HW 5 Solution

# HW 5 Solution - (62E FIGURE on TH‘E LEFT GWEN“ Tap...

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Unformatted text preview: (62E FIGURE on TH‘E LEFT) GWEN“. Tap: 6:616 WEIQHT DF CRETE _ _.-_......—-—.._—-.—._-— 6E1: LEFT—HAND (:0qu m2 pawn-now m: EQleﬁlzumEq; J I 835+ 3.7T“ _ (J erTﬁa+§Tﬁc+gTﬁpﬁw:o I I (2; _ 11' -..: 3 —o.3s Thai-1:, TM, 3:), TM: a - C MA'KIUG‘ TAD-:sm-fb fol) nus soLUnJG- FOR 1"...“b '. T :- 40 61;, T “65.51112.” “3 a.¥9(77)( "3) "5 was-r: wrms FOR TAB mun 7‘; D m (3) AND Sowwt. Fm? Tnca , . :7 - -l - 0.36(666.67)+% Ec— F, (ate) :0, IM- 57(“5'0—‘fé‘ﬂb 5UB§TITUTJNG F01? THE raw sums mm (2): W: 0.8(666.£7{b)+.:—3(%9 lb)+ \$—\$(6l6 21;) -: (565.5 {5 W=1868Ib 4 GIVEN: u) PLATE WEIGHS 6015. (2) mm FORM 30" Aweﬁs wer VERTICAL. FIND: TENSION IN EFHH WIRE, WE no-ra THﬂT 'rH-E VERTICAL FORCE f ﬁf‘Ef-{TED BY THE SUPPDRT 0M inr D. [a EuUHL rN' HHGW'TU DE To THE WEIGHT OF THE PM TE: P: 6015-. “PM; FREE BODY: D Z1350: -" : 0"' 6 ‘ ' ° lwsn?) Smﬁ +Epsm50wsw +1119 551430.:05 50° r. O on : eTRD sin 50' +TeDcm¥D'+Tcgtm€(l§): 0 Z}; = 0: TM, \$in30'cosso‘+'rwsan au‘arnnu‘ 4’” sinso‘ain so“ = o 0R: 7 ‘ . a I . Theo: '30 +TBDS'nﬂ0 - gméo:(§)) MU LT'PLY ('3 BY sin 50:01.) 31’ Cosso', Imp ADJ? : ‘m (0’5 50.035 50'—5 “'50” 5 in M) + 7;”,(6 u‘néa'cos #o' +60560'sinw'):0 REGLLIMG Tum C¢5qmsp—Sindsin/5 =cas(a[.m) m Sin arcs/s +cosa(€f“/5 = sincaup w: HM: 7nw(uo)+-rbpsin(mo‘) ‘ 53:0,397zﬂ TAD (3) suamn‘urt mm (1') map sow; Foe 1gp: T” (- Sin 55+ 0.31:2!“ cos-w“) + Baggy: o '0-5 0 T + ‘= ' - -’ 00 m: wasso o Tcpﬁ 1M (19 EF=O-‘-T 3---r °._ - _ J has 0 “coast: \$195,504,604 -0 0“ 19+Tab+129 = 67.252 n: (5) 5095117911? Fan" (5) much) HJT0(5),- 1'5‘1‘72‘76 TAD: 61.18116 T”:2=1.5;6{b (5) suBrn-ru‘rE F-Rull L5) INTO (5) 9ND ('0: TED-‘0.3‘t7276(2‘7.5'lb (A) TED? Lb ‘ '— 'co = Tm; “QB—"215\$ <1 2.123 GWEN: Cow-mm“ o: WaeHT “1:270”: 15 susPEMDED FRaH mm, A. CA’QLE Bnc misses THROUGH- KING 9. FWD: P AND Q Foe EumLIBRIuM puswwwswwn .___.__._-__—_ FREE BODY: RINGﬁ Z::0: T +'_|' +F+Q+VU=O 15 M .. .. ._ WHERE f: P‘: . -n.; _ CSHME Teuuw T m 801'» PomIoNS OF cnsLE) WE, HAVE ms 449%); +(7z;n.) 3' —(16in.)1< HB-BBin (Z'finJi, “may; «ramp; Rc=77in ﬂ _ as . . Inquaﬂa:TH—B—z(-——éb+if-;B)T Sussnwmc FOR I”, 1:“, £9, MD!“ INTO if“) AND FACTDRNG £4,152 .__5_ T +3.— T T P)r: t] 1| 77 q Zé .. ' +('—'T+77ZT Wig '+(‘TIT'77T+Q)5:° 557nm; TH-E Coernu‘ems OF L, 3 l5 gaunt Tb ZERU‘ Mu: REDUCING: " Q) J%T+F=0‘ ' - ' (t) (D gT—Wﬂ) ' (z) (E) q¥T+Q:D _> (3) TIME E1161 Fem—(2) We Emile—i=0». 1-: ‘ _ 77 _ 1-- mam IL) - tsm lb Suasnwurms Foe-r m EQ\$.(I) Rummy; am.“ P:36.DH:>, Q: 59.0 Lb ‘ GIVEN : (I) CDLLRIU A AND B CUNNELTED 3v WIRE or LENGTH 525'mm {2.) 13-.- (3mm); 3: l5"; mm' HM): (4313; (E) Q Foa'eaummuu (Elgmm): (ZOO mm)‘+ (GS-urme £2. a, . «3: H‘éonm A3 :(100 mm): "(lsﬁmmjé +gt+gowmjﬁl 93:53” ?‘ =§—E:1ﬂi._ 553+mk "RB 52‘3- ﬁ 525 .— Pi FREE 300le CDLLDHL ﬂl' (Asf: 9C1+ 31%;: 9 “ if =0: N115 NI}; N13 r?! 4455+ 1:8 2%“) I 315%” suasnTmme FoR a”, MD SETTING TH-E COEFF. (ST-5' mom. TO 25215: 5'25 MMNG 9:391“ mm Somme- Fox 7‘33.- T = \$525 3‘“ N r: (5) FREE BODY; cum». 3 “ET-"=0 4; an, ,7 . A“ “53 * Q5 " Twang“ 3 sues-raw'rwe Fm; g MSETTMG- .. M: NIL» THE coeFF. aF'E EqumLTo 2520: Q5 |' __ “a! - Q (375 '95) ‘ 0 Mﬂ'KING “‘ I” : HSS'N‘ M1: .Souflnlé FOR (.1: __ ‘Hao _ _.5_i§(1l55'~) (340mg ‘ ...
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HW 5 Solution - (62E FIGURE on TH‘E LEFT GWEN“ Tap...

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