{"id":23,"date":"2023-06-27T11:34:58","date_gmt":"2023-06-27T11:34:58","guid":{"rendered":"https:\/\/physigeek.com\/tr\/denge-kosullari\/"},"modified":"2023-06-27T11:34:58","modified_gmt":"2023-06-27T11:34:58","slug":"denge-kosullari","status":"publish","type":"post","link":"https:\/\/physigeek.com\/tr\/denge-kosullari\/","title":{"rendered":"Denge ko\u015fullar\u0131"},"content":{"rendered":"<p>Bu makale denge ko\u015fullar\u0131n\u0131n ne oldu\u011funu a\u00e7\u0131klamaktad\u0131r. Hem denge ko\u015fullar\u0131n\u0131n ger\u00e7ek \u00f6rneklerini bulacaks\u0131n\u0131z, hem de ad\u0131m ad\u0131m \u00e7\u00f6z\u00fclen egzersizlerle antrenman yapabileceksiniz. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFCuales-son-las-condiciones-de-equilibrio\"><\/span> Denge ko\u015fullar\u0131 nelerdir?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Fizikte <strong>denge ko\u015fullar\u0131<\/strong> , bir cismin kendisine uygulanan kuvvetlerin ve momentlerin toplam\u0131n\u0131n s\u0131f\u0131ra e\u015fit olmas\u0131 durumunda dengede oldu\u011funu belirtir.<\/p>\n<p> Dolay\u0131s\u0131yla denge i\u00e7in iki ko\u015ful vard\u0131r: \u0130lk ko\u015ful, bile\u015fke kuvvetin s\u0131f\u0131r olmas\u0131 gerekti\u011fini s\u00f6yler ve ikinci ko\u015ful, bile\u015fke momentin s\u0131f\u0131r olmas\u0131 gerekti\u011fini s\u00f6yler. <\/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\/conditions-dequilibre.png\" alt=\"denge ko\u015fullar\u0131\" class=\"wp-image-416\" width=\"258\" height=\"258\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/conditions-dequilibre-300x300.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/conditions-dequilibre-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/conditions-dequilibre.png 514w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<\/div>\n<p> Unutmay\u0131n ki bir sistemin dengede say\u0131lmas\u0131 i\u00e7in her iki denklemin de kar\u015f\u0131lanmas\u0131 gerekir, sadece bir ko\u015fulun kar\u015f\u0131lanmas\u0131 yeterli de\u011fildir. <\/p>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Primera-condicion-de-equilibrio\"><\/span> Dengenin ilk \u015fart\u0131<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> <strong>Birinci denge ko\u015fulu<\/strong> , bir cisme uygulanan kuvvetlerin toplam\u0131n\u0131n, s\u00f6z konusu cismin \u00f6teleme dengesinde olmas\u0131 i\u00e7in s\u0131f\u0131ra e\u015fit olmas\u0131 gerekti\u011fini s\u00f6yler.<\/p>\n<p> Mant\u0131ksal olarak kuvvetlerin toplam\u0131 \u00fc\u00e7 eksen i\u00e7in de s\u0131f\u0131r olmal\u0131d\u0131r, e\u011fer herhangi bir eksende kar\u015f\u0131lanm\u0131yorsa cisim dengede de\u011fildir.<\/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-9d6873586b63ccddf575a8ee1c7f5137_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\sum\\vv{F_x}=0\\qquad\\sum\\vv{F_y}=0\\qquad\\sum\\vv{F_z}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"319\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<p> Ayr\u0131ca kuvvetlerin toplam\u0131 s\u0131f\u0131rsa bu, cismin do\u011frusal bir ivmesinin olmad\u0131\u011f\u0131 anlam\u0131na gelir. B\u00f6ylece, \u00f6teleme dengesindeki bir cisim hareketsiz olabilir (s\u0131f\u0131r h\u0131zda) veya sabit do\u011frusal h\u0131zda hareket edebilir.<\/p>\n<p> Buradan iki t\u00fcr \u00f6teleme dengesi ay\u0131rt edilebilir:<\/p>\n<ul>\n<li> <strong>Statik \u00f6teleme dengesi<\/strong> : \u0130lk denge ko\u015fulu kar\u015f\u0131land\u0131\u011f\u0131nda ve v\u00fccut da hareketsiz oldu\u011funda.<\/li>\n<li> <strong>Dinamik \u00f6teleme dengesi<\/strong> : \u0130lk denge ko\u015fulu kar\u015f\u0131land\u0131\u011f\u0131nda ve v\u00fccut sabit bir h\u0131za sahip oldu\u011funda (s\u0131f\u0131rdan farkl\u0131). <\/li>\n<\/ul>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Segunda-condicion-de-equilibrio\"><\/span> \u0130kinci denge ko\u015fulu<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> \u0130kinci denge ko\u015fulu birinci denge ko\u015fuluna benzer ancak kuvvetler yerine momentleri kullan\u0131r.<\/p>\n<p> <strong>\u0130kinci denge ko\u015fulu,<\/strong> e\u011fer bir cismin momentlerinin toplam\u0131 s\u0131f\u0131rsa, o zaman cismin d\u00f6nme dengesinde oldu\u011funu s\u00f6yler.<\/p>\n<p> Ayn\u0131 \u015fekilde \u00e7er\u00e7evenin t\u00fcm eksenlerinde momentlerin toplam\u0131 s\u0131f\u0131r olmal\u0131d\u0131r, aksi halde ikinci denge ko\u015fulu do\u011frulanmaz.<\/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-ef043ee3ac4a59374afc86a86f450df6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\sum\\vv{M_x}=0\\qquad\\sum\\vv{M_y}=0\\qquad\\sum\\vv{M_z}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"336\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<p> Bir noktadaki kuvvetin momentinin (veya torkunun), kuvvetin de\u011ferinin, kuvvetten noktaya olan dik mesafeyle \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-765ae97c83695144c85bb65446416345_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"M=F\\cdot d\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"79\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Benzer \u015fekilde, ikinci denge ko\u015fulunun kar\u015f\u0131lanmas\u0131 i\u00e7in cismin a\u00e7\u0131sal ivmesinin s\u0131f\u0131r olmas\u0131 gerekir; bu, bu durumda cismin sabit bir a\u00e7\u0131sal h\u0131zda d\u00f6nmedi\u011fi veya d\u00f6nmedi\u011fi anlam\u0131na gelir. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplos-de-las-condiciones-de-equilibrio\"><\/span> Denge ko\u015fullar\u0131 \u00f6rnekleri<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> \u0130ki denge ko\u015fulunun tan\u0131mlar\u0131n\u0131 g\u00f6rd\u00fckten sonra, kavram\u0131 tam olarak anlamak i\u00e7in a\u015fa\u011f\u0131da g\u00fcnl\u00fck hayattan birka\u00e7 \u00f6rnek g\u00f6rebileceksiniz.<\/p>\n<p> \u00d6rne\u011fin bir cisim tavana as\u0131ld\u0131\u011f\u0131nda sistem tamamen hareketsiz oldu\u011fundan v\u00fccut dengede olur. Sistemin statik dengede oldu\u011funu da s\u00f6yleyebiliriz. <\/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\/premiere-condition-dequilibre.png\" alt=\"dengenin ilk ko\u015fulu\" class=\"wp-image-368\" width=\"307\" height=\"307\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/premiere-condition-dequilibre-300x300.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/premiere-condition-dequilibre-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/premiere-condition-dequilibre-768x766.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/premiere-condition-dequilibre.png 1006w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<\/div>\n<p> G\u00fcnl\u00fck ya\u015famdaki denge ko\u015fullar\u0131n\u0131n bir ba\u015fka \u00f6rne\u011fi de terazidir. Denge kolu sabitlenip d\u00f6nmeyi b\u0131rakt\u0131\u011f\u0131nda sistem hareketsizdir ve dolay\u0131s\u0131yla dengededir. <\/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\/deuxieme-condition-dequilibre.png\" alt=\"ikinci denge ko\u015fulu\" class=\"wp-image-430\" width=\"286\" height=\"286\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/deuxieme-condition-dequilibre-300x300.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/deuxieme-condition-dequilibre-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/deuxieme-condition-dequilibre-768x768.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/deuxieme-condition-dequilibre.png 781w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-las-condiciones-de-equilibrio\"><\/span> \u00c7\u00f6z\u00fclm\u00fc\u015f denge ko\u015fullar\u0131 problemleri<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 \u00e7\u00f6z\u00fclm\u00fc\u015f egzersizi\" 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 yer\u00e7ekimi kuvveti form\u00fcl\u00fcn\u00fc uygulayarak a\u011f\u0131rl\u0131\u011f\u0131n kuvvetini 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-da2fc72dc768050ef84d2a3c9ee4a281_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=\"239\" 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 k\u0131s\u0131tlamalar\u0131n 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-694655a52049a536489cebbaef3bc7a2_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> Calculer le moment que doit faire le support de la poutre suivante pour qu&#8217;elle soit en \u00e9quilibre de rotation : <\/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\/exercice-resolu-deuxieme-condition-dequilibre.png&quot; alt=&quot;Exercice r\u00e9solu de la deuxi\u00e8me condition d'\u00e9quilibre&quot; class=&quot;wp-image-397&quot; width=&quot;237&quot; height=&quot;203&quot; srcset=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-deuxieme-condition-dequilibre-300x257.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-deuxieme-condition-dequilibre.png 643w&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> Pour que la poutre soit en \u00e9quilibre de rotation et que la deuxi\u00e8me condition d&#8217;\u00e9quilibre soit donc remplie, le support doit contrecarrer le moment de torsion g\u00e9n\u00e9r\u00e9 par la force, donc la somme des moments sera nulle. On calcule donc le moment (ou couple) g\u00e9n\u00e9r\u00e9 par la force au niveau de l&#8217;appui : [latex]M_{force}=13\\cdot 9 = 117 \\ Nm&#8221; title=&#8221;Rendered by QuickLaTeX.com&#8221; height=&#8221;343&#8243; width=&#8221;3353&#8243; style=&#8221;vertical-align: 0px;&#8221;><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> \u015eimdi moment dengesi denklemini ifade ediyoruz:<\/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-fdac8512edbbe2c1b6396ee43a776261_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"M_{support}+M_{force}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"170\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Kuvveti olu\u015fturan an ekran\u0131n i\u00e7inden ge\u00e7ti\u011fi i\u00e7in i\u015fareti negatiftir:<\/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-b654ae9d3ab78a23e2f7235a7742a503_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"M_{support}-117=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"145\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ve son olarak denklemdeki bilinmeyeni \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-cfe550481f3747f9121d30c3f35f7f85_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"M_{support}=117\\Nm\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"115\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Elde edilen an\u0131n olumlu bir i\u015fareti vard\u0131r, dolay\u0131s\u0131yla anlam\u0131 ekran\u0131n d\u0131\u015f\u0131ndad\u0131r.<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\"> Al\u0131\u015ft\u0131rma 3<\/h3>\n<p> A\u015fa\u011f\u0131daki \u015fekilde 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. E\u011fer nesne 2&#8217;nin k\u00fctlesi 7 kg ise ve rampan\u0131n e\u011fimi 50\u00b0 ise, t\u00fcm sistem denge ko\u015fullar\u0131nda olacak \u015fekilde nesne 1&#8217;in k\u00fctlesini hesaplay\u0131n. Bu durumda s\u00fcrt\u00fcnme kuvveti ihmal edilebilir. <\/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-dequilibre-des-forces.png\" alt=\"\u00f6teleme dengesi sorunu\" 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>\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-35d7a76d4aead5e24628c76e5f80b4eb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_{1x}=P_1\\cdot \\text{sen}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"131\" 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\"> Bu nedenle t\u00fcm sisteme 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=\"\u00c7\u00f6z\u00fclm\u00fc\u015f \u00e7evirisel denge egzersizi\" 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 form\u00fcle 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-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-9b5e757fb28e9dde3aed458f89a3ed53_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1\\cdot \\text{sen}(\\alpha)=P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"123\" 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-37d57b7e3c4a13f3c4dc4ae981f7d61f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1=9,14\\kg\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"82\" 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 4<\/h3>\n<p> A\u015fa\u011f\u0131daki \u015fekilde g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi 10 m&#8217;lik yatay \u00e7ubuk k\u00fctlesi 8 kg olan bir cismi desteklemektedir. Destekler ile as\u0131l\u0131 cisim aras\u0131ndaki mesafeler bilindi\u011finde, sistem d\u00f6nme ve \u00f6telenme dengesinde ise desteklerin uygulad\u0131\u011f\u0131 kuvvetlerin de\u011feri nedir? <\/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-dequilibre-de-rotation.png\" alt=\"d\u00f6nme dengesi sorunu\" class=\"wp-image-355\" width=\"339\" height=\"120\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-de-rotation-300x107.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-de-rotation.png 643w\" 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\"> \u00d6ncelikle yatay \u00e7ubu\u011fun desteklemesi gereken a\u011f\u0131rl\u0131\u011f\u0131 hesaplamak i\u00e7in yer \u00e7ekimi kuvveti form\u00fcl\u00fcn\u00fc kullan\u0131r\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-eb480f5d32a9e25cefacd5f89d407580_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=m\\cdot g=8\\cdot 9,81 =78,48 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"244\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Buna g\u00f6re sistemin serbest cisim diyagram\u0131 \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-rotation-equilibre.png\" alt=\"\u00e7\u00f6z\u00fcmlenmi\u015f rotasyonel denge egzersizi\" class=\"wp-image-356\" width=\"340\" height=\"297\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-rotation-equilibre-300x261.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-rotation-equilibre.png 654w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> Problem ifadesi bize sistemin kuvvetler dengesinde oldu\u011funu, dolay\u0131s\u0131yla t\u00fcm bu kuvvetlerin toplam\u0131n\u0131n s\u0131f\u0131r olmas\u0131 gerekti\u011fini s\u00f6yler. Bu denge ko\u015fulunu kullanarak a\u015fa\u011f\u0131daki denklemi form\u00fcle 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-635096a57ce10781254f283c9807f64c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_A+F_B-P=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"136\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> \u00d6te yandan bu ifade bize ayn\u0131 zamanda sistemin momentum dengesinde oldu\u011funu da s\u00f6ylemektedir. Yani sistemin herhangi bir noktas\u0131ndaki momentlerin toplam\u0131n\u0131 dikkate al\u0131rsak sonu\u00e7 s\u0131f\u0131r olmal\u0131d\u0131r ve iki destekten birinin referans noktas\u0131n\u0131 al\u0131rsak tek bilinmeyenli bir denklem 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-f1a6d56c3426e8d6c2e890b5e8f4a873_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"M(A)=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"79\" 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-922d5ff929e034db7a9e80d732b0b893_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-P\\cdot 6.5+F_B\\cdot (6.5+3.5)=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"233\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Art\u0131k denklemdeki bilinmeyeni \u00e7\u00f6zerek B deste\u011finin uygulad\u0131\u011f\u0131 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-95596edf27bb6086c35473f55465416d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-78.48\\cdot 6.5+F_B\\cdot 10=0\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"197\" 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-953185a6ad4824b654b8a40e259bbd71_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_B=\\cfrac{78.48\\cdot 6.5}{10}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"125\" 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-c5459080e5aa468fa3d77a16a2b0d9b9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_B=51.01\\N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"87\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ve son olarak, elde edilen de\u011feri d\u00fc\u015fey kuvvetlerin denkleminde yerine koyarak di\u011fer deste\u011fe uygulanan kuvvetin yo\u011funlu\u011funu bilebiliriz: <\/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-635096a57ce10781254f283c9807f64c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_A+F_B-P=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"136\" 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-36325128977806ddaca1436ffb68dcd4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_A+51,01-78,48=0\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"185\" 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-5ecd2d608e61435093289d17c242e21f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_A=27,47\\N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"90\" 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 makale denge ko\u015fullar\u0131n\u0131n ne oldu\u011funu a\u00e7\u0131klamaktad\u0131r. Hem denge ko\u015fullar\u0131n\u0131n ger\u00e7ek \u00f6rneklerini bulacaks\u0131n\u0131z, hem de ad\u0131m ad\u0131m \u00e7\u00f6z\u00fclen egzersizlerle antrenman yapabileceksiniz. Denge ko\u015fullar\u0131 nelerdir? Fizikte denge ko\u015fullar\u0131 , bir cismin kendisine uygulanan kuvvetlerin ve momentlerin toplam\u0131n\u0131n s\u0131f\u0131ra e\u015fit olmas\u0131 durumunda dengede oldu\u011funu belirtir. Dolay\u0131s\u0131yla denge i\u00e7in iki ko\u015ful vard\u0131r: \u0130lk ko\u015ful, bile\u015fke kuvvetin s\u0131f\u0131r olmas\u0131 &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/tr\/denge-kosullari\/\"> <span class=\"screen-reader-text\">Denge ko\u015fullar\u0131<\/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-23","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 Denge ko\u015fullar\u0131<\/title>\n<meta name=\"description\" content=\"Denge ko\u015fullar\u0131n\u0131n ne oldu\u011funu a\u00e7\u0131kl\u0131yoruz. 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