Transcript Monday, September 20, 2004

PHYS 1443 – Section 003
Lecture #8
Monday, Sept. 20, 2004
Dr. Jaehoon Yu
1.
Newton’s Laws of Motion
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•
2.
Application of Newton’s Laws
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•
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Monday, Sept. 20, 2004
Gravitational Force and Weight
Newton’s third law of motion
Free-body diagrams
Application of Newton’s Laws
Motion without friction
Forces of friction
Motion with friction
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
1
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one week from today
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• Miss an exam without pre-approval or a good reason: Your
grade is F.
– Mixture of multiple choice and free style problems
Monday, Sept. 20, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
2
Example for Newton’s 2nd Law of Motion
Determine the magnitude and direction of acceleration of the puck whose
mass is 0.30kg and is being pulled by two forces, F1 and F2, as shown in the
picture, whose magnitudes of the forces are 8.0 N and 5.0 N, respectively.
F  F1 cos q1  8.0  cos  60   4.0N
Components 1x
of F1
F1 y  F1 sin q1  8.0  sin  60   6.9N
F1
q160o
q2-20o
F2
Components F2 x  F2 cos q2  5.0  cos  -20   4.7N
of F2

Components of
total force F
Fx  F1x  F2 x  4.0  4.7  8.7N  max
Fy  F1 y  F2 y  6.9 -1.7  5.2N  ma y
F
8.7
ax  x 
 29m / s 2
m 0.3
Fy
Magnitude and
direction of
a
-1  17 
o
-1  y 
tan
tan

q

acceleration a
 
   30
29
a

Monday, Sept. 20, 2004

F2 y  F2 sin q 2  5.0  sin -20  -1.7N
x


5.2
ay 

 17 m / s 2
m
0.3

Acceleration
Vector a
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
r
2
2
2
a   ax 2   a y    29  17 
 34m / s 2


r




a  ax i  ay j   29 i  17 j  m / s 2


3
Gravitational Force and Weight
Gravitational Force, Fg
The attractive force exerted
on an object by the Earth
F G  ma  m g
Weight of an object with mass M is W  F G  M g  Mg
Since weight depends on the magnitude of gravitational
acceleration, g, it varies depending on geographical location.
By measuring the forces one can determine masses. This is
why you can measure mass using the spring scale.
Monday, Sept. 20, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
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Newton’s Third Law (Law of Action and Reaction)
If two objects interact, the force F21 exerted on object 1 by object 2
is equal in magnitude and opposite in direction to the force F12
exerted on object 2 by object 1.
F21
F12
F 12  - F 21
The action force is equal in magnitude to the reaction force but in
opposite direction. These two forces always act on different objects.
What is the reaction force to the
force of a free fall object?
The gravitational force exerted
by the object to the Earth!
Stationary objects on top of a table has a reaction force (normal force)
from table to balance the action force, the gravitational force.
Monday, Sept. 20, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
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Example of Newton’s 3rd Law
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?
F12
F21=-F12
m
F 12  - F 21
F 12  F 21  F
F 12  ma b
F 12 x  mabx
F 12 y  maby  0
F 21  M a M
M
F 21x  MaMx
F  Ma  0
21 y
Since
F 12  - F 21
Establish the
equation
Monday, Sept. 20, 2004
My
and F 12  - F 21  F
mabx  F  MaMx
Divide by m
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
F M
aMx
abx 

m m
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Example of Newton’s 3rd Law
Man’s velocity
vMxf  vMxi  aMxt  aMx t
Boy’s velocity
vbxf  vbxi  abxt  abxt 
M
M
aMxt 
vMxf
m
m
So boy’s velocity if higher than man’s, if M>m, by the ratio of the masses.
b) Who moves farther while their hands are in contact?
Boy’s displacement
1
M
abx t 2 
aMx t 2
2
2m
M 1
M
2 
xb 
a
t
xM
Mx


m 2

m
xb  vbxi t 
Man’s displacement
Given in the same time interval, since the boy has higher acceleration and thereby higher
speed, he moves farther than the man.
Monday, Sept. 20, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
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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 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
Monday, Sept. 20, 2004
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 1443-003, Fall 2004
Dr. Jaehoon Yu
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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
Which one would you like to select to draw FBD?
What do you think are the forces acting on this object?
FN
Gravitational force
FG  M g
Gravitational force
Me
m
The force pulling the elevator (Tension)
What about the box in the elevator?
Monday, Sept. 20, 2004
FG  M g
A force supporting the object exerted by the floor
Which one would you like to select to draw FBD?
What do you think are the forces acting on this elevator?
FT
FN
F GB  mg
FG  M g
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
Gravitational
force
Normal
force
FT
FG  M g
FN
F9BG  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
T
Fg=Mg
Normal force: n
T
Fg=Mg
Tension force: T
Total force:
F=Fg+n+T=T
If T is a constant
force, ax, is constant
Monday, Sept. 20, 2004
 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 1443-003,
2004
WhatFall
happened
to the motion in y-direction? 10
Dr. Jaehoon Yu
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
Monday, Sept. 20, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
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