Chapter(4

Dynamics:*Newton�s*Laws*of*Motion

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

Static&and&Kinetic&Frictional&Forces

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When%an%object%is%in%contact%with%a%surface%there%is%a%forceacting%on%that%object.%The%component%of%this%force%that%is%parallel&to&the&surface is%called%the%frictional&force.

�cold%welds�

Static&and&Kinetic&Frictional&Forces

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When%the%two%surfaces%are%not%sliding%across%one%anotherthe%friction%is%called%static&friction.

Static&and&Kinetic&Frictional&Forces

The$magnitude$of$the$static$frictional$force$can$have$any$valuefrom$zero$up$to$a$maximum$value.

MAXss ff ≤

NsMAXs Ff µ=

10 << sµ is$called$the$coefficient$of$static$friction.

Not$a$vector$equation!fS is$parallel$to$the$surface,FN is$perpendicular$tothe$surface.

Static&and&Kinetic&Frictional&Forces

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Note%that%the%magnitude%of%the%frictional%force%doesnot%depend%on%the%contact%area%of%the%surfaces.

Static&and&Kinetic&Frictional&Forces

Static&friction&opposes&the&impending relative&motion&betweentwo&objects.

Kinetic&friction&opposes&the&relative&sliding&motion&that&actually&does&occur.

Nkk Ff µ=

10 << sµ is&called&the&coefficient&of&kinetic&friction.k

Static&and&Kinetic&Frictional&Forces

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Usually,(µs > µk

Static&and&Kinetic&Frictional&Forces

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Example. A"sled"and"a"rider"are"moving"at"a"speed"of"4.0"m/s"along"ahorizontal"stretch"of"snow."The"snow"exerts"a"kinetic"frictional"force"on"therunners"of"the"sled,"so"the"sled"slows"down"and"eventually"comes"to"a"stop.The"coefficient"of"kinetic"friction"is"0.050"and"the"mass"of"the"sled"and"rideris"40"kg."Find"the"kinetic"frictional"force"and"the"displacement,"x,"of"the"sled.

Static&and&Kinetic&Frictional&Forces

1. Use&Newton�s&2nd&law&in&x&and&y&directions.

!Fx = -fk = max --> ax = -fk/m = -µkFN/m (since&fk = µkFN)

!Fy = FN - W = FN - mg = may = 0 --> FN = mg

fk = µkFN = µkmg = (0.050)(40)(9.8) = 20 N

Static&and&Kinetic&Frictional&Forces

ax = -µkFN/m = -µkmg/m = -µkg = -(0.050)(9.8) = -0.49 m/s2

2. Solve for x using ax and kinematic equations.

x v0x vx ax t? 4.0 m/s 0 m/s -0.49 m/s2

vx2 = v0x

2 + 2axx --> x = (vx2 - v0x

2)/(2ax)= (02 - 4.02)/(2(-0.49)) = 16 m

independentof)mass)ofsled+rider

fk

Example. Block&1&(mass&m1&=&8.00&kg)&is&moving&on&a&30.0o incline&witha&coefficient&of&kinetic&friction&between&the&block&and&incline&of&0.300.This&block&is&connected&to&block&2&(mass&m2 =&22.0&kg)&by&a&masslesscord&that&passes&over&a&massless&and&frictionless&pulley.&Find&theacceleration&of&each&block&and&the&tension&in&the&cord.

fkfk

m1 =#8.00#kg#####m2 =#22.0#kg#######µk =#0.300

Find#a,#T and#T�

xy

x

y

!a

!a

Block&1:

! Fx = -fk - W1 sin 30.0o + T = m1a

! Fy = FN - W1 cos 30.0o = 0 ! FN = W1 cos 30.0o

Block&2:

! Fy = T� - W2 = m2(-a)

We&also&know:

T = T� since&the&pulley&and&cord&are&massless

fk = µkFN = µkm1g cos 30.0o = (0.300)(8.00)(9.80)(0.866) = 20.4 N

fk

xy

x

y

!a

!a

Equations*we�re*left*with*to*solve*for a and T:

-fk - W1 sin 30.0o + T = m1a

T - W2 = -m2a ! T = W2 - m2a

Substituting*for*T in*the*first*equation:

-fk - W1 sin 30.0o + W2 - m2a = m1a

a = (-fk - W1 sin 30.0o + W2)/(m1 + m2)= (-20.4 - (8.00)(9.80)(0.500) + (22.0)(9.80))/(8.00 + 22.0)= 5.20 m/s2

T = W2 - m2a = (22.0)(9.80) - (22.0)(5.20) = 101 N