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ReviewProblemsWorksheetSoln

ReviewProblemsWorksheetSoln - 3 h n EPA/law PRO ELEMS...

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Unformatted text preview: 3 \\ h) n EPA/law PRO ELEMS A3 .2? 'T g... T TT T ' R x “7, "6v :5 3233' . gg'a- O", a; 4x 50'7- Cu 3+“ 400 . 33 F —o fig 2 ’ x 5‘: “’00)" ( :5 lT‘ISOIé 0 ¢ R Nl‘ [‘50 ‘SP'ISD ‘ 1s 600N and the puileys ar A boy 15 hanging from the puHeys shown. are assumed to be vertical . By using the method of sections, find . . , L . . x’\ T I F \ 2F = O / » ng [451000 (0836a,°l° =O-J? N=8OO JU’ the forces produced in bars I, 2, and 3 of the 2F ’ = 0 .truss shown in Fig- J due to Mic action of a \000 N ‘f 1000 $5,450.90 : O “,7 T: 900 9‘) ha“ 0f\\'eight W = 1,0001%, resting at the JoimL D T — ’ and supported by a cable EF 5:260“. 4 _ / ‘ ‘ ‘ . ; 800 2M - O > ( The cable M 15 mum to AC- 'A 17240) + goo 02530.01 (.30)'Zoo 5‘: r1303 (75) z o A“ ‘9 6 1 A», :5‘10 is AY éVx=O ——-z Ax: 80"0 shaft/J! ——«>Ax=4g0 1|) $00 F 2M = 0 .3 i 7"”‘" . ' ‘ — ' = - _ : LT s 430 \ {Fl ” ~52O (,2,s)—FZ(20) » 0 v? *2. 500 3 \ F2. 5001 cu mm b “3* - — — —F ; 3w) :0 3 - e -300 cosam ( 5w) 5 n . V 3? iFy—O — 3?0 I F; :4]? 15 COM?“ ' ”i : 3 6° - , . . _ . _m /15 fo:o= l’s—‘KD +9005m30fi ECOSB’bfl 2;, E : 333 My Tame“ FIG; J BcHorté use FED-6F EIGHT side. 4-. Determine the qagnitude éf the {areséurg P, in p51“,- necessary to exert a force of 3000 1125 on the material at A. 1/» \ \ \\ K \ l'\ \ F = WM") ' Fm“ S‘tMMW‘ FM" FED ' /7 F ZF1=O = —P(umr)+2T—;Lsmls° BE) 0 —-P81r 1- , / AC sin‘S" \MM“ \\\\\\\§\\\\§§\‘ \ T 5000 - FL C0515 — — P8“ cosl5° =0 ' ' ’ Ts" .-..-. . ._ . ... I T .40 (no , - , ba . sus nded by wires at A and B and supporte l T 5‘ 25th {zxdogvtgggesysfia of egrficafl forces Shaun. Neglect the l — — ” Kl weight ef the frame. . I; The reaction_ et'ifnqport C is: _ . ‘0 V 40'“, 601:» ' 5 2MVM;5 :0 _ 30 ( ' 506.3) +5o(.)-— {(3)30 50 B---y £2 ~'_ 2M :0 _ ‘. c I“) . agar—Sag)-4o(z>—ao(4)—soc«)'flea»)—ro(s>=0\Te 85"” 3 . ZMBX:O I I ‘ if . - Aw 235(4)} 50(4) + 40 (4) 2. 00(1) + so (a) + lo(t)=o @ 0%ch 2p scam-.7Haggai-tweewe—so.|o 6. The derrick shown supports 2 4,0004!» load. M is held by 3: than 'zmd . socket at A and by two cables attached at poi-Ms Hand [-3. In the ' “ position shown, the derrick stands in a vertical plane foru‘ri»:-Ig.‘;‘m angle . _ ~~ «14, = 20° with.thé'xy_plane. Determine the tension mam-h cable and SE- ' v .. ' ‘ ‘ h T" _ —8’1‘-12—5"*0T< 1.3531135 8‘ HY} + (f . ,_ __ . . A «Q. ".t "To TE . - : BE 51 1‘ 2773 —,33 k) ,, A“ 4000 _,\ A A : 2 ‘ U. A2 Ay 0= [@Mx—woomwemxia w?» l. i 4...; ,- ‘A . ”8/; + (03 (5 :ZDZSIHCaovfloSZO’I‘ 52.1201) x 410003] + [113x @569 r7773 —.38 9125 W 7 Given a 200 lb. refrigerator on a ramp inclined at angle 6. The centervof gravity of the refrigerator is located at G. . v 2 Ma. : O ’ - ' .. - , , I z _ If e = 20° and the coefficient of friction between the refrigerator "ZOOIOSZOO'S) *‘ZOQ$7"20(3) +P(4)'O and the ramp is y = 0.5, then the minimum force P (as shown) to ‘ . » . .. move the refrigerator is ‘ . I .- 131'?) N _- ._ 1?: F12- Qb' . 4t V ‘ '. i‘fiéééP+2005mzo—5€—a¥=87.é9 filu'dC’ " 2 ' I Zszo—A; N2 Zoowsw = 1337.9 2 _ ~ and; z 43 Z/AN )’M RH ' 8107.4 (5)0819) i ZF1:O=—ZQOcoSZo+N—‘> 141187.? 57.(;< c3397 4.__YF/$O ‘ E 213:0 = .5(&87.‘1)—zoosm7,o—P ’R‘WMJ (CUF- VZ; ZS.[p £5 fiflwgu’ . =(W ‘ Yéfiv “A ° jaw;— 5w, .L . - Z X ”(W-H47 AA=-<><z-x->dv WA— f"é(@+§)(m-£ 4‘1 z :é(fiq+%z) _, S‘YCAA ’\ b —-.— i 4:", .. Y: m: 5 a __y_1 A \\\\\\\\\\\ ‘* r W: m Y “a I vs?!) Ix: XI X2; x //Il‘1: 568 lechWE mo+es 6. The derrick shown supports a 4.(XX)-ll) load. It is held by a hull and socket at A and by two cables attached al points I) and If. In the position shown. the derrick stands in a vertical plane forming an angle 4» = 20° with the xy plane. Determine the tension hum-ls cable «on! SE. T55 = 4720 12b .G‘Almwl-e :(bcl-ler.’) Fmd 'ch' wmm in table BE gig 13D. C 409015 ..—> = 0 2MA ‘ lie x (TED/”13D + @6339) + is" X—tloomk‘g‘ " O 17f(-31’|‘+.2%f|}7 2: #400033 :0 71440003620715 %(4m)éflb)% =0 Is my x lJBD (1,5: ? «.773 + .981) + T3165” ..77J «3239] +Dzssnw ((05201 453nm) Managua??? +)1/('-77XTW*T;e)W*/'Z/C%Xfig “Ky” 4: O T A ’ cm? .' '53 (TBD'TBe) + 4°°O('2%’>=O "” TED-EE : “fins 12 COMP: ’5' (TED *T333—400'D‘8l) =0 —-———-> TED 4,126 = (05;; ADD 4? 2TB!) = 3255 "'9 TED = ”“8 IL :- 4730 31’ 505112 «4» z TBe =‘H7' AM awhly‘S roumlivj Errand since. I ofllY carried 2 Accimal {AM-€5- ...
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