{"id":29199,"date":"2021-03-25T13:00:30","date_gmt":"2021-03-25T13:00:30","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=29199"},"modified":"2024-09-22T16:16:49","modified_gmt":"2024-09-22T16:16:49","slug":"o-principio-de-pascal","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/pt\/o-principio-de-pascal\/","title":{"rendered":"O Princ\u00edpio de Pascal"},"content":{"rendered":"<p><center><\/p>\n<h1>O Princ\u00edpio de Pascal<\/h1>\n<p><em><strong>Resumo:<\/strong><br \/>\nNesta aula ser\u00e1 explicado o Princ\u00edpio de Pascal, que afirma que a press\u00e3o aplicada em um fluido incompress\u00edvel \u00e9 distribu\u00edda uniformemente. Veremos suas consequ\u00eancias, como o n\u00edvel constante em recipientes e a rela\u00e7\u00e3o entre volumes deslocados. Al\u00e9m disso, ser\u00e3o mostradas aplica\u00e7\u00f5es em m\u00e1quinas hidr\u00e1ulicas e resolveremos exemplos pr\u00e1ticos como o nivelamento de terrenos, diferen\u00e7as de n\u00edvel em l\u00edquidos de diferentes densidades e o funcionamento de um elevador hidr\u00e1ulico.<br \/>\n<\/em><br \/>\n<strong>Objetivos de Aprendizagem:<\/strong><\/p>\n<ol style=\"text-align:left;\">\n<li><strong>Compreender<\/strong> o Princ\u00edpio de Pascal e sua aplica\u00e7\u00e3o em fluidos incompress\u00edveis.<\/li>\n<li><strong>Analisar<\/strong> as consequ\u00eancias do princ\u00edpio, como a distribui\u00e7\u00e3o uniforme da press\u00e3o e o comportamento do n\u00edvel do l\u00edquido.<\/li>\n<li><strong>Aplicar<\/strong> o Princ\u00edpio de Pascal em situa\u00e7\u00f5es pr\u00e1ticas, como o funcionamento de m\u00e1quinas hidr\u00e1ulicas.<\/li>\n<li><strong>Resolver<\/strong> problemas pr\u00e1ticos sobre o nivelamento de terrenos e diferen\u00e7as de n\u00edvel em l\u00edquidos com diferentes densidades.<\/li>\n<li><strong>Explicar<\/strong> o uso do Bar\u00f4metro de Torricelli para medir a press\u00e3o atmosf\u00e9rica e o funcionamento de elevadores hidr\u00e1ulicos.<\/li>\n<\/ol>\n<p><strong>\u00cdNDICE DE CONTE\u00daDOS<\/strong><br \/>\n<a href=\"#1\">Transmiss\u00e3o de for\u00e7as em l\u00edquidos<\/a><br \/>\n<a href=\"#2\">Consequ\u00eancias do Princ\u00edpio de Pascal<\/a><br \/>\n<a href=\"#3\">O Princ\u00edpio de Pascal Aplicado em M\u00e1quinas Hidr\u00e1ulicas<\/a><br \/>\n<a href=\"#4\">Exemplos de Aplica\u00e7\u00e3o<\/a>\n<\/p>\n<p><\/center><\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/9IAFRXx0Cns\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Transmiss\u00e3o de for\u00e7as em l\u00edquidos<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=84s\" target=\"_blank\" rel=\"noopener\"><strong>O Princ\u00edpio de Pascal aplica-se quando lidamos com l\u00edquidos incompress\u00edveis<\/strong><\/a>. As part\u00edculas neste tipo de subst\u00e2ncias exercem for\u00e7as entre si que n\u00e3o preservam a posi\u00e7\u00e3o, mas sim a dist\u00e2ncia relativa m\u00e9dia entre elas. Se isso n\u00e3o ocorresse, as part\u00edculas mudariam sua dist\u00e2ncia relativa m\u00e9dia em algumas regi\u00f5es, alterando o volume total, o que indicaria que estar\u00edamos lidando com um l\u00edquido compress\u00edvel. A maioria dos l\u00edquidos \u00e9, de fato, altamente incompress\u00edvel, e, portanto, quando se aplica uma for\u00e7a em uma determinada regi\u00e3o, essa \u00e9 transmitida integralmente ao longo de todo o fluido.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Consequ\u00eancias do Princ\u00edpio de Pascal<\/h2>\n<h3>O n\u00edvel \u00e9 o mesmo em todos os lugares<\/h3>\n<p style=\"text-align: justify;\">Um exemplo claro disso \u00e9 o mostrado na figura a seguir. Se ela for preenchida com um l\u00edquido, o n\u00edvel subir\u00e1 igualmente em todos os espa\u00e7os dispon\u00edveis, independentemente da forma dos condutos.<\/p>\n<p><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-SD1TKOVRBrw\/YF0upu1TqEI\/AAAAAAAAEtc\/3r34Qw4kXgwJpbXqylLBaKJX45tFH1S9gCLcBGAsYHQ\/s763\/principio%2Bde%2Bpascal%2B-%2Bnivel%2Bde%2Bliquido.png\" alt=\"Princ\u00edpio de Pascal\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-SD1TKOVRBrw\/YF0upu1TqEI\/AAAAAAAAEtc\/3r34Qw4kXgwJpbXqylLBaKJX45tFH1S9gCLcBGAsYHQ\/s763\/principio%2Bde%2Bpascal%2B-%2Bnivel%2Bde%2Bliquido.png\" alt=\"Princ\u00edpio de Pascal\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><\/noscript><\/p>\n<p style=\"text-align: justify;\">Isso se deve ao fato de que, como j\u00e1 vimos, a press\u00e3o \u00e9 uma fun\u00e7\u00e3o da altura.<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">P = \\rho g h<\/span>\n<p style=\"text-align: justify;\">Se houvesse uma diferen\u00e7a de altura entre diferentes partes do l\u00edquido, ent\u00e3o haveria necessariamente uma diferen\u00e7a de press\u00e3o entre essas partes.<\/p>\n<h3>Volume deslocado = Volume cedido<\/h3>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=323s\" target=\"_blank\" rel=\"noopener\"><strong>Se uma press\u00e3o for aplicada em algum ponto do l\u00edquido,<\/strong><\/a> ele se deslocar\u00e1, mantendo sempre o mesmo n\u00edvel em todas as outras partes. O volume cedido <span class=\"katex-eq\" data-katex-display=\"false\">v_1<\/span> seria igual ao volume deslocado <span class=\"katex-eq\" data-katex-display=\"false\">v_2+v_3.<\/span>\n<p><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-nDStNasYWZQ\/YF0upq_20xI\/AAAAAAAAEtg\/bPPlgi4vpug4i70kHoMpkPsvtRtfbQrVACLcBGAsYHQ\/s767\/principio%2Bde%2Bpascalnivel%2Bde%2Bliquido2.png\" alt=\"Princ\u00edpio de Pascal, volume cedido igual ao volume deslocado\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-nDStNasYWZQ\/YF0upq_20xI\/AAAAAAAAEtg\/bPPlgi4vpug4i70kHoMpkPsvtRtfbQrVACLcBGAsYHQ\/s767\/principio%2Bde%2Bpascalnivel%2Bde%2Bliquido2.png\" alt=\"Princ\u00edpio de Pascal, volume cedido igual ao volume deslocado\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><\/noscript><br \/>\n<a name=\"3\"><\/a><\/p>\n<h2>O Princ\u00edpio de Pascal Aplicado em M\u00e1quinas Hidr\u00e1ulicas<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=426s\" target=\"_blank\" rel=\"noopener\"><strong>O Princ\u00edpio de Pascal \u00e9 utilizado na constru\u00e7\u00e3o de prensas hidr\u00e1ulicas.<\/strong><\/a> Acontece que, como a press\u00e3o \u00e9 distribu\u00edda uniformemente em todas as partes de um l\u00edquido, a seguinte express\u00e3o vale:<\/p>\n<p><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-hpJeg_iRsgw\/YF0uplM2GnI\/AAAAAAAAEtY\/tBztNE2eUJQe2uHlrzgZlh_y0QPLGyCnwCLcBGAsYHQ\/s805\/prensahidr%25C3%25A1ulica.PNG\" alt=\"Princ\u00edpio de Pascal, Elevador Hidr\u00e1ulico\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"200\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-hpJeg_iRsgw\/YF0uplM2GnI\/AAAAAAAAEtY\/tBztNE2eUJQe2uHlrzgZlh_y0QPLGyCnwCLcBGAsYHQ\/s805\/prensahidr%25C3%25A1ulica.PNG\" alt=\"Princ\u00edpio de Pascal, Elevador Hidr\u00e1ulico\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"200\" \/><\/noscript><\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">P_2 = P_1<\/span>\n<p style=\"text-align: justify;\">Mas como a press\u00e3o pode ser expressa como for\u00e7a por unidade de \u00e1rea, <span class=\"katex-eq\" data-katex-display=\"false\">P=F\/A<\/span>, ent\u00e3o teremos:<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{F_2}{A_2} = \\frac{F_1}{A_1}<\/span>\n<p style=\"text-align: justify;\">Isso significa que, se aplicarmos uma for\u00e7a <span class=\"katex-eq\" data-katex-display=\"false\">F_1<\/span> no pist\u00e3o 1, no pist\u00e3o 2 sair\u00e1 uma for\u00e7a <span class=\"katex-eq\" data-katex-display=\"false\">F_2<\/span> da forma<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle F_2 = \\frac{A_2}{A_1} \\cdot F_1<\/span>\n<p style=\"text-align: justify;\">O fator <span class=\"katex-eq\" data-katex-display=\"false\">A_2\/A_1<\/span> \u00e9 o que chamaremos de \u00abCoeficiente de amplifica\u00e7\u00e3o ou redu\u00e7\u00e3o da for\u00e7a\u00bb. Amplificar\u00e1 se for maior que 1 e reduzir\u00e1 se for um valor entre 0 e 1.<\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h2>Exemplos de Aplica\u00e7\u00e3o<\/h2>\n<ol>\n<li style=\"text-align: justify;\">Duas pessoas querem nivelar um terreno inclinado. Para isso, pegam uma mangueira e a enchem com \u00e1gua, depois posicionam suas extremidades a uma dist\u00e2ncia horizontal de 275[cm]. O n\u00edvel da \u00e1gua em um ponto \u00e9 de 110[cm] e no outro \u00e9 de 175[cm].<img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-qmCq1t6szPU\/YF0uopFj2HI\/AAAAAAAAEtM\/Bo__H3vPJ9YZCms__pgwk-26VxjBbiYmgCLcBGAsYHQ\/s625\/nivel%2Bde%2Bagua.PNG\" alt=\"Princ\u00edpio de Pascal, N\u00edvel de \u00c1gua\" class=\"aligncenter size-full lazyload\" width=\"625\" height=\"313\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-qmCq1t6szPU\/YF0uopFj2HI\/AAAAAAAAEtM\/Bo__H3vPJ9YZCms__pgwk-26VxjBbiYmgCLcBGAsYHQ\/s625\/nivel%2Bde%2Bagua.PNG\" alt=\"Princ\u00edpio de Pascal, N\u00edvel de \u00c1gua\" class=\"aligncenter size-full lazyload\" width=\"625\" height=\"313\" \/><\/noscript>a) Qual \u00e9 a diferen\u00e7a de n\u00edvel do terreno?\n<p>b) Qual \u00e9 o \u00e2ngulo de inclina\u00e7\u00e3o do terreno?<\/p>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=691s\" target=\"_blank\" rel=\"noopener\"><strong>SOLU\u00c7\u00c3O<\/strong><\/a><\/li>\n<li style=\"text-align: justify;\">Uma mangueira em forma de U cont\u00e9m \u00e1gua em seu interior, com uma densidade de <span class=\"katex-eq\" data-katex-display=\"false\">1000 [kg\/m^3].<\/span> Se em uma das extremidades da mangueira for derramado \u00f3leo at\u00e9 preencher uma altura de <span class=\"katex-eq\" data-katex-display=\"false\">20[cm]<\/span>, com uma densidade de <span class=\"katex-eq\" data-katex-display=\"false\">800[kg\/m^3]<\/span>, qual diferen\u00e7a de n\u00edvel ser\u00e1 produzida entre as extremidades da mangueira?<img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-EYBgWw2b6jY\/YF0uoq4c5AI\/AAAAAAAAEtU\/LCAin5GVeSoRDCIgn3widW9kaTwfJwFRACLcBGAsYHQ\/s596\/nivel-agua-aceite.PNG\" alt=\"Princ\u00edpio de Pascal - Misturas\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-EYBgWw2b6jY\/YF0uoq4c5AI\/AAAAAAAAEtU\/LCAin5GVeSoRDCIgn3widW9kaTwfJwFRACLcBGAsYHQ\/s596\/nivel-agua-aceite.PNG\" alt=\"Princ\u00edpio de Pascal - Misturas\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><\/noscript><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=867s\" target=\"_blank\" rel=\"noopener\"><strong>SOLU\u00c7\u00c3O<\/strong><\/a><\/li>\n<li style=\"text-align: justify;\"><strong>O Bar\u00f4metro de Torricelli:<\/strong> Torricelli projetou o seguinte instrumento para medir a press\u00e3o atmosf\u00e9rica:<img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-hB2h_82Yg6A\/YF0uol0V-OI\/AAAAAAAAEtQ\/1w25pez7rj4LE3-tanK4NMo5NBy_0xJQgCLcBGAsYHQ\/s572\/Principio%2Bde%2Bpascal%2B-%2Btorricelli.PNG\" princ\u00edpio=\"\" de=\"\" pascal=\"\" bar\u00f4metro=\"\" torricelli=\"\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300 alt=\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-hB2h_82Yg6A\/YF0uol0V-OI\/AAAAAAAAEtQ\/1w25pez7rj4LE3-tanK4NMo5NBy_0xJQgCLcBGAsYHQ\/s572\/Principio%2Bde%2Bpascal%2B-%2Btorricelli.PNG\" princ\u00edpio=\"\" de=\"\" pascal=\"\" bar\u00f4metro=\"\" torricelli=\"\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300 alt=\" \/><\/noscript>a) Se a densidade do merc\u00fario \u00e9 de <span class=\"katex-eq\" data-katex-display=\"false\">13.534,0[kg\/m3]<\/span>, qual ser\u00e1 a altura da coluna de merc\u00fario devido \u00e0 press\u00e3o atmosf\u00e9rica? Considere <span class=\"katex-eq\" data-katex-display=\"false\">P_{atm}=1,0[atm]=101.325,0[Pa]<\/span>?\n<p>b) Se em vez de merc\u00fario fosse utilizada \u00e1gua, qual seria a altura da coluna de \u00e1gua para medir a mesma press\u00e3o?<\/p>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=1049s\" target=\"_blank\" rel=\"noopener\"><strong>SOLU\u00c7\u00c3O<\/strong><\/a><\/li>\n<li style=\"text-align: justify;\">Um elevador hidr\u00e1ulico tem dois pist\u00f5es, um com raio de 2[p\u00e9s] e o outro com raio de 30[p\u00e9s]. a) Qual peso deve ser colocado no pist\u00e3o menor se no pist\u00e3o maior se deseja elevar um bloco de uma tonelada? b) Se deseja-se que o pist\u00e3o maior eleve um bloco de concreto a uma altura de <span class=\"katex-eq\" data-katex-display=\"false\">50[cm]<\/span>, qual dist\u00e2ncia o pist\u00e3o menor deve percorrer?\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=1403s\" target=\"_blank\" rel=\"noopener\"><strong>SOLU\u00c7\u00c3O<\/strong><\/a><\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>O Princ\u00edpio de Pascal Resumo: Nesta aula ser\u00e1 explicado o Princ\u00edpio de Pascal, que afirma que a press\u00e3o aplicada em um fluido incompress\u00edvel \u00e9 distribu\u00edda uniformemente. Veremos suas consequ\u00eancias, como o n\u00edvel constante em recipientes e a rela\u00e7\u00e3o entre volumes deslocados. Al\u00e9m disso, ser\u00e3o mostradas aplica\u00e7\u00f5es em m\u00e1quinas hidr\u00e1ulicas e resolveremos exemplos pr\u00e1ticos como o [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":29196,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":14,"footnotes":""},"categories":[637,891],"tags":[],"class_list":["post-29199","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-fisica-pt","category-mecanica-dos-fluidos"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>O Princ\u00edpio de Pascal - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"O Princ\u00edpio de Pascal afirma que, ao aplicar press\u00e3o em um fluido incompress\u00edvel, essa \u00e9 transmitida de maneira uniforme em todas as dire\u00e7\u00f5es. 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