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Newton’s Laws
The Study of Dynamics
Isaac Newton



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

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
Aristotle firmly believed this.
But Physics 1 students know better!
A force diagram illustrating
no net force
A force diagram illustrating
no net force
A force diagram illustrating
no net force
A force diagram illustrating
no net force
Another
example
illustrating
no net force
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.
Newton’s Second Law

∑F = ma
where ∑F is the net force
measured in Newtons (N)
 m is mass (kg)
2
 a is acceleration (m/s )

Units of force

Newton (SI system)



1 N = 1 kg m /s2
1 N is the force required to
accelerate a 1 kg mass at a rate of
1 m/s2
Pound (British system)

1 lb = 1 slug ft /s2
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
Newton’s Third Law
For every action there exists
an equal and opposite reaction.
 If A exerts a force F on B,
then B exerts a force of -F on
A.

Working a Newton’s 2nd Law Problem
Step 1: Draw the problem
Larry pushes a 20 kg
block on a frictionless
floor at a 45o angle
below the horizontal F
L
with a force of 150 N
while Moe pulls the
same block
horizontally with a
force of 120 N. What
is acceleration?
20 kg
FM
Working a Newton’s 2nd Law Problem
Step 2: Diagram

Force diagram
N
N
FL
20 kg
FM
FM
FL
FG
FG

Free Body diagram
Working a Newton’s 2nd Law Problem
Step 3: Set up equations
 F = ma
 Fx = max
 Fy = may
Always resolve two-dimensional problems
into two one-dimensional problems.
Working a Newton’s 2nd Law Problem
Step 4: Substitute
Make a list of givens from
the word problem.
 Substitute in what you
know.

Working a Newton’s 2nd Law Problem
Step 5: Solve
Plug-n-chug.
 Calculate your unknowns.
 Sometimes you’ll need to do
kimematic calculations
following the Newton’s 2nd law
calculations.

Gravity as an accelerating
force
A very commonly used accelerating
force is gravity. Here is gravity in
action. The acceleration is g.
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.
2-Dimensional problem
Larry pushes a 20 kg
block on a frictionless
floor at a 45o angle
below the horizontal
with a force of 150 N
FL
while Moe pulls the
same block horizontally
with a force of 120 N.
a) What is the
acceleration?
b) What is the normal
force?
N
20 kg
FG
FM
Flat surfaces – 1 D
N
mg
N = mg for objects
resting on
horizontal surfaces.
Applied forces affect normal
force.
applied force
normal
friction
weight
N = applied force
V=0
A=0
V>0
A>0
V>0
A=0
Elevator Ride – going up!
V>0
A<0
Heavy feeling
Normal feeling
Normal feeling
Light feeling
N
N
N
N
mg
mg
mg
mg
Ground
floor
Just
starting up
Between
floors
Arriving at
top floor
V<0
A<0
V=0
A=0
V<0
A>0
V<0
A=0
Elevator Ride – going down!
Heavy feeling
Normal feeling
Normal feeling
Light feeling
N
N
N
N
mg
mg
mg
mg
Top
floor
Beginning
descent
Between Arriving at
floors Ground floor
Ramps – 2 D
N = mgcos
N
mgcos
mgsin
The normal force is
perpendicular to angled
ramps as well. It’s always
equal to the component of
weight perpendicular to
the surface.

mg

Ramps – 2 D
N = mgcos
N
How long will it take a 1.0
kg block to slide down a
frictionless 20 m long
ramp that is at a 15o angle
with the horizontal?
mgcos

mgsin
mg

Determination of the
Coefficients of Friction
Coefficient of Static Friction
1)
Set a block of one material on an
incline plane made of the other
material.
2)
Slowly increase angle of plane until
the block just begins to move.
Record this angle.
3)
Calculate s = tan.
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
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Kinetic friction
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exists before sliding occurs
exists after sliding occurs
In general fk <= fs
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.

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
A force diagram illustrating
Static Friction
Normal Force
Bigger
Frictional
Force
Bigger
Applied
Force
Gravity
A force diagram illustrating
Static Friction
The forces on the book are now UNBALANCED!
Normal Force
Frictional
Force
Gravity
Static friction cannot get any larger, and
can no longer completely oppose the
applied force.
Even
Bigger
Applied
Force
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.

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Determination of the
Coefficients of Friction
Coefficient of Kinetic Friction
1)
2)
3)
Set a block of one material on an
incline plane made of the other
material.
Slowly increase angle of plane until
the block just begins to move at
constant speed after giving it a
slight tap. Record this angle.
Calculate k = tan.
Magic Pulleys
N
mg
T
-x
T
m1
mg
m2
x
Pulley problem
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.
m1
m2
Pulley problem
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?
m1
m2
Pulley problem
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?
m1
m2
Tension
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A pulling force.
Generally exists in a rope, string,
or cable.
Arises at the molecular level,
when a rope, string, or cable
resists being pulled apart.
Working a Newton’s Law Problem
Step 1: Identify the body to analyze.

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
This may not be all that easy!
It may be a knot, a nail, a hinge, a
person, an “object” or a “particle”.
It is the focus of your subsequent
analysis.
Working a Newton’s Law Problem
Step 2: Select a reference frame.


This should be an inertial reference
frame which may be moving but not
accelerating.
Think of this as a coordinate system
with a specific origin!
Working a Newton’s Law Problem
Step 3: Make a diagram of forces.

Force diagram
F2
F1

Free Body diagram
F2
F1
Working a Newton’s Law Problem
Step 4: Set up force equations.
F = ma
 Fx = max
 Fy = may

Always resolve two-dimensional problems into
two one-dimensional problems.
Working a Newton’s Law Problem
Step 5: Calculate!


Substitute in what you know into
the second law equations.
Calculate unknown or unknowns.
Ramp (frictionless)
N = mgcos
N
The normal force is
perpendicular to angled
ramps as well. It’s usually
equal to the component
of weight perpendicular
to the surface.
mgcos 
mgsin
mg

Ramp (frictionless)
N = mgcos
N
mgcos 
mgsin
What will acceleration
be in this situation?
F= ma
mgsin = ma
gsin = a
mg

Ramp (frictionless)
N = mgcos
N
How could you keep the
block from accelerating?
T
mgcos 
mgsin
mg

Tension (static 1D)
The horizontal and vertical
components of the tension are equal
to zero if the system is not
accelerating.
T
15 kg
mg
F = 0
T = mg
Tension (static 2D)
The horizontal and vertical
components of the tension are equal
to zero if the system is not
accelerating.
30o
45o
F =
T3
mg
T2
15 kg
T1 0
T3
x
Fy =
0
Tension (elevator)
T
M
Mg
When an
elevator is still,
the tension in
the cable is
equal to its
weight.
Tension (elevator)
T
M
Mg
What about
when the
elevator is just
starting to head
upward from
the ground
floor?
Tension (elevator)
T
M
Mg
What about
when the
elevator is
between floors?
Tension (elevator)
T
M
Mg
What about
when the
elevator is
slowing at the
top floor?
Tension (elevator)
What about if
the elevator
cable breaks?
M
Mg
Pulley problems
Magic pulleys simply bend the coordinate
system.
N
T
T
m1g
-x
m1
F = ma
m2g
= (m1+m2)a
m2g
m2
x
Pulley problems
Tension is determined by examining one
F2 = m2a
block or the other
m2g - T = m2a
T
T
N
m1g

m2
F1 = m1a
T-m1gsin = m1a
m2g