{"id":571,"date":"2021-07-21T14:22:41","date_gmt":"2021-07-21T18:22:41","guid":{"rendered":"http:\/\/pressbooks.library.upei.ca\/statics\/?post_type=chapter&#038;p=571"},"modified":"2025-08-01T16:34:57","modified_gmt":"2025-08-01T20:34:57","slug":"4-5-examples","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.upei.ca\/statics\/chapter\/4-5-examples\/","title":{"raw":"4.5 Examples","rendered":"4.5 Examples"},"content":{"raw":"Here are examples from Chapter 4 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 the author <a href=\"mailto:eosgood@upei.ca\">eosgood@upei.ca<\/a>.\r\n<h1>Example 4.5.1: External Forces, Submitted by Elliott Fraser<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nBilly (160 lbs), Bobby (180 lbs), and Joe (145 lbs) are walking across a small bridge with a length of 11 feet. Both sides of the bridge are supported by rollers. Billy is 2 feet along the bridge, whereas Joe is 9 feet along the bridge. If the maximum force that the left side of the bridge can withstand without failing is 225 lbs, where along the bridge can Bobby stand?\r\n\r\nReal-life scenario:\r\n\r\n[caption id=\"attachment_913\" align=\"aligncenter\" width=\"361\"]<a href=\"https:\/\/flic.kr\/p\/2gAGZ17\"><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-problem-1-1-300x200.jpg\" alt=\"People walking across a bridge.\" class=\"wp-image-913\" width=\"361\" height=\"240\" \/><\/a> Source: https:\/\/www.flickr.com\/photos\/chumlee\/48306801162\/[\/caption]\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\/Elliott-1-draw1-300x203.jpg\" alt=\"A sketch of the problem\" class=\"aligncenter wp-image-916\" width=\"331\" height=\"224\" \/>\r\n\r\n&nbsp;\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\/Elliott-1-draw-3-300x136.jpg\" alt=\"A FBD of the problem.\" class=\"aligncenter wp-image-920\" width=\"394\" height=\"179\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>r<sub>Bi<\/sub> = 2 ft<\/li>\r\n \t<li>r<sub>J<\/sub> = 9 ft<\/li>\r\n \t<li>r<sub>B<\/sub> = 11 ft<\/li>\r\n \t<li>F<sub>Bi<\/sub> = 160 lb<\/li>\r\n \t<li>F<sub>J<\/sub> = 145 lb<\/li>\r\n \t<li>F<sub>Bo<\/sub> = 180 lb<\/li>\r\n \t<li>A<sub>y<\/sub> = 225 lb (since this is the maximum force without failure)<\/li>\r\n<\/ul>\r\nNote: Since the mass of the bridge was not given, we assume it is negligible and ignore it for this question.\r\n\r\nUnknown:\r\n<ul>\r\n \t<li>r<sub>Bo<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nUse equilibrium equations ( $latex \\sum\\underline{F}=0 $ , $latex \\sum\\underline{M}=0 $ ) . Use the sum of forces in y to find By; use the sum of moments to find where Bobby can stand. Solve for x.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\n[latex] \\sum F_y=0=-F_{Bi}-F_{Bo}-F_J+A_y+B_y\\\\\\\\B_y=F_{Bi}+F_{Bo}+F_J-A_y\\\\\\\\B_y=160 lb+180 lb+145 lb-225 lb\\\\\\\\\\\\B_y=260 lb[\/latex]\r\n\r\n[latex]\\sum M_A=0=-(F_{Bi})(r_{Bi})-(F_{Bo})(r_{Bo})-(F_{J})(r_{J})+(B_{y})(r_{B})\\\\r_{bo}=\\frac{-(F_{Bi})(r_{Bi})-(F_{J})(r_{J})+(B_{y})(r_{B})}{F_{Bo}}\\\\r_{Bo}=\\frac{-(160 lb)(2 ft)-(145 lb)(9 ft)+(260 lb)(11 ft)}{180 lb}[\/latex]\r\n\r\n[latex]\\underline{r_{Bo}=6.86 ft}[\/latex]\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nBobby can stand anywhere from 6.8611 ft - 11 ft from A with no problems. If Bobby were to stand between 0 ft and 6.811 ft, the left side of the bridge would fail.\r\n\r\n<\/div>\r\n<h1>Example 4.5.2: Free-Body Diagrams, Submitted by Victoria Keefe<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">A box is sitting on an inclined plane (\u03b8 = 15\u00b0) and is being pushed down the plane with a force of 20 N. Draw the free-body diagram for the box while it is in static equilibrium.<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-130931-300x249.png\" alt=\"Box on a slope with weight, normal force, and a 20\u202fN push downhill.\" class=\"aligncenter wp-image-2388 size-medium\" width=\"300\" height=\"249\" \/>Source:https:\/\/www.omnicalculator.com\/physics\/normal-force<\/div>\r\n<div><\/div>\r\n<div><\/div>\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\/Victoria-1-draw-300x188.jpg\" alt=\"A sketch of the problem\" class=\"aligncenter wp-image-1012\" width=\"215\" height=\"135\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>\u03b8 = 15\u00b0<\/li>\r\n \t<li>F<sub style=\"text-align: initial;background-color: initial\">A<\/sub><\/li>\r\n<\/ul>\r\nUnknown:\r\n<ul>\r\n \t<li>Free-body diagram of the box<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nDraw the box, then draw all forces acting on it\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-132845-274x300.png\" alt=\"FBD of the problem\" class=\"aligncenter wp-image-2393 size-medium\" width=\"274\" height=\"300\" \/>\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nAll forces acting upon the box are drawn, including weight\/gravitational force, normal force, friction, and applied forces.\r\n\r\n<\/div>\r\n<h1>Example 4.5.3: Friction, Submitted by Deanna Malone<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"200\"]<a href=\"https:\/\/static.thenounproject.com\/png\/2745226-200.png\"><img src=\"https:\/\/static.thenounproject.com\/png\/2745226-200.png\" alt=\"Man Kicking Box Icons - Download Free Vector Icons | Noun Project\" class=\"n3VNCb\" data-noaft=\"1\" width=\"200\" height=\"200\" \/><\/a> Source: https:\/\/static.thenounproject.com\/png\/2745226-200.png[\/caption]\r\n\r\nA box is being pushed along level ground with a force of 150 N at an angle of 30\u00b0 with the horizontal. The mass of the box is 12 kg.\r\n\r\na) What is the normal force between the box and the floor?\r\n\r\nb) What is the coefficient of friction between the box and the floor?\r\n\r\n<\/div>\r\n<strong>2. Draw\r\n<\/strong>\r\n\r\nSketch:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-1-300x175.jpg\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-753\" width=\"317\" height=\"185\" \/>\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\/Deanna-1-Draw-2-300x194.jpg\" alt=\"A FBD of the problem.\" class=\"aligncenter wp-image-754\" width=\"324\" height=\"209\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>F<sub>A<\/sub> = 150 N<\/li>\r\n \t<li>\u03b8 = 30\u00b0<\/li>\r\n \t<li>m = 12 kg<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>F<sub>N<\/sub><\/li>\r\n \t<li>\u03bc<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nUse equilibrium equations ( $latex \\sum\\underline{F}=0 $ , $latex \\sum\\underline{M}=0 $ ), SOH CAH TOA, friction equation\r\n\r\n<strong>5. Analysis\r\n<\/strong>\r\n\r\nPart a:\r\n<p class=\"indent\">Find F<sub>g<\/sub>:<\/p>\r\n[latex]Fg=m\\cdot g\\\\Fg=(12kg)(9.81m\/s^2)\\\\\\\\Fg=117.72N[\/latex]\r\n<p class=\"indent\">Find F<sub>N<\/sub> using equilibrium equations:<\/p>\r\n[latex]\\sum Fy=0=F_N-F_g-F_A\\sin 30^{\\circ}\\\\0=F_N-117.72N-150N\\cdot \\sin 30^{\\circ}\\\\F_N=117.72+150N\\cdot\\sin 30^{\\circ}\\\\\\\\\\underline{F_N=192.7 N}[\/latex]\r\n\r\nPart b:\r\n<p class=\"indent\">Find two equations for F<sub>f<\/sub>, set them equal to each other, and solve for \u03bc<\/p>\r\n\r\n<div>[latex]<\/div>\r\n<div>\\begin{align*}\r\n\\sum F_x &amp;= 0 = F_A \\cos 30^\\circ - F_f \\\\\r\n0 &amp;= 150\\,N \\cdot \\cos 30^\\circ - F_f\r\n\\end{align*}<\/div>\r\n<div>[\/latex]<\/div>\r\n<span>[latex]F_f=150N\\cdot\\cos 30^{\\circ}\\\\F_f=\\mu\\cdot F_N\\\\150N\\cdot\\cos 30^{\\circ}=\\mu\\cdot F_N[\/latex]<\/span>\r\n\r\n<span>[latex]\\mu=\\frac{150N\\cdot\\cos 30^{\\circ}}{F_N}\\\\\\mu=\\frac{150N\\cdot\\cos 30^{\\circ}}{192.72N}=0.67405[\/latex]<\/span>\r\n\r\n<span>[latex]\\underline{\\mu=0.67}[\/latex]<\/span>\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nF<sub>N<\/sub>, F<sub>g<\/sub>, and the y component of F<sub>A<\/sub> are the only forces in the y direction so it makes sense that they need to equal zero for the equilibrium equations. F<sub>N<\/sub> is the only positive force in the y direction, so it makes sense that it equals the magnitude of the other two put together.\r\n\r\nThe coefficient found between the box and the floor is reasonable as it is less than 1, and it's reasonable for a box on the floor. For example, if the box was wood and the floor was wood, the coefficient of static friction would be anywhere from 0.5-0.7, so having a coefficient of friction equal to 0.67 makes sense.\r\n\r\n<\/div>\r\n<h1>Example 4.5.4: Friction, Submitted by Dhruvil Kanani<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-1-251x300.jpg\" alt=\"A hand holding a brick against the wall.\" class=\"alignright wp-image-1024\" width=\"188\" height=\"225\" \/>\r\n\r\nA person is trying to prevent a brick from sliding on a rough vertical surface by applying force in the direction of wall. Assuming the coefficient of static friction is 0.49 and the mass of the brick is 5 kg,\r\n<ul>\r\n \t<li>a) Determine the minimum force required to prevent the brick from slipping.<\/li>\r\n \t<li>b) Find the distributed load or intensity if the length of the person\u2019s hands from the tip of his fingers to their wrist is 16 cm.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<strong>2. Draw<\/strong>\r\n\r\nSketch:\r\n\r\nFree-body diagram (box):\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-2-300x230.jpg\" alt=\"FBD of the problem.\" class=\"aligncenter wp-image-1025 size-medium\" width=\"300\" height=\"230\" \/>\r\n\r\nFree-body diagram (distributed load):\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-3-300x140.jpg\" alt=\"FBD of the problem (distributed load)\" class=\"alignnone wp-image-1026 size-medium\" width=\"300\" height=\"140\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>Mass of brick (m) = 5kg<\/li>\r\n \t<li>Coefficient of friction (\u03bc<sub>1<\/sub>) = 0.49<\/li>\r\n \t<li>Acceleration due to gravity (g) = 9.81m\/s2<\/li>\r\n \t<li>Length of the hand (L) = 16 cm<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>Applied force (F<sub>A<\/sub>)<\/li>\r\n \t<li>Intensity (w)<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nUse equilibrium equations ( $latex \\sum\\underline{F}=0 $ , $latex \\sum\\underline{M}=0 $ ), equations for F<sub>g<\/sub> and F<sub>f<\/sub> (see below).\r\n\r\n[latex]F_g=m g\\\\F_f=\\mu F_N[\/latex]\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nPart a:\r\n\r\n[latex]F_g=m\\cdot g\\\\F_g=5kg\\cdot 9.81m\/s^2\\\\F_g=49.05N[\/latex]\r\n\r\n[latex]\\sum F_y=0=-F_g+F_f\\\\F_f=F_g\\\\F_f=49.05N[\/latex]\r\n\r\n[latex]F_f=\\mu_1\u00a0 F_N\\\\F_N=\\frac{F_f}{\\mu_1}\\\\F_N=\\frac{49.05N}{0.49}[\/latex]\r\n\r\n[latex]\\underline{F_N=100.1N}[\/latex]\r\n\r\nPart b:\r\n\r\n[latex]\\sum F_x=0=F_N-F_A\\\\F_A=F_N\\\\F_A=100.1N[\/latex]\r\n\r\n[latex]w=\\frac{F_A}{L}\\\\w=\\frac{100.1N}{(16cm\\times\\frac{1m}{100cm})}[\/latex]\r\n\r\n[latex]\\underline{w=625.64N\/m}[\/latex]\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nIt makes sense that the applied force is larger than the gravitational force. It also makes sense that the normal and applied forces are equal, since they are the only forces in the x direction (the same goes for the friction and gravitational forces).\r\n\r\n<\/div>\r\n&nbsp;\r\n<h1>Example 4.5.5: Friction, Submitted by Emma Christensen<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\n&nbsp;\r\n\r\nA ball is suspended by two ropes and rests on an inclined surface at an angle of 15\u00b0. Rope A pulls on the ball with a force of 200 N, and rope B has a force of 150 N. They each have angles of 20\u00b0 and 60\u00b0, respectively, from the inclined surface plane, as shown in the image below.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio-300x179.png\" alt=\"A 20kg ball on a 15\u00b0 slope with two angled forces: 200N at 20\u00b0 and 150N at 60\u00b0.\" width=\"298\" height=\"178\" class=\"alignnone wp-image-2140\" \/>\r\n\r\na) Draw a free-body diagram of the ball\r\n\r\nb) Find the friction force\r\n\r\n<\/div>\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\/balloninclineplane.drawio-300x179.png\" alt=\"A sketch of the problem\" width=\"300\" height=\"179\" class=\"alignnone wp-image-2140 size-medium\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n<ul>\r\n \t<li>\u03b8<sub>A<\/sub> = 20\u00b0<\/li>\r\n \t<li>\u03b8<sub>B<\/sub> = 60\u00b0<\/li>\r\n \t<li>\u03b8<sub>C<\/sub> = 15\u00b0<\/li>\r\n \t<li>m = 20 kg<\/li>\r\n \t<li>F<sub>A<\/sub> = 200 N<\/li>\r\n \t<li>F<sub>B<\/sub> = 150 N<\/li>\r\n<\/ul>\r\nUnknown:\r\n<ul>\r\n \t<li>F<sub>f<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nDraw the ball, then add forces. Use equilibrium equations ( $latex \\sum\\underline{F}=0 $ , $latex \\sum\\underline{M}=0 $ ) to find the friction force.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nPart a:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/fEmma-2-solve-1-300x209-1.jpg\" alt=\"FBD of the problem.\" width=\"300\" height=\"209\" class=\"aligncenter wp-image-2141 size-full\" \/>\r\n\r\nPart b:\r\n\r\n<em>Step 1: Find F<sub>G<\/sub><\/em>\r\n\r\n[latex]F_G=m g\\\\F_G=20kg\\cdot 9.81m\/s^2\\\\F_G=196.2N[\/latex]\r\n\r\n<em>Step 2: Find the x-component of F<sub>G<\/sub><\/em>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-3-300x211.jpg\" alt=\"Force diagram showing a triangle split into components F_gx and F_gy from force F_g at a 15\u00b0 angle.\" class=\"alignnone wp-image-1135 size-medium\" width=\"300\" height=\"211\" \/>\r\n\r\n[latex]F_{GX}=F_G\\sin(15^{\\circ})\\\\F_{GX}=196.2N\\cdot\\sin(15^{\\circ})\\\\F_{GX}=50.78N[\/latex]\r\n\r\n<em>Step 3: Find the x-component of F<sub>A<\/sub><\/em>\r\n\r\n[latex]\\cos(20^{\\circ})=\\frac{F_AX}{F_{A}}\\\\F_{AX}=F_A\\cdot(cos(20^{\\circ}))\\\\F_{AX}=200N\\cdot(\\cos(20^{\\circ}))\\\\F_{AX}=187.938N[\/latex]\r\n\r\n<em>Step 4: Find the x-component of F<sub>B<\/sub><\/em>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-7-300x152.jpg\" alt=\"Force F_B split into components F_Bx and F_By at a 60\u00b0 angle.\" class=\"alignnone wp-image-1139 size-medium\" width=\"300\" height=\"152\" \/>\r\n\r\n[latex]F_{BX}=F_B\\cdot(\\cos(60^{\\circ}))\\\\F_{BX}=150N\\cdot(\\cos(60^{\\circ}))\\\\F_{BX}=75N[\/latex]\r\n\r\n<em>Step 5: Sum forces in the x-direction to find the frictional force<\/em>\r\n\r\n[latex]\\sum F_x=0=-F_f+F_{BX}-F_{AX}+F_{GX}\\\\F_f=-F_{AX}+F_{BX}+F_{GX}\\\\F_f=-187.938N+75N+50.78N[\/latex]\r\n\r\n[latex]\\underline{F_f=-62.158N}[\/latex]\r\n\r\nBecause the frictional force is negative, that means the frictional force actually acts in the opposite direction, so the friction is keeping the ball from going up the plane.\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nThe units of F<sub>f<\/sub> are newtons, which makes sense because it is a force. It also makes sense that F<sub>Ax<\/sub> is larger than F<sub>Gx<\/sub> and F<sub>Bx<\/sub>.\r\n\r\n<\/div>\r\n<h1>Example 4.5.6: Friction, Submitted by Riley Fitzpatrick<\/h1>\r\n<div class=\"textbox\">\r\n<ol>\r\n \t<li><strong>Problem<\/strong><\/li>\r\n<\/ol>\r\n<div class=\"textbox shaded\">\r\n\r\nA 6m trailer contains 500 kg, which is distributed evenly across its bed, meaning that the weight of the load is acting at the centroid. The trailer has an axle at either end of the bed and a 3m hitch. A truck is towing the load with a force of 5000N.\r\n\r\na. Calculate the weight of the load.\r\n\r\nb. Calculate the total normal force acting on the wheels.\r\n\r\nc. Calculate the frictional force acting on the wheels.\r\n\r\nd. Calculate the coefficient of friction between the wheels and the asphalt.\r\n\r\ne. Given that the static coefficient of friction between the wheels and the asphalt is 0.72. Is the trailer about to slip?\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/riley-chap-3-image-e1667953494391.jpg\" alt=\"Truck unloading logs.\" class=\"alignnone wp-image-1982\" width=\"548\" height=\"269\" \/>\r\n\r\nAn image relatable to the problem. https:\/\/en.wikipedia.org\/wiki\/Sidelifter#\/media\/File:Sidelifter_in_forest.jpg\r\n<div><\/div>\r\n<\/div>\r\n<strong>2. Draw<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-11-e1667948472188.png\" alt=\"FBD of the problem.\" class=\"alignnone wp-image-1980 size-full\" width=\"524\" height=\"304\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>m = 500 kg<\/li>\r\n \t<li>P = 5000 N<\/li>\r\n \t<li>9 = 9.81 m\/s^2<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>w<\/li>\r\n \t<li>N<\/li>\r\n \t<li>F<sub>F <\/sub><\/li>\r\n \t<li><span>\u00b5<\/span><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n<ul>\r\n \t<li>Calculate w using the formula w = mg<\/li>\r\n \t<li>Calculate N using [latex]\\sum{F_{y}} = 0[\/latex]<\/li>\r\n \t<li>Calculate F<sub>F<\/sub> using [latex]\\sum{F_{x}} = 0[\/latex]<\/li>\r\n \t<li>Calculate <span>\u00b5 using [latex]\\mu= \\dfrac{F_{F}}{N}[\/latex]<\/span><\/li>\r\n \t<li>Compare <span>\u00b5 to \u00b5<sub>s<\/sub><\/span><\/li>\r\n<\/ul>\r\n<strong>5.\u00a0 Analysis<\/strong>\r\n\r\n<span style=\"text-decoration: underline\">Calculating w<\/span>\r\n\r\n[latex]w = m\\cdot g [\/latex]\r\n\r\n[latex]\\:\\:\\:=500kg \\cdot 9.81 m\/s^2 = 4905 N[\/latex]\r\n\r\n<strong>w = 4905 N<\/strong><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-11-e1667948472188.png\" alt=\"Free-body diagram of the trailer supported by two rollers with vertical and horizontal forces.\" class=\"alignright wp-image-1980\" width=\"590\" height=\"342\" \/>\r\n\r\n<span style=\"text-decoration: underline\">Calculating N\u00a0<\/span>\r\n\r\n[latex]\\sum {F_y} = N-w =0[\/latex]\r\n\r\n[latex]N=w [\/latex]\r\n\r\n<strong>N = 4905 N<\/strong>\r\n\r\nTherefore, N<sub>A\u00a0<\/sub>= N<sub>B<\/sub> = 4905N\/2 = 2452.5N\r\n\r\n<span style=\"text-decoration: underline\">Finding F<sub>F<\/sub><\/span>\r\n\r\n[latex]\\sum {F_x}= P-F_{F} = 0[\/latex]\r\n\r\n[latex]P = F_{F}[\/latex]\r\n\r\n<strong>F<sub>F\u00a0<\/sub> = 5000 N<\/strong>\r\n\r\n<span style=\"text-decoration: underline\">Finding \u00b5<\/span>\r\n\r\n[latex]\\mu = \\dfrac {F_{F}}{N} = \\dfrac{5000 N }{4905 N}[\/latex]\r\n\r\n<strong>\u00b5 = 1.019\u00a0<\/strong>\r\n\r\n<span style=\"text-decoration: underline\">Comparing \u00b5 and \u00b5<sub>s<\/sub><\/span>\r\n\r\n<strong>1.019&gt; 0.72<\/strong>\r\n\r\n<strong>\u00b5 &gt; \u00b5<sub>s<\/sub><\/strong>\r\n\r\nThus, the trailer is is in motion, i.e. sliding along the surface.\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nFrom initial inspection of the FBD, it is clear that w (acting downwards) is opposite to the normal forces (acting upwards). The sum of the normal forces would not exceed w.\r\n\r\nF<sub>F\u00a0<\/sub>should be equal and opposite to P.\r\n\r\n<\/div>\r\n<h1>Example 4.5.7: Slip or Tip, Submitted by Luke McCarvill<\/h1>\r\n<div class=\"textbox\">\r\n<ol>\r\n \t<li><strong>Problem\u00a0<\/strong><\/li>\r\n<\/ol>\r\n<div class=\"textbox shaded\">\r\n\r\n&nbsp;\r\n\r\n1.8 metres tall Spiderman with strong spider-like webs is attempting to tip over a 10 m tall and 2.5 m wide shipping container sitting upright. Given that the coefficient of static friction between the bottom of the shipping container and the ground is 0.2, and the mass of the container is 2000 kg, how much tension must Spiderman impart on his web in order to barely tip over the container towards him if he is 20 metres away from the bottom of the container?\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_1866\" align=\"alignnone\" width=\"780\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/luke-McCarvil-spiderman.jpg\" alt=\"Spiderman pulling a 2000 kg container using a web slanted upward over 20 m.\" class=\"wp-image-1866\" width=\"780\" height=\"435\" \/> Source:https:\/\/commons.wikimedia.org\/wiki\/File:Container_%E3%80%90_22G1_%E3%80%91_GVTU_201551(1)_%E3%80%90_Container_pictures_taken_in_Japan_%E3%80%91.jpg[\/caption]\r\n\r\n<\/div>\r\n<strong>2. Draw<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/luke-McCarvil-spiderman-1.jpg\" alt=\"FBD of the problem.\" class=\"alignnone wp-image-1868 size-full\" width=\"734\" height=\"451\" \/>\r\n\r\n<strong>3. Knowns and Unknowns\u00a0<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>m = 2000 kg<\/li>\r\n \t<li>h<sub>c<\/sub>\u00a0= 10 m<\/li>\r\n \t<li>h<sub>s<\/sub>\u00a0= 1.8 m<\/li>\r\n \t<li>d = 20 m<\/li>\r\n \t<li>w =2.5<\/li>\r\n \t<li><span>\u03bc\u00a0 = 0.2\u00a0<\/span><\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>T<sub>x<\/sub><\/li>\r\n \t<li>T<sub>y\u00a0<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nComponentizing forces in the x-y direction and using the summation of moments and forces to be equal to zero.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nBecause mass is given force by gravity, [latex]F_{g }= m\\times g=19,620 N[\/latex]\r\n\r\nSince the tension applied by Spider-Man is a force with both x and y components, we would need angle [latex]\\theta[\/latex]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Luke-McCarvill-spidreman-answer.drawio.png\" alt=\"Spiderman pulling a 2000\u202fkg container using web at an angle \u03b8, with a 20\u202fm horizontal distance and height difference (hc \u2212 hS).\" class=\"aligncenter wp-image-1872 size-full\" width=\"561\" height=\"351\" \/>\r\n\r\n[latex]\r\n\\theta = \\tan^{-1} \\left( \\frac{h_c - h_s}{20} \\right) = 22.3^\\circ\r\n[\/latex]\r\n\r\nAnalyzing the force diagram:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-135055-228x300.png\" alt=\"Force diagram.\" class=\"aligncenter wp-image-2396\" width=\"356\" height=\"468\" \/>\r\n\r\nLooking at the force diagram for the shipping container, it is understood that if the container were to tip towards Spiderman, the sum of moments on point B would equal zero. It is noticeable that among the forces on the container( frictional force, normal force, force of gravity, and tension) only force of gravity and x component of tension contribute to the moment at point B. Therefore, the following equation for the moment at B is derived.\r\n\r\n[latex]M_{B} =\\: -h_{c}\\times T_{x}+\\frac{w}{2}\\times F_{g} \\:=0\\\\ \\kern\u00a0 \u00a0 1pc =10m \\times T_{x}+\\frac{2.5}{2}\\times 19,620 N[\/latex]\r\n\r\nthus,\u00a0 \u00a0[latex]T_{x} = 2,452.5 N[\/latex]\r\n\r\nAlso\r\n\r\n[latex]T_{x} = T\\cos{\\theta}[\/latex]\r\n\r\n[latex]\\lvert T\\rvert = \\frac{T_{x}}{\\cos{\\theta}}=2,651 N[\/latex]\r\n\r\nThe required tension to tip the container is 2,651 N\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nWe can ensure that the container will tip, not slip, by examining the role of frictional force in the net force. If the maximum frictional force is greater than the pulling force, the container would tip. To verify this, we analyze the below given force diagram to find the frictional force.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke-McCarvill-spidreman-answer-2.jpg\" alt=\"FBD of the problem solved.\" class=\"alignnone wp-image-1894 size-full\" width=\"762\" height=\"581\" \/>\r\n\r\n[latex]\r\n\\begin{aligned}\r\nF_y &amp;= -T_y + F_n - F_g \\\\\r\n0 &amp;= -T \\sin{\\theta} + F_n - F_g \\\\\r\n0 &amp;= -2651 \\cdot \\sin{22.3^\\circ} + F_n - 19620 \\\\\r\nF_n &amp;= 20626\\,N\r\n\\end{aligned}\r\n[\/latex]\r\n\r\nnow [latex]F_{f}=\\mu \\times F_{n} \\\\ F_{f} = 4,125 N [\/latex]\r\n\r\nBecause F<sub>f<\/sub> &gt; T<sub>x<\/sub>, the container will tip.\r\n\r\n<\/div>\r\n<h1>Example 4.5.8: Friction, Submitted by <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Michael Oppong-Ampomah&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:156,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:1}\">Michael Oppong-Ampomah<\/span><\/h1>\r\n<div class=\"textbox\">\r\n<ol>\r\n \t<li><strong>Problem\u00a0<\/strong><\/li>\r\n<\/ol>\r\n<div class=\"textbox shaded\">\r\n\r\nFind the frictional force between th<span style=\"text-align: initial;font-size: 1em\">e ramp and the cart of mass 70 kg. A person pulls <\/span><span style=\"text-align: initial;font-size: 1em\">the cart with a rope with a tension of\u00a0 300 N, and the ramp is tilted 30 degrees.\u00a0<\/span>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/michael-chapter-4.png\" alt=\"A person unloading boxes from a ramp.\" class=\"aligncenter wp-image-1985\" style=\"margin-bottom: 4.44444em;font-size: 1em\" width=\"368\" height=\"488\" \/>\r\n\r\nA relatable scenario is given in the problem. https:\/\/en.wikipedia.org\/wiki\/Baggage_handler#\/media\/File:BaggageHandlerDetroit7August2006.png\r\n\r\n<\/div>\r\n<strong>2. Draw<\/strong>\u00a0<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Screenshot-12-e1667957018705.png\" alt=\"FBD of the problem.\" class=\"aligncenter wp-image-1986 size-full\" style=\"font-size: 1em\" width=\"464\" height=\"384\" \/>\r\n\r\n<strong>3. Knowns and Unknowns\u00a0<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>m= 70 kg<\/li>\r\n \t<li>P = 300N<\/li>\r\n \t<li>\u03b8 = 30\u00b0<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>F<sub>g<\/sub><\/li>\r\n \t<li>F<sub>F<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach\u00a0<\/strong>\r\n<ul>\r\n \t<li>Find F<sub>g\u00a0<\/sub>by using F<sub>g<\/sub> = m<span>\u2a2fg.<\/span><\/li>\r\n \t<li>Figuring out forces in the x and y components<\/li>\r\n \t<li>Apply the equilibrium equation to figure out the rest of the forces.<\/li>\r\n<\/ul>\r\n<strong>5. Analysis\u00a0<\/strong>\r\n\r\n<span style=\"text-decoration: underline\">Finding F<sub>g<\/sub><\/span>\r\n\r\n[latex]F_{g} = m \\cdot g = 70 kg \\cdot 9.81m\/s^2 = 686.7 N[\/latex]\r\n\r\n<span style=\"text-decoration: underline\">Forces in x and y components<\/span>\r\n\r\nSince P, F<sub>F<\/sub>, and N are in the direction of either x or y, only F<sub>g <\/sub>has x and y components.\r\n\r\n[latex]F_{gx} =686.7\\cdot \\sin{30}[\/latex]\r\n\r\n[latex]F_{gx} =343.35N[\/latex]\r\n\r\n<span style=\"text-decoration: underline\">Equilibrium equation in the x-axis\u00a0\u00a0<\/span>\r\n\r\n\u2211F<sub>x<\/sub> = 300N - F<sub>F<\/sub>\u2212 686.7N sin(30\u00b0) = 0\r\n\r\nF<sub>F<\/sub>=-43.35N\r\n\r\nSince this value is negative, it just means that the drawing has the force of friction acting in the opposite direction.\r\n\r\n<strong>6. Review\u00a0<\/strong>\r\n\r\nThe answer obtained makes sense as the force of friction should be small and be working against the cart sliding down the incline. The force of gravity in the x component is only slightly larger than that of the person pulling the cart so the force of friction being only 43.35N makes sense.\r\n\r\n<\/div>\r\n<h1>Example 4.5.9: Finding the coefficient of friction, 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\nYou are developing an accessible gaming platform for a charity to allow everyone to play their favourite games.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4.png\" alt=\"The accessible gaming platform.\" class=\"alignnone wp-image-1994 size-full\" width=\"613\" height=\"244\" \/>\r\nThe platform can be simplified for the calculations into a box 0.8 m long and 0.15m high. The platform has a mass of 4 kg and rests on the two thighs of a person, each 0.2m wide and each 0.1 m away from the edges of the platform. Say the charity would want to make sure that a force of 10 N (from any of the top corners of the platform, going parallel to the surface) will not push it over. Make proper assumptions if needed and determine a coefficient of friction so that a material for the base can be chosen.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_1.png\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-1995\" width=\"360\" height=\"156\" \/>\r\n\r\n<\/div>\r\n<strong>2. Sketch<\/strong>\r\n\r\n&nbsp;\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_2-1.png\" alt=\"FBD of the problem.\" class=\"alignnone wp-image-1998\" width=\"604\" height=\"215\" \/>\r\n<strong>3. Knows and Unknowns:<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>F<sub>A<\/sub> = 10N<\/li>\r\n \t<li>D<sub>2<\/sub> = 0.2m<\/li>\r\n \t<li>D<sub>3<\/sub> = 0.1m<\/li>\r\n \t<li>L = 0.8m<\/li>\r\n \t<li>H = 0.15m<\/li>\r\n \t<li>m = 4kg<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>D<sub>1<\/sub> = ?<\/li>\r\n \t<li>F<sub>g<\/sub> = ?<\/li>\r\n \t<li>F<sub>fa<\/sub> = ?<\/li>\r\n \t<li>F<sub>fb<\/sub> = ?<\/li>\r\n \t<li>N<sub>A<\/sub> = ?<\/li>\r\n \t<li>N<sub>B<\/sub> = ?<\/li>\r\n \t<li>\u03bc = ?<\/li>\r\n<\/ul>\r\n<strong>4. Approach:<\/strong>\r\n\r\nDefine an origin point to measure all forces from to simplify the problem.\r\nDetermine D<sub>1<\/sub> and the coordinates of all forces measured from the origin O.\r\nFind F<sub>g<\/sub>.\r\nUse the sum of moments around A to find NB.\r\nUse the sum of forces in the Y direction to find N<sub>A<\/sub>.\r\nUse the sum of forces in the X direction to find \u03bc.\r\n\r\n<strong>5. Analysis:<\/strong>\r\nDefining the origin as the bottom left corner of the gaming platform:\r\n[latex] N_A\\ is\\ measured\\ D_3 + \\frac{1}{2} D_2\\ from\\ the\\ origin. \\\\ N_B\\ is\\ measured\\ D_3 + D_2 + D_1 + \\frac{1}{2} D_2 \\ from\\ the\\ origin. \\\\ F_g\\ is\\ measured\\ D_3 + D_2 + \\frac{1}{2} D_1 \\ from\\ the\\ origin.[\/latex]\r\nTo find D<sub>1<\/sub>:\r\n[latex]L = 2D_3 + 2D_2 + D_1 \\\\ D_1 = L - 2D_3 - 2D_2 \\\\ D_1 = 0.8m - 2 \\cdot 0.1m - 2 \\cdot 0.2m \\\\ D_1 = 0.2m[\/latex]\r\nTo find\u00a0F<sub>g<\/sub>:\r\n[latex]F_g = mg = 4kg \\cdot 9.81 m\/s^2 = 39.24N[\/latex]\r\nThe coordinates of these forces can now be shown as:\r\n[latex]N_A\\ is\\ at\\ [0.2, 0]m \\\\ N_B\\ is\\ at\\ [0.6, 0]m \\\\ F_g\\ is\\ at\\ [0.4, 0]m \\\\ F_A\\ is\\ at\\ [0, 0.15]m[\/latex]\r\n\r\nThe sum of moments around point A can now be done to find N<sub>B<\/sub>. The distance between each force and point A is the difference between their coordinates.\r\n[latex]\\sum M_A = 0 = -F_g\u00a0 \\cdot (0.4m - 0.2m) + N_B \\cdot (0.6m - 0.2m) - F_A (0.15m - 0m) \\\\ N_B = \\frac{F_g \\cdot 0.2m + F_A \\cdot 0.15m}{0.4m} \\\\ N_B = \\frac{39.24N \\cdot 0.2m + 10N \\cdot 0.15m}{0.4m} \\\\ N_B = 23.37 N[\/latex]\r\nThe Sum of forces in the Y direction can be used to find N<sub>A<\/sub> now:\r\n[latex]\\sum F_y = 0 = -F_g + N_A + N_B \\\\ N_A = F_g - N_B = 39.24N - 23.37N \\\\ N_A = 15.87N[\/latex]\r\nBefore the sum of forces in the X is done, it must be noted that F<sub>fa<\/sub> and F<sub>fb<\/sub> are equivalent to \u03bcN<sub>A<\/sub> and \u03bcN<sub>B<\/sub>, respectively.\r\n[latex]\\sum F_x = 0 = F_A - F_{fa} - F_{fb} = F_A - \\mu \\cdot N_A - \\mu \\cdot N_B \\\\ -F_A = -\\mu \\cdot N_A - \\mu \\cdot N_B \\\\ \\frac{-F_A}{\\mu} = -N_A - N_B \\\\ \\frac{1}{\\mu} = \\frac{-N_A - N_B}{-F_A} \\\\ \\mu = \\frac{-F_A}{-N_A - N_B} \\\\ \\mu = \\frac{-10N}{-15.87N - 23.37N} \\\\ \\mu = 0.25[\/latex]\r\n\r\n<strong>6. Review:<\/strong>\r\n\r\nTo review this solution, the value for N<sub>A<\/sub> can be solved for by taking the moment around point B.\r\n[latex] \\sum M_B = 0 = F_g \\cdot (0.6m - 0.4m) - N_A \\cdot (0.6m - 0.2m) - F_A \\cdot (0.15m) \\\\ N_A = \\frac{(F_g \\cdot 0.2m ) - (F_A \\cdot 0.15m)}{0.4m} \\\\ N_A = \\frac{(39.24N \\cdot 0.2m) - (10N \\cdot 0.15)}{0.15m}{0.4m} \\\\ N_A = 15.87N[\/latex]\r\nThe value found is the same value originally found, providing evidence that there was no error in the calculation.\r\n\r\n<\/div>","rendered":"<p>Here are examples from Chapter 4 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 the author <a href=\"mailto:eosgood@upei.ca\">eosgood@upei.ca<\/a>.<\/p>\n<h1>Example 4.5.1: External Forces, Submitted by Elliott Fraser<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>Billy (160 lbs), Bobby (180 lbs), and Joe (145 lbs) are walking across a small bridge with a length of 11 feet. Both sides of the bridge are supported by rollers. Billy is 2 feet along the bridge, whereas Joe is 9 feet along the bridge. If the maximum force that the left side of the bridge can withstand without failing is 225 lbs, where along the bridge can Bobby stand?<\/p>\n<p>Real-life scenario:<\/p>\n<figure id=\"attachment_913\" aria-describedby=\"caption-attachment-913\" style=\"width: 361px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/flic.kr\/p\/2gAGZ17\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-problem-1-1-300x200.jpg\" alt=\"People walking across a bridge.\" class=\"wp-image-913\" width=\"361\" height=\"240\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-problem-1-1-300x200.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-problem-1-1-768x512.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-problem-1-1-65x43.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-problem-1-1-225x150.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-problem-1-1-350x233.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-problem-1-1.jpg 1024w\" sizes=\"auto, (max-width: 361px) 100vw, 361px\" \/><\/a><figcaption id=\"caption-attachment-913\" class=\"wp-caption-text\">Source: https:\/\/www.flickr.com\/photos\/chumlee\/48306801162\/<\/figcaption><\/figure>\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\/Elliott-1-draw1-300x203.jpg\" alt=\"A sketch of the problem\" class=\"aligncenter wp-image-916\" width=\"331\" height=\"224\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw1-300x203.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw1-1024x694.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw1-768x520.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw1-1536x1041.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw1-65x44.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw1-225x152.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw1-350x237.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw1.jpg 1550w\" sizes=\"auto, (max-width: 331px) 100vw, 331px\" \/><\/p>\n<p>&nbsp;<\/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\/Elliott-1-draw-3-300x136.jpg\" alt=\"A FBD of the problem.\" class=\"aligncenter wp-image-920\" width=\"394\" height=\"179\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw-3-300x136.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw-3-1024x463.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw-3-768x347.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw-3-1536x694.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw-3-65x29.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw-3-225x102.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw-3-350x158.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Elliott-1-draw-3.jpg 1774w\" sizes=\"auto, (max-width: 394px) 100vw, 394px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>r<sub>Bi<\/sub> = 2 ft<\/li>\n<li>r<sub>J<\/sub> = 9 ft<\/li>\n<li>r<sub>B<\/sub> = 11 ft<\/li>\n<li>F<sub>Bi<\/sub> = 160 lb<\/li>\n<li>F<sub>J<\/sub> = 145 lb<\/li>\n<li>F<sub>Bo<\/sub> = 180 lb<\/li>\n<li>A<sub>y<\/sub> = 225 lb (since this is the maximum force without failure)<\/li>\n<\/ul>\n<p>Note: Since the mass of the bridge was not given, we assume it is negligible and ignore it for this question.<\/p>\n<p>Unknown:<\/p>\n<ul>\n<li>r<sub>Bo<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Use equilibrium equations ( [latex]\\sum\\underline{F}=0[\/latex] , [latex]\\sum\\underline{M}=0[\/latex] ) . Use the sum of forces in y to find By; use the sum of moments to find where Bobby can stand. Solve for x.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>[latex]\\sum F_y=0=-F_{Bi}-F_{Bo}-F_J+A_y+B_y\\\\\\\\B_y=F_{Bi}+F_{Bo}+F_J-A_y\\\\\\\\B_y=160 lb+180 lb+145 lb-225 lb\\\\\\\\\\\\B_y=260 lb[\/latex]<\/p>\n<p>[latex]\\sum M_A=0=-(F_{Bi})(r_{Bi})-(F_{Bo})(r_{Bo})-(F_{J})(r_{J})+(B_{y})(r_{B})\\\\r_{bo}=\\frac{-(F_{Bi})(r_{Bi})-(F_{J})(r_{J})+(B_{y})(r_{B})}{F_{Bo}}\\\\r_{Bo}=\\frac{-(160 lb)(2 ft)-(145 lb)(9 ft)+(260 lb)(11 ft)}{180 lb}[\/latex]<\/p>\n<p>[latex]\\underline{r_{Bo}=6.86 ft}[\/latex]<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>Bobby can stand anywhere from 6.8611 ft &#8211; 11 ft from A with no problems. If Bobby were to stand between 0 ft and 6.811 ft, the left side of the bridge would fail.<\/p>\n<\/div>\n<h1>Example 4.5.2: Free-Body Diagrams, Submitted by Victoria Keefe<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">A box is sitting on an inclined plane (\u03b8 = 15\u00b0) and is being pushed down the plane with a force of 20 N. Draw the free-body diagram for the box while it is in static equilibrium.<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-130931-300x249.png\" alt=\"Box on a slope with weight, normal force, and a 20\u202fN push downhill.\" class=\"aligncenter wp-image-2388 size-medium\" width=\"300\" height=\"249\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-130931-300x249.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-130931-65x54.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-130931-225x187.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-130931-350x291.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-130931.png 585w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/>Source:https:\/\/www.omnicalculator.com\/physics\/normal-force<\/div>\n<div><\/div>\n<div><\/div>\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\/Victoria-1-draw-300x188.jpg\" alt=\"A sketch of the problem\" class=\"aligncenter wp-image-1012\" width=\"215\" height=\"135\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Victoria-1-draw-300x188.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Victoria-1-draw-1024x641.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Victoria-1-draw-768x481.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Victoria-1-draw-65x41.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Victoria-1-draw-225x141.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Victoria-1-draw-350x219.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Victoria-1-draw.jpg 1033w\" sizes=\"auto, (max-width: 215px) 100vw, 215px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>\u03b8 = 15\u00b0<\/li>\n<li>F<sub style=\"text-align: initial;background-color: initial\">A<\/sub><\/li>\n<\/ul>\n<p>Unknown:<\/p>\n<ul>\n<li>Free-body diagram of the box<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Draw the box, then draw all forces acting on it<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-132845-274x300.png\" alt=\"FBD of the problem\" class=\"aligncenter wp-image-2393 size-medium\" width=\"274\" height=\"300\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-132845-274x300.png 274w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-132845-65x71.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-132845-225x247.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-132845-350x384.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-132845.png 694w\" sizes=\"auto, (max-width: 274px) 100vw, 274px\" \/><\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>All forces acting upon the box are drawn, including weight\/gravitational force, normal force, friction, and applied forces.<\/p>\n<\/div>\n<h1>Example 4.5.3: Friction, Submitted by Deanna Malone<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<figure style=\"width: 200px\" class=\"wp-caption alignright\"><a href=\"https:\/\/static.thenounproject.com\/png\/2745226-200.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/static.thenounproject.com\/png\/2745226-200.png\" alt=\"Man Kicking Box Icons - Download Free Vector Icons | Noun Project\" class=\"n3VNCb\" data-noaft=\"1\" width=\"200\" height=\"200\" \/><\/a><figcaption class=\"wp-caption-text\">Source: https:\/\/static.thenounproject.com\/png\/2745226-200.png<\/figcaption><\/figure>\n<p>A box is being pushed along level ground with a force of 150 N at an angle of 30\u00b0 with the horizontal. The mass of the box is 12 kg.<\/p>\n<p>a) What is the normal force between the box and the floor?<\/p>\n<p>b) What is the coefficient of friction between the box and the floor?<\/p>\n<\/div>\n<p><strong>2. Draw<br \/>\n<\/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\/Deanna-1-Draw-1-300x175.jpg\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-753\" width=\"317\" height=\"185\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-1-300x175.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-1-1024x596.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-1-768x447.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-1-1536x894.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-1-65x38.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-1-225x131.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-1-350x204.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-1.jpg 1994w\" sizes=\"auto, (max-width: 317px) 100vw, 317px\" \/><\/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\/Deanna-1-Draw-2-300x194.jpg\" alt=\"A FBD of the problem.\" class=\"aligncenter wp-image-754\" width=\"324\" height=\"209\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-2-300x194.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-2-1024x663.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-2-768x497.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-2-1536x994.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-2-2048x1325.jpg 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-2-65x42.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-2-225x146.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Deanna-1-Draw-2-350x226.jpg 350w\" sizes=\"auto, (max-width: 324px) 100vw, 324px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>F<sub>A<\/sub> = 150 N<\/li>\n<li>\u03b8 = 30\u00b0<\/li>\n<li>m = 12 kg<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>F<sub>N<\/sub><\/li>\n<li>\u03bc<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Use equilibrium equations ( [latex]\\sum\\underline{F}=0[\/latex] , [latex]\\sum\\underline{M}=0[\/latex] ), SOH CAH TOA, friction equation<\/p>\n<p><strong>5. Analysis<br \/>\n<\/strong><\/p>\n<p>Part a:<\/p>\n<p class=\"indent\">Find F<sub>g<\/sub>:<\/p>\n<p>[latex]Fg=m\\cdot g\\\\Fg=(12kg)(9.81m\/s^2)\\\\\\\\Fg=117.72N[\/latex]<\/p>\n<p class=\"indent\">Find F<sub>N<\/sub> using equilibrium equations:<\/p>\n<p>[latex]\\sum Fy=0=F_N-F_g-F_A\\sin 30^{\\circ}\\\\0=F_N-117.72N-150N\\cdot \\sin 30^{\\circ}\\\\F_N=117.72+150N\\cdot\\sin 30^{\\circ}\\\\\\\\\\underline{F_N=192.7 N}[\/latex]<\/p>\n<p>Part b:<\/p>\n<p class=\"indent\">Find two equations for F<sub>f<\/sub>, set them equal to each other, and solve for \u03bc<\/p>\n<div>[latex]<\/div>\n<div>\\begin{align*}  \\sum F_x &= 0 = F_A \\cos 30^\\circ - F_f \\\\  0 &= 150\\,N \\cdot \\cos 30^\\circ - F_f  \\end{align*}<\/div>\n<div>[\/latex]<\/div>\n<p><span>[latex]F_f=150N\\cdot\\cos 30^{\\circ}\\\\F_f=\\mu\\cdot F_N\\\\150N\\cdot\\cos 30^{\\circ}=\\mu\\cdot F_N[\/latex]<\/span><\/p>\n<p><span>[latex]\\mu=\\frac{150N\\cdot\\cos 30^{\\circ}}{F_N}\\\\\\mu=\\frac{150N\\cdot\\cos 30^{\\circ}}{192.72N}=0.67405[\/latex]<\/span><\/p>\n<p><span>[latex]\\underline{\\mu=0.67}[\/latex]<\/span><\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>F<sub>N<\/sub>, F<sub>g<\/sub>, and the y component of F<sub>A<\/sub> are the only forces in the y direction so it makes sense that they need to equal zero for the equilibrium equations. F<sub>N<\/sub> is the only positive force in the y direction, so it makes sense that it equals the magnitude of the other two put together.<\/p>\n<p>The coefficient found between the box and the floor is reasonable as it is less than 1, and it&#8217;s reasonable for a box on the floor. For example, if the box was wood and the floor was wood, the coefficient of static friction would be anywhere from 0.5-0.7, so having a coefficient of friction equal to 0.67 makes sense.<\/p>\n<\/div>\n<h1>Example 4.5.4: Friction, Submitted by Dhruvil Kanani<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-1-251x300.jpg\" alt=\"A hand holding a brick against the wall.\" class=\"alignright wp-image-1024\" width=\"188\" height=\"225\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-1-251x300.jpg 251w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-1-65x78.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-1-225x268.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-1-350x418.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-1.jpg 684w\" sizes=\"auto, (max-width: 188px) 100vw, 188px\" \/><\/p>\n<p>A person is trying to prevent a brick from sliding on a rough vertical surface by applying force in the direction of wall. Assuming the coefficient of static friction is 0.49 and the mass of the brick is 5 kg,<\/p>\n<ul>\n<li>a) Determine the minimum force required to prevent the brick from slipping.<\/li>\n<li>b) Find the distributed load or intensity if the length of the person\u2019s hands from the tip of his fingers to their wrist is 16 cm.<\/li>\n<\/ul>\n<\/div>\n<p><strong>2. Draw<\/strong><\/p>\n<p>Sketch:<\/p>\n<p>Free-body diagram (box):<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-2-300x230.jpg\" alt=\"FBD of the problem.\" class=\"aligncenter wp-image-1025 size-medium\" width=\"300\" height=\"230\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-2-300x230.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-2-1024x784.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-2-768x588.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-2-65x50.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-2-225x172.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-2-350x268.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-2.jpg 1089w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>Free-body diagram (distributed load):<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-3-300x140.jpg\" alt=\"FBD of the problem (distributed load)\" class=\"alignnone wp-image-1026 size-medium\" width=\"300\" height=\"140\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-3-300x140.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-3-1024x479.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-3-768x359.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-3-65x30.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-3-225x105.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-3-350x164.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Dhruvil-1-draw-3.jpg 1031w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>Mass of brick (m) = 5kg<\/li>\n<li>Coefficient of friction (\u03bc<sub>1<\/sub>) = 0.49<\/li>\n<li>Acceleration due to gravity (g) = 9.81m\/s2<\/li>\n<li>Length of the hand (L) = 16 cm<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>Applied force (F<sub>A<\/sub>)<\/li>\n<li>Intensity (w)<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Use equilibrium equations ( [latex]\\sum\\underline{F}=0[\/latex] , [latex]\\sum\\underline{M}=0[\/latex] ), equations for F<sub>g<\/sub> and F<sub>f<\/sub> (see below).<\/p>\n<p>[latex]F_g=m g\\\\F_f=\\mu F_N[\/latex]<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Part a:<\/p>\n<p>[latex]F_g=m\\cdot g\\\\F_g=5kg\\cdot 9.81m\/s^2\\\\F_g=49.05N[\/latex]<\/p>\n<p>[latex]\\sum F_y=0=-F_g+F_f\\\\F_f=F_g\\\\F_f=49.05N[\/latex]<\/p>\n<p>[latex]F_f=\\mu_1\u00a0 F_N\\\\F_N=\\frac{F_f}{\\mu_1}\\\\F_N=\\frac{49.05N}{0.49}[\/latex]<\/p>\n<p>[latex]\\underline{F_N=100.1N}[\/latex]<\/p>\n<p>Part b:<\/p>\n<p>[latex]\\sum F_x=0=F_N-F_A\\\\F_A=F_N\\\\F_A=100.1N[\/latex]<\/p>\n<p>[latex]w=\\frac{F_A}{L}\\\\w=\\frac{100.1N}{(16cm\\times\\frac{1m}{100cm})}[\/latex]<\/p>\n<p>[latex]\\underline{w=625.64N\/m}[\/latex]<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>It makes sense that the applied force is larger than the gravitational force. It also makes sense that the normal and applied forces are equal, since they are the only forces in the x direction (the same goes for the friction and gravitational forces).<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h1>Example 4.5.5: Friction, Submitted by Emma Christensen<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>&nbsp;<\/p>\n<p>A ball is suspended by two ropes and rests on an inclined surface at an angle of 15\u00b0. Rope A pulls on the ball with a force of 200 N, and rope B has a force of 150 N. They each have angles of 20\u00b0 and 60\u00b0, respectively, from the inclined surface plane, as shown in the image below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio-300x179.png\" alt=\"A 20kg ball on a 15\u00b0 slope with two angled forces: 200N at 20\u00b0 and 150N at 60\u00b0.\" width=\"298\" height=\"178\" class=\"alignnone wp-image-2140\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio-300x179.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio-768x457.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio-65x39.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio-225x134.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio-350x208.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio.png 815w\" sizes=\"auto, (max-width: 298px) 100vw, 298px\" \/><\/p>\n<p>a) Draw a free-body diagram of the ball<\/p>\n<p>b) Find the friction force<\/p>\n<\/div>\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\/balloninclineplane.drawio-300x179.png\" alt=\"A sketch of the problem\" width=\"300\" height=\"179\" class=\"alignnone wp-image-2140 size-medium\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio-300x179.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio-768x457.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio-65x39.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio-225x134.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio-350x208.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/balloninclineplane.drawio.png 815w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<ul>\n<li>\u03b8<sub>A<\/sub> = 20\u00b0<\/li>\n<li>\u03b8<sub>B<\/sub> = 60\u00b0<\/li>\n<li>\u03b8<sub>C<\/sub> = 15\u00b0<\/li>\n<li>m = 20 kg<\/li>\n<li>F<sub>A<\/sub> = 200 N<\/li>\n<li>F<sub>B<\/sub> = 150 N<\/li>\n<\/ul>\n<p>Unknown:<\/p>\n<ul>\n<li>F<sub>f<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Draw the ball, then add forces. Use equilibrium equations ( [latex]\\sum\\underline{F}=0[\/latex] , [latex]\\sum\\underline{M}=0[\/latex] ) to find the friction force.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Part a:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/fEmma-2-solve-1-300x209-1.jpg\" alt=\"FBD of the problem.\" width=\"300\" height=\"209\" class=\"aligncenter wp-image-2141 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/fEmma-2-solve-1-300x209-1.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/fEmma-2-solve-1-300x209-1-65x45.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/fEmma-2-solve-1-300x209-1-225x157.jpg 225w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>Part b:<\/p>\n<p><em>Step 1: Find F<sub>G<\/sub><\/em><\/p>\n<p>[latex]F_G=m g\\\\F_G=20kg\\cdot 9.81m\/s^2\\\\F_G=196.2N[\/latex]<\/p>\n<p><em>Step 2: Find the x-component of F<sub>G<\/sub><\/em><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-3-300x211.jpg\" alt=\"Force diagram showing a triangle split into components F_gx and F_gy from force F_g at a 15\u00b0 angle.\" class=\"alignnone wp-image-1135 size-medium\" width=\"300\" height=\"211\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-3-300x211.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-3-768x539.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-3-65x46.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-3-225x158.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-3-350x246.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-3.jpg 931w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>[latex]F_{GX}=F_G\\sin(15^{\\circ})\\\\F_{GX}=196.2N\\cdot\\sin(15^{\\circ})\\\\F_{GX}=50.78N[\/latex]<\/p>\n<p><em>Step 3: Find the x-component of F<sub>A<\/sub><\/em><\/p>\n<p>[latex]\\cos(20^{\\circ})=\\frac{F_AX}{F_{A}}\\\\F_{AX}=F_A\\cdot(cos(20^{\\circ}))\\\\F_{AX}=200N\\cdot(\\cos(20^{\\circ}))\\\\F_{AX}=187.938N[\/latex]<\/p>\n<p><em>Step 4: Find the x-component of F<sub>B<\/sub><\/em><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-7-300x152.jpg\" alt=\"Force F_B split into components F_Bx and F_By at a 60\u00b0 angle.\" class=\"alignnone wp-image-1139 size-medium\" width=\"300\" height=\"152\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-7-300x152.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-7-1024x520.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-7-768x390.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-7-65x33.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-7-225x114.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-7-350x178.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Emma-2-solve-7.jpg 1281w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>[latex]F_{BX}=F_B\\cdot(\\cos(60^{\\circ}))\\\\F_{BX}=150N\\cdot(\\cos(60^{\\circ}))\\\\F_{BX}=75N[\/latex]<\/p>\n<p><em>Step 5: Sum forces in the x-direction to find the frictional force<\/em><\/p>\n<p>[latex]\\sum F_x=0=-F_f+F_{BX}-F_{AX}+F_{GX}\\\\F_f=-F_{AX}+F_{BX}+F_{GX}\\\\F_f=-187.938N+75N+50.78N[\/latex]<\/p>\n<p>[latex]\\underline{F_f=-62.158N}[\/latex]<\/p>\n<p>Because the frictional force is negative, that means the frictional force actually acts in the opposite direction, so the friction is keeping the ball from going up the plane.<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>The units of F<sub>f<\/sub> are newtons, which makes sense because it is a force. It also makes sense that F<sub>Ax<\/sub> is larger than F<sub>Gx<\/sub> and F<sub>Bx<\/sub>.<\/p>\n<\/div>\n<h1>Example 4.5.6: Friction, Submitted by Riley Fitzpatrick<\/h1>\n<div class=\"textbox\">\n<ol>\n<li><strong>Problem<\/strong><\/li>\n<\/ol>\n<div class=\"textbox shaded\">\n<p>A 6m trailer contains 500 kg, which is distributed evenly across its bed, meaning that the weight of the load is acting at the centroid. The trailer has an axle at either end of the bed and a 3m hitch. A truck is towing the load with a force of 5000N.<\/p>\n<p>a. Calculate the weight of the load.<\/p>\n<p>b. Calculate the total normal force acting on the wheels.<\/p>\n<p>c. Calculate the frictional force acting on the wheels.<\/p>\n<p>d. Calculate the coefficient of friction between the wheels and the asphalt.<\/p>\n<p>e. Given that the static coefficient of friction between the wheels and the asphalt is 0.72. Is the trailer about to slip?<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/riley-chap-3-image-e1667953494391.jpg\" alt=\"Truck unloading logs.\" class=\"alignnone wp-image-1982\" width=\"548\" height=\"269\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/riley-chap-3-image-e1667953494391.jpg 228w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/riley-chap-3-image-e1667953494391-65x32.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/riley-chap-3-image-e1667953494391-225x111.jpg 225w\" sizes=\"auto, (max-width: 548px) 100vw, 548px\" \/><\/p>\n<p>An image relatable to the problem. https:\/\/en.wikipedia.org\/wiki\/Sidelifter#\/media\/File:Sidelifter_in_forest.jpg<\/p>\n<div><\/div>\n<\/div>\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\/2022\/02\/Screenshot-11-e1667948472188.png\" alt=\"FBD of the problem.\" class=\"alignnone wp-image-1980 size-full\" width=\"524\" height=\"304\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-11-e1667948472188.png 524w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-11-e1667948472188-300x174.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-11-e1667948472188-65x38.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-11-e1667948472188-225x131.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-11-e1667948472188-350x203.png 350w\" sizes=\"auto, (max-width: 524px) 100vw, 524px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>m = 500 kg<\/li>\n<li>P = 5000 N<\/li>\n<li>9 = 9.81 m\/s^2<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>w<\/li>\n<li>N<\/li>\n<li>F<sub>F <\/sub><\/li>\n<li><span>\u00b5<\/span><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<ul>\n<li>Calculate w using the formula w = mg<\/li>\n<li>Calculate N using [latex]\\sum{F_{y}} = 0[\/latex]<\/li>\n<li>Calculate F<sub>F<\/sub> using [latex]\\sum{F_{x}} = 0[\/latex]<\/li>\n<li>Calculate <span>\u00b5 using [latex]\\mu= \\dfrac{F_{F}}{N}[\/latex]<\/span><\/li>\n<li>Compare <span>\u00b5 to \u00b5<sub>s<\/sub><\/span><\/li>\n<\/ul>\n<p><strong>5.\u00a0 Analysis<\/strong><\/p>\n<p><span style=\"text-decoration: underline\">Calculating w<\/span><\/p>\n<p>[latex]w = m\\cdot g[\/latex]<\/p>\n<p>[latex]\\:\\:\\:=500kg \\cdot 9.81 m\/s^2 = 4905 N[\/latex]<\/p>\n<p><strong>w = 4905 N<\/strong><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-11-e1667948472188.png\" alt=\"Free-body diagram of the trailer supported by two rollers with vertical and horizontal forces.\" class=\"alignright wp-image-1980\" width=\"590\" height=\"342\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-11-e1667948472188.png 524w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-11-e1667948472188-300x174.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-11-e1667948472188-65x38.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-11-e1667948472188-225x131.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-11-e1667948472188-350x203.png 350w\" sizes=\"auto, (max-width: 590px) 100vw, 590px\" \/><\/p>\n<p><span style=\"text-decoration: underline\">Calculating N\u00a0<\/span><\/p>\n<p>[latex]\\sum {F_y} = N-w =0[\/latex]<\/p>\n<p>[latex]N=w[\/latex]<\/p>\n<p><strong>N = 4905 N<\/strong><\/p>\n<p>Therefore, N<sub>A\u00a0<\/sub>= N<sub>B<\/sub> = 4905N\/2 = 2452.5N<\/p>\n<p><span style=\"text-decoration: underline\">Finding F<sub>F<\/sub><\/span><\/p>\n<p>[latex]\\sum {F_x}= P-F_{F} = 0[\/latex]<\/p>\n<p>[latex]P = F_{F}[\/latex]<\/p>\n<p><strong>F<sub>F\u00a0<\/sub> = 5000 N<\/strong><\/p>\n<p><span style=\"text-decoration: underline\">Finding \u00b5<\/span><\/p>\n<p>[latex]\\mu = \\dfrac {F_{F}}{N} = \\dfrac{5000 N }{4905 N}[\/latex]<\/p>\n<p><strong>\u00b5 = 1.019\u00a0<\/strong><\/p>\n<p><span style=\"text-decoration: underline\">Comparing \u00b5 and \u00b5<sub>s<\/sub><\/span><\/p>\n<p><strong>1.019&gt; 0.72<\/strong><\/p>\n<p><strong>\u00b5 &gt; \u00b5<sub>s<\/sub><\/strong><\/p>\n<p>Thus, the trailer is is in motion, i.e. sliding along the surface.<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>From initial inspection of the FBD, it is clear that w (acting downwards) is opposite to the normal forces (acting upwards). The sum of the normal forces would not exceed w.<\/p>\n<p>F<sub>F\u00a0<\/sub>should be equal and opposite to P.<\/p>\n<\/div>\n<h1>Example 4.5.7: Slip or Tip, Submitted by Luke McCarvill<\/h1>\n<div class=\"textbox\">\n<ol>\n<li><strong>Problem\u00a0<\/strong><\/li>\n<\/ol>\n<div class=\"textbox shaded\">\n<p>&nbsp;<\/p>\n<p>1.8 metres tall Spiderman with strong spider-like webs is attempting to tip over a 10 m tall and 2.5 m wide shipping container sitting upright. Given that the coefficient of static friction between the bottom of the shipping container and the ground is 0.2, and the mass of the container is 2000 kg, how much tension must Spiderman impart on his web in order to barely tip over the container towards him if he is 20 metres away from the bottom of the container?<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_1866\" aria-describedby=\"caption-attachment-1866\" style=\"width: 780px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/luke-McCarvil-spiderman.jpg\" alt=\"Spiderman pulling a 2000 kg container using a web slanted upward over 20 m.\" class=\"wp-image-1866\" width=\"780\" height=\"435\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/luke-McCarvil-spiderman.jpg 680w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/luke-McCarvil-spiderman-300x167.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/luke-McCarvil-spiderman-65x36.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/luke-McCarvil-spiderman-225x125.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/luke-McCarvil-spiderman-350x195.jpg 350w\" sizes=\"auto, (max-width: 780px) 100vw, 780px\" \/><figcaption id=\"caption-attachment-1866\" class=\"wp-caption-text\">Source:https:\/\/commons.wikimedia.org\/wiki\/File:Container_%E3%80%90_22G1_%E3%80%91_GVTU_201551(1)_%E3%80%90_Container_pictures_taken_in_Japan_%E3%80%91.jpg<\/figcaption><\/figure>\n<\/div>\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\/2022\/05\/luke-McCarvil-spiderman-1.jpg\" alt=\"FBD of the problem.\" class=\"alignnone wp-image-1868 size-full\" width=\"734\" height=\"451\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/luke-McCarvil-spiderman-1.jpg 734w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/luke-McCarvil-spiderman-1-300x184.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/luke-McCarvil-spiderman-1-65x40.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/luke-McCarvil-spiderman-1-225x138.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/luke-McCarvil-spiderman-1-350x215.jpg 350w\" sizes=\"auto, (max-width: 734px) 100vw, 734px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns\u00a0<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>m = 2000 kg<\/li>\n<li>h<sub>c<\/sub>\u00a0= 10 m<\/li>\n<li>h<sub>s<\/sub>\u00a0= 1.8 m<\/li>\n<li>d = 20 m<\/li>\n<li>w =2.5<\/li>\n<li><span>\u03bc\u00a0 = 0.2\u00a0<\/span><\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>T<sub>x<\/sub><\/li>\n<li>T<sub>y\u00a0<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Componentizing forces in the x-y direction and using the summation of moments and forces to be equal to zero.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Because mass is given force by gravity, [latex]F_{g }= m\\times g=19,620 N[\/latex]<\/p>\n<p>Since the tension applied by Spider-Man is a force with both x and y components, we would need angle [latex]\\theta[\/latex]<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Luke-McCarvill-spidreman-answer.drawio.png\" alt=\"Spiderman pulling a 2000\u202fkg container using web at an angle \u03b8, with a 20\u202fm horizontal distance and height difference (hc \u2212 hS).\" class=\"aligncenter wp-image-1872 size-full\" width=\"561\" height=\"351\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Luke-McCarvill-spidreman-answer.drawio.png 561w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Luke-McCarvill-spidreman-answer.drawio-300x188.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Luke-McCarvill-spidreman-answer.drawio-65x41.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Luke-McCarvill-spidreman-answer.drawio-225x141.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Luke-McCarvill-spidreman-answer.drawio-350x219.png 350w\" sizes=\"auto, (max-width: 561px) 100vw, 561px\" \/><\/p>\n<p>[latex]\\theta = \\tan^{-1} \\left( \\frac{h_c - h_s}{20} \\right) = 22.3^\\circ[\/latex]<\/p>\n<p>Analyzing the force diagram:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-135055-228x300.png\" alt=\"Force diagram.\" class=\"aligncenter wp-image-2396\" width=\"356\" height=\"468\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-135055-228x300.png 228w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-135055-65x86.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-135055-225x296.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-135055-350x461.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-27-135055.png 677w\" sizes=\"auto, (max-width: 356px) 100vw, 356px\" \/><\/p>\n<p>Looking at the force diagram for the shipping container, it is understood that if the container were to tip towards Spiderman, the sum of moments on point B would equal zero. It is noticeable that among the forces on the container( frictional force, normal force, force of gravity, and tension) only force of gravity and x component of tension contribute to the moment at point B. Therefore, the following equation for the moment at B is derived.<\/p>\n<p>[latex]M_{B} =\\: -h_{c}\\times T_{x}+\\frac{w}{2}\\times F_{g} \\:=0\\\\ \\kern\u00a0 \u00a0 1pc =10m \\times T_{x}+\\frac{2.5}{2}\\times 19,620 N[\/latex]<\/p>\n<p>thus,\u00a0 \u00a0[latex]T_{x} = 2,452.5 N[\/latex]<\/p>\n<p>Also<\/p>\n<p>[latex]T_{x} = T\\cos{\\theta}[\/latex]<\/p>\n<p>[latex]\\lvert T\\rvert = \\frac{T_{x}}{\\cos{\\theta}}=2,651 N[\/latex]<\/p>\n<p>The required tension to tip the container is 2,651 N<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>We can ensure that the container will tip, not slip, by examining the role of frictional force in the net force. If the maximum frictional force is greater than the pulling force, the container would tip. To verify this, we analyze the below given force diagram to find the frictional force.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke-McCarvill-spidreman-answer-2.jpg\" alt=\"FBD of the problem solved.\" class=\"alignnone wp-image-1894 size-full\" width=\"762\" height=\"581\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke-McCarvill-spidreman-answer-2.jpg 762w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke-McCarvill-spidreman-answer-2-300x229.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke-McCarvill-spidreman-answer-2-65x50.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke-McCarvill-spidreman-answer-2-225x172.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke-McCarvill-spidreman-answer-2-350x267.jpg 350w\" sizes=\"auto, (max-width: 762px) 100vw, 762px\" \/><\/p>\n<p>[latex]\\begin{aligned}  F_y &= -T_y + F_n - F_g \\\\  0 &= -T \\sin{\\theta} + F_n - F_g \\\\  0 &= -2651 \\cdot \\sin{22.3^\\circ} + F_n - 19620 \\\\  F_n &= 20626\\,N  \\end{aligned}[\/latex]<\/p>\n<p>now [latex]F_{f}=\\mu \\times F_{n} \\\\ F_{f} = 4,125 N[\/latex]<\/p>\n<p>Because F<sub>f<\/sub> &gt; T<sub>x<\/sub>, the container will tip.<\/p>\n<\/div>\n<h1>Example 4.5.8: Friction, Submitted by <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Michael Oppong-Ampomah&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:156,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:{&quot;1&quot;:2,&quot;2&quot;:0}},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:1}\">Michael Oppong-Ampomah<\/span><\/h1>\n<div class=\"textbox\">\n<ol>\n<li><strong>Problem\u00a0<\/strong><\/li>\n<\/ol>\n<div class=\"textbox shaded\">\n<p>Find the frictional force between th<span style=\"text-align: initial;font-size: 1em\">e ramp and the cart of mass 70 kg. A person pulls <\/span><span style=\"text-align: initial;font-size: 1em\">the cart with a rope with a tension of\u00a0 300 N, and the ramp is tilted 30 degrees.\u00a0<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/michael-chapter-4.png\" alt=\"A person unloading boxes from a ramp.\" class=\"aligncenter wp-image-1985\" style=\"margin-bottom: 4.44444em;font-size: 1em\" width=\"368\" height=\"488\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/michael-chapter-4.png 220w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/michael-chapter-4-65x86.png 65w\" sizes=\"auto, (max-width: 368px) 100vw, 368px\" \/><\/p>\n<p>A relatable scenario is given in the problem. https:\/\/en.wikipedia.org\/wiki\/Baggage_handler#\/media\/File:BaggageHandlerDetroit7August2006.png<\/p>\n<\/div>\n<p><strong>2. Draw<\/strong>\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Screenshot-12-e1667957018705.png\" alt=\"FBD of the problem.\" class=\"aligncenter wp-image-1986 size-full\" style=\"font-size: 1em\" width=\"464\" height=\"384\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Screenshot-12-e1667957018705.png 464w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Screenshot-12-e1667957018705-300x248.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Screenshot-12-e1667957018705-65x54.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Screenshot-12-e1667957018705-225x186.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Screenshot-12-e1667957018705-350x290.png 350w\" sizes=\"auto, (max-width: 464px) 100vw, 464px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns\u00a0<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>m= 70 kg<\/li>\n<li>P = 300N<\/li>\n<li>\u03b8 = 30\u00b0<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>F<sub>g<\/sub><\/li>\n<li>F<sub>F<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach\u00a0<\/strong><\/p>\n<ul>\n<li>Find F<sub>g\u00a0<\/sub>by using F<sub>g<\/sub> = m<span>\u2a2fg.<\/span><\/li>\n<li>Figuring out forces in the x and y components<\/li>\n<li>Apply the equilibrium equation to figure out the rest of the forces.<\/li>\n<\/ul>\n<p><strong>5. Analysis\u00a0<\/strong><\/p>\n<p><span style=\"text-decoration: underline\">Finding F<sub>g<\/sub><\/span><\/p>\n<p>[latex]F_{g} = m \\cdot g = 70 kg \\cdot 9.81m\/s^2 = 686.7 N[\/latex]<\/p>\n<p><span style=\"text-decoration: underline\">Forces in x and y components<\/span><\/p>\n<p>Since P, F<sub>F<\/sub>, and N are in the direction of either x or y, only F<sub>g <\/sub>has x and y components.<\/p>\n<p>[latex]F_{gx} =686.7\\cdot \\sin{30}[\/latex]<\/p>\n<p>[latex]F_{gx} =343.35N[\/latex]<\/p>\n<p><span style=\"text-decoration: underline\">Equilibrium equation in the x-axis\u00a0\u00a0<\/span><\/p>\n<p>\u2211F<sub>x<\/sub> = 300N &#8211; F<sub>F<\/sub>\u2212 686.7N sin(30\u00b0) = 0<\/p>\n<p>F<sub>F<\/sub>=-43.35N<\/p>\n<p>Since this value is negative, it just means that the drawing has the force of friction acting in the opposite direction.<\/p>\n<p><strong>6. Review\u00a0<\/strong><\/p>\n<p>The answer obtained makes sense as the force of friction should be small and be working against the cart sliding down the incline. The force of gravity in the x component is only slightly larger than that of the person pulling the cart so the force of friction being only 43.35N makes sense.<\/p>\n<\/div>\n<h1>Example 4.5.9: Finding the coefficient of friction, Submitted by Liam Murdock<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>You are developing an accessible gaming platform for a charity to allow everyone to play their favourite games.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4.png\" alt=\"The accessible gaming platform.\" class=\"alignnone wp-image-1994 size-full\" width=\"613\" height=\"244\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4.png 613w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4-300x119.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4-65x26.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4-225x90.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4-350x139.png 350w\" sizes=\"auto, (max-width: 613px) 100vw, 613px\" \/><br \/>\nThe platform can be simplified for the calculations into a box 0.8 m long and 0.15m high. The platform has a mass of 4 kg and rests on the two thighs of a person, each 0.2m wide and each 0.1 m away from the edges of the platform. Say the charity would want to make sure that a force of 10 N (from any of the top corners of the platform, going parallel to the surface) will not push it over. Make proper assumptions if needed and determine a coefficient of friction so that a material for the base can be chosen.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_1.png\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-1995\" width=\"360\" height=\"156\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_1.png 337w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_1-300x130.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_1-65x28.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_1-225x97.png 225w\" sizes=\"auto, (max-width: 360px) 100vw, 360px\" \/><\/p>\n<\/div>\n<p><strong>2. Sketch<\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_2-1.png\" alt=\"FBD of the problem.\" class=\"alignnone wp-image-1998\" width=\"604\" height=\"215\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_2-1.png 1991w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_2-1-300x107.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_2-1-1024x365.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_2-1-768x273.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_2-1-1536x547.png 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_2-1-65x23.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_2-1-225x80.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/05\/Liam_Chapter4_2-1-350x125.png 350w\" sizes=\"auto, (max-width: 604px) 100vw, 604px\" \/><br \/>\n<strong>3. Knows and Unknowns:<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>F<sub>A<\/sub> = 10N<\/li>\n<li>D<sub>2<\/sub> = 0.2m<\/li>\n<li>D<sub>3<\/sub> = 0.1m<\/li>\n<li>L = 0.8m<\/li>\n<li>H = 0.15m<\/li>\n<li>m = 4kg<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>D<sub>1<\/sub> = ?<\/li>\n<li>F<sub>g<\/sub> = ?<\/li>\n<li>F<sub>fa<\/sub> = ?<\/li>\n<li>F<sub>fb<\/sub> = ?<\/li>\n<li>N<sub>A<\/sub> = ?<\/li>\n<li>N<sub>B<\/sub> = ?<\/li>\n<li>\u03bc = ?<\/li>\n<\/ul>\n<p><strong>4. Approach:<\/strong><\/p>\n<p>Define an origin point to measure all forces from to simplify the problem.<br \/>\nDetermine D<sub>1<\/sub> and the coordinates of all forces measured from the origin O.<br \/>\nFind F<sub>g<\/sub>.<br \/>\nUse the sum of moments around A to find NB.<br \/>\nUse the sum of forces in the Y direction to find N<sub>A<\/sub>.<br \/>\nUse the sum of forces in the X direction to find \u03bc.<\/p>\n<p><strong>5. Analysis:<\/strong><br \/>\nDefining the origin as the bottom left corner of the gaming platform:<br \/>\n[latex]N_A\\ is\\ measured\\ D_3 + \\frac{1}{2} D_2\\ from\\ the\\ origin. \\\\ N_B\\ is\\ measured\\ D_3 + D_2 + D_1 + \\frac{1}{2} D_2 \\ from\\ the\\ origin. \\\\ F_g\\ is\\ measured\\ D_3 + D_2 + \\frac{1}{2} D_1 \\ from\\ the\\ origin.[\/latex]<br \/>\nTo find D<sub>1<\/sub>:<br \/>\n[latex]L = 2D_3 + 2D_2 + D_1 \\\\ D_1 = L - 2D_3 - 2D_2 \\\\ D_1 = 0.8m - 2 \\cdot 0.1m - 2 \\cdot 0.2m \\\\ D_1 = 0.2m[\/latex]<br \/>\nTo find\u00a0F<sub>g<\/sub>:<br \/>\n[latex]F_g = mg = 4kg \\cdot 9.81 m\/s^2 = 39.24N[\/latex]<br \/>\nThe coordinates of these forces can now be shown as:<br \/>\n[latex]N_A\\ is\\ at\\ [0.2, 0]m \\\\ N_B\\ is\\ at\\ [0.6, 0]m \\\\ F_g\\ is\\ at\\ [0.4, 0]m \\\\ F_A\\ is\\ at\\ [0, 0.15]m[\/latex]<\/p>\n<p>The sum of moments around point A can now be done to find N<sub>B<\/sub>. The distance between each force and point A is the difference between their coordinates.<br \/>\n[latex]\\sum M_A = 0 = -F_g\u00a0 \\cdot (0.4m - 0.2m) + N_B \\cdot (0.6m - 0.2m) - F_A (0.15m - 0m) \\\\ N_B = \\frac{F_g \\cdot 0.2m + F_A \\cdot 0.15m}{0.4m} \\\\ N_B = \\frac{39.24N \\cdot 0.2m + 10N \\cdot 0.15m}{0.4m} \\\\ N_B = 23.37 N[\/latex]<br \/>\nThe Sum of forces in the Y direction can be used to find N<sub>A<\/sub> now:<br \/>\n[latex]\\sum F_y = 0 = -F_g + N_A + N_B \\\\ N_A = F_g - N_B = 39.24N - 23.37N \\\\ N_A = 15.87N[\/latex]<br \/>\nBefore the sum of forces in the X is done, it must be noted that F<sub>fa<\/sub> and F<sub>fb<\/sub> are equivalent to \u03bcN<sub>A<\/sub> and \u03bcN<sub>B<\/sub>, respectively.<br \/>\n[latex]\\sum F_x = 0 = F_A - F_{fa} - F_{fb} = F_A - \\mu \\cdot N_A - \\mu \\cdot N_B \\\\ -F_A = -\\mu \\cdot N_A - \\mu \\cdot N_B \\\\ \\frac{-F_A}{\\mu} = -N_A - N_B \\\\ \\frac{1}{\\mu} = \\frac{-N_A - N_B}{-F_A} \\\\ \\mu = \\frac{-F_A}{-N_A - N_B} \\\\ \\mu = \\frac{-10N}{-15.87N - 23.37N} \\\\ \\mu = 0.25[\/latex]<\/p>\n<p><strong>6. Review:<\/strong><\/p>\n<p>To review this solution, the value for N<sub>A<\/sub> can be solved for by taking the moment around point B.<br \/>\n[latex]\\sum M_B = 0 = F_g \\cdot (0.6m - 0.4m) - N_A \\cdot (0.6m - 0.2m) - F_A \\cdot (0.15m) \\\\ N_A = \\frac{(F_g \\cdot 0.2m ) - (F_A \\cdot 0.15m)}{0.4m} \\\\ N_A = \\frac{(39.24N \\cdot 0.2m) - (10N \\cdot 0.15)}{0.15m}{0.4m} \\\\ N_A = 15.87N[\/latex]<br \/>\nThe value found is the same value originally found, providing evidence that there was no error in the calculation.<\/p>\n<\/div>\n","protected":false},"author":60,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"4.5 Examples","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-571","chapter","type-chapter","status-publish","hentry"],"part":56,"_links":{"self":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/571","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":63,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/571\/revisions"}],"predecessor-version":[{"id":2858,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/571\/revisions\/2858"}],"part":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/parts\/56"}],"metadata":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/571\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/media?parent=571"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapter-type?post=571"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/contributor?post=571"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/license?post=571"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}