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Newton’s Laws
Physics 11
Newton’s Laws of Motion
1.
2.
3.
An object at rest or in uniform motion will
remain rest or in uniform motion unless acted
on by an external force.
Acceleration is directly proportional to the
force applied to an object and inversely
proportional to its mass.
For every action, there is an equal and
opposite reaction.
Inertial and Non-Inertial Frames of
Reference




An inertial frame of reference is one in which
Newton’s Laws are valid
An inertial frame is either at rest or in uniform
motion but there can be no acceleration
A non-inertial frame of reference is one in
which Newton’s Laws are not valid
Accelerating frames of reference are always noninertial
Newton’s Second Law

Acceleration is directly proportional to the
force applied to an object and inversely
proportional to its mass.


F  ma
Putting it All Together


Now that we have considered Newton’s Second
Law, you can use that to analyze kinematics
problems with less information than we have
used previously
We can either use dynamics information to then
apply to a kinematic situation or vice versa
Free Body Diagrams


A free body diagram will be
used in most dynamics
problems in order to simplify
the situation
In a FBD, the object is
reduced to a point and forces
are drawn starting from the
point
FN
Fa
Ff
Fg
The Net Force



In most situations, there is more than one force
acting on an object at any given time
When we draw the FBD we should label all
forces that are acting on an object and also
determine which would cancel each other out
Ones that do not completely cancel out will be
used to determine the net force
The Net Force


The net force is a vector sum which means that
both the magnitude and direction of the forces
must be considered
In most situations we consider in Physics 11, the
forces we consider will be parallel or anti-parallel
An Example

A 25kg crate is slid from rest across a floor with
an applied force 72N. If the coefficient of
kinetic friction is .27, determine:
The acceleration of the crate?
 The time it would take to slide the crate 5.0m across
the floor.

FBD
FN=250N
Fa=72N
Ff=?
Fg=-250N
Use the frictional force equation to
determine the magnitude of the
frictional force
F f  FN
F f  (.27)( 250 N )
F f  66 N

F f  66 N
The net force is the sum of the
forces (acting parallel or antiparallel)

Fnet

Fnet

Fnet

Fnet

  Fi


 F f  Fa
 66 N  72 N
 5.8 N
Use Newton’s Second Law to solve
for the acceleration


Fnet  ma

5.8 N  (25kg)a

2
a  0.23m / s
Use kinematics to solve for the time
taken to cross the floor

2



at
d (t ) 
 v0t  d 0
2
2 2
0.23m / s t
5 .0 m 
2
2(5.0m)
t
2
0.23m / s
t  6 .6 s
Problems

Page 168
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Questions 4-8
Page 170
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Page 104
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1-2
Page 108
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1-14
Page 158
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
9, 10 and 13
Physics 20


Page 159
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Conceptual Problems
Section Review
Page 163
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1-3