HW5_sol

# HW5_sol - Prnblam 5.4 5 through 5.8 Locate the centroid...

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Unformatted text preview: Prnblam 5.4 5.] through 5.8 Locate the centroid ﬁfth: piam: ma Show. QIHFWJMM NV 1m, .iﬁrﬁnhmﬂcfgwrmfmé E :I?\$¢f an ‘1 15-4 3’ EH '3 I ElEl/Jildﬁyiz‘lj'5} 3” 3’1) 7" 1'" Wﬁﬁi max-m ? = 5": :h-m 1" Problem 5.5 E {#2315}: 3039.5; say-mg; 5:3 f=aaﬂa Em Problem 5.20 5.1? through 5.2!} A. thin homogeneous wire is bent to form The perimeter of the ﬁgure indicated. Locate the center of gravity of the wire ﬁgure thus formed. 5.20 Fig. P53. 5.1] The homogeneous wire A 3CD is bent as shown and is Eith to a hinge at C. Detemdne the length L for which ponion 3CD ofihe wire is horizontal. my; a. II '. _. . . For pure-r '.':-'J'r :3 _ ._. TL? (in? ﬂﬂf'.Er_JI-'I_ ow“ 1 .rrI--' . rte-r I or argue. .'-.I,-' :5.‘ Tee __’.-1n'j no: wig-r4: . .*.- - r .. Erie-95+ ee. :1! reef/f]- ber'rrw: IFTHHEI'Q. L... i e?‘ or = wager PFIE a;er mﬁ‘ﬂh WI 1: fﬂﬂﬂr“ WI: IKE-0 (ul— WE: Z «if +5 EMC: Q: [/2530 NIKE) + [Fﬂwﬂﬂ - @WXFFL): 0 furioer bmw d Problem 5.4! Problem 5.5!! 5.49 and 5.5:? Dctcrrnme the magnitude and location of the resultant of the distributed load shown. Also calculate the reactions atrt and B. R; ‘ if I200“ 5J= emu N} 32: ﬁlaooaX4.§}-=é?5¢w ' R= 2.700 +532): 9459 y '- ' ' - — x Rn KI HI 1‘ RE RE I = {Illegaglsgﬁaﬂszm} {Frat-trawl; E=Z.5?m .q Hm‘lwﬁ: J EMA = u== 4.53- .3.5? {amigo} 3:549” _E= £40H‘t'l‘d *1 fﬁﬂh Jeoo- 9450 FA a, : oogo M £3: 405ml «I! 5.49 and 5.59 ﬂeten'nine the magnitude and location of the resultant of the distributed load shown. Also calculate the reactions at A and B. 3‘ 33k”) R=3m "‘ +72Mﬂ=t31t32h93 3: :2 will E: aokh’i‘i +1 z'Ej-e o= IE-32+A= Prohiam 5.39 ' 5.4m Detemﬁne the total surface area ofthe body shown. ﬂew: err: W0 guaFHCEE. j H]: 2'3 Qt: Imam-M? ﬂaw Eofﬁﬁﬁ' 7W FIVE paras mun” ﬂutes:de W4 T3 as; FEW MEI? H: N THE :ﬁwﬁ'uﬂtﬂ mi: 2'??? L —.- HfiL -= Féﬂﬂé m: 3:; r41 : 35.33 we)? Wﬁﬁ E 3' 3 i ﬁr : 2.5'99305 {-ﬂﬂwﬂ 7324.; EueFrw ﬂick : a): : Etyﬂﬂgﬁwzi fa: §[500ijrfﬁrz Fe) - 35290 K22. Fa” Hﬂﬂﬂfﬁﬁ Eff-W Wﬁfxﬂl‘) =f?acn.¢ DEM : a : {ﬁgﬂéﬂﬂ‘fﬂmyﬂw +C'(?+"=’/)'_0 5" C = 65%? n5 crzi‘mf *1 .1. Z; :0: g +J§£3fé rja'anfﬁ -2?¢ofé= 5 f 3_ g:-§3J‘E J5 §:\$ﬂn?‘ﬂ’r " 5.32 Detenninr: {a} the distributed land we a: the and C of 1111: bem- ABC for Problem 5.32 . . _ _ . w‘tuch ﬂu: rcactmn at C IS- zcm, 1‘ b] the correspondmg reacuon at B. 5} = gram 25X+9Xx2+w= W15 gpﬂré’rmc, 921%) = 5’45, Far: 54,51 For? C =0: +5 Ewe: a: +{3Amxafmﬂ-(mgxgaj = g a; = 200 Egg £3 = 2m Kai/H 4 Eaﬁﬁgﬁp‘jﬂagﬂé Lp'ﬁLﬂE 9F ﬁg} = {ff-Fun g- ﬂoﬂfﬁrﬁmﬁ=ﬁ . _ 'ib’ﬁ J S=f€m£ﬁa §=msé+ 4 [ah 'thlﬁm 5-33 5.53 Determine the: center ufgravity uflhe machine elm-rant shown. IM- p/wﬂg 7W! MMym: am mm H25? a»: my; @NﬁJF/M cur H mmmyuauwz WEﬁJE {I}; .4 garmgﬂﬂ FEW-E“ (I!) am 29- MLF :vmvoﬁ' E E: mm. - “*3 = Jaf - 0.95.5“: if?! in i “33:59,. iﬂganxHE: 34‘53?th may: Wash-7mm?“ Er :VMEWW i=3 r" ,- “ ...
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HW5_sol - Prnblam 5.4 5 through 5.8 Locate the centroid...

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