Download Friction with no acceleration

Physics 207 – Lecture 8
Lecture 8
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Goals:
v Solve 1D & 2D motion with friction
v Utilize Newton’s 2nd Law
v Differentiate between Newton’s 1st, 2nd and 3rd Laws
v Begin to use Newton’s 3rd Law in problem solving
Physics 207: Lecture 8, Pg 1
Friction with no acceleration
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No net force
So frictional force just cancels applied force
N
FAPPLIED
fFRICTION
j
i
mg
g
Physics 207: Lecture 8, Pg 2
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Physics 207 – Lecture 8
Friction...
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Friction is caused by the “microscopic” interactions between
the two surfaces:
Physics 207: Lecture 8, Pg 3
Friction: Static friction
Static equilibrium: A block with a horizontal force F applied,
As F increases so does fs
FBD
N
F
m
mg
Physics 207: Lecture 8, Pg 4
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fs
Physics 207 – Lecture 8
Friction: Static friction
Static equilibrium: A block with a horizontal force F applied,
Σ Fx = 0 = -F + fs
fs = F
Σ Fy = 0 = - N + mg
N = mg
FBD
N
F
m
fs
FrictionForce
mg
Force applied
Physics 207: Lecture 8, Pg 5
Static friction, at maximum (just before slipping)
fS is proportional to the magnitude of N
fs = µs N
N
F
m
FrictionForce
mg
Force applied
Physics 207: Lecture 8, Pg 6
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fs
Physics 207 – Lecture 8
Model of Static Friction
Magnitude:
f is proportional to the applied forces such that
fs ≤ µs N
µs called the “coefficient of static friction”
Direction:
If just a single “applied” force, friction is in opposite direction
Physics 207: Lecture 8, Pg 7
Kinetic (fk < fs)
Dynamic equilibrium, moving but acceleration is still zero
Σ Fx = 0 = -F + fk
Σ Fy = 0 = - N + mg
fk = F
N = mg
FBD
v
N
F
m
mg
fk = µk N
Physics 207: Lecture 8, Pg 8
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fk
Physics 207 – Lecture 8
Model of Sliding Friction
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Direction: ⊥ to the normal force vector N and
opposite to the velocity.
Magnitude: fk is proportional to the magnitude of N
v fk = µk N
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The constant µk is called the “coefficient of kinetic friction”
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Logic dictates that
µS > µK
for any system
Physics 207: Lecture 8, Pg 9
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 10
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Physics 207 – Lecture 8
Sliding friction (fk < fs) but now |a| > 0
A change in velocity
Σ Fx = -F + fk
FBD
= net Force
Σ Fy = 0 = - N + mg
v
N
N = mg
F
As F increases fk remains nearly constant
(but now there is acceleration)
m
mg
fk = µk N
Physics 207: Lecture 8, Pg 11
Acceleration, Inertia and Mass
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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
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r
| a |∝
1
m
|a|
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Mass is an inherent property of an object
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 12
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fk
Physics 207 – Lecture 8
Mass
We have an idea of what mass is from everyday life.
l In physics:
v 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)
l Mass is an inherent property of an object.
l 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!
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Physics 207: Lecture 8, Pg 13
Exercise
Newton’s 2nd Law
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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).
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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 14
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Physics 207 – Lecture 8
1st: Frictionless experiment (with a ≠ 0)
Two blocks are connected on the table as shown. The
table is frictionless. Find the acceleration of mass 2.
T
Requires two FBDs
N
m2
T
m1
m2g
m1g
Mass 1
Σ Fy = m1ay =
Notice ay = ax = a
Eliminate T
m1a + m2a = m1
T – m1g
Mass 2
Σ Fx = m2ax = -T
Σ Fy = 0 = N – m2g
a = m1 / (m2+m1)g
Physics 207: Lecture 8, Pg 15
Experiment with friction (with a ≠ 0)
Two blocks, of m1 & m2 , are connected on the table as shown.
The table has unknown static and kinetic friction coefficients.
Given an a, find µK.
N
T
Similar but now with friction.
m2
fk
T
m1
m2g
m1g
Mass 1
Σ Fy = m1a =
Mass 2
Σ Fx = m2a = -T + fk
Σ Fy = 0 = N – m2g
T – m1g
T = m1g + m1a = µk m2g – m2a
= -T + µk N
µk = (m1(g+a)+m2a)/m2g
Physics 207: Lecture 8, Pg 16
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Physics 207 – Lecture 8
Experiment with friction (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 µK.
T
Dynamic equilibrium:
Set m2 and adjust m1
to find place when
T
a = 0 and v ≠ 0
fk
m1
Mass 1
m2g
Σ Fx = 0 = -T + ff = -T + µk N
Σ Fy = 0 = N – m2g
T – m1g
T = m1g = µk m2g
m2
m1g
Mass 2
Requires two FBDs
Σ Fy = 0 =
N
µk = m1/m2
To repeat, net force
Physics 207: Lecture 8, Pg 17
acceleration
In physics:
v 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
Physics 207: Lecture 8, Pg 18
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Physics 207 – Lecture 8
Home Exercise
Newton’s 2nd Law
A 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
D
F
G
Physics 207: Lecture 8, Pg 19
Remember: Forces are Conditional
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Notice what happens if we change the direction of the applied
force
The normal force can increase or decrease
Here the normal force exceeds mg
F
F sin θ
θ
F sin θ +mg
N
Let a=0
j
i
fF
mg
Physics 207: Lecture 8, Pg 20
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Physics 207 – Lecture 8
Home Exercise
Newton’s 2nd Law
A constant force is exerted on a cart that is initially at rest on
an air table. The force acts for a short period of time and
gives the cart a certain final speed s.
Force
Cart
Air Track
In a second trial, we apply a force only half as large.
To reach the same final speed, how long must the same force be
applied (recall acceleration is proportional to force if mass fixed)?
A.
B.
C.
D.
4 x as long
2 x as long
1/2 as long
1/4 as long
Physics 207: Lecture 8, Pg 21
Home Exercise Newton’s 2nd Law
Solution
Force
Cart
Air Track
F = ma
Since F2 = 1/2 F1
a2 = 1/2 a1
We know that under constant acceleration, v = a ∆t
So,
a2 ∆t2 = a1 ∆t1 we want equal final velocities
1/2 a1 / ∆t2 = a1 / ∆t1
∆t = 2 ∆t
2
1
(B) 2 x as long
Physics 207: Lecture 8, Pg 22
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Physics 207 – Lecture 8
Home Exercise
Newton’s 2nd Law
A force of 2 Newtons acts on a cart that is initially at rest
on an air track with no air and pushed for 1 second.
Because there is friction (no air), the cart stops
immediately after I finish pushing.
It has traveled a distance, D.
Force
Cart
Air Track
Next, the force of 2 Newtons acts again
but is applied for 2 seconds.
A.
B.
The new distance the cart moves relative
to D is:
C.
D.
8 x as far
4 x as far
2 x as far
1/4 x as far
Physics 207: Lecture 8, Pg 23
Home Exercise
Solution
Force
Cart
Air Track
We know that under constant acceleration,
∆x = a (∆t)2 /2
(when v0=0)
Here ∆t2=2∆t1,
F2 = F1 ⇒ a2 = a1
1
a ∆ t 22 (
∆x2 2
2 ∆ t1 )2
=
=
=4
∆ x1 1 a ∆ t 2
∆ t12
1
2
(B) 4 x as long
Physics 207: Lecture 8, Pg 24
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Physics 207 – Lecture 8
Sample Problem
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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 k for jelloium on steel?
Physics 207: Lecture 8, Pg 25
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 t2
0.80 m = ½ a 4 s2
2
a = 0.40 m/s
µk = (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46
Physics 207: Lecture 8, Pg 26
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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 27
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
Physics 207: Lecture 8, Pg 28
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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 29
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 30
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Physics 207 – Lecture 8
Example (non-contact)
Question: By how much does g change at an altitude of
40 miles? (Radius of the Earth ~4000 mi)
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 31
Recap
Assignment: HW4, (Chapter 6 & 7 due 10/4)
For Monday finish Chapter 7
Physics 207: Lecture 8, Pg 32
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