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Chapter 4 Forces
In One Dimension
Free-body Diagrams
Normal Force

The perpendicular contact force exerted
by a surface on another object.
Friction
Friction Opposes motion
 Ffr=μFN
 Ffr=μ*m*g
 μ - friction constant
 m – mass
 g – gravity
 Friction is related to the perpendicular
component of the weight force, parallel to
the surface.

Friction
μs- static friction coefficient
 μk – kinetic friction coefficient
 Usually refers to objects in direct contact
with each other.


Air resistance is a type of friction
Occurs when an object is moving
 Generally, only appears in free fall

Tension

Tension, the specific name for the force
exerted by a string or rope is an interaction
force.
Section 4.1 Force & Motion
Force - a push or pull acting on an object
that can cause the object to speed up, slow
down, or change direction
 Unit of force = Newton (N) = kg*m/s2
 Forces have both magnitude and direction they are vectors.
 Forces are divided into contact and field
forces

 Fields:
electric, magnetic
Types of Forces






Contact force
 Example: Your book laying on the desk
Field force
 Example: You drop your textbook onto the floor.
System – what we are looking at
Surroundings – everything else
Balanced forces have a net force of zero in every direction
Unbalanced forces have a non-zero net force in at least
one direction
Self Check

Pages 3-10
The Force Is with You
Free-body diagrams

Draw vectors away from objects
Table pushing up on
books
Books pushing down
on table
Determine net force

Add forces acting in the same direction
F1 = 3.0 n E
F2 = 2.0 n E
F net = 5.0 n E

Subtract forces acting in opposite
directions.

3n E + 2n W
1n E
Vectors at Right Angles
Sin q = opp
hyp
Cos q = adj
hyp
Tan q = opp
adj
Newton’s 1st Law (Inertia)

An object has a tendency to resist a change in its
motion unless there is an outside net force acting on
it.

Objects at rest stay at rest, objects in motion stay in
motion unless acted upon by an outside force.

No net force can result in



No motion or constant motion
Known as equilibrium
A net force can result in

Speeding up or slowing down
Newton’s 2nd Law
a = Fnet / m
 The acceleration on an object is equal to
the sum of the forces acting on an object
divided by the mass of the object.
 Sum of all forces acting on an object


Equilibrium if Fnet = 0
Calculating Net Force
Free-body diagrams
Free-body
diagrams are
used to show the
relative
magnitude and
direction of all
forces acting on
an object.
This diagram
shows four
forces acting
upon an
object. There
aren’t always
four forces,
For example,
there could be
one, two, or
three forces.
Forces at angles
Problem 1
A book is at rest on a table top. Diagram the
forces acting on the book.
Problem 1
In this diagram,
there are normal
and gravitational
forces on the
book.
Problem 2
An egg is free-falling from a nest in a tree.
Neglect air resistance. Draw a free-body
diagram showing the forces involved.
Gravity is the
only force
acting on
the egg as it
falls.
Self Check

Pages 17-18
Problem
1. A rock falls freely from a cliff. Draw
vectors and label each.
 +y

V
a
Fnet

A skydiver falling towards earth at a
constant rate…
+y
F air resistance on diver
v
v
v
a=0
F net = 0
F Earth’s mass on diver

A rope pulls a box at a constant speed
across a horizontal surface. The surface
provides a force that resists the box’s
motion.
+x
v
v
v
v
F friction on box
F pull on box
F net = 0
Problem
Two horizontal forces, 255 n and 184 n,
are exerted on a boat. If these forces are
applied in the same direction, find the net
horizontal force on the boat.
 F net = 255n + 184n = 4.39 x 102 n in the
direction of the forces.

Free Fall

The only force acting on the falling ball is Fg.

Fg is the weight force.

Fg is acting down as are the velocity and the
acceleration

Newton’s 2nd law has become:
Fg = mg
A ball in mid-air in free fall has
only the force of gravity acting
on it. Air resistance can be
neglected.
System
V
a
Known:
Fg
a=g
m
Unknown:
Fg
Fnet = ma
Fnet = Fg
Therefore:
Fg = mg
a=g
Newton’s 2nd Law Problem

Two girls are fighting over a stuffed toy
(mass = 0.30 kg). Sally (on left) pulls with
a force of 10.0 n and Susie pulls right with
a force of 11.0 n. What is the horizontal
acceleration of the toy?
Sally
10.0 n
M toy = 0.30 kg
Susie
11.0 n
Solution to Susie’s & Sally’s
dilemma.
Find the net Force:
11.0n R + (-10.0n L)
Fnet = 1.0 n R
Fnet = ma
a = Fnet / mw
a = 1.0 n / 0.30 kg = 3.33 m/s2 Right
Self Check

Pages 12-19
Real and Apparent Weight

Apparent weight is the force an object experiences as a
result of the contact forces acting on it, giving the object
an acceleration

Same when a body is traveling either up or down at a
constant rate, in an elevator, for example.

Apparent weight < real weight when the elevator is
slowing while rising or speeding up while descending.

Apparent weight > real weight when speeding up while
rising or slowing while going down.
Appar
ent
weight
is
Fscale
less
when
…
Fg
Slowly rising or
speeding up while
descending.
Fscale
Fg
Appare
nt
weight
is
greater
when
Speeding up
while rising or
slowing while
going down
Going
Up?
v=0
a=0
v>0
a>0
v>0
a=0
v>0
a<0
Heavy feeling
Normal feeling
Normal feeling
Light feeling
Wapp
Wapp
Wapp
Wapp
W
W
W
W
Ground
floor
Just
starting up
Between
floors
Arriving at
top floor
Going
Down?
v<0
a<0
v=0
a=0
v<0
a>0
v<0
a=0
Heavy feeling
Normal feeling
Normal feeling
Light feeling
Wapp
Wapp
Wapp
Wapp
W
W
W
W
Beginning
descent
Between Arriving at
floors Ground floor
Top
floor
How a bathroom scale works.
When you stand on the
scale the spring exerts an
upward force on you while
are in contact with the
scale.
You are not accelerating,
so the net force acting on
you must be zero.
The spring force, Fsp
upwards must be opposite
and equal to your weight
Fg that is acting downward.
Problem
On Earth, a scale shows that you weigh
585 n.
 A. What is your mass?
 B. What would the scale read on the Moon
where g = 1.60 m/s2?
 C. Back on Earth, what do you weigh in
pounds? (1 kg = 2.2 kg)

A. What is your mass?
m = Fg / g
 m = 585 n /9.8 m/s2
 m = 59.7 kg

B. What would the scale read on the
Moon where g = 1.60 m/s2?
Fg = mgmoon
 Fg= (59.7 kg)(1.60m/s2)
 Fg = 95.5 n

Back on Earth…
m = 59.7 kg x 2.2 lb
1 kg
m = 131 lb
Drag Force and Terminal Velocity

When an object moves through a fluid (liquid or
gas), the fluid exerts a drag force opposite to the
direction of motion of the object.

The force is dependent upon the motion of the
object and the properties of the fluid
(temperature and viscosity - resistance to flow).

As the object’s velocity increases, so does the
drag force.

The terminal velocity is the maximum velocity
reached by the object as it moves through the
fluid. Only occurs if air resistance in present
Terminal Velocity

The constant velocity that is reached when
the drag force equals the force of gravity.

What reaches a terminal velocity faster, a
heavy, compact object or a light object
with larger surface area?
Self Check

Pages 20-21
Identifying Interactive forces

You are on skates and so is your friend. You
push on their arm to move them forward and
they exert an equal and opposite force on you
which causes you to move backwards.

These forces are an interaction pair.

An interaction pair (or action and reaction) is two
forces that have equal magnitude and act in
opposite directions.
Newton’s Third Law

All forces come in pairs and the forces in a
pair act on different objects and are equal
in strength and opposite in direction
The forces simply exist together or not at all.
They result from the contact between the two
of you.
The two forces act on different objects and are
equal and opposite
Numerically, F A on B = - F B on A
A
F B on A
B
F A on B

When a softball of mass 0.18 kg is
dropped, its acceleration toward Earth is g.
What is the force on the Earth due to the
ball and what is Earth’s resulting
acceleration? Earth’s mass is 6.0 x 1024
kg.
Use Newton’s 2nd and 3rd
laws to find a Earth
F Earth on Ball = m ball a
Substitute a = -g
F Earth on Ball = m ball (-g)
Substitute knowns
F Earth on Ball =
(0.18kg)(9.8m/s2)
F Earth on Ball = 1.8 n
Find Earth’s Acceleration
F ball on Earth = - F Earth on ball = - 1.8 n
 a Earth on ball = Fnet/ m Earth
 a Earth on ball = 1.8 n / 6.0 x 1024kg
 a Earth on ball = 2.9 x 10-25 m/s2 toward the
ball

Practice Problem
A 50.0 kg bucket is being lifted by a rope.
The rope will not break if the tension is
525 N or less. The bucket started at rest,
and after being lifted 3.0 m, it is moving at
3.0 m/s. If the acceleration is constant, is
the rope in danger of breaking?
Practice Problem
Poloma hands a 13 kg box to a 61 kg
Stephanie, who stands on a platform. What
is the normal force exerted by the platform
on Stephanie?
Self Check

Pages 23-24