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Transcript
PHYS 1441 – Section 002
Lecture #8
Wednesday, Feb. 25, 2009
Dr. Jaehoon Yu
•
•
•
•
Newton’s Second Law and Third Law
Free Body Diagram
Categories of forces
Application of Newton’s Laws
–
•
Forces of Friction
–
Wednesday, Feb. 25,
2009
Motion without friction
Motion with friction
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
1
Announcements
• Quiz Monday, Mar.2
– Beginning of the class
– Covers Ch4. or what we finish this Wednesday
• Change of Exam Dates
– Mid-term exam scheduled on Mar. 11 now moved to
Wed. Mar. 27
– Second non-comprehensive exam scheduled on Apr. 13
now moved to Wed. Apr. 22
– Final comprehensive exam stays on the same date, May
11
• Colloquium today
Wednesday, Feb. 25,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
2
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.
• Each problem is 10 points each.
• Due is Wednesday, Mar. 4.
• Please be sure to clearly define each variables
used in your derivation.
Wednesday, Feb. 25,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
3
Wednesday, Feb. 25,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
4
Vector Nature of the Force
The direction of force and acceleration vectors can be taken
into account by using x and y components.
r
r
 F  ma
is equivalent to
F
y
Wednesday, Feb. 25,
2009
 may
F
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
x
 max
5
Ex. Stranded man on a raft
A man is stranded on a raft
(mass of man and raft =
1300kg)m as shown in the
figure. By paddling, he
causes an average force P of
17N to be applied to the raft
in a direction due east (the +x
direction). The wind also
exerts a force A on the raft.
This force has a magnitude of
15N and points 67o north of
east. Ignoring any resistance
from the water, find the x and
y components of the rafts
acceleration.
Wednesday, Feb. 25,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
6
First, let’s compute the net force on the raft as follows:
Force
r
P
r
A
r r r
F  P A
Wednesday, Feb. 25,
2009
x component
+17 N
y component
0N
+(15N)cos67o
+(15N)sin67o
+17+15cos67o=
+23(N)
+17+15sin67o=
+14(N)
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
7
Now compute the acceleration components in x and y directions!!
ax 
ay
F
x
m
F


The overall
acceleration is
Wednesday, Feb. 25,
2009
y
m
23 N
2

 0.018 m s
1300 kg
14 N

 0.011 m s 2
1300 kg
r
r
r
a  ax i  a y j 


r
r
0.018i  0.011 j m s 2
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
8
Example for Newton’s 2nd Law of Motion
Determine the magnitude and direction of the 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.
ur
F1x  F 1 cos q1  8.0  cos  60
  4.0N
Components
ur
of F1
F1 y  F 1 sin q1  8.0  sin  60   6.9N
F1
q160o
q2-20o
F2
ur
Components F2 x  F 2 cos q2  5.0  cos  -20   4.7N
ur
of F2
F2 y  F 2 sin q 2  5.0  sin  -20   -1.7N
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

Wednesday, Feb. 25,
2009
x


5.2
ay 

 17 m / s 2
m
0.3

Acceleration
Vector a
PHYS 1441-002, Spring 2009 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


9
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.
F12
ur
2
ur
F21
F 12  - F 21
1
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 falling object?
The gravitational force exerted
by the object to the Earth!
Stationary objects on top of a table has a reaction force (called the normal
force) from table to balance the action force, the gravitational force.
Wednesday, Feb. 25,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
10
Ex. The Accelerations Produced by
Action and Reaction Forces
Which one do you think will
get larger acceleration?
Suppose that the magnitude of the force is 36 N. If the mass
of the spacecraft is 11,000 kg and the mass of the astronaut
is 92 kg, what are the accelerations?
Wednesday, Feb. 25,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
11
Ex. continued
r r
Force exerted on the space craft by the astronaut  F  P.
r
r
Force exerted on the astronaut by the space craft  F  -P.
r
r
P
36 i N
space craft’s r
2
a


 0.0033m s
s
acceleration
ms 11,000 kg
r
r
- P -36 i N
2
astronaut’s r
a

0.39
m
s


A
acceleration
m A 92 kg
Wednesday, Feb. 25,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
12
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
Since
m
ur
ur
F 12  - F 21
Establish the
equation
Wednesday, Feb. 25,
2009
ur
ur
F 12  - F 21
ur
ur
F 12  F 21  F
r
ur
F 12  ma b
ur
r
F 21  M a M
F 12 x  mabx
F 12 y  maby  0
F 21x  MaMx
F  Ma  0
21 y
and
My
ur
ur
F 12  - F 21  F
mabx  F  MaMx
Divide by m
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
F M
aMx
abx 

m m
13
Example of Newton’s 3rd Law, cnt’d
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 is 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
xb  vbxi t  abx t 2 
aMx t 2
2
2m
M 1
M
2 
xb 
a
t
xM
Mx


m 2

m
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.
Wednesday, Feb. 25,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
14
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 action forces 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
Wednesday, Oct. 1, 2008
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.
15
Free Body Diagrams and Solving Problems
•

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
Which one would you like to select to draw FBD?
What do you think are the forces acting on this object?
Gravitational force
FG  M g
Gravitational force
Me
m
The force pulling the elevator (Tension)
What about the box in the elevator?
Wednesday, Oct. 1, 2008
FG  M g
the 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
Gravitational
force
Normal
force
FT
FG  M g
FN
16
F BG  m g
Categories of Forces
• Fundamental Forces: Truly unique forces that
cannot be derived from any other forces
– Total of three fundamental forces
• Gravitational Force
• Electro-Weak Force
• Strong Nuclear Force
• Non-fundamental forces: Forces that can be
derived from fundamental forces
– Friction
– Tension in a rope
– Normal or support forces
Wednesday, Oct. 1, 2008
17
The Normal Force
The normal force is one component of the force that a surface
exerts on an object with which it is in contact – namely, the
component that is perpendicular to the surface.
Monday, Mar. 3, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
18
Some normal force exercises
Case 1: Hand pushing down on the book
FN - 11 N - 15 N= 0
FN  26 N
Case 2: Hand pulling up the book
FN  11 N - 15 N  0
FN  4 N
Monday, Mar. 3, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
19
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
Wednesday, Oct. 1, 2008
 Fx  T  Ma x
F
y
ax 
 - Fg  n  Ma y  0
T
M
ay  0
v xf  vxi  a xt  v xi  
T 
t
M 
1 T  2
x
x

v
t


t
x  f i xi
2M 
What happened to the motion in y-direction?
20