Unit 6 Friction Mechanics Lecture 6, Slide 1 Today’s Concept: Friction UP

Unit 6 Friction

Mechanics Lecture 6, Slide 1

Today’s Concept:Friction UP

Midterm 1


Average=102.1”%”

Excellent Job!You can do the problems!

Unit 5 Homework


Deadline for unit 5 Extended until this Thursday @ 11:30 Pm

Two of the problems were for unit 6….

Video Prelectures/lecture Thoughts


Friction


Friction

Always opposes the relative motion of two surfaces


Friction


Static Friction


Static Friction


Static Friction


Without Friction


sin

sing

m

mg

m

Fa net

Friction


xx maF yy maF

Note that we have rotated the coordinate system!!!

Friction


xx maF yy maF


0ˆ

ˆcosˆ

jma

jmgjNF

y

y

cosmgN

Friction


xx maF


imaifimgF xkxˆˆˆsin

cosmgN

Nf kk cosmgk

cossin gmmgma kx

cossin kx ga

Block


22 2

2

1

t

xaatx

m

fmg

m

Fa knet

sin

2

2sinsin

t

xgmmamgfk

Static Friction


xx maF ifimgF sxˆˆsin

cosmax, mgNf sss

cossin mgmg s

maxmax cossin mgmg s

Starts to Slide

maxmax cossin s

maxmax

max tancos

sin s

Static Friction/Car Skidding


Pushing Blocks


Pushing Blocks


netF23

gmgmmNFynet 14323 2)(

1

1

4321

1

4)( m

F

mmmm

Fa hh

24

22)( 1

1

1114323

hhnet

F

m

FmamammF

x

21

2

1223

22323 2

2gm

FFFF h

netnetnet yx

xnetF23

Force to accelerate twoBlocks with acceleration=a

ynetF23

Normal force = weight of two blocks

Main Points


Main Points


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

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

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 mk. What is the tension in the string?


m2

m2g

T

N

f

m1

m1g

T

m2

m1

g

1) FBD

1) FBD2) SF=ma

add

N = m2g

T – m m2g = m2a m1g – T = m1a

m1g – m m2g = m1a + m2a

a = m1g – m m2g m1 + m2

m2

m2g

T

N

f

m1

m1g

T

m2

m1

g


T is smaller when a is bigger

m1g – T = m1a

T = m1g – m1aa = m1g – m m2g m1 + m2

m2

m2g

T

N

f

m1

m1g

T

m2

m1

g


1) FBD2) SF=ma

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:

Sa

A B C

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.


Sa

56% correct

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


M

f T

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

M

f T

38% correct

Checkpoint A. B. C.


0% 0%0%

gm

f

m

Fa k

net 1

1

11

1

gm

f

m

Fa k

net 2

2

22

2

gaa k 21



00 nettotal FFa


Checkpoint A. B. C.


0% 0%0%

0sin mgfFnet









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


http://astro1.panet.utoledo.edu/~vkarpov/Static_Friction_nature.pdf

http://astro1.panet.utoledo.edu/~vkarpov/Static_Friction_nature.pdf

Climbing Skins for Touring Skis



Accelerating Blocks

21 mm

T

m

Fa

amf 1


gmmT

gm

gm

m

fa

gmf

S

SS

S

)( 21max

2

2

2

2

22

max

max

Accelerating Blocks

gm

gm

m

fa

gmf

kk

k

2

2

2

2

22

21

2

mm

fT

m

Fa


fRRv 2

maFN

RfR

va

22

2

Carnival Ride


Rf

g

Rf

gR

v

gR

Rv

g

a

g

ma

mg

N

mg

mgNf

s

s

s

22

22

22

/

min

min

min

Carnival Ride

min

min

1

s

s

W

N

mgNf


Accelerating Truck

t

va

maFf

ga

mamgNf

S

SS

max

maxmax


Accelerating Truck

gm

mg

m

fa

mgNf

KKsliding

sliding

KKsliding

ga

mamgNf

S

SS

max

maxmax


Mass on Incline

sin

sin

gm

Fa

mgF

x

x

cossin

cos

kx

k

gm

fFa

mgf


Mass on Incline

x

mgk

m

xkg

m

xkfFa

mgf

s

sx

s

cossin

cossin)(

0

cos


cos

)cos)((sin

)(cos)(sin)(0

sin

sin

cos

cos

2

1212

2

22121

2

21

1

2

11

222

min

min12

1

2

min

m

mmm

m

mmgmmg

m

ffFFa

gmF

gmF

gmf

gmf

ss

ssxx

x

x

s

s

Mass on Incline


Mass on Incline 2

sin

sin

gm

Fa

mgF

x

x

cos

sin

cossin

cos

g

ag

gm

fFa

mgf

k

kx

k


Mass on Incline 2

cos

sin

cossin)(

0

cos

gm

xkg

m

xkg

m

xkfFa

mgf

s

sx

s


Mass on Incline 2

sincos

)cos(sin

sincos

cossin

)sincos(cossin

0

)sincos(

s

s

s

s

s

ss

mgmgmgT

TmgTmg

a

TmgNf


)cos(sin

)sincos(

)(cos)(sin

)(cos)(sin)(0

sin

sin

cos

cos

212

22121

2

22121

2

21

1

2

11

222

min

min12

1

2

min

s

s

ss

ssxx

x

x

s

s

mm

mmgmmg

m

mmgmmg

m

ffFFa

gmF

gmF

gmf

gmf

Mass on Incline 2

Documents

Unit 6 Friction Mechanics Lecture 6, Slide 1 Today’s Concept: Friction UP