HW 7 Solution

# HW 7 Solution - Gavan; F5: 435 N, LINE at: Arthur“ or: EB...

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Unformatted text preview: Gavan; F5: 435 N, LINE at: Arthur“ or: EB PHb-SE'b THRMH 0 72mm Club: MOMENT OF E: AWC, 90mm F\Rs‘r NOTE ___ _ doe. I: S “‘95 “My-W MIL-IL =91 wwvx 'n. 5 ma» c038: 9L1 . .3}... sm% Tl m a: a, LQsB=Q489 rm 3% =35 M ii“: :a slug! (4%5 N\k%3= 39° BY oasERVA1\ou..‘ ML=—xc\1 -ﬂr.“ WHERE. I u 0A2 ‘ww-— cows?“ - 6.055 m a ~1=D.cﬂZm-¥O.o§ M: 941.91 m Imam Me, =— mass. “qt-mac N3 - - Las- Hoz Mﬁkszs N3 = -"IZ.4-‘:3 M-m _ Etc“ "ll-'5 N~m§ ‘ Swag: '\I'Ec:r0?é. E, E Am: ; Have: SINK case) 2. ismﬁx +e.') + ism-e.» E = B£Cbseai+ 315.3%;3' g = mama-mam C. C Kmxuia—uumi'j BY bEmhu—Ymm .. ‘. *g\ = BC. Sud KK-‘5\ (“I I '1.Q\ 7. BC stLuH-Eﬁ L23} mammoth) 3.15 CONTINUED Now gxgz BLCuafLL-I-Eubi) 5L Luna“; + magi) = BC. (Case: smut. - smﬁs (33:0; L AND g'xg = ﬁakLohbi—smea;\ 1. Q kcusni+ \$‘Nki\ = E:C.(Lcs& tum K-I- 5w Ba :95 KM}: g4 EQURTRQC. THE maww- Ham-ab 959.5 cc: E95. 03 hub L23) TO ME. MRBN\T~JUES on 1mg. Rxmﬁ— Hum!) was or: Em... V53 hum L43" Rzavawaur. mamas.-- BC. 5m Lok- Pf) :- Bﬂkcosh smm— sauna». cosh} L EL EN Lat-3&3 s BC. Lens 3 smut. + away :35ch Q. aswwa 1h 5m (rt-25+Hukm- Pa) = 1 ms& mun. QR smut 2.05%: = i Swank- (53+ 5?: Smack-£53 ‘ m K E CHM): Gun VECTOR 2; NORMAL 16 THE PLANE becmab 49,— i_\_ hub ‘2 WHEN m Avi+2§-‘5\5 §g4i-1}__sb. (b3 =3i_33_4 2.». BY BEC\N\'\'\UM‘ THE. VELTDR 5‘23 1s NORMAL To THE, MNE DEF-unﬁt: EN 5 mm E}. TH», Evka 2" - lbe‘ i. i E u z_ -s 4 -"I -5 : L-N: -3\$\i+L—2.c+%\i + (-1-‘33‘5 : —4S'3__ \ﬁé- \SE THEN \gng = \S L..3\‘-+ L—q‘; {4}" =\=.3{Tl— Lad HIRV‘E _—z u; —4 = (mm: “Ann; +K\B—k.\\5 '2 8.2+IZE THEN “bed —— 4 Sauna-53‘- = 4 W _ x -— 33(2ng 4 Guam: Amman-\- sues E Ah": (3 o: A ‘PARRLUELDGMM Fmb: AREA 0: 'PARALLELQERKM mnen (on 1: =-."H_+'33-'s¥_._ g a ai+ a} .» SE 0:) E = t.'x__— cal—2E Q a-ZlA-Eihﬁs; HAW“- AREK A:\E-Q\ m '2. L i Exg = -1 a, -3 z z 5 . - L\S+L.31+Q—ka+3s)3 +L—ta—O‘5 =Z\-j_-|v'23;_-ZDE — THE.» A: {(chagzsyug-zonl A: MD 4 (b) i 3* \5 t. 3: 4' ' “:— LS-HoYl-n-{AuuAi +£30_\u\‘_r5" = \Si 4- \oi4-zog ' THEN Fx —_ 5 Km: (2.53 mi A=Zle9 ‘ ...
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## This note was uploaded on 02/17/2012 for the course CEE 240 taught by Professor Chang during the Spring '08 term at UMass (Amherst).

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HW 7 Solution - Gavan; F5: 435 N, LINE at: Arthur“ or: EB...

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