PHYS 1441 – Section 002
Lecture #10
Monday, Oct. 11, 2010
Dr. Jaehoon Yu
•
•
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Force of Friction
Application of Newton’s Laws
Example for applications of Newton’s
Laws
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Example for Motion without friction
Example for Motion with friction
Uniform Circular Motion
Today’s homework is homework #6, due 10pm, Tuesday, Oct. 19!!
Monday, Oct. 11, 2010
PHYS 1441-002, Fall 2010
Dr. Jaehoon Yu
1
Announcements
• Quiz #2 Results
– Class average: 17/30
• Equivalent to: 57/100
– Top score: 30/30
• Evaluation policy
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–
–
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Homework: 30%
Comprehensive final exam: 25%
One better of the two non-comprehensive term exam: 20%
Lab: 15%
Quiz:10%
Extra credit: 10%
• Colloquium this week
Monday, Oct. 11, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Monday, Oct. 11, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
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Reminder: Special Project for Extra Credit
A large man and a small boy stand facing each other on frictionless ice. They put their
hands together and push against each other so that they move apart. a) Who moves away
with the higher speed and by how much? b) Who moves farther in the same elapsed time?
• Derive fomulae for the two problems above in
much more detail than what is shown in this
lecture note.
• Be sure to clearly define each variables used in
your derivation.
• Each problem is 10 points each.
• Due is this Wednesday, Oct. 13.
Monday, Oct. 11, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Applications of Newton’s Laws
Suppose you are pulling a box on frictionless ice, using a rope.
M
What are the forces being
exerted on the box?
T
Gravitational force: Fg
n= -Fg
Free-body
diagram
n= -Fg
Normal force: n
T
Fg=Mg
T
Fg=Mg
Monday, Oct. 11, 2010
Tension force: T
Total force:
F=Fg+n+T=T
If T is a constant
force, ax, is constant
 Fx  T  Max
F
y
ax 
 Fg  n  Ma y  0
v xf  vxi  ax t  vxi  
T
M
ay  0
T 
t
M
1 T  2
x

x

v
t


t
x  f i xi
2 M 
PHYS 1441-002, Fall
2010happened
Dr. Jaehoon
What
to the motion in y-direction?
Yu
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Friction Force
When an object is in contact with a surface there is a force acting on that
object. The component of this force that is parallel to the surface is called
the friction force. This resistive force is exerted on a moving object due to
viscosity or other types of frictional property of the medium in or surface on
which the object moves. Always opposite to the movement!!
Monday, Oct. 11, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Static Friction
When the two surfaces are not sliding across one another
the friction is called static friction. The resistive force exerted
on the object up to the time just before the object starts moving.
Monday, Oct. 11, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Magnitude of the Static Friction
The magnitude of the static friction force can have any value
from zero up to the maximum value.
fs  f
MAX
s
f
0  s  1
MAX
s
 s FN
is called the coefficient of static friction.
What is the unit? None
Once the object starts moving, there is NO MORE static friction!!
Kinetic friction
applies during the move!!
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Monday, Oct. 11, 2010
Yu
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Note that the magnitude of the frictional force does not depend
on the contact area of the surfaces.
f
Monday, Oct. 11, 2010
MAX
s
 s FN
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Kinetic Friction
Static friction opposes the impending relative motion between
two objects.
Kinetic friction opposes the relative sliding motions that is
happening. The resistive force exerted on the object during its
movement. Normally much smaller than static friction!!
f k  k FN
0  k  1
is called the coefficient of kinetic friction.
What is the direction of friction forces?
Monday, Oct. 11, 2010
opposite to the movement
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Coefficient of Friction
What
are
these?
Monday, Oct. 11, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Forces of Friction Summary
Resistive force exerted on a moving object due to viscosity or other types
frictional property of the medium in or surface on which the object moves.
These forces are either proportional to the velocity or the normal force.
Force of static friction, fs: The resistive force exerted on the object until
just before the beginning of its movement
Empirical
Formula
ur
ur
f s  s F N
What does this
formula tell you?
Frictional force increases till it
reaches the limit!!
Beyond the limit, the object moves, and there is NO MORE static
friction but kinetic friction takes it over.
Force of kinetic friction, fk
ur
ur
f k  k F N
The resistive force exerted on the object
during its movement
Which direction does kinetic friction apply?
Monday, Oct. 11, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
Opposite to the motion!
12
Look at this problem again…
Suppose you are pulling a box on a rough surfice, using a rope.
M
What are the forces being
exerted on the box?
T
Gravitational force: Fg
Normal force: n
Tension force: T
Friction force: Ff
n= -Fg
Free-body
diagram
n= -Fg
Ff
T
Ff
Fg=Mg
T
Fg=Mg
Monday, Oct. 11, 2010
Total force:
F=Fg+n+T+Ff
If T is a constant
force, ax, is constant
 Fx  T  Ff  Max ax 
F
y
T  Ff
M
 Fg  n  Ma y  0 a y  0 n  Fg
 T  Ff 
t
 M 
v xf  vxi  ax t  vxi  
x  x f  xi 
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
vxi t 
1  T  Ff  2
t
2  M 
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Example of Using Newton’s Laws
A traffic light weighing 125 N hangs from the cable tied to two other cables
fastened to a support. The upper cables make angles of 37.0o and 53.0o with
the horizontal plane. Find the magnitudes of the tensions in the three cables.
37o
y
53o
T1
Free-body
Diagram
T2
37o
53o
T3
r
ur ur ur ur
F  T 1  T 2  T 3  ma  0
x
Newton’s 2nd law
i 3
x-comp. of
cos  53o 
o
o
T2  0.754T2
Tix  0 T1 cos 37  T2 cos 53  0 T1 
o
net force Fx  
cos
37
 
i 1
 
y-comp. of
net force
i 3
Fy   Tiy  0
i 1
Monday, Oct. 11, 2010
 
 
 
T1 sin 37o  T2 sin 53o  mg  0
 
 
o
o
  1.25T2  125N
T2 
sin
53

0.754

sin
37


T2  100N ; T1  0.754T2  75.4N
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Example w/o Friction
A crate of mass M is placed on a frictionless inclined plane of angle θ.
a) Determine the acceleration of the crate after it is released.
y
ur ur
r
F  F g  n  ma
n
Fx  Ma x  Fgx  Mg sin q
y
n
x

q
Fg
Free-body
Diagram
Supposed the crate was released at the
top of the incline, and the length of the
incline is d. How long does it take for the
crate to reach the bottom and what is its
speed at the bottom?
ax  g sin q
x
F= -Mg F  Ma  n  F  n  mg cosq  0
y
y
gy
2d
g sin q
1 2 1
v
t

ax t  g sin q t 2 t 
d  ix
2
2
v xf  vix  ax t  g sin q
2d
 2dgsinq
g sin q
vxf  2dgsinq
Monday, Oct. 11, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Example w/ Friction
Suppose a block is placed on a rough surface inclined relative to the horizontal. The
inclination angle is increased till the block starts to move. Show that by measuring
this critical angle, qc, one can determine coefficient of static friction, μs.
y
n
n
Fs=μsn
x
F= -Mg


Fg
Free-body
Diagram
Net force
r ur
r ur
ur
F  M a  Fg  n  f s
x comp.
Fx  Fgx  fs  Mg sin q  f s  0
f s   s n  Mg sin q c
y comp.
Fy  Ma y  n  Fgy  n  Mg cosqc  0
n Fgy  Mg cosq c
Mg sin q c
Mg sin q c
 tan q c
s 

Mg cos q c
n
Monday, Oct. 11, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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