{"id":566,"date":"2021-07-21T14:19:33","date_gmt":"2021-07-21T18:19:33","guid":{"rendered":"http:\/\/pressbooks.library.upei.ca\/statics\/?post_type=chapter&#038;p=566"},"modified":"2025-07-31T18:58:01","modified_gmt":"2025-07-31T22:58:01","slug":"1-8-examples","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.upei.ca\/statics\/chapter\/1-8-examples\/","title":{"raw":"1.8 Examples","rendered":"1.8 Examples"},"content":{"raw":"<p style=\"text-align: justify\">Here are examples from Chapter 1 to help you understand these concepts better. These were taken from the real world and supplied by FSDE students in Summer 2021. If you'd like to submit your own examples, please send them to:\r\n<a href=\"mailto:eosgood@upei.ca\"> eosgood@upei.ca<\/a><\/p>\r\n\r\n<h1>Example 1.8.1: Vectors, Submitted by Tyson Ashton-Losee<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nAfter a long day of studying, a student sitting at their computer moves the cursor from the bottom left of the screen to the top right in order to close a web browser. The computer mouse was displaced 6 cm along the x-axis and 3.5 cm along the y-axis. Draw the resultant vector and calculate the distance traveled.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem-2-259x300.jpg\" alt=\"A digital drawing of a computer mouse is shown from above with an arrow labeled r pointing diagonally outward from its center. X and Y axes are displayed in the lower left corner.\" class=\"alignright wp-image-745\" width=\"200\" height=\"232\" \/>\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_743\" align=\"alignnone\" width=\"270\"]<a href=\"https:\/\/www.flickr.com\/photos\/dejankrsmanovic\/33218207918\"><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem-300x200.jpg\" alt=\"A black wired computer mouse is placed on a white surface, viewed from above.\" class=\"wp-image-743\" width=\"270\" height=\"180\" \/><\/a> Source: <a href=\"https:\/\/www.flickr.com\/photos\/155403590@N07\/33218207918\">Black Computer Mouse on Table | Old dusty black computer mou\u2026 | Flickr<\/a>[\/caption]\r\n\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n<strong>2. Draw<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-draw-1.jpg\" alt=\"A right triangle on an XY plane with base 6 cm on the X-axis, height 3.5 cm on the Y-axis, hypotenuse labeled r, and an angle theta (\u03b8) at the origin.\" class=\"alignnone wp-image-746 size-full\" width=\"249\" height=\"137\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnown:\r\n<ul>\r\n \t<li>x = 6 cm<\/li>\r\n \t<li>y = 3.5 cm<\/li>\r\n<\/ul>\r\nUnknown:\r\n<ul>\r\n \t<li>r<\/li>\r\n \t<li>\u03b8<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nUse SOH CAH TOA, first find \u03b8, then r\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\n[latex]\\begin{aligned} &amp;\\tan \\theta=\\frac{y}{x} \\\\ &amp;\\tan \\theta=\\frac{3.5 \\mathrm{~cm}}{6 \\mathrm{~cm}} \\\\ &amp;\\theta=\\tan ^{-1}\\left(\\frac{3.5}{6}\\right) \\\\ &amp;\\theta=30.256^{\\circ} \\\\ &amp;\\sin \\theta=\\frac{y}{r} \\\\ &amp;r=\\frac{y}{\\sin \\theta} \\\\ &amp;r=\\frac{3.5 \\mathrm{~cm}}{\\sin \\left(30.256^{\\circ}\\right)} \\\\ &amp;r=6.946 \\mathrm{~cm} \\\\ &amp;r=6.9 \\mathrm{~cm} \\end{aligned}[\/latex]\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nIt makes sense that the angle is less than 45, because y is smaller than x. Also, if you use the Pythagorean theorem to find r, you get the same answer.\r\n\r\n<\/div>\r\n&nbsp;\r\n<h1>Example 1.8.2: Vectors, Submitted by Brian MacDonald<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nMark is fishing in the ocean with his favourite fishing rod. The distance between the tip of the rod and the reel is 8 ft and the length of the reel handle is 0.25 ft. The angle between the fishing rod and fishing line is 45 degrees. If Mark catches a fish when 25 ft of the fishing line is released while the fish is diving down with a force of 180 N, how much force does Mark need to apply (push down) to the reel handle to bring in the fish? Draw the position vector of the fish relative to the reel.\r\n<p class=\"no-indent\">Assumptions:<\/p>\r\n\r\n<ul>\r\n \t<li>Mark can reel in the fish when he generates more torque with the handle than the amount of torque that the fish is applying to the reel while pulling on the line.<\/li>\r\n \t<li>The fishing line comes out of the reel in a straight line at a 90-degree angle.<\/li>\r\n<\/ul>\r\n[caption id=\"attachment_1394\" align=\"aligncenter\" width=\"1024\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deepsea-1024x768.jpg\" alt=\"A person wearing a cap and life vest is fishing from a boat, holding a bent fishing rod over the sea.\" class=\"wp-image-1394 size-large\" width=\"1024\" height=\"768\" \/> Source: <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Deepsea.JPG\">File:Deepsea.JPG - Wikimedia Commons<\/a>[\/caption]\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-1v2-300x185.jpg\" alt=\"A digital drawing of a fishing rod angled at 45 degrees is 8 ft tall with a 25 ft fishing line. A 180 N fish is hooked at the end. An inset shows a reel with a handle 0.25 ft from its center.\" class=\"aligncenter wp-image-582\" width=\"373\" height=\"230\" \/>\r\n\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n<strong>2. Draw<\/strong>\r\n\r\nSketch:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-1-300x220.jpg\" alt=\"Diagram of a right triangle on grid paper: vertical side AC is 8 ft, horizontal side AB is 0.25 ft, angle at C is 45\u00b0, and the hypotenuse CD measures 25 ft. Points A, B, C, and D are labeled.\" class=\"alignnone wp-image-590\" width=\"345\" height=\"253\" \/>\r\n\r\nFree-body diagram:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-2-300x233.jpg\" alt=\"A right triangle with a green vertical side, angle \u03b8, and two red downward arrows labeled FA and FD. The x and y axes are shown pointing right and up, respectively.\" class=\"alignnone wp-image-589\" width=\"328\" height=\"255\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnown:\r\n<ul>\r\n \t<li>r<sub>AB<\/sub> = 0.25 ft<\/li>\r\n \t<li>r<sub>BC<\/sub> = 8 ft<\/li>\r\n \t<li>r<sub>CD<\/sub> = 25 ft<\/li>\r\n \t<li>F<sub>D<\/sub> = 180 N<\/li>\r\n \t<li>\u03b8 = 45\u00b0<\/li>\r\n<\/ul>\r\nUnknown:\r\n<ul>\r\n \t<li>F<sub>A<\/sub><\/li>\r\n \t<li>vector <span style=\"text-decoration: underline\">r<\/span><sub>AD<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n<p class=\"indent\">Convert inches to meters, then use the equation below.<\/p>\r\n[latex]T=|r| *|F| * \\sin \\theta\\\\[\/latex]\r\n\r\n<strong>5. Analysis<\/strong>\r\n<p class=\"indent\">Step 1: Convert inches to meters<\/p>\r\n[latex]\r\n\\begin{align}\r\n&amp;25 \\mathrm{ft} * \\frac{12 \\mathrm{in}}{1 \\mathrm{ft}} * \\frac{2.54 \\mathrm{cm}}{\\operatorname{1in} } * \\frac{\\operatorname{1m}}{100 \\mathrm{cm}}=7.62 \\mathrm{m}\\\\\\\\\r\n\r\n&amp;\\quad\\mathrm{and}\\\\\\\\\r\n\r\n&amp;0.25 \\mathrm{ft} * \\frac{12\u00a0 \\mathrm{in}}{1\\mathrm{ft}} * \\frac{2.54 \\mathrm{cm}}{\\operatorname{1in} } * \\frac{\\mathrm{1m}}{100 \\mathrm{cm}}=0.0762 \\mathrm{m}\\\\\r\n\r\n\\end{align}\r\n\r\n[\/latex]\r\n\r\n&nbsp;\r\n<p class=\"indent\">Step 2: Solve for T<sub>D<\/sub><\/p>\r\n[latex]\r\n\\begin{aligned}&amp;T_{D}=\\left|r_{C D}\\right| * \\left|F_{D}\\right| * \\sin \\theta\\\\\r\n\r\n&amp;T_{D}=(7.62 m)(180 N) \\sin \\left(45^{\\circ}\\right)\\\\\r\n&amp;T_{D}=969.86766 \\mathrm{Nm}\r\n\r\n\\end{aligned}\r\n[\/latex]\r\n\r\n&nbsp;\r\n<p class=\"indent\">Step 3: Solve for F<sub>A<\/sub><\/p>\r\n[latex]\r\n\\begin{aligned}&amp;T_{A}=\\left|r_{AB}\\right| * \\left|F_{A}\\right| * \\sin \\theta\\\\\r\n\r\n&amp;\\text { Assume } T_{A}=T_{D}\\\\\r\n\r\n&amp;F_{A}=\\frac{T_{D}}{\\left|r_{AB}\\right| * \\sin \\theta} \\\\\r\n&amp;F_{A}=\\frac{969.86766 \\mathrm{ Nm}}{0.0762 \\mathrm{~m} * \\sin \\left(90^{\\circ}\\right)} \\\\\r\n&amp;F_{A}=12 727.9 \\mathrm{N} \\\\\r\n&amp;F_{A}=13 000 \\mathrm{N}\r\n\\end{aligned}\r\n[\/latex]\r\n\r\n&nbsp;\r\n<p class=\"indent\">Vector <span style=\"text-decoration: underline\">r<\/span><sub>AD<\/sub>:<\/p>\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Solve-Brain-1-v4-300x221.jpg\" alt=\"A vector diagram on a grid showing points A, B, C, and D. Vectors are drawn from A to B, B to C (black), and A to D (orange), with labeled distances and axes for x and y shown at the bottom left.\" class=\"alignnone wp-image-628 size-medium\" width=\"300\" height=\"221\" \/>\r\n\r\n[latex]\r\n\\begin{aligned}\r\n&amp;\\vec r_{A D}=\\vec r_{A B}+\\vec r_{B C}+\\vec r_{C D} \\\\\r\n&amp;\\vec {r}_{A D}=\\left[\\begin{array}{c}\r\n0.25 \\\\\r\n0\r\n\\end{array}\\right] f t+\\left[\\begin{array}{l}\r\n0 \\\\\r\n8\r\n\\end{array}\\right] f t+\\left[\\begin{array}{c}\r\n25 \\sin 45^{\\circ} \\\\\r\n-25 \\cos 45^{\\circ}\r\n\\end{array}\\right] f t \\\\\r\n&amp;\\vec r_{A D}=\\left[\\begin{array}{cc}\r\n17.93 \\\\\r\n-9.68\r\n\\end{array}\\right] ft\r\n\\end{aligned}\r\n[\/latex]\r\n\r\n&nbsp;\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nThough yielding a very large number, the answer appears correct from the information given. 13,000 N of force is the amount of force Mark would need to apply to the reel handle to generate the same amount of force that the fish creates. 13,000 N in reality is too much for one to generate, but also in a real scenario, one would not have to generate the same amount of force to reel in the fish, to reel gearing, the amount of torque generated by the fishing rod itself, etc. In other words 13 000 N of force is too high in a real scenario, but with the assumptions given in the problem, the number seems reasonable. The answer also has the correct unit, N.\r\n\r\n<\/div>\r\n<h1>Example 1.8.3: Dot product and cross product, submitted by Anonymous ENGN 1230 Student<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\n[latex]\\underline{a}=[6\\;\\;\\;5\\;\\;\\;3]\\;\\;\\;\\underline{b}=[8\\;\\;\\;1\\;\\;\\;3][\/latex]\r\n\r\na) Find 6a\r\n\r\nb) Find [latex] a\\cdot b[\/latex]\r\n\r\nc) Find [latex] a\\times b[\/latex]\r\n\r\nd) Find [latex]2a\\times b[\/latex]\r\n\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n<strong>2. Draw<\/strong>\r\n<p style=\"text-align: center\">N\/A<\/p>\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnown:\r\n<ul>\r\n \t<li><span style=\"text-decoration: underline\">a<\/span><\/li>\r\n \t<li><span style=\"text-decoration: underline\">b<\/span><\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>6a,<\/li>\r\n \t<li>[latex] a\\cdot b[\/latex]<\/li>\r\n \t<li>[latex] a\\times b[\/latex]<\/li>\r\n \t<li>[latex]2a\\times b[\/latex]<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nUse dot product and cross product equations\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nPart a:\r\n\r\n[latex]6\\underline{a}=6*[6\\;\\;\\;5\\;\\;\\;3]\\\\6\\underline{a}=[36\\;\\;\\;30\\;\\;\\;18][\/latex]\r\n\r\nPart b:\r\n\r\n[latex]\\underline{a}\\cdot\\underline{b}=[6\\;\\;\\;5\\;\\;\\;3]\\cdot[8\\;\\;\\;1\\;\\;\\;3]\\\\=6\\cdot 8+5\\cdot 1+3\\cdot3\\\\=48+5+9\\\\\\underline{a}\\cdot\\underline{b}=62[\/latex]\r\n\r\nPart C:\r\n\r\n[latex]\\underline{a}\\times\\underline{b}=\\begin{bmatrix} \\underline{\\hat{i}} &amp;\\underline{\\hat{j}} &amp; \\underline{\\hat{k}} \\\\ 6 &amp; 5 &amp; 3 \\\\ 8 &amp; 1 &amp; 3 \\end{bmatrix}\\\\(5\\cdot 3-3\\cdot 1)\\underline{\\hat{i}}-(6\\cdot 3-3\\cdot 8)\\underline{\\hat{j}}+(6\\cdot 1-5\\cdot 8)\\underline{\\hat{k}}\\\\\\underline{a}\\times\\underline{b}=12\\underline{\\hat{i}}+6\\underline{\\hat{j}}-34\\underline{\\hat{k}}[\/latex]\r\n\r\nPart D:\r\n\r\n[latex]2\\underline{a}=2*[6\\;\\;\\;5\\;\\;\\;3]=[12\\;\\;\\;10\\;\\;\\;6]\\\\\\underline{b}=[8\\;\\;\\;1\\;\\;\\;3]\\\\2\\underline{a}\\times\\underline{b} = \\begin{bmatrix} \\underline{\\hat{i}} &amp; \\underline{\\hat{j}} &amp; \\underline{\\hat{k}} \\\\ 12 &amp; 10 &amp; 6 \\\\ 8 &amp; 1 &amp; 3 \\end{bmatrix} \\\\=(10\\cdot 3-6\\cdot 1)\\underline{\\hat{i}}-(12\\cdot 3-6\\cdot 8)\\underline{\\hat{j}}+(12\\cdot 1-10\\cdot 8)\\underline{\\hat{k}}\\\\2\\underline{a}\\times\\underline{b}=24\\underline{\\hat{i}}+12\\underline{\\hat{j}}-68\\underline{\\hat{k}}[\/latex]\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nThe answer to part d is double the answer for part c, which makes sense. It also makes sense that the answers to a, c, and d have values in three directions, while b only has magnitude.\r\n\r\n<\/div>\r\n<h1>Example 1.8.4: Torque, Submitted by Luke McCarvill<\/h1>\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox\">\r\n<div class=\"textbox shaded\">\r\n\r\nTo start riding her bicycle, Jane must push down on one of her bike\u2019s pedals, which are on 16-centimetre-long crank arms. Jane can push directly downwards with her legs with a force of 100N. Jane notices that the pedal\u2019s starting position can sometimes make it more or less useful in generating torque.\r\n\r\na) What is the ideal angle that Jane\u2019s bike pedal should be at in order to generate the most torque? Prove this mathematically. (Assume we only care about the very start of her very first push, and choose a reference frame for the angle that makes most sense for you.)\r\n\r\nb) What angle(s) should the bike pedal be at if Jane wants to generate exactly half of the maximum amount of torque?\r\n\r\nc) Is there any position(s) at which the pedal will create zero torque? Where are they and why?\r\n\r\n[caption id=\"attachment_1393\" align=\"aligncenter\" width=\"1024\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Girl_on_a_Bike_Imagicity_116-1024x685.jpg\" alt=\"Girl on a bike\" class=\"wp-image-1393 size-large\" width=\"1024\" height=\"685\" \/> Source: https:\/\/commons.wikimedia.org\/wiki\/File:Girl_on_a_Bike_(Imagicity_116).jpg[\/caption]\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_1395\" align=\"aligncenter\" width=\"1024\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bicycle-3168934_1280-1024x582.png\" alt=\"A green bicycle\" class=\"wp-image-1395 size-large\" width=\"1024\" height=\"582\" \/> Source: https:\/\/pixabay.com\/illustrations\/bicycle-cycle-two-wheeler-pedal-3168934\/[\/caption]\r\n\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n<strong>2. Draw<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-1-Draw--300x231.jpg\" alt=\"A16 cm horizontal bar, with a downward force labeled F\u2090 acting at the right end of the bar. The y- and x-axes are shown in the top left corner.\" class=\"alignnone wp-image-644 size-medium\" width=\"300\" height=\"231\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n\r\n[latex]\r\n\\begin{aligned}\r\n&amp;\\vec r=\\left[\\begin{array}{c}\r\n0.16 \\\\\r\n0 \\\\\r\n0\r\n\\end{array}\\right] m \\\\\r\n&amp;\\vec F_{A}=\\left[\\begin{array}{c}\r\n0 \\\\\r\n-100 \\\\\r\n0\r\n\\end{array}\\right] N\r\n\r\n\\end{aligned}\r\n[\/latex]\r\n\r\nUnknowns:\r\n<ul>\r\n \t<li>position of <span>r<\/span> for maximum torque<\/li>\r\n \t<li>position of <span>r<\/span> for half of the maximum torque<\/li>\r\n \t<li>position of <span>r<\/span> for zero torque, and why<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nFor part a), I will find a general equation for torque based on the given values in terms of \u03b8, then analyze the function for its maxima\r\n\r\nFor part b), I will find the magnitude of 50% of maximum torque and then reverse-engineer the equation to determine what angle(s) the pedal needs to be at to satisfy the equation.\r\n\r\nFor part c), I will look back at my equation and find when the equation equals zero, then try to understand why, given the example problem.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nPart a:\r\n\r\n[latex]\r\n\\begin{aligned}\r\n&amp;T=\\left|\\vec F_{A}\\right| * \\left|\\vec r\\right| * \\sin \\theta \\\\\r\n&amp;T=(100 N) *(0.16 m) * \\sin \\theta \\\\\r\n&amp;T=16 \\sin \\theta \\mathrm{Nm} \\\\\r\n&amp;\\quad\\left\\{90^{\\circ}+360^{\\circ} k ; k \\in \\mathbb{Z}\\right\\}\r\n\\end{aligned}\r\n[\/latex]\r\n\r\n&nbsp;\r\n\r\nThinking about the shape of the sine function in the first period, the maximum occurs at 90 degrees.\r\n\r\nYou could say algebraically that the maximum is at 90, 450, 810, etc., but these angles all represent the same position on the wheel. Therefore, we will use 90.\r\n\r\nPart b:\r\n\r\n[latex]\\begin{aligned}\r\n&amp;T_1=\\left|F_{A}\\right| *|r| * \\sin \\theta \\\\\r\n&amp;T_1=(100 N)*(0.16 m)* \\sin 90^{\\circ} \\\\\r\n&amp;T_1=16 N m \\end{aligned}[\/latex]\r\n\r\nFind 50% of the maximum torque:\r\n\r\n[latex] T_2=\\frac{T_1}{2}[\/latex]\r\n[latex]\\frac{16 Nm}{2}=8 Nm [\/latex]\r\n\r\nRearrange T<sub>2<\/sub> equation: [latex]T_{2}=\\left|F_{A}\\right| *|{r}|\u00a0 \\sin \\theta [\/latex]\r\n[latex]\\begin{aligned}&amp;\\sin \\theta=\\frac{T_{2}}{\\left| F_{A}\\right| * \\left| r \\right|} \\\\\r\n\r\n&amp;\\sin \\theta=\\frac{8 N m}{(100 N)*(0.16 m)} \\\\\r\n&amp;\\sin \\theta=0.5 \\\\\r\n&amp;\\theta=30^{\\circ}, 150^{\\circ}, etc\r\n\\end{aligned}[\/latex]\r\n\r\nTherefore, Jane could push at 30 from vertical, or 150 from vertical to create half the torque.\r\n\r\n*Interesting to note is that half the angle does not yield half the torque; the angle is 30, not 45. This is because the sine function is nonlinear.*\r\n\r\nPart C:\r\n\r\nT = 16 sin\u03b8 tells us that the angles of 0 and 180 will give us zero torque.\r\n\r\nThis makes sense given that pushing straight down on a stable pendulum will not cause the pendulum to rotate!\r\n\r\nLikewise, if you just stand on your pedals, you're providing lots of downward force, but creating zero torque since the crank arm and the direction of the force are parallel (or antiparallel)!\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nThese answers have the correct units (Nm and degrees) and are within a reasonable order of magnitude based on the given information. See logic\/explanations above for more details.\r\n\r\n<\/div>\r\n<h1>Example 1.8.5: Torque, submitted by Hamza Ben Driouech<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA person is pushing on a door with a force of 100 N. The door is at an angle \u03b1 = 45\u00b0 as shown in the sketch below.\r\n\r\na) Calculate the moment when <span style=\"text-decoration: underline\">r<\/span> is 45 cm and 75 cm.\r\n\r\nb) At what angle(s) is the moment zero? Explain why.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-1-300x260.jpg\" alt=\"A sketch of a hand pushing a door\" class=\"aligncenter wp-image-1099\" width=\"287\" height=\"249\" \/>\r\n\r\nAssumptions: Model the force as a single point load acting on the door.\r\n\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n<strong>2. Draw<\/strong>\r\n\r\nSketch:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-2-300x193.jpg\" alt=\"A sketch of the forces acting on the door at certain points\" class=\"alignnone wp-image-1100\" width=\"356\" height=\"229\" \/>\r\n\r\nFree-body diagram:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-3-300x144.jpg\" alt=\"A free-body diagram of the forces acting on the door at certain points\" class=\"alignnone wp-image-1101\" width=\"433\" height=\"208\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>F = 100N<\/li>\r\n \t<li>r<sub>1<\/sub> = 45cm<\/li>\r\n \t<li>r<sub>2<\/sub> = 75cm<\/li>\r\n \t<li>\u03b1 = 45\u00b0<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>M<sub>1<\/sub><\/li>\r\n \t<li>M<sub>2<\/sub><\/li>\r\n \t<li>Angle when M is zero<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nUse the equation below.\r\n\r\n[latex]M=|r|\\cdot|F|\\cdot\\sin\\theta[\/latex]\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nPart a)\r\n\r\nThe angle we were given is not technically the one we should use in the moment equation. The angle should be between r and F. Therefore, we have to find the new angle.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-1-solve-1-1-228x300.jpg\" alt=\"45 degree angle to the door\" class=\"alignnone wp-image-1095\" width=\"142\" height=\"187\" \/>\r\n\r\nAs shown below, the angle we find is also 45\u00b0. Now we can continue and solve for M<sub>1<\/sub> and M<sub>2<\/sub>.\r\n\r\n[latex]\\theta=90^{\\circ}-45^{\\circ}[\/latex]\r\n\r\n[latex]\\theta=45^{\\circ}[\/latex]\r\n\r\n[latex]M_1=|r_1|\\cdot|F|\\cdot\\sin\\theta\\\\M_1=0.45m\\cdot 100N\\cdot\\sin(45^{\\circ})\\\\m_1=31.82Nm\\\\\\\\M_2=|r_2|\\cdot |F|\\cdot\\sin\\theta\\\\M_2=0.75m\\cdot 100N\\cdot\\sin(45^{\\circ})\\\\M_2=53.03Nm[\/latex]\r\n\r\nPart b)\r\n\r\n[latex]M=|r|\\cdot|F|\\cdot\\sin\\theta\\\\if \\sin\\theta=0, M=0\\\\\\sin\\theta=0\\\\\\theta=\\sin^{-1}(0)\\\\\\theta=0^{\\circ}, 180^{\\circ}, 360^{\\circ}[\/latex]\r\n\r\nAnswer: The moment is zero when the angle between the force and the moment arm is 0\u00b0 or 180\u00b0 (360 would represent the same angle as 0\u00b0, as would 540\u00b0, etc.)\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nIt makes sense that the moment is zero when the door is either closed or wide open, because when we apply a force at those positions, no movement of the door is possible.\r\n\r\n<\/div>\r\n<h1>Example 1.8.6: Bonus Vector Material, Submitted by Liam Murdock<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nGeorge travelled a displacement of d<sub>g<\/sub> = [7 \u00a0 0\u00a0 8] m from his car. George\u2019s dog, named Sparky, on the other hand, travelled a displacement of ds = [0 6 6] m from George\u2019s car. George called Sparky's name, and the dog ran to George\u2019s position. It took Sparky four seconds to get there.\r\n<ol>\r\n \t<li>\u00a0What is the displacement from George to his dog?<\/li>\r\n \t<li>What is Sparky\u2019s velocity? (No need to draw.)<\/li>\r\n \t<li>What is Sparky\u2019s speed? (No need to draw.)<\/li>\r\n<\/ol>\r\n[caption id=\"attachment_1392\" align=\"aligncenter\" width=\"910\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/united-kingdom-motherwell-colley-dog.jpg\" alt=\"A dog running in the field\" class=\"wp-image-1392 size-full\" width=\"910\" height=\"607\" \/> Source: https:\/\/www.piqsels.com\/en\/public-domain-photo-oekac[\/caption]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n<strong>2. Draw<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Liam-1-Draw-300x244.jpg\" alt=\"A free-body diagram of the dog and George with dimensions shown\" class=\"alignnone wp-image-633\" width=\"341\" height=\"277\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>d<sub>g<\/sub> = [7 \u00a0 0\u00a0 8] m<\/li>\r\n \t<li>d<sub>s<\/sub> = [0\u00a0 6\u00a0 6] m<\/li>\r\n \t<li>t= 4 seconds<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>v<sub>sg<\/sub>= ?<\/li>\r\n \t<li>d<sub>sg<\/sub>= ?<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nWe are going to use vector operations (both subtraction and division), the velocity-displacement relationship, the velocity-speed relationship, and Pythagoras' theorem to solve this problem.\r\n\r\n&nbsp;\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nPart a:\r\n\r\n[latex]d_{sg}=d_{g}-d_{s}[\/latex]\r\n[latex]d_{sg}=[7\\;\\;\\;0\\;\\;\\;8]m-[0\\;\\;\\;6\\;\\;\\;6]m[\/latex]\r\n[latex]d_{sg}=[7-0\\;\\;\\;0-6\\;\\;\\;8-6]m[\/latex]\r\n[latex]d_{sg}=[7\\;\\;\\;-6\\;\\;\\;2]m[\/latex]\r\n\r\nPart b:\r\n\r\n[latex]v_{sg}=d_{sg}\/t[\/latex]\r\n[latex]v_{sg}=[7\\;\\;\\;-6\\;\\;\\;2]m\/4s[\/latex]\r\n[latex]v_{sg}=[7\/4\\;\\;\\;-6\/4\\;\\;\\;2\/4]m\/s[\/latex]\r\n[latex]v_{sg}=[1.75\\;\\;\\;-1.5\\;\\;\\;0.5]m\/s[\/latex]\r\n\r\nPart C:\r\n\r\n[latex]V_{sg}=\\sqrt{V_{sgx}^2+V_{sgy}^2+V_{sgz}^2}[\/latex]\r\n[latex]V_{sg}=\\sqrt{1.75^2+(-1.5)^2+0.5^2}[\/latex]\r\n[latex]V_{sg}=2.36m\/s[\/latex]\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nPart a:\r\n\r\nOne way to review the question is to walk through the solution verbally. Our solution shows that for Sparky to get to George, he must walk 7 m in the positive x-direction (almost out of the page), 6 m in the negative y-direction (left), and finally 2 m in the positive z-direction (up).\r\n\r\nFirstly, since the dog initially did not go in the x-direction, it makes sense that Sparky would have to copy George's exact x movement. Secondly, since George did not move in the y-direction, it would make sense that Sparky would just have to retrace his steps, and if he initially went 6 m right, he would have to go 6 m left. Thirdly, George and Sparky both went upwards, but George went 2 m higher with an altitude of 8 m compared to Sparky\u2019s 6 m, correlating to Sparky having to go positive 2 m in the z-direction to meet George.\r\n\r\nTherefore, since all the movements make sense for Sparky to meet George (using logic), the answer is proven to be right.\r\n\r\n&nbsp;\r\n\r\nParts b and c:\r\n\r\nSince B and C correlate to the same magnitude, they can be reviewed together. From a quick search, an average dog tops out at a speed of 19 miles per hour. We can convert this to SI units:\r\n\r\n[latex]\r\n\\frac{19 \\text { miles }}{1 h r}\\left(\\frac{1 \\mathrm{~km}}{0.621371 \\text { miles }}\\right)\\left(\\frac{1000 \\mathrm{~m}}{1 \\mathrm{~km}}\\right)\\left(\\frac{1 \\mathrm{hr}}{3600}\\right)=8.49 \\mathrm{~m} \/ \\mathrm{s}\r\n[\/latex]\r\n\r\nThe top speed of an average dog is 8.49 m\/s. So 2.36m\/s is approximately a quarter of the top speed of an average dog. Sparky probably was not sprinting at full speed and he could be a slower dog breed, making\u00a0 2.36m\/s a reasonable answer.\r\n<h1>Example 1.8.7: Cross Product, Submitted By Victoria Keefe<\/h1>\r\n<div class=\"textbox shaded\"><strong>1. Problem:<\/strong>\r\nUsing the cross product, find the torque on point A due to force F. Force F = (5 i - 2 j + k)N. The width of the platform is 1 meter, the height is 0.25 meters, and the length is 5 meters.<\/div>\r\n<strong>2: Sketch<\/strong>\r\n<p style=\"text-align: center\">N\/A<\/p>\r\n<p style=\"text-align: left\"><strong>\r\n3. Knowns and Unknowns<\/strong><\/p>\r\nKnowns:\r\n<ul>\r\n \t<li>F = (5 i - 2 j + k)N<\/li>\r\n \t<li>l = 5m<\/li>\r\n \t<li>w = 1m<\/li>\r\n \t<li>h = 0.25m<\/li>\r\n<\/ul>\r\nUnknown:\r\n<ul>\r\n \t<li>M<\/li>\r\n<\/ul>\r\n<strong>4: Approach<\/strong>\r\n\r\nUse the equation: M = r x F\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\n[latex]r = (-1 \\hat{i} + 0.25 \\hat{j} + 5 \\hat{k})m \\\\ \\bar{F} = (5 \\hat{i} - 2 \\hat{j} + \\hat{k})N \\\\\u00a0 \\bar{M} =[ [(0.25 \\cdot 1) - (-2 \\cdot 5)] \\hat{i} - [(-1 \\cdot 1) - (5 \\cdot 5)] \\hat{j} + [(-1 \\cdot -2) - (5 \\cdot 0.25)] \\hat{k}]Nm \\\\ \\bar{M} = (10.25 \\hat{i} + 26 \\hat{j} +0.75 \\hat{k})Nm[\/latex]\r\n\r\n<strong>6.0 Review<\/strong>\r\n\r\nThe resulting moment has reasonable values and can be verified with a cross product calculator.\r\n\r\n<\/div>","rendered":"<p style=\"text-align: justify\">Here are examples from Chapter 1 to help you understand these concepts better. These were taken from the real world and supplied by FSDE students in Summer 2021. If you&#8217;d like to submit your own examples, please send them to:<br \/>\n<a href=\"mailto:eosgood@upei.ca\"> eosgood@upei.ca<\/a><\/p>\n<h1>Example 1.8.1: Vectors, Submitted by Tyson Ashton-Losee<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>After a long day of studying, a student sitting at their computer moves the cursor from the bottom left of the screen to the top right in order to close a web browser. The computer mouse was displaced 6 cm along the x-axis and 3.5 cm along the y-axis. Draw the resultant vector and calculate the distance traveled.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem-2-259x300.jpg\" alt=\"A digital drawing of a computer mouse is shown from above with an arrow labeled r pointing diagonally outward from its center. X and Y axes are displayed in the lower left corner.\" class=\"alignright wp-image-745\" width=\"200\" height=\"232\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem-2-259x300.jpg 259w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem-2-65x75.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem-2-225x261.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem-2.jpg 307w\" sizes=\"auto, (max-width: 200px) 100vw, 200px\" \/><\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_743\" aria-describedby=\"caption-attachment-743\" style=\"width: 270px\" class=\"wp-caption alignnone\"><a href=\"https:\/\/www.flickr.com\/photos\/dejankrsmanovic\/33218207918\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem-300x200.jpg\" alt=\"A black wired computer mouse is placed on a white surface, viewed from above.\" class=\"wp-image-743\" width=\"270\" height=\"180\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem-300x200.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem-768x512.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem-65x43.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem-225x150.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem-350x233.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-Problem.jpg 1024w\" sizes=\"auto, (max-width: 270px) 100vw, 270px\" \/><\/a><figcaption id=\"caption-attachment-743\" class=\"wp-caption-text\">Source: <a href=\"https:\/\/www.flickr.com\/photos\/155403590@N07\/33218207918\">Black Computer Mouse on Table | Old dusty black computer mou\u2026 | Flickr<\/a><\/figcaption><\/figure>\n<\/div>\n<hr \/>\n<p><strong>2. Draw<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-draw-1.jpg\" alt=\"A right triangle on an XY plane with base 6 cm on the X-axis, height 3.5 cm on the Y-axis, hypotenuse labeled r, and an angle theta (\u03b8) at the origin.\" class=\"alignnone wp-image-746 size-full\" width=\"249\" height=\"137\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-draw-1.jpg 249w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-draw-1-65x36.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Tyson-1-draw-1-225x124.jpg 225w\" sizes=\"auto, (max-width: 249px) 100vw, 249px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Known:<\/p>\n<ul>\n<li>x = 6 cm<\/li>\n<li>y = 3.5 cm<\/li>\n<\/ul>\n<p>Unknown:<\/p>\n<ul>\n<li>r<\/li>\n<li>\u03b8<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Use SOH CAH TOA, first find \u03b8, then r<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>[latex]\\begin{aligned} &\\tan \\theta=\\frac{y}{x} \\\\ &\\tan \\theta=\\frac{3.5 \\mathrm{~cm}}{6 \\mathrm{~cm}} \\\\ &\\theta=\\tan ^{-1}\\left(\\frac{3.5}{6}\\right) \\\\ &\\theta=30.256^{\\circ} \\\\ &\\sin \\theta=\\frac{y}{r} \\\\ &r=\\frac{y}{\\sin \\theta} \\\\ &r=\\frac{3.5 \\mathrm{~cm}}{\\sin \\left(30.256^{\\circ}\\right)} \\\\ &r=6.946 \\mathrm{~cm} \\\\ &r=6.9 \\mathrm{~cm} \\end{aligned}[\/latex]<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>It makes sense that the angle is less than 45, because y is smaller than x. Also, if you use the Pythagorean theorem to find r, you get the same answer.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h1>Example 1.8.2: Vectors, Submitted by Brian MacDonald<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>Mark is fishing in the ocean with his favourite fishing rod. The distance between the tip of the rod and the reel is 8 ft and the length of the reel handle is 0.25 ft. The angle between the fishing rod and fishing line is 45 degrees. If Mark catches a fish when 25 ft of the fishing line is released while the fish is diving down with a force of 180 N, how much force does Mark need to apply (push down) to the reel handle to bring in the fish? Draw the position vector of the fish relative to the reel.<\/p>\n<p class=\"no-indent\">Assumptions:<\/p>\n<ul>\n<li>Mark can reel in the fish when he generates more torque with the handle than the amount of torque that the fish is applying to the reel while pulling on the line.<\/li>\n<li>The fishing line comes out of the reel in a straight line at a 90-degree angle.<\/li>\n<\/ul>\n<figure id=\"attachment_1394\" aria-describedby=\"caption-attachment-1394\" style=\"width: 1024px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deepsea-1024x768.jpg\" alt=\"A person wearing a cap and life vest is fishing from a boat, holding a bent fishing rod over the sea.\" class=\"wp-image-1394 size-large\" width=\"1024\" height=\"768\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deepsea-1024x768.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deepsea-300x225.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deepsea-768x576.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deepsea-1536x1152.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deepsea-65x49.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deepsea-225x169.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deepsea-350x263.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deepsea.jpg 1600w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption id=\"caption-attachment-1394\" class=\"wp-caption-text\">Source: <a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Deepsea.JPG\">File:Deepsea.JPG &#8211; Wikimedia Commons<\/a><\/figcaption><\/figure>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-1v2-300x185.jpg\" alt=\"A digital drawing of a fishing rod angled at 45 degrees is 8 ft tall with a 25 ft fishing line. A 180 N fish is hooked at the end. An inset shows a reel with a handle 0.25 ft from its center.\" class=\"aligncenter wp-image-582\" width=\"373\" height=\"230\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-1v2-300x185.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-1v2-65x40.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-1v2-225x139.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-1v2-350x216.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-1v2.jpg 526w\" sizes=\"auto, (max-width: 373px) 100vw, 373px\" \/><\/p>\n<\/div>\n<hr \/>\n<p><strong>2. Draw<\/strong><\/p>\n<p>Sketch:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-1-300x220.jpg\" alt=\"Diagram of a right triangle on grid paper: vertical side AC is 8 ft, horizontal side AB is 0.25 ft, angle at C is 45\u00b0, and the hypotenuse CD measures 25 ft. Points A, B, C, and D are labeled.\" class=\"alignnone wp-image-590\" width=\"345\" height=\"253\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-1-300x220.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-1-1024x750.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-1-768x563.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-1-65x48.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-1-225x165.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-1-350x257.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-1.jpg 1374w\" sizes=\"auto, (max-width: 345px) 100vw, 345px\" \/><\/p>\n<p>Free-body diagram:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-2-300x233.jpg\" alt=\"A right triangle with a green vertical side, angle \u03b8, and two red downward arrows labeled FA and FD. The x and y axes are shown pointing right and up, respectively.\" class=\"alignnone wp-image-589\" width=\"328\" height=\"255\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-2-300x233.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-2-1024x796.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-2-768x597.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-2-65x51.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-2-225x175.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-2-350x272.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Brian-Draw-2.jpg 1377w\" sizes=\"auto, (max-width: 328px) 100vw, 328px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Known:<\/p>\n<ul>\n<li>r<sub>AB<\/sub> = 0.25 ft<\/li>\n<li>r<sub>BC<\/sub> = 8 ft<\/li>\n<li>r<sub>CD<\/sub> = 25 ft<\/li>\n<li>F<sub>D<\/sub> = 180 N<\/li>\n<li>\u03b8 = 45\u00b0<\/li>\n<\/ul>\n<p>Unknown:<\/p>\n<ul>\n<li>F<sub>A<\/sub><\/li>\n<li>vector <span style=\"text-decoration: underline\">r<\/span><sub>AD<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p class=\"indent\">Convert inches to meters, then use the equation below.<\/p>\n<p>[latex]T=|r| *|F| * \\sin \\theta\\\\[\/latex]<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p class=\"indent\">Step 1: Convert inches to meters<\/p>\n<p>[latex]\\begin{align}  &25 \\mathrm{ft} * \\frac{12 \\mathrm{in}}{1 \\mathrm{ft}} * \\frac{2.54 \\mathrm{cm}}{\\operatorname{1in} } * \\frac{\\operatorname{1m}}{100 \\mathrm{cm}}=7.62 \\mathrm{m}\\\\\\\\    &\\quad\\mathrm{and}\\\\\\\\    &0.25 \\mathrm{ft} * \\frac{12\u00a0 \\mathrm{in}}{1\\mathrm{ft}} * \\frac{2.54 \\mathrm{cm}}{\\operatorname{1in} } * \\frac{\\mathrm{1m}}{100 \\mathrm{cm}}=0.0762 \\mathrm{m}\\\\    \\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p class=\"indent\">Step 2: Solve for T<sub>D<\/sub><\/p>\n<p>[latex]\\begin{aligned}&T_{D}=\\left|r_{C D}\\right| * \\left|F_{D}\\right| * \\sin \\theta\\\\    &T_{D}=(7.62 m)(180 N) \\sin \\left(45^{\\circ}\\right)\\\\  &T_{D}=969.86766 \\mathrm{Nm}    \\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p class=\"indent\">Step 3: Solve for F<sub>A<\/sub><\/p>\n<p>[latex]\\begin{aligned}&T_{A}=\\left|r_{AB}\\right| * \\left|F_{A}\\right| * \\sin \\theta\\\\    &\\text { Assume } T_{A}=T_{D}\\\\    &F_{A}=\\frac{T_{D}}{\\left|r_{AB}\\right| * \\sin \\theta} \\\\  &F_{A}=\\frac{969.86766 \\mathrm{ Nm}}{0.0762 \\mathrm{~m} * \\sin \\left(90^{\\circ}\\right)} \\\\  &F_{A}=12 727.9 \\mathrm{N} \\\\  &F_{A}=13 000 \\mathrm{N}  \\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p class=\"indent\">Vector <span style=\"text-decoration: underline\">r<\/span><sub>AD<\/sub>:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Solve-Brain-1-v4-300x221.jpg\" alt=\"A vector diagram on a grid showing points A, B, C, and D. Vectors are drawn from A to B, B to C (black), and A to D (orange), with labeled distances and axes for x and y shown at the bottom left.\" class=\"alignnone wp-image-628 size-medium\" width=\"300\" height=\"221\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Solve-Brain-1-v4-300x221.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Solve-Brain-1-v4-1024x756.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Solve-Brain-1-v4-768x567.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Solve-Brain-1-v4-65x48.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Solve-Brain-1-v4-225x166.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Solve-Brain-1-v4-350x258.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Solve-Brain-1-v4.jpg 1251w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>[latex]\\begin{aligned}  &\\vec r_{A D}=\\vec r_{A B}+\\vec r_{B C}+\\vec r_{C D} \\\\  &\\vec {r}_{A D}=\\left[\\begin{array}{c}  0.25 \\\\  0  \\end{array}\\right] f t+\\left[\\begin{array}{l}  0 \\\\  8  \\end{array}\\right] f t+\\left[\\begin{array}{c}  25 \\sin 45^{\\circ} \\\\  -25 \\cos 45^{\\circ}  \\end{array}\\right] f t \\\\  &\\vec r_{A D}=\\left[\\begin{array}{cc}  17.93 \\\\  -9.68  \\end{array}\\right] ft  \\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>Though yielding a very large number, the answer appears correct from the information given. 13,000 N of force is the amount of force Mark would need to apply to the reel handle to generate the same amount of force that the fish creates. 13,000 N in reality is too much for one to generate, but also in a real scenario, one would not have to generate the same amount of force to reel in the fish, to reel gearing, the amount of torque generated by the fishing rod itself, etc. In other words 13 000 N of force is too high in a real scenario, but with the assumptions given in the problem, the number seems reasonable. The answer also has the correct unit, N.<\/p>\n<\/div>\n<h1>Example 1.8.3: Dot product and cross product, submitted by Anonymous ENGN 1230 Student<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>[latex]\\underline{a}=[6\\;\\;\\;5\\;\\;\\;3]\\;\\;\\;\\underline{b}=[8\\;\\;\\;1\\;\\;\\;3][\/latex]<\/p>\n<p>a) Find 6a<\/p>\n<p>b) Find [latex]a\\cdot b[\/latex]<\/p>\n<p>c) Find [latex]a\\times b[\/latex]<\/p>\n<p>d) Find [latex]2a\\times b[\/latex]<\/p>\n<\/div>\n<hr \/>\n<p><strong>2. Draw<\/strong><\/p>\n<p style=\"text-align: center\">N\/A<\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Known:<\/p>\n<ul>\n<li><span style=\"text-decoration: underline\">a<\/span><\/li>\n<li><span style=\"text-decoration: underline\">b<\/span><\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>6a,<\/li>\n<li>[latex]a\\cdot b[\/latex]<\/li>\n<li>[latex]a\\times b[\/latex]<\/li>\n<li>[latex]2a\\times b[\/latex]<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Use dot product and cross product equations<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Part a:<\/p>\n<p>[latex]6\\underline{a}=6*[6\\;\\;\\;5\\;\\;\\;3]\\\\6\\underline{a}=[36\\;\\;\\;30\\;\\;\\;18][\/latex]<\/p>\n<p>Part b:<\/p>\n<p>[latex]\\underline{a}\\cdot\\underline{b}=[6\\;\\;\\;5\\;\\;\\;3]\\cdot[8\\;\\;\\;1\\;\\;\\;3]\\\\=6\\cdot 8+5\\cdot 1+3\\cdot3\\\\=48+5+9\\\\\\underline{a}\\cdot\\underline{b}=62[\/latex]<\/p>\n<p>Part C:<\/p>\n<p>[latex]\\underline{a}\\times\\underline{b}=\\begin{bmatrix} \\underline{\\hat{i}} &\\underline{\\hat{j}} & \\underline{\\hat{k}} \\\\ 6 & 5 & 3 \\\\ 8 & 1 & 3 \\end{bmatrix}\\\\(5\\cdot 3-3\\cdot 1)\\underline{\\hat{i}}-(6\\cdot 3-3\\cdot 8)\\underline{\\hat{j}}+(6\\cdot 1-5\\cdot 8)\\underline{\\hat{k}}\\\\\\underline{a}\\times\\underline{b}=12\\underline{\\hat{i}}+6\\underline{\\hat{j}}-34\\underline{\\hat{k}}[\/latex]<\/p>\n<p>Part D:<\/p>\n<p>[latex]2\\underline{a}=2*[6\\;\\;\\;5\\;\\;\\;3]=[12\\;\\;\\;10\\;\\;\\;6]\\\\\\underline{b}=[8\\;\\;\\;1\\;\\;\\;3]\\\\2\\underline{a}\\times\\underline{b} = \\begin{bmatrix} \\underline{\\hat{i}} & \\underline{\\hat{j}} & \\underline{\\hat{k}} \\\\ 12 & 10 & 6 \\\\ 8 & 1 & 3 \\end{bmatrix} \\\\=(10\\cdot 3-6\\cdot 1)\\underline{\\hat{i}}-(12\\cdot 3-6\\cdot 8)\\underline{\\hat{j}}+(12\\cdot 1-10\\cdot 8)\\underline{\\hat{k}}\\\\2\\underline{a}\\times\\underline{b}=24\\underline{\\hat{i}}+12\\underline{\\hat{j}}-68\\underline{\\hat{k}}[\/latex]<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>The answer to part d is double the answer for part c, which makes sense. It also makes sense that the answers to a, c, and d have values in three directions, while b only has magnitude.<\/p>\n<\/div>\n<h1>Example 1.8.4: Torque, Submitted by Luke McCarvill<\/h1>\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox\">\n<div class=\"textbox shaded\">\n<p>To start riding her bicycle, Jane must push down on one of her bike\u2019s pedals, which are on 16-centimetre-long crank arms. Jane can push directly downwards with her legs with a force of 100N. Jane notices that the pedal\u2019s starting position can sometimes make it more or less useful in generating torque.<\/p>\n<p>a) What is the ideal angle that Jane\u2019s bike pedal should be at in order to generate the most torque? Prove this mathematically. (Assume we only care about the very start of her very first push, and choose a reference frame for the angle that makes most sense for you.)<\/p>\n<p>b) What angle(s) should the bike pedal be at if Jane wants to generate exactly half of the maximum amount of torque?<\/p>\n<p>c) Is there any position(s) at which the pedal will create zero torque? Where are they and why?<\/p>\n<figure id=\"attachment_1393\" aria-describedby=\"caption-attachment-1393\" style=\"width: 1024px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Girl_on_a_Bike_Imagicity_116-1024x685.jpg\" alt=\"Girl on a bike\" class=\"wp-image-1393 size-large\" width=\"1024\" height=\"685\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Girl_on_a_Bike_Imagicity_116-1024x685.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Girl_on_a_Bike_Imagicity_116-300x201.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Girl_on_a_Bike_Imagicity_116-768x514.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Girl_on_a_Bike_Imagicity_116-1536x1028.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Girl_on_a_Bike_Imagicity_116-65x44.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Girl_on_a_Bike_Imagicity_116-225x151.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Girl_on_a_Bike_Imagicity_116-350x234.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Girl_on_a_Bike_Imagicity_116.jpg 1600w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption id=\"caption-attachment-1393\" class=\"wp-caption-text\">Source: https:\/\/commons.wikimedia.org\/wiki\/File:Girl_on_a_Bike_(Imagicity_116).jpg<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_1395\" aria-describedby=\"caption-attachment-1395\" style=\"width: 1024px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bicycle-3168934_1280-1024x582.png\" alt=\"A green bicycle\" class=\"wp-image-1395 size-large\" width=\"1024\" height=\"582\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bicycle-3168934_1280-1024x582.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bicycle-3168934_1280-300x171.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bicycle-3168934_1280-768x437.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bicycle-3168934_1280-65x37.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bicycle-3168934_1280-225x128.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bicycle-3168934_1280-350x199.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/bicycle-3168934_1280.png 1280w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption id=\"caption-attachment-1395\" class=\"wp-caption-text\">Source: https:\/\/pixabay.com\/illustrations\/bicycle-cycle-two-wheeler-pedal-3168934\/<\/figcaption><\/figure>\n<\/div>\n<hr \/>\n<p><strong>2. Draw<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-1-Draw--300x231.jpg\" alt=\"A16 cm horizontal bar, with a downward force labeled F\u2090 acting at the right end of the bar. The y- and x-axes are shown in the top left corner.\" class=\"alignnone wp-image-644 size-medium\" width=\"300\" height=\"231\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-1-Draw--300x231.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-1-Draw--1024x790.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-1-Draw--768x592.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-1-Draw--1536x1185.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-1-Draw--65x50.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-1-Draw--225x174.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-1-Draw--350x270.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luke-1-Draw-.jpg 1641w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<p>[latex]\\begin{aligned}  &\\vec r=\\left[\\begin{array}{c}  0.16 \\\\  0 \\\\  0  \\end{array}\\right] m \\\\  &\\vec F_{A}=\\left[\\begin{array}{c}  0 \\\\  -100 \\\\  0  \\end{array}\\right] N    \\end{aligned}[\/latex]<\/p>\n<p>Unknowns:<\/p>\n<ul>\n<li>position of <span>r<\/span> for maximum torque<\/li>\n<li>position of <span>r<\/span> for half of the maximum torque<\/li>\n<li>position of <span>r<\/span> for zero torque, and why<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>For part a), I will find a general equation for torque based on the given values in terms of \u03b8, then analyze the function for its maxima<\/p>\n<p>For part b), I will find the magnitude of 50% of maximum torque and then reverse-engineer the equation to determine what angle(s) the pedal needs to be at to satisfy the equation.<\/p>\n<p>For part c), I will look back at my equation and find when the equation equals zero, then try to understand why, given the example problem.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Part a:<\/p>\n<p>[latex]\\begin{aligned}  &T=\\left|\\vec F_{A}\\right| * \\left|\\vec r\\right| * \\sin \\theta \\\\  &T=(100 N) *(0.16 m) * \\sin \\theta \\\\  &T=16 \\sin \\theta \\mathrm{Nm} \\\\  &\\quad\\left\\{90^{\\circ}+360^{\\circ} k ; k \\in \\mathbb{Z}\\right\\}  \\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Thinking about the shape of the sine function in the first period, the maximum occurs at 90 degrees.<\/p>\n<p>You could say algebraically that the maximum is at 90, 450, 810, etc., but these angles all represent the same position on the wheel. Therefore, we will use 90.<\/p>\n<p>Part b:<\/p>\n<p>[latex]\\begin{aligned}  &T_1=\\left|F_{A}\\right| *|r| * \\sin \\theta \\\\  &T_1=(100 N)*(0.16 m)* \\sin 90^{\\circ} \\\\  &T_1=16 N m \\end{aligned}[\/latex]<\/p>\n<p>Find 50% of the maximum torque:<\/p>\n<p>[latex]T_2=\\frac{T_1}{2}[\/latex]<br \/>\n[latex]\\frac{16 Nm}{2}=8 Nm[\/latex]<\/p>\n<p>Rearrange T<sub>2<\/sub> equation: [latex]T_{2}=\\left|F_{A}\\right| *|{r}|\u00a0 \\sin \\theta[\/latex]<br \/>\n[latex]\\begin{aligned}&\\sin \\theta=\\frac{T_{2}}{\\left| F_{A}\\right| * \\left| r \\right|} \\\\    &\\sin \\theta=\\frac{8 N m}{(100 N)*(0.16 m)} \\\\  &\\sin \\theta=0.5 \\\\  &\\theta=30^{\\circ}, 150^{\\circ}, etc  \\end{aligned}[\/latex]<\/p>\n<p>Therefore, Jane could push at 30 from vertical, or 150 from vertical to create half the torque.<\/p>\n<p>*Interesting to note is that half the angle does not yield half the torque; the angle is 30, not 45. This is because the sine function is nonlinear.*<\/p>\n<p>Part C:<\/p>\n<p>T = 16 sin\u03b8 tells us that the angles of 0 and 180 will give us zero torque.<\/p>\n<p>This makes sense given that pushing straight down on a stable pendulum will not cause the pendulum to rotate!<\/p>\n<p>Likewise, if you just stand on your pedals, you&#8217;re providing lots of downward force, but creating zero torque since the crank arm and the direction of the force are parallel (or antiparallel)!<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>These answers have the correct units (Nm and degrees) and are within a reasonable order of magnitude based on the given information. See logic\/explanations above for more details.<\/p>\n<\/div>\n<h1>Example 1.8.5: Torque, submitted by Hamza Ben Driouech<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A person is pushing on a door with a force of 100 N. The door is at an angle \u03b1 = 45\u00b0 as shown in the sketch below.<\/p>\n<p>a) Calculate the moment when <span style=\"text-decoration: underline\">r<\/span> is 45 cm and 75 cm.<\/p>\n<p>b) At what angle(s) is the moment zero? Explain why.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-1-300x260.jpg\" alt=\"A sketch of a hand pushing a door\" class=\"aligncenter wp-image-1099\" width=\"287\" height=\"249\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-1-300x260.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-1-1024x886.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-1-768x665.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-1-65x56.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-1-225x195.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-1-350x303.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-1.jpg 1355w\" sizes=\"auto, (max-width: 287px) 100vw, 287px\" \/><\/p>\n<p>Assumptions: Model the force as a single point load acting on the door.<\/p>\n<\/div>\n<hr \/>\n<p><strong>2. Draw<\/strong><\/p>\n<p>Sketch:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-2-300x193.jpg\" alt=\"A sketch of the forces acting on the door at certain points\" class=\"alignnone wp-image-1100\" width=\"356\" height=\"229\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-2-300x193.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-2-1024x659.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-2-768x494.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-2-1536x989.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-2-65x42.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-2-225x145.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-2-350x225.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-2.jpg 1807w\" sizes=\"auto, (max-width: 356px) 100vw, 356px\" \/><\/p>\n<p>Free-body diagram:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-3-300x144.jpg\" alt=\"A free-body diagram of the forces acting on the door at certain points\" class=\"alignnone wp-image-1101\" width=\"433\" height=\"208\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-3-300x144.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-3-1024x493.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-3-768x370.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-3-1536x739.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-3-2048x986.jpg 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-3-65x31.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-3-225x108.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-draw-3-350x168.jpg 350w\" sizes=\"auto, (max-width: 433px) 100vw, 433px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>F = 100N<\/li>\n<li>r<sub>1<\/sub> = 45cm<\/li>\n<li>r<sub>2<\/sub> = 75cm<\/li>\n<li>\u03b1 = 45\u00b0<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>M<sub>1<\/sub><\/li>\n<li>M<sub>2<\/sub><\/li>\n<li>Angle when M is zero<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Use the equation below.<\/p>\n<p>[latex]M=|r|\\cdot|F|\\cdot\\sin\\theta[\/latex]<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Part a)<\/p>\n<p>The angle we were given is not technically the one we should use in the moment equation. The angle should be between r and F. Therefore, we have to find the new angle.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-1-solve-1-1-228x300.jpg\" alt=\"45 degree angle to the door\" class=\"alignnone wp-image-1095\" width=\"142\" height=\"187\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-1-solve-1-1-228x300.jpg 228w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-1-solve-1-1-65x86.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-1-solve-1-1-225x297.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-1-solve-1-1-350x461.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Hamza-1-solve-1-1.jpg 490w\" sizes=\"auto, (max-width: 142px) 100vw, 142px\" \/><\/p>\n<p>As shown below, the angle we find is also 45\u00b0. Now we can continue and solve for M<sub>1<\/sub> and M<sub>2<\/sub>.<\/p>\n<p>[latex]\\theta=90^{\\circ}-45^{\\circ}[\/latex]<\/p>\n<p>[latex]\\theta=45^{\\circ}[\/latex]<\/p>\n<p>[latex]M_1=|r_1|\\cdot|F|\\cdot\\sin\\theta\\\\M_1=0.45m\\cdot 100N\\cdot\\sin(45^{\\circ})\\\\m_1=31.82Nm\\\\\\\\M_2=|r_2|\\cdot |F|\\cdot\\sin\\theta\\\\M_2=0.75m\\cdot 100N\\cdot\\sin(45^{\\circ})\\\\M_2=53.03Nm[\/latex]<\/p>\n<p>Part b)<\/p>\n<p>[latex]M=|r|\\cdot|F|\\cdot\\sin\\theta\\\\if \\sin\\theta=0, M=0\\\\\\sin\\theta=0\\\\\\theta=\\sin^{-1}(0)\\\\\\theta=0^{\\circ}, 180^{\\circ}, 360^{\\circ}[\/latex]<\/p>\n<p>Answer: The moment is zero when the angle between the force and the moment arm is 0\u00b0 or 180\u00b0 (360 would represent the same angle as 0\u00b0, as would 540\u00b0, etc.)<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>It makes sense that the moment is zero when the door is either closed or wide open, because when we apply a force at those positions, no movement of the door is possible.<\/p>\n<\/div>\n<h1>Example 1.8.6: Bonus Vector Material, Submitted by Liam Murdock<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>George travelled a displacement of d<sub>g<\/sub> = [7 \u00a0 0\u00a0 8] m from his car. George\u2019s dog, named Sparky, on the other hand, travelled a displacement of ds = [0 6 6] m from George\u2019s car. George called Sparky&#8217;s name, and the dog ran to George\u2019s position. It took Sparky four seconds to get there.<\/p>\n<ol>\n<li>\u00a0What is the displacement from George to his dog?<\/li>\n<li>What is Sparky\u2019s velocity? (No need to draw.)<\/li>\n<li>What is Sparky\u2019s speed? (No need to draw.)<\/li>\n<\/ol>\n<figure id=\"attachment_1392\" aria-describedby=\"caption-attachment-1392\" style=\"width: 910px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/united-kingdom-motherwell-colley-dog.jpg\" alt=\"A dog running in the field\" class=\"wp-image-1392 size-full\" width=\"910\" height=\"607\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/united-kingdom-motherwell-colley-dog.jpg 910w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/united-kingdom-motherwell-colley-dog-300x200.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/united-kingdom-motherwell-colley-dog-768x512.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/united-kingdom-motherwell-colley-dog-65x43.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/united-kingdom-motherwell-colley-dog-225x150.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/united-kingdom-motherwell-colley-dog-350x233.jpg 350w\" sizes=\"auto, (max-width: 910px) 100vw, 910px\" \/><figcaption id=\"caption-attachment-1392\" class=\"wp-caption-text\">Source: https:\/\/www.piqsels.com\/en\/public-domain-photo-oekac<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<\/div>\n<hr \/>\n<p><strong>2. Draw<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Liam-1-Draw-300x244.jpg\" alt=\"A free-body diagram of the dog and George with dimensions shown\" class=\"alignnone wp-image-633\" width=\"341\" height=\"277\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Liam-1-Draw-300x244.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Liam-1-Draw-65x53.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Liam-1-Draw-225x183.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Liam-1-Draw-350x284.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Liam-1-Draw.jpg 543w\" sizes=\"auto, (max-width: 341px) 100vw, 341px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>d<sub>g<\/sub> = [7 \u00a0 0\u00a0 8] m<\/li>\n<li>d<sub>s<\/sub> = [0\u00a0 6\u00a0 6] m<\/li>\n<li>t= 4 seconds<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>v<sub>sg<\/sub>= ?<\/li>\n<li>d<sub>sg<\/sub>= ?<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>We are going to use vector operations (both subtraction and division), the velocity-displacement relationship, the velocity-speed relationship, and Pythagoras&#8217; theorem to solve this problem.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Part a:<\/p>\n<p>[latex]d_{sg}=d_{g}-d_{s}[\/latex]<br \/>\n[latex]d_{sg}=[7\\;\\;\\;0\\;\\;\\;8]m-[0\\;\\;\\;6\\;\\;\\;6]m[\/latex]<br \/>\n[latex]d_{sg}=[7-0\\;\\;\\;0-6\\;\\;\\;8-6]m[\/latex]<br \/>\n[latex]d_{sg}=[7\\;\\;\\;-6\\;\\;\\;2]m[\/latex]<\/p>\n<p>Part b:<\/p>\n<p>[latex]v_{sg}=d_{sg}\/t[\/latex]<br \/>\n[latex]v_{sg}=[7\\;\\;\\;-6\\;\\;\\;2]m\/4s[\/latex]<br \/>\n[latex]v_{sg}=[7\/4\\;\\;\\;-6\/4\\;\\;\\;2\/4]m\/s[\/latex]<br \/>\n[latex]v_{sg}=[1.75\\;\\;\\;-1.5\\;\\;\\;0.5]m\/s[\/latex]<\/p>\n<p>Part C:<\/p>\n<p>[latex]V_{sg}=\\sqrt{V_{sgx}^2+V_{sgy}^2+V_{sgz}^2}[\/latex]<br \/>\n[latex]V_{sg}=\\sqrt{1.75^2+(-1.5)^2+0.5^2}[\/latex]<br \/>\n[latex]V_{sg}=2.36m\/s[\/latex]<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>Part a:<\/p>\n<p>One way to review the question is to walk through the solution verbally. Our solution shows that for Sparky to get to George, he must walk 7 m in the positive x-direction (almost out of the page), 6 m in the negative y-direction (left), and finally 2 m in the positive z-direction (up).<\/p>\n<p>Firstly, since the dog initially did not go in the x-direction, it makes sense that Sparky would have to copy George&#8217;s exact x movement. Secondly, since George did not move in the y-direction, it would make sense that Sparky would just have to retrace his steps, and if he initially went 6 m right, he would have to go 6 m left. Thirdly, George and Sparky both went upwards, but George went 2 m higher with an altitude of 8 m compared to Sparky\u2019s 6 m, correlating to Sparky having to go positive 2 m in the z-direction to meet George.<\/p>\n<p>Therefore, since all the movements make sense for Sparky to meet George (using logic), the answer is proven to be right.<\/p>\n<p>&nbsp;<\/p>\n<p>Parts b and c:<\/p>\n<p>Since B and C correlate to the same magnitude, they can be reviewed together. From a quick search, an average dog tops out at a speed of 19 miles per hour. We can convert this to SI units:<\/p>\n<p>[latex]\\frac{19 \\text { miles }}{1 h r}\\left(\\frac{1 \\mathrm{~km}}{0.621371 \\text { miles }}\\right)\\left(\\frac{1000 \\mathrm{~m}}{1 \\mathrm{~km}}\\right)\\left(\\frac{1 \\mathrm{hr}}{3600}\\right)=8.49 \\mathrm{~m} \/ \\mathrm{s}[\/latex]<\/p>\n<p>The top speed of an average dog is 8.49 m\/s. So 2.36m\/s is approximately a quarter of the top speed of an average dog. Sparky probably was not sprinting at full speed and he could be a slower dog breed, making\u00a0 2.36m\/s a reasonable answer.<\/p>\n<h1>Example 1.8.7: Cross Product, Submitted By Victoria Keefe<\/h1>\n<div class=\"textbox shaded\"><strong>1. Problem:<\/strong><br \/>\nUsing the cross product, find the torque on point A due to force F. Force F = (5 i &#8211; 2 j + k)N. The width of the platform is 1 meter, the height is 0.25 meters, and the length is 5 meters.<\/div>\n<p><strong>2: Sketch<\/strong><\/p>\n<p style=\"text-align: center\">N\/A<\/p>\n<p style=\"text-align: left\"><strong><br \/>\n3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>F = (5 i &#8211; 2 j + k)N<\/li>\n<li>l = 5m<\/li>\n<li>w = 1m<\/li>\n<li>h = 0.25m<\/li>\n<\/ul>\n<p>Unknown:<\/p>\n<ul>\n<li>M<\/li>\n<\/ul>\n<p><strong>4: Approach<\/strong><\/p>\n<p>Use the equation: M = r x F<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>[latex]r = (-1 \\hat{i} + 0.25 \\hat{j} + 5 \\hat{k})m \\\\ \\bar{F} = (5 \\hat{i} - 2 \\hat{j} + \\hat{k})N \\\\\u00a0 \\bar{M} =[ [(0.25 \\cdot 1) - (-2 \\cdot 5)] \\hat{i} - [(-1 \\cdot 1) - (5 \\cdot 5)] \\hat{j} + [(-1 \\cdot -2) - (5 \\cdot 0.25)] \\hat{k}]Nm \\\\ \\bar{M} = (10.25 \\hat{i} + 26 \\hat{j} +0.75 \\hat{k})Nm[\/latex]<\/p>\n<p><strong>6.0 Review<\/strong><\/p>\n<p>The resulting moment has reasonable values and can be verified with a cross product calculator.<\/p>\n<\/div>\n","protected":false},"author":60,"menu_order":8,"template":"","meta":{"pb_show_title":"on","pb_short_title":"1.8 Examples","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-566","chapter","type-chapter","status-publish","hentry"],"part":3,"_links":{"self":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/566","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/users\/60"}],"version-history":[{"count":54,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/566\/revisions"}],"predecessor-version":[{"id":2844,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/566\/revisions\/2844"}],"part":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/566\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/media?parent=566"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapter-type?post=566"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/contributor?post=566"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/license?post=566"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}