Physics 2112 Unit 15 Today’s Concept: Ampere’s Law Unit 15, Slide 1.
Today’s Concept - University of Utahspringer/phys1500/Lectures/PHYS... · 2015. 2. 23. ·...
Transcript of Today’s Concept - University of Utahspringer/phys1500/Lectures/PHYS... · 2015. 2. 23. ·...
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Unit 6 Friction
Today’s Concept:Friction UP
Mechanics Lecture 6, Slide 1
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Midterm 1
Average=102.1”%”
Excellent Job!You can do the problems!
Mechanics Lecture 6, Slide 2
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Unit 5 Homework
Deadline for unit 5 Extended until this Thursday @ 11:30 Pm
Two of the problems were for unit 6….
Mechanics Lecture 6, Slide 3
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Video Prelectures/lecture Thoughts
Mechanics Lecture 6, Slide 4
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Friction
Mechanics Lecture 6, Slide 5
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Friction
Always opposes the relative motion of two surfacesMechanics Lecture 6, Slide 6
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Friction
Mechanics Lecture 6, Slide 7
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Static Friction
Mechanics Lecture 6, Slide 8
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Static Friction
Mechanics Lecture 6, Slide 9
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Static Friction
Mechanics Lecture 6, Slide 10
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Without Friction
θθ sinsin gm
mgm
Fa net ===
Mechanics Lecture 6, Slide 11
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Friction
xx maF =∑ yy maF =∑
Note that we have rotated the coordinate system!!!
Mechanics Lecture 6, Slide 12
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Friction
xx maF =∑ yy maF =∑
Note that we have rotated the coordinate system!!!
0ˆ
ˆcosˆ
=
−=∑jma
jmgjNF
y
y θ
θcosmgN =
Mechanics Lecture 6, Slide 13
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Friction
xx maF =∑
Note that we have rotated the coordinate system!!!
imaifimgF xkxˆˆˆsin =−=∑ θ
θcosmgN =
Nf kk µ= θµ cosmgk=
θµθ cossin gmmgma kx −=
( )θµθ cossin kx ga −=
Mechanics Lecture 6, Slide 14
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Block
22 2
21
txaatx ∆
=⇒=∆
mfmg
mFa knet −
==θsin
∆
−=−= 22sinsint
xgmmamgfk θθ
Mechanics Lecture 5, Slide 15
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Static Friction
xx maF =∑ifimgF sxˆˆsin −=∑ θ
θµµ cosmax, mgNf sss ==
θµθ cossin mgmg s>
maxmax cossin θµθ mgmg s=
Starts to Slide
maxmax cossin θµθ s=
maxmax
max tancossin θ
θθµ ==s
Mechanics Lecture 6, Slide 16
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Static Friction/Car Skidding
Mechanics Lecture 6, Slide 17
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Pushing Blocks
Mechanics Lecture 5, Slide 18
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Pushing Blocks
netF23
gmgmmNFynet 14323 2)( =+==
1
1
4321
1
4)( mF
mmmmFa hh =
+++=
2422)( 1
1
1114323
hhnet
FmFmamammF
x===+=
( )21
212
232
2323 22
gmFFFF hnetnetnet yx
+
=+=
xnetF23
Force to accelerate twoBlocks with acceleration=a
ynetF23
Normal force = weight of two blocks
Mechanics Lecture 5, Slide 19
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Main Points
Mechanics Lecture 6, Slide 20
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Main Points
Mechanics Lecture 6, Slide 21
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CheckPoint
A block slides on a table pulled by a string attached to a hanging weight. In Case 1 the block slides without friction and in Case 2 there is kinetic friction between the sliding block and the table.
In which case is the tension in the string biggest?A) Case 1 B) Case 2 C) Same
68% got this right on first try
Case 2(With Friction)
Case 1(No Friction)
m2
m1
g
m2
m1
g
Mechanics Lecture 6, Slide 22
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In which case is the tension in the string biggest?A) Case 1 B) Case 2 C) Same
B) M1 will not be accelerating as fast in case two than in case one because there is the extra force of friction acting against mass 1. Since the acceleration is smaller in case two, there has to be more of a force acting against gravity, the only other possible force is Tension.
Case 2(With Friction)
Case 1(No Friction)
m2
m1
g
m2
m1
g
Mechanics Lecture 6, Slide 23
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Lets work it out
m2
m1
g
A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is µk. What is the tension in the string?
Mechanics Lecture 6, Slide 24
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m2
m2g
TN
f
m1
m1g
T
m2
m1
g
1) FBD
Mechanics Lecture 6, Slide 27
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1) FBD2) ΣF=ma
add
N = m2gT – µm2g = m2a m1g – T = m1a
m1g – µm2g = m1a + m2a
a = m1g – µm2g
m1 + m2
m2
m2g
TN
f
m1
m1g
T
m2
m1
g
Mechanics Lecture 6, Slide 28
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T is smaller when a is bigger
m1g – T = m1a
T = m1g – m1aa = m1g – µm2g
m1 + m2
m2
m2g
TN
f
m1
m1g
T
m2
m1
g
1) FBD2) ΣF=ma
Mechanics Lecture 6, Slide 29
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CheckpointA.
B.
C.0% 0%0%
A box sits on the horizontal bed of a moving truck. Static friction between the box and the truck keeps the box from sliding around as the truck drives.
If the truck moves with constant acceleration to the left as shown, which of the following diagrams best describes the static frictional force acting on the box:
µS
a
A B C
Mechanics Lecture 6, Slide 30
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CheckPoint
If the truck moves with constant accelerating to the left as shown, which of the following diagrams best describes the static frictional force acting on the box:
A B C
A) In order to keep the box from sliding to the back of the truck as it accelerates, the frictional force needs to pull/push the box forward.
B) Friction always opposes motion/acceleration.
µS
a
56% correct
Mechanics Lecture 6, Slide 31
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Clicker Question
A box of mass M sits on a horizontal table. A horizontal string having tension T applies a force on the box, but static friction between the box and the table keeps the box from moving.
What is the magnitude of the total force acting on the box?
Since acceleration is zero.
A) MgB) mMgC) TD) 0
Mf T
Mechanics Lecture 6, Slide 32
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Clicker QuestionA.
B.
C.
D.
0% 0%0%0%
A box of mass M sits on a horizontal table. A horizontal string having tension T applies a force on the box, but static friction between the box and the table keeps the box from moving.
What is the magnitude of the static frictional force acting on the box?
Since the box is not moving the forces must be equal, otherwise there would be an acceleration.
A) MgB) mMgC) TD) 0
Mf T
38% correct
Mechanics Lecture 6, Slide 33
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Checkpoint A.
B.
C.
0% 0%0%
gmf
mF
a knet µ===
1
1
11
1
gmf
mF
a knet µ===
2
2
22
2
gaa kµ==⇒ 21
28% got this right on first try
Mechanics Lecture 6, Slide 34
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00 ==⇒= nettotal FFa
44% got this right on first try
Mechanics Lecture 6, Slide 35
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Checkpoint A.
B.
C.
0% 0%0%
0sin =−= θmgfFnet
Mechanics Lecture 6, Slide 36
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Mechanics Lecture 6, Slide 37
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Mechanics Lecture 6, Slide 38
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Mechanics Lecture 6, Slide 39
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Mechanics Lecture 6, Slide 40
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Mechanics Lecture 6, Slide 41
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Mechanics Lecture 6, Slide 42
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Mechanics Lecture 6, Slide 43
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Mechanics Lecture 6, Slide 44
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Microscopic explanation of Friction
What is Friction?Why can’t we walk through walls?
Basically the same answer to both questions…
Electron clouds of atoms repel (or bond to) each other
http://www.virneth.co.uk/topFriction/friction0.php http://astro1.panet.utoledo.edu/~vkarpov/Static_Friction_nature.pdf
Mechanics Lecture 6, Slide 45
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Climbing Skins for Touring Skis
Mechanics Lecture 6, Slide 46
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Accelerating Blocks
21 mmT
mFa
+==
amf 1=
Mechanics Lecture 6, Slide 47
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gmmT
gm
gmmf
a
gmf
S
SS
S
µ
µµ
µ
)( 21max
2
2
2
2
22
max
max
+=
===⇒
=
Accelerating Blocks
gm
gmmfa
gmf
kk
k
µµµ
===⇒
=
2
2
2
2
22
21
2
mmfT
mFa
+−
==
Mechanics Lecture 6, Slide 48
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fRRv πω 2==
( )
maFN
RfRva
==
== 22
2π
Carnival Ride
Mechanics Lecture 6, Slide 49
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( )
( ) ( ) Rfg
RfgR
vgR
Rvg
ag
mamg
Nmg
mgNf
s
s
s
22
22
22
/
min
min
min
ππµ
µ
µ
==
=====
==
Carnival Ride
min
min
1
s
s
WN
mgNf
µ
µ
=
==
Mechanics Lecture 6, Slide 50
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Accelerating Truck
tva∆∆
=
maFf ==
gamamgNf
S
SS
µµµ
=⇒===
max
maxmax
Mechanics Lecture 6, Slide 51
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Accelerating Truck
gmmg
mf
a
mgNf
KKsliding
sliding
KKsliding
µµ
µµ
===
==
gamamgNf
S
SS
µµµ
=⇒===
max
maxmax
Mechanics Lecture 6, Slide 52
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Mass on Incline
θ
θ
sin
sin
gmFa
mgF
x
x
==⇒
=
( )θµθ
θµ
cossin
cos
kx
k
gm
fFa
mgf
−=−
=⇒
=
Mechanics Lecture 6, Slide 53
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Mass on Incline
( )
( )x
mgk
mxkg
mxkfFa
mgf
s
sx
s
∆−
=
∆−−=
∆+−==⇒
=
θµθ
θµθ
θµ
cossin
cossin)(0
cos
Mechanics Lecture 6, Slide 54
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θθµθµ
µµθθ
θ
θθµ
θµ
cos)cos)((sin
)(cos)(sin)(0
sin
sincos
cos
2
1212
2
22121
2
21
1
2
11
222
min
min12
1
2
min
mmmm
mmmgmmg
mffFF
a
gmF
gmFgmf
gmf
ss
ssxx
x
x
s
s
−+=
+−+=
−−+==⇒
=
==
=
Mass on Incline
Mechanics Lecture 6, Slide 55
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Mass on Incline 2
θ
θ
sin
sin
gmFa
mgF
x
x
==⇒
=
( )
θθµ
θµθ
θµ
cossin
cossin
cos
gag
gm
fFa
mgf
k
kx
k
−=
−=−
=⇒
=
Mechanics Lecture 6, Slide 56
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Mass on Incline 2
( )
θ
θµ
θµθ
θµ
cos
sin
cossin)(0
cos
gm
xkg
mxkg
mxkfFa
mgf
s
sx
s
∆
−=
∆−−=
∆+−==⇒
=
Mechanics Lecture 6, Slide 57
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Mass on Incline 2
θµθθµθ
θµθθµθ
θθµθθ
θθµµ
sincos)cos(sin
sincoscossin
)sincos(cossin0
)sincos(
s
s
s
s
s
ss
mgmgmgT
TmgTmga
TmgNf
+−
=+−
=
++=⇒=
+==
Mechanics Lecture 6, Slide 58
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)cos(sin)sincos(
)(cos)(sin
)(cos)(sin)(0
sin
sincos
cos
212
22121
2
22121
2
21
1
2
11
222
min
min12
1
2
min
θµθθθµ
µµθθ
µµθθ
θ
θθµ
θµ
s
s
ss
ssxx
x
x
s
s
mm
mmgmmgm
mmgmmgm
ffFFa
gmF
gmFgmf
gmf
−−
=
+=+
+−+=
−−+==⇒
=
==
=
Mass on Incline 2
Mechanics Lecture 6, Slide 59