{"id":298,"date":"2023-06-22T07:14:12","date_gmt":"2023-06-22T07:14:12","guid":{"rendered":"https:\/\/physigeek.com\/pt\/lei-da-alavanca\/"},"modified":"2023-06-22T07:14:12","modified_gmt":"2023-06-22T07:14:12","slug":"lei-da-alavanca","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/lei-da-alavanca\/","title":{"rendered":"Lei da alavanca"},"content":{"rendered":"<p>Neste artigo voc\u00ea descobrir\u00e1 o que \u00e9 a lei da alavancagem. Tamb\u00e9m mostramos um exemplo que explica como a lei da alavanca afeta as for\u00e7as que atuam sobre ela. Al\u00e9m disso, voc\u00ea pode praticar a lei da alavancagem com exerc\u00edcios passo a passo.<\/p>\n<p> Logicamente, antes de ver em que consiste a lei da alavancagem, devemos ter muito claro o que \u00e9 uma alavanca. \u00c9 por isso que recomendamos que voc\u00ea visite o seguinte post antes de continuar com a explica\u00e7\u00e3o: <\/p>\n<div style=\"background-color:#FFFDE7; padding-top: 10px; padding-bottom: 10px; padding-right: 10px; padding-left: 20px; border: 2.5px dashed #FFB74D; border-radius:20px;\"> <span style=\"color:#ff951b\">\u27a4<\/span> <strong>Veja:<\/strong> <a href=\"https:\/\/physigeek.com\/pt\/alavanca\/\">O que \u00e9 uma alavanca?<\/a> <\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-la-ley-de-la-palanca\"><\/span> Qual \u00e9 a lei da alavancagem?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> A <strong>lei da alavanca<\/strong> \u00e9 uma lei que relaciona as diferentes for\u00e7as que atuam sobre uma alavanca. Portanto, a lei da alavanca \u00e9 utilizada para resolver problemas envolvendo alavancas.<\/p>\n<p> Mais especificamente, a lei da alavancagem diz que o produto da pot\u00eancia vezes o comprimento do seu bra\u00e7o \u00e9 equivalente ao produto da resist\u00eancia vezes o comprimento do seu bra\u00e7o.<\/p>\n<p> Assim, a lei da alavanca permite relacionar matematicamente a resist\u00eancia, que \u00e9 a for\u00e7a exercida pela carga sobre a alavanca, com a pot\u00eancia, que \u00e9 a for\u00e7a que deve ser exercida para vencer a carga. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Formula-de-la-ley-de-la-palanca\"><\/span> F\u00f3rmula da lei da alavanca<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> A <strong>f\u00f3rmula da lei da alavanca<\/strong> relaciona matematicamente a pot\u00eancia \u00e0 resist\u00eancia da alavanca. Mais especificamente, a lei da alavanca afirma que a pot\u00eancia vezes o bra\u00e7o de pot\u00eancia \u00e9 igual \u00e0 resist\u00eancia vezes o bra\u00e7o de resist\u00eancia. <\/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\/loi-du-levier.png\" alt=\"f\u00f3rmula da lei da alavancagem\" class=\"wp-image-5155\" width=\"338\" height=\"339\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/loi-du-levier-300x300.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/loi-du-levier-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/loi-du-levier-768x770.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/loi-du-levier.png 779w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p style=\"margin-bottom:6px\"> Ouro:<\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\"><strong>Ponto de apoio ou fulcro (F)<\/strong> : \u00e9 a parte da alavanca sobre a qual permanece. Portanto, ela suporta todo o peso da barra, bem como dos corpos acima dela.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\"><strong>Esfor\u00e7o ou pot\u00eancia (P)<\/strong> : \u00e9 a for\u00e7a aplicada \u00e0 alavanca para neutralizar a carga do outro lado.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\"><strong>Carga ou resist\u00eancia (R)<\/strong> : \u00e9 a for\u00e7a que deve ser superada.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\"><strong>Power Arm (BP)<\/strong> : Esta \u00e9 a dist\u00e2ncia entre o poder e o fulcro.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\"><strong>Bra\u00e7o de Resist\u00eancia (BR)<\/strong> : \u00c9 a dist\u00e2ncia entre a resist\u00eancia e o ponto de apoio.<\/span><\/li>\n<\/ul>\n<p> Observe que a lei da alavanca s\u00f3 \u00e9 verdadeira se a alavanca estiver em equil\u00edbrio, ou seja, se estiver em repouso. Portanto, se a alavanca se mover, a equa\u00e7\u00e3o da alavanca n\u00e3o ser\u00e1 v\u00e1lida. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-de-la-ley-de-la-palanca\"><\/span> Exemplo da lei da alavancagem<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> A t\u00edtulo de exemplo, nesta se\u00e7\u00e3o veremos como o valor da for\u00e7a que deve ser aplicada para neutralizar a resist\u00eancia muda dependendo do comprimento dos bra\u00e7os de alavanca.<\/p>\n<p> Primeiro, veremos o que acontece quando o fulcro est\u00e1 bem no meio do poder e da resist\u00eancia: <\/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\/loi-du-levier-exemple-1.png\" alt=\"lei da alavanca exemplo 1\" class=\"wp-image-5161\" width=\"299\" height=\"214\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/loi-du-levier-exemple-1-300x215.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/loi-du-levier-exemple-1.png 645w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p> Aplicamos a f\u00f3rmula da lei da alavanca para calcular o valor da pot\u00eancia:<\/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-f6834df9d4ef006d7a868d9f1df7af56_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P\\cdot BP=R\\cdot BR\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"133\" 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-02893e0ec4e3c945bb8d92c039e90c2b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=\\cfrac{R\\cdot BR}{BP}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"94\" 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-2b51a9318ca7df21c0442670c1bed97c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=\\cfrac{100\\cdot 150}{150}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"106\" 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-a7d0ccc28cd273b8c44cc50c05f79ffa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=100 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"86\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Assim, se o fulcro estiver exatamente a meio caminho entre a pot\u00eancia e a resist\u00eancia, a for\u00e7a que deve ser exercida na alavanca \u00e9 equivalente \u00e0 resist\u00eancia.<\/p>\n<p> Em segundo lugar, analisaremos o caso em que o ponto de apoio est\u00e1 mais pr\u00f3ximo da resist\u00eancia do que da pot\u00eancia: <\/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\/loi-du-levier-exemple-2.png\" alt=\"lei da alavanca exemplo 2\" class=\"wp-image-5163\" width=\"300\" height=\"218\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/loi-du-levier-exemple-2-300x217.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/loi-du-levier-exemple-2.png 645w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\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-f6834df9d4ef006d7a868d9f1df7af56_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P\\cdot BP=R\\cdot BR\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"133\" 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-02893e0ec4e3c945bb8d92c039e90c2b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=\\cfrac{R\\cdot BR}{BP}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"94\" 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-a96415ed0eee3185bb4337121a191f99_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=\\cfrac{100\\cdot 100}{200}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"106\" 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-325fa07f0a03db4a489592b7fbd47ebc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=50 \\N\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"55\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Portanto, quando o bra\u00e7o de pot\u00eancia \u00e9 maior que o bra\u00e7o de resist\u00eancia, o valor da pot\u00eancia \u00e9 menor que o valor da resist\u00eancia.<\/p>\n<p> Por fim, estudamos o caso em que o ponto de apoio est\u00e1 mais pr\u00f3ximo da pot\u00eancia do que da resist\u00eancia: <\/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\/loi-du-levier-exemple-3.png\" alt=\"exemplo de lei da alavanca 3\" class=\"wp-image-5164\" width=\"300\" height=\"216\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/loi-du-levier-exemple-3-300x216.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/loi-du-levier-exemple-3.png 643w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\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-f6834df9d4ef006d7a868d9f1df7af56_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P\\cdot BP=R\\cdot BR\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"133\" 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-02893e0ec4e3c945bb8d92c039e90c2b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=\\cfrac{R\\cdot BR}{BP}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"94\" 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-6bbe95110f0a90fc6e0442006b5720ad_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=\\cfrac{100\\cdot 220}{80}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"106\" 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-64eb7e33f3c6378d745a24b65c75f3b8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=275 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"86\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Concluindo, quando o fulcro est\u00e1 mais pr\u00f3ximo da pot\u00eancia do que da resist\u00eancia, uma for\u00e7a maior que a resist\u00eancia deve ser exercida para equilibrar a balan\u00e7a. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-ley-de-la-palanca\"><\/span> Exerc\u00edcios resolvidos da lei da alavancagem<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Antes de fazer os exerc\u00edcios, recomendamos que voc\u00ea visite o seguinte link onde explicamos os diferentes tipos de alavancas, pois existe um exerc\u00edcio para cada tipo de alavanca e voc\u00ea deve ter claro o que \u00e9 cada tipo para resolver os problemas. . <\/p>\n<div style=\"background-color:#FFFDE7; padding-top: 10px; padding-bottom: 10px; padding-right: 10px; padding-left: 20px; border: 2.5px dashed #FFB74D; border-radius:20px;\"> <span style=\"color:#ff951b\">\u27a4<\/span> <strong>Veja:<\/strong> <a href=\"https:\/\/physigeek.com\/pt\/tipos-de-alavancas\/\">Tipos de alavancas<\/a><\/div>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 1<\/h3>\n<p> Um corpo de 50 kg \u00e9 colocado pr\u00f3ximo a uma alavanca de primeiro grau feita de uma barra r\u00edgida de 300 cm. Se a dist\u00e2ncia entre a carga e o fulcro \u00e9 de 180 cm, quanto deve pesar o corpo colocado do outro lado da alavanca para que esteja em equil\u00edbrio? <\/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\"> A alavanca neste problema \u00e9 de primeiro grau e conhecemos apenas a resist\u00eancia (50 kg) e o bra\u00e7o de resist\u00eancia (180 cm). No entanto, como sabemos o comprimento da barra, podemos calcular o bra\u00e7o de pot\u00eancia subtraindo o comprimento total da barra menos o comprimento do bra\u00e7o de resist\u00eancia:<\/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-d8009cd38409af0152df783ad89f3f36_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"BP=300-180=120 \\text{ cm}\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"206\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ent\u00e3o, podemos determinar o valor da pot\u00eancia aplicando a regra da alavanca:<\/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-f6834df9d4ef006d7a868d9f1df7af56_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P\\cdot BP=R\\cdot BR\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"133\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Substitu\u00edmos os dados na 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-5959be827849cceb2665451a56d8ff0a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P\\cdot 120=50\\cdot 180\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"134\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E finalmente, resolvemos a inc\u00f3gnita na 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-78a4d183ce7fc344a8b0a34efc858751_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=\\cfrac{50\\cdot 180}{120}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"97\" 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-7ace0a567aab127764f2007e8b8f0b3d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=75 \\text{ kg}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"79\" 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> Num carrinho de m\u00e3o colocamos um objeto de 70 kg a 50 cm do ponto de apoio. Se a parte onde fica o carrinho de m\u00e3o est\u00e1 a 140 cm do fulcro, qual o esfor\u00e7o que devemos fazer para poder transportar o objeto com o carrinho de m\u00e3o? <\/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\"> O carrinho de m\u00e3o \u00e9 uma alavanca de segundo grau, pois a resist\u00eancia est\u00e1 localizada entre o fulcro e a pot\u00eancia. Portanto, para resolver o problema devemos aplicar a lei da alavancagem:<\/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-f6834df9d4ef006d7a868d9f1df7af56_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P\\cdot BP=R\\cdot BR\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"133\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Substitu\u00edmos os dados que conhecemos na 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-e71aa135e71ce25f9237d661b3656cd7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P\\cdot 140=70\\cdot 50\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"125\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E finalmente, resolvemos a inc\u00f3gnita na 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-05e2c7e509587e806745d16426497be4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=\\cfrac{70\\cdot 50}{140}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"88\" 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-bda321058b63b34b706e280791fd6407_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=25 \\text{ kg}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"79\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Voc\u00ea deve, portanto, fazer um esfor\u00e7o equivalente a levantar 25 kg.<\/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> Numa alavanca de terceiro grau, uma for\u00e7a equivalente a 60 N deve ser exercida para neutralizar uma resist\u00eancia de 15 N localizada a 80 cm do fulcro. Calcule a que dist\u00e2ncia do fulcro a energia \u00e9 aplicada. <\/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\"> Neste problema de alavanca de terceiro grau, somos solicitados a determinar o bra\u00e7o de pot\u00eancia. Ent\u00e3o, para resolver o problema, precisamos aplicar a equa\u00e7\u00e3o da alavancagem:<\/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-f6834df9d4ef006d7a868d9f1df7af56_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P\\cdot BP=R\\cdot BR\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"133\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Substitu\u00edmos os dados que conhecemos na 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-1dae5ba7078c412fbe57dd6d8fe0008f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"60\\cdot BP=15\\cdot 80\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"131\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E resolvemos a inc\u00f3gnita na 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-6c20589172f022b76c17eb055f14ab91_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"BP=\\cfrac{15\\cdot 80}{60}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"103\" 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-4b319952f3a57992cb5e4d229af26b5b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"BP=20 \\text{ cm}\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"98\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A pot\u00eancia deve, portanto, ser aplicada a 20 cm do fulcro.<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Neste artigo voc\u00ea descobrir\u00e1 o que \u00e9 a lei da alavancagem. Tamb\u00e9m mostramos um exemplo que explica como a lei da alavanca afeta as for\u00e7as que atuam sobre ela. Al\u00e9m disso, voc\u00ea pode praticar a lei da alavancagem com exerc\u00edcios passo a passo. Logicamente, antes de ver em que consiste a lei da alavancagem, devemos &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/lei-da-alavanca\/\"> <span class=\"screen-reader-text\">Lei da alavanca<\/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-298","post","type-post","status-publish","format-standard","hentry","category-dinamico"],"yoast_head":"<!-- This site is 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