{"id":23,"date":"2023-06-27T11:34:58","date_gmt":"2023-06-27T11:34:58","guid":{"rendered":"https:\/\/physigeek.com\/pt\/condicoes-de-equilibrio\/"},"modified":"2023-06-27T11:34:58","modified_gmt":"2023-06-27T11:34:58","slug":"condicoes-de-equilibrio","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/condicoes-de-equilibrio\/","title":{"rendered":"Condi\u00e7\u00f5es de equil\u00edbrio"},"content":{"rendered":"<p>Este artigo explica o que s\u00e3o condi\u00e7\u00f5es de equil\u00edbrio. Voc\u00ea encontrar\u00e1 exemplos reais de ambas as condi\u00e7\u00f5es de equil\u00edbrio e, al\u00e9m disso, poder\u00e1 treinar com exerc\u00edcios resolvidos passo a passo. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFCuales-son-las-condiciones-de-equilibrio\"><\/span> Quais s\u00e3o as condi\u00e7\u00f5es de equil\u00edbrio?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Na f\u00edsica, as <strong>condi\u00e7\u00f5es de equil\u00edbrio<\/strong> afirmam que um corpo est\u00e1 em equil\u00edbrio se a soma das for\u00e7as e a soma dos momentos aplicados a ele forem iguais a zero.<\/p>\n<p> Portanto, existem duas condi\u00e7\u00f5es para o equil\u00edbrio: a primeira condi\u00e7\u00e3o diz que a for\u00e7a resultante deve ser zero, e a segunda condi\u00e7\u00e3o diz que o momento resultante deve ser zero. <\/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=\"condi\u00e7\u00f5es de equil\u00edbrio\" 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> Tenha em mente que para que um sistema seja considerado em equil\u00edbrio \u00e9 necess\u00e1rio que ambas as equa\u00e7\u00f5es sejam atendidas, n\u00e3o basta que apenas uma condi\u00e7\u00e3o seja atendida. <\/p>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Primera-condicion-de-equilibrio\"><\/span> Primeira condi\u00e7\u00e3o de equil\u00edbrio<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> A <strong>primeira condi\u00e7\u00e3o de equil\u00edbrio<\/strong> diz que a soma das for\u00e7as aplicadas a um corpo deve ser igual a zero para que esse corpo esteja em equil\u00edbrio translacional.<\/p>\n<p> Logicamente, a soma das for\u00e7as deve ser zero para os tr\u00eas eixos, se n\u00e3o for cumprida em nenhum eixo o corpo n\u00e3o est\u00e1 em equil\u00edbrio.<\/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> Al\u00e9m disso, se a soma das for\u00e7as for zero, significa que o corpo n\u00e3o tem acelera\u00e7\u00e3o linear. Assim, um corpo em equil\u00edbrio translacional pode estar em repouso (velocidade zero) ou mover-se com velocidade linear constante.<\/p>\n<p> A partir da\u00ed, dois tipos de equil\u00edbrio translacional podem ser distinguidos:<\/p>\n<ul>\n<li> <strong>Equil\u00edbrio translacional est\u00e1tico<\/strong> : quando a primeira condi\u00e7\u00e3o de equil\u00edbrio \u00e9 atendida e o corpo tamb\u00e9m est\u00e1 em repouso.<\/li>\n<li> <strong>Equil\u00edbrio translacional din\u00e2mico<\/strong> : quando a primeira condi\u00e7\u00e3o de equil\u00edbrio \u00e9 atendida e o corpo tem velocidade constante (diferente de zero). <\/li>\n<\/ul>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Segunda-condicion-de-equilibrio\"><\/span> Segunda condi\u00e7\u00e3o de equil\u00edbrio<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> A segunda condi\u00e7\u00e3o de equil\u00edbrio \u00e9 an\u00e1loga \u00e0 primeira condi\u00e7\u00e3o de equil\u00edbrio, mas utiliza momentos em vez de for\u00e7as.<\/p>\n<p> A <strong>segunda condi\u00e7\u00e3o de equil\u00edbrio<\/strong> diz que se a soma dos momentos de um corpo for zero, ent\u00e3o o corpo est\u00e1 em equil\u00edbrio rotacional.<\/p>\n<p> Da mesma forma, a soma dos momentos deve ser zero em todos os eixos do p\u00f3rtico, caso contr\u00e1rio a segunda condi\u00e7\u00e3o de equil\u00edbrio n\u00e3o se verifica.<\/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> 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> Da mesma forma, para que a segunda condi\u00e7\u00e3o de equil\u00edbrio seja atendida, a acelera\u00e7\u00e3o angular do corpo deve ser zero, o que significa que neste estado o corpo n\u00e3o est\u00e1 girando ou girando a uma velocidade angular constante. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplos-de-las-condiciones-de-equilibrio\"><\/span> Exemplos de condi\u00e7\u00f5es de equil\u00edbrio<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Depois de ver as defini\u00e7\u00f5es das duas condi\u00e7\u00f5es de equil\u00edbrio, voc\u00ea poder\u00e1 ver v\u00e1rios exemplos da vida cotidiana abaixo para compreender completamente o conceito.<\/p>\n<p> Por exemplo, quando um corpo est\u00e1 suspenso no teto, o corpo est\u00e1 em equil\u00edbrio, pois o sistema est\u00e1 completamente em repouso. Tamb\u00e9m podemos dizer que o sistema est\u00e1 em equil\u00edbrio est\u00e1tico. <\/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=\"primeira condi\u00e7\u00e3o de equil\u00edbrio\" 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> Outro exemplo de condi\u00e7\u00f5es de equil\u00edbrio no dia a dia \u00e9 a balan\u00e7a. Quando o bra\u00e7o de equil\u00edbrio se estabiliza e para de girar, o sistema est\u00e1 em repouso e, portanto, tamb\u00e9m em equil\u00edbrio. <\/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=\"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> Problemas de condi\u00e7\u00f5es de equil\u00edbrio resolvidos<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 1<\/h3>\n<p> Dado um corpo r\u00edgido com massa de 12 kg suspenso por duas cordas cujos \u00e2ngulos s\u00e3o mostrados na figura a seguir, calcule a for\u00e7a que cada corda deve exercer para manter o corpo em equil\u00edbrio. <\/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=\"problema da primeira condi\u00e7\u00e3o de equil\u00edbrio\" 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>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> A primeira coisa que precisamos fazer para resolver este tipo de problema \u00e9 desenhar o diagrama de corpo livre da figura: <\/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=\"Exerc\u00edcio resolvido da primeira condi\u00e7\u00e3o de equil\u00edbrio\" 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\"> Observe que na verdade existem apenas tr\u00eas for\u00e7as atuando sobre o corpo suspenso, a for\u00e7a do peso P e as tens\u00f5es das cordas T <sub>1<\/sub> e T <sub>2<\/sub> . As for\u00e7as representadas T <sub>1x<\/sub> , T <sub>1y<\/sub> , T <sub>2x<\/sub> e T <sub>2y<\/sub> s\u00e3o as componentes vetoriais de T <sub>1<\/sub> e T <sub>2<\/sub> respectivamente.<\/p>\n<p class=\"has-text-align-left\"> Assim, como conhecemos os \u00e2ngulos de inclina\u00e7\u00e3o das cordas, podemos encontrar as express\u00f5es para as componentes vetoriais das for\u00e7as de tra\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-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\"> Por outro lado, podemos calcular a for\u00e7a do peso aplicando a f\u00f3rmula da for\u00e7a gravitacional:<\/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\"> A defini\u00e7\u00e3o do problema diz-nos que o corpo est\u00e1 em equil\u00edbrio, portanto a soma das for\u00e7as verticais e a soma das for\u00e7as horizontais deve ser igual a zero. Portanto, podemos estabelecer as equa\u00e7\u00f5es de for\u00e7a e defini-las iguais a zero:<\/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\"> Substitu\u00edmos agora os componentes das restri\u00e7\u00f5es pelas suas express\u00f5es encontradas anteriormente: <\/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\"> E, por fim, resolvemos o sistema de equa\u00e7\u00f5es para obter o valor das for\u00e7as T <sub>1<\/sub> e T <sub>2<\/sub> :<\/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\"> E agora declaramos a equa\u00e7\u00e3o do equil\u00edbrio de 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-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\"> O momento obtido tem sinal positivo, portanto seu significado est\u00e1 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 3<\/h3>\n<p> Conforme mostrado na figura a seguir, dois objetos est\u00e3o conectados por uma corda e uma polia de massas desprez\u00edveis. Se o objeto 2 tem massa de 7 kg e a inclina\u00e7\u00e3o da rampa \u00e9 de 50\u00ba, calcule a massa do objeto 1 para que todo o sistema fique em condi\u00e7\u00f5es de equil\u00edbrio. Neste caso, a for\u00e7a de atrito pode ser desprezada. <\/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=\"problema de equil\u00edbrio translacional\" 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>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> O corpo 1 est\u00e1 em um declive inclinado, ent\u00e3o a primeira coisa a fazer \u00e9 vetorizar a for\u00e7a do seu peso para ter as for\u00e7as nos eixos do declive: <\/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\"> O conjunto de for\u00e7as que atuam em todo o 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-equilibre-des-forces.png\" alt=\"Exerc\u00edcio de equil\u00edbrio translacional resolvido\" 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\"> A defini\u00e7\u00e3o do problema diz-nos que o sistema de for\u00e7as est\u00e1 em equil\u00edbrio, portanto os dois corpos devem estar em equil\u00edbrio. A partir dessas informa\u00e7\u00f5es, podemos formular as equa\u00e7\u00f5es de equil\u00edbrio dos dois corpos: <\/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\"> Agora aplicamos a f\u00f3rmula da for\u00e7a gravitacional e simplificamos a 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-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\"> Finalmente, substitu\u00edmos os dados e resolvemos a massa do corpo 1: <\/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\">Exerc\u00edcio 4<\/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, qual o valor 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. Portanto 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 uma \u00fanica 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 no outro apoio substituindo o valor obtido na equa\u00e7\u00e3o 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 s\u00e3o condi\u00e7\u00f5es de equil\u00edbrio. Voc\u00ea encontrar\u00e1 exemplos reais de ambas as condi\u00e7\u00f5es de equil\u00edbrio e, al\u00e9m disso, poder\u00e1 treinar com exerc\u00edcios resolvidos passo a passo. Quais s\u00e3o as condi\u00e7\u00f5es de equil\u00edbrio? Na f\u00edsica, as condi\u00e7\u00f5es de equil\u00edbrio afirmam que um corpo est\u00e1 em equil\u00edbrio se a soma das for\u00e7as e &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/condicoes-de-equilibrio\/\"> <span class=\"screen-reader-text\">Condi\u00e7\u00f5es 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-23","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 Condi\u00e7\u00f5es de equil\u00edbrio<\/title>\n<meta name=\"description\" content=\"Explicamos quais s\u00e3o as condi\u00e7\u00f5es de equil\u00edbrio. 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