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STUDENT NUMBER:201201
INSTITUTION: UNIVERSITY OF JOHANN
*PLEASE REFER TO THE REFER
AT THE END *
• Born 1642
• Went to University of Cambridge in England as a student
and taught there as a professor after
• Never married
• Gave his attention mostly to physics and mathematics,
but he also gave his attention to religion and alchemy
• Newton was the first to solve three mysteries that
intrigued the scientists
• Laws of Motion
• Laws of Planetary Orbits
• Calculus
NEWTON 1ST LAW
OF MOTION
NEWTON 2ND
LAW OF MOTION

NEWTON 3RD LAW
OF MOTION
4-2 Newton’s First Law of Motion
Newton’s first law is often called the law of
inertia.
Every object continues in its state of rest, or of
uniform velocity in a straight line, as long as no
net force acts on it.
4-2 Newton’s First Law of Motion

Inertial reference frames:
An inertial reference frame is one in which
Newton’s first law is valid.
This excludes rotating and accelerating
frames.
4-3 Mass
Mass is the measure of inertia of an object. In
the SI system, mass is measured in kilograms.
Inertia  The tendency of a body to maintain its
state of rest or motion.
Mass is not weight:
Mass is a property of an object. Weight is the
force exerted on that object by gravity.
If you go to the moon, whose gravitational
acceleration is about 1/6 g, you will weigh much
less. Your mass, however, will be the same.
Newton’s first law

The ball remains in its state of rest till an external
force is acting on it.
Newton’s first law (contd..)
Push the ball

When an external force (push) is acting on a ball (object)
at rest it changes its state (moves over the steps ).
Newton’s first law (contd..)
Push the ball

When external force (Push of the rabbit)
acting on the object (ball) which is in
motion the object changes its position.
• An Object at rest remains at rest, and an object in motion
continues in motion with constant velocity (that is,
constant speed in a straight line), unless it experiences a
net external force.
• The tendency to resist change in motion is called inertia
• People believed that all moving objects would eventually stop
before Newton came up with his laws
• When there is no force
exerted on an object, the
motion of the object remains
the same like described in
the diagram
• Because the equation of
Force is F=ma, the
acceleration is 0m/s². So
the equation is 0N=m*0m/s²
• Therefore, force is not
needed to keep the object
in motion, when
• The object is in equilibrium
when it does not change its
state of motion
• 1st Law: If no net force acts, object remains at rest or
in uniform motion in straight line.
• What if a net force acts? Do Experiments.
• Find, if the net force F  0  The velocity v changes
(in magnitude or direction or both).
• A change in the velocity v (Δv)
 There is an acceleration a = (Δv/Δt)
OR
A net force acting on a body produces an
acceleration!
Fa
Newton’s Second Law

Force
equals
mass times
accelerati
on.
F = ma
• The acceleration of an object is directly proportional to
the net force acting on it and inversely proportional to its
mass

Fnet
Acceleration
• From experiments: The net force F on a body
and the acceleration a of that body are related.
• HOW? Answer by EXPERIMENTS!
• The outcome of thousands of experiments over
hundreds of years:
a  F/m
(proportionality)

• We choose the units of force so that this is not
just a proportionality but an equation:
a  F/m

OR:

(total!)

F = ma
• Newton’s 2nd Law:

F = ma
F = the net (TOTAL!) force acting on mass m
m = the mass (inertia) of the object.
a = acceleration of object. Description of the effect of F
F is the cause of a.

• To emphasize that the F in Newton’s 2nd Law is
the TOTAL (net) force on the mass m, text writes:

∑F = ma
∑ = a math symbol meaning sum (capital sigma)
• Newton’s 2nd Law:
∑F = ma

Based on experiment!
Not derivable
mathematically!!

Force is a vector, so ∑F = ma is true along each coordinate
axis.

∑Fx = max

∑Fy = may

∑Fz = maz

ONE OF THE MOST
FUNDAMENTAL & IMPORTANT
LAWS OF CLASSICAL PHYSICS!!!
4-4 Newton’s Second Law of Motion
Newton’s second law is the relation between
acceleration and force. Acceleration is
proportional to NET force and inversely
proportional to mass.
(4-1)
4-4 Newton’s Second Law of Motion

The unit of force in the SI
system is the newton (N).
Note that the pound is a
unit of force, not of mass,
and can therefore be
equated to newtons but
not to kilograms.
As the man jumps off
the boat, he exerts
the force on the boat
and the boat exerts
the reaction force on
the man.
The man leaps forward
onto the pier, while the
boat moves away from
the pier.
Foil deflected
up

Engine pushed
forward

Flow
backwardpushed backward
Flow
Foil deflected
down deflected
Foil
down
• If two objects interact, the force exerted on object 1 by
object 2 is equal in magnitude but opposite in direction to
the force exerted on object 2 by object 1
• Forces always come in pair when two objects interact
• The forces are equal, but opposite in direction

Fn

Fg
4-5 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.
Law of action-reaction: “Every
action has an equal & opposite
reaction”.
(Action-reaction forces act on
DIFFERENT objects!)
4-5 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.
4-5 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.
4-5 Newton’s Third Law of Motion
Helpful notation: the first subscript is the object
that the force is being exerted on; the second is
the source.
This need not be
done indefinitely, but
is a good idea until
you get used to
dealing with these
forces.

(4-2)
Example 4-11
Fpx  (40.0 N )(cos30.0 0 )  34.6 N
Fpy  (40.0 N )(sin 30.00 )  20.0 N

Fpx  max

Fpx

34.6 N
 ax 

m 10.0 Kg
 3.46 m / s 2

Fy  ma y

 FN  Fpy  mg  ma y

FN  20.0  98.0  0 ; a y  0
FN  78.0 N
Example 4-13
FT  mE g  mE aE   mE a

(1)

FT  mC g  mC aC   mC a

( 2)

subtract
(1) from (2)
(mE  mC ) g  (mE  mC )a
mE  mC
a
g  0.68 m / s 2
mE  mC
FT  mE g  mE a  mE ( g  a)  10500

N
Example 4-14
The advantage of a pulley

2 FT  mg  ma
const. speed  a  0
mg
FT 
2
Example 4-15
Fx  FRC cos  FRB cos  0 ; a  0
 FRB  FRC
Fy  Fp  2 FT sin   0
FT 

Fp
2 sin 

 1700 N
Example 4-16
Example 4-18
4-6 Weight – the Force of Gravity;
and the Normal Force
Weight is the force exerted on
an object by gravity.
Close to the surface of the
Earth, where the gravitational
force is nearly constant, the
weight is:
4-6 Weight – the Force of Gravity;
and the Normal Force
An object at rest must have no net force on it. If
it is sitting on a table, the force of gravity is still
there; what other force is there?
The force exerted perpendicular to a surface is
called the normal force. It is
exactly as large as needed
to balance the force from
the object (if the required
force gets too big,
something breaks!)
m = 10 kg
The normal
force is NOT
always equal
to the weight!!
Example 4-7

m = 10 kg
∑F = ma
FP – mg = ma
Example 4-19
4-7 Solving Problems with Newton’s Laws –
Free-Body Diagrams
1. Draw a sketch.
2. For one object, draw a free-body
diagram (force diagram), showing all
the forces acting on the object. Make
the magnitudes and directions as
accurate as you can. Label each
force. If there are multiple objects,
draw a separate diagram for each one.
3. Resolve vectors into components.
4. Apply Newton’s second law to each
component.
5. Solve.
4-7 Solving Problems with Newton’s Laws –
Free-Body Diagrams

When a cord or rope pulls on
an object, it is said to be under
tension, and the force it exerts
is called a tension force.
4-8 Applications Involving Friction, Inclines
On a microscopic scale,
most surfaces are rough.
The exact details are not
yet known, but the force
can be modeled in a
simple way.

We must account for
Friction to be realistic!
4-8 Applications Involving Friction, Inclines
• Exists between any 2 sliding surfaces.
• Two types of friction:

Static (no motion) friction
Kinetic (motion) friction
• The size of the friction force: Depends on the microscopic details of 2
sliding surfaces.

•
•
•
•

The materials they are made of
Are the surfaces smooth or rough?
Are they wet or dry?
Etc., etc., etc.
4-8 Applications Involving Friction, Inclines

For kinetic (sliding) friction, we write:

k is the coefficient of kinetic friction, and is
different for every pair of surfaces.

k depends on the surfaces & their conditions
k is dimensionless & < 1
4-8 Applications Involving Friction, Inclines

Static friction is the frictional force between two
surfaces that are not moving along each other.
Static friction keeps objects on inclines from
sliding, and keeps objects from moving when a
force is first applied.
4-8 Applications Involving Friction, Inclines
The static frictional force increases as the applied
force increases, until it reaches its maximum.
Then the object starts to move, and the kinetic
frictional force takes over.
4-8 Applications Involving Friction, Inclines
4-8 Applications Involving Friction, Inclines
An object sliding down an incline has three forces acting
on it: the normal force, gravity, and the frictional force.
• The normal force is always perpendicular to the surface.

• The friction force is parallel to it.
• The gravitational force points down.

If the object is at rest,
the forces are the same
except that we use the
static frictional force,
and the sum of the
forces is zero.
4-9 Problem Solving – A General Approach
1. Read the problem carefully; then read it again.
2. Draw a sketch, and then a free-body diagram.
3. Choose a convenient coordinate system.

4. List the known and unknown quantities; find
relationships between the knowns and the
unknowns.
5. Estimate the answer.
6. Solve the problem without putting in any numbers
(algebraically); once you are satisfied, put the
numbers in.

7. Keep track of dimensions.
8. Make sure your answer is reasonable.
 FT1

m1g   FT2
 FT2

Take up as positive!
m1 = m2 = 3.2 kg
m1g = m2g = 31.4 N
a
Bucket 1: FT1 - FT2 - m1g = m1a
Bucket 2: FT2 -m2g = m2a

a

∑F =

 m2 g
ma (y direction), for EACH bucket separately!!!
FT 

Take up positive!
m = 65 kg
mg = 637 N
 FT

FT + FT - mg = ma
2FT -mg = ma
FP = - FT (3rd Law!)

 FP
a

∑F = ma (y direction) on woman + bucket!
 mg
• Kinetic Friction: Experiments determine the
relation used to compute friction forces.
• Friction force Ffr is proportional to the
magnitude of the normal force FN between
two sliding surfaces.
DIRECTIONS of Ffr & FN are  each other!!
Ffr  FN
FN

Ffr

a

Fa
mg
• Static Friction: Experiments are used again.
• The friction force Ffr exists || two surfaces, even if
there is no motion. Consider the applied force Fa:
FN
Ffr

Fa
mg
∑F = ma = 0 & also v = 0

 There must be a friction force Ffr to oppose Fa
Fa – Ffr = 0 or Ffr = Fa

The object is not
moving
• Experiments find that the maximum static friction
force Ffr (max) is proportional to the magnitude (size)
of the normal force FN between the two surfaces.
DIRECTIONS of Ffr & FN are  each other!! Ffr  FN
• Write the relation as Ffr (max) = sFN
s  Coefficient of static friction
• Depends on the surfaces & their conditions
• Dimensionless & < 1
• Always find s > k
 Static friction force:

Ffr  sFN
FN  mA g  5.0  9.8  49 N
F fr   k FN  0.20  49  9.8 N
FBy  mB g  FT  mB a
FAx  FT  F fr  mA a

Example
4-20

 FT  F fr  mA a
 mB g  F fr  mA a  mB a
mB g  F fr 19.6  9.8
a

 1.4 m / s 2
mA  mB 5.0  2.0
FT  F fr  mA a  17 N
Example 4-21

FGx  mg sin 
FGy  mg cos
Fy  FN  mg cos  0
Fx  mg sin    k FN  max

 FN  mg cos
 mg sin    k (mg cos )  max
Problem 48
FN

Ffr

Fp
(m1+m2) g

FN   m1  m2  g



Ffr   k FN   k  m1  m2  g

 Fx  FP  Ffr   m1  m2  a 

a

a = 1.9 m/s2

For contact forces analyze forces on the two blocks separately.
1. What acceleration will result when a 12 N net
force applied to a 3 kg object? A 6 kg object?
2. A net force of 16 N causes a mass to accelerate at
a rate of 5 m/s2. Determine the mass.
3. How much force is needed to accelerate a 66 kg
skier
1 m/sec/sec?

4. What is the force on a 1000 kg elevator that is
falling freely at 9.8 m/sec/sec?
Summary of Chapter 4
• Newton’s first law: If the net force on an object
is zero, it will remain either at rest or moving in a
straight line at constant speed.

• Newton’s second law:
• Newton’s third law:
• Weight is the gravitational force on an object.
• The frictional force can be written:
(kinetic friction) or

(static friction)

• Free-body diagrams are essential for problemsolving
•

Henderson, Tom. Physics. Course home page. 16 May 2008
http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/newtlaws/newtltoc.html>.

•

Serway, Raymond A., and Jerry S. Faughn. "The Laws of Motion." College Physics .
Fifth ed. 1999.
Benson, Tom. Newton’s Third Law applied to Aerodynamics 21 May 2008
http://www.grc.nasa.gov/WWW/K-12/airplane/newton3.html

•

•

Introduction to Rocket Performance. Newton’s Third Law. 12 March 2004
http://www.allstar.fiu.edu/aero/rocket1a.htm

•

Stern, David P. (16) Newton’s Laws of Physics. 1. Force and Inertia. 9 October
http://www-istp.gsfc.nasa.gov/stargaze/Snewton.htm

2004

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Re purposed presentation

  • 1. Re-purposed by: Mr XT MK STUDENT NUMBER:201201 INSTITUTION: UNIVERSITY OF JOHANN *PLEASE REFER TO THE REFER AT THE END *
  • 2. • Born 1642 • Went to University of Cambridge in England as a student and taught there as a professor after • Never married • Gave his attention mostly to physics and mathematics, but he also gave his attention to religion and alchemy • Newton was the first to solve three mysteries that intrigued the scientists • Laws of Motion • Laws of Planetary Orbits • Calculus
  • 3. NEWTON 1ST LAW OF MOTION NEWTON 2ND LAW OF MOTION NEWTON 3RD LAW OF MOTION
  • 4. 4-2 Newton’s First Law of Motion Newton’s first law is often called the law of inertia. Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it.
  • 5. 4-2 Newton’s First Law of Motion Inertial reference frames: An inertial reference frame is one in which Newton’s first law is valid. This excludes rotating and accelerating frames.
  • 6. 4-3 Mass Mass is the measure of inertia of an object. In the SI system, mass is measured in kilograms. Inertia  The tendency of a body to maintain its state of rest or motion. Mass is not weight: Mass is a property of an object. Weight is the force exerted on that object by gravity. If you go to the moon, whose gravitational acceleration is about 1/6 g, you will weigh much less. Your mass, however, will be the same.
  • 7. Newton’s first law The ball remains in its state of rest till an external force is acting on it.
  • 8. Newton’s first law (contd..) Push the ball When an external force (push) is acting on a ball (object) at rest it changes its state (moves over the steps ).
  • 9. Newton’s first law (contd..) Push the ball When external force (Push of the rabbit) acting on the object (ball) which is in motion the object changes its position.
  • 10. • An Object at rest remains at rest, and an object in motion continues in motion with constant velocity (that is, constant speed in a straight line), unless it experiences a net external force. • The tendency to resist change in motion is called inertia • People believed that all moving objects would eventually stop before Newton came up with his laws
  • 11. • When there is no force exerted on an object, the motion of the object remains the same like described in the diagram • Because the equation of Force is F=ma, the acceleration is 0m/s². So the equation is 0N=m*0m/s² • Therefore, force is not needed to keep the object in motion, when • The object is in equilibrium when it does not change its state of motion
  • 12. • 1st Law: If no net force acts, object remains at rest or in uniform motion in straight line. • What if a net force acts? Do Experiments. • Find, if the net force F  0  The velocity v changes (in magnitude or direction or both). • A change in the velocity v (Δv)  There is an acceleration a = (Δv/Δt) OR A net force acting on a body produces an acceleration! Fa
  • 13. Newton’s Second Law Force equals mass times accelerati on. F = ma
  • 14. • The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass Fnet Acceleration
  • 15. • From experiments: The net force F on a body and the acceleration a of that body are related. • HOW? Answer by EXPERIMENTS! • The outcome of thousands of experiments over hundreds of years: a  F/m (proportionality) • We choose the units of force so that this is not just a proportionality but an equation: a  F/m OR: (total!) F = ma
  • 16. • Newton’s 2nd Law: F = ma F = the net (TOTAL!) force acting on mass m m = the mass (inertia) of the object. a = acceleration of object. Description of the effect of F F is the cause of a. • To emphasize that the F in Newton’s 2nd Law is the TOTAL (net) force on the mass m, text writes: ∑F = ma ∑ = a math symbol meaning sum (capital sigma)
  • 17. • Newton’s 2nd Law: ∑F = ma Based on experiment! Not derivable mathematically!! Force is a vector, so ∑F = ma is true along each coordinate axis. ∑Fx = max ∑Fy = may ∑Fz = maz ONE OF THE MOST FUNDAMENTAL & IMPORTANT LAWS OF CLASSICAL PHYSICS!!!
  • 18. 4-4 Newton’s Second Law of Motion Newton’s second law is the relation between acceleration and force. Acceleration is proportional to NET force and inversely proportional to mass. (4-1)
  • 19. 4-4 Newton’s Second Law of Motion The unit of force in the SI system is the newton (N). Note that the pound is a unit of force, not of mass, and can therefore be equated to newtons but not to kilograms.
  • 20. As the man jumps off the boat, he exerts the force on the boat and the boat exerts the reaction force on the man. The man leaps forward onto the pier, while the boat moves away from the pier.
  • 21. Foil deflected up Engine pushed forward Flow backwardpushed backward Flow Foil deflected down deflected Foil down
  • 22. • If two objects interact, the force exerted on object 1 by object 2 is equal in magnitude but opposite in direction to the force exerted on object 2 by object 1 • Forces always come in pair when two objects interact • The forces are equal, but opposite in direction Fn Fg
  • 23. 4-5 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. Law of action-reaction: “Every action has an equal & opposite reaction”. (Action-reaction forces act on DIFFERENT objects!)
  • 24. 4-5 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.
  • 25. 4-5 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.
  • 26. 4-5 Newton’s Third Law of Motion Helpful notation: the first subscript is the object that the force is being exerted on; the second is the source. This need not be done indefinitely, but is a good idea until you get used to dealing with these forces. (4-2)
  • 27.
  • 28. Example 4-11 Fpx  (40.0 N )(cos30.0 0 )  34.6 N Fpy  (40.0 N )(sin 30.00 )  20.0 N Fpx  max Fpx 34.6 N  ax   m 10.0 Kg  3.46 m / s 2 Fy  ma y  FN  Fpy  mg  ma y FN  20.0  98.0  0 ; a y  0 FN  78.0 N
  • 29. Example 4-13 FT  mE g  mE aE   mE a (1) FT  mC g  mC aC   mC a ( 2) subtract (1) from (2) (mE  mC ) g  (mE  mC )a mE  mC a g  0.68 m / s 2 mE  mC FT  mE g  mE a  mE ( g  a)  10500 N
  • 30. Example 4-14 The advantage of a pulley 2 FT  mg  ma const. speed  a  0 mg FT  2
  • 31. Example 4-15 Fx  FRC cos  FRB cos  0 ; a  0  FRB  FRC Fy  Fp  2 FT sin   0 FT  Fp 2 sin   1700 N
  • 34. 4-6 Weight – the Force of Gravity; and the Normal Force Weight is the force exerted on an object by gravity. Close to the surface of the Earth, where the gravitational force is nearly constant, the weight is:
  • 35. 4-6 Weight – the Force of Gravity; and the Normal Force An object at rest must have no net force on it. If it is sitting on a table, the force of gravity is still there; what other force is there? The force exerted perpendicular to a surface is called the normal force. It is exactly as large as needed to balance the force from the object (if the required force gets too big, something breaks!)
  • 36. m = 10 kg The normal force is NOT always equal to the weight!!
  • 37. Example 4-7 m = 10 kg ∑F = ma FP – mg = ma
  • 39. 4-7 Solving Problems with Newton’s Laws – Free-Body Diagrams 1. Draw a sketch. 2. For one object, draw a free-body diagram (force diagram), showing all the forces acting on the object. Make the magnitudes and directions as accurate as you can. Label each force. If there are multiple objects, draw a separate diagram for each one. 3. Resolve vectors into components. 4. Apply Newton’s second law to each component. 5. Solve.
  • 40. 4-7 Solving Problems with Newton’s Laws – Free-Body Diagrams When a cord or rope pulls on an object, it is said to be under tension, and the force it exerts is called a tension force.
  • 41. 4-8 Applications Involving Friction, Inclines On a microscopic scale, most surfaces are rough. The exact details are not yet known, but the force can be modeled in a simple way. We must account for Friction to be realistic!
  • 42. 4-8 Applications Involving Friction, Inclines • Exists between any 2 sliding surfaces. • Two types of friction: Static (no motion) friction Kinetic (motion) friction • The size of the friction force: Depends on the microscopic details of 2 sliding surfaces. • • • • The materials they are made of Are the surfaces smooth or rough? Are they wet or dry? Etc., etc., etc.
  • 43. 4-8 Applications Involving Friction, Inclines For kinetic (sliding) friction, we write: k is the coefficient of kinetic friction, and is different for every pair of surfaces. k depends on the surfaces & their conditions k is dimensionless & < 1
  • 44. 4-8 Applications Involving Friction, Inclines Static friction is the frictional force between two surfaces that are not moving along each other. Static friction keeps objects on inclines from sliding, and keeps objects from moving when a force is first applied.
  • 45. 4-8 Applications Involving Friction, Inclines The static frictional force increases as the applied force increases, until it reaches its maximum. Then the object starts to move, and the kinetic frictional force takes over.
  • 46. 4-8 Applications Involving Friction, Inclines
  • 47. 4-8 Applications Involving Friction, Inclines An object sliding down an incline has three forces acting on it: the normal force, gravity, and the frictional force. • The normal force is always perpendicular to the surface. • The friction force is parallel to it. • The gravitational force points down. If the object is at rest, the forces are the same except that we use the static frictional force, and the sum of the forces is zero.
  • 48. 4-9 Problem Solving – A General Approach 1. Read the problem carefully; then read it again. 2. Draw a sketch, and then a free-body diagram. 3. Choose a convenient coordinate system. 4. List the known and unknown quantities; find relationships between the knowns and the unknowns. 5. Estimate the answer. 6. Solve the problem without putting in any numbers (algebraically); once you are satisfied, put the numbers in. 7. Keep track of dimensions. 8. Make sure your answer is reasonable.
  • 49.  FT1 m1g   FT2  FT2 Take up as positive! m1 = m2 = 3.2 kg m1g = m2g = 31.4 N a Bucket 1: FT1 - FT2 - m1g = m1a Bucket 2: FT2 -m2g = m2a a ∑F =  m2 g ma (y direction), for EACH bucket separately!!!
  • 50. FT  Take up positive! m = 65 kg mg = 637 N  FT FT + FT - mg = ma 2FT -mg = ma FP = - FT (3rd Law!)  FP a ∑F = ma (y direction) on woman + bucket!  mg
  • 51. • Kinetic Friction: Experiments determine the relation used to compute friction forces. • Friction force Ffr is proportional to the magnitude of the normal force FN between two sliding surfaces. DIRECTIONS of Ffr & FN are  each other!! Ffr  FN FN Ffr a Fa mg
  • 52. • Static Friction: Experiments are used again. • The friction force Ffr exists || two surfaces, even if there is no motion. Consider the applied force Fa: FN Ffr Fa mg ∑F = ma = 0 & also v = 0  There must be a friction force Ffr to oppose Fa Fa – Ffr = 0 or Ffr = Fa The object is not moving
  • 53. • Experiments find that the maximum static friction force Ffr (max) is proportional to the magnitude (size) of the normal force FN between the two surfaces. DIRECTIONS of Ffr & FN are  each other!! Ffr  FN • Write the relation as Ffr (max) = sFN s  Coefficient of static friction • Depends on the surfaces & their conditions • Dimensionless & < 1 • Always find s > k  Static friction force: Ffr  sFN
  • 54. FN  mA g  5.0  9.8  49 N F fr   k FN  0.20  49  9.8 N FBy  mB g  FT  mB a FAx  FT  F fr  mA a Example 4-20  FT  F fr  mA a  mB g  F fr  mA a  mB a mB g  F fr 19.6  9.8 a   1.4 m / s 2 mA  mB 5.0  2.0 FT  F fr  mA a  17 N
  • 55. Example 4-21 FGx  mg sin  FGy  mg cos Fy  FN  mg cos  0 Fx  mg sin    k FN  max  FN  mg cos  mg sin    k (mg cos )  max
  • 56. Problem 48 FN Ffr Fp (m1+m2) g FN   m1  m2  g  Ffr   k FN   k  m1  m2  g  Fx  FP  Ffr   m1  m2  a  a a = 1.9 m/s2 For contact forces analyze forces on the two blocks separately.
  • 57. 1. What acceleration will result when a 12 N net force applied to a 3 kg object? A 6 kg object? 2. A net force of 16 N causes a mass to accelerate at a rate of 5 m/s2. Determine the mass. 3. How much force is needed to accelerate a 66 kg skier 1 m/sec/sec? 4. What is the force on a 1000 kg elevator that is falling freely at 9.8 m/sec/sec?
  • 58. Summary of Chapter 4 • Newton’s first law: If the net force on an object is zero, it will remain either at rest or moving in a straight line at constant speed. • Newton’s second law: • Newton’s third law: • Weight is the gravitational force on an object. • The frictional force can be written: (kinetic friction) or (static friction) • Free-body diagrams are essential for problemsolving
  • 59. • Henderson, Tom. Physics. Course home page. 16 May 2008 http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/newtlaws/newtltoc.html>. • Serway, Raymond A., and Jerry S. Faughn. "The Laws of Motion." College Physics . Fifth ed. 1999. Benson, Tom. Newton’s Third Law applied to Aerodynamics 21 May 2008 http://www.grc.nasa.gov/WWW/K-12/airplane/newton3.html • • Introduction to Rocket Performance. Newton’s Third Law. 12 March 2004 http://www.allstar.fiu.edu/aero/rocket1a.htm • Stern, David P. (16) Newton’s Laws of Physics. 1. Force and Inertia. 9 October http://www-istp.gsfc.nasa.gov/stargaze/Snewton.htm 2004