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Forces. Chapter 6 Pages: 116-147. Force. A force is a push or pull upon an object resulting from the object's interaction with another object. Contact Forces Long-Range Forces. Contact Forces.

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Forces

Forces

Chapter 6

Pages: 116-147


Force
Force

A force is a push or pull upon an object resulting from the object's interaction with another object.

  • Contact Forces

  • Long-Range Forces


Contact forces
Contact Forces

Contact forces are types of forces in which the two interacting objects are physically in contact with each other.




AirResistance


Long range forces
Long-Range Forces

Long-Range Forces are types of forces in which the two interacting objects are not in physical contact with each other, but are able to exert a push or pull despite the physical separation.






Force2
Force

  • F for use in equations.

  • Newton is the unit for Force.

  • N abbreviation for Newton.

  • Net Force can accelerate.

  • N = kg m/s2


Force is a vector quantity
Force is a Vector Quantity

  • Magnitude

  • Direction

4000lb


Read pages 118 119 answer question 1 in notebook
Read Pages 118-119Answer Question 1 in Notebook


Homework
Homework

Page: 124

Questions: 7-11


Free body diagrams
Free Body Diagrams

The purpose of a free-body force diagram is to assist you in trying to determine the net force acting on a body.


Net force
Net Force

The purpose of a free-body force diagram is to assist you in trying to determine the net force acting on a body.


Free body diagrams1
Free Body Diagrams

The net force is the vector sum of all the individual forces acting on a system.

Fnet = F1± F2± F3 ± F4 …


Constructing free body force diagram
Constructing “free-body force diagram”

  • Identify the object(s) you will draw a diagram for. 


Constructing free body force diagram1
Constructing “free-body force diagram”

2. Identify all the forces acting directly on the object and the object exerting them.

  • Gravity

  • Table


Constructing free body force diagram2
Constructing “free-body force diagram”

3.Draw a dot to represent the object of interest.


Constructing free body force diagram3

Table

Gravity

Constructing “free-body force diagram”

4. Draw a vector to represent each force.


Constructing free body force diagram4
Constructing “free-body force diagram”

5. If the object is stationary or is moving at a constant velocity, the vectors should graphically add up to zero. 


Constructing free body force diagram5
Constructing “free-body force diagram”

5.   If the object is accelerating, the sum of the vectors should produce a vector in the same direction as the acceleration.


Constructing free body force diagram6

Floor

Gravity

Constructing “free-body force diagram”

Ffloor=Fgravity

Standing on Floor


Constructing free body force diagram7

Muscle

Gravity

Constructing “free-body force diagram”

Fmuscle>Fgravity

Jumping


Constructing free body force diagram8

Gravity

Constructing “free-body force diagram”

Fgravity

In the Air


Types of motion

Road

Brakes

Motor

Gravity

Types of Motion

  • No Motion

FRoad=FGravity

FMotor=FBrakes

No Net Force


Types of motion1

Road

Friction

Motor

Gravity

Types of Motion

  • Constant Velocity

FRoad=FGravity

FMotor=FFriction

No Net Force


Types of motion2

Road

Friction

Motor

Gravity

Types of Motion

  • Speeding Up

FRoad=FGravity

FMotor>FFriction

Net Force


Types of motion3

Road

Friction

Motor

Gravity

Types of Motion

  • Slowing Down

FRoad=FGravity

FMotor<FFriction

Net Force


Finding net force

FM=400N

FR=4000N

FF=400N

FG=4000N

Finding Net Force

VerticalFnet = FR - FG

HorizontalFnet = FF - FM

Fnet=4000N–4000N

Fnet=400N–400N

Fnet= 0N

Fnet= 0N


Finding net force1

FR=4000N

FF=400N

FG=4000N

Finding Net Force

FM=400N

Not Moving OrConstant Velocity


Finding net force2

FM=400N

FR=4000N

FF=100N

FG=4000N

Finding Net Force

VerticalFnet = FR - FG

HorizontalFnet = FF - FM

Fnet=4000N–4000N

Fnet=100N–400N

Fnet= 0N

Fnet= -300N


Finding net force3

FR=4000N

FF=100N

FG=4000N

Finding Net Force

FM=400N

Acceleratingto the Left.


Newton s second law of motion
Newton’s Second Law of Motion

Law

of

Acceleration


Newton s second law of motion1
Newton’s Second Law of Motion

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.


Newton s second law of motion2

FNet = 1N

Accelerates

Ff = 40N

FP = 41N

Newton’s Second Law of Motion


Newton s second law of motion3

a

F

Newton’s Second Law of Motion

m

FNet = ma



Inertia
Inertia

Inertia is the resistance an object has to a change in its state of motion.

Mass




Newton s first law of motion2
Newton’s First Law of Motion

An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.


Mass vs weight
Mass vs. Weight

Mass is the amount of stuff you are made up of. (kg or slugs) Does not change!!!!


Mass vs weight1
Mass vs. Weight

Weight depends on how much gravity is acting on you at the moment; you'd weigh less on the moon than on Earth.

(newtons or pounds)




Weight
Weight

To calculate weight use the acceleration due to gravity (9.8m/s2). This will be called g.

F=ma

Fg=mg


Weight problems
Weight Problems

Mr. Clune has a mass of 110kg. How much does he weight?

Given: m=110kg

g=9.8m/s2

Find: Fg=?

Equation: Fg=mg

=(110kg)(9.8m/s2)

Fg=1078N


F ma problems
F=ma Problems

A boy pulls a sled that has a mass of 5kg across the snow. The sled accelerates at a rate of 0.5m/s2. What is the net force of on the sled?

a=0.5m/s2

Fnet


Given: m=5kg

a=0.5m/s2

Find: Fnet=?

Equation: Fnet=mg

=(5kg)(0.5m/s2)

Fnet=2.5N


F ma problems1
F=ma Problems

A rock with a mass of 10kg fell off a cliff. At a specific time during its’ fall it had an acceleration of 3m/s2, due to air resistance. What is the force of air on this rock at this time?


+

Fnet=Fair + Fg

Fair

a

Fg

Fnet= ma

Fg= mg

Fnet


Fair=Fnet - Fg

Fair=ma- mg

Fair=m(a– g)

Fair=10kg{(-3m/s2)–(-9.8m/s2)}

Fair=10kg{(-3m/s2)+(9.8m/s2)}

Fair=10kg(6.8m/s2)

Fair=68N


Homework1
Homework

Page: 147

Questions: 22, 27,29

Due: 10/25/06


Factors that determine friction
Factors that determine Friction

Weight

Moving

Stationary

Surface


Friction forces

Fg

FN

Friction Forces

FP

Ff

FT


Friction forces1
Friction Forces

  • FN – Normal Force: This force which will affect frictional resistance is the component of applied force which acts perpendicular or "normal" to the surfaces which are in contact and is typically referred to as the normal force.


Friction forces2
Friction Forces

  • FT – Surface Force: This force opposite the normal force which is equal to this force.


Friction forces3
Friction Forces

  • FP – Push or Pull Force: This force is pushing or pulling the object.


Friction forces4
Friction Forces

  • Ff – Friction Force: Frictional resistance to the relative motion of two solid objects.


Friction forces5
Friction Forces

  • Ffs – Static Friction Force: Static frictional forces are non-moving forces between two surfaces. It will increase to prevent any relative motion up until some limit where motion occurs.


Friction forces6
Friction Forces

  • Ffk – Kinetic Friction Force: The force between two surfaces that are moving with respect to one another, the frictional resistance is almost constant over a wide range of low speeds.


Friction forces7
Friction Forces

  • μ –Coefficient of Friction: The ratio of the force of friction (Ff) between two bodies and the force pressing them together (FN).


Coefficient of friction

FN

Ffs

μs=

FN

Coefficient of Friction

Ff


Coefficient of friction1

FN

Ff

Ffk

μk=

FN

Coefficient of Friction


Friction problem

Ffk

FN

Friction Problem

A refrigerator of total weight 400N is pushed at a constant speed across a room by pushing horizontally on one side with a force of 160N. What is the coefficient of kinetic friction?

Ffk = 160N

FN = 400N

μk = ?


140N

μk=

400N

Ffk

μk=

FN

μk= 0.35


Ffs

μs=

= μk

Ffs

FN

FN

If the coefficient of static friction between the floor and the refrigerator was 0.6, how much force would be needed to start the refrigerator moving?

μs = 0.6

FN = 400N

Ffs = ?


= (0.6)

= 240N

(400N)

Ffs

Ffs

= μk

Ffs

FN


Homework2
Homework

Page: 133

Questions: 14,15

Page: 145

Questions: 33-35

Due: 11/2/06


Newton s third law
Newton’s Third Law

"For every action, there is an equal and opposite reaction."


While driving, Anna Litical observed a bug striking the windshield of her car. Obviously, a case of Newton's third law of motion. The bug hit the windshield and the windshield hit the bug. Which of the two forces is greater: the force on the bug or the force on the windshield?


Rockets are unable to accelerate in space because ... windshield of her car. Obviously, a case of Newton's third law of motion. The bug hit the windshield and the windshield hit the bug. Which of the two forces is greater: the force on the bug or the force on the windshield? There is no air in space for the rockets to push off of. There is no gravity is in space. There is no air resistance in space. ... nonsense! Rockets do accelerate in space.


  • A gun recoils when it is fired. The recoil is the result of action-reaction force pairs. As the gases from the gunpowder explosion expand, the gun pushes the bullet forwards and the bullet pushes the gun backwards. The acceleration of the recoiling gun is ...

  • greater than the acceleration of the bullet.

  • smaller than the acceleration of the bullet.

  • the same size as the acceleration of the bullet.


In the top picture, a physics student is pulling upon a rope which is attached to a wall. In the bottom picture, the physics student is pulling upon a rope which is held by the Strongman. In each case, the force scale reads 500 Newtons. The physics student is pulling…


with more force when the rope is attached to the wall. which is attached to a wall. In the bottom picture, the physics student is pulling upon a rope which is held by the Strongman. In each case, the force scale reads 500 Newtons. The physics student is pulling…

with more force when the rope is attached to the Strongman.

the same force in each case.


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