{"id":247,"date":"2023-06-23T09:26:11","date_gmt":"2023-06-23T09:26:11","guid":{"rendered":"https:\/\/physigeek.com\/pt\/forca-normal\/"},"modified":"2023-06-23T09:26:11","modified_gmt":"2023-06-23T09:26:11","slug":"forca-normal","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/forca-normal\/","title":{"rendered":"For\u00e7a normal"},"content":{"rendered":"<p>Este artigo explica o que \u00e9 for\u00e7a normal e como determin\u00e1-la dependendo do tipo de problema. Encontrar\u00e1 assim as caracter\u00edsticas da for\u00e7a normal e, al\u00e9m disso, poder\u00e1 praticar este tipo de for\u00e7a com exerc\u00edcios resolvidos passo a passo. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-la-fuerza-normal\"><\/span>O que \u00e9 for\u00e7a normal?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Na f\u00edsica, <strong>a for\u00e7a normal<\/strong> \u00e9 uma for\u00e7a exercida por uma superf\u00edcie sobre um corpo apoiado nela. Portanto, a dire\u00e7\u00e3o da for\u00e7a normal \u00e9 perpendicular \u00e0 superf\u00edcie e a dire\u00e7\u00e3o da for\u00e7a normal \u00e9 para fora, ou seja, a superf\u00edcie aplica a for\u00e7a normal em dire\u00e7\u00e3o ao corpo.<\/p>\n<p> Em geral, a for\u00e7a normal serve para neutralizar a <a href=\"https:\/\/physigeek.com\/pt\/peso-fisico\/\">for\u00e7a peso<\/a> , que \u00e9 a atra\u00e7\u00e3o gravitacional que a Terra exerce sobre qualquer corpo com massa. Por\u00e9m, quando o corpo repousa sobre uma superf\u00edcie inclinada, o valor da for\u00e7a normal pode n\u00e3o ser suficiente. A seguir veremos como \u00e9 calculada a for\u00e7a normal em um plano inclinado.<\/p>\n<p> Resumindo, as <strong><u style=\"text-decoration-color:#4fd12f\">caracter\u00edsticas da for\u00e7a normal<\/u><\/strong> s\u00e3o:<\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">A for\u00e7a normal \u00e9 uma for\u00e7a de contato, ou seja, s\u00f3 pode ser aplicada se duas superf\u00edcies estiverem em contato.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">A dire\u00e7\u00e3o da for\u00e7a normal \u00e9 perpendicular \u00e0 superf\u00edcie sobre a qual o corpo permanece.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">A dire\u00e7\u00e3o da for\u00e7a normal \u00e9 sempre para fora, pois \u00e9 a superf\u00edcie que aplica a for\u00e7a normal ao corpo.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">Em geral, a magnitude da for\u00e7a normal \u00e9 equivalente \u00e0 proje\u00e7\u00e3o da for\u00e7a resultante na superf\u00edcie de apoio.<\/span><\/li>\n<li> <span style=\"color:#101010;font-weight: normal;\">Normalmente, a for\u00e7a normal \u00e9 geralmente representada pelo s\u00edmbolo N ou F <sub>N.<\/sub><\/span> <\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Como-calcular-la-fuerza-normal\"><\/span> Como calcular a for\u00e7a normal<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Em geral, para <strong>calcular a for\u00e7a normal,<\/strong> deve-se aplicar as equa\u00e7\u00f5es de equil\u00edbrio, que estabelecem que um corpo est\u00e1 em equil\u00edbrio quando a soma das for\u00e7as verticais e a soma das for\u00e7as horizontais s\u00e3o iguais a zero.<\/p>\n<p> Ao aplicar as condi\u00e7\u00f5es de equil\u00edbrio ao problema, seremos capazes de resolver a for\u00e7a normal a partir das equa\u00e7\u00f5es propostas e, portanto, determinar 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-a9333695ba02f6e089d628fe3622a2e5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{c}\\displaystyle\\sum \\vv{F_x}=0\\\\[2ex]\\displaystyle\\sum \\vv{F_y}=0\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"81\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-del-calculo-de-la-fuerza-normal\"><\/span> Exemplo de c\u00e1lculo de for\u00e7a normal<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Agora que conhecemos a defini\u00e7\u00e3o de for\u00e7a normal, vamos ver um exemplo concreto de c\u00e1lculo da for\u00e7a normal.<\/p>\n<ul>\n<li> Um corpo de 8 kg est\u00e1 em repouso sobre um terreno plano. Qual \u00e9 o valor da for\u00e7a normal exercida pelo solo sobre o corpo?<\/li>\n<\/ul>\n<p> Neste problema, o corpo est\u00e1 em repouso sobre uma superf\u00edcie plana, portanto as \u00fanicas for\u00e7as que atuam sobre ele s\u00e3o a for\u00e7a peso e a for\u00e7a normal. <\/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-et-poids-normaux.png\" alt=\"for\u00e7a e peso normais\" class=\"wp-image-4215\" width=\"273\" height=\"297\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-et-poids-normaux-275x300.png 275w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-et-poids-normaux.png 570w\" sizes=\"auto, (max-width: 275px) 100vw, 275px\"><\/figure>\n<p> Portanto, para que um corpo esteja em equil\u00edbrio sobre uma superf\u00edcie plana, a for\u00e7a normal (N) e a for\u00e7a peso (P) devem ser iguais. A normal e o peso t\u00eam, portanto, a mesma dire\u00e7\u00e3o, o mesmo m\u00f3dulo, mas sua dire\u00e7\u00e3o \u00e9 oposta.<\/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> Assim, para determinar o valor da for\u00e7a normal, basta calcular o peso do corpo, que equivale \u00e0 sua massa multiplicada pela acelera\u00e7\u00e3o da gravidade: <\/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-a7355a420808c17875e97713c4bef5ec_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P=m\\cdot g=8 \\cdot 9,81 = 78,48 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"283\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Fuerza-normal-en-un-plano-inclinado\"><\/span> for\u00e7a normal em um plano inclinado<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Nesta se\u00e7\u00e3o derivaremos a f\u00f3rmula da for\u00e7a normal em um plano inclinado, pois seu valor muda dependendo se a superf\u00edcie \u00e9 plana ou inclinada.<\/p>\n<p> Assim, as for\u00e7as que atuam em um corpo apoiado em um plano inclinado s\u00e3o as seguintes: <\/p>\n<figure class=\"wp-block-image aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline.png\" alt=\"for\u00e7a normal em um plano inclinado\" class=\"wp-image-4220\" width=\"308\" height=\"417\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline-222x300.png 222w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline-757x1024.png 757w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline-768x1038.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline.png 770w\" sizes=\"auto, (max-width: 222px) 100vw, 222px\"><\/figure>\n<p> Veja a figura acima: Quando o plano est\u00e1 inclinado, \u00e9 mais conveniente usar a dire\u00e7\u00e3o paralela ao plano (eixo 1) e a dire\u00e7\u00e3o perpendicular ao plano (eixo 2) como eixos. Dessa forma \u00e9 mais f\u00e1cil formular as equa\u00e7\u00f5es de equil\u00edbrio.<\/p>\n<p> Para calcular a <strong>for\u00e7a normal sobre um plano inclinado<\/strong> \u00e9 necess\u00e1rio aplicar a condi\u00e7\u00e3o de equil\u00edbrio no eixo perpendicular ao plano inclinado, pois podemos garantir que o corpo est\u00e1 em equil\u00edbrio neste eixo mas n\u00e3o no eixo paralelo ao 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-badff1735c827c6562a4e074ea4b6bd2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\sum \\vv{F_2}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"80\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<p> Assim, a for\u00e7a normal sobre um plano inclinado \u00e9 equivalente \u00e0 componente do peso do eixo perpendicular ao 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-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> A componente do peso do eixo perpendicular ao plano \u00e9 igual \u00e0 f\u00f3rmula do peso multiplicado pelo cosseno do \u00e2ngulo de inclina\u00e7\u00e3o do 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-efeff10487f285abba9d74ee3eba6b45_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=P\\cdot \\cos(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"117\" 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-9fe1bc1d3a7fbacecb2ce1ccc1dadc67_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=m\\cdot g\\cdot \\cos(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"141\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Resumindo, a <strong>f\u00f3rmula da for\u00e7a normal em um plano inclinado<\/strong> afirma que a for\u00e7a normal \u00e9 igual \u00e0 massa do corpo vezes a gravidade vezes o cosseno do \u00e2ngulo de inclina\u00e7\u00e3o do plano: <\/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\/formule-de-la-force-normale-dans-un-plan-incline.png\" alt=\"f\u00f3rmula para for\u00e7a normal em um plano inclinado\" class=\"wp-image-4232\" width=\"269\" height=\"92\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formule-de-la-force-normale-dans-un-plan-incline-300x102.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formule-de-la-force-normale-dans-un-plan-incline.png 576w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Fuerza-normal-y-fuerza-de-rozamiento\"><\/span> for\u00e7a normal e for\u00e7a de atrito<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Nesta se\u00e7\u00e3o veremos a rela\u00e7\u00e3o entre a for\u00e7a normal e a for\u00e7a de atrito, pois s\u00e3o dois tipos de for\u00e7as ligadas matematicamente. Mas primeiro voc\u00ea precisa saber o que \u00e9 for\u00e7a de atrito.<\/p>\n<p> A for\u00e7a de atrito (ou for\u00e7a de atrito) \u00e9 uma for\u00e7a que ocorre ao tentar mover um corpo sobre uma superf\u00edcie n\u00e3o lisa. A for\u00e7a de atrito \u00e9, portanto, uma for\u00e7a que se op\u00f5e ao movimento de um corpo.<\/p>\n<p> A for\u00e7a de atrito \u00e9 calculada a partir da for\u00e7a normal. Mais precisamente, <strong>a for\u00e7a de atrito \u00e9 igual ao coeficiente de atrito superficial multiplicado pela for\u00e7a normal.<\/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-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 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. <\/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.<\/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 uma resist\u00eancia normal. <\/span><\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-fuerza-normal\"><\/span> Exerc\u00edcios de for\u00e7a normais resolvidos<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 1<\/h3>\n<p> Um corpo de 5 kg est\u00e1 em repouso sobre um terreno plano. Se ent\u00e3o outro corpo de massa 3 kg for adicionado acima do primeiro corpo, qual \u00e9 a for\u00e7a normal exercida pelo solo para sustentar os dois corpos? Dados: g=9,81 m\/ <sup>s2<\/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\"> Como o solo deve suportar ambos os corpos, a for\u00e7a normal ser\u00e1 a soma da for\u00e7a do peso de cada corpo. Portanto, primeiro calcularemos o peso de cada corpo e depois somaremos.<\/p>\n<p class=\"has-text-align-left\"> Lembre-se de que a for\u00e7a do peso \u00e9 calculada multiplicando a massa do corpo pela gravidade.<\/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-c0cdb663ec9f8fe79fbecd960b50fc39_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=m\\cdot g\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"75\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Assim, calculamos o peso de um corpo de 5 kg:<\/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-a15ef1db0a6608fa8e6165ac0e12e925_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1=5\\cdot 9.81=49.05\\N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"160\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Em segundo lugar, determinamos o peso do segundo corpo, cuja massa \u00e9 de 3 kg:<\/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-d57016899db324fc0f785e92341e9f2f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=3\\cdot 9.81=29.43\\N\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"161\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Assim, aplicando a condi\u00e7\u00e3o de equil\u00edbrio vertical, obtemos que a for\u00e7a normal \u00e9 equivalente \u00e0 soma dos dois pesos: <\/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-c65761c8213d33892f422dd6b0a29121_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\sum \\vv{F_y}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"81\" style=\"vertical-align: -8px;\"><\/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-808b8980c7dbb5f2b1cdf14418fea88c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P_1+P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"99\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Concluindo, o valor da for\u00e7a normal exercida pelo solo \u00e9: <\/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-c5819ab4ff951b1edbe6efa0a0111243_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=49,05+29,43=78,48 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"237\" 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> Conforme mostrado na figura a seguir, dois corpos est\u00e3o conectados 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 exercida pelo plano inclinado 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-f258dccd08d6573f74a2261b2192a92f_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> Nous pla\u00e7ons un corps de masse m=2 kg au sommet d&#8217;une rampe avec un angle d&#8217;inclinaison de 30\u00ba. Quel est le coefficient de frottement entre la rampe et le corps si celui-ci est maintenu en \u00e9quilibre ? Donn\u00e9es : g=9,81 m\/s <sup>2<\/sup> <\/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\/probleme-de-force-normale-et-de-force-de-friction.png&quot; alt=&quot;&quot; class=&quot;wp-image-4253&quot; width=&quot;285&quot; height=&quot;176&quot; srcset=&quot;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&quot; sizes=&quot;(max-width: 300px) 100vw, 300px&quot;><\/figure>\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> Comme dans tout probl\u00e8me de physique portant sur les forces, la premi\u00e8re chose \u00e0 faire est de dessiner le diagramme du corps libre du syst\u00e8me. Ainsi, toutes les forces qui agissent dans ce syst\u00e8me sont : <\/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-force-normale-et-friction-force.png&quot; alt=&quot;exercice r\u00e9solu de la force normale et de la force de frottement&quot; class=&quot;wp-image-4254&quot; width=&quot;285&quot; height=&quot;333&quot; srcset=&quot;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&quot; sizes=&quot;(max-width: 256px) 100vw, 256px&quot;><\/figure>\n<p> Ainsi, pour que le syst\u00e8me soit en \u00e9quilibre, la somme des forces sur les axes 1 et 2 doit \u00eatre \u00e9gale \u00e0 z\u00e9ro. Par cons\u00e9quent, les \u00e9quations suivantes sont vraies : [latex]F_R=P_1&#8243; title=&#8221;Rendered by QuickLaTeX.com&#8221; height=&#8221;454&#8243; width=&#8221;7014&#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-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\"> Ent\u00e3o exclu\u00edmos 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","protected":false},"excerpt":{"rendered":"<p>Este artigo explica o que \u00e9 for\u00e7a normal e como determin\u00e1-la dependendo do tipo de problema. Encontrar\u00e1 assim as caracter\u00edsticas da for\u00e7a normal e, al\u00e9m disso, poder\u00e1 praticar este tipo de for\u00e7a com exerc\u00edcios resolvidos passo a passo. O que \u00e9 for\u00e7a normal? Na f\u00edsica, a for\u00e7a normal \u00e9 uma for\u00e7a exercida por uma superf\u00edcie &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/forca-normal\/\"> <span class=\"screen-reader-text\">For\u00e7a normal<\/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-247","post","type-post","status-publish","format-standard","hentry","category-dinamico"],"yoast_head":"<!-- This site is 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