{"id":22,"date":"2023-06-27T12:43:20","date_gmt":"2023-06-27T12:43:20","guid":{"rendered":"https:\/\/physigeek.com\/pt\/segunda-condicao-de-equilibrio\/"},"modified":"2023-06-27T12:43:20","modified_gmt":"2023-06-27T12:43:20","slug":"segunda-condicao-de-equilibrio","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/segunda-condicao-de-equilibrio\/","title":{"rendered":"Segunda condi\u00e7\u00e3o de equil\u00edbrio"},"content":{"rendered":"<p>Este artigo explica o que \u00e9 a segunda condi\u00e7\u00e3o de equil\u00edbrio e em que consiste. Voc\u00ea tamb\u00e9m encontrar\u00e1 exemplos reais da segunda condi\u00e7\u00e3o de equil\u00edbrio e, por fim, poder\u00e1 treinar com exerc\u00edcios resolvidos passo a passo. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFCual-es-la-segunda-condicion-de-equilibrio\"><\/span> Qual \u00e9 a segunda condi\u00e7\u00e3o de equil\u00edbrio?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Na f\u00edsica, a <strong>segunda condi\u00e7\u00e3o de equil\u00edbrio<\/strong> \u00e9 uma regra que diz que um corpo est\u00e1 em equil\u00edbrio rotacional se a soma dos momentos aplicados a ele for igual a zero.<\/p>\n<p> A segunda condi\u00e7\u00e3o de equil\u00edbrio \u00e9, portanto, satisfeita quando o momento resultante \u00e9 zero. Matematicamente, a segunda condi\u00e7\u00e3o de equil\u00edbrio \u00e9 expressa pela seguinte f\u00f3rmula:<\/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-d8a19d84c103481347db47c0b8a6d2a8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\somme \\vv{M}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"52\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Observe que os momentos devem ser somados vetorialmente, pois os momentos que atuam em eixos diferentes n\u00e3o podem ser somados. Esta condi\u00e7\u00e3o n\u00e3o \u00e9 um problema se trabalhar com for\u00e7as coplanares (em duas dimens\u00f5es) j\u00e1 que o momento sempre vai na mesma dire\u00e7\u00e3o, mas deve estar ciente disso quando se trabalha em tr\u00eas dimens\u00f5es.<\/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-ec7e37abbe42c976a3809c9e0f2724f2_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\\qquad\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"336\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<p> Lembre-se de que o momento (ou torque) de uma for\u00e7a em um ponto \u00e9 calculado multiplicando o valor da for\u00e7a pela dist\u00e2ncia perpendicular da for\u00e7a ao ponto.<\/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> Ent\u00e3o, para satisfazer a equa\u00e7\u00e3o da segunda condi\u00e7\u00e3o de equil\u00edbrio, o corpo deve ter acelera\u00e7\u00e3o angular zero, ou em outras palavras, um corpo neste estado n\u00e3o est\u00e1 girando (est\u00e1 em repouso) ou girando com velocidade angular constante.<\/p>\n<p> Assim, podemos distinguir tipos de equil\u00edbrio rotacional:<\/p>\n<ul>\n<li> <strong>Equil\u00edbrio rotacional est\u00e1tico<\/strong> : quando a soma dos momentos \u00e9 zero e a velocidade angular do corpo \u00e9 zero.<\/li>\n<li> <strong>Equil\u00edbrio rotacional din\u00e2mico<\/strong> : quando a soma dos momentos \u00e9 zero e a velocidade angular do corpo \u00e9 constante (diferente de zero). <\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplos-de-la-segunda-condicion-de-equilibrio\"><\/span> Exemplos da segunda condi\u00e7\u00e3o de equil\u00edbrio<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Considerando a defini\u00e7\u00e3o da segunda condi\u00e7\u00e3o de equil\u00edbrio, veremos agora v\u00e1rios exemplos do cotidiano para finalizar a compreens\u00e3o do conceito.<\/p>\n<p> Um exemplo comum da segunda condi\u00e7\u00e3o de equil\u00edbrio \u00e9 uma escala. Quando o sistema se estabiliza, o bra\u00e7o de equil\u00edbrio para de girar e, portanto, a soma dos momentos \u00e9 zero e o sistema est\u00e1 em equil\u00edbrio rotacional. <\/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=\"segunda condi\u00e7\u00e3o de equil\u00edbrio\" class=\"wp-image-430\" width=\"303\" height=\"303\" 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<p> Outro exemplo concreto \u00e9 a Terra. O planeta gira continuamente em seu eixo, mas considera-se que gira a uma velocidade angular constante, portanto satisfaz a segunda condi\u00e7\u00e3o de equil\u00edbrio.<\/p>\n<p> Finalmente, quando suspendemos um objeto no teto e o mantemos em repouso, o objeto cumpre tanto a segunda condi\u00e7\u00e3o de equil\u00edbrio quanto a primeira condi\u00e7\u00e3o de equil\u00edbrio, uma vez que est\u00e1 em equil\u00edbrio translacional e em equil\u00edbrio translacional. rota\u00e7\u00e3o.<\/p>\n<p> Se n\u00e3o entende claramente em que consiste a primeira condi\u00e7\u00e3o de saldo, pode consultar o seguinte artigo onde \u00e9 explicado detalhadamente: <\/p>\n<div style=\"background-color:#FFFDE7; padding-top: 10px; padding-bottom: 10px; padding-right: 10px; padding-left: 20px; border: 2.5px dashed #FFB74D; border-radius:20px;\"> <span style=\"color:#ff951b\">\u27a4<\/span> <strong>Veja:<\/strong> <a href=\"https:\/\/physigeek.com\/pt\/primeira-condicao-de-equilibrio\/\">qual \u00e9 a primeira condi\u00e7\u00e3o de equil\u00edbrio<\/a> <\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-segunda-condicion-de-equilibrio\"><\/span>Exerc\u00edcios resolvidos da segunda condi\u00e7\u00e3o de equil\u00edbrio<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 1<\/h3>\n<p> Calcule o momento que o apoio da viga seguinte deve apresentar para estar em equil\u00edbrio rotacional: <\/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-deuxieme-condition-dequilibre.png\" alt=\"exerc\u00edcio resolvido da segunda condi\u00e7\u00e3o de equil\u00edbrio\" class=\"wp-image-397\" width=\"237\" height=\"203\" srcset=\"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\" 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>veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Para que a viga esteja em equil\u00edbrio rotacional e a segunda condi\u00e7\u00e3o de equil\u00edbrio seja atendida, o apoio deve neutralizar o momento de tor\u00e7\u00e3o gerado pela for\u00e7a, de modo que a soma dos momentos ser\u00e1 zero.<\/p>\n<p class=\"has-text-align-left\"> Calculamos, portanto, o momento (ou torque) gerado pela for\u00e7a ao n\u00edvel do apoio:<\/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-e172a3bbd51bb42bd229d9a88df0a167_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"M_{force}=13\\cdot 9 = 117 \\ Nm\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"202\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E agora propomos a equa\u00e7\u00e3o de equil\u00edbrio dos momentos:<\/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\"> O momento que gera a for\u00e7a passa dentro da tela, ent\u00e3o seu sinal \u00e9 negativo:<\/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\"> E finalmente, resolvemos a inc\u00f3gnita na equa\u00e7\u00e3o:<\/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-772409bbbd945be3fc433e503b43e540_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"M_{support}=117 \\ Nm\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"152\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> O pulso obtido tem sinal positivo, portanto sua dire\u00e7\u00e3o \u00e9 para fora da tela.<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 2<\/h3>\n<p> Como voc\u00ea pode ver na figura a seguir, uma barra horizontal de 10 m sustenta um corpo cuja massa \u00e9 de 8 kg. Conhecendo as dist\u00e2ncias entre os apoios e o corpo suspenso, quais s\u00e3o os valores das for\u00e7as exercidas pelos apoios se o sistema estiver em equil\u00edbrio de rota\u00e7\u00e3o e transla\u00e7\u00e3o? <\/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=\"problema de equil\u00edbrio rotacional\" 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>veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Primeiro, usamos a f\u00f3rmula da for\u00e7a gravitacional para calcular o peso que a barra horizontal deve suportar:<\/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\"> O diagrama de corpo livre do sistema \u00e9, portanto: <\/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=\"exerc\u00edcio de equil\u00edbrio rotacional resolvido\" 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\"> A defini\u00e7\u00e3o do problema diz-nos que o sistema est\u00e1 em equil\u00edbrio de for\u00e7as, portanto a soma de todas estas for\u00e7as deve ser zero. Usando esta condi\u00e7\u00e3o de equil\u00edbrio, podemos formular a seguinte equa\u00e7\u00e3o:<\/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\"> Por outro lado, a afirma\u00e7\u00e3o tamb\u00e9m nos diz que o sistema est\u00e1 em equil\u00edbrio de momento. Ent\u00e3o se considerarmos a soma dos momentos em qualquer ponto do sistema, o resultado deve ser zero, e se tomarmos o ponto de refer\u00eancia de um dos dois apoios, teremos uma equa\u00e7\u00e3o com apenas uma inc\u00f3gnita:<\/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\"> Podemos agora calcular a for\u00e7a exercida pelo suporte B resolvendo a inc\u00f3gnita na equa\u00e7\u00e3o: <\/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\"> E por fim, podemos saber a intensidade da for\u00e7a aplicada ao outro apoio substituindo o valor obtido na equa\u00e7\u00e3o alta das for\u00e7as verticais: <\/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>Este artigo explica o que \u00e9 a segunda condi\u00e7\u00e3o de equil\u00edbrio e em que consiste. Voc\u00ea tamb\u00e9m encontrar\u00e1 exemplos reais da segunda condi\u00e7\u00e3o de equil\u00edbrio e, por fim, poder\u00e1 treinar com exerc\u00edcios resolvidos passo a passo. Qual \u00e9 a segunda condi\u00e7\u00e3o de equil\u00edbrio? Na f\u00edsica, a segunda condi\u00e7\u00e3o de equil\u00edbrio \u00e9 uma regra que diz &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/segunda-condicao-de-equilibrio\/\"> <span class=\"screen-reader-text\">Segunda condi\u00e7\u00e3o de equil\u00edbrio<\/span> Leia mais &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-22","post","type-post","status-publish","format-standard","hentry","category-dinamico"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>\u25b7 Segunda condi\u00e7\u00e3o de equil\u00edbrio<\/title>\n<meta name=\"description\" content=\"Explicamos o que \u00e9 a segunda condi\u00e7\u00e3o de equil\u00edbrio e em que consiste. 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