Chapter 4Forces and Newton’s Laws of

Motion

Why things move the way the do.

Newton’s First Law of Motion

“Maintaining the status quo”

Newton’s First Law of Motion“The Law of Inertia”

“An object at rest will remain at rest and an object in motion will continue to move at a constant velocity, unless acted upon by a net external force.”

The property of an object that causes it to resist changes in its motion is called inertia.

Inertia is proportional to an object’s mass.

Inertia causes motion at a constant velocity.

Fnet =0→ a =0→ constant velocity

Aristotle’s acceptance of a stationary earth…not spinning, was in part due to his lack of a knowledge or understanding of inertia.

In the third century BC the Greek philosopher Aristotle developed a model of the structure and motion of the universe.

In Aristotle’s world it was assumed that the earth was stationary, did not spin or move through space, and was at the center of the universe.

Newton’s Second Law of Motion

“How is Acceleration related to Force?”

Newton’s Second Law of MotionThe law of accelerated motion

“The acceleration of an object is directly proportional to and in the same direction as the net force acting on it and inversely proportional to the object’s mass.”

r a =

r Fnetm

r F net =m⋅

r a

Units of Force :

kg⋅ms2

=kg⋅ms2

=Newton,N

Force is a vector and uses the same sign convention as velocity and acceleration.

⇒ effect on

r v

parallel magnitude increases

opposite magnitude decreases

⎧ ⎨ ⎩

direction of r F net

⇒ direction of r a

r F net =

r F1 +

r F2 +

r F3L

r F 1 ,

r F 2 ,

r F 3 are + or −

.based on their directions

SummaryConstant Velocity

Variables: Relationships:

Constant Acceleration

Variables: Relationships:Caused by Inertia

Fnet

m r F net

m r F net =0

r a =0

Caused by Net Force

r F net =

r Fi∑

a = r Fnetm

constant

s

v

t

s =vt

v=st

t= sv

s

v

vi

vf

a

t

v = st v=

vi +vf2

s= vt t= s v

a=vf −vi

tvf =vi +at

s=vit+12 at2

vf2 =vi

2 + 2as

Describing Forces

All forces between objects can be placed into two broad categories:

Contact forcesAction-at-a-distance forces

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

Frictional ForceTensional ForceAir Resistance ForceSpring ForceBuoyant Force

Action-at-a-distance forces are types of forces in which the two interacting objects are not in physical contact with each other, yet are able to exert a push or pull despite a physical separation.

Gravitational ForceElectric ForceMagnetic Force

Types of ForcesGravitational Force and Weight

Friction

Fluid Resistance

Buoyant Force

Normal Force

Tension

Newton’s Law of Universal Gravitation:

The Gravitational Force

“Every mass in the universe exerts a gravitational force on every other mass. The gravitational force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.”

FG =Gm1m2

r1,22

G = Universal Gravitational Constant

G=6.673×10−11 N⋅m2

kg2

Mass and WeightMass is a measure of the amount of matter in an object and is related to the object’s inertia.

Weight is a measure of the gravitational force acting on an object.

F=m⋅a↓w

↓m= g

↓⋅

The weight of an object equals the product of its mass and the acceleration due to gravity.

force of gravityweight, w

acceleration due to gravity, g

Comparing the equation for weight with the equation for the gravitational force gives:

Gm1m2

r1,22 =m1g⇒ g=G

m2

r1,22

Where :

m2 = mass of the earth

r1,2 = distance of object from

the Earth' s center

r1,2 =rearth+altitude

The Normal ForceWhen two surfaces are in contact, each surface exerts a force on the other. The component of this force which is perpendicular to the surface is called the normal force, FN.

ww

FN FN

FrictionThe force that opposes.

Friction is a force whose direction is ALWAYS opposite to the direction an object is moving or would tend to move if there were no friction.

Friction is force between two surfaces:1. the surface of the object and2. the surface on which it is moving

Friction depends on the characteristics of the two surfaces and the force pressing them together.

Static friction exists between a stationary object and the surface on which it rests.

Friction depends on whether the object is moving or stationary.

Static friction must be overcome before an object can begin moving.

Kinetic friction exists between a moving object and the surface on which it is moving.

Kinetic friction is always opposite to the object’s velocity.

Kinetic friction is usually less than static friction.

Static friction is always opposite to the direction the object would move if there was no friction.

Static friction equals the net applied force up to its maximum value which depends on the mass of the object and the properties of the two surfaces.

μ static > μkinetic

fstatic,max =μstatic⋅FN

fkinetic =μkinetic⋅FN

Consider an object sitting on a stationary horizontal surface.

w

FN

If the object remains in contact with the surface, the acceleration in the vertical, y, direction must be zero.

FN +w=0

FN =−w=−mg=m g

f =μmg

fstatic,max =μsmg

fk =μkm g

Consider an object sitting on a stationary inclined surface.

x

yFN

wθ

θ

wy'

wx'

wy' =wcosθ =mgcosθ

wx' = w sinθ =mg sinθ

If the object remains in contact with the surface, the acceleration in the vertical, y’, direction must be zero.

FN +mgcosθ =0

FN =−mgcosθ=m g cosθ

f =μmg cosθ

FN

wθ

θ

wy'

wx'

The x component of the weight, wx’ tends to cause the object to slide down the incline.The frictional force, f, will be in a direction to oppose this motion.

f

The net force on the object is then:

Fnet,x =wx −f

FN

wθ

θ

wy'

wx'

wx'

Fnet,x' =mg sinθ−μm g cosθ

Fnet,x' =mg sinθ−μcosθ( )

ax' =Fnet,x'

m

ax' = g sinθ−μcosθ( )

The acceleration is then:

For an object to begin sliding down an incline:

wx' ≥fs,max

m g sin θmin( ) ≥μsm g cosθmin( )

sin θmin( ) ≥μscosθmin( )

sin θmin( )cosθmin( )

≥μs

tan θmin( )≥μs

θmin ≥ tan−1 μs

Tension ForceTension, T, is a “pulling” force applied to an object by wire or rope.

Tension always acts along the wire or rope in a direction away from the object on which the tension is applied.

T

When two objects are connected by a wire or rope the tension is the same at each end.

T T

Consider an object with mass m1 sitting on a table and connected by a wire which passes over a pulley to a second mass m2 hanging beside the table.

m 1

m 2

Forces Acting on m1:Weight = m1g

Normal Force, FN

Tension, T

Friction, f

m1g

FN Tf

Forces Acting on m2:

Weight = m2g

m2gTension, T

T

If released m2 will tend to fall pulling m1 to the right.

a2

a1

Newton’s Second Law for m1:

Y-DirectionAssuming object remains on the table.

FN +m1g=0FN =−m1g=m1 g

X-Direction

Newton’s Second Law for m2:

Since m1 and m2 are connected by a wire the magnitudes of their accelerations must be equal. The directions of the accelerations may not be the same.

m1 tends to accelerate to the right (+)

m2 tends to accelerate down (-)

a2 =−a1

T −f =m1a1

T =m1a1 +μm1 g

T −μm1 g =m1a1

T −m2g=m2a2

T =m2a2 +m2g

T =m1a1 +μm1 g

T =m2a2 +m2gSubstitute a2 = -a1

T =m1a1 +μm1 g

T =−m2a1 +m2g

m1a1 +μm1g=−m2a1 +m2g

m1a1 +m2a1 =m2g−μm1g

a1 m1 +m2( ) =m2g−μm1g

a1 =m2g−μm1g

m1 +m2

m2g−μm1g⇒ net force on system

m1 +m2 ⇒ total mass of system

m1 m2 m2g

μm1g

T = m1a1 +μm1g

T = m1 a1 +μg( )

T = m1m2g−μm1g

m1 +m2+μg

⎛ ⎝ ⎜ ⎞

⎠ ⎟

T = m1gm2 −μm1m1 +m2

+μ ⎛ ⎝ ⎜ ⎞

⎠ ⎟

T = m1gm2 −μm1 +μm1 +μm2

m1 +m2

⎛ ⎝ ⎜ ⎞

⎠ ⎟

T = m1gm2 +μm2m1 +m2

⎛ ⎝ ⎜ ⎞

⎠ ⎟

T = m1gm2 1+μ( )m1 +m2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

T = m1g1+μ( )m1m2

+1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Summary

a1 =m2g−μm1g

m1 +m2

a2 =−a1

T = m1g1+μ( )m1m2

+1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Since m1 can not accelerate to the left, a1 > 0, therefore for the system to move

m2g > μsm1g orm2 > μsm1

Fluid ResistanceFluid (liquid or gas) resistance is a frictional force that an object experiences as it moves through a fluid (e.g. air or water).

Fluid resistance depends on the size and shape of the object, density of the fluid, and is directly proportional to the object’s velocity and in the opposite direction.

Example

An object falling through air experiences an upward force of air resistance in addition to the downward force of gravity.

gravitational force

air resistance

r f fluid ∝−

r v

Initially air resistance is small because the velocity is small.air resistance ∝velocity

As the magnitude of the downward velocity increases the force of air resistance increases.

gravitational force > air resistance

net force = gravitational force - air resistanceObject accelerates downward magnitude of

velocity increases

If the object falls far enough the force of air resistance becomes equal to the gravitational force.

At this point:

gravitational force = air resistance

net force, Fnet = 0

a = 0

velocity is constant

This constant velocity reached by an object falling through a fluid is called its terminal velocity.

Buoyant Force

An object immersed in a fluid (liquid or gas) experiences an upward buoyant force.

The buoyant force depends on the volume of the object immersed in the fluid (volume of fluid displaced) and the density of the fluid.

Consider an object dropped into a tank containing a fluid (e.g. oil)

Buoyant Force Only

Buoyant Force & Fluid Resistance

buoyant forcer F B =ρfluid⋅g⋅Vdisplaced

Buoyant Force with Fluid Resistance

The mass of the object is 10kg and the buoyant force is 190N.

If the object starts from rest at a height of 63.5m above the surface of the oil how long will it take for it to reach a depth of 60m in the oil? (Ignore any buoyant force due to air and any fluid resistance.)

Part One of the Motion

Falling from rest a distance of 63.5m to the surface of the oil.

The only force acting on the object is the gravitational force…this part of the motion is Free-Fall.

Part Two of the Motion

Falling a distance of 60m from the surface of the oil.

There are two forces acting on the object:•The gravitational force = mg•The buoyant force = 190N

The object’s velocity as it enters the oil = -35.3m/s

Variables: Relationships:

v =dt v =

vi + vf2

d = v ⋅t t = dv

a =vf −vi

tvf =vi + a⋅t

d =vi ⋅t+12 a⋅t2 v f

2 =vi2 + 2ad

d

v

v i

v f

a

t

m

Fnet

r F net =

r Fi∑

a = r Fnetm

=−63.5m

=0

=10kgFind t=3.6s

find vf=−35.3 ms

=−98N

=−9.8 ms2

=FG = w = mg

=(10kg) × (−9.8 ms2 )

=−98N

=−98N10kg

=−9.8 ms2

d =vi ⋅t+12 a⋅t2

Substituting values w/o units.

−63.5 = 0 ⋅t + 12 (−9.8 m

s2 )t2

−63.5 = −4.9t2

t2 =−63.5−4.9

t = −63.5−4.9 =3.6s

vf =vi + a⋅t

vf =0 +(−9.8 ms2

)⋅(3.6s) =−35.3 ms

Variables: Relationships:

v =dt v =

vi + vf2

d = v ⋅t t = dv

a =vf −vi

tvf =vi + a⋅t

d =vi ⋅t+12 a⋅t2 v f

2 =vi2 + 2ad

d

v

v i

v f

a

t

m

Fnet

r F net =

r Fi∑

a = r Fnetm

=10kgFind t

=−60m

=−35.3 ms

=FG + FB

=−98N + 190N=92N

= 92N10kg

=9.2 ms2

=92N

=9.2 ms2

d =vi ⋅t+12 a⋅t2

Substituting values w/o units.

−60 = −35.3 ⋅t + 12 (9.2)t2

−60 = −35.3 ⋅t + 4.6 ⋅t2

Writing in the General Quadratic Form

4.6⋅t2 −35.3⋅t+60 =0

Comparing to ax2 +bx+ c=0 gives:a = 4.6

b = -35.3

c = 60

Substituting into the Quadratic Formula

x = -b± b2 −4⋅a⋅c2a gives:

t =-(-35.3)± (−35.3)2 −4⋅(4.6)⋅(60)

2(4.6)

Simplifying gives:

t = 35.3±11.929.2

t1 = 35.3 +11.929.2 =5.1s

t2 = 35.3−11.929.2

=2.6s

The time for the object to reach the surface of the oil was 3.6s.

Therefore the object is at a depth of 60m in the oil at two times.

t1 =3.6s+ 5.1s=8.7s

t2 =3.6s+ 2.6s=6.2s

What is the significance of the two times?

What is the maximum depth the object will reachAnd when does it reach this maximum depth?

Variables: Relationships:

v =dt v =

vi + vf2

d = v ⋅t t = dv

a =vf −vi

tvf =vi + a⋅t

d =vi ⋅t+12 a⋅t2 v f

2 =vi2 + 2ad

d

v

v i

v f

a

t

m

Fnet

r F net =

r Fi∑

a = r Fnetm

=10kg

=−35.3 ms

=92N

=9.2 ms2

=0

Find “d”=−67.7m

Find “t”

t =vf −vi

a

v f2 =vi

2 + 2ad

d =vf2 −vi

2

2a =0 − (−35.3 m

s )2

2 ⋅(9.2 ms2 )

=−67.7m

t =vf −vi

a =0 − (−35.3 m

s )

9.2 ms2

=3.84s

The object reaches the surface in 3.6s.

The object is at a depth of 60m moving downward in 6.2s

The object reaches its maximum depth of 67.7m in 7.44s.

Time to reach maximum depth:

3.84s+ 3.6s=7.44s

The object is at a depth of 60m moving upward in 8.7s.

Newton’s Third Law of Motion

“Forces always come in pairs.”

Newton’s Third Law of Motion“Action and Reaction”

“If object A exerts a force on object B, object B must exert an equal and opposite force on object A.”

“For every action there is an equal and opposite reaction.”

Why do the two forces (action and reaction) not cancel each other?

They act on DIFFERENT objects!

r F A→ B =−

r FB→ A

Newton’s Third Law says that action and reaction forces are always equal and opposite.

It does NOT say that their effects are equal!

Consider two objects moving along a horizontal track.

FA→ B ⇒ aB depends on mB

aB =FA→ BmB

FB→ A ⇒ aA depends on mA

aA =FB→ AmA