Today’s Concept - University of Utahspringer/phys1500/Lectures/PHYS... · 2015. 2. 23. ·...

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!


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 surfacesMechanics Lecture 6, Slide 6

Friction


Static Friction


Without Friction

θθ sinsin gm

mgm

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 θ

θcosmgN =


Friction

xx maF =∑


imaifimgF xkxˆˆˆsin =−=∑ θ

θcosmgN =

Nf kk µ= θµ cosmgk=

θµθ cossin gmmgma kx −=

( )θµθ cossin kx ga −=


Block

22 2

21

txaatx ∆

=⇒=∆

mfmg

mFa knet −

==θsin

∆

−=−= 22sinsint

xgmmamgfk θθ


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


Static Friction/Car Skidding


Pushing Blocks


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


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 µk. What is the tension in the string?


m2

m2g

TN

f

m1

m1g

T

m2

m1

g

1) FBD


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


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


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


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


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


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


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



00 ==⇒= nettotal FFa



Checkpoint A.

B.

C.

0% 0%0%

0sin =−= θmgfFnet


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/%7Evkarpov/Static_Friction_nature.pdf

http://astro1.panet.utoledo.edu/%7Evkarpov/Static_Friction_nature.pdf

Climbing Skins for Touring Skis


Accelerating Blocks

21 mmT

mFa

+==

amf 1=


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

+−

==


fRRv πω 2==

( )

maFN

RfRva

==

== 22

2π

Carnival Ride


( )

( ) ( ) 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

=

==


Accelerating Truck

tva∆∆

=

maFf ==

gamamgNf

S

SS

µµµ

=⇒===

max

maxmax


Accelerating Truck

gmmg

mf

a

mgNf

KKsliding

sliding

KKsliding

µµ

µµ

===

==

gamamgNf

S

SS

µµµ

=⇒===

max

maxmax


Mass on Incline

θ

θ

sin

sin

gmFa

mgF

x

x

==⇒

=

( )θµθ

θµ

cossin

cos

kx

k

gm

fFa

mgf

−=−

=⇒

=


Mass on Incline

( )

( )x

mgk

mxkg

mxkfFa

mgf

s

sx

s

∆−

=

∆−−=

∆+−==⇒

=

θµθ

θµθ

θµ

cossin

cossin)(0

cos


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


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

−=

−=−

=⇒

=


Mass on Incline 2

( )

θ

θµ

θµθ

θµ

cos

sin

cossin)(0

cos

gm

xkg

mxkg

mxkfFa

mgf

s

sx

s

∆

−=

∆−−=

∆+−==⇒

=


Mass on Incline 2

θµθθµθ

θµθθµθ

θθµθθ

θθµµ

sincos)cos(sin

sincoscossin

)sincos(cossin0

)sincos(

s

s

s

s

s

ss

mgmgmgT

TmgTmga

TmgNf

+−

=+−

=

++=⇒=

+==


)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


Today’s Concept - University of Utahspringer/phys1500/Lectures/PHYS... · 2015. 2. 23. ·...

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Transcript of Today’s Concept - University of Utahspringer/phys1500/Lectures/PHYS... · 2015. 2. 23. ·...