1. SPH3UW Today’s Agenda Friction What is it?
Systematic catagories of forces
How do we characterize it?
Model of friction
Static & Kinetic friction (kinetic = dynamic in some languages)
Some problems involving friction

2. New Topic: Friction What does it do?
It opposes relative motion of two objects that touch!
How do we characterize this in terms we have learned (forces)?
Friction results in a force in the direction opposite to the direction of relative motion (kinetic friction, static – impending mot) j N
FAPPLIED i ma
fFRICTION
some roughness here mg

3. Surface Friction...
Friction is caused by the “microscopic” interactions between the two surfaces:

4. Surface Friction... Force of friction acts to oppose relative motion:
Parallel to surface.
Perpendicular to Normal force. j N F i ma fF mg

5. Model for Sliding (kinetic) Friction
The direction of the frictional force vector is perpendicular to the normal force vector N.
The magnitude of the frictional force vector |fF| is proportional to the magnitude of the normal force |N |.
|fF| = K | N | ( = K|mg | in the previous example)
The “heavier” something is, the greater the friction will be...makes sense!
The constant K is called the “coefficient of kinetic friction.”
These relations are all useful APPROXIMATIONS to messy reality.

6. Model... Dynamics: i : F  KN = ma j : N = mg so F Kmg = ma
(this works as long as F is bigger than friction, i.e. the left hand side is positive) j N F i ma
K mg mg

7. Lecture 7, Act 1 Forces and Motion
A box of mass m1 = 1.5 kg is being pulled by a horizontal string having tension T = 90 N. It slides with friction (mk = 0.51) on top of a second box having mass m2 = 3 kg, which in turn slides on a frictionless floor. (T is bigger than Ffriction, too.)
What is the acceleration of the second box ?
(a) a = 0 m/s2 (b) a = 2.5 m/s2 (c) a = 3.0 m/s2
Hint: draw FBDs of both blocks – that’s 2 diagrams
slides with friction (mk=0.51 ) T m1
a = ? m2
slides without friction

8. Lecture 7, Act 1 Solution First draw FBD of the top box: N1 m1
f = mKN1 = mKm1g T
m1g

9. Lecture 7, Act 1 Solution
Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2.
As we just saw, this force is due to friction:
= mKm1g m1
f1,2
f2,1 m2

10. Lecture 7, Act 1 Solution Now consider the FBD of box 2: N2
(contact from…)
(friction from…)
f2,1 = mkm1g m2
(contact from…)
(gravity from…)
m1g
m2g

11. Lecture 7, Act 1 Solution
Finally, solve F = ma in the horizontal direction:
mKm1g = m2a
a = 2.5 m/s2
f2,1 = mKm1g m2

12. Inclined Plane with Friction:
Draw free-body diagram: ma
KN j N  mg  i

13. Inclined plane... Consider i and j components of FNET = ma :
i mg sin KN = ma
j N = mg cos 
KN
mg sin Kmg cos  = ma ma j N 
a / g = sin Kcos   mg
mg cos  i
mg sin 

14. Static Friction... j N F i fF mg
So far we have considered friction acting when the two surfaces move relative to each other- I.e. when they slide..
We also know that it acts in when they move together: the ‘static” case.
In these cases, the force provided by friction will depend on the OTHER forces on the parts of the system. j N F i fF mg

15. Static Friction… (with one surface stationary)
Just like in the sliding case except a = 0.
i : F fF = 0
j : N = mg
While the block is static: fF F j N F i fF mg

16. Static Friction…
The maximum possible force that the friction between two objects can provide is fMAX = SN, where s is the “coefficient of static friction.”
So fF  S N.
As one increases F, fF gets bigger until fF = SN and the object starts to move.
If an object doesn’t move, it’s static friction
If an object does move, it’s dynamic friction j N F i fF mg

17. Static Friction...
S is discovered by increasing F until the block starts to slide:
i : FMAX SN = 0
j : N = mg
S FMAX / mg j N
FMAX i
Smg mg

18. Lecture 7, Act 2 Forces and Motion
A box of mass m =10.21 kg is at rest on a floor. The coefficient of static friction between the floor and the box is ms = 0.4.
A rope is attached to the box and pulled at an angle of q = 30o above horizontal with tension T = 40 N.
Does the box move?
(a) yes (b) no (c) too close to call T q m
static friction (ms = 0.4 )

19. Lecture 7, Act 2 Solution y x Pick axes & draw FBD of box:
Apply FNET = ma
y: N + T sin q - mg = maY = 0 N
N = mg - T sin q
= 80 N T
x: T cos q - fFR = maX q
The box will move if T cos q - fFR > 0
fFR m mg

20. Lecture 7, Act 2 Solution y x y: N = 80 N x: T cos q - fFR = maX
The box will move if T cos q - fFR > 0 N
T cos q = 34.6 N T
fMAX = msN
fMAX = msN = (.4)(80N) = 32 N q m
So T cos q > fMAX and the box does move mg
Now use dynamic friction: max = Tcosq - mKN

21. Static Friction:  We can also consider S on an inclined plane.
In this case, the force provided by friction will depend on the angle  of the plane. 

22. (Newton’s 2nd Law along x-axis)
Static Friction...
The force provided by friction, fF , depends on . fF
ma = 0 (block is not moving)
mg sin ff  i j N 
(Newton’s 2nd Law along x-axis) mg 

23. Static Friction...
We can find s by increasing the ramp angle until the block slides:
mg sin ff
In this case, when it starts to slide:
ffSN  Smg cos M
SN i j
mg sin MSmg cos M N
M mg 
Stan M

24. Additional comments on Friction:
Since fF = N , kinetic friction “does not” depend on the area of the surfaces in contact. (This is a surprisingly good rule of thumb, but not an exact relation. Do you see why??)
By definition, it must be true that S ³ K for any system (think about it...).

25. Aside: Graph of Frictional force vs Applied force: fF = SN fF = KN
fF = FA FA

26. Problem: Box on Truck
A box with mass m sits in the back of a truck. The coefficient of static friction between the box and the truck is S.
What is the maximum acceleration a that the truck can have without the box slipping? S m a

27. Problem: Box on Truck Draw Free Body Diagram for box:
Consider case where fF is max... (i.e. if the acceleration were any larger, the box would slip). N j i
fF = SN mg

28. Problem: Box on Truck Use FNET = ma for both i and j components
i SN = maMAX
j N = mg
aMAX = S g N j
aMAX i
fF = SN mg

29. Lecture 7, Act 3 Forces and Motion
An inclined plane is accelerating with constant acceleration a. A box resting on the plane is held in place by static friction. What is the direction of the static frictional force? S a Ff Ff Ff
(a) (b) (c)

30. Lecture 7, Act 3 Solution
First consider the case where the inclined plane is not accelerating. N Ff mg mg Ff N
All the forces add up to zero!

31. Lecture 7, Act 3 Solution
If the inclined plane is accelerating, the normal force decreases and the frictional force increases, but the frictional force still points along the plane: N Ff a mg
All the forces add up to ma!
F = ma
The answer is (a) mg Ff N ma

32. Putting on the brakes
Anti-lock brakes work by making sure the wheels roll without slipping. This maximizes the frictional force slowing the car since S > K .