# Day 2 - Force & N2L

Mechanics

Oct 31, 2013 (4 years and 6 months ago)

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Physics 20

Day 2

Forces and Newton’s 2
nd

Law

3.1

Dynamics

is the branch of mechanics that deals with
forces
:
the cause of motion.

A force is a push or a pull.

Isaac Newton

(1642
-

1727)

symbol:

units:

Force is a vector quantity, and therefore must include direction.

newton (N)

Free
-
Body Diagrams

A
free
-
body diagram

is a vector diagram
that shows all the forces acting on an object.

Smile of the Day!

Daily Objectives:

1)
Apply Newton's second law of motion to explain,
qualitatively, the relationship among net force, mass,
and acceleration
(20 B1.3k)

Some Types of Forces

1)
weight

the force that gravity exerts on an object

An object’s weight can be
measured using a spring scale.

NEVER symbolize weight as
W.

W

=
work

!

2)
normal force

the force that a surface exerts on an object that is
resting on it

“normal” means
perpendicular

3)
applied force

general name given to any push or pull on an
object

= force due to friction

4)
net force

vector sum of
all the forces acting on an
object

Net force

and
unbalanced force

mean
the same thing.

5
)
tension

a force applied through pulling a rope

Types of Forces

Forces come in two basic types:
contact
forces

in which there is contact or touching
between the two objects and
action
-
at
-
a
-
distance forces
, whereby the two objects
are separated from one another, yet they
are still able to exert a push or pull on
each other.

Contact Forces
-

are forces that require
physical contact between interacting
objects. Some examples of these forces
are as follows:

Frictional Forces, Applied Forces, Normal
Forces, Tension
and Compression Forces,
Shearing and Torsion Forces, Spring
Forces

Action
-
at
-
a
-
distance Forces
-

forces that
exert a "push" or "pull" between two
objects despite the fact that the objects
are separated. Examples of these forces
are the four fundamental forces of Nature:

Gravitational Force
, Electromagnetic
Force, Strong Nuclear Force, Weak
Nuclear Force

Newton’s Second Law:

When acted upon by an unbalanced force,
objects accelerate.

The acceleration of an object is directly
proportional to the net force on it and
inversely proportional to its mass

F
net

= ma >

--
> a = F
net
/m

If the forces acting on the skater are
unbalanced
, there is a net force.

This causes her to accelerate in the
direction of the unbalanced force.

There is a linear relationship between acceleration and net force.

Relating Acceleration and Net Force

Relating Acceleration and Mass

There is an inverse relationship between
the acceleration and the mass of a skater.

Newton’s Second Law

When an external non
-
zero net force acts on an object, the object
accelerates in the direction of the net force. The magnitude of the
acceleration is directly proportional to the magnitude of the net
force and inversely proportional to the mass of the object.

= net force (N)

= acceleration (m/s
2
)

m

= mass (kg)

Newton’s Second Law

i) The force used to determine acceleration is the vector
sum of all the forces acting on the object.

Newton’s Second Law:

ii) The acceleration is in the same direction that the net
force is acting.

iii) Weight is the force the earth exerts downwards on a
mass

Newton’s Second Law:

F
g

= mg

iv) The symbol for frictional force is
f
. Friction always
opposes the direction of motion and so is negative in the
net force statement.

Newton’s Second Law:

m

m

Newton’s Second Law and Inertial Mass

Mass is related to the inertia of an object.

The
inertial mass

is a mass measurement based on the ratio of a known
net force on an object to the acceleration of the object.

It takes a larger force to accelerate an object with
a larger mass.

2:07

Example 3.6

Always use a free
-
body diagram
and use it to write an expression
for the net force.

Example 3.7

Example 3.8

Worksheet Package

Newton’s 2
nd

Law

#1, 2, 4, 6, 8

Try….

Atwood Machine

Be able to calculate answers to questions
that involve single forces or multiple forces
acting in a straight line.

(e.g. A person pushing a box when there
is a frictional force on the box, an Atwood
machine (two masses attached over a
pulley), a mass hanging over the edge of a
table and pulling on a second mass on the
table)

Questions will involve the use of kinematics formulas.

i)
Review the question and find a formula which will answer the
question
-

and likely involves acceleration since this concept links
kinematics and dynamics.

ii)
If a variable is missing, find a second formula which would let you
determine its value.

iii)
Continue to work backwards into the problem until all missing
variables are accounted.

iv)
Keep a record of how you thought through the problem. You will
present it in reverse order either showing each step as a separate
calculation or combining the formulas to make a concept statement
that answers the question in one step.

Example 3.9

m

=
30
kg

In any system involving more than one mass, we always find the total
mass and draw a free body diagram.

Example 3.10

m

= 60 kg

Example:

Atwood Machine:

Example 3.10 pg. 154

Two masses are connected by a light rope over a
light, frickionless pulley.
m
1

has a mass of 25 kg,
m
2

has a mass of 35 kg. Determine the motion
(acceleration) of each object once they are released.

Given: m
1

= 25 kg

m
2

= 35 kg

g = 9.81 m/s
2

downwards

Required: acceleration of each object (a
1

and a
2
)

Analysis & Solution:

The difference in mass between 1 and 2 will provide the
net force that will accelerate both objects. Since
m
2

m
1

you would expect
m
2
to accelerate downwards and
m
1

to
accelerate upwards. The rope has negligible mass.

The tension in the rope is the same on both sides of the
pulley, the rope does not stretch.

So the magnitude of accelerate is equal for both objects.

Find the total mass of both objects:

m
T

=
m
2

+
m
1

=
25 kg + 35 kg

=60 kg

Choose an equivalent system in terms of
m
T

to analyze the motion:

m
T

F
1

F
2

F
1

is equal to the gravitational force acting on
m
1

and F
2

is equal to the gravitational force acting
on
m
2

Apply Newton’s second law to find the net force
acting on
m
T

F
net

= F
1
+ F
2

F
net

=
-

m
1
g +
m
2
g

F
net

= (
m
2

m
1
)g

F
net

= (35
-
25 kg)(9.81 m/s
2
)

F
net

= 98.1 kg m/s
2

F
net

= 98.1 N

Use the scalar form of Newton’s second
law to calculate the magnitude of the
acceleration:

F
net

=
m
T
a

a = F
net
/
m
T

=
9.81
N /
60
kg

=
1.6
m/s
2

Mass
1
will
1.6
m/s
2
accelerate upwards,
mass
2
will accelerate
1.6
m/s
2
downward.

Homework:

Newton’s 2
nd

Law Worksheet:

#1, 2, 4, 6, 8, 10, 12, 14