{"id":247,"date":"2023-06-23T09:26:11","date_gmt":"2023-06-23T09:26:11","guid":{"rendered":"https:\/\/physigeek.com\/tr\/normal-kuvvet\/"},"modified":"2023-06-23T09:26:11","modified_gmt":"2023-06-23T09:26:11","slug":"normal-kuvvet","status":"publish","type":"post","link":"https:\/\/physigeek.com\/tr\/normal-kuvvet\/","title":{"rendered":"Normal g\u00fc\u00e7"},"content":{"rendered":"<p>Bu makalede normal kuvvetin ne oldu\u011fu ve sorunun t\u00fcr\u00fcne ba\u011fl\u0131 olarak nas\u0131l belirlenece\u011fi a\u00e7\u0131klanmaktad\u0131r. B\u00f6ylece normal kuvvetin \u00f6zelliklerini bulacak ve ayr\u0131ca ad\u0131m ad\u0131m \u00e7\u00f6z\u00fclen egzersizlerle bu kuvvet t\u00fcr\u00fcn\u00fc uygulayabileceksiniz. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-la-fuerza-normal\"><\/span>Normal kuvvet nedir?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Fizikte <strong>normal kuvvet<\/strong> , bir y\u00fczeyin, \u00fczerine oturan bir cismin \u00fczerine uygulad\u0131\u011f\u0131 kuvvettir. Dolay\u0131s\u0131yla normal kuvvetin y\u00f6n\u00fc y\u00fczeye dik, normal kuvvetin y\u00f6n\u00fc ise d\u0131\u015far\u0131ya do\u011frudur, yani y\u00fczey normal kuvveti cisme do\u011fru uygular.<\/p>\n<p> Genel olarak normal kuvvet, D\u00fcnya&#8217;n\u0131n k\u00fctlesi olan herhangi bir cisme uygulad\u0131\u011f\u0131 yer\u00e7ekimi kuvveti olan <a href=\"https:\/\/physigeek.com\/tr\/fiziksel-agirlik\/\">a\u011f\u0131rl\u0131k kuvvetine<\/a> kar\u015f\u0131 koymaya yarar. Ancak cisim e\u011fimli bir y\u00fczey \u00fczerinde durdu\u011funda normal kuvvetin de\u011feri yeterli olmayabilir. A\u015fa\u011f\u0131da e\u011fimli bir d\u00fczlemdeki normal kuvvetin nas\u0131l hesapland\u0131\u011f\u0131n\u0131 g\u00f6rece\u011fiz.<\/p>\n<p> K\u0131saca <strong><u style=\"text-decoration-color:#4fd12f\">normal kuvvetin \u00f6zellikleri<\/u><\/strong> \u015funlard\u0131r:<\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">Normal kuvvet bir temas kuvvetidir, yani yaln\u0131zca iki y\u00fczey temas halinde oldu\u011funda uygulanabilir.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">Normal kuvvetin y\u00f6n\u00fc cismin \u00fczerinde bulundu\u011fu y\u00fczeye diktir.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">Normal kuvveti cisme uygulayan y\u00fczey oldu\u011fundan normal kuvvetin y\u00f6n\u00fc her zaman d\u0131\u015far\u0131 do\u011frudur.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">Genel olarak normal kuvvetin b\u00fcy\u00fckl\u00fc\u011f\u00fc, ortaya \u00e7\u0131kan kuvvetin destek y\u00fczeyine yans\u0131mas\u0131na e\u015fde\u011ferdir.<\/span><\/li>\n<li> <span style=\"color:#101010;font-weight: normal;\">Normalde normal kuvvet genellikle N veya F <sub>N<\/sub> sembol\u00fcyle temsil edilir.<\/span> <\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Como-calcular-la-fuerza-normal\"><\/span> Normal kuvvet nas\u0131l hesaplan\u0131r<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Genel olarak <strong>normal kuvveti hesaplamak i\u00e7in,<\/strong> dikey kuvvetlerin toplam\u0131 ve yatay kuvvetlerin toplam\u0131 s\u0131f\u0131ra e\u015fit oldu\u011funda bir cismin dengede oldu\u011funu belirleyen denge denklemlerinin uygulanmas\u0131 gerekir.<\/p>\n<p> Denge ko\u015fullar\u0131n\u0131 probleme uygulayarak normal kuvveti \u00f6nerilen denklemlerden \u00e7\u00f6zebilece\u011fiz ve dolay\u0131s\u0131yla normal kuvvetin de\u011ferini belirleyebilece\u011fiz. <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-a9333695ba02f6e089d628fe3622a2e5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{c}\\displaystyle\\sum \\vv{F_x}=0\\\\[2ex]\\displaystyle\\sum \\vv{F_y}=0\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"81\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-del-calculo-de-la-fuerza-normal\"><\/span> Normal kuvvet hesaplama \u00f6rne\u011fi<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Art\u0131k normal kuvvetin tan\u0131m\u0131n\u0131 bildi\u011fimize g\u00f6re, normal kuvvetin hesaplanmas\u0131na ili\u015fkin somut bir \u00f6rnek g\u00f6relim.<\/p>\n<ul>\n<li> 8 kg a\u011f\u0131rl\u0131\u011f\u0131ndaki bir cisim d\u00fcz zeminde hareketsiz duruyor. Yerin cisme uygulad\u0131\u011f\u0131 normal kuvvetin de\u011feri nedir?<\/li>\n<\/ul>\n<p> Bu problemde, v\u00fccut d\u00fcz bir y\u00fczey \u00fczerinde hareketsiz oldu\u011fundan, ona etki eden kuvvetler yaln\u0131zca a\u011f\u0131rl\u0131k kuvveti ve normal kuvvettir. <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-et-poids-normaux.png\" alt=\"normal g\u00fc\u00e7 ve a\u011f\u0131rl\u0131k\" class=\"wp-image-4215\" width=\"273\" height=\"297\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-et-poids-normaux-275x300.png 275w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-et-poids-normaux.png 570w\" sizes=\"auto, (max-width: 275px) 100vw, 275px\"><\/figure>\n<p> Yani bir cismin d\u00fcz bir y\u00fczey \u00fczerinde dengede durabilmesi i\u00e7in normal kuvvet (N) ile a\u011f\u0131rl\u0131k kuvvetinin (P) e\u015fit olmas\u0131 gerekir. Bu nedenle normal ve a\u011f\u0131rl\u0131k ayn\u0131 y\u00f6ne, ayn\u0131 mod\u00fcle sahiptir, ancak y\u00f6nleri z\u0131tt\u0131r.<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-318d3aaff48777c13e5ac24cb775f6b0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"54\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> B\u00f6ylece, normal kuvvetin de\u011ferini belirlemek i\u00e7in, cismin k\u00fctlesinin yer\u00e7ekimi ivmesiyle \u00e7arp\u0131m\u0131na e\u015fde\u011fer olan a\u011f\u0131rl\u0131\u011f\u0131n\u0131 hesaplamak yeterlidir: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-a7355a420808c17875e97713c4bef5ec_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P=m\\cdot g=8 \\cdot 9,81 = 78,48 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"283\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Fuerza-normal-en-un-plano-inclinado\"><\/span> e\u011fik d\u00fczlemde normal kuvvet<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Bu b\u00f6l\u00fcmde e\u011fimli bir d\u00fczlemdeki normal kuvvetin form\u00fcl\u00fcn\u00fc elde edece\u011fiz, \u00e7\u00fcnk\u00fc de\u011feri y\u00fczeyin d\u00fcz veya e\u011fimli olmas\u0131na g\u00f6re de\u011fi\u015fir.<\/p>\n<p> Buna g\u00f6re e\u011fik bir d\u00fczlem \u00fczerinde duran bir cisme etki eden kuvvetler a\u015fa\u011f\u0131daki gibidir: <\/p>\n<figure class=\"wp-block-image aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline.png\" alt=\"e\u011fik d\u00fczlemde normal kuvvet\" class=\"wp-image-4220\" width=\"308\" height=\"417\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline-222x300.png 222w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline-757x1024.png 757w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline-768x1038.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline.png 770w\" sizes=\"auto, (max-width: 222px) 100vw, 222px\"><\/figure>\n<p> Yukar\u0131daki \u015fekle bak\u0131n: D\u00fczlem e\u011fik oldu\u011funda eksen olarak d\u00fczleme paralel y\u00f6n\u00fc (eksen 1) ve d\u00fczleme dik y\u00f6n\u00fc (eksen 2) kullanmak daha uygundur. Bu \u015fekilde denge denklemlerini ifade etmek daha kolayd\u0131r.<\/p>\n<p> <strong>E\u011fik bir d\u00fczlemdeki normal kuvveti<\/strong> hesaplamak i\u00e7in denge ko\u015fulunu e\u011fik d\u00fczleme dik eksene uygulamak gerekir, \u00e7\u00fcnk\u00fc cismin bu eksende dengede oldu\u011funu ancak d\u00fczleme paralel eksende olmad\u0131\u011f\u0131n\u0131 garanti edebiliriz. .<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-badff1735c827c6562a4e074ea4b6bd2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\sum \\vv{F_2}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"80\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<p> Dolay\u0131s\u0131yla e\u011fimli bir d\u00fczlem \u00fczerindeki normal kuvvet, d\u00fczleme dik eksenin a\u011f\u0131rl\u0131\u011f\u0131n\u0131n bile\u015fenine e\u015fde\u011ferdir:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-ea3f790cf878ca23f77405f73a20e7c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"58\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p> D\u00fczleme dik eksenin a\u011f\u0131rl\u0131\u011f\u0131n\u0131n bile\u015feni, a\u011f\u0131rl\u0131\u011f\u0131n form\u00fcl\u00fc ile d\u00fczlemin e\u011fim a\u00e7\u0131s\u0131n\u0131n kosin\u00fcs\u00fcn\u00fcn \u00e7arp\u0131m\u0131na e\u015fittir:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-efeff10487f285abba9d74ee3eba6b45_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=P\\cdot \\cos(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"117\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-9fe1bc1d3a7fbacecb2ce1ccc1dadc67_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=m\\cdot g\\cdot \\cos(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"141\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> K\u0131saca, <strong>e\u011fimli bir d\u00fczlem \u00fczerindeki normal kuvvetin form\u00fcl\u00fc,<\/strong> normal kuvvetin cismin k\u00fctlesi \u00e7arp\u0131 yer\u00e7ekimi \u00e7arp\u0131 d\u00fczlemin e\u011fim a\u00e7\u0131s\u0131n\u0131n kosin\u00fcs\u00fcne e\u015fit oldu\u011funu belirtir: <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formule-de-la-force-normale-dans-un-plan-incline.png\" alt=\"e\u011fik d\u00fczlemde normal kuvvet form\u00fcl\u00fc\" class=\"wp-image-4232\" width=\"269\" height=\"92\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formule-de-la-force-normale-dans-un-plan-incline-300x102.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formule-de-la-force-normale-dans-un-plan-incline.png 576w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Fuerza-normal-y-fuerza-de-rozamiento\"><\/span> normal kuvvet ve s\u00fcrt\u00fcnme kuvveti<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Bu b\u00f6l\u00fcmde normal kuvvet ile s\u00fcrt\u00fcnme kuvveti aras\u0131ndaki ili\u015fkiyi g\u00f6rece\u011fiz \u00e7\u00fcnk\u00fc bunlar matematiksel olarak ba\u011flant\u0131l\u0131 iki t\u00fcr kuvvettir. Ama \u00f6nce s\u00fcrt\u00fcnme kuvvetinin ne oldu\u011funu bilmeniz gerekir.<\/p>\n<p> S\u00fcrt\u00fcnme kuvveti (veya s\u00fcrt\u00fcnme kuvveti), bir cismi p\u00fcr\u00fczs\u00fcz olmayan bir y\u00fczey \u00fczerinde hareket ettirmeye \u00e7al\u0131\u015f\u0131rken ortaya \u00e7\u0131kan bir kuvvettir. Dolay\u0131s\u0131yla s\u00fcrt\u00fcnme kuvveti cismin hareketine kar\u015f\u0131 koyan bir kuvvettir.<\/p>\n<p> S\u00fcrt\u00fcnme kuvveti normal kuvvetten hesaplan\u0131r. Daha kesin olarak <strong>s\u00fcrt\u00fcnme kuvveti, y\u00fczey s\u00fcrt\u00fcnme katsay\u0131s\u0131n\u0131n normal kuvvetle \u00e7arp\u0131m\u0131na e\u015fittir.<\/strong><\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b8e2dc6a1180d664163aeb969b289073_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=\\mu \\cdot N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"86\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p style=\"margin-bottom:5px\"> Alt\u0131n: <\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:8px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-5b005ac29604de5f2904d2da7ade0238_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"22\" style=\"vertical-align: -3px;\"><\/p>\n<p> s\u00fcrt\u00fcnme kuvvetidir. <\/span><\/li>\n<li style=\"margin-bottom:8px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-05d9eae892416bd34247a25207f8b718_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"11\" style=\"vertical-align: -4px;\"><\/p>\n<p> s\u00fcrt\u00fcnme katsay\u0131s\u0131d\u0131r.<\/span><\/li>\n<li><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-7354bae77b50b7d1faed3e8ea7a3511a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: 0px;\"><\/p>\n<p> normal bir diren\u00e7tir. <\/span><\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-fuerza-normal\"><\/span> \u00c7\u00f6z\u00fclm\u00fc\u015f normal kuvvet egzersizleri<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> 1. Egzersiz<\/h3>\n<p> 5 kg a\u011f\u0131rl\u0131\u011f\u0131ndaki bir cisim d\u00fcz zeminde hareketsiz duruyor. Daha sonra ilk cismin \u00fczerine 3 kg k\u00fctleli ba\u015fka bir cisim eklenirse, iki cismi desteklemek i\u00e7in zeminin uygulad\u0131\u011f\u0131 normal kuvvet nedir? Veri: g=9,81 m\/ <sup>s2<\/sup> . <\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#FFF8E1\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>\u00c7\u00f6z\u00fcm\u00fc g\u00f6r\u00fcn<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Zeminin her iki cismi de desteklemesi gerekti\u011finden normal kuvvet, her iki cismin a\u011f\u0131rl\u0131\u011f\u0131n\u0131n kuvvetinin toplam\u0131 olacakt\u0131r. Bu nedenle \u00f6nce her bir cismin a\u011f\u0131rl\u0131\u011f\u0131n\u0131 hesaplay\u0131p sonra bunlar\u0131 toplayaca\u011f\u0131z.<\/p>\n<p class=\"has-text-align-left\"> A\u011f\u0131rl\u0131\u011f\u0131n kuvvetinin, v\u00fccudun k\u00fctlesinin yer\u00e7ekimi ile \u00e7arp\u0131lmas\u0131yla hesapland\u0131\u011f\u0131n\u0131 unutmay\u0131n.<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c0cdb663ec9f8fe79fbecd960b50fc39_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=m\\cdot g\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"75\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> B\u00f6ylece 5 kg&#8217;l\u0131k bir v\u00fccudun a\u011f\u0131rl\u0131\u011f\u0131n\u0131 hesapl\u0131yoruz:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-a15ef1db0a6608fa8e6165ac0e12e925_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1=5\\cdot 9.81=49.05\\N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"160\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> \u0130kinci olarak k\u00fctlesi 3 kg olan ikinci cismin a\u011f\u0131rl\u0131\u011f\u0131n\u0131 belirliyoruz:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-d57016899db324fc0f785e92341e9f2f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=3\\cdot 9.81=29.43\\N\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"161\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> B\u00f6ylece dikey denge ko\u015fulunu uygulayarak normal kuvvetin iki a\u011f\u0131rl\u0131\u011f\u0131n toplam\u0131na e\u015fit oldu\u011funu elde ederiz: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c65761c8213d33892f422dd6b0a29121_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\sum \\vv{F_y}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"81\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-808b8980c7dbb5f2b1cdf14418fea88c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P_1+P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"99\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sonu\u00e7 olarak, zeminin uygulad\u0131\u011f\u0131 normal kuvvetin de\u011feri: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c5819ab4ff951b1edbe6efa0a0111243_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=49,05+29,43=78,48 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"237\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Al\u0131\u015ft\u0131rma 2<\/h3>\n<p> A\u015fa\u011f\u0131daki \u015fekilde g\u00f6sterildi\u011fi gibi, iki cisim k\u00fctleleri ihmal edilebilir bir halat ve bir makara ile birbirine ba\u011flanm\u0131\u015ft\u0131r. E\u011fer 2 cismi <sub>m2<\/sub> =7 kg ise ve rampan\u0131n e\u011fimi 50\u00b0 ise, t\u00fcm sistemin dengede olmas\u0131 i\u00e7in e\u011fik d\u00fczlemin <sub>m1<\/sub> k\u00fctleli cisme uygulad\u0131\u011f\u0131 normal kuvveti hesaplay\u0131n. Egzersiz boyunca s\u00fcrt\u00fcnme kuvvetini ihmal edin. <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-des-forces.png\" alt=\"\u00f6teleme dengesi problemi\" class=\"wp-image-295\" width=\"299\" height=\"240\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-des-forces-300x241.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-des-forces.png 718w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#FFF8E1\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>\u00c7\u00f6z\u00fcm\u00fc g\u00f6r\u00fcn<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Cisim 1 e\u011fimli bir e\u011fim \u00fczerinde oldu\u011fundan yap\u0131lacak ilk \u015fey, kuvvetlerin e\u011fim eksenleri \u00fczerinde olmas\u0131n\u0131 sa\u011flamak i\u00e7in a\u011f\u0131rl\u0131\u011f\u0131n\u0131n kuvvetini vekt\u00f6rle\u015ftirmektir: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c05811c44aa2d58295c811d612a54eee_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_{1x}=P_1\\cdot \\text{sin}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"128\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-1a0b77602980cc17cce9b3baef744df8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_{1y}=P_1\\cdot \\text{cos}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"130\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dolay\u0131s\u0131yla sistemin tamam\u0131na etki eden kuvvetler k\u00fcmesi \u015funlard\u0131r: <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-equilibre-des-forces.png\" alt=\"\u00f6teleme dengesi egzersizi \u00e7\u00f6z\u00fcld\u00fc\" class=\"wp-image-296\" width=\"338\" height=\"272\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-equilibre-des-forces-300x241.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-equilibre-des-forces.png 718w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p class=\"has-text-align-left\"> Problem ifadesi bize kuvvetler sisteminin dengede oldu\u011funu, dolay\u0131s\u0131yla iki cismin de dengede olmas\u0131 gerekti\u011fini s\u00f6yler. Bu bilgiden iki cismin denge denklemlerini \u00f6nerebiliriz: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b62bbb21cbec2be0bba7f8a839b12ba9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"1\\ \\rightarrow \\ \\begin{cases}P_{1x}=T\\\\[2ex]P_{1y}=N\\end{cases} \\qquad\\qquad 2 \\ \\rightarrow \\ T=P_2[\/latex ] Par cons\u00e9quent, la composante vectorielle du poids du corps 1 inclin\u00e9 dans le sens de la pente doit \u00eatre \u00e9gale au poids de l'objet 2. [latex]P_{1x}=P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"83\" width=\"1404\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-4e1b75b6ba5d7bbe88d23e014eb011c5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1\\cdot \\text{sin}(\\alpha)=P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"120\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> \u00d6nceki denklemden cismin 1 k\u00fctlesini hesaplayabiliriz: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-06a53a846ad5bc034f69fa05488404c4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1\\cdot g \\cdot \\text{sin}(\\alpha) =m_2 \\cdot g\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"174\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-802fde26f3388538d766a709d60cf48b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1 \\cdot \\text{sin}(\\alpha) =m_2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"130\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-16ae359d38a8a11d1b1db4988b8eeaf1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1 \\cdot \\text{sin}(50\\text{\u00ba}) =7\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"130\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-4249c6e274233595f50eedc1da64f56f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1 =\\cfrac{7}{\\text{sin}(50\\text{\u00ba})}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"111\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-6e80f0daabb2167ec2f6622b08001a97_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1=9,14 \\ kg\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"106\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> \u00d6te yandan sistemin kuvvet diyagram\u0131na bakt\u0131\u011f\u0131m\u0131zda normal kuvvetin 1 numaral\u0131 cismin a\u011f\u0131rl\u0131\u011f\u0131n\u0131n e\u011fik d\u00fczleme dik vekt\u00f6r bile\u015fenine e\u015fit olmas\u0131 gerekti\u011fini g\u00f6r\u00fcr\u00fcz. <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-82b47c80ab7ef66a41fc4d4425032831_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_{1y}=N\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"66\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-56ad7b690b37b3f53ca20597e165860b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1\\cdot \\text{cos}(\\alpha)=N\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"120\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dolay\u0131s\u0131yla bu denklemden normal kuvvetin de\u011ferini bulabiliriz: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-f258dccd08d6573f74a2261b2192a92f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_1\\cdot \\text{cos}(\\alpha)\\\\[3ex]N=m_1 \\cdot g\\cdot \\text{cos}(\\alpha)\\\\[ 3ex]N=9,14 \\cdot 9,81 \\cdot \\text{cos}(50\\text{\u00ba})\\\\[3ex]N=\\bm{57,63 \\ N}\\end{array}[\/ latex]\n\n<div class=&quot;wp-block-otfm-box-spoiler-end otfm-sp_end&quot;><\/div>\n<h3 class=&quot;wp-block-heading&quot;> Exercice 3<\/h3>\n<p> Nous pla\u00e7ons un corps de masse m=2 kg au sommet d&#8217;une rampe avec un angle d&#8217;inclinaison de 30\u00ba. Quel est le coefficient de frottement entre la rampe et le corps si celui-ci est maintenu en \u00e9quilibre ? Donn\u00e9es : g=9,81 m\/s <sup>2<\/sup> <\/p>\n<figure class=&quot;wp-block-image aligncenter size-full is-resized&quot;><img decoding=&quot;async&quot; loading=&quot;lazy&quot; src=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-force-normale-et-de-force-de-friction.png&quot; alt=&quot;&quot; class=&quot;wp-image-4253&quot; width=&quot;285&quot; height=&quot;176&quot; srcset=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-force-normale-et-de-force-de-friction-300x185.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-force-normale-et-de-force-de-friction.png 702w&quot; sizes=&quot;(max-width: 300px) 100vw, 300px&quot;><\/figure>\n<div class=&quot;wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1&quot; role=&quot;button&quot; tabindex=&quot;0&quot; aria-expanded=&quot;false&quot; data-otfm-spc=&quot;#FFF8E1&quot; style=&quot;text-align:center&quot;>\n<div class=&quot;otfm-sp__title&quot;> <strong>Voir la solution<\/strong><\/div>\n<\/div>\n<p> Comme dans tout probl\u00e8me de physique portant sur les forces, la premi\u00e8re chose \u00e0 faire est de dessiner le diagramme du corps libre du syst\u00e8me. Ainsi, toutes les forces qui agissent dans ce syst\u00e8me sont : <\/p>\n<figure class=&quot;wp-block-image aligncenter size-full is-resized&quot;><img decoding=&quot;async&quot; loading=&quot;lazy&quot; src=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-normale-et-friction-force.png&quot; alt=&quot;exercice r\u00e9solu de la force normale et de la force de frottement&quot; class=&quot;wp-image-4254&quot; width=&quot;285&quot; height=&quot;333&quot; srcset=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-normale-et-friction-force-256x300.png 256w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-normale-et-friction-force.png 702w&quot; sizes=&quot;(max-width: 256px) 100vw, 256px&quot;><\/figure>\n<p> Ainsi, pour que le syst\u00e8me soit en \u00e9quilibre, la somme des forces sur les axes 1 et 2 doit \u00eatre \u00e9gale \u00e0 z\u00e9ro. Par cons\u00e9quent, les \u00e9quations suivantes sont vraies : [latex]F_R=P_1&#8243; title=&#8221;Rendered by QuickLaTeX.com&#8221; height=&#8221;454&#8243; width=&#8221;7014&#8243; style=&#8221;vertical-align: 0px;&#8221;><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-ea3f790cf878ca23f77405f73a20e7c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"58\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Art\u0131k normal kuvvetin de\u011ferini ikinci denklemden hesaplayabiliriz:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-780db8c589b96d398e1400444a11db30_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_2\\\\[3ex]N=P\\cdot \\text{cos}(\\alpha)\\\\[3ex]N=m \\cdot g\\cdot \\text{cos }(\\alpha)\\\\[3ex]N=2 \\cdot 9,81 \\cdot \\text{cos}(30\\text{\u00ba})\\\\[3ex]N=16,99 \\ N\\end{array} \" title=\"Rendered by QuickLaTeX.com\" height=\"196\" width=\"171\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> \u00d6te yandan s\u00fcrt\u00fcnme kuvvetinin de\u011ferini birinci denklemi kullanarak belirliyoruz:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-bef5af0f3a7e907aa90f08435f538cf7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}F_R=P_1\\\\[3ex]N=P\\cdot \\text{sin}(\\alpha)\\\\[3ex]F_R=m \\cdot g\\cdot \\text{sin }(\\alpha)\\\\[3ex]F_R=2 \\cdot 9,81 \\cdot \\text{sin}(30\\text{\u00ba})\\\\[3ex]F_R=9,81 \\ N\\end{array} \" title=\"Rendered by QuickLaTeX.com\" height=\"196\" width=\"175\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Benzer \u015fekilde s\u00fcrt\u00fcnme kuvveti, a\u015fa\u011f\u0131daki form\u00fcl kullan\u0131larak normal kuvvet ve s\u00fcrt\u00fcnme katsay\u0131s\u0131 ile ili\u015fkilendirilebilir:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b8e2dc6a1180d664163aeb969b289073_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=\\mu \\cdot N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"86\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> B\u00f6ylece s\u00fcrt\u00fcnme katsay\u0131s\u0131n\u0131 denklemden silip de\u011ferini hesapl\u0131yoruz: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-2bee3710c7506bf8ff2456662a57f279_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu=\\cfrac{F_R}{N}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"59\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-69da73a9c8ca8ef047563bcb0b957d4b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu=\\cfrac{9,81}{16,99}\" title=\"Rendered by QuickLaTeX.com\" height=\"42\" width=\"80\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-87da99c1b6541f3ad374e4ebb3e9daf1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{\\mu=0.58}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"66\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Bu makalede normal kuvvetin ne oldu\u011fu ve sorunun t\u00fcr\u00fcne ba\u011fl\u0131 olarak nas\u0131l belirlenece\u011fi a\u00e7\u0131klanmaktad\u0131r. B\u00f6ylece normal kuvvetin \u00f6zelliklerini bulacak ve ayr\u0131ca ad\u0131m ad\u0131m \u00e7\u00f6z\u00fclen egzersizlerle bu kuvvet t\u00fcr\u00fcn\u00fc uygulayabileceksiniz. Normal kuvvet nedir? Fizikte normal kuvvet , bir y\u00fczeyin, \u00fczerine oturan bir cismin \u00fczerine uygulad\u0131\u011f\u0131 kuvvettir. Dolay\u0131s\u0131yla normal kuvvetin y\u00f6n\u00fc y\u00fczeye dik, normal kuvvetin y\u00f6n\u00fc ise &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/tr\/normal-kuvvet\/\"> <span class=\"screen-reader-text\">Normal g\u00fc\u00e7<\/span> Devam\u0131 &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"","footnotes":""},"categories":[5],"tags":[],"class_list":["post-247","post","type-post","status-publish","format-standard","hentry","category-dinamik"],"yoast_head":"<!-- This site is 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