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Physics 207 – Lecture 8
Lecture 8
Goals: (Finish Chap. 6 and begin Ch. 7)
Solve 1D & 2D motion with friction
Utilize Newton’s 2nd Law
Differentiate between Newton’s 1st, 2nd and 3rd Laws
Begin to use Newton’s 3rd Law in problem solving
Assignment: HW4, (Chap. 6 & 7, due 10/5)
Finish reading Chapter 7
1st Exam Wed., Oct. 7th from 7:15-8:45 PM Chapters 1-7
in room 2103 Chamberlin Hall
Physics 207: Lecture 8, Pg 1
Net force gives rise to acceleration!
In physics:
A force is an action which causes an object to
accelerate (translational & rotational)
This is Newton’s Second Law
r r
r
∑ F = Fnet = ma ≠ 0
∑ Fx = max
∑ Fy = may
∑ Fz = maz
and mass plays a role
Physics 207: Lecture 8, Pg 2
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Physics 207 – Lecture 8
Mass
We have an idea of what mass is from everyday life.
In physics:
Mass (in Phys 207) is a quantity that specifies
how much inertia an object has
(i.e. a scalar that relates force to acceleration)
(Newton’s Second Law)
Mass is an inherent property of an object.
Mass and weight are different quantities; weight is
usually the magnitude of a gravitational (non-contact)
force.
“Pound” (lb) is a definition of weight (i.e., a force), not
a mass!
Physics 207: Lecture 8, Pg 3
Inertia and Mass
The tendency of an object to resist any attempt to
change its velocity is called Inertia
Mass is that property of an object that specifies how
much resistance an object exhibits to changes in its
velocity (acceleration)
r r
a ∝ Fnet
If mass is constant then
If force constant r
| a |∝
1
m
|a|
Mass is an inherent property of an object
m
Mass is independent of the object’s surroundings
Mass is independent of the method used to measure it
Mass is a scalar quantity
The SI unit of mass is kg
Physics 207: Lecture 8, Pg 4
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Physics 207 – Lecture 8
Exercise
Newton’s 2nd Law
An object is moving to the right, and experiencing
a net force that is directed to the right. The
magnitude of the force is decreasing with time
(read this text carefully).
The speed of the object is
A.
B.
C.
D.
increasing
decreasing
constant in time
Not enough information to decide
Physics 207: Lecture 8, Pg 5
Exercise
Newton’s 2nd Law
A 10 kg mass undergoes motion along a line with
velocities as given in the figure below. In regards to
the stated letters for each region, in which is the
magnitude of the force on the mass at its greatest?
A.
B.
C.
D.
E.
A
B
C
D
G
Physics 207: Lecture 8, Pg 6
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Physics 207 – Lecture 8
Now: Back to forces that oppose motion
Physics 207: Lecture 8, Pg 7
Static and Kinetic Friction
Friction is a model force that exists between objects & it is
conditional
At Static Equilibrium: A block, mass m, with a horizontal force F
applied,
Direction:
1. Force vector ⊥ to the normal force vector N
2. 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 8
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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 9
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 10
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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 11
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 12
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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 13
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 14
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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 15
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 16
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 17
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 18
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Physics 207 – Lecture 8
Home Exercise
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 19
Home exercise
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 20
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Physics 207 – Lecture 8
Forces at different angles
Case1: Downward angled force with friction
Case 2: Upwards angled force with friction
Cases 3,4: Up against the wall
Questions: Does it slide?
What happens to the normal force?
What happens to the frictional force?
Cases 3, 4
Case 2
Case 1
F
N
N
ff
F
F
N
ff
ff
mg
mg
mg
Physics 207: Lecture 8, Pg 21
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
mg cos θ
θ
with mg sin θ = f ≤ µS N
if mg sin θ > µS N, must slide
Critical angle µs = tan θc
θ
θ
Block weight is mg
Page 11
Physics 207: Lecture 8, Pg 22
y
x
Physics 207 – Lecture 8
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
Fx= 0 = mg sin θ – µk mg cos θ
µk = tan θ (only one angle)
y
mg cos θ
θ
fk = µk N = µk mg cos θ
θ
θ
x
mg
Physics 207: Lecture 8, Pg 23
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
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Physics 207: Lecture 8, Pg 24
Physics 207 – Lecture 8
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 25
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 26
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Physics 207 – Lecture 8
Recap
Assignment: HW4, (Chapter 6 & 7 due 10/5)
Finish Chapter 7
Physics 207: Lecture 8, Pg 27
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