Egm2511 hw 3 - FRDELEM 2.4T :1 tum-u is pulh‘ng a partisan and n'ticr at a constant mam ‘ x the ndur weith 55H h and that the resultant fume R

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Unformatted text preview: FRDELEM 2.4T » :1. tum-u is pulh‘ng a partisan] and n'ticr at a constant mam ‘ x the ndur weith 55H h' and that the resultant fume R I - ' partwajl un til: towing yolu: A farm: an anglc Dr 1559 “rim the horizontal Ir . _‘m_ “W? dL'ltJETT‘lIII'Iu {tn the tcmtlnn in the taw rape AB, (45} If": mum DIR ' at? ' SOLUTION Frt-tL-Hud} Diagram: '_I|! — —?-_IRCU'53”n + = a = ———‘— T £05615” w a}; = u: —Imsinfifl” + Rsinfifi” —(55u N} = t1 m3msinfi5°] — 550 = 0 1:05:35” TAH[—sin3fl° + or rfl=4uan _ mam“ ansfi5° [M R {45:} N} marma- nunr MATERIAL I: 2001' Th: racism—Hm can-Imam, Im. an riafll run-trad. Na pfl'l'flil Ink—F '* . avdammrnwflmwbuymmm un' m Mum' . Wmmmwmatflhmmmfihaypfgamjfl . " fl -' '|. SDLUTIGH ' ""_"“'"""l'—‘*- 'l'wu cables. tlL-d together at L" are lauded as she Immjm a. | “'11. nf vuluua‘ at" H' for which thi: the ' cubic. Hudhnig. Diagram M C: BIT PROBLEM 2.55 ( 11' 1'." 11' +—fl" 150 28 3 . [5 15 + 1- n- —— . — .—— = f 51'.” + Uni Wusn [b] n 4 H E I ‘32:}: J1. —T--——15fllb-W=[} 17 Dr 3*r4f | Adding Equations{l}and[2)gives ?rfi.=msu+1w 1 11 . . . l? I Umngfiquanonfl} —?Tfl-+5 25 .=—H-" | W in as 25 ‘anrm T52401b= r£:140=fiw tunaion will nut exceed 24¢} lb in Izmir: =m H) (2} PROBLEM 2.50 Twe- eablea tied together at C are leaded I35 Shem. tint the maximum allowable tension in each cable 13 Wt} N, determm: {a} the magnitude of the largest three P which may be applied at C, {b} the corresponding value af 0:. SOLUTION \‘nte: In pl'LtlJlLIrllh nt'lhis type. P may be direeted aleng nne {tithe eables. with T = flu.“ in that cable and 1" l} in [he nlher, er P may he directed in such a way.r that Tis maximum in both Gables. The aeeend peaaibilily is investigated first. |‘Tee-BUd_t' Diagram it C: Feree Triangle "nice triangle is thtlL‘L‘an with Eli = |Hll"— 85” ,8 = 415” a = elem: N}ens4l5¢' = 121:5 N Sinee P :v H, anlutinn is correct {a} P = 1216 N 4 a = ISIF- 55” — 415“ = 7'15" [:5] a: = 115° 4 ...
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This note was uploaded on 12/15/2008 for the course EGM 2511 taught by Professor Jenkins during the Fall '08 term at University of Florida.

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Egm2511 hw 3 - FRDELEM 2.4T :1 tum-u is pulh‘ng a partisan and n'ticr at a constant mam ‘ x the ndur weith 55H h and that the resultant fume R

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