Homework - Problem#12.18 ME211HomeworkSet#1Solutions 9122008

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Unformatted text preview: Problem#12.18 ME211HomeworkSet#1Solutions 9122008 Twoforcesareappliedattheendofascreweyeinordertoremovethepost. Determinetheangle(090)andthemagnitudeofforceFsothattheresultant forceactingonthepostisdirectedverticallyupwardandhasamagnitudeof750N. Solution 1.)Tobegintheproblem,theforcesshouldberesolvedintodirectionsxandyas shownbelow. Fx = F * sin(30) - 500 * sin( ) Fy = F * cos(30) + 500 * cos( ) 2.)Usingtheconditionsoftheproblem,therewillbetwoequationsandtwo unknowns,whichyieldsenoughinformationtosolvetheproblem 0 = F * sin(30) - 500 * sin( ) 750 = F * cos(30) + 500 * cos( ) 3.)NowyoucansolvetheproblemsofFintermsoftheta 500 * sin( ) = F * sin(30) 500 * cos( ) = 750 - F * cos(30) 4.)Atthispoint,youmustrecallthatsin^2+cos^2=1 500 2 * (sin( )) 2 = (F * sin(30)) 2 500 2 * (cos( )) 2 = (750 - F * cos(30)) 2 500 2 (sin( )) 2 + cos( )) 2 ) = (F * sin(30)) 2 + (750 - F * cos(30)) 2 500 2 = (F * sin(30)) 2 + (750 - F * cos(30)) 2 F = 318.8N,F = 980.23N 5.)DuetothefactthatFresultedinaquadraticequation,therearetwoanswersfor theforce,sobothsolutionswillbetriedintheoriginalequationsforxandytosee whichanswersatisfiesbothequations. F = 318.8N 500 * (sin( )) = (318.8 * sin(30)) = 18.5 500 * (cos( )) = (750 - 318.8 * cos(30)) = 18.5 F = 980.23N 500 * (sin( )) = (980.23* sin(30)) = 0.0171 500 * (cos( )) = (750 - 980.23* cos(30)) = 101.408 6.)AsyoucanseethattheonlysolutionthatiscorrectiswhenF=318.8Nbecauseit yieldsthesameangleinbothequations,18.5. = 18.5 Answer F = 318.8N Answer Problem#22.35 ExpresseachofthethreeforcesactingonthecolumninCartesianvectorformand computethemagnitudeoftheresultantforce. 1.)TowritevectorsinCartesiancoordinates,eachforcehastobebrokendowninto componentsofx,y,andz.Thexcomponentisdictatedwithvector,i;they componentisdictatedbythevector,j;thezcomponentisdictatedbythevectork. TheCartesianvectorsofeachforceareshownbelow. 3 4 F1 = {150 * i -150 * j }lbs 5 5 Answers F2 = {-275 j }lbs F3 = {-75 * cos(60) i - 75 * sin(60) j }lbs 2.)Inordertofindthemagnitudeoftheresultantforce,youmustfirstaddthe componentsoftheforces. 3.)Thenfindthemagnitudeoftheresultantforcebysquaringeachcomponent value,addingthevalues,andtakingthesquarerootofthewholesummation. Answer Problem#32.50 Determinethemagnitudeandcoordinatedirectionanglesoftheresultantforceand sketchthisvectoronthecoordinatesystem. Answer 1.)Todeterminethemagnitudeoftheresultantvector,firsttheCartesian coordinatesofeachforcemustbewrittenandthenaddedtogetherbycomponent. F1 = {(350 * cos(60)) i + (350 * cos(60)) j - (350 * cos(45)) k }N F1 = {(175) i + (175) j - (247.5) k }N 4 4 3 F2 = {(250 * cos(30)) i - (250 * * cos(60)) j + (250 * ) k }N 5 5 5 F2 = {(173.2) i - (100) j + (150) k }N Fresult = {(175 + 173.2) i + (175 -100) j + (-247.5 + 150) k }N Fresult = {348.2 i + 75 j - 97.48 k }N 2.)Thenfindthemagnitudethesamewayaswedidinproblem#2. magnitude = (348.205) 2 + (75) 2 + (-97.48) 2 N = 369.3N Answer 3.)Thedirectionanglesareshownbelow. Answers Problem#42.71 DeterminethelengthsofwiresAD,BD,andCD.TheringatDismidwaybetweenA andB. 1.) FirstyouneedtofindthelocationofD rAD = (1- 2) i + (1- 0) j + (1-1.5) k rAD = (-1) i + (1) j + (-0.5) k rBD = (1- 0) i + (1- 2) j + (1- 0.5) k rBD = (1) i + (-1) j + (0.5)k 2.) rCD = (1- 0) i + (1- 0) j + (1- 2) k rBD = (1) i + (1) j + (-1)k rAD = (-1) 2 + 12 + (-0.5) 2 = 1.5m rBD = (1) 2 + (-1) 2 + (0.5) 2 = 1.5m rCD = (1) 2 + 12 + (-1) 2 = 1.73m Problem#52.91 DeterminetheprojectionoftheforceFalongthepole. 1.) Findthevectorofthepole. U = {(2) i + (2) j + (-1)k } 2.) Findtheunitvectorofthepole. Magnitude = 2 2 + 2 2 + 12 = 3 2 2 -1 u = {( ) i + ( ) j + ( )k } 3 3 3 3.) Findthedotproductoftheforcewiththeunitvectortoyieldtheprojection oftheforceonthepole 2 2 -1 Fprojection = F u = {( * 2) + ( * 4) + ( *10)}kN 3 3 3 2 Fprojection = kN Answer 3 Problem#63.5 DeterminethemagnitudeanddirectionalsenseofthemomentoftheforceatA aboutpointP. 1.) TheequationofamomentisM=rXF;itisacrossproduct.Todeterminethe moment,theforceandmomentarmmustbewritteninCartesian coordinates. r = {-2 i - 8 j + 0 k } F = {400cos(30) i + 400sin(30) j + 0 k } i j k MP = r F = -2 -8 0 400cos(30) 400sin* (30) 0 M P = {400cos(30) * (8) - 400 * sin(30) * (2)}k M p = {2371.28 k }N * m Problem#73.34 TheforceF={600i+300j600k}Nactsattheendofthebeam.Determinethe momentoftheforceaboutpointA. 1.) Similartoproblem#6,thecrossproductoftheleverarmoftheforceand forcemustdone,butfirstweneedtofindtheleverarminCartesian coordinates. r = {0.2 i + 1.2 j + 0k }m F = {600 i + 300 j - 600k }N i j k 2.) M P = r F = 0.2 1.2 0 600 300 -600 M P = {-720 i + 120 j - 660 k }N * m Problem#83.49 ThehoodoftheautomobileissupportedbythestrutAB,whichexertsaforceofF= 24lbonthehood.Determinethemomentofthisforceaboutthehingedaxisy. 1.)Knowingthatwehavea3Dproblemandwearetryingtofindthemoment,your firstthoughtshouldbethatweneedtodoacrossproduct,rXF=M.Tobegin,wefirst needtofindFinCartesianvectorform.Todothat,weneedtofirstdefinethevector uponwhichtheforceacts. ThepositionvectoractsalongthelineAB. rAB = rB - rA = {2,2,4} - {4,0,0} = {(2 - 4),(2 - 0),(4 - 0)} rAB = -2 i + 2 j + 4 k 2.)Thenweneedtomakethatvectoraunitvectorbyfindingthemagnitudeand dividingeachcomponentbythemagnitude. magnitude = (-2) 2 + (2) 2 + (4) 2 -2 + 2 + 4 k i j rAB = 2 2 2 (-2) + (2) + (4) 3.)Thenwewillmultiplythemagnitudeoftheforcebytheunitvectorthatitactsto yieldtheCartesianvectorformoftheforce. -2 + 2 + 4 k i j F = 24 (-2) 2 + (2) 2 + (4) 2 4.)Tocompletethecrossproduct,wemustdefinetheleverarmvector,r.Itdoes notmatterwherealongtheunitvectorthatyouwritetheleverarmtosolongasit startsfromthehingeanditendsalongtheunitvectorofforce.Ichosetohaveitrun tothebaseoftheforcevectoratthebaseofthepole.Youcouldalsowritertorun tothepointofforceonthehoodofthecar. r = {4 i + 0 j + 0 k }m 5.)Atthispoint,youhavebothvectorswritteninCartesianvectorform,andyoucan completethecrossproducttofindthemoment.Itisimportanttonotethatyouonly needthejcomponentbecauseyouarelookingforthatmoment. i j k 4 0 0 -9.79 9.79 19.59 M p = i (0 *19.59 - 0 * 9.79) - j (4 *19.59 - 0 * (-9.79)) + k (4 * 9.79 - 0 * (-9.79)) M P = i (0) - j (78.4) + k (39.16) M hinge = M P j = -78.4lb * ft MP = r F = Problem#93.70 Themeshedgearsaresubjectedtothecouplemomentsshown.Determinethe magnitudeoftheresultantcouplemomentandspecifyitscoordinatedirection angles. 1.) Inordertofindtheresultantmoment,youmustaddthecomponents respectivelyofthemoments;therefore,youmustwritebothmomentsin Cartesianvectorform. M1 = {50 k }N * m 2.) TodeterminethecomponentsofM2youmustdotwocalculationsforthex, andycomponents.Firstyoumustfindthecomponentontothexyplane. Youcanthenlookatthatvector,whichisnowina2Dplane,andfindthe componentonthexandyaxisbybreakingitintocomponentsoncemore. Forexample,forthexcomponentofM2,youmustfirstfindthecomponentof M2ontothexyplaneasshownbelow. M 2xcomponent = -20 * cos(20) Thenyouneedthecomponentonthexaxisofthisvalue. M 2xcomponent = -20 * cos(20) * sin(30) Doitsimilarlyfortheyandzcomponents,andyouyieldaCartesianvectoras follows. M 2 = {(-20cos(20)sin(30)) i - (20cos(20)cos(30)) j + (20sin(20)) k }N * m 3.) Thenaddthembycomponents M1 + M 2 = {(-20cos(20)sin(30)) i - (20cos(20)cos(30)) j + (20sin(20) + 50) k }N * m M1 + M 2 = {-9.39 i -16.28 j + 56.84 k }N * m 3.)Usingtheresultantofthemoment,youcanfindthemagnitude (M1 + M 2 ) magnitude = (-9.39) 2 + (-16.28) 2 + 56.84 2 = 59.866N * m Answer 4.)Thedirectionanglesareshownbelow. = cos-1 ( component i -9.39 ) = cos-1 ( ) = 99.0 magnitude 59.866 component j -16.28 = cos-1 ( ) = cos-1 ( ) = 105.8 Answers magnitude 59.866 component k 56.84 = cos-1 ( ) = cos-1 ( ) = 18.3 magnitude 59.866 ...
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