Unit 6 Friction

Today’s Concept:Friction UP

Mechanics Lecture 6, Slide 1

Midterm 1

Average=102.1”%”

Excellent Job!You can do the problems!

Mechanics Lecture 6, Slide 2

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

Video Prelectures/lecture Thoughts

Mechanics Lecture 6, Slide 4

Friction

Mechanics Lecture 6, Slide 5

Friction

Always opposes the relative motion of two surfacesMechanics Lecture 6, Slide 6

Friction

Mechanics Lecture 6, Slide 7

Static Friction

Mechanics Lecture 6, Slide 8

Static Friction

Mechanics Lecture 6, Slide 9

Static Friction

Mechanics Lecture 6, Slide 10

Without Friction

θθ sinsin gm

mgm

Fa net ===

Mechanics Lecture 6, Slide 11

Friction

xx maF =∑ yy maF =∑

Note that we have rotated the coordinate system!!!

Mechanics Lecture 6, Slide 12

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

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

Block

22 2

21

txaatx ∆

=⇒=∆

mfmg

mFa knet −

==θsin

∆

−=−= 22sinsint

xgmmamgfk θθ

Mechanics Lecture 5, Slide 15

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

Static Friction/Car Skidding

Mechanics Lecture 6, Slide 17

Pushing Blocks

Mechanics Lecture 5, Slide 18

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

Main Points

Mechanics Lecture 6, Slide 20

Main Points

Mechanics Lecture 6, Slide 21

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

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

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

m2

m2g

TN

f

m1

m1g

T

m2

m1

g

1) FBD

Mechanics Lecture 6, Slide 27

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

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

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

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

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

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

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

00 ==⇒= nettotal FFa

44% got this right on first try

Mechanics Lecture 6, Slide 35

Checkpoint A.

B.

C.

0% 0%0%

0sin =−= θmgfFnet

Mechanics Lecture 6, Slide 36

Mechanics Lecture 6, Slide 37

Mechanics Lecture 6, Slide 38

Mechanics Lecture 6, Slide 39

Mechanics Lecture 6, Slide 40

Mechanics Lecture 6, Slide 41

Mechanics Lecture 6, Slide 42

Mechanics Lecture 6, Slide 43

Mechanics Lecture 6, Slide 44

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

Climbing Skins for Touring Skis

Mechanics Lecture 6, Slide 46

Accelerating Blocks

21 mmT

mFa

+==

amf 1=

Mechanics Lecture 6, Slide 47

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

fRRv πω 2==

( )

maFN

RfRva

==

== 22

2π

Carnival Ride

Mechanics Lecture 6, Slide 49

( )

( ) ( ) 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

Accelerating Truck

tva∆∆

=

maFf ==

gamamgNf

S

SS

µµµ

=⇒===

max

maxmax

Mechanics Lecture 6, Slide 51

Accelerating Truck

gmmg

mf

a

mgNf

KKsliding

sliding

KKsliding

µµ

µµ

===

==

gamamgNf

S

SS

µµµ

=⇒===

max

maxmax

Mechanics Lecture 6, Slide 52

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

Mass on Incline

( )

( )x

mgk

mxkg

mxkfFa

mgf

s

sx

s

∆−

=

∆−−=

∆+−==⇒

=

θµθ

θµθ

θµ

cossin

cossin)(0

cos

Mechanics Lecture 6, Slide 54

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

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

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

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

)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