Beer dinamica 9na solucionario 12

PROBLEM 12.1
“The value of gat any lud may be obtained rom the formal
232.0904 0.0053 snp) 4
whic takes ito account the effect of the rotation othe cath ax well a sh ft hat the cath no aly
Sphere. Determine to fou significant Figures (2) he weight in pounds, (#) he mass in pounds ()

alo
msi

SOLUTION

ono

#45

poo
Weihe Wang
PaO MISSION Ws?) = 4.987 Ih «
Das M (0.1858 Ib NG. ME) 5000 fb «
9-0
(0) Mass: Atal lattes:

wm rn

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‘oo tac nay fy sy mo mo pr ari Porn Pa wa jen ee
‘Renan nde pora hi tn ae pw ae aa

PROBLEM 12.2
‘he acceleration due to gravity onthe moon is 1.62 mv”. Determine (a) he weight in newions, (5 he mass
in ilogams, on the moon, ofa god bar, the mass of which hos been officially designated as 2 kg,

| souvnon
a) Weight WF = mg = (2 keX E.62 mis") MEAN A
[o meses nad

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PROBLEM 12.3

A 00 stellte is in circular orbit 1500 km above the surface of Venus, The accelemtion due 10 the
‘gravitational traction of Venus hv alte i 5.32 mi”. Determine the magnitude ofthe near momen,
‘oF te satelite knowing th its bil speed is 254010" kh,

SOLUTION

Fis note

Now

or 130010 kg €

angina aan. DOT ci Cao Al np of ho a nd,
‘Soi ened naps; Ma Tip den cusps Gyno mt

PROBLEM 12.4

A spring sale 4 and lover see having ysl lever ars are
| fastened lo the of ofan elevator, and identical packages are

Athen the scales as shown, Knowing that when the elevator

moves dowmward with un acceleration oF 4 fs the spring sale

indicates a load of 14. I, determine (a) the weight of the

fle estou.) ld ibi y pig sal a ic mss

FEN} needed to balance ho lever wale when Ihe elevaioe mover
ward with an acceleration of 4145

SOLUTION

Assume 2322 16
“

if

, wein 4
lo
A an
4 wren 24
hr
EP)
eu; san a

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PROBLEM 12.4 (Continued)

For the balance system 5,

R
But
and
so tas CEA
ond m m, 0.500 Ibs! À

vt tn

Maen Cnc e, iis ssl Ne ao A y Be ed
1 uo as ea te rer te perso ea, oe ee e
ty trachea mec rs oar ne cod

PROBLEM 12.5

¿mil velocity ofthe pack, (the cuit tion betwocn the puck and the ee

A hockey player hits a puck so hat it comes to rest in 9 sae sliding 30 m on the fs, Detemine (o) the

SOLUTION
(o) Assume unifomiy decelerated motion.
y tal
0% +20)
9

no
PET
o
2 sy 246667 wis ea
wi Eu
o
“CS.
ce NeW omg
sting Fun -nre
“ = mt oe proms 4

‘PROPRIETARY MATERIAL. © 20 The Mra Compu, be A nes o par Sa my e it
tel ad ny rey any mr ioe on fe be edd

‘Kents mr antec pora Mend pe cry as

PROBLEM 12.6

Determine the maximum theoretical speed that sa estomobile stating (im rest can reach after taveing
400 1. Assume thatthe eoefeent of static tion i 080 been the ties and the pavement and hat a) he
automobile has front whee dive and the ut wheels support 62 percent of the atomobil weight, (8) he
automobile has rear-wheel drive and the rear wheos support 43 parent of the automobiles weight.

SOLUTION

(0) Foemuxinum serelestion ya

a he ORI) «Cons
GG 080608
me bro

Now Ej Meme
“ 04% mme
Th a= 4960.81 ms?) 446576 mis
Since cost, we hve

Porto)
Wher Om: gg = 24 mé 00m)
or Vos = (2301 mi
or 2550 4

(6) For maximum sescertion

Now
Then a= O.344(981 ms) = 3.37064 wi?
Since ais constant, we have
v4 2a(¢~0)
EDO mi va = 208.3746 mi 00 m)
lan "51959 me
or Van. UK €

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‘mpfr oy met pr re prea os pa el el ad
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voi pu

PROBLEM 12.7

An anticipation ofa long 7° upgrade, bus driver aceleraes a constant tae wf 3 i while ill ona level
set ofthe gay. Knowing tha the speed of the bus is 60 mih as begin o climb the grude and that
the driver dees nol change the meting ahs vote shit gears, determine the distance traveled by the bas
up the grade when is speed has decreased to 50 ih

‘SOLUTION

Fics wen es on ee een he igh.
” de

wohne sah ome Pa
A

"Now consider wh the hus on Ihe upgrade.

Werne eme PW sin ea”
B

AAA

Suns ter agg W sin = a
£ 8
= dd = 507
(03220 7) 18
09m

or the uniformly decelerated mation

Omak +20 an)
nl

Noting ha 60 ith vo). weave
a (sr)

(an) as common

or pa = 280.167

or Sia 0292 mi 4

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‘raha or ened yom oe rr pri he bee he! dnd
Soares Mr ie rpm scoop a

PROBLEM 12.8

IF am automobile's braking distance from 60 mph is 150 f on level pavement, termine the miles
"racing distance from 60 mph when it i (a) going up a 5° inline, (5) going down a 3-percent incline
‘Assume the braking fore is independent of grade

| sorurion
Assume will deleted mation in al cases

Tor braking on the level surface,

DRS
EU]
25.813 que

BV in se
ee

{080166 HD

Bere

IORI RY aa. mM Cn, me nr of i Mt mb an
nai each amd sated Metron Haft be on ppnow saa Ma

PROBLEM 12.8 (Continued)

oe "
ande joy B= U7I835"

(6) Going dwn 3 percent in

hi Wein Be La
E
LOGS sin 1022)

TT
0-8 PT

ME]

‘rhe a ern à fam hy nym, ta ba porn he ld, ld le o
To chosen y Macia pico dro rer prc ann Aa

PROBLEM 12.9

A 20-4g package sa est on an inline sehen force P is applied
Determine the magnitude of P i 108 is required for the package to
travel Sm up the incline. The stc and Kinti weeiens of tion
beten the pue and the incline are both equal 00,3

SOLUTION

Kineatic: Uniformly cle motion. (x,

1
Kenne)
2 _@xs)
7a
20: 2 Pain SD mg 200-0
in SO" mg cos 20"

or 0.100 mes?

Of WH, = ma: Peas S0— mg sn 20°- UN = ma

or Pr 50° = mg in 20° uk sin 50* + mg cos 20)

For motion impending, set 0 and t=,
9 OD 20.8 IX Sin 20° + 04 cos 20")

ES
N «

100 mt, se =

2040100) + (20K9.814sin 20° #0. cos 20 PP.
à 08 50-03 FIN A

For motion with

‘reed on an n

Fr each section, determine te acceleration ofthe package

ip PROBLEM 12.10
Fat ,
> ‘The aceceation of package sling at Point 4 is 3 ws
A Assuming tht the coeficient of int tion he same
>.
>», m Point À

SOLUTION ]

or any empl 8
{Use and y coordinates as shown, i
aro ms

NER mu: me sin efi

= sin, co
nit

eos

San 305
MENE
022423
Aroma CPE)

4, =9 Gin 18° 0224230 159)

daten

Ar Point A. 920, a,

AH m4 IS 4

PROPRISTARY MATERIAL. © 299 The Coe Cas, be lige sare No ur of i ana wy e in
‘vedo ind nx foro ay mc a a oto rei ote per, med oe ea
‘Reson inde pl Mo fit id ar pri: ne

PROBLEM 12.11

‘Tne two block shown ar orginally at ret. Nepleting the masses of
the pulleys andthe eect of tition inthe ples and bern Mk |
and the horizontal surface, determine (a) the acceleration of etc block,
(the tersion athe cable.

Tin
en |
$ o
wa Zanes em,
isa Tam
2 Hama Baron
Sig
OS
en
J- mi set
À © Mme renogxonnent

LLTAN 4

Compares, Al ie sono. of tht Mod de
‘ep 0 rtd ay fr yy eam, ses or so piso Y E ml, eed Ned

‘ini do anta pomo ae if idler an pe snag he kat

PROBLEM 12.12

“The two blocks shown axe orginally af rest. Neglocng the masses
ofthe palleys andthe effet of friction inthe pulleys and essing
hat the coefficients of ction bewcon block À and the horizontal
face aro 7, 025 and fy =020, dee

‘oF exch loc, (2) the Lion in the cable,

() the aeseection

SOLUTION

la
m
3 of
mis
»
arf
|
Ta Tan

rom the diagram
pty, = ocn

Then DE

and 2300

a. ar o

Find termine ithe blocks will move with a, =a, =0. We lavo

AYRE =O: MN

mil
arme

RO Aie Bo!

Then poster 9.81 mis! 81.75 N

AE MN = or Mamas

Abo, en "Ua a Uda
02830 ex me
sin

Simplis that de looks will move

PROPRISTARE MATERIA. > 20 he Mk MÍ Crane, I. Ai rg of tn Sal wa Be gine,

pre a ated hr e

‘yam to oe mst psf Fear ele ea

‘Wiebe oa ply enters ose port Pps saa a,

PROBLEM 12.12 (Continued)

@ 4 ARE 20 WN ,=0 oF Nyame

‘lising:

Using Bg. (2)
2

oF mg 020 me + IO eme

des"

149%

sin facts
‘Then 4, = 0.098 mis? 4
: tine

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alone leas porel con ifr ha dhl pr eas Beas gl
erent prin

PROBLEM 12.13

“The encens of frition between the lod and the aed
trailer shown are 4, = 0.40 and 4 =0.30. Knowing thet the
peed ofthe rg is ÁS mith, determine the short distance in
‘which the sig cn be brought ta stop ithe ld is nt to hit

SOLUTION
Lea: We asume hat sliding of load relato tier simpending:
FoF, 6
EN

sation o loud is ame as deceleration of tile, which isthe maximum allowable deceleration.

HER m0 Nowe ner

Emm oa

HER at Be

oo

Unis scclerated motion

+20 with

gg, = 2S A
0-46) +2(-12.88)¢

PROPRIETARY MATERLAL 208 The Mean Cong, Is A conve ve of ls Maa nay be plo
Eee ee en o nt,

PROBLEM 12.14

À tactile is tavelng ae 60 mh when the dives
applies his brakes. Knowing that Ihe braking forces of the
tractor and the ae ane 16001 and 13.700 I, respectively,
determine (a) the distance taveled by Me Unctoctrnler
before it comes toa stop, (1) th horizontal component of he
force inthe hitch between the tractor and the Waller while
they me slowing down.

| sorurion

Wen 90:

lo

Ds ls

AB om Edad Ft

or unir decora motion

222.08
(5000+ 17,400)
171982 ist

(3600 613,700) 16

#2) 99 = 60: =88 Més

2 fs Kas) i
nnd

0 (88 Ay? #12

crear

lo Ce

Then HAO «17,1932 5)
ET
e. Ban 4.10 lb (tension) |
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"rh db nfo 6 ym: me rr tes pm of uns ot et ne a

‘Binders o od cor tom Hor hte orc nem coer Hak

PROBLEM 12.15

lock A as a muss of 0 Ag, and block B has a mass
of 8 kg. The coefficient of fiction between all
Sacos of contact are, = 020 and 4 = 0.15, I
20, determine (a) the weccleation of block
(0) the tension in the cord

| sorunon

From the diagam

‘Then 2 ree
and day tags =0

Now

Then

o. aa, w

Fist we determine if the blocks will move for the given value 9. Thus, we seek the value of @ for
rich the Docks ace in impending mali, wih Ihe impending motion 014 down te inci

BAP ER =O) Nyy Hg coed=0 5
or amet
Now Fg Maa

12myeas
NE 20 erg Ming =O

or T=mai02c0s0+sin0) 4

GRE WO MN Msn

e N=, +marecoso
[now LA

0260, mB
A)

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ened a ny ayn a tht aie am ot

PROBLEM 12.45 (Continued)

o T= mgsind 02m, +m) 6060-02 mces8

isin 0.24 +2m,)c060]

Eguating the two expressions for 7
mci02c000 sin) = on, 0020, + 2m, 0050]

or 8024 tnd) =[401an@ 02440 + 2x8)

or tind 04

or 0= 21,8 For impending motion. Since 925%, the Blocks wi
motion af the block.

WER: Nyy My cos25=0

ns
or Na =mgan2s
Sliding: Fa Ha DIES

ar, 2 mas T+ F y + Pan gag

T=mlr(0.15cos25*+sin25")=04)
19.810.15c0s28°-+ sin25*)~ a4)
2 (N)

WERD MN Wy cos28"=0

o DATA ECS

Sir E

Na DIS, + mg) ge0828°

E Esp +H sin 28° gy

1 mg sin 29°~0.15(ong 4 my) co 25°
01m 008 28° m (a)

= el, sin 29°—0.15(m, +2m,J00 3] my
9 $40 sin 25° -0.15(40-+238)e0825°}+ 40a

91182024400 EN)

Eguating the two expressions fr 7
8647982 ay) =91.15202-+ 40a,

or 028575 me

9, = 0986 mis! De 25° 4

@ Werave

8{547952 098575)

SIN €

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PROBLEM 12.16

Block 4 hava mass of 40 kg, and block 8 has a mass of kz.
“The coefficients of cion between al surface a conc ate
1-020 and y <0.15, I P=40N 3, determine (a) the
Acceleration of block B (the tension in he cord.

‘SOLUTION
From the diam

Dy a = constant

Then Dar
and Dpt

Now

Then

e 4 = o

Fira we determine ithe blocks will mows for the given value of P. Ths, ve sek the value of P for which
‘the blocks ae i impending motion, with the impending motion of a down Ihe incline

Be VBE WO: Mag Mons"

or Na

1 00525

Now Fara,
02m 0025

NUE 20 TA Eutin 25000

or 70.2 mg 25° my sn 25°
(ELMO IP (0.2 605254 sin 25%)
4739289 N

WERD: MN cost

Psin25

or a = (+ mag cos 25% Pin 28°

PROPRIETARY MATERIAL © 20) hc Hea Comp e. AN ral Hr of or Mel may Le aa
See in o e rr po ts rn i ed od ed
Sores no pertain

PROBLEM 12.16 (Continued)

Now eux
« EM mx 60825-38251

Nano TE, ty sn 200
« TR rm 00625" Pen 259] 0.2m 0028 me 28 Peu 28"
E
Then P(O.2 sin 25° 4.008 25°) — 47.39249 N + 9.81 m/s? [0.2140 + 2x8)cos 25° - 40 sin 25°] Kg}

@ yg Wa cos 220

o Nog Mig 00250

Shing: Fa

Nas
05m cos

Sas,
« Tama 1299 au]

'93N0,15cos25"+ sin 25]
154799249) (ND

Mag: ~T + Fug + sin 2 ma

4
ERDE My Ny Mason 204 Pin2S 0

or A= Cr + mag 00525" =P sin 28°

Sliding: FUN

0.15. typ ec 28° Pain 28°)

ar ama Poy Pay Mean 28° Peu rm ay

PROPRISTARY MATERIAL. © 200 the ctr Compu. A gis ee pr Mal be dei
PEL u e op for a mene ter wi per MJ, al ae hd
Tchr tr ply a Mfr al Pepe [ro do Mn
Fer lv paar

PROBLEM 12.16 (Continued)

Substituting and using Ex (D)

‘Tay sin 25°—0.14mg # mg) cos 25° Pin 25°]

0.15 mg cos25*+ Peas 25° -m,(-a9)
main 28° 0.1 SU, + 2m)c0825"]
2 POS in 25° 4m 259 ms

BIO in 25° 0.154404 2 BJeu 25°]
2 40(0.15in 25 + x 25) + 40,
9.940042, (N)

Equaing the two expressions for 7
(5479520) =129 940044 400,

or ig = 1.79883 ms?

Weine T-Rsar0s2

eisen]

1.794 mis! 28° 4

Tessand

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(ern hp hr ar e an vaa pnts poms dle pl, ceo ed eed
an che ander portal Meno Mort it corro or ana,

PROBLEM 12.17
cs nd Bae leona cmeya a bat sta
re The hl sky sed san apa ito eo
lit sping cocos Betws te bt and th toes Kein
tate cies of ia Tico bt. the We
Ende box ro Y 050 a (A) 03
SOLUTION
A ALE 20: Mc ISO 4
se
= DE outs mo
ons th
Ya amas Fe Miah,
ce ass o,
=
or au = (32.2 145" X03 eos 15° sin 15°)
A
Bo if BR, =0: Ny -Wycos15*=0 ve
@ Nyasa? "iin
£ e
Siping: Feu
WER mma Fe egy
ce On ist sas Me,
5
o ay =(32.2 Ms” (0.32 cos 15” -sin 15°) 1.619 fs?
p> => orton scot
1997 Mis? x 15° 4

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or ded ay om by yw a we pf ert ed ha
Stender py a Mr del ui ppt Bo acs al

PROBLEM 12.17 (Continued)

inte iti asus tat the bones romain in contact (ay #0, then

0/74, and find (ER, =m)

a OL, cos 1 M sin 1S? Ny

#

» 0320 cos 5" Hy sin 5"1 Way =
«

273102 and Np =-0889 I, which coma te spin.

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‘pe ora myo fy yma ren pen 9 poe, re pd ed
rn e dec cn Mii id asp. oar

we PROBLEM 12.18

nosing ha the sytem sh tar om es
find the velocity at 1= 128 of (a) collar À
(0) cola B. Negleet the masses ofthe pulleys
tid the lc of tin.

SOLUTION

Referring to solution of Problem 11.51 we note that

w

‘where minus sign indicate hata, and ay have opposite some

us
o
SAL meta
la ar
nn
T=10a, 10]
GO Sin from in
| 25-2000)=150,
2527.50, a, 0.909 is! =>
|
PER EN wind

9 Sutstiutin a-0.909 ino Fa (1
= Ce tast=0+ 0455412) wmosismi 4

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"on or ne yom yap an me ft rt pr of all, en en a
ehr au orme cae fre be ce pero ptr asta ck
ova pri

PROBLEM 12.19 |

Eel of the systems shove is initial at re
‘Neglecting and fiction and the masses oft
pulleys, determine. for cach system (0) the
aceelerition of block A, (7) the velocity of block
‘A after it has moved through 10 1, (0) the time
‘required for block to each a velocity of 20 05,

SOLUTION
Lety be postive dowmwar for both block

| Constant of eale: y+ = constan

rere |
or AR

Ford 8, 5m

Block A:

Mock 3

Solving for. &
o
o

ROMA 0) he ee rim a e pf tra
en Be ts rt ea i a et AT
Spectra magenta gn nc
a

| PROBLEM 12.19 (Continued)

(a) Acceleration of lock.

send 1,2001, mom P=0
it 200100, 2!
form (1). (ay = FE fag 10.73 08") €
‘By formula (1), a= 5100 22 (a) = 1073
Sent 1,2001 yO, TS
2000 sone
Bytes. an Go =1610 2)
Sytem 6 220016, Wy=21001, P=0
2200-2100 zone
foma (= 20-210 (az ws
By foul (D, Cops BRB) ooo la
© raya hyde WR scort,
System (I co OTRO baies me
Sen ex = RETIN Gone |
Sen GX CE warn

(‘Time ay, = 20 145 Use formula GS)

Sah as ins 4
sen ei nenne
seno) ‘

San

PROPRISTARY MATERIA © 203 Ye Mime Croquis A eh tf his Mc mb it
wn end nr an a. ionic arity pa of hr nr
nine conde thawte aga opcion ni nt

PROBLEM 12.20

A man standing in un elevator tha is moving wild a constant
acceleration holds ag block betweea two other blocks in such a
‘vay thet he motion of relative to and Cis smpeding. Knowing
{hatte coccion of em Pote all sure ef, 030.
and i, =025, determine (a) the acceleration oF the clovaor its |
moving upward and each of the forces exerted by the man on blocks
Fand Chas e horivoutl component cual ice the weigh OER,
(6) the horizontal components of the forces exerted by the man on |
blocks 4 and € if the acceleration of the elevator is 20 ms"
lowe.

SOLUTION
| Fir we observe tha nse is mot moving relaie to 4 and to C that ay = ge

Werne wena

mgt

For ay, tobe", the nel verte! for st he, sich require ha ihe sion fares be ating as
own. len follows tae the impending motion oF relative to A and Cis down, Thea

VER = meg? 2P My = mye,
26m em = Mada,
.. ou "02981 ws?
“ RT
te
0) Weine vun x dl Ml

[Now we observe that because the tection ofthe impending motion unknown, the directions ofthe
‘ition foves is also unknown (long, mas be daw,

AER, = map: 227 Hz = malay!

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‘Rehan an i ee or ne

PROBLEM 12.20 (Continued)

or Fm lod
PS

¡Since the magnitude of must be positive then follows hat FT, and hat he impendin motion of
telative to and Cis downward, Final

210308)

kex(O8l—2) mie

[OIE IY MA. 209 Can A ar prof Mel at na,
eines pr ni ed pin Pala Mb Mr

PROBLEM 12.21

A page atest om a onseyor bel which i ily at rest, The belt
‘stared and moves tothe rg fr L3 swith a constant acceleration of
2 mi. The bol then moves witha conan deceleration a, and comes.
in a sop afer a tol displacement of 22 m. Knowing thi the
cneiciens of icon betven the package and the Belt are 4, 2 O38
and 44 =025, determine (a) the deceleration a of th belt (A) he

irn uF the package his (0 the bel the Belt comes 198
stop.
SOLUTION
(a) Kinematics ofthe betty, 20
1 Acceleration phase with q = 2 a>
4 =1,+04 OHMS) =26 ms
1 re
Nom tt thal 20604 LOK 2160 m
2. Doceleion pase: 1, =O sine the bel sons
Deren

(6) Motion ofthe package

12. Acceleration phase, Assume so slip. (2) = 2 u

un

A

F ma

ELBA me: eme,

‘The required fiction fro à a

Tes

lable tion fonce s 0.35 0.350

Lean ang (0.399981) 248 me

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‘var hd ans fort me eo aot tr sts ef plo nt

itnnews Me

PROBLEM 1221 (Continued)

Since 2.0 is 3 45m! ie package does sip.
hen 226 mis and 6,

om

2. Deccleation pase: Asume mo sip. (a) =-11.2 8?

ma

LP, =m =

Leo 2-66 ais?

LME og = 343 mt 66 mt
Sins enable ction force ses han the required cion force for op. he

package docs:

SE, He mn

LR ma N an)

an

= 0290080
24925 mu
ah)
2264 (-2.452530.3923)
Am

a» ns

[23

i :
set

CEA

cs

=2321m
Position of package relive to the belt
(shh =2521-22=0321

PROPAUEEARY MATERIAL © 280 Te Me Camps Is At ee tf he Mad yb gta
veh thre nyo by ym ms ‘ren pra ef aa! ed
Teka ce eo Ger ia cae Pr ats.

PROBLEM 12.22

To transport seis of bulls of shingles Ato
fof, a contactor uses a motor-driven lit
Sonsising of a horizontal pltfonn BC which
fides on oi such o the sides of ade,
The if stars fom rest and inially moves with
à constant acceleration a, as shown. The it then
desert a onstrate; de ore
LD, nea the top ofthe lade. Knowing that the
úcoetlcint oF stati cion between a bundle of
shingles and the horizontal plafom is 030,
termine the largest allowable acceleration ay
nd the largest allowable deceleration A, 1 the
Taal soto slide on hepato,

SOLUTION

Accelertion

Arme,

Beaux

Kamm

307%,

ima ta sin 68°)

NM,

Fini so 65°

or ma cos 65° = 030m a + a sin 654)

PRE A
TENTE
= (1.950408)
9.53 ms 953 m4 65° À
Deceleaion ay: Impending sip. 4 =, = 0.085, w
ER, may: NW == din 68"
My may in 68"
LER =m, 1a co 5° Ma Ina,

72

‘PROPRIETARY MATERIAL. © 200 Te Mi Copa Le. A ac No pr of hs Mami may be pie

ml or tal nym i ys

nn aes and pen ce rial ira re pre Fyn ng Hoe

PROBLEM 12.22 (Continued)

e. qd 20565" =030m, a a co 6S")

030%
TS
=

aa 2 69° €

324 me

AOPRIEIANY MATERA. ©2409 e Nea Congr eA nat ee Pe of a na be ma
Ino ated my os Bm ahr en he ace a bone oe
nn das Ge o daa cn rats Bip singe ru

De va parson

PROBLEM 12.23

“To unlond a bound stack of pls fuma track, the diver Fe ts the
ed of the truck and then accelerates from rest Knowing that the
coefficients of fiction between the botom sheet of plsvood and the bad
fe 4, =040 and =030, determine () Ine malle aceceration of
the truck which wil cawie the stick of plywood to slide, (6) the
soseleration of the truck sich causes corse Of the stack o wach the
ed ofthe bad in 005.

SOLUTION
Tata, be aveleation ofthe plywaod, 1, be Ihe acceleration ofthe trek, and py be th accleation
‘ofthe plywood relative to the ck

(@ Find the valu of 2, so that relative motion ofthe ply ith expt the tuck imperia

Ar map and Fi 4, = 0408,

NER, = mas Rae Won 10 mp sin 20

y= mn 20" ana
rl > rn ,
ER =a, Li M Sin 20 mya cos 20° a +
Kama sin 20 14, cos 207) qu mar
mp sin 20" 1006207) = 0.40 me 60820" 4,20)
u. PA 20" = sin 20")
7° 6204 0
= (0031454980)
0209

Lang Lead
O) App Cpe Cr han 20404 Lan
@ Gre Ont Jona 204041

sm +
ze
a

Far

anne = Mr - GX

7 os)

an

pA We 20)

NE

mys My Wry 00820 in 20°

Ny mpl 06 20° a; sin 207)

PROPRIETARY MATERIA. © 298 he Neti Cane A ss
‘eu ost my fon y a oe io or i pe
o amic any ae ier A pari Horo fa Ba

PROBLEM 12.23 (Continued)

AZ BH, Bima, Es Wy Sin 20°= my 8 20 mp
Es mag sin 20% a, cos 29'=0p05)

For siiding wih fotón Au,

30%,

(Sin 20" + 0, 006247 ar

30m (2 00820" + sn 20°)
(03000820 —sin 207) + y
052040208020"

CONSIGO) +(0.9504KE.98)
2407 ECT

PROPRIETURY MATERIA. © 209 Sem Corpus Inc A ote pu of i Meal be tt
Ion erated fot ay a tc rr wien pn of pee! od me
Keren a rao rs pm sim

PROBLEM 12.24

‘The rapelle of a ship of weight ean produce a propusi force Fe they produce fore ofthe same
rmagnitade but of eppositecircetion wien the engines are roversed Knovtag thatthe ship was proceeding
Fra at is maximum speed vp when the engines were put ino reverse, determine the distance the ship
rev before corning lo op. Asim tha the toa rosistango of tne water varies daly ith o
‘ace ofthe velocity,

‘SOLUTION

Atenas spec «=O.

When the propellers are reversed

= ce

a an Av)
a ae
as

mabe

= fing (2g

var. 4
rs

PROPRIETARY MATERIAL. © 240 Tne Nee Cea, hes A gs nd fs Ma a e und,
pent ba oy om ya iv pst rr ft pbs wed pon)
lar non pri cn i a Md ac prin ea nh ee

PROBLEM 12.25

A constant free Ps applied 1 a piston sa od of oil mans m to
‘make ham move in a cylinder filled with il. As the piston moves, the
oi is force through ones in the piston and exerts on the piston a
force of magnitude ke in a dicto oppesite to Ihe motion oF he
piston. Knowing tat the piston stats from est at 7-0 and x=
show thatthe equation relating x, and 1, where is the distance
ave bythe piston rd vi the pod ofthe piston, linear in ech
of ts variables

SOLUTION

onde
En

‘which is line. «

PROPRIETARY sara. © 290 e Mac.

mete: A ua so. rt of Mama my be ae

‘epee dvd e fu Oxy cs, mi pp ma ee i
‘etn aos dir pur on ifm repr Tag Nh

PROBLEM 12.26

A spring AB of constant ki atached 10 a support at À and to
‘alla of mass m. The usted lenge of the spring €.
Knowing that the collar is released ffom rest at + 3, and

ls

+ ink neglecting fiction between the collar and the horizontal ed,
| dci the magie of the velocity of he colar as i
ES passes through Pol
| SOLUTION

| ane ie rig at Point Card ab positive othe gt, Then xs positon corde o the sider B
and isis nal vale. Let he te reichd length ofthe spring. Then, ram the ight range

LVEF
Ear the aiid ul the foe eer y the spring is

‘The elongation ofthe spring is

gabe EE] 4
By gemir, cont: = NG
AER mo: Anden
se ee
AVE HF 0) E < ma.
Vere

Apra (Fra)

RICE ET

ave fe < 0 4

PROPRIETARY MATERIAL © 209 The MC Comi, be A gn sa. No pt fl al e ep,
‘pwned ware y fom or y som. nd je oto premi of pees eed etd
Beane ply Mt Mint pps ar

PROBLEM 12.27

Determine dhe maximum theoretical speed thal a 2700: automobile starting fos rest can rech afer

traveling a quarer of a mile if air resistance is considered. Assume thatthe cuificin of sie fiction

betwee the tes andthe pavement is 070, thatthe automobile has frohe deve, thst he Font Wheels
ort 62, percent af the automobile weight, and that se aerodynamic rag D has a magnitude
where D and y are expressed in pounds and Rs, resp,

some
kg ,
|

i B
dorama Fonda eu Meru

e al 0012)
Eas w-002r)

£ >
=0.002£.017 W= 69
Zen wo)

ones
Now nw e
san
fae pe
Meer 00028 fée [0
i Le Gris
“ Es Sina woo
van Ese Finanz won
1 yy [2170 6}
EN ELLES
Fu TW 6" _ ara
u y

When 20.25 mi 1320

|
75.285 We
or Von “19S mh 4

PROPRIETARY MATERIAL. © 200 The Mini Compas. A ace prt dé Mae me dp
er tn yf oi ae a, wh torus pret of oa, ed bed
‘Bennett oat oe toner Hen a

E PROBLEM 12.28

‘The coeficients of fiction between blocks 4 and Cand the
orizomal surfaces ate 4, =0.24 and, =020. Knowing
Ha my = 3 kg, my TOs, and mo 10%g, detemine
(a) the tension in the cor (6) the acceleration of each
block.

SOLUTION
We frst check tat static equim snot metano:
Ce Te E
0248108
367

Since Wy = mag = 108 > 3:69, equilibrium sot maintained.

tho dk BR: Meme
falls EWN 02m
a) 7-026 =, o
A er 3; Me ames
oe E Fe Ne 0208.
Ten,
ABR mc T 02e o
= Mh HER ma
q m o
wid des,
w
ER: T029g=50, 4027-04 ©)
Eq.t2): 1 02(10)g = | ac =0IT-028 ©
10, ay=8-02F o

Per 10827

PROPRIETARY Maren. © 20 e Mean Cong ne, o A iio ame prt fi Ma me und,
‘eater ay aoe i rpm of en ol e
evar eigen porte “ “

PROBLEM 12.28 (Continued)

Soba into (0
-02r~Lo2r-02¢ +017 029)

2 a

128 = 0387 (O81 mt) T=H6N A

1409 Sutter Tinto (5), (1), and (6:

48520081 rah!)

wii) o

anf) one

en(2te)-aze

142860981 m5?)

org ern. © 00 ser Como a ep! of na et
‘titi eer ac yr M Gun for hr ch re pn faa eg a

PROBLEM 12.29

Solve Problem 12.28, assuming hat a. Sk, my 10 kg, and

me 20 kg.

PROBLEM 12.28 The coefficients of fiction ven blocks À
and C and the horizontal surfaces aro 4, 02% and 4, = 020
Kong dut m,=S kg. ma =10kg, and me = 10 kg, Goom
(o) deter ie card) emo och Sock

SOLUTION
We fst check that sti equilibrium i nt maine
da HI = + mg 0245 +20) = 68

Since Ma mag:
‘We shall assur that ll 3 blocks move and use as. (1 (2) (2), (4) drives in solution of Problem 1228
(2) Tension incon.

He > 6, equilibrium sot maintined

Given dat mr Ske, my=lökg me =20%6
Eg UD: F-02450, ag 2027-024 6
Ex. 7-0200 = 200, ac =00sr-026 6
Ba 10228100) ea m02r m

Subiuaing into (4)

rasan 4

Substtting io (6s

“This means ha ou assumption hat lack C moves was wrong

PROPRIETARY terra. © 20% Re Moi Compare, In. Ar see. fat of ie Mono up ssp
weet or tatty ort) ey mt tr wien perme pe hd de mnt
‘Koto otal MOM i ma ca pp ee rs Ae

PROBLEM 12.29 (Continued)

Assuming u hat Cais ot mo, we have

Suitutng = 0 and from (5) and (ino bg (3)

gwar -hear-02e

Lig=037

| T-360N 4
Subs fr Timo Pes (5) and (7):

coms! |

oran)

Ve also check that 74 y es (2), 0241209) Ag.

‘the value assumed for a thos cect 2.04

ss well as that obtained for 7 T=360N 4

PROPRIETARY MATERIA 208 Vi Main MU Compa ite um pr fh Man ae par,
rr rd o foo ron nt i fs pron of ea at Nat he
‘tno pre fr aa pre ns Me

PROBLEM 12.30

Blocks À and B weigh 20 Ib each, block € weighs 14 Ib, and block D
weighs 16 Ib Knowing hat a downward (ore of mugnltudo 24 1 i
sppled 1 block D, determine (a) the acceleration ofeach block, (9) he
tension in cond ABC: Neglect the weights ofthe pulleys and th eect
ot fiction,

SOLUTION

Note: As shown, the system sin equilibrium
From he dira:

Con 2p, 4299 se = conan
Then Dr Ont
and day autant
Con OIDO == const
ten Bip 000
and Zagsajney=0
“ @ ER mat W254 = hay
yo “
SP Ana
» HER, = Mey: Wy 25 2
up z
in x 20-2547,
a

Note as, and (4) 5 2423
Then Da. (1) warn,

Lo.) > ay =a

Mm

®

°

w

PROPRIETARY MATERIAL. © 209 The Meat Compas be A ng so. po pr ar Mani no e sorgo

hl Mad y fr my pi 0

us sons psi ye mel Roe a a
rams nes pra er or onary he

PROBLEM 12.30 (Continued)

a 4 er,

man or

5

D

Substiing fr; [89 (SY and 7 [Fa 63] ino. (3)

amore) (ana) 3;
A ely

or 322 148 = 2.76 00
aa, ie)
EEE oF a 11.0 BT 4
) Substituting into Fy.)
se)

PROPRIETARY MATERIAL © 20% ie Neem Cong, be, AN its ome pt of Mal ma yd
rs partly ine le hahaha pan: ste
Jang ult pa d

PROBLEM 12.31

locks 4 amd A weigh 20 bench, lock C weighs 16th, and block D
‘weighs 16 Ib, Knowing that a dowmward force of magnitude 10 Ih is
apple o block 2 and tat the system starts Tom rst, determine at

35 the velocity (a) of D relative to, (6) of C relative to D. Neglect
the weights ofthe pes and the effet of ction

Nowe: As showa, the pt equilibrio,

From the digea
Cond 2a Brg ve = constant

then 2,42 He =
and 2a,+20,+0,=0 o
Cond 2: (p> Ya Op va) = constant
Then EU
and Day ajay 20 o

We determine the accelerations of blocks 4, €, uml D, using the
blocks as ee hal

alar,

mas mm Te",
A

or 20-4 a

Bay
HER mi My BAT Pa = Ex
f
« 20-2547, 210-24, w
2
Foming o aa
n

PROPRIETARY MATERIAL. 200 The Ml Compare. gh ri rt of i Ma ma be ena
PR tne ny fo or y a rs mt he rw pores he ps th ae
nd pani N ro ra war ase Ma

PROBLEM 12.31 (Continued)

m

A erage

Then ER 244 12{ AE

« aus
me au festejo

Substtuing for 7 (Eq, Sand 75 LE, (6) in Fo.)

ole

2, 12.2 86) = 690 WE
or TOS
Then c= ACG 90S!) 322008 = 4.60 Ms

DETTE TETE

Not: We have w

omy accelerated motion, so hy
vs 0tar
(Weine You "Yo Va

or Vador Bi 3
CENT

PROPRIETARY MATERIAL. 2909 Te ru Cc. ies ro of hi Howl my Be a
‘eer din $ an oa ar vrei ef Be er, el et Il
‘Reni a iy ne Moti ta sr pst fn ag Hoel

PROBLEM 12.32

“The 15-kg blocks supported by the 25-kg block A and is
attached Yo cord fo which 225-N bovizontal force i anplicd
sahen, Neglecting fiction, determine (a) the aceceration of
block 4, (6) Ihe cocher of block relative 10 À

SOLUTION

(a) int wenote ay =a4 mg here A, is directed along the imlined surface oF 4

» MER, amas Pg sin28°~ me, c0625'E mod À
ga
ns ee
or Nay = 1S{geos25” - a, sin25")
e
” Na (2504-2251 —c0s25°)]/sin25° e
x
peras - |
E

OS c0s25 ain 28° 45000025)

a4=280m € 4

0) From Bg.)
ing “15-8 5iN28°=2 797900025"

or an 832007 D 25° 4

PROPRIETARY MATERIAL © 309 Ta ein Comyn, LA et cad. o port Jal ay e ap
nant fo ny ne D pl en prod pl oe po ead
En pr oe ifr ce pen yn et ge oa

PROBLEM 12.33
lock B of mass 10 kg rests as shown on the upper suis of a 22kg
‘wedge Knowing that te system is tless Tum est and neglecting

fiction, determine (a) the aceleation of 8, (0) the veloc al À
relative tod at ¢=05 3

BD»

14043 Main + nn ng

a

oe a=

Naw wenote: Ay 22, Us where ay, directe along the top surfe of

Y/ EE, =p: May Wy coe? =-my,gsin SOP

or Nye =10(gcos200—a,sins00)

Eguating the two expressions for Ny

fu)
"100 6620040)

1.1 200640 ie

or a 00 mi

NER emo ME am m0, cos

sin 20° a c0850°
=(98Isin 20" + 64061005509 mvs?
274730 mie?

PROPRIETARY MATERIA. 0200 Ve Meinl Cai. Ae ee oe of he owed yb pl
"peo i son a aa, tar ars par ay, oad ee el
‘aa ca adc yay oie sae pps en neg ed

PROBLEM 12.33 “(Continued)

ity Bente Han
Werne OR TATI AORTA RRA
« 9447 ms ee,
sun
“ nr

4"
ag “894 nds? ETS 4

(6) ove: We have uniform aeclerated motion, so hat

Ka ytd ad
Au-0ss van ADO KOSS
vue 288m EI 4

PROPUEEARY stars. (299 Tos Mei Cag, Le, A ight ee No po of i Sl y Be tnt
eel setae xy fom oa sss jr rie mine of ei, ed pd Se ad
‘Sibert ko eke nr rns a

PROBLEM 12.34

A 40 siding panel ie
Support by rollers at Band
ER 234 counterweight À
is tached to a cable m
Show und, in eases nd e,
is italy in contact win a
vertical edge of the panel
Napking Fiction, determine
in cach case shown the
acceleration of the pal tl
ie tem fhe cord
immediatly afer the system
is released fom es.

SOLUTION
@ Past ore een by comercia
Lx
_ FEF
F lve
28, ma

Comienweight 4: Ns accelesion has to comp

ayn ap tap arta

oma: Feu

®

tu

2s

8

ar mas 25
Adding D Eh ant)

PANIER ES,

Substituting foe ita (3)

m

o

a

and

Te 4

PROPRIETARY MATERIA. 209 Ts erh Cp, eA ie ek pu a Al mo be pc
Pen be mom by pmo er wre Fran Dale end on
(rd er or en Bm

[ PROBLEM 12.34 (Continued)

|) Pat E]
Aus: mn
Comcel
AS =e: o

T
Ang (ard (+ #4 fe
as
u, ana €
Be
Subst rin
7-15 4

{Since pane is acc he

Pan

=]

4
Bag 0)

“nterweigh 4: Same ro bly an in Pat

=
er, =ma: 2 T="a 2)
las : a

Since ap. (1) and (2) ar the seo asin 0), we gt the same answers

PS

Re)

[PROPRIETARY MATERIAL. 209 ts Manni Compa, hu. A ge re ip of Moa mae apn
ee he fron o. rie pri ge D ho ee
nsw chon eh per, Ha fi md one wea eu asso Mr.

PROBLEM 12.35

TA S00 rate Bis suspended rom a cable attached to à 401 trolley A
which rides on am inclined beam as shown, Karin that at the
instant shown the trolley has an aoslerion wf 12 MA up and te
Fight, determine (a) the acceleration oF relaie o, (9) he tension in
cable CD.

SOLUTION

©) Fine wenote:

Let gy Where a, is directe perpendicla to cab.
HER,

mas Om, + mye c0828°

or gy “(2 W525"

©) Poreates

A,

(12 Minas

22m
For wiley 4
Y IE, ma Top =Tysim2s"- Wysin2 Ma,
#
1 nage 4 12
or Tay = $0787 loss" 4010) sins.
ae
« STA

PROPRIPCARY MATERIAL 19210 Te Meine SU Cup Js AU SIA He ar ih Mm mre iio
‘abs er nated ny fm vy mn rs es lar, aed So ed
‘Sri rd analy ee fe che ocuparon Fae ka

PROBLEM 12.36

During a hammer thowers pratie swing, the 7.1-kg head 4
the hammer revolves aa constant speed y in horizontal nc
shown, p=0.93 m and 6'= 60" determine (a) the tension in
‘wie BC, (8) the speed of the hammers had.

SOLUTION
| Fest we note aaa a
(o a Oña
ro Lips : |
al ‘sin 60"
EN

GR mans Taco,
© .“ 4

o (80426 Nycosh0.99 m
a Tig.

v,2230m8 €

pede MATA 0 Ni o A rac pa of en or pd,
A D i dv cove ppt Feast Mae,

PROBLEM 12.37

À 450- tetherbal is moving sang horizontal circle path
3 coma speed of 4 mi. Deremine (a) the angle hat the
Sud forms with pole BC, () tr tension inthe cod

SOLUTION

| Pirstvenoe
Ta
where Orne

HR Tuner,

Tyg =e,

u E

»

Substioting fr ad >

mag. 4 2 2
sin=m, sin 0 21 cos
Ex un ent

(my

1 cont = Em con
ES

or eos? + 0.906105¢088-)
Solving
” 0-00 4
U Binden yg = ME 0450 RE DAT ml?
Er
0 Ta =635N 4

PROPRUBTARY SATA © 290 The MG ic he Ags ean A a fh Ma may Be na
‘pro natal mm wy yam tr Fro Y ao oe eet ad
‘Rene dno my ont es ck pros pach Had

PROBLEM 12.38

A single wie ACB of length 80 in. uses rough sing at Cth is
attached (0 a sphere which revolves at a constant spool yn the
hovianlal ice shown, Knowing thal @ = 60 and 0, 30° and that
the tension isthe same in bath pots othe wir, determine the spend:

SOLUTION
est we note

Now

NER, =0% 1yc0030 Ta on

mes.
EE

207208 mg

FB megs Teysin 30° 4 Ton sinn.
>

À

(0.73205 main 30" +60) =m
n

ons 2 {2288 ear a)
(Or:

ve=825 ms 4

PMOPRIGEANY MATERIAL. © 209 The Mu Compro, Inc. A ve. a pare Mal mr be pe

rca il D a

Jove Bah parson

nie fu pe fe ler, nod han ae aed
ier drial compra evar cg he Me

PROBLEM 12.39

A singe wire ACB passes through ring aC that is ached toa 12
Sphere which revolves at a constant speed » in he horzontal cle
sown, Knowing that G80" and dl = 30 ind thatthe tension m |
both portions of the wire Is 2.6 th, terrine a) the api. (8) Ihe
speed

SOLUTION
(0 AER, ~0; Tous #70080, —H =0 BR:
=F const 093616 =
| 16
0 |
©) ana sense aan
tt pd,
na ano, ins

Radius of horizontal cielo

=1645in.= 137000

Emo Tang sind, =

Pd ne)
„ {LION ONO" + 20.5847)

22
31246048 PET

UY MATERIAL. © 209 The Meena Cones fA i roe Yan of Aa ay be dia,
‘ove ha my fr o ly msm ote o peace ft pew jon?
ORI chr evecare Sho ferme oer nera deta mo hana!

PROBLEM 12.40
“Two wies AC an BC ae ti at Ct a Pkg phere which revolves
at Cosa speed» inthe Bora ee bow, Keogh
9.258” and 8,230" and that'd LA m, die the ange uf
fais oF fv hich oth vis rena ut,
SOLUTION
Fist were
were 55 end
Pdo
hen [+ 2 Jango”
ie leer
Lam
“ pa
canso tans?
35680 m
RR among Va "yin SSP E
nas 5 ny
ADR, 202 To 60630 Ty co 9°,
= Ta 08304 Tg 00855 o
Re
By.) 9g =
eb. sinsS ame À
Sub nto Eg, (1) ins ame É
CE = (05690 OA mass"
o [Were rs
ROPRINTAY MATERIAL © 309 he Mean np Inc Apr. qa M br nt

Pie or ned fee an an la os ra pm fe pe er eS ed
‘Koen poy toch be anya rao et

PROBLEM 12.40 (Continued)

owe we form CT")
Fi
TegsinS8" eas 7, cos5S"sin30 = 60530" mc gsin30"
a
B
“e Teg 8028" = me “06308 sino?
me #

us accu when Toy = (dou, Which Occurs wen Ti,

(ebm 7436 ms and wire AC wil be tat Em.

a

pez Y Fa (2) To =

bite in ME um th

A. me em,

er (gag 03500 6) uno"
= (eee "27 ms

Now we fom (asas)

Panama nm Espias

mass”

{au occurs when on iy 0.
wo.

Bosh wires are taut when arme

77 mé ad wire AC wil be tut 1, >2.77 mí,

em €

‘PROPRIETARY MATERIAL. © 290 Ye Mein Cais a ais mB ir Mood mayb igo
vo ara tan fro y cons pet peo of pa ad epee!

PROBLEM 12.41

A100 gsphere Dis rest eave wo dram ARC, which ates
ata constant rate. Neglecting fiction, determino the range of
the allows values ofthe velocity vof the sphere 1 rete of
he normal forces exerted by the sphere onthe incl
surfaces ofthe drum iso exceed 1.1 N

SOLUTION
Fist wenn est |
?
sun p-02m ss |
cos 0620 2m À m
Le {
160 Nysin 20° np =0 |
“ Ming Mia 20 umge &

Case: N is maximum.

La MIN
bq.) (LL N)sin 0-4 sin 20° (0.1 kg) (O81 mí)
or ¿0080958 N

Oo SEIN OK

aa) Bene rr e060" à 098295400420) N
er nn amd

Now we form (sin20°9¢)~(60820302)

160s 60s 20°~ sin 6008202 min 20° mpg cos 20°

e Hsin? y Bin 200— mp coszor
»

(ode occur when 4 (Ma

Onde 1.121 ws

“ornueyany mare. © 299 Tos Met Ni Corsi, eM rat. Ne of at Maa way Be apie
‘em ar dred nay fae ar bp an nD rues porn pul ed pend hd
‘Sinem handen porn Neto Ve Pt er pr Tone soon i A

PROBLEM 12.41 (Continued)

CaseZ: My maximun.

La

moun
10.0) sn 4 ND" ADO e)

| a UN. 0X
sac Bran, = LEP 207
u Cot 168 me

{incor cs

¿cos 20"sin 60 — N, sin Wesen my sine moe co 60°

o. costo sino mpg 28"
a

Ch cas hen Ns Wado

Cou “1663 ms

Wor Ay My LIN

LB ney 1.663 me 4

OPKIETAKO mare. © 20 he Moon Lex AU es sol. q o Mano! be dt
Ice hd e o ir by mma rit pre of he be oe po te ed

‘Roe nha poly io Mra nts Fash

PROBLEM 12.42"

As part an outdoor display, a 12-18 model © of the ar tached
Lo wires AC and UC and revolves at a constant speed the horizontal
ele shown, Determine the range ofthe allowable vale of wi bt
‘ites an 0 remain tat ad ifthe tension in ier ofthe wies is nut 10
exceed 26 1b

SOLUTION
Fist ate a
where gaan

Demat Toi Ts À

a Os 20 o

HER, 0% Teno

Note a (2) piss ha
(2) when Los Cerdas Tea chan
CET Ton Cane Tex “Tera

(se Le Je à maximun,

La Teg 2618

BD CIRT costs? (1218) =0

on Tog 28.1060
OS

(Cog a =]

PROPRIETARY MATERIAL © 20 The Mio Comp, Ic Ah ro ur hs Mol ab pins
‘Renu dl ap fob ae mur. vd e Jr e pane e llo, nl ge fd
‘Mitten pd sor if i cs Hi sas oe Mae

PROBLEM 12.42" (Continued)

Ex)
IN par à 81065
GE = 22080000 +106 359

w Cora 123118

ow oe fom (00159 DHL

Me cost Se Meinst

Fine 154 Tous 015

an

or Tensinsse Me costs 1 sin? © |
A

(a cb To id

Bocas aa cam when Za Codes

Be Ty =0 (me ha wire BC wi nt bt)

mo Taco (2 pn

Ta 186669102610. OK.

oe: EQ) plist when Tey = Teens Ye "cn. Then

Ea.) Baa FERED 5.669 sin
or hag 29.00 0
Tan Teg 2261 wien VOR RU 4

PROPRIETARY eurem. 309 Te Min Cap, I. pt rc. a po Man ay tt
(apa ru yoo yma ie rr ris pmo llo, met so i
‘eink ops i M Bel a poe aw Man.

PROBLEM 12.43"

“The 1.21 yall of sotifgel governor revolve at a constar speed
vin Be horizontal ciel of Gin, radius shown. Neglecting the wells
of links 48, BC. AD, and DE and requring tha the links support only |
terse frees, determine dhe range athe allowable values ul wo hal
the magnitudes ofthe forces inthe Finks do mo exceed 17 Ts

| SOLUTION
| Fist note 2
fe peosn
Een: Ty 020° Tg sing? =
er
A) ©
Me tat, ii
(@) when Toe =(Toedan> Toa = Tre oan
© ven Tes Toa
Cut: hy mus,
a mont
Eg) (17 lb)cos20%—1 y c0530"—(1.2 fb) = 0
soria
0 Ines ea (12 20
$ sanos am

‘PROPRIETARY MATERIA, 02003 Te Min Campus Ic gh en pr a a Mal ab int
‘penned rida up Ron year nen de ror sites pon of pe, a ed ed
‘ribs onic pora, cal hr tia cups Y rca bt Mar

PROBLEM 12.43" (Continued)

Cada “16 944319

Figg 17
Fac ne, = ED 694135020" 117503091
2m
e nora 13.85 Ws
Now for (080°) (1) #23092)

0530” +1 sin 30”

Tg $i, 2070630 Ts 20° in
r

or Tnesinsor °° cox30 1 o
a
ial it
sth
Because (Fy Max OCU when Try = Try as
mo menu
Ho One Tua, Te
num Pha EEE ar nao
M err
ene 242 Miss 951385 Ms À

PROPRIETARY MATERIAL, © 20 Me Neo Can, e. A oe pun of eM na be ee,
rao dae ay ty a x, se th ren pin Fer ma en e ol
‘Titan tatty crs lf abl oni ne asters Mase
pre

PROBLEM 12.44

A child having a mass of 22 kg sits on a swing and is held inthe
position shown by a cum child, Neglecting the mass of the
Ewing, determine the tension in ope AB (2) white the second chi
holde te swing with his anıs outsicched. horizomall
(©) immediacy er the wing es

SOLUTION

Note:The ions of“ as included inthe following iee-body diagrams Because thee are two fopes and
only one i considere.
(0) Porte swing atest,

a
17,20: reas bo de

in Tye BHOSLE w
x 20835
o arto Pa
1 oe
VER: Ty Íwanis=o wey
2 y Ÿ
” rye JOE KOR Meo

onaierany sare
re dant y far le tran a rar por
‘Senator chant pm 1 hocico bb cr pro

© 200 Ve Stn Cn. Ah rl, ep of mal mr eine
= fia, eid

PROBLEM 12.45

A (kg wrecking tall is attached (oa 13:mong sel cable
AR und swings im the vertical are shown. Determine the tension
in the cable (a atthe top Caf the swing, (a dh btn D oF
the swing, hore ne spl Be 42 m.

‘SOLUTION

(0) ALG therwpordeswing, Y =

de

Tas Myc 20°-0

or Tug GUION meso
or TNA
enh »
CES nn en
E
= Pr [panas ER re

PROPRIETARY MATERIA. #2408 oe NCH Can, eA gh Noon of is Ma! ey ead
and arta iy eo a ea Eo tn prs aba tet ed
ia cpm i ifr ered ace gr Puras Mo

SOLUTION |
CET 100 mil 146.667 MV.
| à
| HER =m, À
i j ze
Lo (146.667 Mi"
| 220
= css 1
we peca
À (6) Not ri constant 2 4, =0; 50 mih 273333 1s
la
hh
ESTOS al

PROBLEM 12.46

Dring a highspeed chase, a 2400-6 sports cor teveing ata
cc of 100 mi just loses contact with the road as
reaches the crest A Of a hill (a) Denn the ras of
anatur 2 of the vertical prof ofthe oad a À. (0) Using
the value of p found in para, determino te oros exeried
on à 1604 diver by the seat of his 3100: ear ay the cr,
Utaveling aa constant speed of 0 mi, pasos through A.

622 668.05 1

pio, Aid td pr fh Man ye agape

pve ha na fae hy rca Re pri wien rn he Yair di ie tra
nen under pres ie Mf is nn ene orp er om Ha.

PROBLEM 12.47

The portion of a toboggan vum shown is contained in a vertical

plane, Sections AB and CD have radi of corte as dict, und

scotion BCs straight ard forms an angle of 20" with he horizontal

Knowing thatthe eneMiient ul kinetic friction between slo and

the run i 0.10 and that the speed of the led ix 26 Ws a A,

determine the tangential componen ofthe acest ofthe sled
1 (aus before it coches 8, D) just ater it passes €.

SOLUTION
(a) Nat: Just before By py = 0 R

Sting: Fun

| a+]
Be

ray: Wsin20°— Pg,
®
= a main 0m

Pa
2 es ty?
Then 2622 208 ysin20°—o. toon 20) 0 ERP
a x a

or 2,6951 20° 4

ROPRIETARY SATERLAL. © 30 Ti Mans MN Campes. srs Hp of ne Nad mae po
prec or db e Ju By np mes wh Oe ror wren pay Fae, oat ead eed
‘Ren ened See Mf mB ae pp Den

PROBLEM 12.47 (Continued)

(0) is ft necessary to determine
For section BC
AA DR 202 Myo =H c0s20°=0

or Myc =W eos 20"

Sidigs Fog HN: E 0820"

NER = agg: Wind" Fe =

or ae = in 20° = a 005200)
022 NA ain 20 160820)
9872 ms

oe this uniformly accelerated motion we have
Erd At

SM + 207.9872 EN
o PRET
Now, just fer €, pp =140 1
VER mg Woo” nF
Sin
NE ma;
oe
Tien |

ADA]

ROPRIETARY start. © 290 Ye GM Capas. ira, pra Mas nay e ae
‘oc boda so Ba ae ene te wie eon à de a o ed
‘Sarton onto preci en fr oe rn Pp anh Si,

PROBLEM 12.48

A series of small package, cach with a mass of 0. kg are discharge
fiom a conveyor belt as shown, Knowing thal the fie of sic

m beacon each package and the conveyer belt is 0.0, determine
a) the force exerted y the belt on à package just afer i as passed
Point 4, (D) the angle à éefinag Poin B where the packages ir lip
clive the bal

SOLUTION

Assume package dos not slip.

On the curved portion othe bt

Forany angle @
ER ma,
&
ABE mas E Umgsin@= ma, =0
Ey =mgsino o
(9 ArPoina, .
KORK (018) N=2905N-€
06) ALPoint 8, un

mgsin®= 1, cost mu)
(a)

ren

(ee)

ofen; |
ss

PROPRINTARY MATINAL © 209 The MN Coup I. A gh ee A pr hs Mn ee.
pe dante nym ly nas nth wri rane of epee ct ed he
‘deanna pend cr Mafia cars partes Poren nern i te
se or mgd itn poo

PROBLEM 12.48 (Continued)

‘Squaring and using igerome

densities,
1-00? 0=0.16008°.0- 11304790058 + 26601
1.1608 0-0.130479c0:0-097340=0)
eos =09 7402
‘Check that package does no separate Fem the bel

07 4

nz mins
| aw

PROPRIETARY MATERIAL. © 2509 Ye Nw Care, eA eg st pur Sowa Be pit
ved or aed ay fare easy mae ws rc wn emis pb, ed and ed
Snel tire pro pv enn He

PROBLEM 12.49

A Stig plot file a jt tains in a half vertical Toop of 1200:
radius so that the speed ofthe minor decreases st a const rte
Knowing thatthe pilot's apparent weights at Pointe and (ane
1680 N and 350 N, respectively, determine the force exerted on

© herby te sea ofthe einer when the trainer fat Point

| souution

irs note that the piles apparent weight is equa to the vertical free that she sers on the
sea the je tier

Ava HER m0, fran
ly
nano m "991 e) se
« A O!

2255613742

MO cms He =
+

Sine 4

or 19,5408 mi =25,561.3 PA + 20,6
or 707970 me!

“Then from Ato

era ge

5,5613 wa? +2(-0.79730 mo x1200 m)

2,585 mist

[PROPRIETARY AEATERLAL 2203 1 Nein Cais, I Al rom ot fi Mal mab yr,
‘aw td om th tr rn poi oe milo, a ed
a lah Sn el pee He soda ed

PROBLEM 12.49 (Continued)

au Nyon’ D

: x

ss i
sang SSSI he
o 1014.98 N—
YER ma; WP, =mlal
e. Py =(54 a X0.29730-9.80 ms!
or Der TT
Finally, Froude = PE + BE = (1014.98 Has
Bon

a EN 32564

PROPRIETARY MATERIAL. © 290 Tm Mi Hi Compr, ne. A gs seed. No fr of th am wo eden
Pend ron a fom o E ay ao flr ete pono he pba, a in Sod m
‘Soran car pete Wear Mr tel so poten sa tae a Mae,
‘cena on pn

PROBLEM 12.50

À 250: block B fis ine a small cavity cu in sem Od, which
rotates in he venal plone at constan ste such that v= ve
Knowing that the spring exeta on block A a force of magnitude
PZASN and neglecting the fic of ction, determine the rango of
vales of # fr which block 2 is im contact wich dhe fice of Ue avi
loses othe ais frotan O.

À Y A
Sens encino en e
Y wind Qu” ae.
Hohave contact wilh be ei rs we esd 90, =O
any a”
a = remind Mixa
9-Pemsn0 "7
wm

SN, p-09m
Subsiting at (De

sino = 04075

airs O ss €

PROPRIETARY STERIL. © 20 Tie Mi Copos. Art ot u ds Mol ee
‘ora ed ay rm oy apt Wer crn pron er sd Ae ae
‘in nea ns sin ifn saree What hh

PROBLEM 12.51

A curve in a spa Hack ha a run of 1009 and arated speed of 120 mi,
(See sample Problem 12.6 forthe definition of rated speci), Knowing hat a
facing ea sas skidding onthe curvo when traveling a speed of 130 mi,
‘determine (a) the banking angl 8, (he coelicien of atc Ton between
the ies an the track under the prevailing condon, (2) the minlnum speed at
‘whieh the same car could negotiate ha curv.

souumon

we von D à D=
cleration a ir & da

het MP &

Ekta: F+W sin 0 maces
Fe cos 8m sind o
7

EF ma; NW ews @= ma sind

ne in Bog cos 8 a
in 04m
(a) Bankingangle Rated speed v= 120 mn = 176s. F0 ot rate spe

=I os 0— mg sn 0
»

769

tun À.

096199
pe Coo
ona en.
(6) Siping wa = 180 in = 264s
rope pol 2 ps0-pgins

NT nor pa coso
(268) em 43.89" 1000) 32.2) sin 43991
(264) sn 4389 10002-20648
= 039009 A-0304

w

PROPRIETARY MATERLAL. © 209 The MG rai, Ve it o. pl Mama ay Be pgs
Pe hun e ora e y oy m o per a en fel do e ad
hunde cereal te fir hr tba cep once gt a
ue rag a pen

PROBLEM 12.51 (Continued)

(9 Minimum seo

(1009)09.2) (in 43.89° 0.59009 cos 33.899)
eos 23.89" + 039000 sin STATES

3.369?

15.2 fas P= TBR ith 4

PROPRIETARY MATERIAL. D 206 Th MG il Cp, Me A gt cee sol eed,
‘spe nd io np. ms ih ws porn tar, te ea el
ina er ads pray See Hilfe tb re pra re gt cea,
‘ein aon poro

PROBLEM 12.52

À car
¿ss mot skid. Express your answer in Les ofthe alas fe cure, the banking ange 8, und the
“tai fiction 4, Boton the ties andthe povement

raven on a banked rol constant speed. Determin the range of values of for which the car
ngleof

SOLUTION
Case von
ote: R= +
af E 3 EN
S Aaa Riesen w
+120 Rews(@rg)-W
= Reesor game o
gt a a)
Foming & jee
wo Roos(O+d) me
or Vom © fer Un (8 + 9)
eue?

Notes RoR EN

+ e PS o

Reosi@-9,)-W=0

or Reos(O—4)= me o

Forming {2

Forte ear not to skid

AFRO da 4

[PROPRIETARY SIATENIAL. © 20 "be Men Camps, Ag rm pra as Mal a be pe
rn cd o ah pmo at je mete as pa, La eed
rd Re de spe wa ga

PROBLEM 12.53

Ting tas, such as the American Fler, which will ran om
Washington to New York and Boston, ae designed to travel saely
at high spocl on curved sections of track, which were built for
Slower, convencional tins. As ht enters a cave, each ea i ll
by hydraulic actuators mount an is mucks, The ling ture of
the cas als increases passenger coo by eliminating oF gray
redhcing the side force, (paall to the Moor ofthe cur) o which
passengers fel subject. For inn Anveling at 100 mb on à
‘euved section of tack banked through an angle O 6" and with 2
rated speed of 60 mi, determine (a) the magi oF the sido
force felt by a passenger of weight Y in à standard car vith o.

tie (=, (te uid angle oft 9 1 do poset
ft id foc. (er Sample Posen 12 for dh de ot
mad)
SOLUTION
Rated see On 88 Ms, 100 mi = 14667 ts
Prom Sample Probe 12.6
pue
4 Los
z - ET
Fund” B2un6 À
Lette aus be paa o on ufr
Poms HAsO
90.
[pews
=1 use]
aan
assur pensar 4

PROPRIBTARY MATERIA. © 20 Tne Mini Cpr, Ie, A ee No vf Moa my e ted
‘ep or sed nnn np mos i lr rpm of ly et Ml
Sein cr tty et bb rs rar ee,

PROBLEM 12.53 (Continued)

© tunen
ante 19-00
zp 9
y 04667
an Gap „ea
er ep (32.2)(2288)
Dre
0-16 em...

KOPRISTARS MATER. 209 The Mit Compr be Ay wi Mi prt fr Ama ye ipa
‘ovate rio nyo yma ae pr res poo of si el oe e
“Bonin sete le portly ror pr Pare aca ag Mer,

PROBLEM 12.54

Tests carried outwith the ling tins desribed in Problem 12.53
revealed tht passenger eel quessy when they soe though the car
windows tha the an sounding a curve at high ps, et do no
Feel any side force. Designers therlore, prefer to reduce, Du not

inate that ore. Fr the train of Problem 12.53, determine the
uted angle oft fi passengers ae to fc id rec equal to
10% of thoi weights,

PROBLEM 1253 Ting tins, such es the American Per,
which will un from Waslingin to New York and Boston, are
designed to travel safely thigh speeds on curved sections ol track,
‘which were built Tor slower, conventional tan, As it eters à
cure, cach car is ted hy bye actuators mounled on is
trucks, The tilting Rare of the care alo increases passenger
comfort by eliminating. or sweaty reducing the side Türe F,
(parallel tthe flor ofthe ca) ta which passengers fel subjected
For a tran traveling at 100 mih on a curved section of tack
ranked muugh an angle 0-6” and wih a ned spa ol 60 mh,
determine (a) he magnitude ofthe si Fre fl by a passenger of
weight ina staan cu with o ik ($0), (0) the required
angle of ie ifthe passenger 610 fee no tdo fee, (See Sample
Problem 126 fo the definition of rae spc)

| SOLUTION
Rate spect

From Sample Problem 12.6,

gtd

| Lette axis be parallels the our ofthe car.

fh LASA

1240091049)

o
ACL)

Solving for, | casio 9)~sin(O+ 6)
er

PROPRINTARY MATERIAL. © 209 The Weve Coup Aa sr pr e Neal e py
bs a nn rd a

PROBLEM 12.54 (Continued)

vow PACE 87 at ®
Soit {10 = 10 29199 0100 9-00)
La
Then

190201991 Ww or 0291991 x 0.1044
Sain hides, Boss) 0014024
a 10820102 -047S26-0
¿be positive ot fte na pion a 48700
then, Drbmsnrtun ose

$o1690°-0-10 786° snare <

PROPRIETARY MATERIAL © 200 Ye Mco Cop A pre. por of tum o e e
pr kann Jn Deu ne re ot pts pmo pk, ra ende a
“sneer con pr ee Ht cal cm pp a at

PROBLEM 12.55

A small, 30048 collar) can side on portion AB of u rad which i bent as shown
‘Kuowing that = 40" and that tb rd riss bout the vertical. AC’ a a constant ate oT
S mms, determine the vale o für whic the collar wl ot lige om he a i the fet
of ion between the laa the collars neglect

‘SOLUTION

rst note

v= "bas

oe, o: sinew 0 Y &

91m 1
(mas ana
20468 m

468 mm 4

PROPRIETARY MATERIAL. € 209 he MG Compt, be, AN igh a fof Mol y eds
swe in ns oo” yay ea he ls ar he sc eS ad
En esrb ar mim crepes Pyne oe

5 PROBLEM 12.56

A small 200 colla D can sie on potion AB o rod which i bent as
shown, Knowing that the rod rates about the vertical AC at a constant
fate an tat = 30" and y = GUO nm, determine the range of values of

the speed v for which th cll will wi side on the rod if the cosiicient
] fat Fiction between the rod andthe cllari 030,

NW sin 30" mn" &
a

a enfe

WNBA, =a, =I 0830 =m din 30

. penseur)

1 0ten30

= 0.81 is! #06 my

or 23608

PROPRIETARY MATERIAL 12 Ve MG Cosi a vere o ar fi ne a te a,
‘Rd dal o om By ma the tren pono he er, med ad ed
‘Sateen pd Sie etal appa ahaa st

PROBLEM 12.56 (Continued)

Cane 2: ver, impending motion upd

WER = ma: N -Wsin30” m” -cos30"
if #
m =m gsin30”+*cos30"
À +
Nas amo: FW cm sin 30"

Now Fun
ana) En
Ten poso Fain] 0 02071 "oon?
PET
was
20.6 ny +230
Ba my an
or in =:499 m4

For the colli not 0

236 nv ve 499 mis €

PROPRIREARY MATERIAL. © 20 1e tnt Comp a AD ih ee prof his Me en be do
‘mp be ny form oy a mo mo one re rc ied oi
on era por rn he See

a PROBLEM 12.57

A small, 619 collar D can slide an portion AB of à rod which is bent as
‘howe: Keowing that #=8 in, and dat the od toas about the vertical AC a
{constant mate of 10 rad, determine the amulet allowable valve of the
{velicient of sai ection between the collar and the od if the collars not io
Side whan () ar 1, 0) ar 48°, Indicate for each cae the direction ofthe
Impending motion

SOLUTION

First we note that» ne =
‘lle D sping

JO ra) = 280, and hat requirin 4, = (m impish sing oF

Also, mang,
Now we consider the two possible cases of impending motion,

Case: Impending moin downy.

a/R mas N-Weina-

or Ene

Eine)

Now Fans

O cn)

e Y izátma ton una
PCT wnat tng,

a a
ET)

PROPRIETARY MATFRAL. D 200) 19 Nana Cr, o M one pr Ms Ma ay dat
rch rds nym ae ha De wien poo la, a a eed
‘Sec teen ry eat te pp mn se wk

PROBLEM 12.57 (Continued)

Case 2: Impending motion ype

“ monas À ow)

AIR em: Pava 22 sina

o sue)
q La
or Fone]
,
=$
at)
Hi PEUT) ou
E=)
nm maes ana

and where the "1" eorespand o impending motion downward andthe "o impending ation apa

Was
Weave sete, masse
4200
Then
Gnu 0.1904, motion impending dotan «

{PROPRIETARY MATERIAL. © 469 Ie Mn Ces, AU nd po of Mol e pl
‘em rd ny eB an ae, eb aren pom emt Crd a a
Bitar arto Se ir nah pean a soa

PROBLEM 12.57 (Continued)

o
Wehe use
as we
Tien Gen 0019219"

4, = 0.349 motion impending pat <

rot do

16220 Te ac Cap, AM sige sme eof hi Moma De i
ny fro yay ct. eo sos pro a a el gr ha
Siete er ant ode

PROBLEM 12.58

A semicircular lt of 10. radis seat in a plat which oats about
the vertical AD at const at of 14 md, Asal Ob block is
in the slot as the plate rotates. Knowing that the

in are 4, =03 and 4, 0.25, determine whether he
‘block vil lide inthe sl is relense inthe postion corespondi to
(a) 9807, (6) 0-40. Also detemine the magnitud: and th drei

ofthe fon ore exerted on he block immediately seri release.

Fine pe mme

Ve Puc

Then OS

6-00 Jus na

Sin 8) M

‘Asse thatthe black at rest with espect he plate

VRP, <a Née ino Y
> a
“ wf cos Laie ©
2
AER ma P+ Wen Dem À 06 8
d
= Penser À coxa)
eon)

PROPRIETARY MATERIAL © 200 The se Cup. o AU a cs. tof hs Mol y Bp
‘epee td e sf bn me. a rr pra ya, wed Np ad
diria senna nm Mr fo mia op ir i Mom

PROBLEM 12.58 (Continued)

(o retos oo
e
wen ocre resi ar]
rouen ea sty act]
sun 04
— ou
”

O) exam]

:
a ape 1 à
SU ET
on
canne es sonar ane
sai
pur
+ E
'

lock E wilde i he slot

+ nad op
AR =, the Blok ist et relative 1 he plate, as (ya),
rete ungen to the so

D
Y ER = ma: N-+ eos 40° min 40" >
a : dd o

or rosa ma] Castor

0 2 £0, so tbat Aya musibe

PROPRIETARY MATERIAL © 209 Te Mn Compu AN igh ve Bi et of i Mal e be pn
‘sul or tated yor oy ae ita kr ie prt of he pit joa ted
“arene he i te oo pr a Mm.

PROBLEM 12.58 (Continued)

Sliding: Fon
544.4924 16)
us

[Noting that and ggg ust be ceed as shown if ei directions are reversed hen ZR, is N
while a, Ne te

te oc es donna in the stand Hei Sad
tee solutions
() Assume tht he Blacks rest wi espe tothe ple
rem We R= on,
Then
(om above)
and
Now ungen 4
st eva
026 > Block dos nt land Ra cited as shown
[2
Now FeRsng nd Re
. EX
sino ses
The Fest
ET
2192618
The bock des wld th sot and F=19261h nnd

PROPERTIES sr
esc ede ay a i bla pn Hn son a

PROBLEM 12.58 (Continued)

() Anse that he block at et with spect tothe pate, ays

my ER

Pla Ba a-ssinaor ne

a osos
and es
Now Dan
so tat ome

“The block will sie inthe sot and then

4, where angi m
or Dee
“To determine in which direction he lock wil side, consider he fres-body diagrams fr the 50

es
:
WER, = mas Wcos40"+ cos is
so nn.
me or

PROPRIRTARY SLR. & 2090 Te Mean HE Cis, A ei. np al y e o
‘pono aed nay far by ny nn. ea tr writes pion he ihe rd oo td
(Zon mcr carrera Meer a i eel ep Mpa a St

PROBLEM 12.59

Tee seconds aera polisher is started fiom rs, sal ls of
fleece from along the eircunferense of the 225-mn-diameter
Polishng pad are obsctvad to fly fee ofthe pad. Ihe polisher E
Sate so tht the Meece along the eicurverenee under à
condant tangential acceleration of 4 ms”, determino (0) ihe
‘speed vo tu a it eaves the pad, (9) he magnitad of he force
aquired ofre a tu ifthe average sass af tale Lm

SOLUTION

CE
Ten
au=3s

CEE

Md

Finally,

tn => uniform 3ecleation mation

204 af
ve
v=1200 ms 4

Fj A0 Rex mí) ar
—64x10 N

= 201810

BTE

ENT

oo NY

Ka = 205HI0N 4

PROPRIETARY MAERLL. © 209 1e CN Coms, be A di so. df al ye foe!
(Sa or dd nap frm by a mov me Yn pre 1 de ud, a oe a ad
‘Sinton set permi Moor Mtb emer Draco Mb Dl

PROBLEM 12.60

A tumiable 4 is built into a stage for use in a theatrical
production dv observed during a roll hal tae A
‘Sans to sido on the tumiable 10 after the tuntble begin 10
rotate, Knowing that the unk undergoes a const

seclcration of 0.24 ms, determine the ei
{ston between the rink andthe tumtabl,

SOLUTION

Fist we ote that (a), = constan implies uniformly accelerated motion

my Oat

Mu EN
rem Fonte) rm _
‘Then Fm) Hag
far
sey 20 ve
‘Then mag mp lla ia |
amarre)
1 IA ZETA

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nen sches dsc peed br sot MB bn ca pnp Ppa Hmm,
worn up

PROBLEM 12.61

the paralleklink mechanism ABCD is
cd lo tarsport «component J between
‘manufacturing processes at stations FF,
and G by picking it up a a station wen
= 0 and depositing ita the next tation
sen = 180". Knowing that member
BC remains horizontal throughout is
motion and that links AB and CD voate
at a constant rate in a vertial plano in
uch a way that vp = 22 es, determine
(the minimum valu ofthe eneliient
of ati friction bere Ihe component
and UC if the component is not 1 slide
‘on BC while being uansfencd, (2) the
‘ales of for which ling impending.

SOLUTION

Now

and fr the component not to slide

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‘peel arated any a ra ma rar mien pos he ll, go ed

‘Bont ndo une pony err ina ocean oan ase

PROBLEM 12.61 (Continued)

Th

maine te above exp

We munt determi the valu of 8 whi

o

QE Tue)
a
: men
Bone
Now sine: Ces 180373

A-1008 und 6160609

(2) From above,

er à
das vine noe À
sna w

eine
[77

= tont03915°

er med

(0) We hase impending motion
tothe eft for
tothe right for

9-10 4
9106 4

PROPRIETARY MATERIAL. 248 Ye Nin Mi Cs, nA gs oor a pr fe Ml he ge
Te dect ition Meh sence cc Dn do Ma

Se por

PROBLEM 12.62

Knowing that the coulis of tition hetwcen the component / and member BC ofthe mechanism of
Problem 1261 ave 7, 0,35 and 44, =025, determine (a) the maximum allowable constant speed if te

componen iat to eon BC we sing sonas, le val 0 o rich lg poi
SOLUTION
cones
Ple >
Now
nd compo nt slide
or WA cosas
vr
> hn 1
ER + y nd q

nor equal tothe minimu value

Foca tat hime esti (2) mnt be
of pes 0 3 0) which acs wha ond, sn.) in Pe

a
amos pin 8) ai 041,008 0-0
oO pin)

| or oz,

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PROBLEM 12.62 (Continued)

(9 Te manimum aloe value y, Sethe

(eM na a er ten

fr
em

10

or heu = 29804 À

QD) Fist note thet or 90" = 180% Ba, (1) becomes

ep.

hn E —
Warme

‘where =180° 8. then flows thatthe second value of @ fr which motion i impending is
6180" 192900"
160.7100"

we have impending motion
‘othe for
Lo Me right for
Alerting solution
wem: WH Rm,
Then

Tor impending motion, $9. Also, as shown above, the values of 0 fr which motion is impending

shi lit sa ei = Roses igi mo

‘hat ay is minimum when ma, and Rare perpendicular.

PROPRIETARY MATERIA. 0307 We Sco Cenpan A I wc pa a Maa be en,
a nm yet eet enn ar wld

PROBLEM 12.62 (Continued)

hse,
om te diam
gr %
wand bs
ai $“
# u)
for 9002180, ue ave
from he ig
PS
ont
a Weeping (nahme)

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PROBLEM 12.63

In the eathode-ay tube shown, eletons emitted by the cathode and
atraced by the anede puss though a anal ole in the anode and
thom travel in a tight Tine with a spol nl hey sie the
screen at A. However ifs difference of potential i established
beten the to pale plates, the elctrns willbe subjected fo à
fence F pepondieute tthe plates while hey taal bescen the
plates and wil sie the sercen st Point B, which sat distance 3
from 4. The magnitude ofthe force Wis P=eY/d, where eis the
huge of an electron and ithe dance between the plats.
Derive an expression for the deflection in terms of Y, 4, the
‘barge e and the mass m ofan electron, and the dimensions €,
and

SOLUTION
Consider the motion of one cectron, For the horizontal motion, let x = a the eft edge ofthe pate
and à = E atte ght edge ofthe plate. AU he seren,

,
ante

Horizoial motions There men horizontal Forces ating on he electro tha, 0,

Let 4,0 when the clctron pases the left edge ofthe plate,
‘when it impact onthe seen, For uniform horizontal mation,

Y when de pases the ight edge and (1,

D

so at

ani

Vertical motion The gravity ore ating onthe electro is neglected since we are interested in th deflection
produce bythe electric free, While the eccrun ace pate (05/1) the venia! res om the
electron is Fy eid. Aer it pases the plates (4/1), izo.

For 0=1= hy

se
Ind

raptor la poo

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PROBLEM 12.63 (Continued) |

M a yal
ml aa

Por SIS t a,

Ia

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PROBLEM 12.64

In Problem 12.63, determine the smallest allowable value of the
ratio de in ers Of e, m vy and if at xf, the minimum
permise distance betwen the path of the clés and the
postie plate is 0.054.

PROBLEM 12.63 Inthe choeay tube shown, clectrons emited
by the cathode and atsced by the anode passthrough a small ble
in the anode and then travel in a saigne with a speed y, until
they strike the sereon À Monsever, a diffrence of potential Yi
estolished Between the (mo parallel plates, the electrons will be
‘ubjeted 10 e force F perpendicular Lo the plates wile they travel
‘ete the pito and wil sre he soon at Poin A, which in ata
distance 3 from 4. The magntide of the fore Fis F=ea,
‘where e is the charge of an electron and dis the distance between
the plates Nepketing he ci of gig, drive an expression for
the deflection 6 in tens of Y, m, te charge and the mass m of
an election, and the dimensions, and

SOLUTION

CCamsier the mation of ne cectron Fa the horizontal motion, let = atthe el edge ofthe plate
amd X= atthe right edge ofthe plate. At the screen,

Horizon motion: There reno horizontal forces ating on te electron so that =

Let 4 =0 when the election passes the left coge of the plate, #24, when it passes the right edge, nd 1

o
i tfo
ot mitad E

ES
Vertical maton: The avi, foros acting on the el is neglestod since we ae interested in the deletion
produced by the electric farce, Whi ib electron ds wen de plates (051: 4), th vertical fee on the
lei is F, = eV. Afri passes the plates (y © #43), itis em,

|
|

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‘peel or doy fom up ana mkt rer en ern he pao, ed gon td
Sere cei pale Shea te pan Fc ase th

PROBLEM 12.64 (Continued)

hu
sota case

Bo tet tanh

Ben

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nn nde tados proc ce risks ce ange Ma

= =

SOLUTION

PROBLEM 12.65

‘The current model fa cahodosay tube to be mee ss she
eng ft tube and he spacing betwcen the platos are reduce by
40 percent ad 20 pere, reve. the sze oF the screen 110
remain the some, determino he new Ing © of ho pas, sunning
that all of the other characterises of the tube are to remain
change. (See Problem 12.63 for à description of à eathode ray
tube)

PROBLEM 12.63 tn ie chodcay tube shown, elctons emited
by tho cathode and atracted by the ana ps through a srl hole
in the anode and thea travel in a straight ine with speed va uni
they strike Ihe wren al À. Honor il difference of potential Ys
established Between the Iwo parallel plates, the electrns will be
subject toa force E perpendieua 10 the plates wile they travel
tween dhe plas und will ike the sree at Pont, whic i at à
distance 6 from 4. The magnitud ofthe force Fis P= eV, whee
“e isthe charge of an electron and d isthe disance between the
plates. Negling dhe effets ul gravy, drive an expression for
the deflection din terms of 7 the Charge and he mans an
lec and the dimensions d À, and

Consider he motion of one clecron. For the horizontal mosion, let
and x= E ale cgh edge ofthe lao. At the sereen,

atthe lef edge of the plate

Horizon motion: There te no horizontal forces ating om te electron so tha, =D

Let 4, =0 when the electron passes he Left edge o the plas, #

sacs u te scren, For uni horizontal motion,

1 when it passes the igh eg, and 1

produced by the electric res. While the eloton is betwen the plates (0-57 7), the vota force on the
ecto ik F, =, Aor pases the plates (51 o) LI 270

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eon om mys wi pean ie eo i ed

For

model.

Since e Yo, and yy are constants

Bu

‘Then

PROBLEM 12.65 (Continued)

0)

826, LU L-O40L=04L,

d= d-0.201.

au

1

1-05
a

IS)

083

Let & £ Land db dimensions taken rom the eurent modo! and 8, ©, Land be hose othe modified

rained

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PROBLEM 12.66

Rod 04 routes about O in a horizontal plane. The motion of the
300 g collar is deis by the rolations y = 300 + 100 cos (0.520
and 8 até = 49, where rl express In miners, rin seconds,
and Oi radians, Determine the radial ad transverse compost oF
the force exerted on the collar when (a) 0, {= 038

SOLUTION
‘Rola eoordinates and thi dtivatsas.
F-03401 0m Oman
eos einsam II milh

Pos uns Be

(Forno, 04m 9-0
peo = Sr eats
Bes CES

Components o acceleration

arse aA
eas mst
E27]
O)
Sis eis?

‘omponents of fos.
1,0353

nay (032.513)
© Forr=05s. F=03101m 073927 rad
11107 ms Bears

037447008 danna

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otter poly ht MAA ec rin Pash

PROBLEM 12.66 (Continued)

Components oF aceleran,

Component o fore

haan 4

palin 4

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PROBLEM 12.67

or the ation defined in Problem 12.66, determine the ial
and tansvese components ofthe force exerted on the collar
when = 18

PROBLEM 12.66. Rod 04 rotas about O in a horizontal
plane. The moron of the 300 g collr x define hy 0

fons 300 + 100 cos (0.5) and D = 217 = 30, where r
is express in millimeters, ¢ in second, and @ in radians
Detennine the rail and Irse components of the nec
sxered on the collar when (o)#=0, (0)? = 055,

SOLUTION
Polar coito and ir rites.

03 101000 Bo

Fa -OUSrsin(0Sa)ms AI ds
F2 0257 cos(0.520) me

Hotels 2022929
011107 mis

2037687 más

Componente acceleration
rr 017447 mis?
erh io
Kama =(0:390.17447) SN 4
Fy= may ODIAN Komma 4

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PROBLEM 12.68

Rod O4 oscillates about O ina horizontal plane, The motion ofthe Sb
collar de defined by the relations = Ike +) and 8 = (2 sn ar.
where ris expressed in ft, in seconds, and in nllns Determine the
radial and answers components ofthe once exerted onthe colar when
CIA

SOLUTION
so Eu
Weine rot, (2 sina)
1 zu (sens
Then Ws boss
a Le
20 a
and Fe ns A
os »
(0 20
TT
OT
Now ani
0.1615) -0 M2 ay)?
ET
and wer
AC Me 2 a)
16m
Final ma
sts
LT
ETS :
homey
st a
= nés
Ten a
or Road

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PROBLEM 12.68 (Continued)

CRE ven
7010 Bez
Foon ne ö=0
Now (002 Ris )= (LA ra
and AS
Finally Rama,
ae
mane
or Bond
o. PT

PROBLEM 12.69

A collar of mass m slides en the eetioness arm 44, The arm
is atcheg to um D and rotas about O in horizontal plane at
te rate Aer, where ei à constr Ax he arm drum asembly
Tote, a meckanisin within the drum winds in the core so that
the collar moves toward O with a constant speed. Knovin that
at 4 0, r= y express as a funcion OÙ, c, Ka, and 4
(@ ie tension Tin the cord, (3) the magnitude of the hovizota
force Q exert un By em ddl.

SOLUTION
‘Kinematics

We have:
Arr ni

Año,

Now

and

Kinetics
(4/38 me,
(0) PER Ama

are
aan?
Ane

erh
fe BK) + hen)

oy ye
= =) NN wy

=

Tem À

ten 9]
Damen -9 4

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PROBLEM 12.70

“The 3: colar B sides on the fitness arm AA’. The am is
tached to dram amd rotates about O in a hocizmal plane a the
fate 00,25%, where © and rare expressed in ad and seconds,
respectively, Ax the an drum assembly roles, a rechanism
Sin th drum releases condo thatthe collar moves outa
Fini © with a constant speed of 05 mi Kong tht at 1 0.
12 0, determine the tne at which the tension inthe cord qual
te the magie ofthe brio force exerted on by arm Aa

SOLUTION
Kinetic
Wehare
Arr 0: [ae [ose
= reason
A 0 d= (asin
Bes

an —rÓ 0-10 ms0.759 eas? =—(0.28128 mb
Pr Br

1259 08?
inte

mu 27 = OX 0281250) er
« OCR WO

hax 1259 04? 7

“e 0=0375)N
Now equi hat ro
« (0387s) N=375N
7 Pao

ar 122005 4

rare een. oe Ce be io pe oe
ar en. te Gents sen Pte Mo CS
BEE ita AS

PROBLEM 12.71

‘The 100-2 pin slides slong the lot in the rotating em OC
and along the slot DE which I cut in a fixed horizontal pats

Neglecng Mein and Knowing, tht rod OC rotates at the
constant rate 9, = 12 ral, determine for any ven vale of 9
(a) the null and traer components ofthe reliant lore F
‘exerted on pin 8, (2) the forces P anal Q exited on pin & by
nd OC andthe wall sll DE respectively,

SOLUTION
incmaties
Fram he rowing ofthe system, we have
we |
wo
sno |
‘Then 02220 om 2"
(ajo. dr
and
203 MAN ~sin 200580) q»
CS
Sah

2 fa

ea)
Lie) ye

Now

je

#6) _{ 02
7

(oss!
alas lts
{ (cosa

sin?) yee
| DOTE
(me)

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and

Kinetics
@ Weine
and

(6) Now

PROBLEM 12.71 (Continued)

sin’) go
remota rst

rama

Fy = (576 N) tan 95008 À

PR NV. ee

B
P=S.26 and sec +(5.16 tn sec B) am 0

5.16 tan Osco NO
+f ER: FE =Q0os8

0-626 0

Q= 6.76 Nan de A

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PROBLEM 12.72"

Sls C has» weight of 0.5 Ib and may move in slot eut in arm
AR, which rotates at he consten te A, = 10 rule in a horizontal
plane. The slider auchad o a spring of constant & = 2. IR,
Which is unsriched when r = 0, Knowing that the slider ©
released rom rest with 0 radial vo in he position > 18 in
nd neglecting tito, determine or he position r= 12 in. (a) the |
radial and transverse components of the velocity of the slider,
(6) the radial and transverse components ite secteation (© the
horizontal fore excel onthe slider by arm AB

Data

| SOLUTION

"tu hs the adi coondinate when the spring is unstretched, Force exerted bythe spring.
Beier
ER um: Kremer)
Gr {0)
Bu

erate using the cond

ast
122 mu
=I ds, k=2.5 IAN, de
(=O, ¡IB MISA r=I2inm=1ON

Dat: = 001555 m8

PROPRIETARY MATERIAL. © 200 he creel Camp, I, A igh ch No or of hs Rowe mp e ned
‘pec ia as fo ar y any mnt he rnin pr pln or ed od hd
Beton ecto at nn MJ todo ene sema Hoh

PROBLEM 12.72" (Continued)

(2) Components of slsty when r= 128,
» 25

eo-1sn«o

76.223
WERT

Since rs decreasing,» is negative

i= roc
PET ETUI)

9 Conpnensatacletin
Te
ass
TES

ou =rh + 25-04 NATI

o ponent of
y= meg = (O01SSIK1756)

urn 4

ve 1000s €

610 ae

mo a

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PROBLEM 12.73"

Solve Problem 12.72, assuming that the spring i uneretehel when
slider Cs located 2 in. o the left ofthe midpoin O of arm AB

er

PROBLEM 12.72 Slider C has a weight of Ib and may move in
ot cu in arm AB, which states a he constant ae À = 10 ra
in a horizontal pane. The sider ls attached Lo à spring uf eons
4 = 25 To, whic is unsreiched wien y — 0. Knowing tht the
slider is asc with no radial velocity inthe position = 18
nd negating Fiction, determine or Ih position r= 12 (0) the
radial and transverse components of the velocity of the slider,
(6) the reia and wansverse components of is ecclraion,() the
"viral fore exerted um he

SOLUTION
Let be era coordinate when the spring i umsrechod, Force exerted bythe spring.
B= Alek)
at rem)
(#-4)-4 w

but

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PROBLEM 12.73" (Continued)

w__0sm >
mE SM ous to

> g Mans
bslomds k= 25/8, 4 =0-2 00
TITO neltineish, r=I2in=L0R

(a) Components of velocity when = 12 in,

1
s as;
01 ¿jar Sao
ass) ES
D
=ro210.515
Since ris dereasing, , ie negative
assi 10.1586 4
AT) = 1000 À

(0)

O

“hr ly

2916
ES

IO ISO

PET
(Manswers somponsnt of force,
as 4

1 = (00155342030)

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Teer enel nyt hy te ie os tn ers of he lee Bo ee
‘inn stewed: ony Hoe Maa pas con e Mn

PROBLEM 12.74

A particle of mass m i proc rom Point A with an initial velocity o
perpendiculor o line O4 and moves under a central force Y along à
semiicular path of diameter 04, Observing that r= exe mul u
Lg (12.27), show ha he spec othe pate i» = cas.

SOLUTION
Since the parle moves under a cent fcc, constan.

Using Ey, (1227), her

Radial componsnt of velocity.

a
0)
Zn

‘Transvene componsaL of velocity.
ver = (cost

07

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‘vn of ned np ar ani ih an er pao, tt he
‘een ody ios Meee se rosa

PROBLEM 12.75

path ofthe patil for (a) 9=0, (0) 0245,

SOLUTION

Since the particle moves under central fore, = constant
Usine Ba. 12.27).

asia! compos

‘Tamvene component of veloc.
Dee Eee)

Sneed

Tangentia component of acceleration.

Langer component of force

Any sin 45
arg

For the particle of Problem 12.74, determine the lagen
somponent ofthe central force F along the tangent lo Ihe

N

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PROBLEM 12.76

A particle of mass mi projet fim point A with intl velocity ve
perpendicular to line 04 and moves under a senal fives F directed
vay from the cent of force O, Knowing tht e pue follows pal
defined by the equation = Jcos28 and using Fa. (1227) express
the radial and transverse components ofthe velocity ¥ othe pte as
fasion of 8.

SOLUTION

Since the price aves under ce

Using Fa, (1227)

Radil component of velocity,

, día Ve, ce
ao? tal santa aa
rn) mona ae
eur Kr
Transen component of velo
+= RTE 4

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PROBLEM 12.77

or the particle of Problem 12.76, show (a) thatthe velocity of the
particle and the cental force Fare proportional to the distance y from
the particle 0 the center of force O, (6) hat the rus of cutre al
the ath i proportional 1

PROBLEM 12.26 A partici of mass mis projected from Point with
an int velocity ve perpendicular t ine 04 and moves under a ect
force R dected away Kom the center of force O. Knowing tha the
pate follows a path defined by the equation r= /Yeoe20 and using.
Ea, (12.27, express the radial and tansvers components ofthe velocity
ofthe particle as uch al 8

SOLUTION
Since the parie moves under acetal fore, 4 constant,
Ung a (1227),
heröchenn ur Oele np
irn the expression fr vih respeto ins,
TE) 5020
a a

LA
<7

* c0s20= whe

Ditterentating an,

af, iw, 2008 26 sin 28g vo 26007204 sin? 20
ARS one al

nern ian

TO Are}

ae cree 260 2020
nes

Luce 20460720,

Due

mir nat

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PROBLEM 12.77 (Continued)

e paricl moves under a enual foe, dy =0.
Magnitude of assertion,

an ra ae

Tangs component of assertion.
war
aaa

‘Nona component of acceleration,

Du
eves,

x Pe

CET er

P ma E +

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PROBLEM 12.78

“the raus ofthe orbit of a moon of a given planet is equal to twice the radis ofthat plane. Denoting
by the mean density ofthe plane, show that the time required by the moon to complete on fll revolution
about the planet is (24 2169), where G's the constant of gravitation.

‘SOLUTION
For rota force und crea bit,
Gin m

E
E
2
B
t
2

Solving for

Rp, wens,

al

Tin Ed
Go

| Using r= 28a given eas to

| sm E. BE

BE - 2AxiGpy" À

PROPRES MATERIAL. © 209 the SS Compt e. Ar WIE Mo ur M o
nc indie pod Moe Milf nd ase ono ceo MR oe

PROBLEM 12.79

‘Show thatthe rive of dhe bit of moon ofa given plant cn be determin from the radis of the
planet, the acceleration of gravity at the surface ofthe planes, and Te time Srequired by the mon to complete
one full evolution about the planet. Determine the scccloraion of gravity tthe surface ofthe planet Jit
owing that & = 71,492 km ara that = 3551 days and r= 670.0 x 10 km for is moon Europe!

SOLUTION

Weine roda (64.12.29)
and
Then chan
« unit
Now cnt gt? (4.42301
so that Po rf
der
Yor one ob, a
Nr
or (EE) omo «
Sing fr, ante
eng hat £ 2.551 days = 206406 5 then
de? fie
A
a mo
CNET
= Eng An 4

Nites Kia “2538,

PROPRIETARY MATEXIAL 200 THe Merl Compro be, A gs see No pr of a! ee
‘rhc or rated ps fem i y mes, meer rs pan Y ar, te eed a ad
‘Mon ren pen Uo if taba ane ym Dorta rez
ewe il pon

PROBLEM 12.80

| communication stellte are placed in a gecsynchronous obit i, in a circular orbit such that they completo
| one fll revolution about the cath in une sidreal day (23934 1), nd thas appear stationary with respect to

the ground. Determine (a) the altitude of these satellites above te surface ofthe earth, (9) Ihe velocity with
‘which they describe het ori. Give the answers in both SI and US. customary ais

[ souunon |

For ravie fre and à cie nit,

LE

Let + bei period time to complet one orbit

Bu weder or
wen AT

ar
Da: DETTE
@ St 7981 is, RG

Git = gi? OS? = 39806210" me?
AA
aude he r= 35750 m 123.00 1m €

mu. 82322, R 23960 mi = 2090010 n

GA = gt? 2220090" 14077410 WA?

a rs
aide hr 1222004

PROPRIETARY MATERIAL. © 200 The Mas Compete, In. A Ht Tos. prof is Mal my e dep
Ipoh a el nr rm bay mc i hae pario yal, add e id
‘Bo arenas ety Sotho hn rere a no Sad

PROBLEM 12.80 (Continued)

Ostet
A
esca

iif _ flagrrsad

A ee

S07 kn 4

IO nd

[PROPRIETARY MATERIA. © 200 Vie cio Hil Campa, eA ih oe poof is Maa may Be pe

‘oda el my ot rc, i he wes eo

‘tenis enr panel Ben Half iba a ian Hata Me

PROBLEM 12.81

Determine the miss ofthe earth, Knowing thal the mean rus of the moon's obit about the ar 238,910 mi
and thatthe sun requires 2732 days to complete one fll evolution about he car

SOLUTION

We have:

and

Then

Now

sota

Noting that ays = 2360410

and = DR Om 1261410"

' on
wwe ave Me 21a
' aorta FATAL

or MAA O À

PROPRIETARY MAFIA. © 200 The Mat Cope, lu. oe Ne oof M may e egg
Feen ar ret ny fan 7 o mi ioe tar tes ponen oe bin oe hd la
Rot eden potty tine fennel ro pes ne alos he Med

PROBLEM 12.82

A spaoccaf is placed nt a pole obit abo the planet Mars at an aude of 380 iw. Knowing ha he
ean ey of Mars 2.94 Mg! and that the radi of Mar 3397 kn, determine (a) he timo required
for the spacecraft to complete one fll revolution about Mars, (9) the velocity with which the soccer
describe orbi.

SOLUTION

(Frown the solution to Problem 1278, we have

ON y

“Veo <
here 12 Ra (8397 3) kn
3777 km
Mi at arr
Then a (3,
101955
L Eh 7 min €
9 Wehe 2
LRO
Tos.
or ya 380.0% 4

‘eave att y fora by oy atthe phonon pr of i uba sal on eed
‘Srna oo amén oido Mea if cc cc [rec ag Mec
pun ton poa

ne Comping tA ih oe Se pa e Aa! Be dpa

PROBLEM 12.83

A sel 5 pe in a cer ori aout the plant St at an ale of 2100 mi The satelite
series so wäh a velocity of 8710" má. Kooning ste ai ofthe i abo Sam athe
ende tie of Al, one of Suns moon, are 86.851 mi and 16012 day, respectively, detemine|
(0 the ran ol Sut, (3 the mane uf Str. (The peri ano a alle À ne tim eq to
ample one fll evolton abou the planet)

SOLUTION

elo of At.

where SSI min ASL SHOR

and fy W017 days 51,987
CAUSES, oy serio a

S987
' Mn

Sastre E

fom which

Forest

vere 54.710 mi = 80.267 107 Res

(ISSSTE KS4.S87>10°Y
War
39,600 má

208 84510"

(a) Radius of Saum,

(aide) 39,600 - 2100

Mago Sauer,
LSLGSD NSA STO
EXT

PA

PROPRIETARY MATERIAL © 200 The NG Cas e A ih sted. po in Ma war dope
(red or ded so Do a on omen ps he llo, a! ed i
‘Sin nda en a apt ah

PROBLEM 12.84

‘The peta mes (ee Problem 12.83) ofthe planet Uranus” moon Sulit and Tiana ave han observe
be 0493! days and $706 day, respectively. Knowing thal the ads of let's orbits 64,360 kn, determine
(a) she mass of Uranus, (4) the radis of Tania abi.

‘SOLUTION
Velocity of als De

where 13 = 64360 km 26436410 m
‘and 1, = 0931 days 42,6085

9.490710" ms

Gravitational foxes,

Frm which Gat are? constant
DA

(@) Mass of Urans. at

AO
COTE
= ke

M =86.9%10" ig €

w

[2] mana

io)

36,000 km À

fp 248639108 m

PROPRIETARY MATERIAL. © 26 The Min Ca. igh rm. A por ef the Mama ma he dim,
‘oi se! may a Be wc oh or wien porn of he aio hein eed
‘Sein cider a bt bee pepe Hn ae

‘PROPRIETARY MATERIAL. © 209 The Mew Comps, o Al

PROBLEM 12.85

A 1200: spacer fist is placed into a cirela obit about the ar at an altitude of 2800 mi and then i
transfered 10 circular obi about the moon. Knunving ha the muss of the moon 0.03230 times the muss
ofthe earth und thatthe radius of the moon s 1060 mi, determine (a) he gravitational fore exerted on the
sueca si wa orbiting he th, () the required rau of the orbit othe spacecralt about he moon 4
the perio times (See Problem 12.83) of thet orbits are to be equ, (e the acceleration ol gravity Ihe
surface oF te moon.

SOLUTION

Firs mite 2, =3060 i
Men va = Re the (5960 2000)
= ami
(0) Webwe 1a 0229) $
ad CRE
Then F De)
af
” Fand
(6) From the station do Probe 128
Me
thea
x art _ ert o
= lost, "Joh
fanny’, da
or me se) y = (001230)!°(6760 mi)
nu 1560 mi 4

fa seno. a as a tn y ean,
‘elo nied ony fm or ey os mt ps fe pba at ha
‘Mein eats vn mt Ro Mi ain a Ml

PROBLEM 12.85 (Continued)

0 Wehe GM gh? (8402299)

Subst imo Eu (2)

EEE
BEL
Rede Bulow

ing te esas of Par (8). Then

or SS]

1
Note Some PER

[PROPRIGTARY MATRA, GS The Meike Cayo, A ome Se tof is Mal u be ip
‘seni es addons race I hembra roe opin rar Mea

PROBLEM 12.86

“To place a communications satellite into a gosyachronous
ri {se Problem 12.80) at an alte of 22,240 mi above
the he satelite Fis leased from a
space shut, which is in a circular ob tan aude of
1 thon is propelled by an upper ste booster to
tk. As the satelite passes through A, the
is fd 6 insert satelite nto a ela
transfer orbit The Booster is again fied at 0 insert
into gangrenous oil. Knowing that
the second fing increases the speed of the suelte by
4810 Ma, determine (a) the speed of the satelite as 1
"approaches A om Ih elite nur orbit (8) he nenne
in speed resting rom the rt ring at

SOLUTION
Far eat, R 3960 i= 20.90% 10 R
SR = (02.2920. 900 14077108 0
23080 RS = 1 SEE
y= 3960+ 22,240 26200 mi =135:36x10'N
Soe sila obit tn À
if
Cou» [EE
E E CT
HET

=25302x10° ts

Sesso sinclar oobi show &
ES

Colas =

AT
DETTE
nern

PROPRIETARY MATERIA. © 2 Yo Min Cac, At mel pr of ha Maa mar be Ay
Font ala y et pt u on of eps oh at
doo ono ot nd oni aa ag ae

PROBLEM 12.86 (Continued)

(2) Speed on anse tajsctony at
Oph = 100810 810
sano sox is 4
Conservation of angola momentum u anse mir
TON
rl
CAROL NES)
28RS6x 10
23336210 fa

| © Clon nape a

rs
3330 —

00010

36210

ar = 8000 is 4

PROPRIEDIRY Marta. © 200 he Mans Comes be, gh Rh oe of i Ml ray Ke ep
‘vec oust ins fi hy mena pr we pon of he lobe re oa tt
ene enmity aioe pen coto Va

PROBLEM 12.87

A space vehicle de ju a tear obit of 2200 radius around the
moon. To transfer it toa smaller icular orbit of 2080-km radis, the
etico is fst placed on an elliptic path AB by reducing is sped by
36.3 rvs nl use Ich 4. Knowing that he mass o moon is
7549 x10° kg, dermine (a) the speed othe vehicle A pprvecis A
on the elite path, (9) the amount by which is speed should be
sd spp A inet na the salle circular bi

SOLUTION

Fora circular ori

Ha, (1228)

Then

Conservation of angular moment

(0) Now

e
Demy: Pome
aot

ant

79:10" hy

axé
19301
(u. SEO Ag 73.4910" kp
2080105
Gala #15555 a8
(hn Cae Me
214930 -263) ws
Am
rires tat
a = Fa dr
ee Rx 4657 wis
15913008
Cd = 1551 més 4
dias aden + 8e

Avy = QS385-1551.3) mv
Avy 155 ms 4

2909 Te Mi Coen, A ee of a Mee ma Be tia

or et a fr gam aan Orr nie peris Fe el ed
‘Zines neon ern pres Mr bier pen. Bp oar gah a

PROBLEM 12.88

: Plas fr an nme nin sn onthe plane Mars aid
atada: Tor the curib-return vehicle to first describe a circular orbit at an
Ali y= 220 hy aves Sl of te et it

Vals) of 277 mi A a mg Pol A I ele
/ ee: vn be ine ia ane tr ori by ie te
6h Sie and merca ls speed by Bry 1040 Ak posal

HF wen pin Bt tion = 10,0 km, he whe ve o
te meted io a second wane eit located im ality
diferent pane, by changing the diestion ol is velociy and
reducing ts sped by Ang —-220me Fly, as the vehicle

nettoberai pas ong Pot Caan atado de = 1000 Ko, io speed

o ‘vas to de increased by v= 60 10 inst tin som

uso, Knowing tat the mins of he planet Mars i

3200 km, detemnine tke velociy of the. vehicle alter

completion oF the last maneuver

La = M04 2200 = 5600 km = 5.602.108
Ex = 14004 100,00 = 105,400 kn = 10 4x uf
Fe = M00 1100-4600 an 4.405100

Fis ass ti

ae

Im + 1046 mé 817 ms A

Cameron ang momen: 4,
Amon at
SANA
ve 2067
Socom oc à
201 z
2067-20-18 me € »
ion of sgl momentan: il
Tem 2 me ots
AX 0897) (40210)
venti
Wr 0660 SID <

PROPRISTARY MATERIA © 200 he Mute Cana, A gt OSD Nef ht Haul e yet
“ue so na eB rg m,n er te ptf th yb ged e al
nas más Woon fr kd pre Poca a Doma

PROBLEM 12.89

‘order for the shut o recover the satelite, th sulle fs plac

Ny À À iin stints ath BC by nensing ts spe by Ara 28D a
> passes ihruugh À, As sto apronchen Cha apo is increased

\ A pus shale Sl tli 4 ral ai shown,

by Axe = 26014 to ner ino a second elit tamer bi CP.

À a Knowing thatthe distance from O lo Cs 4289 mu determino the
ef amount by which he spe ofthe shut souk eines as
= Approaches Dto insert mio the cule obit ofthe satel

IN

| SOLUTION

Fistnote

sa

Fora circular orbit,

Erg
An
pu O
por Ca 2S is
Li PR 2
TZ
« Lau si
Werne Pde “ac + By" O57 4280)
5.657

PROPRIETARY MATERIA. © 299 io Mo Cops. A ai ae po of an wap Be pine,
‘rhc rn nay oth yn, ha Be an anf ya, rn Sea ed
ig os ds mts Mtr de en rn por

PROBLEM 12.89 (Continued)

Conservation o angular momentum requires hat
BC: rym meng Fem Worn

AM,

From bg.)

Now

0m

766 Ris

Ning "don. + = (4, 166260) MU

25,0260

4289 mi
Sim

(ae = Cod + Aa

= (24,785 ~24, 732) Vs

425,657 Ms

25,026 Ms

o
o

Avy 53M À

[PROPRIETARY MATERIAL. © 229 Te Mei Comp Ic hs some N of te Moly bp
‘wc or ed rer a eo ithe te rien poo palermo aed
‘Bintan errant pod Iie i orp marcada

a ares

PROBLEM 12.90

A ab collar cn slide on rita ra which à Fre to tte
about a vorical shal. The colar de nally held st by a coed
tache 10 the shall A spring of constant 2 tht is atache tothe
Collar and to the ssf nd i uni when the ella in at À. AS
the fod rotates atthe rate 016 ra, the cord is cu and the ell
moves out along the rod. Neglecting fiction and the mass ofthe
od, determine (a) the mia! and inwene component of the
sceelertion of the collar at 4, (2) the acceleration of the collar
relative 6 the rod at, () the tassverse component ofthe velocity
ofthe coll

‘SOLUTION
Firstnote

(B= Oandat 4

CRE

e

to), =04
Wed

Noting that suis =, WehEVE ALA

(Gin 16nd}]

or Camarada = 153614

(©) Aer he word sc, the only hcizomtal fee acting on the collar ia due to the spring. "Ths, angular
moment about the shaft is conserved,

Then

SEEN
oa ing u he perio “

Dome where oe

cone end

Ch 2.0106 €

PROBLEM 12.91

For the collar of Problem 12.90, assuming thatthe od niall rotes at the rate 9 12 di, tering for
position B ofthe collar (o) the transverse component of he veloiey ofthe cola (2) dh rail und vansverse
‘components oF ts acceleration, (e) acceleration of he Cola relative lo ud

‘SOLUTION
Fist we note Baker)

ms
au di À = |

in
rin
=21b

(2) Ar the cord is cu, the only horizomal fore acting onthe collars due to the spring Thu
moment abou the shal is conserved

runde rato de

ete san,

Then (ede (Gi KZ mst

a sein 4
© Weine a0 4

Now

af (ART,
© Wetwe a

Now Ga

Then at Grade =

or Can 2226 ii €

gyn marl. © Gen em Mp fn ti
(en nnd permed ran Paris bse rg pn De,

PROBLEM 12.92

A 200g ball A and » 400-2 ball B are mounted on a
Horizontal rod which rotates recly about a vertical shaft. The
bulls ae held in th positions shown by pins. The pin hon
‘ is suddenly remove and the hall moves to postion Cas the
fod rotates, Neglecting friction andthe muss of the rod and
owing thatthe initial speed of À is +, = 25m, determi
(a) the radial amd Imravene components ofthe acceletion
fof ball 8 immediately afer the pin is removed, (0) the
‘osslration of bal relative to the rod at that instant, (0) the
speed of al fier hal as reached the stop at C

SOLUTION

[tr and @ be pola coordinates withthe origin ing at esha
ne

Constant of od: By =0, + radins: Os

(a) Companens of aceceraton
‘Sketch the fio body diagrams of hc balls showing th radi and
transverse components ofthe forces acting on them. Owing o.
frictionless sliding of along the rod, (Fy
Radial component of acceleration of

E mala), (ay), 04

Tiansverse components faces

perd à=reû

CEE o

‘Since the od is masles, it st be in equilibrium. Draw is
five body diagram, applying Newton's Third Law.

SEM, <0: Edo tral

0 datum aa =0

ea rem + 20

ed

Atr-0. 70 sota 620
rom Le (D. an

PROPRIETARY MATERIAL. 00 Vi Niro Congas, os. AI em. ar fh Mal o esa
pra dol for sb so ao id ren prin of hc lao, ed os ee
a hrs endo peras Wf hr edn coe prats pr als o,

Beer dinamica 9na solucionario 12

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