{"id":48,"date":"2023-06-26T19:47:34","date_gmt":"2023-06-26T19:47:34","guid":{"rendered":"https:\/\/physigeek.com\/tr\/gerilme-direnci\/"},"modified":"2023-06-26T19:47:34","modified_gmt":"2023-06-26T19:47:34","slug":"gerilme-direnci","status":"publish","type":"post","link":"https:\/\/physigeek.com\/tr\/gerilme-direnci\/","title":{"rendered":"Gerilme direnci"},"content":{"rendered":"<p>Bu makalede fizikte \u00e7ekme kuvvetinin ne oldu\u011fu ve nas\u0131l hesapland\u0131\u011f\u0131 anlat\u0131lmaktad\u0131r. Bir ipin \u00e7ekme kuvvetinin ger\u00e7ek bir \u00f6rne\u011fini bulacaks\u0131n\u0131z ve ayr\u0131ca bu t\u00fcr kuvvetlerin \u00e7\u00f6z\u00fcml\u00fc egzersizleriyle antrenman yapabileceksiniz. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-la-fuerza-de-tension\"><\/span> Gerilme kuvveti nedir?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>\u00c7ekme kuvveti,<\/strong> bir ipin, bir kablonun veya herhangi bir elastik cismin gerilim alt\u0131ndayken, yani b\u00fck\u00fclemedi\u011finde uygulad\u0131\u011f\u0131 kuvvettir.<\/p>\n<p> \u00d6rne\u011fin bir ipin her iki ucuna da kuvvet uyguland\u0131\u011f\u0131nda ip gerginle\u015fir ve dolay\u0131s\u0131yla bir \u00e7ekme kuvveti uygular. Bir sonraki b\u00f6l\u00fcmde bir ipin uygulad\u0131\u011f\u0131 \u00e7ekme kuvvetlerini detayl\u0131 olarak inceleyece\u011fiz.<\/p>\n<p> Germe kuvveti Newton (N) cinsinden \u00f6l\u00e7\u00fcl\u00fcr ve normalde T harfi ile g\u00f6sterilir. Ayr\u0131ca, bir kuvvet t\u00fcr\u00fc oldu\u011fundan, \u00e7ekme kuvvetleri, y\u00f6n\u00fc ipin veya kablonun uzamas\u0131na paralel olan vekt\u00f6rlerdir.<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-de-una-fuerza-de-tension\"><\/span> Gerilme kuvveti \u00f6rne\u011fi<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> \u00c7ekme kuvvetinin tan\u0131m\u0131n\u0131 g\u00f6z \u00f6n\u00fcnde bulundurarak kavram\u0131n daha iyi anla\u015f\u0131lmas\u0131 i\u00e7in bir \u00f6rne\u011fi detayl\u0131 olarak inceleyece\u011fiz.<\/p>\n<p> Gerilme kuvvetinin tipik bir \u00f6rne\u011fi iptir. Bir ipe herhangi bir kuvvet uygulanmad\u0131\u011f\u0131nda halat gev\u015fek kal\u0131r ve dolay\u0131s\u0131yla \u00e7ekme kuvveti olu\u015fmaz. \u00d6te yandan, halat\u0131n her iki ucuna da bir kuvvet uyguland\u0131\u011f\u0131nda gergin kal\u0131r ve dolay\u0131s\u0131yla her iki ucuna da bir \u00e7ekme kuvveti uygular.<\/p>\n<p> Ayr\u0131ca halat\u0131n k\u00fctlesiz ve deforme olmayan bir cisim oldu\u011fu d\u00fc\u015f\u00fcn\u00fcl\u00fcrse halat\u0131n bir ucuna uygulanan kuvvet di\u011fer ucuna iletilir, bunun tersi durumda da ikinci uca uygulanan kuvvet birinci uca iletilir. ipin. IP. .<\/p>\n<p> Soldaki ki\u015finin uygulad\u0131\u011f\u0131 kuvvetin ( <sub>TA<\/sub> ) ipin sa\u011fdaki ki\u015fiye uygulad\u0131\u011f\u0131 kuvvet oldu\u011funu g\u00f6steren a\u015fa\u011f\u0131daki \u00e7izime bak\u0131n. Ve ayn\u0131 \u015fekilde sa\u011fdaki ki\u015finin uygulad\u0131\u011f\u0131 kuvvet (T <sub>B<\/sub> ) soldaki ki\u015fiye iletilir. <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/la-force-de-traction.png\" alt=\"Gerilme kuvveti\" class=\"wp-image-848\" width=\"365\" height=\"219\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/la-force-de-traction-300x179.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/la-force-de-traction.png 741w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<\/div>\n<p> Halat \u00e7ekme oyunu, gerilim kuvvetlerinin bir ip arac\u0131l\u0131\u011f\u0131yla iletildi\u011fi g\u00fcnl\u00fck hayattan somut bir \u00f6rnektir.<\/p>\n<p> Sonu\u00e7 olarak, kuvvetlerin bir v\u00fccuttan di\u011ferine iletilmesi i\u00e7in halatlar, kablolar veya benzeri nesneler kullan\u0131l\u0131r. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Como-calcular-la-fuerza-de-tension\"><\/span> Gerilme kuvveti nas\u0131l hesaplan\u0131r<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Gerilim kuvvetlerini hesaplama ad\u0131mlar\u0131 \u015funlard\u0131r:<\/p>\n<ol style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Ne dikey ne de yatay olan kuvvetleri vekt\u00f6rel olarak ayr\u0131\u015ft\u0131r\u0131n. Bu \u015fekilde t\u00fcm kuvvetler dikey veya yatay olacakt\u0131r.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Sistemin serbest cisim diyagram\u0131n\u0131 \u00e7izin, yani sisteme etki eden t\u00fcm kuvvetlerin grafi\u011fini \u00e7izin.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Sistemin denge denklemlerini kurun. Normalde yatay kuvvetler i\u00e7in bir denklem, d\u00fc\u015fey kuvvetler i\u00e7in ise ba\u015fka bir denklem kurulmal\u0131d\u0131r.<\/span><\/li>\n<li> <span style=\"color:#101010;font-weight: normal;\">Denklemlerden \u00e7ekme kuvvetini \u00e7\u00f6z\u00fcn ve de\u011ferini hesaplay\u0131n.<\/span><\/li>\n<\/ol>\n<p> \u00d6zetle fizikte <strong>\u00e7ekme kuvvetinin hesaplanabilmesi i\u00e7in <span style=\"text-decoration: underline;\"><a href=\"https:\/\/physigeek.com\/tr\/denge-kosullari\/\">denge ko\u015fullar\u0131n\u0131n uygulanmas\u0131 gerekir<\/a><\/span><\/strong> . Denge denklemleri belirlenerek \u00e7ekme kuvveti \u00e7\u00f6z\u00fclebilir ve dolay\u0131s\u0131yla de\u011feri bulunabilir.<\/p>\n<p> A\u015fa\u011f\u0131da bunun nas\u0131l ger\u00e7ekle\u015fti\u011fini g\u00f6rmek i\u00e7in hesaplanan gerilim kuvvetinin ad\u0131m ad\u0131m bir \u00f6rne\u011fi verilmi\u015ftir:<\/p>\n<ul>\n<li> K\u00fctlesi 65 kg olan bir cisim tavana bir iple as\u0131lmaktad\u0131r. \u0130pin v\u00fccudu desteklemek i\u00e7in ne kadar \u00e7eki\u015f g\u00fcc\u00fc uygulamas\u0131 gerekir? Halat\u0131n ihmal edilebilir bir k\u00fctleye sahip oldu\u011fu ve esnemedi\u011fi varsay\u0131lmaktad\u0131r.<\/li>\n<\/ul>\n<p> Her \u015feyden \u00f6nce, D\u00fcnya&#8217;n\u0131n v\u00fccudu \u00e7ekti\u011fi \u00e7ekim kuvvetini belirlemek gerekir. Bunu yapmak i\u00e7in a\u011f\u0131rl\u0131k kuvveti form\u00fcl\u00fcn\u00fc uyguluyoruz:<\/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-10adee55a222c172752922f0d901ea78_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=m\\cdot g=65\\cdot 9,81=637,65 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"261\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p> \u015eimdi serbest cisim diyagram\u0131n\u0131 olu\u015fturuyoruz. Bu durumda elimizde yaln\u0131zca iki dikey kuvvet vard\u0131r: ipin gerilme kuvveti ve a\u011f\u0131rl\u0131\u011f\u0131n kuvveti. <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-de-force-de-tension-resolu.png\" alt=\"Germe kuvvetinin kas\u0131tl\u0131 olarak uygulanmas\u0131\" class=\"wp-image-855\" width=\"243\" height=\"264\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-de-force-de-tension-resolu-276x300.png 276w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-de-force-de-tension-resolu.png 556w\" sizes=\"auto, (max-width: 276px) 100vw, 276px\"><\/figure>\n<\/div>\n<p> \u015eimdi dikey denge ko\u015fulunu ortaya koyal\u0131m. Yukar\u0131ya do\u011fru yaln\u0131zca bir dikey kuvvet ve a\u015fa\u011f\u0131ya do\u011fru bir dikey kuvvet oldu\u011fundan, v\u00fccudun dengede kalabilmesi i\u00e7in iki kuvvetin e\u015fit olmas\u0131 gerekir:<\/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-fe9470d16022905bcce2e03dd9a64bc9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\somme F_y=0\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"52\" 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-1cc7afff797d6c18363325cd43c54b50_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"TP=0\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"59\" 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-994386bc251ba773e73bfb0f81fbbb67_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"T=P\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"50\" 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-fa5442daf69fd2fb31db64f7ca4f0d1c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"T=637,65 \\N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"88\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-fuerza-de-tension\"><\/span> Gerilme kuvveti ile ilgili \u00e7\u00f6z\u00fclm\u00fc\u015f al\u0131\u015ft\u0131rmalar<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> 1. Egzersiz<\/h3>\n<p> A\u015fa\u011f\u0131daki \u015fekilde a\u00e7\u0131lar\u0131 g\u00f6sterilen iki ip ile as\u0131l\u0131 duran, k\u00fctlesi 12 kg olan kat\u0131 bir cisim verildi\u011finde, her bir ipin, cismi dengede tutmak i\u00e7in uygulamas\u0131 gereken kuvveti hesaplay\u0131n. <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-premiere-condition-dequilibre.png\" alt=\"ilk denge ko\u015fulu problemi\" class=\"wp-image-372\" width=\"243\" height=\"243\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-premiere-condition-dequilibre-300x300.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-premiere-condition-dequilibre-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-premiere-condition-dequilibre.png 600w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<\/div>\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\"> Bu t\u00fcr problemleri \u00e7\u00f6zmek i\u00e7in yapmam\u0131z gereken ilk \u015fey, \u015feklin serbest cisim diyagram\u0131n\u0131 \u00e7izmektir: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-de-la-premiere-condition-dequilibre.png\" alt=\"dengenin ilk ko\u015fulunun kararl\u0131 bir \u015fekilde uygulanmas\u0131\" class=\"wp-image-375\" width=\"282\" height=\"335\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-de-la-premiere-condition-dequilibre-252x300.png 252w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-de-la-premiere-condition-dequilibre.png 600w\" sizes=\"auto, (max-width: 252px) 100vw, 252px\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> As\u0131l\u0131 cisme etki eden yaln\u0131zca \u00fc\u00e7 kuvvetin oldu\u011funa dikkat edin: P a\u011f\u0131rl\u0131\u011f\u0131n\u0131n kuvveti ve T <sub>1<\/sub> ve T <sub>2<\/sub> tellerinin gerilmeleri. T <sub>1x<\/sub> , T <sub>1y<\/sub> , T <sub>2x<\/sub> ve T <sub>2y<\/sub> ile temsil edilen kuvvetler s\u0131ras\u0131yla T <sub>1<\/sub> ve T <sub>2&#8217;nin<\/sub> vekt\u00f6r bile\u015fenleridir.<\/p>\n<p class=\"has-text-align-left\"> B\u00f6ylece iplerin e\u011fim a\u00e7\u0131lar\u0131n\u0131 bildi\u011fimiz i\u00e7in \u00e7ekme kuvvetlerinin vekt\u00f6r bile\u015fenleri i\u00e7in ifadeler 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-bc09423d2d10435101c7d6b087add524_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\" T_{1x}=T_1\\cdot \\text{cos}(20\u00ba)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"135\" 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-0603d4b02835532dcefe2290484067fb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\" T_{1y}=T_1\\cdot \\text{sin}(20\u00ba)\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"133\" 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-0b10a6fc64a1a84b9f4f2c47b7990766_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\" T_{2x}=T_2\\cdot \\text{cos}(55\u00ba)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"135\" 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-3e7a1dc2ffa7eb20e5e2d9346f0b96a2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\" T_{2y}=T_2\\cdot \\text{sin}(55\u00ba)\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"133\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> \u00d6te yandan a\u011f\u0131rl\u0131k kuvvetini yer\u00e7ekimi kuvveti form\u00fcl\u00fcn\u00fc uygulayarak 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-dab14bd2fa937f39825c5add90b3ae58_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=m\\cdot g=12\\cdot 9,81 =117,72 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"261\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Problem ifadesi bize v\u00fccudun dengede oldu\u011funu, dolay\u0131s\u0131yla dikey kuvvetlerin toplam\u0131n\u0131n ve yatay kuvvetlerin toplam\u0131n\u0131n s\u0131f\u0131ra e\u015fit olmas\u0131 gerekti\u011fini s\u00f6yler. B\u00f6ylece kuvvet denklemlerini olu\u015fturabilir ve bunlar\u0131 s\u0131f\u0131ra e\u015fitleyebiliriz: <\/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-6532044e76d6b9246f64624159b08c33_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-T_{1x}+T_{2x}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"119\" style=\"vertical-align: -3px;\"><\/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-52aadf04437252b1f9c17107dfc16a84_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"T_{1y}+T_{2y}-P=0\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"140\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> \u015eimdi gerilimlerin bile\u015fenlerini daha \u00f6nce buldu\u011fumuz ifadelerle de\u011fi\u015ftiriyoruz: <\/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-4a4993c55ab7f27b6c0b67793ee5ff8a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-T_1\\cdot\\text{cos}(20\u00ba)+T_2\\cdot \\text{cos}(55\u00ba)=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"239\" 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-204773c167037418680872592d118315_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"T_1\\cdot \\text{sin}(20\u00ba)+T_2\\cdot \\text{sin}(55\u00ba)-117.72=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"293\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ve son olarak T <sub>1<\/sub> ve T <sub>2<\/sub> kuvvetlerinin de\u011ferini elde etmek i\u00e7in denklem sistemini \u00e7\u00f6z\u00fcyoruz: <\/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-c0aa7a8f7fe7234899f77039b42b47d1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left.\\begin{array}{l}-T_1\\cdot 0,94+T_2\\cdot 0,57=0\\\\[2ex]T_1\\cdot 0,34+T_2\\cdot 0,82-117 .72=0\\end{array }\\right\\} \\longrightarrow \\ \\begin{array}{c}T_1=69,56 \\ N\\\\[2ex]T_2=114,74 \\ 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 2<\/h3>\n<p> Comme le montre la figure suivante, deux objets sont reli\u00e9s par une corde et une poulie de masses n\u00e9gligeables. Si l&#8217;objet 2 a une masse de 7 kg et que l&#8217;inclinaison de la rampe est de 50\u00ba, calculez la masse de l&#8217;objet 1 pour que l&#8217;ensemble du syst\u00e8me soit dans des conditions d&#8217;\u00e9quilibre. Dans ce cas, la force de frottement peut \u00eatre n\u00e9glig\u00e9e. <\/p>\n<div class=&quot;wp-block-image&quot;>\n<figure class=&quot;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-dequilibre-des-forces.png&quot; alt=&quot;probl\u00e8me d'\u00e9quilibre translationnel&quot; class=&quot;wp-image-295&quot; width=&quot;299&quot; height=&quot;240&quot; srcset=&quot;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&quot; sizes=&quot;(max-width: 300px) 100vw, 300px&quot;><\/figure>\n<\/div>\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> Le corps 1 est sur une pente inclin\u00e9e, donc la premi\u00e8re chose \u00e0 faire est de vectoriser la force de son poids pour avoir les forces sur les axes de la pente : [latex]P_{1x}=P_1\\cdot \\text{sin}(\\alpha)&#8221; title=&#8221;Rendered by QuickLaTeX.com&#8221; height=&#8221;340&#8243; width=&#8221;2918&#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-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\"> B\u00f6ylece sistemin tamam\u0131na etki eden kuvvetler k\u00fcmesi \u015fu \u015fekildedir: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"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<\/div>\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-ed082b4f064316ab20fb0d26054d3010_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 ] Ainsi, la composante du poids de l'objet 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=\"87\" width=\"1160\" 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\"> \u015eimdi yer \u00e7ekimi kuvveti form\u00fcl\u00fcn\u00fc uygulay\u0131p denklemi basitle\u015ftiriyoruz: <\/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-left\"> Son olarak verileri yerine koyuyoruz ve 1. cismin k\u00fctlesini buluyoruz: <\/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-0457f85ca65afde96b2e575ce54869dd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1 \\cdot \\text{sin}(50\u00ba) =7\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"122\" 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-9a26d132815a0ce878a6ad874c8b40b0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1 =\\cfrac{7}{\\text{sin}(50\u00ba)}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"103\" 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<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Bu makalede fizikte \u00e7ekme kuvvetinin ne oldu\u011fu ve nas\u0131l hesapland\u0131\u011f\u0131 anlat\u0131lmaktad\u0131r. Bir ipin \u00e7ekme kuvvetinin ger\u00e7ek bir \u00f6rne\u011fini bulacaks\u0131n\u0131z ve ayr\u0131ca bu t\u00fcr kuvvetlerin \u00e7\u00f6z\u00fcml\u00fc egzersizleriyle antrenman yapabileceksiniz. Gerilme kuvveti nedir? \u00c7ekme kuvveti, bir ipin, bir kablonun veya herhangi bir elastik cismin gerilim alt\u0131ndayken, yani b\u00fck\u00fclemedi\u011finde uygulad\u0131\u011f\u0131 kuvvettir. \u00d6rne\u011fin bir ipin her iki ucuna da &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/tr\/gerilme-direnci\/\"> <span class=\"screen-reader-text\">Gerilme direnci<\/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-48","post","type-post","status-publish","format-standard","hentry","category-dinamik"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.3 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>\u25b7 Germe kuvveti nas\u0131l hesaplan\u0131r (\u00e7\u00f6z\u00fclm\u00fc\u015f al\u0131\u015ft\u0131rmalar)<\/title>\n<meta name=\"description\" content=\"Germe kuvvetinin ne oldu\u011funu ve nas\u0131l hesapland\u0131\u011f\u0131n\u0131 (form\u00fcl) a\u00e7\u0131kl\u0131yoruz. 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Egzersizlerle gerilim kuvvetleri ad\u0131m ad\u0131m \u00e7\u00f6z\u00fcld\u00fc.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/physigeek.com\/tr\/gerilme-direnci\/","og_locale":"tr_TR","og_type":"article","og_title":"\u25b7 Germe kuvveti nas\u0131l hesaplan\u0131r (\u00e7\u00f6z\u00fclm\u00fc\u015f al\u0131\u015ft\u0131rmalar)","og_description":"Germe kuvvetinin ne oldu\u011funu ve nas\u0131l hesapland\u0131\u011f\u0131n\u0131 (form\u00fcl) a\u00e7\u0131kl\u0131yoruz. 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