{"id":44,"date":"2017-10-08T19:02:23","date_gmt":"2017-10-08T19:02:23","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/?post_type=chapter&#038;p=44"},"modified":"2017-10-08T19:14:20","modified_gmt":"2017-10-08T19:14:20","slug":"newtons-third-law-of-motion-symmetry-in-forces","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/chapter\/newtons-third-law-of-motion-symmetry-in-forces\/","title":{"raw":"Newton\u2019s Third Law of Motion: Symmetry in Forces","rendered":"Newton\u2019s Third Law of Motion: Symmetry in Forces"},"content":{"raw":"<h1 class=\"entry-title\">Newton\u2019s Third Law of Motion: Symmetry in Forces<\/h1>\r\n<div class=\"difficulty\"><\/div>\r\n<div id=\"post-668\" class=\"standard post-668 chapter type-chapter status-publish hentry\">\r\n<div class=\"entry-content\">\r\n<div>\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<div>\r\n<div>\r\n<div>\r\n<div>\r\n<ul>\r\n \t<li>Understand Newton\u2019s third law of motion.<\/li>\r\n \t<li>Apply Newton\u2019s third law to define systems and solve problems of motion.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nThere is a passage in the musical <em>Man of la Mancha<\/em> that relates to Newton\u2019s third law of motion. Sancho, in describing a fight with his wife to Don Quixote, says, \u201cOf course I hit her back, Your Grace, but she\u2019s a lot harder than me and you know what they say, \u2018Whether the stone hits the pitcher or the pitcher hits the stone, it\u2019s going to be bad for the pitcher.\u2019\u201d This is exactly what happens whenever one body exerts a force on another\u2014the first also experiences a force (equal in magnitude and opposite in direction). Numerous common experiences, such as stubbing a toe or throwing a ball, confirm this. It is precisely stated in <em> Newton\u2019s third law of motion<\/em>.\r\n<div>\r\n<div>\r\n<div class=\"textbox shaded\">\r\n<h3><strong>Newton\u2019s Third Law of Motion<\/strong><\/h3>\r\n<section class=\" focusable\">Whenever one body exerts a force on a second body, the first body experiences a force that is equal in\u00a0magnitude and opposite in direction to the force that it exerts.\r\n\r\n<\/section><\/div>\r\n<\/div>\r\nThis law represents a certain <em>symmetry in nature<\/em>: Forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself. We sometimes refer to this law loosely as \u201caction-reaction,\u201d where the force exerted is the action and the force experienced as a consequence is the reaction. Newton\u2019s third law has practical uses in analyzing the origin of forces and understanding which forces are external to a system.\r\n\r\nWe can readily see Newton\u2019s third law at work by taking a look at how people move about. Consider a swimmer pushing off from the side of a pool, as illustrated in Figure 1. She pushes against the pool wall with her feet and accelerates in the direction <em>opposite<\/em> to that of her push. The wall has exerted an equal and opposite force back on the swimmer. You might think that two equal and opposite forces would cancel, but they do not <em>because they act on different systems<\/em>. In this case, there are two systems that we could investigate: the swimmer or the wall. If we select the swimmer to be the system of interest, as in the figure, then F<sub>wall on feet<\/sub> is an external force on this system and affects its motion. The swimmer moves in the direction of F<sub>wall on feet<\/sub>. In contrast, the force F<sub>feet on wall<\/sub> acts on the wall and not on our system of interest. Thus F<sub>feet on wall<\/sub> does not directly affect the motion of the system and does not cancel F<sub>wall on feet<\/sub>. Note that the swimmer pushes in the direction opposite to that in which she wishes to move. The reaction to her push is thus in the desired direction.\r\n<div title=\"Figure 4.12.\">\r\n<div id=\"attachment_12168\" class=\"wp-caption aligncenter\">\r\n\r\n<img class=\"size-full wp-image-12168\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03231158\/Figure_04_04_01.jpg\" alt=\"A swimmer is exerting a force with her feet on a wall inside a swimming pool represented by an arrow labeled as vector F sub Feet on wall, pointing toward the right, and the wall is also exerting an equal force on her feet, represented by an arrow labeled as vector F sub Wall on feet, having the same length but pointing toward the left. The direction of acceleration of the swimmer is toward the left, shown by an arrow toward the left.\" width=\"875\" height=\"329\" \/>\r\n<p class=\"wp-caption-text\">Figure 1. When the swimmer exerts a force <strong>F<\/strong><sub>feet on wall<\/sub> on the wall, she accelerates in the direction opposite to that of her push. This means the net external force on her is in the direction opposite to <strong>F<\/strong><sub>feet<\/sub> on wall. This opposition occurs because, in accordance with Newton\u2019s third law of motion, the wall exerts a force <strong>F<\/strong><sub>wall<\/sub> on feet on her, equal in magnitude but in the direction opposite to the one she exerts on it. The line around the swimmer indicates the system of interest. Note that <strong>F<\/strong><sub>feet<\/sub> on wall does not act on this system (the swimmer) and, thus, does not cancel <strong>F<\/strong><sub>wall on feet<\/sub>. Thus the free-body diagram shows only <strong>F<\/strong><sub>wall<\/sub> on feet, <strong>w<\/strong>, the gravitational force, and <strong>BF<\/strong>, the buoyant force of the water supporting the swimmer\u2019s weight. The vertical forces <strong>w<\/strong> and <strong>BF<\/strong> cancel since there is no vertical motion.<\/p>\r\n\r\n<\/div>\r\nOther examples of Newton\u2019s third law are easy to find. As a professor paces in front of a whiteboard, she exerts a force backward on the floor. The floor exerts a reaction force forward on the professor that causes her to accelerate forward. Similarly, a car accelerates because the ground pushes forward on the drive wheels in reaction to the drive wheels pushing backward on the ground. You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw rocks backward. In another example, rockets move forward by expelling gas backward at high velocity. This means the rocket exerts a large backward force on the gas in the rocket combustion chamber, and the gas therefore exerts a large reaction force forward on the rocket. This reaction force is called <em> thrust<\/em>. It is a common misconception that rockets propel themselves by pushing on the ground or on the air behind them. They actually work better in a vacuum, where they can more readily expel the exhaust gases. Helicopters similarly create lift by pushing air down, thereby experiencing an upward reaction force. Birds and airplanes also fly by exerting force on air in a direction opposite to that of whatever force they need. For example, the wings of a bird force air downward and backward in order to get lift and move forward. An octopus propels itself in the water by ejecting water through a funnel from its body, similar to a jet ski. In a situation similar to Sancho\u2019s, professional cage fighters experience reaction forces when they punch, sometimes breaking their hand by hitting an opponent\u2019s body.\r\n<div>\r\n<div title=\"Example 4.3. Getting Up To Speed: Choosing the Correct System\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 1.\u00a0Getting Up To Speed: Choosing the Correct System<strong>\u00a0<\/strong><\/h3>\r\n<div>\r\n\r\nA physics professor pushes a cart of demonstration equipment to a lecture hall, as seen in Figure 2. Her mass is 65.0 kg, the cart\u2019s is 12.0 kg, and the equipment\u2019s is 7.0 kg. Calculate the acceleration produced when the professor exerts a backward force of 150 N on the floor. All forces opposing the motion, such as friction on the cart\u2019s wheels and air resistance, total 24.0 N.\r\n<div title=\"Figure 4.13.\">\r\n<div>\r\n<div>\r\n<div id=\"attachment_12167\" class=\"wp-caption aligncenter\">\r\n\r\n<img class=\"size-full wp-image-12167\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03231032\/Figure_04_04_02.jpg\" alt=\"A professor is pushing a cart of demonstration equipment. Two systems are labeled in the figure. System one includes both the professor and cart, and system two only has the cart. She is exerting some force F sub prof toward the right, shown by a vector arrow, and the cart is also pushing her with the same magnitude of force directed toward the left, shown by a vector F sub cart, having same length as F sub prof. The friction force small f is shown by a vector arrow pointing left acting between the wheels of the cart and the floor. The professor is pushing the floor with her feet with a force F sub foot toward the left, shown by a vector arrow. The floor is pushing her feet with a force that has the same magnitude, F sub floor, shown by a vector arrow pointing right that has the same length as the vector F sub foot. A free-body diagram is also shown. For system one, friction force acting toward the left is shown by a vector arrow having a small length, and the force F sub floor is acting toward the right, shown by a vector arrow larger than the length of vector f. In system two, friction force represented by a short vector small f acts toward the left and another vector F sub prof is represented by a vector arrow toward the right. F sub prof is longer than small f.\" width=\"1000\" height=\"578\" \/>\r\n<p class=\"wp-caption-text\">Figure 2. A professor pushes a cart of demonstration equipment. The lengths of the arrows are proportional to the magnitudes of the forces (except for <strong>f<\/strong>, since it is too small to draw to scale). Different questions are asked in each example; thus, the system of interest must be defined differently for each. System 1 is appropriate for Example 4.4, since it asks for the acceleration of the entire group of objects. Only <strong>F<\/strong><sub>floor<\/sub> and <strong>f<\/strong> are external forces acting on System 1 along the line of motion. All other forces either cancel or act on the outside world. System 2 is chosen for this example so that <strong>F<\/strong><sub>prof<\/sub> will be an external force and enter into Newton\u2019s second law. Note that the free-body diagrams, which allow us to apply Newton\u2019s second law, vary with the system chosen.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h4><strong>Strategy<\/strong><\/h4>\r\nSince they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System 1 in Figure 2. The professor pushes backward with a force F<sub>foot<\/sub> of 150 N. According to Newton\u2019s third law, the floor exerts a forward reaction force F<sub>floor<\/sub> of 150 N on System 1. Because all motion is horizontal, we can assume there is no net force in the vertical direction. The problem is therefore one-dimensional along the horizontal direction. As noted, f opposes the motion and is thus in the opposite direction of F<sub>floor<\/sub>. Note that we do not include the forces F<sub>prof<\/sub> or F<sub>cart<\/sub> because these are internal forces, and we do not include F<sub>foot<\/sub> because it acts on the floor, not on the system. There are no other significant forces acting on System 1. If the net external force can be found from all this information, we can use Newton\u2019s second law to find the acceleration as requested. See the free-body diagram in the figure.\r\n<h4><strong>Solution<\/strong><\/h4>\r\nNewton\u2019s second law is given by <span class=\"katex\"><span class=\"katex-mathml\">a=Fnetm\\displaystyle a=\\frac{{F}_{\\text{net}}}{m}<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathit\">a<\/span><span class=\"mrel\">=<\/span><span class=\"mord reset-textstyle displaystyle textstyle uncramped\"><span class=\"mfrac\"><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle cramped\"><span class=\"mord textstyle cramped\"><span class=\"mord mathit\">m<\/span><\/span><\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle uncramped\"><span class=\"mord textstyle uncramped\"><span class=\"mord\"><span class=\"mord mathit\">F<\/span><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle scriptstyle cramped\"><span class=\"mord scriptstyle cramped\"><span class=\"text mord scriptstyle cramped\"><span class=\"mord mathrm\">n<\/span><span class=\"mord mathrm\">e<\/span><span class=\"mord mathrm\">t<\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>.\r\n<div>(The SI unit for time, the <em> second\u00a0<\/em>(abbreviated s), has a long history. For many years it was defined as 1\/86,400 of a mean solar day. More recently, a new standard was adopted to gain greater accuracy and to define the second in terms of a non-varying, or constant, physical phenomenon (because the solar day is getting longer due to very gradual slowing of the Earth\u2019s rotation). Cesium atoms can be made to vibrate in a very steady way, and these vibrations can be readily observed and counted. In 1967 the second was redefined as the time required for 9,192,631,770 of these vibrations. (See Figure 1.18.) Accuracy in the fundamental units is essential, because all measurements are ultimately expressed in terms of fundamental units and can be no more accurate than are the fundamental units themselves.<\/div>\r\n<div><\/div>\r\nThe net external force on System 1 is deduced from Figure 2 and the discussion above to be\r\n<div>\u00a0<em>F<\/em><sub> net <\/sub> = <em>F<\/em><sub> floor <\/sub> \u2212 <em>f<\/em> = 150 N \u2212 24 . 0 N = 126 N.<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\nThe mass of System 1 is\r\n\r\nm = (65.0 + 12.0 + 7.0) kg = 84 kg\r\n\r\nThese values of <em>F<\/em><sub> net <\/sub> and <em>m<\/em> produce an acceleration of\r\n\r\n<span class=\"katex\"><span class=\"katex-mathml\">a=Fnetm\\displaystyle a=\\frac{{F}_{\\text{net}}}{m}<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathit\">a<\/span><span class=\"mrel\">=<\/span><span class=\"mord reset-textstyle displaystyle textstyle uncramped\"><span class=\"mfrac\"><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle cramped\"><span class=\"mord textstyle cramped\"><span class=\"mord mathit\">m<\/span><\/span><\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle uncramped\"><span class=\"mord textstyle uncramped\"><span class=\"mord\"><span class=\"mord mathit\">F<\/span><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle scriptstyle cramped\"><span class=\"mord scriptstyle cramped\"><span class=\"text mord scriptstyle cramped\"><span class=\"mord mathrm\">n<\/span><span class=\"mord mathrm\">e<\/span><span class=\"mord mathrm\">t<\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\r\n\r\n<span class=\"katex\"><span class=\"katex-mathml\">a=126\u00a0N84\u00a0kg=1.5\u00a0m\/s2\\displaystyle a=\\frac{126 \\text{ N}}{84\\text{ kg}}=1.5 \\text{ m\/s}^{2}<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathit\">a<\/span><span class=\"mrel\">=<\/span><span class=\"mord reset-textstyle displaystyle textstyle uncramped\"><span class=\"mfrac\"><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle cramped\"><span class=\"mord textstyle cramped\"><span class=\"mord mathrm\">8<\/span><span class=\"mord mathrm\">4<\/span><span class=\"text mord textstyle cramped\"><span class=\"mord mspace\">\u00a0<\/span><span class=\"mord mathrm\">k<\/span><span class=\"mord mathrm\">g<\/span><\/span><\/span><\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle uncramped\"><span class=\"mord textstyle uncramped\"><span class=\"mord mathrm\">1<\/span><span class=\"mord mathrm\">2<\/span><span class=\"mord mathrm\">6<\/span><span class=\"text mord textstyle uncramped\"><span class=\"mord mspace\">\u00a0<\/span><span class=\"mord mathrm\">N<\/span><\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><span class=\"mrel\">=<\/span><span class=\"mord mathrm\">1<\/span><span class=\"mord mathrm\">.<\/span><span class=\"mord mathrm\">5<\/span><span class=\"mord\"><span class=\"text mord displaystyle textstyle uncramped\"><span class=\"mord mspace\">\u00a0<\/span><span class=\"mord mathrm\">m<\/span><span class=\"mord mathrm\">\/<\/span><span class=\"mord mathrm\">s<\/span><\/span><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle scriptstyle uncramped\"><span class=\"mord scriptstyle uncramped\"><span class=\"mord mathrm\">2<\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\r\n<h4><strong>Discussion<\/strong><\/h4>\r\nNone of the forces between components of System 1, such as between the professor\u2019s hands and the cart, contribute to the net external force because they are internal to System 1. Another way to look at this is to note that forces between components of a system cancel because they are equal in magnitude and opposite in direction. For example, the force exerted by the professor on the cart results in an equal and opposite force back on her. In this case both forces act on the same system and, therefore, cancel. Thus internal forces (between components of a system) cancel. Choosing System 1 was crucial to solving this problem.\r\n\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div title=\"Example 4.3. Getting Up To Speed: Choosing the Correct System\">\r\n<div title=\"Example 4.4. Force on the Cart\u2014Choosing a New System\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 2.\u00a0Force on the Cart\u2014Choosing a New System<\/h3>\r\n<div title=\"Example 4.3. Getting Up To Speed: Choosing the Correct System\">\r\n<div title=\"Example 4.4. Force on the Cart\u2014Choosing a New System\">\r\n\r\nCalculate the force the professor exerts on the cart in Figure 2 using data from the previous example if needed.\r\n<h4><strong>Strategy<\/strong><\/h4>\r\nIf we now define the system of interest to be the cart plus equipment (System 2 in Figure 2), then the net external force on System 2 is the force the professor exerts on the cart minus friction. The force she exerts on the cart, F<sub>prof<\/sub>, is an external force acting on System 2. F<sub>prof<\/sub> was internal to System 1, but it is external to System 2 and will enter Newton\u2019s second law for System 2.\r\n<h4><strong>Solution<\/strong><\/h4>\r\nNewton\u2019s second law can be used to find F<sub>prof<\/sub>. Starting with\r\n<div>\r\n<div><span class=\"katex\"><span class=\"katex-mathml\">a=Fnetm\\displaystyle a=\\frac{{F}_{\\text{net}}}{m}<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathit\">a<\/span><span class=\"mrel\">=<\/span><span class=\"mord reset-textstyle displaystyle textstyle uncramped\"><span class=\"mfrac\"><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle cramped\"><span class=\"mord textstyle cramped\"><span class=\"mord mathit\">m<\/span><\/span><\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle uncramped\"><span class=\"mord textstyle uncramped\"><span class=\"mord\"><span class=\"mord mathit\">F<\/span><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle scriptstyle cramped\"><span class=\"mord scriptstyle cramped\"><span class=\"text mord scriptstyle cramped\"><span class=\"mord mathrm\">n<\/span><span class=\"mord mathrm\">e<\/span><span class=\"mord mathrm\">t<\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\r\n<div><\/div>\r\n<div><\/div>\r\nand noting that the magnitude of the net external force on System 2 is\r\n<div><em>F<\/em><sub>net\u00a0<\/sub>=\u00a0<em>F<\/em><sub>prof\u00a0<\/sub>\u2212\u00a0<em>f<\/em>,<\/div>\r\n<div><\/div>\r\nwe solve for <em>F<\/em><sub>prof<\/sub>, the desired quantity:\r\n<div>\u00a0<em>F<\/em><sub> prof \u00a0<\/sub>= <em>F<\/em><sub> net <\/sub> + <em>f<\/em> .<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\nThe value of <em>f<\/em> is given, so we must calculate net <em>F<\/em><sub>net<\/sub>. That can be done since both the acceleration and mass of System 2 are known. Using Newton\u2019s second law we see that\r\n\r\n<span class=\"katex\"><span class=\"katex-mathml\">a=Fnetm\\displaystyle a=\\frac{{F}_{\\text{net}}}{m}<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathit\">a<\/span><span class=\"mrel\">=<\/span><span class=\"mord reset-textstyle displaystyle textstyle uncramped\"><span class=\"mfrac\"><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle cramped\"><span class=\"mord textstyle cramped\"><span class=\"mord mathit\">m<\/span><\/span><\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle uncramped\"><span class=\"mord textstyle uncramped\"><span class=\"mord\"><span class=\"mord mathit\">F<\/span><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle scriptstyle cramped\"><span class=\"mord scriptstyle cramped\"><span class=\"text mord scriptstyle cramped\"><span class=\"mord mathrm\">n<\/span><span class=\"mord mathrm\">e<\/span><span class=\"mord mathrm\">t<\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\r\n\r\nwhere the mass of System 2 is 19.0 kg (<em>m<\/em>= 12.0 kg + 7.0 kg) and its acceleration was found to be a = 1.5 m\/s<sup>2<\/sup>\u00a0in the previous example. Thus,\r\n\r\n<span class=\"katex\"><span class=\"katex-mathml\">a=Fnetm\\displaystyle a=\\frac{{F}_{\\text{net}}}{m}<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathit\">a<\/span><span class=\"mrel\">=<\/span><span class=\"mord reset-textstyle displaystyle textstyle uncramped\"><span class=\"mfrac\"><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle cramped\"><span class=\"mord textstyle cramped\"><span class=\"mord mathit\">m<\/span><\/span><\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle uncramped\"><span class=\"mord textstyle uncramped\"><span class=\"mord\"><span class=\"mord mathit\">F<\/span><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle scriptstyle cramped\"><span class=\"mord scriptstyle cramped\"><span class=\"text mord scriptstyle cramped\"><span class=\"mord mathrm\">n<\/span><span class=\"mord mathrm\">e<\/span><span class=\"mord mathrm\">t<\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\r\n\r\n<em>F<\/em><sub>net\u00a0<\/sub>= (19.0 kg)(1.5 m\/s<sup>2<\/sup>)<sup>\u00a0<\/sup>= 29 N\r\n\r\nNow we can find the desired force:\r\n<div><em>F<\/em><sub>prof\u00a0<\/sub>=\u00a0<em>F<\/em><sub>net\u00a0<\/sub>+\u00a0<em>f<\/em>,<\/div>\r\n<div><\/div>\r\n<div><em>F<\/em><sub>prof\u00a0<\/sub>= 29 N + 24.0 N = 53 N.<\/div>\r\n<div><\/div>\r\n<h4><strong>Discussion<\/strong><\/h4>\r\nIt is interesting that this force is significantly less than the 150-N force the professor exerted backward on the floor. Not all of that 150-N force is transmitted to the cart; some of it accelerates the professor. The choice of a system is an important analytical step both in solving problems and in thoroughly understanding the physics of the situation (which is not necessarily the same thing).\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<div title=\"Example 4.3. Getting Up To Speed: Choosing the Correct System\">\r\n<div title=\"Example 4.4. Force on the Cart\u2014Choosing a New System\">\r\n<h2>PhET Explorations: Gravity Force Lab<\/h2>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>Visualize the gravitational force that two objects exert on each other. Change properties of the objects in order to see how it changes the gravity force.<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"wp-caption aligncenter\">\r\n\r\n<a href=\"http:\/\/phet.colorado.edu\/sims\/html\/gravity-force-lab\/latest\/gravity-force-lab_en.html\" target=\"_blank\" rel=\"noopener\"><img src=\"http:\/\/phet.colorado.edu\/sims\/html\/gravity-force-lab\/latest\/gravity-force-lab-600.png\" alt=\"Gravity Force Lab screenshot\" width=\"300\" height=\"200\" \/><\/a>\r\n<p class=\"wp-caption-text\">Click to run the simulation.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Conceptual Questions<\/h3>\r\n<div id=\"fs-id1572333\" class=\"exercise\">\r\n<div id=\"fs-id1421833\" class=\"problem\">\r\n<p id=\"import-auto-id2689558\">1. When you take off in a jet aircraft, there is a sensation of being pushed back into the seat. Explain why you move backward in the seat\u2014is there really a force backward on you? (The same reasoning explains whiplash injuries, in which the head is apparently thrown backward.)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1595226\" class=\"exercise\">\r\n<div id=\"fs-id1600814\" class=\"problem\">\r\n<p id=\"import-auto-id2674155\">2. A device used since the 1940s to measure the kick or recoil of the body due to heart beats is the \u201cballistocardiograph.\u201d What physics principle(s) are involved here to measure the force of cardiac contraction? How might we construct such a device?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2846557\" class=\"exercise\">\r\n<div id=\"fs-id1266684\" class=\"problem\">\r\n<p id=\"import-auto-id861584\">3. Describe a situation in which one system exerts a force on another and, as a consequence, experiences a force that is equal in magnitude and opposite in direction. Which of Newton\u2019s laws of motion apply?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2661705\" class=\"exercise\">\r\n<div id=\"fs-id1415968\" class=\"problem\">\r\n<p id=\"import-auto-id2631801\">4. Why does an ordinary rifle recoil (kick backward) when fired? The barrel of a recoilless rifle is open at both ends. Describe how Newton\u2019s third law applies when one is fired. Can you safely stand close behind one when it is fired?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2423524\" class=\"exercise\">\r\n<div id=\"fs-id1677679\" class=\"problem\">\r\n<p id=\"import-auto-id1919111\">5. An American football lineman reasons that it is senseless to try to out-push the opposing player, since no matter how hard he pushes he will experience an equal and opposite force from the other player. Use Newton\u2019s laws and draw a free-body diagram of an appropriate system to explain how he can still out-push the opposition if he is strong enough.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2355576\" class=\"exercise\">\r\n<div id=\"fs-id1550500\" class=\"problem\">\r\n<p id=\"import-auto-id2159238\">6. Newton\u2019s third law of motion tells us that forces always occur in pairs of equal and opposite magnitude. Explain how the choice of the \u201csystem of interest\u201d affects whether one such pair of forces cancels.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1008305\" class=\"problems-exercises focusable\">\r\n<div class=\"textbox exercises\">\r\n<h3>Problems &amp; Exercises<\/h3>\r\n<div id=\"fs-id1740619\" class=\"exercise\">\r\n<div id=\"fs-id1407116\" class=\"problem\">\r\n<p id=\"import-auto-id1281048\">1. What net external force is exerted on a 1100-kg artillery shell fired from a battleship if the shell is accelerated at 2.40 \u00d7 10<sup>4<\/sup> m\/s<sup>2<\/sup>? What is the magnitude of the force exerted on the ship by the artillery shell?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1443880\" class=\"solution\">2. A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating at 1.20 m\/s<sup>2<\/sup>. (a) What is the force of friction between the losing player\u2019s feet and the grass? (b) What force does the winning player exert on the ground to move forward if his mass plus equipment is 110 kg? (c) Draw a sketch of the situation showing the system of interest used to solve each part. For this situation, draw a free-body diagram and write the net force equation.<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"import-auto-id1046121\" class=\"definition\">\r\n \t<dt>Newton\u2019s third law of motion:<\/dt>\r\n \t<dd id=\"fs-id1694093\">whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that the first body exerts<\/dd>\r\n<\/dl>\r\n<dl id=\"import-auto-id1552615\" class=\"definition\">\r\n \t<dt>thrust:<\/dt>\r\n \t<dd id=\"fs-id1518123\">a reaction force that pushes a body forward in response to a backward force; rockets, airplanes, and cars are pushed forward by a thrust reaction force<\/dd>\r\n<\/dl>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Selected Solutions to Problems &amp; Exercises<\/h3>\r\n1.\u00a0Force on shell: 2.64 \u00d7 10<sup>7<\/sup>\u00a0N,\u00a0Force exerted on ship = -2.64 \u00d7 10<sup>7<\/sup> N, by Newton\u2019s third law\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>","rendered":"<h1 class=\"entry-title\">Newton\u2019s Third Law of Motion: Symmetry in Forces<\/h1>\n<div class=\"difficulty\"><\/div>\n<div id=\"post-668\" class=\"standard post-668 chapter type-chapter status-publish hentry\">\n<div class=\"entry-content\">\n<div>\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<div>\n<div>\n<div>\n<div>\n<ul>\n<li>Understand Newton\u2019s third law of motion.<\/li>\n<li>Apply Newton\u2019s third law to define systems and solve problems of motion.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>There is a passage in the musical <em>Man of la Mancha<\/em> that relates to Newton\u2019s third law of motion. Sancho, in describing a fight with his wife to Don Quixote, says, \u201cOf course I hit her back, Your Grace, but she\u2019s a lot harder than me and you know what they say, \u2018Whether the stone hits the pitcher or the pitcher hits the stone, it\u2019s going to be bad for the pitcher.\u2019\u201d This is exactly what happens whenever one body exerts a force on another\u2014the first also experiences a force (equal in magnitude and opposite in direction). Numerous common experiences, such as stubbing a toe or throwing a ball, confirm this. It is precisely stated in <em> Newton\u2019s third law of motion<\/em>.<\/p>\n<div>\n<div>\n<div class=\"textbox shaded\">\n<h3><strong>Newton\u2019s Third Law of Motion<\/strong><\/h3>\n<section class=\"focusable\">Whenever one body exerts a force on a second body, the first body experiences a force that is equal in\u00a0magnitude and opposite in direction to the force that it exerts.<\/p>\n<\/section>\n<\/div>\n<\/div>\n<p>This law represents a certain <em>symmetry in nature<\/em>: Forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself. We sometimes refer to this law loosely as \u201caction-reaction,\u201d where the force exerted is the action and the force experienced as a consequence is the reaction. Newton\u2019s third law has practical uses in analyzing the origin of forces and understanding which forces are external to a system.<\/p>\n<p>We can readily see Newton\u2019s third law at work by taking a look at how people move about. Consider a swimmer pushing off from the side of a pool, as illustrated in Figure 1. She pushes against the pool wall with her feet and accelerates in the direction <em>opposite<\/em> to that of her push. The wall has exerted an equal and opposite force back on the swimmer. You might think that two equal and opposite forces would cancel, but they do not <em>because they act on different systems<\/em>. In this case, there are two systems that we could investigate: the swimmer or the wall. If we select the swimmer to be the system of interest, as in the figure, then F<sub>wall on feet<\/sub> is an external force on this system and affects its motion. The swimmer moves in the direction of F<sub>wall on feet<\/sub>. In contrast, the force F<sub>feet on wall<\/sub> acts on the wall and not on our system of interest. Thus F<sub>feet on wall<\/sub> does not directly affect the motion of the system and does not cancel F<sub>wall on feet<\/sub>. Note that the swimmer pushes in the direction opposite to that in which she wishes to move. The reaction to her push is thus in the desired direction.<\/p>\n<div title=\"Figure 4.12.\">\n<div id=\"attachment_12168\" class=\"wp-caption aligncenter\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-12168\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03231158\/Figure_04_04_01.jpg\" alt=\"A swimmer is exerting a force with her feet on a wall inside a swimming pool represented by an arrow labeled as vector F sub Feet on wall, pointing toward the right, and the wall is also exerting an equal force on her feet, represented by an arrow labeled as vector F sub Wall on feet, having the same length but pointing toward the left. The direction of acceleration of the swimmer is toward the left, shown by an arrow toward the left.\" width=\"875\" height=\"329\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. When the swimmer exerts a force <strong>F<\/strong><sub>feet on wall<\/sub> on the wall, she accelerates in the direction opposite to that of her push. This means the net external force on her is in the direction opposite to <strong>F<\/strong><sub>feet<\/sub> on wall. This opposition occurs because, in accordance with Newton\u2019s third law of motion, the wall exerts a force <strong>F<\/strong><sub>wall<\/sub> on feet on her, equal in magnitude but in the direction opposite to the one she exerts on it. The line around the swimmer indicates the system of interest. Note that <strong>F<\/strong><sub>feet<\/sub> on wall does not act on this system (the swimmer) and, thus, does not cancel <strong>F<\/strong><sub>wall on feet<\/sub>. Thus the free-body diagram shows only <strong>F<\/strong><sub>wall<\/sub> on feet, <strong>w<\/strong>, the gravitational force, and <strong>BF<\/strong>, the buoyant force of the water supporting the swimmer\u2019s weight. The vertical forces <strong>w<\/strong> and <strong>BF<\/strong> cancel since there is no vertical motion.<\/p>\n<\/div>\n<p>Other examples of Newton\u2019s third law are easy to find. As a professor paces in front of a whiteboard, she exerts a force backward on the floor. The floor exerts a reaction force forward on the professor that causes her to accelerate forward. Similarly, a car accelerates because the ground pushes forward on the drive wheels in reaction to the drive wheels pushing backward on the ground. You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw rocks backward. In another example, rockets move forward by expelling gas backward at high velocity. This means the rocket exerts a large backward force on the gas in the rocket combustion chamber, and the gas therefore exerts a large reaction force forward on the rocket. This reaction force is called <em> thrust<\/em>. It is a common misconception that rockets propel themselves by pushing on the ground or on the air behind them. They actually work better in a vacuum, where they can more readily expel the exhaust gases. Helicopters similarly create lift by pushing air down, thereby experiencing an upward reaction force. Birds and airplanes also fly by exerting force on air in a direction opposite to that of whatever force they need. For example, the wings of a bird force air downward and backward in order to get lift and move forward. An octopus propels itself in the water by ejecting water through a funnel from its body, similar to a jet ski. In a situation similar to Sancho\u2019s, professional cage fighters experience reaction forces when they punch, sometimes breaking their hand by hitting an opponent\u2019s body.<\/p>\n<div>\n<div title=\"Example 4.3. Getting Up To Speed: Choosing the Correct System\">\n<div class=\"textbox examples\">\n<h3>Example 1.\u00a0Getting Up To Speed: Choosing the Correct System<strong>\u00a0<\/strong><\/h3>\n<div>\n<p>A physics professor pushes a cart of demonstration equipment to a lecture hall, as seen in Figure 2. Her mass is 65.0 kg, the cart\u2019s is 12.0 kg, and the equipment\u2019s is 7.0 kg. Calculate the acceleration produced when the professor exerts a backward force of 150 N on the floor. All forces opposing the motion, such as friction on the cart\u2019s wheels and air resistance, total 24.0 N.<\/p>\n<div title=\"Figure 4.13.\">\n<div>\n<div>\n<div id=\"attachment_12167\" class=\"wp-caption aligncenter\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-12167\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/648\/2016\/11\/03231032\/Figure_04_04_02.jpg\" alt=\"A professor is pushing a cart of demonstration equipment. Two systems are labeled in the figure. System one includes both the professor and cart, and system two only has the cart. She is exerting some force F sub prof toward the right, shown by a vector arrow, and the cart is also pushing her with the same magnitude of force directed toward the left, shown by a vector F sub cart, having same length as F sub prof. The friction force small f is shown by a vector arrow pointing left acting between the wheels of the cart and the floor. The professor is pushing the floor with her feet with a force F sub foot toward the left, shown by a vector arrow. The floor is pushing her feet with a force that has the same magnitude, F sub floor, shown by a vector arrow pointing right that has the same length as the vector F sub foot. A free-body diagram is also shown. For system one, friction force acting toward the left is shown by a vector arrow having a small length, and the force F sub floor is acting toward the right, shown by a vector arrow larger than the length of vector f. In system two, friction force represented by a short vector small f acts toward the left and another vector F sub prof is represented by a vector arrow toward the right. F sub prof is longer than small f.\" width=\"1000\" height=\"578\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. A professor pushes a cart of demonstration equipment. The lengths of the arrows are proportional to the magnitudes of the forces (except for <strong>f<\/strong>, since it is too small to draw to scale). Different questions are asked in each example; thus, the system of interest must be defined differently for each. System 1 is appropriate for Example 4.4, since it asks for the acceleration of the entire group of objects. Only <strong>F<\/strong><sub>floor<\/sub> and <strong>f<\/strong> are external forces acting on System 1 along the line of motion. All other forces either cancel or act on the outside world. System 2 is chosen for this example so that <strong>F<\/strong><sub>prof<\/sub> will be an external force and enter into Newton\u2019s second law. Note that the free-body diagrams, which allow us to apply Newton\u2019s second law, vary with the system chosen.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h4><strong>Strategy<\/strong><\/h4>\n<p>Since they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System 1 in Figure 2. The professor pushes backward with a force F<sub>foot<\/sub> of 150 N. According to Newton\u2019s third law, the floor exerts a forward reaction force F<sub>floor<\/sub> of 150 N on System 1. Because all motion is horizontal, we can assume there is no net force in the vertical direction. The problem is therefore one-dimensional along the horizontal direction. As noted, f opposes the motion and is thus in the opposite direction of F<sub>floor<\/sub>. Note that we do not include the forces F<sub>prof<\/sub> or F<sub>cart<\/sub> because these are internal forces, and we do not include F<sub>foot<\/sub> because it acts on the floor, not on the system. There are no other significant forces acting on System 1. If the net external force can be found from all this information, we can use Newton\u2019s second law to find the acceleration as requested. See the free-body diagram in the figure.<\/p>\n<h4><strong>Solution<\/strong><\/h4>\n<p>Newton\u2019s second law is given by <span class=\"katex\"><span class=\"katex-mathml\">a=Fnetm\\displaystyle a=\\frac{{F}_{\\text{net}}}{m}<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathit\">a<\/span><span class=\"mrel\">=<\/span><span class=\"mord reset-textstyle displaystyle textstyle uncramped\"><span class=\"mfrac\"><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle cramped\"><span class=\"mord textstyle cramped\"><span class=\"mord mathit\">m<\/span><\/span><\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle uncramped\"><span class=\"mord textstyle uncramped\"><span class=\"mord\"><span class=\"mord mathit\">F<\/span><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle scriptstyle cramped\"><span class=\"mord scriptstyle cramped\"><span class=\"text mord scriptstyle cramped\"><span class=\"mord mathrm\">n<\/span><span class=\"mord mathrm\">e<\/span><span class=\"mord mathrm\">t<\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>.<\/p>\n<div>(The SI unit for time, the <em> second\u00a0<\/em>(abbreviated s), has a long history. For many years it was defined as 1\/86,400 of a mean solar day. More recently, a new standard was adopted to gain greater accuracy and to define the second in terms of a non-varying, or constant, physical phenomenon (because the solar day is getting longer due to very gradual slowing of the Earth\u2019s rotation). Cesium atoms can be made to vibrate in a very steady way, and these vibrations can be readily observed and counted. In 1967 the second was redefined as the time required for 9,192,631,770 of these vibrations. (See Figure 1.18.) Accuracy in the fundamental units is essential, because all measurements are ultimately expressed in terms of fundamental units and can be no more accurate than are the fundamental units themselves.<\/div>\n<div><\/div>\n<p>The net external force on System 1 is deduced from Figure 2 and the discussion above to be<\/p>\n<div>\u00a0<em>F<\/em><sub> net <\/sub> = <em>F<\/em><sub> floor <\/sub> \u2212 <em>f<\/em> = 150 N \u2212 24 . 0 N = 126 N.<\/div>\n<div><\/div>\n<div><\/div>\n<p>The mass of System 1 is<\/p>\n<p>m = (65.0 + 12.0 + 7.0) kg = 84 kg<\/p>\n<p>These values of <em>F<\/em><sub> net <\/sub> and <em>m<\/em> produce an acceleration of<\/p>\n<p><span class=\"katex\"><span class=\"katex-mathml\">a=Fnetm\\displaystyle a=\\frac{{F}_{\\text{net}}}{m}<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathit\">a<\/span><span class=\"mrel\">=<\/span><span class=\"mord reset-textstyle displaystyle textstyle uncramped\"><span class=\"mfrac\"><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle cramped\"><span class=\"mord textstyle cramped\"><span class=\"mord mathit\">m<\/span><\/span><\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle uncramped\"><span class=\"mord textstyle uncramped\"><span class=\"mord\"><span class=\"mord mathit\">F<\/span><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle scriptstyle cramped\"><span class=\"mord scriptstyle cramped\"><span class=\"text mord scriptstyle cramped\"><span class=\"mord mathrm\">n<\/span><span class=\"mord mathrm\">e<\/span><span class=\"mord mathrm\">t<\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p><span class=\"katex\"><span class=\"katex-mathml\">a=126\u00a0N84\u00a0kg=1.5\u00a0m\/s2\\displaystyle a=\\frac{126 \\text{ N}}{84\\text{ kg}}=1.5 \\text{ m\/s}^{2}<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathit\">a<\/span><span class=\"mrel\">=<\/span><span class=\"mord reset-textstyle displaystyle textstyle uncramped\"><span class=\"mfrac\"><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle cramped\"><span class=\"mord textstyle cramped\"><span class=\"mord mathrm\">8<\/span><span class=\"mord mathrm\">4<\/span><span class=\"text mord textstyle cramped\"><span class=\"mord mspace\">\u00a0<\/span><span class=\"mord mathrm\">k<\/span><span class=\"mord mathrm\">g<\/span><\/span><\/span><\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle uncramped\"><span class=\"mord textstyle uncramped\"><span class=\"mord mathrm\">1<\/span><span class=\"mord mathrm\">2<\/span><span class=\"mord mathrm\">6<\/span><span class=\"text mord textstyle uncramped\"><span class=\"mord mspace\">\u00a0<\/span><span class=\"mord mathrm\">N<\/span><\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><span class=\"mrel\">=<\/span><span class=\"mord mathrm\">1<\/span><span class=\"mord mathrm\">.<\/span><span class=\"mord mathrm\">5<\/span><span class=\"mord\"><span class=\"text mord displaystyle textstyle uncramped\"><span class=\"mord mspace\">\u00a0<\/span><span class=\"mord mathrm\">m<\/span><span class=\"mord mathrm\">\/<\/span><span class=\"mord mathrm\">s<\/span><\/span><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle scriptstyle uncramped\"><span class=\"mord scriptstyle uncramped\"><span class=\"mord mathrm\">2<\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<h4><strong>Discussion<\/strong><\/h4>\n<p>None of the forces between components of System 1, such as between the professor\u2019s hands and the cart, contribute to the net external force because they are internal to System 1. Another way to look at this is to note that forces between components of a system cancel because they are equal in magnitude and opposite in direction. For example, the force exerted by the professor on the cart results in an equal and opposite force back on her. In this case both forces act on the same system and, therefore, cancel. Thus internal forces (between components of a system) cancel. Choosing System 1 was crucial to solving this problem.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div title=\"Example 4.3. Getting Up To Speed: Choosing the Correct System\">\n<div title=\"Example 4.4. Force on the Cart\u2014Choosing a New System\">\n<div class=\"textbox examples\">\n<h3>Example 2.\u00a0Force on the Cart\u2014Choosing a New System<\/h3>\n<div title=\"Example 4.3. Getting Up To Speed: Choosing the Correct System\">\n<div title=\"Example 4.4. Force on the Cart\u2014Choosing a New System\">\n<p>Calculate the force the professor exerts on the cart in Figure 2 using data from the previous example if needed.<\/p>\n<h4><strong>Strategy<\/strong><\/h4>\n<p>If we now define the system of interest to be the cart plus equipment (System 2 in Figure 2), then the net external force on System 2 is the force the professor exerts on the cart minus friction. The force she exerts on the cart, F<sub>prof<\/sub>, is an external force acting on System 2. F<sub>prof<\/sub> was internal to System 1, but it is external to System 2 and will enter Newton\u2019s second law for System 2.<\/p>\n<h4><strong>Solution<\/strong><\/h4>\n<p>Newton\u2019s second law can be used to find F<sub>prof<\/sub>. Starting with<\/p>\n<div>\n<div><span class=\"katex\"><span class=\"katex-mathml\">a=Fnetm\\displaystyle a=\\frac{{F}_{\\text{net}}}{m}<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathit\">a<\/span><span class=\"mrel\">=<\/span><span class=\"mord reset-textstyle displaystyle textstyle uncramped\"><span class=\"mfrac\"><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle cramped\"><span class=\"mord textstyle cramped\"><span class=\"mord mathit\">m<\/span><\/span><\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle uncramped\"><span class=\"mord textstyle uncramped\"><span class=\"mord\"><span class=\"mord mathit\">F<\/span><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle scriptstyle cramped\"><span class=\"mord scriptstyle cramped\"><span class=\"text mord scriptstyle cramped\"><span class=\"mord mathrm\">n<\/span><span class=\"mord mathrm\">e<\/span><span class=\"mord mathrm\">t<\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\n<div><\/div>\n<div><\/div>\n<p>and noting that the magnitude of the net external force on System 2 is<\/p>\n<div><em>F<\/em><sub>net\u00a0<\/sub>=\u00a0<em>F<\/em><sub>prof\u00a0<\/sub>\u2212\u00a0<em>f<\/em>,<\/div>\n<div><\/div>\n<p>we solve for <em>F<\/em><sub>prof<\/sub>, the desired quantity:<\/p>\n<div>\u00a0<em>F<\/em><sub> prof \u00a0<\/sub>= <em>F<\/em><sub> net <\/sub> + <em>f<\/em> .<\/div>\n<div><\/div>\n<div><\/div>\n<p>The value of <em>f<\/em> is given, so we must calculate net <em>F<\/em><sub>net<\/sub>. That can be done since both the acceleration and mass of System 2 are known. Using Newton\u2019s second law we see that<\/p>\n<p><span class=\"katex\"><span class=\"katex-mathml\">a=Fnetm\\displaystyle a=\\frac{{F}_{\\text{net}}}{m}<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathit\">a<\/span><span class=\"mrel\">=<\/span><span class=\"mord reset-textstyle displaystyle textstyle uncramped\"><span class=\"mfrac\"><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle cramped\"><span class=\"mord textstyle cramped\"><span class=\"mord mathit\">m<\/span><\/span><\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle uncramped\"><span class=\"mord textstyle uncramped\"><span class=\"mord\"><span class=\"mord mathit\">F<\/span><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle scriptstyle cramped\"><span class=\"mord scriptstyle cramped\"><span class=\"text mord scriptstyle cramped\"><span class=\"mord mathrm\">n<\/span><span class=\"mord mathrm\">e<\/span><span class=\"mord mathrm\">t<\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p>where the mass of System 2 is 19.0 kg (<em>m<\/em>= 12.0 kg + 7.0 kg) and its acceleration was found to be a = 1.5 m\/s<sup>2<\/sup>\u00a0in the previous example. Thus,<\/p>\n<p><span class=\"katex\"><span class=\"katex-mathml\">a=Fnetm\\displaystyle a=\\frac{{F}_{\\text{net}}}{m}<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathit\">a<\/span><span class=\"mrel\">=<\/span><span class=\"mord reset-textstyle displaystyle textstyle uncramped\"><span class=\"mfrac\"><span class=\"vlist\"><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle cramped\"><span class=\"mord textstyle cramped\"><span class=\"mord mathit\">m<\/span><\/span><\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><\/span><span class=\"\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle textstyle uncramped\"><span class=\"mord textstyle uncramped\"><span class=\"mord\"><span class=\"mord mathit\">F<\/span><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span><span class=\"reset-textstyle scriptstyle cramped\"><span class=\"mord scriptstyle cramped\"><span class=\"text mord scriptstyle cramped\"><span class=\"mord mathrm\">n<\/span><span class=\"mord mathrm\">e<\/span><span class=\"mord mathrm\">t<\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\">\u200b<\/span>\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"baseline-fix\"><span class=\"fontsize-ensurer reset-size5 size5\"><span class=\"\">\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p><em>F<\/em><sub>net\u00a0<\/sub>= (19.0 kg)(1.5 m\/s<sup>2<\/sup>)<sup>\u00a0<\/sup>= 29 N<\/p>\n<p>Now we can find the desired force:<\/p>\n<div><em>F<\/em><sub>prof\u00a0<\/sub>=\u00a0<em>F<\/em><sub>net\u00a0<\/sub>+\u00a0<em>f<\/em>,<\/div>\n<div><\/div>\n<div><em>F<\/em><sub>prof\u00a0<\/sub>= 29 N + 24.0 N = 53 N.<\/div>\n<div><\/div>\n<h4><strong>Discussion<\/strong><\/h4>\n<p>It is interesting that this force is significantly less than the 150-N force the professor exerted backward on the floor. Not all of that 150-N force is transmitted to the cart; some of it accelerates the professor. The choice of a system is an important analytical step both in solving problems and in thoroughly understanding the physics of the situation (which is not necessarily the same thing).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<div title=\"Example 4.3. Getting Up To Speed: Choosing the Correct System\">\n<div title=\"Example 4.4. Force on the Cart\u2014Choosing a New System\">\n<h2>PhET Explorations: Gravity Force Lab<\/h2>\n<\/div>\n<\/div>\n<div>\n<div>Visualize the gravitational force that two objects exert on each other. Change properties of the objects in order to see how it changes the gravity force.<\/div>\n<\/div>\n<div>\n<div class=\"wp-caption aligncenter\">\n<p><a href=\"http:\/\/phet.colorado.edu\/sims\/html\/gravity-force-lab\/latest\/gravity-force-lab_en.html\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/phet.colorado.edu\/sims\/html\/gravity-force-lab\/latest\/gravity-force-lab-600.png\" alt=\"Gravity Force Lab screenshot\" width=\"300\" height=\"200\" \/><\/a><\/p>\n<p class=\"wp-caption-text\">Click to run the simulation.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Conceptual Questions<\/h3>\n<div id=\"fs-id1572333\" class=\"exercise\">\n<div id=\"fs-id1421833\" class=\"problem\">\n<p id=\"import-auto-id2689558\">1. When you take off in a jet aircraft, there is a sensation of being pushed back into the seat. Explain why you move backward in the seat\u2014is there really a force backward on you? (The same reasoning explains whiplash injuries, in which the head is apparently thrown backward.)<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1595226\" class=\"exercise\">\n<div id=\"fs-id1600814\" class=\"problem\">\n<p id=\"import-auto-id2674155\">2. A device used since the 1940s to measure the kick or recoil of the body due to heart beats is the \u201cballistocardiograph.\u201d What physics principle(s) are involved here to measure the force of cardiac contraction? How might we construct such a device?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2846557\" class=\"exercise\">\n<div id=\"fs-id1266684\" class=\"problem\">\n<p id=\"import-auto-id861584\">3. Describe a situation in which one system exerts a force on another and, as a consequence, experiences a force that is equal in magnitude and opposite in direction. Which of Newton\u2019s laws of motion apply?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2661705\" class=\"exercise\">\n<div id=\"fs-id1415968\" class=\"problem\">\n<p id=\"import-auto-id2631801\">4. Why does an ordinary rifle recoil (kick backward) when fired? The barrel of a recoilless rifle is open at both ends. Describe how Newton\u2019s third law applies when one is fired. Can you safely stand close behind one when it is fired?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2423524\" class=\"exercise\">\n<div id=\"fs-id1677679\" class=\"problem\">\n<p id=\"import-auto-id1919111\">5. An American football lineman reasons that it is senseless to try to out-push the opposing player, since no matter how hard he pushes he will experience an equal and opposite force from the other player. Use Newton\u2019s laws and draw a free-body diagram of an appropriate system to explain how he can still out-push the opposition if he is strong enough.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2355576\" class=\"exercise\">\n<div id=\"fs-id1550500\" class=\"problem\">\n<p id=\"import-auto-id2159238\">6. Newton\u2019s third law of motion tells us that forces always occur in pairs of equal and opposite magnitude. Explain how the choice of the \u201csystem of interest\u201d affects whether one such pair of forces cancels.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1008305\" class=\"problems-exercises focusable\">\n<div class=\"textbox exercises\">\n<h3>Problems &amp; Exercises<\/h3>\n<div id=\"fs-id1740619\" class=\"exercise\">\n<div id=\"fs-id1407116\" class=\"problem\">\n<p id=\"import-auto-id1281048\">1. What net external force is exerted on a 1100-kg artillery shell fired from a battleship if the shell is accelerated at 2.40 \u00d7 10<sup>4<\/sup> m\/s<sup>2<\/sup>? What is the magnitude of the force exerted on the ship by the artillery shell?<\/p>\n<\/div>\n<div id=\"fs-id1443880\" class=\"solution\">2. A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating at 1.20 m\/s<sup>2<\/sup>. (a) What is the force of friction between the losing player\u2019s feet and the grass? (b) What force does the winning player exert on the ground to move forward if his mass plus equipment is 110 kg? (c) Draw a sketch of the situation showing the system of interest used to solve each part. For this situation, draw a free-body diagram and write the net force equation.<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"import-auto-id1046121\" class=\"definition\">\n<dt>Newton\u2019s third law of motion:<\/dt>\n<dd id=\"fs-id1694093\">whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that the first body exerts<\/dd>\n<\/dl>\n<dl id=\"import-auto-id1552615\" class=\"definition\">\n<dt>thrust:<\/dt>\n<dd id=\"fs-id1518123\">a reaction force that pushes a body forward in response to a backward force; rockets, airplanes, and cars are pushed forward by a thrust reaction force<\/dd>\n<\/dl>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Selected Solutions to Problems &amp; Exercises<\/h3>\n<p>1.\u00a0Force on shell: 2.64 \u00d7 10<sup>7<\/sup>\u00a0N,\u00a0Force exerted on ship = -2.64 \u00d7 10<sup>7<\/sup> N, by Newton\u2019s third law<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n","protected":false},"author":51812,"menu_order":13,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-44","chapter","type-chapter","status-publish","hentry"],"part":48,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/wp-json\/pressbooks\/v2\/chapters\/44","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/wp-json\/wp\/v2\/users\/51812"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/wp-json\/pressbooks\/v2\/chapters\/44\/revisions"}],"predecessor-version":[{"id":58,"href":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/wp-json\/pressbooks\/v2\/chapters\/44\/revisions\/58"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/wp-json\/pressbooks\/v2\/parts\/48"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/wp-json\/pressbooks\/v2\/chapters\/44\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/wp-json\/wp\/v2\/media?parent=44"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/wp-json\/pressbooks\/v2\/chapter-type?post=44"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/wp-json\/wp\/v2\/contributor?post=44"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-hccc-generalscience\/wp-json\/wp\/v2\/license?post=44"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}