{"id":21,"date":"2023-06-27T12:59:55","date_gmt":"2023-06-27T12:59:55","guid":{"rendered":"https:\/\/physigeek.com\/pt\/primeira-condicao-de-equilibrio\/"},"modified":"2023-06-27T12:59:55","modified_gmt":"2023-06-27T12:59:55","slug":"primeira-condicao-de-equilibrio","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/primeira-condicao-de-equilibrio\/","title":{"rendered":"Primeira condi\u00e7\u00e3o de equil\u00edbrio"},"content":{"rendered":"<p>Este artigo explica em que consiste a primeira condi\u00e7\u00e3o de equil\u00edbrio. Voc\u00ea tamb\u00e9m encontrar\u00e1 exemplos reais da primeira condi\u00e7\u00e3o de equil\u00edbrio e, por fim, poder\u00e1 praticar com exerc\u00edcios resolvidos sobre este tema. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFCual-es-la-primera-condicion-de-equilibrio\"><\/span> Qual \u00e9 a primeira condi\u00e7\u00e3o de equil\u00edbrio?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Na f\u00edsica, a <strong>primeira condi\u00e7\u00e3o de equil\u00edbrio<\/strong> estabelece que se a soma das for\u00e7as aplicadas a um corpo for igual a zero, esse corpo est\u00e1 em equil\u00edbrio translacional.<\/p>\n<p> Portanto, a primeira condi\u00e7\u00e3o de equil\u00edbrio \u00e9 satisfeita quando a for\u00e7a resultante de um sistema \u00e9 zero. Em outras palavras, a primeira condi\u00e7\u00e3o de equil\u00edbrio \u00e9 atendida quando a seguinte f\u00f3rmula \u00e9 satisfeita:<\/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-b6be8a8a29e6933d96cdf943ceb09119_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\sum \\vv{F}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"75\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<p> Al\u00e9m disso, quando a primeira condi\u00e7\u00e3o de equil\u00edbrio \u00e9 satisfeita, o corpo est\u00e1 em repouso ou movendo-se com velocidade constante. Porque se a soma das for\u00e7as for zero, o corpo n\u00e3o pode ter acelera\u00e7\u00e3o.<\/p>\n<p> Logicamente, para que a primeira condi\u00e7\u00e3o de equil\u00edbrio seja verificada, as for\u00e7as devem ser somadas vetorialmente e n\u00e3o os m\u00f3dulos. Em outras palavras, se a soma das for\u00e7as em cada eixo for zero, ent\u00e3o o corpo r\u00edgido est\u00e1 em equil\u00edbrio mec\u00e2nico.<\/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> Portanto, um m\u00e9todo para verificar se a primeira condi\u00e7\u00e3o de equil\u00edbrio \u00e9 atendida \u00e9 somar todas as for\u00e7as em cada eixo separadamente, e se todas as somas derem zero, o corpo est\u00e1 em equil\u00edbrio translacional. <\/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> Observe que existem dois tipos de equil\u00edbrio translacional:<\/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<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplos-de-la-primera-condicion-de-equilibrio\"><\/span> Exemplos da primeira condi\u00e7\u00e3o de equil\u00edbrio<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Depois de conhecermos a defini\u00e7\u00e3o da primeira condi\u00e7\u00e3o de equil\u00edbrio, voc\u00ea poder\u00e1 ver tr\u00eas exemplos diferentes abaixo para entender completamente o que ela significa.<\/p>\n<p> Os sem\u00e1foros s\u00e3o um exemplo da primeira condi\u00e7\u00e3o de equil\u00edbrio na vida cotidiana. Muitas vezes vemos placas penduradas na rua e elas est\u00e3o sempre em repouso (ficam em p\u00e9 e n\u00e3o caem), portanto em equil\u00edbrio.<\/p>\n<p> Da mesma forma, qualquer objeto que esteja em repouso no solo est\u00e1 em equil\u00edbrio de for\u00e7as, ou em outras palavras, satisfaz a primeira condi\u00e7\u00e3o de equil\u00edbrio. Porque as \u00fanicas for\u00e7as aplicadas ao corpo s\u00e3o o peso e a for\u00e7a normal, e as duas for\u00e7as se op\u00f5em. <\/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\/equilibre-des-forces.png\" alt=\"equil\u00edbrio de poder\" class=\"wp-image-294\" width=\"319\" height=\"252\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/equilibre-des-forces-300x238.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/equilibre-des-forces-768x610.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/equilibre-des-forces.png 852w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<\/div>\n<p> Finalmente, outro exemplo da primeira condi\u00e7\u00e3o de equil\u00edbrio \u00e9 um carro dirigindo a uma velocidade constante em uma rodovia. Qualquer corpo que se mova com velocidade constante implica que sua acelera\u00e7\u00e3o \u00e9 zero e, portanto, a soma das for\u00e7as aplicadas a ele tamb\u00e9m \u00e9 zero. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-primera-condicion-de-equilibrio\"><\/span>Problemas resolvidos da primeira 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> 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-5ce6dd2d7dce6ba9d2e8c66e50e628b6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left.\\begin{array}{l}-T_1\\cdot 0,94+T_2\\cdot 0,57=0\\\\[2ex]T_1\\cdot 0,34+T_2\\cdot 0,82-117 .72=0\\end{array }\\right\\} \\longrightarrow \\ \\begin{array}{c}T_1=69,56 \\ N\\\\[2ex]T_2=114,74 \\ N\\end{array}[\/ latex] \n\n<div class=&quot;wp-block-otfm-box-spoiler-end otfm-sp_end&quot;><\/div>\n<h3 class=&quot;wp-block-heading&quot;> Exercice 2<\/h3>\n<p> Comme le montre la figure suivante, deux objets sont reli\u00e9s par une corde et une poulie de masses n\u00e9gligeables. Si l&#8217;objet 2 a une masse de 7 kg et que la pente de la rampe est de 50\u00ba, calculez la masse de l&#8217;objet 1 pour que l&#8217;ensemble du syst\u00e8me soit dans des conditions d&#8217;\u00e9quilibre. Dans ce cas, la force de frottement peut \u00eatre n\u00e9glig\u00e9e. <\/p>\n<div class=&quot;wp-block-image&quot;>\n<figure class=&quot;aligncenter size-full is-resized&quot;><img decoding=&quot;async&quot; loading=&quot;lazy&quot; src=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-des-forces.png&quot; alt=&quot;probl\u00e8me d'\u00e9quilibre translationnel&quot; class=&quot;wp-image-295&quot; width=&quot;299&quot; height=&quot;240&quot; srcset=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-des-forces-300x241.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-des-forces.png 718w&quot; sizes=&quot;(max-width: 300px) 100vw, 300px&quot;><\/figure>\n<\/div>\n<div class=&quot;wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1&quot; role=&quot;button&quot; tabindex=&quot;0&quot; aria-expanded=&quot;false&quot; data-otfm-spc=&quot;#FFF8E1&quot; style=&quot;text-align:center&quot;>\n<div class=&quot;otfm-sp__title&quot;> <strong>Voir la solution<\/strong><\/div>\n<\/div>\n<p> Le corps 1 est sur une pente inclin\u00e9e, donc la premi\u00e8re chose \u00e0 faire est de d\u00e9composer vectoriellement la force de son poids pour avoir les forces dans les axes de la pente : [latex]P_{1x}=P_1\\cdot \\text{sin}(\\alpha)&#8221; title=&#8221;Rendered by QuickLaTeX.com&#8221; height=&#8221;340&#8243; width=&#8221;2876&#8243; style=&#8221;vertical-align: 0px;&#8221;><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-1a0b77602980cc17cce9b3baef744df8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_{1y}=P_1\\cdot \\text{cos}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"130\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Assim, o conjunto de for\u00e7as que atuam em todo o sistema \u00e9: <\/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 propor 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-4e1b75b6ba5d7bbe88d23e014eb011c5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1\\cdot \\text{sin}(\\alpha)=P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"120\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> 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-6e80f0daabb2167ec2f6622b08001a97_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1=9,14 \\ kg\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"106\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Este artigo explica em que consiste a primeira condi\u00e7\u00e3o de equil\u00edbrio. Voc\u00ea tamb\u00e9m encontrar\u00e1 exemplos reais da primeira condi\u00e7\u00e3o de equil\u00edbrio e, por fim, poder\u00e1 praticar com exerc\u00edcios resolvidos sobre este tema. Qual \u00e9 a primeira condi\u00e7\u00e3o de equil\u00edbrio? Na f\u00edsica, a primeira condi\u00e7\u00e3o de equil\u00edbrio estabelece que se a soma das for\u00e7as aplicadas a &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/primeira-condicao-de-equilibrio\/\"> <span class=\"screen-reader-text\">Primeira 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-21","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 Primeira condi\u00e7\u00e3o de equil\u00edbrio (com exerc\u00edcios resolvidos)<\/title>\n<meta name=\"description\" content=\"Explicamos o que \u00e9 a primeira condi\u00e7\u00e3o de equil\u00edbrio e em que consiste. 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