Baixe Fox e mc donald resolução de exercícios e outras Exercícios em PDF para Engenharia Mecânica, somente na Docsity! j
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| Prabiem “1 2 TA
Biver! Common sutstantes
|
o 7ar Sand
deling dus Totupaste
|
“sita Petregt | Jetto
|
ax |
|
me 0f trese Ssudastanees exhubir characrereshes ot
o lts and fixids under Atterent conditions.
plan and give examples.
É
&
288 GNEETE 3 SQUARE
286 SHEETS 3 SQUARE
Ter, cuolx ane Jelto bthare as Solis ar mom temperature 0º betous
Sesi ar ordidare pressures. 4! high presseres or over long persocs,
; tney extibit Pluta characteristies. A+ higher tempera dures, &é
dé Apre lrguiefsy aro become viscors Fleids.
i
Mode hng clay and sita preta, show Pleid Lenovo” thea sheares
Sloiwty. Flowever, Éhes fracicare under sexeldenta apples stress,
which 15 acharacterisie of solids.
aa PoInpaste behaves AS di So! then At rest 17 tne desde. Inheo the
ese +ube «3 squeeged hard, tootn paste “Bowas * out the Spot, showing
fura behávior. Shaviag cream bebaves smart.
Sand act Solid Usher 1º repose (a sand pre). However, d+ Pleros *
from a Gaout Dr down à Steep meldene.
o
Frotltem (2
va TA
Civen: Tante to contain t0 kg DÊ Op at lema, 35%,
Frad: Tent volume ara diameter nt Spherical.
Solution: Asstume maeal gas benavior,
Basie eguanons: p= pRT
(=
Subbia tu tina, tue obtam p= MET,so
vz
(p= absolute pressure)
%
From Table A.b, R= 259.8 Nunlkg Ke, so
r
y = UR
= 089,25 LENAM (234EDK, mÊ
* kg * Cx rob + ro/x10 JN
W=92.0567 mm?
Fr a sphere, = EmRS = frD3,5
4 L
o-fe8]! fp sonsa et saum
0
«DU
Problem 1/5 | 14 TA
Open-Ended Problem Statement: Consider the physics of “skipping” a stone across the water
surface of a lake. Compare these mechanisms with a stone as it bounces after being thrown
along a roadway.
Discussion: Observation and experience suggest two behaviors when a stone is thrown along a
water surface:
(1) If the angle between the path of the stone and the water surface is steep the stone may
penetrate the water surface. Some momentum pf the stone will be converted to momentum of
the water in the resulting splash. After penetrating the water surface, the high drag" of the
water will slow the stone quickly. Then, because the stone is heavier than water it will sink.
(2) If'the angle between the path of the stone and the water surface is shallow the stone may not
penetrate the water surface. The splash will be smaller than jf the stone penetrated the water
surface. This will transfer less momentum to the water, causing less reduction in speed of the
stone, The only drag force on the stone will bg from friction on the water surface. The drag
will be momentary, causing the stone to lose oply a portion of its kinetic energy. Instead of
sinking, the stone may skip off the surface and|become airborne again.
When the stone is thrown with speed and angle just right, it may skip several times across the
water surface. With each skip the stone loses some forward.speed. After several skips the stone
loses enough forward speed to penetrate the surfacg and sink into the water.
Observation suggests that the shape of the stone significantly affects skipping. Essentially
spherical stones may be made to skip with considetable effort and skill from the thrower. Flatter,
more disc-shaped stones are more likely to skip, prpvided they are thrown with the flat surface(s)
essentially parallel to the water surface; spin may e used to stabilize the stone in flight.
I
By contrast, no stone can ever penetrate the pavembnt of a roadway. Each collision between stone
and roadway will be inelastic; friction between theroad surface and stone will affect the motion of
- the stone only slightly. Regardless of the initial Val between the path of the stone and the
surface of the roadway, the stone may bounce several times, then finally it “will roll to a stop.
I
The shape of the stone is unlikely to affect trajectory of bouncing from a roadway significantly.
5
* Compared to the negligible aerodynamic drag in air.
Problem 1.6 (A TRA
Open-Ended Problem Statement: The barrel of a bicycle tire pump becomes quite warm during
use. Explain the mechanisms responsible for the temperature increase.
Discussion: Two phenomena are responsible for the temperature increase: (1) friction between
the pump piston and barrel and (2) temperature rise of the air as it is compressed in the pump
barrel.
Friction between the pump piston and barrel converts mechanical energy (force on the piston
moving through a distance) into thermal energy as a result of friction. Lubricating the piston helps
to provide a good seal with the pump barrel and reduces friction (and therefore force) between the
piston and barrel.
Temperature of the trapped air rises as it is compressed. The compression is not adiabatic because
it occurs during a finite time interval. Heat is transferred from the warm: compressed air in the
pump barrel to the cooler surroundings. This raises the temperature of the barrel, making its
outside surface warm (or even hot!) to the touch.
Veg E ta |
| Re ottvem NA |
We A Nondord conditions — P= NA Say To sa
ET , RE EU TU PO SE
Ada MR 28 a dm da, Corse sponás Neo AA qa
Tmáo Ac “X sacos A Sa CL Sade.
2 o SUA Me, sJolue .
Gisen +
Ea
“gs * O. 384
A (a ca Sa l&D
8 esc vera Pesa
SA son. a
s& AM WºR A NSAV
p= as E SAN Come &
R= o .O Ns Now tg” R
A
“Ne TULE a Sena, * een REA
a USÊS “0a SS
2 a = *ON B
E na Ee = o=A Mg e tom
— lv tos º
Rã E (EA se =4, ucidi total,
Nes
o a aVh 2 -
“es A (49 Les8N Ml ass + (oc À
| Froblem [10 Veg E Tal
Crven: Pet food can H= loZf Immtzoto!)
D= 73% 4 wmblotar)
ma 397t1g (zo tot)
Find: Magartude and estimated uneertarntes ot pet fovdl deosite,
Souto: Densitg 48 p= mM = 280 = / Es 9” pela, Db)
Frem uncerta inte analyses
“ - 4 no 2, Da
? (G Sm tm)! + (7 up)" + (é a SE Um) Jº
Evaluatag, mf mM
f
anta 24 Bm ss. =
P am 7 py TOM À !5 tem a 0,252 %
Dor DE um €
Eri Colem ccpstpa É!
e f 7 D3k P Tom jp = É LIA
H ep Hen gm /
f e = 1) 4 der. = =!
ee e mpg” 6 F TD 15 Um = o = É 6, 980%
Substrtutras
= flo XocsoJta [Caxtrsn] e Ico a853) j +
ue = 2 2.72 percent us
= E TT, 03*mm 102 m$ -4 43
Vo & Dim E x 102 mm DIna8 tzyuxoTtm
mm. 3a
Ma 30 3
f=% dequio a * 555 = $80 bg lm
Thus p= 930 + Ez kg /m3 (ro to 1) P
10
Problem tl Boo E Tl |
Given: Standard American gotf bas: ms lbttoo! og (20 to 1)
D= Lb$t6,01 1. Coto!)
Find: (a) Desta and speerfee grau ty.
(e) Estimate wuncertaintes 19 cateulates values,
Soluwtron: Density es mass per tunit volume, SO
= m 8 m
p= Da = =b m
do Emrr3 dr) + D$
, 0.453 Jeg 1n.8
- e, fbtoz;—— «DSG DO = HBgk 3
PT Seas los topisyêms” O 9 Jm
and
S6= 2 . H3odka LÊ (jB
Pmo m3” Jo00 kg
The uncertasnta 1m densitg 18 ques ey
fe
us =2[[2 32 “u/Dô Fr
? ls Em tem) “(e ES “p)
m .
SE” 2 = É=/ ; Um et BM = 0.617 percent
D 9 2 3tb m Vos
Cs. D/ St mi Ho k ma
e o Pp 7 pf & m (32) =-ssup=t0598
percent
Thus
=: [Cum + (suo) +
tflo.en)+[-s Cosa") “
E
'
Up = É 1,89 percent (E zug eg los)
Uog = Up = 1:89 percent (t 00214)
Fenaltg,
P= 30 + Z:%4 kg/m? (20 to 1)
S6 = 113? gor (ro dba 1)
!
| |
[ Beam nar | Nes ERA
Gisea, SYondorá VBekcás aee Gas
ma 4341034 Ls No
D= ANSA * O. Bem (es Vo 9
Fá s DenÃa asd Sqeef ve arossias,
&y E XsnokÃe rca Xe rà ves am codes S es podede!
Ed
Veni ds esass SRA Snes SO
5 Ss
<A . BW A
CIT axe am GEN =w 32
3
- N a
to . — = vko A
Fo º casa a, * ASAS vê s Rae
3
end r valo Ra, « ss = do
SG = Qae a * To
Ne. sea sk oasS RS darei e sem a
dq= Us 3 Se “ X 3 SN
A -
t
Os
Re gs Lo sos SB som
ap Dat MN cafe Na
5, EE sas a
uy= + = o não À Ê
aus de NN
de= + (uam * Caos = (oi Teste t
Aç= tia A Cuega Lao
Ve dg= vas (x set)
Soros tina,
p= io 5 ata talo? Go NR
age Nidk E O ora (Go Ns Ss S&
12
Problten Lis | Rep E Ca]
Solution: Laterat acceleratios 15 green by a = VIR.
Given: Lateral aecelerados, a = 0.823 Z, mtastered om Z00—F
chameter Skid pad.
Path deviaton: t2A
, | mta stertentar uncertanta
Vehicle spted: LOS Mph
Find: (a) Estimate unce rtaunty m lateral atce teraton,
(bo) How could experimental procedure te improved?
“4,
From Appendix PF, Up = teu)? + (ua) | ”
From the gue data,
4
Z
ta
vV=aR; V=lar =[0.822, 32.2 88. « 100 +] = 51.5 44 )s
Then . h
u,=t EV sós m Ss A, LO «+ O0ue
- Br ssh mm Bhoo S
and
Up =
h
t+
AR <l
u
ns
+
'
+
s
Q
N
S
Y
"
us *[(2 x 0019272 + (0.02003t] = 2 0.0347
É
Ug = [347 perment
Experimenta! procedure cota be improved by usa à larger
Cirere, ASsumng the Absolute errors sm measurement ar constant.
For D= 400 ft, R=200 tt
t
u
v-/a? = |o0823, 32.244 1008] = TBA/s = 49.bmph
se +
uv =t 3.5 moh
= + 2f
GIL mph
= 9.001 ; Ur
4
4
ua = +[(ex 00101) + (00003) =tp.Dzs Or + 23perent
15
Problem lt
Neg Sd
Given: Dimtnson! DF Soda Cas:
“f
D= bb mm H
| H= HO mm —+
, ot
Find: tuas urement preasios needed to atiow volume to be
eSbinatkd witm an unerrtante DÊ LOS pereeat or kss.
Solutron: Use the methods of Apperdix E:
Computing equadons: = Tem
4
as e Bs ut ESET”
Since yo TOM sen 2% + TD and PE = TDH
4 ed q oD 2
Let up=* de ana um = É =, Ludes PIÃO,
-+[(-Sg me ex) a(o mpH 5º] “= al(E “+ 289]
Lay - + q 2) tapa E 2) (5) >)
Solvrag, , é
uyt= (Ep + (silo +69)
bx d,005
: 1 a Ê
Ux = tus + (eu95] = 2(0.003:64) 2 (0.0078)3] = É 0.00444 v”
If 6x represents hate tre least Count, A minimum reso litros
of about 26x Yo32º mm 8 nteded.
o
= ato MT, = tô
óx = “Ra + (9º wo = ; E z Et b.158 mm
H [Cs =) Tre) <<
. 4 8%, p brsg ma 4 -
Check: Um H To mm * É Jeuy x10
+ 8 t/NSE mM (+ 4x 1978
4o D dé am tesixto
SG
i
!
i
i
i
Edo: Preeisida db twohich D MUS! be measured d eStmase
Problem 1.11
é
Given: American golf ball; m =| b2t0.0/ 03; D = 168 40.
density cofre tocertonta 0f 1 percent.
toluton: ApPIy UnErtanty concepts
Definihos: Density, P = = gy =E7R = zpÊ
bg
Compunoo equation: Up =S[UERE ux)+ =]
From tne detinrhos, P = Mo. tbm el, D)
TO DS
Thus Mal d 2 =
€ am” vao E = 8,80
me
Up = + um) +(3u5)]
um - Um +9U4
L
72
So/viPg, Up = 24 [uu]
Form ne data gen, up = to.nivo
= 150/06 (+ pooh
Lbz 03
4
up = glfo.or0) -(0.006 n*]º -L0.00262 0” & Dzbih
Snee ga o ; then
SD= LDup E É IL 16006 = É 6.004 IA.
<—
| The bai! diameter must bt measured to a precason of
| 000844 10. (bot mm) or beer do estimate density
mito ES perento À mem or Catiptr toutd be Usés,
Seg E Cal po
“
Problem 1.19 (cont'd.) 2/2
Uncertainty in volume of cylinder:
ô= 0.002 in. 0.0508 mm
Y= 1 mm
D(mm) Limm) LVD(-) vo (%) ui (4%) uv Ch)
05 5.09 10.2 19.2 1.00 20.3
0.6 3.54 5.89 8.47 1.44 17.0
07 2.80 3.71 7.26 1.96 14,6
os 1.99 249 8.35 2.55 130
0.9 1.57 1.75 5.84 323 11.7
1,0 127 1.27 5.08 3.99 10.9
141 1.05 0.957 4.62 4.83 10.4
12 0.884 0.737 423 5.75 10.2
1.22 0.855 0.701 416 5.94 10.2
13 0.753 0.580 3.91 6.74 10.3
1.4 0.650 0.464 3.63 7.82 10.7
15 0.566 0.377 3.39 8.98 11.2
16 0.497 0.311 318 10.2 120
17 0.441 0.259 2.99 15 13.0
18 0.393 0.218 2.82 129 14.1
19 0.353" 0.186 2.867 144 154
20 0.318 0.159 2.54 180 16.7
24 0.289 0,137 242 17.6 18.2
22 0.263 0.120 231 19.3 19.9
23 0.241 0.105 2.21 211 216
24 0.221 0.092 212 230 234
2.5 0.204 0.081 203 249 253
Uncertainty vs. Diameter
25
——— L(mm)
s uD (%)
É DI NO ds uL (%)
s uv (%)
És
$
E
z 10
E
s
s
E
>
o
00 05 10 15 20
Cylinder Diameter, D (mm)
e)
Problem 1.20 | sata
Given: Smal! partite accelerating from rest 9 à Fluid,
Nes mwetght 18 W, resistirg Force FDkV, where Vs soccer,
Fmd: Time regured do reach 95 percent of terminar speed, Ve -
Solution: Consider the parriete to be à System.
Apply Mecutons Second lauro.
Gasic equation: Ef = May Particle <p
w
Assumplas: 4) W (5 nek werght
2) Resisting force aets oppostte to V
Form kv
&
Then
= Wo - =emY «May
2a =W-kV Ema, =m “78
º dv j- &
e” = gu DV)
Separariag varabtes,
ev = gde
fe
ty
Integrating, Psting that velocity 15 ger mibaitys
v Voe
dv W E Vl = -
Evo Elm Ev)]= Sgt - gt
vv
or — degt bot
“by ve Wi ,-t8
mEv=e Wo = [4 e % |
But VV as t-200,S0 WU = = | Therefore
kgt
Voe E
We
tg
when É =598, then € W = 00 And car =3. Thus
ke.
+ = 3w/gk
2!
Redem NRA o sa ta o
Gosem 1 Senso ande gsreeNer os Lee cad R a Sed
E A vo NA, Tens A Ot Ca. o Co des A deco À e
Trad Veares coque ad às ceods “5 ves ask É Sereno
Quad Se
SeNNsos: Coma: “das See qatde No er a. saSue, = 4,
Sra Nes astomd Nous p=
. . o ads
Rosie Capelo " 2“ Dea Rsbes—> +
Mentiras 8 Ne net AN “
NEN Resina Sor Bs sqesisÃoA
da sã ES
Ven, Les sa-ÊN = mass E
Sta Sô FO Sa
os 3 a
VA Nessa sua G4= O end
“A =To Rs
VN 2 DN
da” a
Sears ssos valsas 3 3
o NS à
NA 3 “es
a
Neg Es SA “Ss, A dessa > 355º a
= aa. A Gs ONE
er O AE
-osss Cada ss - GS
-AÇ Toas + tnocos)= E SS
a los 4” = 2 8
Ss a
Zz
| Prodam 23 sata
Gsen Verá Necsus A Canas = NEXO TA Woo fera E
astow dé (N= De. Negheà quo cesstance .
Craár Cake É (65 aqued . esa ts) esa , S artes
UR: an cedensa squid | asd 43 sao sas a Laden eh
ENT
NX =utaÃão A Cossissasã) 4
Zeus a Á= -"a te Ato Fe TRA
= na , and fç= 2tço- esula, TE-+& —+
NV
Neo , ves =-"a 5 Nrdr = ada ,; O- 2 =-as
Boo No Soa,
E = não (So U=tiç= cena (on Cs uÃ= Teo
Tosa À a= LS e
S =
Os - As cu:
“Js e 8
Ven - E!
. o z
si sotado Soco aade AL
“= [resta ANDA a Nela NR sh NV = nan “4
se Ta Oss Pas:
Frow Eq.s Se = Seas = A AB O S=Es Va A
v? 4 ? “
= 88 [Cesta ses aa gd Se
PC E ASA (Eq ny and €,= SAY (eq st are qrescieá
Nasa
Eq. 4: Initial Speed vs. Max. Helght Eq. 5: Initial Angle vs. Max. Height
= 8 a 80
E
E q 8
£ $ 40
i “0 É
a < 20
= 2 m
s B
E E
E o so
0 5 10 15 20 25 30 º 5 10 15 20 25 30
Maximum Height, A (m) Maximum Height, A (m)
7
o
| | |
| Vedas, AZ | sata dO
Gac Ma doser AA = gota Ghca Neca dosss.
EN Coca so “y= Bo a ea, 46 Sperd
Tiesd «3 Vesenhad Squad Nes da deem
às Vedrcol LA anca No cas dA V=0.ask
«SD daXicd dass To cen some squaA E Es
AA: EN Squad De SOS, ond id SEMA
EANNINS . Ey= da
Comedor Ms Anser as à sue: so
. a
Rate q ves UE ad = wa, a - “ay N
Sesi. “ENE conÁÃsSss V=o A Neo q8=o -
. = QN as cpgosTe. No SN
Ren TE “ay Yara ano
SA TarençãoS squud a=o amd S=d Rec.
=
RW RAL E) sã Nç= VE «)
e ae «ss . a. 4
Sj= | o via * oa Ne” tab = Sa.a ele Vw]
às To cds. Loc s A dsoas A q toe crendo Os esqrestim
cerca | and a, From Tg d
q Ze A Sd As
Sac VS o= nd or *— Sal a de
Sugertiia dos Conhes arA Ne pokada -
A és = e &s
No Mnt “a, C A
º = * -
- 2 A Qs— OL = Ss er epi Es
os ce PA -IS . EN
Toc Aco asa, Ven o
q: - Ly tota E. QuÃa-(oas) VÊ mSa s
” = GANA S Vaga e
O For Cree Soh Us Red cus Tess once ZE “ae es
JS cn tor PAES “ E as or
açávia, ]
sz
“a tb
I :
[Rem Nota Cola)
Nessa “a, + Voa Seen ap «
s= Y + eras (sam m E E
Nes mar Ce area au
ssesà cas LAN Bases Sai
E N VN Sweo |
venda O gps squa.
From a voe cos varie S= AS as
EA Asi Ta
N- e.
Ta Seia é os expressos for de SAS voe usei
3 Es= mna- Sar =" 84
Segosesaãa, sos çNdes asmá sds SETE voe rose
C SR = CEB 2 z ad
sto ESA NAN
E Na A Se
asd Ae eta
Ee RR
RENAN ES
da 4 GR
e ES Da A ES
Fem
“as A oná 3 ore NASed. Nona
Eq. 4: Speed Ratio vs. Distance Eg. 5: Speed Ratio vs. Time
10 10
3
«Ss
08 08
06 06
04 04
Speed Ratio, V/V. (-—)
02 02
00 Do
9 100 200 300 400 500 600 º 5 10 15
Distance, Y (m) Timo, t (s)
t
20
Problem 1.22 [o AU |
Gen: Unit conversians
pad! Convert (0) 1 Pa to tbé br* (pressure
(b) | gatioa to lrtees (vo treme)
Co (Absinto tofos he? (uiscos tes)
Solution: Use basic detiarheas
(o PRB IN IM, Lt ulo0esa al À pg 107 JE ha
me mu teaE S Zn =
OD ligas = gal, t2/ 08 SD to00L « 379
gal 14.3 mã Te
(> 4 Mestm' = [NS 4 ef ul 2048)tm* = 2.09x197* tofas
mm datas NS Pr fp
qm
320
ds
(b)
(c)
a
Froblem 1.28
Givea! Unet conversteas
pad: Convert (a) | Nro? to def has
be) lho to Men ls
(es 4 Efe lb to nº
Solution: Use taste debarhens
as IM ef
(a) mea a AS
b hp = 55 bh bf, 0.20%
s 2+
(cs 1 Blu + MEL
Tem +
| Blu 2330 Meo fa 2530 1.)
tém kg
31
x (075)?
BM E APIS 2 To AS
ar be * 7%
ER IbÊ Q3098 PO x th PIE
Ibema
| AS
Coressure)
Cpswrer)
mlks (specifaé ererge)
= GE IDT Jet (as
Mu
/a:E
um Cb)
s
bm
«em
Jbf 0/9536 kg
(2)
a
| Frobtemm 1,29 | NEVIS i
Given: Densrtg of mercury 18 PP = lb.B Sig Its,
Aceelerafion DÊ grau nte on mass 13 Gm = 547 fé fa
Find: (a) Specrfie gravitay 0f mereursg.
(6) Specific volume of mercurs, sn mB ug.
€) Specrfé ceght on Farta.
(4) Speeifee evaght on moon.
Solution: ApPly detinitirns: FP = Egor lp, sã ele
ro
( Ne tnGt the mass- based quantirtiés (56 and v) Are independent of
Gravity. .
32
Thus SG = U3slug, fra
” prs " nagatag = Be 56
fre do so4e mB Slug tem “mB
= , = 137x407“m
“es Slug” Ars *2,2 bm OGSBL kg 137n40 ag v
€—————+1—
On Earth,
e
Y = 2 3sug ser te MS gua be /ty3 g
E “as SET Slug fr Per der)
On the moon,
X= 23 SUS surfe Jefçã bt fer3 %,
m Ars * SE Cao dp o et 4 */ , [
| Rede var | NIVA
|
Goes, Sears SL anão Vomiá on dsrentions sf Lora N NR,
& Rome
asd S Loro = Sasidedh = 24
usE Nes = Peca- quites = AS
osà E tece = snáã do = ass
Em 8) comserseos Lados No SI sas
do cdros ás wnks sf noss a Refis supuns -
Grom.
8 Tie cossesseos fados qse
Noca Ne
Peas Quirtes &
N SA dá = “as
à From Nekoss second laro E = a
z .
= (nd RN a oca Be vet sf vossas
NR E E
“Res - eras
“Vo comseà mas do SE
vago Pele. Cos eta 6 É Seegader Raro
Neca od redides GIRA CÊ
ge E “BSO ta
mass (Nei ÃO) qo
enoss (SE vnAS) SO
às
Vodca = Vsodcdloa 4 ZA VASO ÉS 2 SANA
Ts * a
N
N Bicas po es = Recs oder + as —* O dote = sow TA
Eeé
Vedas vas SAO
| Casaes Gen
SL anão Taxá en gas dimensions of
Lea e Nota and Nena
SRA es desta = tocado
VEN SE 6 |
ER SAM = A :
UR É A ENS = “9 “e
Fred) conserta o ta SE us
de as
neo a e psd, ade E anão
(O ceNdisci Iagas ST ora) Nº Bda sf moss
Sais
Se comaer sen Cadiers ose
vraselnad = a itasaisoM « SAS, « NE camada do vmazad
& Seas, NeE
Lx
V Nsmemass = Ceraron » SA cida = wa o
à God = 8 Sad « On “e ESA Gamito us 22
To Loshoa Secos to a = OáS IN
às USB Isost dosmensens af E RS Ba Te À ONA nosso
ase Garendosu dimensions
. as e
Sea ds To Bs t=T
From tias second Vas Tema = “a
so w =” Ea , a .
= = NT Na > cod Be ani feras 1
= IN
(Lana s
3 To cossesl enoss und Xe SE e
= asedi Aromeran EVA WIN Loo & x Ra
Cadu “ioasdeid Temeras Qu vê >
« TS
W = OAOS ta
3t
| Proa 4.3 |
ATA
Crises, Comgia, comer tochas
as a mass ST ras Ra, unem CNed Co8S,
Ted”, «oo NheiaNt of ússias a Vo esses
NA CA
ao sachasme ams RÉ
Toa <u
SEADE Coe
Ra
E= wma
si = tão indo
hs = Wind = sa Ade
“ao Nas Seg nana See
Ne ssdiare 1S quase
= ae nd - io
s R& R&
From Tonho NAS p= tas Ava Lã? A Ve AS E
Code SaalMs
& s x Sus Rea o a e
vas duo nek vão - edad
37
TAS Neg Snes esmçã 5
ca sf ENA o soda, Sa,
e tas = sa aà NNdz
| A
| |
VeNdem 2d ERECARA
Gin: Sedociy Geda , T=atitay Casbe Na)
Vão Eqpedixem toa Se Las dAreorhnes , and
A Ce presa one direasmunes Lor LZO and se
Sototios :
Ne Nope A Ve DAracelives m Me *e-q Name o ao da
AM V
n Ss q
Fo Na est -tsu3 , Bem um or sr = - Sa, - Nena
BLv.ta
Sm “o cal
To shoe Be AL Tesenliol ea putos
(êa (E Ss
, Sapos che sas velahes ond Sucpa
47056 4
Ang = -D Ay + cond
act
Qn a = AnWã ata ese condats And
Nem est a&
A Ê -& a
or Nena “= E) “ou CAN
Sida, Se condeno a ond do arm Cred.
E . selocê
& ERA y qa spa AS Gere, Jolaes
Vá es resmas asse otiamned 'ea, ass
Ro le condad of tiearatron -. .
“Severo ed e ae Nem avo = 4 ando Re Degomines ase
ANA “a Ne eo NO
== s o as É
as ã a
Fa cre ae fo abs onã t=0 fer ad 4.
A jts Re ese en = S 5 Re esputa em
of a age Ba .
Cosses se Seus la AMCeresk
sodoes fc
Co
“o
| Vedder 2a | azoto
Gases: Elas OS aner O - «Sqónteé. a e Sa
Lda Nec ais oa LS
Ted am a: anal ros Lor a Gas Sreases..
NA a Va E cnsrbeas vs Va Ca apa dra q
Asdar se
Ssesal qo ka SS eo Tae
DNA testa
The Nope € Pe Arenaimes NN Se tea ane. & ae a,
da ot
FATO a
Nes Sa - ENS q Nes = OK and sm Nesse,
MAL Tl BA
dA a Ta
= & Do da .
ara 5 o . VAN cas Na os cd
4 e ENE = cena = ta c
Nina com aa Sie do Nos are que -€
Cor amet Ve AM = o
VMA A Breasses “A Nur &
AM are Soliaas o cesdiast SS ES ASOA
Toe Se Asqoeg ca Be gos (SS Be
cana BE COS arena? e na
For c=o , as o dA and tro Sor sd A.
Ne. NES = che e
Corea A = cast oa ,
Me Gas Ão Qe Coser
as Lee = E “o Ve.
ão ASaa
a
tl
Rede 2S | 222 Ya
Gases Fios Sick . “qa (GR Aa DA
—. adere = 46 Gel. o aê % Ev, coarbarlas en.
Met: a fers Areganlimes seduda Oo a RA qosses |
Rrcrg o qo ED Ras |
Soc. Snes A ARS anã = Ás do Pe same
a da Nos . à s
Sicepanhmes ase qosada ko QB sda so Deca s
TN - 3
RS RES É Aa Ha
Nine da e SA
S AR q =º
E Seacia,
dnad De LENA er (ua “a = (ua Sar cen
tara Recuad tas O (as (Cat = = 2d
= (e 2 4 tendoÃL = do
- SD uea coma = ão
|
1 Be-o36, ]
Cato do
C=4a
c=2do
C=tu do
C=aA
C= do
|
|
|
|
|
|
|
|
a
l
tz
| Problem 2.8 | ara ta)
tvs,
mato
Gun: Steady, ;ncompressible flows im Ay plane vorte
V AY =
V= t+ 2 É when = Emtls
and coordinvses arm mn meters.
Find: (a) Eguanoa fr Stresmmbine farough (ug) = (1,8).
Cb) Time required for à fieis partie to men Prom
x:lmtdbx: Sm.
. sy >
Soludsa: Tre velociry reta 1 Veuttvf do ur É = 07
Co mpttin. adons: =! =
mputing equadons E) nn ta ST E
Substrhatidg, dz = az - Z So E =
y
Integrar
“grana, lux = bug + E «ly rtwe or qy=ey
Pr por (Xg)=(h3), O =5 -Í
Thus x -5 18 equateos (a)
Por a partict, up = SE = 2 om xdx = Ad
. 24
Integradng, f xdx = xt a So t= tão
2
+= LAB meo (tm, É lts ; co»
m [A
as
| Redes Va | are t!
Cossess + Saci CAS A - SAL vas s WSvere ese N Fa .
Find E Srs Ss sd sm so de ses orá. ay De
Vos asur
“a prior A ç= Ce =
des Sam pane = qc E SN Poeira
A
tem Compare psNeline vos Ssessrce Sender Some vet
Sewlaem
So 8 God veda 1 Ve Ye Sa Lo a N= SA U as vo
os essa “o o = OM NR
ANS & da,
Cs Ve qR O er E = ad sã (E =Çast RS
er CE
e CR da a E soa VA e Lua “id
“LX egrara Egs. 4 oras we SK ass
ESTA
o as A ane, : +
As
os as e e oná dee ass
de as AA + Aa ce o A e ORA =ce
. g, a
às To Has Ve
epa & Go pelve vs Amd À Cory
Ve Porosvels ae tape. R N
A = cel o MNT =
A
A da 2 = A oe Às “e =
ar Cree
E apatia, eAgrestismo So que SR
a SE AA dO o CM T=dn o
Ca
SD e
do
Ce (AO ê
Mt t-o
tm N=C nas val. Crea ass Bem
Ne qe & Ve paid 1& A = 2. RESNS
(Ny Be Aressine à Be Au None Vos à da. 4= Ta
Ras
da a e a zo Ra cos ta Reage do PANPEN
ta 2 Ana - E cs Saà Ane
. |
o SS e Shaus e. =c NOS Noca Qu !
neo ces este (OD o Pes
Lito
| Problem 2uo | arete |
Gren: Velocity Bela, V =-Axi+bg?, where azb= 25)
and Fne coordinates am measeres 10 nters.
Find: (a) Fara mttr re equa dons dr parhele Mmoten.
& Parnine equador fr particte at (ug) = (2,1) at t=0.
(0) Compare wu elrea tre Trroagn same porn.
Solution! Velocity fieis is VEL +VA O Urcax = by
Computing equanors: Up =, =
dt é
das =
dk 'strearmtine UM
up = SE «ax do Hb = - ad and tw = at
Thus x= we
- Za
Vo de -=+by So dE = pbde and bus +tt
Mus y=met = met Parametrie tos
: at g
Sotumg fr CE e “%” = So Xy* A
for particle at (X,%) = (2,1)
Xy = Emyilm = tm? Path tas Cb)
Pe Stremmtina 1s dy «E « &y .44 —-E
dx 2 ax ax x
Separating vanrabies and sntegradang,
Jg + Sd =0 0 buy tlng *CCatol or ug =
Fer pone Cy) = (2,1) the Stredimbne 5 6x mama Zmt
xy = 2m Stramine (0)
Note! AS expected, patinhas and <treamtne corneide for | |
this steads flow, |
“m
| Reddem TAS | art
fes ResACEED WeaiTE
4858 OO RECOLED WAITE
pa
ato"
Greer: Ledoiisa Sida We pa (MAD cul | Raros,
B= Cel Ss S coordmplxs wncasared ms wekgcs .
Vi: Re gofico Re quaiide RA posses Reougin
Se ge NS E Vara Neo + Na SS
Cos OR Re Sremahimes Po e
ey e a es Nronaço Cia Ne
post & Me mãasks Teo,
Soh io:
. de
For o pastiide (um ES omná nr= tala
as x
ds BA LAO = de (de = ( so (ss RAL
,
4 + te e AÊ
An Des sttasat = arabe a de sçePS ES) :
a = Cu = Sal aa , (cas a “. as ap
. s .
Ne qe eso ta qNRies “ea, esa, Los Svcnise hos
Ne Rec oshine ds Sena A (od aysem KN fem da, Ned .
res
da Es E a
O UC 4 B+
Coe da 4 = Chtslne e Ads cg MD
Seo nune, Broa pos GAS ques Css. ss OR
sotoá is via for WB, anã E DAS o
=
mt tro , Aa 3
5.0
aos . ceara:
red
da
Streamiines: t=0s
Pathine
Tets AEW
*= às, <=
30
vm
7
n
o
20
1.0
00
so
Seis ey:
| Recedesm Cada, | rreda
Casas + Seossa, Si Lda Ni= GA ANN + EEN “Vesa arte =, -
p= os (ANE COTiaNoNes GSE ANCasaTEd «sá ds |
Na: Pe GAR Ca
à este org Este
NY e 8 WE Veda REM : 2E NS
Compara SAR JEANS SESA Mesa
a8,
Sasme pork dA tos, and às. =
For a spas teve , = dela end = du fax
Nem Us ca (DS e É E & = (aca
ta É o GRASS and A Ares
ãçãs e A=te ri
1 = exe da , (eat = Ce asd ar “e E
Nie come te ssa sas Ras ron bos
e a N AS ve
Se A censo + ÀS Cousnd (a S. gs Cam 3 AX = “
Sem a cd MMA Cd
SH O qa GO á a A
de - o q
CAM LS a E Ana sâne, s CAD =
tensao MR Goat a. ON ques CA. Ben Ss
ss Aisha, Coe OU 6,4 od € um Kas . |
+e als «285 Sah |
Wui=o ( 4=
Ato, de Ne “9 fareamines: 1=05 t=1s
EE te, d=w
40
30
y tm)
20
1.0
00
E)
Gasem: doi
Bwis
|
| Veces TAS À penta
ba Dedo asE essas. VS des
RIA Ve Eae: ( deco Ro ass otite sa se
TESE e. Ae ER ERAS qavt ( As ud
Ye
END ash e Arterdor qoind Seu
ve DOR ço
Senso:
Cor a gosiide us aro ord = “a
E qe ado lg (e õe |
Aos dn Ãç= LA anã pa A er E 28 ta a,
“se sa Ago (alpes ã euçãs sa ae e Se
Se Ne eng doa Sases. a E cs diodos
Va Necarhuve vã Send (a RS Es dali E q
Sem
da = So ana Be Rea aehise Pena aa E
Ao . Ts d+
ANA Sa (é AS q + or a der SAE =
o Ee EN (Ato ond se ua EMT le
Wux=A
= Nas tn ha
Xar
toa veces / !
= ars Ata
10
Streamiines: t=1s / -Pathline |
t=2s
y tm
52
|
[ Veces EA La | A
e, se Es Bão esta XIS |
=. 2 SÃO
EM pets CAN ouned Voa sms çõe EO om Ve Sair
vos po (ao A tras |
50
Streamlines: t=0s |
|
“0 |
30
> 39
10
00.
e a 2 4 6 8 10 |
x(m) ;
|
|
|
|
& |
| Rede EM | tt TA
gogans E
Ê
|
de
A
Gresess Neceias, Ca Ne GA DT + esa . Vere. arCeis, |
= OF, ond coscdsneies gre veogaredi w enesass |
Ma: Be Acensire Ra qustes Reu Be qui Cao gi
asa Ma leo GERE sue Na
Compete 8 Pa drecanhiras ASSes. Proud Re
Sosve pod A NES Naná US
Sedutor, ;
tre chne Às as connedis qosdes as passed Pesada
po ea A costas turmas Teo vt, ond 55.
For o qostade | as dela ans = Salar
* à; Cas e
ae ate E ara SÁ é - (ecciiar
“= o = Me
A Bos a&a E - aus Blu A
AAA E (q
A= A
so da e a x Vo . - A
= CA = as Va ES ão etasd Ss |
Sad RA das Lor ade, As, Ud Lo vees |
As ARS a AD , us ES 8 (a Ná NA
Re decoder Var ea Na, Ss stora mscluro Coe
om Re vera BETtas os Srs Vando.
Rea Mre coke ds Cound (Ca E Le das +
Ras e 3
- CA day (O d+
Cesbar GALNÇES ora 3 LE ns
0, A “o a É s
ES T ans ds o Sr ar Ao
Solana Suas Cor As e . ads e 5 Vo
“Ar + h
BA ue vs NEM .
="5 , = 3
Eb
RR ,
tro 2 oe As 4 à Ave!
DIMNOS E But CTONQNODE Sepr
DVNOS S ali ENOADÕI OO, Gees
SANDOS 5 dera alas DOR Ger
dons é Soa siena dd
NOS S Bau SISAMSDOS RLL
e
a N
ol o
—| u o
LA - £
E
2 le É
91 Is é o
“4 o , E -
z x ta
nd «
of
4) li .
E
Po lã
[ed ê o
— a o o e o o
2 Ss 2 2 2.8
tu) A
|
ao
faro
aece
peso rena
o
a <
“A E
A E
- ê
. o o
4 E 2
4 * 3
a e “
E
E
4 É
e
70 o bd “ o
(u) 4
aesross 2em axu ate Bee /
a £
ae f
| Problem 2.21 RPA A
em,
Crer: Velocity Bed 19 xy plane;V-ab rbxs, where
a=2ms and bz ls!
Find; (a) Eguahon fr sirearmine tnrough Guy) = (2,5).
(b) At t=Z2S, coordinates vt particie (0,4) at E=0.
CO At erZs esordinates o! partie (L Gas)ar t= ts,
(d) tormpare potniwe, Sirarmine;streaklpe.
Solutroa: For a stramtne de dy
tu vw
Fr V=alrbxs U=a andv=bx, so d dy pm
Oo Ex |
dx =X
dx z dtg
Indtegratina
x* a 4 = b 2
Tr gute or daqui te
Evaluahas c at buyl=(25),
c=y -Ex* Em Ebro Cm) dm
Stream ae Hurongh (x) “(é ge Em u
E
E
To beak. parties, cerne parametrse equahens
=“ <p dx =adt, aod x-% = até-to)
“p dE ,
vp = Sb «by, dy = bxdt =blxotr at —ato)
y-y = butt) + Eleito) iate (t-so)
for tre partick at (Yom) = (OW) at t=0,
Z=0 taé mat t72s, x= Mis «Ym
ga qr E & at tris, gem +gx mB
gy=Bm
(be)
7) <—
[Problem 2,21 Conta.) |
A
re parte at Guy) -(,mic)at t=)s,
X=mWralt-to) = Itadt-)
So at t=35, x = + lmss = Em
= + bxlt-to) + Et) catoler ts)
= astra (MelécD + fu tm(po,)- 2, Is (t-1)
So at t=Bs, gz pa tl += = ot
(cs
Au these pomis lt on tre Same Sirecm ae, às etwn delotu:
C) (Stores)
/0 [
9 T 67 (14,3)
s E
9 L Siam ne
Ylm)
LH
Ss II
4% Re :
st (0,448)
Z IH
1.
õ 1 ! | t !
D / Z 3 % 5
Xe)
For tnis sjeadey flaco, Sireamines, patnhoes, ar Strmakhoes
tornerde, as expected,
bz
HEM SIOSHEEIS FILEF, 6 SOUARE
8 sá
os
teem, Guefhoo , arg its
de As
. , 6
Seta Secad gos GS spa So» = a (A
CAN
apra ES o ares CD trend]
too q A =
te 4 SA-a
Ez uz USAM
Streamiines: t=0s t=1s
vim)
»
Streakline
x(m)
bes
| Problem 22% |
ve netah
given: Vebcite fetd Veaytt bes where aals
torta Streambines at t=0, 8525
and u = =ay cal mst]
so
da
acao
luhere, Mofo AME coord mades of partie te at ty
g=2m+gniSm «MÍst s00m
» 3-9
L=Im+ Ex 2m(2-0)5, Lulu 5, (time +09)
46) = 2ms Lx VE Etsy (23%) s> «3.22Mm
GS Shows on Ine néx page.
The stregminde 13 fauad (at gra €) Lar Lx
Substatuhog U=0y and > &t
Z
Fim tro, EC sat lo m)=(h2), the Co 4
Ei, gt= Entes; AF Cho Go) Ch 2) then Ce
t=2,y*=
Sé = bt, so dy =btde and gym Eur
(At t=35, particle tnar passed (ht) at t=25
(O Plot patnime and sireakline frough (2); compare
Solution: Path ie And strakine are based 01 parametriê
eguations tor a partie, Thus
ve So
É
ud, al yot +8(E- o), 4X =X tOmltoto)+ E, alho
For (23, ty 20, and (Xosfo) =(,2) Thus at tr2s, y=&
j b=08 mó ts
Find: (a)Att-2s, particle nar passes (ni)ax €70<
-to)
=
+
to =p (85
At t=25,(x,b) =
“Sm
For(b), &, = 2 5,00d (hoy) = O, Thus ALto3S, he particie tá ax
607, 3.00) m
At t=38, to=ts &)
Farto, tre sirathoe me be plottes at any é by vagas ts
*
> dk = AM des or yt= Rbt, +
2. zbf bé As
Thus c=U = *o
byte; at Cut) =(h 2), cz 2 ty t=35,€=1
] Ce 9)
- (ays83.25)
x(3) = ime quim tona, dao ((2lur, trça-=))s?a 3.59 mm
(e)
LC
Probtem 214 Contras |
ed
UV = % tt» ae” .
Recall! V=ayt + bt? ,whee a Isib=as IS os Go) * (1, 2) mn.
ess,
Part (a); Pathisme of particle located * ia Pt
met athiine Plot
A€ (eo, ) Qt s*0s: “8
Ez
t 69] tes]xm|y tm 4
[ol 0[1.00 a , e
ê e D:
ú 0
E 0 2 4 6 8 10
é x(m) .
E « , ; ) 8
: Part(b)! Pathine of particle located Pathine Plot
i at (votar t=?S! 8
E
>
sOp top em) 4 es
2 2] 1.00]: 2.00
a y 2
2 4, 7.67] 5.00
em . o
x tm)
ate, Part(c): Srreamtines through pot (rag) At t=0,2,and IS:
Tre 00 1) 2] 3 8
ts) Streamline Plot
c=| 40| 30) 20) 10 6
to ()jx tmb y Cm) | y tm y temo y tm) E 3s
O] 1 2.00] 2.00] 2.00] 2.00 > 4
o” 2/ 200) 224| 245) 2.65 E
O| 3/ 2.00] 2.45] 2.83] 3.16 a
o[| 4] 200] 265 3.16] 361 2 T tos
0" 5/ 2.00] 2.83] 3.46] 4.00
o| 6] 2.00] 3.00] 3.74] 4.36 º
o) 7) 200] 3.16] 4.00] 4.69 o. "emo s 1
of 8[ 200] 3.32) 4.24) 5.00 :
0| 9; 200 3.46] 4.47] 5.20
0] 10[ 200] 3.61] 459] 5.57
Shreakiine at t=33 of sc me em
particdes that passed thru Streakiine Plot
pont txo,go): “8
E
Pos) topm[y(m 7a FRA
[ of 3) 9.25] 4.25]
1º 3] 667 4.00 2
ao 2[* 3[ 3.58| 3.28! o
3] 3, 1.00] 2.00
e a = o 2 4 6 8 1
x(m
€7
| Vedder LIM |
2a
Gisen : Flo cl qudes ese Sáiscas
gas, quiias as Shown. q
Seleção
* NR
+
he O So,
a am
Una = Oro ls
Estas
Findo Sens Acaso om e qe (indicate Atredio
) 4 MAS Pe
Nom A sivaas Syrass across RIR VAR
Jose equidios Ler E
Ee Afeto
= . RETO
do Vivas [EN 24 tone à,
BA agree que | qu + Tso
Tau (eu 3)- » E) %
Few Tolehe Mt, for cut ses, pa ismasE meslati Rus
tm He = MENS Ms ,otom A
wa Ss Ses tm
Lam -2ma nt
Ne spças qiiie so unas susto . Suco Tuto | Re dress
Aress om Re “pas qe IR aà ms Re qias + Avreson.
Re dear Aress sosves Nesta vo sA 4
<= po -- É Str 4
Be ticos res sm Ve tos Loss q js
st Ac Qua tienes mom <
Co quilo, merlaro ve NO
Vivdicoked sm Re A Sb
e
Jo
Problem 2.18 REA TAS
Solution: Apoly defindhois of Necstançan flued, shear stress.
Given: laminar tio betrseen porattel plates.
idos td
tu cpf 4
Umax 1=€ p) A
VYT7er
T=15€, umax = 005 m/s, h= mm, water
Find: Faree on A=03m? setor of fever plate,
. Vs .E dia
Sasc equadoss: T=7, Tl dy
Asscemphons: (1) Mecotonão Praid
. é
From tune gen profile, U e umas[|-(E] »S0 & = Umas C2 XE)
At lowver surface, dy = —hfe = — Fumar S
hE
To liocser) Mm =u - Limas (ht | - re Sema
Tux >0 and surface 45 postos, so PD rega.
Ea max A
h
Faltas
From Appendix A, Table AB) pad x E Nisim at 18º€, 59
—3
F=talyxiO NS posm Bm 1, 1068 com
me 4 S mm mo
P=00!81M (to rights)
-
q!
Problem 2.29 Lo RA TA
ad
Open-Ended Problem Statement: Explain how an ice skate interacts with the ice surface. What
mechanism acts to reduce sliding friction between skate and ice?
Discussion: The normal freezing and melting temperature of ice is 0ºC (32ºF) at atmospheric
pressure. The melting temperature of ice decreases as pressure is increased. Therefore ice can
be caused to melt at a temperature below the normal melting temperature when the ice is
subjected to increased pressure.
A skater is supported by relatively narrow blades with a short contact against the ice. The blade of
a'typical skate is less than 3 mm wide. The length of blade-in contact with the ice may be just ten
or so millimeters. With a 3 mm by 10 mm contact patch, a'75 kg skater is supported by a pressure
between skate blade and ice on the order of tens of megaPascals (hundreds of atmospheres). Such
a pressure is enough to cause ice to melt rapidly.
When pressure is applied to the ice surface by the skater, a thin surface layer of ice melts to
become liquid water and the skate glides on this thin liquid film. Viscous friction is quite small,
so the effective friction coefficient is much smaller than for sliding friction.
The magnitude of the viscous drag force acting on each skate blade depends on the speed of the
skater, the area of contact, and the thickness of the water layer on top of the ice.
The phenomenon of static friction giving way to viscous friction is similar to the hydroplaning of a
pneumatic tire caused by a layer of water on the road surface.
9
RPecihem 2.3 CA val
Gosen! Fodk of vota vos té ; “Om. an each <a, $ porá ap
a None imdined A AS No Re Horas ones o (Na
of smE sentia or. Se aged Tal Be thoM do condád
AS Gio os Me GA LAMA dinero 44 O.9014 R
Seat qrefide vs Lie ts Minos
Force Tecqured.
Se Ae.
Fa
Gunca Bye the ds encima À cordas selori ES, Sem SEaxd
Considar Me Lotes
o
Me áreas é dio qnd lost Ra
Loee Vota danço of Ne od.
E
mt
£ RN
ns
Sw LF =o Vier
, Ef. nsszo
Nos Be Lidia Lorca R Seda
Gere t= pe
“a
Fa sewsl ESP ( Nuseas seo prelio <= K 3
Nasa
as
Setas pLa
ana
Sus
a .
E= pane me
=- Bm st SAB ro
. “ .
Free Fa “2, Neçendra a, Los SME Nei oi QrosF (38), pa=3105 as slot
E = RS A Sp
«2 +
= anus Ne zona bis Mm sm UT ÃO A rosa
has e Ne s Secos Zi
A)
E= 3u.5 NX
75
Rede 2as | TMA RAN
Grao: E 3a ste tuga ia do Vo te
BA o eetA duo sê Mica 4, É] 4
Asmie ms Re mastova La y
E CN RO, 66 Soon + , E
c-ciorim. ts asa izodôn E —
Cadortenrdo: = on dalÃos. «esa Ce Sp
Tecla toe as Nes Ss ?
Nose, ohevoothe Cora e *apo + E=1s WE.
Funã : Moss alovadde Aape speed.
Seven:
TELE MaL
Seca 4 = Xot (Men ZE, = and drisva, Cerca o Daloncad
doq o o Ee va feudo Cora e
EN NPR q
K foca A haras Poe
Or op sur apa, 4 Pesa GAS
Ave T om postuse susfaco meo
Deatide À Tão do Vet
= da = +. est Da N
Or Moto asfalto Sep dg pasa” DE
Qui se L en negeinie Surfoee mea
Ec oÃs to RE
enero ZEco = EV
ds
Eç eta were T=p ma
Es Et a ala (ral
c= PER TR = 2pER
Solana, fer *
v
> . &. A
a Res (A Ca
“= Ce = 15 %K+ o. ne "oa * sons WE SE “O
apa . “a
& NS
N= 343 “ a
To
I Rea Ra
BD DD
Grace à Vos É eras A dNdas om PR coa
a Nei Ca SL SA of EE |
nus Acme ds. Coros SIE Oil film |
tviscosity, 1) ;
SC Tio ta RA, AX tive T=o,
TNASE vm AS celeese d Ceesen ca. mas
Fusár md Expressos, for “3a ais gre oe od
tas O CONS A a
RS) Viene cha “ares US anda a Boi,
sq. as o DA seas A mel
(O Exqresatem for Visa qua N= SC 4 NS A.
Ny
Se RSS a In »
da 0 a 4 Ee 3
E E
Emas , .
: Rose capela + Cuz pa Sa, EE=ma TR
i Wesusgisons O Neon Na BoA
=) Assvegs eso a, qretde NR Sa CAs. “a,
Fr
k
Vea da
Roca cplino pitas ÃO
Cor Ve Node, E= EE = A se SS
O Tor Be Lola, emass 2a = ma - Fe =
. s de &
Sta Ag= EN , Vier asa N Ra Bom E So vo 4S as
oo, ES as
DO RATE A E MA go 15
E
deb; ar Rs = Gra) ED Pes. |
arde sorvetes asd ks cd
Sa
te ÇaeS SEDA DR into ES,
fe
“Ne todae voe
ta EMDDA A (an PS 4
à dedo
= “as - E
Sama Sor 4, *
a - ABL .
Ser q Wan (UI — o a : q
an
Pa
a sos Mece otas esgonetnsdu No Naa Pa
9
| Vedder LI | mma RA
Coen: ve , donder à, to de .—
ars at secdos > à te — —
q LS Barao:
PR
“AD n.
mota poa e PL
Ver rad pow As Re epa Wimsesm tose and die
NE nus Ema A speed, Ee Sonhe-
Fwd: Fosco ceqpired No qui Ve voe
Selo
ZE =mas
Sysce À e = cesso , ApQNea: Lora end Voe esta Ne
tona frios fora v Fe
das
Fo E <A ubhese Te p &e and ns <a
Tssusmma a Niveos sichocs So clodaes vs ser mish
Via Sa, SN
er) “Pos o Asa
Ve Shoe Ao os Qoshase f eusfors sá o s reapinee
4 Bsedaas
= Fe =
Entas EN te
Ps
je
E= “Oca an « 2X, SOm 4 O Bens Oem DA mr «Aa, RAS:
165 un E ce E Om Arm vescan ha
= tora
Po
Vrehem “BE | UMA Rei
d
Guaen Concanhire senvindar instalar
| RENS m. d=o cera alem, | |
“Xones Vindas solcias SL 25o cen q |
| |
i
Es
Gap LNea. o cosas 64 dA ar.
— |
Verne Norge SEM pired fo rotode Ne |
wner eindar CD
Ne ceara Nora cad Tabanea Nye ceseara Serge & Me Eras Sá
Ve Seos Lora 4 ae AS TA abere = URI
Fec o Hessniar fina Ts + Sa
Fer sema E Cear qrefdoy A = x +
USere N= Nomgg nos. sstocdas of waner espdas = Rus
caça — tus LeRd = 2x Rua
4
&
ond Ye Nocque T = pra 2x nê ud
a N
Fon Fa fes cndos Sh À ade (32), pe Io no mM stvd
Soda, nomenind ssoves .
<.
a
TJ. 2x nE vs = 24x dioso de, prosas! “e = eb «BY S Ro res dona
.&. a
a R<rod ma q
tes bas R$
To s3 4-6
—
g
$1
Veotham 2.3% | vma Cê
MN
Gien:, Comerc cmfndas Srsesdies & ETSD
Tras eptndas cas A úseweo ren, N
Cund Voce, ef Najia chessonçe a
Seslios
Re “enpoted ergue mad bloco Re candura Norge SL Re eos Cora |.
Ra Swen | AOC, INE tSOmm
Nora > “Tm O. im
Ne Seas Corce ae da Esta Ber Weix An
Fer
o Nesdoni Roud Cm ja É
N
Suvia Be sda, rede & astumeá da tr licor 4 e» a
Rus,
abrace A as Re Nongeia socios, Ra vrer condes Ne Rio
N Re W
Estas nã acto = Expe tem
= asp ess
aná Ve Tosca pa Ta RF = SEP eta
Sohanra, fer +
pe
pe
Ná . oe tm om, Lido amo (A
es Ria 2 (AS at O roses É Ema
v o,
* tes «es 4 Seso) em
Re coã. ma E
Gorvas! muslor
Pz
[ Veda 2a | tema Ta) |
Gram: Tem oder cdndar Cmoss t, and todios VI EL a
comcarkse Eilendas is tader 15 Árigem . Va Cody
OSS EDTVR Ta Ns ara
Solienesa mmer cicinias do 6. LESTE
aus CEO NEED & «est Na paid é Be Sie cssetioas
“oa Na redpe Sed
Tião. qo adere A é, Se STE oe ad às Vejo
des CERTAS A s o ES “E Std
€N expressem Ley ya
Sã NS,
. . San
Vas en polises + fz p» Sá,
EV=sra EN TTA
Wesuame, (9 NasKoraas, Guia
en Neseoe motoca, qref de
+
P
TR Pe
U e ENO am pago Ro E
é Te tuto P(A m
To UBV o : ao
Pusóa, asealesol tom, ta Re Tens vs Pe cora de Fe
Vos So caldas sas ER Tete ad e «MN
ns c For Re enoat TR a E qa SE ne IR
E= Pap WA dee
fre EA A “Se o 48 às
| 14 WaR E eo ss = (a E
“a Mto OS ='s pistas e Ca NNE = =€
Ea
5 'acio =f te = (E a fem .
Thesgiiha , dt-= Ele = às NE = Ml
Ga = ES, = o GS) 5 ae Bda
Susana fes B < a RAN o
“tg e SR AD
vas ps de G- Seia Be - aii
sDosarivsa às cuuss À LS
Droit ER ia
e
| Problem 2.43 | aaa TO
Solutioa: Appiy summatos of torgues and Nestor! sond late.
Crea! Cireulor altumintm shatt 19 journal.
Symmetrie elearaner gap
filled With SME Ioo-30 ar 30%.
Sheatt tumed by mass qnd Cord.
Find: (a) Deve lop and solve a drtfererhas
equa fios for angular speed as a funetron of time.
bo) Calcelate maximum angular speed,
(6) Estrias fame needes da reach 95 percent of maximum Speed,
Basie equations ET =I ça ZF = m V= Rus
“ . am =
Por the mass é Z6, emg-t em = mk da O
+
gv ma
For the shafr: ET > ER Ncous = (e
9
to Tuiscous = TRA = my RarRL eme we
+ A a
Assume: (NNewtonian gu te, (U) Smat gap, (5) Linear propfe
Then EG. 2 becomes tR - UAM = Teto Time ds
Must iplenia Ext bg R and dombindg turfa 69.3 grves
mg R - meto - Er aaEt 1 = da or mar Ita ty «(Tema do
Th mas be curte A-Bi co where AzmaR, Ro = Itm
Separahas variables dm = E
ds [mu t
Invegradng $ o - “Blula-eu] =-Shult 88) -/ dt
. “ a . !
Simplfgrg = Se -€ de or uu = gli-e-*%] 6s>
=
8)
we)
Me maximum angular sos (t nO) “w- AB.
, Niogã 8
A = mgR = ODio kg, 9.81 4 x 0.025 ia = 2.45 210 É Nim
zum REL 09S 3 O sê “P unants
B= Ze = EU DOS de, aos Ja 1050 m O esasn Bd "UBE Mi
pr
| Problem 1.43 (cont'd.) |
t)
Evaldo fog, Comas 8 2.82X%0 Yume
Thus
Umas * “estad , re bos 25.1 rpm
=A «Zumo Lo 242 radis.
ET limrod ma Mimas
From E4.8, w= 0.98 WUmox eshen 6 PL = po,” Flo 2 S;t m Se
C=Ttmbta imetemee =limeme
Ma WR(LSL ALI CSTRÊL SS for
M = USTdb sm, dose muco ka D.b4B 9
3
C (ix D.LYE kg , Doro kg lb .ors mt , 09x 107? kg um*
Thus
-y 45%
ts 3w L.oPxp 0 Dt = 075 tas
The terminal speed conta awe bile Computes from Ep. bg
setting deo [dt DD, initnoud Lo lug Ft drferenha! eguateon.
f9
| Recdemn Cho VA Cai
Graam: Cone asd qioie stands Bios .
Ber SL cone juá Noucdhes Re quite
SE cera nto e no po
Cica. as Persas on expression Cor Ve Snes ERA po
case do Re Irapaado Adi Che Re age E E,
& Emalu die Se Ea on Re árden de ae Ts Sample? —
cone a Teses sf Segs dress
aná agonia + Ve sudo
SAS Rey
Seca Ne enoie 8“ Sera swod, Ve Ones age See casta, a
dão sera Emo.
“Eis Cebsonaide Ko sósume o Auseas evoca qefde
across Ce GER anã No efe VA eMRÃo
Ne Seas (doeeeisa) ride ds Fr
qse da dao ER E. 1 = cho
NR “o
Wu Sana, vadias (S >
Se aveia, AS = us Gead
Ne SER va W= vÃsns
. Us
2 TO AS
*= Trans É Ian
Sce Ds ses Senolh AonB= O and
: “a
=
o
de: Ne dneos colo às esdagandosl & u a Ns
Sasçõe 15 supresa, to Ve some Sens CO
Ne Nerape a Pe Arusem cone a EV
= (rar Bee SF a Ta 88
Sea 4 é a condos (Ls a aysen “5 Ver Tons cerdas
ONA R,
= ( TSE = (eta = EA < Cervo
à o
— .< q?
T= ã W Ta TY
Go
| Problem 2H | MARS
Gen: VistoUs Chetch made from pair of closely spaces dusts.
Input Speed, toy
c N
butput speed, tas . SS - ES —
Vistous Dil 1A gap, pt bi 4 Do
Find algebras expressos in derons vE Ms, RA ti, And Loo e
(a) Torque transmita; T
(8 Pover transmitida
tes Ship raio, a = Blur, in decems of T
(a) Eftierênco, 7) 19 terms FA, Wk, dna T
Solution: Aprlg Nevstons lar of viscose
Bosiceguanoas: =ude dpetdAa dr =rdF
q T 4 la F
Asstumphons: (1) Newotonian gua RA
() Marcou sap so velocity profile 16 lençar
donsider o segment of plates! fo sz
pa arte “a Eça
o du su DU CO) E ” ==
Tt Am dg e dy = E) def Plyo
de .
da = rdrdo End Vitus Boto View
de «da = MODO args fab pa dedo ; dr Are o fabio dns
& E : a
Tt, = E ad Ceower transmettes)
à = At) ( tar
dk TUR bo
Pocuer out Tido op, But Ub-ty “AM, So
Power in Teuy ty ,
td Aty
= LOC.i- SD =tImw
9 7 ri
Ertiuênca 15 7) =
Os
[Peres vas I 2a TE)
a
Cosas Comcadime, ca estelar Ses ti
LSrem neves Nam das e NA a SL
e
dos E 9) SNS umas See 1
Ata Gras GSGUÁ CSS Rae
Sl Fer Colsnãos esa chdas
TS. «em esspressasm Ses EE Nara pre doa 25 [
So o end o f
às cagfed em fer rsss do err so
Vosiam, de À e > ese of 's
to
(A Tre AA Soo Re oq
SG AVONL -
SE a R
seio SEES é ON S4 1% th RO
Tor q = tem vo Ma = e. aos =D “a
OS Te o ques ddr of ela, Ve assess *b Cass e
ÇÃO E oesnieá Volootra os Bu per
A ds Door Ny O da tunes am
(For Saio, tos A “Bio cas grondsi
Ia de 6
G) tp Er. Aos?
en ASR COSAGAS COM o ce cerrenã?
Sds. Rose dai Rua PM
Wasudmçiions «S Neas ae do Nes Somar Ng pu
(8 a andor as
& 2 RE S ss
Longe Sepp Be pao
E qu E RE ETA = RuE E test do agua | a
A a too Ep Le DE
e. a Date da o s Ps.
o “o 5 5 re ads SRA
Nem” Car = Ss E = (Las = a ç Eme dr
“T+ = TEM 3 e o) Ve a
Tora = pio (O ta o ESSAS = RR by
A
For TRestom [Tarado É vos , Sem ns
“Te E pad a Mono Rea,
END
FA ú ao
SA TO *SO a
Problem 2.51 am ATSA
Crer: Sphericas Incust bearing showa: h
cu Do
Find: Obtam and plot ar algebrare a yo
expression tre the torque 09 fhe i
Em
Sphericas mem ber,as à funchimo E»
of &.
o
Oil film (viscosity, a) Lo
dolution: ApPly defrartoss
= Computing equatrens: Copa T=hrtda
Assumptions: E) Mecotpniaa Elec, (0) Marnw GP; (8) Lacuna” flow
From the figure, = Rans wet =nkS108
- - [E 8
7 pndie = p(ÇO) qro cpa tas
JA zimrêdao = UTRE sinta do
Thus
e T =P Rsins( es Ss) py pt smo do = trtctof Jsunsado
o õ
8
T = cuca ft osip css)” = mesto R[auia su +53]
ô
. 3 .z
+ ” =| Costa E E
D plot, nwmatize to T/2 to e + os a. tosa E
Plo trsng! 48 ' , f Ê
0.
T
j
T
Norma tiged 2.4
Torque,
T/ 2a RV pa
h
(--—) ;
to So So 0 40
Spherical Member Angle,& (eg)
T
1
check ctimensans: | parto R$] a ELLA A = FL vo
h Za t t
as
| Problem 2,52 MA Ta
séc,
Gen: Rotating bearms shown:
Narro gap filed wrth Vistous 8
Dil, pues (250 cp.
. . é = 60rpm [|| RS 7Smm
Find: (8) Algebraie expressos for Sheor o=0.5mm dl
Stress 009 Sphertas mentes, 1 qe Eds Olin gap
bb) Find maximamn Shea” Sress. ! [
(e) Algeterese expressa fr videos . Fo= 20mm
ee on spherreal mese.
(dy Eva luate Jorque,
Solution: Apply definrtmoas
Comp vting equa Boss: € = pu rh rTdA
Assumpteas: (1) Newtençãs fluid, (1) Marrocos sap 65) Liam nar mohon
From the figure, P= RnB, Us ur a tuR Sind, die - 40 de
h=a+R(i-cose) dA = emrdr = 27Ran8R cosas)
Thus € = MR SAD
AFRCI- case)
Fm the table below, Tmax = 67.9 Nim* ar & = los” (not at 8)
Tmax
, Bmax es
Torque 18 Taj Pashnd SB COB ta
o A FREI-Coss)
This must be evaluated numercaty 00 graphicaltgi From Appedis
6, 1 Poise = 0.1 kglmis. Thus ja = Es KGlmis = LIS NS] am»
T = SNIS, Em tOS, DOIS, Sin b.SR /
Mm s 2.0005 + bd ti-cosb Sn
Tatulating resutis of smitar coletados gues!
= 47.9 Mfon*
. , 2
a Here "turcos" is S0Êtosa
memenaa cnoncnos com ArRLI- Cosa)
0.s 10.2 11
1.5 29.3 3.858-05 3.968-05 Ap = | deg 20,0075 Pad
2.5 45.6 SLO7E-05 1.308-04 anel = !etg
vo au SMEOA 5.178.04 ua
. . +. 29E- «17E- a
5.5 66.6 2.95E-04 B.12E-04 for the numeral integration.
65 67.9 3.538-04 1.178-03
5 li aa LER ko
. . «ASE- «OLE- =t -,
9.5 63.6 4:798-04 2.498-03 Lomax" sn É =sm 428) = 48.59
10:5 611 5.07B-D4 3.008-03 . R as
1.5 58.5 5.298-04 3.538-03
12.5 56.0 S.ATE-04 4.098-03
13.5 53.5 5.61E-O4 4.64E-03
14.5 51.0 S-728-04 5.218-03
5.018-04 5.798-03
Integrated torque = 5.508-03 N-m
Probjem 2,52 | EINE
ren: Smau gas tublboles form ,a toda tule opened; D=0.! mm.
End: Esfimade pressure difference fr saside to outside
Such & bubble.
Solutroa: Consider à free- body dagram of half a bubble:
Tivo frias act: f
Pressure: p= Mhrg D —>
Surface tensor: ge STD 2 +
Surmming farees far equterimm
> -E = 4p 2
ZA Fo -& = 4rEE-swD =0
so 4p D — =0 om At = dá
7 o 2-5
Assuming Soda-gas nberdacê IS Sicnrtar do tuater ar; tes
= 78 mim, and
Ap = texto 2 t = "2014103 A 2 2,4) PA
m
Sp
x =
ro ox t0"2m
23