{"id":253,"date":"2023-06-23T06:39:35","date_gmt":"2023-06-23T06:39:35","guid":{"rendered":"https:\/\/physigeek.com\/pt\/plano-inclinado\/"},"modified":"2023-06-23T06:39:35","modified_gmt":"2023-06-23T06:39:35","slug":"plano-inclinado","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/plano-inclinado\/","title":{"rendered":"Plano inclinado"},"content":{"rendered":"<p>Este artigo explica o que s\u00e3o planos inclinados na f\u00edsica e como problemas desse tipo s\u00e3o resolvidos. Voc\u00ea encontrar\u00e1 as f\u00f3rmulas das for\u00e7as que atuam em um plano inclinado e, al\u00e9m disso, poder\u00e1 treinar com exerc\u00edcios resolvidos passo a passo no plano inclinado. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-un-plano-inclinado\"><\/span> O que \u00e9 um plano inclinado?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Um <strong>plano inclinado<\/strong> \u00e9 uma superf\u00edcie inclinada em um determinado \u00e2ngulo. Na f\u00edsica, o plano inclinado \u00e9 usado para praticar problemas de for\u00e7a.<\/p>\n<p> Por exemplo, uma rampa ou uma estrada inclinada s\u00e3o planos inclinados.<\/p>\n<p> O plano inclinado permite transportar um objeto usando menos for\u00e7a. J\u00e1 que empurrar um objeto em um plano inclinado requer menos for\u00e7a do que levant\u00e1-lo verticalmente.<\/p>\n<p> Al\u00e9m disso, o plano inclinado \u00e9 considerado uma das seis m\u00e1quinas simples cl\u00e1ssicas. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Formulas-del-plano-inclinado\"><\/span> F\u00f3rmulas de plano inclinado<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Agora que sabemos a defini\u00e7\u00e3o de plano inclinado, vamos ver quais f\u00f3rmulas atuam em um plano inclinado e quais equa\u00e7\u00f5es as conectam.<\/p>\n<p> O primeiro problema que encontramos nos exerc\u00edcios de plano inclinado \u00e9 que a maioria das for\u00e7as atua numa dire\u00e7\u00e3o paralela ou perpendicular ao plano inclinado. Portanto, os eixos coordenados t\u00edpicos (um eixo vertical e um eixo horizontal) n\u00e3o s\u00e3o muito \u00fateis para estes tipos de problemas. \u00c9 por isso que, em geral, em planos inclinados trabalhamos com um sistema de coordenadas diferente: <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/plan-incline.png\" alt=\"plano inclinado\" class=\"wp-image-4369\" width=\"391\" height=\"368\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/plan-incline-300x283.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/plan-incline-768x724.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/plan-incline.png 1010w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p> <strong>Em f\u00edsica, para resolver um problema de plano inclinado, utilizamos dois eixos diferentes:<\/strong> um primeiro eixo cuja dire\u00e7\u00e3o \u00e9 paralela ao plano inclinado e, por outro lado, um segundo eixo cuja dire\u00e7\u00e3o \u00e9 perpendicular ao plano inclinado.<\/p>\n<p> Al\u00e9m disso, como voc\u00ea pode ver na imagem, <strong>geralmente tr\u00eas for\u00e7as diferentes atuam em um plano inclinado<\/strong> (se houver atrito): a for\u00e7a peso, a for\u00e7a normal e a for\u00e7a de atrito (ou for\u00e7a de atrito). Mas logicamente, se n\u00e3o houver atrito no plano inclinado, a for\u00e7a de atrito \u00e9 desprezada.<\/p>\n<p> Por\u00e9m, a for\u00e7a do peso \u00e9 decomposta vetorialmente em duas componentes: uma componente paralela ao plano inclinado e outra componente perpendicular ao plano inclinado. Desta forma todas as for\u00e7as podem ser expressas nos eixos de trabalho do plano inclinado. Assim, os dois componentes do peso do corpo apoiado no plano inclinado s\u00e3o calculados pelo seno e pelo cosseno do \u00e2ngulo de inclina\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-a26edbf89d563f1351d0ec9771f7e7bc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1=m\\cdot g\\cdot \\text{sen}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"142\" 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-f5f191747feef04cf0a63f61a6b56cfd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=m\\cdot g\\cdot \\text{cos}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"141\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Finalmente, as for\u00e7as que atuam num plano inclinado podem ser relacionadas pelas duas f\u00f3rmulas seguintes: <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-du-plan-incline.png\" alt=\"f\u00f3rmulas de plano inclinado\" class=\"wp-image-4388\" width=\"491\" height=\"151\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-du-plan-incline-300x93.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-du-plan-incline-768x237.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-du-plan-incline.png 1022w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p> Observe que, se a defini\u00e7\u00e3o do problema n\u00e3o disser o contr\u00e1rio, o corpo no plano inclinado poderia deslizar pela encosta, raz\u00e3o pela qual uma poss\u00edvel acelera\u00e7\u00e3o est\u00e1 inclu\u00edda na equa\u00e7\u00e3o do eixo paralelo ao plano. Por outro lado, o corpo n\u00e3o pode se mover na dire\u00e7\u00e3o do eixo perpendicular ao plano inclinado, portanto a soma das for\u00e7as \u00e9 zero. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-resuelto-del-plano-inclinado\"><\/span> Exemplo resolvido do plano inclinado<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Para que voc\u00ea possa ver como os problemas de planos inclinados s\u00e3o resolvidos na f\u00edsica, voc\u00ea pode ver abaixo um exemplo resolvido passo a passo.<\/p>\n<ul>\n<li> Colocamos um corpo de massa m=6 kg no topo de um plano inclinado de 45\u00ba. Se o corpo desliza sobre o plano inclinado com uma acelera\u00e7\u00e3o de 4 m\/s <sup>2<\/sup> , qual \u00e9 o coeficiente de atrito din\u00e2mico entre a superf\u00edcie do plano inclinado e a do corpo? Dados: g=10 m\/s <sup>2<\/sup> . <\/li>\n<\/ul>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique.png\" alt=\"problema do coeficiente de atrito ou atrito din\u00e2mico\" class=\"wp-image-4281\" width=\"203\" height=\"205\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique-298x300.png 298w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique.png 479w\" sizes=\"auto, (max-width: 298px) 100vw, 298px\"><\/figure>\n<p class=\"has-text-align-left\"> A primeira coisa que precisamos fazer para resolver qualquer problema de f\u00edsica relacionado \u00e0 din\u00e2mica \u00e9 desenhar o diagrama de corpo livre. Ent\u00e3o, todas as for\u00e7as que atuam no sistema s\u00e3o: <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-dynamique.png\" alt=\"exerc\u00edcio resolvido do coeficiente de atrito ou atrito din\u00e2mico\" class=\"wp-image-4282\" width=\"248\" height=\"301\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-dynamique-247x300.png 247w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-dynamique.png 572w\" sizes=\"auto, (max-width: 247px) 100vw, 247px\"><\/figure>\n<p class=\"has-text-align-left\"> No sentido do eixo 1 (paralelo ao plano inclinado) o corpo apresenta uma acelera\u00e7\u00e3o, por\u00e9m, no sentido do eixo 2 (perpendicular ao plano inclinado) o corpo est\u00e1 em repouso. A partir dessas informa\u00e7\u00f5es, estabelecemos as equa\u00e7\u00f5es das for\u00e7as do sistema:<\/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-d87a1ef6aaa3476891df5da8334cbc49_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1-F_R=m\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"124\" 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-6bdf90ed250934bf6cffbb110bc792a4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2-N=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"90\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ent\u00e3o, podemos calcular a for\u00e7a normal a partir da segunda 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-59341555fe3d5fe315ceb1864547873b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_2\\\\[3ex]N=m\\cdot g\\cdot \\text{cos}(\\alpha) \\\\[3ex] N=6 \\cdot 10 \\cdot \\ text{cos}(45\u00ba)\\\\[3ex]N=42,43 \\ N\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"151\" width=\"185\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Por outro lado, calculamos o valor da for\u00e7a de atrito (ou for\u00e7a de atrito) a partir da primeira equa\u00e7\u00e3o apresentada:<\/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-d8f2aff2a81d98ddcea04b1988282fda_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}P_1-F_R=m\\cdot a\\\\[3ex]F_R=P_1-m\\cdot a\\\\[3ex]F_R=m\\cdot g\\cdot \\text{sin} (\\alpha)-m\\cdot a\\\\[3ex]F_R=6\\cdot 10\\cdot \\text{sin}(45\u00ba)-6\\cdot 4\\\\[3ex]F_R=18.43 \\ N\\end{ array} \" title=\"Rendered by QuickLaTeX.com\" height=\"195\" width=\"204\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E uma vez conhecido o valor da for\u00e7a normal e da for\u00e7a de atrito, podemos determinar o coeficiente de atrito din\u00e2mico usando sua f\u00f3rmula correspondente: <\/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-b783c9e37bcf4d077d9496489fc5d7d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu_d=\\cfrac{F_R}{N}=\\cfrac{18.43}{43.43}=0.42\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"187\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-del-plano-inclinado\"><\/span> Exerc\u00edcios resolvidos no plano inclinado<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 1<\/h3>\n<p> Colocamos um corpo de massa m=2 kg no topo de um plano inclinado com \u00e2ngulo de inclina\u00e7\u00e3o de 30\u00ba. Qual \u00e9 o coeficiente de atrito entre a rampa e o corpo se este permanecer em equil\u00edbrio? Dados: g=9,81 m\/s <sup>2<\/sup> <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-force-normale-et-de-force-de-friction.png\" alt=\"\" class=\"wp-image-4253\" width=\"285\" height=\"176\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-force-normale-et-de-force-de-friction-300x185.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-force-normale-et-de-force-de-friction.png 702w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\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\"> Como em qualquer problema de f\u00edsica que envolva for\u00e7as, a primeira coisa a fazer \u00e9 desenhar o diagrama de corpo livre do sistema. Ent\u00e3o, todas as for\u00e7as que atuam neste sistema s\u00e3o: <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-normale-et-friction-force.png\" alt=\"resolver o exerc\u00edcio de for\u00e7a normal e for\u00e7a de atrito\" class=\"wp-image-4254\" width=\"285\" height=\"333\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-normale-et-friction-force-256x300.png 256w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-normale-et-friction-force.png 702w\" sizes=\"auto, (max-width: 256px) 100vw, 256px\"><\/figure>\n<p class=\"has-text-align-left\"> Assim, para que o sistema esteja em equil\u00edbrio, a soma das for\u00e7as nos eixos 1 e 2 deve ser igual a zero. Portanto, as seguintes equa\u00e7\u00f5es s\u00e3o verdadeiras:<\/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-a692b08b4d7c08a2c55556233dc56651_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=P_1\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"63\" 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-ea3f790cf878ca23f77405f73a20e7c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"58\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Agora podemos calcular o valor da for\u00e7a normal a partir da segunda 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-780db8c589b96d398e1400444a11db30_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_2\\\\[3ex]N=P\\cdot \\text{cos}(\\alpha)\\\\[3ex]N=m \\cdot g\\cdot \\text{cos }(\\alpha)\\\\[3ex]N=2 \\cdot 9,81 \\cdot \\text{cos}(30\\text{\u00ba})\\\\[3ex]N=16,99 \\ N\\end{array} \" title=\"Rendered by QuickLaTeX.com\" height=\"196\" width=\"171\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Por outro lado, determinamos o valor da for\u00e7a de atrito usando a primeira 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-bef5af0f3a7e907aa90f08435f538cf7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}F_R=P_1\\\\[3ex]N=P\\cdot \\text{sin}(\\alpha)\\\\[3ex]F_R=m \\cdot g\\cdot \\text{sin }(\\alpha)\\\\[3ex]F_R=2 \\cdot 9,81 \\cdot \\text{sin}(30\\text{\u00ba})\\\\[3ex]F_R=9,81 \\ N\\end{array} \" title=\"Rendered by QuickLaTeX.com\" height=\"196\" width=\"175\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Da mesma forma, a for\u00e7a de atrito pode ser relacionada \u00e0 for\u00e7a normal e ao coeficiente de atrito usando a seguinte f\u00f3rmula:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b8e2dc6a1180d664163aeb969b289073_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=\\mu \\cdot N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"86\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Portanto, resolvemos o coeficiente de atrito da equa\u00e7\u00e3o e calculamos seu valor: <\/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-2bee3710c7506bf8ff2456662a57f279_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu=\\cfrac{F_R}{N}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"59\" 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-69da73a9c8ca8ef047563bcb0b957d4b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu=\\cfrac{9,81}{16,99}\" title=\"Rendered by QuickLaTeX.com\" height=\"42\" width=\"80\" style=\"vertical-align: -16px;\"><\/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-87da99c1b6541f3ad374e4ebb3e9daf1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{\\mu=0.58}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"66\" 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 2<\/h3>\n<p> Como vemos no seguinte sistema formado por um plano inclinado e uma polia, dois corpos est\u00e3o ligados por uma corda e uma polia de massas desprez\u00edveis. Se o corpo 2 tem massa m <sub>2<\/sub> = 7 kg e a inclina\u00e7\u00e3o da rampa \u00e9 de 50\u00ba, calcule a for\u00e7a normal que o plano inclinado exerce sobre o corpo de massa m <sub>1<\/sub> para que todo o sistema fique em equil\u00edbrio. Despreze a for\u00e7a de atrito durante todo o exerc\u00edcio. <\/p>\n<figure class=\"wp-block-image 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 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-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<figure class=\"wp-block-image 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<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-b62bbb21cbec2be0bba7f8a839b12ba9_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 ] Par cons\u00e9quent, la composante vectorielle du poids du corps 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=\"83\" width=\"1404\" 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\"> A partir da equa\u00e7\u00e3o anterior, podemos calcular 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-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-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-16ae359d38a8a11d1b1db4988b8eeaf1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1 \\cdot \\text{sin}(50\\text{\u00ba}) =7\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"130\" 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-4249c6e274233595f50eedc1da64f56f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1 =\\cfrac{7}{\\text{sin}(50\\text{\u00ba})}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"111\" 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<p class=\"has-text-align-left\"> Por outro lado, se olharmos o diagrama de for\u00e7as do sistema, observamos que a for\u00e7a normal deve ser igual \u00e0 componente vetorial do peso do corpo 1 perpendicular ao plano inclinado. <\/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-82b47c80ab7ef66a41fc4d4425032831_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_{1y}=N\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"66\" 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-56ad7b690b37b3f53ca20597e165860b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1\\cdot \\text{cos}(\\alpha)=N\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"120\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ent\u00e3o, a partir desta equa\u00e7\u00e3o podemos encontrar o valor da for\u00e7a normal: <\/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-4a591e86900fca256cfa079da1bd0461_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_1\\cdot \\text{cos}(\\alpha)\\\\[3ex]N=m_1 \\cdot g\\cdot \\text{cos}(\\alpha)\\\\[ 3ex]N=9,14 \\cdot 9,81 \\cdot \\text{cos}(50\\text{\u00ba})\\\\[3ex]N=\\bm{57,63 \\ 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 3<\/h3>\n<p> Un tra\u00eeneau de 70 kg glisse sur une pente de 30\u00ba avec une vitesse initiale de 2 m\/s. Si le coefficient de frottement dynamique entre le tra\u00eeneau et la neige est de 0,2, calculez la vitesse que le tra\u00eeneau acquerra apr\u00e8s avoir parcouru 20 m\u00e8tres. Donn\u00e9es : g=10 m\/s <sup>2<\/sup> . <\/p>\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> Tout d&#8217;abord, nous r\u00e9alisons le sch\u00e9ma corporel libre du tra\u00eeneau : <\/p>\n<figure class=&quot;wp-block-image 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-plan-incline.png&quot; alt=&quot;exercice r\u00e9solu de la force de frottement ou de frottement sur un plan inclin\u00e9&quot; class=&quot;wp-image-4345&quot; width=&quot;305&quot; height=&quot;355&quot; srcset=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-plan-incline-258x300.png 258w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-plan-incline.png 706w&quot; sizes=&quot;(max-width: 258px) 100vw, 258px&quot;><\/figure>\n<p> Le tra\u00eeneau a une acc\u00e9l\u00e9ration dans la direction de l&#8217;axe 1 (parall\u00e8le au plan inclin\u00e9) mais reste au repos dans la direction de l&#8217;axe 2 (perpendiculaire au plan inclin\u00e9), donc les \u00e9quations des forces sont : [latex]P_1-F_R=m\\cdot a&#8221; title=&#8221;Rendered by QuickLaTeX.com&#8221; height=&#8221;213&#8243; width=&#8221;8731&#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-6bdf90ed250934bf6cffbb110bc792a4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2-N=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"90\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A partir da segunda equa\u00e7\u00e3o podemos calcular a for\u00e7a normal que atua no tren\u00f3<\/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-90b32b903f8be520ec73748b3de9b8b3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_2\\\\[3ex]N=m\\cdot g\\cdot \\text{cos}(\\alpha) \\\\[3ex] N=70 \\cdot 10 \\cdot \\ text{cos}(30\u00ba)\\\\[3ex]N=606,22 \\ N\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"151\" width=\"194\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Como agora sabemos o valor da for\u00e7a normal e o coeficiente de atrito din\u00e2mico, podemos calcular a for\u00e7a de atrito aplicando a f\u00f3rmula correspondente:<\/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-e0a32cc7650b33325233258788c218d4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=\\mu\\cdot N=0,2 \\cdot 606,22=121,24 \\ N \" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"298\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Assim, para determinar a velocidade final, devemos primeiro encontrar a acelera\u00e7\u00e3o do tren\u00f3, e esta pode ser calculada a partir da primeira equa\u00e7\u00e3o de for\u00e7a apresentada: <\/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-d87a1ef6aaa3476891df5da8334cbc49_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1-F_R=m\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"124\" 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-fa13e0490f51e32ac03b455043f6f32d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{P_1-F_R}{m}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"99\" 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-a6274d836af5618f7ef99e7f179c3902_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{m\\cdot g\\cdot \\text{sin}(\\alpha)-F_R}{m}\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"177\" 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-a3a9db70462cd187d50b851ede83983f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{70\\cdot 10\\cdot \\text{sin}(30\u00ba)-121.24}{70}\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"221\" 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-ba0d7325efa059351cc3d9aef838a9e2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=3,27 \\ \\cfrac{m}{s^2}\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"92\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Uma vez conhecida a acelera\u00e7\u00e3o do tren\u00f3, calculamos o tempo que leva para percorrer os 20 metros com a equa\u00e7\u00e3o do movimento retil\u00edneo em acelera\u00e7\u00e3o constante: <\/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-69f632cd171007df0f5bd6f0fa458a5c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x=v_0\\cdot t +\\cfrac{1}{2}\\cdot a \\cdot t^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"150\" 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-b97ac72bf22d70273fece0cce195f4ca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"20=2\\cdot t +\\cfrac{1}{2}\\cdot 3.27 \\cdot t^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"172\" 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-11a99cb686bf68cbcca594d0d60f801b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"0=1,64t^2+2t-20\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"158\" 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-7aa7e01e70b4199d597d05e261c970df_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle t=\\cfrac{-2\\pm \\sqrt{2^2-4\\cdot 1.64\\cdot (-20)}}{2\\cdot 1.64}=\\cfrac{-2\\ pm 11.63}{ 3.28}=\\begin{cases}2.94\\\\[2ex]-4.15 \\ \\color{red}\\bm{\\times}\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"507\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Logicamente, exclu\u00edmos a solu\u00e7\u00e3o negativa, pois o tempo \u00e9 uma quantidade f\u00edsica que n\u00e3o pode ser negativa.<\/p>\n<p class=\"has-text-align-left\"> Finalmente, calculamos a velocidade final usando a f\u00f3rmula de acelera\u00e7\u00e3o constante: <\/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-5ace98bfb166f5b813f593760fcfa048_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{v_f-v_0}{t_f-t_0}\\quad \\longrightarrow \\quad v_f=a\\cdot (t_f-t_0)+v_0\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"330\" style=\"vertical-align: -18px;\"><\/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-817a054f54a84b5a88667fc794feeb4d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v_f=3.27\\cdot (2.94-0)+2=\\bm{11.61} \\ \\cfrac{\\bm{m}}{\\bm{s}}\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"280\" style=\"vertical-align: -12px;\"><\/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 planos inclinados na f\u00edsica e como problemas desse tipo s\u00e3o resolvidos. Voc\u00ea encontrar\u00e1 as f\u00f3rmulas das for\u00e7as que atuam em um plano inclinado e, al\u00e9m disso, poder\u00e1 treinar com exerc\u00edcios resolvidos passo a passo no plano inclinado. O que \u00e9 um plano inclinado? Um plano inclinado \u00e9 uma superf\u00edcie &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/plano-inclinado\/\"> <span class=\"screen-reader-text\">Plano inclinado<\/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-253","post","type-post","status-publish","format-standard","hentry","category-dinamico"],"yoast_head":"<!-- This site is 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