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