Capitulo 5 estatica solucionario Beer 9 edicion

[CHAPTER 5 |

PROBLEM 5.1

‘Locate the centri ofthe plane are su,

SOLUTION

CRE Hmm)
| ü 00 | tos, 15 OSO | sad
CES 130 Zas ECT

E | is MENOS ETS

865.0"
Ea 15300

LA 14055010"

15300

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PROBLEM 5.2

Lee he setoid the plane area show,

SOLUTION
“how
Am) Fam
1 1200 360 |
2 300 san |
2 | mo 35440
‘Then a. F=1621mn 4
za mo
y 04, 550
za van

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PROBLEM 5.3

aca the enim of the plane area shown,

| sorurion
Aint sain [pin [aint pain?
' pas. |g 5 7 450
ncis=aıs 25 | 2s | ms mes
z 0500 mors 28125
75095 y:
1 F-10280,
The ae in. €
F4 28125

E 634,
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PROBLEM 5.4

Locate the centroid ofthe plane are shown.

SOLUTION
|
Ain? int | pain
a aos x À 146 1
FT eo » [rs | 1
Te a 5 M
The PE]
Festy=306 Festina
CET
1609-20

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PROBLEM 5.5

sate the cena ofthe plane tea show,

SOLUTION
! 4 ,
« -
EN dm
À
RE x
| Ain? zin | pin ETS Faia”
1 | oma | 7 | w | 10 | 20
i aa 6 2 301.59 603.19
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he Pen
20
DA, 21068
EMS

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PROBLEM 5.6

Locate he onto ofthe plane aca shown,

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y “oh
15 pe
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Ei
pers a
Fam | sam"
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Poke dotted oro by ny one rw pono 80
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PROBLEM 5.7

“Locate the conoid ofthe plane ane shown,

SOLUTION
a y
E x
F
Y ain Fin | ein | Pain?
1 Jon =22082 0 | 161277 o 36581 |
2 ace 320 en x 3000 | 2560
B masa 320 | 39021
Then EA F-1643in. 4
Ea sn

21746. 4

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PROBLEM 5.8

Lacate the entra the plane area show,

sorunon
y ste
5
List -
im 1.
on
L E
is.
| ma in [y im] rain To aim
| 30x50 21500 15 25 22500 3500
2 qa 35343 men | 30 | Bo 106029
5 me | CNE
x uso ;
: „uud Fu 4
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PROBLEM 5.9

‘Locate the cenroid of the plan ne shown,

SOLUTION
Y
- /%
Hope
x
2A
Am? zan | Fam

1 | con

6x0" ET

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3 sas | 25468 | noue | 7200
Y 7200 ase ET
men PES ET 1000m 4
IDA TODA Fam 4

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PROBLEM 5.11

Locate he centroid ofthe plans ars shown

SOLUTION
note hat symmen implies o.

ETS
cat

Fain?
sus |
usos
5 125659 [soc
25 in! E
„224 1066 int oc Fe6asin. 4

TO

Then

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"| PROBLEM 5.12
SEI A ce tei ath pn as
e
SOLUTION
15
A. mm" FAm
E (05x80) 1200 EST | Den
a] loe | a | m | a | ao
| 2503 IE
Then
roma ao Fas05mm 4
»
Faso Felsen 4

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PROBLEM 5.13

Locate the centroid of de plane area sown

SOLUTION
hy tate
= u
= Bot yam
Es
am
mn | Hm Fam?
1 9 15 2500 | 5000
2 moss | re | 9000 | 2137
z 90686 CRETE
10500 É
The osos: am
Sexe * 1
26137 =
an F-288mm 4

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PROBLEM 5.14

Locate the centroid of the plane arc shown,

Dimensions i in

A, in
pr
1 2000
2 0
z 1200 i
Th F=900in. 4
7-90. 4

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PROBLEM 5.15
“ Locate the cui ole pla ra show.

EVE

à
Am ma | y mm | mm | Sm
"| ose as | a so | am
>| os a |» | soso | sao
Y 3750 u 112500 243000
a rama
X(3750 mm?) = 112500 mm? or F=300mm 4
= ras
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PROBLEM 5.16 |

etc the yevonlinate ofthe certo ofthe shaded area in |
terms of ry rand

SOLUTION

ia dbz,

Fin, determine the location ofthe cent.

From Figure 5

Similarly

Then

and

PROBLEM 5.17

Show tha as 13 approaches 3, the location of the cent
approaches tat for an are alle oF rio (5 +72

SOLUTION

Dan és

Fir, termine the losstion oF the eno

rom Figure 584:

Sia (Edo

ele

then

en)

PROBLEM 5.17 (Continued)

Using Few 58, ¥en fs ra

Lee yO,
yet

Dale

e Gold tans

Now
let
‘Then he
Mer
PN qe rara
3) PGA
HAT
“Or

een as
a.
an

Soda

‘Which sprees with Equation (1),

w

a Foun |

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PROBLEM 5.18.

For the arca shown, dtr the ai a for which 7

sorunon
s 3 a
ES sa
[27
- fe
36
“i.
[ 4 z va FA
; PRE 2
es 3
à 2
=| lab e
: ca
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E

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1 PROBLEM 5.19

3 or be sena area of Problem 5.1, determine the ratio rr 50
Lx |. a
est La si

DRE
SOLUTION

‘hen
lat pe
pr +4)
lO +
o 16" +(16-9mp+(16-91)=0

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PROBLEM 5.19 (Continued)

Ha PRAG EN CE ALOE 9)
ETC]
oe pa -05726
parar

Taking the postive rot

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PROBLEM 5.20

A composite eam is constructed by baling four
plates o Tour 60 x 60 1m angles as shown.
Te bolts are equally spaced along te Bean, and

fe beam suppors a vertcal Toad. Ax proved

mechanics of materials, the shearing forces
exerted on the bolis st and are proportional 10
the fst moment with rec 1 the comida
‘is ofthe rl shaded anexo shown, respective,
in Parts a and ofthe figure. Knowing Mat the
force exerted on the bo a Aix 280 N, determine
the force exerted on the bolts,

SOLUTION

| Prom de problem statement: is proportionate D,

hasta. Fen Fa = Qda
(Oda (Ode [22
(ose 2
Forte st moments: (@.,-(225+42Joaosiay
821600 mun?
(Ode ~(O,)442[ 25 |(48x12) + 2(225- 30X12x60)
1364688 mm?
Thea ALPEN] or Fi, =459N 4

EU

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PROBLEM 5.21

“The horizontal aid Brough

ti Cafe ars shown, and
it divides the area int (wo component areas A and Ay. Detormins he fst
‚moment of each component aca with respect Io de axis, and explain the
resus bined,

‘SOLUTION u
u
| pty
TEN
foo PONE
=)
Note that sey
Pi conca] a
Voz lio?
due Vile aja
A O
sr @>--s0ù
pon

00) HQ) =0
Hi results expected since x is cetridal axis (thus
and

CEST

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, PROBLEM 5.22

‘The horizontal x axis is drawa tough de ceaoid€ ofthe area shown,
and vides the rea ino two component ais A and a, Determine
‘ths fr moment of each componen area With respecto the x axis, and
explain the results obs,

SOLUTION
fs eine the locate o De cet C We ave
Ain? ¡EA
,
| eus [es | as
CR CR | ss
u 45x29 65 =
5 ms | dan
“then Franz
F(20.25) = 82.6875
o Fan
Now os
Then an |jos ausm [e.srss-s
116.5 - 4.0833)in [4 SKDTin” or (Qh=233 10 4 |
and a duos Jasmin |

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PROBLEM 5.22 (Continued)

Now 2 =@ +0,» 20
‘This reali peut sine iv contwidal ans this 7-0)

and Grp Fra F.020,-0

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PROBLEM 5.23

‘The Sint moment ofthe shaded area with respect othe x axis is denoted by 0,

n (a) press Q, in terms fc, and te distance rom the base of the shaded
Are 10 he ani, (9) For what alar of y de O, ax, and ha Ui
‘exam value?

SOLUTION
Stade ce <
| A
@ <
CET
For yO:

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si PROBLEM 5.24

F

A thin, bomogeneous wire is bent 1 Form the perimeter of the figure
Indicated. Locate he center of gravity ofthe wire igure ths Fun.

essa

SOLUTION
ru Fr“
E sm
4
2 m? Slam |

il x vo se]

Tp ao [es] EI BETT
CEE CTN aassxi0"
O] 7 ers] ant
UNC INEA INES Jun [mer | #0
CONEA CA ET 5

= [m CET

Pen
(1080 mm) = 186.3x 10° mm? 21725 mn 4
FIL = 35.

E40 mm) 108.3507 m Ÿ-975 mn 4

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PROBLEM 5.25

A thin, homogeneous wire i bento form the pernos of the figure
indicated. Locate the center of gravity of the wire igure thus foma. |

SOLUTION
Ftd atc shoes rer ov il ici wih condo
amen
| EA A, min Fon | Fm | Ln Hmm?
oe ele N
AE a
6 CNE 4
ieee m
oe fes Le CT
la ED D HN en
6 sw | 0 30 0 FT
AE wie) |
Then EL Km |
FEL=EÿL F(008 F2298 mm 4

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PROBLEM 5.26

A in, hamogeneons wir bento form the porter ofthe figure
indicated, Locate the enter of gravity ofthe wire figure ts formed,

‘SOLUTION
Fst note hat because the wies homogeneous, is entr of gravity will vinci with he coma o the
creping ine
4 o
6 lo
Tin y
me u 165 o o
| s » 35 ms
os] 2 ns 15 425 ns |
3 | yes | 6 23 | 15256 | 10m
3 su wears | sms
then EFL
868.209) 162726 or F=1845in, €
snd 222073

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PROBLEM 5.27

A tha, homogeneous wir is bento form the pereter ofthe
figure indicesed. Locate the contr of gravity ofthe wre Figure
ns ema

SOLUTION
Fist note that because the wi
corresponding ie.
AL
©
o lo
—
Lin Bin Lyin?
[a Is » o = o
3 16 ae) $ E ee
3 20 io io 00 ve
4 16 o 8 o 128
s ss 19 o m 0
6 | mann | e 24192 o zu
E EE CET
Tien rind
Fastin €

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PROBLEM 5.28

A uniform circula od of weight § Ib and roe 10 in, itched o pin st Cand

to the able AB. Determine (2) he tension in the cable, (0) te reaction at
SOLUTION 1
Por quater ence rake

som

mm)

50

C,=81o]

C2948 575° 4

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PROBLEM 5.29

Mens ARCDE i component oa mail and is formed thm a single
piece of aluminum tubing. Knowing thatthe members supported at Card

that (= 21m, determine the distance d o thot potion BCD ofthe member |
horizon,

SOLUTION

Find note thal for equim, dh enter of gravity ofthe compact amt is o a veda ine dough €

Fuster, because te tubing is uniform, he enter o gravity ofthe component wil coincide withthe croi
ofthe corresponding line. Thus, X 20

Sei: su
na
3
02s nas)

Jos {hanche

or (07511512 =

Lam her uns

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PROBLEM 5.30

Member ARCDE is a component of a mobile and is formed hom a
single piece of aluminum wbing. Knowing thal the member le
supported a Can dat 0,50 m, determine te length fam DE
shat this porion ofthe member horton

SOLUTION

Fist not ha for equilibrium, Ih center of gravity of the component muse le on & vital line then €.
ure, because te tubing is wifi, tbe center of gravity of the component wil coincido wi the cetrold
of the corresponding in, Tha,

So that ted

x 23a ins) 750 he
(a Jor AA
(0,25 mins ®.
{rene patine

o car fus” 4-0
ae “ig

ih -Lnistenisase0

. Lens JETA 80,

bi 2

u men «ros

Nowe that sin 3522 0 fr both values oso bo vales ar acceptable,

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PROBLEM 5.31

‘The homens wie ABC is hont int a semicitenlar ar anda taht
as shown and is atsched to a hinge at, Determine Ne val für shi
‘wie isin equilibrium forte indicated position,

SOLUTION

Fit one hat Fr equis, he center of gravity ofthe wire must ie ona vertical ie through 4. Fares
because the wire is homogeneous, is enter ol gray will cinco with the gend ofthe eoresponding
Tine. Thus,

So that

Then

or 02567 4

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PROBLEM 5.32

Determine the distance A fr which the centroid ofthe shade.
ea is as far above lice BB" as posible when (u) & = 0.10,
(a= 080.

SOLUTION
EA a ge)
a EZ
1 Los
|
2 ln |
3
»
beam
Teen en
» PR
man be a
e
or à i
eh) u
amt A en |
de 3 “kin |
Fe A) o!
Sim a.) ici
Ai Zah ra’ =O

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PROBLEM 5.32 (Continued)

aaa
. y A
The a =
fen]
Note the only the negative oa is acceptable since <a Then
© 010
| er ba 05m 4
o 4030
1-0] où no 4

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PROBLEM 5.33

ww thal the distance / hy bac selected 1 maxi the
distance y fom line BEF tothe cetrid of the shaded ares,
Show tht F= 201,

o

Haan) ah? =0 o
Rearanging Eq (2) (which defines he value o which maximizes 7) yield
20)

“Then substitating into Eg. (1) (hich di 7)

2a Mi

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SOLUTION

PROBLEM 5.34

Deen by dies integration the enr ot the area shawn, Express your
answer in teams ofan A

yok il

vee x
A VE ,

Auer 3

Fast +1 ne

a bet

dra 2

Ar [re (alae Loi

¡PUTOS OS

e Mam Cope Inc Ad tre of Ml a ed,
th Jona pe ft pro
ani er ae me A,

PROBLEM 5.35

Determine by iret integration the ci
our answer in tem oF a aa

‘SOLUTION

Aa)
None
Ps)
and by, ye
2h tax je
Then fes [Aeon ae
sd ¡Zee

neo ade [HO = jar

if

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PROBLEM 5.35 (Continued)

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PROBLEM 5.36

Determine by dives tegration the centro of the ara shown. Express your
answer in ms of «and a

SOLUTION >
Forte 4

Se
a =

y
then de

Ex *

Now
Then
and
ence
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" PROBLEM 5.37

Deering by diet integration he conti o the area shown,

SOLUTION

ax
Forte coment (ED) sho AE all

thea

To integre, it

‘Then

i fom oy

PROBLEM 5.37 (Continued)

Fous PP oA Haar

" PROBLEM 5.38

Determine by dret integration necesi ofthe aca show

SOLUTION |

Fit mote tpm imp

Fer helene (son 4
Fre =2 Gigme sam) x
rom 4

Then

ond

so

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PROBLEM 5.39

Determine by dice integration the centoid of he ara shown

SOLUTION
est note that symmety pics

pa

-04

‘hen ar für [aso
„an Eo

sand tte [ato fre Lora}

“él = Loir,
ee)

= (a) =a! 2 since] oe
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PROBLEM 5.40

"eterno by direct integration the ceri ofthe are dwn. xpress
our ansner in terms far and 0.

SOLUTION
a sat yeh A
Fo at
CORRE
Thea y Lota? IE
Now x
ooo Todo Ge
ant ata ieh
then a fate [hara bts a
and fase [ Le-sal-s [le 20? rar
{2 2543):
(Er ) dan
Ho salas
tres =
yoko
10

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Y PROBLEM 5.41

Determine by diet integration the centroid of the arc show, Expres
your ser erm a a

SOLUTION

nahe Ph

dope À

1
IEA)

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PROBLEM 5.41 (Continued)

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PROBLEM 542
semis nt ington e cena src stan
SOLUTION
Weine
Then
ant
u,
Mar
Hence a> feu Fain
va raat

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fam e sys

Fr

SOLUTION

E
B

à

ess

Java

PROBLEM 5.43

DDatemine by dires integration the centroid ofthe area shown. Express

‘your amer ons ol wand.

120 3=b az or

hae
ne

sat

PROBLEM 5.43 (Continued)

" PROBLEM 5.44

Determine by dret iteration the certo afte arc shown, Express
our answer in terms of amd

SOLUTION
Nor nat
Te
¿eno 2)
Byabsenaion ne termes E
Now ias
dr O20 Fact nk ma dan yee Ba
Rues
Then as fur [tars Lea
ay
ee

aod fuu- Dare): f “bes

=¿ooro([as or] [aer],

alah
6

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© PROBLEM 5.44 (Continued)

Grada fhe) Marae (20

tel

ty
Er

Hemos

fia a] Ler ru

Par

pa frase (Jaja a

Je Al Sis ran. pr i al y ee
pol arte pum Me plo, tt el?
tnd cre ran Poio Ma,

» PROBLEM 5.45

3 homogeneous wire is bom nto dhe shape shown. Determine by iret
ren the x cording of is nto

ae

SOLUTION

is not hat because the wire is homogeneous, is center of gravity cocidos with he centroid of the
somesponding Ine

Now Br acos!O and den ale vy

Where 10000: duree Osin 040
aan 8: dy= Rain” cos al

Then dl = [Ca cos? sin BAO? + (a sin? Boos ado] Y
= Ses sin Boos Bsn? 8)
Sacos Ain Bld

Le

and

Hence

«Son

Alternat

pasito uvo-(2]

PROPRISTARY MATERIAL. © 200 The Me Coi A gh sal, no hs Mn o eds
or ite ca fr or y y en‘ nr sl pomo pt a mud Rod a
cedido Mo Mio ci ro pp oma dci nt

PROBLEM 5.45 (Continued)

u PY

18)
ae

ejac:
fau [4

Se finde

y PROBLEM 5.46

A homogeneous wie is Bent ino te shape shown. Determine ly dicct
w A of is eet,

SOLUTION

Fist mote Ma because 1
corresponding line

mogeneous, its center of gravity coincides with he centros of the

New Ta 1500 né dnd
Then fa (lue nas - En
e
ani A
¡A
Ts «

PROPRIETARY MATERLU. $200 The Mim Comrie be A igh sone o Ma a be de
“em tn e ia ein a of in wl Sl

PROBLEM 5.47"

A iomagenesus wit het into the shape shown. Determine by
veut integration the coordinate of Cenrcd. Express Your
Anserin ens of

‘SOLUTION

rs not hat because the wie is homogeneous, is center of rity wil coincide with he conti of the
caesar I.

We have at soa, yea di
a
Then 1

and

PROPRIRTARY MATERIAL & 70 To eine orga, A om Re Mo pt of i nl may e pa
reco ar na enor by ng me hr wi pn of palsy a en
nern preci Som Pf ade re rept ences Mr

PROBLEM 5.47° (Continued)

À Use integration by pas w

wor dena ae

dads

> Mu

2

‘hen De [Zeenat

2 pa
(40290
glace [

Jura]

PROPRIETARY MATERIAL > 200 The Meer Cana, be A ph ee ur ke ha mo, ed
Fete uta rm pay en pre mi pm fhe pee ed hed
‘Rees MC hic ca pana cn oat

PROBLEM 5.48"

eerie by dec integration the cernoid ofthe aa shown,

SOLUTION

Wehne

and

en =
a]
La

amd Jue Jaco]

Use iron by ars th wer boom
dende va 2h in SE

Then fro sin sn

fut PE =] N”

= 010637402

COPRITARY MATERIAL + 20 Ti Men Cos La. A an. pr of i Mal wb ged
‘rt ent fm ry yo ted ties pen Jato, ud e nd
‘Serre oy Ato nl po pana abs Mah

PROBLEM 5.48 (Continued)

A A oo Ma)

Ja Jrad Ber) osos oe 70450 €

PROPRIETARY MATERIAL © 20 He Momo Corpus IA gh He pur Mma ap dp
‘mo er tated ny A r ata ie hr wien roe pro po ed
in nti pad ar Ml ol ar Venn

PROBLEM 5.40" ]

Detennine by ic integration the centroid of the ar shown,

souron
veine
ss
mu
1336230
mi fie [Fat cacas)

Lee moan
Topo coat it

wee and dado

eo aa +8
en Jen fan 0040)

Now tet wae then deco

rssindtd, then vus

mon Pension rino-i[-eto-

si

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‘pel dd many oo vt ena th re pros Bardo, cece i e ol
‘Rian nar ac os pr Hei o mb om pactos D y a
api

PROBLEM 5.49* (Continued)

So that (in som

arco]

GOA ye am

año O!

ar fet sinaate
E

wae and dur 08

finado and 000

Men orsinas 0% ont fo
sorta Jo neo?" ets)
fat more)
ut
13.094"
Here

FRE) 120260

(230!) =413.094" er

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‘wh or id fr 7 a. et rs pen flea on ed nd

‘Bowron tl atic pss Hs eh Ma

PROBLEM 5.50

‘Determine th cntoid ofthe are shown when a Zi

SOLUTION

Wehave

ind: Sond F whew a=2in.

| Le -24)

Weine rue! Koran
gras Lin 4
2-2m2-4

wi 7 oe Fais
22-12-10 su «

ol Ne of ht Ha may Be ied

mer wou posi Free

PROBLEM 5.51

Determine the vale of fur which the ratio REF 89,

serle)
a area
ma afan lo
sg
and fau re Nadel a T

ae

dora
“fs ame 1

ama er

oparsrany MATERLL. ©

| PROBLEM 5.51 (Continued)

iad: a so tat

À Using trial and rro or musical methods and ienring th

tiva soon à

in. ind

a=ISOtin, and amd 4

PROPRIBYARY SANZ. © 200 Tue Mani Cop A gh note. N ri Mama m and,
{eye dn afm Du mos an pao Fake a en ee
Te nt or oe TE El al

, PROBLEM 5.52
Determine the volume
by rotating the are o
Gy they

the sora are of the solid obtained
blem 5.1 about (a) the line = 240 mun,

PROBLEM SA Locate the corso of he plane are shown,

Le

SOLUTION

From the solution to Problem $1 wo have

es’ dome
ee) a
EPA = 1.448510 mm Seow
Applying tb sors of Paps idos we have ca

Volume «2/40 -x)4 pie] *
IOMA ETA)
= IAS 3X10") 2] Volume 619106 me? À
E
SRG Ly +l + Fla +h)

Where Ts are mensured with espe to ine x = 240 mm
Arca = Daft 203240 + (15130) + (0270)
15210) + (24030) nes = 85810 nan? €

(©) Rotation aboat the y ais
Volume = Zr, 2H)
ORIO mn)

Arca = 27 RE 288 Ve
25d + Role +R Ra + RL)
= 2402012204 (24030300)
(22530) CIO +(10SK2109] Ares 171 en? À

ROPRISTARY MATERIAL. © 210 The Cen Cp, te Al ph eee Nf Ma np ep
wc dle a form eon et ‘ta rar tine pmo fu a ape ead
‘Keilor Bot dn pes sar cm th

PROBLEM 5.53

Determine the volume and the surface ares of the solid obtained by sota
‘he ares of Problem 52 about (a) the ine y =60 mu, the ais,

PROBLEM $2 Locate the conoid ofthe plane are shown,

SOLUTION ,
From the solution 1 Problem 52 we have E Lion
Aero _
Zinn ne
Si 5840 el
Applying th theorems of Pappus-Gudims we have Lu
(€) Ronan about ne y=6omm x
Volume 2560 DA
250604250)
2740) 55640] Volume 308 10 es €

Arch
ern
ZZ Fala Fila Fabel)

Where Je ae measured with respect to ie y= 60m,

aaa» [conan «ca Rom +c +00! +04)

Aves 32x10) 4
(0) Rotation abot the rs

Volume 24. Am 2408.84) 22

200) Volume 177.2546 mm?
Arca 20 Xd = 20 Eig ML 2 Hy EN TL RI)
= 2x don + 29294 05120) GS En 1 00° + 000

Arca 24x10 mnt 4

PROPRIETARY MATERIA. © 200 Me Mio Copos im. An eB fr of is Mowe pho
‘ars ed my fot a ae, wih rt esc th ea ah de ce
ital i Md coco on ams Ppa oe

PROBLEM 5.54

Determine the volume ar the surface a of the solid obtained y rotating the
area of Problen about (a) the x ans, (0) hey axis

PROBLEM 5.8 Loca the conti othe plane aca sown,

SOLUTION
4
rom the solution to Problem 5.8 we have ©
11057 in a @
FR —141370
Bot 220897 in! @

Applying the coreme oF Puppen. we have
() Rotation about th x axis:

Volume = 29h Au 2424
(26897 ia") a

Volune=1620x10/in7 4
er
a]
20 + Tola al HL)
240.515) AAA
SONO) = 25H50) or Ares = 2845100 in? 4

(0) Rotation sh th mn
Volume Kim 2AEEA

2204470) or Volume 889410 in 4

A 2 Le BEE
25 Hale bla HE)
zur

acaso roscas)

or ‘Area =15.48%108 in! 4

PROPRISTANY MANERA 210 The Celi Conan be A ag sor. or of Sal a be ed
oma yy ano te rect fu a et
Eee ate pte het il acpi ar een en

PROBLEM 5.55

Determine the volume of the solid generated by trang the public
area shawn about (a) Ihe. xi, () dis AA

SOLUTION

Fin, from Figure $ we have

Applying. the second theorem of Pappus lis we have
(a) Rotation about the x ais:

or Volume not 4

(0) Rotation about se line 444

Er) ve vane av 4

PROPRIETARY MATERIA. © 2010 Te Mae Congres bn, AN ge re of Mama a aed
‘Eonar ome ye sie we en

PROBLEM 5.56

Deceriine the volume and the surface area of the cha link
showa, whichis made Go a G=randimete ay of = 10 mm
and L= 30mm.

SOLUTION

“The are À and circumference C ofthe cross section ofthe bar are

Alo, the semicircular ends ft ink can be obtained by rotas the ros section through a horizontal

semicircular are of rive A Now, pls icons of Pappu-Guldius, we have forthe volume 7
PA od

ADD MAR

SUL aR A

or Y 2130 mun + 10 0m)

= 3670 731700" 4
A= WA +A)

UCI) + AAC)
ALADO

or the rea

or 42130 mun + (10 Le

2520 ma or d= 232000 4

PROPRIETARY MATERIAL. D 2010 The Neil ergo, A gs ome fh Maal ay be dd
[ere sordas uh psa a wien pain of he per ed oa re
‘Eien cenafan A oe ao oh Jn

| PROBLEM 5.57

‘Verify that the expressions for the volumes ofthe fics our shapes in Figure 521 0

age 253 ar ona,

SOLUTION

Following the second theorem of Pappus<ildimus, in ech case a specii
generating sea À il be ou about the x axis to produce the given sap,
Values of 7 ave from Figure 3.83,

9) Hemiphere: the generating are ia quer cle

E) «ee

weine Pie

42) Semilliroït u evolution: the gene

in en ia quaner lie

Mehare Y =2aF4 wind

(©) Farboloid fevolutin he generating ren is a quer perabols

wer rail) a ranted

(4) Come: he generating ora va mgt

vom ze) we robin à

Iced hun nr farm yo na tee

Nee ta ay Be sand

ren porn a aß ma on ee

‘Traitor rasan pry ee ee ones pn arent Mae,

PROBLEM 5.58

A Bine bole i ile in apiece of Lán-abick stk the hole
is thes countrsink a shown. termine he volume of steel removed
‘uring the eousersnking process.

‘SOLUTION

he required volume ca he generated by rotating the area shown abou the yan. pling the second

‘core of Paps Galdins, we have

Y
Hu.
+
rl Y 00050? 4

ROPRISTARY SLATERLA, 2 301 Vie MeN Compa tA gh foc or of th Mel de
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PROBLEM 5.59

Determine the capaci in is, fe punch ove
shown if = 250 rm,

‘SOLUTION

The Yume can be gnc y ring ange ad shear ch sh abo the ya App he
second theorem of Pappus-Guldinus and using Figure Sn, we have ja

ren)
A) gf 2Rsin3@ Vx a}
EEES 5
le CRE K xe ke)
ee
a Bee
(ists)
36,
u
Wi
7 OB
om
Since 10
vom. vad

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pr din In ou rm by ny nae ven pono he oo ma dea a
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PROBLEM 5.60

Three disent drive bell profiles are 10 he
sue IP any given time each belt makes
onset with ome al of the circumference of
its pulley, determine the conta aro betwee
the elt and he pulley for each design.

SOLUTION

it,

the ih

Al’ Ve.

I

Jue
hi

SOLUTION

Applying the frst theorem of Pappus Ouldins, Ihe contac arc de of abet
ds given by

Ac = ALAN
‘where the individual lengths are the eng ofthe blt ros section that are
in ontne wih the pale.

fa) Ae = A204) 4 Sola)
(aus) [ae]. ore
of) [ue arret

m aran
ES

(O AGAN
al (3-282) Joss

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‘pected oy ory mo ie rion os of pla ea pod td
a ne pray Nn ei dr repo oracle Mea

See he

PROBLEM 5.61

‘The sum shade forthe smal high neni lamp shown has a unir ticks of} men. Koning that
the density of lumina s 2800 kr determine the mas of the shod,

SOLUTION

Y
2
el
“The mass fie lamp a given by
m= pr

Where is the surcar ands the thickness ofthe shade. The asc canbe generated by rotting the lino
‘sen about the was. Applying the fie theoren of Paps Gulli e have

Av TAS 2a EF.
Tr

BLUE mm /G2 + m)
¿james Emmy

[sm (o
ar ams 325 mn
A:

TO +02 mm)
(Se ee]

BA 3+ 40608431729 1867.50)

109034 sm?
Then mp
= (2800 Kay 10905410 Pm 0.001 1)
or m= 0.0008 iy À

PROPRIEYARYSEATERLAL © 200 Mts Wi Cyan Ang ve yr of hs Monel me pe,
‘eras intel my a E pr ar pmo Fale Ro ee
eric teenie and Satin ann cee

PROBLEM 5.62

‘The escuicheun (a decorativo plate placed on a pipe whee the
Pipe ents from a wall) shown sent om brace Knowing that
the density of bras is 8470 Kim determino the mas 6 the
cscutcioon

SOLUTION

The mass ofthe esate i given by m
rotating Ihe arca A about the a

Fr te ie me 950
ré om
1229324 mm

EI our 0143168 04

Ares 4 can be obiined by combining the following four anc:

T. ~ Bem
¿Pa CSc Es Be
he + T
en

ES i im
1 | tos, Tongans non.
2 | cms | Mirage | om
3 sas 029-4167 iss
‘ - loss 2 |
Loasy=62 2
z 2507

ROPMISTARY MATERIAL. © 700 The Ci Con, Ve Alpe sn. Ne pt of m ny be pit
‘rd ry o as rs ed nd

PROBLEM 5.62 (Continued)

then v~ 20374
= 24035997 10
225180
mens)?
BET Kg K22.618%10 m0)
0191528 kg

io 4

PROPRIBTARY MATERIA. © 200 ne Met Compre, In. A mp stan wa ri
D a e) na y my cea el tr von on oe rable soto Son
‘tev dr ne ee tn pr ibe cr rre Hr gs

PROBLEM 5.63

A manufacturer is planing to proce 20,000 wooden pers
having the shape shown, Determine how many gallon a
pain sho be ordered, knowing that each peg, wll be given
Fo cons oF pitt and tht une allow of pain covers 100

SOLUTION
“The number of gallons of pint needed is aven ty
11 gallon
serfs umi pepsi ar pe cc
iles rat en Ear Je cos)
o Number of gators 4004, (4,80)
where, is the surface are of one eg. A, canbe generated by rotating the ine shawn about the avi.

Using the fr theorem of Pappus-Culdinus and Figures Sb,
We have

or Zara ana
22272277

T Wa m?

035 | 0328 cons

: os - ons

a] 9.0625 00175781
Er eae onesies ous Le

5 Promo os PIE au | ain
6 20579 LUS ros | tsa

ISLA 12532 in?

ROPRICTANY MATERIAL. © 210 Vie Menu np Ic is rae pr Moral o be dont
ean mo ta Joa a of ep a ie

tn renege ee,

PROBLEM 5.63 (Continued)

me

then = Pah 25812 i.)
4, = 2x x
0054678 1

Fully Nunber of gallons = 4000, 054678
2223.87 gallons Osler 22 pone 4

PROPRIBTARY MATERIA.) 200 he Mn HE Copo o. i mre pr of Maa be nt
‘rhe orem a or. gm rt wie pif ele ken a
‘Set on pls pa a

avers PROBLEM 5.64

The wooden peg shown is timed fiom a dowel 1 iin
lume and 4 in. ong, Determine the percentage of te
‘nil volume oF te dowel hat becomes est

SOLUTION

“To bain he ston rt necessary I determine the volume ofthe pes, That volume ca be generated
‘by rotating the are shown about the xa

‘|

[eee 1}

“The yenersiag ars x net divided into six components as indicated

o AS
Applying the second ore of Pappus Guldinus and then using Figure 5.89, we have
Vaca = 28422985 A

Ain? >, in. FA, in.
i ren ons oise
3 Game) aT as sa
(O28) = 0.82987 =
3 AAA E DES
4] Sos um DT ET
E ZXOSTSSUITA2S es

5 os

aint 749° | 001005

0020020

à ones am

me am te cnt ri At rl
EEE

PROBLEM 5.64 (Continued)

Thea

Now Yan = (diameten)? length)
Ed in Gin
Zin) Gin)
TO

Then

7
Set Yo og
(es)

4159)

100% oF % Waste = 06.5% 4

PROPRIRTARY MATESLAL. © 210 Te Mute oe, Alp Rm No ptf hi Mal my de land,
‘avon er set ns a Bn ont Be wien pam of fe ma hn ee
Fan ner rj bl ean eee Yea Mama,

PROBLEM 5.65"

“The shade for wallanounted ight i formed rom iy co
twasluceat plastic, Determine the surface area ofthe ouside of
tado, Mowing that has the parabolic cos section show.

| sorurion
Fint aoe that he required surface area À can be generate by rotating the parabolic cross section through a
radians about the y ais. Applying he fs theorem of Pappus-Culdins we have
Anat
Now at 00m y 250
2502 AMOO? or £20025 mm

wd PER
de
ala)“
Do
whee Be
‘Then adi de
Wehe al fade (ra)
Los
fee]
Ao
e RS WA
175433 mm”

uy AROS) or 4255110 mt À

PROPRIRTARY MATERIA. & 710 The Mun Ce, A pt Mer Mar! he ph
Pared or Sri cay mer a mio. es i ft mt pence of i ptt od pone haa
Pl rm i ih lars: Yoh wm

nee rr a encore
+

and location of he resultant of the ditu ad, (2) the
ot reactions the hen spent

SOLUTION
a satan
120 1) ptr
EN vt
A A ake tte
xe
non so
AS
FSU EOI) aa
(a) Sib, F-A33f €
(0) Resetons DEM, =0: BOR -a2smMEI 0

1 585.0015

sn

BE 20. Asso SR 0

2090 1

PROPRIETARY MATERIAL © 200 he Mani Compare A e amd Ma md dt,
ch mr nin mr mm ia ll, a on et

‘yr PROBLEM 5.67

For the boum and loading shown, determine (2) the magnitude and

oom Rare tonte parses we
SE sapo
te
SOLUTION
= at wth
i ea ie
ol, te =
nh Tih
(0) Weine 4-0) (200 Nes} = 800 8
Ram =1600N
thea EEE RRR
or Re 001600 2400
en Late “E2400 = som 2501600)
R-20N] À -233m 4
0) Reocions
20: 4, 41400-21000
o Aion

Amon! mio]

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‘pe trl nan By a ar aha I ory po of ala, et oe
‘Sinton ca nd pared Sige drid Corren aoe o Ho

qa PROBLEM 5.68

Ls Detenmine the reactions atthe beam supports for he given

Ar 20: A-12IN-20KN=0
azote

EM, 20: M,- (ARE -CIN 1) =O

a)

PROPRIETARY MATERIAL. © 200 Ye Medi Coni. gt ra on Mad ar Sp
‘me ar hd a or ar By nn SD Jr ne ann of paseo Sone hee?
‘indo ep Baa Ma

Al
cane
Lat anf |
N
rima MG

Pr y m PROBLEM 5.69

Detemnine the reactions athe bean apport ithe given
loading.

SOLUTION
Gion onto) E

Lemon) non

fig = Géo = 120016
“hea

SYEM, 0: 2.NY720 O) (4 MC ISOO lo) (6 AC, = (7 AY(L200 16) = 0

‘i cama
or By = 136016 Boot

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Cepas bet yo ry ay am lc rr pee of pero ca han eed
Kessler Re hiram ann ic

PROBLEM 5.70

Determine the rections at the beam sopor Fr the given loading

SOLUTION

= COMAS A)
y= 30001

Leone)

~ (4000 41.5 8) (LE 28) + HIS 8) =0
Dto

ap, =0: 44 7401-30001 6001

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cazar aun sy jor oy man mt ten psn Je rd e le tn
‘Sony fein portly Shoe ce

PROBLEM 5.71

Determine the reactions the beam support fir the ven Ling,

SOLUTION

ist replace the given King with the loading shown below. The two Jondings are sale because both
ae defined by linear relation between load and distance aa the values at (he end points aro th sam

24m
oh |
à

LS
ee I

Weave 8, = 18.6 my2200 Nim) = 3960

=

16m 200 Nn) = 4520 N
Wen AERO 1,0
DEM, 20: Com, HAN)

1. 8mxs2209)=0

« 4-80 Anson
ER =0: sons N 14RD 18, =0

a a, =-8eON moss]

eal Cop, o A th ir o o Neal o de int
Frl ar rae nn form y m at lp io pio pte eke he)
‘occas Ao ti ri mace das na

PROBLEM 5.72

Determine ths reasons he bu supports far te given
heading

‘SOLUTION
prie q pumas
PTS re
nate
bend
veine cames
Léna
ten une 4-0
hy, mo: 4,-800m 2000-0
Per Actua!
DAM¿20: M, GUISO -C6S MOI) 0
u sm

PROPRIETARY SaTERDA. © 200 Tue Mein Cops, Ic Arc. pon of eM! par een
‘ead ray o ol uy rc. oor separa ale rm ond ae
‘es ren enrol Wn in rein Fee nega

“an PROBLEM 5.73

ermine te eations atte ba supports fr the given aang.

SOLUTION

Fit rela the given losing with the loading shown below. The two loadings ae va ocn both
are defined by a parabole relation between fe! and dsc und ie vals at the end point ave the same

unre

x

Weine
Then Er
AIR, <0: 4, +100 Neon 20
= 4 =20mN
Drau: 26x00 0) miooo
a“ Mean 2008 m}4

PROPRISTARENIATERLAL. © 28 ie Ma MI Cn, I A igs ree pr fie Man may a
ot arn nc mu err rm kr nal nd

Wehe

hen

Alo

Fui Las (and (2)
Thon

Now

PROBLEM 5.74
Determine (a) the distance a so tha the ver

suppers 4 and ase equal, (2) he correspond
the suppons.

A

18 =1200 1 6000

40,24, =D, ~ 600+ O(N)

Dey

am, fe a oman)

(am fogs PEN
4400-7050?

00 +3 4004 7004 Ste
se+420

©

e

746410

056 4

er iia

Ie oy jm a pom of i feral po hed
adap seat Momo

PROBLEM 5.74 (Continued)

© Werne La,
3)

= 600+ 006083590)
761 AsB=WiN] 4

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‘echt or raed ey fam ob on us ieee wea per tyler cred hoe ee
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fever ng ten pent

PROBLEM 5.75

Determine (o) the distance a o thatthe rection a support 8 is
minimum, 0) the corresponding reactions atthe por.

SOLUTION
e
tm
itm {im sooo {nr oem eo
we DS o
na atta 10000 re
z
i) Fa) 8, = SUKI —100(1) + 800 B-750N! 4
ee
HER, <0: 4, KON - 304 -DN +750N=0
or 4, =1050N A=1050N] 4

PROPRIETARY MATERIAL, © 010 De Mean Cops, o A te sr. opt of is Maa ey Br dt
A ora yma ante rr win pao th ple tl end
cor sates rad ne ar ia Comerc urease a
er

PROBLEM 5.76
Danish esos eas he hn lig
a
SOLUTION 7
Mund
|
Weine
Ten Karo
FDEM, =0: (44,100 kip-() (2 1) (4050 1)
ams aan,
or = 2525010] 4
alas, <0: 8, 4801350852900
ie an n-is00n 4

PROPRUETART MATERIAL © 200 ne Mon Wi Cog e o vd por 2 ant ur det,
‘corel fn en cas tat hi esp or on et
nn ica oes a kek

PROBLEM 5.77
Dsemine (he dti nd aus theo ofthe
team ACD fr lt ato (0)
on rot at
SOLUTION |
@ |
Were
Men EM, =O: (LION. My tO AKEEEOM) ER I=
« mann 4
o 0 G20

PROPRIETARY MATI. ©3010 Ve Sin Congas Inc A ce Ba pr fi Ml y, ep
(ae he fo Ir ana mh rr en peta fe io, ed od le m
rinden ses ante Fame He ar nin mr ran ro ts Mo

mn CH, PROBLEM 8.78

‘he beam 48 supports tv consentatd Id ad res on oi
exon a Paris diribuie upward lend as shown, Determine the

a E
HI. e es

D |
SOLUTION :
pira pren
zum en
» A
° E “
| oem pain sn
a Lots)
Melon,
DE, 0: AAMO.2-0)- BONE I) =(090, 40m) 20 o
For 405m: 2412-06) (0903)-U54@, 20
14409 05t0,=0 1000 4
ARE, =0: HAN ON OSLO =0 ay SOU À

‘PROPRIETARY MATERIA. € 3010 Me Mei Cm be Ag ans A pf Mama ee
‘ett a nated fora roy e. he lar at cn 9 bon rd ped
“Kintera eds patty Ue Mier peter orn Seed,

win

Cda

quero Tr.

1 —

SOLUTION
À Wehe
© ann
a six mo

— 7 |

PROBLEM 5.79

For the beam aa lod of Problem 578, determine) the distance a
for hich «= 20 Ni, (6) the corresponding vale oF 0.

PROBLEM 878 The can AR suppors tw concent feds and
rests on soil that exer linearly distributed upward xd as shown
Detennine the values o o, and x costesponding to equilibrium.

= juameom-
= mk Nin 090 18

(12-024 KN 0.6 mx IS RN 0.3 mxIO KN =D

a-035m €

24K FIRKN à (9) EN -3DEN 20

Ders)

‘erat ory hepa esse ae

PROBLEM 5.80

‘the cross section ofa concrete dam is as shown, For a amide
dam section determine (2) resi ul the sion Forces
exerted by the ground on the baso oe dam, (9) the point of
application ef the resultant of Pasta, (e) Ihe relat 07 the
pressure forces exert ys water the fase AC of the da

SOLUTION

Consider fee body mado af don an impor section ul water shown. (Thickness = 1m)

“o
= [Th .
i"

»
de
1 ORI AG YO St)
24.2 mk KR)

ren
CAZA MA iO)

EN
O)

EN

Far sam ES)

EN
ETES mes
Age <0) yoo 48-0

Forn vom]

PROPRIECANY MATERIA. © 2010 Me Matin Comp A vo cs. pf Ama y ern
edad tae nt fo ya ch tr is pen Mb, do ed
‘Rn ere pst Me Mc abd on ppt now Me,

PROBLEM 5.80 (Continued)

o

5
= Stss)=3m
j= ¿(8

zw ı pam

(830.7) ~(G {542.5 KN) ~ (5.6 m 20 EN)
(6 MNBA SAN) +2 25431) 0

(830.7) -1627.5-11390- 842.74 610.3=0

1830.7) 26989 =0

2325 m (lorightof 4) €
(e) Resaltant on face BC
Dirt computation
P= pgh= (U0 kei TEE
PTOS NE

accio
som
os
est
5
= ones? xs mm) |

tea computo: as fe dy ol water soon RCD.

REN IRA
Ren 718430 4

Septic hey fom ur by u means. tho te pr rt pono th uber, wed pod te in
‘Rob centro poly Sn era ep rosa ube Hd
Sree ake poi

PROBLEM 5.81

“The cross section ofa concrete dam i a sh or a
dam section determine (a) the resultat of the reaction forces
exerted by the ground on te Base AB ofthe dam, (8)
pplication of tho resultant of Pet, () the ves
presse faces exerted by the water an the face HC Of the dam.

SOLUTION

‘The fa body shown consists of Il hick section of the dam and the quater icular section of water above
the dam

Boten

mn
Ciegas

ET

or area 3 fst ose,

Then

Lo.

Mairead ns sn cannon Vpn tah

PROBLEM 5.81 (Continued)

& Now wer
So that u eje wun]-sıosin
a = SO WN VEAL MO 19] =25,200 10
mw, asomar? y 2x2 Jp x 19,19
soma) [ar Gor joa | 151966
nam] jervan aa
Also 2453101 AXE IL. MERKE Ny} 13,759 1b
Tron BAO: #000 ce Hatem 4
ib 25.200 1196121619 =O
o rana veusokips 4
© Weine ane, =0: 411.969 my ACT YS 954B) —C Nasa)
AI 1496) (17 was
ra)
o 112,968r-627,980- 60,00-478280- 3,0101 VAE
or remand
(©) Consider ater sen BCD si ebay
Wehe Em
Loan
‘hen

‘ded oda nao a, wi
Eo npr eh ld postin pen ene Ma

-R=256kips <2 575° 0 REG A

cst: HO rc o per ft Ma ar ra

PROBLEM 5.82
“The Ax dose AP oa ki bot and Win plc by»
‘hin od BC: The mao ese fore the 10 en stn wou bring.
1320 AN, and desen pets quie for in he od oto exer
Do pan hs wal rc slowly filed it wae, dem he
min lobe do water nthe tank

SOLUTION

Comite the ech dng of i

ete eo ig de x
1.
Wehne pelage dined)
2073 ro
am

| Now Dem, 20: CE if

|

Whee ER

Tht or da

CRE 0
“ 120 Rem 54, Nin 20
“ du 2h 4

ROBAR rau. 3 Te M Ca AE an Nw ef a a és
‘Sino sen endet frelon ao bel er man poate te eo

PROBLEM 5.83

‘The 3° m side of an apn tank i hinged at its boom and ie eld in
Place bya hi rdf. The tank st be fled wäh giycrine, whose density
18 1263 kg. Determine the fore Tin the rod and he reachons st he Hinge
fee he tank filed to sath of 29m,

‘SOLUTION

We have

“Then

Consider the ice body dingrm uf the side,

Mesa)

Am) (26m JO ia PS
oi HA
470 yeh *
5 fs
Sauce am ( m)aman-o
roer Tem
FA 24D AN ASTRO
AS AAN 4

PROPRIETARY MATERIA © 200 ne Mean Com. ig rare, e nn hn Mod ma be

tet
ies teri ant pron ef te pale are Spas fus]
RG D M

PROBLEM 5.84

“The tion fore Between u 6% 6 square sluic gate 48 and ts guides ise
to 10 percent ofthe ecullan ofthe pressure forces ext by th water onthe fee
‘ofthe gate. Determine the intl fee needed ot the gate Tit weh 1000 Ib.

SOLUTION
‘Consider the free-body diagram of the gate. TT
m PRE
ET ant
ron oa
ein
ha Pacino en)
Zoos san
07m
Final Se, <0: 726957 101000 10-0

‘PROPRIETARY MATERIA. 200 Ihe Moot Compr, be, A ig Fr der Ma y he die
A ne ow ery ye i tr wen ern of he plier ed bon eed
nino Ref AAA ga ro o a

PROBLEM 5.85

A freshwater nash is dine tothe ocean though an automatic de
‘ate that is 4 N wide and (ug. The gate is eld by Hing une
lon isp edge at 4 and boats on a sil t,he water evel ithe

mars is #6 determine the ocean level for which the ato will
open, (Specific weigh of salt water 6€ It!)

Since gate is 4 wie
Thos:

w= (4 Mp Ad
ih

area
Ber
slot
Lomeru-0-a=-ea—a gun
Mitin

Ga)

DOUTE 7

aaa: OB Ae)

Lenmar
jer-m-3ar-n

Aorta 61-9,
2d 22h Hard ar
Ya

AM ie ec prof Mind eb pie
‘pra ar ed of my yma ea te oc ay pa er tet eo

tr ae or Be D nibs A Dre sun
‘mana ipo ”

PROBLEM 5.85 (Continued)

win Boo amd hman: ONDA
x
1-52
7
Data: SOS
pam
LA
E
00058 asin

da

ROPRISTARY MATERLAL. € 200 ue ke Comore. Al gt Ea of ds Ml y, ere
“esi! racy wry ng en i nn ofa te nd

PROBLEM 5.86

Te dam fora lake i designe to withstand the additional force caused
by sit chal has seed om the fake bottom. Assuming thal il is
«suivant to a quid of density, =1.76x10 km” and considering a
Ewido section of dam, determine the percentage increase 1 the
ce acting on the dam face for sil accumlation o depth? m.

SOLUTION

| Fira, determine the force on te dam fae with the it

We here

,
Lao
= Laos

bs
LG mK ID" Km O. ls 6.6 ml T

Com

13.06 KN
Fire on the dam fae wih sit

Next determine 1

Wehe A= LG MO km OST AE mt

03.290 kN
PE. m OO kar KOT? HAG Wd
90.252 iN
De wy KC1.76%10" Kn NOT ws? X20 my = +,
CR
Then FLA) + Ge = SST AN
“The percentage increase, Ys ins,

698740 thine. 658% 4

PROPHUELARY MATERIAL. © Jo Ta Me Cages, tA fs a at Ha mae pc
Ieee nut mj oh a a rl mr porros fhe Yaa on ee
‘Berntsen cen pr ee fete lidad open peasant A

“

PROBLEM 5.87

“The beso oda for la l designed to resist ap 10 120 percent of
the horizontal force of the water, Aer commetion, à 5 ound tha
tht equivalent to a quid o density 2, = 1 76X10" kee)
Setting om the Take bottom at de rate of 12 inver. Considering u
Y mewido section of dam, determine the number of years until te
da becomes neal

SOLUTION
From Problem 8.85, the force on the dam Face before the sik e deposite, à
alle ie Pa, où he dam is the:

Pan 1321 66 XN} =25639 KN

13.65 KN. The maximum

Next determi the force Pom the dam fie alter depth of silt as sete

al

ooh Wn

Wehne 166. dpa O kn X91 ms 40.6 dm]

mar
(Py = 1d TCO? Kgim KOST mL en)
are
n= comas ape ant om)
“som
PB) | (Pa =14.905(43,560—13.2000d +d)
ona
on

‘Now required shat PP, Lo determine dhe maximum vale Of
IDA + HN BORN 256398

Lo 33856

282 yours 4

Finally, 3.3856 = 12107 wes ar

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md cad sh où ay men in DJ cn pr à pas ord een ed
‘Gani cheney I oil Id cre re Pca ana Se

PROBLEM 5.88

A 05 x am gate AB is looted atthe boston of tank Fille
‘vi wate. Th gate hingod ong its op edge À and reis on a
Ietonless stop at Determine the reetions a 4 and 2 when
cable HCD is slack.

bose!

SOLUTION

Fis conser the force ofthe wate on the gate

We have

so that = 05108 AO AT mi AS)

29 N à
OS NOS MISCO Kim KO ms? 093 ot

24,06 N

Reactions a and B when T=0

7

Weine
pee
Jam, o 408 m2 0 12008 masas (03 mi 5 J
ar mision PR
u RAISINS 4
Paro: 4 415I074N-89N-180466N=0
u Ans

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! PROBLEM 5.89

JA 0 x am ga AR ds located at the bottom of a tank filled ith
‘water. The gate hinged slog is op dye À and set on a Sctioniess

‘open he ee

a =.

fy jus m0. mIDACIO Km OST ONE
152466

Tioopen

First note that when the gute begins to open, the reaction at # — 0. dera ah
a
Tin emo: Loro ecos miissen Wwe
3 3 one
045-027 7) de
iG

or 238.464 978.152 0.33882

or TEN

Ingrid or cise ny frm or by ay m. mühe the rir tien person uf be publi. or ed Aya a eed
‘Kiet le oferta pri no Mom

PROBLEM 5.90

A long tough is supported by continuous hinge along its
Tower edge and by a series of horzomal cables attached to.
li up eg: Deter he tio i cach ofthe che,
ata time when the tough is completely fl of water.

SOLUTION

Com eg eg ha ghana
at ny ps
Dee tie or
fan
nern
PAR
us
a

Pans €

rome urea © 0 ne cn Cc be A a ah Me pot of ht aa e A,
‘Recently i bn can rg otk

PROBLEM 5.91

A 4x2 gate fe hing aA ani hod in position by rod CD.
End D ess agains spring whose constant i 828 Ib. The spring
is undeformed wien the gate is vertical. Assuming thet the force
cecertal by rod CD on the gale remain Normal, determine
‚minimum depth of water for which the brtom B of the gate
move to the end of the cvlndicl potion a the Moo.

SOLUTION

Fist determine the forces exerted on the gate bythe spring and he water when B is atthe end of te

marca porton tte sr
We have. mol gas
4
en ke = GR
ai Fete
TT
ET
an
We have Pal apa! ar) sd
243 ny
i O |
| sad i
| Sl)
2044-0390
For a soga opens, =
À vaig above fico body grams oft as, we have
| En à
Sem, <0: E
ora -0
or (03264 -191.2)+ (66564-35670) 4002
a 46008
48 seston cora! anon 4

[PROPRIETARY MATERIAL. © 2019 Ye Bin Cay as A ih ome pt of as Mae! ay gar,
Te ne M aeg Pes Ma
‘err sar

PROBLEM 5.92
Solve Problem 5.91 if the gate weighs 10001.

PROBLEM 591 A 492-1 que is ngs at 4 and i held in
posi hy rod CD. End D vests agains a spring whose const is
98 Ih. The spring is undefonved when the gute is verucal.
‘Assuming thatthe fore exert hy rod CD othe yal remains
ron, determine the minimum depth of war fr which the
Soto Bote gat wi mec othe end of sti potion
of te ler.

SOLUTION

First determine the frees exerted onthe gate by the ping ad ie water when Ihe end the
eindicl portion ofthe or

Wehave sn 9-30"
Then an |
ind Fag = Brgy SINS tind"

sais
Ass dean

1

Weave La

Prom
then A)

49 6

RD NUS 60030)
eu as
ct

Tor dya 80 that gate pes, W
Using the above feesboay diagrams ofthe gato, we have

EM, =0: ($e }osaca—ay-[$a amet ossssoiy
Gaia da en

or 68284 151.2) + (668.64 35620) 43024 1000=0
or oon
424 = assumption comet ara d

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(mad und ns fm yt m we pr ne enon oh Fler ei
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PROBLEM 5.93

A prismatially shape gue place at the end ofa esate cane) iy

Suppor by a pin and bracket andá ess na Ftionss support
lanos di 0.10 m bel the ete of gravity C

ofthe te Determine the depth of ward for which he git will open.

Froumon

Fin mot that wen the gate fs sbout to open (lock
action of the allan P ofthe prenne for esse Ih
at hamogencous, then center gravity Coincido with he corto

ad
Now a:
ris
otha Losa.
09-15
SimplDing pete
w
Alternative solution
Considera fee ty consigo Lam hick Seton o he stead We Wangular section BE of water
ove the ite,
Now

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Sts ol rociado Mma mba pp Y ri o a

PROBLEM 5:93 (Continued)

| Then with A, =Ofes explained above), we have

200 SalI pa
demo [Foal Sa (Soe)

00250) Loe }-0

ntm

2 EN
ing Psoe dran 4
Santana o uso 4

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es rn fo pat aa he tar rps fo er, oe so ed
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PROBLEM 5.94

‘A prsmaticaly shaped gate placed atthe end ot
issüpponil ya pin and becker a À wl ess a I
AU. Determine the distance i he gate e o open wh

SOLUTION

Firs woe tha wher dhe gate i about o open Ceci rio is pening), B, > 0 and the line of
ston af the voit Pte reso res pases ugh the in at 4 In adit, it is aimed tat the
dnl ofthe tsar area, Then

te is homogen, thn is comer u gai coincide ml

dann As
Fr
u nn tan sen
Nor
sota or

309-500)

‘Simplifying pies

789 gy 706
eis= a
as = 7 w
Atom slain
Consider tise had consiga Lan thik eto ol the a andthe agli econ HDE of water
pa
o Per)
Now Pai compro sq] E
ir
l
"te,
Sos

‘reel did oy ro Dr
rine ite heen

tic e i oc Mo he Mm ant
reno ee, ed in

«se

PROBLEM 5.94 (Continued)

“Then with B, =0(as explained above), wehave

> EA] AA CS (A
dao [joo for [osa]
sss
sunny Marois on

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Pads tint rer a a mt rr oes emi of ft ool jn head
ne fiesta ae caos don

PROBLEM 5.95

A SSegilon 2-im meer drum i place oni side to act as a
dam ina 30m de freshwater anne. Keow ng that te dram i
anchored othe sides of the ebm, terne the resaltan of the
rosa ess ang om the drum

SOLUTION

CConsidr the elements of water shown. The rsullant o the weights of water above each section ofthe dm
and the resullani o the pressure Forces acting on the vertical surface a the elements cs ete esl
Prostate ince acia on he dra, Then

Aye) >

Ale] A
En »
A

en Es

ACL (US)

Then HEN: Re 86542-70635) 2 2149118
Jer, + 255.80 112528) 337.581
Filly Ra JET ano

RB 575° 4

see po a i mr ei
pnt mba, a mo nd a
econ Procenet se

PROBLEM 5.96

Determine the location ofthe cent oF the compte body shown
when (a) #29, (0) = 2.56,

Cynder neo |

(o For

lated
a [Le om

Connie of hae of cone À

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‘hr trl e fo wh np mses o pr ven wma of he ahr ota eon ie
‘SoReal ond ee partly cr Mtb ss pre Ppa M,

PROBLEM 5.96 (Continued)

() For h= 25h: r srelpeja sure

ern] 1
erratas Les
E zes ¿Osa

WAS
Ss

ART ENT 2

11368
Centos 0.011360 sight oF ase of come €
ote: Cen ia ese oP em oe = V5 = 2 449

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PROBLEM 5.97

Patrie the yconrinate ofthe centroid of the body shown.

SOLUTION

Fit note thatthe values of Pl eth same forthe piven body ad the boy shown below. Then

oe
tine bea
z ¿ca Gear)
| then ea] Eaton wy on ya 4

DPROPRUEEARY MATIAS. © 208 He Metin Cops, A gle ses N ptf Mura on dt
Speen ir ifn ae ion tary pai of i a

PROBLEM 5.98

Determine the 2 eoondinate ofthe eno ofthe body shown. (in: Use
‘the slt of Sample Problem 5.13)

nder” from a “alcone," a shown,

HaitCone

Maley

From Sample Problem 5.13

Weave

hen

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‘ian ren crap He ri ae Pepe Hn ack te Mo,

PROBLEM 5.99

“The composite body show i formed by removing semiclipeci uf
revolution of semimjor aus A and semiminor aus a2 fom a
hemisphere uri termine a) they coordina af the cent
‘when b= a, (D) th aio a for which = 0a

ana
; T7
2 far 1 3
sono Er
hen # à La (aa =
pin qu
zw Bauen
| 1
or Yu 7-00

LE

PROPRIETARY MATERIAL © 010 Te Mein Corpa, nM ie rc a at of le Mol ey be tnt
pra dono) yo by amar, ror its pom pale ar and e
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PROBLEM 5.99 (Continued)

o won Yee
Sainte.)
sofa 222)
comet]
. (dran

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PROBLEM 5.100

Tor the stop bracket shows, locate the reontinate ofthe center

oran

SOLUTION

sum that she brackets homogeneous 3 tha is center of gravity coincides with the centroid oF te
volume,

1
2
3
a | Less | 3615 | 180060
[= 209190 [9786600

zur _ 9786600,

PROPRIRERY MATERLA. © 2010 ie Mtr Compu. Ang NM raf Ml a ean
‘puto ae yan et a er ind DJs mel ro fe pb eet fp ed
To Seay Uca cogi

jem PROBLEM 5.101

For the sap esc show, locate he coordiate ofthe enter of
ity

SOLUTION

‘Assume that the brackets humos so that center of gravity olnides with the centro of the von

z Wot
ü : 5
2 | compose | mens | one
3 | eso | arena un
A | sta | iso
y En EIN
seno h
en MEL w Zonas

protean go. © 200 o Bs Cno, aa po a sn,
En ende pony Sane ar tate evar a ham

Ae PROBLEM 5.102
LA. |
urn For the: machine element shown, locate the y coordinate ofthe |
A cotes

A

see

re

SOLUTION

First assume that the machine element is homogencous so
eto ofthe coresponding volume

is comer of gravity wil coincido with the

F Fm Fm m vw | mnt
ACT ° soon 1458000,

a docs = A800 | 2 48 4416000 2304000
: (12910) 45239 us se 495010 244290

Ww | ass 28 9 267590 EN

Y 2069672 i BE
Wehe 22

204967 2m) 3920280 m“ a.

PROPRIETARY MATERIAL © 200 he Michi Cans Inc. Al gh rca. por Ama wo eed,
posed rd as fm ol np meson esr pf yaar ar a ea a ae
‘icra red Keren forth ib cre rt pro asa tr es
sow gon paar

PROBLEM 5.103

For the machine element shown, leat te y coordinar ofthe
center of gravity.

SOLUTION
For half ylindrica hole:

For half eyndcal pte

| sn Pi
1 | Rectongutrpine | oxao7)=2.0 | ass | as | tars | m
1 ETS ET to | 2 | x00 [tno
m [antoine | Gaza | um | 2 | ser | am
ty | Halteplinder | S(2(0.28)=4.712 | -0375 | -285 | 1767 | 3699
v| singen [oser | ons | 7 | um | 257 |
Y EX) EN
22
een Shin y = 3 Rés in ann 4

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‘moc ented yom ator he rien prison le. a 0 jo ee
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PROBLEM 5.104

For the mach
ster of gravity

len shown, lote the coordinat of the

SOLUTION

or half lin hole:

F2
4025
ae
21470

Im 2m

or hal eylindricl plate

u Tan [zu [mn
esta pats | Tons [as | ae
CTE 10 | 2 | 00
ui {Half cylinder) 28) (= 2454 1470 2 3607 | 4908
Ww | marins | eroza=amz | os | ras | 1767 | 5699
v | im see ons | 7 Lin | sm
Y sus | os]
Now ar |
OPEN ze |

Soong ttn pr

the comer oF eit.

PROBLEM 5.105

For the machine element shown, lor te x coordinate of

SOLUTION

Firat sue that
‘eentoid ofthe corresponding volume

€ machine element is homogencoss so ha its center af gravis wil coince withthe

Pm" mm | fon | mm mt

1 | caasy0)= 162000 | so y 101000 158000

m | asno 4 416000 2304000

m | rastas Us an IE

w | una | 2 y 267590 on

z 2049672 [Um EE

Weave

F(2049672 mom) 12723420 mm A]

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pre oreo ya ia Jrs ics pe of lle, et a ae
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See wargames

j PROBLEM 5.106

an Locate the center of gravity ofthe sheet metal form shown

a

RA
EU
PT

SOLUTION

"Vist, assume hat he sheet metal is homogeneous o ha he cont of ravi Of form wil cosido with
the centroid of th comesponding ars.

=» [zeae a

15 j 4 12 324 | -0864 | 259

u | "asanzen [om Le DEE

m| Eosyesum | | on | am Passe | sons loués |
Weine Aaron. 230m oe X-0295m 4
FEV =EFW: 73.3694 m?)=5.0585 m0 or Ÿ=0421m 4

PRY REV: 70133684 m") = 22.769 0" or Z-1703m 4

PROPRIBTARY MATERIAL. 1 ue Ml Copa, De A is sare No ut fi an we pt
epee ar a erm 7 a er be an A leo a jo e ea
‘Seatac met owe M mb acne Pee nt

PROBLEM 5.107

ity of the shest metal fom sown,

SOLUTION

Fat assume tar the sheet meals homogeneous so Hat he center ol gaviy ofthe form will coincide with
ho oeil of the eoresponing ares. Now note ut snc imp

x N = a
2 50.930 mm
R

1250 4

de mo Aus
\
> el
an zm | RT ]
1 | comes [7s | 0 | su | immo |
T 1
a | Zooesy=ss | 20095 | som | esoo | só |
m | Zas) ns o | cmo o
= 60 sour | Goo
were FRA EG A GO) 1647100000? Herman 4
ZEA=ErA: Z(98460 mm’) =3.300%10°men? or Z=335mm 4

PROPRUETARY MATERIA. © 20 Yue MC Ces inc Aa ne. Jr fe Mama be opie
Ip sent if hy msn Dept mi ne bier aero ol
‘etc md o pci ae rs bro om paran Tom svn Cs Ml

PROBLEM 5.108

A srastchasket, designe 1
and hos ise inthe shape oF
the center of gravity of te wastebasket, knowing that it is mado of
shock metal of nfo thickness.

Ihe corner of rom, 16 in i

SOLUTION
v
By symmetry: we Love
ac (Cylindical surface) r <a
5 Dur
tee ete Ws
4
Ain? E
1 | omas 5 800 1280
MIRTA o o 1280,
| u 25133 63002 10000 | 20106
DE ass 15333 o
| 649.7 m3 | 43706
AU STIn 2730.30
E 42059 im HoZoa2iin 4

EST in!) ~4890.6in?

Ferostin

PROPRIETARY SERIAL. © 200 Va Nan Compas, > A as o pn of aa a a
‘ee nt ony e ra ha jr mien en fh dr a boa de Bed
Din near scr lf te havea ne pon Dwar te are,

PROBLEM 5.109

A mung hc for cetro compone fon mn
sect metal of form thickness. Locate the center ol gra
‘ofthe Bracket.

SOLUTION

rst, assume thatthe sheet metal is homogeneous so thatthe enter of gravity o she bracket coincides with
the entoid of the corresponding aca. Then (sce diagram)

Ain En EN iin! | Phi
ı| es@=1s [us] 0 > CO a
n | users | as | -oms | > | ws [es] 25
m | @rso=as | 2875 | as | 3 | ews | ses | ns
w| (ous | 10 | o | as | as) 0 | mous
CRETE vo | 0 OC 0 [ame
| = En] ‘460739 | 103125 | 65.7197

| Were FRA=ETA
HG in?) 400730
Fra=xya
¥(22.6364in?)=-103125 in!
PRATIEA
2026") = 65.1197

2000. 4

PROPRIETARY mem. © 200 be Mari Compre be, AN gh eo ef of Hp et
‘ee sed yf ot e a tr tn pms he so ho ee
“ain ero pata nn rset os rat Ye user
‘nar price

PROBLEM 5.110

A thin shec of plaie of uniform thickness is bent 10
form a desk organize, Locate the center of gravity of the
organizer.

SOLUTION

First assure that dhe plastic is homogeneous so tha the center of gravity ofthe organize wil coincide with
| the cenit ofthe corresponding anc Now not ha smc ities

Ferm €

226 2.1803 mm

5, 364 2 239.220 mm

5, =50-2% 54.180 mm

m+ PL 136820 0m

26-2282. 1803 mm

75925 18.103 mm

660 56549 mar

A EAS 204248 om?

PROPRIETARY MATERIAL. © 200 he ni Copo, be, re ron. ps Bo ep,
(mh han is far ar y ye td he ro pom of e li lord
enr ae ft id rm pte yc ah Shrek

PROBLEM 5.110 (Continued)

= Timm | Fa
1 oe o | 10
te ai | im m
37] con «iso | a CN WET a |
4 565.497 39.820 2.1803 22518 1233 i
5 (69K60) = 4140 a 405 173880 167670 Î
ma 9 Tas | an mom
7 | Goo | "aus | ais | ion
a EXT 180 2.1803 30638 | 1233
5 Too | os 0 ao 0

La 565,49 16320 | 2.1803 77370 us

PRET _ CT | os}

We have XxA=YFA: X(2222444 mm’) = 1032766 mn or À =46.5 mm 4

Fan SpA Y(2224:44 mo") 604878 me

UPRLETANI VATE RIAL. © 200 Ts Mc Cent, eA gh oe op of or ol Be gi
PR dns um ny ry noc ma BE os tan pan of ble, rl poe had
‘Bare nr cow Me a ce npn Doma onto Mean

PROBLEM 5.111

A indone eng is fabricted fom sheet metal of if
Fiches. Loate the enter of grit the aa

SOLUTION

Fins, assume hat te she metal is homogeneous so that the center oF gaviy of the awning cocidos with
the ceatoid ofthe eomesponding ars.

259° = 49087 in”
Ten’ msn

des sis

Ets

w | sr cy

TN MODE 2 Ts E sa |
women Dons | 2 [a

v| este 3 ms | m

M am ram | M | ons

Fa TES - m | 3

‘PROPRIETARY MATERIAL > 200 Ye Min I Cyan, wc A ee Sp U Mat u be ee
"Sect nn e e o pvc cil

PROBLEM 5.111 (Continued)

7.0010. €

ows, syrmetry implies x
ur Pots Sin. À

and Press Fl26s29in’)= 41607

Z26528in°) Gin. 4

1868 or 2

On MATERIA. © 200 The Veo Compro, Ve A pi No pf th Al e e dot
poche ordi nae rm ey m ala se pic of Pio Sod td
‘Soares RG iirc br ope Pn Ho

4 PROBLEM 5.112

An cn forthe duct of «ventilating system is made of sheet
‘metal oF uniform hikes, Locate the center of gravity al he
bow.

SOLUTION

First assume hat the sheet metal is bomogencous 0 hat he center gaviy of the duct eins with no
emo of dc eumesponding area. Also, note that the shape ofthe duct impli

Note dat

00 2 (206) 172.676 um

5, 400 À (400) «230.25 mm
ES

3 200) 315.12 mm
0020021512

4
242300 À 200) 2215 12m
wo 3 20

Also note that she corresponding top and Bottom areas will ont equally when dtm

‘Thus Am z,mm A, mm"
MET [use | nase | ose | as
u | Zeon | ass | uno | css | ann
Mi BH] | 15000 | aoc
w | Ej =25u27 | no | non | sen | sc
) i
v 4 Jeu sm | 1502 tes | 1356220
vi | annee oo | so | 350 | 12000000 | nano
E A A sera | oia

PROPRIETARY MARIA. € 710 Me srl Compas, AN gl pi of Mod uy e dpto
‘nc er nae ep forty e mas cs tn 9 pla, o jo el

‘Sinners chosen pray ben fr eta nr Peat rare neo

PROBLEM 5.112 (Continued)

We hase. SA ERA NDS mm?) = 4103760100 or 1802 m0 4

DEAWEE A FOR ma) =44070260 mu or Z 1935 mm 4

PROPRIETARY ana. 209 the Mea Compre, lu. A ii ur Moa up be ra
rc ea ny fra by ny re e te water puros pl hon ad
Beton pele htc Md rt aos ng oa

PROBLEM 5.113

a Reine eyimdrcl duct and a 4 Si. rectangular
ct are 10 be joined as indicted. Knowing thatthe ducs were
fabriatod from he sane sheet met, which is of uniform
Ickes, toate he centr of gravity ofthe sembly.

‘SOLUTION

Assume that he body i homogeneous 0 that is center af gravity ences with th eer oth ara.

Lae Ein’ | Fain
HOD =e NET
ECS 28 on
A Zep aoe a0 | 96%
ES HT ue
5 | unes m | 7
Fay ar Eo Pr si
| Zu mn | mon | m
EN E i | ae] E
s | aaa é no | m | een
53 1566] aa
En 150501 isla
then EN or erat
354 oma, ern

2a"

PROPERTY MATERIAL € 200 Ie Maes Compu, A hs soil a Mol Be ied
‘aed "ated n oy frm oy ym rer tg pan of hbo, a pe mt
oka inant akon Mire bai rro we sein

PROBLEM 5.114

‘A thin tel wire of uniform ross section e bent no the shape shown,
Lee enter of gray

‘SOLUTION

ins asumo thot the wire is homogeneous so thal is comer oF grvity wi cincid wih he cent ofthe
comsponding ine.

pl 0360015 me
À = O3cos60"= 015m

A — ( + ds

ta * 53,09
um Lae [= [#7 [me]
Duo ous | os ois | oases | 0a |
| 1 i +
oe | | oo | 998 Tam |e | osu
ei a I fur
E ICI IEC 3 10 O48 | on
= | soon | 1 Tas NEON EA
We have (3.0283 m) =0.43981 m? or X-0.1452m 4
76.0283 m) =1.20m" or ¥=03%m 4
20289) —1 08 ar

PROPRIETARY MATERIA 120 he Meme Compu. A ih so er of Malo he dor
‘mnt did yf rr. al pr rte pers plo er one m
A RA peavey th Ma,
Sarno pen

PROBLEM 5.115

Lote the entr of gravity ofthe figure shown, knowing thai is
made of hin bras rod of nilo darter.

SOLUTION

Unio od

AB OT

FEL=ERL: F0m)=205m°

19m
im [em [oe [ae zum
we [te] es | os | os os
zo | 06 | 19 | o | 03 as
Loo wo [os [o [o | om |. °
as jo jejej. Jus |
Es fs i [am | ae | os |

VELSEFI: Y 0255 F20510m 4

Zut=zt: Z60m)=0750? 7-04500m 4

PROBIER erst. EN et np eh ore Np of a ar Ana
ru uncer neers perl y Mera Mir ht ind vera reporto pn sens Ma

owt thon pr

PROBLEM 5.116

Tote the centr of gravity ofthe Figure shone, knowing hat it ik
made of thin bres rods of uniform diameter

SOLUTION

De
ANT
KEN.
Ein | in.
ar | ar o |
ap | hors | 15 | s sw | m
ae | | ts o s0 | 0
E ji 1530 | 784,
FEL=ESL: FU0S2265 in)" 15300?

PELAR: ZOS226S in) 784 ia

Sin. €

PROPRIETARY MATERIAL. © 200 The MEC Compt te A ng ni a er fé Mama edo
Ita tnt o ya mts or pomo Wf ee a nd Doe
‘Keys cers Sno ic can nya sc

PROBLEM 5.117

“the fame of a prenne is constricted fom wif slain
tama. Locate the center ol gravity o the portion of the fame
show,

| sorunon Mr

Fin assume lt cel are bomogencons so tht the center of gravity ofthe fame
il coincido with the centroid ofthe correspond line

LA zn In jr] on | uw
i 2 5 o 6 o 2
3 is o 35 ù a
5 3 25 15 | 3 o
5 3 25 5 | ns 10
D en + 9 |» 16
poy 3 5 CO 2
Ds 13 5 35 15 6
s | Peer 2 | 6909 | 0 Û 32.562 0
E AE E
X 2 o s fa a 16 2
2] »ax 6 [ren | sioux
We have. RL: KH) =09 0° ET
| HER or Pasion

Zoomen) ss as

3550 €

‘PROPRIETARY MATERIAL. © 2010 The Mei Cons AM ls aa. Ne at of vu ng be tot
‘cil cd ny or or ms a ivre pres ieee fd o
tn nicer pro Bo rie ans mae ence ha

PROBLEM 5.118

hat the diy pati i 1030

A scsi al In pi ade und is ada shank, Ko
amd of tee is 7860 kg/m’, locate the center of gravity ofthe a

SOLUTION

Fi, te that symmeu implies Y-2-04

;
[= I

—

=

= 2425 mm) 28125 mm
$

9,0000 byw’) Jonas"

À en 22 001250
4.213310" ke

7,825 mm

Wy, (1000 ky SE Joursm oem)
4044810
a 925 mn-25 mm 67.5 mm

mm (Eoonencem
G

=-049510x10 "kg
Sy 192 m7 m2 mm
Re AS

2 1
y= 182.5 mon +210 mm) 185 nm

252071 kg

in, néon Eos me

PROPRIETARY MATERIAL.» 2000 ig Men Copas, in Ao or o par of toma py plat
‘ed da oor yam ec ao fer ne
Sova wh tah prc * di

PROBLEM 5.118 (Continued)

[24 ima | kg

1 CETTE mans | 32000”

u EXT E 22.5107

m | nass rs | same”
cw MET mas | nwrosxio*
y ESTOS us ET

z Er ET

| Wehne ian

55 005310 Mg) = 2360-7210 Pkg

Feelin 4

em the end oF the han)

PROPRIETARY MATERLAL. © 209 The Masia Cone. A gh ae o ek Mol aye ep
“onan ranean by Moria ae epee wera Md

PROBLEM 5.119

A bronze bushing is mounted side a gel ser, Knowing that he
spite weight of bronze is 0,518 In’ and of see is 0264104
détenir locas of the eter of gray o the vel

SOLUTION
“gu
1
y
x
2
Now wet
Frein nn | es 075 Jn" fosin) 0.238801

Din

A ny causant

A jus a ai)} ame

Wohne FEAT
(020100.

1568319) + (0.70 XO 1092
9 10.14 Hy 0.109200

o. F-0.526in. 4

PROPRIETARY MATEREM- © 2010 The Mi Capa, A met pr on Me ed
‘aan a dnt ny oo E ats na ts i rms of fe fle ad Sid re
‘Etnies Hoel al aoa Psa Me

PROBLEM 5.120

of Regih in. Tete

A bess collar, of eng 25 in, is mounted on an aluminum rod

contr of gravity of he compost bod:

(Specie weighs: ass =0.306 Ii, slum = 101 in.)

SOLUTION
ir =]
in. ER + z
TT an.
hs m
Am um,
FE 470.300 win
Ark rod wen
= tiv.) La cin (in.
oo [Etém ei]
En
Brass collar: wer
06m) EG
238000
| Component. any E) Si (on) |
asin 2 ron
3.8693 150
E IET 64612
À VAS 646121 in
Y 13018 F-13804

‘enced dnt nob ep me anh tr ey Pas hear, wa en rd

spent eG to má creo Bp ors wigs as

& PROBLEM 5.121

‘he te legs oa anal lasoppel able ane equally spaced und ne
nad of setting, which has am outside diameter of 24 mn and à
«ross sectional area of 150mm. the diameter and the thickness of
the tabletop are 600 mer and 10 mm, repoctivey. Knowing th the
density a sel is 7860 kg/m" and oF glass is 2190 Kg", cate the
center of gravy of the table

SOLUTION
Pint eat symmetry is -0<
Alsa ofre oe eg, muse ol components tan il at $
li by he
vay 2880
ER x
"atk ous
A 750 Dem To
26m sus
Bm = 284190425005 mn = Au = 200 Em 010)
onu =6.1921 ke
| ern rm | mem]
vom | m [nan |
mn | nossn | uns Sn a
[om EJ 2029
> [aren DE
We have YEm= Em: 8.7478 kg) = 3681.3 kg mm
Be 7-008mn

‘The emer ol gravity 421 mm €

{above the oc)

PROPRIETARY MATERIA. © NE Tie cos Mo Copo. A irc, por of hs Manel way e depto
‘ee a ete ro ay cam, wi ams oft a cc nd re
end pi HE MA hs sl pr a aoe gb so,

PROBLEM 5.122
Determine by diet integration the values of & forthe two volumes obtained by pasing a vertical cun
plane though the give shape of Figure 5.21. The culing plane i paral! io the base ofthe given pe and
Alves the shaping two volumes oF equal height,

A hemisphere

SOLUTION
Choose m he element of volume is of ras and thickaes dt Then

maria, Te

“The equation of e generting curves x? +? a then

av ar pde

Component 1

and

Now

Component 2

PROPRISTAR MATERIA, 200 02 Me Coses, A inh pr of tt Mal nr be dened,
‘hed w ne ny or macs, rr vn pame he Fath ar
(Donna ol er bal cren Pro arn

Now

PROBLEM 5.122 (Continued)

Diner = [le ae eee

feet

DIR

Gre

ae far: qn) dt

er irda
#

PROBLEM 5.123

ect integration the values of for the to volumen sine by pasing a venical cuting
ie given shape Figure 521. The cti lane à parallel the base o ho given shape ne
Aivides th shape ito wo volumes of equal height.

A semilipsoid of revelation

SOLUTION

Y © |
[Choose asthe clement of volume a disk of rads and thickness de. Then ©
Wear Fax

Fue union fe png cures Zu À ota

“Component >

PROPRIETARY MATERIAL. > a 1e Nein Copan AH mc pa of Mama u be psa
(pba or ddd ay fm by y me rer ra porn of eer, eal pe ad
‘aetna oe te compo a av

e

PROBLEM 5.123 (Continued)

an Gruas [ala =]

oles a,

pe e] [ee

Ce
rn

Aral e
sor nar: sE) nos

PROPRIETARY MATERIAL. © 200 To tia Coop. A gh ron Apr fan Ml e rr,
pened dnd y fo by my aos ah ps ei porn fable oat an ed
“ido ter anche Scorn ore parecio Porco reco Mam
Joram hho poro

en =—_

PROBLEM 5.124

Determine by diet integration the vs of © forthe two voies obtsined by passing a vocal cing
ane dough he given shape of Figure $2. Tho eating plane parle othe base ofthe given shape and
‘ives the shape ito two volumes of equ eight

À pad of ew

SOLUTION

‘Choose as the element of volume a disk of radius ran thickness di. Then

A)
and werk nen
Camp nana

and

a qual
Now #0 or sus
W edo 4

Component 2

Brinch enna Bar carp nie al nal

PROBLEM 5.124 (Continued)

and

Now nen
3

more. € A e sine rs e na tf So wb ro,
vane aes ion nan ME lio, ed
er Mfr tbo sre ern Urea nn Mama

PROBLEM 5.125

‘Locate the centroid ofthe volume obtain by rotating ie sd area
about the x axis.

an JUE

SOLUTION
Fist note ht sync implies 3-04
mi

Werne var?

a sO yeas a CA

or be.

Choose athe clement of volume a disk of vs rand ice ds Then

dear, Ry oe eer
B om koch)
so oa
so that det NOS
A PTE
a " à
Th nhs SA (SN
Leth
and Jiu = fire sal
arte [ot at ae ae sate
bat dee se ee]
E
Now sr fur (Eon) on se rula 4

ROFRIGTARY MATERIAL.) 200 The New Compan, A ng rea o por fH Mama me be ip
mdd did ny moe ban mi Jr pm ofthe pide, ed had aed
‘wine rca a pure onan Da.

PROBLEM 5.126

Locate the control ofthe volume obtained by stating the sade area
shout the x as

‘SOLUTION

Fin note dat syn imp 5-04

‘Choose as the element of volume a disk of is rand hicknes dí. Then

der, y

Now ron
0 that dat
ax yea az

Then wer
5

and

Alo

PROPRIETARY MATERIAL. © 200 Tue Mu Ni Cms, os. igi vase pon ef he Mt ay be poe
‘Sonne ended pe eer Ice eed en Jato Mm

PROBLEM 5.127

Locate the cenvoid of the volume vaine y eat
area about he ine

the shaded

SOLUTION

rst, note tat symmetry implies ahd

‘Choose as the clement of volume a disk of rads rad thickness ch Then

dardo,

Then er le)

en rafa er)
Le Hein dran
Then

Elo AF 0) acoso do

LT aa vé nt asd

Kamen. so. eos

T

9,528) 1
2

th)

= 0098470

“af

2

PROPRIEEARY MATERIAL © 200 Me Neem Congr, ic. Ag ne fr of Mos! e dps
‘seins amado prin Seti her pea e ence Mae,

PROBLEM 5.127 (Continued)

E Clare ro
cables dar
[rote ger]
“Lati
i
¡ES Woher FO) aa e pa 4

[PROPRIETARY MATERIAL 2010 Me ii ni ne A th are i mt ina,
rae sated ny oro ay ans fre on peo of peo |
Scanned Mt enn o ep Oo ee an

PROBLEM 5.128"

Locate the centroid ofthe volume generated by revohi
ofthe sine curve hot about the sa,

gti peti

SOLUTION

Hit, noe that sommet implies

Chosen he lement of volumes disk rai and cines de
Thon Mars, en

Now rubio E
a

stat ar

Pi Ede
Es

Ten ro [arta

and

Use integration by ari ide

PROPRIETARY MATERIAL. © 200 Tue Mco Aa ree po o Mam er oa
‘a a tr i a na ee men eon oft pine mol hd lod
‘icici ary ot tl pce mse he es

PROBLEM 5.128" (Continued)

PROBLEM 5.129"

Late the centroid of the volume gener! by revolving the
portion af the sine curve shown about the y an. (a: Use 2

pic shel of radia and thickness the element
lume )

SOLUTION

First note dat symmetry implies

Choose a he element of volume a cylindre shel of radius ad thickness de,

Thon NM Fr

| Now peinte
2
sota

Then

‘Use imeyaton by parts with

Then

254835646

PROPRIETARY MATERIA. © 200 Te Nea Congres eh ace rt of i Mol be a
a ns Ce pay Hr heel eas ere Dona acta Me,

PROBLEM 5.129" (Continued)

Año Praat = $ [3000 2 astron Zar)
writ [ri ar

| Useimegraion yar wih

| te paren? we. à
“che OI Ke a]

2205700"

Now Wa fonts SAS DRE or y 03788 4

CT aaa. cee On pu
Ip on bey form oy hm ir rien prin of pl
ee men a re

|

‘Show that fo a regular pyramid of height hand ide (n=3, 4...) e centroid of the volume ofthe pyramid
located ata distance A above te base.

PROBLEM 5.130"

SOLUTION

‘Choose as he element oa horizontal slice ohickness ab. For ny number X sie
‘he pyramids given by

| AB
whore 2 20) se note below, Using similar trangls, ave

ne mca of th be o

un

+
il

2 der

a

“ ban D

thes Aueh sp

ato

seth A et armer

Now or pain ORD 4
i

hae cam or va

ET

[PROPRIETARY MATERIAL. € 200 Th MG ME Copa, lo. Alpe o o fh Ml ay pt
ehe endet pred eo ifr td ce per yer ar dh

PROBLEM 5.131

Determine by direct integration the locaton ofthe cen of one-half of a
‘hin uniform mier! shell of adi A

| then

SOLUTION
«

inst note hat symmety implies

‘The element of area dA of the shell shown is obtained by cutting the shell with to planes parallel to the
plane. Now

dd = ter Rat)

Wire

that

and

Now

Symmetry implies cord

PROPKIETARY MATERIAL. © 210 The Wn Mi Cris, be A igh ee. Nop oft Smal ay eine
‘Bosh ovary Ses Mile e rec rae ghee,

PROBLEM 5.132

The sides and the base of a punch bowl are of
uniform thikaes 1 <a = 250 m,
¿item the tion ofthe center of gravity of
{the bowl, (6) the punch.

SOLUTION
(a) Bow! |
Fist ote dat symmetry implies

fo the coordinate axes shown her. Now sms tal the Boel may be teed as a shel

{Pony ri bow wl ecke wth i Cnt ofthese Für wa he bon Sea
Src aire y ron he and about yas Th
age = asin dR) s
d On), Roos +
: A a T
m E 2, SR
€. An [ne nate
arte eu al
a
sos Sota" [Ed
= [ere soso,
fons?
LR
4
Py ocn
Now
or 121900 4

PROPRIETARY SLATRBLAL © 200 o Ni Hl Comyn, AL ng eh a i Nel mae ar,
‘ama ore ney hy an lB rr wes paso dl, ar wet ea ee
‘in eal carat Sea Wi i cae ac pen yo ao ha Ml

PROBLEM 5.132 (Continued)

o

|
the punch is oogenesis cetro avi il coin wt the cent of |
the corresponding volume, Choose us he sleet of volume disk adios x and thickres ch. Then

Wands, Fay

Now wer
soto dae Pr
Tien CO Rt

mi |
Now w= Pratt: zoe) ar
s

E R=20mm p=-M2mm À
+]

PROPRIETARY MATERIAL. 2 300 the Compa Inc AB dt med prof he Ma mar be a,
‘dry ao ene mean oft as a ro ied

PROBLEM 5.133

Alter grain à, a ler pes for stas desig the
‘ever o Uh lb füra howe. Vo provide afin evel base for
the sah, the had places a mi ef 3 in. of gavel
Fanci the sb, Delcrnin the vol of vel al und
the eoordinat ofthe centroid ofthe volume of the save,
(int: Te baton surfe ofthe gravel isan oblique plane,
which ean be presented by the euaton y = 0 + br 2)

ny integration, The equation for he bottom of he gravel
no llos:

where the ests. ed ce et
220: p34, and therefore
à
a
iy Sin and therefore

1
4300), or
00), or a

For SON: y= Gin, and theretine

fee(50R), oF

Tele

Now

A volume element can be chosen as
a

PROPRIETARY MATEAIAL. © 200 ue GOA Caen he Al as: Ma o ed
ere a an! Km hry santosh pr ari on fhe bi oa et
‘insect Man M ar pie Po ndo te
oe hemes por

PROBLEM 5.133 (Continued)

hes PANNE

es [a

Lp
¿sorna

1 sar
Mesos ee
[el
anne
Thelma
| Then
The: reese 4
isa

onary mei € en nn e rl he a on et
Lo o warn dran pend iy Grew nah or rn Wear su az an,

PROBLEM 5.134

Detemine by dise integration the locaton o the centroid of
the volume between thee plane and the portion shown ofthe
sue y= 16 Nhe = Ver

SOLUTION

Fürst note that symmetry implies a4

Choose asthe cement of volume lament of base dex and high >. Then

a
a a an = Yin ee
Ar)

Then yo [hr

PROPRIETARY MATERIAL 020% ix com Coop. AN it cod iu Mal ay he ana
(enter dance e foo y wr. ith tc rr in pomos oh leo e o jad e ad
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Dear ein pen

PROBLEM 5.134 (Continued)

wa fool tiles A tre

A

ca
A Ca? ae alge ae

O oo pao]

oF 5 2
sele yp o »
- EN
wel wr be ler] Pat
tw A
nN a> final (3) Re

PROPRIETARY MATERIAL. © 210 Te MeCen ECs, tas I igh sie ofr of a m dd
‘arto onan tno paral Sn lr viel Comerc Qe ete Mr,

PROBLEM 5.135

Locate the cent ofthe seston shown, which ves cut Fin thin eter
pine by two oblique panes,

SOLUTION

Fist note that symmetry implies 5-04 |

Assume hat the pipe has a uniform wall thickness y nd choose as the clement of volume 4 vertical sr al
wit aß and eight (9, 1) Then

and

Then

» A 2 |
and (acos0-+ay+ acoso +20 i

»
sc
ke

ai a
ar = "Me Sa =
2 1

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PROBLEM 5.135 (Continued)

and

Now

and,

rome area. NV Nm nc A ah pt of Mama on
Sp mun! tn fo e ma hous pes ei net
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PROBLEM 5.136"

cat the centroid of he section shown, which wir
‘finder by an obligue plane.

SOLUTION
tnt tat sme imps
À ST
Ss Sb.
‘

(Choose a the element of volume a vertical slice of with 2, thickness, and height. Ts

ar 2, Fun

Now

if
Li

ant od

Then

Las

Then

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PROBLEM 5136" (Continuod)
and
ia
Tin Jin Besos recon a)
[low 026m 0 rte ae
Now
Sou
ay [an sai
hen Gras = La EL ES ee
Lo (2.8028) 2 9, ty u”
abi (Ge) oros ta] .
Saum
Br
Aw LAR Leal
ta a [Load
N en
La Erbin denn
In din D in DN 0940 co 040)
Using
i war ob nf [sno o
P| cod ot ond PO Mea
if foe goed E D

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Now

and

7 PROBLEM 5.136* (Continued)

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‘vc att nym By me, aan pi a pra fb who eh
Bain henna py Re bch pen ipo ako Mal

PROBLEM 5.137

Locate the cen ef the pane rca shown

SOLUTION
2 L
Ac Fon | za ma
METER ° El 61236
2 | hara | 0 68 6700
3 a | 27688
a | METE
Then AEA=ESA
0622 m‘) = 200880 mon
and Fear

(40422) = 270020 mm or Fa266am 4

[PROPRIETARY SEATERIAL © 210 Tim Metin Cine. A ih Mo pf a Ma mar
Taran ons ny fo ey aa ant hfs ore ll on at
Serine Fu hie eal arp a oa Ha

“ PROBLEM 5.138

Late the centroid ofthe pane are show.

4 “La
38 A Lis +
x \
Li ne 7
Oo: 1 Ea
x rx
“hn um [me]
1 15 Less | 2200
> tam | gan
= ee [mao
Then 19707777
(5.3333 in?) = 17.0650 in? or Y 1. 4
Pr Fais
Tis 3333 im?) 106670102 or F-20in €

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‘rere ened op eu a ay mac ptr ie in à ep vd hood de od
ne che anata pry Ken rin Po nidos Ml

PROBLEM 5.139

Ue frame Foca sign is fabricatd fom thin, at sce har sock a mans
por uni cal 473 Ah Th Fame supported by a pin at € and by a
tale 48. Determine (the tension inthe cable (8) the rosction a C

SOLUTION

Fest mote tha becuse the frame sites fom unir bar stock, its cer o gravity will come with
‘the centroid ofthe corresponding Lins.

‘Le Fases
s| le N los

im Em
7 135 ors |
2 0 | 03
E us o o
4 0 2 on}
| s | Samen | coms | 120000 |
ee | ECTS
then Feet R-
Fae2s10)=25106 L Xe
or F 0542470 pu
the re body agra ofthe fame is then tt,
ve vue SJ

2 78m 62810 mx 9 81 mA
SN

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PROBLEM 5.139 (Continued)

isi then equis

o siicn0: ass 3g). 475%
or Msn où SN € |
[0] |
e ras
y mo. 6 faaszany-aurny-o
Men ann

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4 PROBLEM 5.140
ja
TE Deine hy dt gan the eo of tae hav. Expres
È Yor sever ts fond
SD E
| z
oe
SOLUTION
by obsecion
| y
Papa RO yal NCA) or kan ===;
“ . ya0: 0e) ar Cn ! Fute
by
Then de
Now
Then

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PROBLEM 5.140 (Continued)

ste past oa)

PROBLEM 5.141

ye

Determine by direct integration he cet ofthe ssa sven, Express your
N answer in terms oF and

Sauna
s
ne ya
E
sg
E sefafafe-2Jooal 23

ant feats fi

AO 3
Hae fonda x(4ap)-1 ru 4
nat: (jon) $

fs jaja

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N PROBLEM 5.142

‘owing that two equal caps have been removed fon 10-n.diameter wonden
per, determine he (tl surface ars e the remaining portion,

SOLUTION
“The surface ara cn be generted y rotating ho lino shown about the y us. Applying Abe frat core of
Poppus Guldinu, we have

raie y
Now aad Sm
“ mu. JE
Zi
ans
Ps Stns al
(36m.
du = si
92min

3] (4.
[jo mecano]

ind

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pooch ihn pmo

ine PROBLEM 5.143
ai paie terne bm spt ge eg

SOLUTION
2, = Emo Nm)
EN
Ry Uno Nm)
-4soN
Now HER =O: 4-0

o om eo smn (2a}es0-0

ante

AXE, 20: 1300N-2700N 1 8, -:4S0N=0

o B,=1850N CNE

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|... PROBLEM 5.144
Jere

SET. L IT T IN ‘beam i suce 10 ner dite dowaword

Fand ino pat Cm DP ach

AT a E

ml 1 o tome

SOLUTION
Wehave = 6 0 Nm 18008
600m) 650 N

Run Nin) ~ (08, N

op OR NON
Ten eat 20: AKI) (MCG) me N) =D
“ Hog = 50 Nm 4
and HER =0: (Me) N 1800 N 3600 N SON 0
“ Mc =28125 Nm Va = 0 Nm 4

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‘arched dd nay ob np mace ter ve pon eth, foe eon ee
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PROBLEM 5.145

"he square gate A shel in the position shown by hinges along its top edge
and by a shear pin at For a dcp af water 43,50, detain Ihe for
‘exerted on the at by the shear pin,

souuron
Pen ths Metro
En
Lam Ve
me nova) =
ss

Lorca ese

vow e (riens
MIR:
I Laon teens
o E, =2770 4300 4
eww note a cee te

‘rol or aed ey ar ot y mana lr wie pom fhe yw a Rod ea
a DN Vrs nd mt ee

PROBLEM 5.146

Consider the compite boy shown. Determine (a) the value
SER when h= 17, (0) the ratio AL or which

soLumon
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D ee Ya
praia
me
ne Raven
ta

rola) aloe oa

LE. o

@

Keosaı 4

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PROBLEM 5.146 (Continued)

0 Bet ten
siria afte) Lafont)
e ;

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Capitulo 5 estatica solucionario Beer 9 edicion

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