{"id":253,"date":"2023-06-23T06:39:35","date_gmt":"2023-06-23T06:39:35","guid":{"rendered":"https:\/\/physigeek.com\/tr\/egik-duzlem\/"},"modified":"2023-06-23T06:39:35","modified_gmt":"2023-06-23T06:39:35","slug":"egik-duzlem","status":"publish","type":"post","link":"https:\/\/physigeek.com\/tr\/egik-duzlem\/","title":{"rendered":"E\u011fik d\u00fczlem"},"content":{"rendered":"<p>Bu makale fizikte e\u011fik d\u00fczlemlerin ne oldu\u011funu ve bu t\u00fcr problemlerin nas\u0131l \u00e7\u00f6z\u00fcld\u00fc\u011f\u00fcn\u00fc a\u00e7\u0131klamaktad\u0131r. E\u011fik bir d\u00fczleme etki eden kuvvetlerin form\u00fcllerini bulacaks\u0131n\u0131z ve ayr\u0131ca e\u011fimli d\u00fczlemde ad\u0131m ad\u0131m \u00e7\u00f6z\u00fclen egzersizlerle antrenman yapabileceksiniz. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-un-plano-inclinado\"><\/span> E\u011fik d\u00fczlem nedir?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>E\u011fik d\u00fczlem<\/strong> belirli bir a\u00e7\u0131yla e\u011fimli bir y\u00fczeydir. Fizikte e\u011fik d\u00fczlem kuvvet problemlerini \u00e7\u00f6zmek i\u00e7in kullan\u0131l\u0131r.<\/p>\n<p> \u00d6rne\u011fin bir rampa veya e\u011fimli bir yol e\u011fimli d\u00fczlemlerdir.<\/p>\n<p> E\u011fik d\u00fczlem, bir nesneyi daha az kuvvet kullanarak ta\u015f\u0131man\u0131za olanak tan\u0131r. Bir nesneyi e\u011fik bir d\u00fczlemde itmek, onu dikey olarak kald\u0131rmaktan daha az kuvvet gerektirdi\u011finden.<\/p>\n<p> Ayr\u0131ca e\u011fik d\u00fczlem alt\u0131 klasik basit makineden biri olarak kabul edilir. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Formulas-del-plano-inclinado\"><\/span> E\u011fik d\u00fczlem form\u00fclleri<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Art\u0131k e\u011fik d\u00fczlemin tan\u0131m\u0131n\u0131 bildi\u011fimize g\u00f6re, e\u011fik d\u00fczlemde hangi form\u00fcllerin etkili oldu\u011funu ve bunlar\u0131 hangi denklemlerin birbirine ba\u011flad\u0131\u011f\u0131n\u0131 g\u00f6relim.<\/p>\n<p> E\u011fik d\u00fczlem egzersizlerinde kar\u015f\u0131la\u015ft\u0131\u011f\u0131m\u0131z ilk sorun, kuvvetlerin \u00e7o\u011funun e\u011fik d\u00fczleme paralel veya dik y\u00f6nde etki etmesidir. Dolay\u0131s\u0131yla tipik koordinat eksenleri (bir dikey eksen ve bir yatay eksen) bu t\u00fcr problemler i\u00e7in pek kullan\u0131\u015fl\u0131 de\u011fildir. Bu nedenle genel olarak e\u011fik d\u00fczlemlerde farkl\u0131 bir koordinat sistemiyle \u00e7al\u0131\u015f\u0131r\u0131z: <\/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\/plan-incline.png\" alt=\"e\u011fik d\u00fczlem\" class=\"wp-image-4369\" width=\"391\" height=\"368\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/plan-incline-300x283.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/plan-incline-768x724.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/plan-incline.png 1010w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p> <strong>Fizikte e\u011fik d\u00fczlem problemini \u00e7\u00f6zmek i\u00e7in iki farkl\u0131 eksen kullan\u0131r\u0131z:<\/strong> y\u00f6n\u00fc e\u011fik d\u00fczleme paralel olan birinci eksen ve y\u00f6n\u00fc e\u011fik d\u00fczleme dik olan ikinci eksen.<\/p>\n<p> Ayr\u0131ca resimde g\u00f6rebilece\u011finiz gibi <strong>e\u011fik bir d\u00fczlemde genellikle \u00fc\u00e7 farkl\u0131 kuvvet etki eder<\/strong> (e\u011fer s\u00fcrt\u00fcnme varsa): a\u011f\u0131rl\u0131k kuvveti, normal kuvvet ve s\u00fcrt\u00fcnme kuvveti (veya s\u00fcrt\u00fcnme kuvveti). Ancak mant\u0131ksal olarak e\u011fik d\u00fczlemde s\u00fcrt\u00fcnme yoksa s\u00fcrt\u00fcnme kuvveti ihmal edilir.<\/p>\n<p> Ancak a\u011f\u0131rl\u0131\u011f\u0131n kuvveti vekt\u00f6rel olarak iki bile\u015fene ayr\u0131l\u0131r: e\u011fik d\u00fczleme paralel bir bile\u015fen ve e\u011fik d\u00fczleme dik ba\u015fka bir bile\u015fen. Bu \u015fekilde t\u00fcm kuvvetler e\u011fik d\u00fczlemin \u00e7al\u0131\u015fma eksenlerinde ifade edilebilir. B\u00f6ylece, e\u011fimli bir d\u00fczlem \u00fczerinde duran v\u00fccudun a\u011f\u0131rl\u0131\u011f\u0131n\u0131n iki bile\u015feni, e\u011fim a\u00e7\u0131s\u0131n\u0131n sin\u00fcs\u00fc ve kosin\u00fcs\u00fc ile hesaplan\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-a26edbf89d563f1351d0ec9771f7e7bc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1=m\\cdot g\\cdot \\text{sen}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"142\" 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-f5f191747feef04cf0a63f61a6b56cfd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=m\\cdot g\\cdot \\text{cos}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"141\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Son olarak e\u011fik bir d\u00fczleme etki eden kuvvetler a\u015fa\u011f\u0131daki iki form\u00fclle ili\u015fkilendirilebilir: <\/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\/formules-du-plan-incline.png\" alt=\"e\u011fik d\u00fczlem form\u00fclleri\" class=\"wp-image-4388\" width=\"491\" height=\"151\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-du-plan-incline-300x93.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-du-plan-incline-768x237.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-du-plan-incline.png 1022w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p> Problem tan\u0131m\u0131nda aksi belirtilmedi\u011fi takdirde, e\u011fik d\u00fczlemdeki cisim yoku\u015f a\u015fa\u011f\u0131 kayabilir, bu nedenle d\u00fczleme paralel eksen i\u00e7in denklemde olas\u0131 bir ivme yer al\u0131r. \u00d6te yandan cisim e\u011fik d\u00fczleme dik eksen y\u00f6n\u00fcnde hareket edemedi\u011finden kuvvetlerin toplam\u0131 s\u0131f\u0131rd\u0131r. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-resuelto-del-plano-inclinado\"><\/span> E\u011fik d\u00fczlemin \u00e7\u00f6z\u00fclm\u00fc\u015f \u00f6rne\u011fi<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> E\u011fik d\u00fczlem problemlerinin fizikte nas\u0131l \u00e7\u00f6z\u00fcld\u00fc\u011f\u00fcn\u00fc g\u00f6rebilmeniz i\u00e7in a\u015fa\u011f\u0131da ad\u0131m ad\u0131m \u00e7\u00f6z\u00fclm\u00fc\u015f bir \u00f6rne\u011fi g\u00f6rebilirsiniz.<\/p>\n<ul>\n<li> E\u011fimi 45\u00b0 olan bir d\u00fczlemin tepesine m=6 kg k\u00fctleli bir cisim yerle\u015ftiriyoruz. E\u011fer cisim e\u011fik d\u00fczlem \u00fczerinde 4 m\/s <sup>2<\/sup> ivmeyle kay\u0131yorsa, e\u011fik d\u00fczlem y\u00fczeyi ile cismin y\u00fczeyi aras\u0131ndaki dinamik s\u00fcrt\u00fcnme katsay\u0131s\u0131 nedir? Veri: g=10 m\/ <sup>s2<\/sup> . <\/li>\n<\/ul>\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-resolu-coefficient-de-frottement-dynamique.png\" alt=\"S\u00fcrt\u00fcnme katsay\u0131s\u0131 veya dinamik s\u00fcrt\u00fcnme sorunu\" class=\"wp-image-4281\" width=\"203\" height=\"205\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique-298x300.png 298w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique.png 479w\" sizes=\"auto, (max-width: 298px) 100vw, 298px\"><\/figure>\n<p class=\"has-text-align-left\"> Dinamik ile ilgili herhangi bir fizik problemini \u00e7\u00f6zmek i\u00e7in yapmam\u0131z gereken ilk \u015fey serbest cisim diyagram\u0131n\u0131 \u00e7izmektir. Yani sisteme etki eden t\u00fcm kuvvetler \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-force-de-friction-dynamique.png\" alt=\"S\u00fcrt\u00fcnme katsay\u0131s\u0131 veya dinamik s\u00fcrt\u00fcnmenin \u00e7\u00f6z\u00fclm\u00fc\u015f egzersizi\" class=\"wp-image-4282\" width=\"248\" height=\"301\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-dynamique-247x300.png 247w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-dynamique.png 572w\" sizes=\"auto, (max-width: 247px) 100vw, 247px\"><\/figure>\n<p class=\"has-text-align-left\"> Cisim 1. eksen y\u00f6n\u00fcnde (e\u011fik d\u00fczleme paralel) bir ivmeye sahiptir, ancak 2. eksen y\u00f6n\u00fcnde (e\u011fik d\u00fczleme dik) cisim hareketsizdir. Bu bilgiden sistem kuvvetlerinin denklemlerini kurar\u0131z:<\/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-d87a1ef6aaa3476891df5da8334cbc49_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1-F_R=m\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"124\" 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-6bdf90ed250934bf6cffbb110bc792a4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2-N=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"90\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> B\u00f6ylece normal kuvveti 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-59341555fe3d5fe315ceb1864547873b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_2\\\\[3ex]N=m\\cdot g\\cdot \\text{cos}(\\alpha) \\\\[3ex] N=6 \\cdot 10 \\cdot \\ text{cos}(45\u00ba)\\\\[3ex]N=42,43 \\ N\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"151\" width=\"185\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> \u00d6te yandan s\u00fcrt\u00fcnme kuvvetinin (veya s\u00fcrt\u00fcnme kuvvetinin) de\u011ferini sunulan ilk denklemden 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-d8f2aff2a81d98ddcea04b1988282fda_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}P_1-F_R=m\\cdot a\\\\[3ex]F_R=P_1-m\\cdot a\\\\[3ex]F_R=m\\cdot g\\cdot \\text{sin} (\\alpha)-m\\cdot a\\\\[3ex]F_R=6\\cdot 10\\cdot \\text{sin}(45\u00ba)-6\\cdot 4\\\\[3ex]F_R=18.43 \\ N\\end{ array} \" title=\"Rendered by QuickLaTeX.com\" height=\"195\" width=\"204\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Normal kuvvetin ve s\u00fcrt\u00fcnme kuvvetinin de\u011ferini bildi\u011fimizde, buna kar\u015f\u0131l\u0131k gelen form\u00fcl\u00fc kullanarak dinamik s\u00fcrt\u00fcnme katsay\u0131s\u0131n\u0131 belirleyebiliriz: <\/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-b783c9e37bcf4d077d9496489fc5d7d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu_d=\\cfrac{F_R}{N}=\\cfrac{18.43}{43.43}=0.42\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"187\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-del-plano-inclinado\"><\/span> E\u011fik d\u00fczlemde \u00e7\u00f6z\u00fclen al\u0131\u015ft\u0131rmalar<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> 1. Egzersiz<\/h3>\n<p> E\u011fim a\u00e7\u0131s\u0131 30\u00b0 olan e\u011fik bir d\u00fczlemin tepesine m=2 kg k\u00fctleli bir cisim yerle\u015ftiriyoruz. E\u011fer rampa ve g\u00f6vde dengede kal\u0131rsa aras\u0131ndaki s\u00fcrt\u00fcnme katsay\u0131s\u0131 nedir? Veri: g=9,81 m\/s <sup>2<\/sup> <\/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-de-force-normale-et-de-force-de-friction.png\" alt=\"\" class=\"wp-image-4253\" width=\"285\" height=\"176\" srcset=\"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\" 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\"> Kuvvet i\u00e7eren her fizik probleminde oldu\u011fu gibi yap\u0131lacak ilk \u015fey sistemin serbest cisim diyagram\u0131n\u0131 \u00e7izmektir. Bu sisteme etki eden t\u00fcm kuvvetler \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-force-normale-et-friction-force.png\" alt=\"normal kuvvet ve s\u00fcrt\u00fcnme kuvvetinin uygulanmas\u0131n\u0131 \u00e7\u00f6zmek\" class=\"wp-image-4254\" width=\"285\" height=\"333\" srcset=\"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\" sizes=\"auto, (max-width: 256px) 100vw, 256px\"><\/figure>\n<p class=\"has-text-align-left\"> Yani sistemin dengede olabilmesi i\u00e7in 1 ve 2 numaral\u0131 eksenlere etkiyen kuvvetlerin toplam\u0131n\u0131n s\u0131f\u0131ra e\u015fit olmas\u0131 gerekir. Bu nedenle a\u015fa\u011f\u0131daki denklemler do\u011frudur:<\/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-a692b08b4d7c08a2c55556233dc56651_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=P_1\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"63\" 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-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\"> Bu nedenle s\u00fcrt\u00fcnme katsay\u0131s\u0131n\u0131 denklemden \u00e7\u00f6z\u00fcyoruz ve 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<h3 class=\"wp-block-heading\">Al\u0131\u015ft\u0131rma 2<\/h3>\n<p> E\u011fik bir d\u00fczlem ve bir makaradan olu\u015fan a\u015fa\u011f\u0131daki sistemde g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi, iki cisim k\u00fctleleri ihmal edilebilir bir halat ve bir makara ile birbirine ba\u011flanm\u0131\u015ft\u0131r. Cisim 2&#8217;nin k\u00fctlesi m <sub>2<\/sub> = 7 kg ise ve rampan\u0131n e\u011fimi 50\u00b0 ise, t\u00fcm sistemin dengede olmas\u0131 i\u00e7in e\u011fik d\u00fczlemin m <sub>1<\/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-4a591e86900fca256cfa079da1bd0461_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> Un tra\u00eeneau de 70 kg glisse sur une pente de 30\u00ba avec une vitesse initiale de 2 m\/s. Si le coefficient de frottement dynamique entre le tra\u00eeneau et la neige est de 0,2, calculez la vitesse que le tra\u00eeneau acquerra apr\u00e8s avoir parcouru 20 m\u00e8tres. Donn\u00e9es : g=10 m\/s <sup>2<\/sup> . <\/p>\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> Tout d&#8217;abord, nous r\u00e9alisons le sch\u00e9ma corporel libre du tra\u00eeneau : <\/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-plan-incline.png&quot; alt=&quot;exercice r\u00e9solu de la force de frottement ou de frottement sur un plan inclin\u00e9&quot; class=&quot;wp-image-4345&quot; width=&quot;305&quot; height=&quot;355&quot; srcset=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-plan-incline-258x300.png 258w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-plan-incline.png 706w&quot; sizes=&quot;(max-width: 258px) 100vw, 258px&quot;><\/figure>\n<p> Le tra\u00eeneau a une acc\u00e9l\u00e9ration dans la direction de l&#8217;axe 1 (parall\u00e8le au plan inclin\u00e9) mais reste au repos dans la direction de l&#8217;axe 2 (perpendiculaire au plan inclin\u00e9), donc les \u00e9quations des forces sont : [latex]P_1-F_R=m\\cdot a&#8221; title=&#8221;Rendered by QuickLaTeX.com&#8221; height=&#8221;213&#8243; width=&#8221;8731&#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-6bdf90ed250934bf6cffbb110bc792a4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2-N=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"90\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> \u0130kinci denklemden k\u0131za\u011fa etki eden normal kuvveti 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-90b32b903f8be520ec73748b3de9b8b3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_2\\\\[3ex]N=m\\cdot g\\cdot \\text{cos}(\\alpha) \\\\[3ex] N=70 \\cdot 10 \\cdot \\ text{cos}(30\u00ba)\\\\[3ex]N=606,22 \\ N\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"151\" width=\"194\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Art\u0131k normal kuvvetin de\u011ferini ve dinamik s\u00fcrt\u00fcnme katsay\u0131s\u0131n\u0131 bildi\u011fimize g\u00f6re, s\u00fcrt\u00fcnme kuvvetini ilgili form\u00fcl\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-e0a32cc7650b33325233258788c218d4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=\\mu\\cdot N=0,2 \\cdot 606,22=121,24 \\ N \" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"298\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dolay\u0131s\u0131yla son h\u0131z\u0131 belirlemek i\u00e7in \u00f6ncelikle k\u0131za\u011f\u0131n ivmesini bulmal\u0131y\u0131z ve bu, sunulan ilk kuvvet denkleminden hesaplanabilir: <\/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-d87a1ef6aaa3476891df5da8334cbc49_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1-F_R=m\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"124\" 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-fa13e0490f51e32ac03b455043f6f32d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{P_1-F_R}{m}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"99\" 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-a6274d836af5618f7ef99e7f179c3902_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{m\\cdot g\\cdot \\text{sin}(\\alpha)-F_R}{m}\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"177\" 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-a3a9db70462cd187d50b851ede83983f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{70\\cdot 10\\cdot \\text{sin}(30\u00ba)-121.24}{70}\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"221\" 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-ba0d7325efa059351cc3d9aef838a9e2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=3,27 \\ \\cfrac{m}{s^2}\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"92\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> K\u0131za\u011f\u0131n ivmesini bildi\u011fimizde, sabit ivmede do\u011frusal hareket denklemi ile 20 metrelik yolu kat etmek i\u00e7in gereken s\u00fcreyi hesaplar\u0131z: <\/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-69f632cd171007df0f5bd6f0fa458a5c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x=v_0\\cdot t +\\cfrac{1}{2}\\cdot a \\cdot t^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"150\" 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-b97ac72bf22d70273fece0cce195f4ca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"20=2\\cdot t +\\cfrac{1}{2}\\cdot 3.27 \\cdot t^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"172\" 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-11a99cb686bf68cbcca594d0d60f801b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"0=1,64t^2+2t-20\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"158\" style=\"vertical-align: -4px;\"><\/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-7aa7e01e70b4199d597d05e261c970df_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle t=\\cfrac{-2\\pm \\sqrt{2^2-4\\cdot 1.64\\cdot (-20)}}{2\\cdot 1.64}=\\cfrac{-2\\ pm 11.63}{ 3.28}=\\begin{cases}2.94\\\\[2ex]-4.15 \\ \\color{red}\\bm{\\times}\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"507\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Mant\u0131ksal olarak, zaman negatif olamayacak fiziksel bir nicelik oldu\u011fundan negatif \u00e7\u00f6z\u00fcm\u00fc hari\u00e7 tutuyoruz.<\/p>\n<p class=\"has-text-align-left\"> Son olarak, sabit ivme form\u00fcl\u00fcn\u00fc kullanarak son h\u0131z\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-5ace98bfb166f5b813f593760fcfa048_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{v_f-v_0}{t_f-t_0}\\quad \\longrightarrow \\quad v_f=a\\cdot (t_f-t_0)+v_0\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"330\" style=\"vertical-align: -18px;\"><\/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-817a054f54a84b5a88667fc794feeb4d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v_f=3.27\\cdot (2.94-0)+2=\\bm{11.61} \\ \\cfrac{\\bm{m}}{\\bm{s}}\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"280\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Bu makale fizikte e\u011fik d\u00fczlemlerin ne oldu\u011funu ve bu t\u00fcr problemlerin nas\u0131l \u00e7\u00f6z\u00fcld\u00fc\u011f\u00fcn\u00fc a\u00e7\u0131klamaktad\u0131r. E\u011fik bir d\u00fczleme etki eden kuvvetlerin form\u00fcllerini bulacaks\u0131n\u0131z ve ayr\u0131ca e\u011fimli d\u00fczlemde ad\u0131m ad\u0131m \u00e7\u00f6z\u00fclen egzersizlerle antrenman yapabileceksiniz. E\u011fik d\u00fczlem nedir? E\u011fik d\u00fczlem belirli bir a\u00e7\u0131yla e\u011fimli bir y\u00fczeydir. Fizikte e\u011fik d\u00fczlem kuvvet problemlerini \u00e7\u00f6zmek i\u00e7in kullan\u0131l\u0131r. \u00d6rne\u011fin bir rampa veya &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/tr\/egik-duzlem\/\"> <span class=\"screen-reader-text\">E\u011fik d\u00fczlem<\/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-253","post","type-post","status-publish","format-standard","hentry","category-dinamik"],"yoast_head":"<!-- This site is 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bir d\u00fczleme etki eden kuvvetlerin form\u00fcllerini (fizik) ve e\u011fik d\u00fczlemde \u00e7\u00f6z\u00fclm\u00fc\u015f al\u0131\u015ft\u0131rmalar\u0131 bulacaks\u0131n\u0131z.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/physigeek.com\/tr\/egik-duzlem\/\" \/>\n<meta property=\"article:published_time\" content=\"2023-06-23T06:39:35+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/plan-incline.png\" \/>\n<meta name=\"author\" content=\"Jonathan Reynolds\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Yazan:\" \/>\n\t<meta name=\"twitter:data1\" content=\"Jonathan Reynolds\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tahmini okuma s\u00fcresi\" \/>\n\t<meta name=\"twitter:data2\" content=\"6 dakika\" \/>\n<script type=\"application\/ld+json\" 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