static friction

friction-physics

1. Frictional Forces..or lack thereof

2. Friction
Definition: the forcethat
opposes the motion of one
surface past another
Dependent upon: 1) type of
material
2)how tightly pressed they are
together (more weight =more
friction)
It’s time to experience friction,
so rub your hands together.

3. Why would a snowboarder wax the board?
What conditions would change the frictional
force as they move downhill?

4. Characteristics of friction
1)ALWAYS opposes motion
a) it will slow it
b) or prevent it from moving
2)Creates heat
3) Symbol is lowercase ‘f’…Ff

5. 2 types of friction
Static
1) prevents motion from
starting.
2) occurs before motion
Which is bigger…static or
kinetic?
Kinetic friction
1) slows objects down.
2) occurs during motion
It is easier to keep an object
moving than start it from rest

7. New Symbol
•μ–Greek letter “Mu”
•Coefficient of friction.
–μ
s
–coefficient of static friction
–μ
k
–coefficient of kinetic friction
–depends on materials used and their surface
conditions
•Coefficient of friction: Decimal between 0.0 and
1.0, unitless.
•Formula for solving friction:Ff= μ F
N

8. Sand is often placed on an icy road
because the sand:
Decreases the coefficient of friction between the tires of a car
and the road
Increases the coefficient of friction between the tires of a car
and the road
Decrease the gravitational force on a car
Increases the normal force of a car on the road

9. An empty cart is being rolled across a
warehouse floor. If the cart was filled, the
force of kinetic friction between the cart
and the floor would
Decrease
Increase
Remain the same

10. A 24 kg crate initially at rest on a horizontal
floor requires a 75N horizontal force to set it
in motion. Find the coefficient of static
friction between the crate and floor.

11. Once the crate is in motion, a horizontal
force of 53N keeps it moving with constant
velocity. Find the coefficient of kinetic
friction between the crate and floor.

12. A 55kg baseball player slides into third base
with an initial speed of 4.6 m/s. If the
coefficient of kinetic friction between the
player and the ground is 0.46, what is the
player’s acceleration? How far did he slide?
1)Always draw your FBD
2)Fnet= ma= Ff
3)Ff=µFN
=0.46(539N)
= -248 N
4) F=ma
a= F/m
=-248N/55kg
= -4.5 m/s²

13. Where does this apply?
To car tires in
accidents.
By measuring the
length of skid marks,
police can calculate the
speed a car was going
before an accident.

14. Static and Kinetic Frictional Force
Known Variables
Fk= 0.0500
v
o= 4.00 m/s
v
f= 0.00 m/s
Unknown Variables
a
x=
x =
Formula
1) F
N= mg = weight
2) f
k = 
kF
N
3) ΣF= ma
x = 
kF
N
4) ma
x = 
kmg
5) (/m=>) a
x= 
kg
6) a
x= .05 (9.80m/s
2
)
7)v
f
2
=v
o
2
+2a
xx. Find x

15. Static and Kinetic Frictional Force
Known Variables

k= 0.0500
v
o= 4.00 m/s
v
f= 0.00 m/s
Unknown Variables
a
x= 0.49 m/s
2
x =16.3m
Formula
1) F
N= mg = weight
2) ΣF= f
k, f
k = 
kF
N
3) ΣF= ma
x = 
kF
N
4) ma
x = 
kmg
5) (/m=>) a
x= 
kg
6) a
x= .05 (9.80m/s
2
)
7)v
f
2
=v
o
2
+2a
xx. Find x

16. Static and Kinetic Frictional Force
Did we need to know the mass of the
sleder? No.
Why? It cancels out in the a
xequation.
Real Life Ap: This applies to car tires
in accidents. By measuring the length
of skid marks, they can calculate the
speed a car was going before an
accident. 
kof a tire is the same for all
cars since it does not depend on car
mass or surface area of the tires.
Formula
1) F
N= mg = weight
2) ΣF= f
k, f
k = 
kF
N
3) ΣF= ma
x = 
kF
N
4) ma
x = 
kmg
5) (/m=>) a
x= 
kg
6) a
x= .05 (9.80m/s
2
)
7)v
f
2
=v
o
2
+2a
xx. Find x

17. Static and Kinetic Frictional Forces
Static Frictional Force:
Reaction force to anything trying to
start motion.
•Equal and opposite to applied force.
•DOES NOT EXCEED THE
APPLIED FORCE,
but is equal to it.

18. Static and Kinetic Frictional Forces
Static Frictional Force:
Reaction force to anything trying to
start motion.
Equal and opposite to applied force,
until reaches maximum value and motion
starts.
friction “breaks” when F is great enough
and motion begins.

19. Static and Kinetic Frictional Forces
Static Frictional Force Breaks at a certain value:
f
s= 
s F
N
f
s= force of static friction

s = coefficient of static friction
F
N = Normal force

20. Static and Kinetic Frictional Forces
Static Frictional Force Breaks at a certain value:
f
s= 
s F
N
f
s= force of static friction

s = coefficient of static friction
F
N = Normal force

s is a given value. It depends on the object and
the surface.

21. Static and Kinetic Frictional Forces
Static Frictional Force:
f
s= force of static friction

s = coefficient of static friction
F
N = Normal force (usually weight)
Normal force is usuallyjust the
weight of the object.
F
N =Mass* 9.80 m/s
2
IMPORTANT!!!!!
If the surface is not horizontal use trig.
Multiply by cos of the angle of incline.

22. Notes on friction
Almost always:
μ
s
> μ
k
It is easier to keep an object moving than it is to start
from rest. Think about pushing a car.
Both are almost always less than 1. If it was greater
than one, it would be easier to pick the object up
and carry it than it would be to push it across the flat
surface (something like velcro)

23. Problem
Box:
How much force is needed to “budge” this
box?
If we keep pushing that hard, what will the
acceleration be?

24. Problem
Box:
f
s= 
s F
N = .4 (10kg) ( 9.80m/s
2
)
f
s= 39 N (Breaking Force)

25. Problem
Box:
f
k= 
k F
N = .2 (10kg) ( 9.80m/s
2
)
f
k= 19.5 N (Kinetic Force)
Net Force = Pushing Force –Kinetic Friction Force
Net Force = 39N –19.5N = 19.5N
a = F / m = 19.5N / 10kg = 1.95 m/s
2
= 2 m/s
2

26. The Tension Force

27. The Tension Force
Tension is the force balanced
by a rope, cable or wire.
A “simple pulley” changes
direction without affecting
tension.
Tension is the same at every
point in a single rope.

28. Equilibrium Applications of
Newton’s Laws of Motion
An object is in equilibrium when it has zero acceleration
“equilibrium” refers to a lack of change, but in the sense
that the velocity of an object isn’t changing, i.e, there is no
acceleration.
Equilibrium: Constant Speed and Direction.
F
x = 0 and F
y= 0, a
x= 0 m/s
2
and a
y = 0 m/s
2
a
x= 0 m/s
2
and a
y = 0 m/s
2

F
x = 0 and F
y= 0

29. Reasoning Strategy
If F = 0, thenFx = 0 and Fy = 0.
•Draw a free-body diagram the object. Be sure to include only the
forces that act on the object; do not include forces that the object
exerts on its environment.

30. Example
A jet plane is flying with a constant
speed along a straight line at an angle of
30.0
o
above the horizontal. The plane
has a weight W whose magnitude is
W=86,500 N and its engine provide a
forward thrust T of magnitude
T=103,000 N. In addition, the lift force
L (directed perpendicular to the wings)
and the force R of air resistance
(directed opposite to the motion)
act on the plane. Find L and R.

31. List our Forces:
Weight –86,500 @ 270
Thrust -103,000 @ 30
Lift -??? @ 120
Drag -??? @ 210
(drag is a friction)

32. List our Forces:
Weight –86,500 @ 270
Thrust -103,000 @ 30
Lift -??? @ 120
Drag -??? @ 210
(drag is a friction)
This looks really complicated.

33. Shortcut:
Since 3 of our forces are
Perpendicular, lets change
The axes...

34. Now:
Weight –86,500 @ 240
Thrust -103,000 @ 0
Lift -??? @ 90
Drag -??? @ 180
(drag is a friction)
3 nice angles are better than 1.

35. x component
W: 86,500 cos 240
L: 0 (L cos 90)
T: +103,000N
R: -R (R cos 180)
y component
W: 86,500 sin 240
L: +L
T: 0
R: 0
Fx = W cos240.0
o
+ T -R = 0 Fy = -Wcos30.0
o
+ L= 0
R=59,800N and L = 74,900 N

36. Nonequilibriuium Applications
of Newton’s Laws of Motion
Non-equilibriumconditions occur when the object
is accelerating and the forces acting on it are not balanced
so the net force is not zero.
Non-equilibrium:F
x = ma
xand F
y= ma
y
F
x = ma
xand F
y= ma
y

37. Example
A supertanker of mass m = 1.50 X 10
8
kg is being towed by two tugboats. The
tensions in the towing cables apply the forces T1 and T2 at equal angles of 30.0
o
with respect to the tanker’s axis. In addition, the tanker’s engines produce a
forward drive force D whose magnitude is D = 75.0 X 10
3
N. Moreover, the
water applies an opposing force R, whose magnitude is R = 40.0 X 10
3
N. The
tanker moves forward with an acceleration that points along the tanker’s axis and
has a magnitude of 2.00 X10
-3
m/s
2
. Find the magnitudes of T1 and T2.

38. x component
T1: +T1cos30.0
o
T2: +T2cos30.0
o
D: +D = 75.0 X10
3
N
R: -R=-40.0 N
y component
T1: +T1sin30.0
o
T2: -T2sin30.0
o
D: 0
R: 0
Fy = +T
1cos30.0
o
–T
2sin30.0
o
= 0Fx = +T
1cos30.0
o
+ T
2cos30.0
o
+D -R = ma
x
T=1.53 X 10
5
N

39. Example
The figure shows a water skier at four different moments:
a) The skier is floating motionless in the water
b) The skier is being pulled out of the water and up onto the skis
c) The skier is moving at a constant speed along a straight line
d) The skier has let go of the tow rope and is slowing down
For each moment, explain whether the net force acting on the skier
is positive, negative, or zero.

40. Example
A flatbed is carrying a crate up a 10.0
o
hill. the coefficient of
static friction between the truck bed and the crate is 
s= 0.350.
Find the maximum acceleration that the truck can attain before the
crate begins to slip backward relative to the track. (p. 114)

41. x component givens
W: F
wx = -mgsin10.0
o
(gravity pulls backwards at10.0
o
)

s= 0.350
f
s =
sF
N =.35 mgcos10.0
o
F
truck = ma
x = Fx
y component givens
W: -mgsin10.0
o
F
N: F
N = mgsin10.0
o
Equations
Fx = -Gravity+ Friction= Truck Engine Accel
Fx = F
wx + 
sF
N
= ma
x
Fx = -mgsin10.0
o
+ .35 mgcos10.0
o
= ma
x
(/m out)= -9.80sin10.0
o
+.35(9.8)cos10.0
o
=a
x
y comp equations
Fy = -mgsin10.0
o
+ F
N = 0
a
x= 1.68 m/s
2
Who uses this info? Not the driver. The engineers use it to figure out if
they need to add more tie downs to the truck bed design.
Free Body Diagram for X comp
Gravity
Friction
Truck
Engine
10
o

42. Example
Block 1 (mass m1 = 8.00 kg) is moving n a frictionless 30.0
o
incline. This
block is connected to block 2 (mass m2 = 22.0 kg) by a cord that passes over
a massless and frictionless pulley, Find the acceleration of each block and the
tension in the cord. (p. 115)

43. x component
W1: -W1sin30.0
o
T: T
y component
W2: -W2
T: T
Fx = -W
1sin30.0
o
+ T= m
1a Fy = T –W
2
= m
2(-a)