{"id":253,"date":"2023-06-23T06:39:35","date_gmt":"2023-06-23T06:39:35","guid":{"rendered":"https:\/\/physigeek.com\/it\/piano-inclinato\/"},"modified":"2023-06-23T06:39:35","modified_gmt":"2023-06-23T06:39:35","slug":"piano-inclinato","status":"publish","type":"post","link":"https:\/\/physigeek.com\/it\/piano-inclinato\/","title":{"rendered":"Piano inclinato"},"content":{"rendered":"<p>In questo articolo viene spiegato cosa sono i piani inclinati in fisica e come si risolvono problemi di questo tipo. Troverai le formule per le forze che agiscono su un piano inclinato e, inoltre, potrai allenarti con esercizi risolti passo dopo passo sul piano inclinato. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-un-plano-inclinado\"><\/span> Cos&#8217;\u00e8 un piano inclinato?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Un <strong>piano inclinato<\/strong> \u00e8 una superficie inclinata di un certo angolo. In fisica, il piano inclinato viene utilizzato per praticare problemi di forza.<\/p>\n<p> Ad esempio, una rampa o una strada in pendenza sono piani inclinati.<\/p>\n<p> Il piano inclinato consente di trasportare un oggetto utilizzando meno forza. Poich\u00e9 spingere un oggetto su un piano inclinato richiede meno forza che sollevarlo verticalmente.<\/p>\n<p> Inoltre, il piano inclinato \u00e8 considerato una delle sei classiche macchine semplici. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Formulas-del-plano-inclinado\"><\/span> Formule del piano inclinato<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Ora che conosciamo la definizione di piano inclinato, vediamo quali formule agiscono su un piano inclinato e quali equazioni le collegano.<\/p>\n<p> Il primo problema che incontriamo negli esercizi sul piano inclinato \u00e8 che la maggior parte delle forze agisce in una direzione parallela o perpendicolare al piano inclinato. Quindi i tipici assi delle coordinate (un asse verticale e un asse orizzontale) non sono molto utili per questo tipo di problemi. Ecco perch\u00e9, in generale, nei piani inclinati lavoriamo con un sistema di coordinate diverso: <\/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\/plan-incline.png\" alt=\"piano inclinato\" class=\"wp-image-4369\" width=\"391\" height=\"368\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/plan-incline-300x283.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/plan-incline-768x724.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/plan-incline.png 1010w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p> <strong>In fisica, per risolvere un problema del piano inclinato, utilizziamo due assi diversi:<\/strong> un primo asse la cui direzione \u00e8 parallela al piano inclinato e, dall&#8217;altro, un secondo asse la cui direzione \u00e8 perpendicolare al piano inclinato.<\/p>\n<p> Inoltre, come puoi vedere nell&#8217;immagine, <strong>su un piano inclinato (se c&#8217;\u00e8 attrito) agiscono generalmente tre diverse forze<\/strong> : la forza peso, la forza normale e la forza di attrito (o forza di attrito). Ma logicamente, se non c\u2019\u00e8 attrito sul piano inclinato, la forza di attrito viene trascurata.<\/p>\n<p> Tuttavia la forza peso si scompone vettorialmente in due componenti: una componente parallela al piano inclinato ed un&#8217;altra componente perpendicolare al piano inclinato. In questo modo tutte le forze possono essere espresse negli assi di lavoro del piano inclinato. Pertanto le due componenti del peso del corpo appoggiato sul piano inclinato si calcolano dal seno e dal coseno dell&#8217;angolo di inclinazione:<\/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-a26edbf89d563f1351d0ec9771f7e7bc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1=m\\cdot g\\cdot \\text{sen}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"142\" 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-f5f191747feef04cf0a63f61a6b56cfd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=m\\cdot g\\cdot \\text{cos}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"141\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Infine, le forze agenti su un piano inclinato possono essere messe in relazione dalle due formule seguenti: <\/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\/formules-du-plan-incline.png\" alt=\"formule del piano inclinato\" class=\"wp-image-4388\" width=\"491\" height=\"151\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-du-plan-incline-300x93.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-du-plan-incline-768x237.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-du-plan-incline.png 1022w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p> Si noti che, se la formulazione del problema non dice diversamente, il corpo sul piano inclinato potrebbe scivolare lungo il pendio, motivo per cui nell&#8217;equazione per l&#8217;asse parallelo al piano \u00e8 inclusa una possibile accelerazione. D&#8217;altra parte il corpo non pu\u00f2 muoversi nella direzione dell&#8217;asse perpendicolare al piano inclinato, quindi la somma delle forze \u00e8 zero. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-resuelto-del-plano-inclinado\"><\/span> Esempio risolto del piano inclinato<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Per farti vedere come vengono risolti i problemi del piano inclinato in fisica, puoi vedere un esempio risolto passo dopo passo di seguito.<\/p>\n<ul>\n<li> Posizioniamo un corpo di massa m=6 kg sulla sommit\u00e0 di un piano inclinato di 45\u00ba. Se il corpo scivola sul piano inclinato con un&#8217;accelerazione di 4 m\/s <sup>2<\/sup> , qual \u00e8 il coefficiente di attrito dinamico tra la superficie del piano inclinato e quella del corpo? Dati: g=10 m\/s <sup>2<\/sup> . <\/li>\n<\/ul>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique.png\" alt=\"problema del coefficiente di attrito o attrito dinamico\" class=\"wp-image-4281\" width=\"203\" height=\"205\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique-298x300.png 298w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique.png 479w\" sizes=\"auto, (max-width: 298px) 100vw, 298px\"><\/figure>\n<p class=\"has-text-align-left\"> La prima cosa che dobbiamo fare per risolvere qualsiasi problema di fisica riguardante la dinamica \u00e8 disegnare il diagramma di corpo libero. Quindi tutte le forze che agiscono sul sistema sono: <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-dynamique.png\" alt=\"esercizio risolto del coefficiente di attrito o attrito dinamico\" class=\"wp-image-4282\" width=\"248\" height=\"301\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-dynamique-247x300.png 247w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-dynamique.png 572w\" sizes=\"auto, (max-width: 247px) 100vw, 247px\"><\/figure>\n<p class=\"has-text-align-left\"> Nella direzione dell&#8217;asse 1 (parallelo al piano inclinato) il corpo ha un&#8217;accelerazione, invece, nella direzione dell&#8217;asse 2 (perpendicolare al piano inclinato) il corpo \u00e8 fermo. Da queste informazioni stabiliamo le equazioni delle forze del sistema:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-d87a1ef6aaa3476891df5da8334cbc49_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1-F_R=m\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"124\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-6bdf90ed250934bf6cffbb110bc792a4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2-N=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"90\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Quindi, possiamo calcolare la forza normale dalla seconda equazione:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-59341555fe3d5fe315ceb1864547873b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_2\\\\[3ex]N=m\\cdot g\\cdot \\text{cos}(\\alpha) \\\\[3ex] N=6 \\cdot 10 \\cdot \\ text{cos}(45\u00ba)\\\\[3ex]N=42,43 \\ N\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"151\" width=\"185\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Calcoliamo invece il valore della forza di attrito (o forza di attrito) dalla prima equazione presentata:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-d8f2aff2a81d98ddcea04b1988282fda_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}P_1-F_R=m\\cdot a\\\\[3ex]F_R=P_1-m\\cdot a\\\\[3ex]F_R=m\\cdot g\\cdot \\text{sin} (\\alpha)-m\\cdot a\\\\[3ex]F_R=6\\cdot 10\\cdot \\text{sin}(45\u00ba)-6\\cdot 4\\\\[3ex]F_R=18.43 \\ N\\end{ array} \" title=\"Rendered by QuickLaTeX.com\" height=\"195\" width=\"204\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E una volta conosciuto il valore della forza normale e della forza di attrito, possiamo determinare il coefficiente di attrito dinamico utilizzando la formula corrispondente: <\/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-b783c9e37bcf4d077d9496489fc5d7d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu_d=\\cfrac{F_R}{N}=\\cfrac{18.43}{43.43}=0.42\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"187\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-del-plano-inclinado\"><\/span> Esercizi risolti sul piano inclinato<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Esercizio 1<\/h3>\n<p> Posizioniamo un corpo di massa m=2 kg sulla sommit\u00e0 di un piano inclinato con un angolo di inclinazione di 30\u00ba. Qual \u00e8 il coefficiente di attrito tra la rampa e il corpo se quest&#8217;ultimo rimane in equilibrio? Dati: g=9,81 m\/s <sup>2<\/sup> <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-force-normale-et-de-force-de-friction.png\" alt=\"\" class=\"wp-image-4253\" width=\"285\" height=\"176\" srcset=\"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\" 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>Vedi la soluzione<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Come in ogni problema di fisica che coinvolge forze, la prima cosa da fare \u00e8 disegnare il diagramma di corpo libero del sistema. Quindi, tutte le forze che agiscono in questo sistema sono: <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-normale-et-friction-force.png\" alt=\"risolvere l\u2019esercizio della forza normale e della forza di attrito\" class=\"wp-image-4254\" width=\"285\" height=\"333\" srcset=\"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\" sizes=\"auto, (max-width: 256px) 100vw, 256px\"><\/figure>\n<p class=\"has-text-align-left\"> Quindi, affinch\u00e9 il sistema sia in equilibrio, la somma delle forze sugli assi 1 e 2 deve essere uguale a zero. Pertanto sono vere le seguenti equazioni:<\/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-a692b08b4d7c08a2c55556233dc56651_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=P_1\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"63\" 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-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\"> Possiamo ora calcolare il valore della forza normale dalla seconda equazione:<\/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\"> D&#8217;altra parte, determiniamo il valore della forza di attrito utilizzando la prima equazione:<\/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\"> Allo stesso modo, la forza di attrito pu\u00f2 essere correlata alla forza normale e al coefficiente di attrito utilizzando la seguente formula:<\/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\"> Risolviamo quindi il coefficiente di attrito dall&#8217;equazione e calcoliamo il suo valore: <\/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<h3 class=\"wp-block-heading\">Esercizio 2<\/h3>\n<p> Come vediamo nel seguente sistema formato da un piano inclinato e da una carrucola, due corpi sono collegati da una fune e da una carrucola di massa trascurabile. Se il corpo 2 ha massa m <sub>2<\/sub> = 7 kg e l&#8217;inclinazione della rampa \u00e8 50\u00ba, calcolare la forza normale che il piano inclinato esercita sul corpo di massa m <sub>1<\/sub> affinch\u00e9 l&#8217;intero sistema sia in equilibrio. Trascura la forza di attrito durante l&#8217;esercizio. <\/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 dell\u2019equilibrio traslazionale\" 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>Vedi la soluzione<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Il corpo 1 si trova su un pendio inclinato, quindi la prima cosa da fare \u00e8 vettorizzare la forza del suo peso per avere le forze sugli assi del pendio: <\/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\"> Pertanto l\u2019insieme delle forze che agiscono sull\u2019intero sistema sono: <\/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=\"esercizio di equilibrio traslazionale risolto\" 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\"> La formulazione del problema ci dice che il sistema di forze \u00e8 in equilibrio, quindi i due corpi devono essere in equilibrio. Da queste informazioni possiamo proporre le equazioni di equilibrio dei due corpi: <\/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\"> Dall&#8217;equazione precedente possiamo calcolare la massa del 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\"> Se invece guardiamo il diagramma delle forze del sistema, osserviamo che la forza normale deve essere pari alla componente vettoriale del peso del corpo 1 perpendicolare al piano inclinato. <\/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\"> Quindi, da questa equazione possiamo trovare il valore della forza normale: <\/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-4a591e86900fca256cfa079da1bd0461_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> Un tra\u00eeneau de 70 kg glisse sur une pente de 30\u00ba avec une vitesse initiale de 2 m\/s. Si le coefficient de frottement dynamique entre le tra\u00eeneau et la neige est de 0,2, calculez la vitesse que le tra\u00eeneau acquerra apr\u00e8s avoir parcouru 20 m\u00e8tres. Donn\u00e9es : g=10 m\/s <sup>2<\/sup> . <\/p>\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> Tout d&#8217;abord, nous r\u00e9alisons le sch\u00e9ma corporel libre du tra\u00eeneau : <\/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-plan-incline.png&quot; alt=&quot;exercice r\u00e9solu de la force de frottement ou de frottement sur un plan inclin\u00e9&quot; class=&quot;wp-image-4345&quot; width=&quot;305&quot; height=&quot;355&quot; srcset=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-plan-incline-258x300.png 258w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-plan-incline.png 706w&quot; sizes=&quot;(max-width: 258px) 100vw, 258px&quot;><\/figure>\n<p> Le tra\u00eeneau a une acc\u00e9l\u00e9ration dans la direction de l&#8217;axe 1 (parall\u00e8le au plan inclin\u00e9) mais reste au repos dans la direction de l&#8217;axe 2 (perpendiculaire au plan inclin\u00e9), donc les \u00e9quations des forces sont : [latex]P_1-F_R=m\\cdot a&#8221; title=&#8221;Rendered by QuickLaTeX.com&#8221; height=&#8221;213&#8243; width=&#8221;8731&#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-6bdf90ed250934bf6cffbb110bc792a4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2-N=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"90\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dalla seconda equazione possiamo calcolare la forza normale agente sulla slitta<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-90b32b903f8be520ec73748b3de9b8b3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_2\\\\[3ex]N=m\\cdot g\\cdot \\text{cos}(\\alpha) \\\\[3ex] N=70 \\cdot 10 \\cdot \\ text{cos}(30\u00ba)\\\\[3ex]N=606,22 \\ N\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"151\" width=\"194\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Poich\u00e9 ora conosciamo il valore della forza normale e il coefficiente di attrito dinamico, possiamo calcolare la forza di attrito applicando la formula corrispondente:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-e0a32cc7650b33325233258788c218d4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=\\mu\\cdot N=0,2 \\cdot 606,22=121,24 \\ N \" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"298\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Quindi, per determinare la velocit\u00e0 finale, dobbiamo prima trovare l&#8217;accelerazione della slitta, e questa pu\u00f2 essere calcolata dalla prima equazione della forza presentata: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-d87a1ef6aaa3476891df5da8334cbc49_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1-F_R=m\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"124\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-fa13e0490f51e32ac03b455043f6f32d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{P_1-F_R}{m}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"99\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-a6274d836af5618f7ef99e7f179c3902_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{m\\cdot g\\cdot \\text{sin}(\\alpha)-F_R}{m}\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"177\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-a3a9db70462cd187d50b851ede83983f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{70\\cdot 10\\cdot \\text{sin}(30\u00ba)-121.24}{70}\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"221\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-ba0d7325efa059351cc3d9aef838a9e2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=3,27 \\ \\cfrac{m}{s^2}\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"92\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Una volta conosciuta l&#8217;accelerazione della slitta, calcoliamo il tempo impiegato per percorrere i 20 metri con l&#8217;equazione del moto rettilineo ad accelerazione costante: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-69f632cd171007df0f5bd6f0fa458a5c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x=v_0\\cdot t +\\cfrac{1}{2}\\cdot a \\cdot t^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"150\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b97ac72bf22d70273fece0cce195f4ca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"20=2\\cdot t +\\cfrac{1}{2}\\cdot 3.27 \\cdot t^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"172\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-11a99cb686bf68cbcca594d0d60f801b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"0=1,64t^2+2t-20\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"158\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-7aa7e01e70b4199d597d05e261c970df_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle t=\\cfrac{-2\\pm \\sqrt{2^2-4\\cdot 1.64\\cdot (-20)}}{2\\cdot 1.64}=\\cfrac{-2\\ pm 11.63}{ 3.28}=\\begin{cases}2.94\\\\[2ex]-4.15 \\ \\color{red}\\bm{\\times}\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"507\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Logicamente escludiamo la soluzione negativa poich\u00e9 il tempo \u00e8 una grandezza fisica che non pu\u00f2 essere negativa.<\/p>\n<p class=\"has-text-align-left\"> Infine, calcoliamo la velocit\u00e0 finale utilizzando la formula dell&#8217;accelerazione costante: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-5ace98bfb166f5b813f593760fcfa048_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{v_f-v_0}{t_f-t_0}\\quad \\longrightarrow \\quad v_f=a\\cdot (t_f-t_0)+v_0\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"330\" style=\"vertical-align: -18px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-817a054f54a84b5a88667fc794feeb4d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v_f=3.27\\cdot (2.94-0)+2=\\bm{11.61} \\ \\cfrac{\\bm{m}}{\\bm{s}}\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"280\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>In questo articolo viene spiegato cosa sono i piani inclinati in fisica e come si risolvono problemi di questo tipo. Troverai le formule per le forze che agiscono su un piano inclinato e, inoltre, potrai allenarti con esercizi risolti passo dopo passo sul piano inclinato. Cos&#8217;\u00e8 un piano inclinato? Un piano inclinato \u00e8 una superficie &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/it\/piano-inclinato\/\"> <span class=\"screen-reader-text\">Piano inclinato<\/span> Leggi altro &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-253","post","type-post","status-publish","format-standard","hentry","category-dinamico"],"yoast_head":"<!-- This site is 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