Announcements
• CAPA Set #6 due Friday at 10 pm
• This week in Section 
Lab 2: Acceleration due to gravity
Make sure to do the pre-lab and print materials before lab
* Slight mismatch of pre-lab file and web input. Just one
simple question added on web input. Follow web input.
• Read Chapter 5 Sections 5.1-5.5 (Circular Motion)
• Advanced reminder  Exam #2 on Tuesday, October 11
Friction
Why have you been ignoring me all this time?
Force resisting the movement of two objects in physical
contact past each other.
Force of Friction depends on:
1. Characteristics of materials
2. Vertical force pushing the materials together
Coefficient of Friction (m)
When we try to push this
green block sideways…
The Force of Friction acts in the
opposite direction to resist this
pushing force.
Maximum Force of Friction =
coefficient of friction (m) x
Force pushing surfaces together (Normal Force)
Ff(max) = m N
Static versus Kinetic Friction
When an object is not moving (static), molecules can form
more bonds and materials can deform into each other.
Thus, the coefficient of friction can be larger
(call it coefficient of static friction ms).
Once the object is moving, friction is lower
(call it the coefficient of kinetic friction mk).
Also, the force of friction never creates an acceleration,
it always acts to oppose motion
caused by some other effect.
Demonstration with Newton Meter and Blocks
Idealization – Not an Exact Law of Physics
Room Frequency BA
Clicker Question
You are pushing horizontally with a force of 5000 Newtons on a
car that has a weight of 10,000 Newtons.
The car is not moving.
What can you say for certain about the coefficient of friction?
A)
B)
C)
D)
E)
ms = 0
ms = 0.1
ms = 0.5
mk = 0.5
None of the above
But actually, ms could be even larger
since we do not know if we have
reached the maximum.
Fnet = 0 = Fpush – Ffriction
0=Fpush – ms x Normal
ms x Mg = Fpush
ms = Fpush/Mg = 5000/10k
ms= 0.5
Surfaces
µ (static)
µ (kinetic)
Steel on steel
0.74
0.57
Glass on glass
0.94
0.40
Metal on Metal (lubricated)
0.15
0.06
Ice on ice
0.10
0.03
Teflon on Teflon
0.04
0.04
Tire on concrete
1.00
0.80
Tire on wet road
0.60
0.40
Tire on snow
0.30
0.20
A block of mass m is being pushed across a rough
horizontal table. A constant velocity v is maintained
with an external force Fext. What is μK?
-
-
Fnet , y  ma y  N  mg  0
N  mg
Fnet , x  ma x  Fext  m k N  Fext  m k mg  0
Fext
mk 
mg
Clicker Question
Room Frequency BA
fmax = -μSN = -μSMg
An object with mass M is resting
on a rough table whose
coefficients of static and kinetic
friction are μS and μK, respectively.
Which of the following is a
necessary condition to start the
object in motion?
A) M > m
N=Mg
T
M
T
Mg
T=mg (little block)
T>msN (large block)
For motion to begin:
mg > μSMg
m
mg
B) M < m
C) mg > μK Mg
D) Mg > μS mg
E) mg > μs Mg
Room Frequency BA
Clicker Question
A block of mass m is pulled across a rough (μK) flat table
with a constant force Fext at an angle θ.
-
What is the correct expression for the magnitude of the
Normal force to use for calculating the friction force?
A)
B)
C)
D)
E)
N=mg
N=mg sinq
N=mg sinqFextsinq
N=mg –Fextcosq
None of the above
Fnet,y=may=0
0=N-mg+Fext sinq
N=mg-Fext sinq
A block of mass m slides down a rough (μK) incline tilted
at an angle θ from the horizontal.
f = μK N
mg cosθ
mg sinθ
N
θ
+y
Fg = mg
θ
+x
Fnet, y  ma y  N  mg cosq  0  N  mg cosq
Fnet , x  ma x  mg sin q  m k N
ma x  mg sin q  m k mg cosq
Circular Motion