{"id":2103,"date":"2014-12-11T02:29:29","date_gmt":"2014-12-11T02:29:29","guid":{"rendered":"https:\/\/courses.candelalearning.com\/colphysics\/?post_type=chapter&#038;p=2103"},"modified":"2016-11-03T18:21:48","modified_gmt":"2016-11-03T18:21:48","slug":"11-4-variation-of-pressure-with-depth-in-a-fluid","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/chapter\/11-4-variation-of-pressure-with-depth-in-a-fluid\/","title":{"raw":"Variation of Pressure with Depth in a Fluid","rendered":"Variation of Pressure with Depth in a Fluid"},"content":{"raw":"<div>\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<div>\r\n<ul>\r\n\t<li>Define pressure in terms of weight.<\/li>\r\n\t<li>Explain the variation of pressure with depth in a fluid.<\/li>\r\n\t<li>Calculate density given pressure and altitude.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nIf your ears have ever popped on a plane flight or ached during a deep dive in a swimming pool, you have experienced the effect of depth on pressure in a fluid. At the Earth\u2019s surface, the air pressure exerted on you is a result of the weight of air above you. This pressure is reduced as you climb up in altitude and the weight of air above you decreases. Under water, the pressure exerted on you increases with increasing depth. In this case, the pressure being exerted upon you is a result of both the weight of water above you <em>and<\/em> that of the atmosphere above you. You may notice an air pressure change on an elevator ride that transports you many stories, but you need only dive a meter or so below the surface of a pool to feel a pressure increase. The difference is that water is much denser than air, about 775 times as dense. Consider the container in Figure 1.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"225\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20103719\/Figure_12_04_01a.jpg\" alt=\"A container with fluid filled to a depth h. The fluid\u2019s weight w equal to m times g is shown by an arrow pointing downward. A denotes the area of the fluid at the bottom of the container and as well as on the surface.\" width=\"225\" height=\"398\" \/> Figure 1. The bottom of this container supports the entire weight of the fluid in it. The vertical sides cannot exert an upward force on the fluid (since it cannot withstand a shearing force), and so the bottom must support it all.[\/caption]\r\n\r\nIts bottom supports the weight of the fluid in it. Let us calculate the pressure exerted on the bottom by the weight of the fluid. That <em> pressure<\/em> is the weight of the fluid <em>mg\u00a0<\/em>divided by the area <em>A<\/em> supporting it (the area of the bottom of the container):\r\n<p style=\"text-align: center;\">[latex]P=\\frac{mg}{A}\\\\[\/latex].<\/p>\r\nWe can find the mass of the fluid from its volume and density:\r\n<div style=\"text-align: center;\" title=\"Equation 11.14.\"><em>m\u00a0<\/em>=\u00a0<em>\u03c1V<\/em>.<\/div>\r\nThe volume of the fluid <em>V<\/em> is related to the dimensions of the container. It is\r\n<p style=\"text-align: center;\"><em>V = Ah<\/em>,<\/p>\r\nwhere <em>A<\/em> is the cross-sectional area and <em>h<\/em> is the depth. Combining the last two equations gives\r\n<p style=\"text-align: center;\">[latex]m=\\rho Ah\\\\[\/latex].<\/p>\r\nIf we enter this into the expression for pressure, we obtain\r\n<p style=\"text-align: center;\">[latex]P=\\frac{\\left(\\rho{Ah}\\right)g}{A}\\\\[\/latex].<\/p>\r\nThe area cancels, and rearranging the variables yields\r\n<div style=\"text-align: center;\" title=\"Equation 11.18.\"><em>P\u00a0<\/em>=\u00a0<em>h\u03c1g<\/em>.<\/div>\r\n&nbsp;\r\n\r\nThis value is the <em>pressure due to the weight of a fluid<\/em>. The equation has general validity beyond the special conditions under which it is derived here. Even if the container were not there, the surrounding fluid would still exert this pressure, keeping the fluid static. Thus the equation <em>P\u00a0<\/em>=\u00a0<em>h\u03c1g<\/em> represents the pressure due to the weight of any fluid of <em>average density\u00a0<\/em><em>\u03c1<\/em> at any depth <em>h<\/em> below its surface. For liquids, which are nearly incompressible, this equation holds to great depths. For gases, which are quite compressible, one can apply this equation as long as the density changes are small over the depth considered. <em>Example 2: Calculating Average Density: How Dense Is the Air?<\/em>\u00a0illustrates this situation.\r\n<div title=\"Example 11.3. Calculating the Average Pressure and Force Exerted: What Force Must a Dam Withstand?\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 1. Calculating the Average Pressure and Force Exerted: What Force Must a Dam Withstand?<\/h3>\r\n<div>\r\n\r\nIn <a href=\".\/chapter\/11-2-density\/\" target=\"_blank\">Example 1. Calculating the Mass of a Reservoir from Its Volume<\/a><em>,<\/em> we calculated the mass of water in a large reservoir. We will now consider the pressure and force acting on the dam retaining water. (See Figure 2.) The dam is 500 m wide, and the water is 80.0 m deep at the dam. (a) What is the average pressure on the dam due to the water? (b) Calculate the force exerted against the dam and compare it with the weight of water in the dam (previously found to be 1.96 \u00d7 10<sup>13<\/sup> N).\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"300\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20103725\/Figure_12_04_02a.jpg\" alt=\"A two-dimensional view of a dam with dimensions L and h is shown. Force F at h is shown by a horizontal arrow. The force F exerted by water on the dam is F equals average pressure p bar into area A and pressure in turn is average height h bar into density rho into acceleration due to gravity g.\" width=\"300\" height=\"430\" \/> Figure 2. The dam must withstand the force exerted against it by the water it retains. This force is small compared with the weight of the water behind the dam.[\/caption]\r\n<h4><strong>Strategy for (a)<\/strong><\/h4>\r\nThe average pressure [latex]\\overline{P}\\\\[\/latex]\u00a0due to the weight of the water is the pressure at the average depth [latex]\\overline{h}\\\\[\/latex]\u00a0of 40.0 m, since pressure increases linearly with depth.\r\n<h4><strong>Solution for (a)<\/strong><\/h4>\r\nThe average pressure due to the weight of a fluid is\r\n<p style=\"text-align: center;\">[latex]\\overline{P}=\\overline{h}rho g\\\\[\/latex].<\/p>\r\nEntering the density of water from Table 1\u00a0and taking [latex]\\overline{h}\\\\[\/latex]\u00a0to be the average depth of 40.0 m, we obtain\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}\\overline{P}&amp; =&amp; \\left(\\text{40.0 m}\\right)\\left({\\text{10}}^{3}\\frac{\\text{kg}}{{\\text{m}}^{3}}\\right)\\left(9.80\\frac{\\text{m}}{{\\text{s}}^{2}}\\right)\\\\ &amp; =&amp; 3.92\\times {\\text{10}}^{5}\\frac{\\text{N}}{{\\text{m}}^{2}}=\\text{392 kPa.}\\end{array}\\\\[\/latex]<\/p>\r\n\r\n<h4><strong>Strategy for (b)<\/strong><\/h4>\r\nThe force exerted on the dam by the water is the average pressure times the area of contact:\r\n<p style=\"text-align: center;\">[latex]F=\\overline{P}A\\\\[\/latex].<\/p>\r\n\r\n<h4><strong>Solution for (b)<\/strong><\/h4>\r\nWe have already found the value for [latex]\\overline{P}\\\\[\/latex]. The area of the dam is A = 80.0 m x 500 m = 4.00 \u00d7 10<sup>4<\/sup> m<sup>2<\/sup>, so that\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}F&amp; =&amp; \\left(3.92\\times {\\text{10}}^{5}{\\text{N\/m}}^{2}\\right)\\left(4.00\\times {\\text{10}}^{4}{\\text{m}}^{2}\\right)\\ &amp; =&amp; 1.57\\times {\\text{10}}^{\\text{10}}\\text{N.}\\end{array}\\\\[\/latex]<\/p>\r\n\r\n<div title=\"Equation 11.22.\"><\/div>\r\n<strong>Discussion<\/strong>\r\n\r\nAlthough this force seems large, it is small compared with the 1.96 \u00d7 10<sup>13 <\/sup>N\u00a0weight of the water in the reservoir\u2014in fact, it is only 0.0800% of the weight. Note that the pressure found in part (a) is completely independent of the width and length of the lake\u2014it depends only on its average depth at the dam. Thus the force depends only on the water\u2019s average depth and the dimensions of the dam, <em>not<\/em> on the horizontal extent of the reservoir. In the diagram, the thickness of the dam increases with depth to balance the increasing force due to the increasing pressure depth to balance the increasing force due to the increasing pressure.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div title=\"Figure 11.11.\">\r\n<table summary=\"The table shows the value of density in units of kilogram per meter cubed for certain solids, liquids, and gases.\" cellspacing=\"0\" cellpadding=\"0\"><caption>Table 1. Densities of Various Substances<\/caption>\r\n<thead>\r\n<tr>\r\n<th>Substance<\/th>\r\n<th><strong>[latex]\\rho \\left({\\text{10}}^{3}{\\text{kg\/m}}^{3}\\text{or}\\text{g\/mL}\\right)\\\\[\/latex]<\/strong><\/th>\r\n<th>Substance<\/th>\r\n<th>[latex]\\rho \\left({\\text{10}}^{3}{\\text{kg\/m}}^{3}\\text{or}\\text{g\/mL}\\right)\\\\[\/latex]<\/th>\r\n<th>Substance<\/th>\r\n<th>[latex]\\rho \\left({\\text{10}}^{3}{\\text{kg\/m}}^{3}\\text{or}\\text{g\/mL}\\right)\\\\[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td><strong>Solids<\/strong><\/td>\r\n<td><strong>Liquids<\/strong><\/td>\r\n<td><strong>Gases<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Aluminum<\/td>\r\n<td>2.7<\/td>\r\n<td>Water (4\u00baC)<\/td>\r\n<td>1.000<\/td>\r\n<td>Air<\/td>\r\n<td>1.29 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Brass<\/td>\r\n<td>8.44<\/td>\r\n<td>Blood<\/td>\r\n<td>1.05<\/td>\r\n<td>Carbon dioxide<\/td>\r\n<td>1.98 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Copper (average)<\/td>\r\n<td>8.8<\/td>\r\n<td>Sea water<\/td>\r\n<td>1.025<\/td>\r\n<td>Carbon monoxide<\/td>\r\n<td>1.25 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Gold<\/td>\r\n<td>19.32<\/td>\r\n<td>Mercury<\/td>\r\n<td>13.6<\/td>\r\n<td>Hydrogen<\/td>\r\n<td>0.090 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Iron or steel<\/td>\r\n<td>7.8<\/td>\r\n<td>Ethyl alcohol<\/td>\r\n<td>0.79<\/td>\r\n<td>Helium<\/td>\r\n<td>0.18 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Lead<\/td>\r\n<td>11.3<\/td>\r\n<td>Petrol<\/td>\r\n<td>0.68<\/td>\r\n<td>Methane<\/td>\r\n<td>0.72 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Polystyrene<\/td>\r\n<td>0.10<\/td>\r\n<td>Glycerin<\/td>\r\n<td>1.26<\/td>\r\n<td>Nitrogen<\/td>\r\n<td>1.25 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Tungsten<\/td>\r\n<td>19.30<\/td>\r\n<td>Olive oil<\/td>\r\n<td>0.92<\/td>\r\n<td>Nitrous oxide<\/td>\r\n<td>1.98 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Uranium<\/td>\r\n<td>18.70<\/td>\r\n<td>Oxygen<\/td>\r\n<td>1.43 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Concrete<\/td>\r\n<td>2.30\u20133.0<\/td>\r\n<td>Steam (100\u00ba C)<\/td>\r\n<td>0.60 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Cork<\/td>\r\n<td>0.24<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Glass, common (average)<\/td>\r\n<td>2.6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Granite<\/td>\r\n<td>2.7<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Earth\u2019s crust<\/td>\r\n<td>3.3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Wood<\/td>\r\n<td>0.3\u20130.9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Ice (0\u00b0C)<\/td>\r\n<td>0.917<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Bone<\/td>\r\n<td>1.7\u20132.0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<em>Atmospheric pressure<\/em> is another example of pressure due to the weight of a fluid, in this case due to the weight of <em>air<\/em> above a given height. The atmospheric pressure at the Earth\u2019s surface varies a little due to the large-scale flow of the atmosphere induced by the Earth\u2019s rotation (this creates weather \u201chighs\u201d and \u201clows\u201d). However, the average pressure at sea level is given by the <em>standard atmospheric pressure\u00a0<\/em><em>P<\/em><sub>atm<\/sub>, measured to be\r\n<p style=\"text-align: center;\">[latex]\\text{1 atmosphere (atm)}={P}_{\\text{atm}}=1.01\\times {\\text{10}}^{5}{\\text{N\/m}}^{2}=\\text{101 kPa}\\\\[\/latex].<\/p>\r\nThis relationship means that, on average, at sea level, a column of air above 1.00 m<sup>2<\/sup>\u00a0of the Earth\u2019s surface has a weight of 1.01 \u00d7 10<sup>5<\/sup> N, equivalent to 1 atm. (See Figure 3.)\r\n<div title=\"Figure 11.12.\">\r\n<div>\r\n<div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"250\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20103727\/Figure_12_04_03a.jpg\" alt=\"Figure shows a column of air exerting a weight of one point zero one times ten to the power five newtons on a rectangular patch of ground of one square meter cross section.\" width=\"250\" height=\"458\" \/> Figure 3. Atmospheric pressure at sea level averages 1.01 \u00d7 10<sup>5<\/sup> Pa (equivalent to 1 atm), since the column of air over this 1 m<sup>2<\/sup>, extending to the top of the atmosphere, weighs 1.01 \u00d7 10<sup>5<\/sup>.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div title=\"Example 11.4. Calculating Average Density: How Dense Is the Air?\">\r\n<div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2. Calculating Average Density: How Dense Is the Air?<\/h3>\r\n<div>\r\n\r\nCalculate the average density of the atmosphere, given that it extends to an altitude of 120 km. Compare this density with that of air listed in\u00a0Table 1.\r\n<h4><strong>Strategy<\/strong><\/h4>\r\nIf we solve <em>P\u00a0<\/em>=\u00a0<em>h\u03c1g<\/em> for density, we see that\r\n<p style=\"text-align: center;\">[latex]\\overline{\\rho }=\\frac{P}{\\text{hg}}\\\\[\/latex].<\/p>\r\nWe then take <em>P<\/em> to be atmospheric pressure, <em>h<\/em> is given, and <em>g<\/em> is known, and so we can use this to calculate [latex]\\overline{\\rho}\\\\[\/latex].\r\n<h4><strong>Solution<\/strong><\/h4>\r\nEntering known values into the expression for [latex]\\overline{\\rho}\\\\[\/latex]\u00a0yields\r\n<div style=\"text-align: center;\" title=\"Equation 11.25.\">[latex]\\overline{\\rho }=\\frac{1\\text{.}\\text{01}\\times {\\text{10}}^{5}{\\text{N\/m}}^{2}}{\\left(\\text{120}\\times {\\text{10}}^{3}\\text{m}\\right)\\left(9\\text{.}\\text{80}{\\text{m\/s}}^{2}\\right)}=8\\text{.}\\text{59}\\times {\\text{10}}^{-2}{\\text{kg\/m}}^{3}\\\\[\/latex].<\/div>\r\n<h4><strong>Discussion<\/strong><\/h4>\r\nThis result is the average density of air between the Earth\u2019s surface and the top of the Earth\u2019s atmosphere, which essentially ends at 120 km. The density of air at sea level is given in\u00a0Table 1\u00a0as 1.29 kg.m<sup>3<\/sup>\u2014about 15 times its average value. Because air is so compressible, its density has its highest value near the Earth\u2019s surface and declines rapidly with altitude.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div title=\"Example 11.5. Calculating Depth Below the Surface of Water: What Depth of Water Creates the Same Pressure as the Entire Atmosphere?\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 3. Calculating Depth Below the Surface of Water: What Depth of Water Creates the Same Pressure as the Entire Atmosphere?<\/h3>\r\n<div>\r\n\r\nCalculate the depth below the surface of water at which the pressure due to the weight of the water equals 1.00 atm.\r\n<h4><strong>Strategy<\/strong><\/h4>\r\nWe begin by solving the equation <em>P\u00a0<\/em>=\u00a0<em>h\u03c1g<\/em> for depth <em>h<\/em>:\r\n<p style=\"text-align: center;\">[latex]h=\\frac{P}{\\mathrm{\\rho g}}\\\\[\/latex].<\/p>\r\nThen we take <em>P<\/em> to be 1.00 atm and <em>\u03c1<\/em> to be the density of the water that creates the pressure.\r\n<h4><strong>Solution<\/strong><\/h4>\r\nEntering the known values into the expression for <em>h<\/em> gives\r\n<p style=\"text-align: center;\">[latex]h=\\frac{1\\text{.}\\text{01}\\times {\\text{10}}^{5}{\\text{N\/m}}^{2}}{\\left(1\\text{.}\\text{00}\\times {\\text{10}}^{3}{\\text{kg\/m}}^{3}\\right)\\left(9\\text{.}\\text{80}{\\text{m\/s}}^{2}\\right)}=\\text{10}\\text{.}3\\text{m}\\\\[\/latex].<\/p>\r\n\r\n<h4><strong>Discussion<\/strong><\/h4>\r\nJust 10.3 m of water creates the same pressure as 120 km of air. Since water is nearly incompressible, we can neglect any change in its density over this depth.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div title=\"Example 11.5. Calculating Depth Below the Surface of Water: What Depth of Water Creates the Same Pressure as the Entire Atmosphere?\">What do you suppose is the <em>total<\/em> pressure at a depth of 10.3 m in a swimming pool? Does the atmospheric pressure on the water\u2019s surface affect the pressure below? The answer is yes. This seems only logical, since both the water\u2019s weight and the atmosphere\u2019s weight must be supported. So the <em>total<\/em> pressure at a depth of 10.3 m is 2 atm\u2014half from the water above and half from the air above. We shall see in <a href=\".\/chapter\/11-5-pascals-principle\/\" target=\"_blank\">Pascal\u2019s Principle<\/a> that fluid pressures always add in this way.<\/div>\r\n<h2>Section Summary<\/h2>\r\n<ul>\r\n\t<li>Pressure is the weight of the fluid <em>mg<\/em>\u00a0divided by the area <em>A<\/em> supporting it (the area of the bottom of the container):\r\n<div style=\"text-align: center;\">[latex]P=\\frac{\\text{mg}}{A}\\\\[\/latex].<\/div><\/li>\r\n\t<li>Pressure due to the weight of a liquid is given by\r\n<div style=\"text-align: center;\">[latex]P=h\\rho g\\\\[\/latex],<\/div>\r\nwhere<em> P<\/em> is the pressure, <em>h<\/em>\u00a0is the height of the liquid, <em>\u03c1<\/em> is the density of the liquid, and <em>g<\/em>\u00a0is the acceleration due to gravity.<\/li>\r\n<\/ul>\r\n<section>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Conceptual Questions<\/h3>\r\n<div>\r\n<div>\r\n\r\n1. Atmospheric pressure exerts a large force (equal to the weight of the atmosphere above your body\u2014about 10 tons) on the top of your body when you are lying on the beach sunbathing. Why are you able to get up?\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n2. Why does atmospheric pressure decrease more rapidly than linearly with altitude?\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n3. What are two reasons why mercury rather than water is used in barometers?\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n\r\n4. Figure 4 shows how sandbags placed around a leak outside a river levee can effectively stop the flow of water under the levee. Explain how the small amount of water inside the column formed by the sandbags is able to balance the much larger body of water behind the levee.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"292\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03210436\/Figure_12_04_04a.jpg\" alt=\"The figure shows a flooding river on the extreme right, with a levee set up on its left, and sandbags are stacked on the left of the levee. The height of the levee and that of the stacked sandbags is greater than the water level of the flooding river, so the water does not flow over their tops, but a leak under the levee allows some water to flow under it and reach the sandbags.\" width=\"292\" height=\"160\" \/> Figure 4. Because the river level is very high, it has started to leak under the levee. Sandbags are placed around the leak, and the water held by them rises until it is the same level as the river, at which point the water there stops rising.[\/caption]\r\n\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n5. Why is it difficult to swim under water in the Great Salt Lake?\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n6. Is there a net force on a dam due to atmospheric pressure? Explain your answer.\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n7. Does atmospheric pressure add to the gas pressure in a rigid tank? In a toy balloon? When, in general, does atmospheric pressure <em>not<\/em> affect the total pressure in a fluid?\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n8. You can break a strong wine bottle by pounding a cork into it with your fist, but the cork must press directly against the liquid filling the bottle\u2014there can be no air between the cork and liquid. Explain why the bottle breaks, and why it will not if there is air between the cork and liquid.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section>\r\n<div class=\"textbox exercises\">\r\n<h3>Problems &amp; Exercises<\/h3>\r\n<div>\r\n<div>\r\n\r\n1. What depth of mercury creates a pressure of 1.00 atm?\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n2. The greatest ocean depths on the Earth are found in the Marianas Trench near the Philippines. Calculate the pressure due to the ocean at the bottom of this trench, given its depth is 11.0 km and assuming the density of seawater is constant all the way down.\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n3. Verify that the SI unit of <em>h\u03c1g<\/em>\u00a0is N\/m<sup>2<\/sup>.\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n4. Water towers store water above the level of consumers for times of heavy use, eliminating the need for high-speed pumps. How high above a user must the water level be to create a gauge pressure of 3.00 \u00d7 10<sup>5<\/sup> N\/m<sup>2<\/sup>?\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n5. The aqueous humor in a person\u2019s eye is exerting a force of 0.300 N on the 1.10-cm<sup>2<\/sup>\u00a0area of the cornea. (a) What pressure is this in mm Hg? (b) Is this value within the normal range for pressures in the eye?\r\n\r\n<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n6. How much force is exerted on one side of an 8.50 cm by 11.0 cm sheet of paper by the atmosphere? How can the paper withstand such a force?\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n7. What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long gas tank that can hold 50.0 kg of gasoline by the weight of the gasoline in it when it is full?\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n8. Calculate the average pressure exerted on the palm of a shot-putter\u2019s hand by the shot if the area of contact is 50.0 cm<sup>2<\/sup>\u00a0and he exerts a force of 800 N on it. Express the pressure in N\/m<sup>2<\/sup> and compare it with the 1.00 \u00d7 10<sup>6<\/sup> pressures sometimes encountered in the skeletal system.\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n\r\n9. The left side of the heart creates a pressure of 120 mm Hg by exerting a force directly on the blood over an effective area of 15.0 cm<sup>2<\/sup>. What force does it exert to accomplish this?\r\n\r\n<\/div>\r\n<div>\r\n\r\n10. Show that the total force on a rectangular dam due to the water behind it increases with the <em>square<\/em> of the water depth. In particular, show that this force is given by [latex]F=\\rho {\\mathrm{gh}}^{2}L\/2\\\\[\/latex], where [latex]rho\\\\[\/latex] is the density of water,\u00a0<em>h<\/em> is its depth at the dam, and <em>L<\/em> is the length of the dam. You may assume the face of the dam is vertical. (Hint: Calculate the average pressure exerted and multiply this by the area in contact with the water. (See Figure 5.)\r\n\r\n<\/div>\r\n<\/div>\r\n<div><figure>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"283\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03210434\/Figure_12_04_02a.jpg\" alt=\"A two-dimensional view of a dam with dimensions L and h is shown. Force F at h is shown by a horizontal arrow. The force F exerted by water on the dam is F equals average pressure p bar into area A and pressure in turn is average height h bar into density rho into acceleration due to gravity g.\" width=\"283\" height=\"203\" \/> Figure 5.[\/caption]\r\n\r\n<\/figure><\/div>\r\n<\/div>\r\n<\/section>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl><dt>pressure:<\/dt><dd>the weight of the fluid divided by the area supporting it<\/dd><\/dl>\r\n<div class=\"textbox exercises\">\r\n<h3>Selected Solutions to Problems &amp; Exercises<\/h3>\r\n1.\u00a00.760 m\r\n\r\n3.\u00a0[latex]\\begin{array}{lll}{\\left(h\\rho g\\right)}_{\\text{units}}&amp; =&amp; \\left(\\text{m}\\right)\\left({\\text{kg\/m}}^{3}\\right)\\left({\\text{m\/s}}^{2}\\right)=\\left(\\text{kg}\\cdot {\\text{m}}^{2}\\right)\/\\left({\\text{m}}^{3}\\cdot {\\text{s}}^{2}\\right)\\\\ &amp; =&amp; \\left(\\text{kg}\\cdot {\\text{m\/s}}^{2}\\right)\\left({\\text{1\/m}}^{2}\\right)\\\\ &amp; =&amp; {\\text{N\/m}}^{2}\\end{array}\\\\[\/latex].\r\n\r\n5. (a) 20.5 mm Hg\u00a0(b) The range of pressures in the eye is 12\u201324 mm Hg, so the result in part (a) is within that range\r\n\r\n7. 1.09 \u00d7 10<sup>3<\/sup> N\/m<sup>2<\/sup>\r\n\r\n9.\u00a024.0 N\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>","rendered":"<div>\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<div>\n<ul>\n<li>Define pressure in terms of weight.<\/li>\n<li>Explain the variation of pressure with depth in a fluid.<\/li>\n<li>Calculate density given pressure and altitude.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<p>If your ears have ever popped on a plane flight or ached during a deep dive in a swimming pool, you have experienced the effect of depth on pressure in a fluid. At the Earth\u2019s surface, the air pressure exerted on you is a result of the weight of air above you. This pressure is reduced as you climb up in altitude and the weight of air above you decreases. Under water, the pressure exerted on you increases with increasing depth. In this case, the pressure being exerted upon you is a result of both the weight of water above you <em>and<\/em> that of the atmosphere above you. You may notice an air pressure change on an elevator ride that transports you many stories, but you need only dive a meter or so below the surface of a pool to feel a pressure increase. The difference is that water is much denser than air, about 775 times as dense. Consider the container in Figure 1.<\/p>\n<div style=\"width: 235px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20103719\/Figure_12_04_01a.jpg\" alt=\"A container with fluid filled to a depth h. The fluid\u2019s weight w equal to m times g is shown by an arrow pointing downward. A denotes the area of the fluid at the bottom of the container and as well as on the surface.\" width=\"225\" height=\"398\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The bottom of this container supports the entire weight of the fluid in it. The vertical sides cannot exert an upward force on the fluid (since it cannot withstand a shearing force), and so the bottom must support it all.<\/p>\n<\/div>\n<p>Its bottom supports the weight of the fluid in it. Let us calculate the pressure exerted on the bottom by the weight of the fluid. That <em> pressure<\/em> is the weight of the fluid <em>mg\u00a0<\/em>divided by the area <em>A<\/em> supporting it (the area of the bottom of the container):<\/p>\n<p style=\"text-align: center;\">[latex]P=\\frac{mg}{A}\\\\[\/latex].<\/p>\n<p>We can find the mass of the fluid from its volume and density:<\/p>\n<div style=\"text-align: center;\" title=\"Equation 11.14.\"><em>m\u00a0<\/em>=\u00a0<em>\u03c1V<\/em>.<\/div>\n<p>The volume of the fluid <em>V<\/em> is related to the dimensions of the container. It is<\/p>\n<p style=\"text-align: center;\"><em>V = Ah<\/em>,<\/p>\n<p>where <em>A<\/em> is the cross-sectional area and <em>h<\/em> is the depth. Combining the last two equations gives<\/p>\n<p style=\"text-align: center;\">[latex]m=\\rho Ah\\\\[\/latex].<\/p>\n<p>If we enter this into the expression for pressure, we obtain<\/p>\n<p style=\"text-align: center;\">[latex]P=\\frac{\\left(\\rho{Ah}\\right)g}{A}\\\\[\/latex].<\/p>\n<p>The area cancels, and rearranging the variables yields<\/p>\n<div style=\"text-align: center;\" title=\"Equation 11.18.\"><em>P\u00a0<\/em>=\u00a0<em>h\u03c1g<\/em>.<\/div>\n<p>&nbsp;<\/p>\n<p>This value is the <em>pressure due to the weight of a fluid<\/em>. The equation has general validity beyond the special conditions under which it is derived here. Even if the container were not there, the surrounding fluid would still exert this pressure, keeping the fluid static. Thus the equation <em>P\u00a0<\/em>=\u00a0<em>h\u03c1g<\/em> represents the pressure due to the weight of any fluid of <em>average density\u00a0<\/em><em>\u03c1<\/em> at any depth <em>h<\/em> below its surface. For liquids, which are nearly incompressible, this equation holds to great depths. For gases, which are quite compressible, one can apply this equation as long as the density changes are small over the depth considered. <em>Example 2: Calculating Average Density: How Dense Is the Air?<\/em>\u00a0illustrates this situation.<\/p>\n<div title=\"Example 11.3. Calculating the Average Pressure and Force Exerted: What Force Must a Dam Withstand?\">\n<div class=\"textbox examples\">\n<h3>Example 1. Calculating the Average Pressure and Force Exerted: What Force Must a Dam Withstand?<\/h3>\n<div>\n<p>In <a href=\".\/chapter\/11-2-density\/\" target=\"_blank\">Example 1. Calculating the Mass of a Reservoir from Its Volume<\/a><em>,<\/em> we calculated the mass of water in a large reservoir. We will now consider the pressure and force acting on the dam retaining water. (See Figure 2.) The dam is 500 m wide, and the water is 80.0 m deep at the dam. (a) What is the average pressure on the dam due to the water? (b) Calculate the force exerted against the dam and compare it with the weight of water in the dam (previously found to be 1.96 \u00d7 10<sup>13<\/sup> N).<\/p>\n<div style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20103725\/Figure_12_04_02a.jpg\" alt=\"A two-dimensional view of a dam with dimensions L and h is shown. Force F at h is shown by a horizontal arrow. The force F exerted by water on the dam is F equals average pressure p bar into area A and pressure in turn is average height h bar into density rho into acceleration due to gravity g.\" width=\"300\" height=\"430\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The dam must withstand the force exerted against it by the water it retains. This force is small compared with the weight of the water behind the dam.<\/p>\n<\/div>\n<h4><strong>Strategy for (a)<\/strong><\/h4>\n<p>The average pressure [latex]\\overline{P}\\\\[\/latex]\u00a0due to the weight of the water is the pressure at the average depth [latex]\\overline{h}\\\\[\/latex]\u00a0of 40.0 m, since pressure increases linearly with depth.<\/p>\n<h4><strong>Solution for (a)<\/strong><\/h4>\n<p>The average pressure due to the weight of a fluid is<\/p>\n<p style=\"text-align: center;\">[latex]\\overline{P}=\\overline{h}rho g\\\\[\/latex].<\/p>\n<p>Entering the density of water from Table 1\u00a0and taking [latex]\\overline{h}\\\\[\/latex]\u00a0to be the average depth of 40.0 m, we obtain<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}\\overline{P}& =& \\left(\\text{40.0 m}\\right)\\left({\\text{10}}^{3}\\frac{\\text{kg}}{{\\text{m}}^{3}}\\right)\\left(9.80\\frac{\\text{m}}{{\\text{s}}^{2}}\\right)\\\\ & =& 3.92\\times {\\text{10}}^{5}\\frac{\\text{N}}{{\\text{m}}^{2}}=\\text{392 kPa.}\\end{array}\\\\[\/latex]<\/p>\n<h4><strong>Strategy for (b)<\/strong><\/h4>\n<p>The force exerted on the dam by the water is the average pressure times the area of contact:<\/p>\n<p style=\"text-align: center;\">[latex]F=\\overline{P}A\\\\[\/latex].<\/p>\n<h4><strong>Solution for (b)<\/strong><\/h4>\n<p>We have already found the value for [latex]\\overline{P}\\\\[\/latex]. The area of the dam is A = 80.0 m x 500 m = 4.00 \u00d7 10<sup>4<\/sup> m<sup>2<\/sup>, so that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}F& =& \\left(3.92\\times {\\text{10}}^{5}{\\text{N\/m}}^{2}\\right)\\left(4.00\\times {\\text{10}}^{4}{\\text{m}}^{2}\\right)\\ & =& 1.57\\times {\\text{10}}^{\\text{10}}\\text{N.}\\end{array}\\\\[\/latex]<\/p>\n<div title=\"Equation 11.22.\"><\/div>\n<p><strong>Discussion<\/strong><\/p>\n<p>Although this force seems large, it is small compared with the 1.96 \u00d7 10<sup>13 <\/sup>N\u00a0weight of the water in the reservoir\u2014in fact, it is only 0.0800% of the weight. Note that the pressure found in part (a) is completely independent of the width and length of the lake\u2014it depends only on its average depth at the dam. Thus the force depends only on the water\u2019s average depth and the dimensions of the dam, <em>not<\/em> on the horizontal extent of the reservoir. In the diagram, the thickness of the dam increases with depth to balance the increasing force due to the increasing pressure depth to balance the increasing force due to the increasing pressure.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div title=\"Figure 11.11.\">\n<table summary=\"The table shows the value of density in units of kilogram per meter cubed for certain solids, liquids, and gases.\" cellpadding=\"0\" style=\"border-spacing: 0px;\">\n<caption>Table 1. Densities of Various Substances<\/caption>\n<thead>\n<tr>\n<th>Substance<\/th>\n<th><strong>[latex]\\rho \\left({\\text{10}}^{3}{\\text{kg\/m}}^{3}\\text{or}\\text{g\/mL}\\right)\\\\[\/latex]<\/strong><\/th>\n<th>Substance<\/th>\n<th>[latex]\\rho \\left({\\text{10}}^{3}{\\text{kg\/m}}^{3}\\text{or}\\text{g\/mL}\\right)\\\\[\/latex]<\/th>\n<th>Substance<\/th>\n<th>[latex]\\rho \\left({\\text{10}}^{3}{\\text{kg\/m}}^{3}\\text{or}\\text{g\/mL}\\right)\\\\[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td><strong>Solids<\/strong><\/td>\n<td><strong>Liquids<\/strong><\/td>\n<td><strong>Gases<\/strong><\/td>\n<\/tr>\n<tr>\n<td>Aluminum<\/td>\n<td>2.7<\/td>\n<td>Water (4\u00baC)<\/td>\n<td>1.000<\/td>\n<td>Air<\/td>\n<td>1.29 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<tr>\n<td>Brass<\/td>\n<td>8.44<\/td>\n<td>Blood<\/td>\n<td>1.05<\/td>\n<td>Carbon dioxide<\/td>\n<td>1.98 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<tr>\n<td>Copper (average)<\/td>\n<td>8.8<\/td>\n<td>Sea water<\/td>\n<td>1.025<\/td>\n<td>Carbon monoxide<\/td>\n<td>1.25 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<tr>\n<td>Gold<\/td>\n<td>19.32<\/td>\n<td>Mercury<\/td>\n<td>13.6<\/td>\n<td>Hydrogen<\/td>\n<td>0.090 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<tr>\n<td>Iron or steel<\/td>\n<td>7.8<\/td>\n<td>Ethyl alcohol<\/td>\n<td>0.79<\/td>\n<td>Helium<\/td>\n<td>0.18 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<tr>\n<td>Lead<\/td>\n<td>11.3<\/td>\n<td>Petrol<\/td>\n<td>0.68<\/td>\n<td>Methane<\/td>\n<td>0.72 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<tr>\n<td>Polystyrene<\/td>\n<td>0.10<\/td>\n<td>Glycerin<\/td>\n<td>1.26<\/td>\n<td>Nitrogen<\/td>\n<td>1.25 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<tr>\n<td>Tungsten<\/td>\n<td>19.30<\/td>\n<td>Olive oil<\/td>\n<td>0.92<\/td>\n<td>Nitrous oxide<\/td>\n<td>1.98 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<tr>\n<td>Uranium<\/td>\n<td>18.70<\/td>\n<td>Oxygen<\/td>\n<td>1.43 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<tr>\n<td>Concrete<\/td>\n<td>2.30\u20133.0<\/td>\n<td>Steam (100\u00ba C)<\/td>\n<td>0.60 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<tr>\n<td>Cork<\/td>\n<td>0.24<\/td>\n<\/tr>\n<tr>\n<td>Glass, common (average)<\/td>\n<td>2.6<\/td>\n<\/tr>\n<tr>\n<td>Granite<\/td>\n<td>2.7<\/td>\n<\/tr>\n<tr>\n<td>Earth\u2019s crust<\/td>\n<td>3.3<\/td>\n<\/tr>\n<tr>\n<td>Wood<\/td>\n<td>0.3\u20130.9<\/td>\n<\/tr>\n<tr>\n<td>Ice (0\u00b0C)<\/td>\n<td>0.917<\/td>\n<\/tr>\n<tr>\n<td>Bone<\/td>\n<td>1.7\u20132.0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p><em>Atmospheric pressure<\/em> is another example of pressure due to the weight of a fluid, in this case due to the weight of <em>air<\/em> above a given height. The atmospheric pressure at the Earth\u2019s surface varies a little due to the large-scale flow of the atmosphere induced by the Earth\u2019s rotation (this creates weather \u201chighs\u201d and \u201clows\u201d). However, the average pressure at sea level is given by the <em>standard atmospheric pressure\u00a0<\/em><em>P<\/em><sub>atm<\/sub>, measured to be<\/p>\n<p style=\"text-align: center;\">[latex]\\text{1 atmosphere (atm)}={P}_{\\text{atm}}=1.01\\times {\\text{10}}^{5}{\\text{N\/m}}^{2}=\\text{101 kPa}\\\\[\/latex].<\/p>\n<p>This relationship means that, on average, at sea level, a column of air above 1.00 m<sup>2<\/sup>\u00a0of the Earth\u2019s surface has a weight of 1.01 \u00d7 10<sup>5<\/sup> N, equivalent to 1 atm. (See Figure 3.)<\/p>\n<div title=\"Figure 11.12.\">\n<div>\n<div>\n<div style=\"width: 260px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20103727\/Figure_12_04_03a.jpg\" alt=\"Figure shows a column of air exerting a weight of one point zero one times ten to the power five newtons on a rectangular patch of ground of one square meter cross section.\" width=\"250\" height=\"458\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. Atmospheric pressure at sea level averages 1.01 \u00d7 10<sup>5<\/sup> Pa (equivalent to 1 atm), since the column of air over this 1 m<sup>2<\/sup>, extending to the top of the atmosphere, weighs 1.01 \u00d7 10<sup>5<\/sup>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div title=\"Example 11.4. Calculating Average Density: How Dense Is the Air?\">\n<div>\n<div class=\"textbox examples\">\n<h3>Example 2. Calculating Average Density: How Dense Is the Air?<\/h3>\n<div>\n<p>Calculate the average density of the atmosphere, given that it extends to an altitude of 120 km. Compare this density with that of air listed in\u00a0Table 1.<\/p>\n<h4><strong>Strategy<\/strong><\/h4>\n<p>If we solve <em>P\u00a0<\/em>=\u00a0<em>h\u03c1g<\/em> for density, we see that<\/p>\n<p style=\"text-align: center;\">[latex]\\overline{\\rho }=\\frac{P}{\\text{hg}}\\\\[\/latex].<\/p>\n<p>We then take <em>P<\/em> to be atmospheric pressure, <em>h<\/em> is given, and <em>g<\/em> is known, and so we can use this to calculate [latex]\\overline{\\rho}\\\\[\/latex].<\/p>\n<h4><strong>Solution<\/strong><\/h4>\n<p>Entering known values into the expression for [latex]\\overline{\\rho}\\\\[\/latex]\u00a0yields<\/p>\n<div style=\"text-align: center;\" title=\"Equation 11.25.\">[latex]\\overline{\\rho }=\\frac{1\\text{.}\\text{01}\\times {\\text{10}}^{5}{\\text{N\/m}}^{2}}{\\left(\\text{120}\\times {\\text{10}}^{3}\\text{m}\\right)\\left(9\\text{.}\\text{80}{\\text{m\/s}}^{2}\\right)}=8\\text{.}\\text{59}\\times {\\text{10}}^{-2}{\\text{kg\/m}}^{3}\\\\[\/latex].<\/div>\n<h4><strong>Discussion<\/strong><\/h4>\n<p>This result is the average density of air between the Earth\u2019s surface and the top of the Earth\u2019s atmosphere, which essentially ends at 120 km. The density of air at sea level is given in\u00a0Table 1\u00a0as 1.29 kg.m<sup>3<\/sup>\u2014about 15 times its average value. Because air is so compressible, its density has its highest value near the Earth\u2019s surface and declines rapidly with altitude.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div title=\"Example 11.5. Calculating Depth Below the Surface of Water: What Depth of Water Creates the Same Pressure as the Entire Atmosphere?\">\n<div class=\"textbox examples\">\n<h3>Example 3. Calculating Depth Below the Surface of Water: What Depth of Water Creates the Same Pressure as the Entire Atmosphere?<\/h3>\n<div>\n<p>Calculate the depth below the surface of water at which the pressure due to the weight of the water equals 1.00 atm.<\/p>\n<h4><strong>Strategy<\/strong><\/h4>\n<p>We begin by solving the equation <em>P\u00a0<\/em>=\u00a0<em>h\u03c1g<\/em> for depth <em>h<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]h=\\frac{P}{\\mathrm{\\rho g}}\\\\[\/latex].<\/p>\n<p>Then we take <em>P<\/em> to be 1.00 atm and <em>\u03c1<\/em> to be the density of the water that creates the pressure.<\/p>\n<h4><strong>Solution<\/strong><\/h4>\n<p>Entering the known values into the expression for <em>h<\/em> gives<\/p>\n<p style=\"text-align: center;\">[latex]h=\\frac{1\\text{.}\\text{01}\\times {\\text{10}}^{5}{\\text{N\/m}}^{2}}{\\left(1\\text{.}\\text{00}\\times {\\text{10}}^{3}{\\text{kg\/m}}^{3}\\right)\\left(9\\text{.}\\text{80}{\\text{m\/s}}^{2}\\right)}=\\text{10}\\text{.}3\\text{m}\\\\[\/latex].<\/p>\n<h4><strong>Discussion<\/strong><\/h4>\n<p>Just 10.3 m of water creates the same pressure as 120 km of air. Since water is nearly incompressible, we can neglect any change in its density over this depth.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div title=\"Example 11.5. Calculating Depth Below the Surface of Water: What Depth of Water Creates the Same Pressure as the Entire Atmosphere?\">What do you suppose is the <em>total<\/em> pressure at a depth of 10.3 m in a swimming pool? Does the atmospheric pressure on the water\u2019s surface affect the pressure below? The answer is yes. This seems only logical, since both the water\u2019s weight and the atmosphere\u2019s weight must be supported. So the <em>total<\/em> pressure at a depth of 10.3 m is 2 atm\u2014half from the water above and half from the air above. We shall see in <a href=\".\/chapter\/11-5-pascals-principle\/\" target=\"_blank\">Pascal\u2019s Principle<\/a> that fluid pressures always add in this way.<\/div>\n<h2>Section Summary<\/h2>\n<ul>\n<li>Pressure is the weight of the fluid <em>mg<\/em>\u00a0divided by the area <em>A<\/em> supporting it (the area of the bottom of the container):\n<div style=\"text-align: center;\">[latex]P=\\frac{\\text{mg}}{A}\\\\[\/latex].<\/div>\n<\/li>\n<li>Pressure due to the weight of a liquid is given by\n<div style=\"text-align: center;\">[latex]P=h\\rho g\\\\[\/latex],<\/div>\n<p>where<em> P<\/em> is the pressure, <em>h<\/em>\u00a0is the height of the liquid, <em>\u03c1<\/em> is the density of the liquid, and <em>g<\/em>\u00a0is the acceleration due to gravity.<\/li>\n<\/ul>\n<section>\n<div class=\"textbox key-takeaways\">\n<h3>Conceptual Questions<\/h3>\n<div>\n<div>\n<p>1. Atmospheric pressure exerts a large force (equal to the weight of the atmosphere above your body\u2014about 10 tons) on the top of your body when you are lying on the beach sunbathing. Why are you able to get up?<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p>2. Why does atmospheric pressure decrease more rapidly than linearly with altitude?<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p>3. What are two reasons why mercury rather than water is used in barometers?<\/p>\n<\/div>\n<\/div>\n<div>\n<p>4. Figure 4 shows how sandbags placed around a leak outside a river levee can effectively stop the flow of water under the levee. Explain how the small amount of water inside the column formed by the sandbags is able to balance the much larger body of water behind the levee.<\/p>\n<div style=\"width: 302px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03210436\/Figure_12_04_04a.jpg\" alt=\"The figure shows a flooding river on the extreme right, with a levee set up on its left, and sandbags are stacked on the left of the levee. The height of the levee and that of the stacked sandbags is greater than the water level of the flooding river, so the water does not flow over their tops, but a leak under the levee allows some water to flow under it and reach the sandbags.\" width=\"292\" height=\"160\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. Because the river level is very high, it has started to leak under the levee. Sandbags are placed around the leak, and the water held by them rises until it is the same level as the river, at which point the water there stops rising.<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p>5. Why is it difficult to swim under water in the Great Salt Lake?<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p>6. Is there a net force on a dam due to atmospheric pressure? Explain your answer.<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p>7. Does atmospheric pressure add to the gas pressure in a rigid tank? In a toy balloon? When, in general, does atmospheric pressure <em>not<\/em> affect the total pressure in a fluid?<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p>8. You can break a strong wine bottle by pounding a cork into it with your fist, but the cork must press directly against the liquid filling the bottle\u2014there can be no air between the cork and liquid. Explain why the bottle breaks, and why it will not if there is air between the cork and liquid.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section>\n<div class=\"textbox exercises\">\n<h3>Problems &amp; Exercises<\/h3>\n<div>\n<div>\n<p>1. What depth of mercury creates a pressure of 1.00 atm?<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p>2. The greatest ocean depths on the Earth are found in the Marianas Trench near the Philippines. Calculate the pressure due to the ocean at the bottom of this trench, given its depth is 11.0 km and assuming the density of seawater is constant all the way down.<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p>3. Verify that the SI unit of <em>h\u03c1g<\/em>\u00a0is N\/m<sup>2<\/sup>.<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p>4. Water towers store water above the level of consumers for times of heavy use, eliminating the need for high-speed pumps. How high above a user must the water level be to create a gauge pressure of 3.00 \u00d7 10<sup>5<\/sup> N\/m<sup>2<\/sup>?<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p>5. The aqueous humor in a person\u2019s eye is exerting a force of 0.300 N on the 1.10-cm<sup>2<\/sup>\u00a0area of the cornea. (a) What pressure is this in mm Hg? (b) Is this value within the normal range for pressures in the eye?<\/p>\n<\/div>\n<div><\/div>\n<\/div>\n<div>\n<div>\n<p>6. How much force is exerted on one side of an 8.50 cm by 11.0 cm sheet of paper by the atmosphere? How can the paper withstand such a force?<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p>7. What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long gas tank that can hold 50.0 kg of gasoline by the weight of the gasoline in it when it is full?<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p>8. Calculate the average pressure exerted on the palm of a shot-putter\u2019s hand by the shot if the area of contact is 50.0 cm<sup>2<\/sup>\u00a0and he exerts a force of 800 N on it. Express the pressure in N\/m<sup>2<\/sup> and compare it with the 1.00 \u00d7 10<sup>6<\/sup> pressures sometimes encountered in the skeletal system.<\/p>\n<\/div>\n<\/div>\n<div>\n<div>\n<p>9. The left side of the heart creates a pressure of 120 mm Hg by exerting a force directly on the blood over an effective area of 15.0 cm<sup>2<\/sup>. What force does it exert to accomplish this?<\/p>\n<\/div>\n<div>\n<p>10. Show that the total force on a rectangular dam due to the water behind it increases with the <em>square<\/em> of the water depth. In particular, show that this force is given by [latex]F=\\rho {\\mathrm{gh}}^{2}L\/2\\\\[\/latex], where [latex]rho\\\\[\/latex] is the density of water,\u00a0<em>h<\/em> is its depth at the dam, and <em>L<\/em> is the length of the dam. You may assume the face of the dam is vertical. (Hint: Calculate the average pressure exerted and multiply this by the area in contact with the water. (See Figure 5.)<\/p>\n<\/div>\n<\/div>\n<div>\n<figure>\n<div style=\"width: 293px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1322\/2015\/12\/03210434\/Figure_12_04_02a.jpg\" alt=\"A two-dimensional view of a dam with dimensions L and h is shown. Force F at h is shown by a horizontal arrow. The force F exerted by water on the dam is F equals average pressure p bar into area A and pressure in turn is average height h bar into density rho into acceleration due to gravity g.\" width=\"283\" height=\"203\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5.<\/p>\n<\/div>\n<\/figure>\n<\/div>\n<\/div>\n<\/section>\n<div>\n<h2>Glossary<\/h2>\n<dl>\n<dt>pressure:<\/dt>\n<dd>the weight of the fluid divided by the area supporting it<\/dd>\n<\/dl>\n<div class=\"textbox exercises\">\n<h3>Selected Solutions to Problems &amp; Exercises<\/h3>\n<p>1.\u00a00.760 m<\/p>\n<p>3.\u00a0[latex]\\begin{array}{lll}{\\left(h\\rho g\\right)}_{\\text{units}}& =& \\left(\\text{m}\\right)\\left({\\text{kg\/m}}^{3}\\right)\\left({\\text{m\/s}}^{2}\\right)=\\left(\\text{kg}\\cdot {\\text{m}}^{2}\\right)\/\\left({\\text{m}}^{3}\\cdot {\\text{s}}^{2}\\right)\\\\ & =& \\left(\\text{kg}\\cdot {\\text{m\/s}}^{2}\\right)\\left({\\text{1\/m}}^{2}\\right)\\\\ & =& {\\text{N\/m}}^{2}\\end{array}\\\\[\/latex].<\/p>\n<p>5. (a) 20.5 mm Hg\u00a0(b) The range of pressures in the eye is 12\u201324 mm Hg, so the result in part (a) is within that range<\/p>\n<p>7. 1.09 \u00d7 10<sup>3<\/sup> N\/m<sup>2<\/sup><\/p>\n<p>9.\u00a024.0 N<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2103\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Physics. <strong>Authored by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/031da8d3-b525-429c-80cf-6c8ed997733a\/College_Physics\">http:\/\/cnx.org\/contents\/031da8d3-b525-429c-80cf-6c8ed997733a\/College_Physics<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Located at License<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":1,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Physics\",\"author\":\"OpenStax College\",\"organization\":\"\",\"url\":\"http:\/\/cnx.org\/contents\/031da8d3-b525-429c-80cf-6c8ed997733a\/College_Physics\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Located at License\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2103","chapter","type-chapter","status-publish","hentry"],"part":7546,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/chapters\/2103","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/wp\/v2\/users\/1"}],"version-history":[{"count":15,"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/chapters\/2103\/revisions"}],"predecessor-version":[{"id":12043,"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/chapters\/2103\/revisions\/12043"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/parts\/7546"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/chapters\/2103\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/wp\/v2\/media?parent=2103"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/pressbooks\/v2\/chapter-type?post=2103"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/wp\/v2\/contributor?post=2103"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/atd-austincc-physics1\/wp-json\/wp\/v2\/license?post=2103"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}