{"id":28537,"date":"2024-09-16T15:44:16","date_gmt":"2024-09-16T15:44:16","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=28537"},"modified":"2024-09-16T15:44:16","modified_gmt":"2024-09-16T15:44:16","slug":"newtons-laws","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/en\/newtons-laws\/","title":{"rendered":"Newton&#8217;s Laws"},"content":{"rendered":"<p><center><\/p>\n<h1>Newton&#8217;s Laws<\/h1>\n<p><em><strong>Summary:<\/strong><br \/>\n    This class addresses Newton&#8217;s laws and their role in the dynamics of bodies. It explores how mass and velocity determine linear momentum and describes the three laws: inertia, which maintains the state of motion in the absence of external forces, the relationship between force and acceleration, and the action and reaction between bodies. Through examples such as sliding on planes and the motion of pendulums, the application of these laws is illustrated, culminating in practical exercises to consolidate learning.<\/em><\/p>\n<p><strong>Learning Objectives:<\/strong><br \/>\n    By the end of this class, the student will be able to:<\/p>\n<ol style=\"text-align:left;\">\n<li><strong>Understand<\/strong> the three laws of Newton and their application in the dynamics of bodies.<\/li>\n<li><strong>Apply<\/strong> Newton&#8217;s laws to analyze and solve dynamic problems.<\/li>\n<li><strong>Identify<\/strong> the relationship between mass, velocity, and linear momentum.<\/li>\n<li><strong>Analyze<\/strong> the importance of inertial observers in the study of dynamics.<\/li>\n<li><strong>Explain<\/strong> how Newton&#8217;s second law relates force and acceleration.<\/li>\n<li><strong>Describe<\/strong> the concept of inertial mass and how to compare it between different bodies.<\/li>\n<\/ol>\n<p>    <strong>TABLE OF CONTENTS<\/strong><br \/>\n    <a href=\"#1\">Introduction<\/a><br \/>\n    <a href=\"#2\">Newton&#8217;s Laws on the Dynamics of Bodies<\/a><br \/>\n    <a href=\"#3\">How to Use Newton&#8217;s Laws?<\/a><br \/>\n    <a href=\"#4\">Problem Solving Using Newton&#8217;s Laws<\/a><br \/>\n    <\/center><br \/>\n    <a name=\"1\"><\/a><\/p>\n<h2>Introduction<\/h2>\n<p style=\"text-align: justify;\">If the kinematics we have reviewed in the <a href=\"https:\/\/toposuranos.com\/posicion-velocidad-y-aceleracion-cinematica\/\" rel=\"noopener\" target=\"_blank\">previous classes<\/a> allow us to describe the motion of bodies, through Newton&#8217;s laws we obtain the dynamics that allow us to reason about the causes of motion (or changes in the state of motion). Here, the ideas of position and time are important because in terms of these we define velocity and acceleration, but an additional one is added: mass.<\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Tcp_1M1a7H0&amp;t=127s\" target=\"_blank\" rel=\"noopener\"><strong>Mass is important to define the state of motion<\/strong><\/a> of bodies, or linear momentum. It is said that the linear momentum of a body, <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{p}<\/span>, is the product of mass and velocity<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Large \\vec{p}=m\\vec{v}<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">The state of motion is the key idea behind Newton&#8217;s laws.<\/p>\n<p>    <center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/Tcp_1M1a7H0\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p>    <a name=\"2\"><\/a><\/p>\n<h2>Newton&#8217;s Laws on the Dynamics of Bodies<\/h2>\n<h4>First Law (Inertia):<\/h4>\n<p style=\"text-align: center; color: #000080; background-color: #80ff80;\"><strong>In the absence of external agents, all bodies maintain a constant state of motion.<\/strong><\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Tcp_1M1a7H0&amp;t=208s\" target=\"_blank\" rel=\"noopener\"><strong>The first of Newton&#8217;s laws<\/strong><\/a> has the brilliance of establishing two issues of profound importance for physics. The first and most evident: it establishes linear momentum as a conserved quantity; and the second and equally important, but much more implicit, allows us to establish what an <strong>inertial observer<\/strong> is.<\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Tcp_1M1a7H0&amp;t=322s\" target=\"_blank\" rel=\"noopener\"><strong>There are many ways to define an observer,<\/strong><\/a> but among them, there is a special class that we call an inertial observer. The difference lies in that, from the perspective of an inertial observer, in the absence of an external agent, the state of motion of bodies is a conservative quantity.<\/p>\n<h5>What distinguishes an inertial observer from a non-inertial one?<\/h5>\n<p style=\"text-align: justify;\">The difference lies in that, from the perspective of an inertial observer, in the absence of an external agent, the state of motion of bodies is a conservative quantity.<\/p>\n<h4>Second Law (Force and Mass):<\/h4>\n<p style=\"text-align: center; color: #000080; background-color: #80ff80;\"><strong>From the perspective of an inertial observer, the force exerted by an external agent on a body is equivalent to the variation of its state of motion.<\/strong><\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Tcp_1M1a7H0&amp;t=396s\" target=\"_blank\" rel=\"noopener\"><strong>In other words, if a force is applied to a body<\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}<\/span>, then it will have that.<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Large \\displaystyle \\vec{F}=\\frac{d\\vec{p}}{dt}<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">The well-known relationship \u00abforce equals mass times acceleration,\u00bb <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}=m\\vec{a},<\/span><\/span> is just a consequence of Newton&#8217;s second law, which is obtained from the properties of derivatives and the conservation of mass.<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n    \\vec{F} &amp; =\\displaystyle \\frac{d\\vec{p}}{dt} = \\frac{d}{dt}\\left(m\\vec{v} \\right) \\\\ \\\\\n\n    &amp; =\\displaystyle \\underbrace{\\frac{dm}{dt}}_{= 0}\\vec{v} + m \\underbrace{\\frac{d\\vec{v}}{dt}}_{= \\vec{a}} = m\\vec{a}\n\n    \\end{array}<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">In this last step, it has been considered that <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">dm\/dt=0<\/span><\/span> because it is assumed that no mass is being added or removed, and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d\\vec{v}\/dt<\/span><\/span> is the definition of acceleration.<\/p>\n<h5>Inertial Mass<\/h5>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Tcp_1M1a7H0&amp;t=572s\" target=\"_blank\" rel=\"noopener\"><strong>The second of Newton&#8217;s laws<\/strong><\/a> also allows us to specify the concept of mass. Here it appears as a proportionality constant between force and acceleration. The greater the mass, the greater the force required to achieve the same acceleration; as a result, mass is understood as a measure of the inertia of bodies and hence the name <strong>inertial mass.<\/strong> If two bodies at rest relative to an inertial observer are acted upon by the same force (without exchanging matter), then we have that<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m_1 \\vec{a}_1 = \\vec{F} = m_2 \\vec{a}_2<\/span><\/span><\/p>\n<p style=\"text-align:strong>\n    At the end of this class, the student will be able to:<\/p>\n<ol style=\"text-align:left;\">\n<li><strong>Understand<\/strong> Newton&#8217;s three laws and their application in the dynamics of bodies.<\/li>\n<li><strong>Apply<\/strong> Newton&#8217;s laws to analyze and solve dynamics problems.<\/li>\n<li><strong>Identify<\/strong> the relationship between mass, velocity, and linear momentum.<\/li>\n<li><strong>Analyze<\/strong> the importance of inertial observers in the study of dynamics.<\/li>\n<li><strong>Explain<\/strong> how Newton&#8217;s second law relates force and acceleration.<\/li>\n<li><strong>Describe<\/strong> the concept of inertial mass and how to compare it between different bodies.<\/li>\n<\/ol>\n<p>    <strong>CONTENT INDEX<\/strong><br \/>\n    <a href=\"#1\">Introduction<\/a><br \/>\n    <a href=\"#2\">Newton&#8217;s Laws on the Dynamics of Bodies<\/a><br \/>\n    <a href=\"#3\">How to Use Newton&#8217;s Laws?<\/a><br \/>\n    <a href=\"#4\">Problem Solving Using Newton&#8217;s Laws<\/a><br \/>\n    <\/center><br \/>\n    <a name=\"1\"><\/a><\/p>\n<h2>Introduction<\/h2>\n<p style=\"text-align: justify;\">If the kinematics we reviewed in the <a href=\"https:\/\/toposuranos.com\/posicion-velocidad-y-aceleracion-cinematica\/\" rel=\"noopener\" target=\"_blank\">previous classes<\/a> allow us to describe the motion of bodies, Newton&#8217;s laws give us the dynamics that allow us to reason about the causes of motion (or changes in the state of motion). Here, the ideas of position and time are important because, in terms of these, we define velocity and acceleration, but an additional one is added: mass.<\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Tcp_1M1a7H0&amp;t=127s\" target=\"_blank\" rel=\"noopener\"><strong>Mass is important to define the state of motion<\/strong><\/a> of bodies or linear momentum. It is said that the linear momentum of a body, <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{p}<\/span>, is the product of mass times velocity<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Large \\vec{p}=m\\vec{v}<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">The state of motion is the key idea behind Newton&#8217;s laws.<\/p>\n<p>    <center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/Tcp_1M1a7H0\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p>    <a name=\"2\"><\/a><\/p>\n<h2>Newton&#8217;s Laws on the Dynamics of Bodies<\/h2>\n<h4>First Law (Inertia):<\/h4>\n<p style=\"text-align: center; color: #000080; background-color: #80ff80;\"><strong>In the absence of external agents, all bodies maintain a constant state of motion.<\/strong><\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Tcp_1M1a7H0&amp;t=208s\" target=\"_blank\" rel=\"noopener\"><strong>The first of Newton&#8217;s laws<\/strong><\/a> has the genius to establish two issues of profound importance for physics. The first and most obvious: it establishes linear momentum as a conserved quantity; and the second, equally important but much more implicit, allows us to establish what an <strong>inertial observer<\/strong> is.<\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Tcp_1M1a7H0&amp;t=322s\" target=\"_blank\" rel=\"noopener\"><strong>There are many ways to define an observer,<\/strong><\/a> but among them, there is a special class that we call an inertial observer. The difference lies in that, from the perspective of an inertial observer, in the absence of an external agent, the state of motion of bodies is a conservative quantity.<\/p>\n<h5>What distinguishes an inertial observer from one that is not?<\/h5>\n<p style=\"text-align: justify;\">The difference lies in that, from the perspective of an inertial observer, in the absence of an external agent, the state of motion of bodies is a conservative quantity.<\/p>\n<h4>Second Law (Force and Mass):<\/h4>\n<p style=\"text-align: center; color: #000080; background-color: #80ff80;\"><strong>From the perspective of an inertial observer, the force exerted by an external agent on a body is equivalent to the variation of its state of motion.<\/strong><\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Tcp_1M1a7H0&amp;t=396s\" target=\"_blank\" rel=\"noopener\"><strong>In other words, if a force is applied to a body<\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}<\/span> then it will be that<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Large \\displaystyle \\vec{F}=\\frac{d\\vec{p}}{dt}<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">The well-known relationship \u00abforce equals mass times acceleration,\u00bb <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}=m\\vec{a},<\/span><\/span> is just a consequence of Newton&#8217;s Second Law, obtained from the properties of derivatives and mass conservation.<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n    \\vec{F} &amp; =\\displaystyle \\frac{d\\vec{p}}{dt} = \\frac{d}{dt}\\left(m\\vec{v} \\right) \\\\ \\\\\n\n    &amp; =\\displaystyle \\underbrace{\\frac{dm}{dt}}_{= 0}\\vec{v} + m \\underbrace{\\frac{d\\vec{v}}{dt}}_{= \\vec{a}} = m\\vec{a}\n\n    \\end{array}<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">In this last step, it has been considered that <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">dm\/dt=0<\/span><\/span> because it is assumed that no mass is being added or removed, and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d\\vec{v}\/dt<\/span><\/span> is the definition of acceleration.<\/p>\n<h5>Inertial Mass<\/h5>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Tcp_1M1a7H0&amp;t=572s\" target=\"_blank\" rel=\"noopener\"><strong>Newton&#8217;s second law<\/strong><\/a> also allows us to clarify the concept of mass. Here it appears as a constant of proportionality between force and acceleration. The greater the mass, the greater the force must be to achieve the same acceleration; as a result, mass is understood as a measure of the inertia of bodies, hence the name <strong>inertial mass<\/strong>. If the same force (without exchange of matter) acts on two bodies at rest relative to an inertial observer, then it is that<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m_1 \\vec{a}_1 = \\vec{F} = m_2 \\vec{a}_2<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">From this, we can compare the masses of the bodies through the quotient of the magnitudes of the accelerations<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{m_1}{m_2} = \\frac{\\|\\vec{a_2}\\|}{\\|\\vec{a_1}\\|}<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">Therefore, if <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m_2<\/span><\/span> were a \u00abstandard kilogram,\u00bb then it is enough to observe the quotient <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{a}_2\\|\/\\|\\vec{a}_1\\|<\/span><\/span> to know how many kilograms <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m_1<\/span><\/span> has.<\/p>\n<h4>Third Law (Action and Reaction):<\/h4>\n<p style=\"text-align: center; color: #000080; background-color: #80ff80;\"><strong>If body A exerts a \u00abaction\u00bb force on another body B, then B exerts a \u00abreaction\u00bb force on A of equal magnitude but opposite direction.<\/strong><\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=Tcp_1M1a7H0&amp;t=794s\" target=\"_blank\" rel=\"noopener\"><strong>Newton&#8217;s third law not only allows us to speak<\/strong><\/a> with greater precision about forces but also explicitly states that external agents applying a force are also physical objects susceptible to them:<\/p>\n<ul style=\"text-align: justify;\">\n<li>The external agent is a physical object that can be affected by forces.<\/li>\n<li>Forces never occur in isolation but rather in pairs called \u00abaction-reaction pairs,\u00bb and the vector sum of these pairs is always zero.<\/li>\n<li>Action-reaction pairs always occur on different bodies, so the total force on a body is not necessarily zero.<\/li>\n<\/ul>\n<p style=\"text-align: justify;\">Since action-reaction pairs always act on a straight line, this brings what we will see later as the conservation of angular momentum.<\/p>\n<p style=\"text-align: justify;\">Apart from these things, Newton&#8217;s third law implies a few other things:<\/p>\n<ul style=\"text-align: justify;\">\n<li>For a non-zero net force to be applied to a body, at least a second object is necessary.<\/li>\n<li>Action and reaction occur simultaneously. Since two bodies can interact at a distance (through gravitation or electromagnetism), it follows that in Newtonian mechanics, there must be a way to transmit information from one point to another with infinite speed. We know that such a thing is impossible because, according to special relativity, the maximum speed is the speed of light in a vacuum, so we say that this third law is an approximation of reality.<\/li>\n<\/ul>\n<p>    <center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/DuO-cvLNzwQ\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p>    <a name=\"3\"><\/a><\/p>\n<h2>How to Use Newton&#8217;s Laws?<\/h2>\n<p style=\"text-align: justify;\">To understand how to use Newton&#8217;s laws so that their meaning is clear, it is best to resort to examples based on concrete situations and the construction of free-body diagrams.<\/p>\n<h3>Free-Body Diagrams<\/h3>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=DuO-cvLNzwQ&amp;t=96s\" target=\"_blank\" rel=\"noopener\"><strong>A free-body diagram is<\/strong><\/a> a pictorial scheme where we represent the forces acting on a body. Based on what we have reviewed about weight, we can construct the following examples of free-body diagrams.<\/p>\n<h4>A body resting on a horizontal plane<\/h4>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=DuO-cvLNzwQ&amp;t=180s\" target=\"_blank\" rel=\"noopener\"><strong>Due to gravity, all bodies with mass feel a force<\/strong><\/a> directed to the ground. Through Newton&#8217;s second law, we observe that such force is given by the product of the mass and gravitational acceleration <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{g}=-g\\hat{y},<\/span><\/span> where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g=9.81[m\/s^2].<\/span><\/span><\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}_{weight}=m\\vec{g} = -mg\\hat{y}<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">What we understand as \u00abweight\u00bb of a body is actually the magnitude of this weight force that we have just seen.<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">{weight}=\\|\\vec{F}_{weight}\\|= mg<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">When we place a block on a horizontal plane, a pair of action-reaction forces appears: these are the weight and the normal force. Such forces are equal in magnitude but opposite in direction so that the vector sum of the forces on the body is zero, and therefore, its state of motion remains constant over time.<\/p>\n<p>    <center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-ylYcrS1tJbw\/YVfZYBYCPjI\/AAAAAAAAFlU\/0mOIkMhF-Hwrysc_WB5o7MLvrQtxQJu2QCLcBGAsYHQ\/s0\/reposo-plano-horizontal.PNG\" width=\"497\" height=\"280\" alt=\"body resting on a horizontal plane\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-ylYcrS1tJbw\/YVfZYBYCPjI\/AAAAAAAAFlU\/0mOIkMhF-Hwrysc_WB5o7MLvrQtxQJu2QCLcBGAsYHQ\/s0\/reposo-plano-horizontal.PNG\" width=\"497\" height=\"280\" alt=\"body resting on a horizontal plane\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<h4>Sliding on a Horizontal Plane<\/h4>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=DuO-cvLNzwQ&amp;t=343s\" target=\"_blank\" rel=\"noopener\"><strong>Let&#8217;s imagine that the block is now tied to a rope<\/strong><\/a> and we pull it as shown in the following free-body diagram:<\/p>\n<p>    <center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-a2GB2RJ7vTY\/YVgEny43ZkI\/AAAAAAAAFlk\/uZ9_XQ3JwVMiiWjZDNC5Y4ePVZ9GH1bngCLcBGAsYHQ\/s0\/fuerza-cuerpo-en-plano-horizontal.PNG\" width=\"793\" height=\"386\" alt=\"force diagram on an object on a horizontal plane\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-a2GB2RJ7vTY\/YVgEny43ZkI\/AAAAAAAAFlk\/uZ9_XQ3JwVMiiWjZDNC5Y4ePVZ9GH1bngCLcBGAsYHQ\/s0\/fuerza-cuerpo-en-plano-horizontal.PNG\" width=\"793\" height=\"386\" alt=\"force diagram on an object on a horizontal plane\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p style=\"text-align: justify;\">Here we observe the appearance of two action-reaction pairs: on the one hand, we have the pairs associated with the weight and normal forces of the bodies, there is a third action-reaction pair associated with the ends of the rope with which the subject pulls the block, and finally, a pair associated with the applied force <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\vec{F}_1<\/span><\/span> and the friction force, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}_{friction},<\/span><\/span> with a maximum value of <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mu\\|\\vec{F}_\\textnormal{normal}\\|.<\/span><\/span><\/p>\n<h5>Coefficient of Friction and Friction Forces<\/h5>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=DuO-cvLNzwQ&amp;t=458s\" target=\"_blank\" rel=\"noopener\"><strong>Here <span class=\"katex-eq\" data-katex-display=\"false\">\\mu<\/span> is the coefficient of friction<\/strong><\/a> that expresses the resistance to sliding between two surfaces; the coefficient of friction has two versions: a kinetic one (<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mu_c<\/span><\/span>) and a static one (<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mu_e<\/span><\/span>). Static friction appears when the body remains at rest, while kinetic friction appears once the body has begun to slide.<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{lcr}\\mu = \\left\\{\\begin{array}{lll} \\mu_e &amp; ;&amp; \\textnormal{Body at rest} \\\\ \\\\ \\mu_c &amp; ;&amp; \\textnormal{Body in motion} \\end{array}\\right. &amp; ; &amp; \\textnormal{Where } \\mu_c \\leq \\mu_e\\end{array}<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">The friction force opposes the motion of the body experiencing it and can be modeled (simplified) through the following expression<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n        \\vec{F}_\\textnormal{friction} ( \\vec{F}_1 ) = \\left\\{\n\n            \\begin{array}{lll}\n\n            - \\vec{F}_1 &amp; ; &amp; \\|\\vec{F}_1\\| \\leq \\mu_e \\|\\vec{F}_\\textnormal{normal}\\| \\\\ \\\\\n\n            -\\mu_c \\|\\vec{F}_\\textnormal{normal}\\|\\hat{x} &amp; ; &amp; \\mu_e \\|\\vec{F}_\\textnormal{normal}\\| \\lt \\|\\vec{F}\\|\n\n            \\end{array}\n\n    \\right.<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">When the applied force is less than or equal to the maximum static friction, the body remains at rest relative to the ground. If the applied force exceeds static friction, the body begins to move and friction becomes kinetic; the net force on the body is therefore: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}_{net} = \\vec{F}_1 - \\mu_c\\|\\vec{F}_\\textnormal{normal}\\|\\hat{x},<\/span><\/span> and therefore, it moves with an acceleration <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{a} = \\vec{F}_{net}\/M<\/span><\/span>. If once the body is in motion, the applied force equals kinetic friction, then the body moves with a constant velocity.<\/p>\n<h4>Sliding on an Inclined Plane<\/h4>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=DuO-cvLNzwQ&amp;t=753s\" target=\"_blank\" rel=\"noopener\"><strong>When an object slides on an inclined plane<\/strong><\/a> at an angle <span class=\"katex-eq\" data-katex-display=\"false\">\\alpha<\/span> the following force diagram is obtained:<\/p>\n<p>    <center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-vMApZqsSDs0\/YVjUkV187NI\/AAAAAAAAFls\/eZ2DPO6f54AKJDtB7-y1DP3TFjIMZjT7ACLcBGAsYHQ\/s0\/diagrama%2Bde%2Bfuerzas%2Ben%2Bplano%2Binclinado.PNG\" width=\"823\" height=\"554\" alt=\"Force diagram for sliding on an inclined plane\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-vMApZqsSDs0\/YVjUkV187NI\/AAAAAAAAFls\/eZ2DPO6f54AKJDtB7-y1DP3TFjIMZjT7ACLcBGAsYHQ\/s0\/diagrama%2Bde%2Bfuerzas%2Ben%2Bplano%2Binclinado.PNG\" width=\"823\" height=\"554\" alt=\"Force diagram for sliding on an inclined plane\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p style=\"text-align: justify;\">Here, for convenience, a reference system has been chosen oriented so that the horizontal coordinate is aligned with the sliding plane. In this scheme, the weight force is divided into two components: one parallel and the other perpendicular to the motion.<\/p>\n<ul>\n<li><strong>Parallel Component:<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}_{\\textnormal{weight},x}=mg\\sin(\\alpha)\\hat{x}<\/span><\/span><\/li>\n<li><strong>Perpendicular Component:<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}_{weight,y}=-mg\\cos(\\alpha)\\hat{y}<\/span><\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify;\">The friction force appears as a reaction to the parallel component of the weight force, and the normal force as a reaction to the perpendicular component of the weight force. If the horizontal component of the weight force exceeds the maximum static friction, the state of motion of the block will change with an acceleration<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\vec{a} = mg\\left(\\frac{\\sin(\\alpha) - \\mu_c \\cos(\\alpha)}{m}\\right)\\hat{x}<\/span><\/span><\/p>\n<h4>Hanging Mass<\/h4>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=DuO-cvLNzwQ&amp;t=1085s\" target=\"_blank\" rel=\"noopener\"><strong>A mass hanging from a rope attached to a ceiling<\/strong><\/a> and that remains at rest has the following free-body diagram:<\/p>\n<p>    <center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-jWSXd7lQhiw\/YVjpzr8z-pI\/AAAAAAAAFl0\/sZ955vFZz3Y-MtKXQQK8UFPRAaHhgxyPgCLcBGAsYHQ\/s0\/masa-colgante.PNG\" width=\"555\" height=\"521\" alt=\"Free-body diagram of a hanging mass\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-jWSXd7lQhiw\/YVjpzr8z-pI\/AAAAAAAAFl0\/sZ955vFZz3Y-MtKXQQK8UFPRAaHhgxyPgCLcBGAsYHQ\/s0\/masa-colgante.PNG\" width=\"555\" height=\"521\" alt=\"Free-body diagram of a hanging mass\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p style=\"text-align: justify;\">A mass hanging from a rope attached to a ceiling and that remains at rest has the following free-body diagram:<\/p>\n<p style=\"text-align: justify;\">On the rope, there is a pair of forces called \u00abtensions,\u00bb if the rope is inextensible, these forces are equal and opposite. On the block, there is also a pair of forces: weight and rope tension. If the block remains hanging and at rest, then the weight and tension are opposite and of equal magnitude. There is a fourth force not shown here, the one that holds the rope to the ceiling; these four forces form two action-reaction pairs.<\/p>\n<h4>Simple Pendulum Motion<\/h4>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=DuO-cvLNzwQ&amp;t=1158s\" target=\"_blank\" rel=\"noopener\"><strong>A mass attached to an inextensible rope, attached to a ceiling, that oscillates<\/strong><\/a> around an equilibrium position due to its own weight is what we call a simple pendulum. Below is its free-body diagram.<\/p>\n<p>    <center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-KEfWZhZZNl8\/YVkYdDhxLcI\/AAAAAAAAFl8\/CXQTZSVYbxIDw9G_JaVtV9VaG-ruwqHewCLcBGAsYHQ\/s0\/pendulo%2Bsimple.PNG\" width=\"366\" height=\"452\" alt=\"Free-body diagram of the simple pendulum\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-KEfWZhZZNl8\/YVkYdDhxLcI\/AAAAAAAAFl8\/CXQTZSVYbxIDw9G_JaVtV9VaG-ruwqHewCLcBGAsYHQ\/s0\/pendulo%2Bsimple.PNG\" width=\"366\" height=\"452\" alt=\"Free-body diagram of the simple pendulum\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p style=\"text-align: justify;\">Since the rope is inextensible, the radial acceleration is zero and, consequently:<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F_{p,\\parallel} + T = ma_{\\parallel}(t) = 0<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">On the other hand, for the component perpendicular to the rope, it will be<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">F_{p,\\bot}=-mg\\sin(\\theta) = ma_{\\bot}(t)<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">From this last expression, it is possible to infer a differential equation that will allow us to model the angular position <span class=\"katex-eq\" data-katex-display=\"false\">\\theta<\/span> of the simple pendulum over time<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d^2\\theta(t)}{dt^2} + \\frac{g}{l}\\sin(\\theta) = 0<\/span><\/span><\/p>\n<p style=\"text-align: justify;\">But the derivation of this equation and the inferences we can make from it are something we will see in detail later.<\/p>\n<p>    <a name=\"4\"><\/a><\/p>\n<h2>Problem Solving Using Newton&#8217;s Laws<\/h2>\n<p style=\"text-align: justify;\">Use Newton&#8217;s laws to solve the following problems:<\/p>\n<ol style=\"text-align: justify;\">\n<li>A block of <span class=\"katex-eq\" data-katex-display=\"false\">15[kg]<\/span> is placed on a horizontal surface. Between the block and the surface, there is static friction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mu_e=0.55<\/span><\/span> and kinetic friction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mu_c=0.31<\/span><\/span>\n<ol>\n<li type=\"a\">What will be the minimum force necessary to put the block in motion?<\/li>\n<li type=\"a\">Calculate the acceleration of the block when it starts moving due to the force obtained in the previous point.<\/li>\n<\/ol>\n<\/li>\n<li>A block of <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">12[kg]<\/span><\/span> is placed on an inclinable plane. If the static friction coefficient is <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mu_e=0.03,<\/span><\/span> determine the maximum inclination angle for which the block will remain at rest.<\/li>\n<li>A block of <span class=\"katex-eq\" data-katex-display=\"false\">75[kg]<\/span> ascends with a constant velocity on an inclined plane at <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">30^o<\/span><\/span> with respect to the horizontal due to a force applied horizontally to it. If there is a kinetic friction coefficient <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mu_c=0.21,<\/span><\/span> between the block and the surface of the plane, determine the magnitude of the applied force.<\/li>\n<p>    <center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/kQaZdNm-iDQ\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<li>Consider two masses <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m_1<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m_2<\/span><\/span> connected by an inextensible, massless rope that passes over a pulley as shown in the figure. Calculate the acceleration of both masses.\n<p>    <center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEg6YZ3Pc3UlsxceJu3isovO7by4X1DHEtV-82DhHMl7FLepZDp8YQt4DAGxlQGSCemJE1Ai-SS4OdSMoJUc66U2cEKTuSbCbM_MhX-8jTElj2kwGK589fba-JoIcF9fDw_v36cKBD9OPrPJ6ZaXp4tKhK7qvNftjVQDoIyQDgGUlKXHLj_T3LOgT-rNyw\" width=\"298\" height=\"345\" alt=\"ATWOOD MACHINE\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEg6YZ3Pc3UlsxceJu3isovO7by4X1DHEtV-82DhHMl7FLepZDp8YQt4DAGxlQGSCemJE1Ai-SS4OdSMoJUc66U2cEKTuSbCbM_MhX-8jTElj2kwGK589fba-JoIcF9fDw_v36cKBD9OPrPJ6ZaXp4tKhK7qvNftjVQDoIyQDgGUlKXHLj_T3LOgT-rNyw\" width=\"298\" height=\"345\" alt=\"ATWOOD MACHINE\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center>\n    <\/li>\n<li>A flexible rope of mass <span class=\"katex-eq\" data-katex-display=\"false\">M<\/span> hangs between two walls forming an angle <span class=\"katex-eq\" data-katex-display=\"false\">\\alpha<\/span> at the attachment points. Calculate the tension of the rope at the lowest point.\n<p>    <center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEh3rmFCklJ9Z7Hb6M5Nb_lREohLFKZNhujGlrx1m8aUKGvr4RnoB0w3H2wQj1VHo6or-UgVj7_CtuBiL-mIs17CGJT4FF9gVPXPEKR34_6sRLW50L8q8bb5vby23Lby4xLzX92nfDvUnWnJsxtuTzehTRw-N3NJq9R91n-UFXzCgksrGooUWFGX8D1VLA\" width=\"343\" height=\"281\" alt=\"Hanging rope\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEh3rmFCklJ9Z7Hb6M5Nb_lREohLFKZNhujGlrx1m8aUKGvr4RnoB0w3H2wQj1VHo6or-UgVj7_CtuBiL-mIs17CGJT4FF9gVPXPEKR34_6sRLW50L8q8bb5vby23Lby4xLzX92nfDvUnWnJsxtuTzehTRw-N3NJq9R91n-UFXzCgksrGooUWFGX8D1VLA\" width=\"343\" height=\"281\" alt=\"Hanging rope\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center>\n    <\/li>\n<li>A body of mass <span class=\"katex-eq\" data-katex-display=\"false\">m<\/span> moves in circles on the <span class=\"katex-eq\" data-katex-display=\"false\">x,y<\/span> plane with a radius <span class=\"katex-eq\" data-katex-display=\"false\">R<\/span> and a constant angular velocity <span class=\"katex-eq\" data-katex-display=\"false\">\\omega<\/span>. Calculate the force applied to the mass.<\/li>\n<p>    <center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/D3fnVM-HKJ4\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/p>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Newton&#8217;s Laws Summary: This class addresses Newton&#8217;s laws and their role in the dynamics of bodies. It explores how mass and velocity determine linear momentum and describes the three laws: inertia, which maintains the state of motion in the absence of external forces, the relationship between force and acceleration, and the action and reaction between [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":28536,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":8,"footnotes":""},"categories":[651,635],"tags":[],"class_list":["post-28537","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-fundamentals-of-mechanics","category-physics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Newton&#039;s Laws - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Newton&#039;s laws describe the motion of bodies and the forces acting on them. 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