Beer dinamica 9e_manual_de_soluciones_c11a

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About This Presentation

dinamica beer 9 edicion solucionario


Slide Content

|

7 7 u ]
PROBLEM 11.1
‘The motion ofa partie defined by the relation 1.5 = 306 + 4410, where and are expressed in

etes and sols, respectively. Determine the poston, the veliy, and the acceleration of te patel
when 248

SOLUTION

Give or es

Evaluate expressions at 24 5

KEN? ES +10 -66 05 1-60 4

AY 6000-25 =149 més NU]

Ia) 60-228 mu Eon 4

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‘apr rtd yf Ban ine ro wrap sar, mat von ee
Senn tnd thy Sth Hg ht Bane pte a ed

PROBLEM 11.2

“The motion of apatite is define by the relation = 127° 180421 5, where and fare expressed in
‘meters nd condo, espestively. Determine the postion and he velocity when he acceleration of the perte
‘sequal o zer.

SOLUTION
Given: O
Find the timo fora=0.

Substitute nto above expressions.
AOS) 180592 +2103) +5=3 x=300m 4

2603) 36408) 2
rm

mas |

OPE MATERIAL 030 Ts hi Cm, Apr os Mal wy be pen
PT daran 0 Arm y nr te he ir wie pum ne py e rd hdd nt
mi afer mtm oc man Mok

PROBLEM 11.3

“The mation ofa particle is defined by the relation 308, where and aro espresed in

5 2
‘et und seconds, respectively Determine tb time he post, and the acceleration when v=

SOLUTION

AA

We have:
2

Then ve Kase 51-30
a
and

When v=o:

= IS pe a
Aum3e jor Lor es

atom or

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‘cl dnd e e fa ty u ma, ml ps pri se e
‘Bother omer pod ten Mfr bi pe ko lé
ovo ng wn pon

PROBLEM 11.4

he motion of a patch is defined by the relation x = 6/8 + 4004, where x and are expressed in
inches and seconds, respectively. Determine the position, the velocity and the acceleration when /=65.

E u a

SOLUTION

We have x26? 8480081

Then Hata ane sin ae

and ae À 12-40 cost

a

Auta6s 5-0 8 rind
va = 126) 402 sin 6 a nina 4
a 12-40 en Ri

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ae Aha Are yy cae, a Dr pis fe pla, o od Rod de id
‘Snare poh en ne

PROBLEM 11.5

‘The motion ofa particle is dined by the elation x=0Y 2 12 +3143, where x and rate expressed in
ters and second, respectively. Determine the tine, the position, andthe velocity when «0.

SOLUTION

an
" ven
a
28) 8] (Qu coin 4
2Ÿ _{2Ÿ 2
Ge ross 4

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en acer und ers porated Me fr me pp Je ln e Mase.
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PROBLEM 11.6

“The motion of a particle is defied by the relation = 26 187° + 2är+d, where is expressed in meters
and in seconds. Determine (a) when the veoct is zor, (8) he psiton and the foal distance aveled when
Uo acceleration is ero.

‘SOLUTION
Pots aries
gg
des Or
Boop an

a 1-0
a
o 62 3002

arm
© a=0 sen 120-0

00% <

0x or

Ss

209"

SOS) 42425114

find otal distance tod, veto hat

EA ASF ar

For 120.

Diane traveled

From #=0t0¢=1s:

From ¢=tstor

ss nes seine

‘otal distance ave 13.5 2245 m «

PROPRIETARY MATERIAL © 200 the Mes Caps, eI ae pr of Mama o dene
Iu ha ny ey ao ina, on rein Par Flo af yd
TARDE cmon! uso nena one i ceca open Ppt mbr Se,
pm

PROBLEM 11.7

‘The motion of a particle i dena by the relation x=»
| and seconds, respectively. Determine a) hen the vel
Total distncetaveled when x =0,

61-40, where x amd rare expressed in feet
iy is zero, (6) the velocity, the acceleration, and the

SOLUTION

Weave

Then

aná

@ When you,

ar ET
a 25 (Rejet and 12600 4
6) When x Pat 360-400
Freeing Etam 20 or
Nowobsenethat Dean vO
sscrsios v>0
dat 20 27-00
(268: (6) (GP ~36(6) 40.
2560
DT do =1200)=36
wo
EN ES
Then In 6-4 216 8
Fy Fo 0250) 2256
Total dis tele (216 +256) 8472 «

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‘ep ov lista fb ny ose ern prin of yao, ed el i ha
Sein des ty oti en Ferma ag ee

PROBLEM 11.8

“The motion fa article is defined by the elation x=0" 978 +248, where x and fare expressed in inches
and seconds, respectively Determine (a) when the velocity is zero, (9) the poston and the tia distance
eave when the acceleration ero.

SOLUTION
We have
then
and
(When PART
er 209-0
or 122405 and
0) When a=0: Gino où as
Arras y ETT
Fin ohne that 0-128 wo
Peer
Now
amo wo
aunts MO) 124) -8= 12,
#1
Pt,
STE] oar te
o as GS
Then

“Total distance traveled =(20+ 2)

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er A am fre ey ny an ah tr mvs ere ft ble ered on ie nd
Te Ri MA RAS pr msc Ma

PROBLEM 11.9

hat 224 o when y
distance uaveled when smile.

“The acera ofa parle is defined by the relation a = ~8 ms”. Knowing hat x
16 mvs, determine (a) he timo when the velocity is ein, (1) he veloc wa Ub ot

20 m when 1=4 and

[soLurion
We hase

‘hen

Aso

Arr 4x2 20m

When v 216 4 à = 4 m:

Combining

Simplifying

and

(a) When »=0.
(6) Velocity and distance se LL,

on mi

ja frase c

A+ C (ms)

La ferro
2-20 par 4.0%,

42 + Cure 84000)
or

16-84 C2

40 400-8486
OA ONE 4180
41+4=0
Ce ms
vase 32(ms)
CET

“8220

MD
non
«20m

Su SAU +320n-44=-176m

A]

sion 4

nn et ps ae ng a
EEE A Ri]

PROBLEM 11.9 (Continued)

Now observe tit

Then

otal distance taveled=(64-+196) m = 260 me «

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D hare ry fre ay m is swe pers pan a ed Nene Bed
‘inten te a map an Ma

PROBLEM 11.10

‘The acelraton ofa parucl i directly proportional 1 he square of the time. When £0, the pale is
st 5=24m. Knowing that at 36m and y= 18s, express x and io er of

SOLUTION
| Werne an aia
Now ke?
a mi: [o [ira

m QE)
zu an

Lo
or + = 216K)

Kunst
a Ceres
EE
Le
AmO mam - (Ji-tuo-
0 ae [fis jae-noja
“ perenne ee
ano tata)
Now
ıfı
126 296m no tal Mot
nenas molto]
0 ok me
5
yıyı
Then fie
EM alla 210)
“ et eimen «
| me
fi
mi vars (Eee
16) 2
1
e «

PROPRIETARY MATERIAL © 200 Te ene Conga, A ts cd a a his Mon yb tt
Grete ac naher tn RC ea deren Dann aoe

PROBLEM 11.11

“The acceleration of a parce is dice proportional to the tine £ At ¿=0, the velocity of the particle
is #216 ind, Knowing thet 915 in/s and that x= 20 in, when 171 1, determine the velocity, the postion,
std tet distanceteled when #27 5.

SOLUTION
Woh
Now
Are t.e=toins
or wei Lens)
7

avs 15 ind Sims 216 indo Zi”
o Din and vió
ao

saone

o [is
3

1 13
het an

Then

weisen? or y= B30 ins 4

US e 2-20 4

When v=0: or tess

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to ds yf ya nn me ts pra of te oe gon ee
sd pr Oi o e pr m sve nk

PROBLEM 11.11 (Continued)

Now observe that

Aal _
+ 5
Ne is
»
Io -21=12-471=45in.

Prone un. © Ye kn Crono, eM i pl Me rn
ended er ele ero Franc ant righ Mama

PROBLEM 11.12

The wecclration ofa parle ie dined by the elation aK, (a) Knowing that v=-32 Me when 4 = 0
and hat v=+32 (Vs when 14, determine the constant & (9) Waite the equations of motion, owing alse |
hat x=0 when r= 4s. |
wel

SOLUTION
©
©
3.000 4
© sin
an 4
em oe fee
De 3
ven wna
«e
san

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(nr dead ny roy ty rr an ft tk pres fdr or ra pend ed
nn rodean pr Meo Miler iba one pp aren unge Moo
Sooner wou prin

PROBLEM 11.13

| The accleaton of a parce is defined bythe reiiona = AGF, where À à constant. AU 1-0 the
parle stars a a= wi v0. Knowing thal at TA v= 1010, doin () ls me at wi the

[somos 7
| We have

ve o

A Gove fu cera

or v= Ar 20 (més)

Ati=15,v=30 mis: 30 01-207"

Also. e

= «

O Gin

RTE
125 ess — es 995

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[phen ar dt je rl e mo, te rr vts prin pala. red Noe ed
{rao eased eater Go fh india erat rem ata

PROBLEM 11.13 (Continued)

Now observe that

Ten

spurt oe yer Soe teres tes er]
a pa pa fae eg

PROBLEM 11.14

Tes know that ‘of patch is inversely proportional to the cube
ofthe me £ When 125, v=-15 m8, and when #=10 3, 90.36 mis. Knowing at the particle sie
35 far from the origin when #25 as i when 1=10s, determino (a) the postion of the pack
‘when 22 sand when (105, (9) the ttl distance we hy the particle Tun 4

SOLUTION

We have

Now

Ares va 15 a

AL /=10 5, v0.36 mé

a
@ Vene Pan:
In fa fear cso
or Luci
won easy veut)
“ cm
m Kerner

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‘rnb ha a frm by ny mon ine rrr lu, md a ad
‘Brenda poly tin Wf ip ne

PROBLEM 11.14 (Continuod)

avers EC « nome
“ Ñ
seine mure “nern 4

9 When ved.

arate:

Now observe hat 28<7< 88: yO
Secrets: vo

IR
ad num 25)
ES (25)
‘Then By fr 17.2~352)- 18m

os =I76-172=04m
Total distance towel = (18408) m 18.40 m «
Nove: The val distance waved isthe same for bo cases

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‘ern or hated fm ty ar i rr itn pm llo ol ne
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Dee

PROBLEM 11.15

‘he acceleration oF a patil i defined hy the raion hi, Khas been experimentally determined that
wets when +206 1 and that y=9 U6 when x=12 Determine (a) the velocity of te particle
‘when 2215, (9) the position ofthe parce at which ts velociy izo.

SOLUTION
ant,
ae
Separate and integrate using x = 0.6 N, y=15 Rés.
[pra
del = kins]

eo (ez o

doi fe
A Co]

; a
ec
aimee

© Sie BSR ant 13000.

Betas! mE

ess 4

o tere

ots)

"(as

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oval er ib sx former ty ma a fe is rn fe ple tg Id

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Sova aufero

PROBLEM 11.16
A parce sting fiom es at <1 fis accented so that it velocity doubles in manie between
À und À 80. Knight accion ofthe pale dein bythe relation aeäfe- ADI.
Akemi the vals of the ort 4 and ihe parte has a elo 029s when 167
SOLUTION
Weine
When af, v=0: fro [aras
o |
Lo amx-t)
pen aang)
1 E |
zar amz-4]=4(3 ue)
Le |
PAT er oe ee |
ws 1618-48) |
o |
|
“ Antes 4
1 Lao 285 |
Yon]
‘Noting that In(16)=41 |
mass |
Webave sua [236 ama
ee]

15284 |

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‘ree ote ny or ay ta oth tr ete pr faster al eon fee
Wine tae oul actor par DA ea

PROBLEM 11.17

A parie osiltes betwcen the puis x=40 mm and x =160 mun with an acceleration a 4100 - x),
‘whore und x ar expresso in mau" and mm, respectively, and kis constat. The velocity of the particle
is IS mov when x= 100 mm end is zero at bath

(0) the velocity when = 120 mm.

SOLUTION

(0) Wetave
When x=40 mm, v=0: [rare Lruo- ar

=)

Na
Lau vo]

o)

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PROBLEM 11.18

A particle starts from est at the origin and is given an acceleration A+ 4), where and ne expressed
in Ms and m, respectively, and & i a cortan. Krowing thatthe velocity ofthe parce l 4 més when
2x8 m, determine (2) the value of , (9) the position of de partici when +=4,5 m, (e) the maximum
velocity ofthe pace

SOLUTION

(0 Wehe

When x=0,

When x8 m, yd més
o kosa 4

(0) When v245 mis

(©) Notedhat when v= yg = 0.

Now 0 nn shat

190 mus À

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Ton armani ru Mh le proie Pomares mt Be Ma,

PROBLEM 11.19

A piece of electonic cqui
Fuster is dropred so Im
mus. After impact the equips experiences an acceleration
of a=—kx, where Ki a constant and i he compression of
‘he packing material. U the packing malcil experisnect a
maimum compresion of 20 mm, determino the maximum
acceleration of the equipment,

SOLUTION

Separate and iterate

Use 1) = mvs = 002 mand v, =0, Solve ford

= Laon? besos?

‘Maximum aceleracion.
ne = Ain (40, 0004002) = 800 més

a= 800 m7 4

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rar hurt ay aro yma hate se pron of phd pe eed
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PROBLEM 11.20

Based on experimental obewvatons tbe aceleration of: patil defined by the relation a =~(0.1+ sin 8),
‘where a and x are expres in mA and meters, sepoctivey. Knowing that = 0.8 m apd that y més
‘when 4 =U, determino (2) the velocity oF the parce when = 1m, () the position where the ven is
rim, () the maximum veloc

‘SOLUTION

Wehe

Wien x=0,v@1nis

E
© ICM 08e054-03
90328 ms 4
sun) =
m tos i )=0
“ oem 70mm 4
Wien x= 01080154
1 corona
7 0800 OT yy
Yan «0100001394 cu LU 0
nosed ma?
“ Yeu = 0 mis À

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naa dd my font a ear vn he a person el pew Sie ae
Sento hdl Sth Se oa eons Paes Paden a

PROBLEM 11.21

String from x 20 wih ao initial velocity, pace is given an acceleration a = 0.5/0? 249, where a and
‘ae expressed in més and mé, respectively, Determine (a) the position of the particle when v=24 ms,
(0) the speed of te particle when x= 40m.

‘SOLUTION

Weine CT

When 120,90 Et fee
or IES
« Foios

(a) When y= 24 mis: 4 449-7
or send

o

er va ta ms €

morte Maru. 6300 e Mi Conan, nM el pt Hé a os
Eo e talco presio i ig nl concep ena le cg ee,

PROBLEM 11.22

‘The section of a parle i defined by the ation a= À V5, whee sa constant, Knowing that 0
and y= 81 més at 4 = and that v= 36 més when x= 18 m. determine (a) the velocity ofthe pace when
220 m (0) the time required forthe particle t come to rest

| SOLUTION
Le) Webwe
sota

| Wien x=0 valia:

When sitcom Zoe us)
a Er
Finally
When «=20 m: 20° 2-190
A vr
0) Wetave Hanf
fa
as [00
|
Loa Ann |
Loa 20-919
Wien ve: a9

PROBLEM 11.23

The acceleration of panic is defined by the relation a = -O., there a is expressed inn and y in in
Knowing hat at = the velocity i 40 i, determine (a) the distance the partici wll vel before coming

to est, (6) time reputed forthe pate o come 1 es, (che time required or the pale 10 bs reduced
by SO percent fi intial value

‘SOLUTION

@

Separa and integra with

Distance ore,

Tor v=0,

¿Lucio <
10)
mon rauf)
turrones
O tros
a <

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end el noon br np es erry pon oh aay, wa end ee
‘Ren cence panty Shon hel arp Prana Sh

PROBLEM 11.24

A Dowling ball s dropped om a boat so tht it tikes the
surface of à lake with a speed of 25 fs. Assuming the ball
‘experiences downward veelerion of 2100.92 thes
in the water, determine the velocity ofthe bl when it tis
‘the boom of the lake.

SOLUTION
WA 28 MS, rn
210-090 = -
. 10 y
ero 090 and nas
Wh 3-0 3
eo 33083 Me
Since vp > 6, write
4 =e)

he

Integrating,

Pod nn

eee

MIT

11238910" 1

335 €

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u -
PROBLEM 11.25

), where i a constant, Kong tha
= 4m, determine (a) the constant,
ofthe price,

‘The aceceratio of» particle is defined by the reluion a= 0.4 —
1 1=0 the particle starts om ees at x à m and thal when 4
(5) the position of ths article when v= 6 ms, (0) the masinum vlad

| sorurion

Werne

a
Aus (> [osa
1
m Lina sopas
ue
o toto o
Aminen ag 0409
a
Seva sm
10157 4
© Weine
Ten
a k
Wim: A)
aña arena
| osas
“ 12520 4

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‘vec a duet nar by a mn, at th tr wen prs lee od et
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PROBLEM 11.25 (Continued)

€ The maximum velocity occurs when a 0.

VAR

sam 4

Analtertive sohn so begin with a D).

Int Dar
040
ma jose
Thus y,
1
vet

2 store

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ed yo ar ay ma st or wien pas he al. ed aa ed
esas uds ro Ibid one Dorso mt,

PROBLEM 11.26

A ati is projected to the righ rom the position x = with an inti! velocity of ms wecleaton
Of the particle is defined by the relation a=-0.60" where a and y ate express in mv and mis,
respectively, determine (a the distance the particle wil have travel when ls velocity 8 m, (8) the tins
when ¥= 104 () the time raja forthe parte o tuve 6

SOLUTION
Wehe E 06

Wien amt [vn [Losas

or avg 06%

= a w

0-4)

= am
© Werne ow”

When 0.29mi: [rss [ose

m Ae = 06

. foe

Wher v= mis q

=: 172225 4
Weine Klon

Now

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‘pbs or ted 0 Fran a ic rien pao po, ld Te
‘itn rnd aly Me Mf el app Boao ic,

PROBLEM 11.26 (Continued)

9
= ot

When x6 m

rd

An alternative solution i begin with Pa, CD.

“Ten
Now
Pr
io «fa
ne Tears
Mi =
“ CES EE

“Which last the same equation as above.

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pecker hoy moa lh trie pum of ys od de ed
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PROBLEM 11.27

‘Based on observations, the spect of a jogger can be approximated by the
relation v= 75(1~ 0.4)", where wand are expressed in mi and ils.
respectively. Knowing ae à 20 at 2=0, detemino (0) the distance the
Jonger has run whew 21h, (0) he jogger’ acceleration in RU at 2
{6) the time required forthe jogger o un 6 mi

aur-in

0) Wohne

ÉTECEEC
© Font)

When à = 6 mi

100049 =021 w

1 pa
PL Le

1 >
RAT

«= 715 mi 4

PRE)

PIM OREA "1
OS.

om ED

am)

E]

ERAN
A]

pr

va
= 0832205

1303 min À

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ese tet hy om yey oa ae jot cr pene y Flot re

ttn tees ond svar ratty ate Mfrs be cpp Up ambas eee

PROBLEM 11.28

‘Experimental data indicate that in a region downsteearn of a
given louvered supply vet, the velocity of the emit ar is
defined by »=0 IAA, whore y and x are expresced in mis and.
meters, respectively, and x, isthe inal diverge veloc of
the aie For vy =3.6 mis, determine (a) the acceletion ofthe air
atx=2m, (0) sui ore air to flow from x= 10
rim.

SOLUTION
@ Werwe
|
When x 22 mi
© Weine

From xT mw 2

ae

"x

us sus)
v al x

a2-0052 mi 4

Lap a ea
BT mo

20

OO

6079

PROPRIFTARY MATERIAL © 209 ne Men Congres A ih oe ot af Mana nay be pie

rie yn i re of ee en ed

tn aceon Dir ang ahnt

PROBLEM 11.29

‘The accctran due to gravity at an altitude y above the surf of
expressed as

“Toray

where a and y are expressed in "and fect, respectively. Using this expression,
‘compute the height reached by a prject red vertical upward from the trace
‘ofthe earth fit intl velocity i (a) 1500 As, (8) 3000 fe, (0) 36,700 fs,

SOLUTION

Wehave
When

(o »y=1800 N:

5 =3000 ni:

sou?

CR

Y “SOLO À
3000)" |

ae

Yang MOTION À

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‘er otal x mas ion ar ey pon thease, ma en a
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E

PROBLEM 11.29 (Continued)

© 1 =36,700 Ms

or on 23090 €

he velocity 36.00 Ms is approximaly the escape velocity vg fom the earth. Bor 9

Youre 4

PROPRIETARY MATERLAL © 209 The Wem Compre Ine. A hs eee fart of ma ne be dy
‘oral ded ny oa a ns ia DJ mare fe pacers ad Bj te Hed
ria nar sd adas pc Me ford dr nc parao omar acne nant,
lv pr

ry PROBLEM 11.30 Î
| ‘he sen dst iy of «pre tiling ad al
! ines gti ar nd fm coer af eth
|| hep iste ran ee a andr oe ace de
i I
Proce el un ee ete fi ca oo

A Bean (at 30 here)

SOLUTION

Vete

ven

ne

(emana merino mi
* mena vs 4

PROPRIETARY MATERIA. € 209 The Mi Copado, be, A ii remo. por is ol ea,
‘yee st yf sl y mae oa ro rt pra of ea roe ee eo
dende Arme Ptr coh a,

PROBLEM 11.31

‘Tho velocity of « particle is v=y[l-sinGa7J]. Knowing thatthe partite starts fo the origin with an
initial velocity», determine (0) ts position and is acceleration a = 37, () it average velocity dering the
interval 15010 4>7.

SOLUTION

@ Weine

ALTO =

©

mar,

sip 2236040 4

Also

Using Be. (0)

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dame ay om by ay na ps ven pon fh li, ml bd ad
‘Ree ref tly co ht cl app pont

PROBLEM 11.31 (Continued)

Auer: |
er)
fot
(5)
cor
tbe sgt o 0367-0
aro

PROPRIPEARY MATERIA. 200 Te Mine Conan Is AB hs ec at fi Moe arb pa
‘pm or ne lo fe na rar write poof nr et a ed
hance nen eh, ine fr ala o eatin rot are ae ae a,

PROBLEM 11.32
‘The velocity o sider is defino bythe elation v= sin (a). Denoting the velocity and he postion
ofthe adc at 1-0 by amd. espectvch, and knowing thal the man displacement ofthe aides
is 26, show that (a) Y (1 +:300)/200,. (6) the maximum valve of the veloc occurs when
DR.

| SOLUTION
Le Asa += van

hen cp äh Se

exces
Now Lo avncag+9)
At 120.2%) [lam [sincere so
di = r
or sue]

+ ones 14
mt Dar)
Now ober th gg oes ten co 4/46) =. Then

San = 2419 ee CD)
Pa

Sinai or a . =.)

or sv =

Squaring bat ide of his équation

el ~ Dagan, + V9 =v? —
PURA

o OED.

An cs. on of Al Be pc,

PROPRIETARY MATERIAL. 209 ie Mei Coop.
PR did ym ry yes Ra de fe mane oros of ple a gon ed

Mote oxic ray ioe na repr pes

PROBLEM 11.32 (Continued)

(9) Fes observe that nu, occur when au +=. “The comesponding value of is

fer]

Substtting fist for cos and hen for y

ol

aa iaa]

salia Y

met

PROPRIETARY MATERIA. © 30 The aCe Compases. A eh rose rt of Ml a oa
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‘Sib snes ented af a dbl cana pre yoo aan sigs Mon

PROBLEM 11.33

A mori enters a fieeway at 45 Kavi and accelerates

itumly 10 99 kn. rom the ameter in the car, he
‘motors knoe hat she traveled 02 km ile aelerain.
Detenine (a) the acceleration of the ca, (9) the time
ruta rc 9 kh,

SOLUTION
0 jon of the
Data:
exo same
rm]

at to)

NS-125
1500
-10003 a

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[du nr Say am athe rr wien pao ala, on ed
ero tant nda rca Screed te prea Tsk Ma

= PROBLEM 11.34

A truck travels 220 m ig 10 while being decelerated at a
constan rae 006 mA, Dotermine (a) inal velo,
(0) its final velocity, (c) the distance waveled during the
fist 15

‘SOLUTION
CR

201

AO 1.259 ms 4
©) BinaLselciy votar

v=250+ (09400) 00 ms

€ Distance travel daring first LS.

La
ear

esco sain 4

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Em PROBLEM 11.35

Assuming a uniform accleaton of 11 MAT and knowing
A. Aero tte mec fcr a pss 4 $30 mi, dermis
PHARE (a) the time required forthe cart reach 8, () the sped of

km | the car as it passes 8.

| souunion

© Tina onc
mi AA Rs, ya Melt, oat st

soho?
DT
suisse
44+ as)? ICI
u
sans

Rejecting the negative mo.
© Sposa

ug 504 mi À

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‘emake don? nym By a a a DJ e por of the llo. alpen a
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N | PROBLEM 11.36

LE a group af students anne n model rocket inthe vertical direcion. Rad on tacking

T ua, they determine that the altitude of the rocket was 89.6 fat the end of the powered

" ‘parachute failed to deploy sa that the racket fell fieely to the ground after reaching its

DT
: we

an |

ec - ama

aed At
Then 102 (282 MSN +2(-32.2 Ms” gas — 89.6) M

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PROBLEM 11.37

A sprimer in 100-m ace acelerates uniformly forthe Fest 35 m and the runs
‘with constant velocity. he sprinters time for the fst 35 m in 54 , determino
(a) his acceleration, (b) his final velocity (es time for the race,

‘SOLUTION

Given: O=x=35m, u 2 constan
35m<x 100m, »
ALO, 0=0 ven

Find

@a

© ywhen 2 =100.m

9 hen x=100m

@ Weine corola? fr as

1
36m Las 4 9
gas ss

“ a= 24005 ve
20m 4

CET for 38 msi 100,

Now 2042-0) foe 05138

When 235 m ac = 224005 mi )OS m)

or Yay 129628 ms Pis 21296 is 4
© Wehe Kan enl-n) ir 38m< 100m

When x=100.m: 10m 2 35m 4 (129628 mA 5.48

or nal: 4

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and

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428
cap

458
CD,

ase

PROBLEM 11.38

A small package is released from rest al and
‘moves along the skate wheel eoaveyor ABCD.
‘The peckago has uniform pecceration of
D 45 més as moves down Setions AB and CD.
es fa eo am even a CF
2 the veloci of the package a D ix 72 má
ts determine (a) the distanced between C and Da
(0) the time required forthe package o each D.

Pal)

We =0+ 2(4.8 ms )3-0) m
Ame Ce = 52666 ms)
era damen

02m EEE

40m 4

vota

53666 me =0+ 8 m8 Yay
DATE
72 ole 5.3666 mis Ami Ve
Lo = 0381965

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‘Kein sas lap Set ad re ppt wo sing an

PROBLEM 11.38 (Continued)

Kom,
fr Bae,

weave eet ctas

o 3m 2 (82666 male

e hac 055901 à

Final ty Lan az len (LBD 0 58901 +038196) 5

o 120654

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ert calc Sab iu hel ac por en sn i Mm,

PROBLEM 11.39

A police olor ina pat! car parked ia a 70 kn speed zone chere a passing automobile waveling a a
‘ow, consta poet Relieeing that the driver ofthe automobile might be stented, the officer sare hit
far, seceleratesunifornly to 90 km in $ 5, and, maintaining a constat velocity of 90 km, overakes he
motorist 42 safer he automobile passed him. Knowing that 18s elapsed before de lficer began pursuing
the motor, detennine (a Me dite Le air Ice befre overtaking the motors, (4) Uh

speed.

SOLUTION
er à
CR NS GE
Ger Tas as

Bra Gr =90 kav 90 kav = 25 mi

(Parolen

Deren)

a Pe

Abo, AUDE EL

arte Go ess com

pu scies staat 29

ares (oan wets nto

von
rn-asın 4

© temerity Dot

Arten men

“ me

« a. 191mn 4

ovata tans peo

PROBLEM 11.40

As relay cannes 4 enters the 20-u lo exchange vr with w
Speed oF 129 mó, he begins o slow down. tle hands te baton (0
runner 41.82 ltr as they leave the exchange zone withthe

@ turas aorta

ars be

“e 210008 4
| ao (debt
Mans: Ga IRAN
5078 ms
Former: wa
When wer OTE Ws? +20)
Door ig "20603 mt
ae =20 4
(0) Forrummer ET

where 44 th tie at which be begin onu.

Aus: AS
or 107-2398
Runner 8 should stat to run 259 s before À reaches ino exchange one. «

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‘ota sad nye by ng as fear poof fab ow en ea
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PROBLEM 11.41
les À and B ae travcling in adjacent Nighy L

0 ave the posiios und spect shown. Kn
at automobile 4 has a constant acceleration of 18 Ws? and
that B has a constant deceleration of 1.2 4°, determine

(@) when and whore À will ovetake B, (5) the spe u ch
sutomobile a hat tine,

sa
een
TOS

nenne o
Le soessare us

ede taal COUT

0

(o

Kama: 35214096 = 754528 044

126, -75~0
150546

1222 and 4

Dre:

320150903415 05)"

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‘eed nn yf a a rage of per wae

s

PROBLEM 11.41 (Continued)

| ® Yeats sn 15054

Es: 123524180505)
1,2629 0 1,7425 mith €
EG M =82.8-1.215.05)

vp 3474 Ms

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‘ere o aa oar oa of er a od
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PROBLEM 11.42

In a boat race, boat is Fading boat 120
and both boats are traveling ta constant speed
of 1OS mish. At 1=0, the boats accent a
constant rates, Knowing thal when 1 passes 4.
5 and yy =138 mi, determine (a) the
acceleration 04, (8) the accleraion o.

‘SOLUTION

@ Wehave

CRT

and

Aus:

rado ta

(2p =105 ai ECTS

Pet mith =198 Us
198 tvs =154 e+ a (8)

Pe
anne dad? Garon
rent rise
rons ass no eos ms
en
80 eas
ansia

at oo Ms a yb dhe,

‘ral sed ma hy eo i jr te prin oh yar ied joie ae
‘tite each nd tly Mere Trt dl ree pat ng Maw

PROBLEM 11.43

Bones are placed on a chute at uniform intervals af time 1, and
slide down the chute with uniform acceleration. Knowing that as

ox Bis released, the precoding ox À has lead sid 6 m
nd tha ter they’ are 10 m apart, determino () the vale
OF ter (8) the acceleration ofthe boxes

SOLUTION

E Es,

Let 12 = 1 se the time when the boxes are 30 A apar,

ster te

ta na (as A
CE bon
AS
For lao Aa zu)
a ten ein 8 ba o

at Paty tye aga = 30

1 2.2

30= La v PE

D alg ak Hs ©
1

Lera
5 3

a ae o

Diving Fain (y EG
iat nes 4

a

(0 Soin £5.29 er

area |

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‘teenie rt rap Poner no shoe

PROBLEM 11.44

‘Two automobiles 4 and B ace approaching
eich ote in adjacent highway les. At
10, A and Bar I kon apar, hir speeds
fe ¥, =108 kin and vg = 63 ky and
they ine al Points P and Q, respectively.
Knowing that passes Point (240 safer
was there and that B passes Pint 2-42 s
fcr À wae there, determine (0) the
nor accelerations off and (0) wen
the vehicles pass each oer, (e) the speed.
‘of bat ie

SOLUTION

Wem steed? etant

mo, soe Sr
sea osa
“ aos ON:

(6) When ocu pas each other

‘Then O0 hy 0250 a 117.5 md
020005 mig 1000

o 0.05045tjg +98 ~2000

Solving

0822 5 and = 1908s

120-219-2085 4

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‘wc rel nr rae io rr en pena of alada. me oa e ed
Re ps are tn

PROBLEM 11.44 (Continued)

Weine ep Cp yb aat
At va = 17.3 ms +(0.30045 OR.)
2375610
$ 997855 km 4

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PROBLEM 11.45

Car A is parked along the northbound
Jane of gay, and car is tve.
in the soulhbound lane at a constan!
speed of 60 mh. At £=0, A starts nd
ects ata constant rate ay. whl
at r= Ss, B begins to slow down ih a
constant deceleration of magnitude 4/6.
Knowing that when the cas pass each
her x=294 and 9, = vg, determine
(a) the accelention ay, (3 when the
vehicles pas euch oer, () te distance
ete the vehicles st 170.

SOLUTION
kur 20: moral
lie
ort
uses sOy 760 mi = Wis
Au=ss sa = ER MANSO = 440
For ass orto ale,
PO OR =~}
Gas +=) Pre)
A

Assume £5 «when the cars pass eae othe,

ime (uch

mi DAC ET

1

x, 72941: zu

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‘pd dung ay form eb sm, ht prion of aor, ced Net ad
‘Sei cond ater Wr tap tae a oh

PROBLEM 11.45 (Continued)

ES

Then wae
e AA ML +520
Solving La 20.955 and 14-105
CRUE 296 R= 0 Y

er Pen
6 Fromabove La = 7.005 4

ote: An acceptable solution cannot be found fi assumed a 1 = 9

fo) Wehave dare,
= 204 [440 HRS
a

200 0

oe Anand

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PROBLEM 11.46

“Two blocks 4 and B are placed on an inline as shen. At
120, 4 is projected the incline with a initial velocity
OF 23 1 and Bis leased rom ret The blocks pass each
other 1 5 Inter, and A reaches the baton of the inline

‘rom
the bottom of the incline reached by block 41621 fe and
That the acceleration of 4 and (due to gravity and
Seton) ae consant and are directed down the incline,
‘determine a) he acccerations ofA nd (1) the distance
(E) espec A when he block pa each her.

=
SOLUTION
@ Wehe 00 ala 01 wh
a

When DER) La

Then 0-27 by 420,21

“ e, 113

« amet 4

Now norte

sa pdas

2

AL 121, the blocks pas each other

Hood
axé

1s ON

E fer mone rnasm anes]

[Joos fase

or ay 370

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pc ar dred nay frm ya) moos, te rr no be, ad ed de hd
‘Teint cher andar pr BO lf hr mbr sar pe Pp M,
cagon

PROBLEM 11.46 (Continued)

© mitra os Lorna

© Were ra
Amis 9 227 sa

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ind dd x fama yo sna M BS i pon of ef a mal Ed
Beeches

PROBLEM 11.47

Slider block A moves tw the Jef with a constant velocity of 6 mé.
Deren (a) the velocity of block (9) the velocity of poston D of
‘he able, () the relative velocity of potion Co the cable with respect
to pontion D.

rom he diagram, we have
24437 =comata
tren eth 0
und arto, 20 @
(o) Substting ino Eq.(1) ms + 3, 20
er vins 4 |
6) Fromihedigrm 35135 =enmunt
then en
vo=2miel 4

(9 Frum hedingron xy ye “constant

Then PRET

6m) (2 ms) E me

von=smst 4

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PROBLEM 11.48

lock & stats rom reat und moves demand with a constant
secceration. Knowing that fer slider block 4 has moved 400 wm is
velocity e 4 ma, termi (a) he acelerations of and B, (9) the
velocity and the Change in poston PA lr 2s.

‘SOLUTION
From the digran, we have
sg By Senna

Then ro 0)
and ayr3ay=0 @
(a) Fa.@X ag 1309 =0 and ay is constant and

positive =>, is constant and negative

Also, Ea, (1) and (re) =0= (0

Le iO 2e Ga)

When jAr,|=04.m: Em)" «24,004 m)

e. ms 9 4

‘Ten, substtuting imo Eg.)

-20 wis? +30, =0

6676 1 4

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PROBLEM 11.48 (Continued)

9 Webwe Deren

(Balen

or md
1

Alo ve One +04 ba

Au=2s

em |

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PROBLEM 11.49

The elevator shown in the figure moves downward with constan
velocity af 15 Ra. Determine (a) the velocity ofthe cable €, (6) the
velocity of the counterwsight (ete relative velocity ofthe cable C

vith respect 10 the cevator, (4) the relative velocity of the
‘ountensight I respeto e cesto
J
SOLUTION -
Choose the postive direction doused.
Veloce ofeabte €.
Ye +27 =eonstant Tu
+ =0
Bu, PET <
or PET ve 30047 4
© Velocity of commierseieh W.
Yar constat
he rr vy rs wem 4

(6) Relative veloc of C with respect 0 E.
on =e Me =(-30 Ws) (415 Me AS Re

(a) Relative velocity of IP with respect to E.
hog = = Ve CIS 8) 01 8) ==30 Ms

vine 7300 51 4

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rc eo yo yc a ren permi that wa bd
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PROBLEM 11.50

BB an perreo

acceleration. IF the eouncerseight W moves through 30 in $

(a) the secclerations ofthe elevator and he cable €, (5) the velocity of he

ly | q dar

SOLUTION

‘We choose Positive diction donnant for motion a counterweight.
demon

Auı=ss, PET

mn Lao?

pza 0 al
(o) Acedera of and C.
Since WA Ye SCOT Ye yy =O, and ay +
Thus Ae ty = CARS). ag =240 eT 4
Also, Yow Zn = consi, 442,0, and te 42g 20
Tous: eBay ZA) AS, aan 4

(0) Feloity of elevator afer 5.
ms

ve =p gl =0 24 MANS)

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ra erin pol RC ra ld

ar PROBLEM 11.51
Ë that ae 8

of alters.

termine (0) the accelraons of À

Collar À stas from rest and moves pur With constant aeccleaio. Knowing
relative veloety of colar # with respect to colar 4 is 24 ins,

118, (0) the velocity and the change in positon

SOLUTION

Then

and

(a

rom the digan

24+ a+ Ue =) const

1442950 Oo)
4g V2 =0 a
Fa (Hand On

Also, Eg. (2)and a, is consent and negative =, is constant

an positive
then DEP er
Now DRE er
From Eg.)

Sottar

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‘rca ied as fora by mcs se rns mie pte ea ud ha
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PROBLEM 11.51 (Continued)

2in= Sa
o ayediniet 4
an then
o ay stint?

© armés
or Von
how
mimos
or Yeo

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PROBLEM 11.52

Inthe post
(a) he velit
velocity of portion Cof te cable with respect a nr A.

sown, collar 5 moves downward with a velocity of
iy of collar, () the velocity of potion C ofthe cat

12 in. Determine
ble, (e) the relative

SOLUTION
rm hedge
2g tye == constant
Then u o
and ay+ay o
lo) Substring into #4.) vy #20228)
From the diagram By, ye constan
Then Berre
E nds) Hyp 0
er ven dtinal 4
CR
= (48 nds) (2 ind
“ Yon =36 A4 4

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a porn

PROBLEM 11.53

Slider block 8 moves to the right with constr

i ® Pe velocity of 300 mms Determine (a) the velocity
er ser block A. (0) he ey of tion Ca
en } i Sie he oly of no Date

aii cabe (4) the tive velocity of portion Co the
‘aie with resp sider block À
SOLUTION
from the diagram

| Ten o

ES o

Also, weiave

Then o

[a a Br

loa + 200 mn 4

À conte diagam p(y He) cotnt

ma aunt

TS 20 mm

ve = nnd

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PROBLEM 11.53 (Continued)

(0 From the diam Bean constant
“thea De ee
Substituting 600 mms 24200 mans) =

(0 Webave
or ou = 400 mms 4

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PROBLEM 11.54

AL the instant show, sider Block 3 is moving
‘with a consant acceleration, and is speed 5
150 mms, Knowing that air der Dick À
has moved 240 mm to the right its velocity is
0 mm, determine a) the acceleration of À
srl) the aeceention of po he
able, () e velocity and change i position of
Slider block # ar 4 a

SOLUTION
Prom the ig Beta) 2a constant
Then 29,00 o
and 2ay Ba, a
(@) Firstobserve ati block 4 move the right, y, and Fg. (1) = ve >. Then, in

Bq, (Datr=0

2050 mms) ~ 3)

e 00

Also ba. 2) andy = constant “>a = constan

‘Then Vs = SET dd)

When x4 = (y =240 mn:
(ms) = (100 ms? 4 24,240 so)

or a0 nv
E

or a, 53m € 4

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PROBLEM 11.54 (Continued)

‘hen, subsctatng into Ha. 2)

2e, 4

or Dee

(8) Prom the solution 10 Problem 1153

wine
Then antojo
AP
( y
© Wehme Mb rai
Auızes va =150 ss 1-200 Jn NS)
a ven 0 mi 4
A PRE
aus DC = 050 mms
1 :
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LC 200 may
or A!

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PROBLEM 11.55

lock B moves wma with a constant velocity of 20 mays.

nd its Velocity is 30 mm King that at 3 slider Block C
Tas move 87 mm Lo the righ, determine (a) the velocity of sider
block Cat #0, (9) the accelertons of and €, (ey the lame in
positon of block falter 5s.

SOLUTION
rom the digram

Br yg + fe =consant
Then Brg tig bye = o
and Bay tay tac =0 e
Give vo =20 mus bs

(00730 mm?
(a) Sitio ino Fa (Dat 0
rm) +420 =

e =10 ms rl 4

0) Wehave seco Woh ttc?

some (opaca

ac = 6 mis? e muni? à 4

Now

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PROBLEM 11.55 (Continued)

“Then, substuting ito Fg. (2)
32, +440) +6 ms?) =00

Deere or td

| © Wehne DAT

Asse ah =H en,
78m

yasni 4

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PROBLEM 11.56

Bock 1 stars from rest, Block 4 moves with a consta
acceleration, and sides block C moves to the sight with a
‘cantantacelestion ol 75 mm, King that at ¢=28 the
ects of and Care 480 munis dowward and 280 mn io
the right respectively, determine (a) the accelerations of A and,
(6) the int velocities oF À und €, () the change in position of
Sidor Calera.

From the digram
Ba 14) te constant

Then Sr tag Hye 20 ©
ad 30, +4, 20090 o
iver Ge

(00-75

NRK. vq 480 wb

ve = 280 mm +
(2) Le (2)and a, constan and a = constant = ay = consta
Then vy =O at
Atr=2s 480 mms 0,625)

A0 mms? or 2.240 ms? L 4

Substituting ato ba.)
30444240 mas?) +01 mnie?) = 0
45 ms et

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PROBLEM 11.56 (Continued)

@ Wetase Cora
Arz 280 mms (re), + 75 mm s)
Ve 130 mm for A 4

Then, substituting into Eg, (1) at =
36,040) + ((30 mm) 0

wem aint
lc) Weave Al
ai oo osa
Stem ma

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PROBLEM 11.57

Collar A sars fom xt at 1=D and moves downward with a constant
cocon of 7 inde Collar B moves upward with corstentseceleation,
And its ina veloc I 8 ins. Knowing that collar moves through 20 in,
between 1=0 end £=2 5, determine () the sccclertions of collar 8 and
loe €, (ih time at wih the velocity of block Cs ee, () the tance
through which block C wil have moved at tht time,

SOLUTION
tee +0
- thes e ica D
HT
tl ai nu ag o
me Given;
i
J meds
Wehe
wi nes
en te
ents, 6

20 ind) in) ag =0

sin or ame

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PROBLEM 11.57 (Continued)

(8) Substituting ito Eq (1) at 1 =0

20-829 Hr), 0 (dein
Now senna
Wen yo <0 zus
| or 2 8
| 2 ors 4
Lae
© Werne set ours tag
iol? auf
Au seman 2s) hein (2

0667 in. o Ye oh =0667in TA

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PROBLEM 41.58

cece

Collar 4 and sat fom res, and collar A moves wand with an
of Min 2. King a colla A man doer witha

‘mst acceleration and that tz velocity is 8 i

eternine (a) the necleation of block C () the distance rough which

block Call have moved aes 3s

vs efter moving 32 in.

SOLUTION
Frame diagram
1 ea) #2pe A) coma
Then entre o
and aura +40 =0 o
Given: Oe
CE
DITS
a, = contnt L
Wien AL,
CET
(2 Weine Vale Cu]
When yy Co) 2 ins (into =20,(12 0)
or PS
Then, subsitating io La (2)
AP im A) (in) bag 20
oe rind

o
210) ~

Now

‘Substituting nto Lg. (at 20

Wan =0 OF OCI =0

eg acai
te ea lamer
DE + a 0-6)

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PROBLEM 114

(Continued)

Pie [jo-aña

Loa
ÿ Luca!
Loam
hs, reso
m
à
> me eds
‘Theo, ck City moves downy (>) nd hn mores up (<0)
re
Now PTE
Mure [de Lea
we Der
rm,

ES

Sa
tes eV = =D in.
A ON

“Total distance treed

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PROBLEM 11.59

“The system shown sets from wet, ad each component moves with san
acceleration. I the roatve accelerion of block C with respect to collar B

‘paar andthe relative acceleration of tock D with respect o Bock À
un? downward, determine (2) the volociy of Black Cafe 3, (0) he
‘eign positon oF block D ater,

SOLUTION

al ee
SEE

rom the gram
atte Data ye zona
E Then 22,120 11090 o
an Lay Vay ice o
GX Go 79 10m ~ ro) constant
hes 14012930 o
and gy 2up =0 o
Givers At 120, v= alarcon constan
don = 60 mm, de
(O Werne dap Heap
md mu
Sabino E) 0 (8)
IE DEE Deren
æ Wera 0
BR hay M0) (ar +02 0
“e rn 6

Ne os pe a Ms y be rt
ri po of epi, a eed he
a rept oe ang a

PROBLEM 11.59 (Continued)

Solving os. (3) and (6) foray and ap

10 munis?
AUIS 3s: (40 m/s? (3 8)
w ome 4
Lo Weine row
een poco

yor RS nt 4

PROBLEM 11.60

The system shown tarts from rest, andthe length ofthe upper cord is used so
that, 8, and € ae italy atthe sane level, Fach component moves with a
‘consist seecleration, and fier 2 the reaiv change in positon of Dick € with
respecto block 41s 280 mm upward. Anowing that when the relative velocity 07
lla 8 with espec o block is RO mun denn, ih dsplaoements ef A
ud B are 160 mn downwand and 320 mm downward, respectively, determine
(a) he sccleratins o À and 44 ay > 10 mms", (1) the change in position of
block D en te velocity block C's 600 mund pur,

Prom the gra

= y j wen 2y,12v 1 yo = constant
EN mm mern w
Ot om at a
| Cable 2: (yp =34)+0% =>) constant
à nu o
7 Ing +2 ~0 (0)
re

Oh = 0-60

All accelerations constant a 2

on 2280 man À

Vas mis
| Bien
nO 320 me

8 10m

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PROBLEM 11.60" (Continued)

© Wetwe roza

and Per Gen HOH

! 2
Then Dar ever Hee a
AE 28, Ju 280 mus

a

ñ
280 mm 2
a

o ge 704-140 o
Sabin ito Eg.)
2a, +20, +(0, -140)=0

or 6
Now
DEA
CE]
fs

Ako aby MO Lag
When a 80 mm Le 80 (ox ~0,38 o
Then 160
Using En. OY ne más
then e. a= 20 mis À
and o ay 40m 4

Not that Ea, (6) 5 not used; ths, the problem i over determined.

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PROBLEM 17.60" (Continued)

Altematve ston
Weine CETTE

W504 2s Che]

Tien Maa Ha v4 Bars Cdt at, Cool

Wien Veg =80 mens La

50 mons = V2{ yep 320 m) — fa ¿A mr) |

& 20-20, Vo.)

Solving Eqs. (6) and (8) yields a, and ay
€) Subaiuting into Hg (5)
= 20-140 -120 eit

and into Ea.)
20 mms!) 40 nnd?) 20,

or ap 30 mv

Now Deren

Wien ve 000 ns 120 sm

Also

Ares

Le PACA

®

375 mn L 4

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Intro rte e Jara e y es ot i re i ern fbi ml he eit
Terao hanes perl re mh carpe nase a,

PROBLEM 11.61

A panicle moves in righ ine
the igure, Knowing that stars rom the origin with y =
(a) plot the v 2 and x 2 curves for 0. 1. 206, (0) detemine ts
velocity, is posi, and he al distance rele sehr 2 125,

‘SOLUTION
(a

Initial conditions: 120, 162185, gi
Change in v equal sre under amt un
TA

OSI: ETC MAEM 66.8

ancien nen MENO RIE Mg = 130

Deren Ba S WERDE na 120 «

rare vy Hy ETES ION ra 20

PET

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