Download Lecture 8 Static and Kinetic Friction fs ≤ µs N

Physics 207 – Lecture 8
Lecture 8
Goals:
Solve 1D motion with friction
Differentiate between Newton’s 1st, 2nd and 3rd Laws
Begin to use Newton’s 3rd Law in problem solving
Assignment: HW4, (Chapter 6, due 2/17, Wednesday)
Finish Chapter 7
1st Exam Wed., Feb. 17th from 7:15-8:45 PM Chapters 1-6
in room 2103 Chamberlin Hall
Physics 207: Lecture 8, Pg 1
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 µ were
0.
Magnitude: f is proportional to the applied forces such that
fs ≤ µs N
µs called the “coefficient of static friction”
Physics 207: Lecture 8, Pg 2
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Physics 207 – Lecture 8
Case study ... big F
Dynamics:
x-axis i :
y-axis j :
max = F − µKN
may = 0 = N – mg or N = mg
F − µKmg = m ax
so
fk
v
j
N
F
i
ma
ax
fk
µK mg
mg
g
Physics 207: Lecture 8, Pg 3
Case study ... little F
Dynamics:
x-axis i :
y-axis j :
so
max = F − µKN
may = 0 = N – mg or N = mg
F − µKmg = m ax
fk
v
j
N
F
i
ma
ax
fk
µK mg
mg
g
Physics 207: Lecture 8, Pg 4
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Physics 207 – Lecture 8
Friction: Static friction
Static equilibrium: A block with a horizontal force F applied,
Σ Fx = 0 = -F + fs
fs = F
FBD
Σ Fy = 0 = - N + mg N = mg
N
As F increases so does fs
F
m
fs
1
mg
Physics 207: Lecture 8, Pg 5
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 µ were 0.
Magnitude: fS is proportional to the magnitude of N
fs = µs N
N
F
m
mg
Physics 207: Lecture 8, Pg 6
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fs
Physics 207 – Lecture 8
Kinetic or Sliding friction (fk < fs)
Dynamic equilibrium, moving but acceleration is still zero
Σ Fx = 0 = -F + fk
FBD
fk = F
v
Σ Fy = 0 = - N + mg N = mg
N
F
As F increases fk remains nearly constant
(but now there acceleration is acceleration)
m
fk
1
mg
fk = µk N
Physics 207: Lecture 8, Pg 7
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 = µk N
( = µK mg
g in the previous example)
The constant µk is called the “coefficient of kinetic friction”
Logic dictates that
µS > µK
for any system
Physics 207: Lecture 8, Pg 8
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Physics 207 – Lecture 8
Coefficients of Friction
Material on Material
µs = static friction
µk = 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
Physics 207: Lecture 8, Pg 9
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 µS
N
T
Static equilibrium: Set
m2
m2 and add mass to
m1 to reach the
breaking point.
Requires two FBDs
Mass 1
Σ Fy = 0 =
T – m1g
fS
T
m1
m2g
m1g
Mass 2
Σ Fx = 0 = -T + fs = -T + µS N
Σ Fy = 0 = N – m2g
T = m1g = µS m2g µS = m1/m2
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Physics 207: Lecture 8, Pg 10
Physics 207 – Lecture 8
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 µK.
T
Dynamic equilibrium:
Set m2 and adjust m1
to find place when
T
a = 0 and v ≠ 0
fk
m1
Requires two FBDs
Mass 1
Σ Fy = 0 =
T – m1g
N
m2
m2g
m1g
Mass 2
Σ Fx = 0 = -T + ff = -T + µk N
Σ Fy = 0 = N – m2g
T = m1g = µk m2g µk = m1/m2
Physics 207: Lecture 8, Pg 11
An experiment (with a ≠ 0)
Two blocks are connected on the table as shown. The
table has unknown static and kinetic friction coefficients.
N
Design an experiment to find µK.
T
Non-equilibrium: Set m2
T
and adjust m1 to find
regime where a ≠ 0
m1
Requires two FBDs
m1g
Mass 1
Σ Fy = m1a =
fk
m2
m2g
Mass 2
T – m1g
Σ Fx = m2a = -T + fk
Σ Fy = 0 = N – m2g
= -T + µk N
T = m1g + m1a = µk m2g – m2a µk = (m1(g+a)+m2a)/m2g
Physics 207: Lecture 8, Pg 12
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Physics 207 – Lecture 8
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?
Physics 207: Lecture 8, Pg 13
Sample Problem
Σ Fx =ma = F - ff = F - µk N = F - µk mg
Σ Fy = 0 = N – mg
µk = (F - ma) / mg & x = ½ a t 2 0.80 m = ½ a 4 s 2
a = 0.40 m/s2
µk = (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46
Physics 207: Lecture 8, Pg 14
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Physics 207 – Lecture 8
Inclined plane with “Normal” and Frictional Forces
1. Static Equilibrium Case
“Normal” means perpendicular
2. Dynamic Equilibrium (see 1)
Normal
Force
3. Dynamic case with non-zero acceleration
Friction
ΣF=0
f Force
Fx= 0 = mg sin θ – f
mg sin θ
Fy= 0 = –mg cos θ + N
θ
y
mg cos θ
θ
with mg sin θ = f ≤ µS N
if mg sin θ > µS N, must slide
θ
x
Block weight is mg
Critical angle µs = tan θc
Physics 207: Lecture 8, Pg 15
Inclined plane with “Normal” and Frictional Forces
1. Static Equilibrium Case
“Normal” means perpendicular
Normal
Force
2. Dynamic Equilibrium
Friction opposite velocity
v
(down the incline)
ΣF=0
fK
Friction
Force
mg sin θ
Fx= 0 = mg sin θ – fk
Fy= 0 = –mg cos θ + N
θ
mg cos θ
θ
fk = µk N = µk mg cos θ
Fx= 0 = mg sin θ – µk mg cos θ
µk = tan θ (only one angle)
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θ
mg
Physics 207: Lecture 8, Pg 16
y
x
Physics 207 – Lecture 8
Inclined plane with “Normal” and Frictional Forces
3. Dynamic case with non-zero acceleration
Normal
Result depends on direction of velocity
Force
Friction Force
Sliding Down
v
mg sin θ
θ
Fx= max = mg sin θ ± fk
θ
fk
Sliding
Up
Weight of block is mg
Fy= 0 = –mg cos θ + N
fk = µk N = µk mg cos θ
Fx= max = mg sin θ ± µk mg cos θ
ax = g sin θ ± µk g cos θ
Physics 207: Lecture 8, Pg 17
The inclined plane coming and going (not static):
the component of mg along the surface > kinetic friction
Σ Fx = max = mg sin θ ± uk N
Σ Fy = may = 0 = -mg cos θ + N
Putting it all together gives two different accelerations,
ax = g sin θ ± uk g cos θ. A tidy result but ultimately it is the
process of applying Newton’s Laws that is key.
Physics 207: Lecture 8, Pg 18
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Physics 207 – Lecture 8
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.
Physics 207: Lecture 8, Pg 19
Friction in a viscous medium
Drag Force Quantified
With a cross sectional area, A (in m2), coefficient of drag of
1.0 (most objects), ρ sea-level density of air, and velocity, v
(m/s), the drag force is:
D = ½ C ρ A v2 ≅ c A v2
in Newtons
3
c = ¼ kg/m
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
Physics 207: Lecture 8, Pg 20
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Physics 207 – Lecture 8
Fish Schools
Physics 207: Lecture 8, Pg 21
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.
Physics 207: Lecture 8, Pg 22
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Physics 207 – Lecture 8
“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
Physics 207: Lecture 8, Pg 23
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.
Physics 207: Lecture 8, Pg 24
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Physics 207 – Lecture 8
Newton’s 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 = Σ F = ma
a
Law 3:
Forces occur in pairs: FA , B = - FB , A
(For every action there is an equal and opposite reaction.)
Read: Force of B on A
Physics 207: Lecture 8, Pg 25
Newton’s 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:
Newton’s 3rd law concerns force pairs which
act on two different objects (not on the same object) !
Physics 207: Lecture 8, Pg 26
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Physics 207 – Lecture 8
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 Newton’s law of gravity,
as follows:
Physics 207: Lecture 8, Pg 27
Cavendish’s 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!!!
Physics 207: Lecture 8, Pg 28
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Physics 207 – Lecture 8
Example (non-contact)
Consider the forces on an object undergoing projectile motion
FB,E = - mB g
FB,E = - mB g
FE,B = mB g
FE,B = mB g
EARTH
Question: By how much does g change at an altitude of
40 miles? (Radius of the Earth ~4000 mi)
Physics 207: Lecture 8, Pg 29
Example (non-contact)
Consider the forces on an object undergoing projectile motion
FB,E = - mB g
FB,E = - mB g
FE,B = mB g
FE,B = mB g
EARTH
Compare: g = G m2 / 40002
g’ = G m2 / (4000+40)2
g / g’ = / (4000+40)2 / 40002 = 0.98
Physics 207: Lecture 8, Pg 30
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Physics 207 – Lecture 8
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?
Physics 207: Lecture 8, Pg 31
Exercise
Newton’s 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,
A.
B.
C.
greater than
equal to
less than
that exerted by the fly on the bus.
Physics 207: Lecture 8, Pg 32
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Physics 207 – Lecture 8
Exercise
Newton’s 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.
Physics 207: Lecture 8, Pg 33
Exercise 2
Newton’s 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
A. greater than
B. equal to
C. less than
that of the fly.
Physics 207: Lecture 8, Pg 34
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Physics 207 – Lecture 8
Exercise 2
Newton’s Third Law
Solution
By Newton’s third law these two forces form an interaction
pair which are equal (but in opposing directions).
✿
Thus the forces are the same
However, by Newton’s second law Fnet = ma or a = Fnet/m.
So Fb, f = -Ff, b = F0
but |abus | = |F0 / mbus | << | afly | = | F0/mfly |
Answer for acceleration is (C)
Physics 207: Lecture 8, Pg 35
Exercise 3
Newton’s 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?
a
A.
B.
C.
D.
b
2
4
6
Something else
Physics 207: Lecture 8, Pg 36
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Physics 207 – Lecture 8
Exercise 3
Solution:
Fa,f
Fb,a
Ff,a
a
FE,a
Fa,b
bF
Fg,a
Fg,b
Fa,g
Fb,g
Fa,E
E,b
Fb,E
6
Physics 207: Lecture 8, Pg 37
Lecture 8 Recap
Assignment: HW4, (Chapter 6, due 2/17, Wednesday)
Finish Chapter 7
1st Exam Wed., Feb. 17th from 7:15-8:45 PM Chapters 1-6
in room 2103 Chamberlin Hall
Physics 207: Lecture 8, Pg 38
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