{"id":251,"date":"2023-06-23T07:32:36","date_gmt":"2023-06-23T07:32:36","guid":{"rendered":"https:\/\/physigeek.com\/pt\/forca-de-atrito-ou-forca-de-atrito\/"},"modified":"2023-06-23T07:32:36","modified_gmt":"2023-06-23T07:32:36","slug":"forca-de-atrito-ou-forca-de-atrito","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/forca-de-atrito-ou-forca-de-atrito\/","title":{"rendered":"For\u00e7a de atrito (ou for\u00e7a de atrito)"},"content":{"rendered":"<p>Este artigo explica o que \u00e9 for\u00e7a de atrito (ou for\u00e7a de atrito) na f\u00edsica e como ela \u00e9 calculada. Voc\u00ea encontrar\u00e1, portanto, as propriedades da for\u00e7a de atrito, os dois tipos de for\u00e7a de atrito existentes e, al\u00e9m disso, exerc\u00edcios espec\u00edficos para praticar. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-la-fuerza-de-rozamiento\"><\/span> O que \u00e9 for\u00e7a de atrito?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>A for\u00e7a de atrito<\/strong> , tamb\u00e9m chamada <strong>de for\u00e7a de atrito<\/strong> , \u00e9 uma for\u00e7a de contato que ocorre ao tentar mover um corpo atrav\u00e9s da superf\u00edcie de outro corpo.<\/p>\n<p> Mais precisamente, a for\u00e7a de atrito \u00e9 uma for\u00e7a exercida numa dire\u00e7\u00e3o paralela e oposta ao movimento. <\/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\/force-de-friction-ou-force-de-friction.png\" alt=\"\" class=\"wp-image-4319\" width=\"293\" height=\"292\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-de-friction-ou-force-de-friction-300x300.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-de-friction-ou-force-de-friction-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-de-friction-ou-force-de-friction.png 625w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p> Existem dois tipos de for\u00e7as de atrito: for\u00e7a de atrito est\u00e1tico e for\u00e7a de atrito din\u00e2mico. Dependendo do caso, um ou outro atua. Abaixo veremos a diferen\u00e7a entre eles.<\/p>\n<p> Em geral, a for\u00e7a de atrito \u00e9 representada pelo s\u00edmbolo F <sub>R.<\/sub> <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Caracteristicas-de-la-fuerza-de-rozamiento\"><\/span> Caracter\u00edsticas da for\u00e7a de atrito<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Agora que conhecemos a defini\u00e7\u00e3o de for\u00e7a de atrito (ou for\u00e7a de atrito), vamos ver quais s\u00e3o as caracter\u00edsticas desse tipo de for\u00e7a:<\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">A for\u00e7a de atrito \u00e9 uma for\u00e7a de contato, ou seja, s\u00f3 atua se duas superf\u00edcies estiverem em contato.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Al\u00e9m disso, a for\u00e7a de atrito s\u00f3 aparece quando um corpo se move ou tenta mover-se em cima de outro.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">A dire\u00e7\u00e3o da for\u00e7a de atrito \u00e9 paralela \u00e0 dire\u00e7\u00e3o do movimento.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">A dire\u00e7\u00e3o da for\u00e7a de atrito \u00e9 oposta ao movimento.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">A for\u00e7a de atrito n\u00e3o depende da velocidade com que os corpos deslizam.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">A for\u00e7a de atrito n\u00e3o depende do tamanho da superf\u00edcie em contato.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Mas a for\u00e7a de atrito depende dos materiais em contacto, do seu acabamento e da temperatura.<\/span><\/li>\n<li> <span style=\"color:#101010;font-weight: normal;\">A for\u00e7a de atrito \u00e9 diretamente proporcional \u00e0 for\u00e7a normal.<\/span> <\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Formula-de-la-fuerza-de-rozamiento\"><\/span> F\u00f3rmula da for\u00e7a de atrito<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> A for\u00e7a de atrito \u00e9 igual ao coeficiente de atrito multiplicado pela for\u00e7a normal. Portanto, para calcular a for\u00e7a de atrito, deve-se primeiro encontrar a for\u00e7a normal e depois multiplic\u00e1-la pelo coeficiente de atrito entre as duas superf\u00edcies de contato.<\/p>\n<p> A <strong>f\u00f3rmula para a for\u00e7a de atrito<\/strong> (ou for\u00e7a de atrito) \u00e9, portanto, a seguinte:<\/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-a865b1cd2e263b944debf58666ec1269_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 style=\"margin-bottom:5px\"> Ouro: <\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:8px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-5b005ac29604de5f2904d2da7ade0238_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"22\" style=\"vertical-align: -3px;\"><\/p>\n<p> \u00e9 a for\u00e7a de atrito ou atrito, expressa em newtons. <\/span><\/li>\n<li style=\"margin-bottom:8px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-05d9eae892416bd34247a25207f8b718_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"11\" style=\"vertical-align: -4px;\"><\/p>\n<p> \u00e9 o coeficiente de atrito, que n\u00e3o tem unidade.<\/span><\/li>\n<li><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-7354bae77b50b7d1faed3e8ea7a3511a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 a for\u00e7a normal, expressa em newtons. <\/span><\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Fuerza-de-rozamiento-estatico-y-dinamico\"><\/span> For\u00e7a de atrito est\u00e1tica e din\u00e2mica<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> O valor da for\u00e7a de atrito depende se o corpo est\u00e1 em repouso ou em movimento. Por exemplo, com certeza voc\u00ea tentou arrastar um corpo muito pesado e no in\u00edcio foi dif\u00edcil mov\u00ea-lo, mas depois que voc\u00ea conseguiu mover um pouco o corpo fica mais f\u00e1cil continuar arrastando o objeto.<\/p>\n<p> Na verdade, em geral, a for\u00e7a de atrito quando o corpo est\u00e1 parado \u00e9 maior do que quando o corpo est\u00e1 em movimento.<\/p>\n<p> Assim, distinguimos dois tipos de for\u00e7a de atrito (ou for\u00e7a de atrito):<\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\"><strong>For\u00e7a de atrito est\u00e1tico<\/strong> : \u00c9 a for\u00e7a de atrito que atua quando o corpo ainda n\u00e3o est\u00e1 em movimento.<\/span><\/li>\n<li> <span style=\"color:#101010;font-weight: normal;\"><strong>For\u00e7a de atrito din\u00e2mica (ou cin\u00e9tica)<\/strong> : \u00e9 a for\u00e7a de atrito que atua quando o corpo j\u00e1 iniciou o movimento.<\/span><\/li>\n<\/ul>\n<p> Da mesma forma, o coeficiente de atrito est\u00e1tico tamb\u00e9m se distingue do coeficiente de atrito din\u00e2mico, que s\u00e3o usados para determinar a for\u00e7a de atrito est\u00e1tico e a for\u00e7a de atrito din\u00e2mico, respectivamente.<\/p>\n<p> Finalmente, o valor da for\u00e7a de atrito varia conforme mostrado no gr\u00e1fico a seguir: <\/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\/graphique-force-statique-et-dynamique-friction.png\" alt=\"for\u00e7a de atrito est\u00e1tico e din\u00e2mico\" class=\"wp-image-4356\" width=\"381\" height=\"261\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/graphique-force-statique-et-dynamique-friction-300x205.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/graphique-force-statique-et-dynamique-friction.png 725w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p> A for\u00e7a de atrito est\u00e1tico \u00e9 igual \u00e0 for\u00e7a aplicada para tentar mover o corpo, mas sua dire\u00e7\u00e3o \u00e9 oposta. Seu valor m\u00e1ximo \u00e9 o produto entre o coeficiente de atrito est\u00e1tico e a for\u00e7a normal. Quando a for\u00e7a aplicada ultrapassa esse valor, o corpo come\u00e7a a se mover.<\/p>\n<p> Assim, quando o corpo j\u00e1 est\u00e1 em movimento, a for\u00e7a de atrito din\u00e2mico tem valor constante equivalente ao produto entre o coeficiente de atrito din\u00e2mico e a for\u00e7a normal, qualquer que seja o valor da for\u00e7a aplicada. Al\u00e9m disso, este valor \u00e9 ligeiramente inferior ao valor m\u00e1ximo da for\u00e7a de atrito est\u00e1tico. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-fuerza-de-rozamiento\"><\/span> Exerc\u00edcios resolvidos sobre for\u00e7a de atrito<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 1<\/h3>\n<p> Pretende-se mover um bloco de massa m = 12 kg sobre uma superf\u00edcie plana e ele come\u00e7a a se mover exatamente quando uma for\u00e7a de 35 N \u00e9 aplicada. Qual \u00e9 o coeficiente de atrito est\u00e1tico entre o solo e o bloco? Dados: g=10 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-resolu-coefficient-de-frottement-statique.png\" alt=\"problema resolvido de coeficiente de atrito est\u00e1tico\" class=\"wp-image-4302\" width=\"285\" height=\"125\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-statique-300x132.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-statique.png 650w\" 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\"> Primeiro, representamos graficamente todas as for\u00e7as que atuam no bloco: <\/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-statique.png\" alt=\"exerc\u00edcio resolvido sobre o coeficiente de atrito est\u00e1tico ou coeficiente de atrito est\u00e1tico\" class=\"wp-image-4303\" width=\"269\" height=\"359\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-statique-225x300.png 225w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-statique.png 670w\" sizes=\"auto, (max-width: 225px) 100vw, 225px\"><\/figure>\n<p class=\"has-text-align-left\"> Na situa\u00e7\u00e3o limite de equil\u00edbrio, as duas equa\u00e7\u00f5es a seguir s\u00e3o satisfeitas:<\/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-318d3aaff48777c13e5ac24cb775f6b0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"54\" 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-b2d0fc9325264d9c3dceae21b529d2c5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=F\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"60\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Assim a for\u00e7a de atrito ser\u00e1 equivalente \u00e0 for\u00e7a horizontal aplicada ao corpo:<\/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-f433d79addb4c8cfc17c71cf797f4905_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=F=35 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"123\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Por outro lado, podemos calcular o valor da for\u00e7a normal usando a f\u00f3rmula da for\u00e7a peso:<\/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-fdd5a10090733132a78410e57a059c2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P\\\\[3ex] N=m\\cdot g\\\\[3ex] N=12\\cdot 10 \\\\[3ex] N=120 \\ N\\end{array }\" title=\"Rendered by QuickLaTeX.com\" height=\"147\" width=\"88\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Finalmente, uma vez conhecido o valor da for\u00e7a de atrito e da for\u00e7a normal, aplicamos a f\u00f3rmula do coeficiente de atrito est\u00e1tico para determinar o 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-20cdd7c8dcf9b6af55e1fe95eedf8da2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu_e=\\cfrac{F_R}{N}=\\cfrac{35}{120}=0.29\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"174\" style=\"vertical-align: -12px;\"><\/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> 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> . <\/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-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<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 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, propomos 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-31af78ef6e04fa66121d64aa3570f5a6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu_d=\\cfrac{F_R}{N}=\\cfrac{18.43}{43.43}=\\bm{0.42}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"187\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Exerc\u00edcio 3<\/h3>\n<p> Um tren\u00f3 de 70 kg desliza por um declive de 30\u00ba com velocidade inicial de 2 m\/s. Se o coeficiente de atrito din\u00e2mico entre o tren\u00f3 e a neve for 0,2, calcule a velocidade que o tren\u00f3 adquirir\u00e1 ap\u00f3s percorrer 20 metros. Dados: g=10 m\/s <sup>2<\/sup> . <\/p>\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\"> Primeiramente fazemos o diagrama de corpo livre do tren\u00f3: <\/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-plan-incline.png\" alt=\"exerc\u00edcio determinado de uma for\u00e7a de atrito sobre um plano inclinado\" class=\"wp-image-4345\" width=\"305\" height=\"355\" srcset=\"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\" sizes=\"auto, (max-width: 258px) 100vw, 258px\"><\/figure>\n<p class=\"has-text-align-left\"> O tren\u00f3 tem uma acelera\u00e7\u00e3o na dire\u00e7\u00e3o do eixo 1 (paralelo ao plano inclinado), mas permanece em repouso na dire\u00e7\u00e3o do eixo 2 (perpendicular ao plano inclinado), ent\u00e3o as equa\u00e7\u00f5es de for\u00e7a s\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-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\"> 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, descartamos 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-e7b1fffc78a1f3be82ead78bf2635dc0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v_f=3,27\\cdot (2,94-0)+2=11,61 \\ \\cfrac{m}{s}\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"288\" 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 \u00e9 for\u00e7a de atrito (ou for\u00e7a de atrito) na f\u00edsica e como ela \u00e9 calculada. Voc\u00ea encontrar\u00e1, portanto, as propriedades da for\u00e7a de atrito, os dois tipos de for\u00e7a de atrito existentes e, al\u00e9m disso, exerc\u00edcios espec\u00edficos para praticar. O que \u00e9 for\u00e7a de atrito? A for\u00e7a de atrito , &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/forca-de-atrito-ou-forca-de-atrito\/\"> <span class=\"screen-reader-text\">For\u00e7a de atrito (ou for\u00e7a de atrito)<\/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-251","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 For\u00e7a de fric\u00e7\u00e3o (ou for\u00e7a de fric\u00e7\u00e3o)<\/title>\n<meta name=\"description\" content=\"Aqui voc\u00ea descobrir\u00e1 o que \u00e9 for\u00e7a de atrito (ou for\u00e7a de atrito), como ela \u00e9 calculada (f\u00f3rmula) e exerc\u00edcios resolvidos sobre for\u00e7a de atrito.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/physigeek.com\/pt\/forca-de-atrito-ou-forca-de-atrito\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"\u25b7 For\u00e7a de fric\u00e7\u00e3o (ou for\u00e7a de fric\u00e7\u00e3o)\" \/>\n<meta property=\"og:description\" content=\"Aqui voc\u00ea descobrir\u00e1 o que \u00e9 for\u00e7a de atrito (ou for\u00e7a de atrito), como ela \u00e9 calculada (f\u00f3rmula) e exerc\u00edcios resolvidos sobre for\u00e7a de atrito.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/physigeek.com\/pt\/forca-de-atrito-ou-forca-de-atrito\/\" \/>\n<meta property=\"article:published_time\" content=\"2023-06-23T07:32:36+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-de-friction-ou-force-de-friction.png\" \/>\n<meta name=\"author\" content=\"Jonathan Reynolds\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"Jonathan Reynolds\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. tempo de leitura\" \/>\n\t<meta name=\"twitter:data2\" content=\"7 minutos\" \/>\n<script type=\"application\/ld+json\" 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