{"id":17,"date":"2023-06-27T17:03:39","date_gmt":"2023-06-27T17:03:39","guid":{"rendered":"https:\/\/physigeek.com\/pt\/equilibrio-de-poder\/"},"modified":"2023-06-27T17:03:39","modified_gmt":"2023-06-27T17:03:39","slug":"equilibrio-de-poder","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/equilibrio-de-poder\/","title":{"rendered":"Equil\u00edbrio de poder"},"content":{"rendered":"<p>Este artigo explica o que \u00e9 equil\u00edbrio de for\u00e7as e quando um corpo est\u00e1 em equil\u00edbrio. Voc\u00ea tamb\u00e9m aprender\u00e1 sobre equil\u00edbrio de momentos e equil\u00edbrio de for\u00e7as e momentos. Al\u00e9m disso, voc\u00ea poder\u00e1 ver um exemplo e poder\u00e1 praticar com um exerc\u00edcio resolvido sobre equil\u00edbrio de for\u00e7as.<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Equilibrio-de-fuerzas\"><\/span> Equil\u00edbrio de poder<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Um corpo r\u00edgido est\u00e1 em equil\u00edbrio de for\u00e7as quando a soma de todas as for\u00e7as aplicadas a ele \u00e9 igual a zero. Em outras palavras, <strong>um corpo est\u00e1 em equil\u00edbrio de for\u00e7as quando a for\u00e7a resultante \u00e9 zero.<\/strong><\/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-ff1948e7ffb274a2dfda7b4c2bd3cb5a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\somme \\vv{F}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"46\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <span style=\"color:#4fd12f\">\u27a4<\/span> <strong>Veja:<\/strong> <a href=\"https:\/\/physigeek.com\/pt\/forca-resultante\/\">Qual \u00e9 a for\u00e7a resultante?<\/a><\/p>\n<p> Al\u00e9m disso, se um corpo r\u00edgido estiver em condi\u00e7\u00f5es de equil\u00edbrio de for\u00e7as, isso significa que n\u00e3o ter\u00e1 acelera\u00e7\u00e3o. Portanto, o corpo manter\u00e1 sua velocidade ou n\u00e3o se mover\u00e1 se estiver em repouso.<\/p>\n<p> Deve-se levar em conta que para que um corpo esteja em equil\u00edbrio translacional, a soma das for\u00e7as em cada dire\u00e7\u00e3o deve ser zero (tr\u00eas dire\u00e7\u00f5es se trabalharmos no espa\u00e7o e duas dire\u00e7\u00f5es se trabalharmos no plano).<\/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-8173059e9d835ea8000739691bd3c498_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\somme \\vv{F_x}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"53\" 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-f019ed19f677d908d6dc565e03921605_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\somme \\vv{F_y}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"52\" 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-3d5793601c780b47719e5936868096fb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\somme \\vv{F_z}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"52\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p> Se uma das tr\u00eas condi\u00e7\u00f5es anteriores n\u00e3o for atendida, o corpo n\u00e3o estar\u00e1 em equil\u00edbrio de for\u00e7as e, portanto, ter\u00e1 acelera\u00e7\u00e3o.<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Equilibrio-de-momentos\"><\/span> Equil\u00edbrio moment\u00e2neo<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Um corpo r\u00edgido est\u00e1 em equil\u00edbrio de momentos quando a soma de todos os momentos aplicados a ele \u00e9 igual a zero. Em outras palavras, <strong>um corpo est\u00e1 em equil\u00edbrio de momento quando o momento resultante \u00e9 zero.<\/strong><\/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-c54de0f44f7a6549f27abfb7804a2c0c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\somme \\vv{M}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"52\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Portanto, o equil\u00edbrio de momentos \u00e9 an\u00e1logo ao equil\u00edbrio de for\u00e7as, mas a soma deve ser zero em todos os tr\u00eas eixos de rota\u00e7\u00e3o, em vez de em todos os tr\u00eas eixos longitudinais.<\/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-a19ea30e88dea24c5f83bcbeabe34336_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\somme \\vv{M_x}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"58\" 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-e465040a99f8e342adb6376e49a82089_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\somme \\vv{M_y}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"58\" 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-2b60edfb4e54fa1744d266ff1d25c851_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\somme \\vv{M_z}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"58\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p> Basta que uma equa\u00e7\u00e3o anterior n\u00e3o seja cumprida para que o s\u00f3lido r\u00edgido n\u00e3o esteja em equil\u00edbrio de momentos e portanto tenha uma acelera\u00e7\u00e3o rotacional, ou seja, o corpo come\u00e7ar\u00e1 a girar sobre si mesmo (ele partiu do repouso) .<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Equilibrio-de-fuerzas-y-momentos\"><\/span> Equil\u00edbrio de for\u00e7as e momentos<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>Um corpo r\u00edgido est\u00e1 em equil\u00edbrio de for\u00e7as e momentos quando a for\u00e7a resultante e o momento resultante s\u00e3o zero<\/strong> , ou seja, um corpo est\u00e1 em equil\u00edbrio de for\u00e7as e momentos quando a soma de todas as for\u00e7as e todos os momentos s\u00e3o 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-6c79e87aef62838b71a11567cc9a620f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\sum \\vv{F}=0 \\qquad \\sum\\vv{M}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"180\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Logicamente, um corpo s\u00f3 estar\u00e1 em equil\u00edbrio se a soma das for\u00e7as e momentos for zero em todos os eixos.<\/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-0560b8f93d7021c3c8c86531cb30833d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\sum \\vv{F_x}=0\\qquad \\sum \\vv{M_x}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"193\" 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-ea213173f7160398c238be4689cdd784_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\sum \\vv{F_y}=0\\qquad \\sum \\vv{M_y}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"192\" 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-6f64d40d9df66dff51c3fff8b9fbad7f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\sum \\vv{F_z}=0\\qquad \\sum \\vv{M_z}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"191\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Como vimos acima, um corpo n\u00e3o precisa necessariamente estar em equil\u00edbrio de for\u00e7as e momentos ao mesmo tempo, ele tamb\u00e9m pode estar apenas em equil\u00edbrio de for\u00e7as e apresentar algum desequil\u00edbrio nos momentos ou vice-versa.<\/p>\n<p> No entanto, quando um corpo est\u00e1 em equil\u00edbrio de for\u00e7as e de momentos <strong>, diz-se que o corpo est\u00e1 em equil\u00edbrio<\/strong> .<\/p>\n<p> As condi\u00e7\u00f5es de equil\u00edbrio s\u00e3o utilizadas para encontrar o valor de uma for\u00e7a aplicada a um corpo, pois nos permitem formular equa\u00e7\u00f5es e a partir delas podemos resolver as for\u00e7as desconhecidas. Por exemplo, a for\u00e7a normal \u00e9 geralmente calculada declarando a equa\u00e7\u00e3o do equil\u00edbrio vertical.<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-de-equilibrio-de-fuerzas\"><\/span> exemplo de equil\u00edbrio de poder<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Para compreender totalmente o conceito, vejamos um exemplo t\u00edpico de sistema de for\u00e7as em equil\u00edbrio.<\/p>\n<p> Por exemplo, um corpo em repouso localizado no solo est\u00e1 em equil\u00edbrio de for\u00e7as, pois apenas a for\u00e7a do peso e a for\u00e7a normal atuam sobre ele e se op\u00f5em. A soma das for\u00e7as e momentos em todas as dire\u00e7\u00f5es \u00e9, portanto, equivalente a 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\/equilibre-des-forces.png\" alt=\"equil\u00edbrio de poder\" class=\"wp-image-294\" width=\"332\" height=\"264\" 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> Neste caso o corpo tamb\u00e9m est\u00e1 em equil\u00edbrio de momentos, pois n\u00e3o h\u00e1 momento que atue sobre ele. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicio-resuelto-de-equilibrio-de-fuerzas\"><\/span> Exerc\u00edcio de equil\u00edbrio de poder resolvido<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li> 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. <\/li>\n<\/ul>\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 de poder\" 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 decompor vetorialmente 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-c05811c44aa2d58295c811d612a54eee_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_{1x}=P_1\\cdot \\text{sin}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"128\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-1a0b77602980cc17cce9b3baef744df8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_{1y}=P_1\\cdot \\text{cos}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"130\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> 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 de poder 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-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","protected":false},"excerpt":{"rendered":"<p>Este artigo explica o que \u00e9 equil\u00edbrio de for\u00e7as e quando um corpo est\u00e1 em equil\u00edbrio. Voc\u00ea tamb\u00e9m aprender\u00e1 sobre equil\u00edbrio de momentos e equil\u00edbrio de for\u00e7as e momentos. Al\u00e9m disso, voc\u00ea poder\u00e1 ver um exemplo e poder\u00e1 praticar com um exerc\u00edcio resolvido sobre equil\u00edbrio de for\u00e7as. Equil\u00edbrio de poder Um corpo r\u00edgido est\u00e1 em &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/equilibrio-de-poder\/\"> <span class=\"screen-reader-text\">Equil\u00edbrio de poder<\/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-17","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 Equil\u00edbrio de poder<\/title>\n<meta name=\"description\" content=\"O que \u00e9 equil\u00edbrio de for\u00e7as e quando um corpo est\u00e1 em equil\u00edbrio de for\u00e7as e momentos. 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