Download N - Mr. Kleckner`s Class

9/12/2016
Chapter 4 Test
Monday 9/19
Newton’s Laws
The Study of Dynamics
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Isaac Newton
• Arguably the greatest physical genius ever.
• Came up with 3 Laws of Motion to explain the
observations and analyses of Galileo and Johannes
Kepler.
• Invented Calculus.
• Published his Laws in 1687 in the book Mathematical
Principles of Natural Philosophy.
What is Force?
•A force is a push or pull on an object.
•Forces cause an object to accelerate…
•To speed up
•To slow down
•To change direction
Contact forces arise from physical
contact .
Action-at-a-distance forces do not
require contact and include gravity and electrical
forces.
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The net force on an object is the vector sum of
all forces acting on that object.
The SI unit of force is the Newton (N).
kg  m
N 2
s
Arrows are used to represent forces. The length of the arrow
is proportional to the magnitude of the force.
15 N
5N
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Arrows are used to represent forces. The length of the arrow
is proportional to the magnitude of the force.
Individual Forces
4N
Net Force
10 N
6N
Net Force
Individual Forces
5N
37o
3N
4N
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Newton’s First Law
•The Law of Inertia.
•A body in motion stays in motion at constant
velocity and a body at rest stays at rest unless
acted upon by an external force.
•This law is commonly applied to the horizontal
component of velocity, which is assumed not to
change during the flight of a projectile.
The First Law is Counterintuitive
contrary to intuition or to common-sense
expectation (but often nevertheless true).
Aristotle firmly believed this.
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When a train suddenly starts, the passengers tend to fall backwards. This is because
the lower part of the body which is in contact with the train begins to move while the
upper part of the body tends to maintain its position of rest. As result, the upper part
tends to fall backwards
http://physics.tutorvista.com/motion/newton-s-first-law-of-motion.html
A physics book is sitting at rest on a table top.
Are there any forces acting on the book?
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A physics book is sitting at rest on a table top.
Are there any forces acting on the book?
A force diagram illustrating no net force
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A force diagram illustrating no net force
Another example
illustrating no net
force
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Newton’s First Law
•The Law of Inertia.
•A body in motion stays in motion at constant
velocity and a body at rest stays at rest unless
acted upon by an external force.
•If there is no NET force acting on an object there
will be no acceleration.
Newton’s Second Law
•A body accelerates when acted upon by a
net external force.
•The acceleration is proportional to the net
force and is in the direction which the net
force acts.
•This law is commonly applied to the
vertical component of velocity.
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Newton’s Second Law
•∑F = ma
•where ∑F is the net force
measured in Newtons (N)
•m is mass (kg)
•a is acceleration (m/s2)
A free-body-diagram is a diagram that
represents the object and the forces that
act on it.
Ftable
Fg
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Working a Newton’s 2nd Law Problem
F  395  275  560  110 N
F  ma
395 N
110 N  (1850kg)a
560 N
275 N
a  0.059m / s 2
The direction of force and acceleration vectors
can be taken into account by using x and y
components.



F  ma
is equivalent to
F
y
 may
F
x
 max
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Working a Newton’s 2nd Law Problem
17 N
12 N
F
F  12 2  17 2  20.8 N
17 N
F  ma
12 N
20.8 N  (27kg)a
a  0.77m / s 2
Assignment
Read pg. 75-80
Do pg. 97-98
Questions #3,4
Problems #1,2,3,6
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Chapter 4 Test
Monday 9/19
Gravity as an accelerating force
A very commonly used accelerating
force is gravity. Here is gravity in
action. The acceleration is g.
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Gravity as an accelerating force
In the absence of
air resistance,
gravity acts upon
all objects by
causing the same
acceleration…g.
Gravity as an accelerating force
The pulley lets us use gravity as our
accelerating force… but a lot slower
than free fall. Acceleration here is a
lot lower than g.
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The problem of weight
Are weight and mass the same
thing?
•No. Weight can be defined as the force
due to gravitation attraction.
•W = mg
Flat surfaces – 1 D
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.
FN = mg for objects resting
on horizontal surfaces.
FN
mg
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John places a block on the top of a flat table. The
block has a mass of 1.5kg. What is the normal force
the table applies to the block?
FN  mg
FN  (1.5kg)(10m / s 2 )
FN  15 N
John places a block on the top of a flat table. The block has a mass of
1.5kg. He then proceeds to push down on the block with a force of 11N.
What is the normal force the table applies to the block?
FN  15N  11N  26 N
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John places a block on the top of a flat table. The block has a mass of
1.5kg. He then proceeds to pull up on the block with a force of 11N.
What is the normal force the table applies to the block?
FN  15N  11N  4 N
How much tension must a rope withstand in order to accelerate a
1000kg object up with an acceleration of 0.65 m/s2?
Tension  W  ma
T=W+ma
T  10000 N  (1000kg)(.65m / s 2 )
T  10000 N  650 N
T  10650 N
W=mg=10000 N
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Will pushes a 10 kg block on a frictionless floor at a 30o angle below the
horizontal with a force of 150 N.
a) F cos   ma
a) What is the acceleration?
150 cos 30o  (10kg)a
b) What is the normal force?
129.9 N  (10kg)a
a  13m / s 2
Fcos
b) FN  F sin   W
Fsin
FN  150 sin 30o  100 N
F
FN  75  100
FN  175 N
W=mg=100 N
T sin 10o  T sin 10o  W
2T sin 10o  500 N
T sin 10o  250 N
T
Tsin10
250 N
 1440 N
sin 10o
Tsin10
W=mg=500 N
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normal force is
Ramps – 2 D The
perpendicular to angled
ramps as well. It’s always
equal to the component of
weight perpendicular to the
surface.
N = mgcos
N
F=mgsin
mg

What is the normal force for a 5.0 kg block on a
ramp with a 15o angle with the horizontal?
FN  mg cos 
FN  (5.0kg)(10m / s 2 ) cos 15o
FN  48 N
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How long will it take a 5.0 kg block to slide down
a frictionless 20 m long ramp that is at a 15o
angle with the horizontal?
F  mg sin 
ma  mg sin 
a  g sin 
a  (10m / s 2 ) sin 15  2.6m / s 2
1
x  vot  at 2
2
1
20m  0  (2.6m / s 2 )t 2
2
20  1.3t 2
15.38  t 2
t  3.9 s
V>0
V=0
A>0
A=0
Elevator Ride – going
up!
Heavy feeling
Normal feeling
V>0
A=0
V>0
A<0
Normal feeling
Light feeling
N
N
N
N
mg
mg
mg
mg
Ground
floor
Just
starting up
Between
floors
Arriving at
top floor
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V<0
V=0
A<0
A=0
Elevator Ride – going down!
Normal feeling
V<0
A=0
V<0
A>0
Heavy feeling
Normal feeling
Light feeling
N
N
N
N
mg
mg
mg
mg
Top
floor
Beginning
descent
Between
floors
Arriving at
Ground floor
A person weighing 800N rides the elevator up to the 4th
floor from the 1st floor. The elevator has an acceleration
of 1.0m/s2, how heavy does the person feel when the
elevator begins to go up?
W  m( g  a )
W  (80kg)(10  1)
W  880 N
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A person stands on a bathroom scale in a motionless
elevator. When the elevator begins to move, the scale
briefly reads only 0.75 of the person’s regular weight.
Calculate the acceleration of the elevator, and find the
direction of the acceleration?
.75W  m( g  a )
.75mg  mg  ma
.75 g  g  a
.75(10)  10  a
7.5  10  a
 2.5   a
a  2.5m / s 2
Assignment
Read pg. 81-94
Do pg. 98-103
Problems
#10,12,15,16,24,25,72
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Chapter 4 Test
Monday 9/19
Magic Pulleys
N
mg
T
-x
T
m1
mg
m2
x
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Mass 1 (10 kg) rests on a frictionless table connected by a
string to Mass 2 (5 kg). Find (a) the acceleration of each
block and, (b) the tension in the connecting string.
a)T  m2 g  (5kg)(10m / s 2 )  50 N
b) F  ma
50 N  (5kg  10kg)a
50 N  (15kg)a
a  3.3m / s 2
m1
m2
Chapter 4 Test
Monday 9/19
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Friction
•The force that opposes a sliding
motion.
•Enables us to walk, drive a car,
etc.
•Due to microscopic irregularities
in even the smoothest of
surfaces.
There are two types of friction
Static friction
•exists before sliding occurs
•Kinetic friction
•exists after sliding occurs
•In general fk <= fs
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Friction and the Normal
Force
•The frictional force which exists
between two surfaces is directly
proportional to the normal force.
•That’s why friction on a sloping
surface is less than friction on a flat
surface.
Applied forces affect normal force.
friction
applied force
normal
weight
N = applied force
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Static Friction
•fs  sN
• fs : static frictional force (N)
• s: coefficient of static friction
• N: normal force (N)
•Static friction increases as the force trying
to push an object increases… up to a point!
A force diagram illustrating
Static
Friction
Normal Force
Frictional
Force
Applied Force
Gravity
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A force diagram illustrating
Static
Friction
Normal Force
Bigger Applied
Force
Bigger
Frictional
Force
Gravity
A force diagram illustrating
Static
Friction
The forces on the book are now UNBALANCED!
Normal Force
Frictional
Force
Gravity
Even Bigger
Applied Force
Static friction cannot get any larger, and can
no longer completely oppose the applied
force.
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Kinetic Friction
•fk = kN
• fk : kinetic frictional force (N)
• k: coefficient of kinetic friction
• N: normal force (N)
•Kinetic friction (sliding friction) is generally
less than static friction (motionless friction)
for most surfaces.
A force of 48.0N is required to start a 5.0-kg box moving across a
horizontal concrete floor.
a) What is the coefficient of static friction between the box and the
floor?
b) If the 48.0 N force continues, the box accelerates at 0.70 m/s 2.
What is the coefficient of kinetic friction?
a) f s  48 N   s FN
FN  mg  (5kg)(10m / s 2 )  50 N
48 N   s (50 N )
 s  .96
b)netF  ma  (5kg)(0.70m / s 2 )  3.5 N  netF
netF  F  f k
3.5  48  f k
f k  44.5 N   k FN
 k  44.5 / 50  0.89
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What is the maximum acceleration a car can undergo if the
coefficient of static friction between the tires and the road is 0.80?
F  fs
F   s FN
ma   s (mg )
a  s g
a  (0.80)(10m / s 2 )
a  8m / s 2
Chapter 4 Test
Monday 9/19
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Magic Pulleys
N
mg
T
-x
T
m1
mg
m2
x
Mass 1 (10 kg) rests on a frictionless table connected by a
string to Mass 2 (5 kg). Find (a) the acceleration of each
block and, (b) the tension in the connecting string.
a)T  m2 g  (5kg)(10m / s 2 )  50 N
b) F  ma
50 N  (5kg  10kg)a
50 N  (15kg)a
a  3.3m / s 2
m1
m2
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Mass 1 (10 kg) rests on a table connected by a string to Mass
2 (5 kg) as shown. What must the minimum coefficient of
static friction be to keep Mass 1 from slipping?
FN  m1 g  (10kg)(10m / s 2 )  100 N
appliedF  m2 g  50 N
 s FN  appliedF
 s (100 N )  50 N
 s  .5
m1
m2
Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg).
If s = 0.3 and k = 0.2, what is a) the acceleration and b) the tension in
the string?
F  m g  (10kg)(10m / s 2 )  100 N
N
1
 k FN  .2(100)  20 N
appliedF  50 N
NetF  50 N  20 N  30 N
F  ma
30 N  (15kg)a
a  2m / s 2
b)T  m2 g  Fk  50 N  20 N  70 N
m1
m2
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Newton’s Third Law of Motion
Any time a force is exerted on an object, that force is caused by another object.
Newton’s third law:
Whenever one object exerts a force on a second object, the second exerts an
equal force in the opposite direction on the first.
Newton’s Third Law of Motion
A key to the correct application of the
third law is that the forces are exerted
on different objects. Make sure you
don’t use them as if they were acting on
the same object.
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Newton’s Third Law of Motion
Rocket propulsion can also be explained using Newton’s third law: hot gases from
combustion spew out of the tail of the rocket at high speeds. The reaction force is
what propels the rocket.
Note that the rocket does not
need anything to “push” against.
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?
spacecraft
P  ma
36 N  (11000kg)a
a  0.0036m / s 2
astronaut
 P  ma
 36 N  (92kg)a
a  0.39m / s 2
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T sin 10o  T sin 10o  W
2T sin 10o  500 N
T sin 10o  250 N
T
Tsin10
250 N
 1440 N
sin 10o
Tsin10
W=mg=500 N
An automobile engine has a weight W, whose magnitude is W=3150N. This engine is being
positioned above an engine compartment, as Figure 4.29a illustrates. To position the engine, a
worker is using a rope. Find the tension T1 in the supporting cable and the tension T2 in the
positioning rope.
x  components :right  left
T2 sin 80o  T1 sin 10o
T2  .18T1
y  components : up  down
T1 cos 10o  W  T2 cos 80o
.985T1  3150  .174T 2
.985T1  3150  .174(.18T 1)
.985T1  3150  .031T 1
.954T1  3150
T1  3303 N
T2  (.18)(3303)  595 N
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Assignment
Do pg. 100-104
Problems
#27,30,37,41,45,50,78
Chapter 4 Test
Monday 9/19
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Newton’s Third Law of Motion
In nature there are two general types of forces,
fundamental and nonfundamental.
Fundamental Forces
1. Gravitational force
2. Strong Nuclear force
3. Electroweak force
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Newton’s Law of Universal Gravitation
Every particle in the universe exerts an attractive force on every
other particle.
A particle is a piece of matter, small enough in size to be
regarded as a mathematical point.
The force that each exerts on the other is directed along the line
joining the particles.
For two particles that have masses m1 and m2 and are
separated by a distance r, the force has a magnitude
given by
mm
F  G 12 2
r
G  6.673 1011 N  m2 kg 2
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Definition of Weight
The weight of an object on or above the earth is the
gravitational force that the earth exerts on the object.
The weight always acts downwards, toward the center
of the earth.
On or above another astronomical body, the weight is the
gravitational force exerted on the object by that body.
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