1. Homework
P , 33, 34, 35, 37

2. What causes friction?
Why is there Friction? Surface roughness Electronic interactions at the atomic level
Friction is caused by the temporary electrostatic bonds created between two objects in contact with one another.
Examples of Friction - Desirable - Undesirable

3. Examples of Friction. - Desirable. - Walking. - Driving. - Braking
Examples of Friction - Desirable - Walking - Driving - Braking - Undesirable - Engine Efficiency - Coasting - Pushing a heavy object

4. Why would I want to change friction? - How would I do it?

5. Friction & Applying Newton’s 2nd Law
System
Chapter 6.2

6. Friction How does friction affect the motion of objects?
It can slow an object down
like the friction between
the tires and the road.
It is responsible for
increasing the speed of an
object like a car.
It is also responsible
for objects being able to
change direction.

7. Static Friction Static Friction:
The resistive force that keeps an object from moving.
Fforward
Ffriction
Fnet = FAPPLIED – Ffriction
Since the crate is not accelerating, Fnet = 0
FAPPLIED = Ffriction
FAPPLIED
Ffriction
Fground-on-crate
Fgravity
System
Note: As long as the crate does not move, FAPPLIED = Ffriction

8. Fnet = FAPPLIED – Ffriction
Kinetic Friction
Kinetic Friction:
The resistive force that opposes the relative motion of two contacting surfaces that are moving past one another.
Since the crate will initially accelerate, Fnet > 0.
Fforward
Ffriction
Fground-on-crate
Fgravity
Fforward
Ffriction
Fnet = FAPPLIED – Ffriction
Fnet
System
Note: If the crate moves at a constant speed, then FAPPLIED = Ffriction and Fnet = 0.

9. An Important Term APPLIED FORCE Usually whatever is pushing or pulling
NOT the same as Net Force

10. Determining the Frictional Force
For people who had a lot of wrong ideas about Physics the
Greek alphabet sure gets used a lot!
The force of friction is proportional to the normal force and a proportionality constant ( - pronounced mu) called the coefficient of friction.
For static friction:
0 < Ff, static < s FN
For kinetic friction:
Ff, kinetic = k FN
Note: FN = the force normal (perpendicular) to the frictional force on the object.
 is dimensionless
Ff, static > Ff, kinetic Ff FN

11. Frictional Force For static friction: 0 < Ff, static < s FN
For kinetic friction:
Ff, kinetic = k FN

12. Determining the Frictional Force
 (the coefficient of friction) is usually in the range of 0<=  <= 1, but this is not always the case
Material 1
Material 2 
Tire, dry
Road, dry 1
Tire, wet
Road, wet
0.2
Rubber
Steel
1.6
Teflon
0.04
Ice
Wood
0.05
Glass
Metal
Chromium
0.41
Aluminum
1.3

13. Determining the Frictional Force
Sketch a graph of Fs vs applied force
Sketch a graph of Fk versus applied force
Sketch a graph showing the transition from Fs to Fk
Do this as a class exercise

14. Ff versus applied force

15. Ff versus applied force

16. The Normal Force
The normal force is a force that opposes the Earth’s gravitational attraction and is perpendicular to the surface that an object rests or is moving on.
For a horizontal surface, FN = Fg = mg.
For a surface that is not perpendicular to gravity, FN = Fgcos FN 

17. The Normal Force  FN FN Fg Fg FN = Fg = mg FN = Fg cos = mg cos
cos = adj/hyp Fg
FN = Fg = mg
FN = Fg cos = mg cos

18. Example 2: Determining Friction (Balanced Forces)
Assume that the man in the figure is pushing a 25 kg wooden crate across a wooden floor at a constant speed of 1 m/s.
How much force is exerted on the crate?
FAPPLIED Ff FN Fg
System

19. Diagram the Problem FAPPLIED Ff FN Fg FAPPLIED Ff FN Fg+y +x
FAPPLIED Ff FN Fg
FAPPLIED Ff FN Fg
System
y-direction: FN = Fg
x-direction: Fnet = FAPPLIED - Ff
Since the crate is moving with constant speed,
a = 0, Fnet = 0, and FAPPLIED = Ff

20. State the Known and Unknowns
What is known?
Mass (m) = 25 kg
Speed = 1 m/s
Acceleration (a) = 0 m/s2
k = 0.3 (wood on wood)
What is not known?
FAPPLIED = ?

21. Perform Calculations y-direction: x-direction: a = 0 Fg = FN = mg
Fnet = Fforward – Ff
FAPPLIED = Ff
FAPPLIED = kFN; FAPPLIED = kmg
FAPPLIED = (0.3)(25 kg)(9.8 m/s2)
FAPPLIED = 74 N

22. Example 3: Determining Friction (Unbalanced Forces)
Assume that the man in the figure is pushing a 25 kg wooden crate across a wooden floor at a speed of 1 m/s with a force of 74 N.
If he doubled the force on the crate, what would the acceleration be?
Assume Constant speed
FAPPLIED Ff FN Fg
System

23. Diagram the Problem FN FAPPLIED Ff FN Fg Ff FAPPLIED Fg+y +x FN
FAPPLIED Ff FN Fg
System Ff
FAPPLIED Fg
y-direction: FN = Fg
x-direction: Since a > 0, Fnet = Fforward - Ff

24. State the Known and Unknowns
What is known?
Force = 148 N
Mass (m) = 25 kg
Speed = 1 m/s
k = 0.3 (wood on wood)
What is not known?
a ?

25. Perform Calculations y-direction: x-direction: a > 0
Fg = FN = mg
x-direction: a > 0
Fnet = FAPPLIED – Ff
ma = FAPPLIED – Ff
ma = FAPPLIED – kmg
a = FAPPLIED – kmg m
a = (148N)/(25kg) – (0.3)(9.8 m/s2)
a = 2.96 m/s2
Fnet = 148N – 74N
ma = 74N
a = 74N/25kg
a = 2.96 m/s2

26. Key Ideas
Friction is an opposing force that exists between two bodies.
Friction is proportional to the normal force and the coefficient of friction; static or kinetic.
The force required to overcome static friction is greater than that required to overcome kinetic friction.