Newton’s 1st Law of InertiaNewton’s 1st Law of InertiaAny object continues in its state of rest or in its uniform velocity unless it is made to change thatstate by an unbalanced force is acting upon it.

An object does not accelerate itself and it wants to retain a state of zero acceleration.

Every object possesses inertia and the amountdepends on the amount of matter or mass.

The greater the mass, the greater the inertia

or resistance to acceleration. Mass is a measure of the inertia.

Mass is the amount of matter contained in anobject.

Mass is a scalar quantity meaning it has magnitude only. Mass is measured in g, kg, or slugs. For nonrelativistic speeds, the mass of an object remains constant.

Do not use mass and weight interchangeably!

Weight is a measure of the gravitational attraction between an object and the earth.

Weight is a vector quantity because it has both magnitude and direction (the direction is always assumed to be towards the center of the earth).

The weight of an object varies with location as it is dependent on the distance from the center of the earth.

Newton’s 2nd Law of AccelerationNewton’s 2nd Law of AccelerationThe acceleration of an object is directlyproportional to the net force acting on the

objectand is inversely proportional to its mass.

Inertia is the tendency to resist changes inmotion and Newton’s 2nd law expresses thismathematically.

a is directly proportional to the Fnet. By whatever factor Fnet changes, a changes by the same factor.

a α Fnet.

If you double the force, you double the acceleration.

If you decrease the Fnet by 1/3, you decrease the acceleration by 1/3.

A graph of Acceleration vs Force would be a

straight line passing through the origin.

a is inversely proportional to m.

a α 1/m

If you double the mass, the acceleration is reduced by ½.

If you decrease the mass by a factor of 1/3, you would triple the acceleration.

A graph of Acceleration vs Mass would be a hyperbola.

An object always accelerates in the direction of

the net force.

If the net force is applied in the direction of

the object’s motion (velocity), the object accelerates positively (speeds up).

If the net force is applied in the opposite direction of the object motion (velocity),

the object decelerates.

Mathematically, Newton’s 2nd Law is

Fnet = ma

where m is the mass of the object in kg, a is

the acceleration in m/s2, and Fnet is the net force in N.

We now can formally define 1 N of force.

If you have a mass of 1.0 kg and the net force causes it to accelerate at 1.0 m/s/s, then it is by definition 1.0 N of force.

Which Is It?Which Is It?Which is true?

a = Δv/Δt or a = Fnet/m? Acceleration was previously defined to

be the rate at which the velocity changes. Now we are defining acceleration to be Fnet/m. Why the difference?

Both are true!

Previously, we looked at kinematics or how do we

describe motion?

Do objects move at a constant velocity or a constant acceleration?

Newton’s Laws describe the dynamics or why doobjects move as they do?

Is the net force equal to greater than zero?

Newton’s 2Newton’s 2ndnd Law Problems Law ProblemsA car traveling at 32 m/s slows down to a stopand travels a distance of 52 m. If the mass ofthe car is 1375 kg, what net force acted on

thecar?

vi = 32 m/s vf = 0m = 1375 kg Δx = 52 m

vf2 = vi

2 + 2aΔx

0 = (32 m/s)2 + 2 × a × 52 m

a = -9.8 m/s2

Fnet = ma = 1375 kg × -9.8 m/s2 = -1.4 × 104 N

The negative values for a and F make sensebecause the car decelerated in coming to a

stoprequiring a force in the opposite direction to

itsmotion.

A stone weighs 7.4 N. What force must beapplied to make it accelerate upward at 4.2

m/s2?

Fw = 7.4 N g = 9.80 m/s2

a = 4.2 m/s2

•Fw

FT FT = Fnet + Fw

Fnet = ma

Fw = mg

m = 7.4 N/9.80 m/s2 = 0.76 kg

Fnet = 0.76 kg × 4.2 m/s2 = 3.2 NFT = 3.2 N + 7.4 N = 11 N

Some notes from the previous problem:

If the acceleration of the stone is upward,

then the Fnet must also act upwards.

This implies that FT > Fw because the rope

must provide the total force to support the

weight of the object and also provide the net

force.

Free FallFree FallFree fall exists when an object’s weight is theonly force acting on it (straight down, towardsthe center of the earth).

In the absence air resistance, all objectsaccelerate at the same rate.

a = Fnet/m = Fw/m

a = g = 9.80 m/s2 = 980 cm/s2 = 32 ft/s2

Air resistance can usually be ignored for smalldense objects that travel short distances but

therecan be exceptions:

Every baseball fan has heard the expression

that the ball was headed out but the wind knocked it down.

A ping pong ball will never be confused with

a small dense object.

Sometimes air resistance is not what you want:

When throwing a football for distance, a tight

spiral minimizes air resistance.

Long range rifles have grooves in the barrel so the bullets come out spiraling.

If either the football or the bullet started to topple end over end, well …

The reason will be explained in another set of slides discussing angular momentum.

Sometimes you want air resistance.

Just ask any parachutist.

Air resistance depends on both velocity and

surface area.

a = g = Fnet/m = Fw – R/m

At t = 0, R = 0.

As an object accelerates downward, R increases.

a = g = Fnet/m = Fw – R/m

At t = 0, R = 0.

As an object accelerates downward, R increases.

When Fw = R, Fnet = Fw – FR = 0, and a = 0.

When the acceleration equals zero, the object is said to be moving with a terminal velocity.

Two brothers, Pete and Repeat, jump from the

same helicopter and their parachutes are initially

opened. The parachutes are the same size and

Pete weighs 500 N and Repeat weighs 450 N.

Who hits the ground first?

True Weight vs Apparent WeightTrue Weight vs Apparent WeightA man stands on a bathroom scale in anelevator. The scale reads 917 N.

(a) What is the man’s weight?

•Fup

Fw

Fup is the force that the bathroom scalepushes up on the man.

Fw = Fup = 917 N and the man appearsto weigh 917 N.

(b) What is the man’s mass?

Fw = 917 N g = 9.80 m/s2

Fw = mg

m = 917 N/9.80 m/s2 = 93.6 kg

(c) As the elevator moves up, the scale reading

increases to 1017 N. Determine the upward

acceleration of the elevator.

•Fup

Fw

Fnet = ma = Fup - Fw

Fnet = 1017 N – 917 N = 100. N

a = Fnet/m = 100. N/93.6 kg

a = 1.1 m/s2 straight up

(d) As the elevator approaches the 13th floor, the

scale reading decreases to 798 N. What is the

acceleration of the elevator?•

Fup

Fw

Fnet = ma = Fw - Fup

Fnet = 917 N – 798 N = 119 N

a = Fnet/m = 119 N/93.6 kg

a = 1.3 m/s2 straight down

(e) When the elevator reaches the 13th floor it

stops. After about 5 sec the man looks down

and the scale is reading 0. What is going on?•

Fw

Fnet = Fw = mg

The guy is in a heap of trouble ashe is in a state of free fall!

Thoughts To PonderThoughts To PonderIf the elevator was sound proof and there wasno visible connection to the outside world, thereis nothing the man could do to detect uniformmotion.

When the acceleration of a system is zero, there is no experiment that distinguishes between an object at rest (∑F = 0) or an object moving in a straight line at constant speed (∑F = 0).

There are no relativistic speeds involved, so

that the mass of the man remains constant.

Whenever there is an acceleration involved, the

net force will always be in the same direction as

the acceleration.

FrictionFrictionFriction is a force that resists the relative

motionof solid objects that are in contact with eachother.

If the solid is in a fluid (a liquid or a gas),

then it is called viscosity.

Friction is caused by uneven surfaces of touching objects.

Six Principles of FrictionSix Principles of FrictionFriction acts parallel to the surfaces that

are incontact and always opposes motion.

Friction depends on the nature or composition

of the solid surfaces in contact.

Rolling Friction < Sliding Friction < Starting

Friction

Sliding friction is practically independent of

surface area for a given object.

Sliding friction is practically independent of

medium speeds.

Sliding friction is directly proportional to the

force pressing the two surfaces together.

Coefficient of Friction Coefficient of Friction The formula for friction is given by

Ff = µFN

where Ff is the frictional force in N (newtons)and FN is the normal (perpendicular) forcepressing the two surfaces together.

The normal force, FN, will not always equal

the weight of the object!

µ (mu) is the coefficient of friction which isdetermined by what the two solid surfacesconsist of (glass on glass, wood on wood, etc.).

µ = Ff/FN is the ratio of the frictional force to the

normal force.

µs > µk

where µs is the coefficient of starting friction and µk is the coefficient of sliding friction.

Friction Problems Friction Problems A crate weighing 475 N is pulled along a

levelfloor at a uniform speed by a rope which

makesan angle of 30.0° with the floor. The

appliedforce on the rope is 232 N.

(a) Draw a free-body diagram of the box.

Fw

FvFf FH

FN Fθ

(b) Determine the coefficient of friction.

(c) How much force is needed to pull the box?

µ = Ff

FN= FH

FW - FV= F × cosθ

Fw - F × sinθ

µ = 232 N × 0.866475 N – 232 N × 0.500 =

0.560

F × cosθFH = =232 N × 0.866= 201 N

(d) Compare the force in (c) to the weight of the

box.

It is easier to pull the crate, 201 N, than it is

to lift the crate, 475 N.

An Inclined Plane Problem An Inclined Plane Problem A roller coaster reaches the top of a steep

hillwith a speed of 7.0 km/h. It then descends

thehill, which is at an angle of 45° and is 35.0 mlong. If µk is 0.12, what is the speed when itreaches the bottom?

Vi = 7.0 km/h Δx = 35.0 mθ = 45° = 0.12

µk

.

•

θ

FN

Fp

Ff

Fw

FNθ

Fnet = Fp - Ff = Fw sin θ - µk FN

Fnet = mg sinθ - µk mg cos θ

Fnet = ma

.

a = mgsinθ - µmgcosθm

a =9.80 m/s2 × 0.707 – 0.12 × 9.80 m/s2 × 0.707

a =6.10 m/s2

vi =7.0 kmh × 103 m

1 km

× 1 h3600 s =1.9 m/s

vf2 = vi

2 + 2a(x-xi)

vf2 = (1.9 m/s)2 + 2 × 6.10 m/s2 × 35.0 m

vf = 21 m/s

Another Inclined Plane Problem Another Inclined Plane Problem A block is given an initial speed of 4.2 m/s

up a24.0° inclined plane. Ignoring frictional

effects,calculate:

(a) How far up the inclined plane will the block

travel? •FN

FN

Fwθ

Fp θ

vi = 4.2 m/s θ = 24.0°µ = 0 vf = 0

Fnet = ma

a =Fnetm =

-Fpm =

-Fw sinθm = -mg sinθ

m

a =-9.80 m/s2 × 0.407 = - 3.99 m/s2

vf2 = vi

2 + 2a(x-xi)

0 = (4.2 m/s)2 + 2 × (- 3.99 m/s2) × Δx

Δx = 2.2 m

(b) How long does it take before the block returns to its starting point?

vf = vi + aΔt

0 = 4.2 m/s/-3.99 m/s2

Δt = 1.1 s ΔtT = 2 × 1.1 s = 2.2 s

Newton’s 3Newton’s 3rdrd Law Law When one object exerts a force on a secondobject, the second object exerts a force on

thefirst that is equal in magnitude but opposite

indirection.

These two forces are called an action-reaction pair of forces.

F1 = - F2

To apply Newton’s 3rd Law, you mustdistinguish between forces acting on an

objectand forces exerted by the object.

When using Newton’s 3rd Law, you must

have two different objects!

Examples of Newton’s 3Examples of Newton’s 3rdrd Law Law Consider a 10. N ball falling freely in a

vacuumwhere there is no air resistance.

What is the action force?

What is the reaction force?

The action force could be the earth pulling down

on the ball with a force of 10. N.

The reaction force would be the 10. N ballpulling up on the earth.

• Fa = Fw = 10. N

•Fr = Fw = 10. N

It is easy to see why the ball falls toward the

center of the earth.

From Newton’s 2nd Law:

ab = Fbmb

= Fwmb

= mbgmb

=g = 9.80 m/s2

straight down

It is easy to see why the earth remainsstationary.

ae = Fbme

= Fwme

= 10. N5.96 x 1024 kg

ae = 1.68 x 10-24 m/s2 straight up

At The Firing Range At The Firing Range What happens when you fire a long range

rifle?

The action force can be considered to be

the force the gun exerts on the bullet.

The reaction force would be the force the

bullet exerts on the gun.

The acceleration of the bullet and the gun

would be given by:

mg > mb, therefore, ab > ag.

This accounts for the “kickback” or recoil

velocity of the gun.

ab = Fbmb

ag = - Fbmg

Universal Law of Gravitation Universal Law of Gravitation The mutual force of attraction between twoobjects is directly proportional to the product oftheir masses and inversely proportional to thesquare of the distance between their centers.

F α m1 × m2

F α 1/d2

F α (m1 × m2)/d2

The proportionality sign can be replaced with an

equals sign and a constant.F = G × m1 × m2

r2

where G = 6.67 x 10-11 n•m2/kg2.

Newton’s Universal Law of Gravitation is anexample the inverse square law.

If you double the distance, the force decreases by the factor of ¼.

Universal Law of Gravitation Universal Law of Gravitation Problem Problem

What is the mutual force of attraction of a 1.0 kg

mass and the earth if the 1.0 kg mass is resting

on the ground?

m1 = 1.0 kg m2 = me = 5.96 x 1024 kg

re = 6.37 x 106 m G = 6.67 x 10-11 Nm2/kg2

F =G × m1 × m2

r2

.

F = G × m1 × m2

r2

F =6.67 × 10-11 Nm2/kg2 × 5.96 × 1024 kg × 1.0 kg

(6.37 × 106 m)2

F= 9.8 N

Everything Fits!Everything Fits! What is the acceleration due to gravity in

theprevious problem?

re = 6.37 x 106 m me = 5.96 x 1024 kg

G = 6.67 x 10-11 Nm2/kg2

F =G × m1 × m2

r2

.

Fw = G × mb × me

re2

mbg= G × mb × me

re2

g = G × mere

2

=6.67 x 10-11 Nm2/kg2 × 5.96 x 1024 kg(6.37 x 106 m)2

.g=9.80 m/s2

Sound familiar?

g is a constant for a given location!

Relation of Gravity to Weight Relation of Gravity to Weight Gravity describes the force of gravitationalattraction on or near the surface of a planet.

Objects at higher altitudes will weigh less than at sea level.

Masses weigh a little more at either pole than at

the equator.

Going inside the surface of the earth decreasesthe acceleration due to gravity.

Once below the surface of the earth, theattraction of the earth above the object causesthe object to weigh less.

What makes Newton’s Three Laws and theUniversal Law of Gravitation so beautiful is

thatthey work anywhere in the universe.

All of Newton’s Laws are mass dependent.

The only time they break down is at relativistic speeds.

Wrap Up QuestionsWrap Up QuestionsAssess the following statement:

When an object is at rest, there are no external

forces acting on it.

This statement is false because when an object

is at rest, there is no resultant force. The vector

sum of the forces, ΣF = 0.

Two boys pull on a 5.0 m rope each with ahorizontal force of 225 N. If each boy

increasestheir applied force by the same amount,

can therope ever be horizontal?

No, because of the weight of the rope. Nomatter how much force each boy exerts,

thereis no vertical force to cancel the weight of

therope.

You can reduce the force of friction (i.e. sanding

or polishing the surfaces) only so much, beforeit increases again. Why?

By smoothing the surfaces as much aspossible, the separation distance of the atomsor molecules decreases. This makes for astronger attraction.

If the two surfaces are the same material, theforce is cohesion, otherwise the force isadhesion.

Assuming the earth is a perfect sphere and itsmass is evenly distributed, how much would a225 N person weigh at the center of the earth?

The person would weigh 0 N. This answer canbe arrived at either qualitatively orquantitatively.

Qualitatively, the person would an experience aforce of attraction from all directions. But theattractive force would not be downward towardthe center of the earth but rather radially awayfrom the center of the earth.

Quantitatively, one could start with Newton’sUniversal Law of Gravitation.

This formula is applicable when you are on the

surface of earth or above it. However, once you

go below the surface of the earth, the formula

has to be modified.

F = Gm1 × m2

r2

Using the assumptions given in the question,

the mass of the earth is given by:

me = De × Ve = De × 4/3 × π × re3

F = Gmp × De × 4/3 × π × re

3

re2

F = G ×mp × De × 4/3 × π × rwhere mp is the mass of the person and r is thedistance from the center of the earth.