Transcript Chapter 3 Forces and Newton`s Laws

Chapter 3
Forces and Newton’s Laws
• Section 1: Forces
• Section 2: Newton’s Laws of Motion
• Section 3: Using Newton’s Laws
Section 1: Forces
1. Force
Force – a push or a pull that one body exerts on another
• You are constantly exerting forces on objects around you, and
objects around you are exerting forces on you
 What kinds of forces do you exert?
 What kinds of forces are exerted on you?
• A force can cause the motion of an object to change
 Example: hitting a tennis ball with a racket
• Forces do not always change velocity
 Example: aluminum cube on countertop.
• There are two forces acting on the block: a downward
acting force due to gravity, and an upward acting force
due to the countertop
 The two forces are equal in size (strength) and acting
in opposite directions. The aluminum block will never
move under these conditions.
Section 1: Forces
• When two or more forces act on an object at the same time,
the forces combine to form a net force.
• Balanced forces – forces on an object
that are equal size and opposite in
direction. Net force equals zero.
In the case of the aluminum block on the
counter, the two forces are balanced,
so the net (total ) force acting on the
block equals zero (0).
• Unbalanced forces – the result of the net force on an object
being greater than zero.
If we push (apply a force) on
the aluminum block, the net
force is greater than zero,
and the block may move in
the direction of the net force.
• There is a special unit for force called the Newton (N),
𝟏𝑵 = 𝟏
𝒌𝒈 − 𝒎
𝒔𝟐
Section 1: Forces
2. Friction
• Friction – the force that opposes motion between two
surfaces that are touching each other.
Imagine a toy car rolling across the floor. What forces are
acting on the car? Will the car roll forever? What causes the
car to come to a stop?
• The force that allows the car to roll on the floor and finally
brings the car to a stop is the same force--friction.
If there were no friction between the toy car’s wheel and the
floor the car’s wheels would spin but the car would not move.
That same friction between the wheels and the floor causes
the car to stop.
• The amount of friction depends on two things:
1) The kinds of surfaces, and
2) the forces pressing the surfaces together
• Types of friction:
static friction: The force due to microwelds that form
between the two surfaces
sliding friction: The force that opposes the motion of two
surfaces sliding past each other
rolling friction: The friction between a rolling object and
the surface it rolls on
air resistance: The friction between a solid surface and air
Section 1: Forces
It is possible to calculate the amount of friction between two
𝑭𝒇
surfaces. The equation: 𝝁 = 𝑭
𝑵
μ is the symbol for a value call the coefficient of friction; it is a
value with no dimension and is usually a known value. It is always
a value less than 1 and is expressed as a decimal to the
thousandth place.
Imagine you are sliding a
large box across the floor. The
drawing to the right
illustrates the forces acting
on the box.
• FA is the force you are applying on the box.
• FN is the normal force, or the force pressing the box to the floor.
• Ff is the frictional force between the floor and the box. Notice
that friction is acting in a direction opposite of the force you are
applying on the box.
In order to slide the box across the floor you have to apply a force
greater than friction.
Section 1: Forces
Example 1: The normal force of a box on a surface is 5.0 N. If
 = 0.300, what is the frictional force between the box and the
floor?
Solution:
𝐹𝑓
𝝁 = 𝟎. 𝟑𝟎𝟎
𝜇=
𝑭𝑵 = 𝟓. 𝟎 𝑵
𝐹𝑁
𝐹𝑓
𝑭𝒇 = ?
𝜇 𝐹𝑁 =
(𝐹 )
𝐹𝑁 𝑁
𝐹𝑓 = 𝜇(𝐹𝑁 )
𝐹𝑓 = 0.300(5.0𝑁)
𝑭𝒇 = 𝟏. 𝟓𝑵
Example 2: What is the normal force on a box if the frictional
force is 215.0 N and the coefficient of friction between the box
and the floor = 0.275?
Solution:
𝐹𝑓
μ = 0.275
𝜇=
Ff = 215.0 N
𝐹𝑁
𝐹𝑓
FN = ?
𝜇 𝐹𝑁 =
𝐹
𝐹𝑁 𝑁
𝜇(𝐹𝑁 ) 𝐹𝑓
=
𝜇
𝜇
𝐹𝑓
𝐹𝑁 =
𝜇
215.0𝑁
𝐹𝑁 =
0.275
𝑭𝑵 = 𝟕𝟖𝟏. 𝟖𝑵
Section 1: Forces
3. Gravity and Weight
The gravitational force is one of the four fundamental forces in
nature:
1. Gravitational
2. Electromagnetic – electricity, magnetism, and chemical
interaction
3. Strong nuclear force – holds proton and neutrons together
in an atom’s nucleus
4. Weak nuclear force – acts only on the particles in atomic
nuclei during radioactive decay
• Law of Gravitation – Any two masses exert an attractive force
on each other
 The gravitational force depends on two things:
1. The mass of the two objects
2. The distance between the objects
 Newton’s equation for the Law of Gravitation: 𝑭 =
Where:
F = the gravitational force
G = constant
M = mass one object 1
m = mass of object 2
r = the distance between the two objects

𝑮(𝑴)(𝒎)
𝒓𝟐
All the particles in the universe exert a gravitational force
on each other
Section 1: Forces
• Gravitational Acceleration
 Near Earth’s surface the gravitational attraction of Earth
causes all falling objects to have an acceleration of 9.8 m/s2
 The acceleration due to gravity is constant (on Earth) and is
noted using the small letter “g”. So: g = 9.8 m/s2
 Newton’s 2nd Law of Motion relates the size of the force
acting on a body to the body’s mass and its acceleration.
The gravitational force acting on an object can be described
in the same terms:
Gravitational force = mass of object x the acceleration due to gravity
 The gravitational force is known as the weight of an object,
so the equation becomes:
𝑾 = 𝒎𝒈
Where: W = weight (N)
m = mass (kg)
g = 9.8 m/s2
Example 1: What is the weight of a box that has a mass of
10-kg?
Solution
m = 10.0 kg
g = 9.8 m/s2
W=?
𝑊 = 𝑚𝑔
𝑚
𝑊 = 10. 0 𝑘𝑔 9.8 2
𝑠
𝑘𝑔 − 𝑚
𝑾 = 98.0
= 𝟗𝟖. 𝟎 𝑵
𝑠2
Section 1: Forces
Example 2: An object has a weight of 980-N. What is its mass?
Solution
W = 980.0 N
g = 9.8 m/s2
m=?
𝑊 = 𝑚𝑔
𝑊
𝑚𝑔
=
𝑔
𝑔
𝑊
𝑚=
𝑔
980.0 𝑁
𝑚=
𝑚
9.8 2
𝑠
𝑘𝑔 − 𝑚
980.0
𝑠2
𝑚=
𝑚
9.8 2
𝑠
𝒎 = 𝟏𝟎𝟎. 𝟎 𝒌𝒈
Important Relationship:
On a flat, level surface the
weight of an object (W) is equal
to the normal force (FN) between
the object an the surface.
Section 2: Newton’s Laws of Motion
1. Newton’s First Law of Motion: an object moves at a constant
velocity unless an unbalanced force acts on the object
An object moving at a constant velocity keeps moving at
that velocity until a net force act upon it. If an object is at
rest, it stays at rest unless a net force acts upon it.
• The 1st Law is also known as the Law of Inertia
 Inertia – the tendency of an object to resist any change in
motion
 The greater an object’s mass, the greater its inertia
Example: It is easier to change the motion of a toy car
than it is to change the motion of a full-size car
What happens to the driver, who is not wearing a seat belt, of
a car that runs into another vehicle? Why does this happen?
View Video
Section 2: Newton’s Laws of Motion
2. Newton’s 2nd Law of Motion: The net force acting on an
object causes the object to accelerate in the direction of the
net force
• Equation for the 2nd Law:
𝑭
𝜶=
𝒎
Where: α = acceleration (m/s2)
F = force (N)
m = mass (kg)
Example: Megan, riding her bike, applies a net force of 200N
on the pedals. The combined mass of Megan and the bicycle
is 50kg. What is the acceleration of the bike and Megan?
Solution: F = 200.0 N
𝐹
m = 50.0 kg
α=?
𝛼=
𝑚
200.0 𝑁
𝛼=
50.0 𝑘𝑔
𝑘𝑔 − 𝑚
200.0
𝑠2
𝛼=
50.0 𝑘𝑔
𝒎
𝜶 = 𝟒. 𝟎 𝟐
𝒔
The equation for the 2nd law can also be written as:
𝑭 = 𝒎 (𝜶)
View Video
Section 2: Newton’s Laws of Motion
3. Newton’s 3rd Law of Motion: When one object exerts a force
on a second object, the second object exerts a force on the
first that is equal in size and opposite in direction
• Forces always occur in action/reaction pairs.
• Examples of the 3rd Law include:
 walking
 swimming fish
 rocket propulsion
Example: Ann and John are ice-skating. John, who has a mass of 85
kg, pushed Ann, mass = 50 kg, so that her forward velocity went
from 0 m/s to 10 m/s in 5 s. As a result of this push, what is the rate
and direction of John’s acceleration?
Solution:
mJ = 85.0 kg
mA= 50.0kg
Vi = 0.0 m/s
Vf = 10.0 m/s
∆t = 5.0s
αJ = ?
∆𝑉 = 𝑉𝑓 − 𝑉𝑖
𝑚
𝑚
∆𝑉 = 10.0 − 0.0
𝑠
𝑠
𝑚
∆𝑉 = 10.0
𝑠
View Video
∆𝑉
∆𝑡
𝑚
10.0
𝑠
𝛼𝐴 =
5.0𝑠
𝑚
𝛼𝐴 = 2.0 2
𝑠
𝐹𝐴 = 𝑚𝐴 (𝛼𝐴 )
𝛼𝐴 =
𝑚
)
𝑠2
𝑘𝑔 − 𝑚
𝐹𝐴 = 100.0
𝑠2
𝐹𝐴 = 100,0 𝑁
𝐹𝐴 = 50.0𝑘𝑔(2.0
Because of the 3rd Law: FA =-FJ
−𝐹𝐽 = 𝑚𝐽 (𝛼𝐽 )
−𝐹𝐽
𝑚𝐽 (𝛼𝐽 )
=
𝑚𝐽
𝑚𝐽
−𝐹𝐽
𝛼𝐽 =
𝑚𝐽
−100.0𝑁
𝛼𝐽 =
85.0𝑘𝑔
𝑘𝑔 − 𝑚
−100.0
𝑠2
𝛼𝐽 =
85.0 𝑘𝑔
𝒎
𝜶𝑱 = −𝟏. 𝟏𝟖 𝟐
𝒔
The minus sign (-) means that
John/s acceleration is in the
opposite direction of Ann’s.
Section 3: Using Newton’s Laws
• Recall from Chapter 2 that a moving object has a property
called momentum
 A moving object’s inertia and momentum are related in that
the greater the mass of the object the larger the force must
to overcome the object’s inertia and change its momentum
 Equation for momentum: momentum = mass x velocity, or:
Where:
 = momentum (kgm/s)
𝝆=𝒎∙𝒗
m = mass (kg)
v = velocity (m/s)
 An object with a small and high velocity can have a great
deal of momentum, just as an object with a large and low
velocity can have a high momentum.
•
Momentum and the 2nd Law of Motion:
 Recall two equations: 𝜶 =
𝒗𝒇 −𝒗𝒊
𝒕
 By substitution we get: 𝑭 = 𝒎
, and: 𝑭 = 𝒎𝜶
𝒗𝒇 −𝒗𝒊
𝒎𝒗𝒇 −𝒎𝒗𝒊𝒊
=
𝒕
𝒕
mvf = final momentum, and mvi = initial momentum
 The equation becomes: 𝑭 =
∆𝝆
,
𝒕
∆ρ = the change in
momentum
 the net force acting on an object can be calculated by
dividing the change in momentum by the time over which
the change occurs
Section 3: Using Newton’s Laws
• Law of Conservation of Momentum: Momentum of an object
doesn’t change unless there is a change in the object’s mass
or velocity
• Momentum can be transferred from one object to another
Example: playing pool - The momentum lost by the cue ball
equals the momentum gained by the other balls. The total
momentum in the system remains constant, but is
distributed differently.
Example: A cue ball, m = 0.75-kg, is rolling towards the 8-ball, m = 0.5kg, at a velocity of 1.5 m/s. If the cue ball has a velocity of 1 m/s after
hitting the 8-ball, what is the 8-ball’s velocity after the collision?
Solution:
mc = 0.75 kg
m8 =0.50 kg
Vci =1.5 m/s
Vcf = 1.0 m/s
V8 = ?
𝜌𝑖 = 𝜌𝑓
𝜌𝑐𝑖 + 𝜌8𝑖 = 𝜌𝑐𝑓 + 𝜌8𝑓
𝑚𝑐 𝑣𝑐𝑖 + 𝑚8 𝑣8𝑖 = 𝑚𝑐 𝑣𝑐𝑓 + 𝑚8 𝑣8𝑓
Solving for v8f
𝑣8𝑓
𝑚8 𝑣8𝑓 = 𝑚𝑐 𝑣𝑐𝑖 + 𝑚8 𝑣8𝑖 − 𝑚𝑐 𝑣𝑐𝑓
𝑚8 𝑣8𝑓
𝑚𝑐 𝑣𝑐𝑖 + 𝑚8 𝑣8𝑖 − 𝑚𝑐 𝑣𝑐𝑓
=
𝑚8
𝑚8
𝑚𝑐 𝑣𝑐𝑖 + 𝑚8 𝑣8𝑖 − 𝑚𝑐 𝑣𝑐𝑓
𝑣8𝑓 =
𝑚8
𝑚
𝑚
𝑚
0.75𝑘𝑔 1.5
+ 0.5𝑘𝑔 0
− 0.75𝑘𝑔(1.0 )
𝑠
𝑠
𝑠
=
0.5𝑘𝑔
𝑘𝑔𝑚
𝑘𝑔𝑚
1.125
− 0.75
𝑠
𝑠
𝑣8𝑓 =
0.5𝑘𝑔
𝑘𝑔𝑚
0.375
𝑠
𝑣8𝑓 =
0.5𝑘𝑔
𝒎
𝒗𝟖𝒇 = 𝟎. 𝟕𝟓
𝒔