{"id":29263,"date":"2021-03-31T13:00:09","date_gmt":"2021-03-31T13:00:09","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=29263"},"modified":"2024-09-24T19:51:52","modified_gmt":"2024-09-24T19:51:52","slug":"la-flottabilite-et-le-principe-darchimede","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/fr\/la-flottabilite-et-le-principe-darchimede\/","title":{"rendered":"La Flottabilit\u00e9 et le Principe d&#8217;Archim\u00e8de"},"content":{"rendered":"<p><center><\/p>\n<h1>La Flottabilit\u00e9 et le Principe d&#8217;Archim\u00e8de<\/h1>\n<p><em><strong>R\u00e9sum\u00e9 :<\/strong><br \/>\nCette le\u00e7on expliquera le ph\u00e9nom\u00e8ne de la Flottabilit\u00e9 et le Principe d&#8217;Archim\u00e8de, en montrant comment les objets immerg\u00e9s dans un fluide subissent une force de pouss\u00e9e \u00e9gale au poids du fluide d\u00e9plac\u00e9. Ce principe est utilis\u00e9 pour calculer la partie d\u2019un objet qui \u00e9merge au-dessus du fluide, avec des exemples pratiques.<br \/>\n    <\/em><\/p>\n<p>    <strong>Objectifs d&#8217;apprentissage<\/strong><br \/>\n    \u00c0 la fin de cette le\u00e7on, l&#8217;\u00e9tudiant sera capable de :<\/p>\n<ol style=\"text-align:left;\">\n<li><strong>Comprendre<\/strong> le principe d&#8217;Archim\u00e8de et sa relation avec la flottabilit\u00e9.<\/li>\n<li><strong>Calculer<\/strong> la force de pouss\u00e9e sur les objets immerg\u00e9s dans un fluide.<\/li>\n<li><strong>D\u00e9terminer<\/strong> quelle portion d&#8217;un objet flottant \u00e9merge au-dessus du fluide, en se basant sur la densit\u00e9 relative.<\/li>\n<\/ol>\n<p><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/nikSJceTxJ0\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<h2>La Force de Pouss\u00e9e<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=nikSJceTxJ0&amp;t=126s\" target=\"_blank\" rel=\"noopener\"><strong>Lorsque nous parlons de flottabilit\u00e9,<\/strong><\/a> la premi\u00e8re chose \u00e0 laquelle nous pensons est que les objets semblent peser moins lorsqu&#8217;ils sont immerg\u00e9s dans un fluide. Par exemple, une roche qui pourrait \u00eatre soulev\u00e9e avec difficult\u00e9 sous l&#8217;eau serait presque impossible \u00e0 d\u00e9placer hors de l&#8217;eau. Ce ph\u00e9nom\u00e8ne est expliqu\u00e9 par l&#8217;apparition d&#8217;une force appel\u00e9e pouss\u00e9e.<\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=nikSJceTxJ0&amp;t=207s\" target=\"_blank\" rel=\"noopener\"><strong>Lorsqu&#8217;un objet est immerg\u00e9 dans un fluide,<\/strong><\/a> une force de pouss\u00e9e dirig\u00e9e vers le haut appara\u00eet, \u00e9gale au poids du liquide d\u00e9plac\u00e9 par le corps immerg\u00e9. C&#8217;est pourquoi tous les corps immerg\u00e9s dans un fluide semblent perdre une partie de leur poids, car il existe une diff\u00e9rence de forces entre les diff\u00e9rentes r\u00e9gions du corps selon la profondeur. Ainsi, la force de pouss\u00e9e sera :<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">F_{pouss\u00e9e} = F_2 - F_1<\/span>\n<p><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/fuerza-de-empuje.jpg\" alt=\"force de pouss\u00e9e\" width=\"293\" height=\"502\" class=\"aligncenter size-full wp-image-29232 lazyload\" \/><noscript><img decoding=\"async\" src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/fuerza-de-empuje.jpg\" alt=\"force de pouss\u00e9e\" width=\"293\" height=\"502\" class=\"aligncenter size-full wp-image-29232 lazyload\" srcset=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/fuerza-de-empuje.jpg 293w, http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/fuerza-de-empuje-175x300.jpg 175w\" sizes=\"(max-width: 293px) 100vw, 293px\" \/><\/noscript><\/p>\n<p style=\"text-align: justify;\">Puisque <span class=\"katex-eq\" data-katex-display=\"false\">P=F\/A<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">P=\\rho g h<\/span>, nous pouvons d\u00e9duire que <span class=\"katex-eq\" data-katex-display=\"false\">F=\\rho g A h<\/span>, o\u00f9 <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span> est la densit\u00e9 du fluide, <span class=\"katex-eq\" data-katex-display=\"false\">h<\/span> la profondeur, <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> la surface sur laquelle la pression est appliqu\u00e9e, et <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> l&#8217;acc\u00e9l\u00e9ration gravitationnelle. Avec cela, les forces exerc\u00e9es sur les faces sup\u00e9rieure et inf\u00e9rieure sont donn\u00e9es par :<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">F_1 = \\rho g A h_1<\/span>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">F_2 = \\rho g A h_2<\/span>\n<p style=\"text-align: justify;\">Ainsi, nous avons :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rl}\n\n F_{\\text{pouss\u00e9e}} &amp;= \\rho g A h_2 - \\rho g A h_1 \\\\ \\\\\n\n&amp;=\\rho g A \\underbrace{h_2 - h_1}_{\\Delta h} \\\\ \\\\\n\n&amp;= \\rho gV \\\\ \\\\\n\n&amp; =\\text{Poids du volume du liquide d\u00e9plac\u00e9}\n\n\\end{array}\n\n<\/span>\n<p style=\"text-align: justify;\">C&#8217;est ce que l&#8217;on appelle le principe d&#8217;Archim\u00e8de.<\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #008000;\">EXEMPLE :<\/span> <a href=\"https:\/\/www.youtube.com\/watch?v=nikSJceTxJ0&amp;t=482s\" target=\"_blank\" rel=\"noopener\"><strong>Une roche de 70[kg] repose au fond d&#8217;un lac.<\/strong><\/a> Si son volume est de <span class=\"katex-eq\" data-katex-display=\"false\">3\\cdot 10^4 [cm^3]<\/span>, quelle force est n\u00e9cessaire pour la soulever ?<\/p>\n<p style=\"text-align: justify;\">SOLUTION :<\/p>\n<p style=\"text-align: justify;\">La force de pouss\u00e9e sur la roche sera :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rl}\n\nF_{\\text{pouss\u00e9e}} &amp;= \\rho_{\\text{eau}} g V_{roche} \\\\ \\\\\n\n&amp;= 10^3 \\left[\\dfrac{kg}{m^3}\\right] \\cdot 9.81\\left[\\dfrac{m}{s^2}\\right] \\cdot 3 \\cdot 10^4 [cm^3] \\\\ \\\\\n\n&amp;= 10^3 \\left[\\dfrac{kg}{m^3}\\right] \\cdot 9.81\\left[\\dfrac{m}{s^2}\\right] \\cdot 3 \\cdot 10^4 \\left[\\dfrac{m}{100}\\right]^3 = 294[N]\n\\end{array}\n\n<\/span>\n<p style=\"text-align: justify;\">Tandis que la force de gravit\u00e9 de la roche est :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">F_{\\text{poids}} = m_{\\text{roche}}g = 70[kg] \\cdot 9.81 \\left[\\dfrac{m}{s^2}\\right]=686[N]<\/span>\n<p style=\"text-align: justify;\">Par cons\u00e9quent, pour soulever la roche sous l&#8217;eau, une force de <span class=\"katex-eq\" data-katex-display=\"false\">F = 686[N] - 294[N] = 392[N]<\/span> suffira. Sous l&#8217;eau, cette roche peut \u00eatre soulev\u00e9e avec presque la moiti\u00e9 de la force requise hors de l&#8217;eau.<\/p>\n<h2>La Flottabilit\u00e9 et le Principe d&#8217;Archim\u00e8de<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=nikSJceTxJ0&amp;t=858s\" target=\"_blank\" rel=\"noopener\"><strong>Le principe d&#8217;Archim\u00e8de nous aide \u00e0 comprendre pourquoi certains objets flottent<\/strong><\/a> lorsqu&#8217;ils sont immerg\u00e9s dans certains fluides. Par exemple, le bois dans l&#8217;eau. En g\u00e9n\u00e9ral, un objet flotte dans un fluide si la densit\u00e9 du milieu est sup\u00e9rieure \u00e0 celle de l&#8217;objet, et il flottera jusqu&#8217;\u00e0 ce qu&#8217;une partie \u00e9merge \u00e0 la surface. Le corps s&#8217;\u00e9l\u00e8vera jusqu&#8217;\u00e0 atteindre une position d&#8217;\u00e9quilibre. Comment pouvons-nous calculer la portion du corps qui \u00e9merge au-dessus du fluide ? C&#8217;est facile \u00e0 calculer.<\/p>\n<p><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/flotabilidad.jpg\" alt=\"La Flottabilit\u00e9 et le Principe d'Archim\u00e8de\" width=\"543\" height=\"341\" class=\"aligncenter size-full wp-image-29234 lazyload\" \/><noscript><img decoding=\"async\" src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/flotabilidad.jpg\" alt=\"La Flottabilit\u00e9 et le Principe d'Archim\u00e8de\" width=\"543\" height=\"341\" class=\"aligncenter size-full wp-image-29234 lazyload\" srcset=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/flotabilidad.jpg 543w, http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/flotabilidad-300x188.jpg 300w\" sizes=\"(max-width: 543px) 100vw, 543px\" \/><\/noscript><\/p>\n<p style=\"text-align: justify;\">Si nous \u00e9galons la force de gravit\u00e9 avec la force de pouss\u00e9e, nous pouvons calculer quelle portion du corps flottera au-dessus de la surface. Le raisonnement est le suivant :<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n &amp; F_{\\text{poids}} = F_{\\text{pouss\u00e9e}}\\\\ \\\\\n\n\\equiv &amp; m_{\\text{objet}} g = m_{\\text{immerg\u00e9}} g \\\\ \\\\\n\n\\equiv &amp; \\rho_{\\text{objet}}V_{\\text{objet}} g = \\rho_{\\text{immerg\u00e9}} V_{\\text{immerg\u00e9}}  g \\\\ \\\\\n\n\\equiv &amp; \\dfrac{\\rho_{\\text{objet}}}{\\rho_{\\text{immerg\u00e9}}} = \\dfrac{V_{\\text{immerg\u00e9}}}{V_{\\text{objet}} } = \\text{Pourcentage du corps immerg\u00e9}\n\n\\end{array}\n\n<\/span>\n<p style=\"text-align: justify;\"><span style=\"color: #008000;\">EXEMPLE :<\/span> <a href=\"https:\/\/www.youtube.com\/watch?v=nikSJceTxJ0&amp;t=1023s\" target=\"_blank\" rel=\"noopener\"><strong>Un mod\u00e8le simple suppose qu&#8217;un continent est un bloc solide de roche<\/strong><\/a> (avec une densit\u00e9 de <span class=\"katex-eq\" data-katex-display=\"false\">=2800[kg\/m^3]<\/span>) flottant sur le manteau terrestre environnant (avec une densit\u00e9 de <span class=\"katex-eq\" data-katex-display=\"false\">=3300[kg\/m^3]<\/span>). En supposant que le continent a une \u00e9paisseur moyenne de 35[km], calculez la hauteur moyenne qui \u00e9merge au-dessus du manteau.<\/p>\n<p style=\"text-align: justify;\">SOLUTION :<\/p>\n<p style=\"text-align: justify;\">Le pourcentage de la partie immerg\u00e9e du corps sera :<\/p>\n<p style=\"text-align: justify;\">Pourcentage immerg\u00e9 <span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle = \\frac{\\rho_{corps}}{\\rho_{fluide}}<\/span>\n<p style=\"text-align: justify;\">Ainsi, le pourcentage du corps qui n&#8217;est pas immerg\u00e9 et qui \u00e9merge au-dessus du manteau sera :<\/p>\n<p style=\"text-align: justify;\">Pourcentage \u00e9merg\u00e9 <span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle = 1 - \\frac{\\rho_{corps}}{\\rho_{fluide}} = 1 - \\frac{2800}{3300} \\approx 0,15 = 15\\%<\/span>\n<p style=\"text-align: justify;\">\u00c9tant donn\u00e9 que l&#8217;\u00e9paisseur moyenne est de <span class=\"katex-eq\" data-katex-display=\"false\">35[km]<\/span>, la partie \u00e9merg\u00e9e en moyenne sera <span class=\"katex-eq\" data-katex-display=\"false\">15\\% 35[km]\\approx 5.3 [km].<\/span>\n","protected":false},"excerpt":{"rendered":"<p>La Flottabilit\u00e9 et le Principe d&#8217;Archim\u00e8de R\u00e9sum\u00e9 : Cette le\u00e7on expliquera le ph\u00e9nom\u00e8ne de la Flottabilit\u00e9 et le Principe d&#8217;Archim\u00e8de, en montrant comment les objets immerg\u00e9s dans un fluide subissent une force de pouss\u00e9e \u00e9gale au poids du fluide d\u00e9plac\u00e9. Ce principe est utilis\u00e9 pour calculer la partie d\u2019un objet qui \u00e9merge au-dessus du fluide, [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":29250,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":55,"footnotes":""},"categories":[901,647],"tags":[],"class_list":["post-29263","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mecanique-des-fluides","category-physique"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>La Flottabilit\u00e9 et le Principe d&#039;Archim\u00e8de - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"D\u00e9couvrez comment la Flottabilit\u00e9 et le Principe d&#039;Archim\u00e8de expliquent pourquoi les objets flottent ou coulent. 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