Download Monday, February 18, 2013

PHYS 1441 – Section 002
Lecture #9
Monday, Feb. 18, 2013
Dr. Jaehoon Yu
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Free Body Diagram
Newton’s Third Law
Categories of forces
Announcements
• Term exam results
– Class average: 60/103
• Equivalent to 58.2/100
– Top score: 98.5/103
• Evaluation policy
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–
–
–
–
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Homework: 25%
Final comprehensive exam: 23%
Midterm comprehensive exam: 20%
Better of the two term exams: 12%
Lab: 10%
Pop quizzes: 10%
Extra credit: 10%
Monday, Feb. 18, 2013
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
2
Newton’s Second 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.
How do we write the above statement in a mathematical expression?
Newton’s 2nd
Law of Motion
From this
we obtain
m
Since it’s a vector expression, each component must also satisfy:
F
ix
 max
i
Monday, Feb. 18, 2013
F
iy
 may
i
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
F
iz
 maz
i
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Unit of the Force
From the vector expression in the previous page, what do
you conclude the dimension and the unit of the force are?
The dimension of force is
The unit of force in SI is
[m][ a ]  [ M ][ LT 2 ]
[ Force]  [m][a]  [M ][ LT 2 ] 
For ease of use, we define a new
derived unit called, Newton (N)
Monday, Feb. 18, 2013
m
kg
   2   kg  m / s 2
s 
1
1N º1kg × m / s » lbs
4
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
2
4
Free Body Diagram
A free-body-diagram is a diagram that represents
the object and the forces that act on it.
Monday, Feb. 18, 2013
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
5
Ex. Pushing a stalled car
What is the net force in this example?
F= 275 N
Which direction?
Monday, Feb. 18, 2013
+ 395 N
– 560 N =
+110 N
The + x axis of the coordinate system.
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
6
What is the acceleration the car receives?
If the mass of the car is 1850 kg, then by Newton’s
second law, the acceleration is
Since the motion is
in 1 dimension
Now we solve this
equation for a
F


a
Monday, Feb. 18, 2013
m
 F  ma
110 N
 0.059 m s2
1850 kg
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
7
Vector Nature of the Force
The direction of the force and the acceleration vectors can be
taken into account by using x and y components.
is equivalent to
F
y
Monday, Feb. 18, 2013
 may
F
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
x
 max
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Ex. Stranded man on a raft
A man is stranded on a raft
(mass of man and raft =
1300kg) 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.
Monday, Feb. 18, 2013
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
9
First, let’s compute the net force on the raft as follows:
Force
x component
+17 N
+(15N)cos67o
+17+15cos67o=
+23(N)
Monday, Feb. 18, 2013
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
y component
0N
+(15N)sin67o
+15sin67o=
+14(N)
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Now compute the acceleration components in x and y directions!!
ax 
ay
F
x
m
F


y
m
23 N
2

 0.018 m s
1300 kg
14 N

 0.011 m s 2
1300 kg
And put them all
together for the
overall acceleration:
Monday, Feb. 18, 2013
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
11
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.
F1
 220o
F2
total force F
Fx
8.7

Fy
2
y
Monday, Feb. 18, 2013


  4.7N

Fx  F1x  F2 x  4.0  4.7  8.7N  max
Fy  F1 y  F2 y  6.9 1.7  5.2N  ma y
x
x
  6.9N
5.0  sin 20  1.7N
Magnitude and a  m  0.3  29m / s a  m
direction of
a
1  17 
o
1  y 
tan
tan



acceleration a
 
   30
29
a


5.0  cos 20
F2 y 
Components of
  4.0N
8.0  sin 60
Components F2 x 
of F2
 160o

8.0  cos 60
F 
Components 1x
of F1
F1 y 



5.2
 17 m / s 2
0.3
Acceleration
Vector a
PHYS 1441-002, Spring 2013
Dr. Jaehoon Yu
 ax 

2
  a y    29  17 
 34m / s 2
2
2
2




ax i  ay j   29 i  17 j  m / s 2


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