{"id":246,"date":"2017-09-18T18:08:52","date_gmt":"2017-09-18T22:08:52","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/chapter\/4-4-newtons-third-law-of-motion-symmetry-in-forces\/"},"modified":"2021-05-09T16:19:25","modified_gmt":"2021-05-09T20:19:25","slug":"4-4-newtons-third-law-of-motion-symmetry-in-forces","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/chapter\/4-4-newtons-third-law-of-motion-symmetry-in-forces\/","title":{"raw":"4.4 Newton\u2019s Third Law of Motion: Symmetry in Forces","rendered":"4.4 Newton\u2019s Third Law of Motion: Symmetry in Forces"},"content":{"raw":"<div>\n<div class=\"bcc-box bcc-highlight\">\n<h3>Summary<\/h3>\n<div>\n<ul>\n \t<li>Understand Newton's third law of motion.<\/li>\n \t<li>Apply Newton's third law to define systems and solve problems of motion.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"import-auto-id2355102\">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 <strong><span id=\"import-auto-id1260844\">Newton\u2019s third law of motion<\/span><\/strong>.<\/p>\n\n<div id=\"fs-id2688761\" class=\"note\">\n<div class=\"textbox shaded\">\n<div class=\"note\">\n<h3 class=\"title\">NEWTON'S THIRD LAW OF MOTION<span style=\"text-decoration: underline\">\n<\/span><\/h3>\n<p id=\"import-auto-id2639627\">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 it exerts.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<p id=\"import-auto-id1460190\">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 id=\"import-auto-id2301470\">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 <a class=\"autogenerated-content\" href=\"#import-auto-id2338100\">Figure 1<\/a>. 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 <strong><em>F<\/em><sub>wall on feet<\/sub><\/strong> is an external force on this system and affects its motion. The swimmer moves in the direction of <strong><em>F<\/em><sub>wall on feet<\/sub><\/strong>. In contrast, the force <strong><em>F<\/em><sub>feet on wall<\/sub><\/strong> acts on the wall and not on our system of interest. Thus <strong><em>F<\/em><sub>feet on wall<\/sub><\/strong> does not directly affect the motion of the system and does not cancel <strong><em>F<\/em><sub>wall on feet<\/sub><\/strong>. 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\n<figure id=\"import-auto-id2338100\"><figcaption><\/figcaption>\n\n[caption id=\"\" align=\"aligncenter\" width=\"600\"]<img src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2017\/09\/Figure_04_04_01-1.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=\"600\" height=\"329\"> <strong>Figure 1.<\/strong> When the swimmer exerts a force <strong>F<sub>feet on wall<\/sub><\/strong> 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<sub>feet on wall<\/sub><\/strong>. This opposition occurs because, in accordance with Newton\u2019s third law of motion, the wall exerts a force <strong>F<sub>wall on feet<\/sub><\/strong> 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<sub>feet on wall<\/sub><\/strong> does not act on this system (the swimmer) and, thus, does not cancel<strong> F<sub>wall on feet<\/sub><\/strong>. Thus the free-body diagram shows only F<strong><sub>wall on feet<\/sub><\/strong>, <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.[\/caption]<\/figure>\n<p id=\"import-auto-id1772663\">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. Helicopters 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>\nIn another example, rockets move forward by expelling gas backward at high velocity. Rockets range in size from fireworks so small that ordinary people use them to immense Saturn Vs that once propelled massive payloads toward the Moon. The propulsion of all rockets, jet engines, deflating balloons, and even squids and octopuses is explained by the same physical principle\u2014Newton\u2019s third law of motion. 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, as shown in <a href=\"\/douglasphys1107\/chapter\/4-4-newtons-third-law-of-motion-symmetry-in-forces\/#rocket\">Figure 2<\/a>. This reaction force is called <strong><span id=\"import-auto-id1295864\">thrust<\/span><\/strong>. 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.\n\n[caption id=\"\" align=\"aligncenter\" width=\"300\"]<img src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_09_07_01a-1.jpg\" alt=\"Picture a shows a rocket launched into space. It moves upward with velocity v in time t and the burning of fuel is also shown. After time t plus delta t the mass of fuel decreases by delta m and hence the velocity of the rocket increases to v plus delta v. The free body diagram shows the weight W of the rocket downward, reaction force upward and the resultant velocity upward too.\" width=\"300\" height=\"1006\"> <a id=\"rocket\" href=\"\"><\/a><strong>Figure 2.<\/strong> (a) This rocket has a mass<em><strong> m<\/strong><\/em> and an upward velocity <em><strong>v<\/strong><\/em>. The net external force on the system is<strong><em> \u2212mg<\/em><\/strong>, if air resistance is neglected. (b) A time <strong>\u0394<em>t<\/em><\/strong> later the system has two main parts, the ejected gas and the remainder of the rocket. The reaction force on the rocket is what overcomes the gravitational force and accelerates it upward.[\/caption]\n\n<div class=\"textbox shaded\">\n<div id=\"fs-id2355307\" class=\"example\">\n<h3 id=\"import-auto-id1645990\">Example 1: Getting Up To Speed: Choosing the Correct System<\/h3>\nA physics professor pushes a cart of demonstration equipment to a lecture hall, as seen in <a class=\"autogenerated-content\" href=\"#import-auto-id2324690\">Figure 3<\/a>. 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.\n<figure id=\"import-auto-id2324690\"><figcaption><\/figcaption>\n\n[caption id=\"\" align=\"aligncenter\" width=\"550\"]<img src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_04_04_02-1.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=\"550\" height=\"578\"> <strong>Figure 3.<\/strong> 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 <a href=\"#fs-id2092526\">Example 2<\/a>, since it asks for the acceleration of the entire group of objects. Only <strong>F<sub>floor<\/sub><\/strong> 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<sub>prof<\/sub><\/strong> 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.[\/caption]<\/figure>\n<p id=\"fs-id2687234\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1769338\">Since they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System 1 in <a class=\"autogenerated-content\" href=\"#import-auto-id2324690\">Figure 3<\/a>. The professor pushes backward with a force <strong><em>F<\/em><sub>foot<\/sub><\/strong> of 150 N. According to Newton\u2019s third law, the floor exerts a forward reaction force <strong><em>F<\/em><sub>floor<\/sub><\/strong> 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, <em><strong>f<\/strong><\/em> opposes the motion and is thus in the opposite direction of <strong><em>F<\/em><sub>floor<\/sub><\/strong>. Note that we do not include the forces <strong><em>F<\/em><sub>prof<\/sub><\/strong> or <strong><em>F<\/em><sub>cart<\/sub><\/strong> because these are internal forces, and we do not include <strong><em>F<\/em><sub>foot<\/sub><\/strong> 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<p id=\"import-auto-id1333658\"><strong>Solution<\/strong><\/p>\n<p id=\"import-auto-id1443880\">Newton\u2019s second law is given by<\/p>\n\n<div id=\"eip-id1456392\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{a\\:=}[\/latex][latex size=\"2\"]\\boldsymbol{\\frac{F_{\\textbf{net}}}{m}.}[\/latex]<\/div>\n<p id=\"import-auto-id1729170\">The net external force on System 1 is deduced from <a class=\"autogenerated-content\" href=\"#import-auto-id2324690\">Figure 3<\/a> and the discussion above to be<\/p>\n\n<div id=\"eip-id1515829\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{\\textbf{F}_{\\textbf{net}}=\\textbf{F}_{\\textbf{floor}}-\\textbf{f}=150\\textbf{ N}-24.0\\textbf{ N}=126\\textbf{ N}.}[\/latex]<\/div>\n<p id=\"import-auto-id2669812\">The mass of System 1 is<\/p>\n\n<div id=\"eip-id1213154\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{m=(65.0 + 12.0 + 7.0)\\textbf{ kg} = 84\\textbf{ kg}.}[\/latex]<\/div>\n<p id=\"import-auto-id2320233\">These values of <strong><em>F<\/em><sub>net<\/sub><\/strong> and <em><strong>m<\/strong><\/em> produce an acceleration of<\/p>\n\n<div id=\"eip-id1741589\" class=\"equation\" style=\"text-align: center\">$latex \\begin{array}{r @{{}={}}l} \\boldsymbol{a} &amp; \\boldsymbol{\\frac{F_{\\textbf{net}}}{m}} \\\\[1em] \\boldsymbol{a} &amp; \\boldsymbol{\\frac{126 \\;\\textbf{N}}{84 \\;\\textbf{kg}} = 1.5 \\;\\textbf{m} \/ \\;\\textbf{s}^2} \\end{array} $<\/div>\n<p id=\"import-auto-id2685057\"><strong>Discussion<\/strong><\/p>\n<p id=\"import-auto-id2396019\">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\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<div id=\"fs-id2092526\" class=\"example\">\n<h3 id=\"import-auto-id1572504\">Example 2: Force of the Cart\u2014Choosing a New System<\/h3>\nCalculate the force the professor exerts on the cart in <a class=\"autogenerated-content\" href=\"#import-auto-id2324690\">Figure 3<\/a> using data from the previous example if needed.\n<p id=\"import-auto-id2423425\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1745150\">If we now define the system of interest to be the cart plus equipment (System 2 in <a class=\"autogenerated-content\" href=\"#import-auto-id2324690\">Figure 3<\/a>), 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, <strong><em>F<\/em><sub>prof<\/sub><\/strong>, is an external force acting on System 2. <strong><em>F<\/em><sub>prof<\/sub><\/strong> was internal to System 1, but it is external to System 2 and will enter Newton\u2019s second law for System 2.<\/p>\n<p id=\"import-auto-id2682081\"><strong>Solution<\/strong><\/p>\n<p id=\"import-auto-id2693182\">Newton\u2019s second law can be used to find <strong><em>F<\/em><sub>prof<\/sub><\/strong>. Starting with<\/p>\n\n<div id=\"eip-id1949236\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{a\\:=}[\/latex][latex size=\"2\"]\\boldsymbol{\\frac{F_{\\textbf{net}}}{m}}[\/latex]<\/div>\n<p id=\"import-auto-id2147279\">and noting that the magnitude of the net external force on System 2 is<\/p>\n\n<div id=\"eip-id2390884\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{net}}=F_{\\textbf{prof}}-f,}[\/latex]<\/div>\n<p id=\"import-auto-id2401022\">we solve for <strong><em>F<\/em><sub>prof<\/sub><\/strong>, the desired quantity:<\/p>\n\n<div id=\"eip-id1373488\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{prof}}=F_{\\textbf{net}}+f.}[\/latex]<\/div>\n<p id=\"import-auto-id2730714\">The value of <em><strong>f<\/strong><\/em> is given, so we must calculate net <strong><em>F<\/em><sub>net<\/sub><\/strong>. 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\n<div id=\"eip-id2367884\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{net}}=ma,}[\/latex]<\/div>\n<p id=\"import-auto-id2674462\">where the mass of System 2 is 19.0 kg (<em><strong>m<\/strong><\/em>= 12.0 kg + 7.0 kg) and its acceleration was found to be <strong><em>a<\/em> = 1.5 m\/s<sup>2<\/sup><\/strong> in the previous example. Thus,<\/p>\n\n<div id=\"eip-id3477001\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{net}}=ma,}[\/latex]<\/div>\n<div id=\"eip-id1454423\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{net}}=(19.0\\textbf{ kg})(1.5\\textbf{ m\/s}^2)=29\\textbf{ N}.}[\/latex]<\/div>\n<p id=\"import-auto-id860562\">Now we can find the desired force:<\/p>\n\n<div id=\"eip-id1004496\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{prof}}=F_{\\textbf{net}}+f,}[\/latex]<\/div>\n<div id=\"eip-id1197329\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{prof}}=29\\textbf{ N}+24.0\\textbf{ N}=53\\textbf{ N}.}[\/latex]<\/div>\n<p id=\"import-auto-id2655815\"><strong>Discussion<\/strong><\/p>\n<p id=\"import-auto-id2666000\">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.<\/p>\n<p id=\"import-auto-id2335706\">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\n<\/div>\n<\/div>\n<div class=\"example\">\n<div class=\"textbox shaded\">\n<h3>PHET EXPLORATIONS: GRAVITY FORCE LAB<\/h3>\nVisualize 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.\n\n[caption id=\"\" align=\"aligncenter\" width=\"450\"]<a href=\"\/resources\/56ecddb16a348d5796513d8d3cc2616ce2d28271\/gravity-force-lab_en.jar\"><img src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/PhET_Icon-2-1.png\" alt=\"image\" width=\"450\" height=\"147\"><\/a> <strong>Figure 3.<\/strong> <a href=\"https:\/\/phet.colorado.edu\/en\/simulation\/gravity-force-lab\">Gravity Force Lab<\/a>[\/caption]\n\n<\/div>\n<\/div>\n<section id=\"fs-id1446741\" class=\"section-summary\">\n<h1>Section Summary<\/h1>\n<ul id=\"fs-id2332580\">\n \t<li><strong>Newton\u2019s third law of motion<\/strong> represents a basic symmetry in nature. It states: 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.<\/li>\n \t<li>A <strong>thrust <\/strong>is 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.<\/li>\n<\/ul>\n<\/section><section id=\"fs-id1460367\" class=\"conceptual-questions\">\n<div class=\"bcc-box bcc-info\">\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\"><strong>1: <\/strong>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\n<\/div>\n<\/div>\n<div id=\"fs-id1595226\" class=\"exercise\">\n<div id=\"fs-id1600814\" class=\"problem\">\n<p id=\"import-auto-id2674155\"><strong>2: <\/strong>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\n<\/div>\n<\/div>\n<div id=\"fs-id2846557\" class=\"exercise\">\n<div id=\"fs-id1266684\" class=\"problem\">\n<p id=\"import-auto-id861584\"><strong>3: <\/strong>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\n<\/div>\n<\/div>\n<div id=\"fs-id2661705\" class=\"exercise\">\n<div id=\"fs-id1415968\" class=\"problem\">\n<p id=\"import-auto-id2631801\"><strong>4: <\/strong>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\n<\/div>\n<\/div>\n<div id=\"fs-id2423524\" class=\"exercise\">\n<div id=\"fs-id1677679\" class=\"problem\">\n<p id=\"import-auto-id1919111\"><strong>5: <\/strong>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\n<\/div>\n<\/div>\n<div id=\"fs-id2355576\" class=\"exercise\">\n<div id=\"fs-id1550500\" class=\"problem\">\n<p id=\"import-auto-id2159238\"><strong>6: <\/strong>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\n<\/div>\n<\/div>\n<\/div>\n<\/section><section id=\"fs-id1008305\" class=\"problems-exercises\">\n<div class=\"bcc-box bcc-info\">\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\"><strong>1: <\/strong>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\n<\/div>\n<\/div>\n<div id=\"fs-id2300721\" class=\"exercise\">\n<div id=\"fs-id1791532\" class=\"problem\">\n<p id=\"import-auto-id1771007\"><strong>2: <\/strong>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> backward. (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 <span style=\"text-decoration: underline\"><strong>write the net force equation.<\/strong><\/span><\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"import-auto-id1046121\" class=\"definition\">\n \t<dt>Newton\u2019s third law of motion<\/dt>\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>\n<\/dl>\n<dl id=\"import-auto-id1552615\" class=\"definition\">\n \t<dt>thrust<\/dt>\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>\n<\/dl>\n<div class=\"bcc-box bcc-info\">\n<h3>Solutions<\/h3>\n<strong>Problems &amp; Exercises\n<\/strong>\n<p id=\"import-auto-id2297403\"><strong>1: <\/strong>Force on shell: [latex]\\boldsymbol{2.64\\times10^7\\textbf{ N}}[\/latex] Force exerted on ship = [latex]\\boldsymbol{-2.64\\times10^7\\textbf{ N}},[\/latex] by Newton\u2019s third law<\/p>\n2:\u00a0 net force on the losing player = m a = (90.0 kg)(1.20 m\/s<sup>2<\/sup>) = 108 N.\u00a0 The pushing backwards force is 800 N but the net force is only 108 Ns so that means friction = (800-108) = 692 N.\u00a0 \u00a0Without friction the net force would be 800 N and the player would accelerate very quickly backwards.\u00a0 b) net force on the winning player =\u00a0 m a = (110.0 kg)(1.20 m\/s<sup>2<\/sup>) = 132 N.\u00a0 \u00a0He is exerting a force of 800 N so the friction force = ?\n\n<\/div>\n<\/div>","rendered":"<div>\n<div class=\"bcc-box bcc-highlight\">\n<h3>Summary<\/h3>\n<div>\n<ul>\n<li>Understand Newton&#8217;s third law of motion.<\/li>\n<li>Apply Newton&#8217;s third law to define systems and solve problems of motion.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"import-auto-id2355102\">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 <strong><span id=\"import-auto-id1260844\">Newton\u2019s third law of motion<\/span><\/strong>.<\/p>\n<div id=\"fs-id2688761\" class=\"note\">\n<div class=\"textbox shaded\">\n<div class=\"note\">\n<h3 class=\"title\">NEWTON&#8217;S THIRD LAW OF MOTION<span style=\"text-decoration: underline\"><br \/>\n<\/span><\/h3>\n<p id=\"import-auto-id2639627\">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 it exerts.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"import-auto-id1460190\">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 id=\"import-auto-id2301470\">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 <a class=\"autogenerated-content\" href=\"#import-auto-id2338100\">Figure 1<\/a>. 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 <strong><em>F<\/em><sub>wall on feet<\/sub><\/strong> is an external force on this system and affects its motion. The swimmer moves in the direction of <strong><em>F<\/em><sub>wall on feet<\/sub><\/strong>. In contrast, the force <strong><em>F<\/em><sub>feet on wall<\/sub><\/strong> acts on the wall and not on our system of interest. Thus <strong><em>F<\/em><sub>feet on wall<\/sub><\/strong> does not directly affect the motion of the system and does not cancel <strong><em>F<\/em><sub>wall on feet<\/sub><\/strong>. 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<figure id=\"import-auto-id2338100\"><figcaption><\/figcaption><figure style=\"width: 600px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2017\/09\/Figure_04_04_01-1.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=\"600\" height=\"329\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 1.<\/strong> When the swimmer exerts a force <strong>F<sub>feet on wall<\/sub><\/strong> 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<sub>feet on wall<\/sub><\/strong>. This opposition occurs because, in accordance with Newton\u2019s third law of motion, the wall exerts a force <strong>F<sub>wall on feet<\/sub><\/strong> 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<sub>feet on wall<\/sub><\/strong> does not act on this system (the swimmer) and, thus, does not cancel<strong> F<sub>wall on feet<\/sub><\/strong>. Thus the free-body diagram shows only F<strong><sub>wall on feet<\/sub><\/strong>, <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.<\/figcaption><\/figure>\n<\/figure>\n<p id=\"import-auto-id1772663\">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. Helicopters 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<p>In another example, rockets move forward by expelling gas backward at high velocity. Rockets range in size from fireworks so small that ordinary people use them to immense Saturn Vs that once propelled massive payloads toward the Moon. The propulsion of all rockets, jet engines, deflating balloons, and even squids and octopuses is explained by the same physical principle\u2014Newton\u2019s third law of motion. 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, as shown in <a href=\"\/douglasphys1107\/chapter\/4-4-newtons-third-law-of-motion-symmetry-in-forces\/#rocket\">Figure 2<\/a>. This reaction force is called <strong><span id=\"import-auto-id1295864\">thrust<\/span><\/strong>. 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.<\/p>\n<figure style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_09_07_01a-1.jpg\" alt=\"Picture a shows a rocket launched into space. It moves upward with velocity v in time t and the burning of fuel is also shown. After time t plus delta t the mass of fuel decreases by delta m and hence the velocity of the rocket increases to v plus delta v. The free body diagram shows the weight W of the rocket downward, reaction force upward and the resultant velocity upward too.\" width=\"300\" height=\"1006\" \/><figcaption class=\"wp-caption-text\"><a id=\"rocket\" href=\"\"><\/a><strong>Figure 2.<\/strong> (a) This rocket has a mass<em><strong> m<\/strong><\/em> and an upward velocity <em><strong>v<\/strong><\/em>. The net external force on the system is<strong><em> \u2212mg<\/em><\/strong>, if air resistance is neglected. (b) A time <strong>\u0394<em>t<\/em><\/strong> later the system has two main parts, the ejected gas and the remainder of the rocket. The reaction force on the rocket is what overcomes the gravitational force and accelerates it upward.<\/figcaption><\/figure>\n<div class=\"textbox shaded\">\n<div id=\"fs-id2355307\" class=\"example\">\n<h3 id=\"import-auto-id1645990\">Example 1: Getting Up To Speed: Choosing the Correct System<\/h3>\n<p>A physics professor pushes a cart of demonstration equipment to a lecture hall, as seen in <a class=\"autogenerated-content\" href=\"#import-auto-id2324690\">Figure 3<\/a>. 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<figure id=\"import-auto-id2324690\"><figcaption><\/figcaption><figure style=\"width: 550px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/Figure_04_04_02-1.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=\"550\" height=\"578\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 3.<\/strong> 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 <a href=\"#fs-id2092526\">Example 2<\/a>, since it asks for the acceleration of the entire group of objects. Only <strong>F<sub>floor<\/sub><\/strong> 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<sub>prof<\/sub><\/strong> 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.<\/figcaption><\/figure>\n<\/figure>\n<p id=\"fs-id2687234\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1769338\">Since they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System 1 in <a class=\"autogenerated-content\" href=\"#import-auto-id2324690\">Figure 3<\/a>. The professor pushes backward with a force <strong><em>F<\/em><sub>foot<\/sub><\/strong> of 150 N. According to Newton\u2019s third law, the floor exerts a forward reaction force <strong><em>F<\/em><sub>floor<\/sub><\/strong> 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, <em><strong>f<\/strong><\/em> opposes the motion and is thus in the opposite direction of <strong><em>F<\/em><sub>floor<\/sub><\/strong>. Note that we do not include the forces <strong><em>F<\/em><sub>prof<\/sub><\/strong> or <strong><em>F<\/em><sub>cart<\/sub><\/strong> because these are internal forces, and we do not include <strong><em>F<\/em><sub>foot<\/sub><\/strong> 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<p id=\"import-auto-id1333658\"><strong>Solution<\/strong><\/p>\n<p id=\"import-auto-id1443880\">Newton\u2019s second law is given by<\/p>\n<div id=\"eip-id1456392\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{a\\:=}[\/latex][latex]\\boldsymbol{\\frac{F_{\\textbf{net}}}{m}.}[\/latex]<\/div>\n<p id=\"import-auto-id1729170\">The net external force on System 1 is deduced from <a class=\"autogenerated-content\" href=\"#import-auto-id2324690\">Figure 3<\/a> and the discussion above to be<\/p>\n<div id=\"eip-id1515829\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{\\textbf{F}_{\\textbf{net}}=\\textbf{F}_{\\textbf{floor}}-\\textbf{f}=150\\textbf{ N}-24.0\\textbf{ N}=126\\textbf{ N}.}[\/latex]<\/div>\n<p id=\"import-auto-id2669812\">The mass of System 1 is<\/p>\n<div id=\"eip-id1213154\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{m=(65.0 + 12.0 + 7.0)\\textbf{ kg} = 84\\textbf{ kg}.}[\/latex]<\/div>\n<p id=\"import-auto-id2320233\">These values of <strong><em>F<\/em><sub>net<\/sub><\/strong> and <em><strong>m<\/strong><\/em> produce an acceleration of<\/p>\n<div id=\"eip-id1741589\" class=\"equation\" style=\"text-align: center\">[latex]\\begin{array}{r @{{}={}}l} \\boldsymbol{a} & \\boldsymbol{\\frac{F_{\\textbf{net}}}{m}} \\\\[1em] \\boldsymbol{a} & \\boldsymbol{\\frac{126 \\;\\textbf{N}}{84 \\;\\textbf{kg}} = 1.5 \\;\\textbf{m} \/ \\;\\textbf{s}^2} \\end{array}[\/latex]<\/div>\n<p id=\"import-auto-id2685057\"><strong>Discussion<\/strong><\/p>\n<p id=\"import-auto-id2396019\">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<div class=\"textbox shaded\">\n<div id=\"fs-id2092526\" class=\"example\">\n<h3 id=\"import-auto-id1572504\">Example 2: Force of the Cart\u2014Choosing a New System<\/h3>\n<p>Calculate the force the professor exerts on the cart in <a class=\"autogenerated-content\" href=\"#import-auto-id2324690\">Figure 3<\/a> using data from the previous example if needed.<\/p>\n<p id=\"import-auto-id2423425\"><strong>Strategy<\/strong><\/p>\n<p id=\"import-auto-id1745150\">If we now define the system of interest to be the cart plus equipment (System 2 in <a class=\"autogenerated-content\" href=\"#import-auto-id2324690\">Figure 3<\/a>), 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, <strong><em>F<\/em><sub>prof<\/sub><\/strong>, is an external force acting on System 2. <strong><em>F<\/em><sub>prof<\/sub><\/strong> was internal to System 1, but it is external to System 2 and will enter Newton\u2019s second law for System 2.<\/p>\n<p id=\"import-auto-id2682081\"><strong>Solution<\/strong><\/p>\n<p id=\"import-auto-id2693182\">Newton\u2019s second law can be used to find <strong><em>F<\/em><sub>prof<\/sub><\/strong>. Starting with<\/p>\n<div id=\"eip-id1949236\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{a\\:=}[\/latex][latex]\\boldsymbol{\\frac{F_{\\textbf{net}}}{m}}[\/latex]<\/div>\n<p id=\"import-auto-id2147279\">and noting that the magnitude of the net external force on System 2 is<\/p>\n<div id=\"eip-id2390884\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{net}}=F_{\\textbf{prof}}-f,}[\/latex]<\/div>\n<p id=\"import-auto-id2401022\">we solve for <strong><em>F<\/em><sub>prof<\/sub><\/strong>, the desired quantity:<\/p>\n<div id=\"eip-id1373488\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{prof}}=F_{\\textbf{net}}+f.}[\/latex]<\/div>\n<p id=\"import-auto-id2730714\">The value of <em><strong>f<\/strong><\/em> is given, so we must calculate net <strong><em>F<\/em><sub>net<\/sub><\/strong>. 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<div id=\"eip-id2367884\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{net}}=ma,}[\/latex]<\/div>\n<p id=\"import-auto-id2674462\">where the mass of System 2 is 19.0 kg (<em><strong>m<\/strong><\/em>= 12.0 kg + 7.0 kg) and its acceleration was found to be <strong><em>a<\/em> = 1.5 m\/s<sup>2<\/sup><\/strong> in the previous example. Thus,<\/p>\n<div id=\"eip-id3477001\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{net}}=ma,}[\/latex]<\/div>\n<div id=\"eip-id1454423\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{net}}=(19.0\\textbf{ kg})(1.5\\textbf{ m\/s}^2)=29\\textbf{ N}.}[\/latex]<\/div>\n<p id=\"import-auto-id860562\">Now we can find the desired force:<\/p>\n<div id=\"eip-id1004496\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{prof}}=F_{\\textbf{net}}+f,}[\/latex]<\/div>\n<div id=\"eip-id1197329\" class=\"equation\" style=\"text-align: center\">[latex]\\boldsymbol{F_{\\textbf{prof}}=29\\textbf{ N}+24.0\\textbf{ N}=53\\textbf{ N}.}[\/latex]<\/div>\n<p id=\"import-auto-id2655815\"><strong>Discussion<\/strong><\/p>\n<p id=\"import-auto-id2666000\">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.<\/p>\n<p id=\"import-auto-id2335706\">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 class=\"example\">\n<div class=\"textbox shaded\">\n<h3>PHET EXPLORATIONS: GRAVITY FORCE LAB<\/h3>\n<p>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.<\/p>\n<figure style=\"width: 450px\" class=\"wp-caption aligncenter\"><a href=\"\/resources\/56ecddb16a348d5796513d8d3cc2616ce2d28271\/gravity-force-lab_en.jar\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/douglasphys1104\/wp-content\/uploads\/sites\/1393\/2021\/05\/PhET_Icon-2-1.png\" alt=\"image\" width=\"450\" height=\"147\" \/><\/a><figcaption class=\"wp-caption-text\"><strong>Figure 3.<\/strong> <a href=\"https:\/\/phet.colorado.edu\/en\/simulation\/gravity-force-lab\">Gravity Force Lab<\/a><\/figcaption><\/figure>\n<\/div>\n<\/div>\n<section id=\"fs-id1446741\" class=\"section-summary\">\n<h1>Section Summary<\/h1>\n<ul id=\"fs-id2332580\">\n<li><strong>Newton\u2019s third law of motion<\/strong> represents a basic symmetry in nature. It states: 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.<\/li>\n<li>A <strong>thrust <\/strong>is 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.<\/li>\n<\/ul>\n<\/section>\n<section id=\"fs-id1460367\" class=\"conceptual-questions\">\n<div class=\"bcc-box bcc-info\">\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\"><strong>1: <\/strong>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\"><strong>2: <\/strong>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\"><strong>3: <\/strong>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\"><strong>4: <\/strong>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\"><strong>5: <\/strong>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\"><strong>6: <\/strong>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>\n<section id=\"fs-id1008305\" class=\"problems-exercises\">\n<div class=\"bcc-box bcc-info\">\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\"><strong>1: <\/strong>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>\n<div id=\"fs-id2300721\" class=\"exercise\">\n<div id=\"fs-id1791532\" class=\"problem\">\n<p id=\"import-auto-id1771007\"><strong>2: <\/strong>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> backward. (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 <span style=\"text-decoration: underline\"><strong>write the net force equation.<\/strong><\/span><\/p>\n<\/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 class=\"bcc-box bcc-info\">\n<h3>Solutions<\/h3>\n<p><strong>Problems &amp; Exercises<br \/>\n<\/strong><\/p>\n<p id=\"import-auto-id2297403\"><strong>1: <\/strong>Force on shell: [latex]\\boldsymbol{2.64\\times10^7\\textbf{ N}}[\/latex] Force exerted on ship = [latex]\\boldsymbol{-2.64\\times10^7\\textbf{ N}},[\/latex] by Newton\u2019s third law<\/p>\n<p>2:\u00a0 net force on the losing player = m a = (90.0 kg)(1.20 m\/s<sup>2<\/sup>) = 108 N.\u00a0 The pushing backwards force is 800 N but the net force is only 108 Ns so that means friction = (800-108) = 692 N.\u00a0 \u00a0Without friction the net force would be 800 N and the player would accelerate very quickly backwards.\u00a0 b) net force on the winning player =\u00a0 m a = (110.0 kg)(1.20 m\/s<sup>2<\/sup>) = 132 N.\u00a0 \u00a0He is exerting a force of 800 N so the friction force = ?<\/p>\n<\/div>\n<\/div>\n","protected":false},"author":9,"menu_order":4,"comment_status":"closed","ping_status":"closed","template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-246","chapter","type-chapter","status-publish","hentry"],"part":226,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/chapters\/246","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/wp\/v2\/users\/9"}],"replies":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/wp\/v2\/comments?post=246"}],"version-history":[{"count":1,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/chapters\/246\/revisions"}],"predecessor-version":[{"id":247,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/chapters\/246\/revisions\/247"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/parts\/226"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/chapters\/246\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/wp\/v2\/media?parent=246"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/pressbooks\/v2\/chapter-type?post=246"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/wp\/v2\/contributor?post=246"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/douglasphys1104summer2021\/wp-json\/wp\/v2\/license?post=246"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}