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Unformatted text preview: rn rn FEIEET Sample ExumInu‘linns 31' ml and 3?. The disuihuted 10ml Is replaced by 545M100?) mm}=2m:~q
TIE K m The form is applied at {lfajafé ﬁll—ﬁfﬁ m from paint :1. EME:C
_ B I
Jim—E T31)! [HEN] Nil—{ID THJRA =3
HA =146J§TN Answer is D. 38. Divide the area ink) recumgle R and triangle. T. :L'R '—‘ %
ALI = 2
Err = 1
AT =ﬁ
a. : #111111 + MAT __ [23' + {13433}
.4R AT 2 + E
= 0.375 Answer is D. 3:]. First, ﬁnd thc react1mm. REEmy Rugby:
_ {mas N} + (mat) N] + {2}{25 N}
_ 2 322.5 N — Emulation: inf Exam 'I—Muang Sealer: H tint the truss vertically through the eeeenrl seetjen From 4.3. The vertical component in BC is the same 3.5 the
the left. reeetien at E.
25 N 5:: N' a 2 MA = ”
[KM [T mute...) h (3 m}[23l'ﬂ N] —. e
I___[ REA, = 955.? N = .BGH,
2'
32.2.5 N K ADSWIer is A.
I I N m
 t a
l xx 3% I
[ .__\1 44. N = WEBB e = [tale l‘x'jiees {10”} = 601.2 N
' ( 1F :' Ff... = Ne. =[ee1.2 N}{U.2} = 123.2 N
.l' E
L __ [ P;3 = {TEAS N}{etn4ﬂ°] : 504.5 K .
15 :11 15 m F9 3 FL: The hleel: ls. sliding. zen = 0
“(EU IIJJIF + (15 mﬂﬂﬁ N  322.5 = U
F = _223'125 N Anewfer is C. FL“. 2 Np... = (5.11.2 NHDJE] u 9.11? N_ Answer 15 BI 45. The ﬁeebody diagrmn nf the top Huck Fwd ' FEB
5 kg
4D. If there was a force in AH, there would be me other
member at jnint A that. ctmlrl cancel a. vertical mmpo— _
nent. Member A33 is a zeroforce member. FLE = Ill'1'” _ FW = #1739
,_ ,_ x . m
Answer is A. =te.1a](a kg} Keel 8—2]
= “.35 N The freeberiy diagram of the lewer block 55 4].
F55
' F
Fm 20kg
—Iﬂ'—F———i
F = F}; 4 Flt.45,
. . _ u
= tee N + [n.eetteu kg + 5 kg} —;
a x
=8 . N
ZME = e U 9
{P Answer 15 D.
L k3" HBt1 —{]
2
30; = LP’IJQH [mmpmsﬁiom 4E. Since the ﬁtruL'turtz 15 eymmetrjeel.
Angurer is D. REE = EEG
Er. — n
_ 3TH tees1.1”] P = 0
4141. = = N'ﬂl PAC : Answer is D. " Answer is B. measslerwl PI.l:[terinrtsr Int. H FEKEIT Sample Emil“: w 1?. Draw the freshedy of the lower pulley and sprEﬁdﬂ'
her. All forms are the same because the pumaJr's we
ﬁietienleas. F D N
F = ' = 15.5? N
3
AILearner is E. El}.
11D)! LE
43. When she block begins to tip, it will] tip about the
lower right corner. Since the herimn’sal mmpenent ef
the reaction at this comer is: unknown. momenta must
be taken about the corner to determine 3.
Elﬁn“: rigid. cumnu: "' D
4 m
—z[2lf] N]. + {56 N) = 0
z = 5 Tn
Answer is: D.
. For n ideal
as the.
W and
atrepic ically
absu 1t is when:
rludic has a II] gt? — [1} {9.31 —) (13 a}: T1. 5‘ 2 5?
29m 0:: MI... Answer is C. 72. Impaci farce is normal L0 the wall in the 3 direction
unly. $2.; = a v“ r — 0.5: w m 3 1:”: _: ' JG) = 2% mg’s Answer is A. W K r ’.?V///_///L/r9'¥% Prnfessiunui Publimlium. lnz. ﬂ WI hauim'linn: m This is e compound pendulum. The mass moment of
inertia uf the Lube about its end [where it is. Smipended]
is 1 _ 1r . n _ '2 i
ID = 371;; : ﬂﬁﬁliw = 0.01550 kgm The distance from the suspension paint to the center of
the 1113.35 is L 0.3115 In 1
d E. = _2_.._ . Learn The period of eecilletien of e compound pendulum is .' In
T = 2?!" III m—yd
3 2” V" _ : :
\J [0.5 kg) (9.51 {0.1525 In}
=U.9':l4 5
Anewerr is D.
_ 2? _ {EJEED m}
74. a _ v—i ._ E E ma Ex I hr (1 km} been 5
=91] 5 Answer is D. _ 2 a '
__ [1?t + muting
e {10;} [W335 — (eejﬁj + Tln [g]
= 2.34
Answer is C.
T5. Haring : Eprnjectiie
.2
3 ,, e _ m"
ikfﬂ..) — 2 __ . 3 cm; :11
u 1 kg
1111 ‘5.
«5 u — x [ ("11] (lm Em) _= LEE mfﬁ Answer is E.
"H'. _ “lu _ austere—e)
W P_ Lit _ e: {ﬂkgliemozﬂf~em§ﬂ =122.e x 1e” W (12?. GW} Answer is B. 7'3. 3'!
/ [Vssne— yam = —e
a
vein 3e: “ égﬁﬂﬂz = —h Cheese the positive root. . .' _
a: = [Vﬁlnﬂ' + 1...: V3511:2 H + Egh.) Answer is D. 79. The initiai kinetic energy is mara
Bk = —
2
I m‘ 1’
_ {32 kg] (35m 1 km W
2 105 Figmi
32 "r
2 2i}? PETI After launch. some of the kinetic energy will he trams
formed into potential energy; but the total energy will
he unchanged. Answer is C. Pmiﬁﬂﬂml il'ﬂi. w unc—
d be
mod
.16
.ngle
IE: of
.o to
19. The walls are frictionless, 5: there are no vertical
wall components to help support the Spheres. The total
weight of 50 N must be roiistcd by the ﬂoor forum:l F.
Answer is: D.
20. B} itself1 sphere B would require no horizontal sup
_ port. Homeven the horizontal component of sphere A‘s
tum weigh: must. be resisted.
On sphoro E:
‘+ WA 1131 N
‘0' 2W =——:n——=14T"'.32'.\.T
E} I E M” tam? Leuﬁﬂ" Answer is C. Eullﬁnns hr Exam I—Aﬂerlm Section 33 21. Amumc 13'z = D.
2.1;... = —{3.?5}{ﬁﬂl} _ [7.5J(60D}+ [again{g}
n (3.?ajfauﬂj (g) + 25;}. = 0
F]; = 245.] .3 120 the right
Ariswar is D.
24. DiVidLJ the Lanai area. into three shapes: III‘JE'TBU
mutangle, .55., and two ﬁmaJi interim" rectangles. B rimi
C.
I... ism“i .—. (it:E [4}[i213_ 51's
Zia. _ Ag. — SCI — 60C — {300] If —. (i Um: Lin: paralicl LLXiS ihmmm to ﬁnd the mmimrlt nf
, i inertia uf the SIIlliHBI rectangle about the LILaxis.
.411” = 53] [upwade
EMA = {335M501 — {151(st — [imam] 4’3 It? — #313 “ME
+(2H300Hﬁ'1[3.’«'ﬁ}[30rij{g} = a = (1‘1?) {EMF + {and}:
E: 1220.5 {1.4: the right] = 3257 JT 7 L1 —I:1T5 5 f3} 3' 5?ﬁ— [EHEQﬁ‘ﬂ 2F; = A... +1220.6+[3nm[§]=3 4 MT
: ] _ .4511: = ~14ﬁﬁﬂ = ldﬁrﬂﬁ to the left _ Ana'wer is C.
Ansmr is B. 22. AL point E. ?_l'
212:me +Ei=0 15“.
F113. + 1220.15 0'
FED L —13b}3.3 [oumprewiun]:
‘2
25'” = Fan (4'25) + Em = 0
F — 4—1133 3) 2
“E _ ‘ ' 4.25
= I511] (minim)
A1: puim ﬂ;
2
2E) = —FAE ' FADf1 A“. = '0 4.35 FAD — (330 — 55]] 2—) = 339.111 {temion} (Mamba: ABC is horizontal and has no vertical fume
component} Answer is C. 23. A horizontal fume to the right will mutiler the
othcr moments. r _ i. —— Sulul'ium in: Emu l—Afl'emuol Sulfa: 85 — 1'01]. 312:. v=535=12ﬁ+4h1
1’11.
dim {d 2
=—= — 12 4x—1.—24:+4
” m2 Jim): 3+ 3' Al. x = 2.
a = [24}{2} + 4 = 52 Answer is C. 3?. The distance traveled, not. the change in pusitium
is rmquired, so check fur changes in the Sign ﬁr vclﬂcﬂy. If there are any rlwcrsalﬁ in position. then 1.? must pass
through zero. E = =1212+1t—1
V d: v is 23m whan The pﬂsitinns at thu intnrval endpoints :11": III21:31}??? —' {EH—2}“ — [2] + 3 = —19
zt—%}=I:4H—af+[2n—%f— —&}+3=3s
z [+5 = {4: ref T1121 [a‘f — w + a = 2—90? 3 (+2} = (4}[233 H2319}? — 5: + 3 = 41 The diﬁtanne travniﬂd E3
3.5 ~{—19)1+2.9n? — 3.5 +i41 — 2.91m _ 51.: Answer is B. 38. Inhagrate using the average force ﬁver math 1:5an
interval. _ '[ﬂ N5 + 45 m +1335 Ns
_ _ . __. .4 kg = 13.2 mfs Amar is. L“. 39. From the impulse—momentum principle, a impulsae — El momentum Fﬂf.=mﬂv
35
n x 1'
' WI
+(2ﬂN1—25 >[253J
2
=55.U' N5 The mass is irreie'lrant. Answer is B. 4121. Using the impulsemomentum principle and the re—
51.1135 UT Prob. 39, 3.5 .3 3.5
mm: 3;] Fcii=j th+ F'dt
El CI 3 2F T  N
:55 351,5 _lu. {9.5 S} — {SEES Ns Answer is D. # FI'IJIhﬁfDllDI PUbeElliﬂﬂir IﬂL ...
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This note was uploaded on 09/02/2008 for the course BEE 5330 taught by Professor Timmons/aneshansley/ during the Spring '07 term at Cornell.
 Spring '07
 TIMMONS/ANESHANSLEY/

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