Newton's First Law• Newton's first law of motion: An object at rest stays at

rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

The Meaning of Force• A force is a push or pull upon an object resulting from the

object's interaction with another object. • Force is a quantity that is measured using the standard

metric unit known as the Newton.

• All forces (interactions) between objects can be placed into two broad categories

– Contact forces - that result when the two interacting objects are perceived to be physically touching each other.

– Field forces - that result even when the two interacting objects are not in physical contact with each other, yet are able to exert a push or pull despite their physical separation.

Contact Forces Action-at-a-Distance Forces (Field Force)

Applied Force Gravitational Force

Tension Force Electrical Force

Normal Force Magnetic Force

Air Resistance Force

Frictional Force

Spring Force

• The force of gravity is the force with which the earth, moon, or other massively large object attracts another object towards itself. By definition, this is the weight of the object. All objects upon earth experience a force of gravity that is directed "downward" towards the center of the earth. The force of gravity on earth is always equal to the weight of the object as found by the equation:

• Fgrav = m • g• where g = 9.81 N/kg (on Earth) and m = mass (in kg)• Note: g is different at different locations

Gravity Force (Weight) Fgrav

Fg Fg Fg

FgFg

Fg

Practice- indicate Fg on each box with an arrow

Comparing Mass and WeightWeight

• The force of gravity. • Vector, its direction is

downward. • W = mg • The weight of an object

(measured in Newton) will vary according to where in the universe the object is.

Mass • The mass of an object

refers to the amount of matter that is contained by the object;

• Scalar, has no direction• The mass of an object

(measured in kg) will be the same no matter where in the universe that object is located.

• The normal force is the support force exerted upon an object that is in contact with another stable object (usually a surface). The direction of the normal force is perpendicular to the surface, from the surface toward the object and on the object.

Normal Force (FN )

Fg Fg Fg

FgFg

Fg

FN FNFN

FN

FN

FN

Practice- indicate FN on each box with an arrow

• The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. The friction force often opposes the motion of an object.

• Friction results from the two surfaces being pressed together closely, causing intermolecular attractive forces between molecules of different surfaces. Friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together.

Friction Force (Ff)

Ff = μFN

Fg Fg Fg

FgFg

Fg

FN FNFN

FN

FN

FN

Ff

vv

Ff

v

Ff

v

Ff

v

Ff Ff

v

Ff

Practice- indicate Ff on each box with an arrow

• The air resistance is a special type of frictional force that acts upon objects as they travel through the air. The force of air resistance is often observed to oppose the motion of an object. This force will frequently be neglected due to its negligible magnitude.

Air Resistance Force (Fair )

• The tension force is the force that is transmitted through a string, rope, cable or wire when it is pulled tight by forces acting from opposite ends. The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire.

Tension Force (FT )

• The spring force is the force exerted by a compressed or stretched spring upon any object that is attached to it. An object that compresses or stretches a spring is always acted upon by a force that restores the object to its rest or equilibrium position – directed toward equilibrium position.

Spring Force (Fspring )

Balanced and Unbalanced Forces

If two individual forces are of equal magnitude and opposite direction, then the forces are said to be balanced.

When only balanced forces act on an object, the object is said to be at equilibrium.

Unbalanced forces

State of Motion• The state of motion of an object is defined

by its velocity - the speed with a direction.

• Inertia: tendency of an object to resist changes in its velocity.

• Inertia: tendency of an object to resist accelerations.

Newton’s First Law

Inertia

Tendency of an object to maintain its STATE OF MOTION

Also known as the “Law of Inertia”

Forces Don't Keep Objects Moving

Everyday Applications of Newton's First Law

• Blood rushes from your head to your feet while quickly stopping when riding on a descending elevator.

• The head of a hammer can be tightened onto the wooden handle by banging the bottom of the handle against a hard surface.

• A brick is painlessly broken over the hand of a physics teacher by slamming it with a hammer. (CAUTION: do not attempt this at home!)

• To dislodge ketchup from the bottom of a ketchup bottle, it is often turned upside down and thrusted downward at high speeds and then abruptly halted.

• Headrests are placed in cars to prevent whiplash injuries during rear-end collisions.

• While riding a skateboard (or wagon or bicycle), you fly forward off the board when hitting a curb or rock or other object that abruptly halts the motion of the skateboard.

Do these guys have a lot of inertia?

LOTS OF INERTIAhard to…

GET MOVING orSTOP

MORE MASSmeans

MORE INERTIA

Inertia is proportional to MASS

Drawing Free-Body Diagrams• Free-body diagrams are used to show the

relative magnitude and direction of all forces acting upon an object in a given situation.

• The size of the arrow in a free-body diagram reflects the magnitude of the force. The arrow shows the direction that the force is acting.

• Each force arrow in the diagram is labeled to indicate the exact type of force.

• It is generally customary to draw the force arrow from the center of the box outward in the direction that the force is acting.

A block of wood is sitting motionless on a table.What forces are acting on it?

FgWeight

FNNormal

Weight is the force of gravity

pulling an object toward the

CENTER OF THEEARTH

Normal Force is aREACTION

force that any object exerts

when pushed on

practice• A book is at rest on a tabletop. Diagram the

forces acting on the book.

Fg

FN

Determining the Net Force

400 N up

200 N down

20 N leftA B C

40 N

30 NR2 = (30N)2 + (40N)2

θ = tan-1(30/40) = 53.1o

Net force is 50 N at 53.1o West of North

•The net force is the vector sum of all the forces that act upon an object.

Net Force• If there is NO NET FORCE on an object, then it is

at EQUILIBRIUM and either: MOTIONLESS OR MOVING WITH CONSTANT

VELOCITY

• So a “net” or “unbalanced” force will– CHANGE AN OBJECT’S VELOCITY

• Changing velocity means ACCELERATION

A net force (an unbalanced force) causes an accelerationDescription of Motion Net Force: Yes or No?

                                                                                                                                                                                                

                                                                                                                                                                                           

                                                                                                                                                                                       

                                                                               

                                                                              

                                                                           

 

yes

yes

no

no

yes

yes

Force Acceleration• How much acceleration?

• Depends on:– AMOUNT OF FORCE

• MORE FORCE = MORE ACCELERATION • Acceleration is directly related to force

– MASS OF OBJECT• MORE MASS = LESS ACCELERATION• Acceleration is inversely related to mass

Newton’s Second Law“The acceleration of an object is directly proportional tothe net external force acting on the object and inversely

proportional to the mass of the object.”

mFa net

Unit of force is the NEWTON (N)

F

a

mFa net

m

a

Relationships: a ~ F; a ~ 1/m

• If mass is held constant, • doubling of the net force results in …

• a doubling of the acceleration, • halving of the net force results in …

• a halving of the acceleration. • If force is held constant,

• doubling of the mass results in …• a halving of the acceleration

• halving of the mass results in …• a doubling of the acceleration.

mFa net

Example• A 2 kilogram box is pushed with a net,

unbalanced force of 10 newtons.

• What is the acceleration experienced by the box?

a = Fnet / m

a = (10 N) / (2 kg)

a = 5 m/s2

The Big Misconception• The most common misconception is

one that dates back for ages; it is the idea that sustaining motion requires a continued force.

• Newton's laws declare loudly that a net force (an unbalanced force) causes an acceleration;

Are You Infected with the Misconception?• Two students discussing an object that is being acted

upon by two individual forces as shown. During the discussion, Anna Litical suggests to Noah Formula that the object under discussion could be moving.

• Noah Formula objects, arguing that the object could not have any horizontal motion if there are only vertical forces acting upon it.

• Who do you agree with?

FrictionA force that causes surfaces to stick together

and opposes motion.

At the MICROSCOPIC level, most surfaces are very BUMPY and IRREGULAR

Ways to minimize friction

SMOOTH SURFACESLUBRICATION

Coefficient of Friction (μ)

• How much materials STICK TOGETHER

– DIMENSIONLESS (no units)

– The greater the coefficient, the greater the tendency to STICK TOGETHER

– The coefficient is lowered if surfaces are SLIDING past each other

Friction Force• Static Friction

– STATIONARY OBJECTS – cancels out applied force - KEEPS OBJECTS IN PLACE

– CAN CHANGE – increases as the applied force increases until it reaches the maximum quantity for that specific surface.

– ROLLING OBJECTS

• Kinetic Friction– SLIDING OBJECTS– OPPOSES MOTION

Calculating Friction Force

• Amount of friction depends on:– Coefficient of friction

• Static – the object is motionless, rolling, or pushing off from a surface

• Kinetic – the object is sliding across a surface

– Normal Force• Greater normal force HIGHER friction force

Nf FF

Kinetic versus Static Friction• kinetic friction results

when an object moves across a surface.

Ffrict = μ • Fnorm

• The symbol μ represents the coefficient of kinetic friction between the two surfaces. The coefficient value is dependent primarily upon the nature of the surfaces that are in contact with each other. It does not depends on area of contact, the angle of the area, or the temperature, etc.

• Static friction results when the surfaces of two objects are at rest relative to one another and a force exists on one of the objects to set it into motion relative to the other object.

• The static friction force balances the force that you exert on the box such that the stationary box remains at rest.

Ffrict-static ≤ μfrict-static• Fnorm

Finding the unknowns

• Fnet is the vector sum of all the individual forces. The three major equations that will be useful are

– Fnet = m•a,

– Fg = m•g,

– Ff = μ•FN

Example #1• A man pushes a 50 kilogram crate across a

frictionless surface with a constant force of 100 Newtons.

Draw a free-body diagram of the crate.What is the weight of the crate?What is the normal force that pushes on the crate?What is the net force on the crate?What is the crate’s acceleration?

FA

FN

Fg

Fg = mgFg = (50 kg)(9.81 m/s2)

Fg = 490.5 N

FN = Fg

FN = 490.5 N

Fnet will only bethe 100N horizontal

force

a = Fnet / ma = (100 N) / (50 kg)

a = 2 m/s2

Example #2• A horse pulls a 500 kilogram sled with a constant

force of 3,000 Newtons. The force of friction between the sled and the ground is 500 Newtons.

Draw a free-body diagram of the sled.What is the weight of the sled?What is the normal force that pushes on the sled?What is the net force on the sled?What is the sled’s acceleration?

FA

FN

Fg

Ff

Fg = mgFg = (500 kg)(9.81 m/s2)

Fg = 4905 N

FN = Fg

FN = 4905 N

Fnet = ΣFx

Fnet = 3000 N – 500 NFnet = 2500 N

a = Fnet / ma = (2500 N) / (500 kg)

a = 5 m/s2

the object is moving horizontally. Use the diagram to determine the normal force, the net force, the mass, and the acceleration of the object.

8 kg

80 N

40 N right

5 m/s2 right

Example #3

Example #4• Edwardo applies a 4.25-N rightward force to a 0.765-kg

book to accelerate it across a tabletop. The coefficient of friction between the book and the tabletop is 0.410. Determine the acceleration of the book.

   

Example #5•  Lee Mealone is sledding with his friends when he

becomes disgruntled by one of his friend's comments. He exerts a rightward force of 9.13 N on his 4.68-kg sled to accelerate it across the snow. If the acceleration of the sled is 0.815 m/s/s, then what is the coefficient of friction between the sled and the snow?

Free Fall and Air ResistanceFree Fall

• Objects that are said to be undergoing free fall, are

• not encountering air resistance;

• falling under the sole influence of gravity. All objects will fall with the same rate of acceleration, regardless of their mass. This is due to that the acceleration is The ratio of force to mass (Fnet/m)

Falling with air resistance• As an object falls through air, it

usually encounters some degree of air resistance - the result of collisions of the object's leading surface with air molecules.

• The two most common factors that have a direct affect upon the amount of air resistance are– the speed of the object: Increased

speeds result in an increased amount of air resistance.

– the cross-sectional area of the object: Increased cross-sectional areas result in an increased amount of air resistance.

• As an object falls, it picks up speed. The increase in speed leads to an increase in the amount of air resistance. Eventually, the force of air resistance becomes large enough to balances the force of gravity. At this instant in time, the net force is 0 Newton; the object will stop accelerating. The object is said to have reached a terminal velocity.

Falling with air resistance – terminal velocity

Newton's Third Law• For every action, there is an equal and

opposite reaction. • Forces always come in pairs - equal and

opposite action-reaction force pairs. • Examples:

– The propulsion of a fish through the water. – The flying motion of birds. – The motion of a car on the way to school.

Third Law Examples• A firefighter directs a stream of water

from a hose to the east. In what direction is the force on the hose?

• A man getting out of a rowboat jumps north onto the dock. What happens to the boat?

There will be a force on the hose to the WEST

The boat will move to the SOUTH

Identifying Action and Reaction Force Pairs

• Identifying and describing action-reaction force pairs is a simple matter of identifying the two interacting objects and making two statements describing who is pushing on whom and in what direction.

Action/reaction forces vs. equilibrium forces

• Equilibrium forces act on same object

• Action and reactions force act on different objects

Force on the ground

Force on the car FN

Fg

Check Your Understanding1. While driving down the road, a firefly strikes the

windshield of a bus and makes a quite obvious mess in front of the face of the driver. This is a clear case of Newton's third law of motion. The firefly hit the bus and the bus hits the firefly. Which of the two forces is greater: the force on the firefly or the force on the bus?

2. For years, space travel was believed to be impossible because there was nothing that rockets could push off of in space in order to provide the propulsion necessary to accelerate. This inability of a rocket to provide propulsion is because ...

a. ... space is void of air so the rockets have nothing to push off of.

b. ... gravity is absent in space.c. ... space is void of air and so there is no air

resistance in space.d. ... nonsense! Rockets do accelerate in space

and have been able to do so for a long time.

3. Many people are familiar with the fact that a rifle recoils when fired. This recoil is the result of action-reaction force pairs. A gunpowder explosion creates hot gases that expand outward allowing the rifle to push forward on the bullet. Consistent with Newton's third law of motion, the bullet pushes backwards upon the rifle. The acceleration of the recoiling rifle is ...

a. greater than the acceleration of the bullet. b. smaller than the acceleration of the bullet.c. the same size as the acceleration of the bullet. 

Objectives: Forces in Two Dimensions

1. Net Force Problems Revisited 2. Equilibrium and Static3. Inclined Planes

• When forces acting at angles to the horizontal, Newton’s 2nd law still applies:

• Force is a vector quantity. Adding forces in 2 dimensions follows the rules for adding vectors.

∑F = ma

• The two ways for adding vectors are:1. Graphically - Head and tail method2. Mathematically: Add forces by components and

Pythagorean Theorem to determine magnitude and tangent function to determine direction

Net Force Problems Revisited

Determine the Fnet graphically

1. Resolve the vectors at an angle into x and y components.

2. Add all the x components together3. Add all the y components together4. Use Pythagorean Theorem to find the

resultant (hypotenuse)5. Resultant2 = x2 + y2

6. Use trigonometric function to determine the direction: tanθ = opp / adj

Determine the Fnet mathematically

Rx = Ax + Bx + Cx + Dx + Ex

Rx = -14N + 21N + 25N = 32N

A

B C

D

E

Ry = Ay + By + Cy + Dy + Ey

Ry = -14N + 20N + 21N -50 N = -23N

R2 = Rx

2+ Ry2

R = 39.4 N

θ = tan-1(-23/32) = -36o

Determine the Fnet mathematicallyAx = 20cos(225o) = -14 N

Ay = 20sin(225o) = -14 N

Cx = 30cos(45o) = 21 N

Cy = 30sin(45o) = 21 N-14 N

-14 N

21 N

21 N

Example - Pulling on an Angle

30˚

Fg

FN

This applied force (FA)can be broken into

COMPONENTS

FA

FAX

FAY

A block is pushed along a frictionless, horizontal surface with a force of 100 newtons at an angle of

30° above horizontal.

The total vertical force mustbe 0, so

Ry = FN + FAY –Fg = 0FN = Fg – FAY

R = Rx = Fax

Acceleration depends only onFAX

X Y

FAX FAY

Fg

FN

Total = FAX Total = 0

FAx = 100cos(30o) = 87 N

FAy = 100sin(30o) = 50 N

Example • A man pulls a 40 kilogram crate across a

smooth, frictionless floor with a force of 20 N that is 45˚ above horizontal.

What is the net force on the sled?

What is the crate’s acceleration?

Fnet = FA cos θFnet = (20 N)(cos 45°)

Fnet = 14.14 N

a = Fnet / ma = (14.14 N) / (40 kg)

a = 0.35 m/s2

How could the acceleration be increased?

Pushing at a smaller angle will make Fnet greater andtherefore increase acceleration.

Pushing on an Angle

-30˚

Fg

FN

FA

FAX

FAY

This applied force (FA)can be broken into

COMPONENTS

A block is pushed along a frictionless, horizontal surface with a force of 100 newtons at an angle of

30° below horizontal.

The total vertical force mustbe 0, so

FN = Fg + FAY

Acceleration depends only onFAX

X Y

FAX FAY

Fg

FN

Total = FAX Total = 0

Example• A girl pushes a 30 kilogram lawnmower

with a force of 15 Newtons at an angle of 60˚ below horizontal.

Assuming there is no friction, what is the acceleration of the lawnmower?

What could she do to reduce her acceleration?

Fnet = FA cos θFnet = (15 N)(cos 60°)

Fnet = 7.5 N

a = Fnet / ma = (7.5 N) / (30 kg)

a = 0.25 m/s2

Push at an greater angle

Example – find acceleration

• The vertical forces are balanced (Fgrav, Fy, and Fnorm add up to 0 N),• The horizontal forces add up to 29.3 N, right • The net force is 29.3 N, right • a = Fnet / m = 29.3 N / 10 kg = 2.93 m/s2, right

Determine the net force and acceleration

• Fnet = 69.9 N, right • m = (Fgrav / g) = 20 kg • a = (69.9 N) / (20 kg) =3.50 m/s/s, right

Equilibrium and Static• When all the forces that act upon an object are

balanced, then the object is said to be in a state of equilibrium.

• An object at equilibrium is either ...– at rest and staying at rest, or – in motion and continuing in motion with the same

speed and direction.

• "static equilibrium." refers to an object at rest

• A frame is shown with the given tension. Determine the weight of the frame.

Example

Rx = Ax + Bx + Cx = 0

Ax = 50cos(150o) = -43 N

Bx = 50cos(30o) = 43 N

Cx = Rx - Ax - Bx = 0

Ry = Ay + By + Cy = 0

Ay = 50sin(150o) = 25 N

By = 50sin(30o) = 25 N

Cy = Ry - Ay - By = -50 N

A B

C = ?

30o

C2 = Cx

2+ Cy2

R = 50. N

example• A sign is shown with the given mass of 5 kg.

Determine the tension of each cable.

C = Fg

40o 40o

Fg = Tsin40o + Tsin140o

(5 kg)(9.81 m/s2) = 1.286T

T = 38 N

A = T B = T

Tsin40oTsin140o

Tcos40oTcos140o

An important principle• As the angle with the horizontal increases, the amount of

tensional force required to hold the sign at equilibrium decreases.

Fg = 10 N

Down the Slope• A tool used to move objects from one height to

another.• Allows for the movement of an object without

lifting it directly against gravity.

• The object accelerate downward due to the component gravity that is parallel to the plane.

Fg on Inclined Plane

Calculations

• Consider forces:– Perpendicular

• F┴ = Fg cos θ

• Cancel out Normal (FN )

– Parallel• F// = Fg sin θ

• Could be in the same or opposite of Friction (Ff )

Tilt you head method

Essential Knowledge• What happens to the component of weight that is

perpendicular to the plane as the angle is increased?Decreases – Fg perpendicular

• What happens to the component of weight that points ALONG the plane as the angle is increased?

Increases – Fg parallel• What happens to the normal force as the angle is

increased?Decreases – depends on Fg perpendicular

• What happens to the friction force as the angle is increased?

Decreases – depends on normal force

• The net force is the vector sum of all the forces. – All the perpendicular components (including

the normal force) add to 0 N. – All the parallel components (including the

friction force) add together to yield the net force. Which should directed along the incline.

Fnet = F//

mgsinθ = ma

a = gsinθ

In the absence of friction

Fnet = 0

Object is at equilibrium – at rest or moving with constant velocity

Ff

Horizontal:F// = Ff

mgsinθ = μFN

mgsinθ = μ∙mgcosθtanθ = μ

Vertical:

F┴ = FN

mgcosθ = FN

Example

• What is the magnitude of the normal force?FN = Fg perpendicular = Fg cos θ = 43.3 N

• If the box is sliding with a constant velocity, what is the magnitude of the friction force?

Ff = Fg parallel = Fg sin θ = 25 N

Fg = 50N

30°

example• The free-body diagram shows the forces acting upon a 100-kg

crate that is sliding down an inclined plane. The plane is inclined at an angle of 30 degrees. The coefficient of friction between the crate and the incline is 0.3. Determine the net force and acceleration of the crate.

In perpendicular direction:Fnorm = F┴ = 850 N

In parallel direction:Fnet = F// - Ff

Fnet = 500 N - µFnorm

Fnet = 235 Na = Fnet / m = 2.35 m/s2

F┴ = Fgrav∙cos30o = 850 NF// = Fgrav∙sin30o = 500 N

practice

Double Trouble (a.k.a., Two Body Problems)

• Two body-problems can typically be approached using one of two basic approaches. – One approach is the system analysis, the two

objects are considered to be a single object moving (or accelerating) together as a whole.

– Another approach is the individual object analysis, either one of the two objects is isolated and considered as a separate, independent object.

Example - system analysis• A 5.0-kg and a 10.0-kg box are touching each other. A

45.0-N horizontal force is applied to the 5.0-kg box in order to accelerate both boxes across the floor. Ignore friction forces and determine the acceleration of the boxes and the force acting between the boxes.

m = 15 kg

Fnet = 45 N

a = Fnet / m = 3 m/s2

Example - individual analysis

5a = 45 – 10a a = 3 m/s2

In vertical direction: FN = Fg = (5 kg) (9.81 m/s2) = 49 N

In horizontal direction: Fnet = Fapp - Fcontact

(5 kg)a = 45 N - Fcontact

In vertical direction: FN = Fg = (10 kg) (9.81 m/s2) = 98 N

In horizontal direction: Fnet = Fcontact

(10 kg)∙a = Fcontact

Example: system analysis • A 5.0-kg and a 10.0-kg box are touching each other. A 45.0-

N horizontal force is applied to the 5.0-kg box in order to accelerate both boxes across the floor. The coefficient of kinetic friction is 0.200. Determine the acceleration and the contact force. In vertical direction:

FN = Fg = (15 kg) (9.81 m/s2) = 147 N

a = Fnet / m = (15.6 N/15.0 kg) = 1.04 m/s2

In horizontal direction: Fnet = Fapp - Ffrict = 45 N - μ•Fnorm Fnet = 15.6 N

However, in order to find the contact force between the objects, we must make individual analysis.

In vertical direction: FN = Fg = (10 kg) (9.81 m/s2) = 98 N

In horizontal direction: Fnet = Fcontact - Ff

(10 kg)∙(1.04 m/s2) = Fcontact - μ•Fnorm

Example: individual analysis

10.4 = Fcontact – (0.2)(9.8)Fcontact = 8.44 N