Classical Mechanics
Lecture 6
Today’s Concept:
Friction
Mechanics Lecture 6, Slide 1
Midterm Redux
Seems like time is a
big issue!!!
Practice makes
faster…
Average=18.6/20
Mechanics Lecture 6, Slide 2
Friction
Mechanics Lecture 6, Slide 3
Friction
Mechanics Lecture 6, Slide 4
Friction
Mechanics Lecture 6, Slide 5
Friction
Always opposes the relative motion of two surfaces
Mechanics 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
a
Fnet mg sin 

 g sin 
m
m
Mechanics Lecture 6, Slide 11
Friction
Mechanics Lecture 6, Slide 12
Friction
Mechanics Lecture 6, Slide 13
Friction
Mechanics Lecture 6, Slide 14
Friction
Mechanics Lecture 6, Slide 15
Friction
Mechanics Lecture 6, Slide 16
Friction
Mechanics Lecture 6, Slide 17
Friction
Mechanics Lecture 6, Slide 18
Mechanics Lecture 6, Slide 19
Main Points
Mechanics Lecture 6, Slide 20
Main Points
Mechanics Lecture 6, Slide 21
Pre-lecture thoughts
Mechanics Lecture 6, Slide 22
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 23
Mechanics Lecture 6, Slide 24
Mechanics Lecture 6, Slide 25
Mechanics Lecture 6, Slide 26
Mechanics Lecture 6, Slide 27
Mechanics Lecture 6, Slide 28
Mechanics Lecture 6, Slide 29
Mechanics Lecture 6, Slide 30
Mechanics Lecture 6, Slide 31
47%
40%
Checkpoint
A.
13%
B.
C.
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.
S
a
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:
A
B
C
Mechanics Lecture 6, Slide 32
CheckPoint
S
a
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:
56% correct
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.
Mechanics Lecture 6, Slide 33
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?
T
f
A) Mg
B) mMg
C) T
D) 0
M
Since acceleration is zero.
Mechanics Lecture 6, Slide 34
Clicker Question
A.
B.
C.
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.
D.
What is the magnitude of the static frictional force acting on the box?
T
f
A) Mg
B) mMg
C) T
D) 0
M
Since the box is not moving the
forces must be equal, otherwise
38% correct there would be an acceleration.
100%
0%
0%
0%
Mechanics Lecture 6, Slide 35
CheckPoint
f1  k m1 g
f 2  k m1 g
F  ma  a 
F
m
47% got this right on
first try
Mechanics Lecture 6, Slide 36
Checkpoint
A.
B.
C.
a1 
Fnet1
a2 
Fnet2
m1
m2

f1
 k g
m1

f2
 k g
m2
 a1  a2  k g
0%
0%
0%
Mechanics Lecture 6, Slide 37



a  0  Ftotal  Fnet  0
38% got this right on
first try
Mechanics Lecture 6, Slide 38
Checkpoint
A.
B.
C.

Fnet  f  m g sin   0
0%
0%
0%
Mechanics Lecture 6, Slide 39
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.
m2
Case 1
(No Friction)
m2
g
Case 2
m1
(With Friction)
g
m1
In which case is the tension in the string biggest?
A) Case 1 B) Case 2 C) Same
65% got this right on
first try
Mechanics Lecture 6, Slide 40
Resume here
Mechanics Lecture 6, Slide 42
Lets work it out
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 k. What is the tension in the
string?
m2
g
m1
Mechanics Lecture 6, Slide 43
Tension in String
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
k. What is the tension in the string?
T1
T2
m2
What is the relationship between the magnitude
of the tension of the string at block 2 and the
magnitude of the tension in the string at block
1?
A) T1 > T2 B) T1 = T2
g
m1
C) T1 < T2
Mechanics Lecture 6, Slide 44
Acceleration of coupled blocks
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 k. What is the tension in the
string?
m2
g
What is the relationship between the magnitudes of
the acceleration of the two blocks?
m1
A) a1 = a2 B) a1 < a2 C) a1 > a2
Mechanics Lecture 6, Slide 45
1) FBD
m2
N
f
m2
T
g
T
m2g
m1
m1
m1g
Mechanics Lecture 6, Slide 46
1) FBD
2) SF=ma
m2
N
f
T
m2
T
m2g
N = m2g
T –  m2g = m2a
g
m1g – T = m1a
m1
m1
m1g
add
m1g –  m2g = m1a + m2a
m1g –  m2g
a=
m1 + m2
Mechanics Lecture 6, Slide 47
1) FBD
2) SF=ma
m2
N
f
m2
T
g
T
m1
m1
m2g
m1g
m1g –  m2g
a=
m1 + m2
m1g – T = m1a
T = m1g – m1a
T is smaller when a is bigger
Mechanics Lecture 6, Slide 48
Accelerating Blocks
a
F
T

m m1  m2
f  m1a
Mechanics Lecture 6, Slide 49
Accelerating Blocks
f 2 max   S m2 g
a
f 2 max
m2

 S m2 g
m2
 S g
Tmax  (m1  m2 )  S g
f 2   k m2 g
a
a
f 2  k m2 g

 k g
m2
m2
F T  f2

m m1  m2
Mechanics Lecture 6, Slide 50
Carnival Ride
v  R  2fR
v2
2
a
 2f  R
R
N  F  ma
Mechanics Lecture 6, Slide 51
Carnival Ride
f   smin N  m g
s
min
s
min
mg mg g
g
gR


  2
 2
N
ma a v / R
v
gR
g


2Rf 2 2f 2 R


f   smin N  m g
N
1

W  smin
Mechanics Lecture 6, Slide 52
Accelerating Truck
a
v
t
f  F  ma
f max   S N   S m g  m amax
 amax   S g
Mechanics Lecture 6, Slide 53
Accelerating Truck
f sliding   K N   K m g
asliding 
f sliding
m

K m g
m
 K g
f max   S N   S m g  m amax
 amax   S g
Mechanics Lecture 6, Slide 54
Mass on Incline
Fx  m g sin 
a
Fx
 g sin 
m
f   k m g cos
a
Fx  f
 g sin    k cos 
m
Mechanics Lecture 6, Slide 55
Mass on Incline
f   s m g cos
Fx  ( f  kx)
kx
 g sin    s cos  
m
m
m gsin    s cos 
k
x
a0
Mechanics Lecture 6, Slide 56
Mass on Incline
f 2   s 2min m2 g cos
f1   s m1 g cos
Fx2  m2 g sin 
Fx1  m1 g sin 
a0
s2
min

Fx2  ( Fx1  f1 )  f 2
m2

g sin  (m1  m2 )  g cos (  s m1   s 2min m2 )
m2
(sin  (m1  m2 )   s m1 cos )
m2 cos
Mechanics Lecture 6, Slide 57
Mass on Incline 2
Fx  m g sin 
a
Fx
 g sin 
m
f   k m g cos
Fx  f
 g sin    k cos 
m
g sin   a
k 
g cos
a
Mechanics Lecture 6, Slide 58
Mass on Incline 2
f   s m g cos
Fx  ( f  kx)
kx
 g sin    s cos  
m
m
kx 

 g sin  

m 

s 
g cos
a0
Mechanics Lecture 6, Slide 59
Mass on Incline 2
f   s N   s (m g cos  T sin  )
a0
 m gsin   T cos   s (m g cos  T sin  )
T
m g sin    s m g cos m g(sin    s cos )

cos   s sin 
cos   s sin 
Mechanics Lecture 6, Slide 60
Mass on Incline 2
f 2   s 2min m2 g cos
f1   s m1 g cos
Fx2  m2 g sin 
Fx1  m1 g sin 
a0
Fx2  ( Fx1  f1 )  f 2
m2

g sin  (m1  m2 )  g cos (  s m1   s 2min m2 )
m2
g sin  (m1  m2 )  g cos (  s m1   s 2min m2 )
m2  m1
(  s cos  sin  )
(sin    s 2 cos )
Mechanics Lecture 6, Slide 61
Block
1 2
2x
at  a  2
2
t
F
m g sin   f k
a  net 
m
m
2x 

f k  m g sin   m a  m g sin   2 
t 

x 
Mechanics Lecture 5, Slide 62
Pushing Blocks

F23net
a
Fh1
F
 h1
(m1  m2  m3  m4 ) 4m1
F23 netx  (m3  m4 )a  2m1a 
2m1 Fh1 Fh1

4m1
2
F23 net y  N  (m3  m4 ) g  2m1 g
2
F23 net  F
2
23 net x
F
2
23 net y
F 
2
  h1   2m1 g 
 2 
Mechanics Lecture 5, Slide 63