Chapter(4

Dynamics:*Newton�s*Laws*of*Motion

Another(Newton�s(2nd Law(example

Example. The$figure$below$shows$two$forces$with$magnitudes$F1 =$+2000$Nand$F2 =$+3000$N$acting$on$an$object,$the$plus$signs$indicating$that$the$forces$act$along$the$+x$axis.$A$third$force$F3 also$acts$on$the$object$but$is$not$shown$in$the$figure.$The$object$is$moving$with$a$constant$velocity$of$+750$m/s$along$the$x$axis.$Find$the$magnitude$and$direction$of$F3.

+x

F1F2

v =$+750$m/sF3?

Types&of&Forces:&An&Overview

In#nature#there#are#two#general#types#of#forces,fundamental and#nonfundamental.

Fundamental*Forces*66 three#have#been#identified,all#other#forces#are#derived#from#them.

1. Gravitational#force (e.g.#downward#force#on#objects#near#the#Earth�s#surface)

2.#Strong#Nuclear#force (e.g.#force#that#binds#protons#and#neutrons#together#in#the#atomic#nucleus)

3.#Electroweak#force (e.g.#force#between#objects#which#are#electrically#charged)

Types&of&Forces:&An&Overview

Examples)of)Nonfundamental Forces ,,All)of)these)are)derived)from)the)electroweak)force:

normal)or)support)forces

friction

tension)in)a)rope

The$Gravitational$Force

Newton�s(Law(of(Universal(Gravitation

Every(particle(in(the(universe(exerts(an(attractive(force(on(everyother(particle.((

A(particle(is(a(piece(of(matter,(small(enough(in(size(to(be(regarded(as(a(mathematical(point.

The(force(that(each(exerts(on(the(other(is(directed(along(the(linejoining(the(particles.

The$Gravitational$Force

For$two$particles$that$have$masses$m1$and$m2 and$are$separated$by$a$distance$r,$the$force$has$a$magnitude$given$by

221

rmmGF =

2211 kgmN10673.6 ⋅×= −G

The$Gravitational$Force

F =G m1m2

r2

= 6.67×10−11 N ⋅m2 kg2( ) 12 kg( ) 25 kg( )1.2 m( )2

=1.4×10−8 N

Example. Find%the%gravitational%force%between%a%bicycle%of%mass12%kg%and%a%lawnmower%of%mass%25%kg%sitting%1.2%m%away%from%eachother%in%a%garden%shed.

very%small,%considering%it%takes%a%force%ofabout%1%N%to%lift%a%pen!

Gravitational*force*between*two*Nimitz3class*aircraft*carriers

d ="300"m

F =G m1m2

r2 =G m2

d 2 = 6.67×10−11 N•m2

kg2

#

$%

&

'(

9.8×107 kg( )2

300 m( )2 = 7.1 N

=1.6 lb

Find"the"gravitational"forcebetween"the"two"aircraft"carriers"pictured"here."The"mass"of"a"Nimitz>class"aircraft"carrier"is"about"98,000"metric"tons,"and"the"separation"between"the"aircraft"carriers"Is"about"300"m.

1 metric ton =1000 kg98,000 metric tons = 9.8×107 kg

The$Gravitational$Force

Even%though%the%Earth%and%Moon%are%not%point%particles,because%they%are%mostly%spherical,%Newton�s%law%of%Gravitation%may%be%applied%using%the%distance%between%their%centers%as%if%they%were%point%particles.%

Find%the%average%gravitational%force%betweenThe%Earth%(mass%=%5.98%x%1024 kg)%and%Moon(mass%=%7.35%x%1022 kg)%if%their%mean%distanceapart%is%3.85%x%108 m.

F = GMMME/r2

= (6.67 x 10-11)(7.35 x 1022)(5.98 x 1024)/(3.85 x 108)2

= 1.98 x 1020 N very large!

The$Gravitational$Force

The$Gravitational$Force

Definition(of(Weight

The(weight(of(an(object(on(or(above(the(earth(is(the(gravitational(force(that(the(earth(exerts(on(the(object.((The(weight(always(acts(downwards,(toward(the(center(of(the(earth.

On(or(above(another(astronomical(body,(the(weight(is(the(gravitational(force(exerted(on(the(object(by(that(body.(

SI$Unit$of$Weight:$newton((N)

The$Gravitational$Force

The$weight,"W,"of$an$object$can$be$associated$with$its$mass,"m,using$Newton�s$2nd law:

+y

m

W

ay =-.g

!Fy =-.W-=-may =-m(.g)

;;>$$W-=-mg

The$Gravitational$Force

Relation$Between$Mass$and$Weight

2rmM

GW E=

mgW =

2rMGg E=

The$acceleration$of$gravitydepends$on$the$mass$of$the$planetand$the$distance$from$its$center.

The$Gravitational$Force

g =G ME

RE2

= 6.67×10−11 N ⋅m2 kg2( )5.98×1024 kg( )6.38×106 m( )

2

= 9.80 m s2

On#the#earth�s#surface:

Types&of&Forces:&An&Overview

Examples)of)Nonfundamental Forces ,,All)of)these)are)derived)from)the)electroweak)force:

normal)or)support)forces

friction

tension)in)a)rope

The$Normal$Force

Definition(of(the(Normal(ForceThe$normal$force$is$one$component$of$the$force$that$a$surfaceexerts$on$an$object$with$which$it$is$in$contact$– namely,$thecomponent$that$is$perpendicular$to$the$surface.

For$a$block$of$weight$W sitting$at$rest$on$a$table,$from$Newton�s$2nd law:

! Fy =%FN ' W%=%may = 0$$BB>$$FN =%W ,$and$is$directed$upward

Weight is$the$downwardforce$exerted$by$gravityon$an$object.

The$Normal$Force

N 26

0N 15N 11

=

=−−

N

N

F

F

N 4

0N 15N 11

=

=−+

N

N

F

F

! F = FN + FH + W = 0

The$Normal$Force

Apparent(Weight

The$apparent$weight$of$an$object$is$the$reading$of$the$scale.

It$is$equal$to$the$normal$force$the$scale$exerts$on$the$man.

The$Normal$Force

mamgFF Ny =−+=∑

mamgFN +=

apparent'weight

trueweight

From'Newton�s'3rd'law:the'normal'force'exerted'by'the'scale'on'the'man'is'equal'(and'opposite)'to'the'force'the'man'exerts'on'the'scale'>>>'the'man�s'apparent'weight

Sum'of'the'forcesacting'on'the'man: W

FN =mg+ma ⇒ a = FN −mgm

Since W =mg = 700 N ⇒ m =700 N

9.80 m s2 = 71.4 kg

∴a = FN − 70071.4

m s2( )

Find%the%acceleration%in%each%case:

FN = 700 N

a = 700− 70071.4

= 0 ms2

FN =1000 N

a = 1000− 70071.4

= 4.20 ms2

FN = 400 N

a = 400− 70071.4

= −4.20 ms2

FN = 0 N

a = 0− 70071.4

= −9.80 ms2