Physics 207: Lecture 2 Notes

1
Lecture 8

• Goals
• Solve 1D motion with friction
• Differentiate between Newtons 1st, 2nd and 3rd
  Laws
• Begin to use Newtons 3rd Law in problem solving

Assignment HW4, (Chapter 6, due 2/17,
Wednesday) Finish Chapter 7 1st Exam Wed., Feb.
17th from 715-845 PM Chapters 1-6 in room 2103
Chamberlin Hall
2
Static and Kinetic Friction

• Friction exists between objects and its behavior
  has been modeled.
• At Static Equilibrium A block, mass m, with a
  horizontal force F applied,
• Direction A force vector ? to the normal force
  vector N and the vector is opposite to the
  direction of acceleration if m were 0.
• Magnitude f is proportional to the applied
  forces such that
• fs ms N
• ms called the coefficient of static friction

3
Case study ... big F

• Dynamics
• x-axis i max F ? ?KN
• y-axis j may 0 N mg or N mg
• so F ???Kmg m ax

fk
v
N
F
max
fk
?K mg
mg
4
Case study ... little F

• Dynamics
• x-axis i max F ? ?KN
• y-axis j may 0 N mg or N mg
• so F ???Kmg m ax

fk
v
j
N
F
i
max
fk
?K mg
mg
5
Friction Static friction
Static equilibrium A block with a horizontal
force F applied, As F increases so does fs
S Fx 0 -F fs ? fs F S Fy 0 - N
mg ? N mg
6
Static friction, at maximum (just before slipping)
Equilibrium A block, mass m, with a horizontal
force F applied, Direction A force vector ?
to the normal force vector N and the vector is
opposite to the direction of acceleration if m
were 0. Magnitude fS is proportional to the
magnitude of N fs ms N
N
F
fs
m
mg
7
Kinetic or Sliding friction (fk lt fs)
Dynamic equilibrium, moving but acceleration is
still zero As F increases fk remains nearly
constant (but now there acceleration is
acceleration)
FBD
S Fx 0 -F fk ? fk F S Fy 0 - N
mg ? N mg
v
N
F
m1
fk
mg
fk mk N
8
Sliding Friction Modeling

• Direction A force vector ? to the normal force
  vector N and the vector is opposite to the
  velocity.
• Magnitude fk is proportional to the magnitude of
  N
• fk ?k N ( ?K??mg in the previous example)
• The constant ?k is called the coefficient of
  kinetic friction
• Logic dictates that ?S gt ?K for any
  system

9
Coefficients of Friction
Material on Material ?s static friction ?k kinetic friction
steel / steel 0.6 0.4
add grease to steel 0.1 0.05
metal / ice 0.022 0.02
brake lining / iron 0.4 0.3
tire / dry pavement 0.9 0.8
tire / wet pavement 0.8 0.7
10
An experiment

• Two blocks are connected on the table as shown.
  The
• table has unknown static and kinetic friction
  coefficients.
• Design an experiment to find mS

T
Static equilibrium Set m2 and add mass to m1
to reach the breaking point. Requires two FBDs
fS
T
m1
m1g
Mass 1 S Fy 0 T m1g T m1g mS m2g ?
mS m1/m2
Mass 2 S Fx 0 -T fs -T mS N S Fy 0
N m2g
11
A 2nd experiment

• Two blocks are connected on the table as shown.
  The
• table has unknown static and kinetic friction
  coefficients.
• Design an experiment to find mK.

T
Dynamic equilibrium Set m2 and adjust m1 to
find place when a 0 and v ? 0 Requires
two FBDs
fk
T
m1
m1g
Mass 1 S Fy 0 T m1g T m1g mk m2g ?
mk m1/m2
Mass 2 S Fx 0 -T ff -T mk N S Fy 0
N m2g
12
An experiment (with a ? 0)

• Two blocks are connected on the table as shown.
  The
• table has unknown static and kinetic friction
  coefficients.
• Design an experiment to find mK.

T
Non-equilibrium Set m2 and adjust m1 to
find regime where a ? 0 Requires two FBDs
fk
T
m1
m1g
Mass 1 S Fy m1a T m1g T m1g m1a
mk m2g m2a ? mk (m1(ga)m2a)/m2g
Mass 2 S Fx m2a -T fk -T mk N S Fy
0 N m2g
13
Sample Problem

• You have been hired to measure the coefficients
  of friction for the newly discovered substance
  jelloium. Today you will measure the coefficient
  of kinetic friction for jelloium sliding on
  steel. To do so, you pull a 200 g chunk of
  jelloium across a horizontal steel table with a
  constant string tension of 1.00 N. A motion
  detector records the motion and displays the
  graph shown.
• What is the value of µk for jelloium on steel?

14
Sample Problem

• S Fx ma F - ff F - mk N F - mk mg
• S Fy 0 N mg
• mk (F - ma) / mg x ½ a t2 ? 0.80 m
  ½ a 4 s2
• a 0.40 m/s2
• mk (1.00 - 0.20 0.40 ) / (0.20 10.) 0.46

15
Inclined plane with Normal and Frictional
Forces

• Static Equilibrium Case
• Dynamic Equilibrium (see 1)
• Dynamic case with non-zero acceleration

Normal means perpendicular
Normal Force
Friction Force
f
S F 0 Fx 0 mg sin q f Fy 0 mg
cos q N with mg sin q f mS N if mg sin q
gt mS N, must slide Critical angle ms tan qc
mg sin q
q
y
mg cos q
q
q
x
Block weight is mg
16
Inclined plane with Normal and Frictional
Forces

• Static Equilibrium Case
• Dynamic Equilibrium
• Friction opposite velocity
• (down the incline)

Normal means perpendicular
Normal Force
v
Friction Force
fK
S F 0 Fx 0 mg sin q fk Fy 0 mg
cos q N fk mk N mk mg cos q Fx 0 mg
sin q mk mg cos q mk tan q (only
one angle)
mg sin q
q
y
mg cos q
q
q
x
mg
17
Inclined plane with Normal and Frictional
Forces
3. Dynamic case with non-zero acceleration Result
depends on direction of velocity
Fx max mg sin q fk Fy 0 mg cos q
N fk mk N mk mg cos q Fx max mg sin
q mk mg cos q ax g sin q mk g cos q
18
The inclined plane coming and going (not
static)the component of mg along the surface gt
kinetic friction

• Fx max mg sin q uk N
• Fy may 0 -mg cos q N

Putting it all together gives two different
accelerations, ax g sin q uk g cos q. A tidy
result but ultimately it is the process of
applying Newtons Laws that is key.
19
Velocity and acceleration plots
Notice that the acceleration is always down the
slide and that, even at the turnaround point, the
block is always motion although there is an
infinitesimal point at which the velocity of the
block passes through zero. At this moment,
depending on the static friction the block may
become stuck.
20
Friction in a viscous mediumDrag Force Quantified

• With a cross sectional area, A (in m2),
  coefficient of drag of 1.0 (most objects), ?
  sea-level density of air, and velocity, v (m/s),
  the drag force is
• D ½ C ? A v2 ? c A v2 in Newtons
• c ¼ kg/m3
• In falling, when D mg, then at terminal
  velocity
• Example Bicycling at 10 m/s (22 m.p.h.), with
  projected area of 0.5 m2 exerts 30 Newtons
• Minimizing drag is often important

21
Fish Schools
22

• By swimming in synchrony in the correct
  formation, each fish can take advantage of moving
  water created by the fish in front to reduce
  drag.
• Fish swimming in schools can swim 2 to 6 times as
  long as individual fish.

23
Free Fall

• Terminal velocity reached when Fdrag Fgrav (
  mg)
• For 75 kg person with a frontal area of 0.5 m2,
• vterm ? 50 m/s, or 110 mph
• which is reached in about 5 seconds, over 125 m
  of fall

24
Trajectories with Air Resistance

• Baseball launched at 45 with v 50 m/s
• Without air resistance, reaches about 63 m high,
  254 m range
• With air resistance, about 31 m high, 122 m range

Vacuum trajectory vs. air trajectory for 45
launch angle.
25
Newtons Laws

• Law 1 An object subject to no external forces is
  at rest or moves with a constant velocity if
  viewed from an inertial reference frame.
• Law 2 For any object, FNET ??F ma
• Law 3 Forces occur in pairs FA , B -
  FB , A
• (For every action there is an equal and
  opposite reaction.)

26
Newtons Third Law

• If object 1 exerts a force on object 2 (F2,1 )
  then object 2 exerts an equal and opposite force
  on object 1 (F1,2)
• F1,2 -F2,1

For every action there is an equal and opposite
reaction
IMPORTANT Newtons 3rd law concerns force
pairs which act on two different objects (not
on the same object) !
27
Gravity
Newton also recognized that gravity is an
attractive, long-range force between any two
objects. When two objects with masses m1 and m2
are separated by distance r, each object pulls
on the other with a force given by Newtons law
of gravity, as follows
28
Cavendishs Experiment
F m1 g G m1 m2 / r2 g G m2 / r2 If we
know big G, little g and r then will can find m2
the mass of the Earth!!!
29
Example (non-contact)
Consider the forces on an object undergoing
projectile motion
Question By how much does g change at an
altitude of 40 miles? (Radius of the Earth 4000
mi)
30
Example (non-contact)
Consider the forces on an object undergoing
projectile motion
Compare g G m2 / 40002 g G m2 /
(400040)2 g / g / (400040)2 / 40002
0.98
31
The flying bird in the cage

• You have a bird in a cage that is resting on your
  upward turned palm.  The cage is completely
  sealed to the outside (at least while we run the
  experiment!).  The bird is initially sitting at
  rest on the perch.  It decides it needs a bit of
  exercise and starts to fly. Question How does
  the weight of the cage plus bird vary when the
  bird is flying up, when the bird is flying
  sideways, when the bird is flying down?
• Follow up question
• So, what is holding the airplane up in the sky? 

32
Exercise Newtons Third Law
A fly is deformed by hitting the windshield of a
speeding bus.
v
The force exerted by the bus on the fly is,

1. greater than
2. equal to
3. less than

that exerted by the fly on the bus.
33
Exercise Newtons Third Law
A fly is deformed by hitting the windshield of a
speeding bus.
v
The force exerted by the bus on the fly is,

• B. equal to

that exerted by the fly on the bus.
34
Exercise 2Newtons Third Law
Same scenario but now we examine the
accelerations
A fly is deformed by hitting the windshield of a
speeding bus.
v
The magnitude of the acceleration, due to this
collision, of the bus is

1. greater than
2. equal to
3. less than

that of the fly.
35
Exercise 2Newtons Third LawSolution
By Newtons third law these two forces form an
interaction pair which are equal (but in opposing
directions).
?
Thus the forces are the same
However, by Newtons second law Fnet ma or a
Fnet/m. So Fb, f -Ff, b F0 but abus F0
/ mbus ltlt afly F0/mfly
Answer for acceleration is (C)
36
Exercise 3Newtons 3rd Law

• Two blocks are being pushed by a finger on a
  horizontal frictionless floor.
• How many action-reaction force pairs are present
  in this exercise?

1. 2
2. 4
3. 6
4. Something else

37
Exercise 3Solution
a
b
6
38
Lecture 8 Recap
Assignment HW4, (Chapter 6, due 2/17,
Wednesday) Finish Chapter 7 1st Exam Wed., Feb.
17th from 715-845 PM Chapters 1-6 in room 2103
Chamberlin Hall