Transcript Thursday, June 9, 2005

PHYS 1443 – Section 001
Lecture #6
Thursday June 9, 2005
Dr. Nurcan Ozturk for Dr. Brandt
•
•
•
Newton’s Laws/Force
Friction
Circular Motion
Thursday June 9, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
1
Announcements
• Homework:
– HW3 on ch4 due Monday 6/13 at 2pm
– HW4 on ch5 due Tuesday 6/14 at 8pm
Thursday June 9, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
2
Newton’s Laws of Motion
Newton’s 1st law of motion (Law of Inertia): In the absence of external forces, an object at rest
remains at rest and an object in motion continues in motion with a constant velocity.
Newton’s 2nd law of motion: The acceleration of an object is directly proportional
to the net force exerted on it and is inversely proportional to the object’s mass.
Newton’s 3rd law of motion: Whenever one object exerts a force on a second object,
the second exerts an equal and opposite force on the first.
 Fi  ma
i
since Force
is a vector…
For simplicity, we define a new
derived unit called, a Newton (N)
Thursday June 9, 2005
F
ix
 max
i
F
iy
 may
F
iz
i
 maz
i
1
1N  1kg  m / s  lb
4
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
2
3
Some Basic Information
When Newton’s laws are applied, external forces only are of interest!!
Gravitational Force (Weight)
r
r
W  FG  M g  Mg
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 9, 2005
A graphical tool which is a diagram of external forces
on an object and is extremely useful in analyzing forces
and motion!! Drawn only on an object.
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
4
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 9, 2005
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
PHYS 1443-001,
I 2005 in the
Is Summer
there motion
Dr. Andrew Brandt
2M 
y-direction?
5
Example 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
ur ur ur ur
F  T 1  T 2  T 3  ma  0
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 9, 2005
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
T1  I0.754
T2  75.4 N
T2 PHYS
1001443-001,
N ; Summer
2005
Dr. Andrew Brandt
T3  125N
6
Example without 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
b) 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 9, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
7
Forces of Friction
Resistive force exerted on a moving object due to viscosity or other types of
frictional properties 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 is variable and will
increase until it reaches the limit
Beyond the limit, the object moves, and there is NO MORE static
friction but kinetic friction takes over.
Force of kinetic friction, fk
fk  k n
The resistive force exerted on the object
during its movement
Which direction does kinetic friction apply?
Thursday June 9, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
Opposite to the motion!
8
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
Free-body
Diagram
Fg
fs=sn
q
q
Net force
x
F= -Mg
F  M a  F g  n  f s On the verge of sliding, block is stationary,
acceleration is zero, friction is static friction.
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
If block moves at constant speed, acceleration
Mg sin q c
Mg
sin
q
c
s 
 Mg cos q  tan q c is zero, friction is kinetic friction. Coefficient
c
n
of kinetic friction can be calculated as k  tan q
Thursday June 9, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
9
Newton’s Second Law & Uniform Circular Motion
The centripetal acceleration is always perpendicular
to the velocity vector, v, for uniform circular motion.
v2
ar 
r
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 a change in the direction of the velocity
vector. This force is called centripetal force.
v2
 Fr  mar  m r
What do you think will happen to the ball if the string that holds the ball breaks? Why?
Based on Newton’s 1st law, since the external force no longer exist, the
ball will continue its motion without change and will fly away following the
tangential direction to the circle.
Thursday June 9, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
10
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 a 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 9, 2005
v2
 5.00   8.33 N
 0.500 
T m
 
r
1.5
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
2
11
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
x comp.
F
 FN sin q  mar
y comp.
F
 FN cosq  mg  0
x
y
v2
FN sin q  m
r
FN cos q  mg
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 9, 2005
tan q 
142
50  9.8
 0.4
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
q  tan 1 0.4  22o
12
Forces in Non-uniform Circular Motion
The object has both tangential and radial
accelerations.
What does this statement mean?
Fr
F
The object is moving under both
tangential and radial forces.
Ft
F  Fr  Ft
These forces cause not only the velocity but also the speed of the ball to
change. The object undergoes a curved motion under the absence of
constraints, such as a string.
How does the acceleration look?
Thursday June 9, 2005
a  ar2  at2
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
13
Example for Non-Uniform Circular Motion
A ball of mass m is attached to the end of a cord of length R. The ball is moving in a
vertical circle. Determine the tension of the cord at any instant when the speed of
the ball is v and the cord makes an angle q with vertical.
q
T
R
m
What are the forces involved in this motion?
•The gravitational force Fg
•The radial force, T, providing tension.
Fg=mg
tangential
comp.
Radial
comp.
F
t
 mg sin q  mat
2
v
 Fr  T  mg cosq  mar  m R
at  g sin q
 v2

T  m  g cos q 
R

At what angles does the tension become
maximum and minimum. What are the max
and min tensions?
Thursday June 9, 2005
PHYS 1443-001, Summer I 2005
Dr. Andrew Brandt
14