Chapter 4 Dynamics: Newton s Laws of Motion
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