Transcript N e w t o n` s L a w s

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Unit 3
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3.1 Force and Mass
Force – push or pull; required to
change an object’s motion.
Vector – so magnitude and direction
Example of Contact Forces
Examples of Field Forces
Friction
Tension
Gravitational
Applied
Electric
Magnetic
Spring
3.1 Force and mass
Mass – measurement of how difficult it is to
change the objects velocity
Inertia – resistance to change in velocity
So mass is a measurement of an object’s
inertia
3.1 Force and mass
3.2 Newton’s First Law of Motion
1st Law
An object at rest remains at rest as long as no
net force acts on it.
An object moving with constant velocity
continues to move with the same speed and
in the same direction as long as no net force
acts on it.
3.2 Newton’s First Law of Motion
Sometimes called the Law of Inertia
3.2 Newton’s First Law of Motion
3.3 Newton’s Second Law of Motion
2nd Law
A net force causes an acceleration in the
direction of the net force.


F  ma
Simulation
3.3 Newton’s Second Law of Motion
Free body diagrams
Show all the forces acting on an object
For example an object sitting on a table
W – weight = mg
N
N – Normal Force (perpendicular) to the
surface
W
3.3 Newton’s Second Law of Motion
Free body diagrams
If a rope pulls the object toward the right, then
T = Tension
N
Practice Free Body
T
W
3.3 Newton’s Second Law of Motion
Free body diagrams
Steps in problems solving
1.Sketch the forces
2.Isolate the Object
3.Choose a Coordinate System
4.Resolve the Forces into Components
5.Apply Newton’s Second Law of Motion
3.3 Newton’s Second Law of Motion
A 50 kg gopher has a string tied around his
neck and pulled with a force of 80 N at an
angle of 30o to the horizontal. What is his
acceleration?
3.3 Newton’s Second Law of Motion
A 50 kg gopher has a string tied around his
neck and pulled with a force of 80 N at an
angle of 30o to the horizontal. What is his
acceleration?
Free Body
diagram
3.3 Newton’s Second Law of Motion
A 50 kg gopher has a string tied around his
neck and pulled with a force of 80 N at an
angle of 30o to the horizontal. What is his
acceleration?
N
T
Free Body
diagram
W
3.3 Newton’s Second Law of Motion
A 50 kg gopher has a string tied around his
neck and pulled with a force of 80 N at an
angle of 30o to the horizontal. What is his
acceleration?
N
T
Free Body
diagram
Axis
W
3.3 Newton’s Second Law of Motion
A 50 kg gopher has a string tied around his
neck and pulled with a force of 80 N at an
angle of 30o to the horizontal. What is his
T
acceleration?
y
N
T
Free Body
diagram
Tx
Axis
W
3.3 Newton’s Second Law of Motion
A 50 kg gopher has a string tied around his
neck and pulled with a force of 80 N at an
angle of 30o to the horizontal. What is his
T
acceleration?
F  T  ma
y
N
Free Body
x
x
x
Fy  Ty  N  W  may
diagram
Tx
Axis
Equation
W
T cos
 30ma
80 cos
 50a
T sin
a 1.39Nm2 W  0
s
3.3 Newton’s Second Law of Motion
3.4 Newton’s Third Law of Motion
For every force that acts on an object, there is
a reaction force acting on a different object
that is equal in magnitude and opposite in
direction.
If object 1 exerts a force F on object 2, then
object 2 exerts a force –F on object 1.
3.4 Newton’s Third Law of Motion
What are the action reaction pairs in the
following?
3.4 Newton’s Third Law of Motion
A 60 kg man walks off a 3 m long canoe by
walking from one end to the other. He
applies a force of 20 N to the canoe, which
has a mass of 150 kg.
A.What is the acceleration of
the man?
B. What is the acceleration of the canoe?
3.4 Newton’s Third Law of Motion
A 60 kg man walks off a 3 m long canoe by
walking from one end to the other. He
applies a force of 20 N to the canoe, which
has a mass of 150 kg.
Free Body Diagrams
Nc
Nm
P
P
Wm
Wc
3.4 Newton’s Third Law of Motion
A 60 kg man walks off a 3 m long canoe by
walking from one end to the other. He
applies a force of 20 N to the canoe, which
has a mass of 150 kg.
Fx  P  mm a
Equations
Nc
Fx  P  mc a
Fy  N c  Wc  0
Fy  N m  Wm  0
Nm
P
P
Wm
Wc
3.4 Newton’s Third Law of Motion
A 60 kg man walks off a 3 m long canoe by
walking from one end to the other. He
applies a force of 20 N to the canoe, which
has a mass of 150 kg.
P  mm a
20  60a
A-acceleration of man
Nc
Fx  P  mc a
Fy  N c  Wc  0
Nm
a  0.33 sm2
P
P
Wm
Wc
3.4 Newton’s Third Law of Motion
A 60 kg man walks off a 3 m long canoe by
walking from one end to the other. He
applies a force of 20 N to the canoe, which
has a mass of 150 kg.
P  mm a
20  60a
A – acceleration of canoe
Nc
P  mc a
Nm
20  150a
P
a  0.13 sm2
a  0.33 sm2
P
Wm
Wc
3.4 Newton’s Third Law of Motion
Two boxes are tied together with a rope, and
the first one is pulled by a second rope.
Both boxes accelerate at 2.0 m/s2. If the
front box has a mass of 25 kg, and the
second a mass of 50 kg, what is the tension
on each rope?
a
3.4 Newton’s Third Law of Motion
Free body diagrams
Nb
Nf
T2
T1
T2
Wb
Wf
3.4 Newton’s Third Law of Motion
Equations
Fx  T2  mb a
Fx  T1  T2  m f a
Fy  N b  Wb  0
Fy  N f  W f  0
Solve (add)
T1  (m f  mb )a
T2  mb a
T1  T2  m f a
T2  (50kg)2
 100 N
m
Nb
Nf
T1  (25  50) 2
T1  150 N
T2
s2
T1
T2
Wb
Wf
3.4 Newton’s Third Law of Motion
3.5 The Vector Nature of Forces
Forces are vectors, so they can be treated
using vectors rules
3.5 The Vector Nature of Forces
Two men are carrying a 1.3 kg pail of water,
the first dude (Bob) exerts a force of 7N,
and the second one (Leon) exerts a force of
11N @ 28o. What is the angle of Bob’s
force?
Free Body Diagram?
L
B
W
3.5 The Vector Nature of Forces
Two men are carrying a 1.3 kg pail of water,
the first dude (Bob) exerts a force of 7N,
and the second one (Leon) exerts a force of
11N @ 28o. What is the angle of Bob’s
force?
Components?
B
L
By
Bx
W
3.5 The Vector Nature of Forces
Two men are carrying a 1.3 kg pail of water,
the first dude (Bob) exerts a force of 7N,
and the second one (Leon) exerts a force of
11N @ 28o. What is the angle of Bob’s
force?
L
y
Components?
L
By
Bx
Lx
W
3.5 The Vector Nature of Forces
Two men are carrying a 1.3 kg pail of water,
the first dude (Bob) exerts a force of 7N,
and the second one (Leon) exerts a force of
11N @ 28o. What is the angle of Bob’s
force?
L
y
Equations?
Fx  Lx  Bx  max
Fy  Ly  By  W  may
By
Bx
Lx
W
3.5 The Vector Nature of Forces
Two men are carrying a 1.3 kg pail of water,
the first dude (Bob) exerts a force of 7N,
and the second one (Leon) exerts a force of
11N @ 28o. What is the angle of Bob’s
force?
L
y
Values?
Fx  Lx  Bx  max
Fy  Ly  By  W  may
By
Bx
Lx
W
3.5 The Vector Nature of Forces
Two men are carrying a 1.3 kg pail of water,
the first dude (Bob) exerts a force of 7N,
and the second one (Leon) exerts a force of
11N @ 28o. What is the angle of Bob’s
force?
L
y
Values?
L cos   Bx  0
L sin   By  mg  0
By
Bx
Lx
W
3.5 The Vector Nature of Forces
Two men are carrying a 1.3 kg pail of water,
the first dude (Bob) exerts a force of 7N,
and the second one (Leon) exerts a force of
11N @ 28o. What is the angle of Bob’s
force?
L
y
Solve?
By
L cos   Bx  0
L sin   By  mg  0
Bx
Lx
W
3.5 The Vector Nature of Forces
Two men are carrying a 1.3 kg pail of water,
the first dude (Bob) exerts a force of 12.3N,
and the second one (Leon) exerts a force of
11N @ 28o. What is the angle of Bob’s
force?
L
y
Solve?
11cos 28  Bx
11sin 28  (1.3)(9.8)   By
Bx  9.7
By  7.6
By
Bx
Lx
W
 7.6 
o
tan 
  38
3.5 The Vector Natureof Forces
 9.7 
1
3.6 Frictional Forces
Friction – force that opposes motion
Caused by microscopic irregularities of a
surface
Increases as pushing
force increases
3.6 Frictional Forces
Depends on the normal force and the type of
surface
f  mN
f – force of friction (N)
N – normal force
m – coefficient of friction (1 or less)
3.6 Frictional Forces
Three types of friction
1.Static – object at rest
2.Kinetic – object in motion
Surfaces
µ (static)
µ (kinetic)
Steel on steel
0.74
0.57
Glass on glass
0.94
0.4
Metal on Metal (lubricated)
0.15
0.06
0.1
0.03
Teflon on Teflon
0.04
0.04
Tire on concrete
1
0.8
Tire on wet road
0.6
0.4
Tire on snow
0.3
0.2
Ice on ice
3.Rolling – just like it sounds
3.6 Frictional Forces
Example
3.6 Frictional Forces