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Chapter 3 Forces and Newton’s Laws Section 1: Forces Section 2: Newton’s Laws of Motion Section 3: Using Newton’s Laws
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Chapter 3 Forces and Newton’s Laws

Jan 02, 2016

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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 - PowerPoint PPT Presentation
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Page 1: 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

Page 2: Chapter 3  Forces and Newton’s Laws

Section 1: Forces1. ForceForce – 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 countertopThe two forces are equal in size (strength) and acting

in opposite directions. The aluminum block will never move under these conditions.

Page 3: Chapter 3  Forces and Newton’s Laws

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 inthe direction of the net force.

• There is a special unit for force called the Newton (N),

Page 4: Chapter 3  Forces and Newton’s Laws

Section 1: Forces2. 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, and2) 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

Page 5: Chapter 3  Forces and Newton’s Laws

Section 1: Forces

It is possible to calculate the amount of friction between two surfaces. The equation:

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.

Page 6: Chapter 3  Forces and Newton’s Laws

Section 1: ForcesExample 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:

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:

μ = 0.275Ff = 215.0 NFN = ?

Page 7: Chapter 3  Forces and Newton’s Laws

Section 1: Forces3. Gravity and WeightThe gravitational force is one of the four fundamental forces in nature:

1. Gravitational2. Electromagnetic – electricity, magnetism, and chemical

interaction3. Strong nuclear force – holds proton and neutrons together

in an atom’s nucleus4. 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:

All the particles in the universe exert a gravitational force on each other

Where:F = the gravitational forceG = constantM = mass one object 1m = mass of object 2r = the distance between the two objects

Page 8: Chapter 3  Forces and Newton’s Laws

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:

Example 1: What is the weight of a box that has a mass of 10-kg?Solution

Where: W = weight (N) m = mass (kg) g = 9.8 m/s2

m = 10.0 kgg = 9.8 m/s2

W = ?

Page 9: Chapter 3  Forces and Newton’s Laws

Section 1: Forces

Example 2: An object has a weight of 980-N. What is its mass?Solution

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.

W = 980.0 Ng = 9.8 m/s2

m = ?

Page 10: Chapter 3  Forces and Newton’s Laws

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

Page 11: Chapter 3  Forces and Newton’s Laws

Section 2: Newton’s Laws of Motion2. 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:

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:

The equation for the 2nd law can also be written as:

View Video

Where: α = acceleration (m/s2) F = force (N) m = mass (kg)

F = 200.0 Nm = 50.0 kgα = ?

Page 12: Chapter 3  Forces and Newton’s Laws

Section 2: Newton’s Laws of Motion3. 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 kgmA= 50.0kgVi = 0.0 m/sVf = 10.0 m/s∆t = 5.0sαJ = ?

Because of the 3rd Law: FA =-FJ

The minus sign (-) means that John/s acceleration is in the opposite direction of Ann’s.

View Video

Page 13: Chapter 3  Forces and Newton’s Laws

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:

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:, and: By substitution we get:

mvf = final momentum, and mvi = initial momentum The equation becomes: , ∆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

𝝆= 𝒎∙𝒗

Where: = momentum (kgm/s) m = mass (kg) v = velocity (m/s)

Page 14: Chapter 3  Forces and Newton’s Laws

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 anotherExample: 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 = 0.5-kg, 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 kgm8 =0.50 kgVci =1.5 m/sVcf = 1.0 m/sV8 = ?

Solving for v8f