Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Physics 111 Lecture 12

Static Equilibrium

SJ 8th Ed.: Chap 12.1 –

12.3

•Overview -Equilibrium Defined

•Conditions for Equilibrium

•Center of Gravity

–Defined

–Finding it

–When do m

ass center and CG not coincide?

•Static Equilibrium Examples

12

.1 T

he

Rig

id O

bje

ct

in E

qu

ilib

riu

m:

Th

e C

on

dit

ion

s f

or

Eq

uil

ibri

um

12

.2 M

ore

on

th

e C

en

ter

of

Gra

vit

y

12

.3 E

xam

ple

s o

f R

igid

Ob

jec

ts i

n

Sta

tic

Eq

uil

ibri

um

12

.4 E

las

tic

Pro

pe

rtie

s o

f S

oli

ds

(M

od

uli

)

Co

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–S

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2012

The Equilibrium Conditions

•N

o n

ew

ph

ysic

s h

ere

…ap

ply

New

ton

’s s

eco

nd

law

•

Sp

ecia

l case o

f 2

nd

Law

wit

h a

ccele

rati

on

s e

qu

al to

zero

•S

o…

net

forc

es a

nd

to

rqu

es a

re z

ero

as w

ell

{ {{{} }}}

ne

t,e

xt

L =

co

ns

tan

t

IF

.E

Q.

0 T

HE

N

=

0

se

co

nd

co

nd

itio

n m

et

τα

τα

τα

τα

r rrr

r rrr

dtL

dn

et

ext

,iα αααΙ ΙΙΙ

= ==== ===

τ τττ≡ ≡≡≡

τ τττ∑ ∑∑∑

rr

rr

Identify torques, write equations setting their sum = 0

“S

eco

nd

”co

nd

itio

n f

or

eq

uilib

riu

m0

= ===τ τττ

∑ ∑∑∑ext

,ir

{ {{{} }}}

ne

t,e

xt

p =

co

ns

tan

t

IF

F .E

Q.

0 T

HE

N

a =

0

fir

st

co

nd

itio

n m

et

r rrr r rrr

am

dtp

dF

Fn

et

ext

,i

rr

rr

= ==== ===

≡ ≡≡≡∑ ∑∑∑

Identify forces, write equations setting their sum = 0

“F

irst”

co

nd

itio

n f

or

eq

uilib

riu

m:

0= ===

∑ ∑∑∑e

xt

,iFr

Co

pyri

gh

t R

. J

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–S

pri

ng

2012

DYNAMIC VERSUS STATIC EQUILIBRIUM:

STABILITY OF STATIC EQUILIBRIUM

ISOLATED SYSTEM

MECHANICAL EQUILIBRIUM

No

ch

an

ge i

n t

ran

sla

tio

nal o

r ro

tati

on

al

“sta

te”

�co

nsta

nt

mo

men

tum

, z

ero

net

forc

e, c

on

sta

nt

an

gu

lar

mo

men

tum

, zero

net

torq

ue

�zero

lin

ear

accele

rati

on

of

part

icle

or

cm

of

rig

id b

od

y,

zero

an

gu

lar

accele

rati

on

of

rig

id b

od

y

�D

yn

am

ic:

Σ ΣΣΣF

= 0

an

d Σ ΣΣΣ

τ τττ=

0..

..B

UT

... p

an

d/o

r L

are

no

t =

0

�S

tati

c:

In

ad

dit

ion

, p

= 0

an

d L

= 0

�Im

ag

ine d

isp

lacin

g s

ys

tem

a s

mall

am

ou

nt:

do

es i

t re

turn

?

stable

unstable

neutral

EXAMPLES:

boo

k o

n ta

ble

stat

icne

utra

lpu

ck s

lidin

g on

ice

dyn

amic

--ce

ilin

g fa

n –

ondyn

amic

--ce

ilin

g fa

n –

off

stat

icne

utra

lca

r on

str

aigh

t ro

addyn

amic

--ca

r on

cur

veno

t equ

ilib

rium

--la

dder

lean

ing

agai

nst

wal

lst

atic

unst

able

(fr

icti

on)

stab

le (

foot

in

groo

ve)

Co

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t R

. J

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–S

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ng

2012

•T

he a

ctu

al w

eig

ht

can

be r

ep

laced

by t

he

weig

ht

of

a s

ing

le p

art

icle

at

the C

G

•A

sin

gle

no

n-g

ravit

ati

on

al

no

rmal

forc

e a

pp

lie

d a

t th

e C

G c

an

p

rod

uce t

ran

sla

tio

na

l eq

uil

ibri

um

balance a ruler see-saw a box about to tip

m

L

m1

m2 <

m1

L1

L2

•G

ravit

ati

on

al

torq

ues c

an

ce

l ab

ou

t th

e C

G

•A

sin

gle

su

pp

ort

fo

rce

at

the

CG

cre

ate

s z

ero

ad

ded

to

rqu

e (

mo

men

t arm

= 0

)…..

. an

d c

an

can

cel

the w

eig

ht

forc

es

(li

near

eq

uil

ibri

um

)

•C

G o

ften

co

inc

ides w

ith

mass c

en

ter

(CM

) ..

. b

ut

ma

y n

ot

if g

ravit

y v

ari

es a

cro

ss

th

e b

od

y

Definition: The center of gravityof a body is the point

about which all of the torquesdue to gravitational forces

alone add up to zero

Co

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t R

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2012

�12.1. The center of gravity for a sledge hammer lies on the

centerline of the handle, close to the head, at the X mark. Suppose

you saw across the handle through the center of gravity, cuttingthe

ax in two pieces. You then weigh both pieces. Which of the following

will you find?

Center of gravity for a sledge hammer

A)

Th

e h

an

dle

pie

ce is h

ea

vie

r th

an

th

e h

ea

d p

iec

e

B)

Th

e h

ead

pie

ce i

s h

ea

vie

r th

an

th

e h

an

dle

pie

ce

C)

Th

e t

wo

pie

ces a

re e

qu

all

y h

ea

vy

D)

Th

e c

om

para

tive w

eig

hts

dep

en

d o

n m

ore

in

form

ati

on

Definition: The center of gravityof a body

is the point about which all of the torques

due to gravitational forces sum to zero

x

Co

pyri

gh

t R

. J

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ng

2012

Center of gravity: Summary

•If the object has constant density

and is symmetrical, the center of

gravity coincides with its geometric

center.

•The sum of gravitational forces

acting on all the various mass

elements is equivalent to a single

gravitational force (the total

weight) acting through a single point

called the center of gravity (CG)

•The net torque due to gravitational

forces on an object of mass M

equals the force Mg acting at the

center of gravity of the object.

•If g is uniform over the object,

then the center of gravity of the

object coincides with its center of

mass.

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example: Find the balance point (CG) for a level see-saw

To

lo

cate

CG

co

ord

ina

te “

s”

•P

ick l

ikely

CG

lo

cati

on

.

•S

et

su

m o

f to

rqu

es a

rou

nd

it

= 0

, in

clu

din

g o

nly

gra

vit

ati

on

al

forc

es

•S

up

po

rt f

orc

e N

ap

pli

ed

at

CG

exert

s z

ero

to

rqu

e,

pro

vid

es l

inear

eq

uil

ibri

um

FBD

m1

m2

L

xs

balance point = CG

Balance means: Find the CG and put a support force N (= total weight) there:

•Linear acceleration is zero (“static translational equilibrium”)

•Angular acceleration is also zero (“rotational equilibrium”)s

L -

sm

1g

m2g

N

If a

n o

bje

ct

is in

eq

uilib

riu

m, th

en

th

e n

et

torq

ue =

0 f

or

an

y a

xis

ch

osen

(m

ore

belo

w).

•

Fo

r an

axis

no

t at

the C

G, to

rqu

e d

ue t

o N

is n

ot

zero

bu

t sti

ll c

an

cels

th

e g

ravit

ati

on

al to

rqu

es

gm

gm

N

a)m

m(

F

y2

12

10

0− −−−

− −−−= ===

= ===+ +++

⇒ ⇒⇒⇒= ===

∑ ∑∑∑ w

eig

ht

tota

l

g )

m m(

N

2

1+ +++

= ===∴ ∴∴∴

2)

1)

0

N

)s

L(g

m

g

sm

x

cg

+ +++− −−−

− −−−⇒ ⇒⇒⇒

= ===τ τττ

∑ ∑∑∑2

10

cg

lo

cate

s2

L )m

m(

m

s

21

+ +++= ===

∴ ∴∴∴E

x:

Let

m1

= 2

m2

then

s =

1/3

L

Co

pyri

gh

t R

. J

an

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–S

pri

ng

2012

Locating CG’s of rigid bodies

-Pic

k b

od

y u

p (

exte

rnal

forc

e T

) an

d le

t it

sw

ing

fre

ely

,

co

me

to e

qu

ilib

riu

m p

osit

ion

, a

nd

sto

p (

fric

tio

n).

-B

od

y c

om

es

to

rest

wit

h C

G

belo

w o

r ab

ove

su

sp

en

sio

n p

oin

t,

o/w

torq

ue w

ou

ld n

ot

= z

ero

an

d

bo

dy w

ou

ld n

ot

be in

eq

uil

ibri

um

.

-M

ark

a v

ert

ical

lin

e a

t s

usp

en

sio

n p

oin

t-

Rep

eat

pro

ces

s f

or

a d

iffe

ren

t su

sp

en

sio

n p

oin

t-

Th

e C

G is

wh

ere

th

e l

ines

in

ters

ect.

T

pre

vio

us

suspensio

n

poin

t

If gravitation field is uniform, CG & CM coincide but may not beat

geometrical symmetry points unless density is uniform.

•For uniform sphere, cube, disc, rectangle,.. CG at center

•For cone. cylinder, bar, etc CG along axis of symmetry

•For composite objects, break into shapes with symmetry

WEIGHT

ACTS AT

CG

STABILITY

Block tips if CG is to the left of

all support points in base.

Net torque then can not = 0

W

N

Co

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gh

t R

. J

an

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–S

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2012

For Equilibrium Problems: Does it matter w

hich rotation axis you

choose for calculating torques? No. Choose any convenient axis.

O'

ro

O

axes

an

yab

ou

t

sam

e

the

is

net

i

he

nt

0

F

If

τ τττ

= ===∴ ∴∴∴

∑ ∑∑∑v

v

Not

e al

so: if

Pcm

= 0

, th

e an

gula

r m

oment

um is

the

sam

e a

roun

d a

ll p

aral

lel ax

es

O

ax

isa

bo

ut

m

om

en

tum

a

ng

ula

ri

iO

pr

L

vv

r× ×××

= ===∑ ∑∑∑

Compare rotation axis at O with one shifted to O’

for a system of point particles { mi}

ori

gin

O

'

to

by

axis

sh

ift

ii

a

a

'

r

rr

rr

r+ +++

= ===

ii

' ii

' iO

p

a

pr

p ]

a r[

L

vr

vv

vr

vr

∑ ∑∑∑∑ ∑∑∑

∑ ∑∑∑× ×××

+ +++× ×××

= ===× ×××

+ +++= ===

�Take the time derivative of each term, use second law (rotational and linear)

angular momentum

relative to axis O

angular momentum

relative to axis O’

linear momentum

of mass center

cm

P

v≡ ≡≡≡

o'

L

r

= ===

dt

Pd

a

d

t

Ld

dtL

dcm

'O

O

rr

rr

× ×××+ +++

= ===

o'

n

et,

τ τττ= ===

vo

net,

τ τττ= ===

vext

n

et,

F

r

= ===

Rightmost term vanishes

if F

net= Σ ΣΣΣFi= 0

O

O’

irv

i'rv

av

Proof:

mi

IF F

net=

0 T

HE

N τ τττ

netis

th

e s

am

e a

rou

nd

eve

ry p

ara

lle

l ax

is ..s

o...

IFτ τττ n

et=

0 f

or

on

e a

xis

TH

EN

it

is z

ero

fo

r e

ve

ry o

the

r a

xis

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Rules for solving equilibrium problems

Fir

st

Co

nd

itio

n:

Seco

nd

Co

nd

itio

n:

0= ===

ne

tFr

0= ===

τ τττn

et

r

For plane statics (flat world)

�All forces lie in x-y plane. F

z always equals zero

�All torques lie along +/-

z axis. τ τττx,

τ τττ y= 0

00

= ==== ===

= ==== ===

∑ ∑∑∑∑ ∑∑∑

y,iy,

net

x,ix,

net

FF

,F

F

0= ===

τ τττ= ===

τ τττ∑ ∑∑∑

z,iz,

ne

t

Ch

oo

se

an

y c

on

ve

nie

nt

ax

is f

or

torq

ue

ca

lcu

lati

on

.F

irst

co

nd

itio

n im

plie

s t

ha

t re

su

lt w

ill b

e t

he

sam

e (

se

e p

roo

f)fo

r a

ny r

ota

tio

n a

xis

Ca

lcu

late

to

rqu

es

du

e t

o w

eig

ht

of

rig

id b

od

ies

by p

lac

ing

th

eir

w

eig

hts

at

the

ir c

en

ter

of

gra

vit

y l

oc

ati

on

s (

see

pro

of)

Us

e s

tati

c f

ric

tio

n f

orc

e w

here

ne

ed

ed

:N

fs

sµ µµµ

≤ ≤≤≤

Co

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gh

t R

. J

an

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–S

pri

ng

2012

�12.2. The figure shows five sketches of a uniform rod with two

or more forces acting perpendicular to it. The magnitudes of

the forces can be adjusted to any value needed, except zero.

For which of the sketches can the rod be in translational

equilibrium?

When can a system be in equilibrium?

A)

1, 2

,3 ,

4 ,

5

B

) 4

C)

2,

3, 4,

5

D

) 3

, 4

, 5

E)

4, 5

1 2 3

4 5

�12.3. For which of the situations can the rod be in overall

static equilibrium?

Net torque is the same for any

convenient axis when net force = 0

Co

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gh

t R

. J

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2012

Method for Solving Equilibrium Problems

ESSENTIALLY THE SAME AS FOR SOLVING SECOND LAW PROBLEMS

�D

raw

ske

tch, la

bel it

, deci

de w

hat

is

in o

r ou

t of

the “

syst

em

”.�

Dra

w f

ree b

ody

dia

gram

s.

Inc

lude f

orce

s O

Nth

e b

ody

bein

g an

alyz

ed. S

how

poi

nt o

f ap

plic

atio

n to

ind

icat

e t

orqu

es.

�C

hoo

se a

xes

and t

he +

/-se

nse f

or r

otat

ions

. R

epl

ace f

orce

s by

their

x, y,

z c

ompo

nent

s

00

0= ===

τ τττ= ===

= ===∑ ∑∑∑

∑ ∑∑∑∑ ∑∑∑

zy

x

,F

,F

�C

hoo

se a

rot

atio

n ax

is/p

oint

to

use in

calc

ulat

ing

torq

ues.

oA

ll c

hoi

ces

yield

the s

ame n

et

torq

ue, so

lon

g as

the F

irst

equ

ilib

rium

con

dit

ion

appl

ies…

but

...o

Som

e c

hoi

ces

sim

plif

y th

e s

olut

ion.

Loo

k f

or w

ays

to g

ive z

ero

mom

ent

arm

(zero

tor

que)

to irr

ele

vant

or

trou

ble

som

e f

orce

s.�

App

ly

(2 d

imens

ions

)�

Sub

stit

ute t

he a

ctua

lfor

ces

and t

orqu

es

into

the e

quat

ions

.�

Cou

nt t

he u

nkno

wns

; m

ake s

ure t

here

are

N e

quat

ions

when

there

are

N u

nkno

wns

. C

onst

rain

t equ

atio

ns a

re o

ften

needed.

�S

olve

the s

et o

f “s

imul

tane

ous

equ

atio

ns”

alge

bra

ical

ly a

s fa

ras

is

reas

onab

le b

efo

re s

ubst

itut

ing

num

bers

.�

Try

to

inte

rpre

t re

sult

ing

equa

tion

s in

tuit

ively

. C

heck

that

nu

meri

cal an

swers

mak

e s

ens

e, hav

e r

eas

onab

le m

agni

tudes,

phys

ical

uni

ts, etc

.

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example: Mechanical Advantage of a pulley system

Th

e p

ulle

y s

ys

tem

is u

se

d t

o r

ais

e a

weig

ht

slo

wly

at

co

nsta

nt

sp

eed

( a

= 0

).

Use t

he

eq

uil

ibri

um

co

nd

itio

ns a

nd

fre

e b

od

y

dia

gra

ms t

o f

ind

th

e t

en

sio

n in

each

cab

le

an

d t

he l

ifti

ng

fo

rce.

Th

e p

ulle

ys

are

massle

ss

an

d f

ricti

on

less.

Solution requires “first condition”

only

�At each pulley, the sum of upward =

the sum of downward forces

TA

TA

TB

TB

= 2

TA

W =

98

00 N

T1

T2

T2

T3

T3

T3

T4

N

9800

W

T

= ==== ===

∴ ∴∴∴1

N

4900

T

T

T2

21

= ===⇒ ⇒⇒⇒

= ===2

N

2450

T

T

T

23

2= ===

⇒ ⇒⇒⇒= ===

3

N

49

00

T

2T

T4

3= ===

⇒ ⇒⇒⇒= ===

4

Mechanical advantage

(weight lifted/force exerted)

4

2450

9800

TW= ===

= ===3

Co

pyri

gh

t R

. J

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–S

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ng

2012

Example: Angle of the chair lift cable

Th

e c

ha

ir l

ift

is a

t th

e m

idd

le o

f t

he c

ab

le

sp

an

as s

ho

wn

. I

t’s w

eig

ht

cau

ses t

he

cab

le

to d

efl

ec

t b

y a

n a

ng

le θ θθθ

fro

m t

he h

ori

zo

nta

l o

n

bo

th s

ides o

f th

e c

hair

. T

he

pu

lle

ys

are

mass

-

less a

nd

fri

cti

on

less.

Th

e s

kie

r w

eig

hs 7

8 k

g.

Fin

d t

he a

ng

le θ θθθ

wh

en

th

e h

an

gin

g w

eig

ht

is

2200 N

as

sh

ow

n.

m =

78 k

g

W =

2200 N

θ θθθ

T is u

nif

orm

an

d T

= 2

200 N

TT

mg

θ θθθθ θθθ

Fre

e b

od

y d

iag

ram

mg

)

sin

(T

F

y− −−−

θ θθθ= ===

= ===∑ ∑∑∑

20

17

37

02

20

02

89

78

2.

x

.

x

T

mg

)sin

(= ===

= ==== ===

θ θθθ

.

o0

10

= ===θ θθθ

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

PP

10603-0

8:

A c

ar

wh

ose m

ass =

1360 k

g h

as 3

.05 m

betw

een

its

fro

nt

an

d r

ear

axle

s. It

s c

en

ter

of

gra

vit

y i

s lo

cate

d 1

.78 m

beh

ind

th

e f

ron

t axle

. T

he c

ar

is s

tati

on

ary

on

level g

rou

nd

. F

ind

th

e

mag

nit

ud

e o

f th

e f

orc

e f

rom

th

e g

rou

nd

on

each

fro

nt

an

d r

ear

wh

eel (a

ssu

min

g e

qu

al fo

rces o

n

bo

th s

ides o

f th

e c

ar)

.

Example: Distributing weight between front and rear wheels of a car

2F

R2F

F

L=

3.0

5 m

mg

d =

1.7

8 m

Solution: apply “first”and “second”equilibrium

conditions. Calculate torques using axis

through rear wheel contact point with the

ground. m

g =

1360 k

g x

9.8

= 1

3,3

28 N

.

0

F

x= ===

∑ ∑∑∑m

g

F

2 F

2

F

FR

y− −−−

+ +++= ===

= ===∑ ∑∑∑

0

FR

an

d F

Fare

fo

rces o

n e

ac

h w

heel

L

2F

d)

-m

g(L

F 2

x

Fx

Ro

+ +++− −−−

= ==== ===

τ τττ∑ ∑∑∑

00

To

rqu

es a

bo

ut

rear

wh

eels

3.0

5

x 2

1.7

8)

- (3

.05

x

13

,32

8

L2

d)

-m

g(L

F

F= ===

= ===

2

55

50

- 1

3,3

28

2

2F

- m

g

F

FR

= ==== ===

N.

2

77

5

F

F= ===

N.

3

88

9

F R

= ===

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example: A meter stick balances horizontally on a knife-edge at the 50.0

cm mark. When two 5.0 g coins are attached over the 12.0 cm mark

the

stick now balances at the 45.5 cm mark. What is the mass of the meter

stick?

Problem PP10603-11:

m =

2 x

5 g

m =

.01 k

gx

1=

0.1

2 m

x2

=

.455 m

X3

= 0

.5 m

M

= m

ete

r s

tick

mass

0 c

m10

0 c

m

x1

mg

Mg

Nx

2

x3

APPLY EQUILIBRIUM CONDITIONS:

Mg

mg

N

F

y− −−−

− −−−= ===

= ===∑ ∑∑∑

0

Mg

mg

N + +++

= ===

•M

ete

r sti

ck a

lon

e a

cts

lik

e a

mass M

co

ncen

trate

d a

t it

s o

wn

CG

at

x3

= 0

.5 m

•C

ho

ose a

xis

fo

r to

rqu

es a

t 0 c

m (

left

en

d o

f sti

ck).

A

ny o

ther

axis

yie

lds s

am

e n

et

torq

ue

g

xM

Nx

mg

x-

32

1o

− −−−+ +++

= ==== ===

τ τττ∑ ∑∑∑

0

g

xM

Mg

xm

gx

mg

x-

32

1− −−−

+ ++++ +++

= ===2

0

]x

x [M

]x -

xm

[

3

12

− −−−+ +++

= ===2

0

2

3

10

44

74

55

50

0

12

45

50

10

− −−−= ===

− −−−− −−−= ===

− −−−= ===

x.

..

..

. ]

xx [

]x -

xm

[

M

2

12

g

m

.

M

47

4= ===

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

�12.4. The sketches show four overhead views of uniform disks that

can slide or rotate on a frictionless floor. Three forces act on each

disk, either at the rim or at the center. The force vectors rotate

along with the disks, so the force arrangements remain the same.

Which disks are in equilibrium?

Find the forces on the pivot

A)

1, 2, 3, 4

B)

1, 3, 4

C)

3D

)1, 4

E)

3, 4

F

2F

3F

1

FF

2F

2

FF

2F

3

F

2F

4 F

0= ===

net

Fr0

= ===τ τττ

ne

t

r

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example 1: Massless

beam supporting a weight

Th

e 2

.4 m

. lo

ng

weig

htl

ess b

eam

sh

ow

n in

th

e f

igu

re is

su

pp

ort

ed

on

th

e r

igh

t b

y a

cab

le t

hat

makes a

n a

ng

le o

f 5

0o

wit

h t

he h

ori

zo

nta

l b

eam

. A

32 k

g m

ass h

an

gs f

rom

th

e b

eam

1.5

m f

rom

th

e p

ivo

t p

oin

t o

n t

he left

.

a)

Calc

ula

te t

he t

orq

ue c

au

sed

by t

he h

an

gin

g m

ass.

b)

Dete

rmin

e t

he c

ab

le t

en

sio

n t

hat

pro

du

ces e

qu

ilib

riu

m

Choose: rotation axis at O o

T

θ θθθ

mg

Fx

Fy

x

Draw FBD of beam

Part a)

m.N

.

x .

x

mg

x

mass

470

51

89

32

− −−−= ===

− −−−= ===

− −−−= ===

τ τττ

m 2.4

L

m,

1.5

x

,o

= ==== ===

= ===θ θθθ

50

Part b)

meq

uil

ibri

u

for

o

0= ===

τ τττ∑ ∑∑∑

)T

sin

(

T

T

LT

m

gx

-

b

ea

m

to

o

fc

om

po

ne

nt

y

y

θ θθθ= ===

= ===

+ +++= ===

⊥ ⊥⊥⊥

0

N

.

)L

sin

(

mg

x

T

25

6= ===

θ θθθ= ===

m =

32 k

g

x

O

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

For the preceding problem, find the x component of the force at the

pivot point. What equations can you use?

Example 1, continued: Find the forces on the pivot

Fir

st

Co

nd

itio

n:

Seco

nd

Co

nd

itio

n:

0= ===

ne

tFr

0= ===

τ τττn

et

r

o

T

θ θθθ

mg

Fx

Fy

x

m 2.4

L

m,

1.5

x

,o

= ==== ===

= ===θ θθθ

50

N

.

)L

sin

(

mg

x

T

256

= ===θ θθθ

= ===

For the preceding problem, find the y component of the force at the

pivot point. What equations can you use?

F

-T

0

F

xx

x= ===

= ===∑ ∑∑∑

N 1

65

)

Tc

os

(F

x

= ===θ θθθ

= ===

)T

sin

(

m

g -

F

0

F

y

yθ θθθ

+ +++= ===

= ===∑ ∑∑∑

N 1

17

)

Ts

in(

mg

-

F

y

= ===θ θθθ

− −−−= ===

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example 2: As in previous problem but beam has mass now

Th

e 2

.4 m

. lo

ng

beam

sh

ow

n in

th

e f

igu

re is s

up

po

rted

on

th

e r

igh

t b

y a

cab

le t

hat

makes a

n a

ng

le o

f 5

0o

wit

h t

he h

ori

zo

nta

l b

eam

. A

32 k

g m

ass h

an

gs f

rom

th

e b

eam

1.5

m f

rom

th

e p

ivo

t p

oin

t o

n t

he

left

.

a)

Dete

rmin

e t

he c

ab

le t

en

sio

n n

eed

ed

to

pro

du

ce e

qu

ilib

riu

mb

)F

ind

th

e f

orc

es a

t p

oin

t “O

”

MB= mass of beam = 30 kg acts as point mass

at CG of beam.

Choose: rotation axis at O

m =

32 k

g

x

O

o

T

θ θθθ

mg

Fx

Fy

x

MBg

L/2

FBD of beam

L/2

at

C

G

m,

2

.4L

m,

1

.5x

,o

= ==== ===

= ===θ θθθ

50

meq

uil

ibri

u

for

o

0= ===

τ τττ∑ ∑∑∑

)T

sin

(

T

T

L

T g

M

mg

x -

be

am

to

of

co

mp

on

en

t y

y2L

B

θ θθθ= ===

= ===

+ +++− −−−

= ===

⊥ ⊥⊥⊥

0

N

.

)L

sin

(

gM

mg

x

T

LB

44

82

= ===θ θθθ

+ +++= ===

Part a)

Part b)

)co

s(

TF

F

xx

θ θθθ− −−−

= ==== ===

∑ ∑∑∑0

N.

)

co

s(

TF

x2

88

= ===θ θθθ

= ===

gM

mg

)sin

(T

F

FB

yy

− −−−− −−−

θ θθθ+ +++

= ==== ===

∑ ∑∑∑0

N

.

264

)

sin

(T

gM

mg

F B

y= ===

θ θθθ− −−−

+ +++= ===

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example 3:

A u

nif

orm

beam

, o

f le

ng

th L

an

d m

ass m

= 1

.8 k

g, is

at

rest

wit

h i

ts e

nd

s

on

tw

o s

cale

s (

see f

igu

re).

A

un

ifo

rm b

lock, w

ith

mass M

= 2

.7 k

g,

is a

t re

st

on

th

e b

eam

, w

ith

its

cen

ter

a d

ista

nce L

/ 4

fro

m t

he b

eam

's l

eft

en

d.

What do the scales read?

Solve as equilibrium system

Axis at “o”is convenient

Beam weight acts as if it is at

CG of beam (x = L/2)

o

FB

D

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example 3: Solution

2:

1:

3:Scale readings are normal forces and

L Fr

R Fr

mg

F

M

g

Fb

eam

g,

blo

ck

,g

= ==== ===

imp

act

n

ox

0

F

= ===∑ ∑∑∑

mg

Mg

F

F

0

FR

Ly

− −−−− −−−

+ +++= ===

= ===∑ ∑∑∑ Choose axis for torques at left of beam L

F

2L

mg

-

4LM

g -

0

F

0

R

x

L+ +++

= ==== ===

τ τττ∑ ∑∑∑ From 2:

Solve

2

LLm

g

4

LLM

g

FR

+ +++= ===

15

.44

2

mg

4Mg

F

2

9.8

x

1.8

4

9.8

x

2.7

R= ===

= ===+ +++

= ===+ +++

N

.

15

F

R≈ ≈≈≈

mg

Mg

43

2

mg

4Mg

mg

Mg

FL

21+ +++

= ===− −−−

− −−−+ +++

= ===

N

.

29

28.6

8

F L

≈ ≈≈≈= ===

Block and beam

represented as particles

at their CMs(CGs)

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example 4:

A s

afe

wh

ose

ma

ss is

M=

43

0 k

g is

h

an

gin

g b

y a

ro

pe

fro

m a

bo

om

wit

h

dim

en

sio

ns

a=

1.9

m a

nd

b=

2.5

m.

Th

e b

oo

m c

on

sis

ts o

f a

hin

ge

d

be

am

an

d a

ho

rizo

nta

l ca

ble

th

at

co

nn

ec

ts t

he

be

am

to

a w

all. T

he

u

nif

orm

be

am

ha

s a

ma

ss

mo

f 8

5

kg

; th

e m

as

s o

f th

e c

ab

le a

nd

ro

pe

a

re n

eg

lig

ible

.

(a)

Wh

at

is t

he

te

ns

ion

Tc

in t

he

c

ab

le;

i. e

., w

hat

is t

he

ma

gn

itu

de

of

the

fo

rce

Tc

on

th

e b

ea

m f

rom

th

e

ho

rizo

nta

l ca

ble

?

(b)

Wh

at

is t

he

ma

gn

itu

de

of

the

fo

rce

at

the

hin

ge

?

Tr

Tc

mg

Fh

Fv

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example 4 Solution

Tr

Tc

mg

Fh

Fv

The b

eam

is

the s

yste

m t

o an

alyz

eT

he b

eam

’s w

eig

ht

acts

at

it’s

CG c

hc

hx

T

F

T

F

0

F

= ===∴ ∴∴∴

− −−−= ===

= ===∑ ∑∑∑

rv

yT

M

g

Mg

mg

F

0

F= ===

− −−−− −−−

= ==== ===

∑ ∑∑∑ Choose hinge point “O”as axis for torques

-Eliminates Fvand F

hin torque equation

2b

mg

bT

aT

0

r

co

− −−−− −−−

= ==== ===

τ τττ∑ ∑∑∑

)co

t(

ab

θ θθθ= ===

1.7

2.5

x

9.8

]

430

285

[

ab g

]M

2m[

T c

+ +++= ===

+ +++= ===

N

6093

T

c

= ===

N

6093

T

F

c

h= ===

= ===

9.8

)430

(85

M

g

m

g

F

v+ +++

= ===+ +++

= ===

N

5

04

7

F

v

= ===

N

7

91

2

]

F

F

[

|F|

1/2

2 v

2 h= ===

+ +++= ===

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Problem:In the figure, force Fis applied horizontally at the axle of the wheel. The

wheel's radius is rand its mass is m.

Find the minimum force F needed to raise the wheel over a curb of height h.

Example 5: Does the wheel roll over the curb?

Calculate torques around

contact point with curb

Barrier when h > r

r -

h s

mg N

Px

Py

Wh

y s

uch

larg

e w

heels

?

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example 5: Can the wheel roll over the curb?

N

mg

F Px

Py

Geometry:

h r

h

rh)

hr(

rs

as

lo

ng

as

> >>>− −−−

= ===− −−−

− −−−= ===

22

22

Equations for equilibrium:

0= ===

− −−−+ +++

= ===∑ ∑∑∑

mg

PN

Fy

y

0= ===

− −−−= ===

∑ ∑∑∑x

xP

FF

0= ===

− −−−− −−−

− −−−= ===

τ τττ∑ ∑∑∑

)h

r(F

Ns

mg

s

As wheel is about to lift off and roll, N � ���

0

mg

P

y

= ===∴ ∴∴∴

xP

F = ===

22

hrh

mg

mg

sh

)-

F(r

− −−−

= ==== ===

h

rfo

r

mg

h)

-(r

hrh

F

> >>>

− −−−= ===

22

For r > h, F grows as r � ���

h.

F becomes infinite if r=h.

For r < h vehicle is stopped.

F cannot pull it over curb

When does F = mg? Ans: for r = 2h

Take

torques

around

here

r -

h

s

FBD of the W

heel

Problem:In the figure, force Fis applied horizontally at the axle of the wheel. The

wheel's radius is rand its mass is m.

Find the minimum force F needed to raise the wheel over a curb of height h.

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example 6: Climbing a ladder

A ladder

wh

ose length

L=

12

m a

nd m

ass m

= 4

5 k

g leans a

gain

st

a f

riction

less)

wall.

It

s

top is a

t heig

ht

h=

9.3

m a

bove t

he p

avem

en

t

that

the lo

wer

end r

ests

on.

The p

avem

ent

is

not

fric

tionle

ss.

The ladder's o

wn m

ass c

ente

r

is L

/ 3

fro

m t

he lo

wer

end.

A f

irefighte

r of

mass M

= 7

2 k

g c

limbs t

he

ladd

er

until h

er

cente

r of

mass is s

= L

/ 2

from

the lo

wer

end.

Fin

d t

he m

agn

itudes o

f th

e f

orc

es o

n t

he

ladd

er

from

the w

all

an

d t

he p

avem

ent.

Fin

d h

ow

far

up t

he ladd

er

the f

irefighte

r ca

n

clim

b b

efo

re it

slip

s.

The s

tatic f

riction

coeff

icie

nt

betw

een ladd

er

and p

avem

ent

is

0.2

.a =

Lco

s(θ θθθ

)h

= L

sin

(θ θθθ)

S

θ θθθ

1.2

27

)

tan

(

.815

)

co

t(

50.8

.3

/12)

(sin

)

Lh (sin

o1-

1-

= ===θ θθθ

= ===θ θθθ

= ==== ===

= ===θ θθθ

9

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example 5 solution –Climbing a Ladder

S

Sx

Firefighter is a distance S from the base

)co

s(

s

sx

θ θθθ= ===

ba

se

at

forc

e

no

rma

lp

y

F≡ ≡≡≡

py

s

forc

e

fric

tio

n

sta

tic

px

F

Fµ µµµ

≤ ≤≤≤≡ ≡≡≡

we

igh

trs

fire

fig

hte

M

g≡ ≡≡≡

L/3

at

CG

la

dd

er

weig

ht,

sla

dd

er'

m

g≡ ≡≡≡

ho

rizo

nta

l

fr

icti

on

N

o

lad

der

o

n

wall

of

fo

rce

w

F⇒ ⇒⇒⇒

≡ ≡≡≡

mg

Mg

F

F

py

y− −−−

− −−−= ===

= ===∑ ∑∑∑

0

px

wx

F

F

F

− −−−= ===

= ===∑ ∑∑∑

0

Equilibrium conditions

F

F

px

w= ===

9.8

72)

(45

mg

Mg

F

p

y+ +++

= ===+ +++

= ===

N

1146

F

py

= ===

Choose axis for torques

at base of ladder “O”

)sin

(

LF

)co

s(

M

gs

)

(o

sc

3Lm

g

w

o

θ θθθ− −−−

θ θθθ+ +++

θ θθθ= ===

= ===τ τττ

∑ ∑∑∑0

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Ladder Example Solution, continued

Note:

1.

As s grows, so does FW= F

PX(friction) needed to stay in equilibrium.

2.

As

θ θθθ� ���0, frictional force needed becomes huge, ladder slips

)ta

n(g

LsM

mF

Wθ θθθ

+ +++= ===

3

Example:

For S = L/2 (halfway up):

Solve torque equation for F

w

)s

in(

L

)c

os

(M

gs

mg

LF

wθ θθθθ θθθ

+ +++= ===

3

F

N

40

8

).

tan

(

gM

mF

p

xo

w= ===

≈ ≈≈≈

+ +++

= ===8

50

23

Set impending motion condition:

Solve torque equation for s:

How far up the ladder can the firefighter climb before it

slips, assuming µ µµµ

s= 0.2

[ [[[] ]]]

gm

MF

F

F

s

py

sp

xw

+ +++µ µµµ

= ===µ µµµ

= ==== ===

)co

s(

3L

mg

)

sin

(

LF

)co

s(

M

gs

wθ θθθ

− −−−θ θθθ

= ===θ θθθ

3L

Mm

)ta

n(

L

Mg

g)m

M(

s

s− −−−

θ θθθ+ +++

µ µµµ= ===

312

72

45

1.2

27

12

45/7

2)

(1 .2

x

− −−−+ +++

= ===

m

2.3

.

s

≈ ≈≈≈

= ===∴ ∴∴∴

29

2

Getting halfway up

requires

µ µµµs>> 0.2

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Example 7: Rock Climber

In t

he

fig

ure

a r

ock c

lim

ber

wit

h m

ass

m=

55 k

g r

es

ts d

uri

ng

a “

ch

imn

ey c

lim

b,”

pre

ssin

g o

nly

wit

h h

er

sh

ou

lders

an

d f

ee

t ag

ain

st

the

walls o

f a f

issu

re o

f w

idth

w=

1.0

m.

Her

cen

ter

of

mass i

s a

ho

rizo

nta

l d

ista

nce d

= 0

.20 m

fro

m t

he

wall a

gain

st

wh

ich

her

sh

ou

lders

are

pre

ssed

. T

he

co

eff

icie

nt

of

sta

tic

fri

cti

on

betw

een

her

sh

oes a

nd

th

e w

all is µ µµµ

1=

1.1

, an

d

betw

een

her

sh

ou

lders

an

d t

he

wall it

is

µ µµµ2

= 0

.70.

To

res

t, t

he c

lim

ber

wan

ts t

o

min

imiz

e h

er

ho

rizo

nta

l p

ush

on

th

e

walls. T

he m

inim

um

occ

urs

wh

en

her

fee

t an

d h

er

sh

ou

lders

are

bo

th o

n t

he v

erg

e

of

sli

din

g.

(a)

Wh

at

is t

he m

inim

um

ho

rizo

nta

l p

ush

on

th

e w

all

s?

(i

mp

en

din

g m

oti

on

)

(b)

Wh

at

sh

ou

ld t

he v

ert

ical

dis

tan

ce h

be b

etw

een

th

e

sh

ou

lders

an

d f

eet,

in

ord

er

for

her

to r

em

ain

sta

tio

nary

?

µ µµµ1

µ µµµ2

Co

pyri

gh

t R

. J

an

ow

–S

pri

ng

2012

Solution of Example 7: Rock Climber

w =

1.0

m

O

Fin

d t

he m

inim

um

ho

rizo

nta

l p

ush

on

th

e w

alls.

The climber is the system to analyze

N'

N

N'

N

Fx

= ===⇒ ⇒⇒⇒

− −−−= ===

= ===∑ ∑∑∑

0

mg

f

f

F

21

y− −−−

+ +++= ===

= ===∑ ∑∑∑

0

For impending slippage:

N

f

N

f2

2µ µµµ

= ===µ µµµ

= ===1

1

mg

N )

(

21

= ===µ µµµ

+ +++µ µµµ

N

30

0

)

(

mg

N

0.7

1.1

9.8

x

55

21

= ==== ===

µ µµµ+ +++

µ µµµ= ===

+ +++

m 0.7

4

d)

dw(

N

mg

dN

wµ

h

wf

h

N

m

gd

x'N

x

f

1

o

= ===µ µµµ

− −−−− −−−

µ µµµ= ===

− −−−= ===

∴ ∴∴∴

− −−−+ +++

+ ++++ +++

= ==== ===

τ τττ∑ ∑∑∑

21

12

00

0

Th

is c

ho

ice

of

h p

rod

uces i

mp

en

din

g m

oti

on

Wh

at

sh

ou

ld t

he v

ert

ical

dis

tan

ce h

be

be

tween

sh

ou

lders

an

d f

eet?

C

ho

ose a

xis

fo

r to

rqu

es

at

po

int

“O

”