Physics 211: Lecture 7 Todays Agenda

1
Physics 211 Lecture 7Todays Agenda

• Friction
• What is it?
• 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!

j
N
FAPPLIED
i
ma
some roughness here
mg
3
Surface Friction...

• Friction is caused by the microscopic
  interactions between the two surfaces

Friction causes heat !
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 in the
  direction that opposes relative motion between
  the two surfaces.
• 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.

6
Model...

• Dynamics
• i F ? ?KN ma
• j N mg
• so F ???Kmg ma

j
N
F
i
ma
?K mg
mg
7
Lecture 7, Act 0Friction

• A small box of mass is being pulled to the right
  by a horizontal string. It slides with friction
  on top of a bigger box. Assume both boxes were
  initially at rest.
• In which direction does the frictional force act
  on the bottom box?
• A) To the right
• B) To the left
• C) Friction does not affect the bottom box, only
  the top box.

T
m

friction
M
8
Lecture 7, Act 0Solution

• Draw FBD of the top box

N
T
f mKN mKmg
m
mg
9
Lecture 7, Act 0Solution

• 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.

f1,2
m
f2,1
mKmg
M
10
Lecture 7, Act 1More Friction

• A box of mass m 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 M 3 kg, which in turn
  slides on a frictionless floor.
• What is the acceleration of the bottom box ?
• (a) a 0 m/s2 (b) a 2.5 m/s2 (c) a 3.0
  m/s2

T
m


mk0.51
no friction
M
11
Lecture 7, Act 1Solution
f1,2
m
f2,1
mKmg
M
Using FMa for bottom box
a mKmg/M
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

?KN
ma
j
N
?
a / g sin ?????Kcos ?
?
mg
mg cos ??
i
mg sin ??
14
Static Friction...

• 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
  togetherthe 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
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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 doesnt move, its static friction
• If an object does move, its dynamic friction

j
N
F
i
fF
mg
17
Static Friction...
Suitcase

• ?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 2Forces 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 )
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Lecture 7, Act 2Solution

• Pick axes draw FBD of box

• Apply FNET ma

y N T sin q - mg maY 0
N
N mg - T sin q
T
q
m
fFR
x T cos q - fFR maX
ax will be positive (i.e. the box will move) if
T cos q gt fFR
mg
20
Lecture 7, Act 2Solution
y N 80 N
N
T
fMAX msN
q
m
mg
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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.

?
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Static Friction...

• The force provided by friction, fF , depends on ?.

fF
ma 0 (block is not moving)
mg sin ????ff ???
N
?
(Newtons 2nd Law along x-axis)
mg
?
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Static Friction...
Blocks

• 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
mg sin ?M????Smg cos ?M????
N
mg
??M
?S???tan ?M?
?
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Additional comments on Friction

• Since fF ?N , kinetic friction does not
  depend on the area of the surfaces in contact.
• 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
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
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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
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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
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Lecture 7, Act 3Forces 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)
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Lecture 7, Act 3Solution

• First consider the case where the inclined plane
  is not accelerating.

N
Ff
mg
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Lecture 7, Act 3Solution

• 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
32
Problem Putting on the brakes
Wheel

• Anti-lock brakes work by making sure the wheels
  roll without slipping. This maximizes the
  frictional force slowing the car since ?S gt ?K
  .
• The driver of a car moving with speed vo slams on
  the brakes. The coefficient of static friction
  between the wheels and the road is ?S . What is
  the stopping distance D?

33
Problem Putting on the brakes

• Use FNET ma for both i and j components
• i ?SN ma
• j N mg
• a ?S g

N
j
a
i
fF ?SN
mg
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Problem Putting on the brakes

• As in the last example, find ab ?Sg.
• Using the kinematic equation v2 - v02 2a( x
  -x0 )
• In our problem 0 - v02 ? 2ab( D )

vo
ab
v 0
D
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Problem Putting on the brakes

• In our problem 0 - v02 ? 2ab( D )
• Solving for D
• Putting in ab ?Sg

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Recap of todays lecture

• Friction
• What is it?
• How do we characterize it?
• Model of friction.
• Static Kinetic friction.
• Some problems involving friction.
• Box on truck.
• Braking distance.