{"id":117,"date":"2023-02-07T18:30:56","date_gmt":"2023-02-07T23:30:56","guid":{"rendered":"https:\/\/opentextbc.ca\/foundationsofphysics\/?post_type=chapter&#038;p=117"},"modified":"2023-12-20T14:32:02","modified_gmt":"2023-12-20T19:32:02","slug":"equilibrium-newtons-first-law","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/foundationsofphysics\/chapter\/equilibrium-newtons-first-law\/","title":{"raw":"Equilibrium &amp; Newton\u2019s First Law","rendered":"Equilibrium &amp; Newton\u2019s First Law"},"content":{"raw":"<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Key Takeaways<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ul>\r\n \t<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/newtlaws\/Lesson-1\/Newton-s-First-Law\">Newton\u2019s First Law<\/a><\/li>\r\n \t<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/vectors\/Lesson-3\/Equilibrium-and-Statics\">Equilibrium and Statics<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\nEquations Introduced and Used for this Topic: (All equations can be written and solved as both <strong>scalar<\/strong> and <strong>vector <\/strong>and<strong> all equations<\/strong> are generally solved as vectors)\r\n<ul style=\"list-style-type: none;\">\r\n \t<li style=\"text-align: center;\">[latex]\\sum F_{x, y, z} = 0 \\text{ N}[\/latex]<\/li>\r\n \t<li style=\"text-align: center;\">[latex]T = r \\times F[\/latex]<\/li>\r\n \t<li style=\"text-align: center;\">[latex]T_{\\text{clockwise}} = T_{\\text{counter-clockwise}}[\/latex]<\/li>\r\n \t<li style=\"text-align: center;\">[latex](m_1 + m_2) x_{\\text{cm}} = m_1 x_1 + m_2 x_2[\/latex]<\/li>\r\n<\/ul>\r\nWhere...\r\n<ul>\r\n \t<li>[latex]T[\/latex] is the torque, measured in newton-metres (Nm)<\/li>\r\n \t<li>[latex]F[\/latex] is force, measured in newtons (N)<\/li>\r\n \t<li>[latex]r[\/latex] is the length of the lever-arm measured in metres (m)<\/li>\r\n \t<li>[latex]m_1[\/latex] and [latex]m_2[\/latex] are the masses of two bodies, measured in kilograms (kg)<\/li>\r\n \t<li>[latex]x_{\\text{cm}}[\/latex], [latex]x_1[\/latex], and [latex]x_2[\/latex] are the distances, measured in metres (m)<\/li>\r\n \t<li>\u00f8 is the angle given for the triangle and is generally between the adjacent side and the hypotenuse with the angle measured in degrees (be careful to make sure that the calculator is not working in gradients or radians).<\/li>\r\n \t<li><em>Hypotenuse<\/em> is the longest side in a right triangle and is opposite to the 90\u00b0 angle.<\/li>\r\n \t<li><em>Opposite<\/em> is the side that is opposite to the angle that you are given.<\/li>\r\n \t<li><em>Adjacent<\/em> is the side that is beside the angle you are given.<\/li>\r\n<\/ul>\r\n<strong>The Trigonometric Functions:<\/strong>\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]\\text{sine \u00f8 } = \\dfrac{\\text{opposite}}{\\text{hypotenuse}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{cosine \u00f8 } = \\dfrac{\\text{adjacent}}{\\text{hypotenuse}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{tangent \u00f8 } = \\dfrac{\\text{opposite}}{\\text{adjacent}}[\/latex]<\/li>\r\n<\/ul>\r\n<img class=\"wp-image-478 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7-trig-functions.png\" alt=\"\" width=\"557\" height=\"185\" \/>\r\n<h1>7.1 Right Angle Trigonometric Functions<\/h1>\r\n<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Observatory_in_Alexandria_at_the_Time_of_Hipparchus.jpg\"><img class=\"aligncenter wp-image-1208\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-angle-trig.jpeg\" alt=\"\" width=\"667\" height=\"456\" \/><\/a>\r\n\r\nTrigonometry, a term derived from the Greek trigonon (triangle) and metron (measure) has known historical roots dating back 4000 years to writings found in Egyptian and Babylonian mathematics and astronomical records. Indian astronomers were using trigonometry over 2600 years ago and that widespread usage by Islamic and Chinese scientists was common at least 1500 years prior.\r\n\r\nThe study of trigonometry returned to European nations in the Renaissance in translations of Arabic and Greek writings. Modern trigonometry is credited as reaching its current form through the works of Leonard Euler (1707-1783), who is considered to be one of the greatest mathematicians in history and had worked in nearly every area of mathematics and physics. Modern notation used in trigonometry comes directly from Euler\u2019s writings, as are many other currently used notations in mathematics and physics. A list of Euler\u2019s achievements is extensive as well as his writings that have been collected in 92 volumes, making him one of the most prolific writers in mathematics history.\r\n\r\n<strong>Introductory trigonometry<\/strong> is based on identifying the similarities between identical right angled (one angle is 90\u00b0) of different sizes. If the angles of a triangle are identical then all of these triangles are simply larger or smaller copies of each other. In all of the cases shown above, if you take any two sides of any triangle shown and divide them by each other, that number will be exactly the same.\r\n\r\n<img class=\"wp-image-479 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-triangles.png\" alt=\"\" width=\"540\" height=\"288\" \/>\r\n\r\nThese triangle ratios have defined names:\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]\\text{sine } = \\dfrac{\\text{opposite}}{\\text{hypotenuse}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{cosine } = \\dfrac{\\text{adjacent}}{\\text{hypotenuse}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{tangent } = \\dfrac{\\text{opposite}}{\\text{adjacent}}[\/latex]<\/li>\r\n<\/ul>\r\nYou often see these equations shortened to:\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]\\text{sin } = \\dfrac{\\text{opp}}{\\text{hyp}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{cos } = \\dfrac{\\text{adj}}{\\text{hyp}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{tan } = \\dfrac{\\text{opp}}{\\text{adj}}[\/latex]<\/li>\r\n<\/ul>\r\nand memorized as: <strong>SOH - CAH - TOA<\/strong>\r\n\r\nDefining the sides of a triangle follows a set pattern:\r\n\r\n<img class=\"alignnone wp-image-480\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-hypotenuse-triangle.png\" alt=\"\" width=\"243\" height=\"234\" \/>\r\n<ol>\r\n \t<li>The side of a triangle that is opposite to the right angle is called the hypotenuse.<\/li>\r\n \t<li>The opposite and adjacent sides are then defined by the angle you are going to work with.\u00a0 One of the sides will be opposite this angle and the other side will be adjacent to this side.<\/li>\r\n<\/ol>\r\n<strong>For example:<\/strong> The following sides are defined by the right angle and the angle you are going to work with \u00f8.\u00a0 You will have to define the adjacent and opposite sides for every right triangle you work with.\r\n\r\n<img class=\"wp-image-481 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-name-sides-triangle.png\" alt=\"\" width=\"560\" height=\"182\" \/>\r\n\r\nThe other right angled trigonometric rations are the reciprocals of Sine, Cosine and Tangent:\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]\\text{Cosecant } = \\dfrac{1}{\\text{Sine}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{Secant } = \\dfrac{1}{\\text{Cosine}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{Cotangent } = \\dfrac{1}{\\text{Tangent}}[\/latex]<\/li>\r\n<\/ul>\r\nor formally defined as:\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]\\text{Cosecant } = \\dfrac{\\text{hypotenuse}}{\\text{opposite}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{Secant } = \\dfrac{\\text{hypotenuse}}{\\text{adjacent}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{Cotangent } = \\dfrac{\\text{adjacent}}{\\text{opposite}}[\/latex]<\/li>\r\n<\/ul>\r\nYou often see these equations shortened to:\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]\\text{Csc } = \\dfrac{\\text{hyp}}{\\text{opp}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{Sec} = \\dfrac{\\text{hyp}}{\\text{adj}}[\/latex]<\/li>\r\n \t<li>[latex]\\text{Cot } = \\dfrac{\\text{adj}}{\\text{opp}}[\/latex]<\/li>\r\n<\/ul>\r\nThese reciprocal trigonometric functions are commonly used in calculus, specifically in integration and when working with polar coordinates.\u00a0 Anyone taking higher levels of mathematics will encounter these reciprocal trigonometric functions.\r\n\r\nUsing the Pythagorean theorem for 30\u00b0, 45\u00b0 and 60\u00b0 right angle triangles, you can get the exact values of the trigonometric relationship (and the reciprocal values). It is standard to see exams where students are required to draw these 30\u00b0, 45\u00b0 and 60\u00b0 right angle triangles and use the side lengths to generate exact values.\r\n\r\nAnother common sight is to see trigonometric tables being used as approximations of the trig ratios of standard angles from 1\u00b0 to 90\u00b0.\u00a0 For these tables you just choose the value that lines up the trigonometric function you wish to use with the angle that you are using. Basic scientific calculators have essentially made these tables obsolete.\r\n\r\nStandard Reference Angles:\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]\\sin 30^{\\circ} = \\dfrac{1}{2}[\/latex]<\/li>\r\n \t<li>[latex]\\cos 30^{\\circ} = \\dfrac{\\sqrt{3}}{2}[\/latex]<\/li>\r\n \t<li>[latex]\\tan 30^{\\circ} = \\dfrac{1}{\\sqrt{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\sin 45^{\\circ} = \\dfrac{1}{\\sqrt{2}}[\/latex]<\/li>\r\n \t<li>[latex]\\cos 45^{\\circ} = \\dfrac{1}{\\sqrt{2}}[\/latex]<\/li>\r\n \t<li>[latex]\\tan 45^{\\circ} = 1[\/latex]<\/li>\r\n \t<li>[latex]\\sin 60^{\\circ} = \\dfrac{\\sqrt{3}}{2}[\/latex]<\/li>\r\n \t<li>[latex]\\cos 60^{\\circ} = \\dfrac{1}{2}[\/latex]<\/li>\r\n \t<li>[latex]\\tan 60^{\\circ} = \\dfrac{\\sqrt{3}}{1}[\/latex]<\/li>\r\n<\/ul>\r\n<img class=\"aligncenter wp-image-483 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-standard-reference-angles-e1678999268935.png\" alt=\"\" width=\"571\" height=\"147\" \/>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.1.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFind the values that correspond to the following trigonometric functions and angles:\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]\\sin 19^{\\circ} = x[\/latex]<\/li>\r\n \t<li>[latex]\\cos 67^{\\circ} = y[\/latex]<\/li>\r\n \t<li>[latex]\\tan 38^{\\circ} = z[\/latex]<\/li>\r\n<\/ul>\r\n<strong>Solution<\/strong>\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]x = 0.326[\/latex]<\/li>\r\n \t<li>[latex]y = 0.391[\/latex]<\/li>\r\n \t<li>[latex]z = 0.781[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\nIt is also possible to work in reverse, i.e.: given the trigonometric ration of two sides, you can find the angle that you are working with.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.1.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFind the angles that correspond to the following trigonometric values:\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]\\sin \u00f8 = 0.829[\/latex]<\/li>\r\n \t<li>[latex]\\cos \u00f8 = 0.940[\/latex]<\/li>\r\n \t<li>[latex]\\tan \u00f8 = 3.732[\/latex]<\/li>\r\n<\/ul>\r\n<strong>Solution<\/strong>\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>\u00f8 = 56\u00b0<\/li>\r\n \t<li>\u00f8 = 20\u00b0<\/li>\r\n \t<li>\u00f8 = 75\u00b0<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\nSometimes you do not have a value that matches up. For these cases you choose the value that is closest to what you have.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.1.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFind the angles that are closest to the following trigonometric values:\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>[latex]\\sin \u00f8 = 0.297[\/latex]<\/li>\r\n \t<li>[latex]\\cos \u00f8 = 0.380[\/latex]<\/li>\r\n \t<li>[latex]\\tan \u00f8 = 0.635[\/latex]<\/li>\r\n<\/ul>\r\n<strong>Solution<\/strong>\r\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\r\n \t<li>\u00f8 = 17\u00b0<\/li>\r\n \t<li>\u00f8 = 68\u00b0<\/li>\r\n \t<li>\u00f8 = 32\u00b0<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.1.4<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nSolve for the unknown side in the following triangle.\r\n\r\n<img class=\"wp-image-502 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.4.png\" alt=\"\" width=\"505\" height=\"229\" \/>\r\n\r\n<strong>Solution<\/strong>\r\n\r\n[latex]\\begin{align}z &amp;= \\sqrt{10^2+20^2} \\\\ &amp;=22.361 \\\\ &amp;= 22 \\text{ cm} \\end{align}[\/latex]\r\n\r\n[latex]angle \\ \u00f8 = \\tan ^{-1}\\left(\\dfrac{10}{20} \\right) = 27^{\\circ}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.1.5<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nSolve for the unknown sides in the following triangle.\r\n\r\n<img class=\"wp-image-503 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.5.png\" alt=\"\" width=\"501\" height=\"271\" \/>\r\n\r\n<strong>Solution<\/strong>\r\n\r\n[latex]y = (25 \\text{ m})(\\sin 32^{\\circ}) = 13 \\text{ m}[\/latex]\r\n\r\n[latex]x =(25 \\text{ m})(\\cos 32^{\\circ}) = 21 \\text{ m}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\nTo find trigonometric tables, go to <a href=\"https:\/\/opentextbc.ca\/foundationsofphysics\/back-matter\/appendix-a-trigonometric-tables\/\">Appendix A: Trigonometric Tables<\/a>.\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 7.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFind the value of each of the following trigonometric functions using your scientific calculator to 6 digits.\r\n<ol class=\"twocolumn\">\r\n \t<li>[latex]\\sin 48^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]\\sin 29^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]\\cos 25^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]\\cos 61^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]\\tan 11^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]\\tan 57^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]\\sin 11^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]\\cos 57^{\\circ}[\/latex]<\/li>\r\n<\/ol>\r\nUse your scientific calculator to find each angle to the nearest hundredth of a degree.\r\n<ol class=\"twocolumn\" start=\"9\">\r\n \t<li>[latex]\\sin \u00f8 = 0.4848[\/latex]<\/li>\r\n \t<li>[latex]\\sin \u00f8 = 0.6293[\/latex]<\/li>\r\n \t<li>[latex]\\cos \u00f8 =0.6561[\/latex]<\/li>\r\n \t<li>[latex]\\cos \u00f8 =0.6157[\/latex]<\/li>\r\n \t<li>[latex]\\tan \u00f8 = 0.6561[\/latex]<\/li>\r\n \t<li>[latex]\\tan \u00f8 = 0.1562[\/latex]<\/li>\r\n \t<li>[latex]\\sin \u00f8 =0.6561[\/latex]<\/li>\r\n \t<li>[latex]\\cos \u00f8 =0.1562[\/latex]<\/li>\r\n<\/ol>\r\nSolve for all unknowns in the following right triangles.\r\n<ol class=\"twocolumn\" start=\"17\">\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone size-full wp-image-514\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-17.png\" alt=\"\" width=\"880\" height=\"407\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone size-full wp-image-516\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-18.png\" alt=\"\" width=\"820\" height=\"357\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone size-full wp-image-522\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-19.png\" alt=\"\" width=\"765\" height=\"371\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone size-full wp-image-520\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-20.png\" alt=\"\" width=\"788\" height=\"407\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-521\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-21.png\" alt=\"\" width=\"208\" height=\"145\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone size-full wp-image-524\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-22.png\" alt=\"\" width=\"902\" height=\"385\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone size-full wp-image-525\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-23.png\" alt=\"\" width=\"678\" height=\"371\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone size-full wp-image-526\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-24.png\" alt=\"\" width=\"838\" height=\"407\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-527\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-25.png\" alt=\"\" width=\"186\" height=\"203\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-528\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-26.png\" alt=\"\" width=\"215\" height=\"91\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-529\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-27.png\" alt=\"\" width=\"226\" height=\"123\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone size-full wp-image-518\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-28.png\" alt=\"\" width=\"815\" height=\"403\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-517\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-29.png\" alt=\"\" width=\"158\" height=\"196\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone size-full wp-image-519\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-30.png\" alt=\"\" width=\"788\" height=\"348\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-523\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-31.png\" alt=\"\" width=\"234\" height=\"142\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone size-full wp-image-515\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-32.png\" alt=\"\" width=\"788\" height=\"348\" \/><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<h1>7.2 Vector Resolution Into Components<\/h1>\r\n<div class=\"textbox\">Extra Help:<a href=\"https:\/\/www.physicsclassroom.com\/Class\/vectors\/U3L1e.cfm#trig\"> Vector Resolution<\/a><\/div>\r\nThrough practice you will be able to quickly resolve angled vectors using right triangle trigonometry. For example, given a force vector of 100 N angled at 42\u00b0 from the horizontal, you can break it up into vertical and horizontal components using the sine and cosine functions.\u00a0 In this instance the vector components are as follows:\r\n\r\n<img class=\"wp-image-534 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-triangle.png\" alt=\"\" width=\"362\" height=\"275\" \/>\r\n\r\nQuick Resolution:\r\n<ul>\r\n \t<li>[latex]F = 100 \\text{ N}[\/latex]<\/li>\r\n \t<li>[latex]F_x = 100 \\text{ N } \\cos 42^{\\circ}[\/latex] (\u2248 74 N)<\/li>\r\n \t<li>[latex]F_y = 100 \\text{ N } \\sin 42^{\\circ}[\/latex]\u00a0 (\u2248 70 N)<\/li>\r\n<\/ul>\r\nA more detailed resolution is:\r\n<ul>\r\n \t<li>[latex]F_x[\/latex]: [latex]\\cos 42^{\\circ} = \\dfrac{F_x}{100 \\text{ N}}[\/latex] which, when isolated, leaves us with [latex]F_x = 100 \\text{ N } \\cos 42^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]F_y[\/latex]: [latex]\\sin 42^{\\circ} = \\dfrac{F_y}{100 \\text{ N}}[\/latex] which, when isolated, leaves us with [latex]F_y = 100 \\text{ N } \\sin 42^{\\circ}[\/latex]<\/li>\r\n<\/ul>\r\nWhen the vector magnitude is known, then the opposite side vector is the sine of the vector angle and the adjacent side vector is the cosine of the vector angle.\r\n\r\nWhen both vector components are known, the angle of the resultant vector can be found using the tangent function and the magnitude of the vector by using the Pythagorean Theorem.\u00a0 For this, one generally uses a more traditional approach.\u00a0 Consider the following example:\r\n\r\n<img class=\" wp-image-536 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-F_net.png\" alt=\"\" width=\"481\" height=\"366\" \/>\r\n\r\nThe magnitude of the net force is:\r\n<ul>\r\n \t<li>[latex](F_{\\text{net}})^2 = (600 \\text{ N})^2 + (700 \\text{ N})^2[\/latex]<\/li>\r\n \t<li>[latex](F_{\\text{net}})^2 = 360 000 \\text{ N}^2 + 490000 \\text{ N}^2[\/latex]<\/li>\r\n \t<li>[latex](F_{\\text{net}})^2 = 850000 \\text{ N}^2[\/latex]<\/li>\r\n \t<li>[latex]F_{\\text{net}} = 922 \\text{ N } (\\approx 920 \\text{ N})[\/latex]<\/li>\r\n<\/ul>\r\nThe angle this makes is:\r\n<ul>\r\n \t<li>[latex]\\tan \u00f8 = \\dfrac{600 \\text{ N}}{700 \\text{ N}}[\/latex]<\/li>\r\n \t<li>[latex]\\tan \u00f8 = 0.8571[\/latex]<\/li>\r\n \t<li>[latex]\u00f8 = \\tan ^{-1} 0.8571[\/latex]<\/li>\r\n \t<li>[latex]\u00f8 = 40.6^{\\circ}\u00a0 (\\approx 41^{\\circ})[\/latex]<\/li>\r\n<\/ul>\r\nThe main purpose of this section is for you to be able to quickly resolve a vector with an angle into components.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.2.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nResolve the following vectors and angles into components.\r\n\r\n<img class=\"wp-image-537 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-triangle.png\" alt=\"\" width=\"303\" height=\"288\" \/>\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]F_y = 40 \\text{ N } \\sin 38^{\\circ}[\/latex] or 24.6\u00b0 (\u2248 25 N)<\/li>\r\n \t<li>[latex]F_x = 40 \\text{ N } \\cos 38^{\\circ}[\/latex] or 31.5\u00b0 (\u2248 32 N)<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.2.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nResolve the following vectors and angles into components.\r\n\r\n<img class=\"wp-image-538 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-angles.png\" alt=\"\" width=\"308\" height=\"268\" \/>\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]\\vec{v_y} = 48 \\text{ m\/s } \\sin 32^{\\circ}[\/latex] or 25.4 m\/s (\u2248 25 m\/s)<\/li>\r\n \t<li>[latex]\\vec{v_x} = 48 \\text{ m\/s } \\cos 32^{\\circ}[\/latex] or 40.7 m\/s (\u2248 41 m\/s)<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.2.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nResolve the following vectors and angles into components.\r\n\r\n<img class=\"wp-image-539 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-angles.png\" alt=\"\" width=\"303\" height=\"268\" \/>\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]\\vec{F_y} = 32 \\text{ m\/s } \\sin 28^{\\circ}[\/latex] or 15 m\/s<\/li>\r\n \t<li>[latex]\\vec{F_x} = 32 \\text{ m\/s } \\cos 28^{\\circ}[\/latex] or (\u2212) 28 m\/s<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.2.4<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nResolve the following vectors and angles into components.\r\n\r\n<img class=\" wp-image-540 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-angles.png\" alt=\"\" width=\"308\" height=\"269\" \/>\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]\\vec{a_y} = 20.4 \\text{ m\/s}^2 \\text{ } \\sin 42^{\\circ}[\/latex] or (\u2212) 14 m\/s<sup>2<\/sup><\/li>\r\n \t<li>[latex]\\vec{a_x} = 20.3 \\text{ m\/s}^2 \\text{ } \\cos 42^{\\circ}[\/latex] or 15 m\/s<sup>2<\/sup><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 7.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nBreak the following vectors into components (not drawn to scale)\r\n<ol class=\"twocolumn\">\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-543\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-exercise.png\" alt=\"\" width=\"253\" height=\"222\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-546\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-exercise.png\" alt=\"\" width=\"242\" height=\"214\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-548\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-exercise.png\" alt=\"\" width=\"242\" height=\"214\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-550\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-exercise.png\" alt=\"\" width=\"243\" height=\"216\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-551\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.5-exercise.png\" alt=\"\" width=\"247\" height=\"219\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-552\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.6-exercise.png\" alt=\"\" width=\"250\" height=\"221\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-553\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.7-exercise.png\" alt=\"\" width=\"247\" height=\"233\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-554\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.8-exercise.png\" alt=\"\" width=\"249\" height=\"228\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-555\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.9-exercise.png\" alt=\"\" width=\"248\" height=\"236\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-556\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.10-exercise.png\" alt=\"\" width=\"248\" height=\"236\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-557\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.11-exercise.png\" alt=\"\" width=\"245\" height=\"232\" \/><\/li>\r\n \t<li style=\"margin-bottom: 40px;\"><img class=\"alignnone wp-image-558\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.12-exercise.png\" alt=\"\" width=\"250\" height=\"235\" \/><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<h1>7.3 Static Equilibrium (Concurrent Forces)<\/h1>\r\n<div class=\"textbox\">\r\n\r\nIn the News: <a href=\"https:\/\/www.smithsonianmag.com\/smart-news\/\">Scientists Prove Leonardo da Vinci\u2019s 500 Year Old Bridge Design Actually Works<\/a>\r\n\r\n<\/div>\r\nStatic equilibrium is one of the requirements of architecture; engineering tools and techniques have been employed since ancient times to create buildings that are statically safe from collapse. In short, structural design must work to counteract the effects of gravity, storms and earthquakes, to name a few of the forces that can act on a structure.\r\n\r\n<img class=\"alignleft wp-image-560\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Acueducto-de-Tarragona.jpg\" alt=\"Les Ferreres Aqueduct\" width=\"378\" height=\"213\" \/>The bridge to the left was thought to have been built during the reign of Augustus (27 BCE- 14CE) as an aqueduct to supply water to the town of Tarraco in Catalonia, Spain. This structure, 249 m in length, is comprised of a series of free-standing stone arches that have the same radius of 5.9 \u00b1 0.15 m.\u00a0 Of the 931 known Roman bridge structures surviving to date, the largest of these is the Puente Romano bridge in Merida, Spain, having a length of 792 m.\r\n\r\n<img class=\" wp-image-561 alignright\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Azure_Window.jpeg\" alt=\"\" width=\"352\" height=\"264\" \/>The arch design used in Roman bridges is one that is found in nature, such as the Azure Window shown to the right and located in Malta. The success of the strength in the design of arch structures is due to the way in which the force from the weight of the bridge is carried outward along the curve of the arch, and is balanced by the supports anchoring either end.\r\n\r\n<img class=\"wp-image-563 alignleft\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-notre-dame-buttress.jpeg\" alt=\"\" width=\"214\" height=\"287\" \/>The flying buttress design generally known as a design favorite of the architects of the Notre Dame Cathedral consists of half arches placed on either side of the building to support the weight, and wind shears acting on the central dome covering the building. This technique of balancing forces allowed for the walls of the building to be opened up to allow windows to be constructed in the walls, to let in light and support the weight of the roof.\r\n\r\n<img class=\" wp-image-564 alignright\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-VillardButtressReims.jpeg\" alt=\"\" width=\"201\" height=\"291\" \/>The sketch shown to the right is that of Villard de Honnecourt (1225-1235). While of unknown purpose, it is part of a collection of works showing various designs of devices he saw during his travels through Medieval Europe.\r\n\r\n<strong>**The images shown \u00a0below illustrate the forces that act in an arch bridge: force vectors acting in the Action Force (Load) and the Reaction Force (Support for the Load).**<\/strong>\r\n\r\n<strong>Conceptual Origins ...<\/strong>\r\n\r\nThe term equilibrium[footnote]Equilibrium is further classified as being stable, neutral or unstable.[\/footnote] originates from the Latin \u201cacqui\u201d and \u201clibra\u201d, which mean \u201cequal\u201d and \u201cbalance\u201d. When used in physics concepts, \u201cobjects in a state of equilibrium\u201d refers to a balancing of forces acting on the object.\u00a0 This basic concept of equilibrium, i.e.: all forces acting on a point in an object sum to zero, is the embodiment of Newton\u2019s first law.\r\n\r\n<strong>To recall, Newton's first law<\/strong> states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of <strong>inertia.<\/strong> The key point here is that if there is <strong>no net force<\/strong> acting on an object (if all the external forces cancel each other out) then the object will maintain at a <strong>constant velocity<\/strong>. If that velocity is zero, then the object remains at rest. If an external force is applied, the velocity will change because of the force.\r\n\r\nFrom the Latin version of Newton's Principia, his first law reads as...\r\n<p style=\"padding-left: 40px;\"><strong><em>Lex I: Corpus omne perseverare in statu suo quiescendi vel movendi uniformiter in directum, nisi quatenus a viribus impressis cogitur statum illum mutare.<\/em><\/strong><\/p>\r\nTranslated to English, this reads:\r\n<p style=\"padding-left: 40px;\"><strong><em>Law I: Every body persists in its state of being at rest or of moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed<\/em><\/strong><strong>.<\/strong><\/p>\r\nNewton then expands on his first law with:\r\n<p style=\"padding-left: 40px;\"><strong><em>Projectiles persevere in their motions, so far as they are not retarded by the resistance of the air, or impelled downwards by the force of gravity. A top, whose parts by their cohesion are perpetually drawn aside from rectilinear motion, does not cease its rotation, otherwise than as it is retarded by the air. The greater bodies of the planets and comets, meeting with less resistance in more free spaces, preserve their motions both progressive and circular for a much longer time.<\/em><\/strong><\/p>\r\nOne of the challenges to the interpretation of Newton\u2019s laws is that he appears to have written them following the style of earlier Greek philosophers, such as Euclid. As such, historians have suggested that Newton\u2019s first law is simply a natural deduction of his second law since there is nothing about frames of reference included in Newton\u2019s writings.\r\n\r\nEquilibrium is explored in three different ways in this chapter: translational, static and rotational, all relating to the balancing of forces that act on an object or a structure. Equilibrium is a very important concept, and you should encounter a number of different ways that explore equilibrium in specialized fields.\r\n\r\n<strong>Translational equilibrium<\/strong> occurs when no net or resultant force is acting on an object. This means that the sum of all forces that are acting will equal 0 N. Newton\u2019s first law interprets this to mean that if the object is at rest, it will remain at rest and if the object is in motion, it will remain in that exact state of motion.\r\n\r\n<strong>Rotational equilibrium<\/strong> requires that the object is non-rotating. This necessitates that all forces causing it to rotate (torques) will sum to zero. Newton\u2019s first law interprets this to mean that if the object is at rest, not-rotating, it will remain at rest and not-rotating. If the object is rotating, it will remain in that exact state of rotation, neither speeding up nor slowing down.\r\n\r\n<strong>Static equilibrium<\/strong> is a special case of translational and rotational equilibrium, and occurs when the net force or resultant force acting on an object sums to zero and the object is neither moving in any direction nor rotating.\u00a0 Static equilibrium requires that the object in question is at rest and remains at rest.\r\n\r\nWhen we look at the effect of forces acting on a body, there are five commonly observed actions. You will be encountering the first two of these, compression (C) and tension (T), in the problems in this chapter.<strong>\u00a0<\/strong>\r\n\r\n<strong>Free Body Diagrams <\/strong>\r\n<div class=\"textbox textbox--sidebar\">\r\n<ul>\r\n \t<li>Extra Help: <a href=\"https:\/\/physics.info\/newton-first\/\">Forces, Free Body Diagrams and Newton\u2019s Laws of Force<\/a><\/li>\r\n \t<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/newtlaws\/Lesson-2\/Drawing-Free-Body-Diagrams\">Drawing Free Body Diagrams<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\nFree body diagrams are standard tools used to analyze forces in equilibrium. Free body diagrams are used to show the magnitude and direction of all forces that act on a body in a given situation. In many cases these diagrams are drawn to scale, with the length of the arrow in scale to the magnitude of the force and the angle of this force accurately positioned. Learning to draw a free body diagram is important for one to be able to work through questions in statics, dynamics and a few other fields of classical mechanics.\r\n\r\nUsing the free body diagram as a tool, one is able to easily visualize whether or not the forces acting on an object add to zero ([latex]F_{\\text{net}} = 0[\/latex]). If they do add to zero, the object is in static equilibrium and not moving. If the forces acting on the body do not add to zero ([latex]F_{\\text{net}} \\neq 0[\/latex]), it will accelerate.\r\n\r\nHyper physics lists a number of common forces that you can encounter where free body diagrams assist in working out the intricacies of the problem. In Chapter 6, you encountered <strong>weight (<\/strong>[latex]w[\/latex], [latex]F_g[\/latex]<strong>) <\/strong>which is a force vector directed straight down towards the Earth\u2019s centre. In Chapter 7 you will use another force vector called <strong>tension ([latex]T[\/latex], [latex]F_t[\/latex])<\/strong>.\u00a0 Tension is the force that exists in strings, cables, ropes or beams that are acting to apply a force to an external object. Tension for these are the forces transmitted by the cable, rope, string or beam and act in the direction these are pulling or pushing. In the simplest examples that follow, this tension balances the weight. The examples become more complicated when multiple tensions act on an external object.\r\n\r\n<strong>Other common forces<\/strong> that you will encounter are:\r\n<ul>\r\n \t<li><strong>Normal force ([latex]N[\/latex], [latex]F_n[\/latex]), <\/strong>which can be explained as the force exerted by a surface tangentially away from the surface to support an object.<\/li>\r\n \t<li><strong style=\"orphans: 1; text-align: initial; font-size: 14pt;\">Frictional force ([latex]f[\/latex]<\/strong><span style=\"orphans: 1; text-align: initial; font-size: 14pt;\">\u00a0<\/span><strong style=\"orphans: 1; text-align: initial; font-size: 14pt;\">, [latex]F_f[\/latex]),<\/strong><span style=\"orphans: 1; text-align: initial; font-size: 14pt;\"> which is the force acting between two surfaces, and is generally opposite to the direction the object is moving or accelerating, but this is not always true.<\/span><\/li>\r\n \t<li><strong style=\"orphans: 1; text-align: initial; font-size: 14pt;\">Elastic forces ([latex]F_e[\/latex], [latex]F_s[\/latex])<\/strong><span style=\"orphans: 1; text-align: initial; font-size: 14pt;\"> are the forces exerted by springs or rubber bands, etc., when they are stretched (or compressed in some cases). The force in these instances acts in the direction of the stretch or compression.<\/span><\/li>\r\n \t<li><strong style=\"orphans: 1; text-align: initial; font-size: 14pt;\">Buoyancy ([latex]B[\/latex], [latex]F_b[\/latex])<\/strong><span style=\"orphans: 1; text-align: initial; font-size: 14pt;\"> is the force that acts on objects that are immersed or suspended in a fluid.\u00a0 The direction of this force is always upwards and away from the centre of the Earth.<\/span><\/li>\r\n \t<li><strong>Drag or air resistance ([latex]R[\/latex], [latex]D[\/latex], [latex]F_d[\/latex], [latex]F_{\\text{air}}[\/latex]) <\/strong>is the force that acts on any object that is moving through a fluid, be it liquid or gaseous. This force will generally be acting opposite to the direction that the object is moving (opposite to the velocity).<\/li>\r\n \t<li><strong>Lift ([latex]L[\/latex], [latex]F_l[\/latex])<\/strong> is the force that acts on an object that is tangential to its movement through a fluid, meaning that as the object is moving forward through a fluid, the lift will act at 90\u00b0 to the direction of the object\u2019s velocity.<\/li>\r\n \t<li><strong>Thrust ([latex]T[\/latex], [latex]F_t[\/latex])<\/strong> is the action-reaction force that results from pushing a fluid in one direction (generally backwards) that allows the object doing this pushing to move in the opposite direction.<\/li>\r\n<\/ul>\r\nIn all cases, combinations of the above forces can be acting on some object, requiring one to find the net result of forces <strong>([latex]F_{\\text{net}}[\/latex])<\/strong> acting on the object.\u00a0 All these forces are in newtons (N).\r\n\r\nThe following examples will illustrate how one can draw a free body diagram of the forces acting on a body.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.3.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nConsider a parachutist:\r\n<ol type=\"i\">\r\n \t<li>First starts to fall\r\n<img class=\"alignnone wp-image-578 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-i-example.png\" alt=\"\" width=\"212\" height=\"277\" \/><\/li>\r\n \t<li>Reaches terminal velocity\r\n<img class=\"wp-image-579 alignnone\" style=\"orphans: 1; text-align: initial; font-size: 0.9em; word-spacing: normal;\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-ii-example.png\" alt=\"\" width=\"198\" height=\"301\" \/><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.3.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nConsider a person floating in a lake:\r\n\r\n<img class=\" wp-image-580 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.2-example.png\" alt=\"\" width=\"232\" height=\"310\" \/>\r\n\r\n<\/div>\r\n<\/div>\r\nThe principle used to draw these diagrams is the same when you encounter more than two forces acting on an object.\u00a0 Two cables supporting an object at rest or moving at a constant velocity ([latex]F_{\\text{net}} = 0[\/latex]) would show up as two forces acting in unison to balance the weight or force of gravity pulling down on the object.\u00a0 If these cables are acting to accelerate the object either up, down or to one side ([latex]F_{\\text{net}} \\neq 0[\/latex]) then the forces acting on the object would show some imbalance.\u00a0 In all cases, the force arrows will act in the directions that the forces act on the object.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.3.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nSuppose you have a box suspended in air by two cables as shown below.\r\n\r\n<img class=\"aligncenter wp-image-581 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-two-cables.png\" alt=\"\" width=\"296\" height=\"296\" \/>\r\n\r\nIf we were to redraw this using force vectors, it would look like:\r\n\r\n<img class=\"wp-image-582 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-1.png\" alt=\"\" width=\"337\" height=\"345\" \/>\r\n\r\nTo properly analyze this diagram we would need both the magnitudes and directions of all forces acting on the box. (Forces are generally termed tensions when these forces are pulling on the cable.\u00a0 This is compared to a beam that can be experiencing\u00a0 a force that can be acting to compress it.)\r\n\r\nThis diagram to the right can be further reduced to:\r\n\r\n<img class=\"wp-image-583 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-2.png\" alt=\"\" width=\"263\" height=\"403\" \/>\r\n\r\nTrigonometry would be used at this point to break the two Cable Tensions (T1 &amp; T2) into vector components. The rest of the solution would be algebra in balancing the forces acting on the Box.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.3.4<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFind the tension in a cable supporting a single box of 25 kg in Static Equilibrium.\r\n\r\n<strong>Solution<\/strong>\r\n\r\nThe sketch and the free body diagrams for this example are:\r\n\r\n<img class=\"wp-image-635 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example.png\" alt=\"\" width=\"457\" height=\"282\" \/>\r\n\r\nThe solution for this requires that the forces acting on the box must sum to zero.\u00a0 In an equation this looks like:\r\n<p style=\"text-align: center;\">[latex]\\sum \\vec{F_y} = 0 \\text{ N}[\/latex]<\/p>\r\nTo solve this problem, assign cartesian directions as you would on a graph:\r\n\r\nFor y-direction, up is positive and down is negative and for the x-direction, right is positive and left is negative.\r\n\r\n<img class=\"wp-image-636 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-2.png\" alt=\"\" width=\"430\" height=\"305\" \/>\r\n\r\nUsing this vector notation, the solution is as follows:\r\n<ul>\r\n \t<li>[latex]T_1 + w = 0 \\text{ N}[\/latex]<\/li>\r\n \t<li>[latex]T_1 + (25 \\text{ kg})(- 9.8 \\text{ m\/s}^2) = 0 \\text{ N}[\/latex]<\/li>\r\n \t<li>[latex]T_1 - 245 \\text{ N } = 0 \\text{ N}[\/latex]<\/li>\r\n \t<li>[latex]T_1 = 0 \\text{ N } + 245 \\text{ N}[\/latex]<\/li>\r\n \t<li>[latex]T_1 = 245 \\text{ N } (\\approx 250 \\text{ N})[\/latex]<\/li>\r\n<\/ul>\r\nThe direction of this force is upwards.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.3.5<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFind the tension in the two cables supporting a single crate of 100 kg in static equilibrium as shown in the diagram below.\r\n\r\n<img class=\"wp-image-640 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-3.png\" alt=\"\" width=\"373\" height=\"353\" \/>\r\n\r\n<strong>Solution<\/strong>\r\n\r\nThe free body diagram of this problem looks like:\r\n\r\n<img class=\"aligncenter wp-image-642 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example.png\" alt=\"\" width=\"370\" height=\"337\" \/>\r\n\r\nThese forces act in two directions... \u00b1 [latex]x[\/latex] and \u00b1 [latex]y[\/latex], and the solution for this requires balancing these forces in two directions.\r\n<p style=\"text-align: center;\">[latex]\\sum \\vec{F_x} = 0 \\text{ N}[\/latex] and [latex]\\sum \\vec{F_y} = 0 \\text{ N}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{Weight }= (100 \\text{ kg})(-9.8 \\text{m\/s}^2)[\/latex] or [latex]-980 \\text{ N}[\/latex] in the y-direction.<\/p>\r\nSince cable 2 is the only cable supporting the crate, we know that the tension in the cable in the y-direction must be + 980 N.\u00a0 The sketch of this looks as follows:\r\n\r\n<img class=\"wp-image-643 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-2.png\" alt=\"\" width=\"280\" height=\"250\" \/>\r\n\r\nTrigonometry is now used to find the tension of cable 2 in the x direction ([latex]T_{2x}[\/latex]):\r\n<ul>\r\n \t<li>[latex]\\tan 30^{\\circ} = \\dfrac{T_{2x}}{980 \\text{ N}}[\/latex]<\/li>\r\n \t<li>[latex]T_{2x} = 980 \\text{ N } \\tan 30^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]T_{2x} = -566 \\text{ N } (\\approx -570 \\text{ N})[\/latex], negative x direction.<\/li>\r\n<\/ul>\r\nNow, the total tension of cable 2 is:\r\n<ul>\r\n \t<li>[latex]\\cos 30^{\\circ}= \\dfrac{980 \\text{ N}}{T_2}[\/latex]<\/li>\r\n \t<li>[latex]T_2 = \\dfrac{980 \\text{ N}}{\\cos 30^{\\circ}}[\/latex]<\/li>\r\n \t<li>[latex]T_2 = 1132 \\text{ N } (\\approx 1100 \\text{ N})[\/latex]<\/li>\r\n<\/ul>\r\nAnd, the tension of cable 1 is:\r\n<ul>\r\n \t<li>[latex]T_1 = 566 \\text{ N } (\\approx 570 \\text{ N})[\/latex] (equal and opposite to [latex]T_{2x}[\/latex])<\/li>\r\n<\/ul>\r\nPutting all of this in a free body diagram yields:\r\n\r\n<img class=\"wp-image-644 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-3.png\" alt=\"\" width=\"291\" height=\"342\" \/>\r\n\r\nWhen summing up the forces [latex]\\sum \\vec{F_x} = 0 \\text{ N}[\/latex] and [latex]\\sum \\vec{F_y} = 0 \\text{ N}[\/latex]...\r\n\r\n[latex]\\sum \\vec{F_x} = 0 \\text{ N}[\/latex]\r\n<ul>\r\n \t<li>[latex]T_{2y} + w = 0 \\text{ N}[\/latex]<\/li>\r\n \t<li>[latex]980 \\text{ N } - 980 \\text{ N } = 0 \\text{ N}[\/latex]<\/li>\r\n<\/ul>\r\n[latex]\\sum \\vec{F_y} = 0 \\text{ N}[\/latex]\r\n<ul>\r\n \t<li>[latex]T_{2x} + T_1 = 0 \\text{ N}[\/latex]<\/li>\r\n \t<li>[latex]-566 \\text{ N } + 566 \\text{ N } = 0 \\text{ N}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.3.6<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFind the tension in the two cables supporting a single box of 420 kg in static equilibrium as shown in the diagram below and to the right.\r\n\r\n.\r\n\r\n<img class=\"wp-image-647 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example.png\" alt=\"\" width=\"273\" height=\"307\" \/>\r\n\r\n<strong>Solution<\/strong>\r\n\r\nThe free body diagram for this problem looks like:\r\n\r\n<img class=\"wp-image-648 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-2.png\" alt=\"\" width=\"228\" height=\"359\" \/>\r\n\r\nSolving this problem is a little more work.\r\n\r\nFirst, the weight is:\r\n<ul>\r\n \t<li>[latex]w = m g[\/latex]<\/li>\r\n \t<li>[latex]w = (420 \\text{ kg})(9.8 \\text{ m\/s}^2)[\/latex]<\/li>\r\n \t<li>[latex]w = 4116 \\text{ N } (\\approx 4100 \\text{ N})[\/latex]<\/li>\r\n<\/ul>\r\n<img class=\"wp-image-649 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-3.png\" alt=\"\" width=\"268\" height=\"517\" \/>\r\n\r\nSumming up the forces acting on the box:\r\n<ul>\r\n \t<li>In the y direction: [latex]T_{1y} + T_{2y} + w = 0 \\text{ N}[\/latex]<\/li>\r\n \t<li>In the x direction: [latex]T_{1x} + T_{2x} = 0 \\text{ N}[\/latex]<\/li>\r\n<\/ul>\r\nFind what we know about the Tensions.\r\n\r\n<img class=\"aligncenter wp-image-650 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-4.png\" alt=\"\" width=\"168\" height=\"302\" \/>\r\n\r\n[latex]\\sin 30^{\\circ} = T_{1x} \\div T_1[\/latex], or...\r\n<ul>\r\n \t<li>[latex]T_{1x} = T_1 \\sin 30^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]T_{1x} = 0.500 T_2[\/latex]<\/li>\r\n<\/ul>\r\n[latex]\\cos 30^{\\circ} = T_{1y} \\div T_1[\/latex], or...\r\n<ul>\r\n \t<li>[latex]T_{1y} = T_1 \\cos 30^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]T_{1y} = 0.866 T_1[\/latex]<\/li>\r\n<\/ul>\r\n<img class=\"aligncenter wp-image-651 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-5.png\" alt=\"\" width=\"169\" height=\"267\" \/>\r\n\r\n[latex]\\sin 40^{\\circ} = T_{2x} \\div T_2[\/latex], or...\r\n<ul>\r\n \t<li>[latex]T_{2x} = T_2 \\sin 40^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]T_{2x} = 0.643 T_2[\/latex]<\/li>\r\n<\/ul>\r\n[latex]\\cos 40^{\\circ} = T_{2y} \\div T_2[\/latex], or...\r\n<ul>\r\n \t<li>[latex]T_{2y} = T_2 \\cos 40^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]T_{2y} = 0.766 T_2[\/latex]<\/li>\r\n<\/ul>\r\n&nbsp;\r\n\r\nBalancing all forces:\r\n<ul>\r\n \t<li>x direction: [latex]0 \\text{ N } = T_{1x} + T_{2x}[\/latex] or [latex]0 \\text{ N } = 0.500 T_1 - 0.643 T_2[\/latex]<\/li>\r\n \t<li>y direction: [latex]0 \\text{ N } = T_{1y} + T_{2y} + w[\/latex] or [latex]0.866 T_1 + 0.766 T_2 = 4116 \\text{ N}[\/latex]<\/li>\r\n<\/ul>\r\n&nbsp;\r\n\r\nNow, use substitution to solve this system of two equations.\r\n\r\nFirst: Isolate [latex]T_1[\/latex]\r\n<ul>\r\n \t<li>[latex]0 \\text{ N } = 0.643 T_1 - 0.500 T_2[\/latex]<\/li>\r\n \t<li>[latex]0.500 T_1 = 0.643 T_2[\/latex]<\/li>\r\n \t<li>[latex]T_1 = \\dfrac{0.643 T_2}{0.500}[\/latex]<\/li>\r\n \t<li>[latex]T_ 1 = 1.29 T_2[\/latex]<\/li>\r\n<\/ul>\r\nSecond: Substitute for [latex]T_2[\/latex]\r\n<ul>\r\n \t<li>[latex]0.866 T_1 + 0.766 T_2 = 4116 \\text{ N}[\/latex]<\/li>\r\n \t<li>[latex]0.866 (1.29 T_2) + 0.766 T_2 = 4116 \\text{ N}[\/latex]<\/li>\r\n \t<li>[latex]1.12 T_{2} + 0.766 T_{2} = 4116 \\text{ N}[\/latex]<\/li>\r\n \t<li>[latex]1.866 T_2 = 4116 \\text{ N}[\/latex]<\/li>\r\n \t<li>[latex]T_2 = 4116 \\text{ N } \\div 1.462 \\text{ or } 2182 \\text{ N } (\\approx 2200 \\text{ N})[\/latex]<\/li>\r\n<\/ul>\r\nFinally:\r\n<ul>\r\n \t<li>[latex]T_1 = 1.29 T_2[\/latex]<\/li>\r\n \t<li>[latex]T_1 = (1.29)(2182 \\text{ N})[\/latex]<\/li>\r\n \t<li>[latex]T_1 = 2815 \\text{ N } (\\approx 2800 \\text{ N})[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 7.3.7<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFind the tension in the cable and the reactive force from the beam supporting a single crate of 500 kg in static equilibrium as shown in the diagram below.\r\n\r\n<img class=\"wp-image-661 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-6.png\" alt=\"\" width=\"149\" height=\"258\" \/>\r\n\r\n&nbsp;\r\n\r\n<strong>Solution<\/strong>\r\n\r\n[latex]\\text{Weight } = (500 \\text{ kg})(- 9.8 \\text{ m\/s}^2) or -4900 \\text{ N}[\/latex]\r\n\r\nConditions for equilibrium:\r\n<p style=\"text-align: center;\">[latex]\\sum \\vec{F_x} = 0 \\text{ N}[\/latex] or [latex]\\sum \\vec{F_y} = 0 \\text{ N}[\/latex]<\/p>\r\nSince this cable is the only force supporting the crate, we know that the tension in the cable in the y-direction must be + 4900 N.\r\n<ul>\r\n \t<li style=\"text-align: left;\">[latex]T_y = + 4900 \\text{ N}[\/latex]<\/li>\r\n<\/ul>\r\nTrigonometry is now used to find the tension of the cable in the x direction ([latex]T_x[\/latex]).\r\n<ul>\r\n \t<li>[latex]\\tan 42^{\\circ} = \\dfrac{4900 \\text{ N}}{T_x}[\/latex]<\/li>\r\n \t<li>[latex]T_x = \\dfrac{4900 \\text{ N}}{\\tan 42^{\\circ}}[\/latex]<\/li>\r\n \t<li>[latex]T_x = 5442 \\text{ N } (\\approx 5400 \\text{ N})[\/latex]... positive X-direction.<\/li>\r\n<\/ul>\r\nNow, the total tension of the cable is:\r\n<ul>\r\n \t<li>[latex]\\sin 41^{\\circ} = \\dfrac{4900 \\text{ N}}{T}[\/latex]<\/li>\r\n \t<li>[latex]T = \\dfrac{4900 \\text{ N}}{\\sin 42^{\\circ}}[\/latex]<\/li>\r\n \t<li>[latex]T = 7323 \\text{ N } (\\approx 7300 \\text{ N})[\/latex]<\/li>\r\n<\/ul>\r\n<img class=\"wp-image-662 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-example-2.png\" alt=\"\" width=\"256\" height=\"240\" \/>\r\n\r\nThe reactive force from the beam is:\r\n<ul>\r\n \t<li>[latex]C = -5442 \\text{ N } (\\approx - 5400 \\text{ N})[\/latex] (equal and opposite to [latex]T_x[\/latex])<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 7.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>Find the tension in a cable supporting a single box of 100 kg in static equilibrium.\r\n<img class=\"wp-image-664 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-exercise.png\" alt=\"\" width=\"109\" height=\"328\" \/><\/li>\r\n \t<li>Find the tension in a cable supporting a single box weighing 5.0 N in static equilibrium.\r\n<img class=\"wp-image-665 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.2-exercise.png\" alt=\"\" width=\"110\" height=\"331\" \/><\/li>\r\n \t<li>Find the tension in the two cables supporting a single crate of mass 50 kg in static equilibrium as shown in the diagram below to the right.\r\n<img class=\"wp-image-666 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-exercise.png\" alt=\"\" width=\"242\" height=\"233\" \/><\/li>\r\n \t<li>Find the tension in the two cables supporting a single crate of 250 kg in static equilibrium as shown in the diagram below to the right.\r\n<img class=\"aligncenter wp-image-667 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-exercise.png\" alt=\"\" width=\"242\" height=\"233\" \/><\/li>\r\n \t<li>Find the tension in the two cables supporting a single crate of 120 kg in static equilibrium as shown in the diagram below and to the right.\r\n<img class=\"wp-image-668 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-exercise.png\" alt=\"\" width=\"256\" height=\"256\" \/><\/li>\r\n \t<li>Find the tension in the two cables supporting a single crate of 1200 kg in static equilibrium as shown in the diagram below and to the right.\r\n<img class=\"aligncenter wp-image-669 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-exercise.png\" alt=\"\" width=\"268\" height=\"267\" \/><\/li>\r\n \t<li>Find the tension in the cable and the reactive force from the beam supporting a single crate of 420 kg in static equilibrium as shown in the diagram below to the right.\r\n<img class=\"wp-image-671 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-exercise.png\" alt=\"\" width=\"236\" height=\"409\" \/><\/li>\r\n \t<li>If the maximum recommended tension in the cable is 12 000 N, find the reactive force from the beam that would be needed to support this maximum in static equilibrium as shown in the diagram below to the right.\r\n<img class=\"wp-image-672 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.8-exercise.png\" alt=\"\" width=\"217\" height=\"376\" \/><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 7.4: Limits of a cable<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA cable that has a maximum load tension of 50 000 N is slowly lifting a heavy crate straight upwards.\r\n<ol type=\"i\">\r\n \t<li>How close to the top can the crate be safely lifted?<img class=\"wp-image-1187 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/add-mass.png\" alt=\"\" width=\"611\" height=\"252\" \/><\/li>\r\n \t<li>What is the mass that can be lifted to within 10 cm from the top?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<h1>Exercise Answers<\/h1>\r\n<h2>7.1 Right Angle Trigonometric Functions<\/h2>\r\n<ol class=\"twocolumn\">\r\n \t<li>0.743145<\/li>\r\n \t<li>0.484810<\/li>\r\n \t<li>0.906308<\/li>\r\n \t<li>0.484810<\/li>\r\n \t<li>0.194380<\/li>\r\n \t<li>1.53986<\/li>\r\n \t<li>0.190810<\/li>\r\n \t<li>0.544639<\/li>\r\n \t<li>29\u00b0<\/li>\r\n \t<li>39\u00b0<\/li>\r\n \t<li>50\u00b0<\/li>\r\n \t<li>52\u00b0<\/li>\r\n \t<li>33.3\u00b0<\/li>\r\n \t<li>8.9\u00b0<\/li>\r\n \t<li>41.0\u00b0<\/li>\r\n \t<li>81\u00b0<\/li>\r\n \t<li>z = 27.36 (\u2248 27), \u00f8 = 26.6\u00b0 (\u2248 27\u00b0)<\/li>\r\n \t<li>y = 19.6 (\u2248 20),\u00a0\u00f8 = 44.4\u00b0 (\u2248 44\u00b0)<\/li>\r\n \t<li>x = 16, \u00f8 = 53.1\u00b0 (\u2248 53\u00b0)<\/li>\r\n \t<li>x = 21.2 (\u2248 21), y = 13.2 (\u2248 13)<\/li>\r\n \t<li>x = 892 (\u2248 890), y = 803 (\u2248 800)<\/li>\r\n \t<li>z = 242 (\u2248 240),\u00a0 \u00f8 = 24.4\u00b0 (\u2248 24\u00b0)<\/li>\r\n \t<li>x = 9.8, y = 6.9<\/li>\r\n \t<li>x = 37.6 (\u2248 38),\u00a0 y = 42.6 (\u2248 43)<\/li>\r\n \t<li>z = 25, \u00f8 = 53.1\u00b0 (\u2248 53\u00b0)<\/li>\r\n \t<li>y = 75,\u00a0 \u00f8 = 36.9\u00b0 (37\u00b0)<\/li>\r\n \t<li>y = 4, \u00f8 = 53.1\u00b0 (\u2248 53\u00b0)<\/li>\r\n \t<li>y = 11.1 (\u2248 11),\u00a0 z = 27.4 (\u2248 27)<\/li>\r\n \t<li>z = 28.6 (\u2248 29), \u00f8 = 44.4\u00b0 (\u2248 44\u00b0)<\/li>\r\n \t<li>y = 19.6 (\u2248 20),\u00a0 \u00f8 = 44.4\u00b0 (\u2248 44\u00b0)<\/li>\r\n \t<li>y = 8.9, \u00f8 = 41.8\u00b0 (\u2248 42\u00b0)<\/li>\r\n \t<li>x = 35, y = 61<\/li>\r\n<\/ol>\r\n<h2>7.2 Vector Resolution into Components<\/h2>\r\n<ol>\r\n \t<li>[latex]v_y[\/latex] = 25.4 (\u2248 25), [latex]v_x[\/latex] = 40.7 (\u2248 41)<\/li>\r\n \t<li>[latex]v_y[\/latex] = 11, [latex]v_x[\/latex] = 19.1 (\u2248 19)<\/li>\r\n \t<li>[latex]F_y[\/latex] = 15, [latex]F_x[\/latex] = 28.3 (\u2248 28)<\/li>\r\n \t<li>[latex]F_y[\/latex] = 53.7 (\u2248 54), [latex]F_x[\/latex] = 79.6 (\u2248 80)<\/li>\r\n \t<li>[latex]a_y[\/latex] = 13.6, [latex]a_x[\/latex] = 15.2<\/li>\r\n \t<li>[latex]a_y[\/latex] = 22.0, [latex]a_x[\/latex] = 45.1<\/li>\r\n \t<li>[latex]v_y[\/latex] = 25.4 (\u2248 25), [latex]v_x[\/latex] = 40.7 (\u2248 41)<\/li>\r\n \t<li>[latex]v_y[\/latex] = 11, [latex]v_x[\/latex] = 19.1 (\u2248 19)<\/li>\r\n \t<li>[latex]F_y[\/latex] = 15, [latex]F_x[\/latex] = 28.3 (\u2248 28)<\/li>\r\n \t<li>[latex]F_y[\/latex] = 53.7 (\u2248 54), [latex]F_x[\/latex] = 79.6 (\u2248 80)<\/li>\r\n \t<li>[latex]a_y[\/latex] = 13.6, [latex]a_x[\/latex] = 15.2<\/li>\r\n \t<li>[latex]a_y[\/latex] = 22.0, [latex]a_x[\/latex] = 45.1<\/li>\r\n<\/ol>\r\n<h2>7.3 Static Equilibrium (Concurrent Forces)<\/h2>\r\n<ol class=\"twocolumn\">\r\n \t<li>[latex]T_y[\/latex] = 980 N<\/li>\r\n \t<li>[latex]T_y[\/latex] = 5.0 N<\/li>\r\n<\/ol>\r\n<ol start=\"3\">\r\n \t<li>[latex]T_1[\/latex] = 228 N (\u2248 230 N), [latex]T_2[\/latex] = 541 N (\u2248 540 N)<\/li>\r\n \t<li>[latex]T_1[\/latex] = 89.2 N (\u2248 89 N), [latex]T_2[\/latex] = 261 N\u00a0 (\u2248 260 N)<\/li>\r\n \t<li>[latex]T_1[\/latex] = 769 N (\u2248 770 N), [latex]T_2[\/latex] = 663 N (\u2248 660 N)<\/li>\r\n \t<li>[latex]T_1[\/latex] = 8502 N (\u2248 8500 N), [latex]T_2[\/latex] = 5557 N (\u2248 5600 N)<\/li>\r\n \t<li>T = 6685 N ( \u2248 6700 N), [latex]F_C =\u00a0 -5268[\/latex] N\u00a0 (\u2248 [latex]- 5300[\/latex] N)<\/li>\r\n \t<li>[latex]FC = - 14300[\/latex] N (\u2248 [latex]- 14000[\/latex] N)<\/li>\r\n<\/ol>\r\n<h2>7.4 Limits of a Cable<\/h2>\r\n<ol type=\"i\">\r\n \t<li>distance = 0.49 m from the top<\/li>\r\n \t<li>[latex]m = 204 \\text{ kg } = 200\\text{ kg}[\/latex]<\/li>\r\n<\/ol>\r\n<h3>Media Attributions<\/h3>\r\n<ul>\r\n \t<li>\"<a href=\"https:\/\/upload.wikimedia.org\/wikipedia\/commons\/1\/11\/Observatory_in_Alexandria_at_the_Time_of_Hipparchus.jpg\">Observatory in Alexandria at the Time of Hipparchus<\/a>\" Hermann Goll is in the <a href=\"https:\/\/creativecommons.org\/share-your-work\/public-domain\/\">public domain<\/a>.<\/li>\r\n \t<li>\"<a href=\"https:\/\/www.flickr.com\/photos\/calafellvalo\/6969596062\/in\/photolist-fx4tVs-fx4zhN-jkWaky-bQMGU6-bQMH5D-bBT16m-jkSeyR-jkWcsu-bQMGmZ-bQMGAx-jkTUaW-bBSZio-bBT1ym-bQMHt4-bQMHEV-jkU29f-bQMJhZ-jkSfNV-DBe8hg-jkTjNV-bQMHiT-bBSZow-jkTMZ9-jkS1pK-bQMGet-bQMHzk-6fz5LS-bQMFLn-jkTR2G-bBT1aE-jkTYYU-bBT1C9-bBT2fy-jkSdXk-jkTBJ8-jkTjgT-jkToHP-jkTHma-jkSisc-jkTGQF-jkTHXa-bQMJnv-bQMJ9t-bBT2k1-bQMG5g-bBT2r1-bQMKnr-bQMGsz-bQMGwF-bBT3ud\">Acueducto de Tarragona<\/a>\" by <a id=\"yui_3_16_0_1_1678914850254_1764\" class=\"owner-name truncate\" title=\"Go to calafellvalo\u2019s photostream\" href=\"https:\/\/www.flickr.com\/photos\/calafellvalo\/\" data-track=\"attributionNameClick\">calafellvalo<\/a> is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-nc-nd\/2.0\/\">CC BY-NC-ND 2.0 licence<\/a>.<\/li>\r\n \t<li>\"<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Azure_Windownew.JPG\">Azure Window New<\/a>\" by anjab1593 is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by\/3.0\/deed.en\">CC BY 3.0 Unported licence<\/a>.<\/li>\r\n \t<li>\"<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Notre_Dame_buttress.jpg\">Notre Dame buttress<\/a>\" by <a class=\"external text\" href=\"https:\/\/www.flickr.com\/people\/7761867@N06\" rel=\"nofollow\">Jean Lemoine<\/a> is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/2.0\/deed.en\">CC BY-SA 2.0 licence<\/a>.<\/li>\r\n \t<li>\"<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:VillardButtressReims.jpg\">Villard Buttress Reims<\/a>\" uploaded by <a href=\"https:\/\/en.wikipedia.org\/wiki\/User:Wetman\">Wetman<\/a> is in the <a href=\"https:\/\/creativecommons.org\/share-your-work\/public-domain\/\">public domain<\/a>.<\/li>\r\n<\/ul>","rendered":"<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Key Takeaways<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ul>\n<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/newtlaws\/Lesson-1\/Newton-s-First-Law\">Newton\u2019s First Law<\/a><\/li>\n<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/vectors\/Lesson-3\/Equilibrium-and-Statics\">Equilibrium and Statics<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>Equations Introduced and Used for this Topic: (All equations can be written and solved as both <strong>scalar<\/strong> and <strong>vector <\/strong>and<strong> all equations<\/strong> are generally solved as vectors)<\/p>\n<ul style=\"list-style-type: none;\">\n<li style=\"text-align: center;\">[latex]\\sum F_{x, y, z} = 0 \\text{ N}[\/latex]<\/li>\n<li style=\"text-align: center;\">[latex]T = r \\times F[\/latex]<\/li>\n<li style=\"text-align: center;\">[latex]T_{\\text{clockwise}} = T_{\\text{counter-clockwise}}[\/latex]<\/li>\n<li style=\"text-align: center;\">[latex](m_1 + m_2) x_{\\text{cm}} = m_1 x_1 + m_2 x_2[\/latex]<\/li>\n<\/ul>\n<p>Where&#8230;<\/p>\n<ul>\n<li>[latex]T[\/latex] is the torque, measured in newton-metres (Nm)<\/li>\n<li>[latex]F[\/latex] is force, measured in newtons (N)<\/li>\n<li>[latex]r[\/latex] is the length of the lever-arm measured in metres (m)<\/li>\n<li>[latex]m_1[\/latex] and [latex]m_2[\/latex] are the masses of two bodies, measured in kilograms (kg)<\/li>\n<li>[latex]x_{\\text{cm}}[\/latex], [latex]x_1[\/latex], and [latex]x_2[\/latex] are the distances, measured in metres (m)<\/li>\n<li>\u00f8 is the angle given for the triangle and is generally between the adjacent side and the hypotenuse with the angle measured in degrees (be careful to make sure that the calculator is not working in gradients or radians).<\/li>\n<li><em>Hypotenuse<\/em> is the longest side in a right triangle and is opposite to the 90\u00b0 angle.<\/li>\n<li><em>Opposite<\/em> is the side that is opposite to the angle that you are given.<\/li>\n<li><em>Adjacent<\/em> is the side that is beside the angle you are given.<\/li>\n<\/ul>\n<p><strong>The Trigonometric Functions:<\/strong><\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>[latex]\\text{sine \u00f8 } = \\dfrac{\\text{opposite}}{\\text{hypotenuse}}[\/latex]<\/li>\n<li>[latex]\\text{cosine \u00f8 } = \\dfrac{\\text{adjacent}}{\\text{hypotenuse}}[\/latex]<\/li>\n<li>[latex]\\text{tangent \u00f8 } = \\dfrac{\\text{opposite}}{\\text{adjacent}}[\/latex]<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-478 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7-trig-functions.png\" alt=\"\" width=\"557\" height=\"185\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7-trig-functions.png 1686w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7-trig-functions-300x99.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7-trig-functions-1024x340.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7-trig-functions-768x255.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7-trig-functions-1536x509.png 1536w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7-trig-functions-65x22.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7-trig-functions-225x75.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7-trig-functions-350x116.png 350w\" sizes=\"auto, (max-width: 557px) 100vw, 557px\" \/><\/p>\n<h1>7.1 Right Angle Trigonometric Functions<\/h1>\n<p><a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Observatory_in_Alexandria_at_the_Time_of_Hipparchus.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1208\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-angle-trig.jpeg\" alt=\"\" width=\"667\" height=\"456\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-angle-trig.jpeg 800w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-angle-trig-300x205.jpeg 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-angle-trig-768x525.jpeg 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-angle-trig-65x44.jpeg 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-angle-trig-225x154.jpeg 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-angle-trig-350x239.jpeg 350w\" sizes=\"auto, (max-width: 667px) 100vw, 667px\" \/><\/a><\/p>\n<p>Trigonometry, a term derived from the Greek trigonon (triangle) and metron (measure) has known historical roots dating back 4000 years to writings found in Egyptian and Babylonian mathematics and astronomical records. Indian astronomers were using trigonometry over 2600 years ago and that widespread usage by Islamic and Chinese scientists was common at least 1500 years prior.<\/p>\n<p>The study of trigonometry returned to European nations in the Renaissance in translations of Arabic and Greek writings. Modern trigonometry is credited as reaching its current form through the works of Leonard Euler (1707-1783), who is considered to be one of the greatest mathematicians in history and had worked in nearly every area of mathematics and physics. Modern notation used in trigonometry comes directly from Euler\u2019s writings, as are many other currently used notations in mathematics and physics. A list of Euler\u2019s achievements is extensive as well as his writings that have been collected in 92 volumes, making him one of the most prolific writers in mathematics history.<\/p>\n<p><strong>Introductory trigonometry<\/strong> is based on identifying the similarities between identical right angled (one angle is 90\u00b0) of different sizes. If the angles of a triangle are identical then all of these triangles are simply larger or smaller copies of each other. In all of the cases shown above, if you take any two sides of any triangle shown and divide them by each other, that number will be exactly the same.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-479 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-triangles.png\" alt=\"\" width=\"540\" height=\"288\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-triangles.png 1778w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-triangles-300x160.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-triangles-1024x546.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-triangles-768x409.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-triangles-1536x819.png 1536w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-triangles-65x35.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-triangles-225x120.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-right-triangles-350x187.png 350w\" sizes=\"auto, (max-width: 540px) 100vw, 540px\" \/><\/p>\n<p>These triangle ratios have defined names:<\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>[latex]\\text{sine } = \\dfrac{\\text{opposite}}{\\text{hypotenuse}}[\/latex]<\/li>\n<li>[latex]\\text{cosine } = \\dfrac{\\text{adjacent}}{\\text{hypotenuse}}[\/latex]<\/li>\n<li>[latex]\\text{tangent } = \\dfrac{\\text{opposite}}{\\text{adjacent}}[\/latex]<\/li>\n<\/ul>\n<p>You often see these equations shortened to:<\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>[latex]\\text{sin } = \\dfrac{\\text{opp}}{\\text{hyp}}[\/latex]<\/li>\n<li>[latex]\\text{cos } = \\dfrac{\\text{adj}}{\\text{hyp}}[\/latex]<\/li>\n<li>[latex]\\text{tan } = \\dfrac{\\text{opp}}{\\text{adj}}[\/latex]<\/li>\n<\/ul>\n<p>and memorized as: <strong>SOH &#8211; CAH &#8211; TOA<\/strong><\/p>\n<p>Defining the sides of a triangle follows a set pattern:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-480\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-hypotenuse-triangle.png\" alt=\"\" width=\"243\" height=\"234\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-hypotenuse-triangle.png 495w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-hypotenuse-triangle-300x288.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-hypotenuse-triangle-65x63.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-hypotenuse-triangle-225x216.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-hypotenuse-triangle-350x337.png 350w\" sizes=\"auto, (max-width: 243px) 100vw, 243px\" \/><\/p>\n<ol>\n<li>The side of a triangle that is opposite to the right angle is called the hypotenuse.<\/li>\n<li>The opposite and adjacent sides are then defined by the angle you are going to work with.\u00a0 One of the sides will be opposite this angle and the other side will be adjacent to this side.<\/li>\n<\/ol>\n<p><strong>For example:<\/strong> The following sides are defined by the right angle and the angle you are going to work with \u00f8.\u00a0 You will have to define the adjacent and opposite sides for every right triangle you work with.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-481 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-name-sides-triangle.png\" alt=\"\" width=\"560\" height=\"182\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-name-sides-triangle.png 1691w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-name-sides-triangle-300x98.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-name-sides-triangle-1024x333.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-name-sides-triangle-768x250.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-name-sides-triangle-1536x500.png 1536w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-name-sides-triangle-65x21.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-name-sides-triangle-225x73.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-name-sides-triangle-350x114.png 350w\" sizes=\"auto, (max-width: 560px) 100vw, 560px\" \/><\/p>\n<p>The other right angled trigonometric rations are the reciprocals of Sine, Cosine and Tangent:<\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>[latex]\\text{Cosecant } = \\dfrac{1}{\\text{Sine}}[\/latex]<\/li>\n<li>[latex]\\text{Secant } = \\dfrac{1}{\\text{Cosine}}[\/latex]<\/li>\n<li>[latex]\\text{Cotangent } = \\dfrac{1}{\\text{Tangent}}[\/latex]<\/li>\n<\/ul>\n<p>or formally defined as:<\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>[latex]\\text{Cosecant } = \\dfrac{\\text{hypotenuse}}{\\text{opposite}}[\/latex]<\/li>\n<li>[latex]\\text{Secant } = \\dfrac{\\text{hypotenuse}}{\\text{adjacent}}[\/latex]<\/li>\n<li>[latex]\\text{Cotangent } = \\dfrac{\\text{adjacent}}{\\text{opposite}}[\/latex]<\/li>\n<\/ul>\n<p>You often see these equations shortened to:<\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>[latex]\\text{Csc } = \\dfrac{\\text{hyp}}{\\text{opp}}[\/latex]<\/li>\n<li>[latex]\\text{Sec} = \\dfrac{\\text{hyp}}{\\text{adj}}[\/latex]<\/li>\n<li>[latex]\\text{Cot } = \\dfrac{\\text{adj}}{\\text{opp}}[\/latex]<\/li>\n<\/ul>\n<p>These reciprocal trigonometric functions are commonly used in calculus, specifically in integration and when working with polar coordinates.\u00a0 Anyone taking higher levels of mathematics will encounter these reciprocal trigonometric functions.<\/p>\n<p>Using the Pythagorean theorem for 30\u00b0, 45\u00b0 and 60\u00b0 right angle triangles, you can get the exact values of the trigonometric relationship (and the reciprocal values). It is standard to see exams where students are required to draw these 30\u00b0, 45\u00b0 and 60\u00b0 right angle triangles and use the side lengths to generate exact values.<\/p>\n<p>Another common sight is to see trigonometric tables being used as approximations of the trig ratios of standard angles from 1\u00b0 to 90\u00b0.\u00a0 For these tables you just choose the value that lines up the trigonometric function you wish to use with the angle that you are using. Basic scientific calculators have essentially made these tables obsolete.<\/p>\n<p>Standard Reference Angles:<\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>[latex]\\sin 30^{\\circ} = \\dfrac{1}{2}[\/latex]<\/li>\n<li>[latex]\\cos 30^{\\circ} = \\dfrac{\\sqrt{3}}{2}[\/latex]<\/li>\n<li>[latex]\\tan 30^{\\circ} = \\dfrac{1}{\\sqrt{3}}[\/latex]<\/li>\n<li>[latex]\\sin 45^{\\circ} = \\dfrac{1}{\\sqrt{2}}[\/latex]<\/li>\n<li>[latex]\\cos 45^{\\circ} = \\dfrac{1}{\\sqrt{2}}[\/latex]<\/li>\n<li>[latex]\\tan 45^{\\circ} = 1[\/latex]<\/li>\n<li>[latex]\\sin 60^{\\circ} = \\dfrac{\\sqrt{3}}{2}[\/latex]<\/li>\n<li>[latex]\\cos 60^{\\circ} = \\dfrac{1}{2}[\/latex]<\/li>\n<li>[latex]\\tan 60^{\\circ} = \\dfrac{\\sqrt{3}}{1}[\/latex]<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-483\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-standard-reference-angles-e1678999268935.png\" alt=\"\" width=\"571\" height=\"147\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-standard-reference-angles-e1678999268935.png 2053w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-standard-reference-angles-e1678999268935-300x77.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-standard-reference-angles-e1678999268935-1024x263.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-standard-reference-angles-e1678999268935-768x198.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-standard-reference-angles-e1678999268935-1536x395.png 1536w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-standard-reference-angles-e1678999268935-2048x527.png 2048w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-standard-reference-angles-e1678999268935-65x17.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-standard-reference-angles-e1678999268935-225x58.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-standard-reference-angles-e1678999268935-350x90.png 350w\" sizes=\"auto, (max-width: 571px) 100vw, 571px\" \/><\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.1.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Find the values that correspond to the following trigonometric functions and angles:<\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>[latex]\\sin 19^{\\circ} = x[\/latex]<\/li>\n<li>[latex]\\cos 67^{\\circ} = y[\/latex]<\/li>\n<li>[latex]\\tan 38^{\\circ} = z[\/latex]<\/li>\n<\/ul>\n<p><strong>Solution<\/strong><\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>[latex]x = 0.326[\/latex]<\/li>\n<li>[latex]y = 0.391[\/latex]<\/li>\n<li>[latex]z = 0.781[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>It is also possible to work in reverse, i.e.: given the trigonometric ration of two sides, you can find the angle that you are working with.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.1.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Find the angles that correspond to the following trigonometric values:<\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>[latex]\\sin \u00f8 = 0.829[\/latex]<\/li>\n<li>[latex]\\cos \u00f8 = 0.940[\/latex]<\/li>\n<li>[latex]\\tan \u00f8 = 3.732[\/latex]<\/li>\n<\/ul>\n<p><strong>Solution<\/strong><\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>\u00f8 = 56\u00b0<\/li>\n<li>\u00f8 = 20\u00b0<\/li>\n<li>\u00f8 = 75\u00b0<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>Sometimes you do not have a value that matches up. For these cases you choose the value that is closest to what you have.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.1.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Find the angles that are closest to the following trigonometric values:<\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>[latex]\\sin \u00f8 = 0.297[\/latex]<\/li>\n<li>[latex]\\cos \u00f8 = 0.380[\/latex]<\/li>\n<li>[latex]\\tan \u00f8 = 0.635[\/latex]<\/li>\n<\/ul>\n<p><strong>Solution<\/strong><\/p>\n<ul class=\"threecolumn\" style=\"list-style-type: none;\">\n<li>\u00f8 = 17\u00b0<\/li>\n<li>\u00f8 = 68\u00b0<\/li>\n<li>\u00f8 = 32\u00b0<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.1.4<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Solve for the unknown side in the following triangle.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-502 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.4.png\" alt=\"\" width=\"505\" height=\"229\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.4.png 971w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.4-300x136.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.4-768x348.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.4-65x29.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.4-225x102.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.4-350x159.png 350w\" sizes=\"auto, (max-width: 505px) 100vw, 505px\" \/><\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>[latex]\\begin{align}z &= \\sqrt{10^2+20^2} \\\\ &=22.361 \\\\ &= 22 \\text{ cm} \\end{align}[\/latex]<\/p>\n<p>[latex]angle \\ \u00f8 = \\tan ^{-1}\\left(\\dfrac{10}{20} \\right) = 27^{\\circ}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.1.5<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Solve for the unknown sides in the following triangle.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-503 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.5.png\" alt=\"\" width=\"501\" height=\"271\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.5.png 880w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.5-300x162.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.5-768x415.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.5-65x35.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.5-225x122.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/ex-7.1.5-350x189.png 350w\" sizes=\"auto, (max-width: 501px) 100vw, 501px\" \/><\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>[latex]y = (25 \\text{ m})(\\sin 32^{\\circ}) = 13 \\text{ m}[\/latex]<\/p>\n<p>[latex]x =(25 \\text{ m})(\\cos 32^{\\circ}) = 21 \\text{ m}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>To find trigonometric tables, go to <a href=\"https:\/\/opentextbc.ca\/foundationsofphysics\/back-matter\/appendix-a-trigonometric-tables\/\">Appendix A: Trigonometric Tables<\/a>.<\/p>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 7.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Find the value of each of the following trigonometric functions using your scientific calculator to 6 digits.<\/p>\n<ol class=\"twocolumn\">\n<li>[latex]\\sin 48^{\\circ}[\/latex]<\/li>\n<li>[latex]\\sin 29^{\\circ}[\/latex]<\/li>\n<li>[latex]\\cos 25^{\\circ}[\/latex]<\/li>\n<li>[latex]\\cos 61^{\\circ}[\/latex]<\/li>\n<li>[latex]\\tan 11^{\\circ}[\/latex]<\/li>\n<li>[latex]\\tan 57^{\\circ}[\/latex]<\/li>\n<li>[latex]\\sin 11^{\\circ}[\/latex]<\/li>\n<li>[latex]\\cos 57^{\\circ}[\/latex]<\/li>\n<\/ol>\n<p>Use your scientific calculator to find each angle to the nearest hundredth of a degree.<\/p>\n<ol class=\"twocolumn\" start=\"9\">\n<li>[latex]\\sin \u00f8 = 0.4848[\/latex]<\/li>\n<li>[latex]\\sin \u00f8 = 0.6293[\/latex]<\/li>\n<li>[latex]\\cos \u00f8 =0.6561[\/latex]<\/li>\n<li>[latex]\\cos \u00f8 =0.6157[\/latex]<\/li>\n<li>[latex]\\tan \u00f8 = 0.6561[\/latex]<\/li>\n<li>[latex]\\tan \u00f8 = 0.1562[\/latex]<\/li>\n<li>[latex]\\sin \u00f8 =0.6561[\/latex]<\/li>\n<li>[latex]\\cos \u00f8 =0.1562[\/latex]<\/li>\n<\/ol>\n<p>Solve for all unknowns in the following right triangles.<\/p>\n<ol class=\"twocolumn\" start=\"17\">\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-514\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-17.png\" alt=\"\" width=\"880\" height=\"407\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-17.png 880w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-17-300x139.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-17-768x355.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-17-65x30.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-17-225x104.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-17-350x162.png 350w\" sizes=\"auto, (max-width: 880px) 100vw, 880px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-516\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-18.png\" alt=\"\" width=\"820\" height=\"357\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-18.png 820w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-18-300x131.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-18-768x334.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-18-65x28.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-18-225x98.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-18-350x152.png 350w\" sizes=\"auto, (max-width: 820px) 100vw, 820px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-522\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-19.png\" alt=\"\" width=\"765\" height=\"371\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-19.png 765w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-19-300x145.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-19-65x32.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-19-225x109.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-19-350x170.png 350w\" sizes=\"auto, (max-width: 765px) 100vw, 765px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-520\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-20.png\" alt=\"\" width=\"788\" height=\"407\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-20.png 788w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-20-300x155.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-20-768x397.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-20-65x34.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-20-225x116.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-20-350x181.png 350w\" sizes=\"auto, (max-width: 788px) 100vw, 788px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-521\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-21.png\" alt=\"\" width=\"208\" height=\"145\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-21.png 770w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-21-300x209.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-21-768x535.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-21-65x45.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-21-225x157.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-21-350x244.png 350w\" sizes=\"auto, (max-width: 208px) 100vw, 208px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-524\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-22.png\" alt=\"\" width=\"902\" height=\"385\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-22.png 902w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-22-300x128.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-22-768x328.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-22-65x28.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-22-225x96.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-22-350x149.png 350w\" sizes=\"auto, (max-width: 902px) 100vw, 902px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-525\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-23.png\" alt=\"\" width=\"678\" height=\"371\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-23.png 678w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-23-300x164.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-23-65x36.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-23-225x123.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-23-350x192.png 350w\" sizes=\"auto, (max-width: 678px) 100vw, 678px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-526\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-24.png\" alt=\"\" width=\"838\" height=\"407\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-24.png 838w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-24-300x146.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-24-768x373.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-24-65x32.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-24-225x109.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-24-350x170.png 350w\" sizes=\"auto, (max-width: 838px) 100vw, 838px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-527\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-25.png\" alt=\"\" width=\"186\" height=\"203\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-25.png 614w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-25-275x300.png 275w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-25-65x71.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-25-225x245.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-25-350x381.png 350w\" sizes=\"auto, (max-width: 186px) 100vw, 186px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-528\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-26.png\" alt=\"\" width=\"215\" height=\"91\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-26.png 779w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-26-300x127.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-26-768x325.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-26-65x28.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-26-225x95.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-26-350x148.png 350w\" sizes=\"auto, (max-width: 215px) 100vw, 215px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-529\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-27.png\" alt=\"\" width=\"226\" height=\"123\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-27.png 682w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-27-300x163.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-27-65x35.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-27-225x122.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-27-350x190.png 350w\" sizes=\"auto, (max-width: 226px) 100vw, 226px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-518\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-28.png\" alt=\"\" width=\"815\" height=\"403\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-28.png 815w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-28-300x148.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-28-768x380.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-28-65x32.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-28-225x111.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-28-350x173.png 350w\" sizes=\"auto, (max-width: 815px) 100vw, 815px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-517\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-29.png\" alt=\"\" width=\"158\" height=\"196\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-29.png 522w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-29-241x300.png 241w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-29-65x81.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-29-225x280.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-29-350x436.png 350w\" sizes=\"auto, (max-width: 158px) 100vw, 158px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-519\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-30.png\" alt=\"\" width=\"788\" height=\"348\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-30.png 788w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-30-300x132.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-30-768x339.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-30-65x29.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-30-225x99.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-30-350x155.png 350w\" sizes=\"auto, (max-width: 788px) 100vw, 788px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-523\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-31.png\" alt=\"\" width=\"234\" height=\"142\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-31.png 632w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-31-300x183.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-31-65x40.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-31-225x137.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-31-350x213.png 350w\" sizes=\"auto, (max-width: 234px) 100vw, 234px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-515\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-32.png\" alt=\"\" width=\"788\" height=\"348\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-32.png 788w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-32-300x132.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-32-768x339.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-32-65x29.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-32-225x99.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.1-32-350x155.png 350w\" sizes=\"auto, (max-width: 788px) 100vw, 788px\" \/><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h1>7.2 Vector Resolution Into Components<\/h1>\n<div class=\"textbox\">Extra Help:<a href=\"https:\/\/www.physicsclassroom.com\/Class\/vectors\/U3L1e.cfm#trig\"> Vector Resolution<\/a><\/div>\n<p>Through practice you will be able to quickly resolve angled vectors using right triangle trigonometry. For example, given a force vector of 100 N angled at 42\u00b0 from the horizontal, you can break it up into vertical and horizontal components using the sine and cosine functions.\u00a0 In this instance the vector components are as follows:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-534 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-triangle.png\" alt=\"\" width=\"362\" height=\"275\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-triangle.png 1278w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-triangle-300x228.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-triangle-1024x778.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-triangle-768x584.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-triangle-65x49.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-triangle-225x171.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-triangle-350x266.png 350w\" sizes=\"auto, (max-width: 362px) 100vw, 362px\" \/><\/p>\n<p>Quick Resolution:<\/p>\n<ul>\n<li>[latex]F = 100 \\text{ N}[\/latex]<\/li>\n<li>[latex]F_x = 100 \\text{ N } \\cos 42^{\\circ}[\/latex] (\u2248 74 N)<\/li>\n<li>[latex]F_y = 100 \\text{ N } \\sin 42^{\\circ}[\/latex]\u00a0 (\u2248 70 N)<\/li>\n<\/ul>\n<p>A more detailed resolution is:<\/p>\n<ul>\n<li>[latex]F_x[\/latex]: [latex]\\cos 42^{\\circ} = \\dfrac{F_x}{100 \\text{ N}}[\/latex] which, when isolated, leaves us with [latex]F_x = 100 \\text{ N } \\cos 42^{\\circ}[\/latex]<\/li>\n<li>[latex]F_y[\/latex]: [latex]\\sin 42^{\\circ} = \\dfrac{F_y}{100 \\text{ N}}[\/latex] which, when isolated, leaves us with [latex]F_y = 100 \\text{ N } \\sin 42^{\\circ}[\/latex]<\/li>\n<\/ul>\n<p>When the vector magnitude is known, then the opposite side vector is the sine of the vector angle and the adjacent side vector is the cosine of the vector angle.<\/p>\n<p>When both vector components are known, the angle of the resultant vector can be found using the tangent function and the magnitude of the vector by using the Pythagorean Theorem.\u00a0 For this, one generally uses a more traditional approach.\u00a0 Consider the following example:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-536 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-F_net.png\" alt=\"\" width=\"481\" height=\"366\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-F_net.png 1278w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-F_net-300x228.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-F_net-1024x778.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-F_net-768x584.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-F_net-65x49.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-F_net-225x171.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2-F_net-350x266.png 350w\" sizes=\"auto, (max-width: 481px) 100vw, 481px\" \/><\/p>\n<p>The magnitude of the net force is:<\/p>\n<ul>\n<li>[latex](F_{\\text{net}})^2 = (600 \\text{ N})^2 + (700 \\text{ N})^2[\/latex]<\/li>\n<li>[latex](F_{\\text{net}})^2 = 360 000 \\text{ N}^2 + 490000 \\text{ N}^2[\/latex]<\/li>\n<li>[latex](F_{\\text{net}})^2 = 850000 \\text{ N}^2[\/latex]<\/li>\n<li>[latex]F_{\\text{net}} = 922 \\text{ N } (\\approx 920 \\text{ N})[\/latex]<\/li>\n<\/ul>\n<p>The angle this makes is:<\/p>\n<ul>\n<li>[latex]\\tan \u00f8 = \\dfrac{600 \\text{ N}}{700 \\text{ N}}[\/latex]<\/li>\n<li>[latex]\\tan \u00f8 = 0.8571[\/latex]<\/li>\n<li>[latex]\u00f8 = \\tan ^{-1} 0.8571[\/latex]<\/li>\n<li>[latex]\u00f8 = 40.6^{\\circ}\u00a0 (\\approx 41^{\\circ})[\/latex]<\/li>\n<\/ul>\n<p>The main purpose of this section is for you to be able to quickly resolve a vector with an angle into components.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.2.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Resolve the following vectors and angles into components.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-537 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-triangle.png\" alt=\"\" width=\"303\" height=\"288\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-triangle.png 1017w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-triangle-300x285.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-triangle-768x730.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-triangle-65x62.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-triangle-225x214.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-triangle-350x333.png 350w\" sizes=\"auto, (max-width: 303px) 100vw, 303px\" \/><\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]F_y = 40 \\text{ N } \\sin 38^{\\circ}[\/latex] or 24.6\u00b0 (\u2248 25 N)<\/li>\n<li>[latex]F_x = 40 \\text{ N } \\cos 38^{\\circ}[\/latex] or 31.5\u00b0 (\u2248 32 N)<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.2.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Resolve the following vectors and angles into components.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-538 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-angles.png\" alt=\"\" width=\"308\" height=\"268\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-angles.png 847w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-angles-300x261.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-angles-768x668.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-angles-65x57.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-angles-225x196.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-angles-350x305.png 350w\" sizes=\"auto, (max-width: 308px) 100vw, 308px\" \/><\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]\\vec{v_y} = 48 \\text{ m\/s } \\sin 32^{\\circ}[\/latex] or 25.4 m\/s (\u2248 25 m\/s)<\/li>\n<li>[latex]\\vec{v_x} = 48 \\text{ m\/s } \\cos 32^{\\circ}[\/latex] or 40.7 m\/s (\u2248 41 m\/s)<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.2.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Resolve the following vectors and angles into components.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-539 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-angles.png\" alt=\"\" width=\"303\" height=\"268\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-angles.png 834w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-angles-300x265.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-angles-768x679.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-angles-65x57.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-angles-225x199.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-angles-350x309.png 350w\" sizes=\"auto, (max-width: 303px) 100vw, 303px\" \/><\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]\\vec{F_y} = 32 \\text{ m\/s } \\sin 28^{\\circ}[\/latex] or 15 m\/s<\/li>\n<li>[latex]\\vec{F_x} = 32 \\text{ m\/s } \\cos 28^{\\circ}[\/latex] or (\u2212) 28 m\/s<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.2.4<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Resolve the following vectors and angles into components.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-540 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-angles.png\" alt=\"\" width=\"308\" height=\"269\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-angles.png 843w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-angles-300x262.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-angles-768x671.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-angles-65x57.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-angles-225x197.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-angles-350x306.png 350w\" sizes=\"auto, (max-width: 308px) 100vw, 308px\" \/><\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]\\vec{a_y} = 20.4 \\text{ m\/s}^2 \\text{ } \\sin 42^{\\circ}[\/latex] or (\u2212) 14 m\/s<sup>2<\/sup><\/li>\n<li>[latex]\\vec{a_x} = 20.3 \\text{ m\/s}^2 \\text{ } \\cos 42^{\\circ}[\/latex] or 15 m\/s<sup>2<\/sup><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 7.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Break the following vectors into components (not drawn to scale)<\/p>\n<ol class=\"twocolumn\">\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-543\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-exercise.png\" alt=\"\" width=\"253\" height=\"222\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-exercise.png 838w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-exercise-300x264.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-exercise-768x675.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-exercise-65x57.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-exercise-225x198.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.1-exercise-350x308.png 350w\" sizes=\"auto, (max-width: 253px) 100vw, 253px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-546\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-exercise.png\" alt=\"\" width=\"242\" height=\"214\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-exercise.png 838w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-exercise-300x266.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-exercise-768x680.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-exercise-65x58.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-exercise-225x199.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.2-exercise-350x310.png 350w\" sizes=\"auto, (max-width: 242px) 100vw, 242px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-548\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-exercise.png\" alt=\"\" width=\"242\" height=\"214\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-exercise.png 834w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-exercise-300x265.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-exercise-768x679.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-exercise-65x57.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-exercise-225x199.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.3-exercise-350x309.png 350w\" sizes=\"auto, (max-width: 242px) 100vw, 242px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-550\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-exercise.png\" alt=\"\" width=\"243\" height=\"216\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-exercise.png 834w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-exercise-300x267.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-exercise-768x683.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-exercise-65x58.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-exercise-225x200.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.4-exercise-350x311.png 350w\" sizes=\"auto, (max-width: 243px) 100vw, 243px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-551\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.5-exercise.png\" alt=\"\" width=\"247\" height=\"219\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.5-exercise.png 834w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.5-exercise-300x267.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.5-exercise-768x683.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.5-exercise-65x58.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.5-exercise-225x200.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.5-exercise-350x311.png 350w\" sizes=\"auto, (max-width: 247px) 100vw, 247px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-552\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.6-exercise.png\" alt=\"\" width=\"250\" height=\"221\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.6-exercise.png 834w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.6-exercise-300x265.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.6-exercise-768x679.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.6-exercise-65x57.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.6-exercise-225x199.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.6-exercise-350x309.png 350w\" sizes=\"auto, (max-width: 250px) 100vw, 250px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-553\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.7-exercise.png\" alt=\"\" width=\"247\" height=\"233\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.7-exercise.png 792w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.7-exercise-300x283.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.7-exercise-768x724.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.7-exercise-65x61.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.7-exercise-225x212.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.7-exercise-350x330.png 350w\" sizes=\"auto, (max-width: 247px) 100vw, 247px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-554\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.8-exercise.png\" alt=\"\" width=\"249\" height=\"228\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.8-exercise.png 811w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.8-exercise-300x274.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.8-exercise-768x703.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.8-exercise-65x59.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.8-exercise-225x206.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.8-exercise-350x320.png 350w\" sizes=\"auto, (max-width: 249px) 100vw, 249px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-555\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.9-exercise.png\" alt=\"\" width=\"248\" height=\"236\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.9-exercise.png 783w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.9-exercise-300x286.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.9-exercise-768x733.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.9-exercise-65x62.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.9-exercise-225x215.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.9-exercise-350x334.png 350w\" sizes=\"auto, (max-width: 248px) 100vw, 248px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-556\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.10-exercise.png\" alt=\"\" width=\"248\" height=\"236\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.10-exercise.png 783w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.10-exercise-300x286.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.10-exercise-768x733.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.10-exercise-65x62.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.10-exercise-225x215.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.10-exercise-350x334.png 350w\" sizes=\"auto, (max-width: 248px) 100vw, 248px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-557\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.11-exercise.png\" alt=\"\" width=\"245\" height=\"232\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.11-exercise.png 783w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.11-exercise-300x284.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.11-exercise-768x728.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.11-exercise-65x62.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.11-exercise-225x213.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.11-exercise-350x332.png 350w\" sizes=\"auto, (max-width: 245px) 100vw, 245px\" \/><\/li>\n<li style=\"margin-bottom: 40px;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-558\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.12-exercise.png\" alt=\"\" width=\"250\" height=\"235\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.12-exercise.png 788w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.12-exercise-300x282.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.12-exercise-768x723.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.12-exercise-65x61.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.12-exercise-225x212.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.2.12-exercise-350x330.png 350w\" sizes=\"auto, (max-width: 250px) 100vw, 250px\" \/><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h1>7.3 Static Equilibrium (Concurrent Forces)<\/h1>\n<div class=\"textbox\">\n<p>In the News: <a href=\"https:\/\/www.smithsonianmag.com\/smart-news\/\">Scientists Prove Leonardo da Vinci\u2019s 500 Year Old Bridge Design Actually Works<\/a><\/p>\n<\/div>\n<p>Static equilibrium is one of the requirements of architecture; engineering tools and techniques have been employed since ancient times to create buildings that are statically safe from collapse. In short, structural design must work to counteract the effects of gravity, storms and earthquakes, to name a few of the forces that can act on a structure.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignleft wp-image-560\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Acueducto-de-Tarragona.jpg\" alt=\"Les Ferreres Aqueduct\" width=\"378\" height=\"213\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Acueducto-de-Tarragona.jpg 1075w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Acueducto-de-Tarragona-300x169.jpg 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Acueducto-de-Tarragona-1024x576.jpg 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Acueducto-de-Tarragona-768x432.jpg 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Acueducto-de-Tarragona-65x37.jpg 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Acueducto-de-Tarragona-225x127.jpg 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Acueducto-de-Tarragona-350x197.jpg 350w\" sizes=\"auto, (max-width: 378px) 100vw, 378px\" \/>The bridge to the left was thought to have been built during the reign of Augustus (27 BCE- 14CE) as an aqueduct to supply water to the town of Tarraco in Catalonia, Spain. This structure, 249 m in length, is comprised of a series of free-standing stone arches that have the same radius of 5.9 \u00b1 0.15 m.\u00a0 Of the 931 known Roman bridge structures surviving to date, the largest of these is the Puente Romano bridge in Merida, Spain, having a length of 792 m.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-561 alignright\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Azure_Window.jpeg\" alt=\"\" width=\"352\" height=\"264\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Azure_Window.jpeg 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Azure_Window-300x225.jpeg 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Azure_Window-768x576.jpeg 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Azure_Window-65x49.jpeg 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Azure_Window-225x169.jpeg 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-Azure_Window-350x263.jpeg 350w\" sizes=\"auto, (max-width: 352px) 100vw, 352px\" \/>The arch design used in Roman bridges is one that is found in nature, such as the Azure Window shown to the right and located in Malta. The success of the strength in the design of arch structures is due to the way in which the force from the weight of the bridge is carried outward along the curve of the arch, and is balanced by the supports anchoring either end.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-563 alignleft\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-notre-dame-buttress.jpeg\" alt=\"\" width=\"214\" height=\"287\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-notre-dame-buttress.jpeg 765w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-notre-dame-buttress-224x300.jpeg 224w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-notre-dame-buttress-65x87.jpeg 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-notre-dame-buttress-225x301.jpeg 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-notre-dame-buttress-350x468.jpeg 350w\" sizes=\"auto, (max-width: 214px) 100vw, 214px\" \/>The flying buttress design generally known as a design favorite of the architects of the Notre Dame Cathedral consists of half arches placed on either side of the building to support the weight, and wind shears acting on the central dome covering the building. This technique of balancing forces allowed for the walls of the building to be opened up to allow windows to be constructed in the walls, to let in light and support the weight of the roof.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-564 alignright\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-VillardButtressReims.jpeg\" alt=\"\" width=\"201\" height=\"291\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-VillardButtressReims.jpeg 207w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-VillardButtressReims-65x94.jpeg 65w\" sizes=\"auto, (max-width: 201px) 100vw, 201px\" \/>The sketch shown to the right is that of Villard de Honnecourt (1225-1235). While of unknown purpose, it is part of a collection of works showing various designs of devices he saw during his travels through Medieval Europe.<\/p>\n<p><strong>**The images shown \u00a0below illustrate the forces that act in an arch bridge: force vectors acting in the Action Force (Load) and the Reaction Force (Support for the Load).**<\/strong><\/p>\n<p><strong>Conceptual Origins &#8230;<\/strong><\/p>\n<p>The term equilibrium<a class=\"footnote\" title=\"Equilibrium is further classified as being stable, neutral or unstable.\" id=\"return-footnote-117-1\" href=\"#footnote-117-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> originates from the Latin \u201cacqui\u201d and \u201clibra\u201d, which mean \u201cequal\u201d and \u201cbalance\u201d. When used in physics concepts, \u201cobjects in a state of equilibrium\u201d refers to a balancing of forces acting on the object.\u00a0 This basic concept of equilibrium, i.e.: all forces acting on a point in an object sum to zero, is the embodiment of Newton\u2019s first law.<\/p>\n<p><strong>To recall, Newton&#8217;s first law<\/strong> states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of <strong>inertia.<\/strong> The key point here is that if there is <strong>no net force<\/strong> acting on an object (if all the external forces cancel each other out) then the object will maintain at a <strong>constant velocity<\/strong>. If that velocity is zero, then the object remains at rest. If an external force is applied, the velocity will change because of the force.<\/p>\n<p>From the Latin version of Newton&#8217;s Principia, his first law reads as&#8230;<\/p>\n<p style=\"padding-left: 40px;\"><strong><em>Lex I: Corpus omne perseverare in statu suo quiescendi vel movendi uniformiter in directum, nisi quatenus a viribus impressis cogitur statum illum mutare.<\/em><\/strong><\/p>\n<p>Translated to English, this reads:<\/p>\n<p style=\"padding-left: 40px;\"><strong><em>Law I: Every body persists in its state of being at rest or of moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed<\/em><\/strong><strong>.<\/strong><\/p>\n<p>Newton then expands on his first law with:<\/p>\n<p style=\"padding-left: 40px;\"><strong><em>Projectiles persevere in their motions, so far as they are not retarded by the resistance of the air, or impelled downwards by the force of gravity. A top, whose parts by their cohesion are perpetually drawn aside from rectilinear motion, does not cease its rotation, otherwise than as it is retarded by the air. The greater bodies of the planets and comets, meeting with less resistance in more free spaces, preserve their motions both progressive and circular for a much longer time.<\/em><\/strong><\/p>\n<p>One of the challenges to the interpretation of Newton\u2019s laws is that he appears to have written them following the style of earlier Greek philosophers, such as Euclid. As such, historians have suggested that Newton\u2019s first law is simply a natural deduction of his second law since there is nothing about frames of reference included in Newton\u2019s writings.<\/p>\n<p>Equilibrium is explored in three different ways in this chapter: translational, static and rotational, all relating to the balancing of forces that act on an object or a structure. Equilibrium is a very important concept, and you should encounter a number of different ways that explore equilibrium in specialized fields.<\/p>\n<p><strong>Translational equilibrium<\/strong> occurs when no net or resultant force is acting on an object. This means that the sum of all forces that are acting will equal 0 N. Newton\u2019s first law interprets this to mean that if the object is at rest, it will remain at rest and if the object is in motion, it will remain in that exact state of motion.<\/p>\n<p><strong>Rotational equilibrium<\/strong> requires that the object is non-rotating. This necessitates that all forces causing it to rotate (torques) will sum to zero. Newton\u2019s first law interprets this to mean that if the object is at rest, not-rotating, it will remain at rest and not-rotating. If the object is rotating, it will remain in that exact state of rotation, neither speeding up nor slowing down.<\/p>\n<p><strong>Static equilibrium<\/strong> is a special case of translational and rotational equilibrium, and occurs when the net force or resultant force acting on an object sums to zero and the object is neither moving in any direction nor rotating.\u00a0 Static equilibrium requires that the object in question is at rest and remains at rest.<\/p>\n<p>When we look at the effect of forces acting on a body, there are five commonly observed actions. You will be encountering the first two of these, compression (C) and tension (T), in the problems in this chapter.<strong>\u00a0<\/strong><\/p>\n<p><strong>Free Body Diagrams <\/strong><\/p>\n<div class=\"textbox textbox--sidebar\">\n<ul>\n<li>Extra Help: <a href=\"https:\/\/physics.info\/newton-first\/\">Forces, Free Body Diagrams and Newton\u2019s Laws of Force<\/a><\/li>\n<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/newtlaws\/Lesson-2\/Drawing-Free-Body-Diagrams\">Drawing Free Body Diagrams<\/a><\/li>\n<\/ul>\n<\/div>\n<p>Free body diagrams are standard tools used to analyze forces in equilibrium. Free body diagrams are used to show the magnitude and direction of all forces that act on a body in a given situation. In many cases these diagrams are drawn to scale, with the length of the arrow in scale to the magnitude of the force and the angle of this force accurately positioned. Learning to draw a free body diagram is important for one to be able to work through questions in statics, dynamics and a few other fields of classical mechanics.<\/p>\n<p>Using the free body diagram as a tool, one is able to easily visualize whether or not the forces acting on an object add to zero ([latex]F_{\\text{net}} = 0[\/latex]). If they do add to zero, the object is in static equilibrium and not moving. If the forces acting on the body do not add to zero ([latex]F_{\\text{net}} \\neq 0[\/latex]), it will accelerate.<\/p>\n<p>Hyper physics lists a number of common forces that you can encounter where free body diagrams assist in working out the intricacies of the problem. In Chapter 6, you encountered <strong>weight (<\/strong>[latex]w[\/latex], [latex]F_g[\/latex]<strong>) <\/strong>which is a force vector directed straight down towards the Earth\u2019s centre. In Chapter 7 you will use another force vector called <strong>tension ([latex]T[\/latex], [latex]F_t[\/latex])<\/strong>.\u00a0 Tension is the force that exists in strings, cables, ropes or beams that are acting to apply a force to an external object. Tension for these are the forces transmitted by the cable, rope, string or beam and act in the direction these are pulling or pushing. In the simplest examples that follow, this tension balances the weight. The examples become more complicated when multiple tensions act on an external object.<\/p>\n<p><strong>Other common forces<\/strong> that you will encounter are:<\/p>\n<ul>\n<li><strong>Normal force ([latex]N[\/latex], [latex]F_n[\/latex]), <\/strong>which can be explained as the force exerted by a surface tangentially away from the surface to support an object.<\/li>\n<li><strong style=\"orphans: 1; text-align: initial; font-size: 14pt;\">Frictional force ([latex]f[\/latex]<\/strong><span style=\"orphans: 1; text-align: initial; font-size: 14pt;\">\u00a0<\/span><strong style=\"orphans: 1; text-align: initial; font-size: 14pt;\">, [latex]F_f[\/latex]),<\/strong><span style=\"orphans: 1; text-align: initial; font-size: 14pt;\"> which is the force acting between two surfaces, and is generally opposite to the direction the object is moving or accelerating, but this is not always true.<\/span><\/li>\n<li><strong style=\"orphans: 1; text-align: initial; font-size: 14pt;\">Elastic forces ([latex]F_e[\/latex], [latex]F_s[\/latex])<\/strong><span style=\"orphans: 1; text-align: initial; font-size: 14pt;\"> are the forces exerted by springs or rubber bands, etc., when they are stretched (or compressed in some cases). The force in these instances acts in the direction of the stretch or compression.<\/span><\/li>\n<li><strong style=\"orphans: 1; text-align: initial; font-size: 14pt;\">Buoyancy ([latex]B[\/latex], [latex]F_b[\/latex])<\/strong><span style=\"orphans: 1; text-align: initial; font-size: 14pt;\"> is the force that acts on objects that are immersed or suspended in a fluid.\u00a0 The direction of this force is always upwards and away from the centre of the Earth.<\/span><\/li>\n<li><strong>Drag or air resistance ([latex]R[\/latex], [latex]D[\/latex], [latex]F_d[\/latex], [latex]F_{\\text{air}}[\/latex]) <\/strong>is the force that acts on any object that is moving through a fluid, be it liquid or gaseous. This force will generally be acting opposite to the direction that the object is moving (opposite to the velocity).<\/li>\n<li><strong>Lift ([latex]L[\/latex], [latex]F_l[\/latex])<\/strong> is the force that acts on an object that is tangential to its movement through a fluid, meaning that as the object is moving forward through a fluid, the lift will act at 90\u00b0 to the direction of the object\u2019s velocity.<\/li>\n<li><strong>Thrust ([latex]T[\/latex], [latex]F_t[\/latex])<\/strong> is the action-reaction force that results from pushing a fluid in one direction (generally backwards) that allows the object doing this pushing to move in the opposite direction.<\/li>\n<\/ul>\n<p>In all cases, combinations of the above forces can be acting on some object, requiring one to find the net result of forces <strong>([latex]F_{\\text{net}}[\/latex])<\/strong> acting on the object.\u00a0 All these forces are in newtons (N).<\/p>\n<p>The following examples will illustrate how one can draw a free body diagram of the forces acting on a body.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.3.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Consider a parachutist:<\/p>\n<ol type=\"i\">\n<li>First starts to fall<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-578\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-i-example.png\" alt=\"\" width=\"212\" height=\"277\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-i-example.png 435w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-i-example-230x300.png 230w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-i-example-65x85.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-i-example-225x294.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-i-example-350x457.png 350w\" sizes=\"auto, (max-width: 212px) 100vw, 212px\" \/><\/li>\n<li>Reaches terminal velocity<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-579 alignnone\" style=\"orphans: 1; text-align: initial; font-size: 0.9em; word-spacing: normal;\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-ii-example.png\" alt=\"\" width=\"198\" height=\"301\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-ii-example.png 495w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-ii-example-198x300.png 198w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-ii-example-65x99.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-ii-example-225x341.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-ii-example-350x531.png 350w\" sizes=\"auto, (max-width: 198px) 100vw, 198px\" \/><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.3.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Consider a person floating in a lake:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-580 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.2-example.png\" alt=\"\" width=\"232\" height=\"310\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.2-example.png 563w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.2-example-225x300.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.2-example-65x87.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.2-example-350x467.png 350w\" sizes=\"auto, (max-width: 232px) 100vw, 232px\" \/><\/p>\n<\/div>\n<\/div>\n<p>The principle used to draw these diagrams is the same when you encounter more than two forces acting on an object.\u00a0 Two cables supporting an object at rest or moving at a constant velocity ([latex]F_{\\text{net}} = 0[\/latex]) would show up as two forces acting in unison to balance the weight or force of gravity pulling down on the object.\u00a0 If these cables are acting to accelerate the object either up, down or to one side ([latex]F_{\\text{net}} \\neq 0[\/latex]) then the forces acting on the object would show some imbalance.\u00a0 In all cases, the force arrows will act in the directions that the forces act on the object.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.3.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Suppose you have a box suspended in air by two cables as shown below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-581\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-two-cables.png\" alt=\"\" width=\"296\" height=\"296\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-two-cables.png 701w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-two-cables-300x300.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-two-cables-150x150.png 150w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-two-cables-65x65.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-two-cables-225x225.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3-two-cables-350x350.png 350w\" sizes=\"auto, (max-width: 296px) 100vw, 296px\" \/><\/p>\n<p>If we were to redraw this using force vectors, it would look like:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-582 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-1.png\" alt=\"\" width=\"337\" height=\"345\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-1.png 1031w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-1-293x300.png 293w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-1-1002x1024.png 1002w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-1-768x785.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-1-65x66.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-1-225x230.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-1-350x358.png 350w\" sizes=\"auto, (max-width: 337px) 100vw, 337px\" \/><\/p>\n<p>To properly analyze this diagram we would need both the magnitudes and directions of all forces acting on the box. (Forces are generally termed tensions when these forces are pulling on the cable.\u00a0 This is compared to a beam that can be experiencing\u00a0 a force that can be acting to compress it.)<\/p>\n<p>This diagram to the right can be further reduced to:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-583 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-2.png\" alt=\"\" width=\"263\" height=\"403\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-2.png 646w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-2-196x300.png 196w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-2-65x100.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-2-225x345.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-2-350x536.png 350w\" sizes=\"auto, (max-width: 263px) 100vw, 263px\" \/><\/p>\n<p>Trigonometry would be used at this point to break the two Cable Tensions (T1 &amp; T2) into vector components. The rest of the solution would be algebra in balancing the forces acting on the Box.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.3.4<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Find the tension in a cable supporting a single box of 25 kg in Static Equilibrium.<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>The sketch and the free body diagrams for this example are:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-635 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example.png\" alt=\"\" width=\"457\" height=\"282\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example.png 1517w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-300x185.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-1024x631.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-768x473.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-65x40.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-225x139.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-350x216.png 350w\" sizes=\"auto, (max-width: 457px) 100vw, 457px\" \/><\/p>\n<p>The solution for this requires that the forces acting on the box must sum to zero.\u00a0 In an equation this looks like:<\/p>\n<p style=\"text-align: center;\">[latex]\\sum \\vec{F_y} = 0 \\text{ N}[\/latex]<\/p>\n<p>To solve this problem, assign cartesian directions as you would on a graph:<\/p>\n<p>For y-direction, up is positive and down is negative and for the x-direction, right is positive and left is negative.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-636 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-2.png\" alt=\"\" width=\"430\" height=\"305\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-2.png 1517w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-2-300x213.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-2-1024x727.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-2-768x545.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-2-65x46.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-2-225x160.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-2-350x248.png 350w\" sizes=\"auto, (max-width: 430px) 100vw, 430px\" \/><\/p>\n<p>Using this vector notation, the solution is as follows:<\/p>\n<ul>\n<li>[latex]T_1 + w = 0 \\text{ N}[\/latex]<\/li>\n<li>[latex]T_1 + (25 \\text{ kg})(- 9.8 \\text{ m\/s}^2) = 0 \\text{ N}[\/latex]<\/li>\n<li>[latex]T_1 - 245 \\text{ N } = 0 \\text{ N}[\/latex]<\/li>\n<li>[latex]T_1 = 0 \\text{ N } + 245 \\text{ N}[\/latex]<\/li>\n<li>[latex]T_1 = 245 \\text{ N } (\\approx 250 \\text{ N})[\/latex]<\/li>\n<\/ul>\n<p>The direction of this force is upwards.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.3.5<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Find the tension in the two cables supporting a single crate of 100 kg in static equilibrium as shown in the diagram below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-640 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-3.png\" alt=\"\" width=\"373\" height=\"353\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-3.png 907w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-3-300x283.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-3-768x726.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-3-65x61.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-3-225x213.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-example-3-350x331.png 350w\" sizes=\"auto, (max-width: 373px) 100vw, 373px\" \/><\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>The free body diagram of this problem looks like:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-642\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example.png\" alt=\"\" width=\"370\" height=\"337\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example.png 1017w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-300x273.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-768x699.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-65x59.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-225x205.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-350x318.png 350w\" sizes=\"auto, (max-width: 370px) 100vw, 370px\" \/><\/p>\n<p>These forces act in two directions&#8230; \u00b1 [latex]x[\/latex] and \u00b1 [latex]y[\/latex], and the solution for this requires balancing these forces in two directions.<\/p>\n<p style=\"text-align: center;\">[latex]\\sum \\vec{F_x} = 0 \\text{ N}[\/latex] and [latex]\\sum \\vec{F_y} = 0 \\text{ N}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Weight }= (100 \\text{ kg})(-9.8 \\text{m\/s}^2)[\/latex] or [latex]-980 \\text{ N}[\/latex] in the y-direction.<\/p>\n<p>Since cable 2 is the only cable supporting the crate, we know that the tension in the cable in the y-direction must be + 980 N.\u00a0 The sketch of this looks as follows:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-643 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-2.png\" alt=\"\" width=\"280\" height=\"250\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-2.png 838w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-2-300x267.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-2-768x685.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-2-65x58.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-2-225x201.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-2-350x312.png 350w\" sizes=\"auto, (max-width: 280px) 100vw, 280px\" \/><\/p>\n<p>Trigonometry is now used to find the tension of cable 2 in the x direction ([latex]T_{2x}[\/latex]):<\/p>\n<ul>\n<li>[latex]\\tan 30^{\\circ} = \\dfrac{T_{2x}}{980 \\text{ N}}[\/latex]<\/li>\n<li>[latex]T_{2x} = 980 \\text{ N } \\tan 30^{\\circ}[\/latex]<\/li>\n<li>[latex]T_{2x} = -566 \\text{ N } (\\approx -570 \\text{ N})[\/latex], negative x direction.<\/li>\n<\/ul>\n<p>Now, the total tension of cable 2 is:<\/p>\n<ul>\n<li>[latex]\\cos 30^{\\circ}= \\dfrac{980 \\text{ N}}{T_2}[\/latex]<\/li>\n<li>[latex]T_2 = \\dfrac{980 \\text{ N}}{\\cos 30^{\\circ}}[\/latex]<\/li>\n<li>[latex]T_2 = 1132 \\text{ N } (\\approx 1100 \\text{ N})[\/latex]<\/li>\n<\/ul>\n<p>And, the tension of cable 1 is:<\/p>\n<ul>\n<li>[latex]T_1 = 566 \\text{ N } (\\approx 570 \\text{ N})[\/latex] (equal and opposite to [latex]T_{2x}[\/latex])<\/li>\n<\/ul>\n<p>Putting all of this in a free body diagram yields:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-644 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-3.png\" alt=\"\" width=\"291\" height=\"342\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-3.png 880w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-3-255x300.png 255w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-3-871x1024.png 871w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-3-768x903.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-3-65x76.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-3-225x265.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-example-3-350x412.png 350w\" sizes=\"auto, (max-width: 291px) 100vw, 291px\" \/><\/p>\n<p>When summing up the forces [latex]\\sum \\vec{F_x} = 0 \\text{ N}[\/latex] and [latex]\\sum \\vec{F_y} = 0 \\text{ N}[\/latex]&#8230;<\/p>\n<p>[latex]\\sum \\vec{F_x} = 0 \\text{ N}[\/latex]<\/p>\n<ul>\n<li>[latex]T_{2y} + w = 0 \\text{ N}[\/latex]<\/li>\n<li>[latex]980 \\text{ N } - 980 \\text{ N } = 0 \\text{ N}[\/latex]<\/li>\n<\/ul>\n<p>[latex]\\sum \\vec{F_y} = 0 \\text{ N}[\/latex]<\/p>\n<ul>\n<li>[latex]T_{2x} + T_1 = 0 \\text{ N}[\/latex]<\/li>\n<li>[latex]-566 \\text{ N } + 566 \\text{ N } = 0 \\text{ N}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.3.6<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Find the tension in the two cables supporting a single box of 420 kg in static equilibrium as shown in the diagram below and to the right.<\/p>\n<p>.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-647 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example.png\" alt=\"\" width=\"273\" height=\"307\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example.png 811w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-267x300.png 267w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-768x864.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-65x73.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-225x253.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-350x394.png 350w\" sizes=\"auto, (max-width: 273px) 100vw, 273px\" \/><\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>The free body diagram for this problem looks like:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-648 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-2.png\" alt=\"\" width=\"228\" height=\"359\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-2.png 646w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-2-191x300.png 191w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-2-65x102.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-2-225x354.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-2-350x551.png 350w\" sizes=\"auto, (max-width: 228px) 100vw, 228px\" \/><\/p>\n<p>Solving this problem is a little more work.<\/p>\n<p>First, the weight is:<\/p>\n<ul>\n<li>[latex]w = m g[\/latex]<\/li>\n<li>[latex]w = (420 \\text{ kg})(9.8 \\text{ m\/s}^2)[\/latex]<\/li>\n<li>[latex]w = 4116 \\text{ N } (\\approx 4100 \\text{ N})[\/latex]<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-649 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-3.png\" alt=\"\" width=\"268\" height=\"517\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-3.png 678w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-3-155x300.png 155w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-3-530x1024.png 530w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-3-65x126.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-3-225x435.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-3-350x676.png 350w\" sizes=\"auto, (max-width: 268px) 100vw, 268px\" \/><\/p>\n<p>Summing up the forces acting on the box:<\/p>\n<ul>\n<li>In the y direction: [latex]T_{1y} + T_{2y} + w = 0 \\text{ N}[\/latex]<\/li>\n<li>In the x direction: [latex]T_{1x} + T_{2x} = 0 \\text{ N}[\/latex]<\/li>\n<\/ul>\n<p>Find what we know about the Tensions.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-650\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-4.png\" alt=\"\" width=\"168\" height=\"302\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-4.png 421w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-4-167x300.png 167w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-4-65x117.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-4-225x404.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-4-350x629.png 350w\" sizes=\"auto, (max-width: 168px) 100vw, 168px\" \/><\/p>\n<p>[latex]\\sin 30^{\\circ} = T_{1x} \\div T_1[\/latex], or&#8230;<\/p>\n<ul>\n<li>[latex]T_{1x} = T_1 \\sin 30^{\\circ}[\/latex]<\/li>\n<li>[latex]T_{1x} = 0.500 T_2[\/latex]<\/li>\n<\/ul>\n<p>[latex]\\cos 30^{\\circ} = T_{1y} \\div T_1[\/latex], or&#8230;<\/p>\n<ul>\n<li>[latex]T_{1y} = T_1 \\cos 30^{\\circ}[\/latex]<\/li>\n<li>[latex]T_{1y} = 0.866 T_1[\/latex]<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-651\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-5.png\" alt=\"\" width=\"169\" height=\"267\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-5.png 440w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-5-190x300.png 190w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-5-65x103.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-5-225x356.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-5-350x554.png 350w\" sizes=\"auto, (max-width: 169px) 100vw, 169px\" \/><\/p>\n<p>[latex]\\sin 40^{\\circ} = T_{2x} \\div T_2[\/latex], or&#8230;<\/p>\n<ul>\n<li>[latex]T_{2x} = T_2 \\sin 40^{\\circ}[\/latex]<\/li>\n<li>[latex]T_{2x} = 0.643 T_2[\/latex]<\/li>\n<\/ul>\n<p>[latex]\\cos 40^{\\circ} = T_{2y} \\div T_2[\/latex], or&#8230;<\/p>\n<ul>\n<li>[latex]T_{2y} = T_2 \\cos 40^{\\circ}[\/latex]<\/li>\n<li>[latex]T_{2y} = 0.766 T_2[\/latex]<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<p>Balancing all forces:<\/p>\n<ul>\n<li>x direction: [latex]0 \\text{ N } = T_{1x} + T_{2x}[\/latex] or [latex]0 \\text{ N } = 0.500 T_1 - 0.643 T_2[\/latex]<\/li>\n<li>y direction: [latex]0 \\text{ N } = T_{1y} + T_{2y} + w[\/latex] or [latex]0.866 T_1 + 0.766 T_2 = 4116 \\text{ N}[\/latex]<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<p>Now, use substitution to solve this system of two equations.<\/p>\n<p>First: Isolate [latex]T_1[\/latex]<\/p>\n<ul>\n<li>[latex]0 \\text{ N } = 0.643 T_1 - 0.500 T_2[\/latex]<\/li>\n<li>[latex]0.500 T_1 = 0.643 T_2[\/latex]<\/li>\n<li>[latex]T_1 = \\dfrac{0.643 T_2}{0.500}[\/latex]<\/li>\n<li>[latex]T_ 1 = 1.29 T_2[\/latex]<\/li>\n<\/ul>\n<p>Second: Substitute for [latex]T_2[\/latex]<\/p>\n<ul>\n<li>[latex]0.866 T_1 + 0.766 T_2 = 4116 \\text{ N}[\/latex]<\/li>\n<li>[latex]0.866 (1.29 T_2) + 0.766 T_2 = 4116 \\text{ N}[\/latex]<\/li>\n<li>[latex]1.12 T_{2} + 0.766 T_{2} = 4116 \\text{ N}[\/latex]<\/li>\n<li>[latex]1.866 T_2 = 4116 \\text{ N}[\/latex]<\/li>\n<li>[latex]T_2 = 4116 \\text{ N } \\div 1.462 \\text{ or } 2182 \\text{ N } (\\approx 2200 \\text{ N})[\/latex]<\/li>\n<\/ul>\n<p>Finally:<\/p>\n<ul>\n<li>[latex]T_1 = 1.29 T_2[\/latex]<\/li>\n<li>[latex]T_1 = (1.29)(2182 \\text{ N})[\/latex]<\/li>\n<li>[latex]T_1 = 2815 \\text{ N } (\\approx 2800 \\text{ N})[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 7.3.7<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Find the tension in the cable and the reactive force from the beam supporting a single crate of 500 kg in static equilibrium as shown in the diagram below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-661 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-6.png\" alt=\"\" width=\"149\" height=\"258\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-6.png 540w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-6-173x300.png 173w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-6-65x113.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-6-225x390.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-example-6-350x606.png 350w\" sizes=\"auto, (max-width: 149px) 100vw, 149px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>[latex]\\text{Weight } = (500 \\text{ kg})(- 9.8 \\text{ m\/s}^2) or -4900 \\text{ N}[\/latex]<\/p>\n<p>Conditions for equilibrium:<\/p>\n<p style=\"text-align: center;\">[latex]\\sum \\vec{F_x} = 0 \\text{ N}[\/latex] or [latex]\\sum \\vec{F_y} = 0 \\text{ N}[\/latex]<\/p>\n<p>Since this cable is the only force supporting the crate, we know that the tension in the cable in the y-direction must be + 4900 N.<\/p>\n<ul>\n<li style=\"text-align: left;\">[latex]T_y = + 4900 \\text{ N}[\/latex]<\/li>\n<\/ul>\n<p>Trigonometry is now used to find the tension of the cable in the x direction ([latex]T_x[\/latex]).<\/p>\n<ul>\n<li>[latex]\\tan 42^{\\circ} = \\dfrac{4900 \\text{ N}}{T_x}[\/latex]<\/li>\n<li>[latex]T_x = \\dfrac{4900 \\text{ N}}{\\tan 42^{\\circ}}[\/latex]<\/li>\n<li>[latex]T_x = 5442 \\text{ N } (\\approx 5400 \\text{ N})[\/latex]&#8230; positive X-direction.<\/li>\n<\/ul>\n<p>Now, the total tension of the cable is:<\/p>\n<ul>\n<li>[latex]\\sin 41^{\\circ} = \\dfrac{4900 \\text{ N}}{T}[\/latex]<\/li>\n<li>[latex]T = \\dfrac{4900 \\text{ N}}{\\sin 42^{\\circ}}[\/latex]<\/li>\n<li>[latex]T = 7323 \\text{ N } (\\approx 7300 \\text{ N})[\/latex]<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-662 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-example-2.png\" alt=\"\" width=\"256\" height=\"240\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-example-2.png 728w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-example-2-300x281.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-example-2-65x61.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-example-2-225x211.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-example-2-350x328.png 350w\" sizes=\"auto, (max-width: 256px) 100vw, 256px\" \/><\/p>\n<p>The reactive force from the beam is:<\/p>\n<ul>\n<li>[latex]C = -5442 \\text{ N } (\\approx - 5400 \\text{ N})[\/latex] (equal and opposite to [latex]T_x[\/latex])<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 7.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>Find the tension in a cable supporting a single box of 100 kg in static equilibrium.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-664 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-exercise.png\" alt=\"\" width=\"109\" height=\"328\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-exercise.png 311w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-exercise-100x300.png 100w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-exercise-65x195.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.1-exercise-225x676.png 225w\" sizes=\"auto, (max-width: 109px) 100vw, 109px\" \/><\/li>\n<li>Find the tension in a cable supporting a single box weighing 5.0 N in static equilibrium.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-665 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.2-exercise.png\" alt=\"\" width=\"110\" height=\"331\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.2-exercise.png 311w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.2-exercise-100x300.png 100w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.2-exercise-65x195.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.2-exercise-225x676.png 225w\" sizes=\"auto, (max-width: 110px) 100vw, 110px\" \/><\/li>\n<li>Find the tension in the two cables supporting a single crate of mass 50 kg in static equilibrium as shown in the diagram below to the right.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-666 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-exercise.png\" alt=\"\" width=\"242\" height=\"233\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-exercise.png 733w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-exercise-300x289.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-exercise-65x63.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-exercise-225x216.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.3-exercise-350x337.png 350w\" sizes=\"auto, (max-width: 242px) 100vw, 242px\" \/><\/li>\n<li>Find the tension in the two cables supporting a single crate of 250 kg in static equilibrium as shown in the diagram below to the right.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-667\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-exercise.png\" alt=\"\" width=\"242\" height=\"233\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-exercise.png 733w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-exercise-300x289.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-exercise-65x63.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-exercise-225x216.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.4-exercise-350x337.png 350w\" sizes=\"auto, (max-width: 242px) 100vw, 242px\" \/><\/li>\n<li>Find the tension in the two cables supporting a single crate of 120 kg in static equilibrium as shown in the diagram below and to the right.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-668 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-exercise.png\" alt=\"\" width=\"256\" height=\"256\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-exercise.png 701w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-exercise-300x300.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-exercise-150x150.png 150w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-exercise-65x65.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-exercise-225x225.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.5-exercise-350x350.png 350w\" sizes=\"auto, (max-width: 256px) 100vw, 256px\" \/><\/li>\n<li>Find the tension in the two cables supporting a single crate of 1200 kg in static equilibrium as shown in the diagram below and to the right.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-669\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-exercise.png\" alt=\"\" width=\"268\" height=\"267\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-exercise.png 705w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-exercise-300x298.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-exercise-150x150.png 150w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-exercise-65x65.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-exercise-225x224.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.6-exercise-350x348.png 350w\" sizes=\"auto, (max-width: 268px) 100vw, 268px\" \/><\/li>\n<li>Find the tension in the cable and the reactive force from the beam supporting a single crate of 420 kg in static equilibrium as shown in the diagram below to the right.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-671 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-exercise.png\" alt=\"\" width=\"236\" height=\"409\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-exercise.png 540w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-exercise-173x300.png 173w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-exercise-65x113.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-exercise-225x390.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.7-exercise-350x606.png 350w\" sizes=\"auto, (max-width: 236px) 100vw, 236px\" \/><\/li>\n<li>If the maximum recommended tension in the cable is 12 000 N, find the reactive force from the beam that would be needed to support this maximum in static equilibrium as shown in the diagram below to the right.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-672 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.8-exercise.png\" alt=\"\" width=\"217\" height=\"376\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.8-exercise.png 540w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.8-exercise-173x300.png 173w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.8-exercise-65x113.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.8-exercise-225x390.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/7.3.8-exercise-350x606.png 350w\" sizes=\"auto, (max-width: 217px) 100vw, 217px\" \/><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 7.4: Limits of a cable<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A cable that has a maximum load tension of 50 000 N is slowly lifting a heavy crate straight upwards.<\/p>\n<ol type=\"i\">\n<li>How close to the top can the crate be safely lifted?<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1187 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/add-mass.png\" alt=\"\" width=\"611\" height=\"252\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/add-mass.png 2090w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/add-mass-300x124.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/add-mass-1024x422.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/add-mass-768x316.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/add-mass-1536x633.png 1536w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/add-mass-2048x844.png 2048w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/add-mass-65x27.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/add-mass-225x93.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/add-mass-350x144.png 350w\" sizes=\"auto, (max-width: 611px) 100vw, 611px\" \/><\/li>\n<li>What is the mass that can be lifted to within 10 cm from the top?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h1>Exercise Answers<\/h1>\n<h2>7.1 Right Angle Trigonometric Functions<\/h2>\n<ol class=\"twocolumn\">\n<li>0.743145<\/li>\n<li>0.484810<\/li>\n<li>0.906308<\/li>\n<li>0.484810<\/li>\n<li>0.194380<\/li>\n<li>1.53986<\/li>\n<li>0.190810<\/li>\n<li>0.544639<\/li>\n<li>29\u00b0<\/li>\n<li>39\u00b0<\/li>\n<li>50\u00b0<\/li>\n<li>52\u00b0<\/li>\n<li>33.3\u00b0<\/li>\n<li>8.9\u00b0<\/li>\n<li>41.0\u00b0<\/li>\n<li>81\u00b0<\/li>\n<li>z = 27.36 (\u2248 27), \u00f8 = 26.6\u00b0 (\u2248 27\u00b0)<\/li>\n<li>y = 19.6 (\u2248 20),\u00a0\u00f8 = 44.4\u00b0 (\u2248 44\u00b0)<\/li>\n<li>x = 16, \u00f8 = 53.1\u00b0 (\u2248 53\u00b0)<\/li>\n<li>x = 21.2 (\u2248 21), y = 13.2 (\u2248 13)<\/li>\n<li>x = 892 (\u2248 890), y = 803 (\u2248 800)<\/li>\n<li>z = 242 (\u2248 240),\u00a0 \u00f8 = 24.4\u00b0 (\u2248 24\u00b0)<\/li>\n<li>x = 9.8, y = 6.9<\/li>\n<li>x = 37.6 (\u2248 38),\u00a0 y = 42.6 (\u2248 43)<\/li>\n<li>z = 25, \u00f8 = 53.1\u00b0 (\u2248 53\u00b0)<\/li>\n<li>y = 75,\u00a0 \u00f8 = 36.9\u00b0 (37\u00b0)<\/li>\n<li>y = 4, \u00f8 = 53.1\u00b0 (\u2248 53\u00b0)<\/li>\n<li>y = 11.1 (\u2248 11),\u00a0 z = 27.4 (\u2248 27)<\/li>\n<li>z = 28.6 (\u2248 29), \u00f8 = 44.4\u00b0 (\u2248 44\u00b0)<\/li>\n<li>y = 19.6 (\u2248 20),\u00a0 \u00f8 = 44.4\u00b0 (\u2248 44\u00b0)<\/li>\n<li>y = 8.9, \u00f8 = 41.8\u00b0 (\u2248 42\u00b0)<\/li>\n<li>x = 35, y = 61<\/li>\n<\/ol>\n<h2>7.2 Vector Resolution into Components<\/h2>\n<ol>\n<li>[latex]v_y[\/latex] = 25.4 (\u2248 25), [latex]v_x[\/latex] = 40.7 (\u2248 41)<\/li>\n<li>[latex]v_y[\/latex] = 11, [latex]v_x[\/latex] = 19.1 (\u2248 19)<\/li>\n<li>[latex]F_y[\/latex] = 15, [latex]F_x[\/latex] = 28.3 (\u2248 28)<\/li>\n<li>[latex]F_y[\/latex] = 53.7 (\u2248 54), [latex]F_x[\/latex] = 79.6 (\u2248 80)<\/li>\n<li>[latex]a_y[\/latex] = 13.6, [latex]a_x[\/latex] = 15.2<\/li>\n<li>[latex]a_y[\/latex] = 22.0, [latex]a_x[\/latex] = 45.1<\/li>\n<li>[latex]v_y[\/latex] = 25.4 (\u2248 25), [latex]v_x[\/latex] = 40.7 (\u2248 41)<\/li>\n<li>[latex]v_y[\/latex] = 11, [latex]v_x[\/latex] = 19.1 (\u2248 19)<\/li>\n<li>[latex]F_y[\/latex] = 15, [latex]F_x[\/latex] = 28.3 (\u2248 28)<\/li>\n<li>[latex]F_y[\/latex] = 53.7 (\u2248 54), [latex]F_x[\/latex] = 79.6 (\u2248 80)<\/li>\n<li>[latex]a_y[\/latex] = 13.6, [latex]a_x[\/latex] = 15.2<\/li>\n<li>[latex]a_y[\/latex] = 22.0, [latex]a_x[\/latex] = 45.1<\/li>\n<\/ol>\n<h2>7.3 Static Equilibrium (Concurrent Forces)<\/h2>\n<ol class=\"twocolumn\">\n<li>[latex]T_y[\/latex] = 980 N<\/li>\n<li>[latex]T_y[\/latex] = 5.0 N<\/li>\n<\/ol>\n<ol start=\"3\">\n<li>[latex]T_1[\/latex] = 228 N (\u2248 230 N), [latex]T_2[\/latex] = 541 N (\u2248 540 N)<\/li>\n<li>[latex]T_1[\/latex] = 89.2 N (\u2248 89 N), [latex]T_2[\/latex] = 261 N\u00a0 (\u2248 260 N)<\/li>\n<li>[latex]T_1[\/latex] = 769 N (\u2248 770 N), [latex]T_2[\/latex] = 663 N (\u2248 660 N)<\/li>\n<li>[latex]T_1[\/latex] = 8502 N (\u2248 8500 N), [latex]T_2[\/latex] = 5557 N (\u2248 5600 N)<\/li>\n<li>T = 6685 N ( \u2248 6700 N), [latex]F_C =\u00a0 -5268[\/latex] N\u00a0 (\u2248 [latex]- 5300[\/latex] N)<\/li>\n<li>[latex]FC = - 14300[\/latex] N (\u2248 [latex]- 14000[\/latex] N)<\/li>\n<\/ol>\n<h2>7.4 Limits of a Cable<\/h2>\n<ol type=\"i\">\n<li>distance = 0.49 m from the top<\/li>\n<li>[latex]m = 204 \\text{ kg } = 200\\text{ kg}[\/latex]<\/li>\n<\/ol>\n<h3>Media Attributions<\/h3>\n<ul>\n<li>&#8220;<a href=\"https:\/\/upload.wikimedia.org\/wikipedia\/commons\/1\/11\/Observatory_in_Alexandria_at_the_Time_of_Hipparchus.jpg\">Observatory in Alexandria at the Time of Hipparchus<\/a>&#8221; Hermann Goll is in the <a href=\"https:\/\/creativecommons.org\/share-your-work\/public-domain\/\">public domain<\/a>.<\/li>\n<li>&#8220;<a href=\"https:\/\/www.flickr.com\/photos\/calafellvalo\/6969596062\/in\/photolist-fx4tVs-fx4zhN-jkWaky-bQMGU6-bQMH5D-bBT16m-jkSeyR-jkWcsu-bQMGmZ-bQMGAx-jkTUaW-bBSZio-bBT1ym-bQMHt4-bQMHEV-jkU29f-bQMJhZ-jkSfNV-DBe8hg-jkTjNV-bQMHiT-bBSZow-jkTMZ9-jkS1pK-bQMGet-bQMHzk-6fz5LS-bQMFLn-jkTR2G-bBT1aE-jkTYYU-bBT1C9-bBT2fy-jkSdXk-jkTBJ8-jkTjgT-jkToHP-jkTHma-jkSisc-jkTGQF-jkTHXa-bQMJnv-bQMJ9t-bBT2k1-bQMG5g-bBT2r1-bQMKnr-bQMGsz-bQMGwF-bBT3ud\">Acueducto de Tarragona<\/a>&#8221; by <a id=\"yui_3_16_0_1_1678914850254_1764\" class=\"owner-name truncate\" title=\"Go to calafellvalo\u2019s photostream\" href=\"https:\/\/www.flickr.com\/photos\/calafellvalo\/\" data-track=\"attributionNameClick\">calafellvalo<\/a> is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-nc-nd\/2.0\/\">CC BY-NC-ND 2.0 licence<\/a>.<\/li>\n<li>&#8220;<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Azure_Windownew.JPG\">Azure Window New<\/a>&#8221; by anjab1593 is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by\/3.0\/deed.en\">CC BY 3.0 Unported licence<\/a>.<\/li>\n<li>&#8220;<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Notre_Dame_buttress.jpg\">Notre Dame buttress<\/a>&#8221; by <a class=\"external text\" href=\"https:\/\/www.flickr.com\/people\/7761867@N06\" rel=\"nofollow\">Jean Lemoine<\/a> is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/2.0\/deed.en\">CC BY-SA 2.0 licence<\/a>.<\/li>\n<li>&#8220;<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:VillardButtressReims.jpg\">Villard Buttress Reims<\/a>&#8221; uploaded by <a href=\"https:\/\/en.wikipedia.org\/wiki\/User:Wetman\">Wetman<\/a> is in the <a href=\"https:\/\/creativecommons.org\/share-your-work\/public-domain\/\">public domain<\/a>.<\/li>\n<\/ul>\n<hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-117-1\">Equilibrium is further classified as being stable, neutral or unstable. <a href=\"#return-footnote-117-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":125,"menu_order":7,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-117","chapter","type-chapter","status-publish","hentry"],"part":3,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/117","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/users\/125"}],"version-history":[{"count":25,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/117\/revisions"}],"predecessor-version":[{"id":1233,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/117\/revisions\/1233"}],"part":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/117\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/media?parent=117"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapter-type?post=117"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/contributor?post=117"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/license?post=117"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}