Physics II, Pg 1

Physics II Physics II

Today’s AgendaToday’s Agenda

Newton’s 3 laws. How and why do objects move? DynamicsDynamics.

Look at Textbook problemsLook at Textbook problems

Physics II, Pg 2

Sir Issac Newton

Physics II, Pg 3

DynamicsDynamics

Issac Newton (1643-1727) published Principia Mathematica in 1687. In this work, he proposed three “laws” of motion:

Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame.

Law 2: For any object, FFNET = FF = maa

Law 3: Forces occur in pairs: FFA ,B = - FFB ,A.

See text: 5-1 and 5-2

Physics II, Pg 4

Newton’s First LawNewton’s First Law An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frameinertial reference frame.

If no forces act, there is no acceleration.

The above statement can be thought of as the definition of inertial reference frames.An IRF is a reference frame that is not accelerating (or rotating) with respect to the “fixed stars”. If one IRF exists, infinitely many exist since they are related by any arbitrary constant velocity vector!

See text: 5-3

Physics II, Pg 5

Is Cincinnati a good IRF?Is Cincinnati a good IRF? Is Cincinnati accelerating? YES!

Cincinnati is on the Earth.The Earth is rotating.

What is the centripetal acceleration of Cincinnati? T = 1 day = 8.64 x 10 4 sec, R ~ RE = 6.4 x 10 6 meters .

Plug this in: aU = .034 m/s2 ( ~ 1/300 g) Close enough to 0 that we will ignore it. Cincinnati is a pretty good IRF.

av

RR

TRU

2 22

2

Physics II, Pg 6

Newton’s Second LawNewton’s Second Law

For any object, FFNET = FF = maa.

The acceleration aa of an object is proportional to the net force FFNET acting on it.

The constant of proportionality is called “mass”, denoted m.» This is the definition of mass.

» The mass of an object is a constant property of thatobject, and is independent of external influences.

Force has units of [M]x[L/T2] = kg m/s2 = N (Newton)

See text: 5-5

Physics II, Pg 7

Newton’s Second Law...Newton’s Second Law...

What is a force?A Force is a push or a pull.A Force has magnitude & direction (vector).Adding forces is like adding vectors.

FF1 FF2

aaFF1

FF2

aa

FFNET

FFNET = maa

See text: 5-5 and 5-7

Physics II, Pg 8

Newton’s Second Law...Newton’s Second Law...

Components of FF = maa :

FX = maX

FY = maY

FZ = maZ

Suppose we know m and FX , we can solve for aX and apply

the things we learned about kinematics over the last few weeks: v v a t

x x v t a t

x ox x

o ox x

1

22

See text: 5-5, 5-6, and 5-7

Physics II, Pg 9

Example: Pushing a Box on Ice.Example: Pushing a Box on Ice.

A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50N in the ii direction. If the box starts at rest, what is it’s speed v after being pushed a distance d=10m ?

FF

v = 0

m a

ii

See text: 5-5

Physics II, Pg 10

Example: Pushing a Box on Ice.Example: Pushing a Box on Ice.

A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50N in the ii direction. If the box starts at rest, what is it’s speed v after being pushed a distance d=10m ?

d

FF

v

m a

ii

See text: 5-5

Physics II, Pg 11

Example: Pushing a Box on Ice...Example: Pushing a Box on Ice...

Start with F = ma.a = F / m.Recall that v2

2 - v12 = 2a(x2 - x1 ) (lecture 1)

So v2 = 2Fd / m vFd

m

2

d

FF

v

m a

ii

See text: 5-5

Physics II, Pg 12

Example: Pushing a Box on Ice...Example: Pushing a Box on Ice...

Plug in F = 50N, d = 10m, m = 100kg:Find v = 3.2 m/s.

d

FF

v

m a

ii

vFd

m

2

See text: 5-5

Physics II, Pg 13

ForcesForces

Units of force (mks): [F] = [m][a] = kg m s2 = N (Newton) We will consider two kinds of forces:

Contact force:» This is the most familiar kind.

I push on the desk. The ground pushes on the chair...

Action at a distance (a bit mysterious):» Gravity» Electromagnetic, strong & weak nuclear forces.

See text: 5-4

Physics II, Pg 14

Contact forces:Contact forces:

Objects in contact exert forces.

Convention: FFa,b means “the force acting on a due to b”.

So FFhead,thumb means “the force on the head due to the thumb”.

FFhead,thumb

See text: 5-4

Physics II, Pg 15

Gravity...Gravity...

Near the earth’s surface...

But we have just learned that: FFg = maaThis must mean that g is the “acceleration due to

gravity” that we already know!

So, the force on a mass m due to gravity near the earth’s surface is FFg = mgg where gg is 9.8m/s2 “down”.

F GM m

Rm G

M

Rmgg

e

e

e

e

2 2

g GM

Re

e

2and

See text: 9-2

Physics II, Pg 16

Example gravity problem:Example gravity problem:

What is the force of gravity exerted by the earth on a typical physics student?

Typical student mass m = 55kgg = 9.8 m/s2.Fg = mg = (55 kg)x(9.8 m/s2 )

Fg = 539 N

The force that gravity exerts on any object is called its Weight

FFg

See text: 5-6

See text example Mass and Weight.

Physics II, Pg 17

Newtons Third Law:Newtons Third Law:

Forces occur in pairs: FFA ,B = - FFB ,A.

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

In the case of gravity:

F F121 2

122 21 G

m m

R

R12

m1m2

FF12 FF21

See text: 5-8

Physics II, Pg 18

Newtons Third Law...Newtons Third Law...

FFA ,B = - FFB ,A. is true for contact forces as well:

FFm,w FFw,m

FFm,f

FFf,m

See text: 5-8

Physics II, Pg 19

Example of Bad ThinkingExample of Bad Thinking

Since FFm,b = -FFb,m why isn’t FFnet = 0, and aa = 0 ?

a ??a ??FFm,b FFb,m

ice

See text: 5-8

Physics II, Pg 20

Example of Good ThinkingExample of Good Thinking Consider only the box only the box as the system!

FFon box = maabox = FFb,m

Free Body Diagram (next time).

aaboxbox

FFm,b FFb,m

ice

See text: 5-8

Physics II, Pg 21

The Free Body DiagramThe Free Body Diagram

Newtons 2nd says that for an object FF = maa.

Key phrase here is for an objectfor an object..

So before we can apply FF = maa to any given object we isolate the forces acting on this object:

See text: 5-8, 6-1

Physics II, Pg 22

The Free Body Diagram...The Free Body Diagram...

Consider a plank leaning against a wall.What are the forces acting on the plank ?

P = plank F = floor W = wall E = earth

FFPW

FFWP

FFPF FFPE

FFFP FFEP

See text: 5-8, 6-1

Physics II, Pg 23

The Free Body Diagram...The Free Body Diagram...

Consider the previous caseWhat are the forces acting on the plank ?

Isolate the plank from

the rest of the world. FFPW

FFWP

FFPF FFPE

FFFP FFEP

See text: 5-8, 6-1

Physics II, Pg 24

The Free Body Diagram...The Free Body Diagram...

The forces acting on the plank should reveal themselves...

FFPW

FFPF FFPE

See text: 5-8, 6-1

Physics II, Pg 25

Aside...Aside...

In this example the plank is not moving...It is certainly not accelerating!So FFNET = maa becomes FFNET = 0

This is the basic idea behind statics, which we will discuss in a few weeks.

FFPW

FFPF FFPE

FFPW + FFPF + FFPE = 0

See text: 6-1

Physics II, Pg 26

ExampleExample

Example dynamics problem:

A box of mass m = 2kg slides on a horizontal frictionless floor. A force Fx = 10N pushes on it in the ii direction. What is the acceleration of the box?

FF = Fx ii aa = ?

m

j j

i i

See text: 6-1

Physics II, Pg 27

Example...Example...

Draw a picture showing all of the forces

FFFFBF

FFFBFFBE

FFEB

j j

i i

See text: 6-1

Physics II, Pg 28

Example...Example...

Draw a picture showing all of the forces. Isolate the forces acting on the block.

FFFFBF

FFFBFFBE = mgg

FFEB

j j

i i

See text: 6-1

Physics II, Pg 29

Example...Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram.

FFFFBF

mgg

j j

i i

See text: 6-1

Physics II, Pg 30

Example...Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram. Solve Newtons equations for each component.

FX = maX

FBF - mg = maY

FFFFBF

mgg

j j

i i

See text: 6-1

See strategy: Solving Newton’s Law Problems,

Physics II, Pg 31

Example...Example... FX = maX

So aX = FX / m = (10 N)/(2 kg) = 5 m/s 2.

FBF - mg = maY

But aY = 0 So FBF = mg.

The vertical component of the forceof the floor on the object (FBF ) isoften called the Normal Force Normal Force (N).

Since aY = 0 , N = mg in this case.

FX

N

mg

j j

i i

See text: 6-1

Physics II, Pg 32

Example RecapExample Recap

FX

N = mg

mg

aX = FX / m j j

i i

See text: 6-1

Physics II, Pg 33

Tools: Ropes & StringsTools: Ropes & Strings

Can be used to pull from a distance. TensionTension (T) at a certain position in a rope is the magnitude of the

force acting across a cross-section of the rope at that position.The force you would feel if you cut the rope and grabbed the ends.An action-reaction pair.

cut

TT

T

See text: 6-1

Physics II, Pg 34

Tools: Ropes & Strings...Tools: Ropes & Strings...

An ideal (massless) rope has constant tension along the rope.

If a rope has mass, the tension can vary along the rope For example, a heavy rope

hanging from the ceiling...

We will deal mostly with ideal massless ropes.

T = Tg

T = 0

T T

Physics II, Pg 35

Tools: Ropes & Strings...Tools: Ropes & Strings...

The direction of the force provided by a rope is along the direction of the rope:

mg

T

m

Since ay = 0,(not moving)

T = mg

See text: 6-1

Physics II, Pg 36

Scales:Scales:

Springs can be calibrated to tell us the applied force. We can calibrate scales to read Newtons, or...Fishing scales usually read

weight in kg or lbs.

02468

See text: 5-9

Physics II, Pg 37

Tools: Pegs & PulleysTools: Pegs & Pulleys

Used to change the direction of forces.

An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude:

FF1 ideal peg

or pulley

FF2

| FF1 | = | FF2 |

Physics II, Pg 38

Tools: Pegs & PulleysTools: Pegs & Pulleys

Used to change the direction of forces.

An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude:

mg

T

m T = mg

FW,S = mg

Physics II, Pg 39

Recap of today’s lectureRecap of today’s lecture

Newton’s 3 laws:

Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an

inertial reference frame.

Law 2: For any object, FFNET = FF = maa

Law 3: Forces occur in pairs: FFA ,B = - FFB ,A.