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The content summarizes core methods for vector operations, including force vectors, angle calculations, moment computation via cross products, and vector projection. It then reviews equilibrium principles for particles and rigid bodies, equivalent systems for loads and couples, and centroid calculations using composite-part and integration approaches, followed by hydrostatic load and pressure relations and general friction concepts.",{"@graph":69,"@context":122},[70,84,105],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":20,"@type":76,"position":81},"https://docshare.wps.com/document/exam/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/me-270-fall-2019-final-exam-review-material/300773/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/me-270-fall-2019-final-exam-review-material/300773.png","ImageObject",300,407,{"name":92,"@type":93},"Ben ","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-27","2026-09-19",true,{"@type":102,"interactionType":103,"userInteractionCount":39},"InteractionCounter",{"@type":104},"ViewAction",{"@type":106,"mainEntity":107},"FAQPage",[108,114,118],{"name":109,"@type":110,"acceptedAnswer":111},"What does the review material cover under vector operations?","Question",{"text":112,"@type":113},"It covers forces as vectors, angle relationships, moment calculations (including the 3D cross product approach and 2D moment arm idea), and vector projections.","Answer",{"name":115,"@type":110,"acceptedAnswer":116},"How are equivalent systems defined for a loading case?",{"text":117,"@type":113},"Equivalent systems combine loading into an equivalent force and couple such that resultant force and moments match for any chosen reference point, while clarifying this concept is distinct from equilibrium.",{"name":119,"@type":110,"acceptedAnswer":120},"What key formulas are provided for hydrostatic loads and friction?",{"text":121,"@type":113},"Hydrostatic sections give buoyancy FB=ρgV and pressure p=ρgh, with guidance for hydrostatic load construction on flat, vertical, inclined, or curved surfaces. Friction guidance includes conditions for no sliding (f≤µSN) and special handling for impending slip and tipping vs. slipping problems.","https://schema.org",{"og:url":83,"og:type":124,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":126,"canonical":83},"index,follow",{"doc_id":128,"site_id":62},300773,1790508804,{"code":4,"msg":5,"data":131},{"doc_id":128,"user_id":132,"nickname":92,"user_avatar":133,"doc_module":4,"category_id":19,"category_name":20,"doc_title":65,"doc_description":67,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":39,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":139,"language":140,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":67,"update_tm":144,"read_time":145,"preview_page_count":4},2336478951081,"https://ap-avatar.wpscdn.com/davatar_276721f389ce27ea32af1340a28f341c","ME 270 – Fall 2019  \nFinal exam review material  \nME 270: Topics of Questions from Past Final Exams  \n\n|  | TOPIC | Fall 14 | Fall 15 | Fall 16 | Fall 17 | Spring 14 | Spring 15 | Spring 16 | Spring 17 | Spring 18 |\n| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |\n| 1 | vector operations |  |  |  |  |  |  |  |  |  |\n| 2 | equilibrium of particles |  | 1a |  | 1a |  | 1b | 1a | 1a |  |\n| 3 | equilibrium of rigid bodies | 3, 5a | 2, 5a | 2, 4a | 1b, 5a | 1a, 2 | 1a, 2, 4a | 1b, 2, 5a | 1b, 2, 5a | 5a |\n| 4 | equivalent systems | 2c |  | 3a |  | 5a |  |  |  |  |\n| 5 | centroids | 1c |  |  |  |  |  | 4d |  | 3a |\n| 6 | hydrostatic loads | 2a | 1d | 3d |  |  | 1c |  |  |  |\n| 7 | friction | 2b | 1c | 3c | 2 | 1d, 3 | 1d | 1c | 1c | 2a, 2b |\n| 8 | trusses | 4 | 1b |  | 1c | 1b |  | 1d, 3a, 3b | 3 | 1a |\n| 9 | frames and machines |  | 3 | 1 | 1d | 1c | 3 | 3c | 1d | 1b |\n| 10 | SF/BM diagrams | 5b | 4a, 5b | 4b | 5b | 5b | 4b | 4a, 5b | 4a, 5b | 3b, 5b |\n| 11 | axial stress |  |  | 1f | 3c |  | 3e |  | 3e |  |\n|  | shear stress-direct shear | 1d |  | 1h | 3d | 4c | 5a | 3c |  | 1c |\n|  | shear stress-torsion | 1b | 4b | 3e | 4 | 4d | 5b | 4b | 4b | 3c, 4 |\n|  | flexural stress-bending | 5c, 5d | 5c, 5d, 5f | 4c, 4e | 5e | 5d | 4c, 4d, 4e | 5c, 5d | 5c, 5d | 3d, 5c |\n| 12 | 2nd area moments |  | 4c, 4d | 3b, 4e | 3a, 3b, 5d | 4b, 5c | 5c, 5d | 4c, 4d | 4c, 4d | 1d, 3a |\n|  | Other | 1a |  |  |  | 4a |  |  |  |  |\n\n1. VECTOR OPERATIONS  \nForces as vectors  \n!   \nF = F AB = F !AB rAB  \nwhere AB points FROM point A TO point B.  \nAngles betwanb! vectors cosθ = b~~!~~​~~ ~~  \nEquivalently, you could use a cross product to find the angle; however, that is not recommended as cross products require more calculations, in general.  \nCalculations of moments  \nFor 3D forces, ALWAYS use the cross product to find momenM! tsO:= OA × F!A  \nwhere OA points FROM point O TO point A For 2D forces, use either the cross product approach above r use the! moment arm approach:  \nM  = d F  \nO A  \nwith direction found by the right-hand rule.  \n3Dprjection!s !  \nFCD = ( F i CD ) CD = projection of F onto line CD  \n! b  \nO  \n2. EQUILIBRIUM OF PARTICLES  \nFor eFF!D0!: (gives two equations in 2D, and three equations in 3D)  \n3. EQUILIBRIUM OF RIGID BODIES  \nFor eFMF!! D0: 0(g!i(ves two equationsgives two equationinsi2nD2,aD,ndanthdrtehereeequationsequationinsi3nD3)D)  \n4. EQUIVALENT SYSTEMS – Given loading in I. Find equivalent load Fand couple Mat give( Fo!i)nI(.∑ F! )II  \n(∑ M! B )I = (∑ M! B )II  \nfor ANY point B of your choice. This concept is NOT related to equilibrium!  \nSystem I  \nequivalent  \nSystem II  \nLine loads – Given w (x) . Find equivalent force Fand location d.  \nL  \n(∑Fy )II = (∑Fy )I ⇒ F = ∫ w (x )dx  \n0  \nL  \n(∑ MA )II = (∑ MA )I ⇒ Fd = ∫x w(x )dx  \n0  \n⎛ L ⎞  \n⇒ d = ⎝⎜  x w(x )dx ⎠⎟ / F  \nSystem I System II  \n5. CENTROIDS  \nCentroids – composite parts  \nA = ∑ AixC = ~~ ~~1A∑xi Ai y C = ~~ ~~1A∑ yiAi  \nThe coordinates for all components should be referenced to the same origin.  \nCentroids – by integration  \nUsing differential area parallel to y-axis:  \na  \ndA = y1 (x )− y2 (x )dx ⇒ A = ∫ y1 (x )− y2 (x )dx  \n0  \na  \nxC = ~~ ~~1A∫x dA = ~~ ~~1A∫xy1 (x )− y2 (x )dx  \n0  \na y  \ny C = ~~ ~~1A ∫ y dA = ~~ ~~1A ∫ ~~ ~~12 y1 (x )+ y2 (x )y1 (x )− y2 (x )dx y1 (x )  \n0 b   \n= 21A y21 (x )− y22 (x )dx ~~ ~~12 ( y1 + y2 )  y2 (x )  \nx  \nx a  \nAlternately, using differential area parallel to x-axis:  \nb  \ndA = x2 ( y )− x1 ( y )dy ⇒ A = ∫ x2 ( y )− x1 ( y )dy  \n0  \nb  \nxC = ~~ ~~1A ∫x dA = ~~ ~~1A ∫ ~~ ~~12 x2 ( y ) + x1 ( y )x2 ( y )− x1 ( y )dy 0  \nb  \n= 21A ∫ x22 ( y ) − x21 ( y )dy  \n0  \nb  \ny C = ~~ ~~1A∫ y dA = ~~ ~~1A∫ y x1 ( y )− x2 ( y )dy 0  \n12 (x2 + x1 )  \nxa  \n6. HYDROSTATIC LOADS  \nBuoyancy  \nThe buoyancy force on an object in a fluid is equal to the weight of the fluid displaced by the body in the fluid:  \nFB = ρgV  \nwhere Vis the volume of fluid displaced and ρ is the density of the fluid.  \nNote that if the maximum possib","cbCait9tn5f06gY8","https://ap.wps.com/l/cbCait9tn5f06gY8","pdf",7404744,26,"English","# ME 270 - Fall 2019\n## Vector operations\n## Equilibrium of particles\n## Equilibrium of rigid bodies\n## Equivalent systems\n## Centroids\n## Hydrostatic loads\n## Friction","[{\"question\":\"What does the review material cover under vector operations?\",\"answer\":\"It covers forces as vectors, angle relationships, moment calculations (including the 3D cross product approach and 2D moment arm idea), and vector projections.\"},{\"question\":\"How are equivalent systems defined for a loading case?\",\"answer\":\"Equivalent systems combine loading into an equivalent force and couple such that resultant force and moments match for any chosen reference point, while clarifying this concept is distinct from equilibrium.\"},{\"question\":\"What key formulas are provided for hydrostatic loads and friction?\",\"answer\":\"Hydrostatic sections give buoyancy FB=ρgV and pressure p=ρgh, with guidance for hydrostatic load construction on flat, vertical, inclined, or curved surfaces. Friction guidance includes conditions for no sliding (f≤µSN) and special handling for impending slip and tipping vs. slipping problems.\"}]","ME 270 - Fall 2019 - Final exam review material | PDF",1789777192,66]