{"id":19,"date":"2023-06-27T14:57:01","date_gmt":"2023-06-27T14:57:01","guid":{"rendered":"https:\/\/physigeek.com\/pt\/equilibrio-translacional\/"},"modified":"2023-06-27T14:57:01","modified_gmt":"2023-06-27T14:57:01","slug":"equilibrio-translacional","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/equilibrio-translacional\/","title":{"rendered":"Equil\u00edbrio translacional"},"content":{"rendered":"<p>Este artigo explica o que \u00e9 equil\u00edbrio translacional. Voc\u00ea encontrar\u00e1 a rela\u00e7\u00e3o entre o equil\u00edbrio translacional e a primeira condi\u00e7\u00e3o de equil\u00edbrio (ou princ\u00edpio da in\u00e9rcia). Voc\u00ea tamb\u00e9m poder\u00e1 ver v\u00e1rios exemplos de saldos translacionais e, por fim, ter\u00e1 um exerc\u00edcio resolvido sobre o assunto para praticar. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-el-equilibrio-traslacional\"><\/span> O que \u00e9 equil\u00edbrio translacional?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>O equil\u00edbrio translacional<\/strong> \u00e9 um estado f\u00edsico no qual o corpo est\u00e1 em repouso ou em velocidade constante. O equil\u00edbrio translacional ocorre quando a soma das for\u00e7as que atuam no corpo \u00e9 igual 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-022c7518a0834669c82cb9a1e2208ce2_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> Portanto, quando um corpo r\u00edgido est\u00e1 em equil\u00edbrio translacional, isso significa que sua acelera\u00e7\u00e3o \u00e9 zero. Da mesma forma, se um sistema de for\u00e7as est\u00e1 em equil\u00edbrio translacional, a for\u00e7a resultante no sistema \u00e9 zero.<\/p>\n<p> Lembre-se de que, em f\u00edsica, a transla\u00e7\u00e3o \u00e9 um movimento no qual a posi\u00e7\u00e3o de um objeto muda. \u00c9 por isso que o equil\u00edbrio translacional significa que a tradu\u00e7\u00e3o ocorre de forma equilibrada (a uma velocidade constante) ou que simplesmente n\u00e3o h\u00e1 tradu\u00e7\u00e3o.<\/p>\n<p> Existem dois tipos de equil\u00edbrio translacional:<\/p>\n<ul>\n<li> <strong>Equil\u00edbrio translacional est\u00e1tico<\/strong> : corpo cuja soma das for\u00e7as \u00e9 zero e tamb\u00e9m est\u00e1 em repouso.<\/li>\n<li> <strong>Equil\u00edbrio translacional din\u00e2mico<\/strong> : Um corpo cuja soma vetorial de for\u00e7as \u00e9 zero e se move a velocidade constante. <\/li>\n<\/ul>\n<h2 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><\/h2>\n<p> Quando um corpo est\u00e1 em equil\u00edbrio translacional, diz-se que a <strong>primeira condi\u00e7\u00e3o de equil\u00edbrio<\/strong> est\u00e1 satisfeita.<\/p>\n<p> Portanto, a primeira condi\u00e7\u00e3o de equil\u00edbrio \u00e9 satisfeita quando a soma das for\u00e7as de um sistema \u00e9 zero. Tenha em mente que os m\u00f3dulos das for\u00e7as n\u00e3o devem ser somados, mas sim como vetores, ou seja, a soma das for\u00e7as deve ser zero para cada eixo.<\/p>\n<p> Portanto, se estivermos trabalhando com for\u00e7as coplanares (duas dimens\u00f5es), para que um corpo esteja em equil\u00edbrio translacional, as for\u00e7as horizontais (eixo X) e as for\u00e7as verticais (eixo Y) devem ser somadas separadamente, e as duas somas devem dar 0 . <\/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-45b0a8d2ba38c7f5141d9122c85b4363_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\sum \\vv{F_x}=0 \\qquad \\sum\\vv{F_y}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"200\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Equilibrio-traslacional-y-rotacional\"><\/span> Equil\u00edbrio translacional e rotacional<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>Um corpo r\u00edgido est\u00e1 em equil\u00edbrio translacional e rotacional quando a soma das for\u00e7as e a soma dos momentos s\u00e3o iguais a zero.<\/strong> Ou, em outras palavras, um corpo est\u00e1 em equil\u00edbrio translacional e rotacional quando a for\u00e7a resultante e o momento resultante s\u00e3o 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> Nesta situa\u00e7\u00e3o, a velocidade linear do corpo ser\u00e1 zero ou constante e, da mesma forma, a sua velocidade angular ser\u00e1 zero ou constante. Portanto, n\u00e3o ter\u00e1 acelera\u00e7\u00e3o linear nem acelera\u00e7\u00e3o angular.<\/p>\n<p> Al\u00e9m disso, 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<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplos-de-equilibrio-traslacional\"><\/span> Exemplos de equil\u00edbrio translacional<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Considerando a defini\u00e7\u00e3o de equil\u00edbrio translacional, analisaremos tr\u00eas exemplos diferentes para finalizar a compreens\u00e3o do significado deste termo.<\/p>\n<p> Por exemplo, o seguinte objeto suspenso por cordas est\u00e1 em equil\u00edbrio translacional porque todas as for\u00e7as se equilibram. A for\u00e7a do peso \u00e9 compensada pela for\u00e7a T <sub>2<\/sub> e pelas componentes verticais das for\u00e7as T <sub>1<\/sub> e T <sub>3<\/sub> . E por outro lado, as componentes horizontais das for\u00e7as T <sub>1<\/sub> e T <sub>3<\/sub> compensam-se. <\/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-translationnel.png\" alt=\"equil\u00edbrio translacional\" class=\"wp-image-341\" width=\"360\" height=\"360\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/equilibre-translationnel-300x300.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/equilibre-translationnel-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/equilibre-translationnel-768x770.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/equilibre-translationnel.png 968w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<\/div>\n<p> Na verdade, qualquer objeto apoiado no solo em repouso est\u00e1 em equil\u00edbrio de for\u00e7as, uma vez que as \u00fanicas for\u00e7as aplicadas a ele 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 for\u00e7as na tradu\u00e7\u00e3o\" 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> Outro exemplo de equil\u00edbrio translacional \u00e9 um carro movendo-se a uma velocidade constante ao longo da estrada. Qualquer corpo que se mova com velocidade constante implica que sua acelera\u00e7\u00e3o \u00e9 zero e, portanto, a soma das for\u00e7as tamb\u00e9m \u00e9 zero. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicio-resuelto-del-equilibrio-traslacional\"><\/span> Exerc\u00edcio de equil\u00edbrio translacional 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 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 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 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-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 translacional. Voc\u00ea encontrar\u00e1 a rela\u00e7\u00e3o entre o equil\u00edbrio translacional e a primeira condi\u00e7\u00e3o de equil\u00edbrio (ou princ\u00edpio da in\u00e9rcia). Voc\u00ea tamb\u00e9m poder\u00e1 ver v\u00e1rios exemplos de saldos translacionais e, por fim, ter\u00e1 um exerc\u00edcio resolvido sobre o assunto para praticar. O que \u00e9 equil\u00edbrio translacional? O equil\u00edbrio translacional &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/equilibrio-translacional\/\"> <span class=\"screen-reader-text\">Equil\u00edbrio translacional<\/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-19","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 O que \u00e9 equil\u00edbrio translacional? 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