Friction
Friction Problem Situations
Physics
Montwood High School
R. Casao
Friction
• Friction Ff is a force that resists motion
• Friction involves objects in contact
with each other.
• Friction must be overcome before
motion occurs.
• Friction is caused by the uneven
surfaces of the touching objects. As
surfaces are pressed together, they
tend to interlock and offer resistance to
being moved over each other.
Microscopic Friction
Surface Roughness
Magnified section of a
polished steel surface
showing surface bumps
about 5 x 10-7 m (500 nm)
high, which corresponds
to several thousand
atomic diameters.
Adhesion
Computer graphic from
a simulation showing gold
atoms (below) adhering to
the point of a sharp nickel
probe (above) that has
been in contact with the
gold surface.
Friction
• Frictional forces are always in the
direction that is opposite to the
direction of motion or to the net force
that produces the motion.
• Friction acts parallel to the surfaces in
contact.
Types of Friction
• Static friction: maximum frictional force
between stationary objects.
• Until some maximum value is reached and
motion occurs, the frictional force is
whatever force is necessary to prevent
motion.
• Static friction will oppose a force until such
time as the object “breaks away” from the
surface with which it is in contact.
• The force that is opposed is that
component of an applied force that is
parallel to the surface of contact.
Types of Friction
• The magnitude of the static friction force Ffs
has a maximum value which is given by:
Ff s  s  FN
• where μs is the coefficient of static friction
and FN is the magnitude of the normal force
on the body from the surface.
Types of Friction
• Sliding or kinetic friction: frictional force
between objects that are sliding with respect
to one another.
• Once enough force has been applied to the object
to overcome static friction and get the object to
move, the friction changes to sliding (or kinetic)
friction.
• Sliding (kinetic) friction is less than static friction.
• If the component of the applied force on the object
(parallel to the surface) exceeds Ffs then the
magnitude of the opposing force decreases rapidly
to a value Fk given by:
Fk  k  FN
where μk is the coefficient of kinetic friction.
Static Friction
The static frictional force keeps an object from
starting to move when a force is applied. The static
frictional force has a maximum value, but may take on
any value from zero to the maximum, depending on
what is needed to keep
the sum of forces zero.
Types of Friction
• From 0 to the maximum value of the static
frictional force Fs in the figure, the applied
force is resisted by the static frictional force
until “breakaway”.
• Then the sliding (kinetic) frictional force Fk
is approximately constant.
Types of Friction
• Static and sliding friction are
dependent on:
• The nature of the surfaces in contact.
Rough surfaces tend to produce more
friction.
• The normal force (Fn) pressing the
surfaces together; the greater Fn is, the
more friction there is.
Friction vs. Area
Question: Why doesn’t friction depend on contact area?
The microscopic area of contact between a box and the floor is
only a small fraction of the macroscopic area of the box’s bottom
surface.
If the box is turned on its side, the macroscopic area is increased,
but the microscopic area of contact remains the same (because the
contact is more distributed). Therefore the frictional force f is
independent of contact area.
Types of Friction
• Rolling friction: involves one object
rolling over a surface or another
object.
• Fluid friction: involves the
movement of a fluid over an object
(air resistance or drag in water) or
the addition of a lubricant (oil,
grease, etc.) to change sliding or
rolling friction to fluid friction.
Coefficient of Friction
• Coefficient of friction (): ratio of
the frictional force to the normal
force pressing the surfaces together.
 has no units.
• Static:
F
μs 
fs
Fn
• Sliding (kinetic): μ  Ffk
k
Fn
•The maximum frictional force is 50 N. As the applied
force increases from 0 N to 50 N, the frictional force also
increases from 0 N to 50 N and will be equal to the
applied force as it increases.
•Once the static frictional force of 50 N has been
overcome, only a 40 N force is needed to overcome the
40 N kinetic frictional force and produce constant
velocity (a = 0 m/s2).
•As the applied force increases beyond 40 N, the kinetic
frictional force remains at 40 N and the 100 N block will
accelerate.
A Model of Friction
Friction
Static Friction
Kinetic Friction
Fpush  f k  k  FN
Kinetic Friction and Speed
The kinetic frictional force is also
independent of the relative speed of the
surfaces, and of their area of contact.
Rolling Friction
Horizontal Surface – Constant Speed
•Constant speed:
a = O m/s2.
•The normal force
pressing the
surfaces together is
the weight; Fn = Fw
ΣFx  m  a
Fx  Ff  m  a
Ff
Ff
μk 

Fn Fw
Fx  Ff  0 N
Ff  μ k  Fw
Fx  Ff
Fx  Ff  μ k  Fw
Horizontal Surface: a > O m/s2
Fx  Ff
ΣFx  m  a
Fx  Ff  m  a
Fn  Fw
Ff
Ff
μk 

Fn Fw
Ff  μ k  Fw
Horizontal Surface: a > O m/s2
• If solving for:
• Fx: Fx  m  a  Ff
Fx  m  a  μ k  Fw
Fx  m  a  μ k  m  g
• F f:
Ff  Fx  m  a
• a:
Fx  Ff
a
m
Horizontal Surface: Skidding to a
Stop or Slowing Down (a < O m/s2)
• The frictional force is responsible for the
negative acceleration.
• Generally, there is no Fx.
 Ff  m  a
Fn  Fw
Ff
Ff
μk 

Fn Fw
Ff  μ k  Fw
Horizontal Surface: Skidding to a
Stop or Slowing Down (a < O m/s2)
• Most common use involves finding
acceleration with a velocity equation
and finding k:
2
2
v f  v i  (2  a  Δx )
2
Δx  (v i  t )  (0.5  a  t )
v f  v i  (a  t )
• Acceleration will be negative
because the speed is decreasing.
Horizontal Surface: Skidding to a
Stop or Slowing Down (a < O m/s2)
Ff
Ff
m  a a
μk 



Fn
Fw
mg
g
• The negative sign for acceleration a is
dropped because k is a ratio of forces
that does not depend on direction.
• Maximum stopping distance occurs when
the tire is rotating. When this happens,
a = -s·g.
• Otherwise, use a = -k·g to find the
acceleration, then use a velocity equation
to find distance, time, or speed.
Friction, Cars, & Antilock Brakes
The diagram shows forces acting on a car
with front-wheel drive. Typically, Fn > Fn’
because the engine is over the front wheels.
The largest frictional force fs the tire can exert
on the road is µs·Fn. Attempts to make the tire
exert a force larger than this causes the tire to
“burn rubber” and actually reduces the force,
since µk<µs.
Note that while all points on the rolling tire
have the same speed v in the reference frame
of the car, in the reference frame of the road
the bottom of the tire is at rest, while top is
moving forward with a speed of 2·v.
Antilock brakes sense the wheel rotation
and “ease off” if it close to stopping,
maintaining static friction with the road and
allowing better control of steering than if the
wheels were locked.
Antilock Brakes
Example:
The Effect of Antilock Brakes
A car is traveling at 30 m/s along a horizontal road.
The coefficients of friction are ms=0.50 and mk=0.40.
(a) What is the braking distance xa with antilock
brakes?
(b) What is the braking distance xb if the brakes lock?
2
v
v 2  v02  2  a  x and v  0, so x   0
2a
aa  s  g and ab  k  g
v02
(30 m/s)2
xa 

 91.7 m
2
2  s  g 2  (0.5)  (9.81 m/s )
v02
(30 m/s)2
xb 

 114.7 m
2
2  k  g 2  (0.4)  (9.81 m/s )
Example: A Game of Shuffleboard
A cruise-ship passenger uses a
shuffleboard cue to push a shuffleboard disk
of mass 0.40 kg horizontally along the deck,
so that the disk leaves the cue at a speed of
8.5 m/s. The disk then slides a distance of
8.0 m.
f k  k Fn
What is the coefficient of kinetic friction
between the disk and deck?
F
y
 m  ay  0
Fn  m  g  0  Fn  m  g
F
x
 m  ax
 f k   k  m  g  m  ax a x    k g
vx2  v02x  2  ax x  0  v02x  2  k  g  x
v02x
(8.5 m/s)2
k 

 0.46
2
2  g  x 2  (9.81 m/s )  (8.0 m)
Down an Inclined Plane
Down an Inclined Plane
• Resolve Fw into Fx and Fy.
• The angle of the incline is always equal to
the angle between Fw and Fy.
• Fw is always the hypotenuse of the right
triangle formed by Fw, Fx, and Fy.
cos θ 
Fy
Fw
Fx
sin θ 
Fw
Fy  Fw  cos θ
Fx  Fw  sin θ
Down an Inclined Plane
• The force pressing the surfaces
together is NOT Fw, but Fy; Fn = Fy.
ΣF  m  a
or
Fx  Ff  m  a
Fx  Ff
a
m
Ff
Ff
μk 

Fn Fy
Ff  μ k  Fy
m  g  sin       m  g  cos    m  a
mass m cancels out
(g  sin  )  (  g  cos  )  a
(g  sin  )  a

g  cos 
Down an Inclined Plane
• If we place an object on an inclined plane and
increase the tilt angle  to the point at which the
object just begins to slide.
• What is the relation between  and the static
coefficient of friction µs?
y - axis : FN  Fw  cos 
x - axis : f s  s  FN  Fw  sin 
fs
Fw  sin 
s 

Fn Fw  cos 
sin 
s 
 tan 
cos 
Down an Inclined Plane
• If the object slides down the incline at constant speed
(a = 0 m/s2), the relation between  and the kinetic
coefficient of friction µk:
Fx  Ff  m  0
m
s
2
Fx  Ff  0 N
Fx  Ff
Ff
Fx
Fw  sin θ sin θ
μk 



 tan θ
Fn Fy Fw  cos θ cos θ
μ k  tan θ
Down an Inclined Plane
• To determine the angle of the
incline:
• If moving:
1
μk
1
μs
θ  tan
• If at rest:
θ  tan
Example: A Sliding Coin
A hardcover book is resting on a
tabletop with its front cover facing upward.
You place a coin on the cover and very
slowly open the book until the coin starts
to slide. The angle  is the angle of the
cover just before the coin begins to slide.
Find the coefficient of static friction µs
between the coin and book.
F
y
 m  ay
 Fn  m  g  cos  0 or FN  m  g  cos
f s  s  FN at  , so f s  s  m  g  cos 
F
x
 m  ax
 m  g  sin   f s  0 or f s  m  g  sin 
Therefore, s  cos   sin  or s  tan 
Example:
Dumping a file cabinet
Steel on dry steel 
Free-body diagram
A 50.0 kg steel file cabinet is in the back of a dump truck.
The truck’s bed, also made of steel, is slowly tilted. What is
the size of the static friction force when the truck’s bed is
tilted by 20°? At what angle will the file cabinet begin to
slide?
Example:
Dumping a file cabinet
F
F
x
 w  sin   f s  m  g  sin   f s  0;
y
 n  w  cos   n  m  g  cos   0;
f s  m  g  sin   (50.0 kg)  (9.80 m/s 2 )  sin 20  168 N ;
File cabinet will begin to slide when:
f s  f s max   s  n   s  m  g  cos  ;
m  g  sin   f s  m  g  sin   s  m  g  cos   0;
sin 
s 
 tan  ;   arctan  s  arctan(0.80)  38.7
cos 
Non-Parallel Applied Force on Ramp
If an applied force acts on the box
at an angle  above the
horizontal, resolve FA into parallel
and perpendicular components FA
using the angle  +  :
FA ·sin( +  )
N

FA ·cos ( + θ) and FA ·sin ( + θ)
FA ·cos( +  )
fk

m·g ·sin
FA serves to increase acceleration
directly and indirectly: directly by
FA ·cos ( + θ) pulling the box
down the ramp, and indirectly by
FA ·sin ( + θ) lightening the
normal support force with the
ramp (thereby reducing friction).

m·g
m·g ·cos
Non-Parallel Applied Force on Ramp
FA ·sin ( + )
FA
N

FA ·cos( +  )
fk

If FA ·sin( +  ) is not big enough
to lift the box off the ramp, there
is no acceleration in the
perpendicular direction. So,
FA ·sin( +  ) + FN = m·g·cos.
Remember, FN is what a scale
would read if placed under the
box, and a scale reads less if a
force lifts up on the box. So,
FN = m·g ·cos - FA ·sin( +  ),
which means fk = k ·FN
= k ·[m·g ·cos - FA ·sin( +  )].
m·g ·sin

m·g
Non-Parallel Applied Force on Ramp
FA ·sin( +  )
FA
N

FA ·cos( +  )
fk

m·g ·sin

If the combined force of FA ·cos( +  ) +
m·g ·sin is is enough to move the box:
FA ·cos( +  ) + m·g·sin
- k ·[m·g·cos - FA ·sin( +  )] = m·a
m·g ·cos
m·g
Up an Inclined Plane
Up an Inclined Plane
• Resolve Fw into Fx and Fy.
• The angle of the incline is always
equal to the angle between Fw and
Fy.
• Fw is always the hypotenuse of the
right triangle formed by Fw, Fx, and
Fy.
Fy
F
cos θ 
Fw
sin θ 
Fy  Fw  cos θ
Fx  Fw  sin θ
x
Fw
Up an Inclined Plane
• Fa is the force that must be applied
in the direction of motion.
• Fa must overcome both friction and
the x-component of the weight.
• The force pressing the surfaces
together is Fy.
Up an Inclined Plane
Fn  Fy
ΣFx  m  a
Fa  Ff  Fx  m  a
Fa  Ff  Fx
a
m
Ff
Ff
μk 

Fn Fy
Ff  μ k  Fy
•For constant
speed, a = 0 m/s2.
Fa = Fx + Ff
•For a > 0 m/s2.
Fa = Fx + Ff + (m·a)
Pulling an Object on a Flat Surface
Pulling an Object on a Flat Surface
•The pulling force F
is resolved into Fx
and Fy.
Fx
cos θ 
F
Fy
sin θ 
F
Fx  F  cos θ
Fy  F  sin θ
Pulling an Object on a Flat Surface
•Fn is the force that
the ground exerts
upward on the
mass. Fn equals the
downward weight Fw
minus the upward
force Fy from the
pulling force.
•For constant speed,
a = 0 m/s2.
ΣFy  0 N
Fn  Fy  Fw  0 N
Fn  Fw  Fy
Ff
Ff
μk 

Fn Fw  Fy
Ff  μ k  (Fw  Fy )
ΣFx  m  a
Fx  Ff  m  a
Fx  Ff
a
m
Example: Pulling A Sled
Two children sitting on a sled at rest in the
snow ask you to pull them. You pull on the
sled’s rope, which makes an angle of 40° with
the horizontal. The children have a combined
mass of 45 kg, and the sled has a mass of
5.0 kg. The coefficients of static and kinetic
friction are µs=0.20 and µk=0.15, and the sled
is initially at rest.
Find the acceleration of the sled and children if
the rope tension is 100 N.
F
y
 m  ay
 FN  T  sin   m  g  m  0 or FN  m  g  T  sin 
FN  50kg  9.8 m
 100 N  sin 40  425.72 N
s2
 T  cos   f k  m  ax
F
 max
k 
fk
 f k  k  FN  0.15  425.72 N  63.86 N
FN
x
100 N  cos 40  63.86 N  50kg  ax  ax 
100 N  cos 40  63.86 N
 0.2549 m 2
s
50kg
Simultaneous Pulling and Pushing an Object
on a Flat Surface
Simultaneous Pulling and Pushing an Object
on a Flat Surface
Σ Fy  0 N
Fn  Fy  Fw  0 N
Fx
cos θ 
F
Fy
sin θ 
F
Fx  F  cos θ
Fn  Fw  Fy
Fy  F  sin θ
a 
μk
Ff
Ff


Fn
Fw  Fy
Ff  μ k  (Fw  Fy )
ΣFx  m  a
Fx  Fpush  Ff  m  a
Fx  Fpush  Ff
m
Pushing an Object on a Flat Surface
Pushing an Object on a Flat Surface
•The pushing force F
is resolved into Fx
and Fy.
Fx
cosθ 
F
Fy
sinθ 
F
Fx  F  cosθ
Fy  F  sinθ
Pushing an Object on a Flat Surface
•Fn is the force that
the ground exerts
upward on the
mass. Fn equals the
downward weight Fw
plus the upward
force Fy from the
pushing force.
•For constant speed,
a = 0 m/s2.
ΣFy  0 N
Fn  Fy  Fw  0 N
Fn  Fw  Fy
Ff
Ff
μk 

Fn Fw  Fy
Ff  μ k  (Fw  Fy )
ΣFx  m  a
Fx  Ff  m  a
Fx  Ff
a
m
Pulling and Tension
• The acceleration a of both masses is the same.
Pulling and Tension
• For each mass:
Fn1  Fw1
Fn2  Fw2
Ff1  μ k  Fn1
Ff2  μ k  Fn2
• Isolate each mass and examine the
forces acting on that mass.
Pulling and Tension
•m1 = mass
ΣF  m1  a
T1  T2  Ff1  m1  a
•T1 may not be a
tension, but could be
an applied force (Fa)
that causes motion.
Pulling and Tension
•m2 = mass
ΣF  m 2  a
T2  Ff 2  m 2  a
Pulling and Tension
• This problem can often be solved as a
system of equations:
T1  T2  Ff1  m1  a
T2  Ff 2  m 2  a
• See the Solving Simultaneous Equations
notes for instructions on how to solve
this problem using a TI or Casio
calculator.
Revisiting Tension and Friction
Revisiting Tension and Friction
•For the hanging mass, •For the mass on the
table, m1:
m2 :
ΣF  m2  a
ΣF  m  a
Fn1  Fw1
Fw 2  T  m2  a
Ff  μ  Fn1
Fw 2  m 2  g
m2  g  T  m2  a
•The acceleration a of
both masses is the same.
T-Ff  m1  a
Revisiting Tension and Friction
m2  g  m1  a  Ff  m2  a
m2  g  Ff  m2  a  m1  a
m2  g  Ff
a
m2  m1
Example: A Sliding Block
A block of mass m2 = 5.0 kg has been
adjusted so that the block m1 = 7.0 kg
is just on the verge of sliding.
(a) What is the coefficient of static
friction ms between the table and the
block?
F
 FN  m1  g  m1  ay  0 so FN  m1  g
y
F
x
 T  f  m1  ax  0 so f  s  FN  s m1  g  T
F
x'
 m2  g  T  m2  ax '  0 so T  m2  g
m2  g m2 5kg
Therefore,  s  m1  g  m2  g ;  s 


 0.71
m2  g m2 7kg
Example: A Sliding Block
(b) With a slight push, the blocks move with
acceleration a. Find a if µk = 0.54.
F
x
 T  f  m1  ax so T  k  m1  g  m1  ax
T  m1  ax  k  m1  g
F
x'
 m2  g  T  m2  ax '  T  m2  g  m2  ax '
ax  ax '  a
T=T, therefore, m1  a  k  m1  g  m2  g  m2  a
m1  a  m2  a  m2  g  k  m1  g
g  (m2  k  m1 )
a
(m1  m2 )
9.8 m/s 2   5.0 kg  0.54  7.0 kg 
a
 1.0 m/s 2
7.0 kg  5.0 kg
Normal Force Not Associated with Weight.
• A normal force can exist that is
totally unrelated to the weight of an
object.
friction
applied force
normal
weight
FN = applied force
Friction is Always Parallel to
Surfaces….
•In this case, for the block
to remain in position
against the wall without
moving:
• the upward frictional
force Ff has to be equal
and opposite to the
downward weight Fw.
•The rightward applied
force F has to be equal
ad opposite to the
leftward normal force
FN.
Ff
F
FN
FW
(0.20)