Transcript Wednesday, March 5, 2008

PHYS 1441 – Section 002
Lecture #13
Wednesday, Mar. 5, 2008
Dr. Jaehoon Yu
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Wednesday, Mar. 5, 2008
Static and Kinetic Frictional Forces
The Tension Force
Equilibrium Applications of Newton’s Laws
Non-equilibrium Applications of Newton’s Laws
Uniform Circular Motion
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Announcements
• Term exam #2
– Wednesday, March 26, in class
– Will cover CH4.1 – whatever we finish Monday, Mar. 24
• Colloquium today
– 4pm in SH101
– Dr. M. Wiltberger, U. of Colorado, Boulder
– Geospace Modeling
Wednesday, Mar. 5, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Wednesday, Mar. 5, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Special Project Reminder
• Using the fact that g=9.80m/s2 on the Earth’s
surface, find the average density of the Earth.
• 20 point extra credit
• Due: Wednesday, Mar. 12
• You must show your OWN, detailed work to
obtain any credit!!
Wednesday, Mar. 5, 2008
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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.
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PHYS 1441-002, Spring 2008
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Magnitude of Static Friction
The magnitude of the static frictional force can have any value
from zero up to a maximum value.
fs  f
f
MAX
s
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, Spring 2008
Wednesday, Mar. 5, 2008
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Note that the magnitude of the frictional force does not depend
on the contact area of the surfaces.
f
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MAX
s
 s FN
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Kinetic Friction
Static friction opposes the impending relative motion between
two objects.
Kinetic friction opposes the relative sliding motion motions that
actually does occur. The resistive force exerted on the object
during its movement.
f k  k FN
0  s  1
is called the coefficient of kinetic friction.
What is the direction of frictional forces? opposite to the movement
Wednesday, Mar. 5, 2008
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Coefficient of Friction
What
are
these?
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Ex. 10. Sled Riding
A sled is traveling at 4.00m/s along a horizontal stretch of snow.
The coefficient of kinetic friction k=0.0500. How far does the
sled go before stopping?
What are
the forces in
this motion?
The sled comes to a halt because the kinetic frictional force
opposes its motion and causes the sled to slow down.
Wednesday, Mar. 5, 2008
PHYS 1441-002, Spring 2008
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Ex. 10 continued
What is the net force in y direction?
What is the net force in x direction?
0N
Fx   f k  ma
f k  k FN  k mg
So the force equation becomes
fk
 k mg
Solve this for a a  

  k g 
m
m
2
2
0.05  9.80 m s   0.49 m s
Now that we know a and vi, pick the a
kinematic equation to solve for distance
Solve this for x
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x
2
2
v

v
 f i
2a

2ax   v 2f  vi2 
2
0

4.00


2   0.49 
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
 16.3m
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The Tension Force
Cables and ropes transmit forces through tension.
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Tension Force continued
A massless rope will transmit
tension undiminished from one
end to the other.
If the rope passes around a
massless, frictionless pulley, the
tension will be transmitted to the
other end of the rope
undiminished.
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Some Basic Information
When Newton’s laws are applied, external forces are only of interest!!
Why?
Because, as described in Newton’s first law, an object will keep its
current motion unless non-zero net external force is applied.
Normal Force, n:
Reaction force to the net force on a surface due to
the surface structure of an object. Its direction is
always perpendicular to the surface.
Tension, T:
The reactionary force by a stringy object
against an external force exerted on it.
Free-body diagram
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A graphical tool which is a diagram of external
forces on an object and is extremely useful analyzing
forces and motion!! Drawn only on an object.
PHYS 1441-002, Spring 2008
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Equilibrium
An object is in equilibrium when it has zero acceleration.
0
F

ma


Conditions for
Translational
Equilibrium
ma

ma

F

0
F

0
x
y
 x
 y
If an object is not moving at all, the object is in its static equilibrium.
Is an object is moving at a constant velocity in its equilibrium?
Why?
Yes
Because its acceleration is 0.
Wednesday, Mar. 5, 2008
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Strategy for Solving Problems
•
Select an object(s) to which the equations of equilibrium are
to be applied.
• Identify all the forces acting only on the selected object
• Draw a free-body diagram for each object chosen above.
Include only forces acting on the object, not forces the
object exerts on its environment.
• Choose a set of x, y axes for each object and resolve all
forces in the free-body diagram into components that point
along these axes.
• Apply the equations and solve for the unknown quantities
 No matter which object we choose to draw the free body
diagram on, the results should be the same, as long as they
are in the same motion
Wednesday, Mar. 5, 2008
PHYS 1441-002, Spring 2008
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Ex. 11 Traction for the Foot
The weight of the 2.2 kg object
creates a tension in the rope that
passes around the pulleys.
Therefore, tension forces T1 and
T2 are applied to the pulley on the
foot. The foot pulley is kept in
equilibrium because the foot also
applies a force F to it. This force
arises in reaction to the pulling
effect of the forces T1 and T2.
Ignoring the weight of the foot,
fine the magnitude of the force F.
 F  T sin 35 T sin 35  0
 F  T cos 35 T cos 35  F  0
x
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2
1
y
1
2
PHYS 1441-002, Spring 2008
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Ex. 12 Replacing an Engine
An automobile engine has a weight (or gravitational force) W, whose magnitude is
W=3150N. This engine is being positioned above an engine compartment, as shown in
the figure. To position the engine, a worker is using a rope. Find the tension T1 in the
support cabling and the tension T2 in the positioning rope.
Wednesday, Mar. 5, 2008
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First, analyze the forces in x and y
Force
x component
y component
T1
T1 sin10.0
T1 cos10.0
T2
T2 sin 80.0
T2 cos80.0
W
0
W
W  3150 N
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PHYS 1441-002, Spring 2008
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Now compute each force component
F
x
F
y
 T1 sin10.0  T2 sin 80.0  0
 T1 cos10.0 T2 cos80.0 W  0
 sin 80.0
The first equation gives T1  
 sin10.0

 T2

Substitution into the second gives
 sin 80.0

 sin10.0
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
 T2 cos10.0 T2 cos80.0 W  0

PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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T2 
W
 sin 80.0 

 cos10.0  cos80.0
 sin10.0 
3150 N
 sin 80.0 

 cos10.0  cos80.0
 sin10.0 

 582 N
T2  582 N
 sin 80.0
T1  
 sin10.0
Wednesday, Mar. 5, 2008

 T2

T1  3.30 10 N
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Is an accelerating object in its equilibrium? NO
ma

F

0

ma

ma

F

0
F

0
y
 x
 y
x
Wednesday, Mar. 5, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Ex. 14 Towing a Supertanker
A supertanker of mass m=1.50x108kg is being towed by two tugboats. The tension in the towing
cables apply the forces T1 and T2 at equal angles of 30o with respect to the tanker’s axis. In
addition, the tanker’s engines produce a forward drive force D, whose magnitude is D=75.0x103N.
Moreover, the water applies an opposing force R, whose magnitude is R=40.0x103N. The tanker
moves forward with an acceleration that points along the tanker’s axis and has a magnitude of
2.00x10-3m/s2. Fine the magnitude of T1 and T2.
The acceleration is along the x axis so a y  0
Wednesday, Mar. 5, 2008
PHYS 1441-002, Spring 2008
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Figure out X and Y components
Force
x component
y component
T1
T2
T1 cos 30.0
T1 sin 30.0
T2 cos 30.0
T2 sin 30.0
D
0
R
0
D
R
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F
y
 T1 sin 30.0 T2 sin 30.0  0
T1 sin 30.0  T2 sin 30.0  T1  T2  T
F
x
 T1 cos 30.0 T2 cos30.0  D  R
 max
Since T1=T2=T
Solving for T
2T cos 30.0  max  R  D
max  R  D 1.50 10  2.00 10  40.0 10  75.0 10

T
2cos 30.0
2 cos 30.0
5
0
3
3
3
 1.53 10 N
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
Example for Using Newton’s Laws
A traffic light weighing 125 N hangs from a cable tied to two other cables
fastened to a support. The upper cables make angles of 37.0o and 53.0o with
the horizontal. Find the tension in the three cables.
37o
y
53o
T1
Free-body
Diagram
T2
37o
53o
T3
r r r
r
r
T

T

T

ma
0
F 1 2 3
i 3
x-comp. of
Tix  0
net force Fx  
i 1
y-comp. of
net force Fy 
i 3
T
i 1
iy
0
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 
x
Newton’s 2nd law
 
 T1 cos 37  T2 cos 53  0 T1 
 
 
T sin 53   0.754  sin 37   1.25T
 T
cos  37 
cos 53o
o
2

0.754T2
T1 sin 37o  T2 sin 53o  mg  0


2
2
 125N
0.754T2  75.4 N
T2  PHYS
1001441-002,
N; T
1  2008
Spring
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 q.
a) Determine the acceleration of the crate after it is released.
y
r
ur ur
r
F F g  n ma
n
Fx  Ma x  Fgx  Mg sin q
y
n
x
q
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?
Mg
ax  g sin q
Fy  Ma y  n  Fgy  n  mg cos q  0
1 2 1
v
t

a x t  g sin q t 2  t 
d  ix
2
2
v xf  vix  axt g sin q
2d
g sin q
2d
 2dg sin q
g sin q
 vxf  2dg sin q
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PHYS 1441-002, Spring 2008
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, s.
y
n
Free-body
Diagram
Fg
n
fs=kn
x
q
q
Mg
Net force
r ur
r ur
ur
F  M a  Fg n f s
x comp.
Fx  Fgx  f s  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
Wednesday, Mar. 5, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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