Transcript No Slide Title

Chapter 4
Forces and Newton’s Laws of
Motion
Why things move the way the do.
Newton’s First Law of Motion
“Maintaining the status quo”
Newton’s First Law of Motion
“The Law of Inertia”
“An object at rest will remain at rest and an
object in motion will continue to move at a
constant velocity, unless acted upon by a net
external force.”
The property of an object that causes it to resist changes
in its motion is called inertia.
Inertia is proportional to an object’s mass.
Inertia causes motion at a constant velocity.
Fnet  0  a  0  constant velocity
In the third century BC the Greek philosopher Aristotle developed
a model of the structure and motion of the universe.
In Aristotle’s world it was assumed that the earth was stationary,
did not spin or move through space, and was at the center of the
universe.
Aristotle’s acceptance of a stationary earth…not spinning, was in
part due to his lack of a knowledge or understanding of inertia.
Newton’s Second Law of Motion
“How is Acceleration related to Force?”
Newton’s Second Law of Motion
The law of accelerated motion
“The acceleration of an object is directly
proportional to and in the same direction as
the net force acting on it and inversely
proportional to the object’s mass.”
F
a  net
m
Fnet  ma
Units of Force:
kg m
m
kg  2 
 Newton, N
2
s
s
Force is a vector and uses the same sign
convention as velocity and acceleration.
direction of Fnet
 direction of a
parallel magnitude increases
 effect on v 
opposite magnitude decreases
Fnet  F1  F2  F3
F1 , F2 , F3 are + or 
based on their directions.
Summary
Constant Velocity
Caused by Inertia
Variables: Relationships:
s
s  vt
v constant v  st
t
t  vs
m
Fnet  0
Fnet
a0
Constant Acceleration
Caused by Net Force
Variables: Relationships:
v  vf
s
v  st v  i
2
v
s  v t t = vs
vf  vi
vi
a
t
v f  v i  at
v
f
a
s  v it  1 at 2
2
t
m
v 2f  v 2i  2as
Fnet
Fnet   Fi
Fnet
a m
Describing Forces
All forces between objects can be placed into two broad categories:
Contact forces
Action-at-a-distance forces
Contact forces are types of forces in which the two interacting
objects are physically contacting each other.
Frictional Force
Tensional Force
Air Resistance Force
Spring Force
Buoyant Force
Action-at-a-distance forces are types of forces in which the two
interacting objects are not in physical contact with each other, yet
are able to exert a push or pull despite a physical separation.
Gravitational Force
Electric Force
Magnetic Force
Types of Forces
Gravitational Force and Weight
Normal Force
Friction
Tension
Fluid Resistance
Buoyant Force
The Gravitational Force
Newton’s Law of Universal Gravitation:
“Every mass in the universe exerts a gravitational
force on every other mass. The gravitational force is
directly proportional to the product of their masses
and inversely proportional to the square of the
distance between them.”
m1 m2
FG  G 2
r1,2
G  Universal Gravitational Constant
G = 6.673  10
11
Nm
2
kg
2
Mass and Weight
Mass is a measure of the amount of matter in an object and is
related to the object’s inertia.
Weight is a measure of the gravitational force acting on an object.
force of gravity
weight, w
F = ma
  
w = m g
acceleration due
to gravity, g
The weight of an object equals the product of its mass and the
acceleration due to gravity.
Comparing the equation for weight with the equation for the
gravitational force gives:
m 1m 2
m2
G 2  m1g  g  G 2
r1,2
r1,2
Where :
m2  mass of the earth
r1,2  distance of object from
the Earth's center
r1,2  rearth  altitude
The Normal Force
When two surfaces are in contact, each surface exerts a force on
the other. The component of this force which is perpendicular to
the surface is called the normal force, FN.
FN
FN
w
w
Friction
The force that opposes.
Friction is a force whose direction is ALWAYS opposite to
the direction an object is moving or would tend to move if
there were no friction.
Friction is force between two surfaces:
1. the surface of the object and
2. the surface on which it is moving
Friction depends on the characteristics of the two
surfaces and the force pressing them together.
Friction depends on whether the object is moving or
stationary.
Static friction exists between a stationary object and the surface on
which it rests.
Static friction is always opposite to the direction the object would
move if there was no friction.
Static friction must be overcome before an object can begin moving.
Static friction equals the net applied force up to its maximum
value which depends on the mass of the object and the properties
of the two surfaces. fstatic,max   static  FN
Kinetic friction exists between a moving object and the surface on
which it is moving.
Kinetic friction is always opposite to the object’s velocity.
Kinetic friction is usually less than static friction.
fkinetic   kinetic  FN
 static   kinetic
Consider an object sitting on a stationary horizontal surface.
FN
w
If the object remains in contact with the surface, the acceleration
in the vertical, y, direction must be zero.
FN  w  0
FN  w  mg  m g
f  m g
fstatic,max   s m g
fk   km g
Consider an object sitting on a stationary inclined surface.
y y'
FN
x
x'

w y'

w
w x'
w x'  w sin  m g sin
w y'  w cos  mgcos 
FN

w y'

w
w x'
If the object remains in contact with the surface, the acceleration
in the vertical, y’, direction must be zero.
FN  mg cos   0
FN  mg cos   m g cos
f  m g cos
FN
f

w y'
w x'

w
w x'
The x component of the weight, wx’ tends to cause the object to
slide down the incline.
The frictional force, f, will be in a direction to oppose this motion.
The net force on the object is then:
Fnet,x  w x  f
Fnet,x'  m g sin   m g cos
Fnet,x'  m g sin    cos 
The acceleration is then:
Fnet,x'
a x'  m
a x'  g  sin   cos
For an object to begin sliding down an incline:
w x'  fs,max
m g sin min    s m g cos min 
sin min    s cos min 
sinmin 
 s
cosmin 
tan min    s
1
 min  tan  s
Tension Force
Tension, T, is a “pulling” force applied to an object by wire or rope.
Tension always acts along the wire or rope in a direction away
from the object on which the tension is applied.
T
When two objects are connected by a wire or rope the tension is the
same at each end.
T
T
Consider an object with mass m1 sitting on a table and connected
by a wire which passes over a pulley to a second mass m2 hanging
beside the table.
FN
f
m1
T
a1
m1g
Forces Acting on m1:
Weight = m1g
T
m2
a2
Forces Acting on m2:
Weight = m2g
Normal Force, FN
Tension, T
m2 g
Tension, T
If released m2 will tend to fall pulling m1 to the right.
Friction, f
Newton’s Second Law for m1:
Y-Direction
Assuming object remains on the table.
FN  m1g  0
FN  m1g  m1 g
X-Direction
T  f  m 1a1
T  m1 g  m1a 1
T  m1a1  m1 g
Newton’s Second Law for m2:
Since m1 and m2 are connected by a wire the magnitudes of their
accelerations must be equal. The directions of the accelerations may
not be the same.
m1 tends to accelerate to the right (+)
m2 tends to accelerate down (-)
a 2  a1
T  m 2g  m 2 a 2
T  m2a2  m2g
T  m1a1  m1 g
T  m2a2  m2g
Substitute a2 = -a1
T  m1a1  m1 g
T  m 2 a1  m 2g
m1a 1  m1 g  m2 a 1  m 2 g
m1a 1  m2 a 1  m 2 g  m1g
a1  m1  m 2   m 2 g  m1 g
m2 g  m1g
a1  m  m
1
2
m1g
m1
m2
m2 g  m1g  net force on system
m1  m 2  total mass of system
m2 g
T = m1a1  m1g
T = m1 a 1  g 
m 2 g  m1 g

T = m1  m  m
 g


1
2
m 2  m1

T = m1g m  m  
 1

2
m 2  m1  m1  m 2 

T = m1g
m1  m 2


m 2  m 2 
T = m1g m  m 
 1
2 
m 2 1   
T = m1g m  m 
 1
2 


 1   
T = m1g m

 1  1 
 m 2

Summary
m2 g  m1g
a1  m  m
1
2
a 2  a1
Since m1 can not accelerate to the
left, a1 > 0,
therefore for the system to move
m2g > sm1g or
m2 > sm1


 1   
T = m1g m

 1  1 
 m 2

Fluid Resistance
Fluid (liquid or gas) resistance is a frictional force that an object
experiences as it moves through a fluid (e.g. air or water).
Fluid resistance depends on the size and shape of the object,
density of the fluid, and is directly proportional to the object’s
velocity and in the opposite direction.
fluid
Example
An object falling through air experiences an upward force of air
resistance in addition to the downward force of gravity.
f
 v
air resistance
gravitational force
Initially air resistance is small because the velocity
is small. air resistance  velocity
gravitational force > air resistance
net force = gravitational force - air resistance
Object accelerates downward magnitude of
velocity increases
As the magnitude of the downward velocity
increases the force of air resistance increases.
If the object falls far enough the force of air
resistance becomes equal to the gravitational
force.
At this point:
gravitational force = air resistance
net force, Fnet = 0
a=0
velocity is constant
This constant velocity reached by an object falling
through a fluid is called its terminal velocity.
Buoyant Force
An object immersed in a fluid (liquid or gas) experiences an
upward buoyant force.
The buoyant force depends on the volume of the object immersed
in the fluid (volume of fluid displaced) and the density of the fluid.
buoyant force
FB  fluid  g  Vdisplaced
Consider an object dropped into a tank containing a fluid
(e.g. oil)
Buoyant Force
Only
Buoyant Force &
Fluid Resistance
The mass of the object is 10kg and the buoyant force is 190N.
If the object starts from rest at a height of 63.5m above the
surface of the oil how long will it take for it to reach a depth of
60m in the oil? (Ignore any buoyant force due to air and any
fluid resistance.)
Buoyant Force with Fluid Resistance
Part One of the Motion
Falling from rest a distance of 63.5m to the surface of the oil.
The only force acting on the object is the gravitational force…this
part of the motion is Free-Fall.
Part Two of the Motion
Falling a distance of 60m from the surface of the oil.
The object’s velocity as it enters the oil = -35.3m/s
There are two forces acting on the object:
•The gravitational force = mg
•The buoyant force = 190N
Variables:
d  63.5m
v
vi  0
v f find
35.3
vf m
s
a  9.8 m2
s
3.6s
t Find
t
m  10kg
Fnet  98N
Relationships:
vi  vf
d
v
v t
2
d= vt
t= d
v
vf  vi
vf  vi  a  t
a
t
2
2
2
1
d  v i  t  2 a  t v f  v i  2ad
Fnet   Fi  FG  w  mg
 (10kg)  (9.8 m2 )
s
 98N
Fnet 98N  9.8 m
a  m  10kg
s2
d  v i  t  12 a  t 2
Substituting values w/o units.
m )t 2
63.5  0  t  1
(9.8
2
2
s
2
63.5  4.9t
t  63.5
4.9
 3.6s
t  63.5
4.9
2
vf  vi  a  t
m
m

35.3
v f  0  (9.8 2 ) (3.6s)
s
s
Variables:
d  60m
v
v i  35.3 m
s
vf
a  9.2 m2
s
t Find t
m  10kg
Fnet  92N
Relationships:
vi  vf
d
v
v t
2
d= vt
t= d
v
vf  vi
vf  vi  a  t
a
t
2
2
2
1
d  v i  t  2 a  t v f  v i  2ad
Fnet   Fi  FG  FB
 98N  190N
 92N
Fnet  92N  9.2 m
a m
2
10kg
s
d  v i  t  12 a  t 2
Substituting values w/o units.
2
1
60  35.3  t  2 (9.2)t
2
60  35.3  t  4.6  t
Writing in the General Quadratic Form
4.6  t  35.3  t  60  0
2
Comparing to ax  bx  c  0 gives :
a = 4.6
2
b = -35.3
c = 60
Substituting into the Quadratic Formula
x = -b  b  4  a  c gives :
2a
2
-(-35.3)  (35.3)  4 (4.6)  (60)
t=
2(4.6)
2
Simplifying gives:
35.3

11.92
t=
9.2
 11.92  5.1s
t 1 = 35.39.2
 11.92  2.6s
t 2 = 35.39.2
The time for the object to reach the surface of the oil was 3.6s.
Therefore the object is at a depth of 60m in the oil at two times.
t 1  3.6s  5.1s  8.7s
t 2  3.6s  2.6s  6.2s
What is the significance of the two times?
What is the maximum depth the object will reach
And when does it reach this maximum depth?
Variables:
 67.7m
d Find
“d”
v
v i  35.3 m
s
vf  0
a  9.2 m2
s
t Find “t”
m  10kg
Fnet  92N
Relationships:
v  dt
d= vt
vf  vi
a
t
v f  vi
t
a
2
1
d  vi  t  2 a  t
Fnet   Fi
Fnet
a m
vi  vf
v
2
t= d
v
vf  vi  a  t
v 2f  v 2i  2ad
v  v  2ad
2
f
2
i
2
m
0  (35.3 s )
 67.7m

d  2a
2 (9.2 m2 )
s
m)
0

(35.3
v f  vi
s  3.84s
t

a
9.2 m2
s
Time to reach
3.84s  3.6s  7.44s
maximum depth:
v 2f
 v 2i
The object reaches the surface in 3.6s.
The object is at a depth of 60m moving downward in 6.2s
The object reaches its maximum depth of 67.7m in 7.44s.
The object is at a depth of 60m moving upward in 8.7s.
Newton’s Third Law of Motion
“Forces always come in pairs.”
Newton’s Third Law of Motion
“Action and Reaction”
“If object A exerts a force on object B, object
B must exert an equal and opposite force on
object A.”
“For every action there is an equal and
opposite reaction.”
FAB  FBA
Why do the two forces (action and reaction) not cancel each other?
They act on DIFFERENT objects!
Newton’s Third Law says that action and reaction forces are
always equal and opposite.
It does NOT say that their effects are equal!
FAB  a B depends on m B
FAB
aB  m
B
FBA  a A depends on m A
FBA
aA  m
A
Consider two objects moving along a horizontal track.