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# AERO ENGR MECHANICS AERO 211

Texas A&M

GPA 3.54

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This 13 page Class Notes was uploaded by Wilber Bradtke on Wednesday October 21, 2015. The Class Notes belongs to AERO 211 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 28 views. For similar materials see /class/226161/aero-211-texas-a-m-university in Aerospace Engineering at Texas A&M University.

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Date Created: 10/21/15

Friday September 26 2008 727 PM 7317 GOAL Finil thv S1 ipport lozirls at l GIVEN Referring tn the dnulvsis in Smtiun 36 WU are giwn thi39 goonn tr Hf the sysmnl and flu zippliml loads ASSUIVIE W assume gravity arts in the nugativc 241111111011 that the systi39ni is nonplzinzu and that thu 1111 is in r quililn39iuni DRAW 390 draw an FBD of tho towm Tho Anglos 1 and 15 1111151 in lotorninntil t39rmn thi39 gmnn trii39 infurinzitiun given in Tzililn 31 of Smtion I i From this tithh 139 min him that thu listiinm botm39on tho hvight 0139 pin 1 at tho top of thv nurth tawnr and the height of pin L tho rst pin south of rho nurth mwvr in 153 1117 901 111 626 111 From T211110 31 390 Nov that flit horizontal spacing hutm39i n Mljan mit pins is 101 111 HCHH n is given by 160 111 tan n 20 1n 1 686 Similarly the hi ight 11111111 111 antl tho lii39ight 0139 pin X the rst pin north Uftln nort 11 TOWUI is 153 111 7 711 111 816 111 ith 160 111 again thv spitting l h 33911 thi pins 1 is giwn by 16K in 816 111 tan J if 3311 Unfiled Notes Page 1 FORMULATE EQUATIONS AND SOLVE Applying the moment equilibrium 39Ulh x39 at 1 We have 11 7 MM 223927111T13inn 2227111T25i11 f 15 2227 11119531911630 MN 2227 111265 lsin 0311 MN VJ12U MN 111 lition about tllk 1 Applying rlm 1111111 11l qnililu illm umlit i011 211mm thv y nxis at A we have Z Mum 0 JIM l 133111T1lt39osa l 135111T105d 135111Tl rusn 135111T ms i Sinai Tho Hllt39i39ilmriuns from tho 139 s39 form39s in TllU 7111105 in ho 11101110111 abmu39 1110 yaxis at 1 nxaci 1y cam 1 our in This mutation w au v 101139 wth 111M l 39 m 1 wt 11mm Applying tllv 11101110111 Equilibrium miulitiuu about 1110 3 Z 1le 1 VV 135 111TLsi1m 1313 111T7gti11 3 135111Tlsi11rv 135111Tsinii zis the 39rlllll39lllllTiknlS from the Unsilo funM in tho 1is vqlmtiou lmviug us As in the moment lmldnm Almut tho I mlilm m l llK 11101110111 about l lll a391 m A L39xm39ily tunml out in 11 with 411A Tilll k in Vm tor foun WU 11mm Suppo 11707an M A 1 now apply Lhr J39 ltlll39 39Yill funo pxililn iuui equation 2 a 0 RI J Applying Ilil39r393939il1 ful cr vquililirium queuiun m39 haw Fr U I 2715111 7 2T sinJ R 22537si11686 quot MN 7 220511511163a MN I 411329 MN Unfiled Notes Page 2 Finally applying z rliroetiou lioree equilibrium equation We have 251 1 ll 2Tleoso 2Teos739 2 1m nz 196 MN 22537m686 MN 2205 1 15631 MN 1 100 MN 185 MN 21mm 1 63922 MN Thus in veetor form R 0529jt 522k MN 022 MN supme 39m39er of A As one39s intuition might suggest aiul as this simpli ed ana sis of the tower showed the primary l39uuetion ol39 the tower is to resist the downward load nauseil by both the weight ol39 the tower approximately a thir4l of the downward load and the downward eoniponents of the tension in the cables the remaining tmrthirds of the lown load The seeonrlary l39unetion of the tower is to resist the moment about the r axi eaused by the rlil l erenee in the horizontal eoniponents of the tensile i39orees in the eables Though this lil39lerenee is quite sinalli it aets at sueh a large distance from the base of the tower that its nionient about the base is signi cant CHECK e an verify that our answer is orreet by showing that the net moment about any other point is zero In partieular we will verify that the moment about B the top eorner ot the tower is zero Applying the moment equilibrium condition about the r I i is 13 we have 2 U 1 All l 227111Rt 120 MN1112271n70529MN 0 U ll Zl1Ull UJHl 2TinTosnt2T HINT100514135111ll mw 135mm 227 mm J About the y axis at B we have u 27 m2337m686 gtMN27m2lt55 1ms030 MN 135 1nlOG l39 135111 quot mlttll U i Finally about the axis at B we have 2111 n Unfiled Notes Page 3 Friday September 26 2008 727 PM 7257 GOAL a Find an quntiuu fur r vrrirnl furor at A as a funct39iun of tho diver position 1 h Dom11mm tlu forw at A when 1110 diver is standing m 13 v P10 the 39 1391i ELI force at as n fnnr un of r GIVEN Figure 1 57 Wlmrv tht diw r i5 39zki11g along thu lmurd ASSUME VO Msnnw gravity acts in the nogettiw y dirvction and that ho systcm is DIHD AL V w 39 s x d H L FORNIULATE EQUATIONS AND SOLVE Z Muwmc m 0 4mm 7 w r F 77quot 750 RESULTS a T110 Yortiml form at A is RESULTS 1 Vhon tho diw r is at point B r RESULTS 0 CHECK T110 plot shows that when 1 is nogntiw thv diver is Ix39tm vn A and B dn vrtiml form at A is pushing upward whvrms whl n 1110 divor beyond point B I39 is pHMTIV tho wrtital Ehrro at A is pulling dmx39nward which is ronsimonr with Tho problem st quot11011101111 3 Unfiled Notes Page 4 Friday September 26 2008 728 PM 73 H24 GOAL Find 11 rousitm ill 11w 1le md llu Suppon hmds m B GIVEN W0 are given le gvuuwny Hf tlw plum and That llu systvm is in mullihriu uv SSU m tln L39r utm39 uf grm ih39 and might nf w NIB 390 assuuw That tlw sysh m is nunplmml39 DRAW 39139 draw an FED of 11m plate qoo M FORMULATE EQUATIONS AND SOLVE Noting than Hm lliugr39 mum moment about Hug I mm itim ulmnt thv 1 ul N39 1 any m1 11mm JIM U n39 hogiu by applying thv mmm m l qllililn39illm HT 13 Z um u 6 InJT 21u1HUN Z V T 133 N 133N T 133 HMSUH in 39ubh 39l39u Hm Hm suppurt u w m B 139 Apply flw urw equililnilun 39muli nns me inspw un w mm rlom39lv SW 111 B t Bu n Simv llm r39 igt umhing 1n lmlnurv lhrvsu lkn ma 11111 111 1110 ydim39tiun rmv equil lu39ium 39platirm lmwm39m gins E R 0 B 1 400 N 13 133N 100 N B 267 N Thth in wwim mm wo haw FI ICN Mppw ffvr t39r Hf B Unfiled Notes Page 5 i1i1111 111111111 111139 1397e1xis 111 B Z 110111 11 112 11511111111 13111T WV 110 11111111 111v 111n1111 111 11111i1i111 i11111 111111 111 10511141Jl11 15111103 11 11 N010 111111 39 funk 1391113911 111 11311 111139 X2111 ahlo 11f 111139 11111si1111 111 1111111i11 111is 111141111 139111111 139 1112111 11111 1111111111 111191111 111L391l 111111111 hzwv giwu 11 nonzero 11111115411 11139 small 1111111111111 Xis 111 B Hmvvvor 11 1111111 i111111 1i1gt11 111 1111 FED 1111111391 51111111 111111 111011 is 110 1111111111111 211111111 1111 2w139i 211111 11mm F1111111V 1111 11111s1 apply 11111 1111111111111 1111111111139i11111 1 111111i1i1111 111111111 111139 1IB 1 111 3911111391HH11111 51111111111 1111111111111 111 B CHECK 1 1111 111111111111139211 rv1 1111111 1 391 if39ill1 111111 1110 11111 111111110111 11110111 111139 1111101 1111i111 is z139111 I11 1111391i1 1111139 139 will 1111339 111 111139 1111111111111 111111111 1111 111111111 111 g13913911y 39 is 711111 Applying 1111 11101111111 11111111i111 111111 11111111111111 111111111 1110 is 111 C 11 1111 139 11 11 1151103 1 111111 11 115111267 N 1111133N 11 110 2 111 0 2 111113 7 1 111139139 11 211121139T N7 gt111115535N1 11 11 1 11111l 11111 Maxis 211 391 haw Again i1 is 111121139 111211 1110111 is 1111 1111111111111 111111111 11111 139ilt NH 11111 1111111111111 111111111111111111 11111t1i1ilt11111111111111111 11 C39is1111111111211139 39 39 15111111 5111111 11111 11111 1111111111111 111rgt111 F in 211111 in 2111 11mm 13911111 111112111 11i1 1 1 1i1111si 11111 1111511111 1111M 111 1111 1 1211 Unfiled Notes Page 6 Friday September 26 2008 728 PM 7213 GOAL Fiwi HU zuiglv n GIVEN 390 zm given ThziT 11m Vi iglil if The 11mm is ingligihhu Thv gmlnul ry if The a 11101111th10nd 11M 1 w imam is a ilu vwl39rirm vlvmnvul and That thu syslum is in qllimll iluIL 390 assuuw Tho wighi 0T Tlu39 mblt mu Iw miglemi and Tho DRAW 39139 draw an FED of Tim lwmh T01 i is planar W 39 2m FORNIULATE EQUATIONS AND SOLVE 139 imply Th0 lHUlIH m T qniliininm mn riilimi uimut lhi rnxia at B iMFU 2 TUMI I IH OSmFJ 15 1n2 101 15 i RN 11310 v WEI Vv Thmi apply lhv r liu t39Tiun i39m39m L39qnililn39inm 39qlmTiou Zr n J TINWISH FT 511130 n 1 V i q Imansu ms 3 gt1113H uquot H JUL 71 115 Iquot V Nu irurlimi l39um qllilihl iluu umiil km 2 F H i gt 7 WHENquot 39I m sin l 2 3 Suhsliiiui39iug 1 and 2 into 339vngt1uiunu muntirm in only n msfii ltNgtai1m 2 FA Tu 4 how n12pr The 14 r Unfiled Notes Page 7 15 kXJU 211130quot lnun 2 k u L30quot mgh Mn rubh CHECK Our rwdt mak usc shu39t we know thv rin39ulnlmucul nf I39m 111ml art in NW lwgufivv 1 1i1 cquotriltm in unlm39 Tu k ilutt ml tho r39mnp0uun nf I39vuw Tm must ht aiming lvft 0f m riml as mu nuswvr shum39d 39l39u luwk mu nuswm39 murv 39illt hllh39 39 klmw TlmT 11w fours must lw unturnmf sium39 11v lwam is n fluovhu39m 39lvnmuf Using The 1W0 muulh39r tl izmghw L39l39l nh d in tlu diagram bvlmv it 1mm gtr39 11ml 13 m 2111180 r 15111t2116llc Simw thia rms lluv sump nwdl 01 n as that fuuurl in our mluliun hih prm39r s 11le llu furu39s aw 39nuvm39rmn and hmm 011139 umwm39 Imlsi 1w m wvt Unfiled Notes Page 8 Friday September 26 2008 28 PM PROBLEM 8127 The steel disk in E8 27 has an aluminum insert whose faces are ush with the faces of the disk The thickness of the disk is 30 mm Determine a the mass ofthe disk with insert and the location of its center of mass b the centroid of the disk with insert b n mm 2i lilnl lw min i IntMic 7 ill mm GOAL Find the mass and the center of mass CM of an aluminum steel composite disk GIVEN Dimensions of disk and insert and types of materials Lise ASSUME Materials are homogeneous with densities given in Table A2 1 DRAW Create a set of axes for the disk i 3a m duck mm in my FORMULATE EQUATIONS and SOLVE Divide the assembly into three paits Solid steel disk Steel rectangular parallelepiped that is removed from the disk to create a hole Aluminum rectangular parallelepiped that is inserted into the hole la n i r016003 241x10 m p3 7830kgIii3 iii p slquot 1889kg oo41003 480 xio nn pl ZoQOrgIiiquot 39 703769 For center of mass Lise 87 By Syminetiy XM 0mm ZM l 5mm This table summarizes our calculations for YM 7 00247 kgm 1865 kg 7132 Inm Center of mass 0 l32 15 mm b The location ofthe centroid is independent of the mass of the material It only depends on the geometric shape Based on symmetry the centroid occurs at the middle of the disk Centroid O O 15 mitt RESULTS mass 865 kg center of mass O l32 15 mm centroid 0 O lS inm Unfiled Notes Page 9 Friday September 26 2008 728 PM 7228 G AL Fintl tlu Slippm39r lnzuls m39tiug ml The plznw 211 111v li unl vluwl 21ml um I H Hl39 wlwvl GIVEN Vo zn o giw n the L Uigln of 110 plane its H39HH ul39 grzn ily its amnion uml llmf lllt SVSN UJ is in l qllllllu llllll ASSUlVIE 1 mm gravity mix in Tlm nugniiw j lil39 nll it mm 4151 lw s39nnml that 1110 svslvin inmr m39vn tllringli H n39lnlimlly 71w must L39Ullhlllt39l39 lllt mmm nl ulmm lln riaxim l lmvm39m rim to the synuuvn v nf I ll 11 Ull lll llw luml rlno m cutl dl vlivcl is llu mum ennl llIL inmnunt our 1 1m nlwnl llm JquotHVl oxzuIlv lamC15 0111 llic utlm Hi umn the mmuon r lmlnnw ulmm tlu 139 39is is inllmnnlimllv szili 39iml nlnlm flw assumption of synnnvl 139 39 RAW m haw an FBI or llm plenu n 27D mm iiiimam FORMULATE EQUATIONS AND SOLVE WU apply llw nmnioni L qllililn ilnn mu Zia10 lt39 gt 2 1I1F UTm80l Fl 2800 N litinn almm 1119 7 at FJ 2800 N in lln pusiiin yaliimzlinn SWINGl lam u jinn when 390 ilnx r apply 1w 1llll3939k l inn lin m 39qliililn ilnn mnliliun ZR U I gt 2F 17 8000 N 2173 2801 N 801 I B 20m N FD 2000 N in llu pusitivv ytlirm on HunHM out If I M1 Iml39 mw l CHECK 390 ran 39liork run ans m39 illmw by Vi i ih illg ilml ilw 1M nnnnvnl nlmnl any lllm39 mini is zm39u In pzn iii39nlni w will Verify Tlim llll innniont nlwnl l is zvm Applying ilw nmmvni L qllllllll llllu untlilinn nimnl llll39 ziaxia at L w lmw Zn 0 1 Unfiled Notes Page 10 2U21 VU 13 111150N 21112230HN 13110800051 l U 1 Sim0 1110 not mmuvm almut 1 is 4mm our auswm 111ml 1w L39Hl H I39L Unfiled Notes Page 11 Friday September 26 2008 728 PM 7415 GOAL Fiuti Hu h llh itm ill 111v mhh HHi llu Nippon inzuis m A and 39 GIVEN WU zut giwu Tim gmiuvfl39y uf flu39 sysivm 11w Tight 0f Hu39 inn the weight M llw post dlni that IiU sluimr smuul ll uml 11mm rl only normal funvs ASSU IE 0 l5 um g 39 Ads in MN imgntivo ltiil39 iinlii Hiu sfmfmu is pinum and rliv swimu in in 39pliliiii39inm RAW WC draw an FUD 0139 the Imr N010 Til T llJL39 topmost Liilm usinn SilUllid lm l u RNI ULATE EQUATIONS AND SOLVE 390 apply the immile equilibriqu mm s at 1 11mm 211mm llu EM 0 39 1 xi 7 3111 7 075Hi1150 II1T1P57539UH5UDUNIUUUNJ 1 Wu M applv IIIL J39iliil w39iit l him muililu ium L39UIMiiliHH Z F u g 5n 2 Using 2 In l39 i i T in 1 W lim39v 2723110147UTSSiuMY1n5i1150 1875ms50 mil0N 39 2 N HI slum u in FBD s uiim39f him 11 C Using 2 w haw 3 7amp11130 X Unfiled Notes Page 12 139 7100 N Minrm 171 1111 Lusil 1 applv 1111 1vdhwli1111 11111 L Alllililu illlu 39oudilinm Z R 11 11 gt F 11111551 101111 13 5221115 511 N 1111111 N F 1 6039 l N 39 111 111139 I mummf 1m 11 CHECK 1 l ll l1r 139k 1111139 unswvr above by V 11in rvg F is 10111 1g fllal H11 1101 111111110111 21110111 any other Unfiled Notes Page 13

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