Ch-04 Forces Newtons Laws

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4 CHAPTER

Forcesand Newton's Lawsof Motion
PREVIEW
accelerate. why objects you will beginthe studyof dynamics,thatbranchof physicswhich explains In this chapter laws of motion,which applyto all forcesthat of You will be intoduced to the concept force, and studyNewton's subjectto occur in nature.You will leam how to constructfree-bodydiagrams,and use them to analyzesystems such forcesas gravity and friction. You will also apply Newton'slaws of motion to solve a numberof different problems.The applications will includeboth equilibrium andnon-equilibriumproblems.

QUICKREFERENCE lmportantTerms
Force law. It is a second the to The pushor pull required change stateof motionof an object,asdefinedby Newton's (lb). (N), dynes(dyn),or pounds vectorquantifywith unitsof newtons Inertia in line. speed a straight The naturaltendency an objectto remainat restor in uniformmotionat a constant of Mass (kg),grarns or slugs(sl). (g), measure inertia.Units arekilograms of A quantitative Inertial Reference Frame law A reference framein whichNewton's of inertiais valid. Free-Body Diagram all acting anobject. on that A vector diagram represents oftheforces GravitationalForce in exerts everyotherparticle. on that The forceof attraction everyparticleof mass theuniverse Weight body) on an object. by otherlargeastronomical The gravitational forceexerted theearth(or some Normal Force This component is of One component the force that a surfaceexertson an objectwith which it is in contact. directed normal,or perpendicular, the surface. to Friction when it movesor attemptsto move along a surface.It is alwaysdirected The force that an object encounters parallelto thesurface question. in Tension The tendency rope(or similarobject)to be pulledapartdueto the forcesthat areappliedat eitherend. ofa Equilibrium Mathematically, equilibrium meansEF : 0. The statean objectis in if it haszeroacceleration. Apparent Weight The force that an object exerts on the platform of a scale.It may be larger or smaller than the tue weight, depending the acceleration on ofthe object and the scale.

4 Chapter 39

Newton'sLaws of Motion
First Law line,nnless alonga straight speed in An objectcontinues a stateof restor in a stateof motionat a constant that to compelled change stateby a net force.By "net" forcawe meanthe vector sum of all of theforces actingon an object. it from any otherobjector force.If theshipis stationary, isolated in a For exanple,consider spaceship deepspace, line to (its rocketengines shutdown)it will continue movein a straight are will remainso. But if the shipis moving to at with a constant speecl. if the ship weretraveiinginto deepspace say,1C0000 mi/h, it wouldcontinue move So its to forceacted stopor change an outside firing,until line,evenwithoutthe rocketengines in at this speed a straight motion. Second Law to is a m, When a net forceXF actson an objectof mass theacceleration thatresults directlyproportional the net force and has a magnitudethat is inverselyproportionalto the mass.The directionof the is as acceleration the same the directionof thenet force.This statement usuallywrittenas is

XF = ma

or

a: XF/m

( 4.1)

that Thismeans the Thesymbol!F represents net force,that is, thevector sumof all theforcesactingon an object. (4. equation 1)becomes in For mustbe examined. example, two dimensions, thecomponents theforces of

XF*: ma* XFr: ma'
'I'hey are the (4.1)and(4.2) cutbeused detennine unitsof force. to Equations SI CGS BE (N) kg nr/s2= newton = g cm/sz dyne(dyn) (lb) slugftls2= pound

(.2a) 62b)

Third Larv force directed an bodyexer:ts oppositely body,thesecond a one Whenever bodyexerts forceon a second on magnitude thefirstbody. of equal but law, refenedto as the "action-reaction" i.e., "for everyactionthereis an equal, opposite, This is sometimes out ejecthot gases the rearof therocket The firing its rocketengines. engines is reaction". example a spaceship An in is velocity,this is the "action".The "reaction" thattherocketaccelerates theforwarddirection. at very,high

Universal Gravitation
The force acting an Every particle in the universee>ierts atfractiveforce on everyotherparticle in the universe. givenby a magnitude r by m1 between parficles masses andm2 separated a distance has two of

F:G#

m, m. r-

(4.3)

is experimentally) givenby whosevalue(obtained whereG is the universalgravitationalconstant,

G=6.6i3x 1o-11g*24.t2

40 FORCESAND NEWTON'SLAWS OF MOTION

on forcethattheearthexerts an object.Thisforceis always is Theweight of an objecton theearttr thegravitational = (mass Mg, radius= RB) we canwrite of For downward, towardthecenter the earth. theearttr directed

W:G+

M-m R;

be wherethe gravitational acceleration g : GMB/RgZ= 9.80m/s2.Theweightof an objectcantherefore written is

w=mg

(4.s)

SECTIONS OF DISCUSSION SELECTED 4.314.4 Newton's SecondLaw of Motionand the VectorNatureof Newton's SecondLaw of Motion
that (4.1)and(4.2),contains The XF vectors, equation = ma means in Newton's second law, asexpressed equations the that lF, we needto determine net forceactingon an object.Theterminology, means you mustdetermine the that beginby drawinga diagram vectorsum(E = surnmation sign)of all theforcesactingon an object.You should on it, a so-called free-bodydiagram. the acting represents objectandall theforces Diagram Example1 A Free-Body and A 2.0 kg objectis subjected a 6.0N forceactingin thex direction, an 8.0N forceactingin they direction. to (magnitude dilection) olthis object? Whatis theacceleration and where the diagrarn is The first diagram shows mass andforcesactingon it. The first diagram thefree-body the m are actingfrom thispoint. at system the forces shown and mass placed the originof thex-y coordinate is

the In orderto frndtheacceleration theobject,a, we needto find thenetforce,F, actingon the object.Since of perpendicular, canfind thenet forceusing we forcesactingon m are

F_ JF;+F; 0- *'[*)
= tan-l l'l"gN.l -^o Ie.oN]:)r'

t-

Chapter4 41

is is Sincethe acceleration in the samedirection asthe net force, a is directed53oup from the *x axis. Its magnitude a = F/m: (10 NX2.0 kg) = 5 m/s2. The individuai forcesmustbe in In most cases, forcesacting on an object can have any orientation space. the can be found. The next example before the net force, and hencethe acceleration, resolvedinto x and y components illustatesthis. Example2 Finding thenetforceandacceleration of The in to a Suppose 1.8kg objectis subjected forcesasillustrated thefollowingdiagram. magnitudes theforces actingon the object? N, areF1 = 20.0N, F2= 14.0 F3 = 15.0N. Whatis thenet forceandacceleration

(4.2a) and and of we To find EF andtheaccelerations, needto look at the components theforces useequations (EFy).We have (XF*) andy direction (4.2b).Thatis, we needto sumthe forcecomponents the x direction in Force Fl F2
x comDonenl

+(20.0 cos30.0o:+17.3 N N) = - (14.0 cos45.0o - 9.9N N)

F3
EF* = +lf.l N - 9.9N : +7.4N

Force Fl 82 F3

y component +(20.0 sin30.0'= +10.0 N N) +(14.0 sin45.0' = + 9.9N N) -15.0 N

N EF, = +10.0 + 9.9N - l5-o N = +4.9N

42 FORCESAND NEWTON'SLAWS OF MOTION

The accelerations thex andy directions now be obtained in can usingXF" andEFufoundabove:

v ^. = r = - -

XR
m

+7.4N
1.8kg

=4.1 fivS-l

,)

d-.=-=-

Y

XFu +4.9N =
m 1.8kg

z. | ilu"s-

,?

4.7 The Gravitational Force
exerts forceof athaction the a on two of m1 by r. Consider particles masses andm2 separated a distance Eachmass joining theparticles the forcehasa magnitude givenby othermass. The forceis directed alongtheline and

F=G+

m. m^
t'

(4.3)

Example3 Findtheforcebetween objects mass = 1.00x 103kg, rn2=2.00x 103kg, separated r : 3.00m. of by m1

F = Gry f

ro3 * = (6.67x y *r7L*ry'Loo-* kgX2'0010'kg): 1.48 10-5 x tg-rr N (3.00 mf

Example 4 The earth's gravity and weight The weight of an object on earth is defined as the gravitational force exerted by the earth on the object. Find the weight on earthof a 50.0 kg tnass. ntassand radius. tlse MB = 5.98 x 1024kg, l{E = 6.38 x 106 m for t}reeaLth's

w:G + q

M-m

-r l : ( 6 . 6 7 x1 0 " N - z a 9 ( 5 ' 9 8 * r O a k g X (6.38 loom)' x

= 4 . 9 0 x1 0 2 N

Note that accordingto Newton'ssecondlaw, F: ma: mg, sinceg is the acceleration gravity. Therefore, of

g:-'

GM, _ (6.67x 10-llN n 2A<g)15.9g102-a _ kg) = n on *,^2 x : Y . d u I r Y S.

R;

(6.38 10"m)x

So the weight of an object on earth can be expressed W = mg. This expressionwill prove to be quite useful. as

Weight
The weightof an objectis not the same its mass. massof anobjectis thesame as The regardless where object the of is located theuniverse. weight,beingsimplytheforcethatgravityexerts an object,depends howclose in The on on the objectis to another massive objectsuchasthe earth. rq Closeto thesurface theearth,an objectof mass, hasa of weightof mg,where"g" is theacceleration to gravity.The acceleration, closeto thesurface theearth g, due of is

9.80m/s2 or32.2ftls2.

4 Ghapter 43

Forees Statlcand KineticFrictionai 4.814.9The Nornnal ForceaneJ
forcethat a surface exerts anobjectwith p'hichit is in contact.Foran on The norrnalforce,F1q,is theperyendicular (seefigure beiow) we canseethatFN : W. The normalforceis simplythe objectat equilibriumon a flat surface pushing objectagainst surface. the the swface'sreaction theforces to

of If an obiect is at equilibrium aninclinedplane,themagnihrde thenormalforcewill be equalto themagnitude on of a cornponent theobject'sweight,asis shownin the figure. of

w
The magnitude of the normal force is FN = W cos 0, which is the caseeven for a flat surface,i.e., for 0 = 0 since cos 0 = l, thereforeFN = W. When an object is in contactwith a surface,there is anotherforce in addition to the normal force and gravity acting on it. This additional force is friction. When the object slides over the surface, the surfaceexertsa frictional force on the object. This is known as kinetic friction. However, a frictional force exists even if two objects are not 'Ihis in motion. force is static fiiction. An important characteristicof the frictional force is that its magnitudeis proportional to the magnitudeof the normal force. For example, an object sliding on a rough surfaceis subjectto the force ofkinetic friction, fg., accordins to

&: p,kFN
<pk<1.

(4.8)

where pp is the coefficient of kinetic friction. The coefficient of kinetic friction can have values u,hich are usually0

LAWSOF MOTION ANDNEWTON'S 44 FORCES

Example5 at N by surface a 100.0 forcedirected an angleof l4'0" with ilg ,q.S.O objectis pulledovera roughhorizontal ofttre object' ofkinetic friction is 0.40,find theacceleration coefficient (see to respect thesurface figure).Ifthe

force, are whichmustbe included theapplied The for diagram this example. forces a We beginby drawing free-body = mg. If theobjectis movingin the+x direction' r, *r'f,ictional forci, q,, thrnor*riorce, Fy, andtheweight,W we have

in Look attheforces thex andy directions: ydirection: FN*Fy-W=O

so in sincethereis no acceleration they direction, Ftt:W-Fy=-g-Fsin0 "| = N) FN: (5.0kgx9.80ttt/r")- (100.0 sin 14.0o 25N. x direction: F" - f1= ma*

in sincethe objectis accelerating the x direction,so F Fcos0-P1 y=ma* N) (100.0 cos14.0o (0.40X25 = 87N = ma* N)

Chapter4 45

Therefore

) a x : ( 8 7 N ) / m = ( 8 7 N ) / ( 5 . 0 k g= l T m l & .
this force,frictionresists forceand Suppose objectis restingon a flat surface. we apply a smallhorizontal an If force,the object the increase applied the objectremains rest.This is the force of staticfriction,l'r. As we gradually at of will eventually beginto move oncewe have overcome maximumfrictionalforce, frmax. If the coefficient the staticfriction is prr,themagnitude themaximumstaticfrictionalforcecanbewritten of

frmax: FsFN

(4.7)

Example 6 StaticFriction coefficient kinetic whosecoefficient staticfriction is Fs= 0.60andwhose of A 10.0kg block sitson a flat surface of to friction is Fk = 0.a0.(a) Whathorizontal force is required gettheblock to move?(b) If we continue applythe to sarne force asin part (a),whatwill theblock'sacceleration be? (a) We canuseequation (4.7)to find the forceF necessary overcome to friction,that is,

-

F = tmax
F:FsFN=Fsffig

= m/s2) 59N. F: (0.60X10.0 kgx9.80
(b) LJsing1heaboveforce,Nervton'ssccondlaw, )lF .=ma, alrd eqrraticrn (4.8) for kinotic fi:ictiou,w0 liav0

I,F: F-ft XF: F-FtFN XF= F-pkmg = ma
Now

IF mm

F-Pumg

"I("1I"ry{ryg{) 10.0 c k

a

= ),.U lTtlS'-

Example 7 A box weighing 147 N sits on a horizontal surface.The coeffrcient ofstatic friction befweenthe box and the surface is 0.70..If a force P is exertedon the box at an angle directed 37o below the horizontal, what must the magnitude of P be to get the box moving? The box and the associated free body diagram are shown below.

46 FORCESAND NEWTON'S LAWS OF MOTION

(4.2b)and(4.2a). Usethe cornponent equations Newton'ssecond of law, equations In the y direction

XF, = ma,
Thebox hasno acceleration theverticaldirection in so

FN-Py-W:0
which gives

FN:Psin0+W
In the x direction

(1)

XF*= mq
Thebox hasno acceleration thehorizontal in direction so

P*-t=o
Pcosd-FsF1g=0 P cos(l - Fr€ sinQ+'S/):0
(2) Solving for P yields: r r - A s 0 - ll r r Wi n O 'rrs

(2)

( 0.70) 147 ( N) :270 N. cos37"- (0.70) 37o sin

4.10 TheTension Force
Forces are sometimesapplied by ropes, cables,or wiles that ptrll on an object. We usually assume that the rope is massless that the magnitLrde the tension,T, is the samethronghoutthe rope. and of Exanrple 8 Tlrc tensionforce A 12.0kg objectis pulled upwardby a massless rope with an acceleration 3.00 m/s2.What is the tensionin the of

rope? The tension forceis directed upward(+) while theweightof ttre object(mg)is dirpcted downward Sincetheobjectis being O. (+) accelerated upward we canwrite tF: lna as

T-mg=P4 T=m(g+a) 1 = (12.0 kgX9.80 mts2+:.00 m/s2) T:154N.

Chapter4 47

Example9 Suppose objectin theprevious the rvereaccelerating example of downward 3.00m/s2instead upward.What at would thetension the ropebe in thiscase? in The only difference this case in wouldie thattheacceleration downward, takea = -3.00m/s2.We have, is so

T - mg: ma'
so that

T=m(g+a) -3.00m/s2) 1= (12.0 (9.80mts2 kg) T= 82N.
The situations depicted examples and 9 canbe observed in 8 you'reon the whenyou'rein an elevator. Suppose bottom floor and the elevatorstartsaccelerating upward.You feel as if you are beingpusheddownward, the and tension the elevator in cableincreases. Whenyou'reon the top floor andtheelevator accelerating begins downward, you feel as if you are being lifted off the floor andthereis a conesponding in decrease the elevatorcabletension. Whenthe elevator at rest,or movingwitb a steady is justbe y6u speed, feel "nonnal" thc cabletensionwould and equalto the weightof you andtheelevator.

4.11 Equilibrlum Applications Newton's of Lawsqf Motiorr
--/ An object that is at rest or travelingwith a constantveiocify has zero acceieration is said to be in "equilibrium". and T'hisirrrplies that thc net frirceactingon the objectrszr,ro.Intwo dirir,;lsior,s r:irlr u'r: ruiie

t l' x = f l -i

Q.ea) (4.9b)

XFr: 0
There are five stepsthat can be followed in solving equilibriumproblems.They are:

Step 1. Step 2. ' Step 3.

Step 4. Step 5.

Selectan object,the "system",aboutwhich the r:rostinformationis lqtorvn.If two or more objects are connectedit may be necessary hea.teachindividuaily. to , Draw a "free-body"diagramfor the system. discussed As pleviously,lJ:ris a drawingthat is representsthe object and ALL the forces acting on it. Choosea convenient ofx, y axesfor the object and resolvea1lthc forcesinto components set that point along theseaxes.By "convenient" we mean that you should chooseyour axesso that as mauy forces as possiblepoint directly along the x or y axis. This will minimize the number of calculations needed. Apply equations(4.9a) and (4.9b) by setting the sum of the componentsof the forces in the x direction equal to zero and the sum of the componentsia the y direction equal to zero. Solve the two equationsobtarnedin step 4 for the desred unknown quantities.In many cases the equationsmust be solved simultaneouslyfor the two unknowns.

LAWS OF MOTION 48 FORCESANDNEWTON'S just the employing method ouflined. of The following area number examples by Examplel0 A weightsuspended cables in as below.Findthetension eachcable. by A 280 lb block is suspended two cables, shown

30.00

45.

We as and x diagram chosen andy axes, shown. now we Choosing block asour object, havedrawnthefree-body the (4.9a)and(4.9b)usingthe followingtable applyequations

Force

x component y component -T1 cos30.0 +-T2 45.0' cos 0 +Tl sin30.0' +'l'2sin45.0' -2801b

T1 ,T2

w
second law UseNewton's

= + IFx = -T1cos30.0o T2 cos45.0o 0 IFy: +T1sin30.0"+ T2 sin45.0o 280lb : 0
(1), Solve T1 in equation for

(1) (2)

45.0') I cos = Tz T, : Tzt ."-30 0'J 0'816
(2), the Substitute resultinto equation

(3)

:280 lb + (0.816) sin30.0o T2 sin45.0o T2
Solvingfor T2 thenyields,

T= '2

' 2 8 0 1 b, , , ,, - : 2 5 0 1 b (0.816 30.0o+ 45.0) sin sin

(3) And finally, usingthis in equation yields

T 1 : ( 0.816) T2:200Ib.

Ghapter 4 49

Example11 considerthe followingsystem equilibrium. in Deterrnine valueof theweight,w, andthetension theleft wire. the in The free body diagramcan be drawn in two different ways, as shownbelow. In (a), we can make the y axis vertical and draw in the forces. Perhapsa better choice would be to align the coordinateaxesas shownin diagram This is (b). more convenient because of the threeforces two now lie alongthe axes, arethusresolved and into x andy components automatically. , Theequilibrium equations, (4.9a) (4.9b)can and now be written as

IF*=150N-Wsin37o=0
which gives

= YY (150N)(sin 37";= 250*
and

XFr=T-Wcos37o=0
which gives

'l',= :200 N. W cos37o

Noticethat the choice x-and axes (b) madethe equilibrium of y in equations rathereasyto solvefor w andr. If we hadchosen axesasin drawing theequations the (a), *o,rld huu. been bit moredifficult to solve,i.e., a XFx= (150N) cos37o cos53o 0 -T = XFy = (150N) sin37o Tiin 53o W = 0. + Thus,the appropriate choice coordinate of axescanfacilitatethe solution quitea bit in mostcases.

LAWS OF MOTION AND NEWTON'S 50 FORGES

plane on Example12 Object an inclined cord A is = 25.0kg sittingon an inclinedplanewhoseangleof inclination 0 = 30.0o. massless a Consider massM up M m to r frirtionlesspulley and is attached a tnass = 15.0kg' The mass slides the ourr connected M passes to M the Findthecoeflicientof kineticfriction between mass andtheplane. planeat a constant speed.

mass We is velocity (a = 0), the system in equilibrium. needto look at each movewith constant Sincethe masses for diagrams M andm looklike quantities. free-body The the individuallyin orderto determine necessary

on TheForces M

on TheForces m

Mg
For m: In thgy direction so that

XFr:T-m9:0 T =mg = nvs2; t+z N. kgx9.8o T = (15.0 tFy = FN - Mg cos0 :0, Fy: Mg cos0 :212 N. mis2; 30.0o cos FN: (25.0 kgX9.80

ForM: In they direction gives

In the x direction

EF*=T-&-lutgsin0:0 T - p k F N - M g s i n 0 =0

4 Ghapter 51
Solvingfor pl yields

nn:T

T-Mesin0

147 - (25.0kgX9.80 N n/s') sin30.0o=

212N

0.1.2.

4.12 Nonequilibrium Applicationsof Newton'sLawsof Motion
Nonequilibrium occurs whenthe forcesactingon an objectarenot balanced. this case objectaccelerates In the and Newton'ssecond is law

XF*= ma*

g.za)

XF, = ma,

(4.2b)

Example 73 Atwood's Machine Two masses, andm2 (assume > m1), areconnected a massless m1 m2 by sting thatpasses overa massless, frictionfreepulley, asshownin the figure.Find the acceleration themasses thetension thesfring. of and in

We needto look at each mass individuallyin orderto obtainthetwo equations necessary solvefor theunknown to quantities andT. The free-body a diagram eachmass shownabove. for is Since bothmasses connected the are by string,themagnitudes theiraccelerations thesame. of are Sincem2 > ml, m1 hasan upwardacceleration m2 has and a downward acceleration. alsoassume thetension thestringon each We that in sideof thepuiieyis thesame. For m1 the acceleration upward, T ) m1 g andwe cantherefore is so write, T - m1g: rltlt. For m2 the acceleration down,som2 g > T. Therefore, canwrite, is we

mZg'T = m2a.

LAWSOF MOTION ANDNEWTON'S 52 FORCES

gives, into and Solving for T in the first equation substituting the second

mZE-(m1g+ m1a)= m2a.
yields, Finally, solving for the acceleration

a=Gfr6s
thetension' we into for this Substitgting expression the acceleration eitherof the massequations, cansolvefor
T=.,.--ILs ^

(m"- m,)

2m,m"

(mr+rn)'

Example14 of the = with considerthefollowinganangemenl M1 = 15.0kg, M2 10.0kg, andp1= 0.30.Find the acceleration the accompany drawing' diagrams cord.Free-body systemandthe tensionin the connecting

__tts

M1

M2

+
For M1 we have: or

XFx=T-fk:T-pkFg=M1a

T-pkMtB=Mla.
For Me we have:

EFy: MZ E- T = M2 a.
into the first yields: and equation substituting solving for T in the second

MZE-M2a- FtMt g=Ml a

Chapter453

_J

or

(Mt + M2)a: (M2- rp M1)e
The accelerationis therefgre,

u:qry

(ta:tfl,)s

kg rr;111.80 =2.2n/s-. n/s2) . 1ro.o - (0.30)(ir.o ,z

Thetensiolr nowbe foundby substituting valueof a into theequation M2. can this for

T:M2 @-a) -2.2m/s?)= N. 1: (10.0 (9.80mls? kg) 76

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