hw10 tutorial
DESCRIPTION
tutorialTRANSCRIPT
CE
E27
1:A
pplie
dM
echa
nics
II,D
ynam
ics
–Le
ctur
e25
:Ch.
17,S
ec.4
-5–
Pro
f.A
lber
tS.K
im
Civ
ilan
dE
nviro
nmen
talE
ngin
eerin
g,U
nive
rsity
ofH
awai
iatM
anoa
Dat
e:__
____
____
____
____
1/3
6
EQ
UAT
ION
SO
FM
OTI
ON
:RO
TATI
ON
AB
OU
TA
FIX
ED
AX
IS
Toda
y’s
obje
ctiv
es:S
tude
nts
will
beab
leto
1A
naly
zeth
epl
anar
kine
tics
ofa
rigid
body
unde
rgoi
ngro
tatio
nalm
otio
n.
In-c
lass
activ
ities
:•
Rea
ding
Qui
z•
App
licat
ions
•R
otat
ion
Abo
utan
Axi
s•
Equ
atio
nsof
Mot
ion
•C
once
ptQ
uiz
•G
roup
Pro
blem
Sol
ving
•A
ttent
ion
Qui
z
2/3
6
RE
AD
ING
QU
IZ
1In
rota
tiona
lmot
ion,
the
norm
alco
mpo
nent
ofac
cele
ratio
nat
the
body
’sce
nter
ofgr
avity
(G)i
sal
way
s.
(a)
zero
(b)
tang
entt
oth
epa
thof
mot
ion
ofG
(c)
dire
cted
from
Gto
war
dth
ece
nter
ofro
tatio
n(d
)di
rect
edfro
mth
ece
nter
ofro
tatio
nto
war
dG
AN
S:(
c)2
Ifa
rigid
body
rota
tes
abou
tpoi
ntO
,the
sum
ofth
em
omen
tsof
the
exte
rnal
forc
esac
ting
onth
ebo
dyab
out
poin
tOeq
uals
?(a
)IGα
(b)IOα
(c)maG
(d)maO
AN
S:(
b)
3/3
6
AP
PLI
CAT
ION
S
Pin
at th
e ce
nter
of
rota
tion.
•Th
ecr
ank
onth
eoi
l-pum
prig
unde
rgoe
sro
tatio
nab
outa
fixed
axis
,cau
sed
byth
edr
ivin
gto
rque
,M
,fro
ma
mot
or.
•A
sth
ecr
ank
turn
s,a
dyna
mic
reac
tion
ispr
oduc
edat
the
pin.
This
reac
tion
isa
func
tion
ofan
gula
rve
loci
ty,a
ngul
arac
cele
ratio
n,an
dth
eor
ient
atio
nof
the
cran
k.•
Ifth
em
otor
exer
tsa
cons
tant
torq
ueM
onth
ecr
ank,
does
the
cran
ktu
rnat
aco
nsta
ntan
gula
rvel
ocity
?Is
this
desi
rabl
efo
rsuc
ha
mac
hine
?
4/3
6
AP
PLI
CAT
ION
S(c
ontin
ued)
•Th
epe
ndul
umof
the
Cha
rpy
impa
ctm
achi
neis
rele
ased
from
rest
whe
nθ=
0◦.I
tsan
gula
rve
loci
ty(ω
)beg
ins
toin
crea
se.
•C
anw
ede
term
ine
the
angu
lar
velo
city
whe
nit
isin
verti
cal
posi
tion?
•O
nw
hich
prop
erty
(P)o
fthe
pend
ulum
does
the
angu
lar
acce
lera
tion
(α)d
epen
d?•
Wha
tis
the
rela
tions
hip
betw
een
Pan
dα
?
5/3
6
EQ
UAT
ION
SO
FM
OTI
ON
:RO
TATI
ON
AB
OU
TA
FIX
ED
AX
IS(S
ectio
n17
.4)
•W
hen
arig
idbo
dyro
tate
sab
outa
fixed
axis
perp
endi
cula
rto
the
plan
eof
the
body
atpo
intO
,the
body
’sce
nter
ofgr
avity
Gm
oves
ina
circ
ular
path
ofra
dius
rG
.Thu
s,th
eac
cele
ratio
nof
poin
tGca
nbe
repr
esen
ted
bya
tang
entia
lcom
pone
nt(a
G) t=
rGα
and
ano
rmal
com
pone
nt(a
G) n
=rGω2.
•S
ince
the
body
expe
rienc
esan
angu
lara
ccel
erat
ion,
itsin
ertia
crea
tes
am
omen
tofm
agni
tude
,IGα
,equ
alto
the
mom
ento
fthe
exte
rnal
forc
esab
outp
oint
G.T
hus,
the
scal
areq
uatio
nsof
mot
ion
can
best
ated
as:
ΣFn=
m(a
G) n
=mrGω2
ΣFt=
m(a
G) t
=mrGα
MG
=IGα
6/3
6
EQ
UAT
ION
SO
FM
OTI
ON
(con
tinue
d)
•N
ote
that
theΣM
Gm
omen
tequ
atio
nm
aybe
repl
aced
bya
mom
ents
umm
atio
nab
outa
nyar
bitra
rypo
int.
Sum
min
gth
em
omen
tabo
utth
ece
nter
ofro
tatio
nO
yiel
dsΣM
O=
IGα+rGm(a
G) t=
[IG+m(r
G)2]α
•Fr
omth
epa
ralle
laxi
sth
eore
m,I
O=
IG+
m(r
G)2
,th
eref
ore
the
term
inpa
rent
hese
sre
pres
ents
IO
.
•C
onse
quen
tly,w
eca
nw
rite
the
thre
eeq
uatio
nsof
mot
ion
fort
hebo
dyas
:
ΣFn=
m(a
G) n
=mrGω2
ΣFt=
m(a
G) t
=mrGα
MO
=IOα
7/3
6
PR
OC
ED
UR
EFO
RA
NA
LYS
IS
•P
robl
ems
invo
lvin
gth
eki
netic
sof
arig
idbo
dyro
tatin
gab
outa
fixed
axis
can
beso
lved
usin
gth
efo
llow
ing
proc
ess.
1E
stab
lish
anin
ertia
lcoo
rdin
ate
syst
eman
dsp
ecify
the
sign
and
dire
ctio
nof
(aG) n
and(a
G) t
.2
Dra
wa
free
body
diag
ram
acco
untin
gfo
rall
exte
rnal
forc
esan
dco
uple
s.S
how
the
resu
lting
iner
tiafo
rces
and
coup
le(ty
pica
llyon
ase
para
teki
netic
diag
ram
).3
Com
pute
the
mas
smom
ento
fine
rtiaIG
orIO
.4
Writ
eth
eth
ree
equa
tions
ofm
otio
nan
did
entif
yth
eun
know
ns.S
olve
fort
heun
know
ns.
5U
seki
nem
atic
sif
ther
ear
em
ore
than
thre
eun
know
ns(s
ince
the
equa
tions
ofm
otio
nal
low
foro
nly
thre
eun
know
ns).
8/3
6
DE
PE
ND
EN
TM
OTI
ON
EX
AM
PLE
•G
iven
:The
unifo
rmsl
ende
rrod
has
am
ass
of15
kg.
•Fi
nd:
The
reac
tions
atth
epi
nO
and
the
angu
lara
ccel
erat
ion
ofth
ero
dju
staf
tert
heco
rdis
cut.
•P
lan:
Sin
ceth
em
ass
cent
er,G
,mov
esin
aci
rcle
ofra
dius
0.15
m,i
t’sac
cele
ratio
nha
sa
norm
alco
mpo
nent
tow
ardO
and
ata
ngen
tialc
ompo
nent
actin
gdo
wnw
ard
and
perp
endi
cula
rtorG
.App
lyth
epr
oble
mso
lvin
gpr
oced
ure.
9/3
6
EX
AM
PLE
(Sol
utio
n)
•FB
Dan
dK
inet
icD
iagr
am
= r G
•E
quat
ions
ofm
otio
n:
(+ →)Σ
Fn=
man=
mrGω2
⇒O
x=
0N
(+↓)ΣFt=
mat=
mrGα
⇒−O
y+15
(9.81)
=15
(0.15)α
(�+)Σ
MO
=IGα+mrGα(r
G)
⇒(0.15)15
(9.81)
=IGα+m(r
G)2
•U
sing
IG=
(ml2)/12
andrG=
(0.15)
,we
can
writ
e:IGα+m(r
G)2α=
[(15
×0.92)/12
+15
(0.15)
2]α
=1.35α
10/3
6
EX
AM
PLE
(con
tinue
d)
•A
fters
ubst
itutin
g:
22.07=
1.35α⇒
α=
16.4rad/s2
•Fr
omth
ese
cond
equa
tion
inth
ela
stpa
ge:
−O
y+15
(9.81)
=15
(0.15)α
Oy
=15
(9.81)
−15
(0.15)16
.4=
110N
11/3
6
CO
NC
EP
TQ
UIZ
1If
arig
idba
rofl
engt
hl
isre
leas
edfro
mre
stin
the
horiz
onta
lpos
ition
(θ=
0),t
hem
agni
tude
ofits
angu
lar
acce
lera
tion
isat
max
imum
whe
n (a)θ=
0
(b)θ=
90o
(c)θ=
180o
(d)θ=
0oan
dθ=
180o
AN
S:(
d)2
Inth
eab
ove
prob
lem
,whe
nθ=
90o,t
heho
rizon
tal
com
pone
ntof
the
reac
tion
atpi
nA
is.
(a)
zero
(b)mg
(c)
1 2mω2
(d)
Non
eof
the
abov
eA
NS
:(a)
12/3
6
GR
OU
PP
RO
BLE
MS
OLV
ING
•G
iven
:Wsp
here=
30lb
,W
rod=
10lb
•Fi
nd:T
here
actio
nat
the
pinO
just
afte
rthe
cord
AB
iscu
t.•
Pla
n:D
raw
the
free
body
diag
ram
and
kine
ticdi
agra
mof
the
rod
and
sphe
reas
one
unit.
Then
appl
yth
eeq
uatio
nsof
mot
ion.
13/3
6
GR
OU
PP
RO
BLE
MS
OLV
ING
(Sol
utio
n)
•FB
Dan
dki
netic
diag
ram
;30lb
=
OyO
x
10lb
mrod(1.0α)
IG,rodα
mrod(1.0)(0)
2IG,sphereαm
sphere(3)(0)
2
msp
here(3α)
•E
quat
ions
ofm
otio
nfo
rΣFn=
m(a
G) n
:
Ox
=(30/
32.2)(3)(0)2
+(10/
32.2)(1.0)(0)2
⇒O
x=
0lb
14/3
6
GR
OU
PP
RO
BLE
MS
OLV
ING
(con
tinue
d)
30lb
=
OyO
x
10lb
mrod(1.0α)
IG,rodα
mrod(1.0)(0)
2IG,sphereαm
sphere(3)(0)
2
msp
here(3α)
•ΣFt=
m(a
G) t
:⇒O
y=
40−3.10
6αfro
m
−O
y+30
+10
=(30/
32.2)(3α
)+(10/
32.2)(1.0α
)
•M
O=
Ioα
:⇒10
0=
9.17
2αfro
m
30(3.0)+10
(1.0)
=[0.4(30/
32.2)(1)
2+(30/
32.2)(3)
2] sphereα
+[(1/
12)(10/32.2)(2)
2+(10/
32.2)(1)
2] rodα
•Th
eref
ore,
α=
10.9rad/s2
andO
y=
6.14
lb15
/36
CO
NC
EP
TQ
UIZ
(p.4
30)
1A
drum
ofm
assm
isse
tint
om
otio
nin
two
way
s:(a
)by
aco
nsta
nt40
Nfo
rce
(left)
,and
,(b)
bya
bloc
kof
wei
ght4
0N
(rig
ht).
Ifαa
andαb
repr
esen
tthe
angu
lara
ccel
erat
ion
ofth
edr
umin
each
case
,sel
ectt
hetru
est
atem
ent.
(a)
(a)αa>
αb
(b)αa<
αb
(c)αa=
αb
(d)
Non
eof
the
abov
e.
2In
case
(b)a
bove
(rig
htfig
ure)
,wha
tis
the
tens
ionT
inth
eca
ble?
(b)
(a)T
=40
N(b
)T
<40
N(c
)T
>40
N(d
)N
one
ofth
eab
ove.
16/3
6
EQ
UAT
ION
SO
FM
OTI
ON
:GE
NE
RA
LP
LAN
EM
OTI
ON
Toda
y’s
obje
ctiv
es:S
tude
nts
will
beab
leto
1A
naly
zeth
epl
anar
kine
tics
ofa
rigid
body
unde
rgoi
ngge
nera
lpla
nem
otio
n.
In-c
lass
activ
ities
:•
Rea
ding
Qui
z•
App
licat
ions
•E
quat
ions
ofM
otio
n•
Fric
tiona
lRol
ling
Pro
blem
s•
Con
cept
Qui
z•
Gro
upP
robl
emS
olvi
ng•
Atte
ntio
nQ
uiz
17/3
6
RE
AD
ING
QU
IZ
1If
adi
skro
llson
aro
ugh
surfa
cew
ithou
tslip
ping
,the
acce
lera
tion
atth
ece
nter
ofgr
avity
(G)w
illan
dth
efri
ctio
nfo
rce
will
be.
(a)
notb
eeq
ualt
oαr;l
ess
than
µsN
(b)
beeq
ualt
oαr;e
qual
toµkN
(c)
beeq
ualt
oαr;l
ess
than
µsN
(d)
Non
eof
the
abov
eA
NS
:(c)
2If
arig
idbo
dyex
perie
nces
gene
ralp
lane
mot
ion,
the
sum
ofth
em
omen
tsof
exte
rnal
forc
esac
ting
onth
ebo
dyab
out
any
poin
tPis
equa
lto
.(a
)IPα
(b)IPα+
maP
(c)maG
(d)IGα+
rGP×maP
AN
S:(
a)
18/3
6
AP
PLI
CAT
ION
S
•A
sth
eso
ilco
mpa
ctor
acce
lera
tes
forw
ard,
the
front
rolle
rexp
erie
nces
gene
ralp
lane
mot
ion
(bot
htra
nsla
tion
and
rota
tion)
.•
Wha
tare
the
load
sex
perie
nced
byth
ero
llers
haft
orbe
arin
gs?
•Th
efo
rces
show
non
the
rolle
r’sFB
Dca
use
the
acce
lera
tions
show
non
the
kine
ticdi
agra
m.I
sth
epo
intA
theIC
?
19/3
6
AP
PLI
CAT
ION
S(c
ontin
ued)
•Th
ela
wn
rolle
ris
push
edfo
rwar
dw
itha
forc
eof
200N
whe
nth
eha
ndle
isat
45◦ .
•H
owca
nw
ede
term
ine
itstra
nsla
tion
acce
lera
tion
and
angu
lara
ccel
erat
ion?
•D
oes
the
acce
lera
tion
depe
ndon
the
coef
ficie
nts
ofst
atic
and
kine
ticfri
ctio
n?20
/36
AP
PLI
CAT
ION
S(c
ontin
ued)
•D
urin
gan
impa
ct,t
hece
nter
ofgr
avity
ofth
iscr
ash
dum
my
will
dece
lera
tew
ithth
eve
hicl
e,bu
tal
soex
perie
nce
anot
her
acce
lera
tion
due
toits
rota
tion
abou
tpoi
ntA
.•
Why
?
•H
owca
nen
gine
ers
use
this
info
rmat
ion
tode
term
ine
the
forc
esex
erte
dby
the
seat
belt
ona
pass
enge
rdur
ing
acr
ash?
21/3
6
GE
NE
RA
LP
LAN
EM
OTI
ON
(Sec
.17.
5)
•W
hen
arig
idbo
dyis
subj
ecte
dto
exte
rnal
forc
esan
dco
uple
-m
omen
ts,i
tcan
unde
rgo
both
trans
latio
nalm
otio
nan
dro
tatio
nal
mot
ion.
This
com
bina
tion
isca
lled
gene
ralp
lane
mot
ion.
•U
sing
anx−y
iner
tialc
oord
inat
esy
stem
,the
equa
tions
ofm
otio
nsab
outt
hece
nter
ofm
ass,
G,m
aybe
writ
ten
as: ΣFx
=m(a
G) x
(1)
ΣFy
=m(a
G) y
(2)
MG
=IGα
(3) 22
/36
GE
NE
RA
LP
LAN
EM
OTI
ON
(con
tinue
d)
•S
omet
imes
,itm
aybe
conv
enie
ntto
writ
eth
em
omen
tequ
atio
nab
outa
poin
tPot
hert
hanG
.The
nth
eeq
uatio
nsof
mot
ion
are
writ
ten
asfo
llow
s:
ΣFx=
m(a
G) x
(4)
ΣFy=
m(a
G) y
(5)
MP=
Σ(M
k) P
(6)
•In
this
case
,Σ(M
k) P
repr
esen
tsth
esu
mof
the
mom
ents
ofIGα
and
r G/P×
maG
abou
tpoi
ntP
.
23/3
6
FRIC
TIO
NA
LR
OLL
ING
PR
OB
LEM
S
•W
hen
anal
yzin
gth
ero
lling
mot
ion
ofw
heel
s,cy
linde
rs,o
rdi
sks,
itm
ayno
tbe
know
nif
the
body
rolls
with
outs
lippi
ngor
ifit
slid
esas
itro
lls. •
Fore
xam
ple,
cons
ider
adi
skw
ithm
assm
and
radi
usr,s
ubje
cted
toa
know
nfo
rceP
.•
The
equa
tions
ofm
otio
nw
illbe
:ΣFx=
m(a
G) x
⇒P
−F
=maG
ΣFy=
m(a
G) y
⇒N
−mg=
0ΣM
G=
IGα⇒
Fr=
IGα
•Th
ere
are
4un
know
ns(F
,N,a,a
ndaG
)in
thes
eth
ree
equa
tions
.
24/3
6
FRIC
TIO
NA
LR
OLL
ING
PR
OB
LEM
S(c
ontin
ued)
•H
ence
,we
have
tom
ake
anas
sum
ptio
nto
prov
ide
anot
her
equa
tion.
Then
,we
can
solv
efo
rth
eun
know
ns.
•Th
e4t
heq
uatio
nca
nbe
obta
ined
from
the
slip
orno
n-sl
ipco
nditi
onof
the
disk
.
•C
ase
1:A
ssum
eno
slip
ping
and
useaG=
αr
asth
e4t
h
equa
tion
and
DO
NO
Tus
eFf=
µsN
.Afte
rsol
ving
,you
will
need
tove
rify
that
the
assu
mpt
ion
was
corr
ectb
ych
ecki
ngifFf≤
µsN
.•
Cas
e2:
Ass
ume
slip
ping
and
useFf=
µkN
asth
e4t
h
equa
tion.
Inth
isca
se,a
G�=
αr.
25/3
6
PR
OC
ED
UR
EFO
RA
NA
LYS
IS
Pro
blem
sin
volv
ing
the
kine
tics
ofa
rigid
body
unde
rgoi
ngge
nera
lpla
nem
otio
nca
nbe
solv
edus
ing
the
follo
win
gpr
oced
ure.
1.E
stab
lish
thex−y
iner
tialc
oord
inat
esy
stem
.Dra
wbo
thth
efre
ebo
dydi
agra
man
dki
netic
diag
ram
fort
hebo
dy.
2.S
peci
fyth
edi
rect
ion
and
sens
eof
the
acce
lera
tion
ofth
em
ass
cent
er,a
G,a
ndth
ean
gula
racc
eler
atio
n,α
,oft
hebo
dy.I
fnec
essa
ry,c
ompu
teth
ebo
dy’s
mas
sm
omen
tof
iner
tiaIG
.3.
Ifth
em
omen
tequ
atio
nM
P=
Σ(M
k) P
isus
ed,u
seth
eki
netic
diag
ram
tohe
lpvi
sual
ize
the
mom
ents
deve
lope
dby
the
com
pone
ntsm(a
G) x
,m(a
G) y
,and
IGα
.4.
App
lyth
eth
ree
equa
tions
ofm
otio
n.
26/3
6
PR
OC
ED
UR
EFO
RA
NA
LYS
IS(c
ontin
ued)
5.Id
entif
yth
eun
know
ns.I
fnec
essa
ry(i.
e.,t
here
are
four
unkn
owns
),m
ake
your
slip
orno
-slip
assu
mpt
ion
(typi
cally
nosl
ippi
ng,o
rthe
use
ofaG=
αr,i
sas
sum
edfir
st).
6.U
seki
nem
atic
equa
tions
asne
cess
ary
toco
mpl
ete
the
solu
tion.
7.If
asl
ip-n
osl
ipas
sum
ptio
nw
asm
ade,
chec
kits
valid
ity!!!
Key
poin
tsto
cons
ider
1B
eco
nsis
tent
inus
ing
the
assu
med
dire
ctio
ns.T
hedi
rect
ion
ofaG
mus
tbe
cons
iste
ntw
ithα
.2
IfFf=
µkN
isus
ed,F
fm
usto
ppos
eth
em
otio
n.A
sa
test
,as
sum
eno
frict
ion
and
obse
rve
the
resu
lting
mot
ion.
This
may
help
visu
aliz
eth
eco
rrec
tdire
ctio
nof
Ff.
27/3
6
DE
PE
ND
EN
TM
OTI
ON
EX
AM
PLE
•G
iven
:Asp
oolh
asa
mas
sof
200
kgan
da
radi
usof
gyra
tion
(kG
)of0
.3m
.The
coef
ficie
ntof
kine
ticfri
ctio
nbe
twee
nth
esp
oola
ndth
egr
ound
isµk=
0.1.
•Fi
nd:
The
angu
lara
ccel
erat
ion
(α)o
fthe
spoo
land
the
tens
ion
inth
eca
ble.
•P
lan:
Focu
son
the
spoo
l.Fo
llow
the
solu
tion
proc
edur
e(d
raw
aFB
D,e
tc.)
and
iden
tify
the
unkn
owns
.
28/3
6
EX
AM
PLE
(Sol
utio
n)
•Th
efre
ebo
dydi
agra
man
dki
netic
diag
ram
fort
hebo
dyar
e:
I G a
ma G
=
1962
N
•E
quat
ions
ofm
otio
n:
ΣFy=
m(a
G) y
:N
B−
(200
kg)(9.8m/s2)=
0
⇒N
B=
1962
N(7
)
29/3
6
EX
AM
PLE
(con
tinue
d)
•N
ote
that
aG=
(0.4)α
.Why
?•ΣFx=
m(a
G) x
:
T−0.1N
B=
200a
G=
200(0.4)α
T−19
6.2
=80α
•ΣM
G=
IGa:
450−
T(0.4)−0.1N
B(0.6)
=20
(0.3)2α
450−T(0.4)−
196.2(0.6)
=1.8α
•S
olvi
ngth
ese
two
equa
tions
,we
get
α=
7.50
rad/s2
andT
=79
7N30
/36
CO
NC
EP
TQ
UIZ
1A
n80
kgsp
ool(kG=
0.3meter
)is
ona
roug
hsu
rface
and
aca
ble
exer
tsa30
Nlo
adto
the
right
.The
frict
ion
forc
eat
Aac
tsto
the
and
theaG
shou
ldbe
dire
cted
toth
e.
(a)
right
,lef
t
(b)
left,
right
(c)
right
,rig
ht
(d)
left,
left
AN
S:(
c)
0.2
m A
30N
α
0.75
m
G
2Fo
rthe
situ
atio
nab
ove,
the
mom
ente
quat
ion
abou
tGis
?(a
)0.75(F
fA)−0.2(30)=
−(80)(0.3
2)α
(b)−0.2(30)=
−(80)(0.3
2)α
(c)0.75(F
fA)−0.2(30)=
−(80)(0.3
2)α
+80
aG
(d)
Non
eof
the
abov
eA
NS
:(a)
31/3
6
GR
OU
PP
RO
BLE
MS
OLV
ING
•G
iven
:A80
kgla
wn
rolle
rhas
ara
dius
ofgy
ratio
nof
kG=
0.17
5meter
.Iti
spu
shed
forw
ard
with
afo
rce
of20
0N.
•Fi
nd:T
hean
gula
racc
eler
atio
nif
µs=
0.12
andµk=
0.1.
•P
lan:
Follo
wth
epr
oble
mso
lvin
gpr
oced
ure.
•S
olut
ion:
The
mom
ento
fine
rtia
ofth
ero
llera
bout
Gis
IG=
m(k
G)2
=(80)(0.175
)2=
2.45
kg·m
2
32/3
6
GR
OU
PP
RO
BLE
MS
OLV
ING
(con
tinue
d)
•FB
D:
•E
quat
ions
ofm
otio
n:
ΣFx
=m(a
G) x
FA−20
0cos45
=80aG
(8)
ΣFy
=m(a
G) y
NA−78
4.8−20
0sin45
=0
(9)
ΣM
G=
IGα
−0.2F
A=
2.45α
(10)
•W
eha
ve4
unkn
owns
:NA,FA,aG
andα
.•
Ano
ther
equa
tion
isne
eded
toal
low
solv
ing
fort
heun
know
ns.
33/3
6
GR
OU
PP
RO
BLE
MS
OLV
ING
(con
tinue
d)
•Th
eth
ree
equa
tions
we
have
now
are:
FA−20
0cos45
◦=
80aG
NA−78
4.8−20
0sin45
◦=
0
−0.2F
A=
2.45α
•Fi
rst,
assu
me
the
whe
elis
not s
lippi
ng.
Thus
,we
can
writ
eaG=
rα=
0.2α
•N
owso
lvin
gth
efo
ureq
uatio
nsyi
elds
:NA=
926.2N
,FA=
61.4N
,α=
−5.01
rad/s2
,and
aG=
−1.0m/s2
•Th
eno
-slip
assu
mpt
ion
mus
tbe
chec
ked.
IsFA(=
61.4N)≤
µsN
A(=
111.1N
)?
•Ye
s,th
eref
ore,
the
whe
elro
llsw
ithou
tslip
.
34/3
6
CO
NC
EP
TQ
UIZ
1A
slen
der1
00kg
beam
issu
spen
ded
bya
cabl
e.Th
em
omen
tequ
atio
nab
outp
oint
Ais
?
(a)3(10)=
1/12(100)(42)α
(b)3(10)=
1/3(100)(4
2)α
(c)3(10)=
1/12(100)(42)α
+(100
aGx)(2)
(d)
Non
eof
the
abov
e.
AN
S:(
c)
4m3m
10N
A
2S
elec
tthe
equa
tion
that
best
repr
esen
tsth
e‘n
o-sl
ip’
assu
mpt
ion.
(a)Ff=
µsN
(b)Ff=
µkN
(c)aG=
rα
(d)
Non
eof
the
abov
e.A
NS
:(c)
35/3
6
note
36/3
6