{"id":29251,"date":"2021-03-31T13:00:10","date_gmt":"2021-03-31T13:00:10","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=29251"},"modified":"2024-09-24T19:29:22","modified_gmt":"2024-09-24T19:29:22","slug":"buoyancy-and-archimedes-principle","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/en\/buoyancy-and-archimedes-principle\/","title":{"rendered":"Buoyancy and Archimedes&#8217; Principle"},"content":{"rendered":"<p><center><\/p>\n<h1>Buoyancy and Archimedes&#8217; Principle<\/h1>\n<p><em><strong>Summary:<\/strong><br \/>\nThis lesson will explain the phenomenon of Buoyancy and Archimedes&#8217; Principle, showing how submerged objects in a fluid experience an upward force equal to the weight of the displaced fluid. This principle is used to calculate the portion of an object that emerges above the fluid, with practical examples.<br \/>\n    <\/em><\/p>\n<p>    <strong>Learning Objectives<\/strong><br \/>\n    By the end of this lesson, the student will be able to:<\/p>\n<ol style=\"text-align:left;\">\n<li><strong>Understand<\/strong> Archimedes&#8217; principle and its relationship to buoyancy.<\/li>\n<li><strong>Calculate<\/strong> the buoyant force on objects submerged in a fluid.<\/li>\n<li><strong>Determine<\/strong> what portion of a floating object emerges above the fluid, based on relative density.<\/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>Buoyant Force<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=nikSJceTxJ0&amp;t=126s\" target=\"_blank\" rel=\"noopener\"><strong>When we talk about buoyancy,<\/strong><\/a> the first thing we think about is that objects tend to weigh less when submerged in a fluid. For example, a rock that could be lifted with difficulty underwater would be almost impossible to move outside of it. This phenomenon is explained by the appearance of a force called buoyancy.<\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=nikSJceTxJ0&amp;t=207s\" target=\"_blank\" rel=\"noopener\"><strong>When an object is submerged in a fluid,<\/strong><\/a> an upward buoyant force appears, equal to the weight of the liquid displaced by the submerged body. For this reason, all bodies submerged in a fluid seem to lose part of their weight, as there is a difference in forces between different regions of the body according to its depth. Thus, the buoyant force will be:<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">F_{buoy} = 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=\"buoyant force\" 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=\"buoyant force\" 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;\">Since <span class=\"katex-eq\" data-katex-display=\"false\">P=F\/A<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">P=\\rho g h<\/span>, we can infer that <span class=\"katex-eq\" data-katex-display=\"false\">F=\\rho g A h<\/span>, where <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span> is the fluid density, <span class=\"katex-eq\" data-katex-display=\"false\">h<\/span> the depth, <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> the area of the surface where the pressure is applied, and <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> the gravitational acceleration. With this, the forces exerted on the top and bottom faces are given by:<\/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;\">Therefore, we have:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rl}\n\n F_{\\text{buoyancy}} &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{Weight of the displaced fluid volume}\n\n\\end{array}\n\n<\/span>\n<p style=\"text-align: justify;\">This is what is known as Archimedes&#8217; principle.<\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #008000;\">EXAMPLE:<\/span> <a href=\"https:\/\/www.youtube.com\/watch?v=nikSJceTxJ0&amp;t=482s\" target=\"_blank\" rel=\"noopener\"><strong>A 70[kg] rock lies at the bottom of a lake.<\/strong><\/a> If its volume is <span class=\"katex-eq\" data-katex-display=\"false\">3\\cdot 10^4 [cm^3]<\/span>, what force is needed to lift it?<\/p>\n<p style=\"text-align: justify;\">SOLUTION:<\/p>\n<p style=\"text-align: justify;\">The buoyant force on the rock will be:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rl}\n\nF_{\\text{buoyancy}} &amp;= \\rho_{\\text{water}} g V_{rock} \\\\ \\\\\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;\">While the weight force of the rock is:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">F_{\\text{weight}} = m_{\\text{rock}}g = 70[kg] \\cdot 9.81 \\left[\\dfrac{m}{s^2}\\right]=686[N]<\/span>\n<p style=\"text-align: justify;\">Therefore, to lift the rock underwater, a force of <span class=\"katex-eq\" data-katex-display=\"false\">F = 686[N] - 294[N] = 392[N]<\/span> will suffice. Underwater, this rock can be lifted with almost half the force you would need outside of it.<\/p>\n<h2>Buoyancy and Archimedes&#8217; Principle<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=nikSJceTxJ0&amp;t=858s\" target=\"_blank\" rel=\"noopener\"><strong>Archimedes&#8217; principle helps us understand why some objects float<\/strong><\/a> when submerged in certain fluids, such as wood in water. In general, an object floats in a fluid if the medium&#8217;s density is greater than that of the object, and it will float until a portion emerges above the interface. The body will rise until it reaches an equilibrium position. How can we calculate the portion of the body that emerges above the fluid? It&#8217;s easy to calculate.<\/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=\"Buoyancy and Archimedes' Principle\" 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=\"Buoyancy and Archimedes' Principle\" 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;\">If we equate the weight force with the buoyant force, we can calculate what portion of the body will emerge above the interface. The reasoning is as follows:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n &amp; F_{\\text{weight}} = F_{\\text{buoyancy}}\\\\ \\\\\n\n\\equiv &amp; m_{\\text{object}} g = m_{\\text{submerged}} g \\\\ \\\\\n\n\\equiv &amp; \\rho_{\\text{object}}V_{\\text{object}} g = \\rho_{\\text{submerged}} V_{\\text{submerged}}  g \\\\ \\\\\n\n\\equiv &amp; \\dfrac{\\rho_{\\text{object}}}{\\rho_{\\text{submerged}}} = \\dfrac{V_{\\text{submerged}}}{V_{\\text{object}} } = \\text{Percentage of the body submerged}\n\n\\end{array}\n\n<\/span>\n<p style=\"text-align: justify;\"><span style=\"color: #008000;\">EXAMPLE:<\/span> <a href=\"https:\/\/www.youtube.com\/watch?v=nikSJceTxJ0&amp;t=1023s\" target=\"_blank\" rel=\"noopener\"><strong>A simple model assumes a continent as a solid block of rock<\/strong><\/a> (with a density <span class=\"katex-eq\" data-katex-display=\"false\">=2800[kg\/m^3]<\/span>) floating on the surrounding Earth&#8217;s mantle (with a density <span class=\"katex-eq\" data-katex-display=\"false\">=3300[kg\/m^3]<\/span>). Assuming the continent has an average thickness of 35[km], calculate the average height that emerges above the mantle.<\/p>\n<p style=\"text-align: justify;\">SOLUTION:<\/p>\n<p style=\"text-align: justify;\">The percentage of the body submerged will be:<\/p>\n<p style=\"text-align: justify;\">Submerged percentage <span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle = \\frac{\\rho_{body}}{\\rho_{fluid}}<\/span>\n<p style=\"text-align: justify;\">Therefore, the percentage of the body that is not submerged and that emerges above the mantle will be:<\/p>\n<p style=\"text-align: justify;\">Emerging percentage <span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle = 1 - \\frac{\\rho_{body}}{\\rho_{fluid}} = 1 - \\frac{2800}{3300} \\approx 0,15 = 15\\%<\/span>\n<p style=\"text-align: justify;\">Since the average thickness is <span class=\"katex-eq\" data-katex-display=\"false\">35[km]<\/span>, what emerges on average will be <span class=\"katex-eq\" data-katex-display=\"false\">15\\% 35[km]\\approx 5.3 [km].<\/span>\n","protected":false},"excerpt":{"rendered":"<p>Buoyancy and Archimedes&#8217; Principle Summary: This lesson will explain the phenomenon of Buoyancy and Archimedes&#8217; Principle, showing how submerged objects in a fluid experience an upward force equal to the weight of the displaced fluid. This principle is used to calculate the portion of an object that emerges above the fluid, with practical examples. Learning [&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":11,"footnotes":""},"categories":[889,635],"tags":[],"class_list":["post-29251","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-fluid-mechanics","category-physics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Buoyancy and Archimedes&#039; Principle - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Discover how Buoyancy and Archimedes&#039; Principle explain why objects float or sink. Learn to calculate the buoyant force with practical examples that will surprise you. 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