Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Physics 111 Lecture 04

Force and Motion I: The Laws of Motion

SJ 8th Ed.: Ch. 5.1 –5.7

•Dynamics -Some history

•Force Causes Acceleration

•Newton’s First Law: zero net force

•Mass

•Newton’s Second Law

•Free Body Diagrams

•Gravitation

•Newton’s Third Law

•Application to Sample Problems

5.1

Th

e C

on

cep

t o

f a F

orc

e

5.2

New

ton

’s F

irst

Law

an

d

Inert

ial F

ram

es

5.3

Mass

5.4

New

ton

’s S

eco

nd

Law

5.5

Gra

vit

ati

on

al

Fo

rce a

nd

Weig

ht

5.6

New

ton

’s T

hir

d L

aw

5.7

Usin

g N

ew

ton

’s S

eco

nd

Law

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Dynamics -Newton’s Laws of Motion

Kinematics described motion only –no real Physics.

Why does a particle have a certain acceleration?

New concepts (in 17th century):

•Forces -pushes or pulls -

cause acceleration

•Inertia (mass) measures how much matter is being accelerated –

resistance to acceleration

Sir Isaac Newton 1642 –

1727

•Formulated basic laws of mechanics

•Invented calculus in parallel with Liebnitz

•Discovered Law of Universal Gravitation

•Many discoveries dealing with light and optics

•Ran the Royal Mint for many years during old age

•Many rivalries & conflicts, few friends,

no spouse or children, prototype “geek”

Newton’s 3 Laws of M

otion:

•Codified kinematics work by Galileo and other early experimenters

•Introduced mathematics (calculus) as the language of Physics

•Allowed detailed, quantitative prediction and control (engineering).

•Ushered in the “Enlightenm

ent”& “Clockwork Universe”.

•Are accurate enough (with gravity) to predict all com

mon

motions plus those of celestial bodies.

•Fail only for v ~ c and quantum

scale (very small).

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Force: causes any change in the velocities of

particles

A force is that which

causes an acceleration

Newton’s

definition:

Co

nta

ct

forc

es i

nvo

lve “

ph

ysic

al

co

nta

ct”

betw

een

ob

jects

F =

-kx

Fie

ld f

orc

es a

ct

thro

ug

h e

mp

ty s

pac

e w

ith

ou

t p

hysic

al

co

nta

ct

Action at a distance through intervening space? How?

Given atomic physics, Is there any such thing as a real contact force?

The four fundamental forces of nature are:

Gravitation, Electromagnetic, N

uclear, and W

eak force

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Force: what causes any change in the velocities

of particles

Units: 1 N

ewton = force that causes 1 kg to accelerate at 1 m/s

2

1 Pound = force that causes 1 slug to accelerate at 1 ft/s2

Forces are VECTORS

Operate with vector rules

Rep

lace a

fo

rce a

cti

ng

at

a p

oin

t b

y i

ts

co

mp

on

en

ts a

t th

e s

am

e p

oin

t.

iF

x

jF

y

Fr

= ===

Su

perp

osit

ion

: A

set

of

forc

es a

t a p

oin

t h

ave t

he s

am

e e

ffect

that

their

ve

cto

r re

su

ltan

t fo

rce w

ou

ld

= ===

F 1r

3Fr

2Fr

∑ ∑∑∑≡ ≡≡≡

ii

ne

tF

Fr

vN

ota

tio

n:

ne

tFr

Definition: A body is in EQUILIBRIU

M

if the net force applied to it equals zero

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Relative motion -Inertial Reference Frames in 1 Dimension

A f

ram

e o

f re

fere

nce a

mo

un

ts t

o s

ele

cti

ng

a c

oo

rdin

ate

sys

tem

.•

Descri

be p

oin

t P

in

bo

th f

ram

es u

sin

g x

1,

v1, a

1an

d x

2,

v2,

a2.

•O

rig

ins c

oin

cid

e a

t t

= 0

•C

on

sta

nt

v12

= r

ela

tive v

elo

cit

yo

f o

rig

in o

f 1

in

fra

me

2

•x

12

is t

he d

ista

nce b

etw

een

ori

gin

s a

t so

me t

ime t

.

Tra

nsfo

rm t

he c

oo

rdin

ate

s:

)t(x

)t(x

)t(x

12

12

+ +++= ===

Tra

nsfo

rm t

he v

elo

cit

ies

:

dt

dx

dt

dx

dt

dx

12

12

+ +++= ===

12

12

v

v

v + +++

= ===

Th

en

a12

= 0

an

d

Fin

d t

he a

cce

lera

tio

ns:

01

21

2

2

a

d

t

dv

a

d

t

dv

dt

dv

dt

dv

dt

dv

a1

21

21

1= ===

≡ ≡≡≡= ===

+ +++= ===

= ===

a

a 2

1= ===

Th

e a

ccele

rati

on

of

a m

ovin

g o

bje

ct

is t

he s

am

e f

or

a p

air

of

inert

ial

fra

mes.

Example: An accelerating car viewed from a train and the ground,with the train

itself moving at constant velocity

Inert

ial

fra

mes c

an

no

t b

e r

ota

tin

g o

r acce

lera

tin

g r

ela

tive

to

on

e a

no

ther

or

to t

he

fixed

sta

rs.

No

n-i

nert

ial

fra

me

s � ���

fic

titi

ou

s f

orc

es.

x2

y2

x1

y1

x12

x2

x1

P

v1

2

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Newton’s First Law (1686)

Don’t moving objects come to a stop if you stop pushing?

•Stopping implies negative acceleration, due to friction

or other forces opposing motion.

•What effect does inertia have on a curve on an icy road?

“T

he L

aw

of

Inert

ia”:

A b

od

y’s

velo

cit

y is c

on

sta

nt

(i.e

., a

= 0

) if

th

e n

et

forc

e a

cti

ng

on

it

eq

uals

zero

Alt

ern

ate

sta

tem

en

t: A

bo

dy r

em

ain

s in

un

ifo

rm m

oti

on

alo

ng

a

str

aig

ht

lin

e a

t co

nsta

nt

sp

eed

(o

r re

main

s a

t re

st)

un

less it

is

acte

d o

n b

y a

net

exte

rnal fo

rce.

Above assum

e an “inertial reference frame”:

•Equations of physics look simplest in inertial systems.

•Non-inertial frames (e.g., rotating) require fictitious forces &

accelerations in physics equations (e.g., centrifugal and Coriolisforces).

Fir

st

Law

(o

ur

text)

: A

n o

bje

ct

that

do

es n

ot

inte

ract

wit

h o

ther

ob

jects

(n

o n

et

forc

e,

iso

late

d)

can

be p

ut

into

a r

efe

ren

ce f

ram

e i

n w

hic

h t

he

ob

ject

has z

ero

accele

rati

on

(i.

e., i

t’s m

oti

on

can

be t

ran

sfo

rmed

to

an

in

ert

ial

fram

e i

f it

is n

ot

in o

ne a

lread

y).

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Mass: the Measure of Inertia

Apply the same force to different objects.

Different accelerations result. W

hy?

Example: Apply same force to baseball, bowling ball, autom

obile,RR train

Mass m

easu

res in

ert

ia:

•th

e a

mo

un

t o

f “m

att

er”

in a

bo

dy (

i.e., h

ow

man

y a

tom

s o

f each

typ

e)

•re

sis

tan

ce t

o c

han

ges i

n v

elo

cit

y (

i.e.,

accele

rati

on

) w

hen

a f

orc

e a

cts

For a givenforce applied to m1and m

2:

12

21

aa

mm

= ===A mass and resulting acceleration on it

are inversely proportional

Th

e s

am

e “

inert

ial m

ass”

valu

e a

lso

measu

res “

gra

vit

ati

on

al m

ass”

–-

a p

art

icle

’s e

ffect

in p

rod

ucin

g g

ravit

ati

on

al p

ull o

n o

ther

masses

Mass is a

scala

r:2

12

1m

mm

m

to m

w

ith

be

ha

ve

s

res

ult

Th

e

Att

ac

h+ +++

= ===

Mass is in

trin

sic

to

an

ob

ject:

•

it d

oesn

’t d

ep

en

d o

n t

he e

nvir

on

men

t o

r sta

te o

f m

oti

on

(fo

r v <

< c

), o

r ti

me.

Do

n’t

co

nfu

se m

ass w

ith

weig

ht

(a f

orc

e):

mg

W= ===

N.

3.3

Wb

ut

N.

20

W

then

kg

m

If

mo

on

ea

rth

≈ ≈≈≈≈ ≈≈≈

= ===2

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Newton’s Second Law

a

m

F

F i

in

et

rr

r= ===

≡ ≡≡≡∑ ∑∑∑

Su

mm

ari

zin

g:

Vector sum

of forces

acting ON a particle

Inertia of

particle

Acceleration resulting

from

Fn

et

SIMPLE PROPO

RTIONALITY W

HEN IN AN INERTIAL FRAME

Cart

esia

n c

om

po

nen

t eq

uati

on

s:

m

a

F

F x

iix

x,n

et

= ===≡ ≡≡≡∑ ∑∑∑

m

a

F

F y

iiy

y,n

et

= ===≡ ≡≡≡∑ ∑∑∑

m

a

F

F z

iiz

z,n

et

= ===≡ ≡≡≡∑ ∑∑∑

Oth

er

ways t

oW

rite

2n

dL

aw

:

dtr

dm

F

F

2

ii

ne

t2r

rr

= ===≡ ≡≡≡∑ ∑∑∑

vm

p

dtp

d

F w

here

net

rr

rr

≡ ≡≡≡= ===

DIRECTION O

F ACCELERATION AND N

ET FORCE ARE THE SAME

Un

its f

or

Fo

rce:

2T/

L M

]a[ ]

m[

]

F[= ===

= ===

SYSTEM FORCE MASS ACCELERATION

SI Newton (N) Kg m

/s2

CGS Dyne gm

cm/s

2

British Pound (lb) slug ft/s

2

1 d

yn

e =

10

-5N

1 g

m

= 1

0-3

kg

1 lb

=

4.4

5 k

g1 s

lug

=

14.

59

kg

1 k

gW

EIG

HS

2.2

lb

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

4-1

: T

hre

e s

tud

en

ts c

an

all

pu

ll o

n t

he r

ing

(se

e s

ketc

h)

wit

h i

den

tical

forc

es o

f m

ag

nit

ud

e F

, b

ut

in d

iffe

ren

t d

irec

tio

ns w

ith

resp

ec

t to

th

e +

x

axis

. O

ne

of

them

pu

lls a

lon

g t

he +

x a

xis

wit

h f

orc

e F

1as s

ho

wn

.

Wh

at

sh

ou

ld t

he o

ther

two

an

gle

s b

e t

o m

inim

ize t

he

ma

gn

itu

de o

fth

e

rin

g’s

accele

rati

on

?

a)

θ θθθ2 222

= 0

, θ θθθ

3=

0b

) θ θθθ

2 222=

180,

θ θθθ3

= -

180

c)

θ θθθ2 222

= 6

0,

θ θθθ3

= -

60

d)

θ θθθ2 222

= 1

20,

θ θθθ3

= -

120

e)

θ θθθ2 222

= 1

50,

θ θθθ3

= -

150

Tug of war

x1Fr

2Fr

3Fr

θ θθθ3

θ θθθ2

4-2

: W

ha

t sh

ou

ld t

he

oth

er

two

an

gle

s b

e t

o m

axim

ize t

he m

ag

nit

ud

e o

f th

e r

ing

’s

accele

rati

on

?

a)

θ θθθ2 222

= 0

, θ θθθ

3=

0b

) θ θθθ

2 222=

180,

θ θθθ3

= -

180

c)

θ θθθ2 222

= 6

0,

θ θθθ3

= -

60

d)

θ θθθ2 222

= 1

20,

θ θθθ3

= -

120

e)

θ θθθ2 222

= 1

50,

θ θθθ3

= -

150

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example:

A hockey puck whose mass is 0.30 kg is

sliding on a frictionless ice surface. Two

forces act horizontally on it as shown in the

sketch. Find the magnitude and direction of the

puck’s acceleration.

Ap

ply

2n

dL

aw

to

x a

nd

y d

irecti

on

s

)c

os

(F

)c

os

(F

m

a

F

F

F

1x

2x

1x

x,n

et

60

20

2+ +++

− −−−= ===

= ===+ +++

≡ ≡≡≡

)s

in(

F )

sin

(F

m

a

F

F

F

1y

2y

1y

y,n

et

60

20

2+ +++

− −−−= ===

= ===+ +++

≡ ≡≡≡

N

8.7

F x,

ne

t= ===

N

5.2

F y,

ne

t= ===

22

9s/

m

0

.3

8.7

m

F

a

x

ne

t,x

= ==== ===

= ===

2s/

m

17

0

.3

5.2

m

F

a

y

net,

y= ===

= ==== ===

Eva

lua

te:

FB

D

m =

0.3

kg

Co

nvert

to

po

lar

co

ord

ina

tes

:

ox

y-1

2/

2 y2 x

)a/

a( ta

n

m/s

3

4

]a

[a

a

3

12

1= ===

= ===θ θθθ

= ===+ +++

= ===

Th

e n

et

forc

e a

nd

accele

rati

on

ve

cto

rs h

ave t

he s

am

e d

irec

tio

nA

un

it v

ecto

r in

th

at

dir

ec

tio

n i

s:

j 0

.50

i

0.8

5

j3

4

17

i

34

29

j|

a|a

i |a|a

|

a|a

a

y

x+ +++

= ===+ +++

= ===+ +++

= ===≡ ≡≡≡

r

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Measuring Force and Mass

On frictionless surface (e.g., air track): Apply enough horizontal

force F

0to give the standard mass m

0the standard acceleration a

0

m0

= 1

kg

a0

= 1

m/s

2

00

am

F

0≡ ≡≡≡

The (standard) force unit thereby defined = 1N.

No

te:

x-c

om

po

nen

ts o

nly

ab

ove,

F&

aare

in

sam

e d

irecti

on

What about forces in the y –

direction, left out above?

Measuring another mass:

•Apply standard force, record resulting acceleration

m/

F

a0

≡ ≡≡≡a

/ F

m

0= ===

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Gravitational Force, Weight, “Normal”Force

Mass in free fallon Earth’s surface accelerates at

g:

FB

D

m

g h

as t

he s

am

ed

irecti

on

(to

ward

cen

ter

of

Eart

h)

an

d m

ag

nit

ud

e f

or

all

masses (

in a

vo

lum

e l

arg

e

co

mp

are

d t

o a

hu

man

).

jm

g

F g

− −−−= ===

rg

= 9

.8 m

/s2

=

32.2

ft/

s2

at

the E

art

h’s

su

rface

•“Action at a distance”

•The weight is independent of how a mass is

moving (perhaps other forces also act).

w

eig

ht"

the

"

F

g≡ ≡≡≡

r

Mass in contact with a “horizontal”surface (table, air track, …)

may be in equilibrium

for y

:

a

0

F

y

for

m

"E

qu

ilib

riu

"

yy

= ==== ===

↔ ↔↔↔∑ ∑∑∑

FB

D

m

jm

g

F g

− −−−= ===

r

N

ay

= 0

� ���

N =

Fg

(mdoes not accelerate)

su

rface

to

la

rp

erp

en

dic

u

fo

rce"

n

orm

al

"

N

≡ ≡≡≡≡ ≡≡≡

N is a contact force that adjusts to

Fg

Fgpushes on the surface –

-the surface pushesback with N

If a

ynot = 0, N

does not equal

Fg

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Newton’s Third Law

Bodies interact by pushing or pulling on each other

3rd

Law

(an

tiq

ue v

ers

ion

): E

ach

acti

on

has a

n e

qu

al

an

d o

pp

osit

e r

eacti

on

Mo

re m

od

ern

ve

rsio

n:

Wh

en

tw

o b

od

ies in

tera

ct

the f

orc

es t

hat

each

exert

s o

n t

he o

ther

are

alw

ays e

qu

al

in m

ag

nit

ud

e a

nd

op

po

sit

e in

dir

ecti

on

F

F 21

12

rr

− −−−= ===

Example: gravity acting at a distance

2

ob

ject

to

du

e 1

ob

ject

o

n

forc

e

F ≡ ≡≡≡

12

r

1 o

bje

ct

to

du

e

2 o

bje

ct

o

n

forc

e

F ≡ ≡≡≡

21

r

If you ever find a force w/o the 3

rdlaw reaction,

you can build a perpetual motion machine

Example: box on a level surface

•F

gis

th

e p

ull

of

Eart

h o

n t

he b

ox (

weig

ht)

•T

he 3

rdla

w r

eacti

on

is F

e -

the b

ox’s

pu

ll o

n t

he E

art

h•

Nis

th

e s

urf

ace’s

pu

sh

on

th

e b

ox

•F

sis

th

e b

ox’s

pu

sh

on

th

e s

urf

ace

•O

NL

Y f

orc

es

on

th

e b

ox (

Fg

& N

) a

ffect

it’s

mo

tio

n•

Fg

& N

are

NO

T a

3rd

law

pair

(F

sb

& N

are

a p

air

)•

Wh

y t

hen

do

es F

g=

N ?

??

m

jm

g

F g

− −−−= ===

r

N FeF

s

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Free Body Diagrams (FBDs)

Drawing the FBDsis the most important step in analyzing motion.

•DRAW FBDsFIRST –

before you start writing down equations.

•Pictorial sketches are not the same as FBDs.

•Model bodies in your system as point particles. There may be

several. Som

etimes you can treat the system as one object.

•Choose coordinates.

•Include in FBDsonly forces that act O

N your system.

•Exclude forces exerted BY bodies in your system on other bodies.

•Neglect internal forces. When you break up a system for analysis,

INCLUDE formerly internal forces that become external.

•Don’t forget action-at-a-distance forces (fields) such a gravity

m

F g

r

N

F g

r

m

NM

jM

g

F g

− −−−= ===

r

N

m

jm

g

F

' g− −−−

= ===

rIs this a FBD?

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Friction F

ricti

on

less

m

gFNF

Sli

din

g o

r sta

tic

fric

tio

n f

(la

ter)

m

gFNF

f

No

fri

cti

on

para

llel

to s

urf

ace

Fri

cti

on

is a

resis

tive f

orc

e p

ara

llel

to s

urf

ace

m/F

ax

= ===m/)f

F(a

x− −−−

= ===N

f∝ ∝∝∝

Co

nta

ct

forc

e -

alw

ays o

pp

ose

d t

o m

oti

on

Cords

Ten

sio

n o

nly

, n

o c

om

pre

ssio

n

Pu

llin

g c

rea

tes t

en

sio

n T

–th

e f

orc

e t

ran

sm

itte

d b

y t

he

co

rdT

is

th

e s

am

e e

ve

ryw

here

in

a z

ero

-mass

, u

ns

tretc

hab

leco

rd

3rdlaw pair

T

mSu

pp

ort

FB

D f

or

bo

dy

mT Fg

FB

D f

or

co

rd

TT’=

T

3rdlaw pair

FB

D f

or

su

pp

ort

T’

F

Eq

uilib

riu

mF

'T

TF g

= ==== ===

= ===

FBDs: show only forces on bodies

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Newton’s Laws -Summary

Newton’s First Law

A b

ody’s

velo

city is c

onsta

nt (

)

if th

e n

et exte

rnal fo

rce o

n it is

zero

0= ===

ar

-M

otion is a

long a

str

aig

ht lin

e-

Fin

d a

nd u

se a

n inert

ial fr

am

e o

f re

fere

nce

Newton’s Second Law

am

vecto

r

forc

e

net

F

F

ii

net

rr

r= ===

= ==== ===∑ ∑∑∑

In C

art

esia

n c

om

ponents

:

zz

yy

xx

ma

F

ma

F

ma

F

∑ ∑∑∑∑ ∑∑∑

∑ ∑∑∑= ===

= ==== ===

Newton’s Third Law

If b

ody A

exert

s a

foce

on b

ody B

, th

en b

ody B

exert

s a

nd e

qual and

Opposite forc

e o

n b

ody A

.

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Method for solving Newton’s Second Law problems

Systems with several components may have several unknowns….

…and need an equal number of independent equations

•Draw or sketch system. Adopt coordinates. N

ame the variables,

•Draw free body diagrams. Show forces acting on particles. Include

gravity (weights), contact forces, normal forces, friction.

•Apply Second Law to each part

•Make sure there are enough (N) equations; Extra conditions connecting

unknowns (constraint equations) may be applicable

•Simplify and solve the set of (simultaneous) equations.

•Interpret resulting form

ulas. Do they make intuitive sense? Are the

units correct? Refer back to the sketches and original problem

•Calculate num

erical results, and sanity check anwers

(e.g., right order of

magnitude?)

am

FF

in

et

rr

r= ===

= ===∑ ∑∑∑

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example: Traffic Light in Equilibrium

Co

ncep

tuali

ze:

cab

les a

re m

assle

ss

an

d d

on

’t b

reak

no

mo

tio

n

Cate

go

rize:

eq

uil

ibri

um

pro

ble

m –

accele

rati

on

s =

0M

od

el

as p

art

icle

s i

n e

qu

ilib

riu

m2 F

BD

s

T3 Fg

FB

D o

f L

igh

tF

BD

of

Kn

ot

Fg

= 1

22 N

So

lve u

pp

er

for

T2:

FB

D o

f lig

ht

yie

lds T

3=

Fg

( a

lig

ht=

0)

FB

D o

f kn

ot

yie

lds:

)co

s(

T)

co

s(

T0

F

oo

2x

37

53

1− −−−

= ==== ===

∑ ∑∑∑

31

37

53

T)

sin

(T

)sin

(T

0

F

oo

2y

− −−−+ +++

= ==== ===

∑ ∑∑∑

11

53

37

33

1T

.T

T)

oc

os

(

)o

co

s(

2= ===

= ===

N .

)]

sin

(

.)

[sin

(

To

o22

153

33

137

1= ===

+ +++

N

.

T

47

31

= ===

N .

T 4

97

2= ===

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example: Particle M

otion Under a N

et Force

mN

F

W

θ θθθ

x

yM

ass m

is m

ovin

g o

n a

fri

cti

on

less h

ori

zo

nta

l su

rface,

acte

d o

n b

y a

n e

xte

rnal

forc

e o

f m

ag

nit

ud

e F

, m

akin

g a

n a

ng

le θ θθθ

wit

h t

he x

-axis

.

Fin

d e

xp

ressio

ns f

or

the a

ccele

rati

on

alo

ng

th

e h

ori

zo

nta

l an

d v

ert

ical

dir

ecti

on

s

W =

weig

ht

Along x:

xx

ma

)

co

s(

F F

= ===

θ θθθ= ===

∑ ∑∑∑

m

)co

s(

F

a

x

θ θθθ= ===

Set

ay =

0.

Cro

sso

ve

r to

ay >

0w

hen

N a

lso

=0, i.

e, w

hen

Along y: Is

ayzero, or does particle accelerate upward?

yy

ma

W

N )

sin

(F

F

= ===− −−−

+ +++θ θθθ

= ===∑ ∑∑∑

W

N )

sin

(F

0 ⇒ ⇒⇒⇒

− −−−+ +++

θ θθθ= ===

W

)F

sin

(

= ===θ θθθ

When

Fyis > the weight, the

particle accelerates upward

m

W

)sin

(F

a

y

− −−−θ θθθ

= ===

N

0= ===

When

Fyis < the weight, the

particle does not accelerate

a

y0

= === )

sin

(F

W

N θ θθθ

− −−−= ===

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

4-3

: T

he

man

an

d t

he p

latf

orm

weig

h a

to

tal

of

500 N

. H

e p

ulls

up

ward

o

n t

he r

op

e w

ith

fo

rce F

. W

ha

t fo

rce w

ou

ld h

e n

eed

to

exert

in

ord

er

to a

ccele

rate

up

ward

wit

h a

= 0

.1 g

?

Is t

his

po

ssib

le?

a)

F =

50 N

b)

F =

1000 N

c)

F =

550 N

d

) F

= 5

00 N

e)

He

can

no

t li

ft h

imself

by h

is o

wn

bo

ots

trap

s a

t all

.

Bootstraps

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example: Sliding and Hanging Blocks

Blo

ck S

, m

ass

M is

sli

din

g o

n a

fri

cti

on

less

ho

rizo

nta

l su

rfa

ce. B

lock H

, m

ass m

han

gs

fro

m a

m

assle

ss,

un

str

etc

hab

leco

rd w

rap

ped

over

a

massle

ss

pu

lle

y.

Fin

d e

xp

ressio

ns f

or

the

accele

rati

on

s o

f th

e

blo

cks a

nd

th

e t

en

sio

n in

th

e c

ord

.

W’=

mg

Mg

W= ===

NT

T’

No

fri

cti

on

,N

o m

as

s

W =

Mg

W =

Mg

NT

FBD for Block S

a

Ma

T

F

xs

= ==== ===

∑ ∑∑∑0

≡ ≡≡≡= ===

− −−−= ===

∑ ∑∑∑ys

ys

Ma

W

N

F

0= ===

= ===∑ ∑∑∑

xh

xh

ma

F

'm

a

T'

W

'F

yh

= ===− −−−

= ===∑ ∑∑∑

WW’’=

mg

= m

g

T’

FBD for Block H

a’

choose

positive down

Apply 2

ndLaw to blocks S & H separately for x & y

Constraints:

T’=

T,

a’

= a

Eliminate T,T’,a’

ma

Ma

W

'= ===

− −−−

Ma

m

a

mg

+ +++= ===

g

Mm

m

a

+ +++= ===

Find formula for T

g

MmM

m

M

a

T

+ +++

= ==== ===

could also use

system approach

To find this

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example: Block Sliding on a Ramp [“Inclined Plane”]

Mass m

is a

cc

ele

rati

ng

alo

ng

a f

ricti

on

less

incli

ned

su

rface a

s s

ho

wn

, m

akin

g a

n a

ng

le θ θθθ

wit

h t

he h

ori

zo

nta

l.

Fin

d e

xp

ressio

ns f

or

the a

ccele

rati

on

an

d t

he n

orm

al

forc

e

mN

a

θ θθθW

= m

g

Wh

y d

oes t

he b

lock a

ccele

rate

?

Do y

ou e

xpect

N to e

qual W

?Forces acting ON the block are N

and W

•N is norm

al to the surface

•W is vertical as usual

Choose x-yaxes aligned to ramp, for which:

y

x

θ θθθ

N

W

90

90

90

90

− −−−θ θθθ

θ θθθW

y

Wx

rig

ht)

&

do

wn

(p

osit

ive

x

yA

ssu

me

a

a

,a

≡ ≡≡≡= ===

0F

BD

Apply 2nd Law to x and y

) W

sin

( W

)co

s(

WW

xy

θ θθθ= ===

θ θθθ= ===

)sin

(m

gm

a

W

F

x

xx

θ θθθ= ===

= ==== ===

∑ ∑∑∑

0

)m

gco

s(

N

WN

F

yy

= ===θ θθθ

− −−−= ===

− −−−= ===

∑ ∑∑∑

)m

gc

os

(

N θ θθθ

= ===

)g

sin

(

a

a

x

θ θθθ= ===

= ===

Does not depend on m

Check:

What happens as

θ θθθ� ���

90o

as

θ θθθ� ���

0o

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example: “Atwood’s M

achine”with M

assless

Pulley

Bo

th m

asses

ha

ve

th

e s

am

e a

cc

ele

rati

on

a (

co

nstr

ain

t).

Th

e

ten

sio

n T

in

th

e u

ns

tre

tch

ab

lec

ord

is t

he

sam

e o

n b

oth

sid

es

of

the m

assle

ss

pu

lle

y (

an

oth

er

co

ns

train

t).

Fin

d e

xp

ressio

ns f

or

the

accele

rati

on

an

d t

he

ten

sio

nIn

th

e c

ord

.

Fin

d t

he f

orc

e W

totsu

pp

ort

ing

th

e p

ull

ey

a

a

Wto

t

No forces or

motion along x

am

g

mT

F

1

y= ===

− −−−= ===

∑ ∑∑∑1

1

T

m1g

aFBD for m

1

T

m2g

a

FBD for m

2

am

T

gm

F

2y

= ===− −−−

= ===∑ ∑∑∑

22

Add the equations

am

a

m T

g

m

gm

T

2

11

2+ +++

= ===− −−−

+ +++− −−−

g

m

m

mm

a

22

11

+ +++− −−−= ===

a =

0 f

or

m1

= m

2a i

s c

lockw

ise f

or

m2

> m

1 a

= g

fo

r m

1=

0

or

a =

-g

fo

r m

2=

0

Subtract the equations

am

a

m

T

gm

g

m

T

21

12

+ +++− −−−

= ===+ +++

− −−−− −−−

g

m

m

m2m

T

2

1

1

2

+ +++= ===

T =

0 f

or

m1

or

m2

eq

uals

zero

.

T =

m

1g

fo

r m

2=

m1

Wto

t

TT

0

2T

W

tot

= ===− −−−

FBD

for

pulley

g

m

m

m4m

W

2

1to

t12

+ +++= ===

Wto

t=

2m

1g

if

m2

= m

1

Oth

erw

ise n

ot

so

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example: Your Weight in an Elevator

A p

assen

ger

wh

ose m

ass

m

=

72.2

kg

is

sta

nd

ing

on

a

pla

tfo

rm s

cale

in

an

ele

va

tor.

Wh

at

weig

ht

do

es t

he

scale

rea

d f

or

him

as

th

e e

leva

tor

(an

d

him

se

lf)

accele

rate

s u

p a

nd

do

wn

?

No forces or

motion along x

N

mg

F g

= ===FBD for

passenger

+

•F

g=

mg

is the real weight, which doesn’t change

•The building is an inertial frame, as is the elevator

when traveling at constant speed.

•When the elevator and passenger are accelerating

their non-inertial frame � ���

fictitious forces

N

is t

he s

cale

read

ing

= a

pp

are

nt

weig

ht

•Apply Second Law in the building’s reference frame

am

m

gN

F

y= ===

− −−−= ===

∑ ∑∑∑ )

ga(

m

N + +++

= ===

Interpretations:

N

= m

g

for

a

=

0N

orm

al w

eig

ht

for

ele

va

tor

at

rest

or

mo

vin

g w

ith

co

ns

tan

t v

N

> m

g

for

a

>

0In

cre

ased

ap

pare

nt

weig

ht

for

ele

va

tor

accele

rati

ng

up

N

< m

g

for

a

<

0D

ecre

ased

ap

pa

ren

t w

eig

ht

for

ele

va

tor

accele

rati

ng

do

wn

mg

=

fo

rce

exe

rted

by g

ravit

y

N

= 0 f

or

a =

-g

Fre

e f

all

-w

eig

htl

ess

Fic

titi

ou

s f

orc

e a

pp

ears

in

no

n-i

nert

ial

fra

me

ve

locit

y h

as n

o e

ffect

on

N –

ap

pare

nt

weig

ht