{"id":1180,"date":"2021-06-30T17:02:05","date_gmt":"2021-06-30T17:02:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/work-and-force\/"},"modified":"2022-03-19T03:36:54","modified_gmt":"2022-03-19T03:36:54","slug":"work-and-force","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/work-and-force\/","title":{"raw":"Work and Force","rendered":"Work and Force"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate the work done by a variable force acting along a line<\/li>\r\n \t<li>Calculate the work done in pumping a liquid from one height to another<\/li>\r\n \t<li>Find the hydrostatic force against a submerged vertical plate<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Work Done by a Force<\/h2>\r\n<p id=\"fs-id1167794140573\">We now consider work. In physics, <strong>work<\/strong> is related to force, which is often intuitively defined as a push or pull on an object. When a force moves an object, we say the force does work on the object. In other words, work can be thought of as the amount of energy it takes to move an object. According to physics, when we have a constant force, work can be expressed as the product of force and distance.<\/p>\r\n<p id=\"fs-id1167794042466\">In the English system, the unit of force is the pound and the unit of distance is the foot, so work is given in foot-pounds. In the metric system, kilograms and meters are used. One newton is the force needed to accelerate 1 kilogram of mass at the rate of 1 m\/sec<sup>2<\/sup>. Thus, the most common unit of work is the newton-meter. This same unit is also called the <span class=\"no-emphasis\"><em>joule<\/em><\/span>. Both are defined as kilograms times meters squared over seconds squared [latex](\\text{kg}\u00b7{\\text{m}}^{2}\\text{\/}{\\text{s}}^{2}).[\/latex]<\/p>\r\n<p id=\"fs-id1167793514627\">When we have a constant force, things are pretty easy. It is rare, however, for a force to be constant. The work done to compress (or elongate) a spring, for example, varies depending on how far the spring has already been compressed (or stretched). We look at springs in more detail later in this section.<\/p>\r\n<p id=\"fs-id1167793982965\">Suppose we have a variable force [latex]F(x)[\/latex] that moves an object in a positive direction along the [latex]x[\/latex]-axis from point [latex]a[\/latex] to point [latex]b.[\/latex] To calculate the work done, we partition the interval [latex]\\left[a,b\\right][\/latex] and estimate the work done over each subinterval. So, for [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of the interval [latex]\\left[a,b\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026},n,[\/latex] choose an arbitrary point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right].[\/latex] To calculate the work done to move an object from point [latex]{x}_{i-1}[\/latex] to point [latex]{x}_{i},[\/latex] we assume the force is roughly constant over the interval, and use [latex]F({x}_{i}^{*})[\/latex] to approximate the force. The work done over the interval [latex]\\left[{x}_{i-1},{x}_{i}\\right],[\/latex] then, is given by<\/p>\r\n\r\n<div id=\"fs-id1167794065392\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{W}_{i}\\approx F({x}_{i}^{*})({x}_{i}-{x}_{i-1})=F({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793926193\">Therefore, the work done over the interval [latex]\\left[a,b\\right][\/latex] is approximately<\/p>\r\n\r\n<div id=\"fs-id1167794003046\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]W=\\displaystyle\\sum_{i=1}^{n} {W}_{i}\\approx \\displaystyle\\sum_{i=1}^{n} F({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793951829\">Taking the limit of this expression as [latex]n\\to \\infty [\/latex] gives us the exact value for work:<\/p>\r\n\r\n<div id=\"fs-id1167794212535\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]W=\\underset{n\\to \\infty }{\\text{lim}} \\displaystyle\\sum_{i=1}^{n} F({x}_{i}^{*})\\text{\u0394}x={\\displaystyle\\int }_{a}^{b}F(x)dx.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793984329\">Thus, we can define work as follows.<\/p>\r\n\r\n<div id=\"fs-id1167794137488\" class=\"textbox shaded\">\r\n<div class=\"title\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1167794094283\">If a variable force [latex]F(x)[\/latex] moves an object in a positive direction along the [latex]x[\/latex]-axis from point [latex]a[\/latex] to point [latex]b[\/latex], then the <strong>work<\/strong> done on the object is<\/p>\r\n\r\n<div id=\"fs-id1167793879581\" class=\"equation\" style=\"text-align: center;\">[latex]W={\\displaystyle\\int }_{a}^{b}F(x)dx[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1167794122064\">Note that if <em>F<\/em> is constant, the integral evaluates to [latex]F\u00b7(b-a)=F\u00b7d,[\/latex] which is the formula we stated at the beginning of this section.<\/p>\r\n<p id=\"fs-id1167793249357\">Now let\u2019s look at the specific example of the work done to compress or elongate a spring. Consider a block attached to a horizontal spring. The block moves back and forth as the spring stretches and compresses. Although in the real world we would have to account for the force of friction between the block and the surface on which it is resting, we ignore friction here and assume the block is resting on a frictionless surface. When the spring is at its natural length (at rest), the system is said to be at equilibrium. In this state, the spring is neither elongated nor compressed, and in this equilibrium position the block does not move until some force is introduced. We orient the system such that [latex]x=0[\/latex] corresponds to the equilibrium position (see the following figure).<\/p>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213202\/CNX_Calc_Figure_06_05_004.jpg\" alt=\"&quot;This\" \/>\r\n\r\n&nbsp;\r\n<p id=\"fs-id1167793633520\">According to<strong> Hooke\u2019s law<\/strong>, the force required to compress or stretch a spring from an equilibrium position is given by [latex]F(x)=kx,[\/latex] for some constant [latex]k.[\/latex] The value of [latex]k[\/latex] depends on the physical characteristics of the spring. The constant [latex]k[\/latex] is called the <span class=\"no-emphasis\"><em>spring constant<\/em><\/span> and is always positive. We can use this information to calculate the work done to compress or elongate a spring, as shown in the following example.<\/p>\r\n\r\n<div id=\"fs-id1167793298896\" class=\"textbook exercises\">\r\n<h3>Example: The Work Required to Stretch or Compress a Spring<\/h3>\r\nSuppose it takes a force of 10 N (in the negative direction) to compress a spring 0.2 m from the equilibrium position. How much work is done to stretch the spring 0.5 m from the equilibrium position?\r\n<div id=\"fs-id1167794099463\" class=\"exercise\">[reveal-answer q=\"fs-id1167794060082\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794060082\"]\r\n<p id=\"fs-id1167794060082\">First find the spring constant, [latex]k.[\/latex] When [latex]x=-0.2,[\/latex] we know [latex]F(x)=-10,[\/latex] so<\/p>\r\n\r\n<div id=\"fs-id1167793619889\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill F(x)&amp; =\\hfill &amp; kx\\hfill \\\\ \\hfill -10&amp; =\\hfill &amp; k(-0.2)\\hfill \\\\ \\hfill k&amp; =\\hfill &amp; 50\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793432264\">and [latex]F(x)=50x.[\/latex] Then, to calculate work, we integrate the force function, obtaining<\/p>\r\n\r\n<div id=\"fs-id1167794038347\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]W={\\displaystyle\\int }_{a}^{b}F(x)dx={\\displaystyle\\int }_{0}^{0.5}50xdx={25{x}^{2}|}_{0}^{0.5}=6.25.[\/latex]<\/div>\r\nThe work done to stretch the spring is 6.25 J.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSuppose it takes a force of 8 lb to stretch a spring 6 in. from the equilibrium position. How much work is done to stretch the spring 1 ft from the equilibrium position?\r\n<div id=\"fs-id1167793929416\" class=\"exercise\">[reveal-answer q=\"315964\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"315964\"]Use the process from the previous example. Be careful with units.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div class=\"exercise\">[reveal-answer q=\"fs-id1167793281283\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793281283\"]\r\n<p id=\"fs-id1167793281283\">8 ft-lb<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=811&amp;end=903&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.5PhysicalApplications811to903_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.5 Physical Applications\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5667[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1167793455240\" class=\"bc-section section\">\r\n<h2>Work Done in Pumping<\/h2>\r\n<p id=\"fs-id1167793455245\">Consider the work done to pump water (or some other liquid) out of a tank. Pumping problems are a little more complicated than spring problems because many of the calculations depend on the shape and size of the tank. In addition, instead of being concerned about the work done to move a single mass, we are looking at the work done to move a volume of water, and it takes more work to move the water from the bottom of the tank than it does to move the water from the top of the tank.<\/p>\r\n<p id=\"fs-id1167793502586\">We examine the process in the context of a cylindrical tank, then look at a couple of examples using tanks of different shapes. Assume a cylindrical tank of radius 4 m and height 10 m is filled to a depth of 8 m. How much work does it take to pump all the water over the top edge of the tank?<\/p>\r\n<p id=\"fs-id1167793502601\">The first thing we need to do is define a frame of reference. We let [latex]x[\/latex] represent the vertical distance below the top of the tank. That is, we orient the [latex]x\\text{-axis}[\/latex] vertically, with the origin at the top of the tank and the downward direction being positive (see the following figure).<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"481\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213205\/CNX_Calc_Figure_06_05_005.jpg\" alt=\"This figure is a right circular cylinder that is vertical. It represents a tank of water. The radius of the cylinder is 4 m, the height of the cylinder is 10 m. The height of the water inside the cylinder is 8 m. There is also a horizontal line on top of the tank representing the x=0. A line is drawn vertical beside the cylinder with a downward arrow labeled x.\" width=\"481\" height=\"379\" \/> Figure 5. How much work is needed to empty a tank partially filled with water?[\/caption]\r\n<p id=\"fs-id1167793479906\">Using this coordinate system, the water extends from [latex]x=2[\/latex] to [latex]x=10.[\/latex] Therefore, we partition the interval [latex]\\left[2,10\\right][\/latex] and look at the work required to lift each individual \u201clayer\u201d of water. So, for [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of the interval [latex]\\left[2,10\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026},n,[\/latex] choose an arbitrary point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right].[\/latex] (Figure 6) shows a representative layer.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"471\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213208\/CNX_Calc_Figure_06_05_006.jpg\" alt=\"This figure is a right circular cylinder representing a tank of water. Inside of the cylinder is a layer of water with thickness delta x. The thickness begins at xsub(i-1) and ends at xsubi.\" width=\"471\" height=\"251\" \/> Figure 6. A representative layer of water.[\/caption]\r\n<p id=\"fs-id1167793720072\">In pumping problems, the force required to lift the water to the top of the tank is the force required to overcome gravity, so it is equal to the weight of the water. Given that the weight-density of water is 9800 N\/m<sup>3<\/sup>, or 62.4 lb\/ft<sup>3<\/sup>, calculating the volume of each layer gives us the weight. In this case, we have<\/p>\r\n\r\n<div id=\"fs-id1167793506225\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V=\\pi {(4)}^{2}\\text{\u0394}x=16\\pi \\text{\u0394}x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793277647\">Then, the force needed to lift each layer is<\/p>\r\n\r\n<div id=\"fs-id1167793277650\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]F=9800\u00b716\\pi \\text{\u0394}x=156,800\\pi \\text{\u0394}x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793937471\">Note that this step becomes a little more difficult if we have a noncylindrical tank. We look at a noncylindrical tank in the next example.<\/p>\r\n<p id=\"fs-id1167793937476\">We also need to know the distance the water must be lifted. Based on our choice of coordinate systems, we can use [latex]{x}_{i}^{*}[\/latex] as an approximation of the distance the layer must be lifted. Then the work to lift the [latex]i\\text{th}[\/latex] layer of water [latex]{W}_{i}[\/latex] is approximately<\/p>\r\n\r\n<div id=\"fs-id1167794165423\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{W}_{i}\\approx 156,800\\pi {x}_{i}^{*}\\text{\u0394}x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793886516\">Adding the work for each layer, we see the approximate work to empty the tank is given by<\/p>\r\n\r\n<div id=\"fs-id1167793473606\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]W=\\underset{i=1}{\\overset{n}{\\text{\u2211}}}{W}_{i}\\approx \\underset{i=1}{\\overset{n}{\\text{\u2211}}}156,800\\pi {x}_{i}^{*}\\text{\u0394}x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793725962\">This is a Riemann sum, so taking the limit as [latex]n\\to \\infty ,[\/latex] we get<\/p>\r\n\r\n<div id=\"fs-id1167793355091\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill W&amp; =\\underset{n\\to \\infty }{\\text{lim}}\\underset{i=1}{\\overset{n}{\\text{\u2211}}}156,800\\pi {x}_{i}^{*}\\text{\u0394}x\\hfill \\\\ &amp; =156,800\\pi {\\displaystyle\\int }_{2}^{10}xdx\\hfill \\\\ &amp; =156,800\\pi {\\left[\\frac{{x}^{2}}{2}\\right]|}_{2}^{10}=7,526,400\\pi \\approx 23,644,883.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793546878\">The work required to empty the tank is approximately 23,650,000 J.<\/p>\r\nFor pumping problems, the calculations vary depending on the shape of the tank or container. The following problem-solving strategy lays out a step-by-step process for solving pumping problems.\r\n<div id=\"fs-id1167793546887\" class=\"textbox examples\">\r\n<h3>Problem-Solving Strategy: Solving Pumping Problems<\/h3>\r\n<ol id=\"fs-id1167794051212\">\r\n \t<li>Sketch a picture of the tank and select an appropriate frame of reference.<\/li>\r\n \t<li>Calculate the volume of a representative layer of water.<\/li>\r\n \t<li>Multiply the volume by the weight-density of water to get the force.<\/li>\r\n \t<li>Calculate the distance the layer of water must be lifted.<\/li>\r\n \t<li>Multiply the force and distance to get an estimate of the work needed to lift the layer of water.<\/li>\r\n \t<li>Sum the work required to lift all the layers. This expression is an estimate of the work required to pump out the desired amount of water, and it is in the form of a Riemann sum.<\/li>\r\n \t<li>Take the limit as [latex]n\\to \\infty [\/latex] and evaluate the resulting integral to get the exact work required to pump out the desired amount of water.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<p id=\"fs-id1167793505372\">We now apply this problem-solving strategy in an example with a noncylindrical tank.<\/p>\r\n\r\n<div id=\"fs-id1167793505375\" class=\"textbook exercises\">\r\n<h3>Example: A Pumping Problem with a Noncylindrical Tank<\/h3>\r\nAssume a tank in the shape of an inverted cone, with height 12 ft and base radius 4 ft. The tank is full to start with, and water is pumped over the upper edge of the tank until the height of the water remaining in the tank is 4 ft. How much work is required to pump out that amount of water?\r\n<div id=\"fs-id1167793505377\" class=\"exercise\">[reveal-answer q=\"fs-id1167793395586\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793395586\"]\r\n<p id=\"fs-id1167793395586\">The tank is depicted in Figure 7. As we did in the example with the cylindrical tank, we orient the [latex]x\\text{-axis}[\/latex] vertically, with the origin at the top of the tank and the downward direction being positive (step 1).<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"283\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213212\/CNX_Calc_Figure_06_05_007.jpg\" alt=\"This figure is an upside-down cone. The cone has an axis through the center. The top of the cone on the axis is labeled x=0.\" width=\"283\" height=\"486\" \/> Figure 7. A water tank in the shape of an inverted cone.[\/caption]\r\n<p id=\"fs-id1167793385041\">The tank starts out full and ends with 4 ft of water left, so, based on our chosen frame of reference, we need to partition the interval [latex]\\left[0,8\\right].[\/latex] Then, for [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of the interval [latex]\\left[0,8\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026},n,[\/latex] choose an arbitrary point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right].[\/latex] We can approximate the volume of a layer by using a disk, then use similar triangles to find the radius of the disk (see the following figure).<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"746\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213214\/CNX_Calc_Figure_06_05_008.jpg\" alt=\"This figure has two images. The first has the x-axis. Below the axis, on a slant is a line segment extending up to the x-axis. Beside the line segment is a horizontal right circular cylinder. The second image has a triangle. The right triangle mirrors the first image with the hypotenuse the line segment in the first image. The top of the triangle is 4 units. the length of the vertical side is 12 units. The vertical side is also divided into two parts; the first is xsubi, the second is 12-xsubi. It is divided at the level where the first image has the cylinder.\" width=\"746\" height=\"561\" \/> Figure 8. Using similar triangles to express the radius of a disk of water.[\/caption]\r\n<p id=\"fs-id1167793221543\">From properties of similar triangles, we have<\/p>\r\n\r\n<div id=\"fs-id1167793221547\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{{r}_{i}}{12-{x}_{i}^{*}}&amp; =\\hfill &amp; \\frac{4}{12}=\\frac{1}{3}\\hfill \\\\ \\hfill 3{r}_{i}&amp; =\\hfill &amp; 12-{x}_{i}^{*}\\hfill \\\\ \\hfill {r}_{i}&amp; =\\hfill &amp; \\frac{12-{x}_{i}^{*}}{3}\\hfill \\\\ &amp; =\\hfill &amp; 4-\\frac{{x}_{i}^{*}}{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793419284\">Then the volume of the disk is<\/p>\r\n\r\n<div id=\"fs-id1167793419287\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{V}_{i}=\\pi {(4-\\frac{{x}_{i}^{*}}{3})}^{2}\\text{\u0394}x\\text{(step 2).}[\/latex]<\/div>\r\n<p id=\"fs-id1167793287424\">The weight-density of water is 62.4 lb\/ft<sup>3<\/sup>, so the force needed to lift each layer is approximately<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{F}_{i}\\approx 62.4\\pi {(4-\\frac{{x}_{i}^{*}}{3})}^{2}\\text{\u0394}x\\text{(step 3).}[\/latex]<\/div>\r\n<p id=\"fs-id1167793628639\">Based on the diagram, the distance the water must be lifted is approximately [latex]{x}_{i}^{*}[\/latex] feet (step 4), so the approximate work needed to lift the layer is<\/p>\r\n\r\n<div id=\"fs-id1167793929624\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{W}_{i}\\approx 62.4\\pi {x}_{i}^{*}{(4-\\frac{{x}_{i}^{*}}{3})}^{2}\\text{\u0394}x\\text{(step 5).}[\/latex]<\/div>\r\n<p id=\"fs-id1167793250306\">Summing the work required to lift all the layers, we get an approximate value of the total work:<\/p>\r\n\r\n<div id=\"fs-id1167793570048\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]W=\\underset{i=1}{\\overset{n}{\\text{\u2211}}}{W}_{i}\\approx \\underset{i=1}{\\overset{n}{\\text{\u2211}}}62.4\\pi {x}_{i}^{*}{(4-\\frac{{x}_{i}^{*}}{3})}^{2}\\text{\u0394}x\\text{(step 6).}[\/latex]<\/div>\r\n<p id=\"fs-id1167793400817\">Taking the limit as [latex]n\\to \\infty ,[\/latex] we obtain<\/p>\r\n\r\n<div id=\"fs-id1167793400833\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill W&amp; =\\underset{n\\to \\infty }{\\text{lim}}\\underset{i=1}{\\overset{n}{\\text{\u2211}}}62.4\\pi {x}_{i}^{*}{(4-\\frac{{x}_{i}^{*}}{3})}^{2}\\text{\u0394}x\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{8}62.4\\pi x{(4-\\frac{x}{3})}^{2}dx\\hfill \\\\ &amp; =62.4\\pi {\\displaystyle\\int }_{0}^{8}x(16-\\frac{8x}{3}+\\frac{{x}^{2}}{9})dx=62.4\\pi {\\displaystyle\\int }_{0}^{8}(16x-\\frac{8{x}^{2}}{3}+\\frac{{x}^{3}}{9})dx\\hfill \\\\ &amp; =62.4\\pi {\\left[8{x}^{2}-\\frac{8{x}^{3}}{9}+\\frac{{x}^{4}}{36}\\right]|}_{0}^{8}=10,649.6\\pi \\approx 33,456.7\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793454764\">It takes approximately 33,450 ft-lb of work to empty the tank to the desired level.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: A Pumping Problem with a Noncylindrical Tank.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=1355&amp;end=1691&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.5PhysicalApplications1355to1691_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.5 Physical Applications\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1167793553739\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA tank is in the shape of an inverted cone, with height 10 ft and base radius 6 ft. The tank is filled to a depth of 8 ft to start with, and water is pumped over the upper edge of the tank until 3 ft of water remain in the tank. How much work is required to pump out that amount of water?\r\n<div>[reveal-answer q=\"46716\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"46716\"]Use the process from the previous example.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167793553742\" class=\"exercise\">[reveal-answer q=\"fs-id1167793553760\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793553760\"]\r\n<p id=\"fs-id1167793553760\">Approximately 43,255.2 ft-lb<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5684[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Hydrostatic Force and Pressure<\/h2>\r\n<p id=\"fs-id1167793281535\">In this last section, we look at the force and pressure exerted on an object submerged in a liquid. In the English system, force is measured in pounds. In the metric system, it is measured in newtons. Pressure is force per unit area, so in the English system we have pounds per square foot (or, perhaps more commonly, pounds per square inch, denoted psi). In the metric system we have newtons per square meter, also called <span class=\"no-emphasis\"><em>pascals<\/em><\/span>.<\/p>\r\n<p id=\"fs-id1167793630279\">Let\u2019s begin with the simple case of a plate of area [latex]A[\/latex] submerged horizontally in water at a depth [latex]s[\/latex] (Figure 9). Then, the force exerted on the plate is simply the weight of the water above it, which is given by [latex]F=\\rho As,[\/latex] where [latex]\\rho [\/latex] is the weight density of water (weight per unit volume). To find the <strong>hydrostatic pressure<\/strong>\u2014that is, the pressure exerted by water on a submerged object\u2014we divide the force by the area. So the pressure is [latex]p=F\\text{\/}A=\\rho s.[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213218\/CNX_Calc_Figure_06_05_009.jpg\" alt=\"This image has a circular plate submerged in water. The plate is labeled A and the depth of the water is labeled s.\" width=\"487\" height=\"324\" \/> Figure 9. A plate submerged horizontally in water.[\/caption]\r\n<p id=\"fs-id1167793553642\">By <span class=\"no-emphasis\">Pascal\u2019s principle<\/span>, the pressure at a given depth is the same in all directions, so it does not matter if the plate is submerged horizontally or vertically. So, as long as we know the depth, we know the pressure. We can apply Pascal\u2019s principle to find the force exerted on surfaces, such as dams, that are oriented vertically. We cannot apply the formula [latex]F=\\rho As[\/latex] directly, because the depth varies from point to point on a vertically oriented surface. So, as we have done many times before, we form a partition, a Riemann sum, and, ultimately, a definite integral to calculate the force.<\/p>\r\n<p id=\"fs-id1167793553670\">Suppose a thin plate is submerged in water. We choose our frame of reference such that the [latex]x[\/latex]-axis is oriented vertically, with the downward direction being positive, and point [latex]x=0[\/latex] corresponding to a logical reference point. Let [latex]s(x)[\/latex] denote the depth at point [latex]x[\/latex]. Note we often let [latex]x=0[\/latex] correspond to the surface of the water. In this case, depth at any point is simply given by [latex]s(x)=x.[\/latex] However, in some cases we may want to select a different reference point for [latex]x=0,[\/latex] so we proceed with the development in the more general case. Last, let [latex]w(x)[\/latex] denote the width of the plate at the point [latex]x.[\/latex]<\/p>\r\n<p id=\"fs-id1167794127192\">Assume the top edge of the plate is at point [latex]x=a[\/latex] and the bottom edge of the plate is at point [latex]x=b.[\/latex] Then, for [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of the interval [latex]\\left[a,b\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026},n,[\/latex] choose an arbitrary point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right].[\/latex] The partition divides the plate into several thin, rectangular strips (see the following figure).<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"571\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213221\/CNX_Calc_Figure_06_05_010.jpg\" alt=\"This image is the overhead view of a submerged circular plate. The x-axis is to the side of the plate. The plate\u2019s diameter goes from x=a to x=b. There is a strip in the middle of the plate with thickness of delta x. On the axis this thickness begins at x=xsub(i-1) and ends at x=xsubi. The length of the strip in the plate is labeled w(csubi).\" width=\"571\" height=\"350\" \/> Figure 10. A thin plate submerged vertically in water.[\/caption]\r\n<p id=\"fs-id1167793420981\">Let\u2019s now estimate the force on a representative strip. If the strip is thin enough, we can treat it as if it is at a constant depth, [latex]s({x}_{i}^{*}).[\/latex] We then have<\/p>\r\n\r\n<div id=\"fs-id1167793952122\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{F}_{i}=\\rho As=\\rho \\left[w({x}_{i}^{*})\\text{\u0394}x\\right]s({x}_{i}^{*})[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793584536\">Adding the forces, we get an estimate for the force on the plate:<\/p>\r\n\r\n<div id=\"fs-id1167793584539\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]F\\approx \\underset{i=1}{\\overset{n}{\\text{\u2211}}}{F}_{i}=\\underset{i=1}{\\overset{n}{\\text{\u2211}}}\\rho \\left[w({x}_{i}^{*})\\text{\u0394}x\\right]s({x}_{i}^{*})[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794057934\">This is a Riemann sum, so taking the limit gives us the exact force. We obtain<\/p>\r\n\r\n<div id=\"fs-id1167794057937\" class=\"equation\" style=\"text-align: center;\">[latex]F=\\underset{n\\to \\infty }{\\text{lim}}\\underset{i=1}{\\overset{n}{\\text{\u2211}}}\\rho \\left[w({x}_{i}^{*})\\text{\u0394}x\\right]s({x}_{i}^{*})={\\displaystyle\\int }_{a}^{b}\\rho w(x)s(x)dx[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794291708\">Evaluating this integral gives us the force on the plate. We summarize this in the following problem-solving strategy.<\/p>\r\n\r\n<div id=\"fs-id1167794291713\" class=\"textbox examples\">\r\n<h3>Problem-Solving Strategy: Finding Hydrostatic Force<\/h3>\r\n<ol id=\"fs-id1167794291719\">\r\n \t<li>Sketch a picture and select an appropriate frame of reference. (Note that if we select a frame of reference other than the one used earlier, we may have to adjust the equation above accordingly.)<\/li>\r\n \t<li>Determine the depth and width functions, [latex]s(x)[\/latex] and [latex]w(x).[\/latex]<\/li>\r\n \t<li>Determine the weight-density of whatever liquid with which you are working. The weight-density of water is 62.4 lb\/ft<sup>3<\/sup>, or 9800 N\/m<sup>3<\/sup>.<\/li>\r\n \t<li>Use the equation to calculate the total force.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1167793504251\" class=\"textbook exercises\">\r\n<h3>Example: Finding Hydrostatic Force<\/h3>\r\nA water trough 15 ft long has ends shaped like inverted isosceles triangles, with base 8 ft and height 3 ft. Find the force on one end of the trough if the trough is full of water.\r\n<div id=\"fs-id1167793504253\" class=\"exercise\">[reveal-answer q=\"fs-id1167793385052\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793385052\"]\r\n<p id=\"fs-id1167793385052\">Figure 11 shows the trough and a more detailed view of one end.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"578\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213225\/CNX_Calc_Figure_06_05_011.jpg\" alt=\"This figure has two images. The first is a water trough with rectangular sides. The length of the trough is 15 feet, the depth is 3 feet, and the width is 8 feet. The second image is a cross section of the trough. It is a triangle. The top has length of 8 feet and the sides have length 5 feet. The altitude is labeled with 3 feet.\" width=\"578\" height=\"634\" \/> Figure 11. (a) A water trough with a triangular cross-section. (b) Dimensions of one end of the water trough.[\/caption]\r\n<p id=\"fs-id1167793385076\">Select a frame of reference with the [latex]x\\text{-axis}[\/latex] oriented vertically and the downward direction being positive. Select the top of the trough as the point corresponding to [latex]x=0[\/latex] (step 1). The depth function, then, is [latex]s(x)=x.[\/latex] Using similar triangles, we see that [latex]w(x)=8-(8\\text{\/}3)x[\/latex] (step 2). Now, the weight density of water is 62.4 lb\/ft<sup>3<\/sup> (step 3), so applying the force equation from above,\u00a0we obtain<\/p>\r\n\r\n<div id=\"fs-id1167794054206\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill F&amp; ={\\displaystyle\\int }_{a}^{b}\\rho w(x)s(x)dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{3}62.4(8-\\frac{8}{3}x)xdx=62.4{\\displaystyle\\int }_{0}^{3}(8x-\\frac{8}{3}{x}^{2})dx\\hfill \\\\ &amp; =62.4{\\left[4{x}^{2}-\\frac{8}{9}{x}^{3}\\right]|}_{0}^{3}=748.8.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793445714\">The water exerts a force of 748.8 lb on the end of the trough (step 4).<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793445721\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA water trough 12 m long has ends shaped like inverted isosceles triangles, with base 6 m and height 4 m. Find the force on one end of the trough if the trough is full of water.\r\n\r\n[reveal-answer q=\"319826\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"319826\"]Follow the problem-solving strategy and the process from the previous example.[\/hidden-answer]\r\n<div id=\"fs-id1167793445724\" class=\"exercise\">\r\n<div class=\"solution\">\r\n\r\n[reveal-answer q=\"388525\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"388525\"]\r\n<p id=\"fs-id1167793421193\">156,800 N<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1167793421202\" class=\"commentary\">\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=2006&amp;end=2135&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266832\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266832\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.5PhysicalApplications2006to2135_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.5 Physical Applications\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1167794138214\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWhen the reservoir is at its average level, the surface of the water is about 50 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?\r\n<div>[reveal-answer q=\"989393\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"989393\"]Change the depth function, [latex]s(x),[\/latex] and the limits of integration.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167794138217\" class=\"exercise\">[reveal-answer q=\"fs-id1167794138229\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794138229\"]\r\n<p id=\"fs-id1167794138229\">Approximately 7,164,520,000 lb or 3,582,260 t<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\n<p id=\"fs-id1167791543256\"><a href=\"https:\/\/www.history.com\/topics\/great-depression\/hoover-dam\" target=\"_blank\" rel=\"noopener\">To learn more about Hoover Dam, see this article published by the History Channel.<\/a><\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5670[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate the work done by a variable force acting along a line<\/li>\n<li>Calculate the work done in pumping a liquid from one height to another<\/li>\n<li>Find the hydrostatic force against a submerged vertical plate<\/li>\n<\/ul>\n<\/div>\n<h2>Work Done by a Force<\/h2>\n<p id=\"fs-id1167794140573\">We now consider work. In physics, <strong>work<\/strong> is related to force, which is often intuitively defined as a push or pull on an object. When a force moves an object, we say the force does work on the object. In other words, work can be thought of as the amount of energy it takes to move an object. According to physics, when we have a constant force, work can be expressed as the product of force and distance.<\/p>\n<p id=\"fs-id1167794042466\">In the English system, the unit of force is the pound and the unit of distance is the foot, so work is given in foot-pounds. In the metric system, kilograms and meters are used. One newton is the force needed to accelerate 1 kilogram of mass at the rate of 1 m\/sec<sup>2<\/sup>. Thus, the most common unit of work is the newton-meter. This same unit is also called the <span class=\"no-emphasis\"><em>joule<\/em><\/span>. Both are defined as kilograms times meters squared over seconds squared [latex](\\text{kg}\u00b7{\\text{m}}^{2}\\text{\/}{\\text{s}}^{2}).[\/latex]<\/p>\n<p id=\"fs-id1167793514627\">When we have a constant force, things are pretty easy. It is rare, however, for a force to be constant. The work done to compress (or elongate) a spring, for example, varies depending on how far the spring has already been compressed (or stretched). We look at springs in more detail later in this section.<\/p>\n<p id=\"fs-id1167793982965\">Suppose we have a variable force [latex]F(x)[\/latex] that moves an object in a positive direction along the [latex]x[\/latex]-axis from point [latex]a[\/latex] to point [latex]b.[\/latex] To calculate the work done, we partition the interval [latex]\\left[a,b\\right][\/latex] and estimate the work done over each subinterval. So, for [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of the interval [latex]\\left[a,b\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026},n,[\/latex] choose an arbitrary point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right].[\/latex] To calculate the work done to move an object from point [latex]{x}_{i-1}[\/latex] to point [latex]{x}_{i},[\/latex] we assume the force is roughly constant over the interval, and use [latex]F({x}_{i}^{*})[\/latex] to approximate the force. The work done over the interval [latex]\\left[{x}_{i-1},{x}_{i}\\right],[\/latex] then, is given by<\/p>\n<div id=\"fs-id1167794065392\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{W}_{i}\\approx F({x}_{i}^{*})({x}_{i}-{x}_{i-1})=F({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793926193\">Therefore, the work done over the interval [latex]\\left[a,b\\right][\/latex] is approximately<\/p>\n<div id=\"fs-id1167794003046\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]W=\\displaystyle\\sum_{i=1}^{n} {W}_{i}\\approx \\displaystyle\\sum_{i=1}^{n} F({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793951829\">Taking the limit of this expression as [latex]n\\to \\infty[\/latex] gives us the exact value for work:<\/p>\n<div id=\"fs-id1167794212535\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]W=\\underset{n\\to \\infty }{\\text{lim}} \\displaystyle\\sum_{i=1}^{n} F({x}_{i}^{*})\\text{\u0394}x={\\displaystyle\\int }_{a}^{b}F(x)dx.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793984329\">Thus, we can define work as follows.<\/p>\n<div id=\"fs-id1167794137488\" class=\"textbox shaded\">\n<div class=\"title\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<\/div>\n<p id=\"fs-id1167794094283\">If a variable force [latex]F(x)[\/latex] moves an object in a positive direction along the [latex]x[\/latex]-axis from point [latex]a[\/latex] to point [latex]b[\/latex], then the <strong>work<\/strong> done on the object is<\/p>\n<div id=\"fs-id1167793879581\" class=\"equation\" style=\"text-align: center;\">[latex]W={\\displaystyle\\int }_{a}^{b}F(x)dx[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1167794122064\">Note that if <em>F<\/em> is constant, the integral evaluates to [latex]F\u00b7(b-a)=F\u00b7d,[\/latex] which is the formula we stated at the beginning of this section.<\/p>\n<p id=\"fs-id1167793249357\">Now let\u2019s look at the specific example of the work done to compress or elongate a spring. Consider a block attached to a horizontal spring. The block moves back and forth as the spring stretches and compresses. Although in the real world we would have to account for the force of friction between the block and the surface on which it is resting, we ignore friction here and assume the block is resting on a frictionless surface. When the spring is at its natural length (at rest), the system is said to be at equilibrium. In this state, the spring is neither elongated nor compressed, and in this equilibrium position the block does not move until some force is introduced. We orient the system such that [latex]x=0[\/latex] corresponds to the equilibrium position (see the following figure).<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213202\/CNX_Calc_Figure_06_05_004.jpg\" alt=\"&quot;This\" \/><\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793633520\">According to<strong> Hooke\u2019s law<\/strong>, the force required to compress or stretch a spring from an equilibrium position is given by [latex]F(x)=kx,[\/latex] for some constant [latex]k.[\/latex] The value of [latex]k[\/latex] depends on the physical characteristics of the spring. The constant [latex]k[\/latex] is called the <span class=\"no-emphasis\"><em>spring constant<\/em><\/span> and is always positive. We can use this information to calculate the work done to compress or elongate a spring, as shown in the following example.<\/p>\n<div id=\"fs-id1167793298896\" class=\"textbook exercises\">\n<h3>Example: The Work Required to Stretch or Compress a Spring<\/h3>\n<p>Suppose it takes a force of 10 N (in the negative direction) to compress a spring 0.2 m from the equilibrium position. How much work is done to stretch the spring 0.5 m from the equilibrium position?<\/p>\n<div id=\"fs-id1167794099463\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794060082\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794060082\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794060082\">First find the spring constant, [latex]k.[\/latex] When [latex]x=-0.2,[\/latex] we know [latex]F(x)=-10,[\/latex] so<\/p>\n<div id=\"fs-id1167793619889\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill F(x)& =\\hfill & kx\\hfill \\\\ \\hfill -10& =\\hfill & k(-0.2)\\hfill \\\\ \\hfill k& =\\hfill & 50\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793432264\">and [latex]F(x)=50x.[\/latex] Then, to calculate work, we integrate the force function, obtaining<\/p>\n<div id=\"fs-id1167794038347\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]W={\\displaystyle\\int }_{a}^{b}F(x)dx={\\displaystyle\\int }_{0}^{0.5}50xdx={25{x}^{2}|}_{0}^{0.5}=6.25.[\/latex]<\/div>\n<p>The work done to stretch the spring is 6.25 J.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Suppose it takes a force of 8 lb to stretch a spring 6 in. from the equilibrium position. How much work is done to stretch the spring 1 ft from the equilibrium position?<\/p>\n<div id=\"fs-id1167793929416\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q315964\">Hint<\/span><\/p>\n<div id=\"q315964\" class=\"hidden-answer\" style=\"display: none\">Use the process from the previous example. Be careful with units.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793281283\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793281283\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793281283\">8 ft-lb<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=811&amp;end=903&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.5PhysicalApplications811to903_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.5 Physical Applications&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5667\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5667&theme=oea&iframe_resize_id=ohm5667&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1167793455240\" class=\"bc-section section\">\n<h2>Work Done in Pumping<\/h2>\n<p id=\"fs-id1167793455245\">Consider the work done to pump water (or some other liquid) out of a tank. Pumping problems are a little more complicated than spring problems because many of the calculations depend on the shape and size of the tank. In addition, instead of being concerned about the work done to move a single mass, we are looking at the work done to move a volume of water, and it takes more work to move the water from the bottom of the tank than it does to move the water from the top of the tank.<\/p>\n<p id=\"fs-id1167793502586\">We examine the process in the context of a cylindrical tank, then look at a couple of examples using tanks of different shapes. Assume a cylindrical tank of radius 4 m and height 10 m is filled to a depth of 8 m. How much work does it take to pump all the water over the top edge of the tank?<\/p>\n<p id=\"fs-id1167793502601\">The first thing we need to do is define a frame of reference. We let [latex]x[\/latex] represent the vertical distance below the top of the tank. That is, we orient the [latex]x\\text{-axis}[\/latex] vertically, with the origin at the top of the tank and the downward direction being positive (see the following figure).<\/p>\n<div style=\"width: 491px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213205\/CNX_Calc_Figure_06_05_005.jpg\" alt=\"This figure is a right circular cylinder that is vertical. It represents a tank of water. The radius of the cylinder is 4 m, the height of the cylinder is 10 m. The height of the water inside the cylinder is 8 m. There is also a horizontal line on top of the tank representing the x=0. A line is drawn vertical beside the cylinder with a downward arrow labeled x.\" width=\"481\" height=\"379\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. How much work is needed to empty a tank partially filled with water?<\/p>\n<\/div>\n<p id=\"fs-id1167793479906\">Using this coordinate system, the water extends from [latex]x=2[\/latex] to [latex]x=10.[\/latex] Therefore, we partition the interval [latex]\\left[2,10\\right][\/latex] and look at the work required to lift each individual \u201clayer\u201d of water. So, for [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of the interval [latex]\\left[2,10\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026},n,[\/latex] choose an arbitrary point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right].[\/latex] (Figure 6) shows a representative layer.<\/p>\n<div style=\"width: 481px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213208\/CNX_Calc_Figure_06_05_006.jpg\" alt=\"This figure is a right circular cylinder representing a tank of water. Inside of the cylinder is a layer of water with thickness delta x. The thickness begins at xsub(i-1) and ends at xsubi.\" width=\"471\" height=\"251\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. A representative layer of water.<\/p>\n<\/div>\n<p id=\"fs-id1167793720072\">In pumping problems, the force required to lift the water to the top of the tank is the force required to overcome gravity, so it is equal to the weight of the water. Given that the weight-density of water is 9800 N\/m<sup>3<\/sup>, or 62.4 lb\/ft<sup>3<\/sup>, calculating the volume of each layer gives us the weight. In this case, we have<\/p>\n<div id=\"fs-id1167793506225\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V=\\pi {(4)}^{2}\\text{\u0394}x=16\\pi \\text{\u0394}x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793277647\">Then, the force needed to lift each layer is<\/p>\n<div id=\"fs-id1167793277650\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]F=9800\u00b716\\pi \\text{\u0394}x=156,800\\pi \\text{\u0394}x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793937471\">Note that this step becomes a little more difficult if we have a noncylindrical tank. We look at a noncylindrical tank in the next example.<\/p>\n<p id=\"fs-id1167793937476\">We also need to know the distance the water must be lifted. Based on our choice of coordinate systems, we can use [latex]{x}_{i}^{*}[\/latex] as an approximation of the distance the layer must be lifted. Then the work to lift the [latex]i\\text{th}[\/latex] layer of water [latex]{W}_{i}[\/latex] is approximately<\/p>\n<div id=\"fs-id1167794165423\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{W}_{i}\\approx 156,800\\pi {x}_{i}^{*}\\text{\u0394}x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793886516\">Adding the work for each layer, we see the approximate work to empty the tank is given by<\/p>\n<div id=\"fs-id1167793473606\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]W=\\underset{i=1}{\\overset{n}{\\text{\u2211}}}{W}_{i}\\approx \\underset{i=1}{\\overset{n}{\\text{\u2211}}}156,800\\pi {x}_{i}^{*}\\text{\u0394}x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793725962\">This is a Riemann sum, so taking the limit as [latex]n\\to \\infty ,[\/latex] we get<\/p>\n<div id=\"fs-id1167793355091\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill W& =\\underset{n\\to \\infty }{\\text{lim}}\\underset{i=1}{\\overset{n}{\\text{\u2211}}}156,800\\pi {x}_{i}^{*}\\text{\u0394}x\\hfill \\\\ & =156,800\\pi {\\displaystyle\\int }_{2}^{10}xdx\\hfill \\\\ & =156,800\\pi {\\left[\\frac{{x}^{2}}{2}\\right]|}_{2}^{10}=7,526,400\\pi \\approx 23,644,883.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793546878\">The work required to empty the tank is approximately 23,650,000 J.<\/p>\n<p>For pumping problems, the calculations vary depending on the shape of the tank or container. The following problem-solving strategy lays out a step-by-step process for solving pumping problems.<\/p>\n<div id=\"fs-id1167793546887\" class=\"textbox examples\">\n<h3>Problem-Solving Strategy: Solving Pumping Problems<\/h3>\n<ol id=\"fs-id1167794051212\">\n<li>Sketch a picture of the tank and select an appropriate frame of reference.<\/li>\n<li>Calculate the volume of a representative layer of water.<\/li>\n<li>Multiply the volume by the weight-density of water to get the force.<\/li>\n<li>Calculate the distance the layer of water must be lifted.<\/li>\n<li>Multiply the force and distance to get an estimate of the work needed to lift the layer of water.<\/li>\n<li>Sum the work required to lift all the layers. This expression is an estimate of the work required to pump out the desired amount of water, and it is in the form of a Riemann sum.<\/li>\n<li>Take the limit as [latex]n\\to \\infty[\/latex] and evaluate the resulting integral to get the exact work required to pump out the desired amount of water.<\/li>\n<\/ol>\n<\/div>\n<p id=\"fs-id1167793505372\">We now apply this problem-solving strategy in an example with a noncylindrical tank.<\/p>\n<div id=\"fs-id1167793505375\" class=\"textbook exercises\">\n<h3>Example: A Pumping Problem with a Noncylindrical Tank<\/h3>\n<p>Assume a tank in the shape of an inverted cone, with height 12 ft and base radius 4 ft. The tank is full to start with, and water is pumped over the upper edge of the tank until the height of the water remaining in the tank is 4 ft. How much work is required to pump out that amount of water?<\/p>\n<div id=\"fs-id1167793505377\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793395586\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793395586\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793395586\">The tank is depicted in Figure 7. As we did in the example with the cylindrical tank, we orient the [latex]x\\text{-axis}[\/latex] vertically, with the origin at the top of the tank and the downward direction being positive (step 1).<\/p>\n<div style=\"width: 293px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213212\/CNX_Calc_Figure_06_05_007.jpg\" alt=\"This figure is an upside-down cone. The cone has an axis through the center. The top of the cone on the axis is labeled x=0.\" width=\"283\" height=\"486\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. A water tank in the shape of an inverted cone.<\/p>\n<\/div>\n<p id=\"fs-id1167793385041\">The tank starts out full and ends with 4 ft of water left, so, based on our chosen frame of reference, we need to partition the interval [latex]\\left[0,8\\right].[\/latex] Then, for [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of the interval [latex]\\left[0,8\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026},n,[\/latex] choose an arbitrary point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right].[\/latex] We can approximate the volume of a layer by using a disk, then use similar triangles to find the radius of the disk (see the following figure).<\/p>\n<div style=\"width: 756px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213214\/CNX_Calc_Figure_06_05_008.jpg\" alt=\"This figure has two images. The first has the x-axis. Below the axis, on a slant is a line segment extending up to the x-axis. Beside the line segment is a horizontal right circular cylinder. The second image has a triangle. The right triangle mirrors the first image with the hypotenuse the line segment in the first image. The top of the triangle is 4 units. the length of the vertical side is 12 units. The vertical side is also divided into two parts; the first is xsubi, the second is 12-xsubi. It is divided at the level where the first image has the cylinder.\" width=\"746\" height=\"561\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8. Using similar triangles to express the radius of a disk of water.<\/p>\n<\/div>\n<p id=\"fs-id1167793221543\">From properties of similar triangles, we have<\/p>\n<div id=\"fs-id1167793221547\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{{r}_{i}}{12-{x}_{i}^{*}}& =\\hfill & \\frac{4}{12}=\\frac{1}{3}\\hfill \\\\ \\hfill 3{r}_{i}& =\\hfill & 12-{x}_{i}^{*}\\hfill \\\\ \\hfill {r}_{i}& =\\hfill & \\frac{12-{x}_{i}^{*}}{3}\\hfill \\\\ & =\\hfill & 4-\\frac{{x}_{i}^{*}}{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793419284\">Then the volume of the disk is<\/p>\n<div id=\"fs-id1167793419287\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{V}_{i}=\\pi {(4-\\frac{{x}_{i}^{*}}{3})}^{2}\\text{\u0394}x\\text{(step 2).}[\/latex]<\/div>\n<p id=\"fs-id1167793287424\">The weight-density of water is 62.4 lb\/ft<sup>3<\/sup>, so the force needed to lift each layer is approximately<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{F}_{i}\\approx 62.4\\pi {(4-\\frac{{x}_{i}^{*}}{3})}^{2}\\text{\u0394}x\\text{(step 3).}[\/latex]<\/div>\n<p id=\"fs-id1167793628639\">Based on the diagram, the distance the water must be lifted is approximately [latex]{x}_{i}^{*}[\/latex] feet (step 4), so the approximate work needed to lift the layer is<\/p>\n<div id=\"fs-id1167793929624\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{W}_{i}\\approx 62.4\\pi {x}_{i}^{*}{(4-\\frac{{x}_{i}^{*}}{3})}^{2}\\text{\u0394}x\\text{(step 5).}[\/latex]<\/div>\n<p id=\"fs-id1167793250306\">Summing the work required to lift all the layers, we get an approximate value of the total work:<\/p>\n<div id=\"fs-id1167793570048\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]W=\\underset{i=1}{\\overset{n}{\\text{\u2211}}}{W}_{i}\\approx \\underset{i=1}{\\overset{n}{\\text{\u2211}}}62.4\\pi {x}_{i}^{*}{(4-\\frac{{x}_{i}^{*}}{3})}^{2}\\text{\u0394}x\\text{(step 6).}[\/latex]<\/div>\n<p id=\"fs-id1167793400817\">Taking the limit as [latex]n\\to \\infty ,[\/latex] we obtain<\/p>\n<div id=\"fs-id1167793400833\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill W& =\\underset{n\\to \\infty }{\\text{lim}}\\underset{i=1}{\\overset{n}{\\text{\u2211}}}62.4\\pi {x}_{i}^{*}{(4-\\frac{{x}_{i}^{*}}{3})}^{2}\\text{\u0394}x\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{8}62.4\\pi x{(4-\\frac{x}{3})}^{2}dx\\hfill \\\\ & =62.4\\pi {\\displaystyle\\int }_{0}^{8}x(16-\\frac{8x}{3}+\\frac{{x}^{2}}{9})dx=62.4\\pi {\\displaystyle\\int }_{0}^{8}(16x-\\frac{8{x}^{2}}{3}+\\frac{{x}^{3}}{9})dx\\hfill \\\\ & =62.4\\pi {\\left[8{x}^{2}-\\frac{8{x}^{3}}{9}+\\frac{{x}^{4}}{36}\\right]|}_{0}^{8}=10,649.6\\pi \\approx 33,456.7\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793454764\">It takes approximately 33,450 ft-lb of work to empty the tank to the desired level.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: A Pumping Problem with a Noncylindrical Tank.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=1355&amp;end=1691&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.5PhysicalApplications1355to1691_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.5 Physical Applications&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1167793553739\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>A tank is in the shape of an inverted cone, with height 10 ft and base radius 6 ft. The tank is filled to a depth of 8 ft to start with, and water is pumped over the upper edge of the tank until 3 ft of water remain in the tank. How much work is required to pump out that amount of water?<\/p>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q46716\">Hint<\/span><\/p>\n<div id=\"q46716\" class=\"hidden-answer\" style=\"display: none\">Use the process from the previous example.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167793553742\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793553760\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793553760\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793553760\">Approximately 43,255.2 ft-lb<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5684\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5684&theme=oea&iframe_resize_id=ohm5684&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Hydrostatic Force and Pressure<\/h2>\n<p id=\"fs-id1167793281535\">In this last section, we look at the force and pressure exerted on an object submerged in a liquid. In the English system, force is measured in pounds. In the metric system, it is measured in newtons. Pressure is force per unit area, so in the English system we have pounds per square foot (or, perhaps more commonly, pounds per square inch, denoted psi). In the metric system we have newtons per square meter, also called <span class=\"no-emphasis\"><em>pascals<\/em><\/span>.<\/p>\n<p id=\"fs-id1167793630279\">Let\u2019s begin with the simple case of a plate of area [latex]A[\/latex] submerged horizontally in water at a depth [latex]s[\/latex] (Figure 9). Then, the force exerted on the plate is simply the weight of the water above it, which is given by [latex]F=\\rho As,[\/latex] where [latex]\\rho[\/latex] is the weight density of water (weight per unit volume). To find the <strong>hydrostatic pressure<\/strong>\u2014that is, the pressure exerted by water on a submerged object\u2014we divide the force by the area. So the pressure is [latex]p=F\\text{\/}A=\\rho s.[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213218\/CNX_Calc_Figure_06_05_009.jpg\" alt=\"This image has a circular plate submerged in water. The plate is labeled A and the depth of the water is labeled s.\" width=\"487\" height=\"324\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 9. A plate submerged horizontally in water.<\/p>\n<\/div>\n<p id=\"fs-id1167793553642\">By <span class=\"no-emphasis\">Pascal\u2019s principle<\/span>, the pressure at a given depth is the same in all directions, so it does not matter if the plate is submerged horizontally or vertically. So, as long as we know the depth, we know the pressure. We can apply Pascal\u2019s principle to find the force exerted on surfaces, such as dams, that are oriented vertically. We cannot apply the formula [latex]F=\\rho As[\/latex] directly, because the depth varies from point to point on a vertically oriented surface. So, as we have done many times before, we form a partition, a Riemann sum, and, ultimately, a definite integral to calculate the force.<\/p>\n<p id=\"fs-id1167793553670\">Suppose a thin plate is submerged in water. We choose our frame of reference such that the [latex]x[\/latex]-axis is oriented vertically, with the downward direction being positive, and point [latex]x=0[\/latex] corresponding to a logical reference point. Let [latex]s(x)[\/latex] denote the depth at point [latex]x[\/latex]. Note we often let [latex]x=0[\/latex] correspond to the surface of the water. In this case, depth at any point is simply given by [latex]s(x)=x.[\/latex] However, in some cases we may want to select a different reference point for [latex]x=0,[\/latex] so we proceed with the development in the more general case. Last, let [latex]w(x)[\/latex] denote the width of the plate at the point [latex]x.[\/latex]<\/p>\n<p id=\"fs-id1167794127192\">Assume the top edge of the plate is at point [latex]x=a[\/latex] and the bottom edge of the plate is at point [latex]x=b.[\/latex] Then, for [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of the interval [latex]\\left[a,b\\right],[\/latex] and for [latex]i=1,2\\text{,\u2026},n,[\/latex] choose an arbitrary point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right].[\/latex] The partition divides the plate into several thin, rectangular strips (see the following figure).<\/p>\n<div style=\"width: 581px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213221\/CNX_Calc_Figure_06_05_010.jpg\" alt=\"This image is the overhead view of a submerged circular plate. The x-axis is to the side of the plate. The plate\u2019s diameter goes from x=a to x=b. There is a strip in the middle of the plate with thickness of delta x. On the axis this thickness begins at x=xsub(i-1) and ends at x=xsubi. The length of the strip in the plate is labeled w(csubi).\" width=\"571\" height=\"350\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 10. A thin plate submerged vertically in water.<\/p>\n<\/div>\n<p id=\"fs-id1167793420981\">Let\u2019s now estimate the force on a representative strip. If the strip is thin enough, we can treat it as if it is at a constant depth, [latex]s({x}_{i}^{*}).[\/latex] We then have<\/p>\n<div id=\"fs-id1167793952122\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{F}_{i}=\\rho As=\\rho \\left[w({x}_{i}^{*})\\text{\u0394}x\\right]s({x}_{i}^{*})[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793584536\">Adding the forces, we get an estimate for the force on the plate:<\/p>\n<div id=\"fs-id1167793584539\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]F\\approx \\underset{i=1}{\\overset{n}{\\text{\u2211}}}{F}_{i}=\\underset{i=1}{\\overset{n}{\\text{\u2211}}}\\rho \\left[w({x}_{i}^{*})\\text{\u0394}x\\right]s({x}_{i}^{*})[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794057934\">This is a Riemann sum, so taking the limit gives us the exact force. We obtain<\/p>\n<div id=\"fs-id1167794057937\" class=\"equation\" style=\"text-align: center;\">[latex]F=\\underset{n\\to \\infty }{\\text{lim}}\\underset{i=1}{\\overset{n}{\\text{\u2211}}}\\rho \\left[w({x}_{i}^{*})\\text{\u0394}x\\right]s({x}_{i}^{*})={\\displaystyle\\int }_{a}^{b}\\rho w(x)s(x)dx[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794291708\">Evaluating this integral gives us the force on the plate. We summarize this in the following problem-solving strategy.<\/p>\n<div id=\"fs-id1167794291713\" class=\"textbox examples\">\n<h3>Problem-Solving Strategy: Finding Hydrostatic Force<\/h3>\n<ol id=\"fs-id1167794291719\">\n<li>Sketch a picture and select an appropriate frame of reference. (Note that if we select a frame of reference other than the one used earlier, we may have to adjust the equation above accordingly.)<\/li>\n<li>Determine the depth and width functions, [latex]s(x)[\/latex] and [latex]w(x).[\/latex]<\/li>\n<li>Determine the weight-density of whatever liquid with which you are working. The weight-density of water is 62.4 lb\/ft<sup>3<\/sup>, or 9800 N\/m<sup>3<\/sup>.<\/li>\n<li>Use the equation to calculate the total force.<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1167793504251\" class=\"textbook exercises\">\n<h3>Example: Finding Hydrostatic Force<\/h3>\n<p>A water trough 15 ft long has ends shaped like inverted isosceles triangles, with base 8 ft and height 3 ft. Find the force on one end of the trough if the trough is full of water.<\/p>\n<div id=\"fs-id1167793504253\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793385052\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793385052\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793385052\">Figure 11 shows the trough and a more detailed view of one end.<\/p>\n<div style=\"width: 588px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213225\/CNX_Calc_Figure_06_05_011.jpg\" alt=\"This figure has two images. The first is a water trough with rectangular sides. The length of the trough is 15 feet, the depth is 3 feet, and the width is 8 feet. The second image is a cross section of the trough. It is a triangle. The top has length of 8 feet and the sides have length 5 feet. The altitude is labeled with 3 feet.\" width=\"578\" height=\"634\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 11. (a) A water trough with a triangular cross-section. (b) Dimensions of one end of the water trough.<\/p>\n<\/div>\n<p id=\"fs-id1167793385076\">Select a frame of reference with the [latex]x\\text{-axis}[\/latex] oriented vertically and the downward direction being positive. Select the top of the trough as the point corresponding to [latex]x=0[\/latex] (step 1). The depth function, then, is [latex]s(x)=x.[\/latex] Using similar triangles, we see that [latex]w(x)=8-(8\\text{\/}3)x[\/latex] (step 2). Now, the weight density of water is 62.4 lb\/ft<sup>3<\/sup> (step 3), so applying the force equation from above,\u00a0we obtain<\/p>\n<div id=\"fs-id1167794054206\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill F& ={\\displaystyle\\int }_{a}^{b}\\rho w(x)s(x)dx\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{3}62.4(8-\\frac{8}{3}x)xdx=62.4{\\displaystyle\\int }_{0}^{3}(8x-\\frac{8}{3}{x}^{2})dx\\hfill \\\\ & =62.4{\\left[4{x}^{2}-\\frac{8}{9}{x}^{3}\\right]|}_{0}^{3}=748.8.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793445714\">The water exerts a force of 748.8 lb on the end of the trough (step 4).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793445721\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>A water trough 12 m long has ends shaped like inverted isosceles triangles, with base 6 m and height 4 m. Find the force on one end of the trough if the trough is full of water.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q319826\">Hint<\/span><\/p>\n<div id=\"q319826\" class=\"hidden-answer\" style=\"display: none\">Follow the problem-solving strategy and the process from the previous example.<\/div>\n<\/div>\n<div id=\"fs-id1167793445724\" class=\"exercise\">\n<div class=\"solution\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q388525\">Show Solution<\/span><\/p>\n<div id=\"q388525\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793421193\">156,800 N<\/p>\n<\/div>\n<div id=\"fs-id1167793421202\" class=\"commentary\">\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=2006&amp;end=2135&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266832\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266832\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.5PhysicalApplications2006to2135_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.5 Physical Applications&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1167794138214\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>When the reservoir is at its average level, the surface of the water is about 50 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?<\/p>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q989393\">Hint<\/span><\/p>\n<div id=\"q989393\" class=\"hidden-answer\" style=\"display: none\">Change the depth function, [latex]s(x),[\/latex] and the limits of integration.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167794138217\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794138229\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794138229\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794138229\">Approximately 7,164,520,000 lb or 3,582,260 t<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p id=\"fs-id1167791543256\"><a href=\"https:\/\/www.history.com\/topics\/great-depression\/hoover-dam\" target=\"_blank\" rel=\"noopener\">To learn more about Hoover Dam, see this article published by the History Channel.<\/a><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5670\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5670&theme=oea&iframe_resize_id=ohm5670&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/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-1180\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>2.5 Physical Applications. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/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":17533,"menu_order":20,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"2.5 Physical Applications\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1180","chapter","type-chapter","status-publish","hentry"],"part":1160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1180","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1180\/revisions"}],"predecessor-version":[{"id":1651,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1180\/revisions\/1651"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/1160"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1180\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1180"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1180"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1180"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1180"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}