Chapter5Addi-onalApplica-onsofNewton’s

Laws

October1andOctober6,2009

Fric-on

  Fric-on

 Opposesmo-onbetweensystemsincontact Paralleltothecontactsurface Dependsontheforceholdingthesurfacestogether(thenormalforceN)

  Sta-cfric-on  Forcerequiredtomoveasta-onaryobject

 fsislessthanorequaltoµsN

  Kine-cfric-on  Fric-onalforceonanobjectinmo-on

 Canbelessthansta-cfric-on  fk=µkN

Proper-esofSurfaceFric-on

These relations are all useful APPROXIMATIONS to messy reality.

Newton’slaws:example

T=50 N

M = 5 kg

µk = 0.2

Find acceleration of block

Newton’slaws:example

T=50 N

M = 5 kg

µk = 0.2

Find acceleration of block

Frictional force: f = µkMg opposes the tension T Net force: Fnet = T - f = T - µkMg Acceleration: a = Fnet / M = ((50 - 0.2 x 5 x 9.8) / 5) m/s2

Answer: 8.04 m/s2

Newton’slaws:anotherexampleT=50 N

M = 5 kg

µk = 0.2

Find acceleration of block

θ=500

Newton’slaws:anotherexampleT=50 N

M = 5 kg

µk = 0.2

Find acceleration of block

θ=500

Hints: Resolve T in x and y components: Tx=T cosθ, Ty=T sinθ Draw free body diagram Solve for y-component of force, and note that y-acceleration is zero (Obtain relationship between T and N)

Solve for x-component of force, then use ax=Fx/m

Answer: acceleration of block is 6.0 m/s2 in +x direction

€

a =FnetM

=Tx − µkN

M=Tx − µk Mg −Ty( )

M=T cosθ −µk Mg −T sinθ( )

M

RollingFric-on:CarTires  Fric-onkeepsthecarwheelsfromspinninginplace

  Youwantthe-restoroll  Youwantthefric-ontobehigh  Thecontactpointisatrest‐althoughthecarisinmo-on

» WhatmaXersisthecoefficientofsta-cfric-on!

weight Maximum Static friction ( > Fcar on road for car not to spin in place!)

Fcar on road

Froad on car

Consider Newton’s 3rd law:

Froad on car is the actual force ON the car.

Static Friction µsN is its maximum value

SurfaceFric-on…

  Fric-oniscausedbythe“microscopic”interac-onsbetweenthetwosurfaces:

Ques-on1You are pushing a wooden crate across the floor at

constant speed. You decide to turn the crate on end, reducing by half the surface area in contact with the floor. In the new orientation, to push the same crate across the same floor with the same speed, the force that you apply must be about a) four times as great b) twice as great c) equally as great e) one-fourth as great

as the force required before you changed the crate orientation.

Ques-on1You are pushing a wooden crate across the floor at

constant speed. You decide to turn the crate on end, reducing by half the surface area in contact with the floor. In the new orientation, to push the same crate across the same floor with the same speed, the force that you apply must be about a) four times as great b) twice as great c) equally as great e) one-fourth as great

as the force required before you changed the crate orientation.

Frictional force does not depend on the area of contact. It depends only on

the normal force and the coefficient of friction for the contact.

€

fs ≤ µsNfk = µkN

Surfacefric-on…

friction force

InclinedPlanewithFric-on(1)

θ

What is the acceleration of the block?

InclinedPlanewithFric-on(2)

θ mg

N

µKN ma

θ

Draw free-body diagram for block:

InclinedPlanewithFric-on(3)

Consider i and j components of FNET = ma

θ

mg

N

θ

µKN

ma

mg sin θ mg cos θ

€

i component : mg sinθ - µKN = maj component : N = mg cosθ

€

mgsinθ −µKmgcosθ = ma

€

ag

= sinθ −µK cosθ

Sta-cfric-on

•  Previousproblemsconsideredfric-onac-ngwhenthetwosurfacesmoverela-vetoeachother–ie.,whentheslide.– Wealsoknowthatfric-onactswhenthesurfacesarestucktogether:the“sta-ccase.”

•  Whensurfacesdonotmoverela-vetoeachother,thefric-onforcedependsontheOTHERforcesonthepartsofthesystem.

fF

fF

  Themaximumpossibleforcethatthefric-onbetweentwoobjectscanprovideisfMAX=µSN,whereµsisthe“coefficientofsta-cfric-on.”

  SofF≤µSN. AsoneincreasesF,fFgetsbiggerun-lfF=µSNandtheobjectstartstomove.

Sta-cFric-on…(withonesurfacefixed)

Sta-cFric-on…

  Justlikeintheslidingcaseexcepta=0.

i: F- fF=0j: N=mg

fF

  While the block is static: fF = F

Sta-cFric-on…

  The“coefficientofsta8cfric8on,”µS,determinesmaximumsta-cfric-onalforce,µSN,thatthecontactbetweentheobjectscanprovide.

  µSisdiscoveredbyincreasingFun-ltheobjectstartstoslide: FMAX‐µSN=0 N=mg

FMAX=µSmgµS= FMAX/mg

µS N

Sta-cFric-on…

  WecanalsoconsiderµSonaninclinedplane.

  Inthiscase,theforceprovidedbyfric-onwilldependontheangleθoftheplane.

θ

Sta-cFric-on…

θ mg

N

ma = 0 (block is not moving)

  Theforceprovidedbyfric-on,fF,dependsonθ.

θ

fF

mg sin θ - ff = 0

(Newton’s 2nd Law along x-axis)

Sta-cFric-on…

  Wecanfindµsbyincreasingtherampangleun-ltheblockslides:

θM mg N

µSN

In this case:

θ

mg sin θM - µSmg cos θM = 0

µS = tan θM

mg sin θ - ff = 0

ff = µSN = µSmg cos θM

Problem:Sta-cFric-on  Aboxofmassm=10.21kgisatrestonafloor.Thecoefficientof

sta-cfric-onbetweenthefloorandtheboxisµs=0.4.  AropeisaXachedtotheboxandpulledatanangleofθ=30oabove

horizontalwithtensionT=40N.  Doestheboxmove?

(A)yes(B)no(C)tooclosetocall

T

m static friction (µs = 0.4 ) θ

Problem:Sta-cFric-on‐‐Solu-on

  Pickaxes&drawFBDofbox:

T

m θ

N

mg

y

x

  Apply FNET = ma

y: N + T sin θ - mg = maY = 0

N = mg - T sin θ = 80 N

x: T cos θ - fFR = maX

The box will move if T cos θ - fFR > 0

fFR

m =10.21 kg, µs = 0.4, θ = 30o, T = 40 N

Problem:Sta-cFric-on‐‐Solu-ony

x

m =10.21 kg, µs = 0.4, θ = 30o, T = 40 N

T

m

fMAX = µsN

N

mg

x: T cos θ - fFR = maX

y: N = 80 N

The box will move if T cos θ - fFR > 0

T cos θ = 34.6 N

fMAX = µsN = (.4)(80N) = 32 N

So T cos θ > fMAX and the box does move

θ

An-lockBrakes

  An-‐lockbrakesworkbymakingsurethewheelsrollwithoutslipping.Thismaximizesthefric-onalforceslowingthecarsinceµS>µK.

  Thedriverofacarmovingwithspeedvoslamsonthebrakes.Thecoefficientofsta-cfric-onbetweenthewheelsandtheroadisµS=0.5andthecoefficientofkine-cfric-onbetweenthecarandtheroadisµK=0.4.WhatisthestoppingdistanceDfor(a)an-lockbrakesand(b)lockedwheels?

ab

vo

v = 0

D

An-lockBrakes

ab

vo

v = 0

D

If wheels are turning, the tire surface is stationary with respect to the road, and the friction is static friction. So magnitude of maximum friction force here is µSN=0.5mg. If wheels are sliding, the friction is kinetic friction, with magnitude µKN=0.4mg.

Find stopping distance from

Since |ax| = µSg, the ratio of the stopping distances is µS/µK

€

vx2 = v0x

2 + 2axD

When vx = 0, D = - v0x2

2ax

DragForces

•  Objectsmovingthroughafluidsuchasairorwaterexperienceadragforcethatopposesthemo-onoftheobject.Themagnitudeofthedragforceincreasesasthespeedincreases(unlikethekine-cfric-onforce!).Empiricallyitistypicallyfoundthat

where b is a constant and n is typically 1 or 2.

Terminalspeedoffallingobject

•  TheterminalspeedvTisthespeedatwhichthedragforcebvnexactlybalancestheforceofgravitymg.

€

€

bvTn = mg

vT =mgb

1/n

Ques-on–skydiver

Askydiverjumpsoutofanairplaneat5000mal-tude.Shereachesterminalvelocity(duetothedragoftheair)aneraboutsixseconds.

Ifaboxofsteelpartsthathasthesameweightasthediverisdroppedsimultaneously,theboxwillfall:

fasterthanthediver

slowerthanthediver

thesameasthediver

Ques-on–skydiver

Askydiverjumpsoutofanairplaneat5000mal-tude.Shereachesterminalvelocity(duetothedragoftheair)aneraboutsixseconds.

Ifaboxofsteelpartsthathasthesameweightasthediverisdroppedsimultaneously,theboxwillfall:

fasterthanthediver

slowerthanthediver

thesameasthediver

correct

The force of gravity is proportional to the mass of an object, and the drag coefficient is proportional to the cross-section of the object. The box of steel is much more dense than a human body (whose density is about that of water), so the gravitational force on it is much larger compared to the drag force than for the person. (In other words, the terminal velocity vT=(mg/b)1/n, which is larger for the box of steel than it is for the person.)

Mo-onalongacurvedpath

•  Centripetalaccelera-onistheaccelera-onperpendiculartothevelocitythatoccurswhenapar-cleismovingonacurvedpath.

•  Theforceassociatedwiththiscentripetalaccelera-onisdirectedtowardsthecenterofthecircledefinedbytheradiusofcurvature,andhasmagnitudev2/R.

CentripetalAccelera-onDerivation of a=v2/R:

CentripetalAccelera-onDerivation of a=v2/R:

CentripetalForce

  Theforcecausingthecentripetalaccelera-onissome-mescalledthecentripetalforce

  Thisisnotanewforce,itisanewroleforaforce

  Itisaforceac8ngintheroleofaforcethatcausesacircularmo8on

UniformCircularMo-on

•  Ifanobjectismovingalongacurvedpath,itisaccelera-ng.

•  Newton’ssecondlaw⇒aforcemustbeac-ngontheobject.

•  Ifobjectismovingwithconstantspeedonthecircle,

UniformCircularMo-on

  Aforcecausingacentripetalaccelera-onactstowardthecenterofthecircle

  Itcausesachangeinthedirec-onofthevelocityvector

  Iftheforcevanishes,theobjectwouldmoveinastraight‐linepathtangenttothecircle

Mo-oninaHorizontalCircle

  Thespeedatwhichtheobjectmovesdependsonthemassoftheobjectandthetensioninthecord

  Thecentripetalforceissuppliedbythetension

T =mv2

r⇒

Ques-on

AnobjectofmassmissuspendedfromapointintheceilingonastringoflengthL.Theobjectrevolveswithconstantspeedvinahorizontalcircleofradiusr.(Thestringmakesanangleθwiththever-cal).

Thespeedvisgivenbytheexpression:

L θ

Ques-on

AnobjectofmassmissuspendedfromapointintheceilingonastringoflengthL.Theobjectrevolveswithconstantspeedvinahorizontalcircleofradiusr.(Thestringmakesanangleθwiththever-cal).

Thespeedvisgivenbytheexpression:

correct

y

L

€

F net = m

a

x : Tsinθ = m v2

ry : Tcosθ −mg = 0

⇒ tanθ =v2

rg⇒ v = rg tanθ

θ

Horizontal(Flat)Curve

  Theforceofsta-cfric-onsuppliesthecentripetalforce

  Themaximumspeedatwhichthecarcannego-atethecurveis

fs = mv2

rfsmax = m

vmax2

r= µsmg

Note, this does not depend on the mass of the car

BankedCurve

  Thesearedesignedtobenavigablewhenthereisnofric-on

  Thereisacomponentofthenormalforcethatsuppliesthecentripetalforce

ncosθ = mg

nsinθ = m v2

r

Non‐UniformCircularMo-on

  Theaccelera-onandforcehavetangen-alcomponents

  Frproducesthecentripetalaccelera-on

  Ftproducesthetangen-alaccelera-on

  ΣF=ΣFr+ΣFt

Ver-calCircleWithNon‐UniformSpeed

  Thegravita-onalforceexertsatangen-alforceontheobject  LookatthecomponentsofFg

  Thetensionatanypointcanbefound

TopandBoXomofCircle

  ThetensionattheboXomisamaximum

  Thetensionatthetopisaminimum

  IfTtop=0,then

TheCenterofMass

•  Defini-onofcenterofmass:

Where

For a continuous object (e.g., a solid sphere)

CenterofMass(2)

•  Newton’sLawsforacollec-onofobjects:

The acceleration of the center of mass is determined entirely by the net force on the object.

Ahorizontalforceisusedtopushanobjectofmassmupaninclinedplane.Theanglebetweentheplaneandthehorizontalisθ.Thenormalreac-onforceoftheplaneac-ngonthemassmis

A. mgcosθ+Fcosθ.B. mgcosθ.C. mgcosθ+Fsinθ.

E. impossibletodeterminebecausethecoefficientoffric-onisnotgiven.

Ahorizontalforceisusedtopushanobjectofmassmupaninclinedplane.Theanglebetweentheplaneandthehorizontalisθ.Thenormalreac-onforceoftheplaneac-ngonthemassmis

AcargoingaroundacurveofradiusRataspeedVexperiencesacentripetalaccelera-onac.Whatisitsaccelera-onifitgoesaroundacurveofradius3Rataspeedof2V?

A. (2/3)acB. (4/3)acC. (2/9)ac

E. (3/2)ac

AcargoingaroundacurveofradiusRataspeedVexperiencesacentripetalaccelera-onac.Whatisitsaccelera-onifitgoesaroundacurveofradius3Rata

speedof2V?

Thefigureshowsatopviewofaballontheendofastringtravelingcounterclockwiseinacircularpath.Thespeedoftheballisconstant.Ifthestringshouldbreakattheinstantshown,thepaththattheballwouldfollowis

A. 1B. 2C. 3

E. impossibletotellfromthegiveninforma-on.

Thefigureshowsatopviewofaballontheendofastringtravelingcounterclockwiseinacircularpath.Thespeedoftheballisconstant.Ifthestringshouldbreakattheinstantshown,thepaththattheballwould

followis