### Chapter 4 Notes

```Chapter 4
Forces and Newton’s
Laws of Motion
Objective
• We will be able to explain Newton’s 1st
Law of motion
4.1 The Concepts of Force and Mass
A force is a push or a pull.
Contact forces arise from
physical contact .
Action-at-a-distance
forces do not
require contact and
include gravity and
electrical forces.
4.1 The Concepts of Force and Mass
Arrows are used to represent forces. The
length of the arrow is proportional to the
magnitude of the force. Force is a vector
quantity
15 N
5N
4.1 The Concepts of Force and Mass
Mass is a measure of the amount
of “stuff” contained in an object.
(How many atoms are present)
4.2 Newton’s First Law of Motion
Newton’s First Law
An object continues in a state of rest
or in a state of motion at a constant
speed along a straight line, unless
compelled to change that state by a
net force.
4.2 Newton’s First Law of Motion
Newton’s First Law
The net force is the vector sum of all
of the forces acting on an object.
4.2 Newton’s First Law of Motion
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).
Individual Forces
4N
10 N
Net Force
6N
4.2 Newton’s First Law of Motion
Individual Forces
Net Force
5N
64
3N
4N
4.2 Newton’s First Law of Motion
Inertia is the natural tendency of an
object to remain at rest or in motion at
a constant speed along a straight line.
4.2 Newton’s First Law of Motion
The mass of an object is a quantitative
measure of inertia.
SI Unit of Mass: kilogram (kg)
DEMOS
• Cup and Penny demo
• Bottle and Paper Demo
• Weight and String Demo
Question
• Which of the following statements can be
explained by Newton’s first law?
a) When your car suddenly comes to a halt,
you lunge forward
b) When your car rapidly accelerates, you
are pressed backward against the seat
c) Both
d) Neither
Question
• An object is moving at a constant velocity.
All but one of the following statements
could be true. Which one cannot be true?
• (a) No forces act on the object.
• (b) A single force acts on the object.
• (c) Two forces act simultaneously on the
object.
• (d) Three forces act simultaneously on the
object.
Question
• A cup of coffee is sitting on a table in a recreational
vehicle (RV). The cup slides toward the rear of the
RV. According to Newton’s first law, which one or
more of the following statements could describe the
motion of the RV? (A) The RV is at rest, and the
driver suddenly accelerates. (B) The RV is moving
forward, and the driver suddenly accelerates. (C)
The RV is moving backward, and the driver
suddenly hits the brakes.
• (a) A (b) B (c) C (d) A and B (e) A, B, and C
Good Example
• The mechanism for locking a seat belt?
Question
• When we jump into the air why doesn’t the
wall come crashing into us?
• (Earths Rotation ~ 1000 mph)
Mass vs Weight
•
•
•
•
Mass is NOT Weight
Mass is the quantity of matter in an object
Weight is the force of gravity on an object
They are proportional, Not the same!
Weight
• Weight = mass x gravity
• w = m۰g
Question
• Does a 2 kg rock have twice as much
mass of a 1 kg rock? Twice the inertia?
Twice the weight (when weighed in the
same location)?
• Yes, Yes, Yes
Question
• Does a liter of molten lead have the same
volume as a liter of apple juice? Does sit
have the same mass?
• Yes, no
Question
• Why do physicists say mass is more
fundamental than weight?
• Weight changes based on location, mass
doesn’t
Question
• An elephant and a mouse would both have
zero weight in gravity-free space. If they
were moving toward you with the same
speed would they bump into you with the
same force?
• No, the elephant is harder to stop, more
inertia.
Objective
• We will define Newton’s 2nd Law and solve
problems using the equation F = ma.
• We will be able to describe the
relationships between forces for 1 or more
objects using Newton’s 3rd Law
4.3 Newton’s Second Law of Motion
Newton’s Second Law
When a net external force acts on an object
of mass m, the acceleration that results is
directly proportional to the net force and has
a magnitude that is inversely proportional to
the mass. The direction of the acceleration is
the same as the direction of the net force.
Demo
Equation
• The equation for Force


 FNet  ma
  FNet
a
m
4.3 Newton’s Second Law of Motion
SI Unit for Force
m  kg  m
kg  2   2  N
s
s
This combination of units is called a newton (N).

4.3.1. A net force F is required to give an object with mass m an
acceleration a . If a net force 6F is applied to an object with mass
2m, what is the acceleration on this object?
a) a
b) 2a
c) 3a
d) 4a
e) 6a
Question
• The same force is applied to different objects.
Object 1 has a mass m, object 2 has a mass
m/3. Object 2’s acceleration will be ______
times more than object 1.
a) 1/3
b) 1/9
c) 3
d) 6
e) 9
4.3 Newton’s Second Law of Motion
A free-body-diagram is a diagram that
represents the object and the forces that
act on it.
4.3 Newton’s Second Law of Motion
The net force in this case is:
275 N + 395 N – 560 N = +110 N
and is directed along the + x axis of the
coordinate system.
4.3 Newton’s Second Law of Motion
If the mass of the car is 1850 kg then, by
Newton’s second law, the acceleration is
F

a
Net
m
110N
2

 0.059m s
1850 kg
4.4 The Vector Nature of Newton’s Second Law
The direction of force and acceleration vectors
can be taken into account by using x and y
components.


 FNet  ma
is equivalent to
F
y
 may
F
x
 max
4.5 Newton’s Third Law of Motion
Newton’s Third Law of Motion
Whenever one body exerts a force on a
second body, the second body exerts an
oppositely directed force of equal
magnitude on the first body.
Question?
• How come I cannot push myself?
• If I push on myself with a Force (F), then
my body pushes back with a Force (-F).
The forces cancel out
Question
• If I push a book with a force of 20N, and it
pushes back with 20 N, then how come
the book moves? Why don’t I move?
• There are 2 objects. 20 N is enough to
overcome friction for the book, but not
enough for me.
Question?
• What pushes a car forward?
• The ground does. The car pushes back on
the ground. The ground pushes forward on
the car.
Question?
• Scenario: You are on a sail boat and there
is no wind. However you have a big
industrial fan. What do you have to do to
move forward?
• Use the fan to blow out of the boat, not on
the sail. The fan will cause a force on the
air, the air will cause a force on the boat.
Question
• Explain how a rocket in space works if
there is no air to push on?
• Rocket pushes on the exhaust, the
exhaust pushes on the rocket.
Question
• A man on a bike and a man in a truck
have a head on collision. Which person
experiences the greater force? The
greater acceleration?
• Same force for both. Smaller mass =
greater acceleration (bike)
Question
• Does the Earth pull down on a ball, or
does a ball pull up on the Earth?
• Both. However, the Earth is much larger
and experiences a miniscule acceleration.
4.5 Newton’s Third Law of Motion
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?
4.5 Newton’s Third Law of Motion
 
On thespacecraft  F  P.


On theastronaut F  P.


P
 36 N
2
as 

 0.0033m s
ms 11,000kg


 P  36 N
2
aA 

 0.39 m s
mA
92 kg
Objective
• We will be able to draw free body
diagrams that include the force due to
gravity, friction, tension and the normal.
• We will be able to solve problems that
involve friction, tension, gravity and the
normal forces.
4.6 Types of Forces: An Overview
In nature there are two general types of forces,
fundamental and nonfundamental.
Fundamental Forces
1. Gravitational force
2. Strong Nuclear force
3. Electroweak force
4.6 Types of Forces: An Overview
Examples of nonfundamental forces:
friction
tension in a rope
normal or support forces
4.7 The Gravitational Force
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.
4.7 The Gravitational Force
For two particles that have masses m1 and m2 and are
separated by a distance r, the force has a magnitude
given by
m1m2
F G 2
r
G  6.6731011 N  m2 kg2
4.7 The Gravitational Force
m1m2
F G 2
r

11
 6.67 10
8
 1.4 10 N
N  m kg
2
2

12 kg25 kg

2
1.2 m 
4.7 The Gravitational Force
4.7 The Gravitational Force
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.
SI Unit of Weight: newton (N)
4.7 The Gravitational Force
Relation Between Mass and Weight
W G
M Em
r
W  mg
ME
g G 2
r
2
4.7 The Gravitational Force
On the earth’s surface:
ME
g G 2
RE

11
 6.67 10
 9.80 m s
2
Nm

5.9810 kg
kg 
6.3810 m
24
2
2
6
2
Objective
• We will be able to draw free body
diagrams that include the force due to
gravity, friction, tension and the normal.
• We will be able to solve problems that
involve friction, tension, gravity and the
normal forces.
4.8 The Normal Force
Definition of the Normal Force
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.
It is a compression force
4.8 The Normal Force
FN  11 N  15 N  0
FN  26 N
FN  11 N  15 N  0
FN  4 N
4.8 The Normal Force
If the person is standing on a nonaccelerating surface, the normal force is
simply the opposite of the gravitation
force (weight)
F
y
 0N  FN  mg
FN  mg
4.8 The Normal Force
Apparent Weight
The apparent weight of an object is the reading of
the scale.
It is equal to the normal force the man exerts on the
scale.
4.8 The Normal Force
We use this equation if the person or object is
on an accelerating platform.
F
y
  FN  mg  ma
FN  mg  ma
apparent
weight
true
weight
4.9 Static and Kinetic Frictional Forces
When an object is in contact with a surface there is a
force acting on that object. The component of this
force that is parallel to the surface is called the
frictional force.
Friction is always
directed opposite the
direction of motion
4.9 Static and Kinetic Frictional Forces
When the two surfaces
are not sliding across one
another the friction is
called static friction.
4.9 Static and Kinetic Frictional Forces
The magnitude of the static frictional force can
have any value from zero up to a maximum
value.
MAX
s
s
f f
f
MAX
s
0  s  1
 s FN
is called the coefficient of static friction.
4.9 Static and Kinetic Frictional Forces
Note that the magnitude of the frictional force
does not depend on the contact area of the
surfaces.
4.9 Static and Kinetic Frictional Forces
Static friction opposes the impending relative
motion between two objects.
Kinetic friction opposes the relative sliding
motion motions that actually does occur.
f k  k FN
0  s  1
is called the coefficient of kinetic friction.
Question
• Question: A person has a choice of either
pushing or pulling a sled at a constant velocity,
as the drawing illustrates. Friction is present. If
the angle is the same in both cases, does it
require more or less force to push or pull the
sled?
• To pull, because the upward component of the
pulling force reduces the normal force
therefore reducing friction
4.9 Static and Kinetic Frictional Forces
4.9 Static and Kinetic Frictional Forces
The sled comes to a halt because the kinetic frictional force
opposes its motion and causes the sled to slow down.
4.9 Static and Kinetic Frictional Forces
Suppose the coefficient of kinetic friction is 0.05 and the total
mass is 40kg. What is the kinetic frictional force?
f k   k FN   k m g 


0.0540kg 9.80 m s  20kg
2
4.10 The Tension Force
Cables and ropes transmit
forces through tension.
Tension is a stretching
force
4.10 The Tension Force
A massless rope will transmit
tension undiminished from one
end to the other.
If the rope passes around a
massless, frictionless pulley, the
tension will be transmitted to
the other end of the rope
undiminished.
Question
• Question: A rope is used in a tug of war
between two teams of 5 people each. Both
teams are equally strong, so neither team wins.
An identical rope is tied to a tree, and the same
10 people just as hard on the loose end as they
did in the contest. In both cases, the people pull
steadily with no jerking. Which rope sustains the
greater tension, a) the rope tied to the tree or b)
the rope in the tug of war or c) do the ropes
sustain the same tension.
When the angle from vertical
increases, what happens to the
tension force?
• Tension always increases as the angle
away from vertical increases
4.11 Equilibrium Application of Newton’s Laws of Motion
Definition of Equilibrium
An object is in equilibrium when it has zero acceleration.
F
x

0
Fy  0
4.11 Equilibrium Application of Newton’s Laws of Motion
Reasoning Strategy
• Select an object(s) to which the equations of equilibrium are
to be applied.
• Draw a free-body diagram for each object chosen above.
Include only forces acting on the object, not forces the object
exerts on its environment.
• Choose a set of x, y axes for each object and resolve all forces
in the free-body diagram into components that point along these
axes.
• Apply the equations and solve for the unknown quantities.
4.11 Equilibrium Application of Newton’s Laws of Motion
 T1 sin 35  T2 sin 35  0


 T1 cos35  T2 cos35  F  0


4.11 Equilibrium Application of Newton’s Laws of Motion
4.11 Equilibrium Application of Newton’s Laws of Motion
Force

T1

T2

W
x component
y component
 T1 sin 10.0
 T1 cos10.0
 T2 sin 80.0
 T2 cos80.0
0


W
W  3150 N
4.11 Equilibrium Application of Newton’s Laws of Motion

F
Fx   T1 sin 10.0  T2 sin 80.0  0
y
  T1 cos 10 .0  T2 cos 80 .0  W  0
The first equation gives


 sin 80 .0  
T
T1  
  2
 sin 10 .0 
Substitution into the second gives
 sin 80.0 




T cos10.0  T2 cos80.0  W  0
  2
 sin 10.0 
4.11 Equilibrium Application of Newton’s Laws of Motion
T2 
W
 sin 80.0 




cos
10
.
0

cos
80
.
0
 
sin
10
.
0


T2  582 N
T1  3.3010 N
3
4.12 Nonequilibrium Application of Newton’s Laws of Motion
When an object is accelerating, it is not in equilibrium.
F
x

 ma x
Fy  ma y
4.12 Nonequilibrium Application of Newton’s Laws of Motion
The acceleration is along the x axis so
ay  0
4.12 Nonequilibrium Application of Newton’s Laws of Motion
Force

T1

T2

D

R
x component
y component

 T1 cos30.0
 T1 sin 30.0

 T2 cos30.0
 T2 sin 30.0
D
0
R
0


4.12 Nonequilibrium Application of Newton’s Laws of Motion
F
y
  T1 sin 30.0  T2 sin 30.0  0

 T1  T2
F
x
 m ax
  T1 cos30.0  T2 cos30.0  D  R

4.12 Nonequilibrium Application of Newton’s Laws of Motion
T1  T2  T
ma x  R  D
5
T

1
.
53

10
N

2 cos 30.0
```