Transcript Document

Lecture 10

Goals
 Describe Friction
 Problem solving with Newton’s 1st, 2nd and 3rd Laws
 Forces in circular and curvelinear motion
Physics 201: Lecture 10, Pg 1
Another experiment: A modified Atwood’s machine
Two blocks, m1 & m2, are connected by a massless frictionless
string/pulley on the table as shown. The table surface is
frictionless and little g acts downward.
What is the acceleration of the horizontal block?
N
T
Requires two FBDs and
m2
Newton’s 3rd Law.
Mass 1
S Fy = m1a1y= T – m1g
Mass 2
S Fx = m2a2x = -T
S Fy = 0 = N – m2g
T
m1
m2g
m1g
Correlated motion:
|a1y| = | a2x| ≡ a
If m1 moves up moves m2 right
Physics 201: Lecture 10, Pg 2
Another experiment: A modified Atwood’s machine
Two blocks, m1 & m2, are connected by a massless frictionless
string/pulley on the table as shown. The table surface is
frictionless and little g acts downward.
What is the acceleration of the horizontal block.
N
T
Requires two FBDs and
m2
Newton’s 3rd Law.
T
Mass 1
A. S Fy = m1 a= T – m1g
m1
m2g
m1g
Subbing in T from B into A
Mass 2
B. S Fx = m2 a = -T
S Fy = 0 = N – m2g
m1 a= - m2 a – m1g
m1 a + m2 a = m1g
a = m1g / (m1 + m2 )
Physics 201: Lecture 10, Pg 3
A “special” contact force: Friction


What does it do?
 It opposes motion (velocity, actual or that which
would occur if friction were absent!)
How do we characterize this in terms we have learned?
 Friction results in a force in a direction opposite to
the direction of motion (actual or, if static, then
“inferred”)!
j
N
FAPPLIED
ma
fFRICTION
i
mg
Physics 201: Lecture 10, Pg 4
If no acceleration



No net force
So frictional force just equals applied force
Key point: It is conditional!
N
FAPPLIED
fFRICTION
j
i
mg
Physics 201: Lecture 10, Pg 5
Friction...

Friction is caused by the “microscopic” interactions between
the two surfaces:
Physics 201: Lecture 10, Pg 6
Friction: Static friction
Static equilibrium: A block with a horizontal force F applied,
S Fx = 0 = -F + fs  fs = F
S Fy = 0 = - N + mg  N = mg
As F increases so does fs
FBD
N
F
m
mg
Physics 201: Lecture 10, Pg 7
fs
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
m
mg
Physics 201: Lecture 10, Pg 8
fs
Kinetic or Sliding friction (fk < fs)
Dynamic equilibrium, moving but acceleration is still zero
S Fx = 0 = -F + fk
 fk = F
S Fy = 0 = - N + mg  N = mg
As F increases fk remains nearly constant
(but now there acceleration is acceleration)
FBD
v
N
F
m
fk
1
mg
fk = mk N
Physics 201: Lecture 10, Pg 9
Model of Static Friction (simple case)
Magnitude:
f is proportional to the applied forces such that
fs ≤ ms N
ms called the “coefficient of static friction”
Direction:
Opposite to the direction of system acceleration if m were 0
Physics 201: Lecture 10, Pg 10
Sliding Friction

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 = mk N
( = mK mg in the previous example)

The constant mk is called the “coefficient of kinetic friction”

Logic dictates that
mS > mK
for any system
Physics 201: Lecture 10, Pg 11
Coefficients of Friction
Material on Material
ms = static friction
mk = 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 201: Lecture 10, Pg 12
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 201: Lecture 10, Pg 13
Many Forces are Conditional


Notice what happens if we change the direction of the applied
force
The normal force can increase or decrease
N > mg
F
j
i
Let a=0
fF
mg
Physics 201: Lecture 10, Pg 14
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
Mass 1
S Fy = 0 =
T – m1g
fS
T
Mass 2
S Fx = 0 = -T + fs = -T + mS N
S Fy = 0 = N – m2g
m1
N
m2
m2g
m1g
T = m1g = mS m2g  mS = m1/m2
Physics 201: Lecture 10, Pg 15
3rd Law : Static Friction with a bicycle wheel
You are pedaling hard
and the bicycle is
speeding up.
What is the direction of
the frictional force?

You are breaking and
the bicycle is slowing
down
What is the direction of
the frictional force?

Physics 201: Lecture 10, Pg 17
Static Friction with a bicycle wheel
You are pedaling hard and
the bicycle is speeding up.
What is the direction of the
frictional force?

Hint…you are accelerating
to the right
a=F/m
Ffriction, on B from E is to the right
Ffriction, on E from,B is to the left
Physics 201: Lecture 10, Pg 18
Force pairs on an Inclined plane
Forces on the block (static case)
Normal
Force
f= mN
Friction
Force
Forces on the plane by block
y
q
Physics 201: Lecture 10, Pg 23
x
The inclined plane coming and going (not static):
the component of mg along the surface > kinetic friction
S Fx = max = mg sin q ± uk N
S 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 Newton’s Laws that is key.
Physics 201: Lecture 10, Pg 24
Forces with rotation about a fixed axis
Key steps
 Identify
forces (i.e., a FBD)
 Identify
axis of rotation
 Circular
 Apply
motion implies ar & Fr = mar
conditions (position, velocity &
acceleration)
Physics 201: Lecture 10, Pg 29
Example
axis of rotation
The pendulum
Consider a person on a swing:
(A)
(B)
(C)
When is the tension equal to the weight of the person + swing?
(A) At the top of the swing (turnaround point)
(B) Somewhere in the middle
(C) At the bottom of the swing
(D) Never, it is always greater than the weight
(E) Never, it is always less than the weight
Physics 201: Lecture 10, Pg 30
Example
Gravity, Normal Forces etc.
axis of rotation
T
T
y
vt
q
mg
x
mg
at top of swing vt = 0
at bottom of swing vt is max
Fr = m 02 / r = 0 = T – mg cos q
T = mg cos q
Fr = m ac = m vt2 / r = T - mg
T = mg + m vt2 / r
T < mg
T > mg
Physics 201: Lecture 10, Pg 31
Conical Pendulum (Not a simple pendulum)

Swinging a ball on a string of length L around your head
(r = L sin q)
axis of rotation
S Fr = mar = T sin q
L
S Fz = 0 = T cos q – mg
so
T = mg / cos q (> mg)
r
mar = (mg / cos q ) (sin q )
ar = g tan q = vT2/r
 vT = (gr tan q)½
Period:
T= 2p r / vT =2p (r cot q /g)½
= 2p (L cos q /g)½
Physics 201: Lecture 10, Pg 32
Conical Pendulum (very different)

Swinging a ball on a string of length L around your head
axis of rotation
Period:
t = 2p r / vT =2p (r cot q /g)½
= 2p (L cos q / g )½
= 2p (5 cos 5 / 9.8 )½ = 4.38 s
= 2p (5 cos 10 / 9.8 )½ = 4.36 s
= 2p (5 cos 15 / 9.8 )½ = 4.32 s
L
r
Physics 201: Lecture 10, Pg 33
Another example of circular motion
Loop-the-loop 1
A match box car is going to do a loop-the-loop
of radius r.
What must be its minimum speed vt at the top
so that it can manage the loop successfully ?
Physics 201: Lecture 10, Pg 34
Loop-the-loop 1
To navigate the top of the circle its tangential
velocity vT must be such that its centripetal
acceleration at least equals the force due to
gravity. At this point N, the normal force,
goes to zero (just touching).
Fr = mar = mg = mvt2/r
vT
vt = (gr)1/2
mg
Physics 201: Lecture 10, Pg 35
Loop-the-loop 2
The match box car is going to do a loop-the-loop.
If the speed at the bottom is vB, what is the
normal force, N, at that point?
Hint: The car is constrained to the track.
Fr = mar = mvB2/r = N - mg
N = mvB2/r + mg
N
v
mg
Physics 201: Lecture 10, Pg 36
Loop-the-loop 3
Once again the car is going to execute a loop-the-loop.
What must be its minimum speed at the bottom so
that it can make the loop successfully?
This is a difficult problem to solve using just forces. We
will skip it now and revisit it using energy
considerations later on…
Physics 201: Lecture 10, Pg 37
Orbiting satellites vt = (gr)½
Net Force:
ma = mg = mvt2 / r
gr = vt2
vt = (gr)½
The only difference is
that g is less
because you are
further from the
Earth’s center!
Physics 201: Lecture 10, Pg 38
Geostationary orbit

The radius of the Earth is ~6000 km but at 36000 km you are
~42000 km from the center of the earth.

Fgravity is proportional to 1/r2 and so little g is now ~10 m/s2 / 50

vT = (0.20 * 42000000)½ m/s = 3000 m/s
At 3000 m/s, period T = 2p r / vT = 2p 42000000 / 3000 sec =
= 90000 sec = 90000 s/ 3600 s/hr = 24 hrs


Orbit affected by the moon and also the Earth’s mass is
inhomogeneous (not perfectly geostationary)

Great for communication satellites
(1st pointed out by Arthur C. Clarke)
Physics 201: Lecture 10, Pg 39
Recap
Assignment: HW5
For Tuesday: Read Chapter 7.1-7.4
Physics 201: Lecture 10, Pg 40