PHYS 1443 – Section 001
Lecture #7
Thursday, June 7, 2007
Dr. Jaehoon Yu
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Application of Newton’s Laws
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•
•
•
•
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Motion without friction
Forces of Friction
Uniform and Non-uniform Circular Motions
Resistive Forces and Terminal Velocity
Newton’s Law of Universal Gravitation
Kepler’s Laws
Today’s homework is HW #4, due 7pm, Monday, June 11!!
Thursday, June 7, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
1
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 that reacts to gravitational
force due to the surface structure of an object.
Its direction is perpendicular to the surface.
Tension, T:
The reactionary force by a stringy object
against an external force exerted on it.
Free-body diagram
Thursday, June 7, 2007
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 1443-001, Summer 2007
Dr. Jaehoon Yu
2
Free Body Diagrams and Solving Problems
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
1.
2.
3.
4.
5.
6.

Free-body diagram: A diagram of vector forces acting on an object
A great tool to solve a problem using forces or using dynamics
Select a point on an object in the problem
Identify all the forces acting only on the selected object
Define a reference frame with positive and negative axes specified
Draw arrows to represent the force vectors on the selected point
Write down net force vector equation
Write down the forces in components to solve the problems
No matter which one we choose to draw the diagram on, the results should be the same,
as long as they are from the same motion
FN
M
FN
FG  M g
FT
Gravitational force
Me
m
A force supporting the object exerted by the floor
The force pulling the elevator (Tension)
What about the box in the elevator?
Thursday, June 7, 2007
F GB  mg
FG  M g
Which one would you like to select to draw FBD?
What do you think are the forces acting on this elevator?
Gravitational force
FN
FG  M g
Which one would you like to select to draw FBD?
What do you think are the forces acting on this object?
Gravitational
force
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
Normal
force
FT
FG  M g
FN
F3BG  m g
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
Thursday, June 7, 2007
Tension force: T
Total force:
F=Fg+n+T=T
If T is a constant
force, ax, is constant
 Fx  T  Ma x
F
y
ax 
  Fg  n  Ma y  0
T
M
ay  0
v xf  vxi  axt  v xi  
T 
t
M 
1 T  2
x

x

v
t


t
x  f i xi
2M 
PHYS 1443-001,
Summer 2007
What happened
to the motion in y-direction?
Dr. Jaehoon Yu
4
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
Thursday, June 7, 2007
0
 
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
 0.754
T  75.4 N
T2  100
N ; T1 Summer
PHYS 1443-001,
2007 2
Dr. Jaehoon Yu
5
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
n
x
q
F  Fg  n  ma
n
Fx  Ma x  Fgx  Mg sin q
Free-body
Diagram
q
Fg
y
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 cos q  0
y
y
gy
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
Thursday, June 7, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
6
Forces of Friction
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
f s  s 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
f k  k n
The resistive force exerted on the object
during its movement
Which direction does kinetic friction apply?
Thursday, June 7, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
Opposite to the motion!
7
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
n
fs=kn
x
F= -Mg
q
q
Fg
Free-body
Diagram
Net force
F  Ma  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
Thursday, June 7, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
8
Newton’s Second Law & Uniform Circular Motion
The centripetal acceleration is always perpendicular
to velocity vector, v, for uniform circular motion.
v
ar 
r
2
Are there forces in this motion? If so, what do they do?
The force that causes the centripetal acceleration
acts toward the center of the circular path and
causes the change in the direction of the velocity
vector. This force is called the centripetal force.
v2
 Fr  mar  m r
What do you think will happen to the ball if the string that holds the ball breaks?
Based on Newton’s 1st law, since the external force no longer exist.
Therefore, the ball will continue its motion without change and will fly
away following the tangential direction to the circle.
Thursday, June 7, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
9
Example of Uniform Circular Motion
A ball of mass 0.500kg is attached to the end of a 1.50m long cord. The ball is
moving in a horizontal circle. If the string can withstand maximum tension of 50.0 N,
what is the maximum speed the ball can attain before the cord breaks?
Fr m
Centripetal
acceleration:
When does the
string break?
v2
ar 
r
v2
 Fr  mar  m r  T
when the centripetal force is greater than the sustainable tension.
v2
m
 T
r
v  Tr  50.0 1.5  12.2  m / s 
m
0.500
Calculate the tension of the cord
when speed of the ball is 5.00m/s.
Thursday, June 7, 2007
v2
 5.00   8.33 N
 0.500 
T m
 
r
1.5
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
2
10
Example of Banked Highway
(a) For a car traveling with speed v around a curve of radius r, determine a formula
for the angle at which a road should be banked so that no friction is required to keep
the car from skidding.
y
x
mv 2
x comp.  Fx  FN sin q  mar  FN sin q 
0
2
r
mv
FN sin q 
r
FN cos q  mg
y comp.  Fy  FN cos q  mg  0
2
mg
mg sin q
mv
FN sin q 
FN 
 mg tan q 
sin q
cos q
r
v2
tan q 
gr
(b) What is this angle for an expressway off-ramp curve of radius 50m at a design
speed of 50km/h?
v  50km / hr  14m / s
Thursday, June 7, 2007
tan q 
142
50  9.8
 0.4
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
q  tan 1 0.4  22o
11