exm1_s10_solns

# exm1_s10_solns - Name KEY Discussion'SeCtion AU—" l...

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Unformatted text preview: Name KEY \ . Discussion'SeCtion AU— } " l . EMA 201 Exam #1 V March 3'“, 2010 This exam is closed book and closed notes except for-the appended “cheat sheet”. I Each problem is equally weighted. All work must be shown including free body diagrams where appropriate. we PrcoiStEx/x t3 ww1+lio W3 ‘~ g \ 1. Consider the two bars BA and CD )7 and the lines passing through them. In the calculations below, assume positive directions are from B to A and from Cto D. (a) Suppose a line passing through C is parallel to the line through BA (call it CE). What is the angle, between CE and CD? ' (b) Find the projection of res along the line passing through CD. (c) Find the smallest ‘ distance between point A and line CD. (a3 Mace van-02.3 (30mm twin 1 uses oorr venues) So “mm” ‘M’E {WM Gwen) Ct? M10 CD Cw 13C: FEM/vi) vSqu- “Mai v DOT YV/owc/Ti' _ . A F“ 2%-;926Q—2’24rcki4e,’ 73“:th l > Be: A (S A ~w = ”“ -”"—l-("»lc woe," 36 H" II 5 11 h— A A 215' l: 2‘. “re—:2 : [\b ’ in i on D 0/ l. YD - A L . A l_\_A an _Ek W 2 L 3 A A _ u) 0mm "mm mesa m Bo’h—l .UULT wave“, I be .. 1 ﬂ _ a ‘0 nor was 1.1.x -3; 4,1,3. . am a m H I ll lS‘ “ ...more workspace for Problem 1... /,. ' \~ 'I u '1 (QM SMMLGET bugsme Macaw pow-r A M LIL)? cc L) _ 1 \VT thL‘B’c’ 'M‘E VWVQUDLCULﬂ msmmce, we aw 0-HT MS KT WW . ~ Fwme W-moAFc/nmé 011 F3 MUG-MCD- . USmo- 1w: "FR/omme OF- A— mam- wuwaé‘ 7b FMJD A. A '3‘ A 4. A .— A ‘ .— Crir)(rc,¥ c a?) 216% , wily—nook .a I\ r :1 r 0 Ma -~ .‘5 z '. ' ((IPT)( n w: 01 713% __ _(,.3/+2.3_ If t3. 3 L 7.. \ H - 7.13 ea“ (mr)< d3JYZ’W: = "2. \$137 ' 2. The picture shown consists of five l , cables. Cable ABCD supports a drum ﬁ’ , , \ of weight W Cable DF is horizontal, ; r 91 L/ and cable segments AB and CD are vertical. The contact between the drum and cable ABCD is frictionless. lf cable ABCD has 600 lb breaking strength and all other cables having 200 lb breaking strength, determine the largest value W may have. (It is not necessary to incorporate any safety factor.) F‘Bbs 0F wawﬁr N m AomTShﬁ: W A (hm) (MA-Y To \$Wn'2z’l' Time» T If Timer) '1‘“ " ass \m) \/ Ll“ Q IA ' ‘A “D W “Mt; Fr» 0? N) “Rem: Z VT> \ 'l W 1m“ FW'SD OF ’A, We” HAW” A A A l '2. r Qlﬂvit —O is’l‘mE-‘E M) D _. T's—Eu 20.34314 \ l / a - 4 at ‘\z -9. =0 _ l?) vazf— o g rm, JrET/m Z — EN 0.2.7594 ’ ll’Z— u ’ "' P” raw-1m- Pts» or?— D', M7 ‘W Q 0 (\Pﬂi" 0 Tot: 13 \MD §TDG Top: OSQlH QP%2l‘-o éT—m “AT—O ~ 2 w 3 be rs he 2. ~ it“ ~ ...more workspace for Problem 2... to? cm M’Vtx “PA/lwuf cmmm ﬁn; INDIVIDUAL 0&ng To FWD "M? mmr WMZW N: Moor (mo ms 2 ‘3 .' 3) w: lZuo 155 my. Zcm Hos = 0.34% M :3 N 2 :73“ “,5 W: m “as 240.223 w a) H = W715; my. . 2W 11,3 2 0.3mm =>‘ N 2 Zoi’lbs be} m) m —_ Law“ => w H4415 €544“ N '3. A single force F exists in 3D space, with its 0 line of action passing through the point P having coordinates (1,y,z) m. The force is F = 6i + 8] + 10k N and it creates a mOment about point 0 of Mo = —14i + 8] + 2k N-mT NOTE: M0 is NOT a separate load — it is the moment created by F at O. (a) Find the coordinates yand zof point P. (b) Realizing that Mo = Fd, determine the perpendicular distance d from point ,0 to the line of action of F. (0) Considering now the equivalent force system at O, which consists of F and Mo, find where'the line of action of F intersects the xy-plane by matching the moment of F about Oto M i. ' Q A H) ~ 2, Q) VOPKF - V10 ( x L W 12 ‘ ‘ Es A _ B I A L ’ ~ , " A '1, “mm. C) XS: “ ‘3‘ % W‘" "[4L+?b+ \‘c _ /'\ ., (a r 8’ to U 'WM WCQNlM‘. lock- 8’? _._ -14, g t . v .‘ - ummui, (\L‘Hg> '0% ~10 - X’ Na ’ ; m 3’ " V3 ‘ m A- »Xctmg W “A “0 (MS 0‘36; W“: , I SMSPT "M: agar EQM, JUST mm) W MAerNue'S oil t— it H0) Twp 0L: Mo/P .— 2. : Z N.“ T“ “0* l4 + 811;” H" Y Q V ) l: '— of + WMOL = Muir/u CWT“) Q A: : l.l4‘7 w~ QWN k r——————.-—~ -—-——t- / 0 ...more workspace for Problem 3... (Q Show W amount“. SchTHM WMTS 0F IL SING-u; PM a / "In: MUMWT Vin Cam gm NYE: szw 13' Arc,ng .L'm (Wm R WWW marl H1 15 M1 MW” DOA/(9‘13) , TD Mom/E 1m? WWEM‘ WWW-m. "Me QUE?»an {WWW 70 mm W um: dF Ammo 0P 5 WCWW in mm m VOWT 0P wwwumw.w "ma? 5%— FLA-NE m 7mg Vow'r Q (3615‘ ﬂew/W3. A: \ 9 _ q A M :M , (JiVT)>rUQ-XL+U8B (E (S ’L I "L ’0. .J -> .. w‘ A Y “A ‘r-qu F — x L9, 0 .: —\¢u+?g+llo~v~ ‘ u 0 x 3’ lo ' / Q ...
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