mg

““Magic” pulleys simply bend the coordinate Magic” pulleys simply bend the coordinate system.system.

Pulley problemsPulley problems

m1

m2

-x

x

““Magic” pulleys simply bend the coordinate Magic” pulleys simply bend the coordinate system.system.


m1

m2

m1g

NT T

m2g

F = mam2g= (m1+m2)a

All problems should be started from a All problems should be started from a force diagram.force diagram.


m 1 m2

Tension is determined by examining Tension is determined by examining one block or the otherone block or the other


m 1 m2

m1g

NT

T

m2g

Fm1+m2)am2g – T+T – m1gsin

= (m1+m2)a

Tension is determined by examining Tension is determined by examining one block or the otherone block or the other


m 1 m2

m1g

NT

T

m2g

F2m2am2g - T = m2a

F1m1am1gsin = m1a

Atwood machineAtwood machine

A device for measuring g.A device for measuring g. If m1 and m2 are nearly the If m1 and m2 are nearly the

same, slows down freefall such same, slows down freefall such that acceleration can be that acceleration can be measured.measured.

Then, g can be measured.Then, g can be measured.m1

m2

Atwood machineAtwood machine

A device for measuring g.A device for measuring g. If m1 and m2 are nearly the If m1 and m2 are nearly the

same, slows down freefall such same, slows down freefall such that acceleration can be that acceleration can be measured.measured.

Then, g can be measured.Then, g can be measured.m1

m2T

m1g

T

m2gFmam2g-m1g = (m2+m1)a

Thursday, September 20, Thursday, September 20, 20072007

Post-Test ReviewPost-Test Review

Friday, September 21, Friday, September 21, 20072007

WorkdayWorkday

Exam CorrectionsExam Corrections

Homework ReviewHomework Review

Atwood Machine Mini-labAtwood Machine Mini-lab

Draw diagram of apparatus in lab Draw diagram of apparatus in lab book.book.

Record all data.Record all data. Calculate Calculate gg..

Easy ProblemHow fast will the block be sliding at the bottom of the frictionless ramp?

20o

L = 12 m5.0 kg

Easy ProblemHow high up the frictionless ramp will the block slide?

5.0 kg20o

v = 12.0 m/s

Tuesday, September 25, Tuesday, September 25, 20072007

FrictionFriction

Moderate ProblemDescribe acceleration of the 5 kg block. Table and pulley are magic and frictionless.

20o

5.0 kg

1.0 kg

FrictionFriction Friction opposes a sliding motion.Friction opposes a sliding motion. Static friction exists before Static friction exists before

sliding occurssliding occurs (f(fss ssN)N)..

Kinetic friction exists after sliding Kinetic friction exists after sliding occursoccurs ffkk = = kkNN

Friction on flat surfacesFriction on flat surfaces

x

y

Draw a free body diagram for a braking car.

x

y

Draw a free body diagram for a car accelerating from rest.

But we don’t want cars to But we don’t want cars to skid!skid!

Why don’t we?Why don’t we? Let’s use DataStudio to see if we can Let’s use DataStudio to see if we can

detect the difference in magnitude detect the difference in magnitude between static and kinetic friction.between static and kinetic friction.

Wednesday, September 26, Wednesday, September 26, 20072007

Friction LabFriction Lab

Thursday, September 27, Thursday, September 27, 20072007

Friction on a rampFriction on a ramp

Sliding down Sliding up

Friction is always parallel to surfaces….A 1.00 kg book is held against a wall by pressing it against the wall with a force of 50.00 N. What must be the minimum coefficient of friction between the book and the wall, such that the book does not slide down the wall?

F

W

f

N

(0.20)

Problem #1Assume a coefficient of static friction of 1.0 between tires and road. What is the minimum length of time it would take to

accelerate a car from 0 to 60 mph?

Problem #2Assume a coefficient of static friction of 1.0 between tires and road and a coefficient of kinetic friction of 0.80 between tires and road. How far would a car travel after the driver applies the brakes if it skids to a stop?

Centripetal ForceCentripetal Force Inwardly directed force which causes a Inwardly directed force which causes a

body to turn; perpendicular to velocity.body to turn; perpendicular to velocity. Centripetal force always arises from Centripetal force always arises from

other forces, and is not a unique kind other forces, and is not a unique kind of force.of force. Sources include gravity, friction, tension, Sources include gravity, friction, tension,

electromagnetic, normal.electromagnetic, normal. ΣΣF = maF = ma a = va = v22/r/r ΣΣF = m vF = m v22/r/r

Highway CurvesHighway Curves

R

r

z

Friction turns the vehicle

Normal force turns the vehicle

Sample problem• Find the minimum safe turning radius for a car traveling

at 60 mph on a flat roadway, assuming a coefficient of static friction of 0.70.

Sample problem• Derive the expression for the period best banking angle

of a roadway given the radius of curvature and the likely speed of the vehicles.

Friday, September 28, Friday, September 28, 20072007

More on Circular MotionMore on Circular Motion

Conical Pendulum

For conical pendulums, centripetal force is provided by a component of the tension.

z

r

T

mg

LT = 2 L cos

g

Sample problem• Derive the expression for the period of a conical

pendulum with respect to the string length and radius of rotation.

Non-uniform Circular MotionNon-uniform Circular Motion

Consider circular motion in which either speed Consider circular motion in which either speed of the rotating object is changing, or the forces of the rotating object is changing, or the forces on the rotating object are changing.on the rotating object are changing.

If the speed changes, there is a tangential as If the speed changes, there is a tangential as well as a centripetal component to the force.well as a centripetal component to the force.

In some cases, the magnitude of the centripetal In some cases, the magnitude of the centripetal force changes as the circular motion occurs.force changes as the circular motion occurs.

Sample problemSample problemYou swing a 0.25-kg rock in a vertical circle on a 0.80 m You swing a 0.25-kg rock in a vertical circle on a 0.80 m long rope at 2.0 Hz. What is the tension in the rope a) at long rope at 2.0 Hz. What is the tension in the rope a) at the top and b) at the bottom of your swing?the top and b) at the bottom of your swing?

Monday, October 1Monday, October 1

Time-dependent forcesTime-dependent forces

Sample problemSample problemA 40.0 kg child sits in a swing supported by 3.00 m long A 40.0 kg child sits in a swing supported by 3.00 m long chains. If the tension in each chain at the lowest point is chains. If the tension in each chain at the lowest point is 350 N, find a) the child’s speed at the lowest point and b) 350 N, find a) the child’s speed at the lowest point and b) the force exerted by the seat on the child at the lowest the force exerted by the seat on the child at the lowest point.point.

Sample problemSample problemA 900-kg automobile is traveling along a hilly road. If it is to A 900-kg automobile is traveling along a hilly road. If it is to remain with its wheels on the road, what is the maximum speed it remain with its wheels on the road, what is the maximum speed it can have as it tops a hill with a radius of curvature of 20.0 m?can have as it tops a hill with a radius of curvature of 20.0 m?

Non-constant ForcesNon-constant Forces

• Forces can vary with time.Forces can vary with time.• Forces can vary with velocity.Forces can vary with velocity.• Forces can vary with position.Forces can vary with position.

Calculus Concepts forCalculus Concepts forForces that Vary with TimeForces that Vary with Time

• DifferentiationDifferentiation• the tangent (or slope) of a functionthe tangent (or slope) of a function• position -> velocity -> accelerationposition -> velocity -> acceleration

• IntegrationIntegration• the area under a curvethe area under a curve• acceleration -> velocity -> positionacceleration -> velocity -> position

Evaluating IntegralsEvaluating Integrals

If a(t) = tn

then

tn dt = tn+1 / (n+1) + C

tn dt = tn+1 / (n+1) 0

t

0

t

Sample problemSample problem Consider a force that is a function of time:Consider a force that is a function of time:

F(t) = (3.0 t – 0.5 tF(t) = (3.0 t – 0.5 t22)N)N If this force acts upon a 0.2 kg particle at rest for 3.0 If this force acts upon a 0.2 kg particle at rest for 3.0

seconds, what is the resulting velocity and position of seconds, what is the resulting velocity and position of the particle?the particle?

Sample problemSample problem Consider a force that is a function of time:Consider a force that is a function of time:

F(t) = (16 tF(t) = (16 t22 – 8 t + 4)N – 8 t + 4)N If this force acts upon a 4 kg particle at rest for 1.0 If this force acts upon a 4 kg particle at rest for 1.0

seconds, what is the resulting change in velocity of seconds, what is the resulting change in velocity of the particle?the particle?

Drag ForcesDrag Forces

Drag forcesDrag forces slow an object down as it passes slow an object down as it passes

through a fluid.through a fluid. act in opposite direction to velocity.act in opposite direction to velocity. are functions of velocity.are functions of velocity. impose terminal velocity.impose terminal velocity.

Drag as a Function of Drag as a Function of VelocityVelocity

ffDD = bv + cv = bv + cv22

b and c depend uponb and c depend upon shape and size of objectshape and size of object properties of fluidproperties of fluid

b is important at low velocityb is important at low velocity c is important at high velocityc is important at high velocity

Drag Force in Free FallDrag Force in Free Fall

mg mg

fD

mg

fD

mg

fD

when fD equals mg, terminal velocity has been reached

Drag Force in Free FallDrag Force in Free Fall

for fast moving objects

FD = c v2

c = 1/2 D AWhereD = drag coefficient = density of fluidA = cross-sectional area

FD = bv + c v2

for slow moving objects

FD = b v

mg

FD

Sample Problem: Slow moving objectsSample Problem: Slow moving objects Show that vShow that vTT = mg/b = mg/b

mg

FD = bv

Sample Problem: Slow Moving ObjectsSample Problem: Slow Moving Objects Show that v(t) = (mg/b)(1 – e Show that v(t) = (mg/b)(1 – e –bt/m–bt/m))

mg

FD = bv

Sample Problem: Fast moving objectsSample Problem: Fast moving objects Show that vShow that vTT = (2mg / D = (2mg / DA)A)1/21/2

mg

FD = 1/2DAv2

Sample Problem: Fast moving objectsSample Problem: Fast moving objects Derive an expression for the velocity of Derive an expression for the velocity of

the object as a function of timethe object as a function of time

mg

FD = 1/2DAv2

Documents

Newton’s Laws MONDAY, September 14, Re-introduction to Newton’s 3 Laws