{"id":48,"date":"2023-06-26T19:47:34","date_gmt":"2023-06-26T19:47:34","guid":{"rendered":"https:\/\/physigeek.com\/pt\/resistencia-a-traccao\/"},"modified":"2023-06-26T19:47:34","modified_gmt":"2023-06-26T19:47:34","slug":"resistencia-a-traccao","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/resistencia-a-traccao\/","title":{"rendered":"Resist\u00eancia \u00e0 trac\u00e7\u00e3o"},"content":{"rendered":"<p>Este artigo explica o que \u00e9 a for\u00e7a de tens\u00e3o na f\u00edsica e como ela \u00e9 calculada. Voc\u00ea encontrar\u00e1 um exemplo real da for\u00e7a de tens\u00e3o de uma corda e, al\u00e9m disso, poder\u00e1 treinar com exerc\u00edcios resolvidos deste tipo de for\u00e7as. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-la-fuerza-de-tension\"><\/span> O que \u00e9 for\u00e7a de tens\u00e3o?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>A for\u00e7a de tra\u00e7\u00e3o<\/strong> \u00e9 a for\u00e7a exercida por uma corda, um cabo ou qualquer objeto el\u00e1stico quando est\u00e1 tensionado, ou seja, quando n\u00e3o pode ser dobrado.<\/p>\n<p> Por exemplo, quando uma for\u00e7a \u00e9 aplicada a ambas as extremidades de uma corda, esta fica esticada e, portanto, exerce uma for\u00e7a de tens\u00e3o. Abaixo, na pr\u00f3xima se\u00e7\u00e3o, estudaremos detalhadamente as for\u00e7as de tens\u00e3o exercidas por uma corda.<\/p>\n<p> A for\u00e7a de tens\u00e3o \u00e9 medida em newtons (N) e normalmente \u00e9 representada pela letra T. Al\u00e9m disso, por se tratar de um tipo de for\u00e7a, as for\u00e7as de tens\u00e3o s\u00e3o vetores cuja dire\u00e7\u00e3o \u00e9 paralela \u00e0 extens\u00e3o da corda ou cabo.<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-de-una-fuerza-de-tension\"><\/span> Exemplo de for\u00e7a de tens\u00e3o<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Considerando a defini\u00e7\u00e3o de for\u00e7a de tens\u00e3o, analisaremos detalhadamente um exemplo para entender melhor o conceito.<\/p>\n<p> Um exemplo t\u00edpico de for\u00e7a de tens\u00e3o \u00e9 uma corda. Se nenhuma for\u00e7a for aplicada a uma corda, ela permanecer\u00e1 solta e, portanto, n\u00e3o haver\u00e1 for\u00e7a de tens\u00e3o. Por outro lado, se uma for\u00e7a for aplicada em cada extremidade da corda, ela permanece esticada e, portanto, exerce uma for\u00e7a de tens\u00e3o em cada uma das suas extremidades.<\/p>\n<p> Al\u00e9m disso, se a corda for considerada um objeto sem massa e indeform\u00e1vel, a for\u00e7a aplicada a uma extremidade da corda \u00e9 transmitida \u00e0 outra extremidade, e vice-versa, a for\u00e7a exercida \u00e0 segunda extremidade \u00e9 transmitida \u00e0 primeira extremidade. da corda. a corda. .<\/p>\n<p> Observe o desenho a seguir em que a for\u00e7a exercida pela pessoa da esquerda (T <sub>A<\/sub> ) \u00e9 a for\u00e7a exercida pela corda sobre a pessoa da direita. E da mesma forma, a for\u00e7a aplicada pela pessoa da direita (T <sub>B<\/sub> ) \u00e9 transmitida para a pessoa da esquerda. <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/la-force-de-traction.png\" alt=\"For\u00e7a de tens\u00e3o\" class=\"wp-image-848\" width=\"365\" height=\"219\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/la-force-de-traction-300x179.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/la-force-de-traction.png 741w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<\/div>\n<p> O jogo do cabo de guerra \u00e9 um exemplo concreto da vida cotidiana em que as for\u00e7as de tens\u00e3o s\u00e3o transmitidas atrav\u00e9s de uma corda.<\/p>\n<p> Concluindo, cordas, cabos ou objetos semelhantes s\u00e3o utilizados para transmitir for\u00e7as de um corpo para outro. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Como-calcular-la-fuerza-de-tension\"><\/span> Como calcular a for\u00e7a de tens\u00e3o<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> As etapas para calcular as for\u00e7as de tens\u00e3o s\u00e3o:<\/p>\n<ol style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Decompor vetorialmente for\u00e7as que n\u00e3o s\u00e3o verticais nem horizontais. Desta forma todas as for\u00e7as ser\u00e3o verticais ou horizontais.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Desenhe o diagrama de corpo livre do sistema, ou seja, represente graficamente todas as for\u00e7as que atuam no sistema.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Estabele\u00e7a as equa\u00e7\u00f5es de equil\u00edbrio do sistema. Normalmente, uma equa\u00e7\u00e3o deve ser estabelecida para for\u00e7as horizontais e outra equa\u00e7\u00e3o para for\u00e7as verticais.<\/span><\/li>\n<li> <span style=\"color:#101010;font-weight: normal;\">Resolva a for\u00e7a de tens\u00e3o a partir das equa\u00e7\u00f5es e calcule seu valor.<\/span><\/li>\n<\/ol>\n<p> Em resumo, em f\u00edsica para <strong>calcular a for\u00e7a de tens\u00e3o <span style=\"text-decoration: underline;\"><a href=\"https:\/\/physigeek.com\/pt\/condicoes-de-equilibrio\/\">, devem ser aplicadas condi\u00e7\u00f5es de equil\u00edbrio<\/a><\/span><\/strong> . Ao estabelecer as equa\u00e7\u00f5es de equil\u00edbrio, a for\u00e7a de tens\u00e3o pode ser resolvida e, portanto, seu valor pode ser encontrado.<\/p>\n<p> Abaixo est\u00e1 um exemplo passo a passo de for\u00e7a de tens\u00e3o calculada para ver como isso acontece:<\/p>\n<ul>\n<li> Um corpo com massa de 65 kg est\u00e1 suspenso no teto por uma corda. Quanta tra\u00e7\u00e3o a corda deve exercer para sustentar o corpo? Sup\u00f5e-se que a corda tenha massa desprez\u00edvel e n\u00e3o estique.<\/li>\n<\/ul>\n<p> Em primeiro lugar, \u00e9 necess\u00e1rio determinar a for\u00e7a gravitacional com que a Terra atrai o corpo. Para fazer isso, aplicamos 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-10adee55a222c172752922f0d901ea78_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=m\\cdot g=65\\cdot 9,81=637,65 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"261\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p> Agora criamos o diagrama de corpo livre. Neste caso temos apenas duas for\u00e7as verticais: a for\u00e7a de tens\u00e3o da corda e a for\u00e7a do peso. <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-de-force-de-tension-resolu.png\" alt=\"exerc\u00edcio deliberado de for\u00e7a de tens\u00e3o\" class=\"wp-image-855\" width=\"243\" height=\"264\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-de-force-de-tension-resolu-276x300.png 276w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-de-force-de-tension-resolu.png 556w\" sizes=\"auto, (max-width: 276px) 100vw, 276px\"><\/figure>\n<\/div>\n<p> Vamos agora propor a condi\u00e7\u00e3o de equil\u00edbrio vertical. Como existe apenas uma for\u00e7a vertical para cima e uma for\u00e7a vertical para baixo, para que o corpo permane\u00e7a em equil\u00edbrio as duas for\u00e7as devem ser iguais:<\/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-fe9470d16022905bcce2e03dd9a64bc9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\somme F_y=0\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"52\" 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-1cc7afff797d6c18363325cd43c54b50_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"TP=0\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"59\" 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-994386bc251ba773e73bfb0f81fbbb67_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"T=P\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"50\" 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-fa5442daf69fd2fb31db64f7ca4f0d1c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"T=637,65 \\N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"88\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-fuerza-de-tension\"><\/span> Exerc\u00edcios resolvidos sobre for\u00e7a de tens\u00e3o<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Exerc\u00edcio 1<\/h3>\n<p> Dado um corpo r\u00edgido com massa de 12 kg suspenso por duas cordas cujos \u00e2ngulos s\u00e3o mostrados na figura a seguir, calcule a for\u00e7a que cada corda deve exercer para manter o corpo em equil\u00edbrio. <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-premiere-condition-dequilibre.png\" alt=\"problema da primeira condi\u00e7\u00e3o de equil\u00edbrio\" class=\"wp-image-372\" width=\"243\" height=\"243\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-premiere-condition-dequilibre-300x300.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-premiere-condition-dequilibre-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-premiere-condition-dequilibre.png 600w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<\/div>\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 este tipo de problema \u00e9 desenhar o diagrama de corpo livre da figura: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-de-la-premiere-condition-dequilibre.png\" alt=\"exerc\u00edcio resolvido da primeira condi\u00e7\u00e3o de equil\u00edbrio\" class=\"wp-image-375\" width=\"282\" height=\"335\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-de-la-premiere-condition-dequilibre-252x300.png 252w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-de-la-premiere-condition-dequilibre.png 600w\" sizes=\"auto, (max-width: 252px) 100vw, 252px\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> Observe que na verdade existem apenas tr\u00eas for\u00e7as atuando sobre o corpo suspenso, a for\u00e7a do peso P e as tens\u00f5es das cordas T <sub>1<\/sub> e T <sub>2<\/sub> . As for\u00e7as representadas T <sub>1x<\/sub> , T <sub>1y<\/sub> , T <sub>2x<\/sub> e T <sub>2y<\/sub> s\u00e3o as componentes vetoriais de T <sub>1<\/sub> e T <sub>2<\/sub> respectivamente.<\/p>\n<p class=\"has-text-align-left\"> Assim, como conhecemos os \u00e2ngulos de inclina\u00e7\u00e3o das cordas, podemos encontrar as express\u00f5es para as componentes vetoriais das for\u00e7as de tra\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-bc09423d2d10435101c7d6b087add524_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\" T_{1x}=T_1\\cdot \\text{cos}(20\u00ba)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"135\" 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-0603d4b02835532dcefe2290484067fb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\" T_{1y}=T_1\\cdot \\text{sin}(20\u00ba)\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"133\" 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-0b10a6fc64a1a84b9f4f2c47b7990766_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\" T_{2x}=T_2\\cdot \\text{cos}(55\u00ba)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"135\" 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-3e7a1dc2ffa7eb20e5e2d9346f0b96a2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\" T_{2y}=T_2\\cdot \\text{sin}(55\u00ba)\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"133\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Por outro lado, podemos calcular a for\u00e7a do peso aplicando a f\u00f3rmula da for\u00e7a gravitacional:<\/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-dab14bd2fa937f39825c5add90b3ae58_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=m\\cdot g=12\\cdot 9,81 =117,72 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"261\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A defini\u00e7\u00e3o do problema diz-nos que o corpo est\u00e1 em equil\u00edbrio, portanto a soma das for\u00e7as verticais e a soma das for\u00e7as horizontais deve ser igual a zero. Portanto, podemos estabelecer as equa\u00e7\u00f5es de for\u00e7a e defini-las iguais a zero: <\/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-6532044e76d6b9246f64624159b08c33_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-T_{1x}+T_{2x}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"119\" 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-52aadf04437252b1f9c17107dfc16a84_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"T_{1y}+T_{2y}-P=0\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"140\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Substitu\u00edmos agora os componentes das tens\u00f5es pelas suas express\u00f5es encontradas anteriormente: <\/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-4a4993c55ab7f27b6c0b67793ee5ff8a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-T_1\\cdot\\text{cos}(20\u00ba)+T_2\\cdot \\text{cos}(55\u00ba)=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"239\" 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-204773c167037418680872592d118315_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"T_1\\cdot \\text{sin}(20\u00ba)+T_2\\cdot \\text{sin}(55\u00ba)-117.72=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"293\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E, por fim, resolvemos o sistema de equa\u00e7\u00f5es para obter o valor das for\u00e7as T <sub>1<\/sub> e T <sub>2<\/sub> : <\/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-c0aa7a8f7fe7234899f77039b42b47d1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\left.\\begin{array}{l}-T_1\\cdot 0,94+T_2\\cdot 0,57=0\\\\[2ex]T_1\\cdot 0,34+T_2\\cdot 0,82-117 .72=0\\end{array }\\right\\} \\longrightarrow \\ \\begin{array}{c}T_1=69,56 \\ N\\\\[2ex]T_2=114,74 \\ 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 2<\/h3>\n<p> Comme le montre la figure suivante, deux objets sont reli\u00e9s par une corde et une poulie de masses n\u00e9gligeables. Si l&#8217;objet 2 a une masse de 7 kg et que l&#8217;inclinaison de la rampe est de 50\u00ba, calculez la masse de l&#8217;objet 1 pour que l&#8217;ensemble du syst\u00e8me soit dans des conditions d&#8217;\u00e9quilibre. Dans ce cas, la force de frottement peut \u00eatre n\u00e9glig\u00e9e. <\/p>\n<div class=&quot;wp-block-image&quot;>\n<figure class=&quot;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-dequilibre-des-forces.png&quot; alt=&quot;probl\u00e8me d'\u00e9quilibre translationnel&quot; class=&quot;wp-image-295&quot; width=&quot;299&quot; height=&quot;240&quot; srcset=&quot;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&quot; sizes=&quot;(max-width: 300px) 100vw, 300px&quot;><\/figure>\n<\/div>\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> Le corps 1 est sur une pente inclin\u00e9e, donc la premi\u00e8re chose \u00e0 faire est de vectoriser la force de son poids pour avoir les forces sur les axes de la pente : [latex]P_{1x}=P_1\\cdot \\text{sin}(\\alpha)&#8221; title=&#8221;Rendered by QuickLaTeX.com&#8221; height=&#8221;340&#8243; width=&#8221;2918&#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-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<div class=\"wp-block-image\">\n<figure class=\"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<\/div>\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-ed082b4f064316ab20fb0d26054d3010_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 ] Ainsi, la composante du poids de l'objet 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=\"87\" width=\"1160\" 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\"> Agora aplicamos a f\u00f3rmula da for\u00e7a gravitacional e simplificamos a 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-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-left\"> Finalmente, substitu\u00edmos os dados e resolvemos 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-0457f85ca65afde96b2e575ce54869dd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1 \\cdot \\text{sin}(50\u00ba) =7\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"122\" 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-9a26d132815a0ce878a6ad874c8b40b0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1 =\\cfrac{7}{\\text{sin}(50\u00ba)}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"103\" 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<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Este artigo explica o que \u00e9 a for\u00e7a de tens\u00e3o na f\u00edsica e como ela \u00e9 calculada. Voc\u00ea encontrar\u00e1 um exemplo real da for\u00e7a de tens\u00e3o de uma corda e, al\u00e9m disso, poder\u00e1 treinar com exerc\u00edcios resolvidos deste tipo de for\u00e7as. O que \u00e9 for\u00e7a de tens\u00e3o? A for\u00e7a de tra\u00e7\u00e3o \u00e9 a for\u00e7a exercida &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/resistencia-a-traccao\/\"> <span class=\"screen-reader-text\">Resist\u00eancia \u00e0 trac\u00e7\u00e3o<\/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-48","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 Como calcular a for\u00e7a de tens\u00e3o (exerc\u00edcios resolvidos)<\/title>\n<meta name=\"description\" content=\"Explicamos o que \u00e9 for\u00e7a de tens\u00e3o e como ela \u00e9 calculada (f\u00f3rmula). 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