{"id":247,"date":"2023-06-23T09:26:11","date_gmt":"2023-06-23T09:26:11","guid":{"rendered":"https:\/\/physigeek.com\/it\/forza-normale\/"},"modified":"2023-06-23T09:26:11","modified_gmt":"2023-06-23T09:26:11","slug":"forza-normale","status":"publish","type":"post","link":"https:\/\/physigeek.com\/it\/forza-normale\/","title":{"rendered":"Forza normale"},"content":{"rendered":"<p>Questo articolo spiega cos&#8217;\u00e8 la forza normale e come determinarla in base al tipo di problema. Ritroverai cos\u00ec le caratteristiche della forza normale e, inoltre, potrai praticare questo tipo di forza con esercizi risolti passo dopo passo. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-la-fuerza-normal\"><\/span>Cos&#8217;\u00e8 la forza normale?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> In fisica, <strong>la forza normale<\/strong> \u00e8 una forza esercitata da una superficie su un corpo appoggiato su di essa. Pertanto, la direzione della forza normale \u00e8 perpendicolare alla superficie e la direzione della forza normale \u00e8 verso l&#8217;esterno, cio\u00e8 la superficie applica la forza normale verso il corpo.<\/p>\n<p> In generale, la forza normale serve a contrastare la <a href=\"https:\/\/physigeek.com\/it\/peso-fisico\/\">forza peso<\/a> , che \u00e8 l&#8217;attrazione gravitazionale che la Terra esercita su qualsiasi corpo dotato di massa. Tuttavia, quando il corpo poggia su una superficie inclinata, il valore della forza normale potrebbe non essere sufficiente. Di seguito vedremo come si calcola la forza normale su un piano inclinato.<\/p>\n<p> In breve, le <strong><u style=\"text-decoration-color:#4fd12f\">caratteristiche della forza normale<\/u><\/strong> sono:<\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">La forza normale \u00e8 una forza di contatto, cio\u00e8 pu\u00f2 essere applicata solo se due superfici sono in contatto.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">La direzione della forza normale \u00e8 perpendicolare alla superficie su cui rimane il corpo.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">La direzione della forza normale \u00e8 sempre verso l&#8217;esterno, poich\u00e9 \u00e8 la superficie che applica la forza normale al corpo.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">In generale, l&#8217;entit\u00e0 della forza normale equivale alla proiezione della forza risultante sulla superficie di appoggio.<\/span><\/li>\n<li> <span style=\"color:#101010;font-weight: normal;\">Normalmente, la forza normale \u00e8 solitamente rappresentata dal simbolo N o F <sub>N.<\/sub><\/span> <\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Como-calcular-la-fuerza-normal\"><\/span> Come calcolare la forza normale<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> In generale, per <strong>calcolare la forza normale<\/strong> si devono applicare le equazioni di equilibrio, le quali stabiliscono che un corpo \u00e8 in equilibrio quando la somma delle forze verticali e la somma delle forze orizzontali sono pari a zero.<\/p>\n<p> Applicando le condizioni di equilibrio al problema, saremo in grado di risolvere la forza normale dalle equazioni proposte e quindi determinare 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-a9333695ba02f6e089d628fe3622a2e5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{c}\\displaystyle\\sum \\vv{F_x}=0\\\\[2ex]\\displaystyle\\sum \\vv{F_y}=0\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"81\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-del-calculo-de-la-fuerza-normal\"><\/span> Esempio di calcolo della forza normale<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Ora che conosciamo la definizione di forza normale, vediamo un esempio concreto di calcolo della forza normale.<\/p>\n<ul>\n<li> Un corpo di 8 kg \u00e8 fermo su un terreno pianeggiante. Qual \u00e8 il valore della forza normale esercitata dal terreno sul corpo?<\/li>\n<\/ul>\n<p> In questo problema il corpo \u00e8 fermo su una superficie piana, quindi le uniche forze che agiscono su di esso sono la forza peso e la forza normale. <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-et-poids-normaux.png\" alt=\"forza e peso normali\" class=\"wp-image-4215\" width=\"273\" height=\"297\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-et-poids-normaux-275x300.png 275w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-et-poids-normaux.png 570w\" sizes=\"auto, (max-width: 275px) 100vw, 275px\"><\/figure>\n<p> Quindi, affinch\u00e9 un corpo sia in equilibrio su una superficie piana, la forza normale (N) e la forza peso (P) devono essere uguali. La normale e il peso hanno quindi la stessa direzione, lo stesso modulo, ma la loro direzione \u00e8 opposta.<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-318d3aaff48777c13e5ac24cb775f6b0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"54\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Pertanto, per determinare il valore della forza normale, \u00e8 sufficiente calcolare il peso del corpo, che equivale alla sua massa moltiplicata per l&#8217;accelerazione dovuta alla gravit\u00e0: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-a7355a420808c17875e97713c4bef5ec_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P=m\\cdot g=8 \\cdot 9,81 = 78,48 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"283\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Fuerza-normal-en-un-plano-inclinado\"><\/span> forza normale su un piano inclinato<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> In questa sezione ricaveremo la formula della forza normale su un piano inclinato, poich\u00e9 il suo valore cambia a seconda che la superficie sia piana o inclinata.<\/p>\n<p> Pertanto le forze che agiscono su un corpo appoggiato su un piano inclinato sono le seguenti: <\/p>\n<figure class=\"wp-block-image aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline.png\" alt=\"forza normale su un piano inclinato\" class=\"wp-image-4220\" width=\"308\" height=\"417\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline-222x300.png 222w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline-757x1024.png 757w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline-768x1038.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline.png 770w\" sizes=\"auto, (max-width: 222px) 100vw, 222px\"><\/figure>\n<p> Osserva la figura sopra: Quando il piano \u00e8 inclinato, \u00e8 pi\u00f9 conveniente utilizzare come assi la direzione parallela al piano (asse 1) e la direzione perpendicolare al piano (asse 2). In questo modo \u00e8 pi\u00f9 semplice enunciare le equazioni di bilancio.<\/p>\n<p> Per calcolare la <strong>forza normale su un piano inclinato<\/strong> \u00e8 necessario applicare la condizione di equilibrio sull&#8217;asse perpendicolare al piano inclinato, poich\u00e9 possiamo garantire che il corpo \u00e8 in equilibrio su questo asse ma non sull&#8217;asse parallelo al piano .<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-badff1735c827c6562a4e074ea4b6bd2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\sum \\vv{F_2}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"80\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<p> Quindi la forza normale su un piano inclinato \u00e8 equivalente alla componente del peso dell&#8217;asse perpendicolare al piano:<\/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> La componente del peso dell&#8217;asse perpendicolare al piano \u00e8 uguale alla formula del peso moltiplicata per il coseno dell&#8217;angolo di inclinazione del piano:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-efeff10487f285abba9d74ee3eba6b45_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=P\\cdot \\cos(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"117\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-9fe1bc1d3a7fbacecb2ce1ccc1dadc67_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=m\\cdot g\\cdot \\cos(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"141\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> In breve, la <strong>formula della forza normale su un piano inclinato<\/strong> afferma che la forza normale \u00e8 uguale alla massa del corpo moltiplicata per la gravit\u00e0 moltiplicata per il coseno dell&#8217;angolo di inclinazione del piano: <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formule-de-la-force-normale-dans-un-plan-incline.png\" alt=\"formula per la forza normale su un piano inclinato\" class=\"wp-image-4232\" width=\"269\" height=\"92\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formule-de-la-force-normale-dans-un-plan-incline-300x102.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formule-de-la-force-normale-dans-un-plan-incline.png 576w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Fuerza-normal-y-fuerza-de-rozamiento\"><\/span> forza normale e forza di attrito<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> In questa sezione vedremo la relazione tra la forza normale e la forza di attrito, poich\u00e9 sono due tipi di forze legate matematicamente. Ma prima devi sapere cos&#8217;\u00e8 la forza di attrito.<\/p>\n<p> La forza di attrito (o forza di attrito) \u00e8 una forza che si verifica quando si tenta di spostare un corpo su una superficie non liscia. La forza di attrito \u00e8 quindi una forza che si oppone al movimento di un corpo.<\/p>\n<p> La forza di attrito viene calcolata dalla forza normale. Pi\u00f9 precisamente, <strong>la forza di attrito \u00e8 pari al coefficiente di attrito superficiale moltiplicato per la forza normale.<\/strong><\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b8e2dc6a1180d664163aeb969b289073_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=\\mu \\cdot N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"86\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p style=\"margin-bottom:5px\"> Oro: <\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:8px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-5b005ac29604de5f2904d2da7ade0238_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"22\" style=\"vertical-align: -3px;\"><\/p>\n<p> \u00e8 la forza di attrito. <\/span><\/li>\n<li style=\"margin-bottom:8px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-05d9eae892416bd34247a25207f8b718_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"11\" style=\"vertical-align: -4px;\"><\/p>\n<p> \u00e8 il coefficiente di attrito.<\/span><\/li>\n<li><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-7354bae77b50b7d1faed3e8ea7a3511a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e8 una resistenza normale. <\/span><\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-fuerza-normal\"><\/span> Risolti esercizi di forza normale<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Esercizio 1<\/h3>\n<p> Un corpo di 5 kg \u00e8 fermo su un terreno pianeggiante. Se quindi si aggiunge un altro corpo di massa 3 kg sopra il primo corpo, qual \u00e8 la forza normale esercitata dal suolo per sostenere i due corpi? Dati: g=9,81 m\/ <sup>s2<\/sup> . <\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#FFF8E1\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Vedi la soluzione<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Poich\u00e9 il terreno deve sostenere entrambi i corpi, la forza normale sar\u00e0 la somma della forza del peso di ciascun corpo. Pertanto, calcoleremo prima il peso di ciascun corpo e poi li sommeremo insieme.<\/p>\n<p class=\"has-text-align-left\"> Ricorda che la forza del peso si calcola moltiplicando la massa del corpo per la gravit\u00e0.<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c0cdb663ec9f8fe79fbecd960b50fc39_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=m\\cdot g\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"75\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Pertanto, calcoliamo il peso di un corpo di 5 kg:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-a15ef1db0a6608fa8e6165ac0e12e925_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1=5\\cdot 9.81=49.05\\N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"160\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> In secondo luogo, determiniamo il peso del secondo corpo la cui massa \u00e8 3 kg:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-d57016899db324fc0f785e92341e9f2f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=3\\cdot 9.81=29.43\\N\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"161\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Applicando quindi la condizione di equilibrio verticale, otteniamo che la forza normale equivale alla somma dei due pesi: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c65761c8213d33892f422dd6b0a29121_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\sum \\vv{F_y}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"81\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-808b8980c7dbb5f2b1cdf14418fea88c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P_1+P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"99\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> In conclusione il valore della forza normale esercitata dal terreno \u00e8: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c5819ab4ff951b1edbe6efa0a0111243_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=49,05+29,43=78,48 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"237\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Esercizio 2<\/h3>\n<p> Come mostrato nella figura seguente, due corpi sono collegati da una fune e da una puleggia 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 esercitata dal piano inclinato sul corpo di massa m <sub>1<\/sub> in modo che 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-f258dccd08d6573f74a2261b2192a92f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_1\\cdot \\text{cos}(\\alpha)\\\\[3ex]N=m_1 \\cdot g\\cdot \\text{cos}(\\alpha)\\\\[ 3ex]N=9,14 \\cdot 9,81 \\cdot \\text{cos}(50\\text{\u00ba})\\\\[3ex]N=\\bm{57,63 \\ N}\\end{array}[\/ latex]\n\n<div class=&quot;wp-block-otfm-box-spoiler-end otfm-sp_end&quot;><\/div>\n<h3 class=&quot;wp-block-heading&quot;> Exercice 3<\/h3>\n<p> Nous pla\u00e7ons un corps de masse m=2 kg au sommet d&#8217;une rampe avec un angle d&#8217;inclinaison de 30\u00ba. Quel est le coefficient de frottement entre la rampe et le corps si celui-ci est maintenu en \u00e9quilibre ? Donn\u00e9es : g=9,81 m\/s <sup>2<\/sup> <\/p>\n<figure class=&quot;wp-block-image aligncenter size-full is-resized&quot;><img decoding=&quot;async&quot; loading=&quot;lazy&quot; src=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-force-normale-et-de-force-de-friction.png&quot; alt=&quot;&quot; class=&quot;wp-image-4253&quot; width=&quot;285&quot; height=&quot;176&quot; srcset=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-force-normale-et-de-force-de-friction-300x185.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-force-normale-et-de-force-de-friction.png 702w&quot; sizes=&quot;(max-width: 300px) 100vw, 300px&quot;><\/figure>\n<div class=&quot;wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1&quot; role=&quot;button&quot; tabindex=&quot;0&quot; aria-expanded=&quot;false&quot; data-otfm-spc=&quot;#FFF8E1&quot; style=&quot;text-align:center&quot;>\n<div class=&quot;otfm-sp__title&quot;> <strong>Voir la solution<\/strong><\/div>\n<\/div>\n<p> Comme dans tout probl\u00e8me de physique portant sur les forces, la premi\u00e8re chose \u00e0 faire est de dessiner le diagramme du corps libre du syst\u00e8me. Ainsi, toutes les forces qui agissent dans ce syst\u00e8me sont : <\/p>\n<figure class=&quot;wp-block-image aligncenter size-full is-resized&quot;><img decoding=&quot;async&quot; loading=&quot;lazy&quot; src=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-normale-et-friction-force.png&quot; alt=&quot;exercice r\u00e9solu de la force normale et de la force de frottement&quot; class=&quot;wp-image-4254&quot; width=&quot;285&quot; height=&quot;333&quot; srcset=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-normale-et-friction-force-256x300.png 256w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-normale-et-friction-force.png 702w&quot; sizes=&quot;(max-width: 256px) 100vw, 256px&quot;><\/figure>\n<p> Ainsi, pour que le syst\u00e8me soit en \u00e9quilibre, la somme des forces sur les axes 1 et 2 doit \u00eatre \u00e9gale \u00e0 z\u00e9ro. Par cons\u00e9quent, les \u00e9quations suivantes sont vraies : [latex]F_R=P_1&#8243; title=&#8221;Rendered by QuickLaTeX.com&#8221; height=&#8221;454&#8243; width=&#8221;7014&#8243; style=&#8221;vertical-align: 0px;&#8221;><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-ea3f790cf878ca23f77405f73a20e7c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"58\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> 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\"> Quindi eliminiamo 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","protected":false},"excerpt":{"rendered":"<p>Questo articolo spiega cos&#8217;\u00e8 la forza normale e come determinarla in base al tipo di problema. Ritroverai cos\u00ec le caratteristiche della forza normale e, inoltre, potrai praticare questo tipo di forza con esercizi risolti passo dopo passo. Cos&#8217;\u00e8 la forza normale? In fisica, la forza normale \u00e8 una forza esercitata da una superficie su un &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/it\/forza-normale\/\"> <span class=\"screen-reader-text\">Forza normale<\/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-247","post","type-post","status-publish","format-standard","hentry","category-dinamico"],"yoast_head":"<!-- This site is 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