{"id":574,"date":"2021-07-21T14:23:24","date_gmt":"2021-07-21T18:23:24","guid":{"rendered":"http:\/\/pressbooks.library.upei.ca\/statics\/?post_type=chapter&#038;p=574"},"modified":"2025-08-01T16:57:53","modified_gmt":"2025-08-01T20:57:53","slug":"5-5-examples","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.upei.ca\/statics\/chapter\/5-5-examples\/","title":{"raw":"5.5 Examples","rendered":"5.5 Examples"},"content":{"raw":"Here are examples from Chapter 5 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 5.5.1: Method of Sections, Submitted by Riley Fitzpatrick<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA flower cart at a local garden center is being pushed with a force of 500N at joint G. Its back wheels (A) are locked, so it is not moving. There is 1 meter of space between each of the four shelves in height, and each shelf is four meters long.\r\n\r\na) Calculate the reaction forces of the locked wheels and the unlocked wheels.\r\n\r\nb) Calculate the load carried by F<sub>CG<\/sub>, and whether it is in tension or compression\r\n\r\nWhich method did you use, joints or sections? Which is faster for the style of questions? How would your strategy change if you were calculating the load in each member?\r\n\r\n[caption id=\"attachment_790\" align=\"aligncenter\" width=\"398\"]<a href=\"https:\/\/flic.kr\/p\/txjSpP\"><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-problem-300x200.jpg\" alt=\"Colorful flower pots arranged on shelves.\" class=\"wp-image-790\" width=\"398\" height=\"265\" \/><\/a> Source: https:\/\/flic.kr\/p\/txjSpP[\/caption]\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-300x236.jpg\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-800 size-medium\" width=\"300\" height=\"236\" \/>\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\/Riley-1-draw-1-300x236.jpg\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-800 size-medium\" width=\"300\" height=\"236\" \/>\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\/Riley-1-draw-2-1-300x231.jpg\" alt=\"A FBD of the problem.\" class=\"aligncenter wp-image-804 size-medium\" width=\"300\" height=\"231\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li class=\"indent\">P = 500 N<\/li>\r\n \t<li class=\"indent\">Width = 4 m<\/li>\r\n \t<li class=\"indent\">Total height = 3 m<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>R<sub>Ax<\/sub><\/li>\r\n \t<li>R<sub>Ay<\/sub><\/li>\r\n \t<li>R<sub>By<\/sub><\/li>\r\n \t<li>F<sub>CG<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nPart a: Determine reaction forces using equilibrium equations\r\n\r\nPart b: Calculate F<sub>CG<\/sub> using the method of sections. Make a cut, then solve internal forces using equilibrium equations\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nPart a:\r\n\r\nSolving for R<sub>Ax<\/sub>:\r\n\r\n[latex]\\sum F_x=0=P+R_{Ax}\\\\R_{Ax}=-P\\\\R_{Ax}=-500N[\/latex]\r\n\r\nSolving for R<sub>By<\/sub>:\r\n\r\n[latex]\\sum M_A=0=(r_{BA}\\cdot R_{By})-(r_{GA}\\cdot P)\\\\r_{BA}\\cdot R_{By}=r_{GA}\\cdot P\\\\R_{By}=\\frac{r_{GA}\\cdot P}{r_{BA}}\\\\R_{By}=\\frac{2m\\cdot 500N}{4m}\\\\R_{By}=250N[\/latex]\r\n\r\nSolving for R<sub>Ay<\/sub>:\r\n\r\n[latex]\\sum F_y=0=R_{By}+R_{Ay}\\\\R_{Ay}=-R_{By}\\\\R_{Ay}=-250N[\/latex]\r\n\r\nThe answers we got that were negative numbers mean that the direction of the vector is drawn wrong on our original diagram (in reference to our coordinate frame). This makes sense as R<sub>A<\/sub><sub>y<\/sub> and R<sub>B<\/sub><sub>y<\/sub> are the only external forces in the y direction, so they have to cancel each other for the equilibrium equations to be true (Same for the vectors in the x direction.) Therefore, one of them should have a negative direction. We will leave this answer as is for now, but the next time we draw the system, we will change the direction of the arrow.\r\n\r\n[latex]\\underline{R_{Ax}=-500N,\\; R_{Ay}=-250N, \\; R_{By}=250N}[\/latex]\r\n\r\nPart b:\r\n\r\nFirstly, we redraw the diagram, changing the direction of the R<sub>Ay<\/sub> and R<sub>Ax.<\/sub> Then, since we are using the method of sections, we make a cut so that the member F<sub>CG<\/sub> (the one we want to find) is cut.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-31-122912-300x224.png\" alt=\"Truss with external load P and support reactions marked, section cut across the middle.\" class=\"aligncenter wp-image-2423\" width=\"420\" height=\"314\" \/>\r\n\r\nNow we redraw, choosing one of the pieces from the cut. Here, the top half is chosen, but you could also choose the bottom half and get the right answer. The only reason the top half was chosen here is because there are fewer external forces to consider for the top.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-solve-6-300x179.jpg\" alt=\"Free-body diagram of the cut truss section showing applied load P and now internal forces.\" class=\"aligncenter wp-image-815\" width=\"343\" height=\"205\" \/>\r\n\r\nSolve for the member we are looking for:\r\n\r\n<span style=\"text-align: initial;background-color: initial;font-size: 1em\">\u00a0<\/span>[latex]\\sum F_x=0=P+\\frac{4}{\\sqrt{17}}F_{CG}\\\\\\frac{4}{\\sqrt{17}}F_{CG}=-P\\\\F_{CG}=-P(\\frac{\\sqrt{17}}{4})\\\\F_{CG}=-500N(\\frac{\\sqrt{17}}{4})\\\\F_{CG}=-515.388N[\/latex]\r\n\r\nAgain, the number we get is negative. The way we drew F<sub>CG<\/sub> originally was as if the member was in tension. The negative number just means that it is actually compression, not tension.\r\n\r\n[latex]\\underline{F_{CG}=515 \\text{N (Compression)}}[\/latex]\r\n\r\nPart C:\r\n\r\nFor part b, I used the method of sections, as it would be the fastest method. The method of joints would require the lower joints to be solved first, which would be a much slower process, whereas with this method, a simple cut can be made, and the member's load can be quickly solved using equilibrium equations. Had the question asked for all member loads to be solved, however, the method of joints would have been the faster approach.\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nPart a:\r\n\r\nR<sub>Ax<\/sub> is equal and opposite to P so we know it is correct, and the value of R<sub>By<\/sub> should also be correct as its moment about A (250 * 4m = 1000 Nm), is equal and opposite to the moment of P about A, (500 N * 2 m = 1000 Nm). as R<sub>Ay<\/sub> is equal and opposite to R<sub>By<\/sub> it is also correct.\r\n\r\nPart b:\r\n\r\nThe x component of the calculated value of F<sub>CG<\/sub> is equal in magnitude to P (see equation below), and it is the only cut member acting in the x direction. Therefore, it must be correct.\r\n\r\n[latex]\\frac{4}{\\sqrt{17}}(515 N)=500 N[\/latex]\r\n\r\nPart C:\r\n\r\nThe method of sections allows you to solve a very specific area of the system's internal forces (the members that are cut), whereas the method of joints usually requires you to solve most, if not all, of the internal forces of the system. Therefore, the method of sections is the most efficient for finding the internal forces of specific parts of the system, whereas the method of joints is more efficient for solving the whole system.\r\n\r\n<\/div>\r\n<h1>Example 5.5.2: Zero-Force Members, Submitted by Michael Oppong-Ampomah<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA bridge with uneven ground has been built as shown below. Force is applied at three points on the top of the bridge.\r\n\r\na) Find any zero-force members\r\n\r\nb) What purpose do these members serve?\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-draw-1-1-300x257.jpg\" alt=\"A sketch of the problem. \" class=\"alignnone wp-image-1041 size-medium\" width=\"300\" height=\"257\" \/>\r\n\r\n<\/div>\r\n<strong>2. Draw<\/strong>\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\/Michael-1-solve-1-1-300x180.jpg\" alt=\"A FBD of the problem. \" class=\"alignnone wp-image-1042\" width=\"383\" height=\"230\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>All of the known values do not mean anything - we only need to know where the forces exist<\/li>\r\n<\/ul>\r\nUnknown:\r\n<ul>\r\n \t<li>Which members are zero-force<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nLook at each joint and determine how many forces are in each direction. If there is only one force in a direction, that member is zero-force.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nPart a:\r\n\r\nLet's start with joint C. If we think of the forces acting in the x and y directions as shown below by the coordinate frame, we see that there are two forces acting in the y direction, and only one in the x direction. Therefore, assuming the joint is in static equilibrium, member CE is a zero-force.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-3-1-300x190.jpg\" alt=\"Internal forces at joint C.\" class=\"aligncenter wp-image-1044 size-medium\" width=\"300\" height=\"190\" \/>\r\n\r\nIf we do the same type of analysis for the other joints and remove the zero-force members, the structure now looks like this:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-2-1-300x266.jpg\" alt=\"Updated sketch of the problem\" class=\"aligncenter wp-image-1043\" width=\"367\" height=\"325\" \/>\r\n\r\nAfter one more analysis of the joints, we find one more zero-force member, as shown below.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-4-300x190.jpg\" alt=\"New sketch of the problem without any zero force member present.\" class=\"aligncenter wp-image-1045\" width=\"391\" height=\"248\" \/>\r\n\r\nAnswer: CE, DG, and FG are zero-force members.\r\n\r\nPart b:\r\n\r\nZero-force members exist to provide stability to the truss, to keep the shape rigid.\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nAlthough the new truss (without zero-force members) looks strange, there are no joints where there's only one force in one direction; therefore, there are no more zero-force members.\r\n\r\n<\/div>\r\n<h1>Example 5.5.3: Method of Joints, Submitted by Deanna Malone<\/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\nA small truss bridge is over a river, and it has a bucket full of water hanging off the middle. The mass of the water bucket is 15 kg.\r\n\r\na. Solve for the reaction forces.\r\n\r\nb. Use the method of joints to solve each member.\r\n\r\nc. Draw forces and how they act.\r\n\r\nIn the real world, the truss bridge might be similar to this,\r\n\r\n[caption id=\"attachment_1829\" align=\"alignnone\" width=\"761\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Bridge_above_river-truss-deanna-scaled.jpg\" alt=\"A truss bridge.\" class=\"wp-image-1829\" width=\"761\" height=\"403\" \/> Source:https:\/\/commons.wikimedia.org\/wiki\/File:Railroad_Truss_bridge_over_trinity_river_near_Goodrich,_Texas.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\/02\/Screenshot-20-150x137.png\" alt=\"x and y coordinates.\" class=\"alignleft wp-image-1831\" style=\"margin-bottom: 4.44444em\" width=\"94\" height=\"86\" \/><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/deanna-malone-first-truss-image--292x300.png\" alt=\"A sketch of the problem.\" class=\"alignnone wp-image-1698\" width=\"349\" height=\"359\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>F<sub>g\u00a0<\/sub>= 15 kg . 9.81 m\/s<sup>2<\/sup> = 147.15 N<\/li>\r\n \t<li>r<sub>AF<\/sub> = r<sub>BD<\/sub> = r<sub>CE<\/sub> =2 m<\/li>\r\n \t<li>r<sub>DF<\/sub> = r<sub>AB<\/sub> = 6 m<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>R<sub>A<\/sub><sub>x<\/sub><\/li>\r\n \t<li>R<sub>Ay<\/sub><\/li>\r\n \t<li>R<sub>By<\/sub><\/li>\r\n \t<li>F<sub>EF<\/sub><\/li>\r\n \t<li>F<sub>ED<\/sub><\/li>\r\n \t<li>F<sub>BD<\/sub><\/li>\r\n \t<li>F<sub>BE<\/sub><\/li>\r\n \t<li>F<sub>CE<\/sub><\/li>\r\n \t<li>F<sub>AE<\/sub><\/li>\r\n \t<li>F<sub>AC<\/sub><\/li>\r\n \t<li>F<sub>BC<\/sub><\/li>\r\n \t<li>F<sub>A<\/sub><sub>F<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach\u00a0<\/strong>\r\n\r\nFirst, use the equilibrium equations to find the reaction forces. Apply these reaction forces in the method of joints to find the force of each member.\r\n\r\n<strong>5. Analysis\u00a0<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/truss-second-image-deanna-.png\" alt=\"FBD of the problem.\" class=\"alignnone wp-image-1699\" width=\"469\" height=\"381\" \/><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-150x137.png\" alt=\"x and y coordinates.\" class=\"alignleft wp-image-1831\" style=\"margin-bottom: 4.44444em\" width=\"94\" height=\"86\" \/>\r\n\r\na. Calculating the reaction forces using the equilibrium equations and the diagram.\r\n\r\n[latex]\\sum F_x=0=R_{Ax}\\\\R_{Ax}=0N[\/latex]\r\n\r\nSolving for R<sub>By<\/sub>:\r\n\r\n[latex]\\sum M_A=0= -(r_{AC}\\cdot F_g)+(r_{AB}\\cdot R_{By})\\\\r_{AB}\\cdot R_{By}=r_{AC}\\cdot\u00a0 F_g [\/latex]\r\n\r\n[latex]R_{By}=\\frac{r_{AC}\\cdot F_g}{r_{AB}}\\\\R_{By}=\\frac{3m\\cdot 147.15 N}{6m} [\/latex]\r\n\r\n<span style=\"text-align: initial;background-color: initial;font-size: 1em\">[latex]R_{By}=73.575 N[\/latex]<\/span>\r\n\r\nSolving for R<sub>Ay<\/sub>:\r\n\r\n[latex]\\sum F_y=0=R_{By}+R_{Ay} -F_g\\\\R_{Ay}=F_g-R_{By}\\\\R_{By}=73.575N[\/latex]\r\n\r\nb. F<sub>AF<\/sub>, F<sub>EF<\/sub>, F<sub>DE<\/sub>, F<sub>BD<\/sub> are zero force members because for joints F and D, there are only 2 members each, and there is no external load.\r\n\r\nFinding the angle between AC and AE:\r\n\r\ntan <span>\u03b8 = <\/span>r<sub>CE<\/sub>\/r<sub>AC<\/sub>\r\n\r\ntherefore, <span>\u03b8 = 33.7 \u00b0<\/span>\r\n\r\nAnalyzing joint A:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-150x137.png\" alt=\"x and y coordinates.\" class=\"alignleft wp-image-1831\" style=\"margin-bottom: 4.44444em\" width=\"94\" height=\"86\" \/><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-120.png\" alt=\"Sketch of forces at joint A\" class=\"alignnone wp-image-1700 size-full\" width=\"231\" height=\"214\" \/>\r\n\r\n[latex]\\sum F_y=0 = R_{Ay}+ F_{AE}\\cdot sin 33.7 \\\\ =\\frac{-73.575 N}{sin 33.7}[\/latex]\r\n\r\n[latex]F_{AE} = -132.6 N[\/latex]\r\n\r\n[latex]\\sum F_x=0 = F_{AC} + F_{AE} cos 33.7\\\\ F_{AC}=132.6N\\cdot \\cos {33.7}[\/latex]\r\n\r\n[latex]F_{AC} = 110.32 N[\/latex]\r\n\r\nAnalyzing Joint C:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-05-31-140213-300x218.png\" alt=\"A sketch of forces at joint C.\" class=\"alignnone wp-image-2431\" width=\"413\" height=\"300\" \/><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-150x137.png\" alt=\"x and y coordinates.\" class=\"alignleft wp-image-1831\" style=\"margin-bottom: 4.44444em\" width=\"94\" height=\"86\" \/>\r\n\r\nFrom the FBD,\r\n\r\nF<sub>AC<\/sub>= F<sub>BC\u00a0<\/sub>= 110.32 N\r\n\r\nAlso, F<sub>g\u00a0 <\/sub>=F<sub>CE<\/sub> = 147.15 N\r\n\r\nAnalyzing Joint B\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-150x137.png\" alt=\"x and y coordinates.\" class=\"alignleft wp-image-1831\" style=\"margin-bottom: 4.44444em\" width=\"94\" height=\"86\" \/><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/truss-joint-b-deanna-.png\" alt=\"A sketch of forces at joint B.\" class=\"alignnone wp-image-1701 size-full\" width=\"214\" height=\"282\" \/>\r\n\r\n[latex]\\sum F_y=0 = R_{By} +F_{BE}\\cdot sin 33.7[\/latex]\r\n\r\n[latex]F_{BE} = -132.6 N[\/latex]\r\n\r\n[latex]\\sum F_x=0 = F_{BC} -F_{BE} cos 33.7\\\\ F_{BC}=132.6N\\cdot \\cos {33.7}[\/latex]\r\n\r\n[latex]F_{BC} = 110.32 N[\/latex]\r\n\r\nc.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Picture1-300x213.png\" alt=\"Labeled truss diagram showing internal forces in each member and support reactions.\" width=\"394\" height=\"280\" class=\"aligncenter wp-image-2156\" \/>\r\n\r\n<strong>6. Review\u00a0<\/strong>\r\n\r\nThe answers make sense because F<sub>CE<\/sub>is in tension as imagined due to the weight of the bucket. Members AC and BC are under reaction forces, are equally under tension, and members AE and BE are balanced by compression. The identified zero-force members also make sense due to the rules being met.\r\n\r\n<\/div>\r\n<h1><strong>Example 5.5.4\u00a0 Method of Joints, Submitted by Luke McCarvill<\/strong><\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nJeffery is unsuccessfully pushing a 2-dimensional shopping cart shown below with a force of 70N. It is made up of steel (8g\/cm<sup>3<\/sup>) rods with 2cm diameters. The wheel of the cart that is closest to him is stuck (pinned) while the front wheel is free to roll (roller). The total height of the cart is 0.75m, and its total length is 1.5m.\r\n\r\nWhat is the total mass of the cart? (For the purposes of the equilibrium equations, assume the total weight acts only at the centre point of the system, and otherwise the rods are massless.) What are the reaction forces on the wheels?\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke1.png\" alt=\"A digital sketch of the problem. \" class=\"alignnone wp-image-1857\" width=\"538\" height=\"244\" \/>\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\/02\/Luke2.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1875\" width=\"722\" height=\"387\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n\r\n\u2022 [latex]P = 70\\,N[\/latex]\r\n\u2022 [latex]\\sum M = 0[\/latex]\r\n\u2022 [latex]h = 0.75\\,m[\/latex]\r\n\u2022 [latex]\\theta_p = 15^\\circ[\/latex]\r\n\u2022 [latex]d_{\\text{rod}} = 2\\,\\text{cm (diameter)}[\/latex]\r\n\u2022 [latex]\\sum F = 0[\/latex]\r\n\u2022 [latex]\\rho = 8.00\\,\\text{g\/cm}^3[\/latex]\r\n\u2022 [latex]V_{\\text{cylinder}} = \\pi r^2 h[\/latex]\r\n\r\nUnknowns:\r\n\r\n\u2022 [latex]m_{\\text{tot}} = \\, ?[\/latex]\r\n\u2022 [latex]P_x = \\, ?[\/latex]\r\n\u2022 [latex]P_y = \\, ?[\/latex]\r\n\u2022 [latex]R_{Ax} = \\, ?[\/latex]\r\n\u2022 [latex]R_{Ay} = \\, ?[\/latex]\r\n\u2022 [latex]R_{By} = \\, ?[\/latex]\r\n\u2022 [latex]F_g = \\, ?[\/latex]\r\n\r\n<strong>4. Approach<\/strong>\r\n\r\nUse trigonometry and equilibrium equations, as well as the method of joints, to calculate unknowns\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nFinding total mass:\r\n\r\nThere are four 75cm rods, two rods that are [latex]75\\sqrt{2}[\/latex]cm, and one rod that is 1.5m long, all of which have a diameter of 2cm. The total mass is therefore [latex]8g\/cm^3 \\cdot ((4 \\pi \\cdot (1cm)^2\\cdot (75cm) + 2\\pi\\cdot (1cm)^2 \\cdot \\\\ (75 \\sqrt(2)cm) + \\pi\u00a0 (1cm)^2 \\cdot (150cm)) \\\\ \\approx 16,641g = 16.641kg = m_{tot}[\/latex]\r\n\r\nF<sub>g<\/sub> can then be found using [latex]F_g = g \\cdot m_{tot} = -9.81m\/s^2 \\cdot 16.641kg \\approx -163.25N[\/latex]\r\n\r\nThe sum of forces and moments can now be used to determine reaction forces\r\n\r\n[latex]\\sum F_x = P_x - R_{Ax} \\\\ R_{Ax} = P \\cdot cos(\\theta) = 70N cos(15^\\circ) \\\\ R_{Ax} \\approx 67.6N[\/latex]\r\n\r\n[latex]\\sum F_y = -F_g - P_y + R_{Ay} + R_{By} = 0[\/latex]\r\n\r\nWith two unknowns, the sum of moments must be considered. Summing moments around point A finds:\r\n\r\n[latex]\\sum M_A = 0 = -(0.75m \\cdot Pcos(\\theta)) - (0.75m \\cdot F_g) + (1.5m \\cdot R_{By})[\/latex]\r\n\r\n[latex]\\sum M_A = 0 = -(0.75m \\cdot 70N cos(15^\\circ)) - (0.75m \\cdot 163.25N) + (1.5m \\cdot R_{By})[\/latex]\r\n\r\n[latex]R_{By} \\approx 115.43N[\/latex]\r\n\r\nSum of Forces in Y can now be used to find R<sub>Ay<\/sub>\r\n\r\n[latex]F_y = -F_g - P_y + R_{Ay} +R_{By} = 0[\/latex]\r\n\r\n[latex]R_{Ay} \\approx 65.94N[\/latex]\r\n\r\nThe method of joints can now be used to determine if any zero-force members exist in the shopping cart.\r\n\r\nLooking at Point A, it can be found that [latex]F_{AO} = 65.94\\,N[\/latex] in compression and [latex]F_{AC} = 67.6\\,N[\/latex] in tension. Continuing through each joint, it can be found that member CB is a zero-force member as point B only has one component in the X direction, which, in a static system, implies that the force must be zero.\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nGiven the high weight of the cart creating a large F<sub>g<\/sub> and the applied force of 70N, the reaction forces found and the internal forces all seem to be of appropriate magnitude and direction.\r\n\r\n<\/div>\r\n<h1><strong>Example 5.5.5: Method of Sections, Submitted by Liam Murdock<\/strong><\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">You are designing a shelf for a friend, and it can be analyzed as a truss shown below. If you know the forces in the member CD can take a tensile force of 25 N, and GH can take a compressive force of 15 N before breaking. If w = 10 cm and h = 20 cm, what mass can be supported if it is situated exactly on point E?\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Liam_Chapter5_1.png\" alt=\"A digital sketch of the problem.\" class=\"alignnone wp-image-2009 size-full\" width=\"509\" height=\"471\" \/><\/div>\r\n<strong>2. Sketch<\/strong>\r\nDraw a free-body diagram of the truss with a slice through CD, GH, and IH.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Liam_Chapter5_2-1.png\" alt=\"A FBD showing cut sections of the problem.\" class=\"alignnone wp-image-2011 size-full\" width=\"294\" height=\"408\" \/>\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>w = 0.1 m<\/li>\r\n \t<li>h = 0.2 m<\/li>\r\n \t<li>F<sub>CD<\/sub> = -25 N<\/li>\r\n \t<li>F<sub>GH<\/sub> = 15 N<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>F<sub>HI<\/sub> =?<\/li>\r\n \t<li>d<sub>HI<\/sub> = ?<\/li>\r\n \t<li>P =?<\/li>\r\n \t<li>m = ?<\/li>\r\n<\/ul>\r\n<strong>4. Approach\r\n<\/strong>Use the method of sections to show internal forces and analyze as a rigid body.\r\n\r\n<strong>5. Analysis<\/strong>\r\nFirst, the distance d<sub>HI<\/sub> can be found:\r\n[latex]d_{HI} = \\sqrt{(0.1 m)^2 + (0.2 m)^2} \\\\ d_{HI} = 0.2236 m[\/latex]\r\nThis value can be used to find the equivalents to cos\u03b8 and sin\u03b8:\r\n[latex]\\cos{\\theta} = \\frac{0.1 m}{0.2236 m} \\\\ \\sin{\\theta} = \\frac{0.2 m}{0.2236 m}[\/latex]\r\nThe sum of forces in the X direction can be used to determine the value of F<sub>HI<\/sub>:\r\n[latex]\\sum F_x = 0 = -F_{CD} + F_{GH} + F_{HI} \\cdot \\cos{\\theta} \\\\ F_{HI} = \\frac{(F_{CD} - F_{GH})}{\\cos{\\theta}} \\\\ F_{HI} = \\frac{0.2236 m}{0.1 m} \\cdot (25 N - 15 N) \\\\ F_{HI} = 22.36N[\/latex]\r\nThe sum of forces in the Y direction can now be used to determine the value of P:\r\n[latex]\\sum F_y = 0 = -P + F_{HI} \\cdot \\sin{\\theta} \\\\ P = F_{HI} \\cdot \\sin{\\theta} \\\\ P = (22.36 N) \\cdot \\frac{0.2 m}{0.2236 m} \\\\ P = 20 N[\/latex]\r\nThe mass can now be simply found by dividing the force by gravity:\r\n[latex]P = mg \\\\ m = \\frac{P}{g} \\\\ m = \\frac{20 N}{9.81 \\frac{m}{s^2}} \\\\ m = 2.04 kg[\/latex]\r\n\r\n<strong>6. Review\r\n<\/strong>A brief review of this question is the sign and magnitude of the result. The force generated by the mass pushes downwards with gravity, which is expected. The magnitude of the mass, while defining this shelf as one that is particularly flimsy, is a reasonable mass.\r\n\r\n<\/div>","rendered":"<p>Here are examples from Chapter 5 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 5.5.1: Method of Sections, Submitted by Riley Fitzpatrick<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A flower cart at a local garden center is being pushed with a force of 500N at joint G. Its back wheels (A) are locked, so it is not moving. There is 1 meter of space between each of the four shelves in height, and each shelf is four meters long.<\/p>\n<p>a) Calculate the reaction forces of the locked wheels and the unlocked wheels.<\/p>\n<p>b) Calculate the load carried by F<sub>CG<\/sub>, and whether it is in tension or compression<\/p>\n<p>Which method did you use, joints or sections? Which is faster for the style of questions? How would your strategy change if you were calculating the load in each member?<\/p>\n<figure id=\"attachment_790\" aria-describedby=\"caption-attachment-790\" style=\"width: 398px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/flic.kr\/p\/txjSpP\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-problem-300x200.jpg\" alt=\"Colorful flower pots arranged on shelves.\" class=\"wp-image-790\" width=\"398\" height=\"265\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-problem-300x200.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-problem-768x512.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-problem-65x43.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-problem-225x150.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-problem-350x233.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-problem.jpg 1024w\" sizes=\"auto, (max-width: 398px) 100vw, 398px\" \/><\/a><figcaption id=\"caption-attachment-790\" class=\"wp-caption-text\">Source: https:\/\/flic.kr\/p\/txjSpP<\/figcaption><\/figure>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-300x236.jpg\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-800 size-medium\" width=\"300\" height=\"236\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-300x236.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-1024x804.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-768x603.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-1536x1207.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-65x51.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-225x177.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-350x275.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1.jpg 2014w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/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\/Riley-1-draw-1-300x236.jpg\" alt=\"A sketch of the problem.\" class=\"aligncenter wp-image-800 size-medium\" width=\"300\" height=\"236\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-300x236.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-1024x804.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-768x603.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-1536x1207.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-65x51.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-225x177.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1-350x275.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-1.jpg 2014w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/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\/Riley-1-draw-2-1-300x231.jpg\" alt=\"A FBD of the problem.\" class=\"aligncenter wp-image-804 size-medium\" width=\"300\" height=\"231\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-2-1-300x231.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-2-1-1024x787.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-2-1-768x590.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-2-1-1536x1181.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-2-1-2048x1574.jpg 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-2-1-65x50.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-2-1-225x173.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-draw-2-1-350x269.jpg 350w\" 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 class=\"indent\">P = 500 N<\/li>\n<li class=\"indent\">Width = 4 m<\/li>\n<li class=\"indent\">Total height = 3 m<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>R<sub>Ax<\/sub><\/li>\n<li>R<sub>Ay<\/sub><\/li>\n<li>R<sub>By<\/sub><\/li>\n<li>F<sub>CG<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Part a: Determine reaction forces using equilibrium equations<\/p>\n<p>Part b: Calculate F<sub>CG<\/sub> using the method of sections. Make a cut, then solve internal forces using equilibrium equations<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Part a:<\/p>\n<p>Solving for R<sub>Ax<\/sub>:<\/p>\n<p>[latex]\\sum F_x=0=P+R_{Ax}\\\\R_{Ax}=-P\\\\R_{Ax}=-500N[\/latex]<\/p>\n<p>Solving for R<sub>By<\/sub>:<\/p>\n<p>[latex]\\sum M_A=0=(r_{BA}\\cdot R_{By})-(r_{GA}\\cdot P)\\\\r_{BA}\\cdot R_{By}=r_{GA}\\cdot P\\\\R_{By}=\\frac{r_{GA}\\cdot P}{r_{BA}}\\\\R_{By}=\\frac{2m\\cdot 500N}{4m}\\\\R_{By}=250N[\/latex]<\/p>\n<p>Solving for R<sub>Ay<\/sub>:<\/p>\n<p>[latex]\\sum F_y=0=R_{By}+R_{Ay}\\\\R_{Ay}=-R_{By}\\\\R_{Ay}=-250N[\/latex]<\/p>\n<p>The answers we got that were negative numbers mean that the direction of the vector is drawn wrong on our original diagram (in reference to our coordinate frame). This makes sense as R<sub>A<\/sub><sub>y<\/sub> and R<sub>B<\/sub><sub>y<\/sub> are the only external forces in the y direction, so they have to cancel each other for the equilibrium equations to be true (Same for the vectors in the x direction.) Therefore, one of them should have a negative direction. We will leave this answer as is for now, but the next time we draw the system, we will change the direction of the arrow.<\/p>\n<p>[latex]\\underline{R_{Ax}=-500N,\\; R_{Ay}=-250N, \\; R_{By}=250N}[\/latex]<\/p>\n<p>Part b:<\/p>\n<p>Firstly, we redraw the diagram, changing the direction of the R<sub>Ay<\/sub> and R<sub>Ax.<\/sub> Then, since we are using the method of sections, we make a cut so that the member F<sub>CG<\/sub> (the one we want to find) is cut.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-31-122912-300x224.png\" alt=\"Truss with external load P and support reactions marked, section cut across the middle.\" class=\"aligncenter wp-image-2423\" width=\"420\" height=\"314\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-31-122912-300x224.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-31-122912-65x48.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-31-122912-225x168.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-31-122912-350x261.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-05-31-122912.png 739w\" sizes=\"auto, (max-width: 420px) 100vw, 420px\" \/><\/p>\n<p>Now we redraw, choosing one of the pieces from the cut. Here, the top half is chosen, but you could also choose the bottom half and get the right answer. The only reason the top half was chosen here is because there are fewer external forces to consider for the top.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-solve-6-300x179.jpg\" alt=\"Free-body diagram of the cut truss section showing applied load P and now internal forces.\" class=\"aligncenter wp-image-815\" width=\"343\" height=\"205\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-solve-6-300x179.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-solve-6-1024x612.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-solve-6-768x459.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-solve-6-65x39.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-solve-6-225x134.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-solve-6-350x209.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Riley-1-solve-6.jpg 1270w\" sizes=\"auto, (max-width: 343px) 100vw, 343px\" \/><\/p>\n<p>Solve for the member we are looking for:<\/p>\n<p><span style=\"text-align: initial;background-color: initial;font-size: 1em\">\u00a0<\/span>[latex]\\sum F_x=0=P+\\frac{4}{\\sqrt{17}}F_{CG}\\\\\\frac{4}{\\sqrt{17}}F_{CG}=-P\\\\F_{CG}=-P(\\frac{\\sqrt{17}}{4})\\\\F_{CG}=-500N(\\frac{\\sqrt{17}}{4})\\\\F_{CG}=-515.388N[\/latex]<\/p>\n<p>Again, the number we get is negative. The way we drew F<sub>CG<\/sub> originally was as if the member was in tension. The negative number just means that it is actually compression, not tension.<\/p>\n<p>[latex]\\underline{F_{CG}=515 \\text{N (Compression)}}[\/latex]<\/p>\n<p>Part C:<\/p>\n<p>For part b, I used the method of sections, as it would be the fastest method. The method of joints would require the lower joints to be solved first, which would be a much slower process, whereas with this method, a simple cut can be made, and the member&#8217;s load can be quickly solved using equilibrium equations. Had the question asked for all member loads to be solved, however, the method of joints would have been the faster approach.<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>Part a:<\/p>\n<p>R<sub>Ax<\/sub> is equal and opposite to P so we know it is correct, and the value of R<sub>By<\/sub> should also be correct as its moment about A (250 * 4m = 1000 Nm), is equal and opposite to the moment of P about A, (500 N * 2 m = 1000 Nm). as R<sub>Ay<\/sub> is equal and opposite to R<sub>By<\/sub> it is also correct.<\/p>\n<p>Part b:<\/p>\n<p>The x component of the calculated value of F<sub>CG<\/sub> is equal in magnitude to P (see equation below), and it is the only cut member acting in the x direction. Therefore, it must be correct.<\/p>\n<p>[latex]\\frac{4}{\\sqrt{17}}(515 N)=500 N[\/latex]<\/p>\n<p>Part C:<\/p>\n<p>The method of sections allows you to solve a very specific area of the system&#8217;s internal forces (the members that are cut), whereas the method of joints usually requires you to solve most, if not all, of the internal forces of the system. Therefore, the method of sections is the most efficient for finding the internal forces of specific parts of the system, whereas the method of joints is more efficient for solving the whole system.<\/p>\n<\/div>\n<h1>Example 5.5.2: Zero-Force Members, Submitted by Michael Oppong-Ampomah<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A bridge with uneven ground has been built as shown below. Force is applied at three points on the top of the bridge.<\/p>\n<p>a) Find any zero-force members<\/p>\n<p>b) What purpose do these members serve?<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-draw-1-1-300x257.jpg\" alt=\"A sketch of the problem.\" class=\"alignnone wp-image-1041 size-medium\" width=\"300\" height=\"257\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-draw-1-1-300x257.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-draw-1-1-1024x878.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-draw-1-1-768x658.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-draw-1-1-65x56.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-draw-1-1-225x193.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-draw-1-1-350x300.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-draw-1-1.jpg 1416w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<\/div>\n<p><strong>2. Draw<\/strong><\/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\/Michael-1-solve-1-1-300x180.jpg\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1042\" width=\"383\" height=\"230\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-1-1-300x180.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-1-1-1024x614.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-1-1-768x460.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-1-1-1536x921.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-1-1-2048x1227.jpg 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-1-1-65x39.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-1-1-225x135.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-1-1-350x210.jpg 350w\" sizes=\"auto, (max-width: 383px) 100vw, 383px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>All of the known values do not mean anything &#8211; we only need to know where the forces exist<\/li>\n<\/ul>\n<p>Unknown:<\/p>\n<ul>\n<li>Which members are zero-force<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Look at each joint and determine how many forces are in each direction. If there is only one force in a direction, that member is zero-force.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Part a:<\/p>\n<p>Let&#8217;s start with joint C. If we think of the forces acting in the x and y directions as shown below by the coordinate frame, we see that there are two forces acting in the y direction, and only one in the x direction. Therefore, assuming the joint is in static equilibrium, member CE is a zero-force.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-3-1-300x190.jpg\" alt=\"Internal forces at joint C.\" class=\"aligncenter wp-image-1044 size-medium\" width=\"300\" height=\"190\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-3-1-300x190.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-3-1-1024x649.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-3-1-768x487.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-3-1-65x41.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-3-1-225x143.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-3-1-350x222.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-3-1.jpg 1167w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>If we do the same type of analysis for the other joints and remove the zero-force members, the structure now looks like this:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-2-1-300x266.jpg\" alt=\"Updated sketch of the problem\" class=\"aligncenter wp-image-1043\" width=\"367\" height=\"325\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-2-1-300x266.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-2-1-1024x909.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-2-1-768x682.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-2-1-1536x1364.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-2-1-65x58.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-2-1-225x200.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-2-1-350x311.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-2-1.jpg 1698w\" sizes=\"auto, (max-width: 367px) 100vw, 367px\" \/><\/p>\n<p>After one more analysis of the joints, we find one more zero-force member, as shown below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-4-300x190.jpg\" alt=\"New sketch of the problem without any zero force member present.\" class=\"aligncenter wp-image-1045\" width=\"391\" height=\"248\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-4-300x190.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-4-1024x649.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-4-768x486.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-4-1536x973.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-4-2048x1297.jpg 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-4-65x41.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-4-225x143.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Michael-1-solve-4-350x222.jpg 350w\" sizes=\"auto, (max-width: 391px) 100vw, 391px\" \/><\/p>\n<p>Answer: CE, DG, and FG are zero-force members.<\/p>\n<p>Part b:<\/p>\n<p>Zero-force members exist to provide stability to the truss, to keep the shape rigid.<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>Although the new truss (without zero-force members) looks strange, there are no joints where there&#8217;s only one force in one direction; therefore, there are no more zero-force members.<\/p>\n<\/div>\n<h1>Example 5.5.3: Method of Joints, Submitted by Deanna Malone<\/h1>\n<div class=\"textbox\">\n<ol>\n<li><strong>Problem\u00a0<\/strong><\/li>\n<\/ol>\n<div class=\"textbox shaded\">\n<p>A small truss bridge is over a river, and it has a bucket full of water hanging off the middle. The mass of the water bucket is 15 kg.<\/p>\n<p>a. Solve for the reaction forces.<\/p>\n<p>b. Use the method of joints to solve each member.<\/p>\n<p>c. Draw forces and how they act.<\/p>\n<p>In the real world, the truss bridge might be similar to this,<\/p>\n<figure id=\"attachment_1829\" aria-describedby=\"caption-attachment-1829\" style=\"width: 761px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Bridge_above_river-truss-deanna-scaled.jpg\" alt=\"A truss bridge.\" class=\"wp-image-1829\" width=\"761\" height=\"403\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Bridge_above_river-truss-deanna-scaled.jpg 2560w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Bridge_above_river-truss-deanna-300x159.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Bridge_above_river-truss-deanna-1024x543.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Bridge_above_river-truss-deanna-768x407.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Bridge_above_river-truss-deanna-1536x814.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Bridge_above_river-truss-deanna-2048x1085.jpg 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Bridge_above_river-truss-deanna-65x34.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Bridge_above_river-truss-deanna-225x119.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Bridge_above_river-truss-deanna-350x185.jpg 350w\" sizes=\"auto, (max-width: 761px) 100vw, 761px\" \/><figcaption id=\"caption-attachment-1829\" class=\"wp-caption-text\">Source:https:\/\/commons.wikimedia.org\/wiki\/File:Railroad_Truss_bridge_over_trinity_river_near_Goodrich,_Texas.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\/02\/Screenshot-20-150x137.png\" alt=\"x and y coordinates.\" class=\"alignleft wp-image-1831\" style=\"margin-bottom: 4.44444em\" width=\"94\" height=\"86\" \/><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/deanna-malone-first-truss-image--292x300.png\" alt=\"A sketch of the problem.\" class=\"alignnone wp-image-1698\" width=\"349\" height=\"359\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/deanna-malone-first-truss-image--292x300.png 292w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/deanna-malone-first-truss-image--65x67.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/deanna-malone-first-truss-image--225x231.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/deanna-malone-first-truss-image-.png 350w\" sizes=\"auto, (max-width: 349px) 100vw, 349px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>F<sub>g\u00a0<\/sub>= 15 kg . 9.81 m\/s<sup>2<\/sup> = 147.15 N<\/li>\n<li>r<sub>AF<\/sub> = r<sub>BD<\/sub> = r<sub>CE<\/sub> =2 m<\/li>\n<li>r<sub>DF<\/sub> = r<sub>AB<\/sub> = 6 m<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>R<sub>A<\/sub><sub>x<\/sub><\/li>\n<li>R<sub>Ay<\/sub><\/li>\n<li>R<sub>By<\/sub><\/li>\n<li>F<sub>EF<\/sub><\/li>\n<li>F<sub>ED<\/sub><\/li>\n<li>F<sub>BD<\/sub><\/li>\n<li>F<sub>BE<\/sub><\/li>\n<li>F<sub>CE<\/sub><\/li>\n<li>F<sub>AE<\/sub><\/li>\n<li>F<sub>AC<\/sub><\/li>\n<li>F<sub>BC<\/sub><\/li>\n<li>F<sub>A<\/sub><sub>F<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach\u00a0<\/strong><\/p>\n<p>First, use the equilibrium equations to find the reaction forces. Apply these reaction forces in the method of joints to find the force of each member.<\/p>\n<p><strong>5. Analysis\u00a0<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/truss-second-image-deanna-.png\" alt=\"FBD of the problem.\" class=\"alignnone wp-image-1699\" width=\"469\" height=\"381\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/truss-second-image-deanna-.png 372w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/truss-second-image-deanna--300x244.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/truss-second-image-deanna--65x53.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/truss-second-image-deanna--225x183.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/truss-second-image-deanna--350x284.png 350w\" sizes=\"auto, (max-width: 469px) 100vw, 469px\" \/><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-150x137.png\" alt=\"x and y coordinates.\" class=\"alignleft wp-image-1831\" style=\"margin-bottom: 4.44444em\" width=\"94\" height=\"86\" \/><\/p>\n<p>a. Calculating the reaction forces using the equilibrium equations and the diagram.<\/p>\n<p>[latex]\\sum F_x=0=R_{Ax}\\\\R_{Ax}=0N[\/latex]<\/p>\n<p>Solving for R<sub>By<\/sub>:<\/p>\n<p>[latex]\\sum M_A=0= -(r_{AC}\\cdot F_g)+(r_{AB}\\cdot R_{By})\\\\r_{AB}\\cdot R_{By}=r_{AC}\\cdot\u00a0 F_g[\/latex]<\/p>\n<p>[latex]R_{By}=\\frac{r_{AC}\\cdot F_g}{r_{AB}}\\\\R_{By}=\\frac{3m\\cdot 147.15 N}{6m}[\/latex]<\/p>\n<p><span style=\"text-align: initial;background-color: initial;font-size: 1em\">[latex]R_{By}=73.575 N[\/latex]<\/span><\/p>\n<p>Solving for R<sub>Ay<\/sub>:<\/p>\n<p>[latex]\\sum F_y=0=R_{By}+R_{Ay} -F_g\\\\R_{Ay}=F_g-R_{By}\\\\R_{By}=73.575N[\/latex]<\/p>\n<p>b. F<sub>AF<\/sub>, F<sub>EF<\/sub>, F<sub>DE<\/sub>, F<sub>BD<\/sub> are zero force members because for joints F and D, there are only 2 members each, and there is no external load.<\/p>\n<p>Finding the angle between AC and AE:<\/p>\n<p>tan <span>\u03b8 = <\/span>r<sub>CE<\/sub>\/r<sub>AC<\/sub><\/p>\n<p>therefore, <span>\u03b8 = 33.7 \u00b0<\/span><\/p>\n<p>Analyzing joint A:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-150x137.png\" alt=\"x and y coordinates.\" class=\"alignleft wp-image-1831\" style=\"margin-bottom: 4.44444em\" width=\"94\" height=\"86\" \/><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-120.png\" alt=\"Sketch of forces at joint A\" class=\"alignnone wp-image-1700 size-full\" width=\"231\" height=\"214\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-120.png 231w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-120-65x60.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-120-225x208.png 225w\" sizes=\"auto, (max-width: 231px) 100vw, 231px\" \/><\/p>\n<p>[latex]\\sum F_y=0 = R_{Ay}+ F_{AE}\\cdot sin 33.7 \\\\ =\\frac{-73.575 N}{sin 33.7}[\/latex]<\/p>\n<p>[latex]F_{AE} = -132.6 N[\/latex]<\/p>\n<p>[latex]\\sum F_x=0 = F_{AC} + F_{AE} cos 33.7\\\\ F_{AC}=132.6N\\cdot \\cos {33.7}[\/latex]<\/p>\n<p>[latex]F_{AC} = 110.32 N[\/latex]<\/p>\n<p>Analyzing Joint C:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-05-31-140213-300x218.png\" alt=\"A sketch of forces at joint C.\" class=\"alignnone wp-image-2431\" width=\"413\" height=\"300\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-05-31-140213-300x218.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-05-31-140213-65x47.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-05-31-140213-225x163.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-05-31-140213-350x254.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-2025-05-31-140213.png 635w\" sizes=\"auto, (max-width: 413px) 100vw, 413px\" \/><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-150x137.png\" alt=\"x and y coordinates.\" class=\"alignleft wp-image-1831\" style=\"margin-bottom: 4.44444em\" width=\"94\" height=\"86\" \/><\/p>\n<p>From the FBD,<\/p>\n<p>F<sub>AC<\/sub>= F<sub>BC\u00a0<\/sub>= 110.32 N<\/p>\n<p>Also, F<sub>g\u00a0 <\/sub>=F<sub>CE<\/sub> = 147.15 N<\/p>\n<p>Analyzing Joint B<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-150x137.png\" alt=\"x and y coordinates.\" class=\"alignleft wp-image-1831\" style=\"margin-bottom: 4.44444em\" width=\"94\" height=\"86\" \/><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/truss-joint-b-deanna-.png\" alt=\"A sketch of forces at joint B.\" class=\"alignnone wp-image-1701 size-full\" width=\"214\" height=\"282\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/truss-joint-b-deanna-.png 214w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/truss-joint-b-deanna--65x86.png 65w\" sizes=\"auto, (max-width: 214px) 100vw, 214px\" \/><\/p>\n<p>[latex]\\sum F_y=0 = R_{By} +F_{BE}\\cdot sin 33.7[\/latex]<\/p>\n<p>[latex]F_{BE} = -132.6 N[\/latex]<\/p>\n<p>[latex]\\sum F_x=0 = F_{BC} -F_{BE} cos 33.7\\\\ F_{BC}=132.6N\\cdot \\cos {33.7}[\/latex]<\/p>\n<p>[latex]F_{BC} = 110.32 N[\/latex]<\/p>\n<p>c.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Picture1-300x213.png\" alt=\"Labeled truss diagram showing internal forces in each member and support reactions.\" width=\"394\" height=\"280\" class=\"aligncenter wp-image-2156\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Picture1-300x213.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Picture1-65x46.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Picture1-225x160.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Picture1-350x249.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Picture1.png 624w\" sizes=\"auto, (max-width: 394px) 100vw, 394px\" \/><\/p>\n<p><strong>6. Review\u00a0<\/strong><\/p>\n<p>The answers make sense because F<sub>CE<\/sub>is in tension as imagined due to the weight of the bucket. Members AC and BC are under reaction forces, are equally under tension, and members AE and BE are balanced by compression. The identified zero-force members also make sense due to the rules being met.<\/p>\n<\/div>\n<h1><strong>Example 5.5.4\u00a0 Method of Joints, Submitted by Luke McCarvill<\/strong><\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>Jeffery is unsuccessfully pushing a 2-dimensional shopping cart shown below with a force of 70N. It is made up of steel (8g\/cm<sup>3<\/sup>) rods with 2cm diameters. The wheel of the cart that is closest to him is stuck (pinned) while the front wheel is free to roll (roller). The total height of the cart is 0.75m, and its total length is 1.5m.<\/p>\n<p>What is the total mass of the cart? (For the purposes of the equilibrium equations, assume the total weight acts only at the centre point of the system, and otherwise the rods are massless.) What are the reaction forces on the wheels?<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke1.png\" alt=\"A digital sketch of the problem.\" class=\"alignnone wp-image-1857\" width=\"538\" height=\"244\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke1.png 908w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke1-300x136.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke1-768x348.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke1-65x29.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke1-225x102.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke1-350x159.png 350w\" sizes=\"auto, (max-width: 538px) 100vw, 538px\" \/><\/p>\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\/Luke2.png\" alt=\"A FBD of the problem.\" class=\"alignnone wp-image-1875\" width=\"722\" height=\"387\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke2.png 908w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke2-300x161.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke2-768x412.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke2-65x35.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke2-225x121.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Luke2-350x188.png 350w\" sizes=\"auto, (max-width: 722px) 100vw, 722px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<p>\u2022 [latex]P = 70\\,N[\/latex]<br \/>\n\u2022 [latex]\\sum M = 0[\/latex]<br \/>\n\u2022 [latex]h = 0.75\\,m[\/latex]<br \/>\n\u2022 [latex]\\theta_p = 15^\\circ[\/latex]<br \/>\n\u2022 [latex]d_{\\text{rod}} = 2\\,\\text{cm (diameter)}[\/latex]<br \/>\n\u2022 [latex]\\sum F = 0[\/latex]<br \/>\n\u2022 [latex]\\rho = 8.00\\,\\text{g\/cm}^3[\/latex]<br \/>\n\u2022 [latex]V_{\\text{cylinder}} = \\pi r^2 h[\/latex]<\/p>\n<p>Unknowns:<\/p>\n<p>\u2022 [latex]m_{\\text{tot}} = \\, ?[\/latex]<br \/>\n\u2022 [latex]P_x = \\, ?[\/latex]<br \/>\n\u2022 [latex]P_y = \\, ?[\/latex]<br \/>\n\u2022 [latex]R_{Ax} = \\, ?[\/latex]<br \/>\n\u2022 [latex]R_{Ay} = \\, ?[\/latex]<br \/>\n\u2022 [latex]R_{By} = \\, ?[\/latex]<br \/>\n\u2022 [latex]F_g = \\, ?[\/latex]<\/p>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Use trigonometry and equilibrium equations, as well as the method of joints, to calculate unknowns<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Finding total mass:<\/p>\n<p>There are four 75cm rods, two rods that are [latex]75\\sqrt{2}[\/latex]cm, and one rod that is 1.5m long, all of which have a diameter of 2cm. The total mass is therefore [latex]8g\/cm^3 \\cdot ((4 \\pi \\cdot (1cm)^2\\cdot (75cm) + 2\\pi\\cdot (1cm)^2 \\cdot \\\\ (75 \\sqrt(2)cm) + \\pi\u00a0 (1cm)^2 \\cdot (150cm)) \\\\ \\approx 16,641g = 16.641kg = m_{tot}[\/latex]<\/p>\n<p>F<sub>g<\/sub> can then be found using [latex]F_g = g \\cdot m_{tot} = -9.81m\/s^2 \\cdot 16.641kg \\approx -163.25N[\/latex]<\/p>\n<p>The sum of forces and moments can now be used to determine reaction forces<\/p>\n<p>[latex]\\sum F_x = P_x - R_{Ax} \\\\ R_{Ax} = P \\cdot cos(\\theta) = 70N cos(15^\\circ) \\\\ R_{Ax} \\approx 67.6N[\/latex]<\/p>\n<p>[latex]\\sum F_y = -F_g - P_y + R_{Ay} + R_{By} = 0[\/latex]<\/p>\n<p>With two unknowns, the sum of moments must be considered. Summing moments around point A finds:<\/p>\n<p>[latex]\\sum M_A = 0 = -(0.75m \\cdot Pcos(\\theta)) - (0.75m \\cdot F_g) + (1.5m \\cdot R_{By})[\/latex]<\/p>\n<p>[latex]\\sum M_A = 0 = -(0.75m \\cdot 70N cos(15^\\circ)) - (0.75m \\cdot 163.25N) + (1.5m \\cdot R_{By})[\/latex]<\/p>\n<p>[latex]R_{By} \\approx 115.43N[\/latex]<\/p>\n<p>Sum of Forces in Y can now be used to find R<sub>Ay<\/sub><\/p>\n<p>[latex]F_y = -F_g - P_y + R_{Ay} +R_{By} = 0[\/latex]<\/p>\n<p>[latex]R_{Ay} \\approx 65.94N[\/latex]<\/p>\n<p>The method of joints can now be used to determine if any zero-force members exist in the shopping cart.<\/p>\n<p>Looking at Point A, it can be found that [latex]F_{AO} = 65.94\\,N[\/latex] in compression and [latex]F_{AC} = 67.6\\,N[\/latex] in tension. Continuing through each joint, it can be found that member CB is a zero-force member as point B only has one component in the X direction, which, in a static system, implies that the force must be zero.<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>Given the high weight of the cart creating a large F<sub>g<\/sub> and the applied force of 70N, the reaction forces found and the internal forces all seem to be of appropriate magnitude and direction.<\/p>\n<\/div>\n<h1><strong>Example 5.5.5: Method of Sections, Submitted by Liam Murdock<\/strong><\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">You are designing a shelf for a friend, and it can be analyzed as a truss shown below. If you know the forces in the member CD can take a tensile force of 25 N, and GH can take a compressive force of 15 N before breaking. If w = 10 cm and h = 20 cm, what mass can be supported if it is situated exactly on point E?<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Liam_Chapter5_1.png\" alt=\"A digital sketch of the problem.\" class=\"alignnone wp-image-2009 size-full\" width=\"509\" height=\"471\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Liam_Chapter5_1.png 509w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Liam_Chapter5_1-300x278.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Liam_Chapter5_1-65x60.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Liam_Chapter5_1-225x208.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Liam_Chapter5_1-350x324.png 350w\" sizes=\"auto, (max-width: 509px) 100vw, 509px\" \/><\/div>\n<p><strong>2. Sketch<\/strong><br \/>\nDraw a free-body diagram of the truss with a slice through CD, GH, and IH.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Liam_Chapter5_2-1.png\" alt=\"A FBD showing cut sections of the problem.\" class=\"alignnone wp-image-2011 size-full\" width=\"294\" height=\"408\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Liam_Chapter5_2-1.png 294w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Liam_Chapter5_2-1-216x300.png 216w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Liam_Chapter5_2-1-65x90.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Liam_Chapter5_2-1-225x312.png 225w\" sizes=\"auto, (max-width: 294px) 100vw, 294px\" \/><br \/>\n<strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>w = 0.1 m<\/li>\n<li>h = 0.2 m<\/li>\n<li>F<sub>CD<\/sub> = -25 N<\/li>\n<li>F<sub>GH<\/sub> = 15 N<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>F<sub>HI<\/sub> =?<\/li>\n<li>d<sub>HI<\/sub> = ?<\/li>\n<li>P =?<\/li>\n<li>m = ?<\/li>\n<\/ul>\n<p><strong>4. Approach<br \/>\n<\/strong>Use the method of sections to show internal forces and analyze as a rigid body.<\/p>\n<p><strong>5. Analysis<\/strong><br \/>\nFirst, the distance d<sub>HI<\/sub> can be found:<br \/>\n[latex]d_{HI} = \\sqrt{(0.1 m)^2 + (0.2 m)^2} \\\\ d_{HI} = 0.2236 m[\/latex]<br \/>\nThis value can be used to find the equivalents to cos\u03b8 and sin\u03b8:<br \/>\n[latex]\\cos{\\theta} = \\frac{0.1 m}{0.2236 m} \\\\ \\sin{\\theta} = \\frac{0.2 m}{0.2236 m}[\/latex]<br \/>\nThe sum of forces in the X direction can be used to determine the value of F<sub>HI<\/sub>:<br \/>\n[latex]\\sum F_x = 0 = -F_{CD} + F_{GH} + F_{HI} \\cdot \\cos{\\theta} \\\\ F_{HI} = \\frac{(F_{CD} - F_{GH})}{\\cos{\\theta}} \\\\ F_{HI} = \\frac{0.2236 m}{0.1 m} \\cdot (25 N - 15 N) \\\\ F_{HI} = 22.36N[\/latex]<br \/>\nThe sum of forces in the Y direction can now be used to determine the value of P:<br \/>\n[latex]\\sum F_y = 0 = -P + F_{HI} \\cdot \\sin{\\theta} \\\\ P = F_{HI} \\cdot \\sin{\\theta} \\\\ P = (22.36 N) \\cdot \\frac{0.2 m}{0.2236 m} \\\\ P = 20 N[\/latex]<br \/>\nThe mass can now be simply found by dividing the force by gravity:<br \/>\n[latex]P = mg \\\\ m = \\frac{P}{g} \\\\ m = \\frac{20 N}{9.81 \\frac{m}{s^2}} \\\\ m = 2.04 kg[\/latex]<\/p>\n<p><strong>6. Review<br \/>\n<\/strong>A brief review of this question is the sign and magnitude of the result. The force generated by the mass pushes downwards with gravity, which is expected. The magnitude of the mass, while defining this shelf as one that is particularly flimsy, is a reasonable mass.<\/p>\n<\/div>\n","protected":false},"author":60,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"5.5 Examples","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-574","chapter","type-chapter","status-publish","hentry"],"part":58,"_links":{"self":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/574","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":32,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/574\/revisions"}],"predecessor-version":[{"id":2862,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/574\/revisions\/2862"}],"part":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/parts\/58"}],"metadata":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/574\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/media?parent=574"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapter-type?post=574"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/contributor?post=574"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/license?post=574"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}