{"id":29207,"date":"2021-03-25T13:00:07","date_gmt":"2021-03-25T13:00:07","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=29207"},"modified":"2024-09-22T16:33:30","modified_gmt":"2024-09-22T16:33:30","slug":"le-principe-de-pascal","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/fr\/le-principe-de-pascal\/","title":{"rendered":"Le Principe de Pascal"},"content":{"rendered":"<p><center><\/p>\n<h1>Le Principe de Pascal<\/h1>\n<p><em><strong>R\u00e9sum\u00e9 :<\/strong><br \/>\nDans cette le\u00e7on, nous expliquerons le Principe de Pascal, qui stipule que la pression appliqu\u00e9e \u00e0 un fluide incompressible se distribue uniform\u00e9ment. Nous en verrons les cons\u00e9quences, telles que le niveau constant dans les r\u00e9cipients et la relation entre les volumes d\u00e9plac\u00e9s. De plus, nous pr\u00e9senterons des applications dans les machines hydrauliques et r\u00e9soudrons des exemples pratiques, comme le nivellement de terrains, les diff\u00e9rences de niveau dans des liquides de diff\u00e9rentes densit\u00e9s et le fonctionnement d&#8217;un \u00e9l\u00e9vateur hydraulique.<br \/>\n<\/em><br \/>\n<strong>Objectifs d&#8217;apprentissage :<\/strong><\/p>\n<ol style=\"text-align:left;\">\n<li><strong>Comprendre<\/strong> le Principe de Pascal et son application dans les fluides incompressibles.<\/li>\n<li><strong>Analyser<\/strong> les cons\u00e9quences du principe, telles que la distribution uniforme de la pression et le comportement du niveau du fluide.<\/li>\n<li><strong>Appliquer<\/strong> le Principe de Pascal \u00e0 des situations pratiques, comme le fonctionnement des machines hydrauliques.<\/li>\n<li><strong>R\u00e9soudre<\/strong> des probl\u00e8mes pratiques li\u00e9s au nivellement des terrains et aux diff\u00e9rences de niveau dans des liquides de diff\u00e9rentes densit\u00e9s.<\/li>\n<li><strong>Expliquer<\/strong> l&#8217;utilisation du barom\u00e8tre de Torricelli pour mesurer la pression atmosph\u00e9rique et le fonctionnement des \u00e9l\u00e9vateurs hydrauliques.<\/li>\n<\/ol>\n<p><strong>TABLE DES MATI\u00c8RES<\/strong><br \/>\n<a href=\"#1\">Transmission des forces dans les liquides<\/a><br \/>\n<a href=\"#2\">Cons\u00e9quences du Principe de Pascal<\/a><br \/>\n<a href=\"#3\">Le Principe de Pascal appliqu\u00e9 aux machines hydrauliques<\/a><br \/>\n<a href=\"#4\">Exemples d&#8217;application<\/a>\n<\/p>\n<p><\/center><\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/9IAFRXx0Cns\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Transmission des forces dans les liquides<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=84s\" target=\"_blank\" rel=\"noopener\"><strong>Le Principe de Pascal s&#8217;applique lorsqu&#8217;il s&#8217;agit de liquides incompressibles<\/strong><\/a>. Les particules dans ce type de substance exercent des forces entre elles qui ne pr\u00e9servent pas leur position, mais plut\u00f4t la distance relative moyenne entre elles. Si cela ne se produisait pas, les particules modifieraient leur distance relative moyenne dans certaines r\u00e9gions, modifiant ainsi le volume total, ce qui indiquerait que nous traitons avec un liquide compressible. La plupart des liquides sont en fait tr\u00e8s peu compressibles, et donc, lorsque une force est appliqu\u00e9e dans une r\u00e9gion donn\u00e9e, elle est transmise compl\u00e8tement \u00e0 travers tout le fluide.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Cons\u00e9quences du Principe de Pascal<\/h2>\n<h3>Le niveau est le m\u00eame partout<\/h3>\n<p style=\"text-align: justify;\">Un exemple clair de cela est illustr\u00e9 dans la figure suivante. Si ce r\u00e9cipient est rempli d&#8217;un liquide, le niveau montera de mani\u00e8re \u00e9gale dans tous les espaces disponibles, quelle que soit la forme des conduits.<\/p>\n<p><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-SD1TKOVRBrw\/YF0upu1TqEI\/AAAAAAAAEtc\/3r34Qw4kXgwJpbXqylLBaKJX45tFH1S9gCLcBGAsYHQ\/s763\/principio%2Bde%2Bpascal%2B-%2Bnivel%2Bde%2Bliquido.png\" alt=\"Principe de Pascal\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-SD1TKOVRBrw\/YF0upu1TqEI\/AAAAAAAAEtc\/3r34Qw4kXgwJpbXqylLBaKJX45tFH1S9gCLcBGAsYHQ\/s763\/principio%2Bde%2Bpascal%2B-%2Bnivel%2Bde%2Bliquido.png\" alt=\"Principe de Pascal\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><\/noscript><\/p>\n<p style=\"text-align: justify;\">Cela est d\u00fb au fait que, comme nous l&#8217;avons d\u00e9j\u00e0 vu, la pression est une fonction de la hauteur.<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">P = \\rho g h<\/span>\n<p style=\"text-align: justify;\">S&#8217;il y avait une diff\u00e9rence de hauteur entre diff\u00e9rentes parties du liquide, il y aurait alors n\u00e9cessairement une diff\u00e9rence de pression entre ces parties.<\/p>\n<h3>Volume d\u00e9plac\u00e9 = Volume c\u00e9d\u00e9<\/h3>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=323s\" target=\"_blank\" rel=\"noopener\"><strong>Si une pression est appliqu\u00e9e \u00e0 un point du liquide,<\/strong><\/a> celui-ci se d\u00e9placera tout en maintenant le m\u00eame niveau dans toutes les autres parties. Le volume c\u00e9d\u00e9 <span class=\"katex-eq\" data-katex-display=\"false\">v_1<\/span> serait \u00e9gal au volume d\u00e9plac\u00e9 <span class=\"katex-eq\" data-katex-display=\"false\">v_2+v_3.<\/span>\n<p><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-nDStNasYWZQ\/YF0upq_20xI\/AAAAAAAAEtg\/bPPlgi4vpug4i70kHoMpkPsvtRtfbQrVACLcBGAsYHQ\/s767\/principio%2Bde%2Bpascalnivel%2Bde%2Bliquido2.png\" alt=\"Principe de Pascal, volume c\u00e9d\u00e9 \u00e9gal au volume d\u00e9plac\u00e9\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-nDStNasYWZQ\/YF0upq_20xI\/AAAAAAAAEtg\/bPPlgi4vpug4i70kHoMpkPsvtRtfbQrVACLcBGAsYHQ\/s767\/principio%2Bde%2Bpascalnivel%2Bde%2Bliquido2.png\" alt=\"Principe de Pascal, volume c\u00e9d\u00e9 \u00e9gal au volume d\u00e9plac\u00e9\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><\/noscript><br \/>\n<a name=\"3\"><\/a><\/p>\n<h2>Le Principe de Pascal appliqu\u00e9 aux machines hydrauliques<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=426s\" target=\"_blank\" rel=\"noopener\"><strong>Le principe de Pascal est utilis\u00e9 dans la construction des presses hydrauliques.<\/strong><\/a> Comme la pression est uniform\u00e9ment distribu\u00e9e dans toutes les parties d&#8217;un liquide, l&#8217;expression suivante est valable :<\/p>\n<p><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-hpJeg_iRsgw\/YF0uplM2GnI\/AAAAAAAAEtY\/tBztNE2eUJQe2uHlrzgZlh_y0QPLGyCnwCLcBGAsYHQ\/s805\/prensahidr%25C3%25A1ulica.PNG\" alt=\"Principe de Pascal, \u00e9l\u00e9vateur hydraulique\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"200\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-hpJeg_iRsgw\/YF0uplM2GnI\/AAAAAAAAEtY\/tBztNE2eUJQe2uHlrzgZlh_y0QPLGyCnwCLcBGAsYHQ\/s805\/prensahidr%25C3%25A1ulica.PNG\" alt=\"Principe de Pascal, \u00e9l\u00e9vateur hydraulique\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"200\" \/><\/noscript><\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">P_2 = P_1<\/span>\n<p style=\"text-align: justify;\">Mais comme la pression peut \u00eatre exprim\u00e9e comme une force par unit\u00e9 de surface, <span class=\"katex-eq\" data-katex-display=\"false\">P=F\/A<\/span>, nous aurons donc :<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{F_2}{A_2} = \\frac{F_1}{A_1}<\/span>\n<p style=\"text-align: justify;\">Cela signifie que si nous appliquons une force <span class=\"katex-eq\" data-katex-display=\"false\">F_1<\/span> sur le piston 1, une force <span class=\"katex-eq\" data-katex-display=\"false\">F_2<\/span> sera exerc\u00e9e sur le piston 2 selon l&#8217;expression suivante :<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle F_2 = \\frac{A_2}{A_1} \\cdot F_1<\/span>\n<p style=\"text-align: justify;\">Le facteur <span class=\"katex-eq\" data-katex-display=\"false\">A_2\/A_1<\/span> est ce que nous appellerons le \u00abcoefficient d&#8217;amplification ou de r\u00e9duction de la force\u00bb. Il amplifie si sup\u00e9rieur \u00e0 1, et r\u00e9duit s&#8217;il est compris entre 0 et 1.<\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h2>Exemples d&#8217;application<\/h2>\n<ol>\n<li style=\"text-align: justify;\">Deux personnes veulent niveler un terrain en pente. Pour ce faire, ils prennent un tuyau et le remplissent d&#8217;eau, puis placent ses extr\u00e9mit\u00e9s \u00e0 une distance horizontale de 275 cm. Le niveau de l&#8217;eau \u00e0 un point est de 110 cm et \u00e0 l&#8217;autre est de 175 cm.<img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-qmCq1t6szPU\/YF0uopFj2HI\/AAAAAAAAEtM\/Bo__H3vPJ9YZCms__pgwk-26VxjBbiYmgCLcBGAsYHQ\/s625\/nivel%2Bde%2Bagua.PNG\" alt=\"Principe de Pascal, niveau de l'eau\" class=\"aligncenter size-full lazyload\" width=\"625\" height=\"313\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-qmCq1t6szPU\/YF0uopFj2HI\/AAAAAAAAEtM\/Bo__H3vPJ9YZCms__pgwk-26VxjBbiYmgCLcBGAsYHQ\/s625\/nivel%2Bde%2Bagua.PNG\" alt=\"Principe de Pascal, niveau de l'eau\" class=\"aligncenter size-full lazyload\" width=\"625\" height=\"313\" \/><\/noscript>a) Quelle est la diff\u00e9rence de niveau du terrain ?\n<p>b) Quel est l&#8217;angle d&#8217;inclinaison du terrain ?<\/p>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=691s\" target=\"_blank\" rel=\"noopener\"><strong>SOLUTION<\/strong><\/a><\/li>\n<li style=\"text-align: justify;\">Un tuyau en forme de U contient de l&#8217;eau, avec une densit\u00e9 de <span class=\"katex-eq\" data-katex-display=\"false\">1000 [kg\/m^3].<\/span> Si \u00e0 l&#8217;une des extr\u00e9mit\u00e9s du tuyau on verse de l&#8217;huile jusqu&#8217;\u00e0 une hauteur de <span class=\"katex-eq\" data-katex-display=\"false\">20 cm<\/span>, avec une densit\u00e9 de <span class=\"katex-eq\" data-katex-display=\"false\">800[kg\/m^3]<\/span>, quelle diff\u00e9rence de niveau se produira entre les deux extr\u00e9mit\u00e9s du tuyau ?<img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-EYBgWw2b6jY\/YF0uoq4c5AI\/AAAAAAAAEtU\/LCAin5GVeSoRDCIgn3widW9kaTwfJwFRACLcBGAsYHQ\/s596\/nivel-agua-aceite.PNG\" alt=\"Principe de Pascal - M\u00e9langes\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-EYBgWw2b6jY\/YF0uoq4c5AI\/AAAAAAAAEtU\/LCAin5GVeSoRDCIgn3widW9kaTwfJwFRACLcBGAsYHQ\/s596\/nivel-agua-aceite.PNG\" alt=\"Principe de Pascal - M\u00e9langes\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><\/noscript><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=867s\" target=\"_blank\" rel=\"noopener\"><strong>SOLUTION<\/strong><\/a><\/li>\n<li style=\"text-align: justify;\"><strong>Le Barom\u00e8tre de Torricelli :<\/strong> Torricelli a con\u00e7u l&#8217;instrument suivant pour mesurer la pression atmosph\u00e9rique :<img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-hB2h_82Yg6A\/YF0uol0V-OI\/AAAAAAAAEtQ\/1w25pez7rj4LE3-tanK4NMo5NBy_0xJQgCLcBGAsYHQ\/s572\/Principio%2Bde%2Bpascal%2B-%2Btorricelli.PNG\" alt=\"Principe de Pascal, barom\u00e8tre de Torricelli\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-hB2h_82Yg6A\/YF0uol0V-OI\/AAAAAAAAEtQ\/1w25pez7rj4LE3-tanK4NMo5NBy_0xJQgCLcBGAsYHQ\/s572\/Principio%2Bde%2Bpascal%2B-%2Btorricelli.PNG\" alt=\"Principe de Pascal, barom\u00e8tre de Torricelli\" class=\"aligncenter size-full lazyload\" width=\"400\" height=\"300\" \/><\/noscript>a) Si la densit\u00e9 du mercure est de <span class=\"katex-eq\" data-katex-display=\"false\">13.534,0[kg\/m3]<\/span>, quelle sera la hauteur de la colonne de mercure sous la pression atmosph\u00e9rique ? Consid\u00e9rez <span class=\"katex-eq\" data-katex-display=\"false\">P_{atm}=1,0[atm]=101.325,0[Pa]<\/span> ?\n<p>b) Si \u00e0 la place du mercure il y avait de l&#8217;eau, quelle serait la hauteur de la colonne d&#8217;eau pour mesurer la m\u00eame pression ?<\/p>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=1049s\" target=\"_blank\" rel=\"noopener\"><strong>SOLUTION<\/strong><\/a><\/li>\n<li style=\"text-align: justify;\">Un \u00e9l\u00e9vateur hydraulique poss\u00e8de deux pistons, l&#8217;un avec un diam\u00e8tre de 2 pieds et l&#8217;autre avec un diam\u00e8tre de 30 pieds. a) Quel poids doit-on placer sur le plus petit piston si l&#8217;on souhaite soulever un bloc d&#8217;une tonne sur le plus grand piston ? b) Si l&#8217;on souhaite que le grand piston soul\u00e8ve un bloc de b\u00e9ton \u00e0 une hauteur de <span class=\"katex-eq\" data-katex-display=\"false\">50 cm<\/span>, quelle distance le petit piston doit-il parcourir ?\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=9IAFRXx0Cns&amp;t=1403s\" target=\"_blank\" rel=\"noopener\"><strong>SOLUTION<\/strong><\/a><\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Le Principe de Pascal R\u00e9sum\u00e9 : Dans cette le\u00e7on, nous expliquerons le Principe de Pascal, qui stipule que la pression appliqu\u00e9e \u00e0 un fluide incompressible se distribue uniform\u00e9ment. Nous en verrons les cons\u00e9quences, telles que le niveau constant dans les r\u00e9cipients et la relation entre les volumes d\u00e9plac\u00e9s. De plus, nous pr\u00e9senterons des applications dans [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":29196,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":15,"footnotes":""},"categories":[901,647],"tags":[],"class_list":["post-29207","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 v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Le Principe de Pascal - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Le Principe de Pascal stipule que lorsque la pression est appliqu\u00e9e sur un fluide incompressible, elle est transmise uniform\u00e9ment dans toutes les directions. 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